Square-like functions generated by the Laplace-Bessel differential operator

Abstract

We introduce a wavelet-type transform associated with the Laplace-Bessel differential operator ${\mathrm{\Delta }}_B={\sum }_{k=1}^n\frac{{\partial }^2}{\partial x_k}+\frac{2{\nu }_k}{\partial x_k}\frac{\partial }{\partial x_k}$ and the relevant square-like functions. An analogue of the Calderón reproducing formula and the $L_{2,\nu }$ boundedness of the square-like functions are obtained.

MSC:47G10, 42C40, 44A35.

1 Introduction

The classical square functions $f(x)\to S_{\phi }(x)={({\int }_0^{\mathrm{\infty }}|(f\ast {\phi }_t)(x){|}^2\frac{dt}{t})}^{\frac{1}{2}}$, where $\phi \in S$, $S\equiv S({\mathbb{R}}^n)$ is the Schwartz test function space and ${\int }_{{\mathbb{R}}^n}\phi (x)dx=0$, ${\phi }_t(x)=t^{-n}\phi (t^{-1}x)$, $t>0$, play important role in harmonic analysis and its applications; see Stein [1]. There are a lot of diverse variants of square functions and their applications; see Daly and Phillips [2], Jones et al. [3], Pipher [4], Kim [5]. Square-like functions generated by a composite wavelet transform and its $L_2$ estimates are proved by Aliev and Bayrakci [6].

Note that the Laplace-Bessel differential operator ${\mathrm{\Delta }}_B$ is known as an important operator in analysis and its applications. The relevant harmonic analysis, known as Fourier-Bessel harmonic analysis associated with the Bessel differential operator $B_t=\frac{d^2}{d{t}^2}+\frac{2\nu }{t}\frac{d}{dt}$, has been the research area for many mathematicians such as Levitan, Muckenhoupt, Stein, Kipriyanov, Klyuchantsev, Löfström, Peetre, Gadjiev, Aliev, Guliev, Triméche, Rubin and others (see [7–14]). Moreover, a lot of mathematicians studied a Calderón reproducing formula. For example, Amri and Rachdi [15], Guliyev and Ibrahimov [16], Kamoun and Mohamed [17], Pathak and Pandey [18], Mourou and Trimèche [19, 20] and others.

In this paper, firstly we introduce a wavelet-like transform associated with the Laplace-Bessel differential operator,

$${\mathrm{\Delta }}_B=\sum _{k=1}^n\frac{{\partial }^2}{\partial x_k^2}+\frac{2{\nu }_k}{\partial x_k}\frac{\partial }{\partial x_k},\nu =({\nu }_1,{\nu }_2,\dots ,{\nu }_n),\nu >0,$$

and then the relevant square-like function. The plan of the paper is as follows. Some necessary definitions and auxiliary facts are given in Section 2. In Section 3 we prove a Calderón-type reproducing formula and the $L_{2,\nu }$ boundedness of the square-like functions.

2 Preliminaries

${\mathbb{R}}_{+}^n=\{x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n:x_1>0,x_2>0,\dots ,x_n>0\}$ and let $S({\mathbb{R}}_{+}^n)$ be the Schwartz space of infinitely differentiable and rapidly decreasing functions.

$L_{p,\nu }=L_{p,\nu }({\mathbb{R}}_{+}^n)$ ($1\le p<\mathrm{\infty }$, $\nu =({\nu }_1,\dots ,{\nu }_n)$; ${\nu }_1>0,\dots ,{\nu }_n>0$) space is defined as the class of measurable functions f on ${\mathbb{R}}_{+}^n$ for which

$${\parallel f\parallel }_{p,\nu }={\left({\int }_{{\mathbb{R}}_{+}^n}\right|f(x){|}^p{x}^{2\nu }dx)}^{\frac{1}{p}}<\mathrm{\infty },x^{2\nu }dx=x_1^{2{\nu }_1}{x}_2^{2{\nu }_2}\cdots x_n^{2{\nu }_n}d{x}_{1}d{x}_2\cdots d{x}_n.$$

In the case $p=\mathrm{\infty }$, we identify $L_{\mathrm{\infty }}\equiv L_{\mathrm{\infty },\nu }$ with $C_0$ the space of continuous functions vanishing at infinity, and set ${\parallel f\parallel }_{\mathrm{\infty }}={sup}_{x\in {\mathbb{R}}_{+}^n}|f(x)|$.

