# A note on the largest sum-free sets of integers

Yifan Jing  Department of Mathematics, University of Illinois at Urbana-
Champaign, Urbana IL, USA<EMAIL_ADDRESS>and  Shukun Wu  Department of
Mathematics, University of Illinois at Urbana-Champaign, Urbana IL, USA
<EMAIL_ADDRESS>
###### Abstract.

Given $A$ a set of $N$ positive integers, an old question in additive
combinatorics asks that whether $A$ contains a sum-free subset of size at
least $N/3+\omega(N)$ for some increasing unbounded function $\omega$. The
question is generally attacked in the literature by considering another
conjecture, which asserts that as $N\to\infty$,
$\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}(\mathbbm{1}_{(1/3,2/3)}-1/3)(nx)\to\infty$. This conjecture, if true, would
also imply that a similar phenomenon occurs for $(2k,4k)$-sum-free sets for
every $k\geq 1$. In this note, we prove the latter result directly. The new
ingredient of our proof is a structural analysis on the host set $A$, which
might be of independent interest.

###### 2010 Mathematics Subject Classification:

Primary 11B30; Secondary 05D10, 11K70

##  1\. Introduction

Given $X$ a set of positive integers, we say $X$ is _sum-free_ if there does
not exist $x_{1},x_{2},y_{1}$ in $X$ with $x_{1}+x_{2}=y_{1}$. The study of
sum-free sets can be tracked back to Schur [21], where he used extremal
properties of sum-free sets to prove that the Fermat’s last theorem does not
hold in $\mathbb{F}_{p}$.

Let $A$ be a finite subset of an abelian group $G$. A natural question is: How
large can the maximal sum-free subset of $A$ be? We use
$\mathscr{M}_{(2,1)}(A)$ in this note to denote the size of maximal sum-free
subset of $A$. When $A=G$, interest in determining $\mathscr{M}_{(2,1)}(G)$
for finite abelian groups $G$ goes back over 50 years. In 1969, Diananda and
Yap [7] determined the size of the maximal sum-free set in $G$ whenever $|G|$
has a prime factor $p\not\equiv 1\pmod{3}$, and the question is completely
solved for all finite abelian groups recently by Green and Ruzsa [12]. There
is a large body of literature on extremal problems of sum-free sets. For
example, see [1, 3, 11, 18] for counting sum-free sets, and [6, 13, 19] for
the Erdős–Rothschild problems of sum-free sets.

In this note, we are interested in finding the largest sum-free subset of $A$
when $A$ is a set of $N$ integers. More precisely, we define

$\mathscr{M}_{(2,1)}(N)=\inf_{A\subseteq\mathbb{Z}^{>0},|A|=N}\mathscr{M}_{(2,1)}(A).$

The study of determining $\mathscr{M}_{(2,1)}(N)$ originates from a question
asked by Erdős [10]. For the upper bound, in a recent breakthrough by
Eberhard, Green, and Manners [9], they showed that
$\mathscr{M}_{(2,1)}(N)=N/3+o(N)$. While the sharp lower bound on
$\mathscr{M}_{(2,1)}(N)$ is widely open. In particular, the following
conjecture is made in a series of papers [5, 9, 10]; for more background we
refer to the recent survey by Tao and Vu [22].

######  Conjecture 1.

There is a function $\omega(N)\to\infty$ as $N\to\infty$, such that

$\mathscr{M}_{(2,1)}(N)\geq\frac{N}{3}+\omega(N).$

In [10], using a probabilistic argument, Erdős showed that
$\mathscr{M}_{(2,1)}(N)\geq N/3$. This argument is actually not complicated:
Let $\Omega$ be a maximal sum-free subset of $\mathbb{R}/\mathbb{Z}$, then
$(\Omega+\Omega)\cap\Omega=\varnothing$. By Kneser’s inequality [16], we have
an upper bound on $|\Omega|$ that $|\Omega|\leq 1/3$. Now we fix a maximal
sum-free set $\Omega$ of measure $1/3$ (e.g. $(1/3,2/3)$). For any
$x\in\mathbb{R}/\mathbb{Z}$, we let $A_{x}$ be the set of integers $n$ in $A$
such that $nx\in\Omega$. Then clearly $A_{x}$ is sum-free, and hence we have

$\mathscr{M}_{(2,1)}(A)\geq|A_{x}|=\sum_{n\in A}\mathbbm{1}_{\Omega}(nx),$

where $\mathbbm{1}_{\Omega}$ is the characteristic function of $\Omega$. When
$x$ is chosen randomly from $\mathbb{R}/\mathbb{Z}$, the expected size of
$A_{x}$ is $N/3$, which implies that $|A_{x}|\geq N/3$ for some $x$.

The lower bound estimate of $\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}\mathbbm{1}_{\Omega}(nx)$ is later improved to $(N+1)/3$ by Alon and
Kleitman [2], and the best estimate up to date is obtained by Bourgain [5],
where he showed that $\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}\mathbbm{1}_{\Omega}(nx)\geq(N+2)/3$. In particular, the following
conjecture would imply Conjecture 1.

######  Conjecture 2.

There is a function $\omega(N)\to\infty$ as $N\to\infty$, such that for every
set $A$ of $N$ positive integers, there exists a maximal sum-free set
$\Omega\subseteq\mathbb{R}/\mathbb{Z}$ that

$\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}\Big{(}\mathbbm{1}_{\Omega}-\frac{1}{3}\Big{)}(nx)=\omega(N).$

The analogues conjectures for $(k,\ell)$-sum-free sets are also well-studied.
By _$(k,\ell)$ -sum-free_ we mean that for every $k+\ell$ elements
$x_{1},\dots,x_{k},y_{1},\dots,y_{\ell}$, we always have
$\sum_{i=1}^{k}x_{i}\neq\sum_{j=1}^{\ell}y_{j}$ (through out the paper we
always assume that $k<\ell$). We can similar define
$\mathscr{M}_{(k,\ell)}(A)$ to be the size of maximal $(k,\ell)$-sum-free
subset of $A$, and let

$\mathscr{M}_{(k,\ell)}(N)=\inf_{A\subseteq\mathbb{Z}^{>0},|A|=N}\mathscr{M}_{(k,\ell)}(A).$

Recently, Eberhard [8] showed that $\mathscr{M}_{(1,k)}(N)=N/(1+k)+o(N)$. The
result is later generalized by the authors [15] for all $(k,\ell)$, that
$\mathscr{M}_{(k,\ell)}(N)=N/(k+\ell)+o(N)$. It remains to show whether
$\mathscr{M}_{(k,\ell)}(N)\geq N/(k+\ell)+\omega(N)$ for every pair
$(k,\ell)$, where $\omega(N)$ is a function tends to infinity as $N$ tends to
infinity. Using the probabilistic argument by Erdős, one can immediately get
$\mathscr{M}_{(k,\ell)}(N)\geq N/(k+\ell)$. In general, we believe the
following should be true, which is a generalization of Conjecture 2 to all
$(k,\ell)$-sum-free sets.

