Notes for my lecture on multiple recurrence theorem for weakly mixing systems – Part 2

Now we can start to prove the multiple recurrence theorem in the weak mixing case. Again the material is from Furstenberg’s book ‘Recurrence in Ergodic Theory and Combinatorial Number Theory’ which, very unfortunately, is hard to find a copy of.

Definition: A sequence (x_i) \subseteq X converges in density to x if there exists Z \subseteq \mathbb{N} of density 0 s.t. for all neighborhood U of x, \exists N \in \mathbb{N}, \ \{ x_n \ | \ n \geq N and n \notin Z \} \subseteq U.

We denote (x_n) \rightarrow_D \ x.

Theorem: For (X, \mathcal{B}, \mu, T) weakly mixing,

then \forall f_0, f_1, \cdots, f_k \in L^\infty(X), we have

\int f_0(x) f_1(T^n(x)) f_2(T^{2n}(x)) \cdots f_k(T^{kn}(x)) \ d \mu

\rightarrow_D \int f_0 \ d \mu \int f_1 \ d \mu \cdots \int f_k \ d \mu as n \rightarrow \infty

In particular, this implies for any A \in \mathcal{B} with \mu(A)>0, by taking f_0 = f_1 = \cdots = f_k = \chi_A we have
\mu(A \cap T^{-n}(A) \cap \cdots \cap T^{-kn}(A)) \rightarrow_D \mu(A)^k > 0.
Hence we may pick N \in \mathbb{N} for which
\mu(A \cap T^{-N}(A) \cap \cdots \cap T^{-kN}(A)) > 0.

Establishes the multiple recurrence theorem.

To prove the theorem we need the following:

Lemma 1: Let (f_n) be a bounded sequence in Hilbert space \mathcal{H}, if \langle f_{n+m}, f_n \rangle \rightarrow_D a_m as n \rightarrow \infty, a_m \rightarrow_D 0 as m \rightarrow \infty. Then (f_n) converges weakly in density to \overline{0}

In order to prove the lemma 1, we need:

Lemma 2: Given \{ R_q \ | \ q \in Q \} a family of density 1 subsets of \mathbb{N}, indexed by density 1 set Q \subseteq \mathbb{N}. Then for all S \subseteq \mathbb{N} of positive upper density, for all k \geq 1. There exists \{ n_1, n_2, \cdots, n_k \} \subseteq S, n_1 < n_2 \cdots < n_k such that whenever ii \in Q and n_i \in R_{n_j-n_i}.

Proof of lemma 2: We’ll show this by induction. For k=1, since there is no such j>1, the statement is vacant.

We’ll proceed by induction: Suppose for k>1, there exists S_k \subseteq S of positive upper density and integers m_1< \cdots < m_k such that (S_k+m_1) \cup (S_k+m_2) \cup \cdots \cup (S_k+m_k) \subseteq S and for all j>i, m_j - m_i \in Q and S_k+m_i \subseteq R_{m_j-m_i}.

For k+1, we shall find S_{k+1} \subseteq S_k with positive upper density and m_{k+1}>m_k where S_{k+1}+m_{k+1} \subseteq S and for all 1 \leq i \leq k, m_{k+1} - m_i \in Q and S_{k+1}+m_i \subseteq R_{m_{k+1}-m_i}.

Let S_k^* = \{n \ | \ S_k+n \cap S_k has positive upper density \}.

Claim:S_k^* has positive upper density.

Since \overline{D}(S_k) = \epsilon >0, let N = \lceil 1/ \epsilon \rceil.

Hence there is at most N-1 translates of S_k that pairwise intersects in sets of density 0.

Let M < N be the largest number of such sets, let p_1, \cdots, p_M be a set of numbers where (S_k+p_i) \cap (S_k+p_j) has density 0.
i.e. S_k+(p_j-p_i) \cap S_k has density 0.

Therefore for any p>p_M, (S_k+p-p_i) \cap S_k has positive upper density for some i. Hence p-p_i \in S_k^*. S_k^* is syntactic with bounded gap 2 \cdot p_M hence has positive upper density.

Pick \displaystyle m_{k+1} \in S_k^* \cap \bigcap_{i=1}^k(Q+m_i).

(Hence m_{k+1}-m_i \in Q for each 1 \leq i \leq k)

Let \displaystyle S_{k+1} = (S_k - m_{k+1})  \cap S_k \cap \bigcap_{i=1}^k (R_{m_k+1 - m_i}-m_i).

(S_k - m_{k+1}) \cap S_k has positive upper density, \bigcap_{i=1}^k (R_{m_k+1 - m_i}-m_i) has density 1, S_{k+1} has positive upper density.

S_{k+1}, \ m_{k+1} satisfied the desired property by construction. Hence we have finished the induction.

Proof of lemma 1:

Suppose not. We have some \epsilon > 0, \ f \neq \overline{0},

S = \{ n \ | \ \langle f_n, f \rangle > \epsilon \} has positive upper density.

Let \delta = \frac{\epsilon^2}{2||f||^2}, let Q = \{m \ | \ a_m < \delta/2 \} has density 1.

\forall q \in Q, let R_q = \{ n \ | \ \langle f_{n+q}, f_n \rangle < \delta \} has density 1.

Apply lemma 2 to Q, \{R_q \}, S, we get:

For all k \geq 1. There exists \{ n_1, n_2, \cdots, n_k \} \subseteq S, n_1 < n_2 \cdots < n_k such that whenever i<j , n_j - n_i \in Q and n_i \in R_{n_j-n_i}.

i) n_i \in S \ \Leftrightarrow \ \langle f_{n_i}, f \rangle > \epsilon

ii) n_i \in R_{n_j-n_i} \ \Leftrightarrow \ \langle f_{n_i}, f_{n_j} \rangle < \delta

Set g_i = f_{n_i} - \epsilon \cdot \frac{f}{||f||}. Hence

\forall \ 1 \leq i < j \leq k \langle g_i, g_j \rangle = \langle f_{n_i} - \epsilon \frac{f}{||f||}, f_{n_j} - \epsilon \frac{f}{||f||}\rangle < \delta - 2\cdot \frac{\epsilon^2}{||f||^2} + \frac{\epsilon^2}{||f||^2} =  \delta - \frac{\epsilon^2}{||f||^2}= -\delta.

On the other hand, since (f_n) is bounded in \mathcal{H}, \ (g_n) is also bounded (independent of k). Suppose ||g_n||< M for all k,
then we have
\displaystyle 0 \leq || \sum_{i=1}^k g_i ||^2 = \sum_{i=1}^k ||g_i ||^2 + 2 \cdot \sum_{i < j} \langle g_i, g_j \rangle \leq kM - k(k-1) \delta

For large k, \ kM - k^2 \delta<0, contradiction.
Hence S must have density 0.

Proof of the theorem:
By corollary 2 of the theorem in part 1, since T is weak mixing, T^m is weak mixing for all m \neq 0.
We proceed by induction on l. For l=1, the statement is implied by our lemma 2 in part 1.

Suppose the theorem holds for l \in \mathbb{N}, let f_0, f_1, \cdots, f_{l+1} \in L^\infty(X),

Let C = \int f_{l+1} \ d \mu, \ f'_{l+1}(x) = f_{l+1}(x) - C.

By induction hypothesis, \int f_0(x) f_1(T^n(x)) f_2(T^{2n}(x)) \cdots f_l(T^{ln}(x)) \cdot C \ d \mu

\rightarrow_D \int f_0 \ d \mu \int f_1 \ d \mu \cdots \int f_l \ d \mu \cdot C as n \rightarrow \infty

Hence it suffice to show \int f_0(x) f_1(T^n(x)) f_2(T^{2n}(x)) \cdots f_l(T^{ln}(x)) \cdot
f'_{l+1}(T^{(l+1)n}(x)) \ d \mu \rightarrow_D 0

Let \int f_{l+1} \ d\mu = 0

For all n \in \mathbb{N}, set g_n (x)= f_1 \circ T^n(x) \cdot f_2 \circ T^{2n}(x) \cdots f_{l+1} \circ T^{(l+1)n}(x)

For each m \in \mathbb{N}, \ \forall \ 0 \leq i \leq l=1, let F^{(m)}_i (x)= f_i(x) \cdot f_i(T^{im}(x))

\langle g_{n+m}, g_n \rangle = \int (f_1(T^{n+m} (x) \cdots f_{l+1}(T^{(l+1)(n+m)} (x))) \cdot (f_1(T^n(x)) \cdots f_{l+1}(T^{(l+1)n} (x))) \ d\mu
= \int F^{(m)}_1(T^n(x)) \cdots F^{(m)}_{l+1}(T^{(l+1)n}(x)) \ d \mu

Since T^{l+1} is measure preserving, we substitute y = T^{(l+1)n}(x),

= \int F^{(m)}_{l+1}(y) \cdot F^{(m)}_1(T^{-ln}(y)) \cdots F^{(m)}_l(T^{-n}(y)) \ d \mu

Apply induction hypothesis, to the weak mixing transformation T^{-n} and re-enumerate F^{(m)}_i

\langle g_n, g_{n+m} \rangle \rightarrow_D ( \int F^{(m)}_1 \ d\mu) \cdots (\int F^{(m)}_{l+1} \ d\mu) as n \rightarrow \infty.

\int F^{(m)}_{l+1} \ d\mu = \int f_{l+1} \cdot f_{l+1} \circ T^{(l+1)m} \ d\mu

By lemma 2 in part 1, we have \int F^{(m)}_{l+1} \ d\mu \rightarrow_D 0 as m \rightarrow \infty.

We are now able to apply lemma 2 to g_n, which gives (g_n) \rightarrow_D \overline{0} under the weak topology.

i.e. for any f_0, we have \int f_0(x) g_n(x) \ d \mu \rightarrow_D 0.

Establishes the induction step.

Remark: This works for any group of commutative weakly mixing transformations. i.e. if G is a group of measure preserving transformations, all non-identity elements of G are weakly mixing. T_1, \cdots, T_k are distinct elements in G, then \int f_0(x) f_1(T_1^n(x)) f_2(T_2^n(x)) \cdots f_k(T_k^n(x)) \ d \mu \rightarrow_D \int f_0 \ d \mu \int f_1 \ d \mu \cdots \int f_k \ d \mu as n \rightarrow \infty.

Notes for my lecture on multiple recurrence theorem for weakly mixing systems – Part 1

So…It’s finally my term to lecture on the ergodic theory seminar! (background, our goal is to go through the ergodic proof of Szemerédi’s theorem as in Furstenberg’s book). My part is the beginning of the discussion on weak mixing and prove the multiple recurrence theorem in the weak mixing case, the weak mixing assumption shall later be removed (hence the theorem is in fact true for any ergodic system) and hence prove Szemerédi’s theorem via the correspondence principal discussed in the last lecture.

Given two measure preserving systems (X_1, \mathcal{B}_1, \mu_1, T_1) and (X_2, \mathcal{B}_2, \mu_2, T_2), we denote the product system by (X_1 \times X_2, \mathcal{B}_1 \times \mathcal{B}_2, \mu_1 \times \mu_2, T_1 \times T_2) where \mathcal{B}_1 \times \mathcal{B}_2 is the smallest \sigma-algebra on X_1 \times X_2 including all products of measurable sets.

Definition: A m.p.s. (X, \mathcal{B}, \mu, T) is weakly mixing if (X \times X, \mathcal{B} \times \mathcal{B}, \mu \times \mu, T \times T) is ergodic.

Note that weak mixing \Rightarrow ergodic

as for non-ergodic systems we may take any intermediate measured invariant set \times the whole space to produce an intermediate measured invariant set of the product system.

