2. Proofs

In order to prove Proposition 1, we introduce the following stochastic processes as unique strong solutions of the SDEs. For $i=1,2$, we let

\begin{align*}Z^{(i)}_t&=Z^{(i)}_0+\int_0^tg_i(Z^{(i)}_s)\textrm{d} s+\sqrt{2\gamma^2}\int_0^t \int_{0}^{Z^{(i)}_s}W(\textrm{d} s, \textrm{d} u)\nonumber\\[3pt]&\quad +\int_0^t\int_{(0,\infty)} \int_0^{Z^{(i)}_{s-}}z\widetilde{N}(\textrm{d} s, \textrm{d} z, \textrm{d} u)+\int_0^t Z^{(i)}_{s-}\textrm{d} S^{(i)}_s,\end{align*}

where

(9) \begin{equation}S^{(i)}_t= \sigma B^{({\rm e})}_t +\int_0^t \int_{(-1,1)^c} (h_i(z)-1)N^{({\rm e})}({\textrm{d} s},{\textrm{d} z})+\int_0^t \int_{(-1,1)} ({\rm e}^z-1)\widetilde{N}^{({\rm e})}({\textrm{d} s},{\textrm{d} z}),\end{equation}

with $g_1(z)\ge g_2(z)$ for $z\ge 0$, and $h_1(z)\ge h_2(z)\geq 0$ for $z\in \mathbb{R}$. We also assume that, for each ${m\ge 0}$, there is a non-decreasing concave function $z\mapsto r_m(z)$ on $\mathbb{R}_+$ satisfying $\int_{0+} r_m(z)^{-1}\textrm{d} z=\infty$ and, for $i=1,2$,

\begin{equation*} |g_i(x)-g_i(y)|+|x-y|\int_{(-1,1)^c}\big(|h_i(z)-1|\land m\big)\pi(\textrm{d} z)\le r_m(|x-y|) \end{equation*}

for every $0\le x, y\le m$. According to [Reference Palau and Pardo25, Proposition 1], the previous SDEs possess unique positive strong solutions that we denote by $Z^{(i)}$ for $i=1,2$.

Lemma 1. If $Z^{(1)}_0\ge Z^{(2)}_0$, then, almost surely (a.s.),

\[\mathbb{P}\big(Z^{(2)}_t\le Z^{(1)}_t \textrm{ for all } t\ge 0\big)=1.\]

Proof. Let $\tau_m=\inf\{t\ge 0\,{:}\, Z^{(1)}_s\ge m \textrm{ or } Z^{(2)}_s\ge m\}$ for $m\ge 1$. According to the proof of [Reference Palau and Pardo25, Proposition 1], for $i=1, 2$ we have $Z^{(i)}_t=Z^{(i, m)}_t$ for $t<\tau_m$, where $Z^{(i, m)}$ is the unique strong solution to

\begin{align*}Z^{(i, m)}_t&=Z^{(i)}_0+\int_0^tg_i(Z^{(i, m)}_s\land m)\textrm{d} s+\sqrt{2\gamma^2}\int_0^t \int_{0}^{Z^{(i, m)}_{s-}\land m}W(\textrm{d} s, \textrm{d} u)\nonumber\\[3pt] &\quad+\int_0^t\int_{(0,\infty)} \int_0^{Z^{(i, m)}_{s-}\land m}(z\land m)\widetilde{N}(\textrm{d} s, \textrm{d} z, \textrm{d} u)+\int_0^t \big(Z^{(i, m)}_{s-}\land m\big)\textrm{d} S^{(i)}_s,\end{align*}

with $S^{(i)}$ defined by (9). In other words, for $m\ge 1$ we have

\[\mathbb{P}\left(Z^{(1)}_t\ge Z^{(2)}_t, \textrm{ for all} \,\, t<\tau_m\right)=\mathbb{P}\left(Z^{(1, m)}_t\ge Z^{(2, m)}_t, \textrm{ for all } \,\, t<\tau_m\right)\!.\]

Then, a direct application of [Reference Dawson and Li9, Theorem 2.2] implies that the latter probability equals 1. The proof is completed by taking m to $\infty$.