######  Conjecture 3.

There is a function $\omega(N)\to\infty$ as $N\to\infty$, such that for every
set $A$ of $N$ positive integers, there exists a maximal $(k,\ell)$-sum-free
set $\Omega(k,\ell)\subseteq\mathbb{R}/\mathbb{Z}$, and we have

$\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}\Big{(}\mathbbm{1}_{\Omega(k,\ell)}-\frac{1}{k+\ell}\Big{)}(nx)=\omega(N).$

Conjecture 3 is verified by Bourgain [5] for $(1,3)$-sum-free sets, and
recently by the authors [15] for $(k,5k)$-sum-free sets for every $k\geq 1$.
In particular, the authors observed that if Conjecture 3 holds for
$(k,\ell)$-sum-free sets, then it also holds for $(km,\ell m)$-sum-free sets
for every $m\geq 2$. Thus, if Conjecture 2 holds, then this would imply that
Conjecture 3 holds for $(2k,4k)$-sum-free sets for all $k\geq 1$. Hence the
authors believe that the $(2k,4k)$-sum-free problem is one of the most
interesting cases of Conjecture 3. In this paper, we prove the $(2k,4k)$-sum-
free case without assuming Conjecture 2.

######  Theorem 1.1.

For every $k\geq 1$, there is a function $\omega(N)=\log N/\log\log N$, such
that for every set $A$ of $N$ positive integers, there exists a maximal
$(2k,4k)$-sum-free set $\Omega(2k,4k)\subseteq\mathbb{R}/\mathbb{Z}$, and we
have

$\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}\Big{(}\mathbbm{1}_{\Omega(2k,4k)}-\frac{1}{6k}\Big{)}(nx)\gg\omega(N).$

In particular, there is an absolute constant $c>0$, such that

$\mathscr{M}_{(2k,4k)}(N)\geq\frac{N}{6k}+c\,\omega(N).$

We remark that Theorem 1.1 is not enough for the sum-free conjecture. Indeed,
let $\Omega_{1}=(1/6,1/3)$ and $\Omega_{2}=(2/3,5/6)$ be two subsets of
$\mathbb{R}/\mathbb{Z}$. Note that both $\Omega_{1}$ and $\Omega_{2}$ are
$(2,4)$-sum-free in $\mathbb{R}/\mathbb{Z}$. By Theorem 1.1, one can find two
elements $x_{1},x_{2}\in\mathbb{R}/\mathbb{Z}$ such that $A_{x_{1}}:=\\{n\in
A:nx_{1}\in\Omega_{1}\\}$ has at least $N/6+\omega(N)$ elements, and
$A_{x_{2}}:=\\{n\in A:nx_{2}\in\Omega_{2}\\}$ also has at least
$N/6+\omega(N)$ elements. Clearly, both $A_{x_{1}}$ and $A_{x_{2}}$ are
$(2,4)$-sum-free. If $x_{1}=x_{2}$, the above argument implies the sum-free
conjecture since the union of $\Omega_{1}$ and $\Omega_{2}$ is sum-free in
$\mathbb{R}/\mathbb{Z}$. However, it is possible that $x_{1}\neq x_{2}$ and
thus $A_{x_{1}}$ and $A_{x_{2}}$ may share many common elements.

The new ingredients used in proving Theorem 1.1 contain a structural analysis
of the given set $A$. Recall that a Følner sequence in $(\mathbb{N},\cdot)$ is
a collection of sets of integers $\\{F_{n}\\}_{n=1}^{\infty}$, such that for
every $a\in\mathbb{N}^{>0}$,

$\lim_{n\to\infty}\frac{|F_{n}\,\triangle\,(a\cdot F_{n})|}{|F_{n}|}=0.$

Thus, when $A$ is close to a set in a Følner sequence, we expect that
$|A\,\triangle\,(a\cdot A)|$ is small for appropriate $a$. Inspired by the
structure Følner sequences (which is the only known constructive example whose
largest $(k,\ell)$-sum-free subsets have cardinality $N/(k+\ell)+o(N)$, for
all $(k,\ell)$, see [8, 15]), we split our proof into two cases: when
$|A\,\triangle\,(a\cdot A)|$ is small (close to having multiplicative
structures), and when $|A\,\triangle\,(a\cdot A)|$ is large (close to having
additive structures). We mainly consider the case $a=3$ here since the Fourier
coefficients appeared in the later proofs contain a multiplicative character
mod $3$. The first case is resolved by an application of Littlewood–Paley
theorem, and the proof we given also works for sum-free sets. In the second
case, since the main factors in the Fourier coefficients are not
multiplicative, we carefully sieve out small prime factors, and apply a
variant of the weak Littlewood conjecture. The nontrivial lower bound of
$\omega(N)$ eventually comes from the largeness of $|A\,\triangle\,(3\cdot
A)|$. For convenience, we make the following definition.

######  Definition 1.2.

We say a set $A\subseteq\mathbb{Z}^{>0}$ is a _geometric set_ , if
$|A\,\triangle\,(3\cdot A)|\ll|A|^{c}$ for an absolute constant $c<1$.

The paper is organized as follows. In the next section, we deal with the case
when $|A\,\triangle\,(3\\!\cdot\\!A)|$ is small (we actually prove a more
general result there). In Section 3, we prove a generalized version of the
weak Littlewood conjecture, which is used later in the proof. In Section 4, we
prove the case when $|A\,\triangle\,(3\\!\cdot\\!A)|$ is large, and finish the
proof of Theorem 1.1.

### Notation

Given a set $A$ and a positive integer $k$, we use $kA$ to denote the set
$\\{a_{1}+\dots+a_{k}:a_{i}\in A\text{ for }1\leq i\leq k\\},$ and use $k\cdot
A$ to denote the set $\\{ka:a\in A\\}$. For every
$\theta\in\mathbb{R}/\mathbb{Z}$, we write $e(\theta)=e^{2\pi i\theta}$. We
use the standard Vinogradov notation. That is, $f\ll g$ means $f=O(g)$, and
$f\asymp g$ if $f\ll g$ and $f\gg g$.

