2
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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:

theorem Relevant part of the proof: 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 $(5)$?

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  • $\begingroup$ 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. $\endgroup$ – user159517 Apr 9 at 18:31
  • $\begingroup$ @user159517 Why did you delete your answer? $\endgroup$ – 0xbadf00d Apr 9 at 19:51
  • $\begingroup$ 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. $\endgroup$ – user159517 Apr 9 at 19:52
  • $\begingroup$ @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.) $\endgroup$ – 0xbadf00d Apr 9 at 19:53
  • $\begingroup$ I agree that the proof seems a little cryptic. Are you specifically interested in the case $S = \mathbb{R}$? $\endgroup$ – user159517 Apr 9 at 20:04

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