Parabolic equations involving Bessel operators and singular integrals

TL;DR

This paper studies Bessel parabolic equations involving singular integrals, establishing weighted Sobolev inequalities for solutions to these equations, which are relevant for understanding their regularity and behavior.

Contribution

It introduces new weighted Sobolev inequalities for Bessel parabolic equations using singular integral techniques, advancing the analysis of such equations.

Findings

Established weighted Sobolev inequalities for Bessel parabolic equations

Applied singular integral methods in a parabolic context

Provided tools for analyzing regularity of solutions

Abstract

In this paper we consider the evolution equation $\partial_t u=\Delta_\mu u+f$ and the corresponding Cauchy problem, where $\Delta_\mu$ represents the Bessel operator $\partial_x^2+(\frac{1}{4}-\mu^2)x^{-2}$, for every $\mu>-1$. We establish weighted and mixed weighted Sobolev type inequalities for solutions of Bessel parabolic equations. We use singular integrals techniques in a parabolic setting.

Equations

$$\frac{\partial u(t,x)}{\partial t}=\Delta u(t,x)+f(t,x),(t,x)\in {\mathbb{R}}^{n+1}or(t,x)\in (0,\infty )\times {\mathbb{R}}^n,$$

$$u(t,x)={\int }_0^t{\int }_{{\mathbb{R}}^n}\Gamma (t-s,x-y)f(s,y)dyds,(t,x)\in (0,\infty )\times {\mathbb{R}}^n.$$

$${\partial }_{x_i,x_j}^{2}u(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_0^{t-ϵ}{\int }_{{\mathbb{R}}^n}{\partial }_{y_i{y}_j}\Gamma (t-s,x-y)f(s,y)dyds,(t,x)\in (0,\infty )\times {\mathbb{R}}^n,$$

$${\partial }_{t}u(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_0^{t-ϵ}{\int }_{{\mathbb{R}}^n}{\partial }_t\Gamma (t-s,x-y)f(s,y)dyds+f(t,x),(t,x)\in (0,\infty )\times {\mathbb{R}}^n,$$

$$\Vert {\partial }_{x_i,x_j}^{2}u{\Vert }_{L^p((0,\infty )\times {\mathbb{R}}^n)}+\Vert {\partial }_{t}u{\Vert }_{L^p((0,\infty )\times {\mathbb{R}}^n)}\le C\Vert f{\Vert }_{L^p((0,\infty )\times {\mathbb{R}}^n)}.$$

$$u(t,x)={\int }_0^{\infty }{\int }_{{\mathbb{R}}^n}\Gamma (s,y)f(t-s,x-y)dyds,(t,x)\in {\mathbb{R}}^{n+1},$$

$${\partial }_{x_i,x_j}^{2}u(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_{{\Omega }_{ϵ}}{\partial }_{y_i{y}_j}\Gamma (s,y)f(t-s,x-y)dyds-A_{n}f(t,x),(t,x)\in {\mathbb{R}}^{n+1},i,j=1,\dots ,n,$$

$${\partial }_{t}u(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_{{\Omega }_{ϵ}}{\partial }_s\Gamma (s,y)f(t-s,x-y)dyds+(1-A_n)f(t,x),(t,x)\in {\mathbb{R}}^{n+1},$$

$$\frac{\partial u(t,x)}{\partial t}={\Delta }_{\mu }u(t,x)+f(t,x),(t,x)\in \mathbb{R}\times (0,\infty )or(t,x)\in (0,\infty )\times (0,\infty ),$$

$$h_{\mu }(\varphi )(x)={\int }_0^{\infty }\sqrt{xy}{J}_{\mu }(xy)\varphi (y)dy,x\in (0,\infty ),$$

$$h_{\mu }({\Delta }_{\mu }\varphi )(x)=-x^2{h}_{\mu }(\varphi )(x),x\in (0,\infty ).$$

$$W_t^{\mu }(\varphi )(x)={\int }_0^{\infty }{W}_t^{\mu }(x,y)\varphi (y)dy,x\in (0,\infty ).$$

$$u(t,x)={\int }_0^{\infty }{\int }_0^{\infty }{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau ,(t,x)\in \mathbb{R}\times (0,\infty ),$$

$$\frac{\partial u(t,x)}{\partial t}$$

$$=\underset{ϵ\to 0^{+}}{\lim }{\int }_{{\Omega }_{ϵ}(x)}\frac{\partial }{\partial \tau }{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau +Af(t,x),t,x\in (0,\infty ),$$

$$\frac{{\partial }^{2}u(t,x)}{\partial x^2}$$

$$=\underset{ϵ\to 0^{+}}{\lim }{\int }_{{\Omega }_{ϵ}(x)}\frac{{\partial }^2}{\partial x^2}{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau -(1-A)f(t,x)$$

$$(L_{\mu }f)(t,x)={\int }_0^{\infty }{\int }_0^{\infty }{W}_s^{\mu }(x,y)f(t-s,y)dyds,$$

$$R_{\mu }(f)={\delta }_{\mu +1}{\delta }_{\mu }{L}_{\mu }(f)\text{and}\widetilde{R_{\mu }}(f)={\partial }_t{L}_{\mu }(f).$$

$$R_{\mu }(f)(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_{{\Omega }_{ϵ}(t,x)}{K}_{\mu }(t,x;\tau ,y)f(\tau ,y)d\tau dy+f(t,x)\frac{1}{\sqrt{\pi }}{\int }_1^{\infty }{e}^{-s^2/4}ds,(t,x)\in \mathbb{R}\times (0,\infty ),$$

$$\widetilde{R_{\mu }}(f)(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_{{\Omega }_{ϵ}(t,x)}\widetilde{K_{\mu }}(t,x;\tau ,y)f(\tau ,y)d\tau dy+f(t,x)\frac{1}{\sqrt{\pi }}{\int }_0^1{e}^{-s^2/4}ds,(t,x)\in \mathbb{R}\times (0,\infty ),$$

$$K_{\mu }(t,x;\tau ,y)={\delta }_{\mu +1}{\delta }_{\mu }{W}_{t-\tau }^{\mu }(x,y){\chi }_{(0,\infty )}(t-\tau ),x,y\in (0,\infty ),t,\tau \in \mathbb{R},$$

$$\widetilde{K_{\mu }}(t,x;\tau ,y)=-{\partial }_{\tau }{W}_{t-\tau }^{\mu }(x,y){\chi }_{(0,\infty )}(t-\tau ),x,y\in (0,\infty ),t,\tau \in \mathbb{R},$$

$$\left\{\begin{array}[]{c}\partial_{t}u(t,x)=\Delta_{\mu}u(t,x)+f(t,x),\;\;(t,x)\in(0,\infty)\times(0,\infty),\\ u(0,x)=g(x),\;\;x\in(0,\infty).\end{array}\right.$$

$$u(t,x)={\int }_0^t{\int }_0^{\infty }{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau +{\int }_0^{\infty }{W}_t^{\mu }(x,y)g(y)dy,t,x\in (0,\infty ).$$

$$\frac{\partial u(t,x)}{\partial t}$$

$$\frac{{\partial }^{2}u(t,x)}{\partial x^2}=\underset{ϵ\to 0^{+}}{\lim }{\int }_0^{t-ϵ}{\int }_0^{\infty }\frac{{\partial }^2}{\partial x^2}{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau +{\int }_0^{\infty }\frac{{\partial }^2}{\partial x^2}{W}_t^{\mu }(x,y)g(y)dy,t,x\in (0,\infty ).$$

$${\mathbf{R}}_{\mu }(f)(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_0^{t-ϵ}{\int }_0^{\infty }{\delta }_{\mu +1}{\delta }_{\mu }{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau ,t,x\in (0,\infty ),$$

$$\widetilde{{\mathbf{R}}_{\mu }}(f)(t,x)=\underset{ϵ\to 0^{+}}{\lim }{\int }_0^{t-ϵ}{\int }_0^{\infty }{\partial }_{\tau }{W}_{\tau }^{\mu }(x,y)f(t-\tau ,y)dyd\tau ,t,x\in (0,\infty ).$$

$$\left\{\begin{array}[]{c}\frac{\partial}{\partial t}u(t)+Au(t)=f(t),\;\;t\in(0,\infty),\\ u(0)=0\end{array}\right.$$

$${\Vert \frac{\partial }{\partial t}{u}_f\Vert }_{L^p((0,\infty ),X)}+\Vert A{u}_f{\Vert }_{L^p((0,\infty ),X)}\le C\Vert f{\Vert }_{L^p((0,\infty ),X)}.$$

$$u(t)={\int }_0^t{T}_{t-s}(f(s))ds,t\ge 0,$$

Full text

The first author was partially supported by MTM2016-79436-P. The second author was partially supported by MTM2015-66157-C2-1-P

1. Introduction

The model of a parabolic differential equation is the following

[TABLE]

where $\Delta$ denotes the Laplace operator. By $\Gamma$ we denote the fundamental solution of the heat equation $\partial_{t}u-\Delta u=0$, on $(0,\infty)\times\mathbb{{R}}^{n}$, that is, $\Gamma(t,x)=\frac{e^{-\frac{|x|^{2}}{4t}}}{(4\pi t)^{n/2}}$, $x\in\mathbb{{R}}^{n}$, and $t\in(0,\infty)$. Assume that $f$ is a bounded function defined on $(0,\infty)\times\mathbb{{R}}^{n}$ with compact support. We define

[TABLE]

Thus, $u$ is a solution of (1.1) on $(0,\infty)\times\mathbb{{R}}^{n}$ such that $u(0,x)=0$, $x\in\mathbb{{R}}^{n}$. Moreover, Jones [26] proved that

[TABLE]

and there exists a constant $C>0$ such that

[TABLE]

In [27], $L^{p}$-inequalities like (1.2) were established for a more general parabolic singular integral, where the derivatives of the fundamental solution $\Gamma$ are replaced by more general kernels.

Recently, Jones’ results have been extended to weighted and mixed weighted $L^{p}$ spaces by Ping, Stinga and Torrea ([42, Theorem 2.4]). In [42], they also considered the equation (1.1) on the whole space $\mathbb{{R}}^{n+1}$. In [42, Theorem 2.3], it was proved that if $f$ is a $C^{2}$-function defined on $\mathbb{{R}}^{n+1}$ with compact support, when $n\geq 2,$ the function $u$ given by

[TABLE]

is a solution of (1.1) on $\mathbb{{R}}^{n+1}$, where

[TABLE]

and

[TABLE]

where $A_{n}=\frac{1}{n\Gamma(n/2)}\displaystyle\int_{1/4}^{\infty}\omega^{\frac{n-2}{2}}e^{-\omega}d\omega$. Here, for every $\epsilon>0$, $\Omega_{\epsilon}$ denotes the parabolic region $\Omega_{\epsilon}=\{(t,x)\in(0,\infty)\times\mathbb{{R}}:\;\sqrt{t}+|x|>\epsilon\}$.

By using Calderón-Zygmund theory in the parabolic setting in [42, Theorem 2.3, (B)], weighted norm $L^{p}$-inequalities and weighted weak $L^{1}$-estimates as (1.2) were established in this context.

Parabolic equation of (1.1) type, where the Laplace operator $\Delta$ is replaced by the Hermite operator $\mathscr{H}=\Delta-|x|^{2}$, was also studied in [42]. In [42, Theorems 1.3 and 1.4], weighted parabolic Sobolev estimates (like (1.2)) were established in the Hermite setting.

It is well-known that partial differential equations and singular integrals are closely connected (see for instance [12] and [13]). The previous commented examples show this relationship. Although elliptic PDE’s have received a preferential and continuous treatment over time, parabolic PDE’s have been already studied since the sixties in the last century. Apart from Jones’ papers ([26] and [27]), we can find relevant results about parabolic PDE’s and "a priori" Sobolev estimates in [21], [22], and [46]. In the last years, the study of parabolic equations by using harmonic analysis techniques has taken great interest (see, for instance, [1], [16], [17] and [40]).

