Dunkl–Schrödinger Operators

Abstract

In this paper, we consider the Schrödinger operators $L_k=-\Delta _k+V$, where $\Delta _k$ is the Dunkl–Laplace operator and V is a non-negative potential on $\mathbb {R}^d$. We establish that $L_k$ is essentially self-adjoint on $C_0^\infty (\mathbb {R}^d)$. In particular, we develop a bounded $H^\infty$-calculus on $L^p$ spaces for the Dunkl harmonic oscillator operator.

1 Introduction

During the recent decades the ordinary Schrödinger operators $-\Delta +V$ have been generalized in a domains where a family of a Laplace type operators are given, see for example [4, 12, 13]. In a similar way, in the area of harmonic analysis, functional calculus for self-adjoint operators and a number of important applications are developed [9, 10, 14].

In this paper we consider Schrödinger operators $L_k=-\Delta _k+V$ associated to the Dunkl Laplace operator on $\mathbb {R}^d$ given by $\Delta _k=\sum _{j=1}^{^d}T_j^2$ where $T_j$ are a family of differential–difference operators associated to a finite reflection group, which are called Dunkl operators. It is well known that Dunkl theory provides a generalization of the Fourier Analysis. There are many classical results in Fourier analysis that are extended to Dunkl setting and the present work fits within this framework. Arise from the work of Simon [20], we study the problem of essential selfadjointness of $L_k$ and the correspondent heat semi group $e^{-tL_k}$. We investigate the spectral theory, complex analysis and theory of holomorphic functional calculus for the generator of an holomorphic semi group, as present in [9, 10, 14] to develop an $L^p$-boundedness of holomorphic functional calculus for the Dunkl harmonic oscillator $H_k=-\Delta _k+|x|^2$, where the use of the heat kernel being the most powerful tools.

The paper is outline as follows. In the next section we give Backgrounds form Dunkl’s theory. The Sects. 3 and 4 are devoted to study the essential selfadjointness of the operators $\Delta _k$ and $L_k$. The Sect. 5 treat the $L^p$-boundedness of holomorphic functional calculus for $H_k$.

2 Basics of the Dunkl Theory

For details, we refer to [5, 7, 8, 18] and the references cited there.

$\mathbb {R}^d$ is equipped with a scalar product $\langle x,y\rangle =\sum _{j=1}^dx_jy_j$ which induces the Euclidean norm $|x|=\langle x, x\rangle ^{1/2}$.

Let $G\subset \text {O}(\mathbb {R}^d)$ be a finite reflection group associated to a reduced root system R and $k:R\rightarrow [0,+\infty )$ be a G–invariant function (called multiplicity function). Let $R^+$ be a positive root subsystem. The Dunkl operators  $T_\xi$ on $\mathbb {R}^d$ are the following k–deformations of directional derivatives $\partial _\xi$ by difference operators:

$$\begin{aligned} T_\xi f(x)=\partial _\xi f(x) +\sum _{\,\alpha \in R^+} k(\alpha )\,\langle \alpha ,\xi \rangle \, \frac{f(x)-f(\sigma _\alpha .\,x)}{\langle \alpha ,\,x\rangle }, \end{aligned}$$

where $\sigma _\alpha$ denotes the reflection with respect to the hyperplane orthogonal to $\alpha$. The Dunkl operators are antisymmetric with respect to the measure $w_k(x)\,dx$ with density

$$\begin{aligned} w_k(x)=\,\prod _{\,\alpha \in R^+}|\,\langle \alpha ,x\rangle \,|^{\,2\,k(\alpha )}. \end{aligned}$$

The operators $\partial _\xi$ and $T_\xi$ are intertwined by a Laplace–type operator

$$\begin{aligned} V_kf(x)\, =\int _{\mathbb {R}^d}f(y)\,d\nu _x(y), \end{aligned}$$

associated to a family of compactly supported probability measures  $\{\,\nu _x\,|\,x\in \mathbb {R}^d\}$. Specifically,  $\nu _x$ is supported in the the convex hull ${\text {co}}(G.x).$

For every $y\in \mathbb {C}^d$, the simultaneous eigenfunction problem

$$\begin{aligned} T_\xi f=\langle y,\xi \rangle \,f \qquad \forall \;\xi \in \mathbb {R}^d \end{aligned}$$

has a unique solution $f(x)=E_k(x,y)$ such that $E_k(0,y)=1$, called the Dunkl kernel and is given by

$$\begin{aligned} E_k(x,y)\, =\,V(e^{\,\langle \lambda ,.\,\rangle })(x)\, =\int _{\mathbb {R}^d}e^{\,\langle z,y\rangle }\,d\nu _x(z) \qquad \forall \;x\in \mathbb {R}^d. \end{aligned}$$

(2.1)

Furthermore this kernel has a holomorphic extension to $\mathbb {C}^d\times \mathbb {C}^d$ and the following estimate hold: for $\;x, \;y\in \mathbb {C}^d,$

• (ii) $E_k(x,y)=E_k(y,x)$,

• (iii) $E_k(\lambda x,y)=E_k(x,\lambda y)$, for $\lambda \in \mathbb {C}$

• (iv) $E_k(g. x,g.y)=E_k(x, y)$, for $g\in G$.

In dimension $d=1$, these functions can be expressed in terms of Bessel functions. Specifically,

$$\begin{aligned} \textstyle E_k(x,y)= \mathcal {J}_{k-\frac{1}{2}}(xy) +\frac{xy}{2k+1}\, \mathcal {J}_{k+\frac{1}{2}}(xy), \end{aligned}$$

where

$$\begin{aligned} \textstyle \mathcal {J}_\nu (z)\,=\;\Gamma (\nu +1)\; {\displaystyle \sum \limits _{\,n=0}^{+\infty }}\; \frac{(-1)^n}{n!\,\Gamma (\nu +n+1)}\; \left( \frac{z}{2}\right) ^{2n} \end{aligned}$$

are normalized Bessel functions.

For $1\le p\le \infty$, let $L^p(\mathbb {R}^d,w_k(x)dx)$, the space of Lebesgue measurable function $f:\mathbb {R}^d\rightarrow \mathbb {C}$ such that

$$\begin{aligned} \Vert f\Vert _{p,k}=\int _{\mathbb {R}^d}|f(x)|^pw_k(x)dx<\infty , \quad \text {if}\quad 1\le p<\infty , \end{aligned}$$

and

$$\begin{aligned} \Vert f\Vert _{\infty ,k}= ess \sup _{x\in \mathbb {R}^d} |f(x)|<\infty . \end{aligned}$$

The Dunkl transform is defined on $L^1(\mathbb {R}^d,w_k(x)dx)$ by

$$\begin{aligned} \mathcal {F}_kf(\xi )= \frac{1}{c_k}\; \int _{\mathbb {R}^d}f(x)\,E_k(x,-i\,\xi )\,w_k(x)\,dx, \end{aligned}$$

where

$$\begin{aligned} c_k\,=\int _{\mathbb {R}^d}e^{-\frac{|x|^2}{2}}\,w(x)\,dx. \end{aligned}$$

We list some known properties of this transform:

• (i) The Dunkl transform is a topological automorphism of $\mathcal {S}(\mathbb {R}^d)$, the Schwartz space of rapidly decreasing functions on $\mathbb {R}^d$.

• (ii) (Plancherel Theorem) The Dunkl transform extends to an isometric automorphism of $L^2(\mathbb {R}^d,w_k(x)dx)$.

• (iii) (Parseval’s formula). For all $f, g \in L^2(\mathbb {R}^d,w_k(x)dx)$ we have

  $$\begin{aligned} \int _{\mathbb {R}^d}f(x)\overline{g(x)}w_k(x)dx=\int _{\mathbb {R}^d}\mathcal {F}_k(f)(x)\overline{\mathcal {F}_k(g) (x)}w_k(x)dx \end{aligned}$$

• (vi) (Inversion formula) For every $f\in \mathcal {S}(\mathbb {R}^d)$, and more generally for every $f\in L^1(\mathbb {R}^d,w_k(x)dx)$ such that $\mathcal {F}_kf\in L^1(\mathbb {R}^d,w_k(\xi )d\xi )$, we have

  $$\begin{aligned} f(x)=\mathcal {F}_k^2f(-x)\qquad \forall \;x\in \mathbb {R}^d. \end{aligned}$$

• (iv) if f is a radial function in $L^1(\mathbb {R}^d,w_k(\xi )d\xi )$ such that $f(x)=\widetilde{f}(|x|)$, then $\mathcal {F}_k(f)$ is also radial and one has

  $$\begin{aligned} \mathcal {F}_k(f)(x)=b_k\int _0^\infty \widetilde{f}(s) \mathcal {J}_{\gamma _k+d/2-1}(s|x|)s^{2\gamma _k+d}ds. \end{aligned}$$

  (2.2)

  where $b_k= 2^{-(\gamma _k+d/2-1)}/\Gamma (\gamma _k+d/2)$ and $\gamma _k=\sum _{\alpha \in R^+} k(\alpha )$.

