Contents

Preface 2

1 Introduction 31.1 Hardy’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Copson’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 The squence space c(a,p,q) . . . . . . . . . . . . . . . . . . . . . . . . 11

2 The Blocking Technique 182.1 Transformation from Block Form into Section Form . . . . . . . . . . . 182.2 Transformation from Section Form into Block Form . . . . . . . . . . . 312.3 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Conclusion 38

1

Preface

The aims of this essay to present a comprehensive treatment of the so call blockingtechnique, together with application to the study of sequence function spaces, to thestudy of the operator between such spaces, and to classic inequalities. These notes aredivided into two chapter.

• Chapter 1 is given the Hardy’s inequality, Copson’s inequality and some specialcase of sequence spaces. Moreover, we also introduce the definition of sectionform and block form.

• In Chapter 2 we develop the blocking technique. We show that, apart from sometrivial cases, in fact every norm in section form can be transformed into block formand, what is perhaps even more surprising, every norm in block form can be re-translated into section form. Chapter 2 provides a dictionary of transformationsbetween the kinds of norms.

The main materials of the essay were taken from the books “The blocking technique,weighted mean operators and Hardy’s inequality ” of Karl-Goswin Grosse-Erdmann [1].We have also borrowed extensively from lecture notes of G. H. Hardy, J. E. Littlewood,G. Plya [2] and G.Bennett.

2

Chapter 1

Introduction

1.1 Hardy’s inequality

Theorem 1.1.1 (Hardy’s inequality). If p > 1, An = a1 + a2 + · · · + an and an ≥ 0then

∞∑n=1

(Ann

)<

(p

p− 1

)p ∞∑n=1

apn, (1.1.1)

unless all the a are zero. The constant is the best possible.

To prove that the Hardy’s inequality we need some following lemmas

Lemma 1.1.1. If p, q > 1 with 1p

+ 1q

= 1 then for any two non-negative a,b we have

ab ≤ ap

p+bq

q. (1.1.2)

The equality holds iff ap = bq.

Proof of Lemma 1. The inequality is obvious when one of a and b is zero.Assume that a and b are not zero. We consider

f(t) =tp

p+

1

q− t for t ≥ 0.

We have f ′(t) = tp−1 − 1 which implies that f ′(t) = 0 only if t = 1. Note that

f ′′(t) = (p− 1)tp−2 ≥ 0 ∀t ≥ 0,

in the case p ≥ 1. Hence f(t) has a minimum at t = 1. Since f(1) = 0, we find

f(t) ≥ 0 ∀t ≥ 0.

Therefore

f(t) =tp

p+

1

q− t ≥ 0 ∀t ≥ 0.

Taking t = ab1−q, we get the above inequality. The equality holds iff t = 1 that isap = bq.

3

Lemma 1.1.2 (Holder’s inequality). Let ai, bi ≥ 0 and 1p

+ 1q

= 1 for p, q > 1 then

∞∑i=1

aibi ≤

(∞∑i=1

api

)1/p( ∞∑i=1

bqi

)1/q

. (1.1.3)

Moreover, one has equality only when the sequences (ap1, ..., api , ...) and (bq1, ..., b

qi , ...) are

proportional, i.e, ai = C(bi)p−1.

Now, we will prove that Hardy’s inequality.If an = 0 for all n the inequality (1.1.1) holds.If there exists n such that an 6= 0. Let n0 be the minimum index such that a1 = · · · =an0 = 0 and let b1 = an0+1 > 0, b2 = an0+2, . . . so (1.1.1) becomes(

b1n0 + 1

)p+

(b1 + b2n0 + 2

)p+ · · · <

(p

p− 1

)p(bp1 + bp2 + · · · )p ,

an inequality weaker than (1.1.1).We may suppose that a1 > 0 and let α0 = 0, αn = An/n. So

an = An − An−1 = nαn − αn−1.

We have

αpn −p

p− 1αp−1n an = αpn −

p

p− 1αp−1n [αn − (n− 1)αn−1] (1.1.4)

= αn −np

p− 1αpn +

(n− 1)p

p− 1αn−1α

p−1n (1.1.5)

=

(1− np

p− 1

)αpn +

(n− 1)p

p− 1αp−1n αn−1. (1.1.6)

Applying Lemma 1.1.1 for a = αn−1 and b = αp−1n we obtain

αp−1n αn−1 ≤α(p−1)qn

q+αpn−1p

=αpnq

+αpn−1p

. (1.1.7)

From (1.1.6) and (1.1.7)

αpn −p

p− 1αp−1n an ≤

(1− np

p− 1

)αpn +

(n− 1)p

p− 1

[αpnq

+αpn−1p

]=

(1− np

p− 1

)αpn +

n− 1

p− 1

[(p− 1)αpn + αpn−1

]=

1

p− 1

[(n− 1)αpn−1 − nαpn

].

Hence

N∑n=1

αpn −p

p− 1

N∑n=1

αp−1n an ≤N∑n=1

1

p− 1

[(n− 1)αpn−1 − nαpn

]=

1

p− 1[0αp0 − 1αp1 + 1αp1 − 2αp2 + · · · −NαpN ]

=−Np− 1

αpN < 0 (αN =a1 + a2 + · · · aN

N> 0 since a1 > 0).

4

ThusN∑n=1

αpn <p

p− 1

N∑n=1

αp−1n an. (1.1.8)

Using the Holder’s inequality forN∑n=1

αp−1n an we get

N∑n=1

αp−1n an ≤

(N∑n=1

α(p−1)qn

)1/q( N∑n=1

apn

)1/p

=

(N∑n=1

αpn

)1/q( N∑n=1

apn

)1/p

. (1.1.9)

By (1.1.8) and (1.1.9)

N∑n=1

αpn <p

p− 1

(N∑n=1

apn

)1/p( N∑n=1

αpn

)1/q

.

Dividing by the last factor on the right (which is certainly positive), and raising theresult the pth power, we obtain

N∑n=1

αpn <

(p

p− 1

)p N∑n=1

apn.

When we make N tend to infinity we obtain (1.1.1), expect that we have ‘ ≤ ’ in placeof ‘ < ’. There is equality then the equality of (1.1.9) occurs that means (apn) and (αpn)are proportional, i.e, an = Cαn, where C is independent of n, that is

nan = CAn ∀n.

Since 0 < a1 = Ca1 then C must be 1, and then An = nan for all n. This is only

possible is all the a are equal and this is inconsistent with the convergence of∞∑n=1

apn.

To prove that the constant factor the best possible, we will show that for all 0 <

K <(

pp−1

)pthere exist sequence an such that Hardy’s inequality reverses direction.

In fact, we put

ε = 1−K(p− 1

p

)p.

We takean = n−1/p (n ≤ N), an = 0 (n > N).

Then∞∑n=1

apn =N∑n=1

1

n.

We have

An =n∑k=1

k−1/p >

∫ n

1

x−1/pdx =p

p− 1

(n(p−1)/p − 1

)(1 ≤ n ≤ N).

5

So (Ann

)p>

(p

p− 1

)p(n(p−1)/p − 1

n

)p=

(p

p− 1

)p1− εnn

(1 ≤ n ≤ N),

where

εn = 1−(

1− 1

n(p−1)/p

)p.

We will prove that εn → 0 as n→∞.Indeed, we have

un =1

n(p−1)p → 0 as n→∞.

We define the function g(t) = 1− (1− t)p − pt for t ∈ [0, 1]. Note that

g′(t) = p{(1− t)p−1 − 1} < 0,

so g(t) is decreasing with t ∈ [0, 1]. Implying that

g(t) = 1− (1− t)p − pt ≤ g(0) = 0. (1.1.10)

Hence, we obtain0 ≤ 1− (1− un)p ≤ pun,

that mean εn → 0 as n→∞. Therefore,(Ann

)p>

(p

p− 1

)p1− εnn

.

It follows that

∞∑n=1

(Ann

)p>

N∑n=1

(Ann

)p>

(p

p− 1

)p N∑n=1

1− εnn

.

Since limn→∞

εn = 0, then for every ε = ε(K) there is an N1 = N1(ε) = N1(K) then

εn <ε

2for all n > N1.

For N > N1, we have

N∑n=1

1− εnn

=

N1∑n=1

1− εnn

+N∑

n=N1+1

1− εnn

>N∑

n=N1+1

1− εnn

> (1− ε

2)

N∑n=N1+1

1

n.

