# Equivalent Property of a Hilbert-Type Integral Inequality Related to the Beta Function in the Whole Plane

Authors: Dongmei Xin; Bicheng Yang; Aizhen Wang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

By means of the technique of real analysis and the weight functions, a few equivalent statements of a Hilbert-type integral inequality with the nonhomogeneous kernel in the whole plane are obtained. The constant factor related to the beta function is proved to be the best possible. As applications, the case of the homogeneous kernel, the operator expressions, and a few corollaries are considered.


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## Body

## 1. Introduction

Suppose thatp>1,1/p+1/q=1,f(x),g(y)≥0, 0<∫0∞fp(x)dx<∞, and 0<∫0∞gq(y)dy<∞. We have the following well-known Hardy-Hilbert’s integral inequality (see [1]):(1)∫0∞∫0∞fxgyx+ydxdy<πsin⁡π/p∫0∞fpxdx1/p∫0∞gqydy1/q,where the constant factor π/sin⁡(π/p) is the best possible. For p=q=2, (1) reduces to the well-known Hilbert’s integral inequality. By using the weight functions, some extensions of (1) were given by [2, 3]. A few Hilbert-type inequalities with the homogenous and nonhomogenous kernels were provided by [4–7]. In 2017, Hong [8] also gave two equivalent statements between Hilbert-type inequalities with the general homogenous kernel and parameters. Some other kinds of Hilbert-type inequalities were obtained by [9–16].In 2007, Yang [17] gave a Hilbert-type integral inequality in the whole plane as follows:(2)∫-∞∞∫-∞∞fxgy1+ex+yλdxdy<Bλ2,λ2∫-∞∞e-λxf2xdx∫-∞∞e-λyg2ydy1/2,with the best possible constant factor B(λ/2,λ/2) (λ>0, B(u,v) is the beta function) (see [18]). He et al. [19–23] proved a few Hilbert-type integral inequalities in the whole plane with the best possible constant factors.In this paper, by means of the technique of real analysis and the weight functions, a few equivalent statements of a Hilbert-type integral inequality with the nonhomogeneous kernel in the whole plane similar to (2) are obtained. The constant factor related to the beta function is proved to be the best possible. As applications, the case of the homogeneous kernel, the operator expressions, and a few corollaries are considered.


## 2. An Example and Two Lemmas

Example 1.
ForR=(-∞,∞),R+=(0,∞), we set h(u)≔(max{u,1})α+β/|u-1|λ+α(min{u,1})β(u∈R+), and then for a,b≠0,(3)heax+by=max⁡eax+by,1α+βeax+by-1λ+αmin⁡eax+by,1βx,y∈R.For σ,μ>β,σ+μ=λ<1-α(α+2β<1), in view of h(v-1)v1-σ=h(v)vμ-1(0<v<1), we find (4)kλσ≔∫0∞huuσ-1du=∫01huuσ-1+uμ-1du=∫01max⁡u,1α+β1-uλ+αmin⁡u,1βuσ-1+uμ-1du=∫0111-uλ+αuσ-β-1+uμ-β-1du=B1-λ-α,σ-β+B1-λ-α,μ-β∈R+,where B(u,v)≔∫01(1-t)u-1tv-1dt(u,v>0) is the beta function (cf. [18]).In particular, (i) forα=0, we have σ,μ>β,σ+μ=λ<1(β<1/2),h1(u)=(max{u,1})β/|u-1|λ(min{u,1})β(u>0), and (5)kλ1σ=B1-λ,σ-β+B1-λ,μ-β;(ii) forβ=0, we have σ,μ>0,σ+μ=λ<1-α(α<1),h2(u)=(max{u,1})α/|u-1|λ+α(u>0), and (6)kλ2σ=B1-λ-α,σ+B1-λ-α,μ;(iii) forβ=-α, we have σ,μ>-α,σ+μ=λ<1-α(α>-1), h3(u)=(min{u,1})α/|u-1|λ+α(u>0), and (7)kλ3σ=B1-λ-α,σ+α+B1-λ-α,μ+α.In the case of (iii), for α=0, we have σ,μ>0,σ+μ=λ<1,h4(u)=1/|u-1|λ(u>0), and (8)kλ4σ=B1-λ,σ+B1-λ,μ.In the following, we assume thatp>1,1/p+1/q=1,a,b≠0,σ1,σ∈R,σ,μ>β, σ+μ=λ<1-α(α+2β<1), and(9)Kλσ≔1a1/qb1/pkλσ=1a1/qb1/p×B1-λ-α,σ-β+B1-λ-α,μ-β.Forn∈N={1,2,…}, we define two sets Ec≔{t∈R;ct≥0},Fc≔R∖Ec={t∈R;ct<0}(c=a,b), and the following two expressions:(10)I1≔∫Fbeσ1+1/qnby∫Eaheax+byeσ-1/pnaxdxdy,(11)I2≔∫Ebeσ1-1/qnby∫Faheax+byeσ+1/pnaxdxdy.Settingu=eax+by in (10), in view of Fubini theorem (cf. [24]), it follows that(12)I1=1a∫Fbeσ1-σ+1/nby∫eby∞uuσ-1/pn-1dudy=1ab∫01vσ1-σ+1/n-1∫v∞huuσ-1/pn-1dudvv=eby.In the same way, we find that(13)I2=1a∫Ebeσ1-σ-1/nby∫0eyhuuσ+1/pn-1dxdy=1ab∫1∞vσ1-σ-1/n-1∫0vhuuσ+1/pn-1dudv.Lemma 2.
If there exists a constantM, such that for any nonnegative measurable functions f(x) and g(y) in R, the following inequality(14)I≔∫-∞∞∫-∞∞heax+byfxgydxdy≤M∫-∞∞fxeσaxpdx1/p∫-∞∞gyeσ1byqdy1/qholds true, then we have σ1=σ.Proof.
