# Sharp Bounds for Fractional Conjugate Hardy Operator on Higher-Dimensional Product Spaces

Authors: Zequn Wang; Mingquan Wei; Qianjun He; Dunyan Yan
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we obtain the sharp bound for fractional conjugate Hardy operator on higher-dimensional product spaces fromL1ℝn1×⋯×ℝnm to the space wLQℝn1×⋯×ℝnm and Lpℝn1×⋯×ℝnm to the space Lqℝn1×⋯×ℝnm. More generally, the operator norm of the fractional Hardy operator on higher-dimensional product spaces from LPℝn1×⋯×ℝnm to LQIℝn1×⋯×ℝnm is obtained.


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## Body

## 1. Introduction

Letf be a nonnegative integrable function on ℝn1×ℝn2×⋯×ℝnm. Define the fractional conjugate Hardy operator on higher-dimensional product spaces by(1)Hβ1,β2,…,βm∗fx=∫y1>x1⋯∫ym>xmfy1,…,ymB0,y11−β1/n1⋯B0,ym1−βm/nmdy1…dym,where x=x1,…,xm∈ℝn1×ℝn2×⋯×ℝnm, ∏i=1mxi≠0, and 0 ≤ βi < ni with i = 1, 2, …, m.Operator (1) is a generalization of classical Hardy-type operators. In 1925, Hardy first gave the following definition:(2)Hfx=1x∫0xfydy,H∗fx=∫x∞fyydy,for x > 0.Hardy [1] presented the classical Hardy inequalities on G=0,∞:(3)HfLpG≤pp−1fLpG,H∗fLpG≤pfLpG,with 1 < p ≤ ∞, where the constants p/p−1 and p are the best. Later, Christ and Grafakos [2] extended this result to higher-dimensional setting.In 1930, Bliss [3] proved Hardy inequality with power weight as(4)HfLqxα≤CpqfLpG,withα=qp−1.Inequality (4) can be stated as Theorem 1.Theorem 1.
Letf be a nonnegative integrable function on G and 1 < p < q < ∞. The inequality(5)∫0∞∫0xftdtqx−q/p′−1dx1/q≤Cpq∫0∞fpxdx1/p,holds with the sharp constant(6)Cpq≔p′q1/pq−pp1Bp/q−p,pq−1/q−p1/p−1/q,where B(·, ·) is a beta function.Inequality (5) turns into equality with the above constant if and only if(7)fx=caxq/p−1+1q/q−p.The fractional Hardy operator onLqG can be shown as(8)Hβfx=1x1−β∫0xfydy,with0<β<1.Theorem1 shows that(9)HβfLqG≤CpqfLpG,where β = 1/p − 1/q.Lu-Zhao [4] and Persson-Samko [5] extended those one-dimension results to higher-dimension results. For general situations, results could be found in [6–8].Then, it is natural to consider the Hardy-type operator on product spaces, which can be defined as(10)Hnfx=∫y1<x1⋯∫ym<xm1B0,x1⋯1B0,xmfy1,…,ymdy1⋯dym,where x=x1,…,xm∈ℝn1×⋯×ℝnm.He et al. [9] had already proved the boundedness of the fractional Hardy operator on higher-dimensional product spaces.We will discuss the conjugate situation defined as (1), and the state main theorem as follows.Theorem 2.
Let 1 <p < q < ∞. Set 0 < βi < ni and 1/q=1/p−βi/ni with 1 ≤ i ≤ m. Note n = (n1, n2, …, nm). If f∈Lpℝn1×ℝn2×⋯×ℝnm, then we have(11)Hβ1,…,βm∗fLqℝn1×ℝn2×⋯×ℝnm≤Cpqn∗fLpℝn1×ℝn2×⋯×ℝnm,where(12)Cpqn∗=∏i=1mni1/q+1/p′p−1ni−pβi1/q+1/p′Cpq,is sharp and Cpq is the sharp constant in (5).
In 2016, Gao et al. [10] proved the weak-type estimate for Hβ1,…,βm∗ with n1 = ⋯ = nm = n, β1 = ⋯ = βm = β, and m = 1. Their result is as follows.Theorem 3.
Letf be a nonnegative integrable function on ℝn. Set 0 ≤ β < n and 1/q=1/p−β/n. Then, we have(13)Hβ∗L1ℝn⟶Ln/n−β,∞ℝn=1,(14)Hβ∗Lpℝn⟶Lq,∞ℝn=qp′1/p′.
We will extent (13) to higher-dimensional product spaces.Theorem 4.
Set 0 <βi < ni, for i = 1, 2, …, m. Let Q=n1/n1−β1,n2/n2−β2,…,nm/nm−βm. If f∈L1ℝn1×ℝn2×⋯×ℝnm, then we have(15)Hβ1,…,βm∗fwLQℝn1×ℝn2×⋯×ℝnm≤1⋅fL1ℝn1×ℝn2×⋯×ℝnm,where 1 is the sharp bound.Next, we will generalize the results into mixed norm, which will be mentioned in Section4.Theorem 5.
SupposeP = (p1, …, pm) ≥ 1 (i.e., for every i = 1, 2, …, pi ≥ 1) and Q = (q1, …, qm) ≥ 1. Set 1/pi−1/qi=βi/ni with 0 < βi < ni, for i = 1, …, m. We further suppose 1 < pi < ∞ when i ∈ I ⊂ {1, …, m}. Then, operator Hβ1,…,βm∗ is bounded from LPℝn1×⋯×ℝnm to LQIℝn1×⋯×ℝnm. Moreover, there holds(16)Hβ1,…,βm∗LPℝn1×⋯×ℝnm⟶LQIℝn1×⋯×ℝnm=∏i∈ICpiqini∗∏i∉Iqipi′1/pi′,where(17)Cpiqini∗=ni1/qi+1/pi′pi−1ni−piβi1/qi+1/pi′Cpiqi,is sharp and Cpiqi is defined as in (6).


