real harmonic analysis

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Real Harmonic Analysis


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Published by ANU eViewThe Australian National UniversityCanberra ACT 0200, AustraliaEmail: [email protected] title is also available online at http://eview.anu.edu.au

National Library of Australia Cataloguing-in-Publication entry

Author: Auscher, Pascal.

Title: Real harmonic analysis / lectures by Pascal Auscher with the assistance ofLashi Bandara.

ISBN: 9781921934076 (pbk.) 9781921934087 (ebook)

Subjects: Mathematical analysis.Mathematics.

Other Authors/Contributors:Bandara, Lashi.

Dewey Number: 510

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in anyform or by any means, electronic, mechanical, photocopying or otherwise, without the prior permission of the publisher.

Printed by Griffin Press

This edition 2012 ANU eView


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Real Harmonic Analysis

Lectures by Pascal Auscherwith the assistance of Lashi Bandara


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Contents

1 Measure Theory 6

2 Coverings and cubes 8

2.1 Vitali and Besicovitch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Dyadic Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 Whitney coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Maximal functions 18

3.1 Centred Maximal function onRn . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2 Maximal functions for doubling measures . . . . . . . . . . . . . . . . . . . 21

3.3 The Dyadic Maximal function . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.4 Maximal Function on Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Interpolation 30

4.1 Real interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2 Complex interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Bounded Mean Oscillation 39

5.1 Construction and properties of BMO . . . . . . . . . . . . . . . . . . . . . . 39

5.2 John-Nirenberg inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.3 Good inequalities and sharp maximal functions . . . . . . . . . . . . . . . 46

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6 Hardy Spaces 52

6.1 Atoms and H1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.2 H1 BMO Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7 Calderon-Zygmund Operators 59

7.1 Calderon-Zygmund Kernels and Operators. . . . . . . . . . . . . . . . . . . 59

7.2 The Hilbert Transform, Riesz Transforms, and The Cauchy Operator. . . . 61

7.2.1 The Hilbert Transform. . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.2.2 Riesz Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.2.3 Cauchy Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

7.3 Lp boundedness of CZO operators . . . . . . . . . . . . . . . . . . . . . . . 66

7.4 CZO and H1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

7.5 Mikhlin multiplier Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

7.6 Littlewood-Paley Theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

8 Carleson measures and BMO 80

8.1 Geometry of Tents and Cones . . . . . . . . . . . . . . . . . . . . . . . . . . 81

8.2 BMO and Carleson measures . . . . . . . . . . . . . . . . . . . . . . . . . . 84

9 Littlewood-Paley Estimates 95

10 T(1) Theorem for Singular Integrals 100

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Preface

Pascal Auscher recently spent a year at ANU on leave from Universite Paris-Sud, wherehe is Professor of Mathematics. During this time he taught this graduate level course

covering fundamental topics in modern harmonic analysis.

Auscher has made substantial contributions to harmonic analysis and partial differentialequations, in a wide range of areas including functional calculi of operators, heat kernelestimates, Hardy spaces, weighted norm estimates and boundary value problems.

In particular his contributions were essential to the recent solution of a long-standingconjecture known as the Kato square root problem. This involved substantial new devel-opments concerning the so-called Tb theorems and their application to singular integraloperators or more generally to the functional calculus of operators which satisfy Davies-Gaffney estimates. These concepts were fundamental in the solution of the Kato problem

on square roots of elliptic operators by Auscher, Hofmann, Lacey, Tchamitchian and my-self.

Auscher has two published books to his name: Square Root problem for divergence opera-tors and related topics, (with Ph. Tchamitchian) Asterisque vol 249, Soc. Math. de France1998; andOn necessary and sufficient conditions forLp estimates of Riesz transforms as-sociated to elliptic operators onRn and related estimates, Memoirs of the American Math.Society vol. 186, Am. Math. Soc. 2007.

In this lecture course, Auscher provides a basic grounding in the advanced mathematicsrequired for tackling problems in modern harmonic analysis and applying them, for exam-

ple to partial differential equations. He does this from the point of view of an expert whofully understands the significance of the more basic material. Throughout the manuscriptthe material is developed in novel ways including some original proofs, derived from theadvanced outlook of the author.

Thus the book is a mixture of expository material developed from a contemporary per-spective and new work. It is likely to serve two audiences, both graduate students, andestablished researchers wishing to familiarise themselves with modern techniques of har-monic analysis. The lecture notes were taken and written up by an ANU PhD student,Lashi Bandara, thus providing the final polished account of the course material.

Alan McIntosh, Professor of Mathematics, MSI, ANU

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Introduction

This book presents the material covered in graduate lectures delivered at the AustralianNational University in 2010. This material had also been partially presented at the Uni-

versity Paris-Sud before but never published.

Real Harmonic Analysis originates from the seminal works of Zygmund and Calderon,pursued by Stein, Weiss, Fefferman, Coifman, Meyer and many others.

Moving from the classical periodic setting to the real line, then to higher dimensionalEuclidean spaces and finally to, nowadays, sets with minimal structures, the theory hasreached a high level of applicability. This is why it is called real harmonic analysis:the usual exponential functions have disappeared from the picture. Set and functiondecomposition prevail.

This development has serve to solve famous conjectures in the field. The first one is theboundedness of the Cauchy integral operator on Lipschitz curves by Coifman, McIntoshand Meyer thirthy years ago. In the last ten years, there has been the solution of theKato conjecture at the intersection of partial differential equations, functional analysis andharmonic analysis, and of the Painleve conjecture at the border of complex function theoryand geometric analysis. We mention also the breakthroughs on weighted norm inequalitieswith sharp behavior on the weight constants that are occurring at the moment, many ofthe articles being still in submitted form.

Nowadays all these developments bear on variations of boundedness criteria for singularintegral operators and quadratic functionals called Tb theorems after the pionering work

of McIntosh and Meyer followed by David, Journe and Semmes. These are powerful andversatile tools. However, they represent more advanced topics than such lectures couldcover: these would be a follow-up topic. We have chosen to prepare the grounds to moreadvanced reading by presenting the basic material that is considered as well-knownby experts. Nevertheless, anyone who wants to explore this field or apply it to someother problem must know the material we have chosen here. These lecture notes aretherefore introductory to the field and accessible to beginners. They do not pretend tocover everything but a selection of important topics. Even if some lecture notes like thisone exist, there is a need for updates as they do not cover the same material or addressdifferent points.

Let us present it in some more detail. We cover ten chapters: Measure theory, Coveringsand cubes, Maximal functions, Interpolation, Bounded Mean Oscillation, Hardy spaces,

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Calderon-Zygmund Operators, Carleson measures and BMO, Littlewood-Paley estimatesand the T1 theorem for Singular Integrals.

The first chapter is to recall some notation and results from integration theory. The sec-ond chapter presents the fundamental tools of modern real harmonic analysis: coveringlemmas of various sorts (Vitali, Besicovitch, Whitney). We have chosen to mostly restrictour setting to the Euclidean space to emphasize the ideas rather than the technicalities,Except for the Besicovitch covering lemma, of which we give a proof, they all extend tovery general settings and we have tried to give proofs that are not too different from theones in the extended settings. Maximal functions are treated in Chapter 3. These centralobjects can be declined in several forms: centred, uncentred, dyadic. They all serve thesame goal: decompose the space into sets of controlled sizes according to a function. Chap-ter 4 on interpolation is rather standard and we present a simple as possible treatment.Chapter 5 on bounded mean oscillation restricts its attention to the main properties of

BMO functions, the John-Nirenberg inequality and the technique of good lambda inequal-ities to prove the sharp function inequality of Fefferman-Stein. We also cover the recentlydeveloped good lambda inequality with two parameters which is useful for proving in-equalities in restricted ranges of the exponents of the Lebesgue spaces. Chapter 6 presentsatomic Hardy spaces: Coifman-Weiss equivalence of the atomic definitions with p-atomswith 1 < p as a consequence of Calderon-Zygmund decompositions on functions,and applications to the Fefferman duality theorem. Chapter 7 is also an introduction tothe theory of Calderon-Zygmund operators and some applications to multipliers are giventowards Littlewood-Paley theorem. Again more advanced results can be obtained formthe references. Chapter 8 covers the notion of a Carleson measure and Carlesons em-bedding theorem is proved. Then the connection between Carleson measures and BMO

functions is studied in detail with care on the subtle convergence issues that are not oftentreated in the literature. Applications to the Bony-Coifman-Meyer paraproducts are given.Chapter 9 on Littlewood-Paley estimates could also be called T1 theorem for quadraticfunctionals and is extracted with details from an important article of Christ-Journe. Seealso my previous book with Philippe Tchamitchian were not all details are given. Almostorthogonality arguments, in particular the Cotlar-Knapp-Stein lemma, are central here.We finish in Chapter 1 with a proof of the T1 theorem of David-Journe. This proof isnot the original one but is the one given by Coifman-Meyer and deserves more publicity.Again the strategy of this proof adapts to very general settings.

Special thanks go to Alan McIntosh and to the Mathematical Science Institute by inviting

me as a one year visiting fellow at the Australian National University and also to theCNRS for giving me a delegation allowing a leave of absence from my home university.A one year visit that all my family has enjoyed and will remember for years.

Alan McIntosh and I have had and still have a long standing collaboration and it isa pleasure to thank him for sharing so many views on mathematics and life. That heaccepted to preface this manuscript is a great privilege.

Thanks also go to Lashi Bandara whose typesetting genius made scratch notes he tookfrom the lectures a readable manuscript. He also corrected some mistakes in the proofs(the remaining ones are my responsibility). My former student Frederic Bernicot read the

proofs entirely and found some more typos. I am grateful for that.

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Chapter 1

Measure Theory

While we shall focus our attention primarily on Rn, we note some facts about measuresin an abstract setting and in the absence of proofs.

LetXbe a set. The reader will recall that in the literature, a measure on Xis usuallydefined on a -algebra M P(X). This approach is limited in the direction we willtake. In the sequel, it will be convenient to forget about measurability and associate a sizeto arbitrary subsets but still in a meaningful way. We present the Caratheodorys notionof measure and measurability. In the literature, what we call a measure is sometimesdistinguished as anouter measure.

Definition 1.0.1 (Measure/Outer measure). LetX be a set. Then, a map : P(X)[0, +] is called a measure onX if it satisfies:

(i) () = 0,

(ii) (A)

i=1 (Ai) wheneverA

i=1 Ai.

Remark 1.0.2. Note that (ii) includes the statement: ifA B, then(A) (B).

Certainly, the classical measures are additive on disjoint subsets. This is something welose in the definition above. However, we can recover both the measure -algebra and a

notion of measurable set.

Definition 1.0.3 (Measurable). Let be a measure on X (in the sense of Definition1.0.1). We sayA X is-measurable if for allY X,

(A) =(A \ Y) + (A Y).

