## 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 be a probability space with separable. Then the Banach spaces are separable (i.e. have a countable dense subset) for every ; in particular, the Hilbert space , ) is separable. Show that the claim can fail for . (We allow the spaces to be either real or complex valued, unless otherwise specified.)

**Proof：**Suppose there is a countable family of sets such that . Let denote the set of all the linear combination of indicator functions in over rational numbers. One easily verifies that the space is countable. Given a function , we can choose a sequence of such that . This complete the proof.

On the other hand, since could have uncountably many elements, we could choose choose uncountably many set from . If any two of these sets A and have there symmetry difference set with positive measure, then we have that . This means the above claim can fail for .

**Exercise 2**：Prove the following generalization of the recurrence theorem: if is a measure-preserving system and is non-negative, then .

**Proof**：Firstly we observe the fact that . 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 . By Cauchy-schwarz inequality,

The left-hand side can be rearranged as

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 such that whenever . Then

here C is some constant, thus we can get

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

**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 . (Hint: use a Bernoulli system example.)

**Proof：**In the space and define . Let and let is the product measure. For any positive number , we can choose some set E with in the following way.We first represent as the form of

Suppose is the first nonzero term. Let be the open set with the first coordinates “1″ which has measure . Similarly we choose since and finally get a series of non joint open sets satisfying that So we define

Now take any positive number , from the above method, we can stop earlier when the measure of the union set is larger than before we have infinite many open sets. Suppose we have a positive integer such that . Let

One easily verifies , which means , too.

**Exercise 4：**Using the pigeonhole principle instead of the Cauchy-Schwarz inequality (and in particular, the statement that if , then the sets cannot all be disjoint), prove the weaker statement that for any set E of positive measure in a measure-preserving system, the set 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 , we can choose N large enough that . Then from the pegeon-hole

principle, there is some and $T^n(E)$ such that (assuming ), which means . Then take l larger than n-m, and observe for the same N chosen before, we thus have some n such that . Continue this inductively, we get that the set is nonempty for infinite many n.

**Exercise 5：**With the notation and assumptions of Corollary 1, show that the limit exists, is real, and is greater than or equal to . (Hint: the constant function 1 lies in .) Note that this is stronger than the conclusion of Exercise 2.

**Proof：**By Corollary 1, we know that the limit is actually . Suppose the is the orthogonal basis of the closed subspace . Without loss of generality, suppose the constant function. If we define , then , which is obviously real. Since , we are done.

**Exercise 6：**Let be a probability space, and let be a subalgebra. Let .

1. The operator is a bounded self-adjoint projection on . It maps real functions to real functions, it preserves constant functions (and more generally preserves -valued functions), and commutes with complex conjugation.

2. If f is non-negative, then 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 pointwise a. e., then a.e. Similarly, we have the triangle inequality a.e..

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

4. (Contraction) If . (Hint: do the and cases first.) This implies in particular that conditional expectation has a unique continuous extension to for (the case is exceptional, but note that is contained in since is finite).

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

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

If f is a real function, then for any , we have is a real number, so must be a real function on .The rest conclusions are obvious.

2, It is obvious.

3, Note that given any , the equality holds.

4, First consider the situations when and . For the former one, we have

;

for the later one, it suffice to prove that for any large number A, if , then , which is easy to verify.

When , we have that:

**Exercise 7：**Show that if E lies in , then there exists a set which is genuinely invariant () 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 -algebras by null sets.)

**Proof. **Let . One easily sees that both and .

**Exercise 8：**Show that .

**Proof. **““: Suppose there exists a which is not in . It means that there exists some measurable set K such that is not in . Thus , and we draw our conclusion from the fact that on at least a set with positive measure.

““: Suppose there exists a which is not in . Suppose that has positive measure. Since we have that ,we have that . On the other hand, however, T is measure-preserving shift, so should hold, a contradiction.

**Exercise 9.**Show that Corollary 2 continues to hold if is replaced throughout by for any . (Hint: for the case , use that is dense in . For the case , use that is dense in .) What happens when ?

**Proof.**For the case , given any , since is dense in , we can take a sequence converging to f in . Then it suffice to notice the following:

For the case , given any , since is dense in , we can take a sequence converging to f in . Also, one should notice that for a , there exists a sequence of simple functions converging to in. It is obvious that for each simple function, the conclusion holds, hence for .

**Exercise 10：**Let be a countably infinite discrete group. A Flner sequence is a sequence of increasing finite non-empty sets in with with the property that for any given finite set , we have as , where is the product set of and S, denotes the cardinality of , and denotes symmetric difference. (For instance, in the case , the sequence is a Flner sequence.) If acts (on the left) in a measure-preserving manner on a probability space , and , show that converges in to , where is the collection of all measurable sets which are -invariant modulo null sets, and is the function .

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

Let and it suffice to show that . The left-hand side can be written as:

,

where

Now from the triangle inequality we know that the sequence of is uniformly bounded in 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 , we have .

By the Banarch-Alaoglu theorem, there is a further subsequence which converge to some limit . Since c is not zero, G is not zero function. Then from the definition of the the Flner sequence for any , we have that

Suppose , then the right hand side becomes

which converge to 0 from the property of Flner sequence. So on taking limits we get that

On the other hand, we have , and so is 0. By taking limits we have also, leading to a contradiction.

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