## 2. Preliminaries

First, we can use the following lemma to reduce the dimension. This lemma is from [9].Lemma 1.
Supposef∈Lpℝn1×ℝn2×⋯×ℝnm,xγ, with xγ=xγ1xγ2⋯xγm. Define(18)gfx=1ωn11ωn2⋯1ωnm∫ξ1=1∫ξ2=1⋯∫ξm=1fx1ξ1,x2ξ2,…,xmξmdξ1dξ2⋯dξm,where x=x1,x2,…,xm∈ℝn1×ℝn2×⋯×ℝnm and(19)ωni=2πni/2Γni/2,1≤i≤m.
Then, we have(20)Hβ1,β2,…,βm∗fx=Hβ1,β2,…,βm∗gfx,(21)gfLpℝn1×ℝn2×⋯×ℝnm,|x|γ≤fLpℝn1×ℝn2×⋯×ℝnm,xγ.Equality (20) can be proved by basic integral transformation. Inequality (21) is based on Minkowski’s inequality.Remark 1.
Lemma1 provides the idea that we only need to consider Hβ1,β2,…,βm∗ on radial functions. That is also the basic idea when we handle Hardy-type operators.
Notice that Theorem2 with m = 1 recovers part of result in [10]. Next, we will give a simple proof.Theorem 6.
Let 1 <p < q < ∞, 0 < β < n, and 1/q=1/p−β/n. If f∈Lpℝn, then we have(22)Hβ∗fLqℝn≤Cpqn∗fLpℝn,where(23)Cpqn∗=n1/qi+1/p′p−1n−pβ1/qi+1/p′Cpq,is sharp and Cpq is defined as in (6).Proof.
From Lemma1, it suffices to prove the result when f is a nonnegative, radial, smooth function with compact support on ℝn. Using the polar coordinate transformation, inequality (22) is equivalent to(24)n∫0∞n∫s∞frrβ−nrn−1drqsn−1ds1/q≤Cpqn∗n∫0∞fpssn−1ds1/p.
Letgt = f(t−1)t−β−1. Then, (24) can be written as(25)n∫0∞n∫0ugtdtqu−n−1du1/q≤Cpqn∗n∫0∞gpuup1+β−1+ndu1/p.
Defineα0 = −n − 1 and ω0 = p(1 + β) − (1 + n). We have ω0 < p − 1 and α0+1/q=ω0+1/p−1.
Let(26)v=vu=∫0uz−ω0/p−1dz=p−1p−1−ω0,up−1−ω0/p−1,hv=hvu=guuω0/p−1.
So we haveu−ω0/p−1du  = dv and guuω0/p−1=hv so that gudu = hvdv. It follows that(27)∫0∞gpuuω0du=∫0∞gpuuω0p/p−1u−ω0/p−1du=∫0∞hpvdv.
Notice that(28)α0=ω0+1qp−q−1=ω0qp+q1p−1−1=ω0qp−qp′−1,ω0qp−qp′−1+ω0p−1=ω0p−1−1+qp′ω0p′p−1=ω0p−1−1+qp′ω01p−1−1=ω0p−1−1qp′+1=ω0−p+1p−1qp′+1.
We have(29)I≔n∫0∞n∫0ugtdtquα0du1/q=n∫0∞n∫0ugtdtquω0q/p−q/p′−1du1/q=n∫0∞n∫0ugtdtquω0q/p−q/p′−1+ω0/p−1u−ω0/p−1du1/q=n1/q+1/p′p−1p−1−ω01/q+1/p′n1/p∫0∞∫0vhzdzqvp−1/p−1−ω0ω0−p+1/p−1q/p′+1dv1/q=n1/q+1/p′p−1p−1−ω01/q+1/p′n1/p∫0∞∫0vhzdzqv−q/p′−1dv1/q.
Using Bliss’s result, we conclude that(30)I≤n1/q+1/p′p−1n−pβ1/q+1/p′Cpq∫0∞gpuuω0du.
The sharp bound can be reached if and only if(31)fx=cx−ω0/p−1axp−1−ω0/p−1q/p−1+1q/q−p.


