# _Green function and Poisson kernel associated to root systems for annular
regions_

Chaabane REJEB111Université de Tunis El Manar, Faculté des Sciences de Tunis,
Laboratoire d’Analyse Mathématques et Applications LR11ES11, 2092 El Manar I,
Tunis, TUNISIA, Laboratoire de Mathématiques et Physique Théorique CNRS-UMR
7350, Université de Tours, Campus de Grandmont, 37200 Tours, FRANCE and
Université de Sherbrooke, CANADA. Email<EMAIL_ADDRESS>
###### Abstract

Let $\Delta_{k}$ be the Dunkl Laplacian relative to a fixed root system
$\mathcal{R}$ in $\mathbb{R}^{d}$, $d\geq 2$, and to a nonnegative
multiplicity function $k$ on $\mathcal{R}$. Our first purpose in this paper is
to solve the $\Delta_{k}$-Dirichlet problem for annular regions. Secondly, we
introduce and study the $\Delta_{k}$-Green function of the annulus and we
prove that it can be expressed by means of $\Delta_{k}$-spherical harmonics.
As applications, we obtain a Poisson-Jensen formula for
$\Delta_{k}$-subharmonic functions and we study positive continuous solutions
for a $\Delta_{k}$-semilinear problem.

MSC (2020) primary: 31B05, 31B20, 31J05, 35J08; secondary: 31C45, 46F10,
47B39.

Key words: Dunkl-Laplace operator, Poisson kernel, Green function, Dirichlet
problem, spherical harmonics, Newton kernel.

##  1 Introduction

Since the 90’s, extensive studies have been carried out on analysis associated
with Dunkl operators. These are commuting differential-difference operators on
$\mathbb{R}^{d}$ introduced by C. F. Dunkl (see [6]). The Dunkl analysis
includes especially a generalization of the Fourier transform (called the
Dunkl transform) and the Laplace operator known as the Dunkl Laplacian (and
denoted by $\Delta_{k}$).
The Dunkl theory has many applications as well in mathematical physics and
probability theory. In particular, it has been used in the study of the
Calogero-Moser-Sutherland and other integrable systems (see [4, 10]) and in
the study of Markov processes generalizing Brownian motion (see [22]).
Recently, a special interest has been devoted to potential theory associated
with the Dunkl Laplacian. The study focused on $\Delta_{k}$-harmonic functions
(see [2, 11, 12, 17, 19, 20]), on $\Delta_{k}$-Newton potential theory
(including $\Delta_{k}$-subharmonic functions) (see [13]) and on
$\Delta_{k}$-Riesz potentials of Radon measures (see [14]). More recently, by
means of the $\Delta_{k}$-Newton kernel, the Green function of the open unit
ball has been studied in [15]. Note that finding $\Delta_{k}$-Green functions
for other open sets is a rather difficult problem already in the case of the
classical Laplace operator. The aim of this paper is to show that we can
determine the $\Delta_{k}$-Green function for annular regions in
$\mathbb{R}^{d}$ by using $\Delta_{k}$-spherical harmonics as a crucial tool.
Let us assume throughout the paper that $d\geq 2$. Let $A$ be the annulus

$A:=\\{x\in\mathbb{R}^{d},\ \rho<\|x\|<1\\}\quad\text{with}\quad\rho\in(0,1).$

After giving some properties of the $\Delta_{k}$-Green function $G_{k,A}$ of
$A$, we will use it to study the semilinear problem

$\left\\{\begin{array}[]{ll}\Delta_{k}(u\omega_{k})=\phi(.,u)\omega_{k},&\hbox{in
the sense of distributions}\\\ u=f,&\hbox{on}\ \partial{A},\end{array}\right.$

where $\omega_{k}$ is a precise weight function (see (2.6) for its
expression).
More precisely, under some assumptions on the function $\phi$, we will show
that if $f\in\mathcal{C}(\partial{A})$ is nonnegative, this boundary problem
has one and only one positive continuous solution on $A$ which satisfies (see
Theorem 5.2)

$\forall\ x\in A,\quad
u(x)+\int_{A}G_{k,A}(x,y)\phi(y,u(y))\omega_{k}(y)dy=P_{k,A}[f](x).$

Here $P_{k,A}[f]$ is the unique solution in
$\mathcal{C}^{2}(A)\cap\mathcal{C}(\overline{A})$ of the boundary Dirichlet
problem

$\left\\{\begin{array}[]{ll}\Delta_{k}u=0,&\hbox{on}\ A,\\\ u=f,&\hbox{on}\
\partial{A},\end{array}\right.$

that will be given explicitly in Section 3.

This paper is organized as follows. In Section 2, we recall some basics from
Dunkl theory that will be used throughout the paper. In Section 3, we give an
explicit solution of the boundary Dirichlet problem for the annulus. The Green
function $G_{k,A}$ will be introduced and studied in Section 4. Some
applications will be given in the last Section. Precisely, we will obtain a
Poisson-Jensen formula for $\Delta_{k}$-subharmonic functions in the annulus
and we will study positive solutions of the above semilinear problem.

##  2 Basics from Dunkl theory

We start by recalling some useful facts in Dunkl theory. Let $\mathcal{R}$ be
a root system in the Euclidian space $\mathbb{R}^{d}$, in the sense that
$\mathcal{R}$ is a finite set in $\mathbb{R}^{d}\setminus\\{0\\}$ such that
for every $\alpha\in\mathcal{R}$,
$\mathcal{R}\cap\mathbb{R}\alpha=\\{\pm\alpha\\}$ and
$\sigma_{\alpha}(\mathcal{R})=\mathcal{R}$ (where $\sigma_{\alpha}$ is the
reflection w.r.t. the hyperplane $H_{\alpha}$ orthogonal to $\alpha$). The
subgroup $W\subset O(\mathbb{R}^{d})$ generated by the reflections
$\sigma_{\alpha}$, $\alpha\in\mathcal{R}$, is called the Coxeter-Weyl group
associated to $\mathcal{R}$. We refer to ([18]) for more details on root
systems and their Coxeter-Weyl groups.

Let $k$ be a fixed nonnegative multiplicity function on $\mathcal{R}$ (i.e.
$k$ is $W$-invariant). For $\xi\in\mathbb{R}^{d}$, the $\xi$-directional Dunkl
operator associated to $(W,k)$ is defined by

$D_{\xi}f(x):=\partial_{\xi}f(x)+\sum_{\alpha\in\mathcal{R}_{+}}k(\alpha)\mathop{\langle\alpha,\xi\rangle}\frac{f(x)-f(\sigma_{\alpha}.x)}{\mathop{\langle\alpha,x\rangle}},\quad
f\in\mathcal{C}^{1}(\mathbb{R}^{d}),$

where $\partial_{\xi}$ is the usual $\xi$-directional partial derivative and
$\mathcal{R}_{+}$ is a positive subsystem.
Let us denote by $\mathcal{P}(\mathbb{R}^{d})$ (resp.
$\mathcal{P}_{n}(\mathbb{R}^{d})$) the space of polynomial functions on
$\mathbb{R}^{d}$ (resp. the space of homogeneous polynomials of degree
$n\in\mathbb{N}$).
There exists a unique linear isomorphism $V_{k}$ from
$\mathcal{P}(\mathbb{R}^{d})$ onto itself such that
$V_{k}(\mathcal{P}_{n}(\mathbb{R}^{d}))=\mathcal{P}_{n}(\mathbb{R}^{d})$ for
every $n\in\mathbb{N}$, $V_{k}(1)=1$ and

$\forall\ \xi\in\mathbb{R}^{d},\quad D_{\xi}V_{k}=\partial_{\xi}V_{k}.$ (2.1)

The operator $V_{k}$ is known as the Dunkl intertwining operator (see [7, 8]).
It has been extended to a topological isomorphism from
$\mathcal{C}^{\infty}(\mathbb{R}^{d})$ onto itself satisfying (2.1) (see
[26]). Furthermore, according to [23], for each $x\in\mathbb{R}^{d}$, there is
a compactly supported probability measure $\mu_{x}$ on $\mathbb{R}^{d}$ such
that

$\forall\ f\in\mathcal{C}^{\infty}(\mathbb{R}^{d}),\quad
V_{k}(f)(x)=\int_{\mathbb{R}^{d}}f(y)d\mu_{x}(y).$ (2.2)

If $W.x$ denotes the orbit of $x$ under the $W$-action and $Co(x)$ its convex
hull, then

$\text{supp}\ \mu_{x}\subset Co(x)\subset\overline{B}(0,\|x\|).$ (2.3)

The Dunkl-Laplacian is defined as $\Delta_{k}=\sum_{j=1}^{d}D_{e_{j}}^{2}$,
where $(e_{j})_{1\leq j\leq d}$ is the canonical basis of $\mathbb{R}^{d}$. It
can be expressed as follows

$\Delta_{k}f(x)=\Delta f(x)+\sum_{\alpha\in
R_{+}}k(\alpha)\Big{(}2\frac{\mathop{\langle\nabla
f(x),\alpha\rangle}}{\mathop{\langle\alpha,x\rangle}}-\|\alpha\|^{2}\frac{f(x)-f(\sigma_{\alpha}(x))}{\mathop{\langle\alpha,x\rangle}^{2}}\Big{)},\quad
f\in\mathcal{C}^{2}(\mathbb{R}^{d}),$ (2.4)

where $\Delta$ (resp. $\nabla$ ) is the usual Laplace (resp. gradient)
operator (see [6, 8]). Note that if $k$ is the zero function, the Dunkl
Laplacian reduces to the classical one which commutes with the action of
$O(\mathbb{R}^{d})$. For general $k\geq 0$, $\Delta_{k}$ commutes with the
$W$-action (see [24]) i.e.

$\forall\ g\in W,\quad g\circ\Delta_{k}=\Delta_{k}\circ g.$ (2.5)

Let $L^{2}_{k}(S^{d-1})$, $d\geq 2$, be the Hilbert space endowed with the
inner product

${\mathop{\langle
p,q\rangle}}_{k}:=\frac{1}{d_{k}}\int_{S^{d-1}}p(\xi)q(\xi)\omega_{k}(\xi)d\sigma(\xi).$

We denote by $\|.\|_{L^{2}_{k}(S^{d-1})}$ the associated Euclidean norm. Here,
$d\sigma$ is the surface measure on the unit sphere $S^{d-1}$, $\omega_{k}$ is
the weight function given by

$\omega_{k}(x)=\textstyle\prod_{\alpha\in\mathcal{R}_{+}}|\mathop{\langle\alpha,x\rangle}|^{2k(\alpha)}$
(2.6)

and $d_{k}$ is the constant

$d_{k}=\textstyle\int_{S^{d-1}}\omega_{k}(\xi)d\sigma(\xi).$ (2.7)

The function $\omega_{k}$ is $W$-invariant and homogeneous of degree
$2\gamma:=2\textstyle\sum_{\alpha\in\mathcal{R}_{+}}k(\alpha)$.
Let us introduce the constant

$\lambda_{k}:=\frac{d}{2}+\gamma-1\geq 0.$ (2.8)

Let
$\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d}):=\mathcal{P}_{n}(\mathbb{R}^{d})\cap
Ker\Delta_{k}$ be the space of $\Delta_{k}$-harmonic polynomials, homogeneous
of degree $n$ on $\mathbb{R}^{d}$. From [8], we know that if $n\neq m$, then
$\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})\perp\mathcal{H}_{\Delta_{k},m}(\mathbb{R}^{d})$
in $L^{2}_{k}(S^{d-1})$. Moreover, for every $n\in\mathbb{N}$, we have

$\mathcal{P}_{n}(\mathbb{R}^{d})=\textstyle\bigoplus_{j=0}^{\lfloor
n/2\rfloor}\|x\|^{2j}\mathcal{H}_{\Delta_{k},n-2j}(\mathbb{R}^{d}).$ (2.9)

The restriction to the sphere $S^{d-1}$ of an element of
$\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$ is called a
$\Delta_{k}$-spherical harmonic of degree $n$. The space of
$\Delta_{k}$-spherical harmonics of degree $n$ will be denoted by
$\mathcal{H}_{\Delta_{k},n}(S^{d-1})$. This space has a reproducing kernel
$Z_{k,n}$ uniquely determined by the properties (see [5, 8])

i)

for each $x\in S^{d-1}$, $Z_{k,n}(x,.)\in\mathcal{H}_{\Delta_{k},n}(S^{d-1})$,

ii)

for every $f\in\mathcal{H}_{\Delta_{k},n}(S^{d-1})$, we have

$f(x)={\mathop{\langle
f,Z_{k,n}(x,.)\rangle}}_{k}=\frac{1}{d_{k}}\int_{S^{d-1}}f(\xi)Z_{k,n}(x,\xi)\omega_{k}(\xi)d\sigma(\xi),\quad
x\in S^{d-1}.$ (2.10)

From this formula, we can see that

$\forall\ g\in W,\quad\forall\ x,y\in S^{d-1},\quad
Z_{k,n}(gx,gy)=Z_{k,n}(x,y).$ (2.11)

In the classical case (i.e. $k=0$), $Z_{0,n}(x,.)$ is known as the zonal
harmonic of degree $n$ (see [1, 5]). Note that if
$\\{Y_{j,n},j=1,\dots,h(n,d):=dim\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})\\}$
is a real-orthonormal basis of $\mathcal{H}_{\Delta_{k},n}(S^{d-1})$ in
$L^{2}_{k}(S^{d-1})$, then

$Z_{k,n}(x,y)=\sum_{j=1}^{h(n,d)}Y_{j,n}(x)Y_{j,n}(y).$ (2.12)

By means of the Dunkl intertwining operator and Gegenbauer polynomials,
$Z_{k,n}$ is given explicitly by (see [5], Theorem 7.2.6. or [27])

$\forall\ x,y\in S^{d-1},\quad
Z_{k,n}(x,y)=\frac{(n+\lambda_{k})(2\lambda_{k})_{n}}{\lambda_{k}.n!}V_{k}\Big{(}P_{n}^{\lambda_{k}}\big{(}\mathop{\langle.,y\rangle}\big{)}\Big{)}(x),$
(2.13)

where $\lambda_{k}$ is the constant given by (2.8), $P_{n}^{\mu}$, $\mu>-1/2$,
is the normalized Gegenbauer polynomial (see [8] p. 17) defined by

$P_{n}^{\mu}(x):=\frac{(-1)^{n}}{2^{n}(\mu+1/2)_{n}}(1-x^{2})^{1/2-\mu}\frac{d^{n}}{dx^{n}}(1-x^{2})^{n+\mu-1/2},$

and $(x)_{n}:=x(x+1)\dots(x+n-1)$ is the Pochhammer symbol.