##  2\. When $A$ is geometric

In this section, we study the size of the largest sum-free sets when the host
set $A$ is well-structured. Let
$\Omega=(1/3,2/3)\subseteq\mathbb{R}/\mathbb{Z}$, and it is easy to check that
$\Omega$ is sum-free in $\mathbb{R}/\mathbb{Z}$. Define $\mathbbm{1}_{\Omega}$
as the characteristic function of $\Omega$, and let
$f=\mathbbm{1}_{\Omega}-1/3$ be the balanced function of
$\mathbbm{1}_{\Omega}$. By orthogonality of characters we have

$\widehat{f}(n)=\begin{cases}0&\quad\text{if }n=0,\\\
\widehat{\mathbbm{1}_{\Omega}}(n)&\text{otherwise.}\end{cases}$

When $n>0$,

$\displaystyle\widehat{f}(n)$
$\displaystyle=\int_{\mathbb{R}/\mathbb{Z}}\mathbbm{1}_{\Omega}(x)e(-nx)\,\mathrm{d}x$
$\displaystyle=\frac{1}{2\pi
in}\Big{(}-e\Big{(}-\frac{2n}{3}\Big{)}+e\Big{(}-\frac{n}{3}\Big{)}\Big{)}$
$\displaystyle=\frac{1}{\pi
n}e\Big{(}\frac{n}{2}\Big{)}\sin\Big{(}\frac{n\pi}{3}\Big{)}.$

Therefore we obtain

$\displaystyle f(x)$ $\displaystyle=\sum_{n\neq
0}\widehat{f}(n)e(nx)=\sum_{n\neq 0}\frac{1}{\pi
n}e\Big{(}\frac{n}{2}\Big{)}\sin\Big{(}\frac{n\pi}{3}\Big{)}e(nx)$ (1)
$\displaystyle=-\frac{\sqrt{3}}{\pi}\sum_{n\geq 1}\frac{\chi(n)}{n}\cos(2\pi
nx),$

where $\chi(n)$ is a multiplicative character mod $3$, that is

$\chi(n)=\begin{cases}1\quad&\text{when }n\equiv 1\pmod{3},\\\ -1&\text{when
}n\equiv 2\pmod{3},\\\ 0&\text{otherwise.}\end{cases}$

As we mentioned in the introduction,

(2) $\displaystyle\mathscr{M}_{(2,1)}(A)-\frac{N}{3}$
$\displaystyle\geq\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{m\in A}f(mx).$

In the rest of the section, we are going to estimate $\max_{x}\sum_{m\in
A}f(mx)$ when $A$ has certain algebraic constrains.

Let $\mathcal{P}_{3}$ be the collection of intervals
$[3^{k},3^{k+1})\cap\mathbb{N}$, where $k\geq 0$ is an integer. Let $A$ be a
set of $N$ positive integers. We say that $A$ is $(3,c)$-lacunary, if there is
a subset $\mathcal{P}\subset\mathcal{P}_{3}$ with $|\mathcal{P}|\gg N^{c}$,
such that each interval in $\mathcal{P}$ contains at least one element of $A$,
and the intervals in $\mathcal{P}$ form a cover of $A$.

When the set $A$ is $(3,c)$-lacunary, in some sense the distribution of $A$ is
not far away from a union of long geometric progressions, and we expect that
approximately there is a square root cancellation for $\|\sum_{m\in
A}e(mx)\|_{L^{1}(\mathbb{R}/\mathbb{Z})}$. To make this observation rigorous,
we use the Littlewood–Paley theorem.

######  Theorem 2.1 (Littlewood–Paley).

Let $g(x)$ be the trigonometric series

$g(x)=\sum_{n=1}^{\infty}a_{n}e^{inx}.$

For the sequence $\\{a_{n}\\}$, we consider the following auxiliary truncated
function $\Delta_{k}$ defined as

$\Delta_{k}(x)=\sum_{n=n_{k-1}+1}^{n_{k}}a_{n}e^{inx},$

where $n_{0}=0$, $n_{1}=1$, $n_{k+1}/n_{k}\geq\alpha>1$. Then for any
$1<p<\infty$.

$\Bigg{\|}\Big{(}\sum_{k=1}^{\infty}|\Delta_{k}|^{2}\Big{)}^{1/2}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})}\leq
C_{p,\alpha}\|f\|_{L^{p}(\mathbb{R}/\mathbb{Z})}.$

The proof of the Littlewood–Paley theorem can be found in [23, Chapter XV,
Theorem 4.7], and the constant $C_{p,\alpha}$ was calculated in [4] when
$\alpha=2$.

The next lemma gives us a key estimate for lacunary sets.

######  Lemma 2.2.

Assume that $\\{a_{n}\\}_{n=1}^{N}$ is $(3,c)$-lacunary. Define $g(x)$ as

$g(x)=\sum_{n=1}^{N}e^{ia_{n}x}.$

Then $\|g\|_{L^{1}(\mathbb{R}/\mathbb{Z})}\gg N^{c/3}$.

###### Proof.

Let $\phi_{k}$ be the indicator function of the interval $[3^{k},3^{k+1})$. We
denote by $\Delta_{k}(g)$ the Fourier truncation

$\Delta_{k}(g)(x)=\sum_{n=1}^{N}\phi_{k}(a_{n})e^{ia_{n}x}.$

By the Littlewood–Paley theorem (Theorem 2.1), we have for any $1<p<\infty$,

(3) $\|g\|_{L^{p}(\mathbb{R}/\mathbb{Z})}\geq
C_{p}\Bigg{\|}\Big{(}\sum_{k=0}^{\infty}|\Delta_{k}g(x)|^{2}\Big{)}^{1/2}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})}=C_{p}\Bigg{\|}\Big{(}\sum_{k\in
E}|\Delta_{k}g(x)|^{2}\Big{)}^{1/2}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})}.$

Here the set $E$ contains all the positive integers $k$ satisfying
$[3^{k},3^{k+1})\in\mathcal{P}$, so $|E|=|\mathcal{P}|\gg N^{c}$. We bound the
right hand side of equation (3) using Hölder’s inequality so that

$\Bigg{\|}\Big{(}\sum_{k\in
E}|\Delta_{k}g(x)|^{2}\Big{)}^{1/2}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})}\geq|E|^{-1/2}\Bigg{\|}\Big{(}\sum_{k\in
E}|\Delta_{k}g(x)|\Big{)}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})},$

which clearly implies

$\Bigg{\|}\Big{(}\sum_{k\in
E}|\Delta_{k}g(x)|^{2}\Big{)}^{1/2}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})}\geq|E|^{-1/2}\sum_{k\in
E}\bigg{\|}\Delta_{k}g(x)\bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}.$

Using Hölder’s inequality again, we get
$\|\Delta_{k}g\|_{2}^{2}\leq\|\Delta_{k}g\|_{1}\|\Delta_{k}g\|_{\infty}$, and
this implies $\|\Delta_{k}g\|_{L^{1}}\geq 1$ uniformly in $k$. Therefore,

(4) $\Bigg{\|}\Big{(}\sum_{k\in
E}|\Delta_{k}g(x)|^{2}\Big{)}^{1/2}\Bigg{\|}_{L^{p}(\mathbb{R}/\mathbb{Z})}\geq|E|^{1/2}\gg
N^{c/2}.$

Since
$\|\Delta_{k}g\|_{p}^{p}\leq\|\Delta_{k}g\|_{1}\|\Delta_{k}g\|_{\infty}^{p-1}$,
we can bound $\|g\|_{L^{p}}^{p}$ easily by

