# A maximal extension of the Bloom-Maynard bound for sets with no square
differences

Nuno Arala  Mathematics Institute, Zeeman Building, University of Warwick,
Coventry CV4 7AL<EMAIL_ADDRESS>
###### Abstract.

We show that if $h\in\mathbb{Z}[x]$ is a polynomial of degree $k$ such that
the congruence $h(x)\equiv 0\pmod{q}$ has a solution for every positive
integer $q$, then any subset of $\\{1,2,\ldots,N\\}$ with no two distinct
elements with difference of the form $h(n)$, with $n$ positive integer, has
density at most $(\log N)^{-c\log\log\log N}$, for some constant $c$ that
depends only on $k$. This improves on the best bound in the literature, due to
Rice, and generalizes a recent result of Bloom and Maynard.

###### 2010 Mathematics Subject Classification:

Primary: 11B30, 11P55.

###### Contents

  1. 1 Introduction
  2. 2 Notation and conventions
  3. 3 The auxiliary polynomials
    1. 3.1 Construction of the polynomials
    2. 3.2 Sieving good residue classes
  4. 4 Exponential sum estimates
  5. 5 The density increment
  6. 6 Additive Energy bounds and end of the proof
  7. 7 Acknowledgements

##  1\. Introduction

The following question was asked by Lovász: if $A\subseteq\mathbb{N}$ is a set
such that no two different elements of $A$ differ by a square, must $A$ have
upper density $0$? Here, as is customary, we define the upper density of a
subset $A\subseteq\mathbb{N}$ as

$\limsup_{N\to\infty}\frac{|A\cap\\{1,\ldots,N\\}|}{N}\text{.}$

This question was settled by Furstenberg [3] and Sárközy [11], who proved
independently that the answer is affirmative; the former used ergodic theory,
while the latter employed a Fourier-analytic approach, based on the Hardy-
Littlewood circle method. In recent years there has been interest in
strengthening this result by obtaining better bounds for $|A|$, where
$A\subseteq\\{1,\ldots,N\\}$ is a set with no square differences. For example,
Sárközy’s original argument shows directly that, for such $A$,

$\frac{|A|}{N}\ll\frac{(\log\log N)^{2/3}}{(\log N)^{1/3}}\text{.}$ (1.1)

This has since been much improved. For example, Pintz, Steiger and Szemerédi
[8] showed that, under the same assumptions,

$\frac{|A|}{N}\ll(\log N)^{-c\log\log\log\log N}$ (1.2)

for some absolute constant $c$. More recently, Bloom and Maynard [2] devised a
powerful bound for the additive energy of a set of rational numbers with small
denominators, and applied it to improve the above bound to

$\frac{|A|}{N}\ll(\log N)^{-c\log\log\log N}$ (1.3)

for some absolute constant $c$. This is the best bound up to date.

It is only natural to wonder what happens if one changes the set of forbidden
differences, say by replacing it with the set $h(\mathbb{N})$ for some
polynomial $h\in\mathbb{Z}[x]$. Lovász’s problem concerns the case when
$h(x)=x^{2}$. It is quickly seen that the answer to Lovász’s question is no
longer affirmative in general. For example, say we take $h(x)=x^{2}+1$. One
may then consider the subset $3\mathbb{N}$ of $\mathbb{N}$, which has upper
density $1/3>0$, while on the other hand the difference between any two of its
elements is divisible by $3$ and hence not of the form $x^{2}+1$.

The failure of the above polynomial $h$ to yield an affirmative answer to
Lovász’s generalized question is, as seen above, due to the existence of an
integer $q$ (namely, $3$) such that $q$ does not divide $h(n)$ for any integer
$n$. Indeed, if that is the case, then $q\mathbb{N}$ is a subset of
$\mathbb{N}$ with positive upper density such that no two of its elements
differ by an element of $h(\mathbb{N})$. Polynomials for which no such $q$
exists are called _intersective polynomials_ in the literature. We formalize
the definition below.

######  Definition 1.1 (Intersective polynomial).

A polynomial $h\in\mathbb{Z}[x]$ is called _intersective_ if for any positive
integer $q$ the congruence

$h(n)\equiv 0\pmod{q}$

has an integer solution.

From the remarks above, it is clear that, in order for there to be any hope of
extending the Furstenberg-Sarközy’s theorem, or any of its direct descendants,
to the polynomial $h$, one must impose that $h$ is intersective. This turns
out to be the only restriction needed to ensure an affirmative answer to
Lovász’s question, as shown by Kamae and Mendès France [5]. A result of the
quality of (1.1) has been obtained, for general intersective $h$, by Lucier
[6], with improvements by Lyall and Magyar [7] and Rice [10]. Balog, Pelikan,
Pintz and Szemerédi [1] proved the corresponding version of (1.2) for the
polynomial $h(x)=x^{k}$. Recent work of Rice generalizes this, proving a
result of the quality of (1.2) for a general intersective $h$.

The goal of the present work is to generalize (1.3) for arbitrary intersective
polynomials. More precisely, we will prove the following:

######  Theorem 1.2.

Let $k$ be a positive integer. Then there exists a constant $c=c(k)$ such
that, for any intersective polynomial $h\in\mathbb{Z}[x]$ of degree $k$, the
following holds: for any positive integer $N$ and any subset
$A\subseteq\\{1,\ldots,N\\}$ such that the equation $a-b=h(n)$ does not have
any solutions with $a,b\in A$, $a\neq b$ and $n\in\mathbb{N}$, we have

$\frac{|A|}{N}\ll_{h}(\log N)^{-c\log\log\log N}\text{.}$

There are several obstacles that one encounters when trying to extend the
argument from the $h(n)=n^{2}$ case to the general case. In the Fourier-
analytic argument employed by Bloom and Maynard in [2], estimates for
quadratic Gauß sums play a role, and it is vital for the argument that square
root cancellation is available for such sums. More precisely, if $a$ and $q$
are coprime positive integers, one has

$\sum_{s=0}^{q-1}e\left(\frac{as^{2}}{q}\right)\ll q^{1/2}\text{.}$

However, for a general polynomial $h$ of degree $k$, the estimate

$\sum_{s=0}^{q-1}e\left(\frac{ah(s)}{q}\right)\ll q^{1-1/k}$

is best possible. To circumvent this, we adapt an argument of Rice, sieving
out admissible residue classes $s\pmod{q}$ such that square root cancellation
holds for the modified version of the previous sum where we only sum over
these admissible residues.

The other obstacle comes from the density increment strategy that Bloom and
Maynard employ, where the equality $(qn)^{2}=q^{2}n^{2}$ plays an important
role. To get around this, one needs to adapt the density increment argument in
such a way that the intersective polynomial $h$ under consideration changes at
every step of the density increment strategy. This will, of course, require
that the bounds we obtain exhibit some uniformity with respect to $h$.

##  2\. Notation and conventions

Let $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ be the circle group, and let
$\|\cdot\|$ denote the usual norm in $\mathbb{T}$. For $1\leq a\leq q$ with
$\mathrm{gcd}(a,q)=1$, set

$\mathfrak{M}_{a,q}\left(N,K\right)=\left\\{\gamma\in\mathbb{T}:\left\|\gamma-\frac{a}{q}\right\|\leq\frac{K}{N}\right\\}\text{.}$

For a choice of parameter $Q$, we then define the _major arcs_

$\mathfrak{M}(N,K,Q)=\bigcup_{\begin{subarray}{c}(a,q)=1\\\ q\leq
Q\end{subarray}}\mathfrak{M}_{a,q}\left(N,K\right)$

and the _minor arcs_

$\mathfrak{m}(N,K,Q)=\mathbb{T}\setminus\mathfrak{M}_{q}(N,K,Q)\text{.}$

Given a polynomial $h\in\mathbb{Z}[x]$, we say that a set $A$ of integers is
_$h$ -free_ if for any $a\neq b$ in $A$ the difference $a-b$ does not lie in
$h(\mathbb{N})$. We use the notation $e(x)$ for $e^{2\pi ix}$. A summation
sign with an asterisk $\sum^{\ast}$ denotes a sum over primitive residue
classes.

We use the notation $\widehat{f}$ for the Fourier transform of $f$; so if
$f:\mathbb{Z}\to\mathbb{C}$ is a compactly supported function, we define its
Fourier transform $\widehat{f}:\mathbb{T}\to\mathbb{C}$ by

$\widehat{f}(\gamma)=\sum_{n\in\mathbb{Z}}f(n)e(n\gamma)\text{.}$

The convolution of two (compactly supported, say) functions
$f,g:\mathbb{Z}\to\mathbb{C}$ is defined by

$(f\ast g)(n)=\sum_{a+b=n}f(a)g(b)\text{.}$

We will use the notation $\mathbf{1}_{A}$ for the indicator function of a set
$A\subseteq\mathbb{Z}$, so that

$\mathbf{1}_{A}(n)=\begin{cases}1&\text{ if }n\in A\\\ 0&\text{
otherwise.}\end{cases}$

In view of the result we want to prove (Theorem 1.2), for any fixed constant
$c$, given a $h$-free subset $A\subseteq\\{1,\ldots,N\\}$ with density
$\sigma$, one may assume that $\sigma\geq(\log N)^{-c\log\log\log N}$, for
otherwise the result follows. This will be assumed implicitly throughout the
paper. So, for example, the statement of Lemma 4.6 takes the form “Then either
$\sigma<(\log N)^{-\varepsilon\log\log\log N}$, or (…)” but in practice, in
applications of Lemma 4.6, we will always assume automatically that we are in
the second case. We may (and will) also assume that $N$ is larger than any
fixed constant.

The letters $c$ and $C$ will be used to denote arbitrary positive constants;
we make the usual abuse of notation of allowing them to denote different
constants at different times. We hope this will not cause any confusion.

##  3\. The auxiliary polynomials

###  3.1. Construction of the polynomials

As was mentioned the introduction, the simplicity of the polynomial $x^{2}$
makes it particularly suitable for an approach to the problem at hand via the
density increment strategy that has become pervasive in Arithmetic
Combinatorics over the years. If $A\subseteq[N]$ is a set with no square
differences and one manages to locate a progression $q[N^{\prime}]+r$ on which
$A$ has increased density, then by further splitting this progression into
progressions of modulus $q^{2}$ one finds another progression, say
$q^{2}[N^{\ast}]+s$, on which $A$ also has increased density. Defining

$A^{\ast}=\\{x\in[N^{\ast}]:q^{2}x+s\in A\\}$

one sees that $A^{\ast}$ has no square differences, for if $a\neq b$ are in
$A^{\ast}$ and $a-b=n^{2}$, then $q^{2}a+s$ and $q^{2}b+s$ are in $A$ and

$(q^{2}a+s)-(q^{2}b+s)=q^{2}(a-b)=(qn)^{2}\text{,}$

which contradicts the assumption that $A$ has no square differences. Hence we
obtain a new set $A^{\ast}\subseteq[N^{\ast}]$ with no square differences and
increased density.

For a more general intersective polynomial $h$, given a $h$-free set
$A\subseteq[N]$, if we locate a progression on which $A$ of modulus $q$ is
dense we cannot expect to obtain from it another set
$A^{\prime}\subseteq[N^{\prime}]$ of increased density with no nonzero
differences in $h(\mathbb{N})$. What we can do is to use a construction of
Lucier [6], which allows us to obtain a set $A^{\prime}\subseteq[N^{\prime}]$
of increased density with no nonzero differences in $h_{q}(\mathbb{N})$ for
some polynomial $h_{q}\in\mathbb{Z}[x]$. We explain this construction below.
We remark that the very existence of these auxiliary polynomials is precisely
the step in the argument where the intersectivity of $h$ is used.

Fix once and for all an intersective polynomial $h$ of degree $k$, without
loss of generality with positive leading coefficient. The condition that $h$
is intersective is equivalent to the statement that $h$ has a zero
$z_{p}\in\mathbb{Z}_{p}$ for every prime $p$. Let $m_{p}$ be the multiplicity
of $z_{p}$ as a zero of $h$. For each positive integer $d$ define $r_{d}$ to
be the only integer satisfying $-d<r_{d}\leq 0$ and

$r_{d}\equiv z_{p}\pmod{p^{\alpha}}\text{ whenever }p\text{ is a prime and
}p^{\alpha}\mid d\text{.}$

Clearly $r_{d}$ exists and is unique by the Chinese Remainder Theorem. We now
let $\lambda$ be the unique completely multiplicative function that satisfies
$\lambda(p)=p^{m_{p}}$ for every prime $p$. It is clear from the definition
that $d\mid\lambda(d)\mid d^{k}$ for any $d$. We are now ready to define our
auxiliary polynomials.

######  Definition 3.1 (Auxiliary polynomials).

For each $d\in\mathbb{N}$, we define

$h_{d}(x)=\frac{h(r_{d}+dx)}{\lambda(d)}\text{.}$

It is an easy matter to check from the definitions that $h_{d}$ has integer
coefficients. It is also immediate from the definition that
$\lambda(q)h_{q}(\mathbb{N})\subseteq h(\mathbb{N})$, which explains why these
auxiliary polynomials may act as a replacement for the “$q^{2}$ times a square
is a square” property alluded to above. In fact, more is true:

######  Proposition 3.2.

For any positive integers $d,q$,

$\lambda(q)h_{dq}(\mathbb{N})\subseteq h_{d}(\mathbb{N})\text{.}$

###### Proof.

Since by construction we have $r_{dq}\equiv r_{d}\pmod{d}$, we can write
$r_{dq}=r_{d}+ds$ for some $-q<s\leq 0$. For a positive integer $n$, we then
have

$\lambda(q)h_{dq}(n)=\lambda(q)\frac{h(r_{dq}+dqn)}{\lambda(dq)}=\frac{h(r_{dq}+dqn)}{\lambda(d)}=\frac{h(r_{d}+ds+dqn)}{\lambda(d)}=h_{d}(s+qn)\text{.}$

Since $s+qn>0$ (as $s>-q$), the result follows. ∎

We register here a straightforward but essential analytic property of the
auxiliary polynomials.