At the end of this section, in order to simplify notations in the classical
case $k=0$ we will write $L^{2}(S^{d-1})$ for $L^{2}_{0}(S^{d-1})$,
$\mathcal{H}_{\Delta,n}$ for $\mathcal{H}_{\Delta_{0},n}$, $|S^{d-1}|:=d_{0}$
the surface area of $S^{d-1}$ and $Z_{n}:=Z_{0,n}$.

##  3 $\Delta_{k}$-Dirichlet problem for the annulus

In this section, by introducing a Poisson type kernel, we will solve the
Dirichlet problem for the Dunkl Laplacian in annular regions

$A_{R_{1},R_{2}}:=\\{x\in\mathbb{R}^{d}:\ R_{1}<\|x\|<R_{2}\\}.$

Note that from the homogeneity property of $\Delta_{k}$:

$\delta_{r}\circ\Delta_{k}=r^{-2}\Delta_{k}\circ\delta_{r},\quad\text{with}\quad\delta_{r}(f)(x):=f(rx),$

it suffices to do this for the annular region $A=A_{\rho,1}$ with
$\rho\in(0,1)$ fixed.
Recall that the $\Delta_{k}$-Poisson kernel of the unit ball (see [8]) is
given by

$P_{k}(x,y)=\sum_{n=0}^{+\infty}Z_{k,n}(x,y)=\int_{\mathbb{R}^{d}}\frac{1-\|x\|^{2}}{\big{(}1-2\mathop{\langle
x,z\rangle}+\|x\|^{2}\big{)}^{\frac{d}{2}+\gamma}}d\mu_{y}(z),\quad(x,y)\in
B\times{S^{d-1}}.$ (3.1)

From [8], we know that

$\frac{1}{d_{k}}\int_{S^{d-1}}P_{k}(x,\xi)\omega_{k}(\xi)d\sigma(\xi)=1.$
(3.2)

We start by two preliminary useful results. For each $n\in\mathbb{N}$, the
restriction of the Dunkl intertwining operator

$V_{k}:\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\longrightarrow\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$

is a linear isomorphism.
In the first result, we will estimate the matrix-norms of this operator and of
its inverse where the space $\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})$ (resp.
$\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$) is endowed with the
$L^{2}(S^{d-1})$-norm (resp. the $L_{k}^{2}(S^{d-1})$-norm). More precisely,

###### Proposition 3.1

Let $n$ be a nonnegative integer.

1.

For every $f\in\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})$, we have

$\|V_{k}(f)\|_{L^{2}_{k}(S^{d-1})}\leq
dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|f\|_{L^{2}(S^{d-1})}.$ (3.3)

2.

For every $f\in\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$, we have

$\|V_{k}^{-1}(f)\|_{L^{2}(S^{d-1})}\leq\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|f\|_{L^{2}_{k}(S^{d-1})}.$
(3.4)

_Proof:_ 1) Let $f\in\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})$. After rewriting
the reproducing formula (2.10) in the classical case (i.e. $k=0$), applying it
to $f$ and using Fubini’s theorem, we get

$V_{k}(f)(x)=\frac{1}{|S^{d-1}|}\int_{S^{d-1}}f(\xi)V_{k}[Z_{n}(.,\xi)](x)d\sigma(\xi),\quad
x\in\mathbb{R}^{d}.$ (3.5)

But, from [1], Proposition 5.27, we have

$\forall\ z,\xi\in S^{d-1},\quad|Z_{n}(z,\xi)|\leq
dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})$

which implies that

$\forall\ (z,\xi)\in\mathbb{R}^{d}\times
S^{d-1},\quad|Z_{n}(z,\xi)|\leq\Big{(}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\Big{)}\|z\|^{n}.$
(3.6)

Thus, using the relations (2.2), (2.3), (3.5) and (3.6) and the Cauchy-Schwarz
inequality, we obtain

$\forall\ x\in\mathbb{R}^{d},\quad|V_{k}(f)(x)|\leq
dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|f\|_{L^{2}(S^{d-1})}\|x\|^{n}.$
(3.7)

This implies that

$\|V_{k}(f)\|_{L^{2}_{k}(S^{d-1})}\leq
dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|f\|_{L^{2}(S^{d-1})}.$

2) Let $f\in\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$. By applying the
classical case of the formula (2.10) to $V_{k}^{-1}(f)$ and by using (3.6) and
the Cauchy-Schwarz inequality, we deduce that

$\forall\ x\in\mathbb{R}^{d},\quad|V_{k}^{-1}(f)(x)|\leq
dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|V_{k}^{-1}(f)\|_{L^{2}(S^{d-1})}\|x\|^{n}.$

Now, using the following result (see [8], Proposition 5.2.8): for
$p\in\mathcal{P}_{n}(\mathbb{R}^{d})$ and
$q\in\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})$, then

$\frac{1}{|S^{d-1}|}\int_{S^{d-1}}p(\xi)q(\xi)d\sigma(\xi)=\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}d_{k}}\int_{S^{d-1}}p(\xi)V_{k}(q)(\xi)\omega_{k}(\xi)d\sigma(\xi)$

with $p=q=V_{k}^{-1}(f)$, we obtain

$\displaystyle\|V_{k}^{-1}(f)\|_{L^{2}(S^{d-1})}^{2}$
$\displaystyle\leq\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}d_{k}}\int_{S^{d-1}}|V_{k}^{-1}(f)(\xi)f(\xi)|\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle\leq\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}d_{k}}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|V_{k}^{-1}(f)\|_{L^{2}(S^{d-1})}\int_{S^{d-1}}|f(\xi)|\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle\leq\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|V_{k}^{-1}(f)\|_{L^{2}(S^{d-1})}\|f\|_{L^{2}_{k}(S^{d-1})}.$

This proves the desired relation. $\square$

###### Corollary 3.1

The following inequality holds:

$\forall\ x,y\in
S^{d-1},\quad|Z_{k,n}(x,y)|\leq\Big{(}\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}}\Big{)}^{2}\Big{(}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\Big{)}^{5}.$
(3.8)

_Proof:_ Let $\\{Y_{j,n}\\}_{j}$,
$j=1,\dots,h(n,d)=dim\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$, be a real-
orthonormal basis of $\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})$ in
$L^{2}_{k}(S^{d-1})$. Using (3.7) with $f=V_{k}^{-1}(Y_{j,n})$ and (3.4), we
deduce that

$\displaystyle\forall\ x\in S^{d-1},|Y_{j,n}(x)|$ $\displaystyle\leq
dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\|V_{k}^{-1}(Y_{j,n})\|_{L^{2}(S^{d-1})}$
$\displaystyle\leq\frac{(\gamma+\frac{d}{2})_{n}|S^{d-1}|}{(\frac{d}{2})_{n}}\Big{(}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})\Big{)}^{2}.$

Consequently, we obtain the result from (2.12). $\square$

Following the classical case $k=0$ (see [1]), we define the kernel
$\mathbf{P}_{k,1}(.,.)$ on $A\times{S^{d-1}}$ by

$\mathbf{P}_{k,1}(x,\xi):=\sum_{n=0}^{+\infty}a_{k,n}(x)Z_{k,n}(x,\xi),\quad\text{with}\quad
a_{k,n}(x)=\frac{1-\big{(}\frac{\|x\|}{\rho}\big{)}^{-2\lambda_{k}-2n}}{1-\rho^{2\lambda_{k}+2n}}.$
(3.9)

###### Proposition 3.2

The kernel $\mathbf{P}_{k,1}$ satisfies the following properties

i)

For each $\xi\in S^{d-1}$, $\mathbf{P}_{k,1}(.,\xi)$ is a
$\Delta_{k}$-harmonic function on $A$ and $\mathbf{P}_{k,1}(.,\xi)=0$ on
$S(0,\rho)$.

ii)

For every $x\in A$ and $\xi\in S^{d-1}$,

$0\leq\mathbf{P}_{k,1}(x,\xi)\leq P_{k}(x,\xi).$ (3.10)

iii)

Let $x\in A$ and $\xi\in S^{d-1}$ fixed. Then

$\forall\ g\in W,\quad\mathbf{P}_{k,1}(gx,g\xi)=\mathbf{P}_{k,1}(x,\xi).$
(3.11)

_Proof:_ i) Clearly $\mathbf{P}_{k,1}(.,\xi)=0$ on $S(0,\rho)$. On the other
hand, for any $(x,\xi)\in A\times S^{d-1}$ we can write

$a_{k,n}(x)Z_{k,n}(x,\xi)=c_{1,n}Z_{k,n}(x,\xi)-c_{2,n}K_{k}[Z_{k,n}(.,\xi)](x),$

where $c_{1,n},c_{2,n}$ are two nonnegative constants and $K_{k}$ is the
$\Delta_{k}$-Kelvin transform (see [9]) given by

$K_{k}[f](x)=\|x\|^{-2\lambda_{k}}f(x/{\|x\|^{2}})=\|x\|^{2-2\gamma-d}f(x/{\|x\|^{2}})$
(3.12)

and $f$ is a function defined on $\mathbb{R}^{d}\setminus\\{0\\}$. As the
$\Delta_{k}$-Kelvin transform preserves the $\Delta_{k}$-harmonic functions on
$\mathbb{R}^{d}\setminus\\{0\\}$ (see [9]), we deduce that the function
$x\mapsto a_{k,n}(x)Z_{k,n}(x,\xi)$ is $\Delta_{k}$-harmonic on $A$.
According to [8] (see also [1] and [5]), we know that

$dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})=dim\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})={n+d-1\choose{n}}-{n+d-3\choose{n-2}}.$

Hence, we have

$\lim_{n\rightarrow+\infty}n^{2-d}dim\mathcal{H}_{\Delta,n}(\mathbb{R}^{d})=\frac{2}{(d-2)!}.$

Moreover, we have

$\lim_{n\rightarrow+\infty}n^{-\gamma}\frac{(\gamma+\frac{d}{2})_{n}}{(\frac{d}{2})_{n}}=\lim_{n\rightarrow+\infty}n^{-\gamma}\frac{\Gamma(d/2)}{\Gamma(\gamma+d/2)}\frac{\Gamma(d/2+\gamma+n)}{\Gamma(d/2+n)}=\frac{\Gamma(d/2)}{\Gamma(\gamma+d/2)}.$

Consequently, from (3.8), there exists $C=C(d,\gamma)>0$ such that

$\forall\ x\in\mathbb{R}^{d},\quad\forall\ y\in
S^{d-1},\quad|Z_{k,n}(x,y)|\leq Cn^{5d+2\gamma-10}\|x\|^{n}.$ (3.13)

This inequality as well as the fact that $0\leq a_{k,n}(x)<1$ imply that the
series

$\textstyle\sum_{n\geq 0}a_{k,n}(x)Z_{k,n}(x,\xi)$

converges uniformly on $\overline{A}_{\rho,R}\times S^{d-1}$ for every
$R\in(\rho,1)$. Then , by Corollary 3.3 in [11], the function
$\mathbf{P}_{k,1}(.,\xi)$ is $\Delta_{k}$-harmonic on $A$.
ii) For $\varepsilon>0$ small enough and $\xi\in S^{d-1}$, consider the
function

