SOME ELEMENTARY INEQUALITIES OF G. BENNETT · CHAPTER 1. SOME INEQUALITIES 6 Lemma 1.1.1 (Power...

VIETNAM NATIONAL UNIVERSITY

UNIVERSITY OF SCIENCE

FACULTY OF MATHEMATICS, MECHANICS AND INFORMATICS

Luu Quang Bay

THE PAPER

“SOME ELEMENTARY INEQUALITIES”

OF G. BENNETT

Undergraduate Thesis

Advanced Undergraduate Program in Mathematics

Thesis advisor: Dr. Dang Anh Tuan

Hanoi - 2013

2

Acknowledgments

I would like to express my sincere thank to my thesis advisor Dr. Dang Anh Tuan forthe help, support, patience in writing of this thesis.

I would like to show my gratitude to other teachers at the Mathematics Department ofUniversity of Science for their teaching. I also want to thank my friends in K54-AdvancedMath.

Last, I especially thank to my parents for their care, encouragement,and supports.

Contents

Preface 4

1 Some Inequalities 51.1 Notations and some basis inequalities . . . . . . . . . . . . . . . . . . . . 51.2 Hardy’s extension inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Factorable matrices 172.1 Factorable matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Some Corollaries of Theorem 2.1.1 . . . . . . . . . . . . . . . . . . . . . . 192.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Weighted mean matrices 333.1 Weighted mean matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2 Some Corollaries of Theorem 3.1.1 . . . . . . . . . . . . . . . . . . . . . . . 43

4 Littlewood’s problem 514.1 Littlewood’s problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2 Reverse Littlewood’s inequality . . . . . . . . . . . . . . . . . . . . . . . . 54

Conclusion 61

Bibliography 61

3

Preface

The aim of this thesis presents extensive Hardy’s inequalities and some results involves it.Moreover, the thesis gives some conditions such that some special matrices are boundedoperators from lp into lq and we consider Littlewood’s problem. The thesis are dividedinto four chapters:

• Chapter 1 gives some notations and some basis inequalities which will be use toprove some Theorems and Corollaries in the thesis. Besides, we give two extensionsof Hardy’s inequalities .

• In Chapter 2, we consider the factorable matrix and give some conditions to thefactorale matrix is bounded operator from lp into lq. We also give many interestingCorollaries of it. After that, we give an example of factorable matrix.

• Next, we consider the weighted mean matrix and some results involve it in Chapter3. On other hand, we consider the important Theorem in the thesis of J. Cartlidge.[5].

• Finally, in Chapter 4 we consider Littlewood’s problem and we will determine theconstant K of Littlewood’s problem in special case and general case. The reverseLittlewood’s inequality is also considered.

The main materials of the thesis were taken from the paper of G. Bennett [1]. We havealso borrowed extensively from the book of G . H. Hardy, J. E. Littlewood and G. Polya[4]. We also use the papers of D. Sylvain [5] and G. Bennett, K-G. Grosse-Erdmann [2].A note of P. Gao [3] is also used in this thesis .

4

Chapter 1

Some Inequalities

1.1 Notations and some basis inequalities

We shall be concerned with matrix transformations of lp spaces . Here as usual, lp denotesthe space of real-valued sequences x satisfying ‖x‖p ≤ ∞ where

‖x‖pp =∞∑k=1

|xk|p

and‖x‖∞ = sup

k∈N|xk|.

For 1 ≤ p <∞ we denote the conjugate exponent by p∗ so that p∗ = p/(p− 1). The normof matrix A, mapping lp into lq is given by

‖A‖p,q = sup{‖Ax‖q : ‖x‖p ≤ 1},

where

‖Ax‖qq =∞∑n=1

∣∣∣∣∣n∑k=1

ankxk

∣∣∣∣∣q

.

After that, we gives some basis inequalities.

Theorem 1.1.1 (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

. (1.1.1)

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

qi , ...) are

proportional, i.e, ai = Cbp−1i .

Proof. It is classical inequality.

Next, in two following Lemmas, we assume throughout that (an) and (bn) are sequencesof non-negative terms. We consider X and Y are the sum of two non-negative series, wecall X and Y are equivalent iff

K1X ≤ Y ≤ K2X,

where K1, K2 are positive constants and we denote X ∼ Y.

5

CHAPTER 1. SOME INEQUALITIES 6

Lemma 1.1.1 (Power Rule). If p ≥ 1, then, for all n ∈ N

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

. (1.1.2)

The inequalities reverse direction if 0 0).

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

n∑j=1

aj

)p−1

≥

(k∑j=1

aj

)p−1

. (1.1.3)

Then (n∑k=1

ak

)p

=n∑k=1

ak

(n∑k=1

ak

)p−1

=n∑k=1

ak

(n∑j=1

aj

)p−1

≥n∑k=1

ak

(k∑j=1

aj

)p−1

.

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

aj. Because p ≥ 1, then

∫ Ak

Ak−1

xp−1dx ≤ (Ak − Ak−1)Ap−1k = akAp−1k . (1.1.4)

Thus,

Apk − Apk−1 = p

∫ Ak

Ak−1

xp−1dx ≤ pakAp−1k .

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 (1.1.3) and(1.1.4) are reversed and we obtain the inverse inequalities.

Lemma 1.1.2 (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

. (1.1.5)

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 1.1.1.Case 1: p ≥ 1, for left-hand inequality, because k ≥ n and p ≥ 1 so(

∞∑j=k

aj

)p−1

≤

(∞∑j=n

aj

)p−1

. (1.1.6)

Then

∞∑k=n

ak

(∞∑j=k

aj

)p−1

≤∞∑k=n

ak

(∞∑j=n

aj

)p−1

=∞∑k=n

ak

(∞∑k=n

ak

)p−1

=

(∞∑k=n

ak

)p

.

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

aj. Since p ≥ 1 so that we have

Apk − Apk+1 = p

∫ Ak

Ak+1

xp−1dx ≤ p(Ak − Ak+1)Ap−1k = pakA

p−1k . (1.1.7)

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 1, λn > 0, an ≥ 0, and

Λn =n∑k=1

λk, An =n∑k=1

λkak,

then∞∑n=1

λn

(AnΛn

)p≤(

p

p− 1

)p ∞∑n=1

λnapn. (1.2.1)


Proof. We write αn = An/Λn. We consider

λnαpn −

p

p− 1λnanα

p−1n = λnα

pn −

p

p− 1αp−1n [Λnαn − Λn−1αn−1]

=

(λn − Λn

p

p− 1

)αpn +

p

p− 1Λn−1αn−1α

p−1n

≤(λn − Λn

p

p− 1

)αpn +

pΛn−1

p− 1

[αpn−1p

+α(p−1)qn

q

]

=

(λn − Λn

p

p− 1

)αpn +

Λn−1

p− 1

[αpn−1 + (p− 1)αpn

]=

1

p− 1

[(pλn − λn − p(Λn − Λn−1)− Λn−1)α

pn + Λn−1α

pn−1]

=1

p− 1

[(pλn − λn − pλn − Λn−1)α

pn + Λn−1α

pn−1]

=1

p− 1

[−Λnα

pn + Λn−1α

pn−1].

Hence,

N∑n=1

λnαpn −

p

p− 1

N∑n=1

λnanαp−1n ≤ 1

p− 1

N∑n=1

[Λn−1α

pn−1 − Λnα

pn

]=

1

p− 1[0− Λ1α

p1 + Λ1α

p1 − Λ2α

p2 + · · · − ΛNαN ]

= − 1

p− 1ΛNα

pN ≤ 0.

Thus,N∑n=1

λnαpn ≤

p

p− 1

N∑n=1

λnanαp−1n . (1.2.2)

Using Holder’s inequality with indexes p, p/(p − 1) on the right-hand side of (1.2.2) wehave

N∑n=1

λnαpn ≤

p

p− 1

(N∑n=1

λnapn

)1/p( N∑n=1

λnαpn

)(p−1)/p

.

Dividing the above inequality by the last factor on the right-hand side and raising theresult to the pth power, we obtain

N∑n=1

λnαpn ≤

(p

p− 1

)p N∑n=1

λnapn. (1.2.3)

When we make N tend to infinity we obtain (1.2.1).

In the above Theorem, let λk = 1, we get the Hardy’s inequality

∞∑n=1

(1

n

n∑k=1

ak

)p

≤(

p

p− 1

)p ∞∑n=1

apn.


Theorem 1.2.2 (Hardy’s extension inequality II). Let r > s ≥ 1 and u, v, w be N-tupleswith non-negative entries. If

m∑n=1

un

(n∑k=1

vk

)r

≤

(m∑k=1

vk

)s

for m = 1, 2, ..., N (1.2.4)

thenN∑n=1

un

(n∑k=1

vkwk

)r

≤ K(r, s)

(N∑k=1

vkwr/sk

)s

(1.2.5)

where

K(r, s) =

(r

r − s

)r. (1.2.6)

In the above Theorem, we can assume that uN > 0. Indeed, if uN = 0, inequality(1.2.4) becomes

m∑n=1

un

(n∑k=1

vk

)r

≤m∑k=1

vk for m = 1, ..., N − 1.

And (1.2.5) becomes

N−1∑n=1

un

(n∑k=1

vkwk

)r

≤ K(r, s)

(N−1∑k=1

vkwr/sk

)s

.

Thus, we obtain an equivalent with (N − 1)-tuples.To prove Theorem 1.2.2, we need to prove some following Lemmas.

Lemma 1.2.1. Fix N and non-negative N-tuples u,w we say that w is decreasing ifwi ≥ wj whenever i < j both vi > 0, vj > 0 . If Theorem 1.2.2 holds for decreasing abovew then it is true for arbitrary w .

Lemma 1.2.2. If Theorem 1.2.2 is true for s = 1, it holds for s > 1.

Lemma 1.2.3. Let r > 1, s = 1 and u, v, w be N-tuples with non-negative entries. Inaddition,the decreasing w is defined in Lemma 1.2.1 that is satisfied

m∑n=1

un

(n∑k=1

vk

)r

≤m∑k=1

vk for m = 1, ..., N. (1.2.7)

ThenN∑n=1

un

(n∑k=1

vkwk

)r

≤(

r

r − 1

)r N∑k=1

vkwrk, (1.2.8)

Now, we will prove the above Lemmas.

Proof of Lemma 1.2.1. If w is not decreasing that means there exists integers i, j with1 ≤ i < j < N such that wi < wj both vi > 0, vj > 0. We have w = (w1, w2 . . . , wN) withwi ≥ 0 for all i. Take a new N-tuples, w′, as follow

w′ = (w1, w2, . . . , wi−1, w′i, wi+1, . . . , wj−1, w

′j, wj+1 . . . , wn),


where

w′ir/s

= w′jr/s

=viw

r/si + vjw

r/sj

vi + vj. (1.2.9)

We claim thatN∑k=1

vkwr/sk =

N∑k=1

vk(w′k)r/s. (1.2.10)

Indeed,

N∑k=1

vkw′r/sk = v1w

r/s1 + · · ·+ vi(w

′i)r/s + · · ·+ vj(w

′j)r/s + · · ·+ vNw

r/sN

= v1wr/s1 + · · ·+ vi

viwr/si + vjw

r/sj

vi + vj.+ · · ·+ vj

viwr/si + vjw

r/sj

vi + vj+ · · ·+ vNw

r/sN

= v1wr/s1 + · · ·+ viw

r/si + · · ·+ vjw

r/sj · · ·+ vNw

r/sn

=N∑k=1

vkwr/sk .

By (1.2.10), we have the right side of (1.2.5) is unchanged when w is replaced by w′. Now,we will show that the left side of (1.2.5) increases strictly when w is replaced by w′. From(4.1.1), r > s ≥ 1 and by assuming of the hypothesis we have

w′i = w′j =

(viw

r/si + vjw

r/sj

vi + vj

)s/r

>

(viw

r/si + vjw

r/si

vi + vj

)s/r

= wi. (1.2.11)

Similarly, we also have

w′i = w′j =

(viw

r/si + vjw

r/sj

vi + vj

)s/r

<

(viw

r/sj + vjw

r/sj

vi + vj

)s/r

= wj. (1.2.12)

We will comparen∑k=1

vkwk andn∑k=1

vkw′k.

For n < i, we obtainn∑k=1

vkwk =n∑k=1

vkw′k.

For i ≤ n < j, we have

n∑k=1

vkwk = v1w1 + · · ·+ viwi + · · ·+ vnwn

< v1w1 + · · ·+ viw′i + · · ·+ vnwn

=n∑k=1

vkw′k.


