Inequalities for Higher Order Riesz-Laguerre Transforms

Ammar Elobied¹, Abdelgader Siddig², Sabir Widatalla³

¹Department of Physics and Mathematics, Faculty of Education, Sinnar University, Sinnar, Sudan.
²Department of Mathematics, University of Albutana, Ruffa’a, Sudan.
³Department of Mathematics, Faculty of Science, University of Tabuk, Tabuk, Saudi Arabia.

DOI: 10.4236/apm.2022.124025                

Abstract

The weak-type (1, 1) boundedness of the higher order Riesz-Laguerre transforms associated with the Laguerre polynomials and the boundedness for the Riesz-Laguerre transforms of order 2 are considered. We discuss a polynomial weight w that makes the Riesz-Laguerre transforms of order greater than or equal to 2 continuous from L¹ (wdμ_α) into L^1,∞ (dμ_α), under specific value α, where μ_α is the Laguerre measure.

Keywords

Riesz-Laguerre Transform, Polynomial Expansion, Weak-Type, Stein Complex Interpolation Theorem, Calderon-Zygmund-Type, Riesz-Gauss Transform

1. Introduction

The aim of this paper is to discuss the weak-type (1, 1) boundedness of ${\mathcal{R}}_{\propto }^{m+1}$ and the polynomial weight w that makes the Riesz-Laguerre transforms of order greater than or equal to 2 continuous from $L^1\left(wd{\mu }_{\alpha }\right)$ into $L^{1,\infty }\left(d{\mu }_{\alpha }\right)$, under specific value $\alpha$. Following the same notions appear in [1] [2]. The (m+ 1)th Riesz-Laguerre transform with $m\in Z_{\ge 0}^d$ associated with the multidimensional Laguerre operator ${\mathcal{L}}_{\propto }$, where $\propto =\left({\propto }_1,\cdots ,{\propto }_d\right)$ is a multi-index with ${\propto }_i\ge 0,i=1,\cdots ,d$.

The Laguerre operator ${\mathcal{L}}_{\propto }$, is a self-adjoint “Laplacian” on $L^2\left(d{\mu }_{\alpha }\right)$, where ${\mu }_{\alpha }$ is the Laguerre measure of type $\propto =\left({\propto }_1,\cdots ,{\propto }_d\right)$ with ${\propto }_i>-1,i=1,\cdots ,d$ ; defined on $R_{+}^d=\left\{x\in R^d:x_i>0,\text{for}\text{each}i=1,\cdots ,d\right\}$, by

$d{\mu }_{\alpha }\left(x\right)=\underset{i=1}{\overset{d}{{\displaystyle \prod }}}\frac{x_i^{{\propto }_i}{\text{e}}^{-x_i}}{\Gamma \left({\propto }_i+1\right)}dx.$

It is well known that the spectral resolution of ${\mathcal{L}}_{\propto }$ is

${\mathcal{L}}_{\propto }=\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}n{\mathcal{P}}_n^{\propto },$

where ${\mathcal{P}}_n^{\propto }$ is the orthogonal projection on the space spanned by Laguerre polynomials of total degree n and type $\propto$ ind variables [3] [4]. The operator ${\mathcal{L}}_{\propto }$ is the infinitesimal generator of a “heat” semigroup, called the Laguerre semigroup, $\left\{{\text{e}}^{-t{\mathcal{L}}_{\propto }}:t\ge 0\right\}$, defined in the spectral sense as

${\text{e}}^{-t{\mathcal{L}}_{\propto }}=\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}{\text{e}}^{-nt}{\mathcal{P}}_n^{\propto }.$

For any multi-index $m+1=\left(a_1,\cdots ,a_d\right)\in Z_{\ge 0}^d$, the Riesz-Laguerre transforms ${\mathcal{R}}_{\propto }^{m+1}$ of order $\left|m+1\right|=a_1+\cdots +a_d$ are defined by

${\mathcal{R}}_{\propto }^{m+1}={\nabla }_{\alpha }^{m+1}{\left({\mathcal{L}}_{\alpha }\right)}^{-\left|m+1\right|/2}{\mathcal{P}}_0^{\propto \perp },$

where ${\nabla }_{\propto }$ is associated to ${\mathcal{L}}_{\propto }$ defined as ${\nabla }_{\propto }=\left(\sqrt{x_1}{\partial }_{x_1},\cdots ,\sqrt{x_d}{\partial }_{x_d}\right)$, and ${\mathcal{P}}_0^{\propto \perp }$, denotes the orthogonal projection onto the orthogonal complement of the eigenspace corresponding to the eigenvalue 0 of ${\mathcal{L}}_{\propto }$.

In order to use the well-known relationship with the Ornstein-Uhlenbeck context, but not too much exploited in the weak-type inequalities, we are going to perform a change of coordinates in $R_{+}^d$. If $x=\left(x_1,\cdots ,x_d\right)$ is a vector $R_{+}^d$, then $x^2$ will denote the vector $\left(x_1^2,\cdots ,x_d^2\right)$. Let $\Psi :R_{+}^d\to R_{+}^d$ be defined as $\Psi \left(x\right)=x^2$ and let $d{\tilde{\mu }}_{\propto }=d{\mu }_{\propto }o{\Psi }^{-1}$ be the pull-back measure from $d{\mu }_{\propto }$. Then the modified Laguerre measure $d{\tilde{\mu }}_{\propto }$ is the probability measure

$d{\tilde{\mu }}_{\propto }\left(x\right)=2^d\underset{i=1}{\overset{d}{{\displaystyle \prod }}}\frac{x_i^{2{\propto }_i+1}{\text{e}}^{-x_i^2}}{\Gamma \left({\propto }_i+1\right)}dx=2^d\underset{i=1}{\overset{d}{{\displaystyle \prod }}}\frac{x_i^{2{\propto }_i+1}}{\Gamma \left({\propto }_i+1\right)}{\text{e}}^{-{\left|x\right|}^2}dx,$ (1)

on $R_{+}^d$.

The map $f\to {\mathcal{U}}_{\Psi }f=fo\Psi$ is an isometry from $L^q\left(d{\mu }_{\propto }\right)$ onto $L^q\left(d{\tilde{\mu }}_{\propto }\right)$ and from $L^{q,\infty }\left(d{\mu }_{\alpha }\right)$ onto $L^{q,\infty }\left(d{\tilde{\mu }}_{\propto }\right)$, for every q in [1, ∞]. So we may reduce the problem of studying the weak-type boundedness of ${\mathcal{R}}_{\propto }^{m+1}$ to the study of the same boundedness for the modified Riesz-Laguerre transforms ${\tilde{\mathcal{R}}}_{\propto }^{m+1}={\mathcal{U}}_{\Psi }{\mathcal{R}}_{\propto }^{m+1}{\mathcal{U}}_{\Psi }^{-1}$ with respect to the measure $d{\tilde{\mu }}_{\propto }$.

Observe that ${\tilde{\mathcal{R}}}_{\propto }^{m+1}$ coincides, up to a multiplicative constant, with ${\nabla }^{m+1}{\left({\tilde{\mathcal{L}}}_{\propto }\right)}^{-\left|m+1\right|}{\tilde{\mathcal{P}}}_0^{\propto \perp }$, being ${\tilde{\mathcal{L}}}_{\propto }={\mathcal{U}}_{\Psi }{\mathcal{L}}_{\propto }{\mathcal{U}}_{\Psi }^{-1}$, ${\tilde{\mathcal{P}}}_0^{\propto \perp }={\mathcal{U}}_{\Psi }{\mathcal{P}}_0^{\propto \perp }{\mathcal{U}}_{\Psi }^{-1}$ andes Ñ the gradient of $R^d$ associated to the Laplacian operator [5].

