# Higher-Order Riesz Transforms in the Inverse Gaussian Setting and UMD Banach Spaces

Authors: Jorge J. Betancor; Lourdes Rodríguez-Mesa
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study higher-order Riesz transforms associated with the inverse Gaussian measure given byπn/2ex2dx on ℝn. We establish Lpℝn,ex2dx-boundedness properties and obtain representations as principal values singular integrals for the higher-order Riesz transforms. New characterizations of the Banach spaces having the UMD property by means of the Riesz transforms and imaginary powers of the operator involved in the inverse Gaussian setting are given.


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## Body

## 1. Introduction

Our setting isℝn endowed with the measure γ−1 whose density with respect to the Lebesgue measure is πn/2ex2, x∈ℝn. The measure γ−1 is called the inverse Gaussian measure. The study of harmonic analysis operators in ℝn,γ−1 was began by Salogni [1]. The principal motivation for Salogni’s studies was the connection with the Gaussian setting. However, as Bruno and Sjögren [2] pointed out, ℝn,γ−1 can be seen as a model of a variety of settings where a theory of singular integrals has not been developed. Also, the natural Laplacian on ℝn,γ−1, that we will denote by A, can be interpreted as a restriction of the Laplace-Beltrami operator associated with a warped-product manifold whose Ricci tensor is unbounded from below. A complete exposition of the theory of this kind of manifolds can be found in [3].The aim of this paper is to studyLpℝn,γ−1-boundedness properties of higher-order Riesz transforms in the inverse Gaussian setting. Also, we characterize the UMD Banach spaces by using these Riesz transforms.We consider the second-order differential operatorA0 defined by
(1)A0fx=−12Δfx−x·∇fx,x∈ℝn,where f∈Cc∞ℝn, the space of the smooth functions with compact support in ℝn. Here, Δ and ∇ denote the usual Euclidean Laplacian and gradient, respectively.A0 is essentially selfadjoint in L2ℝn,γ−1. A denotes the closure of A0 in L2ℝn,γ−1.For everyk=k1,⋯,kn∈ℕn by Hk, we represent the k-th Hermite polynomial given by Hkx=∏i=1nHkixi, x=x1,⋯,xn∈ℝn, where for every m∈ℕ,
(2)Hmz=−1mez2dmdzme−z2,z∈ℝ.We have that, for everyk=k1,⋯,kn∈ℕn,
(3)AH~k=k+nH~k,where k=k1+k2+⋯+kn and H~kx=e−x2Hkx, x∈ℝn. The spectrum of A in L2ℝn,γ−1 is the discrete set n+mm∈ℕ.The operator−A generates a diffusion semigroup (in the Stein sense [4]) TtAt>0 in ℝn,γ−1 where for every t>0, we have that
(4)TtAfx=∫ℝnTtAx,yfydy,x∈ℝn,for every f∈Lpℝn,γ−1, 1≤p<∞, and being
(5)TtAx,y=e−ntπn/21−e−2tn/2exp−x−e−ty21−e−2t,x,y∈ℝn,t>0.The maximal operatorT∗A defined by
(6)T∗Af=supt>0TtAf,was studied by Salogni ([1]). She proved that T∗A is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. From the general results in [4], it can be deduced that T∗A is bounded from Lpℝn,γ−1 into itself, for every 1<p<∞. Recently, Betancor et al. [5] have characterized the Köethe function spaces with the Hardy-Littlewood property by using the maximal operators
(7)T∗,kAf=supt>0tk∂tkTtf,k∈ℕ.In [1], Lpℝn,γ−1-boundedness properties with 1<p<∞ for some spectral multipliers associated with the operator A were proved. The imaginary power Aiσ, σ∈ℝ\0, of A is a special case of the multipliers studied in [1]. Bruno ([6]) established endpoints results for Aiσ, σ∈ℝ\0, proving that Aiσ is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. Also, he showed that, for λ≥1, the shifted first-order Riesz transform ∇A+λI−1/2 is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. These operators are studied on new Hardy type H1-spaces.Higher-order Riesz transforms associated with the operatorA were studied by Bruno and Sjögren [2]. For every α=α1,⋯,αn∈ℕn\0, the α-th Riesz transform is defined by Rα=∂αA−α/2, where ∂α=∂α/∂x1α1⋯∂xnαn and α=α1+⋯+αn. In ([2], Theorem 1.1), it was established that Rα is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1 if and only if α≤2.In ([6], Remark 2.6), Bruno proved that, for every α∈ℕn with α=1, Rα is bounded from Lpℝn,γ−1 into itself, for every 1<p<∞. In [2], Bruno and Sjögren say that they do not know whether Rα is bounded from Lpℝn,γ−1 into itself for every 1<p<∞ and α∈ℕn, α>1, though they expect so. In our first result, we prove that, as they expected, Rα is bounded from Lpℝn,γ−1 into itself when 1<p<∞ and α∈ℕn\0. We also obtain a representation of Rα as a principal value singular integral. In order to prove our result, we need to use some properties of the negative power A−β, β>0, of A. In Section 2, we analyze A−β, β>0. We obtain that, for every β>0, the operator A−β is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. This result contrasts with the one in ([7], Proposition 6.2) where it is proved that L−β, β>0, is not bounded from L1ℝn,γ into L1,∞ℝn,γ, where L represents the Ornstein-Uhlenbeck operator and γ denotes the Gaussian measure (dγx=π−n/2e−x2dx) on ℝn.Theorem 1.
Letα=α1,⋯,αn∈ℕn\0. For every f∈Cc∞ℝn, the derivative ∂αA−α/2fx exists for almost all x∈ℝn and there exists cα∈ℝ such that
(8)∂xαA−α/2fx=limε⟶0+∫x−y>εRαx,yfydy+cαfx,foralmostallx∈ℝn,where cα=0 if αi is odd for some i=1,⋯,n.Furthermore, whenn=1 and α is even, the last integral is actually absolutely convergent for every x∈ℝ, and in this case, no principal value is needed. Here,
(9)Rαx,y=1Γα/2∫0∞∂xαTtAx,ytα/2−1dt,x,y∈ℝn,x≠y.Letα∈ℕn. Since, for every ℓ∈ℕ,
(10)ddzH~ℓz=−H~ℓ+1z,z∈ℝ,we have that, for every k∈ℕn,
(11)∂xαA−α/2H~kx=−1αk+nα/2H~k+αx,x∈ℝn.Letf∈L2ℝn,γ−1. We can write f=∑k∈ℕnckfH~k in L2ℝn,γ−1, where for every k∈ℕn,
(12)ckf=πn/2H~kL2ℝn,γ−12∫ℝnfyH~kyey2dy.We define(13)Rαf=−1α∑k∈ℕnckfk+nα/2H~k+α.For everyk=k1,⋯,kn∈ℕn,
(14)H~k+αL2ℝn,γ−12=∫ℝnHk+αy2e−y2dy=πn2α+k∏i=1nΓki+αi+1.Then(15)RαfL2ℝn,γ−12=∑k∈ℕnckf2H~k+αL2ℝn,γ−12k+nα=∑k∈ℕnckfH~kL2ℝn,γ−12H~k+αL2ℝn,γ−12k+nαH~kL2ℝn,γ−12≤C∑k∈ℕnckfH~kL2ℝn,γ−12=CfL2ℝn,γ−12,where the constant C depends on n and α. Hence, Rα is bounded from L2ℝn,γ−1 into itself.Iff∈Cc∞ℝn, then Rαfx=∂xαA−α/2fx, x∈ℝn.Theorem 2.
Letα=α1,⋯,αn∈ℕn\0 and 1<p<∞. The Riesz transform Rα can be extended from L2ℝn,γ−1∩Lpℝn,γ−1 to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself. By denoting again Rα to this extension, we have that, there exists cα∈ℝ such that, for every f∈Lpℝn,γ−1,
(16)Rαfx=limε⟶0+∫x−y>εRαx,yfydy+cαfx,foralmostallx∈ℝn,where cα=0 if αi is odd for some i=1,⋯,n.Whenn=1 and α∈ℕ is even the integral defining Rα is absolutely convergent.As it was mentioned, Bruno and Sjögren ([2], Theorem 1.1]) proved that Rα is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1 if and only if 1≤∣α∣≤2. This property also holds in the Gaussian setting (see [8, 9]). Aimar et al. ([10]) introduced Riesz type operators Rα, α∈ℕn\0, related to the Ornstein-Uhlenbeck operator. Rα is bounded from L1ℝn,γ into L1,∞ℝn,γ for every α∈ℕn\0. Motivated by the results in [10], we define Riesz transform in the inverse Gaussian setting whose behavior in L1ℝn,γ−1 is different from the one for Rα.We can writeA0=∑i=1nδi∂xi, where for every i=1,⋯,n, δi=−2−1e−xi2∂xiexi2. We consider the operator A¯0=∑i=1n∂xiδi. We have that A¯0=−n+A0. Let A¯ the closure of A¯0 in L2ℝn,γ−1. For every k∈ℕn, A¯H~k=kH~k, and the spectrum of A¯ in L2ℝn,γ−1 is the set of nonnegative integers. For every ℓ,m∈ℕ, we have that δumH~ℓu=−1mΓℓ+1/Γℓ−m+1H~ℓ−mu, u∈ℝ. Here, we understand H~ℓ=0 when ℓ<0. Then, for every α=α1,⋯,αn∈ℕn, by denoting δα=∏i=1nδiαi, we get
