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 .
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:
where 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:
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:
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 .
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.