\begin{document}

\title[Equicontinuous families of Markov operators]{Equicontinuous families of Markov operators in view of asymptotic stability}
\author{Sander C. Hille}
\address{Mathematical Institute, Leiden University, P.O. Box 9512, 2300 RA Leiden, The Netherlands, (SH,MZ)}
\email{\{shille,m.a.ziemlanska\}@math.leidenuniv.nl}
\author{Tomasz Szarek}
\address{Institute of Mathematics, University of Gda\'nsk, Wita Stwosza
  57, 80-952 Gda\'nsk, Poland, (TS)}
\email{szarek@intertele.pl}
\author{Maria A. Ziemla\'nska}

{\rm d}ate\today

\subjclass[2000]{37A30, 60J05} 
\keywords{Markov operator, asymptotic stability, e-property, Ces\`aro e-property, equicontinuity, tightness}
\thanks{The work of Maja Ziemla\'nska has been partially supported by a Huygens Fellowship of Leiden University. The work of Tomasz Szarek has been supported by the National Science Centre
of Poland, grant number 2016/21/B/ST1/00033.}

\maketitle

\begin{abstract}
Relation between equicontinuity -- the so-called e--property and stability of Markov operators is studied. In particular, it is shown that any asymptotically stable Markov operator with an invariant measure such that the interior of its support is non-empty satisfies the e--property.
\end{abstract}

\section{Introduction}

This paper is centered around two concepts of equicontinuity for Markov operators defined on probability measures on Polish spaces: the {\it e-property} and the {\it Ces\`aro e-property}. Both appeared as a condition (among others) in the study of ergodicity of Markov operators. In particular they are very useful in proving existence of a unique invariant measure and its {\it asymptotic stability}: at whatever probability measure one starts, the iterates under the Markov operator will weakly converge to the invariant measure. The first concept appeared in \cite{Lasota-Szarek:2006,Szarek-Worm:2012} while the second was introduced  in \cite{Worm-thesis:2010} as a theoretical generalisation of the first and allowed the author to extend various results by replacing the e-property condition by apparently weaker the Ces\`aro e-property condition, among others. 

Interest in equicontinuous families of Markov operators existed already before the introduction of the e-property. Jamison \cite{Jamison:1964}, working on compact metric state spaces, introduced the concepts of (dual) Markov operators on the continuous functions that are `uniformly stable' or `uniformly stable in mean' to obtain a kind of asymptotic stability results in this setting. Meyn and Tweedie \cite{Meyn-Tweedie:2009} introduced the so-called `{\it e-chains}' on locally compact Hausdorff topological state spaces, for similar purposes. See also \cite{Zaharopol:2014} for results in a locally compact metric setting.

The above mentioned concepts were used in proving ergodicity for some Markov chains (see \cite{Stettner: 1994, Czapla: 2012, CzaplaHorbacz: 2014, EsS-vonRenesse: 2012, GongLiu: 2015, Komorowksi_ea:2010, WW: 2017}).

It is worth mentioning here that similar concepts appear in  the study of mean equicontinous dynamical systems mainly on compact spaces (see for instance \cite{LiTuJe: 2015}). However it must be stressed here that our space of Borel probability measures defined on some Polish space is non-compact.

Studing the e--property the natural question arose whether any asymptotically stable Markov operator satisfies this property. Proposition 6.4.2 in \cite{Meyn-Tweedie:2009} says that this holds when the phase space is compact. In particular, the authors showed that the stronger e--chain property is satisfied. Unfortunately, the proof contains a gap and it is quite easy to construct an example showing that some additional assumptions must be then added. 

On the other hand, striving to repair the gap of the Meyn-Tweedie result mentioned above, we show that any asymptotically stable Markov operator with an invariant measure such that the interior of its support is nonempty satisfies the e--property.

\section{Preliminaries}

Let $(S, d)$ be a Polish space. By $B(x,r)$ we denote the open ball in $(S,d)$ of radius $r$, centered at $x\in S$ and $\partial B(x,r)$ denotes its boundary. Further $\overline{E}$, ${\rm Int}_S E$ denote the closure of $E\subset S$ and the interior of $E$, respectively. By $C_b(S)$ we denote the vector space of all bounded real-valued continuous functions  on $S$ and by $B_b(S)$ all bounded real-valued Borel measurable functions, both equipped with the supremum norm $|\cdot|$. By $L_b(S)$  we denote the subspace of $C_b(S)$ of all bounded Lipschitz functions (for the metric $d$ on $S$). For $f\in L_b(S)$, ${\rm Lip} f$ denotes the Lipschitz constant of $f$.

By ${\mathcal{M}}(S)$  we denote the family of all finite Borel measures on $S$ and  by $\mathcal{P}(S)$ the subfamily of all probability measures in ${\mathcal{M}} (S)$. For $\mu\in{\mathcal{M}} (S)$, its {\it support} is the set
\[
\supp\mu :=\{x\in S:  \mu(B(x,r))>0\,\,\mbox{for all}\ r>0\}.
\]

An operator $P: {\mathcal{M}} (S)\to {\mathcal{M}} (S)$ is called a {\it Markov operator} (on $S$) if it satisfies the following two conditions:
\begin{enumerate}
\item[({\it i})] {\it (Positive linearity)} $P(\lambda_1\mu_1+\lambda_2\mu_2)=\lambda_1 P\mu_1+\lambda_2 P\mu_2$\\
for $\lambda_1, \lambda_2\geq 0$; $\mu_1, \mu_2\in{\mathcal{M}}(S)$;
\item[({\it ii})] {\it (Preservation of the norm)} $P\mu (S)=\mu (S)$ for $\mu\in{\mathcal{M}}(S)$.
\end{enumerate}
A measure $\mu_*$ is called {\it invariant} if $P\mu_*=\mu_*$. A Markov operator $P$ is {\it asymptotically stable} if there exists a unique invariant measure $\mu_*\in\mathcal{P}(S)$ such that $P^n\mu\to\mu_*$ weakly as $n\to\infty$ for every $\mu\in\mathcal{P}(S)$. 

For brevity we shall use the notation:
$$
\langle f, \mu\rangle :=\int_S f(x)\mu ({\rm d} x)\qquad\mbox{for}\ f\in B_b(S),\ \mu\in{\mathcal{M}} (S).
$$

A Markov operator $P$ is {\it regular} if there exists a linear operator $U: B_b(S)\to B_b(S)$ such that 
\[
\pair{f}{P\mu} = \pair{Uf}{\mu}\qquad \mbox{for all}\ f\in B_b(S),\ \mu\in{\mathcal{M}}(S).
\]
The operator U is called the {\it dual operator} of $P$. A regular Markov operator is a {\it Feller operator} if its dual operator $U$ maps $C_b(S)$ into itself.
Equivalently, $P$ is Feller if it is continuous in the weak topology (cf. \cite{Worm-thesis:2010}, Proposition 3.2.2). 

A Feller operator $P$ satisfies the {\it e--property} at $z\in S$ if for any $f\in L_b(S)$ we have 
\begin{equation}\label{e2_9.04.17}
\lim_{x\to z} \limsup_{n\to\infty}|U^n f (x) - U^n f (z)|=0,
\end{equation}
i.e., if the family of iterates $\{U^n f: n\in\mathbb N\}$ is equicontinuous at $z\in S$. We say that a Feller operator satisfies the {\it e--property} if it satisfies it at any $z\in S$.

D. Worm slightly generalized the e--property introducing the Ces\'aro e--property (see  \cite{Worm-thesis:2010}). Namely,  a Feller operator $P$ will satisfy the {\it Ces\'aro e--property} at $z\in S$ if for any $f\in L_b (S)$ we have 
\begin{equation}\label{e3_9.04.17}
\lim_{x\to z} \limsup_{n\to\infty}\left|\frac{1}{n}\sum_{k=1}^n U^k f (x) - \frac{1}{n}\sum_{k=1}^n U^k f (z)\right|=0.
\end{equation}
Analogously a Feller operator satisfies the {\it Ces\'aro e--property} if it satisfies this property at any $z\in S$.

The following simple example shows that Proposition 6.4.2 in \cite{Meyn-Tweedie:2009} fails.

\begin{example}
Let $S=\{1/n : n\ge 1\}\cup\{0\}$ and let $T: S\to S$ be given by the following formula:
$$
T(0)=T(1)=0\qquad\text{and}\qquad T(1/n)=1/(n-1)\quad\text{for $n\ge 2$}.
$$
The operator $P: \mathcal M(S)\to\mathcal M(S)$ given by the formula $P\mu=T_*(\mu)$ (the pushforward measure) is asymptotically stable but it does not satisfy the e--property at $0$.
\end{example}

Jamison \cite{Jamison:1964} introduced for a Markov operator the property of {\it uniform stability in mean} when $\{U^n f:n\in\mathbb N\}$ is an equicontinuous family of functions in the space of real-vauled continuous function $C(S)$ for every $f\in C(S)$. Here $C$ is a compact metric space. Since the space of bounded Lipschitz functions is dense for the uniform norm in the space of bounded uniform continuous functions, this property coincides with the Ces\'aro e--property for compact metric spaces. Now, if the Markov operator $P$ on the compact metric space is asymptotically stable, with the invariant measure $\mu_*\in\mathcal M_1$, then  
$
\frac{1}{n}\sum_{i=1}^n U^i f\to \langle f, \mu_*\rangle$ pointwise, for every $f\in C(S)$. According to Theorem 2.3 in \cite{Jamison:1964} this implies that $P$ is uniformly stable in mean, i.e., has 
the Ces\'aro e--property.

