Skeleton Decomposition and Law of Large Numbers for Supercritical Superprocesses

Abstract

The goal of this paper is twofold. First, we establish skeleton and spine decompositions for superprocesses whose underlying processes are general symmetric Hunt processes. Second, we use these decompositions to obtain weak and strong law of large numbers for supercritical superprocesses where the spatial motion is a symmetric Hunt process on a locally compact metric space $E$ and the branching mechanism takes the form

$$ \psi _{\beta }(x,\lambda )=-\beta (x)\lambda +\alpha (x)\lambda ^{2}+ \int _{(0,{\infty })}\bigl(e^{-\lambda y}-1+\lambda y\bigr)\pi (x,dy) $$

with $\beta \in \mathcal{B}_{b}(E)$, $\alpha \in \mathcal{B}^{+}_{b}(E)$ and $\pi$ being a kernel from $E$ to $(0,{\infty })$ satisfying $\sup_{x\in E}\int _{(0,{\infty })} (y\wedge y^{2}) \pi (x,dy)<{\infty }$. The limit theorems are established under the assumption that an associated Schrödinger operator has a spectral gap. Our results cover many interesting examples of superprocesses, including super Ornstein-Uhlenbeck process and super stable-like process. The strong law of large numbers for supercritical superprocesses are obtained under the assumption that the strong law of large numbers for an associated supercritical branching Markov process holds along a discrete sequence of times, extending an earlier result of Eckhoff et al. (Ann. Probab. 43(5):2594–2659, 2015) for superdiffusions to a large class of superprocesses. The key for such a result is due to the skeleton decomposition of superprocess, which represents a superprocess as an immigration process along a supercritical branching Markov process.

Appendix

1.1 A.1 Kuznetsov Measure for Superdiffusions

Suppose the operator $L$ is defined by the formula:

$$ L:=\sum_{i,j=1}^{d} \frac{\partial }{\partial x_{j}}\,a_{ij}(x)\frac{ \partial }{\partial x_{j}}+\sum _{i=1}^{d}b_{i}(x)\frac{\partial }{ \partial x_{i}}\quad \mbox{on }\mathbb{R}^{d}, $$

where the matrix $A(x):=(a_{ij}(x))_{i,j}$ is symmetric and positive definite, and all $a_{ij}$ and $b_{i}$ are bounded and (globally) Hölder continuous on $\mathbb{R}^{d}$. It is known (cf. [13, Chap. 2]) that there exists a diffusion process on $\mathbb{R}^{d}$, called the $L$-diffusion, whose transition density function is a fundamental solution of the equation $\partial {u}/ \partial t=Lu$. In this subsection, we assume $E\subset \mathbb{R} ^{d}$ is a nonempty domain and $\xi$ is the subprocess of the $L$-diffusion on $E$. We assume that the branching mechanism $\psi _{\beta }(x,{\lambda })$ is given by (2.1) and all the first partial derivatives of $\psi _{\beta }(x,{\lambda })$ are continuous. This $(\xi ,\psi _{\beta })$-superprocess is also called $(\xi , \psi _{\beta })$-superdiffusion. Let $X$ be a $(\xi ,\psi _{\beta })$-superdiffusion. We are concerned with the set $E_{0}=\{x\in E: \mbox{Kuznetsov measure }\mathbb{N}_{x} \mbox{ exists}\}$. By the argument in Sect. 2.3, we know that (2.13) is a sufficient condition for the existence of $\mathbb{N}_{x}$. We recall that for every open set $D\subset E$ and $t\ge 0$, there is a random measure $\widetilde{X}^{D}_{t}$ concentrated on the boundary of $[0,t)\times D$ such that (2.7) and (2.8) hold for every $\mu \in \mathcal{M}_{F}(E)$ and every $\widetilde{f}\in \mathcal{B}^{+}_{b}([0,t]\times E)$. We also recall that $X_{t}$ is the projection of $\widetilde{X}^{E}_{t}$ on $E$. For every $x\in E$, let $\mathbb{O}_{x}$ be the set of open sets in $E$ that contains $x$. Then condition (2.13) is satisfied if

$$ \mathrm{P}_{\delta _{x}} \bigl( \widetilde{X}^{D}_{t}=0 \bigr) >0 \quad \mbox{for all }t>0\mbox{ and }D\in \mathbb{O}_{x}. $$

(A.1)

It follows from the Markov property of branching exit Markov systems that for any $D_{1} , D_{2}\in \mathbb{O}_{x}$ with $D_{1}\subset D_{2}$, and any $0< s\le t<{\infty }$,

$$ \bigl\{ \widetilde{X}^{D_{1}}_{s}=0\bigr\} \subset \bigl\{ \widetilde{X}^{D_{2}}_{t}=0 \bigr\} \quad \mathrm{P}_{\delta _{x}} \mbox{-a.s.} $$

Thus condition (A.1) is equivalent to that: There is $\delta >0$ sufficiently small such that

$$ \mathrm{P}_{\delta _{x}} \bigl( \widetilde{X}^{ B_{r} }_{t}=0 \bigr) >0 \quad \mbox{for all }t>0\mbox{ and }0< r\le \delta . $$

Here $B_{r}:=B(x,r)$ denotes the ball centered at $x$ with radius $r$.

In the remainder of this subsection, we assume $\psi$ is a spatially independent branching mechanism given by (2.20). Suppose $((Y_{t})_{t\ge 0};\mathbf{P}_{y},y\in \mathbb{R}^{+})$ is a one dimensional CB process with branching mechanism $\psi$ and $\mathbf{P}_{y}(Y_{0}=y)=1$. The process $Y$ is called subcritical, critical or supercritical according to $\psi '(0+)>$, = or $<0$. It is well-known that $\mathbf{P}_{y}( \lim_{t\to \infty }Y_{t}=0)=e^{-y z_{\psi }}$, where $z_{\psi }:= \sup \{{\lambda }\ge 0, \psi ({\lambda })\le 0\}\in [0,{\infty }]$. Moreover, by the Markov property of $Y$, we have

$$ \mathbf{P}_{y} \bigl( e^{-z_{\psi }Y_{t}} \bigr) =e^{-yz_{\psi }} \quad \mbox{for all }y\ge 0\mbox{ and }t\ge 0. $$

In the subcritical and critical cases, $z_{\psi }=0$, while in the supercritical case, $z_{\psi }\in (0,{\infty })$ if $\psi ({\infty })= {\infty }$, and otherwise $z_{\psi }={\infty }$. Given $z_{\psi } \in (0,{\infty })$, conditioned on the event $\{\lim_{t\to \infty }Y _{t}=0\}$, $Y$ is a CB process with branching mechanism

$$ \psi ^{*}({\lambda })=\psi ({\lambda }+z_{\psi })=\psi '(z_{\psi }) {\lambda }+a{\lambda }^{2}+ \int _{(0,{\infty })} \bigl( e^{-{\lambda } x}-1+{\lambda } x \bigr) e^{-xz_{\psi }}\eta (dx). $$

(A.2)

Since $(\psi ^{*})'(0+)=\psi '(z_{\psi })> 0$, the conditioned process is subcritical (cf. Lemma 2 in [2]).

Lemma A.1

Suppose$\psi$given by (2.20) satisfies$z_{\psi }<{\infty }$and that

$$ \int ^{\infty }_{N}\,ds \biggl[ \int ^{s}_{z_{\psi }}\psi (u)\,du \biggr] ^{-1/2}< \infty \quad \textit{for some }N>z_{\psi }. $$

(A.3)

Then for any bounded open set$Q=(t_{1}, t_{2})\times B$, where$0\le t_{1}< t_{2}<\infty$and$B:=B(x_{0}, r)$with$x_{0}\in E$and$r>0$, there is a function$v^{0}(s, x)\in C^{2}(Q)$such that

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \displaystyle \frac{\partial v^{0}}{\partial s}+Lv^{0}\le \psi (v^{0}), & (s,x) \in Q, \\ \displaystyle v^{0}(s, x)\ge z_{\psi },& (s,x)\in Q, \\ \displaystyle \lim_{ Q\ni (s,x)\to z}v^{0}(s, x)=\infty ,& z\in \mathcal{T}_{Q}, \end{array}\displaystyle \right . $$

(A.4)

where$\mathcal{T}_{Q}:=((t_{1},t_{2})\times \partial B)\cup (\{t_{2} \}\times \overline{B})$is a total set of$\partial Q$. By “total” we mean that whenever the$L$-diffusion started from inside$Q$first exits from$Q$, it will exit though the boundary points in$\mathcal{T}_{Q}$with probability one.

Proof

The idea of this proof is similar to that of [13, Theorem 5.3.1]. We only need to prove the result for sufficiently small $t_{2}-t_{1}+r$. Note that (A.3) is equivalent to

$$ \int ^{\infty }_{N}\,ds \biggl[ \int ^{s}_{0}\psi ^{*}(r)\,dr \biggr] ^{-1/2}< \infty \quad \mbox{for some }N>0, $$

where $\psi ^{*}$ is defined in (A.2). By [13, Lemma 5.3.1], for sufficiently small $t_{2}-t_{1}$ there is a nonnegative solution $u^{*}(s)$ of the following problem:

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \frac{\partial u^{*}}{\partial s}\le \psi ^{*}(u^{*}), \quad s\in (t _{1},t_{2}), \\ \displaystyle \lim_{(t_{1},t_{2})\ni s\to t_{2}} u^{*}(s)=\infty . \end{array}\displaystyle \right . $$

Take $u(s):=z_{\psi }+u^{*}(s)$ for $s\in (t_{1}, t_{2})$. Using the fact that $\psi ^{*}({\lambda })=\psi ({\lambda }+z_{\psi })$, it is easy to check that $u$ satisfies (A.5):

$$ \left \{ \textstyle\begin{array}{l} \displaystyle \frac{\partial u}{\partial s}\le \psi (u), \quad s\in (t_{1},t_{2}), \\ \displaystyle u(s)\ge z_{\psi },\quad s\in (t_{1},t_{2}). \\ \displaystyle \lim_{(t_{1},t_{2})\ni s\to t_{2}} u(s)=\infty , \end{array}\displaystyle \right . $$

(A.5)

On the other hand, by [13, Lemma 5.3.4] and the fact that $\psi ^{*}({\lambda })=\psi ({\lambda }+z_{\psi })$, for sufficiently small $r$, there is a nonnegative solution $v(x)$ of the following problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \displaystyle Lv\le \psi (v), & x\in B, \\ \displaystyle \lim_{B\ni x\to z} v(x)=\infty , & z\in \partial B. \end{array}\displaystyle \right . $$

Hence $v^{0}(s,x):=u(s)+v(x)$ is a solution of (A.4). □

Lemma A.2

(Comparison Principle)

Suppose$\psi$is given by (2.20) with$z_{\psi }<{\infty }$and that$Q$is a bounded regular open set in$[0,\infty )\times \mathbb{R}^{d}$. Then$u\le v$in$Q$assuming that

• (a) $u, v\in C^{2}(Q)$;

• (b) $\frac{\partial u}{\partial s}+Lu-\psi (u)\ge \frac{\partial v}{ \partial s}+Lv-\psi (v)$in$Q$;

• (c) $u$is bounded from above and$v\ge z_{\psi }$in$Q$.