Let $x\in \mathbb {R}^d$, the Dunkl translation operator $\tau _x$ is given for $f\in L^2_k(\mathbb {R}^d,w_k(x)dx)$ by

$$\begin{aligned} \mathcal {F}_k(\tau _x(f))(y)= \mathcal {F}_kf(y)\,E_k(x,iy), \quad y\in \mathbb {R}^d. \end{aligned}$$

Trimèche in [23] prove that the operator $\tau _x$ is related to the usual translation by

$$\begin{aligned} \tau _x(f)(y)= (V_k)_x(V_k)_y((V_k)^{-1}(f)(x + y)). \end{aligned}$$

In the case when $f(x)=\widetilde{f}(|x|)$ is a radial function in $\mathcal {S}(\mathbb {R}^d)$, the Dunkl translation is represented by the following integral

$$\begin{aligned} \tau _x(f)(y)= \int _{\mathbb {R}^{n}}\widetilde{f}\left( \sqrt{|y|^2+|x|^2+2<y,\eta >}\right) \;d\nu _x(\eta ). \end{aligned}$$

(2.3)

This formula shows that the Dunkl translation operators can be extended to all radial functions f in $L^p (\mathbb {R}^d,w_k(x)dx)$, $1\le p\le \infty$ and the following holds

$$\begin{aligned} \Vert \tau _x(f)\Vert _{p,k}\le \Vert f\Vert _{p,k }. \end{aligned}$$

(2.4)

We define the Dunkl convolution product for suitable functions f and g by

$$\begin{aligned} f*_kg(x)=\int _{\mathbb {R}^d} \tau _x(f)(-y)g(y)d\mu _k(y),\quad x\in \mathbb {R}^d \end{aligned}$$

We note that it is commutative and satisfies the following property:

$$\begin{aligned} \mathcal {F}_k(f*_kg)=\mathcal {F}_k(f)\mathcal {F}_k(g), \quad \quad f,\;g\in L^2(\mathbb {R}^d,w_k(x)dx). \end{aligned}$$

(2.5)

Moreover, the operator $f \rightarrow f*_kg$ is bounded on $L^p (\mathbb {R}^d,w_k(x)dx)$ provide g is a bounded radial function in $L^1(\mathbb {R}^d,w_k(x)dx)$. In particular we have the the following Young’s inequality:

$$\begin{aligned} \Vert f*_kg\Vert _{p,k}\le \Vert g\Vert _{1,k}\Vert f\Vert _{p,k}. \end{aligned}$$

(2.6)

3 The Dunkl Laplacian Operator

In this section and in what follows, we often use the language of spectral theory for unbounded operators. Our main reference is [16].

Let $(e_1,e_2,\ldots ,e_d)$ be an orthonormal basis of $\mathbb {R}^d$. The Dunkl Laplacian operator is defined by

$$\begin{aligned} \Delta _k=\sum _{j=1}^dT_{j}^2, \end{aligned}$$

where $T_j=T_{e_j}$. We consider $-\Delta _k$ as a densely defined operator on the Hilbert space $L^2(\mathbb {R}^d,w_k(x)dx)$ with domain $D(-\Delta _k)=\mathcal {S}(\mathbb {R}^d)$. By means of the Dunkl transform one can prove that, for $f,g\in \mathcal {S}(\mathbb {R}^d)$,

$$\begin{aligned} -\Delta _k(f)(x)= & {} \mathcal {F}_k^{-1}\left( |x|^2\mathcal {F}_k(f)\right) ,\\ \langle -\Delta _k f, g \rangle= & {} \langle f,- \Delta _k g \rangle ,\\ \langle - \Delta _k f, f \rangle= & {} \sum _{j=1}^d\Vert T_j(f)\Vert _{2,k}^2, \end{aligned}$$

which show that $-\Delta _k$ is a densely defined, symmetric and positive operator on $L^2(\mathbb {R}^d,w_k(x)dx)$. Thus, the Friedrichs extension theorem tells us that there is a positive self-adjoint extension of $-\Delta _k$. Define the linear operator $A_k$ as extension of $-\Delta _k$ by

$$\begin{aligned} D(A_k)= & {} H_k^2(\mathbb {R}^d)=\left\{ f\in L^2\left( \mathbb {R}^d,w_k(x)dx\right) ;\quad |x|^2\mathcal {F}_k(f)\in L^2 \left( \mathbb {R}^d,w_k(x)dx\right) \right\} \\ A_k(f)= & {} \mathcal {F}_k^{-1}\left( |x|^2\mathcal {F}_k(f)\right) ; \quad f \in D(A_k). \end{aligned}$$

Clearly $A_k$ is symmetric and positive.

Theorem 3.1

The operator $A_k$ is self-adjoint and that is the unique positive self-adjoint extension operator of $-\Delta _k$.

Proof

Recall first that the adjoint operator $A_k^*$ is defined on the domain $D(A_k^*)$ consisting of the function $f\in L^2(\mathbb {R}^d,w_k(x)dx)$ for which the functional $g\mapsto \langle A_kg , f\rangle$ is bounded on $D(A_k)$ and by Riesz representation theorem there exists a unique $f^*\in L^2(\mathbb {R}^d,w_k(x)dx)$ such that $\langle A_k(g),f\rangle =\langle g,f^* \rangle$, since $D(A_k)$ is dense in $L^2(\mathbb {R}^d,w_k(x)dx)$. We define $A_k^*$ on $D(A_k^*)$ by $A_k^*(f)=f^*$. Since $A_k$ is symmetric, then $A_k$ is self-adjoint if and only if $D(A_k^*)=D(A_k)$. Noting that $D(A_k)\subset D(A_k^*)$ is obvious. Let $f\in D(A_k^*)$, then there exists a constant $C>0$ such that for all $g \in D(A_k)$ we have

$$\begin{aligned} \langle A_kg,f \rangle \le C\Vert g\Vert _{2,k}. \end{aligned}$$

(3.1)

For $r>0$ define the function $g_r \in L^2(\mathbb {R}^d,w_k(x)dx)$ by

$$\begin{aligned} \mathcal {F}_k(g_r)(x)=|x|^2\mathcal {F}_k(f)(x)\chi _{\{|x|<r\}} \end{aligned}$$

where $\chi$ denotes the characteristic function. Clearly we have $g_r\in D(A_k)$ and in view of (3.1)

$$\begin{aligned} |\langle f, A_kg_r\rangle |= & {} |\langle \mathcal {F}_k(f),\mathcal {F}_k(A_kg_r)\rangle |=\int _{|x|<r}|x|^4|\mathcal {F}_k(f)(x)|^2w_k(x)dx\\\le & {} C \left( \int _{|x|<r}|x|^4|\mathcal {F}_k(f)(x)|^2w_k(x)dx\right) ^{1/2}. \end{aligned}$$

It yields that

$$\begin{aligned} \int _{|x|<r}|x|^4|\mathcal {F}_k(f)(x)|^2w_k(x)dx\le C^2. \end{aligned}$$

Therefore by letting $r\rightarrow \infty$ we deduce that $f\in D(A_k)$ and conclude that $D(A_k^*) \subset D(A_k)$.

Let us now prove that $-\Delta _k$ is essentially self-adjoint, this means that $-\Delta _k$ admits an unique self-adjoint extension and that is equal to $A_k$, since we have proved that $A_k$ is self-adjoint. From the general theory of unbounded operators, see for example the chapter VIII of [16], it suffices to prove that $(-\Delta _k \pm i)D(-\Delta _k)$ is dense, which is equivalent to

$$\begin{aligned} \Big ((-\Delta _k \pm i)D(-\Delta _k)\Big )^\perp =\{0\}. \end{aligned}$$

In fact, let $g\in \Big ((-\Delta _k \pm i)D(-\Delta _k)\Big )^\perp$. Then for any $f\in D(-\Delta _k)=\mathcal {S}(\mathbb {R}^d)$

$$\begin{aligned} 0=\langle (-\Delta _k\pm i)f,g\rangle = \langle (|.|^2\pm i)\mathcal {F}_k(f),\mathcal {F}_k(g)\rangle = \langle (\mathcal {F}_k(f),(|.|^2\pm i)\mathcal {F}_k(g)\rangle \end{aligned}$$

Since $\mathcal {F}_k(\mathcal {S}(\mathbb {R}^d))=\mathcal {S}(\mathbb {R}^d)$, this implies by density argument that $\mathcal {F}_k(g)=0$ and so $g=0$, as desired. $\square$

The operator $A_k$ is generator of strongly continuous one parameter semi group $(e^{-tA_k})_{t\ge 0}$ where the operator $e^{-tA_k}$ is given by

$$\begin{aligned} e^{-tA_k}f= \mathcal {F}_k^{-1} \left( e^{-t|\cdot |^2} \mathcal {F}_k(f)\right) \end{aligned}$$

for all $t>0$ and $f\in L^2(\mathbb {R}^d,w_k(x)dx).$ It follows that $e^{-tA_k}$ is an integral operator given by

$$\begin{aligned} e^{-tA_k}f(x)==k_t*_kf =\int _{\mathbb {R}^d}K_t(x,y)f(y)w_k(y)dy \end{aligned}$$

(3.2)

where

$$\begin{aligned} k_t(x)= \mathcal {F}_k^{-1}(e^{-t|.|^2})(x) = t^{-\gamma _k-d/2}e^{-|x|^2/t} \end{aligned}$$

(3.3)

and from (2.1) and (2.3)

$$\begin{aligned} K_t(x,y)=\tau _x(k_t)(-y)= t^{-\gamma _k-d/2}e^{- (|x|^2+|y|^2)/t}E(2x/t,y). \end{aligned}$$

(3.4)

Corollary 3.2

$e^{-tA_k}$ can be extended to a bounded operator from $L^p(\mathbb {R}^d,w_k(x)dx)$ into $L^\infty (\mathbb {R}^d,w_k(x)dx)$, for $1\le p < \infty$.

Proof

In view of (3.4) and (2.4), we obtain by using Hölder’s inequality

$$\begin{aligned} |k_t*_kf(x)|\le \Vert K_t(x,.)\Vert _{p',k}\Vert f\Vert _{p,k}\le \Vert k_t \Vert _{p',k}\Vert f\Vert _{p,k}, \end{aligned}$$

for $f\in S(\mathbb {R}^d)$ and $1/p+1/p'=1$. Thus one conclude the corollary by density argument. $\square$

4 Dunkl Schrödinger Operator

In this section, we use arguments analogous to those used in [20].