Theorem 1.0.4. Let{Ai}i=1 be a countable set of-measurable sets. Then,

(i)

i=1 Ai and

i=1 Ai are-measurable,

(ii) If the sets{Ai}i=1 are mutually disjoint, then

i=1

Ai = i=1

(Ai),

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(iii) IfAk Ak+1 for allk 1, then

limk

(Ak) =

i=1 Ai ,(iv) IfAk Ak+1 for allk 1, then

limk

(Ak) =

i=1

Ai

,

(v) The setM ={A X :A is measurable} is a-algebra.

A proof of (i) - (iv) can can be found in [GC91, p2]. Then, (v) is an easy consequence and

we leave it as an exercise.

The following result illustrates that we can indeed think about classical measures in thisframework. Recall that a measure space is a triple (X,M, ) where M P(X) is a -algebra and : M [0, +] is a measure in the classicalsense. Measure spaces are alsogiven as the tuple (X, ) by which we mean (X,M, ) where M is the largest -algebracontaining-measurable sets.

Theorem 1.0.5. Let (X,M, ) be a measure space. Then, there exists a measure inthe sense of Definition1.0.1onX such that= onM.

See[G.81,5.2, Theorem 3].

Of particular importance is the following construction of the Lebesgue measure in oursense.

Definition 1.0.6 (Lebesgue Measure). LetA Rn. For a Euclidean ballB, denote thevolume of the ball byvol B. Define the Lebesgue measureL:

L(A) = inf

i=1

vol Bi : A

i=1

Bi and eachBi is an open ball

.

This definition is justified by the following proposition.

Proposition 1.0.7. Let (Rn,M,L) be the classical Lebesgue measure defined on thelargest possible-algebraM and letM ={A X :A isL measurable}. Then,

M = M andL(A) = L(A)

for allA M.

For a more detailed treatment of abstract measure theory, see [GC91, 1] and[G.81,Ch.5].

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Chapter 2

Coverings and cubes

We will consider the setting ofRn with the usual Euclidean norm | | inducing the standardEuclidean metric dE(x, y) =|x y|. We note, however, that some of the material that wediscuss here can be easily generalised to a more abstract setting.

To introduce some nomenclature, we denote the ball centred at x with radius r > 0 byB(x, r). We are intentionally ambiguous as to whether the ball is open or closed. We willspecify when this becomes important.

For a ball B, let rad B denote its radius. For > 0, we denote the ball with the same

centre but radius rad B byB .

This chapter is motivated by the following two questions.

(1) Suppose that =

BBB where B is a family of balls. We wish to extract asubfamily B of balls that do not overlap too much and still cover .

(2) Given a set Rn, how can we select a cover of with a given geometric structure.

2.1 Vitali and Besicovitch

Lemma 2.1.1 (Vitali Covering Lemma). Let {B}I be a family of balls inRn and

suppose thatsupI

rad B< .

Then there exists a subsetI0 I such that

(i) {B}I0 are mutually disjoint.

(ii)IB I05B.Remark 2.1.2. (i) The balls{B}I can be open or closed.

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(ii) This statement only relies on the metric structure ofRn with the euclidean metricgiven by | |. It can be replaced by a metric space(E, d). As an example, we can set(E, d) = (Rn, d), where d is the infinity distance given be by the infinity norm.

This is equivalent to replacing balls by cubes.

(iii) The conditionsupIrad B< is necessary. A counterexample is{Bi = B(0, i) :i N}.

Proof. LetM= supIrad B. For j N, define

I(j) =

I : 2j1M

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Theorem 2.1.4 (Besicovitch Covering Theorem). LetE Rn. For eachx E, letB (x)be a ball centred at x. Assume that E is bounded or that supxErad B(x) < . Then,there exists a countable setE0 Eand a constantC(n) N such that

(i) E

xE0B(x).

(ii)

xE0

B(x) C(n).

Remark 2.1.5. (i) Here, (ii) means that the set{B(x)}xE0 forms a bounded coveringof E with constant C(n). This constant depends only on dimension. The boundedcovering property tells us that the balls are almost disjoint. In fact, we can organiseE0 = E1 . . . ENsuch that each set{B(x)}xEk contain mutually disjoint balls.We will not prove this.

(ii) We can substitute cubes for balls.

(iii) ForEunbounded, the conditionsupxErad B(x) 0, x, z C(y), if |x| , |z| R then|x z| R.

Proof of Besicovitch Covering Theorem. LetM= supxErad B(x) and suppose thatE isbounded and M = . Then, fix an x0 Eand there exists a ball B(x0, R) such thatB(x0, R) E. Then, were done by setting E0= {x0}.

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Now, to show (ii), fix y Rn, let = /6 and let C(y) denote the sector with vertex yand aperture . Define

Ay = {xi E0: y B(xi) and xi C(y)} ,

and letxi be the first element in Ay. Take xj Ay with j > i. Then,xi, xj C(y)

|xi y| rad B(xi)

|xj y| rad B(xj)

and so|xi xj | max {rad B(xi), rad B(xj)} by application of Lemma 2.1.6. But by theordering on E0, xj was selected after xi and so xj B(xi). Consequently rad B(xi) rad B(xi).

Now, suppose thatxiwas selected at generationk, so 2k1M k which implies that rad B(xi)> rad B(xj) which is a contradiction.

Also, note that

B(xj, 2k2M) :j Ay

are mutually disjoint and are all contained in

B(xi, 2k+1M). It follows that,

jAy

L(B(xj, 2k2M)) L(B(xi, 2

k+1M)) = (23)nL(B(0, 2k2M))

and

card Ay =

jAy

1 =

jAy

L(B(xj, 2k2M))

L(B(0, 2k2M)) 23n

and we are done.

We shall not give details for the case that Eis unbounded, but it can be obtained fromthe previous case with some effort. We refer the reader to [GC91, p35].

Corollary 2.1.7 (Sards Theorem). Letf : Rn Rn and let

A= x Rn : lim infr0

L(f(B(x, r)))L(B(x, r))

= 0 .ThenL(f(A)) = 0. (Recall that by our definition ofL, we can measure every subset ofRn).

Proof. Firstly, we note that for all x A and >0, there exists an rx (0, 1] such that

L(f(B(x, rx))) L(B(x, rx)).

Let A0 A be the set of centres given by Besicovitch applied to the set of balls {Bx = B(x, rx)}for which the above measure condition holds. Then,f(A)

xiA0

f(B(xi)) and by the

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subadditivity ofL,

L(f(A)) xiA0L(f(Bxi))

xiA0

L(Bxi)

A+B(0,1)

Bxi (y) dL(y)

C(n)L(A + B(0, 1)).

Now, if A is bounded, then L(A+ B(0, 1)) < and we can obtain the conclusion byletting 0. Otherwise, we replace A byA B(0, k) to obtain L(f(A B(0, k))) = 0and then by taking k we establish L(f(A))) = 0.

It is easy to show that A contains the singular points of f, that is points x at which fis differentiable with vanishing differential df(x) = 0. That is for a differentiable map,almost all points are regular (i.e., non-singular).

2.2 Dyadic Cubes

We begin with the construction of dyadic cubes on Rn

.Definition 2.2.1 (Dyadic Cubes). Let [0, 1)n be the reference cube and let j Z andk= (k1, . . . , kn) Zn. Then define the dyadic cube of generationj with lower left corner2jk

Qj,k =

x Rn : 2jx k [0, 1)n

,

the set of generationj dyadic cubes

Qj ={Qj,k : k Zn} ,

and the set of all cubes

Q = jZ

Qj ={Qj,k :j Z andk Zn} .

We define the length of a cube to be its side length(Qj,k) = 2j.

Remark 2.2.2. If we were to replace[0, 1)n withR =n

i=1[ai, ai + )as a reference cube,then the dyadic cubes with respect to R are constructed via the homothety : Rn Rn

where([0, 1)n) =R.

Theorem 2.2.3 (Properties of Dyadic Cubes). (i) Forj Z,(Qj,k) = 2j, L(Qj,k) =2jn, andQj forms a partition ofRn for eachj Z.

(ii) For all j Z and k Zn, there exists a unique k Zn such thatQj,k Qj1,k.We setQj,k =Qj1,k and it is called theparent ofQj,k.

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(iii) For allj Z andk Zn, the cubesQj+1,k for whichQj,k is the parent are calledthe children ofQj,k.

(iv) For allxR

n, there exists a unique sequence of dyadic cubes (Qj,kj

)jZ

Q suchthatQj+1,kj+1 is a child ofQj,k, Qj,kj is the parent of Qj+1,kj+1 and x Qj,kj forallj Z.

(v) Let E Q such that =QEQ satisfies L() < . Then, there exists a

collectionF Q of mutually disjoint dyadic cubes such that =QFQ

.

Proof. We leave it to the reader to verify (i) - (iv). We prove (v).

Define:

F =

Q Q :Q but

Q

.

We prove that F =. Let Q E, and denote the kth parent byQk. It follows thatL(Qk) = (2n)kL(Q). Now, letk0= maxk:Qk and consequently,Qk0 F.Now, note that

QEQ

QFQ

and by the construction ofF,QFQ

. But

by hypothesis =QEQ and so it follows that =

QFQ

.

To complete the proof, we prove that the cubes in Fare mutually disjoint. Let Q, Q Fand assume that Q = Q. This is exactly that Q Q and Q Q so by (iii),Q Q = .

Remark 2.2.4. (i) The way (v) is formulated does not ensure thatF E. Take forexampleE ={[0, 1/2), [1/2, 1)} which givesF ={[0, 1)}.

(ii) Suppose is an open set withL() < and let E = {Q Q :Q }. Then,F E and we callF the maximal dyadic cubes in E. Maximality here is withrespect to the propertyQ .

(iii) The assumptionL()< cannot be dropped. A counterexample is [0, 2k) R fork N.

2.3 Whitney coverings

The Whitney covering theorems are an important tool in Harmonic Analysis. We initiallygive a dyadic version of Whitney. But first, we recall some terminology from the theory ofmetric spaces. Recall that the diameter of a set EX for a metric space (X, d) is givenby

diam E = supx,yE

d(x, y),

and the distance from Eto another set F Xis given by

dist(E, F) = inf xE,yF

d(x, y).

Then the distance from E to a point y X is simply given by dist(x, E) = dist(E, F)where F ={x}.

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Theorem 2.3.1 (Whitney Covering Theorem for Dyadic cubes). Let O Rn be open.Then, there exists a collection of Dyadic cubesF ={Qi}iI such that

(i)1

30dist(Qi,

cO) diam Qi 1

10dist(Qi,

cO),

(ii)

O=iI

Qi,

(iii) The dyadic cubes inFare mutually disjoint.