For the sequel, it is convenient to express the kernel of ${\tilde{\mathcal{R}}}_{\propto }^{m+1}$ with respect to the Polynomial measure ${\left(m+1\right)}_{\propto }$ defined on $R_{+}^d$ as

$d{\left(m+1\right)}_{\propto }\left(x\right)={\text{e}}^{{\left|x\right|}^2}d{\tilde{\mu }}_{\propto }\left(x\right)\text{.}$ (2)

According to [6] [7], for ${\propto }_i>-1/2,i=1,\cdots ,d$, the kernel of the modified Riesz-Laguerre transforms of order $\left|m+1\right|$ with respect to the polynomial measure ${\left(m+1\right)}_{\propto }$ is defined, off the diagonal, as

${\mathcal{K}}^{m+1}\left(x,s\right)={\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\mathcal{K}}^{m+1}\left(x,s\right){\Pi }_{\alpha }\left(s\right)\left(s\right)ds}$

with

${\mathcal{K}}^{m+1}\left(x,s\right)={\displaystyle \underset{0}{\overset{1}{\int }}{\left(\sqrt{r}\right)}^{\left|m+1\right|-2}}{\left(-\frac{\log r}{1-r}\right)}^{\frac{\left|m+1\right|-2}{2}}\underset{i=1}{\overset{d}{{\displaystyle \prod }}}{H}_{a_i}\left(\frac{x_i\left(\sqrt{r}-s_i\right)}{\sqrt{1-r}}\right)\frac{{\text{e}}^{-\frac{q-\left(r{x}^2,x^2,s\right)}{1-r}}}{{\left(1-r\right)}^{\left|\alpha \right|+d+1}}dr$ (3)

where $H_{a_i}$ is the Hermite polynomial of degree $a_i$ and

$q_{\pm }\left(x,s\right)=\underset{i=1}{\overset{d}{{\displaystyle \sum }}}2x_i\left(1\pm s_i\right),$

${\Pi }_{\alpha }\left(s\right)=\underset{i=1}{\overset{d}{{\displaystyle \prod }}}\frac{\Gamma \left({\alpha }_i+1\right)}{\Gamma \left({\alpha }_i+\frac{1}{2}\right)\sqrt{\pi }}{\left(1-s_i^2\right)}^{{\alpha }_i-1/2},$

$\cos \theta =\cos \theta \left(x,s\right)=\frac{{\displaystyle {\sum }_{i=1}^d{x}_i{s}_i}}{{\left|x\right|}^2},$

$\sin \theta =\sin \theta \left(x,s\right)={\left(1-{\cos }^2\theta \right)}^{1/2}={\left(1-{\left(\frac{{\displaystyle {\sum }_{i=1}^d{x}_i{s}_i}}{{\left|x\right|}^2}\right)}^2\right)}^{1/2}.$

The symbol $a\lesssim b$ means $a\le Cb$ where C is a constant that may be different on each occurrence. And we write a ~ b whenever $a\lesssim b$ and $b\lesssim a$.

2. Main Results

For every multi-index $\propto$ we have the following result see [3].

Theorem 1: The second order Riesz-Laguerre transforms map $L^1\left(d{\mu }_{\alpha }\right)$ continuously into $L^{1,\infty }\left(d{\mu }_{\alpha }\right)$.

Proof: The result follows by splitting the modified Riesz-Laguerre transforms of second order into a local operator and a global one. Let us observe that for a simple covering Lemma, we may pass from estimates with respect to the measure ${\left(m+1\right)}_{\propto }$ on the local part $R_0$ to estimates with respect to the modified Laguerre measure ${\tilde{\mu }}_{\propto }$. Therefore the local operator is equivalent to $T_0^{m+1}$ for $\left|m+1\right|=2$. The global operators bounded weak type (1, 1) and therefore so are the second order modified Riesz-Laguerre transforms.

From [8] and [9], it is known that an upper bound for $\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|$ on G is

$\begin{array}{l}{\tilde{\mathcal{K}}}^{m+1}\left(x,s\right)\\ =\{\begin{array}{ll}{\left({\left|2x\right|}^2\right)}^{\frac{\left|m+1\right|-2}{2}}{\text{e}}^{-{\left|x\right|}^2},\hfill & \cos \theta <0\hfill \\ \frac{1}{2}{\left(4{\left|x\right|}^4{\sin }^2\theta \right)}^{\frac{\left|m+1\right|-2}{4}}{\left(\frac{1+\cos \theta }{1-\cos \theta }\right)}^{\frac{\left|\alpha \right|-d}{2}}{\left(1+(4{\left|x\right|}^4{\sin }^2\theta \right)}^{\frac{1}{4}}{\text{e}}^{-u_0},\hfill & \cos \theta \ge 0\hfill \end{array}\end{array}$ (4)

with

$u_0=\frac{{\left(q_{+}\left(x^2,s\right)q_{-}\left(x^2,s\right)\right)}^{1/2}}{2}$.

Proposition 2: For $\left|m\right|=2$,

$\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\precsim {\text{e}}^{{\left|x\right|}^2}{\mathcal{K}}^{*}\left(x,y,s\right){\text{e}}^{-{\left|y\right|}^2}$

on the global region

$G=R_{+}^d\times {\left[-1,1\right]}^d\backslash R_0=R_1\cup R_2\cup R_3\cup R_4,$

with

$R_0=\left\{\left(y,s\right)\in R_{+}^d\times {\left[-1,1\right]}^d:q-{\left(x^2,y^2,s\right)}^{1/2}\le \frac{C}{1+\left|X\right|}\right\},$

$R_1=\left\{\left(y,s\right)\notin R_0:\cos \theta <0\right\},$

$R_2=\{\left(y,s\right)\notin R_0:\cos \theta \ge 0,\left|y\right|\le \left|x\right|\},$

$R_3=\{\left(y,s\right)\notin R_0:\cos \theta \ge 0,\left|x\right|\le \left|y\right|\le 2\left|x\right|\},$

$R_4=\{\left(y,s\right)\notin R_0:\cos \theta \ge 0,\left|y\right|\le 2\left|x\right|\},$

and

${\mathcal{K}}^{*}\left(x,y,s\right)=\{\begin{array}{lll}{\text{e}}^{-{\left|x\right|}^2}\hfill & \left(y,s\right)\in R_1\hfill & (5)\hfill \\ {\left|x\right|}^{2|\propto |+2d}{\text{e}}^{-c\left|x\right|q_{-}^{1/2}\left(x^2,y^2,s\right)}\hfill & \left(y,s\right)\in R_2\hfill & (6)\hfill \\ {\left|x\right|}^{2|\propto |+2d}\left(1\wedge \frac{{\text{e}}^{-\frac{c{\left|x\right|}^4{\sin }^2\theta }{{\left|y\right|}^2-{\left|x\right|}^2+\sin \theta {\left|x\right|}^2}}}{{\left({\left|y\right|}^2-{\left|x\right|}^2+\sin \theta {\left|x\right|}^2\right)}^{\frac{2|\propto |+2d-1}{2}}}\right)\hfill & \left(y,s\right)\in R_3\hfill & (7)\hfill \\ \left(1+\left|x\right|\right){\text{e}}^{-{\sin }^2\theta {\left|x\right|}^2}\hfill & \left(y,s\right)\in R_4\hfill & (\; 8\; )\hfill \end{array}$

Proof. In this proposition, $\left|m\right|=2$. If $\cos \theta <0$, it is immediate that

$\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\lesssim {\text{e}}^{{\left|x\right|}^2}{\mathcal{K}}^{*}\left(x,y,s\right){\text{e}}^{-{\left|y\right|}^2}.$

Let us then assume that $\cos \theta \ge 0$.

1) First let us consider $\left|x\right|>\left|y\right|$.

