## Szemerédi’s Theorem Part V – Compact extensions

This is the fifth in a series of six posts I am writing about Szemerédi’s theorem. In the previous post I proved that the Sz property lifts through weak mixing extension and in this post I will prove that the Sz property lifts through compact 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 that 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 compact if for every ${f\in L^2({\cal D}\mid{\cal A})}$ we can remove a set of arbitrarily small measure from its support in a way that for every ${\epsilon>0}$ there exists ${r\in{\mathbb N}}$ and functions ${f_1,\cdots,f_r\in L^2({\cal D}\mid{\cal A})}$ such that for every ${x\in X}$ and ${n\in {\mathbb N}}$ there exists ${i\in\{1,\cdots,r\}}$ such that

$\displaystyle \|T^nf-f_i\|_{L^2({\cal D}\mid{\cal A})}<\epsilon$

All this notions are better explained in the previous posts on this series.

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

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

I will present one and a half proofs of this theorem. The first (one half) uses the van der Waerden theorem and is somewhat easier. It has the advantage that it was extended to prove polynomial Szemerédi’s theorem as well as other generalizations. The second proof, obtained by Furstenberg, Katznelson and Ornstein, avoids the use of the coloristic result, while also giving uniform limits.

The first proof is just one half proof because it does not give uniform Cesàro limits, (although one can get uniform limits even when using van der Waerden’s theorem, that is exactly what Furstenberg achieved in his original proof). For this reason, the first proof is not necessary in the proof of Szemerédi’s theorem, or if one is only interested in Szemerédi’s theorem, but not in getting uniform limits, then the second proof is not necessary.

In a vague way, the first proof uses the fact that multiplying an arithmetic progression by a constant gives another arithmetic progression (i.e. arithmetic progressions are invariant under multiplication), and the second proof uses the fact that adding an arithmetic progression with a constant gives another arithmetic progression (i.e. arithmetic progressions are invariant under addition).

— 1. First proof, using van der Waerden’s theorem —

In this section we assume that we are under the assumptions of Theorem 1. To prove this theorem we will need the van der Waerden Theorem to which I presented a proof in a previous post. Recall that ${[K]=\{0,\dots,K-1\}}$.

Theorem 2 (van der Waerden) Let ${r,k\in{\mathbb N}}$. There exists some ${K(r,k)\in{\mathbb N}}$ such that for any coloring of ${[K(r,k)]}$ in ${r}$ colors there exists a monochromatic arithmetic progression of length ${k}$.

Let ${f\in L^\infty({\cal D})}$ such that ${f\geq0}$ and ${\int_Xfd\mu>0}$; and let ${k\in{\mathbb N}}$. We will show that

$\displaystyle \liminf_{N\rightarrow\infty}\frac1N\sum_{n=1}^N\int_XfT^nf...T^{kn}fd\mu>0\ \ \ \ \ (1)$

The first step is to perform some simplifications. Let ${\epsilon>0}$ be such that if ${A\in{\cal A}}$ has ${\mu(A)>1-\epsilon}$ then ${\int_Afd\mu>0}$ and let ${A\in{\cal A}}$ be such that ${\mu(A)>1-\epsilon}$ and ${f1_A}$ is conditionally compact. Note that if (1) holds for ${f1_A}$, since ${f\geq f1_A}$ then it also holds for ${f}$. Therefore we can (and will) assume, without loss of generality that ${f}$ is itself conditionally compact. We can also assume that ${0\leq f\leq1}$ after multiplying it, if necessary, by a constant.

Let ${g=\mathop{\mathbb E}[f\mid\mathcal A]}$ and denote by ${\delta:=\int_Xfd\mu}$. Note that also ${\int_Xgd\mu=\delta}$ and ${0\leq g\leq1}$ a.e.(to see why we have such bound consider the set ${\{x\in X:g\notin[0,1]\}}$ and use the definition of conditional expectation). Let ${\epsilon<\delta^{k+1}/(k+1)2^{k+3}}$.

