if $f$ is weakly mixing then $f^n$ is ergodic? If $f$ is weakly mixing then $f^n$ is ergodic? I think this is false but I can't find a counter-example because I don't know transformations weakly mixing but not mixing. Can you prove or give a counterexample?
 A: That $(f,\mu)$ is weakly mixing means that
$$
\frac{1}{N} \sum_{k=0}^{N-1} |\mu (A \cap f^{-k} B) - \mu(A)\mu(B)| \to 0.
$$
Similarly $(f^n,\mu)$ is weakly mixing if
$$
\frac{1}{N} \sum_{k=0}^{N-1} |\mu (A \cap f^{-kn} B) - \mu(A)\mu(B)| \to 0.
$$
If you compare
$$
\frac{1}{N} \sum_{k=0}^{N-1} |\mu (A \cap f^{-kn} B) - \mu(A)\mu(B)|
$$
and
$$
\frac{1}{nN} \sum_{k=0}^{nN-1} |\mu (A \cap f^{-k} B) - \mu(A)\mu(B)|
$$
you will see that $(f^n,\mu)$ is weakly mixing if $(f,\mu)$ is. Finally, observe that weakly mixing implies ergodicity.
A: You can directly use the definition of weakly mixing as Tomas Persson showed in the answer.
Also, you can do this:
Recall that $f$ is weakly mixing if and only if for every measurable set $ A, B $, there is a set $J_{A, B} \subseteq\mathbb N$ with density zero for which
$$ \mu(A\cap f^{-k}B)\overset{k\to\infty}{\underset{k\notin J_{A,B}}\longrightarrow}\mu(A)\mu(B). $$
For $f^n$, since $f$ is weakly mixing, we have
$$ \mu(A\cap (f^{n})^{-k}B)=\mu(A\cap f^{-nk}B)\overset{k\to\infty}{\underset{nk\notin J_{A,B}}\longrightarrow}\mu(A)\mu(B). $$
 Note that the set $\{k\in\mathbb N\colon nk\in J_{A,B}\}$ has density zero, this is because
\begin{align}
\limsup_{m\to\infty}\frac 1m\left| \{k\in\{k\in\mathbb N\colon nk\in J_{A,B}\}\colon 1\le k\le m\}\right|&\le\limsup_{m\to\infty}\frac{1}{m}|\{k\in J_{A,B}\colon 1\le k\le mn\}|\\
&=n\limsup_{m\to\infty}\frac{1}{mn}|\{k\in J_{A,B}\colon 1\le k\le mn\}|\\
&=n\times 0\\
&=0
\end{align}
and we know that $$\lim_{m\to\infty}\frac 1m\left| \{k\in\{k\in\mathbb N\colon nk\in J_{A,B}\}\colon 1\le k\le m\}\right|=0,$$ i.e., the density of the set $\{k\in\mathbb N\colon nk\in J_{A,B}\}$ is $0$.
Therefore, we conclude that $f^n$ is also weakly mixing. And hence $f^n$ is ergodic.
