# Minimal Gaps and Additive Energy in Real-valued Sequences

Shvo Regavim  School of Mathematical Sciences, Tel Aviv University, Tel Aviv
69978, Israel<EMAIL_ADDRESS>
###### Abstract.

We study the minimal gap statistic for sequences of the form $\left(\alpha
x_{n}\right)_{n=1}^{\infty}$ where $\left(x_{n}\right)_{n=1}^{\infty}$ is a
sequence of real numbers, and its connection to the additive energy of
$\left(x_{n}\right)_{n=1}^{\infty}$. Inspired by a recent paper of
Aistleitner, El-Baz and Munsch we show conditionally on the Lindelöf
Hypothesis that if the additive energy is of lowest possible order then for
almost all $\alpha$, the minimal gap
$\delta_{\min}^{\alpha}(N)=\min\left\\{\alpha x_{m}-\alpha x_{n}\bmod\ 1:1\leq
m\neq n\leq N\right\\}$ is close to that of a random sequence, a result
Rudnick showed for integer-valued sequences. We also show unconditional
results in this direction, as well as some converse theorems about sequences
with large additive energy.

##  1\. Introduction and Statement of Results

Given a sequence of real numbers $(x_{n})_{n\geq 1}$, our goal is to
understand the minimal gap

$\delta_{\min}\left((x_{n})_{n\geq
1},N\right)=\min\left(\norm{x_{m}-x_{n}}:m,n\leq N,m\neq n\right)$

Where
$\norm{x}=\min_{\begin{subarray}{c}k\in\mathbb{Z}\end{subarray}}\absolutevalue{x-k}$
is the distance of $x$ to the nearest integer. In the random case, that is for
$N$ independent uniformly distributed points in the unit interval, the minimal
gap is almost surely of size about $\frac{1}{N^{2}}$. The minimal gap is
always at most $\frac{1}{N}$, as that is the average size of the gap between
two consecutive points.

Computing the minimal gap of a given sequence is difficult, and so we will
mainly discuss the metric theory of minimal gaps. Fix a sequence
$X=(x_{n})_{n\geq 1}$. We will study the properties of the sequence $(\alpha
x_{n})_{n\geq 1}$ for $\alpha\in\mathbb{R}$. We want to understand under which
conditions we can show that for almost all $\alpha$ the minimal gap

$\delta_{\min}^{\alpha}\left(N\right)=\delta_{\min}\left((\alpha x_{n})_{n\geq
1},N\right)$

is of size approximately $\frac{1}{N^{2}}$ as in the random case. It is
relatively easy to show that under very mild spacing conditions on $x_{n}$ the
minimal gap cannot be too small. More specifically, we have

######  Theorem 1.1.

Assume that there is some constant $c>0$ such that
$x_{n+1}-x_{n}\geq\frac{c\log(n)}{n}$. Then for almost all $\alpha$ we have

$\delta_{\min}^{\alpha}\left(N\right)>\frac{1}{N^{2}\log(N)^{1+o(1)}}$

where the implied rate of decay $o(1)$ depends on $\alpha$.

To ensure that the minimal gap is small, we give the following criterion in
terms of the number of solutions to a certain Diophantine inequality and a
mild spacing condition on $x_{n}$. This follows the ideas of Rudnick and
Technau [14], where an analogous condition is developed for the question of
the metric pair correlation of the sequence $x_{n}$.

######  Theorem 1.2.

Fix some $2\geq\beta>1$. Suppose that $x_{n+1}-x_{n}\geq c$ for some $c>0$ and
that for all $\varepsilon>0$ we have

$\sum_{1\leq j_{1},j_{2}<N^{\beta-\varepsilon}}\sum_{\begin{subarray}{c}1\leq
m_{1},m_{2},n_{1},n_{2}\leq N\\\ m_{1}\neq n_{1},m_{2}\neq
n_{2}\end{subarray}}\mathbbm{1}\left(\absolutevalue{j_{1}(x_{m_{1}}-x_{n_{1}})-j_{2}(x_{m_{2}}-x_{n_{2}})}<1\right)\leq
C(\varepsilon)N^{4-\varepsilon}$

for some constant $C(\varepsilon)$. Then for almost all $\alpha$,

$\delta_{\min}^{\alpha}\left(N\right)\ll\frac{1}{N^{\beta-\varepsilon}}$

for all $\varepsilon>0$.

Our main goal in the rest of this paper is to bound this sum for various
sequences. We follow the technique in [1] in order to give a criterion in
terms of the ”additive energy” of the sequence $X$, which we define by

$E^{*}\left(X,N,\gamma\right)=\\#\left\\{(n_{1},n_{2},n_{3},n_{4})\in\left[1,N\right]^{4}:\absolutevalue{x_{n_{1}}+x_{n_{2}}-x_{n_{3}}-x_{n_{4}}}<\gamma\right\\}$

and by convention when we do not write $\gamma$ we let it be equal to 1, that
is

$E^{*}\left(X,N\right)=E^{*}\left(X,N,1\right)$

We always have $E^{*}\left(X,N\right)\geq N^{2}$ and by our spacing conditions
it is easy to see that $E^{*}\left(X,N\right)\ll N^{3}$ (as
$n_{1},n_{2},n_{3}$ determine $n_{4}$ up to a finite number of options). The
additive energy measures in a sense how much additive structure our sequence
has. For example, an arithmetic progression has additive energy $\gg N^{3}$,
of highest possible order, and the sequence $x_{n}=2^{n}$ has additive energy
$\ll N^{2}$, of lowest possible order. In [11] it is shown that if the
additive energy of an integer-valued sequence is at most $N^{2+o(1)}$, then
the minimal gap is for almost all $\alpha$ of size $\frac{1}{N^{2-o(1)}}$. We
will, conditionally, show the corresponding result for real-valued sequences:

######  Theorem 1.3.

Assume the Lindelöf Hypothesis. Take a sequence $X=(x_{n})_{n\geq 1}$ of
positive real numbers such that there exists a constant $c>0$ such that
$x_{n+1}-x_{n}\geq c$, and that $x_{n}$ is of polynomial growth, that is there
exists some constant $C$ such that $x_{n}\ll n^{c}$. Assume furthermore that
for some $\beta$, for all $\varepsilon>0$

$E^{*}\left(X,N\right)\ll N^{\beta+\varepsilon}$

Then for almost all $\alpha$,

$\delta_{\min}^{\alpha}\left(N\right)<\frac{1}{N^{4-\beta-o(1)}}$

where the implied rate of decay $o(1)$ is dependent on $\alpha$

In particular, if the conditions of 1.3 hold with $\beta=2$, that is

$E^{*}\left(X,N\right)=N^{2+\varepsilon}$

for all $\varepsilon>0$ then for almost all $\alpha$ we have

$\delta_{\min}^{\alpha}\left(N\right)=\frac{1}{N^{2+o(1)}}$

For example, if $x_{n}=n^{\theta},\theta\geq 2$ then by Theorem 2 in [10],
$E^{*}\left(X,N\right)=N^{2+o(1)}$ and so the theorem applies in this case.

We will show that this result is optimal, in the sense that for all
$2\leq\beta\leq 3$ there is a sequence $x_{n}$ such that
$E^{*}(X,N)=N^{\beta+o(1)}$ and for almost all $\alpha$ we have
$\delta_{\min}^{\alpha}\left(N\right)<\frac{1}{N^{4-\beta-o(1)}}$.

We have written this result mainly to illustrate the use of the technique. In
fact, we shall prove a more general result, at least when the additive energy
is small:

######  Theorem 1.4.

Take a sequence $X=(x_{n})_{n\geq 1}$ of positive real numbers such that there
exists a constant $c>0$ such that $x_{n+1}-x_{n}\geq c$. Assume furthermore
that

$E^{*}\left(X,N\right)=N^{2+o(1)}$

Assume that for some $\kappa\geq 2$ we have the bound

$\frac{1}{T}\intop_{0}^{T}\absolutevalue{\zeta\left(\frac{1}{2}+it\right)}^{\kappa}\mathrm{d}t\ll
T^{m(\kappa)+\varepsilon}$

on the $\kappa$th moment of the Riemann zeta function. Then for almost all
$\alpha$,

$\delta_{\min}^{\alpha}\left(N\right)<\frac{1}{N^{\frac{2\kappa-4}{\kappa+4\cdot
m(\kappa)}-o(1)}}$

where the implied rate of decay $o(1)$ is dependent on $\alpha$.

For example, suppose we knew that
$\zeta\left(\frac{1}{2}+it\right)=\mathcal{O}(t^{\mu+\varepsilon})$ for all
$\varepsilon>0$. Then, $m(\kappa)\leq\mu\kappa$ and so taking
$\kappa\to\infty$ we would have a minimal gap of size at most
$\frac{1}{N^{\frac{2}{1+4\mu}-\varepsilon}}$. In particular, assuming the
Lindelöf Hypothesis we have $\mu=0$, and thus we get 1.3 as a corollary.

It is interesting to note that the convexity bound
$\zeta\left(\frac{1}{2}+it\right)\ll t^{\frac{1}{4}}$ gives the trivial bound
$\frac{1}{N}$ on the minimal gap (which is obvious, as this is the size of the
average gap), and so any subconvexity bound on the zeta function gives a non-
trivial result.

Using Heath-Brown’s bound on the 12-th moment of the Riemann zeta function111A
tiny improvement comes from using Ivić’s bound which is Theorem 8.3 in [8]
where $\kappa=178/13$ and $m(\kappa)=16/13$, giving an exponent 1 + 31/121 =
1.2562… which in our terms states that $m(12)\leq 1$ we see that under the
assumptions of 1.4 the minimal gap is of size at most
$\frac{1}{N^{\frac{5}{4}-\varepsilon}}$.

Our arguments in Theorems 1.3 and 1.4 are in part an adaptation of those in
Aistleitner, Elbaz and Munsch in [1], where these techniques are used in the
setting of pair correlation.

We will also show using a purely elementary argument that certain sequences
satisfy the metric property we want, specifically rapidly growing sequences
and quadratic polynomials.

######  Theorem 1.5.

Take a sequence $X=(x_{n})_{n\geq 1}$ of positive real numbers such that for
every $\varepsilon>0$, for all $n$ large enough we have

$x_{n+1}>\left(1+\frac{1}{n^{\varepsilon}}\right)x_{n}$

Then for almost all $\alpha$,

$\delta_{\min}^{\alpha}\left(N\right)<\frac{1}{N^{2-o(1)}}$

where the implied rate of decay $o(1)$ is dependent on $\alpha$

Clearly every lacunary sequence satisfies this growth condition and so the
conclusion of the theorem applies in this case. However, this theorem applies
also to some sequences which grow subexponentially such as
$x_{n}=e^{\frac{n}{(\log n)^{\theta}}}$ for any $\theta\in\mathbb{R}$.

We also show this property for quadratic polynomials:

######  Theorem 1.6.

For any $a,b,c\in\mathbb{R}$ such that $a\neq 0$, look at the sequence
$X=(an^{2}+bn+c)_{n\geq 1}$. For almost all $\alpha$,

$\delta_{\min}^{\alpha}\left(N\right)<\frac{1}{N^{2-o(1)}}$

where the implied rate of decay $o(1)$ is dependent on $\alpha$

In the case where $a,b,c$ are integers (or rational numbers) this follows from
[11], but the case where $a,b,c$ are general real numbers is new.

The core idea in all the upper bounds we have given so far is that the random
variables $(x_{n}-x_{m})\alpha$ for $\alpha\in\mathbb{R}$, which are the gaps,
are essentially independent and therefore should behave as in the random case.
In 1.3 the additive energy is essentially a variance bound, in 1.5 we use the
rapid growth of $x_{n}$ to show the independence, and in 1.6 we use the fact
that we are able to compute $x_{n}-x_{m}$ fairly explicitly.

In contrast, we shall also discuss the sequence $\sqrt{n}$, where we do not
show a metric result but a result specifically about the gaps of the sequence
$\sqrt{n}$ itself. Properties of the gaps of $\sqrt{n}$ have been studied in
[7] where their gap distribution is computed and is surprisingly shown to not
be Poissonian (as in the random case), and in [6] where in contrast their pair
correlation is shown to be Poissonian. We will compute the minimal gap of the
sequence:

######  Theorem 1.7.

The minimal gap of the sequence $\left(\sqrt{n}\right)$ is asymptotically
equal to $\frac{1}{2N^{\frac{3}{2}}}$, that is

$\delta_{\min}\left(\left(\sqrt{n}\right)_{k^{2}\neq
n},N\right)\sim\min\left(\norm{\sqrt{m}-\sqrt{n}}:m,n\leq N,m\neq
n,\sqrt{m},\sqrt{n}\not\in\mathbb{Z}\right)\sim\frac{1}{2N^{\frac{3}{2}}}$

Notice that this is in stark contrast to the random case.