Applying the compactness hypothesis we can find ${r\in{\mathbb N}}$ and ${f_1,...,f_r}$ be such that for each ${n\in{\mathbb N}}$ we have

$\displaystyle \left\|\min_{1\leq i\leq r}\|T^nf-f_i\|_{L^2(\mathcal D\mid\mathcal A)}\right\|_{L^\infty}<\epsilon$

Let ${A:=\{x:g(x)>\delta/2\}\in\mathcal A}$. Observe that ${\delta=\int_Agd\mu+\int_{X\setminus A}gd\mu\leq\mu(A)+\frac\delta2}$, hence ${\mu(A)>\delta/2}$. Now apply the van der Waerden Theorem to find ${K\in{\mathbb N}}$ so that each ${r}$-coloring of ${\{1,...,K\}}$ contains an arithmetic progression of length ${k+1}$. By assumption we get that ${1_A}$ satisfies (1) with ${K}$ instead of ${k}$. Let ${c>0}$ be the ${\liminf}$ in (1) with ${1_A}$ and ${K}$, let ${I_n=A\cap T^nA\cap...\cap T^{Kn}A}$ and let ${B:=\{n\in{\mathbb N}:\mu(I_n)>c/2\}}$. We have

$\displaystyle \begin{array}{rcl} c&=&\displaystyle\liminf_{N\rightarrow\infty}\frac1N\sum_{n=1}^N\mu(I_n)\\&=& \displaystyle\liminf_{N\rightarrow\infty}\frac1N\sum_{n=1}^N\left[1_B(n)\mu(I_n)+1_{B^c}(n)\mu(I_n)\right]\\&\leq& \displaystyle\liminf_{N\rightarrow\infty}\frac1N\big|B\cap\{1,\dots,N\}\big|+\frac c2 \end{array}$

and thus

$\displaystyle \liminf_{N\rightarrow\infty}\frac{\big|B\cap\{1,\dots,N\}\big|}N\geq\frac c2$

Interpreting what we have so far, for each ${n\in B}$ and ${x\in I_n}$, we have many (actually ${K}$) points in the orbit of ${x}$ that fall inside ${A}$, and hence ${\mathop{\mathbb E}[f\mid{\cal A}]}$ in that portion of the orbit is bounded away from ${0}$.

Now fix ${n\in B}$, let ${x\in I_n}$ and consider the ${r}$-coloring of ${[K]}$ induced by the orbit ${T^{an}f}$. More precisely the coloring is the map from ${a\in[K]}$ to some ${i\in[r]}$ that satisfies ${\|T^{an}f-f_i\|_{L^2(\mathcal D\mid\mathcal A)}(x)<\epsilon}$. By the choice of ${K}$ we can find some ${a=a(x,n),s=s(x,n)\in{\mathbb N}}$ and ${i=i(x,n)\in[r]}$ such that ${a+ks\leq K}$ and ${\|T^{(a+ts)n}f-f_i\|_{L^2(\mathcal D\mid\mathcal A)}(x)<\epsilon}$ for each ${t=0,...,k}$. Note that for each fixed ${a,s,i}$ (and ${n}$), the set of those ${x\in X}$ such that ${\|T^{(a+ts)n}f-f_i\|_{L^2(\mathcal D\mid\mathcal A)}(x)<\epsilon}$ for all ${t=0,...,k}$ is in ${\mathcal A}$, so we can choose the functions ${a,r,i}$ to be measurable on ${\mathcal A}$.

Since there are only at most ${K^2\cdot r}$ possible ways to choose ${a,s,i}$, we can find a subset ${E_n\subset I_n}$ in ${\mathcal A}$ such that ${\mu(E_n)>\mu(I_n)/(K^2\cdot r)}$ and some constants ${a=a(n),s=s(n)\in[K]}$ and ${i=i(n)\in[r]}$ such that

$\displaystyle (\forall t\in[k])(\forall x\in E_n) \qquad\|T^{(a+ts)n}f-f_i\|_{L^2(\mathcal D\mid\mathcal A)}(x)<\epsilon$

Using the conditional triangular inequality we also have that

$\displaystyle (\forall t\in[k])(\forall x\in E_n) \qquad\|T^{(a+ts)n}f-T^{an}f\|_{L^2(\mathcal D\mid\mathcal A)}(x)<2\epsilon$

Finally note that for each ${x\in E_n\subset I_n\subset T^{-an}A}$ we also have that ${\mathop{\mathbb E}[T^{an}f\mid\mathcal A](x)=T^{an}g(x)>\delta/2}$.