For n ≥ j, we apply the Jensen’s inequality for convex function f(x) = xr/s

vivi + vj

wr/si +

vjvi + vj

wr/sj ≥

(vi

vi + vjwi +

vjvi + vj

wj

)r/s.

So that(vi + vj)w

′i ≥ (vi + vj)

vivi + vj

wi +vj

vi + vjwj = viwi + vjwj.

Hence,n∑k=1

vkwk ≤n∑k=1

vkw′k.

Therefore, we always have for all n ∈ Nn∑k=1

vkwk ≤n∑k=1

vkw′k.

We have shown that the left side of (1.2.5) is increasing strictly when w is replaced by w′.Next, we will point out the above process from arbitrary w to the decreasing w∗ is

defined in Lemma 1.2.1 is finite by induction. In the process, if there exists vi = 0, thedecreasing or nondercreasing of w is not influenced by wi. Thus, we can assume that v ispositive N -tuples.For N = 2, w = (w1, w2). If w1 ≥ w2, we stop.If w1 < w2, we change

w′ = (w′1, w′2) such that w1 < w′1 = w′2 < w2.

Hence, we change once time.We assume that w = (w1, . . . wN−1) and we need at most

(N − 2)(N − 1)

2steps

to give decreasing (N − 1)-tuples w∗.We will show that it holds for n = N, that means after at most

N(N − 1)

2steps,

the process will be stops.We consider w = (w1, w2, . . . , wN , wN−1) which is satisfied

w1 ≥ w2 ≥ · · · ≥ wN−1. (1.2.13)

• If wN−1 ≥ wN , we will stop.

• If wN−1 < wN , we will compare w1 and wN .If w1 < wN , we’ll change w1 and wN to obtain

w′ = (w′1, w2, . . . , wN−1, w′N) such that w2 ≤ w1 < w′1 = w′N .


We continue to change w2 and w′N , we get

w′′ = (w′1, w′2, w3, . . . , wN−1, w

′′N) such that w3 ≤ w2 < w′2 = w′′N < w′1.

So on the process and we need N − 1 times to give a new decreasing N -tuples w∗ .If w1 ≥ wN , we have w1 ≥ wN > wN−1 then there exists

j ∈ {1, 2, . . . , N − 2} such that wj ≥ wN ≥ wj+1.

We also do similarly as the process w1 < wN and after N − j − 1 steps, we get thea decreasing N -tuples w∗.

Therefore, we want to change from w is satisfied (1.2.13) to decreasing N -tuples w∗ thenwe need at most N − 1 steps.Combining the assumption, to change from arbitrary the N−tuples w to the decreasingN -tuples w∗ we need at most

(N − 2)(N − 1)

2+N − 1 =

N(N − 1)

2steps.

We complete the proof of Lemma1.2.1.

Proof of Lemma 1.2.2. Assuming that Theorem 1.2.2 holds for s = 1, that means if

m∑n=1

un

(n∑k=1

vk

)r

≤m∑k=1

vk, (1.2.14)

thenN∑n=1

un

(n∑k=1

vkwk

)r

≤ K(r, 1)N∑k=1

vkwrk, (1.2.15)

where

K(r, 1) =

(r

r − 1

)r.

For s > 1, let x be the a non-negative N-tuple with

‖x‖s∗ =

(N∑n=1

xs∗

)1/s∗

= 1,

wheres∗ =

s

s− 1.

Assuming inequality (1.2.4) holds. Using Holder’s inequality and inequality (1.2.4), weget

m∑n=1

xnu1/sn

(n∑k=1

vk

)r/s

≤ ‖x‖s∗(

m∑n=1

un

(n∑k=1

vk

)r)1/s

≤m∑k=1

vk.


We see that the above inequality is the inequality (1.2.14) with un is replaced by xnu1/sn , r

by r/s. Hence, the inequality (1.2.15) holds with un is replaced by xnu1/sn , r by r/s

N∑n=1

xnu1/sn

(n∑k=1

vkwk

)r/s

≤ K(r/s, 1)N∑n=1

vkwr/sk .

Taking the supermum on the left over all x satisfying ‖x‖s∗ = 1 we get

sup‖x‖s∗=1

N∑n=1

xnu1/sn

(n∑k=1

vkwk

)r/s

≤ K(r/s, 1)N∑n=1

vkwr/sk . (1.2.16)

Let y =

(u1/sn

(n∑k=1

vkwk

)r/s)N

n=1

and x = (xn)Nn=1 . We have

sup‖x‖s∗=1

N∑n=1

xnu1/sn

(n∑k=1

vkwk

)r/s

= sup‖x‖s∗=1

< y, x > .

We claim thatsup‖x‖s∗=1

| < y, x > | = ‖y‖s. (1.2.17)

By Holder’s inequality, we have

| < y, x > | ≤ ‖y‖s‖x‖s∗ = ‖y‖s. (1.2.18)

On the other hand, we choose

|xk| =(|yk|s

‖y‖ss

)1/s∗

is satisfied ‖x‖s∗ = 1. So that

| < y, x > | =∞∑k=1

|yk||xk| =∞∑k=1

|yk||yk|s/s∗

‖y‖s/s∗

s

=

∞∑k=1

|yk|s

‖y‖s/s∗

s

= ‖y‖s. (1.2.19)

From (1.2.18) and (1.2.19) we deduce (1.2.17).By (1.2.16) and (1.2.17), we obtain

‖y‖s =

N∑n=1

un

(n∑k=1

vkwk

)r/s1/s

≤ K(r/s, 1)N∑n=1

vkwr/sk ,

which is equivalent to (1.2.6). This complete the Lemma 1.2.2.

Proof of Lemma 1.2.3. If v = 0, then our problem is trivial. We assume that v 6= 0. Weneed only to show that for v1 > 0. Indeed, if v1 = 0, the inequality (1.2.7) becomes

m∑n=2

un

(n∑k=2

vk

)r

≤m∑k=2

vk for m = 2, ..., N.


And (1.2.8) becomes

N∑n=2

un

(n∑k=2

vkwk

)r

≤(

r

r − 1

)r N∑k=2

vkwrk.

They are equivalent (N − 1)-tuples.For v1 > 0, let

yn =

n∑k=1

vkwk

n∑k=1

vk

.

We consider

yi − yi+1 =

i∑k=1

vkwk

i∑k=1

vk

−

i+1∑k=1

vkwk

i+1∑k=1

vk

=

vi+1

i∑k=1

vkwk − vi+1wi+1

i∑k=1

vk

i∑k=1

vki+1∑k=1

vk

=

i∑k=1

vi+1vk(wk − wi+1)

i∑k=1

vki+1∑k=1

vk

≥ 0.

Thus, y is decreasing. Let

xn = vn − un

(n∑k=1

vk

)r

.

We havem∑n=1

xn =m∑n=1

vn −m∑n=1

un

(n∑k=1

vk

)r

≥ 0.

Let S0 = 0 and Sm =m∑n=1

xn ≥ 0 for m = 1, ..., N. We consider

N∑n=1

xnyrn =

N∑n=1

(Sn − Sn−1)yrn

= S1yr1 + S2y

r2 − S1y

r2 + · · ·+ SNy

rN − SN−1yrN

= S1(yr1 − yr2) + S2(y

r2 − yr3) + · · ·+ SN−1y

rN−1 − yrN + SNy

rN .

Since y is decreasing and Sm ≥ 0 for m = 1, .., N then

N∑n=1

xnyrn ≥ 0.


That means

N∑n=1

[vn − un

(n∑k=1

vk

)r]n∑k=1

vkwk

n∑k=1

vk

r

≥ 0,

or

N∑n=1

vn

n∑k=1

vk

n∑k=1

vk

r

≥n∑n=1

un

(n∑k=1

vkwk

)r

.

Thus, we need only show that

N∑n=1

vn

n∑k=1

vk

n∑k=1

vk

r

≤(

r

r − 1

)r n∑n=1

vkwrk.

It is the inequality of Theorem 1.2.1. Therefore, we complete Lemma 1.2.3.

From three above Lemma, we have shown that Theorem 1.2.2. We complete this sec-tion by showing that Theorem 1.2.2 fails when 0 < r ≤ s and also when 0 < s < 1, r > s.

In case: 0 < r ≤ s, we take v = (1, 1, . . . , 1), w = (1, 0, . . . , 0) and un = Cns−r−1.The inequality (1.2.4) becomes

Cm∑n=1

ns−1 ≤ ms, (1.2.20)

inequality (1.2.5) becomes

CN∑n−1

ns−1 ≤ K(r, s). (1.2.21)

For s ≥ 1, we havem∑n=1

ns−1 ≤ m.ms−1 = ms.

We can choose C = 1 to (1.2.20) is satisfied.For 0 < s < 1, we have

m∑n=1

ns−1 ≤m+1∫1

xs−1dx =(m+ 1)s − 1

s.

We consider(m+ 1)s − 1

sms=

1

s

[(1 +

1

m

)s− 1

ms

]≤ 2s

s.

So that we can choose C = 2−ss to (1.2.20) satisfied.

Since s > 0 then∞∑n=1

ns−1 diverges. Hence, (1.2.21) fails for N large.


In second case: 0 < s < 1 and r > s, we take vk = 2k, wk = v−s/rk , uk = Cvs−rk . So that

(1.2.4) becomes

C

m∑n=1

2n(s−r)

(n∑k=1

2k

)r

≤

(m∑k=1

2k

)s

= 2s (2m − 1)s . (1.2.22)

We have

m∑n=1

2n(s−r)

(n∑k=1

2k

)r

=m∑n=1

2n(s−r)2r (2n − 1)r

≤ 2rm∑n=1

2n(s−r)2nr

= 2rm∑n=1

2ns

=2r+s

2s − 1(2ms − 1)

≤ 2r+s

2s − 12ms

≤ 2r+s

2s − 12s(2m − 1)s.

We can choose C = (2s − 1).2−(r+s) to (1.2.22) is satisfied so that (1.2.5) becomes

CN∑n=1

2(s−r)n

(n∑k=1

2k2−ks/r

)r

≤ K(r, s)

(N∑k=1

2k2−k

)s

= K(r, s)N s. (1.2.23)

We consider

N∑n=1

2(s−r)n

(n∑k=1

2k2−ks/r

)r

=N∑n=1

2(s−r)n(

2(n+1)(1−s/r) − 21−s/r

21−s/r − 1

)r

=

(21−s/r

21−s/r − 1

)r N∑n=1

2(s−r)n (2(1−s/r)n − 1)r.

We consider [2(1−s/r)n − 1

]r2(1−s/r)nr =

[1− 1

2(1−s/r)n

]r≥(

1− 1

21−s/r

)r.

So that

N∑n=1

2(s−r)n

(n∑k=1

2k2−ks/r

)r

≥(

21−s/r

21−s/r − 1

)r (1− 1

21−s/r

)r n∑n=1

2(s−r)n.2nr−ns = N.

Thus, the constant C is chosen so that (1.2.22) is satisfied, as N →∞ the left-hand sideof (1.2.23) grow like N but the right-hand side like N s so that (1.2.23) fails for N large.

Chapter 2

Factorable matrices

2.1 Factorable matrices

In this section we study the mapping properties of matrices A = (ank)∞n,k=1 of the type

ank =

{anbk when 1 ≤ k ≤ n,

0 otherwise .(2.1.1)

such the matrices have been called factorable. The matrices of A may be complex numbers,but the norm of A is unchanged when the entries are replaced by the absolute value. Thuswe shall assume throughout this section that a′s and b′s are non-negative.

Theorem 2.1.1. Let 1 < p ≤ q < ∞, a and b be on non-negative numbers and let A bethe matrix given by (2.1.1). Then the following conditions are equivalent:(i) A is bounded from lp into lq.(ii) There exists K1 such that for m = 1, 2, . . .

m∑n=1

(an

n∑k=1

bp∗

k

)q

≤ K1

(m∑k=1

bp∗

k

)q/p

. (2.1.2)

(iii) There exists K2 such that for m = 1, 2, . . .(∞∑n=m

aqn

)1/q( m∑k=1

bp∗

k

)1/p∗

≤ K2. (2.1.3)

(iv) There exists K3 such that for m = 1, 2, . . .

∞∑k=m

(bk

∞∑n=k

aqn

)p∗

≤ K3

(∞∑n=m

aqn

)p∗/q∗

. (2.1.4)

Proof. Our proof follows the logical cycles

(ii)⇒ (i)⇒ (iii)⇒ (ii) and (iv)⇒ (i)⇒ (iii)⇒ (iv).

We need to prove only the first logical cycle, since both are equivalent. To see why is so,simply replace (a1, a2, . . . , aN) by (bN , bN−1, . . . , b1) and (b1, b2, . . . , bN) by (aN , aN−1, . . . , a1).