The Fourier-Bessel transform and its inverse are defined by

$$f^{\wedge }(x)=F_{\nu }(f)(x)={\int }_{{\mathbb{R}}_{+}^n}f(y)(\prod _{k=1}^n{j}_{{\nu }_k-\frac{1}{2}}(x_k{y}_k))y^{2\nu }dy,$$

(2.1)

$$F_{\nu }^{-1}(f)(x)=c_{\nu }(n)(F_{\nu }f)(x),c_{\nu }(n)={\left[2^{2n}\prod _{k=1}^n{\mathrm{\Gamma }}^2({\nu }_k+\frac{1}{2})\right]}^{-1},$$

(2.2)

where $j_{\nu -\frac{1}{2}}$ is the normalized Bessel function, which is also the eigenfunction of the Bessel operator $B_t=\frac{d^2}{d{t}^2}+\frac{2\nu }{t}\frac{d}{dt}$; $j_{v-\frac{1}{2}}(0)=1$ and $j_{\nu -\frac{1}{2}}^{\prime }(0)=0$ (see [10]).

Denote by $T^y$ ($y\in {\mathbb{R}}_{+}^n$) the generalized translation operator acting according to the law:

$$\begin{array}{rcl}{T}^{y}f(x)& =& {\pi }^{-n/2}\prod _{k=1}^n\mathrm{\Gamma }({\nu }_k+\frac{1}{2}){\mathrm{\Gamma }}^{-1}({\nu }_k){\int }_0^{\pi }\cdots {\int }_0^{\pi }f(\sqrt{x_1^2-2x_1{y}_{1}cos{\alpha }_1+y_1^2},\dots ,\\ \sqrt{x_n^2-2x_n{y}_{n}cos{\alpha }_n+y_n^2})\prod _{k=1}^n{sin}^{2{\nu }_k-1}{\alpha }_{k}d{\alpha }_1\cdots d{\alpha }_n.\end{array}$$

$T^y$ is closely connected with the Bessel operator $B_t$ (see [10]). It is known that (see [11])

$${\parallel T^{y}f\parallel }_{p,\nu }\le {\parallel f\parallel }_{p,\nu }\mathrm{\forall }y\in {\mathbb{R}}_{+}^n,1\le p\le \mathrm{\infty },$$

(2.3)

$${\parallel T^{y}f-f\parallel }_{p,\nu }\to 0,|y|\to 0,1\le p\le \mathrm{\infty }.$$

(2.4)

The generalized convolution ‘B-convolution’ associated with the generalized translation operator is $(f\ast g)(x)={\int }_{{\mathbb{R}}_n^{+}}f(y)(T^{y}g(x))y^{2\nu }dy$ for which

$${(f\ast g)}^{\wedge }=f^{\wedge }{g}^{\wedge }.$$

(2.5)

We consider the B-maximal operator (see [8, 21])

$$M_{B}f(x)=\underset{r>0}{sup}|E_{+}(0,r){|}_{2\nu }^{-1}{\int }_{E_{+}(0,r)}{T}^y|f(x)|y^{2\nu }dy,$$

where $E_{+}(0,r)=\{y\in {\mathbb{R}}_{+}^n:|y|<r\}$ and $|E_{+}(0,r){|}_{2\nu }={\int }_{E_{+}(0,r)}{x}^{2\nu }dx=C{r}^{n+2\nu }$. Moreover, the following inequalities are satisfied (see for details [22]).

• (a) If $f\in L_{1,\nu }({\mathbb{R}}_{+}^n)$, then for every $\alpha >0$,

  $$|\{x:M_{B}f(x)>\alpha \}{|}_{2\nu }\le \frac{c}{\alpha }{\int }_{{\mathbb{R}}_{+}^n}|f(x)|x^{2\nu }dx,$$

where $c>0$ is independent of f.

• (b) If $f\in L_{p,\nu }({\mathbb{R}}_{+}^n)$, $1<p\le \mathrm{\infty }$, then $M_{B}f\in L_{p,\nu }(R_{+}^n)$ and

  $${\parallel M_{B}f\parallel }_{p,\nu }\le C_p{\parallel f\parallel }_{p,\nu },$$

where $c_p$ is independent of f.