######  Proposition 3.3.

For any positive integer $d$, denote by $b_{d}$ and $J_{d}$ the leading
coefficient of $h_{d}$ and the sum of the absolute values of the coefficients
of $h_{d}$, respectively. Then $J_{d}\ll_{h}b_{d}$.

###### Proof.

Write $h(x)=a_{k}x^{k}+\cdots+a_{0}$. A binomial expansion shows that the the
coefficient of $x^{i}$ in $h_{d}(x)$ is

$\frac{1}{\lambda(d)}\sum_{j=i}^{k}a_{j}\binom{j}{i}r_{d}^{j-i}d^{i}\text{.}$
(3.1)

In particular, since $|r_{d}|<d$, it follows from the triangle inequality that

$J_{d}\ll_{h}\frac{d^{k}}{\lambda(d)}\text{.}$

But it also follows from (3.1) that $b_{d}=a_{d}d^{k}/\lambda(d)$, and the
result follows. ∎

###  3.2. Sieving good residue classes

As explained in the introduction, a fundamental obstacle one encounters when
trying to apply ideas inspired by the approach of Bloom and Maynard in [2] is
the lack of square root cancellation exhibited by exponential sums $\pmod{q}$
over values of a polynomial of degree at least $3$. Rice [9] has found an
ingenious way to address this problem. Building on the fact that square root
cancellation always holds for squarefree moduli (by the Chinese Remainder
Theorem and the Weil bound), he then leverages a classical “$p$-adic
stationary phase” argument and shows that if we sum only over a certain large
set of residue classes at which the derivative of the polynomial has
appropriate non-vanishing properties, then we do indeed attain square root
cancellation. Below we explain Rice’s construction of admissible residue
classes, which we will use, and on section §4 we give the relevant square root
cancellation bounds for exponential sums.

######  Definition 3.4.

Let $g\in\mathbb{Z}[x]$ and $Y>0$. For each prime $p$, let $\gamma(g;p)$ be
the smallest positive integer for which the derivative $g^{\prime}$ does not
vanish identically modulo $p^{\gamma(g;p)}$.

  1. (i)

We define

$W(g;Y)=\\{n\in\mathbb{Z}:g^{\prime}(n)\not\equiv
0\pmod{p^{\gamma(g;p)}}\text{ for all }p\leq Y\\}\text{.}$

  2. (ii)

Moreover, given also a positive integer $q$, we define

$W_{q}(g;Y)=\\{n\in\mathbb{Z}:g^{\prime}(n)\not\equiv
0\pmod{p^{\gamma(g;p)}}\text{ for all }p\leq Y\text{ for which
}p^{\gamma(g;p)}\mid q\\}$

which intuitively only captures the restrictions that can be read from
$n\pmod{q}$.

  3. (iii)

Given a prime $p$, we let $j(g;p)$ be the number of roots of $g^{\prime}$
modulo $p^{\gamma(p)}$.

  4. (iv)

In case the polynomial $g$ under consideration is one of the $h_{d}$ we
abbreviate $\gamma(g;p)$, $W(g;Y)$, $W_{q}(g;Y)$ and $j(g;p)$ by
$\gamma_{d}(p)$, $W_{d}(Y)$, $W^{q}_{d}(Y)$, and $j_{d}(p)$ respectively.

If $g(x)=a_{0}+a_{1}x+\cdots+a_{k}x^{k}$, define

$\mathrm{cont}(g)=\mathrm{gcd}(a_{1},\ldots,a_{k})\text{.}$

In [9], section 2.5, Rice proves two key properties of the auxiliary
polynomials $h_{d}$ that will be vital to the application of the exponential
sum estimates in the next section. One is the fact that $\mathrm{cont}(h_{d})$
is absolutely bounded; hence in practice all the constants that occur in the
exponential sum bounds in the next section that depend on $\mathrm{cont}(g)$
are for our purposes absolute constants provided that we only work with the
polynomials $h_{d}$. The other is the fact that the set of moduli $q$ for
which $h_{d}^{\prime}$ vanishes identically modulo $q$ is contained in a
finite set independent of $d$. This implies the following statement.

######  Proposition 3.5.

Let

$w_{d}(Y)=\prod_{p\leq Y}\left(1-\frac{j_{d}(p)}{p^{\gamma_{d}(p)}}\right)$

be the expected proportion of integers in $W_{d}(Y)$. Then we have

$w_{d}(Y)\gg_{h}(\log Y)^{1-k}\text{.}$

###### Proof.

From our previous remark, we deduce that for all but finitely many primes $p$
we have $\gamma_{d}(p)=1$ for every $d$. Moreover, $h_{d}^{\prime}$ has degree
$k-1$ and hence has at most $k-1$ roots modulo $p$. It follows that the factor
$1-j_{d}(p)/p^{\gamma_{d}(p)}$ is at least $1-(k-1)/p$ for all but finitely
many primes. Hence, since all factors are clearly nonzero,

$w_{d}(Y)\gg_{h}\prod_{k\leq p\leq Y}\left(1-\frac{k-1}{p}\right)\gg(\log
Y)^{1-k}\text{,}$

by a classical result of Mertens. ∎

In fact Rice makes precise in [9], via a standard Brun sieve argument, the
intuition that $w_{d}(Y)$ should be close to the proportion of integers in
$[1,X]$ lying in $W_{d}(Y)$, provided that $X$ is sufficiently large in terms
of $Y$. More precisely, he proves the following:

######  Lemma 3.6.

Let $g\in\mathbb{Z}[x]$ and $X,Y>0$ with $c\log X\geq\log Y\log\log Y$. Then,

$|[1,X]\cap W(g;Y)|=X\prod_{p\leq
Y}\left(1-\frac{j(g;p)}{p^{\gamma(g;p)}}\right)+O\left(Xe^{-c\frac{\log
X}{\log Y}}\right)\text{,}$

where $c>0$ depends only on $\mathrm{deg}(g)$ and the collection of moduli for
which $g^{\prime}$ vanishes identically.

###### Proof.

This is [9], Proposition 2.4. ∎

##  4\. Exponential sum estimates

In this section we record some estimates for exponential sums arising from a
polynomial $g$ of degree $k$, with the main goal being to establish that
certain exponential sums are small on appropriate minor arcs. We start by
recalling some results of Rice [9] on which we will build.

######  Lemma 4.1.

Suppose $k\in\mathbb{N}$,
$g(x)=a_{0}+a_{1}x+\cdots+a_{k}x^{k}\in\mathbb{Z}[x]$, and let
$J=|a_{0}|+\cdots+|a_{k}|$. If $X,Y>0$, $a,q\in\mathbb{N}$,
$\alpha=a/q+\beta$, and $c\log(X/q)\geq\log Y\log\log Y$, then

$\displaystyle\sum_{\begin{subarray}{c}n=1\\\ n\in
W(g;Y)\end{subarray}}^{X}g^{\prime}(n)e(g(n)\alpha)$
$\displaystyle=\frac{1}{q}\prod_{\begin{subarray}{c}p\leq Y\\\
p^{\gamma(g;p)}\nmid
q\end{subarray}}\left(1-\frac{j(g;p)}{p^{\gamma(g;p)}}\right)\sum_{\begin{subarray}{c}s=0\\\
s\in
W^{q}(g;Y)\end{subarray}}^{q-1}e\left(\frac{ag(s)}{q}\right)\int_{0}^{X}g^{\prime}(x)e(g(x)\beta)dx$
$\displaystyle+O\left(kJX^{k}e^{-c\frac{\log\left(\frac{X}{q}\right)}{\log
Y}}(1+JX^{k}|\beta|)\right)\text{,}$

where $c=c(k,\mathrm{cont}(g))>0$.

###### Proof.

This is Lemma 4.1 in [9]. ∎

######  Lemma 4.2.

If $g\in\mathbb{Z}[x]$ with $\mathrm{deg}(g)=k\geq 2$, $a,q\in\mathbb{N}$ with
$\mathrm{gcd}(a,q)=1$, and $Y>0$, then

$\left|\sum_{\begin{subarray}{c}s=0\\\ s\in
W^{q}(g;Y)\end{subarray}}^{q-1}e\left(\frac{ag(s)}{q}\right)\right|\ll_{k}\mathrm{gcd}(\mathrm{cont}(g),q)^{3}C^{\omega(q)}\begin{cases}q^{1/2}\text{
if }q\leq Y\\\ q^{1-1/k}\text{ for all }q\end{cases}\text{,}$

where $C=C(k)$ and $\omega(q)$ is the number of distinct prime factors of $q$.

###### Proof.

This is Lemma 4.3 in [9]. ∎

######  Lemma 4.3.

Suppose $k\in\mathbb{N}$,
$g(x)=a_{0}+a_{1}x+\cdots+a_{k}x^{k}\in\mathbb{Z}[x]$ with $a_{k}>0$. Suppose
further that $X,Y,Z\geq 2$, $YZ\leq X$, and $a,q\in\mathbb{N}$ with
$\mathrm{gcd}(a,q)=1$. If $|\alpha-a/q|<1/q^{2}$, then

$\left|\sum_{\begin{subarray}{c}n=1\\\ n\in
W(g;Y)\end{subarray}}^{X}e(g(n)\alpha)\right|\ll_{k}\mathrm{cont}(g)^{6}(\log
Y)^{ek}X\left(e^{-\frac{\log Z}{\log
Y}}+\left(a_{k}\log^{k^{2}}(a_{k}qX)\left(q^{-1}+\frac{Z}{X}+\frac{qZ^{k}}{a_{k}X^{k}}\right)\right)^{2^{-k}}\right)\text{.}$

###### Proof.

This is Lemma 4.5 in [9]. ∎

We will use these to prove the following key estimate. In the following, set
$\rho=1/2^{k+2}k^{2}$.

######  Lemma 4.4.

Suppose $N\gg_{h}1$, and let $r>0$ be arbitrary. Let $d$ be such that $d\ll
N^{\rho}$. Set $M=\left\lfloor(N/b_{d})^{1/k}\right\rfloor$. Then either
$\sigma\leq(\log N)^{-\log\log\log N/4}$, or there exists a constant $\kappa$
(possibly depending on $r$) such that, setting $Y=\sigma^{-(k+2)}$,

$\left|\frac{1}{w_{d}(Y)N}\sum_{\begin{subarray}{c}m=1\\\ n\in
W_{d}(Y)\end{subarray}}^{M}h_{d}^{\prime}(m)e\left(h_{d}(m)\gamma\right)\right|\leq
r\sigma$

whenever $\gamma\in\mathfrak{m}(N,\kappa/\sigma,\kappa/\sigma^{k+1})$.

######  Remark 4.5.

The denominator $w_{d}(Y)N$ is the trivial bound for the exponential sum, so
that the Lemma says something about how much we can improve over the trivial
bound.

###### Proof.

Let $S$ be the expression we wish to bound. Throughout the argument, we
abbreviate $w_{d}(Y)$ by $w$, and let $Z=e^{(\log\log N)^{3}}$. Dirichlet’s
approximation theorem yields a positive integer $q$, with $1\leq q\leq
N/Z^{2k}$, and an integer $a$ with $1\leq a\leq q$ coprime to $q$, such that

$\left|\gamma-\frac{a}{q}\right|<\frac{Z^{2k}}{qN}\text{.}$

Write $\gamma=a/q+\beta$. We split the argument into several cases.

Suppose first that $q\leq N^{1/2k}$. Recalling that $J_{d}\ll b_{d}$
(Proposition 3.3) and that hence $J_{d}M^{d}\ll N$, Lemma 4.1 together with
Proposition 3.5 yields

$S=\frac{w_{q}}{qN}\sum_{\begin{subarray}{c}s=0\\\ s\in
W_{d}^{q}(Y)\end{subarray}}^{q-1}e\left(\frac{ah_{d}(s)}{q}\right)\int_{0}^{M}h_{d}^{\prime}(x)e(h_{d}(x)\beta)dx+O_{h}\left((\log
Y)^{k-1}e^{-c\frac{\log(M/q)}{\log Y}}Z^{2k}\right)\text{,}$

where

$w_{q}=\prod_{\begin{subarray}{c}p\leq Y\\\ p^{\gamma_{d}(p)\mid
q}\end{subarray}}\left(1-\frac{j_{d}(p)}{p^{\gamma_{d}(p)}}\right)^{-1}\text{.}$

Since $q\leq N^{1/2k}$ we have $\log(M/q)\asymp\log(N)$, and hence

$(\log Y)^{k-1}e^{-c\frac{\log(M/q)}{\log
Y}}Z^{2k}\ll(\log(\sigma^{-1}))^{k-1}e^{-c\frac{\log(N)}{\log(\sigma^{-1})}}e^{2k(\log\log
N)^{3}}$

and if $\sigma^{-1}\leq(\log N)^{\log\log\log N/4}$, then the above is
$O(\sigma^{t})$ for any $t>0$, and in particularly is $o(\sigma)$. Hence we
can focus on controlling the main term. For that, we observe that

$\left|\int_{0}^{M}h_{d}^{\prime}(x)e(h_{d}(x)\beta)dx\right|=\left|\int_{h_{d}(0)}^{h_{d}(M)}e(y\beta)dy\right|\ll\min\\{N,|\beta|^{-1}\\}\text{,}$

provided that $h_{d}(0)\ll N$; but since by definition
$h_{d}(0)=h(r_{d})/\lambda(d)\ll d^{k-1}$ (recall that $|r_{d}|<d$), this
condition certainly holds, taking into account that we are assuming $d\ll
N^{\rho}$. Hence we have

$S\ll\frac{w_{q}}{qN}\sum_{\begin{subarray}{c}s=0\\\ s\in
W_{d}^{q}(Y)\end{subarray}}^{q-1}e\left(\frac{ah_{d}(s)}{q}\right)\min\\{N,|\beta|^{-1}\\}+o_{h,N}(\sigma)\text{.}$
(4.1)

We now split into two further sub-cases.