(i) Ifσ1<σ, then for n>1/(σ-σ1)(n∈N), we set two functions (15)fnx≔eσ-1/pnax,x∈Ea0,x∈Fa,gny≔0,y∈Ebeσ1+1/qnby,y∈Fb,and obtain (16)J2≔∫-∞∞e-pσaxfnpxdx1/p∫-∞∞e-qσ1bygnqydy1/q=∫Eae-ax/ndx1/p∫Fbeby/ndy1/q=na1/pb1/q.By (12) and (14), we find(17)1ab∫01vσ1-σ+1/n-1dv∫1∞uα+β+σ-1/pn-1u-1λ+αdu≤I2=∫-∞∞∫-∞∞heax+byfnxgnydxdy≤MJ2=Mna1/pb1/q.For any n>1/σ-σ1(n∈N),σ1-σ+1/n<0, it follows that ∫01vσ1-σ+1/n-1dv=∞.In view of ∫1∞(uα+β+σ-1/pn-1/(u-1)λ+α)du>0, by (17), we find that ∞≤Mn/|a|1/p|b|1/q<∞, which is a contradiction.
(ii) Ifσ1>σ, then for n>(1/(σ1-σ))(n∈N), we set functions (18)f~nx≔0,x∈Eaeσ+1/pnax,x∈Fa,g~ny≔eσ1-1/qnby,y∈Eb0,y∈Fb,and find (19)J~2≔∫-∞∞e-pσaxf~npxdx1/p∫-∞∞e-qσ1byg~nqydy1/q=∫Faeax/ndx1/p∫Ebe-by/ndy1/q=na1/pb1/q.By (13) and (14), we obtain(20)1ab∫1∞vσ1-σ-1/n-1dv∫01uσ-β+1/pn-11-uλ+αdu≤I1=∫0∞∫0∞heax+byf~nxg~nydxdy≤MJ~2=Mna1/pb1/q.For n>1/(σ1-σ)(n∈N), σ1-σ-1/n>0, it follows that ∫1∞yσ1-σ-1/n-1dy=∞. By (20), in view of ∫01(uσ-β+1/pn-1/(1-u)λ+α)du>0, we have ∞≤Mn/|a|1/p|b|1/q<∞, which is a contradiction.
Hence, we conclude thatσ1=σ.
The lemma is proved.Forσ1=σ, we have the following.Lemma 3.
If there exists a constantM, such that for any nonnegative measurable functions f(x) and g(y) in R, the following inequality(21)I=∫-∞∞∫-∞∞heax+byfxgydxdy≤M∫-∞∞fxeσaxpdx1/p∫-∞∞gyeσbyqdy1/qholds true, then we have M≥Kλ(σ)(>0).Proof.
By (12), for σ1=σ, we obtain (22)I1=1ab∫01v1/n-1∫v1huuσ-1/pn-1dudv+1ab∫01v1/n-1∫1∞huuσ-1/pn-1dudv=1ab∫01∫0uv1/n-1dvhuuσ-1/pn-1du+nab∫1∞huuσ-1/pn-1du=nab∫01huuσ+1/qn-1du∫1∞huuσ-1/pn-1du.We use inequality I1≤MJ~2 (for σ1=σ) as follows:(23)a1/pb1/qnI1=1a1/qb1/p∫01Huuσ+1/qn-1du+∫1∞Huuσ-1/pn-1du≤M.By Fatou lemma (cf. [24]) and (23), it follows that (24)Kλσ=1a1/qb1/p×∫01limn→∞⁡huuσ+1/qn-1du+∫1∞limn→∞⁡huuσ-1/pn-1du≤lim_n→∞⁡a1/pb1/qnI1≤M.
The lemma is proved.