On the other hand, the use of Calderón-Zygmund theory in the context of parabolic PDE’s and parabolic singular integrals appears also in [45], where some of Jones’ results are improved, and more recently in [32] where a singular integral approach to the maximal $L^{p}$-regularity is developed (see also [28], [29] and [30]).

In this paper we consider the parabolic equations

[TABLE]

where, for every $\mu>-1$, $\Delta_{\mu}$ represents the Bessel operator defined by $\Delta_{\mu}=\partial_{x}^{2}+(\frac{1}{4}-\mu^{2})x^{-2}$.

Our study is motivated by [42]. The Bessel operator $\Delta_{\mu}$ can be seen as a one dimensional Schrödinger operator with the singular potential $V_{\mu}(x)=(\frac{1}{4}-\mu^{2})x^{-2}$, $x\in(0,\infty)$. Singular integrals associated with parabolic Schrödinger operators $\partial_{t}-\Delta+V$ in $\mathbb{{R}}^{n+1}$ have been investigated in [15], [23], [33] and [41]. Our potentials $V_{\mu}$, $\mu>-1$, are not included in the class of potentials considered in the above mentioned papers. There, the potentials $V$ are nonnegative and in $L^{1}_{\rm loc}(\mathbb{{R}}^{n+1})$ and they belong to the parabolic reverse Hölder classes.

Let $\mu>-1$. For every $\phi\in C^{\infty}_{c}((0,\infty)),$ the space of smooth functions with compact support on $(0,\infty)$, the Hankel transform $h_{\mu}(\phi)$ of $\phi$ is defined by

[TABLE]

where $J_{\mu}$ denotes the Bessel function of the first kind and order $\mu$. $h_{\mu}$ can be extended to $L^{2}((0,\infty))$ as an isometry (see [11] and [50]) and $h_{\mu}^{-1}=h_{\mu}$ on $L^{2}((0,\infty))$. For every $\phi\in C^{\infty}_{c}((0,\infty)),$ we have that (see [52, Lemma 5.4-1]),

[TABLE]

We extend the definition of the operator $\Delta_{\mu}$ as follows. We define the domain of ${\bf\Delta}_{\mu}$, $D(\bf{\Delta}_{\mu})$, by $D({\bf\Delta_{\mu}})=\{\phi\in L^{2}((0,\infty)):\;x^{2}h_{\mu}(\phi)\in L^{2}((0,\infty))\}$ and, for every $\phi\in D(\bf{\Delta_{\mu}})$, ${\bf\Delta}_{\mu}\phi=-h_{\mu}(x^{2}h_{\mu}(\phi))$. According to [52, Theorem 5.4-1], $C^{\infty}_{c}((0,\infty))\subset D(\bf{\Delta}_{\mu})$ and ${\bf\Delta_{\mu}}\phi=\Delta_{\mu}\phi$, $\phi\in C^{\infty}_{c}((0,\infty))$. Note that, for every $\mu\in(-1,1)$, $\Delta_{\mu}\phi=\Delta_{-\mu}\phi$, $\phi\in C^{\infty}_{c}((0,\infty))$, and $\bf{\Delta}_{\mu}\neq\bf{\Delta}_{-\mu}$.

The operator $-\bf{\Delta}_{\mu}$ is positive and selfadjoint on $L^{2}((0,\infty))$. Moreover, $-\bf{\Delta}_{\mu}$ generates a semigroup $\{W_{t}^{\mu}\}_{t>0}$ of operators in $L^{2}((0,\infty))$ where, for every $t>0$ and $\phi\in L^{2}((0,\infty))$,

[TABLE]

Here, $W_{t}^{\mu}(x,y)=\frac{(xy)^{1/2}}{2t}I_{\mu}\left(\frac{xy}{2t}\right)e^{-\frac{x^{2}+y^{2}}{4t}}$, $x,y,t\in(0,\infty)$, where $I_{\mu}$ represents the modified Bessel function of the first kind and order $\mu$. $\{W_{t}^{\mu}\}_{t>0}$ is usually called the heat semigroup associated with the Bessel operator $\bf{\Delta}_{\mu}$.

If, for every $t>0$, $W_{t}^{\mu}$ is given as in (1.4), $\{W_{t}^{\mu}\}_{t>0}$ also defines a semigroup of operators on $L^{p}((0,\infty))$, for each $1<p<\infty$ when $\mu>-1/2$ and for each $1<p<\infty$ such that $-\mu-1/2<\frac{1}{p}<\mu+3/2$, when $-1<\mu\leq-1/2$.

Harmonic analysis associated with Bessel operator (Riesz transforms, maximal operators, Littlewood-Paley functions, fractional Bessel operators, Hardy spaces,..) has been developed in the last years ([4], [5], [6], [7], [8], [9], [19] and [20]) although the first results about this topic had been obtain by Muckenhoupt and Stein ([37]) in the sixties of the last century as the paper about parabolic singular integrals mentioned at the beginning.

Our results concerning to the solutions of (1.3) in the whole space $\mathbb{{R}}\times(0,\infty)$ are the following.

Theorem 1.1**.**

Assume that $f\in L^{\infty}(\mathbb{{R}}\times(0,\infty))$ has compact support on $\mathbb{{R}}\times(0,\infty)$. Then, the function $u(t,x),$ $(t,x)\in\mathbb{{R}}\times(0,\infty)$, given by

[TABLE]

is defined by an absolutely convergent integral, for every $(t,x)\in\mathbb{{R}}\times(0,\infty)$. Moreover, if $f$ is also in $C^{2}(\mathbb{{R}}\times(0,\infty))$, then, for every $(t,x)\in\mathbb{{R}}\times(0,\infty)$, $\frac{\partial u(t,x)}{\partial t}=\Delta_{\mu}u(t,x)+f(t,x),$ being

[TABLE]

*and *

[TABLE]

where, for every $\epsilon,x\in(0,\infty)$, $\Omega_{\epsilon}(x)=\{(\tau,y)\in(0,\infty)^{2}:\;\tau^{1/2}+|x-y|>\epsilon\}$, and $A=\frac{1}{\sqrt{\pi}}\int_{0}^{1}e^{-\frac{w^{2}}{4}}dw.$

The Bessel operator can be written as $\Delta_{\mu}=\delta_{\mu}^{*}\delta_{\mu}$, where $\delta_{\mu}=x^{\mu+1/2}\frac{d}{dx}x^{-\mu-1/2}$, and $\delta_{\mu}^{*}=x^{-\mu-1/2}\frac{d}{dx}x^{\mu+1/2}$ represents the formal adjoint of $\delta_{\mu}$. This decomposition of $\Delta_{\mu}$ suggests, according to Stein’s ideas ([48]), defining the Riesz transform ${\mathfrak{R}}_{\mu}$ associated with $\Delta_{\mu}$ by ${\mathfrak{R}}_{\mu}=\delta_{\mu}\Delta_{\mu}^{-1/2}$. The main $L^{p}$-boundedness properties of ${\mathfrak{R}}_{\mu}$ can be found in [3] and [8].

We now consider the operator $L_{\mu}$ defined by

[TABLE]

being $f$ a measurable complex function defined on $\mathbb{{R}}\times(0,\infty)$, provided that the last integral exists. In Theorem 1.1 we have established that if $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ and has compact support, then $(\partial_{t}-\Delta_{\mu})L_{\mu}(f)=f$. In a similar way we can see that $L_{\mu}((\partial_{t}-\Delta_{\mu})f)=f$, provided that $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ with compact support. Thus, $L_{\mu}$ can be seen as an inverse of $\partial_{t}-\Delta_{\mu}$. Keeping in mind Stein’s ideas ([48]), we define Riesz transforms associated with the parabolic operator $\partial_{t}-\Delta_{\mu}$ as follows: for every $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ with compact support,

[TABLE]

Note that, according to Theorem 1.1, if $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ with compact support, the above definitions of $R_{\mu}(f)$ and $\widetilde{R_{\mu}}(f)$ have sense because the derivatives of $L_{\mu}(f)$ do exist. Moreover, we can write, for every $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ with compact support,

[TABLE]

and

[TABLE]

where

[TABLE]

and $\Omega_{\epsilon}(t,x)=\{(\tau,y)\in(0,\infty)\times(0,\infty):\;\max\{|t-\tau|^{1/2},|x-y|\}>\epsilon\}$, for $\epsilon,x\in(0,\infty)$ and $t\in(0,\infty)$.

Next we establish $L^{p}$-boundedness properties of the Riesz transforms. If $m$ denotes the Lebesgue measure on $\mathbb{{R}}\times(0,\infty)$ and $d$ represents the parabolic metric defined by $d((t,x),(\tau,y))=|t-\tau|^{1/2}+|x-y|$, $t,\tau\in\mathbb{{R}}$ and $x,y\in(0,\infty)$, the triple $(\mathbb{{R}}\times(0,\infty),m,d)$ is a space of homogeneous type in the sense of Coifman and Weiss ([18]). We represent, for every $1\leq p<\infty$, by $A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$ the class of Muckenhoupt weigths in the space of homogeneous type $(\mathbb{{R}}\times(0,\infty),m,d)$. In section 2 we recall the main definitions and results related to Calderón-Zygmund singular integrals on spaces of homogeneous type.

Theorem 1.2**.**

1. (1)

If $\mu>-1$, the Riesz transformations $R_{\mu}$ and $\widetilde{R_{\mu}}$ are bounded from $L^{2}(\mathbb{{R}}\times(0,\infty))$ into itself. 2. (2)

Suppose that $\mu>1/2$ or $\mu=-1/2$. The Riesz transformations $R_{\mu}$ and $\widetilde{R_{\mu}}$ can be extended from $L^{2}(\mathbb{{R}}\times(0,\infty))\cap L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$ to $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$ as bounded operators from $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$

• •

into $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$, for every $1<p<\infty$ and $\omega\in A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$.

• •

into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty),\omega)$, for $p=1$ and $\omega\in A_{1}^{*}(\mathbb{{R}}\times(0,\infty))$. 3. (3)

If $\mu>-1/2$, the Riesz transformations $R_{\mu}$ and $\widetilde{R_{\mu}}$ can be extended from $L^{2}(\mathbb{{R}}\times(0,\infty))\cap L^{p}(\mathbb{{R}}\times(0,\infty))$ to $L^{p}(\mathbb{{R}}\times(0,\infty))$ as bounded operators from $L^{p}(\mathbb{{R}}\times(0,\infty))$

• •

into $L^{p}(\mathbb{{R}}\times(0,\infty))$, for every $1<p<\infty$.

• •

into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$, for $p=1$. 4. (4)

If $-1<\mu\leq-1/2$, then the Riesz transformation $\widetilde{R_{\mu}}$ can be extended from $L^{2}(\mathbb{{R}}\times(0,\infty))\cap L^{p}(\mathbb{{R}}\times(0,\infty))$ to $L^{p}(\mathbb{{R}}\times(0,\infty))$ as a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself, provided that $-\mu-1/2<{1}/{p}<\mu+3/2$ and $1<p<\infty$. 5. (5)

If $-1<\mu\leq-1/2$, then the Riesz transformation ${R_{\mu}}$ can be extended from $L^{2}(\mathbb{{R}}\times(0,\infty))\cap L^{p}(\mathbb{{R}}\times(0,\infty))$ to $L^{p}(\mathbb{{R}}\times(0,\infty))$ as a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself, provided that $p>\frac{1}{\mu+3/2}$ and $1<p<\infty$.

