# On New Extensions of Hilbert's Integral Inequality

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

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## Abstract

It is shown that some new extensions of Hilbert's integral inequality with parameterλ(λ>1/2) can be established by introducing a proper weight function. In particular, when λ=1, a refinement of Hilbert's integral inequality is obtained. As applications, some new extensions of Widder's inequality and Hardy-Littlewood's inequality are given.


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## Body

## 1. Introduction and Lemmas

Letf(x),g(x)∈L2(0,+∞).
It is well known that the inequality of the form
(1.1)∬α∞f(x)g(y)x+y−2αdxdy≤π{∫α∞f2(x)dx}1/2{∫α∞g2(x)dx}1/2
is called Hilbert’s integral inequality, where the coefficient π is
the best possible.In [1], by introducing a parameter λ (λ>1/2), the following extension of (1.1) of the form
(1.2)∬0∞f(x)g(y)xλ+yλdxdy≤(πλsin(π/2λ)){∫0∞x1−λf2(x)dx}1/2{∫0∞x1−λg2(x)dx}1/2
was established.Recently, various improvements and extensions of (1.1) appear in a great deal of papers
(see [2]). The aim of this paper is to
give some new improvements of (1.1) and (1.2) and then present some important
applications.We now introduce some notations
that will be used throughout the paper.Letx−α≥0,y−α≥0,λ>1/2, and let  c(x−α) be an integrable function in [α,+∞). Define E(x,y)=1−c(x−α)+c(y−α) such that E(x,y)≥0 for (x,y)∈(α,+∞)×(α,+∞).
We also define
(1.3)J1=∬α∞f2(x)(x−α)λ+(y−α)λ(x−αy−α)1/2E(x,y)dxdy,J2=∬α∞f2(y)(x−α)λ+(y−α)λ(y−αx−α)1/2E(x,y)dxdy.Lemma 1.1.
With the above-mentioned assumptions, one has(1.4)J1J2=(πλsin(π/2λ))2{(∫α∞(x−α)1−λf2(x)dx)2−(∫α∞k(x)f2(x)dx)2},
where the weight function k(x) is defined by
(1.5)k(x)=(x−α)1−λ{(λsin(π/2λ)π)∫0∞c((x−α)t)t−1/21+tλdt−c(x−α)}.Proof.
By using the substitutiont=(y−α)/(x−α),
it is easy to deduce that
(1.6)J1=∫α∞{∫α∞1(x−α)λ(1+((y−α)/(x−α))λ)(x−αy−α)1/2E(x,y)dy}f2(x)dx=∫α∞{∫0∞11+tλ(1t)1/2(1−c(x−α)+c((x−α)t))dt}(x−α)1−λf2(x)dx=∫α∞{πλsin(π/2λ)+∫0∞c((x−α)t)1+tλ(1t)1/2dt−(πλsin(π/2λ))c(x−α)}(x−α)1−λf2(x)dx=(πλsin(π/2λ)){∫α∞(x−α)1−λf2(x)dx+∫α∞k(x)f2(x)dx},
where k(x) is a function defined by (1.5).
Similarly, we have(1.7)J2=(πλsin(π/2λ)){∫α∞(x−α)1−λf2(x)dx−∫α∞k(x)f2(x)dx}.
From the above equations involving J1 and J2,
(1.4) holds true.Lemma 1.2.
Letx−α≥0 and λ>1/2. Then
(1.8)∫0∞t−1/2(1+tλ)(1+(x−α)λtλ)dt={(πλsin(π/2λ))(x−α)λ−1/2−1(x−α)λ−1,x−α≠1,(πλsin(π/2λ))(1−12λ),x−α=1.Proof.
The casex−α≠1 was studied in [3], or can be obtained by using [4].
Next, consider the case x−α=1.
By the definition and properties of beta function, it is easy to deduce
that
(1.9)∫0∞t−1/2(1+tλ)2dt=1λB(12λ,2−12λ)=1λ(1−12λ)B(12λ,1−12λ)=(1−12λ)πλsin(π/2λ).