\begin{example} Let $(k_n)_{n\ge 1}$ be an increasing sequence of prime numbers. Set
$$
S:=\{(\overbrace{\mathstrut 0, \ldots, 0,}^{k_n^i-1 - times} i/k_n, 0,\ldots )\in l^{\infty}: i\in\{0,\ldots, k_n\}, n\in\mathbb N\}.
$$ 
The set $S$ endowed with the $l^{\infty}$-norm $\|\cdot\|_{\infty}$ is a (noncompact) Polish space. Define $T: S\to S$ by the formula
$$
T((0,\ldots ))=T((\overbrace{\mathstrut 0, \ldots, 0,}^{k_n^{k_n}-1- times} 1, 0,\ldots))=(0,\ldots, 0, \ldots)\qquad\text{for $n\in\mathbb N$}
$$
and
$$
T((\overbrace{\mathstrut 0, \ldots, 0,}^{k_n^i-1- times} i/k_n,  0,\ldots ))=(\overbrace{\mathstrut 0, \ldots, 0,}^{k_n^{i+1}-1- times} (i+1)/k_n, 0,\ldots )\quad\text{for $i\in\{1,\ldots, k_n-1\}, n\in\mathbb N$.}
$$
The operator $P: \mathcal M(S)\to\mathcal M(S)$ given by the formula $P\mu=T_*(\mu)$ is asymptotically stable but it does not satisfy the Ces\'aro  e--property at $0$. Indeed, if we take an arbitrary continuous function $f: S\to\mathbb R_+$ such that $f((0,\ldots, 0, \ldots))=0$ and $f(x)=1$ for $x\in S$ such that $\|x\|_{{\infty}}\ge1/2$ we have
$$
\frac{1}{k_n}\sum_{i=1}^{k_n} U^i f((\overbrace{\mathstrut 0, \ldots, 0}^{k_n-1}, 1/k_n, 0,\ldots ))-\frac{1}{k_n}\sum_{i=1}^{k_n} U^i f((0,\ldots))\ge 1/2.
$$
\end{example} 

We are in a position to formulate the main result of our paper:

\begin{thrm}\label{main_theorem} Let $P$ be an asymptotically stable Feller operator and let $\mu_*$ be its unique invariant measure.
If ${\rm Int}_S(\supp\mu_*){(n)}eq\emptyset$, then $P$ satisfies the e--property.
\end{thrm}

Its proof involves the following lemma:
\begin{lemma}Let $P$ be an asymptotically stable Feller operator and let $\mu_*$ be its unique invariant measure. Let $U$ be dual to $P$. If $\textup{Int}_S(\supp\mu_*){(n)}eq\emptyset$, then for every $f\in C_b(S)$ and any $\varepsilon>0$ there exists a ball $B\subset\supp\mu_*$ and $N\in\mathbb{N}$ such that
\begin{equation}\label{e1_19.11.16}
|U^nf(x)-U^n f(y)|\le\varepsilon\qquad\text{for any $x, y\in B$,\ $n\geq N$}.
\end{equation}
\end{lemma}

\begin{proof} Fix $f\in C_b(S)$ and $\varepsilon>0$. Let $W$ be an open set such that $W\subset\supp\mu_*$. Set $Y={\overline{W}}$ and observe that the subspace $Y$ is a Baire space. Set
$$
Y_n:=\{x\in Y: |U^mf(x)-\langle f, \mu_*\rangle | \le\varepsilon/2\,\,\, \text{for all $m\ge n$}\}
$$
and observe that $Y_n$ is closed and
$$
Y=\bigcup_{n=1}^{\infty} Y_n.
$$
By the Baire category theorem there exist $N\in\mathbb N$ such that $\textup{Int}_Y Y_{N}{(n)}eq\emptyset.$
Thus there exists a set $V\subset Y_N$ open in the space $Y$ and consequently a ball  $B$ in $S$ such that
$B\subset Y_N\subset\supp\mu_*$. Since 
$$
|U^nf(x)-\langle f, \mu_*\rangle | \le\varepsilon/2\qquad\text{for any $x\in B$ and $n\ge N$},
$$
condition (\ref{e1_19.11.16}) is satisfied.
\end{proof}

We are ready to prove Theorem \ref{main_theorem}.
\begin{proof} Assume, contrary to our claim, that $P$ does not satisfy the e--property. Therefore there exist a function $f\in C_b(S)$ and a point $x_0\in S$ such that
$$
\limsup_{x\to x_0} \limsup_{n\to\infty}|U^n f(x)-U^n f(x_0)|>0.
$$
Choose $\varepsilon>0$ such that
$$
\limsup_{x\to x_0} \limsup_{n\to\infty}|U^n f(x)-U^n f(x_0)|\ge 3\varepsilon.
$$
Let $B:=B(z, 2r)$ be a ball such that condition (\ref{e1_19.11.16}) holds. Since $B(z, r)\subset\supp\mu_*$, we have $\gamma:=\mu_*(B(z, r))>0$. Choose $\alpha\in (0, \gamma)$. Since the operator $P$ is asymptotically stable, we have
\begin{equation}\label{e2_19.11.16}
\liminf_{n\to\infty}P^n\mu (B(z, r))>\alpha\qquad\text{for all $\mu\in\mathcal{P}(S)$},
\end{equation}
by the Alexandrov theorem (see \cite{Ethier_Kurtz}).

Let $k\ge 1$ be such that $2(1-\alpha)^k|f|<\varepsilon$. By induction we are going to define two sequences of measures $({(n)}u^{x_0}_i)_{i=1}^k$, 
$(\mu^{x_0}_i)_{i=1}^k$ and a sequence of integers $(n_i)_{i=1}^k$ in the following way: let $n_1\ge 1$ be such that
\begin{equation}
P^{n_1}{\rm d}elta_{x_0}(B(z, r))>\alpha. 
\end{equation}
Choose $r_1<r$ such that $ P^{n_1}{\rm d}elta_{x_0}(B(z, r_1))>\alpha$ and $P^{n_1}{\rm d}elta_{x_0}(\partial B(z, r_1))=0$ and set 
\begin{equation}
{(n)}u^{x_0}_1(\cdot)=\frac{P^{n_1}{\rm d}elta_{x_0}(\cdot\cap B(z, r_1))}{P^{n_1}{\rm d}elta_{x_0}(B(z, r_1))}
\end{equation}
and
\begin{equation}
\mu^{x_0}_1(\cdot)=\frac{1}{1-\alpha}\left(P^{n_1}
{\rm d}elta_{x_0}(\cdot)-\alpha{(n)}u_1^{x_0} (\cdot)\right).
\end{equation}
Assume that we have done it for $i=1,\ldots, l$, for some $l<k$. Now let $n_{l+1}$ be such that
\begin{equation}
P^{n_{l+1}}\mu_l^{x_0}(B(z, r))>\alpha. 
\end{equation}
Choose $r_{l+1}<r$ such that $ P^{n_{l+1}}\mu_l^{x_0}(B(z, r_{l+1}))>\alpha$ and $P^{n_{l+1}}\mu_l^{x_0}(\partial B(z, r_{l+1}))=0$ and set 
\begin{equation}
{(n)}u^{x_0}_{l+1}(\cdot)=\frac{P^{n_{l+1}}\mu_l^{x_0}(\cdot\cap B(z, r_{l+1}))}{P^{n_{l+1}}\mu_l^{x_0}(B(z, r_{l+1}))}
\end{equation}
and
\begin{equation}
\mu^{x_0}_{l+1}(\cdot)=\frac{1}{1-\alpha}\left(P^{n_{l+1}}
\mu^{x_0}_{l}(\cdot)-\alpha{(n)}u_{l+1}^{x_0} (\cdot)\right).
\end{equation}
We are done. We have
$$
\begin{aligned}
P^{n_1+\ldots+n_k}{\rm d}elta_{x_0}(\cdot)&=\alpha P^{n_2+\ldots+n_k}{(n)}u^{x_0}_1(\cdot)
+\alpha (1- \alpha)
P^{n_3+\ldots+n_k}{(n)}u^{x_0}_2(\cdot)
+\ldots + \\
&+\alpha (1- \alpha )^{k-1}
{(n)}u^{x_0}_k (\cdot)+(1-\alpha)^k\mu^{x_0}_k (\cdot).
\end{aligned}
$$
By induction we check that ${(n)}u_i^x-{(n)}u_i^{x_0}\to 0$ and $\mu_i^x-\mu_i^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$.
Indeed, if $i=1$, then ${(n)}u_1^x-{(n)}u_1^{x_0}\to 0$ weakly (as $d(x, x_0)\to 0$), by the fact that $P$ is a Feller operator and $\lim_{d(x, x_0)\to 0}P^{n_1}{\rm d}elta_{x}(B(z, r_1))=P^{n_1}{\rm d}elta_{x_0}(B(z, r_1))$, by the Alexandrov theorem due to the fact that $P^{n_1}{\rm d}elta_{x_0}(\partial B(z, r_1))=0$. On the other hand, the weak convergence $\mu_1^x-\mu_1^{x_0}\to 0$ as $d(x, x_0)\to 0$ follows directly from the definition of $\mu_1^x$. Moreover, observe that for $x$ sufficiently close to $x_0$ we have $P^{n_1}{\rm d}elta_{x}(B(z, r))>\alpha$ and therefore $\mu_1^x\in\mathcal{P}(S)$.