(5) $\|g\|_{L^{1}(\mathbb{R}/\mathbb{Z})}\geq
N^{1-p}\|g\|_{L^{p}(\mathbb{R}/\mathbb{Z})}^{p}$

Finally, we combine estimates (3), (4) and (5) to finish the proof of this
lemma, by choosing $p=1+c/6$. ∎

Now we are going to estimate the right hand side of (2) with assumption that
$A$ is $(3,c)$-lacunary for some constant $c>0$. Let

$F(x)=\sum_{m\in A}f(mx).$

Since $f$ is a balanced function, we have $\int_{\mathbb{R}/\mathbb{Z}}F=0$,
and this implies that

(6)
$\max_{x\in\mathbb{R}/\mathbb{Z}}F(x)\geq\frac{1}{2}\big{\|}F\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}.$

Let $P\asymp N^{2}$ be a prime, and let $\mathcal{M}$ be the collection of
square-free integers generated by primes smaller than $P$. Let $\mu$ be the
Möbius function, so by equation (1),

$\displaystyle\sum_{k\in\mathcal{M}}\frac{\mu(k)\chi(k)}{k}\sum_{m\in
A}f(mkx)$ $\displaystyle=-\frac{\sqrt{3}}{\pi}\sum_{m\in A,\,n\geq
1}\frac{\chi(n)}{n}\cos(2\pi mnx)\sum_{k\in\mathcal{M},\,k\mid n}\mu(k)$
$\displaystyle=-\frac{\sqrt{3}}{\pi}\sum_{\begin{subarray}{c}m\in A\\\ n\geq
1,n\in\mathcal{N}\end{subarray}}\frac{\chi(n)}{n}\cos(2\pi mnx),$

where $\mathcal{N}$ is the set of integers $n$ such that for every $p<P$,
$\gcd(n,p)=1$.

Therefore, by Minkowski’s inequality we have

$\displaystyle\Bigg{\|}\sum_{k\in\mathcal{M}}\frac{\mu(k)\chi(k)}{k}\sum_{m\in
A}f(mkx)\Bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}$
$\displaystyle\gg\Bigg{\|}\sum_{m\in A}\cos(2\pi
mx)\Bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}$
$\displaystyle\quad-\Bigg{\|}\sum_{\begin{subarray}{c}m\in A\\\
n>1,n\in\mathcal{N}\end{subarray}}\frac{\chi(n)}{n}\cos(2\pi
mnx)\Bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}.$

Via Cauchy–Schwarz inequality and Plancherel, the second term is bounded by

$\Bigg{\|}\sum_{\begin{subarray}{c}m\in A\\\
n>1,n\in\mathcal{N}\end{subarray}}\frac{\chi(n)}{n}\cos(2\pi
mnx)\Bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}\leq C|A|P^{-1/2}.$

Note that $P\asymp N^{2}$. By Merten’s estimate, we have

$\displaystyle\Bigg{\|}\sum_{m\in A}\cos(2\pi
mx)\Bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}$
$\displaystyle\ll\sum_{k\in\mathcal{M}}\frac{|\mu(k)|}{k}\big{\|}F(x)\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}+O(1).$
$\displaystyle\ll\prod_{p<P}\Big{(}1+\frac{1}{p}\Big{)}\big{\|}F(x)\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}\asymp\log
N\big{\|}F(x)\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}.$

Since $A$ is $(3,c)$-lacunary, we invoke Lemma 2.2 to get

$\Bigg{\|}\sum_{m\in A}\cos(2\pi
mx)\Bigg{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}\gg N^{c},$

which implies

$\big{\|}F(x)\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}\gg\frac{N^{c/3}}{\log
N}.$

Finally, we use estimate (6) to conclude
$\max_{x\in\mathbb{R}/\mathbb{Z}}F(x)\gg N^{c/4}$. ∎

##  3\. A density estimate

In this section, we prove a generalization of the McGehee–Pigno–Smith theorem
[20], based on the ideas given by Bourgain [5]. Recall that the weak
Littlewood problem [14] is to ask to estimate

$I(N):=\min_{A\subseteq\mathbb{Z},|A|=N}\int_{\mathbb{R}/\mathbb{Z}}\Big{|}\sum_{n\in
A}e^{inx}\Big{|}\,\mathrm{d}x.$

The conjecture, $I(N)\gg\log N$, is resolved by McGehee, Pigno, and Smith
[20], and independently by Konyagin [17].

Let $\mathcal{N}_{1}$ be the set of natural numbers that only contains prime
factors at least $Q$, where $Q\asymp(\log N)^{100}$ is a prime. We will use
the following lemma from [5, Section 5].

######  Lemma 3.1.

Let $A$ be a finite subset of $\mathbb{Z}^{+}$ with $|A|=N$. For all $R\geq
1$, we define

$A_{R}=\\{m\in A:m<R\\}.$

Also, we use ${\rm Proj}_{R}\sum a_{k}e^{ikx}$ to denote the truncated sum
$\sum_{|k|\leq R}a_{k}e^{ikx}$. Assume $|a_{n}|\leq 1$ and $P>(\log N)^{20}$.
Then there is an absolute big constant $C$, such that

$\Bigg{\|}{\rm Proj}_{R}\sum_{\begin{subarray}{c}n\in\mathcal{N}_{1},m\in
A\end{subarray}}\frac{a_{n}}{n}e^{imnx}\Bigg{\|}_{2}<CP^{-1/15}|A_{R}|^{1/2}.$

Now we are able to prove our technical lemma. The proof basically follows the
arguments used in [15] (which is also based on the ideas in [20] and [5]), but
with more explanations and a stronger conclusion.

######  Lemma 3.2.

Let $B=\\{m_{1},\ldots,m_{M}\\}$ be a finite subset of $\mathbb{N}^{>0}$ and
let $P>(\log N)^{100}$. Assume that $w:\mathbb{N}^{>0}\to\mathbb{C}$ is a
weight. Then there exists a function $\Phi(x)$ with $\|\Phi\|_{\infty}<10$
such that

(7)
$\bigg{|}\bigg{\langle}\sum_{j=1}^{M}e^{im_{j}x}w(m_{j}),\Phi(x)\bigg{\rangle}\bigg{|}\gg\sum_{j=1}^{M}\frac{|w(m_{j})|}{j};$

while for any $\beta\in\mathbb{Z}$,

(8) $\bigg{|}\bigg{\langle}\sum_{\begin{subarray}{c}n\in\mathcal{N}_{1},m\in
B\end{subarray}}\frac{a_{n}}{n}e^{i\beta
mnx},\Phi(x)\bigg{\rangle}\bigg{|}\leq C(\log M)^{-2}.$

Here $c,C$ are two absolute constants.

###### Proof.

Let $k_{0}$ be the largest natural number that $10^{6k_{0}}<M$. We group $B$
into disjoint subsets $\\{B_{k}\\}_{k=0}^{k_{0}}$ such that for $0\leq k\leq
k_{0}-1$, $|B_{k}|=10^{6k}$. Here $B_{0}=\\{m_{1}\\}$,
$B_{1}=\\{m_{2},\ldots,m_{10^{6}+1}\\},\cdots,$ and
$B_{k_{0}}=A\setminus(\bigcup_{k\leq k_{0}-1}B_{k})$. From the construction we
know $|B_{k_{0}}|\asymp 10^{6k_{0}}$. Let
$\tau:\mathbb{N}^{>0}\to\mathbb{S}^{1}$ be the argument function that
$\tau(m)w(m)\geq 0$. For each $B_{k}$, we define

$\widetilde{P}_{k}=\frac{1}{|B_{k}|}\sum_{m\in B_{k}}e^{imx}\tau(m).$

Let $I_{k}=[a_{k},b_{k}]$ be the interval with $a_{k}=\min\\{m:m\in B_{k}\\}$,
$b_{k}=\max\\{m:m\in B_{k}\\}$, and let $\xi_{k}$ be the center of $I_{k}$. We
also define

$P_{k}=\widetilde{P}_{k}\ast\left(e^{i\xi_{k}x}F_{|I_{k}|}\right),$

where $F_{C}=\sum_{|m|\leq C}\frac{C-|m|}{C}e^{imx}$ is the $C$-Féjer kernel.
Consequently,

(9) ${\rm supp}(\widehat{P}_{k})={\rm
supp}(\widehat{\widetilde{P}_{k}})\subset I_{k},$

and for any $m\in B_{k}$

$w(m)\widehat{P_{k}}(m)>10^{-6k-1}|w(m)|.$

This shows that the functions $P_{k}$ are good test functions. However, the
function $\sum_{k}P_{k}(x)$ has one drawback: It is not distributed evenly on
the torus. That is, the $L^{\infty}$-norm $\sum_{k}P_{k}(x)$ is comparably
large.

To overcome this difficulty, for each $P_{k}$, we construct a function $Q_{k}$
serving as a “compensator”. Specifically, let $\mathcal{H}$ be the Hilbert
transform in $L^{2}(\mathbb{R}/\mathbb{Z})$ that
$\widehat{\mathcal{H}{f}}(n)=-i{\rm sgn}(n)\widehat{f}(n)$, so that when $f$
is a real-valued function, $\mathcal{H}f$ is also real-valued. We define

(10)
$Q_{k}=\Big{(}e^{-(|\tilde{P}_{k}|-i\mathcal{H}[|\tilde{P}_{k}|])}\Big{)}\ast
F_{|I_{k}|}.$

Since the Fourier series of
$e^{-(|\tilde{P}_{k}|-i\mathcal{H}[|\tilde{P}_{k}|])}$ is supported in non-
positive integers,

(11) ${\rm supp}\big{(}\widehat{Q}_{k}\big{)}\subset[-|I_{k}|,0].$

Using the inequality that $|e^{-z}-1|\leq|z|$ if $z\in\mathbb{C}$ and ${\rm
Re}(z)\geq 0$, we can easily prove

(12)
$\|1-Q_{k}\|_{2}\leq\|\tilde{P}_{k}\|_{2}+\big{\|}\mathcal{H}[|\tilde{P}_{k}|]\big{\|}_{2}<2|B_{k}|^{-1/2}.$

Thus, $Q_{k}$ is approximately the identical function. In fact, $|Q_{k}|$ is
relatively small when $|P_{k}|$ is relatively large, so $Q_{k}$ can help us
“mollify” the function $P_{k}$.

We will use the functions $P_{k},Q_{k}$ to construct our test function $\Phi$.
In specific, we set $\Phi_{0}=P_{0}$ and set

(13) $\Phi_{k}=Q_{k}\Phi_{k-1}+P_{k},\hskip 28.45274pt1\leq k\leq k_{0}.$

Define $\Phi=\Phi_{k_{0}}$, which has the explicit formula

(14)
$\Phi=P_{k_{0}}+P_{k_{0}-1}Q_{k_{0}}+P_{k_{0}-2}Q_{k_{0}-1}Q_{k_{0}}+\cdots+P_{0}Q_{1}\cdots
Q_{k_{0}}.$

We claim $\|\Phi\|_{\infty}<10$. To see this, we first recall the basic
inequality: $\frac{a}{10}+e^{-a}\leq 1$ if $a\geq 0$. Then, observing
$|P_{0}|=1$ and

$\Big{\|}\frac{1}{10}|P_{k}|+|Q_{k}|\Big{\|}_{\infty}\leq\Big{\|}\big{(}\frac{1}{10}|\widetilde{P}_{k}|+e^{-|\widetilde{P}_{k}|}\big{)}\ast
F_{|I_{k}|}\Big{\|}_{\infty}\leq\Big{\|}\frac{1}{10}|\widetilde{P}_{k}|+e^{-|\widetilde{P}_{k}|}\Big{\|}_{\infty}\leq
1,$

we argue inductively using (13) to conclude our claim.

Next, we will verify (7). We will prove that for any $m\in B_{k}$,

(15) $\big{|}\widehat{\Phi}(m)-\widehat{P}_{k}(m)\big{|}\leq
10^{-1}\big{|}\widehat{P_{k}}(m)\big{|}=\frac{1}{10|B_{k}|}.$

In fact, using the support condition (11) and the equation (14), we have

$\widehat{\Phi}(m)-\widehat{P}_{k}(m)=\widehat{P}_{k_{0}}(m)+\widehat{P_{k_{0}-1}}\ast\widehat{Q_{k_{0}}}(m)+\cdots+\widehat{P_{k}}\ast(1-(Q_{k_{0}}\cdots
Q_{k+1})^{\wedge})(m),$

which, combining the support condition of $\widehat{P_{k}}$ in (9), equals to

$\sum_{j=k}^{k_{0}-1}\widehat{P_{j}}\ast(1-(Q_{k_{0}}\cdots
Q_{j+1})^{\wedge})(m).$

We estimate the above quantity using the equality

$1-Q_{k+1}\cdots
Q_{k_{0}}=(1-Q_{k+1})+Q_{k+1}(1-Q_{k+2})+\cdots+(1-Q_{k_{0}})Q_{k+1}\ldots
Q_{k_{0}-1},$

so that

$|\widehat{\Phi}(m)-\widehat{P}_{k}(m)|=\bigg{|}\sum_{j=k}^{k_{0}-1}\widehat{P_{j}}\ast(1-(Q_{k_{0}}\cdots
Q_{j+1})^{\wedge})(m)\bigg{|}\leq\sum_{j=k}^{k_{0}-1}\|P_{j}\|_{2}\sum_{l=j}^{k_{0}-1}\|1-Q_{l}\|_{2}.$

Since $\|P_{j}\|_{2}\leq|B_{j}|^{-1/2}$ and since (12), the right hand side of
the above inequality can be bounded as