$h_{\varepsilon}(x):=\textstyle\sum_{n\geq
0}a_{k,n}(x)Z_{k,n}((1-\varepsilon)x,\xi).$

As above, from the inequality (3.13) and the homogeneity of $Z_{k,n}(.,\xi)$,
we see that $h_{\varepsilon}$ defines a $\Delta_{k}$-harmonic function in the
annular region $A_{\rho,R}$ with $R=(1-\varepsilon)^{-1}$. Furthermore,
$h_{\varepsilon}=0$ on $S(0,\rho)$ and if $x\in S^{d-1}$, then

$h_{\varepsilon}(x)=\textstyle\sum_{n\geq
0}Z_{k,n}((1-\varepsilon)x,\xi)=P_{k}((1-\varepsilon)x,\xi).$ (3.14)

where $P_{k}$ is the $\Delta_{k}$-Poisson kernel of the unit ball (see [8]).
In particular, $h_{\varepsilon}\geq 0$ on $S^{d-1}$. Consequently, by the weak
minimum principle for $\Delta_{k}$-harmonic functions (see [11] or [21]), we
deduce that

$\forall\ x\in A,\quad h_{\varepsilon}(x)\geq 0.$

On the other hand, for each fixed $(x,\xi)$ in $A\times S^{d-1}$, we have

$\displaystyle|\mathbf{P}_{k,1}(x,\xi)-h_{\varepsilon}(x)|$
$\displaystyle\leq\sum_{n\geq
1}(1-(1-\varepsilon)^{n})a_{k,n}(x)|Z_{k,n}(x,\xi)|$ $\displaystyle\leq
C\sum_{n\geq 1}(1-(1-\varepsilon)^{n})n^{5d+2\gamma-10}\|x\|^{n}.$

Hence, by the monotone convergence theorem we have
$\mathbf{P}_{k,1}(x,\xi)=\lim_{\varepsilon\rightarrow 0}h_{\varepsilon}(x)$.
Finally, we obtain $\mathbf{P}_{k,1}\geq 0$ on $A\times S^{d-1}$.
$\bullet$ For $\xi\in S^{d-1}$ fixed, the function $x\mapsto
P_{k}((1-\varepsilon)x,\xi)-h_{\varepsilon}(x)$ is $\Delta_{k}$-harmonic on
$A$. Moreover, since $P_{k}$ is a nonnegative kernel, we have

$\forall\ x\in S(0,\rho),\quad
P_{k}((1-\varepsilon)x,\xi)-h_{\varepsilon}(x)=P_{k}((1-\varepsilon)x,\xi)\geq
0.$

By (3.14), $x\mapsto P_{k}((1-\varepsilon)x,\xi)-h_{\varepsilon}(x)$ is the
zero function on $S^{d-1}$. So, the weak maximum principle implies that

$\forall\ x\in A,\quad P_{k}((1-\varepsilon)x,\xi)\geq h_{\varepsilon}(x).$

Letting $\varepsilon\longrightarrow 0$, we obtain
$P_{k}(.,\xi)\geq\mathbf{P}_{k,1}(.,\xi)$ on $A$.

iii) The result follows immediately from (2.11). $\square$

###### Proposition 3.3

Let $f$ be a continuous function on $S^{d-1}$. Then the function

$\mathbf{P}_{k,1}[f](x)=\frac{1}{d_{k}}\int_{S^{d-1}}\mathbf{P}_{k,1}(x,\xi)f(\xi)\omega_{k}(\xi)d\sigma(\xi)$
(3.15)

is the unique solution in $\mathcal{C}^{2}(A)\cap\mathcal{C}(\overline{A})$ of
the boundary Dirichlet problem

$\left\\{\begin{array}[]{ll}\Delta_{k}u=0,&\hbox{on}\ A;\\\ u=f,&\hbox{on}\
S^{d-1}\\\ u=0,&\hbox{on}\ S(0,\rho).\end{array}\right.$

_Proof:_ The uniqueness follows from the weak maximum principle for
$\Delta_{k}$-harmonic functions (see [11] or [21]). The inequality (3.13)
allowed us to write for any $x\in A$ that

$\mathbf{P}_{k,1}[f](x)=\sum_{n=0}^{+\infty}u_{n}(x),\quad\text{with}\quad
u_{n}(x)=\frac{a_{k,n}(x)}{d_{k}}\int_{S^{d-1}}Z_{k,n}(x,\xi)f(\xi)\omega_{k}(\xi)d\sigma(\xi).$

By differentiation theorem under integral sign, the functions $u_{n}$ are
$\Delta_{k}$-harmonic on $A$. Moreover, by (3.13) we have

$\forall\ n,\quad|u_{n}(x)|\leq C\|f\|_{\infty}n^{5d+2\gamma-10}\|x\|^{n}.$

This proves that the series $\sum_{n\geq 0}u_{n}$ converges uniformly on each
closed annular region $\overline{A}_{\rho,R}$ whenever $R\in(\rho,1)$. Then,
we conclude that $\mathbf{P}_{k,1}[f]$ is $\Delta_{k}$-harmonic on $A$.
On the other hand, it is easy to see that $\mathbf{P}_{k,1}[f]=0$ on
$S(0,\rho)$.
It remains to prove that for every $\xi\in S^{d-1}$,
$\lim_{x\rightarrow\xi}\mathbf{P}_{k,1}[f](x)=f(\xi)$.
\- If $f\in\mathcal{H}_{\Delta_{k},m}(\mathbb{R}^{d})$, then $u_{n}=0$ if
$n\neq m$ and $u_{m}(x)=a_{k,m}(x)f(x)=\mathbf{P}_{k,1}[f](x)$. Therefore,
$\mathbf{P}_{k,1}[f]=f$ on $S^{d-1}$.
\- If $f\in\mathcal{P}_{m}(\mathbb{R}^{d})$, then by (2.9), there exist
$f_{1},\dots,f_{m}$, with
$f_{j}\in\mathcal{H}_{\Delta_{k},n-2j}(\mathbb{R}^{d})$ such that

$f(x)=\textstyle\sum_{j=0}^{[m/2]}\|x\|^{2j}f_{j}(x).$

This implies that $\mathbf{P}_{k,1}[f]=f$ on $S^{d-1}$.
\- If $f$ is an arbitrary polynomial function, the result also holds.
\- Suppose that $f$ is a continuous function on $S^{d-1}$ and let $p$ be a
polynomial function.
By (3.10) and (3.2) we have

$\displaystyle|\mathbf{P}_{k,1}[f](x)-f(x)|$
$\displaystyle\leq|\mathbf{P}_{k,1}[f](x)-\mathbf{P}_{k,1}[p](x)|+|\mathbf{P}_{k,1}[p](x)-p(x)|+|p(x)-f(x)|$
$\displaystyle\leq 2\|f-p\|_{\infty}+|\mathbf{P}_{k,1}[p](x)-p(x)|.$

This inequality as well as the Stone-Weierstrass theorem show that
$\lim_{x\rightarrow\xi}\mathbf{P}_{k,1}[f](x)=f(\xi)$ for every $\xi\in
S^{d-1}$. This completes the proof. $\square$

Now, for $x\in A$ and $\xi\in S^{d-1}$, consider the functions

$b_{k,n}(x)=\|x\|^{-n}\big{(}\frac{\|x\|}{\rho}\big{)}^{-2\lambda_{k}-n}\frac{1-\|x\|^{2\lambda_{k}+2n}}{1-\rho^{2\lambda_{k}+2n}}=\rho^{-n}(1-a_{k,n}(x))$

and

$\mathbf{P}_{k,2}(x,\xi):=\sum_{n=0}^{+\infty}b_{k,n}(x)Z_{k,n}(x,\xi).$
(3.16)

By means of the Poisson kernel of the unit ball, we can write

$\mathbf{P}_{k,2}(x,\rho\xi)=P_{k}(x,\xi)-\mathbf{P}_{k,1}(x,\xi).$ (3.17)

This relation as well as the properties of $P_{k}$ and $\mathbf{P}_{k,1}$
prove that $\mathbf{P}_{k,2}(.,\rho\xi)$ is a nonnegative
$\Delta_{k}$-harmonic function on $A$ with $\mathbf{P}_{k,2}(.,\rho\xi)=0$ on
$S^{d-1}$.

Let $f$ be a continuous function on $S(0,\rho)$ and define the function

$\mathbf{P}_{k,2}[f](x)=\frac{1}{d_{k}}\int_{S^{d-1}}\mathbf{P}_{k,2}(x,\rho\xi)f(\rho\xi)\omega_{k}(\xi)d\sigma(\xi),\quad
x\in A.$ (3.18)

Using (3.17), we can write

$\displaystyle\mathbf{P}_{k,2}[f](x)$
$\displaystyle=\frac{1}{d_{k}}\int_{S^{d-1}}\big{(}P_{k}(x,\xi)-\mathbf{P}_{k,1}(x,\xi)\big{)}f(\rho\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=P_{k}[\delta_{\rho}.f](x)-\mathbf{P}_{k,1}[\delta_{\rho}.f](x).$

Here, $P_{k}[\phi]$ denotes the Poisson integral of $\phi$ and
$\delta_{\rho}.f(x)=f(\rho{x})$.
Then, using Proposition 3.3 and theorem A in [19], we obtain immediately the
following result:

###### Proposition 3.4

Let $f$ be a continuous function on $S(0,\rho)$. Then $\mathbf{P}_{k,2}[f]$ is
the unique solution in $\mathcal{C}^{2}(A)\cap\mathcal{C}(\overline{A})$ of
the boundary Dirichlet problem

$\left\\{\begin{array}[]{ll}\Delta_{k}u=0,&\hbox{on}\ A;\\\ u=0,&\hbox{on}\
S^{d-1}\\\ u=f,&\hbox{on}\ S(0,\rho).\end{array}\right.$

###### Definition 3.1

Let $f$ be a continuous function on $\partial A$. We define the
$\Delta_{k}$-Poisson integral of $f$ for the annulus $A$ by

$P_{k,A}[f](x):=\mathbf{P}_{k,1}[f](x)+\mathbf{P}_{k,2}[f](x)$ (3.19)

###### Remark 3.1

1.

Obviously $P_{k,A}[1]=1$.

2.

Using (3.11) and a similar relation for the kernel $\mathbf{P}_{k,2}$, we
obtain

$g.P_{k,A}[f]=P_{k,A}[g.f],\quad\text{with}\quad g.f(x):=f(g^{-1}x).$ (3.20)

From Propositions 3.3 and 3.4, we deduce the following main result:

###### Theorem 3.1

Let $f\in\mathcal{C}(\partial{A})$. Then the function $P_{k,A}[f]$ is the
unique solution in $\mathcal{C}^{2}(A)\cap\mathcal{C}(\overline{A})$ of the
boundary Dirichlet problem

$\left\\{\begin{array}[]{ll}\Delta_{k}u=0,&\hbox{on}\ A;\\\ \\\ u=f&\hbox{on}\
\partial A.\end{array}\right.$

From this theorem and the weak maximum principle for $\Delta_{k}$-harmonic
function (see [11]), we obtain the following result:

###### Corollary 3.2

Let $h$ be a $\Delta_{k}$-harmonic function on $A$ and continuous on
$\overline{A}$. Then,

$\forall\ x\in A,\quad h(x)=P_{k,A}[h](x).$

##  4 $\Delta_{k}$-Green function of the annulus

Our aim in this section is to introduce and study the Green function of the
annular region $A=\\{x\in\mathbb{R}^{d},\ \rho<\|x\|<1\\}$ for the Dunkl-
Laplace operator. In the sequel, we will assume that $d+2\gamma>2$ i.e.
$\lambda_{k}>0$ where $\lambda_{k}$ is the constant (2.8).
Let us first recall that the $\Delta_{k}$-Newton kernel, introduced in [13],
is given by

$N_{k}(x,y):=\int_{0}^{+\infty}p_{k}(t,x,y)dt,$ (4.1)

with $p_{k}$ the Dunkl heat kernel (see [21, 24])

$p_{k}(t,x,y)=\frac{1}{(2t)^{d/2+\gamma}c_{k}}\int_{\mathbb{R}^{d}}e^{-(\|x\|^{2}+\|y\|^{2}-2\mathop{\langle
x,z\rangle})/{4t}}d\mu_{y}(z)$ (4.2)

and $c_{k}$ the Macdonald-Mehta constant given by

$c_{k}:=\int_{\mathbb{R}^{d}}\exp(-\frac{\|x\|^{2}}{2})\omega_{k}(x)dx.$

According to [13], the positive and symmetric kernel $N_{k}$ takes the
following form