Assume now that we have proved that ${(n)}u_i^x-{(n)}u_i^{x_0}\to 0$ and $\mu_i^x-\mu_i^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$ for $i=1,\ldots, l$. We show that ${(n)}u_{l+1}^x-{(n)}u_{l+1}^{x_0}\to 0$ and $\mu_{l+1}^x-\mu_{l+1}^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$ too. Analogously, $\lim_{d(x, x_0)\to 0}P^{n_{l+1}}\mu_{l}^{x}(B(z, r_{l+1}))=P^{n_{l+1}}\mu_{l}^{x_0}(B(z, r_{l+1}))$, by the Alexandrov theorem due to the fact that $P^{n_{l+1}}\mu_{l}^{x_0}(\partial B(z, r_{l+1}))=0$ and from the definition of ${(n)}u_{l+1}^x$ we obtain that ${(n)}u_{l+1}^x-{(n)}u_{l+1}^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$. The weak convergence $\mu_{l+1}^x-\mu_{\l+1}^{x_0}\to 0$ as $d(x, x_0)\to 0$ follows now directly from the definition of $\mu_{l+1}^x$ and for $x$ sufficiently close to $x_0$ we have $P^{n_{l+1}}\mu_{l}^{x}(B(z, r))>\alpha$ and therefore $\mu_{l+1}^x\in\mathcal{P}(S)$. We are done.

Observe that for any $x$ sufficiently close to $x_0$ and  all $n\ge n_1+\ldots+n_k$ we have
$$
\begin{aligned}
P^{n}{\rm d}elta_{x}(\cdot)&=\alpha P^{n-n_1}{(n)}u^{x}_1(\cdot)
+\alpha (1- \alpha)
P^{n-n_1-n_2}{(n)}u^{x}_2(\cdot)
+\ldots \\
&+\alpha (1- \alpha )^{k-1}
P^{n-n_1-\ldots-n_k}{(n)}u^{x}_k (\cdot)+(1-\alpha)^k P^{n-n_1-\ldots-n_k}\mu^{x}_k (\cdot),
\end{aligned}
$$
where $\supp{(n)}u^x_i\subset B(z, r)$ for all $i=1,\ldots, k$. Thus
\begin{align}
\limsup_{n\to\infty}|\langle f, P^n{(n)}u^x_i\rangle-\langle f, P^n{(n)}u^{x_0}_i\rangle|&=\limsup_{n\to\infty} |\langle U^nf-\langle f, \mu_*\rangle, {(n)}u^x_i\rangle-\langle U^nf -\langle f, \mu_*\rangle, {(n)}u^{x_0}_i\rangle|{(n)}onumber\\
&\le \varepsilon/2+\varepsilon/2=\varepsilon\label{e_20.11.16} 
\end{align}
for all $i=1,\ldots, k$ and $x$ sufficiently close to $x_0$. Hence
$$
\begin{aligned}
3\varepsilon<\limsup_{x\to x_0} \limsup_{n\to\infty}&|U^n f(x)-U^n f(x_0)|=
\limsup_{x\to x_0} \limsup_{n\to\infty}|\langle f, P^n{\rm d}elta_x\rangle - \langle f, P^n{\rm d}elta_{x_0}\rangle|\\
&\le \varepsilon (\alpha+\alpha (1-\alpha)+\ldots \alpha (1-\alpha)^{k-1}) +2(1-\alpha)^k |f|\\
&\le \varepsilon+\varepsilon=2\varepsilon,
\end{aligned} 
$$
which is impossible. This completes the proof.
\end{proof}
{\bf Acknowledgements.}\ We thank Klaudiusz Czudek for providing us with Example 1.

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\end{document}

\section{Equicontinuous families of Markov operators and tightness} 
\label{sec:equicont and tightness}

Let $P$ be a regular Markov operator with dual operator $U$. The operator $P$ has the {\it e-property} for the admissible metric $d$ on $S$ if for every $f\in\mathrm{BL}(S,d)$, 
\[
\{ U^nf : n\in\mathbb{N}\} \subset C_b(S)
\] 
is an equicontinuous family (cf. \cite{Lasota-Szarek:2006,Szarek-Worm:2012}). Similarly, one defines the Ces\`aro e--property, following \cite{Worm-thesis:2010}, Chapter 7. To that end, one first defines the {\it Ces\`aro averages} $P^{(n)}$ of the Markov operator $P$:
\[
P^{(n)}\mu := \frac{1}{n} \sum_{k=1}^n P^k\mu.
\]
Further $U^{(n)}$ is the dual operator of $P^{(n)}$: $U^{(n)} = \frac{1}{n}\sum_{k=1}^n U^k$.
Now, $P$ has the {\it Ces\`aro e-property} for the admissible metric $d$ on $S$ if for every $f\in\mathrm{BL}(S,d)$, 
\[
\{ U^{(n)} f : n\in\mathbb{N}\} \subset C_b(S)
\] 
is an equicontinuous family. 

Obviously a Markov operator with the e-property has the Ces\`aro e-property. In Section \ref{sec:asympt stability} we shall provide an example of a Markov operator that has the Ces\`aro e-property, but not the e-property. Moreover, Theorem \ref{thrm:asympt stability implies Cesaro} will exhibit a substantial class of Markov operators of interest that have the Ces\`aro e-property.

If $S$ is compact, then any continuous function on $S$ is uniformly continuous, and bounded. Since $\mathrm{BL}(S,d)$ is dense in the space $U_b(S)$ of uniformly continuous and bounded functions for the supremum norm topology (for any Polish space, cf. \cite{Dudley:1966}, Lemma 8), one obtains:
\begin{prop}\label{prop:e-prop compact admissible}
A Markov operator on a compact space $K$ has the e-property for an admissible metric $d$ if and only if $\{U^nf:n\in\mathbb{N}\}$ is equicontinuous in $C(K)$ for every $f\in C(K)$. In particular, it has the e-property for any admissible metric.
\end{prop}
\begin{proof}
Let $g\in C(K)$ and $\varepsilon>0$. Because $K$ is compact there exists $f\in\mathrm{BL}(K,d)$ such that $\|f-g\|_\infty<\frac{1}{3}\varepsilon$. There exists an open neighbourhood $U_0$ of $x_0$ in $K$ such that 
$$
|U^nf(x)-U^nf(x_0)|<\frac{1}{3}\varepsilon\qquad\text{for all $x\in U_0$ and $n\in\mathbb{N}$.}
$$
Then for any $x\in x_0$ and $n\in \mathbb{N}$ we have
\begin{align*}
|U^ng(x) -U^ng(x_0)| &\leq |U^ng(x)-U^nf(x)| + |U^nf(x)-U^nf(x_0)|+ |U^nf(x_0)-U^ng(x_0)|\\
& < \|U^ng-U^nf\|_\infty + \smfrac{1}{3}\varepsilon + \|U^ng-U^nf\|_\infty\\
 & \leq 2\|g-f\|_\infty + \smfrac{1}{3}\varepsilon < \varepsilon. 
\end{align*}
For any $d'\in{\mathcal{D}}_S$, $\mathrm{BL}(K,d')\subset C(K)$. So the last statement follows immediately by restriction. 
\end{proof}

\begin{remark}
A similar statement as Proposition \ref{prop:e-prop compact admissible} holds for the Ces\`aro e-property, with similar proof. 
\end{remark} 

If $S$ is non-compact, $\mathrm{BL}(S,d)$ is no longer dense in $C_b(S)$. The argument put forward above for compact spaces obviously cannot be applied. The following question then arises:
{\it\begin{quote}
Under what conditions on the Markov operator $P$ or the space $S$ does it hold that $\{U^nf:n\in\mathbb{N}\}$ is equicontinuous for all $f\in C_b(S)$?
\end{quote}}
(and a similar question for the Ces\`aro e-property). The following observation gives a hint: 
\begin{lemma} We have
\[ C_b(S) = \bigcup_{d\in{\mathcal{D}}_S} \mathrm{BL}(S,d). \]
\end{lemma}
\begin{proof}
Let $f\in C_b(S)$. Define the metric $d'$ on $S$ by
\begin{equation}\label{eq:def new metric}
d'(x,y) := d(x,y) + |f(x)-f(y)|.
\end{equation}
Then, by construction, $f\in\mathrm{BL}(S,d')$. We claim that $d'$ is an admissible metric on $S$. To prove this statement we have to show that the identity map is a topological isomorphism between $S$ equipped with the metric topology of $d$ and $S$ with the metric topology of $d'$. Working with metric spaces, we use a sequence argument: let $x_0\in S$ and $(x_n)\subset S$ be such that $d(x_n,x_0)\to 0$ as $n\to\infty$. According to \eqref{eq:def new metric}, $d'(x_n,x_0)\to 0$, using the continuity of $f$. On the other hand, if $d'(x_n,x_0)\to 0$ as $n\to\infty$, then $d(x_n,x_0)\to 0$ too.
\end{proof}
\begin{clry}\label{clry:char question}
The family $\{U^nf:n\in\mathbb{N}\}$ is equicontinuous in $C_b(S)$ for any $f\in C_b(S)$ if and only if $P$ has the e-property for any admissible metric $d$ on $S$.
\end{clry}

There exist examples of Markov operators that have the e-property for one admissible metric, but not for some other metric:

\begin{example}\label{ex:one metric eprop} {\it (cf. \cite{Worm-thesis:2010}, Section 7.2, p.145 -- unpublished).} Let $S=[0,\infty)$ with the Euclidean metric $d$. Define $\Phi:S\to S$ by $\Phi(x)=x+1$. The map $\Phi$ defines a Markov operator $P_\Phi$ by means of
\[
P_\Phi\mu(A) := \mu\bigl(\Phi^{-1}(A)\bigl).
\]
The operator $P_\Phi$ has the e-property for $d$. Now define $d'(x,y):=d(e^x,e^y)$. Obviously $d'$ is an admissible metric on $S$. For all $n\in\mathbb{N}$ and $x>0$ one has
\[
d'\bigl( \Phi^n(0),\Phi^n(x) \bigr) = e^{n+x} - e^n \geq xe^n.
\]
Thus, the family of maps $(\Phi^n)_{n\in\mathbb{N}}$ is not equicontinuous at $0$ in $(S,d')$. Consequently, $P_\Phi$ cannot have the e-property for $d'$ (argument similar to proof of \cite{Hille_ea:2016}, Proposition 4.1).
\end{example}

Note that our question of interest is {\it purely topological}. There is no role for any particular truly metric property, which is consistent with Corollary \ref{clry:char question}. In what follows we provide a sufficient condition that answers the question positively. As it turns out, the conditions on $P$ as formulated below in Theorem \ref{thrm:e-prop for one is e-prop for any} can be transformed in a purely topological condition (as it should). 

