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 nα

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 nα T, n α

T

≤ ε. Setting q :=

|n − n|, we thus have qα

T

≤ 2ε. We may assume that qα = 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 qα

T

≤ 2ε ≤

0.02, we see that this set consists of intervals of length at most 2ε/ qα T,

punctuated by gaps of length at least 0.9/ qα

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/ qα

T

≥ δN/q, and the claim follows.