Let $E$ be a locally compact separable metric with countable base. We consider a sequence of Hunt processes $\{X^{(n)}\}_{n \in \mathbb{N}}$ on $E$. That is, each $X^{(n)}=(\{X_t^{(n)}\}_{t \in [0,\infty]},\{P_x^{(n)}\}_{x \in E})$ is a cad-lag Markov process on $E$ with the strong Markov property (and the quasi-left-continuity). Let $m$ be a probability measure on $E$. For $n \in \mathbb{N}$, we write $P_{m}^{(n)}$ for the law of $X^{(n)}$ with initial distribution $m$.

We now assume the following condition.

${\rm \bf(A)}$ The laws $\{P_{m}^{(n)}\}_{n\in \mathbb{N}}$ of $\{X^{(n)}\}_{n \in \mathbb{N}}$ are tight in $D([0,\infty),E)$. Here, $D([0,\infty),E)$ is the space of $E$-valued right continuous functions on $[0,\infty)$ with finite left limits.

Under the condition ${\rm \bf(A)}$, we have a subsequential limit $(X,P)$ of $\{X^{(n)},P_m^{(n)}\}_{n \in \mathbb{N}})$. For a bounded measurable function $f\colon E \to \mathbb{R}$, we set $(P_t f)(x)=E[f(X_t) \mid X_0=x]$ for $m$-a.e. $x$. Then, each $P_t$ is extended to a contraction operator on $L^\infty(E,m)$ (the extended operator is denoted by the same symbol).

Can we show that $P_t(P_sf)=P_{t+s}f$ for every $t,s>0$ and $f \in L^\infty(E,m)$?

That is, I would like to know that $(X,P)$ is a time-homogeneous Markov process on $E.$

  • $\begingroup$ Haven't gone through the details, but what if we work on $E = \mathbb{R}$, and define $X^{(n)}$ with the following transitions: from $0$ it jumps to either $1/n$ or $-1$ at rate 1. The states $1/n$ and $1$ are absorbing. It seems that in the limit, the state $1/n$ merges with $0$, so the process has probability $1/2$ to stay at 0 forever, which is impossible for a Markov process. $\endgroup$ May 11, 2022 at 15:50
  • $\begingroup$ @NateEldredge Thank you for your comment. The state $0$ is a cemetery point for your limit process? If so, I feel that it does not necessarily contradict the Markov property. Am I making a mistake? $\endgroup$
    – sharpe
    May 11, 2022 at 16:04
  • $\begingroup$ Maybe I have to think some more. I guess the problem is that "initial state" becomes inconsistent in the limit. $\endgroup$ May 11, 2022 at 16:10
  • $\begingroup$ @NateEldredge Thank you very much. If you know sufficient conditions to reach an affirmative conclusion, please let me know. $\endgroup$
    – sharpe
    May 11, 2022 at 16:19
  • $\begingroup$ I think this is often studied in terms of convergence of the associated Dirichlet forms. I don't know the exact results, but "Mosco convergence" comes up a lot. doi.org/10.1006/jfan.1994.1093 seems to be the original paper $\endgroup$ May 11, 2022 at 16:55

1 Answer 1


If your semigroups $\{P_t^{(n)}\}$ and $\{P_t\}$ are Feller then Ethier and Kurtz Theorem 4.2.5 give that your limiting process $X$ exists and is a Markov process (indeed, a Feller process).

This may seem obvious (since clearly $X$ is Markov if its semigroup is Feller) but without the Feller property, it's not always clear that $X$ exists. Ethier and Kurtz Corollary 4.2.8 gives an if and only if guarantee: if your semigroup $\{P_t\}$ is strongly continuous (and it is already a positive contraction semigroup) then in order for $X$ to exist, $\{P_t\}$ must also have a conservative generator, and thus be Feller.

In light of that, you might want to know when a convergent sequence of Feller semigroups has a limit that is also Feller. I'm afraid I don't know the answer to that. But there are at least several ways to check, see Theorem 6.3 in Swart and Winter's lecture notes. Chapter 6 of those notes also contains a more readable treatment of Ethier and Kurtz's convergence theorems (see in particular Corollary 6.11).


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