If $q>\kappa/\sigma^{k+1}$, then Lemma 4.2, together with the observation that
$w_{q}\ll_{h}2^{\omega(q)}$, guarantees that the exponential sum is bounded
above by $C^{\omega(q)}q^{1-1/k}$ for some constant $C$, which implies that

$\displaystyle S$
$\displaystyle=O\left(\frac{C^{\omega(q)}}{q^{1/k}}\right)+o_{h,N}(\sigma)$
$\displaystyle=O\left(\frac{1}{q^{1/(k+1)}}\right)+o_{h,N}(\sigma)$

by using the trivial bound $C^{\omega(q)}\ll q^{1/k-1/(k+1)}$. Using
$q>\kappa/\sigma^{k+1}$ yields now

$S=O\left(\sigma/\kappa^{1/(k+1)}\right)+o_{h,N}(\sigma)$

which gives the desired result in this case for an appropriate choice of
$\kappa$.

If instead $q\leq\kappa/\sigma^{k+1}$, then the assumption that
$\gamma\in\mathfrak{m}(N,\kappa/\sigma,\kappa/\sigma^{k+1})$ guarantees that
$|\beta|\geq\frac{\kappa}{\sigma N}$. Bounding the exponential sum trivially
this time and bounding $\min\\{N,|\beta|^{-1}\\}$ from above by
$|\beta|^{-1}$, we obtain

$S=O\left(\frac{|\beta|^{-1}}{N}\right)+o_{h,N}(\sigma)=O\left(\frac{\sigma}{\kappa}\right)+o_{h,N}(\sigma)\text{.}$

This gives the desired result in this case, by choosing an appropriate value
of $\kappa$.

We now tackle the case when $q>N^{1/2k}$. Using partial summation, we may
write

$S=\frac{1}{wN}\left(h_{d}^{\prime}(M)\sum_{\begin{subarray}{c}n=1\\\ n\in
W_{d}(Y)\end{subarray}}^{M}e(h_{d}(n)\gamma)-\int_{1}^{M}h_{d}^{\prime\prime}(t)\left(\sum_{\begin{subarray}{c}1\leq
n\leq t\\\ n\in
W_{d}(Y)\end{subarray}}e(h_{d}(n)\gamma)\right)dt\right)\text{.}$

The assumptions

$\frac{N}{Z^{2k}}>q>N^{1/2k},\quad d<N^{\rho},\quad\sigma>(\log
N)^{-\log\log\log N/4}\text{,}$

together with the fact that $b_{d}\ll d^{k}$ are easily seen to imply that

$(\log Y)^{ek}X\left(e^{-\frac{\log Z}{\log
Y}}+\left(b_{d}\log^{k^{2}}(b_{d}qX)\left(q^{-1}+\frac{Z}{X}+\frac{qZ^{k}}{b_{d}X^{k}}\right)\right)^{2^{-k}}\right)\ll
wM\sigma$ (4.2)

for any $X$ with $\sigma M<X\leq M$. This is the step in the argument that
justifies the choice of the exponent $\rho$; it is chosen so that
$b_{d}q^{-2^{-k}}$ is bounded above by a negative power of $N$, namely
$N^{k\rho-1/(k\cdot 2^{k+1})}$, and hence yields a power saving against which
logarithmic factors and powers of $\sigma^{-1}$ (which are smaller than any
positive power of $N$) cannot compete. It also justifies the exponent of $3$
(in fact any exponent bigger than $2$ would suffice) in our choice of $Z$; it
makes $e^{\log Z/\log Y}$ bigger than any power of $\sigma^{-1}$. Observe also
that $YZ\leq\sigma M$ if $N$ is large enough. Applying this with $X=M$,
together with Lemma 4.3 and observing that $h_{d}^{\prime}(t)\ll
b_{d}t^{k-1}$, yields immediately

$h_{d}^{\prime}(M)\sum_{\begin{subarray}{c}n=1\\\ n\in
W_{d}(Y)\end{subarray}}^{M}e(h_{d}(n)\gamma)\ll b_{d}M^{d}w\sigma\asymp
wN\sigma\text{.}$

We now decompose

$\displaystyle\int_{1}^{M}h_{d}^{\prime\prime}(t)\left(\sum_{\begin{subarray}{c}1\leq
n\leq t\\\ n\in W_{d}(Y)\end{subarray}}e(h_{d}(n)\gamma)\right)dt$
$\displaystyle=\int_{1}^{\sigma
M}h_{d}^{\prime\prime}(t)\left(\sum_{\begin{subarray}{c}1\leq n\leq t\\\ n\in
W_{d}(Y)\end{subarray}}e(h_{d}(n)\gamma)\right)dt+\int_{\sigma
M}^{M}h_{d}^{\prime\prime}(t)\left(\sum_{\begin{subarray}{c}1\leq n\leq t\\\
n\in W_{d}(Y)\end{subarray}}e(h_{d}(n)\gamma)\right)dt\text{.}$

For the first integral, we observe that $h_{d}^{\prime\prime}(t)\ll
b_{d}M^{k-2}$ for $1\leq t\leq\sigma M$ and bound the exponential sum
trivially by $\sigma M$, yielding that

$\int_{1}^{\sigma
M}h_{d}^{\prime\prime}(t)\left(\sum_{\begin{subarray}{c}1\leq n\leq t\\\ n\in
W_{d}(Y)\end{subarray}}e(h_{d}(n)\gamma)\right)dt\ll b_{d}M^{d}\sigma^{2}\ll
wN\sigma\text{.}$

Finally, we use (4.2), Lemma 4.3 and the bound $h_{d}^{\prime\prime}(t)\ll
b_{d}M^{k-2}$ to conclude that

$\int_{\sigma
M}^{M}h_{d}^{\prime\prime}(t)\left(\sum_{\begin{subarray}{c}1\leq n\leq t\\\
n\in W_{d}(Y)\end{subarray}}e(h_{d}(n)\gamma)\right)dt\ll M\cdot
b_{d}M^{k-2}\cdot wM\sigma\asymp wN\sigma\text{.}$

Together, these imply that $S=O(\sigma)$ in this case, finishing the proof. ∎

######  Lemma 4.6.

There exists $\varepsilon=\varepsilon(k)>0$ for which the following statement
is true. Keep the notation and assumptions of Lemma 4.4. Fix
$q\leq\kappa/\sigma^{k+1}$ and some $a$ with $1\leq a\leq q$ and
$\mathrm{gcd}(a,q)=1$. Then either $\sigma<(\log N)^{-\varepsilon\log\log\log
N}$, or the following hold:

  1. (i)

For any $\gamma\in\mathfrak{M}_{a,q}(N,\kappa/\sigma)$,

$\left|\frac{1}{wN}\sum_{\begin{subarray}{c}m=1\\\ n\in
W_{d}(Y)\end{subarray}}^{M}h_{d}^{\prime}(m)e\left(h_{d}(m)\gamma\right)\right|\ll(\log
N)^{1/4}q^{-1/2}\min\left\\{1,(N|\beta|)^{-1}\right\\}\text{.}$

  2. (ii)

We have

$\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}\left|\frac{1}{wN}\sum_{\begin{subarray}{c}m=1\\\
n\in
W_{d}(Y)\end{subarray}}^{M}h_{d}^{\prime}(m)e\left(h_{d}(m)\gamma\right)\right|^{2}d\gamma\ll\frac{(\log
N)^{1/2}}{qN}\text{.}$

###### Proof.

We recover the first part of the proof of Lemma 4.4, namely (4.1). Since
$q\leq Y$, we can bound the exponential sum by $C^{\omega(q)}q^{1/2}$,
according to Lemma 4.2. Using the bounds $w_{q}\ll 2^{\omega(q)}$ and
$\omega(q)\ll\log q/\log\log q$, we obtain

$S=O\left(\frac{C^{\log q/\log\log
q}}{q^{1/2}}\min\\{1,(N|\beta|)^{-1}\\}\right)+O_{t}(\sigma^{t})$

(note that we had remarked earlier that the error term $o(\sigma)$ was
actually $O(\sigma^{t})$ for any $t>0$). The error term can hence be absorbed
into the main term (recall that both $q$ and $\beta$ are bounded by powers of
$\sigma^{-1}$) and hence

$S=O\left(\frac{C^{\log q/\log\log
q}}{q^{1/2}}\min\\{1,(N|\beta|)^{-1}\\}\right)\text{.}$

We now choose $\varepsilon>0$ so that the assumption that $\sigma\geq(\log
N)^{-\varepsilon\log\log\log N}$ implies

$C^{\log q/\log\log q}\ll(\log N)^{1/4}\quad\text{for any
}q\leq\kappa/\sigma^{k+1}\text{.}$

It can easily be checked that such an $\varepsilon$ can be found; indeed,
since $\log q/\log\log q$ is increasing, the assumption that
$q\ll\sigma^{-(k+1)}$ implies that

$\displaystyle\frac{\log q}{\log\log q}$
$\displaystyle\ll\frac{(k+1)\log(\sigma^{-1})}{\log\log(\sigma^{-1})}+O(1)$
$\displaystyle\leq\frac{(k+1)\varepsilon\log\log N\cdot\log\log\log
N}{\log(\varepsilon)+\log\log\log N+\log\log\log\log
N}+O(1)\leq\varepsilon\log\log N+O(1)$

if $N$ is large enough, which in turn yields

$C^{\log q/\log\log q}\ll C^{(k+1)\varepsilon\log\log N}=(\log
N)^{(k+1)\varepsilon\log(C)}$

and so it suffices to choose $\varepsilon$ so that
$(k+1)\varepsilon\log(C)\leq 1/4$. This is, as a matter of fact, perhaps the
part where the argument is the closest to breaking down; in order for this
strategy to work we need to be able to obtain a small power of $\log N$ in our
estimates in this Lemma, and assuming the negation of what we want to prove is
“just enough” to obtain precisely that.

Then either $\sigma<(\log N)^{-\varepsilon\log\log\log N}$, or

$S\ll(\log N)^{1/4}q^{-1/2}\min\\{1,(N|\beta|)^{-1}\\}\text{,}$

thereby proving (i). Now (ii) follows directly by applying (i) pointwise.
Indeed, we have

$\displaystyle\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}\left|\frac{1}{wN}\sum_{\begin{subarray}{c}m=1\\\
n\in
W_{d}(Y)\end{subarray}}^{M}h_{d}^{\prime}(m)e\left(h_{d}(m)\gamma\right)\right|^{2}d\gamma$
$\displaystyle\ll(\log
N)^{1/2}q^{-1}\int_{0}^{\infty}(\min\\{1,(N|\beta|)^{-1}\\})^{2}d\beta$
$\displaystyle=(\log
N)^{1/2}q^{-1}\left(\int_{0}^{1/N}1d\beta+\int_{1/N}^{\infty}(N|\beta|)^{-2}d\beta\right)$
$\displaystyle=\frac{2(\log N)^{1/2}}{qN}\text{,}$

which implies what we wanted. ∎

##  5\. The density increment

The purpose of this section is twofold. First we prove a variation (Lemma 5.1)
of the $L^{2}$ density increment lemma used in [2], adapted to our setup.
Secondly we prove (Lemma 5.2) that if $A\subseteq[N]$ is a $h_{d}$-free set of
density $\sigma$ then
$\widehat{\mathbf{1}_{A}}-\sigma\widehat{\mathbf{1}_{N}}$ has big $L^{2}$ mass
around certain rationals with small denominator, at which the Fourier
coefficients of $A$ are large. We then put the two together in Lemma 5.3,
where we establish that, in the absence of a good density increment, each
denominator appears a relatively small number of times among the rational
numbers in the conclusion of Lemma 5.2. This will pave the way to an
application of the additive energy bound of Bloom and Maynard (Theorem 2 in
[2]).