(17)δαH~k=−1α∏i=1nΓki+1Γki−αi+1H~k−α,k=k1,⋯,kn∈ℕn,kr≥αr,r=1,⋯,n.Ifα=α1,⋯,αn∈ℕn\0 and k=k1,⋯,kn∈ℕn\0, with kr≥αr, r=1,⋯,n,
(18)δαA¯−α/2H~k=−1αkα/2∏i=1nΓki+1Γki−αi+1H~k−α.In other cases,δαA¯−α/2H~k=0 (see Section 4 for details).Letα=α1,⋯,αn∈ℕn\0. We define the Riesz transform R¯α on L2ℝn,γ−1 as follows
(19)R¯αf=−1α∑k=k1,⋯,kn∈ℕnkr≥αr,r=1,⋯,n1kα/2∏i=1nΓki+1Γki−αi+1ckfH~k−α.Thus,R¯α is bounded from L2ℝn,γ−1 into itself. If f∈Cc∞ℝn and c0f=0 then R¯αf=δαA¯−α/2f.Theorem 3.
Letα=α1,⋯,αn∈ℕn\0. The Riesz transform R¯α can be extended from L2ℝn,γ−1∩Lpℝn,γ−1 to Lpℝn,γ−1 as a bounded operator from
(i)
Lpℝn,γ−1 into itself, for every 1<p<∞(ii)
L1ℝn,γ−1 into L1,∞ℝn,γ−1, provided that n=1 or ∣α∣>n, when n>1
By denoting againR¯α to the extension we have that, for every f∈Lpℝn,γ−1, 1≤p<∞,
(20)R¯αfx=limε⟶0+∫x−y>εR¯αx,yfydy+cαfx,foralmostallx∈ℝn,where cα=0 when αi is odd for some i=1,⋯,n. Here
(21)R¯αx,y=1Γα/2∫0∞δxαTtA¯x,ytα/2−1dt,x,y∈ℝn,x≠y.LetX be a Banach space. Suppose that Mrr=1m is a X-valued martingale. The sequence dr=Mr−Mr−1r=1m, where M0 is understood as 0, is called the martingale difference associated with Mrr=1m. We say that drr=1m is a Lp-martingale difference sequence when it is the difference sequence associated with a Lp-martingale. If 1<p<∞, X is said to be a UMDp-space when there exists β>0 such that for all X-valued Lp-martingale difference sequence drr=1m and for all εrr=1m∈−1,1m,
(22)E∑r=1mεrdrp≤βE∑r=1mdrp.UMD is an abbreviation of unconditional martingale difference. If X is UMDp for some 1<p<∞, then X is UMDp for every 1<p<∞. This fact justifies to call UMD to the property without any reference to p. Burkholder [11] and Bourgain [12] proved that the UMD property of X is necessary and sufficient for the boundedness of the Hilbert transform in Lpℝ,X, 1<p<∞. The UMD property is a central notion in the development of the harmonic analysis when the functions are taking values in infinite-dimensional spaces. UMD Banach spaces have been characterized by using other singular integrals that can be seen as Riesz transforms associated to orthogonal systems (see [13–16], for instance). In the following result, we characterize the Banach spaces with the UMD property by using Riesz transforms in the inverse Gaussian setting. For every i=1,⋯,n, we define ei=e1i,⋯,eni where eji=0, i≠j, and eii=1.Theorem 4.
LetX be a Banach space. The following assertions are equivalent.
(i)
X is UMD(ii)
For everyi=1,⋯,n, Rei can be extended from Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself, for every 1<p<∞(iii)
For everyi=1,⋯,n, Rei can be extended from Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself, for some 1<p<∞(iv)
For everyi=1,⋯,n, Rei can be extended from L1ℝn,γ−1⊗X to L1ℝn,γ−1,X as a bounded operator from L1ℝn,γ−1,X into L1,∞ℝn,γ−1,X
Also, the equivalences hold when in the properties (ii), (iii), and (iv), we replaceRei by the maximal operator Rei∗ defined by
(23)Rei∗fx=supε>0∫x−y>εReix,yfydy,x∈ℝnandi=1,⋯,n.Theorem 5.
LetX be a Banach space. The following assertions are equivalent.
(i)
X is UMD(ii)
For every1≤p<∞ there exists, for each i=1,⋯,n, the limit
(24)limε⟶0+∫x−y>εReix,yfydy,foralmostallx∈ℝn,for every f∈Lpℝn,γ−1,X.(iii)
For some1≤p<∞, there exists, for every i=1,⋯,n, the limit(25)limε⟶0+∫x−y>εReix,yfydy,foralmostallx∈ℝn,for each f∈Lpℝn,γ−1,X.
(iv)
For every1≤p<∞, f∈Lpℝn,γ−1,X and i=1,⋯,n, Rei∗fx<∞, for almost all x∈ℝn(v)
For some1≤p<∞ and for every f∈Lpℝn,γ−1,X and i=1,⋯,n, Rei∗fx<∞, for almost all x∈ℝnThe properties stated in Theorems4 and 5 can also be established when we replace R-Riesz transforms by R¯-Riesz transforms.Next aim is to state characterizations of the Banach spaces with theUMD property by using imaginary powers Aiσ, σ∈ℝ\0, of A. Salogni ([1], Theorem 3.4.3) proved that, for every 1<p<∞ and σ∈ℝ\0,
(26)AiσLpℝn,γ−1⟶Lpℝn,γ−1~eϕp∗σ,as σ⟶∞, when ϕp=arcsin2/p−1. Actually, Aiσ is a Laplace transform type multiplier associated with A defined by the function ϕσt=t−iσ/Γ1−iσ, t>0, for every σ∈ℝ\0. Then, since TtAt>0 is a Stein diffusion semigroup, the Lpℝn,γ−1-boundedness of Aiσ, σ∈ℝ\0, follows from the general results established in ([4], Chapter III). Recently, Bruno ([6], Theorem 4.1) proved that Aiσ, σ∈ℝ\0, is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1.Letσ∈ℝ\0. We have that
(27)Aiσf=∑k∈ℕnk+niσckfH~k,f∈L2ℝn,γ−1.It is immediate to see thatAiσ is bounded from L2ℝn,γ−1 into itself. For every f∈L2ℝn,γ−1⊗X, we define Aiσ in the obvious way when X is a Banach space. In order to the operator Aiσ is bounded from L2ℝn,γ−1⊗X into itself as subspace of L2ℝn,γ−1,X, we need to impose some additional property to the Banach space X. For instance, if X is isomorphic to a Hilbert space, then Aiσ can be extended from L2ℝn,γ−1⊗X to L2ℝn,γ−1,X as a bounded operator from L2ℝn,γ−1,X into itself. We are going to characterize the UMD Banach spaces as those Banach spaces for which Aiσ can be extended from L2ℝn,γ−1∩Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself, when 1<p<∞, and from L1ℝn,γ−1,X into L1,∞ℝn,γ−1,X. Our result is motivated by the one in ([17], p. 402) where UMD Banach spaces are characterized by the Lpℝ,dx-boundedness properties of the imaginary power −d2/dx2iσ, σ∈ℝ\0, of −d2/dx2. Guerre-Delabriére’s result was extended to higher dimensions by considering imaginary powers of the Laplacian in ([18], Proposition 1). In [19], this kind of characterization for UMD Banach spaces is obtained in Hermite and Laguerre settings. As far as we know, this property has not been proved for the Ornstein-Uhlenbeck operator in the Gaussian framework.Theorem 6.
Letσ∈ℝ\0. For every f∈L2ℝn,γ−1, we have that
(28)Aiσfx=limε⟶0+∫x−y>εKσAx,yfydy+αεfx,foralmostallx∈ℝn,where
(29)KσAx,y=−∫0∞ϕσt∂tTtAx,ydt,x,y∈ℝn,x≠y,and
(30)αε=1Γn/2∫0∞ϕσε24ue−uun/2−1du,ε∈0,∞,being ϕσt=t−iσ/Γ1−iσ, t∈0,∞.
LetX be a Banach space. The following assertions are equivalent.
(i)
X is UMD(ii)
For every1<p<∞, Aiσ can be extended from L2ℝn,γ−1∩Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself(iii)
For some1<p<∞, Aiσ can be extended from L2ℝn,γ−1∩Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself(iv)
Aiσ can be extended from L2ℝn,γ−1∩L1ℝn,γ−1⊗X to L1ℝn,γ−1,X as a bounded operator from L1ℝn,γ−1,X into L1,∞ℝn,γ−1,X
We define the maximal operatorA∗iσ by
(31)A∗iσfx=supε>0∫x−y>εKσAx,yfydy,f∈⋃p≥1Lpℝn,γ−1,X.
The following assertions are equivalent to (i).(v)
For every1<p<∞, A∗iσ is bounded from Lpℝn,γ−1,X into itself(vi)
For some1<p<∞, A∗iσ is bounded from Lpℝn,γ−1,X into itself(vii)
A∗iσ is bounded from L1ℝn,γ−1,X into L1,∞ℝn,γ−1,X(viii)
For every1≤p<∞ and every f∈Lpℝn,γ−1,X there exists the limit(32)limε⟶0+∫x−y>εKσAx,yfydy+αεfx,foralmostallx∈ℝn.(ix)
For some1≤p<∞ and every f∈Lpℝn,γ−1,X, there exists the limit(33)limε⟶0+∫x−y>εKσAx,yfydy+αεfx,foralmostallx∈ℝn.(x)
For every1≤p<∞ and every f∈Lpℝn,γ−1,X, A∗iσfx<∞, for almost all x∈ℝn(xi)
For some1≤p<∞ and every f∈Lpℝn,γ−1,X, A∗iσfx<∞, for almost all x∈ℝnThis paper is organized as follows. In Section2 we study the negative power A−β, β>0, of A. Higher order Riesz transforms in the inverse Gauss setting are considered in Section 3 where we prove Theorems 1 and 2. Theorem 3 is established in Section 4 and Theorems 4 and 5 in Section 5. Finally, Section 6 is devoted to show the proof of Theorem 6.Throughout this paper,C and c denote positive constants that can change in each occurrence.


