# Equivalent Conditions of a Hilbert-Type Multiple Integral Inequality Holding

Authors: Jianquan Liao; Yong Hong; Bicheng Yang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

Let∑i=1n1/pi=1 pi>1, in this paper, by using the method of weight functions and technique of real analysis; it is proved that the equivalent parameter condition for the validity of multiple integral Hilbert-type inequality ∫R+nKx1,⋯,xn∏i=1nfixidx1⋯dxn≤M∏i=1nfipi,αi with homogeneous kernel Kx1,⋯,xn of order λ is ∑i=1nαi/pi=λ+n−1, and the calculation formula of its optimal constant factor is obtained. The basic theory and method of constructing a Hilbert-type multiple integral inequality with the homogeneous kernel and optimal constant factor are solved.


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## Body

## 1. Preliminary

Assuming thatr>1, fx≥0, and α∈R, define
(1)Lr,α0,+∞=fx≥0:∥f∥r,α=∫0+∞xαfrxdx1/r<+∞.Particularly, denoteLr0,+∞=Lr,00,+∞ and fr=fr,0. If 1/p+1/q=1 p>1, f∈Lp0,+∞, and g∈Lq0,+∞, then there holds the well-known Hilbert’s integral inequality [1]
(2)∫0+∞∫0+∞fxgyx+ydxdy<πsinπ/pfpgq,where the constant factor π/sinπ/p is optimal.In general, let∑i=1n1/pi=1 pi>1, αi∈Ri=1,2,⋯,n, fixi∈Lpi,αi0,+∞, Kx1,⋯,xn be a nonnegative measurable function, and M be a constant; we call
(3)∫R+nKx1,⋯,xn∏i=1nfixidx1⋯dxn≤M∏i=1nfipi,ai,the Hilbert-type multiple integral inequality.Fruitful results have been obtained for the Hilbert-type inequality [2–12], but the results of the multi-integral form are much less, especially for the parameters’ conditions and the best constant factor for the Hilbert-type multi-integral inequality. The research for these problems is natural and important. However, the related references are less. In this paper, we will discuss the cases for the homogeneous integral kernel.Lemma 1.
Letn≥2 be an integer, αi∈Ri=1,2,⋯,n, Kx1,⋯,xn be a homogeneous nonnegative measurable function of order λ, ∑i=1n1/pi=1 pi>1, and ∑i=1nαi/pi=λ+n−1. Denote
(4)ωjxj=∫R+n−1Kx1,⋯,xn∏i=1i≠jnxi−αi+1/pidx1⋯dxj−1dj+1⋯dxn,Wj=∫R+n−1Ku1,⋯,uj−1,1,uj+1,⋯,un∏i=1i≠jnui−αi+1/pidu1⋯duj−1duj+1⋯⋯dun,where j=1,2,⋯,n. Then, we have
(5)ωjxj=xjαj+1/pj−1Wj,and W1=W2=⋯=Wn.Proof.
SinceKx1,⋯,xn is a homogeneous function of order λ, we have
(6)Kx1,⋯,xn=xjλKx1xj,⋯,xj−1xj,1,xj+1xj,⋯,xnxj,and then
(7)ωjxj=xjλ∫R+n−1Kx1xj,⋯,xj−1xj,1,xj+1xj,⋯,xnxj×∏i=1i≠jnxi−αi+1/pidx1⋯dxj−1dxj+1⋯dxn.Settingxi/xj=uii=1,⋯,j−1,j+1,⋯,n, then we find
(8)∂x1,⋯,xj−1,xj+1,⋯,xn∂u1,⋯,uj−1,uj+1,⋯,un=xjn−1,ωjxj=xjλ+n−1−∑i=1i≠jnαi+1/pi∫R+n−1Ku1,⋯,uj−1,1,uj+1,⋯,un×∏i=1i≠jnui−αi+1/pidu1⋯duj−1duj+1⋯dun=xjαj+1/pj−1Wj.Thus, (5) holds, because for any j, we have
(9)Wj=∫R+n−1u1λK1,u2u1,⋯,uj−1u1,1u1,uj+1u1,⋯,unu1×∏i=1i≠jnui−αi+1/pidu1⋯duj−1duj+1⋯dun.Settingu2/u1=v2,⋯,uj−1/u1=vj−1,1/u1=vj,uj+1/u1=vj+1,⋯, and un/u1=vn, then it follows that u1=1/vj,u2=1/vjv2,⋯,uj−1=1/vjvj−1,uj+1=1/vjvj+1,⋯, and un=1/vjvn. So we get
(10)Wj=∫R+n−11vjλK1,v2,⋯,vn1vj−α1+1/p1v2vj−α2+1/p2⋯vj−1vj−αj−1+1/pj−1×vj+1vj−αj+1+1/pj+1⋯vnvj−αn+1/pnvj−ndv2⋯dvn=∫R+n−1K1,v2,⋯,vnv2−α2+1/p2⋯vj−1−αj−1+1/pj−1vj−λ−n+∑i=1i≠jnαi+1/pi×vj+1−αj+1+1/pj+1⋯vn−αn+1/pndv2⋯dvn=∫R+n−1K1,v2,⋯,vn∏i=2nvi−αi+1/pidv2⋯dvn=W1.Hence, we obtainW1=W2=⋯=Wn.Lemma 2.
Let0<a<2, then
(11)∫0+∞min1,tmax1,tt−adt=12−a+1a.Proof.
Since0<a<2, we have a−1<1 and a+1>1, so
(12)∫0+∞min1,tmax1,tt−adt=∫01t1t−adt+∫1+∞1tt−adt=∫011ta−1dt+∫1+∞1ta+1dt=12−a+1a.