Let V be a nonnegative measurable function on $\mathbb {R}^d$ that is finite almost everywhere. In the Hilbert space $L^2(\mathbb {R}^d,w_k(x)dx)$ we consider the operator

$$\begin{aligned} \mathcal {L}_k=A_k+V \end{aligned}$$

with domain $D(\mathcal {L}_k)=H_k^2(\mathbb {R}^d)\cap D(V)$ where

$$\begin{aligned} D(V)=\left\{ f\in L^2\left( \mathbb {R}^d,w_k(x)dx\right) ;\quad Vf\in L^2\left( \mathbb {R}^d,w_k(x)dx\right) \right\} . \end{aligned}$$

We call this operator the Dunkl Schrödinger operator. We have already proven that $A_k$ is positive self-adjoint operator, we should add here that the multiplication operator $f\rightarrow Vf$ is a positive self-adjoint, see [16, VIII.3, Proposition 1]. The important fact that we shall use comes from the theory of the quadratic form. Define the form $q_k$ by

$$\begin{aligned} D(q_k)= & {} \left\{ f\in L^2\left( \mathbb {R}^d,w_k(x)dx\right) ; \left( \sum _{j=1}^n|T_jf|^2\right) ^{1/2}, V^{1/2}f \in L^2\left( \mathbb {R}^d,w_k(x)dx\right) \right\} \\ q_k(f)= & {} \sum _{j=1}^n\Vert T_jf\Vert _{2,k}^2+\Vert V^{1/2}f\Vert ^2_{2,k} \end{aligned}$$

Clearly $C_0^\infty (\mathbb {R}^d)\subset D(q_k)$, so $q_k$ is densely defined.

Lemma 4.1

The quadratic form $q_k$ is closed.

Proof

We shall prove that if a sequence $(\varphi _n)_n\in D(q_k)$ and $\varphi \in L^2(\mathbb {R}^d,w_k(x)dx)$ such that $\Vert \varphi _n-\varphi \Vert _{2,k}\rightarrow 0$ and $q_k(\varphi _n-\varphi _m)\rightarrow 0$ then $\varphi \in D(q_k)$ and $q_k(\varphi _n-\varphi )\rightarrow 0$.

In fact, since $q_k(\varphi _n-\varphi _m)\rightarrow 0$ then $(V^{1/2}\varphi _n)_n$ and $(T_j\varphi _n)_n$ are Cauchy sequences in $L^2(\mathbb {R}^d,w(x)dx)$ and so are convergent. Let $g_j=\lim T_j\varphi _n$ and $h=\lim V^{1/2}\varphi _n$. For any function $\psi \in C_0^\infty$,

$$\begin{aligned} \langle g_j,\psi \rangle =\lim \langle T_j \varphi _n, \psi \rangle =-\lim \langle \varphi _n, T_j\psi \rangle =-\langle \varphi ,T_j \psi \rangle . \end{aligned}$$

This yields that $T_j\varphi =g_j \in L^2(\mathbb {R}^d,w(x)dx)$. Similarly,

$$\begin{aligned} \langle h,\psi \rangle =\lim \langle V^{1/2}\varphi _n, \psi \rangle =\lim \langle \varphi _n, V^{1/2}\psi \rangle =\langle \varphi ,V^{1/2}\psi \rangle =\langle V^{1/2}\varphi ,\psi \rangle \end{aligned}$$

and so $V^{1/2}\varphi =h\in L^2(\mathbb {R}^d,w(x)dx)$. Therefore $\varphi \in D(q_k)$ and $q_k(\varphi _n-\varphi )\rightarrow 0$

$\square$

Let $B_{q_k}$ be the associated sesquilinear form on $D(q_k)$. It follows that $D(q_k)$ is a Hilbert space with the inner product:

$$\begin{aligned} \langle \varphi ,\psi \rangle _{q_k}=\langle \varphi ,\psi \rangle + B_{q_k}(\varphi ,\psi ),\quad \varphi ,\;\psi \;\in D(q_k). \end{aligned}$$

The corresponding norm is given by

$$\begin{aligned} \Vert \varphi \Vert _{q_k}= \sqrt{\Vert \varphi \Vert _{2,\;k}^2+ q_k(\varphi ) }. \end{aligned}$$

The important consequence of the above lemma is that there exist a unique positive self adjoint operator $L_k$ such that.

$$\begin{aligned} q_k(\varphi )=\langle L_k(\varphi ),\varphi \rangle ,\qquad \varphi \in D(q_k), \end{aligned}$$

and is defined as follows

$$\begin{aligned} D(L_k)= & {} \{\varphi \in D(q_k)/ \exists \; \widetilde{\varphi }\in L^2(\mathbb {R}^d,w(x)dx),\; B_{q_k}( \varphi ,\psi )= \langle \tilde{\varphi }, \psi \rangle \;\forall \;\psi \in D(q_k)\}\nonumber \\ L_k(\varphi )= & {} \widetilde{\varphi },\qquad \forall \;\varphi \in D(L_k). \end{aligned}$$

(4.1)

Moreover,

$$\begin{aligned} D(q_k)=D\left( L_k^{1/2}\right) \quad \text {and}\quad q_k(\varphi )=\Vert L_k^{1/2}(\varphi )\Vert _{2,k}. \end{aligned}$$

Theorem 4.2

Assume that $V\in L^1_{loc}(\mathbb {R}^d,w_k(x)dx)$ and $V\ge 0$, then $C_0^\infty (\mathbb {R}^d)$ is dense in $D(q_k)$ in the norm $\Vert \varphi \Vert _{q_k}= \sqrt{\Vert \varphi \Vert _{2,\;k}^2+ q_k(\varphi ) }.$

For the proof we will need the following lemmas.

Lemma 4.3

The range of $e^{-L_k}$ is dense in the Hilbert space $( D(q_k), \Vert .\Vert _{q_k})$.

Proof

Let $\varphi \in D(q_k)$ such that

$$\begin{aligned} \langle e^{-L_k}\psi , \varphi \rangle +B_{q_k}( e^{-L_k}\psi , \varphi )=\langle e^{-L_k}\psi , \varphi \rangle + \langle L_k e^{-L_k}\psi , \varphi \rangle =0 ,\quad \forall \psi \in L^2(\mathbb {R}^d,w_k(x)dx). \end{aligned}$$

This implies that

$$\begin{aligned} \langle (L_k+1) e^{-L_k}\varphi ,\psi \rangle =0,\quad \forall \psi \in L^2(\mathbb {R}^d,w_k(x)dx). \end{aligned}$$

Since $L_k+1$ is invertible, then $e^{-L_k}\varphi =0$ which implies that $\varphi =0$. The density of $Ran(e^{-L_k})$ follows. $\square$

Lemma 4.4

$L^\infty (\mathbb {R}^d,w_k(x)dx)\cap D(q_k)$ is dense in the Hilbert space $( D(q_k), \Vert .\Vert _{q_k})$.

Proof

Using the Kato’s stong Trotter product formula, see Theorem S.21 of [16], we have in the strong convergence

$$\begin{aligned} \lim \left( e^{-tA_k/n} e^{-tV/n}\right) ^n=e^{-tL_k}, \quad t\ge 0. \end{aligned}$$

(4.2)

By the fact that $|e^{-tV/n}f|\le |f|$ and $|e^{-tA_k}(f)|\le e^{-tA_k}(|f|)$, which can be seen from (3.2), it follows that

$$\begin{aligned} |e^{-tL_k}(f)|\le e^{-tA_k}(|f|). \end{aligned}$$

(4.3)

In particular we have

$$\begin{aligned} \Vert e^{-L_k}(f)\Vert _{\infty }\le \Vert e^{-A_k}(|f|)\Vert _{\infty }\le c \Vert f\Vert _{2,k}, \end{aligned}$$

(4.4)

where the second inequality follows from (3.2), by using Cauchy-Schwarz Inequality and (2.4). From (4.4) we have that

$$\begin{aligned} Ran(e^{-L_k})\subset L^\infty \left( \mathbb {R}^d,w_k(x)dx\right) . \end{aligned}$$

(4.5)

Since the function $\lambda \rightarrow \lambda ^{1/2}e^{-\lambda }$ is bounded on $(0,\infty )$ then by spectral theorem the operator $L_k^{1/2}e^{-L_k}$ is bounded on $L^2(\mathbb {R}^d, w_k(x)dx)$. We deduce that

$$\begin{aligned} Ran\left( e^{-L_k}\right) \subset D\left( L_k^{1/2}\right) =D(q_k). \end{aligned}$$

(4.6)

Similarly we have

$$\begin{aligned} Ran(e^{-L_k})\subset D(L_k) \end{aligned}$$

and in view of (4.5) and (4.6) we have that

$$\begin{aligned} Ran(e^{-L_k})\subset L^\infty (\mathbb {R}^d,w_k(x)dx) \cap D(q_k). \end{aligned}$$

We then conclude Lemma 4.4 from the result of Lemma 4.3. $\square$

Proof of Theorem 4.2

Let

$$\begin{aligned} S=\{\varphi \in L^\infty (\mathbb {R}^d,w_k(x)dx) \cap D(q_k);\; supp(\varphi ) \;is\; compact\}. \end{aligned}$$

We claim that S is dense in $( D(q_k), \Vert .\Vert _{q_k})$. Observe that for $\varphi \in L^\infty (\mathbb {R}^d,w_k(x)dx) \cap D(q_k)$ and $\psi \in C_0^\infty (\mathbb {R}^d)$ be a radial function, we have that $V^{1/2}\varphi \psi \in L^2(\mathbb {R}^d,w_k(x)dx)$ and in the distributional sense

$$\begin{aligned} T_j(\varphi \psi ) =T_j(\varphi ) \psi +\varphi T_j(\psi ) ,\quad j=1,2\ldots d \end{aligned}$$

(4.7)

which gives that $T_j(\varphi \psi )\in L^2(\mathbb {R}^d,w_k(x)dx)$. Hence $\varphi \psi \in S$. From this fact if we choose a radial function $\psi \in C_0^\infty (\mathbb {R}^d)$ with $\psi (x)=1$ near 0 and we set $\varphi _n=\psi (./n)\varphi$ then by Lebesgue dominated convergence theorem and (4.7) the following hold

• $\Vert \varphi _n- \varphi \Vert _{2,k}\rightarrow 0$,

• $\Vert V^{1/2}\varphi _n- V^{1/2}\varphi \Vert _{2,k}\rightarrow 0$,

• $\Vert T_j\varphi _n- T_j\varphi \Vert _{2,k} \rightarrow 0$.