$N_{k}(x,y)=\frac{1}{2d_{k}\lambda_{k}}\int_{\mathbb{R}^{d}}\big{(}\|x\|^{2}+\|y\|^{2}-2\mathop{\langle
x,z\rangle}\big{)}^{-\lambda_{k}}d\mu_{y}(z).$ (4.3)

Note that if $y=0$, then $\mu_{y}=\delta_{0}$ (with $\delta_{x_{0}}$ the Dirac
measure at $x_{0}\in\mathbb{R}^{d}$) and so

$N_{k}(x,0)=\frac{1}{2d_{k}\lambda_{k}}\|x\|^{-2\lambda_{k}}.$

In addition, for each fixed $x\in\mathbb{R}^{d}$, the function $N_{k}(x,.)$ is
$\Delta_{k}$-harmonic and of class $C^{\infty}$ on $\mathbb{R}^{d}\setminus
W.x$ (where $W.x$ is the $W$-orbit of $x$), $\Delta_{k}$-superharmonic (see
below for precise definition) on the whole space $\mathbb{R}^{d}$ and
satisfies

$-\Delta_{k}[N_{k}(x,.)\omega_{k}]=\delta_{x},\quad\text{in}\quad\mathcal{D}^{\prime}(\mathbb{R}^{d}),$

where
-for a $W$-invariant open set $\Omega\subset\mathbb{R}^{d}$, $\mathcal{D}(\Omega)$ and $\mathcal{D}^{\prime}(\Omega)$ denote respectively the space of $C^{\infty}$-functions on $\Omega$ with compact support and the space of Schwartz distributions on $\Omega$.
-for $f\in L^{1}_{loc}(\Omega,\omega_{k}(x)dx)$, $\Delta_{k}(f\omega_{k})$ is the Schwartz distribution on $\Omega$ defined by

$\mathop{\langle\Delta_{k}(f\omega_{k}),\varphi\rangle}=\mathop{\langle
f\omega_{k},\Delta_{k}\varphi\rangle},\quad\varphi\in\mathcal{D}(\Omega).$

Moreover, for any $x\in\mathbb{R}^{d}$, $N_{k}(x,x)=+\infty$. For more details
about the $\Delta_{k}$-Newton kernel, we refer to ([13], Section 6).
Let $\Omega$ be a $W$-invariant open subset of $\mathbb{R}^{d}$. Recall that a
function $u:\Omega\longrightarrow[-\infty,+\infty[$ is
$\Delta_{k}$-subharmonic if (see [13])

1.

$u$ is upper semi-continuous on $\Omega$,

2.

$u$ is not identically $-\infty$ on each connected component of $\Omega$,

3.

$u$ satisfies the volume sub-mean property i.e. for each closed ball
$\overline{B}(x,r)\subset\Omega$, we have

$u(x)\leq
M_{B}^{r}(u)(x):=\frac{1}{m_{k}[B(0,r)]}\int_{\mathbb{R}^{d}}u(y)h_{k}(r,x,y)\omega_{k}(y)dy.$
(4.4)

Here $m_{k}$ is the measure $dm_{k}(x):=\omega_{k}(x)dx$ and
$y\mapsto{h}_{k}(r,x,y)$ is the nonnegative compactly supported measurable
function given by

$h_{k}(r,x,y):=\int_{\mathbb{R}^{d}}\textbf{1}_{[0,r]}(\sqrt{\|x\|^{2}+\|y\|^{2}-2\mathop{\langle
x,z\rangle}})d\mu_{y}(z).$ (4.5)

We refer to [11] for more details on the kernel $h_{k}$.

The following result gives some useful facts about the Poisson integral of the
$\Delta_{k}$-Newton kernel:

###### Proposition 4.1

i)

For each $x\in A$, the function $P_{k,A}[N_{k}(x,.)]$ is the solution of the
Dirichlet problem

$\left\\{\begin{array}[]{ll}\Delta_{k}u=0,&\hbox{on}\ A;\\\
u=N_{k}(x,.)&\hbox{on}\ \partial A.\end{array}\right.$

ii)

The function $(x,y)\mapsto P_{k,A}[N_{k}(x,.)](y)$ is continuous on
$A\times\overline{A}$.

iii)

For each fixed $y\in A$, the function $x\mapsto P_{k,A}[N_{k}(x,.)](y)$ is
$\Delta_{k}$-harmonic in $A$.

We need the following lemma:

###### Lemma 4.1

The function $(x,y)\mapsto N_{k}(x,y)$ is continuous on
$\mathbb{R}^{d}\times\mathbb{R}^{d}\setminus\\{(x,gx),\quad
x\in\mathbb{R}^{d},g\in W\\}$.

_Proof:_ Using the following inequality (see [24] Lemma 4.2)

$\displaystyle\forall\ t>0,\quad
p_{k}(t,x,y)\leq\frac{1}{c_{k}(2t)^{d/2+\gamma}}\max_{g\in
W}e^{-\|x-gy\|^{2}/{4t}},$

we can apply the Lebesgue dominated convergence theorem in formula (4.1) to
obtain the result of the lemma.

$\square$

_Proof of Proposition 4.1:_ i) If $x\in A$, then the function $N_{k}(x,.)$ is
continuous on $\partial A$ and by Theorem 3.1, we obtain the first assertion.

ii) From the first assertion, for each $x\in A$, $P_{k,A}[N_{k}(x,.)]$ is
extendable to a continuous function on $\overline{A}$ with
$P_{k,A}[N_{k}(x,.)]=N_{k}(x,.)$ on $\partial{A}$.
Let $(x_{0},y_{0})\in A\times\overline{A}$. For every $(x,y)\in
A\times\overline{A}$ we have

$\displaystyle\Big{|}P_{k,A}[N_{k}(x,.)](y)-P_{k,A}[N_{k}(x_{0},.)](y_{0})\Big{|}$
$\displaystyle\leq\Big{|}P_{k,A}[N_{k}(x,.)](y)-P_{k,A}[N_{k}(x_{0},.)](y)\Big{|}$
$\displaystyle+\Big{|}P_{k,A}[N_{k}(x_{0},.)](y)-P_{k,A}[N_{k}(x_{0},.)](y_{0})\Big{|}$
$\displaystyle\leq\mathbf{P}_{k,1}\Big{[}\big{|}K_{x_{0}}(x,.)\big{|}\Big{]}(y)+\mathbf{P}_{k,2}\Big{[}\big{|}K_{x_{0}}(x,.)\big{|}\Big{]}(y)$
$\displaystyle+\Big{|}P_{k,A}[N_{k}(x_{0},.)](y)-P_{k,A}[N_{k}(x_{0},.)](y_{0})\Big{|},$

where $K_{x_{0}}(x,y):=N_{k}(x,y)-N_{k}(x_{0},y)$.
We already know that

$\lim_{y\rightarrow
y_{0}}P_{k,A}[N_{k}(x_{0},.)](y)=P_{k,A}[N_{k}(x_{0},.)](y_{0}).$

Now, let $\varepsilon>0$ and $R>0$ be such that
$\overline{B}(x_{0},R)\subset{A}$. Since $(x,\xi)\longmapsto N_{k}(x,\xi)$ is
uniformly continuous on $\overline{B}(x_{0},R)\times S^{d-1}$, we deduce that
there exists $\eta>0$ such that

$\forall\ (x,\xi)\in B(x_{0},\eta)\times
S^{d-1},\quad|K_{x_{0}}(x,\xi)|=\big{|}N_{k}(x,\xi)-N_{k}(x_{0},\xi)\big{|}<\varepsilon.$

Then, using (3.15) as well as the inequalities (3.2) and (3.10), we get for
every $x\in B(x_{0},\eta)$ and every $y\in A$

$\mathbf{P}_{k,1}\big{[}\big{|}K_{x_{0}}(x,.)\big{|}\big{]}(y)\leq\frac{1}{d_{k}}\int_{S^{d-1}}\mathbf{P}_{k,1}(y,\xi)|K_{x_{0}}(x,\xi)|\omega_{k}(\xi)d\sigma(\xi)\leq\varepsilon.$

The same idea works if we replace the kernel $\mathbf{P}_{k,1}$ by
$\mathbf{P}_{k,2}$. Finally, we obtain

$\lim_{(x,y)\rightarrow(x_{0},y_{0})}P_{k,A}[N_{k}(x,.)](y)=P_{k,A}[N_{k}(x_{0},.)](y_{0}).$

That is the function $(x,y)\mapsto P_{k,A}[N_{k}(x,.)](y)$ is continuous on
$A\times\overline{A}$ as desired.

iii) According to Corollary 4.6 in [12], it is enough to show that the
functions $x\mapsto u_{y}(x):=\textbf{P}_{k,1}[N_{k}(x,.)](y)$ and $x\mapsto
v_{y}(x):=\textbf{P}_{k,2}[N_{k}(x,.)](y)$ satisfy the volume-mean property.
Let then $x_{0}\in A$ and $R>0$ such that $\overline{B}(x_{0},R)\subset A$. As
the kernels $N_{k}$, $h_{k}$ __and $\textbf{P}_{k,1}$ are nonnegative, we can
use Fubini’s theorem to obtain

$\displaystyle
M_{B}^{R}(u_{y})(x_{0})=\frac{1}{d_{k}}\int_{S^{d-1}}\textbf{P}_{k,1}(y,\xi)M_{B}^{R}[N_{k}(.,\xi)](x_{0})\omega_{k}(\xi)d\sigma(\xi)$

But for any $\xi\in S^{d-1}$, the function $N_{k}(.,\xi)$ is
$\Delta_{k}$-harmonic on $A$. Hence, it satisfies the volume-mean property
i.e. $M_{B}^{R}[N_{k}(.,\xi)](x_{0})=N_{k}(x_{0},\xi)$. Therefore, we obtain

$\displaystyle M_{B}^{R}(u_{y})(x_{0})$
$\displaystyle=\frac{1}{d_{k}}\int_{S^{d-1}}\textbf{P}_{k,1}(y,\xi)N_{k}(x_{0},\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\textbf{P}_{k,1}[N_{k}(x_{0},.)](y)=u_{y}(x_{0}).$

We prove similarly that $x\mapsto v_{y}(x):=\textbf{P}_{k,2}[N_{k}(x,.)](y)$
is also a $\Delta_{k}$-harmonic function in $A$. This shows the desired
result.

$\square$

###### Definition 4.1

For $x\in A$, the function $G_{k,A}(x,.)$ defined by

$G_{k,A}(x,y):=N_{k}(x,y)-P_{k,A}[N_{k}(x,.)](y),\quad y\in A,$ (4.6)

is called the $\Delta_{k}$-Green function of $A$ with pole $x$.

The $\Delta_{k}$-Green function $G_{k,A}$ has the following properties:

###### Proposition 4.2

Let $x\in A$. Then

1.

The function $G_{k,A}(x,.)$ is $\Delta_{k}$-harmonic on $A\setminus W.x$, is
$\Delta_{k}$-superharmonic on $A$ and satisfies

$-\Delta_{k}[G_{k,A}(x,.)\omega_{k}]=\delta_{x}\quad\text{in}\quad\mathcal{D}^{\prime}(A).$
(4.7)

2.

$G_{k,A}(x,x)=+\infty$ and $G_{k,A}(x,y)<+\infty$ whenever $y\notin W.x$ .

3.

For every $\xi\in\partial{A}$, $\lim_{y\rightarrow\xi}G_{k,A}(x,y)=0$.

4.

For every $y\in A$, $G_{k,A}(x,y)>0$.

5.

For every $x,y\in A$, $G_{k,A}(x,y)=G_{k,A}(y,x)$.

6.

For every $x,y\in A$ and $g\in W$, $G_{k,A}(gx,gy)=G_{k,A}(x,y)$.

7.

The zero function is the greatest $\Delta_{k}$-subharmonic minorant of
$G_{k,A}(x,.)$ on $A$.

8.

The function $(x,y)\mapsto G_{k,A}(x,y)$ is continuous on
$A\times\overline{A}\setminus\\{(x,gx):\ x\in A,g\in W\\}$.

_Proof:_ The first and the second assertions follow from the properties of the
$\Delta_{k}$-Newton kernel previously mentioned. In addition, by Proposition
4.1, we easily obtain the third statement.

4) As $\lim_{y\rightarrow\xi\in\partial{A}}G_{k,A}(x,y)=0$, the weak minimum
principle for $\Delta_{k}$-superharmonic functions (see [13], Theorem 3.1)
implies that $G_{k,A}(x,.)\geq 0$ on $A$.
If $G_{k,A}(x,y_{0})=0$ for some $y_{0}\in A$, it follows from the strong
maximum principle (see [13]) that $G_{k,A}(x,.)$ is the zero function on $A$
which is impossible because $G_{k,A}(x,x)=+\infty$. Thus, $G_{k,A}$ is a
positive kernel on $A\times{A}$.