Fundamental in our approach is an alternative characterisation of the e-property that was shown in \cite{Worm-thesis:2010}, Theorem 7.2.2 and \cite{Hille_ea:2016}, Theorem 4.1 and formulated for families $(P_\lambda)_{\lambda\in\Lambda}$ of regular Markov operators. In the formulation below of these results we added an additional equivalent condition (statement ({\it iii})) that is convenient in what follows:
\begin{thrm}\label{thrm:char e-property}
Let $(P_\lambda)_{\lambda\in\Lambda}$ be a family of regular Markov operators on a complete separable metric space $(S,d)$ and let $U_\lambda$ be the dual operator of $P_\lambda$. The following statements are equivalent:
\begin{enumerate}
\item[(i)] The family $\{U_\lambda f: \lambda\in\Lambda\}$ is equicontinuous in $C_b(S)$ for all $f\in\mathrm{BL}(S,d)$.
\item[(ii)] The family of maps $\{P_\lambda: \lambda\in\Lambda\}$ is equicontinuous in $C({\mathcal{M}}^+(S)_{\mathrm{BL},d},{\mathcal{M}}^+(S)_{\mathrm{BL},d})$.
\item[(iii)] The family of maps $\{P_\lambda: \lambda\in\Lambda\}$ is equicontinuous in $C(\mathcal{P}(S)_{\mathrm{weak}},\mathcal{P}(S)_{\mathrm{BL},d})$.
\end{enumerate}
\end{thrm}
\begin{proof}
{\it (i)}$\Rightarrow${\it(ii)}.\ This is the difficult (and remarkable) part of the proof of \cite{Hille_ea:2016}, Theorem 4.1.\\
{\it (ii)}$\Rightarrow${\it(iii)}.\ Follows immediately by restriction of the Markov operators $P^n$ to $\mathcal{P}(S)$.\\
{\it (iii)}$\Rightarrow${\it(i)}.\ Upon closer inspection, the proof in \cite{Hille_ea:2016}, Theorem 4.1 that ({\it ii}) implies ({\it i}) actually shows this step.
\end{proof}
Statement ({\it ii}) means
{\it\begin{enumerate} 
\item[(ii')] For every $\mu_0\in{\mathcal{M}}^+(S)$ and $\varepsilon>0$ there exists an open neighbourhood $U_0$ of $\mu_0$ in ${\mathcal{M}}^+(S)_{\mathrm{BL},d}$ such that $\|P_\lambda\mu_0- P_\lambda\mu\|_{\mathrm{BL},d}^*<\varepsilon$ for all $\mu\in U_0$ and $\lambda\in\Lambda$.
\end{enumerate}}

By taking $\Lambda=\mathbb{N}$ and either $P_n= P^n$ or $P_n:=P^{(n)}$ for a regular Markov operator $P$ one obtains characterisations of the e-property and Ces\`aro e-property, respectively. We call a family $(P_\lambda)$ tight if $\{P_\lambda\mu:\lambda\in\Lambda\}$ is uniformly tight for every $\mu\in{\mathcal{M}}^+(S)$. The characterisations of Theorem \ref{thrm:char e-property} allow us to derive the following result:

\begin{thrm}\label{thrm:e-prop for one is e-prop for any}
Let $(P_\lambda)_{\lambda\in\Lambda}$ be a tight family of regular Markov operator on $S$. If $(P_\lambda)_{\lambda\in\Lambda}$ is equicontinuous for one admissible metric on $S$, then it is equicontinuous for any admissible metric.
\end{thrm}

The key point in the proof of Theorem \ref{thrm:e-prop for one is e-prop for any} is a series of results on characterisation of compact sets in the space of continuous maps when equipped with the compact-open topology.
These can be stated in quite some generality, originating in \cite{Kelley:1955,Bargley-Yang:1966,Engelking:1977}.

A topological space $X$ is a {\it $k$-space} according to \cite{Kelley:1955}, p. 230, if for any $A\subset X$ holds that if the intersection of $A$ with each closed compact set in $X$ is closed, then $A$ is closed. According to \cite{Engelking:1977}, Theorem 3.3.20, every first countable Hausdorff space is a $k$-space. In particular, every metric space is a $k$-space.

The following is a rephrasing -- to our particular setting -- of \cite{Kelley:1955}, Theorem 7.18, p. 234, which is a kind of generalized Arzela-Ascoli theorem:
\begin{thrm}\label{thrm:char compactness}
Let $X$ be a Hausdorff $k$-space and $Y$ a metric space. Let $\mathcal{C}= C(X,Y)$, equipped with the compact-open topology. A subset $\mathcal{F}$ of $\mathcal{C}$ is compact if and only if the following three conditions hold:
\begin{enumerate}
\item[(a)] $\mathcal{F}$ is closed in $\mathcal{C}$,
\item[(b)] The closure of $\mathcal{F}[x]:=\{f(x): f\in\mathcal{F}\}$ in $Y$ is compact for every $x\in X$.
\item[(c)] $\mathcal{F}$ is equicontinuous on every compact subset of $X$.
\end{enumerate}
\end{thrm}

In \cite{Bargley-Yang:1966}, Theorem 3, it is shown that (c) in Theorem \ref{thrm:char compactness} can be replaced by the condition 
{\it \begin{quote}
(c')\quad $\mathcal{F}$ is equicontinuous on $X$,
\end{quote}}
because $Y$ is a metric space. Note that the compact-open topology is a toplogical construct. The specific metric on $Y$ is relevant for its definition only through the induced metric topology. Moreover, Theorem \ref{thrm:char compactness} can be rephrased for a family $\mathcal{F}$ that is relatively compact in $\mathcal{C}$, meaning that its (compact-open) closure is compact:

\begin{thrm}\label{thrm:char rel compactness}
Let $X$ be a Hausdorff $k$-space and $Y$ a metric space. Let $\mathcal{C}= C(X,Y)$, equipped with the compact-open topology. A subset $\mathcal{F}$ of $\mathcal{C}$ is relatively compact iff:
\begin{enumerate}
\item[(a)] The closure of $\mathcal{F}[x]:=\{f(x): f\in\mathcal{F}\}$ in $Y$ is compact for every $x\in X$.
\item[(b)] $\mathcal{F}$ is equicontinuous on every compact subset of $X$.
\end{enumerate}
Statement (b) can be replaced by
\begin{enumerate}
\item[(b')] $\mathcal{F}$ is equicontinuous on $X$.
\end{enumerate}
\end{thrm}

\begin{proof}
Let $\overline{\mathcal{F}}$ be the closure of $\mathcal{F}$ in $\mathcal{C}$. Assume it is compact, then according to Theorem \ref{thrm:char compactness}, the closure of $\overline{\mathcal{F}}[x]$ in $Y$ is compact for every $x\in X$. Hence the closure of $\mathcal{F}[x]$, which is contained in the closure of $\overline{\mathcal{F}}[x]$, will be compact too. The family $\mathcal{F}$ is equicontinuous on $X$ for every compact subset of $X$, because it is a subset of $\overline{\mathcal{F}}$ that has this property.\\
On the other hand, if $\mathcal{F}$ satisfies (a) and (b), or (b'), then $\overline{\mathcal{F}}$ obviously satisfies condition (a) in Theorem \ref{thrm:char compactness}. Now let $f\in\overline{\mathcal{F}}$. Then there exists a net $(f_{(n)}u)\subset \mathcal{F}$ such that $f_{(n)}u\to f$. Point evaluation at $x$ is continuous for the compact-open topology, so $f_{(n)}u(x)\to f(x)$ in $Y$. Since $f_{(n)}u(x)$ is contained in a compact set in $Y$ for every ${(n)}u$, $f(x)$ will be contained in this compact set too. So (b) holds in Theorem \ref{thrm:char compactness} for $\overline{\mathcal{F}}$. In a similar way one can show (c) in Theorem \ref{thrm:char compactness}. Let $K\subset X$ be compact. The compact-open topology on $C(X,Y)$ is identical to the topology of uniform convergence on compact subsets (cf. \cite{Kelley:1955}, Theorem 7.11). So if $f_*\in \overline{\mathcal{F}}$ and $(f_{(n)}u)\subset \mathcal{F}$ is a net such that $f_{(n)}u\to f_*$, then $f_{(n)}u|_K\to f_*|_K$ uniformly. If $x_0\in K$, then for every $\varepsilon>0$ there exists an open neighbourhood $U$ of $x_0$ in $K$ such that
\[
d_Y(f(x),f(x_0))<\smfrac{1}{2}\varepsilon\qquad \mbox{for all}\ f\in\mathcal{F},\ x\in U.
\]
Consequently, 
\[
d_Y(f_*(x),f_*(x_0)) = \lim_{(n)}u d_Y(f_{(n)}u(x),f_{(n)}u(x_0)) \leq \smfrac{1}{2}\varepsilon <\varepsilon
\]
for all $x\in U$. So $\overline{\mathcal{F}}$ is equicontinuous on $K$ too. Theorem \ref{thrm:char compactness} then yields the compactness of $\overline{\mathcal{F}}$ in $\mathcal{C}$, hence the relative compactness of $\mathcal{F}$.
\end{proof}

