# On Hilbert's Inequality for Double Series and  Its Applications

Authors: Zhou Yu; Gao Mingzhe
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

---

## Abstract

This study shows that a refinement of the Hilbert inequality for double series can be established by introducing a real functionu(x) and a parameter λ. In particular, some sharp results of the classical Hilbert inequality are obtained by means of a sharpening of the Cauchy inequality. As applications, some refinements of both the Fejer-Riesz inequality and Hardy inequality in Hp function are given.


---

## Body

## 1. Introduction

Let{an} and {bn} be two sequences of complex numbers. If λ=0,1, then
(1.1)|∑m=1−λ∞∑n=1−λ∞amb¯nm+n+λ|≤π(∑n=1−λ∞|an|2)1/2(∑n=1−λ∞|bn|2)1/2,(1.2)|∑m=1−λ∞∑n=1−λ∞m≠namb¯nm−n|≤π(∑m=1−λ∞|an|2)1/2(∑n=1−λ∞|bn|2)1/2,
where the constant factor π is the best possible. It is all known that the inequalities (1.1) and (1.2) are called the Hilbert theorem for double series.
The two forms (1.1) and (1.2) on the Hilbert inequality were combined
into one similar form in some  papers (e.g., [1, 2], etc.),  that is,
(1.3)|∑m=1−λ∞∑n=1−λ∞ambnm+n+λ|2+|∑m=1−λ∞∑n=1−λ∞m≠nambnm−n|2≤π2∑n=1−λ∞|an|2∑n=1−λ∞|bn|2.
Recently, the various extensions on (1.1) appeared in some papers (e.g., [3, 4], etc.).
They focalize on changing the denominator of the function of the left-hand side of (1.1). Such as the denominator (m+n+λ) is replaced by (αm+βn)μ in [3], and the denominator (m+n+λ) is replaced by (mu(m)+nv(n))μ in [4]. Some new results in these papers were yielded. The inequality (1.3) is obviously a significant refinement of (1.1) and (1.2). However, both extensions and refinements on (1.2) and (1.3) do not commonly appear in previous papers. The main
purpose of the present paper is to establish both an extension and a
significant refinement on (1.3). Explicitly, let u(x)>0x∈[0,+∞) be a real function and let limx→∞u(x)=+∞.
If the denominator (m+n+λ) of the first
term of the  left-hand side of (1.3) is
replaced by u(m)+u(n), and the denominator  (m−n) of the second term of the left-hand side of (1.3) is replaced by u(m)−u(n),
then a new inequality established is significant in theory and applications; and as applications, we will give both extensions and refinements on
Fejer-Riesz’s inequality and Hardy’s inequality. For convenience, we introduce some notations and functions as follows:
(1.4)V(a,b)=∑m=0∞∑n=0∞u(m)≠v(n)ambn¯u(m)−u(n),Uk(a,b)=∑m=0∞∑n=0∞ambn¯(u(m)+u(n))k,k=1,2,∥x∥2=∑n=0∞|xn|2,Tk(x)=∑n=0∞|xn|uk(n),k=1,2,(f,g)=∫02πf(t)g(t)¯dt,∥α∥2=∫02π|α|2dt,whereα=f,g,h.
In particular, when b=a¯,
the notations Uk(a,a) and V(a,a) are denoted, respectively, by Uk(a) and V(a). Throughout this paper, we will frequently use these notations, and we stipulate that Z denotes integer set and that u(n)=Zn+λ/2, where Zn∈Z, n∈N0,λ is an integer or 0<λ<1.