5) Since $N_{k}$ is a symmetric kernel, we have to prove that

$\forall\ x,y\in A,\quad P_{k,A}[N_{k}(x,.)](y)=P_{k,A}[N_{k}(y,.)](x).$

Let $y\in A$ and consider the function

$\displaystyle H_{y}(x):=P_{k,A}[N_{k}(x,.)](y)-P_{k,A}[N_{k}(y,.)](x).$

From Proposition 4.1 , i) and iii), $H_{y}$ is a $\Delta_{k}$-harmonic
function in $A$. On the other hand, writing

$H_{y}(x)=N_{k}(x,y)-G_{k,A}(x,y)-P_{k,A}[N_{k}(y,.)](x)$

and using the positivity of $G_{k,A}$ as well as the symmetry property of the
kernel $N_{k}$, we conclude that

$\limsup_{x\mapsto\xi\in\partial{A}}H_{y}(x)\leq N_{k}(\xi,y)-N_{k}(y,\xi)=0.$

Then the weak maximum principle yields that $H_{y}\leq 0$ on $A$. That is, we
have

$\forall\ x,y\in A,\quad P_{k,A}[N_{k}(x,.)](y)\leq P_{k,A}[N_{k}(y,.)](x).$

By interchanging the role of $x$ and $y$, we also get the reverse inequality.
Finally, we obtain the desired equality.

6) The result follows immediately from (3.20) and from the relation
$N_{k}(gx,gy)=N_{k}(x,y)$, $x,y\in\mathbb{R}^{d}$, $g\in W$ (see [13]).

7) As $G_{k,A}(x,.)$ is positive on $A$, we know that the zero function is a
$\Delta_{k}$-subharmonic minorant of $G_{k,A}(x,.)$.
Now, let $s$ be a $\Delta_{k}$-subharmonic function on $A$ such that $s\leq
G_{k,A}(x,.)$ on $A$. Using the statement 3), we obtain
$\limsup_{z\rightarrow\xi\in\partial A}s(z)\leq 0$. Thus the weak maximum
principle for $\Delta_{k}$-subharmonic functions (see [13]) yields that $s\leq
0$ on $A$.

8) The result follows immediately from the statement ii) of Proposition 4.1
and Lemma 4.1. $\square$

In the following result, we will express the Green function $G_{k,A}$ in terms
of the $\Delta_{k}$-spherical harmonics. More precisely, we have

###### Theorem 4.1

The $\Delta_{k}$-Green function in $A$ is given by

$G_{k,A}(x,y)=N_{k}(x,y)-\sum_{n=0}^{+\infty}\frac{a_{k,n}(y)\|x\|^{n}+b_{k,n}(y)\|x\|^{-n-2\lambda_{k}}\rho^{n}}{d_{k}(2\lambda_{k}+2n)}Z_{k,n}\big{(}\frac{x}{\|x\|},y\big{)}.$
(4.8)

We need the following result:

###### Proposition 4.3

For $x,y\in\mathbb{R}^{d}$ such that $\|y\|<\|x\|$, we have

$N_{k}(x,y)=\sum_{n=0}^{+\infty}\frac{\|x\|^{-2\lambda_{k}}}{d_{k}(2\lambda_{k}+2n)}\|y\|^{n}\|x\|^{-n}Z_{k,n}\big{(}\frac{x}{\|x\|},\frac{y}{\|y\|}\big{)}.$
(4.9)

_Proof:_ Let $\|y\|<\|x\|$. From (4.3), we have

$\displaystyle N_{k}(x,y)$
$\displaystyle=\frac{\|x\|^{-2\lambda_{k}}}{2d_{k}\lambda_{k}}\int_{\mathbb{R}^{d}}\big{(}1-\frac{2\mathop{\langle
x,z\rangle}}{\|x\|^{2}}+\frac{\|y\|^{2}}{\|x\|^{2}}\big{)}^{-\lambda_{k}}d\mu_{y}(z)$
$\displaystyle=\frac{\|x\|^{-2\lambda_{k}}}{2d_{k}\lambda_{k}}\int_{\mathbb{R}^{d}}\sum_{n=0}^{+\infty}\|y\|^{n}\|x\|^{-n}\frac{(2\lambda_{k})_{n}}{n!}P_{n}^{\lambda_{k}}\Big{(}\mathop{\langle\frac{x}{\|x\|},\frac{z}{\|y\|}\rangle}\Big{)}d\mu_{y}(z)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{\|x\|^{-2\lambda_{k}}}{d_{k}(2\lambda_{k}+2n)}\|y\|^{n}\|x\|^{-n}Z_{k,n}\big{(}\frac{x}{\|x\|},\frac{y}{\|y\|}\big{)};$

where in the second line, we have used the relation (2.3) and the generating
relation (see for example [8], p. 18)

$(1-2ar+r^{2})^{-\mu}=\sum_{n=0}^{+\infty}\frac{(2\mu)_{n}}{n!}P_{n}^{\mu}(a)r^{n},\quad\mu>0,\quad|r|<1,\
|a|\leq 1$

and in the last line, we have used
\- the inequality $\underset{x\in[-1,1]}{\sup}|P_{n}^{\mu}(x)|\leq
P_{n}^{\mu}(1)$ (see [25], Theorem 7.32.1) and the above generation relation
with $a=1$ which allows us to permute the symbols $\sum$ and $\int$,
\- the fact that $\mu_{\frac{y}{\|y\|}}$ is the image measure of $\mu_{y}$ by
the dilation $\xi\mapsto\frac{\xi}{\|y\|}$
\- the relations (2.2) and (2.13). $\square$

_Proof of Theorem 4.1:_ By Theorem 3.1, we have

$\displaystyle
P_{k,A}[N_{k}(x,.)](y)=\mathbf{P}_{k,1}[N_{k}(x,.)](y)+\mathbf{P}_{k,2}[N_{k}(x,.)](y):=I_{1}+I_{2}.$

$\bullet$ We have

$\displaystyle I_{1}$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{a_{k,n}(y)}{d_{k}}\int_{S^{d-1}}Z_{k,n}(y,\xi)N_{k}(x,\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{a_{k,n}(y)}{d_{k}}\int_{S^{d-1}}\sum_{m=0}^{+\infty}\frac{\|x\|^{m}}{d_{k}(2\lambda_{k}+2m)}Z_{k,m}\big{(}\frac{x}{\|x\|},\xi\big{)}Z_{k,n}(y,\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\sum_{n=0}^{+\infty}a_{k,n}(y)\sum_{m=0}^{+\infty}\frac{\|x\|^{m}}{d_{k}(2\lambda_{k}+2m)}\frac{1}{d_{k}}\int_{S^{d-1}}Z_{k,m}\big{(}\frac{x}{\|x\|},\xi\big{)}Z_{k,n}(y,\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{a_{k,n}(y)\|x\|^{n}}{d_{k}(2\lambda_{k}+2n)}Z_{k,n}\big{(}\frac{x}{\|x\|},y\big{)},$

where, we have used
-the relation (4.9) in the second line;
-the inequalities (3.13) and $\|x\|<1$ in order to interchange the symbols $\int$ and $\sum$ in the third line;
-the fact that $\mathcal{H}_{\Delta_{k},n}(\mathbb{R}^{d})\perp\mathcal{H}_{\Delta_{k},m}(\mathbb{R}^{d})$ whenever $m\neq n$ and the reproducing formula (2.10) in the last line.

Notice that if $\|x\|>\rho$, then (4.9) yields that

$\forall\ \xi\in S^{d-1},\quad
N_{k}(x,\rho\xi)=\sum_{n=0}^{+\infty}\frac{\|x\|^{-2\lambda_{k}}}{d_{k}(2\lambda_{k}+2n)}\rho^{n}\|x\|^{-n}Z_{k,n}\big{(}\frac{x}{\|x\|},\xi\big{)}.$

Hence we obtain similarly

$\displaystyle I_{2}$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{b_{k,n}(y)}{d_{k}}\int_{S^{d-1}}Z_{k,n}(y,\xi)N_{k}(x,\rho\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{b_{k,n}(y)}{d_{k}}\int_{S^{d-1}}\sum_{m=0}^{+\infty}\frac{\|x\|^{-2\lambda_{k}-m}\rho^{m}}{d_{k}(2\lambda_{k}+2m)}Z_{k,m}\big{(}\frac{x}{\|x\|},\xi\big{)}Z_{k,n}(y,\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\sum_{n=0}^{+\infty}b_{k,n}(y)\sum_{m=0}^{+\infty}\frac{\|x\|^{-2\lambda_{k}-m}\rho^{m}}{d_{k}(2\lambda_{k}+2m)}\frac{1}{d_{k}}\int_{S^{d-1}}Z_{k,m}\big{(}\frac{x}{\|x\|},\xi\big{)}Z_{k,n}(y,\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\sum_{n=0}^{+\infty}\frac{b_{k,n}(y)\|x\|^{-2\lambda_{k}-n}\rho^{n}}{d_{k}(2\lambda_{k}+2n)}Z_{k,n}\big{(}\frac{x}{\|x\|},y\big{)}.$

This gives the desired formula (4.8). $\square$

###### Remark 4.1

Using (4.9) and replacing the functions $a_{k,n}$ and $b_{k,n}$ by their
expressions, if $x,y\in A$ with $\|y\|<\|x\|$ we can write

$\displaystyle
G_{k,A}(x,y)=\sum_{n=0}^{+\infty}\frac{\Big{(}\|y\|^{2\lambda_{k}+2n}-\rho^{2\lambda_{k}+2n}\Big{)}(1-\|x\|^{2\lambda_{k}+2n}\Big{)}}{d_{k}(2\lambda_{k}+2n)(1-\rho^{2\lambda_{k}+2n})(\|x\|\|y\|)^{2\lambda_{k}+n}}Z_{k,n}\big{(}\frac{x}{\|x\|},\frac{y}{\|y\|}\big{)}.$

This formula generalizes the classical case (if $k=0$, $2\lambda_{0}=d-2$)
proved in [16].

##  5 Applications

###  5.1 Poisson-Jensen formula for $\Delta_{k}$-subharmonic functions

Our goal now is to prove an analogue of the Poisson-Jensen formula for
$\Delta_{k}$-subharmonic functions on $\Omega\supset\overline{A}$. Note that a
Poisson-Jensen formula has been proved when $u$ is a
$C^{2}-\Delta_{k}$-subharmonic function on $\Omega$ which contains the closed
unit ball (see [15]).

###### Theorem 5.1

Let $u$ be a $\Delta_{k}$-subharmonic function on a $W$-invariant open set
$\Omega\supset\overline{A}$. Then,

$u(x)=P_{k,A}[u](x)-\int_{A}G_{k,A}(x,y)d\nu_{u}(y),\quad x\in A,$ (5.1)

where $\nu_{u}:=\Delta_{k}(u\omega_{k})$ is the $\Delta_{k}$-Riesz measure of
$u$ (see [13]).

_Proof:_ Let $O$ be a bounded $W$-invariant open set such that
$\overline{A}\subset O\subset\overline{O}\subset\Omega$ . Using the Riesz
decomposition theorem for $\Delta_{k}$-subharmonic functions (see [13]), we
deduce that there exists a $\Delta_{k}$-harmonic function $h$ on $O$ such that

$\forall\ x\in O,\quad u(x)=h(x)-\int_{O}N_{k}(x,y)d\nu_{u}(y):=h(x)-s(x).$

Then, we have

$\forall\ x\in A,\quad P_{k,A}[u](x)=P_{k,A}[h](x)-P_{k,A}[s](x).$

From Corollary 3.2, we have $P_{k,A}[h]=h$ on $A$. Moreover, for $x\in A$, we
have

$P_{k,A}[s](x)=\mathbf{P}_{k,1}[s](x)+\mathbf{P}_{k,2}[s](x).$

The crucial part here is to show that

$\forall\ x\notin A,\quad P_{k,A}[N_{k}(x,.)]=N_{k}(x,.)\quad\text{on}\ A.$
(5.2)

Assume this relation for the moment. By Fubini’s theorem, we have

$\displaystyle\mathbf{P}_{k,1}[s](x)$
$\displaystyle:=\frac{1}{d_{k}}\int_{S^{d-1}}\mathbf{P}_{k,1}(x,\xi)s(\xi)\omega_{k}(\xi)d\sigma(\xi)$
$\displaystyle=\frac{1}{d_{k}}\int_{O}\int_{S^{d-1}}\mathbf{P}_{k,1}(x,\xi)N_{k}(\xi,y)\omega_{k}(\xi)d\sigma(\xi)d\nu_{u}(y)$
$\displaystyle=\int_{O}\mathbf{P}_{k,1}[N_{k}(y,.)](x)d\nu_{u}(y).$

By the same way, we also have

$\mathbf{P}_{k,2}[s](x)=\int_{O}\mathbf{P}_{k,2}[N_{k}(y,.)](x)d\nu_{u}(y).$

The above relations as well as (5.2) imply that

$\displaystyle P_{k,A}[s](x)$
$\displaystyle=\int_{O}P_{k,A}[N_{k}(y,.)](x)d\nu_{u}(y)$
$\displaystyle=\int_{A}P_{k,A}[N_{k}(y,.)](x)d\nu_{u}(y)+\int_{O\setminus
A}P_{k,A}[N_{k}(y,.)](x)d\nu_{u}(y)$
$\displaystyle=\int_{A}\big{(}N_{k}(x,y)-G_{k,A}(x,y)\big{)}d\nu_{u}(y)+\int_{O\setminus
A}N_{k}(x,y)d\nu_{u}(y)$
$\displaystyle=\int_{O}N_{k}(x,y)d\nu_{u}(y)-\int_{A}G_{k,A}(x,y)d\nu_{u}(y).$

This implies the desired Poisson-Jensen formula. Now, it remains to prove
(5.2). We will distinguish three cases.