## 2. Theorem and Its Corollary

Theorem 2.1.
Letf(x) and g(x) be two real functions such
that 0<∫α∞(x−α)1−λf2(x)dx<+∞ and 0<∫α∞(x−α)1−λg2(x)dx<+∞,
where λ>1/2. Then
(2.1)(∬α∞f(x)g(y)(x−α)λ+(y−α)λdxdy)4≤(πλsin(π/2λ))4{(∫α∞(x−α)1−λf2(x)dx)2−(∫α∞ωλ(x)f2(x)dx)2}×{(∫α∞(x−α)1−λg2(x)dx)2−(∫α∞ωλ(x)g2(x)dx)2},
where the weight function ωλ(x) is defined by
(2.2)ωλ(x)={(x−α)1−λ{(x−α)λ−1/2−1(x−α)λ−1−11+(x−α)λ},x−α≠1,12−12λ,x−α=1.Proof.
First, assumef=g.
Let F(x,y)=f(x)f(y)/((x−α)λ+(y−α)λ), E(x,y)=1−c(x−α)+c(y−α).
Then the following holds:(2.3)∬α∞F(x,y)dxdy=∬α∞F(x,y)E(x,y)dxdy.
In fact, it is obvious that
(2.4)∬α∞F(x,y)E(x,y)dxdy=∬α∞F(x,y)dxdy−∬α∞F(x,y)c(x−α)dxdy+∬α∞F(x,y)c(y−α)dxdy.
We need only to show that
(2.5)∬α∞F(x,y)c(x−α)dxdy=∬α∞F(x,y)c(y−α)dxdy.
Let φ(x)=∫α∞f(t)/((x−α)λ+(t−α)λ)dt.
Then
(2.6)∬α∞F(x,y)c(x−α)dxdy=∫α∞(∫α∞f(y)(x−α)λ+(y−α)λdy)f(x)c(x−α)dx=∫α∞(∫α∞f(t)(x−α)λ+(t−α)λdt)f(x)c(x−α)dx=∫α∞φ(x)f(x)c(x−α)dx=∫α∞φ(y)f(y)c(y−α)dy=∫α∞(∫α∞f(t)(y−α)λ+(t−α)λdt)f(y)c(y−α)dy=∫α∞(∫α∞f(x)(y−α)λ+(x−α)λdx)f(y)c(y−α)dy=∬α∞F(x,y)c(y−α)dxdy.
Noting that E(x,y)≥0 and applying Schwarz’s inequality, we
have
(2.7)(∬α∞F(x,y)dxdy)2=(∬α∞F(x,y)E(x,y)dxdy)2=(∬α∞{f(x)((x−α)λ+(y−α)λ)1/2(x−αy−α)1/4(E(x,y))1/2}×{f(y)((x−α)λ+(y−α)λ)1/2(y−αx−α)1/4(E(x,y))1/2}dxdy)2≤∬α∞{f2(x)(x−α)λ+(y−α)λ(x−αy−α)1/2E(x,y)}dxdy×∬α∞{f2(y)(x−α)λ+(y−α)λ(y−αx−α)1/2E(x,y)}dxdy=J1J2.
It follows from (1.4) that
(2.8)(∬α∞F(x,y)dxdy)2≤(πλsin(π/2λ))2{(∫α∞(x−α)1−λf2(x)dx)2−(∫α∞k(x)f2(x)dx)2},
where the weight function k(x) is defined by (1.5).
Letc(x)=1/(1+xλ), where x≥α and λ>1/2. It is obvious that E(x,y)≥0.
By Lemma
1.2, it is easy to deduce that
(2.9)k(x)=(x−α)1−λ{(λsin(π/2λ)π)∫0∞c((x−α)t)t−1/21+tλdt−c(x−α)}=(x−α)1−λ{(λsin(π/2λ)π)∫0∞t−1/2(1+tλ)(1+(x−α)λtλ)dt−c(x−α)}=ωλ(x).
Substitute k(x)=ωλ(x) into (2.8) to obtain
(2.10)(∬α∞F(x,y)dxdy)2≤(πλsin(π/2λ))2{(∫α∞(x−α)1−λf2(x)dx)2−(∫α∞ωλ(x)f2(x)dx)2}.
Next, consider the casef≠g.
By Schwarz’s inequality, we have
(2.11)(∬α∞f(x)g(y)(x−α)λ+(y−α)λdxdy)4={(∫01(∫α∞t(x−α)λ−1/2f(x)dx∫α∞t(y−α)λ−1/2g(y)dy)dt)2}2≤{∫01(∫α∞t(x−α)λ−1/2f(x)dx)2dt}2{∫01(∫α∞t(y−α)λ−1/2g(y)dy)2dt}2={∬α∞f(x)f(y)(x−α)λ+(y−α)λdxdy}2{∬α∞g(x)g(y)(x−α)λ+(y−α)λdxdy}2.
Based on (2.10), it follows from (2.11) that the inequality
(2.1) is valid at once. Theorem is proved.The special caseλ=1 in
Theorem 2.1 yields the following Hilbert’s integral inequality.Corollary 2.2.
If0<∫α∞f2(x)dx<+∞ and 0<∫α∞g2(x)dx<+∞,
then
(2.12)(∬α∞f(x)g(y)x+y−2αdxdy)4≤π4{(∫α∞f2(x)dx)2−(∫α∞ω1(x)f2(x)dx)2}×{(∫α∞g2(x)dx)2−(∫α∞ω1(x)g2(x)dx)2},
where the weight function ω1(x) is defined by
(2.13)ω1(x)=1x−α+1−1x−α+1.Proof.
It follows directly from the proof of Theorem2.1 and
so the details are omitted.Remark 2.3.
By settingf=g in
Corollary 2.2, (2.12) yields
(2.14)(∬α∞f(x)f(y)x+y−2αdxdy)2≤π2{(∫α∞f2(x)dx)2−(∫α∞ω1(x)f2(x)dx)2},
where the weight function ω1(x) is defined by (2.13).Remark 2.4.
For the caseλ=2 in Theorem 2.1, (2.1) becomes
(2.15)(∬α∞f(x)g(y)(x−α)2+(y−α)2dxdy)4≤π44{(∫α∞(x−α)−1f2(x)dx)2−(∫α∞ω2(x)f2(x)dx)2}×{(∫α∞(x−α)−1g2(x)dx)2−(∫α∞ω2(x)g2(x)dx)2},
where the weight function ω2(x) is defined by
(2.16)ω2(x)={(x−α)−1{(x−α)3/2−1(x−α)2−1−11+(x−α)2},x−α≠1,14,x−α=1.


