csymbol=c

# Heat and Martin kernel estimates for Schrödinger operators with critical
Hardy potentials

G. Barbatis  G. Barbatis, Department of Mathematics, National and Kapodistrian
University of Athens, 15784 Athens, Greece<EMAIL_ADDRESS>,  K.T.
Gkikas  K.T. Gkikas, Department of Mathematics, National and Kapodistrian
University of Athens, 15784 Athens, Greece<EMAIL_ADDRESS>and  A.
Tertikas  A. Tertikas, Department of Mathematics & Applied Mathematics,
University of Crete, 70013 Heraklion, Greece and Institute of Applied and
Computational Mathematics, FORTH, 71110 Heraklion, Greece<EMAIL_ADDRESS>
###### Abstract.

Let $\Omega$ be a bounded domain in ${\mathbb{R}}^{N}$ with $C^{2}$ boundary
and let $K\subset\partial\Omega$ be either a $C^{2}$ submanifold of the
boundary of codimension $k<N$ or a point. In this article we study various
problems related to the Schrödinger operator $L_{\mu}=-\Delta-\mu d_{K}^{-2}$
where $d_{K}$ denotes the distance to $K$ and $\mu\leq k^{2}/4$. We establish
parabolic boundary Harnack inequalities as well as related two-sided heat
kernel and Green function estimates. We construct the associated Martin kernel
and prove existence and uniqueness for the corresponding boundary value
problem with data given by measures. Next we apply the results to the study of
$L_{\mu}u+g(u)=0$ and establish existence and uniqueness under suitable
assumptions on the function $g$. To prove our results we introduce among other
things a suitable notion of boundary trace. This trace is different from the
one used by Marcus and Nguyen [MN1] thus allowing us to cover the whole range
$\mu\leq k^{2}/4$.

Keywords: heat kernel; Martin kernel; Green function; Hardy potential;
boundary trace

Mathematics Subject Classification: Primary 35K08; 35J25; 35J75; 35J08.
Secondary 46E35; 31C35

###### Contents

  1. 1 Introduction
  2. 2 Main results
    1. 2.1 Heat kernel and boundary Harnack inequality
    2. 2.2 Martin Kernels and boundary value problems
  3. 3 Hardy-Sobolev type inequalities
  4. 4 Heat Kernel Estimates for small time
    1. 4.1 Doubling Property
    2. 4.2 Density of $C^{\infty}_{c}(\Omega)$ functions
    3. 4.3 Poincaré inequality
    4. 4.4 Moser inequality
    5. 4.5 Harnack inequality
  5. 5 Heat kernel estimates for large time
    1. 5.1 Weighted logarithmic Sobolev inequality
    2. 5.2 Green function estimates
  6. 6 The linear elliptic problem
    1. 6.1 Subsolutions and Supersolutions
    2. 6.2 Existence and uniqueness
  7. 7 Martin kernel
    1. 7.1 $L_{\mu}$-harmonic measure
    2. 7.2 The Poisson kernel of $L_{\mu}$
  8. 8 Boundary value problem for linear equations
    1. 8.1 Boundary trace
    2. 8.2 Boundary value problem for linear equations
  9. 9 The Nonlinear Problem
    1. 9.1 Weak $L^{p}$ estimates
    2. 9.2 Existence and Uniqueness
  10. A Pointwise estimates on eigenfunctions

##  1\. Introduction

The study of linear Schrödinger operators with singular potentials is central
in the theory of parabolic and elliptic partial differential equations. In
recent years in particular there has been an intense study of operators with
Hardy potentials, see e.g. [BM, MMP, BMS, FT, BFT1, DD2, GM, C, FM, BEL,
BFT3].

Throughout this work we assume that $\Omega$ is a bounded $C^{2}$ domain; we
note however that some of the results presented in this introduction are valid
under weaker regularity assumptions.

Consider the problem

$\left\\{\begin{array}[]{ll}u_{t}=\Delta u+V(x)u,&x\in\Omega,\;t>0,\\\
u=0,&x\in\partial\Omega,\;t>0,\\\
u(0,x)=u_{0}(x),&x\in\Omega\,,\end{array}\right.$ (1.1)

where $V\in L^{1}_{\rm loc}(\Omega)$ and set

$\lambda^{*}=\inf_{C^{\infty}_{c}(\Omega)}\frac{\int_{\Omega}|\nabla
w|^{2}dx-\int_{\Omega}Vw^{2}dx}{\int_{\Omega}w^{2}dx}.$

Cabré and Martel [CM] have established that if $\lambda^{*}>-\infty$ then for
regular enough initial data there exists a global in time weak solution of
(1.1) which in addition satisfies an exponential in time bound. Conversely,
the existence of a weak solution which satisfies an exponential bound implies
that $\lambda^{*}>-\infty$. In the prototype case of the Hardy potential
$V(x)=c|x|^{-2}$ this has already been studied by Baras and Goldstein [BG].

Given the existence of a weak solution one natural question is the existence
and asymptotic behaviour of the heat kernel and Green function. If the
potential is not too singular then the asymptotic behaviour of the heat kernel
for small time is the same as that of the Laplacian, namely

$\displaystyle
C^{-1}\Big{(}\frac{d(x)d(y)}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}t^{-\frac{N}{2}}\exp\Big{(}-C\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\leq h(t,x,y)\leq
C\Big{(}\frac{d(x)d(y)}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}t^{-\frac{N}{2}}\exp\Big{(}-C^{-1}\frac{|x-y|^{2}}{t}\Big{)},$

where $d(x)={\rm dist}(x,\partial\Omega)$ denotes the distance to the
boundary, see e.g. [Z].

In the case of a more singular potential such as a Hardy potential, the
problem has been studied in [DS, D2, VZ, MS, LS, BFT2, MT1, MT2, FMT1, FMT2].

A distinction that plays an important role in this context is whether the
singularity of the Hardy potential occurs in the interior or on the boundary
of the domain. For the potential $\mu|x|^{-2}$,
$0\leq\mu\leq(\frac{N-2}{2})^{2}$, where $0\in\Omega$, for small time we have

$\displaystyle
C^{-1}\Big{(}\frac{d(x)d(y)}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}\Big{(}\frac{|x|\;|y|}{(|x|+\sqrt{t})(|y|+\sqrt{t})}\Big{)}^{\theta_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\leq$ $\displaystyle h(t,x,y)$ $\displaystyle\leq$
$\displaystyle
C\Big{(}\frac{d(x)d(y)}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}\Big{(}\frac{|x|\;|y|}{(|x|+\sqrt{t})(|y|+\sqrt{t})}\Big{)}^{\theta_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C^{-1}\frac{|x-y|^{2}}{t}\Big{)},$

where $\theta_{+}$ is the largest solution to the equation
$\theta^{2}+(N-2)\theta+\mu=0$; see [FMT1]. This estimate was generalized in
[GN3] in case where the distance is taken from a closed surface
$\Sigma\subset\Omega$ of codimension $k$, $2\leq k\leq N$; see also [GN1, GN2]
for more results within this framework.

On the other hand, when the distance is taken from the boundary
$\partial\Omega$ the following small time estimate is valid for the heat
kernel of the operator $-\Delta-\mu d(x)^{-2}$, $0\leq\mu\leq\frac{1}{4}$,

$\displaystyle
C^{-1}\Big{(}\frac{d(x)d(y)}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}^{1+\theta_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\leq$ $\displaystyle h(t,x,y)\;\leq
C\Big{(}\frac{d(x)d(y)}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}^{1+\theta_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C^{-1}\frac{|x-y|^{2}}{t}\Big{)},$

where $\theta_{+}$ is the largest solution to the equation
$\theta^{2}+\theta+\mu=0$, see [FMT1, FMT2].

Another function that is important in the study of this type of problems is
the Martin kernel [A, HW, M]. Ancona proved the existence of the Martin kernel
$K_{\mu,\partial\Omega}(x,y)$ of
$L_{\mu}^{\partial\Omega}=-\Delta-\frac{\mu}{d^{2}}$, $\mu<\frac{1}{4}$, with
pole at $y$, which is unique up to a normalization (see [A, Theorem 3]). He
showed that for any positive solution $u$ of $L_{\mu}^{\partial\Omega}u=0$
there exists a unique nonnegative Radon measure $\nu$ on $\partial\Omega$ such
that

$\displaystyle u(x)=\int_{\partial\Omega}K_{\mu,\partial\Omega}(x,y)d\nu(y).$
(1.2)

The case $\mu=\frac{1}{4}$ was treated by Gkikas and Véron in [GV]. In
particular, they showed that the representation formula (1.2) holds true
provided the bottom of the spectrum of $L_{\mu}^{\partial\Omega}$ is positive.

When $K\subset\Omega$ is a closed smooth surface of codimension
$k\in\\{3,\ldots,N\\}$, analogous results where obtained in [GN3] for the
operator $L_{\mu}^{K}=-\Delta-\frac{\mu}{d_{K}^{2}}$,
$\mu\leq\frac{(k-2)^{2}}{4}$, under the assumption that the bottom of the
spectrum of $L_{\mu}^{K}$ is positive.

Our aim in this article is to study such problems in the case where the Hardy
potential involves the distance to a smooth submanifold of the boundary,
including the case of a boundary point. In this direction:

$\bullet$ We establish parabolic boundary Harnack inequalities as well as
related two-sided heat kernel estimates. For small time, our approach is based
on the ideas of Grigoryan and Saloff–Coste [GS-C] (see also [S-C]), while for
large time, we exploit the work of Davies in [D1, D2] to obtain sharp- two
sided heat kernel estimates; see also [FMT1, FMT2].

$\bullet$ In the spirit of [CFMS, HW] (see also [GN3, GV]), we construct the
Martin kernel of $L_{\mu}$ in $\Omega$ and we prove the uniqueness also in the
critical case. Using the heat kernel estimates, we obtain sharp pointwise
estimates for the Green function as well as the Martin kernel. We also show
that every nonnegative $L_{\mu}$-harmonic function (i.e. solution of
$L_{\mu}u=0$ in $\Omega$ in the sense of distributions) can be represented as
the integral of the Martin kernel with respect to a finite measure on
$\partial\Omega$.

$\bullet$ Using the properties of the Green function and Martin kernel we
study the boundary value problem with data given by measures. Following
Marcus-Véron [MV1] we prove existence, uniqueness as well as a representation
formula for any solution of this problem. As an application, we establish
existence and uniqueness results for solutions of the equation
$L_{\mu}u+g(u)=0$ under suitable assumptions on the function $g$.

##  2\. Main results

Throughout this article we consider a bounded $C^{2}$ domain
$\Omega\subset\mathbb{R}^{N}$, $N\geq 3$ and a $C^{2}$ compact submanifold
without boundary $K\subset\partial\Omega$ of codimension $k$, $1\leq k\leq N$.
For the extreme cases $k=N$ and $k=1$ we assume that $K=\\{0\\}$ and
$K=\partial\Omega$ respectively. We set $d_{K}(x)=\text{dist}(x,K)$ and define
the operator

$L_{\mu}=-\Delta-\frac{\mu}{d_{K}^{2}}\;,\qquad\mbox{ in }\Omega,$

where $\mu$ is a parameter; we shall always assume that
$\mu\leq\frac{k^{2}}{4}$ so that $L_{\mu}$ is bounded from below. The study of
the parabolic equation $u_{t}+L_{\mu}u=0$ with Dirichlet boundary conditions
is strongly related with the minimization problem,

$\displaystyle C_{\Omega,K}=\inf_{u\in
H_{0}^{1}(\Omega)\setminus\\{0\\}}\frac{\int_{\Omega}|\nabla
u|^{2}dx}{\int_{\Omega}\frac{|u|^{2}}{d_{K}^{2}}dx}.$

It is well known that $0<C_{\Omega,K}\leq\frac{k^{2}}{4}$ (see, e.g., [FM]).

For $\mu\leq\frac{k^{2}}{4}$ the infimum

$\displaystyle\lambda_{\mu}:=\inf_{u\in
H_{0}^{1}(\Omega)\setminus\\{0\\}}\frac{\int_{\Omega}|\nabla
u|^{2}dx-\mu\int_{\Omega}\frac{u^{2}}{d_{K}^{2}}dx}{\int_{\Omega}u^{2}dx}$
(2.1)

is finite. Moreover, if $\mu<\frac{k^{2}}{4},$ then there exists a minimizer
$\phi_{\mu}\in H_{0}^{1}(\Omega)$ of (2.1); see [FM] for more details. In
addition, by [MN1, Lemma 2.2] the eigenfunction $\phi_{\mu}$ satisfies

$\phi_{\mu}(x)\asymp d(x)d_{K}^{\gamma_{+}}(x),\quad\quad\mbox{ in }\Omega\,,$
(2.2)

provided $\mu<C_{\Omega,K}$. Here and below we denote by $\gamma_{+}$ (resp.
$\gamma_{-}$) the largest (resp. the smallest) solution of the equation
$\gamma^{2}+k\gamma+\mu=0$.

On the other hand, if $\mu=\frac{k^{2}}{4}$ then there is no
$H_{0}^{1}(\Omega)$ minimizer. However, there exists a function $\phi_{\mu}\in
H^{1}_{loc}(\Omega)$ such that $L_{\mu}\phi_{\mu}=\lambda_{\mu}\phi_{\mu}$ in
$\Omega$ in the sense of distributions. In the Appendix we follow ideas of
[BMS, DD1, DD2, FMT2] and prove that estimate (2.2) is valid for any
$\mu\leq\frac{k^{2}}{4}$.

###  2.1. Heat kernel and boundary Harnack inequality

Let $u\in C^{1}((0,\infty):C^{2}(\Omega)),$ then by setting
$u=e^{-\lambda_{\mu}t}\phi_{\mu}v,$ we can easily see that

$\frac{u_{t}+L_{\mu}u}{\phi_{\mu}}=v_{t}-\phi_{\mu}^{-2}\,\text{div}\left(\phi_{\mu}^{2}\nabla
v\right)=:v_{t}+{\mathcal{L}}_{\mu}v.$ (2.3)

Hence, instead of studying the properties of the operator $L_{\mu},$ it is
more convenient to study the operator $\frac{\partial}{\partial
t}+{\mathcal{L}}_{\mu}.$ In this direction, we introduce the weighted Sobolev
space $H^{1}(\Omega;\phi_{\mu}^{2})$.

######  Definition 2.1.

Let $D\subset\Omega$ be an open set. We denote by $H^{1}(D;\phi_{\mu}^{2})$
the weighted Sobolev space

$\displaystyle H^{1}(D;\phi_{\mu}^{2}):=\\{u\in
H^{1}_{loc}(D):\;|u|\phi_{\mu}+|\nabla u|\phi_{\mu}\in L^{2}(D)\\}$

endowed with the norm

$\displaystyle\left\|u\right\|^{2}_{H^{1}(D;\phi_{\mu}^{2})}=\int_{D}u^{2}\phi_{\mu}^{2}dx+\int_{D}|\nabla
u|^{2}\phi_{\mu}^{2}dx.$

We also denote by $H^{1}_{0}(D;\phi_{\mu}^{2})$ the closure of
$C_{c}^{\infty}(D)$ in the norm
$\left\|\cdot\right\|_{H^{1}(D;\phi_{\mu}^{2})}$. It is worth mentioning here
that $H^{1}_{0}(\Omega;\phi_{\mu}^{2})=H^{1}(\Omega;\phi_{\mu}^{2})$ (see
Theorem 4.5).

Next, we normalize $\phi_{\mu}$ so that $\int_{\Omega}\phi_{\mu}^{2}dx=1$. We
define the bilinear form $Q:H_{0}^{1}(\Omega;\phi_{\mu}^{2})\times
H_{0}^{1}(\Omega;\phi_{\mu}^{2})\to{\mathbb{R}}$ by

$\displaystyle Q(u,v)=\int_{\Omega}\nabla u\cdot\nabla v\,\phi_{\mu}^{2}dx.$

The associated operator is the operator ${\mathcal{L}}_{\mu}$ defined in (2.3)
and generates a contraction semigroup $T(t):L^{2}(\Omega;\phi_{\mu}^{2})\to
L^{2}(\Omega;\phi_{\mu}^{2})$, $t\geq 0$, denoted also by
$e^{-{\mathcal{L}}_{\mu}t}$. This semigroup is positivity preserving and by
[D2, Lemma 1.3.4] we can easily show that satisfies the conditions of [D2,
Theorems 1.3.2 and 1.3.3]. Using the logarithmic Sobolev inequality (Theorem
5.1) and some ideas of Davies [D1, D2], we shall show that
$e^{-{\mathcal{L}}_{\mu}t}$ is ultracontractive and therefore has a kernel
$k(t,x,y).$ More precisely, we prove the following large time estimates:

######  Theorem 2.2.

Let $\mu\leq\frac{k^{2}}{4}$ and $T>0.$ Then there holds

$\displaystyle k(t,x,y)\asymp 1$

for any $t\geq T$ and $x,y\in\Omega$. The implicit constants depend only on
$\Omega,K,\mu$ and $T.$

For small time the two-sided heat kernel estimate is different. A pivotal
ingredient in the proof of this estimate is the boundary Harnack inequality.
However, in order to state the boundary Harnack inequality, we first need to
give the following definition of weak solution.

######  Definition 2.3.

Let $D\subset\Omega$ be an open set. We say that $v\in
C^{1}((0,T):H^{1}(D;\phi_{\mu}^{2}))$ is a weak solution of
$v_{t}+{\mathcal{L}}_{\mu}v=0$ in $(0,T)\times D$ if for each $\Phi\in
C^{1}_{c}((0,T):C_{c}^{\infty}(D)),$ we have

$\displaystyle\int_{0}^{T}\int_{D}(v_{t}\Phi+\nabla
v\cdot\nabla\Phi)\phi_{\mu}^{2}\,dy\,dt=0.$

######  Theorem 2.4 (Boundary Harnack inequality).

Let $\mu\leq k^{2}/4$ and $v$ be a non-negative solution of
$v_{t}+{\mathcal{L}}_{\mu}v$ in $(0,r^{2})\times\mathcal{B}(x,r)\cap\Omega$.
There exists $\beta_{1}>0$ and a positive constant
$C=C(\Omega,K,\beta_{1},\mu)$ such that for all $r<\beta_{1}$ there holds

$\sup_{(\frac{r^{2}}{4},\frac{r^{2}}{2})\times\mathcal{B}(x,\frac{r}{2})\cap\Omega}v\leq
C\inf_{(\frac{3r^{2}}{4},r^{2})\times\mathcal{B}(x,\frac{r}{2})\cap\Omega}v.$
(2.4)

Here $\mathcal{B}(x,r)$ are suitably defined “balls” (see Definition 4.1). Let
us briefly explain the proof of the above theorem. We first prove the doubling
property for the “balls” $\mathcal{B}(x,r)$ (Lemma 4.2), the Poincaré
inequality (Theorem 4.9) and the Moser inequality (Theorem 4.21). The last
three results along with the density Theorem 4.5 allow us to apply a Moser
iteration argument similar to the one in [GS-C, S-C] so that we reach the
desired result. Due to the fact that $K\subset\partial\Omega,$ the proof of
the above theorem is more complicated than the one in [FMT1, FMT2] and new
essential difficulties arise which should be handled in a very delicate way.

Proceeding as in the proof of [S-C, Theorem 5.4.12], we may deduce that the
boundary Harnack inequality (2.4) implies the following sharp two-sided heat
kernel estimate for small time.

######  Theorem 2.5.

Let $\mu\leq\frac{k^{2}}{4}$ and $T>0$. Then there exists $C>1$ which depends
on $\Omega,K,\mu$ and $T$ such that

$\displaystyle
C^{-1}\left((d(x)+\sqrt{t})(d(y)+\sqrt{t})\right)^{-1}\left((d_{K}(x)+\sqrt{t})(d_{K}(y)+\sqrt{t})\right)^{-\gamma_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\leq\,k(t,x,y)$ $\displaystyle\leq
C\\!\left((d(x)+\sqrt{t})(d(y)+\sqrt{t})\right)^{-1}\\!\\!\left((d_{K}(x)+\sqrt{t})(d_{K}(y)+\sqrt{t})\right)^{-\gamma_{+}}\\!\\!t^{-\frac{N}{2}}\exp\\!\Big{(}\\!-C^{-1}\frac{|x-y|^{2}}{t}\Big{)}$

for any $0<t\leq T$ and $x,y\in\Omega$.

Let $h(t,x,y)$ denote the Dirichlet heat kernel of $L_{\mu}$. It is then
immediate that
$h(t,x,y)=(\phi_{\mu}(x)\phi_{\mu}(y))e^{-\lambda_{\mu}t}k(t,x,y)$. Hence, by
Theorems 2.2 and 2.5, we obtain the following theorem.

######  Theorem 2.6.

Let $\mu\leq\frac{k^{2}}{4}$. There exist $T=T(\Omega,K,\mu)>0,$
$C_{1}=C_{1}(\Omega,K,\mu,T,\lambda_{\mu})>1$ and $C_{2}=C(\Omega,K,\mu,T)>1$
such that

$\displaystyle{\rm(i)}$ $\displaystyle
C^{-1}_{1}\Big{(}\frac{d(x)}{d(x)+\sqrt{t}}\Big{)}\Big{(}\frac{d(y)}{d(y)+\sqrt{t}}\Big{)}\Big{(}\frac{d_{K}(x)}{d_{K}(x)+\sqrt{t}}\Big{)}^{\gamma_{+}}\Big{(}\frac{d_{K}(y)}{d_{K}(y)+\sqrt{t}}\Big{)}^{\gamma_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C_{1}\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\leq h(t,x,y)\leq$ $\displaystyle
C_{1}\Big{(}\frac{d(x)}{d(x)+\sqrt{t}}\Big{)}\Big{(}\frac{d(y)}{d(y)+\sqrt{t}}\Big{)}\Big{(}\frac{d_{K}(x)}{d_{K}(x)+\sqrt{t}}\Big{)}^{\gamma_{+}}\Big{(}\frac{d_{K}(y)}{d_{K}(y)+\sqrt{t}}\Big{)}^{\gamma_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C^{-1}_{1}\frac{|x-y|^{2}}{t}\Big{)},$

for any $0<t<T$ and $x,y\in\Omega.$

$\displaystyle{\rm(ii)}\hskip
56.9055ptC^{-1}_{2}\phi_{\mu}(x)\phi_{\mu}(y)e^{-\lambda_{\mu}t}\leq
h(t,x,y)\leq C_{2}\phi_{\mu}(x)\phi_{\mu}(y)e^{-\lambda_{\mu}t},$

for any $\ t>T$ and $x,y\in\Omega.$

If $\lambda_{\mu}>0,$ then by the above theorem we can obtain the existence of
a minimal Green function $G_{\mu}(x,y)$ of $L_{\mu}$ as well as precise
asymptotic for $G_{\mu}(x,y)$ (see Subsection 5.2 for more details).

###  2.2. Martin Kernels and boundary value problems

If $\mu<C_{\Omega,K}$ then the operator
$L_{\mu}=-\Delta-\frac{\mu}{d_{K}^{2}}$ is coercive in $H_{0}^{1}(\Omega)$.
Hence, taking into account the discussion on the first eigenfunction
$\phi_{\mu}$ of (2.1), we may apply Ancona’s results in [A] to deduce that any
positive solution $u$ of $L_{\mu}u=0$ in $\Omega$ can be represented like
(1.2). If $\mu=C_{\Omega,K}<\frac{k^{2}}{4}$ then there exists an $H^{1}_{0}$
minimiser of the Hardy quotient and therefore there is no Green function and
the operator is not coercive. In the remaining case
$\mu=C_{\Omega,K}=\frac{k^{2}}{4}$, the operator $L_{\mu}$ clearly is not
coercive and this case is not covered by Ancona’s results in [A]. One of the
main goals of this work is to prove that the assumption $\lambda_{\mu}>0$
suffices to have a respective representation formula, also in the case
$\mu=\frac{k^{2}}{4}$.