_First case:_ $x\notin\overline{A}$. As $N_{k}(x,.)$ is $\Delta_{k}$-harmonic
on $A$ and continuous on $\overline{A}$, we deduce by Corollary 3.2 that
$P_{k,A}[N_{k}(x,.)]=N_{k}(x,.)$ on $A$.

_Second case:_ $x\in S^{d-1}$. For $\varepsilon>0$ small enough, the function
$N_{k}\big{(}(1+\varepsilon)x,.\big{)}$ is $\Delta_{k}$-harmonic in the open
ball $B(0,1+\varepsilon)\supset\overline{A}$. Therefore, again by Corollary
3.2, we obtain

$\forall\ y\in A,\quad
N_{k}\big{(}(1+\varepsilon)x,y\big{)}=P_{k,A}[N_{k}\big{(}(1+\varepsilon)x,.\big{)}](y).$
(5.3)

Clearly we have $\lim_{\varepsilon\rightarrow
0}N_{k}\big{(}(1+\varepsilon)x,y\big{)}=N_{k}(x,y)$ for every fixed $y\in A$.
Moreover, using (4.3) and the fact that supp$\
\mu_{y}\subset\overline{B}(0,\|y\|)$ we can see that

$N_{k}\big{(}(1+\varepsilon)x,y\big{)}\leq
N_{k}(x,y),\quad\text{whenever}\quad\|y\|\leq\|x\|.$

Consequently, we can use the Lebesgue dominated convergence theorem to obtain

$\forall\ y\in A,\quad\lim_{\varepsilon\rightarrow
0}P_{k,A}[N_{k}\big{(}(1+\varepsilon)x,.\big{)}](y)=P_{k,A}[N_{k}(x,.)](y).$

Hence, letting $\varepsilon\longrightarrow 0$ in the relation (5.3), we get
the result in this case.

_Third case:_ $x\in S(0,\rho)$. Let $0<\varepsilon<1/2$. In this case, the
function $N_{k}\big{(}(1-\varepsilon)x,.\big{)}$ is $\Delta_{k}$-harmonic in
$\mathbb{R}^{d}\setminus\overline{B}\big{(}0,(1-\varepsilon)\rho\big{)}\supset\overline{A}$
and then from Corollary 3.2 we deduce that

$\forall\ y\in A,\quad
N_{k}\big{(}(1-\varepsilon)x,y\big{)}=P_{k,A}[N_{k}\big{(}(1-\varepsilon)x,.\big{)}](y).$

Note that from (2.3), we can write

$N_{k}(x,y)=\frac{1}{2d_{k}\lambda_{k}}\int_{\mathbb{R}^{d}}\Big{(}\sum_{g\in
W}\lambda_{g}(z)\|x-gy\|^{2}\Big{)}^{-\lambda_{k}}d\mu_{y}(z),$

where for every $z\in\text{supp}\ \mu_{y}$, the nonnegative numbers
$\lambda_{g}(z)$ are such that $\sum_{g\in W}\lambda_{g}(z)=1$. Using the
above relation we easily see that

$N_{k}\big{(}(1-\varepsilon)x,y\big{)}\leq
2^{2\lambda_{k}}N_{k}(x,y),\quad\text{whenever}\quad\|x\|\leq\|y\|,$

Finally by same way as in the second case we obtain the result.

$\square$

###  5.2 Positive solution of $\Delta_{k}$-nonlinear elliptic problem on the
annulus

In this section, we will investigate the positive continuous solutions of the
semilinear problem

$\Delta_{k}(u\omega_{k})=\phi(.,u)\omega_{k}\quad\text{in}\quad\mathcal{D}^{\prime}(A),$

in the sense that

$\forall\
\varphi\in\mathcal{D}(A),\quad\int_{A}u(x)\Delta_{k}\varphi(x)\omega_{k}(x)dx=\int_{A}\varphi(x)\phi(x,u(x))\omega_{k}(x)dx.$

We assume that $\phi(x,u(x))=\phi_{1}(x)\phi_{2}(u(x))$ where
$\bullet$ $\phi_{1}$ is a nonnegative bounded measurable function on $A$.
$\bullet$ $\phi_{2}$ is a nonnegative and nondecreasing continuous function on
$[0,+\infty[$ with $\phi_{2}(0)=0$.

In [3], by using some tools from probabilistic potential theory, the authors
have studied the positive solution on the unit ball $B$ of the semilinear
problem

$\Delta_{k}(u)=\varphi(u)\quad\text{in}\
\mathcal{D}^{\prime}(B)\quad\text{and}\quad u=f\quad\text{on}\ \partial{B}.$

Let us denote by $\mathcal{C}^{+}(\overline{A})$ the convex cone of
nonnegative and continuous functions on $A$.

###### Theorem 5.2

Let $\phi$ be as above. Then, for every $f\in\mathcal{C}^{+}(\partial{A})$,
the semilinear Dirichlet problem

$\left\\{\begin{array}[]{ll}\Delta_{k}(u\omega_{k})=\phi(.,u)\omega_{k},&\mbox{
in }\ \mathcal{D}^{\prime}(A)\\\ \\\ u=f,&\mbox{ on }\ \partial
A\end{array}\right.$ (5.4)

admits one and only one solution $u\in\mathcal{C}^{+}(\overline{A})$.
Furthermore, we have

$\forall\ x\in A,\quad
u(x)+\int_{A}G_{k,A}(x,y)\phi(y,u(y))\omega_{k}(y)dy=P_{k,A}[f](x).$

We begin by showing the uniqueness of the solution. This fact follows
immediately from the following maximum principle type result:

###### Lemma 5.1

Let $u,v\in\mathcal{C}(A)$ and let $\phi$ be a function satisfying the above
conditions. If

$\left\\{\begin{array}[]{ll}\Delta_{k}(u\omega_{k})-\phi(.,u)\omega_{k}\leq\Delta_{k}(v\omega_{k})-\phi(.,v)\omega_{k},&\hbox{in}\
\mathcal{D}^{\prime}(A),\\\ \\\ \limsup\limits_{x\to
y\in\partial{A}}(v-u)(x)\leq 0,&\end{array}\right.$

then $v\leq u$ in $A$.

_Proof:_ Let $U$ be the upper semi-continuous function defined by

$U(x)=\left\\{\begin{array}[]{ll}v(x)-u(x),&\hbox{if}\ x\in A;\\\
\limsup\limits_{y\to x\in\partial{A}}(v-u)(y),&\hbox{if}\
x\in\partial{A}\end{array}\right.$

and $x_{0}\in\overline{A}$ be such that $U(x_{0})=\max_{\overline{A}}U$.
We have to prove that $U(x_{0})\leq 0$. We suppose the contrary i.e.
$U(x_{0})>0$. As $U\leq 0$ on $\partial{A}$, this implies that
$x_{0}\notin\partial{A}$.
Let $O$ be the nonempty open set given by

$O:=\\{x\in A:\quad U(x)>0\\}$

and $\Omega$ be the connected component of $x_{0}$ in $O$ which is also an
open set of $\mathbb{R}^{d}$.
To get a contradiction, we claim that it is suffices to establish that

$U=U(x_{0})\quad\text{on}\quad\Omega.$ (5.5)

Indeed,
$\bullet$ If $\Omega=O$ (i.e. $O$ is connected), then (5.5) holds on
$O=\Omega$. But

$\partial{O}\subset\partial{A}\cup(A\setminus{O})=\\{x\in\overline{A},\
U(x)\leq 0\\}.$

Consequently, using the fact that $U$ is upper semi-continuous and (5.5), we
get

$\forall\ x\in\partial{O},\quad
U(x)=\limsup_{y\rightarrow{x},y\in{O}}U(y)=U(x_{0}).$

Thus, we obtain a contradiction.
$\bullet$ If $\Omega\neq O$, then as $\Omega$ is a connected component of $O$
we have $\partial{\Omega}\cap O=\emptyset$. Therefore, we have
$\partial{\Omega}\subset\partial{A}\cup A\setminus{O}$ and as above we get a
contradiction.

Now, our aim is to prove that $U=U(x_{0})$ on $\Omega$. For this, we introduce
the nonempty closed set

$\Omega_{0}:=\\{x\in\Omega:\quad U(x)=U(x_{0})\\}.$

Note that

$\Omega=\big{(}\Omega\cap\cup_{\alpha\in\mathcal{R}}H_{\alpha}\big{)}\cup\big{(}\Omega\setminus\cup_{\alpha\in\mathcal{R}}H_{\alpha}\big{)}=\big{(}\Omega\cap\cup_{\alpha\in\mathcal{R}}H_{\alpha}\big{)}\cup\big{(}\cup_{g\in
W}\Omega\cap{g}.\mathbf{C}\big{)},$

where $\mathbf{C}$ is a fixed Weyl chamber and $g\mathbf{C}$, $g\in W$, are
the connected components of
$\mathbb{R}^{d}\setminus\cup_{\alpha\in\mathcal{R}}H_{\alpha}$.
Fix $\xi\in\Omega_{0}$ and $R>0$ such that the open ball $B(\xi,R)$ is
contained in $\Omega$. We will distinguish three possible locations of $\xi$
depending on the sets $E_{1}:=\\{\alpha\in\mathcal{R},\quad
U(\sigma_{\alpha}\xi)=U(\xi)\\}$ and
$E_{2}:=\\{\alpha\in\mathcal{R},\quad\xi\in H_{\alpha}\\}\subset E_{1}$.

First case: $E_{1}=\mathcal{R}$. This implies that $U(g\xi)=U(\xi)>0$ for all
$g\in W$. Moreover, clearly there exists $r\in(0,R]$ such that

$\forall\ g\in W,\ \forall\ x\in B(g\xi,r),\quad U(x)\geq 0.$ (5.6)

Consider the $W$-invariant continuous function $U^{W}$ defined on $A$ by

$U^{W}(x):=\frac{1}{|W|}\sum_{g\in W}g.U(x)=\frac{1}{|W|}\sum_{g\in
W}U(g^{-1}x).$

We easily see that $U^{W}$ has a maximum at the point $\xi$ with
$U^{W}(\xi)=U(\xi)=U(x_{0})$. Furthermore, using the $W$-invariance property
of $\Delta_{k}$ (i.e. $g\circ\Delta_{k}=\Delta_{k}\circ g$) and the hypothesis
of the lemma, we obtain

$\displaystyle\Delta_{k}(U^{W}\omega_{k})$
$\displaystyle=\frac{1}{|W|}\sum_{g\in
W}g.[\Delta_{k}(U\omega_{k})]\geq\frac{1}{|W|}\sum_{g\in
W}g.[\big{(}\phi(.,v)-\phi(.,u)\big{)}\omega_{k}]\quad\text{in}\quad\mathcal{D}^{\prime}(A).$

Now, since $U\geq 0$ on $B^{W}(\xi,r)=\cup_{g\in W}B(g\xi,r)$ (from (5.6)) and
$\phi_{2}$ is nondecreasing, we deduce that

$\Delta_{k}(U^{W}\omega_{k})\geq
0\quad\text{in}\quad\mathcal{D}^{\prime}(B^{W}(\xi,r)).$

That is $U$ is weakly $\Delta_{k}$-subharmonic on $B^{W}(\xi,r)$. But the
continuity of $U$ and the Weyl lemma for $\Delta_{k}$-subharmonic functions
(see [13], Theorem 5.2) imply that $U^{W}$ is strongly
$\Delta_{k}$-subharmonic on the open $W$-invariant set $B^{W}(\xi,r)$.
Now, if we follow the proof of the strong maximum principle in [13] for the
$W$-invariant $\Delta_{k}$-subharmonic function $U^{W}$, then we conclude that

$U=U(\xi)=U(x_{0}),\quad\text{on}\quad B(\xi,r).$

Hence, we have $B(\xi,r)\subset\Omega_{0}$.

