# Equivalent condition for mixing

Let $(X,\mathscr{A},\mu)$ be a measurable space and $T$ measure preserving map and define:

$T$ is mixing iff $\lim_{n\to\infty}\mu(A\cap T^{-n}B)=\mu(A)\mu(B)$ for all $A,B\in\mathscr{A}$.

How to prove that mixing property is equivalent with:

$\lim_{n\to\infty}\langle U_T^n f,g\rangle=\langle f,1\rangle\langle 1,g\rangle$ for all $f,g$ in dense subset of $L^2(\mu)$?

Here, $(U_Tf)(x)=f(Tx)$.

One implication is direct (take characteristic functions for $f$ and $g$), but I'm having trouble to prove the second one.

Any help or hint is welcome. Thanks in advance.

• Fist prove that it holds for characteristic functions, then step functions, and finally use a density argument. – D. Thomine May 13 '17 at 13:21
• @ D. Thomine What is density argument? – alans May 13 '17 at 13:29
• Step functions are dense in $\mathbb{L}^2$. Hence, given general $f$ and $g$, you can find step functions $f_\varepsilon$ and $g_\varepsilon$ which are $\varepsilon$-close to $f$ and $g$. Then see what you can say about the convergence. – D. Thomine May 13 '17 at 13:58

## 1 Answer

We denote the two statements we are interested in as follows,

Mixing: A measure space $(X,\mathscr{A},\mu)$ with measure preserving map $T:X\rightarrow X$ is said to be mixing if for every $A,B\in \mathscr{A}$, $$\lim_{n\to\infty}\mu(A\cap T^{-n}B)=\mu(A)\mu(B).$$

Alternate Mixing: A measure space $(X,\mathscr{A},\mu)$ with measure preserving map $T:X\rightarrow X$ is said to be alternate mixing if for every $f,g\in D\subseteq L^2(\mu)$, $$\lim_{n\to\infty}\langle U_T^n f,g\rangle=\lim_{n\to\infty}\langle f(T^n),g\rangle=\langle f,1\rangle\langle 1,g\rangle=\mathbb{E}(f)\mathbb{E}(g).$$ Where $D$ is a dense subset of $L^2(\mu)$.

As stated by the OP, the implication (Alternate Mixing)$\implies$(Mixing) follows easily by considering indicator functions.

For the other implication, we will need the following lemma that follows from Lemma $3.13$ of Rudin's Real and Complex Analysis, $3$rd Ed, p. $69$.

Lemma Let $(X,\mathscr{A},\mu)$ be any measure space. Then the set $S$ of all simple functions with finite support are dense in $L^2(\mu)$.

(Mixing)$\implies$(Alternate Mixing)

Assume that the statement of the mixing definition holds true. Take any $r,t\in S$. We define, $$r(x)=\sum^n_{i=1}a_i\mathbf{1}_{A_i}(x)\qquad t(x)=\sum^m_{j=1}b_j\mathbf{1}_{B_j}(x)$$ Where $\bigcup_{i=1}^n{A_i}\subseteq X$ and $\bigcup_{j=1}^m{B_j}\subseteq X$ and also $\{a_i\}_{i=1}^n\cup\{b_j\}_{j=1}^m\subseteq \mathbb{R}$.

Consider then, $$\langle r(T^n),t\rangle=\int_X \left(\sum_{i}a_i\mathbf{1}_{T^{-n}A_i}\right) \left(\sum_{j}b_j\mathbf{1}_{B_j}\right)d\mu$$ $$= \int_X \sum_{i,j}a_ib_j\ \mathbf{1}_{T^{-n}A_i\cap B_j}\ d\mu.$$ By the linearity of the integral, $$\langle r(T^n),t\rangle=\sum_{i,j}a_ib_j \int_X\mathbf{1}_{T^{-n}A_i\cap B_j}\ d\mu=\sum_{i,j}a_ib_j\ \mu(T^{-n}A_i\cap B_j).$$ Therefore, $$\lim_{n\rightarrow\infty}\langle r(T^n),t\rangle=\lim_{n\rightarrow\infty}\sum_{i,j}a_ib_j\ \mu(T^{-n}A_i\cap B_j)=\sum_{i,j}a_ib_j\ \lim_{n\rightarrow\infty}\mu(T^{-n}A_i\cap B_j).$$ By our assumption, we have that, $$\lim_{n\to\infty}\langle U_T^n f,g\rangle=\lim_{n\rightarrow\infty}\langle r(T^n),t\rangle=\sum_{i,j}a_ib_j\mu(A_i)\mu(B_j)=\left(\sum_{i}a_i\mu(A_i)\right)\left(\sum_{j}b_j\mu(B_j)\right)$$ $$=\mathbb{E}(r)\mathbb{E}(t)=\langle r,1\rangle\langle 1,t\rangle.$$

And the required result follows.

We can use the result we have just proven to prove that the alternate mixing definition holds true on all of $L^2(\mu)$.

This will follow from the fact that any $f,g\in L^2(\mu)$ can be approximated arbitrarily well by two sequences of simple measurable functions in $S$. Using these sequences, the above argument can be modified to prove the general result.