Ergodic Theory (Weak Mixing)

Question

If $T$ is weak mixing then we know that $T\times T \times \ldots \times T$ is also weak mixing. Does anyone know if this is true for $T\times T \times \ldots$?

Answer 1

It seems to me that the answer is YES, and that it is generally the case that weak mixing of all finite products $T_1 \times T_2 \times \dots \times T_k$ (for all $k$) implies weak mixing of the infinite product $T_1 \times T_2 \times \dots \times T_k \times \dots $.

Let $X$ be the space on which $T$ acts. I am assuming $X$ is a compact metrizable space with a Borel probability measure $\mu$. Then, $S = T\times T \times \dots $ acts on $X^\infty = X \times X \times \dots$. Denote the product measure on $X^\infty$ by $\mu^\infty$ (note that existence of $\mu^\infty$ is not a trivial problem, but there is a theorem that assures it exists).

$\newcommand{\eps}{\varepsilon}$ Let $A,B \subset X^\infty$ be two measurable sets with positive densities. Let $A_n$ denote the portion of these sequences $x_1,x_2, \dots$ for which there exist $y_{n+1},\dots$ with $(x_1,x_2,\dots,x_n,y_{n+1},\dots) \in A$. These are measurable, $A_n \supseteq A_{n+1}$ and $\bigcap_n A_n = A$. It follows that $\lim_n\mu^\infty(A_n) = \mu^\infty(A)$. For a fixed $\varepsilon$, we can select $M$ so large that $\mu^\infty(A \triangle A_M) < \eps$. Likewise, we may assume $\mu^\infty(B \triangle B_M) < \eps$.

Consider the expression $a_n = |\mu^\infty(A\cap S^{-n}B) - \mu(A) \mu(B)|$. The definition of weak mixing is that $a_n$ tends to $0$ in the Cesaro sense. Because $S$ is measure preserving and $A_M,B_M$ approximate $A,B$, we can approximate $a_n$: $$ a_n = b_n + O(\eps), \qquad b_n = |\mu^\infty(A_M\cap S^{-n}B_M) - \mu(A_M) \mu(B_M)|$$ (here, $O(\eps)$ is something smaller than, say, $100 \eps$).

It follows that $\frac{1}{N} \sum_{n=1}^N a_n = \frac{1}{N} \sum_{n=1}^N b_n + O(n)$. But $\mu^\infty(A_M) = \mu^M(\tilde{A}_M)$ where $\tilde{A}_M \in X^M$ is the projection of $A_M$; and likewise for other sets. We thus have: $$b_n = |\mu^\infty(\tilde{A}_M\cap S^{-n}\tilde{B}_M) - \mu(\tilde{A}_M) \mu(\tilde{B}_M)| $$ Because $T^M$ was mixing, $\lim_N \frac{1}{N} \sum_{n=1}^N b_n = 0$. This means that $\limsup_N \frac{1}{N} \sum_{n=1}^N a_n = O(\eps)$. However $\eps > 0$ was chosen arbitrarily, so $\limsup_N \frac{1}{N} \sum_{n=1}^N a_n = 0 = \lim_N \frac{1}{N} \sum_{n=1}^N a_n$. Thus, $S$ is weakly mixing, as desired.

I thing the same reasoning should go for other notions of mixing.