A natural question to ask is given a sequence $(x_{n})_{n\geq 1}$ what can we
say in general about the set of $\alpha\in\mathbb{R}$ such that the minimal
gap of $(\alpha x_{n})_{n\geq 1}$ is on order of $\frac{1}{N^{2+o(1)}}$, that
is random. We will show that for integer-valued sequences we have a 0-1 law.

######  Theorem 1.8.

Let $(x_{n})_{n\geq 1}$ be a sequence of integers. Then the set

$A=\left\\{\alpha\in\mathbb{R}:\delta_{\min}^{\alpha}\left(N\right)<\frac{1}{N^{2-o(1)}}\right\\}$

is either of zero measure or of full measure.

The proof is fairly general and works for most metric statistics of integer-
valued sequences.

In the last section we discuss sequences with large additive energy. We will
try to understand how true the converse to 1.4 is: given a sequence with large
additive energy, what can we say about its minimal gap? It is easy to generate
a sequence with large additive energy, but small metric minimal gap, simply by
interlacing the terms of a sequence with large additive energy and a sequence
with small metric minimal gap. Therefore, the best one can hope for is given a
sequence with large energy, to find a relatively dense subsequence with large
minimal gap. In the case where the sequence is integer-valued we establish the
following:

######  Theorem 1.9.

Let $x_{n}$ be a monotone increasing sequence of positive integers such that
for some $r>0$

$E_{N}:=\\#\left\\{(a,b,c,d)\in[1,N]^{4}\ \big{\rvert}\
x_{a}+x_{b}=x_{c}+x_{d}\right\\}\geq rN^{3}$

Then there exist increasing sequences of positive integers
$n_{k},a_{k}\to\infty$ such that $n_{k}$ is of density bounded from below in
$\left\\{1,\dots,a_{N}\right\\}$ and for almost all $\alpha$ we have

$\delta_{\min}^{\alpha}(x_{n_{k}},a_{N})\gg\frac{1}{a_{N}\log(a_{N})^{1+\varepsilon}}$

The key is the Balog-Szemerédi-Gowers theorem, which essentially says that
given a set with large additive energy there is a large subset with a
relatively small difference set, and we shall show that an integer sequence
with small difference set has large minimal gap. For the case of real-valued
sequences, the answer is more subtle and depends on the covering number of the
difference set with respect to different scales. Define the covering number of
a set $A\in\mathbb{R}$ with respect to a scale $\gamma$ to be

$\mathrm{cov}(A,\gamma)=\inf_{B\subset\mathbb{R}}\left\\{\absolutevalue{B}\
\big{\rvert}\ \forall a\in A\ \exists b\in
B:\absolutevalue{a-b}<\gamma\right\\}$

that is, the smallest number of balls with radius $\gamma$ that are needed to
cover $A$. A natural question is what are the conditions on $\gamma$ required
in order to deduce that a sequence whose difference set has a small covering
number on scale $\gamma$ has a large minimal gap. In the case where $\gamma$
is small we prove

######  Theorem 1.10.

For any sequence of real numbers $x_{n}$ such that
$x_{n+1}-x_{n}\geq\frac{c\log n}{n}$ for some constant $c>0$, if for some
$\varepsilon_{0}>0$ we have

$\mathrm{cov}\left(X_{N}-X_{N},\gamma\right)\ll N$

where $\gamma=\frac{1}{N\log(N)^{1+\varepsilon_{0}}}$, then for every
$\varepsilon>0$ and for almost all $\alpha\in\mathbb{R}$ that

$\delta_{\min}^{\alpha}(x_{n})\gg\frac{1}{N\log(N)^{1+\varepsilon}}$

From this we will deduce the following corollary, which is an analogue of
Theorem 1.9 in the real-valued case:

######  Corollary 1.11.

Let $x_{n}$ be a sequence of real numbers such that
$x_{n+1}-x_{n}\geq\frac{c\log n}{n}$ for some constant $c>0$, and for some
constant $r>0$ we have

$E\left(X,N,\gamma\right)\geq rN^{3}$

Then there exist increasing sequences of positive integers
$n_{k},a_{k}\to\infty$ such that $n_{k}$ is of density bounded from below in
$\left\\{1,\dots,a_{N}\right\\}$ and for almost all $\alpha$ we have

$\delta_{\min}^{\alpha}(x_{n_{k}},a_{N})\gg\frac{1}{a_{N}\log(a_{N})^{1+\varepsilon}}$

If $\gamma$ is slightly larger, we show:

######  Theorem 1.12.

There exists a sequence $\varepsilon_{n}\downarrow 0$ such that the sequence
$x_{n}=n-\frac{1}{n^{1-\varepsilon_{n}}}$ satisfies for all $\varepsilon>0$
and for almost all $\alpha\in\mathbb{R}$ that

$\delta_{\min}^{\alpha}(N)\ll\frac{1}{N^{2-\varepsilon}}$

where the implied constant depends on $\alpha$ and $\varepsilon$.

Notice that this sequence satisfies that

$\mathrm{cov}\left(X_{N}-X_{N},\gamma\right)\ll N$

for $\gamma=\frac{1}{N^{1-\varepsilon}}$

For converse results on the absence of almost sure Poisson pair correlation if
the additive energy is large, see [2, 4].

Acknowledgements: This work is part of the author’s M.Sc. thesis written under
the supervision of Zeev Rudnick at Tel Aviv University. This research was
supported by the European Research Council (ERC) under the European Union’s
Horizon 2020 research and innovation programme (Grant agreement No. 786758).

##  2\. Preliminaries

We will use the notation $\mathbbm{1}_{[-\frac{1}{M},\frac{1}{M}]}$ for the
indicator function of the interval $[-1/M,1/M]$ extended with period 1. Let us
look at the function

$X_{N,M}(\alpha)=\sum_{1\leq m\neq n\leq
N}\mathbbm{1}_{[-\frac{1}{M},\frac{1}{M}]}\left(\alpha x_{m}-\alpha
x_{n}\right)$

By definition, $X_{N,M}(\alpha)$ is positive if and only if
$\delta_{min}^{\alpha}(x_{n})\leq\frac{1}{M}$, and so our question is for what
size of $M$ can we ensure that $X_{N,M}(\alpha)$ is almost always positive.
The general idea is to find the expectation of $X_{N,M}$, bound its variance
and then finish by a standard argument using Chebyshev’s inequality and Borel-
Cantelli. The variance will turn out to be essentially the number of solutions
to a certain Diophantine inequality. We will first bound this in the general
case using the Riemann zeta function, and in two specific cases (quadratic
polynomials and rapidly growing sequences) we will bound the sum directly in
an elementary way.

In the integer-valued case, we can simply integrate with respect to the
Lebesgue integral on the interval $[0,1]$, and use the fact that
${(\alpha+1)x_{n}}={\alpha x_{n}}$. In the real-valued case we cannot do this
as the Lebesgue measure is not a probability measure on $\mathbb{R}$, and so
we must integrate with respect to a different measure. A good choice is the
measure $\mu$ which is defined by

$d\mu(x)=a(x)\mathrm{d}x:=\frac{\sin^{2}\left(\pi x\right)}{\left(\pi
x\right)^{2}}\mathrm{d}x$

Given a function $F$, our definition of the Fourier transform will be

$\hat{F}(\xi)=\int_{\mathbb{R}}e^{-i\xi t}F(t)\mathrm{d}t$

The Fourier transform of the function $a(x)=\frac{\sin\left(\pi
x\right)^{2}}{\pi^{2}x^{2}}$ is

$\hat{a}(y)=\max\left(\frac{2\pi-\absolutevalue{y}}{2\pi},0\right)$

Notice that we have normalized $a$ so that

$\int_{\mathbb{R}}a(x)\mathrm{d}x=1$

which is compactly supported on the interval $(-2\pi,2\pi)$, and uniformly
bounded by $1$.

To make the analysis cleaner it will be useful to replace
$\mathbbm{1}_{[-\frac{1}{M},\frac{1}{M}]}$ with a smoother function. A nice
choice are the Selberg polynomials $f_{M-1,1,M}^{\pm}$, which are
trigonometric polynomials of degree at most $M-1$ that satisfy

$f_{M-1,1,M}^{-}(x)\leq\mathbbm{1}_{[-\frac{1}{M},\frac{1}{M}]}(x)\leq
f_{M-1,1,M}^{+}(x)$

for all $x$, and also that

$f_{M-1,1,M}^{\pm}(x)=\sum_{\absolutevalue{j}<M}c_{j}^{\pm}e^{2\pi ijx}$

where

$\absolutevalue{c_{j}^{\pm}}\leq\min\left(\frac{2}{M},\frac{1}{\pi\absolutevalue{j}}\right)+\frac{1}{M}\leq\frac{3}{M}$

and also

$c_{0}^{-}=\intop_{0}^{1}f_{M-1,1,M}^{-}(x)\mathrm{d}x=\frac{1}{M},c_{0}^{+}=\intop_{0}^{1}f_{M-1,1,M}^{+}(x)\mathrm{d}x=\frac{3}{M}$

Now we can define

$X_{N,M}^{\pm}(\alpha)=\sum_{1\leq m\neq n\leq N}f_{M-1,1,M}^{\pm}\left(\alpha
x_{m}-\alpha x_{n}\right)$

Notice that

$X_{N,M}^{-}(\alpha)\leq X_{N,M}(\alpha)\leq X_{N,M}^{+}(\alpha)$

Therefore bounding $X_{N,M}^{+}$ from above corresponds to bounding the
minimal gap from below, and bounding $X_{N,M}^{-}$ from below corresponds to
bounding the minimal gap from above. For clarity, we will write
$f^{\pm}:=f_{M-1,1,M}^{\pm}$

##  3\. Minimal Gaps are not too Small

In this section we shall prove Theorem 1.1. We shall do this by computing the
expectation of $X_{N,M}^{+}$ and showing that it is small. Notice that

$\absolutevalue{\mathbbm{E}\left[X_{N,M}^{+},\mu\right]-\frac{3N(N-1)}{M}}=\absolutevalue{\sum_{1\leq
m\neq n\leq N}\int_{\mathbb{R}}\left(f^{+}\left(\alpha x_{m}-\alpha
x_{n}\right)-\frac{3}{M}\right)a(\alpha)\mathrm{d}\alpha}=$
$=\absolutevalue{\sum_{0<\absolutevalue{j}<M}c_{j}^{+}\sum_{1\leq m\neq n\leq
N}\int_{\mathbb{R}}e^{2\pi ij\left(\alpha x_{m}-\alpha
x_{n}\right)}a(\alpha)\mathrm{d}\alpha}\leq$
$\leq\frac{3}{M}\sum_{0<\absolutevalue{j}<M}\sum_{1\leq m\neq n\leq
N}\hat{a}(2\pi j(x_{n}-x_{m}))\leq$ $\leq\frac{3}{M}\sum_{1\leq m\neq n\leq
N}\sum_{0<\absolutevalue{j}<M}\mathbbm{1}\left(\absolutevalue{j(x_{n}-x_{m})}<1\right)$

Assume without loss of generality that $n>m$. By our spacing conditions, for a
pair $j,n,m$ to contribute to the sum we must have

$1>\absolutevalue{j(x_{n}-x_{m})}>\absolutevalue{\frac{j(n-m)\log(n)}{n}}$

and so we must have

$\frac{n}{\log(n)}>j(n-m)$

Fixing $n$, the number on the right is the number of integer points under the
hyperbola $xy=\frac{n}{\log(n)}$ and this is easily seen to be $\ll n$.
Summing over $n$ from $1$ to $N$ we get that the error term contributes
$\ll\frac{N^{2}}{M}$ and so

$\mathbbm{E}\left[X_{N,M}^{+},\mu\right]\ll\frac{N^{2}}{M}$

Fix some constant $\varepsilon>0$, and for every
$\gamma\in\mathbb{R},N\in\mathbb{N}$ define

$A(N,\gamma)=\left\\{\alpha\in\mathbb{R}:a(\alpha)\geq\gamma,\delta_{\min}^{\alpha}(N)\leq\frac{1}{N^{2}\log(N)^{1+\varepsilon}}\right\\}$

Using our bound on the expectation we get

$\gamma\cdot\absolutevalue{A(N,\gamma)}\leq\mathbbm{E}\left[X_{N,N^{2}\log(N)^{1+\varepsilon}}^{+},\mu\right]\ll\frac{1}{\log(N)^{1+\varepsilon}}$

and so

$\absolutevalue{A(N,\gamma)}\ll\frac{1}{\gamma\log(N)^{1+\varepsilon}}$

Using Borel-Cantelli and the fact that the series
$\sum_{n=1}^{\infty}\frac{1}{n^{1+\varepsilon}}$ converges, we see that the
set of $\alpha$ that are in $A(2^{N},\gamma)$ for infinitely many $N$ is of
measure 0. Take $\alpha$ such that for all $N$ large enough, $\alpha\not\in
A(2^{N},\gamma)$. For such an $\alpha$, for any large enough $n\in\mathbb{N}$,
taking $N$ to be the unique positive integer such that

$2^{N-1}<n\leq 2^{N}$

we see that

$\delta_{\min}^{\alpha}(n)\geq\delta_{\min}^{\alpha}(2^{N})\geq\frac{1}{2^{4N}\log(2^{N})^{1+\varepsilon}}\geq\frac{1}{5n^{2}\log(n)^{1+\varepsilon}}$

Therefore, for any $\gamma>0$, the set of $\alpha$ that satisfy

$a(\alpha)\geq\gamma$

and also do not satisfy

$\delta_{\min}^{\alpha}(n)\geq\frac{1}{5n^{2}\log(n)^{1+\varepsilon}}$

is of measure 0. As the set of zeroes of $a(\alpha)$ is countable, and
therefore of measure 0, we see that for almost every $\alpha$, for all $n$
large enough we have

$\delta_{\min}^{\alpha}(n)\geq\frac{1}{5n^{2}\log(n)^{1+\varepsilon}}$

for every $\varepsilon>0$, as required.

