SOME ELEMENTARY INEQUALITIES OF G. BENNETT · CHAPTER 1. SOME INEQUALITIES 6 Lemma 1.1.1 (Power...
Transcript of 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 < p < 1 (and a1 > 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.
CHAPTER 1. SOME INEQUALITIES 7
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 < p < 1 the inequalities (1.1.6) and (1.1.7) are reversed. We completethe Lemma Power Rule for Tails.
1.2 Hardy’s extension inequalities
Theorem 1.2.1 (Hardy’s extension inequality I). Let p > 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)
CHAPTER 1. SOME INEQUALITIES 8
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.
CHAPTER 1. SOME INEQUALITIES 9
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),
CHAPTER 1. SOME INEQUALITIES 10
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.
CHAPTER 1. SOME INEQUALITIES 11
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 .
CHAPTER 1. SOME INEQUALITIES 12
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.
CHAPTER 1. SOME INEQUALITIES 13
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.
CHAPTER 1. SOME INEQUALITIES 14
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.
CHAPTER 1. SOME INEQUALITIES 15
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.
CHAPTER 1. SOME INEQUALITIES 16
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∗
.
CHAPTER 2. FACTORABLE MATRICES 19
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.
CHAPTER 2. FACTORABLE MATRICES 20
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.
CHAPTER 2. FACTORABLE MATRICES 21
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)
CHAPTER 2. FACTORABLE MATRICES 22
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,
CHAPTER 2. FACTORABLE MATRICES 23
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.
CHAPTER 2. FACTORABLE MATRICES 24
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)
The constant is best possible.
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).
CHAPTER 2. FACTORABLE MATRICES 25
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)
The constant is best possible.
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)
The constant is best possible.
Corollary 2.2.7. If 1 < p ≤ q <∞ and c > 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 ).
CHAPTER 2. FACTORABLE MATRICES 26
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 < p ≤ q <∞),
(iv) α >1
qand α + β >
1
q+
1
p∗(p > q).
Moreover, ‖A‖1,∞ = 1. And in the case 1 < p <∞, q =∞, α = 0, β > 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.
CHAPTER 2. FACTORABLE MATRICES 27
(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.
CHAPTER 2. FACTORABLE MATRICES 28
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 < p <∞limn→∞
lnn
(lnn)1/p=∞.
Therefore, A is not bounded from lp in l∞.
(⇐) We will prove that if α > 0 and α + β ≥ 1/p∗, A is bounded from lp in l∞ with1 < p <∞.
We consider
|(Ax(n))m| =
∣∣∣∣∣m−αm∑k=1
k−βxk
∣∣∣∣∣ ≤m∑k=1
k−(α+β)|xk|
≤
(m∑k=1
k−(α+β)p∗
)1/p∗ ( m∑k=1
|xk|p)1/p
≤ K‖x‖p.
Then A is bounded.
CHAPTER 2. FACTORABLE MATRICES 29
We continue to prove that if α = 0 and β > 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 < p ≤ q <∞.(⇒) We choose x = (1, 0, . . .), then
‖Ax‖qq =∞∑m=1
m−αq
(m∑k=1
k−βxk
)q
=∞∑m=1
m−αq.
Since Ax ∈ lq then∞∑m=1
m−αq < +∞.
Hence, αq > 1. We assume that
α + β <1
q+
1
p∗. (2.3.6)
CHAPTER 2. FACTORABLE MATRICES 30
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
.
CHAPTER 2. FACTORABLE MATRICES 31
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 < p <∞.(⇒) Similarly part (c), we also take x = (1, 0, . . .) and we obtain α > 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.
CHAPTER 2. FACTORABLE MATRICES 32
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,
0 otherwise .(3.1.1)
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 35
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 36
(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.
CHAPTER 3. WEIGHTED MEAN MATRICES 37
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 38
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 39
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 40
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 < p <∞. Let A be given by (3.1.1)and suppose that an > 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.
CHAPTER 3. WEIGHTED MEAN MATRICES 41
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)
CHAPTER 3. WEIGHTED MEAN MATRICES 42
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 43
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)
CHAPTER 3. WEIGHTED MEAN MATRICES 44
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,
CHAPTER 3. WEIGHTED MEAN MATRICES 45
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 46
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 47
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 48
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,
0 otherwise .(3.2.11)
CHAPTER 3. WEIGHTED MEAN MATRICES 49
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.
CHAPTER 3. WEIGHTED MEAN MATRICES 50
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.
CHAPTER 4. LITTLEWOOD’S PROBLEM 53
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.
CHAPTER 4. LITTLEWOOD’S PROBLEM 54
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)
CHAPTER 4. LITTLEWOOD’S PROBLEM 55
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)
CHAPTER 4. LITTLEWOOD’S PROBLEM 56
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.
CHAPTER 4. LITTLEWOOD’S PROBLEM 57
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)
].
CHAPTER 4. LITTLEWOOD’S PROBLEM 58
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.
CHAPTER 4. LITTLEWOOD’S PROBLEM 59
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
.
CHAPTER 4. LITTLEWOOD’S PROBLEM 60
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 < p ≤ q < ∞ or p > 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.
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