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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.

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  • $\begingroup$ "The following expression is maximised": it would be helpful to clarify what is fixed and what is allowed to vary. Are you fixing $f$ and asking whether the function $T\mapsto \lim\dots$ attains its maximum? $\endgroup$
    – YCor
    Feb 2, 2022 at 8:03
  • $\begingroup$ Yes i will clarify. $\endgroup$
    – Nate River
    Feb 2, 2022 at 8:11
  • $\begingroup$ I'm sure that probably I'm missing something, but isn't the integral in your sum the same for all values of k by the fact that T is measure-preserving? If this is true, then you're just asking for T which maximizes the integral of |Tf - f|. This is still maybe an interesting question, but is ostensibly much simpler. $\endgroup$ Feb 2, 2022 at 20:04
  • $\begingroup$ @RonniePavlov Oh damn you’re right, it reduces to a single term.. $\endgroup$
    – Nate River
    Feb 2, 2022 at 21:47
  • $\begingroup$ One more question: why did you say something about entropy in the question? Is there some connection from your quantity to entropy? $\endgroup$ Feb 4, 2022 at 14:09

1 Answer 1

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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.

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  • $\begingroup$ Very nice, essentially by taking them to be on opposite sides of the median, you prevent cancellation effects and attain the highest possible difference. And then compose with an ergodic map on one end to get overall ergodicity. Brilliant! $\endgroup$
    – Nate River
    Feb 4, 2022 at 7:20

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