## Szemerédi’s Theorem Part IV – Weak mixing extensions

This is the fourth in a series of six posts I am writing about Szemerédi’s theorem. In the first three posts, besides setting up the notation and definitions necessary, I reduced Szemerédi’s theorem to three facts. Those three facts are proved in the last three posts in this series. In this post I will prove the fact that the Sz property lifts through weak mixing extensions.

To briefly recall our setup, ${(X,{\cal B},\mu,T)}$ is a measure preserving system, a ${\sigma}$-subalgebra ${{\cal D}\subset{\cal B}}$ is called a factor if for every ${D\in{\cal D}}$ also ${T^{-1}D:=\{x\in X:Tx\in D\}\in{\cal D}}$; and we say a factor ${{\cal D}}$ is Sz if for every nonnegative bounded function ${f\in L^\infty({\cal D})}$ measurable in ${{\cal D}}$ satisfying ${\int_Xfd\mu>0}$ and every ${k\in{\mathbb N}}$ we have

$\displaystyle \liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\int_Xf(x)f(T^nx)f(T^{2n}x)\cdots f(T^{kn}x)d\mu(x)>0$

If ${{\cal A}\subset{\cal D}}$ are factors of ${(X,{\cal B},\mu,T)}$ we call the pair ${({\cal D}\mid{\cal A})}$ an extension. This extension is called weak mixing if every ${f\in L^2({\cal D}\mid{\cal A})}$ such that ${\mathop{\mathbb E}[f\mid{\cal A}]=0}$ satisfies

$\displaystyle \forall g\in L^2({\cal D}\mid{\cal A})\qquad\lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\left|\langle T^nf,g\rangle_{L^2({\cal D}\mid{\cal A})}\right|=0\quad\text{ in }L^2({\cal A}) \ \ \ \ \ (1)$

All this notions are better explained in the previous posts on this series, in particular the definition of the Hilbert module ${L^2({\cal D}\mid{\cal A})}$ and the notation of uniform Ceàro limits.

Remark 1 If an extension is weak mixing for ${T}$, then it is also weak mixing for ${T^i}$ for each ${i\in{\mathbb N}}$. To see this note that

$\displaystyle \frac1{N-M}\sum_{n=N}^M|\langle T^{in}f,g\rangle_{(\mathcal D\mid\mathcal A)}|\leq i\frac1{Ni-Mi}\sum_{n=Mi}^{Ni}|\langle T^nf,g\rangle_{(\mathcal D\mid\mathcal A)}|$

The purpose of this post is to prove the following theorem:

Theorem 1 If ${({\cal D}\mid{\cal A})}$ is a weak mixing extension of a measure preserving system ${(X,{\cal B},\mu,T)}$ and ${{\cal A}}$ is Sz, then also ${{\cal D}}$ is Sz.

— 1. Sz lifts through weak mixing extensions —

In this section we assume that ${({\cal D}\mid{\cal A})}$ is a weak mixing extension of a measure preserving system ${(X,{\cal B},\mu,T)}$ and ${{\cal A}}$ is Sz. First we present a useful lemma that helps explain why we consider factors of ${\mathcal B}$ instead of arbitrary ${\sigma}$-subalgebra:

Lemma 2 Let ${f\in L^2(\mathcal B)}$ and let ${\mathcal A\subset\mathcal B}$ be a factor. Then ${T\mathop{\mathbb E}[f\mid\mathcal A]=\mathop{\mathbb E}[Tf\mid\mathcal A]}$.

Proof: Note that ${T}$ is a unitary operator on ${L^2(\mathcal B)}$ and the conditional expectation is the orthogonal projection onto ${L^2({\cal A})}$. Since ${{\cal A}}$ is a factor, it is easy to see that ${T\mathop{\mathbb E}[f\mid{\cal A}]}$ is also in ${L^2({\cal A})}$. We will use the Riesz representation theorem, thus it suffices to show that for every ${g\in L^2(\mathcal A)}$ we have ${\langle T\mathop{\mathbb E}[f\mid\mathcal A],g\rangle=\langle\mathop{\mathbb E}[Tf\mid\mathcal A],g\rangle}$. Indeed

$\displaystyle \left\langle T\mathop{\mathbb E}[f\mid\mathcal A],g\right\rangle=\left\langle \mathop{\mathbb E}[f\mid\mathcal A],T^{-1}g\right\rangle=\left\langle f,T^{-1}g\right\rangle=\langle Tf,g\rangle=\langle\mathop{\mathbb E}[Tf\mid\mathcal A],g\rangle$

The fact that ${\mathcal A}$ is a factor, which implies that also ${T^{-1}g\in L^2(\mathcal A)}$, was used in the second equality. $\Box$

The next lemma asserts that weak mixing extensions are also ergodic extensions:

Lemma 3 Let ${f\in L^2(\mathcal D)}$ be such that ${\mathop{\mathbb E}[f\mid\mathcal A]=0}$. Then

$\displaystyle \lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^NT^nf=0\qquad\text{ in }L^2(\mathcal D)$

Proof: By the ergodic theorem we know that this limit is the projection of ${f}$ onto the space of invariant functions. Thus it suffices to show that any ${T}$-invariant function ${g\in L^2(\mathcal D)}$ is measurable with respect to ${\mathcal A}$.