17

CHAPTER 2. FACTORABLE MATRICES 18

The resulting factorable matrix, which we denote by As, is the one obtained from A byreflection in the sinister diagonal. It follows, therefore, that

‖As‖q∗,p∗ = ‖A‖p,q. (2.1.5)

If, further, we replace p by q∗ and q by p∗ we see that the hypothesis 1 < p ≤ q <∞ andthe condition (i), (iii) are unchanged , while (ii) and (iv) are interchanged. Therefore, weneed only to prove the equivalence of conditions (i), (ii) and (iii).

(ii) ⇒ (i). Let x = (x1, x2, . . . , xN) with xi ≥ 0. If (ii) holds, we apply the Theorem1.2.2 with

un =aqnK1

, vk = bp∗

k , r = q, s = q/p.

Defining w by

wk =

{xkb

1/(1−p)k if bk > 0,

0 otherwise .

Inequality (1.2.5) gives

N∑n=1

aqnK1

(n∑k=1

bp∗

k xkb1/(1−p)k

)q

≤ K(q, q/p)

[N∑k=1

bp∗

k

(xkb

1/(1−p)k

)p]q/p,

orN∑n=1

(n∑k=1

anbkxk

)q

≤ K1K(q, q/p)

(N∑k=1

xpk

)q/p

.

So that‖Ax‖qq ≤ K1K(q, q/p)‖x‖qp = K1(p

∗)q‖x‖qp.Thus, (i) holds and

‖A‖p,q ≤ p∗K1/q1 . (2.1.6)

(i) ⇒ (iii). Let m be a fixed 1 ≤ m ≤ N and suppose that (i) holds. If x, y areN-tuples with non-negative. We have by Holder’s inequality∣∣∣∣ N∑

n=1

anyn

n∑k=1

bkxk

∣∣∣∣ ≤[

N∑n=1

(an

n∑k=1

bkxk

)q]1/q( N∑n=1

yq∗

n

)1/q∗

≤ ‖Ax‖q‖y‖q∗≤ ‖A‖p,q‖x‖p‖y‖q∗ . (2.1.7)

Setting

xk =

{b1/(p−1)k when 1 ≤ k ≤ m,

0 otherwise ,

yn =

{aq−1n when m ≤ n ≤ N,

0 otherwise .

So that (2.1.7) gives

N∑n=m

aqn

m∑k=1

bp∗

k ≤ ‖A‖p,q

(m∑k=1

bq∗

k

)1/p( N∑n=m

aqn

)1/q∗

.


Then (N∑

n=m

aqn

)1/q( m∑k=1

bp∗

k

)1/p∗

≤ ‖A‖p,q,

so that (iii) holds with K2 ≤ ‖A‖p,q.

(iii) ⇒ (ii). Let B0 = 0 and Bn =n∑k=1

bp∗

k for n = 1, 2, . . . , N. Applying the Power

Rule, for q ≥ 1 we get

Bqn =

(n∑k=1

bp∗

k

)q

≤ q

n∑k=1

bp∗

k

(k∑

m=1

bp∗

m

)q−1

= q

n∑k=1

bp∗

k Bq−1k .

Thenm∑n=1

aqnBqn ≤ q

m∑n=1

aqn

n∑k=1

bp∗

k Bq−1k

= qm∑k=1

bp∗

k Bq−1k

m∑n=k

aqn

≤ qm∑k=1

bp∗

k Bq−1k

N∑n=k

aqn.

Since (iii) holds, then

N∑n=k

aqn ≤ Kq2

(k∑

m=1

bp∗

k

)−q/p∗= Kq

2B−q/p∗k .

Therefore,

m∑n=1

aqnBqn ≤ qKq

2

m∑k=1

bp∗

k Bq−1k B

−q/p∗k

= qKq2

m∑k=1

bp∗

k Bq/p−1k

≤ qKq2B

q/pm .

So that (ii) holds with K1 ≤ qKq2 . We complete Theorem 2.1.1.

2.2 Some Corollaries of Theorem 2.1.1

Corollary 2.2.1. Let p be the fixed such that 1 < p <∞. If

an

n∑k=1

bp∗

k ≤ Kb1/(p−1)n (2.2.1)

for n = 1, 2, ..., then A is bounded on lp and

‖A‖p,p ≤ Kp∗. (2.2.2)

The constant p∗ is the best possible.


Proof. From inequality (2.2.1), we have

m∑n=1

(an

n∑k=1

bp∗

k

)p

≤ K

m∑n=1

bp∗

n .

Applying Theorem 2.1.1 (ii)⇒ (i) with p = q, we obtain A is bounded on lp and

‖A‖p,p ≤ Kp∗.

To prove that the p∗ is best constant we will find an, bk is satisfied (2.2.1) but

‖A‖p,p > C,

where 0 < C < Kp∗.Let an = 1/n, bk = 1. We have (2.2.1) always holds with K = 1. Thus, A is bounded lp

that means

‖Ax‖p =∞∑n=1

(1

n

n∑k=1

xk

)p

≤ p∗∞∑n=1

xpn.

This is Hardy’s inequality. We will prove that the best constant in this part after we provethe Corollary 2.2.3.

By using the implication (iv) ⇒ (i) of Theorem 2.1.1 as similarly argument, we alsogive the following Corollary.

Corollary 2.2.2. Let p be the fixed and 1 < p <∞. If

bk

∞∑n=k

apn ≤ Kap−1k (2.2.3)

for n = 1, 2, ..., then A is bounded on lp and

‖A‖p,p ≤ Kp. (2.2.4)

The constant p is the best possible.

Now, we begin some inequalities of Hardy, Copson and Leinder. For these results wesuppose that

xn ≥ 0, λ > 0,Λn = λ1 + · · ·+ λn.

Corollary 2.2.3. If 1 < c ≤ p, then

∞∑n=1

λnΛ−cn

(n∑k=1

λkxk

)p

≤(

p

c− 1

)p ∞∑n=1

λnΛp−cn xpn. (2.2.5)

The constant is best possible.

Proof. Using the substitution

an =

(λnΛcn

)1/p

, bk =λ1/p∗

k

Λ1−c/pn

, ypn = λnΛp−cn xpn.


The inequality (2.2.5) becomes

∞∑n=1

apn

(n∑k=1

bkyk

)p

≤(

p

c− 1

)p ∞∑n=1

ypn.

By hypothesis 1 < c ≤ p so

0 <c− 1

p− 1< 1.

Applying the Power Rule Lemma, we have

n∑k=1

bp∗

k =n∑k=1

λkΛ(c−p)/(p−1)k

=n∑k=1

λk

(k∑j=1

λk

)(c−1)/(p−1)−1

≤ p− 1

c− 1Λ(c−1)/(p−1)n .

We also have

a−1n b1/(p−1)n =

(λnΛcn

)−1/p(λ1/p∗n

Λ1−c/pn

)1/(p−1)

=λ−1/pn λ

1/pn

Λ−c/pn Λ

(1−c/p)/(p−1)n

= Λ(c−1)/(p−1)n .

Therefore,n∑k=1

bp∗

k ≤p− 1

c− 1a−1n b1/(p−1)n .

Thus, Corollary (2.2.1) may be apply with K = (p − 1)/(c − 1) to give A is bounded onlp and

‖A‖p,p ≤ Kp∗ =p

c− 1.

Now, we’ll prove that the constant is best possible.We take

xn =

{n(c−p−1)/p when 1 ≤ n ≤ N,

0 otherwise.

And

λn =

{1 when 1 ≤ n ≤ N,

0 otherwise.

Inequality (2.2.5) becomes

N∑n=1

n−c

(n∑k=1

n(c−p−1)/p

)p

≤(

p

c− 1

)p N∑n=1

1

n. (2.2.6)


And we have

n∑k=1

k(c−p−1)/p >

∫ n

1

x(c−p−1)/pdx

=p

c− 1

(n(c−1)/p − 1

).

Therefore,

n−c

(n∑k=1

n(c−p−1)/p

)p

>

(p

c− 1

)pn−cnc−1

(1− 1

n(c−1)/p

)p=

(p

c− 1

)pn−1

(1− p

n(c−1)/p

).

Summing on n, we obtain

N∑n=1

n−c

(n∑k=1

n(c−p−1)/p

)p

>

(p

c− 1

)p N∑n=1

(1− p

n(c−1/p

) 1

n.

ThenN∑n=1

n−c(

n∑k=1

n(c−p−1)/p)p

N∑n=1

n−1>

(p

c− 1

)p 1−pN∑n=1

n(1−c−p)/p

N∑n=1

n−1

.Since 1 < c ≤ p so (c+ p− 1)/p > 1. Thus,

N∑n=1

n(1−c−p)/p converges as N →∞.

In addition, we haveN∑n=1

n−1 → ∞ as N → ∞. Hence, the right hand side of the above

inequality tend to(

pc−1

)pas N tend to infinite . We complete Corollary 2.2.3.

Now, we come back the Corollary 2.2.1 to show that the best constant. From theCorollary 2.2.3 we find the K = (p− 1)/(c− 1). We give

c =p− 1

K+ 1.

Thus, we can choose an, bk

an =

{n(1−K−p)/(Kp) when 1 ≤ n ≤ N,

0 otherwise,

bn =

{n(K+p−1)/(Kp)−1 when 1 ≤ n ≤ N,

0 otherwise.

And

xn =

{n−1/p when 1 ≤ n ≤ N,

0 otherwise,


where N is choose later. We have an and bk are satisfied the hypothesis of Corollary 2.2.1.Indeed,

an

n∑k=1

bp∗

k = n(1−K−p)/(Kp)n∑k=1

k[(K+p−1)/(Kp)−1]p∗

= n(1−K−p)/(Kp)n∑k=1

k1/K−1

< n(1−K−p)/(Kp)∫ n

0

x1/K−1dx

= n(1−K)/(Kp)n−1/KKx1/K

= Kb1/pn .

We have

n∑k=1

bkxk =n∑k=1

n(K+p−1)/(Kp)−1n−1/p

>

∫ n

1

x(K+p−1)/(Kp)−1−1/p

=Kp

p− 1

[n(p−1)/(Kp) − 1

].

Thus,

‖Ax‖pp =N∑n=1

apn

(n∑k=1

bkxk

)p

=N∑n=1

n(1−K−p)/K

(n∑k=1

n(K+p−1)/(Kp)−1n−1/p

)p

>N∑n=1

n(1−K−p)/K[Kp

p− 1

(n(p−1)/(Kp) − 1

)]p> (Kp∗)p

N∑n=1

n(1−K−p)/Kn(p−1)/K[1− p

n(p−1)/(Kp)

]= (Kp∗)p

N∑n=1

1

n

[1− p

n(p−1)/(Kp)

]

= (Kp∗)p

1−p

N∑n=1

n−1/(Kp∗)+1

N∑k=1

1n

N∑n=1

1

n

= (Kp∗)pI‖x‖pp.

Thus,‖A‖p > Kp∗I1/p.


We haveN∑n=1

n−1/(Kp∗)+1 converges as N →∞ and

N∑k=1

n−1 is divergent as N →∞. There-

fore,

I =

1−p

N∑n=1

n−1/(Kp∗)+1

N∑k=1

1n

→ 1 as N →∞.

Thus, 0 < C = Kp∗I1/p < Kp∗. We have shown that the best possible constant inCorollary 2.2.1.

Corollary 2.2.4. If 0 ≤ c < 1, then

∞∑n=1

λnΛ−cn

(∞∑k=n

λkxk

)p

≤(

p

1− c

)p ∞∑n=1

λnΛp−cn xpn. (2.2.7)


Proof. We also substitute ypn = λnΛp−cn xpn and take A′ give by

an =λ1/p∗n

Λ1−c/pn

, bk =

(λkΛck

)1/p

.

By Power Rule inequality, we have

n∑k=1

bpk =n∑k=1

λkΛ−ck ≤

1

1− cΛ1−cn .

In addition,

a−1n b1/(p∗−1)

n =λ−1/p∗n

Λ−1+c/pn

(λnΛcn

)(p−1)/p

=λ−1/p∗n

Λ−1+c/pn

λ1/p∗n

Λc(p−1)/pn

= Λ1−cn .

Hence,n∑k=1

bpk ≤1

1− ca−1n b1/(p

∗−1)n .

By Corollary 2.2.1, we get A′ is bounded on lp∗

and

‖A′‖p∗,p∗ ≤p

1− c.

Therefore,

‖A‖p,p ≤p

1− c.

We complete Corollary (2.2.4).