## 2. Negative Powers ofA

In this section, we proveLpℝn,γ−1-boundedness properties of the negative powers A−β, β>0, of A. These properties are different than the ones of the negative powers of the Ornstein-Uhlenbeck operator L=−Δ/2+x∇. We prove that, for every β>0, A−β defines a bounded operator from L1ℝn,γ−1 into L1,∞ℝn,γ−1. However, in ([7], Proposition 6.2), it was proved that if β>0, L−β is not bounded from L1ℝn,γ into L1,∞ℝn,γ.Letβ>0. We define
(34)A−βf=∑k∈ℕnckfk+nβH~k,f∈L2ℝn,γ−1.A−β is bounded from L2ℝn,γ−1 into itself. Moreover, when β>1, the series also converges pointwisely in ℝn. Indeed, let f∈L2ℝn,γ−1. We have that
(35)ckf≤πn/2fL2ℝn,γ−1H~kL2ℝn,γ−1−1,k∈ℕn.Also, for everyk∈ℕn, (see (14)),
(36)H~kL2ℝn,γ−1=πn/22k/2∏j=1nΓkj+11/2,and according to ([20], p. 324),
(37)Hjz≤2Γj+12j/2ez2/2,z∈ℝandj∈ℕ.Then,(38)ckfH~kx≤2ne−x2/2fL2ℝn,γ−1,x∈ℝn,k∈ℕn,and if β>1 it follows that
(39)∑k∈ℕnckfk+nβH~kx≤Ce−x2/2fL2ℝn,γ−1,x∈ℝn.The series in (34) converges pointwise absolutely for each β>0 when f∈Cc∞ℝn. Indeed, let f∈Cc∞ℝn. Partial integration allows us to see that, for every r∈ℕ, there exists C=Cf,r>0 such that
(40)ckf≤Ck+nrH~kL2ℝn,γ−1,k∈ℕn.Then,(41)∑k∈ℕnckfk+nβH~kx≤Ce−x2/2∑k∈ℕn1k+nβ+r<∞.We also consider the operator defined, for everyf∈L2ℝn,γ−1, by
(42)Sβf=1Γβ∫0∞TtAftβ−1dt,where the integral is understood in the L2ℝn,γ−1-Bochner sense.For eacht>0, we can write
(43)TtAf=∑k∈ℕne−k+ntckfH~k,f∈L2ℝn,γ−1.We obtain(44)TtAfL2ℝn,γ−12=∑k∈ℕne−2k+ntckf2H~kL2ℝn,γ−12≤e−2nt∑k∈ℕnckf2H~kL2ℝn,γ−12=e−2ntfL2ℝn,γ−12,f∈L2ℝn,γ−1,t>0.Then,(45)SβfL2ℝn,γ−1≤1Γβ∫0∞TtAfL2ℝn,γ−1tβ−1dt≤fL2ℝn,γ−1Γβ∫0∞e−nttβ−1dt≤fL2ℝn,γ−1nβ,f∈L2ℝn,γ−1.Suppose now thatf∈Cc∞ℝn. By using (40), we obtain that there exists C=Cf>0 such that
(46)TtAfx≤Ce−x2/2e−nt,x∈ℝn,t>0.We can write(47)Sβfx=1Γβ∑k∈ℕnckfH~kx∫0∞e−k+nttβ−1dt=∑k∈ℕnckfk+nβH~kx=A−βfx,x∈ℝn.SinceCc∞ℝn is dense in L2ℝn,γ−1, A−βf=Sβf, f∈L2ℝn,γ−1.According to ([6], Theorem 2.5), we have that, for every f∈Cc∞ℝn(48)A−βfx=∫ℝnMβx,yfydy,for all x outside the support of f, where
(49)Mβx,y=1Γβ∫0∞TtAx,ytβ−1dt,x,y∈ℝn,x≠y.Proposition 7.
Letβ>0. The operator A−β can be extended from Lpℝn,γ−1∩L2ℝn,γ−1 to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself, when 1<p<∞, and from L1ℝn,γ−1 into L1,∞ℝn,γ−1.Proof.
We use the method consisting in decomposing the operatorA−β in two parts called local and global parts. This procedure of decomposition was employed by Muckenhoupt ([21, 22]) in the Gaussian setting.
From now on, we consider the functionm given by mx=min1,1/x2, x∈ℝn\0, and m0=1 and the region N defined by
(50)N=x,y∈ℝn×ℝn:x−y≤nmx.
We decomposeA−β as follows
(51)A−β=Aloc−β+Aglob−β,where Aloc−βfx=A−βfχNx,·x, x∈ℝn.
According to ([1], Lemma 3.3.1), we get
(52)Mβx,y≤C∫0mxe−cx−y2/ttn/2tβ−1dt+∫mx∞e−nt1−e−2tn/2tβ−1dt,x,y∈N.
By choosing0<ε<min2β,n, we obtain
(53)Mβx,y≤C∫0mxe−cx−y2/ttn−ε/2tβ−ε/2−1dt+∫mx∞dttn−ε/2+1≤C1x−yn−ε∫01tβ−ε/2−1dt+mx−n−ε/2≤Cx−yn−ε,x,y∈N.
In the last inequality, we have taken into account thatmx~1/1+x, x∈ℝn, and that 1+xx−y≤C, provided that x,y∈N.
We have that(54)supx∈ℝn∫ℝnMβx,yχNx,ydy≤Csupx∈ℝn∫x−y≤nmxdyx−yn−ε≤Csupx∈ℝn∫0nmxrε−1dr≤Csupx∈ℝnmxε/2<∞.
Also, sincemx~1/1+x~1/1+y~my, x,y∈N,
(55)supy∈ℝn∫ℝnMβx,yχNx,ydx<∞.
Hence, the operatorMβ,loc defined by
(56)Mβ,locfx=∫ℝnMβx,yχNx,yfydy,x∈ℝn,is bounded from Lpℝn,dx into itself, for every 1≤p<∞. Since Mβ,loc is a local operator, by using ([1], Proposition 3.2.5), we deduce that Mβ,loc is bounded from Lpℝn,γ−1 into itself, for every 1≤p<∞.
Suppose thatf,g∈Cc∞ℝn. As above, we take 0<ε<min2β,n and we get
(57)∫ℝnfx∫ℝn∫0∞tβ−1TtAx,yχNx,ygydtdydx≤C∫ℝnfx∫ℝngyχNx,yx−yn−εdydx<∞.
Then,(58)∫ℝnfxAloc−βgxdx=∫ℝnfxSβ,locgxdx=∫ℝnfxMβ,locgxdx.
We deduce thatAloc−βgx=Mβ,locgx, for almost all x∈ℝn. Since Aloc−β and Mβ,loc are bounded from L2ℝn,γ−1 into itself, Aloc−βf=Mβ,locf, f∈L2ℝn,γ−1. It follows that, for every 1≤p<∞, Aloc−β can be extended from L2ℝn,γ−1∩Lpℝn,γ−1 to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself.
We now study the operatorMβ,glob defined by
(59)Mβ,globfx=∫ℝnMβx,yχNcx,yfydy,x∈ℝn.
By making the change of variabless=1−e−2t, t∈0,∞, and taking into account that x−ry2=y−rx2+1−r2x2−y2, x,y∈ℝn, r∈ℝ, we obtain
(60)Mβx,y=π−n/2Γβ∫0∞e−nt1−e−2tn/2e−x−e−ty2/1−e−2ttβ−1dt=π−n/22βΓβey2−x2∫01e−y−x1−s2/ssn/21−sn/2−1−log1−sβ−1ds,x,y∈ℝn.
We now use some notations that were introduced in [23] and proceed as in the proof of ([23], Proposition 2.2). For every x,y∈ℝn, we define
(61)a=x2+y2,b=2x,y,s0=2a2−b2a+a2−b2andus=y−x1−s2s,s∈0,1.
Assume thatx,y∉N. Suppose first that b≤0. In this case, us≥a/s−x2, s∈0,1. Furthermore, s≤C−log1−s≤C1−s−1/4β, s∈0,1. Then,
(62)Mβx,y≤Cey2∫01e−a/ssn/2+11−s3/4ds.
By making the change of variabler=1/s−1, s∈0,1, and taking into account that a≥1/2, we obtain
(63)Mβx,y≤Ce−x2∫0∞e−r/21+rn/2−1/4r3/4dr≤Ce−x2.
Suppose nowb>0. We write u0=us0=y2−x2/2+x+yx−y/2. By using that (see [24], p. 850)
(64)sups∈0,1e−ussn/2~e−u0s0n/2,we have that
(65)Mβx,y≤Cey2−x2sups∈0,1e−ussn/2∫011−sn/2−1−log1−sβ−1ds≤Cey2−x2e−u0s0n/2.
Sinces0~a2−b2/a we conclude that, when x,y∉N,
(66)Mβx,y≤Ce−x2,ifx,y≤0,x+yx−yn/2expy2−x22−x−yx+y2,ifx,y>0.
Let1<q<∞. Since y2−x2≤x+yx−y, x,y∈ℝn, and x−yx+y≥n, when x,y∈Nc and b>0, as in ([25], p. 501), we obtain
(67)∫ℝnex2/q−y2/qMβx,yχNcx,ydy≤C∫ℝne−x21−1/qe−y2/qdy+∫ℝnx+yne−x+yx−y1/2−1/q−1/2dy≤C,x∈ℝn.
Also, we have that(68)supy∈ℝn∫ℝnex2/q−y2/qMβx,yχNcx,ydx<∞.
We deduce thatMβ,glob is bounded from Lpℝn,γ−1 into itself, for every 1<p<∞.
Next, we are going to see thatMβ,glob is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. We decompose Mβx,y, x,y∈ℝn, as follows
(69)Mβx,y=π−n/22βΓβey2−x2∫01/2+∫1/21e−y−x1−s2/ssn/21−sn/2−1−log1−sβ−1ds=I1x,y+I2x,y.
For everyx,y∈ℝn\0, we denote by θx,y∈0,π the angle between x and y (we understand θx,y=0, when n=1). By using ([1], Lemma 3.3.3), we get that, for every x,y∈Nc, x,y≠0,
(70)I1x,y≤Cey2−x2sups∈0,1e−y−x1−s2/s1−sn/2sn/2∫01/2−log1−sβ−11−sds≤Cey2−x2supr∈0,1e−1+ry−1−rx2/r1−rnrn/2≤Cey2−x2min1+xn,xsinθx,y−n.
On the other hand, by proceeding as in ([2], Proposition 5.1), we estimate I2x,y, x,y∈Nc. We first observe that
(71)I2x,y≤Cey2−x2∫1/21e−y−x1−s21−sn/2−1−log1−sβ−1ds,xandy∈ℝn.
Ify≥2x then y−x1−s≥3y/4, s∈1/2,1, and it follows that
(72)I2x,y≤Ce−cy2ey2−x2∫1/211−sn/2−1−log1−sβ−1ds≤Cey2−x2yn−1≤Cey2−x2xn−1,x,y≠0.
Ifx,y∈ℝn\0 we define r0=yx−1cosθx,y and we write y=yx+y⊥, where yx is parallel to x and y⊥ is orthogonal to y.
By making the change of variablesr=1−s and since y−rx2=r−r02x2+y⊥2, it follows that, if y≤2x,
(73)I2x,y≤Ce−cy⊥2ey2−x2∫01/2r−r0n−1+r0n−1−logrβ−1e−cr−r02x2dr≤Ce−cy⊥2ey2−x21xn−1+yxn−1∫01/2−logrβ−1dr≤Ce−cy⊥2ey2−x21xn−1+xyxn−1,x,y∈Nc.
In the last inequality, we have used thatx≥C when x,y∈Nc and 2x≥y.
By combining the above estimates, we obtain(74)Mβx,y≤Cey2−x2min1+xn,xsinθx,y−n+1xn−1+e−cy⊥2xyxn−1χy≤2xx,y,x,y∈Nc.
According to ([1], Lemma 3.3.4) and ([2], Lemmas 4.2 and 4.3), we can see that the operator Mβ,glob is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1.
The estimation (74) allows us to prove that, for every g∈Cc∞ℝn, Aglob−βfx=Mβ,globfx, for almost all x∈ℝn. Then, since Aglob−β and Mβ,glob are bounded from L2ℝn,γ−1 into itself, Aglob−βf=Mβ,globf, f∈L2ℝn,γ−1. Hence, Aglob−β can be extended from L2ℝn,γ−1∩Lpℝn,γ−1 to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself, when 1<p<∞, and from L1ℝn,γ−1 into L1,∞ℝn,γ−1.
Thus, the proof is finished.


