Can anyone help me out with proofs/counterexamples? I'm working on an operator-valued multiplicative ergodic theorem and need what may(?) be a well-known fact. This fact (if true) would help me get rid of an annoying asymmetry in the conclusion of a theorem.


I'm assuming that $T$ is an invertible ergodic transformation of a probability space $(X,\mathcal B,\mu)$ and that $f$ is a measurable (but not necessarily integrable) function on $X$.



> 
> Is it true that $f(T^nx)/n \to 0$ a.e. if and only if $f(T^{-n}x)/n\to 0$ a.e.?
> 


Comments:


(1) In probability language if you define $X\_n=f(T^nx)$ this is a stationary sequence of random variables. Borel-Cantelli 1 shows that if $\mathbb E |X\_0|<\infty$ (i.e. $f\in L^1$) then $X\_n/n\to 0$ as $n\to\pm\infty$: The probability that $|X\_n|/n > 1/k$ is $\mathbb P(|X\_0| > n/k)$. The sum of this series is over-estimated by $k\mathbb E|X\_0| < \infty$. Hence almost surely $|X\_n|/n < 1/k$ for all large $n$. Since this is true for all $k$ you get $X\_n/n\to 0$.


If the $X\_n$ are i.i.d. random variables, then the converse holds by Borel-Cantelli 2. This shows that for i.i.d. random variables, the boxed question has an affirmative answer.


(2) In the case where the $X\_n$ are not i.i.d. I believe there are examples where $X\_n/n\to 0$ almost surely even though $\mathbb E|X\_0|=\infty$.


In the case that $f\in L^1$, both sides of the implication in the main question are true. The unresolved case is $f\not\in L^1$.



Non-integrable ergodic theory
$\newcommand{\R}{\mathbb R}$
$\newcommand{\P}{\mathbf P}$
$\newcommand{\Z}{\mathbb Z}$


I found this question very interesting and gave it much thought this week. I believe I have a proof now. I think it would be interesting to see what generalizations one can get from this argument. Below I use probabilistic notation, which I'm more used to.


Let $\{X\_n\}\_{n\in \Z}$ be a stationary stochastic process, taking values in $\R$. Let $L=\limsup\_{n\to -\infty} \frac{X\_n}{|n|}$ and $R=\limsup\_{n\to \infty} \frac{X\_n}{|n|}$.


**Theorem:** $L=R$, almost surely.


**Proof:**
It is enough to prove the theorem for ergodic processes. We may also assume, WLOG, that all the values are nonnegative integers.


Let $A$ be the event that for some $n<0$ we have $X\_n\ge |n|$ and let $B$ be the event that for some $n>0$ we have $X\_n\ge |n|$.


**Lemma:** $\P(A) \le 2 \P(B) .$


**Proof of lemma:**
For a given realization of $X\_n$, let $I$ be all indices $i$ for which $T^i X$ is in $A$ and let $J$ be all the indices for which $T^i X$ is in $B$. We claim that the density of $J$ is no more than twice the density of $I$, which then implies the conclusion.


$I$ can be written as $\cup\_{n \in \Z} \{n,\ldots,n+X\_n\}$, while $J=\cup\_{n \in \Z} \{n-X\_n,\ldots,n\}$. In particular, $J$ is contained in $\bar{J}=\cup\_{n \in \Z} \{n-X\_n,\ldots,n+X\_n\}$. But if we write $I$ as a union of *disjoint* intervals, then in $\bar{J}$ each of these intervals is extended to the left by at most the length of the interval. Hence, the density of $\bar{J}\supset J$ is at most twice that of $I$. $\blacksquare$


Of course, by symmetry, we also have $\P(B) \le 2 \P(A)$. Let $A\_K$ be the event that for some $n<0$ we have $X\_n \ge \max(|n|,K)$ and define $B\_K$ analogously. By applying the lemma to the process $\{Y\_n\}\_{n\in \Z}$ defined by $Y\_n=X\_n$ if $X\_n\ge K$ and $Y\_n=0$ otherwise, we get that $\P(A\_K) \le 2 \P(B\_K)$ and vice verse.


In particular, $\lim\_{K\to \infty} \P(A\_K) =0$ if and only if $\lim\_{K\to \infty} \P(B\_K) = 0$. These limits necessary exists, since these are monotone decreasing events.


For ergodic processes, we have that $L$ and $R$ are a.s. constant. Now, if $L<1$ a.s. then we have $\P(A\_K)\to 0$. In the other direction, if $\P(A\_K)\to 0$ then a.s. $L\le 1$. Similar implications hold for $R$ and $B\_K$. Hence, we get that if $L<1$ then $R\le 1$ and vice verse. By applying these to a rescaled process $X\_n / \alpha$, we get that for any $\alpha$ we have $L<\alpha$ implies $R\le \alpha$ (and vice verse), so we must have $L=R$. $\blacksquare$