## 3. Main Results and Some Corollaries

Theorem 4.
IfM is a constant, then the following statements (i), (ii), and (iii) are equivalent:
(i) For any nonnegative measurable functionf(x) in R, we have the following inequality:(25)J≔∫-∞∞epσ1by∫-∞∞heax+byfxdxpdy1/p≤M∫-∞∞fxeσaxpdx1/p.
(ii) For any nonnegative measurable functionsf(x) and g(y) in R, we have the following inequality:(26)I=∫-∞∞∫-∞∞heax+byfxgydxdy≤M∫-∞∞fxeσaxpdx1/p∫-∞∞gyeσ1byqdy1/q.
(iii)σ1=σ, and M≥Kλ(σ)(>0).Proof.
( i ) = > ( i i ). By Hölder’s inequality (see [25]), we have(27)I=∫-∞∞eσ1by∫-∞∞heax+byfxdxe-σ1bygydy≤J∫-∞∞gyeσ1byqdy1/q.Then by (25), we have (26).
( i i ) = > ( i i i ). By Lemma 2, we have σ1=σ. Then by Lemma 3, we have M≥Kλ(σ)(>0).
( i i i ) = > ( i ). Setting u=eax+by, we obtain the following weight functions: for y,x∈R,(28)ωσ,y≔eσby∫-∞∞heax+byeσaxdx=1a∫0∞huuσ-1du=1akλσ,(29)ϖσ,x≔eσax∫-∞∞heax+byeσbydy=1bkλσ.
By Hölder’s inequality with weight and (28), we have(30)
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Forσ1=σ, by Fubini theorem (see [24]) and (29), we have (31)J≤1akλσ1/q∫-∞∞∫-∞∞heax+byeσbyeσaxp/qfpxdxdy1/p=1akλσ1/q∫-∞∞∫-∞∞heax+byeσbyeσaxp/qdyfpxdx1/p=1akλσ1/q∫-∞∞ωσ,xe-pσaxfpxdx1/p=Kλσ∫-∞∞e-pσaxfpxdx1/p.For Kλ(σ)≤M, we have (25).
Therefore, the statements (i), (ii), and (iii) are equivalent.
The theorem is proved.Theorem 5.
The following statements (i) and (ii) are valid and equivalent:
(i) For anyf(x)≥0, satisfying 0<∫-∞∞f(x)/eσaxpdx<∞, we have the following inequality: (32)J1=∫-∞∞epσby∫-∞∞max⁡eax+by,1α+βfxdxeax+by-1λ+αmin⁡eax+by,1βpdy1/p<Kλσ∫-∞∞fxeσaxpdx1/p.
(ii) For anyf(x)≥0, satisfying 0<∫-∞∞f(x)/eσaxpdx<∞ and g(y)≥0, satisfying 0<∫-∞∞g(y)/eσbyqdy<∞, we have the following inequality:(33)I=∫-∞∞∫-∞∞max⁡eax+by,1α+βfxgyeax+by-1λ+αmin⁡eax+by,1βdxdy<Kλσ∫-∞∞fxeσaxpdx1/p∫-∞∞gyeσbyqdy1/q.Moreover, the constant factorKλ(σ) in (32) and (33) is the best possible.In particular, forα=β=0,σ,μ>0,σ+μ=λ<1(34)K~λσ≔1a1/qb1/pB1-λ,σ+ζ1-λ,μ,we have the following equivalent inequalities with the best possible constant factor K~λ(σ):(35)∫-∞∞epσby∫-∞∞fxeax+by-1λdxpdy1/p<K~λσ∫-∞∞fxeσaxpdx1/p,(36)∫-∞∞∫-∞∞fxgyeax+by-1λdxdy<K~λσ∫-∞∞fxeσaxpdx1/p∫-∞∞gyeσbyqdy1/q.Proof.
We first prove that (32) is valid. If (30) takes the form of equality for a y∈R, then (see [25]), there exist constants A and B, such that they are not all zero, and (37)Aeσbyeσaxp/qfpx=Beσaxeσbyq/pa.e.inR.We suppose that A≠0 (otherwise B=A=0). Then it follows that (38)fxeσaxp=e-qσbyBAa.e.inR,which contradicts the fact that 0<∫-∞∞f(x)/eσaxpdx<∞. Hence, (30) takes the form of strict inequality. For σ1=σ by the proof of Theorem 4, we obtain (32).
( i ) = > ( i i ). By (27) (for σ1=σ) and (32), we have (33).
