research article martingale morrey-hardy and campanato-hardy...

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Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 690258, 14 pages http://dx.doi.org/10.1155/2013/690258 Research Article Martingale Morrey-Hardy and Campanato-Hardy Spaces Eiichi Nakai, 1 Gaku Sadasue, 2 and Yoshihiro Sawano 3 1 Department of Mathematics, Ibaraki University, Mito, Ibaraki 310-8512, Japan 2 Department of Mathematics, Osaka Kyoiku University, Kashiwara, Osaka 582-8582, Japan 3 Department of Mathematics and Information Sciences, Tokyo Metropolitan University, Minami-Osawa 1-1, Hachioji-shi, Tokyo 192-0397, Japan Correspondence should be addressed to Gaku Sadasue; [email protected] Received 10 June 2013; Accepted 17 August 2013 Academic Editor: Natasha Samko Copyright © 2013 Eiichi Nakai et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We introduce generalized Morrey-Campanato spaces of martingales, which generalize both martingale Lipschitz spaces introduced by Weisz (1990) and martingale Morrey-Campanato spaces introduced in 2012. We also introduce generalized Morrey-Hardy and Campanato-Hardy spaces of martingales and study Burkholder-type equivalence. We give some results on the boundedness of fractional integrals of martingales on these spaces. 1. Introduction Lebesgue spaces and Hardy spaces play an important role in martingale theory and in harmonic analysis as well. Morrey-Campanato spaces are very useful to know more pre- cise properties of functions and martingales. It is known that Morrey-Campanato spaces contain , BMO, and Lip as special cases; see, for example, [1, 2]. In martingale theory, Weisz [3] introduced martingale Lipschitz spaces for general filtrations {F } ≥0 and proved the duality between martingale Hardy spaces and martingale Lip- schitz spaces. is result was extended to generalized martin- gale Campanato spaces and martingale Orlicz-Hardy spaces in [4]. Recently, martingale Morrey-Campanato spaces were introduced in [5], where each sub--algebra F is generated by countable atoms. In this paper, we introduce martingale Morrey-Hardy and Campanato-Hardy spaces based on square functions and unify Hardy, Lipschitz, and Morrey-Campanato spaces in [35]. To do this, we first introduce generalized martingale Morrey-Campanato spaces by using subfamilies {B } ≥0 of the filtration {F } ≥0 with B F for each ≥0. We estab- lish Burkholder-type equivalence and discuss equivalence between martingale Morrey spaces and martingale Cam- panato spaces in a suitable condition. We also establish a John-Nirenberg-type theorem for generalized martingale Campanato-Hardy spaces; see eorem 15. On these martingale spaces, we introduce generalized fractional integrals as martingale transforms and prove their boundedness. Our result extends several results in [57] to these spaces. e fractional integrals are very useful tools to analyse function spaces in harmonic analysis. Actually, on the Euclidean space, Hardy and Littlewood [8, 9] and Sobolev [10] investigated the fractional integrals to establish the the- ory of Lebesgue spaces and Lipschitz spaces. Stein and Weiss [11], Taibleson and Weiss [12], and Krantz [13] also inves- tigated the fractional integrals to establish the theory of Hardy spaces; see also [14]. e - boundedness of the fractional integrals is well known as the Hardy-Littlewood- Sobolev theorem derived from [810]. is boundedness has been extended to Morrey-Campanato spaces by Peetre [1] and Adams [15]; see also Chiarenza and Frasca [16]. In martingale theory, based on the result on the Walsh multiplier by Watari [7, eorem 1.1], Chao and Ombe [6] proved the bounded- ness of the fractional integrals for , , BMO, and Lipschitz spaces of the dyadic martingale. e boundedness of the fractional integrals for martingale Morrey-Campanato spaces was established in [5]. For other types of operators for martingales, see the recent work by Tanaka and Terasawa [17]. At the end of this section, we make some conventions. roughout this paper, we always use to denote a positive constant that is independent of the main parameters involved but whose value may differ from line to line. Constants with

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Page 1: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Hindawi Publishing CorporationJournal of Function Spaces and ApplicationsVolume 2013 Article ID 690258 14 pageshttpdxdoiorg1011552013690258

Research ArticleMartingale Morrey-Hardy and Campanato-Hardy Spaces

Eiichi Nakai1 Gaku Sadasue2 and Yoshihiro Sawano3

1 Department of Mathematics Ibaraki University Mito Ibaraki 310-8512 Japan2Department of Mathematics Osaka Kyoiku University Kashiwara Osaka 582-8582 Japan3Department of Mathematics and Information Sciences Tokyo Metropolitan University Minami-Osawa 1-1Hachioji-shi Tokyo 192-0397 Japan

Correspondence should be addressed to Gaku Sadasue sadasueccosaka-kyoikuacjp

Received 10 June 2013 Accepted 17 August 2013

Academic Editor Natasha Samko

Copyright copy 2013 Eiichi Nakai et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We introduce generalizedMorrey-Campanato spaces ofmartingales which generalize bothmartingale Lipschitz spaces introducedby Weisz (1990) and martingale Morrey-Campanato spaces introduced in 2012 We also introduce generalized Morrey-Hardy andCampanato-Hardy spaces of martingales and study Burkholder-type equivalence We give some results on the boundedness offractional integrals of martingales on these spaces

1 Introduction

Lebesgue spaces 119871119901 and Hardy spaces 119867119901 play an importantrole in martingale theory and in harmonic analysis as wellMorrey-Campanato spaces are very useful to knowmore pre-cise properties of functions and martingales It is known thatMorrey-Campanato spaces contain 119871119901 BMO and Lip

120572as

special cases see for example [1 2]In martingale theory Weisz [3] introduced martingale

Lipschitz spaces for general filtrations F119899119899ge0 andproved theduality betweenmartingaleHardy spaces andmartingale Lip-schitz spacesThis result was extended to generalizedmartin-gale Campanato spaces and martingale Orlicz-Hardy spacesin [4] Recently martingale Morrey-Campanato spaces wereintroduced in [5] where each sub-120590-algebraF119899 is generatedby countable atoms

In this paper we introduce martingale Morrey-Hardyand Campanato-Hardy spaces based on square functions andunify Hardy Lipschitz and Morrey-Campanato spaces in[3ndash5] To do this we first introduce generalized martingaleMorrey-Campanato spaces by using subfamilies B119899119899ge0 ofthe filtration F119899119899ge0 withB119899 sub F119899 for each 119899 ge 0We estab-lish Burkholder-type equivalence and discuss equivalencebetween martingale Morrey spaces and martingale Cam-panato spaces in a suitable condition We also establisha John-Nirenberg-type theorem for generalized martingaleCampanato-Hardy spaces see Theorem 15

On these martingale spaces we introduce generalizedfractional integrals as martingale transforms and prove theirboundedness Our result extends several results in [5ndash7] tothese spaces The fractional integrals are very useful tools toanalyse function spaces in harmonic analysis Actually on theEuclidean space Hardy and Littlewood [8 9] and Sobolev[10] investigated the fractional integrals to establish the the-ory of Lebesgue spaces and Lipschitz spaces Stein andWeiss[11] Taibleson and Weiss [12] and Krantz [13] also inves-tigated the fractional integrals to establish the theory ofHardy spaces see also [14] The 119871119901-119871119902 boundedness of thefractional integrals is well known as the Hardy-Littlewood-Sobolev theorem derived from [8ndash10] This boundedness hasbeen extended toMorrey-Campanato spaces by Peetre [1] andAdams [15] see also Chiarenza and Frasca [16] Inmartingaletheory based on the result on theWalsh multiplier byWatari[7 Theorem 11] Chao and Ombe [6] proved the bounded-ness of the fractional integrals for119867119901119871119901 BMO andLipschitzspaces of the dyadic martingale The boundedness of thefractional integrals formartingaleMorrey-Campanato spaceswas established in [5] For other types of operators formartingales see the recent work byTanaka andTerasawa [17]

At the end of this section we make some conventionsThroughout this paper we always use 119862 to denote a positiveconstant that is independent of themain parameters involvedbut whose value may differ from line to line Constants with

2 Journal of Function Spaces and Applications

subscripts such as119862119901 are dependent on the subscripts If119891 le

119862119892 we then write 119891 ≲ 119892 or 119892 ≳ 119891 and if 119891 ≲ 119892 ≲ 119891 we thenwrite 119891 sim 119892

2 Definitions and Notation

Let (Ω Σ 119875) be a probability space and F = F119899119899ge0 anondecreasing sequence of sub-120590-algebras of Σ such that Σ =

120590(⋃119899 F119899) For the sake of simplicity let Fminus1 = F0 The set119861 isin F119899 is called atom more precisely (F119899 119875)-atom if any119860 sub 119861 119860 isin F119899 satisfying 119875(119860) = 119875(119861) or 119875(119860) = 0 Denoteby 119860(F119899) the set of all atoms inF119899

The expectation operator and the conditional expectationoperators relative toF119899 are denoted by119864 and119864119899 respectivelyIt is known from the Doob theorem that if 119901 isin (1 infin)then any 119871119901-boundedmartingale converges in 119871119901 Moreoverif 119901 isin [1 infin) then for any 119891 isin 119871119901 its correspondingmartingale (119891119899)119899ge0 with 119891119899 = 119864119899119891 is an 119871119901-boundedmartingale and converges to 119891 in 119871119901 (see eg [18]) For thisreason a function 119891 isin 1198711 and the corresponding martingale(119891119899)119899ge0 will be denoted by the same symbol 119891

Let M be the set of all martingales such that 1198910 = 0 For119901 isin [1 infin] let 119871

0

119901be the set of all 119891 isin 119871119901 such that 1198640119891 = 0

For any 119891 isin 1198710

119901 its corresponding martingale (119891119899)119899ge0 with

119891119899 = 119864119899119891 is an 119871119901-boundedmartingale inM For this reasonwe regard 119871

0

119901as a subset ofM

Let B = B119899119899ge0 be subfamilies of F = F119899119899ge0 withB119899 sub F119899 for each 119899 ge 0 We denote byB sub F this relationofB andF

In this paper we always postulate the following conditiononB

There exists a countable subsetB1015840

0sub B0

such that 119875 ( ⋃

119861isinB10158400

119861) = 1

(1)

We first define generalized martingale Morrey-Campanatospaces with respect toB as follows

Definition 1 Let B sub F 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) For 119891 isin 1198711 let1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

=1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817L119901120601

=1003817100381710038171003817119891

1003817100381710038171003817L119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

=1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(2)

and define

119871119901120601 = 119871119901120601 (B) = 119891 isin 1198710

1199011003817100381710038171003817119891

1003817100381710038171003817119871119901120601

lt infin

L119901120601 = L119901120601 (B) = 119891 isin 1198710

1199011003817100381710038171003817119891

1003817100381710038171003817L119901120601

lt infin

Lminus

119901120601= L

minus

119901120601(B) = 119891 isin 119871

0

1199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

lt infin

(3)

Remark 2 By the condition (1) the functionals 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

are norms on 1198710

119901

Remark 3 Let 119891 isin 1198710

1 Then 119891 isin L119901120601 if and only if its

corresponding martingale (119891119899)119899ge0 is L119901120601-bounded that issup

119899ge0119891119899L

119901120601

lt infin The same conclusion holds for Lminus

119901120601

Furthermore if each sub-120590-algebraF119899 is generated by count-able atoms B = 119860(F119899)

119899ge0and 120601 is almost decreasing

then the same conclusion holds for 119871119901120601 More precisely seeProposition 8

Remark 4 In general 119891L119901120601

le 2119891119871119901120601

and hence 119871119901120601 sub

L119901120601 Actually for any 119861 isin B119899

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

+ (int119861

10038161003816100381610038161198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

(4)

Similarly 119891L119901120601

le 2119891Lminus119901120601

andLminus

119901120601sub L119901120601

Definition 5 For 120601 equiv 1 denoteL119901120601 andLminus

119901120601by BMO119901 and

BMOminus

119901 respectively For 120601(119903) = 119903

120572 120572 gt 0 denote L119901120601 andLminus

119901120601by Lip

119901120572and Lipminus

119901120572 respectively

If 120601(119903) = 119903120582 120582 isin (minusinfin infin) then we simply denote 119871119901120601

L119901120601 andLminus

119901120601by 119871119901120582L119901120582 andLminus

119901120582 respectively

A function 120601 (0 1] rarr (0 infin) is said to be almostincreasing (resp almost decreasing) if there exists a positiveconstant 119862 such that120601 (119904) le 119862120601 (119905) (resp 120601 (119905) le 119862120601 (119904)) for 0 lt 119904 le 119905 le 1

(5)For the caseB = F the spaces BMO119901(F) and Lip

119901120572(F)

with 120572 ge 0 were introduced by Weisz [3]Recall that119860(F119899) is the set of all atoms inF119899 and letA =

119860(F119899)119899ge0 Suppose that each sub-120590-algebraF119899 is generatedby countable atoms for the time being Then BMO119901(F) =

BMO119901(A) and Lip119901120572

(F) = Lip119901120572

(A) see [5] In general if120601 is almost increasing then

119871119901120601 (F) = 119871119901120601 (A)

L119901120601 (F) = L119901120601 (A)

Lminus

119901120601(F) = L

minus

119901120601(A)

(6)

Journal of Function Spaces and Applications 3

with equivalent norms respectively However if 120601 is notalmost increasing then these equalities fail in general see [5]

In this paper we do not always assume that each sub-120590-algebraF119899 is generated by countable atoms Let

119860(F119899)perp

= 119861 isin F119899 119875 (119861 cap 119860) = 0 forall119860 isin 119860 (F119899) (7)

and let

B119899 = 119860 (F119899) cup 119860(F119899)perp

(119899 ge 0) (8)

In this case if 120601 is almost increasing then we will show that

119871119901120601 (F) = 119871119901120601 (B)

L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B)

(9)

with equivalent norms respectively (see Proposition 9)Moreover ifF0 is nonatomic thenB119899 = F119899 for all 119899 ge 0 Ifeach sub-120590-algebraF119899 is generated by countable atoms thenB119899 = 119860(F119899) for all 119899 ge 0 Therefore our definition gener-alizes those in [3ndash5]

Next we define Morrey-Hardy and Campanato-Hardyspaces based on square functions with respect to B asfollows For 119891 isin M we denote by 119878119899(119891) and 119878(119891) the squarefunction of 119891

119878119899 (119891) = (

119899

sum

119896=0

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

119878 (119891) = (

infin

sum

119896=0

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

(10)

where 119889119896119891 = 119891119896 minus 119891119896minus1 (119899 ge 0 with convention 1198890119891 = 0 and119878minus1(119891) = 0) We further define

119878(119899)

(119891) = (119878(119891)2

minus 119878119899(119891)2)12

= (

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

(11)

Definition 6 Let B sub F 119901 isin (0 infin) and 120601 (0 1] rarr

(0 infin) For 119891 = (119891119899)119899ge0 isin M let

10038171003817100381710038171198911003817100381710038171003817119867119878119901120601

=1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

119878(119891)119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

=1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119878(119899)

(119891)119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

=1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(12)

and define

119867119878

119901120601= 119867

119878

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817119867119878119901120601

lt infin

H119878

119901120601= H

119878

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

lt infin

H119878minus

119901120601= H

119878minus

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

lt infin

(13)

By (1) the functionals 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

arequasinorms onM

Remark 7 If we take 120601 equiv 1 and B = F then the norm119891H119878minus

119901120601

coincides with the norm 119891BMO119878119901

in [19 Definition245] In this point our notation is different from the one in[19]

In the end of this section we present the definitionof regularity on F and the doubling condition on 120601 Thefiltration F = F119899119899ge0 is said to be regular if there existsa constant 119877 ge 2 such that

119891119899 le 119877119891119899minus1 (14)

holds for all nonnegative martingales (119891119899)119899ge0 We say thesmallest constant 119877 satisfying (14) the regularity constant ofF A function 120601 (0 1] rarr (0 infin) is said to satisfy thedoubling condition if there exists a positive constant 119862120601 suchthat

119862minus1

120601le

120601 (119904)

120601 (119905)le 119862120601 forall119904 119905 isin (0 1] with 1

2le

119904

119905le 2 (15)

The smallest constant 119862120601 satisfying (15) is called the doublingconstant of 120601

3 Properties of Morrey-Hardy andCampanato-Hardy Spaces

In this section we investigate the properties ofMorrey-Hardyand Campanato-Hardy spaces The proofs of the results inthis section will be given in Section 6

First we state basic properties of the norms

Proposition 8 Let B sub F 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Let 119891 isin 1198711 and let (119891119899)119899ge0 be its correspondingmartingale 119891119899 = 119864119899119891 Then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

le sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817119871119901120601

10038171003817100381710038171198911003817100381710038171003817L119901120601

= sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817L119901120601

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

= sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817Lminus119901120601

(16)

Moreover if each sub-120590-algebra F119899 is generated by countableatomsB = A and120601 is almost decreasing that is there exists apositive constant 1198620 such that 120601(119905) le 1198620120601(119904) for 0 lt 119904 le 119905 le 1then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

le sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817119871119901120601

le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(17)

