Hence, by Theorem 2.1, for all $\mu$ such that $\mu(\varphi_2) > 0$ and $\mu(\varphi_1) < +\infty$,

$$
\begin{align*}
|\mu(\eta_{n+m}) - \mu(\eta_n)| &= \mu(\eta_n) |\mathbb{E}_{\mu}(\eta_m(X_n) | n < \tau_{\partial}) - 1| \\
&= \mu(\eta_n) |\mathbb{E}_{\mu}(\eta_m(X_n) | n < \tau_{\partial}) - \nu_{QSD}(\eta_m)| \\
&\le C\mu(\varphi_1)\alpha^n \frac{\mu(\varphi_1)}{\mu(\varphi_2)}.
\end{align*}
$$

For any $x \in E$, applying this result to $\mu = (\delta_x + v_{QSD})/2$, we deduce that

$$
|\eta_{n+m}(x) - \eta_n(x)| \le C\varphi_1(x)^2\alpha^n.
$$

This shows that $(\eta_n(x))_{n \ge 0}$ is a Cauchy sequence and hence that, for all $x \in E$,

$$
\eta(x) = \lim_{n \to +\infty} \theta_0^{-n} \mathbb{P}_x(n < \tau_{\partial})
$$

and, by (10.1), that $\eta \in L^\infty(\varphi_1)$.

Then, since $\eta_n$ is bounded in $L^\infty(\varphi_1)$, we deduce by dominated convergence
that $v_{QSD}(\eta) = 1$ and that, for all $x \in E$,

$$
\delta_x P_1 \eta = \lim_{n \to +\infty} \delta_x P_1 \eta_n = \lim_{n \to +\infty} \theta_0 \eta_{n+1}(x) = \theta_0 \eta(x). \quad (10.2)
$$

The fact that $\eta$ is lower bounded away from 0 on $K$ is an immediate consequence of Lemma 9.9 (integrating (9.6) with respect to $v_{QSD}(dx)$) and the fact that $v_{QSD}(\varphi_1) < +\infty$.

It only remains to prove that $\eta \in L^\infty(\varphi_1^{\log\theta_0/\log\theta_1})$. To prove this, we use the operator $R_n$ introduced in (9.19). By Lemma 9.12 and using the fact that $\eta \in L^\infty(\varphi_1)$, for all $x \in E$,

$$
\begin{align*}
\eta(x) &= \theta_0^{-n} P_n \eta(x) \le C \theta_0^{-n} [R_n \varphi_1(x) + (P_n - R_n) \varphi_1(x)] \\
&\le C \bar{R} \theta_0^{-n} + C \left(\frac{\theta_1}{\theta_0}\right)^n \varphi_1(x).
\end{align*}
$$

Applying this inequality for $n = \lfloor -\log \varphi_1(x) / \log \theta_1 \rfloor$, we deduce

$$
\eta(x) \le C \exp \left( \frac{\log \varphi_1(x)}{\log \theta_1} \log \theta_0 \right) \le C \varphi_1(x)^{\log \theta_0 / \log \theta_1},
$$

which concludes the proof.
## 10.2 Proof of Theorem 2.3

We start with Point (i). We introduce $\Gamma_n = \mathbb{1}_{n<\tau_\partial}$ and define for all $x \in E'$ and $n \ge 0$ the probability measure

$$Q_n^{\Gamma, x} = \frac{\Gamma_n}{\mathbb{E}_x(\Gamma_n)} \mathbb{P}_x,$$

so that the Q-process exists if and only if $Q_n^{\Gamma, x}$ admits a proper limit when $n \to \infty$. For all $0 \le k \le n$, we have by the Markov property

$$\frac{\mathbb{E}_x(\Gamma_n | \mathcal{F}_k)}{\mathbb{E}_x(\Gamma_n)} = \frac{\mathbb{1}_{k<\tau_\partial} \mathbb{P}_{X_k}(n-k<\tau_\partial)}{\mathbb{P}_x(n<\tau_\partial)}.$$

By the pointwise convergence in (2.2) (proved in Subsection 10.1), this converges almost surely as $n \to +\infty$ to

$$M_k := \mathbb{1}_{k<\tau_\partial} \theta_0^{-k} \frac{\eta(X_k)}{\eta(x)} = \theta_0^{-k} \frac{\eta(X_k)}{\eta(x)},$$

and $\mathbb{E}_x(M_k) = \theta_0^{-k} \frac{P_k \eta(x)}{\eta(x)} = 1$. These two properties allow to apply the penalization's theorem of Roynette, Vallois and Yor [94, Theorem 2.1], which implies that $M$ is a martingale under $\mathbb{P}_x$ and that $Q_n^{\Gamma, x}(A)$ converges to $\mathbb{E}_x(M_k \mathbb{1}_A)$ for all $A \in \mathcal{F}_k$ when $n \to \infty$. This means that $\mathbb{Q}_x$ is well defined and

$$\left. \frac{d\mathbb{Q}_x}{d\mathbb{P}_x} \right|_{\mathcal{F}_k} = M_k.$$

Note that the fact that $\eta(x) = 0$ for all $x \in E \setminus E'$ implies that $(X_n, n \ge 0)$ is $E'$-valued $\mathbb{Q}_x$-almost surely for all $x \in E'$. The fact that $X$ is Markov under $(\mathbb{Q}_x)_{x \in E'}$ and Point (ii) can be easily deduced from the last formula (see e.g. [20, Section 6.1]).

It remains to prove Point (iii). We define the function $\psi = \varphi_1/\eta \times \|\eta\|_{L^\infty(\varphi_1)}$ on $E'$. Note that, since $\eta \in L^\infty(\varphi_1)$, $\psi$ is uniformly lower bounded. Moreover, for all $x \in E'$,

$$\tilde{P}_1 \psi(x) = \frac{\theta_0^{-1} \|\eta\|_{L^\infty(\varphi_1)}}{\eta(x)} P_1 \varphi_1(x) \le \frac{\theta_1}{\theta_0} \psi(x) + \frac{c_2 \|\eta\|_{L^\infty(\varphi_1)}}{\theta_0 \eta(x)} \mathbb{1}_K(x) \le \tilde{\theta} \psi(x) + \tilde{c},$$

where $\tilde{\theta} = \theta_1/\theta_0$ and

$$\tilde{c} = \frac{c_2 \|\eta\|_{L^\infty(\varphi_1)}}{\theta_0 \inf_K \eta}.$$

Hence, for all $x \in E$ and all $n \ge 1$,

$$\tilde{P}_n \psi(x) \le \tilde{\theta} \tilde{P}_{n-1} \psi(x) + \tilde{c} \le \dots \le \tilde{\theta}^n \psi(x) + \frac{\tilde{c}}{1-\tilde{\theta}}. \quad (10.3)$$
Using Lemma 9.5, we have that, for all $x \in E'$,

$$ \mathbb{Q}_x(T_K > n) = \mathbb{E}_x \left( \theta_0^{-n} \frac{\varphi_1(X_n)}{\eta(X_n)} \mathbb{1}_{T_K > n} \mathbb{1}_{X_n \in E'} \right) \le \tilde{\theta}^n \psi(x) \quad (10.4) $$

Now, choosing $m_K$ large enough so that $\sup_{x \in K} \tilde{\theta}^{m_K} [\sup_K \psi + \tilde{c}/(1-\tilde{\theta})] \le 1/2$, we deduce that, for all $x \in K$ and all $n_0 \ge 0$,

$$ \mathbb{Q}_x(\exists n \in \{n_0, \dots, n_0 + m_K\}, X_n \in K) \ge 1 - \tilde{\theta}^{m_K} \tilde{P}_{n_0} \psi(x) \ge 1/2. \quad (10.5) $$

Now, let $n_K \ge 1$ be such that $\inf_{x \in K} \mathbb{P}_x(X_n \in K) > 0$ for all $n \ge n_K$ (such a $n_K$ exists by (E1) and (E4), see the proof of Lemma 9.7) and let

$$ a := \inf_{n \in \{n_K, \dots, n_K + m_K\}} \inf_{x \in K} \mathbb{P}_x(X_n \in K) > 0. $$

so that, for all $x \in K$, all $n \in \{n_K, \dots, n_K + m_K\}$ and all $A \subset E$ measurable,

$$ \begin{aligned} \mathbb{Q}_x(X_{n+n_1} \in A) &\ge \mathbb{Q}_x(X_n \in K, X_{n+n_1} \in A) = \frac{\theta_0^{-n-n_1}}{\eta(x)} \mathbb{E}_x(\mathbb{1}_{X_n \in K} \mathbb{E}_{X_n}(\eta(X_{n_1}) \mathbb{1}_{X_{n_1} \in A})) \\ &\ge \frac{\theta_0^{-n-n_1}}{\eta(x)} ac_1 v(\eta \mathbb{1}_A) = \frac{\theta_0^{-n-n_1} v(\eta)}{\eta(x)} ac_1 v_\eta(A) \ge \frac{ac_1}{c_3} v_\eta(A), \end{aligned} $$

where we used (E1) and (E3) and defined $v_\eta(dx) := \frac{\eta(x)v(dx)}{v(\eta)}$. We deduce from the last inequality and (10.5) that, for all $n_0 \ge 0$,

$$ \mathbb{Q}_x(X_{n_0+n_1+n_K+m_K} \in \cdot) \ge \sum_{n=n_0}^{n_0+m_K} \mathbb{Q}_x [\mathbb{1}_{X_n \in K} \mathbb{Q}_{X_n} (X_{n_0+n_1+n_K+m_K-n} \in \cdot)] \ge \frac{ac_1}{2c_3} v_\eta. $$

Hence, for all $n \ge n_K + m_K + n_1$,

$$ \mathbb{Q}_x(X_n \in \cdot) \ge \frac{ac_1}{2c_3} v_\eta, \quad \forall x \in K. $$

For all $x \in E'$, setting $k_x = \lceil \frac{\log(2\psi(x))}{-\log\tilde{\theta}} \rceil$, it follows from (10.4) that $\mathbb{Q}_x(T_K \le k_x) \ge \frac{1}{2}$, and hence

$$ \mathbb{Q}_x(X_{k_x+n_K+m_K+n_1} \in \cdot) \ge \frac{ac_1}{4c_3} v_\eta. $$

In particular, for all $R > 0$, setting $k_R = \lceil \frac{\log(2R)}{-\log\tilde{\theta}} \rceil + n_K + m_K + n_1$, we have, for all $x, y \in E'$ such that $\psi(x) + \psi(y) < R$,

$$ \| \delta_x P_{k_R} - \delta_y P_{k_R} \|_{TV} \le 1 - \frac{ac_1}{4c_3}. \quad (10.6) $$
By [57, Thm 3.9], together with (10.3), the last assertion implies that there
exist constants $C > 0$ and $\tilde{\alpha}_1 \in (0, 1)$ such that, for all real function $h$ on $E'$ such
that $\|\|h\|$ < $\infty$,

$$
\|\| \tilde{P}_n h \|\| \le C \tilde{\alpha}_1^n \|\| h \|\|, \quad (10.7)
$$

where

$$
\|h\| = \sup_{x,y \in E'} \frac{|h(x) - h(y)|}{2 + \psi(x) + \psi(y)}.
$$

This implies (2.4). In particular, for all $x \in E'$,

$$
\| \delta_x \tilde{P}_n - \beta \|_{TV} \xrightarrow[n \to +\infty]{} 0.
$$

Hence, (2.5) is a consequence of Lebesgue's dominated convergence theorem.
This ends the proof of Theorem 2.3.