After that, we also have some similarly inequalities with Λ∗n =∞∑k=n

λn for n = 1, 2, . . .

Corollary 2.2.5. If 0 ≤ c < 1, then

∞∑n=1

λn(Λ∗n)−c

(∞∑k=n

λkxk

)p

≤(

p

1− c

)p ∞∑n=1

λn(Λ∗n)p−cxpn. (2.2.8)


Corollary 2.2.6. If 1 < c ≤ p,

∞∑n=1

λn(Λ∗n)−c

(∞∑k=n

λkxk

)p

≤(

p

c− 1

)p ∞∑n=1

λn(Λ∗n)p−cxpn. (2.2.9)


Corollary 2.2.7. If 1 1, then

∞∑n=1

λnΛ(1−c)q/p−1n

(n∑k=1

λkxk

)q

≤ K

(∞∑n=1

λnΛp−cn xpn

)q/p

. (2.2.10)

Proof. Letaqn = λnΛ(1−c)q/p−1

n , bk = λ1/p∗

k Λc/p−1k , ypn = λnΛp−c

n xpn.

Inequality (2.2.10) becomes

∞∑n=1

(an

n∑k=1

bkyk

)q

≤ K

(∞∑n=1

ypn

)q/p

.

Applying Theorem 2.1.1 (ii)⇒ (i), we need only to point out

m∑n=1

(an

n∑k=1

bp∗

k

)q

≤ K1

(m∑n=1

bp∗

k

)q/p

.

We consider

n∑k=1

bp∗

k =n∑k=1

λkΛ(c−p)/(p−1)k

≤ K

(n∑k=1

λk

)(c−1)/(p−1)

(By Power Rule Lemma ).


Thus,

m∑n=1

(an

n∑k=1

bp∗

k

)q

≤ Kq

m∑n=1

aqnΛ(c−1)q/(p−1)n

= Kq

m∑n=1

λnΛ(1−c)q/p−1n Λ(c−1)q/(p−1)

n

= Kq

m∑n=1

λnΛ(c−1)q/[p(p−1)]−1n (By Power Rule Lemma )

≤ C1

(Λ(c−1)/(p−1)m

)q/p≤ C2

(m∑n=1

bp∗

k

)q/p

,

where C1, C2 are constants. We complete Corollary 2.2.7.

2.3 Example

Our next result gives a complete description of the mapping properties of factorable ma-trices of the form

an = n−α, bk = k−β. (2.3.1)

Corollary 2.3.1. Let 1 ≤ p, q ≤ ∞ and let A be the matrix given by (2.3.1). Then A isbounded from lp into lq if and only if

(i) α ≥ 0 and α + β ≥ 0 (p = 1, q =∞),

(ii) α > 0 and α + β ≥ 1

p∗or α = 0 and β >

1

p∗(p > 1, q =∞),

(iii) α >1

qand α + β ≥ 1

q+

1

p∗(1 1

qand α + β >

1

q+

1

p∗(p > q).

Moreover, ‖A‖1,∞ = 1. And in the case 1 1/p∗ we have

‖A‖p,∞ = ‖k−β‖p∗ .

Before proving, we have a notation is used in proof of the above Corollary. We estimatethe following sum

n∑k=1

n−α ∼

{n1−α if 0 < α < 1,

lnn if α = 1.

Proof. We also have

(Ax)m = ym = m−αm∑k=1

k−βxk.


(a) First, we will show that A is bounded from l1 to l∞ ⇔ (i).(⇒) Let x = (1, 0, . . .) so that ym = m−α. Since A is bounded from l1 in l∞ so that Ax ∈ lp.Thus,

supm∈N

m−α < +∞.

That means α ≥ 0. Now, we take x(n) = (0, . . . 0, 1︸︷︷︸n

, 0, . . .) so

(Ax(n))m =

m−αn−β for m > n,

n−(α+β) for m = n,

0 for m < n.

Since α ≥ 0 , we get‖Ax(n)‖∞ = n−(α+β).

Because A is bounded, we have for all n

n−(α+β) = ‖Ax(n)‖∞ ≤ ‖A‖1,∞‖x(n)‖1 = ‖A‖1,∞.

Then α + β ≥ 0.(⇐) For every m ∈ N,∣∣∣∣∣m−α

m∑k=1

k−βxk

∣∣∣∣∣ ≤m∑k=1

k−(α+β)|xk| (Since α ≥ 0)

≤m∑k=1

|xk| (Since α + β ≥ 0)

≤ ‖x‖1.

Hence, ‖Ax‖∞ ≤ ‖x‖1 and|A‖1,∞ ≤ 1. (2.3.2)

We claim that‖A‖1,∞ = 1.

Indeed, we choose x0 = (1, 0, · · · ) so that

‖Ax0‖∞ = supm∈N

m−α = 1

and ‖x0‖1 = 1. Since‖Ax0‖∞ ≤ ‖A‖1,∞‖x0‖1.

Then ‖A‖1,∞ ≥ 1 with choosing above x0.Combining (2.3.2), we complete the claim.(b) Next, we will show that A is bounded from lp to l∞ ⇔ (ii) with 1 < p <∞.(⇒) Similarly part (a), we also let x = (1, 0 · · · ) and we give α ≥ 0.For α ≥ 0, we choose x(n) = (1, 2, . . . , n, 0 . . .) then

(Ax(n))m =

m−α

n∑k=1

k−β+1 ∼ m−αn2−β for m ≥ n,

m−αm∑k=1

k−β+1 ∼ m−αm2−β for m < n.


We havesupm≥n

m−αn2−β = n2−α−β ≤ ‖Ax‖∞. (2.3.3)

We also have A is bounded so

‖Ax(n)‖∞ ≤ ‖A‖p,∞‖x(n)‖p = ‖A‖p,∞

(n∑k=1

kp

)1/p

≤ ‖A‖p,∞n(p+1)/p. (2.3.4)

From (2.3.3) and (2.3.4), we obtain 2− α− β ≤ (p+ 1)/p that means α + β ≥ 1/p∗.For α = 0, we have β ≥ 1/p∗ and

(Ax(n))m =m∑k=1

k−βxk.

We need only to point out that if β = 1/p∗, A is not bounded from lp into l∞ .Let x(n) = (1−1/p, . . . , n−1/p, 0, . . .) so

(Ax(n))m =

n∑k=1

k−1 ∼ lnn for m ≥ n,

m∑k=1

k−1 = lnm for m < n.

Thus, ‖Ax(n)‖∞ ∼ lnn and ‖x(n)‖p =

(n∑k=1

k−1)1/p

∼ (lnn)1/p.

Since for 1 0 and α + β ≥ 1/p∗, A is bounded from lp in l∞ with1 1/p∗, A is bounded lp to l∞ with p > 1.

|(Ax(n))m| =

∣∣∣∣∣m∑k=1

k−βxk

∣∣∣∣∣≤

m∑k=1

k−β|xk|

≤

(m∑k=1

k−βp∗

)1/p∗ ( m∑k=1

xpk

)1/p

≤ ‖(k−β)‖p∗‖x‖p.

Then A is bounded and ‖A‖p,∞ ≤ ‖(k−β)‖p∗ . In this case, we can show that

‖A‖p,∞ = ‖k−β‖p∗ . (2.3.5)

By takingx(n) = (1−β(p

∗−1), . . . , n−β(p∗−1), 0, . . .).

We have (Ax(n))m =n∑k=1

k−βp∗. Hence, we obtain

‖Ax(n)‖∞ =n∑k=1

k−βp∗

=

(n∑k=1

k−βp∗

)1/p∗ ( n∑k=1

k−βp∗

)1/p

=

(n∑k=1

k−βp∗

)1/p∗ ( n∑k=1

k−β(p∗−1)p

)1/p

= ‖(k−β)‖p∗‖x(n)‖p.

In addition, ‖A‖p,∞ ≤ ‖(k−β)‖p∗ . We get (2.3.5).

(c) After that, we will prove the equivalence of (iii) and A is bounded from lp intolq with 1 1. We assume that

α + β <1

q+

1

p∗. (2.3.6)


Since α > 1/q, we deduce

β +1

p< 1.

Let x(N) = (1−1/p, . . . , N−1/p, 0, . . .). We get

(Ax(N))m =

m−α

m∑k=1

k−(1/p+β) for m < N,

m−αN∑k=1

k−(1/p+β) for m ≥ N.

Then,

‖Ax(N)‖qq =N∑m=1

m−αq

(m∑k=1

k−(1/p+β)

)q

+∞∑

m=N+1

m−αq

(N∑k=1

k−(1/p+β)

)q

≥N∑m=1

m−αq

(m∑k=1

k−(1/p+β)

)q

∼N∑m=1

m−αqm(1−1/p−β)q (Since 0 < β +1

p< 1).

Since

αq + (β +1

p− 1)q < 1.

Then‖Ax(N)‖qq ∼ N1−−αqm(1−1/p−β)q.

On other hand,

‖x(N)‖qp =

(N∑m=1

m−1

)q/p

∼ (lnN)q/p.

Thus,

limN→∞

‖Ax(N)‖qq‖x(N)‖qp

= +∞.

Hence, A is not bounded that this is opposite the hypothesis. Hence we get the proof.(⇐) Applying Theorem 2.1.1, so we need only show that

m∑n=1

n−αq

(n∑k=1

k−βp∗

)q

≤ K

(m∑n=1

n−βp∗

)q/p

.


Since α + β ≥ 1/q + 1/p∗ so βp∗ ≥ 1 + p∗(1/q − α). Therefore,

m∑n=1

n−αq

(n∑k=1

n−βp∗

)q

≤m∑n=1

n−αq

(n∑k=1

kp∗(α−1/q)−1

)q

∼m∑n=1

n−αqnp∗(α−1/q)q (Since p∗(α− 1/q)− 1 < −1)

=m∑n=1

n−αq+p∗(αq−1)

∼ m1−αq+p∗(αq−1) (Since αq − p∗(αq − 1) < 1)

= m(αq−1)/(p−1).

On the other hand,(m∑k=1

n−βp∗

)q/p

≥ Kmp∗(α−1/q)q/p = Kmp∗(αq−1)/p = Km(αq−1)/(p−1),

where K is constant. Thus, A is bounded from lp into lq.(d) Finally, we will show that A is bounded form lp into lq ⇔ (iv) with 1 < q 1/q.Next, we suppose that

α + β ≤ 1

q+

1

p∗.

We use inequality| < Ax, y > | ≤ ‖A‖p,q‖x‖p‖y‖q∗ . (2.3.7)

Let x(N) = (1−1/p, . . . , N−1/p, 0, . . .) and y(N) = (1−1/q∗, . . . , N−1/q

∗, 0, . . .).

‖x(N)‖p‖y(N)‖q∗ =

(N∑k=1

k−1

)1/p( N∑k=1

k−1

)1/q∗

∼ (lnN)1/p+1/q∗ .

And

< Ax(N), y(N) >=N∑m=1

m−(α+1/q∗)m∑k=1

k−(β+1/p) ∼N∑m=1

m−(α+1/q∗)m1−β−1/p = I2.

If α + β = 1/q + 1/p∗, we have

1− α− β − 1

p− 1

q∗= −1.

Thus| < Ax(N), y(N) > | > I2 ∼ (lnN).

Since p > q so that

0 ≤ 1

p+

1

q∗=

1

p+q − 1

q= 1 +

1

p− 1

q< 1.


We have lnN > (lnN)1/p+1/q∗ . Therefore,

limN→∞

| < Ax(N), y(N) > |‖x(N)‖p‖y(N)‖q∗

= +∞.

. Thus, (2.3.7) fails as N tend to infinite . That means A is not bounded.If α + β < 1/q + 1/p∗, we deduce

1− α− β − 1

p− 1

q∗> −1.

Then I2 ∼ N−α−β−1/q−1/p∗. Thus,

limN→∞

| < Ax(N), y(N) > |‖x(N)‖p‖y(N)‖q∗

= +∞.

Therefore, (2.3.7) is not true as N →∞.Summary, if

α >1

qand α + β ≤ 1

q+

1

p∗,

then we get A is not bounded which it is opposite the hypothesis. We complete this part.

(⇐) Let γ = 1− α. Applying part (c) with p = q to the factorable matrix A′ where

a′n = n−α, b′k = k−γ.

We obtain A′ is bounded on lq. On the other hand, the matrix A may be factored :

A = A′D,

where D is the diagonal matrix with diagonal entries :

dk = kγ−β.