Proof of Proposition 1. The first statement follows directly from Lemma 1 by taking

\[g_1(z)=g_2(z)=bz-g(z) \textrm{ for } z\ge 0\quad \textrm{and}\quad h_1(z)=h_2(z)={\rm e}^{z} \textrm{ for } z\in \mathbb{R}.\]

For the second statement, we recall that the competition mechanism g is non-negative and non-decreasing, implying that we can take $g_1(z)=b z$, $g_2(z)=bz-g(z)$, and $h_1(z)=h_2(z)={\rm e}^{z}$. Again from Lemma 1, we deduce that the process Z(x) is stochastically dominated by $ Z^\sharp(y)$ for $y\ge x$.

Proof of Corollary 1. The first part of this result follows from the previous proposition and [Reference He, Li and Xu13, Theorem 4.1], where it was deduced that $Z^\sharp(x)$ becomes extinct with positive probability.

It remains to deduce that the process Z(x) is strong Feller under Grey’s condition (6). With this purpose in mind, we introduce, for $0\leq x\leq y$,

\[T^{(x,y)}=\inf\{t\ge 0\,{:}\, Z_t(x)=Z_t(y)\}.\]

From the Markov property, $Z_t(x)=Z_t(y)$ a.s. for any $t\geq T^{(x,y)}$. Thus, for any $f\in \mathcal{B}_b([0,\infty))$, we have

(10) \begin{equation}\begin{split}\big|\texttt{P}_t\, f(y)-\texttt{P}_t\, f(x)\big|&\le \mathbb{E}\left[\big|f(Z_t(y))-f(Z_t(x))\big|\textbf{1}_{\{t<T^{(x,y)}\}}\right]\\&\le2 \|\,f\|_{[0,\infty)} \mathbb{P}(t<T^{(x,y)}), \end{split}\end{equation}

where $(\texttt{P}_t, t\geq 0)$ denotes the semigroup associated with the process Z and $\|\cdot\|_{[0,\infty)}$ denotes the supremum norm on $[0,\infty)$. Next, we define $\Gamma_t^{(x,y)}\coloneqq Z_t(y)-Z_t(x)$ for $ t\ge 0$. From Proposition 1, it is clear that $\Gamma_t^{(x,y)}\geq 0$ a.s. for any $t\geq 0$. Moreover, it satisfies

(11) \begin{align}\Gamma^{(x,y)}_t&= y-x+b\int_0^t \Gamma^{(x,y)}_s\textrm{d} s +\sqrt{2\gamma^2} \int_0^t \int_0^{\Gamma^{(x,y)}_s}W^\prime(\textrm{d} s, \textrm{d} u)\nonumber\\[3pt] &\quad-\int_0^t G_{Z_t(x)}(\Gamma^{(x,y)}_s) \textrm{d} s+\int_0^t\int_{(0,\infty)} \int_0^{\Gamma^{(x,y)}_{s-}}z\widetilde{N}^{\prime}(\textrm{d} s, \textrm{d} z, \textrm{d} u)+\int_0^t \Gamma^{(x,y)}_{s-}\textrm{d} S_s,\end{align}

where $G_x(z)\coloneqq g(x+z)-g(x)$ is a non-decreasing continuous and non-negative function on $[0,\infty)$,

\[W^\prime(\textrm{d} s, \textrm{d} u)\coloneqq W(\textrm{d} s, \textrm{d} u +Z_s(x))\]

is a Gaussian white noise with intensity $\textrm{d} s \textrm{d} u$, and

\[\widetilde{N}^{\prime}(\textrm{d} s, \textrm{d} z, \textrm{d} u)\coloneqq \widetilde{N}(\textrm{d} s, \textrm{d} z, \textrm{d} u+Z_s(x))\]

is a Poisson random measure with intensity $\textrm{d} s\mu(\textrm{d} z) \textrm{d} u$. Again using Proposition 1, we deduce that

(12) \begin{equation}\mathbb{P}\left(\Gamma^{(x,y),\sharp}_t\ge \Gamma^{(x,y)}_t \textrm{ for all } t\ge 0\right)=1,\end{equation}

where the process $\Gamma^{(x,y), \sharp}$ is the unique strong solution of (11) but with $G_x(z)\equiv 0$ for any $x,z\in [0,\infty)$. Observe that $\Gamma^{(x,y), \sharp}$ has the same law as $ Z^\sharp(y-x)$, and [Reference He, Li and Xu13, Theorem 4.1] thus implies that the former hits 0 with positive probability and that 0 is an absorbing boundary. In addition to (10) and (12), we have

\[\begin{split}\big|\texttt{P}_t\, f(y)-\texttt{P}_t\, f(x)\big|&\le 2\|\,f\|_{{[0,\infty)}} \mathbb{P}^\sharp_{y-x}(T_0>t)\\[3pt]&\le 2\|\,f\|_{{[0,\infty)}} \mathbb{E}\left[1-{\rm e}^{-(y-x)v_t(0,\infty, K)}\right]\!,\end{split}\]

where $v_t(0,\infty,K)$ is a functional of the auxiliary Lévy process K which is defined by

\begin{multline*}K_t=bt-\left( \frac{\sigma^2}{2}+\int_{(-1,1)} ({\rm e}^z-1-z)\pi(\textrm{d} z)\right) t+ \sigma B^{({\rm e})}_t \\+\int_0^t \int_{(-1,1)^c} zN^{({\rm e})}({\textrm{d} s},{\textrm{d} z})+\int_0^t \int_{(-1,1)} z\widetilde{N}^{({\rm e})}({\textrm{d} s},{\textrm{d} z}), \qquad t\ge 0.\end{multline*}

Moreover, we have

\[\mathbb{P}^\sharp_{y}(Z_t=0 \mid K)={\rm e}^{-yv_t(0,\infty,K)}>0 \]

(see, for instance, [Reference He, Li and Xu13, Theorems 4.1, 4.3]). In other words, $|\texttt{P}_t\,f(y)-\texttt{P}_t\,f(x)|$ goes to 0 as $|x-y|$ goes to 0, implying that $\texttt{P}_t\, f$ is a continuous bounded function on $[0, \infty)$, i.e. the semigroup $(\texttt{P}_t, t\ge 0)$ is strong Feller.

Finally, it is clear that the strong Feller property and the right continuity of the unique strong solution (1) implies (i) and (ii). The proof is now complete.

Finally, we prove Theorem 1.

Proof of Theorem 1. First of all, from Lemma 1, it is enough to prove our result for a process with a random environment which has no downward jumps larger than $1-{\rm e}^{-1}$. Hence, we assume throughout this proof that Z(x) is a solution to (1) with

\begin{equation*} S_t= \sigma B^{({\rm e})}_t +\int_0^t \int_{(1,\infty)} ({\rm e}^z-1)N^{({\rm e})}({\textrm{d} s},{\textrm{d} z})+\int_0^t \int_{(-1,1)} ({\rm e}^z-1)\widetilde{N}^{({\rm e})}({\textrm{d} s},{\textrm{d} z}).\end{equation*}

Our first step is to prove (8). To do so, we observe that a similar Lyapunov function to the one used in [Reference Le19] will work in our case. Indeed, recall that $T_M$ denotes the first passage time for the process Z below a level $M>0$, i.e. $T_M\coloneqq \inf \{t \geq 0, Z_t\leq M\}$. As we will see below, the finiteness of the first moment of such random times will be useful for deducing our result. Our first step is to show that there exists $M>0$ such that

(13) \begin{equation} \sup_{x\geq 0} \mathbb{E}_x\big[ T_M\big] = \sup_{x\geq M} \mathbb{E}_x\big[ T_M\big] <\infty.\end{equation}