Proof. We define Eas the collection of dyadic cubes Q such that:

(a) Q ,

(b) diam Q 110dist(Q,cO),

and let F be the maximal subcollection ofEas in Remark2.2.4. This collection F is welldefined: letx O and (Qj,kj)jZ the dyadic sequence which contains x. So, there existsa Qj,kj and since x is fixed, we can impose the condition (b). This proves (ii), and

(iii) and by construction diam Q 110dist(Q,cO) for every Q F. It remains to check

the lower bound.

So, take Q F. Then, by maximality, eitherQ or dist(Q, cO)< 10 diamQ. Thus,in either case,

dist(Q, cO) 10 diamQ.Combining this with the fact that diamQ = 2 diam Q, we find

dist(Q, cO) 10 diamQ + diamQ (20 + 2) diam Q 30 diam Q,and this completes the proof.

Remark 2.3.2. In (i), the constant 1/10 could be replaced by 1/2 for all > 0.However, this would change the constant1/30 in the lower bound.

The Whitney dyadic cubes introduced in the preceding theorem satisfy some importantproperties.

Proposition 2.3.3. The Whitney Dyadic cubes ofO satisfy:

(i) For all i I, 3Qi O,

(ii) For all i, j I, if3Qi 3Qj = , then

1

4

diam Qidiam Qj

4,

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(iii) There exists a constant only depending on dimension,C(n), such that

iI3Qi C(n).

Proof. (i) Suppose there exists an z cO 3Qi. So, there exists a y Qi such thatd(y, z) diam Qi. But,

10 diam Qi dist(Qi,cO) dist(y, cO) dist(y, z) diam Qi

which is a contradiction.

(ii) By symmetry, it suffices to prove that

diam Qi

diam Qj4.

Let y 3Qi 3Qj . We note that dist(y, Qi) diam Qi since y 3Qi, and by thetriangle inequality,

10 diam Qi dist(Qi,cO) dist(y, cO) + dist(y, Qi) dist(y,

cO) + diam Qi

which shows that dist(y, cO) 9 diam Qi.

Also, there existsz Qj such thatd(y, z) diam Qj and dist(y, cO) dist(z, cO) +diam Qj . We estimate dist(z, cO). Fix w Qj and we find

dist(z, cO) d(z, w) + dist(w, cO) diam Qj+ dist(w,cO).

By taking an infimum over all w Qj and using the fact that dist(Qj, cO) 30 diam Qj, we find that dist(z, cO) 31 diam Qj and consequently dist(y, cO) 32 diam Qj.

Putting these estimates together, we get that

diam Qidiam Qj

32

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and this completes the proof.

Theorem 2.3.4 (Whitney Covering Theorem for metric spaces). Let (E, d) be a metricspace, and letO Ebe open. Then there exists a set of ballsE ={B}Iand a constantc1< independent ofO such that

(i) The balls inEare mutually disjoint,

(ii) O=

Ic1B,

(iii) 4c1BO.

Proof. Let(x) = dist(x, cO), and fix 0<

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By the triangle inequality,

(x) = dist(x,cO) dist(z, cO) + d(x, z) dist(z,

cO) +1

2

(x)

and we conclude1

2dist(z, cO).

Furthermore,

dist(z, cO) dist(x,cO) + d(z, x) (x) +

1

2(x) =

3

2(x).

Combining these two estimates, and by symmetry we conclude that

1

3

(x)

(x)

3.

Let E =

B(x,3(x)) : A

. This collection of balls are mutually disjoint by the

previous inequality. Now,

d(x, x) d(x, z) + d(x, z)1

2(x) +

1

2(x)

1

2(x) +

3

2(x) 2(x).

Now, set

C=3+ 2

3

,

and combining this with the estimate above, B (x, 3(x)) B(x, C3(x)). So again,the volumes of the balls can be compared since with constant Csince is fixed, and usinga volume argument as in the proof of Besicovitch (Theorem 2.1.4) we attain a bound oncard A depending only on dimension.

Remark 2.3.6. We note that the preceding proposition can be proved for a metric spacehaving the following structural property: There exists a constant0 < C 0, the number of mutually disjoint balls of radiusR contained in a ball of radius2R is bounded byC.

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Chapter 3

Maximal functions

3.1 Centred Maximal function onRn

We begin with the introduction of the classical Hardy-Littlewood maximal function.

Definition 3.1.1 (Hardy-Littlewood Maximal function). Let be a reference measure,Borel, positive, and locally finite. Let be a second, positive, Borel measure. Define:

M()(x) = sup

r>0

1

(B(x, r))

B(x,r)

d= sup

r>0

(B(x, r))

(B(x, r))

for eachx Rn. IffL1loc(d), then setd(x) =|f(x)|d(x) and define

M(f)(x) =M()(x) = supr>0

1

(B(x, r))

B(x,r)

|f| d.

Remark 3.1.2. By convention, we take 00 = 0.

The first and fundamental question is to ask the size ofM() in terms of.

Theorem 3.1.3 (Maximal Theorem). Suppose that is a finite measure. Then there

exists a constant depending only on the dimensionC=C(n)> 0 such that for all >0,

{x Rn :M()(x)> } C

(Rn).

Remark 3.1.4. (i) The functionM()may not be a Borel function but it is a Lebesguemeasurable function. This a reason we wanted our measures to be defined on arbitrarysets. The proof however does illustrate that{x Rn :M()(x)> } B whereBis a Borel set with(B) C (R

n).

(ii) With some regularity assumptions on , M() becomes lower semi-continuous.That is, for all >0, the set{x Rn :M()(x)> } is open.

For example, suppose the map (x, r) (B(x, r)) is continuous (or equivalently(Sn1(x, r)) = 0 (Exercise)). Then, M() is lower semi-continuous (Exercise).

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Proof of Maximal Theorem. LetO= {x Rn :M()(x)> }. So, for allx O, thereexistsBx= B(x, rx) such that (Bx)> (Bx). Fix R >0, and apply Besicovitch to thesetO B(0, R). So, there is an E O B(0, R) at most countable such that

O B(0, R)

xE

Bx andxE

Bx C(n).

Then,

(O B(0, R))xE

(Bx)xE

1

(Bx) =

xE

1

Rn

Bx d

= 1

Rn

xE

Bx d C(n)

(Rn).

The sets O B(0, R) are increasing as R , so

(O) = limR

(O B(0, R)) C(n)

(Rn)

which completes the proof.

Corollary 3.1.5. For allfL1(d) and for all >0,

{x Rn :M(f)(x)> } C(n)

Rn

|f| d.

Theorem 3.1.6(Lebesgue Differentiation Theorem). LetfL1loc(d). Then, there existsanLf Rn such that(cLf) = 0 and

limr0

1

(B(x, r))

B(x,r)

|f(x) f(y)| d(y) = 0

for allx Lf. In particular,

f(x) = limr0

1

(B(x, r))

B(x,r)

f(y)d(y)

for allx Lf.

Remark 3.1.7. 1. This is a local statement, and so we can replace f by fK whereKis any compact set. Consequently we can assume thatfL1(d).

2. Iffis continuous, then we can takeLf = . That is, the Theorem holds everywhere.

Proof. Define:

f(x) = lim supr0

1

(B(x, r))

B(x,r)

|f(x) f(y)| d(y)

and we remark that the measurability off is the same as the measurability ofM(f).Note that fis subadditive, that is, f+g f+ g. Also, f |f| + M(f).

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Let > 0, and since C0c(Rn) is dense in L1(d) (since is locally finite), there exists ag C0c(R

n) such that

Rn

|f g| d < .

Now, observe that from the previous remark (ii), g = 0 and so it follows that

ffg+ g |f g| + M(f g).

Fix >0 and note that

{x Rn :f(x)> }

x Rn :|(f g)| (x)>

2

x Rn :M(f g)(x)>

2

.

From this, it follows that

{x Rn :f(x)> }

x Rn :|(f g)| (x)>

2

+

x Rn :M(f g)(x)>

2

12

Rn

|f g| d +C(n)

2

Rn

|f g| d

1 + C(n)

2

.

Now letting 0, we find that {x Rn :f(x)> }= 0 and {x Rn :f(x)> 0}=limn

x Rn :f(x)> 1n

= 0. To complete the proof, we set Lf ={x Rn :f(x)> 0}.

Remark 3.1.8 (On the measurability off). Let

f; 1k

(x) = lim supr0, r 1

k

1

(B(x, r))

B(x,r)

|f(x) f(y)| d(y)

and

M1k (f)(x) = sup

r>0, r 1k

1

(B(x, r))

B(x,r)

|f| d.

Then, note that asn , f; 1k

f. Now, for all Q + Q,

|f(x) f(y)| |f(x) | + | f(y)|

and so it follows that

f; 1k

(x) |f(x) | + M1k ( f)(x)

and also that

f, 1k

(x) infQ+Q

|f(x) | + M1k ( f)(x)

.

By the density ofQ + Q, equality holds forx F whereF ={x Rn :f(x) C}. SincefL1(d), we have(cF) = 0.

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Cheap proof. We use Cheap Trick3.2.3coupled with the Maximal theorem for M(f).

The preceding proof has the disadvantage that it is inherently tied up with Rn. Thefollowing is a better proof.

Better proof. Define: O=

x Rn :M(f)(x)>

,

Mm ()(x) = supBx, radBm

(B)

(B),

and O m = x Rn :Mm (f)(x)> . Then, O

= m=1 Om .

Fix m. Then, for all x O m , there exists a ball Bx with x Bx with rad Bx m suchthat

1

(Bx)

Bx

|f| d > .

By a repetition of the argument in Lemma 3.2.4, B Om makingM(f) Borel.

Let B = {Bx}xOm, and apply the Vitali Covering Lemma 2.1.1. So, there exists a

countable subset of centres C Om and mutually disjoint

Bxj

xjC Bsatisfying

Om xjC

5Bxj .

Therefore,

(Om )

xjC

(5Bxj ) C3xjC

(Bxj )

C3

xjC

Bxj

|f| d=C3

xjC

Bxj

|f| dC3

Om

|f| d C3

O

|f| d

Now, O m is an increasing sequence of sets and so we obtain the desired conclusion lettingm .

Remark 3.2.6. This proof is not only better because it frees itself from Besicovitch to themore general Vitali, but we get a better estimate since the integral is overO rather thanRn.

Corollary 3.2.7 (Lebesgue Differentiation Theorem for doubling measures). We have

f(x) = limBx, radB0

1

(B)

B

f(y) d(y)

for-almost everywherex Rn.

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3.3 The Dyadic Maximal function

Let be a positive, locally finite Borel measure.

Definition 3.3.1(Dyadic Maximal function). LetQ0 Q and letD(Q0)be the collectionof dyadic subcubes ofQ0. Define:

MQ(f)(x) = supQD(Q0), Qx

1

(Q)

Q

|f| d

forx Q0 andfL1loc(Q0; d).

Theorem 3.3.2 (Maximal theorem for the dyadic maximal function). Suppose f L1(Q0; d) and >0. Then,

x Q0: MQ(f)(x)>

1

Q0

|f| d.

Remark 3.3.3. Notice here that the bounding constant here is 1. It is independent ofdimension and.

We give two proofs.

Proof 1. Let =

x Q0: MQ(f)(x)>

. If = and x , there exists a

Q D(Q0) such that x Q and

1

(Q)

Q

|f| d > . ()

Let Cbe the collection of all such cubes. Since Q is countable, Cis also countable. Then,for every Q C, we have that Q so therefore, =

QCQ . LetM ={Qi}iI C

be themaximal subcollection. Then, M is a partition ofQ0. So,

() =iI

(Qi) 1

iI

Qi

|f| d 1

|f| d.

Remark 3.3.4. (i) The uniqueness ofM is a consequence of the disjointness of thedyadic cubes at each generation.

(ii) The proof gives us an even sharper inequality since we are only integrating on the set rather thanQ0.

Proof 2. If1

(Q0)

Q0

|f| d > ,

then = {Q0} and theres nothing to do. So assume the converse. We construct amutually disjoint subset F by the following procedure. Consider a dyadic child of Q0,say,Q . IfQ satisfies (), we stop and putQ in F. Otherwise, we apply this procedure to

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Q in place ofQ0. That is, we consider whether a given dyadic child ofQ to satisfies ().The collection F is called the stopping cubes for the property (). 1

We claim that:=

QF

Q.

Clearly, Q F implies that Q (Proof 1). Suppose that there exists x butxQFQ . But we know thatM

Q(f)(x)> so there exists a dyadic cube Q

D(Q0)such thatx Q satisfying the property (). The stopping time did not stop before Q byhypothesisx

QFQ , butQ

satisfies (). By the construction ofF, we have Q Fwhich is a contradiction.

The estimate then follows by the same calculation as in Proof 1.

Remark 3.3.5. The collectionFis the maximal collectionM from Proof 1.Definition 3.3.6 (Dyadic maximal function onRn). Define:

MQ(f)(x) = supQQ, Qx

1

(Q)

Q

|f| d

forx Rn andfL1loc(d).

Corollary 3.3.7. SupposefL1(Rn; d) and >0. Then,

x Rn :MQ(f)(x)>

1

Rn

|f| d.

Proof. Since we are considering dyadic cubes, we begin by splitting Rn into quadrants,and let g = f(R+)n. We note that it suffices to prove the statement for one quadrantsince the argument is unchanged for the others.

For each k N, letQk = [0, 2k)n. Then, note that:

MQ(f)(x) = supkN

gk(x)

where

gk(x) = supQD(Qk), QQk, Qx

1

(Q)

Q

|f| d.

Now, let k=

x Qk :gk(x)>

. We compute:

{x (R+)n :gk(x)> }= (k) 1

Qk

|f| d 1

(R+)n

|f| d.

The desired conclusion is achieved by letting k and summing over all quadrants.

Corollary 3.3.8. ForfL1loc(d), we have

f(x) = limQQ, L(Q)0,Qx

1

(Q)

Q

f(y) d(y)

for-almost everywherex Rn.1This is a stopping time argument, a typical technique in probability.

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Proof. Exercise.

Remark 3.3.9. For a fixedx, the set{Q Q : L(Q) 0, Q x} is really a sequence.

Consequently, the limit is really the limit of a sequence.

We have the following important Application.

Proposition 3.3.10. ForfL1loc(d),

|f|

MQ(f)

M(f)

M(f)

for-almost everywherex Rn.

Remark 3.3.11. The centred maximal function uses Besicovitch and is confined to Rn.The others do not and can be generalised to spaces with appropriate structure. For instance,in the case of the Dyadic maximal function, we must be able to at least perform a dyadicdecomposition of the space.

A consequence of the Lebesgue differentiation is:

M(f) M(f).

Exercise 3.3.12.Try to compareM

Q

withM (andM

). For simplicity, take=L

(but note that this has nothing to do with the measure but rather the geometry).

3.4 Maximal Function on Lp spaces

In this section, we let M denote either M, M or M

Q . We firstly note that for every

fL(d),Mf(x) f

for allxR

n. So,Mfis a bounded operator on L(d). We investigate the boundednesson Lp(d) forp } 0, we can restrict to a -algebra (depending on g) where it is thecase. Then this reduces to the assumption is finite by approximating g via gN,R =g{xRn:g(x)N}B(0,R) and then applying the Monotone Convergence Theorem.

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We compute and apply Fubini:

p

0

{x Rn :g(x)> }p1 d

=p

0

{xRn:g(x)>}

1d(x)p1 d

=

Rn

g(x)0

pp1 d d(x)

=

Rn

gp d(x).

Theorem 3.4.2 (Boundedness of the Maximal function on Lp). Let 1 < p 0 such that wheneverfLp(d)thenMfLp(d)

andMfLp(d) CfLp(d).

For the case ofM = M, we assume that is doubling andCmay also depend on theconstants in the doubling condition.

Proof. Assume that fL1(d) Lp(d). Let >0 and define

f(x) =

f(x) |f(x)| 20 |f(x)|< 2

and

|f| |f| + 2

.

By the subadditivity of the supremum,

Mf Mf+

2

and

{x Rn :Mf(x)> }

x Rn :Mf(x)>

2

.

Therefore,

{x Rn :Mf(x)> } x Rn :Mf(x)> 2 C Rn |f| d.Then,

p

0

{x Rn :Mf(x)> } p1 d p 0

C

Rn

|f| p2 dd

C p

Rn

2f(x)0

|f| p2 dd

C2p1 p

p 1

Rn

|f|p d

Now, for a general f Lp(d) by the density of L1(d) Lp(d) in Lp(d), we can take

a sequence (for example simple functions) fk L1

(d) Lp

(d) satisfying |fk| |f|.Then, Mfk Mf and the proof is complete by invoking the Monotone ConvergenceTheorem.

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Remark 3.4.3. (i) The constantC(n, p) satisfies:

C(n, p) C2p1 p

p 11p

()

So, if we letp , then we find thatC2p1

p

p 1

1p

1.

But asp 1, C pp1 and blows up!

In fact it is true that if0 =fL1(dL), thenMfL1(dL)(Exercise). Forp = 1,the best result is the Maximal Theorem.

(ii) We remark on the optimality ofC(n, p). Consider the inequality (), and note that

the C is the Maximal Theorem constant. This depends on n and the measure (unless we considerMQ).

Suppose = L. Then the best upper bound for C with respect to n is n log n.Consider the operator norm ofM:

Mp,n= supfLp(dL), fLp(dL)=1

MfLp(dL).

IfM= M(ie, the centred maximal function), thenMp,ncan be shown to be boundedinn for any fixedp >1. This is important in stochastic analysis.

Now, supposeM= MQ. Then,

Mp,n= pp 1

.

To see this, first note that

L{x Rn :Mf(x)> } 1

{xRn:Mf(x)>}

|f| dL

and

MfpLp(dL)=

Rn

|Mf| dL

=p

0

L{x Rn :Mf(x)> } p1 d

p

0

{xRn:Mf(x)>}

|f| dLp1 d

=p

0

Mf0

p1 d

|f| dL

= p

p 1

Rn

(Mf)p1 |f| dL

p

p 1Mfp1

Lp(dL)fp

Lp(dL)

which shows that

MfLp(dL)

p

p 1 fLp(dL).

Optimality can be shown via martingale techniques. We will not prove this.

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We give the following important application of Maximal function theory.

Theorem 3.4.4 (Hardy-Littlewood Sobolev inequality). Let0 < < n,1 p < nn andq satisfying

1

p+

n= 1 +

1

q.

Then, with = 1||

,

(i) Whenp= 1, there exists a constantC(, n) such that for all >0,

L{x Rn :| u(x)|> } C

n

un

L1(dL),

(ii) Whenp > 1, there exists a constantC(p, , n) such that

uLp(dL) CuLp(dL),

where

( u)(x) =

Rn

1

|x y|u(y) dL(y)

is the convolution ofu with.

Remark 3.4.5 (Motivation). Such a arises when trying to integrate functions inRn. Such potentials are one way to anti-derive. Formally, in the case of= n 2,

1|x|n2 u() =C(n) u()||2where denotes theFourier Transform. See [Ste71a].

Proof (Hedbergs inequality). We prove (i). Letu Cc (Rn), and set w = u. Wetake >0 to be chosen later. Then,

w(x)

|xy|

1

|x y||u(y)| dL(y) +

k=0

2k1|xy|2k

1

|x y||u(y)| dL(y).

First, whenever|x y| ,|xy|

1

|x y||u(y)| dL(y)

1

uL1(dL).

So, fix k 0. Then,

2k1|xy|2k

1

|x y||u(y)| dL(y)

(2k1)L(B(x, 2k))

L(B(x, 2k))|xy|2k

|u(y)| dL(y)

bn2(2k)nML(u)(x)

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where bn= L(B(0, 1)), the volume of the n ball. It follows then that

k=0 2k1|xy|2k1

|x y||u(y)| dL(y)

k=0

bn2(2k)nML(u)(x)

n bn2

1 2(n)ML(u)(x).

Now, let

A= A(n, ) = bn2

1 2(n)

and solve forsuch thatnA(n, )ML(u)(x) =uL1(dL).Since ifu0 then MLu(x)=0 almost everywhere (which we leave as an exercise),

=

uL1(dL)

A(n, )ML(u)(x)

1n

.

Then, for almost everywherex Rn,

w(x) uL1(dL)+ nA(n, )MLu(x)

=2uL1(dL)

= 2 uL1(dL)

A(n, )ML(u)(x)

nuL1(dL)

= 2

A(n, )ML(u)(x)

uL1(dL)

1q

uL1(dL)

=A(n, )1q ML(u)(x)

1q u

1 1q

L1(dL).

From this, it follows that,

L{x Rn :w(x)> } Lx Rn :ML(x)> qu11qL1(dL)q

C(n)

q

u

1 1q

L1(dL)

quL1(dL)

= C(n)

q uq

L1(dL).

The density of Cc (Rn) in L1(dL) completes the proof.

We leave (ii) as an exercise.

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Chapter 4

Interpolation

4.1 Real interpolation

Suppose that (M, ) and (N, ) are measure spaces and the measures and are-finitemeasures.

Recall that when 0 }=0. This is a Banach space and the norm is then given by

f= esssup |f|= inf{ >0 : {x M :|f(x)|> }= 0} .

We define a generalisation of these spaces called the WeakLp spaces.

Definition 4.1.1 (Weak L

p

space). Let 0 0

p {x M :|f(x)|> }< .

For a functionfLp,(M, ), we define a quasi-norm:

fp,=

sup>0

p {x M :|f(x)|> }

1p

.

Whenp= , we letLp,(M, ) = L.