Sine α + β > 1/q + 1/p∗, we get

(γ − β)pq

p− q=

(1− α− β)pq

p− q< (1− 1

p− 1

q∗)pq(p− q) = −1 (2.3.8)

then d ∈ lpq/(p−q). It follows from Holder’s inequality, that D is bounded lp into lq andthus,

‖Ax‖q = ‖A′Dx‖q ≤ ‖A′‖q, q‖Dx‖q ≤ ‖A′‖q,q‖D‖p,q‖x‖p.

Hence A is bounded from lp into lq.

Chapter 3

Weighted mean matrices

3.1 Weighted mean matrices

In this section we obtain simple necessary and sufficient conditions for a weighted meanmatrix to map lp into lq (1 ≤ p, q ≤ ∞). We recall that a weighted matrix, A is an infinitematrix of the form

ank =

{ak/An when 1 ≤ k ≤ n,


where an’s is non-negative numbers with partial sums An = a1 + a2 + · · ·+ an.We claim that

‖A‖p,∞ = 1 (1 ≤ p <∞), (3.1.2)

‖A‖1,q = supkak

(∞∑n=k

A−qn

)1/q

(1 ≤ q <∞). (3.1.3)

First, we will prove that 3.1.2. We have

(Ax)m =

A−1m

m∑k=1

akxk when 1 ≤ m < n,

A−1nn∑k=1

akxk when m = n,

0 otherwise.

Then,

‖Ax‖∞ = A−1n

n∑k=1

akxk

≤ A−1n

(n∑k=1

ap∗

k

)1/p∗ ( n∑k=1

xpk

)1/p

(By Holder’s inequality)

≤ A−1n

(n∑k=1

ap∗

k

)1/p∗

‖x‖p. (3.1.4)

Since p ≥ 1, then p∗ ≥ 1 and we have

n∑k=1

ap∗

k ≤

(n∑k=1

ak

)p∗

= Ap∗

n .

33

CHAPTER 3. WEIGHTED MEAN MATRICES 34

So that

A−1n

(n∑k=1

ap∗

k

)1/p∗

≤ 1 (3.1.5)

From (3.1.4) and (3.1.5) we give‖Ax‖∞ ≤ ‖x‖p.

Thus,‖A‖p,∞ ≤ 1. (3.1.6)

Let x0 = (1, 0, . . .), we have Ax0 = (1, A−12 a1, . . . , A−1n a1, 0, . . .)

T . Hence we obtain

‖Ax0‖p,∞ = 1 = ‖x0‖p.

Therefore, ‖A‖p,∞ = 1.Next, we will prove that (3.1.3).

Let x(k) = (0, . . . 0, 1︸︷︷︸k

, 0, . . .), we have

(Ax(k))m =

0 for 1 ≤ m < k,

A−1m ak for k ≤ m ≤ n,

0 otherwise.

Thus,

‖Ax‖q =

[n∑

m=k

(A−1m ak

)q]1/q

= ak

(n∑

m=k

A−qm

)1/q

≤ ak

(∞∑m=k

A−qm

)1/q

≤ supk∈N

ak

(∞∑m=k

A−qm

)1/q

.

And we also have ‖x(k)‖1 = 1, we give

‖A‖1,q ≤ supk∈N

ak

(∞∑n=k

A−qn

)1/q

. (3.1.7)

We take

f (k)n =

{A−1n akxk when 1 ≤ k ≤ n,

0 when k > n.


We have

‖f (k)‖q =

(∞∑n=1

|f (k)n |q

)1/q

=

(k−1∑n=1

|f (k)n |q +

∞∑n=k

|f (k)n |q

)1/q

=

[∞∑n=k

(A−1n ak|xk|

)q]1/q

= ak|xk|

(∞∑n=k

A−qn

)1/q

.

On other hand,∞∑k=1

f (k) has a coordinate

∞∑k=1

f (k) = (∞∑k=1

f(k)1 ,

∞∑k=1

f(k)2 , . . .).

So that

‖∞∑k=1

f (k)‖q =

[∞∑n=1

(∞∑k=1

f (k)n

)q]1/q

=

[∞∑n=1

(n∑k=1

A−1n akxk

)q]1/q= ‖Ax‖q.

We aslo have

‖∞∑k=1

f (k)‖q ≥∞∑k=1

‖f (k)‖q.

Thus,

‖Ax‖q ≥∞∑k=1

ak|xk|

(∞∑n=k

A−qn

)1/q

≥ supk∈N

ak

(∞∑n=k

A−qn

)1/q ∞∑k=1

|xk|.

Hence,

‖A‖1,q ≥ supk∈N

ak

(∞∑n=k

A−qn

)1/q

. (3.1.8)

From (3.1.7) and (3.1.8) we obtain (3.1.3).

Theorem 3.1.1. Let A be the weighted mean matrix given by (3.1.1).(a) If 1 < p ≤ q <∞ the following conditions are equivalent :

(i) A maps lp into lq.


(ii) For some constant K1 and all m = 1, 2, . . . such that

m∑n=1

(A−1n

n∑k=1

ap∗

k

)q

≤ K1

(m∑k=1

ap∗

k

)q/p

.

(iii) For some constant K2 and all m = 1, 2, . . . such that(∞∑n=m

A−qn

)1/q( m∑k=1

ap∗

k

)1/p∗

≤ K2.

(iv) For some constant K3 and all m = 1, 2, . . . such that

∞∑k=m

(ak

∞∑n=k

A−qn

)p∗

≤ K3

(∞∑n=m

A−qn

)p∗/q∗

.

(b) If 1 ≤ q < p <∞ the A does not map lp into lq.

Proof. Part (a) is special case of Theorem 2.1.1 to weighted mean matrix with an isreplaced by A−1n , bk by ak. Part (b) is a consequence of Proposition 3.1.1 below.

Proposition 3.1.1. Let A be a lower triangular matrix with

ρ = lim infn→∞

|n∑k=1

ank| > 0. (3.1.9)

If 0 < q < p <∞ then A is not bounded lp in lq.

To prove the Proposition 3.1.1, we need to prove that Pitt’s Theorem [5].

Theorem 3.1.2 (Pitt’s Theorem). If 1 ≤ q < p <∞. Then every bounded linear operatorfrom lq into lp is compact.

To prove Pitt’s Theorem we need to show the following Lemma that is proved in paperof D. Sylvain [5]

Lemma 3.1.1. If z ∈ lp, 1 ≤ p <∞ and for every weakly null sequence (w(i)) in lp, then

lim supi→∞

‖z + z(i)‖pp = ‖z‖pp + lim supi→∞

‖z(i)‖pp. (3.1.10)

Proof. Let z(k) = (z1, z2, . . . , zk, 0, . . .) and z(k) → z in lp. We also have

‖z(k) − z‖p =

(∞∑

j=k+1

|zj|p)1/p

.

First, we will prove that equality is true for (z(k)). Since (w(i)) is weakly null sequence sothat

w(i)n → 0 as i→∞,∀n.


We have

lim supi→∞

‖w(i)‖pp = lim supi→∞

[k∑j=1

|w(i)j |p +

∞∑j=k+1

|w(i)j |p]

= lim supi→∞

∞∑j=k+1

|w(i)j |p.

Fixing k, we have

lim supi→∞

k∑j=1

|z(k)j + w(i)j |p =

k∑j=1

|z(k)j |p.

Therefore,

lim supi→∞

‖z(k) + z(i)‖pp = lim supi→∞

∞∑j=1

|z(k)j + w(i)j |p

= lim supi→∞

[k∑j=1

|z(k)j + w(i)j |p +

∞∑j=k+1

|z(k)j + w(i)|p]

=k∑j=1

|z(k)j |p + lim supi→∞

∞∑j=k+1

|w(i)j |p

= ‖z(k)‖pp + lim supi→∞

‖w(i)‖pp.

Next, we will prove that Lemma 3.1.1 is true for z ∈ lp. We use the inequality∣∣‖x‖pp − ‖y‖pp∣∣ ≤ p‖x− y‖p (‖x‖p + ‖y‖p)p−1 .

Thus,∣∣‖z + w(i)‖pp − ‖z(k) + w(i)‖pp∣∣ ≤ p‖z − z(k)‖p

(‖z + w(i)‖p + ‖z(k) + w(i)‖p

)p−1≤ p‖z − z(k)‖p

(2‖z‖p + 2‖w(i)‖p

)p−1≤ pCp‖z − z(k)‖p.

Hence,

‖z(k) + w(i)‖pp − pCp‖z − z(k)‖p ≤ ‖z + w(i)‖pp ≤ ‖z(k) + w(i)‖pp + pCp‖z − z(k)‖p.

That means

‖z(k)‖pp+lim supi→∞

‖w(i)‖pp−pCp‖z−z(k)‖p ≤ ‖z+w(i)‖pp ≤ ‖z(k)‖pp+lim supi→∞

‖w(i)‖pp+pCp‖z−z(k)‖p.

Since ‖z − z(k)‖p → 0, then

‖z(k)‖pp + lim supi→∞

‖w(i)‖pp = ‖z + w(i)‖pp.

Hence, we complete Lemma 3.1.1.


Now, we come back Pitt’s Theorem that is shown in the paper of D. Sylvain [5].

Proof of Pitt’s Theorem. Let T : lq → lp be a norm-one operator. Every bounded sequencein lp has a weakly Cauchy sequence. Thus, for proving the compactness of T, it is enoughto show that T is weak-to-norm continuous. So, let us consider a weakly sequence (hn) inlp. We have to show that

limn→∞

‖T (hn)‖ = 0.

Fix 0 < ε < 1. By definition of the norm of T, we have ‖T‖p,q = 1 and there exists xε ∈ lpsuch that ‖xε‖p = 1 and 1− ε ≤ ‖T (xε)‖ ≤ 1. Moreover, for all n ∈ N and for all t > 0

‖T (xε) + T (thn)‖q ≤ ‖T‖p,q‖xε + thn‖p = ‖xε + thn‖p. (3.1.11)

In the left-hand side of (3.1.11), we apply Lemma 3.1.1 in lq, with z = T (xε) and theweakly null sequence (T (thn)). We have

lim supn→∞

‖T (xε) + T (thn)‖qq = ‖T (xε)‖qq + lim supn→∞

‖T (thn)‖qq.

Similarly we also apply Lemma 3.1.1 to the right-hand side of (3.1.11) with z = xε andthe weakly null sequence (thn) to obtain

lim supn→∞

‖xε + thn‖pp = ‖xε‖pp + lim supn→∞

‖thn‖pp.

Thus, [‖T (xε)‖qq + tq lim sup

n→∞‖Thn‖qq

]1/q=

[lim supn→∞

‖T (xε) + T (thn)‖qq]1/q

≤[lim supn→∞

‖xε + thn‖pp]1/p

=

[‖xε‖pp + lim sup

n→∞‖thn‖pp

]1/p=

[‖xε‖pp + tp lim sup

n→∞‖hn‖pp

]1/p.

Recall that ‖xε‖ = 1, 1 − ε ≤ ‖T (xε)‖ ≤ 1 and that (hn) is weakly convergent so it isbounded by some M > 0. This gives

lim supn→∞

‖Thn‖qq ≤1

tq

[(1 + tpMp)q/p − (1− ε)q

].

Taking t = ε1/p here, we get

lim supn→∞

‖Thn‖qq ≤1

εq/p[(1 + εMp)q/p − (1− ε)q

]≤ 1

εq/p

[1 +

q

pMpε− (1− qε)

].

Because (1 − ε)q ≤ 1 − qε with 0 ≤ ε < 1 and q ≥ 1. Letting ε → 0 we obtainlim supn→∞

‖Thn‖qq ≤ 0, and therefore the sequence (T (hn)) norm-converges to 0.


We come back Proposition 3.1.1

Proof of Proposition 3.1.1. We assume that A is satisfied the hypothesis of Proposition3.1.1 and A is bounded from lp to lq. By Pitt’s Theorem, we have A must be compactoperator. On the other hand,

ρ = lim infn→∞

|n∑k=1

ank| > 0.

So there exists N0, for all n ≥ N0 ∣∣∣∣∣n∑k=1

ank

∣∣∣∣∣ > ρ

21/(2q).

For m > N0, we havem∑n=1

∣∣∣∣∣n∑k=1

ank

∣∣∣∣∣q

> (m−N0)ρq√

2.

Hence, we can choose m > 4N0 to

m∑n=1

∣∣∣∣∣n∑k=1

ank

∣∣∣∣∣q

>3m

4

ρq√2> ρq

m

2. (3.1.12)

Let x(n) ∈ lq such that

x(n)k =

{n−1/q if 1 ≤ k ≤ n,

0 if k > n.(3.1.13)

We have to point out x(n) → 0 in weakly in lq that means

| < x(n), y > | = n−1/pn∑k=1

|yk| → 0 as n→∞.

where y ∈ lq∗ . Since y ∈ lq∗ so∞∑k=1

|yk|q∗< +∞. Thus,

limn→∞

∞∑k=n+1

|yk|p∗

= 0.