• (d) for every$\tilde{z}\in \mathcal{T}_{Q}$, $\limsup_{Q\ni z\to \tilde{z}}[u(z)-v(z)]\le 0$.

Proof

The proof is similar to that of [13, Theorem 5.2.3] by using the fact that $\psi ({\lambda })$ is an increasing function in ${\lambda }\in [z_{\psi }, \infty )$. We omit the details here. □

Lemma A.3

If$x\in E$satisfies the following condition: there exist an open set$D\in \mathbb{O}_{x}$and a function$\psi$in the form of (2.20) with$z_{\psi }<{\infty }$such that (A.3) holds and

$$ \psi _{\beta }(x, {\lambda })\ge \psi ({\lambda })\quad \textit{for all }x \in D \textit{ and } {\lambda }\ge 0, $$

(A.6)

then$\mathrm{P}_{\delta _{x}} ( X_{t}=0 ) >0$for all$t>0$and hence$x\in E_{0}$.

Proof

By the argument in the beginning of this subsection, it suffices to prove that for $\delta >0$ small enough such that $B_{\delta }:=B(x, \delta )\subset D$, we have

$$ \mathrm{P}_{\delta _{x}} \bigl( \widetilde{X}^{ B_{r} }_{t}=0 \bigr) >0 \quad \mbox{for all }t>0\mbox{ and } B_{r} :=B(x,r)\mbox{ with }r \in (0,\delta ]. $$

(A.7)

Fix $r\in (0,\delta ]$ and an arbitrary $T>0$. For any $\lambda >0$, let

$$ \widetilde{u}^{B_{r}}_{\lambda }(s,y):=-\log \mathrm{P}_{\delta _{y}} \bigl( \exp \bigl( -\bigl\langle \lambda ,\widetilde{X}^{B_{r}}_{s} \bigr\rangle \bigr) \bigr) \quad \mbox{for }y\in B_{r}\mbox{ and }s \in [0,T]. $$

(A.8)

Since all the first partials of $\psi _{\beta }(x,\lambda )$ are continuous, it follows by [13, Theorem 5.2.2] that $v_{\lambda }(s,y):=\widetilde{u}^{B_{r}}_{\lambda }(T-s,y)$ for $(s,y)\in (0,T) \times B_{r}$ is a solution of the following boundary problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} \displaystyle \frac{\partial v_{\lambda }}{\partial s}+Lv_{\lambda }=\psi _{\beta }(y,v _{\lambda }(s,y)), & (s,y)\in (0,T)\times B_{r}, \\ \displaystyle \lim_{(0,T)\times B_{r}\ni (s,y)\to z}v_{\lambda }(s,y)=\lambda , & z\in \partial ((0,T)\times B_{r}). \end{array}\displaystyle \right . $$

Since $\psi _{\beta }(y,\lambda )\ge \psi (\lambda )$ for $y\in B_{r}$ and $\lambda \ge 0$, we have

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \displaystyle \frac{\partial v_{\lambda }}{\partial s}+Lv_{\lambda }\ge \psi (v_{ \lambda }), & (s,y)\in (0,T)\times B_{r}, \\ \displaystyle \lim_{(0,T)\times B_{r}\ni (s,y)\to z}v_{\lambda }(s,y)=\lambda , & z\in \partial ((0,T)\times B_{r}). \end{array}\displaystyle \right . $$

By Lemma A.1, there is a solution $v^{0}(s,y)$ of the following problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \displaystyle \frac{\partial v^{0}}{\partial s}+Lv\le \psi (v^{0}), & (s,y) \in (0,T)\times B_{r}, \\ \displaystyle v^{0}(s,x)\ge z_{\psi },& (s,y)\in (0,T)\times B_{r}, \\ \displaystyle \lim_{ (0,T)\times B_{r} \ni (s,y)\to z}v^{0}(s, y)={\infty },& z \in \mathcal{T}_{(0,T)\times B_{r}}, \end{array}\displaystyle \right . $$

where $\mathcal{T}_{(0,T)\times B_{r}}:=((0,T)\times \partial B_{r}) \cup (\{T\}\times \overline{B_{r}})$ is a total set of $\partial ((0,T) \times B_{r})$. By Lemma A.2, for any $\lambda >0$,

$$ v_{\lambda }(s, y)\le v^{0}(s,y)\quad \mbox{for all } (s,y)\in (0,T) \times B_{r}. $$

Letting $\lambda \uparrow \infty$ in (A.8), we get that for any $t\in (0,T)$

$$ -\log \mathrm{P}_{\delta _{x}}\bigl(\widetilde{X}^{B_{r}}_{t}=0 \bigr)= \lim_{\lambda \to \infty }v_{\lambda }(T-t, x)\le v^{0}(T-t, x)< {\infty }. $$

Since $T>0$ is arbitrary, we get (A.7) for all $t>0$ and $r\in (0,\delta ]$. □

Note that condition (A.3) is stronger than $\int ^{{\infty }} \psi ({\lambda })^{-1}d{\lambda }<{\infty }$, which is usually called Grey’s condition. Grey’s condition is a necessary and sufficient condition for a CB process to become extinct in a finite time with positive probability. Lemma A.3 says that for a superdiffusion with spatially dependent branching mechanism, if the branching mechanism is locally dominated by a spatially independent branching mechanism which satisfies a condition stronger than the Grey’s condition, then this superdiffusion become extinct in a finite time with positive probability.

Recall that $\alpha (x)$ is the non-negative bounded Borel measurable function in (2.1) and $E_{+}=\{x\in E:\ \alpha (x)>0\}$.

Proposition A.1

9.1-A.1 For a$(\xi ,\psi _{\beta })$-superdiffusion, it holds that$E_{+} \subset E_{0}$.

Proof

Since all the first partials of $\psi _{\beta }$ are continuous, $\alpha$ and $\beta$ in $\psi _{\beta }$ are continuous functions. Thus for any $x\in E_{+}$, there exists a neighborhood $D$ of $x$ such that $\sup_{y\in D}\beta (y)\le \beta ^{*}<{\infty }$ and $\inf_{y\in D} \alpha (y)\ge \alpha ^{*}>0$. We have

$$ \psi _{\beta }(y,\lambda )\ge -\beta ^{*}\lambda + \alpha ^{*}\lambda ^{2}=: \psi ^{*}(\lambda )\quad \mbox{for }y\in D\mbox{ and }\lambda \ge 0. $$

It is easy to verify that $\psi ^{*}$ satisfies (A.3). Hence $x\in E_{0}$ by Lemma A.3. □

In the rest of this subsection, we assume Condition 1′ holds. We will establish the existence of $\mathbb{N}^{*}_{x}$ for all $x\in E_{+}$ under Condition 1′.

First we note that there exists a superdiffusion $((X_{t})_{t\ge 0}; \mathrm{P}^{*}_{\mu },\mu \in \mathcal{M}_{F}(E))$ with $\mathrm{P} ^{*}_{\mu }(X_{0}=\mu )=1$ such that for all $\mu \in \mathcal{M}_{F}(E)$, $f\in \mathcal{B}^{+}_{b}(E)$ and $t\ge 0$,

$$ \mathrm{P}^{*}_{\mu } \bigl[ e^{-\langle f,X_{t}\rangle } \bigr] =e ^{-\langle u^{*}_{f}(t,\cdot ),\mu \rangle }, $$

where $u^{*}_{f}(t,x)$ is the unique nonnegative locally bounded solution to the integral equation given by (2.17). The branching mechanism of this superdiffusion is $\psi ^{*}_{\beta ^{*}}(x,\lambda ):=- \beta ^{*}(x)\lambda +\psi ^{*}_{0}(x,\lambda )$ where $\beta ^{*}$ and $\psi ^{*}_{0}$ are given in Proposition 2.2. Since $w$ is only locally bounded on $E$, $\beta ^{*}$ is bounded from above but may not be bounded from below. Hence the branching mechanism $\psi ^{*}_{\beta ^{*}}(x,\lambda )$ does not satisfy the usual assumptions in Sect. 2.1. Nevertheless, one may consider the $(\xi ^{D},\psi ^{*} _{\beta ^{*}})$-superdiffusion in every bounded open domain $D$, where the underlying spatial motion is killed upon hitting the boundary of $D$. Then by using an increasing sequence of compactly embedded domains to approximate $E$, the process $(X,\mathrm{P}^{*}_{\mu })$ can be defined as a distributional limit of these $(\xi ^{D},\psi ^{*}_{\beta ^{*}})$-superdiffusions. (See, for example, the argument before Theorem 2.4 in [15].) We remark that this method of construction may fail in general when the spatial motion has discontinuous sample paths since the process can potentially jump everywhere when it exits from the domain $D$.