##  4\. Minimal Gaps can be Small

From here on we assume that for some constant $c>0$ we have $x_{n+1}-x_{n}\geq
c$. In fact, rescaling $x_{n}$ we can assume that $x_{n+1}-x_{n}\geq 1$, as
this obviously does not affect metric properties. In this case we have

$\mathbbm{E}\left[X_{N,M}^{+},\mu\right]=\frac{3N(N-1)}{M}$
$\mathbbm{E}\left[X_{N,M}^{-},\mu\right]=\frac{N(N-1)}{M}$

as looking at our computations in the previous section we see that
$\absolutevalue{2\pi j(x_{n}-x_{m})}\geq 2\pi$, and so $\hat{a}\left(2\pi
j(x_{n}-x_{m})\right)=0$ which implies this result.

Now we will bound the variance of $X_{N,M}^{-}$. By definition, we have

$\mathrm{Var}\left(X_{N,M}^{-},\mu\right)=\mathrm{Var}\left(X_{N,M}^{-}-\frac{N(N-1)}{M},\mu\right)=\mathbbm{E}\left[\left(X_{N,M}^{-}-\frac{N(N-1)}{M}\right)^{2},\mu\right]=$
$=\sum_{0<\absolutevalue{j_{1}},\absolutevalue{j_{2}}<M}\sum_{\begin{subarray}{c}m_{1}\neq
n_{1},m_{2}\neq n_{2}\\\ 1\leq m_{1},n_{1},m_{2},n_{2}\leq
N\end{subarray}}c_{j_{1}}^{-}c_{j_{2}}^{-}\int_{\mathbb{R}}e^{2\pi
i\alpha\left(j_{1}\left(x_{m_{1}}-x_{n_{1}}\right)-j_{2}\left(x_{m_{2}}-x_{n_{2}}\right)\right)}a(\alpha)\mathrm{d}\alpha\ll$
$\ll\frac{1}{M^{2}}\sum_{0<j_{1},j_{2}<M}\sum_{\begin{subarray}{c}m_{1}<n_{1},m_{2}<n_{2}\\\
1\leq m_{1},n_{1},m_{2},n_{2}\leq
N\end{subarray}}\mathbbm{1}\left(\absolutevalue{j_{1}\left(x_{n_{1}}-x_{m_{1}}\right)-j_{2}\left(x_{n_{2}}-x_{m_{2}}\right)}<1\right)$

For technical reasons, it will be more convenient to use the Cauchy-Schwarz
inequality to separate $j_{1},j_{2}$ into dyadic intervals. Writing
$U=\lceil\log_{2}(M)\rceil$, so that $2^{U}>M$ we get

$\mathrm{Var}\left(X_{N,M}^{-},\mu\right)\ll\int_{\mathbb{R}}\left(\sum_{u=1}^{U}\sum_{1\leq
m\neq n\leq N}\sum_{2^{u-1}\leq j<2^{u}}c_{j}^{-}e^{2\pi
ij\alpha(x_{m}-x_{n})}\right)^{2}\mathrm{d}\mu<$
$<\int_{\mathbb{R}}\left(\sum_{k=1}^{U}1\right)\sum_{u=1}^{U}\absolutevalue{\sum_{2^{u-1}\leq\absolutevalue{j}<2^{u}}\sum_{1\leq
m\neq n\leq N}c_{j}^{-}e^{2\pi
ij\alpha(x_{m}-x_{n})}}^{2}a(\alpha)\mathrm{d}\alpha\ll$
$\ll\frac{\log(N)}{M^{2}}\sum_{u=1}^{U}\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{\begin{subarray}{c}m_{1}<n_{1},m_{2}<n_{2}\\\ 1\leq
m_{1},n_{1},m_{2},n_{2}\leq
N\end{subarray}}\mathbbm{1}\left(\absolutevalue{j_{1}\left(x_{n_{1}}-x_{m_{1}}\right)-j_{2}\left(x_{n_{2}}-x_{m_{2}}\right)}<1\right)$

Recall that we have assumed that this sum is $\ll N^{4-\varepsilon}$ if
$M=N^{\beta-\varepsilon}$, and therefore

$\mathrm{Var}\left(X_{N,M}^{-},\mu\right)\ll\frac{N^{4-\varepsilon}}{M^{2}}$

In the next section we will show how a small variance implies a relatively
small minimal gap which will complete the proof of Theorem 1.2, and then we
shall bound the number of solutions to this Diophantine inequality in various
cases.

##  5\. Variance to Minimal Gap

This argument is standard but is included for the sake of completeness. In
general, suppose we have a series of random variables $Z_{n}$ such that
$\mathbbm{E}\left[Z_{n}\right]>0$ and we wish to show that almost all $Z_{n}$
are positive almost always. Notice that by Chebyshev’s inequality we have

$\mathbb{P}\left(Z_{n}\leq
0\right)\leq\mathbb{P}\left(\absolutevalue{Z_{n}-\mathbbm{E}\left[Z_{n}\right]}\geq\mathbbm{E}\left[Z_{n}\right]\right)\leq\frac{\mathrm{Var}(Z_{n})}{\mathbbm{E}\left[Z_{n}\right]^{2}}$

and so by Borel-Cantelli it is enough that

$\sum_{n}\frac{\mathrm{Var}(Z_{n})}{\mathbbm{E}\left[Z_{n}\right]^{2}}<\infty$

In our case, taking $M=N^{\beta-\varepsilon}$ and using what we know about the
variance and the expectation we have

$\frac{\mathrm{Var}\left(X_{N,M}^{-},\mu\right)}{\mathbbm{E}\left[X_{N,M}^{-},\mu\right]^{2}}\ll\frac{1}{N^{\varepsilon}}$

Now taking the random variables

$Z_{n}=X_{2^{n},2^{n(\beta-\varepsilon)}}$

we see that

$\sum_{n}\frac{\mathrm{Var}(Z_{n})}{\mathbbm{E}\left[Z_{n}\right]^{2}}\ll\sum_{n}\frac{1}{2^{\varepsilon
n}}<\infty$

and so for almost all $\alpha$, for all $n$ large enough we have

$\delta_{\min}^{\alpha}\left(2^{n}\right)<\frac{1}{2^{n(\beta-\varepsilon)}}$

Now, taking any $N$ large enough, and choosing $n$ to be the unique integer
such that $2^{n}\leq N<2^{n+1}$ we have

$\delta_{\min}^{\alpha}(N)\ll\frac{1}{N^{\beta-\varepsilon}}$

for almost all $\alpha$. Now, taking $\varepsilon=\frac{1}{n}$ and taking a
countable intersection of sets of full measure we see that for almost all
$\alpha$,

$\delta_{\min}^{\alpha}(N)<\frac{1}{N^{\beta-o(1)}}$

as required.

##  6\. Bounding the Riemann zeta function

We shall now prove Theorems 1.3 and 1.4, following the method in Aistleitner,
El-Baz and Munsch [1].
From the computation in the previous sections we see that it is enough to
bound the sum

$\sum_{2^{u-1}\leq j_{1},j_{2}<2^{u}}\sum_{1\leq m,n\leq
N}\mathbbm{1}(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1)$

Notice that $\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1$ is equivalent to
$\absolutevalue{\frac{j_{1}}{j_{2}}-\frac{z_{n}}{z_{m}}}<\frac{1}{j_{2}z_{m}}$
and so by the triangle inequality
$\frac{z_{n}}{z_{m}}<\frac{j_{1}+1}{j_{2}}\leq 2$. In the same way we get
$\frac{1}{2}<\frac{z_{n}}{z_{m}}$.

Therefore dividing the $z_{n}$ into $\log N$ intervals of the form
$[2N^{\beta},4N^{\beta}]$ where $N^{\beta}\in[\frac{1}{2},C]$, we see it is
enough to bound

$\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\mathbbm{1}(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1)$

Set $T=2^{u-1}N^{\beta}$. The inequality
$\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1$ implies

$\absolutevalue{\frac{j_{1}z_{m}}{j_{2}z_{n}}-1}<\frac{1}{j_{2}z_{n}}\leq\frac{1}{2^{u-1}N^{\beta}}=\frac{1}{T}$

There exist $\beta_{1}>\beta_{2}>0$ such that if
$\absolutevalue{x-1}<\frac{1}{T}$ then $\absolutevalue{\log
x}<\frac{\beta_{1}}{T}$ and if $\absolutevalue{\log x}<\frac{\beta_{2}}{T}$
then $\absolutevalue{x-1}<\frac{1}{T}$. Now define the Dirichlet polynomials

$D(t)=\sum_{2^{u-1}\leq j<2^{u}}j^{it}$
$P(t)=\sum_{z_{n}\in[N^{\beta},8N^{\beta}]}z_{n}^{it}$

We see that

$\absolutevalue{D(t)}^{2}\absolutevalue{P(t)}^{2}=\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\left(\frac{j_{1}z_{m}}{j_{2}z_{n}}\right)^{it}$

Take $\Phi$ to be a non-negative Schwarz function, whose Fourier transform is
compactly supported in $(-\beta_{1},\beta_{1})$ and satisfies $\hat{\Phi}\geq
1$ in $[-\beta_{2},\beta_{2}]$. Then,

$\int_{\mathbb{R}}\absolutevalue{D(tT)}^{2}\absolutevalue{P(tT)}^{2}\Phi(t)=\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\hat{\Phi}\left(T\log\left(\frac{j_{1}z_{m}}{j_{2}z_{n}}\right)\right)\geq$
$\geq\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\mathbbm{1}(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}<1})$

where the last inequality follows from the fact that $\hat{\Phi}\geq 1$ inside
$[-\beta_{2},\beta_{2}]$. Therefore it is enough to bound the integral. Notice
that we have the pointwise bounds

$\absolutevalue{D(t)}^{2}\leq D(0)^{2}\leq 2^{2u}$
$\absolutevalue{P(t)}^{2}\leq
P(0)^{2}=\sum_{k\in[N^{\beta},8N^{\beta}],z_{n}}\mathbbm{1}(z_{n}\in[k,k+1])\ll\min(N^{4},N^{\beta}E^{*}(X,N))$

where $N^{4}$ is trivial and $N^{\beta}E^{*}(X,N)$ follows from Cauchy-
Schwarz. This allows us to trivially bound the contribution of
$\absolutevalue{t}\leq\frac{1}{T}$ up to a constant by

$\frac{2^{2u}N^{\beta}E^{*}(X,N)}{T}=2^{u}E^{*}(X,N)$

For $\absolutevalue{t}\geq\frac{1}{T}$ we will show connect
$\absolutevalue{D(t)}^{2}$ to the Riemann zeta function via a convolution
formula. This lemma is a special case of Lemma 5.3 in [5].

######  Lemma 6.1.

Let $F$ be a holomorphic function in the strip
$y=\imaginary(z)\in\left[-\frac{3}{2},0\right]$ such that

$\sup_{-\frac{3}{2}\leq y\leq 0}\absolutevalue{F(x+iy)}\ll\frac{1}{x^{2}+1}$

Then for all $s=\frac{1}{2}+it,t\neq 0$ we have

$\sum_{k,\ell\geq 1}\frac{\hat{F}(\log
k\ell)}{k^{s}\ell^{\bar{s}}}=\int_{\mathbb{R}}\zeta(s+iu)\overline{\zeta(s-iu)}F(u)\mathrm{d}u+2\pi\zeta(1-2it)F(is-i)+2\pi\zeta(1+2it)F(i\bar{s}-i)$

Sketch of proof: Integrate the function
$w\mapsto\zeta(s+w)\zeta(\bar{s}+w)F(-iw)$ on the vertical line $\real(w)=0$.
Shift the contour to $\real(w)=\frac{3}{2}$, and the residues that are
collected correspond to the terms
$2\pi\zeta(1-2it)F(is-i)+2\pi\zeta(1+2it)F(i\bar{s}-i)$. On this line, the
series for $\zeta$ converges absolutely, and we can interchange the order of
summation and integration. Then, shifting termwise each integral back to
$\real(w)=0$ we get exactly our identity.

