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 , where is a probability space and preserves the measure, i.e. for each we have .

Theorem 1Let be a m.p.s. and let have positive measure. Let be a polynomial such that . Then for any , the setis 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 2Let be a Hilbert space, let be a unitary operator on and let be a polynomial such that . Let be an idempotent ultrafilter. Then there exists a subspace such that for eachwhere denotes the orthogonal projection onto .

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

Lemma 3Let be an idempotent ultrafilter and let be a bounded sequence in such that for each we have –. Then also – weakly.

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

For each we have that

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:

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

Since was chosen arbitrarily we conclude that .

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 , for the general case, a more sophisticated choice of is required. Let

The -lim in the definition of is in the weak topology, but because the norm does not depend on it is actually a -lim in the strong topology as well. To describe the orthogonal space of , we note that for there is some such that:

It turns out that actually:

Now each arbitrary vector decomposes uniquely as the sum of its projection onto plus the orthogonal . Thus it suffices to prove that for each and that for each . We start by the second claim, and we will use the lemma 3 with :

so taking – and using the hypothesis on we conclude that

so by the lemma 3 we conclude that indeed when .

To prove the other claim, let and be such that . First I show that for any we have . We have by the triangular inequality that

so for each we have

The intersection is in because both sets are (the fact that the first is in uses the first part of the Corollary 9 from the first post.) Now this implies by the definition that .

Recall from the previous post the notation – and the second part of the Corollary 9. We have

Now by the previous observation that is the same as

and because the operator is continuous we can pass it to the inside and get:

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

This concludes the proof. Note that the image space is not necessarily the space , we just proved that .

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

*Proof:* (of Theorem 1) Let and let be the Koopman operator defined by . We apply theorem 2 to this setting (with an arbitrary idempotent ultrafilter ). Note that the constant function is invariant under , so , and thus . Thus for any we have . Now make be the characteristic function of , we have

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

Thus for each , the set

Note also that this happens for every idempotent ultrafilter . Finally we use the fact that a set belonging to every idempotent ultrafilter is an IP set, and hence a syndetic set.

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