I have a question about operators of Markov processes.
Let $(E,\mu)$ be a locally compact separable metric measure space. Let $X=(X_t,P_x)$ be a $\mu$-symmetric Markov process on $E$. For any bounded $f:E \to \mathbb{R}$, we define $p_tf(x)$ by \begin{equation*} p_{t}f(x)=E_{x}[f(X_t)],\, x \in E,\ t>0. \end{equation*} By the Markov property of $X$, $p_t$ is extended linear operators on $L^{p}(E,\mu)$ for any $1\le p \le \infty$. The extensions are also denoted by $p_t$. It is known that $\{p_t\}_{t>0}$ is a strongly continuous contraction semigroup on $L^{p}(E,\mu)$ for any $1\le p<\infty$. If $p=\infty$, in general, $\{p_t\}$ does not have strong continuity: $\lim_{t \to 0}\|p_tf-f\|_{L^{p}(E,\mu)}=0$. Even if $p=\infty$, $\{p_t\}_{t>0}$ is contractive: $\|p_tf\|_{L^{p}(E,\mu)} \le \|f\|_{L^{p}(E,\mu)}$ for any $f \in L^{p}(E,\mu)$ and $t>0$.
For $p \in [1,\infty]$ and $f \in L^{p}(E,\mu)$, we can define \begin{equation*} R_{\alpha}f=\int_{0}^{\infty}\exp(-\alpha t)p_tf\,dt,\, \alpha>0. \end{equation*}
It is known that each $R_{\alpha}$ is a compact operator on $L^{2}(E,\mu)$ if and only if each $p_t$ is a compact operator.
Question
The following assertions are true?
- each $R_{\alpha}$ is a compact operator on $L^{1}(E,\mu)$ if and only if each $p_t$ is a compact operator on $L^{1}(E,\mu)$.
- If each $R_{\alpha}$ is a compact operator on $L^{1}(E,\mu)$, $\{p_t\}_{t>0}$ is eventually compact. Namely, for some $t>0$, $p_t$ is a compact operator on $L^{1}(E,\mu)$.