Let $E^{*}_{0}$ be the set of points in $E$ where the Kuznetsov measure $\mathbb{N}^{*}_{x}$ corresponding to the $(\xi ,\psi ^{*}_{\beta ^{*}})$-superdiffusion exists. It follows that for every $x\in E^{*}_{0}$, every $t>0$ and $f\in \mathcal{B}^{+}_{b}(E)$,

$$ \mathbb{N}^{*}_{x} \bigl( 1-e^{-\langle f,X_{t}\rangle } \bigr) =u ^{*}_{f}(t,x). $$

According to [11, Theorem I.1.1 and Theorem I.1.2], for any bounded open set $D\Subset E$ and $t\ge 0$, there exists a random measure $\widetilde{X}^{*D}_{t}$ concentrated on the boundary of $[0,t)\times D$, such that for every $\mu \in \mathcal{M}_{F}(E)$ and $\widetilde{f}\in \mathcal{B}^{+}_{b}([0,t]\times E)$,

$$ \mathrm{P}^{*}_{\mu } \bigl[ e^{-\langle \widetilde{f},\widetilde{X} ^{*D}_{t}\rangle } \bigr] =e^{-\langle \widetilde{u}^{*D}_{ \widetilde{f}}(t, \cdot ),\mu \rangle }, $$

where $\widetilde{u}^{*D}_{\widetilde{f}}(t,x)$ is the unique nonnegative locally bounded solution to the following integral equation:

$$ \widetilde{u}^{*D}_{\widetilde{f}}(t, x)=\varPi _{x} \bigl[ \widetilde{f}(t \wedge \tau _{D}, \xi _{t\wedge \tau _{D}}) \bigr] - \varPi _{x} \biggl[ \int _{0}^{t\wedge \tau _{D}}\psi ^{*}_{\beta ^{*}} \bigl( \xi _{s}, \widetilde{u}^{*D}_{\widetilde{f}}(t-s, \xi _{s}) \bigr) \,ds \biggr] . $$

(A.9)

Define $\widetilde{w}(t,x):=w(x)$ for $(t,x)\in [0,{\infty })\times E$. Using the local boundedness of $w$ on $E$, we can deduce that for any bounded open set $D\Subset E$,

$$ \widetilde{w}(t,x)=\varPi _{x} \bigl( \widetilde{w}(t\wedge \tau _{D}, \xi _{t\wedge \tau _{D}}) \bigr) -\varPi _{x} \biggl( \int _{0}^{t\wedge \tau _{D}}\psi _{\beta }\bigl(\xi _{s},\widetilde{w}(t-s,\xi _{s})\bigr)\,ds \biggr) . $$

(A.10)

The next proposition shows that $\{\widetilde{X}^{*D}_{t};t\ge 0\}$ is a Doob’s $h$-transformed process of $\{\widetilde{X}^{D}_{t};t\ge 0\}$ via the function $\widetilde{w}$.

Proposition A.2

Let$\mathcal{M}^{w}_{F}(E):=\{\mu \in \mathcal{M}_{F}(E):\langle w, \mu \rangle <{\infty }\}$. Suppose$D\Subset E$is a bounded open set and$\mu \in \mathcal{M}^{w}_{F}(E)$. Then for every$t\ge 0$and$\widetilde{f}\in \mathcal{B}^{+}_{b}([0,t]\times E)$,

$$ e^{-\langle w,\mu \rangle } \mathrm{P}_{\mu } \bigl[ e^{-\langle \widetilde{f}+\widetilde{w},\widetilde{X}^{D}_{t}\rangle } \bigr] = \mathrm{P}^{*}_{\mu } \bigl[ e^{-\langle \widetilde{f},\widetilde{X} ^{*D}_{t}\rangle } \bigr] . $$

(A.11)

Proof

Since the random measure $\widetilde{X}^{D}_{t}$ is supported on the boundary of $[0,t)\times D$ and $\widetilde{w}$ is locally bounded on $[0,{\infty })\times E$, it follows by (2.7) and (2.8) that

$$ e^{-\langle w,\mu \rangle } \mathrm{P}_{\mu } \bigl[ \exp \bigl( - \bigl\langle \widetilde{f}+\widetilde{w},\widetilde{X}^{D}_{t}\bigr\rangle \bigr) \bigr] =\exp \bigl( -\bigl\langle \widetilde{u}^{D}_{\widetilde{w}+ \widetilde{f}}(t, \cdot )-w(\cdot ),\mu \bigr\rangle \bigr) , $$

where $\widetilde{u}^{D}_{\widetilde{w}+\widetilde{f}}(t,x)$ is the unique nonnegative locally bounded solution to the following integral equation

$$ \widetilde{u}^{D}_{\widetilde{w}+\widetilde{f}}(t,x)=\varPi _{x} \bigl[ ( \widetilde{w}+\widetilde{f}) (t\wedge \tau _{D},\xi _{t\wedge \tau _{D}}) \bigr] -\varPi _{x} \biggl( \int _{0}^{t\wedge \tau _{D}}\psi _{\beta }\bigl(\xi _{s}, \widetilde{u}^{D}_{\widetilde{w}+\widetilde{f}}(t-s,\xi _{s})\bigr)\,ds \biggr) $$

(A.12)

Using (A.12) and (A.10), it is straightforward to check that $\widetilde{u}^{*D}_{\widetilde{f}}(t,x):=\widetilde{u}^{D}_{ \widetilde{f}+\widetilde{w}}(t,x)-w(x)$ is the unique nonnegative locally bounded solution to (A.9). Thus we get (A.11). □

Lemma A.4

If$x\in E$satisfies the condition of LemmaA.3, then$x\in E^{*}_{0}$.

Proof

Based on the argument in the beginning of this subsection, it suffices to prove that there exists $\delta >0$ such that

$$ \mathrm{P}^{*}_{\delta _{x}}\bigl(\widetilde{X}^{*B}_{t}=0 \bigr)>0 \quad \mbox{for all } t>0\mbox{ and }B=B(x,r)\mbox{ with }r\le \delta . $$

We choose $\delta$ sufficiently small such that $B\subset D$, and $\psi _{\beta }(y, \lambda )\ge \psi (\lambda )$ for all $y\in B$ and $\lambda \ge 0$, where $D$ and $\psi$ satisfy the conditions of Lemma A.3. It follows from Proposition A.2 that

$$\begin{aligned} \mathrm{P}^{*}_{\delta _{x}}(\widetilde{X}^{*B}_{t}=0)&= \lim_{\lambda \to {\infty }}\mathrm{P}_{\delta _{x}}(e^{-\langle \lambda , \widetilde{X}^{*B}_{t}\rangle }) \\ &= e^{-w(x)}\lim_{\lambda \to {\infty }}\mathrm{P}_{\delta _{x}}(e^{- \langle \widetilde{w}+\lambda , \widetilde{X}^{B}_{t}\rangle }) \\ &= e^{-w(x)}\lim_{\lambda \to \infty }e^{-\widetilde{u}^{B}_{ \widetilde{w}+\lambda }(t,x)}, \end{aligned}$$

where $\widetilde{u}^{B}_{\widetilde{w}+\lambda }(t,x)$ is the unique nonnegative solution of (2.8) with initial condition $\widetilde{w}+\lambda$. Then $v(s, y):=\widetilde{u}^{B}_{ \widetilde{w}+\lambda }(t-s, y)$ for $(s,y)\in [0,t]\times B$, is a solution to the following boundary problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \displaystyle \frac{\partial v}{\partial s}+Lv=\psi _{\beta }(y,v(s,y)), & (s,y) \in (0,t)\times B \\ \displaystyle \lim_{(0,t)\times B\ni (s,y)\to z}v(t, y)=\widetilde{w}(z)+\lambda , & z\in \partial ((0,t)\times B). \end{array}\displaystyle \right . $$

Since $\psi _{\beta }(y,\lambda )\ge \psi (\lambda )$ for $y\in B$ and $\lambda \ge 0$, applying similar arguments as in the proof of Lemma A.3, we get $\lim_{\lambda \to {\infty }}\widetilde{u}^{B} _{\widetilde{w}+\lambda }(t,x)<{\infty }$. Therefore $\mathrm{P}^{*} _{\delta _{x}}(\widetilde{X}^{*B}_{t}=0)>0$ for $t>0$. Hence $x\in E^{*}_{0}$. □

Proposition A.3

For a$(\xi ,\psi _{\beta })$-superdiffusion that satisfies Condition 1′, it holds that$E_{+}\subset E^{*}_{0}$.

Proof

This proposition follows from Lemma A.4 (instead of Lemma A.3) in the same way as the proof of Proposition A.1. We omit the details here. □

1.2 A.2 Verifications of Examples

Example 1

It is easy to check that Assumptions 1–4 are satisfied with $\lambda _{1}=-\beta$, $h(x)\equiv 1$ and $w(x)\equiv z_{\psi }$. Then the $h$-transformed semigroup $P^{h}_{t}$ is given by

$$ P^{h}_{t}f(x):=\frac{e^{-\beta t}}{h(x)}\,\varPi _{x} \bigl[ e^{\beta t}h( \xi _{t})f(\xi _{t}) \bigr] =P_{t}f(x)\quad \mbox{for }f\in \mathcal{B}^{+}_{b} \bigl(\mathbb{R}^{d}\bigr). $$

This implies that the $h$-transformed process $\xi ^{h}$ is still an inward OU process with generator ℒ. It is known that the transition density of $\xi ^{h}$ with respect to $m$ is given by

$$ \begin{aligned} &p^{h}(t,x,y)= \bigl( 1-e^{-2ct} \bigr) ^{-d/2}\exp \biggl( -\frac{c}{e ^{2ct}-1} \bigl( |x|^{2}+|y|^{2}-2e^{ct}x \cdot y \bigr) \biggr)\quad \mbox{for }t>0 \\ &\quad \mbox{and }x,y\in \mathbb{R}^{d}. \end{aligned} $$

