$\newcommand{\set}[1]{\\{#1}\\}$ $\newcommand{\ab}[1]{\langle #1\rangle}$ $\newcommand{\mc}{\mathcal}$ $\newcommand{\Z}{\mathbb Z}$ $\newcommand{\C}{\mathbf C}$

Exercise 2.8.4 in Einsiedler and Ward's Ergodic Theory with a View Towards Number Theory asks the following.

Exercise. Give a different proof of the mean ergodic theorem as follows. For a measure preserving system $(X, \mc X, \mu, T)$ and a function $f\in L^2_\mu$, show that the map $n\mapsto \ab{T^nf, f}:\Z\to \C$ is positive definite and apply the Herglotz-Bochner Theorem to translate the problem into one concerned with functions on $S^1$, and there use the fact that $\frac{1}{N} \sum_{n=0}^{N-1} z^n$ converges for each $z\in S^1$.

First let me assume that the system is invertible (otherwise I was facing trouble showing the positive-definiteness of the said map). Applying the Herglotz-Bochner theorem we can find a positive measure $\rho$ on $S^1$ such that $\hat \rho(n) = \ab{T^nf, f}$ for all $n$. Let $f_N = \frac{1}{N}\sum_{n=0}^{N-1}T^nf$.

I am only able to show the weak$^\ast$ convergence of $(f_N)_{N\geq 1}$ using the approach mentioned in the problem.

Note that $\ab{f_N, f} = \frac{1}{N} \sum_{n=0}^{N-1} \hat\rho(n)$. Using dominated convergence theorem we see that the latter converges to $\rho(1)$. It follows, by similar reasoning, that $\ab{f_N, T^kf}$ converges to $\hat\rho(1)$ for all $k$. Now define $S = \overline{\text{Span}\{T^nf:\ n\in \Z}\}$. It follows that $\ab{f_N, g}$ converges whenever $g\in S$. Also, if $g\in S^\perp$, then $\ab{f_N, g}$ clearly converges to $0$. Thus for any $g\in L^2_\mu$ we have that $\ab{f_N, g}$ converges. This shows the weak* convergence of $f_N$.

How do I upgrade to norm convergence?

  • $\begingroup$ Not sure, but Rudin 'Functional analysis' Theorem 12.44 might help. $\endgroup$ Commented Apr 25, 2023 at 5:02

1 Answer 1


First notice that $$||f_N||_2=\langle f_N, f_N \rangle = \frac{1}{N^2} \sum_{n,m=0}^{N-1} \langle T^nf,T^mf \rangle,$$ then use the fact that $T$ is measure preserving to obtain $\langle T^nf,T^mf \rangle = \langle T^{n-m}f,f \rangle$ and then use the Bochner-Herglotz theorem. Indeed, by the theorem there is a finite and positive measure $\sigma_f$, so that $\langle T^{n-m}f,f \rangle = \int_{\mathbb{S}} z^{n-m}\ d \sigma_{f}(z)$. Therefore, $||f_N||_2$ becomes $$\frac{1}{N^2} \sum_{n,m=0}^{N-1} \int_{\mathbb{S}} z^n \cdot \bar{z}^m\ d \sigma_{f}(z) = \big|\big| \frac{1}{N} \sum_{n=0}^{N-1} z^n \big|\big|_{L^2(\sigma_f)}.$$

In the same manner one sees that

$$\big| \big| f_N-f_M \big| \big|_2 = \big|\big| \frac{1}{N} \sum_{n=0}^{N-1} z^n - \frac{1}{M} \sum_{m=0}^{M-1} z^m\big|\big|_{L^2(\sigma_f)}.$$

Then you can use the hint and the dominated convergence theorem to conclude that $(f_N)$ is a Cauchy sequence and thus converges.

  • $\begingroup$ I can elaborate further if you like, although it's fairly straightforward. $\endgroup$
    – User
    Commented Jun 22, 2023 at 17:15
  • $\begingroup$ Thanks for the response. Though I do not see your idea. Can you elaborate. Thank you. $\endgroup$ Commented Jun 22, 2023 at 17:20
  • 1
    $\begingroup$ Thanks for adding details. What this (to me) shows is that the $L^2$-norm of $f_N$ converges as $N\to \infty$. I don't see why this shows that $f_N$ itself converges in the $L^2$-sense as $N\to \infty$. $\endgroup$ Commented Jun 24, 2023 at 2:57
  • $\begingroup$ Oh, yes, you want convergence in norm! You can repeat the same process to show that $(f_N)$ is Cauchy. $\endgroup$
    – User
    Commented Jun 24, 2023 at 12:11
  • $\begingroup$ I'll try it. But usually it is very hard to establish that such averages are Cauchy. A useful tool is here is the van der Corput trick. $\endgroup$ Commented Jun 25, 2023 at 3:32

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