Remark 4.1.2. (i) The WeakL

p

spaces are really a special case ofLorentz spaces L

p,q

.They truly generalise theLp spaces for they are equal to Lp whenq= p. See [Ste71b]for a detailed treatment.

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(ii) We emphasise that for 0 }

x M :|f|>

2

x M :|g|>

2

.

(iii) When1< p }

fp d(x) 1

pfpp.

In particular, Lp(M, ) Lp,(M, ).

Remark 4.1.5. The inclusion is strict in general. For instance,

1|x|

L n ,(Rn, dL) \ Ln (Rn, dL),

when0< .

Remark 4.1.6 (-finiteness of). The above definitions do not require-finiteness of.We will, however, require this in the interpolation theorem to follow.

Definition 4.1.7(Sublinear operator). LetK = R orK = C, and letFMdenote the spacemeasurable functions f : M K. LetDM be a subspace ofFMand letT :DM FN.We say thatT is sublinear if

|T(f1+ f2)(x)| |T f1(x)| + |T f2(x)|

for-almost allx N.

Example 4.1.8. (i) T :f MLf whereD= L1loc(R

n, dL).

(ii) Any linearT is sublinear.

Definition 4.1.9 (Weak/Strong type). LetT :DM FN be sublinear and let1 p, q. Then,

(i) Tis of strong type(p, q)ifT :DMLp(M, ) Lq(N, )is a bounded map. That is,

there exists aC >0 such that wheneverf DM Lp(M, ) we haveT fLq(N, )

andT fq Cfp.

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(ii) Tis of weak type(p, q)(forq 0 such that wheneverf DM Lp(M, ) we have

T fLq,(N, ) and

T fq, Cfp.

(iii) T is of weak type (p, ) if it is of strong type(p, ).

Remark 4.1.10. Note that ifT is of strong type(p, q) then it is of weak type(p, q).

Example 4.1.11. (i) The maximal operatorT :f Mfis of weak type(1, 1)and ofstrong type(p, p) for1< p .

(ii) The operatoru u where is the potential in Theorem3.4.4is of weak type(1, n ) if0< < n on(R

n,L).

Theorem 4.1.12 (Marcinkiewicz Interpolation Theorem). Given (M, ) and (N, ) let

DMbe stable under multiplication by indicator functions. That is, iff DM thenXfDM. Let1 p1 < p2 , 1 q1 < q2 , withp1q1 andp2q2. Furthermore, letT :DM FN be a sublinear map that is of weak type (p1, q1) and(p2, q2). Then, for all

p (p1, p2), T is of strong type(p, q) withq satisfying

1

p =

1

p1+

p2and

1

q =

1

q1+

q2.

Proof. We prove the case when q1 = p1, q2 = p2 < and leave the general case as anexercise.

First, by the weak type (pi, pi) hypothesis for i = 1, 2 we have constants Ci > 0 such that

T fpi, Cifpi

for all f DM Lpi(M, ). Fix such an falong with p (p1, p2) andN

|T f|p d= p

0

{x N :|T f(x)|> } p1 d.

Fix >0 and define

g(x) =

f(x) |f(x)|>

0 otherwise

andh(x) =f(x) g(x).

So, g = f{xM:|f(x)|>} DM and h= f{xM:|f(x)|} DMby the stability hypoth-esis onDM.

Now,

M

|g|p1 d=

{xM:|f(x)|>}

|f|p1 d

M

|f|p1fpp1 d fpppp1

and by similar calculation,

M

|h|p2 =

{xM:|f(x)|}

|f|p2 d p2p

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Since T is sublinear, for-almost all x N,

|T f(x)| |T g(x)| + |T h(x)|

and so

{x N :|T f(x)|> }

x N :|T g(x)|>

2

+

x N :|T h(x)|>

2

.

We compute,

0

x N :|T g(x)|>

2

p1 d

0

Cp11

2

p1 gp1p1

p1 d

(2C1)p1

0pp11

{xM:|f(x)|}

|f|p1 dd

= (2C1)p1

M

|f|0

pp11 d

d

=(2C1)

p1

p p1

M

|f|p d.

By a similar calculation, but this time integrating from |f| to with for > 0, weget

0 x N :|T h(x)|> 2 d (2C2)p2

p2 p

M |f|

p

d.

Putting these estimates together,

N

|T f|p p

(2C1)

p1

p p1+

(2C2)p2

p2 p

M

|f|p d

which completes the proof.

Remark 4.1.13. If the definition ofg was changed to g = f{xM:|f(x)|>a} witha R+, then this leads to better bounds by optimising a. In fact, we can get a log convexcombination ofC1, C2. That is:

T fp C(p, p1, p2)C11 C

2fp.

For a functionh(), Log convex here means thatlog h() is convex.

4.2 Complex interpolation

We begin by considering a baby version of complex interpolation. Leta CN. Then,for 1 p

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and for p =

a= sup1iN

|ai|

are norms. Let M be an N Nmatrix, then

M(L(CN),p)= Mp,p= sup

aCN,ap=0

M apap

is the associated operator norm. We leave it as an (easy) exercise to verify that

M,= supi

Nj=1

|mij|

where M = (mij). Then,

M1,1= M, = sup

j

Ni=1

|mij|

where we have used the fact that 1 and are dual norms onCN. For 1< p

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This is really a special case of more general interpolation results of the form

Mq,q M1p,p M

r,r

for 1 p < q < r and1

q =

1

p +

r.

A proof of this statement cannot be accessed easily via elementary techniques since thereis no explicit characterisation ofMq,q andMr,r. This was the essential ingredient ofthe preceding proof. This more general statement comes as a consequence of the powerfulcomplex interpolation method.

We prove this by employing the following 3 lines theorem of Hadamard. First note thatin the following statement, Sdenotes the topological interior ofS and H(S) denotes theholomorphic functions on S.

Lemma 4.2.2 (3 lines Theorem of Hadamard). Let S = { C : 0 Re 1}. LetF H(S) andF C0(S). Further suppose thatF< onSand let

C0= suptR

|F(t)| and C1= suptR

|F(1 + t)| .

Then forx (0, 1), andt R,

|F(x + t)| C1x0 Cx1 .

Proof. Assume that C0, C1 > 0. If not, prove the theorem with C0+ , C1+ in place ofC0, C1 for > 0 to conclude that |F()| (C0+)1Re (C1+)Re and letting 0conclude that

|F()|= 0 =C1Re 0 CRe 1

for S.

Fix > 0 and set G() = F()C10 C1 e

2 for S. Then, G H(S), G C0(S).Now, fix R >0 and letQR= S { C : ImR}. By the maximum principle,

supQR

|G()|= supQR

|G()| .

We consider each part of QR. For =t with|t| R,

|G()|= |F(t)| C10 e

(t)2 1

and for = 1 + t with|t| R,

|G()|= |F(1 + t)| C11 e

Re (1t2+2t) e.

Then, for = x + R with|t| R,

|G()| |F(x + R)| Cx10 C

x1 e

Re (x2R2+2xR) CeRe (1R2)

where

C=

supS

|F()|

sup0x1

Cx10 Cx1

35

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and lastly, when =x R with|t| R,

|G()| CeRe (1R2)

by the same calculation since Re (x2 R2 + 2xR) = (x2 R2). So, for fixed, thereexists an R0 such that whenever R R0,

CeR2

1

|G(x)| e

S, R R0 such that QR and |G()| e

and so for all Sand all >0,

|F()| C10 C

1e2e .

The proof is then completed by fixing and letting 0.

Remark 4.2.3. This lemma can be proved assuming some growth onF for |Im | (rather thanFbounded). However, there needs to be some control on the growth.

Lemma 4.2.4 (Phragmen-Lindelof). Let C be the closed subset between the linesR 2 (obtained from the conformal map (

12)). SupposeF H(), C

0() andassume that F is bounded on the lines x 2 . Suppose there exists an A > 0, [0, 1)such that for all,

|F()| exp(A exp(|Re |)).

Then, |F()| 1 for.

Proof. The proof is the same as the proof of the preceding lemma, except e()2

= e2

term needs to be replaced by something decreasing faster to 0 as |Re | . We leavethe details as an exercise (or see [Boa54]).

We are now in a position to state and prove the powerful complex interpolation theoremof Riesz-Thorin.

Theorem 4.2.5 (Riesz-Thorin). Let 1 p < r and let (M, ), (N, ) be -finite

measure spaces. Let DM denote the space of simple, integrable functions on M, andT :DM FN beC-linear. Furthermore, assume thatT is of strong type(p, p) and(r, r)with bounds Mp and Mr respectively. Then, T is of strong type (q, q) for any q (p, r)with boundMq satisfying

MqM1

p Mr ,

where1

q =

1

p +

r

with0<

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Corollary 4.2.6. With the above assumptions, Thas a continuous extension to a boundedoperatorLq(M,d) Lq(N,d) for eachq (p, r).

Proof. Fix q (p, r) and since q < , DM is dense in Lq(M,d). Then, define theextension by the usual density argument.

Proof of Riesz-Thorin. Fix p < q < r and note that DM is dense in Lq(M,d), DN isdense in Lq

(N,d) and by duality, Lq(N,d) = Lq

(N,d). Thus, it suffices to show that

A= supfDM, fq=1, gDN, gq=1

N

Tf g d


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Note thata() = 1 if and only ifb() = 1, and f=f and g = g. Set

F() =

N

T f g d

and note that it is well defined for each by the same reasoning as previously for f andg in place off and g. Furthermore,

F() =

k

l

|k|a() |k|

b() k|k|

l|l|

N

TAk Bl d

soF H(C) H(S), C0(S) and

F() =

N

Tf g d.

In order to apply Lemma 4.2.2we estimate suptR |F(t)| , suptR |F(1 + t)|. So, by thestrong type (p, p) property ofT,

suptR

|F(t)|= suptR

N

T ft gt d

sup

tRT ftpgtp

MpsuptR

ftpgtp .

Using the fact that Ak are mutually disjoint, we can write

ft

p

p= k |k|a(t)

p(A

k)

and since a(t) = qp + t

qr

qp

,

ftpp =

k

|k|q (Ak) =f

qq.

By a similar calculation,gtp

p =gq

q ifp


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Chapter 5

Bounded Mean Oscillation

In this chapter, the framework we work within is Rn with the metricd= dand Lebesguemeasure L. By Q(x, r), we always denote a ball with respect to d of radius r centred atx. That is, Q= Q(x, r) represents an arbitrary cube in Rn.

We remark that for the theory, there is nothing special about Rn and d. It is justconvenient to work in this setting. The following material could be defined and studiedsimilarly on spaces ofhomogeneous type.

5.1 Construction and properties ofBMO

We introduce some notation. Let mXfdenote the mean of the function fon the set X.That is

mXf= 1

L(X)

X

f dL.

