#  New Lower Bounds for the Schur-Siegel-Smyth Trace Problem

Bryce Joseph Orloski and Naser Talebizadeh Sardari  Penn State department of
Mathematics, McAllister Building, Pollock Rd, State College, PA 16802 USA
nzt5208<EMAIL_ADDRESS>
###### Abstract.

We derive and implement a new way to find lower bounds on the smallest
limiting trace-to-degree ratio of totally positive algebraic integers and
improve the previously best known bound to 1.798249. Our method adds new
constraints to Smyth’s linear programming method to decrease the number of
variables required in the new problem of interest. This allows for faster
convergence recovering Schur’s bound in the simplest case and Siegel’s bound
in the second simplest case of our new family of bounds. We also prove the
existence of a unique optimal solution to our newly phrased problem and
express the optimal solution in terms of polynomials. Lastly, we solve this
new problem numerically with a gradient descent algorithm to attain the new
bound 1.798249.

We would like to thank Professors Sarnak, Zargar, and Zikatanov for their
comments on the earlier version of this work. The authors also thank
Pennsylvania State University’s Institute for Computational and Data Sciences
for allowing us to use their ROAR servers for our numerical experiments.

##  1\. Introduction

###  1.1. The Schur-Siegel-Smyth trace problem.

For an algebraic integer $\alpha$ with $\deg(\alpha)=n$ and the set of Galois
conjugates $[\alpha]:=\\{\alpha_{1},\dots,\alpha_{n}\\}$, define

$\overline{\text{tr}}(\alpha):=\frac{\sum_{i=1}^{n}\alpha_{i}}{n},$
$\text{Nr}(\alpha)=\prod_{i=1}^{n}\alpha_{i},\text{ and }$
$\Delta(\alpha)=\prod_{i<j}(\alpha_{i}-\alpha_{j})^{2}\in\mathbb{Z}^{+}.$

The algebraic integer $\alpha$ is called totally positive if $\alpha_{i}>0$
for $i=1,\dots,n$. Let $\mathcal{A}$ be the set of totally positive algebraic
integers. The Schur-Siegel-Smyth trace problem is to compute the value of the
number
$\lambda^{SSS}:=\liminf_{\alpha\in\mathcal{A}}\overline{\text{tr}}(\alpha)$
for a fixed ordering of $\mathcal{A}$. Schur [Sch18] in 1918 proved that

$\lambda^{SSS}\geq\sqrt{e}.$

In other words, this means that for any $\lambda<\sqrt{e}$, there are only
finitely many $\alpha\in\mathcal{A}$ for which
$\overline{\text{tr}}(\alpha)<\lambda$. Siegel [Sie45] improved this to

$\lambda^{SSS}\geq e(1+\nu^{-1})^{-\nu}\approx 1.7336105,$

where $\nu$ is the positive root of the transcendental equation

$(1+\nu)\log(1+\nu^{-1})+\frac{\log\nu}{1+\nu}=1.$

Siegel used only that $\text{Nr}(\alpha)\geq 1$ and $\Delta(\alpha)\geq 1$ for
$\alpha\in\mathcal{A}$, and he solved a similar optimization problem as Schur
using Lagrange multipliers.

The previous best lower bound [WWW21] was achieved numerically by using a
linear programming method initiated first by Smyth [Smy84]. In what follows,
we explain Smyth’s method. Smyth showed that if $\mu$ is the limit of uniform
probability measures attached to distinct conjugate algebraic integers all of
whose conjugates lie in a compact set $K$ of the complex plane, then

$\int\log|Q(x)|d\mu(x)\geq 0$

for every $Q(x)\in\mathbb{Z}[x]$. In a recent breakthrough, Smith [Smi21]
proved that these necessary conditions are sufficient when $K$ is a subset of
$\mathbb{R}$ under some assumptions on $K$. We [OT23] recently established
this result for compact subset $K$ of the complex plane under some assumptions
and our new method gives a different treatment of Smith’s Theorem. In
particular, we introduced some new constraints on the limiting measures of
conjugate algebraic integers. For a limiting measure $\mu$, we proved in
[OT23] that

(1) $\int\log(|Q(x_{1},\dots,x_{n})|)d\mu(x_{1})\dots d\mu(x_{n})\geq 0$

for every $Q(x_{1},\dots,x_{n})\in\mathbb{Z}[x_{1},\dots,x_{n}]$. The goal of
this paper is to improve Smyth’s lower bound method by using new constraints
in (1).

