Posted by: yglima | October 7, 2009

## ERT1: Poincaré’s Recurrence Theorem and Von Neumann’s Theorems

Let’s define the settings: we will always consider a measure-preserving system ${(X,\mathcal B,\mu,T)}$, meaning that ${(X,\mathcal B,\mu)}$ is a probability space and ${T}$ is a measurable transformation that preserves ${\mu}$:

$\displaystyle T_*\mu=\mu\ \iff\ \mu(T^{-1}A)=\mu(A)\,,\ \forall\,A\in\mathcal B.$

If ${\mu(A)>0}$, then there exists ${n>0}$ such that ${\mu(A\cap T^{-n}A)>0}$. This is easy to see because the family ${A_n=T^{-n}A}$, ${n\ge 0}$, satisfies a stationary condition

$\displaystyle\mu(T^{-n}A\cap T^{-m}A)=\mu\left(A\cap T^{-(m-n)}A\right),\ \forall\,m\geq n\ge 0.$

So, if ${m=\lfloor 1/\mu(A)\rfloor+1}$, two of the sets ${A_0,\ldots,A_m}$ have positive-measure intersection. In fact, if this is not the case, then

$\displaystyle \begin{array}{rcl} 1&\ge& \mu\left(\bigcup_{n=0}^{m}A_n\right)\\ &=&\sum_{n=0}^{m}\mu(A_n)\\ &=&(m+1)\cdot\mu(A)\\ &>&1, \end{array}$

a contradiction. We then get the original Poincaré’s Recurrence Theorem:

Theorem 1 (PRT) If ${\mu(A)>0}$, then there exists ${n\in\mathbb N}$ such that ${\mu(A\cap T^{-n}A)>0}$.

Remark 1 The modern statements of PRT are: if ${\mu(A)>0}$, then a.e. point ${x\in A}$ returns to ${A}$. This means that ${\mu(A\cap(\cup_{n\ge 1}T^{-n}A))=\mu(A)}$, which obviously implies the above theorem.

This proves more: call a set ${S\subseteq\mathbb N}$ syndetic if it has bounded gaps, i.e., if there exists ${n_0>0}$ such that ${S\cap\{n,n+1,\ldots,n+n_0\}\not=\emptyset,\ \forall\,n\in\mathbb N}$.

Exercise 1 Prove that if ${\mu(A)>0}$, then the set ${\{n\in\mathbb N\,;\,\mu(A\cap T^{-n}A)>0\}}$ is syndetic.

For further discussions about PRT, the reader may consult this paper of Vitaly Bergelson. Altought its simplicity, this is a remarkable result. It implies, for example, that almost every ${\alpha\in{\mathbb R}}$ has infinitely many ${7}$‘s in its decimal representation, and the same happens for any finite sequence of digits.

As ${T}$ preserves ${\mu}$, it defines a unitary operator ${U_T}$ on the Hilbert space ${L^2(X,\mathcal B,\mu)}$ by ${U_Tf=f\circ T}$, for simplicity denoted from now on as ${T:L^2\rightarrow L^2}$. With this notation, if ${f=\chi_A}$, then

$\displaystyle \mu\left(A\cap T^{-n}A\right)=\int_X \chi_A\cdot\chi_{T^{-n}A}d\mu=\int_X f\cdot T^nfd\mu=\left\langle f,T^nf\right\rangle,$

where ${\left\langle f,g\right\rangle=\int_X fgd\mu}$ is the inner product in ${L^2}$, so

$\displaystyle \begin{array}{rcl} \displaystyle\sum_{n=1}^N\mu\left(A\cap T^{-n}A\right)&=&\displaystyle\sum_{n=1}^N\left\langle f,T^nf\right\rangle\\ &=&\left\langle f,\displaystyle\sum_{n=1}^NT^nf\right\rangle,\\ \end{array}$

such that the sequence ${f_N=N^{-1}\cdot\sum_{n=1}^NT^nf}$, ${N\ge 1}$, may give more general results than PRT. This is what happens.

Theorem 2 (Von Neumann) If ${f\in L^2}$, then the sequence ${f_N=N^{-1}\cdot\sum_{n=1}^NT^nf}$, ${N\ge 1}$, converges in ${L^2}$.