6

So that (Ann

)p>

(p

p− 1

)p(1− ε

2)

N∑n=N1+1

1

n

=

(1− ε2)

N∑n=N1+1

1n

N∑n=1

1n

(p

p− 1

)p N∑n=1

1

n

= (1− ηN)

(p

p− 1

)p ∞∑n=1

apn,

where

ηN = 1−(1− ε

2)

N∑n=N1+1

1n

N∑n=1

1n

→ ε

2as N →∞,

becauseN∑

n=N1+1

1N

N∑n=1

1n

→ 1 as N →∞.

Hence (Ann

)p> (1− ε)

(p

p− 1

)p ∞∑n=1

apn.

Therefore any inequality of the type

∞∑n=1

(Ann

)p< (1− ε)

(p

p− 1

)p ∞∑n=1

apn

is false if an is chosen as above and N is sufficiently large.

Another way to prove the best possible constant, we can select

an = n−1/p−ε, ∀n = 1, 2, . . . and 0 < ε < 1− 1/p.

We have

An =n∑k=1

k−1/p−ε >

∫ n

1

x−1/p−εdx =1

1− 1/p− ε(n1−1/p−ε − 1)

=p

p− 1− εp(n1−1/p−ε − 1)

>p

p− 1(n1−1/p−ε − 1).

7

Implying that(Ann

)p>

(p

p− 1

)pn−εp−1

(1− 1

n1−1/p−ε

)p≥

(p

p− 1

)pn−εp−1

(1− p

n1−1/p−ε

)(since (1.1.10))

=

(p

p− 1

)p (n−εp−1 − p

n2−1/p−ε+εp

).

Futhermore,

N∑n=1

(Ann

)p>

(p

p− 1

)p( N∑n=1

n−εp−1 −N∑n=1

p

n2−1/p−ε+εp

)

=

(p

p− 1

)p( N∑n=1

apn − pN∑n=1

1

n2+εp−1/p−ε

)

>

(p

p− 1

)p( N∑n=1

apn − pN∑n=1

1

n2−1/p

)(since 0 < ε < 1− 1

p, p > 1)

>

(p

p− 1

)p( N∑n=1

apn − pC

),

where C =∞∑n=1

1n2−1/p . Thus

N∑n=1

(Ann

)pN∑n=1

apn

>

(p

p− 1

)p 1− CN∑n=1

apn

. (1.1.11)

SinceN∑n=1

an =N∑n=1

n−1−εp → ∞ as N → ∞ and ε → 0+ so that the right hand side

of inequality (1.1.11) tends to(

pp−1

)pas N tend to infinite. Therefore, we get the

constant is best possible. Therefore, we complete the Hardy’s inequality.

1.2 Copson’s inequality

We consider the dual version of Hardy’s inequality - Copspn’s inequality.

Lemma 1.2.1 (Power Rule). If p ≥ 1 then

n∑k=1

ak

(k∑j=1

aj

)p−1

≤

(n∑k=1

ak

)p

≤ p

n∑k=1

ak

(k∑j=1

aj

)p−1

. (1.2.1)

The inequalities reverse direction if 0 < p ≤ 1 (and a1 > 0).

8

Proof. Case 1 : p ≥ 1First, we will prove the left-hand inequality. Because k ≤ n so(

n∑k=1

ak

)p

=n∑k=1

ak

(n∑k=1

ak

)p−1

(1.2.2)

=n∑k=1

aj

(n∑j=1

aj

)p−1

(1.2.3)

≥n∑k=1

ak

(k∑j=1

aj

)p−1

. (1.2.4)

Second, we prove that the right-hand inequality. Let Ak =k∑j=1

aj, we consider

Apk − Apk−1 = p

∫ Ak

Ak−1

xp−1dx ≤ p|Ak − Ak−1|Ap−1k = pakAp−1k . (1.2.5)

Take summation on k we get

n∑k=1

(Apk − A

pk−1)

= Apn ≤ pn∑k=1

ak

(k∑j=1

ak

)p−1

.

Hence, we get the right-hand inequality .

The same argument applies when 0 < p ≤ 1, which case the inequality (1.2.4) and(1.2.5) are reversed.

Lemma 1.2.2 (Power Rule for Tails). If p ≥ 1, then

∞∑k=n

ak

(∞∑j=k

aj

)p−1

≤

(∞∑k=n

ak

)p

≤ p∞∑k=n

ak

(∞∑j=k

aj

)p−1

. (1.2.6)

The inequality reverse direction when 0 < p < 1 if the term of a are positive.

Proof. This may be proved directly an argument similarly to that used in Lemma 1.2.1.Case 1: p ≥ 1, for left-hand inequality

∞∑k=n

ak

(∞∑j=k

aj

)p−1

≤∞∑k=n

ak

(∞∑j=n

aj

)p−1

(1.2.7)

=∞∑k=n

ak

(∞∑k=n

ak

)p−1

(1.2.8)

=

(∞∑k=n

ak

)p

. (1.2.9)

For right-hand inequality: We also let Ak =∞∑j=k

aj. We consider

Apk − Apk+1 = p

∫ Ak

Ak+1

xp−1dx ≤ p|Ak − Ak+1|Ap−1k = pakAp−1k . (1.2.10)

9

So∞∑k=n

(Apk − A

pk+1

)= Apn ≤ p

∞∑k=n

ak

(∞∑j=k

aj

)p−1

.

(∞∑k=n

ak

)p

≤ p∞∑k=n

ak

(∞∑j=k

aj

)p−1

.

Similarly, when 0 < p ≤ 1 the inequalities (1.2.9) and (1.2.10) are reversed.

Theorem 1.2.1 (Copson’s inequality). Suppose that a1 > 0 and Ak = a1+a2+· · ·+ak.If p ≥ 1, then

∞∑n=1

an

(∞∑k=n

akxkAk

)p

≤ pp∞∑n=1

anxpn. (1.2.11)

for every sequence x of non-negative term. The inequality change direction when 0 <p ≤ 1.

Proof. The theorem is obvious when p = 1. Indeed, the inequality becomes

∞∑n=1

an

∞∑k=n

akxkAk≤

∞∑n=1

anxn.

we have∞∑k=1

akxkAk

k∑n=1

an =∞∑k=1

akxkAk

Ak =∞∑k=1

akxk.

Let us consider p > 1, we must to show that

L ≤ ppR, (1.2.12)

where

L =∞∑n=1

an

(∞∑k=n

akxkAk

)p

(1.2.13)

and

R =∞∑n=1

anxpn. (1.2.14)

10

Applying the Power Rule for Tails (Lemma 1.2.2 ) to (1.2.13), we have

L =∞∑n=1

an

(∞∑k=n

akxkAk

)p

≤ p

∞∑n=1

an

∞∑k=n

akxkAk

(∞∑j=k

ajxjAk

)p−1

= p

∞∑k=1

akxkAk

k∑n=1

an

(∞∑j=k

ajxjAk

)p−1

= p

∞∑k=1

akxk

(∞∑j=k

ajxjAk

)p−1

= p∞∑k=1

a1/pk xka

1/qk

(∞∑j=k

ajxjAk

)p−1 (1

p+

1

q= 1

)

≤ p

(∞∑k=1

akxpk

) 1p

∞∑k=1

ak

(∞∑j=k

ajxjAk

) p−1q

1q

.

The last step courtesy of Holder’s inequality. If follows that

L ≤ pR1pL

1q .

Rasing the pth power both two sides, we obtain

Lp ≤ ppRLpq .

Dividing by the last factor on the right which is certainly positive, we get the inequalityis equivalent to (1.2.12).

The same argument applies when 0 < p < 1, provided that a and x are sequences ofpositive terms. It is obvious that nothing is lost here because Lemma 1.2.2, as stated,follows from the restricted version in which a and x are assumed to the positive.

1.3 The squence space c(a,p,q)

Let a=(an) be any sequence of non-negative terms. Then, for 0 < p, q ≤ ∞, we definethe following space:

c(a, p, q) =

{x :

∞∑n=1

an( n∑k=1

|xk|p)1/p

q <∞} (1.3.1)

with the usual modifications if p or q is infinite, with the following norms respectively

‖x‖ =

∞∑n=1

an( n∑k=1

|xk|p)1/p

q1/q

(1.3.2)

and its companion

‖x‖ =

∞∑n=1

an( ∞∑k=n

|xk|p)1/p

q1/q

. (1.3.3)

11

These are said to be norms in section.If 0 < q <∞, p =∞ so norm in section form becomes

‖x‖ =

[∞∑n=1

(an sup

k≤n|xk|)q]1/q

.

