# Hilbert transforms and the
equidistribution of zeros of polynomials

Emanuel Carneiro, Mithun Kumar Das, Alexandra Florea, Angel V. Kumchev,
Amita Malik, Micah B. Milinovich, Caroline Turnage-Butterbaugh, and Jiuya Wang
ICTP - The Abdus Salam International Centre for Theoretical Physics
Strada Costiera, 11, I - 34151, Trieste, Italy and IMPA - Instituto de
Matemática Pura e Aplicada
Rio de Janeiro - RJ, Brazil, 22460-320<EMAIL_ADDRESS><EMAIL_ADDRESS>Indian Institute of Science Education and Research Berhampur, Engg. School
Road, Berhampur, India-760010<EMAIL_ADDRESS>University of
California, Irvine, Department of Mathematics, Irvine, CA 92697.
<EMAIL_ADDRESS>Department of Mathematics, Towson University, 8000 York Road,
Towson, MD 21252, USA<EMAIL_ADDRESS>American Institute of Mathematics,
600 East Brokaw Road, San Jose, CA 95112-1006, USA and Max-Planck-Institut für
Mathematik, Vivatsgasse 7, 53111 Bonn, Germany<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Mathematics, University of Mississippi,
University, MS 38677, USA<EMAIL_ADDRESS>Department of Mathematics and
Statistics, Carleton College, Northfield, MN 55057, USA.
<EMAIL_ADDRESS>Department of Mathematics, Duke University, 120 Science
Dr, Durham, NC, 27708, USA<EMAIL_ADDRESS>
###### Abstract.

We improve the current bounds for an inequality of Erdős and Turán from 1950
related to the discrepancy of angular equidistribution of the zeros of a given
polynomial. Building upon a recent work of Soundararajan, we establish a novel
connection between this inequality and an extremal problem in Fourier analysis
involving the maxima of Hilbert transforms, for which we provide a complete
solution. Prior to Soundararajan (2019), refinements of the discrepancy
inequality of Erdős and Turán had been obtained by Ganelius (1954) and
Mignotte (1992).

###### Key words and phrases:

Polynomials, Erdős-Turán inequality, equidistribution, discrepancy, Hilbert
transform, extremal problems.

###### 2010 Mathematics Subject Classification:

42A05, 42A50

##  1\. Introduction

###  1.1. Background

Following the elegant treatment of Soundararajan [18], we revisit the
classical work of Erdős and Turán [9] on the distribution of zeros of
polynomials in the complex plane. In particular, we establish a connection
between the upper bound for the discrepancy of the angles of the zeros of a
given polynomial and an extremal problem in Fourier analysis involving the
maxima of Hilbert transforms. Before describing this extremal problem, which
is solved completely in this paper, we first describe our application in
number theory.

Let

$P(z)=\prod_{j=1}^{N}\big{(}z-\alpha_{j}\big{)}=z^{N}+a_{N-1}z^{N-1}+\cdots+a_{0}$

be a monic polynomial of degree $N$, with $a_{0}\neq 0$ and roots
$\alpha_{j}=\rho_{j}\,e^{2\pi i\theta_{j}}$. Roughly speaking, Erdős and Turán
proved that if the size of $P(z)$ on the unit circle is small, and $a_{0}$ is
not too small, then its roots cluster around the unit circle and the angles
$2\pi\theta_{j}$ become equidistributed as $N\to\infty$. Two notions of size,
or height, of a polynomial that have been considered in this problem are

$H(P)=\max_{|z|=1}\,\frac{|P(z)|}{\sqrt{|a_{0}|}}\ \ \ {\rm and}\ \ \
h(P)=\int_{0}^{1}\log^{+}\\!\left(\frac{\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}}{\sqrt{|a_{0}|}}\right)\,\text{\rm d}\theta,$

where $\log^{+}\\!x=\max\\{\log x,0\\}$. By Parseval’s identity, we have

$\int_{0}^{1}|P\big{(}e^{2\pi i\theta}\big{)}\big{|}^{2}\,\text{\rm
d}\theta=1+|a_{N-1}|^{2}+\ldots+|a_{0}|^{2},$

from which it follows easily that $H(P)\geq 1$ and therefore $h(P)\leq\log
H(P)$. Hence, the assumption that $h(P)$ is small is weaker than the
assumption that $H(P)$ is small. Let us also define the quantity

$\mathcal{M}(P)=\prod_{j=1}^{N}\max\left\\{\rho_{j},\frac{1}{\rho_{j}}\right\\}.$

The observation that the zeros cluster around the unit circle is given by the
inequality [18, Theorem 1]

$\log\mathcal{M}(P)\leq 2\,h(P),$

that follows by an interesting application of Jensen’s formula in complex
analysis.

We focus on the study of the equidistribution of the angles $2\pi\theta_{j}$.
Given an interval $I$ on $\mathbb{R}/\mathbb{Z}$, we let $N(I;P)$ denote the
number of zeros $\alpha_{j}=\rho_{j}\,e^{2\pi i\theta_{j}}$ for which
$\theta_{j}\in I$. A convenient way to measure the distribution of the
sequence $\\{\theta_{j}\\}_{j=1}^{N}$ is by means of its discrepancy, defined
by

$\mathcal{D}(P):=\sup_{I}\Big{|}N(I;P)-|I|N\Big{|},$

where $|I|$ denotes the length of the interval $I$. We list a few notable
results in estimating the discrepancy $\mathcal{D}(P)$. Erdős and Turán, in
their original paper [9] of 1950, proved that

$\mathcal{D}(P)\leq C\sqrt{N\log H(P)},$ (1.1)

with $C=16$. In 1954, Ganelius [10] established (1.1) with the constant
$C=\sqrt{2\pi/k}=2.5619\ldots$, where
$k=1/1^{2}-1/3^{2}+1/5^{2}-\ldots=0.9159\ldots$ denotes Catalan’s constant.
Amoroso and Mignotte [2] have produced examples that show that the constant
$C$ in (1.1) must be at least $\sqrt{2}$. In 1992, Mignotte [14] refined
Ganelius’s result by establishing the stronger inequality

$\mathcal{D}(P)\leq C\sqrt{N\,h(P)},$ (1.2)

with the same constant $C=\sqrt{2\pi/k}=2.5619\ldots$. Only recently, in 2019,
Soundararajan [18] improved this result by establishing (1.2) with the
constant

$C=\frac{8}{\pi}=2.5464\ldots.$

Our goal is to provide an improvement of the admissible value of $C$ in (1.2).
We follow the general outline of proof of Soundararajan in [18] up to a
certain point, then we diverge and introduce a novel ingredient: the
connection to a certain extremal problem in Fourier analysis involving the
maxima of Hilbert transforms. As a direct consequence of Theorems 2 and 3
below, we prove that the constant

$C=\frac{\ \ 4}{\sqrt{\pi}}=2.2567\ldots$

is admissible in (1.2) and show that this constant is the best possible with
our particular strategy.

######  Theorem 1.

If $P$ is a monic polynomial of degree $N$ with $P(0)\neq 0$, then

$\mathcal{D}(P)\leq\frac{\ \ 4}{\sqrt{\pi}}\,\sqrt{N\,h(P)}.$

Remark. In their original paper [9], Erdős and Turán were also interested in
estimating the number $\mathcal{R}(P)$ of real roots of a polynomial $P$. In
particular, the notion of discrepancy can be used towards this goal. From the
definition, letting $I$ denote either the point $0$ or $\frac{1}{2}$, it
plainly follows that $\mathcal{R}(P)\leq 2\,\mathcal{D}(P)$.

We also note that, for the example $P(z)=(z-1)^{N}$, one has
$\mathcal{D}(P)=N$ and

$h(P)=N\int_{0}^{1}\log^{+}\\!\big{|}e^{2\pi i\theta}-1\big{|}\,\text{\rm
d}\theta=N\,\frac{3\sqrt{3}\,L(2,\chi_{3})}{4\pi},$ (1.3)

where $\chi_{3}$ denotes the quadratic character modulo $3$. This last
identity was observed by C. J. Smyth in a slightly different context, see [5,
Appendix 1]. Hence, the constant $C$ in (1.2) cannot be smaller than

$\sqrt{\frac{4\pi}{3\sqrt{3}\,L(2,\chi_{3})}}=1.75936\ldots.$ (1.4)

###  1.2. Fourier optimization

Throughout this paper we consider functions in two different environments: the
ones defined on $\mathbb{R}$ (usually denoted here with capital letters) and
the ones defined on $\mathbb{R}/\mathbb{Z}$ (usually denoted here with lower
case letters).

For $F\in L^{1}(\mathbb{R})$ we define its Fourier transform
$\widehat{F}:\mathbb{R}\to\mathbb{C}$ by

$\widehat{F}(t)=\int_{-\infty}^{\infty}e^{-2\pi itx}\,F(x)\,\text{\rm d}x.$

By Plancherel’s theorem one can extend the Fourier transform to an isometry on
$L^{2}(\mathbb{R})$. The Hilbert transform $\mathcal{H}$ is another classical
operator in harmonic analysis that has a few (equivalent) interpretations. As
a singular integral it is defined by

$\mathcal{H}(F)(x)={\rm
p.v.}\,\frac{1}{\pi}\int_{\mathbb{R}}F(x-t)\,\frac{1}{t}\,\text{\rm d}t\,,$
(1.5)

where the notation p.v. here means that such integral should be understood as
a Cauchy principal value. The classical theory of singular integrals
guarantees that the Hilbert transform is a well-defined operator on
$L^{p}(\mathbb{R})$ for $1\leq p<\infty$, being a bounded operator if
$1<p<\infty$, and satisfying a weak-type-$(1,1)$ estimate when $p=1$. See, for
instance [19, Chapters V and VI] or [13, Chapter 4] for proofs of these facts
and the connections with the theory of conjugate harmonic functions. In
particular, the appropriate limiting process in (1.5) converges a.e. for $F\in
L^{p}(\mathbb{R})$, $1\leq p<\infty$. The operator
$\mathcal{H}:L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$ is an isometry that can
be alternatively defined on the Fourier space by the relation111Recall that
${\rm sgn}:\mathbb{R}\to\mathbb{R}$ is defined by ${\rm sgn}(t)=1$, if $t>0$;
${\rm sgn}(0)=0$; and ${\rm sgn}(t)=-1$, if $t<0$.

