# Equivalent Parameter Conditions for the Validity of Half-Discrete Hilbert-Type Multiple Integral Inequality with Generalized Homogeneous Kernel

Authors: Qiang Chen; Bing He; Yong Hong; Zhen Li
Journal: Journal of Function Spaces
 (2020)

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## Abstract

LetGu,v be a homogeneous nonnegative function of order λ,Kn,xm,ρ=Gnλ1,xm,ρλ2. By using the weight coefficient method, the equivalent parameter conditions and best constant factors for the validity of the following half-discrete Hilbert-type multiple integral inequality ∫ℝ+m∑n=1∞Kn,xm,ρanfxdx≤Ma~p,αfq,β are discussed. Finally, its applications in operator theory are discussed.


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## Body

## 1. Preliminaries

Suppose thatm∈ℕ+,ℝ+m=x=x1,x2,⋯,xm:xi≥0,i=1,2,⋯,m. In this paper, it is always assumed that x∈ℝ+m, xm,ρ=x1ρ+x2ρ+⋯+xmρ1/ρρ>0. For any given r>0,α∈ℝ, we write(1)lrα=a~=an:a~r,α=∑n=1∞nαanr1/r<+∞,an≥0,Lrαℝ+m=fx≥0:fr,α=∫ℝ+mxm,ραfrxdx1/r<+∞.LetGu,v be a homogeneous nonnegative function of order λ. We call Ku,v=Guλ1,vλ2 a generalized homogeneous function. Obviously, Ku,v satisfies(2)Ku,v=uλλ1K1,u−λ1/λ2v=vλλ2Kv−λ2/λ1u,1.We will discuss the equivalent parameter conditions for the validity of the half-discrete Hilbert-type multiple integral inequality with generalized homogeneous kernel and the optimal constant factors of the inequality under certain special conditions. Finally, the boundedness and norm of the corresponding series operator and integral operator are discussed. The relevant literature can be found in [1–16].Lemma 1 (see [17]).
Letm∈ℕ+,ρ>0,r>0,ψu be measurable. xm,ρ≤r represents the region Ωr=x=x1,x2,⋯,xm:xm,ρ≤r,xi≥0. Then,(3)∫∥x∥m,ρ≤rψxm,ρdx=Γm1/ρρm−1Γm/ρ∫0rψuum−1du,∫∥x∥m,ρ≥rψxm,ρdx=Γm1/ρρm−1Γm/ρ∫r+∞ψuum−1du,where Γt is the Gamma function.
In view of Lemma1, one has(4)∫ℝ+mψxm,ρdx=Γm1/ρρm−1Γm/ρ∫0+∞ψuum−1du.Lemma 2.
Assume that1/p+1/q=1p>1,m∈ℕ+,ρ>0,λ1λ2>0, Gu,v is a homogeneous nonnegative measurable function of order λ, Kn,xm,ρ=Gnλ1,xm,ρλ2, 1/λ2αλ2−mλ1/p+βλ1−λ2/q−λλ1λ2=c, Kt,1t−a+1/p+c is monotonically decreasing on 0,+∞. Denote(5)W1=∫0+∞K1,tt−b+m/q+m−1dt,W2=∫0+∞Kt,1t−a+1/p+cdt.
Then,(6)λ1W2=λ2W1,ω1n=∫ℝ+mKn,xm,ρxm,ρ−β+m/qdx=Γm1/ρρm−1Γm/ρnα+1/p−1−cW1,ω2x=∑n=1∞Kn,xm,ρn−α+1/p+c≤xm,ρβ+m/q−mW2.Proof.
According to1/λ2αλ2−mλ1/p+βλ1−λ2/q−λλ1λ2=c, after a simple calculation, we have −λ1/λ2λλ2−β+m/q+m−1−λ1/λ2−1=−α+1/p+c. Then,(7)W1=∫0+∞Kt−λ2/λ1,1tλλ2−β+m/q+m−1dt=λ1λ2∫0+∞Ku,1u−λ1/λ2λλ2−β+m/q+m−1−λ1/λ2−1du=λ1λ2∫0+∞Ku,1u−α+1/p+cdu=λ1λ2W2.
Hence,λ1W2=λ2W1.
It follows from Lemma1 that(8)ω1n=Γm1/ρρm−1Γm/ρ∫0+∞Kn,tt−β+m/q+m−1dt=Γm1/ρρm−1Γm/ρ∫0+∞nλλ1K1,t·n−λ1/λ2t−β+m/q+m−1dt=Γm1/ρρm−1Γm/ρnλλ1−λ1/λ2β+m/q−m+1+λ1/λ2∫0+∞K1,uu−β+m/q+m−1du=Γm1/ρρm−1Γm/ρnα+m/p−1−cW1.
Notice thatKt,1t−α+1/p+c is monotonically decreasing on 0,+∞;we have(9)ω2x=∑n=1∞xm,ρλλ2Kn·xm,ρ−λ2/λ1,1n−α+1/p+c=xm,ρλλ2−λ2/λ1α+1/p−c∑n=1∞Kn·xm,ρ−λ2/λ1,1n·xm,ρ−λ2/λ1−α+1/p+c≤xm,ρλλ2−λ2/λ1α+1/p−c∫0+∞Kt·xm,ρ−λ2/λ1,1t·xm,ρ−λ2/λ1−α+1/p+cdt=xm,ρλλ2−λ2/λ1α+1/p−1−c∫0+∞Ku,1u−α+1/p+cdu=xm,ρβ+m/q−mW2.