## 3. Some Applications

As applications, we will give some extensions and
refinements of Widder’s inequality and Hardy-Littlewood’s inequality.Letan≥0(n=0,1,2,…),A(x)=∑n=0∞anxn,A*(x)=∑n=0∞anxn/n!. Then
(3.1)∫01A2(x)dx≤π∫0∞(e−xA*(x))2dx.
Inequality (3.1) is called Widder’s inequality (see [5]).We will give an extension of (3.1) below.Theorem 3.1.
Under the above assumptions, iff(x)=e−(x−α)A*(x−α),
then
(3.2)(∫01A2(x)dx)2≤π2{(∫α∞f2(x)dx)2−(∫α∞ω1(x)f2(x)dx)2},
where ω1(x) is defined by (2.13).Proof.
First, observe that the following holds:(3.3)∫0∞e−tA∗(tx)dt=∫0∞e−t∑n=0∞an(xt)nn!dt=∑n=0∞anxnn!∫0∞tne−tdt=∑n=0∞anxn=A(x). Let tx=s−α.
Then we have
(3.4)∫01A2(x)dx=∫01{∫0∞e−tA*(tx)dt}2dx=∫01(∫α∞e−(s−α)/xA∗(s−α)ds)21x2dx=∫1∞(∫α∞e−(s−α)yA*(s−α)ds)2dy=∫0∞(∫α∞e−(s−α)u−(s−α)A*(s−α)ds)2du=∫0∞(∫α∞e−(s−α)uf(s)ds)2du=∬α∞f(s)f(t)s+t−2αdsdt,
where f(x)=e−(x−α)A∗(x−α). Using
Remark 2.3
, the inequality (3.2) follows from (3.4) at once.
In particular, whenα=0,
we obtain a refinement of (3.1).Corollary 3.2.
With the assumptions as Theorem3.1,
if f(x)=e−xA*(x),
then
(3.5)(∫01A2(x)dx)2≤π2{(∫0∞f2(x)dx)2−(∫0∞ω1(x)f2(x)dx)2},
where ω1(x) is defined by (2.13).Letf(x)∈L2(0,1). If an=∫01xnf(x)dx, n=0,1,2,…,
then we have Hardy-Littlewood’s inequality (see [6])
of the form
(3.6)∑n=0∞an2≤π∫01f2(x)dx,
where π is the best constant that keeps (3.6) valid.
In [7], the inequality (3.6) was extended to the following
inequality:
(3.7)∫0∞f2(x)dx≤π∫01h2(x)dx,
where f(x)=∫01txh(x)dx, x∈[0,+∞).The inequality (3.7) is called Hardy-Littlewood’s integral
inequality. Afterwards, the inequality (3.7) was refined into the
following form (see [8]):
(3.8)∫0∞f2(x)dx≤π∫01th2(t)dt.We will give a new extension of (3.8) here.Theorem 3.3.
Letλ>1/2, h(t)∈L2(0,1), and h(t)≠0.
Define a function by f(x)=∫01t(x−α)λ|h(t)|dt. If 0<∫α+∞(x−α)1−λf2(x)dx<+∞,
then
(3.9)(∫α∞f2(x)dx)4≤(πλsin(π/2λ))2{(∫α∞(x−α)1−λf2(x)dx)2−(∫α∞ωλ(x)f2(x)dx)2}(∫01th2(t)dt)2,
where the weight function   ωλ(x) is defined by (2.2).Proof.
By writingf2(x) in the form
(3.10)f2(x)=∫01f(x)t(x−α)λ|h(t)|dt
and applying Schwarz’s inequality, we obtain
(3.11)(∫α+∞f2(x)dx)4={∫α∞(∫01f(x)t(x−α)λ|h(t)|dt)dx}4={∫01(∫α+∞f(x)t(x−α)λ−1/2dx)t1/2|h(t)|dt}4≤{∫01(∫α+∞f(x)t(x−α)λ−1/2dx)2dt∫01th2(t)dt}2={∫01(∬α+∞f(x)f(y)t(x−α)λ+(y−α)λ−1dxdy)dt∫01th2(t)dt}2=(∬α+∞f(x)f(y)(x−α)λ+(y−α)λdxdy)2(∫01th2(t)dt)2.
By (2.10), the inequality (3.9) follows from (3.11) at once.Remark 3.4.
By settingλ=1 and α=0,
we obtain the following refinement of (3.8):
(3.12)(∫0∞f2(x)dx)4≤π2{(∫0∞f2(x)dx)2−(∫0∞ω1(x)f2(x)dx)2}(∫01th2(t)dt)2,
where the weight function ω1(x) is defined by (2.13).



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