$\widehat{\mathcal{H}(F)}(t)=-i\,{\rm sgn}(t)\,\widehat{F}(t).$ (1.6)

Similarly, in the periodic setting, if $f\in L^{1}(\mathbb{R}/\mathbb{Z})$ we
define its Fourier transform $\widehat{f}:\mathbb{Z}\to\mathbb{C}$ by

$\widehat{f}(k)=\int_{\mathbb{R}/\mathbb{Z}}e^{-2\pi
ik\theta}\,f(\theta)\,\text{\rm d}\theta.$

The periodic Hilbert transform is the singular integral operator defined by

$\mathcal{H}(f)(\theta)={\rm
p.v.}\,\int_{\mathbb{R}/\mathbb{Z}}f(\theta-\alpha)\,\cot(\pi\alpha)\,\text{\rm
d}\alpha.$ (1.7)

Again, the appropriate limiting process in (1.7) converges a.e. if $f\in
L^{p}(\mathbb{R}/\mathbb{Z})$ for $1\leq p<\infty$, defining a bounded
operator on $L^{p}(\mathbb{R}/\mathbb{Z})$ if $1<p<\infty$, and verifying a
weak-type-$(1,1)$ estimate when $p=1$. In particular,
$\mathcal{H}:L^{2}(\mathbb{R}/\mathbb{Z})\to L^{2}(\mathbb{R}/\mathbb{Z})$ can
be alternatively defined via the Fourier coefficients

$\widehat{\mathcal{H}(f)}(k)=-i\,{\rm sgn}(k)\,\widehat{f}(k).$ (1.8)

Although we use the same notation for the Fourier transforms and Hilbert
transforms on $\mathbb{R}$ and $\mathbb{R}/\mathbb{Z}$, it will be clear from
the context which one we are referring to. We consider below some sharp
inequalities for the Hilbert transform. Classical works in this theme include
the ones of Pichorides [15], in which he finds the operator norm
$\|\mathcal{H}\|_{L^{p}\to L^{p}}$ for $1<p<\infty$ (see also [12] for a
simplified proof), and of Davis [8], in which he finds the weak-type-$(1,1)$
operator norm (such works consider both the situation in the real line and in
the periodic setting).

Throughout the paper we let $\mathcal{A}$ be the following class of real-
valued functions:

$\mathcal{A}=\left\\{\begin{array}[]{l}F:\mathbb{R}\to\mathbb{R}\ \ {\rm
even,continuous\ and\ non\\!-\\!negative};\\\ {\rm
supp}(F)\subseteq[-\tfrac{1}{2},\tfrac{1}{2}];\\\ \widehat{F}\in
L^{1}(\mathbb{R}).\end{array}\right.$

For each $F\in\mathcal{A}$ we define its periodization
$f_{F}:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ by

$f_{F}(\theta):=\sum_{k\in\mathbb{Z}}F(\theta+k).$

One can verify that $f_{F}\in L^{1}(\mathbb{R}/\mathbb{Z})$ and that
$\widehat{f_{F}}(k)=\widehat{F}(k)$ for all $k\in\mathbb{Z}$. Moreover, in
this situation, by a classical result of Plancherel and Pólya (see [16] or
[21, eq. (3.1)]), for any $\delta>0$ we have

$\sum_{k\in\mathbb{Z}}\big{|}\widehat{F}(\delta
k)\big{|}\ll_{\delta}\big{\|}\widehat{F}\big{\|}_{L^{1}(\mathbb{R})}.$

In particular, for $F\in\mathcal{A}$, both $\mathcal{H}(F)$ defined by (1.6)
and $\mathcal{H}(f_{F})$ defined by (1.8) via Fourier inversion are bounded
and continuous functions. We consider the following optimization problem
involving the $L^{\infty}$-norms of these Hilbert transforms.

Extremal Problem 1 (EP1). With notations as above, find the infimum:

$\displaystyle{\bf C}:=\inf_{0\neq
F\in\mathcal{A}}\frac{\max\big{\\{}\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\,,\,\|\mathcal{H}(f_{F})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}\big{\\}}}{\|F\|_{L^{1}(\mathbb{R})}}.$
(1.9)

This problem is the main theme of study in this paper. Without necessarily
knowing the precise value of the constant ${\bf C}$, our first main result
gives a non-obvious theoretical connection between this optimization problem,
purely in analysis, and the angular discrepancy $\mathcal{D}(P)$ of a
polynomial $P$.

######  Theorem 2.

Let ${\bf C}$ be given by (1.9). If $P$ is a monic polynomial of degree $N$
with $P(0)\neq 0$, then

$\mathcal{D}(P)\leq\frac{4\sqrt{\bf C}}{\sqrt{\pi}}\,\sqrt{N\,h(P)}.$

We prove this result in Sections 2 and 3. From the observations leading to
(1.4) and Theorem 2 we automatically have a lower bound coming from the number
theory side:

${\bf
C}\geq\frac{\pi}{16}\left(\frac{4\pi}{3\sqrt{3}\,L(2,\chi_{3})}\right)=0.6077\ldots.$

In Theorem 2, we go much further in our understanding of this problem. Before
stating this result, we set up a second optimization problem, somewhat related
to the first one. Let $\mathcal{A}^{*}$ be the following class of real-valued
functions (slightly larger than $\mathcal{A}$):

$\mathcal{A^{*}}=\left\\{\begin{array}[]{l}F\in L^{1}(\mathbb{R}),\ \ F\geq
0;\\\ {\rm supp}(F)\subseteq[-\tfrac{1}{2},\tfrac{1}{2}].\end{array}\right.$

Consider the following problem:

Extremal Problem 2 (EP2). With notations as above, find the infimum:

$\displaystyle{\bf C^{*}}:=\inf_{0\neq
F\in\mathcal{A^{*}}}\frac{\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}}{\|F\|_{L^{1}(\mathbb{R})}}.$
(1.10)

Since $\mathcal{A}\subseteq\mathcal{A^{*}}$ and
$\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\leq\max\big{\\{}\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\,,\,\|\mathcal{H}(f_{F})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}\big{\\}}$,
it is clear from the definitions of (EP1) and (EP2) that ${\bf C^{*}}\leq{\bf
C}.$ Our second main result establishes a complete solution for both of these
extremal problems at once.

######  Theorem 3.

For ${\bf C}$ given by (1.9) and ${\bf C^{*}}$ given by (1.10), we have

${\bf C^{*}}={\bf C}=1.$

Moreover, there are no extremal functions $F\in\mathcal{A}$ for the problem
${\rm(EP1)}$, and the unique (modulo multiplication by a positive constant)
extremal function for the problem ${\rm(EP2)}$ is

$\ \ \ \ \ \ \ \ \ \ \ \ \ \
\qquad\qquad\mathfrak{F}(x):=\frac{2}{\pi}\log\left(\frac{1+\sqrt{1-4x^{2}}}{2|x|}\right)\
\ \ \ \ \big{(}{\rm for}\ -\tfrac{1}{2}\leq x\leq\tfrac{1}{2}\big{)}.$ (1.11)

The proof of this theorem is given in Section 5. The proof relies on the
serendipitous existence of two magic functions: the even function given in
(1.11) and an odd function given in (5.2). For a different perspective on
lower bounds for the Hilbert transform over intervals (mostly in $L^{2}$), see
[1].

Theorem 1 now follows directly from Theorems 2 and 3.

##  2\. Soundararajan’s proof revisited

We now prepare for the proof of Theorem 2. At first, we closely follow
Soundararajan’s strategy of proof for the inequality (1.2) in [18], which we
briefly review for the convenience of the reader. At a certain stage of the
argument (discussed in §2.4 below), we make a crucial change of direction that
leads to our optimization problem in analysis. This is discussed in full
detail in the next section, where we complete the proof of Theorem 2.

###  2.1. Schur’s observation

First note that we can assume without loss of generality that the zeros of the
polynomial are all in the unit circle, an observation due to Schur [17]. In
fact, letting $P(z)=\prod_{j=1}^{N}\big{(}z-\rho_{j}\,e^{2\pi
i\theta_{j}}\big{)}$ as above, we may define
$Q(z)=\prod_{j=1}^{N}\big{(}z-e^{2\pi i\theta_{j}}\big{)}$ and observe, for
$|z|=1$, that

$\left|\frac{z}{\sqrt{\rho_{j}}}-\sqrt{\rho_{j}}\,e^{2\pi
i\theta_{j}}\right|^{2}\geq\big{|}z-e^{2\pi i\theta_{j}}\big{|}^{2}.$

By multiplying over $j$, we find that $|P(z)|/\sqrt{|a_{0}|}\geq|Q(z)|$ for
$|z|=1$, and therefore $h(P)\geq h(Q)$. Hence, from now on we assume that
$\rho_{j}=1$ for $j=1,2,\ldots,N$.

