## Solutions to Ergodic Theory：Lecture Eight

Lecture eight：The mean ergodic theorem

((note:：Professor Terence Tao began posting his lecture notes about the course “ergodic theory” early in 2008. I try doing all the exercises. This is the ten exercises in lecture eight. Here is the the page of this course，and here is the page of this lecture. ）)

Exercise 1：Let $(X, {\mathcal X}, \mu)$ be a probability space with ${\mathcal X}$ separable. Then the Banach spaces $L^p(X, {\mathcal X}, \mu)$ are separable (i.e. have a countable dense subset) for every $1 \leq p < \infty$; in particular, the Hilbert space $L^2(X, {\mathcal X}$, $\mu$) is separable. Show that the claim can fail for $p = \infty$. (We allow the $L^p$ spaces to be either real or complex valued, unless otherwise specified.)

Proof：Suppose there is a countable family of sets $\mathscr{G}$ such that $\mathcal X = \sigma(\mathscr{G})$. Let $\Bbb Q (\mathscr{G})$ denote the set of all the linear combination of indicator functions in $\mathscr{G}$ over rational numbers. One easily verifies that the space $\Bbb Q (\mathscr{G})$ is countable. Given a function $f\in L^p(X, {\mathcal X}, \mu)$, we can choose a sequence of $f_i \in \Bbb Q (\mathscr{G})$ such that $f_i \stackrel {L^p}{\to} f$. This complete the proof.

On the other hand, since $\mathcal X$ could have uncountably many elements, we could choose choose uncountably many set from $\mathcal X$. If any two of these sets A and $A'$ have there symmetry difference set with positive measure, then we have that $\|I_A-I_{A'}\|_\infty =1$. This means the above claim can fail for $p = \infty$.$\diamond$

Exercise 2：Prove the following generalization of the $Poincar\acute{e}$ recurrence theorem: if $(X, {\mathcal X}, \mu, T)$ is a measure-preserving system and $f \in L^1(X, {\mathcal X},\mu)$ is non-negative, then $\limsup_{n \to +\infty} \int_X f T^n f \geq (\int_X f\ d\mu)^2$.

Proof：Firstly we observe the fact that $\int_X T^nf(x) d\mu =\int_X f(x) d\mu$. It can be easily established by considering the indicator function firstly, the simple function and then general function f as the limit of simple functions. Thus we have $\int_X sum_{n=1}^N T^n f(x) d\mu =N\int_X f(x)d\mu$. By Cauchy-schwarz inequality,

$\int_X( \sum_{n=1}^N T^n f(x))^2 d\mu \geq N^2 (\int_X f(x)d\mu) ^2$

The left-hand side can be rearranged as

$\sum_{n=1}^N \sum_{m=1}^N \int_X T^nf(x) T^m f(x)d\mu$

and $latex\int_X T^mf(x) T^n f(x)d\mu =\int_X f(x) T^{n-m}(x)d\mu$.

For any $latex\varepsilon >0$ and any m a positive integer, there exists some $M >0$ such that $\int f(x) T^n f(x)d \mu< \limsup_{n\to \infty}\int_X f(x) T^n f(x)d\mu +\varepsilon/2$ whenever $n\geq M$. Then

$\sum_{n=1}^N\int_Xf(x)T^{n-m}f(x)d\mu \leq(N-2M) (\limsup_{n\to \infty}\int_X f(x) T^n f(x) d\mu +\varepsilon ) +2M C$

here C is some constant, thus we can get

$\sum_{n=1}^N \sum_{m=1}^N \int_X T^nf(x) T^m f(x)d\mu \leq N^2 (\limsup_{n\to \infty}\int_X f(x) T^n f(x) d\mu +\varepsilon/2)$

for all large N, the final conclusion follows from the arbitrariness of $\varepsilon$.

Exercise 3：Give examples to show that the quantity $\mu(X)^2$ in the conclusion of Theorem 1 cannot be replaced by any smaller quantity in general,
regardless of the actual value of $\mu(X)$. (Hint: use a Bernoulli system example.)