If q =∞, 0 < p <∞ we get norm in section

‖x‖ = supnan

(n∑k=1

|xk|

)1/p

and for p, q =∞ it becomes

‖x‖ = supnan sup

k≤n|xk|.

To define norms in block form we will consider index sequence, that is, any sequenceof intergers with m0,mν+1 ≥ mν for all ν and mν → ∞. The block associated with mare defined as

Iν = [mν ,mν+1) = {n ∈ N : mν ≤ n < mν+1}.

Let α ∈ R and 0 < p, q ≤ ∞. Then, for any sequence x = (xn) , we consider

‖x‖ =

∞∑ν=1

1

2να

(∑k∈Iν

|xk|p)1/p

q1/q

. (1.3.4)

again with modifications if p or q is infinite. We call this a norm in block form.Now, one gives some special cases of the sequence spaces c(a, p, q). Let x = (xk) be

real or complex sequence and p > 1 we define the Cesaro sequence space

ces(p) =

{x = (xk) :

∞∑n=1

(1

n

n∑k=1

|xk|

)p

<∞

}. (1.3.5)

under the section norm

‖x‖ces(p) =

[∞∑n=1

(1

n

n∑k=1

|xk|

)p]1/p(1.3.6)

By definition of the sequence space and ces(p) space, we have

ces(p) = c({1/n}n, 1, p).

We are known

lp =

{x = (xk) :

∞∑n=1

|xk|p <∞

}is Banach space. Immediately, we imply the inclusion

lp ⊂ ces(p).

12

Indeed, let x = (xn) ∈ lp so∞∑n=1

|xk|p < ∞. By Hardy’s inequality we have for any

p > 1∞∑n=1

(1

n

n∑k=1

|xk|

)p

<

(p

p− 1

)p ∞∑n=1

|xn|p <∞.

Now, we will prove that ces(p) is Banach space under norm (1.3.6). We will check theaxioms of norm and the completeness of ces(p).It is easy to see that ‖x‖ces(p) ≥ 0; ‖x‖ces(p) = 0⇔ x = 0 and

‖λx‖ces(p) = |λ| ‖x‖ces(p) ∀λ ∈ C,∀x ∈ ces(p).

We check the triangle inequality

‖x+ y‖ces(p) ≤ ‖x‖ces(p) + ‖y‖ces(p) ,∀x ∈ ces(p).

For p > 1, we can choose q > 1 such that 1p

+ 1q

= 1. Applying the Holder’s inequalitywe have

∞∑n=1

(1

n

n∑k=1

|xk|

)(1

n

n∑k=1

|xk + yk|

)p−1

≤

[∞∑n=1

(1

n

n∑k=1

|xk|

)p]1/p ∞∑n=1

(1

n

n∑k=1

|xk + yk|

)(p−1)q1/q

.

That is

∞∑n=1

(1

n

n∑k=1

|xk|

)(1

n

n∑k=1

|xk + yk|

)p−1

≤ ‖x‖ces(p) ‖x+ y‖p−1ces(p) . (1.3.7)

Similarly, we also have

∞∑n=1

(1

n

n∑k=1

|yk|

)(1

n

n∑k=1

|xk + yk|

)p−1

≤ ‖y‖ces(p) ‖x+ y‖p−1ces(p) . (1.3.8)

From (1.3.7) and (1.3.8) imply that

‖x+ y‖pces(p) ≤(‖x‖ces(p) + ‖y‖ces(p)

)‖x+ y‖p−1ces(p) .

Hence we obtain the triangle inequality.Next, we show that the completeness of ces(p). Let {(x(m)

k )k}m is Cauchy sequence inces(p) so

∀ε > 0,∃m0, ∀m, l ≥ m0 :∥∥x(m) − x(l)

∥∥ces(p)

< ε. (1.3.9)

That mean∞∑n=1

(1

n

n∑k=1

|x(m)k − x(l)k |

)p

< εp, (1.3.10)

13

then for each n ∈ N

|x(m)n − x(l)n | < nε.

Implying that {(x(m)n )}∞m=1 is Cauchy in C. Thus, there exists

xk = limm→∞

x(m)k .

We consider x = (xk)k∈N and we need to show that

x ∈ ces(p) and x(m) → x w.p.t the norm (1.3.6).

From (1.3.10), we have for all m, l > m0[N∑n=1

(1

n

n∑k=1

|x(m)k − x(l)k |

)p]1/p< ε (N = 1, 2, . . .).

Fix N ≥ 1 and m ≥ m0 take l→∞ we get[N∑n=1

(1

n

n∑k=1

|x(m)k − xk|

)p]1/p< ε.

Let N →∞; then for all m > m0[∞∑n=1

(1

n

n∑k=1

|x(m)k − xk|

)p]1/p< ε, (1.3.11)

that mean x(m) − x ∈ ces(p). We also have x(m) ∈ ces(p) then

x = x(m) − (x(m) − x) ∈ ces(p)

and by (1.3.11) implies ∥∥∥x(m)k − xk

∥∥∥ces(p)

< ε ∀m ≥ m0.

Thus, we have shown the completeness of ces(p). Therefore, ces(p) is Banach spaceunder the section norm (1.3.6).

Now, we define

Ces(p) =

{x :

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|

)p

<∞}.

with the norm

‖x‖ =

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|

)p1/p

. (1.3.12)

14

We claim that Ces(p) is Banach space under norm (1.3.12). Checking the propertiesof norm is easy, we need only to show that completeness of Ces(p) .

Let {ξm = (ξ(mk )k}m be any Cauchy sequence in Ces(p). That means for every ε > 0

there is an k0 such that for all m,n > k0, ∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|ξ(m)k − ξ(n)k |

)p1/p

< ε. (1.3.13)

So that (2ν+1−1∑k=2ν

|ξ(m)k − ξ(n)k |

)p

<1

2ν(1−p)εp

or2ν+1−1∑k=2ν

|ξ(m)k − ξ(n)k | <

ε

2ν(1−p)/p,∀ν = 0, 1, 2 . . . . (1.3.14)

So that for k ∈ N there exists ν ∈ Z+ such that 2ν ≤ k < 2ν+1 then

|ξ(m)k − ξ(n)k | <

ε

2ν(1−p)/p.

That mean {ξ(m)k }∞m=1 is Cauchy sequence of number. It also converges, that is

ξ(m)k → ξk as m→∞.

Let ξ = (ξk)k∈N we will point out ξ ∈ Ces(p) and ξm → ξ with the norm (1.3.12). By(1.3.13), we have for all m,n > k0, N = 1, 2 . . .

N∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|ξ(m)k − ξ(n)k |

)p

< εp.

Let n→∞, we get for m > k0

N∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|ξ(m)k − ξk|

)p

< εp

Let N →∞ then for m > k0

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|ξ(m)k − ξk|

)p

< εp. (1.3.15)

Thus, ξm − ξ ∈ Ces(p). Moreover, ξm ∈ Ces(p) so that

ξ = ξm + (ξ − ξm) ∈ Ces(p).

From (1.3.15), one implies that ξm → ξ. Therefore, we have shown the completenessof Ces(p) under the norm (1.3.12).

Remark 1.3.1. If {x(m) = (x(m)k )k}m converges in ces(p) or Ces(p) respectively norm

(1.3.6), (1.3.12) then for each k ∈ N the {x(m)k }m is also convergent.

15

Another special sequence space is Lim′s space which is defined

Lim’s space =

{x :

∞∑n=1

1

n

(1

n

n∑k=1

|xk|

)p

<∞

}. (1.3.16)

It follows section norm is

‖x‖ =

[∞∑n=1

1

n

(1

n

n∑k=1

|xk|

)p]1/pBy definition of c(a, p, q) and Lim′s space we have

Lim’s space = C({1/n1+1/p}n, 1, p).

We will prove that the Lim′sspace is Banach space under this norm is defined above.

Let {x(m) = (x(m)k )k}m is Cauchy sequence in the Lim′s space. Then for any ε > 0

there is an N0 such that for all m, l > N0,

∞∑n=1

1

n

(1

n

n∑k=1

|x(m)k − x(l)k |

)p

< εp. (1.3.17)

It follows that for every n = 1, 2, . . . we have

|x(m)n − x(l)n | < εn1+1/p (m, l > N0). (1.3.18)

We choose a fixed k. From (1.3.18) we see that {(x(m)k )m}∞k=1 is Cauchy sequence of

complex number. It converges since C are complete, say,

x(m)k → xk as k →∞.