In view of this, one can easily check that $\xi ^{h}$ has the Feller property, that is, $P^{h}_{t}$ maps $C_{0}(\mathbb{R}^{d})$ to $C_{0}(\mathbb{R}^{d})$ and $\lim_{t\to 0}\|P^{h}_{t}f-f\|_{\infty }=0$ for all $f\in C_{0}(\mathbb{R}^{d})$. It remains to show that Assumption 5 holds for this example. Let $Z$ be the skeleton process. We need to show that for any $\mu \in \mathcal{M}_{c}(E)$, $\sigma >0$ and $f\in \mathcal{B}^{+}(E)$ with $\frac{fw}{h}$ bounded,

$$ \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma }\langle f,Z_{n\sigma } \rangle =\langle f,wh\rangle W^{h/w}_{\infty }(Z)\quad \mathbb{P}_{ \mu } \mbox{-a.s.} $$

(A.13)

For $t\ge 0$, $Z_{t}=\sum_{i=1}^{N_{0}}Z^{i,0}_{t}$, where $N_{0}= \langle 1,Z_{0}\rangle$ and $Z^{i,0}$ denotes the independent subtree of the skeleton initiated by the $i$th particle at time 0. Recall that under $\mathbb{P}_{\mu }$ for $\mu \in \mathcal{M}_{c}(E)$, $N_{0}$ is a Poisson random variable with mean $\langle w,\mu \rangle$. Moreover, given $Z_{0}$, $Z^{i,0}$ follows the same distribution as $(Z;\mathbb{P}_{\cdot ,\delta _{z_{i}(0)}})$. Thus we have

$$\begin{aligned} &\mathbb{P}_{\mu } \Bigl( \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma } \langle f,Z_{n\sigma }\rangle =\langle f,wh\rangle W^{h/w}_{\infty }(X) \Bigr) \\ &\quad \ge \mathbb{P}_{\mu } \Biggl( \bigcap _{i=1}^{N_{0}}\Bigl\{ \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma } \bigl\langle f,Z^{i,0}_{n \sigma }\bigr\rangle =\langle f,wh\rangle W^{h/w}_{\infty }\bigl(Z^{i,0}\bigr)\Bigr\} \Biggr) \\ &\quad =\mathbb{P}_{\mu } \Biggl[ \prod_{i=1}^{N_{0}} \mathbb{P}_{\cdot , \delta _{z_{i}(0)}} \Bigl( \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma } \langle f,Z_{n\sigma }\rangle =\langle f,wh\rangle W^{h/w}_{\infty }(Z) \Bigr) \Biggr] . \end{aligned}$$

Hence to prove (A.13), it suffices to prove that for every $x\in E$,

$$ \mathbb{P}_{\cdot ,\delta _{x}} \Bigl(\, \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma } \langle f,Z_{n\sigma }\rangle =\langle f,wh\rangle W^{h/w} _{\infty }(Z) \Bigr) =1. $$

(A.14)

Sufficient conditions for (A.14) are given in [15] (see conditions (2.23)–(2.26), (2.34) and (2.35) there). In this example, conditions (2.23)–(2.26) in [15] hold for $\sigma _{1}=\sigma _{3}=p$, $\sigma _{2}=2$ and $\varphi _{1}(x)=\varphi _{2}(x)\equiv 1$, condition (2.34) holds for $a(t)=\sqrt{(\frac{-\lambda _{1}}{c}+ \sigma )t}$ and condition (2.35) in [15] holds for $K=1$. Hence we prove (A.13). □

Example 2

Assumptions 1–4 are satisfied with $\lambda _{1}=cd-\beta$, $h(x)= ( \frac{c}{\pi } ) ^{d/2}e ^{-c|x|^{2}}$ and $w(x)\equiv z_{\psi }$. The transition density of $\xi$ with respect to $m$ is given by

$$ \begin{aligned} &p(t,x,y)= \bigl( e^{2ct}-1 \bigr) ^{-d/2} \exp \biggl( -\frac{c}{1-e ^{-2ct}} \bigl( |x|^{2}+|y|^{2}-2e^{-ct}x \cdot y \bigr) \biggr) \\ &\quad \mbox{for }t>0\mbox{ and }x,y\in \mathbb{R}^{d}. \end{aligned} $$

The $h$-transformed process $\xi ^{h}$ is symmetric with respect to the measure $\widetilde{m}(dx)=h^{2}m(dx)= ( \frac{c}{\pi } ) ^{d/2}e^{-c|x|^{2}}dx$, and the transition density of $\xi ^{h}$ with respect to $\widetilde{m}$ is given by

$$ \begin{aligned} &p^{h}(t,x,y)= \bigl( 1-e^{-2ct} \bigr) ^{-d/2}\exp \biggl( -\frac{c}{e ^{2ct}-1} \bigl( |x|^{2}+|y|^{2}-2e^{ct}x \cdot y \bigr) \biggr) \\ &\quad \mbox{for }t>0\mbox{ and }x,y\in \mathbb{R}^{d}. \end{aligned} $$

This implies that $\xi ^{h}$ is an inward OU process on $\mathbb{R} ^{d}$ with infinitesimal generator $\mathcal{L}^{h}:=\frac{1}{2} \Delta -cx\cdot \nabla$. Thus $\lim_{t\to 0}\|P^{h}_{t}f-f\|_{\infty }=0$ for any $f\in C_{0}(\mathbb{R}^{d})$. In view of Example 4.2 in [15], Assumption 5 is also satisfied. □

Example 3

Suppose $\widehat{\psi }_{\beta }(x, \lambda ) := -\beta (x)\lambda +\alpha (x)\lambda ^{2}$. Let $\mathbb{X}$ be a $(Y,\psi _{\beta })$-superdiffusion and $\widehat{\mathbb{X}}$ be a $(Y,\widehat{\psi }_{\beta })$-superdiffusion. For $x\in \mathbb{R} ^{d}$, let $w_{\mathbb{X}}(x):=-\log \mathrm{P}_{\delta _{x}} ( \exists t\ge 0: \langle 1,\mathbb{X}_{t}\rangle =0 )$ and $w_{\widehat{\mathbb{X}}}(x):=-\log \mathrm{P}_{\delta _{x}} ( \exists t\ge 0: \langle 1,\widehat{\mathbb{X}}_{t}\rangle =0 )$. We know from [17] that the function $w_{\widehat{\mathbb{X}}}$ is continuous on $\mathbb{R}^{d}$ and solves the equation $\mathcal{L}u-\widehat{\psi }_{\beta }(x,u)=0$ on $\mathbb{R}^{d}$. Since $\psi _{\beta }\ge \widehat{\psi }_{\beta }$ pointwise, it follows by [15, Lemma 4.5] that $w_{\mathbb{X}}\le w_{\widehat{\mathbb{X}}}$ pointwise. Thus $w_{\mathbb{X}}$ is locally bounded on $\mathbb{R} ^{d}$ and hence is continuous on $\mathbb{R}^{d}$ by [15, Lemma 2.1]. Recall that $X$ is a $(\xi ,\psi _{\beta })$-superdiffusion where $\xi =Y^{E}$ is the subprocess of $Y$ killed upon leaving $E$. We mentioned in Sect. 2.2 that one may think of $X_{t}$ describing the mass in $\mathbb{X}_{t}$ which historically avoids exiting $E$.

Since the function $\rho$ in the generator is bounded between two positive constants, we know from [22] that $\xi$ has a positive continuous transition density $p_{E}(t,x,y)$ with respect to $m$, and for each $T>0$, there exist positive $c_{i}$, $i=1,\ldots ,4$ such that for every $(t,x,y)\in (0,T]\times E\times E$,

$$\begin{aligned} c_{1}f_{E}(t,x,y) t^{-d/2}\exp \biggl( - \frac{c_{2}|x-y|^{2}}{t} \biggr) &\le p_{E}(t,x,y) \\ &\le c_{3}f_{E}(t,x,y) t^{-d/2}\exp \biggl( -\frac{c _{4}|x-y|^{2}}{t} \biggr) , \end{aligned}$$

(A.15)

where $f_{E}(t,x,y):= ( 1\wedge \frac{\delta _{E}(x)}{\sqrt{t}} ) ( 1\wedge \frac{ \delta _{E}(y)}{\sqrt{t}} )$ and $\delta _{E}(x)$ denotes the Euclidean distance between $x$ and the boundary of $E$. Since $\beta$ is bounded, it follows that the Feynman-Kac semigroup $P^{\beta }_{t}$ admits an integral kernel $p_{E}^{\beta }(t,x,y)$ which is positive, symmetric and continuous in $(x,y)$ for each $t>0$, and

$$ e^{-\|\beta \|_{\infty }t}p_{E}(t,x,y)\le p_{E}^{\beta }(t,x,y) \le e ^{\|\beta \|_{\infty }t}p_{E}(t,x,y)\quad \mbox{for all } x,y\in E \mbox{ and }t>0. $$

Thus $p^{\beta }_{E}(t,x,y)$ satisfies the same two-sided estimates (A.15) with possibly different constants $c_{i}>0$, $i=4,\ldots ,8$. By this estimate, $\int _{E\times E}p_{E}^{\beta }(t,x,y)^{2}m(dx)m(dy)= \int _{E}p_{E}^{\beta }(2t,x,x)m(dx)<{\infty }$ for every $t\in (0,T]$. Thus $P^{\beta }_{t}$ is a Hilbert-Schimidt operator in $L^{2}(E,m)$ and hence is compact. The infinitesimal generator of $P^{\beta }_{t}$ is $\mathcal{L}^{(\beta )}:= ( \mathcal{L}+\beta ) |_{E}$ with zero Dirichlet boundary condition. We use $\sigma (\mathcal{L}^{ \beta })$ to denote the spectrum set of $\mathcal{L}^{\beta }$. It then follows from Jentzch’s theorem that $\lambda _{1}:=\inf \{-\lambda :\ \lambda \in \sigma (\mathcal{L}^{(\beta )})\}$ is a simple eigenvalue and a corresponding eigenfunction $h$ can be chosen to be nonnegative with $\int _{E}h(x)^{2}m(dx)=1$. We assume $\lambda _{1}<0$. Since $h(x)=e^{\lambda _{1}t}\int _{E}p_{E}^{\beta }(t,x,y)h(y)m(dy)$, $h$ is continuous and positive on $E$. Moreover, by the estimate (A.15), we have

$$ c_{9} \bigl( 1\wedge \delta _{E}(x) \bigr) \le h(x)\le c_{10} \bigl( 1 \wedge \delta _{E}(x) \bigr) \quad \mbox{for all } x\in E. $$

(A.16)

Here $c_{9},\ c_{10}$ are positive constants independent of $x$. Therefore, Assumptions 2–3 hold.