## 2. The Equivalent Conditions for a Hilbert-Type Multiple Integral Inequality Holding

Theorem 3.
Letn≥2 be an integer, ∑i=1n1/pi=1pi>1, αi∈Ri=1,2,⋯,n, Kx1,⋯,xn be a homogeneous nonnegative measurable function of order λ, and
(13)Wn=∫R+n−1Ku1,⋯,un−1,1∏i=1n−1ui−αi+1/pidu1⋯dun−1,is convergent, such that
(14)Wn1=∫01⋯∫01Ku1,⋯,un−1,1∏i=1n−1ui−αi+1/pidu1⋯dun−1>0,Wn2=∫1∞⋯∫1∞Ku1,⋯,un−1,1∏i=1n−1ui−αi+1/pidu1⋯dun−1>0.
Then(i)
For allfixi∈Lpi,αi0,+∞i=1,2,⋯,n, there exists a constant M, such that the Hilbert-type multiple integral inequality
(15)∫R+nKx1,⋯,xn∏i=1nfixidx1⋯dxn≤M∏i=1nfipi,αi,holds true if and only if ∑i=1nαi/pi=λ+n−1.
(ii)
If (15) holds, then the best constant factor is infM=Wn.Proof.
(i) Suppose that there exists a constantM such that (15) holds. Denote c=∑i=1nαi/pi−λ+n−1.
Ifc>0, then for 0<ε<c, we set
(16)fixi=xi−αi−1+ε/pi,0<xi≤1,0,xi>1,where i=1,2,⋯,n. We find
(17)∏i=1n∫0+∞xiαifipixidxi1/pi=∏i=1n∫01xi−1+εdxi1/pi=1ε,∫R+nKx1,,⋯,xn∏i=1nfixidx1⋯dxn=∫01xn−αn−1+ε/pn∫01⋯∫01Kx1,⋯,xn·∏i=1n−1xi−αi−1+ε/pidx1⋯dxn−1dxn=∫01xn−αn−1+ε/pn+λ∫01⋯∫01Kx1xn,⋯,xn−1xn,1·∏i=1n−1xi−αi−1+ε/pidx1⋯dxn−1dxn=∫01xn−αn−1+ε/pn+λ×∫01/xn⋯∫01/xnKt1,⋯,tn−1,1xnn−1∏i=1n−1xnti−αi−1+ε/pi·dt1⋯dtn−1dxn=∫01xn−αn−1+ε/pn+λ+n−1+∑i=1n−1−αi−1+ε/pi×∫01/xn⋯∫01/xnKt1,⋯,tn−1,1∏i=1n−1ti−αi−1+ε/pidt1⋯dtn−1dxn=∫01xn−1−c+ε∫01/xn⋯∫01/xnKt1,⋯,tn−1,1·∏i=1n−1ti−αi−1+ε/pidt1⋯dtn−1dxn≥∫01xn−1−c+εdxn·∫01⋯∫01Kt1,⋯,tn−1,1∏i=1n−1ti−αi−1+εpidt1⋯dtn−1.Thus, by (15), we get
(18)∫01xn−1−c+εdxn∫01⋯∫01Kt1,⋯,tn−1,1∏i=1n−1ti−αi−1+ε/pidt1⋯dtn−1≤Mε.
SinceWn1>0 and ε−c<0, ∫01xn−1−c+εdxn is divergent to +∞. So we get a contradiction that +∞≤M/ε, namely, c>0, cannot be held.
Ifc<0, then for 0<ε<−c, we set
(19)fixi=xi−αi−1−ε/pi,xi≥1,0,0<xi<1,where i=1,2,⋯,n. Similarly, we get
(20)∫1+∞xn−1−c−εdxn∫1+∞⋯∫1+∞Kt1,⋯,tn−1,1·∏i=1n−1ti−αi−1−ε/pidt1⋯dtn−1≤Mε.
SinceWn2>0 and −c−ε>0, ∫1+∞xn−1−c−εdxn is divergent to +∞; also, we get a contradiction that +∞≤M/ε, namely, c<0, cannot be held.