## 3. Main Result

In this section, we will give the proof of Theorems2 and 4.Proof of Theorem 2.
Without loss of generality, we only need to discuss the case form = 2. The case for m > 2 is also true by the same method.
Assume thatf is a nonnegative integrable function. Fixing variable x1, we define(32)Fy2,x1=∫y1>x1fy1,y2B0,y11−β1/n1dy1.
So(33)Hβ1,β2∗fx1,x2q=∫y2>x2Fy2,x1B0,y21−β2/n2dy2q.
By simple integral estimate, we have(34)∫ℝn1∫ℝn2Hβ1,β2∗fx1,x2qdx1dx2=∫ℝn1∫ℝn2∫y2|>|x2Fy2,x1B0,y21−β2/n2dy2qdx1dx2≤∫ℝn1Cpqn2∗∫ℝn2Fy2,x1pdy21pqdx1=Cpqn2∗q∫ℝn1∫ℝn2Fy2,x1pdy2q/pdx1≕I.
Applying generalized Minkowski’s inequality, we have(35)I=Cpqn2∗q∫ℝn1∫ℝn2Fy2,x1pdy2q/pdx1q/p⋅p/q≤Cpqn2∗q∫ℝn2∫ℝn1Fy2,x1qdx1p/qdy2q/p=Cpqn2∗q∫ℝn2∫ℝn1∫y1>x1fy1,y2B0,y11−β1/n1dy1qdx1p/qdy2q/p≤Cpqn2∗q∫ℝn2Cpqn1∗q∫ℝn1fpy1,y2dy1q/pp/qdy2q/p=Cpqn1∗Cpqn2∗∫ℝn2∫ℝn1fpy1,y2dy1dy21/pq.
It implies that(36)Hβ1,β2∗fLqℝn1×ℝn2≤Cpqn1∗Cpqn2∗fLpℝn1×ℝn2.
On the other hand, by Lemma1, we only need to prove the situation when f is a nonnegative radial smooth function with compact support so that we can separate the variable. Using the similar method of Theorem 6, it is easy to find that the best constant can be reached when(37)fx=cx−ωn1/p−1axp−1−ωn1/p−1q/p−1+1q/q−pcx−ωn2/p−1axp−1−ωn2/p−1q/p−1+1q/q−p,where ωni=pi1+βi−1+ni with i = 1, 2. We finish the Proof of Theorem 2.Proof of Theorem 4.
Without loss of generality, we only discuss the casem = 2. The case m > 2 is the same.
Whenm = 2, operator Hβ1β2∗ can be written as(38)Hβ1β2∗fx1,x2=∫y1>x1∫y2>x2fy1,y2B0,y11−β1/n1B0,y21−β2/n2dy1dy2.
Using Theorem3 and Fubini’s theorem, it implies that(39)Hβ1β2∗f⋅,x2Ln1/n1−β1,∞ℝn1=supλ1>0λ1x1:Hβ1β2∗fx1,x2>λ1n1/n1−β1≤1⋅∫y2>x2f⋅,y2B0,y21−β2/n2dy2L1ℝn1≤1⋅∫y2>x21B0,y21−β2/n2∫ℝn1fy1,y2dy1dy2.