With this goal in mind, we introduce $A>b+\overline{\pi}(1)$, where $\overline{\pi}(x)=\pi([x,\infty))$. According to the integral condition in (7) and [Reference Le and Pardoux20, Lemma 2.3], we have that

\begin{equation*} \lim_{y\to +\infty} \frac{g(y)}{y}=\infty. \end{equation*}

In other words, there exists a constant $M>(z_0+1){\rm e}$ such that

(14) \begin{equation}\int_{M{\rm e}^{-1}}^{\infty}\frac{\textrm{d} w}{g(w)}\leq \frac{1}{A} \quad \text{and} \quad \frac{1}{g(y)} \leq \frac{y}{g(y)}\leq \frac{1}{A}, \qquad \text{for } y\geq M{\rm e}^{-1}.\end{equation}

The constant M will be our threshold. For our purposes, we define the function G in $C^2(\mathbb{R})$ as

\[ G(y)=\left\{ \begin{aligned} &\int_{z_0}^y\frac{\textrm{d} w}{g(w)} &&\text{ if } y \geq z_0+1,\\& 0 && \text{ if } y\leq z_0, \end{aligned} \right. \]

and such that G is non-negative and non-decreasing. Thus, applying Itô’s formula to $G(Z_{t\land T_M})$, we find

(15) \begin{eqnarray} &&\hspace*{-110pt} G(Z_{t \wedge T_M})- G(Z_0)= -t \wedge T_M + b\int_0^{t\wedge T_M} \frac{{Z_s}}{g(Z_s)} \textrm{d} s\end{eqnarray}

(16) \begin{eqnarray} &&\hspace*{-84pt}- \int_0^{t\wedge T_M} \frac{g'(Z_s) }{g(Z_s)^2}\left( \gamma^2 Z_s+\frac{\sigma^2}{2}Z_s^2\right){\textrm{d} s} \end{eqnarray}

(17) \begin{eqnarray} && +\,\sqrt{2\gamma^2} \int_0^{t\wedge T_M}\int_0^{Z_s} \frac{1}{g(Z_s)}W(\textrm{d} s, \textrm{d} u)+ \int_0^{t\wedge T_M} \frac{\sigma{Z_s}}{g(Z_s)} \textrm{d} B^{({\rm e})}_s \end{eqnarray}

(18) \begin{eqnarray} &&+\,\int_0^{t \wedge T_M}\int_{(0,\infty)} Z_s\left(G(Z_{s}+z)-G(Z_{s})-\frac{z}{g(Z_{s})}\right) \mu({\textrm{d} z}){\textrm{d} s}\end{eqnarray}

(19) \begin{eqnarray}&& +\,\int_0^{t \wedge T_M}\int_{(0,\infty)}\int_0^{Z_{s-}} [G(Z_{s-}+z)-G(Z_{s-})]\widetilde{N}({\textrm{d} s},{\textrm{d} z},{\textrm{d} u})\end{eqnarray}

(20) \begin{eqnarray}&&\hspace*{-40pt} +\,\int_0^{t \wedge T_M}\int_{(1,\infty)} [G({\rm e}^z Z_{s-})-G(Z_{s-})]{N}^{({\rm e})}({\textrm{d} s},{\textrm{d} z})\end{eqnarray}

(21) \begin{eqnarray}&& \hspace*{-40pt}+\int_0^{t \wedge T_M}\int_{(-1,1)} [G({\rm e}^z Z_{s-})-G(Z_{s-})]\widetilde{N}^{({\rm e})}({\textrm{d} s},{\textrm{d} z})\end{eqnarray}

(22) \begin{eqnarray}&&+\int_0^{t \wedge T_M}\int_{(-1,1)} \left(G({\rm e}^z Z_{s})-G(Z_{s})-\frac{({\rm e}^z-1)Z_s}{g(Z_s)}\right) \pi(\textrm{d} z){\textrm{d} s}.\end{eqnarray}

We take expectations, assuming that Z starts from x, on both sides of the previous identity and we study each term separately. Our aim is to prove that the expectation of $G(Z_{t \wedge T_M})- G(x)$ may be upper-bounded by $\mathbb{E}_x[t\wedge T_M]$.