## 2. Main Results

Theorem 3.
Suppose that1/p+1/q=1p>1, m∈ℕ+,ρ>0,λ1λ2>0, Gu,v is a homogeneous nonnegative measurable function of order λ, 1/λ2αλ2−mλ1/p+βλ1−mλ2/q−λλ1λ2=c, Gtλ1,1t−α+1/p and Gtλ1,1t−α+1/p+c are monotonically decreasing on 0,+∞, and(10)W0=λ2∫0+∞G1,tλ2t−β+m/q+m−1dtis convergent. Then,(i)
The necessary and sufficient condition for the validity of inequality(11)∫ℝ+m∑n=1∞Gnλ1,xm,ρλ2anfxdx≤Ma~p,αfq,β,for some constant M>0 is c≥0, where a~=an∈lpα,fx∈Lqβℝ+m(ii)
Whenc=0, that is, αλ2−mλ1/p+βλ1−mλ2/q−λλ1λ2=0, the best constant factor of (11) is(12)infM=W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/p.Proof.
DenoteKn,xm,ρ=Gnλ1,xm,ρλ2. (i) Suppose that (11) holds. We prove that c≥0 by using reduction to absurdity. If c<0, for ε=−c/2∣λ1∣>0, take(13)an=n−α−1−∣λ1∣ε/p,n=1,2,⋯,n,fx=xm,ρ−β−m−λ2ε/q,xm,ρ≥1,0,0<xm,ρ<1.
It follows from Lemma1 that(14)Ma~p,αfq,β=M∑n=1∞n−1−∣λ1∣ε1/p∫xm,ρ≥1xm,ρ−m−∣λ2∣εdx1/q=M1+∑k=2∞n−1−∣λ1∣ε1/pΓm1/ρρm−1Γm/ρ∫1+∞t−1−∣λ2∣εdt1/q≤M1+∫1+∞t−1−λ1εdt1/pΓm1/ρρm−1Γm/ρ·1λ2ε1/q=Mε·λ11/pλ21/qΓm1/ρρm−1Γm/ρ1/q1+λ1ε1/p=2M−cλ1λ21/q1−c21/pΓm1/ρρm−1Γm/ρ1/q.
SinceGtλ1,1t−α+1/p=Kt,1t−α+1/p is monotonically decreasing on 0,+∞, we have(15)∫ℝ+m∑n=1∞Gnλ1,xm,ρλ2anfxdx=∫∥x∥m,ρ≥1xm,ρ−β+m+∣λ2∣ε/q∑n=1∞Kn,xm,ρn−α+1+∣λ1∣ε/pdx=∫∥x∥m,ρ≥1xm,ρλλ2−β+m+∣λ2∣ε/q−λ2/λ1α+1+∣λ1∣ε/p∑n=1∞Kn·xm,ρ−λ2/λ1,1n·xm,ρ−λ2/λ1−α+1+∣λ1∣ε/pdx≥∫∥x∥m,ρ≥1xm,ρλλ2−β+m+λ2ε/q−λ2/λ1α+1+λ1ε/p×∫1+∞Ku·xm,ρ−λ2/λ1,1u·xm,ρ−λ2/λ1−α+1+λ1ε/pdudx=∫∥x∥m,ρ≥1xm,ρλλ2−β+m+λ2ε/q−λ2/λ1α+1+λ1ε/p+λ2/λ1∫∥x∥m,ρ−λ2/λ1+∞Kt,1t−α+1+λ1ε/pdtdx=∫xm,ρ≥1xm,ρ−m−λ2/λ1c−λ2ε∫xm,ρ−λ2/λ1+∞Kt,1t−α+1+λ1ε/pdtdx≥∫xm,ρ≥1xm,ρ−m−λ2/λ1c−λ2εdx∫1+∞Kt,1t−α+1+λ1ε/pdt=Γm1/ρρm−1Γm/ρ∫1+∞t−1−λ2/λ1c−λ2εdt∫1+∞Kt,1t−α+1+λ1ε/pdt=Γm1/ρρm−1Γm/ρ∫1+∞t−1−λ2/2λ1cdt∫1+∞Kt,1t−α+1+λ1ε/pdt.
It follows from (11), (14), and (15) that(16)∫1+∞t−1−λ2/2λ1cdt∫1+∞Kt,1t−α+1+λ1ε/pdt≤2M−cλ1λ21/q1−c21/pΓm1/ρρm−1Γm/ρ−1/ρ<+∞.