Let ${g}$ be a ${T}$-invariant function and let ${g_0=g-\mathop{\mathbb E}[g\mid\mathcal A]}$. By Lemma 2, ${Tg_0=Tg-\mathop{\mathbb E}[Tg\mid\mathcal A]=g-\mathop{\mathbb E}[g\mid\mathcal A]=g_0}$, so ${g_0}$ is also ${T}$-invariant. Moreover ${g_0}$ is weak mixing and so

$\displaystyle 0=\lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N|\langle T^ng_0,g_0\rangle_{({\cal D}\mid{\cal A})}|=\|g_0\|^2_{({\cal D}\mid{\cal A})}$

Thus ${g_0=0}$ and hence ${g=\mathop{\mathbb E}[g\mid\mathcal A]}$. Therefore every ${T}$-invariant function is indeed measurable with respect to ${\mathcal A}$ and this proves the result. $\Box$

We need a technical lemma (which is, in some way or the other, behind all results in recurrence or convergence):

Lemma 4 (van der Corput trick) Let ${(u_n)}$ be a sequence taking values in a Hilbert space. If

$\displaystyle \lim_{D_1-D_2\rightarrow\infty}\frac1{D_1-D_2}\sum_{d=D_1}^{D_2}\limsup_{N-M\rightarrow\infty}\left|\frac1{N-M}\sum_{n=M}^N\langle u_{n+d},u_n\rangle\right|=0 \ \ \ \ \ (2)$

then also

$\displaystyle \lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=N}^Mu_n=0\qquad\text{ in the strong topology}$

I have used the van der Corput trick several times before in this blog, I never quite proved this version here, but the proof from this post can easily be adapted to deal with uniform limits.

The following lemma is the key to show that weak mixing extensions of Sz systems are Sz. The proof is essentially the same proof that weak mixing systems satisfy the multiple recurrence property, properly relativized.

Lemma 5 Assume that the extension is weak mixing. Let ${f_1,...,f_k\in L^\infty(\mathcal D)}$ and assume that for some ${i\in\{1,\dots,k\}}$ we have ${\mathop{\mathbb E}[f_i\mid{\cal A}]=0}$. Then

$\displaystyle \lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=1}^kT^{in}f_i=0\qquad\text{ in }L^2(\mathcal D)$

To get this lemma, we need to induct on some stronger hypothesis, and hence we will instead prove the following general case:

Lemma 6 Assume that the extension is weak mixing. Let ${f_1,...,f_k\in L^\infty(\mathcal D)}$ and let ${a_1,\cdots,a_k\in{\mathbb Z}}$ be distinct and non-zero. Assume that, for some ${i\in\{1,\dots,k\}}$, we have ${\mathop{\mathbb E}[f_i\mid{\cal A}]=0}$. Then

$\displaystyle \lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=1}^kT^{a_in}f_i=0\qquad\text{ in }L^2(\mathcal D)$

Proof: We can (and will) assume, without loss of generality, that ${\mathop{\mathbb E}[f_k\mid{\cal A}]=0}$, since otherwise we can just permutate the values of ${a_i}$. We proceed by induction on ${k}$. For ${k=1}$ we can use Remark 1 to deduce that the extension is weak mixing for ${T^{a_1}}$, and hence we can assume that ${a_1=1}$. But now, this case reduces to Lemma 3.

Now let ${k>1}$ and let ${u_n=\prod_{i=1}^kT^{a_in}f_i}$. We are going to apply the van der Corput trick, so it suffices to show that (2) is satisfied. We have

$\displaystyle \begin{array}{rcl} \displaystyle\langle u_{n+d},u_n\rangle&=& \displaystyle\int_X\prod_{i=1}^kT^{a_i(n+d)}f_i\prod_{i=1}^kT^{a_in}f_id\mu\\&=&\displaystyle \int_X\prod_{i=1}^kT^{a_in}\left(T^{a_id}f_i.f_i\right)d\mu\\&=& \displaystyle\int_XT^{a_1n}\left[T^{a_1d}f_1.f_1\prod_{i=2}^kT^{(a_i-a_1)n}\left(T^{a_id}f_i.f_i\right)\right]d\mu\\&=& \displaystyle\int_XT^{a_1d}f_1.f_1\prod_{i=2}^kT^{(a_i-a_1)n}\left(T^{a_id}f_i.f_i\right)d\mu\end{array}$

Let ${g_d=T^{a_kd}f_k.f_k}$ and ${g'_d=g_d-\mathop{\mathbb E}[g_d\mid\mathcal A]}$. By the induction hypothesis we have