######  Lemma 5.1.

Let $A\subseteq[N]$ have density $\sigma=|A|/N$, and suppose that $A$ is
$h_{d}$-free. Let $\nu\in[0,1)$ and let $K,q\geq 1$ be such that $\nu\sigma
N\gg K\lambda(q)$, and suppose that

$\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\int_{\mathfrak{M}_{a,q}(N,K)}\left|\widehat{\mathbf{1}_{A}}(\gamma)-\sigma\widehat{\mathbf{1}_{[N]}}(\gamma)\right|^{2}d\gamma\geq\nu\sigma|A|\text{.}$

Then there exists $N^{\ast}\gg\nu\sigma N/K\lambda(q)$ and a $h_{qd}$-free
subset $A^{\ast}\subseteq[N^{\ast}]$, the density
$\sigma^{\ast}=|A^{\ast}|/N^{\ast}$ of which satisfies

$\sigma^{\ast}\geq\left(1+\frac{\nu}{73}\right)\sigma\text{.}$

###### Proof.

Let

$\mathcal{P}=\\{q,2q,\ldots,N^{\prime}q\\}$

where $N^{\prime}$ will be chosen later according to some constraints that we
will find along the argument. Suppose $\gamma\in\mathfrak{M}_{a,q}(N,K)$, and
observe that, for any $n^{\prime}\in[N^{\prime}]$,

$\left|1-e(\gamma qn^{\prime})\right|\ll\|\gamma qn^{\prime}\|=\|\gamma
qn^{\prime}-an^{\prime}\|=\left\|q\left(\gamma-\frac{a}{q}\right)n^{\prime}\right\|\leq\frac{qKN^{\prime}}{N}\text{.}$

The above implies that $\left|1-e(\gamma qn^{\prime})\right|<1/2$ if
$N^{\prime}\ll N/qK$, for some sufficiently small value of the implied
constant. Assuming this, we obtain that, for
$\gamma\in\mathfrak{M}_{a,q}(N,K)$,

$\left|N^{\prime}-\widehat{\mathbf{1}_{\mathcal{P}}}(\gamma)\right|=\left|\sum_{n\in\mathcal{P}}(1-e(\gamma
n))\right|\leq\frac{|\mathcal{P}|}{2}=\frac{N^{\prime}}{2}\text{,}\text{ and
therefore
}\left|\widehat{\mathbf{1}_{\mathcal{P}}}(\gamma)\right|\geq\frac{N^{\prime}}{2}\text{.}$

Let $g=\mathbf{1}_{A}-\sigma\mathbf{1}_{[N]}$. By Plancherel’s Theorem and the
fact that
$\widehat{\mathbf{1}_{\mathcal{P}}*g}=\widehat{\mathbf{1}_{\mathcal{P}}}\widehat{g}$,

$\displaystyle\|\mathbf{1}_{\mathcal{P}}*g\|_{2}^{2}$
$\displaystyle=\int_{\mathbb{T}}\left|\widehat{\mathbf{1}_{\mathcal{P}}}(\gamma)\right|^{2}\left|\widehat{g}(\gamma)\right|^{2}d\gamma$
$\displaystyle\geq\int_{\mathfrak{M}_{q}(N,K)}\left|\widehat{\mathbf{1}_{\mathcal{P}}}(\gamma)\right|^{2}\left|\widehat{g}(\gamma)\right|^{2}d\gamma$
$\displaystyle\geq\frac{N^{\prime
2}}{4}\int_{\mathfrak{M}_{q}(N,K)}\left|\widehat{g}(\gamma)\right|^{2}d\gamma\geq\frac{\nu\sigma
N^{\prime 2}|A|}{4}\text{.}$

Writing
$\mathbf{1}_{\mathcal{P}}*g=\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A}-\sigma\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}$
and expanding the inner product, we arrive at

$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A}\right\|_{2}^{2}\geq\frac{\nu\sigma
N^{\prime
2}|A|}{4}+2\sigma\langle\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A},\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\rangle-\sigma^{2}\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\right\|_{2}^{2}\text{.}$
(5.1)

We now observe that
$\langle\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A},\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\rangle$
counts solutions $(p,a,p^{\prime},n)\in\mathcal{P}\times
A\times\mathcal{P}\times[N]$ to the equation $p+a=p^{\prime}+n$. Rewriting the
equation as $a=p^{\prime}-p+n$, we see that

$\langle\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A},\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\rangle=\sum_{x\in
A}\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}}*\mathbf{1}_{[N]}(x)\text{.}$

It follows that

$\displaystyle\left|\langle\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A},\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\rangle-|A|N^{\prime
2}\right|$ $\displaystyle=\left|\sum_{x\in
A}(\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}}*\mathbf{1}_{[N]}(x)-N^{\prime
2}\mathbf{1}_{[N]}(x))\right|$
$\displaystyle\leq\sum_{x\in\mathbb{Z}}\left|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}}*\mathbf{1}_{[N]}(x)-N^{\prime
2}\mathbf{1}_{[N]}(x)\right|$
$\displaystyle\leq\sum_{x\in\mathbb{Z}}\left|\sum_{y\in\mathbb{Z}}(\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}})(y)\mathbf{1}_{[N]}(x-y)-\sum_{y\in\mathbb{Z}}(\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}})(y)\mathbf{1}_{[N]}(x)\right|$
$\displaystyle\leq\sum_{y\in\mathbb{Z}}(\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}})(y)\sum_{x\in\mathbb{Z}}\left|\mathbf{1}_{[N]}(x-y)-\mathbf{1}_{[N]}(x)\right|$
$\displaystyle\ll\sum_{y\in\mathbb{Z}}(\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}})(y)|y|$
$\displaystyle\ll
qN^{\prime}\sum_{y\in\mathbb{Z}}(\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-\mathcal{P}})(y)=qN^{\prime
3}\text{.}$

It follows that
$\langle\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A},\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\rangle=|A|N^{\prime
2}+O(qN^{\prime 3})$. On the other hand, since
$\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}(x)$ counts solutions
$(p,n)\in\mathcal{P}\times[N]$ to $p+n=x$, and is hence bounded by
$|\mathcal{P}|=N^{\prime}$, we have

$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\right\|_{2}^{2}\leq
N^{\prime}\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{[N]}\right\|_{1}=N^{\prime
2}N=\frac{|A|N^{\prime 2}}{\sigma}\text{.}$

Using these estimates in (5.1), we arrive at

$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A}\right\|_{2}^{2}\geq\sigma|A|N^{\prime
2}+\frac{\nu\sigma N^{\prime 2}|A|}{4}+O(\sigma qN^{\prime 3})\text{.}$

If $N^{\prime}\ll\nu\sigma N/q$ for some sufficiently small value of the
implied constant, it follows that the term $O(\sigma qN^{\prime 3})$ is at
most $\nu\sigma N^{\prime 2}|A|/8$, and therefore

$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A}\right\|_{2}^{2}\geq\left(1+\frac{\nu}{73}\right)\sigma
N^{\prime 2}|A|\text{.}$

We now observe that
$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A}\right\|_{2}^{2}$ counts
solutions $(p,a,p^{\prime},a^{\prime})\in\mathcal{P}\times
A\times\mathcal{P}\times A$ to $p+a=p^{\prime}+a^{\prime}$. Rewriting the
equation as $p-a^{\prime}=p^{\prime}-a$, we see that
$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{A}\right\|_{2}^{2}=\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}\right\|_{2}^{2}$.
We therefore obtain

$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}\right\|_{1}\cdot\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}\right\|_{\infty}\geq\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}\right\|_{2}^{2}\geq\left(1+\frac{\nu}{73}\right)\sigma
N^{\prime 2}|A|\text{.}$

On the other hand, it is clear that
$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}\right\|_{1}=|\mathcal{P}|\times|A|=N^{\prime}|A|$,
and therefore we conclude that

$\left\|\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}\right\|_{\infty}\geq\left(1+\frac{\nu}{73}\right)\sigma
N^{\prime}\text{.}$

Letting $x$ be an integer at which $\mathbf{1}_{\mathcal{P}}*\mathbf{1}_{-A}$
attains its maximum, this means precisely that
$|\mathcal{P}\cap(A+x)|\geq(1+\nu/73)\sigma N^{\prime}$.

We recall that the constraints on $N^{\prime}$ we found along the way, in
order to obtain the conclusion above, were that $N^{\prime}\ll N/qK$ and
$N^{\prime}\ll\nu\sigma N/q$. It is therefore clear that we can find
$N^{\prime}\gg\nu\sigma N/qK$ for which the conclusion holds. Now partition
$[N^{\prime}]$ into residue classes modulo $\lambda(q)/q$. Each class contains
$\gg\nu\sigma N/\lambda(q)K$ elements, and by the Pigeonhole Principle there
is a class $C$, with $|C|=N^{\ast}$, such that

$|qC\cap(A+x)|\geq\left(1+\frac{\nu}{73}\right)\sigma N^{\ast}\text{.}$

Let $qC=\\{\lambda(q)n+r:1\leq n\leq N^{\ast}\\}$ for some $r\in\mathbb{Z}$,
and let

$A^{\ast}=\\{n\in[N^{\ast}]:\lambda(q)n+r\in A+x\\}\text{.}$

Note that $|A^{\ast}|=|qC\cap(A+x)|$. So it remains only to prove that
$A^{\ast}$ is $h_{dq}$-free. Assume it is not, so that there exist $a\neq b$
in $A^{\ast}$ and a positive integer $n$ such that $a-b=h_{dq}(n)$. Then
$\lambda(q)a+r-x$ and $\lambda(q)b+r-x$ are in $A$, and

$(\lambda(q)a+r-x)-(\lambda(q)b+r-x)=\lambda(q)(a-b)=\lambda(q)h_{dq}(n)\in
h_{d}(\mathbb{N})\text{,}$

by Proposition 3.2, contradicting the assumption that $A$ is $h_{d}$-free.
This finishes the proof. ∎

######  Lemma 5.2.

Let $N\gg_{h}1$, and suppose $A\subseteq[N]$ is $h_{d}$-free, where $d\leq
N^{\rho}$, with $\sigma=|A|/N>(\log N)^{-\log\log\log N}$. Then there exist
$B,Q$, with $\sigma^{O(1)}\ll B\ll\sigma^{-O(1)}$ and $1\leq
Q\ll\sigma^{-O(1)}$, and a set

$\Gamma\subseteq\\{(a,q):q\leq Q,1\leq a\leq q,\mathrm{gcd}(a,q)=1\\}$

such that:

  1. (i)

There is a choice of frequencies
$\gamma_{a,q}\in\mathfrak{M}_{a,q}(N,\kappa/\sigma)$ for each $(a,q)\in\Gamma$
for which

$\sum_{(a,q)\in\Gamma}|\widehat{\mathbf{1}_{A}}(\gamma_{a,q})|\gg\frac{B|A|Q^{1/2}}{(\log
N)^{1/4}(\log(1/\sigma))^{2}}\text{;}$

  2. (ii)

Setting $g=\mathbf{1}_{A}-\sigma\mathbf{1}_{[N]}$, we have, for each
$(a,q)\in\Gamma$,

$\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\gg\frac{\sigma|A|}{B^{2}}\text{.}$

###### Proof.

Let $M=\left\lfloor(N/b_{d})^{1/k}\right\rfloor$. For $\gamma\in\mathbb{T}$,
define

$S(\gamma)=\frac{1}{wN}\sum_{\begin{subarray}{c}m=1\\\ m\in
W_{d}(Y)\end{subarray}}^{M}h_{d}^{\prime}(m)e(h_{d}(m)\gamma)\text{.}$

We then observe that, by orthogonality,

$\int_{\mathbb{T}}\widehat{\textbf{1}_{A}}(\gamma)\overline{\widehat{\textbf{1}_{A}}(\gamma)}S(\gamma)d\gamma=\frac{1}{wN}\sum_{\begin{subarray}{c}1\leq
m\leq M\\\ m\in W_{d}(Y)\\\ a,b\in A\\\
b-a=h_{d}(m)\end{subarray}}h_{d}^{\prime}(m)\text{.}$

The assumption that $A$ is $h_{d}$-free implies that only the terms with
$h_{d}(m)=0$ (and hence $a=b$) contribute to the sum above. Note that
$h_{d}(m)=0$ implies, by definition of $h_{d}$, that $h(r_{d}+dm)=0$; since
$h$ has only finitely many zeros we must have $r_{d}+dm\ll_{h}1$, and since
$|r_{d}|<d$ by assumption, it follows that $m\ll_{h}1$. Since $b_{d}\leq
d^{k}<N^{k\rho}$, it follows that each term $h_{d}^{\prime}(m)$ in the sum
above contributes at most $O_{h}(N^{k\rho})$, and hence

$\int_{\mathbb{T}}\widehat{\textbf{1}_{A}}(\gamma)\overline{\widehat{\textbf{1}_{A}}(\gamma)}S(\gamma)d\gamma=\frac{1}{wN}\sum_{\begin{subarray}{c}1\leq
m\leq M\\\ m\in W_{d}(Y)\\\ a,b\in A\\\
b-a=h_{d}(m)\end{subarray}}h_{d}^{\prime}(m)=O_{h}\left(\frac{1}{wN^{1-k\rho}}\right)\text{.}$
(5.2)

Recall that $g=\mathbf{1}_{A}-\sigma\mathbf{1}_{[N]}$. We claim that

$\int_{\mathbb{T}}|\widehat{g}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)S(\gamma)|d\gamma\gg\sigma|A|\text{.}$
(5.3)

To see this, we split into two cases, according to whether
$|A\cap[N/2,N]|>|A|/2$ or $|A\cap[1,N/2]|>|A|/2$. Assuming the former, we use
(5.2) to derive that

$\displaystyle-\int_{\mathbb{T}}\widehat{g}(\gamma)\overline{\widehat{\mathbf{1}_{A}}(\gamma)}S(\gamma)d\gamma$
$\displaystyle=\sigma\int_{\mathbb{T}}\mathbf{1}_{[N]}(\gamma)\overline{\widehat{\mathbf{1}_{A}}(\gamma)}S(\gamma)d\gamma-O\left(\frac{1}{wN^{1-k\rho}}\right)$
$\displaystyle=\frac{\sigma}{wN}\sum_{\begin{subarray}{c}m\in
W_{d}(Y)\cap[1,M]\\\ a\in[N],b\in A\\\
a-b+h_{d}(m)=0\end{subarray}}h_{d}^{\prime}(m)-O\left(\frac{1}{wN^{1-k\rho}}\right)\text{.}$

We now estimate the contribution to the above sum of those values of $m$ for
which $h_{d}^{\prime}(m)<0$. Recalling the definition of $h_{d}$, we see that

$h_{d}^{\prime}(m)=\frac{d}{\lambda(d)}h^{\prime}(r_{d}+dm)\text{.}$

Since $\lambda(d)\geq d$, $|r_{d}|<d$ and the leading coefficient of $h_{d}$
is assumed to be positive, the condition $h_{d}^{\prime}(m)<0$ implies that
$m\ll_{h}1$ and also that $|h_{d}^{\prime}(m)|\ll_{h}1$. Therefore the values
of $m$ for which $h_{d}^{\prime}(m)<0$ subtract at most
$O_{h}(|A|)=O_{h}(\sigma N)$ from the sum above, and hence