###  2.2. Smoothed sums and $h(P)$

If we define $\psi(\theta)=\log|2\sin(\pi\theta)|$, then its Fourier
coefficients are given by $\widehat{\psi}(0)=0$ and
$\widehat{\psi}(k)=-\tfrac{1}{2|k|}$ for $k\neq 0$ (e.g. [11, §1.441, eq. 2]).
Hence, for $P(z)=\prod_{j=1}^{N}\big{(}z-e^{2\pi i\theta_{j}}\big{)}$ and
$k\in\mathbb{Z}\setminus\\{0\\}$, we have

$\displaystyle\begin{split}\int_{\mathbb{R}/\mathbb{Z}}e^{2\pi
ik\theta}&\log\big{|}P\big{(}e^{2\pi i\theta}\big{)}\big{|}\,\text{\rm
d}\theta=\sum_{j=1}^{N}\int_{\mathbb{R}/\mathbb{Z}}e^{2\pi
ik\theta}\log\big{|}e^{2\pi i\theta}-e^{2\pi i\theta_{j}}\big{|}\,\text{\rm
d}\theta\\\ &=\sum_{j=1}^{N}e^{2\pi
ik\theta_{j}}\int_{\mathbb{R}/\mathbb{Z}}e^{2\pi ik\alpha}\log\big{|}e^{2\pi
i\alpha}-1\big{|}\,\text{\rm d}\alpha\\\ &=\sum_{j=1}^{N}e^{2\pi
ik\theta_{j}}\int_{\mathbb{R}/\mathbb{Z}}\psi(\alpha)\,e^{2\pi
ik\alpha}\,\text{\rm d}\alpha=-\frac{1}{2|k|}\sum_{j=1}^{N}e^{2\pi
ik\theta_{j}}.\end{split}$ (2.1)

Identity (2.1) is essentially contained in [18, Lemma 2]. Let
$g:\mathbb{R}/\mathbb{Z}\to\mathbb{C}$ be a continuous and integrable function
such that $\\{k\,\widehat{g}(k)\\}_{k\in\mathbb{Z}}$ is absolutely summable,
and set

$\mathcal{G}:=\max_{\theta}\left|\sum_{k\neq 0}2|k|\,\widehat{g}(k)\,e^{2\pi
ik\theta}\right|.$

By expanding $g$ into its Fourier series, and using (2.1), we get

$\displaystyle\left|\sum_{j=1}^{N}g(\theta_{j})-N\int_{\mathbb{R}/\mathbb{Z}}g(\theta)\,\text{\rm
d}\theta\right|=\left|\sum_{k\neq 0}\widehat{g}(k)\sum_{j=1}^{N}e^{2\pi
ik\theta_{j}}\right|=\left|-\int_{\mathbb{R}/\mathbb{Z}}\log\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}\left(\sum_{k\neq 0}2|k|\widehat{g}(k)e^{2\pi
ik\theta}\right)\text{\rm d}\theta\right|$ $\displaystyle\ \ \ \
\leq\mathcal{G}\int_{\mathbb{R}/\mathbb{Z}}\Big{|}\log\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}\Big{|}\,\text{\rm
d}\theta=\mathcal{G}\left(\int_{\mathbb{R}/\mathbb{Z}}2\log^{+}\\!\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}\,\text{\rm
d}\theta-\int_{\mathbb{R}/\mathbb{Z}}\log\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}\,\text{\rm d}\theta\right)$ (2.2) $\displaystyle\ \ \ \
=2\,\mathcal{G}\,h(P).$

In the last passage above, note the use of Jensen’s formula in the identity

$\int_{\mathbb{R}/\mathbb{Z}}\log\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}\,\text{\rm d}\theta=0.$

Inequality (2.2) is the content of [18, Proposition 1].

###  2.3. Majorizing the characteristic function of an interval

Having established the preliminaries in §2.1 and §2.2 above, we now move on to
the proof itself. First observe that if we can prove the upper bound

$\displaystyle N(I;P)-|I|\,N\leq C\sqrt{N\,h(P)}\,,$ (2.3)

for a certain universal constant $C$ and all intervals
$I\subset\mathbb{R}/\mathbb{Z}$, we may use the identity

$N(I;P)-|I|\,N=|I^{c}|\,N-N(I^{c};P),$

where $I^{c}$ denotes the complementary interval to $I$, to obtain the
corresponding lower bound. Therefore, it suffices to obtain the upper bound
(2.3).

Let $0\neq F\in\mathcal{A}$, normalized so that
$\|F\|_{L^{1}(\mathbb{R})}=\int_{\mathbb{R}}F(x)\,\text{\rm d}x=1$. For each
$0<\delta\leq 1$, let

$F_{\delta}(x):=\tfrac{1}{\delta}F\big{(}\tfrac{x}{\delta}\big{)}$

so that ${\rm
supp}(F_{\delta})\subset\big{[}-\tfrac{\delta}{2},\tfrac{\delta}{2}\big{]}$.
We let

$f_{\delta}(\theta):=\sum_{k\in\mathbb{Z}}F_{\delta}(\theta+k)$

be the periodization of $F_{\delta}$. Note that
$\int_{\mathbb{R}/\mathbb{Z}}f_{\delta}(\theta)\,\text{\rm d}\theta=1$ and,
more generally, that

$\widehat{f_{\delta}}(k)=\widehat{F_{\delta}}(k)=\widehat{F}(\delta k)$

for all $k\in\mathbb{Z}$. For each interval $I\subset\mathbb{R}/\mathbb{Z}$,
let $I_{\delta}$ be the interval obtained by widening $I$ on either side by
$\delta/2$; if $|I|+\delta\geq 1$, then we just consider $I_{\delta}$ to be
all of $\mathbb{R}/\mathbb{Z}$. Let $\chi_{I_{\delta}}$ be the characteristic
function of the interval $I_{\delta}$ and let $g_{\delta}$ be the convolution
of $\chi_{I_{\delta}}$ and $f_{\delta}$, that is

$\displaystyle
g_{\delta}(\theta)=\int_{\mathbb{R}/\mathbb{Z}}\chi_{I_{\delta}}(\alpha)\,f_{\delta}(\theta-\alpha)\,\text{\rm
d}\alpha.$ (2.4)

Note that $g_{\delta}$ is a continuous and non-negative function that
majorizes the characteristic function of the original interval $I$. We then
write

$\displaystyle
N(I;P)-|I|\,N\leq\sum_{j=1}^{N}g_{\delta}(\theta_{j})-|I|\,N=\left(\sum_{j=1}^{N}g_{\delta}(\theta_{j})-N\int_{\mathbb{R}/\mathbb{Z}}g_{\delta}(\theta)\,\text{\rm
d}\theta\\!\right)\\!+N\\!\left(\int_{\mathbb{R}/\mathbb{Z}}g_{\delta}(\theta)\,\text{\rm
d}\theta-|I|\\!\right).$ (2.5)

Our goal now is to bound the two terms appearing on the right-hand side of
(2.5). For the second term, we use the definition (2.4) and Fubini’s theorem
to get

$\displaystyle 0\leq
N\left(\int_{\mathbb{R}/\mathbb{Z}}g_{\delta}(\theta)\,\text{\rm
d}\theta-|I|\right)=N\left(\int_{\mathbb{R}/\mathbb{Z}}\chi_{I_{\delta}}(\alpha)\,\text{\rm
d}\alpha-|I|\right)\leq N\Big{(}(|I|+\delta)-|I|\Big{)}=N\delta.$ (2.6)

Now, if $I_{\delta}=[\alpha,\beta]$, for all $k\in\mathbb{Z}\setminus\\{0\\}$
we have

$\widehat{\chi_{I_{\delta}}}(k)=\frac{e^{-2\pi ik\alpha}-e^{-2\pi
ik\beta}}{2\pi ik}.$ (2.7)

Recall that
$\widehat{g_{\delta}}(k)=\widehat{\chi_{I_{\delta}}}(k)\,\widehat{f_{\delta}}(k)$
for all $k\in\mathbb{Z}$, hence the sequence
$\\{k\widehat{g_{\delta}}(k)\\}_{k\in\mathbb{Z}}$ is absolutely summable.
Letting

$\mathcal{G}_{\delta}:=\max_{\theta}\left|\sum_{k\neq
0}2|k|\widehat{g_{\delta}}(k)e^{2\pi ik\theta}\right|\,,$

we have seen in (2.2) that the first term on the right-hand side of (2.5)
satisfies

$\displaystyle\left|\sum_{j=1}^{N}g_{\delta}(\theta_{j})-N\int_{\mathbb{R}/\mathbb{Z}}g_{\delta}(\theta)\,\text{\rm
d}\theta\right|\leq 2\,\mathcal{G}_{\delta}\,h(P).$ (2.8)

###  2.4. Understanding the cancellation

We now need to bound the quantity $\mathcal{G}_{\delta}$ and this is where we
diverge from Soundararajan’s original proof [18]. From (2.7) we have

$\displaystyle\sum_{k\neq 0}2|k|\widehat{g_{\delta}}(k)\,e^{2\pi
ik\theta}=\frac{1}{\pi}\sum_{k\neq 0}-i\,{\rm
sgn}(k)\widehat{f_{\delta}}(k)\left(e^{2\pi ik(\theta-\alpha)}-e^{2\pi
ik(\theta-\beta)}\right),$ (2.9)

and hence

$\displaystyle\left|\sum_{k\neq 0}2|k|\widehat{g_{\delta}}(k)\,e^{2\pi
ik\theta}\right|$ $\displaystyle\leq\frac{1}{\pi}\left|\sum_{k\neq 0}-i\,{\rm
sgn}(k)\widehat{f_{\delta}}(k)\,e^{2\pi
ik(\theta-\alpha)}\right|+\frac{1}{\pi}\left|\sum_{k\neq 0}-i\,{\rm
sgn}(k)\widehat{f_{\delta}}(k)\,e^{2\pi ik(\theta-\beta)}\right|$
$\displaystyle=\frac{1}{\pi}\big{|}\mathcal{H}(f_{\delta})(\theta-\alpha)\big{|}+\frac{1}{\pi}\big{|}\mathcal{H}(f_{\delta})(\theta-\beta)\big{|}.$

This plainly yields

$\mathcal{G}_{\delta}\leq\frac{2}{\pi}\|\mathcal{H}(f_{\delta})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}.$
(2.10)

Equality is actually attained if one considers the maximum over all intervals
$[\alpha,\beta]$, so there is no loss in this use of the triangle inequality.

Remark. In the corresponding step in [18], Soundararajan is working in the
restricted subclass of $\mathcal{A}$ for which $\widehat{F}\geq 0$, and at the
end he chooses $F$ to be a triangular graph. He couples the terms $k$ and $-k$
in (2.9) and uses the triangle inequality, further moving the absolute values
inside the sum, to get

$\mathcal{G}_{\delta}\leq\frac{4}{\pi}\max_{\theta}\sum_{k\geq
1}\widehat{f_{\delta}}(k)\,|\sin(2\pi k\theta)|.$

This particular extra step of moving the absolute values inside disregards
some cancellation in the sum. This is precisely the point where our analysis
diverges from [18].

We now state a relation that is fundamental for our purposes, which
essentially says that the supremum over this one-parameter family (for
$0<\delta\leq 1$) of $L^{\infty}$-norms of Hilbert transforms in (2.10), when
properly normalized, occurs at one of the endpoints $\delta=0^{+}$ or
$\delta=1$.

######  Proposition 4.

Let $F\in\mathcal{A}$ and $0<\delta\leq 1$. With notations as above, we have

$\displaystyle\sup_{0<\delta\leq
1}\delta\,\|\mathcal{H}(f_{\delta})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}=\max\\!\left\\{\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\,,\,\|\mathcal{H}(f_{F})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}\right\\}.$

We postpone the proof of this result until the next section.

###  2.5. Conclusion

Assume for a moment that we have established Proposition 4. Let us simplify
the notation by writing

$\mathcal{C}(F):=\max\\!\left\\{\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\,,\,\|\mathcal{H}(f_{F})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}\right\\}.$

It then follows from (2.10) and Proposition 4 that

$\mathcal{G}_{\delta}\leq\frac{2}{\pi\delta}\,\mathcal{C}(F),$ (2.11)

and from (2.5), (2.6), (2.8), and (2.11) we get

$N(I;P)-|I|\,N\leq N\delta+\frac{4}{\pi\delta}\,\mathcal{C}(F)\,h(P).$

The choice of

$\delta=\sqrt{\frac{4\,\mathcal{C}(F)\,h(P)}{\pi N}}$ (2.12)

minimizes the right-hand side of the expression above and leads to the bound

$N(I;P)-|I|\,N\leq\frac{4\sqrt{\mathcal{C}(F)}}{\sqrt{\pi}}\,\sqrt{N\,h(P)}.$

Note that this is independent of the interval $I$. Minimizing over
$F\in\mathcal{A}$ we arrive at the desired conclusion

$\mathcal{D}(P)\leq\frac{4\sqrt{\bf C}}{\sqrt{\pi}}\sqrt{N\,h(P)}.$

Therefore, Theorem 2 follows from Proposition 4.

Remark. From the fact that $\log^{+}\\!xy\leq\log^{+}\\!x\,+\,\log^{+}\\!y$
for any $x,y>0$, if $P(z)=\prod_{j=1}^{N}\big{(}z-e^{2\pi i\theta_{j}}\big{)}$
we get

$\displaystyle h(P)=\int_{0}^{1}\log^{+}\big{|}P\big{(}e^{2\pi
i\theta}\big{)}\big{|}\,\text{\rm d}\theta\leq
N\int_{0}^{1}\log^{+}\big{|}e^{2\pi i\theta}-1\big{|}\,\text{\rm
d}\theta=N\,\frac{3\sqrt{3}\,L(2,\chi_{3})}{4\pi}\,,$

as remarked in (1.3). Hence, the choice of $\delta$ in (2.12) indeed falls in
the interval $(0,1]$ if

$\mathcal{C}(F)\leq\frac{\pi^{2}}{3\sqrt{3}\,L(2,\chi_{3})}=2.43107\ldots.$
(2.13)

In Section 4, we observe that there are functions $F\in\mathcal{A}$ that
verify this bound. For example, the triangle function
$F_{\blacktriangle}(x)=2\,\max\big{\\{}1-2|x|,0\big{\\}}$ has
$\mathcal{C}(F_{\blacktriangle})=1.12219\ldots$. Hence, without loss of
generality, we may assume that from the start we are working under the
threshold (2.13).

##  3\. Maxima of Hilbert transforms

The purpose of this section is to prove the key Proposition 4, hence
concluding the proof of Theorem 2. Recall that we have been using the
definition of the Hilbert transforms via the multipliers (1.6) and (1.8) and
Fourier inversion (hence all Hilbert transforms here are bounded and
continuous functions). In this section, the alternative representations of the
Hilbert transforms as singular integrals will be particularly useful.
Throughout this section we continue to assume that $0\neq F\in\mathcal{A}$ is
normalized so that $\|F\|_{L^{1}(\mathbb{R})}=\int_{\mathbb{R}}F(x)\,\text{\rm
d}x=1$, and for each $0<\delta\leq 1$ we let
$F_{\delta}(x):=\tfrac{1}{\delta}F\big{(}\tfrac{x}{\delta}\big{)}$ and
$f_{\delta}(\theta):=\sum_{k\in\mathbb{Z}}F_{\delta}(\theta+k)$. For
$x\in\mathbb{R}$, let

$\|x\|:=\min\\{|x-n|\,:\,n\in\mathbb{Z}\\}$

be the distance of $x$ to the nearest integer.

###  3.1. Hilbert transforms as singular integrals

For each $0<\delta\leq 1$, since $\mathcal{H}(f_{\delta})$ is an odd and
continuous function in $\mathbb{R}/\mathbb{Z}$, we have
$\mathcal{H}(f_{\delta})(0)=\mathcal{H}(f_{\delta})(\pm\tfrac{1}{2})=0$. We
start by establishing the following useful relation between the periodic
Hilbert transforms $\mathcal{H}(f_{\delta})$ and the Hilbert transform
$\mathcal{H}(F)$.

######  Lemma 5.

Let $0<\delta\leq 1$ and $-\frac{1}{2}<\theta<\frac{1}{2}$. Then

$\displaystyle\delta\,\mathcal{H}(f_{\delta})(\theta)=\mathcal{H}(F)\left(\tfrac{\theta}{\delta}\right)+\frac{\delta}{\pi}\sum_{k\geq
1}\int_{0}^{\frac{\delta}{2}}f_{\delta}(\alpha)\frac{4\theta(\theta^{2}-\alpha^{2}-k^{2})}{\big{(}(\theta-\alpha)^{2}-k^{2}\big{)}\big{(}(\theta+\alpha)^{2}-k^{2}\big{)}}\,\text{\rm
d}\alpha.$ (3.1)

###### Proof.

Let $\Gamma\subset\mathbb{R}$ be the set of full measure (i.e.
$\mathbb{R}\setminus\Gamma$ has measure zero) such that for every $x\in\Gamma$
the limit

$\lim_{\varepsilon\to
0}\frac{1}{\pi}\int_{\varepsilon\leq|t|}F(x-t)\,\frac{1}{t}\,\text{\rm d}t$
(3.2)

exists and is equal to $\mathcal{H}(F)(x)$. Similarly, for a fixed
$0<\delta\leq 1$, let $\Gamma_{\delta}\subset\mathbb{R}/\mathbb{Z}$ be the set
of full measure such that for every $\theta\in\Gamma_{\delta}$ the limit

$\lim_{\varepsilon\to
0}\int_{\varepsilon\leq|\alpha|\leq\frac{1}{2}}f_{\delta}(\theta-\alpha)\,\cot(\pi\alpha)\,\text{\rm
d}\alpha$ (3.3)

exists and is equal to $\mathcal{H}(f_{\delta})(\theta)$.