( i i ) = > ( i ). We set the following function: (39)gy≔epσby∫-∞∞heax+byfxdxp-1y∈R.If J1=∞, then it is impossible since (32) is valid; if J1=0, then (32) is trivially valid. In the following, we suppose that 0<J1<∞. By (33), we have (40)0<∫-∞∞gyeσbyqdy=J1p=I<Kλσ∫-∞∞fxeσaxpdx1/p∫-∞∞gyeσbyqdy1/q<∞,J1=∫-∞∞gyeσbyqdy1/p<Kλσ∫-∞∞fxeσaxpdx1/p;namely, (32) follows, which is equivalent to (33).
Hence, Statements (i) and (ii) are valid and equivalent.
If there exists a constantM≤Kλ(σ), such that (33) is valid when replacing Kλ(σ) by M, then by Lemma 3, we have Kλ(σ)≤M. Hence, the constant factor M=Kλ(σ) in (33) is the best possible.
The constant factorKλ(σ) in (32) is still the best possible. Otherwise, by (27) (for σ1=σ), we would reach a contradiction that the constant factor Kλ(σ) in (33) is not the best possible.
The theorem is proved.Forg(y)=e-λbyG(y), and μ1=λ-σ1 in Theorems 4 and 5, then replacing b (G(y)) by -b (g(y)), setting(41)kλeax,eby≔max⁡eax,ebyα+βeax-ebyλ+αmin⁡eax,ebyβx,y∈R,we have the following corollaries.Corollary 6.
IfM is a constant, then the following statements (i), (ii), and (iii) are equivalent:
(i) For any nonnegative measurable functionf(x) in R, we have the following inequality:(42)∫-∞∞epμ1by∫-∞∞kλeax,ebyfxdxpdy1/p≤M∫-∞∞fxeσaxpdx1/p.
(ii) For any nonnegative measurable functionsf(x) and g(y) in R, we have the following inequality:(43)∫-∞∞∫-∞∞kλeax,ebyfxgydxdy≤M∫-∞∞fxeσaxpdx1/p∫-∞∞gyeμ1byqdy1/q.
(iii)μ1=μ, and M≥Kλ(σ)(>0).Corollary 7.
The following statements (i) and (ii) are valid and equivalent:
(i) For anyf(x)≥0, satisfying 0<∫-∞∞f(x)/eσaxpdx<∞, we have the following inequality:(44)∫-∞∞epbμy∫-∞∞max⁡eax,ebyα+βfxeax-ebyλ+αmin⁡eax,ebyβdxpdy1/p<Kλσ∫-∞∞fxeσaxpdx1/p.
(ii) For anyf(x)≥0, satisfying 0<∫-∞∞f(x)/eσaxpdx<∞, and g(y)≥0, satisfying 0<∫-∞∞g(y)/eμbyqdy<∞, we have the following inequality:(45)∫-∞∞∫-∞∞max⁡eax,ebyα+βfxgyeax-ebyλ+αmin⁡eax,ebyβdxdy<Kλσ∫-∞∞fxeσaxpdx1/p∫-∞∞gyeμbyqdy1/q.
Moreover, the constant factorKλ(σ) in (44) and (45) is the best possible.
In particular, forα=β=0,σ,μ>0, σ+μ=λ<1, we have the following equivalent inequalities with the best possible constant factor K~λ(σ):(46)∫-∞∞epμby∫-∞∞fxeax-ebyλdxpdy1/p<K~λσ∫-∞∞fxeσaxpdx1/p,(47)∫-∞∞∫-∞∞fxgyeax-ebyλdxdy<K~λσ∫-∞∞fxeσaxpdx1/p∫-∞∞gyeμbyqdy1/q.In (35) and (36), setting F(x)=e(λa/2)xf(x),G(y)=e(λb/2)yg(y), then replacing back F(x)(G(y)) by f(x)(g(y)), and introducing the hyperbolic sine function as sinh(s)=(es-e-s)/2, we haveCorollary 8.
Ifσ,μ>0, σ+μ=λ<1, then the following statements (i) and (ii) are valid and equivalent:
(i) For anyf(x)≥0, satisfying 0<∫-∞∞e(λ/2-σ)axf(x)pdx<∞, we have the following inequality:(48)∫-∞∞epσ-λ/2by∫-∞∞fxsinh⁡ax+by/2λdxpdy1/p<2K~λσ∫-∞∞eλ/2-σaxfxpdx1/p.
(ii) For anyf(x)≥0, satisfying 0<∫-∞∞e(λ/2-σ)axf(x)pdx<∞ and g(y)≥0, satisfying 0<∫-∞∞e(λ/2-σ)byg(y)qdy<∞, we have the following inequality:(49)∫-∞∞∫-∞∞fxgysinh⁡ax+by/2λdxdy<2K~λσ∫-∞∞eλ/2-σaxfxpdx1/p∫-∞∞eλ/2-σbygyqdy1/q.
Moreover, the constant factor2K~λ(σ) in (48) and (49) is the best possible.