Moreover, when $\mu>-1/2$ in all these cases the extensions of the operators $R_{\mu}$ and $\widetilde{R_{\mu}}$ are defined by (1.5) and (1.6), respectively, where the limit exist a.e. $(t,x)\in\mathbb{{R}}\times(0,\infty)$ and the equalities are understood also in a.e. $(t,x)\in\mathbb{{R}}\times(0,\infty)$.

$L^{p}$-boundedness properties for the Riesz transforms established in Theorem 1.2 can be seen as Sobolev estimates in our parabolic Bessel setting. Note that the auxiliar operator $\delta_{\mu}$ plays the role of derivatives to define correct Sobolev spaces in the Bessel setting (see [10]). On the other hand, $(4)$ and $(5)$ in Theorem 1.2 remember the so called pencil phenomenon that appears related to the $L^{p}$-boundedness properties of harmonic analysis operators in Laguerre settings (see [25], [34], [35], and [38]).

As an application of vector-valued Calderón-Zygmund theory (see [45]), we establish the following mixed weighted norm inequalities for Riesz transforms $R_{\mu}$ and $\widetilde{R_{\mu}}$. For every $1\leq p<\infty$, we denote the classical classes of Muckenhoupt weights by $A_{p}(\Omega)$, where $\Omega=(0,\infty)$ or $\Omega=\mathbb{{R}}$.

Theorem 1.3**.**

Assume that $\mu>1/2$ or $\mu=-1/2$. If $1<p<\infty$ and $v\in A_{p}((0,\infty))$, then the Riesz transforms $R_{\mu}$ and $\widetilde{R_{\mu}}$ can be extended from $L^{2}(\mathbb{{R}}\times(0,\infty))\cap L^{q}(\mathbb{{R}},u,L^{p}((0,\infty),v))$ to $L^{q}(\mathbb{{R}},u,L^{p}((0,\infty),v))$ as a bounded operator from $L^{q}(\mathbb{{R}},u,L^{p}((0,\infty),v))$ into itself, provided that $1<q<\infty$ and $u\in A_{q}(\mathbb{{R}})$; and, for every $u\in A_{1}(\mathbb{{R}})$, from $L^{2}(\mathbb{{R}}\times(0,\infty))\cap L^{1}(\mathbb{{R}},u,L^{p}((0,\infty),v))$ to $L^{1}(\mathbb{{R}},u,L^{p}((0,\infty),v))$ as a bounded operator from $L^{1}(\mathbb{{R}},u,L^{p}((0,\infty),v))$ into $L^{1,\infty}(\mathbb{{R}},u,L^{p}((0,\infty),v))$.

Note that from Theorem 1.3 we can deduce that, if $\mu>1/2$ or $\mu=-1/2$, $R_{\mu}$ and $\widetilde{R_{\mu}}$ define bounded operators from $L^{p}(\mathbb{{R}}\times(0,\infty),uv)$ into itself, for every $1<p<\infty$, $u\in A_{p}(\mathbb{{R}})$ and $v\in A_{p}((0,\infty))$. Moreover, $uv\in A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$ provided that $u\in A_{p}(\mathbb{{R}})$ and $v\in A_{p}((0,\infty))$, but $A_{p}^{*}(\mathbb{{R}}\times(0,\infty))\neq A_{p}(\mathbb{{R}})\cdot A_{p}((0,\infty))$, when $1<p<\infty$. Hence, Theorem 1.2 (2) is not a special case of strong type results in Theorem 1.3.

We now consider the following Cauchy problem associated with (1.3):

[TABLE]

Theorem 1.4**.**

Let $\mu>-1$. Assume that $f\in L^{\infty}((0,\infty)\times(0,\infty))$ with compact support and $g\in L^{\infty}((0,\infty))$ with compact support. We define

[TABLE]

Then, the last integrals are absolutely convergent for every $t,x\in(0,\infty)$. Moreover, if $f$ is also in $C^{2}((0,\infty)\times(0,\infty))$, then the function $u$ defined by (1.8) is a classical solution of (1.7) and

[TABLE]

*and *

[TABLE]

For every $f\in C^{2}((0,\infty)\times(0,\infty))$ with compact support we define

[TABLE]

and

[TABLE]

Note that the above limits do exist.

Theorem 1.5**.**

1. (1)

Suppose that $\mu>1/2$ or $\mu=-1/2$. The Riesz transformations ${\bf R}_{\mu}$ and $\widetilde{{\bf R}_{\mu}}$ can be extended to $L^{p}((0,\infty)\times(0,\infty),\omega)$ as bounded operators from $L^{p}((0,\infty)\times(0,\infty),\omega)$

• •

into $L^{p}((0,\infty)\times(0,\infty),\omega)$, for every $1<p<\infty$ and $\omega\in A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$.

• •

into $L^{1,\infty}((0,\infty)\times(0,\infty),\omega)$, for $p=1$ and $\omega\in A_{1}^{*}(\mathbb{{R}}\times(0,\infty))$. 2. (2)

If $\mu>-1/2$, the Riesz transformations ${\bf R}_{\mu}$ and $\widetilde{{\bf R}_{\mu}}$ can be extended to $L^{p}((0,\infty)\times(0,\infty))$ as bounded operators from $L^{p}((0,\infty)\times(0,\infty))$

• •

into $L^{p}((0,\infty)\times(0,\infty))$, for every $1<p<\infty$.

• •

into $L^{1,\infty}((0,\infty)\times(0,\infty))$, for $p=1$.

Moreover, the extensions of ${\bf R}_{\mu}$ and $\widetilde{{\bf R}_{\mu}}$ to $L^{p}((0,\infty)\times(0,\infty),\omega)$ are defined as the principal value integral operators in (1.9) and (1.10), respectively, where the limits exist a.e. $(t,x)\in(0,\infty)\times(0,\infty)$.

Suppose that $X$ is a Banach space and $A:D(A)\subset X\to X$ is an operator. If $1<p<\infty$, we say that $A$ has maximal $L^{p}$- regularity when there exists a constant $C>0$ such that, for every $f\in L^{p}((0,\infty),X)$ there exists a unique $u_{f}\in L^{p}((0,\infty),D(A))$ solution of the Cauchy problem

[TABLE]

satisfying

[TABLE]

If the operator $-A$ generates a semigroup $\{T_{t}\}_{t\geq 0}$ of operators on $X$, the solution of (1.11) can be written as

[TABLE]

and $A$ has maximal $L^{p}$-regularity when the operator

[TABLE]

is bounded from $L^{p}((0,\infty),X)$ into itself. Note that $\frac{\partial}{\partial t}T_{t}=-AT_{t}$, $t>0$. This fact leads, from the point of view of harmonic analysis, to replace the property of maximal $L^{p}$-regularity by the $L^{p}$-boundedness of certain Banach space valued singular integrals. If suitable Gaussian bounds hold for the semigroup generated by $-A$, then $A$ has maximal $L^{p}$-regularity (see [14] and [24]).

Theorem 1.6**.**

Let $\mu>-1/2$. Assume that $1<p,q<\infty$. Then, the Bessel operator $\Delta_{\mu}$ has maximal $L^{p}$-regularity on $L^{q}((0,\infty))$.

Note that Theorem 1.6 actually establishes mixed norm estimates for $\widetilde{{\bf R}_{\mu}}$.

In the next sections we will prove our Theorems. In order to show our results we use two different ways. On the one hand, we employ scalar and vectorial Calderón-Zygmund theory in the parabolic context. Here we need to get estimates involving the kernels of the integral operators. In order to do this, the properties of the Bessel function $I_{\mu}$ plays a crucial role. On the other hand, we use a comparative approach. In this second way we take advantages that Bessel operators $\Delta_{\mu}$ are nice (in a suitable sense) perturbations of the Laplacian. Then, it is possible to deduce the properties of our integral operators from the corresponding ones associated to the Laplace operator established in [42].

Throughout this paper we will denote by $C$ and $c$ positive constants, not necessarily the same in each occurrence.

Acknowledgements. The authors thank Professor José Luis Torrea (UAM, Madrid) for posing the problems studied in this paper and for reading a first version of the manuscript. His comments have allowed us to improve Theorem 1.5 and its proof.

2. Proof of Theorem 1.1.

In this section and in the following ones we use some properties of the modified Bessel function $I_{\nu}$ that can be found in the Lebedev’s monograph ([31]) and we recall now. For every $\nu>-1$, the modified Bessel function $I_{\nu}$ is defined by

[TABLE]

The following properties hold

[TABLE]

[TABLE]

where $[\nu,0]=1$ and

[TABLE]

and

[TABLE]

Suppose that $f\in L^{\infty}(\mathbb{{R}}\times(0,\infty))$ is a complex function such that supp$f$ is compact on $\mathbb{{R}}\times(0,\infty)$. Since $\int_{0}^{\infty}W_{\tau}^{\mu}(x,y)y^{\mu+1/2}dy=x^{\mu+1/2}$, $\tau,x\in(0,\infty)$, we can write

[TABLE]

Here $C>0$ depends on the support of $f$. Hence, the integral defining

[TABLE]

is absolutely convergent.

Assume now that $f\in C^{1}(\mathbb{{R}}\times(0,\infty))$ and it has compact support. By proceeding as above we can prove that

[TABLE]

where the integrals are absolutely convergent.

We can write

[TABLE]

Here and in the sequel we denote by

[TABLE]

the classical heat kernel.

According to (2.12) and (2.13) ([7, Lemma 3.1]), we have that

[TABLE]

Then,

[TABLE]

By partial integration we get

[TABLE]

Also, we have that

[TABLE]

By using (2.13), it follows that

[TABLE]

Partial integration leads to

[TABLE]

We are going to see that the integrals on the right hand side of (2.17), (2.18) and (2) are absolutely convergent. By (2.12), (2.13), and (2.14) ([7, pages 128-131]) we have that, for every $0<y<x/2$,

[TABLE]

Let $x\in(0,\infty)$ and $t\in\mathbb{R}$. Since supp$f$ is compact, there exist $0<a<x/2$, $2x<b$ and $c>0$, such that $f(t-\tau,y)=0$, $(\tau,y)\notin(-\infty,c)\times(a,b)$. Then,

[TABLE]

In a similar way we can see that

[TABLE]

Again, according to (2.12), (2.13) and (2.14) ([7, pages 128-130]) we have that

[TABLE]

Then

[TABLE]

On the other hand,

[TABLE]

In the last equality we have taken into account that

[TABLE]

Indeed, let $t\in\mathbb{R}$ and $x\in(0,\infty)$. Since $f\in C^{1}(\mathbb{R}\times(0,\infty))$ with compact support, by using mean value theorem we deduce that $|f(t-s,y)-f(t,y)|\leq Cs$, $s,y\in(0,\infty)$. Then, we can write

[TABLE]

On the other hand, for a certain $a>0$ such that $2/a<x<a/2$ and $f(t,y)=0$, $y\notin(1/a,a)$. It follows, with the obvious extension of $f$, that

[TABLE]

and

[TABLE]

Moreover, it is well known that

[TABLE]

Putting together all the above estimates we obtain

[TABLE]

We conclude that

[TABLE]

From (2.15), (2.17), (2.18), (2) and (2) we deduce that

[TABLE]

We conclude that

[TABLE]

Assume now that $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ and it has compact support. We consider the function

[TABLE]

Note that there exists $0<a<b<\infty$ such that

[TABLE]

Let $t\in\mathbb{{R}}$. There exists $\tau_{0}\in(0,\infty)$ for which

[TABLE]

By (2.14) we have that

[TABLE]

Then, from (2.12) it follows that

[TABLE]

and (2.13) implies that

[TABLE]

From (2) and (2) it follows that

[TABLE]

Hence,

[TABLE]

and the last integral is absolutely convergent.

On the other hand, for every $x,y\in(0,\infty)$,

[TABLE]

By proceeding as above we get that

[TABLE]

and the last integral is absolutely convergent.