## 2. Lemmas

In order to prove our assertions, we need the following  lemmas.Lemma 2.1.
If both∑n=0∞an and ∑n=0∞bn are absolute convergent, then
(i)
∑m=0∞∑n=0∞ambn is absolute convergent,(ii)
∥a∥2 and ∥b∥2 are convergent.The proof of it has been given in the paper [2]. Hence, it is omitted here.Lemma 2.2.
Letf,g,h∈L2[0,2π],
and let h be a variable unit-vector. Then,
(2.1)|(f,g)|2≤∥f∥2∥g∥2(1−(|(f,h)|∥f∥−|(g,h)|∥g∥)2).
In particular, when h is orthogonal to f, we have
(2.2)|(f,g)|2≤∥f∥2∥g∥2(1−(|(g,h)|∥g∥)2);
and the equality in (2.2) holds if and only if f,g and h are linearly dependent.The proof of these results has been given in [5, 6].Define a function by(2.3)r(x)=πU1(x)sin2λπ−U2(x)sin2λπ.Lemma 2.3.
Let{an} and {bn} be two sequences of complex numbers. If λ is an integer or 1/2≤λ<1, then
(2.4)r(a)r(b)≥0.Proof.
Whenλ is an integer, it is clear that r(a)r(b)=0. So we consider only the case for 1/2≤λ<1. It is easy to deduce that
(2.5)U1(x)=∫01|∑m=0∞xmtu(m)−1/2|2dt>0,U2(x)=∫01dss∫0s|∑n=0∞xntu(n)−1/2|2dt>0.
When 1/2≤λ<1, it follows from (2.3) that r(x)<0 for any x∈C. Hence, we have r(a)r(b)>0.Lemma 2.4.
Letf(z)=∑n=0∞anzu(n)−1/2. If f(z) is analytic in the unit-disc |z|<1, then
(2.6)|12π∫−ππt|f(−eit)|2dt|=|V(a)|.Proof.
(2.7)|∫−ππt|f(−eit)|2dt|=|∫−ππtf(−eit)f(−eit)¯dt|=|∫−ππt∑m=0∞∑n=0∞aman¯(cos(π+t)+isin(π+t))u(m)−1/2(cos(π−t)+isin(π−t))u(n)−1/2dt|=2π|(∑m=0∞∑n=0∞u(m)≠u(n)aman¯u(m)−u(n))sinλπ+i(∑m=0∞∑n=0∞u(m)≠u(n)aman¯u(m)−u(n))cosλπ|=2π|V(a)|.
Thereby, the relation (2.6) holds.