$\sum_{j=k}^{k_{0}-1}\|P_{j}\|_{2}\sum_{l=j}^{k_{0}-1}\|1-Q_{l+1}\|_{2}\leq
2\sum_{j=k}^{k_{0}-1}10^{-3j}\sum_{l=j}^{k_{0}-1}10^{-3(l+1)},$

which implies what we need that

$|\widehat{\Phi}(m)-\widehat{P}_{k}(m)|\leq 10^{-3k-2}\leq
10^{-1}|\widehat{P_{k}}(m)|.$

As a consequence of (15), for any $m\in B_{k}$,

${\rm Re}(w\widehat{\Phi})(m)>\frac{1}{2}w(m)\widehat{P}_{k}(m)\geq
10^{-6k-1}|w(m)|.$

We use the above inequality to sum up all $m\in B$ to get

$\bigg{|}\bigg{\langle}\sum_{j=1}^{M}e^{im_{j}x}w(m_{j}),\Phi(x)\bigg{\rangle}\bigg{|}\geq\sum_{j=1}^{M}{\rm
Re}(w\widehat{\Phi})(m_{j})\gg\sum_{j=1}^{M}\frac{|w(m_{j})|}{j},$

and this gives (7).

Finally, we remark that the proof of (8) is given in [15]. At this point, we
complete the proof of the lemma. ∎

As an application of Lemma 3.2, we have the following corollary:

######  Corollary 3.3.

Let $B=\\{m_{1},\ldots,m_{M}\\}$ be a finite subset of $\mathbb{N}^{>0}$ and
let $P>(\log N)^{100}$. Recall that $\mathcal{N}$ is the set of natural
numbers that only contains prime factors at least $P$. Assume $|a_{n}|\leq 1$.
Then for any $\Gamma\subset\mathbb{Z}$ with $|\Gamma|\leq\log N$, we have

$\Bigg{\|}\sum_{j=1}^{M}e^{im_{j}x}w(m_{j})+\sum_{\begin{subarray}{c}n\in\mathcal{N}_{1},m\in
B\end{subarray}}\Big{(}\sum_{\beta\in\Gamma}\frac{a_{n}}{n}e^{i\beta
mnx}\Big{)}\Bigg{\|}_{1}\geq c\sum_{j=1}^{M}\frac{|w(m_{j})|}{j}-o(1).$

###### Proof.

We apply Lemma 3.2 to obtain a function $\Phi(x)$ satisfying (7) and (8). Then

$\displaystyle\Bigg{\|}\sum_{j=1}^{M}e^{im_{j}x}w(m_{j})+\sum_{\begin{subarray}{c}n\in\mathcal{N}_{1},m\in
B\end{subarray}}\Big{(}\sum_{\beta\in\Gamma}\frac{a_{n}}{n}e^{i\beta
mnx}\Big{)}\Bigg{\|}_{1}\|\Phi\|_{\infty}$
$\displaystyle\geq\Big{|}\Big{\langle}\sum_{j=1}^{M}e^{im_{j}x}w(m_{j}),\Phi(x)\Big{\rangle}\Big{|}-\sum_{\beta\in\Gamma}\Big{|}\Big{\langle}\sum_{\begin{subarray}{c}n\in\mathcal{N}_{1},m\in
B\end{subarray}}\frac{a_{n}}{n}e^{i\beta mnx},\Phi(x)\Big{\rangle}\Big{|}$
$\displaystyle>c\sum_{j=1}^{M}\frac{|w(m_{j})|}{j}-o(1),$

as desired. ∎

##  4\. When $A$ is not geometric

In this section, we consider the case when the host set $A$ is uniformly
distributed, in the sense that $|A\,\triangle\,3\cdot A|\gg N^{c}$ for some
positive constant $c>0$. We will focus on finding the largest $(2,4)$-sum-free
in $A$. Let $\Omega_{1}=(1/6,1/3)\subseteq\mathbb{R}/\mathbb{Z}$, and let
$\Omega_{2}=(2/3,5/6)\subseteq\mathbb{R}/\mathbb{Z}$. It is clear that both
$\Omega_{1}$ and $\Omega_{2}$ are $(2,4)$-sum-free in $\mathbb{R}/\mathbb{Z}$.
Let $\mathbbm{1}_{\Omega_{t}}$ be the indicator function of $\Omega_{t}$ for
$t=1,2$. Given $A\subseteq\mathbb{N}^{>0}$ of size $N$, let
$\mathscr{M}_{(2,4)}(A)$ be the size of the maximum $(2,4)$-sum-free subset of
$A$. Again we have

$\displaystyle\mathscr{M}_{(2,4)}(A)\geq\max_{x\in\mathbb{R}/\mathbb{Z}}\sum_{n\in
A}\mathbbm{1}_{\Omega_{t}}(nx),$

for every $t=1,2$. We introduce the balanced function
$f_{t}:\mathbb{R}/\mathbb{Z}\to\mathbb{C}$ defined by
$f_{t}=\mathbbm{1}_{\Omega_{t}}-\frac{1}{6}$. Hence,

$\widehat{f_{t}}(n)=\begin{cases}0\quad&\text{ if }n=0,\\\
\widehat{\mathbbm{1}_{\Omega_{t}}}(n)&\text{ otherwise}.\end{cases}$

When $n>0$, the Fourier coefficient $\widehat{f}_{t}(n)$ is

$\displaystyle\widehat{f_{t}}(n)$
$\displaystyle=\int_{\mathbb{T}}\mathbbm{1}_{\Omega_{t}}(x)e(-nx)d\mu(x)$
$\displaystyle=\frac{1}{2\pi
in}\Big{(}-e\big{(}-\frac{(t-1)n}{2}-\frac{n}{3}\big{)}+e\big{(}-\frac{(t-1)n}{2}-\frac{n}{6}\big{)}\Big{)}$
$\displaystyle=\frac{1}{\pi
n}e\Big{(}\frac{(2t-1)n}{4}\Big{)}\sin\Big{(}\frac{n\pi}{6}\Big{)}.$

Hence, for every $t=1,2$ we have

$\displaystyle f_{t}(x)=\sum_{n\neq 0}\widehat{f_{t}}(n)e(nx)=\sum_{n\neq
0}\frac{1}{\pi
n}e\Big{(}\frac{(2t-1)n}{4}\Big{)}\sin\Big{(}\frac{n\pi}{6}\Big{)}e(nx).$

We will prove that either $\|f_{1}\|_{1}\gg\log N/\log\log N$ or
$\|f_{2}\|_{1}\gg\log N/\log\log N$. However, it seems hard to estimate
$\|f_{t}\|_{1}$ directly. In order to get around this difficult, we consider
their sum $f_{1}+f_{2}$ and difference $f_{1}-f_{2}$. Let
$\Gamma(x):=f_{1}(x)+f_{2}(x)$ be the sum so that