We now consider the function

[TABLE]

Note that, by extending $f$ in the obvious way,

[TABLE]

Then,

[TABLE]

and the last integral is absolutely convergent.

Partial integration leads to

[TABLE]

In a similar way we can see that

[TABLE]

We conclude that, for $i=1,2$,

[TABLE]

By combining (2.23) and (2.28) we obtain

[TABLE]

The other representations of the derivatives of $u$ as principal values can be proved by proceeding as above and by taking into account [42, Theorem 1.3,(A)].

3. Proof of Theorem 1.2, (1) and (2).

Assume that $\mu>-1$. If $f$ is a measurable complex function defined on $\mathbb{{R}}\times(0,\infty)$, we define $L_{\mu}f$ as follows

[TABLE]

provided that the last integral exists with $(t,x)\in\mathbb{{R}}\times(0,\infty)$.

In Theorem 1.1 we established that if $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ and it has compact support, then $(\partial_{t}-\triangle_{\mu})L_{\mu}f=f$. Also, in a similar way we can see that if $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ and it has compact support, then $L_{\mu}(\partial_{t}-\triangle_{\mu})f=f$. In other words, we can see that for good enough functions, $L_{\mu}=(\partial_{t}-\triangle_{\mu})^{-1}$.

Suppose that $f\in C^{2}(\mathbb{{R}}\times(0,\infty))$ and it has compact support. According to [31, page 134] we have that $|\sqrt{z}J_{\nu}(z)|\leq C$, $z\in(1,\infty)$, and $|\sqrt{z}J_{\nu}(z)|\leq Cz^{\nu+1/2}$, $z\in(0,1)$, when $\nu>-1$. Let $z\in(0,\infty)$ and $t\in\mathbb{R}$. There exist $0<a<b<\infty$ and $c>0$ such that

[TABLE]

Let $\mu>-1/2$. From (2.12) and (2.13) we deduce that

[TABLE]

We also have that $|\sqrt{xz}J_{\mu}(xz)|\leq C$, $x,z\in(0,\infty)$. Then, we get

[TABLE]

Hence,

[TABLE]

Assume that now $-1<\mu\leq-1/2$. By using again (2.12) and (2.13) we obtain that

[TABLE]

Then, it follows that

[TABLE]

Hence,

[TABLE]

This fact justifies the interchanges in the orders of integration to get

[TABLE]

because (see [51, p. 195])

[TABLE]

Also, we have that, for certain $0<a<b<+\infty$ and $-\infty<c<d<+\infty$,

[TABLE]

Note that, fixed $z\in(0,\infty)$, $|\sqrt{yz}J_{\mu}(yz)|\leq C$, $y\in(a,b)$.

We denote, as usual, by $\mathscr{F}$ the Fourier transformation defined by, for every $\phi\in L^{1}(\mathbb{R})$, by

[TABLE]

Then,

[TABLE]

We define the space of functions $S_{\mu}$ as follows. A smooth function $f$ on $\mathbb{{R}}\times(0,\infty)$ is in $S_{\mu}$ if and only if, for every $m,k,l\in\mathbb{{N}}$,

[TABLE]

By proceeding as above we can see that if $f\in S_{\mu}$ then the integral defining $L_{\mu}(f)(t,x)$ is absolutely convergent, for every $x\in(0,\infty)$ and $t\in\mathbb{{R}}$, and

[TABLE]

We consider the function space $C_{c,0}^{\infty}(\mathbb{{R}})$ that consists of all those $C^{\infty}(\mathbb{{R}})$-functions $\phi$ such that supp$\,\phi$ is compact and $\phi(t)=0$, $t\in(-r,r)$, for some $r>0$. $C_{c,0}^{\infty}(\mathbb{{R}})$ is a dense subspace of $L^{2}(\mathbb{{R}})$. We define $Z=\mathscr{F}(C_{c,0}^{\infty}(\mathbb{{R}}))$ and

[TABLE]

Since the Fourier transform $\mathscr{F}$ is an isometry on $L^{2}(\mathbb{{R}})$, $Z$ is a dense subspace of $L^{2}(\mathbb{{R}})$. Then $Z\otimes C_{c}^{\infty}(0,\infty)$ is a dense subset of $L^{2}(\mathbb{{R}}\times(0,\infty))$. If $\alpha\in Z$ and $\beta\in C_{c}^{\infty}(0,\infty)$, for a certain $r>0$,

[TABLE]

and hence

[TABLE]

It follows that, for every $f\in Z\otimes C_{c}^{\infty}(0,\infty)$,

[TABLE]

According to [52], we have that, for every $\beta\in H_{\mu}$, $\delta_{\mu}h_{\mu}(\beta)=-h_{\mu+1}(z\beta),$ where $\delta_{\mu}=x^{\mu+1/2}\frac{d}{dx}x^{-\mu-1/2}$. Here $H_{\mu}$ denotes the space introduced by Zemanian [52, Chapter 5] consisting of all those $\phi\in C^{\infty}(0,\infty)$ such that, for every $m,k\in\mathbb{{N}}$,

[TABLE]

Since $z\beta\in H_{\mu+1}$, for every $\beta\in H_{\mu}$, we can write

[TABLE]

for each $f\in Z\otimes C_{c}^{\infty}(0,\infty)$.

We define the Riesz transformation $R_{\mu}$ by $R_{\mu}f=\mathscr{F}^{-1}h_{\mu+2}\left(\frac{z^{2}}{z^{2}+i\rho}\mathscr{F}h_{\mu}(f)\right),\;\;f\in L^{2}(\mathbb{{R}}\times(0,\infty)).$

Thus, $R_{\mu}f=\delta_{\mu+1}\delta_{\mu}L_{\mu}f$, $f\in Z\otimes C_{c}^{\infty}(0,\infty)$, and $R_{\mu}$ is bounded from $L^{2}(\mathbb{{R}}\times(0,\infty))$ into itself.

Also, for every $f\in Z\otimes C_{c}^{\infty}(0,\infty)$, we have that

[TABLE]

We define the Riesz transformation $\widetilde{R_{\mu}}$ by $\widetilde{R_{\mu}}f=\mathscr{F}^{-1}h_{\mu}\left(\frac{-i\rho}{z^{2}+i\rho}\mathscr{F}h_{\mu}(f)\right)$, $f\in L^{2}(\mathbb{{R}}\times(0,\infty))$. Thus, $\widetilde{R_{\mu}}f=\partial_{t}L_{\mu}f$, $f\in Z\otimes C_{c}^{\infty}(0,\infty)$, and $\widetilde{R_{\mu}}$ is bounded from $L^{2}(\mathbb{{R}}\times(0,\infty))$ into itself.

Suppose that $f(t,x)=\alpha(t)\beta(x)$, $t\in\mathbb{{R}}$ and $x\in(0,\infty)$, where $\alpha\in Z$ and $\beta\in C_{c}^{\infty}(0,\infty)$. We have that

[TABLE]

and the last integral is absolutely convergent. Then, we can write, for every $t\in\mathbb{R}$ and $x\in(0,\infty)$,

[TABLE]

where $\Omega_{\varepsilon}=\{(\tau,y)\in(0,\infty)\times\mathbb{R}:|y|+\sqrt{\tau}>\varepsilon\}$. By partial integration as in the proof of [42, Theorem 2.3, (B)] we obtain

[TABLE]

By proceeding as in the proof of Theorem 1.1 we can see that, for every $f\in Z\otimes C_{c}^{\infty}(0,\infty)$,

[TABLE]

where $\Omega_{\varepsilon}(x)=\{(\tau,y)\in(0,\infty)\times(0,\infty):\,|y-x|+\sqrt{\tau}>\varepsilon\}$, for every $x\in(0,\infty)$.

In a similar way we can show that, for every $f\in Z\otimes C_{c}^{\infty}(0,\infty)$,

[TABLE]

In order to prove Theorem 1.2, (ii), we use Calderón-Zygmund theory on the space of homogeneous type $(\mathbb{{R}}\times(0,\infty),m,d)$, where $m$ and $d$ denote the Lebesgue measure and the parabolic distance, respectively, on $\mathbb{{R}}\times(0,\infty)$. We now recall the definitions and results that will be useful in the sequel. We describe now Calderón-Zygmund theory in the more general vectorial setting because we will use it in the proof of Theorem 1.3.

Suppose that $X$ and $Y$ are Banach spaces. By $\mathscr{L}(X,Y)$ we denote the space of bounded operators from $X$ to $Y$. If $1\leq p<\infty$ we represent by $L^{p}(\mathbb{{R}}\times(0,\infty),X)$ and $L^{p,\infty}(\mathbb{{R}}\times(0,\infty),X)$ the Bochner Lebesgue $L^{p}$-space and weak Bochner Lebesgue $L^{p,\infty}$-space. Assume that $T$ is a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty),X)$ into $L^{p}(\mathbb{{R}}\times(0,\infty),Y)$, for some $1<p<\infty$, satisfying that

[TABLE]

for every $f\in S$, where $S$ represents a linear space that is dense in $L^{q}(\mathbb{{R}}\times(0,\infty),X)$, for every $1\leq q<\infty$. Here

[TABLE]

is a strongly measurable function, being

[TABLE]

We say that $K$ is a standard Calderón-Zygmund kernel in $(\mathbb{{R}}\times(0,\infty),m,d)$ when the following properties hold

(a)$\|K(t,x;s,y)\|_{\mathscr{L}(X,Y)}\leq\frac{C}{d((t,x),(s,y))^{3}}$, $(t,x)\neq(s,y)$.

(b) provided that $d((t,x),(s_{0},y_{0}))>d((s,y),(s_{0},y_{0}))$,

[TABLE]

If $1<p<\infty$, a weight $w$ on $\mathbb{{R}}\times(0,\infty)$ is in the Muckenhoupt class $A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$ when there exists $C>0$ such that

[TABLE]

for every ball (with respect to $d$) in $\mathbb{{R}}\times(0,\infty)$.

A weight $w$ is in $A_{1}^{*}(\mathbb{{R}}\times(0,\infty))$ when there exists $C>0$ such that, for a.e. $(t,x)\in\mathbb{{R}}\times(0,\infty)$,

[TABLE]

for every ball $B$ (with respect to $d$) containing $(t,x)$.

The Calderón-Zygmund Theorem says that if $T$ satisfies the above properties where $K$ in (3.29) is a standard Calderón-Zygmund kernel, then the operator $T$ can be extended,

(a) for every $1<q<\infty$ and $w\in A_{q}^{*}(\mathbb{{R}}\times(0,\infty))$, from $L^{p}(\mathbb{{R}}\times(0,\infty),X)\cap L^{q}(\mathbb{{R}}\times(0,\infty),w,X)$ to $L^{q}(\mathbb{{R}}\times(0,\infty),w,X)$ as a bounded operator from $L^{q}(\mathbb{{R}}\times(0,\infty),w,X)$ into $L^{q}(\mathbb{{R}}\times(0,\infty),w,Y)$;

(b) for every $w\in A_{1}^{*}(\mathbb{{R}}\times(0,\infty))$, from $L^{p}(\mathbb{{R}}\times(0,\infty),X)\cap L^{1}(\mathbb{{R}}\times(0,\infty),w,X)$ to $L^{1}(\mathbb{{R}}\times(0,\infty),w,X)$ as a bounded operator from $L^{1}(\mathbb{{R}}\times(0,\infty),w,X)$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty),w,Y)$.

Moreover, the maximal operator given by

[TABLE]

defines a bounded operator from

(a) $L^{q}(\mathbb{{R}}\times(0,\infty),w,X)$ into $L^{q}(\mathbb{{R}}\times(0,\infty),w)$, for every $1<q<\infty$ and $w\in A^{*}_{q}(\mathbb{{R}}\times(0,\infty))$;

(b) $L^{1}(\mathbb{{R}}\times(0,\infty),w,X)$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty),w)$, for every $w\in A^{*}_{1}(\mathbb{{R}}\times(0,\infty))$.

