# The Generalizations of Hilbert's Inequality

Authors: Liubin Hua; Huazhong Ke; Yongjin Li
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

By introducing some parameters and estimating
the weight functions, we establish a new Hilbert-type inequalities with best
constant factors. The equivalent inequalities are also considered.


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## Body

## 1. Introduction

Iff,g are real functions such that 0<∫0∞f2(x)dx<∞ and 0<∫0∞g2(x)dx<∞,
then we have (see [1])(1.1)∫0∞∫0∞f(x)g(y)x+ydxdy<π{∫0∞f2(x)dx∫0∞g2(x)dx}1/2,where the constant factor π is the best possible. Inequality (1.1) is the well-known
Hilbert's inequality. Inequality (1.1) had been generalized by Hardy-Riesz (see [2]) in 1925 as if f,g are real functions such that 0<∫0∞fp(x)dx<∞ and 0<∫0∞gq(x)dx<∞,
then(1.2)∫0∞∫0∞f(x)g(y)x+ydxdy<πsin(π/p){∫0∞fp(x)dx}1/p{∫0∞gq(x)dx}1/q,where the constant factor c=π/sin(π/p) is the best possible. When p=q=2,
(1.2) reduces to (1.1). Inequality (1.2) is named after Hardy-Hilbert's
integral inequality, which is important in the analysis and its applications
(see [3]), it has been studied and generalized in many directions by a number of mathematicians (see [4–8]).Under the same condition of (1.2), we have Hardy-Hilbert's type inequality (see [1, Theorems 341 and 342]):(1.3)∫0∞∫0∞f(x)g(y)max{x,y}dxdy<4{∫0∞f2(x)dx∫0∞g2(x)dx}1/2,∫0∞∫0∞lnx−lnyx−yf(x)g(y)dxdy<π2{∫0∞f2(x)dx∫0∞g2(x)dx}1/2,where the constant factors 4 and π2 are both the best possible.Recently, Li et al. [9], by introducing the function |lnx−lny|/(x+y+|x−y|),
establish new inequalities similar to Hilbert-type inequality for
integrals.Theorem 1.1.
Iff(x),g(x)≥0,0<∫0∞f2(x)dx<∞,0<∫0∞g2(x)dx<∞.
Then, one has(1.4)∫0∞∫0∞|lnx−lny|x+y+|x−y|f(x)g(y)dxdy<4(∫0∞f2(x)dx)1/2(∫0∞g2(x)dx)1/2,where the constant factor 4 is
the best possible.In this paper, we give further analogs of Hilbert-type
inequality and its applications. The main result unifies and generalizes the
classical results as follows.Assume thatr>0,s>−min{1,r},t⩾0,p,q>1,
and 1/p+1/q=1. If f,g⩾0, such that 0<∫0∞fp(x)dx<∞,0<∫0∞gq(x)dx<∞, then(1.5)∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|f(x)g(y)dxdy<K1(∫0∞fp(x)dx)1/p(∫0∞gq(x)dx)1/q,where the constant factor(1.6)K1=∫0∞|lnu|t1+ru+s|1−u|u−1/qduis the best possible.