(16) $\Gamma(x)=\frac{1}{2\pi}\sum_{n\geq
1}\frac{(-1)^{n}}{n}\sin\Big{(}\frac{n\pi}{3}\Big{)}\cos(4\pi nx).$

Also, we let $\Lambda(x)=f_{1}(x)-f_{2}(x)$ be the difference and let

$\gamma(n)=\begin{cases}1&\text{ when }n\equiv 1\pmod{4},\\\ -1&\text{ when
}n\equiv 3\pmod{4},\\\ 0&\text{ otherwise,}\end{cases}$

so that we can express $\Lambda(x)$ as

$\displaystyle\Lambda(x)$ $\displaystyle=\frac{2}{\pi}\sum_{n\geq
1}\frac{\gamma(n)}{n}\sin\Big{(}\frac{n\pi}{6}\Big{)}\sin(2\pi nx)$ (17)
$\displaystyle=\frac{2}{\pi}\Bigg{(}\sum_{\begin{subarray}{c}n\geq 1\\\ 3\nmid
n,2\nmid n\end{subarray}}\frac{1}{2n}\sin(2\pi
nx)-\sum_{\begin{subarray}{c}n\geq 1\\\ 3\mid n,2\nmid
n\end{subarray}}\frac{1}{n}\sin(2\pi nx)\Bigg{)}.$

We first deal with the function $\Gamma(x)$. Recall that $\mathcal{N}_{1}$ is
the set of positive integers $m$ such that $m$ only contains prime factor
larger than $Q\asymp(\log N)^{100}$. We also define $\mathcal{N}_{2}$ be the
set of square-free integers generated by primes that at most $Q$. Since
$(-1)^{n}\sin(n\pi/3)=-\sqrt{3}\chi(n)$ where $\chi(n)$ is a multiplicative
character mod $3$, we can sieve out the small prime factors in (16) by

(18) $\sum_{t\in\mathcal{N}_{2}}\frac{\mu(t)\chi(t)}{t}\sum_{m\in
A}\Gamma(mtx)=-\frac{\sqrt{3}}{2\pi}\sum_{m\in A}\Bigg{(}\cos(4\pi
mx)+\sum_{n\in\mathcal{N}_{1}}\frac{\chi(n)}{n}\cos(4\pi nmx)\Bigg{)},$

where $\mu$ is the Möbius function.

Next, we consider $\Lambda(x)$. Since the coefficients $\gamma(n)\sin(n\pi/6)$
is not multiplicative, $\Lambda(x)$ is more difficult to handle. As shown in
equation (4), $\Lambda(x)$ can be partition into two parts according to the
divisibility of the number 3. This motivates us to first sieve out those
integers $n$ that $3\\!\mid\\!n$, by a restricted Möbius function defined only
on integers divisible by $3$. In this way, except for the first term, all
other terms with significant contribution in the second part cancel out, while
the first part remains unchanged. Then, we use another sieve for the first
part in a similar fashion. It turns out that we can combine these two steps to
one by using the Möbius function directly as our sieve. In fact,

$\displaystyle\sum_{m\in\mathcal{N}_{2}}\frac{\mu(m)}{m}\Lambda(mx)=\frac{2}{\pi}\sum_{n\geq
1}\frac{1}{2n}\sin(2\pi nx)\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid
n\end{subarray}}\mu(m)\gamma\Big{(}\frac{n}{m}\Big{)}\sin\Big{(}\frac{n\pi}{6m}\Big{)}.$

The equation above and be further partitioned as

$\displaystyle\,\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid
n\end{subarray}}\mu(m)\gamma\Big{(}\frac{n}{m}\Big{)}\sin\Big{(}\frac{n\pi}{6m}\Big{)}$
$\displaystyle=$
$\displaystyle\,\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid n\\\ 3\nmid
n\end{subarray}}\frac{1}{2}\mu(m)+\sum_{9\nmid n,3\mid
n}\Bigg{(}\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid n\\\ 3\nmid
m\end{subarray}}-\mu(m)+\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid
n\\\ 3\mid
m\end{subarray}}\frac{1}{2}\mu(m)\Bigg{)}+\sum_{\begin{subarray}{c}9\mid
n,m\mid n\\\ m\in\mathcal{N}_{2}\end{subarray}}-\mu(m)$ $\displaystyle=$
$\displaystyle\\!:\,I_{1}(n)+I_{2}(n)+I_{3}(n).$

By the inclusive-exclusive principle, for $n\notin\mathcal{N}_{1}\cup
3\cdot\mathcal{N}_{1}$, $I_{1}(n)$ is always $0$ unless $n=1$, and $I_{3}(n)$
is always $0$. In $I_{2}(n)$,

$\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid n\\\ 3\mid
m\end{subarray}}\frac{1}{2}\mu(m)=\sum_{\begin{subarray}{c}m\in\mathcal{N}_{2},m\mid
n\\\ 3\nmid m\end{subarray}}-\frac{1}{2}\mu\Big{(}\frac{m}{3}\Big{)},$

which implies that $I_{2}(n)$ is $0$ unless $n=3$. Therefore, we get

(19) $\displaystyle\Lambda_{1}(x):=\sum_{t\in\mathcal{N}_{2}}\sum_{m\in
A}\frac{\mu(t)}{t}\Lambda(tmx)$ $\displaystyle=$
$\displaystyle\,\frac{2}{\pi}\sum_{m\in A}\Bigg{(}\frac{1}{2}\sin(2\pi
mx)-\frac{1}{2}\sin(6\pi mx)+\sum_{n\in\mathcal{N}_{1}\cup
3\cdot\mathcal{N}_{1}}\frac{\eta(n)}{n}\sin(2\pi nmx)\Bigg{)},$

where $\eta$ is defined as

$\eta(n)=\begin{cases}\frac{1}{2}&\text{ when }n\in\mathcal{N}_{1},\\\
-\frac{3}{2}&\text{ when }n\in 3\cdot\mathcal{N}_{1}.\end{cases}$

Note that $\Lambda_{1}$ indeed has the expression

$\displaystyle\Lambda_{1}(x)=\frac{1}{\pi}\sum_{m\in A}\Bigg{(}\sin(2\pi
mx)-\sin(6\pi mx)+\sum_{n\in\mathcal{N}}\frac{1}{n}\Big{(}\sin(2\pi
nmx)-\sin(6\pi nmx)\Big{)}\Bigg{)}.$

Let $B=A\,\triangle\,3\cdot A$, so by our assumption on the host set $A$,
$|B|\gg N^{c}$. For any number $m\in A\cup(3\cdot A)$, we define

$\epsilon(m)=\begin{cases}1&\text{ when }m\in A,\\\ -1&\text{ when }m\in
3\cdot A.\end{cases}$

We can simplify $\Lambda_{1}(x)$ as

(20) $\Lambda_{1}(x)=\frac{1}{\pi}\sum_{m\in B}\Big{(}\epsilon(m)\sin(2\pi
mx)+\sum_{n\in\mathcal{N}_{1}}\frac{\epsilon(m)}{n}\sin(2\pi nmx)\Big{)}.$

