Existence of the limit of periodic measures

Question

Let $T: X \to X$ be a continuous map over a compact metric space. We say that a measure $\mu$ is $T$-invariant if $T_{\ast} \mu= \mu$. We denote by $M(X, T)$ the space of all $T$-invariant Borel probability measures.

Assume that $f: M(X, T) \to \mathbb{R}$ is upper semi-continuous and affine (you can think the map look like $\mu \mapsto \int \varphi d\mu$, where $\varphi$ is $\mu$-a.e bounded). Suppose that $\nu_p$ is an $T^p$-invariant Borel probability measure. We denote an $T$-invariant measure $\mu_{p}:=\frac{1}{p}\sum_{i=0}^{p-1}T_{\ast}^{i} \nu$.

Does the limit of the sequence $(\mu_p)$ exist as $p \to \infty$? If so, what is it?

I think when $p \to \infty$, it is also affected on $\nu$ as it is an $T^p$-invariant.

By compactness, there is a subsequence $\mu_{p_{i}}$ that converges to $\nu$. Is it true that $$f(\mu_{p_{i}}) \to f(\nu)?$$

Answer 1

I may be misunderstanding the question, but in full generality, the answers are "no," i.e. the $\mu_p$ can fail to converge, and even when they converge to a limit $\mu$, $f(\mu_p)$ can fail to converge to $\mu$.

Consider the shift map $T$ on $\{0,1\}^{\mathbb{Z}}$ (with the product topology), defined by $(T x)(n) = x(n+1)$ for all $x,n$. In other words, $T$ just shifts a sequence one unit left.

Note that $0^\mathbb{Z}$ and $1^{\mathbb{Z}}$ are fixed points under $T$, and so their delta measures $\delta_0 := \delta_{0^\mathbb{Z}}$ and $\delta_1 := \delta_{1^\mathbb{Z}}$ are $T$-invariant. Therefore, they're also $T^p$-invariant for all $p$.

So, if you choose $\nu_p = \delta_0$ for even $n$ and $\nu_p = \delta_1$ for odd $n$, then $\mu_p = \nu_p$ for all $p$ and clearly doesn't converge.

For your other question, take $h$ to be measure-theoretic entropy. (I can't give a full definition here, but there are lots of great references, such as Walters's "Ergodic Theory"). It is known to be affine and upper semi-continuous (for the shift map) with respect to the weak-star topology on $M(X,T)$ (I assume this is the topology you're dealing with, but you didn't say...)

By the way, this also means that your claim about the general form of such $f$ is not quite correct; entropy $h(\mu)$ can't be written as the integral of a single function w.r.t. $\mu$.

Anyway, if you take your $\nu_p$ to be the average over delta-measures of ALL sequences with period $p$, then $\mu_p = \nu_p$. (Example: $\mu_2 = \nu_2 = \frac{1}{4}(\delta_{\ldots 0000 \ldots} + \delta_{\ldots 0101 \ldots} + \delta_{\ldots 1010 \ldots} + \delta_{\ldots 1111 \ldots})$.) It's easily checked that $\mu_p \rightarrow \mu$, where $\mu$ is the uniform i.i.d. Bernoulli measure, i.e. $\mu$ gives probability $1/2$ to seeing any letter at any location and different locations are independent.

But all periodic measures have entropy $0$, so all $\mu_p$ do as well, and $\mu$ has entropy $\log 2 > 0$. So $h(\mu_p)$ does not converge to $h(\mu)$.

The final note I would make is that if you redefine $\mu_p$ to be the average of delta measures over all $p$-periodic points (as in the example I just wrote), then there are conditions under which $\mu_p$ must converge and is known to converge to the so-called measure of maximal entropy of $(X,T)$. The simplest is when $(X,T)$ has the specification property, which is a very strong sort of topological mixing property; this was proved by Bowen in "Some systems with unique equilibrium states."

Answer 2

The notations are contradictory. Once $p$ is fixed, and then it varies.

Do you set $\mu_{n}:=\frac{1}{n}\sum_{i=0}^{n-1}T_{\ast}^{i}\nu$ for ALL $n \ge 1$ and assume that $T_{\ast}^{p} \nu = \nu$ for SOME fixed $p$ ?

If yes, $\mu_{p}$ is $T$-invariant and for every $n \ge 1$, calling $q_n$ and $r_n$ the quotient and the remainder of the Euclidean division by $p$, we get by $p$-periodicity of $(T_{\ast}^{i}\nu)_{i \ge 0}$ $$\sum_{i=0}^{n-1} T_{\ast}^{i}\nu = \sum_{i=0}^{pq_n-1} T_{\ast}^{i}\nu + \sum_{i=pq_n}^{n-1} T_{\ast}^{i}\nu = q_n\sum_{i=0}^{p-1} T_{\ast}^{i}\nu + \sum_{i=0}^{r_n-1} T_{\ast}^{i}\nu.$$ Dividing by $n$ we get $$\mu_n = \frac{pq_n}{n} \mu_p + \frac{r_n}{n} \mu_{r_n},$$ with the convention $\mu_0=\nu$ (actually, the choice of $\mu_0$ can be arbitrary and plays no role). If $f : M(X,T) \to \mathbb{R}$ is affine, we get $$f(\mu_n) = \frac{pq_n}{n} f(\mu_p) + \frac{r_n}{n} f(\mu_{r_n}).$$ The sequence $(f(\mu_{r_n}))_{n \ge 1}$ is periodic hence bounded, so $$f(\mu_n) \to f(\mu_p) \text{ as } n \to +\infty.$$ No other property of $f$ is required.