$\displaystyle-\int_{\mathbb{T}}\widehat{g}(\gamma)\overline{\widehat{\mathbf{1}_{A}}(\gamma)}S(\gamma)d\gamma$
$\displaystyle=\frac{\sigma}{wN}\sum_{\begin{subarray}{c}m\in
W_{d}(Y)\cap[1,M]\\\ a\in[N],b\in A\\\ a-b+h_{d}(m)=0\\\
h_{d}^{\prime}(m)>0\end{subarray}}h_{d}^{\prime}(m)-O\left(\frac{\sigma^{2}}{w}\right)-O\left(\frac{1}{wN^{1-k\rho}}\right)\text{.}$
$\displaystyle=\frac{\sigma}{wN}\sum_{\begin{subarray}{c}m\in
W_{d}(Y)\cap[1,M]\\\ a\in[N],b\in A\\\ a-b+h_{d}(m)=0\\\ h_{d}^{\prime}(m)\geq
0\end{subarray}}h_{d}^{\prime}(m)-O\left(\frac{\sigma^{2}}{w}\right)\text{.}$

Now all the summands are non-negative, and it is clear that if $M\ll m\ll M$
and $b\in A\cap[N/2,N]$ then $b-h_{d}(m)\in[N]$, for some value of the implied
constants. This means that at least $w\sigma|A|M$ summands are positive (here
we are using Lemma 3.6), and the contribution of each of those is $\gg
b_{d}M^{k-1}$. Therefore,

$-\int_{\mathbb{T}}\widehat{g}(\gamma)\overline{\widehat{\mathbf{1}_{A}}(\gamma)}S(\gamma)d\gamma\gg\frac{1}{wN}\cdot
w\sigma|A|M\cdot
b_{d}M^{k-1}-O\left(\frac{\sigma^{2}}{w}\right)\gg\sigma|A|\text{.}$

This proves (5.3), under the assumption that $|A\cap[N/2,N]|>|A|/2$. If
instead $|A\cap[1,N/2]|>|A|/2$, we derive in a completely analogous way that

$-\int_{\mathbb{T}}\widehat{\mathbf{1}_{A}}(\gamma)\overline{\widehat{g}(\gamma)}S(\gamma)d\gamma\geq\frac{\sigma}{wN}\sum_{\begin{subarray}{c}m\in
W_{d}(Y)\cap[1,M]\\\ a\in A,b\in[N]\\\ a-b+h_{d}(m)=0\\\ h_{d}^{\prime}(m)\geq
0\end{subarray}}h_{d}^{\prime}(m)-O\left(\frac{\sigma^{2}}{w}\right)\text{.}$

Now, if $M\ll m\ll M$ and $a\in A\cap[1,N/2]$, for some value of the implied
constants, then $a+h_{d}(m)\in[N]$, and we derive as before that

$-\int_{\mathbb{T}}\widehat{\mathbf{1}_{A}}(\gamma)\overline{\widehat{g}(\gamma)}S(\gamma)d\gamma\gg\sigma|A|\text{,}$

establishing (5.3) in this case as well.

We now claim that there exists a constant $\kappa>0$ such that (5.3) remains
true if we integrate only over the major arcs
$\mathfrak{M}(N,\kappa/\sigma,\kappa/\sigma^{k+1})$, i.e., such that

$\int_{\mathfrak{M}(N,\kappa/\sigma,\kappa/\sigma^{k+1})}|\widehat{g}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)S(\gamma)|d\gamma\gg\sigma|A|\text{.}$
(5.4)

To prove this, let $r$ be any positive real number such that $2r$ is smaller
than the implied constant in (5.3). We apply Lemma 4.4, yielding $\kappa>0$
such that

$|S(\gamma)|\leq r\sigma\text{ whenever
}\gamma\in\mathfrak{m}\left(N,\frac{\kappa}{\sigma},\frac{\kappa}{\sigma^{k+1}}\right)\text{.}$

This implies

$\displaystyle\int_{\mathfrak{m}(N,\kappa/\sigma,\kappa/\sigma^{k+1})}|\widehat{g}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)S(\gamma)|d\gamma$
$\displaystyle\leq
r\sigma\int_{\mathbb{T}}|\widehat{g}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)|d\gamma$
$\displaystyle\leq
r\sigma\left(\int_{\mathbb{T}}|\widehat{\mathbf{1}_{A}}(\gamma)|^{2}d\gamma+\sigma\int_{\mathbb{T}}|\widehat{\mathbf{1}_{[N]}}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)|d\gamma\right)\text{.}$

By orthogonality we have

$\int_{\mathbb{T}}|\widehat{\mathbf{1}_{A}}(\gamma)|^{2}d\gamma=|A|\text{.}$

On the other hand, by the Cauchy-Schwarz inequality and orthogonality,

$\displaystyle\int_{\mathbb{T}}|\widehat{\mathbf{1}_{[N]}}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)|d\gamma$
$\displaystyle\left(\int_{\mathbb{T}}|\widehat{\mathbf{1}_{[N]}}(\gamma)|^{2}d\gamma\right)^{1/2}\left(\int_{\mathbb{T}}|\widehat{\mathbf{1}_{A}}(\gamma)|^{2}d\gamma\right)^{1/2}$
$\displaystyle=(|A|N)^{1/2}\leq N\text{.}$

We therefore obtain that

$\int_{\mathfrak{m}(N,\kappa/\sigma,\kappa/\sigma^{k+1})}|\widehat{g}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)S(\gamma)|d\gamma\leq
2r\sigma|A|\text{,}$

which together with (5.3), and recalling the choice of $r$, implies (5.4).

Now (5.4) can be rewritten as

$\sum_{q\leq\kappa/\sigma^{k+1}}{\sum_{1\leq a\leq
q}}^{\ast}\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)\widehat{\mathbf{1}_{A}}(\gamma)S(\gamma)|d\gamma\gg\sigma|A|$

and this implies by the second part of Lemma 4.6 and Cauchy-Schwarz that

$\sum_{q\leq\kappa/\sigma^{k+1}}{\sum_{1\leq a\leq
q}}^{\ast}\frac{1}{q^{1/2}}\left(\underset{\gamma\in\mathfrak{M}_{a,q}(N,\kappa/\sigma)}{\max}\widehat{\mathbf{1}_{A}(\gamma)}\right)\left(\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\right)^{1/2}\gg\frac{\sigma|A|N^{1/2}}{(\log
N)^{1/4}}\text{.}$ (5.5)

We observe that those pairs $(a,q)$ for which

$\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\leq\frac{\sigma^{3k+5}N}{\log
N}$

contribute at most

$\sum_{q\leq\kappa/\sigma^{k+1}}q^{1/2}|A|\frac{\sigma^{(3k+5)/2}N^{1/2}}{(\log
N)^{1/2}}\leq\left(\frac{\kappa}{\sigma^{k+1}}\right)^{3/2}|A|\frac{\sigma^{(3k+5)/2}N^{1/2}}{(\log
N)^{1/2}}\ll\frac{\sigma|A|N^{1/2}}{(\log N)^{1/2}}\text{.}$

Here we used the trivial inequality $\widehat{\mathbf{1}_{A}(\gamma)}\leq|A|$.
It follows that the main contribution to (5.5) comes from those pairs $(a,q)$
in the set

$\Omega=\left\\{(a,q):1\leq q\leq\kappa/\sigma^{k+1},1\leq a\leq
q,\mathrm{gcd}(a,q)=1,\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma>\frac{\sigma^{3k+5}N}{\log
N}\right\\}\text{,}$

or in other words,

$\sum_{(a,q)\in\Omega}\frac{1}{q^{1/2}}\left(\underset{\gamma\in\mathfrak{M}_{a,q}(N,\kappa/\sigma)}{\max}\widehat{\mathbf{1}_{A}(\gamma)}\right)\left(\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\right)^{1/2}\gg\frac{\sigma|A|N^{1/2}}{(\log
N)^{1/4}}\text{.}$

We note that $|\widehat{g}(\gamma)|\leq 2N$ for every $\gamma\in\mathbb{T}$,
and since the arc $\mathfrak{M}_{a,q}(N,\kappa/\sigma)$ has length
$O(1/\sigma)$, we have

$\frac{\sigma^{(3k+5)/2}N^{1/2}}{(\log
N)^{1/2}}<\left(\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\right)^{1/2}\ll\frac{N^{1/2}}{\sigma^{1/2}}\quad\text{for
every $(a,q)\in\Omega$.}$

Since $\log N=O(1/\sigma)$, the range determined by the previous inequalities
can be covered by $O(\log(1/\sigma))$ intervals of the form $[\sigma
N^{1/2}/B,2\sigma N^{1/2}/B]$, for some values of $B$ with $\sigma^{O(1)}\ll
B\ll\sigma^{-O(1)}$. Similarly, the interval $[1,\kappa/\sigma^{k+1}]$ can be
covered by $O(\log(1/\sigma))$ intervals of the form $[Q,2Q]$, with
$Q\ll\sigma^{-O(1)}$. By the Pigeonhole Principle, it follows that there
exists a subset $\Gamma\subseteq\Omega$ and values of $B,Q$ with the indicated
properties such that, if $(a,q)\in\Gamma$, then $Q\leq q\leq 2Q$ and

$\sigma
N^{1/2}/B\leq\left(\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\right)^{1/2}\leq
2\sigma N^{1/2}/B\text{,}$

and moreover

$\sum_{(a,q)\in\Gamma}\frac{1}{q^{1/2}}\left(\underset{\gamma\in\mathfrak{M}_{a,q}(N,\kappa/\sigma)}{\max}\widehat{\mathbf{1}_{A}(\gamma)}\right)\left(\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\right)^{1/2}\gg\frac{\sigma|A|N^{1/2}}{(\log
N)^{1/4}(\log(1/\sigma))^{2}}\text{.}$

The previous two inequalities put together imply that

$\frac{1}{Q^{1/2}}\sum_{(a,q)\in\Gamma}\left(\underset{\gamma\in\mathfrak{M}_{a,q}(N,\kappa/\sigma)}{\max}\widehat{\mathbf{1}_{A}(\gamma)}\right)\frac{\sigma
N^{1/2}}{B}\gg\frac{\sigma|A|N^{1/2}}{(\log
N)^{1/4}(\log(1/\sigma))^{2}}\text{,}$

which simplifies to the desired result. ∎

We will now deduce that, if the density increment condition of Lemma 5.1 does
not hold, so that the $L^{2}$ mass of $\widehat{g}$ around rationals with
small denominators is very small, then the set $\Gamma$ constructed in Lemma
5.2 has few pairs $(a,q)$ with a given value of $q$.

######  Lemma 5.3.

Keep the assumptions of Lemma 5.2, and let $\nu\geq N^{-1/2}$. Then one of the
following holds:

  1. (i)

There exists $N^{\ast}\gg\nu\sigma^{O(1)}N$, an integer $q\ll\sigma^{-O(1)}$
and a $h_{qd}$-free subset $A^{\ast}\subseteq[N^{\ast}]$ with density

$\sigma^{\ast}\geq\left(1+\frac{\nu}{73}\right)\sigma\text{;}$

  2. (ii)

There exist $B$, $Q$ and a set $\Gamma$ as in the conclusion of Lemma 5.2 with
the additional property that, for every $q\leq Q$,

$|\\{a:(a,q)\in\Gamma\\}|\ll\nu B^{2}\text{.}$

###### Proof.

We assume that condition (i) does not hold; note that, given that
$\lambda(q)\leq q^{d}$, the condition $\nu\sigma N\gg K\lambda(q)$ holds
automatically whenever $q,K\ll\sigma^{-O(1)}$, so that Lemma 5.1 implies that,
for such $q$ and $K$,

$\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\int_{\mathfrak{M}_{a,q}(N,K)}\left|\widehat{g}(\gamma)\right|^{2}d\gamma<\nu\sigma|A|\text{.}$

We now consider values $B$ and $Q$ and a set $\Gamma$ as in the conclusion of
Lemma 5.2, and for a value of $q\leq Q$ we add up the inequality

$\int_{\mathfrak{M}_{a,q}(N,\kappa/\sigma)}|\widehat{g}(\gamma)|^{2}d\gamma\gg\frac{\sigma|A|}{B^{2}}$

over all values of $a$ for which $(a,q)\in\Gamma$. This yields

$|\\{a:(a,q)\in\Gamma\\}|\frac{\sigma|A|}{B^{2}}\ll\nu\sigma|A|\text{,}$

which simplifies to (ii). ∎

##  6\. Additive Energy bounds and end of the proof

We quote the following result from [2], which essentially says that a set of
frequencies at which the Fourier coefficients of a set $A$ are large has high
additive energy.