\begin{proof} {\it (Theorem \ref{thrm:e-prop for one is e-prop for any}).}\ Let $d$ be the admissible metric on $S$ for which $(P_\lambda)$ is equicontinuous in $\mathcal{C}_d:=C(\mathcal{P}(S)_{\mathrm{weak}}, \mathcal{P}(S)_{\mathrm{BL},d})$. Let $d'$ be any other admissible metric on $S$. We must show that $(P_\lambda)$ is an equicontinuous family in $\mathcal{C}_{d'}:=C(\mathcal{P}(S)_{\mathrm{weak}}, \mathcal{P}(S)_{\mathrm{BL},d'})$.\\
By assumption, $\{P_\lambda\mu:\lambda\in\Lambda\}$ is tight for every $\mu\in\mathcal{P}(S)$. By Prokhorov's Theorem, it is relatively compact in $\mathcal{P}(S)_{\mathrm{BL},d}$, because the $\|\cdot\|_{\mathrm{BL},d}$-norm topology coincides with the weak topology on ${\mathcal{M}}^+(S)$. Because $(P_\lambda)$ is equicontinuous in $\mathcal{C}_d$, Theorem \ref{thrm:char rel compactness} yields that $(P_\lambda)$ is relatively compact in $\mathcal{C}_d$, for the compact-open topology. Since the topologies on $\mathcal{P}(S)$ defined by the norms $\|\cdot\|_{\mathrm{BL},d'}$, $d'$ admissible, all coincide with the weak topology, $(P_\lambda)$ is relatively compact in $\mathcal{C}_{d'}$ for any admissible metric $d'$. Again application of Theorem \ref{thrm:char rel compactness}, but now in opposite direction, yields that $(P_\lambda)$ is equicontinuous in $\mathcal{C}_{d'}$.
\end{proof}

\section{Equicontinuity and asymptotic stability}
\label{sec:asympt stability}

In this section we discuss the results that we obtained by attempting to repair a gap in the proof of \cite{Meyn-Tweedie:2009}, Proposition 6.4.2, which asserted that an asymptotically stable Markov-Feller operator should have the e-property, in our wording. The following example shows that this cannot be expected to hold in this generality:

\begin{example}\label{ex:asympt stable MO without eprop}
Let $S:=\{1/n:n\in\mathbb{N}\}\cup\{0\}$, equipped with the relative topology of $\mathbb{R}$. The space  $S$ is a compact Polish space. Define $T:S\to S$ by $T(1/n)=1/(n-1)$ for $n>1$, $T(0)=0$ and $T(1)=0$. Then $T$ is continuous and one can easily check that the operator $P_T$ given by the formula
\[
P_T\mu(A) := \mu(T^{-1}(A))\qquad \mbox{for}\ A\subset S
\]
is a Feller operator that is asymptotically stable, with invariant measure $\mu_*={\rm d}elta_0$. However, $P_T$ does not satisfy the e-property, because the family of maps $\{T^n:n\in\mathbb{N}\}$ is not equicontinuous at $0$.
\end{example}

Indeed, Proposition \ref{prop:simple attractor} and Theorem \ref{main_theorem} provide additional conditions that do allow one to conclude that the asymptotically stable Markov-Feller operator will have the e-property. 
Recall the Ces\`aro averages $P^{(n)}$ of a Markov operator $P$ and the concept of Ces\`aro e-property (Section \ref{sec:equicont and tightness}). The following result shows that asymptotically stability of a Feller operator implies the {\it Ces\`aro e-property}, without additional conditions. Moreover, combined with the example above, it implies that the set of Markov operators with the e-property is a proper subset of those with the Ces\`aro e-property.

\begin{thrm}\label{thrm:asympt stability implies Cesaro}
Let $P$ be an asymptotically stable Markov-Feller operator on the Polish space $S$ with invariant measure $\mu_*$. Then for any admissible metric $d$ on $S$,
\begin{equation}\label{eq:uniform conv global}
\sup_{\mu\in\mathcal{P}(S)} \| P^{(n)}\mu - \mu_*\|_{\mathrm{BL},d}^* \to 0 \qquad \mbox{as}\ n\to\infty.
\end{equation} 
Moreover, $P$ has the Ces\`aro e-property for any admissible metric on $S$.
\end{thrm}

Its proof -- as we present it -- builds on some general results for dynamical systems in complete metric spaces that we will introduce first.
These are versions for discrete time dynamical systems of those in \cite{Sell-You:2002}, Chapter 2. 

Let $(X,\rho)$ be a complete metric space and $\phi:X\to X$ a continuous map. The {\it $\omega$-limit set} of $E\subset X$ is
\[
\omega(E):= \{x\in X: \mbox{there exist}\ (x_k)\subset E,\ n_k\in\mathbb{N}, n_k\to\infty,\ \phi^{n_k}(x_k)\to x\ \mbox{as}\ k\to\infty\}.
\]
One writes $\omega(x)$ instead of $\omega(\{x\})$. Let $E,F\subset X$, non-empty. The set $E$ is called {\it invariant} (under $\phi$) if $\phi(E)=E$. Moreover, $E$ {\it attracts} $F$ if
\[
{\rm d}elta(\phi^n[F], E) := \sup_{x\in F} \rho(\phi^n(x), E) \to 0 \qquad \mbox{as}\ n\to \infty.
\]

The function $\phi$ is {\it asymptotically compact} on $E\subset X$ if for every sequence $(x_k)\subset E$ and $(n_k)\subset\mathbb{N}$ for which $n_k\to\infty$, there exists subsequences $(x_{k_m})$ and $(n_{k_m})$ such that the sequence $(\phi^{n_{k_m}}(x_{k_m}))$ converges in $X$. The importance of asymptotic compactness for the theory of dynamical systems is indicated by the following lemma:

\begin{lemma}[based on \cite{Sell-You:2002}, Lemma 23.1]
\label{lem:asympt compactness implications}
If $\phi$ is asymptotically compact on the non-empty set $E\subset X$, then $\omega(E)$ is a non-empty compact invariant set, and $\omega(E)$ attracts $E$.
\end{lemma}
\begin{proof} The proof is similar to that of Lemma 23.1 in \cite{Sell-You:2002}.
\end{proof}

The following proposition gives useful conditions for establishing asymptotic compactness:
\begin{prop}\label{prop:asympt compactness cond}
Let $\phi:X\to X$ be continuous. If the orbits $\{\phi^n(x): n\in\mathbb{N}\}$ are precompact in $(X,\rho)$ for every $x\in X$ and $\mathcal{A}_w:=\bigcup_{x\in X}\omega(x)$ is precompact, then $\phi$ is asymptotically compact on $X$. Moreover, $\mathcal{A}_s:=\omega(X)$ is a non-empty compact invariant set that attracts $X$.
\end{prop}
\begin{proof}
Let $(x_k)\subset X$ and $(n_k)\subset \mathbb{N}$ such that $n_k\to\infty$. Since orbits are precompact, there exists a subsequence $(n^1_k)$ of $n_k$ such that $\phi^{n^1_k}(x_1)$ converges, say to $x^*_1$, as $k\to\infty$. Inductively, one obtains a subsequence $(n^l_k)$ of $(n^{l-1}_k)$ such that
$\phi^{n^l_k}(x_l)$ converges to $x^*_l$. The sequence $(x^*_l)$ is contained in $\mathcal{A}_w$, which is precompact by assumption. Hence there exists a subsequence $(x^*_{l_m})$ that converges to $x^*$ in $X$.\\

Let $\varepsilon>0$. For every $m\in\mathbb{N}$ there exists $N_m\in\mathbb{N}$ such that
\[
\rho\bigl( \phi^{n^{l_m}_k}(x_{l_m}),x^*_{l_m} \bigr) < \smfrac{1}{2}\varepsilon \qquad \mbox{for all}\ k\geq N_m.
\]
Define $m_1:=n_k^{l_1}$ with $k=N_1$. Then $m_{i+1}:=n_k^{l_{i+1}}$ with $k\geq N_{i+1}$ so large that $m_{i+1}> m_i$. Thus, $(m_i)$ is a subsequence of $(n_k)$ such that $m_i\to\infty$. Next, there exists $N\in\mathbb{N}$ such that $\rho(x^*_{l_i},x^*)<\varepsilon/2$ for all $i\geq N$. Then finally,
\[
\rho\bigl( \phi^{m_i}(x_{l_i}), x^*\bigr)) \leq \rho\bigl(\phi^{m_i}(x_{l_i}), x^*_{l_i}\bigr) + \rho(x^*_{l_i}, x^*)\ <\ \smfrac{1}{2}\varepsilon + \smfrac{1}{2}\varepsilon\ =\ \varepsilon
\]
for all $i\geq N$. This proves the asymptotic compactness of $\phi$ on $X$. The last statement follows from Lemma \ref{lem:asympt compactness implications}.
\end{proof}