Sinceλ2/2λ1c<0, then ∫1+∞t−1−λ2/2λ1cdt=+∞, which contradicts (16). Therefore, c≥0.
Suppose thatc≥0. It follows from Hölder’s inequality and Lemma 2 that(17)∫ℝ+m∑n=1∞Gnλ1,xm,ρλ2anfxdx=∫ℝ+m∑n=1∞nα+1−cp/pqxm,ρβ+m/pq·anxm,ρβ+m/pqnα+1−cp/pq·fxKn,xm,ρdx≤∫ℝ+m∑n=1∞nα+1−cp/qxm,ρ−β+m/qanpKn,xm,ρdx1/p×∫ℝ+m∑n=1∞xm,ρβ+m/pn−α+1−cp/pfqxKn,xm,ρdx1/q=∑n=1∞nα+1−cp/qanpω1n1/p∫ℝ+mxm,ρβ+m/pfqxω2xdx1/q≤W11/pW21/qΓm1/ρρm−1Γm/ρ1/p∑n=1∞nα+1−cp/q+α+1/p−1−canp1/p×∫ℝ+mxm,ρβ+m/p+β+m/q−mfqxdx1/q=W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/p∑n=1∞nα−pcanp1/p∫ℝ+mxm,ρβfqxdx1/q≤W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/p∑n=1∞nαanp1/p∫ℝ+mxm,ρβfqxdx1/q=W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/pa~p,αfq,β.
TakeM≥W0/λ11/qλ21/pΓm1/ρ/ρm−1Γm/ρ1/p arbitrarily; one can get (11).(ii)
Whenc=0, if (12) does not hold, then, there exists constant M0>0, such that(18)M0<W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/p,(19)∫ℝ+m∑n=1∞Gnλ1,xm,ρλ2anfxdx≤M0a~p,αfq,β.
Forε>0 and δ are sufficiently small, take(20)an=n−α+1+∣λ1∣ε/p,n=1,2,⋯,n,fx=xm,ρ−β+m+λ2ε/q,xm,ρ≥δ,0,0<xm,ρ<δ.
Then,(21)M0a~p,αfq,β≤M0ελ11/pλ21/qΓm1/ρρm−1Γm/ρ1/q1+∣λ1∣ε1/pδ−∣λ2∣ε/q,(22)∫ℝ+m∑n=1∞Gnλ1,xm,ρλ2anfxdx=∑n=1∞n−α+1+∣λ1∣ε/p∫xm,ρ≥δxm,ρ−β+m+∣λ2∣ε/qKn,xm,ρdx=Γm1/ρρm−1Γm/ρ∑n=1∞n−α+1+∣λ1∣ε/p∫δ+∞Kn,uu−β+m+λ2ε/q+m−1du=Γm1/ρρm−1Γm/ρ∑n=1∞nλλ1−α+1+∣λ1∣ε/p∫δ+∞K1,u·n−λ1/λ2u−β+m+λ2ε/q+m−1du=Γm1/ρρm−1Γm/ρ∑n=1∞nλλ1−α+1+∣λ1∣ε/p−λ1/λ2β+m+λ2ε/q−m+1+λ1/λ2×∫δ·n−λ1/λ2+∞K1,tt−β+m+λ2ε/q+m−1dt≥Γm1/ρρm−1Γm/ρ∑n=1∞n−1−∣λ1∣ε∫δ+∞K1,tt−β+m+λ2ε/q+m−1dt≥Γm1/ρ∣λ1∣ερm−1Γm/ρ∫δ+∞K1,tt−β+m+λ2ε/q+m−1dt.
It follows from (19), (21), and (22) that(23)λ11/pλ21/q∣λ1∣Γm1/ρρm−1Γm/ρ1/p∫δ+∞K1,tt−β+m+λ2ε/q+m−1dt≤M01+∣λ1∣ε1/pδ−∣λ2∣ε/q.
Letε→0+; then,(24)λ11/pλ21/q∣λ1∣Γm1/ρρm−1Γm/ρ1/p∫δ+∞K1,tt−β+m/q+m−1dt≤M0.
In addition, letδ→0+; we have(25)W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/p≤M0.
This contradicts (18). Hence, (12) holds.