######  Lemma 6.1.

Let $A\subseteq[N]$ be a set of density $\sigma=|A|/N$. Let set
$S\subseteq\mathbb{T}$ be a finite set. Then, for every $m\geq 1$,

$\sum_{\gamma\in
S}|\widehat{\mathbf{1}_{A}}(\gamma)|\ll|A|\sigma^{-1/2m}E_{2m}(S;1/2N)^{1/2m}\text{,}$

where we define

$E_{2m}(S;\delta)=|\\{(b_{1},\ldots,b_{2m})\in
S^{2m}:\|b_{1}+\cdots+b_{m}-b_{m+1}-\cdots-b_{2m}\|\leq\delta\\}|\text{.}$

###### Proof.

This is ([2], Lemma 7), but we provide a proof for completeness. For each
$\gamma$ let $\omega_{\gamma}$ be a complex number with absolute value $1$ for
which
$\omega_{\gamma}\widehat{\mathbf{1}_{A}(\gamma)}=|\widehat{\mathbf{1}_{A}(\gamma)}|$.
By Hölder’s inequality,

$\displaystyle\sum_{\gamma\in S}|\widehat{\mathbf{1}_{A}}(\gamma)|$
$\displaystyle=\sum_{\gamma inS}\omega_{\gamma}\sum_{a\in A}e(a\gamma)$
$\displaystyle\leq|A|^{1-1/2m}\left(\sum_{a\in A}\left|\sum_{\gamma\in
S}\omega_{\gamma}e(a\gamma)\right|^{2m}\right)^{1/2m}\text{.}$

We now consider the function $\psi$ on $\mathbb{R}$ defined by
$\psi(t)=\sin(\pi t)^{2}/(\pi t)^{2}$, whose Fourier transform
$\widehat{\psi}$ satisfies

$\widehat{\psi}(\xi)=\begin{cases}1-|\xi|&\text{ if }|\xi|\leq 1\\\ 0&\text{
otherwise. }\\\ \end{cases}$

Since $\psi$ is non-negative and $\psi(t)\gg 1$ for $t\in[-1/2,1/2]$, it
follows that

$\displaystyle\sum_{a\in A}\left|\sum_{\gamma\in
S}\omega_{\gamma}e(a\gamma)\right|^{2m}$
$\displaystyle\ll\sum_{n\in\mathbb{Z}}\psi\left(\frac{n}{2N}\right)\left|\sum_{\gamma\in
S}\omega_{\gamma}e(a\gamma)\right|^{2m}$
$\displaystyle\ll\sum_{(b_{1},\ldots,b_{2m})\in
S^{2m}}\left|\sum_{n\in\mathbb{Z}}\psi\left(\frac{n}{2N}\right)e\left(n(b_{1}+\cdots+b_{m}-b_{m+1}-\cdots-
b_{2m})\right)\right|\text{.}$

We now apply the Poisson summation formula ([4], Theorem 4.4) to the inner
sum, which, together with the fact that $\widehat{\psi}$ has support $[-1,1]$,
implies that the above is

$\displaystyle\ll\sum_{(b_{1},\ldots,b_{2m})\in
S^{2m}}N\left|\sum_{h\in\mathbb{Z}}\widehat{\psi}\left(2N(b_{1}+\cdots+b_{m}-b_{m+1}-\cdots-
b_{2m})\right)\right|$ $\displaystyle\ll N|\\{(b_{1},\ldots,b_{2m})\in
S^{2m}:\|b_{1}+\cdots+b_{m}-b_{m+1}-\cdots-b_{2m}\|\leq 1/2N\\}|\text{.}$

Substituting this into our application of Hölder finishes the proof. ∎

Lemma 6.1 runs into some tension against the following additive energy bound
of Bloom and Maynard if the large Fourier coefficients occur at rationals with
different denominators, which we know will be the case in the absence of a
good density increment, according to Lemma 5.3.

######  Lemma 6.2.

Let $Q\geq 4$ and $m\geq 2$. Let $S$ be a set of positive rational frequencies
in $\mathbb{T}$ with denominator at most $Q$ contained in an interval of
length $1/8m$ and assume that $S$ contains at most $n$ frequencies with
denominator $q$ for every $q\leq Q$. Then we have

$E_{2m}(S;0)\leq(\log Q)^{C^{m}}(Qn)^{m}$

for some absolute constant $C$.

###### Proof.

Let $S^{\prime}$ be a set of representatives on the real line of the elements
of $S$, contained in an interval of length $1/8m$. The number $E_{2m}(S;0)$
counts $2m$-tuples $(b_{1},\ldots,b_{2m})\in S^{\prime 2m}$ for which

$b_{1}+\cdots+b_{m}-b_{m+1}-\cdots-b_{2m}\in\mathbb{Z}\text{.}$

But the above expression lies in $[-1/4,1/4]$, so if it is an integer then it
must equal $0$. Therefore the result follows from Theorem 2 of [2]. ∎

We are now ready to state and prove the next result which essentially
determines the quality of the density increment that the previous additive
energy bounds allow us to obtain. It is similar in spirit to Lemma 8 of [2],
with the differences arising from our slightly worse bound for the sum in
Lemma 5.2(i) (Bloom and Maynard manage to avoid the power of $\log N$ in the
denominator). However this slight worsening of the quality of the density
increment, as measured by the parameter $\nu$, turns out not to be detrimental
to the quality of the final bound.

######  Lemma 6.3.

Let $N\gg_{h}1$, and suppose as usual that $d\leq N^{\rho}$. Then, for some
absolute constant $c>0$, if we set

$\nu=\frac{1}{(\log
N)^{1/2}}\exp\left(-c\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\right)\text{,}$

then there exist $N^{\ast}\gg\sigma^{O(1)}N/(\log N)^{1/2}$, an integer
$q\ll\sigma^{-O(1)}$ and a $h_{qd}$-free subset $A^{\ast}\subseteq[N^{\ast}]$
with density

$\sigma^{\ast}\geq\left(1+\frac{\nu}{73}\right)\sigma\text{.}$

###### Proof.

Let $m\geq 2$ be an integer to be chosen later. We apply Lemma 5.3; our goal
is to prove that case (i) must hold for the presented choice of $\nu$. Assume
it does not; observe that, by an application of the Pigeonhole Principle,
there exists a subset $\Gamma^{\prime}\subseteq\Gamma$ such that the
frequencies $\gamma_{a,q}$ for $(a,q)\in\Gamma^{\prime}$ lie on an interval of
length $1/8m$ and such that

$\sum_{(a,q)\in\Gamma}|\widehat{\mathbf{1}_{A}}(\gamma_{a,q})|\gg\frac{B|A|Q^{1/2}}{m(\log
N)^{1/4}(\log(1/\sigma))^{2}}\text{.}$

Let $S=\\{\gamma_{a,q}:(a,q)\in\Gamma^{\prime}\\}$ and let
$T=\\{a/q:(a,q)\in\Gamma\\}$. It follows from Lemma 6.1 that

$\frac{B|A|Q^{1/2}}{m(\log
N)^{1/4}(\log(1/\sigma))^{2}}\ll|A|\sigma^{-1/2m}E_{2m}(S;1/2N)^{1/2m}\text{.}$

Suppose $(a_{1},q_{1}),\ldots,(a_{2m},q_{2m})$ are such that

$\|\gamma_{a_{1},q_{1}}+\cdots+\gamma_{a_{m},q_{m}}-\gamma_{a_{m+1},q_{m+1}}-\cdots-\gamma_{a_{2m},q_{2m}}\|\leq\frac{1}{2N}\text{.}$
(6.1)

Since $\gamma_{a,q}$ differs from $a/q$ by at most $\kappa/\sigma N$, it
follows that

$\left\|\frac{a_{1}}{q_{1}}+\cdots+\frac{a_{m}}{q_{m}}-\frac{a_{m+1}}{q_{m+1}}-\cdots-\frac{a_{2m}}{q_{2m}}\right\|\leq\frac{1}{2N}+\frac{\kappa
m}{\sigma N}\ll\frac{m}{\sigma N}\text{.}$

On the other hand, the sum above, if nonzero, has denominator at most
$Q^{2m}$. Since $Q\ll\sigma^{-O(1)}$, it would follow that $\sigma N\ll
m\sigma^{-O(m)}$; but, as we will see, our final choice of $m$ will be such
that $m\ll\log\log(1/\sigma)$, so that these would contradict our standing
assumptions on the relative sizes of $N$ and $\sigma$. It follows that,
whenever (6.1) holds, we have

$\frac{a_{1}}{q_{1}}+\cdots+\frac{a_{m}}{q_{m}}-\frac{a_{m+1}}{q_{m+1}}-\cdots-\frac{a_{2m}}{q_{2m}}=0\text{.}$

Therefore $E_{2m}(S;1/2N)\leq E_{2m}(T;0)$, and it follows that

$\frac{B|A|Q^{1/2}}{m(\log
N)^{1/4}(\log(1/\sigma))^{2}}\ll|A|\sigma^{-1/2m}E_{2m}(T;0)^{1/2m}\text{.}$

We now apply Lemma 6.2 to bound the additive energy, noting that according to
the conclusion of Lemma 5.3 we can take $n\ll\nu B^{2}$. It follows that

$\frac{B|A|Q^{1/2}}{m(\log
N)^{1/4}(\log(1/\sigma))^{2}}\ll|A|\sigma^{-1/2m}E_{2m}(T;0)^{1/2m}1\leq|A|\sigma^{-1/2m}(\log
Q)^{C^{m}/2m}\nu^{1/2}Q^{1/2}B\text{,}$

and now - and this is the key reason why we needed to obtain square root
cancellation in our exponential sums! - we can cancel the powers of $Q$ on
both sides (as well as some other factors) and we obtain

$\nu\gg\frac{\sigma^{1/m}}{(\log N)^{1/2}m^{2}\log(1/\sigma)^{4}(\log
Q)^{C^{m}/m}}\text{.}$

Recalling that $Q\ll\sigma^{-O(1)}$, we see that, if $c^{\prime}>0$ is
sufficiently small, then setting $m=\lceil
c^{\prime}\log\log(1/\sigma)\rceil$, this yields

$\nu\gg\frac{1}{(\log
N)^{1/2}}\exp\left(-O\left(\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\right)\right)\text{.}$

This means that case (ii) in Lemma 5.3 is not compatible with the given choice
of $\nu$ if $c$ is sufficiently small, so that case (i) must hold, which is
precisely what we want. ∎

We now apply an iterative density increment argument to deduce our main
result, Theorem 1.2.

###### Proof.

Assume that $A\subseteq[N]$ is $h$-free with density $\sigma=|A|/N$. We may as
well assume that $N\gg_{h}1$. We set

$\nu=\frac{1}{(\log
N)^{1/2}}\exp\left(-c\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\right)\text{,}$

as given by Lemma 6.3. Applying Lemma 6.3 iteratively $t$ times produces:

  * •

A sequence of positive integers $N_{0},N_{1},\ldots,N_{t}$ with $N_{0}=N$ and
$N_{i+1}\geq\sigma^{O(1)}N_{i}/(\log N)^{1/2}$;

  * •

A sequence of positive integers $d_{0},\ldots,d_{t}$ with $d_{0}=1$ and
$d_{i+1}\leq\sigma^{-O(1)}d_{i}$;

  * •

A sequence of subsets $A_{i}\subseteq[N_{i}]$ such that $A_{i}$ is
$h_{d_{i}}$-free and has density $\sigma_{i}$, with

$\sigma_{i+1}\geq\left(1+\frac{\nu}{73}\right)\sigma_{i}\text{.}$

We perform this iteration until we find $t$ such that $N_{t}<N^{1/2}$. Until
then the conditions of Lemma 6.3 remain in place; for expository purposes we
will defer the verification that the technical condition $d_{t}\leq N^{\rho}$
remains true when that holds to the end of the proof.

We now observe that

$1\geq\sigma_{t}\geq\left(1+\frac{\nu}{73}\right)^{t}\sigma\text{,}$

whence

$t\leq\frac{\log(1/\sigma)}{\log(1+\nu/73)}\ll\nu^{-1}\log(1/\sigma)\text{.}$
(6.2)

On the other hand, one has

$N^{1/2}>N_{t}\geq\frac{\sigma^{O(t)}}{(\log N)^{t/2}}N\text{,}$

which implies, after canceling $N^{1/2}$ and taking logarithms,

$\log N\ll t\log(1/\sigma)+t\log\log N\text{.}$

Now using the bound for $t$ given by (6.2), it follows that

$\log N\ll\nu^{-1}\log(1/\sigma)^{2}+\nu^{-1}\log(1/\sigma)\log\log N\text{,}$

whence

$\log N\ll\nu^{-1}\log(1/\sigma)^{2}\text{ or }\log
N\ll\nu^{-1}\log(1/\sigma)\log\log N\text{.}$ (6.3)

If the former holds, then unraveling the definition of $\nu$ yields

$(\log
N)^{1/2}\ll\exp\left(c\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\right)\log(1/\sigma)^{2}$

and taking logarithms from both sides we obtain

$\log\log
N\ll\frac{\log(1/\sigma)}{\log\log(1/\sigma)}+\log\log(1/\sigma)\ll\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\text{.}$

The above is easily seen to imply that $\log(1/\sigma)\gg\log\log
N\cdot\log\log\log N$, which is a restatement of the desired result.

If, on the other hand, the latter holds in (6.3), then similarly unraveling
the definition of $\nu$ yields

$(\log
N)^{1/2}\ll\exp\left(c\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\right)\log(1/\sigma)\log\log
N\text{,}$

and taking logarithms gives

$\log\log
N\ll\frac{\log(1/\sigma)}{\log\log(1/\sigma)}+\log\log(1/\sigma)+\log\log\log
N\text{,}$

and since $\log\log\log N=o(\log\log N)$, it follows that

$\log\log N\ll\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\text{,}$

which, as we have seen, implies the result.

It remains only to check that $d_{t}\leq N^{\rho}$, which it suffices to do
under the assumption that $\sigma\geq(\log N)^{-\log\log\log N/4c}$, where $c$
is the constant appearing in the statement of Lemma 6.3. Under that
assumption, it is easily seen that

$\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\leq\frac{1}{4c}\log\log N$

(if $N$ is large enough). Now observe that $d_{t}\leq\sigma^{-O(t)}$, and

$\displaystyle t$ $\displaystyle\leq(\log
N)^{1/2}\exp\left(c\frac{\log(1/\sigma)}{\log\log(1/\sigma)}\right)\log(1/\sigma)$
$\displaystyle\leq(\log N)^{1/2}\exp\left(\frac{1}{4}\log\log
N\right)\log(1/\sigma)$ $\displaystyle=(\log N)^{3/4}\log(1/\sigma)$
$\displaystyle\ll(\log N)^{3/4}\log\log N\log\log\log N\text{.}$

This implies that

$d_{t}\leq(\log N)^{O((\log N)^{3/4}\log\log N(\log\log\log
N)^{2})}=\exp(O((\log N)^{3/4}(\log\log N)^{2}(\log\log\log N)^{2}))\text{,}$

and the above is smaller than $N^{\rho}$ if $N$ is sufficiently large. ∎

##  7\. Acknowledgements

The author was funded through the Engineering and Physical Sciences Research
Council Doctoral Training Partnership at the University of Warwick. Special
thanks are due to Sam Chow for bringing this problem to the attention of the
author.

