Vitaly Bergelson, Florian Richter and I have recently uploaded to the arXiv our new paper “Single and multiple recurrence along non-polynomial sequences”. In this paper we address the question of what combinatorial structure is present in the set of return times of a non-polynomial sequence, i.e., sets of the form
where is a measure preserving system, , and is some non-polynomial function.
The departure point for this paper was the observation that when is a polynomial with integer coefficients and , set in (1) is syndetic, i.e. has bounded gaps (for the special case this is the content of Khintchine’s theorem, and the general case was obtained by Furstenberg, see a proof in my earlier post), but the same is no longer true in general for non-polynomial sequences .
Our first theorem shows that nevertheless, the set has an interesting (and a-priori surprising) combinatorial property, it is thick, i.e., contains arbitrarily long intervals of integers. One reason why this is surprising is that the notions of thick and syndetic sets are dual in the sense that a set is thick if and only if its complement is not syndetic and vice-versa. See my previous post for some discussion of the basic properties of thick and syndetic sets.
Theorem 1 Let be a measure preserving system, let , and , . Then the set defined in (1) for is thick.
Theorem 1 implies in particular that for every there exists some such that each of the sets , , … , have a large intersection with . Our second theorem proves that in fact all of these sets intersect together.
In the paper we deal with a considerably more general class of non-polynomial functions (see Definition 4 below) but for now I will restrict attention to the example .
— 1. Well distributed and thickly dense sequences —
can be studied using the theory of uniform distribution of sequences modulo . For instance, the fact that has positive density when is a polynomial is a corollary of Weyl’s theorem that the sequence is uniformly distributed in for every irrational .
In fact, in this situation, the sequence is well distributed, i.e., for every non-empty interval and for every sequence of intervals of integers with lengths we have
This fact, together with the spectral theorem implies that the set
is syndetic when satisfies .
It is not hard to see the relation between well distribution and syndeticity: if a sequence is well distributed modulo then it is syndetically dense, i.e., it visits every interval in a syndetic set of times. More precisely, a sequence in a topological space is syndetically dense if for every non-empty open set , the set is syndetic.
On the other hand, when for , , the fact that the set has positive density was derived by Bergelson and Håland from the fact that (any multiple of) the sequence is uniformly distributed on . However, the sequence is not well distributed on . This is easy to see for . Indeed in this case, grows to infinity but slower and slower, so that there are arbitrarily long intervals of integers where barely moves at all. For instance, for in the interval , the function only changes by which clearly prevents it from becoming well distributed modulo .
What we observed is that in fact exhibits the opposite phenomenon of well distribution: it is thickly dense, i.e. it visits every interval in a thick set of times. More precisely, a sequence in a topological space is thickly dense if for every non-empty open set , the set is thick.
It turns out that for every non-integer positive , the sequence is thickly dense in !
This stands in stark contrast with the syndetic denseness of polynomials. Indeed observe that if for any two disjoint subintervals and of , if for a sequence the set is thick, then the set , which is contained in the complement of can not be syndetic. Similarly, if is syndetic then can not be thick. Therefore, for any nonempty open set and irrational , we obtain the following bizarre behaviour for the set : it is syndetic when is an integer but thick when is not.
We don’t actually prove Theorem 3 in the paper, but here is a sketch of the proof.
Proof of Theorem 3: To prove that the set is thick it suffices to show that for every there exists such that . We can rewrite this condition in terms of the carthesian product of as
Therefore it will suffice to show that for every , the closure
contains the diagonal of . This will be achieved by showing that the sequence
is in fact uniformly distributed in some sub-torus of which contains the diagonal. When we can in fact take to be all of .
Let and observe that is a subgroup of . Now using standard tools from the theory of uniform distribution (namely the Weyl’s criterion, Fejér theorem on uniform distribution and the van der Corput trick) one can easily show that the sequence is uniformly distributed in the closed subgroup .
Finally, it suffices to show that the diagonal of is contained in . Using the generalized binomial theorem, it is easy to see that for every we have
and hence if then . This implies that any point in the diagonal of satisfies and hence .
Unfortunately this is not quite enough to establish Theorem 1. One of the difficulties is that, as the more astute reader will have noticed, we conveniently started talking about the sequence instead of or even for some .
— 2. Well distribution with respect to Riesz means —
As mentioned earlier, in the paper we deal with a more general class of functions than . To describe this class we will make use of the discrete derivative operator on functions defined as .
For each let and let .
- For each non integer, the function is in where .
- The functions , or are all in .
Theorem 1 holds for any function arising from .
In view of the discussion in the previous section (or just by considering a rotation on the torus) there is a close connection between Theorem 5 and the fact that for any the sequence is thickly dense in . However, the proof we used to establish thick denseness of the sequence in the previous section will not work for functions such as , because the triple is not uniformly distributed in any subgroup of .
It is true that the sequence is dense in whenever , but one can not prove this using the usual theory of uniform distribution. Instead we resort to the concept of Riesz means (or weighted means, as discussed in this previous post), and indeed of uniform Riesz means:
exists. We say that the uniform limit with respect to exists if
Note that for the coincides with the Cesàro limit. However, for different weights we get more general modes of convergence. For instance, if , the – is what is usually called the logarithmic averages.
It makes sense to define uniform distribution with respect to Riesz means. We say that a sequence taking values in a compact group is uniformly distributed with respect to a weight if for every continuous function the limit – exists and equals , where is the Haar measure on . If moreover the uniform limit – exists and equals , then we say that is well distributed with respect to .
It turns out that every function is well distributed with respect to some Riesz mean, and moreover, the sequence is well distributed with respect to some Riesz mean in some subgroup of .
One can also extend the notion of syndetic sets to that of -syndetic sets. Recall that a set is syndetic if it has non-empty intersection with every large enough interval. In other terms, if there exists a length such that for every interval with we have . For Riesz means , we replace the cardinality of a finite set with the quantity . Observe that letting we recover the cardinality of in this way.
It turns out that the set in Theorem 5 is not only thick (in the usual sense) but also -syndetic for some . This follows easily from the facts mentioned about and the observation that if a sequence is well distributed with respect to on some compact group , then for every non-empty open set of , the set is -syndetic. This in turn follows directly from the next proposition.
Proposition 8 Let and let . Then is -syndetic if and only if for every sequence of intervals in with
The statement is still true if one replaces with .
Proof: Assume first that is not -syndetic and let be an interval of length such that but . Then it is clear that
Next assume that is -syndetic and let be given by Definition 7. Then every interval with can be partitioned into disjoint intervals , with and so we have
which implies that