Proof：In the space $\{0,1\}$ and define $P\{0\} = P\{1\}=\frac{1}{2}$. Let $X=\{0,1\}^\Bbb Z$ and let $\mu$ is the product measure. For any positive number $\varepsilon >0$, we can choose some set E with $\mu(E)=\varepsilon$ in the following way.We first represent $\varepsilon$ as the form of

$\sum_{n=1}^\infty \frac{a_n}{2^n}, a_n=0,1$

Suppose $a_{n_1}$ is the first nonzero term. Let $U_1$ be the open set with the first $n_1$ coordinates “1” which has measure $\frac{1}{2^n}$. Similarly we choose $U_i$ since and finally get a series of non joint open sets satisfying that $\mu(\cup_nU_n)=\varepsilon.$ So we define $E=\cup_{n=1}^\infty U_n$

Now take any positive number $\delta < \varepsilon =\mu (E)$, from the above method, we can stop earlier when the measure of the union set is larger than $\delta$ before we have infinite many open sets. Suppose we have a positive integer $n_0$ such that $\mu(\cup_{i=1}^{n_0}U_i)>\delta$. Let $E_0=\cup_{i=1}^{n_0}U_i)$

One easily verifies $\mu(T^nE_0\cap E_0)=\mu(E_0)^2>\delta^2$, which means $\mu(T^nE\cap E)>\delta^2$, too.

Exercise 4：Using the pigeonhole principle instead of the Cauchy-Schwarz inequality (and in particular, the statement that if $\mu(E_1) + \ldots + \mu(E_n) > 1$, then the sets $E_1,\ldots,E_n$ cannot all be disjoint), prove the weaker statement that for any set E of positive measure in a measure-preserving system, the set $E \cap T^n E$ is non-empty for infinitely many n. (This exercise illustrates the general point that the Cauchy-Schwarz inequality can be viewed as a quantitative strengthening of the pigeonhole principle.)

Proof：Since $\int_X\sum_{n=1}^N 1_{T^nE}d\mu =N\mu(E)$, we can choose N large enough that $N\mu(E) > 1$. Then from the pegeon-hole
principle, there is some $T^m (E)$ and $T^n(E)$ such that $T^m(E)\cap T^n(E) \neq \phi$ (assuming $n>m$), which means $E\cap T^{n-m}(E) \neq \phi$. Then take l larger than n-m, and observe $\int_X\sum_{n=1}^N 1_{T^{nl}E}d\mu =N\mu(E)$ for the same N chosen before, we thus have some n such that $E \cap T^{ln}(E) \neq \phi$. Continue this inductively, we get that the set $E \cap T^n(E)$ is nonempty for infinite many n.

Exercise 5：With the notation and assumptions of Corollary 1, show that the limit $\lim_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} \int_X T^n f \overline{f}\ d\mu$ exists, is real, and is greater than or equal to $|\int_X f|^2$. (Hint: the constant function 1 lies in $L^2(X, {\mathcal X}, \mu)^T$.) Note that this is stronger than the conclusion of Exercise 2.

Proof：By Corollary 1, we know that the limit is actually $\int_X\pi(f)\overline {f}d\mu=<\pi(f),f>$. Suppose the $\{x_n\}_{n=1}^\infty$ is the orthogonal basis of the closed subspace $L^2(X, {\mathcal X}, \mu)^T$. Without loss of generality, suppose $x_1=1$ the constant function. If we define $a_n=$, then $<\pi(f),f>=\sum_{n=1}^\infty |x_n|^2$, which is obviously real. Since $|x_1|^2=||^2=|\int_Xfd\mu|^2$, we are done.

Exercise 6：Let $(X, {\mathcal X}, \mu)$ be a probability space, and let ${\mathcal X}'$ be a sub$-\sigma-$algebra. Let $f \in L^2(X, {\mathcal X}, \mu)$.

1. The operator $f \mapsto {\Bbb E}( f|{\mathcal X'})$ is a bounded self-adjoint projection on $L^2(X,{\mathcal X},\mu)$. It maps real functions to real functions, it preserves constant functions (and more generally preserves ${\mathcal X'}$-valued functions), and commutes with complex conjugation.

2. If f is non-negative, then ${\Bbb E}( f|{\mathcal X'})$ is non-negative (up to sets of measure zero, of course). More generally, we have a comparison principle: if f, g are real-valued and $f \leq g$ pointwise a. e., then ${\Bbb E}( f|{\mathcal X'}) \leq {\Bbb E}( g|{\mathcal X'})$ a.e. Similarly, we have the triangle inequality $|{\Bbb E}( f|{\mathcal X'})| \leq {\Bbb E}( |f||{\mathcal X'})$ a.e..

3. (Module property) If $g \in L^\infty(X, {\mathcal X}', \mu)$, then ${\Bbb E}( f g|{\mathcal X’}) = {\Bbb E}( f|{\mathcal X’}) g$ a.e..

4. (Contraction) If $f \in L^2(X, {\mathcal X},\mu) \cap L^p(X, {\mathcal X},\mu) for some 1 \leq p \leq \infty, then \|{\Bbb E}(f|{\mathcal X'})\|_{L^p} \leq \|f\|_{L^p}$. (Hint: do the $p=1$ and $p=\infty$ cases first.) This implies in particular that conditional expectation has a unique continuous extension to $L^p(X, {\mathcal X},\mu)$ for $1 \leq p \leq \infty$ (the $p=\infty$ case is exceptional, but note that $L^\infty$ is contained in $L^2$ since $\mu$ is finite).