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# A Degree Six Inequality on Convex Quadrilaterals

Mazen Bou Khuzam
Dept. of Mathematics and Natural Science
The American University of Iraq at Sulaimani
Email<EMAIL_ADDRESS>
###### Abstract

We prove a degree-six inequality on convex quadrilaterials. This inequality
originated from work on the Atiyah-Sutcliffe conjectures on configurations of
points in $\mathbb{R}^{3}$.

$\bf{Keywords}$: Atiyah determinant ; Atiyah-Sutcliffe conjectures

##  1 Introduction

†† I would like to thank Ms. Shams Watha for her help in writing numerous
codes to verify some of the inequalities associated with this work
$Z_{1}$$Z_{2}$$W^{\prime}$$W$$c$$d$$f$$b$$a$$e$$Z_{4}$$Z_{3}$$X$$Y$$\alpha_{1}$$\beta_{1}$
Figure 1: Convex Quadrilateral

The inequality we introduce in this paper has its origin from the Atiyah
determinant. M. F. Atiyah constructed this determinant in [1] (see also
related work in [2]) in an attempt to answer a geometric question asked by M.
V. Berry and J. M. Robbins in [4] based on their work on the Spin Statistics
Theorem. He conjectured that the determinant should be non-zero in order to
finalize one of his original constructions. Together with P. Sutcliffe, he
added two new conjectures in [3] implying the original conjecture. However,
proving these conjectures turned out to be notoriously difficult, as described
by him. The linear case and the case of three points were rather
straightforward, but the problem started to become increasingly resistent when
we reach the case of four points. M. Eastwood and P. Norbury proved the first
conjecture in [6] for the case of four points, and other proofs for special
cases have appeared in [5] , [8], and [7]. Nevertheless, the main reason or
geometric reason behind the difficulty of the $n=4$ case and the general case
remained in question. In this paper, we introduce a new geometric inequality
obtained from work on the Atiyah determinant. This degree-six inequality seems
to lie at the heart of the problem and will be used in a subsequent paper to
prove the second Atiyah-Sutcliffe conjecture on convex quadrilaterals. We hope
that this work would help in finding similar inequalities for $n\geq 5$ in
order to solve the general conjecture.

To present this inequality, we start by assuming we are working in the complex
plane and our four points are located at the pairwise distinct points $z_{1}$,
$z_{2}$, $z_{3}$, and $z_{4}$. The case of a convex quadrilateral is shown in
figure 1. In the triangle $z_{1}z_{2}z_{3}$, the lengths of the sides are
assigned as follows: The side facing the vertex at $z_{1}$ has length $a$, the
side facing the vertex at $z_{2}$ has length $b$, and the side facing the
vertex at $z_{3}$ has length $c$. The other three sides have lengths
$|z_{4}z_{1}|=d$, $|z_{4}z_{2}|=e$, and $|z_{4}z_{3}|=f$. We define the
following degree-six expression corresponding to the side $z_{1}z_{2}$ by

$E_{12}=f\,A_{123}\,A_{124}\,(a+b+e+d-2c)$

This expression includes the area of the triangle $z_{1}z_{2}z_{3}$, denoted
by $A_{123}$, multiplied by the area of the triangle $z_{1}z_{2}z_{4}$. These
are the two triangles sharing the side $z_{1}z_{2}$. This is multiplied with
$(a+b-c)+(e+d-c)$ which is the sum of the two triangle inequalities at the
side $z_{1}z_{2}$ which has length $c$. Finally, the whole product is
multiplied by $f$, the length of the free (unused) side. It is obvious that
$E_{12}$ is of degree six and is positive by construction. One can define five
similar quantities, one for each of the remaining five sides of the
quadrilateral. For example: $E_{13}=e\,A_{123}\,A_{134}\,(c+a+d+f-2b)$. Using
these six expressions we introduce and prove the following inequality

###### Theorem 1

For a convex quadrilateral, the following inequality is satisfied:

$E_{12}+E_{23}+E_{34}+E_{41}\geq E_{13}+E_{24}$

In section 2 we simplify the expression of the multiplicity-one terms. In
section 3, we simplify the multiplicity-two terms and then we use these two
simplifications to prove the inequality in section 4.

##  2 The Multiplicity-One Terms

Each of the $E_{ij}$’s consists of 5 terms which means that our inequality
comprise of 30 terms. We start by displaying these thirty terms in order to be
able to select from them based on symmetric considerations that will be used
in the proof of the inequality. Here is the full inequality:

$\displaystyle{f\,A_{123}\,A_{124}\,(d+e+a+b-2c)+d\,A_{123}\,A_{234}\,(c+b+e+f-2a)}$

$\displaystyle{+c\,A_{134}\,A_{234}\,(d+b+e+a-2f)+a\,A_{124}\,A_{134}\,(c+e+b+f-2d)}$

$\displaystyle{-e\,A_{123}\,A_{134}\,(c+a+d+f-2b)-b\,A_{124}\,A_{234}\,(c+d+a+f-2e)\geq
0}$

(2)

The first important note about inequality (2) is that the terms of multipicity
one happen to have a natural decomposition into three disjoint classes: The
$X$-terms, the $Y$-terms, and the $W$-terms, where $X$, $Y$, and $W$ are the
angles shown in figure 1. We will start with the $X$-terms first: These are
multiplicity-one terms in (2) that do not have an $a$ or $d$ factor. We call
these the $X$-terms since $a$ and $d$ are the side-lengths of the angle $X$.
Accordingly, the $X$-terms are:

$\displaystyle{f(e+b)A_{123}A_{124}+c(b+e)A_{134}A_{234}-e(c+f)A_{123}A_{134}-b(c+f)A_{124}A_{234}}$

Rearranging these terms gives us:

$\displaystyle{bc\,A_{234}(A_{134}-A_{124})+ce\,A_{134}\,(A_{234}-A_{123})-bfA_{124}\,(A_{234}-A_{123})}$

$\displaystyle{-ef\,A_{123}\,(A_{134}-A_{124})}$

We note that the last factor in each of the above four terms is actually the
same. It is easy to see that this factor is equal to
$\displaystyle{\frac{1}{2}ad\sin X}$. Substituting this back in the last
expression, we see that the $X$-terms can be written as:

$\left(bc\,A_{234}+ce\,A_{134}-bfA_{124}-ef\,A_{123}\right)(\frac{1}{2}ad\sin
X)$

Replacing the areas of the triangles using the formulae
$A_{234}=\frac{1}{2}ef\sin\beta_{4}$, $A_{134}=\frac{1}{2}bf\sin\alpha_{3}$,
$A_{124}=\frac{1}{2}ce\sin\beta_{2}$, and
$A_{123}=\frac{1}{2}bc\sin\alpha_{1}$ allows us to decompose the $X$-terms as
a product of a scalar quantity with an angular one:

$\frac{1}{4}(abcdef)(\sin\beta_{4}+\sin\alpha_{3}-\sin\beta_{2}-\sin\alpha_{1})\sin
X$

The angular part can be simplified further by using that
$\displaystyle{\sin\alpha+\sin\beta=}$

$2\sin(\frac{\alpha+\beta}{2})\,\cos(\frac{\alpha-\beta}{2})$ on the first two
terms and the last two terms to obtain

$\frac{1}{2}(abcdef)\left(\sin(\frac{\alpha_{3}+\beta_{4}}{2})\cos(\frac{\alpha_{3}-\beta_{4}}{2})-\sin(\frac{\alpha_{1}+\beta_{2}}{2})\cos(\frac{\alpha_{1}-\beta_{2}}{2})\right)\sin
X$

Since $W^{\prime}=\alpha_{3}+\beta_{4}=\alpha_{1}+\beta_{2}$, the answer
becomes:

$\frac{1}{2}(abcdef)\left(\cos(\frac{\alpha_{3}-\beta_{4}}{2})-\cos(\frac{\alpha_{1}-\beta_{2}}{2})\right)\sin\frac{W^{\prime}}{2}\sin
X$

Finally, using
$\displaystyle{\cos\alpha-\cos\beta=-2\sin(\frac{\alpha+\beta}{2})\,\sin(\frac{\alpha-\beta}{2})}$,
and the fact that

$\alpha_{3}-\beta_{2}=\alpha_{1}-\beta_{4}$ and
$\alpha_{3}-\alpha_{1}=\beta_{2}-\beta_{4}=-Y$, the $X$-terms take the
factorized form:

$(abcdef)\sin
X\sin\frac{W^{\prime}}{2}\sin\frac{Y}{2}\sin(\frac{\alpha_{1}-\beta_{4}}{2})$

Similar factorrization for the $Y$-terms and $W$-terms can be obtained in
exactly the same way and are summarized in the following lemma:

###### Lemma 1

: The multiplicity-one terms of our inequality can be simplified as follows:

The $X$-terms add up to: $\displaystyle{(abcdef)\sin
X\sin\frac{W^{\prime}}{2}\sin\frac{Y}{2}\sin(\frac{\alpha_{1}-\beta_{4}}{2})}$

The $Y$-terms add up to: $\displaystyle{(abcdef)\sin
Y\sin\frac{W}{2}\sin\frac{X}{2}\sin(\frac{\beta_{1}-\alpha_{2}}{2})}$

The $W$-terms add up to: $\displaystyle{(abcdef)\sin
W\cos\frac{X}{2}\cos\frac{Y}{2}\sin(\frac{\gamma_{1}+\gamma_{3}}{2})}$

##  3 The Multiplicity-Two Terms

We will see that these multiplicity-two terms have a natural dichotomy
requiring different ways for simplifying their expressions. First, let us
collect them from the expression in (2) and display them in front of us:

$\displaystyle{-2adA_{123}A_{234}-2adA_{124}A_{134}-2cfA_{124}A_{123}-2cfA_{134}A_{234}}$

$\displaystyle{+2beA_{123}A_{134}+2beA_{124}A_{234}}$

Similar to what we did for multiplicity-one terms, we write this expression as
a scalar part multiplied with an angular part as follows:

$\displaystyle{\frac{1}{2}(abcdef)(-\sin\alpha_{1}\sin\beta_{4}-\sin\alpha_{3}\sin\beta_{2}-\sin\alpha_{4}\sin\beta_{3}-\sin\alpha_{2}\sin\beta_{1}}$

$\displaystyle{+\sin\gamma_{1}\sin\gamma_{3}-\sin\gamma_{2}\sin\gamma_{4})}$

We linearize each term using
$\displaystyle{\sin\alpha\sin\beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))}$
and regroup their angular parts as follows:

The first part of the angular expression is

$\displaystyle{P_{1}=\frac{1}{4}(\cos(\alpha_{1}+\beta_{4})+\cos(\alpha_{3}+\beta_{2})+\cos(\alpha_{4}+\beta_{3})+\cos(\alpha_{2}+\beta_{1})}$

$\displaystyle{+\cos(\gamma_{1}-\gamma_{3})+\cos(\gamma_{2}-\gamma_{4}))}$

The second part of the angular expression is

$\displaystyle{P_{2}=\frac{1}{4}(-\cos(\alpha_{1}-\beta_{4})-\cos(\alpha_{3}-\beta_{2})-\cos(\alpha_{4}-\beta_{3})-\cos(\alpha_{2}-\beta_{1})}$

$\displaystyle{-\cos(\gamma_{1}+\gamma_{3})-\cos(\gamma_{2}+\gamma_{4}))}$

We notice that all angles appearing in the expression of $P_{1}$ can be
expressed in terms of $X$, $Y$, and $W$ (or $W^{\prime}=\pi-W$). Namely,
$\alpha_{1}+\beta_{4}=W^{\prime}+Y$, $\alpha_{3}+\beta_{2}=W^{\prime}-Y$,
$\alpha_{2}+\beta_{1}=W+X$, $\alpha_{4}+\beta_{3}=W-X$,
$\gamma_{1}-\gamma_{3}=X+Y$, and $\gamma_{2}-\gamma_{4}=X-Y$. Accordingly,
$P_{1}$ can be simplified as follows:

$\displaystyle{P_{1}=\frac{1}{2}(\cos X\cos W+\cos Y\cos W^{\prime}+\cos X\cos
Y)}$

This can be simplified further by using the half-angle formula to produce

$\displaystyle{P_{1}=\frac{1}{2}(1-2\sin^{2}\frac{X}{2})(1-2\sin^{2}\frac{W}{2})+\frac{1}{2}(1-2\sin^{2}\frac{Y}{2})(1-2\sin^{2}\frac{W^{\prime}}{2})}$

$+\frac{1}{2}(1-2\sin^{2}\frac{X}{2})(1-2\sin^{2}\frac{Y}{2})$

Using $\displaystyle{\sin^{2}\frac{W}{2}+\sin^{2}\frac{W^{\prime}}{2}=1}$ we
can simplify $P_{1}$ to become

$\displaystyle{P_{1}=\frac{1}{2}-2\sin^{2}\frac{X}{2}\cos^{2}\frac{W}{2}-2\sin^{2}\frac{Y}{2}\cos^{2}\frac{W^{\prime}}{2}+2\sin^{2}\frac{X}{2}\sin^{2}\frac{Y}{2}}$

Inserting $\displaystyle{\sin^{2}\frac{Y}{2}+\cos^{2}\frac{Y}{2}}$ as a factor
in the second term, $\displaystyle{\sin^{2}\frac{X}{2}+\cos^{2}\frac{X}{2}}$
as a factor in the third term, and
$\displaystyle{\sin^{2}\frac{W}{2}+\cos^{2}\frac{W}{2}}$ as a factor in the
fourth term allows us to write $P_{1}$ as