Let ε > 0,∃k0,∀n ≥ k0,∞∑

k=n+1

|yk|q∗<( ε

2

)q∗.

So we get∞∑

k=k0+1

|yk|q∗<( ε

2

)q∗.

For n > k0,n∑

k=k0+1

|yk| ≤

(n∑

k=k0+1

1q

)1/q( n∑k=k0+1

|yk|q∗

)1/q∗

< n1/p ε

2.


Hence,

n−1/pn∑k=1

|yk| =

k0∑k=1

|yk|+n∑

k=k0+1

|yk|

n1/p

≤

k0∑k=1

|yk|

n1/p+ε

2

= C1

n1/p+ε

2.

We also have

limn→∞

1

n1/p= 0.

Then for ε is chosen above, ∃n0 : ∀n ≥ n0

1

n1/p<

ε

2C.

Therefore,

n−1/pn∑k=1

|yk| < ε.

We have shown x(n) → 0 is weakly in lq. Since A is compact so that

‖Ax(n)‖q → 0.

But if m ≥ 4N0, we have

‖Ax(m)‖qq ≥m∑n=

∣∣∣∣∣n∑k=1

ankx(m)k

∣∣∣∣∣q

=m∑n=1

m−1

∣∣∣∣∣n∑k=1

ank

∣∣∣∣∣q

≥ ρq

2.

This is contradiction (3.1.12) as m→∞

Theorem 3.1.1 seems to new event in the special q = p. The best previously knownresults in this case are given in a unpublished Thesis of Jame Cartlidge. Cartlidge’s mainresult from which he derives many interesting corollaries, is the following

Theorem 3.1.3 (Theorem C). Let p be the fixed , 1 0 for n = 1, 2, ... If

L = supn

(An+1

an+1

− Anan

)< p, (3.1.14)

then A is bounded on lp and‖A‖p,q ≤

p

p− L.


Proof. We use the proof of P. Gao [3] to show that the Theorem C.To prove the theorem we will show that

∞∑n=1

Bpn ≤

p

p− L

∞∑n=1

anBp−1n (3.1.15)

where

Bn =n∑k=1

akxkAn

.

We consider the function

f(xn) =

(Anan

+ p− 1

)Bpn − pxnBp−1

n .

For n ≥ 2, xn ≥ 0, we have

f ′(xn) = p(p− 1)anAn

(Bp−1n − xnBp−2

n

).

Hence f ′(xn) = 0 where Bn = xn. That means

n−1∑k=1

akxk + anxn

An= an.

or Bn−1 = xn. We also have

f(xn) = Bp−1n

[(Anan

+ p− 1

)Bn − pxn

]= Bp−1

n

[(Anan

+ p− 1

) n−1∑k=1

akxkAn

+

(Anan

+ p− 1

)anxnAn− pxn

]

= Bp−1n

[(Anan

+ p− 1

) n−1∑k=1

akxkAn

+ xn(p− 1)

(n

An− 1

)].

Since an/An ≤ 1 so thatlimxn→∞

f(xn) ≤ 0.

That means ∃y such that f(y) < M and ∃x ∈ [Bn−1, y] satisfied

f ′(x) =f(y)− f(Bn−1)

y −Bn−1< 0.

Simllarly, we have

f(0) =

(Anan

+ p− 1

)[n−1∑k=1

akxkAn−1

]p≥ 0.

Thus f(xn) ≤ f(Bn−1). Therefore(Anan

+ p− 1

)Bpn − pxnBp−1

n ≤(anAn− 1

)Bpn−1. (3.1.16)


By defining B0 = 0 the above inequality also holds for n = 1. Summing (3.1.16) fromn = 1 to N gives

N∑n=1

(Anan

+ p− 1

)Bpn − p

N∑n=1

xnBp−1n ≤

N∑n=1

(anAn− 1

)Bpn−1.

SoN∑n=1

(Anan

+ p− 1

)Bpn −

N−1∑n=0

(an+1

An+1

− 1

)Bpn ≤ p

N∑n=1

xnBp−1n .

N∑n=1

(Anan

+ p− An+1

an+1

)Bpn +

N∑n=1

(1 +

An+1

an+1

)Bpn −

N−1∑n=0

(an+1

An+1

− 1

)Bpn ≤ p

N∑n=1

xnBp−1n .

Hence

N∑n=1

(Anan

+ p− An+1

an+1

)Bpn +

(aN+1

AN+1

− 1

)BpN ≤ p

N∑n=1

xnBp−1n . (3.1.17)

By condition (3.1.14) we have

p+Anan− An+1

an+1

≥ p− L.

We get the inequality (3.1.15) follwos (3.1.17). Apply the Holder inequality we have

N∑n=1

xnBp−1n ≤

(N∑n=1

xpn

)p( N∑n=1

Bpn

)(p−1)/p

. (3.1.18)

Combinating (3.1.15) and (3.1.18) we obtain(N∑n=1

Bpn

)1/p

≤ p

p− L

(N∑n=1

xpn

)1/p

.

This completes the proof.

Cartlidge’s result is less satisfactory than Theorem 3.1.1 since his hypothesis (3.1.14)is highly susceptible to changes individual a′n s. On the other hand, Theorem 3.1.3 covermany case of partial interest, wherein the a′n s be have in a ” regular ” fashion. Aparticularly interesting consequence of Theorem 3.1.3, again due to Cartlidge is

Corollary 3.1.1 (Corollary C). If (an)∞n=1 is an increasing sequence then A is boundedon lp where 1 < p <∞.

Proof. Since (an)∞n=1is an increasing sequence so that

An+1

an+1

− Anan≤ An+1

an− Anan

= 1.

Apply Theorem (3.1.3), we get the proof.


3.2 Some Corollaries of Theorem 3.1.1

Corollary 3.2.1. Let p be the fixed, 1 < p <∞ if

ap∗

1 + · · ·+ ap∗

n = O(a1/(p−1)n An

), (3.2.1)

then A is bounded on lp.

Proof. From the hypothesis (3.2.1) we get

ap∗

1 + · · ·+ ap∗n

Ana1/(p−1)n

≤ K,

then (n∑k=1

ap∗

k

)p

A−pn ≤ Kpap∗

n .

So thatm∑n=1

(A−1n

n∑k=1

ap∗

k

)p

≤ Kp

m∑n=1

ap∗

n .

The above inequality is a part (ii) of Theorem 3.1.1 (a) with p = q then A is bounded onlp.

Corollary 3.2.2. Let p be the fixed and 1 < p <∞. If

an

∞∑k=n

A−pk = O(A1−pn ),

then A is bounded on lp.

Proof. By the hypothesis we have

an

∞∑k=n

A−pk < KA1−pn .

So that

ap∗

n

(∞∑k=n

A−pk

)p∗

< Kp∗A−pn .

Thus,∞∑n=m

(an

∞∑k=n

A−pk

)p∗

< Kp∗∞∑n=m

A−pn .

By the part (iv) ⇒ (i) of Theorem 3.1.1 (a) with p = q we point out A is bounded onlp.

After that, we give a counterexample to point out the inverse direction of Corollary3.2.1 does not hold that means A is bounded but

lim supn→∞

n∑j=1

ap∗

j

a1/(p−1)n An

=∞. (3.2.2)


Let

an =

{1 if n 6= 2k,

2−k if n = 2k.

For 2k ≤ n < 2k+1, we have

An =n∑j=1

aj =k∑r=0

2−r + n− k = 2− 1

2k+ n− k,

n∑j=1

ap∗

j =k∑r=0

2−rp∗

+ n− k =2p

∗

2p∗ − 1

[1− 1

2(k+1)p∗

]+ n− k.

We compute

I =

2k∑j=1

ap∗

j

a1/(p−1)2k

A2k

>2k − k

2−k/(p−1)(2k + 2− k)=

(2k − k)2k/(p−1)

2k + 2− k.

Thus k →∞, we get I →∞. Therefore, we have point out (3.2.2).Next, we will check the boundedness of A. By Corollary 3.2.2 we need only to show that

am

∞∑n=m

A−pn = O(A1−pm ).

We have

An = 2− 1

2k+ n− k ≥ 2− 2

n+ n− log2 n ≥

n

2.

So

A−pn ≤2p

np.

Thus,

Ap−1m am

∞∑n=m

A−pn ≤ mp−1.∞∑n=m

2p

np∼ mp−12pm1−p = 2p.

Therefore, A is bounded.

We continue to give a example to point out the inverse direction of Corollary 3.2.2 isnot true that means A is bounded but

lim supm→∞

am∞∑n=m

A−pn

A1−pm

= +∞. (3.2.3)

Taking

an =

{1 if n 6= 2k,

k if n = 2k.(3.2.4)

For 2k ≤ n < 2k+1, we have

An =2k∑j=1

aj +n∑

j=2k+1

aj =k∑r=0

r + (2k − k) + (n− 2k) =k(k + 1)

2+ n− k,


n∑j=1

ap∗

j =k∑r=0

rp∗ + n− k ≤ kp∗+1 + n− k.

We have

An =k(k + 1)

2+ n− k ≥ (log2 n− 1) log2 n

2+ n− log2 n ≥

n

2,

An =k(k + 1)

2+ n− k ≤ log2 n(log2 n+ 1)

2+ n ≤ 2n.

We estimate

Ap−12k

a2k

∞∑n=2k

A−pn ≥(

2k

2

)p−1k∞∑

n=2k

1

(2n)p

∼ k

22p−12k(p−1)2k(1−p)

=k

22p−1 .

Hence, we get (3.2.3).To check that the boundedness of A, we use Corollary 3.2.1. Thus, we need only to showthat

n∑j=1

ap∗

j = O(a1/(p−1)n An).

We have

n∑j=1

ap∗

j

a1/(p−1)n An

≤ kp∗+1 + n− k

k(k + 1)/2 + n− k

≤ (log2 n)p∗+1 + n

n/2

=2(log2 n)p

∗+1

n+ 1/2.

We have

limn→∞

2(log2 n)p∗+1

n= 0.

Thus, there exists C > 0 such that

2(log2 n)p∗+1

n< C.

Therefore,n∑j=1

ap∗

j

a1/(p−1)n An

≤ C +1

2.

So that A is bounded.


Now, we give an other sequence an such that A is bounded but an is not satisfied

am

∞∑n=m

A−pn = O(A1−pm ) and

n∑j=1

ap∗

j = O(a1/(p−1)n An).

Let

an =

1 if n 6= 2k,

2−k if n = 2k, k is odd,

k if n = 2k, k is even.

For 2k ≤ n < 2k+1, we get

An =2k∑j=0

aj + n− 2k.

If k = 2m, we have

A2k = A22m =2m∑r=0

a2r + 22m − 2m

=m−1∑i=0

a22i+1 +m∑i=1

a22i + 22m − 2m

=m−1∑i=0

(2i+ 1) +m∑i=1

2−2i + 22m − 2m

= m(m− 1) +m+1

3

(1− 1

4m

)+ 22m − 2m

= m2 + 22m +1

3

(1− 1

4m

)− 2m.

Similarly, if k = 2m+ 1, we also have

A2k =m∑i=0

(2i+ 1) +m∑i=1

2−2i + 22m+1 − 2m− 1

= m(m+ 1) +m+1

3

(1− 1

4m

)+ 22m+1 − 2m− 1

= m2 + 22m+1 +1

3

(1− 1

4m

)− 1.

Thus,

An =

{m2 + 1

3

(1− 1

4m

)− 2m+ n if n = 22m,

m2 + 13

(1− 1

4m

)− 1 + n if n = 22m+1.

(3.2.5)

We can see that2n > An ≥ A2k > 2k >

n

2.


If k is even, we have

Ap−122ma22m∞∑

n=22m

A−pn ≥ (22m − 2m)p−12m∞∑

n=22m

(1

2n

)p≥ (22m − 2m)p−12m

22m(1−p)

2p

≥ m

2p−1

[1− 2m

22m

].

So the left hand side tends to infinite where m tends to infinite.If k is odd , we have

22m+1∑j=1

ap∗

j

a1/(p−1)22m+1 A22m+1

≥ 22m+12(2m+1)/(p−1)

2(m2 + 22m+1 + 13)

=2(2m+1)p∗

2(m2 + 22m+1 + 13)

We see that the left hand side is also tends to infinite as m to infinite. We check theboundedness of A

l∑n=1

(A−1n

n∑j=1

ap∗

j

)p

≤ Kl∑

n=1

ap∗

n .

