The mean ergodic theorem for weakly mixing extension

Question

I asked this question in https://math.stackexchange.com/q/4236870/528430, but did not get any help.

I got stuck with the following while going through the proof of Lemma 3.21 from the book 'Ergodic Theory: Independence and Dichotomies' by Kerr and Li.

Problem: If $(X,\mu,T)$ is a weakly mixing extension of $(Y,\nu,S)$, then $$\{f\in L^2(X|Y):f\circ T=f\}\subseteq L^{\infty}(Y).$$

Here I recall the following definitions which are used for the above problem.

Definition 1: Let $(X,\mu)$ and $(Y,\nu)$ be two probability measure space. Let $T:X\rightarrow X$ and $S:Y\rightarrow Y$ be two invertible measure preserving transformations. We say that $(X,\mu,T)$ is an extension of $(Y,\nu,S)$ if there is a $T$ invariant conull set $X'\subseteq X$ and an equivariant measurable map $\pi : X'\rightarrow Y$ such that $\mu (\pi^{-1}(A))=\nu (A)$ for all measurable $A\subseteq Y$.

Definition 2: Given an extension $(X,\mu,T)\rightarrow (Y,\nu,S)$, let $\mathbb{E}_Y:L^2(X)\rightarrow L^2(Y)$ be the conditional expectation. Then $L^2(X|Y)$ is the completion of $L^{\infty}(X)$ with respect to the norm $\|f\|:=\|\mathbb{E}_Y(f\cdot\overline{f})\|^{1/2}$.

Definition 3: An element $f\in L^2(X|Y)$ is said to be conditionally weakly mixing if the mean $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{s=-n}^{s=n}\|\mathbb{E}_Y((f\circ T^s)\cdot\overline{f})\|=0.$$

Definition 4: The extension $(X,\mu,T)\rightarrow (Y,\nu,S)$ is said to be weakly mixing if every element in $L^2(X|Y)$ orthogonal to $L^{\infty}(Y)$ is conditionally weakly mixing.

Thanks in advance for any help or suggestion.

Uri Bader · Accepted Answer · 2021-09-11 12:22:14Z

I suspect that your confusion stems from misinterpreting the norm on $L^2(X|Y)$ in your definition 2. You should note that $\mathbb{E}_Y$ takes $L^\infty(X)$ to $L^\infty(Y)$ and the new norm taken on $L^\infty(X)$ is with respect to the $L^\infty$-norm on $L^\infty(Y)$, not the $L^2$-norm. That is $\|f\|:=\|\mathbb{E}_Y(f\cdot\overline{f})\|_\infty^{1/2}$ for $f\in L^\infty(X)$.

It is now easy to see that $L^2(X|Y)$ decomposes equivariantly as an $L^\infty(Y)$-Hilbert module to $L^\infty(Y)\oplus L^\infty(Y)^\perp$. Accordingly, you get a corresponding decomposition of $L^2(X|Y)^T$ and you are done by observing that a $T$-invariant weakly mixing element must be 0.

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