We recall this elementary identity, already used in the previous post:

$\displaystyle \begin{array}{lcr} \displaystyle\prod_{t=1}^{k+1}(a_t+b_t)&=&\displaystyle({a_1}+{b_1})\prod_{t=2}^{k+1}(a_t+b_t)= {a_1\cdot\prod_{t=2}^{k+1}(a_t+b_t)}+{b_1\cdot\prod_{t=2}^{k+1}(a_t+b_t)}\\&=& \displaystyle {a_1\cdot a_2\prod_{t=3}^{k+1}(a_t+b_t)+a_1\cdot b_2\prod_{t=3}^{k+1}(a_t+b_t)}+{b_1\cdot\prod_{t=2}^{k+1}(a_t+b_t)}\\&=&...= \displaystyle\prod_{t=1}^{k+1}a_t+\sum_{j=0}^k\left(\prod_{t=1}^ja_t\cdot b_{j+1}\cdot\prod_{t=j+2}^{k+1}(a_t+b_t)\right)\end{array} \ \ \ \ \ (2)$

This time we will take ${a_t=T^{an}f}$ (yes, this is independent of ${t}$) and ${b_{t+1}=T^{(a+ts)n}f-T^{an}f}$. Now using the Conditional Cauchy-Schwartz inequality we can bound (the conditional expectation of) each of the products inside the sum. Recall that ${\|f\|_{L^\infty}\leq1}$ and that the norm ${\|\cdot\|_{L^2({\cal D}\mid{\cal A})}}$ is bounded by the ${L^\infty}$ norm. Let ${j\in\{0,\dots,k\}}$ and let ${x\in E_n}$.

$\displaystyle \begin{array}{rcl} &&\displaystyle\left|\mathop{\mathbb E}\left[\left.\prod_{t=1}^ja_t\cdot b_{j+1}\cdot\prod_{t=j+2}^{k+1}(a_t+b_t)\right|{\cal A}\right](x)\right|\\ &=&\displaystyle\left|\mathop{\mathbb E}\left[\left.\left(T^{an}f\right)^j.\left(T^{(a+js)n}f-T^{an}f\right).\prod_{t=j+1}^kT^{(a+ts)n}f\right| {\cal A}\right](x)\right|\\&=&\displaystyle\left|\left\langle T^{(a+js)n}f-T^{an}f,\left(T^{an}f\right)^j.\prod_{t=j+1}^kT^{(a+ts)n}f\right\rangle_{L^2(\mathcal D\mid\mathcal A)}(x)\right|\\ &\leq&\displaystyle\left\|T^{(a+js)n}f-T^{an}f\right\|_{L^2(\mathcal D\mid\mathcal A)}(x)\left\|\left(T^{an}f\right)^j.\prod_{t=j+1}^kT^{(a+ts)n}f\right\|_{L^2(\mathcal D\mid\mathcal A)}(x)\\&\leq& 2\epsilon \end{array}$

Using Jensen inequality for conditional expectation we can get a lower bound on the first product in the identity (2):

$\displaystyle \mathop{\mathbb E}\left[\left.\left(T^{nr}f\right)^{k+1}\right|\mathcal A\right](x)=\mathop{\mathbb E}\left[\left.f^{k+1}\right|\mathcal A\right](T^{nr}x)\geq\left(\mathop{\mathbb E}\left[\left.f\right|\mathcal A\right](T^{nr}x)\right)^{k+1}\geq\left(\frac\delta2\right)^{k+1}$

We now put together the estimates into (2). Using the triangular inequality and recalling the condition on ${\epsilon}$ we get

$\displaystyle \mathop{\mathbb E}\left[\left.\prod_{t=0}^kT^{(a+ts)n}f\right|{\cal A}\right](x) \geq\left(\frac\delta2\right)^{k+1}-2(k+1)\epsilon >\delta^{k+1}2^{-k-2}$

Since ${E_n\in \mathcal A}$ and ${f\geq0}$ we deduce from this that

$\displaystyle \int_X\prod_{t=0}^kT^{(a+ts)n}fd\mu\geq\int_{E_n}\prod_{t=0}^kT^{(a+ts)n}fd\mu>\frac{\delta^{k+1}}{2^{k+2}}\mu(E_n)\geq \frac{\delta^{k+1}\mu(I_n)}{2^{k+2}K^2r}$

Recall that ${B\subset{\mathbb N}}$ was defined such that ${n\in B\Rightarrow \mu(I_n)>c/2}$. Thus for every ${n\in B}$ there exists some ${s\in[K]}$ such that

$\displaystyle \int_X\prod_{t=0}^kT^{tsn}fd\mu=\int_X\prod_{t=0}^kT^{(a+ts)n}fd\mu\geq\frac{\delta^{k+1}\mu(I_n)}{2^{k+2}K^2r}\geq \frac{\delta^{k+1}c}{2^{k+3}K^2r}=:d \ \ \ \ \ (3)$

Note that while ${s}$ depends on ${n}$, there are only finitely many choices for ${s}$.