4 Journal of Function Spaces and Applications

Proposition 9 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge 0)If 120601 is almost increasing that is there exists a positive constant1198620 such that 120601(119904) le 1198620120601(119905) for 0 lt 119904 le 119905 le 1 then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B)le

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(F)le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B) (18)

and the same conclusions hold for sdot L119901120601

sdot Lminus119901120601

sdot 119867119878119901120601

sdot H119878

119901120601

and sdot H119878minus119901120601

Consequently

119871119901120601 (F) = 119871119901120601 (B) L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B) 119867119901120601 (F) = 119867119901120601 (B)

H119878

119901120601(F) = H

119878

119901120601(B) H

119878minus

119901120601(F) = H

119878minus

119901120601(B)

(19)

with equivalent norms respectively

For 119901 isin (0 infin) let 119867119878

119901be the set of all 119891 isin M such that

119878(119891)119871119901

lt infin Let 119891119867119878119901

= 119878(119891)119871119901

Note that if 120601(119903) =

119903minus1119901 and Ω isin B0 then 119867

119878

119901120601= 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

The following is well known as Burkholderrsquos inequality

Theorem 10 (Burkholder [20]) If 119901 isin (1 infin) then there existpositive constants 119888119901 and 119862119901 that depend only on 119901 such that

1198881199011003817100381710038171003817119891

1003817100381710038171003817119871119901

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901

le 119862119901

10038171003817100381710038171198911003817100381710038171003817119871119901

(20)

for all 119891 isin 1198710

1sub M

For expressions of the constants 119888119901 and 119862119901 see forexample [21ndash23] See also [24] for Burkholderrsquos inequality onBanach functions spaces

Our first result is the following which is an extension ofBurkholderrsquos inequality to martingale Campanato spaces

Theorem 11 Let B sub F 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Then

1198881199011003817100381710038171003817119891

1003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901

10038171003817100381710038171198911003817100381710038171003817L119901120601

(21)

119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(22)

for all 119891 isin 1198710

1sub M where 119888119901 and 119862119901 are the constants in

Theorem 10

Next we give the relations between L119901120601 and Lminus

119901120601and

betweenH119878

119901120601andH119878minus

119901120601We consider the following condition

onB

120596 isin Ω 119864119899minus1 [120594119861] (120596) gt 0 isin B119899minus1 forall119861 isin B119899 (119899 ge 1)

(23)

Theorem 12 LetB sub F and 120601 (0 1] rarr (0 infin) Then

L119901120601 sup Lminus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817L119901120601

le 21003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(24)

H119878

119901120601sup H

119878minus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(25)

Conversely if F is regular B satisfies (23) and 120601 satisfiesthe doubling condition then there exists a positive constant 119862dependent only on 119901 the regularity constant of F and thedoubling constant of 120601 such that

L119901120601 sub Lminus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817L119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(26)

H119878

119901120601sub H

119878minus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(27)

We give a relation betweenmartingaleMorrey spaces andmartingale Campanato spaces in the following form

Theorem 13 Suppose that every 120590-algebraF119899 is generated bycountable atoms Let B = A 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (28)

Then there exists a positive constant 119862 such that

1

2

10038171003817100381710038171198911003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

forall119891 isin 1198710

1

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

forall119891 isin M

(29)

Moreover ifF is regular then 119891119871119901120601

119891L119901120601

and 119891Lminus119901120601

areequivalent to each other and 119891

119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

areequivalent to each other

Using Theorems 11ndash13 we have Burkholder-type equiva-lence for generalized martingale Morrey spaces

Corollary 14 Suppose that every 120590-algebra F119899 is generatedby countable atoms LetB = A 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (30)

IfF is regular then there exist positive constants 119888 and 119862 suchthat for all 119891 isin 119871

0

1

1198881003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817119871119901120601

(31)

For the martingale BMO spaces based on square func-tions the John-Nirenberg-type equivalence was establishedbyWeisz [25] and [19Theorem250]We extend this theoremto the spacesH119878

119901120601andH119878minus

119901120601

Theorem 15 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899 (119899 ge 0)119901 isin (0 infin) and 120601 (0 1] rarr (0 infin) Assume that 120601 is almostincreasing and satisfies the doubling conditionThen 119891H119878

119901120601

le

119862119901119902120601119891H119878minus119902120601

for all 119902 isin (0 infin) If we further assume thatF isregular then 119891H119878

119901120601

and 119891H119878minus119901120601

are equivalent to 119891H1198782120601

Journal of Function Spaces and Applications 5

4 Fractional Integrals

In this section we state the results on the boundedness offractional integrals as martingale transforms The proofs ofthe results in this section will be given in Section 7

Let (120574119899)119899ge0 be a sequence of nonnegative bounded func-tions adapted to F = F119899119899ge0 that is 120574119899 is F119899-measurablefor every 119899 ge 0 Let 119868120574 be the martingale transform associateto (120574119899)119899ge0 that is

(119868120574119891)119899

=

119899

sum

119896=0

120574119896minus1119889119896119891 (32)

with convention 120574minus11198890119891 = 0 Note that if 119891 = (119891119899)119899ge0 isin Mthen 119868120574119891 = ((119868120574119891)119899)119899ge0 isin M

We now define a generalized fractional integral 119868120588 formartingales as a special case of 119868120574 under the assumption thatevery 120590-algebra F119899 is generated by countable atoms Ourdefinition generalizes the fractional integral for dyadic mar-tingales introduced in [6 7] The idea of 119868120588 comes from [26]

Suppose that every 120590-algebra F119899 is generated by count-able atoms Let 119887119899 be anF119899-measurable function such that

119887119899 (120596) = 119875 (119861) for as 120596 isin 119861 with 119861 isin 119860 (F119899) (33)

that is

119887119899 = sum

119861isin119860(F119899)

119875 (119861) 120594119861 as (34)

For a bounded function 120588 (0 1] rarr (0 infin) we define a gen-eralized fractional integral 119868120588119891 = ((119868120588119891)119899)119899ge0 of119891 = (119891119899)119899ge0 isin

M by

(119868120588119891)119899

=

119899

sum

119896=0

120588 (119887119896minus1) 119889119896119891 (35)

The generalized fractional integral 119868120588 is obtained by taking120574119899 = 120588(119887119899) in (32) If 120588(119903) = 119903

120572 120572 gt 0 then we simply denote

119868120588 by 119868120572For quasinormed spaces 1198721 and 1198722 of martingales

we denote by 119861(1198721 1198722) the set of all bounded martingaletransforms from1198721 to1198722 that is119879 isin 119861(1198721 1198722)means thatthere exists a positive constant 119862 such that

100381710038171003817100381711987911989110038171003817100381710038171198722

le 1198621003817100381710038171003817119891

10038171003817100381710038171198721

(36)

for all martingales 119891 = (119891119899)119899ge0 isin 1198721We first study the boundedness on the spaces H119878

119901120601

On martingale Campanato-Hardy spaces we consider thefractional integral as a martingale transform associated withmonotone multipliers We say a sequence of nonnegativemeasurable functions 120574 = (120574119899)119899ge0 is almost decreasing if thereexists a positive constant 119862 such that

120574119896 (120596) le 119862120574ℓ (120596) as forall119896 ge ℓ (37)

For an almost decreasing sequence 120574 = (120574119899)119899ge0 we define 119860120574

by

119860120574 = inf 119862 gt 0 119862 satisfies (37) (38)

InTheorem 16 below we do not need any assumption onF119899119899ge0

Theorem 16 Let B sub F 119901 isin (0 infin) and 120601 120595 (0 1] rarr

(0 infin) Let (120574119899)119899ge0 be a sequence of nonnegative boundedalmost decreasing adapted functions and let 119868120574 be the martin-gale transform defined by (32) Assume that

119862120574120601120595 = sup119899ge0

sup119861isinB119899

120601 (119875 (119861))

120595 (119875 (119861))

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817119871infin

lt infin (39)

Then

119868120574 isin 119861 (H119878

119901120601H

119878

119901120595) (40)

with10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(41)

If every 120590-algebra F119899 is generated by countable atomsthen we can apply Theorem 16 to the generalized fractionalintegral 119868120588 The following corollary extends [5 Theorem 58]to the spacesH119878

119901120601

Corollary 17 Assume that every 120590-algebra F119899 is generatedby countable atoms and B = A Let 119901 isin (0 infin) and 120588 120601 120595

(0 1] rarr (0 infin) Suppose that 120588 is almost increasing and that

sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin (42)

Then

119868120588 isin 119861 (H119878

119901120601H

119878

119901120595) (43)

If one further assumes that F119899119899ge0 is regular and that 120595 isalmost increasing and satisfies the doubling condition then

119868120588 isin 119861 (H119878

119901120601H

119878

119902120595) (119901 119902 isin (0 infin)) (44)

We next study the boundedness on martingale Morrey-Hardy spaces 119867

119878

119901120601and martingale Hardy spaces 119867

119878

119901

Recall that119860(F119899)perp

= F119899 for all 119899 ge 0 ifF0 is nonatomic

Proposition 18 Let B sub F 0 lt 119901 lt 119902 lt infin and 120601

(0 1] rarr (0 infin) Let (120574119899)119899ge0 be a sequence of adapted func-tions Suppose that F0 is nonatomic and that B = FAssume in addition that 120601 is almost decreasing that 119905

1119901120601(119905)

is almost increasing and that lim119905rarr0120601(119905) = infin Then 119868120574 notin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

According to Proposition 18 to consider the bounded-ness on 119867

119878

119901120601and 119867

119878

119901 we suppose that every 120590-algebra F119899

is generated by countable atoms and thatB = AIn this case if 120601(119903) = 119903

minus1119901 and F0 = Ω 0 then119867

119878

119901120601coincides with 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

However ifF0 = Ω 0 then 119867

119878

119901120601does not coincide with 119867

119878

119901in general

We do not always assume thatF0 = Ω 0

Theorem 19 Suppose that every 120590-algebraF119899 is generated bycountable atoms thatB = A and that F119899119899ge0 is regular Let

6 Journal of Function Spaces and Applications

0 lt 119901 lt 119902 lt infin and 120601 (0 1] rarr (0 infin) and let (120574119899)119899ge0 bea sequence of nonnegative bounded adapted functions Assumethat 120601 satisfies the doubling condition and that there exists apositive constant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902 as

(45)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then

119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (46)

Furthermore if 120601(119905) = 119905minus1119901 then

119868120574 isin 119861 (119867119878

119901 119867

119878

119902) (47)

As a consequence of Theorem 19 we have the followingcorollary which gives an extension of [5 Corollary 57] to thespaces 119867

119878

119901120601and gives a martingale Morrey-Hardy version of

Gunawan [27 Theorem B]

Corollary 20 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and 120588 120601 (0 1] rarr (0 infin) Assume that120588 is bounded that both 120588 and 120601 satisfy the doubling conditionand that there exists a positive constant 119862 such that

120601 (119903) int

119903

0

120588 (119905)

119905119889119905 + int

1

119903

120601 (119905) 120588 (119905)

119905119889119905 le 119862120601(119903)

119901119902

(0 lt 119903 lt 1)

(48)

Then

119868120588 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (49)

The following extends the results for dyadic martingalesin [6 7] and the result for 0 lt 119901 le 1 in [28]

Corollary 21 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and minus1119901 + 120572 = minus1119902 Then

119868120572 isin 119861 (119867119878

119901 119867

119878

119902) (50)

5 Lemmas

We prepare some lemmas to prove the results in Sections 3and 4

Lemma 1 LetB119899 satisfy 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge

0) Suppose that 120601 (0 1] rarr (0 infin) is almost increasing that

is120601(119903) le 1198620120601(119904) for all 0 lt 119903 le 119904 le 1Then for all nonnegativefunctions 119865

sup119861isinF119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

le 1198620 sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

(51)

Proof Let

N = sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875) (52)

For any 119861 isin F119899 we can choose the sets 119861119895 119895 = 0 1 2

(finite or infinite) such that

119861 = cup119895119861119895 1198610 isin 119860(F119899)perp

119861119895 isin 119860 (F119899)

119895 = 1 2

119875 (119861) = sum

119895

119875 (119861119895)

(53)

In this case 119861119895 isin B119899 119895 = 0 1 2 since119860(F119899)cup119860(F119899)perp

sub

B119899 Then

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

=1

120601 (119875 (119861))

1

119875 (119861)sum

119895

(int119861119895

119865 119889119875)

= sum

119895

119875 (119861119895)

119875 (119861)(

1

120601 (119875 (119861)) 119875 (119861119895)

int119861119895

119865 119889119875)

le sum

119895

119875 (119861119895)

119875 (119861)(

1198620

120601 (119875 (119861119895)) 119875 (119861119895)

int119861119895

119865 119889119875)

le 1198620sum

119895

119875 (119861119895)

119875 (119861)N

= 1198620N

(54)

This shows the conclusion

Lemma 2 (see [5 Lemma 33]) Suppose that every 120590-algebraF119899 is generated by countable atoms and that F119899119899ge0 is regularThen every sequence

1198610 sup 1198611 sup sdot sdot sdot sup 119861119899 sup sdot sdot sdot 119861119899 isin 119860 (F119899) (55)

has the following property for each 119899 ge 1

119861119899 = 119861119899minus1 119900119903 (1 +1

119877) 119875 (119861119899) le 119875 (119861119899minus1) le 119877119875 (119861119899)

(56)

where 119877 is the constant in (14)

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Page 2: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

2 Journal of Function Spaces and Applications

subscripts such as119862119901 are dependent on the subscripts If119891 le

119862119892 we then write 119891 ≲ 119892 or 119892 ≳ 119891 and if 119891 ≲ 119892 ≲ 119891 we thenwrite 119891 sim 119892

2 Definitions and Notation

Let (Ω Σ 119875) be a probability space and F = F119899119899ge0 anondecreasing sequence of sub-120590-algebras of Σ such that Σ =

120590(⋃119899 F119899) For the sake of simplicity let Fminus1 = F0 The set119861 isin F119899 is called atom more precisely (F119899 119875)-atom if any119860 sub 119861 119860 isin F119899 satisfying 119875(119860) = 119875(119861) or 119875(119860) = 0 Denoteby 119860(F119899) the set of all atoms inF119899

The expectation operator and the conditional expectationoperators relative toF119899 are denoted by119864 and119864119899 respectivelyIt is known from the Doob theorem that if 119901 isin (1 infin)then any 119871119901-boundedmartingale converges in 119871119901 Moreoverif 119901 isin [1 infin) then for any 119891 isin 119871119901 its correspondingmartingale (119891119899)119899ge0 with 119891119899 = 119864119899119891 is an 119871119901-boundedmartingale and converges to 119891 in 119871119901 (see eg [18]) For thisreason a function 119891 isin 1198711 and the corresponding martingale(119891119899)119899ge0 will be denoted by the same symbol 119891

Let M be the set of all martingales such that 1198910 = 0 For119901 isin [1 infin] let 119871

0

119901be the set of all 119891 isin 119871119901 such that 1198640119891 = 0

For any 119891 isin 1198710

119901 its corresponding martingale (119891119899)119899ge0 with

119891119899 = 119864119899119891 is an 119871119901-boundedmartingale inM For this reasonwe regard 119871

0

119901as a subset ofM

Let B = B119899119899ge0 be subfamilies of F = F119899119899ge0 withB119899 sub F119899 for each 119899 ge 0 We denote byB sub F this relationofB andF

In this paper we always postulate the following conditiononB

There exists a countable subsetB1015840

0sub B0

such that 119875 ( ⋃

119861isinB10158400

119861) = 1

(1)

We first define generalized martingale Morrey-Campanatospaces with respect toB as follows

Definition 1 Let B sub F 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) For 119891 isin 1198711 let1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

=1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817L119901120601

=1003817100381710038171003817119891

1003817100381710038171003817L119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

=1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(2)

and define

119871119901120601 = 119871119901120601 (B) = 119891 isin 1198710

1199011003817100381710038171003817119891

1003817100381710038171003817119871119901120601

lt infin

L119901120601 = L119901120601 (B) = 119891 isin 1198710

1199011003817100381710038171003817119891

1003817100381710038171003817L119901120601

lt infin

Lminus

119901120601= L

minus

119901120601(B) = 119891 isin 119871

0

1199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

lt infin

(3)

Remark 2 By the condition (1) the functionals 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

are norms on 1198710

119901

Remark 3 Let 119891 isin 1198710

1 Then 119891 isin L119901120601 if and only if its

corresponding martingale (119891119899)119899ge0 is L119901120601-bounded that issup

119899ge0119891119899L

119901120601

lt infin The same conclusion holds for Lminus

119901120601

Furthermore if each sub-120590-algebraF119899 is generated by count-able atoms B = 119860(F119899)

119899ge0and 120601 is almost decreasing

then the same conclusion holds for 119871119901120601 More precisely seeProposition 8

Remark 4 In general 119891L119901120601

le 2119891119871119901120601

and hence 119871119901120601 sub

L119901120601 Actually for any 119861 isin B119899

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

+ (int119861

10038161003816100381610038161198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

(4)

