• $\left(\kappa^{(n)}_t\right)_{t\ge0}$ 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

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 $(5)$?

  • $\begingroup$ 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 '19 at 18:31
  • $\begingroup$ @user159517 Why did you delete your answer? $\endgroup$ – 0xbadf00d Apr 9 '19 at 19:51
  • 1
    $\begingroup$ 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 '19 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 '19 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 '19 at 20:04

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 $t$.

  • $\begingroup$ But what if the given $X$ and $X^n$ don't satisfy $X_0=x$ and $X^n_0=x_n$? The only connection between $T$ and $X$ (resp. $T^n$ and $X^n$) is $$\operatorname E\left[f(X_t)\mid (X_r)_{r\le s}\right]=(T(t-s)f)(X_s)\;\;\;\text{almost surely}$$ (resp. $\operatorname E\left[f(X^n_t)\mid (X^n_r)_{r\le s}\right]=(T_n(t-s)f)(X^n_s)$ almost surely) for all $f\in C_0(E)$ and $t\ge s\ge0$. So, if we change the initial distribution of $X$ (resp. $X^n$) don't we loose this connection? $\endgroup$ – 0xbadf00d May 25 '19 at 11:55
  • $\begingroup$ This is the same thing which is worrying me in this question: math.stackexchange.com/q/3234403/47771. (And this is related to: math.stackexchange.com/q/3235408/47771.) $\endgroup$ – 0xbadf00d May 25 '19 at 11:56
  • $\begingroup$ I would like to accept your answer, but could you tell me what I'm missing (even it is obvious to you)? $\endgroup$ – 0xbadf00d May 26 '19 at 14:02
  • $\begingroup$ Well you've assumed that if $X_n^0 \overset{d}{\to} X^0$ then you have $X_n \overset{d}{\to} X$ in an appropriate skorohod space. So just take $X_n$ to be the markov process with transition kernel $\kappa^{(n)}$ and initial value $x_n$. The operators $\{T_n(t)\}_{t \geq 0}$ are determined entirely by the corresponding transition kernels since $T_n(t)f(x) = \int \kappa^{(n)}(t,x,dy) f(y)$ (in the first edition of kallenberg this is at the bottom of the first page of the chapter on Feller processes and semigroups). This guarantees you are still talking about the right operators. $\endgroup$ – Rhys Steele May 28 '19 at 13:11
  • $\begingroup$ From your answer it is still not clear why we've got $X_t^n \to X_t$ in distribution. As noted by user159517, $$\pi_t:D([0,\infty),E)\to E\;,\;\;\;x\mapsto x(t)$$ is only continuous at points $x\in D([0,\infty),E)$ such that $t$ is a continuity point of $x$. And It's not clear to me why Kallenberg is claiming that $X$ is almost surely continuous. $\endgroup$ – 0xbadf00d Jun 2 '19 at 6:08

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