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

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

$$$$\mathbf{P}_{t}f(x) = \mathbb E [f(X_t)|X_0=x].$$$$ on bounded measurable functions $$\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 $$\Phi$$ $$$$\mathbf{P}_{t}(\Phi(f)) \geq \Phi\left(\mathbf{P}_{t} f\right)$$$$

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.

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.
• Thank you that's very helpful. However, my main question is how to $\textbf{derive}$ strong continuity in $L^2(\mu)$ from the Markov process. I am not assuming this property in the first place. In other words, I am trying to obtain the definition from BGL. Commented Feb 11, 2021 at 9:31
• Derive based on what hypotheses? If the underlying Markov process is a strong Markov process with right-continuous sample paths, then the strong continuity in $L^2$ holds. Something like this is touched on in BGL, pp. 11-12. Commented Feb 11, 2021 at 15:06
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 $$L^2(\mu)$$.