## Convergence along ultrafilters – part II

This is the second of a series of two post whose aim is to prove the following recurrence theorem. Recall that a measure preserving system (shortened to m.p.s.) is a quadruple ${(X,{\cal B},\mu, T)}$, where ${(X,{\cal B},\mu)}$ is a probability space and ${T:X\rightarrow X}$ preserves the measure, i.e. for each ${A\in {\cal B}}$ we have ${\mu(T^{-1}A)=\mu(A)}$.

Theorem 1 Let ${(X,{\cal B},\mu, T)}$ be a m.p.s. and let ${A\in {\cal B}}$ have positive measure. Let ${q\in {\mathbb Z}[x]}$ be a polynomial such that ${q(0)=0}$. Then for any ${\lambda<1}$, the set

$\displaystyle \{n\in {\mathbb Z}:\mu(T^{-q(n)}A\cap A)>\lambda\mu(A)^2\}$

is syndetic, i.e. has bounded gaps.

In the first post I developed some theory of ultrafilters that will be needed for the proof.

As is the case in many single (or even double) recurrence theorems, it is possible to establish a general statement about Hilbert spaces which imply the desired recurrence. In this case, the Hilbert space result is the following:

Theorem 2 Let ${H}$ be a Hilbert space, let ${U}$ be a unitary operator on ${H}$ and let ${q}$ be a polynomial such that ${q(0)=0}$. Let ${p}$ be an idempotent ultrafilter. Then there exists a subspace ${S\subset H}$ such that for each ${f\in H}$

$\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{p(n)}f=P_Sf$

where ${P_S}$ denotes the orthogonal projection onto ${S}$.

To prove this we will need a version of the van der Corput trick for convergence along an idempotent ultrafilter:

Lemma 3 Let ${p}$ be an idempotent ultrafilter and let ${\{x_n\}_n}$ be a bounded sequence in ${H}$ such that for each ${d\in {\mathbb N}}$ we have ${p}$${\displaystyle\lim_{n\rightarrow\infty}\langle x_{n+d},x_n\rangle=0}$. Then also ${p}$${\displaystyle\lim_{n\rightarrow\infty}x_n=0}$ weakly.

Proof: I found this nice proof on this paper of Schnell (it’s the Lemma 4 there).

For each ${N\in{\mathbb N}}$ we have that

$\displaystyle p\text{-}\lim_{n\rightarrow\infty}x_n=p\text{-}\lim_{n_1\rightarrow\infty}p\text{-}\lim_{n_2\rightarrow\infty}...p\text{-}\lim_{n_N\rightarrow\infty}\frac1N \sum_{k=1}^Nx_{n_k+...+n_N}$

Taking norms and using the Cauchy-Schwartz inequality (which implies that the norm of a weak limit is at most the liminf of the norms of the sequence) we get:

$\displaystyle \begin{array}{rcl} \displaystyle\left\|p\text{-}\lim_{n\rightarrow\infty}x_n\right\|^2&\leq&\displaystyle p\text{-}\lim_{n_1\rightarrow\infty}p\text{-}\lim_{n_2\rightarrow\infty}\cdots p\text{-}\lim_{n_N\rightarrow\infty}\frac1{N^2} \left\|\sum_{k=1}^Nx_{n_k+\cdots +n_N}\right\|^2\\ &=&\displaystyle p\text{-}\lim_{n_1\rightarrow\infty}p\text{-}\lim_{n_2\rightarrow\infty}\cdots p\text{-}\lim_{n_N\rightarrow\infty}\frac1{N^2} \left\langle\sum_{k=1}^Nx_{n_k+\cdots +n_N},\sum_{l=1}^Nx_{n_l+\cdots +n_N}\right\rangle\\ &=&\displaystyle\frac1{N^2}\sum_{k,l=1}^N p\text{-}\lim_{n_1\rightarrow\infty}p\text{-}\lim_{n_2\rightarrow\infty}\cdots p\text{-}\lim_{n_N\rightarrow\infty}\left \langle x_{n_k+\cdots +n_N},x_{n_l+\cdots +n_N}\right\rangle \end{array}$

Now we use the Proposition 7 from my last post, together with the fact that ${p}$ is an idempotent ultrafilter to get:

$\displaystyle \begin{array}{rcl} &=&\displaystyle\frac1{N^2}\sum_{k=1}^Np\text{-}\lim_{n_k\rightarrow\infty}\|x_{n_k}\|^2+ \frac2{N^2}\sum_{k

Since ${N}$ was chosen arbitrarily we conclude that ${\displaystyle\left\|p\text{-}\lim_{n\rightarrow\infty}x_n\right\|^2=0}$.