Proof.1, The operator is bounded from the inequality:

$\int_X|{\Bbb E}( f|{\mathcal X'})|^2d\mu \leq \int_X {\Bbb E}( f|{\mathcal X'}) \overline{{\Bbb E}( f|{\mathcal X'})} d\mu =\int_X f\overline{{\Bbb E}( f|{\mathcal X'})} d\mu$ $\leq (\int_X |f|^2d\mu)^{\frac{1}{2}} (\int_X|{\Bbb E}( f|{\mathcal X’})|^2d\mu)^{\frac{1}{2}}$

If f is a real function, then for any $A\in \mathcal X'$, we have $\int_A {\Bbb E}( f|{\mathcal X'})d\mu=\int_Afd\mu$ is a real number, so ${\Bbb E}( f|{\mathcal X'})$ must be a real function on $\mathcal X'$.The rest conclusions are obvious.

2, It is obvious.

3, Note that given any $A\in \mathcal X'$, the equality $\int_A\Bbb E(f|\mathcal X')gd\mu=\int_fgd\mu$ holds.

4, First consider the situations when $p=1$ and $p=\infty$. For the former one, we have

$\int_X\Bbb |E(f|\mathcal X')|d\mu=\int_{E(f|\mathcal X')>0}fd\mu+\int_{E(f|\mathcal X')\leq 0}-fd\mu\leq \int_X|f|d\mu$;

for the later one, it suffice to prove that for any large number A, if $\mu(|\Bbb E(f|\mathcal X')|\geq A)>0$, then $\mu(|f|\geq A)>0$, which is easy to verify.

When $\infty >p>1$, we have that:

$\int_X |\Bbb E (f|\mathcal X')|^p d\mu=$

$\int_X |\Bbb E(f|\mathcal X')| |\Bbb E (f|\mathcal X')|^{p-1}d\mu \leq (\int_X|f|^pd\mu)^{\frac{1}{p}}(\int_X|E(f|\mathcal X')|^p)^{\frac{p-1}{p}}$

Exercise 7：Show that if E lies in ${\mathcal X}^T$, then there exists a set $F \in {\mathcal X}$ which is genuinely invariant ($TF=F$) and which differs from E only by a set of measure zero. Thus it does not matter whether we deal with shift-invariance or essential shift-invariance here. (More generally, it will not make any significant difference if we modify any of the sets in our $\sigma$-algebras by null sets.)

Proof. Let $F= \bigcap_{-\infty}^{\infty} T^n E$. One easily sees that both $TF=F$ and $m(F)=m(E)$.

Exercise 8：Show that $L^2( X, {\mathcal X}, \mu)^T = L^2( X, {\mathcal X}^T, \mu)$.

Proof. $\subseteq$“: Suppose there exists a $f \in L^2( X, {\mathcal X}, \mu)^T$ which is not in $L^2( X, {\mathcal X}^T, \mu)$. It means that there exists some measurable set K such that $E= f^{-1}(K)$ is not in ${\mathcal X}^T$. Thus $E \neq T(E)$, and we draw our conclusion from the fact that $Tf \neq f$ on at least a set with positive measure.

$\supseteq$“: Suppose there exists a $f \in L^2( X, {\mathcal X}^T, \mu)$ which is not in $L^2( X, {\mathcal X}, \mu)^T$. Suppose that $E= \{x; f(T^{-1}(x))> f(x)\}$ has positive measure. Since we have that $T^{-1}(E)=E$,we have that $\int_E Tf > \int_E f$. On the other hand, however, T is measure-preserving shift, so $\int_E Tf = \int_E f$ should hold, a contradiction.

Exercise 9.Show that Corollary 2 continues to hold if $L^2$ is replaced throughout by $L^p$ for any $1 \leq p < \infty$. (Hint: for the case $p<2$, use that $L^2$ is dense in $L^p$. For the case $p>2$, use that $L^\infty$ is dense in $L^p$.) What happens when $p = \infty$?