Using these limits, we define x = (xk) and show that x belongs to Lim′s space andx(m) → x. From (1.3.17), we have for all m, l > N0[

N∑n=1

1

n

(1

n

n∑k=1

|x(m)k − x(l)k |

)p]1/p< ε (N = 1, 2, . . .).

Let l→∞, we obtain for m > N0[N∑n=1

1

n

(1

n

n∑k=1

|x(m)k − xk|

)p]1/p< ε (N = 1, 2, . . .).

We may now let N →∞; then for m > N0[∞∑n=1

(1

n

n∑k=1

|x(m)k − xk|

)p]1/p< ε. (1.3.19)

This show that x(m) − x belongs to Lim′s space. Since x(m) ∈ Lim′s space then

x = x(m) − (x(m) − x) ∈ Lim’s space.

16

Furthermore, (1.3.19) implies that x(m) → x, this proves completeness of Lim′s space,where 1 ≤ p < +∞.We defined an space

Lim(p) =

{x :

∞∑ν=0

2ν−p

(2ν+1−1∑k=2ν

|xk|

)p

<∞}.

with the norm

‖x‖ =

∞∑ν=0

2ν−p

(2ν+1−1∑k=2ν

|xk|

)p1/p

. (1.3.20)

Remark 1.3.2. We also prove that Lim(p) is Banach space with the norm (1.3.20) as

the proof of Ces(p) space. Moreover, if {x(m) = (x(m)k )k}m converges in Lim’s space or

Lim(p) then for each k ∈ N the {x(m)k }m is also convergent respectively.

We are known that from Hardy’s inequality we implies that

lp ⊂ ces(p).

Can we get Hardy’s inequality from the inclusion lp ⊂ ces(p)? This answer is yes.Before explaining the above answer, we need to repeat a closed graph theorem whichis among fundamental theorem of functional analysis.

Theorem 1.3.1 (Closed graph theorem). Let X and Y be Banach spaces then everyclosed linear operator from X to Y is continuous.

Comeback the answer, from the inclusion lp ⊂ ces(p), we have identity operator

Id : (lp, ‖.‖ces(p) → (lp, ‖.‖Ces(p))

is linear. By Remark 1.3.1, it is easy to imply that the map Id is also closed operator.Applying the closed graph theorem we obtain the map Id is continuous, i.e,

‖x‖lp ≤ C‖x‖ces(p),∀x ∈ lp.

We get the Hardy inequality but we do not point out the constant C.

17

Chapter 2

The Blocking Technique

2.1 Transformation from Block Form into Section

Form

Give index sequence m = (mν) and positive monotonic sequence s = (sn) with sn → 0as n→∞. We say that m and s are correlated if

1

2ν≥ sn >

1

2ν+1if mν ≤ n < mν+1

sn >1

2if n < m1.

(2.1.1)

Given any such s we see that there is a unique index sequence m correlated to it; it isdefined by m0 = 1 and, for ν ≥ 1,

mν = min

{n : sn ≤

1

2ν

}. (2.1.2)

Conversely, to any index sequence m there are infinitely many sequence s correlatedto it, for instance the one defined by

sn =1

2νfor mν ≤ n < mν+1, ν ≥ 0.

Example 2.1.1. Let sn = 1/n. Using (2.1.1) we have

1

2ν≥ 1

n>

1

2ν+1

or2ν ≤ n < 2ν+1.

Hence, the index sequence is mν = 2ν .

Throughout this section, let m= (mν) be a fixed index sequence and s= (sn) afixed positive monotonic sequence converging to 0 that is correlated to m,

Theorem 2.1.1. Let 0 < p, q < ∞ and α ∈ R. Then, for any sequence x = (xn), thecondition

∞∑ν=1

1

2να

(∑k∈Iν

|xk|p)1/p

q <∞. (2.1.3)

18

is equivalent to any of the following conditions, β 6= 0, γ 6= 0 and δ are real numberwith γ/q + δ/p = α :

∞∑n=1

(sβn − s

βn+1

)sγ−βn

(n∑k=1

sδk|xk|p)q/p

<∞ if β > 0, γ > 0, (2.1.4)

∞∑n=1

(sβn+1 − sβn

)sγns−βn+1

(n∑k=1

sδk|xk|p)q/p

<∞ if β < 0, γ > 0, (2.1.5)

∞∑n=1

(sβn − s

βn−1

)sγ−βn

(∞∑k=n

sδk|xk|p)q/p

<∞ if β < 0, γ < 0, (2.1.6)

∞∑n=1

(sβn−1 − sβn

)sγns−βn−1

(∞∑k=n

sδk|xk|p)q/p

<∞ if β > 0, γ < 0. (2.1.7)

In addition, for any number γ 6= 0 and δ with γ/q + δ/p+ α, (2.1.3) is equivalent to

∞∑n=1

sγ+δn |xn|p(

n∑k=1

sδk|xk|p)q/p−1

<∞ if γ > 0, (2.1.8)

∞∑n=1

sγ+δn |xn|p(∞∑k=n

sδk|xk|p)q/p−1

<∞ if γ < 0. (2.1.9)

Remark 2.1.1. In conditions (2.1.6) and (2.1.7) the coefficient for n = 1 is undefined.Obviously, its value has no influence on the validity of the theorem. However, it turnsout that the most natural choice is to take its value as sγ1 , that is, to choose s0 = ∞,if one likes.

Proof of Theorem 2.1.1. Replacing |xk|p by |yk|, q/p by q1 and αp by α1

so (2.1.3) becomes∞∑ν=1

1

2να1

(∑k∈Iν

|yk|

)q1

<∞;

(2.1.4) becomes

∞∑n=1

(sβn − s

βn+1

)sγ−βn

(n∑k=1

sδk|yk|

)q1

<∞ if β > 0, γ > 0;

(2.1.5) becomes

∞∑n=1

(sβn+1 − sβn

)sγns−βn+1

(n∑k=1

sδk|yk|

)q1

<∞ if β < 0, γ > 0;

(2.1.6) becomes

∞∑n=1

(sβn − s

βn−1

)sγ−βn

(∞∑k=n

sδk|yk|

)q1

<∞ if β < 0, γ < 0;

19

(2.1.7) becomes

∞∑n=1

(sβn−1 − sβn

)sγns−βn−1

(∞∑k=n

sδk|yk|

)q1

<∞ if β > 0; γ < 0.

(2.1.8) becomes

∞∑n=1

(sγn − s

γn+1

)( n∑k=1

sδk|yk|

)q1−1

if γ > 0

and (2.1.9) becomes

∞∑n=1

(sγn − s

γn+1

)( ∞∑k=n

sδk|yk|

)q1−1

if γ < 0.

So we need only consider the case p = 1.(2.1.3)⇒(2.1.4). By (2.1.3) we have 1/2ν+1 < sn ≤ 1/2ν for n ∈ Iν and ν ≥ 1.

If γ − β ≥ 0 we get

sγ−βn ≤ 1

2ν(γ−β).

Conversely, if γ − β < 0 we obtain

sγ−βn <1

2γ−β1

2ν(γ−β).

Taking K1 = max{

1, 1/2γ−β}. So that

sn ≤ K11

2ν(γ−β). (2.1.10)

We consider

Aν =∑n∈Iν

(sβn − s

βn+1

)sγ−βn

(n∑k=1

sδk|xk|

)q

≤ K1

∑n∈Iν

(sβn − s

βn+1

) 1

2ν(γ−β)

(n∑k=1

sδk|xk|

)q

≤ K1

(sβmν − s

βmν+1

) 1

2ν(γ−β)

ν∑µ=0

∑k∈Iµ

1

2µδ|xk|

q

.

By hypothesis we have δ = α− γ/q, then

Aν ≤ K1

(sβmν − s

βmν+1

) 1

2ν(γ−β)

ν∑µ=0

∑k∈Iµ

1

2µ(α−γ/q)|xk|

q

= K1

(sβmν − s

βmν+1

) 2νβ

2νγ

ν∑µ=0

∑k∈Iµ

2µγ/q

2µα|xk|

q

20

Since 0 < sβmν − sβν+1 < sβν ≤ 1/2νβ so that

Aν ≤ K11

2νγ

ν∑µ=0

2µγ/q

2µα

∑k∈Iµ

|xk|

q

. (2.1.11)

Now, we consider the case q > 1. We have

Aν ≤ K11

2νγ

ν∑µ=0

2µγ/q

Cν

Cν2µα

∑k∈Iµ

|xk|

q

. (2.1.12)

where Cν =∑ν

µ=0 2µγ/q. Applying Jensen’s inequality we obtain ν∑µ=0

2µγ/q

Cν

Cν2µα

∑k∈Iµ

|xk|

q

≤ν∑

µ=0

2µγ/q

Cν

Cν2µα

∑k∈Iµ

|xk|

q

. (2.1.13)

Combine (2.1.12) and (2.1.13)

Aν ≤ K1Cq−1ν

2νγ

ν∑µ=0

2µγ/q

1

2µα

∑k∈Iµ

|xk|

q

.