Remark 22. As noted in [57, Remark 3.10], it is possible to obtain explicit constants $\tilde{C}$ and $\tilde{\alpha}_1$ in (10.7) from the parameters in (10.3) and (10.6) (note that a slight modification of the proof of Lemma 9.9 entails that one can actually take $\tilde{c} = \frac{1/\theta_1}{1-\theta_1/\theta_0} \le \frac{1/\theta_1}{1-\theta_1/\theta_2}$). More precisely, setting $\alpha = \frac{ac_1}{4c_3}$ and $K = \tilde{c}/(1-\tilde{\theta})$ and $\gamma = \tilde{\theta}$, then taking any $\alpha_0 \in (0, \alpha)$ and $R > \frac{2K}{1-\gamma}$, and setting $b = \frac{2\alpha_0}{\gamma R+2K}$,

$$
\alpha_R = (1 - \alpha + \alpha_0) \vee \frac{2 + b\gamma R + b\gamma K}{2 + bR} \in (0, 1),
$$

and $C_R = \frac{2(b+1+K+K/(1-\gamma))}{\alpha_R}$, we obtain that, for all $f \in L^\infty(\varphi_1/\eta)$,

$$
|\tilde{P}_n f - \beta(f)| \le C_R \alpha_R^{n/k_R} \|f\|_{L^\infty(\varphi_1/\eta)}.
$$

**10.3 Proof of Corollary 2.4 and end of the proof of Theorem 2.2**

Let $|g| \le \varphi_1$ and set $h = g/\eta$. Then (2.4) entails that, for all $x \in E'$,

$$
|\theta_0^{-n} \mathbb{E}_x(g(X_n) \mathbb{1}_{X_n \in E'}) - \eta(x) \nu_{QSD}(g \mathbb{1}_{E'})| \le C \bar{\alpha}^n \varphi_1(x).
$$

In what follows, we set $v' = v_{QSD}(\cdot \cap E')$ and, for all $k \ge 1$,

$$
g_k(x) = \mathbb{1}_{x \in E'} \mathbb{E}_x (\mathbb{1}_{X_1 \notin E'} g(X_k) \mathbb{1}_{k < \tau_\delta}).
$$

Note that, defining $E'' := E \setminus E'$, $E'' \cup \{\emptyset\}$ is an absorbing set. Since $K \subset E'$, we thus
have

$$
g_k(x) \le \mathbb{1}_{x \in E'} \mathbb{E}_x (\mathbb{1}_{k < T_K \wedge \tau_\delta} \varphi_1(X_k)) \le \theta_1^k \varphi_1(x).
$$
We also define the measure $v''$ on $E''$ by

$$v'' = \sum_{\ell \ge 1} \theta_0^{-\ell} \mathbb{E}_{v'} (1_{X_1 \notin E'} 1_{X_\ell \in \cdot}) = \sum_{\ell \ge 1} \theta_0^{-\ell} v'(g_\ell).$$

Hence we have for all $n \ge 1$ and all $x \in E'$

$$\begin{aligned}
|\theta_0^{-n} \mathbb{E}_x(g(X_n) 1_{X_n \in E''}) - \eta(x) v''(g)| &\le \sum_{\ell=1}^n |\theta_0^{-n} \mathbb{E}_x(g_\ell(X_{n-\ell})) - \theta_0^{-\ell} \eta(x) v'(g_\ell)| \\
&\le \sum_{\ell=1}^n \theta_0^{-\ell} |\theta_0^{-(n-\ell)} \mathbb{E}_x(g_\ell(X_{n-\ell})) - \eta(x) v'(g_\ell)| \\
&\le \sum_{\ell=1}^n \theta_0^{-\ell} \varphi_1(x) C \bar{\alpha}^{n-\ell} \|g_\ell\|_{L^\infty(\varphi_1)} \\
&\le C \varphi_1(x) \sum_{\ell=1}^n \left(\frac{\theta_1}{\theta_0}\right)^\ell \bar{\alpha}^{n-\ell}.
\end{aligned}$$

We thus proved that, setting $v_0 = v' + v''$, and up to a change in the constants C and $\bar{\alpha}$, for all $x \in E'$,

$$|\theta_0^{-n} \mathbb{E}_x(g(X_n) 1_{n<\tau_\theta}) - \eta(x) v_0(g)| \le C \bar{\alpha}^n \varphi_1(x).$$

Now, by Lemma 9.5, for all $x \in'', |\mathbb{E}_x g(X_n) 1_{n<\tau_\theta}| \le \theta_1^n \varphi_1(x)$. This, we have proved that, up to a change in $\bar{\alpha}$, for all $x \in E$,

$$|\theta_0^{-n} \mathbb{E}_x(g(X_n)) - \eta(x) v_0(g)| \le C \bar{\alpha}^n \varphi_1(x). \quad (10.8)$$

Integrating with respect to $v_{QSD}$ shows that $v_0 = v_{QSD}$. Thus, we have proved (2.6).

To conclude, we can now end the proof of Theorem 2.2. Indeed, taking $g = 1$ immediately entails that the convergence (2.2) is geometric in $L^\infty(\varphi_1)$.

## 10.4 Proof of Corollary 2.5

If $\eta(x) > 0$, it follows from Corollary 2.4 and the fact that $v_{QSD}(\varphi_2) > 0$ that there exists $k \ge 0$ such that $\delta_x P_k \varphi_2 > 0$. Hence $E' \subset \{x \in E : \exists k \ge 0, P_k \varphi_2(x) > 0\}$. Conversely, if $P_k \varphi_2(x) > 0$, we apply Theorem 2.1 to $\mu = \frac{\delta_x P_k}{\delta_x P_k 1_E}$. Since $v_{QSD}(\eta) > 0$, there exists $n \ge 0$ such that $0 < \frac{\mu P_n \eta}{\mu P_n 1_E} = \frac{\delta_x P_{n+k} \eta}{\delta_x P_{n+k} 1_E} = \frac{\theta_0^{n+k}}{\delta_x P_{n+k} 1_E} \eta(x)$. Hence we have proved that $E' = \{x \in E : \exists k \ge 0, P_k \varphi_2(x) > 0\}$.

The fact that any $\mu$ such that $\mu(E') > 0$ and $\mu(\varphi_1^{1/p}) < +\infty$ for some $p < \log\theta_1 / \log\theta_2$ belongs to the domain of attraction of $v_{QSD}$ follows from Remark 1 and Corollary 2.4.
In the case where $\varphi_1$ is bounded, the domain of attraction contains all measures $\mu$ such that $\mu(E') > 0$. If $\mu(\eta) = 0$, then $\mu P_k \eta = 0$ for all $k \ge 0$, which means that $\mu P_k$ gives no mass to $E'$. Hence the convergence of conditional distributions to $v_{QSD}$ cannot hold true. The uniqueness of the quasi-stationary distribution follows immediately.

## 10.5 Proof of Corollary 2.6

Applying (2.5) with $\mu(\eta \cdot) / \mu(\eta)$ instead of $\mu$ and recalling that $\mu(E \setminus E') = 0$, we obtain

$$ \sup_{f:E' \to \mathbb{R}, \|f\|_{\infty} \le 1} \left| \theta_0^{-n} \frac{\mu P_n(\eta f)}{\mu(\eta)} - \beta(f) \right| \underset{n \to +\infty}{\longrightarrow} 0. $$

This entails the convergence result (2.7).

Assume from now on that $\eta$ is positive on $E$. Let $v$ be a quasi-stationary distribution on $E$ such that $v(\eta) < +\infty$ and denote by $\tilde{\theta}_0 \in (0, 1]$ the associated decay parameter, such that $P_v(X_n \in \cdot) = \tilde{\theta}_0^n v$ for all $n \ge 0$. Then, according to (2.7), we have, for all $g \in L^\infty(\eta)$,

$$ \left| \theta_0^{-n} \tilde{\theta}_0^n v(g) - v(\eta) v_{QSD}(g) \right| \underset{n \to +\infty}{\longrightarrow} 0. $$

This entails that $\tilde{\theta}_0 = \theta_0$ and that $v$ is proportional to $v_{QSD}$. Since they both are probability measures, we deduce that $v = v_{QSD}$, which concludes the proof of the second claim of Corollary 2.6.

Finally, assuming that $\eta$ is lower bounded away from 0 on $E$, we deduce from (2.7) with $g \equiv 1$ that, for all probability measure $\mu$ on $E$ such that $\mu(\eta) < +\infty$,

$$ \theta_0^{-n} \mu P^n 1_E \underset{n \to +\infty}{\longrightarrow} \mu(\eta) > 0. $$

This and (2.7) imply that

$$ \sup_{g:E \to \mathbb{R}, \|g\|_{L^\infty(\eta)} \le 1} \left| \mathbb{E}_\mu(g(X_n) | n < \tau_\partial) - v_{QSD}(g) \right| \underset{n \to +\infty}{\longrightarrow} 0, $$

hence (2.8) holds true and the proof of Corollary 2.6 is completed.

## 11 Proof of the results of Section 3

In this section are proved Lemma 3.1 in Subsection 11.1, Lemma 3.2 in Subsection 11.2, Proposition 3.3 in Subsection 11.3 and Lemma 3.4 in Subsection 11.4. Then we prove Theorem 3.5 in Subsection 11.5, Lemma 3.6 in Subsection 11.6 and Proposition 3.7 in Subsection 11.7.
## 11.1 Proof of Lemma 3.1

The function $\varphi_2$ defined in the statement satisfies, for all $x \in E$, $\varphi_2(x) \in [0, 1]$ and, for all $x \in K$, $\varphi_2(x) \ge \frac{\theta_2^{-1}-1}{\theta_2^{-\ell}-1} > 0$. Moreover, we have, for all $x \in E$,

$$P_1\varphi_2(x) = \theta_2\varphi_2(x) - \frac{\theta_2^{-1}-1}{\theta_2^{-\ell}-1} (\theta_2 \mathbb{1}_K(x) - \theta_2^{-\ell+1} P_\ell \mathbb{1}_K(x)) \ge \theta_2\varphi_2(x)$$

since $\ell$ is chosen such that $\theta_2^{-\ell} P_\ell \mathbb{1}_K(x) \ge \mathbb{1}_K(x)$ for all $x \in E$.

Our assumption also implies that there exists $n_0$ such that, for all $n \ge n_0$, $\theta_2^{-n} \inf_{x \in K} \mathbb{P}_x(X_n \in K) \ge 1$. Choosing $n_4(x) = n_0$ for all $x \in K$ entails (E4), which concludes the proof of Lemma 3.1.

## 11.2 Proof of Lemma 3.2

Assume that

$$\mathbb{E}_x(\theta_1^{-T_K \wedge \tau_\partial}) < +\infty \quad \forall x \in E \text{ and } \sup_{y \in K} \mathbb{E}_y(\mathbb{E}_{X_1}(\theta_1^{-T_K \wedge \tau_\partial}) \mathbb{1}_{1<\tau_\partial}) < +\infty$$

and set $\varphi_1(x) = \mathbb{E}_x(\theta_1^{-T_K \wedge [\tau_\partial]})$ for all $x \in E$. Then, for all $x \in E \setminus K$, using Markov's property at time 1,

$$P_1\varphi_1(x) = \mathbb{E}_x(\mathbb{E}_{X_1}(\theta_1^{-T_K \wedge [\tau_\partial]}) \mathbb{1}_{1<\tau_\partial}) \le \mathbb{E}_x(\theta_1^{-(T_K \wedge [\tau_\partial]-1)}) = \theta_1\varphi_1(x).$$

Moreover, for all $x \in K$, $P_1\varphi_1(x) \le \theta_1^{-1} \sup_{y \in K} \mathbb{E}_y(\mathbb{E}_{X_1}(\theta_1^{-T_K \wedge \tau_\partial}) \mathbb{1}_{1<\tau_\partial})$, and hence the first part of the lemma is proved.

Assume now that there exist two constants $C > 0$, $\theta_1 > 0$ and a function $\varphi_1: E \to [1, +\infty)$ such that $\sup_K \varphi_1 < +\infty$ and $P_1\varphi_1 \le \theta_1\varphi_1 + C \mathbb{1}_K$. Then, for all $n \ge 1$ and all $x \in E \setminus K$,

$$\mathbb{E}_x(\varphi_1(X_n) \mathbb{1}_{n<T_K \wedge \tau_\partial}) \le \theta_1^n \varphi_1(x).$$

Thus we deduce that, for all $x \in E$,

$$\mathbb{P}_x(n < T_K \wedge \tau_\partial) \le \theta_1^n \varphi_1(x).$$

In particular, for all $\theta > \theta_1$ and all $x \in E$,

$$\mathbb{E}_x(\theta^{-T_K \wedge \tau_\partial}) \le \frac{1}{\theta - \theta_1} \varphi_1(x) < +\infty.$$

We also deduce that

$$\sup_{x \in K} \mathbb{E}_x (\mathbb{E}_{X_1} (\theta^{-T_K \wedge \tau_\partial})) \le \frac{1}{\theta - \theta_1} \sup_{x \in K} P_1 \varphi_1(x) < +\infty.$$

This concludes the proof of Lemma 3.2.
11.3 Proof of Proposition 3.3

Condition (E4) implies that there exists $x_0 \in E$ such that $\mathbb{P}_{x_0}(X_{n_0} \in K) > 0$. We then deduce from our assumption (3.2) that Condition (E1) is satisfied with the probability measure $\nu$ on $K$ defined by

$$ \nu(\cdot) = \frac{\mathbb{P}_{x_0}(X_{n_0} \in \cdot \cap K)}{\mathbb{P}_{x_0}(X_{n_0} \in K)} $$

and the constants $c_1 = \mathbb{P}_{x_0}(X_{n_0} \in K)/C > 0$ and $n_1 = m_0$.