A complete study about vector valued Calderón-Zygmund theory on spaces of homogeneous type can be encountered in [43], [44] and [45].

We have that, for every $f\in Z\otimes C_{c}^{\infty}(0,\infty)$,

[TABLE]

and

[TABLE]

We consider the kernel functions defined as follows

[TABLE]

and

[TABLE]

It is clear that, for every $f\in Z\otimes C^{\infty}_{c}(0,\infty)$, $R_{\mu}f(t,x)=\int_{\mathbb{{R}}}\int_{0}^{\infty}K_{\mu}(x,t;y,\tau)f(\tau,y)dyd\tau,$ and $\widetilde{R_{\mu}}f(t,x)=\int_{\mathbb{{R}}}\int_{0}^{\infty}\widetilde{K_{\mu}}(x,t;y,\tau)f(\tau,y)dyd\tau,$ $(t,x)\not\in\text{supp}f$.

We remark that $d((t,x),(s,y))=|x-y|+\sqrt{|t-s|},$ $t,s\in\mathbb{{R}}$ and $x,y\in(0,\infty)$.

Proposition 3.7**.**

Let $\mu>1/2$ or $\mu=-1/2$. The kernels $K_{\mu}$ and $\widetilde{K_{\mu}}$ are standard Calderón-Zygmund with respect to the homogeneous type space $(\mathbb{{R}}\times(0,\infty),m,d)$.

Proof.

Firstly we analyze $K_{\mu}$. We consider the function,

[TABLE]

According to (2.14) we have that

[TABLE]

By taking into account (2.12) we obtain

[TABLE]

On the other hand, we can write

[TABLE]

By (2.13) we deduce that

[TABLE]

Hence, if $s,x,y\in(0,\infty),$ and $xy\geq s$, then

[TABLE]

and

[TABLE]

We conclude that

[TABLE]

According to (3) we have that $\displaystyle\lim_{s\to 0^{+}}\mathbb{K}_{\mu}(x,y,s)=0$, $x,y\in(0,\infty)$, $x\neq y$. Hence, $K_{\mu}$ is a continuous function on $[((0,\infty)\times\mathbb{{R}})\times((0,\infty)\times\mathbb{{R}})]\setminus D$.

By using that $\frac{\partial}{\partial x}=\delta_{\mu+2}+\frac{\mu+5/2}{x}$ and (2.14), from (3) it follows that

[TABLE]

and since $\partial_{y}=\delta_{\mu}+\frac{\mu+1/2}{y}$,

[TABLE]

It is clear that

[TABLE]

Now, we estimate $\partial_{x}\mathbb{K}_{\mu}(x,y,s)$. By (2.12), (3) leads that

[TABLE]

On the other hand, (2.13) allows us to deduce from (3) that

[TABLE]

Then, $|\partial_{x}\mathbb{K}_{\mu}(x,y,s)|\leq C\frac{e^{-c\frac{(x-y)^{2}}{s}}}{s^{5/2}}\left(\frac{(x-y)^{2}}{x}+\frac{|x-y|^{3}}{s}\right)$, $x,y,s\in(0,\infty)$ and $\frac{xy}{s}\geq 1$. We have that, for every $x,y,s\in(0,\infty)$, and $xy\geq s$,

[TABLE]

Hence, we get

[TABLE]

We conclude that

[TABLE]

We now estimate $\partial_{y}\mathbb{K}_{\mu}(x,y,s)$. By (2.12), from (3) we deduce that

[TABLE]

Note that in the last estimate we use that $\mu>1/2$ or $\mu=-1/2$.

By (3) we can write

[TABLE]

Then, (2.13) leads to

[TABLE]

Hence, by (3.38) we obtain

[TABLE]

We conclude that

[TABLE]

Now we estimate $\partial_{s}\mathbb{K}_{\mu}(x,y,s)$. Let $x,y\in(0,\infty)$. We define the function $\varphi_{x,y}(z)=\mathbb{K}(x,y,z),\,\,\,z\in\mathbb{C},\,\,\,\text{Re}\,z>0$. Thus, $\varphi_{x,y}$ is an holomorphic function in $\{z\in\mathbb{C}:Re\,z>0\}$. Note that, if $a>0$, $Arg\,\frac{a}{z}=-Arg(z)$ and Re\Bigg{(}\frac{a}{z}\Bigg{)}=\frac{a}{|z|^{2}}Re\,z, $z\in\mathbb{C}$. Note that $Rez\geq\frac{\sqrt{2}}{2}|z|$ provided that $|Argz|\leq\frac{\pi}{4}.$ Hence, $\left|e^{-\frac{(x-y)^{2}}{4z}}\right|=e^{\frac{-Rez(x-y)^{2}}{4|z|^{2}}}\leq e^{-\frac{\sqrt{2}}{8}\frac{(x-y)^{2}}{|z|}},$ and $\left|e^{-\frac{x^{2}+y^{2}}{4z}}\right|\leq e^{-\frac{\sqrt{2}}{8}\frac{x^{2}+y^{2}}{|z|}}$, when $|Argz|\leq\frac{\pi}{4}.$

According (2.12) and (2.13) as in (3) and (3), we can obtain

[TABLE]

By using Cauchy integral formula we get

[TABLE]

Here $C$ and $c$ do not depend on $x,y\in(0,\infty)$. Then, we obtain

[TABLE]

Hence, for every $x,y\in(0,\infty)$, $x\neq y$, $\displaystyle\lim_{s\to 0^{+}}\partial_{s}\mathbb{K}_{\mu}(x,y,s)=0$.

Then, we deduce that $\mathbb{K}_{\mu}$ is in $C^{1}((0,\infty)\times(0,\infty)\times\mathbb{{R}}\setminus\{(x,x,0):\;x\in(0,\infty)\})$, and $K_{\mu}$ is in $C^{1}((\mathbb{{R}}\times(0,\infty)\times\mathbb{{R}}\times(0,\infty))\setminus D)$.

According to (3.39), (3.40) and (3.41), we have that

[TABLE]

[TABLE]

for every $(x,t;y,\tau)\in[((0,\infty)\times\mathbb{{R}})\times((0,\infty)\times\mathbb{{R}})]\setminus D$, where $D=\{(x,t;x,t):\;x\in(0,\infty)\;\;\text{and}\;\;t\in\mathbb{{R}}\}.$

Let now $x,y,y_{0}\in(0,\infty)$ and $t,\tau,\tau_{0}\in\mathbb{{R}}$ such that $d((x,t);(y_{0},\tau_{0}))=\sqrt{|t-\tau_{0}|}+|x-y_{0}|>2(\sqrt{|\tau-\tau_{0}|}+|y-y_{0}|)=2d((y,\tau);(y_{0},\tau_{0})).$ Then, $s(x,t;y_{0},\tau_{0})+(1-s)(x,t;y,\tau)\not\in D$, for every $s\in(0,1)$. Indeed, suppose that $s\in(0,1)$ and that $s(x,t;y_{0},\tau_{0})+(1-s)(x,t;y,\tau)\in D$. We have that $x=sy_{0}+(1-s)y$ and $t=s\tau_{0}+(1-s)\tau$. It follows that

[TABLE]

and this is not possible.

By using the mean value theorem, we can write

[TABLE]

for certain $s\in(0,1)$. Then,

[TABLE]

Note that $|\tau-\tau_{0}|\leq\sqrt{|\tau-\tau_{0}|}(\sqrt{|\tau-\tau_{0}|}+|y-y_{0}|)<\frac{1}{2}\sqrt{|\tau-\tau_{0}|}(\sqrt{|t-\tau_{0}|}+|x-y_{0}|),$ and

[TABLE]

We get $|K_{\mu}(x,t;y,\tau)-K_{\mu}(x,t;y_{0},\tau_{0})|\leq C\frac{|y-y_{0}|+\sqrt{|\tau-\tau_{0}|}}{(\sqrt{|t-\tau_{0}|}+|x-x_{0}|)^{4}}.$

By proceeding in a similar way we also obtain $|K_{\mu}(y,\tau;x,t)-K_{\mu}(y_{0},\tau_{0};x,t)|\leq C\frac{|y-y_{0}|+\sqrt{|\tau-\tau_{0}|}}{(\sqrt{|t-\tau_{0}|}+|x-x_{0}|)^{4}}.$ We have just proved that $K_{\mu}$ is a standard Calderón-Zygmund kernel with respect to the homogeneous type space $(\mathbb{{R}}\times(0,\infty),m,d)$.

Now we prove that $\widetilde{K_{\mu}}$ is a standard Calderón-Zygmund kernel with respect to the homogeneous type space $(\mathbb{{R}}\times(0,\infty),m,d)$.

We define

[TABLE]

By (2.14) we get

[TABLE]

According to (2.12), we can write

[TABLE]

Note that $\mu\geq-1/2$.

Also, by (2.13), we have that

[TABLE]

Then,

[TABLE]

We conclude that

[TABLE]

By proceeding in a similar way, (2.12) and (2.13) lead to

[TABLE]

By using Cauchy integral formula we deduce that

[TABLE]

Then, we obtain

[TABLE]

On the other hand, by using (2.14) (see (2)), we have that

[TABLE]

Then, (2.12) implies that

[TABLE]

provided that $\mu>1/2$ or $\mu=-1/2$. Also, (2.13) leads to

[TABLE]

We obtain,

[TABLE]

and, by taking $z=\max\{x,y\}$,

[TABLE]

We conclude that

[TABLE]

The same arguments, by using again (2.12), (2.13) and (2.14), allows us to obtain

[TABLE]

and Cauchy integral formula leads to

[TABLE]

Symmetries imply that

[TABLE]

We get

[TABLE]

Putting together the above estimates and proceeding as in the $K_{\mu}$-case we can prove that $\widetilde{K_{\mu}}$ is a standard Calderón-Zygmund kernel with respect to the homogeneous type space $(\mathbb{{R}}\times(0,\infty),m,d)$.

Thus, the proof of this proposition is finished.

∎

Now the statements in Theorem 1.2, (ii), follows from Calderón-Zygmund theorem.

4. Proof of Theorem 1.2, (3), (4) and (5)

In order to prove the parts (3), (4) and (5) in Theorem 1.2 we use a procedure which is different from the one employed in Section 3 to prove Theorem 1.2, (2). As it was mentioned in the introduction, Ping, Stinga and Torrea [42] investigated $L^{p}$-boundedness properties of the Riesz transformations associated with the parabolic equation (1.1). They studied, when a one dimensional spatial variable is considered, the following two operators

[TABLE]

and

[TABLE]

where, for every $\varepsilon>0$, $\Omega_{\varepsilon}=\{(s,y)\in(0,\infty)\times\mathbb{R},\sqrt{s}+y>\varepsilon\}$. Our procedure consists, roughly speaking, in studying the $L^{p}$-boundedness properties of the difference operators $R_{\mu}-R$ and $\widetilde{R}_{\mu}-\widetilde{R}$. Then, $L^{p}$-boundedness properties of $R_{\mu}$ and $\widetilde{R}_{\mu}$ are deduced from the corresponding ones of $R$ and $\widetilde{R}$, respectively, established in [42, Theorem 2.3, (B)].

Calderón-Zygmund theorem employed in the proof of Theorem 1.2, (ii), in the previous section allows us to consider weighted $L^{p}$-spaces but the parameter $\mu$ is restricted to $\mu=-1/2$ or $\mu>1/2$. This comparative approach applies to the full range of values of $\mu>-1$.

4.1. Riesz transformation $\widetilde{R_{\mu}}$

We consider the operator

[TABLE]

where we name $\mathscr{K}_{\mu}(s,x,y)=\partial_{s}W_{s}^{\mu}(x,y),$ $s,x,y\in(0,\infty)$.

