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I have a question regarding a confusion from reading the Princeton Companion to Mathematics on the topic of Ergodics Theorems. It is about proving a stronger version of Poincare Recurrence Theorem using Neumann's Mean Ergodic Theorem. I apologize if this question is too easy for this site.

Let $A$ be a subset of positive measure, $T$ a measure preserving transformation and $U$ a unitary operator defined by $Uf(x):=f(Tx)$. Define the ergodic average $A_{N,M}$ to be $$ A_{N,M}(f):=\frac 1{N-M}\sum_{n=M}^{N-1}U^nf. $$

It is not hard to verify that for $f:=1_A$, $\langle f,U^nf\rangle = \mu(A\cap T^{-n}A)$. Here's the passage that confuses me:

It follows that $$ \langle f,A_{N,M}(f)\rangle = \frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A). $$ If we let $N − M$ tend to infinity, then $A_{N,M}f$ tends to a $U$-invariant function $g$. Since $g$ is $U$-invariant, $\langle f,g\rangle = \langle U^nf,g\rangle$ for every $n$ and therefore $\langle f,g\rangle=\langle A_{N,M}(f),g\rangle$ for every $N$ and $M$ and finally $\langle f,g\rangle=\langle g,g\rangle$. By the Cauchy–Schwarz inequality, this is at least $$ \left(\int g(x)d\mu\right)^2 = \left(\int f(x)d\mu\right)^2=\mu(A)^2. $$ Therefore we deduce $$ \lim_{N-M\to\infty}\frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A)\ge \mu(A)^2. $$

How does one arrive at the final result from Cauchy–Schwarz inequality?

I understand everything prior to the statement "By the Cauchy-Schwarz ..." but not after that. Using Cauchy–Schwarz I could deduce $||g||\le ||f||$ and get $$ \lim_{N-M\to\infty}\frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A)\le \mu(A)^2. $$ which is the complete opposite of what was stated.

Here is the link to the same question posted on MSE with no answer.

Edit: As pointed out by John Griesmer, the inequality I derived should read $$ \lim_{N-M\to\infty}\frac 1{N-M}\sum_{n=M}^{N-1}\mu(A\cap T^{-n}A)\le ||f||^2 = \mu(A). $$ instead. A silly mistake there by me.

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    $\begingroup$ When $\mu$ is a probability measure, setting $h\equiv 1$, Cauchy-Schwarz implies $\int g \cdot \bar{h}\, d\mu \leq \Bigl(\int |g|^2\, d\mu\, \int |h|^2\, d\mu\Bigr)^{1/2} $. Squaring both sides, simplifying using $h\equiv 1$ and the fact that $\mu$ is a probability measure produces the desired inequality. $\endgroup$ – John Griesmer Jan 13 '17 at 14:44
  • $\begingroup$ @JohnGriesmer How do we get $\int gd\mu=\mu(A)$ though? Do we need to apply Birkhoff's theorem for convergence in $L^1$ or is it much simpler than that? $\endgroup$ – BigbearZzz Jan 13 '17 at 14:56
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    $\begingroup$ Evaluate $\int A_{N,M}f\, d\mu$ for all $N$ and $M$ and take a limit. $\endgroup$ – Nik Weaver Jan 13 '17 at 15:26
  • $\begingroup$ Thank you everyone very much. Lastly, if you would be so kind, I want to know if we actually have an equality there since I've also shown the reverse inequality (which I hope it is correct). $\endgroup$ – BigbearZzz Jan 13 '17 at 15:39
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    $\begingroup$ Note that $\|f\|^2= \int |f|^2\, d\mu = \int (1_A)^2\, d\mu=\mu(A)$, not $\mu(A)^2$. $\endgroup$ – John Griesmer Jan 13 '17 at 17:35
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$A_{N,M}(f)$ converges to some $U$-invariant function $g$ that satisfies $\langle g, f\rangle$ = $\langle g, g\rangle$.

We also have $\langle g, 1\rangle = \lim \langle A_{N,M}(f), 1\rangle = \langle f, 1\rangle$ since $\langle A_{N,M}(f), 1\rangle = \langle f, 1\rangle$ for all $N, M$.

So we have $\lim \ \langle A_{N,M}(f) , f \rangle = \langle g, f \rangle \ =\langle g, g \rangle = \langle g, g \rangle \langle 1, 1 \rangle \geq \langle g , 1 \rangle^2$

the last inequality following from the Cauchy-Schwarz inequality. For $f = {\bf 1}_A$, we have $\langle f, 1 \rangle = \int {\bf 1}_A d\mu = \mu(A).$

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    $\begingroup$ Isn't it $\langle f,g\rangle=\langle g,g\rangle$, not $\langle f,g\rangle=\langle f,f\rangle$? $\endgroup$ – BigbearZzz Jan 13 '17 at 20:11
  • $\begingroup$ What I understand from the other comments is that we have $\lim \ \langle A_{N,M}(f) , f \rangle = \langle g, f \rangle \ =\langle g,g \rangle = \langle g,g \rangle \langle 1, 1 \rangle \geq \langle g , 1 \rangle^2=\langle f , 1 \rangle^2$. Surely if we have $\langle f,g\rangle=\langle f,f\rangle$ that would be great but I still can't see how. $\endgroup$ – BigbearZzz Jan 13 '17 at 21:17
  • $\begingroup$ @BigbearZzz. Indeed, I have corrected the answer. $\endgroup$ – coudy Jan 14 '17 at 11:13

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