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Let $X$ be a compact metric space, $T$ a homeomorphism on $X$ and $\mu$ a $T$-invariant probability measure. Let $\phi:X\to\mathbb{R}$ be a continuous function and $\phi_n(x)=\phi(x)+\cdots+\phi(T^{n-1}x)$ be the induced cocycle.

A point $x\in X$ is said to be $\phi$-transient if $\phi_n(x)\to\infty$ as $n\to\infty$. Otherwise $x$ is said to be $\phi$-recurrent.

It may be more appropriate to define these notations in the skew-product system $(X\times\mathbb{R},\mu\times m)$, with $(T,\phi):(x,t)\mapsto (Tx,t+\phi(x))$. But above definitions also looks natural :)

Two propositions strengthen the $\phi$-recurrence property:

(1). For $\mu$-a.e. $\phi$-recurrent point $x$, $\displaystyle \liminf_{n\to\infty}|\phi_{n}(x)|=0$.

(2). Assume $(T,\mu)$ is ergodic and $\int\phi d\mu=0$. Then $\mu$-a.e. point $x$ is $\phi$-recurrent. In particular $\displaystyle \liminf_{n\to\infty}|\phi_{n}(x)|=0$ $\mu$-a.e. $x$.

My question is, how to prove these two properties?

Difference with Birkhoff Ergodic Theorem: $\frac{\phi_n(x)}{n}\to0$, but doesn't tell how $\phi_n(x)$ behaves.

Thank you!


Just recalled how to prove the recurrence in (2):

Let $C=\max_{x\in X}|\phi(x)|>0$. Then we divide $X$ into three subsets:

  • (recurrent part) $x\in X_r$ if $\phi_n(x)\in[-C,C]$ infinitely often;

  • (positive part) $x\in X_+$ if $\phi_n(x)\ge C$ for all $n$ large;

  • (negative part) $x\in X_-$ if $\phi_n(x)\le -C$ for all $n$ large.

Moreover both three are invariant. We need to show $X_{\pm}$ are of zero measure. If $\mu(X_+)>0$, then $\mu(X_+)=1$ (by ergodicity). So

$$0=\int\phi d\mu=\int\phi_n d\mu=\lim_{n\to\infty}\int \phi_n d\mu\ge\int\liminf_{n\to\infty}\phi_n d\mu\ge C>0,$$ contradicts in itself and hence $\mu(X_+)=0$. Similarly we have $\mu(X_-)=0$ and hence $\mu(X_r)=1$.

So (2) follows from (1).

share|improve this question
    
What is $f$? Is this $T$? –  Qiaochu Yuan Apr 17 '13 at 7:31
    
Classical Gottschalk-Hedlund's Theorem (see Theorem 2.9.4 in Katok-Hasselblatt) tells you that if you have a recurrent point then your cocycle is a coboundary: $\phi_n(x)=u(T^n(x))-u(x)$, with $u$ continuous function on $X$. –  Michele Triestino Apr 17 '13 at 8:29
2  
What is the question? –  Ian Morris Apr 17 '13 at 8:31
    
@Qiaochu Yes. This is a typo. @Ian I revised my question. I want to know how to prove these two propositions. –  Pengfei Apr 17 '13 at 21:06
    
Perhaps I am missing something, but how is the inequality $\lim_{n \to \infty} \int \phi_n d\mu \geq \int \liminf_{n \to \infty}\phi_n d\mu$ justified? Fatou's lemma does not work here, for example, because the functions $\phi_n$ might fail to be uniformly bounded below. –  Ian Morris Apr 18 '13 at 10:26
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1 Answer

I am not quite sure what the question is, but the following result might be a helpful starting point:

Theorem (Giles Atkinson, 1976): Let $T$ be an ergodic invertible measure-preserving transformation of a probability space $(X,\mathcal{F},\mu)$ and let $\phi \colon X \to \mathbb{R}$ be integrable. Then the following statements are equivalent:

  1. $\int \phi\,d\mu=0$

  2. For all $A \in \mathcal{F}$ such that $\mu(A)>0$, and all $\varepsilon>0$, there exists a nonzero integer $n$ such that $\mu\left(A \cap T^{-n}A \cap \{x \colon |\phi_n(x)|<\varepsilon\}\right)>0$.

Here we define $\phi_n(x)=0$ if $n=0$ and $\phi_n(x)=\phi_{-n}(T^nx)$ if $n$ is negative. Note that by taking $A=X$ and considering a sequence of values of $\varepsilon$ tending to zero we obtain $\liminf_{|n|\to \infty} \phi_n(x)=0$ a.e.

This falls slightly short of proving the statement (2) in the original post since a priori the sequence of integers $n$ along which convergenence to zero takes place might not take infinitely many positive values. I am fairly sure that the sequence can be taken so as to tend to $+\infty$ (indeed it is not unusual to see Atkinson's theorem cited in that form) but Atkinson's original paper unfortunately does not include that statement. I seem to remember that Aaronson's book on infinite ergodic theory includes a proof and so might shed light on this question, but I appear to have mislaid my copy.

(That a.e. $x \in X$ is recurrent in the topological sense (when $X$ is a compact metric space) when $\mu$ is an invariant Borel probability measure is a standard result: see for example Proposition 4.1.18 in Katok and Hasselblatt.)

share|improve this answer
    
Proposition 4.1.18 there is about the base dynamics, and doesn't evolve the cocycle $\phi$. –  Pengfei Apr 17 '13 at 21:18
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