This implies that $\varphi _n\rightarrow \varphi$ in norm $\Vert .\Vert _{q_k}$.

We now claim that $C_0^\infty (\mathbb {R}^d)$ is dense in $( D(q_k), \Vert .\Vert _{q_k})$. Take a radial function $\rho \in C_0^\infty (\mathbb {R}^d)$ with

$$\begin{aligned} \int _{\mathbb {R}^d}\rho (x)w_k(x)dx=1. \end{aligned}$$

For $\varphi \in S$ we define $(\varphi _n)_n$ by $\varphi _n= \rho _n*_k\varphi$ where $\rho _n=n^{-2\gamma _k-d}\rho (x/n)$. Let us observe that $\varphi _n\in C_0^\infty (\mathbb {R}^d)$ and

$$\begin{aligned} T_j\varphi _n=\rho _n*_kT_j\varphi ,\quad j=1,2\ldots d, \end{aligned}$$

which can be seen as following: for $\psi \in C_0^\infty (\mathbb {R}^d)$,

$$\begin{aligned} \int _{\mathbb {R}^d}\varphi _n(x) T_j\psi (x)w_k(x)dx= & {} \int _{\mathbb {R}^d} \int _{\mathbb {R}^d}\varphi (y)\tau _x(\rho _n)(-y)T_j(\psi )(x)w_k(y)w_k(x)dydx\\= & {} \int _{\mathbb {R}^d} \left( \int _{\mathbb {R}^d}\tau _{-y}(\rho _n)(x)T_j(\psi )(x)w_k(x)dx\right) \varphi (y)w_k(y)dy\\= & {} - \int _{\mathbb {R}^d} \left( \int _{\mathbb {R}^d}\tau _{-y}(T_j(\rho _n))(x) \psi (x)w_k(x)dx\right) \varphi (y)w_k(y)dy\\= & {} -\int _{\mathbb {R}^d} \left( \int _{\mathbb {R}^d}\tau _{x}(T_j (\rho _n))(-y)\varphi (y)w_k(y)dy\right) \psi (x)w_k(x)dx\\= & {} -\int _{\mathbb {R}^d} \left( \int _{\mathbb {R}^d}T_j\tau _{x} (\rho _n)(-y)\varphi (y)w_k(y)dy\right) \psi (x)w_k(x)dx\\= & {} \int _{\mathbb {R}^d} \left( \int _{\mathbb {R}^d}T_j \Big (\tau _{x}(\rho _n)(-.)\Big )(y)\varphi (y)w_k(y)dy\right) \psi (x)w_k(x)dx\\= & {} -\int _{\mathbb {R}^d} \left( \int _{\mathbb {R}^d}\tau _{x} (\rho _n)(-y)T_j(\varphi )(y)w_k(y)dy\right) \psi (x)w_k(x)dx\\= & {} - \int _{\mathbb {R}^d}\rho _n*_kT_j\varphi (x)\psi (x)w_k(x)dx. \end{aligned}$$

Therefore as convergent in $L^2(\mathbb {R}^d,w_k(x)dx)$ we obtain

• $\varphi _n\rightarrow \varphi$,

• $V^{1/2}\varphi _n\rightarrow V^{1/2}\varphi$,

• $T_j\varphi _n\rightarrow T_j\varphi$

and thus $\varphi _n\rightarrow \varphi$ in the norm $\Vert .\Vert _{q_k}$. This conclude the proof of the density of $C_0^\infty (\mathbb {R}^d)$.

$\square$

Next we define in the distributional way $\mathcal {L}_{k,dist}= A_k+V$, that is for $\varphi \in L^2(\mathbb {R}^d,w_k(x)dx)$,

$$\begin{aligned} \int _{\mathbb {R}^d} \mathcal {L}_{k,dist}\varphi (x)\psi (x)w_k(x)dx=\int _{\mathbb {R}^d} \varphi (x)(A_k +V)\psi (x)w_k(x)dx, \quad \forall \;\psi \in C_0^\infty (\mathbb {R}^d) . \end{aligned}$$

Clearly $\mathcal {L}_{k,dist}= \mathcal {L}_{k}$ on $C_0^\infty (\mathbb {R}^d)$.

Corollary 4.5

We have that

$$\begin{aligned} D(L_k)=\{ \varphi \in D(q_k);\; \mathcal {L}_{k,dist}(\varphi )\in L^2(\mathbb {R}^d,w_k(x)dx)\}. \end{aligned}$$

Proof

For $\varphi \in D(q_k)$ and $\psi \in C_0^\infty (\mathbb {R}^d)$ we have

$$\begin{aligned} B_{q_k}(\varphi ,\psi )= & {} \sum _{j=1}^d \langle T_j(\varphi ),T_j(\psi )\rangle + \langle V\varphi , \psi \rangle \\= & {} -\sum _{j=1}^d \langle \varphi ,T_j^2(\psi )\rangle + \langle \varphi , V \psi \rangle \\= & {} \langle \varphi , (A_k +V)\psi \rangle \\= & {} \langle \mathcal {L}_{k,dist} \varphi , \psi \rangle . \end{aligned}$$

We conclude the corollary by the definition (4.1) of the domain $D(L_k)$ and the density of $C_0^\infty (\mathbb {R}^d)$. We add here that when $\mathcal {L}_{k,dist}(\varphi )\in L^2(\mathbb {R}^d,w_k(x)dx)$,

$$\begin{aligned} \mathcal {L}_{k}(\varphi )=\mathcal {L}_{k,dist}(\varphi ). \end{aligned}$$

(4.8)

This fact will be used in the proof of the next theorem. $\square$

Theorem 4.6

Assume that $V\in L^2_{loc}(\mathbb {R}^d,w_k(x)dx)$ and $V\ge 0$, then $\mathcal {L}_k$ is essentially self-adjoint on $C^\infty _0(\mathbb {R}^d)$ and its closure is $L_k$.

The proof of this theorem is the same as the proof of Theorem 4.2.

Now recall that $D\subset D(L_k)$ is a core of $L_k$ if for all $\varphi \in D(L_k)$ there exist in D a sequence $(\varphi _n)_n$ such that $\Vert \varphi _n-\varphi \Vert _{2,k}\rightarrow 0$ and $\Vert L_k(\varphi _n)-L_k(\varphi )\Vert _{2,k}\rightarrow 0$.

Lemma 4.7

$D(L_k)\cap L^\infty (\mathbb {R}^d,w_k(x)dx)$ is a core of $L_k$.

Proof

Let $\varphi \in D(L_k)$, by the spectral theorem

$$\begin{aligned} \Vert e^{-tL_k}(\varphi ) -\varphi \Vert _{2,k}^2=\int _0^\infty |e^{-t\lambda }-1|^2\;d(<P_\lambda \varphi ,\varphi >), \end{aligned}$$

where $P_\lambda$ is the projection-valued measure with respect to $L_k$. So using the dominated convergence theorem we obtain that $\Vert e^{-tL_k}(\varphi )-\varphi \Vert _{2,k}\rightarrow 0$ when $t\rightarrow 0$. Similarly,

$$\begin{aligned} \Vert L_ke^{-tL_k}(\varphi ) -L_k(\varphi )\Vert _{2,k}= \Vert e^{-tL_k }L_k(\varphi )-L_k(\varphi )\Vert _{2,k}\rightarrow 0 \end{aligned}$$

when $t\rightarrow 0$. Now according to the proof of Lemma 4.4 one can see that

$$\begin{aligned} Ran(e^{-tL_k})\subset D(L_k)\cap L^\infty (\mathbb {R}^d,w_k(x)dx). \end{aligned}$$

We thus conclude the proof by taking the sequence $\varphi _n=e^{-\frac{1}{n}\;L_k}(\varphi )$. $\square$

Proof of Theorm 4.6

We first show for $\varphi \in D(L_k)$ and $\psi \in C^\infty _0(\mathbb {R}^d)$ be a radial function we have that $\varphi \psi \in D(L_k)$. Indeed, since $D(L_k)\subset D(q_k)$ then we already have $\varphi \psi \in D(q_k)$ and by a direct calculation

$$\begin{aligned} \mathcal {L}_{k,dist}(\varphi \psi )= & {} \mathcal {L}_{k,dist}(\varphi )\psi -2\sum _{j=1}^dT_j\varphi T_j\psi -\varphi \Delta _k\psi \nonumber \\&+ \sum _{j=1}^d\sum _{\alpha \in R_+}k(\alpha ) \alpha _j \frac{\Big (\varphi (x)-\varphi (\sigma _\alpha \cdot x)\Big ) \Big (T_j(\psi )(x)-T_j(\psi )(\sigma _\alpha \cdot x)\Big )}{\langle x,\alpha \rangle }.\nonumber \\ \end{aligned}$$

(4.9)

This is proven by showing that both sides have the same inner product with a function in $C_0^\infty$ and using the density of $C_0^\infty$ in form norm. Since the function

$$\begin{aligned} x\rightarrow \frac{ T_j(\psi )(x)-T_j(\psi )(\sigma _\alpha .x) }{\langle x,\alpha \rangle } \end{aligned}$$

is in $C^\infty _0(\mathbb {R}^d)$, it follows that $\mathcal {L}_{k,dist}(\varphi \psi )\in L^2(\mathbb {R}^d,w_k(x)dx)$ and from Corollary 4.5 we have $\varphi \psi \in D(L_k)$.