## 2. Some Lemmas

Our results will be based on the following results. In
the following lemmas, assume thatr>0, s>−min{1,r}, t⩾0,p,q>1,
and 1/p+1/q=1.Lemma 2.1.
Define the
following weight functions:(2.1)ω1(x)=∫0∞|lnx−lny|tx+ry+s|x−y|(xy)1/qdy,x>0,ω2(y)=∫0∞|lnx−lny|tx+ry+s|x−y|(yx)1/pdx,y>0.Then, ω1=ω2=K1, where K1 is defined by (1.1).Proof.
Forω1(x), let u=y/x, and we have(2.2)ω1=∫0∞|lnu|t1+ru+s|1−u|u−1/qdu=K1.
Forω2(y), first let v=x/y, and then let u=1/v. Thus, we have(2.3)ω2=∫0∞|lnv|t1+rv+s|v−1|v−1/pdv=∫0∞|lnu|t1+ru+s|1−u|u−1/qdu=K1.Hence, the lemma is proved.Lemma 2.2.
Assume thatε>0, then(2.4)∫0∞|lnu|t1+ru+s|1−u|u−(1+ε)/qdu=K1+o(1)(ε→0+).Proof.
Since(2.5)limu→0u1/2p|lnu|t=0,limu→+∞u1/2p|lnu|t=+∞,it follows that there exists δ1∈(0,1),
such that if u∈(0,δ1), then u−1/2p>|lnu|t.
Moreover, there exists δ2∈(1,+∞),
such that if u∈(δ2,+∞), then u1/2q>|lnu|t.
Using the expression of K1,
if s⩽0,
then(2.6)|∫0∞|lnu|t1+ru+s|1−u|u−(1+ε)/qdu−K1|=|∫0∞|lnu|t1+ru+s|1−u|(u−(1+ε)/q−u−1/q)du|⩽(∫0δ1+∫δ1δ2+∫δ2∞)|lnu|t1+ru+s|1−u||u−(1+ε)/q−u−1/q|du<∫0δ1|lnu|t1+s(u−(1+ε)/q−u−1/q)du+∫δ1δ2|lnu|t1+ru+s|1−u||u−(1+ε)/q−u−1/q|du+∫δ2∞|lnu|t(r+s)u(u−1/q−u−(1+ε)/q)du<∫0δ1u−1/2p1+s(u−(1+ε)/q−u−1/q)du+∫δ1δ2|lnu|t1+ru+s|1−u||u−(1+ε)/q−u−1/q|du+∫δ2∞u1/2q(r+s)u(u−1/q−u−(1+ε)/q)du=11+s(11−1/2p−(1+ε)/qδ11−1/2p−(1+ε)/q−11−1/2p−1/qδ11−1/2p−1/q)+∫δ1δ2|lnu|t1+ru+s|1−u||u−(1+ε)/q−u−1/q|du+1r+s(2qδ2−1/2q−2q1+2εδ2−(1+2ε)/2q)=o(1)(ε→0+).Therefore, Lemma 2.2 is proved
for s⩽0.
If s>0,
then we can replace the right-hand side of the first strict inequality above
with(2.7)∫0δ1|lnu|t1(u−(1+ε)/q−u−1/q)du+∫δ1δ2|lnu|t1+ru+s|1−u||u−(1+ε)/q−u−1/q|du+∫δ2∞|lnu|tru(u−1/q−u−(1+ε)/q)du.By the same way, we can show
that the lemma is valid for s>0.
Hence, the lemma is proved.Lemma 2.3.
Assume thatε>0, then(2.8)∫1∞x−1−εdx∫0x−1|lnu|t1+ru+s|1−u|u−(1+ε)/qdu=O(1)(ε→0+).Proof.
Usingδ1 introduced in the proof of Lemma 2.2, if s⩽0,
then(2.9)0<∫1∞x−1−εdx∫0x−1|lnu|t1+ru+s|1−u|u−(1+ε)/qdu=∫1∞x−1−εdx∫0x−1|lnu|t1+ru+s(1−u)u−(1+ε)/qdu<∫1∞x−1−εdx∫0x−1|lnu|t1+su−(1+ε)/qdu=(∫11/δ1+∫1/δ1∞)x−1−εdx∫0x−1|lnu|t1+su−(1+ε)/qdu=∫11/δ1x−1−εdx(∫0δ1+∫δ1x−1)|lnu|t1+su−(1+ε)/qdu+∫1/δ1∞x−1−εdx∫0x−1|lnu|t1+su−(1+ε)/qdu=∫11/δ1x−1−εdx∫0δ1|lnu|t1+su−(1+ε)/qdu+∫11/δ1x−1−εdx∫δ1x−1|lnu|t1+su−(1+ε)/qdu+∫1/δ1∞x−1−εdx∫0x−1|lnu|t1+su−(1+ε)/qdu<∫11/δ1x−1−εdx∫0δ1u−1/2p1+su−(1+ε)/qdu+∫11/δ1x−1−εdx∫δ1x−1|lnu|t1+su−(1+ε)/qdu+∫1/δ1∞x−1−εdx∫0x−1u−1/2p1+su−(1+ε)/qdu=(1−δ1)δ11−1/2p−(1+ε)/qε(1+s)(1−1/2p−(1+ε)/q)+∫11/δ1x−1−εdx∫δ1x−1|lnu|t1+su−(1+ε)/qdu+2pδ1(1+2ε)/2p(1+2ε)(1+s)(1−1/2p−(1+ε)/q)=O(1)(ε→0+).Therefore, Lemma 2.3 is proved
for s⩽0.
If s>0,
then we can replace the right-hand side of the second strict inequality
mentioned above with(2.10)∫1∞x−1−εdx∫0x−1|lnu|t1u−(1+ε)/qdu.By the same way, we can show that
the lemma is valid for s>0.
Therefore, the lemma is proved.


