**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](https://drive.google.com/file/d/1FqoY7rZhEV6miICQVGX-5a9M5b3SoTIB/view?usp=sharing) 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.



The mean ergodic theorem for weakly mixing extension
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.