This theorem, also known as Mean Ergodic Theorem, is in fact a spectral theoretical result and a more general version holds, given by

Theorem 3 If ${T:\mathcal H\rightarrow\mathcal H}$ is a unitary operator on a Hilbert space ${\mathcal H}$, then the sequence of operators ${T_N=N^{-1}\cdot\sum_{n=1}^NT^n}$, ${N\ge 1}$, converges pointwise in norm to the orthogonal projection ${P:\mathcal H\rightarrow\mathcal M}$ onto the subspace of ${T}$-fixed elements ${\mathcal M=\{x\in\mathcal H\,;\,Tx=x\}}$.

Proof: When $T$ is unitary, ${{\rm Ker}(T-I)={\rm Ker}(T^*-I)}$. From the general orthogonal decomposition

$\displaystyle \mathcal H={\rm Ker}(T^*-I)\oplus \overline{{\rm Im}(T-I)},$

we obtain

$\displaystyle \mathcal H={\rm Ker}(T-I)\oplus \overline{{\rm Im}(T-I)}.$

For ${x\in{\rm Ker}(T-I)}$, the convergence is obvious. If ${x=Ty-y}$, then

$\displaystyle \left\|\dfrac{1}{N}\sum_{n=1}^NT^nx\right\|=\left\|\dfrac{T^{N+1}y-y}{N}\right\|\le\dfrac{2\left\|y\right\|}{N}\longrightarrow 0\ \text{ as }N\rightarrow+\infty.$

By approximation and applying the triangle inequality, the same happens in ${\overline{{\rm Im}(T-I)}}$, which concludes the proof.$\Box$

Remark 2 Being, as we said, Hilbertian in nature, Theorem 2 also holds when ${\mu(X)=+\infty}$.

Exercise 2 Under the same conditions of Theorem 3, prove that the same conclusion happens for a sequence ${(M-N)^{-1}\cdot\sum_{n=N+1}^{M}T^n}$ such that $M-N\rightarrow+\infty$.

Let’s show how to use these convergences to obtain recurrence results.

Proposition 4 Let ${f\in L^2\backslash\{0\}}$ be such that ${{f\ge 0}}$. Then ${Pf\ge 0}$ and ${\left\|Pf\right\|>0}$.

Note that ${f=\chi_A}$ satisfies the above conditions.

Proof: Consider the function ${g=\max\{Pf,0\}}$. Then ${g\in\mathcal M}$ and ${{\left\|f-g\right\|\le\left\|f-Pf\right\|}}$. Because ${Pf}$ minimizes the distance of ${f}$ to ${\mathcal M}$, we have ${Pf=g\ge0}$. In addition, if we had ${\left\|Pf\right\|=0}$, then ${f\in\mathcal M^\perp}$, such that ${N^{-1}\sum_{n=1}^{N}T^nf\rightarrow0}$. Integrating, we conclude

$\displaystyle \int_X fd\mu=\int_X\left(\dfrac{1}{N}\sum_{n=1}^{N}T^nf\right)\rightarrow0\ \Longrightarrow\ \int_X fd\mu=0\ \Longrightarrow\ f=0,$

a contradiction. $\Box$

Exercise 3 Using the above proposition, prove that if ${\mu(A)>0}$, then the set ${\{n\in\mathbb N\,;\,\mu(A\cap T^{-n}A)>0\}}$ is syndetic. (Hint: if ${f=\chi_A}$, then ${\left\langle f,(M-N)^{-1}\cdot\sum_{n=N+1}^{M}T^{n}\right\rangle}$ converges to ${\left\langle f,Pf\right\rangle=\left\|Pf\right\|^2}$ as ${M-N\rightarrow+\infty}$.)

So, expressions of the type ${(M-N)^{-1}\cdot\sum_{n=N+1}^{M}T^nf}$, from now on called ergodic averages, are important when dealing with recurrence. This will be our main interest in the next posts. For another perspective on Von Neumann’s Theorem and related results, the reader is referred to this Terence Tao’s lecture.

Previous posts: ERT0.

1. I think in Remark1 you mean to say $\mu(A\cap (\cup_{n \geq 1} T^{-n} A))= 1$ ?
2. sorry not 1 but $\mu (A)$ ? [You’re absolutely right! The remark was changed accordingly – M.]