Similarly 119891L119901120601

le 2119891Lminus119901120601

andLminus

119901120601sub L119901120601

Definition 5 For 120601 equiv 1 denoteL119901120601 andLminus

119901120601by BMO119901 and

BMOminus

119901 respectively For 120601(119903) = 119903

120572 120572 gt 0 denote L119901120601 andLminus

119901120601by Lip

119901120572and Lipminus

119901120572 respectively

If 120601(119903) = 119903120582 120582 isin (minusinfin infin) then we simply denote 119871119901120601

L119901120601 andLminus

119901120601by 119871119901120582L119901120582 andLminus

119901120582 respectively

A function 120601 (0 1] rarr (0 infin) is said to be almostincreasing (resp almost decreasing) if there exists a positiveconstant 119862 such that120601 (119904) le 119862120601 (119905) (resp 120601 (119905) le 119862120601 (119904)) for 0 lt 119904 le 119905 le 1

(5)For the caseB = F the spaces BMO119901(F) and Lip

119901120572(F)

with 120572 ge 0 were introduced by Weisz [3]Recall that119860(F119899) is the set of all atoms inF119899 and letA =

119860(F119899)119899ge0 Suppose that each sub-120590-algebraF119899 is generatedby countable atoms for the time being Then BMO119901(F) =

BMO119901(A) and Lip119901120572

(F) = Lip119901120572

(A) see [5] In general if120601 is almost increasing then

119871119901120601 (F) = 119871119901120601 (A)

L119901120601 (F) = L119901120601 (A)

Lminus

119901120601(F) = L

minus

119901120601(A)

(6)

Journal of Function Spaces and Applications 3

with equivalent norms respectively However if 120601 is notalmost increasing then these equalities fail in general see [5]

In this paper we do not always assume that each sub-120590-algebraF119899 is generated by countable atoms Let

119860(F119899)perp

= 119861 isin F119899 119875 (119861 cap 119860) = 0 forall119860 isin 119860 (F119899) (7)

and let

B119899 = 119860 (F119899) cup 119860(F119899)perp

(119899 ge 0) (8)

In this case if 120601 is almost increasing then we will show that

119871119901120601 (F) = 119871119901120601 (B)

L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B)

(9)

with equivalent norms respectively (see Proposition 9)Moreover ifF0 is nonatomic thenB119899 = F119899 for all 119899 ge 0 Ifeach sub-120590-algebraF119899 is generated by countable atoms thenB119899 = 119860(F119899) for all 119899 ge 0 Therefore our definition gener-alizes those in [3ndash5]

Next we define Morrey-Hardy and Campanato-Hardyspaces based on square functions with respect to B asfollows For 119891 isin M we denote by 119878119899(119891) and 119878(119891) the squarefunction of 119891

119878119899 (119891) = (

119899

sum

119896=0

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

119878 (119891) = (

infin

sum

119896=0

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

(10)

where 119889119896119891 = 119891119896 minus 119891119896minus1 (119899 ge 0 with convention 1198890119891 = 0 and119878minus1(119891) = 0) We further define

119878(119899)

(119891) = (119878(119891)2

minus 119878119899(119891)2)12

= (

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

(11)

Definition 6 Let B sub F 119901 isin (0 infin) and 120601 (0 1] rarr

(0 infin) For 119891 = (119891119899)119899ge0 isin M let

10038171003817100381710038171198911003817100381710038171003817119867119878119901120601

=1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

119878(119891)119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

=1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119878(119899)

(119891)119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

=1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(12)

and define

119867119878

119901120601= 119867

119878

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817119867119878119901120601

lt infin

H119878

119901120601= H

119878

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

lt infin

H119878minus

119901120601= H

119878minus

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

lt infin

(13)

By (1) the functionals 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

arequasinorms onM

Remark 7 If we take 120601 equiv 1 and B = F then the norm119891H119878minus

119901120601

coincides with the norm 119891BMO119878119901

in [19 Definition245] In this point our notation is different from the one in[19]

In the end of this section we present the definitionof regularity on F and the doubling condition on 120601 Thefiltration F = F119899119899ge0 is said to be regular if there existsa constant 119877 ge 2 such that

119891119899 le 119877119891119899minus1 (14)

holds for all nonnegative martingales (119891119899)119899ge0 We say thesmallest constant 119877 satisfying (14) the regularity constant ofF A function 120601 (0 1] rarr (0 infin) is said to satisfy thedoubling condition if there exists a positive constant 119862120601 suchthat

119862minus1

120601le

120601 (119904)

120601 (119905)le 119862120601 forall119904 119905 isin (0 1] with 1

2le

119904

119905le 2 (15)

The smallest constant 119862120601 satisfying (15) is called the doublingconstant of 120601

3 Properties of Morrey-Hardy andCampanato-Hardy Spaces

In this section we investigate the properties ofMorrey-Hardyand Campanato-Hardy spaces The proofs of the results inthis section will be given in Section 6

First we state basic properties of the norms

Proposition 8 Let B sub F 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Let 119891 isin 1198711 and let (119891119899)119899ge0 be its correspondingmartingale 119891119899 = 119864119899119891 Then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

le sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817119871119901120601

10038171003817100381710038171198911003817100381710038171003817L119901120601

= sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817L119901120601

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

= sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817Lminus119901120601

(16)

Moreover if each sub-120590-algebra F119899 is generated by countableatomsB = A and120601 is almost decreasing that is there exists apositive constant 1198620 such that 120601(119905) le 1198620120601(119904) for 0 lt 119904 le 119905 le 1then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

le sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817119871119901120601

le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(17)

4 Journal of Function Spaces and Applications

Proposition 9 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge 0)If 120601 is almost increasing that is there exists a positive constant1198620 such that 120601(119904) le 1198620120601(119905) for 0 lt 119904 le 119905 le 1 then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B)le

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(F)le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B) (18)

and the same conclusions hold for sdot L119901120601

sdot Lminus119901120601

sdot 119867119878119901120601

sdot H119878

119901120601

and sdot H119878minus119901120601

Consequently

119871119901120601 (F) = 119871119901120601 (B) L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B) 119867119901120601 (F) = 119867119901120601 (B)

H119878

119901120601(F) = H

119878

119901120601(B) H

119878minus

119901120601(F) = H

119878minus

119901120601(B)

(19)

with equivalent norms respectively

For 119901 isin (0 infin) let 119867119878

119901be the set of all 119891 isin M such that

119878(119891)119871119901

lt infin Let 119891119867119878119901

= 119878(119891)119871119901

Note that if 120601(119903) =

119903minus1119901 and Ω isin B0 then 119867

119878

119901120601= 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

The following is well known as Burkholderrsquos inequality

Theorem 10 (Burkholder [20]) If 119901 isin (1 infin) then there existpositive constants 119888119901 and 119862119901 that depend only on 119901 such that

1198881199011003817100381710038171003817119891

1003817100381710038171003817119871119901

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901

le 119862119901

10038171003817100381710038171198911003817100381710038171003817119871119901

(20)

for all 119891 isin 1198710

1sub M

For expressions of the constants 119888119901 and 119862119901 see forexample [21ndash23] See also [24] for Burkholderrsquos inequality onBanach functions spaces

Our first result is the following which is an extension ofBurkholderrsquos inequality to martingale Campanato spaces

Theorem 11 Let B sub F 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Then

1198881199011003817100381710038171003817119891

1003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901

10038171003817100381710038171198911003817100381710038171003817L119901120601

(21)

119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(22)

for all 119891 isin 1198710

1sub M where 119888119901 and 119862119901 are the constants in

Theorem 10

Next we give the relations between L119901120601 and Lminus

119901120601and

betweenH119878

119901120601andH119878minus

119901120601We consider the following condition

onB

120596 isin Ω 119864119899minus1 [120594119861] (120596) gt 0 isin B119899minus1 forall119861 isin B119899 (119899 ge 1)

(23)

Theorem 12 LetB sub F and 120601 (0 1] rarr (0 infin) Then

L119901120601 sup Lminus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817L119901120601

le 21003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(24)

H119878

119901120601sup H

119878minus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(25)

Conversely if F is regular B satisfies (23) and 120601 satisfiesthe doubling condition then there exists a positive constant 119862dependent only on 119901 the regularity constant of F and thedoubling constant of 120601 such that

L119901120601 sub Lminus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817L119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(26)

H119878

119901120601sub H

119878minus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(27)

We give a relation betweenmartingaleMorrey spaces andmartingale Campanato spaces in the following form

Theorem 13 Suppose that every 120590-algebraF119899 is generated bycountable atoms Let B = A 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (28)

Then there exists a positive constant 119862 such that

1

2

10038171003817100381710038171198911003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

forall119891 isin 1198710

1

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

forall119891 isin M

(29)

Moreover ifF is regular then 119891119871119901120601

119891L119901120601

and 119891Lminus119901120601

areequivalent to each other and 119891

119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

areequivalent to each other

Using Theorems 11ndash13 we have Burkholder-type equiva-lence for generalized martingale Morrey spaces

Corollary 14 Suppose that every 120590-algebra F119899 is generatedby countable atoms LetB = A 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (30)

IfF is regular then there exist positive constants 119888 and 119862 suchthat for all 119891 isin 119871

0

1

1198881003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817119871119901120601

(31)

For the martingale BMO spaces based on square func-tions the John-Nirenberg-type equivalence was establishedbyWeisz [25] and [19Theorem250]We extend this theoremto the spacesH119878

119901120601andH119878minus

119901120601

Theorem 15 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899 (119899 ge 0)119901 isin (0 infin) and 120601 (0 1] rarr (0 infin) Assume that 120601 is almostincreasing and satisfies the doubling conditionThen 119891H119878

119901120601

le

119862119901119902120601119891H119878minus119902120601

for all 119902 isin (0 infin) If we further assume thatF isregular then 119891H119878

119901120601

and 119891H119878minus119901120601

are equivalent to 119891H1198782120601

Journal of Function Spaces and Applications 5

4 Fractional Integrals

In this section we state the results on the boundedness offractional integrals as martingale transforms The proofs ofthe results in this section will be given in Section 7

Let (120574119899)119899ge0 be a sequence of nonnegative bounded func-tions adapted to F = F119899119899ge0 that is 120574119899 is F119899-measurablefor every 119899 ge 0 Let 119868120574 be the martingale transform associateto (120574119899)119899ge0 that is

(119868120574119891)119899

=

119899

sum

119896=0

120574119896minus1119889119896119891 (32)

with convention 120574minus11198890119891 = 0 Note that if 119891 = (119891119899)119899ge0 isin Mthen 119868120574119891 = ((119868120574119891)119899)119899ge0 isin M

We now define a generalized fractional integral 119868120588 formartingales as a special case of 119868120574 under the assumption thatevery 120590-algebra F119899 is generated by countable atoms Ourdefinition generalizes the fractional integral for dyadic mar-tingales introduced in [6 7] The idea of 119868120588 comes from [26]

Suppose that every 120590-algebra F119899 is generated by count-able atoms Let 119887119899 be anF119899-measurable function such that

119887119899 (120596) = 119875 (119861) for as 120596 isin 119861 with 119861 isin 119860 (F119899) (33)

that is

119887119899 = sum

119861isin119860(F119899)

119875 (119861) 120594119861 as (34)

For a bounded function 120588 (0 1] rarr (0 infin) we define a gen-eralized fractional integral 119868120588119891 = ((119868120588119891)119899)119899ge0 of119891 = (119891119899)119899ge0 isin

M by

(119868120588119891)119899

=

119899

sum

119896=0

120588 (119887119896minus1) 119889119896119891 (35)

The generalized fractional integral 119868120588 is obtained by taking120574119899 = 120588(119887119899) in (32) If 120588(119903) = 119903

120572 120572 gt 0 then we simply denote

119868120588 by 119868120572For quasinormed spaces 1198721 and 1198722 of martingales

we denote by 119861(1198721 1198722) the set of all bounded martingaletransforms from1198721 to1198722 that is119879 isin 119861(1198721 1198722)means thatthere exists a positive constant 119862 such that

100381710038171003817100381711987911989110038171003817100381710038171198722

le 1198621003817100381710038171003817119891

10038171003817100381710038171198721

(36)

for all martingales 119891 = (119891119899)119899ge0 isin 1198721We first study the boundedness on the spaces H119878

119901120601

On martingale Campanato-Hardy spaces we consider thefractional integral as a martingale transform associated withmonotone multipliers We say a sequence of nonnegativemeasurable functions 120574 = (120574119899)119899ge0 is almost decreasing if thereexists a positive constant 119862 such that

120574119896 (120596) le 119862120574ℓ (120596) as forall119896 ge ℓ (37)

For an almost decreasing sequence 120574 = (120574119899)119899ge0 we define 119860120574

by

119860120574 = inf 119862 gt 0 119862 satisfies (37) (38)

InTheorem 16 below we do not need any assumption onF119899119899ge0

Theorem 16 Let B sub F 119901 isin (0 infin) and 120601 120595 (0 1] rarr

(0 infin) Let (120574119899)119899ge0 be a sequence of nonnegative boundedalmost decreasing adapted functions and let 119868120574 be the martin-gale transform defined by (32) Assume that

119862120574120601120595 = sup119899ge0

sup119861isinB119899

120601 (119875 (119861))

120595 (119875 (119861))

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817119871infin

lt infin (39)

Then

119868120574 isin 119861 (H119878

119901120601H

119878

119901120595) (40)

with10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(41)

If every 120590-algebra F119899 is generated by countable atomsthen we can apply Theorem 16 to the generalized fractionalintegral 119868120588 The following corollary extends [5 Theorem 58]to the spacesH119878

119901120601

Corollary 17 Assume that every 120590-algebra F119899 is generatedby countable atoms and B = A Let 119901 isin (0 infin) and 120588 120601 120595

(0 1] rarr (0 infin) Suppose that 120588 is almost increasing and that

sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin (42)

Then

119868120588 isin 119861 (H119878

119901120601H

119878

119901120595) (43)

If one further assumes that F119899119899ge0 is regular and that 120595 isalmost increasing and satisfies the doubling condition then

119868120588 isin 119861 (H119878

119901120601H

119878

119902120595) (119901 119902 isin (0 infin)) (44)

We next study the boundedness on martingale Morrey-Hardy spaces 119867

119878

119901120601and martingale Hardy spaces 119867

119878

119901

Recall that119860(F119899)perp

= F119899 for all 119899 ge 0 ifF0 is nonatomic

Proposition 18 Let B sub F 0 lt 119901 lt 119902 lt infin and 120601

(0 1] rarr (0 infin) Let (120574119899)119899ge0 be a sequence of adapted func-tions Suppose that F0 is nonatomic and that B = FAssume in addition that 120601 is almost decreasing that 119905

1119901120601(119905)

is almost increasing and that lim119905rarr0120601(119905) = infin Then 119868120574 notin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

According to Proposition 18 to consider the bounded-ness on 119867

119878

119901120601and 119867

119878

119901 we suppose that every 120590-algebra F119899

is generated by countable atoms and thatB = AIn this case if 120601(119903) = 119903

minus1119901 and F0 = Ω 0 then119867

119878

119901120601coincides with 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

However ifF0 = Ω 0 then 119867

119878

119901120601does not coincide with 119867

119878

119901in general

We do not always assume thatF0 = Ω 0

Theorem 19 Suppose that every 120590-algebraF119899 is generated bycountable atoms thatB = A and that F119899119899ge0 is regular Let

6 Journal of Function Spaces and Applications

0 lt 119901 lt 119902 lt infin and 120601 (0 1] rarr (0 infin) and let (120574119899)119899ge0 bea sequence of nonnegative bounded adapted functions Assumethat 120601 satisfies the doubling condition and that there exists apositive constant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902 as

(45)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then

119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (46)

Furthermore if 120601(119905) = 119905minus1119901 then

119868120574 isin 119861 (119867119878

119901 119867

119878

119902) (47)

As a consequence of Theorem 19 we have the followingcorollary which gives an extension of [5 Corollary 57] to thespaces 119867

119878

119901120601and gives a martingale Morrey-Hardy version of

Gunawan [27 Theorem B]

Corollary 20 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and 120588 120601 (0 1] rarr (0 infin) Assume that120588 is bounded that both 120588 and 120601 satisfy the doubling conditionand that there exists a positive constant 119862 such that

120601 (119903) int

119903

0

120588 (119905)

119905119889119905 + int

1

119903

120601 (119905) 120588 (119905)

119905119889119905 le 119862120601(119903)

119901119902

(0 lt 119903 lt 1)

(48)

Then

119868120588 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (49)

The following extends the results for dyadic martingalesin [6 7] and the result for 0 lt 119901 le 1 in [28]

Corollary 21 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and minus1119901 + 120572 = minus1119902 Then

119868120572 isin 119861 (119867119878

119901 119867

119878

119902) (50)

5 Lemmas

We prepare some lemmas to prove the results in Sections 3and 4

Lemma 1 LetB119899 satisfy 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge

0) Suppose that 120601 (0 1] rarr (0 infin) is almost increasing that

is120601(119903) le 1198620120601(119904) for all 0 lt 119903 le 119904 le 1Then for all nonnegativefunctions 119865

sup119861isinF119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

le 1198620 sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

(51)

Proof Let

N = sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875) (52)

For any 119861 isin F119899 we can choose the sets 119861119895 119895 = 0 1 2

(finite or infinite) such that

119861 = cup119895119861119895 1198610 isin 119860(F119899)perp

119861119895 isin 119860 (F119899)

