When does weak convergence of a sequence imply almost everywhere convergence of Cesaro averages? Fix $(X,\mu)$ a nonatomic probability space,
$U\colon L^2(X,\mu) \to L^2(X,\mu)$ a unitary operator and $(r_n)$ some fixed sequence such that $U^{r_n}\to V$ in the weak operator topology: $\langle U^{r_n}f,g \rangle \to \langle Vf,g \rangle$ for every $f,g\in L^2(X,\mu)$.
For $f\in L^2(X,\mu)$ denote $f_k:= U^{r_k}f$.
Is the following true:


*

*For every $f\in L^2(X,\mu)$, Cesaro averages $\frac{1}{n}\sum_{k=1}^{n}f_k$ converge almost everywhere?
or, at least,


*For any $\varepsilon>0$ there is $\delta>0$ such that $ \mu(\{ x\in X : \limsup_n |\frac{1}{n} \sum_{k=1}^n f_k(x)| > \varepsilon \}) < \varepsilon $ whenever $\|f\|_{L^2}< \delta$?
For an arbitrary bounded sequence $(f_n)\subset L^2(X,\mu)$ (not necessarily of the form $f_n= U^{r_n}f$) there is only a subsequence whose Cesaro averages converge almost everywhere (by Komlòs and Révész), and Banach-Saks say that weak convergence of $f_n$ implies strong convergence of Cesaro averages again for a subsequence only.
But is there any way to say something about Cesaro averages of the whole sequence, assuming that it is of the form $f_n= U^{r_n}f$ and converges weakly?

No, there is nothing to be said about the Cesàro averages of the whole sequence. They can be as bad as the sequence itself. 
Indeed, given any weakly convergent sequence $\{f_n\}$, we can consider another sequence $\{g_n\}$ defined as 
$$f_1,f_1, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_3, \dots $$
where the term $f_n$ appears $2^{n^2}$ times. Note that $\{g_n\}$ also converge weakly.
Due to the super-exponential number of repetition, the Cesàro mean $h_N = N^{-1}\sum_{k=1}^N g_k$ is getting close to $g_N=f_n$ when $N$ is of the form $2^{n^2}$. Thus, the sequence of Cesàro means mimics the potentially bad behavior of the original sequence.