The following lemma is similar to Lemma 2 in \cite{Komorowksi_ea:2010} (for Markov semigroups):
\begin{lemma}\label{lem:repeated Cesaro averages}
For $l,n\in\mathbb{N}$, $n\geq l+1$, and $\mu\in\mathcal{P}(S)$,
\begin{equation}
\bigl\|P^{(n)} P^{(l)}\mu - P^{(n)}\mu\bigr\|_\mathrm{TV} \leq \frac{l+1}{n}.
\end{equation}
\end{lemma}
\begin{proof}
One has
\begin{align*}
\bigl\|P^{(n)} P^{(l)}\mu - P^{(n)}\mu\bigr\|_\mathrm{TV} & = \bigl\| \frac{1}{nl}\cdot \sum_{k=1}^n\sum_{m=1}^l P^{m+k}\mu - \frac{1}{n}\sum_{k=1}^n P^k\mu\bigr\|_\mathrm{TV}\\
& = \bigl\| \frac{1}{nl}\cdot \sum_{m=1}^l \left(\sum_{k'=m+1}^{n+m} P^{k'}\mu - \sum_{k'=1}^n P^{k'}\mu\right) \bigr\|_\mathrm{TV}\\
& = \bigl\| \frac{1}{nl}\cdot \sum_{m=1}^l \left( \sum_{k'=n+1}^{n+m} P^{k'}\mu - \sum_{k'=1}^m P^{k'}\mu\right) \bigr\|_\mathrm{TV}\\
& \leq \frac{1}{nl}\cdot \sum_{m=1}^l 2m = \frac{l+1}{n}.
\end{align*}
\end{proof}

We can now proceed to the proof of Theorem \ref{thrm:asympt stability implies Cesaro}.

\begin{proof}{\it (Theorem \ref{thrm:asympt stability implies Cesaro}).}\ 
Fix $d\in{\mathcal{D}}_S$. To simplify notation, we write $\|\cdot\|$ instead of $\|\cdot\|_{\mathrm{BL},d}^*$. By asymptotic stability, the invariant measure $\mu_*\in\mathcal{P}(S)$ is unique and $P^n\mu\to\mu_*$ as $n\to\infty$, for every $\mu\in\mathcal{P}(S)$.  Hence all orbits $\{P^n\mu:n\in\mathbb{N}\}$ for $\mu\in\mathcal{P}(S)$ are precompact in $\mathcal{P}(S)_{\mathrm{BL},d}$. That is, $P$ is tight. Moreover, the weak attractor of the dynamical system defined by $P$ in $\mathcal{P}(S)$ by iteration,
\[
\mathcal{A}_w := \bigcup_{\mu\in\mathcal{P}(S)} \omega(\mu) = \{\mu_*\},
\]
is compact. According to Proposition \ref{prop:asympt compactness cond}, $P$ is asymptotically compact on $\mathcal{P}(S)$ and $\mathcal{A}_s:=\omega(\mathcal{P}(S))$ is an attractor in a strong sense: it is a non-empty compact $P$-invariant set that attracts $\mathcal{P}(S)$ uniformly:
\begin{equation}\label{eq:uniform attraction}
{\rm d}elta\bigl(P^n[\mathcal{P}(S)],\mathcal{A}_s\bigr) := \sup_{\mu\in \mathcal{P}(S)} d(P^n\mu,\mathcal{A}_s) \to 0\quad\mbox{as}\ n\to\infty.
\end{equation}
Clearly, $\mu_*\in\mathcal{A}_s$. 

First we claim:
{\it\begin{enumerate}
\item[(C1)] The family ${\mathcal{E}}:=\bigl\{ \frac{1}{n}\sum_{k=1}^n (P|_{\mathcal{A}_s})^n : n\in\mathbb{N}\bigr\}$ is equicontinuous in $C(\mathcal{A}_s,\mathcal{P}(S)_{\mathrm{BL},d})$.
\end{enumerate}}
Suppose that ${\mathcal{E}}$ is not equicontinuous in $C(\mathcal{A}_s,\mathcal{P}(S)_{\mathrm{BL},d})$. Then there exists $\mu_0\in\mathcal{A}_s$ at which the familiy is not equicontinuous: there exists $\varepsilon_0>0$ such that for every open neighbourhood $U_0$ of $\mu_0$ in $\mathcal{A}_s$, there exists $\mu\in U_0$ such that for some $n\in\mathbb{N}$
\begin{equation}\label{eq:negation equicont}
\|P|_{\mathcal{A}_s}^{(n)}\mu - P|_{\mathcal{A}_s}^{(n)}\mu_0\| = \|P^{(n)}\mu - P^{(n)}\mu_0\|\geq \varepsilon_0.
\end{equation}
Thus, we can find a sequence $(\mu_m)\subset\mathcal{A}_s$ such that $\mu_m\to\mu_0$, and $n_m\in\mathbb{N}$, $n_m\to\infty$ as $m\to\infty$, such that 
\begin{equation}\label{eq:selection sequence}
\|P^{(n_m)}\mu_m - P^{(n_m)}\mu_0\| \geq \varepsilon_0.
\end{equation}
In fact, for each $k$ one can select $n_k\in \mathbb{N}$ and $\mu_k\in B(\mu_0,\frac{1}{k})$ such that \eqref{eq:selection sequence} holds. If the sequence $(n_k)$ were bounded, then there exists a subsequence $(n_{k_m})$ that is constant, say $n'_m:=n_{k_m}=n'$ for all $m$. Then $\mu_{k_m}\to \mu_0$, while 
\[
\|P^{(n'_m)}\mu_{k_m} - P^{(n'_m)}\mu_0\| = \|P^{(n')}\mu_{k_m} - P^{(n')}\mu_0\|\geq \varepsilon_0.
\]
This contradicts the continuity of $P^{(n')}$. So there exists a subsequence of $(n_{k_m})$ of $(n_k)$ with $n_{k_m}\to\infty$ as $m\to\infty$. We now write $\mu_m:=\mu_{k_m}$ and $n_m:=n_{k_m}$.

Since $\mathcal{A}_s$ is compact, its closed convex hull, ${\overline{\mathrm{co}}}(\mathcal{A}_s)$, is compact too (e.g. use \cite{Rudin:1973}, Theorem 3.25). Since $P^{(n_m)}\mu_m\in{\overline{\mathrm{co}}}(\mathcal{A}_s)$ for all $m$, upon going to a subsequence, we may assume that $P^{(n_m)}\mu_m\to\mu^*$ as $m\to\infty$. Since $P$ is a Feller operator, $PP^{(n_m)}\mu_m \to P\mu^*$. On the other hand
\[
PP^{(n_m)}\mu_m  = \frac{1}{n_m} \sum_{k=1}^{n_m} P^k\mu_m + \frac{1}{n_m} P^{n_m+1} \mu_m  - \frac{1}{n_m} P\mu_m,
\]
from which we derive that $PP^{(n_m)}\mu_m\to\mu^*$. So $P\mu^*=\mu^*$. Because aymptotic stability implies that the invariant measure is unique, $\mu^*=\mu_*$. Then
\[
\| P^{(n_m)}\mu_m - P^{(n_m)}\mu_0 \| \leq \| P^{(n_m)}\mu_m - \mu_* \| + \| \mu_* - P^{(n_m)}\mu_0 \| \to 0,
\]
which contradicts \eqref{eq:selection sequence}. Thus, (C1) holds.

From (C1) we obtain
\begin{equation}\label{eq:uniform conv on attractor}
\sup_{\mu\in\mathcal{A}_s} \| P^{(n)}\mu - \mu_*\| \to 0 \qquad \mbox{as}\ n\to\infty.
\end{equation}
Indeed, let $\varepsilon>0$. According to (C1), for each $\mu\in\mathcal{A}_S$ there exists and open neighbourhood $U_\mu$ of $\mu$ in $\mathcal{A}_s$ such that 
\[
\| P^{(n)} \mu - P^{(n)} {(n)}u\| < \smfrac{\varepsilon}{2}\qquad \mbox{for every}\ {(n)}u\in U_\mu,\ n\in\mathbb{N}.
\]
$\mathcal{A}_s$ is compact, so there exists finitely many $U_\mu$ that cover $\mathcal{A}_s$, say $\mathcal{A}_s =\bigcup_{i=1}^m U_{\mu_i}$. For each $i$ there exists $N_i\in\mathbb{N}$ such that 
\[
\| P^{(n)} \mu_i - \mu_*\| < \smfrac{\varepsilon}{2}\qquad \mbox{for all}\ n\geq N_i.
\]
Take $N:=\max_{1\leq i\leq m}(N_i)$. Note that the choice of $N$ is independent of $\mu$. For any $\mu\in\mathcal{A}_s$ there exists $i$ such that $\mu\in U_{\mu_i}$. Hence for all $n\geq N$,
\[
\| P^{(n)}\mu - \mu_*\| \leq \|P^{(n)}\mu - P^{(n)}\mu_i\| + \|P^{(n)} \mu_i - \mu_*\| < \smfrac{\varepsilon}{2} + \smfrac{\varepsilon}{2} =\varepsilon,
\]
which proves \eqref{eq:uniform conv on attractor}.

