# A Hilbert Integral-Type Inequality with Parameters

Authors: Shang Xiaozhou; Gao Mingzhe
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

A Hilbert-type integral inequality with parametersα and  (α,λ>0) can be  established by introducing a nonhomogeneous kernel function. And the constant factor is
proved to be the best possible. And then some important and especial results are enumerated.
As applications, some equivalent forms are studied.


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## Body

## 1. Introduction

Letf(x),g(x)∈L2(0,+∞). Then  (1.1)∬0∞f(x)g(y)x+ydxdy≤π{∫0∞f2(x)dx}1/2{∫0∞g2(x)dx}1/2,
where the constant factor π is the best possible, and the equality in (1.1) holds if and only if f(x)=0, or g(x)=0.This is the famous Hilbert integral inequality (see [1, 2]). Owing to the importance of the Hilbert inequality in analysis and applications, some mathematicians have been studying them. Recently, various improvements and extensions of (1.1) appear in a great deal of papers (see [3–11], etc.). Specially, Gao and Hsu enumerated the research articles to more than 40 in the paper [6]. It is obvious that the integral kernel function of the left hand side of (1.1) is a homogeneous form of degree -1. In general, the Hilbert type integral inequality with a homogeneous kernel of degree -1 has been studied in the paper [1]. The purpose of the present paper is to establish a Hilbert integral type inequality with a non-homogeneous kernel. And the constant factor is proved to be the best possible. And then some important and especial results are enumerated and some equivalent forms are studied.For convenience, we need to introduce the Catalan constant and define a real function.The Catalan constant is defined by(1.2)G=12∫01Kdk=∑n=0∞(-1)n(2n+1)2,
where K is the complete elliptic integral, namely, (1.3)K=K(k)=∫0π/2dθ1-k2sin2θ.
And the approximation of G is that (1.4)G=0,915965594⋯.
These results can be found in the paper [12, page 503].Letλ>0. Define a real function by(1.5)S(λ)=∑n=0∞(-1)n1(2n+1)λ+1.In particular,S(1)=G, where G is the Catalan constant.In order to prove our main results, we need the following lemmas.Lemma 1.1.
Leta be a positive number and b>-1. Then
(1.6)∫0∞xbe-axdx=Γ(b+1)ab+1.Proof.
According to the definition of the gamma function, we obtain immediately (1.6). This result can be also found in the paper [12, page 226, formula (1053)].Lemma 1.2.
Letα,λ>0. Define a function by
(1.7)C(α,λ)=2(2α)λ+1Γ(λ+1)S(λ),
where Γ(z) is the gamma function, and S(λ) is a function defined by (1.5). Then
(1.8)∫0∞k(1,u)uα/2-1du=C(α,λ),
where  k(1,u)=|ln(1/u)|λ/(1+uα).Proof.
It is easy to deduce that(1.9)∫0∞k(1,u)uα/2-1du=∫01(ln(1/u))λ1+uαuα/2-1du+∫1∞(lnu)λ1+uαuα/2-1du=∫01(ln(1/u))λ1+uαu(α/2)-1du+∫01(ln(1/v))λ1+vαu(α/2)-1du=2∫01(ln(1/u))λ1+uαu(α/2)-1du=2∫0∞tλe-(α/2)t1+e-αtdt.
Expanding 1/(1+e-αt) into power series of e-αtand then using Lemma 1.1, we have
(1.10)∫0∞tλe-(α/2)t1+e-αtdt=∫0∞tλe-(α/2)t(1-e-αt+e-2αt+e-3αt-e-4αt+⋯)dt=∫0∞(tλe-(α/2)t-tλe-(3/2)αt+tλe-(5/2)αt-tλe-(7/2)αt+⋯)dt=(2α)λ+1Γ(λ+1)∑n=0∞(-1)n1(2n+1)λ+1=(2α)λ+1Γ(λ+1)S(λ).
It follows from (1.9), (1.10), and (1.7) that the equality (1.8) holds.