Taking the Cesàro average we conclude that

$\displaystyle \begin{array}{rcl} \displaystyle\liminf_{N\rightarrow\infty}\frac1N\sum_{n=1}^N\int_X\prod_{t=0}^kT^{(a+ts)n}fd\mu&\geq& \displaystyle\liminf_{N\rightarrow\infty}\frac1N\sum_{n=1}^N\frac{\delta^{k+1}\mu(I_n)}{2^{k+2}K^2r}\\&\geq& \displaystyle\liminf_{N\rightarrow\infty}\frac1N\sum_{n=1}^N1_B(n)\frac{\delta^{k+1}\mu(I_n)}{2^{k+2}K^2r}\\&\geq& \displaystyle\frac{\delta^{k+1}c}{2^{k+3}K^2r}\liminf_{N\rightarrow\infty}\frac{|B\cap[1,N]|}N\\&>& \displaystyle\frac{\delta^{k+1}c^2}{2^{k+4}K^2r}>0 \end{array}$

— 2. Second proof, with uniform limits —

In this section I present another proof of Theorem 1. This proof was obtained by Furstenberg, Katznelson and Ornstein and avoids the use of van der Waerden’s theorem. While this doesn’t seem to be very important, since van der Waerden’s theorem is significantly easier to prove that Szemerédi’s theorem, this allows more flexibility when trying to mimic this proof to other situations where a coloring result may not be available. Besides, this second proof can be applied to uniform Cesàro limits.

Let ${f\in L^\infty({\cal D})}$ be nonnegative and such that ${\int_Xfd\mu>0}$. We can assume without loss of generality that ${f}$ is itself conditionally compact, as done in the first proof. Thus we know that the orbit ${T^nf}$ is totally bounded in the ${:L^2({\cal D}\mid{\cal A})}$ norm. In particular, for any fixed ${\epsilon>0}$, almost every ${x\in X}$ and ${i\in[k]}$, if ${F\subset{\mathbb Z}}$ is large enough subset, than for some distinct ${a,b\in F}$ we have ${\|T^{ia}f-T^{ib}f\|_{({\cal D}\mid{\cal A})}(x)<\epsilon}$. We will carefully choose some finite subset ${F\subset{\mathbb Z}}$ which does not satisfy this property and is maximal. More precisely, given ${F\subset{\mathbb Z}}$ (and keeping ${\epsilon>0}$ fixed), let

$\displaystyle \begin{array}{rcl} S(F)&=&\displaystyle\big\{x\in X:(\forall a\neq b\in F)(\exists i\in[k])\text{ s.t. }\|T^{ia}f-T^{ib}f\|_{({\cal D}\mid{\cal A})}(x)>\epsilon\big\}\\&=&\displaystyle\bigcap_{a\neq b\in F}\bigcup_{i\in[k]}\big\{x\in X:\|T^{ia}f-T^{ib}f\|_{({\cal D}\mid{\cal A})}(x)\in[\epsilon,\infty)\big\} \end{array}$

For instance, if ${F}$ is a singleton, then ${S(F)=X}$. Note that ${S(F)\in{\cal A}}$. Intuitively speaking, ${S(F)}$ is the set of points for which ${F}$ is not “good enough” to cover the orbit of ${f}$. We want a set of points for which ${F}$ is not good enough, but if we add any other integer to ${F}$, than the new set is “good”. Thus we define:

$\displaystyle Q(F)=\{x\in S(F):x\notin S(E)\text{ for all }E\subset{\mathbb Z}\text{ with }|E|>|F|\}$

Lemma 3 Almost every ${x\in X}$ is in some ${Q(F)}$

Proof: We now will make precise the claims above. Let ${r\in{\mathbb N}}$ and ${f_1,...,f_r}$ be such that for each ${n\in{\mathbb N}}$ and almost every ${x\in X}$ there is some ${t\in[r]}$ such that