In our case, take $F(x)=\frac{\sin^{2}(Kx)}{\pi Kx^{2}}$ where
$K=(1+\varepsilon)\log(2^{u})$. Notice that
$\hat{F}(y)=\min\left(0,1-\frac{\absolutevalue{y}}{2K}\right)$, and also that
$F\left(t\pm\frac{i}{2}\right)\ll\frac{2^{u(1+\varepsilon)}}{t^{2}+1}$. Then
we have

$\sum_{j_{1},j_{2}\geq 1}\frac{\hat{F}(\log
j_{1}j_{2})}{(j_{1}j_{2})^{\frac{1}{2}}}\left(\frac{j_{1}}{j_{2}}\right)^{it}=$
$=\int_{\mathbb{R}}\zeta\left(\frac{1}{2}+it+iu\right)\zeta\left(\frac{1}{2}-it+iu\right)F(u)\mathrm{d}u+2\pi\zeta(1-2it)F\left(-t-\frac{i}{2}\right)+2\pi\zeta(1+2it)F\left(t-\frac{i}{2}\right)$

Notice that since the Fourier transform of $\Phi$ is positive, then increasing
the coefficients of $\absolutevalue{D(t)}^{2}$ can only increase our integral,
and so we can replace $\absolutevalue{D(t)}^{2}$ with

$2^{u}\sum_{j_{1},j_{2}\geq 1}\frac{\hat{F}(\log
j_{1}j_{2})}{(j_{1}j_{2})^{\frac{1}{2}}}\left(\frac{j_{1}}{j_{2}}\right)^{it}$

Now using the convolution identity we get

$\int_{\absolutevalue{t}\geq\frac{1}{T}}\absolutevalue{D(tT)}^{2}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t\ll
2^{u}(\mathrm{Int_{1}}+\mathrm{Int_{2}})$

where

$\mathrm{Int_{1}}=\int_{\absolutevalue{t}\geq\frac{1}{T}}\int_{\mathbb{R}}\zeta\left(\frac{1}{2}+itT+iu\right)\zeta\left(\frac{1}{2}-itT+iu\right)F(u)\mathrm{d}u\
\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$
$\mathrm{Int_{2}}=\int_{t\geq\frac{1}{T}}\zeta(1+2itT)F\left(tT-\frac{i}{2}\right)\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$

Using the easy bound $\zeta(1+it)\ll\log(t+1)$ for $t\geq 1$ we have

$\mathrm{Int_{2}}\ll\frac{N^{\beta}E^{*}(X,N)2^{u(1+\varepsilon)}}{T}\int_{t\geq
1}\frac{\log(t+1)}{t^{2}+1}\Phi\left(\frac{t}{T}\right)\mathrm{d}t\ll
E^{*}(X,N)2^{u\varepsilon}$

Now we turn to bounding $\mathrm{Int_{1}}$. From here on we shall assume the
Lindelöf Hypothesis, that is
$\absolutevalue{\zeta\left(\frac{1}{2}+it\right)}\ll t^{\varepsilon}$ for all
$\varepsilon>0$. Therefore,

$\absolutevalue{\zeta\left(\frac{1}{2}+it+iu\right)}\absolutevalue{\zeta\left(\frac{1}{2}-it+iu\right)}\ll(\absolutevalue{t}^{\frac{\varepsilon}{2}}+\absolutevalue{u}^{\frac{\varepsilon}{2}})^{2}\ll\absolutevalue{t}^{\varepsilon}+\absolutevalue{u}^{\varepsilon}$

This gives us that

$\mathrm{Int_{1}}\ll\mathrm{Int_{a}}+T^{\varepsilon}\mathrm{Int_{b}}$

where

$\mathrm{Int_{a}}=\int_{\mathbb{R}}\int_{\mathbb{R}}u^{\varepsilon}F(u)\mathrm{d}u\
\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t\ll\int_{\mathbb{R}}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$

and

$\mathrm{Int_{b}}=\int_{\mathbb{R}}\int_{\mathbb{R}}F(u)\mathrm{d}u\
t^{\varepsilon}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t=\int_{\mathbb{R}}t^{\varepsilon}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$

Due to the rapid decay of $\Phi$ one can see that the contribution of the
$\absolutevalue{t}\geq N$ to $\mathrm{Int_{b}}$ is $\mathcal{O}(1)$, and we
can bound the integral for $\absolutevalue{t}\leq N$ simply by

$N^{\varepsilon}\int_{\mathbb{R}}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$

and so in both cases it suffices to get a good bound for

$\int_{\mathbb{R}}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t=\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\hat{\Phi}\left(T\log\left(\frac{z_{m}}{z_{n}}\right)\right)\ll$
$\ll\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\mathbbm{1}(\absolutevalue{z_{m}-z_{n}}<1)\leq
E^{*}(X,N)$

Combining our bounds together we get

$\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{z_{m},z_{n}\in[N^{\beta},8N^{\beta}]}\mathbbm{1}(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1)\ll
2^{u}E^{*}(X,N)N^{\varepsilon}$

for all $\varepsilon>0$.

Over all, we have shown that

$\mathrm{Var}\left(X_{N,M}^{-},\mu\right)\ll\frac{1}{M}E^{*}(X,N)N^{\varepsilon}$

##  7\. Tightness of Bounds

Taking the sequence $x_{n}=\lfloor n^{4-\beta}\rfloor$ for some
$2\leq\beta\leq 3$, by Theorem 2 of [10] we have that the number of solutions
to the Diophantine inequality

$\absolutevalue{m_{1}^{\beta}+m_{2}^{\beta}-m_{3}^{\beta}-m_{4}^{\beta}}<\delta
N^{\beta}$

where $1\leq m_{1},m_{2},m_{3},m_{4}\leq N$ is at most
$N^{2+\varepsilon}+\frac{N^{4+\varepsilon}}{\delta}$. Choosing
$\delta=\frac{1}{4N^{\beta}}$ we see that in our case we have

$E^{*}(\lfloor n^{4-\beta}\rfloor,N)\ll N^{\beta+\varepsilon}$

and so by Theorem 1.3, under the Lindelöf Hypothesis we have for almost
$\alpha$

$\delta_{\min}^{\alpha}(N)\ll\frac{1}{N^{4-\beta-\varepsilon}}$

On the other hand, clearly we have

$\delta_{\min}^{\alpha}(N)\geq\min_{1\leq n\leq N^{4-\beta}}\norm{\alpha n}$

and it is well known that for almost all $\alpha$ we have $\min_{1\leq n\leq
M}\norm{\alpha n}\gg\frac{1}{M^{1+\varepsilon}}$, therefore for almost all
$\alpha$ we have

$\delta_{\min}^{\alpha}(N)\gg\frac{1}{N^{4-\beta+\varepsilon}}$

which shows that the bound is indeed optimal.

##  8\. Unconditional Results

In the previous sections we have translated the variance to an integral over
the Riemann zeta function, and used the Lindelöf Hypothesis to bound it. In
this section we will prove a weaker, but unconditional result. The method is
essentially the same, but we need to introduce some extra technicalities.
Following the argument here we can remove the requirement of polynomial growth
in Theorem 1.3.

First, let us assume that $z_{m},z_{n}\geq L$ where $L$ is some number
depending on $N,u$ that we shall choose later.

For any integer $k\geq 1$, we set $b_{k}$ to be the number of $z_{m}$’s in the
interval $[k,k+1)$, that is

$b_{k}=\sum_{m=1}^{L}\mathbbm{1}(z_{m}\in[k,k+1))$

By our spacing assumption we have $z_{n}\geq 1$ and so
$\sum_{k=1}^{\infty}b_{k}=\frac{N(N-1)}{2}$. Notice that

$\sum_{k=1}^{\infty}b_{k}^{2}=\sum_{k=1}^{\infty}\sum_{m,n}\mathbbm{1}(z_{m},z_{n}\in[k,k+1))\ll
E^{*}(X,N)$

where we recall that

$E^{*}\left(X,N\right)=\\#\left\\{(n_{1},n_{2},n_{3},n_{4})\in\left[1,N\right]^{4}:\absolutevalue{x_{n_{1}}+x_{n_{2}}-x_{n_{3}}-x_{n_{4}}}<1\right\\}$

is the additive energy of $X$. We split the interval $[1,\infty)$ into
disjoint intervals

$I_{h}=\left[e^{h/T},e^{(h+1)/T}\right)$

for each $h\geq 0$ and set

$a_{h}=\left(\sum_{k\in I_{h}}b_{k}^{2}\right)^{\frac{1}{2}}$

where $T=2^{u-1}L$.

Fix $j_{1},j_{2}$ and assume without loss of generality that $j_{2}\geq
j_{1}$. Assume that $\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1$ where $z_{m}\in
I_{h_{1}},z_{n}\in I_{h_{2}}$ for some $h_{1},h_{2}\geq 0$. Then,

$\absolutevalue{\frac{j_{2}}{j_{1}}-\frac{z_{m}}{z_{n}}}<\frac{1}{j_{1}z_{n}}$

implies that $\frac{z_{m}}{z_{n}}\leq 2$. We clearly have

$e^{(h_{1}-h_{2}-1)/T}\leq\frac{z_{m}}{z_{n}}\leq e^{(h_{1}-h_{2}+1)/T}$

and as $\frac{z_{m}}{z_{n}}\leq 2$ we have

$e^{(h_{1}-h_{2})/T}\leq\frac{z_{m}}{z_{n}}e^{1/T}<2e^{1/T}$

which implies

$\frac{z_{m}}{z_{n}}\leq
e^{(h_{1}-h_{2})/T}\left(1+\left(e^{1/T}-1\right)\right)\leq
e^{(h_{1}-h_{2})/T}+2e^{1/T}\left(e^{1/T}-1\right)<e^{(h_{1}-h_{2})/T}+\frac{3}{T}$

Similarly, one can show that

$e^{(h_{1}-h_{2})/T}-\frac{3}{T}\leq\frac{z_{m}}{z_{n}}$

and so by the triangle inequality

$\absolutevalue{\frac{j_{2}}{j_{1}}-e^{(h_{1}-h_{2})/T}}<\frac{1}{2^{u-1}L}+\frac{3}{T}=\frac{4}{T}$

where the last equality is true by our choice of $T$, and so we now know some
condition that $h_{1},h_{2}$ must satisfy in order to contribute to the sum.
Now fix $h_{1},h_{2}$, and take some $z_{n}\in[k,k+1)$ where $k\in I_{h_{2}}$.
In order for $\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1$ to hold we must have
$\absolutevalue{\left\lceil\frac{j_{2}k}{j_{1}}\right\rceil-z_{m}}<4$, because
$1\leq\frac{j_{2}}{j_{1}}<2$. As $\frac{j_{2}}{j_{1}}\geq 1$, the mapping
$k\mapsto\left\lceil\frac{j_{2}k}{j_{1}}\right\rceil$ is injective, and so

$\sum_{z_{m}\in I_{h_{1}},z_{n}\in
I_{h_{2}}}\mathbbm{1}\left(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1\right)\ll\sum_{k\in
I_{h_{2}}}b_{k}\sum_{\begin{subarray}{c}z_{m}\in I_{h_{1}}\\\
\absolutevalue{\left\lceil\frac{j_{2}k}{j_{1}}\right\rceil-
z_{n}}<4\end{subarray}}1\ll$ $\ll\sum_{k\in
I_{h_{2}}}\sum_{\begin{subarray}{c}\ell=-4\\\
\left\lceil\frac{j_{2}k}{j_{1}}\right\rceil+\ell\in
I_{h_{1}}\end{subarray}}^{3}b_{k}b_{\left\lceil\frac{j_{2}k}{j_{1}}\right\rceil+\ell}\ll\left(\sum_{k\in
I_{h_{2}}}b_{k}^{2}\right)^{\frac{1}{2}}\left(\sum_{k\in
I_{h_{1}}}b_{k}^{2}\right)^{\frac{1}{2}}=a_{h_{1}}a_{h_{2}}$

by Cauchy-Schwarz.