###  1.2. Main Results

Smyth gave a lower bound on the trace problem by minimizing $\int xd\mu(x)$
subject to

$\int\log|Q(x)|d\mu(x)\geq 0$

for every $Q(x)\in A$, where $A\subset\mathbb{Z}[x]$ is a fixed finite subset.
In fact, Smyth worked with the equivalent associated dual linear programming
problem [BV04, Chapter 5] by finding maximal $\lambda_{A}$ such that

(2) $x\geq\lambda_{A}+\sum_{Q\in A}\lambda_{Q}\log|Q(x)|$

holds for every $x\in\mathbb{R}^{+}$ and some choice of scalars
$\lambda_{Q}\geq 0$. In this paper, we minimize $\int xd\mu(x)$ subject to
inequalities (1) for every $Q\in\mathbb{Z}[x_{1},\dots,x_{n}]$. Following
Smyth’s method, we consider maximizing $\lambda_{B}$ such that

(3) $\frac{x+y}{2}\geq\lambda_{B}+\sum_{Q\in B}\lambda_{Q}\log|Q(x,y)|$

for every $x,y\in\mathbb{R}^{+}$ and some $\lambda_{B},\lambda_{Q}\geq 0$ and
a finite subset $B\subset\mathbb{Z}[x,y]$. By integrating both sides of (3)
with respect to $d\mu(x)d\mu(y)$, it follows that

$\int xd\mu(x)\geq\lambda_{B}.$

We note that the bounds in (3) are at least as good as Smyth’s method which
uses (2). We verify this next. Suppose that (2) holds for some choices of
$A\subset\mathbb{Z}[x]$ and $\lambda_{A},\lambda_{Q}\geq 0$. This implies

$\begin{cases}x\geq\lambda_{A}+\sum_{Q\in A}\lambda_{Q}\log|Q(x)|,\\\
y\geq\lambda_{A}+\sum_{Q\in A}\lambda_{Q}\log|Q(y)|.\end{cases}$

By taking the mean of the above inequalities, we have

$\frac{x+y}{2}\geq\lambda_{A}+\sum_{Q\in
A}\lambda_{Q}\frac{\log|Q(x)|+\log|Q(y)|}{2}$

for every $x,y\in\mathbb{R}^{+}$. This is a special case of (3) with

$B:=\\{Q(x),Q(y):Q\in A\\}.$

This implies that $\lambda_{A}\leq\lambda_{B}$. Moreover, our main theorem
shows that Smyth’s bound $\lambda_{A}$ is strictly smaller than
$\lambda_{B^{+}}$, where $B^{+}=B\cup\\{x-y\\}$.

There is a loss in working with the dual problem (3) as the constraints in (1)
are not linear in $\mu$. Next, we improve upon this dual bound by working
directly with (1). The main difficulty in working directly with (1) is that,
as stated before, this problem is no longer a linear programming problem and
the associated dual problem in (3) is not equivalent in general. Our main idea
is to find a particular $Q(x_{1},\dots,x_{n})\in\mathbb{Z}[x_{1},\dots,x_{n}]$
such that the functional
$\mu\mapsto\int\log(|Q(x_{1},\dots,x_{n})|)d\mu(x_{1})\dots d\mu(x_{n})$ is
concave, and then apply the strong duality identity which holds for the convex
optimization problems. We state our main theorem for
$Q(x_{1},x_{2})=x_{1}-x_{2}$. A similar method works for
$Q(x_{1},\dots,x_{4})=x_{1}-x_{2}-x_{3}+x_{4}$ which is a possible object of
future research. In the case of $Q(x_{1},x_{2})=x_{1}-x_{2}$, we define

$I(\mu):=\int\log|x_{1}-x_{2}|d\mu(x_{1})d\mu(x_{2})$

to be the logarithmic energy of $\mu$.

