# If $X^{(n)},X$ are càdlàg and $X^{(n)}\to X$ in distribution, do the corresponding transition semigroups strongly converge?

Let

• $$\left(\kappa^{(n)}_t\right)_{t\ge0}$$ and $$(\kappa_t)_{t\ge0}$$ be Markov semigroups on $$(\mathbb R,\mathcal B(\mathbb R))$$ for $$n\in\mathbb N$$
• $$(T_n(t))_{t\ge0}$$ and $$(T(t))_{t\ge0}$$ be strongly continuous contraction semigroups on $$C_0(\mathbb R)$$ (continuous functions $$\mathbb R\to\mathbb R$$ vanishing at infinity equipped with the supremum norm) with $$T_n(t)f=\int\kappa^{(n)}_t(\;\cdot\;,{\rm d}y)f(y)\tag1$$ and $$T(t)f=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)\tag2$$ for all $$f\in C_0(\mathbb R)$$ and $$t\ge0$$
• $$X^{(n)}$$ and $$X$$ be real-valued càdlàg Markov processes with transition semigroups $$\left(\kappa^{(n)}_t\right)_{t\ge0}$$ and $$(\kappa_t)_{t\ge0}$$, respectively, for $$n\in\mathbb N$$

Assume $$X^{(n)}_0\xrightarrow{n\to\infty}X_0\tag3$$ in distribution and $$X^{(n)}\xrightarrow{n\to\infty}X\tag4$$ in distribution (with respect to the Skorohod topology). Are we able to conclude $$\left\|T_n(t)f-T(t)f\right\|_\infty\xrightarrow{n\to\infty}0\tag5$$ for all $$f\in C_0(\mathbb R)$$ and $$t\ge0$$?

The desired claim is part of the following theorem in the book of Kallenberg, but I don't understand his proof:

I don't know how he's arguing that $$X$$ is almost surely continuous at $$t$$. Is this really true? In any case, if we assume that $$X$$ is almost surely continuous, it is at least clear to me that $$(T_n(t)f)(x_n)\xrightarrow{n\to\infty}(T(t)f)(x)$$ for all $$(x_n)_{n\in\mathbb N}\subseteq\mathbb R$$ and $$x\in\mathbb R$$ with $$x_n\xrightarrow{n\to\infty}x$$ and $$t\ge0$$. But why is that sufficient for $$(5)$$?

• You do not have that $X_t^{(n)} \to X_t$ as $n \to \infty$ in distribution. Note that the functional $$\pi_t : D(\mathbb{R}) \to \mathbb{R}, x \mapsto x(t)$$ is only guaranteed to be continuous at points $x_0 \in D(\mathbb{R})$ for which $t$ is a continuity point. – user159517 Apr 9 at 18:31
• @user159517 Why did you delete your answer? – 0xbadf00d Apr 9 at 19:51
• I figured out the mistake: in my example, the value of $X^{(n)}$ at $t = 0$ was not equal to $X_0$. Lol! Sorry for that, time to go to sleep. – user159517 Apr 9 at 19:52
• @user159517 Do you've got an idea how to prove the claim? I don't really understand Kallenberg's argument. (And he's assuming compactness, which I don't want to assume.) – 0xbadf00d Apr 9 at 19:53
• I agree that the proof seems a little cryptic. Are you specifically interested in the case $S = \mathbb{R}$? – user159517 Apr 9 at 20:04