Furthermore, if $f\in L_{p,\nu }({\mathbb{R}}_{+}^n)$, $1\le p\le \mathrm{\infty }$, then

$$\underset{r\to 0}{lim}|E_{+}(0,r){|}_{2\nu }^{-1}{\int }_{E_{+}(0,r)}{T}^{y}f(x)y^{2\nu }dy=f(x).$$

Now, we will need the generalized Gauss-Weierstrass kernel defined as

$$g_{\nu }(x,t)=F_{\nu }^{-1}\left(e^{-t|\cdot {|}^2}\right)(x)=\sqrt{c_{\nu }(n)}{t}^{\frac{-(n+2|\nu |)}{2}}{e}^{\frac{-x^2}{4t}},x\in {\mathbb{R}}_{+}^n,t>0$$

(2.6)

$c_{\nu }(n)$ being defined by (2.2) and $|\nu |={\nu }_1+{\nu }_2+\cdots +{\nu }_n$.

The kernel $g_{\nu }(x,t)$ possesses the following properties:

$$(\mathrm{a})F_{\nu }(g_{\nu }(\cdot ,t))(x)=e^{-t|x{|}^2}(t>0);$$

(2.7)

$$(\mathrm{b}){\int }_{{\mathbb{R}}_{+}^n}{g}_{\nu }(y,t)dy=1(t>0).$$

(2.8)

Given a function $f:{\mathbb{R}}_n^{+}\to \mathbb{C}$, the generalized Gauss-Weierstrass semigroup, $G_{t}f(x)$ is defined as

$$G_{t}f(x)={\int }_{{\mathbb{R}}_{+}^n}{g}_{\nu }(y,t)(T^{y}f(x))y^{2\nu }dy,t>0.$$

(2.9)

This semigroup is well known and arises in the context of stable random processes in probability, in pseudo-differential parabolic equations and in integral geometry; see Koldobsky, Landkof, Fedorjuk, Aliev, Rubin, Sezer and Uyhan (see [23–26]).

The following lemma contains some properties of the semigroup ${\{G_{t}f\}}_{t\ge 0}$. (Compare with the analogous properties of the classical Gauss-Weierstrass integral [1, 27, 28].)

Lemma 2.1 If $f\in L_{p,\nu }$, $1\le p\le \mathrm{\infty }$ ($L_{\mathrm{\infty }}\equiv C_0$), then

$$(\mathrm{a}){\parallel G_{t}f\parallel }_{p,\nu }\le c{\parallel f\parallel }_{p,\nu },$$

(2.10)

$$(\mathrm{b})\underset{t\to 0}{lim}{G}_{t}f(x)=f(x).$$

(2.11)

The limit is understood in $L_{p,\nu }$ norm and pointwise almost all $x\in {\mathbb{R}}_{+}^n$. If $f\in C_0$, then the limit is uniform on ${\mathbb{R}}_{+}^n$.

$$(\mathrm{c})\underset{t>0}{sup}|G_{t}f(x)|\le c{M}_{B}f(x),$$

(2.12)

where $M_{B}f$ is the well-known Hardy-Littlewood maximal function.

Moreover, let $h(z)$ be an absolutely continuous function on $[0,\mathrm{\infty })$ and

$$\alpha ={\int }_0^{\mathrm{\infty }}\frac{h(z)}{z}dz<\mathrm{\infty }.$$

(2.13)

If we denote $w(z)=h^{\prime }(z)$, we have from (2.13)

$$h(0)=0\text{and}h(\mathrm{\infty })=0$$

(2.14)

(see for details [29]).

Now, we define the following wavelet-like transform:

$$V_{t}f(x)=\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}{G}_{tz}f(x)w(z)dz,$$

(2.15)

where $w(z)$ is known as ‘wavelet function’, ${\int }_0^{\mathrm{\infty }}w(z)dz=0$, and the function $G_{tz}f(x)$ is the generalized Gauss-Weierstrass semigroup.

Using wavelet-like transform (2.15), we define the following square-like functions:

$$(Sf)(x)={\left({\int }_0^{\mathrm{\infty }}\right|V_{t}f(x){|}^2\frac{dt}{t})}^{\frac{1}{2}}.$$

(2.16)

3 Main theorems and proofs

Theorem 3.1

• (a) Let $f\in L_{p,\nu }$, $1\le p\le \mathrm{\infty }$ ($L_{\mathrm{\infty }}\equiv C_0$), $\nu >0$. We have

  $${\parallel V_{t}f\parallel }_{p,\nu }\le c_1{c}_2{\parallel f\parallel }_{p,\nu }(\mathrm{\forall }t>0),$$

  (3.1)

where $c_1=2^{2|\nu |-n}$, $|\nu |={\nu }_1+{\nu }_2+\cdots +{\nu }_n$, $c_2=\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}|w(z)|dz<\mathrm{\infty }$.