For 2s ≤ l < 2s+1, we havel∑

n=1

ap∗

n ≥2s∑n=1

ap∗

n ≥ 2s − s. (3.2.6)

On other hand,

n∑j=1

ap∗

j ≤2k+1−1∑j=1

ap∗

j ≤k+1∑r=0

rp∗

+ 2k+1 − k ≤ (k + 1)p∗+1 + 2k+1.

Then

l∑n=1

(A−1n

n∑j=1

ap∗

j

)p

≤2s+1−1∑n=1

(A−1n

n∑j=1

ap∗

j

)p

=s∑

k=0

∑2k≤n<2k+1

(A−1n

n∑j=1

ap∗

j

)p

≤s∑

k=0

2k(

(k + 1)p∗+1 + 2k+1

2k

)p.

Since

limk→∞

(k + 1)p∗+1

2k= 0,

then there exists K > 0 such that

(k + 1)p∗+1 + 1

2k< K.


Hence,l∑

n=1

(A−1n

n∑j=1

ap∗

j

)p

≤s∑

k=0

2k (K + 2)p = (K + 2)p2s+1. (3.2.7)

From (3.2.6) and (3.2.7), we get

l∑n=1

(A−1n

n∑j=1

ap∗

j

)p

l∑n=1

ap∗n

≤ (K + 2)p2s+1

2s − s≤ 4(K + 2)p.

We deduce A is bounded.

Corollary 3.2.3. Let p be the fixed ,1 < p <∞. If A is bounded on lp, then

∞∑k=n

A−pk = O(nA−pn ). (3.2.8)

Proof. Since A is bounded on lp so that we apply Theorem 3.1.1 (a): (i) ⇒ (iii) withp = q we obtain (

∞∑k=n

A−pk

)1/p( n∑m=1

ap∗

m

)1/p∗

≤ K.

That means∞∑k=n

A−pk

(n∑

m=1

ap∗

m

)p/p∗

≤ Kp. (3.2.9)

Using the Holder’s inequality we also have

n∑m=1

am ≤ n1/p

(n∑

m=1

ap∗

k

)1/p∗

. (3.2.10)

By the inequalities (3.2.9) and (3.2.10) we get

1

n

∞∑k=n

A−pk

(n∑

m=1

am

)p

≤ Kp.

We deduce∞∑k=n

A−pk ≤ KpnA−pn .

The question : Is A bounded on lp when inequality (3.2.8) holds ? We give the exampleto point out the answer of the above question is no. We choose

ak =

{k when m ∈ Z+ for k = 2m,



For 2k ≤ n < 2k+1 and let a0 = 1 then

An = 1 +n∑j=1

aj = 1 +k∑l=0

2l = 2k+1,

n∑j=0

ap∗

i = 1 +k∑l=0

2lp∗

= 1 +2(k+1)p∗ − 1

2p∗ − 1.

We calculate∞∑j=n

A−pj =2k+1−1∑j=n

A−pj +∞∑

j=2k+1

A−pj

=2k+1−1∑j=n

2−(k+1)p +∞∑

l=k+1

∑2l≤k<2l+1

A−pk

= 2−(k+1)p(2k+1 − n

)+

∞∑l=k+1

∑2l≤k<2l+1

2−(l+1)p

= 2−(k+1)p(2k+1 − n

)+

∞∑l=k+1

2l2−(l+1)p

= 2−(k+1)p(2k+1 − n

)+

2−p2(k+1)(1−p)

1− 21−p

= 2−(k+1)p(2k+1 − n

)+

2(k+1)(1−p)

2p − 2

=

[(2k+1 − n)(2p − 2) + 2k+1

]2−(k+1)p

2p − 2.

We have∞∑j=n

A−pj

nA−pn=

(2k+1 − n)(2p − 2) + 2k+1

n(2p − 2)<

2p

2p − 2.

We need only to check that

supn

(∞∑j=n

A−pj

)1/p( n∑j=0

ap∗

j

)1/p∗

=∞.

(∞∑j=n

A−pj

)1/p( n∑j=0

ap∗

j

)1/p∗

=

([(2k+1 − n)(2p − 2) + 2k+1

]2−(k+1)p

2p − 2

)1/p(1 +

2(k+1)p∗ − 2p∗

2p∗ − 1

)1/p∗

>

[(2k+1 − n)(2p − 2) + 2k+1

]1/p2−(k+1)

2

2k

2

>n1/p

8.

We have shown the inverse of Corollary 3.2.3 is not true.


Proposition 3.2.1. If A is bounded on lp for some 1 ≤ p <∞, then

n∑k=1

Akk

= O(An).

Proof. We fix n and define x ∈ lp by

xk =

{a1/pk for k = 1, ..., n,

0 otherwise .

We use the Holder’s inequality

Ak ≤ k1/(p+1)

(k∑j=1

a1+1/pj

)p/(p+1)

.

We see that A is bounded on lp, then

n∑k=1

Akk

=n∑k=1

A−pkAp+1k

k

≤∞∑k=1

A−pk

(k∑j=1

a1+1/pj

)p

= ‖Ax‖pp≤ ‖A‖pp,pAn.

Chapter 4

Littlewood’s problem

4.1 Littlewood’s problem

In this section we study a class of inequalities formulated by Littlewood in connection withsome work on the general theory of orthogonal series. This simplest (non-trivial) examplesare the inequality

∞∑n=1

a3n

n∑m=1

a2mAm ≤ K∞∑n=1

a4nA2n (4.1.1)

and a companion result

∞∑n=1

anA2n

(∞∑m=n

a3/2m

)2

≤ K∞∑n=1

a2nA4n. (4.1.2)

We recall that a′s are arbitrary non-negative numbers with partial sums An = a1 + · · ·+anand K is an absolute constant. Now we find the constant K of (4.1.1). The special case of(4.1.1), in which a′s are decreasing is worth considering separately for then a particularlysimple and natural proof is available.

We suppose that we have shown that the inequality (4.1.1) is true for decreasingsequence. We assume that a = (a1, . . . , ai, ai+1, . . .) is new sequence that is not decreasingthat means there exists i ∈ N such that ai < ai+1. We rearrange a new sequence a′ =(a1, . . . , ai+1, ai, . . .). We calculate

∞∑n=1

a3n

n∑m=1

a2mAm −∞∑n=1

a3n

n∑m=1

a′2mA′m

= a3i

i∑m=1

a2mAm + a3i+1

i+1∑m=1

a2mAm − a3i+1

i∑m=1

a′2mA′m − a3i

i+1∑m=1

a′2mA′m

= a5iAi + a3i+1a2iAi + a5i+1Ai+1 − a5i+1(Ai + ai+1 − ai)− a3i a2i+1(Ai + ai+1 − ai)− a5iAi+1

= (a5i − a5i+1)(Ai − Ai+1) + a2i+1a2iAi(ai+1 − ai)− a3i a2i+1(ai+1 − ai)− a5i+1(ai+1 − ai)

= (ai+1 − ai)[ai+1(a

4i+1 + a3i+1ai + a2i+1a

2i + ai+1a

3i + a4i ) + a2i+1a

2iAi − a5i+1 − a3i a2i+1

]= (ai+1 − ai)

[a4i+1ai + a3i+1a

2i + ai+1a

4i + a2i+1a

2iAi]

> 0.

51

CHAPTER 4. LITTLEWOOD’S PROBLEM 52

We also have

∞∑n=1

a4nA2n −

∞∑n=1

a′4nA′2n

= a4iA2i + a4i+iA

2i+1 − a4i+1(Ai + ai−1 − ai)2 − a4iA2

i+1

= (a4i+1 − a4i )(A2i+1 − A2

i )− 2a4i+1Ai(ai+1 − ai)− a4i+1(ai+1 − ai)2

= (ai+1 − ai)[(a3i+1 + a2i+1ai + ai+1a

2i + a3i )ai+1(2Ai + ai+1)− 2a4i+1Ai − a4i+1(ai+1 − ai)

]= (ai+1 − ai)

[(a2i+1ai + ai+1a

2i + a3i )ai+1(2Ai + ai+1) + a4i+1ai

]= (ai+1 − ai)

[2a4i+1ai + a3i+1a

2i + 2ai+1a

3iAi + 2a2i+1a

2iAi + 2a3i+1aiAi + a2i+1a

3i

]> 0.

We see that

∞∑n=1

a3n

n∑m=1

a2mAm −∞∑n=1

a3n

n∑m=1

a′2mA′m <

∞∑n=1

a4nA2n −

∞∑n=1

a′4nA′2n.

Thus, after each steps rearranges the non-negative sequence a to give the decreasing se-quence, the left hand side of (4.1.1) is less some times than right hand side of (4.1.1).Therefore, the inequality (4.1.1) holds.

Now, we will find a constant K of inequality (4.1.1) with a decreasing sequence. Wemay assume that a1 > 0. Indeed, If a1 = 0, inequality (4.1.1) becomes

∞∑n=2

a3n

n∑m=1

a2mAm ≤ K∞∑n=2

a4nA2n,

is weakly the inequality (4.1.1). The left hand side of (4.1.1) may be rewritten as

L =∞∑n=1

a2nAnanAn

n∑m=1

a2mAm.

Since the a’s is decreasing, we haveanAn≤ 1

n.

Thus,

L ≤∞∑n=1

a2nAn1

n

n∑m=1

a2mAm.

Applying the Holder’s inequality, we get

∞∑n=1

a2nAn1

n

n∑m=1

a2mAm ≤

(∞∑n=1

a4nA2n

)1/2 ∞∑n=1

(1

n

n∑m=1

a2mAm

)21/2

.

Using the Hardy’s inequality, we also have

∞∑n=1

(1

n

n∑m=1

a2mAm

)2

≤ 2∞∑n=1

a4nA2n.


Therefore

L ≤ 2∞∑n=1

a4nA2n.

So that (4.1.1) holds with K = 2.To handle the unrestried version of (4.1.1) a substitute for Hardy’s inequality is called

for, and this is what led to study weighted mean and factorable matrices. To illustratethese ideas we now prove a very general of version (4.1.1).

Theorem 4.1.1. Let p, q, r ≥ 1. If (an)∞n=1 is a sequence of non-negative numbers withpartial sum An = a1 + · · ·+ an, then

∞∑n=1

apnAqn

(∞∑m=n

a1+p/qm

)r

≤(p(q + r)− q

p

)r ∞∑n=1

(apnAqn)1+r/q . (4.1.3)

Proof. We also assume that a1 > 0 so An > 0. Let λn = anAq/pn , yn = a

p/qn An.

We have

Λn = λ1 + · · ·+ λn

= a1Aq/p1 + · · ·+ anA

q/pn

≤ a1Aq/pn + · · ·+ anA

q/pn

= A1+q/pn .

Then

1 ≤ A1+q/pm

Λm

.

Therefore,

a1+p/qm ≤ a1+p/qm A

1+q/pm

Λm

=λmym

Λ. (4.1.4)

Letθ =

r

p(q + r)− q.

It is easy to see that 0 ≤ θ < 1. Using (4.1.4), we estimate left hand side of (4.1.3) weobtain

LHS ≤∞∑n=1

λpn

(∞∑m=n

λmymΛm

)r

.

Applying the Holder’s inequality, we have

∞∑n=1

λp(1−θ)n λpθn

(∞∑m=n

λmymΛm

)r

≤

(∞∑n=1

λp(1−θ)/(1−pθ)n

)1−pθ ∞∑n=1

λn

(∞∑m=n

λmymΛm

)r/pθpθ .(4.1.5)

We see that

1− θ1− pθ

=p(q + r)− q − rp(q + r)− q − pr

=(p− 1)(q + r)

q(p− 1)= 1 +

r

p.


So thatλp(1−θ)/(1−pθ)n = (apnA

qn)1+r/p . (4.1.6)

Using Corollary 2.2.4 with c = θ, p = r/(pθ), xm = ym/Λm we have ∞∑n=1

λn

(∞∑m=n

λmymΛm

)r/(pθ)pθ ≤ [(

r

(pθ)

)r/pθ ∞∑n=1

λnΛr/(pθ)n

(ynΛn

)r/(pθ)]pθ

=

[r

pθ

]r( ∞∑n=1

λnyr/(pθ)m

)pθ

=

[p(q + r)− q

p

]r( ∞∑n=1

anAq/pn

(ap/qn An

)q+r−q/p)pθ

=

[p(q + r)− q

p

]r( ∞∑n=1

ap(q+r)/qn Aq+rn

)pθ

.