## 4. Operator Expressions

We set the following functions:φ(x)≔e-pσax,ψ(y)≔e-qσby,ϕ(y)≔e-qμby, wherefrom, ψ1-p(y)=epσby,ϕ1-p(y)=epμby(x,y∈R), and define the following real normed linear spaces:(50)Lp,φR≔f:fp,φ≔∫-∞∞φxfxpdx1/p<∞,wherefrom, (51)Lq,ψR=g:gq,ψ≔∫-∞∞ψygyqdy1/q<∞,Lq,ϕR=g:gq,ϕ≔∫-∞∞ϕygyqdy1/q<∞,Lp,ψ1-pR=h:hp,ψ1-p=∫-∞∞ψ1-pyhypdy1/p<∞,Lq,ϕ1-pR=h:hp,ϕ1-p=∫-∞∞ϕ1-pyhypdy1/p<∞.(a) In view of Theorem5, for f∈Lp,φ(R), setting (52)h1y≔∫-∞∞heax+byfxdxy∈R,by (34), we have(53)h1p,ψ1-p=∫-∞∞ψ1-pyH1pydy1/p≤Kλσfp,φ<∞.Definition 9.
Define a Hilbert-type integral operator with the nonhomogeneous kernelT(1):Lp,φ(R)→Lp,ψ1-p(R) as follows: for any f∈Lp,φ(R), there exists a unique representation T(1)f=h1∈Lp,ψ1-p(R), satisfying for any y∈R,T(1)f(y)=h1(y).
In view of (53), it follows that (54)T1fp,ψ1-p=h1p,ψ1-p≤Kλσfp,φ,and then the operator T(1) is bounded satisfying (55)T1=supf≠θ∈Lp,φR⁡T1fp,ψ1-pfp,φ≤Kλσ.
If we define the formal inner product ofT(1)f and g as follows: (56)T1f,g≔∫-∞∞∫-∞∞heax+byfxdxgydy,then we can rewrite Theorem 5 as follows.Theorem 10.
The following statements (i) and (ii) are valid and equivalent:
(i) For anyf(x)≥0,f∈Lp,φ(R), satisfying fp,φ>0, we have the following inequality:(57)T1fp,ψ1-p<Kλσfp,φ.
(ii) For anyf(x),g(y)≥0,f∈Lp,φ(R),g∈Lq,ψ(R), satisfying fp,φ>0, and gq,ψ>0, we have the following inequality:(58)T1f,g<Kλσfp,φgq,ψ.
Moreover, the constant factorKλ(σ) in (57) and (58) is the best possible, namely, (59)T1=Kλσ.(b) In view of Corollary7, for f∈Lp,φ(R), setting (60)h2y≔∫-∞∞kλeax,ebyfxdxy∈R,by (44), we have(61)h2p,ϕ1-p=∫-∞∞ϕ1-pyh2pydy1/p≤Kλσfp,φ<∞.Definition 11.
Define a Hilbert-type integral operator with the homogeneous kernelT(2):Lp,φ(R)→Lp,ϕ1-p(R) as follows: for any f∈Lp,φ(R), there exists a unique representation T(2)f=h2∈Lp,ϕ1-p(R), satisfying for any y∈R,T(2)f(y)=h2(y).
In view of (61), it follows that (62)T2fp,ϕ1-p=h2p,ϕ1-p≤Kλσfp,φ, and then the operator T(2) is bounded satisfying (63)T2=supf≠θ∈Lp,φR⁡T2fp,ϕ1-pfp,φ≤Kλσ.If we define the formal inner product ofT(2)f and g as follows: (64)T2f,g≔∫-∞∞∫-∞∞kλeax,ebyfxdxgydy,then we can rewrite Corollary 7 as follows.Corollary 12.
The following statements (i) and (ii) are valid and equivalent:
(i) For anyf(x)≥0,f∈Lp,φ(R), satisfying fp,φ>0, we have the following inequality:(65)T2fp,ϕ1-p<Kλσfp,φ.
(ii) For anyf(x),g(y)≥0,f∈Lp,φ(R),g∈Lq,ϕ(R), satisfying fp,φ>0, and gq,ϕ>0, we have the following inequality:(66)T2f,g<Kλσfp,φgq,ϕ.
Moreover, the constant factorKλ(σ) in (65) and (66) is the best possible, namely, (67)T2=Kλσ.



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