We shall also fix our attention in the operator

[TABLE]

where $\mathscr{K}(s,x,y)=\partial_{s}W_{s}(x-y)$, $s\in(0,\infty)$ and $x,y\in\mathbb{{R}}$.

Ping, Stinga and Torrea in [42] studied $L^{p}$-boundedness properties for the operator $\widetilde{R}$. Our objective is to prove $L^{p}$-boundedness properties for the operator $\widetilde{R_{\mu}}$ by using the corresponding ones for $\widetilde{R}$.

According to (2.21) we have that

[TABLE]

By (2.20) we obtain

[TABLE]

From (4.47) and (4.1) we deduce that

[TABLE]

By proceeding as in (2.20) we get

[TABLE]

Also, from (3.43), it follows that

[TABLE]

Then,

[TABLE]

and when $\mu>-1/2$,

[TABLE]

Observe that this estimate can not be improved for $-1<\mu\leq-1/2$.

We now suppose that $g$ is a complex valued continuous function with compact support in $(0,\infty)$. We define $g_{0}$ as the odd extension of $g$ to $\mathbb{{R}}$. We can write

[TABLE]

This fact and the following estimate will be useful in the sequel.

Note that

[TABLE]

Then,

[TABLE]

Let $f\in C_{c}^{\infty}(\mathbb{{R}}\times(0,\infty))$. We define

[TABLE]

Thus, $f_{0}\in C_{c}^{\infty}(\mathbb{{R}}^{2})$. We can write

[TABLE]

This fact suggests the following analysis. From (4.1) we deduce that

[TABLE]

and

[TABLE]

Finally, we have that

[TABLE]

Fix $\mu>-1/2$. Let $f\in C_{c}^{\infty}(\mathbb{{R}}\times(0,\infty))$. We define $f_{0}$ as above. According to (4.1) and [42, Theorem 2.3, (B)], we can write, for every $t\in\mathbb{R}$ and $x\in(0,\infty)$,

[TABLE]

where $W_{\varepsilon}(x)=\{(\tau,y)\in(0,\infty)\times\mathbb{R}:\,\sqrt{\tau}+|x-y|>\varepsilon\}$, for every $x\in\mathbb{R}$ and $\varepsilon>0$.

Then,

[TABLE]

Note that from (4.1), (4.1), (4.1), (4.1), (4.1) and (4.1) we deduce that the last six integrals are absolutely convergent. We get

[TABLE]

where $T_{\mu}(t,x,y)=\mathscr{K}_{\mu}(t,x,y)-\mathscr{K}(t,x,y)+\mathscr{K}(t,x,-y)$, $t\in\mathbb{R}$ and $x,y\in(0,\infty)$.

Note that

[TABLE]

We consider the operator $\widetilde{T_{\mu}}(f)(t,x)=\int_{0}^{\infty}\int_{0}^{\infty}T_{\mu}(\tau,x,y)f(t-\tau,y)d\tau dy$, $t\in\mathbb{{R}}$, $x\in(0,\infty)$.

According to (4.56) we obtain

• •

There exists a constant $C>0$ such that, for every $f\in L^{1}(\mathbb{{R}}\times(0,\infty))$,

[TABLE]

We have used that $T_{\mu}(u,x,y)=T_{\mu}(u,y,x)$, $u,x,y\in(0,\infty)$.

• •

There exists a constant $C>0$ such that, for every $f\in L^{\infty}(\mathbb{{R}}\times(0,\infty))$,

[TABLE]

Marcinkiewicz interpolation theorem allows us to conclude that $\widetilde{T_{\mu}}$ is a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself, for every $1\leq p\leq\infty$.

By [42, Theorem 2.3, (B)] we deduce that, for every $1\leq p<\infty$, the operator $\widetilde{R_{\mu}}$ can be extended to $L^{p}(\mathbb{{R}}\times(0,\infty))$ as a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself when $1<p<\infty$ and from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$.

As it was established in [42, Theorem 2.3, (B)], the maximal opertor

[TABLE]

is bounded from $L^{p}(\mathbb{{R}}^{2})$ into itself, for every $1<p<\infty,$ and from $L^{1}(\mathbb{{R}}^{2})$ into $L^{1,\infty}(\mathbb{{R}}^{2})$. Then, the above results imply that the maximal operator

[TABLE]

is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself, for every $1<p<\infty$, and from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$.

Since the principal value $\displaystyle\lim_{\epsilon\to 0^{+}}\int_{\Omega_{\epsilon}(x)}\widetilde{R_{\mu}}(\tau,x,y)f(t-\tau,y)d\tau dy$ exists, for every $f\in C_{c}^{\infty}(\mathbb{{R}}\times(0,\infty))$ and $(t,x)\in\mathbb{{R}}\times(0,\infty)$, and as $C_{c}^{\infty}(\mathbb{{R}}\times(0,\infty))$ is a dense subspace of $L^{p}(\mathbb{{R}}\times(0,\infty))$, $1\leq p<\infty$, we have that, for every $f\in L^{p}(\mathbb{{R}}\times(0,\infty))$, $1\leq p<\infty$, there exists the principal value $\displaystyle\lim_{\epsilon\to 0^{+}}\int_{\Omega_{\epsilon}(x)}\widetilde{R_{\mu}}(\tau,x,y)f(t-\tau,y)d\tau dy$ for almost all $(t,x)\in\mathbb{{R}}\times(0,\infty)$. Also, the operator $\widehat{R_{\mu}}$ defined on $L^{p}(\mathbb{{R}}\times(0,\infty))$, $1\leq p<\infty$, as follows

[TABLE]

is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself, for every $1<p<\infty$, and from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$.

Our objective is to study the $L^{p}$-boundedness properties for $\widetilde{R_{\mu}}$ when $-1<\mu\leq-1/2$. We have that

[TABLE]

where $S_{\mu}=h_{\mu}h_{\mu+2}$ and $S_{\mu}^{*}=h_{\mu+2}h_{\mu}$.

The composition operators $h_{\mu}h_{\nu}$ are named transplantation operators associated with Hankel transforms.

According to [39, Theorem 2.1], if $\mu>-1$, $1<p<\infty$ and $v$ is a nonnegative measurable function such that

[TABLE]

then, the operator $S_{\mu}$ can be extended to $L^{p}(v)$ as a bounded operator from $L^{p}(v)$ into itself. Here, as usual, $p^{\prime}$ denotes the conjugated of $p$, that is, $p^{\prime}=\frac{p}{p-1}$.

Since

[TABLE]

provided that $p(\mu+1/2)+1>0$ and $1-p^{\prime}(\mu+3/2)<0$, $S_{\mu}$ defines a bounded operator from $L^{p}((0,\infty))$ into itself when $1<p<\infty$ and $\mu>-1/2-1/p$ and $\mu>-1$. Then $S_{\mu}^{*}$ is bounded from $L^{p}((0,\infty))$ into itself when $1<p<\infty$, $\mu>-1$ and $\mu>-1/2-(1-1/p)=-3/2+1/p$. Hence, since $\widetilde{R_{\mu+2}}$ is bounded from $L^{p}((0,\infty))$ into itself when $1<p<\infty$ and $\mu>-1$, $\widetilde{R_{\mu}}$ defines a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself provided that $-1<\mu\leq 1/2$ and $-1/2-\mu<1/p<3/2+\mu$.

Our results about $\widetilde{R_{\mu}}$ can be summarized as follows:

• •

$\widetilde{R_{\mu}}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself when

1. (1)

$\mu>-1/2$ and $1<p<\infty$. 2. (2)

$-1<\mu\leq-1/2$ and $-1/2-\mu<1/p<3/2+\mu$.

• •

$\widetilde{R_{\mu}}$ is bounded from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$, when $\mu>-1/2$.

\mu$$\frac{1}{p}weakstrong\mu=-\frac{1}{2}-\frac{1}{p}$$\mu=-\frac{3}{2}+\frac{1}{p}$$\frac{1}{2}$${1}$$\frac{-1}{2}$${-1}

4.2. Riesz transformation $R_{\mu}$

We consider the operator $R_{\mu}$ defined by

[TABLE]

for every $f\in C_{c}^{\infty}((\mathbb{{R}}\times(0,\infty))$. We recall that

[TABLE]

We also write

[TABLE]

According to (3) we have that

[TABLE]

Also, we get

[TABLE]

We define $T_{\mu}(x,y,s)=\mathbb{K}_{\mu}(x,y)-\mathbb{K}(x,y,s)+\mathbb{K}(x,-y,s),\;\;s,x,y\in(0,\infty).$

Our first objective is to study when

[TABLE]

is true.

We decompose the proof of (4.57) in several steps. According to (2.12), we can write

[TABLE]

By (2.13) we get

[TABLE]

From (4.2) it follows that, for every $s,x,y\in(0,\infty)$ such that $xy\geq s$,

[TABLE]

By using (4.2) and (4.60) we obtain

[TABLE]

and

[TABLE]

We have that

[TABLE]

Then,

[TABLE]

and

[TABLE]

From (4.63) and (4.64) we deduce that

[TABLE]

and

[TABLE]

On the other hand, we have that

[TABLE]

We can write

[TABLE]

Finally we are going to estimate \displaystyle\int_{x/2}^{3x/2}\int_{0}^{\infty}\big{|}\mathbb{K}_{\mu}(x,y,s)-\mathbb{K}(x,y,s)\big{|}dsdy.

According to (4.2) we obtain

[TABLE]

By using (3) we get

[TABLE]

but at this moment it is not sufficient. We need to improve the last estimates.

Since $I_{\nu}(z)=\frac{e^{z}}{\sqrt{2\pi z}}\left(1+\frac{1-4\nu^{2}}{8z}+O\left(\frac{1}{z^{2}}\right)\right)$, for $z>0$ and $\nu>-1$ (see (2.13)), we can write

[TABLE]

We get

[TABLE]

By using the last equality we deduce that

[TABLE]

By combining (4.2), (4.2), (4.65), (4.2), (4.2), (4.2) and (4.2) we obtain

[TABLE]

This property implies that the operator

[TABLE]

is bounded from $L^{\infty}(\mathbb{{R}}\times(0,\infty))$ into itself, for every $\mu>-1$.

The operator $R_{\mu}$ is bounded from $L^{2}(\mathbb{{R}}\times(0,\infty))$ into itself and the operator $R$ is bounded from $L^{2}(\mathbb{{R}}^{2})$ into itself, see [42, Theorem 2.3, (B)]. From these facts we deduce that $T_{\mu}$ is bounded from $L^{2}(\mathbb{{R}}\times(0,\infty))$ into itself. Hence, interpolation theorem implies that $T_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself, for every $2\leq p\leq\infty$. By using again [42, Theorem 2.3, (B)] we conclude that $R_{\mu}$ defines a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself for every $2\leq p<\infty$ and $\mu>-1$.

It is remarkable that the operator $R_{\mu}$ is not selfadjoint in $L^{2}(\mathbb{{R}}\times(0,\infty))$.

To simplify we now consider the function

[TABLE]

We are going to see that $\displaystyle\sup_{x\in(0,\infty)}\int_{0}^{\infty}\int_{0}^{\infty}|M_{\mu}(x,y,s)|dyds<\infty.$

We define

[TABLE]

From (4.2), (4.2) and (4.60) we deduce that

[TABLE]

[TABLE]

and, for every $x,y,s\in(0,\infty)$ and $xy\geq s$,

[TABLE]

By (4.70) and (4.72) it follows that

[TABLE]

and

[TABLE]

provided that $\mu>-1/2$.

Also, for $x\in(0,\infty)$, we have

[TABLE]

and, as in (4.2),

[TABLE]

From (4.2), (4.2), (4.2) and (4.2) and by taking into account the other above estimates, we deduce that

[TABLE]

Then, the operator $T_{\mu}$ is bounded from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into itself, provided that $\mu>-1/2$.