In order to state the main results we first need to give some notations and
definitions. For $\beta>0$ we set

$K_{\beta}=\\{x\in\mathbb{R}^{N}\setminus
K:\;d_{K}(x)<\beta\\},\qquad\Omega_{\beta}=\\{x\in\Omega:\;d(x)<\beta\\}.$

We assume that $\beta$ is small enough so that for any $x\in\Omega_{\beta}$
there exists a unique $\xi_{x}\in\partial\Omega,$ which satisfies
$d(x)=|x-\xi_{x}|.$ Now set

$\tilde{d}_{K}(x)=\sqrt{|\mathrm{dist}^{\partial\Omega}(\xi_{x},K)|^{2}+|x-\xi_{x}|^{2}}\,,\qquad
x\in K_{\beta},$ (2.5)

where $\mathrm{dist}^{\partial\Omega}(\xi_{x},K)$ denotes the distance of
$\xi_{x}$ to $K$ measured on $\partial\Omega$.

Let $\beta_{0}>0$ (this will be determined in Lemma 6.1). We consider a smooth
cut-off function $0\leq\eta_{\beta_{0}}\leq 1$ with compact support in
$K_{\frac{\beta_{0}}{2}}$ such that $\eta_{\beta_{0}}=1$ in
$\overline{K}_{\frac{\beta_{0}}{4}}$. We define

$W(x)=\left\\{\begin{aligned}
&(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{\gamma_{-}},\qquad&&\text{if}\;\mu<\frac{k^{2}}{4},\\\
&(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{-\frac{k}{2}}(x)|\ln\tilde{d}_{K}(x)|,\qquad&&\text{if}\;\mu=\frac{k^{2}}{4},\end{aligned}\right.\quad
x\in\Omega\cap K_{\beta_{0}},$

and

$\tilde{W}(x):=(1-\eta_{\beta_{0}}(x))+\eta_{\beta_{0}}(x)W(x),\quad
x\in\Omega.$

Let $h\in C(\partial\Omega)$ and $u\in H^{1}_{loc}(\Omega)\cap C(\Omega)$. We
write $\widetilde{\rm tr}(u)=h$ whenever

$\lim_{x\in\Omega,\;x\rightarrow
y\in\partial\Omega}\frac{u(x)}{\tilde{W}(x)}=h(y)\qquad\text{uniformly for
}y\in\partial\Omega.$ (2.6)

In Section 6 we prove that for any $h\in C(\partial\Omega)$ the problem

$\displaystyle\left\\{\begin{aligned}
L_{\mu}v&=0,\qquad\mathrm{in}\;\;\Omega,\\\ \widetilde{\rm
tr}(v)&=h,\qquad\mathrm{on}\;\;\partial\Omega,\end{aligned}\right.$

has a unique solution $v=v_{h}\in H^{1}_{loc}(\Omega)\cap C(\Omega)$. From
this and the accompanying estimate follows that for any $x_{0}\in\Omega$ the
mapping $h\mapsto v_{h}(x_{0})$ is a linear positive functional on
$C(\partial\Omega)$. Thus there exists a unique Borel measure on
$\partial\Omega$, called $L_{\mu}$-harmonic measure in $\Omega,$ denoted by
$\omega^{x_{0}}$, such that

$\displaystyle v_{h}(x_{0})=\int_{\partial\Omega}h(y)d\omega^{x_{0}}(y).$

Thanks to the Harnack inequality the measures $\omega^{x}$ and
$\omega^{x_{0}},$ $x_{0},\;x\in\Omega$, are mutually absolutely continuous.
Therefore, the Radon-Nikodyn derivative exists and we set

$\displaystyle
K_{\mu}(x,y):=\frac{dw^{x}}{dw^{x_{0}}}(y)\qquad\mathrm{for}\;\omega^{x_{0}}\text{-
almost all }y\in\partial\Omega.$

######  Definition 2.7.

Fix $\xi\in\partial\Omega.$ A function $\mathcal{K}$ defined in $\Omega$ is
called a kernel function for $L_{\mu}$ with pole at $\xi$ and basis at
$x_{0}\in\Omega$ if

$\displaystyle{\rm(i)}$ $\mathcal{K}(\cdot,\xi)$ is $L_{\mu}$-harmonic in
$\Omega$, $\displaystyle{\rm(ii)}$ $\displaystyle\mbox{
$\frac{\mathcal{K}(\cdot,\xi)}{\tilde{W}(\cdot)}\in
C(\overline{\Omega}\setminus\\{\xi\\})$ and for any
$\eta\in\partial\Omega\setminus\\{\xi\\}$ we have
}\lim_{x\in\Omega,\;x\to\eta}\frac{\mathcal{K}(x,\xi)}{\tilde{W}(x)}=0,$
$\displaystyle{\rm(iii)}$ $\mathcal{K}(x,\xi)>0$ for each $x\in\Omega$ and
$\mathcal{K}(x_{0},\xi)=1.$

Using the ideas in [CFMS], we show the existence and uniqueness of a kernel
function with pole at $\xi$ and basis at $x_{0}$ (see Proposition 7.3). As a
result we obtain the existence of the Martin kernel and moreover

$\displaystyle
K_{\mu}(x,\xi)=\lim_{y\in\Omega,\;y\to\xi}\frac{G_{\mu}(x,y)}{G_{\mu}(x_{0},y)},\quad\forall\xi\in\partial\Omega.$

In addition, by the estimates on Green function $G_{\mu}(x,y)$ of $L_{\mu}$
(see Proposition 5.3) we obtain the following result.

######  Theorem 2.8.

Assume that $\mu\leq\frac{k^{2}}{4}$ and $\lambda_{\mu}>0$. We then have:
$({\rm i})$ If $\mu<\frac{k^{2}}{4}$ or $\mu=\frac{k^{2}}{4}$ and $k<N$ then

$K_{\mu}(x,\xi)\asymp\frac{d(x)}{|x-\xi|^{N}}\left(\frac{d_{K}(x)}{\left(d_{K}(x)+|x-\xi|\right)^{2}}\right)^{\gamma_{+}},\quad\mbox{
in }\Omega\times\partial\Omega$ (2.7)

$({\rm ii})$ If $\mu=\frac{N^{2}}{4}$ (so $k=N$), then

$K_{\mu}(x,\xi)\asymp\frac{d(x)}{|x-\xi|^{N}}\left(\frac{|x|}{\left(|x|+|x-\xi|\right)^{2}}\right)^{-\frac{N}{2}}+\frac{d(x)}{|x|^{\frac{N}{2}}}\big{|}\ln|x-\xi|\big{|},\quad\mbox{
in }\Omega\times\partial\Omega$ (2.8)

When $K=\partial\Omega$, Filippas, Moschini and Tertikas [FMT1] derived sharp
two-sided estimate on the associated heat kernel. These estimates where then
used in order to obtain sharp estimates on $G_{\mu}(x,y).$ Chen and Véron
[CV2] studied the operator $L_{\mu}$ with $K=\\{0\\}\subset\partial\Omega$ and
they constructed the corresponding Martin kernel. The case $K\subset\Omega$
was thoroughly studied by Gkikas and Nguyen in [GN3]. Estimates on the Green
kernel of $L_{\mu V}=-\Delta-\mu V,$ where $V$ is a singular potential such
that $|V(x)|\leq cd^{-2}(x)$ in $\Omega$, have been given by Marcus [M1, M2].
Marcus and Nguyen [MN1] used Ancona’s result to show that the Martin kernel
$K_{\mu}(x,y)$ is well defined and they applied the results in [M2] to the
model case $L_{\mu}$ in order to obtain estimates on the Green kernel
$G_{\mu}(x,y)$ and the Martin kernel $K_{\mu}(x,y)$. However, their results do
not cover the critical case $\mu=\frac{k^{2}}{4}.$

In this work, we follow a different approach which does not use Ancona’s
result [A] and allows us to study the critical case. In particular our work is
inspired by the articles [FMT1, GV, GN3]. The main difference here is that
$K\subset\partial\Omega,$ which has an effect on the value of the optimal
Hardy constant $C_{\Omega,K}$ as well as on the behaviour of the eigenfunction
$\phi_{\mu}.$ As a result, this fact yields substantial difficulties and
reveals new aspects of the study of $L_{\mu}.$

We are now ready to state the representation formula.

######  Theorem 2.9.

Assume that $\mu\leq\frac{k^{2}}{4}$ and $\lambda_{\mu}>0$. Let $u$ be a
positive $L_{\mu}$-harmonic function in $\Omega.$ Then $u\in
L^{1}(\Omega;\phi_{\mu})$ and there exists a unique Radon measure $\nu$ on
$\partial\Omega$ such that

$\displaystyle
u(x)=\int_{\partial\Omega}K_{\mu}(x,\xi)d\nu(\xi)=:\mathbb{K}_{\mu}[\nu].$

In order to study the corresponding boundary value problem, we should first
introduce the notion of the boundary trace. We will define it in a dynamic
way. In this direction, let $\\{\Omega_{n}\\}$ be a smooth exhaustion of
$\Omega$, that is an increasing sequence of bounded open smooth domains such
that $\overline{\Omega_{n}}\subset\Omega_{n+1}$, $\cup_{n}\Omega_{n}=\Omega$
and
$\mathcal{H}^{N-1}(\partial\Omega_{n})\to\mathcal{H}^{N-1}(\partial\Omega)$.
The operator $L_{\mu}^{\Omega_{n}}$ defined by

$L_{\mu}^{\Omega_{n}}u=-\Delta u-\frac{\mu}{d^{2}_{K}}u$

is uniformly elliptic and coercive in $H^{1}_{0}(\Omega_{n})$ and its first
eigenvalue $\lambda_{\mu}^{\Omega_{n}}$ is larger than $\lambda_{\mu}$. For
$h\in C(\partial\Omega_{n})$ the problem

$\left\\{\begin{array}[]{ll}L_{\mu}^{\Omega_{n}}v=0,&\qquad\text{in
}\Omega_{n}\\\ v=h,&\qquad\text{on }\partial\Omega_{n},\end{array}\right.$

admits a unique solution which allows to define the
$L_{\mu}^{\Omega_{n}}$-harmonic measure on $\partial\Omega_{n}$ by

$v(x_{0})={\displaystyle\int_{\partial\Omega_{n}}}h(y)d\omega^{x_{0}}_{\Omega_{n}}(y).$

######  Definition 2.10 ($L_{\mu}$-boundary trace).

A function $u\in W^{1,p}_{loc}(\Omega),\;p>1$, possesses an
$L_{\mu}$-_boundary trace_ if there exists a measure
$\nu\in\mathfrak{M}(\partial\Omega)$ such that for any smooth exhaustion
$\\{\Omega_{n}\\}$ of $\Omega$, there holds

$\lim_{n\rightarrow\infty}\int_{\partial\Omega_{n}}\phi
u\,d\omega_{\Omega_{n}}^{x_{0}}=\int_{\partial\Omega}\phi\,d\nu,\qquad\forall\phi\in
C(\overline{\Omega}).$

The $L_{\mu}$-boundary trace of $u$ will be denoted by $\mathrm{tr}_{\mu}(u)$.

Let $\mathfrak{M}(\partial\Omega)$ denote the space of bounded Borel measures
on $\partial\Omega$ and $\mathfrak{M}(\Omega;\phi_{\mu})$ the space of Borel
measures $\tau$ on $\Omega$ such that

$\int_{\Omega}\phi_{\mu}d|\tau|<\infty.$

Arguing as in [MV2] we obtain in Lemma 8.1 that for any
$\nu\in\mathfrak{M}(\partial\Omega)$ we have
$\mathrm{tr}_{\mu}(\mathbb{K}_{\mu}[\nu])=\nu$.

Assume now that $\tau\in\mathfrak{M}(\Omega;\phi_{\mu})$ and let

$u=\mathbb{G}_{\mu}[\tau]:=\int_{\Omega}G_{\mu}(x,y)d\tau(y).$

Then $u\in W^{1,p}_{loc}(\Omega)$ for every $1<p<\frac{N}{N-1}$ and
$\mathrm{tr}_{\mu}(u)=0$ (see Lemma 8.2).

Next, we give the definition of weak solutions of the following boundary value
problem.

######  Definition 2.11.

Let $\tau\in\mathfrak{M}(\Omega;\phi_{\mu})$ and
$\nu\in\mathfrak{M}(\partial\Omega)$. We say that $u\in
L^{1}(\Omega;\phi_{\mu})$ is a weak solution of

$\left\\{\begin{array}[]{ll}L_{\mu}u=\tau,&\qquad\text{in }\;\Omega,\\\
\mathrm{tr}_{\mu}(u)=\nu,&\end{array}\right.$ (2.9)

if

$\int_{\Omega}u\,L_{\mu}\zeta\,dx=\int_{\Omega}\zeta\,d\tau+\int_{\Omega}\mathbb{K}_{\mu}[\nu]L_{\mu}\zeta\,dx\;,\qquad\forall\zeta\in\mathbf{X}_{\mu}(\Omega,K),$

where

${\bf X}_{\mu}(\Omega,K)=\Big{\\{}\zeta\in
H_{loc}^{1}(\Omega):\phi_{\mu}^{-1}\zeta\in
H^{1}(\Omega;\phi_{\mu}^{2}),\,\phi_{\mu}^{-1}L_{\mu}\zeta\in
L^{\infty}(\Omega)\Big{\\}}.$ (2.10)

Let us state our main result for problem (2.9).

######  Theorem 2.12.

Let $\tau\in\mathfrak{M}(\Omega;\phi_{\mu})$ and
$\nu\in\mathfrak{M}(\partial\Omega)$. There exists a unique weak solution
$u\in L^{1}(\Omega;\phi_{\mu})$ of (2.9),

$u=\mathbb{G}_{\mu}[\tau]+\mathbb{K}_{\mu}[\nu]$ (2.11)

Furthermore there exists a positive constant $C=C(\Omega,K,\mu)$ such that

$\displaystyle\|u\|_{L^{1}(\Omega;\phi_{\mu})}\leq\frac{1}{\lambda_{\mu}}\|\tau\|_{\mathfrak{M}(\Omega;\phi_{\mu})}+C\|\nu\|_{\mathfrak{M}(\partial\Omega)}.$
(2.12)

If in addition $d\tau=fdx+d\rho$ where $f\in L^{1}(\Omega;\phi_{\mu})$ and
$\rho\in\mathfrak{M}(\Omega;\phi_{\mu})$, then for any
$\zeta\in\mathbf{X}_{\mu}(\Omega,K)$ with $\zeta\geq 0$, there hold

$\int_{\Omega}|u|L_{\mu}\zeta\,dx\leq\int_{\Omega}\mathrm{sign}(u)f\zeta\,dx+\int_{\Omega}\zeta
d|\rho|+\int_{\Omega}\mathbb{K}_{\mu}[|\nu|]L_{\mu}\zeta\,dx,$ (2.13)
$\int_{\Omega}u_{+}L_{\mu}\zeta\,dx\leq\int_{\Omega}\mathrm{sign}_{+}(u)f\zeta\,dx+\int_{\Omega}\zeta\,d\rho_{+}+\int_{\Omega}\mathbb{K}_{\mu}[\nu_{+}]L_{\mu}\zeta\,dx.$
(2.14)

It is worth mentioning here that Marcus and Nguyen [MN1] studied problem (2.9)
by introducing an alternative normalized boundary trace $\text{tr}^{*}(u)$
(see [MN1, Definition 1.2]). However this normalized boundary trace is well
defined only if $\mu<\min(C_{\Omega,K},\frac{2k-1}{4}).$ As a consequence they
showed that the boundary value problem

$\left\\{\begin{array}[]{ll}L_{\mu}u=\tau,&\qquad\text{in
}\;\Omega,\\\\[2.84544pt] \text{tr}^{*}(u)=\nu,&\end{array}\right.$

admits a unique solution provided $\mu<\min(C_{\Omega,K},\frac{2k-1}{4}).$

Finally, we study a semilinear problem involving the operator $L_{\mu}$ and a
subcritical absorption (for relevant works see [CV2, CV1, DD1, DN, GV, MN1,
MV1, MN2, MM, GN1, GN2]). In particular we are concerned with the problem

$\left\\{\begin{array}[]{ll}L_{\mu}u+g(u)=0,&\qquad\text{in
}\;\Omega,\\\\[2.84544pt] \mathrm{tr}_{\mu}(u)=\nu,&\end{array}\right.$ (2.15)

where $g:\mathbb{R}\rightarrow\mathbb{R}$ is a nondecreasing continuous
function such that $g(0)=0$. The above problem was treated by Marcus and
Nguyen [MN1], where they consider the normalized boundary trace
$\text{tr}^{*}(u)$ instead of $\mathrm{tr}_{\mu}(u).$ In the case where the
boundary trace is defined in a dynamic way, we have the following result.

######  Theorem 2.13.

Let $\mu\leq k^{2}/4$ and let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a
nondecreasing continuous function such that $g(0)=0$. Assume that for some
$p>1$ there holds

$\int_{1}^{\infty}t^{-1-p}(g(t)-g(-t))dt<+\infty.$ (2.16)

Let $\nu\in\mathfrak{M}(\partial\Omega)$. Then

(a) If (2.16) holds true with
$p=\min\left(\frac{N+1}{N-1},\frac{N+{\gamma_{+}}+1}{N+{\gamma_{+}}-1}\right)$
then there exists a unique weak solution $u$ of (2.15).

(b) Assume that either $k<N$ or $k=N$ and $\mu<N^{2}/4$. If $\nu$ has support
in $K$ and (2.16) holds true with
$p=\frac{N+{\gamma_{+}}+1}{N+{\gamma_{+}}-1}$ then there exists a unique weak
solution $u$ of (2.15).

(c) If $\nu$ has compact support in $\partial\Omega\setminus K$ and (2.16)
holds true with $p=\frac{N+1}{N-1}$ then there exists a unique weak solution
$u$ of (2.15).

Moreover in all three cases the weak solution $u$ satisfies

$u+\mathbb{G}_{\mu}[g(u)]=\mathbb{K}_{\mu}[\nu],\quad
a.e.\;\text{in}\;\Omega.$

When $k=N$ and $\mu=N^{2}/4$, the subcritical condition (2.16) must be
modified by a logarithmic factor in order to obtain the desired result, that
is

######  Theorem 2.14.

Let $K=\\{0\\}$ and $\mu=\frac{N^{2}}{4}$. Assume that $g$ satisfies

$\displaystyle\int_{1}^{\infty}s^{-\frac{N+2}{N-2}-1}(\ln
s)^{\frac{N+2}{N-2}}\big{(}g(s)-g(-s)\big{)}ds<\infty$

and let $\alpha>0$. Then there exists a unique solution $u$ of (2.15) with
$\nu=\alpha\delta_{0}$. Moreover the function $u$ satisfies

$\displaystyle
u+\mathbb{G}_{\mu}[g(u)]=\alpha\mathbb{K}_{\mu}[\delta_{0}],\quad
a.e.\;\text{in}\;\Omega.$

##  3\. Hardy-Sobolev type inequalities

In this section we shall prove various Hardy-Sobolev type inequalities that
will be essential for our analysis. We start by recalling the following
result:

######  Proposition 3.1.

[FM, Lemma 2.1] There exists $\beta_{0}=\beta_{0}(K,\Omega)$ small enough such
that, for any $x\in\Omega\cap K_{\beta_{0}},$ the following estimates hold:

a) $\displaystyle\tilde{d}_{K}^{2}(x)=d^{2}_{K}(x)(1+g(x))$ b)
$\displaystyle\nabla
d(x)\cdot\nabla\tilde{d}_{K}(x)=\frac{d(x)}{\tilde{d}_{K}(x)}$ c)
$\displaystyle|\nabla\tilde{d}_{K}(x)|^{2}=1+h(x)$ d)
$\displaystyle\tilde{d}_{K}(x)\Delta\tilde{d}_{K}(x)=k-1+f(x),$

where the functions $g,\;h$ and $f$ satisfy

$|g(x)|+|h(x)|+|f(x)|\leq C_{1}(\beta_{0},N)\tilde{d}_{K}(x),\quad\forall
x\in\Omega\cap K_{\beta_{0}}.$ (3.1)

######  Lemma 3.2.

Assume that $\alpha\neq 0$ and $\gamma+\alpha+k-1\neq 0$. There exists
$\beta_{0}>0$ and $C=C(\gamma,\alpha,k,\beta_{0},N)$ such that for any open
$V\subset K_{\beta_{0}}\cap\Omega$ and for any $u\in C^{\infty}_{c}(V)$ there
holds

$\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx+\int_{V}d^{\alpha-1}\tilde{d}_{K}^{\gamma}|u|dx\leq
C\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|\nabla u|dx.$

###### Proof.

By Proposition 3.1 we have

$\displaystyle\gamma\int_{V}$ $\displaystyle
d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx+\gamma\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}h|u|dx=\int_{V}d^{\alpha}\nabla\tilde{d}_{K}^{\gamma}\cdot\nabla\tilde{d}_{K}|u|dx$
$\displaystyle=-\alpha\int_{V}d^{\alpha-1}\tilde{d}_{K}^{\gamma}\nabla
d\cdot\nabla\tilde{d}_{K}|u|dx-\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}\Delta\tilde{d}_{K}|u|dx-\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}\nabla\tilde{d}_{K}\cdot\nabla|u|dx$
$\displaystyle=-\alpha\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx-\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}(k-1+f)|u|dx-\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}\nabla\tilde{d}_{K}\cdot\nabla|u|dx\,,$

that is

$(\gamma+\alpha+k-1)\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx=-\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}(f+\gamma
h)|u|dx-\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}\nabla\tilde{d}_{K}\cdot\nabla|u|dx\,.$

By the above equality, Proposition 3.1 and (3.1), we can easily prove that

$\displaystyle\big{(}|\gamma+\alpha+k-1|-C(C_{1},\gamma)\beta_{0}\big{)}\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx\leq(1+C_{1}\sqrt{\beta_{0}})\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|\nabla
u|dx,$

where $C_{1}=C_{1}(\beta_{0},N)$ is the constant in inequality (3.1). Choosing
$\beta_{0}$ small enough, we obtain

$\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx\leq
C\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|\nabla u|dx.$ (3.2)

By (3.2) and Proposition 3.1 we have

$\displaystyle\left|\alpha\int_{V}d^{\alpha-1}\tilde{d}_{K}^{\gamma}|u|dx\right|$
$\displaystyle=\left|\int_{V}(\nabla d^{\alpha}\cdot\nabla
d)\tilde{d}_{K}^{\gamma}|u|dx\right|$ $\displaystyle\leq
C\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma-1}|u|dx+\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|\nabla
u|dx,$

provided $\beta_{0}$ is small enough. The result now follows. ∎

######  Lemma 3.3.

Assume that $a\neq 0$ and $c+a+k-1\neq 0$. Let $1\leq q\leq\frac{N}{N-1}$ and
$b=a-1+N\frac{q-1}{q}$. If $\beta_{0}$ is small enough then there exists
$C=C(a,c,k,\beta_{0},q,N)$ such that for any open $V\subset\Omega\cap
K_{\beta_{0}}$ and for any $u\in C^{\infty}_{c}(V)$ the following inequality
is valid

$\Big{(}\int_{V}d^{qb}\tilde{d}_{K}^{qc}|u|^{q}dx\Big{)}^{\frac{1}{q}}\leq
C\int_{V}d^{a}\tilde{d}_{K}^{c}|\nabla u|dx.$ (3.3)

