A counterexample for the Mean Ergodic Theorem in $L_\infty$

Question

The so-called Mean Ergodic Theorem goes back to von Neumann for Hilbert spaces. Later on, versions of this result in reflexive Banach spaces have also appeared (see, e.g., the book by Krengel, Ergodic Theorems, section 2.1. At the beginning of the same book (end of p.4), Krengel mentions that such a result is not true for $L_\infty$.

More precisely, he claims that letting $\Omega=[0,1[$, endowed with the Lebesgue measure, and considering the measure-preserving transformation $\tau\omega:=\omega+\alpha$ (mod 1) where $\alpha$ is irrational, one can find a suitable (highly discontinuous) function $f\in L_\infty(\Omega)$ such that $A_nf:=(f+f\circ\tau+f\circ\tau^2+\ldots+f\circ\tau^{n-1})/n$ does not converge to $\bar f:=\int_0^1f$ in the $L_\infty(\Omega)$ norm. He leaves this fact as an exercise.

Does anyone know how to show this?

Answer 1

Let $I_n$ be a collection of rapidly-shrinking intervals with $|I_n|=1/n!$, say. Let $f_0=0$. Inductively define $$ f_n(x)=\begin{cases} (-1)^n&\text{if $x\in\bigcup_{i=0}^{n-1}\tau^i(I_n)$;}\\ f_{n-1}(x)&\text{otherwise.} \end{cases} $$ Let $f(x)=\lim_{n\to\infty}f_n(x)$. This limit exists by the first Borel-Cantelli lemma (as the set of points that switch infinitely often has measure 0).

Notice that the intervals $I_n$ decrease so rapidly that there exists a positive measure subset, $J_n$ of $I_n$ so that $\bigcup_{i=0}^{n-1}\tau^i(J_n)$ is disjoint from $\bigcup_{j=n+1}^\infty \bigcup_{i=0}^{k-1}\tau^i(I_j)$.

Now for each $N$, $A_Nf$ takes the value $-1$ on each odd $J_n$ with $n\ge N$ and $1$ on each even $J_n$ with $n\ge N$. On the other hand, by the Birkhoff ergodic theorem, $A_Nf$ converges almost everywhere to a constant, $c$, so that $\|A_Nf-c\|_\infty\ge 1$ for each $N$.

By the way, there is context for this kind of question in the work of Jones, Bellow, Wierdl, Rosenblatt and collaborators, where they introduce the concept of "strong sweeping out" in subsequence ergodic theorems. In the background of my answer is the Rokhlin lemma, which can be used if one needs more precise control on how the bad parts fill out the space.