3
$\begingroup$

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 $L^{1}(E,\mu)$.
$\endgroup$
8
  • 2
    $\begingroup$ Two remarks: (i) The implication "All $p_t$ are compact $\Rightarrow$ all $R_\alpha$ are compact" is true for every strongly continuous semigroup on every Banach space, so the question is actually about the converse implication. (ii) For the converse implication to be true on $L_2$ the symmetry assumption on the Markov process is essential, since this is not true for general Markov processes which have $\mu$ as an invariant measure. $\endgroup$ Jul 25, 2018 at 13:23
  • $\begingroup$ Do you have a counterexample? $\endgroup$
    – sharpe
    Jul 26, 2018 at 0:54
  • 1
    $\begingroup$ Let $\mathbb{T}$ denote the complex unit circle and consider the Markov process on $\mathbb{T}$ which maps any point $x \in \mathbb{T}$ to the point $e^{it}x$ with probability $1$ after time $t$. The associated transition semigroup is the shift semigroup which leaves the Lebegues measure $\lambda$ on $\mathbb{T}$ invariant; the restriction of the transition semigroup to $L^p(\mathbb{T},\lambda)$ for any $p \in [1,\infty)$ is a strongly continuous semigroup with compact resolvent, though none of its operators is compact on $L^p(\mathbb{T},\lambda)$. $\endgroup$ Jul 26, 2018 at 6:24
  • $\begingroup$ Is this Markov process $\lambda$-symmetric? $\endgroup$
    – sharpe
    Jul 26, 2018 at 8:51
  • $\begingroup$ No, the semigroup operators are shift operators whose adjoint operators are shifts into the converse direction. That's why I wrote in my first comment that "the symmetry assumption is essential". The above counterexample shows that your assertion on $L^2$ does not remain true without the symmetry assumption. Still, this does not answer your question, of course. $\endgroup$ Jul 26, 2018 at 9:02

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.