Since $\cos \theta \ge 0,q_{+}^{1/2}\ge \left|x\right|$ and since $\left|x\right|>\left|y\right|$, then $q_{+}^{1/2}\le 2\left|x\right|$. Therefore $q_{+}^{1/2}~\left|x\right|$. On the other hand, since $q_{-}^{1/2}\ge \frac{C}{1+\left|x\right|}$ then $\left|x\right|\ge c$. Thus

$\begin{array}{c}\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\lesssim \left[{\left(\frac{q_{+}}{q_{-}}\right)}^{\frac{\left|\alpha \right|+d}{2}}+{\left(\frac{q_{+}}{q_{-}}\right)}^{\frac{\left|\alpha \right|+d}{2}}{\left(q_{+}{q}_{-}\right)}^{1/4}\right]{\text{e}}^{-u_0}\\ \lesssim \left[{\left|x\right|}^{\left|\alpha \right|+d}{\left(1+\left|x\right|\right)}^{\left|\alpha \right|+d}+\frac{{\left(q_{+}^{1/2}\right)}^{\left|\alpha \right|+d+1/2}}{{\left(q_{-}^{1/2}\right)}^{\left|\alpha \right|+d-1/2}}\right]{\text{e}}^{-u_0}\\ \lesssim {\left|x\right|}^{2\left|\alpha \right|+2d}{\text{e}}^{-u_0}={\text{e}}^{{\left|x\right|}^2}{\left|x\right|}^{2\left|\alpha \right|+2d}{\text{e}}^{-\frac{{\left(q_{+}{q}_{-}\right)}^{1/2}}{2}{\text{e}}^{\frac{{\left|y\right|}^2-{\left|x\right|}^2}{2}}{\text{e}}^{-{\left|y\right|}^2}}\\ \lesssim {\text{e}}^{{\left|x\right|}^2}{\left|x\right|}^{2\left|\alpha \right|+2d}{\text{e}}^{-\frac{\left|x\right|q_{-}^{1/2}\left(x^2,y^2,s\right)}{2}}{\text{e}}^{-{\left|y\right|}^2}={\text{e}}^{{\left|x\right|}^2}{\mathcal{K}}^{*}\left(x,y,s\right){\text{e}}^{-{\left|y\right|}^2}.\end{array}$

2) Now let us assume $\left|y\right|\ge \left|x\right|$ and rewrite $u_0$ in the following way:

$\begin{array}{c}{u}_0=\frac{{\left|y\right|}^2-{\left|x\right|}^2}{2}+\frac{{\left(q_{+}\left(x^2,y^2,s\right)q_{-}\left(x^2,y^2,s\right)\right)}^{1/2}}{2}\\ ={\left|y\right|}^2-{\left|x\right|}^2+\frac{{\left(q_{+}\left(x^2,y^2,s\right)q_{-}\left(x^2,y^2,s\right)\right)}^{1/2}-\left({\left|y\right|}^2-{\left|x\right|}^2\right)}{2}\\ ={\left|y\right|}^2-{\left|x\right|}^2\frac{q_{+}{q}_{-}-{\left({\left|y\right|}^2-{\left|x\right|}^2\right)}^2}{2\left({\left|y\right|}^2-{\left|x\right|}^2+{\left(q_{+}{q}_{-}\right)}^{1/2}\right)}\\ ={\left|y\right|}^2-{\left|x\right|}^2+\frac{2{\sin }^2\theta {\left|x\right|}^2-{\left|y\right|}^2}{{\left|y\right|}^2-{\left|x\right|}^2+{\left(q_{+}{q}_{-}\right)}^{1/2}}.\end{array}$ (9)

Since

$q_{+}{q}_{-}={\left({\left|x\right|}^2+{\left|y\right|}^2\right)}^2-4{\left|x\right|}^2{\left|y\right|}^2{\cos }^2\theta ={\left({\left|y\right|}^2-{\left|x\right|}^2\right)}^2+4{\left|x\right|}^2{\left|y\right|}^2{\sin }^2\theta$,

and taking into account that sinθ is non-negative, we obtain that

${\left(q_{+}{q}_{-}\right)}^{1/2}~{\left|y\right|}^2-{\left|x\right|}^2+\left|x\right|\left|y\right|\sin \theta \ge {\left|y\right|}^2-{\left|x\right|}^2+{\left|x\right|}^2\sin \theta .$ (10)

Thus, from (9) together with (10) we get

$u_0\ge {\left|y\right|}^2-{\left|x\right|}^2+\frac{c{\left|x\right|}^4{\sin }^2\theta }{{\left|y\right|}^2-{\left|x\right|}^2+\left|x\right|\left|y\right|\sin \theta }.$ (11)

Claim 3: $\max \left({\left|y\right|}^2-{\left|x\right|}^2,{\left|x\right|}^2\sin \theta \right)\ge 1$.

Proof. If $\sin \theta \ge \frac{1}{{\left|x\right|}^2}$, the inequality is immediate.

If $\sin \theta \le \frac{1}{{\left|x\right|}^2}$, then ${\left|y\right|}^2\ge {\left|x\right|}^2+1$. This inequality is immediate when $\left|x\right|\le 1$

by adjusting conveniently the constant C in the definition of the global zone and it is also immediate for d = 1 and $\left|x\right|>1$. Now let us assume that $d\ge 2$ and $\left|x\right|>1$.

$\begin{array}{c}\frac{{\left(C/2\right)}^2}{{\left|x\right|}^2}\le \frac{C^2}{{\left(1+\left|x\right|\right)}^2}\le q_{-}\left(x^2,y^2,s\right)={\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|\left|y\right|\sqrt{1-{\sin }^2\theta }\\ \le {\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|\left|y\right|\sqrt{1-\frac{1}{{\left|x\right|}^4}}.\end{array}$

Hence

${\left|x\right|}^2-2\left|x\right|\sqrt{1-\frac{1}{{\left|x\right|}^4}}\left|y\right|-{\left|x\right|}^2-\frac{{\left(C/2\right)}^2}{{\left|x\right|}^2}\ge 0$

for all $\left|y\right|\ge \left|x\right|$, then

$\left|y\right|\ge \left|x\right|\sqrt{1-\frac{1}{{\left|x\right|}^4}}+\frac{\sqrt{{\left(C/2\right)}^2-1}}{\left|x\right|}$

which implies that

${\left|y\right|}^2\ge {\left|x\right|}^2+2\sqrt{{\left(C/2\right)}^2-1}\sqrt{1-\frac{1}{{\left|x\right|}^4}}+\frac{{\left(C/2\right)}^2-2}{{\left|x\right|}^2}\ge {\left|x\right|}^2+1.$

Therefore by applying this claim to inequality (10) we obtain that $q_{+}{q}_{-}\ge c$ in this context. If $\left|x\right|\le \left|y\right|\le 2\left|x\right|$, by taking into account (11), we get

$u_0\ge {\left|y\right|}^2-{\left|x\right|}^2+\frac{c{\left|x\right|}^4{\sin }^2\theta }{{\left|y\right|}^2-{\left|x\right|}^2+{\left|x\right|}^2\sin \theta },$ (12)

then

$\begin{array}{c}\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\lesssim {\left(\frac{q_{+}}{q_{-}}\right)}^{\frac{\left|\alpha \right|+d}{2}}{\left(q_{+}{q}_{-}\right)}^{1/4}{\text{e}}^{-u_0}\lesssim \frac{q_{+}^{\frac{\left|\alpha \right|+d}{2}}}{{\left(q_{+}{q}_{-}\right)}^{\frac{2|\propto |+2d}{4}}}{\left(q_{+}{q}_{-}\right)}^{1/4}{\text{e}}^{-u_0}\\ \lesssim \frac{{\left|x\right|}^{2\left|\alpha \right|+2d}}{{\left[{\left(q_{+}{q}_{-}\right)}^{1/2}\right]}^{\frac{2|\propto |+2d-1}{2}}}{\text{e}}^{-u_0}\\ \lesssim {\text{e}}^{{\left|x\right|}^2}\frac{{\left|x\right|}^{2\left|\alpha \right|+2d}}{{\left({\left|y\right|}^2-{\left|x\right|}^2+{\left|x\right|}^2\sin \theta \right)}^{\frac{2|\propto |+2d-1}{2}}}{\text{e}}^{-\frac{c{\left|x\right|}^4{\sin }^2\theta }{{\left|y\right|}^2-{\left|x\right|}^2+{\left|x\right|}^2\sin \theta }{\text{e}}^{-{\left|y\right|}^2}}.\end{array}$

To get the last inequality we have used (10) and (12). On the other hand, since $q_{+}{q}_{-}\ge c$ it is immediate the following inequality

$\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\lesssim {\left|x\right|}^{2\left|\alpha \right|+2d}{\text{e}}^{{\left|x\right|}^2-{\left|y\right|}^2}.$

Thus

$\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\lesssim {\text{e}}^{{\left|x\right|}^2}{K}^{*}\left(x,y,s\right){\text{e}}^{-{\left|y\right|}^2}.$