Let us now check Condition (E3) and the last part of Proposition 3.3. We define $T_K^{(n_0)} = \inf\{n \ge n_0 \text{ s.t. } X_n \in K\}$. Lemma 9.5 (which only makes use of Condition (E2)) implies that, for all $x \in E$, $\mathbb{P}_x(n < T_K \wedge \tau_\partial) \le \theta_1^n \varphi_1(x)$. Hence, for all $x \in E$ and all $n \ge n_0$,

$$ \begin{aligned} \mathbb{P}_x(n < \tau_\partial \wedge T_K^{(n_0)}) &= \mathbb{E}_x(\mathbb{1}_{n_0 < \tau_\partial} \mathbb{P}_{X_{n_0}}(n - n_0 < \tau_\partial \wedge T_K)) \\ &\le \theta_1^{n-n_0} \mathbb{E}_x(\mathbb{1}_{n_0 < \tau_\partial} \varphi_1(X_{n_0})) \\ &\le (\theta_1 + c_2)^{n_0} \theta_1^{n-n_0} \varphi_1(x). \end{aligned} $$

Since $\varphi_1 \ge 1$, we also have $\mathbb{P}_x(n < \tau_\partial) \le C\theta_1^n \varphi_1(x)$ for all $n < n_0$. Hence we proved that, for all $x \in E$ and $n \ge 0$,

$$ \mathbb{P}_x(n < \tau_\partial \wedge T_K^{(n_0)}) \le C\theta_1^n \varphi_1(x). \quad (11.1) $$

Therefore, for some constant $C > 0$,

$$ \begin{aligned} \mathbb{P}_x(n < \tau_\partial) &\le \mathbb{P}_x(n < \tau_\partial \wedge T_K^{(n_0)}) + \mathbb{P}_x(T_K^{(n_0)} \le n < \tau_\partial) \\ &\le C \varphi_1(x) \theta_1^n + \sum_{k=n_0}^{n} \mathbb{E}_x(\mathbb{1}_{T_K^{(n_0)}=k} \mathbb{P}_{X_k}(n-k < \tau_\partial)). \end{aligned} \quad (11.2) $$

Now, for all $x \in E$, all $y \in K$ and all $k \in \{n_0, \dots, n\}$, (3.2) and (11.1) entail

$$ \begin{aligned} \mathbb{E}_x(\mathbb{1}_{T_K^{(n_0)}=k} \mathbb{P}_{X_k}(n-k < \tau_\partial)) &\le \mathbb{E}_x(\mathbb{1}_{k-n_0 < T_K^{(n_0)} \wedge \tau_\partial} \mathbb{E}_{X_{k-n_0}}(\mathbb{1}_{X_{n_0} \in K} \mathbb{P}_{X_{n_0}}(n-k < \tau_\partial))) \\ &\le \mathbb{E}_x(\mathbb{1}_{k-n_0 < T_K^{(n_0)} \wedge \tau_\partial} C \mathbb{P}_y(n+m_0-k < \tau_\partial)) \\ &\le \theta_1^{k-n_0} \varphi_1(x) C \mathbb{P}_y(n-k < \tau_\partial), \end{aligned} $$

where the constant $C$ may change from line to line. Using Lemma 9.8, which only makes use of (E1), (E2) and (E4), there exists $n_6 \in \mathbb{Z}_+$ such that, for all $y \in K$ and for all $n, k \in \mathbb{Z}_+$ such that $n-k \ge n_6$,

$$ \begin{aligned} \mathbb{P}_y(n < \tau_\partial) &\ge \mathbb{P}_y(X_{n-k} \in K) \inf_{z \in K} \mathbb{P}_z(k < \tau_\partial) \\ &\ge \mathbb{P}_y(n-k < \tau_\partial) \inf_{T \ge n_6} \inf_{z \in K} \mathbb{P}_z(X_T \in K | T < \tau_\partial) \inf_{z \in K} P_k \varphi_2(z) \\ &\ge C'' \theta_2^k \mathbb{P}_y(n-k < \tau_\partial), \end{aligned} $$
where $C'' > 0$. Hence,

$$ \mathbb{E}_x \left( \mathbb{1}_{T_K^{(n_0)} = k} \mathbb{P}_{X_k} (n-k < \tau_\partial) \right) \le \varphi_1(x) \left( \frac{\theta_1}{\theta_2} \right)^k \frac{\theta_1^{-n_0} C}{C''} \mathbb{P}_y (n < \tau_\partial). $$

Now, we deduce from (11.2) and (11.1) that, for all $x \in E$ and all $y \in K$,

$$ \begin{aligned} \mathbb{P}_x(n < \tau_\partial) &\le C \varphi_1(x) \left[ \theta_1^n + \mathbb{P}_y(n < \tau_\partial) \sum_{k=1}^{n-n_6} \left( \frac{\theta_1}{\theta_2} \right)^k \right] + \mathbb{P}_x(T_K^{(n_0)} \wedge \tau_\partial \ge n-n_6) \\ &\le C \varphi_1(x) \left[ \theta_1^n + \mathbb{P}_y(n < \tau_\partial) + \theta_1^{n-n_6} \right] \\ &\le C \varphi_1(x) \mathbb{P}_y(n < \tau_\partial) \end{aligned} $$

since $\mathbb{P}_y(n < \tau_\partial) \ge \theta_2^n \inf_K \varphi_2$. This implies (E3) since $\sup_K \varphi_1 < \infty$.

## 11.4 Proof of Lemma 3.4

Combining Theorem 2.2 and the fact that $\inf_K \eta > 0$, we deduce that

$$ \liminf_{n \to +\infty} \inf_{x \in K} \theta_0^{-n} \mathbb{P}_x(n < \tau_\partial) > 0. $$

Let $\theta'_2 < \theta_0$. Using Lemma 9.8,

$$ \lim_{n \to +\infty} (\theta'_2)^{-n} \inf_{x \in K} \mathbb{P}_x(X_n \in K) = +\infty. $$

Hence the result follows from Lemma 3.1.

## 11.5 Proof of Theorem 3.5

We assume that Assumption (F) is satisfied. In Subsection 11.5.1, we prove that Assumption (E) holds true for the sub-Markovian semigroup $(P_n)_{n \ge 0}$ of the absorbed Markov process $(X_{nt_2}, n \in \mathbb{Z}_+)$.
In Subsection 11.5.2, we prove the existence of a quasi-stationary distribution for $(X_t)_{t \in I}$ with the claimed properties and in Subsection 11.5.3, we prove the convergence of $e^{\lambda_0 t} \mathbb{P}_x(t < \tau_\partial)$ to $\eta(x)$ for $t \in I$, $t \to +\infty$.

### 11.5.1 Proof of (E)

We fix $\theta_1 \in (\gamma_1^{t_2}, \gamma_2^{t_2})$ and set $\theta_2 = \gamma_2^{t_2}$. Let us first remark that the last line of Condition (F2) implies that $\gamma_2^{-t} \mathbb{P}_v(X_t \in L) \to +\infty$ when $t \to +\infty$. Hence, using Condition (F1), we deduce that

$$ \gamma_2^{-t} \inf_{x \in L} \mathbb{P}_x(X_t \in L) \xrightarrow{t \to +\infty} +\infty. \quad (11.3) $$
We consider a number $n_0 \in \mathbb{N}^*$ large enough so that $\gamma_2^{-t} \inf_{x \in L} \mathbb{P}_x(X_t \in L) \ge 1 \lor \frac{c_2}{\theta_1 - \gamma_1^{t_2}}$, for all $t \ge (n_0 - 1)t_2$ and we set

$$ \varphi_1 = \psi_1 \quad \text{and} \quad \varphi_2 = \frac{\gamma_2^{-t_2} - 1}{\gamma_2^{-n_0 t_2} - 1} \sum_{k=0}^{n_0-1} \gamma_2^{-k t_2} P_k \mathbb{1}_L. $$

**Step 1.** *Proof of (E2), (E4) and (E1) for $(P_n)_{n \in \mathbb{Z}_+}$.*

For all $x \in E \setminus L$, it follows from (F0) and the second line of (F2) that

$$ P_1 \psi_1(x) = \mathbb{E}_x \left( \psi_1(X_{t_2}) \mathbb{1}_{t_2 < \tau_L \wedge \tau_\partial} \right) + \mathbb{E}_x \left( \mathbb{1}_{\tau_L \le t_2 \wedge \tau_\partial} \mathbb{E}_{X_{\tau_L}} \left( \mathbb{1}_{t_2-s < \tau_\partial} \psi_1(X_{t_2-s}) \right) \Big|_{s=\tau_L} \right) \\ \le \gamma_1^{t_2} \psi_1(x) + \mathbb{P}_x(\tau_L \le t_2) c_2. $$

We define

$$ K = \{ y \in E, \mathbb{P}_y(\tau_L \le t_2) / \psi_1(y) \ge (\theta_1 - \gamma_1^{t_2}) / c_2 \}. $$

The second line of (F2) at time $t=0$ and the fact that $\theta_1 - \gamma_1^{t_2} < 1$ imply that $L \subset K$. Moreover, we have, for all $x \notin K$,

$$ P_1 \psi_1(x) \le \theta_1 \psi_1(x). \quad (11.4) $$

Hence, for all $x \in E$,

$$ P_1 \psi_1(x) \le \theta_1 \psi_1(x) + c_2 \mathbb{1}_K(x). \quad (11.5) $$

Note that it immediately follows from the definition of $K$ that $\sup_{x \in K} \psi_1(x) < \infty$. In particular, the first and third lines of (E2) are proved.

Moreover, using the Markov property provided by (F0) and the definition of $n_0$, we deduce that, for all $t \ge n_0 t_2$,

$$ \inf_{x \in K} \gamma_2^{-t} \mathbb{P}_x(X_t \in L) \ge \inf_{x \in K} \mathbb{P}_x(\tau_L \le t_2) \inf_{s \in [0, t_2]} \inf_{y \in L} \gamma_2^{-t} \mathbb{P}_y(X_{t-s} \in L) \ge 1, \quad (11.6) $$

where we used the fact that, for all $x \in K$, $\mathbb{P}_x(\tau_L \le t_2) \ge \frac{\theta_1 - \gamma_1^{t_2}}{c_2}$. In particular,

$$ P_1 \varphi_2 = \gamma_2^{t_2} \varphi_2 + \frac{\gamma_2^{-t_2} - 1}{\gamma_2^{-n_0 t_2} - 1} \left( \gamma_2^{-(n_0-1)t_2} P_{n_0} \mathbb{1}_L - \gamma_2^{t_2} \mathbb{1}_L \right) \ge \gamma_2^{t_2} \varphi_2 = \theta_2 \varphi_2. $$

In addition, for all $x \in K$,

$$ \varphi_2(x) \ge \frac{\gamma_2^{-t_2} - 1}{\gamma_2^{-n_0 t_2} - 1} \gamma_2^{-(n_0-1)t_2} \mathbb{P}_x(X_{n_0 t_2} \in L) \ge \frac{1 - \gamma_2^{t_2}}{\gamma_2^{-n_0 t_2} - 1}. $$
Hence (E2) is proved. Moreover, (11.6) also entails that (E4) holds true.

Fix $n_1 > n_0$ such that $n_1 t_2 - t_1 \ge n_0 t_2$. Condition (F1) and then (11.6) imply that, for all $x \in K$,

$$ \mathbb{P}_x(X_{n_1 t_2} \in \cdot \cap K) \ge \mathbb{P}_x(X_{n_1 t_2 - t_1} \in L) c_1 \nu(\cdot \cap L) \ge \gamma_2^{n_1 t_2 - t_1} c_1 \nu(\cdot \cap L). $$

Extending $\nu$ as a probability measure on $K$, we obtain (E1).

