Let's consider $\{a_{n} \} -$ a bounded sequence of real numbers. Is it true that $\frac{1}{n} \sum_{k=0}^{n-1}{|a_{k}|^{p}}$ (Cesaro sums) converges or diverges for all $p \geq 1$? (more presicely: if it converges for some $p_{0} \geq 1$, will it converge for any other $p_{1} \geq 1$?) Also, it would be great to establish, how the situation changes, if $0 < p <1$.
My assumption is that for $p \geq 1$ the convergence/divergence saves, whereas for $0 < p <1$ it does not (still i'm not sure). A bit more general fact about Cesaro summation is that if $b_{n} \to b$, then $\frac{1}{n} \sum_{k=0}^{n-1}{b_{k}} \to b$ but this does not help too much, since we are considering quite a more general case.
The problem naturally rises while considering the property of strong and weak mixing of measure preserving transformations.
Are there any hints that might help?
Certainly not. Here's how to construct a counterexample - I'll let you insert the epsilons:
The sequence $(a_n)$ is going to consist of blocks that look like $1,1,1,\dots,1$ alternating with blocks that look like $2,0,2,0,\dots,2,0$. Now regardless of how we adjust the length of those blocks we have $$\frac1n\sum_{j=1}^na_j\to1.$$
But the average of $2^2,0^2$ is $2$, while the average of $1^2,1^2$ is $1$. So: Start with a $1,1$ block. At this point the average of $a_j^2$ is $1$. Now insert a $2,0$ block. If that block is long enough you get an $n$ so that $\frac1n\sum_1^na_j^2$ is close to $2$. Then an even longer $1,1$ block brings the average of $a_j^2$ close to $1$ again, and a yet longer $2,0$ block brings it back up close to $2$... You get Cesaro means with lim sup $2$ and lim inf $1$.