######  Theorem 1.1.

Suppose that $A\subset\mathbb{R}[x]$ is a finite subset of polynomials with
real roots. Let $\lambda_{A}=\inf_{\mu}\int xd\mu(x)$, where $\mu$ varies
among probability measures supported on $\mathbb{R}^{+}$ with $I(\mu)\geq 0$
and $\int\log|Q(x)|d\mu\geq 0$ for every $Q\in A$. There exists a unique
measure $\mu_{A}$ supported on a finite union of intervals
$\Sigma\subset[0,18]$ such that

$\begin{array}[]{cc}\int xd\mu_{A}(x)=\lambda_{A},\\\ I(\mu_{A})=0,\\\
x\geq\lambda_{A}+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}U_{\mu_{A}}(x),\text{ and}\\\
\int\log|Q(x)|d\mu_{A}=0\text{ if $\lambda_{Q}\neq 0$,}\par\end{array}$

where in the third line the inequality only holds for non-negative $x$ and
equality holds for every $x\in\Sigma$ with some scalars $\lambda_{Q}\geq 0$
and $\lambda_{0}>0$. Moreover if $\Sigma:=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$
and $\\{x\\}\subset A$, then $\mu_{A}$ has the following density function up
to a scalar

$\frac{|p(x)|\sqrt{|H(x)|}}{\prod_{\lambda_{Q}\neq 0}|Q(x)|},$

where $H(x)=\prod_{i=0}^{l}(x-a_{2i})(x-a_{2i+1})$ and $p(x)$ is a polynomial
with degree $\sum_{Q\in A}\deg(Q)-l-1$. Furthermore, every gap interval
$(a_{2i+1},a_{2i+2})$ contains a root of some $Q\in A$.

######  Corollary 1.2.

Let $\lambda_{A}$ be as in Theorem 1.1. We have

$\lambda_{A}\leq\lambda^{SSS}.$

###### Proof.

Suppose that $\alpha_{n}$ is a sequence of totally positive algebraic integers
such that

$\lim_{n\to\infty}\overline{\text{tr}}(\alpha_{n})=\lambda^{SSS}.$

Let

$\mu_{\alpha_{n}}:=\frac{1}{\deg(\alpha_{n})}\sum_{\alpha_{n,j}\in[\alpha_{n}]}\delta_{\alpha_{n,j}}$

be the uniform discrete probability measure on the set of Galois conjugates of
$\alpha_{n}$. Since $\mu_{\alpha_{n}}$ is a tight family of probability
measures, Prokhorov’s theorem implies that there exists a convergent
subsequence. Suppose that $\mu$ is a limiting measure of a subsequence. As we
mentioned before, $\mu$ satisfies all the constraints of Theorem 1.1 for any
finite set $A$. Hence,

$\lambda_{A}\leq\lambda^{SSS}$

for any finite set $A$. ∎

######  Remark 4.

Note that Corollary 5.5 of Smith [Smi21] directly implies that the least upper
bound of such $\lambda_{A}$ is $\lambda^{SSS}$.

We recover Schur’s lower bound by taking $A=\emptyset$ [Sch18] and Siegel’s
lower bound by taking $A=\\{x\\}$ [Sie45]. We state this in the following
corollaries.

######  Corollary 1.3.

If $A=\emptyset$, then $\lambda_{A}=\sqrt{e}$. Moreover,
$d\mu_{A}=\frac{1}{2\pi}\sqrt{\frac{(4\sqrt{e}-x)}{ex}}dx$.

######  Corollary 1.4.

If $A=\\{x\\}$, then

$\lambda_{A}=e(1+\nu^{-1})^{-\nu}=1.7336105\dots,$

where $\nu$ is the positive root of the transcendental equation

(5) $(1+\nu)\log(1+\nu^{-1})+\frac{\log\nu}{1+\nu}=1.$

Moreover,

$\mu_{A}=\frac{2\sqrt{(b-x)(x-a)}}{\pi(a+b-2\sqrt{ab})x}dx$

where

$a=e(1+\nu^{-1})^{-\nu}\left(1-\frac{1}{\sqrt{\nu+1}}\right)^{2}\text{ and
}b=e(1+\nu^{-1})^{-\nu}\left(1+\frac{1}{\sqrt{\nu+1}}\right)^{2}.$

###  1.3. Gradient Descent

We developed an algorithm which given a finite $A\subset\mathbb{Z}[x]$,
returns a distribution approximating $\mu_{A}$. Our algorithm is based on a
gradient descent algorithm.