Let $\psi \in C_0^\infty (\mathbb {R}^d)$ be a radial function with $\psi (x)=1$ near 0 and set $\varphi _n=\psi (./n)\varphi$. In view of (4.8) and (4.9) we get that $\Vert H(\varphi _n)-H(\varphi )\Vert _{2,k}\rightarrow 0$. This yields that

$$\begin{aligned} S'=\{\varphi \in D(L_k)\cap L^\infty /\; supp(\varphi ) \;is\; compact \} \end{aligned}$$

is a core for $L_k$.

Now we proceed as follows. Let $\varphi \in S'$ then $A_k\varphi +V\varphi \in L^2(\mathbb {R}^d,w_k(x)dx)$ and $V\varphi \in L^2(\mathbb {R}^d,w_k(x)dx)$, since $V\in L^2_{loc}(\mathbb {R}^d,w_k(x)dx)$ and $\varphi \in L^\infty$. It follows that $A_k\varphi \in L^2(\mathbb {R}^d,w_k(x)dx)$. However, if $\varphi _n= \rho _n*_k\varphi \in C^\infty _0(\mathbb {R}^d)$, where $(\rho _n)_n$ is defined in the proof of Theorem 3.1 and as $\varphi \in L^\infty$ and $supp(\varphi )$ is compact then $\Vert V\varphi _n - V\varphi \Vert _{2,k}\rightarrow 0$. But by means of Dunkl transform we see that the

$$\begin{aligned} A_k(\varphi _n)=\rho _n*_kA_k(\varphi ) \end{aligned}$$

and thus $\Vert A_k(\varphi _n) -A_k(\varphi )\Vert _{2,k}\rightarrow 0$. This yields that

$$\begin{aligned} \Vert \varphi _n - \varphi \Vert _{2,k}\rightarrow 0,\quad \text {and}\quad \Vert \mathcal {L}_k(\rho _n)-A_k\varphi -V\varphi \Vert _{2,k}\rightarrow 0 \end{aligned}$$

and conclude that $L_k$ is the closure of $\mathcal {L}_k$ on $C^\infty _0(\mathbb {R}^d)$. $\square$

We closed this section by showing that the semi-group corresponding to the Dunkl Schrödinger operator $L_k$ has an integral kernel.

Theorem 4.8

$W_t= e^{-tL_k}$, $t>0$ is an integral operator given by

$$\begin{aligned} W_t(f)(x)= \int _{\mathbb {R}^d}\mathcal {W}_t(x,y)f(y)w_k(y)dy \end{aligned}$$

with

$$\begin{aligned} 0\le \mathcal {W}_t(x,y) \le K_t(x,y). \end{aligned}$$

(4.10)

Proof

Recall that from (4.3)

$$\begin{aligned} |W_t(f)|=|e^{-tL_k}(f)|\le e^{-tA_k}(|f|),\quad f\in C_0^\infty (\mathbb {R}^d). \end{aligned}$$

Thus by Corollary 3.2, $W_t$ is bounded from $L^p(\mathbb {R}^d,w_k(x)dx)$ to $L^\infty (\mathbb {R}^d,w_k(x)dx)$ for all $1\le p\le \infty$. The theorem of Dunford and Pettis (see for example Theorem 4.2 of [19]) asserts that such operator is a kernel operator. Since $e^{tA_k}$ is an integral operator with positive kernel, it is positivity preserving. Thus using the Trotter product formula (4.2) we have that $W_t$ is positivity preserving which implies that $\mathcal {W}_t(x,y)\ge 0$. The second inequality of (4.10) follows from (4.3) and from Theorem 2.2 of [19]. $\square$

5 Dunkl Harmonic Oscillator

We first recall some known facts about the Dunkl harmonic oscillator. The reader is referred to [2, 3, 15, 17].

The Dunkl harmonic oscillator is the Dunkl Schrödinger operator $H_k = - \Delta _k +|x|^2$. It can be expressed in terms of generalized Hermite functions $h_n^k$,

$$\begin{aligned} H_k(f)=\sum _{n\in \mathbb {N}^d}\left( 2|n|+\gamma _k+d\right) \langle f, h_n^k\rangle \; h_n^k \end{aligned}$$

where $|n|=n_1+\cdots +n_d$. The functions $h_n^k$ are eigenfunctions of $H_k$ with

$$\begin{aligned} H_k\left( h_n^k\right) =\left( 2|n|+\gamma _k+d\right) \; h_n^k \end{aligned}$$

and form an orthonormal basis of $L^2(\mathbb {R}^d,w_k(x)dx.$

The holomorphic Hermite semi-group $e^{-zH_k}$, $Re(z) >0$ is given by

$$\begin{aligned} e^{-zH_k}(f)=\sum _{n\in \mathbb {N}^d}e^{-z(2|n|+\gamma _k+d)}\langle f, h_n^k\rangle \; h_n^k. \end{aligned}$$

(5.1)

It has the following integral representation

$$\begin{aligned} e^{-zH_k}(f)(x)=\int _{\mathbb {R}^d}\mathcal {H}_z(x,y)f(y)w_k(y)dy, \end{aligned}$$

where from the generalized Mehler-formula,

$$\begin{aligned} \mathcal {H}_z(x,y)= & {} \sum _{n\in \mathbb {N}^d}e^{-z(2|n|+\gamma _k+d)} h_n^k(x) h_n^k(y) \\= & {} c_k\left( \frac{\sinh (2z)}{2}\right) ^{-\gamma _k-\frac{d}{2} } E_k\left( \frac{x}{\sinh (2z)},y\right) \;e^{-\frac{\coth (2z)}{2}(|x|^2+|y|^2)}. \end{aligned}$$

It can be written as

$$\begin{aligned} \mathcal {H}_z(x,y) =c_k \sinh (2z)^{-\gamma _k } \int _{\mathbb {R}^d} \mathcal {H}_z^0(\eta ,y) \; e^{-\frac{\coth (2z)}{2}(|x|^2-|\eta |^2)}d\nu _x(\eta ). \end{aligned}$$

where $\mathcal {H}_z^0$ is the kernel of the classical Hermite semi-group given by

$$\begin{aligned} \mathcal {H}_z^0(x,y)= & {} (2\pi \sinh (2z))^{-\frac{d}{2}}\;e^{-\frac{\coth (2z)}{2}(|x-y|^2-\tanh (t)\;\langle x,y \rangle }\nonumber \\= & {} (2\pi \sinh (2z))^{-\frac{d}{2}}\;e^{-\frac{1}{4} (\coth (z) (|x-y|^2+\tanh (z) |x-y|^2) }. \end{aligned}$$

(5.2)

Proposition 5.1

For all $z\in \mathbb {C}$, $0\le \arg (z)\le \omega < \pi /2$ there exist $c>0$ such that

$$\begin{aligned} |\mathcal {H}_z(x,y)|\le \mathcal {H}_{Re(z)}\;(cx,cy). \end{aligned}$$

Let us first prove the following lemma

Lemma 5.2

If $0\le \arg (z)\le \omega <\pi /2$ then there exists $c>0$ such that

$$\begin{aligned} c\;\coth (Re(z)) \le Re(\coth (z))\le \coth (Re(z)). \end{aligned}$$

Proof

Note first that for $z=t+iu$

$$\begin{aligned} Re(\coth (z))=\frac{e^{4t}-1}{(e^{2t}-1)^2+2e^{2t}(1-\cos (2u))}\le \frac{e^{2t}+1}{e^{2t}-1}=\coth (t). \end{aligned}$$

Now if $0\le \arg (z)\le \omega < \pi /2$, then

$$\begin{aligned} |u|\le \tan (\omega ) \; t . \end{aligned}$$

Choosing $a= \pi /(4\tan (\omega )$, it follows that for $t\in (0,a]$, we have $2|u|\le \pi /2$ and $\cos (2u) \ge \cos (2\tan (\omega ) \; t)$. Then we get

$$\begin{aligned} Re(\coth (z))\ge \frac{e^{4t}-1}{(e^{2t}-1)^2+2e^{2t}(1- \cos (2\tan (\omega )t)}. \end{aligned}$$

If we take the function

$$\begin{aligned} \varphi (t)= \left( \frac{e^{4t}-1}{(e^{2t}-1)^2+2e^{2t}(1- \cos (2\tan (\omega )t)}\right) \; \left( \frac{e^{2t}-1}{e^{2t}+1}\right) \end{aligned}$$

we see that

$$\begin{aligned} \lim _{t\rightarrow 0}\varphi (t)= 2\cos ^2(\omega )>0 \end{aligned}$$

and $\varphi$ define a positive continuous function on the interval [0, a], thus $\inf _{y\in (0,a]}\varphi (t)=c>0$. Therefore, for $0<t\le a$

$$\begin{aligned} Re(\coth (z))\ge c\coth (t) . \end{aligned}$$

For $t\ge a>0$ there exit $c>0$ so that

$$\begin{aligned} 2e^{2t}(1-\cos (2u))\le 4e^{2t}\le c (e^{2t}-1)^2. \end{aligned}$$

It follows that,

$$\begin{aligned} Re(\coth (z))\ge \frac{e^{4t}-1}{(1+c)(e^{2t}-1)^2}= c'\coth (t) . \end{aligned}$$