Combining what we know, we get that for fixed $j_{1},j_{2}$ we have

$\sum_{\begin{subarray}{c}1\leq m,n\leq N\\\
z_{m},z_{n}>MN^{\varepsilon}\end{subarray}}\mathbbm{1}(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1)\ll\sum_{\begin{subarray}{c}h_{1},h_{2}\geq
0\\\
\absolutevalue{e^{(h_{1}-h_{2})/T}-\frac{j_{2}}{j_{1}}}\leq\frac{4}{T}\end{subarray}}a_{h_{1}}a_{h_{2}}$

and so summing over all $j_{1},j_{2}$ in $\left[2^{u-1},2^{u}\right)$ we get
that

$\sum_{2^{u-1}\leq j_{1},j_{2}<2^{u}}\sum_{\begin{subarray}{c}1\leq m,n\leq
N\\\
z_{m},z_{n}>L\end{subarray}}\mathbbm{1}(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1)\ll$
$\ll\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{\begin{subarray}{c}h_{1},h_{2}\geq 0\\\
\absolutevalue{e^{(h_{1}-h_{2})/T}-\frac{j_{2}}{j_{1}}}\leq\frac{4}{T}\end{subarray}}a_{h_{1}}a_{h_{2}}$

##  9\. Unconditional zeta bounds

Let $\Phi(t)$ be a Schwarz function with a non-negative compactly supported
Fourier transform. Defining

$D(t)=\sum_{2^{u-1}\leq j<2^{u}}j^{it}$ $P(t)=\sum_{h\geq 0}a_{h}e^{iht/T}$

we see that

$\sum_{2^{u-1}\leq j_{1},j_{2}<2^{u}}\sum_{\begin{subarray}{c}h_{1},h_{2}\geq
0\\\
\absolutevalue{e^{(h_{1}-h_{2})/T}-\frac{j_{2}}{j_{1}}}\leq\frac{4}{T}\end{subarray}}a_{h_{1}}a_{h_{2}}\leq\int_{\mathbb{R}}\absolutevalue{D(tT)}^{2}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$

This is very similar to what we had in previous sections, however notice that
this time we have a concrete upper bound on the size of $T$. This is why we
have defined $P(t)$ as we did: dividing the $z_{n}$ into intervals of
increasing size allows a kind of ”orthogonalization”, which reduces the
contribution of very large $z_{m},z_{n}$.

We can use a simple pointwise bound to see that

$\int_{\absolutevalue{t}\leq\frac{1}{T}}\absolutevalue{D(tT)}^{2}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t\ll\frac{2^{2u}N^{4}}{T}$

Now to bound the rest of the integral, as before we can replace $D(t)$ with

$2^{u}\sum_{j_{1},j_{2}}\frac{\hat{F}\left(\log(j_{1}j_{2})\right)}{(j_{1}j_{2})}^{\frac{1}{2}}\left(\frac{j_{1}}{j_{2}}\right)^{it}$

Now by the convolution identity we have

$\int_{\absolutevalue{t}\geq\frac{1}{T}}\absolutevalue{D(tT)}^{2}\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t\ll
2^{u}\left(\mathrm{Int_{1}}+\mathrm{Int_{2}}\right)$

where

$\mathrm{Int_{1}}=\int_{\absolutevalue{t}\geq\frac{1}{T}}\int_{\mathbb{R}}\zeta\left(\frac{1}{2}+itT+iu\right)\zeta\left(\frac{1}{2}-itT+iu\right)F(u)\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}u\mathrm{d}t$
$\mathrm{Int_{2}}=\int_{\absolutevalue{t}\geq\frac{1}{T}}\zeta\left(1+2itT\right)F\left(tT-\frac{i}{2}\right)\absolutevalue{P(tT)}^{2}\Phi(t)\mathrm{d}t$

Using the easy bound $\zeta(1+it)\ll\log(t+1)$ for $t\geq 1$ we have

$\mathrm{Int_{2}}\ll\frac{2^{u(1+\varepsilon)}N^{4}}{T}\int_{t\geq
1}\frac{\log(1+t)}{t^{2}+1}\mathrm{d}t\ll\frac{2^{u}N^{4+\varepsilon}}{T}$

Now we shall turn to bounding $\mathrm{Int_{1}}$. Notice that

$\mathrm{Int_{1}}\ll\frac{1}{T}\int_{\mathbb{R}}\int_{\mathbb{R}}\absolutevalue{\zeta\left(\frac{1}{2}+it+iu\right)\zeta\left(\frac{1}{2}-it+iu\right)F(u)}\mathrm{d}u\
\absolutevalue{P(t)}^{2}\Phi\left(\frac{t}{T}\right)\mathrm{d}t\ll$
$\ll\frac{1}{T}\int_{\mathbb{R}}\frac{1}{u^{2}+1}\int_{\mathbb{R}}\absolutevalue{\zeta\left(\frac{1}{2}+it+iu\right)}\absolutevalue{\zeta\left(\frac{1}{2}-it+iu\right)}\absolutevalue{P(t)}^{2}\Phi\left(\frac{t}{T}\right)\mathrm{d}t\
\mathrm{d}u$

Using Hölder’s inequality where $\frac{2}{\kappa}+\frac{1}{B}=1$ we get

$\int_{\mathbb{R}}\absolutevalue{\zeta\left(\frac{1}{2}+it+iu\right)}\absolutevalue{\zeta\left(\frac{1}{2}-it+iu\right)}\absolutevalue{P(t)}^{2}\Phi\left(\frac{t}{T}\right)\ll$
$\ll\left(\int_{\mathbb{R}}\absolutevalue{\zeta\left(\frac{1}{2}+it+iu\right)}^{\kappa}\Phi\left(\frac{t}{T}\right)\mathrm{d}t\right)^{\frac{1}{\kappa}}\left(\int_{\mathbb{R}}\absolutevalue{\zeta\left(\frac{1}{2}-it+iu\right)}^{\kappa}\Phi\left(\frac{t}{T}\right)\mathrm{d}t\right)^{\frac{1}{\kappa}}\crossproduct$
$\crossproduct\left(\int_{\mathbb{R}}\absolutevalue{P(t)}^{2(B-1)}\absolutevalue{P(t)}^{2}\Phi\left(\frac{t}{T}\right)\right)^{\frac{1}{B}}\ll
T^{\frac{2}{\kappa}+\frac{2m(\kappa)}{\kappa}+\varepsilon}\absolutevalue{P(0)}^{\frac{4}{\kappa}}T^{1-\frac{2}{\kappa}}E^{*}(X,N)^{1-\frac{2}{\kappa}}\ll$
$\ll T\cdot
T^{\frac{2m(\kappa)}{\kappa}}N^{\frac{8}{\kappa}}E^{*}(X,N)^{1-\frac{2}{\kappa}}N^{\varepsilon}$

Furthermore, by the rapid decay of $\Phi$ we see that the contribution $t\geq
T\log T$ is negligible. Using the simple convexity bound
$\zeta\left(\frac{1}{2}+it\right)\ll t^{\frac{1}{4}}$ we can bound the
remaining integral by

$T^{\frac{1}{2}+\varepsilon}\int_{\absolutevalue{t}\leq T\log
T}\int_{\absolutevalue{u}\geq
T}\frac{1}{u^{2}}\absolutevalue{P(t)}^{2}\Phi\left(\frac{t}{T}\right)\mathrm{d}u\
\mathrm{d}t\ll T^{\frac{1}{2}+\varepsilon}E^{*}(X,N)$

Combining these we get that our sum is bounded by

$2^{u}\frac{1}{T^{\frac{1}{2}-\varepsilon}}+\frac{2^{2u}N^{4+\varepsilon}}{T}+2^{u}T^{\frac{2m(\kappa)}{\kappa}}N^{\frac{8}{\kappa}}E^{*}(X,N)^{1-\frac{2}{\kappa}}N^{\varepsilon}$

Choosing $L=2^{u}N^{\varepsilon}$ then $T=2^{2u}N^{\varepsilon}$ and our sum
is bounded by

$N^{4-\varepsilon}+(2^{u})^{1+\frac{4m(\kappa)}{\kappa}}N^{\frac{8}{\kappa}}E^{*}(X,N)^{1-\frac{2}{\kappa}}N^{\varepsilon}$

In the case where $E^{*}(X,N)\ll N^{2+\varepsilon}$ an easy computation shows
that in order for our sum to be $\ll N^{4-\varepsilon}$ we must have $2^{u}\ll
N^{\frac{2\kappa-4}{\kappa+4m(\kappa)}-\varepsilon}$, which is sufficient to
prove Theorem 1.4.

We have only discussed $z_{m},z_{n}\geq L$, but for $z_{m},z_{n}<L$ we can
divide into dyadic intervals of the form $[N^{\beta},2N^{\beta}]$ and use the
same method of proof. Everything remains the same, except we have a better
pointwise bound $\absolutevalue{P(t)}^{2}<N^{\beta}E^{*}(X,N)$ instead of
$\absolutevalue{P(t)}^{2}\leq N^{4}$.

##  10\. Rapidly Growing Sequences

In this section we will prove Theorem 1.5. We assume that
$x_{n+1}>x_{n}\left(1+\frac{1}{n^{o(1)}}\right)$. Fix $\delta>0$. We will show
that there is some small constant $\varepsilon>0$ such that for every $u$ that
satisfies $2^{u}<N^{2-\delta}$, we have

$\sum_{2^{u-1}\leq
j_{1},j_{2}<2^{u}}\sum_{m,n}\mathbbm{1}\left(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1\right)\ll
N^{4-\varepsilon}$

which, using the method we have used in the previous section, is sufficient to
deduce that for almost all $\alpha$ we have
$\delta_{\min}^{\alpha}(N)<\frac{1}{N^{2-o(1)}}$.

Call a pair $(m,n)$ admissible if there exist $2^{u-1}\leq j_{1},j_{2}<2^{u}$
such that $\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1$, or equivalently
$\absolutevalue{\frac{z_{m}}{z_{n}}-\frac{j_{2}}{j_{1}}}<\frac{1}{j_{1}z_{n}}$.
Recall that $j_{2},j_{1}$ are in a dyadic interval, and so
$\frac{j_{2}}{j_{1}}\in\left[\frac{1}{2},2\right]$, and in particular any
admissible pair satisfies $\frac{z_{m}}{z_{n}}\in\left[\frac{1}{3},3\right]$.

By our growth conditions, for any constant $C>0$ there are at most
$N^{\frac{\delta}{2}}\ n$’s such that $z_{n}<N^{C}$, and so there are at most
$N^{\delta}$ admissible pairs $(m,n)$ such that $z_{m}<N^{C}$ or
$z_{n}<N^{C}$. Each admissible pair contributes at most $2^{2u}<N^{4-2\delta}$
to the sum (as that is the number of pairs $j_{1},j_{2}$ in the allowed
range), and so the contribution of $m,n$ such that $z_{m}<N^{C}$ or
$z_{n}<N^{C}$ is at most $N^{4-\delta}$.

From here on we assume that $z_{m},z_{n}>N^{C}$ for some arbitrarily large
constant $C$. Take an admissible pair $(m,n)$, and take two pairs
$(j_{1},j_{2}),(j_{1}^{{}^{\prime}},j_{2}^{{}^{\prime}})$ which satisfy the
relevant inequality, that is

$\absolutevalue{\frac{z_{m}}{z_{n}}-\frac{j_{2}}{j_{1}}}<\frac{1}{j_{1}z_{n}}$
$\absolutevalue{\frac{z_{m}}{z_{n}}-\frac{j_{2}^{{}^{\prime}}}{j_{1}^{{}^{\prime}}}}<\frac{1}{j_{1}^{{}^{\prime}}z_{n}}$

By the triangle inequality,

$\absolutevalue{\frac{j_{2}}{j_{1}}-\frac{j_{2}^{{}^{\prime}}}{j_{1}^{{}^{\prime}}}}<\frac{1}{z_{n}}$

However, the LHS is a fraction with denominator $<N^{4-2\delta}$, and as
$z_{n}>N^{C}$ we must have

$\frac{j_{2}}{j_{1}}=\frac{j_{2}^{{}^{\prime}}}{j_{1}^{{}^{\prime}}}$

.

In particular, this implies that each admissible pair $(m,n)$ contributes at
most $N^{2-\delta}$ to our sum, and so if we show that there are at most
$N^{2+\frac{\delta}{2}}$ admissible pairs our sum will be of size at most
$N^{4-\frac{\delta}{2}}$, which will prove our claim.