Definition 5.1.1 (Bounded Mean Oscillation (BMO)). LetfL1loc(Rn), real or complex

valued. We say thatf hasbounded mean oscillation if

f = supQ

mQ |f mQ|= supQ

1L(Q)

Q

|f mQ f| dL

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forf, g BMO and K. Furthermore, f = 0 if and only iff constant almosteverywherex Rn.

(iii) BMO/K is a normed space with normf+ K= f,

making BMO/K a Banach space. BMO convergence is often called convergencemodulo constant.

(iv) ForfL(Rn), x0 Rn andt >0 the function defined by

ft,x0(x) =f

x x0

t

L(Rn).

Similarly, forfBMO, ft,x0 BMO andft,x0= f.

Proof. We prove that f = 0 if and only iff constant almost everywhere x Rn (ii)

and leave the rest as an exercise.

Note that the if direction is trivial. To prove the only if direction, assume thatf = 0. Then, for every cube Q R

n, f = mQ f almost everywhere x Q. Let Qj bean exhaustion ofRn by increasing cubes. That is, letQj = [2j, 2j]n for j = 0, 1, 2, . . . .Then, f = mQjfalmost everywhere on Qj and hence mQ0f = mQjf for all j. Letting

j we establish that fis constant almost everywhere onRn.

Exercise 5.1.3. 1. LetA GLn(K)andx0 Rn. WritefA,x0(x) =f(Axx0). Showthat iffBMO thenfA,x0 BMO and

fA,x0 2An,det A

1f.

(Hint: Use the next Lemma). This exercise illustrates that we can indeed change theshape of cubes.

2. Show that

f= supB

mB(|f mBf|)<

whereB are Euclidean balls defines an equivalent semi-norm onBMO.

Lemma 5.1.4. LetfL1loc(Rn) and suppose there exists aC >0 such that for all cubes

Q, there exists acQ K such thatmQ |f cQ| C. Then, fBMO andf 2C.

Proof. We write

f mQ f=f cQ+ cQ mQ f=f cQ+ mQ(cQ f)

and it follows that

mQ |f mQ f| mQ |f cQ| + mQ |cQ f| 2 mQ |f cQ| 2C.

40


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Example 5.1.5 (ln |x| BMO). In particular, this implies thatL(Rn) BMO.

We show that this is true forn= 1. Letf(x) = ln |x|. Fort >0,

ft(x) = lnx

t

= ln |x| ln |t|= f(x) + c.So, forIan interval of lengtht,

mI |f mIf|= mI |ft mIft| .

By change of variables, y= xt J whereJis of unit length and

mI |ft mIft|= mJ|f mJf|

and so we are justified in assuming thatIhas unit length. So, letI= [x0 12 , x0+

12 ] and

by symmetry, assumex0 0.

Consider the case0 x0 3. Then,

mI |f|=

I

|f(x)| dx

72

12

|ln |x|| dx 0

0 x 0.

Then, f BMO and consequently, BMO is not stable under multiplication by indicatorfunctions.

41

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Proof. LetI= [, ] for >0 and small. We compute,

mI |f mIf|= 1

2

0

|ln |x| mIf| dL(x) + 1

2

0

|mIf| dL(x).

Then,

1

2

0

|mIf| dL(x) = 1

2

0

12 0

ln |x| dLx

dL(y)=

1

2

1

2| ln + |

= 1

4|ln + 1| .

But this is notbounded as 0.

Proposition 5.1.7 (Further properties of BMO). (i) Whenever f BMO, then |f|

BMO and |f| 2f.

(ii) Suppose f, g BMO are real valued. Then, f+, f, max(f, g), min(f, g) BMO.Furthermore,

max(f, g), min(f, g)3

2(f+ g) .

(iii) LetfBMO real valued. Then we have the followingApproximation by truncation.Let

fN(x) =

N f(x)> N

f(x) Nf(x) N

N f(x)< N

for N R+. Then, fN L(Rn), fN 2f andfN f almost everywhereinRn.

(iv) Assumef is complex valued. ThenfBMO if and only ifIm f, Re fBMO and

Im f, Re f f Im f+ Re f.

Proof. (i) Let CQ = |mQ f|. Then, ||f| CQ| |f mQ f| and so mQ ||f| CQ| mQ |f mQ f| f. Then, apply Lemma5.1.4.

(ii) Apply (i). Exercise.

(iii) Pick Q a cube, and let x, y Q. Then,|fN(x) fN(y)| |f(x) f(y)| and

fN(x) mQ fN =

Q

(fN(x) fN(y)) dL(y).

So, Q

|fN(x) mQ fN|

Q

Q

|fN(x) fN(y)| dL(x)dL(y)

Q

Q

|f(x) mQ f+ mQ f f(y)| dL(x)dL(y)

Q

Q|f(x) mQ f| + |mQ f f(y)| dL(x)dL(y)

2f.

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(iv) Exercise.

Exercise 5.1.8. Find a functionfL1loc(Rn) with|f| BMO, butf /BMO.

Proposition 5.1.9. LetQ, R be cubes withQ R. LetfBMO. Then,

|mQ f mR f| L(R)

L(Q)f.

Proof. We compute,

|mQ f mR f| mQ |f mR f|

= 1L(Q)

Q

|f mR f|

1

L(Q)

R

|f mR f|

L(R)

L(Q)

1

L(R)

R

|f mR f|

L(R)

L(Q)f.

Corollary 5.1.10. (i) Suppose thatL(R) 2L(Q)withQ R. Then|mR f mQ f| 2f.

(ii) Suppose thatQ, Rare arbitrary cubes (not necessarilyQ R). Then,|mR f mQ f| Cnf (Q, R) where

(Q, R) = ln

2 +

L(R)

L(Q)+L(Q)

L(R)+

dist(Q, R)

(L(Q) L(R))1n

.

Proof. We leave the proof as an exercise but note that the proof of (i) is easy. The proof

of (ii) requires a telescoping argument.

5.2 John-Nirenberg inequality

BMO was invented by John-Nirenberg for use in partial differential equations.

Theorem 5.2.1 (John-Nirenberg inequality). There exist constants C = C(n) 0 and = (n)> 0 depending on dimension such that for allf BMO withf = 0 and forall cubesQ and >0,

L{x Q :|f(x) mQ f|> } Ce fL(Q)

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Remark 5.2.2 (On the exponential decay). Note that by the definition of theBMO normcombined with Tchebitchev-Markov inequality (Proposition4.1.4), we get decay in 1 since

L {x Q :|f(x) mQ f|> }

1

{xQ:|f(x)mQf|>}

|f mQ f| dL 1

fL(Q).

It is natural to ask the question why we get extra gain into exponential decay. The reasonis as follows. The expression above is for a single cube. But the exponential decay comes

from the fact that we have scale and translation invariant estimates. This is typical inharmonic analysis.

Proof of the John-Nirenbeg inequality. First, note that it is enough to assume that Q =Q0= [0, 1)n by the scale and translation invariance of BMO. Furthermore, we can assume

that mQ0f = 0 since f = f mQ f. By multiplying by a constant coupled withthe fact that is a semi-norm, we need to only consider f= 1.

Let F = {x Q :|f(x)|> } . We show that L(F) Ce. We prove this for fN

(the truncation off) and letN to establish the claim via the monotone convergencetheorem. So, without loss of generality, assume thatfL(Rn).

Consider the case when > 1. As before, let D(Q) denote the dyadic subcubes ofQ. Let E =

x Q : MQf(x)>

and we have that F E up to a set of null

measure (since |f| MQ

f almost everywhere). Hence, L(F) L(E) and we showthat L(E) Ce. Then, by definition

supQD(Q), Qx

mQ |f|= MQf(x)

andmQ |f|= mQ |f mQ f| f= 1.

Coupled with this and the assumption that >1, we have that E Q. Let C ={Qi,}be the maximal collection of subcubes Q D(Q) such thatE = Qi,. So,

mQi, |f|> and mQi,|f|

withQi, Q.For each i, we estimate

mQi,f mQi, f . Let {Rk}nk=0 be a set of cubes such thatQi,= R0 R1 Rn=Qi,

andL(Rk+1)

L(Rk) 2.

It is trivial that such a collection exists. By this we have that for all k ,mRk+1f mRkf 2f= 244

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and summing overk , mQi,f mQi, f

2n.

Therefore, mQi,f 2n + mQi, f 2n + .Now, pick a > 2n+ 1 to be chosen later. Let C = {Qj,+} be the maximal disjointcovering for E+. Then, E+ E and for each j, there exists a unique i such thatQj,+ Qi,. Fix i, and estimate:

L(E+ Qi,) = L

{j:Qj,+Qi,}Qj,+

= {j:Qj,+Qi,}L(Qj,+)

1

+

{j:Qj,+Qi,}

Qj,+

|f| dL

1

+

E+Qi,

|f| dL

1

+

Qi,

f mQi,f dL + 1 + mQi,fL(E+ Qi,)

1

+ |Qi,| f+

2n +

+ L(E+ Qi,).

Therefore,

L(E+ Qi,) 1

2nL(Qi,)

and summing overi,

L(E+) 1

2nL(E).

Now, fix such that 1< 2n and observe that

L(E+k) 1 2nkL(E)for k N. If 2, then there exists a unique k N such that 2 + k

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Definition 5.2.3 (BMOp). ForfLploc(R

n), define

f,p= sup

Q

(mQ |f mQ f|p)

1p

and define

BMOp =

fLploc(Rn) :f,p<

.

Remark 5.2.4. Note thatBMOp BMO andf f,p.

Corollary 5.2.5. For all10 such that

supQ

Q

exp(|f mQ f|) dL 0

Q(x,r)

|f(y)| dL(y)

and the uncentred cubic maximal function:

Mf(x) = sup

Qx

Q

|f(y)| dL(y).

Proposition 5.3.2. There existC1, C2> 0 such that

C1MLf MfC2MLf

andC1MLf M

fC2MLf

46

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Proof. The proof follows easily noting that there exist constants A and B such that forevery cube Q, there exist balls B1, B2 with the same centre such that B1 Q B2 andAL(B1) = L(Q) =BL(B2).

Remark 5.3.3. In particular, this means that we can simply substituteM andM inplace ofML andM

L

in Theorems and obtain the same conclusions.

We now introduce a new type of maximal function which will be the primary tool of thissection.

Definition 5.3.4 (Sharp maximal function). ForfL1loc(Rn) andx Rn, define

Mf(x) = supQx

mQ |f mQ f| .

Remark 5.3.5. (i) f BMO if and only if M

f L

(Rn

). Furthermore, f =Mf.

(ii) Mf2Mf.

In particular (ii) means that iffLp(Rn) with1 } (I)

wheref, g are measurable, >0, > 1, (0, 1) and(, )> 0.