We see that

Cν =2(ν+1)γ/q − 1

2γ/q − 1

≤ 2(ν+1)γ/q

2γ/q − 1

= 2νγ/q2γ/q

2γ/q − 1,

so that

Cq−1ν 2νγ ≤ 2νγ(q−1)/q

2νγ

(2γ/q

2γ/q − 1

)q−1=

1

2νγ/q

(2γ/q

2γ/q − 1

)q−1.

Thus

Aν ≤ K1

2νγ/q

ν∑µ=0

2µγ/q

1

2µα

∑k∈Iµ

|xk|

q

,

where K = K1

(2γ/q

2γ/q−1

)q−1.

21

Since the constants K are independent of ν, so that summing on ν gives

∞∑n=m1

(sβn − s

βn+1

)sγ−βn

(n∑k=1

sδk|xk|

)q

≤ K

∞∑ν=1

1

2νγ/q

ν∑µ=0

2µγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K∞∑µ=0

∞∑ν=µ

2µγ/q

2νγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K∞∑µ=0

∞∑ν=0

1

2νγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K

∞∑µ=0

(1

1− 12γ/q

) 1

2µα

∑k∈Iµ

|xk|

q

= K2γ/q

2γ/q − 1

∞∑µ=0

1

2µα

∑k∈Iµ

|xk|

q

= K1

(2γ/q

2γ/q − 1

)q ∞∑µ=0

1

2µα

∑k∈Iµ

|xk|

q

.

For the case q ≤ 1: We replace Jensen’s inequality by the inequality(∞∑k=1

ck

)q

≤∞∑k=1

cqk. (2.1.14)

with non-negative ck. We will show that above the inequality by induction:

• For k = 1 we have cq1 ≤ cq1 the inequality holds.

• For k = 2 the inequality (2.1.14) becomes

(c1 + c2)q ≤ cq1 + cq2,

or

1 ≤ cq1 + cq2(c1 + c2)q

=

(c1

c1 + c2

)q+

(c2

c1 + c2

)q.

Since q ≤ 1 so (c1

c1 + c2

)q+

(c2

c1 + c2

)q≥ c1c1 + c2

+c2

c1 + c2= 1.

Thus, for k = 2 the inequality (2.1.14) is true.

• Now, we assume the inequality (2.1.14) is holds for k = n and we need to show

22

that it is true for k = n+ 1(n+1∑k=1

ck

)q

=

(n∑k=1

ck + cn+1

)q

≤

(n∑k=1

ck

)q

+ cqn+1

≤n∑k=1

cqk + cqn+1 =n+1∑k=1

cqk.

Let n tend to infinity we get the inequality (2.1.14).

Applying the inequality (2.1.14) we have ν∑µ=0

2µγ/q

2µα

∑k∈Iµ

|xk|

q

≤ν∑

µ=0

2µγ/q

2µα

∑k∈Iµ

|xk|

q

. (2.1.15)

From (2.1.11) and (2.1.15) we obtained

Aν ≤ K11

2νγ

ν∑µ=0

2µγ/q

2µα

∑k∈Iµ

|xk|

q

.

We also sum on ν gives

∞∑ν=1

Aν ≤ K1

∞∑ν=1

1

2νγ

ν∑µ=0

2µγ

1

2µα

∑k∈Iµ

|xk|

q

= K1

∞∑µ=0

∞∑ν=µ

2µγ

2νγ

1

2µα

∑k∈Iµ

|xk|

q

= K1

∞∑µ=0

(1

1− 12γ

) 1

2µα

∑k∈Iµ

|xk|

q

= K12γ

2γ − 1

∞∑µ=0

1

2µα

∑k∈Iµ

|xk|

q

<∞.

(2.1.4)⇒(2.1.3) Fix ν ≥ 1 and assume that Iν 6= ∅. Then mν+1 − 1 6= 0, so that forβ > 0 we deduce from (2.1.1)

sβmν+1−1 − sβmν+2

≥ 1

2(ν+1)β− 1

2(ν+3)β= M1

1

2νβ

where

M1 =1

2νβ− 1

23νβ.

23

Hence, for δ = α− γ/q(1

2να

∑k∈Iν

|xk|

)q

=

(1

2ν(δ+γ/q)

∑k∈Iν

|xk|

)q

=1

2να1

2ν(γ−β)

(1

2νδ

∑k∈Iν

|xk|

)q

≤ 1

M1

(sβmν+1−1 − s

βmν+2

) 1

2ν(γ−β)

(∑k∈Iν

sδk|xk|

)q

=1

M1

mν+2−1∑n=mν+1−1

(sβn − s

βn+1

) 1

2ν(γ−β)

(∑k∈Iν

sδk|xk|

)q

.

We have used Iν 6= ∅ one more: It implies that mν ≤ mν+1− 1, so that for mν+1− 1 ≤n ≤ mν+1 − 1 we have

1/2ν+2 < sn ≤ 1/2ν .

Hence,

sγ−βn >1

2(ν+2)(γ−β) if γ − β ≥ 0

and

sγ−βn ≥ 1

2ν(γ−β)if γ − β < 0.

Therefore, we put M2 = max{1, 22(γ−β)} we obtain

1

2ν(γ−β)≤M2s

γ−βn .

Thus, we have(1

2να

∑k∈Iν

|xk|

)q

≤ M2

M1

mν+2−1∑n=mν+1−1

(sβn − s

βn+1

)sγ−βn

(n∑k=1

sδk|xk|

)q

. (2.1.16)

Take summation on all ν ≥ 1 with Iν 6= ∅ we get (2.1.4) implies (2.1.3) for any q > 0.(2.1.3)⇒(2.1.6). By (2.1.10) and (2.1.1)

Bν =∑n∈Iν

(sβn − s

βn−1

)sγ−βn

(∞∑k=n

sδk|xk|

)q

≤ K1

∑n∈Iν

(sβn − s

βn−1

) 1

2ν(γ−β)

(∞∑k=n

sδk|xk|

)q

≤ K1

(sβmν − s

βmν−1

) 1

2ν(γ−β)

∞∑µ=ν

∑k∈Iµ

1

2µδ|xk|

q

≤ K11

2νγ

∞∑µ=ν

1

2µδ

∑k∈Iµ

|xk|

q

.

24

Now using δ = α− γ/q,

Bν ≤ K11

2νγ

(∞∑µ=ν

2µγ/q

2µα

)q

. (2.1.17)

Since γ < 0, we have

Dν =∞∑µ=ν

2µγ/q = 2νγ/q∞∑µ=0

2µγ/q =2νγ/q

1− 2γ/q.

For the case p > 1. From (2.1.17) and applying Jensen’s inequality we get

Bν ≤ K11

2νγ

∞∑µ=ν

2µγ/q

Dν

Dν

2µα

∑k∈Iµ

|xk|

q

≤ K11

2νγ

∞∑µ=ν

2νγ/q

Dν

Dν

2µα

∑k∈Iµ

|xk|

q

≤ K1Dq−1ν

2νγ

∞∑µ=ν

2νγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K1

(2νγ/q

1− 2γ/q

)p−11

2νγ

∞∑µ=ν

2νγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K1

2νγ/q

∞∑µ=ν

2νγ/q

1

2µα

∑k∈Iµ

|xk|

q

,

where

K = K1

(1

1− 2γ/q

)q−1.

Summing on ν gives

∞∑ν=1

Bν ≤ K∞∑ν=1

1

2νγ/q

∞∑µ=ν

2µγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K∞∑µ=1

2µγ/qµ∑ν=0

1

2νγ/q

1

2µα

∑k∈Iµ

|xk|

q

= K∞∑µ=1

2µγ/q2−(µ+1)γ/q − 1

2−γ/q − 1

1

2µα

∑k∈Iµ

|xk|

q

≤ K2−µγ/q

2−γ/q − 1

∞∑µ=1

1

2µα

∑k∈Iµ

|xk|

q

25

For case p ≤ 1 : From (2.1.14) and (2.1.17) imply that

Bν ≤ K11

2νγ

∞∑µ=ν

2µγ/q

2µα

∑k∈Iµ

|xk|

q

.