## 2. Main Results

In the section we will formulate our main results.Theorem 2.1.
Letfand g be two real functions, and let α and λ be arbitrary two positive numbers. If ∫0∞x1-αf2(x)dx<+∞ and ∫0∞x1-αg2(x)dx<+∞, then
(2.1)∬0∞|lnxy|λf(x)g(y)1+(xy)αdxdy≤C(α,λ){∫0∞x1-αf2(x)dx}1/2{∫0∞x1-αg2(x)dx}1/2,
where C(α,λ) is defined by (1.7). And the constant factor C(α,λ) in (2.1) is the best possible. And the equality in (2.1) holds if and only if f(x)=0, or g(x)=0.Proof.
We may apply Hardy’s technique and Cauchy-Schwarz’s inequality to estimate the left-hand side of (2.1)  as follows:
(2.2)∬0∞|lnxy|λf(x)g(y)1+(xy)αdxdy=∬0∞|lnxy|λ/2f(x)(1+(xy)α)1/2(xy)(2-α)/4|lnxy|λ/2f(x)(1+(xy)α)1/2(yx)(2-α)/4g(y)dxdy≤{∬0∞|lnxy|λ1+(xy)α(xy)(2-α)/2f2(x)dxdy}1/2{∬0∞|lnxy|λ1+(xy)α(yx)(2-α)/2g2(y)dxdy}1/2={∫0∞ω(α,λ,x)f2(x)dx}1/2{∫0∞ω(α,λ,x)g2(x)dx}1/2,
where ω(α,λ,x)=∫0∞(|lnxy|λ/(1+(xy)α))(x/y)(2-α)/2dy.
By substitutingu for xy, it is easy to deduce that
(2.3)ω(α,λ,x)=∫0∞|lnxy|λ1+(xy)α(xy)(2-α)/2dy=x1-α∫0∞|lnu|λ1+uα(1u)(2-α)/2du=x1-α∫0∞|ln(1/u)|λ1+uαu(α/2)-1du.
Based on  (1.8), we obtain
(2.4)ω(α,λ,x)=x1-αC(α,λ).
It is known from (2.2) and (2.4) that the inequality (2.1) is valid.
Iff(x)=0, or g(x)=0, then the equality in (2.1) obviously holds. If f(x)≠0 and g(x)≠0, then
(2.5)0<∫0∞x1-αf2(x)dx<+∞,0<∫0∞x1-αg2(x)dx<+∞.
If (2.2) takes the form of the equality, then there exists a pair of non-zero constants c1 and c2 such that
(2.6)c1|lnxy|λ1+(xy)αf2(x)(xy)1-α/2=c2|lnxy|λ1+(xy)αg2(y)(yx)1-α/2a.e.on(0,+∞)×(0,+∞).
Then we have
(2.7)c1x2-αf2(x)=c2y2-αg2(y)=C0.(constant)a.e.on(0,+∞)×(0,+∞).
Without losing the generality, we suppose that c1≠0; then
(2.8)∫0∞x1-αf2(x)dx=C0c1∫0∞x-1dx.
This contradicts that 0<∫0∞x1-αf2(x)dx<+∞. Hence it is impossible to take the equality in (2.2). It shows that it is also impossible to take the equality in (2.1).
It remains to need only to show thatC(α,λ) in (2.1) is the best possible. Below we will apply Yang’s technique (see [13]) to verify our assertion.
For alln∈N, define two functions by
(2.9)fn(x)={x(α/2)-1+1/2nx∈(0,1),0x∈[1,∞),gn(x)={0x∈(0,1],x(α/2)-1-1/2nx∈(1,∞).
Then we have
(2.10)(∫01x1-αfn2(x)dx)1/2=(∫1∞x1-αgn2(x)dx)1/2=n.
Let0≤k≤C(α,λ) such that the inequality (2.1) is still valid when C(α,λ) is replaced by k. Specially, we have
(2.11)1n∬0∞|lnxy|λfn(x)gn(y)1+(xy)αdxdy≤k(1n){∫0∞x1-αfn2(x)dx}1/2{∫0∞x1-αgn2(x)dx}1/2=k.
Let k(1,xy)=|lnxy|λ/(1+(xy)α). By Fubini’s theorem, we have
(2.12)k≥1n∬0∞k(1,xy)fn(x)gn(y)dxdy=1n∫1∞y(α/2)-1-1/2n(∫01k(1,xy)x(α/2)-1+1/2ndx)dy=1n∫1∞y-1-1/n(∫0yk(1,u)u(α/2)-1+1/2ndu)dy=1n{∫1∞y-1-1/n(∫01k(1,u)u(α/2)-1+1/2ndu)dy+∫1∞y-1-1/n(∫1yk(1,u)u(α/2)-1+1/2ndu)dy}=1n{∫1∞n(∫01k(1,u)u(α/2)-1+1/2ndu)+∫1∞k(1,u)u(α/2)-1+1/2n(∫u∞y-1-1/ndy)du}=∫01k(1,u)u(α/2)-1+1/2ndu+∫1∞k(1,u)u(α/2)-1-1/2ndu.
By Fatou’s lemma and (1.8), we have