Recall that, for $\|\alpha\|\geq\varepsilon>0$, we have the absolutely
convergent expansion (e.g. [11, §1.421 eq. 3])

$\cot(\pi\alpha)=\frac{1}{\pi}\left(\frac{1}{\alpha}+\sum_{k\geq
1}\frac{2\alpha}{\alpha^{2}-k^{2}}\right).$ (3.4)

Assume that $\theta\in\Gamma_{\delta}$ and $\frac{\theta}{\delta}\in\Gamma$.
Let $\varepsilon$ be small and write
$X_{\varepsilon}=\big{\\{}\alpha\in\big{[}-\frac{\delta}{2},\frac{\delta}{2}]\,:\,\|\theta-\alpha\|\geq\varepsilon\big{\\}}$
and
$Y_{\varepsilon}=\big{\\{}\beta\in\big{[}-\frac{1}{2},\frac{1}{2}]\,:\,\|\theta-\delta\beta\|\geq\varepsilon\big{\\}}$.
Using (3.4), and with a change of variables $\alpha=\delta\beta$, we note that

$\displaystyle\delta\int_{\varepsilon\leq|\alpha|\leq\frac{1}{2}}$
$\displaystyle f_{\delta}(\theta-\alpha)\,\cot(\pi\alpha)\,\text{\rm
d}\alpha=\delta\int_{X_{\varepsilon}}f_{\delta}(\alpha)\,\cot(\pi(\theta-\alpha))\,\text{\rm
d}\alpha$
$\displaystyle=\frac{\delta}{\pi}\int_{X_{\varepsilon}}\frac{f_{\delta}(\alpha)}{(\theta-\alpha)}\,\text{\rm
d}\alpha\ +\frac{\delta}{\pi}\sum_{k\geq
1}\int_{X_{\varepsilon}}f_{\delta}(\alpha)\frac{2(\theta-\alpha)}{(\theta-\alpha)^{2}-k^{2}}\,\text{\rm
d}\alpha$
$\displaystyle=\frac{1}{\pi}\int_{Y_{\varepsilon}}\frac{F(\beta)}{\big{(}\frac{\theta}{\delta}-\beta\big{)}}\,\text{\rm
d}\beta+\frac{\delta}{\pi}\sum_{k\geq
1}\int_{X_{\varepsilon}}f_{\delta}(\alpha)\frac{2(\theta-\alpha)}{(\theta-\alpha)^{2}-k^{2}}\,\text{\rm
d}\alpha.$

Passing to the limit as $\varepsilon\to 0$, and using the fact that
$f_{\delta}$ is even (to combine $\alpha$ and $-\alpha$ in the integral
below), we get

$\displaystyle\begin{split}\delta\,\mathcal{H}(f_{\delta})(\theta)&=\mathcal{H}(F)\left(\tfrac{\theta}{\delta}\right)+\frac{\delta}{\pi}\sum_{k\geq
1}\int_{-\frac{\delta}{2}}^{\frac{\delta}{2}}f_{\delta}(\alpha)\frac{2(\theta-\alpha)}{(\theta-\alpha)^{2}-k^{2}}\,\text{\rm
d}\alpha\\\
&=\mathcal{H}(F)\left(\tfrac{\theta}{\delta}\right)+\frac{\delta}{\pi}\sum_{k\geq
1}\int_{0}^{\frac{\delta}{2}}f_{\delta}(\alpha)\frac{4\theta(\theta^{2}-\alpha^{2}-k^{2})}{\big{(}(\theta-\alpha)^{2}-k^{2}\big{)}\big{(}(\theta+\alpha)^{2}-k^{2}\big{)}}\,\text{\rm
d}\alpha.\end{split}$ (3.5)

In principle, (3.5) holds for $\theta$ in the set of full measure
$(-\tfrac{1}{2},\tfrac{1}{2})\cap\Gamma_{\delta}\cap\delta\Gamma$. Since the
functions in (3.5) are continuous functions of
$\theta\in(-\tfrac{1}{2},\tfrac{1}{2})$ we conclude that the identity is valid
for all $\theta$ in this range. ∎

###  3.2. Proof of Proposition 4

We start by observing that, for $0<\delta\leq 1$ and
$-\tfrac{1}{2}\leq\theta\leq-\tfrac{\delta}{2}$, we have

$\mathcal{H}(f_{\delta})(\theta)\leq 0.$ (3.6)

In fact, if $\delta<1$ and
$\theta\in(-\tfrac{1}{2},-\tfrac{\delta}{2})\cap\Gamma_{\delta}$, using that
$f_{\delta}$ is even, non-negative and supported in
$[-\tfrac{\delta}{2},\tfrac{\delta}{2}]$ along with the singular integral
representation (3.3), we get

$\displaystyle\mathcal{H}(f_{\delta})(\theta)$
$\displaystyle=\lim_{\varepsilon\to
0}\int_{X_{\varepsilon}}f_{\delta}(\alpha)\,\cot(\pi(\theta-\alpha))\,\text{\rm
d}\alpha=\int_{-\frac{\delta}{2}}^{\frac{\delta}{2}}f_{\delta}(\alpha)\,\cot(\pi(\theta-\alpha))\,\text{\rm
d}\alpha$
$\displaystyle=2\int_{0}^{\frac{\delta}{2}}f_{\delta}(\alpha)\,\big{(}\cot(\pi(\theta-\alpha))+\cot(\pi(\theta+\alpha))\big{)}\,\text{\rm
d}\alpha\leq 0.$

A similar argument shows that if $x\leq-\tfrac{1}{2}$ then
$\mathcal{H}(F)(x)\leq 0.$

Since $\mathcal{H}(f_{\delta})$ is an odd and continuous function in
$[-\tfrac{1}{2},\tfrac{1}{2}]$ (not identically zero), its maximum in absolute
value coincides with the positive maximum, and we investigate the latter.
Recall that
$\mathcal{H}(f_{\delta})(0)=\mathcal{H}(f_{\delta})(\pm\tfrac{1}{2})=0$. We
split our analysis into two cases.

####  3.2.1. Case 1

Assume that $0<\theta<\frac{1}{2}$ is such that
$\mathcal{H}(f_{\delta})(\theta)>0$. In this case, the sum on the right-side
of (3.1) is clearly non-positive and it plainly follows by Lemma 5 that

$\delta\,\mathcal{H}(f_{\delta})(\theta)\leq\mathcal{H}(F)\left(\tfrac{\theta}{\delta}\right).$
(3.7)

####  3.2.2. Case 2

Assume that $-\tfrac{1}{2}<\theta<0$ is such that
$\mathcal{H}(f_{\delta})(\theta)>0$. As observed in (3.6), we must have
$-\frac{\delta}{2}<\theta<0$ in this situation. Using Lemma 5, letting
$\theta^{\prime}=\frac{\theta}{\delta}$ (hence
$-\frac{1}{2}<\theta^{\prime}<0$) and changing variables $\alpha=\delta\beta$
in the integral, we rewrite (3.1) as

$\displaystyle\delta\,\mathcal{H}(f_{\delta})(\theta)$
$\displaystyle=\mathcal{H}(F)(\theta^{\prime})+\frac{1}{\pi}\sum_{k\geq
1}\int_{0}^{\frac{1}{2}}F(\beta)\frac{4\theta^{\prime}\left(\theta^{\prime
2}-\beta^{2}-\left(\tfrac{k}{\delta}\right)^{2}\right)}{\left((\theta^{\prime}-\beta)^{2}-\left(\tfrac{k}{\delta}\right)^{2}\right)\left((\theta^{\prime}+\beta)^{2}-\left(\tfrac{k}{\delta}\right)^{2}\right)}\,\text{\rm
d}\beta.$ (3.8)

The important observation now is that, for each $k\geq 1$, the term

$\frac{4\theta^{\prime}\left(\theta^{\prime
2}-\beta^{2}-\left(\tfrac{k}{\delta}\right)^{2}\right)}{\left((\theta^{\prime}-\beta)^{2}-\left(\tfrac{k}{\delta}\right)^{2}\right)\left((\theta^{\prime}+\beta)^{2}-\left(\tfrac{k}{\delta}\right)^{2}\right)}$

is positive and, for fixed $-\frac{1}{2}<\theta^{\prime}<0$ and
$0\leq\beta\leq\frac{1}{2}$, the function

$h(x):=\frac{4\theta^{\prime}\big{(}\theta^{\prime
2}-\beta^{2}-x^{2}\big{)}}{\big{(}(\theta^{\prime}-\beta)^{2}-x^{2}\big{)}\big{(}(\theta^{\prime}+\beta)^{2}-x^{2}\big{)}}$

verifies $h^{\prime}(x)<0$ for $x\geq 1$. This is a routine calculation. The
conclusion is that we could replace $\delta$ on each summand on the right-hand
side of (3.8) by its maximum value $\delta=1$ and do better, i.e.

$\displaystyle\begin{split}\delta\,\mathcal{H}(f_{\delta})(\theta)&\leq\mathcal{H}(F)(\theta^{\prime})+\frac{1}{\pi}\sum_{k\geq
1}\int_{0}^{\frac{1}{2}}F(\beta)\frac{4\theta^{\prime}\left(\theta^{\prime
2}-\beta^{2}-k^{2}\right)}{\big{(}(\theta^{\prime}-\beta)^{2}-k^{2}\big{)}\big{(}(\theta^{\prime}+\beta)^{2}-k^{2}\big{)}}\,\text{\rm
d}\beta\\\ &=\mathcal{H}(f_{1})(\theta^{\prime})\,,\end{split}$ (3.9)

where the last identity follows from another application of Lemma 5.

####  3.2.3. Conclusion

From (3.7) and (3.9), we plainly arrive at the conclusion that

$\displaystyle\sup_{0<\delta\leq
1}\delta\,\|\mathcal{H}(f_{\delta})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}\leq\max\big{\\{}\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\,,\,\|\mathcal{H}(f_{F})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}\big{\\}}=:\mathcal{C}(F).$
(3.10)

If
$\mathcal{C}(F)=\|\mathcal{H}(f_{F})\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}$,
then (3.10) is obviously an equality (recall that $f_{F}=f_{1}$ in this
notation). On the other hand, if
$\mathcal{C}(F)=\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}$, let
$x_{0}\in\mathbb{R}$ be such that

$\mathcal{C}(F)=\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}=\mathcal{H}(F)(x_{0})$

(note that $\mathcal{H}(F)$ goes to zero at infinity, hence such $x_{0}$
indeed exists). Let $\theta(\delta)=\delta x_{0}$, for $\delta$ sufficiently
small so that $-\tfrac{1}{2}<\theta(\delta)<\tfrac{1}{2}$. We apply Lemma 5
once more, by changing variables $\alpha=\delta\beta$ in the integral and
rewriting (3.1) in the form

$\displaystyle\delta\,\mathcal{H}(f_{\delta})(\delta x_{0})$
$\displaystyle=\mathcal{H}(F)(x_{0})+\frac{1}{\pi}\sum_{k\geq
1}\int_{0}^{\frac{1}{2}}F(\beta)\frac{4\delta^{2}x_{0}\big{(}\delta^{2}x_{0}^{2}-\delta^{2}\beta^{2}-k^{2}\big{)}}{\big{(}\delta^{2}(x_{0}-\beta)^{2}-k^{2}\big{)}\big{(}\delta^{2}(x_{0}+\beta)^{2}-k^{2}\big{)}}\,\text{\rm
d}\beta.$ (3.11)

An application of the dominated convergence theorem on the right-hand side of
(3.11) guarantees that

$\lim_{\delta\to 0^{+}}\delta\,\mathcal{H}(f_{\delta})(\delta
x_{0})=\mathcal{H}(F)(x_{0})\,,$

and we have equality in (3.10) as desired. This concludes the proof of
Proposition 4.

##  4\. A brief interlude

Before moving to the final section, where we present the proof of Theorem 3,
let us briefly make some remarks to highlight a few important elements in our
discussion. Throughout this section let $f=f_{F}$.

###  4.1. Dichotomy

In the definition of $\mathcal{C}(F)$ we have a maximum between two
$L^{\infty}$-norms. One may wonder if one of these is always dominated by the
other. Our first observation is that this is not always the case. In
principle, there are examples of functions for which either $L^{\infty}$-norm
can be maximal.

If the maximum value of $\mathcal{H}(f)(\theta)$ occurs at a certain
$0<\theta<\tfrac{1}{2}$, then

$\mathcal{C}(F)=\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}.$ (4.1)

This follows directly from (3.7) with $\delta=1$. This is the case, in
particular, if $F$ is radial decreasing. In fact, under such assumption, for
a.e. $-\tfrac{1}{2}<\theta<0$ we have

$\mathcal{H}(f)(\theta)=\lim_{\varepsilon\to
0}\int_{\varepsilon\leq|\alpha|\leq\frac{1}{2}}f(\theta-\alpha)\,\cot(\pi\alpha)\,\text{\rm
d}\alpha=\lim_{\varepsilon\to
0}\int_{\varepsilon}^{\frac{1}{2}}\big{(}f(\theta-\alpha)-f(\theta+\alpha)\big{)}\,\cot(\pi\alpha)\,\text{\rm
d}\alpha\leq 0.$

Since $\mathcal{H}(f)$ is continuous, this inequality is valid for all
$-\tfrac{1}{2}<\theta<0$. Note that above we used the fact that
$\|\theta+\alpha\|\leq\|\theta-\alpha\|$ in our range to argue that
$f(\theta-\alpha)\leq f(\theta+\alpha)$.

On the other hand, if the maximum value of $\mathcal{H}(F)(x)$ occurs at a
certain $-\frac{1}{2}<x<0$ (recall that we have seen that
$\mathcal{H}(F)(x)\leq 0$ for $x\leq-\tfrac{1}{2}$), then

$\mathcal{C}(F)=\|\mathcal{H}(f)\|_{L^{\infty}(\mathbb{R}/\mathbb{Z})}.$ (4.2)

This follows from (3.1) with $\delta=1$. There are indeed functions $F$ with
such behaviour, for instance the piecewise linear function, normalized so that
$\int_{\mathbb{R}}F(x)\,\text{\rm d}x=1$,

$\displaystyle F(x)=\begin{cases}0,\ {\rm if}\ 0\leq|x|\leq\frac{1}{4};\\\
64|x|-16,\ {\rm if}\ \frac{1}{4}\leq|x|\leq\frac{5}{16};\\\
\frac{1}{3}(32-64|x|),\ {\rm if}\
\frac{5}{16}\leq|x|\leq\frac{1}{2}.\end{cases}$ (4.3)

See Figure 1 for the plots of the Hilbert transforms of this example. In a
certain sense, the cases for which (4.2) holds are slightly unusual, and
produce large $L^{\infty}$-norms. We prove in the next section that functions
$F$ such that $\mathcal{C}(F)$ is very close to the infimum ${\bf C}$ tend to
like option (4.1) better.

Figure 1. For the function $F$ defined in (4.3), on the left we have the graph
of $\mathcal{H}(F)$ and on the right we have the graph of
$\mathcal{H}(f_{F})$.

###  4.2. The triangle function

Consider the triangle function $F_{\blacktriangle}$ given by

$F_{\blacktriangle}(x)=2\,\max\big{\\{}1-2|x|,0\big{\\}}.$

Note that $\|F_{\blacktriangle}\|_{L^{1}(\mathbb{R})}=1$. An application of
integration by parts in (3.2) shows that

$\displaystyle\mathcal{H}(F_{\blacktriangle})(x)=\frac{1}{\pi}\int_{0}^{\frac{1}{2}}F_{\blacktriangle}^{\prime}(t)\,\log\left(\frac{|x-t|}{|x+t|}\right)\,\text{\rm
d}t=\frac{4}{\pi}\int_{0}^{\frac{1}{2}}\log\left(\frac{|x+t|}{|x-t|}\right)\,\text{\rm
d}t.$

We seek the global maximum of $\mathcal{H}(F_{\blacktriangle})(x)$ when $x>0$.
One can check that this function is decreasing if $x>\tfrac{1}{2}$, simply
because $\tfrac{|x+t|}{|x-t|}<\tfrac{|y+t|}{|y-t|}$ if $x>y>\tfrac{1}{2}$ for
all $0<t<\tfrac{1}{2}$. For $0<x<\tfrac{1}{2}$, we may write

$\displaystyle\mathcal{H}(F_{\blacktriangle})(x)=\frac{4}{\pi}\left(\int_{x}^{x+\frac{1}{2}}\log|y|\,\text{\rm
d}y-\int_{x-\tfrac{1}{2}}^{x}\log|y|\,\text{\rm d}y\right).$

Hence, by the fundamental theorem of calculus, we have

$\displaystyle\mathcal{H}(F_{\blacktriangle})^{\prime}(x)=\log|x+\tfrac{1}{2}|-2\log|x|+\log|x-\tfrac{1}{2}|\,,$

and for $0<x<\tfrac{1}{2}$ we find that
$\mathcal{H}(F_{\blacktriangle})^{\prime}(x)=0$ if and only if

$\frac{\big{(}x+\tfrac{1}{2}\big{)}\big{(}\tfrac{1}{2}-x\big{)}}{x^{2}}=1\,,$

which yields $x=1/(2\sqrt{2})$. This is the global maximum and by (4.1) we get

$\displaystyle\|\mathcal{H}(F_{\blacktriangle})\|_{L^{\infty}(\mathbb{R})}=\mathcal{H}(F_{\blacktriangle})\big{(}\tfrac{1}{2\sqrt{2}}\big{)}=\frac{4}{\pi}\left(\int_{\frac{1}{2\sqrt{2}}}^{\frac{1}{2\sqrt{2}}+\frac{1}{2}}\log
y\,\text{\rm
d}y-\int_{\frac{1}{2\sqrt{2}}-\tfrac{1}{2}}^{\frac{1}{2\sqrt{2}}}\log|y|\,\text{\rm
d}y\right)=\frac{4}{\pi}\log(1+\sqrt{2}).$

This shows that the constant ${\bf C}$ in (EP1) satisfies

${\bf
C}\leq\mathcal{C}(F_{\blacktriangle})=\frac{4}{\pi}\log(1+\sqrt{2})=1.12219\ldots\,,$

and, as a consequence of Theorem 2, for monic polynomials $P$ of degree $N$
with $P(0)\neq 0$ we deduce that

$\mathcal{D}(P)\leq C\sqrt{N\,h(P)}\quad\text{with}\quad
C=\frac{8}{\pi}\sqrt{\log(1+\sqrt{2})}=2.3906\ldots.$ (4.4)

Remark. In [18], Soundararajan works with the triangle test function
$F_{\blacktriangle}$ as above, establishing a bound in (1.2) with
$C=8/\pi=2.54\ldots$. Later, it came to our attention that, in unpublished
notes222Personal communication., he independently arrived at the refined
inequality in (4.4) by further studying the situation with this particular
test function.

##  5\. Magic functions

In this section we prove Theorem 3. Ultimately, our proof relies on the
existence of two magic functions. The first one, mentioned in the statement of
the theorem, is the even function, supported in $[-\frac{1}{2},\frac{1}{2}]$,

$\mathfrak{F}(x):=\frac{2}{\pi}\log\left(\frac{1+\sqrt{1-4x^{2}}}{2|x|}\right)\
\ \ \ \ \big{(}{\rm for}\ -\tfrac{1}{2}\leq x\leq\tfrac{1}{2}\big{)}.$ (5.1)

The second one is the odd function, also supported in
$[-\frac{1}{2},\frac{1}{2}]$, given by

$\qquad\qquad\ \ \ \mathfrak{G}(x):=\frac{2x}{\sqrt{1-4x^{2}}}\ \ \ \ \
\qquad\big{(}{\rm for}\ -\tfrac{1}{2}<x<\tfrac{1}{2}\big{)}.$ (5.2)

We first treat the extremal problem (EP1), to find the value of the sharp
constant ${\bf C}$. Later, with some of the main ingredients already laid out,
we discuss the details that lead to the solution of the extremal problem (EP2)
and the sharp constant ${\bf C}^{*}$.