## 3. Integral Case

In this section, we will state our main results.Proposition 3.1.
Assume thatr>0,s>−min{1,r},t⩾0,p,q>1,
and 1/p+1/q=1. If f,g⩾0, such that 0<∫0∞fp(x)dx<∞,0<∫0∞gq(x)dx<∞, then(3.1)∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|f(x)g(y)dxdy<K1(∫0∞fp(x)dx)1/p(∫0∞gq(x)dx)1/q,where the constant factor(3.2)K1=∫0∞|lnu|t1+ru+s|1−u|u−1/qduis the best possible.Proof.
Using
Hölder's inequality, we have(3.3)∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|f(x)g(y)dxdy=∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|[(xy)1/pqf(x)][(yx)1/pqg(y)]dxdy⩽(∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|(xy)1/qfp(x)dxdy)1/p×(∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|(yx)1/pgq(y)dxdy)1/q=(∫0∞ω1(x)fp(x)dx)1/p(∫0∞ω2(y)gq(y)dy)1/q=K1(∫0∞fp(x)dx)1/p(∫0∞gq(x)dx)1/q.
If (3.3) takes the form of equality, then there are
constants a and b,
such that they are not all zero, and(3.4)a|lnx−lny|tx+ry+s|x−y|(xy)1/qfp(x)=b|lnx−lny|tx+ry+s|x−y|(yx)1/pgq(y),a.e.  in(0,∞)×(0,∞).Hence,(3.5)axfp(x)=bygq(y),a.e.  in(0,∞)×(0,∞).Therefore, there is a constant c,
such that(3.6)axfp(x)=bygq(y)=c,a.e.  in(0,∞)×(0,∞).
We claim that a=0.
In fact, if a≠0,
then fp(x)=c/ax, a.e., in (0,∞),
which contradicts the fact that 0<∫0∞fp(x)dx<∞.
In the same way, we claim that b=0.
This is a contradiction. Hence, by (3.3), we have (3.1).
If the constant factorK1 in (3.1) is not the best possible factor, then
there exists a positive constant H (with H<K1), such that (3.1) is still valid if we
replace K1 with H.
For ε>0 sufficiently small, construct the following functions:(3.7)fε(x)={0,x∈(0,1),x−(1+ε)/p,x∈(0,+∞),gε(x)={0,x∈(0,1),x−(1+ε)/q,x∈(0,+∞),Thus, we obtain(3.8)H(∫0∞fεp(x)dx)1/p(∫0∞gεq(x)dx)1/q=Hε,I:=∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|fε(x)gε(y)dxdy=∫1∞∫1∞|lnx−lny|tx+ry+s|x−y|x−(1+ε)/py−(1+ε)/qdxdy=∫1∞x−(1+ε)/pdx∫1∞|lnx−lny|tx+ry+s|x−y|y−(1+ε)/qdy.Setting u=y/x,