Thus, ∞∑n=1

λn

(∞∑m=n

λmymΛm

)r/(pθ)pθ ≤ [p(q + r)− q

p

]r( ∞∑n=1

(apnAqn)1+r/q

)pθ

. (4.1.7)

From (4.1.5),(4.1.6) and (4.1.7), we obtain

LHS ≤[p(q + r)− q

p

]r ∞∑n=1

(apnAqn)1+r/q .

We complete Theorem (4.1.1).

Inequalities (4.1.1) and (4.1.2) are also special cases. For p = 2, q = 1, r = 1 and byinterchanging the order of summation on the left we get the inequality (4.1.1) and forq = r = 2, p = 1 we have the inequality (4.1.2).

4.2 Reverse Littlewood’s inequality

Does there exist a constant K such that

∞∑n=1

a4nA2n ≤ K

∞∑n=1

a3n

n∑m=1

a2mAm ? (4.2.1)

Unfortunately, the inequality (4.2.1) is false. This can be seen by taking

an =

0 when n > 2N .

1 when n = 2r for r = 1, . . . , N

ε otherwise .

(4.2.2)


For 2r ≤ n < 2r+1 so that

An =r∑s=1

a2s +∑

1≤m<nm 6=2s

am = r + (n− r)ε.

We compute

LHS =N∑r=1

a42rA22r +

∑1≤n<2N

n6=2r

a4nA2n

>N∑r=1

[r + (2r − r)ε]2

> ε2N∑r=1

22r

> ε222N = N4 with taking ε =N2

2N.

We also have

RHS =∞∑n=1

a3n

n∑m=1

a2mAm

=N∑r=1

a32r

2r∑m=1

a2mAm +∑

1≤n<2N

n6=2r

a3n

n∑m=1

a2nAm

=N∑r=1

2r∑m=1

a2mAm + ε3∑

1≤n<2N

n 6=2r

n∑m=1

a2mAm

<N∑r=1

r∑s=1

a22sA2s +N∑r=1

∑1≤n<2r

n6=2s

a2nAm + ε32N2N∑m=1

a2mAm

<

N∑r=1

r∑s=1

[s+ (2s − s)ε] + ε2N∑r=1

2rA2r + ε322N[N + 2Nε

]< (1− ε)

N∑r=1

r(r + 1)

2+ ε

N∑r=1

(2r+1 − 2) + ε2N∑r=1

2r(r + 2rε) + ε322N[N + 2Nε

]< N3 + εN2N+1 + ε2N2N(N + 2Nε) + ε322N(N + 2Nε)

< 6N3 with ε =N2

2N.

Thus, LHS > RHS with choosing above an. We have pointed out there dose not exist Kinequality (4.2.2). Now, we consider for general case

∞∑n=1

(apnAqn)1+r/p ≤ K

∞∑n=1

apnAqn

(∞∑m=n

a1+p/qm

)r

. (4.2.3)


We also take

an =

0 if n > 2N ,

1 if n = 2r for r = 1, 2, .., N,

ε otherwise .

We estimate the left hand side

LHS =2N∑n=1

(apnAqn)1+r/q

=N∑s=1

ap2sAq2s +

∑1≤n<2N

n6=2s

(apnAqn)1+r/q

>

N∑s=1

[s+ (2s − s)ε]q+r

>N∑s=1

[(1− ε)q+rsq+r + 2s(q+r)εq+r

]= (1− ε)q+r

N∑s=1

sq+r + εq+rN∑s=1

2s(q+r)

> (1− ε)q+rN q+r + εq+r2N(q+r)

> N q+r +Nα(q+r) with ε =Nα

2N.

We also have

RHS =2N∑n=1

apnAqn

2N∑m=n

a1+p/qm

r

=N∑s=1

ap2sAq2s

2N∑m=2s

a1+p/qm

r

+∑

1≤n<2N

n 6=2s

apnAqn

2N∑m=n

a1+p/qm

r

= I1 + εpI2.


We continue to compute I1.

I1 =N∑s=1

[s+ (2s − s)ε]q[N + 1 + (2N − 2s)ε1+p/q

]r<

N∑s=1

(s+ 2sε)q(N + 1 + 2Nε1+p/q

)r<

N∑s=1

2q (sq + 2sqεq) 2r[(N + 1)r + 2Nrεr(1+p/q)

]< 2q+r

[(2N)r + 2Nrεr(1+p/q)

] N∑s=1

(sq + 2sqεq)

= 2q+r[(2N)r + 2Nrεr(1+p/q)

]N

(N∑s=1

N q + εqN∑s=1

2sq

)

< 2q+r[(2N)r + 2Nrεr(1+p/q)

]N

(N q+1 + εq

2(N+1)q − 2q

2q − 1

)< 2q+r

[(2N)r + 2Nrεr(1+p/q)

]N

(N q+1 + εq

2(N+1)q

2q − 1

)= 2q+rN

[2rN r + 2Nr

(Nα

2N

)r(1+p/q)] [N q+1 +

2qNαq

2q − 1

]with ε =

Nα

2N

< 2q+rN

[2rN r + 2Nr

(Nα

2N

)r(1+p/q)]2q+1Nαq

2q − 1

<22q+2r+2

2q − 1Nαq+r.

We also have

I2 =∑

1≤n<2N

n6=2s

Aqn

2N∑m=n

a1+p/qm

r

=N−1∑s=0

∑2s<n<2s+1

Aqn

2N∑m=n

a1+p/qm

r

<N−1∑s=0

2sAq2s+1

2N∑m=1

a1+p/qm

r

<N−1∑s=0

2s[s+ 1 + 2s+1ε

]q [N + 2Nε1+p/q

]r<

N−1∑s=0

2s2q[(s+ 1)q + 2q(s+1)εq

]2r[N r + 2Nrεr(1+p/q)

].


I2 < 2q+r[N r + 2Nrεr(1+p/q)

] [N−1∑s=0

2s(s+ 1)q + εqN−1∑s=0

2s2q(s+1)

]< 2q+r

[N r + 2Nrεr(1+p/q)

] (2NN q + εq2N(q+1)

)< 2q+r

[N r + 2Nr

(Nα

2N

)r(1+p/q)] (2NN q + 2NNαq

)with ε =

Nα

2N

< 2q+r2N r2(2NNαq)

< 2q+r+12NNαq+r.

If p > 1, r > 1, then

εpI2 = 2q+r+1Nαp+αq+r

2N(p−1) → 0 as N →∞.

Therefore,

RHS <22q+2r+2

2q − 1Nαq+r.

Thus, we need to choose α > 1.If p > 1, r = 1, then

LHS > N q+1 +Nα(q+1),

RHS <22q+4

2q − 1Nαq+1.

Hence, we need to choose α > 1.If p = 1, r > 1, then

εpI2 = 2q+r+1Nα+αq+r.

AndLHS > N q+r +Nα(q+r),

RHS <22q+2r+2

2q − 1Nαq+r + 2q+r+1Nα+αq+r

Thus, we choose

α >r

r − 1.

If r = 1, p = 1, then

LHS =∞∑n=1

(anAqn)1+1/q =

∞∑m=1

a1+1/qm Aq+1

m .

And

RHS =∞∑n=1

anAqn

∞∑m=n

a1+1/qm =

∞∑m=1

a1+1/qm

m∑n=1

anAqn.

So we need only to compare Aq+1m and

m∑n=1

anAqn. By Power Rule Lemma, we always have

m∑n=1

anAqn ≤ Aq+1

m ≤ qm∑n=1

anAqn.

Thus we always find a constant K in inequality (4.2.3) with q = 1, r = 1. Summary, wehave not found K to the Littlewood’s inequality reserve direction in general case. And inthe last part, we will give the reverse Littlewood’s inequality with a decreasing sequence.


Lemma 4.2.1. Let s, t ≥ 0. If a is an N−tuple of non-negative number with partial sumsAn = a1 + · · ·+ an, then

N∑n=1

anAsn

(N∑

m=n

am

)t

≥ β(s+ 1, t+ 1)As+t+1N (4.2.4)

where

β(s+ 1, t+ 1) =

∫ 1

0

ys(1− y)tdy.

Proof. Let A0 = 0. We have An−1 ≤ x ≤ An so that

AN − x ≤ AN − An−1 =N∑

m=n

am.

Thus, ∫ An

An−1

xs(AN − x)tdx ≤ Asn

(N∑

m=n

am

)t ∫ An

An−1

dx = anAsn

(N∑

m=n

am

)t

.

Therefore,

n∑n=1

anAsn

(N∑

m=n

am

)t

≥∫ An

0

xs(AN − x)tdx

=

∫ 1

0

(yAN)s(AN − yAn)td(yAN)

= As+t+1N

∫ 1

0

ys(1− y)tdy.

We are done.

Theorem 4.2.1. Let p, q, r ≥ 1. If (an)∞n=1 is a decreasing sequence of non-negativenumber with partial sum An = a1 + · · ·+ an, then

∞∑n=1

(apnAqn)1+r/q ≤ 1

β(q + 1, r)

∞∑n=1

apnAqn

(∞∑m=n

a1+p/qm

)r

. (4.2.5)

Proof. Applying Lemma 4.2.1 with s = q, t = r − 1 we obtain for m = 1, 2, ..

β(q + 1, r)Aq+rm ≤m∑n=1

anAqn

(m∑j=n

aj

)r−1

.

So that

β(q + 1, r)ap+pr/qm Aq+rm ≤ ap+pr/qm

m∑n=1

anAqn

(m∑j=n

aj

)r−1

.


Since a’s are decreasing sequence, then

ap+pr/qm

m∑n=1

anAqn

(m∑j=n

aj

)r−1

= a1+p/qm ap−1m a(r−1)p/qm

m∑n=1

anAqn

(m∑j=n

aj

)r−1

≤ a1+p/qm ap−1n a(r−1)p/qj

m∑n=1

anAqn

(m∑j=n

aj

)r−1

= a1+p/qm

m∑n=1

apnAqn

(m∑j=n

a1+p/qj

)r−1

Therefore,

β(q + 1, r)ap+pr/qm Aq+rm ≤ a1+p/qm

m∑n=1

apnAqn

(m∑j=n

a1+p/qj

)r−1

.

Summing on m, we get

β(q + 1, r)∞∑m=1

(apmAqm)1+r/q ≤

∞∑m=1

a1+p/qm

m∑n=1

apnAqn

(m∑j=n

a1+p/qj

)r−1

=∞∑n=1

apnAqn

∞∑m=n

a1+p/qm

(m∑j=n

a1+p/qj

)r−1

≤∞∑n=1

apnAqn

(∞∑m=n

a1+p/qm

)r

.

We are done.

Conclusion

In this thesis, we have shown that extensive Hardy’s inequality II in case r > s ≥ 1, butin other cases 0 < r ≤ s and 0 < s < 1, r > s it is not true. In Chapter 1, we have notknown the best constants of two Hardy’s extensive inequalities.

In Chapter 2 and 3, we have applied extensive Harydy’s inequality II to give some con-ditions such that the factorable matrix and weighted mean matrix are bounded operatorsfrom lp into lq. In addition, many Corollaries are involved the factorable matrices and theweighted mean matrices. Besides, we have given an special factor matrix in section 2.3and have found the conditions to it is bounded operator. Moreover, we have pointed outthe norm of it for p = 1, q = ∞, but we have not computed the norm of this factorablematrix case p > 1, q = ∞ or 1 q. In case p = 1 and 1 ≤ q < ∞, wedo not find the condition of α and β to the factorable matrix in section 2.3 is boundedoperator. Next, for the weighted mean matrix, we have determined the norm of it in casep = 1, q = ∞ and in case p > 1, q = ∞. In other cases, we have not computed the normof the weighted mean matrix. In last section of Chapter 3, we gave some counterexamplesto point out the inverse of some Corollaries does not hold.

Finally, in Chapter 4 we consider the Littlewood’s problem and found a constant Kbut it is not the best constant of Littlewood’s problem. We have proved the general caseof Littlewood’s inequality and we give account example for reverse Littlewood’s inequality.We also have shown that the reverse Littewood’s inequality for the decreasing sequence.Can we add other conditions to the reverse Littlewood’s inequality ?

61

Bibliography

[1] Bennett, G., Some elementary inequalities, Quart. J. Math. Oxford Ser. (2) 38 (1987),401-425.

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

[3] Gao, P., On a result of Cartlidge, Department of Mathematics, University of Michigan,Ann Arbor. (2003) .

[4] Hardy, G. H., Littlewood, J. E., Polya, G. Inequalities. Cambridge Univ. Press, (1965).

[5] Sylvain, D., A short proof of Pitt’s compactness Theorem, Proceeding of the AmericanMathematical Society . Volume 137, No. 4, ( 2009), 1371-1372.

62

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