## 3. Higher-Order Riesz Transforms Associated with the OperatorA

In this section, we prove Theorems1 and 2 concerning to the higher-order Riesz transforms in the inverse Gaussian setting.Proof of Theorem 1.
Letf∈Cc∞ℝn. If ℓ∈ℕn, we have that
(75)∂xℓTtAx,y=−1ℓe−nte−x−e−ty2/1−e−2tπn/21−e−2tn+ℓ/2Hℓx−e−ty1−e−2t,x,y∈ℝn,t>0,and then
(76)∂xℓTtAx,y≤Ce−nte−cx−e−ty2/1−e−2t1−e−2tn+ℓ/2,x,y∈ℝn,t>0.
Suppose thatk∈ℕ and ℓ∈ℕn such that ℓ<k. Then,
(77)∫ℝnfy∫0∞∂xℓTtAx,ytk/2−1dtdy<∞,x∈ℝn.Indeed, by considering the functionm defined in Proposition 7 and using (76) and ([1], Lemma 3.3.1), we obtain, for ε∈0,1,
(78)∫0∞∂xℓTtAx,ytk/2−1dt≤C∫0mxe−cx−y2/ttn+ℓ−k/2+1dt+∫mx∞e−nt1−e−2tn+ℓ/2tk/2−1dt≤C1x−yn−ε∫01tk−∣ℓ∣−ε/2−1dt+∫mx∞dttn−ε/2+1≤C1x−yn−ε+1mxn−ε/2≤Cx−yn−ε,x,y∈N.On the other hand, by reading the proof of ([1], Lemma 3.3.3), we deduce that
(79)∫0∞∂xℓTtAx,ytk/2−1dt≤Cey2−x2supt∈0,∞e−cy−e−tx2/1−e−2t1−e−2tn+ℓ/2∫0∞e−nttk/2−1dt≤Cey2−x21+xn+ℓ,x,y∈Nc.Note that this estimation also holds whenℓ=k.Sincef∈Cc∞ℝn, (77) holds. Hence, according to ([15], Lemma 4.2), we have that, when k∈ℕ, ℓ∈ℕn and ℓ<k,
(80)∂xℓA−k/2fx=1Γk/2∫ℝnfy∫0∞∂xℓTtAx,ytk/2−1dtdy,foralmostallx∈ℝn.We assumeα=α1,⋯,αn∈ℕn\0, so we can suppose without loss of generality that α1≥1. Let us take ℓ=α1−1,α2,⋯,αn. According to (80), we can write
(81)∂xαA−α/2fx=∂x11Γα/2∫ℝnfy∫0∞∂xℓTtAx,ytα/2−1dtdy,foralmostallx∈ℝn.Assume firstn>1 and write
(82)∫ℝnfy∫0∞∂xℓTtAx,ytα/2−1dtdy=Fx+Gx,x∈ℝn,where
(83)Fx=∫ℝnfy∫0∞∂xℓTtAx,y−Wtx−ytα/2−1dtdy,x∈ℝn,and
(84)Gx=∫ℝnfy∫0∞∂xℓWtx−ytα/2−1dtdy,x∈ℝn.Here,Wt, t>0, denotes the classical heat kernel Wtz=e−z2/2t/2πtn/2, z∈ℝn.Next, we show that(85)∂x1Fx=∫ℝnfy∫0∞∂xαTtAx,y−Wtx−ytα/2−1dtdy,foralmostallx∈ℝn.By taking into account (76), we get
(86)∫mx∞∣∂xαTtAx,y−Wtx−y∣t∣α∣/2−1dt≤C∫mx∞e−ntt∣α∣/2−11−e−2tn+∣α∣/2+e−cx−y2/ttn/2+1dt≤C∫mx∞dttn/2+1≤Cmxn/2≤C1+∣x∣n,x,y∈ℝn.Also by using (79), we obtain
(87)∫0mx∂xαTtAx,y−Wtx−yt∣α∣/2−1dt≤Cey2−x21+xn+∣α∣+∫0mxe−cx−y2/ttn/2+1dt≤Cey2−x21+xn+α+1x−yn+α≤Cey21+xn+α,x,y∈Nc.Now we are going to estimate(88)Ix,y=∫0mx∂xαTtAx,y−Wtx−yt∣α∣/2−1dt,x,y∈N.We can write(89)∂xαTtAx,y−Wtx−y=−1απn/2e−nt1−e−2tn+α/2H~αx−e−ty1−e−2t−12tn+α/2H~αx−y2t=−1απn/2e−nt−11−e−2tn+α/2H~αx−e−ty1−e−2t+11−e−2tn+α/2−12tn+α/2H~αx−e−ty1−e−2t+12tn+α/2H~αx−e−ty1−e−2t−H~αx−y2t=∑j=13Ijt,x,y,x,y∈ℝn.Then(90)Ix,y≤∑j=13∫0mxIjt,x,ytα/2−1dt=∑j=13Ijx,y,x,y∈ℝn.By proceeding as in the proof of ([1], Lemma 3.3.1), we can see that
(91)e−cx−e−ty2/1−e−2t1−e−2tn/2≤Ce−cx−y2/ttn/2,x,y∈N,t∈0,1.Then, by taking into account that(92)e−nt−1≤Ctand11−e−2tn+α/2−12tn+α/2≤Ctn+α/2−1,t∈0,1,it follows by using (91) that
(93)I1x,y+I2x,y≤C∫0mxe−cx−y2/ttn/2dt≤Cx−yn−1/2∫01dtt1/4≤Cx−yn−1/2,x,y∈N.Let us analyze the termI3x,y, x,y∈N. For every z=z1,⋯,zn, w=w1,⋯,wn∈ℝn, we write
(94)H~αz−H~αw=∏j=1nH~αjzj−∏j=1nH~αjwj=∑k=1n∏j=1k−1H~αjwjH~αkzk−H~αkwk∏j=k+1nH~αjzj.Letx,y∈N, t∈0,1, and consider z=x−e−ty/1−e−2t and w=x−y/2t. By taking into account that 1+x+yx−y≤C, it follows that e−cz2≤Ce−cw2 and then, for each k=1,⋯,n,
(95)∏j=1k−1H~αjwj∏j=k+1nH~αjzj≤Cexp−c∑j=1,j≠knxj−yj2t,and, by considering also (1) and using the mean value theorem, we get
(96)H~αkzk−H~αkwk≤Ce−cxk−yk2/tzk−wk≤Ce−cxk−yk2/t1−e−tyk1−e−2t+xk−yk11−e−2t−12t≤Ce−cxk−yk2/ttyk+xk−yk≤Ce−cxk−yk2/ttxk+xk−yk≤Ce−cxk−yk2/tt1+x.Then,(97)H~αx−e−ty1−e−2t−H~αx−y2t≤Ce−cx−y2/tt1+x,x,y∈N,t∈0,1,and thus,
(98)I3x,y≤C1+x∫0mxe−cx−y2/ttn+1/2dt≤C1+xx−yn−1/2∫0mxt−3/4dt=C1+xmx1/4x−yn−1/2≤C1+xx−yn−1/2,x,y∈N.We deduce that(99)Ix,y≤C1+xx−yn−1/2,x,y∈N.This estimation, jointly with (86) and (87), leads to
(100)∫ℝnfy∫0∞∂xαTtAx,y−Wtx−yt∣α∣/2−1dtdy≤C∫ℝnfyey21+xn+α+1+xx−yn−1/2dy≤C1+xn+α<∞,x∈ℝn,where we have used thatf has compact support. According to ([15], Lemma 4.2), (85) is then established.We are going to evaluate∂x1Gx. We write Gx=∫ℝnfx−yΦydy, x∈ℝn, where
(101)Φz=∫0∞∂zℓWtzt∣α∣/2−1dt,z∈ℝn.Sincen>1, we have that
(102)Φz≤C∫0∞e−cz2/ttn+1/2dt≤Czn−1,z∈ℝn\0.According to ([15], Lemma 4.2), we can derivate under the integral sign obtaining
(103)∂x1Gx=∫ℝn∂x1fx−yΦydy=−∫ℝn∂y1fx−yΦydy,foralmostallx∈ℝn,where the last integral is absolutely convergent.For everyz=z1,⋯,zn∈ℝn, we define z¯=z2,⋯,zn. Partial integration leads to
(104)∫ℝn∂y1fx−yΦydy=limε⟶0+∫y>ε∂y1fx−yΦydy=limε⟶0+∫y¯<ε∫−∞−ε2−y¯2+∫y¯<ε∫ε2−y¯2+∞+∫y¯>ε∫ℝ∂y1fx−yΦydy1dy¯=limε⟶0+−∫y>εfx−y∂y1Φydy+∫y¯<εfx−yΦyy1=−∞y1=−ε2−y¯2dy¯+∫y¯<εfx−yΦyy1=ε2−y¯2y1=+∞dy¯=limε⟶0+−∫y¯>εfx−y∂y1Φydy+Jεx,x∈ℝn,where(105)Jεx=∫y¯<εfx1+ε2−y¯2,x¯−y¯Φ−ε2−y¯2,y¯dy¯−∫y¯<εfx1−ε2−y¯2,x¯−y¯Φε2−y¯2,y¯dy¯,x∈ℝn.Let us estimatelimε⟶0+Jεx, x∈ℝn. We recall that Φ can be written as follows
(106)Φz=−1α−12ℓ/22πn/2∫0∞H~ℓz2tdttn+1/2,z∈ℝn.Suppose now thatα1 is odd. Then,
(107)Φ−ε2−y¯2,y¯=Φε2−y¯2,y¯,y¯∈ℝn−1,y¯<ε.We have that, for everyx∈ℝn and ε>0,
(108)Jεx=∫y¯<εfx1+ε2−y¯2,x¯−y¯−fx1−ε2−y¯2,x¯−y¯Φε2−y¯2,y¯dy¯=εn−1∫z¯<1fx1+ε1−z¯2,x¯−εz¯−fx1−ε1−z¯2,x¯−εz¯Φε1−z¯2,εz¯dz¯.On the other hand, by performing the change of variables=ε2/2t, we get
(109)Φε1−z¯2,εz¯=−1α−12ℓ/22πn/2∫0∞H~α1−1ε1−z¯22t∏i=2nH~αiεzi2tdttn+1/2=−1α−1ε1−n2α/2πn/2∫0∞H~α1−1s1−z¯2∏i=2nH~ℓizissn−3/2ds,ε>0,z¯∈ℝn−1,z<1.It follows that(110)Jεx=−1α−12α/2πn/2∫z¯<1fx1+ε1−z¯2,x¯−εz¯−fx1−ε1−z¯2,x¯−εz¯×∫0∞Hα1−1s1−z¯2∏i=2nHℓizise−ssn−3/2dsdz¯,x∈ℝn,ε>0.Then, by using the dominated convergence theorem, we obtain(111)limε⟶0+Jεx=0,x∈ℝn.Suppose now thatα1 is even. Then,
(112)Φ−ε1−y¯2,y¯=−Φε1−y¯2,y¯,ε>0,y¯∈ℝn−1,y¯<ε,and proceeding as above, we get
(113)Jεx=−1α2α/2πn/2∫z¯<1fx1+ε1−z¯2,x¯−εz¯+fx1−ε1−z¯2,x¯−εz¯×∫0∞Hα1−1s1−z¯2∏i=2nHαizise−ssn−3/2dsdz¯,x∈ℝn,ε>0.It follows that(114)limε⟶0+Jεx=−1α2α/2−1πn/2fx∫z¯<1∫0∞Hα1−1s1−z2∏i=2n−1Hαizise−ssn−3/2dsdz¯=−cαfx,x∈ℝn.We note that ifαi is odd for some i=2,⋯,n, then cα=0. Thus, we conclude that
(115)∂x1Gx=limε⟶0+∫x−y>εfy∂x1Φx−ydy+cαfx=limε⟶0+∫x−y>εfy∫0∞∂xαWtx−ytα/2−1dtdy+cαfx,foralmostallx∈ℝn,wherecα=0 when αi is odd for some i=1,⋯,n.By putting together (85) and (115), we obtain
(116)∂xαA−α/2fx=1Γα/2∂x1Fx+Gx=1Γα/2∫ℝnfy∫0∞∂xαTtAx,y−Wtx−ytα/2−1dtdy+limε⟶0+∫x−y>εfy∫0∞∂xαWtx−ytα/2−1dtdy+cαfx=1Γα/2limε⟶0+∫x−y>εfy∫0∞∂xαTtAx,ytα/2−1dtdy+cαfx,foralmostallx∈ℝn.We now deal with the casen=1. We have α∈ℕ, α≥1. According to (80), we can write
(117)dαdxαA−α/2fx=1Γα/2ddx∫ℝnfy∫0∞∂xα−1TtAx,ytα/2−1dtdy=1Γα/2ddxF¯x+G¯x,x∈ℝ,where
(118)F¯x=∫ℝfy∫0∞∂xα−1TtAx,y−dα−1dxα−1Wtx−y−dα−1dxα−1Wt0χ1,∞ttα/2−1dtdy,x∈ℝ,and
(119)G¯x=∫ℝfy∫0∞dα−1dxα−1Wtx−y−dα−1dxα−1Wt0χ1,∞ttα/2−1dtdy,x∈ℝ.Note that ifα is even, then dα−1/dxα−1Wt0=0.By proceeding as in the case ofn>1 and taking into account that
(120)dα−1dxα−1Wtx−y−dα−1dxα−1Wt0≤Ctα/2H~α−1x−y2t−H~α−10≤Cx−ytα+1/2,x,y∈ℝ,t>1,we can see that the integral defining F¯x is absolutely convergent for every x∈ℝ and
(121)ddxF¯x=∫ℝfy∫0∞∂xαTtAx,y−Wtx−ytα/2−1dtdy,being also this integral absolutely convergent for almost x∈ℝ.On the other hand, by considering(122)Φ¯y=∫0∞dα−1dxα−1Wty−dα−1dxα−1Wt0χ1,∞ttα/2−1dt=−1α−12α/2π∫0∞H~α−1y2t−H~α−10χ1,∞tdtt,y∈ℝ\0,we can write
(123)G¯x=∫ℝfx−yΦ¯ydy,x∈ℝn,and according to ([15], Lemma 4.2), we get
(124)ddxG¯x=−∫ℝ∂yfx−yΦ¯ydy=−limε⟶0+∫y>ε∂yfx−yΦ¯ydy,foralmostallx∈ℝ.By partial integration, we obtain(125)∫y>ε∂yfx−yΦ¯ydy=−∫y>εfx−yΦ¯′ydy+fx+εΦ¯−ε−fx−εΦ¯ε,ε>0,x∈ℝn.Then(126)ddxG¯x=limε⟶0+∫y>εfx−yΦ¯′ydy+J¯εx,foralmostallx∈ℝ,where
(127)J¯εx=fx−εΦ¯ε−fx+εΦ¯−ε,ε>0,x∈ℝ.By taking into account (10) and that H~αz≤Ce−cz2, z∈ℝ, it follows that
(128)Φ¯′y=−1α2α+1/2π∫0∞H~αy2tt−3/2dt=∫0∞dαdxαWtyta/2−1dt,y∈ℝ\0.On the other hand, we have that(129)Φ¯y=−1α−12a/2π∫01H~α−1y2tdtt+∫1∞H~α−1y2t−H~α−10dtt=−1α−12α/2π∫y22∞H~α−1sdss+∫0y2/2H~α−1s−H~α−10dss,y>0,and,