(2.13)k≥lim̲n→∞∫01k(1,u)uα/2-1+1/2ndu+lim̲n→∞∫1∞k(1,u)uα/2-1-1/2ndu≥∫01lim̲n→∞k(1,u)uα/2-1+1/2ndu+∫1∞lim̲n→∞k(1,u)uα/2-1-1/2ndu=∫01k(1,u)uα/2-1du+∫1∞k(1,u)uα/2-1du=∫0∞k(1,u)uα/2-1du=C(α,λ).
It follows thatk=C(α,λ) in (2.1) is the best possible. Thus the proof of theorem is completed.Based on Theorem2.1, we may build some important and interesting inequalities.Theorem 2.2.
Letn be a nonnegative integer and α>0. If ∫0∞x1-αf2(x)dx<+∞ and ∫0∞x1-αg2(x)dx<+∞, then
(2.14)∬0∞(lnxy)2nf(x)g(y)1+(xy)αdxdy≤C(α,2n){∫0∞x1-αf2(x)dx}1/2{∫0∞x1-αg2(x)dx}1/2,
where C(α,2n)=(π/α)2n+1En,theEn′s are the Euler numbers, namely, E0=1,E1=1,E2=5,E3=61,E4=1385, and so forth, and the constant factor C(α,2n)in (2.14) is the best possible. And the equality in (2.14) holds if and only if  f(x)=0, or g(x)=0.Proof.
We need only to verify the constant factorC(α,2n) in (2.14). When λ=2n, it is known from (1.10) that
(2.15)C(α,2n)=2(2α)2n+1Γ(2n+1)∑k=0∞(-1)k1(2k+1)2n+1.
It is known from the paper [14] that
(2.16)∑k=0∞(-1)k1(2k+1)2n+1=π2n+122n+2(2n)!En,
where the En′s are the Euler numbers, namely, E0=1,E1=1,E2=5,E3=61,E4=1385, and so forth.
It follows that the constant factorC(α,2n) in (2.14) is correct.In particular, based on Theorem2.2, the following results are gotten.Corollary 2.3.
If∫0∞f2(x)dx<+∞ and ∫0∞g2(x)dx<+∞, then
(2.17)∬0∞f(x)g(y)1+xydxdy≤π{∫0∞f2(x)dx}1/2{∫0∞g2(x)dx}1/2,
where the constant factor π in (2.17) is the best possible. And the equality in (2.17) holds if and only if f(x)=0, or g(x)=0.Corollary 2.4.
If∫0∞f2(x)dx<+∞ and ∫0∞g2(x)dx<+∞, then
(2.18)∬0∞(lnxy)2f(x)g(y)1+xydxdy≤π3{∫0∞f2(x)dx}1/2{∫0∞g2(x)dx}1/2,
where the constant factor π3 in (2.18) is the best possible. And the equality in (2.18) holds if and only if f(x)=0, or g(x)=0.Theorem 2.5.
Letfand g be two real functions, and α a positive number. If ∫0∞x1-αf2(x)dx<+∞ and ∫0∞x1-αg2(x)dx<+∞, then
(2.19)∬0∞|lnxy|f(x)g(y)1+(xy)αdxdy≤8Gα2{∫0∞x1-αf2(x)dx}1/2{∫0∞x1-αg2(x)dx}1/2,
where G is the Catalan constant, that is, G=0,915965594⋯.And the constant factor 8G/α2 in (2.19)  is the best possible. And the equality in (2.19) holds if and only if f(x)=0, or g(x)=0.Proof.
We need only to verify the constant factor in (2.19).  For case λ=1, based on (1.7), (1.5) and (1.2), it is easy to deduce that the constant factor is that C(α,1)=8G/α2.In particular, whenα=λ=1, the following result is obtained．Corollary 2.6.
Letf and g be two real functions. If ∫0∞f2(x)dx<+∞ and ∫0∞g2(x)dx<+∞, then
(2.20)∬0∞|lnxy|f(x)g(y)1+xydxdy≤8G{∫0∞f2(x)dx}1/2{∫0∞g2(x)dx}1/2,
where G is the Catalan constant, that is, G=0,915965594⋯.And the constant factor 8G in (2.20) is the best possible. And the equality in (2.20) holds if and only iff(x)=0, or g(x)=0.A great deal of the Hilbert type inequalities can be established provided that the parameters are properly selected．