Finally, we combine (18) and (20) to get

$\displaystyle-\frac{2\sqrt{3}\pi}{3}\sum_{t\in\mathcal{N}_{2}}\frac{\mu(t)\chi(t)}{t}\sum_{m\in
A}\Gamma(mtx)+\frac{2\sqrt{3}\pi}{3}\sum_{t\in\mathcal{N}_{2}}\frac{\mu(t)\chi(t)}{t}\sum_{m\in
A}\Gamma(3mtx)$
$\displaystyle+i\pi\sum_{t\in\mathcal{N}_{2}}\frac{\mu(t)}{t}\sum_{m\in
A}\Lambda(2tmx)$ $\displaystyle=\ $ $\displaystyle\sum_{m\in
B}\Bigg{(}\epsilon(m)\cos(4\pi
mx)+\sum_{n\in\mathcal{N}_{1}}\frac{\chi(n)\epsilon(m)}{n}\cos(4\pi
nmx)\Bigg{)}$ $\displaystyle+i\sum_{m\in B}\Bigg{(}\epsilon(m)\sin(4\pi
mx)+\sum_{n\in\mathcal{N}_{1}}\frac{\epsilon(m)}{n}\sin(4\pi nmx)\Bigg{)}$
(21) $\displaystyle=\ $ $\displaystyle\sum_{m\in B}e^{4\pi
imx}\epsilon(m)+\sum_{m\in
B,n\in\mathcal{N}_{1}}\frac{\epsilon(m)}{n}\Big{(}(\chi(n)+1)e^{4\pi
nmx}+(\chi(n)-1)e^{-4\pi nmx}\Big{)}.$

Now we can employ Corollary 3.3 and the triangle inequality to (21), to obtain

$\displaystyle
2\sum_{t\in\mathcal{N}_{2}}\frac{|\mu(t)|}{t}\big{\|}\Gamma\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}+\sum_{t\in\mathcal{N}_{2}}\frac{|\mu(t)|}{t}\big{\|}\Lambda\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}\gg\log
N.$

Merten’s estimate tells us

$\sum_{t\in\mathcal{N}_{2}}\frac{1}{t}\ll\prod_{p<Q}\Big{(}1+\frac{1}{p}\Big{)}\asymp\log\log
N.$

Hence we have

$\max\big{\\{}\big{\|}\Gamma\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})},\big{\|}\Lambda\big{\|}_{L^{1}(\mathbb{R}/\mathbb{Z})}\big{\\}}\gg\frac{\log
N}{\log\log N}.$

This implies that there is $t\in\\{1,2\\}$, such that $\|f_{t}\|\gg\log
N/\log\log N$, and since $f_{t}$ is balanced, we get
$\max_{x\in\mathbb{R}/\mathbb{Z}}f_{t}(x)\gg\log N/\log\log N$.

With all tools in hand we are going to prove our main theorem.

###### Proof of Theorem 1.1.

Fix $k\geq 1$. We first assume $|A\,\triangle\,3\cdot A|\gg N^{1/2}$. By the
result proved earlier in this section, we may assume that for
$\Omega_{1}=(1/6,1/3)$, there is $x_{0}\in\mathbb{R}/\mathbb{Z}$ such that

$\sum_{n\in A}\mathbbm{1}_{\Omega_{1}}(nx_{0})\gg\frac{N}{6}+c\frac{\log
N}{\log\log N},$

where $c>0$ is an absolute constant. Consider the continuous group
homomorphism $\chi:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ with
$\chi(x)=kx$ for every $x$. Then the Bohr set $\chi^{-1}(\Omega_{1})$ is a
union of $k$ disjoint open intervals $I_{1},\dots,I_{k}$ in
$\mathbb{R}/\mathbb{Z}$, each of which has measure $1/6k$. It is also easy to
see that $I_{t}$ is $(2k,4k)$-sum-free for every $1\leq t\leq k$. Indeed,
suppose that $I_{1}$ is not $(2k,4k)$-sum-free, then there are $6k$ elements
$a_{1},\dots,a_{2k},b_{1},\dots,b_{4k}$ in $I_{1}$ such that
$\sum_{j=1}^{2k}a_{j}=\sum_{j=1}^{4k}b_{j}$. We may assume $a_{1}\leq\dots\leq
a_{2k}$ and $b_{1}\leq\dots\leq b_{4k}$. Define
$\alpha_{r}=\tfrac{1}{k}\sum_{j=rk+1}^{(r+1)k}a_{j}$, and
$\beta_{s}=\tfrac{1}{k}\sum_{j=sk+1}^{(s+1)k}b_{j}$ for all $r\in\\{0,1\\}$
and $s\in\\{0,1,2,3\\}$. Since $I_{1}$ is an interval, $\alpha_{0},\alpha_{1}$
and $\beta_{0},\dots,\beta_{3}$ all belong to $I_{1}$, and
$\sum_{j=0}^{1}\alpha_{j}=\sum_{j=0}^{3}\beta_{j}$. Now, using the fact that
$\chi$ is a group homomorphism, we have
$\sum_{j=0}^{1}\chi(\alpha_{j})=\sum_{j=0}^{3}\chi(\beta_{j})$, and this
contradicts the fact that $\Omega_{1}$ is $(2,4)$-sum-free.

Let $x_{1}=x_{0}/k$. By pigeonhole principle, there is
$t_{0}\in\\{1,\dots,k\\}$ such that

$\sum_{n\in A}\mathbbm{1}_{I_{t}}(nx_{1})\gg\frac{N}{6k}+\frac{c}{k}\frac{\log
N}{\log\log N},$

this finishes the proof of the first case.

Now let us assume $|A\,\triangle\,3\cdot A|\ll N^{1/2}$. Then $A$ is at least
a $(3,1/2)$-lacunary set. Let $\Omega=(1/3,2/3).$ By the result proved in
Section 2, there is $y_{0}\in\mathbb{R}/\mathbb{Z}$, such that

$\sum_{n\in A}\mathbbm{1}_{\Omega}(ny_{0})\gg\frac{N}{3}+cN^{1/8},$

for some constant $c>0$. Let $y_{1}=y_{0}/2k$, then similarly there is an open
interval $I\subseteq\mathbb{R}/\mathbb{Z}$, such that $I$ has length $1/6k$,
$I$ is $(2k,4k)$-sum-free, and

$\sum_{n\in A}\mathbbm{1}_{I}(ny_{1})\gg\frac{N}{6k}+\frac{c}{2k}N^{1/8},$

this finishes the proof. ∎

## Acknowledgements

The research is partially supported by the Arnold O. Beckman Research Award
(Campus Research Board RB2101). The authors would like to thank George Shakan
for carefully reading the first draft of the paper and for making many useful
comments.

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