## 3. Applications

Define series operatorT1 and singular integral operator T2 with kernel Gnλ1,xm,ρλ2 by, respectively,(26)T1a~x=∑n=1∞Gnλ1,xm,ρλ2an,T2fn=∫ℝ+mGnλ1,xm,ρλ2fxdx.According to the basic theory of Hilbert-type inequality (11) can be equivalently written as the following two expressions:(27)T1a~p,β1−p≤Ma~p,α,T2fq,α1−q≤Mfq,β.Thus, Theorem 3 is equivalent to the following theorem.Theorem 4.
Assume that1/p+1/q=1p>1,m∈ℕ+,ρ>0,λ1λ2>0, Gu,v is a homogeneous nonnegative measurable function of order λ, 1/λ1αλ2−mλ1/p+βλ1−λ2/q−λλ1λ2=c,Gtλ1,1t−α+1/p and Gtλ1,1t−α+1/p+c are monotonically decreasing on 0,+∞, operators T1 and T2 are as defined in (26), and(28)W0=∣λ2∣∫0+∞G1,tλ2t−β+m/q+m−1dtis convergent. Then,(i)
T1:lpα→Lpβ1−pℝ+m and T2:Lqβℝ+m→lqα1−q are bounded operators if and only if c≥0(ii)
Whenc=0, the operator norms of T1 and T2 are(29)T1=T2=W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/p.Corollary 5.
Suppose thatm∈ℕ+,ρ>0,1/p+1/q=1p>1,a>0,λ1>0,λ2>0,1/λ2αλ2−mλ1/p+βλ1−λ2/q+aλ1λ2=c, α≥maxλ1pb−1,−1,λ1pb+pc−1,pc−1, mq−1−λ2qa+b<β<mq−1. Denote(30)W0=∫0111+tat1/λ2m−β+m/q−1+ta+b−1/λ2m−β+m/q−1dt.
Define operatorsT1 and T2 by, respectively,(31)T1a~x=∑n=1∞min1,nλ1/xm,ρλ2bnλ1+xm,ρλ2a·an,a~=an∈lpα,T2fn=∫ℝ+mmin1,nλ1/xm,ρλ2bnλ1+xm,ρλ2a·fxdx,fx∈Lqβℝ+m.
Then,(i)
T1 is a bounded operator from lpα to Lpβ1−pℝ+m, and T2 is a bounded operator from Lqβℝ+m to lqα1−q if and only if c≥0(ii)
Whenc=0, the operator norms of T1 and T2 are(32)T1=T2=W0λ11/qλ21/pΓm1/ρρm−1Γm/ρ1/pProof.
Denote(33)Gu,v=min1,u/vbu+va.
Then,Gu,v is a homogeneous nonnegative function of λ=−a order.