We start with the integral on the right-hand side of (15). From the second inequality in (14), we see that

(23) \begin{equation}b\mathbb{E}_x\left[\int_0^{t\wedge T_M} \frac{{Z_s}}{g(Z_s)} \textrm{d} s\right]\le\frac{b}{A} \mathbb{E}_x\left[t\wedge T_M\right].\end{equation}

Next, since $Z_s \geq M{\rm e}^{-1}>z_0+1$ for $s \leq t \wedge T_M$, and since g is non-negative and non-decreasing, the expectation of (16) is non-positive. The same assumptions imply that, for any $x\ge z_0+1$,

(24) \begin{equation} G(x+z)-G(x)-\frac{z}{g(x)}\le 0, \qquad \textrm{for }z> 0.\end{equation}

In other words, the expectations of (18) and (22) are also non-positive.

For (17), studying the quadratic variation of both continuous local martingales together with the second formula in (14), we observe that both processes are real martingales. Therefore their expectations are equal to 0.

For (19), we split the interval $(0,\infty)$ into (0, 1] and $(1,\infty)$. Using the inequality in (24) together with the second formula of (14), we deduce that

\[\mathbb{E}_x\left[ \int_0^{t \wedge T_M}\int_{(0,1)}Z_{s} \Big(G(Z_{s}+z)-G(Z_{s})\Big)^2 \mu({\textrm{d} z}){\textrm{d} s} \right]\le \frac{t}{A^2}\int_{(0,1)}z^2\mu({\textrm{d} z})<\infty. \]

Similarly, we get that

\[\mathbb{E}_x\left[ \int_0^{t \wedge T_M}\int_{[1,\infty)}Z_{s} \Big|G(Z_{s}+z)-G(Z_{s})\Big| \mu({\textrm{d} z}){\textrm{d} s} \right]\le \frac{t}{A}\int_{[1,\infty)}z\mu({\textrm{d} z})<\infty. \]

In other words, the integral of (19) restricted to (0, 1] is a square integrable martingale and the integral restricted to $(1, \infty)$ is a martingale. Finally, the expectation of (19) equals 0.

For (20), we use Fubini’s theorem and the first inequality in (14) to deduce that

\begin{multline*}\mathbb{E}_x\Bigg[ \int_0^{t \wedge T_M}\int_{(1,\infty)} |G({\rm e}^z Z_{s-})-G(Z_{s-})|\pi(\textrm{d} z) \textrm{d} s\Bigg] \\[3pt]\leq \overline{\pi}(1) \mathbb{E}_x\left[ \int_0^{t \wedge T_M} \int_{Z_{s}}^{\infty} \frac{\textrm{d} w}{g(w)} \textrm{d} s \right] \leq t \overline{\pi}(1) \left(\int_{M{\rm e}^{-1}}^{\infty} \frac{\textrm{d} w}{g(w)}\right)\leq \frac{t}{A} \overline{\pi}(1).\end{multline*}

In other words, (20) may be written as the sum of a martingale and a finite variation process, and, since the martingale has null expectation, we deduce that

(25) \begin{align}&\mathbb{E}_x\left[\int_0^{t \wedge T_M}\int_{(1,\infty)} [G({\rm e}^z Z_{s-})-G(Z_{s-})]N^{({\rm e})}({\textrm{d} s},{\textrm{d} z})\right ]\nonumber\\[3pt]&\qquad =\mathbb{E}_x\left[\int_0^{t \wedge T_M}\int_{(1,\infty)} \left[G({\rm e}^z Z_{s-})-G(Z_{s-})\right]\pi(\textrm{d} z) \textrm{d} s \right]\le\frac{\overline{\pi}(1)}{A}\mathbb{E}[t\wedge T_M].\end{align}

