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 Theorem 2:*

To simplify the technical details, I will restrict the proof to the case when , for the general case, a similar but more sophisticated choice of is required. Let

Observe that the expression can be interpreted both in the weak topology or the strong topology. Indeed, if it holds in the weak topology, then and hence, using the fact that the norm does not depend on ,

showing that in norm. To see why is a subspace, observe that if , then for any ,

because , being an idempotent ultrafilter, contains the set . Therefore if and , it follows that , and hence is closed under sums and thus it is a subspace.

Next we need to describe the orthogonal space of . Since is idempotent, observe that for each , the operator is an orthogonal projection. Indeed

Therefore for each , the space can be decomposed into the orthogonal components

This implies that the space contains for every , and therefore that

We claim that this inclusion is actually an equality. To prove this claim, let and . For every there exists and such that and . Therefore

Since was arbitrarily close to we conclude that also . We remark that we used the fact that in the strong topology, but only that in the weak topology.

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 Lemma 3 with :

so taking – and using the hypothesis on we conclude that

so by Lemma 3 we conclude that indeed when .

To prove the other claim, let . By an approximation argument we can assume that there exists such that . Using again the fact that is idempotent we have

Since , it suffices now to show that when is a multiple of (say ) . We do this using (1). Indeed,

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.

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