###### Proof.

Let $0\leq\theta_{i}\leq 1,i=1,2$, be such that $\theta_{1}+\theta_{2}=1$ and
$\frac{N-1}{N}\theta_{1}+\theta_{2}=\frac{1}{q}$. By Hölder inequality we have

$\displaystyle\int_{V}d^{qb}\tilde{d}_{K}^{qc}|u|^{q}dx$
$\displaystyle=\int_{V}\Big{(}d^{qa\theta_{1}}\tilde{d}_{K}^{qc\theta_{1}}|u|^{\theta_{1}q}\Big{)}\left(d^{q(a-1)\theta_{2}}\tilde{d}_{K}^{qc\theta_{2}}|u|^{\theta_{2}q}\right)dx$
$\displaystyle\leq\|d^{a}\tilde{d}^{c}_{K}u\|_{L^{\frac{N}{N-1}}(V)}^{\theta_{1}q}\|d^{a-1}\tilde{d}^{c}_{K}u\|_{L^{1}(V)}^{\theta_{2}q},$

and therefore

$\|d^{b}\tilde{d}_{K}^{c}u\|_{L^{q}(V)}\leq\|d^{a}\tilde{d}^{c}_{K}u\|_{L^{\frac{N}{N-1}}(V)}+\|d^{a-1}\tilde{d}^{c}_{K}u\|_{L^{1}(V)}.$
(3.4)

By the $L^{1}$ Sobolev inequality and Lemma 3.2 we have

$\displaystyle\|d^{a}\tilde{d}^{c}_{K}u\|_{L^{\frac{N}{N-1}}(V)}$
$\displaystyle\leq
C\left(|c|\int_{V}d^{a}\tilde{d}_{K}^{c-1}|u|dx+|a|\int_{V}d^{a-1}\tilde{d}_{K}^{c}|u|dx+\int_{V}d^{a}\tilde{d}_{K}^{c}|\nabla
u|dx\right)$ $\displaystyle\leq C\int_{V}d^{a}\tilde{d}_{K}^{c}|\nabla u|dx.$

Combining this with Lemma 3.2 and (3.4) concludes the proof. $\hfill\Box$

######  Lemma 3.4.

Assume that $a\neq 0$ and $c+a+k-1\neq 0$. Let $2<Q\leq\frac{2N}{N-2}$ and
$b=a-1+N\frac{Q-2}{2Q}$. If $\beta_{0}$ is small enough then there exists
$C=C(c,a,k,\beta_{0},Q,N)$ such that for any open $V\subset\Omega\cap
K_{\beta_{0}}$ and for any $v\in C^{\infty}_{c}(V)$ there holds

$\bigg{(}\int_{V}(d^{b}\tilde{d}_{K}^{c})^{\frac{2Q}{Q+2}}|v|^{Q}dx\bigg{)}^{\frac{2}{Q}}\leq
C\int_{V}d^{2a-\frac{2Qb}{Q+2}}\tilde{d}_{K}^{\frac{4c}{Q+2}}|\nabla
v|^{2}dx.$

###### Proof.

Let $s=\frac{Q}{2}+1$ and write $Q=qs$. Applying (3.3) to the function
$u=|v|^{s}$ we obtain

$\displaystyle\left(\int_{V}\big{(}d^{b}\tilde{d}_{K}^{c}\big{)}^{\frac{2Q}{Q+2}}|v|^{Q}dx\right)^{\frac{Q+2}{2Q}}$
$\displaystyle\leq C\int_{V}d^{a}\tilde{d}_{K}^{c}|v|^{\frac{Q}{2}}|\nabla
v|dx.$ (3.5)

Now, by Schwarz inequality, we have

$\displaystyle\int_{V}d^{a}\tilde{d}_{K}^{c}|v|^{\frac{Q}{2}}|\nabla v|dx$
$\displaystyle=\int_{V}d^{b{\frac{Q}{Q+2}}}\tilde{d}_{K}^{c{\frac{Q}{Q+2}}}|v|^{\frac{Q}{2}}\,d^{a-b{\frac{Q}{Q+2}}}\tilde{d}_{K}^{c(1-{\frac{Q}{Q+2}})}|\nabla
v|dx$
$\displaystyle\leq\left(\int_{V}\big{(}d^{b}\tilde{d}_{K}^{c}\big{)}^{\frac{2Q}{Q+2}}|v|^{Q}dx\right)^{\frac{1}{2}}\left(\int_{V}d^{2a-\frac{2Qb}{Q+2}}\tilde{d}_{K}^{c(2-\frac{2Q}{Q+2})}|\nabla
v|^{2}dx\right)^{\frac{1}{2}}.$

The result follows by (3.5) and the last inequality. ∎

######  Corollary 3.5.

Let $\alpha\neq 0$ and assume that $(\alpha+\gamma)\frac{N-1}{N-2}+k-1\neq 0$.
There exists $\beta_{0}$ small enough and $C>0$ such that for any open
$V\subset\Omega\cap K_{\beta_{0}}$ and for all $u\in C^{\infty}_{c}(V)$ there
holds

$\left(\int_{V}\big{(}d^{\frac{\alpha}{2}}\tilde{d}_{K}^{\frac{\gamma}{2}}|u|\big{)}^{\frac{2N}{N-2}}dx\right)^{\frac{N-2}{N}}\leq
C\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|\nabla u|^{2}dx.$

###### Proof.

We apply Lemma 3.4 with $Q=\frac{2N}{N-2},$ $a=\alpha\frac{N-1}{N-2},$
$c=\gamma(\frac{N-1}{N-2})$. ∎

######  Corollary 3.6.

Let $\alpha>0$ and $\gamma\geq 0$. There exists $\beta_{0}>0$ and $C>0$ such
that for any open $V\subset\Omega\cap K_{\beta_{0}}$ and all $u\in
C^{\infty}_{c}(V),$ the following inequality is valid

$\left(\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|u|^{\frac{2(N+\alpha+\gamma)}{N+\alpha+\gamma-2}}dx\right)^{\frac{N+\alpha+\gamma-2}{N+\alpha+\gamma}}\leq
C\int_{V}d^{\alpha+\frac{2\gamma}{N+a+\gamma}}\tilde{d}_{K}^{\gamma-\frac{2\gamma}{N+a+\gamma}}|\nabla
u|^{2}dx.$

###### Proof.

This follows by Lemma 3.4 with
$Q=\frac{2(N+\alpha+\gamma)}{N+\alpha+\gamma-2},$ $c=\frac{\gamma}{q}$,
$b=\frac{\alpha}{q},$ where $q=\frac{2Q}{Q+2}$. ∎

######  Corollary 3.7.

Let $\alpha>0$, $\gamma<0$ and assume that
$\alpha+\gamma\frac{N+\alpha-1}{N+\alpha}+k-1\neq 0$. There exists
$\beta_{0}>0$ and $C>0$ such that for any open $V\subset\Omega\cap
K_{\beta_{0}}$ and all $u\in C^{\infty}_{c}(V)$ there holds

$\left(\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma}|u|^{\frac{2(N+\alpha)}{N+\alpha-2}}dx\right)^{\frac{N+\alpha-2}{N+\alpha}}\leq
C\int_{V}d^{\alpha}\tilde{d}_{K}^{\gamma\frac{N+\alpha-2}{N+\alpha}}|\nabla
u|^{2}dx\,.$

###### Proof.

The proof follows from Lemma 3.4, with $Q=\frac{2(N+\alpha)}{N+\alpha-2},$
$c=\gamma\frac{N+\alpha-1}{N+\alpha}$ and
$b=\alpha\frac{N+\alpha-1}{N+\alpha}$. ∎

##  4\. Heat Kernel Estimates for small time

We are now going to introduce some notation and tools that will be useful for
our local analysis near $K$ and $\partial\Omega$; see e.g. [K].

Let $x=(x^{\prime},x^{\prime\prime})\in{\mathbb{R}}^{N}$,
$x^{\prime}=(x_{1},..,x_{k})\in{\mathbb{R}}^{k}$,
$x^{\prime\prime}=(x_{k+1},...,x_{N})\in{\mathbb{R}}^{N-k}$. For $\beta>0$, we
denote by $B_{\beta}^{k}(x^{\prime})$ the ball in ${\mathbb{R}}^{k}$ with
center $x^{\prime}$ and radius $\beta$. For any $\xi\in K$ we also set

$V_{K}(\xi,\beta)=\Big{\\{}x=(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta,\;|x_{i}-\Gamma_{i,K}^{\xi}(x^{\prime\prime})|<\beta,\;\forall
i=1,...,k\Big{\\}},$

for some functions $\Gamma_{i,K}^{\xi}:{\mathbb{R}}^{N-k}\to{\mathbb{R}}$,
$i=1,...,k$.

Since $K$ is a $C^{2}$ compact submanifold in $\mathbb{R}^{N}$ without
boundary, there exists $\beta_{0}>0$ such that

  * •

For any $x\in K_{6\beta_{0}}$, there is a unique $\xi\in K$ satisfying
$|x-\xi|=d_{K}(x)$.

  * •

$d_{K}\in C^{2}(K_{4\beta_{0}})$, $|\nabla d_{K}|=1$ in $K_{4\beta_{0}}$ and
there exists $g\in L^{\infty}(K_{4\beta_{0}})$ such that

$\Delta d_{K}(x)=\frac{k-1}{d_{K}(x)}+g(x),\qquad\text{in }K_{4\beta_{0}}.$

(See [V, Lemma 2.2] and [DN, Lemma 6.2].)

  * •

For any $\xi\in K$, there exist $C^{2}$ functions $\Gamma_{i,K}^{\xi}\in
C^{2}({\mathbb{R}}^{N-k};{\mathbb{R}})$, $i=1,...,k$, such that defining

$V_{K}(\xi,\beta):=\Big{\\{}x=(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta,\;|x_{i}-\Gamma_{i,K}^{\xi}(x^{\prime\prime})|<\beta,\;i=1,...,k\Big{\\}},$

we have (upon relabelling and reorienting the coordinate axes if necessary)

$V_{K}(\xi,\beta)\cap
K=\Big{\\{}x=(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta,\;x_{i}=\Gamma_{i,K}^{\xi}(x^{\prime\prime}),\;i=1,...,k\Big{\\}}.$

  * •

There exist $\xi^{j}$, $j=1,...,m_{0}$, ($m_{0}\in{\mathbb{N}}$) and
$\beta_{1}\in(0,\beta_{0})$ such that

$K_{2\beta_{1}}\subset\bigcup_{i=1}^{m_{0}}V_{K}(\xi^{i},\beta_{0}).$ (4.1)

Now set

$\delta_{K}^{\xi}(x):=\Big{(}\sum_{i=1}^{k}|x_{i}-\Gamma_{i,K}^{\xi}(x^{\prime\prime})|^{2}\Big{)}^{\frac{1}{2}},\qquad
x=(x^{\prime},x^{\prime\prime})\in V_{K}(\xi,4\beta_{0}).$

Then there exists a constant $C=C(N,K)$ such that

$d_{K}(x)\leq\delta_{K}^{\xi}(x)\leq C\|K\|_{C^{2}}d_{K}(x),\quad\forall x\in
V_{K}(\xi,2\beta_{0}),$ (4.2)

where $\xi^{j}=((\xi^{j})^{\prime},(\xi^{j})^{\prime\prime})\in K$,
$j=1,...,m_{0}$, are the points in (4.1) and

$\|K\|_{C^{2}}:=\sup\\{\|\Gamma_{i,K}^{\xi^{j}}\|_{C^{2}(B_{5\beta_{0}}^{N-k}((\xi^{j})^{\prime\prime}))}:\;i=1,...,k,\;j=1,...,m_{0}\\}<\infty.$

For simplicity we shall write $\delta_{K}$ instead of $\delta_{K}^{\xi}$.
Moreover, $\beta_{1}$ can be chosen small enough so that for any $x\in
K_{\beta_{1}}$,

$B(x,\beta_{1})\subset V_{K}(\xi,\beta_{0}),$

where $\xi\in K$ satisfies $|x-\xi|=d_{K}(x)$.

When $K=\partial\Omega$ we assume that

$V_{\partial\Omega}(\xi,\beta)\cap\Omega=\Big{\\{}x:\sum_{i=2}^{N}|x_{i}-\xi_{i}|^{2}<\beta^{2},\;0<x_{1}-\Gamma_{1,\partial\Omega}^{\xi}(x_{2},...,x_{N})<\beta\Big{\\}}.$

Thus, when $x\in K\subset\partial\Omega$ is a $C^{2}$ compact submanifold in
$\mathbb{R}^{N}$ without boundary, of co-dimension $k$, $1<k\leq N,$ we have
that

$\Gamma_{1,K}^{\xi}(x^{\prime\prime})=\Gamma_{1,\partial\Omega}^{\xi}(\Gamma_{2,K}^{\xi}(x^{\prime\prime}),...,\Gamma_{k,K}^{\xi}(x^{\prime\prime}),x^{\prime\prime}).$
(4.3)

Let $\xi\in K$. For any $x\in V_{K}(\xi,\beta_{0})\cap\Omega,$ we define

$\delta(x)=x_{1}-\Gamma_{1,\partial\Omega}^{\xi}(x_{2},...,x_{N}),$

and

$\delta_{2,K}(x)=\Big{(}\sum_{i=2}^{k}|x_{i}-\Gamma_{i,K}^{\xi}(x^{\prime\prime})|^{2}\Big{)}^{\frac{1}{2}}.$

Then by (4.3), there exists a constant $A>1$ which depends only on $\Omega$,
$K$ and $\beta_{0}$ such that

$\frac{1}{A}(\delta_{2,K}(x)+\delta(x))\leq\delta_{K}(x)\leq
A(\delta_{2,K}(x)+\delta(x)),$ (4.4)

Thus by (4.2) and (4.4) there exists a constant $C=C(\Omega,K,\gamma)>1$ which
depends on $k,N,\Gamma_{i,K}^{\xi},\Gamma_{1,\partial\Omega}^{\xi},\gamma$
such that

$C^{-1}\delta^{2}(x)(\delta_{2,K}(x)+\delta(x))^{\gamma}\leq
d^{2}(x)d_{K}^{\gamma}(x)\leq
C\delta^{2}(x)(\delta_{2,K}(x)+\delta(x))^{\gamma}.$ (4.5)

We set

$\displaystyle\mathcal{V}_{K}(\xi,\beta_{0})=\left\\{(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta_{0},\;|\delta(x)|<\beta_{0},\;|\delta_{2,K}(x)|<\beta_{0}\right\\}.$

We may then assume that

$\displaystyle\mathcal{V}_{K}(\xi,\beta_{0})\cap\Omega=\left\\{(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta_{0},0<\delta(x)<\beta_{0},\;|\delta_{2,K}(x)|<\beta_{0}\right\\},$

$\displaystyle\mathcal{V}_{K}(\xi,\beta_{0})\cap\partial\Omega=\left\\{(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta_{0},\delta(x)=0,\;|\delta_{2,K}(x)|<\beta_{0}\right\\},$

and

$\displaystyle\mathcal{V}_{K}(\xi,\beta_{0})\cap
K=\left\\{(x^{\prime},x^{\prime\prime}):|x^{\prime\prime}-\xi^{\prime\prime}|<\beta_{0},\delta(x)=0,\;\delta_{2,K}=0\right\\}.$

Let $\beta_{1}>0$, $1<\gamma<2,$ and $0<r<\beta_{1}.$ For any $x\in
V_{\partial\Omega}(\xi,\frac{\beta_{0}}{16})$ with $d(x)\leq\gamma r,$. Taking
$\beta_{1}$ small enough we have

$\mathcal{D}(x,r):=\Big{\\{}y:\sum_{i=2}^{N}|y_{i}-x_{i}|^{2}<r^{2},\;|\delta(y)|<r+d(x)\Big{\\}}\subset\subset
V_{\partial\Omega}(\xi,\frac{\beta_{0}}{16}).$

and there exist $C_{\xi}=C(\Gamma^{\xi},\Omega)>1,$ such that

$\mathcal{D}(x,r)\subset B(x,C_{\xi}r).$ (4.6)

Also,

$\mathcal{D}(x,r)\cap\Omega=\\{y:\sum_{i=2}^{N}|y_{i}-x_{i}|^{2}<r^{2},\;0<\delta(y)<r+d(x)\\}.$

######  Definition 4.1.

Let $\beta_{1}>0$ be small enough, $r\in(0,\beta_{1}),$ $b\in(1,2),$ $\xi\in
K$ and $x\in V(\xi,\frac{\beta_{0}}{16}).$ We define

(i) $\displaystyle\mathcal{B}(x,r)=B(x,r),\mbox{ if }d(x)>br$ (ii)
$\displaystyle\mathcal{B}(x,r)=\mathcal{D}(x,r),\mbox{ if $d(x)\leq br$ and
}d_{K}(x)>bC_{\xi}r$ (iii)
$\displaystyle\mathcal{B}(x,r)=\\{y=(y^{\prime},y^{\prime\prime}):|y^{\prime\prime}-x^{\prime\prime}|<r,\;|\delta_{2,K}(y)|<r+d_{K}(x),\;|\delta(y)|<r+d(x)\\},$
if $d(x)\leq br$ and $d_{K}(x)\leq bC_{\xi}r$.

Finally, we set

$\overline{\mathcal{M}}(x,r)=\int_{\mathcal{B}(x,r)\cap\Omega}d^{2}(y)d^{\gamma}_{K}(y)dy.$

###  4.1. Doubling Property

######  Lemma 4.2.

Let $\gamma\geq-k$. Let $\xi\in\partial\Omega$ and $x\in
V(\xi,\frac{\beta_{0}}{16}).$ Then, there exist $\beta_{1}>0$ and
$C=C(\Omega,K,\gamma,\beta_{0})>1$ such that

$\frac{1}{C}(r+d(x))^{2}(r+d_{K}(x))^{\gamma}r^{N}\leq\overline{\mathcal{M}}(x,r)\leq
C(r+d(x))^{2}(r+d_{K}(x))^{\gamma}r^{N},$ (4.7)

for any $0<r<\beta_{1}.$

###### Proof.

We will consider three cases.

Case 1. $d(x)>br$ Since $d_{K}(x)\geq d(x),$ we can easily show that for any
$y\in B(x,r)$ we have $\frac{b-1}{b}d(x)\leq d(y)\leq\frac{b+1}{b}d(x)$ and
$\frac{b-1}{b}d_{K}(x)\leq d_{K}(y)\leq\frac{b+1}{b}d_{K}(x)$. Thus the proof
of (4.7) follows easily in this case.

Case 2. $d(x)\leq br$ and $d_{K}(x)>bC_{\xi}r.$ By (4.6), we again have that
$\frac{b-1}{b}d_{K}(x)\leq d_{K}(y)\leq\frac{b+1}{b}d_{K}(x).$ Using the last
inequality and proceeding as the proof of [FMT1, Lemma 2.2], we obtain the
desired result.

Case 3. $d(x)\leq br$ and $d_{K}(x)\leq bC_{\xi}r.$

Let $\overline{y}=(y_{2},...,y_{k})\in\mathbb{R}^{k-1}.$ By (4.5) and the
definition of $\mathcal{B}(x,r)$ , we have

$\displaystyle\overline{\mathcal{M}}(x,r)$
$\displaystyle=\int_{\mathcal{B}(x,r)\cap\Omega}d^{2}(y)d^{\gamma}_{K}(y)dy\leq\int_{\mathcal{B}(x,r)\cap\Omega}C\delta^{2}(y)(\delta_{2,K}(y)+\delta(y))^{\gamma}dy$
$\displaystyle\leq
C\int_{B^{N-k}(x^{\prime\prime},r)}\int_{0}^{d(x)+r}\int_{|\overline{y}|<d_{K}(x)+r}(|\overline{y}|+y_{1})^{\gamma}y_{1}^{2}d\overline{y}\,dy_{1}\,dy^{\prime\prime}$
$\displaystyle=CC(k,N)r^{N-k}\int_{0}^{d(x)+r}\int_{0}^{d_{K}(x)+r}s^{k-2}(s+y_{1})^{\gamma}y_{1}^{2}ds\,dy_{1}.$
(4.8)

Now, if $\gamma>0$ then

$\displaystyle\int_{0}^{d(x)+r}$
$\displaystyle\int_{0}^{d_{K}(x)+r}s^{k-2}(s+y_{1})^{\gamma}y_{1}^{2}ds\,dy_{1}$
$\displaystyle\leq\frac{1}{k-1}(2r+d(x)+d_{K}(x))^{\gamma}(d_{K}(x)+r)^{k-1}(d(x)+r)^{3}$
$\displaystyle\leq\frac{(b+2)^{\gamma}(bC_{\xi}+1)^{k-1}(b+1)}{k-1}(r+d_{K}(x))^{\gamma}(d(x)+r)^{2}r^{k}.$

If $-k\leq\gamma\leq 0,$ then

$\displaystyle\int_{0}^{d(x)+r}$
$\displaystyle\int_{0}^{d_{K}(x)+r}s^{k-2}(s+y_{1})^{\gamma}y_{1}^{2}ds\,dy_{1}$
$\displaystyle\leq\int_{0}^{d(x)+r}\int_{0}^{d_{K}(x)+r}s^{k-2}(s+y_{1})^{\gamma+2}ds\,dy_{1}$
$\displaystyle\leq\int_{0}^{d(x)+r}\int_{0}^{d_{K}(x)+r}(s+y_{1})^{\gamma+k}ds\,dy_{1}$
$\displaystyle\leq(d_{K}(x)+r)(d(x)+r)(2r+d(x)+d_{K}(x))^{\gamma+k}$
$\displaystyle\leq(2C_{\xi}+2)(d(x)+r)^{2}(d_{K}(x)+r)^{\gamma}(2r+d(x)+d_{K}(x))^{k}$
$\displaystyle\leq(2C_{\xi}+2)(2+b+bC_{\xi})^{k}(d(x)+r)^{2}(d_{K}(x)+r)^{\gamma}r^{k}.$

Similarly, for the opposite inequality, we have

$\displaystyle\int_{0}^{d(x)+r}$
$\displaystyle\int_{0}^{d_{K}(x)+r}s^{k-2}(s+y_{1})^{\gamma}y_{1}^{2}ds\,dy_{1}$
$\displaystyle\geq\int_{\frac{d(x)+r}{2}}^{d(x)+r}\int_{\frac{d_{K}(x)+r}{2}}^{d_{K}(x)+r}s^{k-2}(s+y_{1})^{\gamma}y_{1}^{2}dsdy_{1}$
$\displaystyle\geq
C(b,C_{\xi},k,\gamma)(d(x)+r)^{2}(d_{K}(x)+r)^{\gamma}r^{k}.$ (4.9)

The desired result follows by (4.8)-(4.9). ∎

From (2.2) and Lemma 4.2, we have the following corollary.