Fix $z_{n}=x_{n_{1}}-x_{n_{2}}$. We will show that there are at most
$N^{\frac{\delta}{2}}\ m$’s such that the pair $(m,n)$ is admissible, and as
there are $<N^{2}$ options for $n$, this will prove what we need. Take $m$
such that $(m,n)$ is admissible, and write $z_{m}=x_{m_{1}}-x_{m_{2}}$. By our
growth conditions, for some arbitrarily small $\varepsilon$ we have

$\frac{x_{n_{1}}}{n_{1}^{\varepsilon}}\ll z_{n}<x_{n_{1}}$
$\frac{x_{m_{1}}}{m_{1}^{\varepsilon}}\ll z_{m}<x_{m_{1}}$

And so remembering that

$1\gg\frac{z_{m}}{z_{n}}\gg 1$

we get that

$\frac{x_{n_{1}}}{n_{1}^{\varepsilon}}\ll x_{m_{1}}\ll
n_{1}^{\varepsilon}\cdot x_{n_{1}}$

By the fact that
$\left(1+\frac{1}{n^{\varepsilon}}\right)^{n^{\varepsilon}\log(n)}\gg
n>n^{\varepsilon}$ we see easily that $\absolutevalue{m_{1}-n_{1}}\ll
n_{1}^{3\varepsilon}<N^{2\varepsilon}$, and so there are at most
$N^{2\varepsilon}$ options for $m_{1}$. Now fix $m_{1}$, and we shall show
that there are at most $N^{2\varepsilon}$ options for $m_{2}$. Assume by
contradiction that this is not the case. Then there exist
$m_{2}<m_{2}^{{}^{\prime}}<m_{1}-N^{2\varepsilon}$ such that
$(m,n),(m^{\prime},n)$ are both admissible pairs, where

$z_{m}=x_{m_{1}}-x_{m_{2}},\ z_{m^{\prime}}=x_{m_{1}}-x_{m_{2}^{{}^{\prime}}}$

That is there exist $2^{u-1}\leq j_{1},j_{2},k_{1},k_{2}<2^{u}$ such that

$\absolutevalue{\frac{z_{m}}{z_{n}}-\frac{j_{2}}{j_{1}}}<\frac{1}{j_{1}z_{n}}$
$\absolutevalue{\frac{z_{m^{\prime}}}{z_{n}}-\frac{k_{2}}{k_{1}}}<\frac{1}{k_{1}z_{n}}$

By using the triangle inequality, we see that

$\absolutevalue{\left(\frac{k_{2}}{k_{1}}-\frac{j_{2}}{j_{1}}\right)-\frac{z_{m^{\prime}}-z_{m}}{z_{n}}}<\frac{1}{z_{n}}$

Now using the reverse triangle inequality
$\absolutevalue{z-w}\geq\absolutevalue{z}-\absolutevalue{w}$ we get that

$\absolutevalue{\frac{k_{2}}{k_{1}}-\frac{j_{2}}{j_{1}}}<\frac{x_{m_{2}^{{}^{\prime}}}-x_{m_{2}}}{z_{n}}+\frac{1}{z_{n}}$

As we have shown,

$x_{m_{1}}\ll m_{1}^{\varepsilon}\cdot z_{n}$

and so

$x_{m_{2}}\ll\frac{x_{m_{1}}}{\left(1+\frac{1}{m_{1}^{\varepsilon}}\right)^{N^{2\varepsilon}}}<\frac{z_{n}}{2N^{4}}$

Combining this information we get

$\absolutevalue{\frac{k_{2}}{k_{1}}-\frac{j_{2}}{j_{1}}}<\frac{1}{N^{4}}$

However the LHS is a fraction with denominator smaller than $N^{4}$, which
means that it is $0$. Using our triangle inequality from the beginning again
we get

$\frac{x_{m_{2}^{{}^{\prime}}}-x_{m_{2}}}{z_{n}}<\frac{1}{z_{n}}$

and so

$x_{m_{2}^{{}^{\prime}}}-x_{m_{2}}<1$

a clear contradiction to our growth conditions. All in all, we have seen that
for each $n$ there are at most $N^{3\varepsilon}\ m$’s such that $(n,m)$ is an
admissible pair. Choosing $\varepsilon=\frac{\delta}{6}$ we get our desired
bound. $\blacksquare$

##  11\. Quadratic Polynomials

In this section we will prove Theorem 1.6. The gaps of quadratic polynomials
are well studied, see for example [12, 13]. In the case where $a,b,c$ are
integers (or more generally, rationals) this follows from [11], and in fact
Theorem 1.2 in [3] shows a stronger statement: for almost all $\alpha$, not
only do there exist $1\leq m\neq n\leq N$ and some integer $k$ such that
$\absolutevalue{\alpha(m^{2}-n^{2})-k}<1$, but we can choose $k=k_{1}k_{2}$
where $c(\alpha)N\leq k_{1},k_{2}\leq C(\alpha)N$. However, it appears that
for general real numbers $a,b,c$ this is a new result.

First of all, notice that we can assume without loss of generality that
$a=1,c=0$ by rescaling and translating the sequence, and so we can work with
the sequence $n^{2}+\beta n$ for some $\beta\in\mathbb{R}$. The gaps of this
sequence are of the form $n^{2}+\beta n-m^{2}-\beta m=(n-m)(n+m+\beta)$ and so
it is enough to bound

$\sum_{1\leq
j_{1},j_{2}<M}\sum_{z_{m},z_{n}}\mathbbm{1}\left(\absolutevalue{j_{1}z_{m}-j_{2}z_{n}}<1\right)\ll\sum_{1\leq
j_{1},j_{2}<M}\sum_{1\leq x<y,z<w\leq
2N}\mathbbm{1}\left(\absolutevalue{j_{1}x(y+\beta)-j_{2}z(w+\beta)}<1\right)$

We need to show that this sum is at most
$\frac{N^{4}}{\log(N)^{1+\varepsilon}}$. Notice that

$j_{1}x(y+\beta)-j_{2}z(w+\beta)=j_{1}xy-j_{2}zw+\beta(j_{1}x-j_{2}z)$

Fix $k=j_{1}x-j_{2}z$. Then there are at most 2 options for $j_{1}xy-j_{2}zw$,
which we will denote by $C$. Then we have

$j_{1}xy-j_{2}zw=j_{2}z(y-w)+ky=C$

and so

$j_{2}z(y-w)=C-ky$

Fixing $y$, there are at most $d(C-ky)^{2}$ options for $z,y-w,j_{2}$, but
using $w=y-(y-w),\ j_{1}x=j_{2}z+k$ we see that we can find $w,j_{1}x$ as
well. Now, there are $d(j_{2}z+k)$ options for $j_{1},x$ and so over all using
the standard divisor bound $d(n)\ll n^{\varepsilon}$ we have shown that fixing
$k,y$ there are $\ll N^{\varepsilon}$ options for $j_{1},j_{2},x,y,z,w$. As
$k\leq MN,\ y\leq 2N$, running over all pairs of $k,y$, the sum is at most
$MN^{2+\varepsilon}$, and choosing $M=N^{2-2\varepsilon}$ we get what we
desire.

Of course, a slightly stronger result could be achieved by using a stronger
divisor bound like

$d(n)\ll 2^{(1+\varepsilon)\frac{\log N}{\log\log N}}$

In the case where our sequence is a quadratic polynomial with integer (or
rational) coefficients, we can divide the sum according to the value of $ab$
and use the fact that $\sum_{n\leq N^{2}}d(n)^{2}\ll N^{2}\log(N)^{3}$ to get
a minimal gap of size

$\frac{\log(N)^{4+\varepsilon}}{N^{2}}$

which is optimal up to a power of $\log N$.

##  12\. Zero-One Law for Integer Valued Sequences

Given a sequence $x_{n}$ which perhaps does not satisfy the metric minimal gap
condition, one can ask about the nature of the set
$X=\\{\alpha\in\mathbb{R}:\delta_{min}^{\alpha}(x_{n})\ll\frac{1}{N^{2-o(1)}}\\}$.
In this section we will show that in the case where $x_{n}$ is an integer
valued sequence, $X$ satisfies a zero-one law: $X$ is either of full measure
or of zero measure. The argument is general and works for the majority of
metric statistics (e.g. metric Poissonian gap distribution, metric Poissonian
pair correlation…). It would be interesting to generalize this to the case
where $(x_{n})$ is real valued.

Clearly $X$ is measurable. Notice that $\alpha$ is in $X$ if and only if
$\alpha+1$ is in $X$, because $\alpha x_{n}=(\alpha+1)x_{n}\mod 1$.
Furthermore, for every $k\in\mathbb{Z}$, if $\alpha$ is in $X$ then so is
$k\alpha$, because $\norm{k\alpha x_{m}-k\alpha x_{n}}\leq k\norm{\alpha
x_{m}-\alpha x_{n}}$ and so $\delta_{min}^{k\alpha}(x_{n})\leq
k\delta_{min}^{\alpha}(x_{n})\ll\frac{1}{N^{2-o(1)}}$. The claim now follows
from the following elementary lemma:

######  Lemma 12.1.

Every measurable set $X$ of real numbers which satisfies the following
conditions:

(i) $x\in X$ if and only if $x+1\in X$

(ii) For every $k\in\mathbb{Z}$, if $x\in X$ then $kx\in X$
is either of full measure or of zero measure

Let us prove the lemma: if $X$ is of zero measure then we are done, so we can
assume that $X$ has positive measure. We will show that for every
$\varepsilon>0$, $\mu(X\cap[0,1])\geq 1-\varepsilon$, which shows that
$\mu(X^{c}\cap[0,1])=0$, and by the 1-periodicity of $X$ we will get that
$\mu(X^{c})=0$.

Now fix $\varepsilon>0$. Because $\mu(X)>0$, there exists an interval $[a,b]$
such that $\mu(X\cap[a,b])>\left(1-\frac{\varepsilon}{2}\right)(b-a)$. By
property (ii), if $[a,b]$ is such an interval then so is $[ka,kb]$ for every
$k\in\mathbb{Z}$, and so we can take $b-a$ to be arbitrarily large. Notice
that

$\mu(X\cap[\lfloor a\rfloor,\lceil
b\rceil])\geq\mu(X\cap[a,b])>\left(1-\frac{\varepsilon}{2}\right)(b-a)>(1-\varepsilon)(\lceil
b\rceil-\lceil a\rceil)$

where the last inequality is true if $a,b$ are large enough (we use the
obvious inequality $b-a+2>\lceil b\rceil-\lfloor a\rfloor$). But now by the
1-periodicity of $X$ we have $\mu(X\cap[k,k+1])=\mu(X\cap[0,1]$ for any
$k\in\mathbb{Z}$, and so

$\mu(X\cap[0,1])=\frac{\mu(X\cap[\lfloor a\rfloor,\lceil b\rceil])}{\lceil
b\rceil-\lfloor a\rfloor}>1-\varepsilon$

which is exactly what we needed to prove.

Alternatively, it is well known that the map $x\mapsto 2x$ for
$x\in\mathbb{R}/\mathbb{Z}$ preserves the Lebesgue measure and is ergodic with
respect to it. We can view $X$ as a set inside $\mathbb{R}/\mathbb{Z}$ which
satisfies $X\subset F^{-1}X$. Therefore by ergodicity, $\mu(X)=0,1$ as
required.

The reason that this proof fails to work in the case of real-valued sequences
is that we lose the 1-periodicity of the set $X$, and the lemma is not true
without it (for example, $\\{x\in\mathbb{R}\ \big{\rvert}\
\absolutevalue{x}>1\\}$ is a counterexample).

##  13\. Minimal Gap in $(\sqrt{n})_{n\geq 1}$

In this section we will discuss the fractional parts of the sequence
$\sqrt{n}$ and prove Theorem 1.7. Of course, technically the minimal gap is
$0$ as $\sqrt{4}-\sqrt{1}=1$, and so we throw away all of the integers in this
sequence, that is square roots of perfect squares. Then, defining

$\delta_{\min}^{2}(N)=\min\left(\norm{\sqrt{m}-\sqrt{n}}:m\neq n\leq
N,\sqrt{m},\sqrt{n}\not\in\mathbb{Z}\right)$

We shall show that

$\delta_{\min}^{2}(N)\sim\frac{1}{2N^{\frac{3}{2}}}$

This is in stark contrast to the random case where we expect
$\delta_{\min}^{2}(N)$ to be on order of $\frac{1}{N^{2}}$.

Take $n>m$ achieving the minimum in $\delta_{N}$. Then there exists
$k\in\mathbb{Z}$ and $\varepsilon\in\mathbb{R}$ such that

$\norm{\sqrt{n}-\sqrt{m}}=\absolutevalue{\varepsilon}$

and

$\sqrt{n}-\sqrt{m}=k+\varepsilon$

Squaring both sides we get

$n+m-\sqrt{4mn}=k^{2}+2k\varepsilon+\varepsilon^{2}$

or equivalently

$\sqrt{4mn}-\left(m+n-k^{2}\right)=2k\varepsilon+\varepsilon^{2}$

Notice that clearly the right hand side is non zero, as $\varepsilon\neq
0,-2k$ and so the left hand side is non zero as well. Then, we have

$\absolutevalue{\sqrt{4mn}-\left(m+n-k^{2}\right)}\geq\absolutevalue{\sqrt{4mn}-\sqrt{4mn+1}}=\frac{1}{\sqrt{4mn}+\sqrt{4mn+1}}=\frac{1+o(1)}{4\sqrt{mn}}$

and so we have

$2k\absolutevalue{\varepsilon}=(1+o(1))\absolutevalue{2k\varepsilon+\varepsilon^{2}}\geq\frac{1+o(1)}{4\sqrt{m}\sqrt{n}}$

and therefore using the arithmetic mean-geometric mean inequality we get

$\delta_{\min}^{2}(N)=\absolutevalue{\varepsilon}\geq\frac{1+o(1)}{8\sqrt{n}\sqrt{m}(\sqrt{n}-\sqrt{m})}\geq\frac{1+o(1)}{2n^{\frac{3}{2}}}\geq\frac{1+o(1)}{2N^{\frac{3}{2}}}$

which proves the lower bound.

