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

Question

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:

Relevant part of the proof:

EDIT: What's bothering me most: Why are we allowed to assume $X_0=x$ and $X^n_0=x_n$? Can the general case somehow be reduced to that case?

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 $(5)$?

Answer 1

You know that $T_n(t)f(x_n) \to T(t)f(x)$ as $n \to \infty$ whenever $x_n \to x$ since this is precisely what is given by the convergence in distribution of $X_t^n \to X_t$ for $X_0 = x, X_0^n = x_n$.

In my answer to your other question, I proved the following result:

Lemma: If $E$ is a compact metric space (a locally compact and separable metric space) and $g_n,g$ are continuous functions (that vanish at infinity) on $E$ such that whenever $x_n \to x$ in $E$, $|g_n(x_n) - g(x)| \to 0$ as $n \to \infty$ then $\|g_n - g\|_\infty \to 0$ as $n \to \infty$.

The proof is somewhat tedious so I won't repost it here but you can find it in my linked answer. Since you have reduced to the case where $S$ is compact, the lemma implies that $\|T_n(t)f - T(t)f\|_\infty \to 0$ as desired by taking $g_n = T_n(t)f, g = T(t)f$ for arbitrary fixed $f$ and $t$.