Let $w(x):=-\log \mathrm{P}_{\delta _{x}} ( \exists t\ge 0:\ \langle 1,X_{t}\rangle =0 )$. Since $h$ is bounded on $E$, it is easy to see that condition (2.27) (or, equivalently, Assumption 4(i)) holds. Thus by Theorem 2.1 and Remark 2.5, $\mathrm{P}_{\delta _{x}} ( W^{h}_{\infty }(X)>0 ) >0$ for every $x\in E$. It follows that

$$ \begin{aligned} w(x)&\ge -\log \Bigl( 1-\mathrm{P}_{\delta _{x}} \Bigl( \limsup_{t\to {\infty }}\langle 1,X_{t}\rangle >0 \Bigr) \Bigr) \\ &\ge -\log \bigl( 1-\mathrm{P}_{\delta _{x}} \bigl( W^{h}_{\infty }(X) >0 \bigr) \bigr) >0\quad \mbox{for }x\in E. \end{aligned} $$

On the other hand, since $X_{t}$ describes only part of the mass in $\mathbb{X}_{t}$, we have

$$ \mathrm{P}_{\delta _{x}} \bigl( \exists t\ge 0:\ \langle 1,X_{t} \rangle =0 \bigr) \ge \mathrm{P}_{\delta _{x}} \bigl( \exists t\ge 0:\ \langle 1,\mathbb{X}_{t}\rangle =0 \bigr) $$

and hence $w(x)\le w_{\mathbb{X}}(x)$ on $E$. This together with the continuity of $w_{\mathbb{X}}$ on $\mathbb{R}^{d}$ implies that $w$ is bounded on $E$ and hence is continuous on $E$ by [15, Lemma 2.1]. This shows that Condition 1 holds, and, consequently, Assumption 1 is satisfied.

We now show that Assumption 4(ii) holds for this example. By condition (8.1) and the boundedness of $w$, one can easily prove that

$$ \frac{\psi _{\beta }(x,w(x))}{w(x)}=-\beta (x)+\alpha (x)w(x)+ \int _{(0,{\infty })}\frac{e^{-w(x)y}-1+w(x)y}{w(x)}\,\pi (x,dy) $$

is a bounded function on $E$. Let $\{D_{n}:n\ge 1\}$ be a sequence of bounded domains with smooth boundaries such that $D_{n}\Subset D_{n+1} \Subset E$ for $n\ge 1$ and $\bigcup_{n=1}^{{\infty }}D_{n}=E$. We know from the argument in the beginning of Sect. 5 that for $t>0$, $x\in E$ and $n$ sufficiently large so that $x\in D_{n}$,

$$\begin{aligned} w(x) =&\varPi _{x} \biggl[ w(\xi _{t\wedge \tau _{D_{n}}})\exp \biggl( - \int _{0}^{t\wedge \tau _{D_{n}}}\frac{\psi _{\beta }(\xi _{s},w(\xi _{s}))}{w( \xi _{s})}\,ds \biggr) \biggr] . \end{aligned}$$

Consequently,

$$\begin{aligned} w(x) \ge &\exp \biggl( -t\sup_{y\in E} \frac{\psi _{\beta }(y,w(y))}{w(y)} \biggr) \varPi _{x} \bigl( w( \xi _{t\wedge \tau _{D_{n}}}) \bigr) \\ =&e^{-c_{13}t} \bigl( \varPi _{x} \bigl[ w(\xi _{t});t< \tau _{D_{n}} \bigr] +\varPi _{x} \bigl[ w(\xi _{\tau _{D_{n}}});t\ge \tau _{D_{n}} \bigr] \bigr) \\ \ge &e^{-c_{13}t}\varPi _{x} \bigl[ w(\xi _{t});t< \tau _{D_{n}} \bigr] =e ^{-c_{13}t}\varPi _{x} \bigl[ w(Y_{t});t< \tau _{D_{n}} \bigr] . \end{aligned}$$

Thus by letting $n\to {\infty }$, we get $w(x)\ge e^{-c_{13}t}\varPi _{x} [ w(Y_{t});t<\tau _{E} ] =e^{-c_{13}t}\varPi _{x} [ w(Y ^{E}_{t}) ]$. Using this and the heat kernel estimates in (A.15), we have for $x\in E$

$$\begin{aligned} w(x) \ge &e^{-c_{13}}\varPi _{x} \bigl[ w \bigl(Y^{E}_{1}\bigr) \bigr] \\ =&e^{-c_{13}} \int _{E}p_{E}(1,x,y)w(y)m(dy) \\ \ge &c_{1}e^{-c_{13}} \int _{E}\bigl(1\wedge \delta _{E}(x)\bigr) \bigl(1\wedge \delta _{E}(y)\bigr)e^{-c_{2}|x-y|^{2}}w(y)m(dy) \\ \ge &c_{1}e^{-c_{13}-c_{2}\operatorname{diam}(E)^{2}}\bigl(1\wedge \delta _{E}(x) \bigr) \int _{E}\bigl(1\wedge \delta _{E}(y) \bigr)w(y)m(dy) \\ \ge &c_{14}\bigl(1\wedge \delta _{E}(x)\bigr). \end{aligned}$$

(A.17)

Here $c_{14}>0$ is constant and the last inequality comes from the fact that $(1\wedge \delta _{E}(y))w(y)$ is positive everywhere on $E$. By (A.17) and (A.16), we have for $x\in E$,

$$\begin{aligned} & \int _{(0,{\infty })}r^{2}e^{-w(x)r}\pi (x,dr) \\ &\quad \le \int _{(0,{\infty })}r ^{2}e^{-c_{14}(1\wedge \delta _{E}(x))r}\pi (x,dr) \\ &\quad \le \int _{(0,{\infty })}r^{2}e^{-c_{15}h(x)r}\pi (x,dr) \\ &\quad =\frac{1}{h(x)} \int _{(0,{\infty })}r\log ^{*}r \biggl( \frac{rh(x)}{ \log ^{*}(rh(x))}e^{-c_{15}h(x)r} \biggr) \frac{\log ^{*}(rh(x))}{\log ^{*}r}\,\pi (x,dr) \\ &\quad \le \frac{1}{h(x)} \int _{(0,{\infty })}r\log ^{*}r \biggl( \frac{rh(x)}{ \log ^{*}(rh(x))}e^{-c_{15}h(x)r} \biggr) \frac{\log ^{*}(r\|h\|_{ \infty })}{\log ^{*}r}\,\pi (x,dr), \end{aligned}$$

where $c_{15}$ is a positive constant. It then follows from condition (8.1), and the fact that functions $y\mapsto \frac{y}{\log ^{*}y}e^{-c_{15}y}$ and $y\mapsto \frac{\log ^{*}(y\|h\| _{\infty })}{\log ^{*}y}$ are bounded on $(0,{\infty })$ that $\int _{(0,{\infty })}r^{2}e^{-w(x)r}\pi (x, dr)\le c_{16}h(x)^{-1}$ on $E$. Immediately we get $\langle ( \int _{(0,{\infty })}r^{2}e ^{-w(\cdot )r}\pi (\cdot ,dr) ) ^{2},1\wedge h^{4}\rangle \le c_{16}^{2}\int _{E}(h(x)^{-2}\wedge h(x)^{2})m(dx)<{\infty }$. Hence Assumption 4(ii) holds.

We next show Assumption 6 holds. Let $P^{h}_{t}$ be the $h$-transformed semigroup from $P^{\beta }_{t}$ given by (2.25). $P^{h}_{t}$ admits an integral kernel $p_{E}^{h}(t,x,y)$ with respect to the measure $\widetilde{m}(dx):=h(x)^{2}m(dx)$, which is related to $p_{E}^{ \beta }(t,x,y)$ by

$$ p_{E}^{h}(t,x,y)=e^{\lambda _{1}t} \frac{p_{E}^{\beta }(t,x,y)}{h(x)h(y)}\quad \mbox{for }x,y\in E \mbox{ and }t\ge 0. $$

Define $\widetilde{a}_{t}(x):=p_{E}^{h}(t,x,x)$ for $x\in E$ and $t>0$. Clearly by (A.15) and (A.16), we have $\sup_{x\in E} \widetilde{a}_{t}(x)<{\infty }$ for every $t>0$. Let $e_{\beta }(t):= \int _{0}^{t}\beta (\xi _{s})\,ds$. Suppose $f\in C_{0}(E)$. Then for any given ${\varepsilon }>0$, there is $\delta >0$ so that $|f(x)-f(y)|< {\varepsilon }$ whenever $|x-y|<\delta$. By the two-sided estimate of $p_{E}^{\beta }(t,x,y)$, for sufficiently small $t\in (0,1]$,

$$\begin{aligned} \sup_{x\in E} \bigl\vert P^{h}_{t}f(x)-f(x) \bigr\vert =&\sup_{x\in E}\frac{ \vert e^{\lambda _{1}t}P^{\beta }_{t}(hf)(x)-h(x)f(x) \vert }{h(x)} \\ =&\sup_{x\in E}e^{\lambda _{1}t}\frac{ \vert \varPi _{x} [ e_{ \beta }(t)h(\xi _{t})(f(\xi _{t})-f(\xi _{0})) ] \vert }{h(x)} \\ \le &{\varepsilon }+\sup_{x\in E}e^{\lambda _{1}t} \frac{ \vert \varPi _{x} [ e_{\beta }(t)h(\xi _{t})(f(\xi _{t})-f(\xi _{0}));|\xi _{t}- \xi _{0}|\ge \delta ] \vert }{h(x)} \\ \le &{\varepsilon }+e^{(\lambda _{1}+\|\beta \|_{\infty })t}\|h\|_{ \infty }\|f\|_{\infty } \sup_{x\in E}\frac{\varPi _{x} ( |\xi _{t}-\xi _{0}|\ge \delta ) }{h(x)} \\ \le &{\varepsilon }+c_{11}\sup_{x\in E} \frac{\delta _{E}(x)}{1\wedge \delta _{E}(x)} \int _{y\in E:\ |y-x|\ge \delta }t^{-(d+1)/2}\exp \bigl( -c_{4}|x-y|^{2}/t \bigr) dy \\ \le &{\varepsilon }+c_{12}\bigl(1+\operatorname{diam}(E)\bigr) \int _{|z|\ge \delta }t ^{-(d+1)/2}e^{-c_{4}|z|^{2}/t}\,dz \\ =&{\varepsilon }+c_{12}\bigl(1+\operatorname{diam}(E)\bigr)t^{-1/2} \int _{\delta t ^{-1/2}}^{{\infty }}r^{d-1}e^{-c_{4}r^{2}}\,dr. \end{aligned}$$

It follows that $\lim_{t\to 0}\|P^{h}_{t}f-f\|_{\infty }=0$ for all $f\in C_{0}(E)$.