######  Corollary 1.5.

Let $A=\\{x,x-1\\}$. Then

$|\lambda_{A}-1.7773797|\leq 10^{-7}.$

Moreover, $\mu$ is supported on the union of two intervals
$[a_{0},a_{1}]\cup[a_{2},a_{3}]$ with the following numerical values:

$a_{0}=0.0362736\dots,a_{1}=0.828301\dots,a_{2}=1.190973\dots,\text{ and
}a_{3}=5.707091\dots.$

Furthermore, it has the following density function:

$d\mu_{A}(x)=\frac{\sqrt{|\left(x-a_{0}\right)\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)|}}{c|x\left(x-1\right)|}dx$

where

$c=6.420592\dots.$

######  Corollary 1.6.

If $A=\\{x,x-1,x^{2}-3x+1\\}$, then

$|\lambda_{A}-1.793023|\leq 10^{-6}$

where $\Sigma$ is a union of four intervals approximated by

$[0.0409275,0.34114487]\cup[0.4252603,0.811681]\cup[1.211488,2.4844644]\cup[2.7580631,5.52512172].$

######  Corollary 1.7.

If $A=\\{x,x-1,x^{2}-3x+1,x^{3}-5x^{2}+6x-1\\}$, then

$|\lambda_{A}-1.798249|\leq 10^{-6}.$

Figure 1. Density function of $\mu_{\\{x,x-1,x^{2}-3x+1,x^{3}-5x^{2}+6x-1\\}}$

This beats the currently best known lower bound on the problem. For extra
details about the measures used in these corollaries, see Figure 1 and
Appendix B.

Below is a table of the history of the best known lower bound on the problem
with the last line including the newest bound which is the main topic of this
paper. This is the largest improvement on the lower bound since Smyth [Smy84]
in 1984.

Authors | Year | Bound Estimate
---|---|---
Schur [Sch18] | 1918 | 1.6487
Siegel [Sie45] | 1945 | 1.7336
Smyth [Smy84] | 1984 | 1.7719
Flammang, et al. [FGR99] | 1997 | 1.7735
McKee and Smyth [MS04] | 2004 | 1.7783
Aguirre, Bilbao, and Peral [ABP06] | 2006 | 1.78002
Aguirre and Peral [AP07] | 2007 | 1.7836
Aguirre and Peral [AP08] | 2008 | 1.7841
Flammang [Fla09] | 2009 | 1.78702
Liang and Wu [LW11] | 2011 | 1.7919
Flammang [Fla19] | 2016 | 1.7928
Wang, Wu, and Wu [WWW21] | 2021 | 1.7931
Our new lower bound | 2023 | 1.79824

The proof of the previously best known lower bound by Wang, Wu, and Wu uses
Smyth’s method with $130$ polynomials whereas with our method, we reduce the
number of polynomials to 4. There is, however, a computational trade-off. The
difficulty in the former method is finding good candidate polynomials with low
trace, whereas the largest difficulty in the authors’ method is computing the
support of the optimal distribution given by their linear programming problem.

###  1.4. Method of Proof

It is well-known that $I(\mu)$ is concave which means that for all probability
measures $\mu_{1}$ and $\mu_{2}$,

(6) $I(\mu_{1})+I(\mu_{2})-2I(\mu_{1},\mu_{2})\leq 0,$

where
$I(\mu_{1},\mu_{2}):=\int\log|x_{1}-x_{2}|d\mu_{1}(x_{1})d\mu_{2}(x_{2})$
[Ahl10, Lemma 2.1]. It follows from Lemma 2.4 that

(7) $I(\mu)\geq 0$

is equivalent to the following family of linear inequalities

(8) $2I(\mu,\mu_{i})-I(\mu_{i})=\int(2U_{\mu_{i}}(x)-I(\mu_{i}))d\mu\geq 0$

for every measure $\mu_{i}$ with non-negative energy $I(\mu_{i})\geq 0$, where

$U_{\mu}(x)=\int\log|x-z|d\mu(z)$

is the logarithmic potential of $\mu$. This implies that the following
optimization problem:

$\begin{array}[]{cc}\min\int xd\mu(x)\\\ \int\log|Q|d\mu(x)\geq 0\text{ for
every }Q\in A\\\ I(\mu)\geq 0,\end{array}$

is equivalent to the following linear programming problem:

$\begin{array}[]{cc}\min\int xd\mu(x)\\\ \int\log|Q|d\mu(x)\geq 0\text{ for
every }Q\in A\\\ \int(2U_{\mu_{i}}(x)-I(\mu_{i}))d\mu\geq 0,\end{array}$

for every measure $\mu_{i}$ with non-negative energy $I(\mu_{i})\geq 0$. We
denote the above optimization problem by the primal problem. Let $\Lambda_{A}$
be the minimum of the primal problem. Suppose that $\mu_{A}$ is the
probability measure that achieves this minimum, then