119895 = 1 2

119875 (119861) = sum

119895

119875 (119861119895)

(53)

In this case 119861119895 isin B119899 119895 = 0 1 2 since119860(F119899)cup119860(F119899)perp

sub

B119899 Then

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

=1

120601 (119875 (119861))

1

119875 (119861)sum

119895

(int119861119895

119865 119889119875)

= sum

119895

119875 (119861119895)

119875 (119861)(

1

120601 (119875 (119861)) 119875 (119861119895)

int119861119895

119865 119889119875)

le sum

119895

119875 (119861119895)

119875 (119861)(

1198620

120601 (119875 (119861119895)) 119875 (119861119895)

int119861119895

119865 119889119875)

le 1198620sum

119895

119875 (119861119895)

119875 (119861)N

= 1198620N

(54)

This shows the conclusion

Lemma 2 (see [5 Lemma 33]) Suppose that every 120590-algebraF119899 is generated by countable atoms and that F119899119899ge0 is regularThen every sequence

1198610 sup 1198611 sup sdot sdot sdot sup 119861119899 sup sdot sdot sdot 119861119899 isin 119860 (F119899) (55)

has the following property for each 119899 ge 1

119861119899 = 119861119899minus1 119900119903 (1 +1

119877) 119875 (119861119899) le 119875 (119861119899minus1) le 119877119875 (119861119899)

(56)

where 119877 is the constant in (14)

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Journal of Function Spaces and Applications 3

with equivalent norms respectively However if 120601 is notalmost increasing then these equalities fail in general see [5]

In this paper we do not always assume that each sub-120590-algebraF119899 is generated by countable atoms Let

119860(F119899)perp

= 119861 isin F119899 119875 (119861 cap 119860) = 0 forall119860 isin 119860 (F119899) (7)

and let

B119899 = 119860 (F119899) cup 119860(F119899)perp

(119899 ge 0) (8)

In this case if 120601 is almost increasing then we will show that

119871119901120601 (F) = 119871119901120601 (B)

L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B)

(9)

with equivalent norms respectively (see Proposition 9)Moreover ifF0 is nonatomic thenB119899 = F119899 for all 119899 ge 0 Ifeach sub-120590-algebraF119899 is generated by countable atoms thenB119899 = 119860(F119899) for all 119899 ge 0 Therefore our definition gener-alizes those in [3ndash5]

Next we define Morrey-Hardy and Campanato-Hardyspaces based on square functions with respect to B asfollows For 119891 isin M we denote by 119878119899(119891) and 119878(119891) the squarefunction of 119891

119878119899 (119891) = (

119899

sum

119896=0

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

119878 (119891) = (

infin

sum

119896=0

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

(10)

where 119889119896119891 = 119891119896 minus 119891119896minus1 (119899 ge 0 with convention 1198890119891 = 0 and119878minus1(119891) = 0) We further define

119878(119899)

(119891) = (119878(119891)2

minus 119878119899(119891)2)12

= (

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2)

12

(11)

Definition 6 Let B sub F 119901 isin (0 infin) and 120601 (0 1] rarr

(0 infin) For 119891 = (119891119899)119899ge0 isin M let

10038171003817100381710038171198911003817100381710038171003817119867119878119901120601

=1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875(119861)int119861

119878(119891)119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

=1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119878(119899)

(119891)119901119889119875)

1119901

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

=1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(B)

= sup119899ge0

sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(12)

and define

119867119878

119901120601= 119867

119878

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817119867119878119901120601

lt infin

H119878

119901120601= H

119878

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

lt infin

H119878minus

119901120601= H

119878minus

119901120601(B) = 119891 isin M

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

lt infin

(13)

By (1) the functionals 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

arequasinorms onM

Remark 7 If we take 120601 equiv 1 and B = F then the norm119891H119878minus

119901120601

coincides with the norm 119891BMO119878119901

in [19 Definition245] In this point our notation is different from the one in[19]

In the end of this section we present the definitionof regularity on F and the doubling condition on 120601 Thefiltration F = F119899119899ge0 is said to be regular if there existsa constant 119877 ge 2 such that

119891119899 le 119877119891119899minus1 (14)

holds for all nonnegative martingales (119891119899)119899ge0 We say thesmallest constant 119877 satisfying (14) the regularity constant ofF A function 120601 (0 1] rarr (0 infin) is said to satisfy thedoubling condition if there exists a positive constant 119862120601 suchthat

119862minus1

120601le

120601 (119904)

120601 (119905)le 119862120601 forall119904 119905 isin (0 1] with 1

2le

119904

119905le 2 (15)

The smallest constant 119862120601 satisfying (15) is called the doublingconstant of 120601

3 Properties of Morrey-Hardy andCampanato-Hardy Spaces

In this section we investigate the properties ofMorrey-Hardyand Campanato-Hardy spaces The proofs of the results inthis section will be given in Section 6

First we state basic properties of the norms

Proposition 8 Let B sub F 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Let 119891 isin 1198711 and let (119891119899)119899ge0 be its correspondingmartingale 119891119899 = 119864119899119891 Then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

le sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817119871119901120601

10038171003817100381710038171198911003817100381710038171003817L119901120601

= sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817L119901120601

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

= sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817Lminus119901120601

(16)

Moreover if each sub-120590-algebra F119899 is generated by countableatomsB = A and120601 is almost decreasing that is there exists apositive constant 1198620 such that 120601(119905) le 1198620120601(119904) for 0 lt 119904 le 119905 le 1then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

le sup119899ge0

1003817100381710038171003817119891119899

1003817100381710038171003817119871119901120601

le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(17)

4 Journal of Function Spaces and Applications

Proposition 9 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge 0)If 120601 is almost increasing that is there exists a positive constant1198620 such that 120601(119904) le 1198620120601(119905) for 0 lt 119904 le 119905 le 1 then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B)le

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(F)le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B) (18)

and the same conclusions hold for sdot L119901120601

sdot Lminus119901120601

sdot 119867119878119901120601

sdot H119878

119901120601

and sdot H119878minus119901120601

Consequently

119871119901120601 (F) = 119871119901120601 (B) L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B) 119867119901120601 (F) = 119867119901120601 (B)

H119878

119901120601(F) = H

119878

119901120601(B) H

119878minus

119901120601(F) = H

119878minus

119901120601(B)

(19)

with equivalent norms respectively

For 119901 isin (0 infin) let 119867119878

119901be the set of all 119891 isin M such that

119878(119891)119871119901

lt infin Let 119891119867119878119901

= 119878(119891)119871119901

Note that if 120601(119903) =

119903minus1119901 and Ω isin B0 then 119867

119878

119901120601= 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

The following is well known as Burkholderrsquos inequality

Theorem 10 (Burkholder [20]) If 119901 isin (1 infin) then there existpositive constants 119888119901 and 119862119901 that depend only on 119901 such that

1198881199011003817100381710038171003817119891

1003817100381710038171003817119871119901

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901

le 119862119901

10038171003817100381710038171198911003817100381710038171003817119871119901

(20)

for all 119891 isin 1198710

1sub M

For expressions of the constants 119888119901 and 119862119901 see forexample [21ndash23] See also [24] for Burkholderrsquos inequality onBanach functions spaces

Our first result is the following which is an extension ofBurkholderrsquos inequality to martingale Campanato spaces

Theorem 11 Let B sub F 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Then

1198881199011003817100381710038171003817119891

1003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901

10038171003817100381710038171198911003817100381710038171003817L119901120601

(21)

119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(22)

for all 119891 isin 1198710

1sub M where 119888119901 and 119862119901 are the constants in

Theorem 10

Next we give the relations between L119901120601 and Lminus

119901120601and

betweenH119878

119901120601andH119878minus

119901120601We consider the following condition

onB

120596 isin Ω 119864119899minus1 [120594119861] (120596) gt 0 isin B119899minus1 forall119861 isin B119899 (119899 ge 1)

(23)

Theorem 12 LetB sub F and 120601 (0 1] rarr (0 infin) Then

L119901120601 sup Lminus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817L119901120601

le 21003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(24)

H119878

119901120601sup H

119878minus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(25)

Conversely if F is regular B satisfies (23) and 120601 satisfiesthe doubling condition then there exists a positive constant 119862dependent only on 119901 the regularity constant of F and thedoubling constant of 120601 such that

L119901120601 sub Lminus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817L119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(26)

H119878

119901120601sub H

119878minus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(27)

We give a relation betweenmartingaleMorrey spaces andmartingale Campanato spaces in the following form

Theorem 13 Suppose that every 120590-algebraF119899 is generated bycountable atoms Let B = A 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (28)

Then there exists a positive constant 119862 such that

1

2

10038171003817100381710038171198911003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

forall119891 isin 1198710

1

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

forall119891 isin M

(29)

Moreover ifF is regular then 119891119871119901120601

119891L119901120601

and 119891Lminus119901120601

areequivalent to each other and 119891

119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

areequivalent to each other

Using Theorems 11ndash13 we have Burkholder-type equiva-lence for generalized martingale Morrey spaces

Corollary 14 Suppose that every 120590-algebra F119899 is generatedby countable atoms LetB = A 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (30)

IfF is regular then there exist positive constants 119888 and 119862 suchthat for all 119891 isin 119871

0

1

1198881003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817119871119901120601

(31)

For the martingale BMO spaces based on square func-tions the John-Nirenberg-type equivalence was establishedbyWeisz [25] and [19Theorem250]We extend this theoremto the spacesH119878

119901120601andH119878minus

119901120601

Theorem 15 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899 (119899 ge 0)119901 isin (0 infin) and 120601 (0 1] rarr (0 infin) Assume that 120601 is almostincreasing and satisfies the doubling conditionThen 119891H119878

119901120601

le

119862119901119902120601119891H119878minus119902120601

for all 119902 isin (0 infin) If we further assume thatF isregular then 119891H119878

119901120601

and 119891H119878minus119901120601

are equivalent to 119891H1198782120601

Journal of Function Spaces and Applications 5

4 Fractional Integrals

In this section we state the results on the boundedness offractional integrals as martingale transforms The proofs ofthe results in this section will be given in Section 7

Let (120574119899)119899ge0 be a sequence of nonnegative bounded func-tions adapted to F = F119899119899ge0 that is 120574119899 is F119899-measurablefor every 119899 ge 0 Let 119868120574 be the martingale transform associateto (120574119899)119899ge0 that is

(119868120574119891)119899

=

119899

sum

119896=0

120574119896minus1119889119896119891 (32)

with convention 120574minus11198890119891 = 0 Note that if 119891 = (119891119899)119899ge0 isin Mthen 119868120574119891 = ((119868120574119891)119899)119899ge0 isin M

We now define a generalized fractional integral 119868120588 formartingales as a special case of 119868120574 under the assumption thatevery 120590-algebra F119899 is generated by countable atoms Ourdefinition generalizes the fractional integral for dyadic mar-tingales introduced in [6 7] The idea of 119868120588 comes from [26]

Suppose that every 120590-algebra F119899 is generated by count-able atoms Let 119887119899 be anF119899-measurable function such that

119887119899 (120596) = 119875 (119861) for as 120596 isin 119861 with 119861 isin 119860 (F119899) (33)

that is

119887119899 = sum

119861isin119860(F119899)

119875 (119861) 120594119861 as (34)

For a bounded function 120588 (0 1] rarr (0 infin) we define a gen-eralized fractional integral 119868120588119891 = ((119868120588119891)119899)119899ge0 of119891 = (119891119899)119899ge0 isin

M by

(119868120588119891)119899

=

119899

sum

119896=0

120588 (119887119896minus1) 119889119896119891 (35)

The generalized fractional integral 119868120588 is obtained by taking120574119899 = 120588(119887119899) in (32) If 120588(119903) = 119903

120572 120572 gt 0 then we simply denote

119868120588 by 119868120572For quasinormed spaces 1198721 and 1198722 of martingales

we denote by 119861(1198721 1198722) the set of all bounded martingaletransforms from1198721 to1198722 that is119879 isin 119861(1198721 1198722)means thatthere exists a positive constant 119862 such that

100381710038171003817100381711987911989110038171003817100381710038171198722

le 1198621003817100381710038171003817119891

10038171003817100381710038171198721

(36)

for all martingales 119891 = (119891119899)119899ge0 isin 1198721We first study the boundedness on the spaces H119878

119901120601

On martingale Campanato-Hardy spaces we consider thefractional integral as a martingale transform associated withmonotone multipliers We say a sequence of nonnegativemeasurable functions 120574 = (120574119899)119899ge0 is almost decreasing if thereexists a positive constant 119862 such that

120574119896 (120596) le 119862120574ℓ (120596) as forall119896 ge ℓ (37)

For an almost decreasing sequence 120574 = (120574119899)119899ge0 we define 119860120574

by

119860120574 = inf 119862 gt 0 119862 satisfies (37) (38)

InTheorem 16 below we do not need any assumption onF119899119899ge0

Theorem 16 Let B sub F 119901 isin (0 infin) and 120601 120595 (0 1] rarr

(0 infin) Let (120574119899)119899ge0 be a sequence of nonnegative boundedalmost decreasing adapted functions and let 119868120574 be the martin-gale transform defined by (32) Assume that

119862120574120601120595 = sup119899ge0

sup119861isinB119899

120601 (119875 (119861))

120595 (119875 (119861))

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817119871infin

lt infin (39)

Then

119868120574 isin 119861 (H119878

119901120601H

119878

119901120595) (40)

with10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(41)

If every 120590-algebra F119899 is generated by countable atomsthen we can apply Theorem 16 to the generalized fractionalintegral 119868120588 The following corollary extends [5 Theorem 58]to the spacesH119878

119901120601

Corollary 17 Assume that every 120590-algebra F119899 is generatedby countable atoms and B = A Let 119901 isin (0 infin) and 120588 120601 120595

(0 1] rarr (0 infin) Suppose that 120588 is almost increasing and that

sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin (42)

Then

119868120588 isin 119861 (H119878

119901120601H

119878

119901120595) (43)

If one further assumes that F119899119899ge0 is regular and that 120595 isalmost increasing and satisfies the doubling condition then

119868120588 isin 119861 (H119878

119901120601H

119878

119902120595) (119901 119902 isin (0 infin)) (44)

We next study the boundedness on martingale Morrey-Hardy spaces 119867

119878

119901120601and martingale Hardy spaces 119867

119878

119901

Recall that119860(F119899)perp

= F119899 for all 119899 ge 0 ifF0 is nonatomic

Proposition 18 Let B sub F 0 lt 119901 lt 119902 lt infin and 120601

(0 1] rarr (0 infin) Let (120574119899)119899ge0 be a sequence of adapted func-tions Suppose that F0 is nonatomic and that B = FAssume in addition that 120601 is almost decreasing that 119905

1119901120601(119905)

is almost increasing and that lim119905rarr0120601(119905) = infin Then 119868120574 notin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

According to Proposition 18 to consider the bounded-ness on 119867

119878

119901120601and 119867

119878

119901 we suppose that every 120590-algebra F119899

is generated by countable atoms and thatB = AIn this case if 120601(119903) = 119903

minus1119901 and F0 = Ω 0 then119867

119878

119901120601coincides with 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

However ifF0 = Ω 0 then 119867

119878

119901120601does not coincide with 119867

119878

119901in general

We do not always assume thatF0 = Ω 0

Theorem 19 Suppose that every 120590-algebraF119899 is generated bycountable atoms thatB = A and that F119899119899ge0 is regular Let

6 Journal of Function Spaces and Applications

0 lt 119901 lt 119902 lt infin and 120601 (0 1] rarr (0 infin) and let (120574119899)119899ge0 bea sequence of nonnegative bounded adapted functions Assumethat 120601 satisfies the doubling condition and that there exists apositive constant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902 as

(45)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then

119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (46)

Furthermore if 120601(119905) = 119905minus1119901 then

119868120574 isin 119861 (119867119878

119901 119867

119878

119902) (47)

As a consequence of Theorem 19 we have the followingcorollary which gives an extension of [5 Corollary 57] to thespaces 119867

119878

119901120601and gives a martingale Morrey-Hardy version of

Gunawan [27 Theorem B]

Corollary 20 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and 120588 120601 (0 1] rarr (0 infin) Assume that120588 is bounded that both 120588 and 120601 satisfy the doubling conditionand that there exists a positive constant 119862 such that

120601 (119903) int

119903

0

120588 (119905)

119905119889119905 + int

1

119903

120601 (119905) 120588 (119905)

119905119889119905 le 119862120601(119903)

119901119902

(0 lt 119903 lt 1)

(48)

Then

119868120588 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (49)

The following extends the results for dyadic martingalesin [6 7] and the result for 0 lt 119901 le 1 in [28]

Corollary 21 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and minus1119901 + 120572 = minus1119902 Then

119868120572 isin 119861 (119867119878

119901 119867

119878

119902) (50)

5 Lemmas

We prepare some lemmas to prove the results in Sections 3and 4

Lemma 1 LetB119899 satisfy 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge

0) Suppose that 120601 (0 1] rarr (0 infin) is almost increasing that

is120601(119903) le 1198620120601(119904) for all 0 lt 119903 le 119904 le 1Then for all nonnegativefunctions 119865

sup119861isinF119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

le 1198620 sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

(51)

Proof Let

N = sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875) (52)

For any 119861 isin F119899 we can choose the sets 119861119895 119895 = 0 1 2

(finite or infinite) such that

119861 = cup119895119861119895 1198610 isin 119860(F119899)perp

119861119895 isin 119860 (F119899)

119895 = 1 2

119875 (119861) = sum

119895

119875 (119861119895)

(53)

In this case 119861119895 isin B119899 119895 = 0 1 2 since119860(F119899)cup119860(F119899)perp

sub

B119899 Then

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

=1

120601 (119875 (119861))

1

119875 (119861)sum

119895

(int119861119895

119865 119889119875)

= sum

119895

119875 (119861119895)

119875 (119861)(

1

120601 (119875 (119861)) 119875 (119861119895)

int119861119895

119865 119889119875)

le sum

119895

119875 (119861119895)

119875 (119861)(

1198620

120601 (119875 (119861119895)) 119875 (119861119895)

int119861119895

119865 119889119875)

le 1198620sum

119895

119875 (119861119895)

119875 (119861)N

= 1198620N

(54)

This shows the conclusion

Lemma 2 (see [5 Lemma 33]) Suppose that every 120590-algebraF119899 is generated by countable atoms and that F119899119899ge0 is regularThen every sequence

1198610 sup 1198611 sup sdot sdot sdot sup 119861119899 sup sdot sdot sdot 119861119899 isin 119860 (F119899) (55)

has the following property for each 119899 ge 1

119861119899 = 119861119899minus1 119900119903 (1 +1

119877) 119875 (119861119899) le 119875 (119861119899minus1) le 119877119875 (119861119899)

(56)

where 119877 is the constant in (14)

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

4 Journal of Function Spaces and Applications

Proposition 9 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge 0)If 120601 is almost increasing that is there exists a positive constant1198620 such that 120601(119904) le 1198620120601(119905) for 0 lt 119904 le 119905 le 1 then

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B)le

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(F)le 1198620

10038171003817100381710038171198911003817100381710038171003817119871119901120601

(B) (18)

and the same conclusions hold for sdot L119901120601

sdot Lminus119901120601

sdot 119867119878119901120601

sdot H119878

119901120601

and sdot H119878minus119901120601

Consequently

119871119901120601 (F) = 119871119901120601 (B) L119901120601 (F) = L119901120601 (B)

Lminus

119901120601(F) = L

minus

119901120601(B) 119867119901120601 (F) = 119867119901120601 (B)

H119878

119901120601(F) = H

119878

119901120601(B) H

119878minus

119901120601(F) = H

119878minus

119901120601(B)

(19)

with equivalent norms respectively

For 119901 isin (0 infin) let 119867119878

119901be the set of all 119891 isin M such that

119878(119891)119871119901

lt infin Let 119891119867119878119901

= 119878(119891)119871119901

Note that if 120601(119903) =

119903minus1119901 and Ω isin B0 then 119867

119878

119901120601= 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

The following is well known as Burkholderrsquos inequality

Theorem 10 (Burkholder [20]) If 119901 isin (1 infin) then there existpositive constants 119888119901 and 119862119901 that depend only on 119901 such that

1198881199011003817100381710038171003817119891

1003817100381710038171003817119871119901

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901

le 119862119901

10038171003817100381710038171198911003817100381710038171003817119871119901

(20)

for all 119891 isin 1198710

1sub M

For expressions of the constants 119888119901 and 119862119901 see forexample [21ndash23] See also [24] for Burkholderrsquos inequality onBanach functions spaces

Our first result is the following which is an extension ofBurkholderrsquos inequality to martingale Campanato spaces

Theorem 11 Let B sub F 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Then

1198881199011003817100381710038171003817119891

1003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901

10038171003817100381710038171198911003817100381710038171003817L119901120601

(21)

119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(22)

for all 119891 isin 1198710

1sub M where 119888119901 and 119862119901 are the constants in

Theorem 10

Next we give the relations between L119901120601 and Lminus

119901120601and

betweenH119878

119901120601andH119878minus

119901120601We consider the following condition

onB

120596 isin Ω 119864119899minus1 [120594119861] (120596) gt 0 isin B119899minus1 forall119861 isin B119899 (119899 ge 1)

(23)

Theorem 12 LetB sub F and 120601 (0 1] rarr (0 infin) Then

L119901120601 sup Lminus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817L119901120601

le 21003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(24)

H119878

119901120601sup H

119878minus

119901120601119908119894119905ℎ

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(25)

Conversely if F is regular B satisfies (23) and 120601 satisfiesthe doubling condition then there exists a positive constant 119862dependent only on 119901 the regularity constant of F and thedoubling constant of 120601 such that

L119901120601 sub Lminus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817L119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

119891119900119903 119901 isin [1 infin)

(26)

H119878

119901120601sub H

119878minus

119901120601119908119894119905ℎ 119862

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

ge1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

119891119900119903 119901 isin (0 infin)

(27)

We give a relation betweenmartingaleMorrey spaces andmartingale Campanato spaces in the following form

Theorem 13 Suppose that every 120590-algebraF119899 is generated bycountable atoms Let B = A 119901 isin [1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (28)

Then there exists a positive constant 119862 such that

1

2

10038171003817100381710038171198911003817100381710038171003817L119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

forall119891 isin 1198710

1

10038171003817100381710038171198911003817100381710038171003817H119878minus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

forall119891 isin M

(29)

Moreover ifF is regular then 119891119871119901120601

119891L119901120601

and 119891Lminus119901120601

areequivalent to each other and 119891

119867119878119901120601

119891H119878119901120601

and 119891H119878minus119901120601

areequivalent to each other

Using Theorems 11ndash13 we have Burkholder-type equiva-lence for generalized martingale Morrey spaces

Corollary 14 Suppose that every 120590-algebra F119899 is generatedby countable atoms LetB = A 119901 isin (1 infin) and 120601 (0 1] rarr

(0 infin) Assume that120601 satisfies the doubling condition and thereexists a positive constant 119862

1015840

120601such that

int

1

119903

120601 (119905)

119905119889119905 le 119862

1015840

120601120601 (119903) (0 lt 119903 lt 1) (30)

IfF is regular then there exist positive constants 119888 and 119862 suchthat for all 119891 isin 119871

0

1

1198881003817100381710038171003817119891

1003817100381710038171003817119871119901120601

le1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

le 1198621003817100381710038171003817119891

1003817100381710038171003817119871119901120601

(31)

For the martingale BMO spaces based on square func-tions the John-Nirenberg-type equivalence was establishedbyWeisz [25] and [19Theorem250]We extend this theoremto the spacesH119878

119901120601andH119878minus

119901120601

Theorem 15 Let 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899 (119899 ge 0)119901 isin (0 infin) and 120601 (0 1] rarr (0 infin) Assume that 120601 is almostincreasing and satisfies the doubling conditionThen 119891H119878

119901120601

le

119862119901119902120601119891H119878minus119902120601

for all 119902 isin (0 infin) If we further assume thatF isregular then 119891H119878

119901120601

and 119891H119878minus119901120601

are equivalent to 119891H1198782120601

Journal of Function Spaces and Applications 5

4 Fractional Integrals

In this section we state the results on the boundedness offractional integrals as martingale transforms The proofs ofthe results in this section will be given in Section 7

Let (120574119899)119899ge0 be a sequence of nonnegative bounded func-tions adapted to F = F119899119899ge0 that is 120574119899 is F119899-measurablefor every 119899 ge 0 Let 119868120574 be the martingale transform associateto (120574119899)119899ge0 that is

(119868120574119891)119899

=

119899

sum

119896=0

120574119896minus1119889119896119891 (32)

with convention 120574minus11198890119891 = 0 Note that if 119891 = (119891119899)119899ge0 isin Mthen 119868120574119891 = ((119868120574119891)119899)119899ge0 isin M

We now define a generalized fractional integral 119868120588 formartingales as a special case of 119868120574 under the assumption thatevery 120590-algebra F119899 is generated by countable atoms Ourdefinition generalizes the fractional integral for dyadic mar-tingales introduced in [6 7] The idea of 119868120588 comes from [26]

Suppose that every 120590-algebra F119899 is generated by count-able atoms Let 119887119899 be anF119899-measurable function such that

119887119899 (120596) = 119875 (119861) for as 120596 isin 119861 with 119861 isin 119860 (F119899) (33)

that is

119887119899 = sum

119861isin119860(F119899)

119875 (119861) 120594119861 as (34)

For a bounded function 120588 (0 1] rarr (0 infin) we define a gen-eralized fractional integral 119868120588119891 = ((119868120588119891)119899)119899ge0 of119891 = (119891119899)119899ge0 isin

M by

(119868120588119891)119899

=

119899

sum

119896=0

120588 (119887119896minus1) 119889119896119891 (35)

The generalized fractional integral 119868120588 is obtained by taking120574119899 = 120588(119887119899) in (32) If 120588(119903) = 119903

120572 120572 gt 0 then we simply denote

119868120588 by 119868120572For quasinormed spaces 1198721 and 1198722 of martingales

we denote by 119861(1198721 1198722) the set of all bounded martingaletransforms from1198721 to1198722 that is119879 isin 119861(1198721 1198722)means thatthere exists a positive constant 119862 such that

100381710038171003817100381711987911989110038171003817100381710038171198722

le 1198621003817100381710038171003817119891

10038171003817100381710038171198721

(36)

for all martingales 119891 = (119891119899)119899ge0 isin 1198721We first study the boundedness on the spaces H119878

119901120601

On martingale Campanato-Hardy spaces we consider thefractional integral as a martingale transform associated withmonotone multipliers We say a sequence of nonnegativemeasurable functions 120574 = (120574119899)119899ge0 is almost decreasing if thereexists a positive constant 119862 such that

120574119896 (120596) le 119862120574ℓ (120596) as forall119896 ge ℓ (37)

For an almost decreasing sequence 120574 = (120574119899)119899ge0 we define 119860120574

by

119860120574 = inf 119862 gt 0 119862 satisfies (37) (38)

InTheorem 16 below we do not need any assumption onF119899119899ge0

Theorem 16 Let B sub F 119901 isin (0 infin) and 120601 120595 (0 1] rarr

(0 infin) Let (120574119899)119899ge0 be a sequence of nonnegative boundedalmost decreasing adapted functions and let 119868120574 be the martin-gale transform defined by (32) Assume that

119862120574120601120595 = sup119899ge0

sup119861isinB119899

120601 (119875 (119861))

120595 (119875 (119861))

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817119871infin

lt infin (39)

Then

119868120574 isin 119861 (H119878

119901120601H

119878

119901120595) (40)

with10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(41)

If every 120590-algebra F119899 is generated by countable atomsthen we can apply Theorem 16 to the generalized fractionalintegral 119868120588 The following corollary extends [5 Theorem 58]to the spacesH119878

119901120601

Corollary 17 Assume that every 120590-algebra F119899 is generatedby countable atoms and B = A Let 119901 isin (0 infin) and 120588 120601 120595

(0 1] rarr (0 infin) Suppose that 120588 is almost increasing and that

sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin (42)

Then

119868120588 isin 119861 (H119878

119901120601H

119878

119901120595) (43)

If one further assumes that F119899119899ge0 is regular and that 120595 isalmost increasing and satisfies the doubling condition then

119868120588 isin 119861 (H119878

119901120601H

119878

119902120595) (119901 119902 isin (0 infin)) (44)

We next study the boundedness on martingale Morrey-Hardy spaces 119867

119878

119901120601and martingale Hardy spaces 119867

119878

119901

Recall that119860(F119899)perp

= F119899 for all 119899 ge 0 ifF0 is nonatomic

Proposition 18 Let B sub F 0 lt 119901 lt 119902 lt infin and 120601

(0 1] rarr (0 infin) Let (120574119899)119899ge0 be a sequence of adapted func-tions Suppose that F0 is nonatomic and that B = FAssume in addition that 120601 is almost decreasing that 119905

1119901120601(119905)

is almost increasing and that lim119905rarr0120601(119905) = infin Then 119868120574 notin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

According to Proposition 18 to consider the bounded-ness on 119867

119878

119901120601and 119867

119878

119901 we suppose that every 120590-algebra F119899

is generated by countable atoms and thatB = AIn this case if 120601(119903) = 119903

minus1119901 and F0 = Ω 0 then119867

119878

119901120601coincides with 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

However ifF0 = Ω 0 then 119867

119878

119901120601does not coincide with 119867

119878

119901in general

We do not always assume thatF0 = Ω 0

Theorem 19 Suppose that every 120590-algebraF119899 is generated bycountable atoms thatB = A and that F119899119899ge0 is regular Let

6 Journal of Function Spaces and Applications

0 lt 119901 lt 119902 lt infin and 120601 (0 1] rarr (0 infin) and let (120574119899)119899ge0 bea sequence of nonnegative bounded adapted functions Assumethat 120601 satisfies the doubling condition and that there exists apositive constant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902 as

(45)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then

119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (46)

Furthermore if 120601(119905) = 119905minus1119901 then

119868120574 isin 119861 (119867119878

119901 119867

119878

119902) (47)

As a consequence of Theorem 19 we have the followingcorollary which gives an extension of [5 Corollary 57] to thespaces 119867

119878

119901120601and gives a martingale Morrey-Hardy version of

Gunawan [27 Theorem B]

Corollary 20 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and 120588 120601 (0 1] rarr (0 infin) Assume that120588 is bounded that both 120588 and 120601 satisfy the doubling conditionand that there exists a positive constant 119862 such that

120601 (119903) int

119903

0

120588 (119905)

119905119889119905 + int

1

119903

120601 (119905) 120588 (119905)

119905119889119905 le 119862120601(119903)

119901119902

(0 lt 119903 lt 1)

(48)

Then

119868120588 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (49)

The following extends the results for dyadic martingalesin [6 7] and the result for 0 lt 119901 le 1 in [28]

Corollary 21 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and minus1119901 + 120572 = minus1119902 Then

119868120572 isin 119861 (119867119878

119901 119867

119878

119902) (50)

5 Lemmas

We prepare some lemmas to prove the results in Sections 3and 4

Lemma 1 LetB119899 satisfy 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge

0) Suppose that 120601 (0 1] rarr (0 infin) is almost increasing that

is120601(119903) le 1198620120601(119904) for all 0 lt 119903 le 119904 le 1Then for all nonnegativefunctions 119865

sup119861isinF119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

le 1198620 sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

(51)

Proof Let

N = sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875) (52)

For any 119861 isin F119899 we can choose the sets 119861119895 119895 = 0 1 2

(finite or infinite) such that

119861 = cup119895119861119895 1198610 isin 119860(F119899)perp

119861119895 isin 119860 (F119899)

119895 = 1 2

119875 (119861) = sum

119895

119875 (119861119895)

(53)

In this case 119861119895 isin B119899 119895 = 0 1 2 since119860(F119899)cup119860(F119899)perp

sub

B119899 Then

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

=1

120601 (119875 (119861))

1

119875 (119861)sum

119895

(int119861119895

119865 119889119875)

= sum

119895

119875 (119861119895)

119875 (119861)(

1

120601 (119875 (119861)) 119875 (119861119895)

int119861119895

119865 119889119875)

le sum

119895

119875 (119861119895)

119875 (119861)(

1198620

120601 (119875 (119861119895)) 119875 (119861119895)

int119861119895

119865 119889119875)

le 1198620sum

119895

119875 (119861119895)

119875 (119861)N

= 1198620N

(54)

This shows the conclusion

Lemma 2 (see [5 Lemma 33]) Suppose that every 120590-algebraF119899 is generated by countable atoms and that F119899119899ge0 is regularThen every sequence

1198610 sup 1198611 sup sdot sdot sdot sup 119861119899 sup sdot sdot sdot 119861119899 isin 119860 (F119899) (55)

has the following property for each 119899 ge 1

119861119899 = 119861119899minus1 119900119903 (1 +1

119877) 119875 (119861119899) le 119875 (119861119899minus1) le 119877119875 (119861119899)

(56)

where 119877 is the constant in (14)

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Journal of Function Spaces and Applications 5

4 Fractional Integrals

In this section we state the results on the boundedness offractional integrals as martingale transforms The proofs ofthe results in this section will be given in Section 7

Let (120574119899)119899ge0 be a sequence of nonnegative bounded func-tions adapted to F = F119899119899ge0 that is 120574119899 is F119899-measurablefor every 119899 ge 0 Let 119868120574 be the martingale transform associateto (120574119899)119899ge0 that is

(119868120574119891)119899

=

119899

sum

119896=0

120574119896minus1119889119896119891 (32)

with convention 120574minus11198890119891 = 0 Note that if 119891 = (119891119899)119899ge0 isin Mthen 119868120574119891 = ((119868120574119891)119899)119899ge0 isin M

We now define a generalized fractional integral 119868120588 formartingales as a special case of 119868120574 under the assumption thatevery 120590-algebra F119899 is generated by countable atoms Ourdefinition generalizes the fractional integral for dyadic mar-tingales introduced in [6 7] The idea of 119868120588 comes from [26]

