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.**



Do Lyapunov functions imply exponential integrability of hitting times?
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 $E = \mathbb{R}$.


