Let $T:X\rightarrow X$ be a continuous map on a compact metric space $(X,d)$. Suppose that $\mu$ is ergodic with respect to $T$ and for every $x\in X$ there exists a constant $C=C(x)$ such that for every $f \in C(X), f \geq 0$, \begin{align*} \limsup_{N \rightarrow \infty} \frac{1}{N} \sum_{n=0}^{N-1} f (T^nx) \leq C \int f d\mu. \end{align*}

Show that $T$ is uniquely ergodic.

I knew the following theorem

$\textbf{Theorem}$ the following properties are equivalent.

(i) $T$ is uniquely ergodic

(ii) For every $f\in C(X)$, \begin{align*} A_N^f:=\frac{1}{N} \sum_{n=0}^{N-1} f(T^nx) \rightarrow C_f, \end{align*}
where $C_f$ is a constant independent of $x$.

I don't know how to induce the convergence of $A_N^f$ from the assumption in the problem.

Any help is appreciated...

Thank you!!

  • $\begingroup$ please see my answer below $\endgroup$ – mathworker21 Nov 3 at 23:06

Lemma: Let $\nu,\mu$ be mutually singular probability measures on a metric space $X$. Then $$\sup_{\substack{f \in C(X,\mathbb{R}^{\ge 0}) \\ f \not \equiv 0}} \frac{\int f d\nu}{\int f d\mu} = +\infty.$$

Proof: Take disjoint $A,B$ with $\nu(A) = 1, \mu(B) = 1$. Take $\epsilon > 0$. Since finite Borel measures on a metric space are inner regular, there are closed $A' \subseteq A$ and $B' \subseteq B$ with $\nu(A') \ge 1-\epsilon$ and $\mu(B') \ge 1-\epsilon$. By Urysohn's Lemma, there is $f \in C(X,[0,1])$ with $f \equiv 1$ on $A'$ and $f \equiv 0$ on $B'$. For this $f$, we have $\frac{\int fd\nu}{\int fd\mu} \ge \frac{1-\epsilon}{\epsilon}$.


Main Problem:

Suppose $T$ is not uniquely ergodic. Let $\nu$ be another ergodic measure w.r.t. $T$. By the ergodic theorem, for any $f \in C(X,\mathbb{R})$, $\frac{1}{N}\sum_{n \le N} f(T^nx) \to \int fd\nu$ for $\nu$-a.e. $x \in X$. Since $X$ is compact, $C(X,\mathbb{R})$ is separable (w.r.t. $||\cdot||_\infty$ topology), so in fact for $v$-a.e. $x \in X$, all $f \in C(X,\mathbb{R})$ satisfy $\frac{1}{N}\sum_{n \le N} f(T^nx) \to \int fd\nu$. Take one such $x$. Then, by hypothesis, there is some $C$ with $\int fd\nu \le C \int fd\mu$ for every $f \in C(X,\mathbb{R}^{\ge 0})$. This contradicts the Lemma.


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