$\displaystyle \|T^nf-f_t\|_{L^2(\mathcal D\mid\mathcal A)}<\frac\epsilon2 \ \ \ \ \ (4)$

Let ${x\in X}$ and ${F\subset{\mathbb Z}}$ be a finite subset. For each ${a\in F}$ and ${i\in[k]}$ there exists some ${t\in[r]}$ that satisfies (4) with ${n=ai}$ (at least for almost every ${x}$, we will ignore the set of measure ${0}$ where this may not occur). Thus we can associate each ${a\in F}$ with a sequence ${t_0,t_1,\dots t_k}$ in ${[r]}$. Since there are only ${r^{k+1}}$ possible such sequences, if ${|F|>r^{k+1}}$ there must be some pair ${a,b\in F}$ with the same sequence. Using the conditional triangular inequality we conclude that for any ${i\in[k]}$ we have ${\|T^{ia}f-T^{ib}f\|_{L^2(\mathcal D\mid\mathcal A)}(x)<\epsilon}$.

This means that, if ${|F|>r^{k+1}}$, the set ${S(F)}$ has measure ${0}$. Since ${S(F)=X}$ whenever ${F}$ is a singleton, we deduce that every ${x\in X}$ is in some ${Q(F)}$. $\Box$

Let ${g=\mathop{\mathbb E}[f\mid\mathcal A]}$ and denote by ${\delta:=\int_Xfd\mu}$. Also, let ${\epsilon<\delta^{k+1}/(k+1)2^{k+2}}$. Let ${A:=\{x:g(x)>\delta/2\}\in\mathcal A}$. Observe that ${\delta=\int_Agd\mu+\int_{X\setminus A}gd\mu\leq\mu(A)+\frac\delta2}$, hence ${\mu(A)>\delta/2}$. Since there are only countably many finite subsets of ${{\mathbb Z}}$, we can find a finite set ${F\subset{\mathbb Z}}$ such that ${\mu\big(Q(F)\cap A\big)>0}$. Since ${Q(F)\subset S(F)}$, for all ${x\in Q(F)}$ and each pair ${a\neq b\in F}$ there exists some ${i=i(x,a,b)\in[k]}$ such that ${\|T^{ia}f-T^{ib}f\|_{({\cal D}\mid{\cal A})}(x)>\epsilon}$. Thus we can define a function ${w:Q(F)\rightarrow[k]^{F^2}}$ such that ${w(x)\in[k]^{F^2}=\{q:F^2\rightarrow[k]\}}$ is such that ${w(x)(a,b)=i(y,a,b)}$. The point here is that ${w}$ has a finite image, and hence there exists a set o positive measure ${E\subset Q(F)\cap A}$ and a map ${q:F^2\rightarrow[k]}$ such that for all ${x\in E}$ we have ${i(x,a,b)=q(a,b)}$ and hence is independent of ${x}$.

Lemma 4 Let ${n\in{\mathbb Z}}$ and ${x\in E\cap T^{-n}E\cap T^{-2n}E\cap\cdots\cap T^{-kn}E}$ (in particular assume this intersection is non-empty). Then there exists ${a=a(n,x)\in F}$ s.t. ${\forall i\in[k]}$ we have ${\|T^{i(n+a)}f-f\|_{({\cal D}\mid{\cal A})}(x)<\epsilon}$.

Proof: Since for each ${t\in[k]}$ we have ${T^{tn}x\in E}$, we have that for any pair ${a\neq b\in F}$ there exists ${i=q(a,b)}$ independent of ${t}$ such that ${\|T^{ia}f-T^{ib}f\|_{({\cal D}\mid{\cal A})}(T^{tn}x)>\epsilon}$ and hence ${\|T^{tn+ia}f-T^{tn+ib}f\|_{({\cal D}\mid{\cal A})}(x)>\epsilon}$. In particular we can take ${t=i}$ and we have

$\displaystyle \|T^{i(n+a)}f-T^{i(b+n)}f\|_{({\cal D}\mid{\cal A})}(x)>\epsilon \ \ \ \ \ (5)$

On the other hand, ${E\subset Q(F)}$, so ${x\notin S(G)}$ for any ${G\subset{\mathbb Z}}$ with ${|G|>F}$. We will make ${G=(F+n)\cup\{0\}}$. If ${0\in F+n}$, then ${-n\in F}$ and hence, taking ${a=-n}$, the conclusion of the lemma is trivially true. Otherwise ${|G|>|F|}$ and hence there exists some pair ${a,b\in G}$ such that for all ${i\in[k]}$ we have

$\displaystyle \|T^{ia}f-T^{ib}f\|_{({\cal D}\mid{\cal A})}(x)<\epsilon$

If neither ${a}$ nor ${b}$ equal ${0}$, then ${a=a'+n}$ and ${b=b'+n}$ with ${a',b'\in F}$. But this would contradict (5). Thus either ${a=0}$ or ${b=0}$ and we are done.