Summing on ν are given

∞∑ν=1

Bν ≤ K1

∞∑ν=1

1

2νγ

∞∑µ=ν

2µγ

1

2µα

∑k∈Iµ

|xk|

q

= K1

∞∑µ=1

2µγµ∑ν=0

1

2νγ

1

2µα

∑k∈Iµ

|xk|

q

= K1

∞∑µ=1

2µγ2−µ+1γ − 1

2−γ − 1

1

2µα

∑k∈Iµ

|xk|

q

≤ K11

2−γ − 1

∞∑µ=1

1

2µα

∑k∈Iµ

|xk|

q

.

The proof of the (2.1.6)⇒(2.1.3) is similarly and is therefore omitted as (2.1.4)⇒(2.1.3).We only remark estimate 1/2νβ by sβmν − s

βmν−1−1.

The equivalence of (2.1.5) with (2.1.3) is obvious once it is noted (2.1.5) is simplyobtained from (2.1.4) by substituting β by −β. The same is true for (2.1.6) and (2.1.7).

To complete the proof of the theorem it suffices to show that in the special case ofβ = γ the conditions (2.1.4) and (2.1.8) are equivalent.

(2.1.4)⇒(2.1.8). For β = γ then (2.1.4) becomes

A =∞∑n=1

(sγn − s

γn+1

)( n∑k=1

sδk|xk|

)q

<∞ if γ > 0. (2.1.18)

By apply the Power Rule (Lemma 1.2.1) we have(n∑k=1

sδk|xk|

)q

≥n∑k=1

sδk|xk|

(k∑j=1

sδj |xj|

)q−1

. (2.1.19)

From (2.1.34) and (2.1.35) we get

A ≥∞∑n=1

(sγn − s

γn+1

) n∑k=1

sδk|xk|

(k∑j=1

sδj |xj|

)q−1

=∞∑k=1

sδk|xk|∞∑n=k

(sγn − s

γn+1

)( k∑j=1

sδj |xj|

)q−1

=∞∑k=1

sδk|xk|sγk

(k∑j=1

sδj |xj|

)q−1

=∞∑k=1

sδ+γk |xk|

(k∑j=1

sδj |xj|

)q−1

.

26

so∞∑k=1

sδ+γk |xk|

(k∑j=1

sδj |xj|

)q−1

<∞.

(2.1.4)⇐(2.1.8). Applying the Power Rule (Lemma 1.2.1) we point out

A =∞∑n=1

(sγn − s

γn+1

)( n∑k=1

sδk|xk|

)q

≤ q

∞∑n=1

(sγn − s

γn+1

) n∑k=1

sδk|xk|

(k∑j=1

sδj |xj|

)q−1

= q∞∑k=1

sδ+γk |xk|

(k∑j=1

sδj |xj|

)q−1

<∞.

Similarly, we also apply the Power Rule for Tails (Lemma 1.2.2) so we show that(2.1.6) and (2.1.9) are equivalent.

The next two theorems treat the case where p or q is finite.

Theorem 2.1.2. Let 0 < q < ∞ and α ∈ R. Then, for any sequence x = (xn)thecondition

∞∑ν=0

(1

2ναsupk∈Iν|xk|)q

<∞ (2.1.20)

is equivalent to each of the flowing conditions, where β 6= 0, γ 6= 0 and δ are real numberwith γ/q + δ = α

∞∑n=1

(sβn − s

βn+1

)sγ−βn

(supk≤n

sδk|xk|)q

<∞ if β > 0, γ > 0, (2.1.21)

∞∑n=1

(sβn+1 − sβn

)sγns−βn+1

(supk≤n

sδk|xk|)q

<∞ if β < 0, γ > 0, (2.1.22)

∞∑n=1

(sβn − s

βn−1

)sγ−βn

(supk≥n

sδk|xk|)q

<∞ if β < 0, γ < 0, (2.1.23)

∞∑n=1

(sβn+1 − sβn

)sγns−βn−1

(supk≥n

sδk|xk|)q

<∞ if β > 0, γ < 0. (2.1.24)

Proof. As in the proof of Theorem 2.1.1 we need only consider the case q = 1, and weneed only prove the equivalence of (2.1.20) with (2.1.21) and (2.1.23).

(2.1.20)⇒(2.1.22). Fix ν > 1. Using (2.1.10) we obtain∑n∈Iν

(sβn − s

βn+1

)sγ−βn sup

k≥nsδk|xk| ≤ K1

∑n∈Iν

(sβn − s

βn+1

) 1

2ν(γ−β)supk≥mν

sδk|xk|

≤ K1

(sβmν+1−1 − s

βmν−1

) 1

2ν(γ−β)supµ≥ν

supk∈Iµ

sδk|xk|

≤ K11

2νγsupµ≥ν

1

2µδsupk∈Iµ|xk|

= K1 supµ≥ν

1

2(ν−µ)γXµ,

27

where

Xµ =1

2µδsupk∈Iµ|xk|.

Summing on ν we have

∞∑n=m1

(sβn − s

βn+1

)sγ−βn sup

k≥nsδk|xk| ≤ K1 sup

µ≤ν

1

2(ν−µ)γXµ. (2.1.25)

For any N ∈ N we observe that

N∑ν=0

supµ≥ν

1

2(ν−µ)γXµ =N−1∑ν=0

supµ≥ν

1

2(ν−µ)γXµ + supµ≥N

1

2(N−µ)γXµ

≤N−1∑ν=0

(supµ≥ν+1

1

2(ν−µ)γXµ + supµ=ν

1

2(ν−µ)γXµ

)+ sup

µ≥N

1

2(N−µ)γXµ

= 2γN−1∑ν=0

supµ≥ν+1

1

2(ν+1−µ)γXµ +N−1∑ν=0

Xν + supµ≥N

1

2(N−µ)γXµ.

Since γ < 0 so

N∑ν=0

supµ≥ν

1

2(ν−µ)γXµ ≤ 2γN∑ν=0

supµ≥ν

1

2(ν−µ)γXµ +N−1∑ν=0

Xν + supµ≥N

Xµ (2.1.26)

From the hypothesis∞∑µ=0

Xµ =∞∑µ=0

1

2µδsupk∈Iµ|xk|

converges thenlim supN→∞

XN = limN→∞

supµ≥N

Xµ = 0. (2.1.27)

From (2.1.26) and (2.1.27) we deduce after letting N →∞ that

∞∑ν=0

supµ≥ν

1

2(ν−µ)γXµ ≤ 2γ∞∑ν=0

supµ≥ν

1

2(ν−µ)γXµ +∞∑ν=0

Xν .

So that∞∑ν=0

supµ≥ν

1

2(ν−µ)γXµ ≤1

1− 2γ

∞∑ν=0

Xν . (2.1.28)

Therefore, (2.1.22) follows from (2.1.25) and (2.1.28) under the assumption (2.1.20).(2.1.22)⇒(2.1.20). With the obvious changes the proof is the same as that of (2.1.4)

⇒(2.1.3). We also using (2.1.1) then

sβmν+1−1 − sβmν−1 ≥ K

1

2νβ.

28

We consider

Xν =1

2ναsupk∈Iν|xk| =

1

2ν(δ+γ)supk∈Iν|xk|

=1

2νβ1

2ν(γ−β)

(1

2νδsupk∈Iν|xk|)

≤ K(sβmν − s

βmν−1−1

) 1

2ν(γ−β)supk∈Iν

sδk|xk|

≤ K

mν−1−2∑n=mν

(sβn − s

βn−1

)sγ−βn sup

k∈Iνsδk|xk|

≤ K

mν−1−2∑n=mν

(sβn − s

βn−1

)sγ−βn sup

k≥νsδk|xk|

Summing on ν and by assumption of (2.1.22) we obtain (2.1.20).The proof of equivalence (2.1.20)⇔(2.1.21) is similar and therefore omitted.