we have(3.9)I=∫1∞x−1−εdx∫x−1∞|lnu|t1+ru+s|1−u|u−(1+ε)/qdu=∫1∞x−1−εdx∫0∞|lnu|t1+ru+s|1−u|u−(1+ε)/qdu−∫1∞x−1−εdx∫0x−1|lnu|t1+ru+s|1−u|u−(1+ε)/qdu=1ε(K1+o(1))+O(1)(ε→0+).Since(3.10)I<Hε,it follows that(3.11)K1+o(1)<H(ε→0+).Hence,(3.12)K1⩽H.This contradicts the fact that H<K1.
So the constant factor K1 in (3.1) is the best possible.
Then Proposition 3.1 is proved.Proposition 3.2.
Under the same assumptions of Proposition3.1, one has(3.13)∫0∞[∫0∞|lnx−lny|tx+ry+s|x−y|f(x)dx]pdy<K1p∫0∞fp(x)dx,where the constant factor K1p is the best possible. Inequalities
(3.1) and (3.13) are equivalent.Proof.
Setting(3.14)g(y)=[∫0∞|lnx−lny|tx+ry+s|x−y|f(x)dx]p−1,y∈(0,∞),by (3.1), we have(3.15)∫0∞gq(y)dy=∫0∞[∫0∞|lnx−lny|tx+ry+s|x−y|f(x)dx]pdy=∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|f(x)g(y)dxdy<K1(∫0∞fp(x)dx)1/p(∫0∞gq(x)dx)1/q.As a result,(3.16)(∫0∞gq(x)dx)1/p<K1(∫0∞fp(x)dx)1/p.Hence,(3.17)∫0∞gq(x)dx<K1p∫0∞fp(x)dx.Then, we have (3.13).
Conversely, by Hölder's inequality, we
have(3.18)∫0∞∫0∞|lnx−lny|tx+ry+s|x−y|f(x)g(y)dxdy=∫0∞[∫0∞|lnx−lny|tx+ry+s|x−y|f(x)dx]g(y)dy⩽(∫0∞[∫0∞|lnx−lny|tx+ry+s|x−y|f(x)dx]pdy)1/p(∫0∞gq(x)dx)1/q<K1(∫0∞fp(x)dx)1/p(∫0∞gq(x)dx)1/q.Then, by (3.13), we have (3.1).
Hence inequalities (3.1) and (3.13) are equivalent.
If the constant factorK1p in (3.13) is not the best possible, then by
(3.18) we can get a contradiction that the constant factor K1 in (3.1) is not the best possible. Hence Proposition 3.2 is proved.Remark 3.3.
In (3.1), let r=1,s=0,t=0,p=q=2,
and we have Hilbert's integral inequality(3.19)∫0∞∫0∞f(x)g(y)x+ydxdy<π{∫0∞f2(x)dx∫0∞g2(x)dx}1/2.Let r=1,s=1,t=0,p=q=2;
we have Hardy-Hilbert's classical inequality(3.20)∫0∞∫0∞f(x)g(y)2max{x,y}dxdy<2{∫0∞f2(x)dx∫0∞g2(x)dx}1/2.Let r=1,s=1/3,t=0,p=q=2,
and we can combine the above two inequalities as
follows:(3.21)∫0∞∫0∞f(x)g(y)(2/3)(x+y+max{x,y})dxdy<K1{∫0∞f2(x)dx∫0∞g2(x)dx}1/2.Let r=1,s=−1/3,t=0,
and we can get the following inequality:(3.22)∫0∞∫0∞f(x)g(y)(2/3)(x+y+min{x,y})dxdy<K1{∫0∞f2(x)dx∫0∞g2(x)dx}1/2.