The lemma follows. $\square$

Proof of Proposition 5.1

This follows from (5.2), Lemma 5.2 and the fact that

$$\begin{aligned} |\sinh (z)|= \sqrt{\sinh ^2(t)+\sin ^2(u)}\ge \sinh (t), \end{aligned}$$

for $z=t+iu$. Indeed,

$$\begin{aligned} |\mathcal {H}_z^0(x,y)|= & {} | (2\pi \sinh (2z))^{-\frac{d}{2}}\;e^{-\frac{1}{4} (\coth (z) (|x-y|^2+\tanh (z) |x-y|^2) }|\\\le & {} (2\pi \sinh (2t))^{-\frac{d}{2}}\;e^{-\frac{c}{4} (\coth (t) (|x-y|^2+\tanh (t) |x-y|^2)} \end{aligned}$$

and

$$\begin{aligned} |\mathcal {H}_z(x,y)|\le & {} c_k \sinh (2t)^{-\gamma _k-d/2 } \int _{\mathbb {R}^d} e^{-\frac{c}{4} (\coth (t) (|\eta -y|^2+\tanh (t) |\eta -y|^2 -\frac{\coth (2t)}{2}(|x|^2-|\eta |^2)}d\nu _x(\eta )\\= & {} c_k \sinh (2t)^{-\gamma _k-d/2 } \int _{\mathbb {R}^d} e^{-\frac{c}{2} (\coth (2t) (|\eta -y|^2+ |x|^2-|\eta |^2) -2\tanh (t) \langle y,\eta \rangle )}d\nu _x(\eta )\\= & {} c_k \sinh (2t)^{-\gamma _k-d/2 } e^{-\frac{c\coth (2t)}{2} (|x|^2+|y|^2)} \int _{\mathbb {R}^d} e^{ \frac{c}{\sinh (2t)}\langle y,\eta \rangle } d\nu _x(\eta )\\= & {} c_k \sinh (2t)^{-\gamma _k-d/2 } e^{-\frac{c\coth (2t)}{2} \left( |x|^2+|y|^2\right) } E_k\left( \frac{c}{\sinh (2t)}x,y\right) \\= & {} \mathcal {H}_{t}\left( \sqrt{c}\;x,\sqrt{c}\;y\right) , \end{aligned}$$

which is the desired inequality. $\square$

Now, since from (4.10)

$$\begin{aligned} 0\le \mathcal {H}_t(x,y)\le K_t(x,y),\quad t>0, \end{aligned}$$

(5.3)

then we can state

Corollary 5.3

For all $z\in \mathbb {C}$, $0\le \arg (z)\le \omega < \pi /2$ there exists $c>0$ such that

$$\begin{aligned} |\mathcal {H}_z(x,y)|\le K_{Re(z)}\;(cx,cy). \end{aligned}$$

5.1 $H^\infty$-Functional Calculus on $L^p(\mathbb {R}^d, w_k(x)dx)$ for Dunkl Oscillator Operator

We briefly recall the definition of sectorial operators and their holomorphic functional calculus. More on basic properties of sectorial operators can be found in [1, 11, 14].

A closed operator T on complex Hilbert space is said to be sectorial of type $\omega \in [0,\pi [$ if the following hold

• (i) The spectrum $\sigma (T)\subset S_\omega =\{z\in \mathbb {C}^*,\; |Arg(z)|<\omega \}\cup \{0\}$

• (ii) For each $\mu >\omega$ there exists $C_\mu$ such that $\Vert (T-zI)^{-1}\Vert \le C_\mu |z|^{-1}$ for $z\notin S_\mu$.

A non-negative self-adjoint operator in a Hilbert space is an operator of type $S_\omega$ for all $\omega >0$.

Let $H^\infty (S_\mu ^o)$ be the space of bounded holomorphic functions in the open sector $S_\mu ^o$, the interior of $S_\mu$ equipped with the norm $\Vert f\Vert _{\infty }= \sup _{z \in S_\mu ^o}|f(z)|$ and let

$$\begin{aligned} \Psi \left( S_\mu ^o\right) =\left\{ \xi \in H^\infty (S_\mu ^o);\;\exists \; s>0;\; |\xi (z)|\le C |z|^s/(1+|z|)^{2s} \right\} . \end{aligned}$$

For $\xi \in \Psi (S_\mu ^o)$ we define the operator $\xi (T)$

$$\begin{aligned} \xi (T)=\frac{1}{2 \pi i}\int _{\gamma }\xi (z)(T-zI)^{-1}dz \end{aligned}$$

(5.4)

where $\gamma$ is the unbounded contour

$$\begin{aligned} \gamma (t)=\left\{ \begin{array}{ll} te^{i\theta }, &{}\quad t\ge 0\\ -te^{-i\theta }, &{}\quad t\le 0. \end{array}\right. \end{aligned}$$

(5.5)

When T is a one–one operator of type $\omega$ then one can define a holomorphic functional calculus as follows: Let $\psi$ the function defined on $\mathbb {C}{\setminus }\{-1\}$ by $\psi (z) = z/(1+z)^2$. For each $\mu > \omega$ and for each $f\in H^\infty (S_\mu ^o)$ we have that $\psi$, $f\psi \in \Psi (S_\mu ^o)$ and $\psi (T)$ is one–one. So $f\psi (T)$ is a bounded operator and $\psi (T)^{-1}$ is a closed operator. Define f(T) by

$$\begin{aligned} f(T)=\psi (T)^{-1}(f\psi )(T)=(I+T)^2T^{-1}(f\psi )(T). \end{aligned}$$

(5.6)

The definitions given by (5.4) and (5.6) are consistent with the usual definition of polynomials of an operator.

We say that T has bounded $H^\infty$-functional calculus if further for all $f\in H^\infty (S_\mu ^o)$ the operator f(T) is bounded and

$$\begin{aligned} \Vert f(T)\Vert \le c_\mu \Vert f\Vert _\infty , \end{aligned}$$

for some constant $c_\mu$. An interesting result is given by the following

Proposition 5.4

[1] If T is a positive self-adjoint operator then it has a bounded $H^\infty$-functional calculus for all $\mu > 0$ and

$$\begin{aligned} \Vert \xi (T)\Vert \le \Vert \xi \Vert _\infty , \end{aligned}$$

for all $\xi \in H^\infty (S_\mu ^o)$.

Theorem 5.5

[1] Let T be a one–one operator of type $\omega$ on a Hilbert space H and $\{\xi _s \}$ be a uniformly bounded sequence in $H^\infty (S_\mu ^o)$, $\mu >\omega$, which converges to a function $\xi \in H^\infty (S_\mu ^o)$ uniformly on compact subsets of $S_\mu ^o$, such that $\{\xi _s(T)\}$ is a uniformly bounded set of the Banach algebra $\mathcal {L}(H)$ of all bounded operators . Then $\xi (T)\in \mathcal {L}(H)$, $\xi _s(T)(u)\rightarrow \xi (T)(u)$ for all $u\in H$, and $\Vert \xi (T)\Vert \le \sup _{s}\Vert \xi _s(T)\Vert$.

Now Assume that an operator T has a bounded $H^\infty$-functional calculus on the Hilbert space $L^2(\mathbb {R}^d,w_k(x)dx)$. We say that T has a bounded $H^\infty$-functional calculus on $L^p(\mathbb {R}^d,w_k(x)dx)$ for $1<p<\infty$, if for all $\xi \in H^\infty (S_\mu ^o)$ the operator $\xi (T)$ can be extended to a bounded operator on $L^p(\mathbb {R}^d,w_k(x)dx)$ that is,

$$\begin{aligned} \Vert \xi (T)(u)\Vert _p \le c \Vert u\Vert _p, \end{aligned}$$

for some constant $c>0$ and for all $u\in L^p(\mathbb {R}^d,w_k(x)dx)$.

We conclude with the following remark.

Remark 5.6

A function $\xi \in H^\infty (S_\mu ^o)$ is the limit of a uniformly bounded sequence of functions in $\Psi (S_\mu ^o)$ in the sense of uniform convergence on compact subsets of $S_\mu ^o$.

5.2 The Main Result

Our main result in this section is the following theorem.

Theorem 5.7

The Dunkl harmonic oscillator operator $H_k$ has a bounded $H^\infty$-functional calculus on $L^p(\mathbb {R}^d,w_k(x)dx)$ for $1<p<\infty$. Moreover, for each $\xi \in H^\infty (S_\mu ^0)$, $0<\mu <\pi$, the operator $\xi (L_k)$ is of weak type (1, 1).

We mention here that the operator $-\Delta _k$ has a $H^\infty$-functional calculus on $L^p(\mathbb {R}^d,w_k(x)dx)$, see [6].

By Theorem 5.7 we recover and extend the result of [3] where a particular case is dealt with $\xi (z)=z^{ia}, a\in \mathbb {R}$. For the proof we follow the elegant approach of [9].

Let $k_t$ and $K_t$ the kernels given by (3.3) and (3.4). Recall that

$$\begin{aligned} K_t(x,y)=\tau _x(k_t)(-y). \end{aligned}$$

Lemma 5.8

There exists $c>0$ such that for all $t>0$ and $z\in \mathbb {R}^d$,

$$\begin{aligned} \sup _{y\in B(z,\sqrt{t})} K_t(x,y)\le c \;\inf _{y\in B(z,\sqrt{t})}K_{2t}(x,y) \end{aligned}$$

(5.7)

Proof

Recall that

$$\begin{aligned} \tau _x(k_t)(-y)=t^{-\gamma _k-d/2}\int _{\mathbb {R}^d}e^{-(|y-\eta |^2+|x|^2-|\eta |^2)/t} d\nu _x(\eta ). \end{aligned}$$

Let $y_1,y_2\in B(z,\sqrt{t})$ and $\eta \in \mathbb {R}^d$ we have

$$\begin{aligned} \frac{|y_2-\eta |^2}{2t}\le \frac{|y_1-\eta |^2+|y_2-y_1|^2}{t}\le \frac{|y_1-\eta |^2}{t}+2. \end{aligned}$$

It follows that

$$\begin{aligned} t^{-\gamma _k+d/2}e^{-|y_1-\eta |^2/t}\le e^{2}2^{\gamma _k+d/2}(2t)^{-\gamma _k-d/2}e^{-|y_2-\eta |^2/2t} . \end{aligned}$$

Now, since we have

$$\begin{aligned} e^{-( |x|^2-|\eta |^2)/t}\le e^{-( |x|^2-|\eta |^2)/2t} \end{aligned}$$

thus for all $y_1,y_2\in B(z,\sqrt{t})$

$$\begin{aligned} \tau _x(k_t)(-y_1)\le c \tau _x(k_{2t})(-y_2), \end{aligned}$$

which yields the desired inequality. $\square$

For $f\in L^1_{loc}(\mathbb {R}^d,w_k(x)dx)$, the Dunkl maximal function $M_kf$ is defined by

$$\begin{aligned} M_kf(x)=\sup _{r>0}\frac{1}{d_kr^{2\gamma _k+d}}|f*_k\chi _{B_r}| \end{aligned}$$

where $\chi _{B_r}$ is the characteristic function of the ball $B_r$ of radius r centered at 0. According to the Theorem 6.2 of [22].