$\Box$

Now we can finish the proof. By assumption we get that ${1_E}$ satisfies (1). Let ${c>0}$ be the ${\liminf}$ in (1) with ${1_E}$, let ${I_n=E\cap T^{-n}E\cap...\cap T^{-kn}E}$ and let ${B:=\{n\in{\mathbb N}:\mu(I_n)>c/2\}}$. We have

$\displaystyle \begin{array}{rcl} c&=&\displaystyle\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\mu(I_n)\\&=& \displaystyle\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{n=M}^N\left[1_B(n)\mu(I_n)+1_{B^c}(n)\mu(I_n)\right]\\&\leq& \displaystyle\liminf_{N-M\rightarrow\infty}\frac1{N-M}\big|B\cap\{M,\dots,N\}\big|+\frac c2 \end{array}$

and thus

$\displaystyle \liminf_{N-M\rightarrow\infty}\frac{\big|B\cap\{M,\dots,N\}\big|}{N-M}\geq\frac c2$

For each ${n\in B}$ and ${x\in I_n}$ we can apply Lemma 4 and find ${a(n,x)\in F}$. We now use identity 2 with ${a_t=f}$ and ${b_t=T^{t(n+a)}f}$, and arguing as above we have

$\displaystyle \mathop{\mathbb E}\left[\left.\prod_{i=0}^kT^{i(n+a)}f\right| {\cal A}\right](x)\geq\left(\frac\delta2\right)^{k+1}-(k+1)\epsilon>\frac{\delta^{k+1}}{2^{k+2}}$

While ${a}$ depends on ${x}$, it can only take finitely many possible values. Thus this dependency is lost if we simply add over all ${a\in F}$. Since ${I_n\in{\cal A}}$ and ${f\geq0}$ we can then integrate and deduce that

$\displaystyle \int_X\sum_{a\in F}\prod_{i=0}^kT^{i(n+a)}fd\mu\geq\int_{I_n}\sum_{a\in F}\prod_{i=0}^kT^{i(n+a)}fd\mu> \frac{\delta^{k+1}}{2^{k+2}}\mu(I_n)=:c_2>0$

Thus we showed that for each ${n\in B}$ there is some ${a\in F}$ such that ${n+a}$ “gives” multiple recurrence for ${f}$. Let ${a_1=\min F}$ and ${a_2=\max F}$. Let ${M be such that ${N-M>a_2-a_1}$. Then we have

$\displaystyle \begin{array}{rcl} \displaystyle\sum_{m=M}^N\int_X\prod_{i=0}^kT^{im}fd\mu&=&\displaystyle\frac1{|F|}\sum_{a\in F}\sum_{n=M-a}^{N-a}\int_X\prod_{i=0}^kT^{i(n+a)}fd\mu\\&\geq&\displaystyle\frac1{|F|}\sum_{a\in F}\sum_{n=M-a_1}^{N-a_2}\int_X\prod_{i=0}^kT^{i(n+a)}fd\mu\\&=&\displaystyle\frac1{|F|}\sum_{m=M-a_1}^{N-a_2}\int_X\sum_{a\in F}\prod_{i=0}^kT^{i(n+a)}fd\mu\\&\geq&\displaystyle\frac{c_2}{|F|}|B\cap[M-a_1,N-a_2]| \end{array}$

Finally, taking the uniform Cesàro limit we conclude:

$\displaystyle \begin{array}{rcl} \displaystyle\liminf_{N-M\rightarrow\infty}\frac1{N-M}\sum_{m=M}^N\int_X\prod_{i=0}^kT^{im}fd\mu&\geq& \displaystyle\liminf_{N-M\rightarrow\infty}\frac{|B\cap[M-a_1,N-a_2]|}{N-M}\frac{c_2}{|F|}\\&=&\displaystyle c\frac{c_2}{|F|}>0 \end{array}$

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