Second case: $E_{1}\neq\mathcal{R}$ and $E_{2}=\emptyset$ i.e.
$\xi\notin\cup_{\alpha\in\mathcal{R}}H_{\alpha}$. So there is a unique
$g_{0}\in W$ such that $\xi\in\Omega\cap g_{0}\mathbf{C}$. Clearly, we can
suppose that $B(\xi,R)\subset\Omega\cap g_{0}\mathbf{C}$.
Let $U^{W}$ be the W-invariant continuous function defined on
$B^{W}(\xi,R):=\cup_{g\in W}B(g\xi,R)$ by

$U^{W}(x)=g.U(x):=U({g}^{-1}.x)\quad\text{whenever}\quad x\in B(g\xi,R).$

We are going to establish that the function $U^{W}$ is
$\Delta_{k}$-subharmonic on $B^{W}(\xi,r)$ for some $r>0$ will be chosen
later. Again from the continuity of $U^{W}$ and the Weyl lemma, it is enough
to show that $U^{W}$ is $\Delta_{k}$-subharmonic in sense of distributions.
$\bullet$ Firstly, we have the following decomposition

$U^{W}=\sum_{i=1}^{n}U^{W}\textbf{1}_{B(g_{i}\xi,R)}=\sum_{i=1}^{n}g_{i}.[U\textbf{1}_{B(\xi,R)}],$

where $g_{1}=id,g_{2},\dots,g_{n}\in W$ are such that
$\textbf{1}_{B^{W}(\xi,R)}=\sum_{i=1}^{n}\textbf{1}_{B(g_{i}\xi,R)}$.
$\bullet$ Secondly, for $f\in\mathcal{C}^{2}(B^{W}(\xi,R))$, we can write
$\Delta_{k}=L_{k}-A_{k}$ where

$L_{k}f(x)=\Delta
f(x)+2\sum_{\alpha\in\mathcal{R}_{+}}k(\alpha)\frac{\mathop{\langle\nabla
f(x),\alpha\rangle}}{\mathop{\langle\alpha,x\rangle}}$

and

$A_{k}f(x)=\sum_{\alpha\in\mathcal{R}_{+}}k(\alpha)\|\alpha\|^{2}\frac{f(x)-f(\sigma_{\alpha}(x))}{\mathop{\langle\alpha,x\rangle}^{2}}.$

$\bullet$ For $\varphi\in\mathcal{D}(B^{W}(\xi,R))$ nonnegative, we have

$\displaystyle\mathop{\langle\Delta_{k}(U^{W}\omega_{k}),\varphi\rangle}$
$\displaystyle=\sum_{i=1}^{n}\mathop{\langle\Delta_{k}\big{(}g_{i}.[U\textbf{1}_{B(\xi,R)}]\omega_{k}\big{)},\varphi\rangle}=\sum_{i=1}^{n}\mathop{\langle
g_{i}.[U\omega_{k}\textbf{1}_{B(\xi,R)}],\Delta_{k}\varphi\rangle}$
$\displaystyle=\sum_{i=1}^{n}\mathop{\langle
g_{i}.[U\omega_{k}\textbf{1}_{B(\xi,R)}],L_{k}\varphi-A_{k}\varphi\rangle}$
$\displaystyle=\sum_{i=1}^{n}\mathop{\langle
U\omega_{k},(L_{k}[g_{i}^{-1}.\varphi])\textbf{1}_{B(\xi,R)}\rangle}-\mathop{\langle\sum_{i=1}^{n}g_{i}.[U\textbf{1}_{B(\xi,R)}]\omega_{k},A_{k}(\varphi)\rangle}$
$\displaystyle=\sum_{i=1}^{n}\mathop{\langle
U\omega_{k},L_{k}\big{(}[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\big{)}\rangle}-\underbrace{\mathop{\langle
U^{W}\omega_{k},A_{k}(\varphi)\rangle}}_{=0}$
$\displaystyle=\sum_{i=1}^{n}\mathop{\langle
U\omega_{k},\Delta_{k}\big{(}[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\big{)}+A_{k}([g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\big{)}\rangle}$
$\displaystyle=\sum_{i=1}^{n}\mathop{\langle\Delta_{k}(U\omega_{k}),[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\rangle}+\sum_{i=1}^{n}\mathop{\langle
U\omega_{k},A_{k}\big{(}[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\big{)}\rangle}$
$\displaystyle\geq\sum_{i=1}^{n}\mathop{\langle[\phi(.,v)-\phi(.,u)]\omega_{k},[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\rangle}+\sum_{i=1}^{n}\mathop{\langle
U\omega_{k},A_{k}\big{(}[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\big{)}\rangle}$
$\displaystyle\geq\sum_{i=1}^{n}\mathop{\langle
U\omega_{k},A_{k}\big{(}[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\big{)}\rangle},$

where we have used
\- the fact that $L_{k}$ commutes with the $W$-action i.e. $L_{k}\circ
g=g\circ L_{k}$, $g\in W$, in the third line and the fact that it preserves
the support in the forth line,
\- the $W$-invariance property of $U^{W}$ which implies that $\mathop{\langle
U^{W}\omega_{k},A_{k}(\varphi)\rangle}=0$ in the forth line,
\- the decomposition $L_{k}=\Delta_{k}+A_{k}$ in the fifth line,
\- the fact that
$[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\in\mathcal{D}(B(\xi,R))$ in the
sixth line,
\- the hypothesis of the lemma in the seventh line,
\- the nondecreasing property of $\phi_{2}$ in the last line.
$\bullet$ As $A_{k}$ is a symmetric operator in the sense that
$\mathop{\langle A_{k}(f)\omega_{k},\psi\rangle}=\mathop{\langle
f\omega_{k},A_{k}(\psi)\rangle}$, it yields that

$\mathop{\langle\Delta_{k}(U^{W}\omega_{k}),\varphi\rangle}\geq\sum_{i=1}^{n}\mathop{\langle
A_{k}(U)\omega_{k},[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\rangle},$

Clearly $A_{k}(U)(\xi)\geq 0$. But, since $E_{1}\neq\mathcal{R}$, we must have
$A_{k}(U)(\xi)\geq 0$. Hence, there exists $r>0$ such that $A_{k}(U)>0$ on
$B(\xi,r)$.
Thus, $\Delta_{k}(U^{W}\omega_{k})\geq 0$ in
$\mathcal{D}^{\prime}(B^{W}(\xi,r))$ i.e. $U^{W}$ is weakly
$\Delta_{k}$-subharmonic on $B^{W}(\xi,r)$ as desired.
Now, again, if we follow the proof of the strong maximum principle in [13] for
the $W$-invariant $\Delta_{k}$-subharmonic function $U^{W}$, then we conclude
that

$U=U(\xi)=U(x_{0})\quad\text{on}\quad B(\xi,r).$

That is $B(\xi,r)\subset\Omega_{0}$.

Third case: $E_{1}\neq\mathcal{R}$ and $E_{2}\neq\emptyset$. Let
$W^{\prime}\subsetneq W$ be the isotropy group of $\xi$. Here, we choose $R>0$
under the further following assumption

$3R\leq\min_{\widetilde{g}\in\nicefrac{{W}}{{W^{\prime}}},\ g\neq
id}\|\xi-g\xi\|,\quad\text{with}\quad\nicefrac{{W}}{{W^{\prime}}}:=\\{\widetilde{g}=gW^{\prime},\
g\in W\\}.$

Let $S$ be the $W^{\prime}$-invariant continuous function defined on $A$ by
$S=\frac{1}{|W^{\prime}|}\sum_{g^{\prime}\in W^{\prime}}g^{\prime}.U$.
$\bullet$ Clearly, $S$ has a maximum at the point $\xi$ with
$S(\xi)=U(\xi)=U(x_{0})$. Furthermore, using the $W$-invariance property of
$\Delta_{k}$ as well as the hypothesis the lemma , we get

$\Delta_{k}(S\omega_{k})\geq\frac{1}{|W^{\prime}|}\sum_{g^{\prime}\in
W^{\prime}}g^{\prime}.(\phi(.,v)-\phi(.,u))\omega_{k}\quad\text{in}\quad\mathcal{D}^{\prime}(A).$
(5.7)

$\bullet$ Now, consider the $W$-invariant continuous function $S^{W}$ defined
on

$B^{W}(\xi,R):=\cup_{g\in
W}B(g\xi,R)=\cup_{\widetilde{g}\in\nicefrac{{W}}{{W^{\prime}}}}B(g\xi,R)$

by

$S^{W}(x):=g.S(x)=S(g^{-1}.x)\quad\text{whenever}\quad x\in
B(g\xi,R)\quad\text{and}\ \widetilde{g}\in\nicefrac{{W}}{{W^{\prime}}}.$

Note that thanks to the previous condition on $R$, the function $S^{W}$ is
well defined.
$\bullet$ Let $\widetilde{g_{1}}=\widetilde{id}$ and
$\widetilde{g_{2}},\dots,\widetilde{g_{m}}\in\nicefrac{{W}}{{W^{\prime}}}$
such that $\textbf{1}_{B^{W}(\xi,R)}=\sum_{i=1}^{m}\textbf{1}_{B(g_{i}\xi,R)}$
and then we can write

$S^{W}=\sum_{i=1}^{m}S^{W}\textbf{1}_{B(g_{i}\xi,R)}=\sum_{i=1}^{m}g_{i}.[S\textbf{1}_{B(\xi,R)}].$

$\bullet$ Let $\varphi\in\mathcal{D}(B^{W}(\xi,R))$ be nonnegative. Following
the same idea as in the second case (where we replace $U$ by $S$ and $U^{W}$
by $S^{W}$) and using (5.7) we see that we can obtain

$\mathop{\langle\Delta_{k}(S^{W}\omega_{k}),\varphi\rangle}\geq\sum_{i=1}^{m}\mathop{\langle
S\omega_{k},A_{k}(\psi_{i})\rangle},\quad\text{with}\quad\psi_{i}=[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}.$
(5.8)

On the other hand, the $W^{\prime}$-invariance property of the function $S$
implies that

$\forall\ i=1,\dots,m,\ \forall\ \alpha\in
E_{2},\quad\textstyle\int_{B^{W}(\xi,R)}S(x)\frac{\psi_{i}(x)-\psi_{i}(\sigma_{\alpha}x)}{\mathop{\langle\alpha,x\rangle}^{2}}\omega_{k}(x)dx=0.$

Hence, for every $i=1,\dots,m$ we have

$\displaystyle\mathop{\langle S\omega_{k},A_{k}(\psi_{i})\rangle}$
$\displaystyle=\textstyle\sum_{\alpha\in\mathcal{R}_{+}\setminus
E_{2}}k(\alpha)\|\alpha\|^{2}\int_{B^{W}(\xi,R)}S(x)\frac{\psi_{i}(x)-\psi_{i}(\sigma_{\alpha}x)}{\mathop{\langle\alpha,x\rangle}^{2}}\omega_{k}(x)dx$
$\displaystyle=\textstyle\sum_{\alpha\in\mathcal{R}_{+}\setminus
E_{2}}k(\alpha)\|\alpha\|^{2}\int_{B(\xi,R)}\frac{S(x)-S(\sigma_{\alpha}x)}{\mathop{\langle\alpha,x\rangle}^{2}}[g_{i}^{-1}.\varphi](x)\omega_{k}(x)dx$
$\displaystyle=\mathop{\langle
A_{k}(S)\omega_{k},[g_{i}^{-1}.\varphi]\textbf{1}_{B(\xi,R)}\rangle}$

As $E_{1}\neq\mathcal{R}$ and

$S(\xi)-S(\sigma_{\alpha}\xi)=\frac{1}{|W^{\prime}|}\sum_{g^{\prime}\in
W^{\prime}}\big{(}U(\xi)-U(\sigma_{g^{\prime-1}.\alpha}\xi)\big{)}\geq 0,$

we deduce that $A_{k}(S)(\xi)>0$. Consequently, there exists $r>0$ such that
$A_{k}(S)\geq 0$ on $B(\xi,r)$. This fact, (5.8), the continuity of $S$ and
the Weyl lemma show that $S^{W}$ is $\Delta_{k}$-subharmonic on
$B^{W}(\xi,r)$.
Now, by the strong maximum principle, we obtain

$S=S(\xi)=U(x_{0})\quad\text{on}\quad B(\xi,r).$

Thus, we get

$U=U(\xi)=U(x_{0})\quad\text{on}\quad B(\xi,r).$

This completes the proof of the lemma. $\square$

The main tool to establish the existence of a solution of the boundary problem
(5.4) is the Schauder fixed point theorem. In order to apply this theorem, we
will prove the following intermediate result:

###### Proposition 5.1

Let $f$ be a bounded function on $A$ and $G_{k,A}[f]$ be $\Delta_{k}$-Green
potential of $f$ on A given by

$G_{k,A}[f](x):=\int_{A}G_{k,A}(x,y)f(y)\omega_{k}(y)dy,\quad x\in A.$ (5.9)

Then $G_{k,A}[f]$ belongs to $\mathcal{C}_{0}(A)$. Moreover, we have

$-\Delta_{k}\big{(}G_{k,A}[f]\omega_{k}\big{)}=f\omega_{k}\quad\text{in}\quad\mathcal{D}^{\prime}(A).$
(5.10)

Before proving this result, we need to show the following lemma:

###### Lemma 5.2

We have

$\lim_{r\rightarrow 0}\sup_{x\in
A}\eta_{x,r}=0,\quad\text{with}\quad\eta_{x,r}:=\int_{B^{W}(x,r)}N_{k}(x,y)\omega_{k}(y)dy$
(5.11)

and $B^{W}(x,r):=\cup_{g\in W}B(gx,r)$.