It follows frommq−1−λ2qa+b<β<mq−1 that 1/λ2m−β+m/q>0 and a+b−1/λ2m−β+m/q>0. Hence, W0 is convergent and(34)∣λ2∣∫0+∞G1,tλ2t−β+m/q+m−1dt=∫0+∞G1,uu1/λ2m−β+m/q−1du=∫0+∞min1,u−1b1+ua·u1/λ2m−β+m/q−1du=∫01u1/λ2m−β+m/q−11+uadu+∫1+∞u−b+1/λ2m−β+m/q−11+uadu=∫0111+tat1/λ2m−β+m/q−1+ta+b−1/λ2m−β+m/q−1dt.
In view ofα≥maxλ1pb+pc−1,pc−1, we obtain λ1b−α+1/p+c<0 and −α+1/p+c<0. Therefore,(35)Gtλ1,1t−α+1/p+c=11+tλ1a·tλ1b−α+1/p+c,0<t≤1,11+tλ1a·t−α+1/p+c,t>1is monotonically decreasing on 0,+∞.
According toα≥maxλ1pb−1,−1, it is also known that Gtλ1,1t−α+1/p is monotonically decreasing on 0,+∞.
In summary, it follows from Theorem4 that Corollary 5 holds.
Takeb=0 in Corollary 5, by virtue of the properties of Beta function; we get(36)W0=∫0111+tat1/λ2m−β+m/q−1+ta−1/λ2m−β+m/q−1dt=B1λ2m−β+mq,a−1λ2m−β+mqand the following result.Corollary 6.
Assume thatm∈ℕ+,ρ>0,1/p+1/q=1p>1,a>0,λ1>0,λ2>0,1/λ2αλ2−mλ1/p+βλ1−λ2/q+aλ1λ2=c, α≥max−1,pc−1, mq−1−λ2qa<β<mq−1. Define operators T1 and T2 by, respectively,(37)T1a~x=∑n=1∞annλ1+xm,ρλ2a,a~=an∈lpα,T2fn=∫ℝ+mfxnλ1+xm,ρλ2adx,fx∈Lqβℝ+m.
Then,(i)
T1:lpα→Lpβ1−pℝ+m and T2:Lqβℝ+m→lqα1−q are bounded operators if and only if c≥0(ii)
Whenc=0, the operator norms of T1 and T2 are(38)T1=T2=1λ11/qλ21/pB1λ2m−β+mq,a−1λ2m−β+mqΓm1/ρρm−1Γm/ρ1/p
In Corollary6, let α=λ1/λ21−p/raλ2,β=λ2/λ11−q/saλ1, m=1, 1/r+1/sr>1; then,(39)αλ2−λ1p+βλ1−λ2q+aλ1λ2=0.
The following results can be obtained.Corollary 7.
Assume that1/p+1/q=1p>1, 1/r+1/sr>1, a>0,λ1>0,λ2>0, α≥−1, q1−λ2a−1<β<q−1, α=λ1/λ21−p/raλ2,β=λ2/λ11−q/sq/saλ1. Define operators T1 and T2 by, respectively,(40)T1a~x=∑n=1∞annλ1+xλ2a,a~=an∈lpα,T2fn=∫0+∞fxnλ1+xλ2adx,fx∈Lqβ0,+∞.
Then,T1:lpα→Lpβ1−p0,+∞ and T2:Lqβ0,+∞→lqα1−q are bounded operators, and the operator norms of T1 and T2 are(41)T1=T2=1λ11/qλ21/pBar+1λ1q−1λ2p,as+1λ2p−1λ1q.



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