Having established \eqref{eq:uniform conv on attractor}, select $l\in\mathbb{N}$ such that $\|P^{(l)}\mu-\mu_*\|<\varepsilon/3$ for all $\mu\in\mathcal{A}_s$. The operator $P^{(l)}$ is continuous on $\mathcal{P}(S)_{\mathrm{BL},d}$, so for each $\mu\in\mathcal{A}_s$ there exists an open neighbourhood $U_\mu$ of $\mu$ in $\mathcal{P}(S)$ such that $\|P^{(l)}{(n)}u-P^{(l)}\mu\|<\varepsilon/3$ for all ${(n)}u\in U_\mu$. Because $U:=\bigcup_{\mu\in\mathcal{A}_s} U_\mu$ is open and $\mathcal{A}_s\subset U$ is compact, there exists ${\rm d}elta>0$ such that
\begin{equation}
\mathcal{A}_s^{\rm d}elta := \{{(n)}u\in\mathcal{P}(S): d({(n)}u,\mathcal{A}_s)<{\rm d}elta\} \subset U.
\end{equation}
According to \eqref{eq:uniform attraction}, there is $N_0\in\mathbb{N}$ such that $P^n[\mathcal{P}(S)]\subset \mathcal{A}_s^{\rm d}elta$ for all $n\geq N_0$. Hence, for each ${(n)}u\in\mathcal{P}(S)$ and $n\geq N_0$, there exists $\mu_{{(n)}u,n}\in\mathcal{A}_s$ such that $P^n{(n)}u \in U_{\mu_{{(n)}u,n}}$. Consequently, for any ${(n)}u\in\mathcal{P}(S)$, $\mu\in\mathcal{A}_s$ and $n\geq N_0$,
\begin{align*}
\|P^{(l)}P^n{(n)}u - P^{(l)}\mu\| &\leq \|P^{(l)}P^n{(n)}u - P^{(l)}\mu_{{(n)}u,n}\| \\
&\qquad +\|P^{(l)}\mu_{{(n)}u,n} - \mu_*\| + \|\mu_* - P^{(l)}\mu\| \\
& < \smfrac{\varepsilon}{3} + \smfrac{\varepsilon}{3} + \smfrac{\varepsilon}{3} = \varepsilon, 
\end{align*}
according to the choice of $l$ in view of \eqref{eq:uniform conv on attractor} and the sets $U_{\mu}$. Take $N_0'\in\mathbb{N}$, $N_0'\geq N_0$,  such that $\varepsilon N_0'>2(N_0-1)$. Then for all ${(n)}u\in\mathcal{P}(S)$, $\mu\in\mathcal{A}_s$ and $n\geq N'_0$:
\begin{align*}
\|P^{(n)} P^{(l)}{(n)}u - P^{(l)}\mu\|\ & \leq\ \bigl\| \frac{1}{n}\sum_{k=1}^{N_0-1} \bigl(P^kP^{(l)}{(n)}u - P^{(l)}\mu\bigr)\bigr\| + \bigl\| \frac{1}{n} \sum_{k=N_0}^n \bigl(P^kP^{(l)}{(n)}u - P^{(l)}\mu\bigr) \bigr\|\\
& \leq\ \smfrac{2(N_0-1)}{n} +  \frac{1}{n} \sum_{k=N_0}^n \bigl\|P^kP^{(l)}{(n)}u - P^{(l)}\mu\bigr\|\\
& \leq\ \varepsilon + \smfrac{n-N_0+1}{n}\cdot \varepsilon\ <\ 2\varepsilon.
\end{align*}
Let $N\in \mathbb{N}$ be such that $N\geq \max(N'_0,l+1, \frac{l+1}{\varepsilon})$. Then, by noting that $\|\mu'\|\leq\|\mu'\|_\mathrm{TV}$ holds for any $\mu'\in{\mathcal{M}}(S)$ and using Lemma \ref{lem:repeated Cesaro averages}, for any choice of $\mu\in\mathcal{A}_s$, one obtains for all ${(n)}u\in\mathcal{P}(S)$ and $n\geq N$:
\begin{align*}
\|P^{(n)}{(n)}u - \mu_*\|\ &\leq\ \|P^{(n)}{(n)}u - P^{(n)} P^{(l)}{(n)}u\| + \|P^{(n)} P^{(l)}{(n)}u - P^{(l)}\mu\| + \|P^{(l)}\mu -\mu_*\|\\
& \leq\ \varepsilon + 2\varepsilon + \smfrac{1}{3}\varepsilon\ <\ 4\varepsilon. 
\end{align*}
Thus we have proven \eqref{eq:uniform conv global}.

To complete the proof of the Ces\`aro e-property, fix $\mu_0\in\mathcal{P}(S)$ and let $\varepsilon>0$. According to \eqref{eq:uniform conv global} there exists $N\in\mathbb{N}$ such that 
\begin{equation}\label{eq:unform est}
\sup_{\mu\in\mathcal{P}(S)} \|P^{(n)}\mu-\mu_*\|<\smfrac{\varepsilon}{2}\quad \mbox{for all}\ n\geq N.
\end{equation}
Let $\bar{\mu}_n:=P^{(n)}\mu_0$ for $n=1,2,{\rm d}ots, N-1$. The preimage $V_n$ of $B(\bar{\mu}_n,\varepsilon)$ under $P^{(n)}$ is an open neighbourhood of $\mu_0$ by the Feller property of $P$. Put $U:=\bigcap_{n=1}^{N-1} V_n$. Then $U$ is an open neighbourhood of $\mu_0$ in $\mathcal{P}(S)$ and for every $\mu\in U$ and $n=1,2,{\rm d}ots, N-1$ one has $\|P^{(n)}\mu_0-P^{(n)}\mu\|<\varepsilon$ by construction. For $n\geq N$,
\[
\|P^{(n)}\mu_0-P^{(n)}\mu\|\leq \|P^{(n)}\mu_0-\mu_*\|+ \|\mu_*-P^{(n)}\mu\|< \smfrac{\varepsilon}{2} +\smfrac{\varepsilon}{2} =\varepsilon,
\]
because of \eqref{eq:unform est}. So $\{P^{(n)} :n\in\mathbb{N}\}$ is equicontinuous at $\mu_0$ for the metric $d$. Since $P$ is tight, Theorem \ref{thrm:e-prop for one is e-prop for any} yields that $P$ has the Ces\`aro e-property for every admissible metric. The proof is complete.
\end{proof}

\begin{remark} 
Much of the proof of Theorem \ref{thrm:asympt stability implies Cesaro} would be simplified if one knew {\it a priori} that $\mathcal{A}_s=\{\mu_*\}$. However, in principle $\mathcal{A}_s$ could be strictly larger than $\mathcal{A}_w$. In fact, in various situations, $\mathcal{A}_s{(n)}eq\{\mu_*\}$, e.g. Example \ref{ex:asympt stable MO without eprop}, because of the following result.
\end{remark}

\begin{prop}\label{prop:simple attractor}
Let $P$ be a Markov-Feller operator that is asymptotically stable, with invariant measure $\mu_*$. If $\mathcal{A}_s = \omega(\mathcal{P}(S)) =\{\mu_*\}$, then $P$ has the e-property for any admissible metric on $S$.
\end{prop}
\begin{proof}
As in the start of the proof of Theorem \ref{thrm:asympt stability implies Cesaro} one gets that $\mathcal{A}_s = \omega(\mathcal{P}(S))$ attracts $\mathcal{P}(S)$ uniformly, equation \eqref{eq:uniform attraction}. Since it is assumed that $\mathcal{A}_s=\{\mu_*\}$, we immediately get
\begin{equation}
\sup_{\mu\in\mathcal{P}(S)} \|P^n-\mu_*\|_{\mathrm{BL},d}^* \to 0\qquad \mbox{as}\ n\to\infty.
\end{equation}
From this one obtains equicontinuity of the family $\{P^n:n\in\mathbb{N}\}$ for the metric $d$, similarly as in the last part of the proof of Theorem \ref{thrm:asympt stability implies Cesaro}, now with $P^{(n)}$ replaced by $P^n$.
\end{proof}

The result presented in Theorem \ref{thrm:asympt stability implies Cesaro} can be strengthened to the conclusion that the Markov operator has the e-property also under an additional condition on the invariant measure:
\begin{thrm}\label{main_theorem} Let $P$ be an asymptotically stable Markov-Feller operator and let $\mu_*$ be its unique invariant measure.
If $\textup{Int}_S(\supp\mu_*){(n)}eq\emptyset$, then $P$ satisfies the e--property.
\end{thrm}

Its proof involves the following lemma:
\begin{lemma}Let $P$ be an asymptotically stable Markov-Feller operator and let $\mu_*$ be its unique invariant measure. Let $U$ be its dual. If $\textup{Int}_S(\supp\mu_*){(n)}eq\emptyset$, then for every $f\in C_b(S)$ and any $\varepsilon>0$ there exists a ball $B\subset\supp\mu_*$ and $N\in\mathbb{N}$ such that
\begin{equation}\label{e1_19.11.16}
|U^nf(x)-U^n f(y)|\le\varepsilon\qquad\text{for any $x, y\in B$,\ $n\geq N$}.
\end{equation}
\end{lemma}

\begin{proof}. Fix $f\in C_b(S)$ and $\varepsilon>0$. Let $W$ be an open set such that $W\subset\supp\mu_*$. Set $Y={\overline{W}}$ and observe that the subspace $Y$ is a Baire space. Denote by 
$$
Y_n:=\{x\in Y: |U^mf(x)-\langle f, \mu_*\rangle | \le\varepsilon/2\,\,\, \text{for all $m\ge n$}\}
$$
and observe that $Y_n$ is closed and
$$
Y=\bigcup_{n=1}^{\infty} Y_n.
$$
By the Baire category theorem there exist $N\in\mathbb N$ such that $\textup{Int}_Y Y_{N}{(n)}eq\emptyset.$
Thus there exists a set $V\subset Y_N$ open in the space $Y$ and consequently a ball  $B$ in $S$ such that
$B\subset Y_N\subset\supp\mu_*$. Since 
$$
|U^mf(x)-\langle f, \mu_*\rangle | \le\varepsilon/2\qquad\text{for any $x\in B$ and $m\ge N$},
$$
condition (\ref{e1_19.11.16}) is satisfied.
\end{proof}

