Do Lyapunov functions imply exponential integrability of hitting times?

Question

I have a question of some integrability of hitting times.

Let $X=(\{X_t\}_{t \ge0},\{P_x\}_{x \in E})$ be a diffusion process on a locally compact separable metric space $E$.

We assume that there exist Borel measurable functions $f\colon E \to [1,\infty)$ and $g \colon E \to \mathbb{R}$ such that $\left\{f(X_t)-f(x)-\int_{0}^{t}g(X_s)\,ds\right\}$ is a local martingale in $P_x$ for any $x \in E$. By convention, we will write $\mathcal{L}f$ for $g$. We further assume that $f$ is bounded, and that there exist $\alpha,\beta \in (0,\infty)$ and compact subset $K \subset E$ such that \begin{align*} \mathcal{L}f(x) \le -\alpha f(x) +\beta\textbf{1}_{K}(x),\quad x \in E. \end{align*}

Under the conditions stated above, we can prove that \begin{align*} \sup_{x \in E}E_x[\sigma_K]<\infty, \end{align*} where $\sigma_K$ is the first hitting time of $K$. Indeed, fix $x \in E \setminus K$ and a localizing sequence $\{\tau_l\}_{l \ge 1}$ . Then, for any $l \ge 1$, $\left\{ f(X_{t\wedge \sigma_K \wedge \tau_l})-f(x)-\int_{0}^{t\wedge \sigma_K \wedge \tau_l}\mathcal{L}f(X_s)\,ds \right\}_{t \ge 0}$ is a $P_x$-martingale. Therefore, we obtain that for any $t>0$ and $l \ge 1$, \begin{align*} -E_{x} \left[\int_{0}^{t \wedge \tau_l \wedge \sigma_K} \mathcal{L}f(X_s)\,ds\right] &= f(x)-E_{x}[f(X_{t \wedge \tau_l \wedge \sigma_K})] \le f(x). \end{align*} Because $\mathcal{L}f \le -\alpha f$ on $E \setminus K$, it follows that \begin{align*} E_{x}[t \wedge \sigma_K \wedge \tau_l] \le f(x)/\alpha. \end{align*} Because $f$ is bounded, Fatou's lemma shows that $\sup_{x \in E \setminus K}E_x[\sigma_K]<\infty$. It clearly holds that $\sup_{x \in K}E_x[\sigma_K]=0.$

My question

We can take increasing compact subsets $\{K_n\}_{n=1}^{\infty}$ of $E$ such that $E=\bigcup_{n=1}^{\infty}K_n$. In this situation, I would want to expect that \begin{align*} (1)\quad \lim_{n \to \infty}\sup_{x \in E}E_{x}[\sigma_{K_n}]=0. \end{align*} Then, there exists $N \in \mathbb{N}$ such that for any $n \ge N$, $\sup_{x \in E}E_x[e^{\sigma_{K_n}}]<\infty$.

Can we prove $(1)$? If not, please tell me a counterexample.

Answer 1

It may depend on what exactly you mean by a diffusion on a general metric space.

Here is a counterexample for a discontinuous process. Take $E = \mathbb{R}$, and let $(X_t)$ be the process started at $x \in \mathbb{R}$ and jumping to the origin after a unit exponential time $\tau$. In other words, $(X_t)$ jumps to the origin at rate $1$, and then stays there forever. Take $f(x) = I \{x \ne 0 \}$. Then $g(x) = -f$. The conditions on $f$ are satisfied with $K = \{ 0\}$.

In this situation (1) fails because $\sup_{x \in E}E_{x}[\sigma_{K_n}] = \sup_{x \in E}E_{x}[\sigma_{K}] = E_{x} \tau = 1$.

EDIT. Here is a set-up that should give a counterexample with a continuous process. Take $E = [0, \infty) \times [0,1]$, and consider the system $$ dX _t = h(X_t, Y_t), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$ $$ dY _t = g(X_t, Y_t), \ \ \ \ \ \ \ (X_0, Y_0) \in E $$ with $h$ and $g$ satisfying the following conditions:

1. $h([0, \infty) \times (0, 1]) = \{0\}$.
2. $g([0, \infty) \times (0, 1] ) = \{-1\}$
3. $h(x, 0) = - \varphi (x)$, where $\varphi > 0$ is a function such that the ODE $ z' = \varphi (z), \ z(0) = 0$ escapes to infinity in a finite time (for example, $\varphi (x) = 1 + x ^2$) .
4. $g(x, 0 ) = 0$.

For $(x,y) \in E$, take now $f(x,y) = (y + t_x) \vee 1$, where $t_x$ is the time when the solution to the ode $$z' = \varphi (z), \ z(0) = 0$$ reaches $x$. It holds that $t_x\leq t_{expl}$, where $t_{expl}$ is the explosion time for $z$. Hence $f$ is bounded. Furthermore, $f(X_t, Y_t) = (f(X_0, Y_0) - t) \vee 1$, hence $f$ satisfy the inequality \begin{align*} \mathcal{L}f \le -\alpha f +\beta\textbf{1}_{K}, \end{align*}

with $\alpha = \frac{1}{\|f \|_{\infty}}$ and $K = \{ (0,0)\}$.

However, (1) is not satisfied because any compact $\mathcal{K} \subset E$ is bounded, and for $x \in (0,\infty)$ such that $\{x\} \times [0,1] \cap \mathcal{K} = \varnothing$ we have $E_{(x,1)}[\sigma_{\mathcal{K}}] \geq 1$ because $\sigma_{\mathcal{K}} \geq 1$ under $P_{(x,1)}$.

Remark. The example can be modified without much trouble to make $h$ and $g$ continuous. I have a feeling that (1) might actually be true if $E = \mathbb{R}$.