# Cesaro summation

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?

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 $1$.