Theorem 2.1.3. Let 0 < p < ∞ and α ∈ R. Then, for any sequence x = (xn)thecondition

supν

1

2να

(∑k∈Iν

|xk|

)1/p

<∞ (2.1.29)

is equivalent to each of the flowing conditions, where γ 6= 0 and δ are real number withγ + δ/p = α

supnsγn

(n∑k=1

sδk|xk|p)1/p

<∞ if γ > 0, (2.1.30)

supnsγn

(∞∑k=n

sδk|xk|p)1/p

<∞ if γ < 0. (2.1.31)

In addition, for real numbers γ 6= 0, δ, µ and σ with γ + δ/q = α and ρ > 0, (2.1.29) isequivalent to

n∑k=1

(sργk − s

ργk+1

)( k∑j=1

sδj |xj|p)(γ+σ)/p

= O

(n∑k=1

sδk|xk|p)σ/p

if γ > 0, (2.1.32)

n∑k=1

(sργk − s

ργk−1)( ∞∑

j=k

sδj |xj|p)(γ+σ)/p

= O

(∞∑k=n

sδk|xk|p)σ/p

if γ < 0. (2.1.33)

Proof. We note that by an obvious change of variable |xk|p = yk and parametersδ = δ1, γ = γ1/p, α = α1/p so that we need only to show that p=1.

(2.1.29)⇒(2.1.30). Let ν ≥ 1. Then we have for n ∈ Iv so

1

2ν+1< sn ≤

1

2ν.

29

We consider

sγn

n∑k=1

sδk|xk| ≤1

2νγ

mν+1−1∑k=1

sδk|xk|

≤ 1

2νγ

ν∑µ=0

∑k∈Iµ

1

2µδ|xk|

=ν∑

µ=0

1

2να1

2γ(ν−µ)

∑k∈Iµ

|xk| (since γ + δ = α)

≤ supµ

1

2να

∑k∈Iµ

|xk|

ν∑µ=0

1

2γ(ν−µ)

=1

2γν2(ν+1)γ − 1

2γ − 1supµ

1

2να

∑k∈Iµ

|xk|

(since γ > 1).

This implies (2.1.30).(2.1.30)⇒(2.1.29). For ν ≥ 1 with Iν 6= ∅ we obtain, again using (2.1.1):

1

2να

∑k∈Inu

|xk| =1

2νγ

∑k∈Inu

1

2νδ|xk|

< 2γsn∑k∈Iν

sδk|xk|.

Taking sup on ν both two sides we get

supν

1

2να

∑k∈Inu

|xk| ≤ 2γ supνsn∑k∈Iν

sδk|xk|.

Condition (2.1.29) now follows.(2.1.29)⇔(2.1.31) is prove similarly.(2.1.30)⇔(2.1.32) We certain new sequences (Xn) and (Sn) and new parameters

Γ,∆ and A. Let γ, ρ, σ > 0 and δ ∈ R with γ + δ = α. We define, for n ≥ 1,

Tn =n∑k=1

sδk|xk|

andXn = sργn − s

ργn+1.

If (Tn) is bounded, then (2.1.30) and (2.1.32) both hold, so that nothing is to be proved.Hence we assume that Tn → ∞. Also, after possibly shifting indices, we may assumethat Tn 6= 0 for all n. We then set

Sn = T−1n ,

which gives us a monotonic positive sequence with Sn → 0. With these sequences(Sn), (Xn) and the parameters Γ = σ,∆ = −(ρ+σ) and A = −ρ is replaced respectively

30

by xn, sn and γ, δ, α so that in the case p = 1 the conditions (2.1.30) becomes

supn

(n∑k=1

sδk|xk|

)−σ n∑k=1

(k∑j=1

skj |xj|

)ρ+σ (sργk − s

ργk+1

)<∞. (2.1.34)

From (2.1.32) with p = 1 implies that(n∑k=1

sδk|xk|

)−σ n∑k=1

(sργk − s

ργk+1

)( k∑j=1

skj |xj|

)ρ+σ

< C. (2.1.35)

for some constant C. By the equivalent of (2.1.34) and (2.1.35) we complete the proofof (2.1.30)⇔(2.1.32).

(2.1.31)⇔(2.1.33) is prove similarly.

For the sake completeness we include the almost trivial case of p = q = ∞. Theproof follows the same line as that Theorem (2.1.3).

Theorem 2.1.4. Let α ∈ R. Then, for any sequence x = (xn), the condition

supν

1

2ναsupk∈Inu

|xk| <∞ (2.1.36)

is equivalent to the following conditions, where γ and δ are real number with γ+ δ = α

supνsγn sup

k≤n|xk| <∞ if γ > 0 (2.1.37)

supνsγn sup

k≥n|xk| <∞ if γ ≤ 0. (2.1.38)

2.2 Transformation from Section Form into Block

Form

These transformation are now easily found by reading the results of the previous section“backwards”. Throughout this section, let a = (an) be fixed sequence of non negativenumber. In order to avoid trivialities we assume that a is not a finite sequence.

Theorem 2.2.1. Let x = (xn) be any sequence.(a) Let 0 < p, q <∞. Assume that

∑∞k=1 a

qk <∞, and define an index sequence m by

letting mν be the least n such that (∑∞

k=n aqk)

1/q ≤ 1/2ν (ν ≥ 1). Then

∞∑n=1

an( n∑k=1

|xk|p)1/p

q <∞ (2.2.1)

is equivalent to

∞∑ν=0

1

2ν

(∑k∈Iν

|xk|p)1/p

q <∞. (2.2.2)

31

(b) Let 0 < q <∞. Assume that∑∞

k=1 aqk <∞, and define m as in (a). Then

∞∑n=1

(an sup

k≤n|xk|)q

<∞ (2.2.3)

is equivalent to

∞∑ν=0

(1

2νsupk∈Iν|xk|)q

<∞. (2.2.4)

(c) Let 0 < p < ∞. Assume that ak → 0, and define an index sequence m by lettingmν be the least n such that sup

k≥nak ≤ 1/2ν (ν ≥ 1). Then

supnan

(n∑k=1

|xk|p)1/p

<∞ (2.2.5)

is equivalent to

supν

1

2ν

(∑k∈Iν

|xk|p)1/p

<∞. (2.2.6)

(d) Assume that ak → 0, and define m as in (c). Then

supnan sup

k≤n|xk| <∞ (2.2.7)

is equivalent to

supν

1

2νsupk∈Iν|xk| <∞. (2.2.8)

Proof. (a) Let sn = (∑∞

k=n aqk)

1/q. By hypothesis, we see that m and s are correlated

so we apply the Theorem (2.1.1) with α = 1, β = γ = q and δ = 0. We get (2.1.3)becomes

∞∑ν=0

1

2ν

(∑k∈Iν

|xk|p)1/p

q <∞.and (2.1.4) becomes

∞∑n=1

(sqn − s

qn+1

)s0n

(n∑k=1

s0k|xk|p)q/p

<∞

or∞∑n=1

(∞∑k=n

aqn −∞∑k=n

aqn+1

)(n∑k=1

|xk|p)q/p

<∞.

So that∞∑n=1

an( n∑k=1

|xk|p)1/p

q <∞.32

Therefore we complete part (a).(b) follows in the same way from Theorem (2.1.2), we also take sn as part (a) with

α = 1, β = γ = q and δ = 0.We also (c) and (d) can be deduced from Theorem (2.1.3) and (2.1.4) when we set

sn = supk≥n

ak. Indeed, we apply theorem (2.1.3) for δ = 0, α = γ = 1 we have (2.1.29)

becomes

supν

1

2ν

(∑k∈Iν

|xk|p)1/p

<∞.

and (2.1.30) becomes

supnsn

(n∑k=1

s0k|xk|p)1/q

<∞,

or

supn

supk≥n

ak

(n∑k=1

|xk|p)1/q

<∞.

To complete the proof of part (c), we need only show that

supn

(supk≥n

ak)cn = supnancn, (2.2.9)

where cn = (∑n

k=1 |xk|p)1/q. Indeed, we have an ≤ sup

k≥nak implies that

ancn ≤ supk≥n

akcn.

Thus, we obtainsupnancn ≤ sup

n(supk≥n

ak)cn. (2.2.10)

On the other hand, we see that (cn) is non-negative, non-decreasing sequence then fork ≥ n we have ck ≥ cn so akck ≥ akcn we get

supkakck ≥ sup

kakcn ≥ sup

k≥nakcn.

Taking sup both sides we get

supnancn ≥ sup

n(supk≥n

ak)cn. (2.2.11)

From (2.2.10) and (2.2.11) can be deduced (2.2.9).

We turn to the companion result for tail section.