By invoking interpolation theorem we infer that $T_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty)$ into itself, for every $1\leq p\leq\infty$ when $\mu>-1/2$. By using again [42, Theorem 2.3, (B)] we conclude that $R_{\mu}$ defines, for every $1<p<\infty$, a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself and from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$ provided that $\mu>-1/2$.

By following the same argument as in the previous section, the use of the maximal operator associated to the singular integral $R_{\mu}$ and [42, Theorem 2.3, (B)] allow us to conclude that, for every $f\in L^{p}(\mathbb{{R}}\times(0,\infty))$, $1\leq p<\infty$, the limit

[TABLE]

exists, for a.e. $(t,x)\in\mathbb{{R}}\times(0,\infty)$, when $\mu>-1/2$. Moreover, the operator $\mathbb{R}_{\mu}$ defined by

[TABLE]

is bounded, for every $1<p<\infty$, from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself and from $L^{1}(\mathbb{{R}}\times(0,\infty)$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty)$ provided that $\mu>-1/2$.

Our next objective is to complete the study of the boundedness of the operator $R_{\mu}$ when $-1<\mu\leq-1/2$.

We have that, for every $f\in L^{2}(\mathbb{{R}}\times(0,\infty))$, $R_{\mu}f=\mathscr{F}^{-1}h_{\mu+2}\left(\frac{z^{2}}{z^{2}+i\rho}\mathscr{F}h_{\mu}(f)\right)$.

Then, the adjoint $R_{\mu}^{*}$ of $R_{\mu}$ is given by

[TABLE]

where $\tilde{f}(t,x)=f(-t,x)$, $t\in\mathbb{{R}}$ and $x\in(0,\infty)$.

We consider the operator $\mathscr{H}_{\mu}f=\mathscr{F}^{-1}h_{\mu}\left(\frac{z^{2}}{z^{2}+i\rho}h_{\mu+2}\mathscr{F}{f}\right)$, $f\in L^{2}(\mathbb{{R}}\times(0,\infty))$. Since $h_{\alpha}^{2}=I$ in $L^{2}((0,\infty))$, for every $\alpha>-1$ we can write

[TABLE]

where $S_{\mu}=h_{\mu}h_{\mu+2}$.

According to [39, Theorem 2.1], $S_{\mu}$ defines a bounded operator from $L^{p}((0,\infty))$ into itself when $1<p<\infty$, $\mu>-1/2-1/p$ and $\mu>-1$. By taking into account that $R_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself when $2\leq p<\infty$, we conclude that $R_{\mu}^{*}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself provided that $\mu>-1/2-1/p$ and $2\leq p<\infty$. Duality implies that $R_{\mu}$ defines a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself provided that $1<p\leq 2$ and $\mu>1/p-3/2$.

Our results about $R_{\mu}$ can be summarized as follows:

• •

$R_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$ into itself when

1. (1)

$\mu>-1/2$ and $1<p<\infty$. 2. (2)

$-1<\mu\leq-1/2$ and $p>\frac{1}{\mu+3/2}$.

• •

$R_{\mu}$ is bounded from $L^{1}(\mathbb{{R}}\times(0,\infty))$ into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$, when $\mu>-1/2$.

\mu$$\frac{1}{p}weakstrong\mu=-\frac{3}{2}+\frac{1}{p}$$\frac{1}{2}$${1}$$\frac{-1}{2}$${-1}

5. Proof of Theorem 1.3

We firstly consider the Riesz transform $R_{\mu}$ defined by

[TABLE]

for every $f\in L^{p}(\mathbb{{R}}\times(0,\infty))$ $1\leq p<\infty$, where $\mathbb{K}_{\mu}(s,x,y)=\delta_{\mu+1}\delta_{\mu}W_{s}^{\mu}(x,y)$, $s,x,y\in(0,\infty)$. We have that, for each $f\in L^{p}(\mathbb{{R}}\times(0,\infty))$, $1\leq p<\infty$,

[TABLE]

Proposition 5.8**.**

Let $\mu>-1/2$ and $1<p<\infty$. The operator $R_{\mu}$ can be extended to $L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ as a bounded operator from $L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ into itself, for $1<q<\infty$ and from $L^{1}(\mathbb{{R}},L^{p}((0,\infty)))$ into $L^{1,\infty}(\mathbb{{R}},L^{p}((0,\infty)))$.

Proof.

We define, for every $s\in(0,\infty)$,

[TABLE]

for every $F\in L^{p}((0,\infty))$.

Note that, according to (3), (3.33) and (3),

[TABLE]

Then,

[TABLE]

and also

[TABLE]

It follows that, for every $s\in(0,\infty)$, $\mathbb{T}_{s}$ defines a bounded operator from $L^{p}((0,\infty))$ into itself and $\|\mathbb{T}_{s}\|_{p\to p}\leq\frac{C}{s}$.

For every $t,s\in\mathbb{{R}}$, we define

[TABLE]

for $F\in L^{p}((0,\infty))$.

Thus, for every $t,s\in\mathbb{{R}}$, $\mathbb{H}(t,s)\in\mathscr{L}(L^{p}((0,\infty)))$ and $\|\mathbb{H}(t,s)\|_{p\to p}\leq\frac{C}{|t-s|}$.

We consider, for every $g\in C_{c}^{\infty}(\mathbb{R}\times(0,\infty))\subset L^{p}(\mathbb{R},L^{p}((0,\infty)))$,

[TABLE]

Note that if $g\in C_{c}^{\infty}(\mathbb{R}\times(0,\infty))$ and $t\not\in\text{supp}(g)$, then

[TABLE]

and the integral $\displaystyle\int_{\mathbb{{R}}}\mathbb{H}(t,s)(g(s))ds$ converges in the $L^{q}(\mathbb{{R}})$-Bochner sense.

We established in Theorem 1.2 that $R_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))=L^{p}(\mathbb{{R}},L^{p}((0,\infty)))$ into itself.

Let $g\in C^{\infty}_{c}(\mathbb{{R}}\times(0,\infty))$ and $\ell\in(L^{p}((0,\infty)))^{\prime}=L^{p^{\prime}}((0,\infty))$. According to the well-known properties of Bochner integrals we have that

[TABLE]

The interchange of the order of integration is justified by (5). Note that

[TABLE]

We conclude that, for every $t\not\in\text{supp}(g),$

[TABLE]

Suppose that $t_{1},t_{2},s\in(0,\infty)$, being $|t_{1}-s|>2|t_{1}-t_{2}|$. Then, $s>\max\{t_{1},t_{2}\}$ or $s<\min\{t_{1},t_{2}\}$. Let $g\in L^{p}((0,\infty))$. Assume that $s<\min\{t_{1},t_{2}\}$. We have that

[TABLE]

Note that $\mathbb{H}(t_{1},s)(g)-\mathbb{H}(t_{2},s)(g)=0$ when $s>\max\{t_{1},t_{2}\}$. According to (3.41) we get

[TABLE]

for some $u\in(\min\{t_{1},t_{2}\}-s,\max\{t_{1},t_{2}\}-s)$. Suppose that $t_{1}<t_{2}$. Then, $u>t_{1}-s$ and $t_{2}-s=t_{2}-t_{1}+t_{1}-s<3(t_{1}-s)/2<3u/2$. We obtain

[TABLE]

It follows that

[TABLE]

Also,

[TABLE]

We conclude that $\mathbb{H}(t_{1},s)-\mathbb{H}(t_{2},s)\in\mathscr{L}(L^{p}((0,\infty)))$ and

[TABLE]

By using vector valued Calderón-Zygmund theory we deduce that the operator $R_{\mu}$ can be extended to $L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ as a bounded operator from

• •

$L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ into itself for every $1<q<\infty$,

• •

$L^{1}(\mathbb{{R}},L^{p}((0,\infty)))$ into $L^{1,\infty}(\mathbb{{R}},L^{p}((0,\infty)))$.

∎

Let now $s\in(0,\infty)$. We consider again the operator

[TABLE]

According to (5.77) we get

[TABLE]

Also by (3.39) and (3.40) we obtain

[TABLE]

provided that $\mu>1/2$ or $\mu=-1/2$.

Since $\mathbb{T}_{s}$ is bounded from, for instance, $L^{2}((0,\infty))$ into itself and $\|\mathbb{T}_{s}\|_{2\to 2}\leq\frac{C}{s}$, Calderón-Zygmund theory implies that, for every $1<p<\infty$ and $\omega\in A_{p}((0,\infty))$, the operator $\mathbb{T}_{s}$ can be extended to $L^{p}((0,\infty),\omega)$ as a bounded operator from $L^{p}((0,\infty),\omega)$ into itself and $\|\mathbb{\mathbb{T}}_{s}\|_{L^{p}((0,\infty),\omega)\to L^{p}((0,\infty),\omega)}\leq\frac{C}{s},$ provided that $\mu>1/2$ or $\mu=-1/2$.

In the sequel we assume that $\mu>1/2$ or $\mu=-1/2$.

Let $1<p<\infty$ and $\omega\in A_{p}((0,\infty))$. We define as above, for every $t,s\in\mathbb{{R}}$,

[TABLE]

for $F\in L^{p}((0,\infty))$. Also, for every $g\in L^{p}(\mathbb{R},L^{p}((0,\infty),\omega))$, we consider

[TABLE]

We have that, for every $1<p<\infty$ and $\omega\in A_{p}((0,\infty))$,

[TABLE]

Then, we infer that, for every $g\in C_{c}^{\infty}(\mathbb{R}\times(0,\infty))$ and $t\notin supp\,g$,

[TABLE]

According Theorem 1.2, (2), $R_{\mu}$ is bounded from $L^{p}(\mathbb{R}\times(0,\infty),W)=L^{p}(\mathbb{R},L^{p}((0,\infty),\omega))$ into itself, where $W(t,x)=\omega(x)$, $(t,x)\in\mathbb{R}\times(0,\infty)$.

Suppose that $t_{1},t_{2},s\in(0,\infty)$, being $|t_{1}-s|>2|t_{1}-t_{2}|$. We are going to see that,

[TABLE]

We have proved that,

[TABLE]

From (5.79) we deduce that

[TABLE]

Our objective is to see that

[TABLE]

and

[TABLE]

Assume that $s<\min\{t_{1},t_{2}\}$. We can write

[TABLE]

for some $u\in(\min\{t_{1},t_{2}\}-s,\max\{t_{1},t_{2}\}-s)$.

We have that (see (3))

[TABLE]

Let $x,y\in(0,\infty)$. We define the function

[TABLE]

$F_{x,y}$ is holomorphic in $\mathbb{{C}}\setminus(-\infty,0]$.

According to (2.12) and (2.13), by proceeding as in the proof of (3.39), we obtain

[TABLE]

By using Cauchy integral formula we obtain that

[TABLE]

By proceeding as above we get

[TABLE]

and

[TABLE]

for $x,y\in(0,\infty)$, $x\neq y$.

Calderón-Zygmund theory leads to

[TABLE]

In a similar way we can see that

[TABLE]

provided $|s_{1}-t|\geq 2|s_{1}-s_{2}|$.

Calderón-Zygmund theory implies that, for every $\omega\in A^{p}((0,\infty))$, $1<p<\infty$, $R_{\mu}$ defines a bounded operator

• •

from $L^{q}(\mathbb{{R}},v,L^{p}((0,\infty),\omega))$ into itself, for every $1<q<\infty$ and $v\in A^{q}(\mathbb{{R}})$,

• •

from $L^{1}(\mathbb{{R}},v,L^{p}((0,\infty),\omega))$ into $L^{1,\infty}(\mathbb{{R}},v,L^{p}((0,\infty),\omega))$, for every $v\in A^{1}(\mathbb{{R}})$,

provided that $\mu=-1/2$ and $\mu>1/2$.