Now if $\left|y\right|\ge 2\left|x\right|$ then ${\left(q_{+}{q}_{-}\right)}^{1/4}\le {\left({\left|x\right|}^2+{\left|y\right|}^2\right)}^{1/2}\lesssim {\left|y\right|}^2$, and thus

$\left(1+{\left(q_{+}{q}_{-}\right)}^{1/4}\frac{\left|x\right|\left|y\right|}{{\left|x\right|}^2+{\left|y\right|}^2}\right)\precsim \left(1+\left|x\right|\right).$

Besides $q_{-}\ge {\left(\left|y\right|-\left|x\right|\right)}^2\ge c{\left|y\right|}^2$ and $q_{+}\le C{\left|y\right|}^2$ therefore $\frac{q_{+}}{q_{-}}\le C$. On the other

hand, from (9) together with ${\left(q_{+}{q}_{-}\right)}^{1/2}\le {\left|y\right|}^2-{\left|x\right|}^2+2\left|x\right|\left|y\right|\sin \theta$ and ${\left|y\right|}^2-{\left|x\right|}^2+\left|x\right|\left|y\right|\sin \theta \le 2{\left|y\right|}^2$ we get

$u_0\ge {\left|y\right|}^2-{\left|x\right|}^2+\frac{{\sin }^2\theta {\left|x\right|}^2{\left|y\right|}^2}{{\left|y\right|}^2-{\left|x\right|}^2+\left|x\right|\left|y\right|\sin \theta }\ge {\left|y\right|}^2-{\left|x\right|}^2+\frac{{\sin }^2\theta }{2}{\left|x\right|}^2.$

Therefore

$\left|{\mathcal{K}}^m\left(x,y,s\right)\right|\lesssim {\text{e}}^{{\left|x\right|}^2}{\mathcal{K}}^{*}\left(x,y,s\right){\text{e}}^{-{\left|y\right|}^2}.$

Proposition 4: The operator ${\mathcal{K}}^{*}$ defined as

${\mathcal{K}}^{*}f\left(x\right)={\text{e}}^{{\left|x\right|}^2}{\displaystyle {\int }_{R_{+}^d}{\displaystyle {\int }_{{\left[-1,1\right]}^d}{\chi }_G}}\left(x,s\right){\mathcal{K}}^{*}\left(x,s\right){\Pi }_{\propto }\left(s\right)\left|f\left(x\right)\right|d{\tilde{\mu }}_{\propto }\left(x\right),$

is of weak type (1, 1) with respect to the measure ${\tilde{\mu }}_{\propto }$.

Proof. The method of proof used in [1] is an adaptation to our context of the techniques developed in [10] [11] [12] which allows us to get rid of the classical one called “forbidden regions technique”.

The kernels (5) and (6) define strong type (1, 1) operators. Indeed,

${\text{e}}^{{\left|x\right|}^2}{\displaystyle {\int }_{R_{+}^d}{\displaystyle {\int }_{{\left[-1,1\right]}^d}{\chi }_{R_1}\left(x,s\right){\text{e}}^{-{\left|x\right|}^2}{\Pi }_{\propto }\left(s\right)\left|f\left(x\right)\right|d{\tilde{\mu }}_{\propto }\left(x\right)}}\le C{\Vert f\Vert }_1.$

Moreover, for semi-integer values of the parameter α, by [6]

${\left|x\right|}^{2\left(|\propto |+d\right)}{\text{e}}^{-C{\left|x\right|}^2{\left(2\left(1-\cos \theta \right)\right)}^{1/2}}$ (13)

is in $L^1\left(d{\mu }_{\alpha }\right)$ uniformly in x and s and so the operator is of strong type with respect to ${\tilde{\mu }}_{\propto }$ on $R_2$. Finally the result for the other values of $\propto$ is obtained via the multidimensional Stein’s complex interpolation Theorem. So to get the weak-type (1, 1) inequality for the operator ${\mathcal{K}}^{*}$ it suffices to prove that the operators

$S_{i}f\left(x\right)={\text{e}}^{{\left|x\right|}^2}{\displaystyle {\int }_{R_{+}^d}{\displaystyle {\int }_{{\left[-1,1\right]}^d}{\chi }_{R_{i+3}}}}\left(x,s\right){\mathcal{K}}^{*}\left(x,s\right){\Pi }_{\propto }\left(s\right)ds\left|f\left(x\right)\right|d{\tilde{\mu }}_{\propto }\left(x\right),i=0,1$

map $L^1\left(d{\tilde{\mu }}_{\propto }\right)$ continuously into $L^{1,\infty }\left(d{\tilde{\mu }}_{\propto }\right)$.

Without loss of generality, we may assume that $f\ge 0$. Fix $\lambda >0$ and let

$E_i=\left\{x\in R_{+}^d:S_{i}f\left(x\right)>\lambda \right\},$

for $i=0,1$. We must prove that ${\tilde{\mu }}_{\propto }\left(E_i\right)\le C\frac{{\Vert f\Vert }_1}{\lambda }$. Let $r_0$ and $r_1$ be the positive roots of the equations

$r_0^{2\left(|\propto |+d\right)}{\text{e}}^{r_0^2}{\Vert f\Vert }_1=\lambda \text{and}{r}_1{\text{e}}^{r_1^2}{\Vert f\Vert }_1=\lambda .$

We may observe that indeed, if $E_i\cap \left\{x\in R_{+}^d:\left|x\right|<r_i\right\}=\varnothing$ : indeed, if $\left|x\right|<r_i$, we have

$S_{0}f\left(x\right)\le {\left|x\right|}^{2\left(|\propto |+d\right)}{\text{e}}^{{\left|x\right|}^2}{\Vert f\Vert }_1<\lambda ,$

$S_{1}f\left(x\right)\le \left|x\right|{\text{e}}^{{\left|x\right|}^2}{\Vert f\Vert }_1<\lambda .$

On the other hand, we may take $\lambda >K\Vert f\Vert$ in [1], and by choosing K large enough we may assume that both $r_0$ and $r_1$ are larger that one. Hence

${\tilde{\mu }}_{\propto }\left\{x\in R_{+}^d:\left|x\right|<2r_i\right\}\le {\displaystyle {\int }_{\left|x\right|<2r_i}\underset{j=1}{\overset{d}{{\displaystyle \prod }}}{x}_j^{2{\propto }_j+1}{\text{e}}^{-{\left|x\right|}^2}dx}\lesssim r_i^{2\left|\alpha \right|}{\text{e}}^{-4r_i^2}\le C\frac{{\Vert f\Vert }_1}{\lambda }.$

Thus we only need to estimate ${\tilde{\mu }}_{\propto }\left\{x\in R_{+}^d:r_i\le \left|x\right|\le 2r_i\right\}$.

We let ${E^{\prime }}_i$ denote the set of $x^{\prime }\in S^{d-1}$ for which there exists a $\rho \in \left[r_i,2r_i\right]$ with $\rho x^{\prime }\in E$. For each $x^{\prime }\in {E^{\prime }}_i$ we let $\rho \left(x^{\prime }\right)$ be the smallest such $\rho$. Observe that

$\sin \theta \left(x,s\right)=\sin \theta \left(x^{\prime },s\right)=\sin \theta .$

Then $S_{i}f\left(\rho \left(x^{\prime }\right)x^{\prime }\right)=\lambda$, by continuity. This implies for $i=0$ and $x^{\prime }\in {E^{\prime }}_0$,