(130)Φ¯y=12α/2π∫y2/2∞H~α−1sdss+∫0y2/2H~α−1s−H~α−10dss,y<0.Letx∈ℝ. Since f∈Cc∞ℝn, we get
(131)fx+ε−fx−ε≤Cε,ε>0.Ifα is odd, then Φ¯ is even and, for every ε∈0,1,
(132)J¯εx=fx+ε−fx−εΦ¯ε≤Cε∫ε2/2∞H~α−1sdss+∫0ε2/2H~α−1s−H~α−10dss≤Cε∫1∞e−sds+∫ε2/21dss+∫0ε2/2sdss≤Cε1+logε+ε.Hence, ifα is odd, limε⟶0+J¯εx=0.Ifα is even, then Φ¯ is odd, Hα−10=0, and
(133)Φ¯ε=−12α/2π∫0∞H~α−1sdss,ε>0.We obtain(134)limε⟶0+J¯εx=−fx2α/2−1π∫0∞H~α−1sdss,provided that α is even.We conclude that, for certaincα∈ℝ(135)ddxGx=limε⟶0+∫x−y>εfy∫0∞dαdxαWtx−ytα/2−1dtdy+cαfx,foralmostallx∈ℝ.Then, we get(136)dαdxαA−α/2fx=1Γα/2limε⟶0+∫x−y>εfy∫0∞∂xαTtAx,ytα/2−1dtdy+cαfx,foralmostallx∈ℝ,where cα=0 when α is odd.Finally, we observe that whenα is even
(137)∫0∞dαdzαWtztα/2−1dt=12α+1/2π∫0∞e−z2/2tHαz2tt−3/2dt=12α/2−1π1z∫0∞e−s2Hαsds=0.By taking into account the arguments developed in this proof, we can see that(138)∫ℝfy∫0∞∂xαTtAx,ytα/2−1dtdy=∫ℝfy∫0∞∂xαTtAx,y−Wtx−ytα/2−1dtdy<∞,for every x∈ℝ. Then,
(139)dαdxαA−α/2fx=1Γα/2∫ℝfy∫0∞∂xαTtAx,ytα/2−1dtdy+cαfx,foralmostallx∈ℝ.The proof of Theorem1 is completed.Proof of Theorem 2.
For everyf∈Cc∞ℝn, we have that
(140)A−α/2fx=∑k∈ℕnckfk+nα/2H~kx,x∈ℝn,and according to (36), (37), and (40), the last series is pointwise absolutely convergent, it defines a smooth function on ℝn and
(141)∂xαA−α/2fx=∑k∈ℕnckfk+nα/2∂xαH~kx=−1α∑k∈ℕnckfk+nα/2H~k+αx,x∈ℝn.
Then, according to Theorem1, for each f∈Cc∞ℝn,
(142)Rαfx=limε⟶0+∫x−y>εRαx,yfydy+cαfx,foralmostallx∈ℝn.
Here,cα=0 when αi is odd for some i=1,⋯,n.
To establish our result, it is sufficient to show that for everyf∈Lpℝn,γ−1 the limit
(143)limε⟶0+∫x−y>εRαx,yfydy,exists for almost all x∈ℝn and the operator Lα defined by
(144)Lαfx=limε⟶0+∫x−y>εRαx,yfydy+cαfx,
is bounded fromLpℝn,γ−1 into itself. Thus, Lα is the unique extension of Rα from L2ℝn,γ−1∩Lpℝn,γ−1 to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself.
For everyβ>0 we define the set
(145)Nβ=x,y∈ℝn×ℝn:x−y≤βnmin1,1x.
Observe thatN1=N and that if β>0, a∈0,1 and x,y∈Nβc, then
(146)ax−ay≥aβnmin1,1x≥a2βnmin1,1ax,that is, ax,ay∈Na2βc. In particular, we have that if a∈0,1 then ax,ay∈Nc provided that x,y∈N1/a2c.
Letβ>0. We consider the operators Rα,loc and Rα,glob defined on Cc∞ℝn by
(147)Rα,locfx=RαfχNβx,·x,Rα,globfx=RαfχNβcx,·x,x∈ℝn.We recall that(148)Rαx,y=−1απn/2Γα/2∫0∞e−nt1−e−2tn+α/2H~ℓx−e−ty1−e−2ttα/2−1dt,x,y∈ℝn,x≠y.Then,(149)Rαx,y≤C∫0∞e−nte−ηx−e−ty2/1−e−2t1−e−2tn+α/2tα/2−1dt,x,y∈ℝn,x≠y,for everyη∈0,1. From now on we consider 1/p<η<1 and β=1/η. Since x−ry2=y−rx2+1−r2x2−y2, x,y∈ℝn, r∈ℝ, by making the change of variables s=1−e−2t in the last integral, we get
(150)Rαx,y≤Ceηy2−x2∫01e−ηy−x1−s2/ssn+α/21−sn/2−1−log1−sα/2−1ds,x,y∈ℝn,x≠y.Assume first thatx,y∈Nβc. Then, ηx,ηy∈Nc. By proceeding as in the proof of (66), it follows that, when x,y≤0,
(151)Rαx,y≤Ce−ηx2∫0∞e−r/21+rn+α/2−1/4r3/4dr≤Ce−ηx2.Suppose now thatx,y>0. Again, as in the estimation in (66), since ηx,ηy∈Nc, we obtain
(152)∫1/21e−ηy−x1−s2/ssn+α/21−sn/2−1−log1−sα/2−1ds≤Cx+yx−yn/2expη2y2−x2−x+yx−y.On the other hand, proceeding as in ([23], p. 862) and considering the notation in (61) and ([23], Lemma 2.3), we have
(153)∫01/2e−ηy−x1−s2/ssn+α/21−sn/2−1−log1−s∣α∣/2−1ds≤Csups∈0,1e−ηussn/21−1/n∫01/2e−ηus/ns−log1−s∣α∣/2−1s∣α∣/2ds≤Ce−ηu0s0n/21−1/n∫01/2e−us/ns3/2ds≤Ce−ηu0s0n/21−1/n∫01e−us/ns3/21−sds≤Ce−ηu0s0n/21−1/ne−ηu0/ns0≤Ce−ηu0s0n/2≤Cx+yx−yn/2expη2y2−x2−x+yx−y.From the above estimates, we conclude that, whenx,y∈Nβc,
(154)Rαx,y≤Ce−ηx2,x,y≤0,x+yx−yn/2expηy2−x22−x−yx+y2,x,y>0.On the other hand, we consider the kernel(155)ℝαx,y=1Γα/2∫0∞∂xαWtx−ytα/2−1dt,x,y∈ℝn,x≠y.Let us show that(156)Rαx,y−ℝαx,y≤C1+xx−yn−1/2,x,y∈Nβ,x≠y.For everyx∈ℝn, we have that
(157)Rαx,y−ℝαx,y=1Γα/2∫0mx+∫mx∞∂xαTtAx,y−Wtx−ytα/2−1dt=Ix,y+Jx,y,x,y∈ℝn,x≠y.The same proof of (99) allows us to obtain that
(158)Ix,y≤C1+xx−yn−1/2,x,y∈Nβ,x≠y.Also, we get(159)Jx,y≤C∫mx∞e−nt1−e−2tn+α/2H~αx−ye−t1−e−2ttα/2−1dt+∫mx∞H~αx−y2tdttn/2+1≤C∫mx∞e−nte−cx−e−ty2/1−e−2t1−e−2tn+α/2tα/2−1dt+∫mx∞e−cx−y2/2ttn/2+1dt≤C∫mx∞dttn/2+1=Cmxn/2≤C1+xn≤C1+xx−yn−1/2,x,y∈Nβ,x≠y,and thus, (156) is established.From (154), we obtain that Rα,glob can be extended to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself, and the extension is given by (147). Indeed, when x,y∈Nβc, we have that x−yx+y≥C. By taking into account also that y2−x2≤x+yx−y, we get
(160)∫ℝnRαx,yex2−y2/pχNβcx,ydy≤C∫ℝne−η−1/px2−y2/pdy+∫ℝnx+ynexpη2−1py2−x2−η2x−yx+ydy≤Ce−η−1/px2∫ℝne−y2/pdy+∫ℝnx+ynexp−x−yx+yη2−1p−η2dy,x∈ℝn.Sinceη>1/p it follows that
(161)supx∈ℝn∫ℝnRαx,yex2−y2/pχNβcx,ydy<∞,and in a similar way
(162)supy∈ℝn∫ℝnRαx,yex2−y2/pχNβcx,ydx<∞,from which we deduce that Rα,glob is bounded on Lpℝn,γ−1.On the other hand, by using (156) and that mx~1/1+x, x∈ℝn, we have that
(163)∫ℝnRαx,y−ℝαx,yχNβx,ydy≤C1+x∫ℝn1x−yn−1/2χNβx,ydy=C1+x∫0βnmxr−1/2dr=C1+xmx≤C,x∈ℝn.Also, sincemx~my, x,y∈Nβ, we get
(164)supy∈ℝn∫ℝnRαx,y−ℝαx,yχNβx,ydx<∞.Hence, the operatorSα defined by
(165)Sαfx=∫ℝnRαx,y−ℝαx,yχNβx,yfydy,x∈ℝn,is a bounded operator from Lpℝn,dx into itself. Since Sα is a local operator, by ([1], Proposition 3.2.5) Sα is bounded from Lpℝn,γ−1 into itself.We now observe that the kernelℝα is a standard Calderón-Zygmund kernel. Indeed, we get
(166)ℝαx,y≤C∫0∞Hαx−y2te−x−y2/2ttn/2+1dt≤C∫0∞e−cx−y2/ttn/2+1dt≤Cx−yn,x,y∈ℝn,x≠y.Leti=1,⋯,n and denote αi=α1,⋯,αi+1,⋯,αn. We have that
(167)∂xiℝαx,y=−1α+12πn/22α+1/2Γα/2∫0∞Hαix−y2te−x−y2/2ttn+3/2dt,x,y∈ℝn,x≠y.Then,(168)∂xiℝαx,y≤Cx−yn+1,x,y∈ℝn,x≠y.The Euclideanα-order Riesz transform ℝα is bounded from Lqℝn,dx into itself, for every 1<q<∞. According to ([1], Proposition 3.2.5), the operator ℝα,loc defined by
(169)ℝα,locfx=ℝαfχNβx,·x,is bounded from Lqℝn,γ−1 into itself, for every 1<q<∞.We can writeRα,loc=Sαf+ℝα,locf on Cc∞ℝn. Then, Rα,loc can be extended from Cc∞ℝn to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself.SinceRα=Rα,loc+Rα,glob, we conclude that Rα can be extended from Cc∞ℝn to Lpℝn,γ−1 as a bounded operator from Lpℝn,γ−1 into itself.Let us consider the maximal operator(170)Rα,∗fx=supε>0∫x−y>εRαx,yfydy,f∈Lpℝn,γ−1,1<p<∞.Letf∈Lpℝn,γ−1, 1<p<∞. For every ε>0, by using the above estimates, we can write
(171)∫x−y>εRαx,yfydy≤∫x−y>εRαx,y−ℝαx,yχNβx,yfydy+∫x−y>εRαx,yχNβcx,yfydy+∫x−y>εℝαx,yχNβx,yfydy<∞,x∈ℝn.We also have that(172)Rα,∗fx≤C∫ℝn1+xx−yn−1/2χNβx,yfydy+∫ℝnRαx,yχNβcx,yfydy+supε>0∫x−y>εℝαx,yχNβx,yfydy,x∈ℝn.Since the maximal operator(173)ℝα,∗fx=supε>0∫x−y>εℝαx,yfydy,is bounded from Lpℝn,dx into itself, by using a vector-valued version of ([1], Proposition 3.2.5) (see ([16], Proposition 2.3)), we deduce that the local maximal operator
(174)ℝα,loc,∗fx=supε>0∫x−y>εℝαx,yχNβx,yfydy,is bounded from Lpℝn,γ−1 into itself.By using the same arguments as above, we conclude thatRα,∗ is bounded from Lpℝn,γ−1 into itself.From (142) and since Cc∞ℝn is dense in Lpℝn,γ−1 and ℝα,∗ is bounded in Lpℝn,γ−1, by using a standard procedure, we can conclude that the limit
(175)limε⟶0+∫x−y>εRαx,yfydy,exists for almost all x∈ℝn and Lα (defined in (144)) is a bounded operator from Lpℝn,γ−1 into itself.Remark 10.
TheLpℝn,γ−1-boundedness of the local part Rα,loc of Rα can be proved also by using Calderón-Zygmund theory. We have preferred to do it by comparing Rα,loc with the classical local Riesz transform ℝα,loc because in this way we can know how the singularity of Rα,loc is. Furthermore, these comparative results will be useful in the proof of Theorems 1.4 and 1.5.


