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<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}.

\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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1 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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