Suppose $X=\{X_t,t\in \mathbb{R}_+\}$ is a right-continuous Feller process in $\mathbb{R}^d$,$D$ is an open set in $\mathbb{R}^d$, denote $\tau=inf\{t>0,X_t\notin D\}$. Assume that it exists two constants $T>0$, $\delta>0$ such that for any $x\in D$, we have $P_x(\tau<T)>\delta$. Then how to prove that $P_x(\tau>nT)<(1-\delta)^n$ and $E_x(\tau)\leq \frac{T}{\delta}$?
It is trivial that $\tau$ is a stopping time. Then should I use the strong Markov property of $X_t$? Any help is welcomed!
Start with some $n$ and work backwards. Assume $n \geq 2$ since the $n=1$ case is assumed. Condition on the state of the system at time $(n-1)T$ and apply the Markov Property: $$ P_x(\tau>nT) = E_x[P_x(\tau>nT|{\cal F}_{(n-1)T}^+)] = E_x[1_{\{\tau > (n-1)T\}} P_{X_{(n-1)T}} (\tau'>T)], $$ where $\tau'$ is for a new process started at $X_{(n-1)T}$ (you might want to make the notation more rigorous here). On $\{\tau>(n-1)T\}$, $X_{(n-1)T} \in D$ and so $P_{X_{(n-1)L}}(\tau'>T) \leq 1-\delta$ by assumption. Therefore, $$ P_x(\tau>nT) \leq (1-\delta) P_x(\tau>(n-1)T). $$ By induction, you get the desired result.
As for the second claim, you can write the expectation as $$ E[\tau] = \int_0^\infty P(\tau>t)dt. $$ Split this into integrals from $0$ to $T$, $T$ to $2T$, and so on, upper bound each integral by a term of the form $T \cdot P(\tau>nT)$, and use $P(\tau>nT) \leq (1-\delta)^n$ and the fact that $\sum_{n=0}^\infty (1-\delta)^n = \frac1\delta$ to get the result.