• (b) Let $f\in L_{p,\nu }$, $1<p\le \mathrm{\infty }$ ($L_{\mathrm{\infty }}\equiv C_0$). We have

  $${\int }_0^{\mathrm{\infty }}{V}_{t}f(x)\frac{dt}{t}\equiv \underset{\rho \to \mathrm{\infty }}{\underset{ϵ\to 0}{lim}}{\int }_{ϵ}^{\rho }{V}_{t}f(x)\frac{dt}{t}=f(x),$$

  (3.2)

where limit can be interpreted in the $L_{p,\nu }$ norm and pointwise for almost all $x\in {\mathbb{R}}_{+}^n$. If $f\in C_0$, the convergence is uniform on ${\mathbb{R}}_{+}^n$.

Theorem 3.2 If $f\in L_{2,\nu }$, then

$${\parallel Sf\parallel }_{2,\nu }\le \frac{1}{2}{\parallel f\parallel }_{2,\nu }.$$

(3.3)

Proof of Theorem 3.1 (a) By using the Minkowski inequality, we have

$$\begin{array}{c}\begin{array}{rl}{\parallel V_{t}f\parallel }_{p,\nu }& =\frac{1}{\alpha }{\left({\int }_{{\mathbb{R}}_{+}^n}\right|{\int }_0^{\mathrm{\infty }}{G}_{tz}f(x)w(z)dz{|}^p{x}^{2\nu }dx)}^{\frac{1}{p}}\\ \le \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}|w(z)|{\parallel G_{tz}f\parallel }_{p,\nu }dz,\end{array}\hfill \\ \begin{array}{rl}{\parallel G_{tz}f\parallel }_{p,\nu }& ={\left({\int }_{{\mathbb{R}}_{+}^n}\right|{\int }_{{\mathbb{R}}_{+}^n}{g}_{\nu }(y,tz)T^{y}f(x)y^{2\nu }dy{|}^p{x}^{2\nu }dx)}^{\frac{1}{p}}\\ \le {\int }_{{\mathbb{R}}_{+}^n}|g_{\nu }(y,tz)|{\left({\int }_{{\mathbb{R}}_{+}^n}\right|T^{y}f(x){|}^p{x}^{2\nu }dx)}^{\frac{1}{p}}{y}^{2\nu }dy\\ \le {\parallel f\parallel }_{p,\nu }{\int }_{{\mathbb{R}}_{+}^n}|g_{\nu }(y,tz)|y^{2\nu }dy=c_1{\parallel f\parallel }_{p,\nu }.\end{array}\hfill \end{array}$$

Taking into account the following equality for $Re\mu >0$, $Re\nu >0$, $p>0$ (see [[30], p.370])

$${\int }_0^{\mathrm{\infty }}{x}^{\nu -1}{e}^{-\mu x^p}dx=\frac{1}{p}{\mu }^{-\frac{\nu }{p}}\mathrm{\Gamma }\left(\frac{\nu }{p}\right),$$

we have

$${\int }_0^{\mathrm{\infty }}{x}^{2\nu }{e}^{-x^2}dx=\frac{1}{2}\mathrm{\Gamma }(\nu +\frac{1}{2}),\nu >0$$