$\displaystyle \lim_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=2}^{k-1} T^{(a_i-a_1)n}\left(T^{a_id}f_i.f_i\right)\cdot T^{(a_k-a_1)n}g'_d=0 \ \ \ \ \ (3)$

On the other hand we have the trivial bound

$\displaystyle \left\|T^{a_1d}f_1.f_1\prod_{i=2}^{k-1}T^{(a_i-a_1)n}\left(T^{a_id}f_i.f_i\right)\right\|_{L^\infty}\leq \prod_{i=1}^{k-1}\|f_i\|_{L^\infty}^2=:b$

Taking the uniform Cesàro limit of the inner product ${\langle u_{n+d},u_n\rangle}$ and using the previous estimate and (3) (and the triangular inequality) we have

$\displaystyle \begin{array}{rcl} &&\displaystyle\limsup_{N-M\rightarrow\infty}\left|\frac1{N-M}\sum_{n=M}^N\langle u_{n+d},u_n\rangle\right|\\&\leq& \displaystyle\limsup_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^Nb\int_X\left|T^{(a_k-a_1)n}\mathop{\mathbb E}[g_d\mid\mathcal A]\right|d\mu\\&=&\displaystyle b\limsup_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\int_X\left|\mathop{\mathbb E}[g_d\mid\mathcal A]\right|d\mu\\&=&\displaystyle b\int_X\left|\mathop{\mathbb E}[g_d\mid{\cal A}]\right|d\mu \end{array}$

Finally we have that

$\displaystyle \begin{array}{rcl} &&\displaystyle\limsup_{D\rightarrow\infty}\frac1D\sum_{d=1}^D\limsup_{N\rightarrow\infty}\left|\frac1N\sum_{n=1}^N\langle u_{n+d},u_n\rangle\right|\\&\leq&\displaystyle b\limsup_{D\rightarrow\infty}\frac1D\sum_{d=1}^D\int_X\left|\mathop{\mathbb E}[g_d\mid\mathcal A]\right|d\mu\\&=&\displaystyle b\limsup_{D\rightarrow\infty}\int_X\frac1D\sum_{d=1}^D\left|\langle T^{kd}f_k,f_k\rangle_{L^2(\mathcal D\mid\mathcal A)}\right|d\mu\\&=&0 \end{array}$

where in the last line we used the definition of weak mixing extension (1). $\Box$

We can now prove Theorem 1. Observe the following elementary identity:

$\begin{array}{rcl} \displaystyle\prod_{i=1}^k(a_i+b_i)&=&\displaystyle(a_1+b_1)\prod_{i=2}^k(a_i+b_i)\\ &=& \displaystyle a_1\cdot\prod_{i=2}^k(a_i+b_i)b_1\cdot\prod_{i=2}^k(a_i+b_i)\\&=& \displaystyle a_1\cdot a_2\prod_{i=3}^k(a_i+b_i)+a_1\cdot b_2\prod_{i=3}^k(a_i+b_i)b_1\cdot\prod_{i=2}^k(a_i+b_i)\\&=&\displaystyle...= \prod_{i=1}^ka_i+\sum_{i=0}^{k-1}\left[\prod_{j=1}^ia_j\cdot b_{i+1}\cdot\prod_{j=i+2}^k(a_j+b_j)\right] \end{array}$

The idea here is that the initial product was transformed into a sum of products, and the first product on this sum only has ${a_i}$‘s and each of the other products in this sum contain at one term ${b_i}$.

Let ${f\in L^\infty(\mathcal D)}$ be a non-negative function such that ${\int_Xfd\mu>0}$. Let ${g=\mathop{\mathbb E}[f\mid\mathcal A]}$ and note that ${\int_Xgd\mu=\int_Xfd\mu>0}$ and ${g\geq0}$ a.e. (to see this, consider the set ${\{x\in X:g(x)<0\}\in{\cal A}}$ and use the definition of conditional expectation). Also let ${h=f-g}$.

We will use the identity with ${a_i=T^{in}g}$ and ${b_i=T^{in}h}$. Then, with exception of the first product, which consists only of ${g}$‘s, each other product in the sum contains some ${h}$. Taking the uniform Cesàro limit and using Lemma 5 we get

$\displaystyle \liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=1}^kT^{in}f\geq \liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=1}^kT^{in}g$

Finally we can get multiple recurrence for ${f}$:

$\displaystyle \begin{array}{rcl} &&\displaystyle\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\int_XfT^nf...T^{kn}fd\mu\\&=&\displaystyle\left\langle f,\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=1}^kT^{in}f\right\rangle\\&\geq&\displaystyle\left\langle f,\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\prod_{i=1}^kT^{in}g\right\rangle\\ &=&\displaystyle\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\int_XgT^ng...T^{kn}gd\mu\\&>&0 \end{array}$

where the inequality on the last line follows from the hypothesis that ${{\cal A}}$ is Sz. Since ${f\in L^\infty}$ was arbitrary (within the conditions ${f\geq0}$ and ${\int_Xfd\mu>0}$) we showed that ${{\cal D}}$ is Sz and this proves Theorem 1.