For any A, B \in \mathcal{B}, let N(A, B) = \{ n \in \mathbb{N} \ | \ \mu(A \cap T^{-n}(B)) > 0 \}.

Ergodic \Leftrightarrow for all A,B with positive measure, N(A,B) \neq \phi

Weakly mixing \Rightarrow for all A,B,C,D with positive measure, N(A \times C, B \times D) \neq \phi.

Since n \in N(A \times C, B \times D)

\Leftrightarrow \ \mu \times \mu(A \times C \cap T^{-n}(B \times D)) > 0

\Leftrightarrow \ \mu \times \mu(A \cap T^{-n}(B) \times C \cap T^{-n}(D)) > 0

\Leftrightarrow \ \mu(A \cap T^{-n}(B)) > 0 and \mu(C \cap T^{-n}(D)) > 0

\Leftrightarrow \ n \in N(A,B) and n \in N(C,D)

Hence T is weakly mixing \Rightarrow for all A,B,C,D with positive measure, N(A,B) \cap N(C,D) \neq \phi. We’ll see later that this is in fact \Leftrightarrow but let’s say \Rightarrow for now.

As a toy model for the later results, let’s look at the proof of following weak version of ergodic theorem:

Theorem 1: Let (X, \mathcal{B}, \mu, T) be ergodic m.p.s., f, g \in L^2(X) then

\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int f \cdot g \circ T^n \ d\mu = \int f \ d\mu \cdot \int g \ d\mu.

Proof: Let \mathcal{U}: f \mapsto f \circ T, \ \mathcal{U} is unitary on L^2(X).

Hence \{ \frac{1}{N+1} \sum_{n=0}^N g \circ T^n \ | \ N \in \mathbb{N} \} \subseteq \overline{B( \overline{0}, ||g||)}

Any weak limit point of the above set is T-invariant, hence ergodicity implies they must all be constant functions.

Suppose \lim_{i \rightarrow \infty} \frac{1}{N_i+1} \sum_{n=0}^N g \circ T^n \equiv c

then we have c = \int c \ d\mu = \lim_{i \rightarrow \infty} \frac{1}{N_i+1} \sum_{n=0}^N \int g \circ T^n \ d\mu = \int g \ d \mu

Therefore the set has only one limit point under the weak topology.

Since the closed unit ball in Hilbert space is weakly compact, hence \frac{1}{N+1} \sum_{n=0}^N g \circ T^n converges weakly to the constant valued function \int g \ d \mu.

Therefore \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int f \cdot g \circ T^n \ d\mu

= \int f \cdot ( \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g \circ T^n) \ d\mu

= \int f \cdot (\int g \ d\mu) d\mu = \int f \ d\mu \cdot \int g \ d\mu.

Next, we apply the above theorem on the product system and prove the following:

Theorem 2: For (X, \mathcal{B}, \mu, T) weakly mixing,

\lim_{N \rightarrow \infty} \frac{1}{1+N} \sum_{n=0}^N (\int f \cdot (g \circ T^n) \ d \mu - \int f \ d \mu \int g \ d \mu)^2

= 0

Proof: For f_1, f_2: X \rightarrow \mathbb{R}, let f_1 \otimes f_2: X \times X \rightarrow \mathbb{R} where f_1 \otimes f_2 (x_1, x_2) = f_1(x_1) f_2(x_2)

By theorem 1, we have

\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N(\int f \cdot g \circ T^n \ d \mu)^2

= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int (f \otimes f) \cdot ((g \otimes g) \circ (T \times T)^n) \ d \mu \times \mu

= (\int f \otimes f \ d\mu \times \mu) \cdot (\int g \otimes g \ d\mu \times \mu)

= (\int f \ d\mu)^2 (\int g \ d\mu)^2 \ \ \ \ ( \star )

Set a_n = \int f \cdot (g \circ T^n) \ d \mu, \ a = \int f \ d \mu \int g \ d \mu hence by theorem 1, we have

\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n = a;

By ( \star ), we have

\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n^2 = a^2

Hence \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N (a_n-a)^2

= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N (a_n^2 - 2a \cdot a_n + a^2)

= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n^2 - 2a \cdot  \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n + a^2

= a^2 - 2 a \cdot a + a^2 = 0

This establishes theorem 2.