## 3. Some Equivalent Forms

As applications, we will build some new inequalities.Theorem 3.1.
Letf be a real function, and let α and λ be arbitrary two positive numbers. If ∫0∞x1-αf2(x)dx<+∞, then
(3.1)∫0∞yα-1{∫0∞|lnxy|λ1+(xy)αf(x)dx}2dy≤(C(α,λ))2∫0∞x1-αf2(x)dx,
where C(α,λ) is defined by (1.7) and the constant factor (C(α,λ))2 in (3.1)  is the best possible.  And the equality in (3.1) holds if and only if f(x)=0. And the inequality (3.1) is equivalent to (2.1).Proof.
First, we assume that the inequality (2.1) is valid.  Define a function g(y) by
(3.2)g(y)=yα-1∫0∞|lnxy|λ1+(xy)αf(x)dx,y∈(0,+∞).
By using (2.1), we have
(3.3)∫0∞yα-1{∫0∞|lnxy|λ1+(xy)αf(x)dx}2dy=∬0∞|lnxy|λ1+(xy)αf(x)g(y)dxdy≤C(α,λ){∫0∞x1-αf2(x)dx}1/2{∫0∞y1-αg2(y)dy}1/2=C(α,λ){∫0∞x1-αf2(x)dx}1/2{∫0∞yα-1(∫0∞|lnxy|λ1+(xy)αf(x)dx)2dy}1/2.
It follows from (3.3) that the inequality (3.1) is valid after some simplifications.
On the other hand, assume that the inequality (3.1) keeps valid; by applying in turn Cauchy’s inequality and (3.1), we have
(3.4)∬0∞|lnxy|λ1+(xy)αf(x)g(y)dxdy=∫0∞y(α-1)/2{∫0∞|lnxy|λ1+(xy)αf(x)dx}y(α-1)/2g(y)dy≤{∫0∞yα-1(∫0∞|lnxy|λ1+(xy)αf(x)dx)2dy}1/2{∫0∞y1-αg2(y)dy}1/2≤{(C(α,λ))2∫0∞x1-αf2(x)dx}1/2{∫0∞y1-αg2(y)dy}1/2=C(α,λ){∫0∞x1-αf2(x)dx}1/2{∫0∞y1-αg2(y)dy}1/2.
Therefore, the inequality (3.1) is equivalent to (2.1).
If the constant factor(C(α,λ))2 in (3.1) is not the best possible, then it is known from (3.4) that the constant factor C(α,λ) in (2.1) is also not the best possible. This is a contradiction. The theorem is proved.Theorem 3.2.
Letn be a nonnegative integer and α>0. If ∫0∞x1-αf2(x)dx<+∞, then
(3.5)∫0∞yα-1{∫0∞(lnxy)2n1+(xy)αf(x)dx}2dy≤(C(α,2n))2∫0∞x1-αf2(x)dx,
where C(α,2n)=(π/α)2n+1En, the En's are the Euler numbers,namely, E0=1,E1=1,E2=5,E3=61,E4=1385, and so forth, and the constant factor C(α,2n) is the best possible. And the equality in (3.5) holds if and only if f(x)=0.  And the inequality (3.5) is equivalent to (2.14).Theorem 3.3.
Letfand g be two real functions, and let α be a positive numbers. If ∫0∞x1-αf2(x)dx<+∞ and ∫0∞x1-αg2(x)dx<+∞, then
(3.6)∫0∞yα-1{∫0∞|lnxy|1+(xy)αf(x)dx}2dy≤8Gα2{∫0∞x1-αf2(x)dx}1/2{∫0∞x1-αg2(x)dx}1/2,
where G is the Catalan constant, that is, G=0,915965594⋯. And the constant factor 8G/α2 in (3.6)  is the best possible. And the equality in (3.6) holds if and only if f(x)=0. And the inequality (3.6) is equivalent to (2.19).The proofs of Theorems3.2 and 3.3 are similar to one of Theorem 3.1; they are omitted here.Similarly, we can establish also some new inequalities which they are, respectively, equivalent to the inequalities (2.17), (2.18) and (2.20). These are omitted here.



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