Proposition 5.3.7. Suppose there exists ap0 (0, ) such thatfp0 0. Then, for allp [p0, ) satisfying

1

Cp= sup

>1, (0,1)

(1 p(, ))> 0

we havefp (Cp)

1p

gp

for some >1 and } d.

47


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Also,

{x Rn :|f(x)|> } {x Rn :|f(x)|> , |g(x)| } {x Rn :|g(x)|> }

and so it follows that

L{x Rn :|f(x)|> } L{x Rn :|f(x)|> , |g(x)| }+L{x Rn :|g(x)|> }

(, )L{x Rn :|f(x)|> }+L{x Rn :|g(x)|> }

by invoking (I). Therefore, it follows that

pp N

0p1L{x Rn :|f(x)|> } d

pp(, )

Nk

0p1L{x Rn :|f(x)|> } d

+pp N

k

0p1L{x Rn :|g(x)|> } d

pp(, )

N0

p1L{x Rn :|f(x)|> } d

+p

p 0

p1L {x Rn :|g(x)|> } d.

By the assumption thatfp0 } d

pp

0

p1L{x Rn :|g(x)|> } d.

Then, apply the monotone convergence theorem to obtain the conclusion.

The goal is to prove the following important inequality.

Theorem 5.3.8 (Fefferman-Stein inequality). Letp0 (0, )andfL1loc(R

n)such that

M

fp0 0 (independent off) suchthatM

fp CpM

fp.

Corollary 5.3.9. Let p (1, ). Then, there exists a C(p, n) > 0 such that for allfLp(Rn),

Mfp C(p, n)M

fp.

In particular, fp Mfp M

fp onLp(Rn).

Proof. Apply the theorem withp0= p sincefLp(Rn) if and only ifMfLp(Rn).

To prove the Fefferman-Stein inequality, by Proposition5.3.7,it suffices to prove (I) withf replaced with M

f and g replaced withMf. First, we need two key Lemmas.

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Lemma 5.3.10 (Localisation for maximal functions). There exists 0 =0(n)> 1 suchthat for allfL1loc(R

n), for all cubesQ, and all >0 if there existsC >1 andx C Q

withMf(x) , then for all 0

Q

x Rn :Mf(x)>

x Rn :M(fMQ)(x)>

0

withM=C+ 2.

Proof. We know that Mf0Mf for some 0= 0(n)> 1. Let

x Q

x Rn :Mf(x)>

and so

Mf(x)> 0

>

since > 0. So, there exists an r >0 such that

1

|Q(x, r)|

Q(x,r)

|f(y)| dL(y)>

0.

First, x Q(x, r) since Mf(x) . This implies that x x r. Secondly, by

hypothesis, x C Q and letting xQ be the centre ofQ,

x x x xQ+ xQ x rad Q + Crad Q (C+ 1) radQ.

So, for any y Q(x, r)

y xQ y x+ x xQ r+ rad Q (C+ 1) rad Q + rad Q Mrad Q

Thus,

M(fMQ)(x)

Q(x,r)

|fMQ| dL =

Q(x,r)

|f| dL >

0

and completes the proof.

Lemma 5.3.11 (Proving (I)). Fixq (1, ] and 1. LetFL1loc(R

n), F0 such

that for all cubesQ there existsGQ, HQ: Rn R+ measurable with

(i) F GQ+ HQ almost everywherex Q,

(ii) For allx Rn,

MF(x)

supQxffl

Q HqQ

1q

q


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Then, there existsC= C(q, n)> 0 and0= 0(, n) 1 such that for all >0, >

0,

and (0, 1], (I) holds forMF andG in place off andg respectively. That is,

Lx Rn : MF(x) > , |G(x)| (, ) Lx Rn : MF(x) > where

(, ) =C

q+

.

Remark 5.3.12. Whenq= ,

is replaced by0.

Proof. Let >0 andE =

x Rn :MF(x)>

is open by the lower semi-continuity

ofMF.

IfE = Rn, theres nothing to do. So, suppose thatE= Rn and use a Whitney decompo-sition with dyadic cubes (Theorem2.3.1). So, E =Qi, mutually disjoint with diam Qicomparable with dist(Qi, cE). In particular, there exists a constant C = C(n) > 1 suchthat for all i, CQi cE = . That is, for each i, there exists a xi CQi such thatMf(xi) . SetDi= Qi

x Rn :Mf(x)> , G

, and so

i

L(Di) = L

x Rn :Mf(x)> , G

since 1.

We estimate each Di for each i. Assume Di =. So, there exists a yi Qi such thatG(yi) . So, by the Localisation Lemma 5.3.10

L(Di) L(Qi

x Rn :Mf(x)>

)

L

x Rn :M(fMQi)(x)>

0

A + B

where

A= Lx Rn :M(GMQiMQi)(x)> 20and

B = L

x Rn :M

(HMQiMQi)(x)>

20

.

We estimate A by invoking the Maximal theorem (weak type (1, 1)):

A 2C0

Rn

GMQiMQi dL 2C0

MQi

GMQi dL

2C0L(MQi)

G(yi)

2C

0L(MQi).

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By the Maximal theorem (weak type (q, q)) for q 20, thenx Rn :M(HMQiMQi)(x)>

0

= .

We are now in a position to prove the Fefferman-Stein inequality.

Proof of the Fefferman-Stein inequality. By hypothesis,fL1loc(Rn) such that M

fp0 0 for fixed and small .

Thus, we conclude that for some Cp> 0,

Mfp= M

Fp CpGp= CpMfp

and the proof is complete.

We have the following Corollary to the Fefferman-Stein inequality.

Corollary 5.3.13 (Stampacchia). Suppose thatT is sublinear onDRn, a subspace of thespace of measurable functions stable under multiplication by indicator functions. Suppose

further that T : Lp(Rn) Lp(Rn) for some p [1, ) and T : DRn L(Rn) BMOare bounded. Then for al l q (p, ), T is strong type (q, q) with log convex control ofoperators norms.

f

51

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Chapter 6

Hardy Spaces

6.1 Atoms and H1

Hardy spaces are function spaces designed to be better suited to some applications thanL1. We consider atomic Hardy spaces.

Definition 6.1.1 (-atom). LetQbe a cube inRn. A measurable functiona : Q C iscalled an-atom onQ if

(i) spt a Q,

(ii) a 1L(Q) ,

(iii)

Q a dL = 0.

We denote the collection of-atoms onQ byAQ andA =QA

Q .

Remark 6.1.2. Note that (i) along with (ii) implies thata1 1.

Definition 6.1.3 (p-atom). LetQ be a cube inRn. A measurable functiona: Q C iscalled anp-atom onQ if

(i) spt a Q,

(ii) ap 1

L(Q)1 1p

,

(iii)

Q a dL = 0. We denote the collection ofp-atoms onQ byApQ andA

p =QApQ.

Definition 6.1.4 (H1,p). Let1< p andf L1(Rn). We say thatf H1,p if thereexistp-atoms{ai}iN and(i)iN

1(N) such that

f=

i=0 iaialmost everywhere.

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Also,

bip Qi

|a|p

1p

+ |mQia|L(Qi)1p =

1

L(Qi)1 1

p

i

where i = 2 |mQi |a|p|

1p L(Qi). We note that i = 0. For otherwise, if i = 0,

then a = 0 on Qi but we assume that (mQi |a|p)

1p > >0 which is a contradiction.

Therefore, ai = 1i

bi is a p-atom.

Now, by Holders inequality and the Maximal Theorem,

i=1

i = 2

i=1

Qi

|a|p 1p L(Qi)1 1p 2

i=1

Qi

|a|p 1

pi=1

L(Qi)

1 1p

2Q |a|p1p

L(E

)11p 2Q |a|p Cp1 2 1L(Q)

p1

.

Now, we choose such that

2

1

L(Q)

p1=

1

2

to find=

cpL(Q)

,

with cp= 41

p1 . It then follows thatbH1,p 12 .

Now, consider g ,

g=

a on Q \ E

mQia onQi for eachi.

Certainly, onQ \ E,|a|p MQ(|a|p) p almost everywhere and so|g| almost

everywhere. OnQi, by maximality and Holders inequality,

|mQia| mQi |a| 2n mQi |a| 2n.

Hence,|g| 2n. It then follows that

g 2n

= 2

n cp

L(Q) .

We also have

Q g =

Q a = 0, so 12ncp

g AQ which implies thatg H1, with

gH1,2ncp.

(ii) Fixf0 H1,p with f0= 0. We show there exists a decomposition f0= f1+ g0 with

f1H1,p 2

3f0H1,p and g

0H1, 4

32ncpf0H1,p .

For every >0 there exists an atomic decomposition f0=

i=1 jaj with

i=1

|j | f0H1,p+ .

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We apply (i) to each aj to find a decomposition aj = bj +gj with bj H1,p 12 .

Certainly, f1 =

j=1 jbj exists since H

1,p is a Banach space (Proposition 6.1.7),and

f1H1,p 12

j=1

|j| 12f0H1,p+ .

So, choose = 13f0H1,p . Setting g0 =

j=0 jgj (the sum converging in H1,

because it is a Banach space andgjH1, 2ncp), we find

g0H1,2ncp

f0H1,p+

=

4

32ncpf0H1,p .

(iii) We iterate

f0= f1+ g0

f1= f2+ g1

f2= f3+ g2

... = ...

and so for each k ,

f0= fk+ g0 + g1 + + gk1.

Certainly fk 0 in H1,p, and g0 + g1 + + gk converges to g in H1, with

gH1, 43

2ncp

j=0

23j f0H1,p .

By Proposition6.1.7, convergence in H1, implies convergence in H1,p and sof0= gand f0 H

1,.

This motivates the following definition.

Definition 6.1.9 (Hardy space H1). We defineH1 to be anyH1,p for1 < p with thecorresponding norm.

6.2 H1 BMO Duality

We show that the dual space of H1 and BMO are isomorphic with equivalent norms. Thisrelationship was first established by C. Fefferman but using a different characterisation ofH1.

Theorem 6.2.1 (H1 BMO duality). The dual ofH1 is isomorphic to BMO with equiv-alent norms.

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Not quite a correct proof. We work with H1 = H1,2 and BMO = BMO2 with corre-sponding norms H1,2 and . Letb BMO2 and a A

2. Set

Lb(a) =

Rnb(x)a(x) dL(x).

Then, Lb(a) is well defined and since for all cubes Q we have spta Q and

Q a= 0,

|Lb(a)|

Q

|b mQ b|

12

a2 b.

By linearity, Lb is defined on VectA2. Let fVectA2. Then, f =

mi=1 iai and

Lb(f) b

m

i=1 iai.By density of VectA2 in H1,2, Lb can be defined on all of H

1,2.