Suppose that every 120590-algebra F119899 is generated by count-able atoms Let 119887119899 be anF119899-measurable function such that

119887119899 (120596) = 119875 (119861) for as 120596 isin 119861 with 119861 isin 119860 (F119899) (33)

that is

119887119899 = sum

119861isin119860(F119899)

119875 (119861) 120594119861 as (34)

For a bounded function 120588 (0 1] rarr (0 infin) we define a gen-eralized fractional integral 119868120588119891 = ((119868120588119891)119899)119899ge0 of119891 = (119891119899)119899ge0 isin

M by

(119868120588119891)119899

=

119899

sum

119896=0

120588 (119887119896minus1) 119889119896119891 (35)

The generalized fractional integral 119868120588 is obtained by taking120574119899 = 120588(119887119899) in (32) If 120588(119903) = 119903

120572 120572 gt 0 then we simply denote

119868120588 by 119868120572For quasinormed spaces 1198721 and 1198722 of martingales

we denote by 119861(1198721 1198722) the set of all bounded martingaletransforms from1198721 to1198722 that is119879 isin 119861(1198721 1198722)means thatthere exists a positive constant 119862 such that

100381710038171003817100381711987911989110038171003817100381710038171198722

le 1198621003817100381710038171003817119891

10038171003817100381710038171198721

(36)

for all martingales 119891 = (119891119899)119899ge0 isin 1198721We first study the boundedness on the spaces H119878

119901120601

On martingale Campanato-Hardy spaces we consider thefractional integral as a martingale transform associated withmonotone multipliers We say a sequence of nonnegativemeasurable functions 120574 = (120574119899)119899ge0 is almost decreasing if thereexists a positive constant 119862 such that

120574119896 (120596) le 119862120574ℓ (120596) as forall119896 ge ℓ (37)

For an almost decreasing sequence 120574 = (120574119899)119899ge0 we define 119860120574

by

119860120574 = inf 119862 gt 0 119862 satisfies (37) (38)

InTheorem 16 below we do not need any assumption onF119899119899ge0

Theorem 16 Let B sub F 119901 isin (0 infin) and 120601 120595 (0 1] rarr

(0 infin) Let (120574119899)119899ge0 be a sequence of nonnegative boundedalmost decreasing adapted functions and let 119868120574 be the martin-gale transform defined by (32) Assume that

119862120574120601120595 = sup119899ge0

sup119861isinB119899

120601 (119875 (119861))

120595 (119875 (119861))

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817119871infin

lt infin (39)

Then

119868120574 isin 119861 (H119878

119901120601H

119878

119901120595) (40)

with10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(41)

If every 120590-algebra F119899 is generated by countable atomsthen we can apply Theorem 16 to the generalized fractionalintegral 119868120588 The following corollary extends [5 Theorem 58]to the spacesH119878

119901120601

Corollary 17 Assume that every 120590-algebra F119899 is generatedby countable atoms and B = A Let 119901 isin (0 infin) and 120588 120601 120595

(0 1] rarr (0 infin) Suppose that 120588 is almost increasing and that

sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin (42)

Then

119868120588 isin 119861 (H119878

119901120601H

119878

119901120595) (43)

If one further assumes that F119899119899ge0 is regular and that 120595 isalmost increasing and satisfies the doubling condition then

119868120588 isin 119861 (H119878

119901120601H

119878

119902120595) (119901 119902 isin (0 infin)) (44)

We next study the boundedness on martingale Morrey-Hardy spaces 119867

119878

119901120601and martingale Hardy spaces 119867

119878

119901

Recall that119860(F119899)perp

= F119899 for all 119899 ge 0 ifF0 is nonatomic

Proposition 18 Let B sub F 0 lt 119901 lt 119902 lt infin and 120601

(0 1] rarr (0 infin) Let (120574119899)119899ge0 be a sequence of adapted func-tions Suppose that F0 is nonatomic and that B = FAssume in addition that 120601 is almost decreasing that 119905

1119901120601(119905)

is almost increasing and that lim119905rarr0120601(119905) = infin Then 119868120574 notin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

According to Proposition 18 to consider the bounded-ness on 119867

119878

119901120601and 119867

119878

119901 we suppose that every 120590-algebra F119899

is generated by countable atoms and thatB = AIn this case if 120601(119903) = 119903

minus1119901 and F0 = Ω 0 then119867

119878

119901120601coincides with 119867

119878

119901and 119891

119867119878119901120601

= 119891119867119878119901

However ifF0 = Ω 0 then 119867

119878

119901120601does not coincide with 119867

119878

119901in general

We do not always assume thatF0 = Ω 0

Theorem 19 Suppose that every 120590-algebraF119899 is generated bycountable atoms thatB = A and that F119899119899ge0 is regular Let

6 Journal of Function Spaces and Applications

0 lt 119901 lt 119902 lt infin and 120601 (0 1] rarr (0 infin) and let (120574119899)119899ge0 bea sequence of nonnegative bounded adapted functions Assumethat 120601 satisfies the doubling condition and that there exists apositive constant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902 as

(45)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then

119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (46)

Furthermore if 120601(119905) = 119905minus1119901 then

119868120574 isin 119861 (119867119878

119901 119867

119878

119902) (47)

As a consequence of Theorem 19 we have the followingcorollary which gives an extension of [5 Corollary 57] to thespaces 119867

119878

119901120601and gives a martingale Morrey-Hardy version of

Gunawan [27 Theorem B]

Corollary 20 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and 120588 120601 (0 1] rarr (0 infin) Assume that120588 is bounded that both 120588 and 120601 satisfy the doubling conditionand that there exists a positive constant 119862 such that

120601 (119903) int

119903

0

120588 (119905)

119905119889119905 + int

1

119903

120601 (119905) 120588 (119905)

119905119889119905 le 119862120601(119903)

119901119902

(0 lt 119903 lt 1)

(48)

Then

119868120588 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (49)

The following extends the results for dyadic martingalesin [6 7] and the result for 0 lt 119901 le 1 in [28]

Corollary 21 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and minus1119901 + 120572 = minus1119902 Then

119868120572 isin 119861 (119867119878

119901 119867

119878

119902) (50)

5 Lemmas

We prepare some lemmas to prove the results in Sections 3and 4

Lemma 1 LetB119899 satisfy 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge

0) Suppose that 120601 (0 1] rarr (0 infin) is almost increasing that

is120601(119903) le 1198620120601(119904) for all 0 lt 119903 le 119904 le 1Then for all nonnegativefunctions 119865

sup119861isinF119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

le 1198620 sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

(51)

Proof Let

N = sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875) (52)

For any 119861 isin F119899 we can choose the sets 119861119895 119895 = 0 1 2

(finite or infinite) such that

119861 = cup119895119861119895 1198610 isin 119860(F119899)perp

119861119895 isin 119860 (F119899)

119895 = 1 2

119875 (119861) = sum

119895

119875 (119861119895)

(53)

In this case 119861119895 isin B119899 119895 = 0 1 2 since119860(F119899)cup119860(F119899)perp

sub

B119899 Then

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

=1

120601 (119875 (119861))

1

119875 (119861)sum

119895

(int119861119895

119865 119889119875)

= sum

119895

119875 (119861119895)

119875 (119861)(

1

120601 (119875 (119861)) 119875 (119861119895)

int119861119895

119865 119889119875)

le sum

119895

119875 (119861119895)

119875 (119861)(

1198620

120601 (119875 (119861119895)) 119875 (119861119895)

int119861119895

119865 119889119875)

le 1198620sum

119895

119875 (119861119895)

119875 (119861)N

= 1198620N

(54)

This shows the conclusion

Lemma 2 (see [5 Lemma 33]) Suppose that every 120590-algebraF119899 is generated by countable atoms and that F119899119899ge0 is regularThen every sequence

1198610 sup 1198611 sup sdot sdot sdot sup 119861119899 sup sdot sdot sdot 119861119899 isin 119860 (F119899) (55)

has the following property for each 119899 ge 1

119861119899 = 119861119899minus1 119900119903 (1 +1

119877) 119875 (119861119899) le 119875 (119861119899minus1) le 119877119875 (119861119899)

(56)

where 119877 is the constant in (14)

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

6 Journal of Function Spaces and Applications

0 lt 119901 lt 119902 lt infin and 120601 (0 1] rarr (0 infin) and let (120574119899)119899ge0 bea sequence of nonnegative bounded adapted functions Assumethat 120601 satisfies the doubling condition and that there exists apositive constant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902 as

(45)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then

119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (46)

Furthermore if 120601(119905) = 119905minus1119901 then

119868120574 isin 119861 (119867119878

119901 119867

119878

119902) (47)

As a consequence of Theorem 19 we have the followingcorollary which gives an extension of [5 Corollary 57] to thespaces 119867

119878

119901120601and gives a martingale Morrey-Hardy version of

Gunawan [27 Theorem B]

Corollary 20 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and 120588 120601 (0 1] rarr (0 infin) Assume that120588 is bounded that both 120588 and 120601 satisfy the doubling conditionand that there exists a positive constant 119862 such that

120601 (119903) int

119903

0

120588 (119905)

119905119889119905 + int

1

119903

120601 (119905) 120588 (119905)

119905119889119905 le 119862120601(119903)

119901119902

(0 lt 119903 lt 1)

(48)

Then

119868120588 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) (49)

The following extends the results for dyadic martingalesin [6 7] and the result for 0 lt 119901 le 1 in [28]

Corollary 21 Suppose that every 120590-algebra F119899 is generatedby countable atoms thatB = A and that F119899119899ge0 is regularLet 0 lt 119901 lt 119902 lt infin and minus1119901 + 120572 = minus1119902 Then

119868120572 isin 119861 (119867119878

119901 119867

119878

119902) (50)

5 Lemmas

We prepare some lemmas to prove the results in Sections 3and 4

Lemma 1 LetB119899 satisfy 119860(F119899) cup 119860(F119899)perp

sub B119899 sub F119899(119899 ge

0) Suppose that 120601 (0 1] rarr (0 infin) is almost increasing that

is120601(119903) le 1198620120601(119904) for all 0 lt 119903 le 119904 le 1Then for all nonnegativefunctions 119865

sup119861isinF119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

le 1198620 sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

(51)

Proof Let

N = sup119861isinB119899

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875) (52)

For any 119861 isin F119899 we can choose the sets 119861119895 119895 = 0 1 2

(finite or infinite) such that

119861 = cup119895119861119895 1198610 isin 119860(F119899)perp

119861119895 isin 119860 (F119899)

119895 = 1 2

119875 (119861) = sum

119895

119875 (119861119895)

(53)

In this case 119861119895 isin B119899 119895 = 0 1 2 since119860(F119899)cup119860(F119899)perp

sub

B119899 Then

1

120601 (119875 (119861))(

1

119875 (119861)int119861

119865 119889119875)

=1

120601 (119875 (119861))

1

119875 (119861)sum

119895

(int119861119895

119865 119889119875)

= sum

119895

119875 (119861119895)

119875 (119861)(

1

120601 (119875 (119861)) 119875 (119861119895)

int119861119895

119865 119889119875)

le sum

119895

119875 (119861119895)

119875 (119861)(

1198620

120601 (119875 (119861119895)) 119875 (119861119895)

int119861119895

119865 119889119875)

le 1198620sum

119895

119875 (119861119895)

119875 (119861)N

= 1198620N

(54)

This shows the conclusion

Lemma 2 (see [5 Lemma 33]) Suppose that every 120590-algebraF119899 is generated by countable atoms and that F119899119899ge0 is regularThen every sequence

1198610 sup 1198611 sup sdot sdot sdot sup 119861119899 sup sdot sdot sdot 119861119899 isin 119860 (F119899) (55)

has the following property for each 119899 ge 1

119861119899 = 119861119899minus1 119900119903 (1 +1

119877) 119875 (119861119899) le 119875 (119861119899minus1) le 119877119875 (119861119899)

(56)

where 119877 is the constant in (14)

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Journal of Function Spaces and Applications 7

Lemma 3 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular For 119861 isin 119860(F119898)let 119861119895 isin 119860(F119895) be

119861 = 119861119898 sub 119861119898minus1 sub sdot sdot sdot sub 1198610 (57)

Let 120601 (0 1] rarr (0 infin) Suppose that 120601 satisfies the doublingconditionThen there exists a positive constant119862 that dependsonly on 120601 and the regularity constant 119877 such that

119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

le 119862 int

119887ℓ

119887119898

120601 (119905)

119905119889119905 119900119899 119861 (58)

where 119887119895 is the function defined by (33)

Proof Let 119869 = 119895 119887119895 = 119887119895minus1 Then by Lemma 2 we have119898

sum

119895=ℓ+1

120601 (119887119895) 120594119887119895

= 119887119895minus1

= sum

119895isin119869ℓlt119895le119898

120601 (119887119895)

= sum

119895isin119869ℓlt119895le119898

1

log (119887119895minus1119887119895)

int

119887119895minus1

119887119895

120601 (119887119895)

119905119889119905

≲ sum

119895isin119869ℓlt119895le119898

int

119887119895minus1

119887119895

120601 (119905)

119905119889119905

= int

119887ℓ

119887119898

120601 (119905)

119905119889119905

(59)

In Theorem 13 we do not assume that F119899119899ge0 is regularHence we need the following lemma

Lemma 4 Let 120601 (0 1] rarr (0 infin) Suppose that every 120590-algebraF119899 is generated by countable atoms For119861 isin 119860(F119899) let119861119895 isin 119860(F119895) be

119861 = 119861119899 sub 119861119899minus1 sub sdot sdot sdot sub 1198610 (60)

For the sequence 119861119896119899

119896=0above one defines a decreasing seq-

uence of integers 119899119895 = 119899119895(119861119896119899

119896=0) inductively by

1198991 = sup 119896 isin [0 119899] cap Z 119875 (119861119896) ge 2119875 (119861)

119899119895 = sup 119896 isin [0 119899119895minus1] cap Z 119875 (119861119896) ge 2119875 (119861119899119895minus1

)

(119895 ge 2)

(61)

where one uses the convention sup 0 = minus1 One further defines

119869 = 119895 119899119895 ge 0 119899+

119895= 1 + 119899119895 (62)

Suppose that 120601 satisfies the doubling condition Then thereexists a positive constant 119862 that depends only on 120601 such that

sum

119895isin119869

120601 (119887119899+119895

) le 119862 int

1

119887119899

120601 (119905)

119905119889119905 119900119899 119861 (63)

where 119887119895 is the function defined by (33)

Note that this lemma is the counterpart to the techniquein [29 page 1104 line 5]

Proof By the definition of 119899119895 if 119895 isin 119869 then

119887119899119895minus1

le 119887119899+119895

lt 2119887119899119895minus1

le 119887119899119895

on 119861 (64)

where we use the convention 1198990 = 119899Using the doubling condition on 120601 we have

sum

119895isin119869

120601 (119887119899+119895

) ≲ sum

119895isin119869

int

2119887119899119895minus1

119887119899119895minus1

120601 (119905)

119905119889119905 le int

1

119887119899

120601 (119905)

119905119889119905 (65)

because the intervals (119887119899119895minus1

2119887119899119895minus1

) are disjointed by (64)

In the proof of Theorem 19 we need the following esti-mates for the square function of 119868120574119891

Lemma 5 Suppose that every 120590-algebra F119899 is generated bycountable atoms and that F119899119899ge0 is regular Let 119901 119902 isin (0 infin)

with 119901 lt 119902 Let (120574119899)119899ge0 be a sequence of nonnegative boundedadapted functions Suppose that 120601 (0 1] rarr (0 infin) satisfiesthe doubling condition Assume that there exists a positiveconstant 119862 such that

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

times

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

le 119862120601(119887119899)119901119902

119886119904

(66)

for all 119899 ge 0 where 119887119896 is the measurable function defined by(33) Then for 119891 isin M with 119891

119867119878119901120601

= 1

119878119899 (119868120574119891) le 119862120601(119887119899minus1)119901119902

119878(119899)

(119868120574119891) le 119862120601(119887119899)119901119902minus1

119878 (119891)

(67)

for all 119899 ge 0 where 119862 is a positive constant independent of 119891

Proof Let 119891 isin M such that 119891119867119878119901120601

= 1 We first show that

10038161003816100381610038161198891198961198911003816100381610038161003816 le 119862120601 (119887119896minus1) (68)

where 119862 is a positive constant that depends only on 120601 andthe regularity constant 119877 Let 119861 isin 119860(F119896) Then on the set 119861keeping in mind that

10038161003816100381610038161198891198961198911003816100381610038161003816 = (

1

119875 (119861)int119861

10038161003816100381610038161198891198961198911003816100381610038161003816

119901119889119875)

1119901

(69)

we have

10038161003816100381610038161198891198961198911003816100381610038161003816 le (

1

119875 (119861)int119861

119878(119891)119901119889119875)

1119901

le 120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

≲ 120601 (119887119896minus1)

(70)

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

8 Journal of Function Spaces and Applications

Wehave obtained (68)We now show (67) Using (68) and theassumption (66) we have

119878119899(119868120574119891)2

=

119899

sum

119896=0

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

119899

sum

119896=0

1205742

119896minus1120601(119887119896minus1)