$\begin{array}{c}\lambda =S_{0}f\left({\rho }_0\left(x^{\prime }\right)x^{\prime }\right)\\ ={\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\text{x}}_{R_3}{\text{e}}^{{\left|x\right|}^2}{x}^{2\left(\left|\alpha \right|+d\right)}}}\left(1\wedge \frac{{\text{e}}^{-\frac{c{\left|x\right|}^2{\sin }^2\theta }{\sin \theta {\left|x\right|}^2}}}{{\left(\sin \theta {\left|x\right|}^2\right)}^{\frac{2\left(\left|\alpha \right|+d\right)-1}{2}}}\right){\Pi }_{\alpha }\left(s\right)dsf\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right)\\ \lesssim {\text{e}}^{p_0^2{\left(x^{\prime }\right)}^2}{r}_0^{2\left(\left|\alpha \right|+d\right)}{\Pi }_{\alpha }\left(s\right)dsf\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right)\\ \times {\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d\left\{\left|x\right|\ge r_0\right\}}{\int }\text{x}\left(x\right)}}\left(1\wedge \frac{{\text{e}}^{-\frac{c{r}_0^4{\sin }^2\theta }{{\left|x\right|}^2-r_0^2+\sin \theta r_0^2}}}{{\left({\left|x\right|}^2-r_0^2+\sin \theta r_0^2\right)}^{\frac{2\left(\left|\alpha \right|+d\right)-1}{2}}}\right),\end{array}$ (14)

and for $i=1$ and $x^{\prime }\in {E^{\prime }}_1$,

$\begin{array}{c}\lambda =S_{1}f\left(p_i\left(x^{\prime }\right)x^{\prime }\right)\\ ={\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\text{x}}_{R_5}}}\left(x\right){\text{e}}^{{\left|x\right|}^2}\left(1+\left|x\right|\right){\text{e}}^{-{\sin }^2\theta {\left|x\right|}^2}{\Pi }_{\alpha }\left(s\right)dsf\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right)\\ \lesssim {\text{e}}^{p_0^2{\left(x^{\prime }\right)}^2}{r}_1{\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d\left\{\left|x\right|>r_1\right\}}{\int }}}\text{x}\left(x\right){\text{e}}^{-c{\sin }^2\theta r_1^2}{\Pi }_{\alpha }\left(s\right)dsf\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right)\end{array}$ (15)

Clearly, since r₀ and r₁ are greater than one, we have

$\begin{array}{c}{\tilde{\mu }}_{\alpha }\left\{x\in E_i:r_i\le \left|x\right|\le 2r_i\right\}\le {\displaystyle \underset{{E^{\prime }}_i}{\int }d\sigma \left(x^{\prime }\right)}\underset{{\rho }_i\left(x^{\prime }\right)}{\overset{2r_i}{{\displaystyle \int }}}{\text{e}}^{-{\rho }^2}{\rho }^{2\left(|\propto |+d\right)-1}d\rho \\ \lesssim {\displaystyle \underset{{E^{\prime }}_i}{\int }{\text{e}}^{-{\rho }_i{\left(x^{\prime }\right)}^2}{r}_i{}^{2\left(|\propto |+d-1\right)}d\sigma (\; x\; \prime \; )}\end{array}$

combining this estimate for $i=0$ with (14), we get

${\tilde{\mu }}_{\alpha }\left\{x\in E_i:r_i\le \left|x\right|\le 2r_i\right\}\le \frac{C}{\lambda }{\displaystyle \underset{{E^{\prime }}_0}{\int }{r}_0^{2\left(\left(2|\propto |+2d\right)-1\right)}d\sigma \left(x^{\prime }\right)}\left(I_0+I{I}_0\right),$ (16)

with

$I_0={\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{\sin \theta r_0^2\le c\right\}}{\int }\frac{{\text{e}}^{-\frac{c{r}_0^4{\sin }^2\theta }{{\left|x\right|}^2-r_0^2+\sin \theta r_0^2}}}{{\left({\left|x\right|}^2-r_0^2+\sin \theta r_0^2\right)}^{\frac{2\left(|\propto |+d\right)-1}{2}}}f}}\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right){\Pi }_{\alpha }\left(s\right)ds,$

and

$I{I}_0={\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{\left|x\right|\ge r_1\sin \theta r_0^2\le c\right\}}{\int }f}}\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right){\Pi }_{\alpha }\left(s\right)ds.$

Similarly for $i=1$ with (15), we obtain

${\tilde{\mu }}_{\alpha }\left\{x\in E_1:r_1\le \left|x\right|\le 2r_1\right\}\le \frac{C}{\lambda }{\displaystyle \underset{{E^{\prime }}_1}{\int }{r}_1^{2\left(|\propto |+d\right)-1}d\sigma }\left(\bar{x}\right)\left(I_1+I{I}_1\right),$ (17)

with

$I_1={\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{\left|x\right|\ge r_1\sin \theta r_1^2\ge c\right\}}{\int }{\text{e}}^{-c{\sin }^2\theta r_1^2}f}}\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right){\Pi }_{\alpha }\left(s\right)ds,$

and

$I{I}_1={\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{\sin \theta r_1^2\le c\right\}}{\int }f}}\left(x\right)d_{{\tilde{\mu }}_{\alpha }}\left(x\right){\Pi }_{\alpha }\left(s\right)ds.$

It is immediate to verify that

$r_0^{2\left(2\left(|\propto |+d\right)-1\right)}{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{x^{\prime }:\sin \theta r_0^2\le c\right\}}{\int }\sigma \left(x^{\prime }\right){\Pi }_{\alpha }\left(s\right)ds}}\le C$

and

$r_1^{2\left(|\propto |+d\right)-1}{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{x^{\prime }:\sin \theta r_0^2\le c\right\}}{\int }d\sigma \left(x^{\prime }\right){\Pi }_{\alpha }\left(s\right)ds}}\le C.$

Which give, after changing the order of integration in (16) and (17), the desired estimate for the terms involving II₀ and II₁, respectively as in [5]. Now let us prove that for $\left|x\right|\ge r_0$

$r_0^{2\left(2\left(|\propto |+d\right)-1\right)}{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{x^{\prime }:\sin \theta r_0^2\le c\right\}}{\int }\frac{{\text{e}}^{-\frac{c{r}_0^4{\sin }^2\theta }{{\left|x\right|}^2-r_0^2+\sin \theta r_0^2}}}{{\left({\left|x\right|}^2-r_0^2+\sin \theta r_0^2\right)}^{\frac{2\left(|\propto |+d\right)-1}{2}}}d\sigma \left(x^{\prime }\right){\Pi }_{\alpha }\left(s\right)ds}}\le C$

and for $\left|x\right|>r_1$

$r_1^{2\left(|\propto |+d\right)-1}{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\displaystyle \underset{\left\{x^{\prime }:\sin \theta r_1^2\ge c\right\}}{\int }{\text{e}}^{-c{\sin }^2\theta r_1^2}d\sigma \left(x^{\prime }\right){\Pi }_{\alpha }\left(s\right)ds}}\le C.$

Firstly, one considers the case where $\propto =\left(\frac{n_1}{2},-1,\cdots ,\frac{n_d}{2}-1\right)$ with $n_i\in N$ and

$n_i>1$ for each $i=1,\cdots ,d$. In this case the inner integrals can be interpreted as integrals over $S^{\left|n\right|-1}$ with respect to the Lebesgue measure, expressed in polyradial

coordinates in [11]. The same estimates are obtained also for $\propto \in \frac{N^d}{2}-1+i{R}^d$.

Finally the result for the other values of $\propto$ are obtained via the multidimensional Stein’s complex interpolation Theorem. Indeed, let $F:C^d\to C$ the function defined by

$F\left(\xi \right)=r_0^{2\left(2\xi +2d-1\right)}{\displaystyle \underset{\left\{\sin \theta r_0^2\le c\right\}}{\int }\frac{{\text{e}}^{-\frac{c{r}_0^4{\sin }^2\theta }{{\left|x\right|}^2-r_0^2+\sin \theta r_0^2}}}{{\left({\left|x\right|}^2-r_0^2+\sin \theta r_0^2\right)}^{\frac{2\left(\left|\xi \right|+d\right)-1}{2}}}{\Pi }_{\xi }}\left(s\right)ds.$

We have seen that $\left|F\left(\frac{n}{2}-1\right)\right|\le C$ and it is easy to prove that $\left|F\left(\frac{n}{2}-1+i\zeta \right)\right|\le \left|F\left(\frac{n}{2}-1\right)\right|$, whenever n is a integer vector and $\zeta \in R^d$.

