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Let $(X,\mathcal{A},\mu)$ be a probability space and $f:X\to X$ be a measurable map that preserves $\mu$. Fix $n\in \mathbb{Z}^+$.

It's not hard to see that $f$ ergodic does not necessarily imply $f^n$ ergodic. For example, take $X=\mathbb{Z}/4\mathbb{Z}$ with the uniform probability measure, $f:x\mapsto x+1$ and $n=2$.

Is it true that if $f$ is (strong) mixing then $f^n$ is mixing?

EDIT: Yes, it is, since a subsequence of a real convergent sequence is convergent and has the same limit. What I meant to ask is:

Is it true that if $f^n$ is (strong) mixing then $f$ is mixing?

The result is easily seen to be true if we replace "mixing" by "ergodic".

Davide Giraudo
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Bruno Stonek
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1 Answers1

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"The map $f$ is strong mixing" means that for each $A,B\in\mathcal A$, we have $$\lim_{N\to\infty}\mu(A\cap T^{-N}(B))=\mu(A)\mu(B).$$ Since the sequence $(\mu(A\cap (T^{n})^{-k}(B))-\mu(A)\mu(B),k\geqslant 1)$ is a subsequence of a sequence which converges to $0$, it converges itself to $0$.

For the converse, fix $A,B\in\mathcal A$ and define $$c_k:=\mu(A\cap T^{-k}(B))-\mu(A)\mu(B).$$ We have that $c_{nk}\to 0$ by assumption. Using $T^{-i}B$ instead of $B$ in the , we get $c_{nk+i}\to 0$ as $k\to\infty$, for each $0\leqslant i\leqslant n-1$.

Davide Giraudo
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    Thanks, I've got it. A couple of silly remarks: to get $c_{nk+i}\to 0$, we use the fact that $f$ preserves the measure: $\mu(f^{-i}(A))=\mu(A)$. To conclude, we use the fact that if we got a partition of $\mathbb{N}$ into infinite subsets, and we have a sequence such that it converges to $l$ when the indices are restricted to each subset, then it converges to $l$. – Bruno Stonek Oct 31 '13 at 17:19