It remains to prove that Assumption 5 holds. Fix $\phi \in \mathcal{B}^{+}_{b}(E)$, $\mu \in \mathcal{M}_{c}(E)$ and $\sigma >0$. For $m,n\in \mathbb{N}$, we have

$$ e^{\lambda _{1}(m+n)\sigma }\biggl\langle \frac{h}{w}\phi ,Z_{(m+n)\sigma } \biggr\rangle -\bigl\langle \phi ,h^{2}\bigr\rangle W^{h/w}_{\infty }(Z)=I(m,n)+ \mathit{II}(m,n)+\mathit{III}(n), $$

(A.18)

where $I(m,n):=e^{\lambda _{1}(m+n)\sigma }\langle \frac{h}{w}\phi ,Z _{(m+n)\sigma }\rangle -\mathbb{P}_{\mu } ( e^{\lambda _{1}(m+n)\sigma }\langle \frac{h}{w}\phi ,Z_{(m+n)\sigma }\rangle \vert \mathcal{F}_{n\sigma } )$,

$$ \mathit{II}(m,n):=\mathbb{P}_{\mu } \biggl( e^{\lambda _{1}(m+n)\sigma} \biggl\langle \frac{h}{w}\phi ,Z_{(m+n)\sigma }\biggr\rangle \vert \mathcal{F}_{n\sigma } \biggr) -\bigl\langle \phi ,h^{2}\bigr\rangle W^{h/w} _{n\sigma }(Z), $$

and $\mathit{III}(n):=\langle \phi ,h^{2}\rangle ( W^{h/w}_{n\sigma }(Z)-W ^{h/w}_{\infty }(Z) )$. Note that by the Markov property of $Z$,

$$\begin{aligned} \mathbb{P}_{\mu } \biggl( e^{\lambda _{1}n\sigma }\biggl\langle \frac{h}{w}\phi ,Z_{(m+n)\sigma }\biggr\rangle \vert \mathcal{F}_{n \sigma } \biggr) =&e^{\lambda _{1}(m+n)\sigma }\mathbb{P}_{\cdot ,Z _{n\sigma }} \biggl( \biggl\langle \frac{h}{w}\phi ,Z_{m\sigma }\biggr\rangle \biggr) \\ =&e^{\lambda _{1}(m+n)\sigma }\biggl\langle \frac{1}{w}P^{\beta }_{m\sigma }(h \phi ),Z_{n\sigma }\biggr\rangle =e^{\lambda _{1}n\sigma }\biggl\langle \frac{h}{w}P^{h}_{m\sigma }\phi ,Z_{n\sigma }\biggr\rangle . \end{aligned}$$

Thus $\mathit{II}(m,n)=e^{\lambda _{1}n\sigma }\langle \frac{h}{w}P^{h}_{m \sigma }g,Z_{n\sigma }\rangle$, where $g(x):=\phi (x)-\langle \phi ,h ^{2}\rangle$. Since $\langle g,h^{2}\rangle =0$, $\vert P^{h}_{m \sigma }g(x) \vert \le e^{-\lambda _{h}(m\sigma -\frac{1}{2})} \tilde{a}_{1}(x)^{1/2}\|g\|_{L^{2}(E,h^{2}m)}$ for $m\sigma >1/2$. Here $\lambda _{h}>0$ denotes the spectral gap in $\sigma (\mathcal{L}^{ ( \beta )})$. Thus we have for $m\sigma >1/2$

$$ |\mathit{II}(m,n)|= \biggl\vert e^{\lambda _{1}n\sigma }\biggl\langle \frac{h}{w}P^{h}_{m \sigma }g,Z_{n\sigma }\biggr\rangle \biggr\vert \le c_{14}e^{-\lambda _{h}m \sigma }\sup_{x\in E} \tilde{a}_{1}(x)^{1/2}W^{h/w}_{n\sigma }(Z). $$

It is shown in the proof of the equivalence between Assumption 5 and Assumption 5′ that for every $m\in \mathbb{N}$, $\lim_{n\to {\infty }}I(m,n)=0$$\mathbb{P}_{\mu }$-a.s. Thus by letting $n\to {\infty }$ and then $m\to {\infty }$ in (A.18), we obtain that

$$ \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma }\biggl\langle \frac{h}{w}\phi ,Z _{n\sigma }\biggr\rangle -\bigl\langle \phi ,h^{2}\bigr\rangle W^{h/w}(Z)=0\quad \mathbb{P}_{\mu }\mbox{-a.s.} $$

 □

Example 4

Let $\lambda _{h}:=\lambda _{2}-\lambda _{1}$ be the spectral gap in $\sigma (\mathcal{L}^{(\beta ) })$, where $\lambda _{2}:=\inf \{ \lambda \in \sigma (- \mathcal{L}^{(\beta ) }): \lambda \neq \lambda _{1} \}$. Since $P^{\beta }_{t}h(x)=e ^{-\lambda _{1}t}h(x)$, it follows by property (ii) and Hölder inequality that $h\in L^{4}(E,m)$. Let $g_{0}(\theta ):=\log ^{*} \theta /\theta$ for $\theta \in (0,+\infty )$. Then

$$\begin{aligned} \biggl\langle \int _{(0,+\infty )}r\log ^{*}\bigl(rh(\cdot )\bigr)\pi (\cdot ,dr),h^{2} \biggr\rangle =&\biggl\langle \int _{(0,+\infty )}g_{0}\bigl(rh(\cdot ) \bigr)r^{2}\pi ( \cdot ,dr),h^{3}\biggr\rangle \\ \le &\|g_{0}\|_{\infty }\biggl\langle \int _{(0,+\infty )}r^{2}\pi (\cdot ,dr),h ^{3} \biggr\rangle \\ < &+\infty . \end{aligned}$$

The last inequality comes from the fact that $h\in L^{4}(E,m)$ and condition (8.2). Thus we have by Theorem 2.1 and Remark 2.5 that $\mathrm{P}_{\delta _{x}} ( W^{h}_{\infty }(X)>0 ) >0$ for each $x\in E$. Hence

$$ w(x)\ge -\log \Bigl( 1-\mathrm{P}_{\delta _{x}} \Bigl( \limsup _{t\to +\infty }\langle 1,X_{t}\rangle >0 \Bigr) \Bigr) \ge - \log \bigl( 1-\mathrm{P}_{\delta _{x}} \bigl( W^{h}_{\infty }(X)>0 \bigr) \bigr) >0. $$

In view of Remark 2.3, $w(x)$ is a bounded function on $E$ under our assumptions. Using the boundedness of $w$ and (8.2), it is easy to verify that Assumptions 1–4 and Assumption 6 are satisfied by this example. Next we will show that Assumption 5′ is also satisfied, that is, for all $\mu \in \mathcal{M}_{c}(E)$, $\phi \in \mathcal{B} ^{+}_{b}(E)$, $\sigma >0$ and some $m\in \mathbb{N}$,

$$ \lim_{n\to +\infty }e^{\lambda _{1}n\sigma }\biggl\langle \frac{h}{w}P^{h} _{m\sigma }\phi ,Z_{n\sigma }\biggr\rangle =\bigl\langle \phi ,h^{2}\bigr\rangle W ^{h/w}_{\infty }(Z)\quad \mathbb{P}_{\mu } \mbox{-a.s.} $$

(A.19)

It is showed in the proof of Lemma 7.4 that

$$ e^{\lambda _{1}n\sigma }\biggl\langle \frac{h}{w}\,P^{h}_{m\sigma } \phi ,Z _{n\sigma }\biggr\rangle =e^{\lambda _{1}(m+n)\sigma } \sum _{i=1}^{N_{n \sigma }}\mathbb{P}_{\mu } \bigl( \bigl\langle \phi h,I^{i,n\sigma }_{m \sigma }\bigr\rangle \bigm| \mathcal{F}_{n\sigma } \bigr) +e^{\lambda _{1}n \sigma }\biggl\langle \frac{h}{w}\,\theta ^{*}_{\phi h}(m\sigma , \cdot ),Z _{n\sigma }\biggr\rangle . $$

Thus by Lemma 5.1, Lemma 7.3 and Theorem 2.2, under Assumptions 1–3 and Assumption 4(i), (A.19) is equivalent to that

$$ \lim_{n\to +\infty }e^{\lambda _{1}(m+n)\sigma }\sum_{i=1}^{N_{n\sigma }} \bigl\langle \phi h,I^{i,n\sigma }_{m\sigma }\bigr\rangle =\bigl\langle \phi ,h ^{2}\bigr\rangle W^{h}_{\infty }(X)\quad \mathbb{P}_{\mu } \mbox{-a.s.} $$

(A.20)

Since $e^{\lambda _{1}(m+n)\sigma }\langle \phi h, X_{(m+n)\sigma } \rangle \ge e^{\lambda _{1}(m+n)\sigma }\sum_{i=1}^{N_{n\sigma }} \langle \phi h,I^{i,n\sigma }_{m\sigma }\rangle$ and $\phi \in \mathcal{B}^{+}_{b}(E)$, (A.20) (or equivalently (A.19)) is equivalent to

$$ \lim_{n\to +\infty }e^{\lambda _{1}n\sigma }\langle \phi h,X_{n\sigma } \rangle =\bigl\langle \phi ,h^{2}\bigr\rangle W^{h}_{\infty }(X) \quad \mathbb{P}_{\mu } \mbox{-a.s.} $$

