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.
Best Answer
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.