$\Box$

We can now proceed to prove the Hilbert space version

Proof of Theorem 2:

To simplify the technical details, I will restrict the proof to the case when ${q(n)=n^2}$, for the general case, a similar but more sophisticated choice of ${W}$ is required. Let

$\displaystyle W:=\overline{\bigcup_{a\in{\mathbb N}}\{f\in H:p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f\}}$

Observe that the expression ${p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f}$ can be interpreted both in the weak topology or the strong topology. Indeed, if it holds in the weak topology, then ${p\text{-}\lim_{n\rightarrow\infty}\langle U^{an}f,f\rangle=\|f\|^2}$ and hence, using the fact that the norm ${\|U^{an}f\|}$ does not depend on ${n}$,

$\displaystyle p\text{-}\lim_{n\rightarrow\infty}\big\|U^{an}f-f\big\|^2=p\text{-}\lim_{n\rightarrow\infty}\|U^{an}f\|^2+\|f\|^2-2\text{Re}\langle U^{an}f,f\rangle=0,$

showing that ${p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f}$ in norm. To see why ${W}$ is a subspace, observe that if ${p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f}$, then for any ${d\in{\mathbb N}}$,

$\displaystyle p\text{-}\lim_{n\rightarrow\infty}\big\|U^{adn}f-f\big\|=0, \ \ \ \ \ (1)$

because ${p}$, being an idempotent ultrafilter, contains the set ${d{\mathbb N}}$. Therefore if ${p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f}$ and ${p\text{-}\lim_{n\rightarrow\infty}U^{bn}g=g}$, it follows that ${p\text{-}\lim_{n\rightarrow\infty}U^{abn}(f+g)=f+g}$, and hence ${W}$ is closed under sums and thus it is a subspace.

Next we need to describe the orthogonal space ${W^\perp}$ of ${W}$. Since ${p}$ is idempotent, observe that for each ${a\in{\mathbb N}}$, the operator ${U_a:f\mapsto p\text{-}\lim_{n\rightarrow\infty}U^{an}f}$ is an orthogonal projection. Indeed

$\displaystyle U_a^2f=p\text{-}\lim_{n\rightarrow\infty}U^{an}p\text{-}\lim_{m\rightarrow\infty}U^{am}f= p\text{-}\lim_{n\rightarrow\infty}p\text{-}\lim_{m\rightarrow\infty}U^{a(n+m)}f=(p+p)\text{-}\lim_{n\rightarrow\infty}U^{an}f=U_af.$

Therefore for each ${a\in{\mathbb N}}$, the space ${H}$ can be decomposed into the orthogonal components

$\displaystyle H=\Big\{f:p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f\Big\}\oplus\Big\{g:p\text{-}\lim_{n\rightarrow\infty}U^{an}g=0\Big\}.$

This implies that the space ${\Big\{g:p\text{-}\lim_{n\rightarrow\infty}U^{an}g=0\Big\}}$ contains ${W^\perp}$ for every ${a\in{\mathbb N}}$, and therefore that

$\displaystyle W^\perp\subset\bigcap_{a\in{\mathbb N}}\Big\{g\in H:p\text{-}\lim_{n\rightarrow\infty}U^{an}g=0\Big\}.$

We claim that this inclusion is actually an equality. To prove this claim, let ${f\in W}$ and ${g\in\bigcap_{a\in{\mathbb N}}\Big\{g\in H:p\text{-}\lim_{n\rightarrow\infty}U^{an}g=0\Big\}}$. For every ${\epsilon>0}$ there exists ${a\in{\mathbb N}}$ and ${f_1\in H}$ such that ${\|f-f_1\|<\epsilon}$ and ${p\text{-}\lim_{n\rightarrow\infty}U^{an}f_1=f_1}$. Therefore

$\displaystyle \langle f_1,g\rangle = p\text{-}\lim_{n\rightarrow\infty}\langle U^{an}f_1,U^{an}g\rangle = p\text{-}\lim_{n\rightarrow\infty}\langle f_1,U^{an}g\rangle = 0$

Since ${f_1}$ was arbitrarily close to ${f}$ we conclude that also ${\langle f,g\rangle=0}$. We remark that we used the fact that ${p\text{-}\lim_{n\rightarrow\infty}U^{an}f_1=f_1}$ in the strong topology, but only that ${p\text{-}\lim_{n\rightarrow\infty}U^{an}g=0}$ in the weak topology.