On the other hand, choose $d$ to be the unique positive integer such that

$4(d+1)^{2}>N\geq 4d^{2}$

and take

$n=\sqrt{4d^{2}-2},m=\sqrt{d^{2}-1}$

We will use the Taylor series expansion of $\sqrt{x}$ around $1$ to the second
order, which is

$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^{2}+O(x^{3})$

and get that

$\sqrt{m}=d\sqrt{1-\frac{1}{d^{2}}}=d-\frac{1}{2d}-\frac{1}{8d^{3}}+O\left(\frac{1}{d^{5}}\right)$
$\sqrt{n}=2d\sqrt{1-\frac{1}{2d^{2}}}=2d-\frac{1}{2d}-\frac{1}{16d^{3}}+O\left(\frac{1}{d^{5}}\right)$

and so

$\sqrt{n}-\sqrt{m}=d+\frac{1}{16d^{3}}+O\left(\frac{1}{d^{5}}\right)$

and therefore

$\delta_{\min}^{2}(N)\leq\frac{1+o(1)}{16d^{3}}=\frac{1+o(1)}{2N^{\frac{3}{2}}}$

which is our required lower bound. Combining these, we have shown that

$\delta_{\min}^{2}(N)\sim\frac{1+o(1)}{2N^{\frac{3}{2}}}$

as requested.

Note that this exact argument shows in fact that if $a,b,c\leq N$ then

$\min_{\begin{subarray}{c}1\leq a,b,c\leq N\\\
\sqrt{a}-\sqrt{b}-\sqrt{c}\not\in\mathbb{Z}\end{subarray}}\left(\absolutevalue{\sqrt{a}-\sqrt{b}-\sqrt{c}}\right)\sim\frac{1}{2N^{\frac{3}{2}}}$

which is a strictly stronger statement (1.7 is this statement when we only let
$c$ range over squares).

One can also ask about the minimal gap in $n^{\frac{1}{k}}$ for
$1<k\in\mathbb{Z}$, that is looking at

$\delta_{\min}^{k}(N)=\min\left(\norm{\sqrt[k]{n}-\sqrt[k]{m}}:m\neq n\leq
N,\sqrt[k]{n},\sqrt[k]{m}\not\in\mathbb{Z}\right)$

The case $k=2$ is the case we have just considered. Our example easily
generalizes via

$n=2^{k}d^{k}-2^{k-1},m=d^{k}-1$

to show that

$\delta_{\min}^{k}(N)\ll\frac{1}{N^{2-\frac{1}{k}}}$

However our method to achieve a lower bound generalizes as follows:

$1\leq
N_{\mathbb{Q}\left(\sqrt[k]{m},\sqrt[k]{n}\right)/\mathbb{Q}}\left(\sqrt[k]{m}-\sqrt[k]{n}-\ell\right)\ll\left(\sqrt[k]{m}-\sqrt[k]{n}-\ell\right)N^{k-\frac{1}{k}}$

which gives just

$\delta_{\min}^{k}(N)\gg\frac{1}{N^{k-\frac{1}{k}}}$

which for $k>2$ does not match our lower bound. It would be interesting to
know the answer for general $k$.

##  14\. Sequences with large additive energy

In this section we will discuss sequences with large additive energy. For
simplicity, we will assume that our sequences are integer-valued, though under
some strong assumptions we can work with real-valued sequences as well. For a
set $A$ of integers we recall that its additive energy is defined by

$E(A)=\\#\left\\{(a,b,c,d)\in A^{4}\ \big{\rvert}\ a+b=c+d\right\\}$

In [11] it is shown that a sequence with additive energy of $\ll
N^{2+\varepsilon}$ (the lowest possible energy) has a metric minimal gap of
size at most $\frac{1}{N^{2-\varepsilon}}$. One can ask the opposite question:
does a large additive energy imply that the sequence has a large minimal gap?
The answer to this is easily seen to be no. For example, take the sequence
defined by

$x_{2k}=k^{2},\ x_{2k+1}=k$

Then this sequence has a small metric minimal gap, because $k^{2}$ has a small
metric minimal gap, but its additive energy is of order $N^{3}$.

Let us look more carefully at our argument for Theorem 1.1. Denoting the set
$X_{N}=\\{x_{1},\dots,x_{N}\\}$ We used the random variable

$\sum_{m,n\in X_{N},m\neq
n}\mathbbm{1}_{[-\frac{1}{M},\frac{1}{M}]}\left(\alpha x_{m}-\alpha
x_{n}\right)$

However, there may be multiplicities in the values of $x_{m}-x_{n}$, and then
we are summing over some indicators multiple times, and so it would be more
economical to work with the random variable

$\sum_{0\neq t\in
X_{N}-X_{N}}\mathbbm{1}_{[-\frac{1}{M},\frac{1}{M}]}\left(\alpha t\right)$

Then we see that the expectation of this variable is

$\frac{2\absolutevalue{X_{N}-X_{N}}}{M}$

and the same type of argument as in section 3 shows that in this case the
minimal gap is at least of order

$\frac{1}{\absolutevalue{X_{N}-X_{N}}\log(N)^{1+\varepsilon}}$

Therefore, if $\absolutevalue{X_{N}-X_{N}}$ is small then the minimal gap is
large. By Cauchy-Schwarz one can see that
$E(X_{N})\geq\frac{\absolutevalue{X_{N}}^{4}}{\absolutevalue{X_{N}-X_{N}}}$,
and so $\absolutevalue{X_{N}-X_{N}}$ small implies that $E(X_{N})$ is large.
We can use the following theorems from additive combinatorics to give a
partial converse to this.

######  Theorem 14.1.

(Balog-Szemerédi-Gowers Theorem)

Let $A$ be a finite subset of an abelian group. If for some $K$ we have
$E(A)\geq\frac{A^{3}}{K}$ (where $E(A)$ is the additive energy) then there is
a subset $A^{\prime}\subset A$ such that $\absolutevalue{A^{\prime}}\geq
K^{-\mathcal{O}(1)}\absolutevalue{A}$ and
$\absolutevalue{A^{\prime}+A^{\prime}}\leq
K^{\mathcal{O}(1)}\absolutevalue{A^{\prime}}$

######  Theorem 14.2.

(Plünnecke-Ruzsa inequality)

If $A,B$ are finite subsets of an abelian group and $K$ is a constant such
that $\absolutevalue{A+B}\leq K\absolutevalue{A}$, then for all non-negative
integers $m,n$ we have $\absolutevalue{mB-nB}\leq K^{m+n}\absolutevalue{A}$

Using $A=B=A^{\prime},m=n=1$ in 14.2 together with 14.1 we see that if
$E(A)\geq\frac{A^{3}}{K}$ there exists a subset $A^{\prime}\subset A$ such
that $\absolutevalue{A^{\prime}}\geq K^{-\mathcal{O}(1)}\absolutevalue{A}$ and
$\absolutevalue{A^{\prime}-A^{\prime}}\leq
K^{\mathcal{O}(1)}\absolutevalue{A^{\prime}}$. With this, we can prove Theorem
1.9 which we restate here for convenience:

######  Theorem 14.3.

Let $x_{n}$ be a monotone increasing sequence of positive integers such that
for some $r>0$

$E_{N}:=\\#\left\\{(a,b,c,d)\in[1,N]^{4}\ \big{\rvert}\
x_{a}+x_{b}=x_{c}+x_{d}\right\\}\geq rN^{3}$

Then there exist increasing sequences of positive integers
$n_{k},a_{k}\to\infty$ such that $n_{k}$ is of density bounded from below in
$\left\\{1,\dots,a_{N}\right\\}$ and for almost all $\alpha$ we have

$\delta_{\min}^{\alpha}(x_{n_{k}},a_{N})\gg\frac{1}{a_{N}\log(a_{N})^{1+\varepsilon}}$

Using the Cantelli argument as above it is enough to show that for the
aforementioned $a_{k},x_{n_{k}}$ we have a small difference set, that is

$\\#\left\\{x_{n_{a}}-x_{n_{b}}\ \big{\rvert}\ 1\leq n_{a},n_{b}\leq
a_{N}\right\\}\leq C(r)a_{N}\log(a_{N})^{\varepsilon}$

Let $a_{0}=0,a_{n+1}=2^{2^{a_{n}}}$, and divide the positive integers into
intervals $I_{h}=\left(a_{h},a_{h+1}\right]$ for $h\geq 0$. As we have stated,
there exists a subset $J_{h}\subset I_{h}$ of density at least $c(r)>0$ such
that $A_{h}=\\{x_{k}\ \big{\rvert}\ k\in J_{h}\\}$ satisfies
$\absolutevalue{A_{h}-A_{h}}\leq C(r)\absolutevalue{I_{h}}$. The desired
subsequence $n_{k}$ be the union of all the $J_{h}$.

Splitting the set of differences into differences where both elements are
contained in $I_{N-1}$ (which are the majority) and differences where this is
not the case we get that

$\\#\left\\{x_{n_{a}}-x_{n_{b}}\ \big{\rvert}\ 1\leq n_{a},n_{b},\leq
a_{N}\right\\}\leq\\#\left\\{x_{n_{a}}-x_{n_{b}}\ \big{\rvert}\
a_{N-1}<n_{a},n_{b}\leq a_{N}\right\\}+2a_{N-1}a_{N}\ll$ $\ll
C(r)a_{N}+a_{N}\log\log(a_{N})\ll a_{N}\log(a_{N})^{\varepsilon}$

as required.

Of course, under weaker assumptions on the additive energy in Theorem 14.3 we
can get a correspondingly weaker result regarding the subsequence, using
exactly the same techniques.

A natural question is how to extend this result to real-valued sequences.
Firstly, an important concept will be the covering number of a set. Given a
set $A\subset\mathbb{R}$ and a scale $\gamma>0$, define the covering number of
a set to be

$\mathrm{cov}\left(A,\gamma\right)=\inf_{B\subset\mathbb{R}}\left\\{\absolutevalue{B}\
\big{\rvert}\ \forall a\in A\ \exists b\in
B:\absolutevalue{a-b}<\gamma\right\\}$

that is, the smallest number of balls with radius $\gamma$ that are needed to
cover $A$. This number essentially measures the size of the set
$\absolutevalue{A}$ if we say that two numbers which are of difference
$\ll\gamma$ are equivalent, so this is a ”fuzzy” notion of size, and $\gamma$
is the scale of fuzziness which we permit. We can define the additive energy
of a set with respect to a certain scale $\gamma$ by

$E^{*}(A,\gamma)=\\#\\{(a,b,c,d)\in A^{4}\ \big{\rvert}\
\absolutevalue{a+b-c-d}<\gamma\\}$

Replacing each element $a\in A$ with $\gamma\lfloor\frac{a}{\gamma}\rfloor$
and then using the above argument we see that given a set $A$ such that
$E^{*}(A,\gamma)\gg N^{3}$ there is a subset $A^{\prime}\subset A$ such that
$\mathrm{cov}\left(A^{\prime}-A^{\prime},2\gamma\right)\leq
K^{\mathcal{O}(1)}\absolutevalue{A}$. Using the same argument we did to prove
Theorem 14.3 we see that as before given a sequence $x_{n}$ of real numbers
with large additive energy of scale $\gamma$, there is a subsequence
$x_{n_{k}}$ such that along a certain subsequence $N_{k}\to\infty$ the
difference set has a small size of scale $2\gamma$.

Now, we can ask the following question: given a sequence $x_{n}$ whose
difference set is of small size (that is,
$\mathrm{cov}\left(X_{N}-X_{N},\gamma\right)\ll N$), can we say that the
metric minimal gap of $x_{n}$ is large? This is exactly the content of
Theorems 1.10 and 1.12, which we will now prove, and once again restate them
for convenience:

######  Theorem 14.4.

For any sequence of real numbers $x_{n}$ such that
$x_{n+1}-x_{n}\geq\frac{c\log n}{n}$ for some constant $c>0$, if for some
$\varepsilon_{0}>0$ we have

$\mathrm{cov}\left(X_{N}-X_{N},\gamma\right)\ll N$

where $\gamma=\frac{1}{N\log(N)^{1+\varepsilon_{0}}}$, we have that for every
$\varepsilon>0$ and for almost all $\alpha\in\mathbb{R}$ that

$\delta_{\min}^{\alpha}(x_{n})\gg\frac{1}{N\log(N)^{1+\varepsilon}}$

######  Theorem 14.5.

There exists a sequence $\varepsilon_{n}\downarrow 0$ such that the sequence
$x_{n}=n-\frac{1}{n^{1-\varepsilon_{n}}}$ satisfies for all $\varepsilon>0$
and for almost all $\alpha\in\mathbb{R}$ that

$\delta_{\min}^{\alpha}(N)\ll\frac{1}{N^{2-\varepsilon}}$

where the implied constant depends on $\alpha$ and $\varepsilon$.

