5
$\begingroup$

I'm trying to solve the following problem:

Let $T\colon X\to X$ be a continuous map on a compact metric space $X$, uniquely ergodic. Let $Y\neq \emptyset$ be an open set. Show that $t(x) = \min\left\{n\ge 1 : T^n x\in Y\right\}$ is well defined and bounded.

I was able to prove it assuming that the measure has full support:

Thanks to unique ergodicity, for every continous map $f$, $\frac 1 n \sum_{k=0}^{n-1} f(T^k x) \to c=\int_X f$ uniformly. If $t$ isn't bounded, then for each $n$ I can find a $x_n$ such that $\frac 1 n \sum_{k\le n-1} f(T^k x) = 0$, and $c=0$. In particular if I take $f\ge0$ to be 1 on an open set inside $Y$ and $0$ outside of $Y$, then I have a contradiction assuming that the measure has full support!

Do I need to assume that? Or is there another way? Taking such a $f$ was a hint given in the exercise, but for example if $\mu(Y) =0$ I don't know what to do. In fact I don't know any example where there is unique ergodicity and where some open sets have 0 measure, I'm also interested in such examples.

Thanks for any help

$\endgroup$
2
$\begingroup$

Here's a counter-example for what you are trying to prove. As your question hints at, this is a very simple example where unique ergodicity holds yet some nonempty open set has measure zero.

Take $X = \mathbb R \cup \{P\}$ to be the one-point compactification of the real line $\mathbb R$. Define $f : X \to X$ by the formula $$f(x) = \begin{cases} P & \quad\text{if $x=P$} \\ x+1 & \quad\text{if $x \in \mathbb R$} \end{cases} $$ The only invariant probability measure for this example is the Dirac measure on the point $P$, so $f$ is uniquely ergodic, but $\mathbb R = X - \{P\}$ is a nonempty invariant open subset.

Now define $Y = (-1,\infty) \subset \mathbb R$, which is an open subset of $X$. Note that $t(P) = +\infty$, and perhaps this might be what you would mean by saying that $t(P)$ is "not well-defined". And note also that while $t(x)$ is indeed well-defined for $x \in \mathbb R$, and while $t(x)=0$ for $x > -1$, on the other hand for any natural number $n$ any value of $x \in (-\infty,-n]$ satisfies $t(x) \ge n$, and so $t(x)$ has no upper bound.

My guess as to what is going on here is that you need to simply add some kind of topological hypothesis to your problem, for example that $f$ is topologically minimal, meaning that there is no proper invariant closed subset.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! I'm ashamed that I didn't think of it! Yes, by well-defined I mean exactly what you said $\endgroup$ – Tom Feb 8 '19 at 16:53
  • $\begingroup$ @Tom A usual definition of first return time to $Y$ assumes that you take a point $x \in Y$, not just any point of phase space. Does your problem define it differently? $\endgroup$ – Evgeny Feb 8 '19 at 17:34
  • $\begingroup$ No, in my problem this terminology isn't used. I agree that this is not very right to talk about the first return time of a point that wasn't there... $\endgroup$ – Tom Feb 8 '19 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.