in one dimension. By using this equality, we get

$$\begin{array}{rcl}{c}_1& =& {\int }_{{\mathbb{R}}_{+}^n}|g_{\nu }(y,t)|y^{2\nu }dy\\ =& 2^{-n}\prod _{k=1}^n{\mathrm{\Gamma }}^{-1}({\nu }_k+\frac{1}{2})t^{\frac{{}^{-(n+2|\nu |)}2}{}}{\int }_{{\mathbb{R}}_{+}^n}{e}^{-\frac{|y{|}^2}{4t}}{y}^{2\nu }dy(y=2\sqrt{t}y,dy=2^n{t}^{\frac{n}{2}}dy)\\ =& 2^{-n}\prod _{k=1}^n{\mathrm{\Gamma }}^{-1}({\nu }_k+\frac{1}{2})t^{\frac{{}^{-(n+2|\nu |)}2}{}}{\int }_{{\mathbb{R}}_{+}^n}{e}^{-|y{|}^2}{2}^{2|\nu |}{t}^{|\nu |}{2}^n{t}^{\frac{n}{2}}{y}^{2\nu }dy\\ =& 2^{2|\nu |}\prod _{k=1}^n{\mathrm{\Gamma }}^{-1}({\nu }_k+\frac{1}{2}){\int }_{{\mathbb{R}}_{+}^n}{e}^{-|y{|}^2}{y}^{2\nu }dy\\ =& 2^{2|\nu |}\prod _{k=1}^n{\mathrm{\Gamma }}^{-1}({\nu }_k+\frac{1}{2})\prod _{k=1}^n\mathrm{\Gamma }({\nu }_k+\frac{1}{2})2^{-n}\\ =& 2^{2|\nu |-n}.\end{array}$$

So we have ${\parallel G_{tz}f\parallel }_{p,\nu }\le 2^{2|\nu |-n}{\parallel f\parallel }_{p,\nu }$, and then inequality (3.1).

• (b) Let $(A_{ϵ,\rho }f)(x)={\int }_{ϵ}^{\rho }{V}_{t}f(x)\frac{dt}{t}$, $0<ϵ<\rho <\mathrm{\infty }$. Applying Fubini’s theorem, we get

  $$\begin{array}{rcl}(A_{ϵ,\rho }f)(x)& =& \frac{1}{\alpha }{\int }_{ϵ}^{\rho }({\int }_0^{\mathrm{\infty }}{G}_{tz}f(x)w(z)dz)\frac{dt}{t}\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}w(z)({\int }_{ϵ}^{\rho }{G}_{tz}f(x)\frac{dt}{t})dz\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}w(z)({\int }_{ϵz}^{\rho z}{G}_{t}f(x)\frac{dt}{t})dz\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}({\int }_{\frac{t}{\rho }}^{\frac{t}{ϵ}}w(z)dz)G_{t}f(x)\frac{dt}{t}\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}\frac{1}{t}[h\left(\frac{t}{ϵ}\right)-h\left(\frac{t}{\rho }\right)]G_{t}f(x)dt\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}\frac{h(t)}{t}{G}_{ϵt}f(x)dt-\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}\frac{h(t)}{t}{G}_{\rho t}f(x)dt\\ =& (A_{ϵ}f)(x)-(A_{\rho }f)(x).\end{array}$$

By Theorem 1.15 in [[28], p.3], if $1<p\le \mathrm{\infty }$ ($L_{\mathrm{\infty }}\equiv C_0$), then

$$\underset{\rho \to \mathrm{\infty }}{lim}{\parallel G_{\rho t}f\parallel }_{p,\nu }=0.$$

Therefore, by the Minkowski inequality and the Lebesgue dominated convergence theorem, taking into account Lemma 2.1, we have

$$\begin{array}{rcl}{\parallel A_{\rho }f\parallel }_{p,\nu }& =& \frac{1}{\alpha }{\left({\int }_{{\mathbb{R}}_n^{+}}{({\int }_{{}^0\mathrm{\infty }}^{}\frac{h(t)}{t}{G}_{\rho t}f(x)dt)}^p{x}^{2\nu }dx\right)}^{\frac{1}{p}}\\ \le & \frac{1}{\alpha }{\int }_{{}^0\mathrm{\infty }}^{}\frac{h(t)}{t}{\parallel G_{\rho t}f\parallel }_{p,\nu }dt\\ =& \frac{1}{\alpha }{\int }_{{}^{}{}_0{}^{}\mathrm{\infty }}^{}\frac{h(\frac{t}{\rho })}{\frac{t}{\rho }}{\parallel G_{\rho t}f\parallel }_{p,\nu }\frac{1}{\rho }dt\to 0,\rho \to \mathrm{\infty }\end{array}$$