Theorem 2.2.2. Let x = (xn) be any sequence.(a) Let 0 < p, q <∞. Assume that

∑∞k=1 a

qk =∞, and define an index sequence m by

letting mν be the least n such that (∑∞

k=n aqk)

1/q ≥ 2ν (ν ≥ 1). Then

∞∑n=1

an( ∞∑k=n

|xk|p)1/p

q <∞ (2.2.12)

33

is equivalent to

∞∑ν=0

2ν

(∑k∈Iν

|xk|p)1/p

q <∞. (2.2.13)

(b) Let 0 < q <∞. Assume that∑∞

k=1 aqk =∞, and define m as in (a). Then

∞∑n=1

(an sup

k≤n|xk|)q

<∞ (2.2.14)

is equivalent to

∞∑ν=0

(2ν sup

k∈Iν|xk|)q

<∞. (2.2.15)

(c) Let 0 < p < ∞. Assume that sup ak = ∞, and define an index sequence m byletting mν be the least n such that supk≥n ak ≥ 2ν (ν ≥ 1). Then

supnan

(∞∑k=n

|xk|p)1/p

<∞ (2.2.16)

is equivalent to

supν

2ν

(∑k∈Iν

|xk|p)1/p

<∞. (2.2.17)

(d) Assume that sup ak =∞, and define m as in (c). Then

supnan sup

k≤n|xk| <∞ (2.2.18)

is equivalent to

supν

2ν supk∈Iν|xk| <∞. (2.2.19)

Proof. (a) and (b) follow from Theorems (2.1.1) and (2.1.2) upon setting

sn =

(n∑k=1

aqk

)−1/pand α = −1, β = γ = −q, δ = 0. In the same way (c) and (d) follows Theorem (2.1.3)

and (2.1.4) with sn =(supk≤n ak

)−1if one notes in addition that

supn

(supk≤n

ak)cn = supnancn, (2.2.20)

for any non-negative, non-increasing sequence (cn). To prove that (2.2.20) we also dosimilarly as prove that (2.2.9).

34

2.3 Comments

The main results in the previous two sections state that for certain norm ‖.‖block and‖.‖section in block and section form we have

∀x ‖x‖block <∞ ⇔ ‖x‖section <∞. (2.3.1)

Now, if s0 = ∞ in Theorem 2.1.1 and 2.1.2 and if a1 > 0 in Theorem 2.2.2, thenthese norm are positive definite. In that case , by a well known reasoning based on theclosed graph theorem, (2.3.1) is equivalent to the quantitative statement that there areconstants K and K ′ such that

∀x K‖x‖block < ‖x‖section < K ′‖x‖block. (2.3.2)

Now, we spell out the most prominent particular case of our results, the renorming ofthe ces(p) sequence space.

Theorem 2.3.1. For 1 < p <∞, the space ces(p) conists of all sequence x in Ces(p)space and the norm in ces(p) is equivalent the norm in Ces(p).

Proof. We apply Theorem 2.1.1 by taking sn = 1/n with α = (p − 1)/p, β = −1, γ =p− 1 and δ = 0. Since sn = 1/n the we get the index sequence mν = 2ν so that (2.1.3)becomes

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|

)p

<∞.

And β = −1 < 0, γ = p− 1 > 1 then (2.1.5) becomes

∞∑n=1

1

n+ 1

1

np−1

(n∑k=1

|xk|

)p

<∞.

We deduce that ‖x‖ <∞ if and only if

∞∑n=1

1

n+ 1

1

np−1

(n∑k=1

|xk|

)p

<∞.

We also have

∞∑n=1

1

n+ 1

1

np−1

(n∑k=1

|xk|

)p

≥ 1

2

∞∑n=1

1

np

(n∑k=1

|xk|

)p

.

And∞∑n=1

1

n+ 1

1

np−1

(n∑k=1

|xk|

)p

<

∞∑n=1

1

np

(n∑k=1

|xk|

)p

.

Therefore x ∈ ces(p).Next, we will prove that the equivalence of the ces(p) and Ces(p) that means there

exists constants C1, C2 > 0 such that

∀x ∈ ces(p) C1‖x‖Ces(p) ≤ ‖x‖ces(p) ≤ C2‖x‖Ces(p).

35

Thus, we prove this we need to point out two identity linear operators

Id1 : (ces(p), ‖.‖ces(p)) ↪→ (Ces(p), ‖.‖Ces(p)).

andId2 : (Ces(p), ‖.‖Ces(p)) ↪→ (ces(p), ‖.‖ces(p)).

are continuous.Indeed, in chapter 1 we have shown ces(p) and Ces(p) are Banach space with respec-tively norm

‖x‖ces(p) =

(∞∑n=1

1

np

(n∑k=1

|xk|

)p

)

)1/p

and

‖x‖Ces(p) =

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|

)p1/p

.

We haveId1(x

(m)) = x(m).

If {x(m) = (x(m)k )k}m converges at x = {xk} in ces(p) under ‖.‖ces(p) and {x(m) =

(x(m)k )k}m is convergent at y = {yk} under ‖.‖Ces(p). By Remark 1.3.1 we have for each

k ∈ N the x(m)k converges xk and x

(m)k converges yk . Hence, x = y that means the

map Id is closed operator. Using the closed graph theorem, we implies that the linearoperator Id1 is continuous. Similarly, we obtain the identity linear Id2 is continuousoperator. We complete Theorem 2.3.1.

We have the above similar for Lim′s space and Lim(p), i.e,

Lim’s space ≡ Lim(p).

Moreover, two norms of its are equivalent. We also applying the Theorem 2.1.1 bytaking sn = 1/n with with α = p, β = −1, γ = p and δ = 0. Since sn = 1/n the we getthe index sequence mν = 2ν so that (2.1.3) becomes

∞∑ν=0

2−νp

(2ν+1−1∑k=2ν

|xk|

)p

<∞.

And β = −1 < 0, γ = p > 1 then (2.1.5) becomes

∞∑n=1

1

n+ 1

1

np

(n∑k=1

|xk|

)p

<∞.

We deduce that ‖x‖Lim(p) <∞ if and only if

∞∑n=1

1

n+ 1

1

np

(n∑k=1

|xk|

)p

<∞.

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We also have

∞∑n=1

1

n+ 1

1

np

(n∑k=1

|xk|

)p

≥ 1

2

∞∑n=1

1

np+1

(n∑k=1

|xk|

)p

.

And∞∑n=1

1

n+ 1

1

np

(n∑k=1

|xk|

)p

<

∞∑n=1

1

np+1

(n∑k=1

|xk|

)p

.

Therefore x ∈ Lim′s space.To prove that the equivalent of two norms we also do the same as the proof of

Theorem 2.3.1.In end this chapter we point out a simple result - a new proof of Hardy’s inequality.

To prove the inequality we need only to show that

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|

)p

≤ C∞∑k=1

|xk|p.

Indeed, we use Holder’s inequality

2ν+1−1∑k=2ν

|xk|.1 ≤

(2ν+1−1∑k=2ν

|xk|p)1/p(

2ν+1−1∑k=2ν

1q

)1/q

(where1

p+

1

q= 1)

=

(2ν+1−1∑k=2ν

|xk|p)1/p

2ν/p.

So that

∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|

)p

≤∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|p)

2νp/q

=∞∑ν=0

2ν(1−p)

(2ν+1−1∑k=2ν

|xk|p)

2ν(q−1)

=∞∑ν=0

2ν(1−p)2ν+1−1∑k=2ν

|xk|p

=∞∑k=1

|xk|p.

We have shown the new proof of Hardy’s inequality.

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Conclusion

In these essay we have shown that the two inequalities which are Hardy’s inequalityand the dual of Hardy’s inequality is Coposon’s inequality in chapter 1. Moreover,we point out the best constant possible of Hardy’s inequality. Next, we have provedthat some sequence space is Banach space which are ces(p), Ces(p), Lim′s space andLim(p). In chapter 2 we concentrate the relationship between section norm and blocknorm. We have transformed from block into section form and opposite direction. Inparticular, by using the closed graph theorem we imply that the equivalence of sectionform and block form of ces(p) and Lim′s space.

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Bibliography

[1] Grosse-Erdmann and Karl-Goswin, (1998). The blocking technique, weighted meanoperators and Hardy’s inequality (Springer-Verlag)

[2] G. H. Hardy, J. E. Littlewood, G. Polya, (1965). Inequalities (Cambridge Univer-sity Press, Cambridge).

[3] G. Bennett and K-G. Grosse-Erdmann, On series of positive terms, Houston Journalof Mathematics. Volume 31, No. 2 (2005), 541-586.

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