From the above discussions, we getc=0; that is, ∑i=1nαi/pi=λ+n−1.
Conversely, assume that∑i=1nαi/pi=λ+n−1 holds. Note that
(21)∏j=1nxjαj+1/pj∏i=1nxiαi+1/pi−1/pj=∏j=1nxjαj+1/pj∏i=1nxi−αi+1/pi∑k=1n1/pk=∏j=1nxjαj+1/pj∏i=1nxi−αi+1/pi=1.
By Ho¨lder’s inequality and Lemma 1, we find
(22)∫R+nKx1,,⋯,xn∏i=1nfixidx1⋯dxn=∫R+nKx1,,⋯,xn∏j=1nxjαj+1/pj∏i=1nxiαi+1/pi−1/pjfjxj·dx1⋯dxn≤∏j=1n∫R+nxjαj+1∏i=1nxi−αi+1/pifjpjxjK·x1,,⋯,xndx1⋯dxn1/pj=∏j=1n∫0+∞xjαj+1−αj+1/pjfjpj·xj∫R+n−1∏i=1i≠jnxi−αi+1/piKx1,⋯,xn×dx1⋯dxj−1dxj+1⋯dxndxj1/pj=∏j=1n∫0+∞xjαj+1−αj+1/pjfjpjxjωjxjdxj1/pj=∏j=1n∫0+∞xjαj+1−αj+1/pjfjpjxjxjαj+1/pj−1Wjdxj1/pj=∏j=1nWj1/pj∏j=1n∫0+∞xjαjfjpjxjdxj1/pj=Wn∏i=1nfipi,αi.
So, for allM≥Wn, (15) holds.
(ii) Next, we prove that when the equality (15) holds, infM=Wn. Otherwise, there exists a constant M0<Wn, such that
(23)∫R+nKx1,⋯,xn∏i=1nfixidx1⋯dxn≤M0∏i=1nfipi,αi.
For a sufficient smallε>0 and δ>0, let
(24)f1x1=x1−α1−1−ε/p1,x1≥1,0,0<x1<1,and when i=2,3,⋯,n, we let
(25)fixi=xi−αi−1−ε/pi,xi≥δ,0,0<xi<δ..
Thus, we get(26)∏i=1nfipi,αi=∫1+∞x1−1−εdx11/p1∏i=2n∫δ+∞xi−1−εdxi1/pi=1ε∏i=2n1δε1/pi.
We still have(27)∫R+nKx1,⋯,xn∏i=1nfixidx1⋯dxn=∫1+∞x1−α1+1+ε/p1∫δ+∞⋯∫δ+∞Kx1,⋯,xn·∏i=2nxi−αi+1+ε/pidx2⋯dxndx1=∫1+∞x1λ−α1+1+ε/p1·∫δ+∞⋯∫δ+∞K1,x2x1x2/x1,⋯,xnx1·∏i=2nxi−αi+1+ε/pidx2⋯dxndx1=∫1+∞x1λ−α1+1+ε/p1·∫δ/x1+∞⋯∫δ/x1+∞K1,u2,⋯,un∏i=2nx1ui−αi+1+ε/pix1n−1·du2⋯dundx1=∫1+∞x1−1−ε∫δ/x1+∞⋯∫δ/x1+∞K1,u2,⋯,un·∏i=2nui−αi+1+ε/pidu2⋯dundx1≥∫1+∞x1−1−ε∫δ+∞⋯∫δ+∞K1,u2,⋯,un·∏i=2nui−αi+1+ε/pidu2⋯dundx1=1ε∫δ+∞⋯∫δ+∞K1,u2,⋯,un∏i=2nui−αi+1+ε/pidu2⋯dun.
Combining this with (23) and (26), we obtain
(28)∫δ+∞⋯∫δ+∞K1,u2,⋯,un∏i=2nui−αi+1+ε/pidu2⋯dun≤M0∏i=2n1δε1/pi.
Letε⟶0+, and then by the Lebesgue dominated convergence theorem, we have
(29)∫δ+∞⋯∫δ+∞K1,u2,⋯,un∏i=2nui−αi+1/pidu2⋯dun≤M0.
Takingδ⟶0+, we get
(30)Wn=W1=∫R+n−1K1,u2,⋯,un∏i=2nui−αi+1/pidu2⋯dun≤M0.
This is a contradiction compared withM0<Wn, and then infM=Wn.