**Step 3. Estimation of the survival probability.**

Our goal here is to prove a version of Lemma 9.9, where (9.6) is replaced by

$$ \mathbb{P}_x(nt_2 < \tau_\partial) \le C \frac{\varphi_1(x)}{1 - \theta_1/\theta_2} \inf_{y \in L} \mathbb{P}_y(nt_2 < \tau_\partial), \quad \forall x \in E, \forall n \in \mathbb{N}. \quad (11.7) $$

Since the proof is similar, we only highlight the main differences. First, Lemma 9.8 only uses (E1), (E2) and (E4), so that there exist $n_6 \ge 1$ and $\zeta_1 > 0$ such that, for all $x \in K$ and all $n \ge n_6$,

$$ \delta_x P_n 1_K \ge \zeta_1 \delta_x P_n 1_E. $$

Hence, for all $x \in K$ and all $N \ge n_0 + n_6$, using (11.6),

$$ \delta_x P_N 1_L \ge \gamma_2^{n_0 t_2} \delta_x P_{N-n_0} 1_K \ge \zeta_1 \gamma_2^{n_0 t_2} \delta_x P_{N-n_0} 1_E \ge \zeta_1 \gamma_2^{n_0 t_2} \delta_x P_N 1_E. $$

Hence,

$$ \inf_{N \ge n_0 + n_6} \inf_{x \in K} \mathbb{P}_x(X_{Nt_2} \in L | Nt_2 < \tau_\partial) > 0. \quad (11.8) $$

Third, it follows from (F2) that, for all $x \in E \setminus L$,

$$ \mathbb{P}_x(nt_2 < \tau_L \wedge \tau_\partial) \le \gamma_1^{nt_2} \psi_1(x) = \theta_1^n \varphi_1(x). \quad (11.9) $$

and from (E2) that, for all $x \in E$,

$$ \mathbb{P}_x(nt_2 < \tau_\partial) \ge \gamma_2^{nt_2} \varphi_2(x). \quad (11.10) $$

Therefore, following the same lines as in (9.9) (replacing $K$ with $L$), we deduce from (11.9) and (11.10) that, for all $x \in E$

$$ \begin{aligned} \mathbb{P}_x(nt_2 < \tau_\partial) &\le \theta_1^n \varphi_1(x) + c_3 \int_0^{nt_2} \inf_{y \in L} \mathbb{P}_y((n - \lceil s/t_2 \rceil) t_2 < \tau_\partial) \mathbb{P}_x(\tau_L \wedge \tau_\partial \in ds) \\ &\le C \inf_{z \in L} \mathbb{P}_z(nt_2 < \tau_\partial) \varphi_1(x) + \frac{c_3 \gamma_2^{-t_2}}{c} \inf_{z \in L} \mathbb{P}_z(nt_2 < \tau_\partial) \mathbb{E}_x(\gamma_2^{-\tau_L \wedge \tau_\partial}), \end{aligned} $$

which entails (11.7), where we used in the second inequality the fact that

$$ \mathbb{P}_x(nt_2 < \tau_\partial) \ge c \gamma_2^{kt_2} \inf_{y \in L} \mathbb{P}_y((n-k)t_2 < \tau_\partial), \quad \forall x \in L, $$
which is deduced from (11.8) exactly as in Lemma 9.9.

Step 4. *Proof of (E3).*

Using (11.7) and the fact that $\sup_{x \in K} \varphi_1(x) < +\infty$, we deduce that there exists a constant $C > 0$ such that, for all $n \in \mathbb{N}$,

$$ \sup_{x \in K} \mathbb{P}_x(nt_2 < \tau_\partial) \le C \inf_{y \in L} \mathbb{P}_y(nt_2 < \tau_\partial). $$

Moreover, using the Markov property at time $n_0 t_2$ and (11.6), we have that, for all $t \ge 0$,

$$ \inf_{x \in K} \mathbb{P}_x(t < \tau_\partial) \ge \inf_{x \in K} \mathbb{P}_x(t + n_0 t_2 < \tau_\partial) \ge \gamma_2^{n_0 t_2} \inf_{y \in L} \mathbb{P}_y(t < \tau_\partial). $$

These inequalities imply (E3).

### 11.5.2 Existence of a quasi-stationary distribution for $(X_t)_{t \in I}$

Subsection 11.5.1 and Theorem 2.1 imply that there exists a probability measure $v_{QSD}$ on $E$ such that

$$ \mathbb{P}_{v_{QSD}}(X_{nt_2} \in \cdot | nt_2 < \tau_\partial) = v_{QSD}, \forall n \in \mathbb{Z}_+, $$

such that $v_{QSD}(\varphi_1) < \infty$ and $v_{QSD}(\varphi_2) > 0$, which is equivalent to $v_{QSD}(L) > 0$ because of the quasi-stationarity and the form of $\varphi_2$. For all $t \in [0, t_2]$, let us define the probability measure $v_t$ on $E$ by

$$ v_t = \mathbb{P}_{v_{QSD}}(X_t \in \cdot | t < \tau_\partial). $$

For all $n \in \mathbb{Z}_+$, we have, using the Markov property and the fact that $v_{QSD}$ is a quasi-stationary distribution for $(X_{nt_2})_{n \ge 0}$,

$$ \mathbb{P}_{v_t}(X_{nt_2} \in \cdot | nt_2 < \tau_\partial) = \mathbb{E}_{v_{QSD}}(\mathbb{P}_{X_{nt_2}}(X_t \in \cdot | t < \tau_\partial) | nt_2 < \tau_\partial) = \mathbb{P}_{v_{QSD}}(X_t \in \cdot | t < \tau_\partial), $$

hence $v_t$ is a quasi-stationary distribution for $(P_n)_{n \ge 0}$. Moreover, the third line of (F2) and the quasi-stationarity of $v_t$ imply that $v_t(L)$ is positive.

Fix $\rho_1 \in (\theta_1^{1/t_2}, \gamma_2)$. It follows from (11.9) that there exists a constant $C > 0$ such that, for all $x \in E$,

$$ \varphi'_1(x) := \mathbb{E}_x(\rho_1^{-\tau_L \wedge \tau_\partial}) \le C \varphi_1(x). $$

We also have that, for all $x \in E \setminus L$,

$$ \begin{aligned} \mathbb{E}_x(\mathbb{1}_{t_2 < \tau_L \wedge \tau_\partial} \varphi'_1(X_{t_2})) &= \rho_1^{t_2} \mathbb{E}_x(\mathbb{1}_{t_2 < \tau_L \wedge \tau_\partial} \rho_1^{-\tau_L \wedge \tau_\partial}) \\ &\le \rho_1^{t_2} \varphi'_1(x) \end{aligned} \quad (11.11) $$
and the inequality is trivial for $x \in L$. In addition, for all $t \in [0, t_2]$ and all $x \in L$, $\mathbb{E}_x(\varphi'_1(X_t) \mathbb{1}_{t<\tau_\partial}) \le C \mathbb{E}_x(\psi_1(X_t) \mathbb{1}_{t<\tau_\partial}) \le C c_2$. Hence Condition (F) is satisfied replacing $\gamma_1$ with $\rho_1$ and $\psi_1$ with $\varphi'_1$. Therefore, we can apply Step 1 to prove that (E) is satisfied with $\varphi'_1$ and $\varphi'_2$ where

$$ \varphi'_2 = \frac{\gamma_2^{-t_2} - 1}{\gamma_2^{-n'_0 t_2} - 1} \sum_{k=0}^{n'_0 - 1} \gamma_2^{-kt_2} P_k \mathbb{1}_L $$

for an integer $n'_0$ that can be chosen larger than $n_0$. We also deduce as in the beginning of Step 2 that $v_{QSD}$ is the unique quasi-stationary distribution of $(P_n)_{n \ge 0}$ such that $v_{QSD}(\varphi'_1) < \infty$ and $v_{QSD}(L) > 0$.

Moreover, by Markov's property at time $t$ we have for all $x \in E$ and $t \ge 0$,

$$ \begin{aligned} \varphi'_1(x) &= \mathbb{E}_x [\mathbb{1}_{t<\tau_L \wedge \tau_\partial} \rho_1^{-\tau_L \wedge \tau_\partial}] + \mathbb{E}_x [\mathbb{1}_{t \ge \tau_L \wedge \tau_\partial} \rho_1^{-\tau_L \wedge \tau_\partial}] \\ &\le \rho_1^{-t} \mathbb{E}_x [\mathbb{1}_{t<\tau_L \wedge \tau_\partial} \varphi'_1(X_t)] + \rho_1^{-t} \mathbb{P}_x (t \ge \tau_L \wedge \tau_\partial) \\ &\le \rho_1^{-t} (\mathbb{E}_x [\mathbb{1}_{t<\tau_\partial} \varphi'_1(X_t)] + 1) \end{aligned} \quad (11.12) $$

so that, for all $t \in [0, t_2]$,

$$ \begin{aligned} v_t(\varphi'_1) &\le \rho_1^{-(t_2-t)} [\mathbb{E}_{v_{QSD}} (\mathbb{1}_{t_2<\tau_\partial} \varphi'_1(X_{t_2})) / \mathbb{P}_{v_{QSD}} (t<\tau_\partial) + 1] \\ &\le \rho_1^{-(t_2-t)} [\mathbb{E}_{v_{QSD}} (\mathbb{1}_{t_2<\tau_\partial} \varphi'_1(X_{t_2})) / \mathbb{P}_{v_{QSD}} (t_2<\tau_\partial) + 1] \\ &= \rho_1^{-(t_2-t)} (v_{QSD}(\varphi'_1) + 1) < \infty. \end{aligned} $$

Since we observed that $v_t(L) > 0$, we deduce that $v_t = v_{QSD}$ for all $t \in I \cap [0, t_2]$.

Using the Markov property, we deduce that $v_t = v_{QSD}$ for all $t \in I$ and hence that $v_{QSD}$ is a quasi-stationary distribution for $(X_t)_{t \in I}$. Since any quasi-stationary distribution for $(X_t)_{t \in I}$ is also a quasi-stationary distribution for $(P_n)_{n \ge 0}$, we deduce that $v_{QSD}$ is the unique quasi-stationary distribution for $(X_t)_{t \in I}$ such that $v_{QSD}(\varphi_1) < +\infty$ and $v_{QSD}(L) > 0$.

Let $t \ge t_2$ be fixed and define $k \in \mathbb{N}$ such that $0 \le t - kt_2 < t_2$. It follows from the fact that $P_1 \varphi'_1 \le \bar{C} \varphi'_1$ and from (11.12) that

$$ \begin{aligned} \mathbb{E}_x [\mathbb{1}_{t<\tau_\partial} \varphi'_1(X_t)] &\le \bar{C}^k \mathbb{E}_x [\mathbb{1}_{t-kt_2<\tau_\partial} \varphi'_1(X_{t-kt_2})] \\ &\le \bar{C}^k \rho_1^{-(k+1)t_2+t} \mathbb{E}_x [\mathbb{1}_{t_2<\tau_\partial} \varphi'_1(X_{t_2}) + \mathbb{1}_{t-kt_2<\tau_\partial}] \\ &\le C \bar{C}^k \rho_1^{-(k+1)t_2+t} \mathbb{E}_x [\mathbb{1}_{t_2<\tau_\partial} \varphi_1(X_{t_2}) + 1] \\ &\le C \bar{C}^k \rho_1^{-(k+1)t_2+t} (\theta_1 + c_2 + 1) \varphi_1(x). \end{aligned} \quad (11.13) $$

Note that a similar inequality may not hold true with $\varphi'_1$ replaced by $\varphi_1$ under our assumptions. This explains why we need to introduce $\varphi'_1$.
Now, let $\mu$ be a probability measure such that $\mu(\varphi_1) < \infty$ and $\mu(\varphi_2) > 0$. Then, for all $t \ge n_0 t_2$, it follows from (11.6) that, for all $k \ge 0$,

$$ \mathbb{P}_{\mu}(X_{t+kt_2} \in L) \geq \mathbb{P}_{\mu}(X_{kt_2} \in L) \inf_{y \in L} \mathbb{P}_{y}(X_t \in L) \geq \gamma_2^{t_2} \mathbb{P}_{\mu}(X_{kt_2} \in L). $$

Therefore, for all $t \in [n_0 t_2, (n_0 + 1) t_2]$,

$$ \begin{aligned} \mathbb{E}_{\mu}(\varphi_2(X_t)) &= \frac{\gamma_2^{-t_2} - 1}{\gamma_2^{-n_0 t_2} - 1} \sum_{k=0}^{n_0-1} \gamma_2^{kt_2} \mathbb{P}_{\mu}(X_{t+kt_2} \in L) \\ &\geq \frac{\gamma_2^{-t_2} - 1}{\gamma_2^{-n_0 t_2} - 1} \gamma_2^{(n_0+1)t_2} \sum_{k=0}^{n_0-1} \gamma_2^{kt_2} \mathbb{P}_{\mu}(X_{kt_2} \in L) = \gamma_2^{(n_0+1)t_2} \mu(\varphi_2). \end{aligned} $$

This and inequality (11.13) imply that (using that $n_0' \ge n_0$), for all $t \in [n_0 t_2, (n_0 + 1) t_2]$ and for a constant $C > 0$ that may change from line to line,

$$ \frac{\mu_t(\varphi'_1)}{\mu_t(\varphi'_2)} \le C \frac{\mu_t(\varphi'_1)}{\mu_t(\varphi_2)} \le C \frac{\mu(\varphi_1)}{\mu(\varphi_2)}, $$

where $\mu_t := \mathbb{P}_\mu(X_t \in \cdot | t < \tau_\partial)$. It then follows the fact that (E) is satisfied by $(P_n, n \ge 0)$ with the functions $\varphi'_1$ and $\varphi'_2$ that there exist constants $\alpha < 1$ and $C > 0$ such that, for all $t \in [n_0 t_2, (n_0 + 1) t_2]$,

$$ \left\| \frac{\mu_t P_n}{\mu_t P_n \mathbb{1}_E} - v_{QSD} \right\|_{TV} \le C \alpha^n \frac{\mu(\varphi_1)}{\mu(\varphi_2)}, $$

Using Markov property, we deduce that

$$ \| \mathbb{P}_{\mu}(X_{nt_2+t} \in \cdot | nt_2 + t < \tau_{\partial}) - v_{QSD} \|_{TV} \le C \alpha^n \frac{\mu(\varphi_1)}{\mu(\varphi_2)}. $$

This ends the proof of (3.4).