Remark 6.2.2. This proof is correct if we know that for all f VectA2, fH1,2 inf

finite|i|. Note that we automatically have fH1,2 inf

finite|i|. This subtletywent unnoticed for some time, and it has only been recently thatfH1,2 inf

finite|i|

was proved by an intricate argument. Also this equivalence is not true for all atoms - thereexists a counter example for-atoms.

We take a different approach in proving the theorem.

Proof of theH1 BMO duality. We work with H1 = H1,2 and BMO = BMO2 with corre-sponding norms H1,2 and .

(i) Takeb L(Rn) and fH1,2. Define,

Lb(a) =

Rn

b(x)a(x) dL(x)

and note that it is well defined since H1,2 L1(Rn). Iff=i=1 iai, we can applyDominated Convergence since

Rn

i=1

|b(x)iai(x)| dL(x) b

i=1

|i| .

Also,

|Lb(a)|=

Rn

(b(x) mQ b) ai(x) dL(x)

b

if sptai Q. This implies that |Lb(f)| bi=1 |i| and taking an infimum over

all possible i,|Lb(f)| bfH1,2 .

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(ii) Now take b BMO2 and f VectA2. Let (bk)k=1 be the truncation of b. So,

|bk| |b| almost everywhere, bk b almost everywhere, and bk 2b. 1 Now,

suppose f =mi=1 iai and Lbk(f) =

mi=1 iLbk(ai). Since b BMO2, we have

that b L2loc(Rn) which implies b L2(sptai). So, certainly|bkai| |bai| L1(Rn)almost everywhere. Thus,

Rn

bkai dL

Rn

bai dL,

ie., Lbk(ai) Lb(a). This implies that

|Lb(f)| supk

|Lbk(f)| 2bfH1,2 .

(iii) We now apply a density argument and extendLb to the whole of H1,2. Let Lb denote

this extension.So we have shown that whenever b BMO2 we have Lb (H

1,2). LetT: BMO2(H1,2) denote the map b Lb.

(iv) We leave it as an exercise to show thatT is linear.

(v) We show that T is injective. Let b BMO2 such that Lb = 0. We show thatb isconstant.

Fix a cube Q and let fL2(Q) with

Q f dL= 0. Then, there exists a Ksuch

that f A2

Q which implies thatfVectA2. So,

0 = Lb(f) =Lb(f) =

Qbf dL

and since we assume

Q f= 0, it follows that b|Q is constant. By exhaustion ofRn

by increasing Q, we deduce that b is constant.

(vi) Lastly, we show that T is surjective. Let L (H1,2) and fix a cube Q. Let L20(Q) =fL2(Q) :

Q f dL = 0

and note that L20(Q) H

1,2. Take f L20(Q) and

=f2L(Q)12 . Then,

f

2

1

L(Q)12

so f is a 2-atom. Thus,

fH1,2 =f

H1,2

f2L(Q)12 .

Furthermore, L|L20(Q) (L20(Q)) and so by the Riesz Representation Theorem, there

exists a bQ L20(Q) such that for all fL

20(Q),

L(f) =

Q

bQf dL

andbQ2 L(H1,2)L(Q)1

2 .1Here we assume b real valued. For complex valued b, separate real and imaginary parts.

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LetQ, Q denote two cubes with Q Q. Then, wheneverfL20(Q),

L(f) = Q

bQf dL = Q

bQf dL

and so bQ bQ is constant almost everywhere in Q. Define b as follows:

b(x) =

b[1,1]n(x) x [1, 1]

n

b[2j ,2j ]n(x) + cj x [2j, 2j]n \ [2j1, 2j1]n, j1

wherecj is the constant such thatb [2j ,2j ]n b[1,1]n =cj on [1, 1]n.

We show that b BMO2, b L(H1,2) and L =Lb. Fix Q, and let j N

such that Q [2j, 2j]n. Let k be such that 2 k j. Then, ck ck1 =b[2k,2k]n b[2k1,2k1]n which is constant on [2

k1, 2k1] and in particular on

[2k1, 2k1] \ [2k2, 2k2]. Therefore,b(x) =b[2j ,2j ]n(x) +cj on all of [2j, 2j]nand in particular on Q. Also, there exists a constantc such that b[2j ,2j ]n bQ =con the cube Q and so b = bQ+ c + cj on Q. Then,

b mQ b= bQ+ c + cj mQ bQ c cj =bQ

since mQ bQ = 0. Therefore,

Q

|b mQ b|2 dL =

Q

|bQ|2 dL L2(H1,2)L(Q).

The fact that L = Lb

follows from the fact that L(a) =Lb(a) for all a A2.

Remark 6.2.3. BMOand atomicH1 can be defined on any space of homogeneous type.The results of this section go through in such generality. See [CR77].

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Chapter 7

Calderon-Zygmund Operators

7.1 Calderon-Zygmund Kernels and Operators

We denote the diagonal ofRn Rn by ={(x, x) :x Rn}.

Definition 7.1.1 (Calderon-Zygmund Kernel). Let 0 < 1. A Calderon-ZygmundKernel of order is a continuous functionK : c K such that there exist aC >0 andsatisfies:

(i) For all(x, y) c,

|K(x, y)| C

|x y|n,

(ii) For allx, y, y Rn satisfying |y y| 12|x y| whenx=y,K(x, y) K(x, y) C |y y||x y|

1|x y|n

,

(iii) For allx, x, y Rn satisfying|x x| 12|x y| whenx=y,

K(x, y) K(x, y) C |x x|

|x y| 1

|x y|n ,

We writeKCZK and norm it viaK= inf{C :(i) to (iii) hold}.

Remark 7.1.2. (i) The constant 12 can be replaced by any (0, 1). Then, the constantC changes.

(ii) The Euclidean norm | | can be changed to any other norm. Again, C changes.

(iii) When= 1, yK(x, y) exists almost everywhere and satisfies:

|yK(x, y)| C

|x y|n+1

for all(x, y) c.

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(iv) When= 1, xK(x, y) exists almost everywhere and satisfies:

|xK(x, y)| C

|x y|n+1

for all(x, y) c.

(v) DefineK(x, y) =K(y, x). Then, KCZK impliesK CZK.

Definition 7.1.3 (Kernel associated to an operator). LetT L(L2(Rn)). We say that akernelK : c K is associated to T if for allfL2(Rn), withspt f compact,

T f(x) =

Rn

K(x, y)f(y) dL(y)

for almost everyx c(spt f).

Remark 7.1.4. This integral is a Lebesgue integral for allx c(spt f). Moreover, thissays thatT fcan be represented by this integral away from the support off.

Definition 7.1.5(Calderon-Zygmund Operator). A Calderon-Zygmund Operator of order is an operator T L(L2(Rn)) that is associated to a K CZK. We define CZOto be the collection of all Calderon-Zygmund operators of order . Also, TCZO =TL(L2(Rn))+ K.

Remark 7.1.6. (i) T CZO if and only ifT CZO.

Also, letf, g L2(Rn) withspt f, spt g compact andspt f spt g= . Then,

Tg, f= g , T f =

Rn

g(x)

Rn

K(x, y)f(y) dL(y)

dL(x)

=

Rn

f(y)

Rn

K(x, y)g(x) dL(x)

dL(y)

and since fwas arbitrary, Tg(y) =

RnK(x, y)g(x) dL(x) for almost every y c(spt g). That is, T has associated kernelK.

(ii) T CZO if and only if Ttr CZO, where Ttr is the real transpose of T. Theassociated kernel to Ttr isKtr(x, y) =K(y, x).

(iii) The map T K, where T CZO and K CZK the associated kernel, is notinjective. Consequently, one cannot define a CZO uniquely given a kernel K CZO. The following is an important illustration.

Let m L(Rn) and let Tm be the map f mf. It is easy to check that this isa bounded operator on L2(Rn). Let K = 0 on c and let f L2(Rn) with spt fcompact. Then, wheneverxspt f, Tmf(x) =m(x)f(x) = 0. Therefore,

Tmf(x) =

Rn

K(x, y)f(y) dL(y)

wheneverx c(spt f), which shows the associated kernel to Tm is0.

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7.2 The Hilbert Transform, Riesz Transforms, and The Cauchy

Operator

We discuss three important examples that have motivated the theory.

7.2.1 The Hilbert Transform

Definition 7.2.1 (Hilbert Transform). We define a map H : S(Rn) S(Rn) by

H() = p. v.

1

x

.

That is, p. v. 1

x

(y) = lim

0

{x:|x|>}

1

x(y x) dL(x).

Proposition 7.2.2. Hextends to a bounded operator onL2(R).

Proof. We can analyse this convolution via the Fourier Transform. For a function S(Rn), the Fourier transform is given by

() =

Rn

ex dL(x).

We can extend this naturally to T S(Rn) by defining T via T , = T, for every S(Rn). So, when S(R),

p. v.

1

x

, = p. v.

1

x,

= lim0

{x:|x|>}

1

x(x) dL(x)

= lim0

{x:1>|x|>}

1

x(x) dL(x)

= lim0

Rn

(){x:1>|x|>}

1x ex dL(x) dL().

Now, fix Rn. Then,{x:1>|x|>}

1

xex dL(x) =

{x:1>|x|>}

1

xsin(x ) dL(x)

=2

{x:1>x>}

sin(x )dL(x)

=2

{x:1>x>}

sin(x ||) sgn() dL(x)

= 2

sgn() 1

||

||

sin uu

dL(u).

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The integral appearing on the right hand side is uniformly bounded on and . Thus, byDominated Convergence,

p. v. 1x , = R sgn()()dL()and so for all S(R),H() = sgn()(). Since the Fourier transform is boundedon L2(R), we extend H to the whole of L2(R) by definingHf() = sgn()f() almosteverywhere in R. Then, this extension agrees on S(R) and by Plancherel Theorem,Hf2= f2.

Proposition 7.2.3. HCZO1.

Proof. LetKCZK1 be defined by

K(x, y) = 1

(x y),

when x = y. Fix f L2(R) with spt f compact. Then, fix x c(sptf) and chooser such that B(x, r) spt f = . So, there exists a sequence n Cc (R) such thatsptn B(x, r) = and n f in L2(Rn). Then, for every z B(x, r),

Hn(z) =

R

K(z, y)n(y)dL(y)

R

K(z, y)f(y)dL(y)

andHn H fin L2(Rn). Covering c(sptf) with countably many such balls, we concludethat

Hf(x) =

Rn

K(x, y)f(y) dL(y)

almost everywhere x c(sptf). Therefore HCZO1.

The Hilbert Transform comes from Complex Analysis. Let f Cc (R) and take theCauchy extension f toC \ R. That is,

F(z) = 1

2

R

f(t)

z t dL(t)

where z =x+y, y = 0. It is an easy fact that F is holomorphic onC \ R. ButC \ R isnot connected. So,

limy0+

F(x y) = 1

2(f(x) Hf(x))

and consequently

Hf(x) = limy0+

F(x + y) F(x y)

,

and

real harmonic analysis

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