Finally, for (21), the fact that $g(w) \geq Aw$ for $w\geq M{\rm e}^{-1}$ implies that

\[\begin{split}\mathbb{E}&\left[ \int_0^{t \wedge T_M}\int_{(-1, 1)} \left(G({\rm e}^zZ_s)-G(Z_s)\right)^2 \pi({\textrm{d} z}){\textrm{d} s} \right]\\[3pt]& \leq \mathbb{E}\left[ \int_0^{t \wedge T_M}\int_{(-1,1)} \left(\int_{Z_{s}}^{{\rm e}^z Z_{s}} \frac{\textrm{d} w}{Aw} \right)^2 \pi({\textrm{d} z}){\textrm{d} s} \right]\leq \frac{t}{A^2} \int_{(-1,1)} z^2 \pi({\textrm{d} z}) <\infty.\end{split}\]

In other words, (21) is a square integrable martingale and its expectation is equal to 0.

Thus, putting all the pieces together (i.e. inequalities (23) and (25), the fact that the expectations of (16), (18), and (22) are non-positive, and the fact that the expectations of (17), (19), and (21) are null), we deduce that

\begin{equation*}\begin{aligned}\mathbb{E}_x\left[ \int_x^{Z_{t\wedge T_M}} \frac{\textrm{d} w}{g(w)} \right] \leq -\left(1-\frac{b}{A}-\frac{\overline{\pi}(1)}{A}\right) \mathbb{E}_x[t\wedge T_M]. \end{aligned}\end{equation*}

Since $A>b+\overline{\pi}(1)$, we have, for any $x,t\geq 0$, that

(26) \begin{equation} \mathbb{E}_x\Big[t\wedge T_M\Big] \leq \frac{1}{C(A)} \mathbb{E}_x\left[ \int^x_{Z_{t\wedge T_M}} \frac{\textrm{d} w}{g(w)} \right] \leq \frac{1}{C(A)} \int_{M{\rm e}^{-1}}^{\infty} \frac{\textrm{d} w}{g(w)}, \end{equation}

where $C(A)\coloneqq (1-\frac{b\,{+}\,\overline{\pi}(1)}{A})>0$. Hence, using the monotone convergence theorem, as t goes to $\infty$, we deduce (13).

In order to prove that the process becomes extinct with positive probability, we first show that the time to extinction for the process Z starting from M is not almost surely infinite. We recall that we assumed that the environment has no negative jumps larger than $1-{\rm e}^{-1}$. Using Proposition 1 (both processes with the same restriction on negative jumps of the environment), we observe that, for any $x\leq M$, the process Z(x) is stochastically dominated by a process $ Z^\sharp(x)$ which is a CB process in a Lévy random environment (without competition) with branching mechanism $\psi$. Since $\psi$ satisfies Grey’s condition, [Reference He, Li and Xu13, Theorem 4.1] guarantees that there is $t_0>0$ for which

\[ 0<\mathbb{P}^\sharp_M\big( Z^\sharp_{t_0}=0\big)\leq \inf_{x \leq M}\mathbb{P}_x\big(Z_{t_0}=0\big)\coloneqq p. \]

Reasoning by recurrence and using the Markov property, we can prove that the extinction time of Z is stochastically dominated from above by an integrable random variable. To be more precise, the algorithm is as follows.

We start from $Z_0=x$; we wait a random time $T_M$ until the process is below the level M. Then, the process becomes extinct before a time interval of size $t_0$ with probability p. If the process is not extinct after $T_M+t_0$, we set $X_1\coloneqq Z_{T_M+t_0}$ and we repeat the procedure with a process starting from $X_1$ according to the Markov property. Finally, we obtain an a.s. finite sequence $(X_i)_{i\geq 1}$ such that Z is stochastically dominated by the random variable $ {\sum_{i=1}^{\xi} \big(T^{(X_i)}_{M}+t_0\big)}, $ where $\xi$ is a geometric random variable that counts the number of random steps before Z becomes extinct, $\big\{T^{(X_i)}_{M}\big\}_{i\geq 0}$ are independent of $\xi$, and, conditionally on the event $\{X_i=x\}$, $T^{(X_i)}_{M}$ has the same distribution as $T_{M}$ under $\mathbb{P}_x$. Hence,

\[ \sup_{x \geq 0}\mathbb{E}_x[T_0] \leq \frac{1}{p}\left({t_0}+\sup_{x\geq 0} \mathbb{E}_x\big[T_M\Big]\right)<\infty,\]

and in particular the process becomes extinct a.s.