We conclude that(40)∫y2>x21B0,y21−β2/n2∫ℝn1fy1,y2dy1dy2Ln2/n2−β2,∞ℝn2=supλ2>0λ2x2:∫y2>x21B0,y21−β2/n2∫ℝn1fy1,y2dy1dy2<λ2n2/n2−β2≤1⋅∫ℝn2∫ℝn1fy1,y2dy1dy2=1⋅fL1ℝn1×ℝn2.
Therefore, we have(41)Hβ1β2∗fwLn1/n1−β1,n2/n2−β2ℝn1×ℝn2≤1⋅fL1ℝn1×ℝn2.
On the other hand, define(42)fϵx=x−β+n/ϵ,x≥1,0,x<1.
Let(43)Fx1,x2=fϵ1x1fϵ2x2=fϵ1x1fϵ2x2,FL1ℝn1×ℝn2=fϵ1L1ℝn1fϵ2L1ℝn2=n1νn1β1+n1/ϵ−n1n2νn2β2+n2/ϵ−n2.
We have(44)Hβi∗fixi=νβi/ni−1∫yi>xiyi−βi+ni/ϵi−ni−βiχyi≥1yidyi.
If |xi| ≥ 1, we have(45)Hβi∗fixi=niνniβi/nixiβi−βi+ni/ϵiβi+ni/ϵi−βi.
If |xi| < 1, we have(46)Hβi∗fixi=niνniβi/niβi+ni/ϵi−βi.
Since that when |xi| < 1 and ϵ is small enough,(47)Hβi∗fixi≤niνniβi/niβi+ni/ϵi−βi.
We obtain that(48)x1∈ℝn1:Hβ1,β2∗Fx1,x2>λ1=B0,1∩x1∈ℝn1:n1νn1β1/n1β1+n1/ϵ1−β1Hβ2∗f2x2>λ1+B0,1c∩x1∈ℝn1:n1νn1β1/n1xiβ1−β1+n1/ϵ1β1+n1/ϵ1−β1Hβ2∗f2x2>λ1x1∈ℝn1:x1β1+n1/ϵ1−β1<n1νn1β1/n1λ1β1+n1/ϵ1−β1Hβ2∗f2x2.
Set(49)Cϵi=niνniβi/niβi+ni/ϵi−βi.
Notice that whenλi>Cϵi, xi∈ℝn1:Hβi∗fixi>λi=Ø. If ϵi is small enough, Cϵi tends to zero. Hence, when ϵ1 is small enough, we obtain that(50)I0≔sup0<λ1<Hβ2∗f2x2λ1x1∈ℝn1:Hβ1,β2∗Fx1,x2>λ1n1/n1−β1=Cϵ1Hβ2∗f2x2n1−β1/β1+n1/ϵ1−β1n1νn1n1−β1/n1sup0<λ1<minCϵ1,Hβ2∗f2x2λ11−n1−β1/β1+n1/ϵ1−β1=CϵiHβ2∗f2x2n1−β1/β1+n1/ϵ1−β1n1νn1n1−β1/n1=Hβ2∗f2x2n1−β1/β1+n1/ϵ1−β1β1+n1/ϵ1−n1β1+n1/ϵ1−β1fϵ1L1ℝn1.
Using the same method forx2, we obtain that(51)sup0<λ2λ2x2∈ℝn2:I0>λ2n2/n2−β2=Cϵ2β1+n1/ϵ1−n1β1+n1/ϵ1−β1fϵ1L1ℝn1n2−β2/n1−β1β1+n1/ϵ1−n1/β1+n1/ϵ1−β1×n2νn2n2−β2/n2sup0<λ2<Cϵ2λ21−n2−β2/n1−β1β1+n1/ϵ1−n1/β1+n1/ϵ1−β1=β1+n1/ϵ1−n1β1+n1/ϵ1−β1fϵ1L1ℝn1n2−β2/n1−β1β1+n1/ϵ1−n1/β1+n1/ϵ1−β1β2+n2/ϵ2−n2β2+n2/ϵ2−β2fϵ2L1ℝn2.
Letϵ1 ⟶ 0+ and ϵ2 ⟶ 0+, it implies that(52)Hβ1β2∗FwLn1/n1−β1,n2/n2−β2ℝn1×ℝn2≥1⋅FL1ℝn1×ℝn2.
We finish the Proof of Theorem 4.