## 4. Riesz Transform Associated with the OperatorA¯

Our objective in this section is to prove Theorem3.We defineL02ℝn,γ−1 as the space that consists of all those f∈L2ℝn,γ−1 such that c0f=∫ℝnfxdx=0. Let β>0. For every f∈L02ℝn,γ−1, A¯−βf is defined by
(176)A¯−βf=∑k∈ℕn\0ckfkβH~k.We have that(177)A¯−βfL2ℝn,γ−12=∑k∈ℕn\0ckf2H~kL2ℝn,γ−12k2β≤fL2ℝn,γ−12,f∈L02ℝn,γ−1.We introduce the operatorS¯β defined by
(178)S¯βf=1Γβ∫0∞TtA¯f−c0fe·2tβ−1dt,f∈L2ℝn,γ−1.Letf∈L2ℝn,γ−1. We have that
(179)TtA¯f−c0fe−·2=∑k∈ℕne−ktckfH~k−e·2c0f=∑k∈ℕn\0e−ktckfH~k,t>0.Then,(180)TtA¯f−c0fe−·2L2ℝn,γ−1≤e−tfL2ℝn,γ−1,t>0.Hence, the integral definingS¯βf converges in the L2ℝn,γ−1-Bochner sense.Letf∈Cc∞ℝn. According to (36), (37), and (40), we get
(181)∑k∈ℕn\0e−ktckfH~kx≤Ce−te−x2/2∑k∈ℕn\01k2≤Ce−te−x2/2,t>0,x∈ℝn.It follows that the series that definesTtA¯fx−c0fe−x2 converges pointwisely and absolutely and
(182)S¯βfx=1Γβ∑k∈ℕn\0ckfH~kx∫0∞e−kttβ−1dt=∑k∈ℕn\0ckfkβH~kx=A¯−βf0x,x∈ℝn,where f0x=fx−c0fe−x2, x∈ℝn.On the other hand, sincesuppf is compact, we have that
(183)∫mx∞∫ℝnTtA¯x,y−e−x2fydytβ−1dt≤C∫mx∞∫ℝne−cx−e−ty2/1−e−2t1−e−2t−n/2−1fydy+∫ℝne−cx−e−ty2/1−e−2t−e−x−e−ty2fydy+∫ℝne−x−e−ty2−e−x2fydytβ−1dt≤C∫0∞e−ttβ−1dt,x∈ℝn,and, taking 0<ε<min2β,n,
(184)∫0mxtβ−1∫ℝnTtA¯x,y−e−x2fydydt≤C∫0mxtβ−1∫ℝnecxe−cx−y2/ttn/2+e−x2fydydt≤Cecx∫ℝnfyx−yn−εdy+mxβe−x2∫ℝnfydy≤Cecx+mxβe−x2,x∈ℝn.Then, we obtain(185)S¯βfx=∫ℝnK¯βx,yfydy,x∈ℝn,where
(186)K¯βx,y=1Γβ∫0∞TtA¯x,y−e−x2tβ−1dt,x,y∈ℝn,and
(187)TtA¯x,y=entTtAx,y=ey2−x2πn/21−e−2tn/2exp−y−e−tx21−e−2t,x,y∈ℝn,t>0.By denotingΠ0 the projection from L2ℝn,γ−1 to L02ℝn,γ−1, we have proved that, for every f∈Cc∞ℝn,
(188)A¯−βΠ0fx=∫ℝnK¯βx,yfydy,x∈ℝn.Letf∈Cc∞ℝn. Next, we show that
(189)δxαA¯−α/2Π0fx=limε⟶0+1Γα/2∫x−y>εfy∫0∞δxαTtA¯x,ytα/2−1dtdy+cαfx,for almost all x∈ℝn. Here, cα∈ℝ and, when αi is odd for some i=1,⋯,n, cα=0.For everyℓ∈ℕn,
(190)δxℓTtA¯x,y=−1ℓe−x22ℓ∂xℓex2TtA¯x,y=−1ℓey2−x22ℓπn/21−e−2tn+ℓ/2e−ℓtH~ℓy−e−tx1−e−2t,x,y∈ℝn,t>0,Then, for eachℓ∈ℕn,
(191)δxℓTtA¯x,y≤Cey2−x2e−ℓte−cy−e−tx2/1−e−2t1−e−2tn+ℓ/2,x,y∈ℝn,t>0.Sinceδxℓe−x2=0, x∈ℝn, when ℓ∈ℕn\0, by proceeding as in the proof of (80), we get that for every ℓ∈ℕn\0 and k∈ℕ being ℓ<k,
(192)δxℓA¯−k/2Π0fx=1Γk/2∫ℝnfy∫0∞δxℓTtA¯x,ytk/2−1dtdy,foralmostallx∈ℝn.Without loss of generality we can assume thatα1≥1 and consider ℓ=α1−1,α2,⋯,αn. When ℓ∈ℕn\0, we can proceed as in the proof of Theorem 1. For n>1, we write
(193)δxαA¯−α/2Π0fx=1Γα/2δx1∫ℝnfy∫0∞δxℓTtA¯x,y−ey2−x2Wty−xtα/2−1dtdy+∫ℝnfy∫0∞δxℓey2−x2Wty−xtα/2−1dtdy=1Γα/2δx1Fx+Gx,x∈ℝn.We observe that, for everyr∈ℕn,
(194)δxrey2−x2Wty−x=−1r2rey2−x2∂xrWty−x=−1r2rπn/2ey2−x22tn+r/2H~ry−x2t,x,y∈ℝn,t>0.By considering the decomposition(195)δxαTtA¯x,y−ey2−x2Wty−x=−1∣α∣2απn/2ey2−x2e−αt1−e−2tn+α/2H~ℓy−e−tx1−e−2t−12tn+α/2H~αy−x2t=−1α2απn/2ey2−x2e−αt−11−e−2tn+α/2H~αy−e−tx1−e−2t+11−e−2tn+α/2−12tn+α/2H~αy−e−tx1−e−2t+12tn+α/2H~αy−e−tx1−e−2t−H~αy−x2t,x,y∈ℝn.We can argue as in the proof of (85) to obtain that
(196)δx1Fx=∫ℝnfy∫0∞δxαTtA¯x,y−ey2−x2Wty−xtα/2−1dtdy,foralmostallx∈ℝn.On the other hand, to deal withδx1Gx, we consider gx=fxex2 and Ψx=∫0∞∂xℓWt−xt∣α∣/2−1dt, x∈ℝn, and write
(197)Gx=e−x22ℓ∫ℝngx−yΨydy,x∈ℝn.We proceed as in the proof of (115) to get
(198)δx1Gx=limε⟶0+∫x−y>εfy∫0∞δxαey2−x2Wtx−yt∣α∣/2−1dtdy+cαfx,foralmostallx∈ℝn,where cα∈ℝ and cα=0 if αi is odd for some i=1,⋯,n. Thus, (189) is established when n>1.Ifn=1, we can also follow the proof of Theorem 1 by using the decomposition
(199)δxαA¯−a/2Π0fx=1Γα/2δxF¯x+G¯x,x∈ℝn,where
(200)F¯x=∫ℝnfy∫0∞δxℓTtA¯x,y−δxℓey2−x2Wty−x−12ℓdℓdxℓWt0χ1,∞ttα/2−1dtdy,x∈ℝn,and
(201)G¯x=∫ℝnfy∫0∞δxℓey2−x2Wty−x−12ℓdℓdxℓWt0χ1,∞ttα/2−1dtdy,x∈ℝn.Whenℓ=0, that is, α=1,0,⋯,0, we can replace F in (193) and F¯ in (199) by
(202)Fx=∫ℝnfy∫0∞TtA¯x,y−e−x2−ey2−x2Wty−xdtdyt,x∈ℝn,and
(203)F¯x=∫ℝnfy∫0∞TtA¯x,y−e−x2−ey2−x2Wty−x−12ℓdℓdxℓWt0χ1,∞tdtdyt,x∈ℝn,respectively, and proceed as above to obtain (189).According to (36), (37), and (40), we get
(204)R¯αfx=δαA−α/2Π0fx,x∈ℝn,and then,
(205)R¯αfx=limε⟶0+1Γα/2∫x−y>εfy∫0∞δxαTtA¯x,ytα/2−1dtdy+cαfx,foralmostallx∈ℝn,with cα=0, when αi is odd for some i=1,⋯,n.We are going to show theLpℝn,γ−1-boundedness properties of R¯α. We recall that
(206)R¯αx,y=−1α2απn/2Γα/2ey2−x2∫0∞e−αt1−e−2tn+α/2H~αy−e−tx1−e−2tt∣α∣/2−1dt,x,y∈ℝn,x≠y.Consider first1<p<∞ and choose 1−1/p<η<1. By making the change of variables s=1−e−2t, t∈0,∞, we obtain
(207)R¯αx,y≤Cey2−x2∫01e−ηy−x1−s2/ssn+α/21−s∣α∣/2−1−log1−s∣α∣/2−1ds,x,y∈ℝn,x≠y.Letβ=η−1 and consider the local and global operators defined on Cc∞ℝn by
(208)R¯α,locfx=R¯αfχNβx,·x,andR¯α,globfx=R¯αfχNβcx,·x,x∈ℝn.By proceeding as in the proof of (154) it follows that, for each x,y∈Nβc,
(209)R¯αx,y≤Ce1−ηy2−x2,x,y≤0,x+yx−ynexp1−η2y2−x2−η2x+yx−y,x,y>0.We have that(210)∫ℝne−y2/p+x2/pR¯αx,yχNβcx,ydy≤C∫ℝney21−η−1/pe1/p−1x2dy+∫ℝnx+ynexp−x+yx−yη2−1−1p−η2dy,x∈ℝn.Then, sinceη>1−1/p,
(211)supx∈ℝn∫ℝne−y2/p+x2/pR¯αx,yχNβcx,ydy<∞.Also, we get(212)supy∈ℝn∫ℝne−y2/p+x2/pR¯αx,yχNβcx,ydx<∞.We conclude thatR¯α,glob is bounded from Lpℝn,γ−1 into itself.We are going to study the operatorR¯α,loc. We write
(213)R¯αx,y=1Γα/2∫0∞δxαTtA¯x,y−ey2−x2Wty−xtα/2−1dt+∫0∞δxαey2−x2Wty−xtα/2−1dt=Ix,y+Jx,y,x,y∈ℝn,x≠y.By taking into account that(214)δxαTtA¯x,y−ey2−x2Wty−x≤Cey2−x2e−αt1−e−2tn+α/2+e−cx−y2/ttn+α/2,x,y∈ℝn,t>0,we get (see (86))
(215)∫mx∞δxαTtA¯x,y−ey2−x2Wty−xtα/2−1dt≤Cey2−x21+xn,x,y∈ℝn.Also, from (195) and proceeding as in the proof of (99), we can see that
(216)∫0mxδxαTtA¯x,y−ey2−x2Wty−xtα/2−1dt≤Cey2−x21+xx−yn−1/2,x,y∈Nβ.Sincey2−x2≤C when x,y∈Nβ, we obtain that
(217)Ix,y≤C1+xx−yn−1/2,x,y∈Nβ,x≠y.On the other hand, we observe that(218)Jx,y=ey2−x22αΓα/2∫0∞∂yαWty−xtα/2−1dt=2−αey2−x2ℝαy,x,x,y∈ℝn,x≠y,where ℝα·,· is the classical kernel considered in (155). The result can be established by proceeding as in the proof of Theorem 2 by taking into account that the Euclidean Riesz transform ℝα is bounded from Lpℝn,dx into itself.To deal with the casep=1, we consider the local and global operators R¯α,loc and R¯α,glob defined above with N instead of Nβ. Since the classical Riesz transform ℝα is bounded from L1ℝn,dx into L1,∞ℝn,dx, we can use (217) and argue as in the proof of Theorem 2 to obtain that R¯α,loc defines a bounded operator from L1ℝn,γ−1 into L1,∞ℝn,γ−1.In order to prove thatR¯α,glob defines a bounded operator from L1ℝn,γ−1 into itself, we make the change of variables r=e−t, t∈0,∞, and write R¯αx,y=R¯α,1x,y+R¯α,2x,y, where
(219)R¯α,1x,y=−1α2απn/2Γα/2ey2−x2∫012rα−11−r2n+α/2H~αy−rx1−r2−logrα/2−1dr,x,y∈ℝn,x≠y.Suppose thatn=1 or α≥n+1 when n>1. By using ([1], Lemma 3.3.3), it follows that
(220)R¯α,1x,y≤Cey2−x2∫01/2rα−1e−cy−rx2/1−r21−r2n/2−logr∣α∣/2−1dr≤Cey2−x2supr∈0,1rne−cy−rx2/1−r21−r2n/2∫01−logr∣α∣/2−1dr≤Cey2−x2min1+xn,xsinθx,y−n,x,y∈Nc.On the other hand, by proceeding as in the estimation ofK20x,y in ([2], proof of Proposition 5.1), we obtain, for every x,y∈Nc,
(221)R¯α,2x,y≤Cey2−x2∫1/21e−cy−rx2/1−r21−r2n+2/2dr≤Cey2−x2x−n+min1+xn,xsinθx,y−n.From ([1], Lemma 3.3.4) and ([2], Lemma 4.2), we deduce that R¯α,glob defines a bounded operator from L1ℝn,γ−1 into L1,∞ℝn,γ−1.Thus, we conclude thatRα can be extended to L1ℝn,γ−1 as a bounded operator from L1ℝn,γ−1 into L1,∞ℝn,γ−1.