Let us prove that the point $\infty$ is a continuous entrance point. From Chebyshev’s inequality, the monotone convergence theorem, and (26), we deduce that, for any $t>0$,

\[\begin{aligned}\lim_{m\to \infty}\lim_{x\to \infty}\mathbb{P}_x(T_m>t)&\le \lim_{m\to \infty}\lim_{x\to \infty}\frac{\mathbb{E}_x[T_m]}{t}\\[3pt]&\leq \lim_{m\to \infty}\frac{1}{C(A)t}\int_{m{\rm e}^{-1}}^{\infty} \frac{\textrm{d} w}{g(w)}=0.\end{aligned}\]

In other words, the point $\infty$ is a continuous entrance point.

It remains to prove the extension to a $C_b$ Feller process on $[0,\infty]$. Let $C_b([0,\infty])$ denote the set of continuous bounded functions on $[0,\infty)$ with a limit at $\infty$, and recall that $(\texttt{P}_t, t\geq 0)$ denotes the semigroup associated with the process Z which is $C_b$ Feller on $[0,\infty)$.

Let $t> 0$ be fixed. Recall, from Proposition 1, that, for any non-decreasing sequence $\{x_n\}_{n\ge 1}$ of strictly positive real numbers, the sequence of random variables $Z_t(x_n)$ is non-decreasing. Hence, it converges a.s. to a limit that we denote by $Z_t(\infty) \in [0,\infty]$. Then, for any $f\in C_b([0,\infty])$, f is bounded, and from the dominated convergence theorem we deduce that

\[\texttt{P}_t\,f(x)=\mathbb{E}_x[\,f(Z_t)]\underset{x\to \infty}{\longrightarrow}\mathbb{E}[\,f(Z_t(\infty))].\]

We denote this limit by $\texttt{P}_t\,f(\infty)$. Let us prove that such an extension on $[0,\infty]$ gives a semigroup satisfying the $C_b$ Feller property on $[0,\infty]$. The definition through a limit guarantees that $\texttt{P}$ remains a semigroup on $[0,\infty]$ and that $\texttt{P}_t\,f\in C_b([0,\infty])$ for $t\ge 0$. Thus, it remains to prove condition (ii) of Corollary 1; precisely, it remains to prove that, for any $f\in C_b([0,\infty])$, $|\texttt{P}_t\,f(\infty)-f(\infty)|\to 0 \textrm{ as } t\to 0.$ Observe that

\[|\texttt{P}_t\,f(\infty)-f(\infty)|\le \limsup_{x\to \infty}\mathbb{E}_x\left[\left|f(Z_t)-f(\infty)\right|\right].\]

Moreover, we have

\[\begin{split}\mathbb{E}_x\left[\left|f(Z_t)-f(\infty)\right|\right]&=\mathbb{E}_x\left[\left|f(Z_t)-f(\infty)\right|\textbf{1}_{\{T_{ x/2}\le t\}}\right]+\mathbb{E}_x\left[\left|f(Z_t)-f(\infty)\right|\textbf{1}_{\{T_{x/2}\ge t\}}\right]\\[4pt]&\le 2||f||_{[0,\infty]} \mathbb{P}_x(T_{x/2}\le t)+\sup_{y\ge \frac{x}{2}}|f(y)-f(\infty)|,\end{split}\]

where $||\cdot||_{[0,\infty]}$ is the supremum norm over $[0,\infty]$. The second term is smaller as soon as x is large enough, and the first part is small as soon as t is small enough, which completes the proof of Theorem 1.