######  Corollary 4.3.

Let $x\in V(\xi,\frac{\beta_{0}}{16})$ and

$\mathcal{M}(x,r)=\int_{\mathcal{B}(x,r)\cap\Omega}\phi_{\mu}^{2}(y)dy.$

Then, there exist $\beta_{1}>0$ and $C=C(\Omega,K,\beta_{0})>1$ such that

$\frac{1}{C}(r+d(x))^{2}(r+d_{K}(x))^{2\gamma_{+}}r^{N}\leq\mathcal{M}(x,r)\leq
C(r+d(x))^{2}(r+d_{K}(x))^{2\gamma_{+}}r^{N},$

for any $0<r<\beta_{1}.$

###  4.2. Density of $C^{\infty}_{c}(\Omega)$ functions

######  Lemma 4.4.

Let $k\leq N$, $\gamma\geq-k,$
$x=(x_{1},x_{2},...,x_{k},x_{k+1},...,x_{N})=(x_{1},\overline{x},x^{\prime\prime})$.
Let

$O=(0,1)\times B^{\mathbb{R}^{k-1}}(0,1)\times B^{\mathbb{R}^{N-k}}(0,1)$

and $u\in H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})$. Assume that
there exists $0<\varepsilon_{0}<1$ such that $u(x)=0$ if either
$x_{1}>\varepsilon_{0}$ or
$|\overline{x}|^{2}+|x^{\prime\prime}|^{2}>\varepsilon_{0}^{2}$. Then there
exists a sequence $\\{u_{n}\\}_{n=1}^{\infty}\subset C^{\infty}_{c}(O)$ such
that

$u_{n}\rightarrow
u,\quad\text{in}\;H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})$

###### Proof.

Let $m\in\mathbb{N}$. Set

$v_{m}(x)=\left\\{\begin{aligned} &m,\qquad&&\text{if }u(x)>m,\\\
&u(x),&&\text{if }-m\leq u(x)\leq m,\\\ &-m&&\text{if
}u(x)<-m.\end{aligned}\right.$

Then we can easily prove that $v_{m}\rightarrow u$ in
$H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})$.

Let $\varepsilon>0$. There exists $m_{0}\in\mathbb{N},$ such that

$\left\|v_{m_{0}}-u\right\|_{H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})}=\left(\int_{O}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}(|v_{m_{0}}-u|^{2}+|\nabla
v_{m_{0}}-\nabla u|^{2})dx\right)^{\frac{1}{2}}<\frac{\varepsilon}{3}.$ (4.10)

For any $0<h<1,$ we consider the function

$\eta_{h}(x_{1})=\left\\{\begin{aligned} &1\qquad&&\text{if }x_{1}>h,\\\
&1-(\ln h)^{-1}\ln\left(\frac{x_{1}}{h}\right)&&\text{if }h^{2}\leq x_{1}\leq
h,\\\ &0&&\text{if }x_{1}<h^{2},\end{aligned}\right.$

We will show that $z_{h}:=\eta_{h}v_{m_{0}}\rightarrow v_{m_{0}}$ in
$H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})$, as $h\rightarrow 0^{+}.$
We can easily show that $z_{h}\rightarrow v_{m_{0}}$ in
$L^{2}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})$. Also,

$\displaystyle\begin{aligned}
\int_{O}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}|\nabla(v_{m_{0}}(1-\eta_{h}))|^{2}dx\leq&\;2\int_{O}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}|\nabla
v_{m_{0}}|^{2}|(1-\eta_{h})|^{2}dx\\\
&+2\int_{O}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}|v_{m_{0}}|^{2}|\nabla\eta_{h}|^{2}dx\\\
\leq&\;2\int_{O}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}|\nabla
v_{m_{0}}|^{2}|(1-\eta_{h})|^{2}dx\\\ &+C(N,k)m_{0}^{2}(\ln
h)^{-2}\int_{h^{2}}^{h}\int_{0}^{1}(x_{1}+r)^{\gamma}r^{k-2}drdx_{1}\rightarrow
0,\end{aligned}$

since $\gamma\geq-k.$ Thus there exists $h_{0}\in(0,1)$ such that

$\left\|v_{m_{0}}-z_{h_{0}}\right\|_{H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})}<\frac{\varepsilon}{3}.$
(4.11)

Note that $z_{h_{0}}$ vanishes outside $\tilde{O}_{\sigma}=(\sigma,1)\times
B^{\mathbb{R}^{k-1}}(0,1)\times B^{\mathbb{R}^{N-k}}(0,1),$ for some
$\sigma=\sigma(h_{0})\in(0,1).$ Thus $z_{h_{0}}\in
H^{1}_{0}(\tilde{O}_{\sigma}),$ which implies the existence of a sequence
$\\{u_{n}\\}\subset C_{c}^{\infty}(\tilde{O}_{\sigma})$ such that
$u_{n}\rightarrow z_{h_{0}}$ in $H^{1}_{0}(\tilde{O}_{\sigma})$. Hence, there
exists $n_{0}\in\mathbb{N}$ such that

$\left\|z_{h_{0}}-u_{n}\right\|_{H^{1}(O;x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma})}<\frac{\varepsilon}{3},\quad\forall
n\geq n_{0}.$ (4.12)

The desired result follows by (4.10), (4.11) and (4.12). $\Box$

We write a point $x\in{\mathbb{R}}^{N}$ as
$x=(x_{1},x_{2},...,x_{k},x_{k+1},...,x_{N})=(x_{1},\overline{x},x^{\prime\prime})$.
Given $r_{1},r_{2},r_{3}>0$ we denote

$O_{r_{1},r_{2},r_{3}}=(0,r_{1})\times B^{\mathbb{R}^{k-1}}(0,r_{2})\times
B^{\mathbb{R}^{N-k}}(0,r_{3}).$

######  Theorem 4.5.

Assume that $\gamma\geq-k$. Then $C^{\infty}_{c}(\Omega)$ is dense in
$H^{1}(\Omega;d^{2}d_{K}^{\gamma}).$

###### Proof.

Let $u\in H^{1}(\Omega;d^{2}d_{K}^{\gamma})$ and $\beta_{0}>0$ be the constant
in Lemma 4.2. Let $\xi\in K$ and $0\leq\phi_{\xi}\leq 1$ be a smooth function
with $\text{supp}(\phi_{\xi})\subset\mathcal{V}_{K}(\xi,\frac{\beta_{0}}{8}),$
and $\phi=1$ in $\mathcal{V}_{K}(\xi,\frac{\beta_{0}}{16})$. Then the function
$v=u\phi_{\xi}$ belongs in $H^{1}(\Omega;d^{2}d_{K}^{\gamma})$.

By (4.5) we have

$\displaystyle\int_{\Omega}d^{2}(x)d_{K}^{\gamma}(x)(|v|^{2}+|\nabla
v|^{2})dx$ $\displaystyle\asymp
C(\Omega,K)\int_{\mathcal{V}_{K}(\xi,\frac{\beta_{0}}{8})}\delta^{2}(x)(\delta_{2,K}(x)+\delta(x))^{\gamma}(|v|^{2}+|\nabla
v|^{2})dx$ $\displaystyle\asymp
C(\Omega,K)\int_{O_{1,\frac{\beta_{0}}{8},\frac{\beta_{0}}{8}}}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}(|\overline{v}|^{2}+|\nabla_{y}\overline{v}|^{2})dy,$

where $\overline{y}=(y_{2},...,y_{k})$ and

$\displaystyle\overline{v}(y)=v\left(y_{1}+\Gamma^{\xi}_{1,\partial\Omega}\left(y_{2}+\Gamma^{\xi}_{2,K}(y^{\prime\prime}),...,y_{k}+\Gamma^{\xi}_{k,K}(y^{\prime\prime}),y^{\prime\prime}\right),y_{2}+\Gamma^{\xi}_{2,K}(y^{\prime\prime}),...,y_{k}+\Gamma^{\xi}_{k,K}(y^{\prime\prime}),y^{\prime\prime}\right).$

The desired result follows by Lemma 4.4 and a partition of unity argument. ∎

By Corollaries 3.6 and 3.7, Theorem 4.5 and using a partition of unity
argument, we obtain the following two results.

######  Corollary 4.6.

Let $\gamma\geq 0$. There exists $C=C(\Omega,K,\gamma)$ such that

$\left(\int_{\Omega}d^{2}d_{K}^{\gamma}|u|^{\frac{2(N+2+\gamma)}{N+\gamma}}dx\right)^{\frac{N+\gamma}{N+2+\gamma}}\leq
C\left(\int_{\Omega}d^{2}d_{K}^{\gamma}|\nabla
u|^{2}dx+\int_{\Omega}d^{2}d_{K}^{\gamma}u^{2}dx\right),$

for any $u\in H^{1}(\Omega;d^{2}d_{K}^{\gamma})$.

######  Corollary 4.7.

Let $-k\leq\gamma<0$. There exists $C=C(\Omega,K,\gamma)$ such that

$\left(\int_{\Omega}d^{2}d_{K}^{\gamma}|u|^{\frac{2(N+2)}{N}}dx\right)^{\frac{N}{N+2}}\leq
C\left(\int_{\Omega}d^{2}d_{K}^{\gamma\frac{N}{N+2}}|\nabla
u|^{2}dx+\int_{\Omega}d^{2}d_{K}^{\gamma}u^{2}dx\right),$

for any $u\in H^{1}(\Omega;d^{2}d_{K}^{\gamma})$.

###  4.3. Poincaré inequality

######  Lemma 4.8.

Let $1\leq k\leq N$ and $\gamma\geq-k$. Assume that
$0<c_{0}r_{2}<r_{3}<r_{1}<r_{2},$ for some constant $0<c_{0}<1.$ Then there
exists a positive constant $C=C(c_{0},N,K,\gamma)$ such that

$\inf_{\zeta\in\mathbb{R}}\int_{O_{r_{1},r_{2},r_{3}}}|f(x)-\zeta|^{2}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}dx\leq
Cr^{2}_{2}\int_{O_{r_{1},r_{2},r_{3}}}|\nabla
f(x)|^{2}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}dx,$

for any $f\in C^{1}(\overline{O}_{r_{1},r_{2},r_{3}}).$

###### Proof.

Let $\zeta\in\mathbb{R}$ and $y_{1}=\frac{x_{1}}{2r_{1}},$
$\overline{y}=\frac{\overline{x}}{2r_{2}}$ and
$y^{\prime\prime}=\frac{x^{\prime\prime}}{2r_{3}}.$ Set
$\overline{f}(y)=f(2r_{1}y_{1},2r_{2}\overline{y},2r_{3}y^{\prime\prime}).$
Then

$\displaystyle\int_{O_{r_{1},r_{2},r_{3}}}$
$\displaystyle|f(x)-\zeta|^{2}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}dx$
$\displaystyle\asymp
C(c_{0},N,k,\gamma)r_{2}^{N+\gamma+2}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\overline{f}(y)-\zeta|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy.$
(4.13)

Let

$\zeta_{\overline{f}}=\Big{(}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\Big{)}^{-1}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}\overline{f}(y)y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy.$

We assert that there exists a positive constant $C>0$ such that

$\displaystyle\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\overline{f}(y)-\zeta_{\overline{f}}|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\leq
C\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla\overline{f}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy,$
(4.14)

for any $\overline{f}\in
C^{1}(\overline{O}_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}).$

We will prove this by contradiction. Let $\\{\overline{f}_{n}\\}\subset
C^{1}(\overline{O}_{\frac{1}{2},\frac{1}{2},\frac{1}{2}})$ be a sequence such
that

$\displaystyle\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\overline{f}_{n}(y)-\zeta_{\overline{f}_{n}}|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy>n\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla\overline{f}_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy.$
(4.15)

Setting

$g_{n}(y)=(\overline{f}_{n}(y)-\zeta_{\overline{f}_{n}})\Big{(}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\overline{f}_{n}(y)-\zeta_{\overline{f}_{n}}|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\Big{)}^{-1},$

(4.15) becomes

$\displaystyle
1=\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy>n\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla
g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy$

and we also have $\zeta_{g_{n}}=0.$

Let $\varepsilon>0$. There exists an extension $\overline{g}_{n}$ of $g_{n}$
such that $\overline{g}_{n}=g_{n}$ in
$\overline{O}_{\frac{1}{2},\frac{1}{2},\frac{1}{2}},$ $\overline{g}_{n}\in
C^{1}(\overline{O}_{1,1,1}),$ $\overline{g}_{n}=0$ if $y_{1}>\frac{2}{3}$ or
$|\overline{y}|>\frac{2}{3}$ or $|y^{\prime\prime}|>\frac{2}{3}$ and there
exists a positive constant $C_{1}=C_{1}(N,k,q)$ such that

$\displaystyle\int_{O_{1,1,1}}|\overline{g}_{n}(y)|^{q}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\leq\;$
$\displaystyle
C_{1}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|g_{n}(y)|^{q}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy$
$\displaystyle\int_{O_{1,1,1}}|\nabla\overline{g}_{n}(y)|^{q}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\leq\;$
$\displaystyle
C_{1}\bigg{(}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla
g_{n}(y)|^{q}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy$
$\displaystyle+\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|g_{n}(y)|^{q}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\bigg{)},$

for any $q>1.$ Assume first that $-k\leq\gamma<0$. Given $\sigma\in(0,1/2)$,
by Corollary 3.7 we have that for some $C=C(\gamma,N,k)$,

$\displaystyle\int_{O_{\sigma,\frac{1}{2},\frac{1}{2}}}|g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy$
$\displaystyle\leq
C\sigma^{\frac{6}{N+2}}\Big{(}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\overline{g}_{n}(y)|^{\frac{2(N+2)}{N}}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\Big{)}^{\frac{N}{N+2}}$
$\displaystyle\leq
C\sigma^{\frac{6}{N+2}}\Big{(}\int_{O_{1,1,1}}|\overline{g}_{n}(y)|^{\frac{2(N+2)}{N}}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\Big{)}^{\frac{N}{N+2}}$
$\displaystyle\leq
C\sigma^{\frac{6}{N+2}}\int_{O_{1,1,1}}|\nabla\overline{g}_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy$
$\displaystyle\leq
C\sigma^{\frac{6}{N+2}}\Big{(}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla
g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy+\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\Big{)}$
$\displaystyle\leq C\sigma^{\frac{6}{N+2}}(1+\frac{1}{n}).$ (4.16)

Similarly in case $\gamma\geq 0$, by Corollary 3.6 we can show

$\displaystyle\int_{O_{\sigma,\frac{1}{2},\frac{1}{2}}}|g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\leq
C(\gamma,N,k)\sigma^{\frac{2(3+\gamma)}{N+2+\gamma}}(1+\frac{1}{n})$ (4.17)

Since $(g_{n})$ is bounded in $H^{1}((\sigma,\frac{1}{2})\times
B^{\mathbb{R}^{k-1}}(0,\frac{1}{2})\times
B^{\mathbb{R}^{N-k}}(0,\frac{1}{2}))$ uniformly in $\sigma\in(0,\frac{1}{2})$,
by (4.16) and (4.17), we can easily show that there exists a subsequence
$(g_{n_{k}})$ such that $g_{n_{k}}\rightarrow g$ in
$L^{2}(O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}};y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma})$.

But

$\lim_{n\rightarrow\infty}\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla
g_{n}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy=0,$

which implies that $\nabla g=0$ a.e. in
$O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}$. Hence there exists constant $c$
such that $g=c$ a.e. in $O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}.$ But
$\zeta_{g_{n_{k}}}=0$ and $g_{n_{k}}\rightarrow g$ in
$L^{2}(O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}),$ thus $c=0,$ which is clearly
a contradiction since

$\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|g(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy=1.$

Since

$\displaystyle\int_{O_{\frac{1}{2},\frac{1}{2},\frac{1}{2}}}|\nabla\overline{f}(y)|^{2}y_{1}^{2}(y_{1}+|\overline{y}|)^{\gamma}dy\asymp
C(N,k,\gamma)\int_{O_{r_{1},r_{2},r_{3}}}r^{-N-\gamma}|\nabla
f(x)|^{2}x_{1}^{2}(x_{1}+|\overline{x}|)^{\gamma}dx,$ (4.18)

the result follows by (4.13), (4.14) and (4.18). ∎

######  Theorem 4.9.

Assume that $\gamma\geq-k$. Let $\xi\in K$, $x\in V(\xi,\frac{\beta_{0}}{16})$
and let $\beta_{1}$ be the constant in Lemma 4.2. Then there exists a positive
constant $C=C(C_{\xi},\Omega,K,\gamma,b)>0$ such that

$\inf_{\zeta\in\mathbb{R}}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)-\zeta|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy\leq
Cr^{2}\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla
f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy,$ (4.19)

for any $0<r<\beta_{1}$ and $f\in
C^{1}(\overline{\mathcal{B}(x,r)\cap\Omega})$.

###### Proof.

Case 1. $d(x)\geq br$ Since $d_{K}(x)\geq d(x),$ we can easily show that for
any $y\in B(x,r)$ $\frac{b-1}{b}d(x)\leq d(y)\leq\frac{b+1}{b}d(x)$ and
$\frac{b-1}{b}d(x)\leq d_{K}(y)\leq\frac{b+1}{b}d_{K}(x)$. Thus the proof of
(4.19) follows easily in this case.

Case 2. $d(x)\leq br$ and $d_{K}(x)>bC_{\xi}r.$ By (4.6), we again have that
$\frac{b-1}{b}d_{K}(x)\leq d_{K}(y)\leq\frac{b+1}{b}d_{K}(x)$. Using the last
inequality and proceeding as the proof of [FMT1, Theorem 2.5], we obtain the
desired result.

Case 3. $d(x)\leq br$ and $d_{K}(x)\leq bC_{\xi}r.$ By (4.5), it is enough to
prove the following inequality

$\inf_{\zeta\in\mathbb{R}}\int_{\mathcal{B}(x,r)\cap\Omega}|f-\zeta|^{2}\delta^{2}(\delta_{2,K}+\delta)^{\gamma}dy\leq
Cr^{2}\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla
f|^{2}\delta^{2}(\delta_{2,K}+\delta)^{\gamma}dy.$

This is a consequence of Lemma 4.8. ∎

By (2.2) and the above theorem, we can easily prove the following result.

######  Corollary 4.10.

Let $\mu\leq k^{2}/4$ and let $\beta_{1}$ be the constant in Lemma 4.2. Then
there exists a constant $C=C(\Omega,K,\gamma,b)>0$ such that for any
$0<r<\beta_{1}$ any $f\in C^{1}(\overline{\mathcal{B}(x,r)\cap\Omega})$ and
all $x\in\Omega$ there holds

$\inf_{\zeta\in\mathbb{R}}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)-\zeta|^{2}\phi_{\mu}^{2}(y)dy\leq
Cr^{2}\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla
f(y)|^{2}\phi_{\mu}^{2}(y)dy\,.$

###### Proof.

If $\mathrm{dist}(x,K)<\beta_{0}/16$ the result follows from Theorem 4.9. In
case $\mathrm{dist}(x,K)>\beta_{0}/16$ the result is well known. ∎

In view of the proof of Lemma 4.8, Corollaries 4.6 and 4.7 and (2.2), we can
prove the following Poincaré inequality in $\Omega.$

######  Theorem 4.11.

Let $\mu\leq k^{2}/4$. There exists a positive constant $C=C(\Omega,K,\mu)$
such that

$\inf_{\zeta\in\mathbb{R}}\int_{\Omega}|f(y)-\zeta|^{2}\phi_{\mu}^{2}(y)dy\leq
C\int_{\Omega}|\nabla f(y)|^{2}\phi_{\mu}^{2}(y)dy,$ (4.20)

for any $f\in C^{1}(\overline{\Omega}).$

###  4.4. Moser inequality

######  Theorem 4.12.

Let $\xi\in K,$ $\gamma\geq-k,$ $x\in V(\xi,\frac{\beta_{0}}{16})$ and let
$\beta_{1}$ be the constant in Lemma 4.2. Then for any $\nu\geq
N+\max\\{2,2+\gamma\\},$ there exists $C=C(\Omega,K,\nu,\beta_{1})$ such that

$\displaystyle\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2(1+\frac{2}{\nu})}d^{2}(y)d^{\gamma}_{K}(y)dy$
$\displaystyle\leq
Cr^{2}\overline{\mathcal{M}}(x,r)^{-\frac{2}{\nu}}\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla
f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy$
$\displaystyle\times\Big{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy\Big{)}^{\frac{2}{\nu}},$
(4.21)

for any $0<r<\beta_{1}$ and all $f\in
C^{\infty}_{c}(\mathcal{B}(x,r)\cap\Omega)$.

###### Proof.

The cases $[d(x)>br]$ and $[d(x)\leq br\mbox{ and }d_{K}(x)>bC_{\xi}r]$ are
proved as in [FMT2, Theorem 3.5] and [FMT1, Theorem 2.6] respectively, using
also the inequalities already obtained in the proof of Lemma 4.2.

So, let us assume that $d(x)\leq br$ and $d_{K}(x)\leq bC_{\xi}r$. We consider
first the case where $-k\leq\gamma<0.$ By Hölder inequality, we have

$\displaystyle\bigg{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy\bigg{)}^{\frac{2(\nu-N-2)}{\nu(N+2)}}$
$\displaystyle\leq\overline{\mathcal{M}}(x,r)^{\frac{4(\nu-N-2)}{\nu(N+2)(N+4)}}\bigg{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2(1+\frac{2}{N+2})}d^{2}(y)d^{\gamma}_{K}(y)dy\bigg{)}^{\frac{2(\nu-N-2)}{\nu(N+4)}}.$
(4.22)

Moreover

$\displaystyle\int_{\mathcal{B}(x,r)\cap\Omega}$
$\displaystyle|f(y)|^{2(1+\frac{2}{\nu})}d^{2}(y)d^{\gamma}_{K}(y)dy$
$\displaystyle\leq\overline{\mathcal{M}}(x,r)^{1-\frac{(\nu+2)(N+2)}{\nu(N+4)}}\bigg{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2(1+\frac{2}{N+2})}d^{2}(y)d^{\gamma}_{K}(y)dy\bigg{)}^{\frac{(\nu+2)(N+2)}{\nu(N+4)}}$
$\displaystyle=\overline{\mathcal{M}}(x,r)^{1-\frac{(\nu+2)(N+2)}{\nu(N+4)}}\bigg{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2(1+\frac{2}{N+2})}d^{2}(y)d^{\gamma}_{K}(y)dy\bigg{)}^{1-\frac{2(\nu-N-2)}{\nu(N+4)}}$
$\displaystyle\leq\overline{\mathcal{M}}(x,r)^{\frac{2}{N+2}-\frac{2}{\nu}}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2(1+\frac{2}{N+2})}d^{2}(y)d^{\gamma}_{K}(y)dy$
$\displaystyle\qquad\bigg{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy\bigg{)}^{-\frac{2(\nu-N-2)}{\nu(N+2)}},$
$\displaystyle\leq\overline{\mathcal{M}}(x,r)^{\frac{2}{N+2}-\frac{2}{\nu}}\Big{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{\frac{2(N+2)}{N}}d^{2}(y)d^{\gamma}_{K}(y)dy\Big{)}^{\frac{N}{N+2}}$
$\displaystyle\qquad\Big{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy\Big{)}^{\frac{2}{\nu}},$
(4.23)

where in the second to last inequality we have used (4.22). By Corollary 3.7
and Proposition 3.1, we have

$\displaystyle\Big{(}\int_{\mathcal{B}(x,r)\cap\Omega}|f(y)|^{\frac{2(N+2)}{N}}d^{2}(y)d^{\gamma}_{K}(y)dy\Big{)}^{\frac{N}{N+2}}\leq
C\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla f(y)|^{2}d^{2}(y)d^{\frac{\gamma
N}{N+2}}_{K}(y)dy$ $\displaystyle\leq
Cr^{-\frac{2\gamma}{N+2}}\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla
f(y)|^{2}d^{2}(y)d^{\gamma}_{K}(y)dy$ (4.24)

Now, by Lemma 4.2

$\overline{\mathcal{M}}(x,r)\asymp
C(\Omega,K,\gamma,N,C_{\xi},\beta_{0})r^{N+\gamma+2}.$ (4.25)

The desired result follows by (4.23), (4.24) and (4.25).