and

$$\begin{array}{rcl}{\parallel A_{ϵ}f-f\parallel }_{p,\nu }& =& {\left({\int }_{{\mathbb{R}}_n^{+}}{(\frac{1}{\alpha }{\int }_{{}^0\mathrm{\infty }}^{}\frac{h(t)}{t}{G}_{ϵt}f(x)dt-f(x))}^p{x}^{2\nu }dx\right)}^{\frac{1}{p}}\\ \stackrel{(\text{2.13})}{=}& {\left({\int }_{{\mathbb{R}}_n^{+}}{(\frac{1}{\alpha }{\int }_{{}^0\mathrm{\infty }}^{}\frac{h(t)}{t}{G}_{ϵt}f(x)dt-\frac{1}{\alpha }{\int }_{{}^0\mathrm{\infty }}^{}\frac{h(t)}{t}f(x)dt)}^p{x}^{2\nu }dx\right)}^{\frac{1}{p}}\\ \le & \frac{1}{\alpha }{\int }_{{}^0\mathrm{\infty }}^{}\frac{h(t)}{t}{\parallel G_{ϵt}f-f\parallel }_{p,\nu }dt\to 0,ϵ\to 0.\end{array}$$

Finally, for $1<p\le \mathrm{\infty }$ ($L_{\mathrm{\infty }}\equiv C_0$), we get

$${\parallel A_{ϵ,\rho }f-f\parallel }_{p,\nu }={\parallel A_{ϵ}f-f\parallel }_{p,\nu }+{\parallel A_{\rho }f\parallel }_{p,\nu }\to 0,ϵ\to 0,\rho \to \mathrm{\infty }.$$

The a.e. convergence is based on the standard maximal function technique (see [[31], p.60], [29] and [32]). □

Proof of Theorem 3.2 Firstly, let $f\in S({\mathbb{R}}_{+}^n)$. By making use of the Fubini and Plancherel (for Fourier-Bessel transform) theorems, we get

$$\begin{array}{rl}{\parallel Sf\parallel }_{2,\nu }^2& ={\int }_{{\mathbb{R}}_{+}^n}\left({\int }_0^{\mathrm{\infty }}\right|V_{t}f(x){|}^2\frac{dt}{t})x^{2\nu }dx\\ ={\int }_0^{\mathrm{\infty }}\left({\int }_{{\mathbb{R}}_n^{+}}\right|V_{t}f(x){|}^2{x}^{2\nu }dx)\frac{dt}{t}\\ ={\int }_0^{\mathrm{\infty }}\left({\int }_{{\mathbb{R}}_n^{+}}\right|{(V_{t}f)}^{\wedge }(x){|}^2{x}^{2\nu }dx)\frac{dt}{t}\end{array}$$

and

$$\begin{array}{rcl}{(V_{t}f)}^{\wedge }(x)& =& F_{\nu }(V_{t}f)(x)=\frac{1}{\alpha }{\int }_{{\mathbb{R}}_n^{+}}({\int }_0^{\mathrm{\infty }}{G}_{tz}f(y)w(z)dz)\prod _{k=1}^n{j}_{{\nu }_k-\frac{1}{2}}(x_k{y}_k)y^{2\nu }dy\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}w(z)({\int }_{{\mathbb{R}}_n^{+}}{G}_{tz}f(y)\prod _{k=1}^n{j}_{{\nu }_k-\frac{1}{2}}(x_k{y}_k)y^{2\nu }dy)dz\\ =& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}w(z){(G_{tz}f)}^{\wedge }(x)dz\\ \stackrel{(\text{2.5})}{=}& \frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}w(z)f^{\wedge }(x)e^{-tz|x{|}^2}dz.\end{array}$$

Now, by using Fubini’s theorem, we have

$$\begin{array}{rcl}{\parallel Sf\parallel }_{2,\nu }^2& =& \frac{1}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}\left[{\int }_{{\mathbb{R}}_n^{+}}{(f^{\wedge }(x))}^2{({\int }_0^{\mathrm{\infty }}w(z)e^{-tz|x{|}^2}dz)}^2{x}^{2\nu }dx\right]\frac{dt}{t}\\ =& \frac{1}{{\alpha }^2}{\int }_{{\mathbb{R}}_n^{+}}{(f^{\wedge }(x))}^2{\int }_0^{\mathrm{\infty }}\frac{dt}{t}{({\int }_0^{\mathrm{\infty }}w(z)e^{-tz|x{|}^2}dz)}^2{x}^{2\nu }dx\\ (t=\tau |x{|}^{-2},dt=|x{|}^{-2}d\tau )\\ =& \frac{1}{{\alpha }^2}{\int }_{{\mathbb{R}}_n^{+}}{(f^{\wedge }(x))}^2{\int }_0^{\mathrm{\infty }}\frac{d\tau }{\tau }{({\int }_0^{\mathrm{\infty }}w(z)e^{-\tau z}dz)}^2{x}^{2\nu }dx\\ =& C^2\frac{1}{{\alpha }^2}{\parallel f\parallel }_{2,\nu }^2,\end{array}$$

where

$$C={\left({\int }_0^{\mathrm{\infty }}\frac{d\tau }{\tau }{({\int }_0^{\mathrm{\infty }}{e}^{-\tau z}w(z)dz)}^2\right)}^{1/2}.$$