## 4. Discrete Case

We also give results for the discrete case.Proposition 4.1.
Assume thatr>0,s∈(−min{1,r},min{1,r}],p,q>1 and 1/p+1/q=1.
If am,bn⩾0,
such that(4.1)0<∑m=1∞amp<∞,0<∑n=1∞bnq<∞,then one has(4.2)∑n=1∞∑m=1∞ambnm+rn+s|m−n|<K1(∑m=1∞amp)1/p(∑n=1∞bnq)1/q,(4.3)∑n=1∞[∑m=1∞amm+rn+s|m−n|]p<K1p∑m=1∞amp,where the constant factor(4.4)K1=∫0∞11+ru+s|1−u|u−1/qduis the best possible, and inequalities (4.2) and (4.3) are
equivalent.Proof.
Define
the following weight functions:(4.5)ϖ1(m)=∑n=1∞1m+rn+s|m−n|(mn)1/q,m=1,2,…,ϖ2(n)=∑m=1∞1m+rn+s|m−n|(nm)1/p,n=1,2,…,then,(4.6)ϖ1(m)<ω1(m)=K1,ϖ2(n)<ω2(n)=K1,where ω1,ω2 are defined in Lemma 2.1. By Hölder's inequality, we have(4.7)∑n=1∞∑m=1∞ambnm+rn+s|m−n|=∑n=1∞∑m=1∞1m+rn+s|m−n|[(mn)1/pqam][(nm)1/pqbn]≤(∑n=1∞∑m=1∞1m+rn+s|m−n|(mn)1/qamp)1/p(∑n=1∞∑m=1∞1m+rn+s|m−n|(nm)1/pbnq)1/q=(∑m=1∞ϖ1(m)amp)1/p(∑n=1∞ϖ2(n)bnq)1/q<K1(∑m=1∞amp)1/p(∑n=1∞bnq)1/q.The last strict inequality holds
because both series {am} and {bn} have positive terms. Thus, we have (4.2).
If the constant factorK1 in (4.2) is not the best possible, then there
exists a positive constant H (with H<K1), such that (4.2) is still valid if we
replace K1 by H.
For ε>0 small enough, construct series(4.8)a˜m=m−(1+ε)/p,b˜n=n−(1+ε)/q.Then, we have(4.9)∑n=1∞∑m=1∞a˜mb˜nm+rn+s|m−n|>∫0∞∫0∞fε(x)gε(y)dxdyx+ry+s|x−y|=1ε(K1+o(1))+O(1),(4.10)(∑m=1∞a˜mp)1/p(∑n=1∞b˜nq)1/q=∑n=1∞n−1−ε<1+∫1∞x−1−εdx=1+1ε,where fε(x) and gε(x) are defined in the proof of
Proposition 3.1. Since(4.11)1ε(K1+o(1))+O(1)<H(1+1ε),it follows that(4.12)K1+o(1)<H.Hence,(4.13)K1⩽H.This contradicts with the fact
that H<K1.
So the constant factor K1 in (4.2) is the best possible one.
Setting(4.14)bn=[∑m=1∞amm+rn+s|m−n|]p−1,n=1,2,…,we find(4.15)∑n=1∞[∑m=1∞amm+rn+s|m−n|]p=∑n=1∞bnq=∑n=1∞∑m=1∞ambnm+rn+s|m−n|.By the same argument used in the
proof of Proposition 3.2, we can show that (4.3)
is valid, the constant factor K1p in (4.3) is the best possible one and
inequalities (4.2) and (4.3) are equivalent.Remark 4.2.
Proposition4.1 is the corresponding series form
of Propositions 3.1 and 3.2 for s∈(−min{1,r},min{1,r}],t=0,
and it is also a generalization of Hilbert's inequality. Here, we restrict the
constants s,t so that we can use the monotony of functions
to obtain (4.6) and (4.9).



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