### 11.5.3 Convergence to $\eta$

To finish the proof of Theorem 3.5, it remains to prove that the convergence (3.5) is exponential in $L^\infty(\psi_1^{1/p})$ and that $P_t\eta = e^{-\lambda_0 t}\eta$. Because of Remark 1, it is enough to prove this for $p = 1$. Since we proved that (E) holds true for the semi-group $(P_n)_{n \ge 0}$ and for the functions $\varphi'_1$ and $\varphi'_2$, it follows from Theorem 2.2 that there exist constants $\lambda_0 \in [0, \log(1/\gamma_2)]$, $\alpha \in (0, 1)$ and $C > 0$ such that, for all $y \in E$,

$$ |e^{\lambda_0 nt_2} \mathbb{P}_y(nt_2 < \tau_\partial) - \eta(y)| \le C \alpha^n \varphi'_1(y). $$
For any $t \in [t_2, 2t_2]$, integrating this inequality with respect to $\mathbb{P}_x(X_t \in dy; t < \tau_\partial)$, we deduce from (11.13) that

$$ \left| e^{\lambda_0 nt_2} \mathbb{P}_x(n t_2 + t < \tau_\partial) - \mathbb{E}_x(\eta(X_t) \mathbb{1}_{t<\tau_\partial}) \right| \le C \alpha^n \varphi_1(x) $$

for a constant C independent of $t \in [t_2, 2t_2]$. Setting $\eta_t(x) = \mathbb{E}_x[e^{\lambda_0 t} \eta(X_t) \mathbb{1}_{t<\tau_\partial}]$, we obtain for all $t \in [t_2, 2t_2]$

$$ \left| e^{\lambda_0 (nt_2+t)} \mathbb{P}_x(n t_2 + t < \tau_\partial) - \eta_t(x) \right| \le C e^{2\lambda_0 t_2} \alpha^n \varphi_1(x). $$

Proceeding as in (10.2), we deduce, letting $n \to +\infty$, that $P_1\eta_t = e^{-\lambda_0 t_2}\eta_t$. It then follows from Corollary 2.4 that $\eta_t(x) = \eta(x)v_{QSD}(\eta_t)$ for all $x \in E$. Since we proved above that $v_{QSD}$ is a quasi-stationary distribution with decay parameter $\lambda_0$, by definition of $\eta_t$, $v_{QSD}(\eta_t) = 1$ and thus $P_t\eta = e^{-\lambda_0 t}\eta$. This ends the proof of Theorem 3.5.

## 11.6 Proof of Lemma 3.6

Proceeding as in (11.11) and (11.12), we have that, for all $x \in E$ and $t \in I$,

$$ \mathbb{E}_x(\psi_1(X_{t_2}) \mathbb{1}_{t_2 < \tau_L \wedge \tau_\partial}) \le \gamma_1^{t_2} \psi_1(x) \quad \text{and} \quad \psi_1(x) \le \gamma_1^{-t} (\mathbb{E}_x[\mathbb{1}_{t < \tau_\partial} \psi_1(X_t)] + 1). $$

Therefore, for all $t \le t_2$ and all $x \in L$,

$$ \begin{aligned} \mathbb{E}_x[\mathbb{1}_{t < \tau_\partial} \psi_1(X_t)] &\le \gamma_1^{-(t_2-t)} \mathbb{E}_x\{[\mathbb{E}_{X_t}(\mathbb{1}_{t_2-t < \tau_\partial} \psi_1(X_{t_2-t})) + 1] \mathbb{1}_{t < \tau_\partial}\} \\ &\le \gamma_1^{-(t_2-t)} [\mathbb{E}_x(\mathbb{1}_{t_2 < \tau_\partial} \psi_1(X_{t_2})) + 1] \\ &\le c_2 := \gamma_1^{-t_2} \left[ \sup_{y \in L} \mathbb{E}_y(\mathbb{1}_{t_2 < \tau_\partial} \psi_1(X_{t_2})) + 1 \right]. \end{aligned} $$

This concludes the proof of Lemma 3.6.

## 11.7 Proof of Proposition 3.7

Let us first assume that (E) is satisfied with $\varphi_1$ bounded and (3.8) and prove that (3.7) holds true. Theorem 2.1 and Remark 1 entail that, for all $n \ge n'_4$,

$$ \begin{aligned} \left\| \frac{\mu P_n}{\mu P_n \mathbb{1}_E} - v_{QSD} \right\|_{TV} &\le \alpha^{n-n'_4} \frac{\|\varphi_1\|_{\infty}}{\inf_{x \in K} \varphi_2(x)} \frac{\mu P_{n'_4} \mathbb{1}_E}{\mu P_{n'_4} \mathbb{1}_K} \\ &\le \alpha^{n-n'_4} \frac{\|\varphi_1\|_{\infty}}{\underline{c} \inf_{x \in K} \varphi_2(x)}. \end{aligned} $$
Hence the convergence is uniform.

Let us now assume that (3.7) holds true. It was proved in [20] that this is equivalent to the following condition.

**Condition (A).** There exist positive constants $c_1, c_2$, a positive integer $k_0$ and a probability measure $v$ on $E$ such that

(A1) (Conditional Dobrushin coefficient) For all $x \in E$,

$$ \mathbb{P}_x(X_{k_0} \in \cdot | k_0 < \tau_{\partial}) \geq c_1 v. $$

(A2) (Global Harnack inequality) We have

$$ \sup_{k \in \mathbb{Z}_+} \frac{\sup_{y \in E} \mathbb{P}_y(k < \tau_{\partial})}{\mathbb{P}_v(k < \tau_{\partial})} \leq c_2. $$

Several consequences of Condition (A) were deduced in [20], among which the fact that the convergence (2.2) in Theorem 2.2 holds true with respect to the $L^\infty$ norm on $E$ with $\eta(x) > 0$ for all $x \in E$. In particular, $\eta$ is bounded, $P_1\eta = \theta_0\eta$ and there exists a constant $C'$ such that, for all $n \geq 0$,

$$ \sup_{x \in E} \mathbb{P}_x(n < \tau_{\partial}) \leq C' \theta_0^n. \qquad (11.14) $$

We fix $\varepsilon \in (0, 1/(4C'))$. Since $\eta$ is positive on $E$, there exists $\delta > 0$ such that the set $K := \{x \in E : \eta(x) \geq \delta\}$ satisfies $v_{QSD}(K) \geq 1 - \varepsilon$ and $v(K) > 0$. Setting $\varphi_2 = \eta/\|\eta\|_\infty$, the part of (E2) dealing about $\varphi_2$ is satisfied with $\theta_2 = \theta_0$. Since the convergence in Theorem 2.2 holds true with respect to the $L^\infty$ norm, we deduce from the choice of $K$ that there exists $k \geq k_0$ such that

$$ c := \inf_{x \in K} \mathbb{P}_x(k_0 < \tau_{\partial}) \geq \inf_{x \in K} \mathbb{P}_x(n < \tau_{\partial}) > 0. $$

It follows from (A1) and (A2) that, for all $n \geq 0$,

$$ \inf_{x \in K} \mathbb{P}_x(n < \tau_{\partial}) \geq \inf_{x \in K} \mathbb{P}_x(n + k_0 < \tau_{\partial}) \geq c_1 c \mathbb{P}_v(n < \tau_{\partial}) \geq \frac{c_1 c}{c_2} \sup_{y \in E} \mathbb{P}_y(n < \tau_{\partial}). $$

This implies (E3) and that $\inf_{x \in K} \mathbb{P}_x(k_0 < \tau_{\partial}) > 0$. Hence, (E1) follows from (A1) with the probability measure $\frac{v(\cdot \cap K)}{v(K)}$. Moreover, for any $n$ large enough to have $C\alpha^n \leq 1/2$ where the constants $C$ and $\alpha$ are those of (3.7), we have $\mathbb{P}_x(X_n \in K | t < \tau_{\partial}) \geq v_{QSD}(K) - C\alpha^n \geq 1/2 - \varepsilon > 0$ and hence (E4) is satisfied. The last computation also entails (3.8) with $n'_4 = n$.
It remains to construct a function $\varphi_1$ satisfying (E2) with $\theta_1 < \theta_0$. For all $x \in E$,

$$ \mathbb{P}_x(X_n \in E \setminus K | n < \tau_\partial) \leq \nu_{QSD}(E \setminus K) + C\alpha^n \leq \varepsilon + C\alpha^n. $$

Using (11.14), we deduce that

$$ \mathbb{P}_x(X_n \in E \setminus K) \leq C'(\varepsilon + C\alpha^n)\theta_0^n, $$

so that there exists $n_0$ large enough such that

$$ \mathbb{P}_x(n_0 < T_K \wedge \tau_\partial) \leq \frac{1}{3}\theta_0^{n_0} = \left(\frac{\theta_0}{3^{1/n_0}}\right)^{n_0}. $$

From this follows that, for all $k \in \mathbb{N}$ and all $x \in E$,

$$ \mathbb{P}_x(kn_0 < T_K \wedge \tau_\partial) \leq \left( \frac{\theta_0}{3^{1/n_0}} \right)^{kn_0}. $$

In particular, for $\theta_1 := \theta_0/2^{1/n_0}$,

$$ \varphi_1(x) := \mathbb{E}_x \left( \theta_1^{-T_K \wedge [\tau_\partial]} \right), \quad \forall x \in E, $$

is a bounded function on E and Lemma 3.2 implies that, for all $x \in E$,

$$ P_1 \varphi_1(x) \leq \theta_1 \varphi_1(x) + \|\varphi_1\|_{\infty} \mathbb{1}_K(x). $$

Since $\theta_1 < \theta_0$, (E2) is proved.

# 12 Proof of the results of Section 4.1

In order to prove Theorem 4.1, we check Condition (F). The goal of Subsection 12.1 is to give the construction of the process X and to check (F0) with $L = K_k$ for any $k \geq 1$. In Subsection 12.2, we explain how (F1) and (F3) can be deduced from general Harnack inequalities. Finally, Subsection 12.3 completes the proof of Theorem 4.1. The proof of Corollary 4.2 is then given in Subsection 12.4.

## 12.1 Construction of the diffusion process X and Markov property

The goal of this section is to construct a weak solution X to the SDE (4.1) with absorption out of D, and prove that it is Markov and satisfies a strong Markov
property at appropriate stopping times, enough to entail Condition (F0) for $L = K_k$ for any $k \ge 1$. We introduce the natural path space for the process $X$ as

$$ \mathcal{D} := \left\{ w : \mathbb{R}_+ \to D \cup \{\partial\} : \forall k \ge 1, w \text{ is continuous on } [0, \tau_k(w)] \right\} $$
$$ \text{and } w(t) = \partial, \forall t \ge \sup_{k \ge 1} \tau_k(w) $$

where $\tau_k(w) := \inf\{t \ge 0 : w_t \in D \setminus K_k\}$. Note that $\mathcal{D}$ contains functions which are not càdlàg since they may not have a left limit at $\tau_\partial$- and, indeed, it is easy to construct examples where $X$ is not càdlàg $\mathbb{P}$-a.s.<fnref n="2" /> Note also that this definition means that we are looking for a process $X$ such that

$$ \tau_\partial := \sup_{k \ge 1} \tau_{D \setminus K_k}, $$

which is the natural definition of $\tau_\partial$ when the left limit of $X$ at time $\tau_\partial$ does not exist.