Now we introduce the possible roots of the equations mentioned in the following Remark see [13]

Remark 5: 1) a) if $r_0=r_1$ then we have

$r_0^{2\left|\alpha \right|+2d-1}=1,$

and

$2\left|\alpha \right|+2d-1=0,$

which implies that

$\left|\alpha \right|=\frac{1}{2}\left(1-2d\right)$,

b) if $r_0\ne r_1$ we have the quadratic equation

$\left(2\left|\alpha \right|+2d\right)\ln r_0=\ln r_1+r_1^2-r_0^2$

we assume, for simplicity, that $r_0={\text{e}}^n$ and $r_1={\text{e}}^{2n}$ we can find

$\left(2\left|\alpha \right|+2d\right)\ln {\text{e}}^n=\ln {\text{e}}^{2n}+{\text{e}}^{4n}-{\text{e}}^{2n}$

${\left({\text{e}}^{2n}\right)}^2-{\text{e}}^{2n}+2n\left(1-\left|\alpha \right|-d\right)=0$

so that

${\text{e}}^{2n}=\frac{1\pm \sqrt{1-8n\left(1-\left|\alpha \right|-d\right)}}{2},$

where $n\ge 1$, we can easily find $r_0$.

2) $S_0$ and $S_1$ aremonotone.

3) Since $\frac{{\left|x\right|}^{2\left(|\propto |+d\right)}}{\left|x\right|}<1$, then ${\left|x\right|}^{2\left(|\propto |+d\right)}\le \left|x\right|<C$.

Proposition 6: For all m, the operator

$T_0^{m+1}f\left(x\right)=p.v.{\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\chi }_{R_0}}}\left(x,s\right){\mathcal{K}}^{m+1}\left(x,s\right){\Pi }_{\propto }\left(s\right)dsf\left(x\right)d{\left(m+1\right)}_{\propto }\left(x\right),$

which is the modified Riesz-Laguerre transform restricted to the local regionR₀, is of weak type (1, 1) with respect to the measure ${\tilde{\mu }}_{\propto }$

Proof. The proof of this result follows the same steps like the proof of the weak-type boundedness on the local zone of the first order Riesz-Laguerre transforms done in [4] [9]. For the former we have the Calderon–Zygmund-type estimates for the kernel $K^{m+1}$.

Lemma 7: There exists a constant C such that

$\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|\phi \left(x,s\right)\le C{\left(2{\left|x\right|}^2\left(1-\cos \theta \right)\right)}^{-\left(|\propto |+d\right)},$

$\left|{\nabla }_{\left(x,x\right)}\left({\mathcal{K}}^{m+1}\left(x,s\right)\phi \left(x,s\right)\right)\right|\le C{\left(2{\left|x\right|}^2\left(1-\cos \theta \right)\right)}^{-\left(|\propto |+d+1/2\right)}$

being $\phi \left(x,s\right)$ acut-off function defined in [3] and $\left(x,s\right)\in R_0$ .

Proof. Since $\left|x_i\left(\sqrt{r}-s_i\right)\right|\le q_{-}^{1/2}\left(r{x}^2,x^2,s\right)$, then

$\begin{array}{l}\left|\underset{i=1}{\overset{d}{{\displaystyle \prod }}}{H}_{a_i}\left(\frac{x_i\left(\sqrt{r}-s_i\right)}{\sqrt{1-r}}\right)\right|{\text{e}}^{-\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}}\\ \lesssim \underset{k=0}{\overset{\left|m+1\right|}{{\displaystyle \sum }}}{\left(\frac{q_{-}^{1/2}\left(r{x}^2,x^2,s\right)}{1-r}\right)}^{k/2}{\text{e}}^{-\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}}\end{array}$

$\lesssim {\text{e}}^{-\frac{q_{-}\left(r{x}^2,x^2,s\right)}{2\left(1-r\right)}}\lesssim {\text{e}}^{-c\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}},$

where last inequality follows from this one:

$q_{-}\left(r{x}^2,x^2,s\right)\ge {\left(2{\left|x\right|}^2\left(1-\cos \theta \right)\right)}^{1/2}-2C\left(1-r^{1/2}\right)$

when $\left(x,s\right)\in R_0$ in [6]. Thus on $R_0$

$\begin{array}{c}\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|\phi \left(x,s\right)\lesssim {\displaystyle {\int }_{1/2}^1{\left(\sqrt{r}\right)}^{\left|m+1\right|-1}{\left(-\frac{\log r}{1-r}\right)}^{\frac{\left|m+1\right|-2}{2}}\frac{{\text{e}}^{-c\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}}}{{\left(1-r\right)}^{|\propto |+d+1}}dr}\\ \lesssim {\displaystyle {\int }_0^{1/2}{\left(\sqrt{r}\right)}^{\left|m+1\right|-2}{\left(-\log r\right)}^{\frac{\left|m+1\right|-2}{2}}dr}+{\displaystyle {\int }_{1/2}^1\frac{{\text{e}}^{-c\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}}}{{\left(1-r\right)}^{|\propto |+d+1}}dr}\\ \lesssim 1+{\left(2{\left|x\right|}^2\left(1-\cos \theta \right)\right)}^{-\left(|\propto |+d\right)}.\end{array}$

In computing the gradient to f the kernel with respect to x we are going to have integrals such as ${\mathcal{K}}^{m+1}\left(x,s\right){\partial }_{x_j}\left(x,s\right)$,

$\begin{array}{l}{\displaystyle {\int }_0^1{\left(\sqrt{r}\right)}^{\left|m+1\right|-1}}{\left(-\frac{\log r}{1-r}\right)}^{\frac{\left|m+1\right|-2}{2}}\underset{i\ne j}{{\displaystyle \prod }}{H}_{a_i}\left(\frac{x_i\left(\sqrt{r}-s_i\right)}{\sqrt{1-r}}\right)\\ \times H_{a_i-1}\left(\frac{x_i\left(\sqrt{r}-s_i\right)}{\sqrt{1-r}}\right)\frac{{\text{e}}^{-\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}}}{{\left(1-r\right)}^{|\propto |+d+3/2}}dr,\end{array}$

with $H_{-1}\equiv 0$ and

${\displaystyle {\int }_0^1{\left(\sqrt{r}\right)}^{\left|m+1\right|-2}}{\left(-\frac{\log r}{1-r}\right)}^{\frac{\left|m+1\right|-2}{2}}{H}_{a_i}\left(\frac{x_i\left(\sqrt{r}-s_i\right)}{\sqrt{1-r}}\right)\times \frac{x_j\left(\sqrt{r}-s_j\right)}{\sqrt{1-r}}\frac{{\text{e}}^{-\frac{q_{-}\left(r{x}^2,x^2,s\right)}{1-r}}}{{\left(1-r\right)}^{|\propto |+d+3/2}}dr.$

In order to estimate these three integrals we use the same estimates described at the beginning of this proof. For the first one we have to use

$2\left|{\nabla }_x\phi \left(x,s\right)\right|\lesssim \frac{1}{{\left(2{\left|x\right|}^2\left(1-\cos \theta \right)\right)}^{1/2}}$

The gradient with respect to x is treated similarly.

For the latter we have the following Theorem regarding the $L^{p}d{\tilde{\mu }}_{\propto }$ -boundedness for $1<p<\infty$ of the modified Riesz-Laguerre transform of any order on G.

Theorem8: The operator

${\mathcal{R}}_g^{m+1}f\left(x\right)={\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\chi }_G\left(x,s\right){\mathcal{K}}^{m+1}\left(x,s\right){\Pi }_{\propto }\left(s\right)dsf\left(x\right)d{\mu }_{\propto }(\; x\; )}}$

is strong-type $\left(p,p\right)$ for $1<p<\infty$ with respect to the measure ${\tilde{\mu }}_{\propto }$.