$\int xd\mu_{A}(x)=\Lambda_{A}.$

The associated dual problem is maximizing $\lambda$ such that

$x\geq\lambda+\sum_{Q_{i}\in
A}\lambda_{Q}\log|Q(x)|+\sum_{\mu_{i}}\lambda_{i}(2U_{\mu_{i}}(x)-I(\mu_{i}))$

for $x\geq 0$, where $\mu_{i}$ varies among probability measures. Let
$\lambda_{A}$ be the maximum of the dual problem. It follows that
$\Lambda_{A}\geq\lambda_{A}$. As part of the proof of our main theorem, we
show that strong duality holds which means $\Lambda_{A}=\lambda_{A}$. By
integrating the above inequality with respect to $d\mu_{A}$, it follows that
the above inequality is equality on the support of $\mu_{A}$ and
$\mu_{i}=\mu_{A}$. Then

(9) $x\geq\lambda_{A}+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}(2U_{\mu_{A}}(x)-I(\mu_{A}))$

for $x\geq 0$,

$\int\log|Q(x)|d\mu_{A}=0$

for $Q\in A$ if $\lambda_{Q}\neq 0$, and

$I(\mu_{A})=0.$

In fact, the above conditions identify $\mu_{A}$ uniquely. It follows under
some assumptions that $\mu_{A}$ is supported on the finite union of some
intervals $\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$ and has the following density
function up to a scalar:

$\frac{|p(x)|\sqrt{|H(x)|}}{\prod_{\lambda_{Q}\neq 0}|Q(x)|},$

where $H(x)=\prod_{i=0}^{l}(x-a_{2i})(x-a_{2i+1})$ and $p(x)$ is a polynomial
with degree $\sum_{Q\in A}\deg(Q)-l-1$.

###  1.5. Organization of the Paper

In section 2, we formulate a convex optimization problem and prove that strong
duality holds for this problem. In section 3, we give an expression for the
optimal solution of the optimization problem in terms of polynomials. In
section 4, we give a proof of Theorem 1.1. In section 5, we give proofs of
Corollary 1.3 and Corollary 1.4. Finally in section 6, we discuss our gradient
descent algorithm and present our numerical results.

##  2\. Convex optimization and strong duality

In this section, we give a proof of the following proposition.

######  Proposition 2.1.

Let $A=\\{Q_{1},\dots,Q_{n}\\}$ be any finite subset of polynomials. Define

$\Lambda_{A}=\inf\int xd\mu(x)$

subjected to $I(\mu)\geq 0$ and the following constraints:

$\int\log|Q_{i}(x)|d\mu(x)\geq 0$

for every $Q_{i}\in A$. Then there exists a unique probability measure
$\mu_{A}$ and constants $\lambda_{Q},\lambda_{0}\geq 0$ such that $\int
xd\mu(x)=\Lambda_{A}$, $I(\mu)=0$,

$x\geq\Lambda_{A}+\sum\lambda_{Q}\log|Q_{i}(x)|+\lambda_{0}U_{\mu_{A}}(x)$

for every non-negative $x$ with equality for every $x$ in the support of
$\mu_{A}$, and

$\int\log|Q_{i}(x)|d\mu_{A}(x)=0$

if $\lambda_{Q}\neq 0$.

The proof of the above proposition follows from several auxiliary lemmas. We
begin by showing the support of $\mu_{A}$ is inside a fixed compact set
independent of $A$. Our proof is based on the Principle of Domination for
potentials that we cite below from [ST97, Theorem 3.2].

######  Theorem 2.2.

[ST97, Theorem 3.2] Let $\mu$ and $\nu$ be two positive finite Borel measures
with compact support on $\mathbb{C}$ and suppose that the total mass of $\nu$
does not exceed that of $\mu$. Assume further that $\mu$ has finite
logarithmic energy. If the inequality

$U_{\mu}(z)\geq U_{\nu}(z)+c$

holds $\mu$-almost everywhere for some constant $c$, then it holds for all
$z\in\mathbb{C}$.

First, we cite the following lemma from [Ahl10, Lemma 2.1].

######  Lemma 2.3.

[Ahl10, Lemma 2.1] Suppose that $\mu_{1},\dots,\mu_{n}$ are probability
measures with finite logarithmic energy. Let $\sum_{i=1}^{n}c_{i}=1$ and
$c_{i}>0$. We have

$I\left(\sum_{i}c_{i}\mu_{i}\right)\geq\sum_{i}c_{i}I(\mu_{i})$

with equality if and only if all $\mu_{i}$ are identical.

