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<l}p\text{-}\lim_{n_k\rightarrow\infty}p\text{-}\lim_{n_l\rightarrow\infty}\langle x_{n_k+n_l},x_{n_l}\rangle\\&=& \displaystyle\frac1Np\text{-}\lim_{n\rightarrow\infty}\|x_n\|^2 \end{array}

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


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


About Joel Moreira

PhD Student at OSU in Mathematics. I'm portuguese.
This entry was posted in Combinatorics, Ergodic Theory, Ramsey Theory and tagged , , , , . Bookmark the permalink.

One Response to Convergence along ultrafilters – part II

  1. Pingback: Applications of the coloring trick | I Can't Believe It's Not Random!

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