We are ready to prove Theorem \ref{main_theorem}.
\begin{proof} {\it (Theorem \ref{main_theorem}).}\  Assume, contrary to our claim, that $P$ does not satisfy the e--property. Therefore there exist a function $f\in C_b(S)$ and a point $x_0\in S$ such that
$$
\limsup_{x\to x_0} \limsup_{n\to\infty}|U^n f(x)-U^n f(x_0)|>0.
$$
Choose $\varepsilon>0$ such that
$$
3\varepsilon\le \limsup_{x\to x_0} \limsup_{n\to\infty}|U^n f(x)-U^n f(x_0)|.
$$
Let $B:=B(z, 2r)$ be a ball such that condition (\ref{e1_19.11.16}) holds. Since $B(z, r)\subset\supp\mu_*$, we have $\gamma:=\mu_*(B(z, r))>0$. Choose a positive constant $\alpha$: $0<\alpha<\gamma$. Since the operator $P$ is asymptotically stable, we have
\begin{equation}\label{e2_19.11.16}
\liminf_{n\to\infty}P^n\mu (B(z, r))>\alpha\qquad\text{for all $\mu\in\mathcal{P}(S)$},
\end{equation}
by the Alexandrov theorem.

Let $k\ge 1$ be such that $2(1-\alpha)^k\|f\|_\infty<\varepsilon$. By induction we are going to define two sequences of measures $({(n)}u^{x_0}_i)_{i=1}^k$, 
$(\mu^{x_0}_i)_{i=1}^k$ and a sequence of integers $(n_i)_{i=1}^k$ in the following way: let $n_1\ge 1$ be such that
\begin{equation}
P^{n_1}{\rm d}elta_{x_0}(B(z, r))>\alpha. 
\end{equation}
Choose $r_1<r$ such that $ P^{n_1}{\rm d}elta_{x_0}(B(z, r_1))>\alpha$ and $P^{n_1}{\rm d}elta_{x_0}(\partial B(z, r_1))=0$ and set 
\begin{equation}
{(n)}u^{x_0}_1(\cdot)=\frac{P^{n_1}{\rm d}elta_{x_0}(\cdot\cap B(z, r_1))}{P^{n_1}{\rm d}elta_{x_0}(B(z, r_1))}
\end{equation}
and
\begin{equation}
\mu^{x_0}_1(\cdot)=\frac{1}{1-\alpha}\left(P^{n_1}
{\rm d}elta_{x_0}(\cdot)-\alpha{(n)}u_1^{x_0} (\cdot)\right).
\end{equation}
Assume that we have done it for $i=1,\ldots, l$, for some $l<k$. Now let $n_{l+1}$ be such that
\begin{equation}
P^{n_{l+1}}\mu_l^{x_0}(B(z, r))>\alpha. 
\end{equation}
Choose $r_{l+1}<r$ such that $ P^{n_{l+1}}\mu_l^{x_0}(B(z, r_{l+1}))>\alpha$ and $P^{n_{l+1}}\mu_l^{x_0}(\partial B(z, r_{l+1}))=0$ and set 
\begin{equation}
{(n)}u^{x_0}_{l+1}(\cdot)=\frac{P^{n_{l+1}}\mu_l^{x_0}(\cdot\cap B(z, r_{l+1}))}{P^{n_{l+1}}\mu_l^{x_0}(B(z, r_{l+1}))}
\end{equation}
and
\begin{equation}
\mu^{x_0}_{l+1}(\cdot)=\frac{1}{1-\alpha}\left(P^{n_{l+1}}
\mu^{x_0}_{l}(\cdot)-\alpha{(n)}u_{l+1}^{x_0} (\cdot)\right).
\end{equation}
We are done. We have
$$
\begin{aligned}
P^{n_1+\ldots+n_k}{\rm d}elta_{x_0}(\cdot)&=\alpha P^{n_2+\ldots+n_k}{(n)}u^{x_0}_1(\cdot)
+\alpha (1- \alpha)
P^{n_3+\ldots+n_k}{(n)}u^{x_0}_2(\cdot)
+\ldots + \\
&+\alpha (1- \alpha )^{k-1}
{(n)}u^{x_0}_k (\cdot)+(1-\alpha)^k\mu^{x_0}_k (\cdot).
\end{aligned}
$$
By induction we check that ${(n)}u_i^x-{(n)}u_i^{x_0}\to 0$ and $\mu_i^x-\mu_i^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$.
Indeed, if $i=1$, then ${(n)}u_1^x-{(n)}u_1^{x_0}\to 0$ weakly (as $d(x, x_0)\to 0$), by the fact that $P$ is a Feller operator and $\lim_{d(x, x_0)\to 0}P^{n_1}{\rm d}elta_{x}(B(z, r_1))=P^{n_1}{\rm d}elta_{x_0}(B(z, r_1))$, by the Alexandrov theorem due to the fact that $P^{n_1}{\rm d}elta_{x_0}(\partial B(z, r_1))=0$. On the other hand, the weak convergence $\mu_1^x-\mu_1^{x_0}\to 0$ as $d(x, x_0)\to 0$ follows directly from the definition of $\mu_1^x$. Moreover, observe that for $x$ sufficiently close to $x_0$ we have $P^{n_1}{\rm d}elta_{x}(B(z, r))>\alpha$ and therefore $\mu_1^x\in\mathcal{P}(S)$.

Assume now that we have proved that ${(n)}u_i^x-{(n)}u_i^{x_0}\to 0$ and $\mu_i^x-\mu_i^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$ for $i=1,\ldots, l$. We show that ${(n)}u_{l+1}^x-{(n)}u_{l+1}^{x_0}\to 0$ and $\mu_{l+1}^x-\mu_{l+1}^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$ too. Analogously, $\lim_{d(x, x_0)\to 0}P^{n_{l+1}}\mu_{l}^{x}(B(z, r_{l+1}))=P^{n_{l+1}}\mu_{l}^{x_0}(B(z, r_{l+1}))$, by the Alexandrov theorem due to the fact that $P^{n_{l+1}}\mu_{l}^{x_0}(\partial B(z, r_{l+1}))=0$ and from the definition of ${(n)}u_{l+1}^x$ we obtain that ${(n)}u_{l+1}^x-{(n)}u_{l+1}^{x_0}\to 0$ weakly as $d(x, x_0)\to 0$. The weak convergence $\mu_{l+1}^x-\mu_{\l+1}^{x_0}\to 0$ as $d(x, x_0)\to 0$ follows now directly from the definition of $\mu_{l+1}^x$ and for $x$ sufficiently close to $x_0$ we have $P^{n_{l+1}}\mu_{l}^{x}(B(z, r))>\alpha$ and therefore $\mu_{l+1}^x\in\mathcal{P}(S)$. We are done.

Observe that for any $x$ sufficiently close to $x_0$ and  all $n\ge n_1+\ldots+n_k$ we have
$$
\begin{aligned}
P^{n}{\rm d}elta_{x}(\cdot)&=\alpha P^{n-n_1}{(n)}u^{x}_1(\cdot)
+\alpha (1- \alpha)
P^{n-n_1-n_2}{(n)}u^{x}_2(\cdot)
+\ldots \\
&+\alpha (1- \alpha )^{k-1}
P^{n-n_1-\ldots-n_k}{(n)}u^{x}_k (\cdot)+(1-\alpha)^k P^{n-n_1-\ldots-n_k}\mu^{x}_k (\cdot),
\end{aligned}
$$
where $\supp{(n)}u^x_i\subset B(z, r)$ for all $i=1,\ldots, k$. Thus
\begin{align}
\limsup_{n\to\infty}|\langle f, P^n{(n)}u^x_i\rangle-\langle f, P^n{(n)}u^{x_0}_i\rangle|&=\limsup_{n\to\infty} |\langle U^nf-\langle f, \mu_*\rangle, {(n)}u^x_i\rangle-\langle U^nf -\langle f, \mu_*\rangle, {(n)}u^{x_0}_i\rangle|{(n)}onumber\\
&\le \varepsilon/2+\varepsilon/2=\varepsilon\label{e_20.11.16} 
\end{align}
for all $i=1,\ldots, k$ and $x$ sufficiently close to $x_0$. Hence
$$
\begin{aligned}
3\varepsilon<\limsup_{x\to x_0} \limsup_{n\to\infty}&|U^n f(x)-U^n f(x_0)|=
\limsup_{x\to x_0} \limsup_{n\to\infty}|\langle f, P^n{\rm d}elta_x\rangle - \langle f, P^n{\rm d}elta_{x_0}\rangle|\\
&\le \varepsilon (\alpha+\alpha (1-\alpha)+\ldots \alpha (1-\alpha)^{k-1}) +2(1-\alpha)^k \|f\|\\
&\le \varepsilon+\varepsilon=2\varepsilon,
\end{aligned} 
$$
which is impossible. This completes the proof.
\end{proof}
{\bf Acknowledgements.}\ We thank Daniel Worm for pointing out Example \ref{ex:one metric eprop} and Klaudiusz Czudek for providing Example \ref{ex:asympt stable MO without eprop}.

\end{document}