In particular, for every $1<p<\infty$, $v\in A^{p}(\mathbb{{R}})$ and $\omega\in A^{p}((0,\infty))$, $R_{\mu}$ defines a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty),v\omega)$ into itself, when $\mu=-1/2$ or $\mu>1/2$.

We had proved that, for every $1<p<\infty$ and $W\in A^{p}_{*}(\mathbb{{R}}\times(0,\infty))$, $R_{\mu}$ defines a bounded operator from $L^{p}(\mathbb{{R}}\times(0,\infty),W)$ into itself.

The proof of Theorem 1.3 for the Riesz transformation $R_{\mu}$ is finished.

Our next objective is to get mixed norm inequalities for the Riesz transform $\widetilde{R_{\mu}}$ defined by, for every $f\in L^{p}(\mathbb{{R}}\times(0,\infty))$, $1\leq p<\infty$,

[TABLE]

where $\mathscr{K}_{\mu}(s,x,y)=\frac{\partial}{\partial s}(W_{s}^{\mu}(x,y))$, $s,x,y\in(0,\infty)$.

Proposition 5.9**.**

Let $\mu>-1/2$ and $1<p<\infty$. The operator $\widetilde{R_{\mu}}$ can be extended to $L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ as a bounded operator from $L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ into itself, for $1<q<\infty$ and from $L^{1}(\mathbb{{R}},L^{p}((0,\infty)))$ into $L^{1,\infty}(\mathbb{{R}},L^{p}((0,\infty)))$.

Proof.

This result can be proved by proceeding as in the proof of Proposition 5.8. We define, for every $s\in(0,\infty)$,

[TABLE]

By (3.43), we have that

[TABLE]

Then, for every $s\in(0,\infty)$, $\mathbb{T}_{s}$ defines a bounded operator from $L^{p}((0,\infty))$ into itself and $\|\mathbb{T}_{s}\|_{L^{p}((0,\infty))\to L^{p}((0,\infty))}\leq\frac{C}{s}.$

For every $t,s\in\mathbb{{R}}$, we define

[TABLE]

for $F\in L^{p}((0,\infty))$.

Thus, for every $t,s\in\mathbb{{R}}$, $\mathscr{H}(t,s)\in\mathscr{L}(L^{p}((0,\infty)))$ and $\|\mathscr{H}(t,s)\|_{p\to p}\leq\frac{C}{|t-s|}$.

Let $1<q<\infty$, We consider, for every $g\in L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$,

[TABLE]

In Theorem 1.2 we proved that the Riesz transformation $\widetilde{R_{\mu}}$ is bounded $L^{p}(\mathbb{{R}}\times(0,\infty))=L^{p}(\mathbb{{R}},L^{p}((0,\infty)))$.

We have that, for every $t\not\in\text{supp}(g),$

[TABLE]

Also, if $t_{1},t_{2},s\in(0,\infty)$, being $|t_{1}-s|>2|t_{1}-t_{2}|$, $\mathscr{H}(t_{1},s)-\mathscr{H}(t_{2},s)\in\mathscr{L}(L^{p}((0,\infty)))$.

[TABLE]

By invoking vector valued Calderón-Zygmund theory we prove that the operator $\widetilde{R_{\mu}}$ can be extended to $L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ as a bounded operator from

• •

$L^{q}(\mathbb{{R}},L^{p}((0,\infty)))$ into itself for every $1<q<\infty$,

• •

$L^{1}(\mathbb{{R}},L^{p}((0,\infty)))$ into $L^{1,\infty}(\mathbb{{R}},L^{p}((0,\infty)))$.

∎

Suppose now that $\mu>1/2$ or $\mu=-1/2$.

By (5.80), we get

[TABLE]

From (3.44), we have that

[TABLE]

By symmetry, we also have that

[TABLE]

By using Calderón-Zygmund theory we deduce that, for every $1<p<\infty$ and $\omega\in A^{p}((0,\infty))$, $\mathscr{T}_{s}$ defines a bounded operator from $L^{p}((0,\infty),\omega)$ into itself and $\|\mathscr{T}_{s}\|_{L^{p}((0,\infty),\omega)\to L^{p}((0,\infty),\omega)}\leq\frac{C}{s}$, for each $s\in(0,\infty)$.

We consider again, for every $t,s\in\mathbb{{R}}$,

[TABLE]

Let $1<p<\infty$ and $\omega\in A^{p}((0,\infty))$. For every $t,s\in\mathbb{{R}}$, $\mathscr{H}(t,s)$ defines a bounded operator from $L^{p}((0,\infty),\omega)$ into itself and $\|\mathscr{H}(t,s)\|_{L^{p}((0,\infty),\omega)\to L^{p}((0,\infty),\omega)}\leq\frac{C}{|t-s|}$.

Let $1<q<\infty$, We consider, as above, for every $g\in L^{q}(\mathbb{{R}},L^{p}((0,\infty)),\omega))$,

[TABLE]

The last integral is absolutely convergent in the $L^{p}((0,\infty),\omega)$-Bochner sense. Moreover, by proceeding as above we get that, if $g\in L^{q}(\mathbb{{R}},L^{p}((0,\infty)),\omega)$ and $t\not\in\text{supp}(g)$, then

[TABLE]

Suppose that $t_{1},t_{2},s\in\mathbb{{R}}$, being $|t_{1}-s|>2|t_{1}-t_{2}|$. Then, we have that $\mathscr{H}(t_{1},s)-\mathscr{H}(t_{2},s)$ defines a bounded operator from $L^{p}((0,\infty),\omega)$ into itself and

[TABLE]

Moreover, according to Theorem 1.2, (2), the operator $\widetilde{R_{\mu}}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty),W)=L^{p}(\mathbb{R},L^{p}((0,\infty),\omega))$, where $W(t,x)=\omega(x)$, $(t,x)\in\mathbb{{R}}\times(0,\infty)$, because $W\in A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$.

Again, according to Calderón-Zygmund theory we deduce that $\widetilde{R_{\mu}}$ defines, for every $1<q<\infty$ and $v\in A^{q}(\mathbb{{R}})$, a bounded operator from $L^{q}(\mathbb{{R}},v,L^{p}((0,\infty),\omega))$ into itself and from $L^{1}(\mathbb{{R}},v,L^{p}((0,\infty),\omega))$ into $L^{1,\infty}(\mathbb{{R}},v,L^{p}((0,\infty),\omega))$, for every $v\in A^{1}(\mathbb{{R}})$.

The same remark at the end of the study of the mixed norm inequalities for $R_{\mu}$ is now in order with respect to $\widetilde{R_{\mu}}$.

6. Proof of Theorems 1.4, 1.5, and 1.6

6.1. Proof of Theorem 1.4

Let $\mu>-1$. We consider the following Cauchy problem

[TABLE]

Suppose initially that $f\in L_{c}^{\infty}((0,\infty)\times(0,\infty))$ and $g\in L^{\infty}_{c}((0,\infty))$. We define

[TABLE]

According to (2.16) the integrals defining $u_{1}$ and $u_{2}$ are absolutely convergent for every $t,x\in(0,\infty)$.

Assume now that $f\in C_{c}^{2}((0,\infty)\times(0,\infty))$. By (2.20), (2), and (2.28), we have that $\partial_{t}u_{2}(t,x)=\Delta_{\mu}u_{2}(t,x)$, $(t,x)\in(0,\infty)\times(0,\infty)$. Moreover,

[TABLE]

and

[TABLE]

By [7, Theorem 2.1] we have that $\displaystyle\lim_{t\to 0}u_{2}(t,x)=g(x)$, for a.e. $x\in(0,\infty)$ and for every $x\in(0,\infty)$ provided that $g$ is continuous on $(0,\infty)$.

By proceeding as in the first section and by taking into account [42, (2.17)] we can obtain that, for $i=1,2$,

[TABLE]

By using parametric derivation we get

[TABLE]

Putting together the above equalities we get

[TABLE]

Moreover, by (2.16) since $f$ has compact support, we can find $a>0$ such that, for every $x\in(0,\infty)$, there exists $C>0$ for which

[TABLE]

Then, $\displaystyle\lim_{t\to 0^{+}}u_{1}(t,x)=0,\;\;x\in(0,\infty).$ Thus, we prove that the function $u$ is a classical solution of (6.81).

6.2. Proof of Theorem 1.5

Assume now that $\mu>-1/2$. It was established in (3), (3, (3.33), and (3 that

[TABLE]

We can write

[TABLE]

By proceeding as in [42, page 11] we get

[TABLE]

where $\psi(z)=\chi_{(\epsilon,\infty)}(z)z^{-5}$, $z\in(0,\infty)$.

Suppose that $f$ is a measurable function on $\mathbb{{R}}\times(0,\infty)$ such that $f(s,y)=0$, $s<0$. We can write

[TABLE]

where $\mathscr{M}$ denotes the centered maximal function in our parabolic setting.

We consider the maximal operator $T_{*}$ defined by

[TABLE]

According to the boundedness properties of the maximal function $\mathscr{M}$ we deduce that the operator $T_{*}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$

• •

into $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$, for every $1<p<\infty$ and $\omega\in A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$.

• •

into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty),\omega)$, for $p=1$ and $\omega\in A_{1}^{*}(\mathbb{{R}}\times(0,\infty))$.

Let $f\in{L^{p}((0,\infty)^{2},\omega)}$ with $1\leq p<\infty$ and $\omega\in A^{p}_{*}(\mathbb{{R}}\times(0,\infty))$. We define $\tilde{f}$ by

[TABLE]

If $\mu>1/2$ or $\mu=-1/2$, by Theorem 1.2, we know that there exists the limit

[TABLE]

Then, a wellknown argument allows us to obtain the existence of the limit

[TABLE]

By using again Theorem 1.2 we can also prove that the above limit exists for almost everywhere $(t,x)\in(0,\infty)^{2}$, when $\mu\in(-1/2,1/2]$ and $f\in L^{p}((0,\infty)^{2})$.

From Theorem 1.2 and the boundedness properties of the operator $T_{*}$ we deduce that

1. (1)

If $\mu>1/2$ or $\mu=-1/2$. The operator $S_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$

• •

into $L^{p}(\mathbb{{R}}\times(0,\infty),\omega)$, for every $1<p<\infty$ and $\omega\in A_{p}^{*}(\mathbb{{R}}\times(0,\infty))$.

• •

into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty),\omega)$, for $p=1$ and $\omega\in A_{1}^{*}(\mathbb{{R}}\times(0,\infty))$. 2. (2)

If $\mu>-1/2$, The operator $S_{\mu}$ is bounded from $L^{p}(\mathbb{{R}}\times(0,\infty))$

• •

into $L^{p}(\mathbb{{R}}\times(0,\infty))$, for every $1<p<\infty$.

• •

into $L^{1,\infty}(\mathbb{{R}}\times(0,\infty))$, for $p=1$.

According to (2.20), we have that

[TABLE]

By proceeding as above we can prove that the desired properties for the Riesz transforms $\widetilde{{\bf R}_{\mu}}$.

6.3. Proof of Theorem 1.6

We now consider the Cauchy problem

[TABLE]

with $\mu>-1/2$ and $f\in L^{p}((0,\infty),L^{q}((0,\infty)))$, being $1<p,q<\infty$. The operator $\Delta_{\mu}$ generates the semigroup $\{W_{t}^{\mu}\}_{t>0}$ in $L^{q}((0,\infty))$, where, for every $t>0$,

[TABLE]

By (2.12) and (2.13) we have that

[TABLE]

Then, according to [36, Theorem 3.3], $\Delta_{\mu}$ has the maximal $L^{p}$-regularity property in $L^{q}((0,\infty))$, that is, there exists $C>0$ such that

[TABLE]

where

[TABLE]

Note that (6.82) is a mixed-norm inequality.