We endow the path space $\mathcal{D}$ with its natural filtration

$$ \mathcal{F}_t = \sigma(w_s, s \le t) = \bigvee_{n \ge 1, 0 \le t_1 < t_2 < \dots < t_n \le t} \sigma(w_{t_1}, w_{t_2}, \dots, w_{t_n}) $$

and we follow the usual method which consists in constructing for all $x \in D$ a probability measure $P_x$ on $\mathcal{D}$ and a stochastic process $(B_t, t \ge 0)$ on $\mathcal{D} \times \mathcal{C}(\mathbb{R}_+, \mathbb{R}^r)$, such that $B$ is a standard $r$-dimensional Brownian motion under $P_x \otimes W^r$, where $W^r$ is the $r$-dimensional Wiener measure and such that $w_0 = x \ P_x \otimes W^r$-almost surely and the canonical process $(w_t, t \ge 0)$ solves the SDE (4.1) for this Brownian motion $B$ on the time interval $[0, \sup_k \tau_k(w))^3$.

For this construction, we use the fact that $b$ and $\sigma$ can be extended out of $K_k$ to $\mathbb{R}^d$ as globally Hölder and bounded functions $b_k$ and $\sigma_k$ and such that $\sigma_k$ is uniformly elliptic on $\mathbb{R}^d$. Hence (see e.g. [66, Rk. 5.4.30]) the martingale problem is well-posed for the SDE

$$ dX_t^k = b_k(X_t^k)dt + \sigma_k(X_t^k)dB_t. $$

<footnote n="2">²For example, one may consider $D$ the open disc of radius 1 centered at 0 in $\mathbb{R}^2$, $\sigma = \text{Id}$ and $b(x) = (-x_2\beta(|x|), x_1\beta(|x|))$ where $x = (x_1, x_2) \in D$. Decomposing the process in polar coordinates $(R_t, \theta_t) := (|X_t|, \arctan(X_t^{(1)} / X_t^{(2)}))$, the radius $R_t$ is a 2-dimensional Bessel process, and $X_t$ is sent to $\partial$ when $R_t$ hits 1 (in a.s. finite time). The angle $\theta_t$ is solution to $d\theta_t = R_t^{-1}dW_t - \beta(R_t)dt$ before $\tau_\partial$, for some Brownian motion $W$. Hence, if $\beta(r)$ converges sufficiently fast to $+\infty$ when $r \to 1$, $\theta_t$ a.s. converges to $-\infty$ when $t \to \tau_\partial$, so $X$ does not admit a left limit at time $\tau_\partial$.</footnote>

<footnote n="3">³Since $\sigma(x)$ is non-degenerate for all $x \in D$, the space $\mathcal{C}(\mathbb{R}_+, \mathbb{R}^r)$ equipped with the Wiener measure $W^r$ is only used to construct the Brownian path $B_t$ after time $\sup_k \tau_k(w)$ and could be omitted for our purpose since we only need to construct the process $B$ up to time $\sup_k \tau_k(w)$.</footnote>
Let us denote by $\mathbb{P}_x^k$ the solution to this martingale problem for the initial condition $x \in \mathbb{R}^d$. This is a probability measure on $\mathcal{C} := \mathcal{C}(\mathbb{R}_+, \mathbb{R}^d)$, equipped with its canonical filtration $(\mathcal{G}_t)_{t \ge 0}$.

For all $k \ge 1$, we define $\tau_k'(w) = \inf\{t \ge 0, w_t \notin \text{int}(K_k)\}$, where $\text{int}(K_k)$ is the interior of $K_k$. Since the paths $w \in \mathcal{D}$ or $\mathcal{C}$ are continuous at time $\tau_k'$ and $\mathbb{R}^d \setminus \text{int}(K_k)$ is closed, it is standard to prove that $\tau_k'$ is a stopping time for the canonical filtration $(\mathcal{F}_t)_{t \ge 0}$ on $\mathcal{D}$ and for the canonical filtration $(\mathcal{G}_t)_{t \ge 0}$ on $\mathcal{C}$. We define as usual the stopped $\sigma$-fields $\mathcal{F}_{\tau_k'}$ and $\mathcal{G}_{\tau_k'}$, and we define for all $x \in \text{int}(K_k)$ the restriction of $\mathbb{P}_x$ to $\mathcal{F}_{\tau_k'}$ as the restriction of $\mathbb{P}_x^k$ to $\mathcal{G}_{\tau_k'}$, where we can identify the events of the two filtrations since they both concern continuous parts of the paths. This construction is consistent for $k$ and $k+1$ (meaning that if $x \in K_k$, they give the same probability to events of $\mathcal{F}_{\tau_k}$) by uniqueness of the solutions $\mathbb{P}_x^k$ and $\mathbb{P}_x^{k+1}$ to the above martingale problems. Hence there exists a unique extension $\mathbb{P}_x$ of the above measures to $\bigvee_{k \ge 1} \mathcal{F}_{\tau_k'}$. Note that, because of the specific structure of the path space $\mathcal{D}$, we have

$$ \bigvee_{k \ge 1} \mathcal{F}_{\tau_k'} = \mathcal{F}_{\infty}. \qquad (12.1) $$

To check this, it suffices to observe that, for all $t \ge 0$ and all measurable $A \subset D \cup \{\partial\}$,

$$ \begin{aligned} \{w_t \in A\} &= \{t < \tau_{\partial}, w_t \in A \cap D\} \cup \{\tau_{\partial} \le t, \partial \in A\} \\ &= \left( \bigcup_{k \ge 1} \{t < \tau_k', w_t \in A \cap D\} \right) \cup \left( \bigcap_{k \ge 1} \{\tau_k' \le t, \partial \in A\} \right), \end{aligned} \qquad (12.2) $$

hence $\{w_t \in A\} \in \bigvee_{k \ge 1} \mathcal{F}_{\tau_k'}$, and, proceeding similarly, the same property holds for events of the form $\{w_{t_1} \in A_1, \dots, w_{t_n} \in A_n\}$.

We recall (see [66, Section 5.4]) that $(\mathbb{P}_x^k)_{x \in \mathbb{R}^d}$ forms a strong Markov family on the canonical space $\mathcal{C}$. Our goal is now to prove that the family of probability measures $(\mathbb{P}_x)_{x \in D \cup \{\partial\}}$, where $\mathbb{P}_{\partial}$ is defined as the Dirac measure on the constant path equal to $\partial$, forms a Markov kernel of probability measures, for which the strong Markov property applies at well-chosen stopping times.

We first need to prove that $(\mathbb{P}_x)_{x \in D}$ defines a kernel of probability measures, i.e. that $x \mapsto \mathbb{P}_x(\Gamma)$ is measurable for all events $\Gamma$ of $\mathcal{F}_{\infty}$. We prove it for an event of the form $\{w_t \in A\}$, the extension to events of the form $\{w_{t_1} \in A_1, \dots, w_{t_n} \in A_n\}$, and hence to all events of $\mathcal{F}_{\infty}$, being easy. This follows from (12.2):

$$ \begin{aligned} \mathbb{P}_x(w_t \in A) &= \lim_{k \to +\infty} \mathbb{P}_x(t < \tau_k', w_t \in A \cap D) + \mathbb{1}_{\partial \in A} \lim_{k \to +\infty} \mathbb{P}_x(\tau_k' \le t) \\ &= \lim_{k \to +\infty} \mathbb{P}_x^{k+1}(t < \tau_k', w_t \in A \cap D) + \mathbb{1}_{\partial \in A} \lim_{k \to +\infty} \mathbb{P}_x^{k+1}(\tau_k' \le t). \end{aligned} $$
Since all the probabilities in the right-hand side are measurable functions of $x$, so is $x \mapsto \mathbb{P}_x(w_t \in A)$.

Now, let us prove that $(X_t, t \ge 0)$ is Markov. It is well-known that this is implied by the following property: for all $n \ge 1$ and $0 \le t_1 \le \dots \le t_{n+1}$ and $A_1, \dots, A_{n+1}$ measurable subsets of $D \cup \{\partial\}$,

$$ \mathbb{P}_x(w_{t_1} \in A_1, \dots, w_{t_{n+1}} \in A_{n+1}) = \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, \dots, w_{t_n} \in A_n} \mathbb{P}_{w_{t_n}}(w_{t_{n+1}} - t_n \in A_{n+1}) \right]. $$

We prove this property only for $n=1$. It is easy to extend the proof to all values of $n \ge 1$. We have

$$ \mathbb{P}_x(w_{t_1} \in A_1, w_{t_2} \in A_2) = \mathbb{P}_x(w_{t_1} \in A_1, w_{t_2} \in A_2, \tau_\partial > t_2) \\ + \mathbb{P}_x(w_{t_1} \in A_1, t_1 < \tau_\partial \le t_2) \mathbb{1}_{\partial \in A_2} + \mathbb{P}_x(\tau_\partial \le t_1) \mathbb{1}_{\partial \in A_1 \cap A_2}. $$

Now, using that $(\mathbb{P}_x^k)_{x \in \mathbb{R}^d}$ is a Markov family for all $k \ge 1$,

$$ \begin{aligned} \mathbb{P}_x(w_{t_1} \in A_1, w_{t_2} \in A_2, \tau_\partial > t_2) &= \lim_{k \to \infty} \mathbb{P}_x(w_{t_1} \in A_1, w_{t_2} \in A_2, \tau_k > t_2) \\ &= \lim_{k \to \infty} \mathbb{P}_x^k(w_{t_1} \in A_1, w_{t_2} \in A_2, \tau_k > t_2) \\ &= \lim_{k \to \infty} \mathbb{E}_x^k \left[ \mathbb{1}_{w_{t_1} \in A_1, t_1 < \tau_k} \mathbb{P}_{w_{t_1}}^k (w_{t_2-t_1} \in A_2, \tau_k > t_2 - t_1) \right] \\ &= \lim_{k \to \infty} \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, t_1 < \tau_\partial} \mathbb{P}_{w_{t_1}} (w_{t_2-t_1} \in A_2, \tau_k > t_2 - t_1) \right] \\ &= \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, t_1 < \tau_\partial} \mathbb{P}_{w_{t_1}} (w_{t_2-t_1} \in A_2, \tau_\partial > t_2 - t_1) \right] \end{aligned} $$

and similarly

$$ \begin{aligned} \mathbb{P}_x(w_{t_1} \in A_1, t_1 < \tau_\partial \le t_2) \mathbb{1}_{\partial \in A_2} &= \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, t_1 < \tau_\partial} \mathbb{P}_{w_{t_1}} (\tau_\partial \le t_2 - t_1) \right] \mathbb{1}_{\partial \in A_2} \\ &= \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, t_1 < \tau_\partial} \mathbb{P}_{w_{t_1}} (\tau_\partial \le t_2 - t_1, w_{t_2-t_1} \in A_2) \right]. \end{aligned} $$

Since

$$ \mathbb{P}_x(\tau_\partial \le t_1) \mathbb{1}_{\partial \in A_1 \cap A_2} = \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, \tau_\partial \le t_1} \mathbb{P}_{w_{t_1}} (w_{t_2-t_1} \in A_2) \right], $$

we have proved that $\mathbb{P}_x(w_{t_1} \in A_1, w_{t_2} \in A_2) = \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1} \mathbb{P}_{w_{t_1}}(w_{t_2-t_1} \in A_2) \right]$. This ends the proof of the Markov property.