If $\gamma>0$, the proof of (4.21) is similar, the only difference is that we
use Corollary 3.6 instead of Corollary 3.7. ∎

By (2.2) and the above theorem, we have

######  Corollary 4.13.

Let $\mu\leq k^{2}/4$ and let $\beta_{1}$ be the constant in Lemma 4.2. Then
for any $\nu\geq N+\max\\{2,2+\gamma\\},$ there exists
$C=C(\Omega,K,\nu,\beta_{1})$ such that for any $x\in\Omega$, any
$r\in(0,\beta_{1})$ and any $f\in
H^{1}_{0}(\mathcal{B}(x,r)\cap\Omega;\phi_{\mu}^{2})$ there holds

$\displaystyle\int_{\mathcal{B}(x,r)\cap\Omega}|f|^{2(1+\frac{2}{\nu})}\phi_{\mu}^{2}dy$
$\displaystyle\leq
Cr^{2}\mathcal{M}(x,r)^{-\frac{2}{\nu}}\Big{(}\int_{\mathcal{B}(x,r)\cap\Omega}|\nabla
f|^{2}\phi_{\mu}^{2}dy\Big{)}$
$\displaystyle\times\Big{(}\int_{\mathcal{B}(x,r)\cap\Omega}f^{2}\phi_{\mu}^{2}dy\Big{)}^{\frac{2}{\nu}}.$

###  4.5. Harnack inequality

We consider the problem

$(\partial_{t}+L_{\phi_{\mu}})u:=u_{t}-\phi_{\mu}^{-2}\text{div}(\phi_{\mu}^{2}\nabla
u)=0,\quad\text{in}\;(0,T)\times\mathcal{B}(x,r)\cap\Omega,$ (4.26)

for any $T>0$ and $r<\frac{\beta_{1}}{4}$ where $\beta_{1}$ is the constant in
Lemma 4.2. Similarly with Definition 2.3 we have

######  Definition 4.14.

Let $D\subset\Omega$ be an open set. A function $v\in
C^{1}((0,T):H^{1}(D;\phi_{\mu}^{2}))$ is a weak subsolution of
$v_{t}+{\mathcal{L}}_{\mu}v=0$ in $(0,T)\times D$ if for any non-negative
$\Phi\in C^{1}_{c}((0,T):C_{c}^{\infty}(D))$ we have

$\displaystyle\int_{0}^{T}\int_{D}(v_{t}\Phi+\nabla
v\cdot\nabla\Phi)\phi_{\mu}^{2}\,dy\,dt\leq 0.$

We now set

$Q=(s-r^{2},s)\times\mathcal{B}(x,r)\cap\Omega$ $Q_{\delta}=(s-\delta
r^{2},s)\times\mathcal{B}(x,\delta r)\cap\Omega.$

Now we are ready to apply the Moser iteration argument in order to prove the
Harnack inequality for nonnegative weak solutions. The proof is based on the
ideas in the proof of Harnack inequality in noncompact smooth manifold (see
[S-C, Chapter 5]). Let us note here that Theorem 4.5 allows to us to consider
test functions in $C_{c}^{\infty}({\mathcal{B}(x,r)}))$ instead of
$C_{c}^{\infty}({\mathcal{B}(x,r)\cap\Omega}))$. Thus we are able to prove
boundary Harnack inequalities.

Let us first state the $L^{p}$ mean value inequality for nonnegative
subsolutions of the operator $\partial_{t}+L_{\phi_{\mu}}.$

######  Theorem 4.15.

Let $\mu\leq k^{2}/4$, $\nu\geq N+\max\\{2,2+2\gamma_{+}\\}$ and $p>0$. There
exists a constant $C(\nu,\lambda,\beta_{1},p,\Omega,K)$ such that for any
$x\in\Omega$ and for any positive subsolution $v$ of (4.26) in $Q$ we have the
estimate

$\sup_{Q_{\delta}}|v|^{p}\leq\frac{C}{(\delta^{\prime}-\delta)^{\nu+2}r^{2}\overline{\mathcal{M}}(x,r)}\int_{Q_{\delta^{\prime}}}|v|^{p}\phi_{\mu}^{2}\,dy\,dt,$

for each $0<\delta<\delta^{\prime}\leq 1.$

The proof of the above theorem is similar to the proof of [S-C, Theorem 5.2.9]
and we omit it (see also [FMT1, Theorem 2.12]). Similarly one can establish
the proof of the parabolic Harnack inequality up to the boundary of Theorem
2.4.

Let $k(t,x,y)$ be the heat kernel of the problem

$\displaystyle{\left\\{\begin{array}[]{lll}{v_{t}=-L_{\mu}v,}&{\mathrm{in}\;\;(0,T]\times\Omega,}\\\
{v=0,}&{\mathrm{on}\;\;(0,T]\times\partial\Omega,}\\\
{v(0,x)=v_{0}(x),}&{\mathrm{in}\;\;\Omega.}\end{array}\right.}$

By the parabolic Harnack inequality (2.4), and following the methods of
Grigoryan and Saloff-Coste (see for example [GS-C, Theorem 2.7] and [S-C,
Theorem 5.4.12]) we obtain the following sharp two-sided heat kernel estimate
for small time:

######  Theorem 4.16.

Let $\beta_{1}$ be the constant of Lemma 4.2. Then there exist positive
constants $A_{1},\;A_{2},$ $C_{1}$ and $C_{2},$ such that for all
$x,\;y\in\Omega$ and all $0<t<\frac{\beta_{1}^{2}}{4}$ the heat kernel
$k(t,x,y)$ satisfies

$\displaystyle\frac{C_{1}}{\mathcal{M}^{\frac{1}{2}}(x,\sqrt{t})\mathcal{M}^{\frac{1}{2}}(y,\sqrt{t})}\exp\Big{(}-A_{1}\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\leq k(t,x,y)$
$\displaystyle\leq\frac{C_{2}}{\mathcal{M}^{\frac{1}{2}}(x,\sqrt{t})\mathcal{M}^{\frac{1}{2}}(y,\sqrt{t})}\exp\Big{(}-A_{2}\frac{|x-y|^{2}}{t}\Big{)}.$

######  Theorem 4.17.

Let $\mu\leq\frac{k^{2}}{4}$. There exists positive constants
$T=T(\Omega,K,\mu)>0$ and $C=C(\Omega,K,\mu)>1$ such that the heat kernel
$k(t,x,y)$ satisfies

$\displaystyle({\rm i})$ $\displaystyle
C^{-1}\Big{(}\frac{1}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}\Big{(}\frac{1}{(d_{K}(x)+\sqrt{t})(d_{K}(y)+\sqrt{t})}\Big{)}^{\gamma_{+}}t^{-\frac{N}{2}}\exp\Big{(}-C\frac{|x-y|^{2}}{t}\Big{)}$
$\displaystyle\hskip 85.35826pt\leq k(t,x,y)\leq$ $\displaystyle
C\Big{(}\frac{1}{(d(x)+\sqrt{t})(d(y)+\sqrt{t})}\Big{)}\Big{(}\frac{1}{(d_{K}(x)+\sqrt{t})(d_{K}(y)+\sqrt{t})}\Big{)}^{\gamma_{+}}t^{-\frac{N}{2}}\exp\Big{(}\\!\\!-C^{-1}\frac{|x-y|^{2}}{t}\Big{)}$
for any $0<t\leq T$ and $x,y\in\Omega$. $\displaystyle({\rm ii})$
$\displaystyle C^{-1}\leq k(t,x,y)\leq C\,\;\;\mbox{for any $t>T$ and
$x,y\in\Omega$.}$

_Proof of Theorem 4.17 (i)._ This follows easily from Theorem 4.16 and
Corollary 4.3. $\Box$

##  5\. Heat kernel estimates for large time

###  5.1. Weighted logarithmic Sobolev inequality

######  Theorem 5.1.

Let $\mu\leq k^{2}/4$. There exists a positive constant $C=C(\Omega,K,\mu)$
such that for any $\epsilon>0$ there holds

$\int_{\Omega}u^{2}\ln\frac{|u|}{\left\|u\right\|_{L^{2}(\Omega;\phi_{\mu}^{2})}}\phi_{\mu}^{2}dx\leq\varepsilon\int_{\Omega}|\nabla
u|^{2}\phi_{\mu}^{2}dx+b(\varepsilon)\int_{\Omega}u^{2}\phi_{\mu}^{2}dx,$
(5.1)

for all $u\in H^{1}(\Omega;\phi_{\mu}^{2})$; here
$b(\varepsilon)=C-\frac{N+2+\max(\gamma_{+},0)}{4}\min(\ln\varepsilon,0)$.

###### Proof.

We may assume that $\left\|u\right\|_{L^{2}(\Omega;\phi_{\mu}^{2})}=1$. Assume
first that $-\frac{k}{2}\leq\gamma_{+}<0$. Then

$\displaystyle\int_{\Omega}|u|^{2}\ln|u|\phi_{\mu}^{2}dx$
$\displaystyle=\frac{N}{4}\int_{\Omega}|u|^{2}\ln|u|^{\frac{4}{N}}\phi_{\mu}^{2}dx$
$\displaystyle\leq\frac{N}{4}\ln\bigg{(}\int_{\Omega}|u|^{\frac{2(N+2)}{N}}\phi_{\mu}^{2}dx\bigg{)}$
$\displaystyle=\frac{N+2}{4}\ln\bigg{(}\Big{(}\int_{\Omega}|u|^{\frac{2(N+2)}{N}}\phi_{\mu}^{2}dx\Big{)}^{\frac{N}{N+2}}\bigg{)}$
$\displaystyle\leq\frac{N+2}{4}\ln\left(C_{0}\left(\int_{\Omega}|\nabla
u|^{2}\phi_{\mu}^{2}dx+\int_{\Omega}|u|^{2}\phi_{\mu}^{2}dx\right)\right),$

where in the last inequality, we used Corollary 4.7 and (2.2). Using the fact
that
$\frac{N+2}{4}\log\theta=\frac{N+2}{4}\ln\frac{4\varepsilon\theta}{C_{0}(N+2)}+\frac{N+2}{4}\ln\frac{C_{0}(N+2)}{4\varepsilon},\;\forall\varepsilon,\theta>0,$
we obtain the desired result with $b(\varepsilon)=1+\frac{N+2}{4}(\ln
C_{0}+\ln\frac{N+2}{4}-\ln\varepsilon),$ if $0<\varepsilon\leq 1$

Similarly, if $\varepsilon\geq 1$ and $-\frac{k}{2}\leq\gamma_{+}<0,$ we
obtain the desired result with $b(\varepsilon)=1+\frac{N+2}{4}(\ln
C_{0}+\ln\frac{N+2}{4}).$

If $\gamma_{+}>0$ we proceed as above and we use Corollary 4.6 instead of
Corollary 4.7, in order to obtain (5.1) with
$b(\varepsilon)=1+\frac{N+2+2\gamma_{+}}{4}(\ln
C_{1}+\ln\frac{N+2+2\gamma_{+}}{4}-\ln\varepsilon),$ where $C_{1}$ is the
constant in Corollary 4.6. ∎

######  Theorem 5.2.

Let $\mu\leq k^{2}/4$ and let $u\in H^{1}(\Omega;\phi_{\mu}^{2})$ be such that
$\int_{\Omega}u\,\phi_{\mu}^{2}dx=0.$ There exists a positive constant
$C=C(\Omega,K,\mu)$ such that for any $\epsilon>0$ there holds

$\int_{\Omega}u^{2}\ln\frac{|u|}{\left\|u\right\|_{L^{2}(\Omega;\phi_{\mu}^{2})}}\phi_{\mu}^{2}dx\leq\varepsilon\int_{\Omega}|\nabla
u|^{2}\phi_{\mu}^{2}dx+b(\varepsilon)\int_{\Omega}u^{2}\phi_{\mu}^{2}dx,$

where $b(\varepsilon)=C-\frac{N+2+\max(2\gamma_{+},0)}{4}\ln\varepsilon$.

###### Proof.

By (4.20) and in view of the proof of (5.1) we obtain the desired result. ∎

Proof of Theorem 4.17 (ii). We normalize $\phi_{\mu}$ so that
$\int_{\Omega}\phi_{\mu}^{2}dx=1$. We define the bilinear form
$Q:H_{0}^{1}(\Omega;\phi_{\mu}^{2})\times
H_{0}^{1}(\Omega;\phi_{\mu}^{2})\to{\mathbb{R}}$ by

$Q(u,v)=\int_{\Omega}\nabla u\cdot\nabla v\,\phi_{\mu}^{2}dx.$

We recall here that
$H^{1}(\Omega;\phi_{\mu}^{2})=H^{1}_{0}(\Omega;\phi_{\mu}^{2})$ by (2.2) and
Theorem 4.5.

Let ${\mathcal{L}}_{\mu}$ denote the self-adjoint operator on
$L^{2}(\Omega;\phi_{\mu}^{2})$ associated to the form $Q$, so that formally we
may write

${\mathcal{L}}_{\mu}u=-\phi_{\mu}^{-2}\,\text{div}\left(\phi_{\mu}^{2}\nabla
u\right).$

The operator ${\mathcal{L}}_{\mu}$ generates a contraction semigroup
$T(t):L^{2}(\Omega;\phi_{\mu}^{2})\to L^{2}(\Omega;\phi_{\mu}^{2})$, $t\geq
0$, denoted also by $e^{-{\mathcal{L}}_{\mu}t}$. This semigroup is positivity
preserving and by [D2, Lemma 1.3.4] we can easily show that satisfies the
conditions of [D2, Theorems 1.3.2 and 1.3.3]. Thus, by (5.1), we can apply
[D2, Corollary 2.2.8] to deduce that

$\|e^{-{\mathcal{L}}_{\mu}t}u\|_{L^{\infty}(\Omega)}\leq
C_{t}\|u\|_{L^{2}(\Omega;\phi_{\mu}^{2})},\quad\quad t>0,\;\;u\in
L^{2}(\Omega;\phi_{\mu}^{2}),$ (5.2)

where

$C_{t}=e^{\frac{1}{t}\int_{0}^{t}b(\varepsilon)d\varepsilon}.$

Hence, by [D2, Lemma 2.1.2], $e^{-{\mathcal{L}}_{\mu}t}$ is ultracontractive
and has a kernel $k(t,x,y)$ such that

$0\leq k(t,x,y)\leq C_{t}.$

By the last inequality, the upper estimate in Theorem 4.17 (ii) follows
easily.

For the lower estimate in 4.17 (ii) we will give two proofs. One using the
boundary Harnack inequality (2.4) and the other one proceeding as the proof of
[D1, Theorem 6].

First proof (as in the proof of [D1, Theorem 6]). First we note that since
$H^{1}(\Omega;\phi_{\mu}^{2})$ is compactly embedded in
$L^{2}(\Omega;\phi_{\mu}^{2})$, the operator ${\mathcal{L}}_{\mu}$ has compact
resolvent. In addition, we have that ${\mathcal{L}}_{\mu}1=0$ and hence, by
(4.20),

${\rm sp}({\mathcal{L}}_{\mu})\subset\\{0\\}\cup[\lambda,\infty),$

for some $\lambda>0.$ Thus, using the spectral theorem, we can easily show
that for any $f\in L^{2}(\Omega;\phi_{\mu}^{2})$ such that
$\int_{\Omega}f\phi_{\mu}^{2}dx=0$ we have

$\|e^{-{\mathcal{L}}_{\mu}t}f\|_{L^{2}(\Omega;\phi_{\mu}^{2})}\leq e^{-\lambda
t}\left\|f\right\|_{L^{2}(\Omega;\phi_{\mu}^{2})},\qquad\forall t\geq 0.$
(5.3)

Now, let $f\in L^{1}(\Omega;\phi_{\mu}^{2})$ and
$\int_{\Omega}f\phi_{\mu}^{2}dx=0.$ By (5.2) and (5.3), we have

$\displaystyle\|e^{-{\mathcal{L}}_{\mu}t}f\|_{L^{\infty}(\Omega)}=\|e^{-{\mathcal{L}}_{\mu}\frac{t}{3}}\big{(}e^{-{\mathcal{L}}_{\mu}\frac{2t}{3}}f\big{)}\|_{L^{\infty}(\Omega)}\leq
C_{\frac{t}{3}}\|e^{-{\mathcal{L}}_{\mu}\frac{2t}{3}}f\|_{L^{2}(\Omega;\phi_{\mu}^{2})}\leq
e^{-\frac{\lambda
t}{3}}C_{\frac{t}{3}}\|e^{-{\mathcal{L}}_{\mu}\frac{t}{3}}f\|_{L^{2}(\Omega;\phi_{\mu}^{2})}.$

Taking adjoints we have

$\|e^{-{\mathcal{L}}_{\mu}\frac{t}{3}}f\|_{L^{2}(\Omega;\phi_{\mu}^{2})}\leq
C_{\frac{t}{3}}\left\|f\right\|_{L^{1}(\Omega;\phi_{\mu}^{2})},$

hence

$\|e^{-{\mathcal{L}}_{\mu}t}f\|_{L^{\infty}(\Omega)}\leq e^{-\frac{\lambda
t}{3}}C_{\frac{t}{3}}^{2}\left\|f\right\|_{L^{1}(\Omega;\phi_{\mu}^{2})}.$

Let now $f\in L^{1}(\Omega;\phi_{\mu}^{2})$. The function
$g:=f-\int_{\Omega}f\phi_{\mu}^{2}dx$ satisfies
$\int_{\Omega}g\phi_{\mu}^{2}dx=0$, thus

$e^{-{\mathcal{L}}_{\mu}t}g=e^{-{\mathcal{L}}_{\mu}t}f-{\langle{f},{1}\rangle}_{L^{2}(\Omega;\phi_{\mu}^{2})}.$

Hence the operator

$\tilde{T}(t)f=e^{-{\mathcal{L}}_{\mu}t}f-{\langle{f},{1}\rangle}_{L^{2}(\Omega;\phi_{\mu}^{2})}$

satisfies

$\|\tilde{T}(t)f\|_{L^{\infty}(\Omega)}=\|e^{-{\mathcal{L}}_{\mu}t}g\|_{L^{\infty}(\Omega)}\leq
e^{-\frac{\lambda
t}{3}}C_{\frac{t}{3}}^{2}\left\|g\right\|_{L^{1}(\Omega;\phi_{\mu}^{2})}\leq
2e^{-\frac{\lambda
t}{3}}C_{\frac{t}{3}}^{2}\left\|f\right\|_{L^{1}(\Omega;\phi_{\mu}^{2})}.$

Therefore the integral kernel $\tilde{k}(t,x,y)$ of $\tilde{T}(t)$ satisfies
$\tilde{k}(t,x,y)=k(t,x,y)-1$ and

$|\tilde{k}(t,x,y)|\leq 2e^{-\frac{\lambda t}{3}}C_{\frac{t}{3}}^{2}.$

The desired result follows if we choose $t$ large enough.

Second proof (using the boundary Harnack inequality (2.4)). Let
$x_{0}\in\Omega.$ Then by (2.4) we can show that

$k(t-1,x,y)\leq C(\Omega,K)k(t,x,x_{0}),$

for all $t\geq 2$ and $x,y\in\Omega.$ Thus,

$\displaystyle 1=\int_{\Omega}k(t-1,x,y)\phi_{\mu}^{2}(y)dy$
$\displaystyle\leq C(\Omega,K)\int_{\Omega}k(t,x,x_{0})\phi_{\mu}^{2}(y)dy$
$\displaystyle=C(\Omega,K)k(t,x,x_{0}),\quad\forall t\geq 2.$

The desired result follows. ∎

###  5.2. Green function estimates

In this subsection we prove the existence of the Green kernel of $L_{\mu}$
along with sharp two-sided estimates.

######  Proposition 5.3.

Let $\mu\leq k^{2}/4$ and assume that $\lambda_{\mu}>0$. For any $y\in\Omega$
there exists a minimal Green function $G_{\mu}(\cdot,y)$ of the equation

$L_{\mu}u=\delta_{y}\quad\text{in }\Omega,$

where $\delta_{y}$ denotes the Dirac measure at $y$. Furthermore, the
following estimates hold

$\displaystyle G_{\mu}(x,y)$ $\displaystyle\asymp\left\\{\begin{aligned}
&|x-y|^{2-N}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\frac{d_{K}(x)d_{K}(y)}{\left(d_{K}(x)+|x-y|\right)\left(d_{K}(y)+|x-y|\right)}\right)^{\gamma_{+}},\\\
&\hskip 227.62204pt\text{if }{\gamma_{+}}>-\frac{N}{2},\\\
&|x-y|^{2-N}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\Big{(}\frac{|x|\,|y|}{\left(|x|+|x-y|\right)\left(|y|+|x-y|\right)}\Big{)}^{-\frac{N}{2}}\\\
&\quad\quad+\frac{d(x)d(y)}{(|x||y|)^{\frac{N}{2}}}\left|\ln\Big{(}\min\Big{\\{}\frac{1}{|x-y|^{2}},\frac{1}{d(x)d(y)}\Big{\\}}\Big{)}\right|,\quad\text{if
}\;{\gamma_{+}}=-\frac{N}{2}.\end{aligned}\right.$ (5.4)

###### Proof.

First, let $C_{1}>0$ and $T$ be as in Theorem 2.6. We note that

$\left(\Big{(}\frac{\sqrt{t}}{d(x)}+1\Big{)}\Big{(}\frac{\sqrt{t}}{d(y)}+1\Big{)}\right)^{-1}=\frac{d(x)d(y)}{(\sqrt{t}+d(x))(\sqrt{t}+d(y))}\leq\min\\{1,\frac{d(x)d(y)}{t}\\}$
(5.5)

and

$\displaystyle\left(\Big{(}\frac{\sqrt{t}}{d(x)}+1\Big{)}\right.$
$\displaystyle\left.\Big{(}\frac{\sqrt{t}}{d(y)}+1\Big{)}\right)^{-1}e^{-\frac{C_{1}|x-y|^{2}}{t}}=\frac{d(x)d(y)}{(\sqrt{t}+d(x))(\sqrt{t}+d(y))}e^{-\frac{C_{1}|x-y|^{2}}{t}}$
(5.6) $\displaystyle\geq
C\min\\{1,\frac{d(x)d(y)}{t}\\}e^{-\frac{(1+C_{1})|x-y|^{2}}{t}}$

for all $x,y\in\Omega$ and $0<t<T$, where $C=C(C_{1},T)>0$.

By Theorem 2.6, (2.2) and estimates (5.5)–(5.6), there exist
$C_{i}=C_{i}(\Omega,K,\mu)>0$, $i=1,2$ and $T=T(\Omega,K,\mu)>0$ such that for
$t\in(0,T)$ and $x,y\in\Omega$,

$\displaystyle\begin{aligned}
C_{1}\min\Big{\\{}1,\frac{d(x)d(y)}{t}\Big{\\}}\Big{(}\frac{d_{K}(x)}{d_{K}(x)+\sqrt{t}}\Big{)}^{\gamma_{+}}\Big{(}\frac{d_{K}(y)}{d_{K}(y)+\sqrt{t}}\Big{)}^{\gamma_{+}}t^{-\frac{N}{2}}&e^{-\frac{C_{2}|x-y|^{2}}{t}}\leq
h(t,x,y)\\\ \leq
C_{2}\min\Big{\\{}1,\frac{d(x)d(y)}{t}\Big{\\}}\Big{(}\frac{d_{K}(x)}{d_{K}(x)+\sqrt{t}}\Big{)}^{\gamma_{+}}\Big{(}\frac{d_{K}(y)}{d_{K}(y)+\sqrt{t}}\Big{)}^{\gamma_{+}}&t^{-\frac{N}{2}}e^{-\frac{C_{1}|x-y|^{2}}{t}},\end{aligned}$
(5.7)

while

$C_{1}\leq\frac{h(t,x,y)}{d(x)d(y)d_{K}^{{\gamma_{+}}}(x)d_{K}^{{\gamma_{+}}}(y)e^{-\lambda_{\mu}t}}\leq
C_{2},\qquad\forall t\geq T,\;x,y\in\Omega.$ (5.8)

By (5.7) and (5.8), we deduce the existence of the minimal Green kernel
$G_{\mu}$ of $L_{\mu}$, given by

$G_{\mu}(x,y)=\int_{0}^{\infty}h(t,x,y)dt=\int_{0}^{T}h(t,x,y)dt+\int_{T}^{\infty}h(t,x,y)dt.$
(5.9)

Using (5.8) we easily see that the second integral in (5.9) satisfies the
required upper estimate in both cases considered (i.e.
$\gamma_{+}>-\frac{N}{2}$ or $\gamma_{+}=-\frac{N}{2}$). We now concentrate on
the first integral in (5.9).