Lemma 5.9

There exists $c>0$ such that

$$\begin{aligned} \sup _{t>0}|k_t*f(x)|\le c M_kf(x). \end{aligned}$$

Lemma 5.10

For all $\delta >0$ there exists a constant $c>0$ such that for all $r>0$

$$\begin{aligned} \sup _{x\in \mathbb {R}^d}\int _{\min _{g\in G}| y-g.x| > r }K_t(x,y)w_k(y)dy\le c(1+r^2t^{-1})^{-\delta }. \end{aligned}$$

Proof

Let us note that for all $\eta \in co(G.x)$

$$\begin{aligned} \min _{g\in G}| y-g.x|^2\le |x|^2+|y|^2-2\langle y,\eta \rangle \le \max _{g\in G}| y-g.x|^2. \end{aligned}$$

From which

$$\begin{aligned} e^{-(|x|^2+|y|^2-2\langle y,\eta \rangle )/t}\le e^{(-\min _{g\in G}| y-g.x|^2)/2t}e^{-(|x|^2+|y|^2-2\langle y,\eta \rangle )/2t}. \end{aligned}$$

Then in view of (2.3) and (3.3) we have that

$$\begin{aligned} K_t(x,y)=\tau _x(k_t)(-y) \le 2^{\gamma _k+d/2}e^{(-\min _{g\in G}| y-g.x|^2)/2t}\tau _{x}(k_{2t})(-y) \end{aligned}$$

and from (2.4)

$$\begin{aligned} \int _{\min _{g\in G}| y-g.x| > r }K_t(x,y)w_k(y)dy\le & {} 2^{\gamma _k+d/2}\; e^{-r^2/2t}\; \Vert \tau _{x}(k_{2t})\Vert _{1,k}\\\le & {} 2^{\gamma _k+d/2}\;e^{-r^2/2t}\; \Vert k_{2t}\Vert _{1,k} \\\le & {} c \;(1+r^2t^{-1})^{-\delta } \end{aligned}$$

which is the desired inequality. $\square$

Lemma 5.11

Let T be a bounded operator on $L^2(\mathbb {R}^d,w_k(x)dx )$ . Suppose that $(T_n)_n$ is a sequence of bounded operators satisfy the condition: there exists $c>0$ such that for each $f \in L^2(\mathbb {R}^d,w_k(x)dx )\cap L^1(\mathbb {R}^d,w_k(x)dx )$ and $\lambda >0$

$$\begin{aligned} \mu _k\{x\in \mathbb {R}^d,\;\; |T_n(f)(x)|> \lambda \}\le c\; \frac{\Vert f\Vert _{k,1}}{\lambda },\quad n\in \mathbb {N} \end{aligned}$$

where the measure $d\mu _x=w_k(x)dx$. Assume that for each $f \in L^2(\mathbb {R}^d,w_k(x)dx )\cap L^1(\mathbb {R}^d,w_k(x)dx )$ there is a subsequence $(T_{n_j})_j$ such that

$$\begin{aligned} T(f)(x)=\lim _{j\rightarrow \infty } T_{n_j}(f)(x); \quad a.e.\; x\in \mathbb {R}^d . \end{aligned}$$

(5.8)

Then T is of weak type (1, 1), i.e., there exists $c>0$ such that

$$\begin{aligned} \mu _k\{x\in \mathbb {R}^d,\;\; |T(f)(x)|> \lambda \}\le c\; \frac{\Vert f\Vert _{k,1}}{\lambda } \end{aligned}$$

for each $f \in L^2(\mathbb {R}^d,w_k(x)dx )\cap L^1(\mathbb {R}^d,w_k(x)dx )$ and $\lambda >0$.

Proof

Let $f \in L^2(\mathbb {R}^d,w_k(x)dx )\cap L^1(\mathbb {R}^d,w_k(x)dx )$ and $\lambda >0$. Put

$$\begin{aligned} A_j=\{x\in \mathbb {R}^d,\;\; T_{n_j}(f)(x)> \lambda \} \end{aligned}$$

and

$$\begin{aligned} C_j=\bigcap _{\ell \ge j}A_{n_\ell }. \end{aligned}$$

Then, we have $C_j\subset C_{j+1}$. From (5.8) we see that

$$\begin{aligned} \mu _k(\{x\in \mathbb {R}^d,\;\; | T(f)(x)|> \lambda \})\le \mu _k\left( \bigcup _{j}C_j \right) =\lim _{j\rightarrow \infty }\mu _k\left( C_j \right) \end{aligned}$$

But,

$$\begin{aligned} \mu _k(C_j) \le \mu _k(A_{n_j}) \le c\; \frac{\Vert f\Vert _{k,1}}{\lambda }. \end{aligned}$$

Therefore

$$\begin{aligned} \mu _k(\{x\in \mathbb {R}^d,\;\; |T(f)(x)|> \lambda \})\le c\; \frac{\Vert f\Vert _{k,1}}{\lambda } \end{aligned}$$

which is the required inequality. $\square$

The next lemma show that the Euclidian $\mathbb {R}^d$ endowed with measure $\mu _k$ is a space of homogeneous type. Recall that a metric measure space $(X,d, \varrho )$ is said to be of homogeneous type if $\varrho$ is a doubling measure on X, i.e., there exists $c> 0$ such that such that for all $x_0\in \mathbb {R}^d$ and $r>0$,

$$\begin{aligned} \varrho (B(x_0,2r))\le c \varrho (B(x_0,r)). \end{aligned}$$

Where $B(x_0,r)=\{x\in X,\; d(x,x_0)<r\}$.

Lemma 5.12

$\mu _k$ is a doubling measure.

Proof

It is enough to check that the weight $w_k$ belongs to a Muckenhaupt class $A_p$ for some $p > 1$. Indeed, it is known [21, Ch V, 6.5] that when P is a polynomial on $\mathbb {R}^d$ having degree $\ell$ then $|P|^a$ belongs to $A_p$ whenever $-1<\ell a<p-1$. Applying this fact to the polynomial $P_\alpha (x) = \langle x,\alpha \rangle$ for $\alpha \in R^+$ and taking $p > 2N\gamma _k + 1$ where N is the cardinality of $R^+$, we see that $|P_\alpha |^{2Nk(\alpha )}\in A_p$. Then according to [21, Ch V, 6.1] and the fact that

$$\begin{aligned} w_k=\left( \prod _{\alpha \in R^+}|P_\alpha |^{2Nk(\alpha )}\right) ^{\frac{1}{N}} \end{aligned}$$

we obtain, with this choice of p, that $w_k \in A_p$. $\square$

We are now able to prove Theorem 5.7.

Proof of the Theorem 5.7

Let $\xi \in H^\infty (S_\mu ^o)$, $\mu >0$. it suffices by Marcinkiewicz interpolation and duality to prove that $\xi (H_k)$ is a weak-type (1, 1). Before beginning we note first that in view of the convergence Theorem 5.5 and Lemma 5.11 one can assume that $\xi \in \Psi (S_\mu ^o)$.

Let $f\in L^2(\mathbb {R}^d), w_k(x)dx)\cap L^1(\mathbb {R}^d), w_k(x)dx)$ and $\lambda >0$. From the Calderon-Zygmund decomposition, there exist a functions g and $f_j$ and balls $B_j=B(x_j , r_j)$ such that

• (i) $f=g+h$ with $h= \sum _j f_j$,

• (ii) $\Vert g\Vert _\infty \le c \lambda$,

• (iii) $supp(f_j)\subset B_j$, and

• (iv) $\Vert f_j\Vert _{1,k}\le c \mu _k(B_j)$,

• (v) $\sum _j\mu _k(B_j)\le \dfrac{c}{\lambda }\;\Vert f\Vert _{1,k}$,

• (vi) Each point of $\mathbb {R}^d$ is contained in at most a finite number M of the balls $B_j$.

Note that (iv) and (v) imply that $\Vert h\Vert _{1,k}\le c\Vert f\Vert _{1,k}$. Hence

$$\begin{aligned} \Vert g\Vert _{1,k}\le (1+c)\Vert f\Vert _{1,k}. \end{aligned}$$

(5.9)

Let $S_t=e^{-tH_k}$, we split $h= h_1+h_2$ with

$$\begin{aligned} h_1= \sum _j S_{t_j}(f_j); \quad h_2= h-h_1=\sum _j(I-S_{t_j})f_j \end{aligned}$$

where $t_j=r_j^2$. Then

$$\begin{aligned} \mu _k\{x;\; |\xi (H_k)(f)(x)|>\lambda \}\le & {} \mu _k\left\{ x;\; |\xi (H_k)(g)(x)|>\frac{\lambda }{3}\right\} \nonumber \\&+\mu _k\left\{ x;\; |\xi (H_k)(h_1)(x)|>\frac{\lambda }{3}\right\} \nonumber \\&+\mu _k\left\{ x;\; |\xi (H_k)(h_2)(x)|>\frac{\lambda }{3}\right\} . \end{aligned}$$

(5.10)

For the first term of the right hand side of (5.10) we use the $L^2$-boundedness of $\xi (H_k)$, (5.9) and (ii) to get

$$\begin{aligned} \mu _k\left\{ x;\; \xi (H_k)(g)>\frac{\lambda }{3}\right\} \le \frac{9}{\lambda ^2}\Vert \xi (H_k)(g)\Vert _{2,k}\le \frac{c}{\lambda ^2}\Vert g\Vert _{2,k}\le \frac{c}{\lambda ^2}\Vert g\Vert _{1,k}\Vert g\Vert _{\infty }\le \frac{c}{\lambda }\Vert f\Vert _{1,k}. \end{aligned}$$

