# How to prove that $P^n(x,\cdot)\stackrel{w}{\rightarrow} \pi$ for all $x$, implies the equicontinuity of $\{P^n f\}_n$ on compact sets?

Let $X$ be a separable and locally compact metric space and let $P\colon X\times \mathcal{B}(X) \rightarrow \mathbb{R}$ be the transition probabability kernel of a homogeneus Markov chain on $X$. Define $$Pf(x)=\int_{X} f(y)\,P(x,dy)\;\;\; (f\in C(X), x\in X),$$ where $C(X)$ is the space of bounded continuous functions $f:X\rightarrow \mathbb{R}$. We say that $P$ has the Feller property if $Pf \in C(X)$ for all $f\in C(X)$.

My question is: how to prove the following theorem (Proposition 6.4.2, Meyn and Tweedie, Markov chains and stochastic stability: http://probability.ca/MT/Chap6.pdf):

Assume that $P$ has the Feller property, and that there exists a unique probility measure $\pi$ on $\mathcal{B}(X)$ such that for every $x\in X$: $$P^n(x,\cdot)\stackrel{w}{\rightarrow} \pi\;\;\;(n\to\infty).$$ Then the family $\{P^n f: n\in\mathbb{N}\}$ is equicontinuous on compact subsets of $X$ whenever $f\in C(X)$ ($\stackrel{w}{\rightarrow}$ denotes the weak convergence).

The authors claim that it follows directly from Ascoli's Theorem. However I can not see this. In order to be able to apply Ascoli's theorem we need to know that for each compact set $C$ at least some subseqence $(P^{k_n} f)_{n}$ is uniformly convergent on $C$. Since $C$ is compact we may choose $x_n \in C$ such that $$|P^n f(x_n)-\pi(f)|=\max_{x\in C} |P^n f(x)-\pi(f)| \;\;\;(n\in\mathbb{N}),$$ where $\pi(f)=\int_X f(y)\,\pi(dy)$. So it suffices to show that there exists a strictly increasing sequence $(k_n)_{n\in\mathbb{N}}$ of natural numbers such that $$P^{k_n}f(x_{k_n})\to \pi(f)$$ but I have a problem in this place. I would be grateful for any ideas.

• To say it another way, it's clear that $P^n f \to \pi f$ pointwise. What's asserted is that the convergence is uniform on compact sets. I agree that I don't see why that is. – Nate Eldredge Oct 25 '13 at 20:07
• Thanks for the reply. Now, at least, I know it is not obvious. Maybe it should be assumed that the weak convergence of $P^n(x,\cdot)$ is uniform with respect to $x$ on compact sets. – Dawid C. Oct 25 '13 at 21:06
• The authors write "Since the limit in (6.20) is continuous (and in fact constant) it follows from Ascoli’s Theorem (...)". This sentence sounds a little strange for me. – Dawid C. Oct 25 '13 at 21:14
• The Dini Theorem was the first thing I have tried to use in poving this. – Dawid C. Oct 26 '13 at 1:20

This doesn't answer the question, but the result does hold if we assume the strong Feller property ($Pg \in C(X)$ for every bounded measurable $g$).
Namely, suppose $f \in C(X)$. Then \begin{align*}|P^{n+1} f - \pi f| &= |P(P^n f - \pi f)|\\ &\le P |P^n f - \pi f| \\ &\le P \sup_{m \ge n} |P^m f - \pi f|.\end{align*} Set $F_n = \sup_{m \ge n} |P^m f - \pi f|$. By construction, $F_n$ is monotone decreasing, so by the monotonicity of $P$, $P F_n$ is also monotone decreasing. Moreover, $\|F_n\|_\infty \le 2 \|f\|_\infty$. And since $P^m f \to \pi f$ pointwise, we have $F_n \to 0$ pointwise. So by monotone convergence, we also have $P F_n \to 0$ pointwise. Finally, by the strong Feller property, each $P F_n$ is continuous. So $P F_n$ is a monotone decreasing sequence of continuous functions which converges to 0 pointwise. By Dini's theorem, we have $P F_n \to 0$ uniformly on compact sets. This gives us that $P^{n+1} f \to \pi f$ uniformly on compact sets; in particular, $\{P^n f\}$ is equicontinuous on compact sets.
• Under the strong Feller property, you have shown that $(P^n f)_n$ is uniformly convergent on compacts sets, which is even stronger that we need. In general, I have tried something similar to this. Namely I have tried to show that a the subsequence of $(|P^n f - \pi f|)_n$ is decreasing on the set $K_0=\{x_n: n\in\mathbb{N}\} \cup \{x_0\}$, where $x_0\in K$ and $x_n$ is an arbitrary sequence of the set $K$ convergent to $x_0$. The we would have the uniform convergence on the set $K_0$ by the Dini Theorem, and therefore, on the set $K$. – Dawid C. Oct 26 '13 at 1:39
• I mean the Arzela-Ascoli theorem says that if $(f_n)$ is a sequence of continuos functions on a compact set then each subsequence of $(f_n)$ has a unformly convergent subsequence iff $\{f_n\}$ is equicontinous and pointwise bounded. So, you are right, if $(f_n)$ is pointwise convergent it equivalent to the uniform convergence of $(f_n)$. – Dawid C. Oct 26 '13 at 2:37