Since $w(z)=h^{\prime }(z)$, $h(z)\ge 0$, $h(\mathrm{\infty })=h(0)=0$, it follows that

$$\begin{array}{rcl}C& =& {\left({\int }_0^{\mathrm{\infty }}\frac{d\tau }{\tau }{({\int }_0^{\mathrm{\infty }}{e}^{-\tau z}w(z)dz)}^2\right)}^{1/2}\\ =& {\left({\int }_0^{\mathrm{\infty }}{({\int }_0^{\mathrm{\infty }}\sqrt{\tau }{e}^{-\tau z}h(z)dz)}^{2}d\tau \right)}^{1/2}\\ \le & {\int }_0^{\mathrm{\infty }}h(z){\left({\int }_0^{\mathrm{\infty }}\tau e^{-2\tau z}d\tau \right)}^{1/2}dz(2z\tau =t,2zd\tau =dt)\\ =& {\int }_0^{\mathrm{\infty }}h(z){\left({\int }_0^{\mathrm{\infty }}\frac{t}{2z}{e}^{-t}\frac{1}{2z}dt\right)}^{1/2}dz\\ =& {\int }_0^{\mathrm{\infty }}\frac{h(z)}{2z}{\left({\int }_0^{\mathrm{\infty }}t{e}^{-t}dt\right)}^{1/2}dz=\frac{1}{2}\alpha .\end{array}$$

Finally, we get

$${\parallel Sf\parallel }_{2,\nu }\le \frac{1}{2}{\parallel f\parallel }_{2,\nu }.$$

For arbitrary $f\in L_{2,\nu }({\mathbb{R}}_{+}^n)$, the result follows by density of the class $S({\mathbb{R}}_{+}^n)$ in $L_{2,\nu }({\mathbb{R}}_{+}^n)$. Namely, let $(f_n)$ be a sequence of functions in $S({\mathbb{R}}_{+}^n)$, which converge to f in $L_{2,\nu }({\mathbb{R}}_{+}^n)$-norm. That is, ${lim}_{n\to \mathrm{\infty }}$ ${\parallel f_n(x)-f(x)\parallel }_{2,\nu }=0$, $\mathrm{\forall }x\in {\mathbb{R}}_{+}^n$.

From the ‘triangle inequality’ (${({\parallel u\parallel }_{2,\nu }-{\parallel v\parallel }_{2,\nu })}^2\le {\parallel u-v\parallel }_{2,\nu }^2$), we have

$$\begin{array}{rcl}|(S{f}_n)(x)-(S{f}_m)(x){|}^2& =& {[{\left({\int }_0^{\mathrm{\infty }}\right|V_t{f}_n(x){|}^2\frac{dt}{t})}^{\frac{1}{2}}-\left({\left({\int }_0^{\mathrm{\infty }}\right|V_t{f}_m(x){|}^2\frac{dt}{t})}^{\frac{1}{2}}\right)]}^2\\ \le & {\int }_0^{\mathrm{\infty }}|V_t{f}_n(x)-V_t{f}_m(x){|}^2\frac{dt}{t}\\ =& {\int }_0^{\mathrm{\infty }}|V_t(f_n-f_m){|}^2\frac{dt}{t}\\ =& {(S(f_n-f_m)(x))}^2.\end{array}$$

Hence

$${\parallel S{f}_n-S{f}_m\parallel }_{2,\nu }\le {\parallel S(f_n-f_m)\parallel }_{2,\nu }\le \frac{1}{2}{\parallel f_n-f_m\parallel }_{2,\nu }.$$

This shows that the sequence $(S{f}_n)$ converges to Sf in $L_{2,\nu }({\mathbb{R}}_{+}^n)$-norm. Thus

$${\parallel Sf\parallel }_{2,\nu }\le \frac{1}{2}{\parallel f\parallel }_{2,\nu },\mathrm{\forall }f\in L_{2,\nu }\left({\mathbb{R}}_{+}^n\right)$$

and the proof is complete. □