(A.21)

Fix $\phi \in \mathcal{B}^{+}_{b}(E)$ and $\mu \in \mathcal{M}_{c}(E)$. Let $g(x):=\phi (x)-\langle \phi ,h^{2}\rangle$ for $x\in E$. We have by (2.6) that for $t>0$,

$$\begin{aligned} &\mathbb{P}_{\mu } \bigl[ \bigl( e^{\lambda _{1}t}\langle gh,X_{t} \rangle \bigr) ^{2} \bigr] \\ &\quad =\bigl\langle P^{h}_{t}g,h\mu \bigr\rangle ^{2}+ \int _{0}^{t}e^{\lambda _{1}s} \biggl\langle P^{h}_{s} \biggl[ \biggl( 2\alpha + \int _{(0,+\infty )}y^{2} \pi (\cdot ,dy) \biggr) h \bigl( P^{h}_{t-s}g \bigr) ^{2} \biggr] ,h \mu \biggr\rangle \,ds \\ &\quad =:I(t)+\mathit{II}(t). \end{aligned}$$

Since $g\in L^{2}(E,\widetilde{m})$ and $\langle g,h^{2}\rangle =0$, we have by (3.6) for $r>t_{0}/2$ and $x\in E$,

$$ \bigl|P^{h}_{\lambda }g(x)\bigr|\le e^{-\lambda _{h}(r-\frac{t_{0}}{2})} \widetilde{a}_{t_{0}}(x)^{1/2}\|g\|_{L^{2}(E,\widetilde{m})}=:c_{1}e ^{-\lambda _{h}r}\widetilde{a}_{t_{0}}(x)^{1/2}. $$

(A.22)

Thus $I(t)\le c_{2}e^{-2\lambda _{h}t}$ for all $t>t_{0}/2$ and some $c_{2}>0$. On the other hand, for $r\in (0,t_{0}/2]$ and $x\in E$,

$$ \bigl|P^{h}_{\lambda }g(x)\bigr|\le \|g\|_{\infty } \le \|g\|_{\infty }e^{\lambda _{h}(\frac{t_{0}}{2}-r)}=:c_{3}e^{-\lambda _{h}r}. $$

(A.23)

Let ${\varepsilon }\in (0,-\lambda _{1}\wedge 2\lambda _{h})$, we have by (A.22) and (A.23) that

$$\begin{aligned} \mathit{II}(t) \le &\biggl\| 2\alpha + \int _{(0,+\infty )}y^{2}\pi (\cdot ,dy)\biggr\| _{ \infty } \int _{0}^{t}e^{\lambda _{1}s}\bigl\langle P^{h}_{s} \bigl( h \bigl( P^{h}_{t-s}g \bigr) ^{2} \bigr) ,h\mu \bigr\rangle \,ds \\ \le &c_{4} \int _{0}^{t}e^{\lambda _{1}s-2\lambda _{h}(t-s)}\bigl\langle P ^{h}_{s} \bigl( h ( 1\vee \widetilde{a}_{t_{0}} ) \bigr) ,h\mu \bigr\rangle \,ds \\ \le &c_{4}e^{-{\varepsilon } t} \int _{0}^{+\infty }e^{(\lambda _{1}+ {\varepsilon })s}\bigl\langle P^{h}_{s} \bigl( h ( 1\vee \widetilde{a}_{t_{0}} ) \bigr) ,h\mu \bigr\rangle \, ds. \end{aligned}$$

(A.24)

Since $h\in L^{4}(E,m)$, it follows by property (ii) that $h ( 1 \vee \widetilde{a}_{t_{0}} ) \in L^{2}(E,\widetilde{m})$. Thus by (3.6), $\langle P^{h}_{s} ( h ( 1\vee \widetilde{a} _{t_{0}} ) ) ,h\mu \rangle$ is bounded from above for $s$ sufficiently large. Therefore the integral in the right hand side of (A.24) is finite. It implies that $\sum_{n=1}^{+\infty } \mathbb{P}_{\mu } [ ( e^{\lambda _{1}n\sigma }\langle gh,X _{n\sigma }\rangle ) ^{2} ] <+\infty$, and hence we obtain (A.21) by Borel-Cantelli Lemma. □

Example 5

By a similar argument as that for [3, IV.5], we have

$$ h(x)\ge c_{0}|x|^{-d-\alpha }\quad \mbox{for }|x|\ge 1 , $$

(A.25)

for some positive constant $c_{0}$. Let $\lambda _{2}$ be the second bottom of the spectrum of $\sigma (\mathcal{E}^{(\beta ) })$, that is $\lambda _{2}:=\inf \{\mathcal{E}^{(\beta ) }(u,u):\ u\in \mathcal{F}\mbox{ with }\int _{\mathbb{R}^{d}}u(x)h(x)\,dx=0\mbox{ and }\int _{\mathbb{R}^{d}}u(x)^{2}\,dx=1 \}$. Then the spectral gap $\lambda _{h}:=\lambda _{2}-\lambda _{1}>0$. Recall that the $h$-transformed semigroup $P^{h}_{t}$ is an $\widetilde{m}$-symmetric semigroup with $\widetilde{m}(dx)=h(x)^{2}\,dx$. $P^{h}_{t}$ admits an integral kernel $p^{h}(t,x,y)$, given by (3.5), with respect to the measure $\widetilde{m}(dx)$. Let $p^{\beta }(t,x,y)$ be the integral kernel of $P^{\beta }_{t}$ with respect to the Lebesgue measure, which satisfies the estimates in (2.22). For any $f\in C_{c}(\mathbb{R}^{d})$, $t>0$ and $x\in \mathbb{R}^{d}$, we have by (8.3) and (2.22) that

$$\begin{aligned} \bigl\vert P^{h}_{t}f(x)-f(x) \bigr\vert &\le \int _{\mathbb{R}^{d}}p^{h}(t,x,y) \bigl\vert f(y)-f(x) \bigr\vert \widetilde{m}(dy) \\ &=\frac{e^{\lambda _{1}t}}{h(x)} \int _{\mathbb{R}^{d}}p^{\beta }(t,x,y)h(y) \bigl\vert f(y)-f(x) \bigr\vert \,dy \\ &\le c_{1} \frac{e^{(\lambda _{1}+\|\beta \|_{\infty })t}}{h(x)} \int _{\mathbb{R}^{d}} \biggl( t^{-d/\alpha }\wedge \frac{t}{|x-y|^{d+ \alpha }} \biggr) h(y)\bigl|f(y)-f(x)\bigr|\,dy.\quad \end{aligned}$$

(A.26)

Let $f_{t}(x):=\frac{1}{h(x)}\int _{\mathbb{R}^{d}} ( t^{-d/ \alpha }\wedge \frac{t}{|x-y|^{d+\alpha }} ) h(y)|f(y)-f(x)|\,dy$. Suppose $\mbox{supp}(f)\subset B(0,R)$ for some $R>1$. For $x\in B(0,2R)$ and $z\in \mathbb{R}^{d}$, $h(z)/h(x)\le \|h\|_{\infty }/\inf_{y\in B(0,2R)}h(y)<{\infty }$, and we have

$$\begin{aligned} f_{t}(x) \le &c_{2} \int _{\mathbb{R}^{d}} \biggl( t^{-d/\alpha }\wedge \frac{t}{|x-y|^{d+\alpha }} \biggr) \bigl|f(y)-f(x)\bigr|\,dy \\ =&c_{2} \int _{\mathbb{R}^{d}} \biggl( t^{-d/\alpha }\wedge \frac{t}{|z|^{d+ \alpha }} \biggr) \bigl|f(x+z)-f(x)\bigr|\,dz. \end{aligned}$$

(A.27)

Since $f\in C_{c}(\mathbb{R}^{d})$ is uniformly continuous on $\mathbb{R}^{d}$, for any ${\varepsilon }>0$, there exists $\delta >0$, such that $|f(z_{1})-f(z_{2})|\le {\varepsilon }$ whenever $|z_{1}-z _{2}|<\delta$. Thus by (A.27),

$$\begin{aligned} f_{t}(x) \le &c_{2}{\varepsilon } \int _{|z|< \delta }t^{-d/\alpha } \wedge \frac{t}{|z|^{d+\alpha }} \,dz+2c_{2}\|f\|_{\infty } \int _{|z|\ge \delta }t^{-d/\alpha }\wedge \frac{t}{|z|^{d+\alpha }}\,dz \\ =&c_{2}{\varepsilon } \int _{|u|< \delta t^{-1/\alpha }}1\wedge \frac{1}{|u|^{d+ \alpha }}\,du+2c_{2}\|f \|_{\infty } \int _{|u|\ge \delta t^{-1/\alpha }}1\wedge \frac{1}{|u|^{d+\alpha }}\,du. \end{aligned}$$

(A.28)

Since $\int _{\mathbb{R}^{d}}1\wedge \frac{1}{|u|^{d+\alpha }}\,du< {\infty }$, by letting $t\to 0$ and ${\varepsilon }\to 0$ in (A.28), we get $\sup_{x\in B(0,2R)}f_{t}(x)\to 0$. On the other hand, for $x\notin B(0,2R)$, by the fact $\mbox{supp}(f)\subset B(0,R)$, the boundedness of $h$ and (A.25), we have

$$\begin{aligned} f_{t}(x) =&\frac{1}{h(x)} \int _{|y|\le R} \biggl( t^{-d/\alpha }\wedge \frac{t}{|x-y|^{d+\alpha }} \biggr) h(y)\bigl|f(y)\bigr|\,dy \\ \le &c_{3} \int _{|y|\le R} \biggl( t^{-d/\alpha }\wedge \frac{t}{|x-y|^{d+ \alpha }} \biggr) |x|^{d+\alpha }\,dy \\ \le &c_{3} \int _{|y|\le R}\frac{t|x|^{d+\alpha }}{(|x|-|y|)^{d+\alpha }}\,dy \\ \le &c_{3} \int _{|y|\le R} \frac{t|x|^{d+\alpha }}{(|x|-R)^{d+\alpha }}\,dy \\ \le &2^{d+\alpha }c_{3}t. \end{aligned}$$

Thus $\sup_{x\notin B(0,2R)}f_{t}(x)\to 0$ as $t\to 0$. Hence by (A.26) we conclude that

$$ \lim_{t\to 0}\bigl\| P^{h}_{t}f-f \bigr\| _{\infty }=0 $$

(A.29)

for all $f\in C_{c}(\mathbb{R}^{d})$. Since $C_{c}(\mathbb{R}^{d})$ is dense in $(C_{0}(\mathbb{R}^{d}),\|\cdot \|_{\infty })$, (A.29) is true for all $f\in C_{0}(\mathbb{R}^{d})$.