We now prove that the following definition of weak mixing is equivalent to the original definition.

Theorem 3: (X, \mathcal{B}, \mu, T) weakly mixing iff for all (Y, \mathcal{D}, \nu, S) ergodic, (X \times Y, \mathcal{B} \times \mathcal{D}, \mu \times \nu, T \times S) is ergodic.

proof:\Leftarrow” is obvious as if (X, \mathcal{B}, \mu, T) has the property that its product with any ergodic system is ergodic, then (X, \mathcal{B}, \mu, T) is ergodic since we can take its product with the one point system.
This implies that the product of the system with itself (X \times X, \mathcal{B} \times \mathcal{B}, \mu \times \mu, T \times T) is ergodic, which is the definition of being weakly mixing.

\Rightarrow” Suppose (X, \mathcal{B}, \mu, T) weakly mixing.

T \times S is ergodic iff all invariant functions are constant a.e.

For any g_1, g_2 \in L^2(X), \ h_1, h_2 \in L^2(Y), let C = \int g_1 \ d \mu, let g_1' = g_1-C; hence \int g_1' \ d \mu = 0.

\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g_1 \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu

= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int C \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu

+ \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g_1' \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu

Since \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int C \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu

= C \cdot \int g_2 \ d \mu \cdot \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int h_1 \cdot (h_2 \circ S^n) \ d \nu

By theorem 1, since S is ergodic on Y,

=\int g_1 \ d \mu \cdot \int g_2 \ d \mu \cdot \int h_1 \ d \nu \cdot \int h_2 \ d \nu

On the other hand, let a_n = \int g_1' \cdot (g_2 \circ T^n) \ d \mu, \ b_n = \int h_1 \cdot (h_2 \circ S^n) \ d \nu

By theorem 2, since T is weak mixing \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N (a_n - 0 \cdot \int g_2 \ d \mu)^2 = 0 hence \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n^2 = 0 \ \ \ (\ast)

(\sum_{n=0}^N a_n \cdot b_n)^2 \leq ( \sum_{n=0}^N a_n^2) \cdot ( \sum_{n=0}^N b_n^2) by direct computation i.e. subtract the left from the right and obtain a perfect square.

Therefore \lim_{N \rightarrow \infty} (\frac{1}{N+1} \sum_{n=0}^N a_n \cdot b_n)^2

\leq (\frac{1}{N+1} \sum_{n=0}^N a_n^2) \cdot (\frac{1}{N+1} \sum_{n=0}^N b_n^2)

Approaches to 0 as N \rightarrow \infty by (\ast).

Therefore, \lim \frac{1}{N+1} \sum_{n=0}^N \int g_1' \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu = 0

Combining the two parts we get

\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g_1 \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu

= \int g_1 \ d \mu \cdot \int g_2 \ d \mu \cdot \int h_1 \ d \nu \cdot \int h_2 \ d \nu.

Since the linear combination of functions of the form f(x, y) = g(x)h(y) is dense in L^2(X \times Y) (in particular the set includes all characteristic functions of product sets and hence all simple functions with basic sets being product sets)

We have shown that for a dense subset of f \in L^2(X \times Y) the sequence of functions \frac{1}{N+1}\sum_{n=0}^N f(T^n(x), S^n(y)) converge weakly to the constant function. (Since it suffice to check convergence a dense set of functional in the dual space)

Hence for any f \in L^2(X \times Y), the average weakly converges to the constant function \int f \ d \mu \times \nu.

For any T \times S-invariant function, the average is constant, hence this implies all invariant functions are constant a.e. Hence we obtain ergodicity of the product system.

Establishes the theorem.