2120594119887119896

= 119887119896minus1

le (

119899

sum

119896=0

120574119896minus1120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902

≲ 120601(119887119899minus1)2119901119902

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1205742

119896minus1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

le 119878(119899)

(119891)2

infin

sum

119896=119899+1

1205742

119896minus1120594119887119896

= 119887119896minus1

le 119878(119899)

(119891)2(

infin

sum

119896=119899+1

120574119896minus1120594119887119896

= 119887119896minus1

)

2

≲ 120601(119887119899)2119901119902minus2

119878(119899)

(119891)2

(71)

Remark 22 In the course of the proof the embedding ℓ1

997893rarr

ℓ2 is used If one does not use the embedding then

119899

sum

119896=0

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899minus1)119901119902

infin

sum

119896=119899+1

120574119896minus11003816100381610038161003816119889119896119891

1003816100381610038161003816 ≲ 120601(119887119899)119901119902minus1

119878(119899)

(119891)

(72)

6 Proofs of the Results in Section 3

In this section we prove the results in Section 3Proposition 8 can be proved in the sameway as [5 Propo-

sition 22] so we omit the proof Proposition 9 is a directconsequence of Lemma 1 Then we will prove Theorems 1112 and 13

Recall that 119878(119899)

(119891) is defined by (11)

61 Proof of Theorem 11 We first show Theorem 11 Burk-holderrsquos inequality on generalized martingale Campanatospaces

Proof of Theorem 11 Let 119891 isin 1198710

1and 119861 isin B119899 Then 119891120594119861 minus

119864119899[119891120594119861] isin 1198710

1sub M and

119889119896 (119891120594119861 minus 119864119899 [119891120594119861]) = 0 if 119896 le 119899

(119889119896119891) 120594119861 if 119896 gt 119899(73)

Therefore we have 119878(119891120594119861 minus 119864119899[119891120594119861]) = 119878(119899)

(119891)120594119861 HenceusingTheorem 10 we have

119888119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

= 1198881199011003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]

1003817100381710038171003817119871119901

le1003817100381710038171003817119878 (119891120594119861 minus 119864119899 [119891120594119861])

1003817100381710038171003817119871119901

= (int119861

119878(119899)

(119891)119901119889119875)

1119901

le 119862119901

1003817100381710038171003817119891120594119861 minus 119864119899 [119891120594119861]1003817100381710038171003817119871119901

= 119862119901(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

(74)

We have obtained (21)We next show (22) Using (74) we have

(int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

le (int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119888minus1

119901(int

119861

119878(119899minus1)

(119891)119901119889119875)

1119901

+ (int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

(75)

Therefore119888119901

1 + 119888119901

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

le1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(76)

For the converse part using the inequality

(int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

le 2(int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875)

1119901

(77)

which we have mentioned in Remark 4 we obtain

(int119861

119878(119899minus1)

(119891)119901119889119875)

1119901

le (int119861

119878(119899)

(119891)119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le 119862119901(int119861

|119891 minus 119864119899119891|119901119889119875)

1119901

+ (int119861

|119889119899119891|119901119889119875)

1119901

le (2119862119901 + 1) (int119861

|119891 minus 119864119899minus1119891|119901119889119875)

1119901

(78)

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

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Page 9: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Journal of Function Spaces and Applications 9

That is1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

le (2119862119901 + 1)1003817100381710038171003817119891

1003817100381710038171003817H119878minus119901120601

(79)

62 Proof of Theorem 12 We next show Theorem 12 a rela-tion betweenL119901120601 andLminus

119901120601H119878

119901120601 andH119878minus

119901120601

Proof of Theorem 12 Inequality (24) was mentioned inRemark 4 Inequality (25) is deduced from the inequality119878(119891)

2minus 119878119899(119891)

2le 119878(119891)

2minus 119878119899minus1(119891)

2We now show (26) Let 119861 isin B119899 and 119861

1015840= 120596 isin Ω

119864119899minus1[120594119861](120596) gt 0 Since 1198611015840

isin F119899minus1 we have

119864 [120594Ω1198611015840120594119861] = 119864 [120594Ω1198611015840119864119899minus1 [120594119861]] = 0 (80)

that is 119861 sub 1198611015840

Suppose that F119899119899ge0 is regular To show (26) we firstprove

1198611015840

= 120596 isin Ω 119864119899minus1 [120594119861] (120596) ge1

119877 (81)

where 119877 is the regularity constant By the definition of 1198611015840 we

have 1198611015840

sup 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 We will show theconverse By the regularity we have 120594119861 le 119877119864119899minus1[120594119861] Thisimplies 119861 sub 120596 isin Ω 119864119899minus1[120594119861](120596) ge 1119877 or equivalently

120594119861 le 120594119864119899minus1

[120594119861]ge1119877 (82)

Operating 119864119899minus1 we have

119864119899minus1 [120594119861] le 120594119864119899minus1

[120594119861]ge1119877 (83)

We have obtained (81)From (81) we deduce that

119875 (1198611015840) = 119864 [120594119864

119899minus1[120594119861]ge1119877] le 119864 [119877119864119899minus1 [120594119861]] = 119877119875 (119861)

(84)

Hence using the assumption (23) and the doubling conditionon 120601 with (84) we have

1

119875 (119861)int119861

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le1

119875 (119861)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

≲1

119875 (1198611015840)int1198611015840

1003816100381610038161003816119891 minus 119864119899minus11198911003816100381610038161003816

119901119889119875

le 120601(119875 (1198611015840))

11990110038171003817100381710038171198911003817100381710038171003817

119901

L119901120601

≲ 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

L119901120601

(85)

We have obtained (26)By the same way as above we have (27) The proof is

completed

63 Proof of Theorem 13 We now prove Theorem 13 a rela-tion betweenmartingaleMorrey spaces andmartingale Cam-panato spaces

Proof of Theorem 13 The part 119891L119901120601

le 2119891119871119901120601

was shownin Remark 4 and the part 119891H119878minus

119901120601

le 119891119867119878119901120601

is obvious Wenow show the part 119891

119871119901120601

≲ 119891Lminus119901120601

Let 119861 isin 119860(F119899) We take 119861119896 isin 119860(F119896) such that 119861 = 119861119899 sub

119861119899minus1 sub sdot sdot sdot sub 1198610 Let 119899119895 be the decreasing sequence of integersdefined in Lemma 4 with convention 1198990 = 119899 Since 119899119895minus1 ge 119899119895the function 119864119899

119895minus1

119891 minus 119864119899119895

119891 is constant on 119861119899119895minus1

Therefore onthe set 119861 we have

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

= (1

119875 (119861119899119895minus1

)

int119861119899119895minus1

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 21119901

(1

119875 (119861119899+119895

)

int119861119899+

119895

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861119899+119895

))

(86)

where 119899+

119895is the same as in (62)

Let 119869 be the same as in Lemma 4 and let119898 = max 119869 UsingLemma 4 and the assumption (28) we have

10038161003816100381610038161003816119864119899119891 minus 119864119899

119898

11989110038161003816100381610038161003816

le sum

119895isin119869

100381610038161003816100381610038161003816119864119899119895minus1

119891 minus 119864119899119895

119891100381610038161003816100381610038161003816

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sum

119895isin119869

120601 (119875 (119861119899+119895

))

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

int

1

119875(119861)

120601 (119905)

119905119889119905

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861

(87)

For |119864119899119898

119891| we may assume that 119899119898 gt 0 By the definition of119899119898 we have 119875(1198611) le 119875(1198610) lt 2119875(119861119899

119898

) le 2119875(1198611) le 2119875(1198610)Therefore

120601 (119875 (1198611)) ≲ int

119875(1198610)

119875(1198610)2

120601 (119905)

119905119889119905

≲ int

1

119875(119861)

120601 (119905)

119905119889119905 ≲ 120601 (119875 (119861))

(88)

Hence on the set 119861 the constant 119864119899119898

119891 has the followingbound

10038161003816100381610038161003816119864119899119898

11989110038161003816100381610038161003816

=10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

= (1

119875(119861119899119898

)int119861119899119898

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

10 Journal of Function Spaces and Applications

le 21119901

(1

119875(1198611)int1198611

10038161003816100381610038161003816119864119899119898

119891 minus 119864011989110038161003816100381610038161003816

119901

119889119875)

1119901

le 211199011003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (1198611)) ≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

120601 (119875 (119861))

(89)

Combining (87) and (89) we have

10038161003816100381610038161198641198991198911003816100381610038161003816 ≲

10038171003817100381710038171198911003817100381710038171003817Lminus119901120601

120601 (119875 (119861)) on 119861 (90)

Using (24) inTheorem 12 we have

(int119861

10038161003816100381610038161198911003816100381610038161003816

119901119889119875)

1119901

le (int119861

1003816100381610038161003816119891 minus 1198641198991198911003816100381610038161003816

119901119889119875)

1119901

+ 119875(119861)1119901 1003816100381610038161003816119864119899119891

1003816100381610038161003816

≲ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

+ 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

sim 119875(119861)1119901

120601 (119875 (119861))1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(91)

that is

10038171003817100381710038171198911003817100381710038171003817119871119901120601

≲1003817100381710038171003817119891

1003817100381710038171003817Lminus119901120601

(92)

We can show 119891119867119878119901120601

≲ 119891H119878minus119901120601

by the same way Indeed in(86) we can replace |119864119899

119895minus1

119891minus119864119899119895

119891| and 119891Lminus119901120601

by 119878119899119895minus1

(119891)2minus

119878119899119895

(119891)212 and 119891H119878minus

119901120601

respectively The rest is similar andwe can obtain 119891

119867119878119901120601

≲ 119891H119878minus119901120601

IfB = A thenB satisfies (23)Therefore ifF is regular

we can applyTheorem 12 to obtain the equivalence of 119891119871119901120601

119891L

119901120601

and 119891Lminus119901120601

and the equivalence of 119891119867119878119901120601

119891H119878119901120601

and 119891H119878minus

119901120601

64 Proof of Theorem 15 We will now proveTheorem 15 theJohn-Nirenberg-type theorem for martingale Campanato-Hardy spaces Following Weisz [19 Definition 245] wedefine

10038171003817100381710038171198911003817100381710038171003817BMO119878

119901

= sup119899ge0

100381710038171003817100381710038171003817100381710038171003817

(119864119899 [119878(119891)2

minus 119878119899minus1(119891)21199012

])

1119901100381710038171003817100381710038171003817100381710038171003817119871infin

(93)

for 119891 isin M and 119901 isin (0 infin)

Proof of Theorem 15 We may assume that B = F byProposition 9

By Holderrsquos inequality and Theorem 12 we only need toshow that 119891H119878

119901120601

le 119862119901119902120601119891H119878minus119902120601

for 0 lt 119902 le 1 lt 119901

Recall the notation 119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2 Let 119891 isin

H119878minus

119902120601cap 119871

0

1sub M 119860 isin F119899 and 119898 ge 119899 + 1 By (73) we have

119878(119898minus1)

(119891120594119860 minus 119864119899[119891120594119860]) = 119878(119898minus1)

(119891)120594119860 Hence for 119861 isin F119898119898 ge 119899 + 1 we have

1

119875 (119861)int119861

119878(119898minus1)

(119891120594119860 minus 119864119899 [119891120594119860])119902119889119875

=1

119875 (119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le1

119875 (119860 cap 119861)int119860cap119861

119878(119898minus1)

(119891)119902119889119875

le 120601(119875 (119860 cap 119861))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

≲ 120601(119875 (119860))1199021003817100381710038171003817119891

1003817100381710038171003817

119902

H119878minus119902120601

(94)

Therefore for the F119898119898ge119899-martingale (119864119898[119891120594119860] minus

119864119899[119891120594119860])119898ge119899 we have10038171003817100381710038171003817(119864119898 [119891120594119860] minus 119864119899 [119891120594119860])

119898ge119899

10038171003817100381710038171003817BMO119878119902

≲ 120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(95)

By [19 Theorem 250] there exists a positive constant 119862119901119902120601

that depends only on 119901 119902 and 120601 such that

(119864119899+1 [119878(119899)

(119891120594119860 minus 119864119899 [119891120594119860])119901])

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(96)

Combining (96) and the fact that 119878(119899)

(119891120594119860 minus 119864119899[119891120594119860]) =

119878(119899)

(119891)120594119860 we have

(119864119899+1 [119878(119899)

(119891)119901])

1119901

120594119860 le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(97)

Therefore for 119860 isin F119899 we have

(1

119875(119860)int119860

119878(119899)

(119891)119901119889119875)

1119901

= (1

119875(119860)int119860

119864119899+1 [119878(119899)

(119891)119901] 119889119875)

1119901

le 119862119901119902120601120601 (119875 (119860))1003817100381710038171003817119891

1003817100381710038171003817H119878minus119902120601

(98)

that is1003817100381710038171003817119891

1003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(99)

for 119891 isin H119878minus

119902120601cap 119871

0

1 For general 119891 isin H119878minus

119902120601 applying (99) to the

martingale 119891(119898)

= (119891min(119898119899))119899ge0 we have

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878119901120601

le 119862119901119902120601

10038171003817100381710038171003817119891

(119898)10038171003817100381710038171003817H119878minus119902120601

le 119862119901119902120601

10038171003817100381710038171198911003817100381710038171003817H119878minus119902120601

(100)

Taking 119901 = 2 in (100) we have that 119891 is an 1198712-bounded

martingale Therefore we have (99) for all 119891 isin H119878minus

119902120601 The

proof is completed

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Journal of Function Spaces and Applications 11

7 Proofs of the Results in Section 4

In this section we prove the results in Section 4

71 Proofs of Theorem 16 and Corollary 17 Recall that119878(119899)

(119891)2

= 119878(119891)2

minus 119878119899(119891)2

Proof of Theorem 16 Using the assumption that (120574119899)119899ge0 isalmost decreasing we have

119878(119899)

(119868120574119891)2

=

infin

sum

119896=119899+1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816

2

le 1198602

1205741205742

119899

infin

sum

119896=119899+1

10038161003816100381610038161198891198961198911003816100381610038161003816

2

= 1198602

1205741205742

119899119878(119899)

(119891)2

(101)

Then for 119861 isin B119899 using the assumption (39) we have

int119861

119878(119899)

(119868120574119891)119901

119889119875 le 119860119901

120574int119861

120574119901

119899119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

int119861

119878(119899)

(119891)119901119889119875

le 119860119901

120574

1003817100381710038171003817120574119899120594119861

1003817100381710038171003817

119901

119871infin

119875 (119861) 120601(119875 (119861))1199011003817100381710038171003817119891

1003817100381710038171003817

119901

H119878119901120601

le 119860119901

120574119862

119901

120574120601120595119875 (119861) 120595(119875 (119861))

11990110038171003817100381710038171198911003817100381710038171003817

119901

H119878119901120601

(102)

Therefore we have

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817H119878119901120595

le 119860120574119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(103)

and 119868120574 isin 119861(H119878

119901120601H119878

119901120595) The proof is completed

Proof of Corollary 17 Let 120574119899 = 120588(119887119899) Then we have that(120574119899)119899ge0 is almost decreasing and that 120574119899120594119861

119871infin

= 120588(119875(119861)) for119861 isin 119860(F119899) Hence

119862120574120601120595 = sup119899ge0

sup119861isin119860(F

119899)

120588 (119875 (119861)) 120601 (119875 (119861))

120595 (119875 (119861))

le sup0lt119905le1

120588 (119905) 120601 (119905)

120595 (119905)lt infin

(104)

Therefore we can apply Theorem 16 to obtain 119868120574 isin

119861(H119878

119901120601H119878

119901120595) If we further assume that F119899119899ge0 is regular

and that 120595 is almost increasing and satisfies the doublingcondition then by the John-Nirenberg-type equivalence(Theorem 15) we have 119868120574 isin 119861(H119878

119901120601H119878

119902120595) for all 119902 isin

(0 infin)

Remark 23 Assume that (39) holds Suppose further thatthere exists a positive number119862

1015840 such thatsuminfin

119896=119899+1120574119896minus1 le 119862

1015840120574119899

as for all 119899 ge 0 Then in the light of Remark 22 we see that

sup119899ge0

sup119861isinB119899

1

120595 (119875 (119861))(

1

119875(119861)int119861

(

infin

sum

119896=119899+1

|120574119896minus1119889119896119891|)

119901

119889119875)

1119901

le 1198621015840119862120574120601120595

10038171003817100381710038171198911003817100381710038171003817H119878119901120601

(105)

72 Proofs of Theorem 19 and Corollaries 20 and 21

Proof of Theorem 19 We first show the part 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) Assume (45) Let 119891 = (119891119899)119899ge0 isin M such that

119891119867119878119901120601

= 1 We need only to show that there exists 119862 gt 0

independent of 119891 such that

10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 119862 (106)

To obtain (106) we first show that

119878 (119868120574119891) le 119862119878(119891)119901119902

(107)

Let 119873 = suminfin

119899=0120594120601(119887

119899)le119878(119891) We define measurable subsets Ω1

Ω2 and Ω3 by

Ω1 = 119873 = infin

Ω2 = 119873 = 0

Ω3 = 0 lt 119873 lt infin

(108)