By the change of variable $s=\frac{|x-y|^{2}}{t}$, we obtain for $i=1,2$,

$\displaystyle\int_{0}^{T}\min\Big{\\{}1,\frac{d(x)d(y)}{t}\Big{\\}}\Big{(}\frac{d_{K}(x)}{d_{K}(x)+\sqrt{t}}\Big{)}^{\gamma_{+}}\Big{(}\frac{d_{K}(y)}{d_{K}(y)+\sqrt{t}}\Big{)}^{\gamma_{+}}t^{-\frac{N}{2}}e^{-\frac{C_{i}|x-y|^{2}}{t}}dt=|x-y|^{2-N}$
$\displaystyle\int_{\frac{|x-y|^{2}}{T}}^{\infty}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(x)}+1\Big{)}\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(y)}+1\Big{)}\right)^{-{\gamma_{+}}}s^{\frac{N}{2}-2}e^{-C_{i}s}ds$
$\displaystyle=:|x-y|^{2-N}S_{i}(x,y)\,.$

By (5.7) we therefore have for some $c_{1},c_{2}>0$ that

$c_{1}|x-y|^{2-N}S_{2}(x,y)\leq\int_{0}^{T}h(t,x,y)dt\leq
c_{2}|x-y|^{2-N}S_{1}(x,y)\;,\qquad x,y\in\Omega.$ (5.10)

In the sequel, we assume that $\frac{|x-y|^{2}}{T}<\frac{1}{2}.$ The proof in
the case $\frac{|x-y|^{2}}{T}>\frac{1}{2}$ is similar, indeed simpler. We
write

$\displaystyle\begin{aligned}
&S_{1}=\int_{\frac{|x-y|^{2}}{T}}^{1}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(x)}+1\Big{)}\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(y)}+1\Big{)}\right)^{-{\gamma_{+}}}s^{\frac{N}{2}-2}e^{-C_{1}s}ds\\\
&\quad\quad+\int_{1}^{\infty}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(x)}+1\Big{)}\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(y)}+1\Big{)}\right)^{-{\gamma_{+}}}s^{\frac{N}{2}-2}e^{-C_{1}s}ds.\end{aligned}$
(5.11)

Concerning the second term in the RHS of (5.11) we have

$\displaystyle\int_{1}^{\infty}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\bigg{(}\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(x)}+1\Big{)}\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(y)}+1\Big{)}\bigg{)}^{-{\gamma_{+}}}s^{\frac{N}{2}-2}e^{-C_{1}s}ds$
$\displaystyle\qquad\leq
C\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\Big{(}\frac{|x-y|}{d_{K}(x)}+1\Big{)}\Big{(}\frac{|x-y|}{d_{K}(y)}+1\Big{)}\right)^{-{\gamma_{+}}},$

and therefore the required estimate is satisfied.

Let $\gamma_{+}\leq 0$. For the first term in the RHS of (5.11) we have

$\displaystyle\int_{\frac{|x-y|^{2}}{T}}^{1}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(x)}+1\Big{)}\Big{(}\frac{|x-y|}{\sqrt{s}d_{K}(y)}+1\Big{)}\right)^{-{\gamma_{+}}}s^{\frac{N}{2}-2}e^{-C_{1}s}ds$
$\displaystyle=\int_{\frac{|x-y|^{2}}{T}}^{1}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\Big{(}\frac{|x-y|}{d_{K}(x)}+\sqrt{s}\Big{)}\Big{(}\frac{|x-y|}{d_{K}(y)}+\sqrt{s}\Big{)}\right)^{-{\gamma_{+}}}s^{\frac{N}{2}+{\gamma_{+}}-2}e^{-C_{1}s}ds$
$\displaystyle\leq
C\left(|x-y|^{-2{\gamma_{+}}}\big{(}d_{K}(x)d_{K}(y)\big{)}^{{\gamma_{+}}}\int_{\frac{|x-y|^{2}}{T}}^{1}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}s^{\frac{N}{2}+{\gamma_{+}}-2}e^{-C_{1}s}ds\right.$
$\displaystyle\;\;\left.+|x-y|^{-{\gamma_{+}}}\big{(}d_{K}(x)d_{K}(y)\big{)}^{\\!{\gamma_{+}}}\\!\\!\\!\\!\\!\int_{\frac{|x-y|^{2}}{T}}^{1}\\!\\!\\!\min\Big{\\{}\\!1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}(d_{K}(x)+d_{K}(y))^{-{\gamma_{+}}}s^{\frac{N}{2}+\frac{{\gamma_{+}}}{2}-2}e^{-C_{1}s}ds\right.$
$\displaystyle\quad\left.+\int_{\frac{|x-y|^{2}}{T}}^{1}\min\Big{\\{}1,s\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}s^{\frac{N}{2}-2}e^{-C_{1}s}ds\right)$
$\displaystyle=:C(J_{1}+J_{2}+J_{3})$ (5.12)

It is easily seen that

$J_{3}\leq C\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\,.$

Concerning $J_{1}$ and $J_{2}$ we consider two cases.

Case I. $-\frac{N}{2}<\gamma_{+}\leq 0$. In view of (5.10) and (5.12), it is
enough to establish that for $i=1,2$ we have

$J_{i}\leq\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}\left(\frac{d_{K}(x)d_{K}(y)}{\left(d_{K}(x)+|x-y|\right)\left(d_{K}(y)+|x-y|\right)}\right)^{\gamma_{+}}\\!\\!,\quad
i=1,2.$ (5.13)

In order to prove (5.13) we shall need to consider additional cases.

Case Ia. $\frac{d(x)d(y)}{|x-y|^{2}}\leq 1$. In this case it is immediate that

$J_{1}=C|x-y|^{-2-2{\gamma_{+}}}\big{(}d_{K}(x)d_{K}(y)\big{)}^{{\gamma_{+}}}d(x)d(y).$

and

$J_{2}=C|x-y|^{-2-{\gamma_{+}}}(d_{K}(x)d_{K}(y))^{{\gamma_{+}}}\big{(}d_{K}(x)+d_{K}(y)\big{)}^{-{\gamma_{+}}}d(x)d(y).$

Hence inequality (5.13) is satisfied.

Case Ib. $\frac{d(x)d(y)}{|x-y|^{2}}>1$. In this case we have
$\frac{1}{4}d_{K}(y)\leq d_{K}(x)\leq 4d_{K}(y)$. Indeed, suppose that
$d_{K}(x)>4d_{K}(y)$. Then, since $d_{K}(x)\leq|x-y|+d_{K}(y)$, we easily
obtain that $d_{K}(y)\leq\frac{1}{3}|x-y|$ and $d_{K}(x)\leq\frac{4}{3}|x-y|$;
hence $d(x)d(y)\leq\frac{4}{9}|x-y|^{2}$, a contradiction.

To proceed we first note that

$\displaystyle J_{1}$ $\displaystyle\leq$
$\displaystyle|x-y|^{-2{\gamma_{+}}}\big{(}d_{K}(x)d_{K}(y)\big{)}^{{\gamma_{+}}}\left(\frac{d(x)d(y)}{|x-y|^{2}}\int_{0}^{\frac{|x-y|^{2}}{d(x)d(y)}}s^{\frac{N}{2}+{\gamma_{+}}-1}e^{-C_{1}s}ds\right.$
(5.14) $\displaystyle\hskip
99.58464pt\left.+\int_{\frac{|x-y|^{2}}{d(x)d(y)}}^{1}s^{\frac{N}{2}+{\gamma_{+}}-2}e^{-C_{1}s}ds\right)$

and similarly

$\displaystyle J_{2}$ $\displaystyle\leq$
$\displaystyle|x-y|^{-{\gamma_{+}}}\Big{(}\frac{d_{K}(x)d_{K}(y)}{d_{K}(x)+d_{K}(y)}\Big{)}^{{\gamma_{+}}}\left(\frac{d(x)d(y)}{|x-y|^{2}}\int_{0}^{\frac{|x-y|^{2}}{d(x)d(y)}}s^{\frac{N}{2}+\frac{{\gamma_{+}}}{2}-1}e^{-C_{1}s}ds\right.$
(5.15) $\displaystyle\hskip
113.81102pt\left.+\int_{\frac{|x-y|^{2}}{d(x)d(y)}}^{1}s^{\frac{N}{2}+\frac{{\gamma_{+}}}{2}-2}e^{-C_{1}s}ds\right)$

We now consider different subcases.

Case 1. $-\frac{N}{2}<{\gamma_{+}}<-N+2$. From (5.14) and (5.15) we obtain

$J_{1}\leq c,\qquad J_{2}\leq c.$

It follows that (5.13) is satisfied.

Case 2. ${\gamma_{+}}=-N+2>-\frac{N}{2}$. In this case (5.14) and (5.15) give

$J_{1}\leq c$

and

$J_{2}\leq
c|x-y|^{-{\gamma_{+}}}\Big{(}\frac{d_{K}(x)d_{K}(y)}{d_{K}(x)+d_{K}(y)}\Big{)}^{{\gamma_{+}}}\bigg{(}1+\ln\Big{(}\frac{d(x)d(y)}{|x-y|^{2}}\Big{)}\bigg{)}\leq
c$

Again it is easily seen that (5.13) is satisfied.

Case 3. $\max\\{-\frac{N}{2},-N+2\\}<{\gamma_{+}}<-\frac{N-2}{2}$. In this
case we obtain

$J_{1}\leq c,\qquad J_{2}\leq
c|x-y|^{-{\gamma_{+}}}\Big{(}\frac{d_{K}(x)d_{K}(y)}{d_{K}(x)+d_{K}(y)}\Big{)}^{{\gamma_{+}}}\leq
c$

and (5.13) once again follows.

Case 4. ${\gamma_{+}}=-\frac{N-2}{2}<0$. In this case we obtain

$\displaystyle J_{1}$ $\displaystyle\leq
c|x-y|^{-2{\gamma_{+}}}\big{(}d_{K}(x)d_{K}(y)\big{)}^{{\gamma_{+}}}\left(1+\ln\Big{(}\frac{d(x)d(y)}{|x-y|^{2}}\Big{)}\right)\leq
c,$ $\displaystyle J_{2}$ $\displaystyle\leq
c|x-y|^{-{\gamma_{+}}}\Big{(}\frac{d_{K}(x)d_{K}(y)}{d_{K}(x)+d_{K}(y)}\Big{)}^{{\gamma_{+}}}\leq
c$

and (5.13) once again follows.

Case 5. $-\frac{N-2}{2}<{\gamma_{+}}\leq 0$. In this case we obtain

$J_{1}\leq
c|x-y|^{-2{\gamma_{+}}}\big{(}d_{K}(x)d_{K}(y)\big{)}^{{\gamma_{+}}}\leq
c,\qquad J_{2}\leq
c|x-y|^{-{\gamma_{+}}}\Big{(}\frac{d_{K}(x)d_{K}(y)}{d_{K}(x)+d_{K}(y)}\Big{)}^{{\gamma_{+}}}\leq
c$

and (5.13) once again follows.

Case II. $\gamma_{+}=-\frac{N}{2}$. The proof is very similar to the previous
case and for the sake of brevity we shall only consider $J_{1}$, where the
main difference appears. We note that in this case we have $d_{K}(x)=|x|$.

We assume that $\frac{|x-y|^{2}}{T}\leq\frac{1}{2}$. The proof in the case
$\frac{|x-y|^{2}}{T}>\frac{1}{2}$ is similar, indeed simpler.

Case IIa. $\frac{d(x)d(y)}{|x-y|^{2}}\leq 1$. In this case we easily obtain

$J_{1}\leq
c\,|x-y|^{N-2}d(x)d(y)(|x|\,|y|)^{-\frac{N}{2}}\log\Big{(}\frac{T}{|x-y|^{2}}\Big{)},$

and this is estimated using the second term in the RHS of (5.4).

Case IIb. $\frac{d(x)d(y)}{|x-y|^{2}}\geq 1$. We may assume that
$\frac{|x-y|^{2}}{d(x)d(y)}>\frac{|x-y|^{2}}{T}$, otherwise we need only
consider the second of the two integrals below.

We have

$\displaystyle J_{1}$ $\displaystyle=$
$\displaystyle|x-y|^{N}(|x|\,|y|)^{-\frac{N}{2}}\bigg{(}\frac{d(x)d(y)}{|x-y|^{2}}\int_{\frac{|x-y|^{2}}{T}}^{\frac{|x-y|^{2}}{d(x)d(y)}}s^{-1}e^{-C_{1}s}ds+\int_{\frac{|x-y|^{2}}{d(x)d(y)}}^{1}s^{-2}e^{-C_{1}s}ds\bigg{)}$
$\displaystyle\leq$ $\displaystyle
c|x-y|^{N-2}d(x)d(y)\big{(}|x|\,|y|\big{)}^{-\frac{N}{2}}\log\Big{(}\frac{T}{d(x)d(y)}\Big{)},$

which satisfies the upper bound in (5.4). Hence the upper bound has been
established in all cases.

This concludes the proof of the upper estimate when ${\gamma_{+}}\leq 0$. If
${\gamma_{+}}>0$ then the proof is essentially similar, indeed simpler, and is
omitted.

The proof of the lower bound is much simpler. For example, in case
$\gamma_{+}\leq 0$ we have from (5.10)

$G_{\mu}(x,y)\geq c_{1}|x-y|^{2-N}S_{2}(x,y)\geq c|x-y|^{2-N}J_{1}(x,y),$

where $J_{1}$ is as above, the only difference being that the exponential
factor in the integrand is $e^{-C_{2}s}$ instead of $e^{-C_{1}s}$. The result
then follows easily. ∎

##  6\. The linear elliptic problem

###  6.1. Subsolutions and Supersolutions

We recall the definition of the function $\tilde{d}_{K}$ from (2.5). Given
parameters $\epsilon>0$ and $M\in{\mathbb{R}}$ we define the functions

$\begin{array}[]{ll}\eta_{\gamma_{+},\varepsilon}=e^{-Md}(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{\gamma_{+}}-d\tilde{d}_{K}^{\gamma_{+}+\varepsilon}&\\!\\!\zeta_{{\gamma_{+}},\varepsilon}=e^{Md}(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{\gamma_{+}}+d\tilde{d}_{K}^{\gamma_{+}+\varepsilon}\\\
\eta_{\gamma_{-},\varepsilon}=e^{-Md}(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{\gamma_{-}}+d\tilde{d}_{K}^{\gamma_{-}+\varepsilon}&\\!\\!\zeta_{\gamma_{-},\varepsilon}=e^{Md}(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{\gamma_{-}}-d\tilde{d}_{K}^{\gamma_{-}+\varepsilon}\\\
\zeta_{\scaleto{+}{3pt},\varepsilon}=\\!e^{-Md}(-\ln\tilde{d}_{K})(d\\!+\tilde{d}^{2}_{K})\tilde{d}_{K}^{-\frac{k}{2}}\\!-\\!d\tilde{d}_{K}^{-\frac{k}{2}+\varepsilon}&\\!\\!{\zeta_{\scaleto{-}{3pt},\varepsilon}}=\\!e^{Md}(-\ln\tilde{d}_{K})(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{-\frac{k}{2}}\\!+\\!d\tilde{d}_{K}^{-\frac{k}{2}+\varepsilon}\end{array}$

######  Lemma 6.1.

Let $\mu\leq k^{2}/4$ and $0<\varepsilon<1$. There exist positive constants
$\beta_{0}=\beta_{0}(\Omega,K,\mu,\varepsilon)$ and
$M=M(\Omega,K,\mu,\varepsilon)$ such that the following hold in
$K_{\beta_{0}}\cap\Omega$:
$({\rm i})$ The functions $\eta_{\gamma_{+},\varepsilon}$ and
$\zeta_{{\gamma_{+}},\varepsilon}$ are non-negative in
$K_{\beta_{0}}\cap\Omega$ and satisfy

$L_{\mu}\eta_{\gamma_{+},\varepsilon}\geq 0,\qquad\quad
L_{\mu}\zeta_{{\gamma_{+}},\varepsilon}\leq 0,\quad\mbox{ in
}K_{\beta_{0}}\cap\Omega.$

$({\rm ii})$ If $\mu<k^{2}/4$ and $\varepsilon<\min\\{1,\sqrt{k^{2}-4\mu}\\})$
then $\eta_{\gamma_{-},\varepsilon}$ and $\zeta_{\gamma_{-},\varepsilon}$ are
non-negative in $K_{\beta_{0}}\cap\Omega$ and satisfy

$L_{\mu}\eta_{\gamma_{-},\varepsilon}\geq 0,\qquad\quad
L_{\mu}\zeta_{\gamma_{-},\varepsilon}\leq 0,\quad\mbox{ in
}K_{\beta_{0}}\cap\Omega.$ (6.1)

$({\rm iii})$ The functions ${\zeta_{\scaleto{+}{3pt},\varepsilon}}$ and
${\zeta_{\scaleto{-}{3pt},\varepsilon}}$ are non-negative in
$K_{\beta_{0}}\cap\Omega$ and satisfy

$L_{\frac{k^{2}}{4}}{\zeta_{\scaleto{+}{3pt},\varepsilon}}\geq 0,\qquad\quad
L_{\frac{k^{2}}{4}}{\zeta_{\scaleto{-}{3pt},\varepsilon}}\leq 0,\quad\mbox{ in
}K_{\beta_{0}}\cap\Omega.$

###### Proof.

Let $M\in\mathbb{R}$. By Proposition 3.1 we have in $\Omega\cap
K_{\beta_{0}}$,

$\displaystyle\Delta(d^{a}\tilde{d}_{K}^{b})$
$\displaystyle=d^{a-2}\tilde{d}_{K}^{b}\Big{(}a(a-1)+ad\Delta d\Big{)}$
$\displaystyle\quad+d^{a}\,\tilde{d}_{K}^{b-2}\Big{(}2ab+b(k-1+f)+b(b-1)(1+h)\Big{)}$
$\displaystyle\nabla e^{Md}\cdot\nabla(d^{a}\tilde{d}_{K}^{b})$
$\displaystyle=Me^{Md}(ad^{a-1}\tilde{d}_{K}^{b}+bd^{a+1}\tilde{d}_{K}^{b-2})$
$\displaystyle\Delta e^{Md}$ $\displaystyle=e^{Md}(M^{2}+M\Delta d)$

Thus

$\displaystyle
L_{\mu}(e^{Md}d^{a}\tilde{d}_{K}^{b})=-e^{Md}d^{a-1}\tilde{d}_{K}^{b}\left(M^{2}d+Md\Delta
d+2aM+a\Delta d+a(a-1)d^{-1}\right)$
$\displaystyle-e^{Md}d^{a}\tilde{d}_{K}^{b-1}\Big{(}\frac{2Mbd+bf+b(b-1)h+\mu
g}{\tilde{d}_{K}}\Big{)}-\big{(}b(k-1)+b(b-1)+2ab+\mu\big{)}e^{Md}d^{a}\tilde{d}_{K}^{b-2}$

Now let $M\in\mathbb{R}$ and $0<\varepsilon<1$. Using the above formulas we
find

$\displaystyle L_{\mu}$
$\displaystyle(e^{Md}(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{{\gamma_{+}}})-L_{\mu}(d\tilde{d}_{K}^{{\gamma_{+}}+\varepsilon})$
$\displaystyle=-e^{Md}\tilde{d}_{K}^{{\gamma_{+}}}\Big{(}(M^{2}d+Md\Delta
d+2M+\Delta d)+(M^{2}+M\Delta d)\tilde{d}_{K}^{2}\Big{)}$
$\displaystyle\quad-e^{Md}d\tilde{d}_{K}^{{\gamma_{+}}-2}\Big{(}2M{\gamma_{+}}d+{\gamma_{+}}f+{\gamma_{+}}({\gamma_{+}}-1)h+\mu
g\Big{)}$
$\displaystyle\quad-e^{Md}\tilde{d}_{K}^{{\gamma_{+}}}\Big{(}2({\gamma_{+}}+k)+({\gamma_{+}}+2)\big{(}({\gamma_{+}}+1)h+f+2Md\big{)}\Big{)}$
$\displaystyle\quad+\epsilon(2{\gamma_{+}}+k+\epsilon)d\tilde{d}_{K}^{{\gamma_{+}}+\epsilon-2}$
$\displaystyle\quad+d\tilde{d}_{K}^{{\gamma_{+}}+\epsilon-2}\Big{(}({\gamma_{+}}+\epsilon)({\gamma_{+}}+\epsilon-1)h+({\gamma_{+}}+\epsilon)f+\mu
g\Big{)}$ $\displaystyle\quad+(\Delta
d)\tilde{d}_{K}^{{\gamma_{+}}+\epsilon}.$

The RHS in the last equality consists of six terms. We now choose $\beta_{0}$
small enough and $M<0$ so that the sum of the first, third and sixth terms is
non-negative in $K_{\beta_{0}}\cap\Omega$. The fourth term is clearly
positive, and by taking $\beta_{0}$ smaller if necessary it may also control
the second and the fifth terms. Hence
$L_{\mu}\eta_{\gamma_{+},\varepsilon}\geq 0$ in $K_{\beta_{0}}\cap\Omega$.

