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Let $\tau:\Omega\to \Omega$ be a measure-preserving transformation with $\mu(\Omega)<\infty$. Define $T:L_p(\Omega)\to L_p(\Omega)$ as $Tf:=f\circ \tau$. I want to prove that for all $1\leq p<\infty$, given $f\in L_p$ there exists $\bar{f}\in L_p$ such that $\|\bar{f}\|_p\leq \|f\|_p,$ $\bar{f}\circ \tau=\bar f$, and $\|\frac{1}{n}\sum_{k=0}^{n-1}T^kf-\bar{f}\|_p\to 0$ as $n\to \infty$.

For $p=2$ this is just von Neumann's mean ergodic theorem. Using this One can easily prove the result for $1\leq p<2$. But how to prove the statement for $2<p<\infty$? Also is it true for $p=\infty$?

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  • $\begingroup$ Something's odd about the properties of $\overline{f}$: you probably mean $\overline{f} \circ \tau = \overline{f}$ instead of $\overline{f} \circ \tau = f$, right? $\endgroup$
    – Jochen Glueck
    Commented May 30, 2020 at 20:06
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    $\begingroup$ @Jochen. Yes. Edited. $\endgroup$
    – A beginner mathmatician
    Commented May 30, 2020 at 20:11

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False for p infinite. True for finite. See e.g. the book by Krengel, Ergodic theorems. Other sources (that also go further) are [1, Sec. I.2.1] or [2, Theorem 8.8].

[1] T. Eisner, Stability of operators and operator semigroups, Operator Theory: Advances and Applications, vol. 209, Birkh¨auser Verlag, Basel, 2010.

[2] T. Eisner, B. Farkas, M. Haase, and R. Nagel, Operator theoretic aspects of ergodic theory, Graduate Texts in Mathematics, vol. 272, Springer, Cham, 2015. http://www.math.uni-leipzig.de/~eisner/book-EFHN.pdf

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  • $\begingroup$ From Krengel only I am reading this. There is no proof of this fact. He only stated this. As I mentioned I could prove for $1\leq p<2$ but not for $p>2$. Also now I see that he has given only a hint for $p=\infty$. But after lot of try I could not solve that too. $\endgroup$
    – A beginner mathmatician
    Commented May 30, 2020 at 18:59
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    $\begingroup$ @Abeginnermathmatician: It is indeed in Krengel's book: For $p \in (1,\infty)$ it is an immediate consequence of Theorem 2.1.2 on page 73. Right after the theorem, Krengel explains why it is also true for $p = 1$ (as a consequence of Theorem 2.1.1 on page 72). $\endgroup$
    – Jochen Glueck
    Commented May 30, 2020 at 20:14
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    $\begingroup$ @Abeginnermathmatician: For a $p=\infty$ counterexample, consider the shift on 2 symbols with the Bernoulli (1/2,1/2) measure and $f(x)=x_0$. Then the pointwise limit of the averages is 1/2 almost everywhere; for each $n$, there is a set of positive measure (namely the set of $x$’s with $x_0=...=x_{n-1}=0$) where the difference between the average and the limit is 1/2. $\endgroup$
    – Anthony Quas
    Commented May 30, 2020 at 20:15

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