Now each arbitrary vector ${v\in H}$ decomposes uniquely as the sum of its projection ${f_v\in W}$ onto ${W}$ plus the orthogonal ${g_v\in W^\perp}$. Thus it suffices to prove that ${\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{n^2}(p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f)=p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f}$ for each ${f\in W}$ and that ${\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{n^2}g=0}$ for each ${g\in W^\perp}$. We start by the second claim, and we will use Lemma 3 with ${x_n=U^{n^2}g}$:

$\displaystyle \langle x_{n+d},x_n\rangle=\langle U^{n^2+2nd+d^2}g,U^{n^2}g\rangle=\langle U^{2dn}g,U^{-d^2}g\rangle$

so taking ${p}$${\lim}$ and using the hypothesis on ${g}$ we conclude that

$\displaystyle p\text{-}\lim_{n\rightarrow\infty}\langle x_{n+d},x_n\rangle=\left\langle p\text{-}\lim_{n\rightarrow\infty}U^{2dn}g,U^{-d^2}g\right\rangle=0$

so by Lemma 3 we conclude that indeed ${p\text{-}\lim_{n\rightarrow\infty}U^{n^2}g=0}$ when ${g\in W^\perp}$.

To prove the other claim, let ${f\in W}$. By an approximation argument we can assume that there exists ${a\in{\mathbb N}}$ such that ${\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f}$. Using again the fact that ${p}$ is idempotent we have

$\displaystyle p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f = p\text{-}\lim_{n\rightarrow\infty}p\text{-}\lim_{m\rightarrow\infty}U^{(n+m)^2}f = p\text{-}\lim_{n\rightarrow\infty}U^{n^2}p\text{-}\lim_{m\rightarrow\infty}U^{m^2+2nm}f$

Since ${a{\mathbb N}\in p}$, it suffices now to show that when ${n}$ is a multiple of ${a}$ (say ${n=ad}$) ${p\text{-}\lim_{m\rightarrow\infty}U^{m^2+2nm}f=p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f}$. We do this using (1). Indeed,

$\displaystyle p\text{-}\lim_{m\rightarrow\infty}U^{m^2+2nm}f-p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f = p\text{-}\lim_{m\rightarrow\infty}U^{m^2}\big(U^{2nm}f-f\big) = 0$

$\Box$

Now we can finally deduce theorem 1 from the Hilbert space version we just proved:

Proof of Theorem 1: Let ${H=L^2(X,\mu)}$ and let ${U:H\rightarrow H}$ be the Koopman operator defined by ${(Uf)(x)=f(Tx)}$. We apply theorem 2 to this setting (with an arbitrary idempotent ultrafilter ${p}$). Note that the constant function ${1}$ is invariant under ${U}$, so ${1\in S}$, and thus ${P_S1=1}$. Thus for any ${f\in H}$ we have ${\langle f,1\rangle=\langle f, P_S1\rangle=\langle P_Sf,1\rangle}$. Now make ${f=1_A}$ be the characteristic function of ${A}$, we have

$\displaystyle 0<\mu(A)=\int_Xfd\mu=\langle f,1\rangle=\langle P_Sf,1\rangle\leq \|P_Sf\|$

where in the last inequality we used Cauchy-Schwartz. On the other hand, since ${P_S}$ is a projection we have

$\displaystyle \|P_Sf\|^2=\langle P_Sf,f\rangle=p\text{-}\lim_{n\rightarrow\infty}\langle U^{p(n)}f,f\rangle$

Thus for each ${\lambda>0}$, the set

$\displaystyle \{n:|\langle U^{p(n)}f,f\rangle|>\lambda\mu(A)^2\}\in p$

Note also that this happens for every idempotent ultrafilter ${p}$. Finally we use the fact that a set belonging to every idempotent ultrafilter is an IP${^*}$ set, and hence a syndetic set. $\Box$

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