## 4. Case for Mixed Norm

The mixed norm space was first defined in [11] by Benedek and Panzone and received much concern such as [12]. In 2018, Wei and Yan [7] defined a more general mixed norm space called as weak and strong mixed-norm space. We list its definition for completeness.Definition 1.
Let (Xi, Si, μi) be n given, totally σ–finite measure spaces, for 1 ≤ i ≤ n. P = x(p1, p2, …, pn) is a given n–tuple with 1 ≤ pi ≤ ∞. The set I satisfies I ⊂ {1, …, n}. A function f(x1, x2, …, xn) measurable in the product spaces X,S,μ=∏i=1nXi,∏i=1nSi,∏i=1nμi is said to belong to the space LPIX if the number obtained after subsequently taking successfully the mixed norm where we take pi–norm for i ∈ I and weak pj–norm for j ∈{1, …, n}\I. In natural order, it is finite. The number so obtained will be denoted by fPI, finite or not.
We give some necessary remarks for the spaceLPIX:(1)
If the setI = {1, …, n}, we call LPIX strong mixed norm space, which is also denoted by LP(X) or Lp1,…,pnX(2)
If the setI is empty, we call LPIX weak mixed norm space, which is also denoted by wLP(X) or wLp1,…,pnX(3)
The spacesLPIX is a quasi-normed space for P ≥ 1
For more properties, we refer readers to [7].
There is a basic lemma which plays an important role in the proof of our main theorems.Lemma 2.
Let (X, S, μ) be defined as in the above definitions. If pn ≥ ⋯ ≥ p1 ≥ 1 and f∈Lpn,…,p1X, then f∈Lp1,…,pnX. Moreover, there holds(53)fLp1,…,pnX≤fLpn,…,p1X.Lemma2 is a direct generalization of Minkowski’s inequality. For the proof of the lemma, readers can refer to [7]. It is not hard to see that Fubini’s theorem is a special case of Lemma 2. With all these, we can describe Hardy conjugate operator’s mixed norm.Theorem 7.
SupposeP = (p1, …, pm) ≥ 1 and Q = (q1, …, qm) ≥ 1. Set 0 < βi < n and 1/pi−1/qi=βi/ni, for i = 1, …, m. We further suppose 1 < pi < ∞ when i ∈ I ⊂{1, …, m}. Then, operator Hβ1,…,βm∗ is bounded from LPℝn1×⋯×ℝnm to LQIℝn1×⋯×ℝnm. Moreover, there holds(54)Hβ1,…,βm∗LPℝn1×⋯×ℝnm⟶LQIℝn1×⋯×ℝnm=∏i∈ICpiqini∗∏i∉Iqipi′1/pi′,where(55)Cpiqini∗=ni1/qi+1/pi′pi−1ni−piβi1/qi+1/pi′Cpiqi,is sharp and Cpiqi is defined as in (6).Proof.
Without loss of generality, we only discuss the casem = 2 and I = {1}.
Notice thatf∈Lp1,p2ℝn1×ℝn2. Applying Theorem 2, we obtain that(56)Hβ1,β2∗f⋅,x2Lq1ℝn1≤Cp1q1n1∗∫y2>x2f⋅,y2B0,y21−β2/n2dy2Lp1ℝn1.
Using Fubini’s theorem, it is easy to find(57)∫y2>x2f⋅,y2B0,y21−β2/n2dy2Lp1ℝn1=∫ℝn1∫y2>x2fy1,y2B0,y21−β2/n2dy2p1dy11/p1≤∫y2>x2∫ℝn1fy1,y2p1dy11/p1B0,y21−β2/n2dy2.
Using Theorem3, the following inequality holds:(58)∫y2|>|x2f⋅,y2B0,y21−β2/n2dy2Lp1ℝn1Lq1,∞ℝn1≤q2p2′1/p2′fLp1,p2ℝn1×ℝn1.
Combining above estimates, we obtain the desired result.
On the other hand, we takef1 the sharp function on ℝn1 given in Theorem 2 and f2 the sharp function on ℝn2 given in [10]. Define(59)Fx1,x2=f1x1f2x2.
By the definition of norm of general operator, we have(60)Hβ1,β2∗FLp1,p2ℝn1×ℝn2⟶LQIℝn1×ℝn2=supfLp1,p2ℝn1×ℝn2≠0Hβ1,β2∗fLQIℝn1×ℝn2fLp1,p2ℝn1×ℝn2≥Hβ1,β2∗FLQIℝn1×ℝn2fLp1,p2ℝn1×ℝn2=Hβ1,β2∗f1f2LQIℝn1×ℝn2f1f2Lp1,p2ℝn1×ℝn2=Hβ1∗f1Lq1ℝn1f1Lp1ℝn1Hβ2∗f2wLq2ℝn2f2Lp2ℝn2=Cp1q1n1∗q2p2′1/p2′.
Combining those, we finish the proof of Theorem7.



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