$P_{1}=\frac{1}{2}-2\sin^{2}\frac{X}{2}\cos^{2}\frac{W}{2}\cos^{2}\frac{Y}{2}-2\cos^{2}\frac{X}{2}\cos^{2}\frac{W^{\prime}}{2}\sin^{2}\frac{Y}{2}$

As for $P_{2}$, we notice that $\alpha_{1}-\beta_{4}=\alpha_{3}-\beta_{2}$,
$\alpha_{4}-\beta_{3}=\alpha_{2}-\beta_{1}$, and
$\gamma_{1}+\gamma_{3}=2\pi-(\gamma_{2}+\gamma_{4})$ which allows us to write
it as

$\displaystyle{P_{2}=\frac{1}{2}\left(-\cos(\beta_{4}-\alpha_{1})-\cos(\alpha_{2}-\beta_{1})-\cos(\gamma_{3}+\gamma_{1})\right)}$.
By realizing that the sum of the angles in this last expression is $\pi$ and
using the identity

$\cos\alpha+\cos\beta+\cos\gamma=1+4\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})\sin(\frac{\gamma}{2})\hskip
17.07164pt;\hskip 8.53581pt\alpha+\beta+\gamma=\pi$

(3)

we can see that
$\displaystyle{P_{2}=-\frac{1}{2}-2\sin(\frac{\alpha_{1}-\beta_{4}}{2})\sin(\frac{\beta_{1}-\alpha_{2}}{2})\sin(\frac{\gamma_{1}+\gamma_{3}}{2})}$

We add $P_{1}$ and $P_{2}$ and summarize these computations in the following
lemma

###### Lemma 2

: The multiplicity-two terms of our inequality can be writen as $(abcdef)$
times

$-2\sin^{2}\frac{X}{2}\cos^{2}\frac{W}{2}\cos^{2}\frac{Y}{2}-2\cos^{2}\frac{X}{2}\cos^{2}\frac{W^{\prime}}{2}\sin^{2}\frac{Y}{2}$

$-2\sin(\frac{\alpha_{1}-\beta_{4}}{2})\sin(\frac{\beta_{1}-\alpha_{2}}{2})\sin(\frac{\gamma_{1}+\gamma_{3}}{2})$

##  4 Proof of The Inequality

Now that we have simplified each group of terms in the previous two sections,
we turn to the proof of inequality (1). We first prove the that the sum of the
$X$, $Y$, $W$-terms, and $P_{1}-\frac{1}{2}$ (which is the first line of Lemma
2) is non-negative. In particular, using the simplifications obtained in the
previous two sections, we prove the following theorem:

###### Theorem 2

: For the convex quadrilateral presented in figure 1 where
$\gamma_{2}+\gamma_{3}\leq\pi$ and $\gamma_{3}+\gamma_{4}\leq\pi$, we have

$\begin{array}[]{cccc}\sin
X\sin\frac{W^{\prime}}{2}\sin\frac{Y}{2}\sin(\frac{\alpha_{1}-\beta_{4}}{2})&&&\\\
+\sin
Y\sin\frac{W}{2}\sin\frac{X}{2}\sin(\frac{\beta_{1}-\alpha_{2}}{2})&\begin{array}[]{c}-2\sin^{2}\frac{X}{2}\cos^{2}\frac{W}{2}\cos^{2}\frac{Y}{2}\\\
\\\
-2\cos^{2}\frac{X}{2}\cos^{2}\frac{W^{\prime}}{2}\sin^{2}\frac{Y}{2}\end{array}&\geq&0\\\
+\sin
W\cos\frac{X}{2}\cos\frac{Y}{2}\sin(\frac{\gamma_{1}+\gamma_{3}}{2})&&&\end{array}$

(4)

The first column in this inequality is nothing but the simplified expressions
of the sum of the $X$, $Y$, and $W$-terms, whereas the second column is the
first line in Lemma 2 (which is $P_{1}-\frac{1}{2}$). To prove Theorem 2, we
first need the following corollary:

###### Corollary 1

: On the convex quadrilateral depicted in figure 1, we have the following 3
inequalities:

$\left|\sin\left(\frac{\alpha_{1}-\beta_{4}}{2}\right)\right|\leq\sin\left(\frac{W^{\prime}-Y}{2}\right)$

$\left|\sin\left(\frac{\beta_{1}-\alpha_{2}}{2}\right)\right|\leq\sin\left(\frac{W-X}{2}\right)$

$\sin\left(\frac{\gamma_{1}+\gamma_{3}}{2}\right)\geq\sin\left(\frac{X+Y}{2}\right)$

Proof: By using
$\sin(\frac{\alpha_{3}-\beta_{2}}{2})+\sin(\frac{\alpha_{3}+\beta_{2}}{2})=2\sin\frac{\alpha_{3}}{2}\cos\frac{\beta_{2}}{2}$
and noticing that

$\alpha_{3}-\beta_{2}=\alpha_{1}-\beta_{4}$ and
$\alpha_{3}+\beta_{2}=W^{\prime}-Y$, we conclude:

$\sin(\frac{\alpha_{1}-\beta_{4}}{2})=\sin(\frac{Y-W^{\prime}}{2})+2\sin\frac{\alpha_{3}}{2}\cos\frac{\beta_{2}}{2}\geq\sin(\frac{Y-W^{\prime}}{2})$

In addition,
$\sin(\frac{\alpha_{3}-\beta_{2}}{2})\leq\sin(\frac{\alpha_{3}+\beta_{2}}{2})$
implies that
$\sin(\frac{\alpha_{1}-\beta_{4}}{2})\leq\sin(\frac{W^{\prime}-Y}{2})$ which
concludes the first inequality in this corollary.

The other two inequalities can be proved in a similar way by using

$\sin(\frac{\beta_{3}-\alpha_{4}}{2})+\sin(\frac{\beta_{3}+\alpha_{4}}{2})=2\sin\frac{\beta_{3}}{2}\cos\frac{\alpha_{4}}{2}$
for the second one and

$\sin(\frac{\gamma_{1}+\gamma_{3}}{2})-\sin(\frac{\gamma_{1}-\gamma_{3}}{2})=2\cos\frac{\gamma_{1}}{2}\sin\frac{\gamma_{3}}{2}$
for the third one.

We also have to use that $\beta_{3}-\alpha_{4}=\beta_{1}-\alpha_{2}$ and
$\beta_{3}+\alpha_{4}=W-X$ for the second inequality and
$\gamma_{1}-\gamma_{3}=X+Y$ for the last one.

$\bf{Proof\,\,of\,\,Theorem\,2}$: Based on Corollary 1, the left hand side of
(4) is not less than:

$\begin{array}[]{ccc}\sin
X\sin\frac{W^{\prime}}{2}\sin\frac{Y}{2}\sin(\frac{Y-W^{\prime}}{2})&&\\\
+\sin
Y\sin\frac{W}{2}\sin\frac{X}{2}\sin(\frac{X-W}{2})&\begin{array}[]{c}-2\sin^{2}\frac{X}{2}\cos^{2}\frac{W}{2}\cos^{2}\frac{Y}{2}\\\
-2\cos^{2}\frac{X}{2}\cos^{2}\frac{W^{\prime}}{2}\sin^{2}\frac{Y}{2}\end{array}&\\\
+\sin W\cos\frac{X}{2}\cos\frac{Y}{2}\sin(\frac{X+Y}{2})&&\end{array}$

Expanding $\displaystyle{\sin(X+Y)/2=\sin X/2\cos Y/2+\cos X/2\sin Y/2}$ in
the third line of the first column and combining the terms with the ones in
the second column yields:

$\begin{array}[]{cc}2\sin\frac{X}{2}\cos\frac{X}{2}\sin\frac{W^{\prime}}{2}\sin\frac{Y}{2}\sin(\frac{Y-W^{\prime}}{2})&\\\
+2\sin\frac{Y}{2}\cos\frac{Y}{2}\sin\frac{W}{2}\sin\frac{X}{2}\sin(\frac{X-W}{2})&\\\
+2\cos\frac{W}{2}\sin\frac{X}{2}\cos^{2}\frac{Y}{2}\left(\sin\frac{W}{2}\cos\frac{X}{2}-\cos\frac{W}{2}\sin\frac{X}{2}\right)&\\\
+2\sin\frac{W}{2}\cos^{2}\frac{X}{2}\sin\frac{Y}{2}\left(\sin\frac{W^{\prime}}{2}\cos\frac{Y}{2}-\cos\frac{W^{\prime}}{2}\sin\frac{Y}{2}\right)&\end{array}$

Note that the last two quatities in parentheses are nothing but
$\displaystyle{\sin(W-X)/2}$ and $\displaystyle{\sin(W^{\prime}-Y)/2}$.
Combined with the first two rows gives us

$2\cos\frac{X}{2}\sin\frac{Y}{2}\sin\left(\frac{W^{\prime}-Y}{2}\right)\left(-\sin\frac{X}{2}\cos\frac{W}{2}+\sin\frac{W}{2}\cos\frac{X}{2}\right)$

$+2\sin\frac{X}{2}\cos\frac{Y}{2}\sin(\frac{W-X}{2})\left(-\sin\frac{Y}{2}\cos\frac{W^{\prime}}{2}+\cos\frac{Y}{2}\sin\frac{W^{\prime}}{2}\right)$

$=2\cos\frac{X}{2}\sin\frac{Y}{2}\sin\left(\frac{W^{\prime}-Y}{2}\right)\sin\left(\frac{W-X}{2}\right)+2\sin\frac{X}{2}\cos\frac{Y}{2}\sin\left(\frac{W-X}{2}\right)\sin\left(\frac{W^{\prime}-Y}{2}\right)$

$=2\sin\left(\frac{W^{\prime}-Y}{2}\right)\sin\left(\frac{W-X}{2}\right)\sin\left(\frac{X+Y}{2}\right)\,\geq
0$

This last expression is nonnegative because
$W^{\prime}-Y=\alpha_{3}+\beta_{2}$, $W-X=\alpha_{4}+\beta_{3}$, and these
angles will be between $0$ and $\pi$ when divided by 2 (in addition to $X+Y$),
which conlcudes the proof of Theorem 2.

Remark: Now that Theorem 2 has been established, let us go back and collect
all remaining terms on the left hand side of (4). We obviously have the final
answer which is
$2\sin\frac{W^{\prime}-Y}{2}\sin\frac{W-X}{2}\sin\frac{X+Y}{2}$. However, we
have some additional terms remaining from corollary 1. Namely, we have the
following three terms from the inequalities used in Corollary 1 (substituted
in the left column of Theorem 2) which remain on the left hand side of (4):

$\,\,\,\,\,2\sin
X\sin\frac{W^{\prime}}{2}\sin\frac{Y}{2}\sin\frac{\alpha_{3}}{2}\cos\frac{\beta_{2}}{2}$

$+2\sin
Y\sin\frac{W}{2}\sin\frac{X}{2}\sin\frac{\beta_{3}}{2}\cos\frac{\alpha_{4}}{2}$
(5)

$+2\sin
W\cos\frac{X}{2}\cos\frac{Y}{2}\cos\frac{\gamma_{1}}{2}\sin\frac{\gamma_{3}}{2}$

In order to finish the proof of the quadilaterial inequality (1), we need to
show that the remaining terms give a positive answer when added to the second
line of Lemma 2, which is
$-2\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\sin\frac{\gamma_{3}+\gamma_{1}}{2}$.
Well, if the quadrilateral at hand satisfies
$\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\leq 0$,
then our result is immediate. Accordingly, we still need to prove our
inequality (1) under the assumption that
$\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\geq 0$.
Let us start by adding the resulting positive term
$2\sin\frac{W^{\prime}-Y}{2}\sin\frac{W-X}{2}\sin\frac{X+Y}{2}$ from Theorem 2
to the second line of Lemma 2:

$\displaystyle{2\sin\frac{W^{\prime}-Y}{2}\sin\frac{W-X}{2}\sin\frac{X+Y}{2}-2\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\sin\frac{\gamma_{3}+\gamma_{1}}{2}}$

$\displaystyle{\geq
2\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\left(\sin\frac{X+Y}{2}-\sin\frac{\gamma_{3}+\gamma_{1}}{2}\right)}$

$\displaystyle{=-4\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\cos\frac{\gamma_{1}}{2}\sin\frac{\gamma_{3}}{2}}$

where the first inequality follows from Corollary 1 and the last equality
comes from the fact that $X+Y=\gamma_{1}-\gamma_{3}$. We can see that, under
our assumption
$\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\geq 0$,
our outcome is still negative. Accordingly, we take this answer and add to it
the third line of (5) to get

$\displaystyle{4\sin\frac{W}{2}\cos\frac{W}{2}\cos\frac{X}{2}\cos\frac{Y}{2}\cos\frac{\gamma_{1}}{2}\sin\frac{\gamma_{3}}{2}}$

$\displaystyle{-4\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\cos\frac{\gamma_{1}}{2}\sin\frac{\gamma_{3}}{2}}$

This is easily seen to be positive: We factor out
$4\cos\frac{\gamma_{1}}{2}\sin\frac{\gamma_{3}}{2}$ which is positive. We then
note that $\sin\frac{W}{2}\cos\frac{X}{2}\geq\sin\frac{W-X}{2}$ and
$\sin\frac{W^{\prime}}{2}\cos\frac{Y}{2}\geq\sin\frac{W^{\prime}-Y}{2}$,
which, by Corollary (1), are not less than
$\sin\frac{\alpha_{1}-\beta_{4}}{2}$ and $\sin\frac{\beta_{1}-\alpha_{2}}{2}$
respectively. Since
$\sin\frac{\beta_{4}-\alpha_{1}}{2}\sin\frac{\alpha_{2}-\beta_{1}}{2}\geq 0$,
we can see that we end up with a non-negative answer, which finally finishes
the proof of inequality (1).

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