1.1. Equidistribution in tori 15
Exercise 1.1.19. Let α, β
Td,
let N 1, and let 0 δ 1. Suppose that
the linear sequence (αn + β)n=1
N
is not totally δ-equidistributed. Show that
there exists a non-zero k
Zd
with |k|
d
δ−Od(1)
such that k · α
T
d
δ−Od(1)/N.
Next, we give an analogue of Corollary 1.1.7:
Exercise 1.1.20 (Single-scale van der Corput lemma). Let x1,x2,...,xN
Td
be a sequence which is not totally δ-equidistributed for some 0 δ 1/2.
Let 1 H
δ−Cd
N for some sufficiently large Cd depending only on d.
Then there exists at least
δCd
H integers h [−H, H] such that the sequence
(xn+h xn)n=1
N
is not totally
δCd
-equidistributed (where we extend xn by
zero outside of {1,...,N}). (Hint: Apply Lemma 1.1.6.)
Just as in the asymptotic setting, we can use the van der Corput lemma
to extend the linear equidistribution theory to polynomial sequences. To
get satisfactory results, though, we will need an additional input, namely
the following classical lemma, essentially due to Vinogradov:
Lemma 1.1.14. Let α T, 0 ε 1/100, 100ε δ 1, and N 100/δ.
Suppose that
T
ε for at least δN values of n [−N, N]. Then there
exists a positive integer q = O(1/δ) such that αq
T
εq
δN
.
The key point here is that one starts with many multiples of α being
somewhat close (O(ε)) to an integer, but concludes that there is a single
multiple of α which is very close (O(ε/N), ignoring factors of δ) to an
integer.
Proof. By the pigeonhole principle, we can find two distinct integers n, n
[−N, N] with |n n | 1/δ such that T, n α
T
ε. Setting q :=
|n n|, we thus have
T
2ε. We may assume that = 0 since we
are done otherwise. Since N 100/δ, we have N/q 10 (say).
Now partition [−N, N] into q arithmetic progressions {nq + r : −N/q +
O(1) n N/q + O(1)} for some r = 0,...,q 1. By the pigeonhole
principle, there must exist an r for which the set
{−N/q + O(1) n N/q + O(1) : α(nq + r)
T
ε}
has cardinality at least δN/q. On the other hand, since
T

0.02, we see that this set consists of intervals of length at most 2ε/ T,
punctuated by gaps of length at least 0.9/
T
(say). Since the gaps are
at least 0.45/ε times as large as the intervals, we see that if two or more of
these intervals appear in the set, then the cardinality of the set is at most
100εN/q δN/q, a contradiction. Thus at most one interval appears in the
set, which implies that 2ε/
T
δN/q, and the claim follows.
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