## 3. Theorems and Their Corollaries

In order to prove our assertions, we need also to introduce the following functions:(3.1)s1(x)=2|T1(x)cosλπ−(1/2π)T2(x)sinλπ|(π2∥x∥2−r(x))1/2,s2(x)=2|T1(x)sinλπ−(1/π)T2(x)sin2(λπ/2)|(π2∥x∥2+r(x))1/2.Theorem 3.1.
Letr(x) be a function defined by (2.3), let {an} and {bn} be two nonzero sequences of complex numbers,
and let both ∑n=0∞an and ∑n=0∞bn be absolute convergent. Then,(i) ifλ is an integer, then
(3.2)|U1(a,b)|2+|V(a,b)|2≤(π2∥a∥2∥b∥2)(1−R2),
where
(3.3)R2={1π2{(T1(a)∥a∥−T2(b)π∥b∥)2+(T1(b)∥b∥−T2(a)π∥a∥)2}ifλisodd,1π2((T1(a)∥a∥)2+(T1(b)∥b∥)2)ifλiseven;(ii) if0<λ<1, then
(3.4)|U1(a,b)cos2λπ−12πU2(a,b)sin2λπ|2+|V(a,b)|2≤{π2∥a∥2∥b∥2−1π2r(a)r(b)}(1−R2),
where R2=min{(s1(a)−s2(b))2,(s1(b)−s2(a))2}, si(x)(i=1,2) is defined by (3.1). In particular, when 1/2≤λ<1, we have r(a)r(b)>0.Proof.
Define two functions by(3.5)f(a,t)=∑m=0∞amtsin(u(m))t,g(b,t)=∑n=0∞bntcos(u(n))t,t∈[0,2π].
Since both ∑n=0∞an and ∑n=0∞bn are absolute convergent by Lemma 2.1, the double series ∑m=0∞∑n=0∞ambn is absolute convergent. Accordingly, f(a,t)g(b,t) is uniformly convergent in the interval [0,2π].
Thereby, the interchange in order of summation and integration can be made. In what follows, we stipulate that the interchanges in order of summation and integration are justified. It is easy to deduce that
(3.6)∥f∥2=π2∥a∥2−r(a),∥g∥2=π2∥b∥2+r(b),|(f,g)|=|∫02πf(a,t)g(b,t)¯dt|=π|V(a,b)+(U1(a,b)cos2λπ−12πU2(a,b)sin2λπ)|,
where r(x) is a function defined by (2.3). By Lemma 2.2, we have
(3.7)|V(a,b)+(U1(a,b)cos2λπ−12πU2(a,b)sin2λπ)|2=1π2|(f,g)|2≤1π2∥f∥2∥g∥2(1−r)=1π2{π2∥a∥2−r(a)}{π2∥b∥2+r(b)}(1−r),
where r=(|(f,h)|/∥f∥−|(g,h)|/∥g∥)2,  h is a variable unit-vector, it can be properly chosen
in accordance with our requirement.(i) Whenλ is an integer, it is known from (2.3) that r(x)=0.We selecth1=2t/2π， it is easy to deduce that ∥h1∥=1,  and
(3.8)|(f,h1)|=|2(∑m=0∞amu(m))|. Since the series ∑n=0∞an is
absolute convergent, it is justified that the complex number an is replaced by |an| in (3.8). Hence, we have
(3.9)|(f,h1)|=2T1(a). Similarly
(3.10)|(g,h1)|={2πT2(b)ifλisodd,0ifλiseven.We therefore obtain that(3.11)r1=(|(f,h1)|∥f∥−|(g,h1)|∥g∥)2={2π2(T1(a)∥a∥−T2(b)π∥b∥)2ifλisodd,2π2(T1(a)∥a∥)2ifλiseven.
Hence, the inequality (3.7) can be
reduced to
(3.12)|V(a,b)+U1(a,b)|2≤{π2∥a∥2∥b∥2}(1−r1).Notice thatU1(b,a)=U1(a,b)¯ and V(b,a)=−V(a,b)¯. If we still select the unit-vector
h2=2t/2π, then, interchanging a and b in (3.12), we have
(3.13)|−V(a,b)¯+U1(a,b)¯|2≤(π2∥a∥2∥b∥2)(1−r2),
where r2 is defined by
(3.14)r2=(|(f,h2)|∥f∥−|(g,h2)|∥g∥)2={2π2(T1(b)∥b∥−T2(a)π∥a∥)2ifλisodd,2π2(T1(b)∥b∥)2ifλiseven.Adding (3.12) and (3.13), then the inequality (3.2) follows after simplifications.(ii) When0<λ<1,
we firstly consider h in (3.7). We still select unit-vector h1=2t/2π. It is easy to deduce that