Proof.The proof of this result in [1] is an adaptation to our context of the same result for the higher order Riesz–Gauss transform s done in [6]. Taking into account that on G, $q_{+}\left(x^2,s\right)q_{-}\left(x^2,s\right)\ge c$ when $\cos \theta \ge 0$, an upperbound for $\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|$ is

${\widetilde{\stackrel{˜}{\mathcal{K}}}}^{m+1}\left(x,s\right)=\{\begin{array}{l}{\left(2{\left|x\right|}^2\right)}^{\frac{\left|m+1\right|-2}{2}}{\text{e}}^{-{\left|x\right|}^2}\text{if}\cos \theta <0,\\ {\left(2{\left|x\right|}^2\left(1+\cos \theta \right)\right)}^{\left|\alpha \right|+d}{\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left|m+1\right|-1}{2}}{\text{e}}^{-{\left|x\right|}^2\sin \theta }\text{if}\cos \theta \ge 0.\end{array}$

Thus

$\begin{array}{l}{\displaystyle {\int }_{R_{+}^d}{\left|{\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\mathcal{K}}^{m+1}}\left(x,s\right){\Pi }_{\propto }\left(s\right)dsf\left(x\right)d{\mu }_{\propto }\left(x\right)\right|}^{p}d{\tilde{\mu }}_{\propto }}\\ \lesssim {\displaystyle {\int }_{R_{+}^d}{\left({\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\widetilde{\stackrel{˜}{\mathcal{K}}}}^{m+1}}\left(x,s\right){\Pi }_{\propto }\left(s\right)ds\left|f\left(x\right)\right|d{\mu }_{\propto }\left(x\right)\right)}^{p}d{\tilde{\mu }}_{\propto }}\\ \lesssim {\displaystyle {\int }_{R_{+}^d}{\left({\displaystyle {\int }_{R_{+}^d}{\left(2{\left|x\right|}^2\right)}^{\frac{p^{\prime }\left(\left|m+1\right|-2\right)}{2}}d{\tilde{\mu }}_{\propto }}\right)}^{p-1}}d{\tilde{\mu }}_{\propto }\left(x\right){\Vert f\Vert }_{L^p\left(d{\tilde{\mu }}_{\propto }\right)}^p.\end{array}$

For the region $G\cap \left\{\cos \theta \ge 0\right\}$ we are going to use the following estimates:

$2x^2\le q_{+}\le {\left|2x\right|}^2,q_{-}\ge 0,{\left(q_{+}{q}_{-}\right)}^{\frac{1}{2}}\ge 0,$

$0\le \left|\frac{1}{p}-\frac{1}{2}\right|{\left|x\right|}^2\sin \theta <{\left|x\right|}^2\sin \theta ,\text{since}p>1,$

$\begin{array}{l}{\displaystyle {\int }_{R_{+}^d}{\left|{\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\mathcal{K}}^{m+1}}\left(x,s\right){\Pi }_{\propto }\left(s\right)dsf\left(x\right)d{\mu }_{\propto }\left(x\right)\right|}^{p}d{\tilde{\mu }}_{\propto }}\\ \lesssim {\displaystyle {\int }_{R_{+}^d}{\left({\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\widetilde{\stackrel{˜}{\mathcal{K}}}}^{m+1}}\left(x,s\right){\Pi }_{\propto }\left(s\right)ds\left|f\left(x\right)\right|d{\mu }_{\propto }\left(x\right)\right)}^{p}d{\tilde{\mu }}_{\propto }}\\ \lesssim {\displaystyle {\int }_{R_{+}^d}{\left({\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\left|2x\right|}^{2\left(|\propto |+d\right)}}{\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left(m+1\right)-1}{2}}{\text{e}}^{-\frac{2{\left|x\right|}^2\sin \theta }{2}}{\Pi }_{\propto }\left(s\right)ds\times \left|f\left(x\right)\right|{\text{e}}^{-\frac{|x{|}^2}{p}}d{\mu }_{\propto }\left(x\right)\right)}^{p}d{\mu }_{\propto }\left(x\right)}\\ \lesssim {\displaystyle {\int }_{R_{+}^d}{\left({\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\left|2x\right|}^{2\left(|\propto |+d\right)}}{\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left(m+1\right)-1}{2}}{\text{e}}^{-\left(\frac{1}{2}-\left|\frac{1}{p}-\frac{1}{2}\right|\right){\left|x\right|}^2\sin \theta }{\Pi }_{\propto }\left(s\right)ds\times \left|f\left(x\right)\right|{\text{e}}^{-\frac{|x{|}^2}{p}}d{\mu }_{\propto }\left(x\right)\right)}^{p}d{\mu }_{\propto }\left(x\right)}\\ \lesssim {\displaystyle {\int }_{R_{+}^d}{\left({\displaystyle {\int }_{G\cap \left\{\cos \theta <0\right\}}{\left|2x\right|}^{2\left(|\propto |+d\right)}}{\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left(m+1\right)-1}{2}}{\text{e}}^{-c{\left|x\right|}^2\sin \theta }{\Pi }_{\propto }\left(s\right)ds\times \left|f\left(x\right)\right|{\text{e}}^{-\frac{|x{|}^2}{p}}d{\mu }_{\propto }\left(x\right)\right)}^{p}d{\mu }_{\propto }}\left(x\right).\end{array}$

To finish the proof we just need to check that the kernel

$H\left(x,s\right):={\left|2x\right|}^{2\left(|\propto |+d\right)}{\text{e}}^{-c_{p}2{\left|x\right|}^2\sin \theta }\chi \mathfrak{G}\cap \left\{\cos \theta \ge 0\right\},$

for $\mathfrak{G}=\left\{\left(x,s\right):q_{-}{}^{\frac{1}{2}}\left(x^2,s\right)\ge \frac{c}{1+2\left|x\right|}\right\}$ is in $L^1\left(d{\left(m+1\right)}_{\propto }(\; x\; )\right)$

and independently of the remaining variables. Due to the symmetry of the kernel we are going to check only the first Claim given in [1].

$\begin{array}{l}{\displaystyle {\int }_{R_{+}^d}H\left(x,s\right)d{\left(m+1\right)}_{\propto }\left(x\right)}\\ \lesssim {\displaystyle {\int }_{0\le 1}{\left|x\right|}^{2\left(|\propto |+d\right)}{\text{e}}^{-c_p\left|x\right|{\left(2{\left|x\right|}^2\left(1-\cos \theta \right)\right)}^{1/2}}d{\left(m+1\right)}_{\propto }\left(x\right)}\\ +{\displaystyle {\int }_{0>1}2{\left|x\right|}^{2\left(|\propto |+d\right)}{\text{e}}^{-{\tilde{c}}_p\left(2\left|x\right|\right)}d{\left(m+1\right)}_{\propto }}\left(x\right).\end{array}$

It is clear that the second integral is bounded independentl y of x and s, for the first one see (13) for any x.

It is known that the first order Riesz-Laguerre transforms are weak-type (1, 1). Furthermore, we also know from that the the Riesz-Laguerre transforms of order higher than 2 need not be weak-type (1, 1) with respect to ${\mu }_{\alpha }$. However, we can prove the following result that has to do with certain kind of weights we can add on the domain of these transforms to make them satisfy a weak-type inequality.

Let us mention that in the Gaussian context something quite similar occur with the higher order Riesz-Gauss transforms. Perez proved that for $\left|m+1\right|>2$, the Riesz-Gauss transforms of order $\left|m+1\right|$ associated to the Ornstein-Uhlenbeck semigroup, map $L^1\left(\left(1+{\left|x\right|}^{\left|m+1\right|-2}\right)d\gamma \right)$ continuously into $L^{1,\infty }\left(d\gamma \right)$, with $d\gamma \left(x\right)={\text{e}}^{-{\left|x\right|}^2}dx$. Regarding the weights for the Riesz-Laguerre transforms of order higher than 2, then [1] proved the following

Theorem 9: The Riesz-Laguerre transforms order $\left|m+1\right|$ with $\left|m+1\right|>2$, map $L^1\left(wd{\mu }_{\alpha }\right)$ continuously into $L^{1,\infty }\left(d{\mu }_{\alpha }\right)$. Where

$w\left(x\right)={\left(1+\sqrt{\left|x\right|}\right)}^{\left|m+1\right|-2}$

Proof. As we mention in the preliminaries to prove this theorem is equivalent to prove that the modified Riesz-Laguerre transforms of order higher than 2 map $L^1\left({\tilde{w}}_{\epsilon }d{\tilde{\mu }}_{\propto }\right)$ continuously into $L^{1,\infty }\left(d{\tilde{\mu }}_{\propto }\right)$, with $\tilde{w}\left(x\right)={\left(1+\left|x\right|\right)}^{\left|m+1\right|-2}$. For each $x\in R_{+}^d$. Let us write

$R_{+}^d\times {\left[-1,1\right]}^d={\displaystyle \underset{i=0}{\overset{4}{\cup }}{R}_i}.$

Therefore, in order to get the result, it will be enough to prove that each of the following operators

$T_i^{m+1}f\left(x\right)={\displaystyle \underset{R_{+}^d}{\int }{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\chi }_{R_i}}}\left(x,s\right){\mathcal{K}}^{m+1}\left(x,s\right){\Pi }_{\propto }\left(s\right)dsf\left(x\right)d{\mu }_{\propto }\left(x\right),$

for $i=0,\cdots ,4$ maps $L^1\left({\tilde{w}}_{\epsilon }d{\tilde{\mu }}_{\propto }\right)$ continuously into $L^{1,\infty }\left(d{\tilde{\mu }}_{\propto }\right)$.