## 5. UMD Spaces and Riesz Transforms in the Inverse Gaussian Setting

Proof of Theorem 4.
For everyi=1,⋯,n, by ℝei, we denote the i-th Euclidean Riesz transform defined, for every f∈Lpℝn,dx, 1<p<∞, by
(222)ℝeifx=limε⟶0+∫x−y>εℝeix−yfydy,foralmostallx∈ℝn,where
(223)ℝeiz=1π∫0∞∂xiWtzdtt,z∈ℝn,z≠0.
Observe that(224)ℝeiz=−Γn+1/2πn+1/2zizn+1,z∈ℝn,z≠0,i=1,⋯,n.LetX be a Banach space. For every i=1,⋯,n, we define ℝei on Lpℝn,dx⊗X, 1≤p<∞, in the obvious way.The UMD-property forX can be characterized by using ℝei, i=1,⋯,n. The properties stated in Theorems 4 and 5 hold when Rei is replaced by ℝei, i=1,⋯,n. The estimations established in the proofs of Theorems 1 and 2 allow us to pass from ℝei to Rei, i=1,⋯,n.Leti=1,⋯,n and 1<p<∞. We are going to see that the following two assertions are equivalent:
(i)
Rei can be extended from L2ℝn,γ−1⋂Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself(ii)
ℝei can be extended from (L2ℝn,dx⋂Lpℝn,dx⊗X to Lpℝn,dx,X as a bounded operator from Lpℝn,dx,X into itselfWe choose1/p<η<1 and consider the global and local operators as in the previous sections according to the region Nβ, with β=η−1.Suppose that (ii) holds. We can writeRei=Rei,loc−ℝei,loc+ℝei,loc+Rei,glob. Since ℝei is a Calderón-Zygmund operator, by using a vectorial version of ([1], Proposition 3.2.5) (see ([16], Proposition 2.3)), we deduce that ℝei,loc can be extended from L2ℝn,γ−1∩Lpℝn,γ−1⊗X to Lpℝn,γ−1,X as a bounded operator from Lpℝn,γ−1,X into itself.According to (154) and (156), we have that
(225)Reix,y−ℝeix−y≤Lix,y,x,y∈Nβ,andReix,y≤Mix,y,x,y∈Nβc,and the integral operators(226)Lifx=∫ℝnLix,yχNβx,yfydy,andMifx=∫ℝnMix,yχNβcx,yfydy,are bounded from Lpℝn,γ−1 into itself. Then, Li and Mi define bounded operators from Lpℝn,γ−1,X into itself, and the same property holds for the operators Rei,loc−ℝei,loc and Rei,glob. We conclude that (i) holds.Suppose now that (i) holds. By (149), we get
(227)Reix,y≤C∫0∞e−nte−cx−e−ty2/1−e−2t1−e−2tn+1/2dtt≤C∫0mxe−cx−y2/ttn/2+1dt+∫mx∞dttn/2+1≤C1x−yn+1mxn/2≤Cx−yn,x,y∈Nβ,x≠y.In a similar way, we get, for eachk=1,⋯,n,
(228)∂xkReix,y≤C∫0∞e−nte−cx−e−ty2/1−e−2t1−e−2tn/2+1dtt≤Cx−yn+1,x,y∈Nβ,x≠y.Then, according to a vector-valued version of ([1], Propositions 3.2.5 and 3.2.7) (see ([16], Propositions 2.3 and 2.4), we deduce that Rei,loc defines a bounded operator from Lpℝn,dx,X into itself. Also, Rei,loc−ℝei,loc defines a bounded operator from Lpℝn,dx,X into itself. We conclude that ℝei,loc defines a bounded operator from Lpℝn,dx,X into itself. Since ℝei is dilatation invariant, by proceeding as in the proof of ([16], Theorem 1.10, (ii)⟹(i)), it follows that ℝei can be extended from L2ℝn,dx⋂Lpℝn,dx⊗X to Lpℝn,dx,X as a bounded operator from Lpℝn,dx,X into itself.The same arguments allow us to prove that the following assertions are equivalent.(i)
Rei can be extended from L1ℝn,γ−1⋂L2ℝn,γ−1⊗X to L1ℝn,γ−1,X as a bounded operator from L1ℝn,γ−1,X into L1,∞ℝn,γ−1,X(ii)
ℝei can be extended from L1ℝn,dx⋂L2ℝn,dx⊗X to L1ℝn,dx,X as a bounded operator from L1ℝn,dx,X into L1,∞ℝn,dx,XFurthermore, in a similar way, we can see that (i)⟺(ii) and (iii)⟺(iv) when Rei and ℝei are replaced by Rei,∗ and ℝei,∗, respectively.The proof of Theorem4 is thus finished.Proof of Theorem 5.
Let1<p<∞ and i=1,⋯,n. We are going to see that the following two assertions are equivalent.
(a)
For everyf∈Lpℝn,γ−1,X, there exists(229)limε⟶0+∫x−y>εReix,yfydy,foralmostallx∈ℝn.(b)
For everyf∈Lpℝn,dx,X, there exists(230)limε⟶0+∫x−y>εℝeix−yfydy,foralmostallx∈ℝn.We consider again1/p<η<1 and β=η−1. Suppose that a is true. Let f∈Lpℝn,dx,X. We can write
(231)∫x−y>εℝeix−yfydy=∫x−y>εℝeix−y−Reix−yχNβx,yfydy+∫x−y>εReix,yχNβx,yfydy+∫x−y>εℝeix−yχNβcx,yfydy,x∈ℝn,ε>0.Since (225) holds and the operator Li is bounded from Lpℝn,dx,X into itself, there exists the limit
(232)limε⟶0+∫x−y>εReix,y−ℝeix−yχNβx,yfydy,foralmostallx∈ℝn.On the other hand, we get(233)∫ℝnℝeix−yχNβcx,yfydy≤C∫ℝn1x−ynχNβcx,yfydy≤C∫x−y>βnmxdyx−ynp′1/p′fLpℝn,dx,X≤C∫βnmx∞drrnp′−1+11/p′fLpℝn,dx,X≤Cmxn/2pfLpℝn,dx,X,x∈ℝn.Then, there exists the limit(234)limε⟶0+∫ℝnℝeix−yχNβcx,yfydy,x∈ℝn.Suppose thatg∈Lpℝn,γ−1,X. It was seen in the proof of Theorem 2 that
(235)∫ℝnReix,yχNβcx,ygydy∈Lpℝn,γ−1.Then,(236)limε⟶0+∫ℝnReix,yχNβcx,ygydy=∫ℝnReix,yχNβcx,ygydy,foralmostallx∈ℝn.Since (a) holds, there also exists the limit(237)limε⟶0+∫x−y>εReix,yχNβx,ygydy,foralmostallx∈ℝn.Letk∈ℕ. We have that y≤βn+k provided that x≤k and x−y≤βnmin1,x−1. Then, for every ε>0,
(238)∫x−y>εReix,yχNβx,yfydy=∫x−y>εReix,yχNβx,yχB0,βn+kyfydy,x≤k.Sincef∈Lpℝn,dx,X, χB0,βn+kf∈Lpℝn,γ−1,X and then there exists
(239)limε⟶0+∫x−y>εReix,yχNβx,yfydy,foralmostallx∈B0,k.Hence, we get that there exists(240)limε⟶0+∫x−y>εReix,yχNβx,yfydy,foralmostallx∈ℝn.We conclude that there exists(241)limε⟶0+∫x−y>εℝeix−yfydy,foralmostallx∈ℝn.In a similar way, we can see that (a) holds provided that (b) is true. Note thatLpℝn,γ−1,X⊂Lpℝn,dx,X.As it was proved in [2] the operator Si defined by
(242)Sifx=∫ℝnReix,yχNβcx,yfydy,x∈ℝn,is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. Then, for every f∈L1ℝn,γ−1, there exists
(243)limε⟶0+∫x−y>εReix,yχNβcx,yfydy=∫ℝnReix,yχNβcx,yfydy,foralmostallx∈ℝn.The same arguments allow us to prove that (a)⟺(b) when p=1.By using an-dimensional version of ([26], Theorem D), we deduce that the properties (i), (ii), and (iii) in Theorem 5 are equivalent.By proceeding as above, we can see that the properties in (a) and (b) continue being equivalent when we replace the principal values by the maximal operatorsRei,∗ and ℝei,∗ in (a) and (b), respectively. Then, the property (b) is equivalent to the property UMD for X (see the comments before the proof of ([16], Theorem 1.10, p. 19).Thus, the proof of Theorem5 is finished.