(3.15)|(f,h1)|=2|(∑m=0∞amu(m))cosλπ−12π(∑m=0∞am(u(m))2)sinλπ|,|(g,h1)|=2|(∑n=0∞bnu(n))sinλπ−1π(∑n=0∞bn(u(n))2)sin2λπ2|.
Since the series ∑n=0∞an and ∑n=0∞bn are
absolute convergent, it is justified that the complex numbers an and bn are replaced, respectively, by |an| and |bn| in the above relations. Let s1(x)=|(f,h1)|/∥f∥),s2(x)=|(g,h1)|/∥g∥.  By using (3.1), we find s1(a),  s2(b).
Let R12=(s1(a)−s2(b))2. We obtain from (3.7)
(3.16)|V(a,b)+(U1(a,b)cos2λπ−12πU2(a,b)sin2λπ)|2≤{π2∥a∥2∥b∥2−(∥b∥2r(a)−∥a∥2r(b))−1π2r(a)r(b)}(1−R12).Notice thatU1(b,a)=U1(a,b)¯, U2(b,a)=U2(a,b)¯, and V(b,a)=−V(a,b)¯. If we still select the unit-vector h2=2t/2π, then, interchanging a and b in (3.16), we have
(3.17)|−V(a,b)¯+(U1(a,b)¯cos2λπ−12πU2(a,b)¯sin2λπ)|2≤{π2∥a∥2∥b∥2−(∥a∥2r(b)−∥b∥2r(a))−1π2r(b)r(a)}(1−R22),
where R22=(s1(b)−s2(a))2.
Let R2=min{R12,R22}. Adding (3.16) and (3.17), the inequality (3.4) can be gotten after simplifications. In particular, when 1/2≤λ<1,
by Lemma 2.3, we have r(a)r(b)≥0. The proof of Theorem 3.1 is completed.Corollary 3.2 .
Let∑n=0∞an be absolute convergent. Then,
(i)
ifλ is an integer, then
(3.18)|U1(a)|2+|V(a)|2≤(π2∥a∥4)(1−R2),
where
(3.19)R2={2π2∥a∥2(T1(a)−T2(a)π)2ifλisodd,2π2∥a∥2(T1(a))2ifλiseven;(ii)
if0<λ<1, then
(3.20)|U1(a)cos2λπ−12πU2(a)sin2λπ|2+|V(a)|2≤{π2∥a∥4−1π2r2(a)}(1−R2),
where R2=(s1(a)−s2(a))2,si(x)(i=1,2) is defined by
(3.1). In particular, when u(n)=n+λ/2,
according to (3.2), one obtains a refinement of
(1.3) immediately.Corollary 3.3.
Ifλ=0,1, then
(3.21)|∑m=1−λ∞∑n=1−λ∞ambn¯m+n+λ|2+|∑m=1−λ∞∑n=1−λ∞m≠nambn¯m−n|2≤(π2∑n=1−λ∞|an|2∑n=1−λ∞|bn|2)(1−R2),
where
(3.22)R2={1π2{(T1(a)∥a∥−T2(b)π∥b∥)2+(T1(b)∥b∥−T2(a)π∥a∥)2}ifλ=1,1π2((T1(a)∥a∥)2+(T1(b)∥b∥)2)ifλ=0,
where
(3.23)Tk(x)=∑n=1−λ∞|xn|(n+λ/2)k,k=1,2.Corollary 3.4.
Ifu(n)=n+1/4, then
(3.24)|(∑m=0∞∑n=0∞amb¯nm+n+1/2)|2+|∑m=0∞∑n=0,m≠n∞amb¯nm−n|2≤{π2∥a∥2∥b∥2−1π2r(a)r(b)}(1−R2),
where
(3.25)r(a)r(b)>0,R2=min{(s1(a)−s2(b))2,(s1(b)−s2(a))2},s1(x)=|−(1/2π)T2(x)|(π2∥x∥2−r(x))1/2,s2(x)=2|T1(x)−(1/2π)T2(x)|(π2∥x∥2+r(x))1/2,Tk(x)=∑n=0∞|xn|(n+1/4)k,k=1,2. Since λ=1/2,
it is known from Lemma 2.3 that r(a)r(b)>0.
Ifr(a)r(b) and R in (3.24) are replaced by zero, then the inequality (3.24) can be reduced to
(3.26)|(∑m=0∞∑n=0∞amb¯nm+n+1/2)|2+|∑m=0∞∑n=0,m≠n∞amb¯nm−n|2<π2∥a∥2∥b∥2.
The inequalities (3.24) and (3.26) are refinements of the Hilbert-Ingham inequality
(3.27)|∑m=0∞∑n=0∞amb¯nm+n+1/2|≤π∥a∥∥b∥.One has yet a new inequality according to Theorem3.1(ii).Theorem 3.5.
With the assumptions as Theorem3.1, if λ=1/4, then
(3.28)|12πU2(a,b)|2+|V(a,b)|2≤{π2∥a∥2∥b∥2−1π2r(a)r(b)}(1−R2),
where
(3.29)R2=min{(s1(a)−s2(b))2,(s1(b)−s2(a))2},s1(x)=|T1(x)−(1/2π)T2(x)|(π2∥x∥2−r(x))1/2,s2(x)=|T1(x)−(1/2π)T2(x)(2−1)|(π2∥x∥2+r(x))1/2,Tk(x)=∑n=0∞|xn|(n+1/8)k,k=1,2.