0

$\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|\lesssim \{\begin{array}{l}{\left(2{\left|x\right|}^2\right)}^{\frac{\left|m+1\right|-2}{2}}{\mathcal{K}}^{*}\left(x,s\right),\text{if}\cos \theta <0,\\ {\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left|m+1\right|-2}{2}}{\mathcal{K}}^{*}\left(x,s\right),\cos \theta \ge 0\end{array}$

If $\left(x,s\right)\in R_i$, $\left|k^{m+1}\left(x,s\right)\right|$ is controlled by $C{\left(1+\left\{\left|x\right|\right\}\right)}^{\left|m+1\right|-2}{\text{e}}^{-{\left|x\right|}^2}$ and there for it is immediate to prove that $T_1^{m+1}$ maps $L^1\left({\tilde{w}}_{\epsilon }d{\tilde{\mu }}_{\propto }\right)$ into $L^1\left(d{\tilde{\mu }}_{\propto }\right)$.

Now if $\left(x,s\right)\in R_i$, with $i=2,3,4$, weclaim that

$\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|\lesssim \tilde{w}\left(x\right){\mathcal{K}}^{*}\left(x,s\right)$

If $\left(x,s\right)\in R_2$ since

$q_{+}\le {\left(2\left|x\right|\right)}^2\lesssim {\left|x\right|}^2,$

then

$\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|\lesssim {\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left|m+1\right|-2}{2}}{\text{e}}^{-C{\left({\left|x\right|}^4\left(1-\cos \theta \right)\right)}^{1/2}}\lesssim \tilde{w}\left(x\right){\text{e}}^{-C{\left({\left|x\right|}^4\left(1-\cos \theta \right)\right)}^{1/2}}.$

Also

$q_{+}{q}_{-}=4{\left|x\right|}^4{\sin }^2\theta .$

Thus

$\left|{\mathcal{K}}^{m+1}\left(x,s\right)\right|\lesssim {\left(2{\left|x\right|}^2\sin \theta \right)}^{\frac{\left|m+1\right|-2}{2}}{\mathcal{K}}^{*}\left(x,s\right)\lesssim \tilde{w}\left(x\right){\mathcal{K}}^{*}\left(x,s\right).$

And this concludes the proof of the Theorem.

It should be noted that there is another proof of Theorem 9 for multi-indices of half-integer type by taking $f_w$ as the function f in [3] [7].

Now we introduce a sharp estimate forw.

Corollary 10: The Riesz-Laguerre transforms of order $\left|m+1\right|$ with $\left|m+1\right|>2$, map $L^1\left(wd{\mu }_{\alpha }\right)$ continuously into $L^{1,\infty }\left(d{\mu }_{\alpha }\right)$. Where

$|\propto |=\frac{8n\left(1-d\right)-1-{\left(2{\text{e}}^{2n}-1\right)}^2}{8n}$

and

$w\left(x\right)\le {\left(1+\sqrt{\left|C\right|}\right)}^{\left|m+1\right|-2}=K^{\left|m+1\right|-2}$

Proof. From Theorem 9 and Remark 5: We can directly see that

$w\left(x\right)={\left(1+\sqrt{\left|x\right|}\right)}^{\left|m+1\right|-2}\le {\left(1+\sqrt{\left|C\right|}\right)}^{\left|m+1\right|-2}\le K^{\left|m+1\right|-2},$

where $m\ge 2$.

Theorem 11: The weight w is the optimal polynomial weight needed to get the weak type (1, 1) inequality for the Riesz-Laguerre transforms of order $\left|m+1\right|$.

Proof. This proof follows essentially in [9]. With the notation of that Theorem 1 One takes $\eta \in {ℝ}_{+}^d$ with $\left|\eta \right|$ sufficiently large, away from the axis and obtains the following lower bound for ${\mathcal{K}}^{m+1}\left(x,\eta \right)$

${\mathcal{K}}^{m+1}\left(x,\eta \right)=C{\displaystyle \underset{{\left[-1,1\right]}^d}{\int }{\mathcal{K}}^{m+1}\left(x,\eta ,s\right){\Pi }_{\propto }\left(s\right)ds}\ge C{\left|\eta \right|}^{\left|m+1\right|-2|\propto |-d-1}{\text{e}}^{{\xi }^2-{\left|\eta \right|}^2}$ (18)

for $x\in J=\left\{\xi \frac{\eta }{\left|\eta \right|}+v:v\perp \eta ,\left|v\right|<1,\frac{1}{2}\left|\eta \right|<\xi <\frac{3}{2}\left|\eta \right|\right\}$.

Now if we assume that the Riesz-Laguerre transforms of order $\left|m+1\right|>2$ map $L^1\left({\tilde{w}}_{\epsilon }d{\tilde{\mu }}_{\propto }\right)$ continuously into $L^{1,\infty }\left(d{\tilde{\mu }}_{\propto }\right)$ with ${\tilde{w}}_{\epsilon }={\left(1+\left|x\right|\right)}^{\epsilon }$ and

$0<\epsilon <\left|m+1\right|-2$ then by taking $f\ge 0$ in $L^1\left({\tilde{w}}_{\epsilon }d{\tilde{\mu }}_{\propto }\right)$ close to an approxima tion of a point mass at η, with ${\Vert f\Vert }_{L^1\left({\tilde{w}}_{\epsilon }d{\tilde{\mu }}_{\propto }\right)}=1$ we have that ${\mathcal{R}}_{\propto }^{m+1}f\left(x\right)$ is close

to ${\text{e}}^{{\left|\eta \right|}^2}{\mathcal{K}}^{m+1}\left(x,\eta \right){\left|\eta \right|}^{-\epsilon }$ and by applying in equality (18) we get that

$e\left|n\right|k^{m+1}\left(x,\eta \right){\left|\eta \right|}^{-ϵ}\ge {\left|\eta \right|}^{\left|m+1\right|-2\left|\alpha \right|-d-1-ϵ}{\text{e}}^{-{\left(\frac{\left|n\right|}{2}\right)}^2}$. Therefore setting

$\lambda ={\left|\eta \right|}^{\left|m+1\right|-2\left|\alpha \right|-d-1-ϵ}{\text{e}}^{-{\left(\frac{\left|n\right|}{2}\right)}^2}$

we obtain

$\begin{array}{l}{\text{e}}^{-{\left(\frac{\left|\eta \right|}{2}\right)}^2}{\left|\eta \right|}^{2\left|\alpha \right|+d-1}\lesssim {\tilde{\mu }}_{\alpha }\left(J\right)\le {\tilde{\mu }}_{\alpha }\left\{x\in R_{+}^d:R_{\alpha }^{m+1}f\left(x\right)>\lambda \right\}\\ \lesssim \frac{1}{\lambda }=C{\left|\eta \right|}^{2\left|\alpha \right|+d-\left|m+1\right|+1+ϵ}{\text{e}}^{-{\left(\frac{\left|\eta \right|}{2}\right)}^2}.\end{array}$

Hence ${\left|\eta \right|}^{\left|m+1\right|-2-ϵ}$ must be bounded which is a contradiction. Therefore the conclusion of Theorem 11 holds.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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