## 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:

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

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 the 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 more sophisticated choice of ${W}$ is required. Let

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

The ${p}$-lim in the definition of ${W}$ is in the weak topology, but because the norm ${\|U^{an}f\|}$ does not depend on ${n}$ it is actually a ${p}$-lim in the strong topology as well. To describe the orthogonal space ${W^\perp}$ of ${W}$, we note that for ${f\in W}$ there is some ${a\in{\mathbb N}}$ such that:

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

It turns out that actually:

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

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 the 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 the 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 ${f\in W}$ and ${a\in{\mathbb N}}$ be such that ${\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{an}f=f}$. First I show that for any ${d\in{\mathbb N}}$ we have ${\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{n^2+adn}f=p\text{-}\lim_{n\rightarrow\infty}U^{n^2}f=:f_0}$. We have by the triangular inequality that

$\displaystyle \|U^{n^2+adn}f-f_0\|\leq\|U^{n^2+adn}f-U^{n^2}f\|+\|U^{n^2}f-f_0\|= \|U^{adn}f-f\|+\|U^{n^2}f-f_0\|$

so for each ${\epsilon>0}$ we have

$\displaystyle \{n:\|U^{n^2+adn}f-f_0\|<\epsilon\}\supset \left\{n:\|U^{adn}f-f\|<\frac\epsilon2\right\}\cap \left\{n:\|U^{n^2}f-f_0\|<\frac\epsilon2\right\}\in p$

The intersection is in ${p}$ because both sets are (the fact that the first is in ${p}$ uses the first part of the Corollary 9 from the first post.) Now this implies by the definition that ${\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{n^2+adn}f=p\text{-}\lim_{n\rightarrow\infty}U^{n^2}f}$.

Recall from the previous post the notation ${p}$${\displaystyle\lim_{n\rightarrow\infty;n\in A}}$ and the second part of the Corollary 9. We have

$\displaystyle p\text{-}\lim_{n\rightarrow\infty}U^{n^2}\left(p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f\right)= p\text{-}\lim_{n\rightarrow\infty; n\in a{\mathbb N}}U^{n^2}\left(p\text{-}\lim_{m\rightarrow\infty}U^{m^2}f\right)$

Now by the previous observation that is the same as

$\displaystyle p\text{-}\lim_{n\rightarrow\infty; n\in a{\mathbb N}}U^{n^2}\left(p\text{-}\lim_{m\rightarrow\infty}U^{m^2+2nm}f\right)$

and because the operator ${U}$ is continuous we can pass it to the inside and get:

$\displaystyle p\text{-}\lim_{n\rightarrow\infty; n\in a{\mathbb N}}\left(p\text{-}\lim_{m\rightarrow\infty}U^{n^2}U^{m^2+2nm}f\right)= p\text{-}\lim_{n\rightarrow\infty}\left(p\text{-}\lim_{m\rightarrow\infty}U^{(n+m)^2}f\right)$

Using the Proposition 7 from the first post we conclude that:

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

This concludes the proof. Note that the image space ${S}$ is not necessarily the space ${W}$, we just proved that ${S\subset W}$. $\Box$

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. $\Box$