Proving strong continuity of semigroup in $L^2(\mu)$ ($\mu$ invariant measure) from non-cadlag Markov process

Question

I am trying to derive the definition of a Markov semigroup from Section 1.1 Bakry, Gentil, Ledoux Analysis and Geometry of Markov Diffusion Operators from a Markov process. I am unsure about my proof of strong continuity of the semigroup on $L^2(\mu)$ for an invariant measure $\mu$. Here is my work:

Suppose we are given a Markov process $X_t, t \geq 0$ on a Polish space $\mathcal X$. The semigroup is defined as

\begin{equation} \mathbf{P}_{t}f(x) = \mathbb E [f(X_t)|X_0=x]. \end{equation} on bounded measurable functions $\mathcal M_b (\mathcal X)$.

The semigroup property, mass conservation $\mathbf{P}_{t} 1 = 1$ and positivity preservation $f \geq 0 \Rightarrow \mathbf{P}_{t} f \geq 0$ are straightforward to show from the definition.

Also, these properties allow to show a Jensen's inequality for the semigroup, for convex functions $\Phi$ \begin{equation} \mathbf{P}_{t}(\Phi(f)) \geq \Phi\left(\mathbf{P}_{t} f\right) \end{equation}

A $\sigma$-finite measure $\mu$ is invariant under the semigroup if \begin{equation*} \int_\mathcal X \mathbf{P}_{t}f \: d \mu = \int_\mathcal X f \: d \mu, \quad \forall f \in \mathcal M_b (\mathcal X). \end{equation*}

With this, we can extend the semigroup to a contraction of $L^p(\mu), 1 \leq p\leq\infty$

\begin{align*} \| f \|_{L^p(\mu)}^p= \int_\mathcal X| f|^p \:d \mu =\int_\mathcal X \mathbf{P}_{t} |f|^p \:d \mu \geq \int_\mathcal X |\mathbf{P}_{t} f|^p \:d \mu = \|\mathbf{P}_{t} f \|_{L^p(\mu)}^p \end{align*}

Finally, for any bounded, continuous function $f \in \mathcal C_b(\mathcal X), 1 \leq p< \infty$,

\begin{equation*} \int_\mathcal X |P_t f(x)- f(x)|^p \: d \mu(x) \to 0 \quad \text{as}\quad t \to 0 \end{equation*} by the DCT since $|P_t f- f|^p(x) \leq 2^{p-1}(\|P_t f\|_\infty +\|f\|_\infty)\leq 2^{p}\|f\|_\infty< \infty$. Since $C_b(\mathcal X)$ is dense in $L^p(\mathcal X, \mu)$ we are done.

Now this argument with the DCT seems only to work for finite measures. For $\sigma$-finite measures, should I split the state-space into finite regions and apply the argument?

Also I am not sure what licenses me to say that $P_t f(x) \to f(x)$ as $t \to 0$ for $\mu$-a.e. $x$ and bounded continuous $f$. This works when the process is right continuous but I am not sure about the general case.

Answer 1

The strong continuity (in $L^2(\mu)$) of $({\bf P}_t)$ is part of the definition of Markov semigroup in BGL (condition (vi) on page 11). On the other hand, the strong continuity on $L^p(\mu)$ for $1\le p<\infty$ then follows automatically because $\|{\bf P}_t\|_p\le 1$: By a well-known result, strong continuity of a semigroup like $({\bf P}_t)$ follows from weak continuity (see, for example, Theorem IX.1 in Yosida'a Functional Analysis). The weak continuity, that is the continuity of $t\mapsto \int g\cdot {\bf P}_tf d\mu$ for $g\in L^q(\mu)$ and $f\in L^p(\mu)$, follows because (i) $g$ and $f$ can be taken to be in $L^1(\mu)\cap L^\infty(\mu)\subset L^2(\mu)$ by the $L^p(\mu)$ contraction property of $({\bf P}_t)$ and the denseness of $L^1(\mu)\cap L^\infty(\mu)$ in $L^p(\mu)$ ($\mu$ is $\sigma$-finite!) and (ii) continuity of $t\mapsto \int g\cdot {\bf P}_tf d\mu$ for $f,g\in L^1(\mu)\cap L^\infty(\mu)$ follows from $L^2(\mu)$ continuity of the semigroup.

Answer 2

Added note: As far as I can tell, BGL asserts strong continuity in $L^2$ of the semigroup of a right-continuous Markov process that is symmetric with respect to an invariant measure.

All the arguments that I know of need enough functions $f\in L^1(\mu)\cap L^\infty(\mu)$ such that $t\mapsto P_tf(x)$ is right continuous for $\mu$-a.e. $x$. If the process has right-continuous paths, if the state space is nice (locally compact, second countable), and if $\mu$ is a Radon measure, then there are plenty of bounded continuous functions $f$ in $L^1(\mu)$. More generally, if $\mu$ is $\sigma$-finite and the process is a "right Markov process" (essentially, right continuous and strong Markov) then the set of bounded integrable $f$ such that $t\mapsto P_tf(x)$ is right continuous for $\mu$-a.e. $x$, is dense in $L^2(\mu)$.