###  5.1. Lower bound via duality

The map $\mathcal{H}:L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$ is an isometry
and verifies $\mathcal{H}^{2}=-I$ (therefore the inverse of $\mathcal{H}$ is
$-\mathcal{H}$). Hence, whenever $F_{1},F_{2}\in L^{2}(\mathbb{R})$ we have

$\int_{\mathbb{R}}\mathcal{H}(F_{1})(x)\,\overline{F_{2}(x)}\,\text{\rm
d}x=-\int_{\mathbb{R}}F_{1}(x)\,\overline{\mathcal{H}(F_{2})(x)}\,\text{\rm
d}x.$ (5.3)

Since $\mathcal{H}:L^{p}(\mathbb{R})\to L^{p}(\mathbb{R})$ is a bounded
operator for $1<p<\infty$, identity (5.3) extends to the situation where
$F_{1}\in L^{p}(\mathbb{R})$ and $F_{2}\in L^{p^{\prime}}(\mathbb{R})$, where
$\tfrac{1}{p}+\tfrac{1}{p^{\prime}}=1$ and $1<p,p^{\prime}<\infty$. The odd
function $\mathfrak{G}$ belongs to $L^{p}(\mathbb{R})$ for $1\leq p<2$ but not
to $L^{2}(\mathbb{R})$. It verifies

$\|\mathfrak{G}\|_{L^{1}(\mathbb{R})}=\int_{\mathbb{R}}|\mathfrak{G}(x)|\,\text{\rm
d}x=1.$ (5.4)

The Hilbert transform of $\mathfrak{G}$ can be explicitly computed and is
given by

$\mathcal{H}(\mathfrak{G})(x)=\begin{cases}-1,\ {\rm if}\ |x|<\tfrac{1}{2};\\\
-1+\frac{2|x|}{\sqrt{4x^{2}-1}},\ {\rm if}\ |x|>\tfrac{1}{2}.\end{cases}$
(5.5)

We refer the reader to [3, p. 248, eq. (25)] for this computation333Letting
$L(x)$ be the function on the left-hand side of [3, p. 248, eq. (25)] with
$a=\frac{1}{2}$, we have
$\mathfrak{G}(x)=\tfrac{1}{2}\big{(}L(x)-L(-x)\big{)}$. Note also that the
Hilbert transform in [3] is defined with a multiplying factor of $-1$.. We
shall see in a moment that the fact that $\mathfrak{G}$ has
$L^{1}(\mathbb{R})$-norm equal to $1$ and that its Hilbert transform is
constant (equal to $-1$) in the interval $[-\tfrac{1}{2},\tfrac{1}{2}]$ is
precisely what makes it magical. The graphs of $\mathfrak{G}$ and
$\mathcal{H}(\mathfrak{G})$ are plotted in Figure 2.

Take any $0\neq F\in\mathcal{A}$, normalized so that
$\|F\|_{L^{1}(\mathbb{R})}=\int_{\mathbb{R}}F(x)\,\text{\rm d}x=1$. Note that
$F\in L^{p}(\mathbb{R})$ for all $1\leq p\leq\infty$. Using (5.3) with
$F_{1}=F$ and $F_{2}=\mathfrak{G}$, together with (5.4), (5.5), and the fact
that ${\rm supp}(F)\subset[-\tfrac{1}{2},\tfrac{1}{2}]$, we get the following
relation

$\displaystyle\begin{split}\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}&=\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}\int_{\mathbb{R}}|\mathfrak{G}(x)|\,\text{\rm
d}x\geq\int_{\mathbb{R}}\mathcal{H}(F)(x)\,\overline{\mathfrak{G}(x)}\,\text{\rm
d}x\\\
&=-\int_{\mathbb{R}}F(x)\,\overline{\mathcal{H}(\mathfrak{G})(x)}\,\text{\rm
d}x=\int_{-\frac{1}{2}}^{\frac{1}{2}}F(x)\,\text{\rm d}x=1.\end{split}$ (5.6)

Since (5.6) holds for any such normalized $F\in\mathcal{A}$, we plainly get
the lower bound

${\bf C}\geq 1.$

In addition, once we establish in the next subsection that ${\bf C}$ is
actually equal to $1$, relation (5.6) also tells us that there are no
extremizers for the problem (EP1) in the class $\mathcal{A}$. In fact,
equality in (5.6) could only be attained if

$\mathcal{H}(F)(x)={\rm
sgn}(x)\,\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}={\rm sgn}(x)$

for a.e. $-\tfrac{1}{2}<x<\tfrac{1}{2}$, which cannot occur since
$\mathcal{H}(F)$ is odd and continuous when $F\in\mathcal{A}$.

Figure 2. On the left, the graph of the magic function $\mathfrak{G}$. On the
right, the graph of the Hilbert transform $\mathcal{H}(\mathfrak{G})$.

###  5.2. A rogue extremal function

We now turn our attention to the function $\mathfrak{F}$ defined in (5.1).
Observe first that $\mathfrak{F}\geq 0$ in $[-\tfrac{1}{2},\tfrac{1}{2}]$,
$\mathfrak{F}$ is continuous and radial decreasing on
$\mathbb{R}\setminus\\{0\\}$ (with a logarithmic singularity at the origin),
and $\mathfrak{F}$ is smooth on $\mathbb{R}\setminus\\{0,\pm\tfrac{1}{2}\\}$.
Moreover, $\mathfrak{F}\in L^{p}(\mathbb{R})$ for $1\leq p<\infty$, and we
note that

$\|\mathfrak{F}\|_{L^{1}(\mathbb{R})}=\int_{\mathbb{R}}\mathfrak{F}(x)\,\text{\rm
d}x=1.$ (5.7)

####  5.2.1. The Hilbert transform of $\mathfrak{F}$

The Hilbert transform $\mathcal{H}(\mathfrak{F})$ is an odd function. For
almost every $x\in\mathbb{R}$ it is given by its singular integral
representation. We may carefully apply integration by parts (excluding the
singularities and then passing to the limit) to get, for a.e. $x>0$,

$\displaystyle\mathcal{H}(\mathfrak{F})(x)$ $\displaystyle={\rm
p.v.}\,\frac{1}{\pi}\int_{\mathbb{R}}\mathfrak{F}(x-t)\,\frac{1}{t}\,\text{\rm
d}t=\frac{1}{\pi}\int_{\mathbb{R}}\mathfrak{F}^{\prime}(x-t)\,\log|t|\,\text{\rm
d}t=\frac{1}{\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}\mathfrak{F}^{\prime}(t)\log|x-t|\,\text{\rm
d}t$
$\displaystyle=-\frac{1}{\pi}\int_{0}^{\frac{1}{2}}\mathfrak{F}^{\prime}(t)\log\left(\frac{|x+t|}{|x-t|}\right)\,\text{\rm
d}t$
$\displaystyle=\frac{2}{\pi^{2}}\int_{0}^{\frac{1}{2}}\frac{1}{t\,\sqrt{1-4t^{2}}}\log\left(\frac{|x+t|}{|x-t|}\right)\,\text{\rm
d}t.$

This last integral can be evaluated explicitly, see [11, §4.297 eqs. 8 and
10], yielding

$\displaystyle\mathcal{H}(\mathfrak{F})(x)=\begin{cases}{\rm sgn}(x),\ {\rm
if}\ |x|\leq\tfrac{1}{2};\\\ \frac{2}{\pi}\arcsin\left(\frac{1}{2x}\right),\
{\rm if}\ |x|>\tfrac{1}{2}.\end{cases}$ (5.8)

From (5.8) we see that
$\|\mathcal{H}(\mathfrak{F})\|_{L^{\infty}(\mathbb{R})}=1$, and given that
$\mathfrak{F}$ has the correct normalization (5.7), it is essentially an
extremizer for our problem. We say ‘essentially’ because $\mathfrak{F}$ does
not exactly belong to our class $\mathcal{A}$, but it is almost there (hence
the rogue in the title of this subsection). The graphs of $\mathfrak{F}$ and
$\mathcal{H}(\mathfrak{F})$ are plotted in Figure 3.