## 3. Applications

Taking some different integral kernels and different parameters, we can get a great deal of Hilbert-type inequalities in former literatures and other some new equalities. Moreover, the necessary and sufficient conditions for the existence of these inequalities are obtained.Corollary 4.
Let the integern≥2, ∑i=1n1/pi=1 pi>1, αi∈R, fixi∈Lpi,αi0,+∞ i=1,2,⋯,n, and
(31)Wn=∫R+n−1minu1,⋯,un−1,1maxu1,⋯,un−1,1∏i=1n−1ui−αi+1/pidu1⋯dun−1,convergence. Then, there exists a constant M, such that the necessary and sufficient condition for the equality hold
(32)∫R+nminx1,⋯,xnmaxx1,⋯,xn∏i=1nfixidx1⋯dxn≤M∏i=1nfipi,αi,is ∑i=1nαi/pi=n−1. And when equation (32) holds, the best constant factor is infM=Wn.Proof.
LetKx1,⋯,xn=minx1,⋯,xn/maxx1,⋯,xn; then, Kx1,⋯,xn certainly is a homogeneous nonnegative measurable function of order λ=0. By Theorem 3, the corollary holds.Corollary 5.
Let1/p+1/q=1p>1, −1<α<2p−1, −1<β<2q−1, fx∈Lp,α0,+∞, and gy∈Lq,β0,+∞. Then, there exists a constant M, such that the necessary and sufficient condition for the equality hold
(33)∫0+∞∫0+∞minx,ymaxx,yfxgydxdy≤Mfp,αgq,β,is α/p+β/q=1. And when equation (33) holds, the best constant factor is infM=p/α+1+q/β+1.Proof.
Since0<β+1/q<2, by Lemma 2, we have
(34)∫0+∞min1,u2max1,u2u2−β+1/qdu2=2−β+1q−1+β+1q−1=pα+1+qβ+1.
Combining this with the casen=2 of Corollary 4, the proof is completed.Remark 6.
(i) Lettingαi=pi−1i=1,⋯,n in Corollary 5, we can get
(35)∫R+nminx1,⋯,xnmaxx1,⋯,xn∏i=1nfixidx1⋯dxn≤n!∏i=1nfipi,pi−1,where the constant factor n! is the best possible. The above equality is the main result in [2].
(ii) Lettingα=β=1 in Corollary 5, we get
(36)∫0+∞∫0+∞minx,ymaxx,yfxgydxdy≤pq2fp,1gq,1,where the constant factor pq/2 is optimal.
(iii) Lettingα=p/2 and β=q/2 in Corollary 5, then we have
(37)∫0+∞∫0+∞minx,ymaxx,yfxgydxdy≤8pqp+2q+2fp,p/2gq,q/2,where the constant factor 8pq/p+2q+2 is the best possible.
(iv) Lettingα=3/2p−2 and β=3/2q−2 in Corollary 5, we obtain
(38)∫0+∞∫0+∞minx,ymaxx,yfxgydxdy≤8pq3p−23q−2fp,3/2p−2gq,3/2q−2,where the constant factor 8pq/3p−23q−2 is optimal.



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