The equidistribution theorem asserts that if is an irrational phase, then the sequence is equidistributed on the unit circle, or equivalently that
for any continuous (or equivalently, for any smooth) function . By approximating uniformly by a Fourier series, this claim is equivalent to that of showing that
for any non-zero integer (where ), which is easily verified from the irrationality of and the geometric series formula. Conversely, if is rational, then clearly fails to go to zero when is a multiple of the denominator of .
One can then ask for more quantitative information about the decay of exponential sums of , or more generally on exponential sums of the form for an arithmetic progression (in this post all progressions are understood to be finite) and a polynomial . It will be convenient to phrase such information in the form of an inverse theorem, describing those phases for which the exponential sum is large. Indeed, we have
for some , then there exists a subprogression of of size such that varies by at most on (that is to say, lies in a subinterval of of length at most ).
Proof: By a linear change of variable we may assume that is of the form for some . We may of course assume that is non-zero in , so that ( denotes the distance from to the nearest integer). From the geometric series formula we see that
and so . Setting for some sufficiently small absolute constant , we obtain the claim.
Thus, in order for a linear phase to fail to be equidistributed on some long progression , must in fact be almost constant on large piece of .
As is well known, this phenomenon generalises to higher order polynomials. To achieve this, we need two elementary additional lemmas. The first relates the exponential sums of to the exponential sums of its “first derivatives” .
Proof: Squaring (1), we see that
We write and conclude that
where is a subprogression of of the same spacing. Since , we conclude that
for values of (this can be seen, much like the pigeonhole principle, by arguing via contradiction for a suitable choice of implied constants). The claim follows.
The second lemma (which we recycle from this previous blog post) is a variant of the equidistribution theorem.
or else there is a natural number such that
Proof: We may assume that and , since we are done otherwise. Then there are at least two with , and by the pigeonhole principle we can find in with and . By the triangle inequality, we conclude that there exists at least one natural number for which
If then we are done, so suppose that . Suppose that are elements of such that and . Writing for some , we have
By hypothesis, ; note that as and we also have . This implies that and thus . We then have
We conclude that for fixed with , there are at most elements of such that . Iterating this with a greedy algorithm, we see that the number of with is at most ; since , this implies that
and the claim follows.
Now we can quickly obtain a higher degree version of Lemma 1:
for some , then there exists a subprogression of with such that varies by at most on .
Proof: We induct on . The cases are immediate from Lemma 1. Now suppose that , and that the claim had already been proven for . To simplify the notation we allow implied constants to depend on . Let the hypotheses be as in the proposition. Clearly cannot exceed . By shrinking as necessary we may assume that for some sufficiently small constant depending on .
By rescaling we may assume . By Lemma 3, we see that for choices of such that
for some interval . We write , then is a polynomial of degree at most with leading coefficient . We conclude from induction hypothesis that for each such , there exists a natural number such that , by double-counting, this implies that there are integers in the interval such that . Applying Lemma 3, we conclude that either , or that
We partition into arithmetic progressions of spacing and length comparable to for some large depending on to be chosen later. By hypothesis, we have
so by the pigeonhole principle, we have
for at least one such progression . On this progression, we may use the binomial theorem and (4) to write as a polynomial in of degree at most , plus an error of size . We thus can write for for some polynomial of degree at most . By the triangle inequality, we thus have (for large enough) that
and hence by induction hypothesis we may find a subprogression of of size such that varies by most on , and thus (for large enough again) that varies by at most on , and the claim follows.
This gives the following corollary (also given as Exercise 16 in this previous blog post):
for some , then there is a natural number such that for all .
One can obtain much better exponents here using Vinogradov’s mean value theorem; see Theorem 1.6 this paper of Wooley. (Thanks to Mariusz Mirek for this reference.) However, this weaker result already suffices for many applications, and does not need any result as deep as the mean value theorem.
Proof: To simplify notation we allow implied constants to depend on . As before, we may assume that for some small constant depending only on . We may also assume that for some large , as the claim is trivial otherwise (set ).
Applying Proposition 4, we can find a natural number and an arithmetic subprogression of such that and such that varies by at most on . Writing for some interval of length and some , we conclude that the polynomial varies by at most on . Taking order differences, we conclude that the coefficient of this polynomial is ; by the binomial theorem, this implies that differs by at most on from a polynomial of degree at most . Iterating this, we conclude that the coefficient of is for , and the claim then follows by inverting the change of variables (and replacing with a larger quantity such as as necessary).
For future reference we also record a higher degree version of the Vinogradov lemma.
for all .
Proof: We induct on . For this follows from Lemma 3 (noting that if then ), so suppose that and that the claim is already proven for . We now allow all implied constants to depend on .
For each , let denote the number of such that . By hypothesis, , and clearly , so we must have for choices of . For each such , we then have for choices of , so by induction hypothesis, either (5) or (6) holds, or else for choices of , there is a natural number such that
for , where are the coefficients of the degree polynomial . We may of course assume it is the latter which holds. By the pigeonhole principle we may take to be independent of .
Since , we have
We can again assume it is the latter that holds. This implies that modulo , so that
for choices of . Arguing as before and iterating, we obtain the claim.
The above results also extend to higher dimensions. Here is the higher dimensional version of Proposition 4:
Proposition 7 (Multidimensional Weyl exponential sum estimate, inverse form) Let and , and let be arithmetic progressions of length at most for each . Let be a polynomial of degrees at most in each of the variables separately. If
for some , then there exists a subprogression of with for each such that varies by at most on .
A much more general statement, in which the polynomial phase is replaced by a nilsequence, and in which one does not necessarily assume the exponential sum is small, is given in Theorem 8.6 of this paper of Ben Green and myself, but it involves far more notation to even state properly.
Proof: We induct on . The case was established in Proposition 5, so we assume that and that the claim has already been proven for . To simplify notation we allow all implied constants to depend on . We may assume that for some small depending only on .
By a linear change of variables, we may assume that for all .
We write . First suppose that . Then by the pigeonhole principle we can find such that
and the claim then follows from the induction hypothesis. Thus we may assume that for some large depending only on . Similarly we may assume that for all .
By the triangle inequality, we have
for (the claim also holds for but we discard it as being trivial). By the pigeonhole principle, there thus exists a natural number such that
for all and for choices of . If we write
where is a polynomial of degrees at most , then for choices of we then have
Applying Lemma 6 in the and the largeness hypotheses on the (and also the assumption that ) we conclude (after enlarging as necessary, and pigeonholing to keep independent of ) that
whenever for , with nonzero. Permuting the indices, and observing that the claim is trivial for , we in fact obtain (8) for all , at which point the claim easily follows by taking for each .
An inspection of the proof of the above result (or alternatively, by combining the above result again with many applications of Lemma 6) reveals the following general form of Proposition 4, which was posed as Exercise 17 in this previous blog post, but had a slight misprint in it (it did not properly treat the possibility that some of the could be small) and was a bit trickier to prove than anticipated (in fact, the reason for this post was that I was asked to supply a more detailed solution for this exercise):
Proposition 8 (Multidimensional Weyl exponential sum estimate, inverse form, II) Let be an natural number, and for each , let be a discrete interval for some . Let
Again, the factor of is natural in this bound. In the case, the option (10) may be deleted since (11) trivially holds in this case, but this simplification is no longer available for since one needs (10) to hold for all (not just one ) to make (11) completely trivial. Indeed, the above proposition fails for if one removes (10) completely, as can be seen for instance by inspecting the exponential sum , which has size comparable to regardless of how irrational is.