To conclude this subsection, let us prove that the strong Markov property holds for all stopping times $\tau_F$ where $F \subset D$ is closed in $D$. Note that $\tau_F$ is indeed a stopping time for the filtration $\mathcal{F}_t$ since $\tau_F = \sup_k \tau_F \wedge \tau'_k = \sup_k \tau_{(F \cup D^c) \cup \text{int}(K_k)^c}$,
where the complement is understood in $\mathbb{R}^d$, $(F \cup D^c) \cup \text{int}(K_k)^c$ is a closed subset of $\mathbb{R}^d$ and all $w \in \mathcal{D}$ is continuous at time $\tau_{(F \cup D^c) \cup \text{int}(K_k)^c}$. Let $x \in D$, $t_1, t_2, s \ge 0$ and $A, B \subset D$ be measurable sets. We proceed as above: first, observe that

$$ egin{aligned} \{w_{t_1} \in A, t_1 < \tau_F \le t_2, w_{\tau_F+s} \in B\} &= \bigcup_{\ell \ge 1} \{w_{t_1} \in A, t_1 < \tau_F \le t_2, w_{\tau_F+s} \in B, w_r \in K_\ell \ \forall r \in [0, \tau_F + s]\} \\ &= \bigcup_{\ell \ge 1} \{w_{t_1} \in A, t_1 < \tau_F \wedge \tau'_\ell \le t_2, w_{\tau_F \wedge \tau'_\ell + s} \in B, \tau'_\ell > \tau_F + s\}. \end{aligned} $$

Since $\tau_F \wedge \tau'_\ell$ is a $\mathcal{G}_t$-stopping time on $\mathcal{C}(\mathbb{R}_+, \mathbb{R}^d)$ and using the strong Markov property under $\mathbb{P}^\ell$, we deduce that

$$ egin{aligned} \mathbb{P}_x(w_{t_1} \in A, t_1 < \tau_F \le t_2, w_{\tau_F+s} \in B) &= \lim_{\ell \to +\infty} \mathbb{P}_x^\ell(w_{t_1} \in A, t_1 < \tau_F \wedge \tau'_\ell \le t_2, w_{\tau_F \wedge \tau'_\ell + s} \in B, \tau'_\ell > \tau_F + s) \\ &= \lim_{\ell \to +\infty} \mathbb{E}_x^\ell \left[ \mathbb{1}_{w_{t_1} \in A, t_1 < \tau_F \wedge \tau'_\ell \le t_2} \mathbb{P}_{w_{\tau_F \wedge \tau'_\ell}}^\ell (w_s \in B, s < \tau'_\ell) \right] \\ &= \lim_{\ell \to +\infty} \mathbb{E}_x^\ell \left[ \mathbb{1}_{w_{t_1} \in A, t_1 < \tau_F \le \tau'_\ell \wedge t_2} \mathbb{P}_{w_{\tau_F}}^\ell (w_s \in B, s < \tau'_\ell) \right] \\ &= \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A, t_1 < \tau_F \le \tau_\partial \wedge t_2} \mathbb{P}_{w_{\tau_F}} (w_s \in B, s < \tau_\partial) \right]. \end{aligned} $$

Similarly,

$$ egin{aligned} \mathbb{P}_x(w_{t_1} \in A, t_1 < \tau_F \le t_2, w_{\tau_F+s} = \partial) &= \lim_{\ell \to +\infty} \mathbb{P}_x^\ell(w_{t_1} \in A, t_1 < \tau_F \le t_2 \wedge \tau'_\ell, \tau'_\ell \le \tau_F + s) \\ &= \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A, t_1 < \tau_F \le t_2 \wedge \tau_\partial} \mathbb{P}_{w_{\tau_F}} (w_s = \partial) \right] \end{aligned} $$

and thus

$$ \mathbb{P}_x(w_{t_1} \in A, t_1 < \tau_F \le t_2, w_{\tau_F+s} \in B) = \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A, t_1 < \tau_F \le t_2 \wedge \tau_\partial} \mathbb{P}_{w_{\tau_F}} (w_s \in B) \right] $$

for all $A, B \subset D \cup \{\partial\}$ measurable. The previous computation extends without difficulty to prove

$$ egin{aligned} \mathbb{P}_x(w_{t_1} \in A_1, \dots, w_{t_n} \in A_n, t_n < \tau_F \le t_{n+1}, w_{\tau_F+s_1} \in B_1, \dots, w_{\tau_F+s_m} \in B_m) &= \mathbb{E}_x \left[ \mathbb{1}_{w_{t_1} \in A_1, \dots, w_{t_n} \in A_n, t_n < \tau_F \le t_{n+1}} \mathbb{P}_{w_{\tau_F}} (w_{s_1} \in B_1, \dots, w_{s_m} \in B_m) \right] \quad (12.3) \end{aligned} $$

for all $n, m \ge 1$, $0 \le t_1 \le \dots \le t_{n+1}$, $0 \le s_1 \le \dots \le s_m$ and $A_1, \dots, A_n, B_1, \dots, B_m \subset D \cup \{\partial\}$ measurable. This implies the strong Markov property at time $\tau_F$, in the sense that, for all $k \ge 1$, all $x \in E$ and all $\Gamma \in \mathcal{F}_\infty$,

$$ \mathbb{P}_x(w^{\tau_F} \in \Gamma | \mathcal{H}_{\tau_F}) = \mathbb{P}_{w_{\tau_F}}(\Gamma), \quad \mathbb{P}_x\text{-almost surely,} $$
where $w^{\tau_F} = (w_{\tau_F+s}, s \ge 0)$ and

$$ \mathcal{H}_{\tau_F} = \sigma\left( \begin{array}{l} \{w_{t_1} \in A_1, \dots, w_{t_n} \in A_n, t_n < \tau_F \le t_{n+1}\}, n \in \mathbb{N}, \\ 0 \le t_1 \le \dots \le t_{n+1}, A_1, \dots, A_n \in D \text{ measurable} \end{array} \right). $$

This form of strong Markov property at time $\tau_F$ is enough for our purpose, since it entails (F0) for $L = K_k$ for all $k \ge 1$.

## 12.2 Harnack inequalities

Our goal here is to check Conditions (F1) and (F3) for the diffusion process constructed above. We will make use of general Harnack inequalities of Krylov and Safonov [72].

**Proposition 12.1.** There exist a probability measure $v$ on $D$ and a constant $t_v > 0$ such that, for all $k \ge 1$, there exists a constant $b_k > 0$ such that

$$ \mathbb{P}_x(X_{t_v} \in \cdot) \ge b_k v(\cdot), \quad \forall x \in K_k. \qquad (12.4) $$

Moreover, for all $k \ge 1$ such that $K_k$ is non-empty,

$$ \inf_{t \ge 0} \inf_{x \in K_k} \frac{\mathbb{P}_x(t < \tau_\partial)}{\sup_{x \in K_k} \mathbb{P}_x(t < \tau_\partial)} > 0. \qquad (12.5) $$

*Proof.* Consider a bounded measurable function $f: D \to \mathbb{R}_+$ with $\|f\|_{\infty} \le 1$ and define the application $u: (t, x) \in \mathbb{R}_+ \times E \mapsto \mathbb{E}_x[1_{\{t < \tau_\partial\}} f(X_t)]$. It is proved in [26] using [72] that, for all $k \ge 1$, there exist two constants $N_k > 0$ and $\delta_k > 0$, which do not depend on $f$ (provided $\|f\|_{\infty} \le 1$), such that

$$ u(\delta_k^2, x) \le N_k u(2\delta_k^2, y), \quad \text{for all } x, y \in K_k \text{ such that } |x - y| \le \delta_k/2. \qquad (12.6) $$

Note that the proof given in [26] makes use of the following strong Markov property: for all open ball $B$ such that $B \subset K_k$ for some $k \ge 1$, all $x \in B$, $t \ge 0$ and all measurable $f: D \cup \{\partial\} \to \mathbb{R}_+$,

$$ \mathbb{E}_x[f(X_t)1_{\tau_{D\setminus B} \le t < \tau_\partial}] = \mathbb{E}_x[1_{\tau_{D\setminus B} \le t} \mathbb{E}_{X_{\tau_{D\setminus B}}} [f(X_{t-u})1_{\{t-u < \tau_\partial\}}]|_{u=\tau_{D\setminus B}}]. $$

This property follows from (12.3).

*Step 1: Proof of (12.4)*
Fix $x_1 \in D$ and $k_1 \ge 1$ such that $x_1 \in \text{int}(K_{k_1})$. Let $\nu$ denote the conditional law $\mathbb{P}_{x_1}(X_{\delta_{k_1}^2} \in \cdot | \delta_{k_1}^2 < \tau_\partial)$. Then, for all measurable $A \subset D \cup \{\partial\}$, Harnack's inequality (12.6) with $f = 1_A$ entails that, for all $x \in D$ such that $|x - x_1| < \frac{\delta_{k_1}}{2} \wedge d(x_1, D \setminus K_{k_1})$,

$$ \mathbb{P}_x(2\delta_{k_1}^2 \in A) \ge \frac{\mathbb{P}_{x_1}(\delta_{k_1}^2 < \tau_\partial)}{N_{k_1}} \nu(A). $$

Since the diffusion is locally elliptic and $D$ is connected, for all $k \ge 1$, there exists a constant $d_k > 0$ such that

$$ \inf_{x \in K_k} \mathbb{P}_x(X_1 \in B(x_1, (\delta_{k_1}/2) \wedge d(x_1, D \setminus K_{k_1}))) \ge d_k. $$

This and Markov's property entail that, for all $x \in K_k$,

$$ \mathbb{P}_x(X_{1+2\delta_{k_1}^2} \in \cdot) \ge d_k \frac{\mathbb{P}_{x_1}(\delta_{k_1}^2 < \tau_\partial)}{N_{k_1}} \nu. $$

This implies the first part of Proposition 12.1.

### Step 2: Proof of (12.5)

Fix $k \ge 1$ such that $K_k$ is non-empty and consider $\ell > k$ such that $K_k$ is included in one connected component of $\text{int}(K_\ell)$. For all $t \ge 2\delta_\ell^2$, the inequality (12.6) applied to $f(x) = \mathbb{P}_x(t - 2\delta_\ell^2 < \tau_\partial)$ and the Markov property entail that

$$ \mathbb{P}_x(t - \delta_\ell^2 < \tau_\partial) \le N_\ell \mathbb{P}_y(t < \tau_\partial), \text{ for all } x, y \in K_\ell \text{ such that } |x - y| \le \delta_\ell/2. $$

Since $s \mapsto \mathbb{P}_x(s < \tau_\partial)$ is non-increasing, we deduce that

$$ \mathbb{P}_x(t < \tau_\partial) \le N_\ell \mathbb{P}_y(t < \tau_\partial), \text{ for all } x, y \in K_\ell \text{ such that } |x - y| \le \delta_\ell/2. $$

Since $K_k$ has a finite diameter and is included in a connected component of $K_\ell$, we deduce that there exists $N'_k$ equal to some power of $N_\ell$ such that, for all $t \ge 2\delta_\ell^2$,

$$ \mathbb{P}_x(t < \tau_\partial) \le N'_k \mathbb{P}_y(t < \tau_\partial), \text{ for all } x, y \in K_k. $$

Now, for $t \le 2\delta_\ell^2$, we simply use the fact that, for all $x \in K_k$, $\mathbb{P}_x(2\delta_\ell^2 < \tau_\partial) \ge \mathbb{P}_x(2\delta_\ell^2 < \tau_B)$ where $B = (x, 1/2k)$ and hence $x \mapsto \mathbb{P}_x(2\delta_\ell^2 < \tau_\partial)$ is uniformly bounded from below on $K_k$ by a constant $1/N_k'' > 0$. In particular,

$$ \mathbb{P}_x(t < \tau_\partial) \le 1 \le N_k'' \mathbb{P}_y(2\delta_\ell^2 < \tau_\partial) \le N_k'' \mathbb{P}_y(t < \tau_\partial), \text{ for all } x, y \in K_k. $$

This concludes the proof of Proposition 12.1. □
12.3 Proof of Theorem 4.1

Our aim is to prove that Condition (F) holds true with $L = K_k$ for some $k \ge 1$
large enough. We have already proved (F0), (F1) and (F3) with $L = K_k$ for any
$k \ge 1$. Hence we only have to check (F2). Fix $\rho_1 \in (\lambda_0, \lambda_1)$, $\rho_2 \in (\lambda_0, \rho_1)$ and
$p \in (1, \lambda_1 / \rho_1)$ and define

$$
\psi_1(x) = \varphi(x)^{1/p}, \quad \forall x \in D. \tag{12.7}
$$

Fix $\rho'_1 \in (\rho_1, \lambda_1/p)$ and

$$
t_2 \geq \frac{2s_1(C + \lambda_1)}{\lambda_1 - p\rho'_1} \lor \frac{\log 2}{\rho'_1 - \rho_1},
$$

where the constant $C$ comes from (4.5). Set $L = K_{k_0}$ with $k_0$ large enough so that
$v(K_{k_0}) > 0$ and, using (4.6),

$$
\mathbb{P}_x(s_1 < \tau_{K_{k_0}} \wedge \tau_\partial) \le e^{-(\rho'_1+C/p)t_2}
$$

for all $x \in D_0$.

From the definition of $\lambda_0$ and applying the same argument as in Step 2 of the
proof of Proposition 12.1 with $f(x) = \mathbb{P}_x(X_{t-2\delta_l^2} \in L)$ with $\ell$ large enough to have
$K_{k_0}$ included in one connected component of $K_\ell$, we deduce that

$$
\liminf_{t \to +\infty} e^{\rho_2 t} \inf_{x \in L} \mathbb{P}_x (X_t \in L) = +\infty,
$$

and hence the last line of (F2) is proved with $\gamma_2 = e^{-\rho_2}$.

