Entropy maximising ergodic transformation

Question

Let $(\Omega, \mathcal F, \mu)$ be a standard probability space.

Question: For each $f \in L^\infty (\Omega)$, does there exist an ergodic measure preserving transformation $T: \Omega \to \Omega$ such that the following expression is maximised?

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^n\int_\Omega |T^k f - T^{k-1} f| \, d\mu$$

Where for $k \geq 1$, $T^k f (x) := f(T^k x)$ and $T^0 f = f$ by convention.

To be unambiguous, the expression is maximized for fixed $f$, over all ergodic measure preserving transformations $T$.

Remark: The limit exists due to the pointwise ergodic theorem.

Answer 1

I don't know if you're still interested now that the average collapses, but I think that there is such a maximizing invertible ergodic $T$ as long as you assume that your probability space is Lebesgue.

Under these assumptions, if you normalize so that the median of $f$ is $0$ (i.e. so that $\mu(f < 0) = \mu(f > 0)$), then I think the maximum value of your quantity $\int |f - Tf|$ is given by $2 \int |f| dx$.

It's clear from the triangle inequality that for every measure-preserving $T$, $\int |f - Tf| \leq \int (|f| + |Tf|) = 2\int |f|$.

Now we want to find $T$ for which $\int |f - Tf| = 2\int |f|$. Define $Z = \{x \ : \ f(x) = 0\}$, $N = \{x \ : \ f(x) < 0\}$, and $P = \{x \ : \ f(x) > 0\}$. If $Z$ has positive measure, split it arbitrarily into $Z^+$ and $Z^-$ of equal measure (since we assumed $\mu$ is nonatomic). Define $P' = P \cup Z^+$ and $N' = N \cup Z^-$. Then $\mu(P') = \mu(N') = 1/2$, and $f$ is nonnegative on $P'$ and nonpositive on $N'$.

I claim that any measure-preserving $T$ with $T(N') = P'$ and $T(P') = N'$ works.

For such $T$, and for every $x \in N'$, $Tx \in P'$, meaning that $Tf(x) - f(x)$ is nonnegative. Similarly, for every $x \in P'$, $Tf(x) - f(x)$ is nonpositive. Therefore, $\int |f - Tf|$ is

$\int_{N'} (Tf - f) + \int_{P'} (f - Tf) = \int_{N'} Tf - \int_{N'} f + \int_{P'} f - \int_{P'} Tf = 2 (\int_{P'} f - \int_{N'} f) = 2 \int |f|$. (The second-to-last equation follows from the fact that $T$ is measure-preserving.)

Now we just need to justify that there exists an invertible ergodic $T$ with $T(P') = N'$ and $T(N') = P'$, but this should be easy. Just find a measure-preserving bijection $S: P' \rightarrow N'$ ($P'$ and $N'$ are positive measure subsets of a Lebesgue space, and so are also Lebesgue and thus isomorphic as probability spaces) and any invertible ergodic map $R: P' \rightarrow P'$. Finally, define

$T(x) = \begin{cases} S(x) & n \in P' \\ RS^{-1}(x) & n \in N' \end{cases}$

Since $T^2$ restricted to $P'$ is just $R$, for every positive measure subset $A \subset P'$, $\bigcup_n T^{2n} A = \bigcup_n R^n A = P'$ by ergodicity. Ergodicity of $T$ should follow almost immediately.