####  5.2.2. Approximating the rogue extremal function

We need to make a small correction to $\mathfrak{F}$ via a standard
approximation argument. Let $\varphi\in C^{\infty}_{c}(\mathbb{R})$ be a non-
negative radial decreasing function supported in
$[-\tfrac{1}{2},\tfrac{1}{2}]$, with $\int_{\mathbb{R}}\varphi(x)\,\text{\rm
d}x=1$ and Fourier transform $\widehat{\varphi}$ also non-negative. To
construct such a function, we can just take $\varphi=\psi*\psi$, where $\psi$
is a smooth, radial decreasing, non-negative function supported in
$[-\tfrac{1}{4},\tfrac{1}{4}]$. Recall that the convolution of two radial
decreasing functions is still radial decreasing (for a beautiful proof of this
fact we refer the reader to [4, p. 171]). For $\varepsilon>0$ small, let

$\varphi_{\varepsilon}(x):=\tfrac{1}{\varepsilon}\,\varphi\\!\left(\tfrac{x}{\varepsilon}\right)\quad{\rm
and}\quad\mathfrak{F}_{1-\varepsilon}(x):=\tfrac{1}{1-\varepsilon}\,\mathfrak{F}\\!\left(\tfrac{x}{1-\varepsilon}\right),$

and define

$F^{\varepsilon}:=\mathfrak{F}_{1-\varepsilon}*\varphi_{\varepsilon}.$

Observe that $F^{\varepsilon}$ is a smooth, radial decreasing and non-negative
function, supported in $[-\tfrac{1}{2},\tfrac{1}{2}]$ with
$\int_{\mathbb{R}}F^{\varepsilon}(x)\,\text{\rm d}x=1$. Moreover,
$\widehat{F^{\varepsilon}}(t)=\widehat{\mathfrak{F}}\big{(}(1-\varepsilon)t\big{)}\,\widehat{\varphi}(\varepsilon
t)\in L^{1}(\mathbb{R})$ (recall that $\widehat{\mathfrak{F}}$ is bounded
since $\mathfrak{F}\in L^{1}(\mathbb{R})$). Hence
$F^{\varepsilon}\in\mathcal{A}$. At the level of the Hilbert transform, we
have

$\displaystyle\mathcal{H}(F^{\varepsilon})(x)=\big{(}\mathcal{H}(\mathfrak{F}_{1-\varepsilon})*\varphi_{\varepsilon}\big{)}(x)=\left(\tfrac{1}{1-\varepsilon}\mathcal{H}(\mathfrak{F})\left(\tfrac{{}^{\centerdot}}{1-\varepsilon}\right)*\varphi_{\varepsilon}\right)\\!(x),$

and we see from (5.8) that

$\|\mathcal{H}(F^{\varepsilon})\|_{L^{\infty}(\mathbb{R})}=\frac{1}{1-\varepsilon}.$

As we have argued in §4.1, since $F^{\varepsilon}$ is radial decreasing we do
have
$\mathcal{C}(F^{\varepsilon})=\|\mathcal{H}(F^{\varepsilon})\|_{L^{\infty}(\mathbb{R})}$.
Sending $\varepsilon\to 0$ we conclude that

${\bf C}=1.$

Figure 3. On the left, the graph of the magic function $\mathfrak{F}$. On the
right, the graph of the Hilbert transform $\mathcal{H}(\mathfrak{F})$.

###  5.3. The extremal problem (EP2)

Note that $\mathfrak{F}\in\mathcal{A}^{*}$ and that ${\bf C}^{*}\leq{\bf
C}=1$. We now verify the lower bound. Let $0\neq F\in\mathcal{A}^{*}$ be a
given function, normalized so that
$\|F\|_{L^{1}(\mathbb{R})}=\int_{\mathbb{R}}F(x)\,\text{\rm d}x=1$. We may
assume without loss of generality that
$\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}<\infty$. Since
$\mathcal{H}(F)(x)=1/(\pi x)+O\big{(}1/|x|^{2}\big{)}$ for $|x|$ large, we
find that $\mathcal{H}(F)\in L^{p}(\mathbb{R})$ for any $1<p\leq\infty$. This
implies that $F$ must have been in $L^{p}(\mathbb{R})$, for any $1\leq
p<\infty$, from the start.

This last claim deserves a brief justification. An argument of Calderón and
Capri [6, Lemma 4]444This lemma is stated for the situation when the singular
integral operator belongs to $L^{1}$, but the proof works for $L^{p}$
($1<p<\infty$) as well. One simply applies Minkowski’s inequality for
integrals to arrive at eq. (17) with $1<p<\infty$. shows that whenever $F\in
L^{1}(\mathbb{R})$ and $\mathcal{H}(F)\in L^{p}(\mathbb{R})$, for some
$1<p<\infty$, and $\Psi$ is a continuous function of compact support, for a.e.
$x\in\mathbb{R}$ we have

$\mathcal{H}(F*\Psi)(x)=\big{(}\mathcal{H}(F)*\Psi\big{)}(x).$ (5.9)

Letting $\varphi$ be a smooth function of compact support, with
$\int_{\mathbb{R}}\varphi(x)\,\text{\rm d}x=1$, and setting
$\varphi_{\varepsilon}(x):=\tfrac{1}{\varepsilon}\varphi\left(\tfrac{x}{\varepsilon}\right)$
as usual, identity (5.9) holds with $\Psi$ replaced by
$\varphi_{\varepsilon}$. Since $F*\varphi_{\varepsilon}$ and
$\mathcal{H}(F)*\varphi_{\varepsilon}$ belong to $L^{p}(\mathbb{R})$ we may
apply the Hilbert transform on both sides of (5.9), using the fact that
$\mathcal{H}^{2}=-I$ on $L^{p}(\mathbb{R})$, to arrive at

$-(F*\varphi_{\varepsilon})(x)=\mathcal{H}\big{(}\mathcal{H}(F)*\varphi_{\varepsilon}\big{)}(x)$
(5.10)

for a.e. $x\in\mathbb{R}$. Letting $\varepsilon\to 0$, since
$\mathcal{H}(F)*\varphi_{\varepsilon}\to\mathcal{H}(F)$ in $L^{p}(\mathbb{R})$
and the Hilbert transform is bounded on $L^{p}(\mathbb{R})$, the right-hand
side of (5.10) converges to $\mathcal{H}\big{(}\mathcal{H}(F)\big{)}$ in
$L^{p}(\mathbb{R})$. The left-hand side of (5.10) converges to $-F$ a.e. The
conclusion is that we must indeed have
$-F=\mathcal{H}\big{(}\mathcal{H}(F)\big{)}$, and therefore $F\in
L^{p}(\mathbb{R})$ as well. An alternative way to argue when $F$ has compact
support and $\int_{\mathbb{R}}F(x)\,\text{\rm d}x=1$, is by observing that
$F-\chi_{[-\frac{1}{2},\frac{1}{2}]}$ belongs to the Hardy space
$H^{1}(\mathbb{R})=\\{G\in L^{1}(\mathbb{R})\ ;\mathcal{H}(G)\in
L^{1}(\mathbb{R})\\}$. This is a Banach space with norm
$\|G\|_{H^{1}(\mathbb{R})}:=\|G\|_{L^{1}(\mathbb{R})}+\|\mathcal{H}(G)\|_{L^{1}(\mathbb{R})}$
(see e.g. [13, Theorem 6.7.4]) in which $\mathcal{H}$ is an isometry with
$\mathcal{H}^{2}=-I$.

Having gone through these considerations, the application of (5.6) is
justified and we arrive at the conclusion that ${\bf C}^{*}\geq 1$, and hence
${\bf C}^{*}=1$. Let us now discuss the uniqueness of the extremizer. Equality
happens in (5.6) if and only if

$\mathcal{H}(F)(x)={\rm
sgn}(x)\,\|\mathcal{H}(F)\|_{L^{\infty}(\mathbb{R})}={\rm sgn}(x)$

for a.e. $-\tfrac{1}{2}<x<\tfrac{1}{2}$. This implies that

$\mathcal{H}(\mathfrak{F}-F)(x)=0$ (5.11)

for a.e. $-\tfrac{1}{2}<x<\tfrac{1}{2}$. We are now in position to invoke a
suitable uniqueness result, first established in a classical paper by Tricomi
[20], and revisited recently by Coifman and Steinerberger [7, Theorem 1].

######  Lemma 6 (cf. [20] and [7]).

Let $G$ be a real-valued function such that ${\rm
supp}(G)\subset[-\tfrac{1}{2},\tfrac{1}{2}\big{]}$ and
$G(x)(1-4x^{2})^{1/4}\in L^{2}\big{(}-\tfrac{1}{2},\tfrac{1}{2}\big{)}$. If
$\mathcal{H}(G)\equiv 0$ on $\big{(}-\tfrac{1}{2},\tfrac{1}{2}\big{)}$ then,
for some $c\in\mathbb{R}$, we have

$\qquad\qquad\qquad\qquad G(x)=\frac{c}{\sqrt{1-4x^{2}}}\qquad\ \ \big{(}{\rm
for}\ -\tfrac{1}{2}<x<\tfrac{1}{2}\big{)}.$

From (5.11) and Lemma 6 we arrive at

$\mathfrak{F}(x)-F(x)=\frac{c}{\sqrt{1-4x^{2}}}$

for a.e. $-\tfrac{1}{2}<x<\tfrac{1}{2}$. Since
$\int_{\mathbb{R}}\mathfrak{F}(x)\,\text{\rm
d}x=\int_{\mathbb{R}}F(x)\,\text{\rm d}x=1$, we conclude that $c=0$ and
$F=\mathfrak{F}$, as proposed.

## Acknowledgements

This project started at the workshop Arithmetic statistics, discrete
restriction, and Fourier analysis at the American Institute of Mathematics
(AIM) in 2021. We thank Theresa Anderson, Frank Thorne, and Trevor Wooley for
the organization of the workshop, and the staff of AIM for providing a superb
scientific atmosphere. We are thankful to William Beckner, Tiago Picon, Ruiwen
Shu, Mateus Sousa, and the referee for enlightening remarks. We also thank
Kannan Soundararajan for sharing some of his unpublished notes on the theme.
EC acknowledges support from FAPERJ - Brazil, AF was supported by NSF
DMS-2101769 and the NSF Postdoctoral Fellowship DMS-1703695, AM was supported
by NSF DMS-1854398 FRG, MBM was supported by NSF DMS-2101912 and a Simons
Foundation Collaboration Grant for Mathematicians, and CT-B was supported by
NSF DMS-1902193 and NSF DMS-1854398 FRG.

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