## 6. UMD Spaces and the Imaginary Powers ofA

In this section, we prove Theorem6.According to ([18], (11)), we have that, for every f∈Cc∞ℝn,
(244)−Δ2iσfx=limε⟶0+αεfx+∫x−y>εKσx−yfydy,foralmostallx∈ℝn,where
(245)Kσz=−∫0∞ϕσt∂tWtzdt,z∈ℝn\0,(246)αε=1Γn/2∫0∞ϕσε24ue−uun/2−1du,ε∈0,∞,and
(247)ϕσt=t−iσΓ1−iσ,t∈0,∞.Note that the limitlimt⟶0+ϕσt does not exist.Letf,g∈Cc∞ℝn. By proceeding as in ([18], p. 213), we can see that
(248)Aiσf,gL2ℝn,γ−1=∫0∞ϕσt−ddt∫ℝn∫ℝnTtAx,yfydygx¯γ−1xdxdt=∫0∞ϕσt−ddt∫ℝn∫ℝnWtx−yfydygx¯γ−1xdxdt+∫0∞ϕσt−ddt∫ℝn∫ℝnTtAx,y−Wtx−yfydygx¯γ−1xdxdt.We have that(249)∂tTtAx,y=−n−2e−t∑i=1nyixi−e−tyi+2e−2tx−e−ty21−e−2tTtAx,y1−e−2t=−n−2e−tx,y−e−ty2+2e−2tx−e−ty21−e−2tTtAx,y1−e−2t,x,y∈ℝn,t>0,and writing2e−tx,y=x2+e−2ty2−x−e−ty2, we get
(250)∂tTtAx,y=−n+e−2ty2−x2−1−e−2tx2+1+e−2t1−e−2tx−e−ty2TtAx,y1−e−2t,x,y∈ℝn,t>0.Also, we have that(251)∂tWtx−y=−n+x−y2tWtx−y2t,x,y∈ℝn,t>0.We can write, for everyx,y∈ℝn and t>0,
(252)∂tTtAx,y−∂tWtx−y=−n+x−y2tTtAx,y−Wtx−y2t+−n11−e−2t−12t+e−2t1−e−2ty2−x2−x2−x−e−ty21−e−2t+2x−e−ty21−e−2t2−x−y22t2TtAx,y.The derivative under the integral sign is justified and we get(253)Aiσf,gL2ℝn,γ−1=−∫0∞ϕσt∫ℝn∫ℝn∂tWtx−yfydygx¯γ−1xdxdt−∫0∞ϕσt∫ℝn∫ℝn∂tTtAx,y−Wtx−yfydygx¯γ−1xdxdt.To ensure the change on the order of integration, we are going to see that(254)Jx=∫0∞∫ℝn∂tTtAx,y−Wtx−yfydydt<∞,x∈ℝn,and
(255)∫0∞∫ℝn∫ℝn∂tTtAx,y−Wtx−yfygxγ−1xdydxdt<∞.We observe that, sinceg∈Cc∞ℝn, it is sufficient to see that Jx≤hx, x∈ℝn, for certain continuous function h.We consider the decomposition(256)Jx=∫0mx+∫mx∞∫ℝn∂tTtAx,y−Wtx−yfydydt=J1x+J2x,x∈ℝn.Sincef∈Cc∞ℝnby (252), we have that
(257)J2x≤C1+x2∫mx∞∫suppf1tn/2+1dtdy≤C1+x2mxn/2,x∈ℝn.We now estimateJ1x, x∈ℝn. We take into account that, if suppf⊂B0,rf, with rf>0, then
(258)x−e−ty2≥x−y2+1−e−t2y2−2x−yy1−e−t≥x−y2−dx1−e−t,x∈ℝn,y∈suppf,t>0,wheredx=2rfrf+x, x∈ℝn. By considering the decomposition (89) for α=0, the estimations (92) and the proof of (97), we can see that
(259)TtAx,y−Wtx−y≤Cedxe−cx−y2/t1tn/2−1+y+x−yn,x∈ℝn,y∈suppf,t∈0,1.Then, according to (252), we have that
(260)∂tTtAx,y−Wtx−y≤Cedxe−cx−y2/t1+x2tn/2+y+x−ynt+x−yy+xtn/2+1+y+x−ytn/2≤Cedxe−cx−y2/t1+x2tn/2+1+xnt+1+xtn+1/2,x∈ℝn,y∈suppf,t∈0,1.Sincemx~1+x−2, x∈ℝn, we get when t∈0,mx,
(261)∂tTtAx,y−Wtx−y≤Cedxe−cx−y2/t1+xtn+1/2+1+xnt,x∈ℝn,y∈suppf.We deduce that(262)J1x≤Cedx∫0mx∫suppf1+xt3/4+1+xntn/2−5/4dyx−yn−1/2≤Cedx1+xmx1/4+1+xnmxn/2−1/4≤Cedx1+x,x∈ℝn.Then,Jx≤C1+xn+2+edx1+x, x∈ℝn and (254) and (255) are established.Note that the estimations that we have just proved are depending onf.By interchanging the order of integration, we get(263)Aiσf,gL2ℝn,γ−1=−Δ2iσf,gL2ℝn,γ−1−∫ℝn∫ℝnfy∫0∞ϕσt∂tTtAx,y−∂tWtx−ydtdyg¯xγ−1xdx.It follows that(264)Aiσfx=−Δ2iσfx−∫ℝnfy∫0∞ϕσt∂tTtAx,y−∂tWtx−ydtdy=−Δ2iσfx−limε⟶0+∫x−y>εfy∫0∞ϕσt∂tTtAx,y−∂tWtx−ydtdy,for almost allx∈ℝn.We conclude that(265)Aiσfx=limε⟶0+∫x−y>εKσAx,yfydy+αεfx,foralmostallx∈ℝn,where
(266)KσAx,y=−∫0∞ϕσt∂tTtAx,ydt,x,y∈ℝn,t>0.Salogni ([1], Theorem 3.4.3) proved that Aiσ is bounded from Lpℝn,γ−1 into itself, for every 1<p<∞, and Bruno ([6], Theorem 4.1, (i)) established that Aiσ is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1. In order to extend Aiσ to functions taking values in a Banach space, we need to prove these results in a different way by making a comparation between Aiσ and −Δ/2iσ.Letβ>0. We define the local and global part as follows
(267)Alociσfx=limε⟶0+αεfx+∫x−y>εKσAx,yχNβx,yfydy,x∈ℝn,and
(268)Aglobiσfx=∫x−y>εKσAx,yχNβcx,yfydy,x∈ℝn.The operators−Δ/2lociσ and −Δ/2globiσ are defined in analogous way.We are going to see that(269)KσAx,y−Kσx−y≤C1+xx−yn−1/2,x,y∈Nβ.We can write(270)KσAx,y−Kσx−y≤C∫0mx∂tTtAx,y−Wtx−ydt+∫mx∞∂tTtAx,ydt+∫mx∞∂tWtx−ydt=∑i=13Iix,y,x,y∈ℝn.First, we observe that(271)I3x,y≤C∫mx∞e−cx−y2/ttn/2+1dt≤Cmxn/2≤C1+xn≤C1+xx−yn−1/2,x,y∈Nβ.By using (250) and since y2−x2≤x−yx+y≤C when x,y∈Nβ, we have that
(272)I2x,y≤C1+x2∫mx∞e−n/2ttn/2+1dt≤C1+x2mxn/2+1≤C1+xn≤C1+xx−yn−1/2,x,y∈Nβ.Finally, from (252), proceeding as above for the estimation of J1, but now by taking into account that x,y∈Nβ, we obtain that
(273)I1x,y≤C∫0mxe−cx−y2/t1+xtn+1/2+1+xntdt≤C1+xmx1/4+1+xnmxn/2−1/4x−yn−1/2≤C1+xx−yn−1/2,x,y∈Nβ.Thus, (269) is proved.As it was established in Section 3.2,(274)supx∈ℝn∫ℝn1+xx−yn−1/2χNβx,ydy<∞andsupy∈ℝn∫ℝn1+xx−yn−1/2χNβx,ydx<∞.Then, the operatorLβ defined by
(275)Lβfx=∫ℝnKσAx,y−Kσx,yχNβx,yfydy,x∈ℝn,is bounded from Lpℝn,dx into itself for every 1≤p<∞. From ([1], Proposition 3.2.5), it follows that Lβ is bounded from Lpℝn,γ−1 into itself, for every 1≤p<∞.Note that(276)Alociσf−−Δ2lociσf≤CLβf.We now study the global operatorAglobiσ. We recall that Aglobiσ is the integral operator defined by
(277)Aglobiσfx=−∫ℝn∫0∞ϕσt∂tTtAx,ydtχNβcx,yfydy,x∈ℝn.We have, by makingr=e−t, t∈0,∞, that
(278)∫0∞ϕσt∂tTtAx,ydt≤C∫01∂rrne−x−ry2/1−r21−r2n/2dr≤C∫01n1−r2+2r1−r2∑i=1nyixi−ryi−2r2x−ry21−r22rn−1e−x−ry2/1−r21−r2n/2dr=C∫011−r2n+x2−r2y2−1+r2x−ry21−r22rn−1e−x−ry2/1−r21−r2n/2dr.For everyx,y∈ℝn, there exists a polynomial Px,y with degree 4 and a positive function Qx,y such that
(279)∂rrne−x−ry2/1−r21−r2n/2=Px,yrQx,yr,r∈0,1.Then, for everyx,y∈ℝn, the function
(280)∂rrne−x−ry2/1−r21−r2n/2changes the sign at most four times in 0,1. We deduce that
(281)∫0∞ϕσt∂tTtAx,ydt≤C∫01∂rrne−x−ry2/1−r21−r2n/2dr≤Csupr∈0,1rne−x−ry2/1−r21−r2n/2,x,y∈ℝn.Ifu,v∈Nλ, with λ>0, we have that
(282)11+λn1+v≤11+u≤1+λn1+v.Also,min1,x−1/2≤1+x−1≤min1,x−1, x∈ℝn. Then, if y,x∈N, then x,y∈N21+n.We takeβ=21+n. Since y,x∉N when x,y∉Nβ, according to ([24], Proposition 2.1), we get, for x,y∉Nβ,
(283)supr∈0,1e−x−ry2/1−r21−r2n/2≤Ce−x2,x,y≤0,x+yx−yn/2expy2−x22−x−yx+y2,x,y>0.By proceeding as in the proof of Theorem2, we can see that, for every 1<p<∞,
(284)supx∈ℝn∫ℝnex2−y2/pχNβcx,y∫0∞ϕσt∂tTtAx,ydtdy<∞,and
(285)supy∈ℝn∫ℝnex2−y2/pχNβcx,y∫0∞ϕσt∂tTtAx,ydtdy<∞.Hence, the operatorLβ defined by
(286)Lβfx=∫ℝnfyχNβcx,y∫0∞ϕσt∂tTtAx,ydtdy,x∈ℝn,is bounded from Lpℝn,γ−1 into itself, for every 1<p<∞.On the other hand, according to ([1], Lemma 3.3.3)
(287)supt>0TtAx,y≤Ce−x2min1+xn,xsinθx,y−n,x,y∈Nβc,x,y≠0.We recall thatθx,y∈0,π represents the angle between the nonzero vectors x and y when n>1 and θx,y=0, x,y∈ℝn, when n=1. By ([1], Lemma 3.3.4) the operator Lβ is bounded from L1ℝn,γ−1 into L1,∞ℝn,γ−1.Note thatAglobiσf≤Lβf.By taking into account that−Δ/2iσ is a Calderón-Zygmund operator, the arguments developed in the proofs of Theorems 2 and 3 allow us to finish the proof of this theorem.



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*Source: 6899603-2021-11-10.xml*