The proofs of the other cases of the lemma are similar and are omitted. For
(iii) we also use the relations

$\displaystyle\Delta\ln\tilde{d}_{K}$
$\displaystyle=\frac{\Delta\tilde{d}_{K}}{\tilde{d}_{K}}-\frac{|\nabla\tilde{d}_{K}|^{2}}{\tilde{d}_{K}^{2}}$
$\displaystyle\nabla\ln\tilde{d}_{K}\cdot\nabla(e^{Md}d\tilde{d}_{K}^{b})$
$\displaystyle=\tilde{d}_{K}^{b-2}e^{Md}\Big{(}Md^{2}+d+b\,d|\nabla\tilde{d}_{K}|^{2}\Big{)}$

and

$\displaystyle-
L_{\mu}\big{(}(-\ln\tilde{d}_{K})e^{Md}d\tilde{d}_{K}^{b}\big{)}=(-\ln\tilde{d}_{K})e^{Md}\tilde{d}_{K}^{b}\left(M^{2}d+Md\Delta
d+2M+\Delta d\right)$
$\displaystyle\qquad+(-\ln\tilde{d}_{K})e^{Md}d\tilde{d}_{K}^{b-1}\Big{(}\frac{2Mbd+bf+b(b-1)h+\mu
g}{\tilde{d}_{K}}\Big{)}$
$\displaystyle\qquad+(-\ln\tilde{d}_{K})\big{(}\,b(k+1)+b(b-1)+\mu\,\big{)}e^{Md}d\tilde{d}_{K}^{b-2}$
$\displaystyle\qquad+e^{Md}d\tilde{d}_{K}^{b-2}\Big{(}-2Md-f+(1-2b)h-2b-k\Big{)}.$

∎

######  Lemma 6.2.

Let $\beta_{0}>0$ be the constant in Lemma 6.1, $\xi\in\partial\Omega$ and
$0<r<\frac{\beta_{0}}{16}.$ We assume that $u\in
H^{1}_{loc}(B_{r}(\xi)\cap\Omega)\cap C(B_{r}(\xi)\cap\Omega)$ is
$L_{\mu}$-harmonic in $B_{r}(\xi)\cap\Omega$ and

$\lim_{\mathrm{dist}(x,F)\to
0}\frac{u(x)}{\tilde{W}(x)}=0,\quad\forall\;\text{compact}\;F\subset
B_{r}(\xi)\cap\partial\Omega.$ (6.2)

Then there exists $C=C(u,\Omega,K,r)>0$ such that

$|u|\leq C\phi_{\mu}\,,\qquad x\in B_{\frac{r}{4}}(\xi)\cap\Omega\,.$ (6.3)

Moreover, if $0\leq\eta_{r}\leq 1$ is a smooth function with compact support
in $B_{\frac{r}{2}}(\xi)$ with $\eta_{r}=1$ on $B_{\frac{r}{4}}(\xi)$, then

$\frac{\eta_{r}u}{\phi_{\mu}}\in H^{1}_{0}(\Omega;\phi_{\mu}^{2}).$ (6.4)

Furthermore, if $u$ is nonnegative there exists
$\Cl[c]{charnack}=\Cr{charnack}(\Omega,K)>0$ such that

$\frac{u(x)}{\phi_{\mu}(x)}\leq\Cr{charnack}\frac{u(y)}{\phi_{\mu}(y)},\quad\forall
x,y\in B_{\frac{r}{16}}(\xi)\cap\Omega.$ (6.5)

###### Proof.

We will only consider the case $\mu<k^{2}/4$ and $\xi\in
K_{\frac{\beta}{16}}\cap\partial\Omega$; the proof of the other cases is very
similar and we omit it.

Since $u$ is $L_{\mu}$\- harmonic in $B_{r}(\xi)\cap\Omega$, by standard
elliptic estimates we have that $u\in C^{2}(B_{r}(\xi)\cap\Omega)$. Set
$w_{l}=\max\\{u-l\eta_{{\gamma_{-}},\varepsilon},0\\}$ where $l>0$ and
$\eta_{{\gamma_{-}},\varepsilon}$ is the supersolution in (6.1). Then by
Kato’s formula we have

$L_{\mu}w_{l}\leq 0\,,\qquad\text{in }B_{r}(\xi)\cap\Omega.$

Setting $v_{l}=\frac{w_{l}}{\phi_{\mu}}$, by straightforward calculations we
have

$-\mathrm{div}(\phi_{\mu}^{2}\nabla
v_{l})+\lambda_{\mu}\phi_{\mu}^{2}v_{l}\leq 0\,,\qquad\text{in
}B_{r}(\xi)\cap\Omega.$ (6.6)

We note here that $v_{l}=0$ if $u\leq
l\eta_{{\alpha_{\scaleto{+}{3pt}}},\varepsilon}$, thus by the assumptions we
can easily obtain that $v_{l}\in H^{1}(B_{\frac{r}{2}}(\xi);\phi_{\mu}^{2})$.

By Theorem 4.15, we can prove the existence of a constant $r_{\beta_{0}}$ and
$C=C(K)>0$ such that for any
$r^{\prime}\leq\min\\{\frac{r}{2},r_{\beta_{0}}\\}$ and $p\geq 1$ the
following inequality holds

$\sup_{x\in B_{\frac{r^{\prime}}{2}}(\xi)\cap\Omega}v_{l}\leq
C\bigg{(}\Big{(}\int_{B_{r^{\prime}}(\xi)\cap\Omega}\phi_{\mu}^{2}dx\Big{)}^{-1}\int_{B_{r^{\prime}}(\xi)\cap\Omega}|v_{l}|^{p}\phi_{\mu}^{2}dx\bigg{)}^{\frac{1}{p}}.$
(6.7)

From (6.2) and the definition of $w_{l}$, we have

$w_{l}\leq u_{+}\leq
C\tilde{W}=C(d+\tilde{d}^{2}_{K})\tilde{d}_{K}^{\gamma_{-}},\quad\text{in
}B_{\frac{r}{2}}(\xi)\cap\Omega.$

This and (2.2) imply that

$\displaystyle\int_{B_{r^{\prime}}(\xi)\cap\Omega}|v_{l}|\phi_{\mu}^{2}dx$
$\displaystyle=\int_{B_{\frac{r}{2}}(\xi)\cap\Omega}|w_{l}|\phi_{\mu}dx$
$\displaystyle\leq
C\int_{B_{\frac{r}{2}}(\xi)\cap\Omega}(d+\tilde{d}^{2}_{K})d\tilde{d}_{K}^{-k}dx\leq
C\int_{B_{\frac{r}{2}}(\xi)\cap\Omega}d_{K}^{2-k}dx<\infty.$

Thus by (6.7) and the last inequality we deduce that

$\sup_{B_{\frac{r^{\prime}}{2}}(\xi)\cap K}v_{l}<C_{1}$

for some constant $C_{1}>0$ which does not depend on $l$. Thus

$w_{l}\leq C_{1}\phi_{\mu}\;,\qquad\text{in
}B_{\frac{r^{\prime}}{2}}(\xi)\cap\Omega.$

By letting $l\to 0$, we derive

$u_{+}\leq C_{1}\phi_{\mu}\;,\qquad\text{in
}B_{\frac{r^{\prime}}{2}}(\xi)\cap\Omega.$

Thus by a covering argument we can find a constant $C_{2}>0$ such that

$u_{+}\leq C_{2}\phi_{\mu}\;,\qquad\text{in }B_{\frac{r}{2}}(\xi)\cap\Omega.$
(6.8)

This implies $v_{0}:=\frac{u_{+}}{\phi_{\mu}}<C_{2}$ in
$B_{\frac{r}{2}}(\xi)\cap\Omega$.

Using $\eta^{2}_{r}v_{l}$ as a test function in (6.6) we can easily obtain

$\int_{B_{\frac{r}{2}}(\xi)\cap\Omega}|\nabla(\eta_{r}v_{l})|^{2}\phi_{\mu}^{2}dx+\lambda_{\mu}\int_{B_{\frac{r}{2}}(\xi)\cap\Omega}|\eta_{r}v_{l}|^{2}\phi_{\mu}^{2}dx\leq\frac{C}{r^{2}}\int_{B_{\frac{r}{2}}(\xi)\cap\Omega}|v_{l}|^{2}\phi_{\mu}^{2}dx.$

By (6.8) and by letting $l\to 0$ we obtain that $\eta_{r}v_{0}\in
H^{1}(\Omega;\phi_{\mu}^{2})$, which in turn implies that
$\frac{\eta_{r}u_{+}}{\phi_{\mu}}\in H^{1}(\Omega;\phi_{\mu}^{2})$. Applying
the same argument to $-u$ we obtain

$u_{-}\leq C_{2}\phi_{\mu}\quad\text{in }B_{\frac{r}{2}}(\xi)\cap\Omega,$

and $\frac{\eta_{r}u_{-}}{\phi_{\mu}}\in H^{1}(\Omega;\phi_{\mu}^{2})$. By
using the fact that $u=u_{+}-u_{-}$, we obtain (6.4) and (6.3).

We next prove the boundary Harnack inequality (6.5). Let $u$ be a nonnegative
$L_{\mu}$-harmonic function and put $v=\frac{u}{\phi_{\mu}}$. Then $v\in
H^{1}(B_{\frac{r}{4}}(\xi);\phi_{\mu}^{2})$ and $v$ satisfies

$-\phi_{\mu}^{-2}\mathrm{div}(\phi_{\mu}^{2}\nabla
v)+\lambda_{\mu}v=0,\qquad\mbox{ in }B_{\frac{r}{4}}(\xi)\cap\Omega.$

The function $\hat{v}(x,t):=e^{\lambda_{\mu}t}v(x)$ then satisfies

$\partial_{t}\hat{v}-\phi_{\mu}^{-2}\mathrm{div}(\phi_{\mu}^{2}\nabla\hat{v})=0,\quad\text{in
}B_{\frac{r}{4}}(\xi)\cap\Omega\times(0,\frac{r^{2}}{16}).$

By the Harnack inequality (2.4),

ess sup
$\displaystyle\Big{\\{}\hat{v}(t,x):(t,x)\in(\frac{r^{2}}{64},\frac{r^{2}}{32})\times{\mathcal{B}}(\xi,\frac{r}{8})\cap\Omega\Big{\\}}$
$\displaystyle\leq C\,\textrm{ess
inf}\Big{\\{}\hat{v}(t,x):(t,x)\in(\frac{3r^{2}}{64},\frac{r^{2}}{16})\times{\mathcal{B}}(\xi,\frac{r}{8})\cap\Omega\Big{\\}}.$

This implies (6.5). ∎

######  Lemma 6.3.

Let $\mu\leq k^{2}/4$ and assume that $\lambda_{\mu}>0$. Let $u\in
H^{1}_{loc}(\Omega)\cap C(\Omega)$ be $L_{\mu}$-subharmonic in $\Omega$.
Assume that

$\limsup_{\mathrm{dist}(x,F)\to 0}\frac{u(x)}{\tilde{W}(x)}\leq
0,\quad\forall\;\text{compact}\;F\subset\partial\Omega.$ (6.9)

Then $u\leq 0$ in $\Omega.$

###### Proof.

First we note that $u_{+}=\max(u(x),0)$ is a nonnegative $L_{\mu}$-subharmonic
function in $\Omega$. Let $v=\frac{u_{+}}{\phi_{\mu}}$. In view of the proof
of (6.4), $v\in H^{1}_{0}(\Omega;\phi_{\mu}^{2})$; moreover by a
straightforward calculation we have

$-\mathrm{div}(\phi_{\mu}^{2}\nabla v)+\lambda_{\mu}\phi_{\mu}^{2}v\leq
0\quad\text{in }\Omega.$ (6.10)

Since $v\in H^{1}_{0}(\Omega;\phi_{\mu}^{2})$, we can use it as a test
function for (6.10) and obtain

$\int_{\Omega}|\nabla
v|^{2}\phi_{\mu}^{2}dx+\lambda_{\mu}\int_{\Omega}|v|^{2}\phi_{\mu}^{2}dx\leq
0.$

Hence $v=0$ and the result follows. ∎

###  6.2. Existence and uniqueness

The aim of this subsection is to prove existence and uniqueness of the
solution of $L_{\mu}u=f,$ with smooth boundary data. We also prove the
boundary Harnack inequalities and maximum principle for the operator
$L_{\mu}.$ Let us first define the notion of a weak solution.

######  Definition 6.4.

Let $f\in L^{2}(\Omega)$. We say that $u$ is a weak solution of

$L_{\mu}u=f\,,\quad\text{ in }\Omega$ (6.11)

if $\frac{u}{\phi_{\mu}}\in H^{1}_{0}(\Omega;\phi_{\mu}^{2})$ and

$\int_{\Omega}\nabla
u\cdot\nabla\psi\,dx-\mu\int_{\Omega}\frac{u\psi}{d^{2}_{K}}dx=\int_{\Omega}f\psi\,dx,\qquad\forall\psi\in
C_{c}^{\infty}(\Omega).$

In the next lemma we give the first existence and uniqueness result.

######  Lemma 6.5.

Let $\mu\leq k^{2}/4$ and assume that $\lambda_{\mu}>0$. For any $f\in
L^{2}(\Omega)$ there exists a unique weak solution $u$ of (6.11). Furthermore
there holds

$\int_{\Omega}u^{2}dx\leq C\int_{\Omega}f^{2}dx,$ (6.12)

where $C=C(\lambda_{\mu})>0.$

###### Proof.

We first observe that $u$ is a weak solution of (6.11) if and only if
$v=\frac{u}{\phi_{\mu}}$ satisfies

$\int_{\Omega}\phi_{\mu}^{2}\nabla v\cdot\nabla\zeta
dx+\lambda_{\mu}\int_{\Omega}\phi_{\mu}^{2}v\zeta
dx=\int_{\Omega}\phi_{\mu}f\zeta dx\;,\qquad\forall\zeta\in
H^{1}_{0}(\Omega;\phi_{\mu}^{2}).$ (6.13)

We define on $H_{0}^{1}(\Omega;\phi_{\mu}^{2})$ the inner product

$\langle\psi,\zeta\rangle_{\phi_{\mu}^{2}}=\int_{\Omega}\phi_{\mu}^{2}(\nabla\psi\cdot\nabla\zeta+\lambda_{\mu}\psi\,\zeta)dx$

and consider the bounded linear functional $T_{f}$ on
$H_{0}^{1}(\Omega;\phi_{\mu}^{2})$ given by

$T_{f}(\zeta)=\int_{\Omega}\phi_{\mu}f\zeta dx.$

Then (6.13) becomes

$\langle v,\zeta\rangle_{\phi_{\mu}^{2}}=T_{f}(\zeta)\quad\forall\zeta\in
H_{0}^{1}(\Omega;\phi_{\mu}^{2}).$ (6.14)

By Riesz representation theorem there exists a unique function $v\in
H_{0}^{1}(\Omega;\phi_{\mu}^{2})$ satisfying (6.14). Furthermore, by choosing
$\zeta=v$ in (6.13) and then using Young’s inequality, we obtain

$\int_{\Omega}\phi_{\mu}^{2}|\nabla
v|^{2}dx+\frac{\lambda_{\mu}}{2}\int_{\Omega}\phi_{\mu}^{2}v^{2}dx\leq
C(\lambda_{\mu})\int_{\Omega}f^{2}dx.$ (6.15)

By putting $u=\phi_{\mu}v$, we deduce that $u$ is the unique weak solution of
(6.11). Moreover, (6.12) follows from (6.15). ∎

The next lemma will be useful in order to prove existence and uniqueness of
solution for the equation $L_{\mu}u=f$ with zero boundary data.

######  Lemma 6.6.

[GN3, Lemma 5.3] Let $\gamma<N$ and $\alpha\in(0,\min\\{k,\gamma\\})$. There
exists a positive constant $C=C(\alpha,\gamma,\Omega,K)$ such that

$\sup_{x\in\Omega}\int_{\Omega}|x-y|^{-N+\gamma}d_{K}^{-\alpha}(y)dy<C.$

In the following lemma we prove the existence of solution for the equation
$L_{\mu}u=f$ with zero boundary data, as well as pointwise estimates.

######  Lemma 6.7.

Let $\mu\leq k^{2}/4$ and assume that $\lambda_{\mu}>0$, $\gamma_{-}-1<b<0$
and $f\in L^{\infty}(\Omega)$. Then there exists a unique $u\in
H^{1}_{loc}(\Omega)\cap C(\Omega)$ which satisfies $L_{\mu}u=fd_{K}^{b}$ in
the sense of distributions as well as (6.9). Moreover, for any
$\gamma\in(-\infty,{\gamma_{+}}]\cap(-\infty,b+1)\cap(-\infty,0]$ there exists
a positive constant $C=C(\Omega,K,b,\mu,\gamma)$ such that

$|u(x)|\leq C\|f\|_{L^{\infty}(\Omega)}d(x)d_{K}^{\gamma}(x),\qquad
x\in\Omega.$ (6.16)

###### Proof.

We assume first that $f\geq 0.$ Set $f_{n}=\min\\{fd_{K}^{b},n\\}$. By Lemma
6.5, there exists a unique solution $u_{n}$ of $L_{\mu}v=f_{n}$ in $\Omega$.
Furthermore, a standard argument yields the representation formula

$u_{n}(x)=\int_{\Omega}G_{\mu}(x,y)f_{n}(y)dy.$

We assume first that $0<\mu<\frac{k^{2}}{4}.$ By (5.4) we have

$\displaystyle 0$ $\displaystyle\leq\int_{\Omega}G_{\mu}(x,y)f_{n}(y)dy$
$\displaystyle\leq
C_{1}\int_{\Omega}\min\left\\{\frac{1}{|x-y|^{N-2}},\frac{d(x)d(y)}{|x-y|^{N}}\right\\}\left(\frac{d_{K}(x)d_{K}(y)}{\left(d_{K}(x)+|x-y|\right)\left(d_{K}(y)+|x-y|\right)}\right)^{\gamma_{+}}f_{n}(y)dy$
$\displaystyle\leq
Cd_{K}^{\gamma_{+}}(x)\int_{\Omega}|x-y|^{-N+2-2{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}d_{K}^{\gamma_{+}}(y)f_{n}(y)dy$
$\displaystyle\quad+C\int_{\Omega}|x-y|^{-N+2-{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}d_{K}^{\gamma_{+}}(y)f_{n}(y)dy$
$\displaystyle\quad+Cd_{K}^{\gamma_{+}}(x)\int_{\Omega}|x-y|^{-N+2-{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}f_{n}(y)dy$
$\displaystyle\quad+C\int_{\Omega}|x-y|^{-N+2}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}f_{n}(y)dy$
$\displaystyle=C(I_{1}+I_{2}+I_{3}+I_{4}).$

First we note that if $d_{K}(y)\leq\frac{1}{4}d_{K}(x)$ then
$|x-y|\geq\frac{3}{4}d_{K}(x).$ Thus for $\gamma\leq{\gamma_{+}}$, we have

$\displaystyle I_{1}$
$\displaystyle=d_{K}^{\gamma_{+}}(x)\int_{\Omega\cap\\{d_{K}(y)\leq\frac{1}{4}d_{K}(x)\\}}|x-y|^{-N+2-2{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}d_{K}^{\gamma_{+}}(y)f_{n}(y)dy$
$\displaystyle\quad+d_{K}^{\gamma_{+}}(x)\int_{\Omega\cap\\{d_{K}(y)>\frac{1}{4}d_{K}(x)\\}}|x-y|^{-N+2-2{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}d_{K}^{\gamma_{+}}(y)f_{n}(y)dy$
$\displaystyle\leq
C\|f\|_{L^{\infty}(\Omega)}d_{K}^{\gamma}(x)\int_{\Omega\cap\\{d_{K}(y)\leq\frac{1}{4}d_{K}(x)\\}}|x-y|^{-N+2-\gamma-{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}d_{K}^{b+{\gamma_{+}}}(y)dy$
$\displaystyle\quad+C\|f\|_{L^{\infty}(\Omega)}d_{K}^{\gamma}(x)\int_{\Omega\cap\\{d_{K}(y)>\frac{1}{4}d_{K}(x)\\}}|x-y|^{-N+2-2{\gamma_{+}}}\min\Big{\\{}1,\frac{d(x)d(y)}{|x-y|^{2}}\Big{\\}}d_{K}^{b-\gamma+2{\gamma_{+}}}(y)dy$
$\displaystyle\leq
C\|f\|_{L^{\infty}(\Omega)}d_{K}^{\gamma}(x)d(x)\int_{\Omega\cap\\{d_{K}(y)\leq\frac{1}{4}d_{K}(x)\\}}|x-y|^{-N-\gamma-{\gamma_{+}}}d_{K}^{b+{\gamma_{+}}+1}(y)dy$
$\displaystyle\quad+C\|f\|_{L^{\infty}(\Omega)}d_{K}^{\gamma}(x)d(x)\int_{\Omega\cap\\{d_{K}(y)>\frac{1}{4}d_{K}(x)\\}}|x-y|^{-N-2{\gamma_{+}}}d_{K}^{b-\gamma+2{\gamma_{+}}+1}(y)dy$
$\displaystyle\leq C\|f\|_{L^{\infty}(\Omega)}d_{K}^{\gamma}(x)d(x)$

where in the last inequalities we have used Lemma 6.6.

Similarly we can prove that

$\displaystyle I_{1}+I_{2}+I_{3}+I_{4}\leq
C\|f\|_{L^{\infty}(\Omega)}d_{K}^{\gamma}(x)d(x)$

Combining the above estimates, we deduce that for any
$\gamma\in(-\infty,{\gamma_{+}}]\cap(-\infty,b+1),$ there exists a positive
constant $C=C(\Omega,K,\mu,b,\gamma)$ such that

$|u_{n}(x)|\leq C\|f\|_{L^{\infty}(\Omega)}d(x)d_{K}^{\gamma}(x),\qquad
x\in\Omega.$ (6.17)

If we choose $\gamma\in({\gamma_{-}},{\gamma_{+}}]\cap({\gamma_{-}},b+1),$
then we can show that

$\lim_{\mathrm{dist}(x,F)\to
0}\frac{d(x)d_{K}^{\gamma}(x)}{\tilde{W}(x)}=0,\quad\forall\;\text{compact}\;F\subset\partial\Omega.$
(6.18)

Thus by the above inequality, (6.17) and applying Lemma 6.3, we can easily
show that $u_{n}\nearrow u$ locally uniformly in $\Omega$ and in
$H^{1}_{loc}(\Omega).$ Furthermore, by standard elliptic theory $u\in
C^{1}(\Omega)$ and, by (6.17),

$|u(x)|\leq C\|f\|_{L^{\infty}(\Omega)}d(x)d_{K}^{\gamma}(x),\qquad
x\in\Omega.$ (6.19)

The uniqueness follows by (6.19), (6.18) and Lemma 6.3.

For the general case, we set $u=u_{+}-u_{-}$ where $u_{\pm}$ are the unique
solutions of $L_{\mu}v=f_{\pm}d_{K}^{-b}$ in $\Omega\setminus K$ respectively,
which satisfy (6.16). Thus $u$ satisfies (6.16) and the result follows in the
case $0<\mu<\frac{k^{2}}{4}$.

The proof in the cases $\mu=\frac{k^{2}}{4}$ and $\mu\leq 0$ is similar and is
omitted. ∎

The following lemma is the main result of this subsection.

######  Lemma 6.8.

Let $\mu\leq k^{2}/4$ and assume that $\lambda_{\mu}>0$. For any $h\in
C(\partial\Omega)$ there exists a unique $L_{\mu}$-harmonic function $u\in
H^{1}_{loc}(\Omega)\cap C(\Omega)$ satisfying

$\lim_{x\in\Omega,\;x\rightarrow
y\in\partial\Omega}\frac{u(x)}{\tilde{W}(x)}=h(y)\qquad\text{uniformly in
}y\in\partial\Omega.$

Furthermore there exists a constant $c=c(\Omega,K)>0$

$\left|\\!\left|\frac{u}{\tilde{W}}\right|\\!\right|_{L^{\infty}(\Omega)}\leq
c\|h\|_{C(\partial\Omega)}.$

###### Proof.

Uniqueness is a consequence of Lemma 6.3.