_Proof:_ Let $x\in A=A_{\rho,1}$ and $r\in(0,\rho)$. Since $N_{k}(x,.)$ is
$\Delta_{k}$-harmonic on $\mathbb{R}^{d}\setminus W.x$ and
$\Delta_{k}$-superharmonic in $\mathbb{R}^{d}$, by the (super-) mean volume
property (4.4) we deduce that

$\displaystyle 0\leq\eta_{x,r}$ $\displaystyle\leq\int_{B(0,r+\|x\|)\setminus
B(0,\|x\|-r)}N_{k}(x,y)\omega_{k}(y)dy$
$\displaystyle=\int_{B(0,r+\|x\|)}N_{k}(x,y)\omega_{k}(y)dy-\int_{B(0,\|x\|-r)}N_{k}(x,y)\omega_{k}(y)dy$
$\displaystyle\leq
m_{k}[B(0,r+\|x\|)]N_{k}(x,0)-m_{k}[B(0,\|x\|-r)]N_{k}(x,0)$ $\displaystyle=C\
N_{k}(x,0)\big{[}(\|x\|+r)^{d+2\gamma}-(\|x\|-r)^{d+2\gamma}\big{]}$
$\displaystyle\leq
C\frac{\rho^{-2\lambda_{k}}}{2d_{k}\lambda_{k}}\big{[}(\|x\|+r)^{d+2\gamma}-(\|x\|-r)^{d+2\gamma}\big{]}.$

This shows that $\lim_{r\rightarrow 0}\sup_{x\in A}\eta_{x,r}=0$ as desired.
$\square$

_Proof of Proposition_ 5.1: We can suppose that $f$ is nonnegative. Let
$\varepsilon>0$. From (5.11), there exists $r>0$ such that

$\forall\ x\in A,\quad\eta_{x,2r}<\varepsilon.$ (5.12)

$\bullet$ First, we will prove that $G_{k,A}[f](x)\longrightarrow 0$ when $x$
tends to $\partial{A}$. Let $x\in A$. By (5.12) we have

$\displaystyle G_{k,A}[f](x)$
$\displaystyle=\int_{A}G_{k,A}(x,y)f(y)\omega_{k}(y)dy$
$\displaystyle=\int_{A\cap
B^{W}(x,r)}G_{k,A}(x,y)f(y)\omega_{k}(y)dy+\int_{A\setminus{B}^{W}(x,r)}G_{k,A}(x,y)f(y)\omega_{k}(y)dy$
$\displaystyle\leq\|f\|_{\infty}\eta_{x,r}+\|f\|_{\infty}\int_{A\setminus{B}^{W}(x,r)}G_{k,A}(x,y)\omega_{k}(y)dy$
$\displaystyle\leq\varepsilon\|f\|_{\infty}+\|f\|_{\infty}\int_{A\setminus{B}^{W}(x,r)}G_{k,A}(x,y)\omega_{k}(y)dy.$

Since for every $z\in\text{supp}\ \mu_{y}\subset Co(y)$, we can write

$\|x\|^{2}+\|y\|^{2}-2\mathop{\langle x,z\rangle}=\textstyle\sum_{g\in
W}\lambda_{g}(z)\|x-gy\|^{2},$

with $\lambda_{g}(z)\geq 0$ and $\sum_{g\in W}\lambda_{g}(z)=1$, we deduce
that

$\forall\ y\in A\setminus{B}^{W}(x,r),\quad 0\leq G_{k,A}(x,y)\leq
N_{k}(x,y)\leq\frac{r^{-2\lambda_{k}}}{2d_{k}\lambda_{k}}.$

Hence, we can apply the Lebesgue dominated convergence theorem to obtain

$\lim_{x\rightarrow\xi\in\partial{A}}\int_{A\setminus{B}^{W}(x,r)}G_{k,A}(x,y)\omega_{k}(y)dy=0.$

$\bullet$ Now, we will prove that $G_{k,A}[f]$ is continuous on $A$. Fix
$x_{0}\in A$ and assume that $\overline{B}(x_{0},2r)\subset A$. Since $f$ is
bounded, it is enough to prove that

$\lim_{x\rightarrow
x_{0}}\int_{A}|G_{k,A}(x,y)-G_{k,A}(x_{0},y)|\omega_{k}(y)dy=0$ (5.13)

For any $x\in B(x_{0},r)$, we have

$\displaystyle\int_{A}|G_{k,A}(x,y)-G_{k,A}(x_{0},y)|\omega_{k}(y)dy$
$\displaystyle\leq\int_{B^{W}(x_{0},r)}|G_{k,A}(x,y)-G_{k,A}(x_{0},y)|\omega_{k}(y)dy$
$\displaystyle+\int_{A\setminus
B^{W}(x_{0},r)}|G_{k,A}(x,y)-G_{k,A}(x_{0},y)|\omega_{k}(y)dy$
$\displaystyle=I_{1}(x,x_{0})+I_{2}(x,x_{0}).$

As $B^{W}(x_{0},r)\subset B^{W}(x,2r)$, by (5.12) we have

$\displaystyle I_{1}(x,x_{0})$
$\displaystyle\leq\int_{B^{W}(x_{0},r)}N_{k}(x,y)\omega_{k}(y)dy+\int_{B^{W}(x_{0},r)}N_{k}(x_{0},y)\omega_{k}(y)dy$
$\displaystyle\leq\eta_{x,2r}+\eta_{x_{0},r}\leq 2\varepsilon.$

In addition, by item 8) in Proposition 4.2, we know that the function
$(x,y)\longmapsto G_{k,A}(x,y)$ is continuous on the compact set
$\overline{B}^{W}(x_{0},r)\times\big{(}\overline{A}\setminus
B^{W}(x_{0},r)\big{)}$. Thus, there exists $\theta>0$ such that for every
$x\in B(x_{0},\theta)$ and every $y\in A\setminus B^{W}(x_{0},r)$, we have

$|G_{k,A}(x,y)-G_{k,A}(x_{0},y)|\leq\varepsilon.$

This implies that

$\displaystyle\forall\ x\in B(x_{0},\theta),\quad
I_{2}(x,x_{0})\leq\varepsilon\int_{A}\omega_{k}(y)dy.$

Finally, we conclude that $G_{k,A}[f]\in\mathcal{C}_{0}(A)$.
$\bullet$ Let $\varphi\in\mathcal{D}(A)$. Using Fubini’s theorem, the symmetry
property of the Green function and (4.7) we get

$\displaystyle-\mathop{\langle\Delta_{k}\big{(}G_{k,A}[f]\omega_{k}\big{)},\varphi\rangle}$
$\displaystyle=-\int_{A}f(y)\mathop{\langle\Delta_{k}\big{(}G_{k,A}(.,y)\omega_{k}\big{)},\varphi\rangle}\omega_{k}(y)dy$
$\displaystyle=\int_{A}f(y)\varphi(y)\omega_{k}(y)dy.$

This completes the proof. $\square$

_Proof of Theorem_ 5.2: Fix $f\in\mathcal{C}^{+}(\partial{A})$ and

$\displaystyle c_{1}$
$\displaystyle:=\inf_{x\in{\overline{A}}}\big{(}P_{k,A}[f](x)-G_{k,A}[\phi(.,c_{2})](x)\big{)}$
$\displaystyle=\inf_{x\in{\overline{A}}}\big{(}P_{k,A}[f](x)-\phi_{2}(c_{2})G_{k,A}[\phi_{1}](x)\big{)},\quad\text{with}\quad
c_{2}:=\max_{\overline{A}}P_{k,A}[f].$

Let us consider the bounded, closed and convex set

$\mathcal{M}:=\\{u\in\mathcal{C}(\overline{A}):\quad c_{1}\leq u\leq c_{2}\\}$

endowed the uniform topology and the map
$T:\mathcal{C}(\overline{A})\longrightarrow\mathcal{C}(\overline{A})$ defined
by

$T(u):=P_{k,A}[f]-G_{k,A}\big{(}\phi(.,u)\big{)}.$

Note that since $\phi(x,u(x))=\phi_{1}(x)\phi_{2}(u(x))$ is bounded, by
Proposition 5.1, $G_{k,A}\big{(}\phi(.,u)\big{)}\in\mathcal{C}_{0}(A)$ and
then $T$ is well defined. Moreover, as $\phi_{2}$ is nondecreasing, for every
$u\in\mathcal{M}$ and every $x\in\overline{A}$, we have

$c_{1}\leq P_{k,A}[f](x)-G_{k,A}[\phi(.,c_{2})](x)\leq T(u)(x)\leq
P_{k,A}[f](x)\leq c_{2}.$ (5.14)

Hence, we have $T(\mathcal{M})\subset\mathcal{M}$.
Now, we want to establish that $T$ has a unique fixed point in $\mathcal{M}$
by using the Schauder theorem.
$\bullet$ Firstly, we will prove that $T(\mathcal{M})$ is relatively compact.
For this, we will use the Arzelà-Ascoli theorem. From (5.14), $T(\mathcal{M})$
is pointwise bounded.
Let $x_{0}\in A$. For every $u\in\mathcal{M}$ we have

$\displaystyle|T(u)(x)-T(u)(x_{0})|$
$\displaystyle\leq\big{|}P_{k,A}[f](x)-P_{k,A}[f](x_{0})\big{|}+\big{|}G_{k,A}[\phi(.,u)](x)-G_{k,A}[\phi(.,u)](x_{0})\big{|}$
$\displaystyle\leq\big{|}P_{k,A}[f](x)-P_{k,A}[f](x_{0})\big{|}$
$\displaystyle+\int_{A}\big{|}G_{k,A}(x,y)-G_{k,A}(x_{0},y)\big{|}\phi_{1}(y)\phi_{2}(u(y))\omega_{k}(y)dy$
$\displaystyle\leq\big{|}P_{k,A}[f](x)-P_{k,A}[f](x_{0})\big{|}$
$\displaystyle+\phi_{2}(c_{2})\|\phi_{1}\|_{\infty}\int_{A}\big{|}G_{k,A}(x,y)-G_{k,A}(x_{0},y)\big{|}\omega_{k}(y)dy.$

Therefore, from (5.13) and the continuity of the function $P_{k,A}[f]$, we
conclude that $T(\mathcal{M})$ is equicontinuous. Finally, $T(\mathcal{M})$ is
relatively compact as desired.
$\bullet$ Secondly, we will prove that
$T:\mathcal{M}\longrightarrow\mathcal{M}$ is continuous. Let then $(u_{n})$ be
sequence in $\mathcal{M}$ which converges uniformly to $u\in\mathcal{M}$. We
have

$\displaystyle|T(u_{n})(x)-T(u)(x)|$
$\displaystyle\leq\int_{A}G_{k,A}(x,y)\phi_{1}(y)\big{|}\phi_{2}(u_{n}(y))-\phi_{2}(u(y))\big{|}\omega_{k}(y)dy.$

But

$0\leq
G_{k,A}(x,y)\phi_{1}(y)\big{|}\phi_{2}(u_{n}(y))-\phi_{2}(u(y))\big{|}\leq
2\phi_{2}(c_{2})\|\phi_{1}\|_{\infty}G_{k,A}(x,y).$

Thus, we can use the Lebesgue dominated convergence theorem to obtain that
$T(u_{n})\longrightarrow T(u)$ pointwise. Hence, by equicontinuity, we get the
uniform convergence.
Consequently, there exists $u\in\mathcal{M}$ such that

$u=T(u)=P_{k,A}[f]-G_{k,A}\big{(}\phi(.,u)\big{)}.$

Finally, note that from the properties of $P_{k,A}$ as well as (5.10), $u$ is
a solution of (5.4). This finishes the proof of the theorem.

$\square$

_Acknowledgement_

It is a pleasure to thank the referee for the valuable suggestions which
improved the presentation of the paper.

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