If $f_n (\omega) = \sum_{i=1}^n f_1 (T^i \omega)$ and $T$ is an ergodic action with respect to the measure $\mu$ then it is know as Birkhoff's theorem that

$$ \lim_{n \rightarrow \infty} \frac{f_n}{n} = \int_{\Omega} f_1(\omega) d\mu. $$

I was wondering what happens if one studies $h_n (\omega) = \sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega)$ is it still true that

$$ \lim_{n \rightarrow \infty} \frac{h_n}{n} = \int_{\Omega} f(\omega)g(\omega) d\mu? $$

If we assume that $g(T^{2i}\omega)$ form a family of i i d random variables.

  • $\begingroup$ I think the integral you have on the right is not the correct limit. The integral of $f\circ T^{i}.g\circ T^{2i}$ is $\int_{\Omega}f(\omega)g\left(T^i\omega\right)d\mu$ by the $T$-invariance of the measure. Assuming that all of the functions are bounded so that we are allowed to swap limit and integral, the integral of the limit $\lim_{n \rightarrow \infty} \frac{h_n}{n} $ then should be $$\lim_{n\to\infty}\int_{\Omega}f(\omega)g_n(\omega)d\mu=\int_{\Omega}f(\omega)d\mu.\int_{\Omega}g(\omega)d\mu$$ since the Birkhoff averages $g_n$ of $g$ tend to $\int_{\Omega}g(\omega)d\mu$ $\mu$-a.e. $\endgroup$ – KhashF Feb 6 '19 at 20:17
  • $\begingroup$ Having that said, I think it is better to ask whether $\lim_{n \rightarrow \infty} \frac{h_n}{n}=\int_{\Omega}f(\omega)d\mu.\int_{\Omega}g(\omega)d\mu$ $\mu$-a.e.? This is true if you make your assumptions about $T$ stronger, e.g. suppose $T$ is weak mixing. Because then, $T\times T^2$ would be an ergodic transformation of the product space $\left(\Omega\times\Omega,\mu\otimes\mu\right)$ and it suffices to apply the Birkhoff Ergodic Theorem to the function $(\omega,\omega')\mapsto f(\omega).g(\omega')$ on this space. $\endgroup$ – KhashF Feb 6 '19 at 20:32
  • $\begingroup$ I@KhashF : I had thought of something like what you are suggesting. However, with your outer-product function, do you really get $h_n=\sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega)$? Or do you get $\sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega')$ instead? $\endgroup$ – Iosif Pinelis Feb 6 '19 at 23:15

$\newcommand{\om}{\omega} \newcommand{\Om}{\Omega} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb{Z}} \newcommand{\de}{\delta}$ Let $\Om:=\{-1,1\}^{\N_0}$ and $\mu:=(\frac12\de_{-1}+\frac12\de_1)^{\otimes\N_0}$, where $\N_0:=\{0,1,\dots\}$ and $\de_a$ denotes the Dirac point measure at $a$. For $\om\in\Om$, suppose that $f(\om)=g(\om)=\om_0$, where $\om_j:=\om(j)$ for $j\in\N_0$. For $\om\in\Om$ and $j\in\N_0$, let $(T\om)_j:=\om_{j+1}$, so that $X_j(\om):=f(T^j\om)=\om_j$.

Then $X_0,X_1,\dots$ are independent Rademacher random variables, with $P(X_j=\pm1)=1/2$. Let $S_n:=\sum_1^n X_i X_{2i}$. Then the question becomes the following: is it true that \begin{equation} A_n:=\tfrac1n\,S_n\to EX_0^2 \end{equation} almost surely (or at least in probability), as $n\to\infty$?

We have \begin{equation} E S_n^2=\sum_1^n EX_i^2 X_{2i}^2+2\sum_{1\le i<j\le n}EX_i X_{2i}X_j\, EX_{2j}=n, \end{equation} so that $EA_n^2=1/n\to0$ and hence $A_n\to0$ in probability. On the other hand, $EX_0^2=1\ne0$.
So, the answer to your question is no in general.


These averages converge almost surely, although as the previous comment shows it is not clear how to identify the limit. The almost sure convergence is a theorem of Bourgain from the following reference.

"Double recurrence and almost sure convergence", J. Reine Angew. Math. 404, pp 140--161, 1990.

It is unknown whether averages of the form $\frac{1}{n} \sum_{k=0}^{n-1} f_1(T^kx)f_2(T^{2k}x)f_3(T^{3k}x)$ converge almost surely.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy