# 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, $n_j - n_i \in Q$ 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 $n \rightarrow \infty$.