On pair correlation and discrepancy

We say that a sequence $\{x_n\}_{n \geq 1}$ in $[0,1)$ has Poissonian pair correlations if \begin{equation*} \lim_{N \rightarrow \infty} \frac{1}{N} \# \left\{ 1 \leq l \neq m \leq N \, : \, \left\lVert x_l-x_m \right\rVert<\frac{s}{N} \right\} = 2s \end{equation*} for all $s>0$. In this note we show that if the convergence in the above expression is - in a certain sense - fast, then this implies a small discrepancy for the sequence $\{x_n\}_{n \geq 1}$. As an easy consequence it follows that every sequence with Poissonian pair correlations is uniformly distributed in $[0,1)$.


Introduction
The concept of Poissonian pair correlations for a sequence (x n ) n≥1 in [0, 1) was introduced by Rudnick and Sarnak in [5], and has been intensively studied by several authors over the last years (see for instance [2,3,6,7,8]). Let · denote distance to the nearest integer. We say that a sequence (x n ) n≥1 of real numbers in [0, 1) has Poissonian pair correlations if for every s > 0.
In this note we are concerned with the relation between the Poissonian pair correlation property and the notion of uniform distribution. We say that the sequence (x n ) n≥1 is uniformly distributed, or equidis- It is well-known that uniform distribution does not necessarily imply Poissonian pair correlations. One example confirming this is the Kronecker sequence ({nα}) n≥1 , which is uniformly distributed for every irrational α, but does not have Poissonian pair correlations for any value of α. Whether the converse implication holds has until recently remained an open question: Is every sequence in [0, 1) with Poissonian pair correlations uniformly distributed? We answer this question in the affirmative by establishing a quantitative result connecting the speed of convergence in (1.1) to the star-discrepancy D * N of the sequence. We recall that the star-discrepancy D * N of (x n ) n≥1 is defined as where A N ([0, a)) := #{1 ≤ n ≤ N : x n ∈ [0, a)}, and that (x n ) n≥1 is uniformly distributed in [0, 1) if and only if lim N →∞ D * N = 0 (see for example [4]).
The main result of this paper is the following.
Theorem 1. Let (x n ) n≥1 be a sequence in [0, 1), and suppose that there exists a function F : N × N → R + which is monotonically increasing in its first argument, and which satisfies max s=1,...,K 1 2s 2) for all N ∈ N and all K ≤ N/2. One can then find an integer N 0 > 0 such that for N ∈ N, N ≥ N 0 , and arbitrary K satisfying N is the star-discrepancy of (x n ) n≥1 . The next result is an easy consequence of Theorem 1.
Proof. Suppose that (x n ) n≥1 has Poissonian pair correlations, and fix any ε > 0. We then have for all sufficiently large N ≥ N(ε). Hence, we may construct a function Without loss of generality, we may assume that N(ε) ≥ 1/ε 5 . If we fix and accordingly K satisfies (1.3). By Theorem 1 it thus follows that

Proof of Theorem 1
For a fixed pair of integers (N, K), where K satisfies (1.3), we introduce the notation Aiming for a proof by contradiction, we assume that ND * N > H(N, K) for infinitely many pairs (N, K). That is, there exist integers 1 < N 1 < N 2 < . . . and corresponding integers K 1 , K 2 , . . . satisfying (1.3), as well as real numbers B 1 , B 2 , . . . ∈ (0, 1), such that either for every j. We assume in what follows that (2.1) holds (the case when (2.2) holds is treated analogously). Note that (2.1) implies Let N := N j , K := K j , B := B j and H := H(N j , K j ) for some fixed j. We now consider the distribution of the points x n into subintervals of [0, 1) of length K/N. Let for i = 0, 1, . . . , ⌊N/K⌋ − 1, and let Moreover, for arbitrary positive integers l, let A l := A l mod(⌊N/K⌋+1) .

If we introduce the notation
We have that .
Now consider Γ K := min where by min x 1 ,...,x N we mean the minimum over all configurations of the points x 1 , . . . , x N satisfying (2.1). If we define and thus We have To see this, assume to the contrary that Z 1 and Z L − (L − 1)Z L−1 /L are all less than 2Z K /(K + 1). Then by succesive insertions we get the contradiction Z K < Z K . Hence, we have Let us now estimate min where the minimum on the right hand side is taken over all possible values of A 0 , A 1 , . . . , A ⌊N/K⌋ provided that the points x 1 , . . . , x N satisfy (2.1). By definition, we have A 0 + · · · + A ⌊N/K⌋ = N. Introducing the notation G i = A i + A i+1 + · · · + A i+K−1 , we thus get Moreover, by invoking condition (2.1) on the distribution of x 1 , . . . , x N , we have and consequently where the minimum on the right hand side is taken over all positive reals G 0 , G 1 , . . . , G ⌊N/K⌋ satisfying (2.8) -(2.10). It is an easy exercise to verify that this minimum is attained when Note that since K ≤ N 2/5 and H ≥ 5N 4/5 , we have K 2 ≤ H/5, and hence by (2.3) both the numerator and the denominator of these G i are positive. Thus, we get

2LKN
H This implies that which is a contradiction. Thus, our assumption (2.1) must be incorrect, and the proof of Theorem 1 is complete. (Note that the last inequality above is trivially true if N 2 /K ≤ NF (K 2 , N); In the opposite case we have K < N/F (K 2 , N), and by the condition (1.3) imposed on K we then get K ≥ N 2/5 /2, and consequently N 2 /K ≤ 2N 8/5 .)