_Existence._ We will only consider the case $0<\mu<\frac{k^{2}}{4},$ the proof
in the other cases is very similar. First we assume that $h\in
C^{2}(\overline{\Omega})$. Then a function $u\in C^{2}(\Omega)$ is
$L_{\mu}$-harmonic if and only if $v:=\tilde{W}h-u$ is a solution of

$L_{\mu}v=L_{\mu}(\tilde{W}h)=h(L_{\mu}\tilde{W})-2\nabla\tilde{W}\cdot\nabla
h-\tilde{W}\Delta h,\qquad\text{in }\Omega\,;$ (6.20)

Arguing as in the proof of Lemma 6.1 we see that there exists
$C=C(\Omega,K,\mu,\beta_{0})$ such that

$|L_{\mu}\tilde{W}|\leq Cd_{K}^{{\gamma_{-}}},\qquad\text{in }\Omega.$

Hence (6.20) can be written as

$L_{\mu}v=fd_{K}^{{\gamma_{-}}},\qquad\text{in }\Omega,$

with $\left\|f\right\|_{L^{\infty}(\Omega)}\leq
C({\gamma_{-}},\Omega,K)\left\|h\right\|_{C^{2}(\overline{\Omega})}.$

By Lemma 6.7 there exists a unique solution $v$ of (6.20) that satisfies

$|v(x)|\leq C\|h\|_{C^{2}(\overline{\Omega})}d(x)d_{K}^{\gamma}(x),\qquad
x\in\Omega,$

for any
$\gamma\in({\gamma_{-}},{\gamma_{+}}]\cap({\gamma_{-}},{\gamma_{-}}+1)$. Thus

$\left|\frac{u(x)}{\tilde{W}(x)}-h(x)\right|\leq
C\|h\|_{C^{2}(\overline{\Omega})}\frac{d(x)d_{K}^{\gamma}(x)}{\tilde{W}(x)},\qquad
x\in\Omega,$ (6.21)

and the desired result follows in this case, since

$\lim_{\mathrm{dist}(x,F)\to
0}\frac{d(x)d_{K}^{\gamma}(x)}{\tilde{W}(x)}=0,\qquad\forall\;\text{compact}\;F\subset\partial\Omega.$

for any
$\gamma\in({\gamma_{-}},{\gamma_{+}}]\cap({\gamma_{-}},{\gamma_{-}}+1).$

Suppose now that $h\in C(\partial\Omega)$. We can then find a sequence
$\\{h_{n}\\}_{n=1}^{\infty}$ of smooth functions in $\partial\Omega$ such that
$h_{n}\rightarrow h$ in $L^{\infty}(\partial\Omega).$ Then there exist
$H_{n}\in C^{2}(\overline{\Omega})$ with value $H_{n}|_{\partial\Omega}=h_{n}$
and $\|H_{n}\|_{L^{\infty}(\overline{\Omega})}\leq
C\|h_{n}\|_{L^{\infty}(\partial\Omega)}$ where $C$ does not depend on $n$ or
$h_{n}$. By the previous case there exists a unique weak solution $u_{n}$ of
$L_{\mu}u=0$ satisfying

$\displaystyle\left|\frac{u_{n}(x)}{\tilde{W}(x)}-H_{n}(x)\right|\leq
C\|H_{n}\|_{C^{2}(\overline{\Omega})}\frac{d(x)d_{K}^{\gamma}(x)}{\tilde{W}(x)},\qquad\forall
x\in\Omega,$

for some $C$ which does not depend on $n$ and $h_{n}$.

By (6.21) and Lemma 6.3, we can easily show that

$\left|\frac{u_{n}(x)-u_{m}(x)}{\tilde{W}(x)}\right|\leq
C\|h_{n}-h_{m}\|_{L^{\infty}(\partial\Omega)},\qquad x\in\Omega;$

thus $u_{n}\rightarrow u$ locally uniformly in $\Omega$.

Now, let $y\in\partial\Omega.$ Then

$\left|\frac{u(x)}{\tilde{W}(x)}-h(y)\right|\leq\left|\frac{u(x)-u_{n}(x)}{\tilde{W}(x)}\right|+\left|\frac{u_{n}(x)}{\tilde{W}(x)}-h_{n}(y)\right|+\left|h_{n}(y)-h(y)\right|$

and the result follows by letting successively $x\to y$ and $n\to\infty$. ∎

##  7\. Martin kernel

###  7.1. $L_{\mu}$-harmonic measure

Let $x_{0}\in\Omega,$ $h\in C(\partial\Omega)$ and denote
$L_{\mu,x_{0}}(h):=v_{h}(x_{0})$ where $v_{h}$ is the solution of the
Dirichlet problem (see Lemma 6.8)

$\left\\{\begin{aligned} L_{\mu}v&=0,\qquad\mathrm{in}\;\;\Omega,\\\
\widetilde{\rm
tr}(v)&=h,\qquad\mathrm{in}\;\;\partial\Omega,\end{aligned}\right.$

where $\widetilde{\rm tr}(v)=h$ is understood in the sense of Lemma 6.8 (cf.
also (2.6)). By Lemma 6.3, the mapping $h\mapsto L_{\mu,x_{0}}(h)$ is a
positive linear functional on $C(\partial\Omega).$ Thus there exists a unique
Borel measure on $\partial\Omega$, called $L_{\mu}$-harmonic measure in
$\Omega,$ denoted by $\omega^{x_{0}}$, such that

$v_{h}(x_{0})=\int_{\partial\Omega}h(y)d\omega^{x_{0}}(y).$

Thanks to the Harnack inequality the measures $\omega^{x}$ and
$\omega^{x_{0}}$, $x_{0},\,x\in\Omega$, are mutually absolutely continuous.
For every fixed $x$ we denote the Radon-Nikodyn derivative by

$K_{\mu}(x,y):=\frac{dw^{x}}{dw^{x_{0}}}(y),\qquad\mathrm{for}\;\omega^{x_{0}}\text{-
almost all }y\in\partial\Omega.$ (7.1)

Let $\xi\in\partial\Omega$. We set $\Delta_{r}(\xi)=\partial\Omega\cap
B_{r}(\xi)$ and denote by $x_{r}=x_{r}(\xi)$ the point in $\Omega$ determined
by $d(x_{r})=|x_{r}-\xi|=r$. We recall here that
$\beta_{0}=\beta_{0}(\Omega,K,\mu)>0$ is small enough and has been defined in
Lemma 6.1.

######  Lemma 7.1.

Let $\mu\leq k^{2}/4$ and assume that $\lambda_{\mu}>0$. Let
$0<r\leq\beta_{0}$. We assume that $u$ is a positive $L_{\mu}$-harmonic
function in $\Omega$ such that

$\displaystyle({\rm i})$ $\displaystyle\frac{u}{\tilde{W}}\in
C(\overline{\Omega\setminus B_{r}(\xi)}),$ $\displaystyle({\rm ii})$
$\displaystyle\lim_{x\in\Omega,x\to
x_{0}}\frac{u(x)}{\tilde{W}(x)}=0,\quad\forall
x_{0}\in\partial\Omega\setminus\overline{B_{r}(\xi)},\text{uniformly with
respect to $x_{0}$.}$

Then

$\displaystyle\begin{aligned}
c^{-1}\frac{u(x_{r}(\xi))}{G_{\mu}(x_{r}(\xi),x_{\frac{r}{16}}(\xi))}&G_{\mu}(x,x_{\frac{r}{16}}(\xi))\leq
u(x)\\\ &\leq
c\frac{u(x_{r}(\xi))}{G_{\mu}(x_{r}(\xi),x_{\frac{r}{16}}(\xi))}G_{\mu}(x,x_{\frac{r}{16}}(\xi)),\qquad\forall
x\in\Omega\setminus\overline{B_{2r}(\xi)},\end{aligned}$ (7.2)

with $c>1$ depending only $\Omega$, $K$ and $\mu$.

###### Proof.

It follows from Lemma 6.5 that there exists $c>1$ such that

$\displaystyle\begin{aligned}
c^{-1}\frac{u(x_{2r}(\xi))}{G_{\mu}(x_{2r}(\xi),x_{\frac{r}{16}}(\xi))}&G_{\mu}(x,x_{\frac{r}{16}}(\xi))\leq
u(x)\\\ &\leq
c\frac{u(x_{2r}(\xi))}{G_{\mu}(x_{2r}(\xi),x_{\frac{r}{16}}(\xi))}G_{\mu}(x,x_{\frac{r}{16}}(\xi)),\qquad\forall
x\in\Omega\cap\partial B_{2r}(\xi),\end{aligned}$

Applying Harnack inequality between $x_{2r}(\xi)$ and $x_{r}(\xi)$ we obtain

$\displaystyle\begin{aligned}
c^{-1}\frac{u(x_{r}(\xi))}{G_{\mu}(x_{r}(\xi),x_{\frac{r}{16}}(\xi))}&G_{\mu}(x,x_{\frac{r}{16}}(\xi))\leq
u(x)\\\ &\leq
c\frac{u(x_{r}(\xi))}{G_{\mu}(x_{r}(\xi),x_{\frac{r}{16}}(\xi))}G_{\mu}(x,x_{\frac{r}{16}}(\xi)),\qquad\forall
x\in\Omega\cap\partial B_{2r}(\xi).\end{aligned}$

For $\varepsilon>0$ let

$u_{\varepsilon}(x)=u(x)-c\frac{u(x_{r}(\xi))}{G_{\mu}(x_{r}(\xi),x_{\frac{r}{16}}(\xi))}G_{\mu}(x,x_{\frac{r}{16}}(\xi))-\varepsilon
v_{1}(x),$

where $c$ is as above. Then $u_{\varepsilon}$ is $L_{\mu}$-harmonic and the
function $u_{\varepsilon}^{+}=\max(u_{\varepsilon},0)$ has compact support in
$\Omega\setminus\overline{B_{2r}(\xi)}$. Set
$v_{\varepsilon}=\frac{u_{\varepsilon}}{\phi_{\mu}}$ and
$v_{\varepsilon}^{+}=\frac{u_{\varepsilon}^{+}}{\phi_{\mu}}$. Using
$u_{\varepsilon}^{+}$ as a test function we obtain

$\int_{\Omega\setminus\overline{B_{2r}(\xi)}}\nabla v_{\varepsilon}\cdot\nabla
v_{\varepsilon}^{+}\phi_{\mu}^{2}dx+\lambda_{\mu}\int_{\Omega\setminus\overline{B_{2r}(\xi)}}v_{\varepsilon}v_{\varepsilon}^{+}\phi_{\mu}^{2}dx=0.$

Letting $\varepsilon\to 0$ in the above equation we get

$\lambda_{\mu}\int_{\Omega}|v^{+}|^{2}\phi_{\mu}^{2}dx\leq 0,$

hence
$u(x)-c\frac{u(x_{r}(\xi))}{G_{\mu}(x_{r}(\xi),x_{\frac{r}{16}}(\xi))}G_{\mu}(x,x_{\frac{r}{16}}(\xi))\leq
0$ for all $x\in\Omega\setminus\overline{B_{2r}(\xi)}$. The proof of the lower
estimate in (7.2) is similar and we omit it. ∎

###  7.2. The Poisson kernel of $L_{\mu}$

In this section we establish some properties of the Poisson kernel associated
to $L_{\mu}$.

######  Definition 7.2.

A function $\mathcal{K}$ defined in $\Omega$ is called a kernel function for
$L_{\mu}$ with pole at $\xi\in\partial\Omega$ and basis at $x_{0}\in\Omega$ if

$\displaystyle{\rm(i)}$ $\mathcal{K}(\cdot,\xi)$ is $L_{\mu}$-harmonic in
$\Omega$, $\displaystyle{\rm(ii)}$ $\displaystyle\mbox{
$\frac{\mathcal{K}(\cdot,\xi)}{\tilde{W}(\cdot)}\in
C(\overline{\Omega}\setminus\\{\xi\\})$ and for any
$\eta\in\partial\Omega\setminus\\{\xi\\}$ we have
}\lim_{x\in\Omega,\;x\to\eta}\frac{\mathcal{K}(x,\xi)}{\tilde{W}(x)}=0,$
$\displaystyle{\rm(iii)}$ $\mathcal{K}(x,\xi)>0$ for each $x\in\Omega$ and
$\mathcal{K}(x_{0},\xi)=1.$

######  Proposition 7.3.

Assume that $\lambda_{\mu}>0$. There exists a unique kernel function for
$L_{\mu}$ with pole at $\xi$ and basis at $x_{0}.$

###### Proof.

The proof is similar to that of [CFMS, Theorem 3.1] and we include it for the
sake of completeness.

Existence. We shall prove that the function $K_{\mu}(x,\xi)$ defined by (7.1)
has the required properties.

Fix $\xi\in\partial\Omega$. Set

$u_{n}(x)=\frac{\omega^{x}(\Delta_{2^{-n}}(\xi))}{\omega^{x_{0}}(\Delta_{2^{-n}}(\xi))},\qquad\forall
n\geq n_{0}.$

Clearly $u_{n}(x)\to K_{\mu}(x,\xi)$, $x\in\Omega$. Since $u_{n}\geq 0$,
$L_{\mu}u_{n}=0$ in $\Omega$ and $u_{n}(x_{0})=1$ the sequence $\\{u_{n}\\}$
is locally bounded in $\Omega$ by Harnack inequality. Hence we can find a
subsequence, again denoted by $\\{u_{n}\\},$ which converges to
$K_{\mu}(\cdot,\xi)$ locally uniformly in $\Omega$.

Let $\eta\in\partial\Omega\setminus\\{\xi\\}$ and let $n_{1}\in\mathbb{N}$ be
such that
$\eta\in\partial\Omega\setminus\overline{B_{2^{-n+1}}(\xi)},\;\forall n\geq
n_{1}$. By Lemma 7.1 we have

$u_{n}(x)\leq
c\frac{u_{n}(x_{2^{-n_{1}}}(\xi))}{G_{\mu}(x_{2^{-n_{1}}},x_{2^{-n_{1}-4}}(\xi))}G_{\mu}(x,x_{2^{-n_{1}-4}}(\xi)),\qquad\forall
x\in\Omega\setminus\overline{B_{2^{-n_{1}+1}}(\xi)},$

which implies

$K_{\mu}(x,\xi)\leq
c\frac{u_{n}(x_{2^{-n_{1}}}(\xi))}{G_{\mu}(x_{2^{-n_{1}}},x_{2^{-n_{1}-4}}(\xi))}G_{\mu}(x,x_{2^{-n_{1}-4}}(\xi)),\qquad\forall
x\in\Omega\setminus\overline{B_{2^{-n_{1}+1}}(\xi)}.$

It follows that

$\lim_{x\in\Omega,\;x\to\eta}\frac{K_{\mu}(x,\xi)}{\tilde{W}(x)}=0,$

hence $K_{\mu}(x,\xi)$ is a kernel function for $L_{\mu}$ with pole at $\xi$
and basis at $x_{0}$.

Uniqueness. Assume $f$ and $g$ are two kernel functions for $L_{\mu}$ in
$\Omega$ with pole at $\xi$ and basis at $x_{0}$. Let $0<r<\beta_{0}$. By
Lemma 7.1 and the properties of $f$ and $g$ there holds

$\frac{1}{c^{\prime}}\frac{f(x_{r}(\xi))}{g(x_{r}(\xi))}\leq\frac{f(x)}{g(x)}\leq
c^{\prime}\frac{f(x_{r}(\xi))}{g(x_{r}(\xi))}\;,\qquad\forall
x\in\Omega\setminus\overline{B_{2r}(\xi)}.$

In particular we can obtain if we take $x=x_{0}$

$\frac{f(x_{r}(\xi))}{g(x_{r}(\xi))}\leq c^{\prime},$

and hence

$\frac{f(x)}{g(x)}\leq c^{\prime 2}=:c\,,\qquad\forall x\in\Omega.$

We derive that for any two kernel functions $f$ and $g$ for $L_{\mu}$ with
pole at $\xi$ and basis at $x_{0}$ there holds

$f(x)\leq cg(x)\leq c^{2}f(x)\,,\qquad\quad x\in\Omega.$

Obviously $c\geq 1.$ If $c=1$ the result is proved. If $c>1$ then we set
$A=\frac{1}{c-1}$ and $f+A(f-g)$ is also a kernel function for $L_{\mu}$ with
pole at $\xi$ and basis at $x_{0}$. Repeating the argument for the functions
$f+A(f-g)$ and $g$ we obtain that

$f+A(f-g)+A\big{(}f-g+A(f-g)\big{)},$

is also a kernel function with pole at $\xi$ and basis at $x_{0}$. Proceeding
in this manner we conclude that for each positive integer $k$ there exist
nonnegative numbers $a_{1k},...,a_{kk}$ such that

$f+\Big{(}kA+\sum_{i=1}^{k}a_{ik}\Big{)}(f-g)$

is a kernel function with pole at $\xi$ and basis at $x_{0}$. Hence

$f+\Big{(}kA+\sum_{i=1}^{k}a_{ik}\Big{)}(f-g)\leq cf.$

This last inequality can hold for all $k$ only if $f\equiv g$. ∎

######  Proposition 7.4.

Assume that $\lambda_{\mu}>0$. For any $x\in\Omega$, the function $\xi\mapsto
K_{{\mu}}(x,\xi)$ is continuous on $\partial\Omega$.

###### Proof.

The proof is an adaptation of that of [CFMS, Corollary 3.2]. Suppose that
$\\{\xi_{n}\\}$ is a sequence converging to $\xi$. Then the sequence
$\\{K_{\mu}(\cdot,\xi_{n})\\}$ of positive solutions of $L_{\mu}u=0$ in
$\Omega$ has a subsequence which converges locally uniformly in $\Omega$ to a
positive $L_{\mu}$-harmonic function. Moreover, for any $r>0,$
$\frac{K_{\mu}(x,\xi_{n})}{\tilde{W}(x)}$ converges to zero uniformly in $n$
as $x\rightarrow\eta\in\partial\Omega\setminus B_{r}(\xi)$. Hence the limit
function of the subsequence is the kernel function $K_{{\mu}}(x,\xi)$. By the
uniqueness of the kernel function we conclude that the convergence

$K_{{\mu}}(x,\xi_{n})\rightarrow K_{{\mu}}(x,\xi)$

holds for the entire sequence $\\{\xi_{n}\\}$. ∎

We can now identify the Martin boundary and topology with their classical
analogues. We begin by recalling the definitions of the Martin boundary and
related concepts.

Let $x_{0}\in\Omega$ be fixed. For $x,\;y\in\Omega$ we set

$\mathcal{K}_{\mu}(x,y):=\frac{G_{L_{\mu}}(x,y)}{G_{L_{\mu}}(x_{0},y)}.$

Consider the family of sequences $\\{y_{k}\\}_{k\geq 1}$ of points of $\Omega$
without cluster points in $\Omega$ for which $\mathcal{K}_{\mu}(x,y_{k})$
converges in $\Omega$ to a $L_{\mu}$-harmonic function, denoted by
$\mathcal{K}_{\mu}(x,\\{y_{k}\\})$. Two such sequences ${y_{k}}$ and
${y_{k}^{\prime}}$ are called equivalent if
$\mathcal{K}_{\mu}(x,\\{y_{k}\\})=\mathcal{K}_{\mu}(x,\\{y_{k}^{\prime}\\})$
and each equivalence class is called an element of the Martin boundary
$\Gamma$. If $Y$ is such an equivalence class (i.e., $Y\in\Gamma$) then
$\mathcal{K}_{\mu}(x,Y)$ will denote the corresponding harmonic limit
function. Thus each $Y\in\Omega\cup\Gamma$ is associated with a unique
function $\mathcal{K}_{\mu}(x,Y).$ The Martin topology on $\Omega\cup\Gamma$
is given by the metric

$\rho(Y,Y^{\prime})=\int_{A}\frac{|\mathcal{K}_{\mu}(x,Y)-\mathcal{K}_{\mu}(x,Y^{\prime})|}{1+|\mathcal{K}_{\mu}(x,Y)-\mathcal{K}_{\mu}(x,Y^{\prime})|}dx,\qquad
Y,Y^{\prime}\in\Omega\cup\Gamma,$

where $A$ is a small enough neighbourhood of $x_{0}$. The function
$\mathcal{K}_{\mu}(x,Y)$ is a $\rho$-continuous function of
$Y\in\Omega\cup\Gamma$ for any fixed $x\in\Omega$. Moreover $\Omega\cup\Gamma$
is compact and complete with respect to $\rho$, $\Omega\cup\Gamma$ is the
$\rho$-closure of $\Omega$ and the $\rho$-topology is equivalent to the
Euclidean topology in $\Omega$.

######  Proposition 7.5.

Assume that $\lambda_{\mu}>0$. There is a one-to-one correspondence between
the Martin boundary of $\Omega$ and the Euclidean boundary $\partial\Omega.$
If $Y\in\Gamma$ corresponds to $\xi\in\partial\Omega$ then
$\mathcal{K}_{\mu}(x,Y)=K_{\mu}(x,\xi).$ The Martin topology on
$\Omega\cup\Gamma$ is equivalent to the Euclidean topology on
$\Omega\cup\partial\Omega.$

###### Proof.

The proof is similar as the one of Theorem 4.2 in [HW] and we include it for
the sake of completeness. By uniqueness of the kernel function we have that

$\mathcal{K}_{\mu}(x,\\{y_{k}\\})=K_{\mu}(x,\xi),$

where $\\{y_{k}\\}$ is a sequence in $\Omega$ such that
$y_{k}\rightarrow\xi\in\partial\Omega.$ It follows that each point of $\Gamma$
may be associated with a point of $\partial\Omega.$ Lemma 7.1 clearly shows
that $K_{\mu}(\cdot,\xi)\neq K_{\mu}(\cdot,\xi^{\prime})$ if
$\xi\neq\xi^{\prime}$. Hence, the functions $\mathcal{K}_{\mu}(x,y_{k})$
cannot converge if the sequence $\\{y_{k}\\}$ has more than one cluster point
on $\partial\Omega$ and different points of $\partial\Omega$ must be
associated with different points of $\Gamma.$ This gives a one-to-one
correspondence between $\partial\Omega$ and $\Gamma$ with
$\mathcal{K}_{\mu}(x,Y)=K_{\mu}(x,\xi)$ when $Y\in\Gamma$ corresponds to
$\xi\in\partial\Omega.$ If $\xi_{k}\rightarrow\xi$ in the Euclidean topology
then $\mathcal{K}_{\mu}(x,Y_{k})\rightarrow\mathcal{K}_{\mu}(x,Y)$ and,
therefore, $Y_{k}\rightarrow Y$ in the $\rho$-topology by Lebesgue’s dominated
convergence theorem. On the other hand suppose that $Y_{k}\rightarrow Y$ in
the $\rho$-topology. If $\xi_{k}$ does not converge to $\xi$ in the Euclidean
topology there is a subsequence $\xi_{k_{j}}$ such that
$\xi_{k_{j}}\rightarrow\xi^{\prime}\neq\xi$ in the Euclidean topology. Then
$Y_{k_{j}}\rightarrow Y^{\prime}$ and $Y_{k_{j}}\rightarrow Y$ in the $\rho$\-
topology with $Y\neq Y^{\prime},$ which is impossible. Therefore, the Martin
$\rho$-topology on $\Omega\cup\Gamma$ is equivalent to the Euclidean topology
on $\Omega\cup\partial\Omega.$ ∎

Proof of Theorem 2.8. The result follows immediately by Proposition 7.5 and
Proposition 5.3. $\hfill\Box$

The next lemma will be used to prove the representation formula of Theorem
2.9.

######  Lemma 7.6.

Assume that $\lambda_{\mu}>0$. Let $F\subset\partial\Omega$ and $D$ be an open
smooth neighbourhood of $F.$ We assume $\Omega\cap D\subset\Omega_{\beta}$ for
some $\beta>0.$ Let $u$ be a positive $L_{\mu}$-harmonic function in $\Omega.$
There exists a $L_{\mu}$-superharmonic function $V$ such that

$V(x)=\left\\{\begin{array}[]{lll}v(x),&\text{in }\;\Omega\setminus
D,\\\\[2.84526pt] u(x),&\text{in }\;\Omega\cap\overline{D},\end{array}\right.$

where $v$ satisfies

$\left\\{\begin{array}[]{ll}L_{\mu}v=0,&\text{in
}\,\Omega\setminus\overline{D},\\\\[2.84526pt]
\lim_{x\in\Omega\setminus\overline{D},\;x\to y}v(x)=u(y),&\forall y\in\partial
D\cap\Omega,\\\\[2.84526pt] \lim_{x\in\Omega\setminus\overline{D},\;x\to
y}\frac{v(x)}{\tilde{W}(x)}=0,&\forall