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=<f,x_n>, then <\pi(f),f>=\sum_{n=1}^\infty |x_n|^2, which is obviously real. Since |x_1|^2=|<f,1>|^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 inL^\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\philner 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\philner 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:

<G_n,f_U>=\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 <G_n,F_U> 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 <F_{n_j},f_U>\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\philner 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\philner 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.

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