Let us now check that the first line of Assumption (F2) holds true for all $x \in D_0$ and then for all $x \in D \setminus D_0$. For all $x \in D_0$, we have $\psi_1(x) \le \sup_{x \in D_0} \varphi^{1/p}(x) < +\infty$, and hence, for all $t \in [s_1, t_2]$, using Hölder's inequality and the definition of $k_0$,

$$
\begin{align}
\mathbb{E}_x \left( \psi_1(X_t) \mathbb{1}_{t < \tau_L \wedge \tau_\partial} \right) &\le \mathbb{E}_x \left( \mathbb{1}_{t < \tau_\partial} \varphi(X_t) \right)^{1/p} \mathbb{P}_x \left( t < \tau_L \wedge \tau_\partial \right)^{\frac{p-1}{p}} \nonumber \\
&\le \varphi(x)^{1/p} e^{C t_2 / p} \mathbb{P}_x \left( s_1 < \tau_L \wedge \tau_\partial \right)^{\frac{p-1}{p}} \tag{12.8} \\
&\le e^{-\rho_1 t_2} \le e^{-\rho_1 t_2} \psi_1(x). \nonumber
\end{align}
$$

To prove (12.8), we used the fact that $\mathcal{L}\varphi \le C \le C\varphi$ and Itô's formula to obtain
$P_t\varphi \le e^{Ct}\varphi$. Since this argument is used repeatedly in the sequel, we give it in
details for sake of completeness. It follows from Itô's formula that, for all $k \ge 1$,
$\mathbb{P}_x$-almost surely,

$$
e^{-C(t \wedge \tau_{K_k^c})} \varphi(X_{t \wedge \tau_{K_k^c}}) = \varphi(x) + \int_0^t \mathbb{1}_{s \le \tau_{K_k^c}} e^{-Cs} (\mathcal{L}\varphi(X_s) - C\varphi(X_s)) \mathrm{d}s \\
+ \int_0^t \mathbb{1}_{s \le \tau_{K_k^c}} e^{-Cs} \nabla\varphi(X_s)^* \sigma(X_s) \mathrm{d}B_s.
$$
Since $\nabla\varphi(x)$ and $\sigma(x)$ are uniformly bounded on $K_k$, the last term has zero expectation, and thus

$$ \mathbb{E}_x \left[ e^{-C(t \wedge \tau_{K_k}^c)} \varphi(X_{t \wedge \tau_{K_k}^c}) \right] \leq \varphi(x). $$

Letting $k \to +\infty$, we deduce form Fatou's lemma that

$$ \mathbb{E}_x \left[ e^{-Ct} \mathbb{1}_{t < \tau_{\partial}} \varphi(X_t) \right] \leq \varphi(x) \quad (12.9) $$

as claimed.

This proves the second line of (F2) for all $x \in D_0$ and $\gamma_1 = e^{-\rho_1}$.

Now, for all $x \in D \setminus D_0$, since $D_0$ is closed in $D$, it follows from the strong Markov property (12.3) at time $\tau_{D_0}$ that

$$ \begin{aligned} \mathbb{E}_x \left( \psi_1(X_{t_2}) \mathbb{1}_{t_2 < \tau_L \wedge \tau_\partial} \right) &= \mathbb{E}_x \left( \mathbb{1}_{t_2-s_1 < \tau_L \wedge \tau_\partial \wedge \tau_{D_0}} \mathbb{E}_{X_{t_2-s_1}} \left( \psi_1(X_{s_1}) \mathbb{1}_{s_1 < \tau_L \wedge \tau_\partial} \right) \right) \\ &\quad + \mathbb{E}_x \left( \mathbb{1}_{\tau_{D_0} \le t_2 - s_1} \mathbb{E}_{X_{\tau_{D_0}}} \left( \psi_1(X_{t_2-u}) \mathbb{1}_{t_2-u < \tau_\partial \wedge \tau_L} \right) \Big|_{u=\tau_{D_0}} \right). \end{aligned} \quad (12.10) $$

Using Hölder's inequality and (12.9), we deduce that, for all $y \in D$,

$$ \mathbb{E}_y \left( \psi_1(X_{s_1}) \mathbb{1}_{s_1 < \tau_L \wedge \tau_\partial} \right) \le \mathbb{E}_y \left( \varphi(X_{s_1}) \mathbb{1}_{s_1 < \tau_\partial} \right)^{1/p} \le e^{\frac{s_1 C}{p}} \varphi(y)^{1/p} = e^{\frac{s_1 C}{p}} \psi_1(y). $$

Hence, the first term in the right-hand side of (12.10) satisfies

$$ \mathbb{E}_x \left( \mathbb{1}_{t_2-s_1 < \tau_L \wedge \tau_\partial \wedge \tau_{D_0}} \mathbb{E}_{X_{t_2-s_1}} \left( \psi_1(X_{s_1}) \mathbb{1}_{s_1 < \tau_L \wedge \tau_\partial} \right) \right) \le e^{\frac{s_1 C}{p}} \mathbb{E}_x \left( \mathbb{1}_{t_2-s_1 < \tau_L \wedge \tau_\partial \wedge \tau_{D_0}} \psi_1(X_{t_2-s_1}) \right). $$

As a consequence, using again Hölder's inequality and applying as above Itô's formula using that $\mathcal{L}\varphi(x) \le -\lambda_1\varphi(x)$ for all $x \notin D_0$, one has

$$ \begin{aligned} \mathbb{E}_x \left( \mathbb{1}_{t_2-s_1 < \tau_L \wedge \tau_\partial \wedge \tau_{D_0}} \mathbb{E}_{X_{t_2-s_1}} \left( \psi_1(X_{s_1}) \mathbb{1}_{s_1 < \tau_L \wedge \tau_\partial} \right) \right) &\le e^{-\lambda_1 \frac{t_2-s_1}{p}} e^{\frac{s_1 C}{p}} \varphi(x)^{1/p} \\ &\le e^{-t_2 \frac{\rho_1 + \lambda_1}{2}} \psi_1(x), \end{aligned} $$

where we used in the last inequality that $t_2 \ge \frac{2s_1(C+\lambda_1)}{\lambda_1 - p\rho_1}$. Moreover, using (12.8), we obtain that the second term in the right-hand side of (12.10) satisfies

$$ \begin{aligned} \mathbb{E}_x \left( \mathbb{1}_{\tau_{D_0} \le t_2 - s_1} \mathbb{E}_{X_{\tau_{D_0}}} \left( \psi_1(X_{t_2-u}) \mathbb{1}_{t_2-u < \tau_\partial \wedge \tau_L} \right) \Big|_{u=\tau_{D_0}} \right) &\le e^{-\rho_1' t_2} \mathbb{P}_x (\tau_{D_0} \le t_2 - s_1) \le e^{-\rho_1' t_2} \psi_1(x). \end{aligned} $$
We finally deduce from (12.10) that, for all $x \in D \setminus D_0$,

$$ \mathbb{E}_x(\psi_1(X_{t_2}) \mathbb{1}_{t_2 < \tau_L \wedge \tau_\partial}) \le 2e^{-\rho_1' t_2} \psi_1(x) \le e^{-\rho_1 t_2} \psi_1(x), $$

where we used that $t_2 \ge \log 2 / (\rho_1' - \rho_1)$. This concludes the proof that the first line of (F2) holds true with $\gamma_1 = e^{-\rho_1}$.

Since $\varphi$ is locally bounded, $\sup_L \varphi < \infty$, and hence, using again (12.9), we deduce that, for all $t \ge 0$,

$$ \sup_{x \in L} \mathbb{E}_x(\psi_1(X_t) \mathbb{1}_{t < \tau_\partial}) \le \sup_{x \in L} \mathbb{E}_x(\varphi(X_t) \mathbb{1}_{t < \tau_\partial}) \le e^{Ct} \sup_{x \in L} \varphi(x) < \infty, $$

which implies the second line of Assumption (F2).

In addition, because of the local uniform ellipticity of the diffusion $X$, for all $n_0 \ge 1$, $\psi_2 := \sum_{k=0}^{n_0} P_k \mathbb{1}_L$ is uniformly bounded away from zero on all compact subsets of $D$. This and Theorem 3.5 concludes the proof of Theorem 4.1.

## 12.4 Proof of Corollary 4.2

Using Theorem 3.5, there exists $\lambda'_0$ such that, for all $x \in D$,

$$ \eta(x) = \lim_{t \to +\infty} e^{\lambda'_0 t} \mathbb{P}_x(t < \tau_\partial). $$

We choose in the definition of $\lambda_0$ a ball $B$ such that $v_{QSD}(B) > 0$ (recall that $\lambda_0$ is independent of the choice of $B$). Given $x \in D$ such that $\eta(x) > 0$,

$$ \lim_{t \to +\infty} e^{\lambda'_0 t} \mathbb{P}_x(X_t \in B) = \eta(x) v_{QSD}(B) \in (0, +\infty). $$

Hence, $\lambda_0 = \lambda'_0$ and the infimum in the definition of $\lambda_0$ is a minimum. The facts that $\mathbb{P}_{v_{QSD}}(t < \tau_\partial) = e^{-\lambda_0 t}$ and $P_t \eta = e^{-\lambda_0 t} \eta$ are then direct consequences of Theorem 3.5.

Let us now prove that $\eta$ is $\mathcal{C}^2$. First, it follows from [97, Theorem 7.2.4] that $x \mapsto e^{\lambda_0 t} \mathbb{P}_x(t < \tau_\partial)$ is continuous for all $t \ge 0$ (see e.g. [26] for a detailed proof). Hence the uniform convergence in Theorem 2.2 implies that $\eta$ is continuous on $D$.

Now, let $B$ be any non-empty open ball such that $\bar{B} \subset D$. We consider the following initial-boundary value problem (in the terminology of [50]) associated to the differential operator $\mathcal{L}$ defined in (4.3)

$$ \begin{cases} \partial_t u(t,x) - \mathcal{L}u(t,x) - \lambda_0 u(t,x) = 0 & \text{for all } (t,x) \in (0, T] \times B, \\ u(0,x) = \eta(x) & \text{for all } x \in B, \\ u(t,x) = \eta(x) & \text{for all } (t,x) \in (0, T] \times \partial B. \end{cases} $$
Since the coefficients of $\mathcal{L}$ are Hölder and uniformly elliptic in $\bar{B}$ and since $\eta$ is continuous, we can apply Corollary 1 of Chapter 3 of [50] to obtain the existence and uniqueness of a solution $u$ to the above problem, continuous on $[0, T] \times \bar{B}$ and $\mathcal{C}^{1,2}((0, T] \times B)$. Now, we can apply Itô's formula to $e^{\lambda_0 s} u(T - s, X_s)$: for all $s < \tau_{B^c} \wedge T$ and all $x \in B$, $\mathbb{P}_x$-almost surely,

$$e^{\lambda_0 s} u(T - s, X_s) = u(T, x) + \int_0^s e^{\lambda_0 r} \left( -\frac{\partial u}{\partial t} + \mathcal{L}u + \lambda_0 u \right) (T - r, X_r) dr + \int_0^s e^{\lambda_0 r} \nabla u(T - r, X_r) \sigma(X_r) dB_r.$$

Since $u$ is bounded and continuous on $[0, T] \times \bar{B}$ and $\nabla u(t, x)$ is locally bounded in $(0, T] \times B$, it follows from standard localization arguments that

$$\begin{aligned} u(T, x) &= \mathbb{E}_x \left[ e^{\lambda_0 (T \wedge \tau_{B^c})} u(T - (T \wedge \tau_{B^c}), X_{T \wedge \tau_{B^c}}) \right] \\ &= \mathbb{E}_x \left[ e^{\lambda_0 (T \wedge \tau_{B^c})} \eta(X_{T \wedge \tau_{B^c}}) \right]. \end{aligned}$$

Now, the Markov property and the fact that $P_t \eta = e^{-\lambda_0 t} \eta$ entail that $e^{\lambda_0 t} \eta(X_t)$ is a martingale on $(\mathcal{D}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P}_x)$, hence

$$\eta(x) = \mathbb{E}_x \left[ e^{\lambda_0 (T \wedge \tau_{B^c})} \eta(X_{T \wedge \tau_{B^c}}) \right] = u(T, x).$$

Therefore, $\eta \in \mathcal{C}^2(D)$ and $\mathcal{L}\eta(x) = -\lambda_0 \eta(x)$ for all $x \in D$.

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