Recall that $w(x)=-\log \mathrm{P}_{\delta _{x}} ( \exists t \ge 0,\ \langle 1,X_{t}\rangle =0 )$. The argument in Remark 2.3 shows that $w(x)$ is bounded on $E$ and so Condition 1 is satisfied. Let $W^{h}_{t}(X):=e^{\lambda _{1}t}\langle h,X_{t}\rangle$. Then $W^{h}_{t}(X)$ is a nonnegative $\mathrm{P}_{\mu }$-martingale with respect to $\mathcal{F}_{t}:=\sigma \{X_{s}:s\in [0,t]\}$ for all $\mu \in \mathcal{M}_{F}(E)$. Let $W^{h}_{\infty }(X):= \lim_{t\to {\infty }}W^{h}_{t}(X)$. Note that by (2.6),

$$\begin{aligned} \mathrm{P}_{\delta _{x}} \bigl( W^{h}_{t}(X)^{2} \bigr) =&h(x)^{2}+e ^{2\lambda _{1}t} \int _{0}^{t}P^{\beta }_{s} \bigl[ \widehat{\alpha } \bigl( P^{\beta }_{t-s}h \bigr) ^{2} \bigr] (x)\,ds \\ =&h(x)^{2}+ \int _{0}^{t}e^{2\lambda _{1}s}P^{\beta }_{s} \bigl[ \widehat{\alpha }h^{2} \bigr] (x)\,ds \\ \le &h(x)^{2}+\|\widehat{\alpha }h\|_{\infty } \int _{0}^{t}e^{2\lambda _{1}s}P^{\beta }_{s}h(x)\,ds \\ =&h(x)^{2}+\|\widehat{\alpha }h\|_{\infty }h(x) \int _{0}^{t}e^{\lambda _{1}s}\,ds \\ \le &h(x)^{2}+c_{4}h(x). \end{aligned}$$

(A.30)

Here $\widehat{\alpha }(x):=2\alpha (x)+\int _{(0,{\infty })}y^{2} \pi (x,dy)$ for $x\in E$ and $c_{4}$ is a positive constant independent of $t$. Thus for any $\mu \in {\mathcal{M}}^{h}_{F}(E)$, $\{W^{h}_{t}(X): \ t\ge 0\}$ is an $L^{2}$-bounded nonnegative martingale and hence $W^{h}_{t}(X)$ converges to $W^{h}_{\infty }(X)$$\mathrm{P}_{\mu }$-a.s. and in $L^{2}(P_{\mu })$. In particular $\mathrm{P}_{\mu } ( W^{h}_{\infty }(X)>0 ) >0$. Thus $\mathrm{P}_{\delta _{x}} ( \exists t>0,\ \langle 1,X_{t}\rangle =0 ) \le 1-\mathrm{P}_{\delta _{x}} ( W^{h}_{\infty }(X)>0 ) <1$, and so $w(x)>0$ for all $x\in \mathbb{R}^{d}$. Furthermore, by Cauchy-Schwartz inequality and (A.30), we have

$$ \mathrm{P}_{\delta _{x}} \bigl( W^{h}_{\infty }(X)>0 \bigr) \ge \frac{ ( \mathrm{P}_{\delta _{x}} [ W^{h}_{\infty }(X) ] ) ^{2}}{\mathrm{P}_{\delta _{x}} [ W^{h}_{\infty }(X)^{2} ] }= \frac{h(x)^{2}}{\lim_{t\to {\infty }}\mathrm{P}_{\delta _{x}} [ W^{h}_{t}(X)^{2} ] } \ge \frac{h(x)}{h(x)+c _{4}} . $$

Thus we get

$$ w(x)\ge -\log \bigl( 1-\mathrm{P}_{\delta _{x}} \bigl( W^{h}_{\infty }(X)>0 \bigr) \bigr) \ge \log \bigl(h(x)+c_{4}\bigr)-\log c_{4}. $$

Using this, (A.25) and the fact that $h$ is bounded away from 0 and ${\infty }$ on compact sets, we conclude that the function $h(x)/w(x)$ is bounded from above on $\mathbb{R}^{d}$. We see from the above arguments that Assumptions 1–4 and Assumption 6 are satisfied by this example.

Finally we will show that Assumption 5 is also satisfied. Let $Z$ be the skeleton process. We need to show that for all $\mu \in \mathcal{M} _{c}(E)$, $\sigma >0$ and $f\in \mathcal{B}^{+}(E)$ with $\frac{fw}{h}$ bounded,

$$ \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma }\langle f,Z_{n\sigma } \rangle =\langle f,wh\rangle W^{h/w}_{\infty }(Z)\quad \mathbb{P}_{\mu }\mbox{-a.s.} $$

(A.31)

Fix $f\in \mathcal{B}^{+}(E )$ with $fw/h$ bounded. Let $g(x):=f(x)- \langle f,wh\rangle \frac{h}{w}(x)= ( \frac{fw}{h}(x)-\langle f,wh \rangle ) \frac{h}{w}(x)$ for $x\in E$. Through the same argument as in Example 1, to prove (A.31), it suffices to prove that for every $x\in E$,

$$ \mathbb{P}_{\cdot ,\delta _{x}} \Bigl( \lim_{n\to {\infty }}e^{\lambda _{1}n\sigma } \langle g,Z_{n\sigma }\rangle =0 \Bigr) =1. $$

(A.32)

We observe that $g$, $\frac{gw}{h}$ are bounded functions on $\mathbb{R}^{d}$ by the boundedness of $\frac{h}{w}$ and $\frac{fw}{h}$. For every $x\in \mathbb{R}^{d}$,

$$ \mathbb{P}_{\cdot ,\delta _{x}} \bigl[ \bigl( e^{\lambda _{1}t}\langle g,Z_{t}\rangle \bigr) ^{2} \bigr] =\frac{h(x)}{w(x)} \bigl( I(t,x)+\mathit{II}(t,x) \bigr) , $$

where $I(t,x):=e^{\lambda _{1}t}P^{h}_{t} ( \frac{w}{h}g^{2} ) (x)$, $\mathit{II}(t,x):=\int _{0}^{t}e^{\lambda _{1}s}P^{h}_{s} [ (2\alpha +b)\frac{h}{w}P^{h}_{t-s} ( \frac{w}{h}g ) ^{2} ] (x)\,ds$. For $I$, we have

$$ I(t,x)\le \bigl\| \frac{w}{h}g\bigr\| _{\infty }\|g\|_{\infty }e^{\lambda _{1}t}. $$

(A.33)

For $\mathit{II}$, we have

$$ \mathit{II}(t,x)\le \|2\alpha +b\|_{\infty }\biggl\| \frac{h}{w} \biggr\| _{\infty }\| \frac{w}{h}g\|_{\infty } \int _{0}^{t}e^{\lambda _{1}s}P^{h}_{s} \biggl[ \biggl\vert P^{h}_{t-s} \biggl( \frac{w}{h}g \biggr) \biggr\vert \biggr] (x)\,ds. $$

(A.34)

Since $\langle \frac{w}{h}g,h^{2}\rangle =0$, by (2.26) and Hölder inequality for any $t>s>0$

$$\begin{aligned} P^{h}_{s} \biggl[ \biggl\vert P^{h}_{t-s} \biggl( \frac{w}{h}g \biggr) \biggr\vert \biggr] (x) =& \int _{\mathbb{R}^{d}}p^{h}(s,x,y) \biggl\vert P ^{h}_{t-s} \biggl( \frac{w}{h}g \biggr) \biggr\vert \widetilde{m}(dy) \\ \le & \biggl( \int _{\mathbb{R}^{d}}p^{h}(s,x,y)^{2} \widetilde{m}(dy) \biggr) ^{1/2} \biggl\| P^{h}_{t-s} \biggl( \frac{w}{h}g \biggr) \biggr\| _{L^{2}(\mathbb{R}^{d}, \widetilde{m})} \\ \le &e^{-\lambda _{h}(t-s)}p^{h}(2s,x,x)^{1/2}\biggl\| \frac{w}{h}g\biggr\| _{L^{2}( \mathbb{R}^{d},\widetilde{m})} \end{aligned}$$

(A.35)

Let ${\varepsilon }\in (0,-\lambda _{1}\wedge \lambda _{h})$. Note that by (3.6), for each $x$, $p^{h}(s,x,x)$ is bounded from above when $s$ is suffciently large. This together with (A.34) and (A.35) yield that for every $x\in \mathbb{R}^{d}$,

$$ \mathit{II}(t,x) \le c_{5} \int _{0}^{t}e^{\lambda _{1}s-\lambda _{h}(t-s)}\,ds \le c_{5}e^{-{\varepsilon } t} \int _{0}^{t}e^{(\lambda _{1}+{\varepsilon })s}\,ds \le c_{6}e^{-{\varepsilon } t}. $$

(A.36)

Thus by (A.33) and (A.36), we get $\sum_{n=1}^{{\infty }} \mathbb{P}_{\cdot ,\delta _{x}} [ ( e^{\lambda _{1}n\sigma } \langle g,Z_{n\sigma }\rangle ) ^{2} ] <{\infty }$. Consequently, (A.32) follows by Borel-Cantelli lemma. □