I have a question about Feller processes.

In this paper "On the doubly Feller property of resolvent" the transition semigroup $(P_t)_{t \ge 0}$ of a Hunt process (on a metric space $E$) is said to have the Feller property if the following two conditions are satisfied.

  1. For every $f \in C_{\infty}(E)$ and $t >0$, we have $P_{t}f \in C_{\infty}(E)$.
  2. For every $f \in C_{\infty}(E)$ and $x \in E$, $\lim_{t \to 0}P_{t}f(x)=f(x)$.

Here, $C_{\infty}(E)$ is the family of continuous functions on $E$ vanishing at infinity.

My question

I think the transition semigroup $(P_t)_{t \ge 0}$ of a Hunt process $X$ always satisfies the condition 2. Since $f(X_t) \to f(X_0)$ as $t \to 0$ from the right continuity of the sample path of $X$. This implies \begin{align*} \lim_{t \to 0}P_{t}f(x)=\lim_{t \to 0}E_{x}[f(X_t)]=E_{x}[f(X_0)]=E_{x}[f(x)]=f(x) \end{align*} by the dominated convergence theorem and the normal property of $X$.

Why they need the condition 2?


2 Answers 2


The Feller property is sufficient (bot not necessary) for the existence of a Hunt process associated with $(P_t)$. As you observe, condition (2) in the Feller property is (in isolation) also a necessary consequence of the Hunt property.

  • 1
    $\begingroup$ Yes: If $f\in C_\infty(E)$ then $t\mapsto f(X_t(\omega))$ is bounded and continuous for each $\omega$, with limit $f(X_0(\omega))$ as $t\downarrow 0$. Property (2) follows by dominated convergence, as you noted. $\endgroup$ Commented Jul 22, 2017 at 15:30
  • $\begingroup$ Are you saying that (2) is not necessary in definition of Hunt process...but it is a consequence of Hunt process? I am confused since in that case why the paper needs (2) in def? $\endgroup$
    – Henry.L
    Commented Jul 23, 2017 at 0:45
  • 1
    $\begingroup$ (2) is not in the definition of Hunt process (a strong Markov process with right-continuous and quasi-left-continuous sample paths). On the other hand, (2) is a necessary consequence for the semigroup of a Hunt process. $\endgroup$ Commented Jul 23, 2017 at 14:56
  • $\begingroup$ Yes I missed sth and corrected it now. $\endgroup$
    – Henry.L
    Commented Jul 24, 2017 at 13:17

Sorry I probably missed "$(P_t)_{t \ge 0}$ of a Hunt process" when I first composed this answer and that is why confusion arise.

I think it is most beneficial if we start from and stick to definition.

Hunt process is a strong Markov process with some regularity on sample functions, usually càdlàg sample functions but subject to some alternations. Suppose $X$ is on a (locally) compact space $E$ and there is a one-to-one correspondence between Hunt process and Feller-Dynkin semigroups.

The paper you cited provided (1)(2) as a set of equivalent conditions that allows you to regard the transition semigroup $(P_t)_{t \ge 0}$ as a Feller-Dynkin semigroup as well. A Hunt process may admits a transition semigroup satisfying (2) of course but it also needs (1) to make the operations within the transition semingroup closed.

I hope I made myself clear now.

  • $\begingroup$ @Henry.L: Two points: (1) the sample paths of a Hunt process are not merely "left-limited" but more stringently "quasi-left-continuous"; (2) the transition semigroup of a Hunt process is not necessarily Feller-Dynkin. $\endgroup$ Commented Jul 24, 2017 at 17:36
  • $\begingroup$ @JohnDawkins (1) $X$ is compact so what is the difference between these two notions? I think they are the same. One step further if Hunt proc is also feller process then just take its cadlag version. (2) Hunt process must also posses Feller-Dynkin property as in OP. $\endgroup$
    – Henry.L
    Commented Jul 24, 2017 at 22:37
  • $\begingroup$ @Henry.L: (1) A Hunt process is cadlag, but is also quasi-left continuous (meaning that if $(T_n)$ is an increasing sequence of stopping times with limit $T$, then $\lim_nX_{T_n}=X_T$ on the event $\{T_n<T,\,\forall n\}$.) For example, if $B$ is a standard one-dimensional reflecting Brownian motion (on $[0,\infty)$) and if $T_1:=\inf\{t>0:B_t=1\}$, define $X_t=B_t$ for $0\le t<T_1$ but $X_t=-1$ for $t\ge T_1$. Then $X$ is a cadlag strong Markov process (with state space $[0,1)\cup(1,\infty)\cup\{-1\}$) but $X$ is not a Hunt process, because q-l-c fails at $T_1$. $\endgroup$ Commented Jul 25, 2017 at 13:43
  • $\begingroup$ @Henry.L: (2) I can only repeat that a Hunt process need not be a Feller-Dynkin process. I can supply an example if you wish. $\endgroup$ Commented Jul 25, 2017 at 13:43
  • $\begingroup$ (1) I mean when $X$ is defined on a compact space $E$, if you can find a counter example I would be interested to know. When $E$ is not compact sure. (2) Yes. I said a Hunt process is a F-D if if must possess F-D property then there is a one-to-one correspondence in my last comment. $\endgroup$
    – Henry.L
    Commented Jul 25, 2017 at 17:25

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