Theorem 14.4 follows from what we have said about integer-valued sequences:
take a set $B_{N}$ of size $\ll N$ such that for every $m,n\leq N$ we have
$\absolutevalue{x_{m}-x_{n}-b}<\frac{1}{N\log(N)^{1+\varepsilon_{0}}}$ for
some $b\in B$. Then, suppose that

$\delta_{\min}^{\alpha}(x_{n},N)<\frac{1}{N\log(N)^{1+\varepsilon}}$

that is, there exist some $m\neq n\leq N$ such that
$\norm{\alpha(x_{m}-x_{n})}<\frac{1}{N\log(N)^{1+\varepsilon}}$, and so
$\norm{\alpha b}<\frac{\absolutevalue{\alpha}+1}{N\log(N)^{1+\varepsilon}}$.
Now using the same arguments as in sections 2 and 3 shows that this happens
for a set of $\alpha$ of measure 0.

Theorem 14.4 combined with our argument above immediately proves Corollary
1.11.

Now we will prove Theorem 14.5. Obviously for any such sequence where
$\varepsilon_{n}\to 0$, the set of differences is
$\frac{1}{N^{1-\varepsilon}}$ close to the set of differences of $x_{n}=n$,
which is of linear size, and so we just need to show that the minimal gap is
sufficiently small. In our sequence we will let $\varepsilon_{n}=\eta_{k}$ be
constant on intervals of the form $\left(4^{k-1},4^{k}\right]$ and $\eta_{k}$
will decay slowly as $k\to\infty$. Throughout, we will assume that $\eta_{k}$
is small. We will show that $\delta_{\min}^{\alpha}(N)$ is small for almost
all $\alpha\in[-C,C]$ for any $C>0$, and therefore it is small for almost all
$\alpha\in\mathbb{R}$. Notice that
$\delta_{\min}^{\alpha}(N)\leq\delta_{\min}^{\alpha}\left(4^{\lfloor\log_{4}(N)\rfloor}\right)$.
Therefore, to show that $\delta_{\min}^{\alpha}(N)$ is small for all $N$ it is
enough to look at the case where $N=4^{k}$. Clearly,

$\delta_{\min}^{\alpha}(N)=\min\left\\{\norm{\alpha(x_{n}-x_{m})}\
\big{\rvert}\ 1\leq m\neq n\leq
N\right\\}\leq\min\left\\{\norm{\alpha(x_{n}-x_{m})}\ \big{\rvert}\
\frac{N}{4}<m\neq n\leq N\right\\}=$
$\min\left\\{\norm{\alpha\left(n-m-\frac{1}{n^{1-\eta_{k}}}+\frac{1}{m^{1-\eta_{k}}}\right)}\
\big{\rvert}\ \frac{N}{4}<m\neq n\leq N\right\\}$

Next we will state a useful lemma:

######  Lemma 14.6.

For all $\eta>0$ small enough (say $\eta<\frac{1}{10}$), for any $\alpha$ such
that there exist coprime integers $a,q,\ 0<q<\frac{N}{3}$ that satisfy

$\absolutevalue{\alpha-\frac{a}{q}}\leq\frac{3}{qN}$

and

$\absolutevalue{\alpha-\frac{a}{q}}\geq\frac{9}{2q^{1+\frac{1}{1-\eta}}}$

we have

$\min\left\\{\norm{\alpha\left(n-m-\frac{1}{n^{1-\eta}}+\frac{1}{m^{1-\eta}}\right)}\
\big{\rvert}\ \frac{N}{4}<m\neq n\leq N\right\\}\leq\frac{8}{N^{2-\eta}}$

For now assume this lemma, and we will prove it later. By Dirichlet’s
approximation theorem,, for all $\alpha$ there exist some coprime integers
$a,q,\ 0<q<\frac{N}{3}$ such that

$\absolutevalue{\alpha-\frac{a}{q}}\leq\frac{3}{qN}$

By Khinchin’s theorem, for almost all $\alpha$, for all $q$ large enough we
have $\absolutevalue{\alpha-\frac{a}{q}}\geq\frac{9}{2q^{2}\log(q)^{2}}$, and
so if $\absolutevalue{\alpha-\frac{a}{q}}<\frac{9}{2q^{1+\frac{1}{1-\eta}}}$
we must have

$q^{\frac{\eta}{1-\eta}}<\log(q)^{2}$

Taking logarithms this is equivalent to

$\frac{\log q}{\log\log q}<\frac{2(1-\eta)}{\eta}$

and therefore

$\log q<\frac{4}{\eta}\log\left(\frac{2}{\eta}\right)$

which means that

$q<\left(\frac{2}{\eta}\right)^{\frac{4}{\eta}}$

However, notice that the set of $\alpha\in[-C,C]$ such that
$\absolutevalue{\alpha-\frac{a}{q}}\leq\frac{1}{qN}$ for some
$q<\left(\frac{2}{\eta}\right)^{\frac{4}{\eta}}$ is of measure at most

$\frac{4C\left(\frac{2}{\eta}\right)^{\frac{4}{\eta}}}{N}$

simply by bounding the measure independently for each $q$. That is, for all
$\alpha\in[-C,C]$ up to a set of measure at most
$\frac{4C\left(\frac{2}{\eta}\right)^{\frac{4}{\eta}}}{N}$ we have

$\min\left\\{\norm{\alpha\left(n-m-\frac{1}{n^{1-\eta}}+\frac{1}{m^{1-\eta}}\right)}\
\big{\rvert}\ \frac{N}{4}<m\neq n\leq N\right\\}\leq\frac{8}{N^{2-\eta}}$

Now we return to our case, where $\eta=\eta_{k}$. Applying Lemma 14.6 we see
that for all $\alpha\in[-C,C]$ up to a set of measure at most
$\frac{2C\left(\frac{2}{\eta_{k}}\right)^{\frac{4}{\eta_{k}}}}{4^{k}}$ we have

$\delta_{\min}^{\alpha}\left(4^{k}\right)\leq\frac{8}{4^{k(2-\eta_{k})}}$

If we let $\eta_{k}$ tend to 0 sufficiently slowly so that

$\sum_{k=1}^{\infty}\frac{\left(\frac{2}{\eta_{k}}\right)^{\frac{4}{\eta_{k}}}}{4^{k}}<\infty$

then by Borel-Cantelli, for almost all $\alpha\in[-C,C]$ we will have

$\delta_{\min}^{\alpha}(4^{k})\ll\frac{1}{4^{k(2-o(1))}}$

and therefore for almost all $\alpha\in[-C,C]$ we have

$\delta_{\min}^{\alpha}(N)\ll\frac{1}{N^{2-o(1)}}$

as $N\to\infty$, exactly as required. $\blacksquare$.

Proof of Lemma 14.6:

Take some number $\frac{N}{3}\leq k\leq\frac{2N}{3}$, and let us look at all
pairs of $\frac{N}{4}\leq m<n\leq N$ such that $n-m=k$. If $n,m$ is such a
pair then obviously so is $n+1,m+1$. Notice that

$\left(x_{n}-x-m\right)-\left(x_{n+1}-x_{m+1}\right)=\left(\frac{1}{m^{1-\eta}}-\frac{1}{(m+1)^{1-\eta}}\right)-\left(\frac{1}{n^{1-\eta}}-\frac{1}{(n+1)^{1-\eta}}\right)$

This is a positive number which is at most

$\frac{1}{m^{1-\eta}}-\frac{1}{(m+1)^{1-\eta}}\leq\frac{1-\eta}{m^{2-\eta}}<\frac{16}{N^{2-\eta}}$

Therefore, the points $x_{n}-x_{m}$ for $n-m=k$ fixed are contained in the
interval

$\left[k+\left(\frac{1}{N-k}\right)^{1-\eta}-\left(\frac{1}{N}\right)^{1-\eta},k+\left(\frac{4}{N}\right)^{1-\eta}-\left(\frac{1}{k+\frac{N}{4}}\right)^{1-\eta}\right]$

Notice that for some $t$ in this interval we have $\alpha t\in\mathbb{Z}$,
then there exist some $m,n$ such that
$\absolutevalue{x_{n}-x_{m}-t}<\frac{8}{N^{2-\eta}}$ and therefore
$\norm{\alpha(x_{n}-x_{m})}<\frac{8\absolutevalue{\alpha}}{N^{2-\eta}}$, which
is a small gap. Now we will find such a $t$.

Notice that for $\frac{N}{3}\leq k\leq\frac{2N}{3}$ (and for $\eta$ close
enough to 0) the above interval contains the interval

$\left[k+\frac{\left(\frac{3}{2}\right)^{1-\eta}-1}{N^{1-\eta}},k+\frac{4^{1-\eta}-\left(\frac{12}{11}\right)^{1-\eta}}{N^{1-\eta}}\right]$

Notice that for $\eta$ small enough this contains the interval

$\left[k+\frac{1}{3N^{1-\eta}},k+\frac{7}{3N^{1-\eta}}\right]$

We want $\alpha t$ to be small for some $t$ in one of these intervals, or
equivalently $\alpha
k\in\left[-\frac{\alpha(c_{1}+c_{2})}{N^{1-\eta}},-\frac{\alpha
c_{1}}{N^{1-\eta}}\right]$ for some $k$, where this interval is taken $\bmod\
1$. Writing $k=\frac{N}{3}+\ell$ for some $0\leq\ell\leq\frac{N}{3}$ then what
we need is

$\alpha\ell\in\left[-\frac{N}{3}\alpha-\frac{\alpha}{3N^{1-\eta}},-\frac{N}{3}\alpha-\frac{7\alpha}{3N^{1-\eta}}\right]$

The exact structure of the right hand side does not matter, what matters is
the fact that it is an interval of length
$\frac{2\absolutevalue{\alpha}}{N^{1-\eta}}$ for some absolute constant
$c_{2}$. To emphasize this, let us write that interval as

$\left[y,y+\frac{2}{N^{1-\eta}}\right]$

for some $y$. Recall that we are given coprime integers $a,q$ such that
$0<q<\frac{N}{3}$ and

$\absolutevalue{\alpha-\frac{a}{q}}\leq\frac{3}{qN}$

For all $\ell\leq\frac{N}{3}$ we have

$\absolutevalue{\alpha\ell-\frac{a\ell}{q}}\leq\frac{3\ell}{qN}\leq\frac{1}{q}$

and so if for some $0\leq\ell\leq\frac{N}{3}$ we have

$\frac{a\ell}{q}\in\left[y+\frac{1}{q},y+\frac{2}{N^{1-\eta}}-\frac{1}{q}\right]$

then we would finish. Because $a,q$ are coprime, as $\ell$ runs from $0$ to
$\frac{N}{3},\ \frac{a\ell}{q}$ runs over all fractions with denominator $q$.
Therefore, if the interval
$\left[y+\frac{1}{q},y+\frac{2}{N^{1-\eta}}-\frac{1}{q}\right]$ is of length
at least $\frac{1}{q}$ it contains such a fraction which is what we need to
prove. Therefore it is sufficient to show that

$q\geq\frac{3N^{1-\eta}}{2}$

or equivalently

$\left(\frac{2q}{3}\right)^{\frac{1}{1-\eta}}\geq\frac{2q^{\frac{1}{1-\eta}}}{3}\geq
N$

Assume to the contrary that this is not the case. Then we have

$\absolutevalue{\alpha-\frac{a}{q}}\leq\frac{3}{qN}<\frac{9}{2q^{1+\frac{1}{1-\eta}}}$

a contradiction to what we know about $q$. $\blacksquare$.

Notice that Theorem 1.10 is not quite a full converse to 1.11: in order to
prove Corollary 1.11 we used the fact that every sequence with large additive
energy has a relatively dense subsequence with a small difference set, and
then that every sequence with a small difference set has a large minimal gap.
In Theorem 1.10 we showed that the second part of this proof fails in the case
where $\gamma=\frac{1}{N^{1-\varepsilon}}$, as we showed that there exists a
sequence with small difference set and large minimal gap. However,
theoretically our sequence $n-\frac{1}{n^{1-\varepsilon_{n}}}$ could have a
subsequence such as described in 1.11, just we would have to choose the
subsequence more carefully. We conjecture that this is not the case.
Specifically,

Conjecture: Every subsequence of $n-\frac{1}{n^{1-\varepsilon_{n}}}$ with
lower positive density has a metric minimal gap of size
$\frac{1}{N^{2-o(1)}}$.

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