Proof.For the case $p<2$, given any $f\in L^p$, since $L^2$ is dense in $L^p$, we can take a sequence $g_k\in L^2$ converging to f in $L^p$. Then it suffice to notice the following:

$\|\frac{1}{N}\sum_{n=0}^{N-1}T^nf-\Bbb E(f| \mathcal X^T)\|_{L^p}$

$\leq \|\frac{1}{N}\sum_{n=0}^{N-1}T^nf-\frac{1}{N}\sum_{n=0}^{N-1}T^ng_k\|_{L^p} +$

$\|\frac{1}{N}\sum_{n=0}^{N-1}T^ng_k-\Bbb E(g_k| \mathcal X^T)\|_{L^p}+\|\Bbb E(g_k| \mathcal X^T)-\Bbb E(f| \mathcal X^T)\|_{L^p}$

For the case $p>2$, given any $f\in L^p$, since $L^\infty$ is dense in $L^p$, we can take a sequence $g_k\in L^\infty$ converging to f in $L^p$. Also, one should notice that for a $g_k\in L^\infty$, there exists a sequence of simple functions $g_{ki}$ converging to $g_k$ in$L^\infty$. It is obvious that for each simple function, the conclusion holds, hence for $f\in L^p$.

Exercise 10：Let $(\Gamma,\cdot)$ be a countably infinite discrete group. A F$\phi$lner sequence is a sequence of increasing finite non-empty sets $F_n$ in $\Gamma$ with $\bigcup_n F_n = \Gamma$ with the property that for any given finite set $S \subset \Gamma$, we have $|(F_n \cdot S) \Delta F_n|/|F_n| \to 0$ as $n \to \infty$, where $F_n \cdot S := \{ fs: f \in F_n, s \in S\}$ is the product set of $F_n$ and S, $|F_n|$ denotes the cardinality of $F_n$, and $\Delta$ denotes symmetric difference. (For instance, in the case $\Gamma = {\Bbb Z}$, the sequence $F_n := \{-n,\ldots,n\}$ is a F$\phi$lner sequence.) If $\Gamma$ acts (on the left) in a measure-preserving manner on a probability space $(X, {\mathcal X}, \mu)$, and $f \in L^2(X, {\mathcal X}, \mu)$, show that $\frac{1}{|F_n|} \sum_{\gamma \in F_n} f \circ \gamma^{-1}$ converges in $L^2$ to ${\Bbb E}(f|{\mathcal X}^\Gamma)$, where ${\mathcal X}^\Gamma$ is the collection of all measurable sets which are $\Gamma$-invariant modulo null sets, and $f \circ \gamma^{-1}$ is the function $x \mapsto f(\gamma^{-1} x)$.

Proof. (Basically the same procedure of the proof of Corollary 2)

Let $f_U=f-\Bbb E(f|\mathcal X^\Gamma)$ and it suffice to show that $\|\frac{1}{|F_n|} \sum_{\gamma \in F_n} f_U \circ \gamma^{-1}\|_{L^2} \to 0$. The left-hand side can be written as:

$=\int_XG_n\overline {f_U} d\mu$,

where $G_n=\frac{1}{|F_n|^2}\sum_{\gamma_1\in F_n}\sum_{\gamma_2\in F_n}f_U\circ \gamma_1^{-1}\gamma_2$

Now from the triangle inequality we know that the sequence of $G_n$ is uniformly bounded in $L^2$ and so by Cauchy-Schwarz we know that the inner products $$ are bounded. If they converge to zero, we are done. otherwise, by the Bolzano-weierstrass theorem, we have for some subsequence $G_{n_j}$, we have $\to c$.

By the Banarch-Alaoglu theorem, there is a further subsequence $G_{n_{j_k}}$ which converge to some limit $G\in L^2(X,\mathcal X,\mu)$. Since c is not zero, G is not zero function. Then from the definition of the the F$\phi$lner sequence for any $\gamma \in \Gamma$, we have that

$\|G_n\gamma- G_n\|_{L^2}$

$=\|\frac{1}{|F_n|^2}\sum_{\gamma_1\in F_n}\sum_{\gamma_2\in F_n}f_U\circ \gamma_1^{-1}\gamma_2\gamma$

$-\frac{1}{|F_n|^2}\sum_{\gamma_1\in F_n}\sum_{\gamma_2\in F_n}f_U\circ \gamma_1^{-1}\gamma_2\|_{L^2}$

Suppose $K=F_n\Gamma\cap F_n$, then the right hand side becomes

$\|\frac{1}{|F_n|^2}\sum_{\gamma_1\in F_n}\sum_{\gamma_2\notin K}( f_U\circ \gamma_1^{-1}\gamma_2\gamma -f_U\circ \gamma_1^{-1}\gamma_2\|_{L^2}\leq C\frac{|F_n\cdot\gamma \Delta F_n|}{|F_n|}$

which converge to 0 from the property of F$\phi$lner sequence. So on taking limits we get that $G\in L^2(X,\mathcal X^\Gamma,\mu )$

On the other hand, we have $\Bbb E(f_U|\mathcal X^\Gamma)=0$, and so $\Bbb E(G_n|\mathcal X^\Gamma)$ is 0. By taking limits we have $\Bbb E(G|\mathcal X^\Gamma)=0$ also, leading to a contradiction.