## 4. Applications toHP Function

Letf(z) be analytic in the unit-disc |z|<1.
If f(z)=∑n=0∞anzn∈Hp with p>0,  then
(4.1)∫01|f(t)|pdt≤12∫02π|f(eit)|pdt,
where the coefficient 1/2 is the best possible. It is called the Fejer-Riesz inequality in Hp function [7].We will give both an extension and a refinement of (4.1) in what follows.Theorem 4.1.
Letf(z)=∑n=0∞anzu(n)−1/2∈Hp, where p>0 and u(n)=Zn+(λ/2)(Zn∈Z,λ∈N0).
If f(z) is analytic in the unit-disc |z|<1, then
(4.2)|∫01|f(t)|pdt|2+|12π∫−ππt|f(−eit)|pdt|2≤(12∫02π|f(eit)|pdt)2(1−R2),
where R2>0.Proof.
At first, we prove the theorem for casep=2. Let f(z)=∑m=0∞amzu(m)−1/2.
It is easy to deduce that
(4.3)∫01|f(t)|2dt=∑m=0∞∑n=0∞u(m)+u(n)≠0aman¯u(m)+u(n)=U1(a),12π∫02π|f(eit)|2dt=∥a∥2.
By Lemma 2.4, we have
(4.4)|12π∫−ππt|f(−eit)|2dt|=|∑m=0∞∑n=0∞u(m)≠u(n)aman¯u(m)−u(n)|=|V(a)|.
Since the series ∑n=0∞an is absolutely convergent, it is justified that
the complex number an is replaced by |an|.
According (3.18), we have
(4.5)|∫01|f(t)|2dt|2+|12π∫−ππt|f(−eit)|2dt|2≤(12∫02π|f(eit)|2dt)2(1−R2),
where R2 is defined by (3.19). It is easy to deduce that
(4.6)2π2∥a∥2=4π(∫02π|f(eit)|2dt)−1,T1(a)=∫01t−1/2|f(t)|dt,T2(a)=∫01dss∫0st−1/2|f(t)|dt.
Because u(n)=Zn+λ/2≠1/π, T1(a)−T2(a)/π≠0. It shows that R2>0.
Hence, the inequality (4.2)
is valid when p=2.
If p≠2, then by the Blaschke decomposition theorem, it holds that f(z)=B(z)G(z),
where B(z) is Blaschke function and G(z)≠0,|B(z)|≤1 in  |z|<1 and |B(eit)|=1.
LetF(z)=(G(z))p/2∈H2. According to the above result for p=2, we have
(4.7)∫01|f(t)|pdt|2+|12π∫−ππt|f(−eit)|pdt|2=|∫01|F(t)|2dt|2+|12π∫−ππt|F(−eit)|2dt|2≤{12∫02π|F(eit)|2dt}2(1−RF2)={12∫02π|G(eit)|pdt}2(1−RG2)={12∫02π|f(eit)|pdt}2(1−R2).
Based on the case for p=2,
we have RF2>0.
Hence, R2>0. The proof of Theorem 4.1 is completed.
Let(4.8)f(z)=∑m=0∞cmzm∈H1.
Then,
(4.9)∑n=0∞|cn|n+1≤12∫−ππ|f(eit)|dt.
It is called the Hardy inequality in Hp function [7].We will give both an extension and a refinement of (4.9) as follows.Theorem 4.2.
Letf(z)=∑m=0∞cmzu(m) be analytic in the unit-disc |z|<1,
where u(m)=Zm+λ/2
(with Zm∈Z and λ∈N0) and f∈H1.
Then,
(4.10)(∑n=0∞|cn|u(n))2+|12π∫−ππt|f(−eit)|2dt|2≤(12∫−ππ|f(eit)|dt)2(1−R2),
where
(4.11)R2={4π(∫01t−1|f(t)|dt−1π∫01dss∫0st−1|f(t)|dt)2(∫02π|f(eit)|dt)−1ifλisodd,4π(∫01t−1|f(t)|dt)2(∫02π|f(eit)|dt)−1ifλiseven.Proof.
By Blaschke decomposition theorem, we have(4.12)f(z)=B(z)G(z)=B(z)G1/2(z)G1/2(z)=f1(z)f2(z),
where B(z) is Blaschke function, f1,f2∈H2.
Let f1(z)=B(z)G1/2(z)=∑m=0∞amzu(m),f2(z)=G1/2(z)=∑n=0∞bnzu(n).
It is easy deduce that
(4.13)∥a∥2=∥b∥2=∑m=0∞|am|2=∑n=0∞|bn|2=12π∫−ππ|G1/2(eit)|2dt=12π∫−ππ|B(eit)G1/2(eit)|2dt=12π∫02π|f(eit)|dt.
Owing to f(z)=f1(z)f2(z), it holds that ∫01t−1|f(t)|dt=∫01t−1|f1(t)|2dt.
It is easy to deduce that
(4.14)∑n=0∞|cn|u(n)≤∑n=0∞∑r+s=n|ar||as|u(r)+u(s)=∑m=0∞∑n=0∞|am||an|u(m)+u(n).
By Lemma 2.4, we find that
(4.15)|12π∫−ππt|f(−eit)|dt|=|12π∫−ππt|f1(−eit)|2dt|=|∑m=0∞∑n=0∞u(m)≠u(n)|am||an|u(m)−u(n)|. It follows from (4.13), (4.14), (4.15), and Corollary 3.2 that
(4.16)(∑n=0∞|cn|u(n))2+|12π∫−ππt|f(−eit)|dt|2≤(12∫02π|f(eit)|dt)2(1−R2),
where R2 is defined by (3.19). It is easy to deduce that
(4.17)2π2∥a∥2=4π(∫02π|f(eit)|2dt)−1,T1(a)=∫01t−1|f1(t)|2dt=∫01t−1|f(t)|dt,T2(a)=∫01dss∫0st−1|f1(t)|2dt=∫01dss∫0st−1|f(t)|dt.
These show that the inequality (4.10) is valid.



---

*Source: 165089-2008-04-16.xml*