For the second term we have

$$\begin{aligned} \mu _k\left\{ x;\; |\xi (H_k)(h_1)(x)|>\frac{\lambda }{3}\right\} \le \frac{9}{\lambda ^2}\Vert \xi (H_k)(h_1)\Vert _{2,k}\le \frac{9}{\lambda ^2} \sum _j\Vert S_{t_j}(f_j)\Vert _{2,k}^2. \end{aligned}$$

Using (5.3) and Lemma 5.8 we get

$$\begin{aligned} |S_{t_j}(f_j)(x)|\le & {} \int _{\mathbb {R}^d}\tau _x(k_{t_j})(-y)|f_j(y)|w_k(y)dy\\\le & {} \sup _{y\in B(z,\sqrt{t_j})}\tau _x(k_{t_j})(-y)\Vert f_j\Vert _{1,k}\\\le & {} c \lambda \mu (B_j)\inf _{y\in B(z,\sqrt{t_j})}\tau _x(k_{2t_j})(-y)\\\le & {} c\lambda \int _{\mathbb {R}^d}\tau _x(k_{2t_j})(-y) \chi _{B_j}w_k(y)dy \end{aligned}$$

and in view of Lemma 5.9 it follows that for any $\varphi \in L^2(\mathbb {R}^d,w_k(x)dx)$,

$$\begin{aligned} |\langle S_{t_j}(f_j),\varphi \rangle _{k}|\le c\lambda \langle k_{2t_j}*_k|\varphi |, \chi _{B_j}\rangle _{k}\le c\lambda \langle M_k|\varphi |, \chi _{B_j}\rangle _{k}. \end{aligned}$$

Thus the $L^2$-boundedness of $M_k$ yield

$$\begin{aligned} \left\| \sum _jS_{t_j}(f_j)\right\| _{2,k}= & {} \sup \left\{ \left| \left\langle \sum _jS_{t_j}(f_j),\varphi \right\rangle _{k}\right| ,\; \Vert \varphi \Vert _{2,k}\le 1\right\} \\\le & {} c\lambda \sup \left\{ \left\langle M_k|\varphi |,\sum _j \chi _{B_j}\right\rangle _{k},\; \Vert \varphi \Vert _{2,k}\le 1\right\} \\\le & {} c\lambda \left\| \sum _j\chi _{B_j}\right\| _{2,k}. \end{aligned}$$

We now use properties (vi) of the Calderon–Zygmund decomposition to obtain the estimate

$$\begin{aligned} \left\| \sum _j\chi _{B_j}\right\| _{2,k}\le M \left( \sum _j\mu (B_j)\right) ^{\frac{1}{2}}. \end{aligned}$$

Thus in view (v)

$$\begin{aligned} \left\| \sum _jS_{t_j}(f_j)\right\| _{2,k}^2\le c\lambda \Vert f\Vert _{1,k} \end{aligned}$$

and we conclude that

$$\begin{aligned} \mu _k\left\{ x;\; |\xi (H_k)(h_1)(x)|>\frac{\lambda }{3}\right\} \le \frac{c}{\lambda }\Vert f\Vert _{1,k}. \end{aligned}$$

Now consider the third term of the right hand side of (5.10). Putting

$$\begin{aligned} \mathcal {B}_j= \bigcup _{g\in G}g\big (B(y_j,2r_j)\Big )=\bigcup _{g\in G}B(gy_j,2r_j) \end{aligned}$$

and write

$$\begin{aligned} \mu _k\left\{ x;\; |\xi (H_k)(h_2)(x)|>\frac{\lambda }{3}\right\} \le \sum _j\mu _k (\mathcal {B}_j)+\mu _k\left\{ x\notin \bigcup _{ j}\mathcal {B}_j ;\;|\xi (H_k)(h_2)(x)|>\frac{\lambda }{3}\right\} . \end{aligned}$$

By the doubling property of the measure $\mu _k$ and (iii) we have

$$\begin{aligned} \sum _j\mu _k ( \mathcal {B}_j ) \le |G|\sum _j\mu _k ( B(y_j,2r_j)) \le |G|\sum _j\mu _k ( B_j)\le \frac{c}{\lambda }\Vert f\Vert _{1,k}, \end{aligned}$$

since $w_k$ is a G-invariant function. So it remains to prove that

$$\begin{aligned} \mu _k\left\{ x\notin \bigcup _{ j}\mathcal {B}_j ;\;|\xi (H_k)(h_2)(x)|>\frac{\lambda }{3}\right\} \le \frac{c}{\lambda }\;\Vert f\Vert _{1,k}. \end{aligned}$$

(5.11)

As in [9] define $\xi _j(v)=\xi (v)(1-e^{-t_jv})$ and write

$$\begin{aligned} \xi (H_k)(h_2)=\sum _j\xi _j(H_k)f_j. \end{aligned}$$

We represent the operator $\xi _j(H_k)$ by

$$\begin{aligned} \xi _j(H_k)=\frac{1}{2i\pi }\int _{\gamma }\xi _j(v)(H_k-vI)^{-1}dv =\xi _j^+(H_k)+\xi _j^-(H_k) \end{aligned}$$

where

$$\begin{aligned} \xi _j^{\pm }(H_k)=\frac{1}{2i\pi }\int _{\gamma ^{\pm }}\xi _j(v)(H_k-vI)^{-1}dv \end{aligned}$$

and the contour $\gamma =\gamma ^+\cup \gamma ^-$ with $\gamma ^+(t)=te^{i\theta }$ for $t\ge 0$, $\gamma ^-(t)=-te^{i\theta }$ for $t< 0$ and whith $0<\theta <\pi /2$. Consider $\xi _j^{+}(H_k)$, for $v\in \gamma ^+$ we substitute

$$\begin{aligned} (H_k-vI)^{-1}=\int _{\Gamma ^+} e^{vz}e^{-zH_k}dz \end{aligned}$$

where the curve $\Gamma ^+$ is defined by $\Gamma ^+=te^{i\beta }$ with $\pi /2-\theta<\beta <\pi /2$. It follows that

$$\begin{aligned} \xi _j^{+}(H_k)= \int _{\Gamma ^+} n^+(z)e^{-zH_k} dz, \end{aligned}$$

where

$$\begin{aligned} n^+(z)=\int _{\gamma ^+}\xi _j(v)e^{vz} dv. \end{aligned}$$

Here we have used Fubini’s theorem to change the order of integration. Therefore we have

$$\begin{aligned}&\displaystyle { \int _{x\in \mathcal {B}_j^c} |\xi _j^{+}(H_k)(f_j)(x)|w_k(x)dx}\\&\quad \le c \int _0^{+\infty } \int _0^{+\infty } |1-e^{-t_jv}|\;e^{-|v||z|\sigma }\\&\qquad \left( \int _{\mathbb {R}^d}\int _{x\in \mathcal {B}_j^c}|\mathcal {H}_z(x,y)||f_j(y)|w_k(x)w_k(y)dxdy\right) d|v|d|z| \end{aligned}$$

where $v=|v|e^{i\theta }$ and $\sigma =\cos (\theta +\beta )$. Let us noting that $x\in \mathcal {B}_j^c$ is equivalent to the condition

$$\begin{aligned} \min _{g\in G}|x-gy|>2r_j. \end{aligned}$$

Then using Proposition 5.1 and Lemma 5.10

$$\begin{aligned} \int _{x\in \mathcal {B}_j^c}|\mathcal {H}_z(x,y)|w_k(x)dx\le & {} \int _{\min _{g\in G}|x-gy|>2r_j}K_{Re(z)}(cx,cy)w_k(x)dx\\= & {} c^{2\gamma _k} \int _{\min _{g\in G}|x-g.cy|>2r_j}K_{Re(z)}(x,cy)w_k(x)dx\\\le & {} C \left( 1+\frac{r_j}{Re(z)}\right) ^{-\delta }\le C \left( 1+\frac{r_j}{|z|}\right) ^{-\delta }, \end{aligned}$$

from which it follows that

$$\begin{aligned}&\displaystyle { \int _{x\in \mathcal {B}_j^c} |\xi _j^{+}(H_k)(f_j)(x)|w_k(x)dx}\\&\quad \le C\Vert f_j\Vert _{1,k} \int _0^{+\infty } \int _0^{+\infty } |1-e^{-t_jv}|\;e^{-|v||z|\sigma }\left( 1+ r_j |z|^{-1} \right) ^{-\delta } d|v|d|z|. \end{aligned}$$

This integral is treated in [9] by splitting it into two parts, $I_1$ and $I_2$, corresponding to integration over $t_j|v|>1$ and $t_j|v|\le 1$ and gives

$$\begin{aligned} \int _{x\in \mathcal {B}_j^c} |\xi _j^{+}(H_k)(f_j)(x)|w_k(x)dx\le C\Vert f_j\Vert _{1,k}. \end{aligned}$$

We also obtain similar estimates for $\xi _j^{-}(H_k)$ by the same argument. Therefore

$$\begin{aligned} \mu _k\left\{ x\notin \bigcup _{ j}\mathcal {B}_j ;\;|\xi (H_k)(h_2)(x)|>\frac{\lambda }{3}\right\}\le & {} \frac{3}{\lambda } \sum _j\int _{x\notin \cup _i\mathcal {B}_i } |\xi _j (H_k)(f_j)(x)|w_k(x)dx\\\le & {} \frac{3}{\lambda }\sum _j\int _{x\notin \mathcal {B}_j } |\xi _j (H_k)(f_j)(x)|w_k(x)dx\\\le & {} c\sum _j\Vert f\Vert _{1,k}\le \frac{c}{\lambda }\;\Vert f\Vert _{1,k}. \end{aligned}$$

This archives the proof of the weak type estimates (1, 1) for $\xi (H_k)$. $\square$