Let 120596 isin Ω1 Then we can take infinitely many integers 119899 suchthat 120601(119887119899minus1(120596)) le 119878(119891)(120596) For such 119899 we have

119878119899 (119868120574119891) (120596) le 119862120601(119887119899minus1 (120596))119901119902

le 119862119878 (119891) (120596)119901119902 (109)

by Lemma 5 Letting 119899 rarr infin along 119899 that satisfies120601(119887119899minus1(120596)) le 119878(119891)(120596) we have (107) on Ω1

On Ω2 again by Lemma 5 we have

119878 (119868120574119891) le 119862120601(1198870)119901119902minus1

119878 (119891) le 119862119878(119891)119901119902minus1

119878 (119891) = 119862119878(119891)119901119902

(110)

Let 120596 isin Ω3 Then we can take an integer 119899 such that

120601 (119887119899minus1 (120596)) le 119878 (119891) (120596) 120601 (119887119899 (120596)) gt 119878 (119891) (120596) (111)

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

12 Journal of Function Spaces and Applications

Hence by Lemma 5 we have

119878 (119868120574119891) (120596) le 119878119899 (119868120574119891) (120596)

+ 119878(119899)

(119868120574119891) (120596)

≲ 120601(119887119899minus1 (120596))119901119902

+ 120601(119887119899 (120596))119901119902minus1

119878 (119891) (120596)

le 119878 (119891) (120596)119901119902

+ 119878 (119891) (120596)119901119902minus1

119878 (119891) (120596)

≲ 119878 (119891) (120596)119901119902

(112)

We have obtained (107)We now show (106) Let 119861 isin cup119899119860(F119899) Using (107) we

have

(int119861

119878(119868120574119891)119902

119889119875)

1119902

≲ (int119861

119878(119891)119901119889119875)

1119901

119901119902

le 119875(119861)1119902

120601(119875 (119861))1199011199021003817100381710038171003817119891

1003817100381710038171003817

119901119902

119867119878119901120601

= 119875(119861)1119902

120601(119875 (119861))119901119902

(113)

We have obtained (106)We now show the part 119868120574 isin 119861(119867

119878

119901 119867

119878

119902) Let 120601(119905) = 119905

minus1119901We simply denote 119867

119878

119901120601by 119867

119878

119901minus1119901 Let 119891 = (119891119899)119899ge0 isin M

such that 119891119867119878119901

= 1 Observe that 119891119867119878119901minus1119901

le 119891119867119878119901

By

the assumption that (45) holds for 120601(119905) = 119905minus1119901 we can apply

(107) to 119891119891119867119878119901minus1119901

and we have

119878 (119868120574119891) le 119862119878(119891)1199011199021003817100381710038171003817119891

1003817100381710038171003817

1minus119901119902

119867119878119901minus1119901

le 119862119878(119891)119901119902

(114)

Hence we obtain

(intΩ

119878(119868120574119891)119902

119889119875)

1119902

≲ (intΩ

119878 (119891)119901119889119875)

1119901

119901119902

= 1 (115)

The proof is completed

Proof of Corollary 20 Let 120574119899 = 120588(119887119899) We only have to verify(45) Using Lemma 3 and the assumption (48) we have

119899

sum

119896=0

120588 (119887119896minus1) 120601 (119887119896minus1) 120594119887119896

= 119887119896minus1

+ 120601 (119887119899)

infin

sum

119896=119899+1

120588 (119887119896minus1) 120594119887119896

= 119887119896minus1

≲ int

1

119887119899

120588 (119905) 120601 (119905)

119905119889119905

+ 120601 (119887119899) int

119887119899

0

120588 (119905)

119905119889119905 ≲ 120601(119887119899)

119901119902

(116)

ByTheorem 19 we have the conclusion

Proof of Corollary 21 If 120588(119903) = 119903120572 and 120601(119905) = 119905

minus1119901 then

int

1

119903

120588 (119905) 120601 (119905)

119905119889119905 + 120601 (119903) int

119903

0

120588 (119905)

119905119889119905 sim 119903

120572minus1119901

= 119903minus1119902

= 120601(119903)119901119902

(117)

Observing (116) and applying Theorem 19 to 119868120572 we have theconclusion

Remark 24 In the light of Remark 22 we see that

sup119899ge0

sup119861isin119860(F

119899)

1

120601(119875 (119861))119901119902

times (1

119875 (119861)int119861

(

infin

sum

119896=1

1003816100381610038161003816120574119896minus11198891198961198911003816100381610038161003816)

119902

119889119875)

1119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(118)

which is similar to (105)In words of harmonic analysis this corresponds to the

embedding 119865119904+119899119901

1minus1198991199012

11990111199021

997893rarr 119865119904

11990121199022

for 0 lt 1199011 lt 1199012 lt infin0 lt 1199021 1199022 le infin and 119904 isin R see [30 Section 23] and [30 page129] for the definition of the space 119865

119904

119901119902and the above embed-

ding respectively It may be interesting to observe that thisembedding is translated into the fact that 119868120572 makes functionshave bounded variation

73 Proof of Proposition 18 In this subsection we proveProposition 18

Proof of Proposition 18 To prove 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) 0

we only need to show the following for any 119891 = (119891119899)119899ge0 isin M

if 119868120574 isin 119861 (119867119878

119901120601 119867

119878

119902120601119901119902) then 120594|120574

119896minus1|gt0119889119896119891 = 0

(119896 = 1 2 )

(119)

We now show (119) for 119891 = (119891119899)119899ge0 isin M We may assumethat 119875(|120574119896minus1| gt 0) gt 0 For 120574119896minus1 define

F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0)

= 119861 isin F119896 F119896minus1 119875 (1003816100381610038161003816120574119896minus1

1003816100381610038161003816 gt 0 cap 119861) gt 0

(120)

If F+

119896(|120574119896minus1| gt 0) = 0 then the function 120594|120574

119896minus1|gt0119889119896119891 is

F119896minus1-measurable Therefore we have

120594|120574119896minus1

|gt0119889119896119891 = 119864119896minus1 [120594|120574119896minus1

|gt0119889119896119891]

= 120594|120574119896minus1

|gt0119864119896minus1 [119889119896119891] = 0

(121)

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

Journal of Function Spaces and Applications 13

To complete the proof of (119) we only have to show thefollowing

if F+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) = 0 then 119868120574 notin 119861 (119867

119878

119901120601 119867

119878

119902120601119901119902)

(122)

Assume thatF+

119896(|120574119896minus1| gt 0) = 0 We fix 119861 isin F119896 F119896minus1 and

120575 gt 0 such that 119875(|120574119896minus1| ge 120575 cap 119861) gt 0 Let 1198611 = |120574119896minus1| ge

120575 and 1198611015840

1= |120574119896minus1| ge 120575 cap 119861 Note that 1198611 isin F119896minus1 119861

1015840

1isin

F+

119896(|120574119896minus1| gt 0) and 119861

1015840

1sub 1198611

To prove (122) we define two decreasing sequences ofmeasurable sets 119861119899

infin

119899=1and 119861

1015840

119899infin

119899=1that satisfy

119861119899 isin F119896minus1 119875 (119861119899) =119875 (1198611)

2119899minus1

1198611015840

119899isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899sub 119861119899

(123)

for every 119899 ge 1 inductively as followsSuppose that we can choose 119861119899minus1 and 119861

1015840

119899minus1that satisfy

119861119899minus1 isin F119896minus1 119875 (119861119899minus1) =119875 (1198611)

2119899minus2

1198611015840

119899minus1isin F

+

119896(

1003816100381610038161003816120574119896minus11003816100381610038161003816 gt 0) 119861

1015840

119899minus1sub 119861119899minus1

(124)

By the assumption that F0 is nonatomic F119896minus1 is alsononatomic Hence there exists 119861119899 isin F119896minus1 such that 119861119899 sub

119861119899minus1 119875(119861119899) = 119875(119861119899minus1)2 and 119875(119861119899 cap 1198611015840

119899minus1) gt 0 Let 119861

1015840

119899= 119861119899 cap

1198611015840

119899minus1 Then we have 119861119899 and 119861

1015840

119899with 119861119899 sub 119861119899minus1 and 119861

1015840

119899sub 119861

1015840

119899minus1

that satisfy (123)For the set 119861

1015840

119899defined above let 119892119899 = 1205941198611015840

119899

minus 119864119896minus1[1205941198611015840119899

]Since 119861119899 isF119896minus1-measurable we have

119892119899120594Ω119861119899

= (1205941198611015840119899

minus 119864119896minus1 [1205941198611015840119899

]) 120594Ω119861119899

= 1205941198611015840119899

120594Ω119861119899

minus 119864119896minus1 [1205941198611015840119899

120594Ω119861119899

]

= 0

(125)

By (125) and the assumption that 1199051119901

120601(119905) is almost increasingwe have for any 119860 isin cup

infin

119899=0F119899

1

120601 (119875 (119860))(

1

119875 (119860)int119860

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860)) 119875(119860)1119901

(int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

≲1

120601 (119875 (119860 cap 119861119899)) 119875(119860 cap 119861119899)1119901

times (int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

=1

120601 (119875 (119860 cap 119861119899))

times (1

119875 (119860 cap 119861119899)int119860cap119861119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

(126)

We now show that 119868120574 notin 119861(119867119878

119901120601 119867

119878

119902120601119901119902) Suppose that 119868120574 isin

119861(119867119878

119901120601 119867

119878

119902120601119901119902) that is there exists a positive number 119862 such

that10038171003817100381710038171003817119868120574119891

10038171003817100381710038171003817119867119878119902120601119901119902

le 1198621003817100381710038171003817119891

1003817100381710038171003817119867119878119901120601

(127)

for all 119891 isin 119867119878

119901120601

Since 1198611015840

119899isin F119896 F119896minus1 we have 119889119896119892119899 = 119892119899 = 0 and 119889119895119892119899 =

0 for 119895 = 119896Therefore we have

119878 (119892119899) = 119892119899 119878 (119868120574119892119899) = 120574119896minus1119892119899 (128)

For 119892119899 we take 119863119899 isin cupinfin

119899=0F119899 such that

1

120601 (119875 (119863119899))(

1

119875(119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

ge1

2

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

(129)

By (126) we may assume that 119863119899 sub 119861119899 As a consequence wehave119863119899 sub 1198611 = |120574119896minus1| ge 120575 and lim119899rarrinfin119875(119863119899) = 0 by (123)Then using (127) with (128) we have

1

120601(119875 (119863119899))119901119902

(1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

1

120601(119875 (119863119899))119901119902

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816120574119896minus1119892119899

1003816100381610038161003816

119902119889119875)

1119902

le1

120575

10038171003817100381710038171003817119868120574119892119899

10038171003817100381710038171003817119867119878119902120601119901119902

le119862

120575

1003817100381710038171003817119892119899

1003817100381710038171003817119867119878119901120601

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119901119889119875)

1119901

le2119862

120575

1

120601 (119875 (119863119899))

times (1

119875 (119863119899)int119863119899

1003816100381610038161003816119892119899

1003816100381610038161003816

119902119889119875)

1119902

(130)

Therefore we have

120601(119875 (119863119899))1minus119901119902

le2119862

120575 (131)

However this contradicts 119901 lt 119902 and lim119905rarr0120601(119905) = infin Wehave (122) and hence have (119)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

14 Journal of Function Spaces and Applications

Acknowledgments

The authors are thankful to Professor Masaaki Fukasawa inOsaka University for his kind hint about Remarks 22 23and 24 The first author was supported by Grant-in-Aid forScientific Research (C) (no 24540159) Japan Society for thePromotion of Science The second author was supportedby Grant-in-Aid for Scientific Research (C) (no 24540171)Japan Society for the Promotion of Science The third authorwas supported by Grant-in-Aid for Young Scientists (B) (no24740085) Japan Society for the Promotion of Science

References

[1] J Peetre ldquoOn the theory of L119901120582 spacesrdquo Journal of FunctionalAnalysis vol 4 no 1 pp 71ndash87 1969

[2] W Yuan W Sickel and D Yang Morrey and Campanato MeetBesov Lizorkin and Triebel Lecture Notes in MathematicsSpringer Berlin Germany 2010

[3] F Weisz ldquoMartingale Hardy spaces for 0 lt 119875 le 1rdquo ProbabilityTheory and Related Fields vol 84 no 3 pp 361ndash376 1990

[4] T Miyamoto E Nakai and G Sadasue ldquoMartingale Orlicz-Hardy spacesrdquoMathematische Nachrichten vol 285 no 5-6 pp670ndash686 2012

[5] E Nakai and G Sadasue ldquoMartingale Morrey-Campanatospaces and fractional integralsrdquo Journal of Function Spaces andApplications vol 2012 Article ID 673929 29 pages 2012

[6] J-A Chao and H Ombe ldquoCommutators on dyadic martin-galesrdquoProceedings of the JapanAcademy vol 61 no 2 pp 35ndash381985

[7] C Watari ldquoMultipliers for Walsh-Fourier seriesrdquo The TohokuMathematical Journal vol 16 pp 239ndash251 1964

[8] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals Irdquo Mathematische Zeitschrift vol 27 no 1 pp 565ndash606 1928

[9] G H Hardy and J E Littlewood ldquoSome properties of fractionalintegrals IIrdquo Mathematische Zeitschrift vol 34 no 1 pp 403ndash439 1932

[10] S L Sobolev ldquoOn a theorem in functional analysisrdquo Matem-aticheskii Sbornik vol 4 pp 471ndash497 1938

[11] E M Stein and GWeiss ldquoOn the theory of harmonic functionsof several variables I The theory of 119867

119901-spacesrdquo Acta Mathe-matica vol 103 pp 25ndash62 1960

[12] M H Taibleson and G Weiss ldquoThe molecular characterizationof certain Hardy spacesrdquo in Representation theorems for Hardyspaces vol 77 ofAsterisque pp 67ndash149 SocMath France Paris1980

[13] S G Krantz ldquoFractional integration on Hardy spacesrdquo StudiaMathematica vol 73 no 2 pp 87ndash94 1982

[14] E Nakai ldquoRecent topics of fractional integralsrdquo Sugaku Exposi-tions vol 20 no 2 pp 215ndash235 2007

[15] D R Adams ldquoA note on Riesz potentialsrdquo Duke MathematicalJournal vol 42 no 4 pp 765ndash778 1975

[16] F Chiarenza and M Frasca ldquoMorrey spaces and Hardy-Littlewoodmaximal functionrdquo Rendiconti di Matematica e dellesue Applicazioni vol 7 no 3-4 pp 273ndash279 1987

[17] H Tanaka and Y Terasawa ldquoPositive operators and maximaloperators in a filtered measure spacerdquo Journal of FunctionalAnalysis vol 264 no 4 pp 920ndash946 2013

[18] JNeveuDiscrete-ParameterMartingales North-HollandAms-terdam The Netherlands 1975

[19] F Weisz Martingale Hardy Spaces and Their Applications inFourier Analysis vol 1568 of Lecture Notes in MathematicsSpringer Berlin Germany 1994

[20] D L Burkholder ldquoMartingale transformsrdquo Annals of Mathe-matical Statistics vol 37 pp 1494ndash1504 1966

[21] D L Burkholder ldquoSharp inequalities for martingales andstochastic integrals Colloque Paul Levy sur les ProcessusStochastiques (Palaiseau 1987)rdquo Asterisque no 157-158 pp 75ndash94 1988

[22] D L Burkholder ldquoExplorations in martingale theory and itsapplicationsrdquo in Ecole drsquoEte de Probabilites de Saint-Flour XIX1989 vol 1464 of Lecture Notes in Mathematics pp 1ndash66Springer Berlin Germany 1991

[23] R L Long Martingale Spaces and Inequalities Peking Univer-sity Press Beijing China 1993

[24] M Kikuchi ldquoCharacterization of Banach function spaces thatpreserve the Burkholder square-function inequalityrdquo IllinoisJournal of Mathematics vol 47 no 3 pp 867ndash882 2003

[25] F Weisz ldquoInterpolation between martingale Hardy and BMOspaces the real methodrdquo Bulletin des Sciences Mathematiquesvol 116 no 2 pp 145ndash158 1992

[26] E Nakai ldquoOn generalized fractional integralsrdquo Taiwanese Jour-nal of Mathematics vol 5 no 3 pp 587ndash602 2001

[27] H Gunawan ldquoA note on the generalized fractional integraloperatorsrdquo Journal of the Indonesian Mathematical Society vol9 no 1 pp 39ndash43 2003

[28] G Sadasue ldquoFractional integrals on martingale Hardy spacesfor 0 lt 119875 le 1rdquoMemoirs of Osaka Kyoiku University vol 60 no1 pp 1ndash7 2011

[29] Y Sawano and H Tanaka ldquoSharp maximal inequalities andcommutators on Morrey spaces with non-doubling measuresrdquoTaiwanese Journal of Mathematics vol 11 no 4 pp 1091ndash11122007

[30] H Triebel Theory of Function Spaces Birkhauser BaselSwitzerland 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article Martingale Morrey-Hardy and Campanato-Hardy …downloads.hindawi.com/journals/jfs/2013/690258.pdf · 2019. 7. 31. · unify Hardy, Lipschitz, and Morrey-Campanato

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