If $(κ_t)_{t≥0}$ is the transition semigroup of a continuous Markov process, is $t↦(κ_tf)(x)$ continuous for all bounded continuous $f$ and fixed $x$?

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$(\mathcal F_t)_{t\ge0}$$ be a filtration on $$(\Omega,\mathcal A)$$
• $$E$$ be a metric space
• $$(X_t)_{t\ge0}$$ be an $$E$$-valued right-continuous time-homogeneous Markov process on $$(\Omega,\mathcal A,\operatorname P)$$
• $$\kappa_t$$ be a regular version of the conditional probability of $$X_t$$ given $$X_0$$, i.e. $$\kappa_t$$ is a Markov kernel on $$(E,\mathcal B(E))$$ with $$\operatorname P\left[X_{s+t}\in B\mid\mathcal F_s\right]=\operatorname P\left[X_t\in B\mid X_0\right]=\kappa_t(X_0,B)\;\;\;\text{almost surely for all }B\in\mathcal B(E)\tag1$$ for all $$s,t\ge0$$
• $$f:E\to\mathbb R$$ be bounded and continuous

Are we able to show that $$[0,\infty)\ni t\mapsto(\kappa_t f)(x):=\int\kappa_t(x,{\rm d}y)f(y)\tag1$$ is continuous for all $$x\in\mathbb R$$?

Let's take a look: Let $$(t_n)_{n\in\mathbb N}\subseteq[0,\infty)$$ and $$t\ge0$$ with $$t_n\xrightarrow{n\to\infty}t$$. For simplicity, assume $$t=0$$. By the dominated convergence theorem, $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0\right]\xrightarrow{n\to\infty}\operatorname E\left[f\left(X_0\right)\mid X_0\right]=f(X_0)\;\;\;\text{almost surely}\tag2$$ and hence $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0=\;\cdot\;\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\mid X_0=\;\cdot\;\right]=f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag3$$ By $$(1)$$, $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0=\;\cdot\;\right]=\kappa_{t_n}f\;\;\;\text{for all }n\in\mathbb N\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag4$$ Thus, $$\kappa_{t_n}f\xrightarrow{n\to\infty}f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag4$$

The problem with $$(4)$$ is the dependence of the null set on $$(t_n)_{n\in\mathbb N}$$.

However, $$\left|\kappa_{t_n}\right|\le\left\|f\right\|_\infty$$ for all $$n\in\mathbb N$$ and hence another application of the dominated convergence theorem yields $$\operatorname E\left[\left(\kappa_{t_n}f\right)(X_0)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\right]\tag5.$$

If we fix an $$x\in E$$ and assume $$\operatorname P\circ\:X_0^{-1}=\delta_x$$, the continuity of $$(1)$$ is obvious from $$(5)$$.

Now, Mars Plastic wrote in his answer that the assumption $$\operatorname P\circ\:X_0^{-1}=\delta_x$$ is no restriction.

Why?

I have some crazy idea: Assume $$E$$ is separable. Let $$D([0,\infty),E)$$ denote the space of càdlàg functions $$[0,\infty)\to E$$ equipped with the Skorohod topology and $$\pi_t:D([0,\infty),E)\to E\;,\;\;\;x\mapsto x(t)$$ for $$t\ge0$$. By separability, $$\mathcal B\left(D([0,\infty),E)\right)=\sigma(\pi_t:t\ge0)\tag6.$$ Now, assume that there is a Markov kernel $$\kappa$$ with source $$(E,\mathcal E)$$ and target $$(\tilde \Omega,\tilde{\mathcal A}):=\left(D([0,\infty),E),\mathcal B\left(D([0,\infty),E)\right)\right)$$ with $$\kappa(x,\;\cdot\;)\circ\left(\pi_{t_0},\ldots,\pi_{t_n}\right)^{-1}=\delta_x\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\;\;\;\text{for all }n\in\mathbb N\text{ and }0=t_0<\cdots for all $$x\in E$$. Then, if $$\mu$$ is any probability measureon $$(E,\mathcal E)$$ and $$\tilde{\operatorname P}[\tilde A]:=(\mu\kappa)(\tilde A)=\int\mu({\rm d}x)\kappa(x,\tilde A)\;\;\;\text{for }\tilde A\in\tilde{\mathcal A},$$ then it's easy to see that $$(\pi_t)_{t\ge0}$$ is a càdlàg Markov process on $$(\tilde\Omega,\tilde{\mathcal A},\tilde{\operatorname P})$$ with transition semigroup $$(\kappa_t)_{t\ge0}$$ and initial distribution $$\operatorname P\circ\:\pi_0^{-1}=\mu$$.

Clearly, we could choose $$\mu=\delta_x$$ for some fixed $$x\in E$$ and immediately obtain the continuity of $$(1)$$ by the same argumentation as before.

First of all, I have no idea if we can prove the existence of $$\kappa$$. As you may guess, my idea was inspired by the usual existence proof of a Markov process with a given transition semigroup and initial distribution. Therein, $$E$$ is assumed to be a Polish space, $$\pi_t$$ is replaced by $$E^{[0,\:\infty)}\ni x\mapsto x(t)$$ and $$(\tilde\Omega,\tilde{\mathcal A})$$ is replaced by $$\left(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)}\right)$$. The definition of $$\tilde{\operatorname P}$$ is the same.

So, if this is the correct approach, the concrete Markov process $$(X_t)_{t\ge0}$$ in the question is of no use. All we need is the transition semigroup $$(\kappa_t)_{t\ge0}$$ satisfying the usual consistency conditions (Chapman-Kolmogorov equations) and then hope that $$E$$ is "nice enough" to admit the existence of $$\kappa$$.

So, the question is: Is this the correct approach? And if so: When is $$E$$ is "nice enough"?

Your presentation is slightly bulky, but the general argument is correct. Note that you can fix a point $$x\in \mathbb R$$ and that it suffices to consider $$P\circ X_0^{-1}=\delta_{x}$$. Then your reasoning yields that for any $$(t_n)\subset\mathbb [0,\infty)$$ converging to $$t\in[0,\infty)$$ we have $$\begin{equation} (\kappa_{t_n}f)(x)=E[f(X_{t_n})|X_0=x] \to E[f(X_t)|X_0=x]=(\kappa_tf)(x), \quad n\to\infty, \end{equation}$$ and that's all you need. There are no null sets which depend on anything that bothers us.
PS: Since you are in a time-homogeneous setting, it is indeed irrelevant if you consider $$t=0$$ or any other $$t\in [0,\infty)$$.
• Why can we assume $\operatorname P\circ X_0^{-1}=\delta_x$? – 0xbadf00d Jan 21 at 15:51
• Assume $t=0$. In general, it's clear to me that $$\kappa_{t_n}f\xrightarrow{n\to\infty}f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag5$$ However, the null set in $(5)$ should depend on $(t_n)_{n\in\mathbb N}$. Applying the bounded convergence theorem yields $$\operatorname E\left[\kappa_{t_n}f(X_0)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\right]\tag6$$ and hence the claim is clear to me, if $\operatorname P\circ\:X_0^{-1}=\delta_x$ for a fixed $x\in E$. But how does the general case follow? – 0xbadf00d Jan 22 at 13:07