Define

$I(\eta,\mu):=\int U_{\eta}(x)d\mu(x).$

######  Lemma 2.4.

Suppose that $\mu$ is a probability measure with finite energy. If $I(\mu)\geq
0$, then

$I(\eta,\mu)-\frac{I(\eta)}{2}\geq 0$

for every probability measure $\eta$ with finite energy. On the other hand, if
for every $\eta$ with $I(\eta)=0$,

$I(\eta,\mu)\geq 0,$

then $I(\mu)\geq 0.$

###### Proof.

Suppose that $I(\mu)\geq 0$. By Lemma 2.3, we have

$I\left(\frac{\mu+\eta}{2}\right)\geq\frac{I(\mu)+I(\eta)}{2}.$

The above implies

$I(\eta,\mu)-\frac{I(\eta)}{2}\geq\frac{I(\mu)}{2}\geq 0$

for every probability measure $\eta$ with finite energy. On the other hand,
suppose that

$I(\eta,\mu)-\frac{I(\eta)}{2}\geq 0$

for every $\eta$ with $I(\eta)=0$. Suppose to the contrary that $I(\mu)<0$.
Let $\sigma$ be the equilibrium measure of the interval $[0,5]$. We note that
$I(\sigma)=\log(5/4)>0.$ So there exists a parameter $0<\alpha<1$ such that

$I(\alpha\mu+(1-\alpha)\sigma)=\alpha^{2}I(\mu)+2\alpha(1-\alpha)I(\mu,\sigma)+(1-\alpha)^{2}I(\sigma)=0.$

Take $\eta:=\alpha\mu+(1-\alpha)\sigma$. Note that $I(\eta)=0$ and

(10)
$-\frac{\alpha}{2}I(\mu)-\frac{(1-\alpha)^{2}}{2\alpha}I(\sigma)=(1-\alpha)I(\mu,\sigma).$

By our assumption, we have

$I(\eta,\mu)=\alpha I(\mu)+(1-\alpha)I(\mu,\sigma)\geq 0.$

By substituting (10), we obtain

$I(\eta,\mu)=\alpha
I(\mu)+(1-\alpha)I(\mu,\sigma)=\frac{\alpha}{2}I(\mu)-\frac{(1-\alpha)^{2}}{2\alpha}I(\sigma)\geq
0$

This is a contradiction as $I(\mu)<0$ and $I(\sigma)>0$. So we must have
$I(\mu)\geq 0$, concluding our lemma. ∎

######  Lemma 2.5.

Suppose that $\mu$ and $\nu$ are probability measures with finite energy, and
for some $\alpha>0$,

$U_{\mu}(z)\geq\alpha U_{\nu}(z)$

for every $z\in\mathbb{C}$. Then

$I(\mu)\geq\alpha^{2}I(\nu).$

###### Proof.

We have

$\displaystyle I(\mu)=\int U_{\mu}(z)d\mu(z)$ $\displaystyle\geq\alpha\int
U_{\nu}(z)d\mu(z)$ $\displaystyle=\alpha\int\int\log|x-z|d\nu(x)d\mu(z)$
$\displaystyle=\alpha\int U_{\mu}(x)d\nu(x)$ $\displaystyle\geq\alpha^{2}\int
U_{\nu}(x)d\nu(x)=\alpha^{2}I(\nu).$

This completes the proof of our lemma. ∎

######  Lemma 2.6.

Let $A=\\{Q_{1},\dots,Q_{n}\\}$ be any finite subset of integer polynomials.
Suppose that $\mu$ is a probability measure on $\mathbb{R}^{+}$ such that
$I(\mu)\geq 0$ and $\int\log|Q_{i}(x)|d\mu(x)\geq 0$ for every $Q_{i}\in A$.
There exists a probability measure $\mu^{\prime}$ supported on $[0,18]$ such
that $I(\mu^{\prime})\geq 0$, $\int\log|Q_{i}(x)|d\mu(x)\geq 0,$ and

$\int xd\mu^{\prime}(x)\leq\int xd\mu(x),$

where the above is equality only if $\mu$ is supported on $[0,18]$.

###### Proof.

Our method is to truncate the support of $\mu$ with a parameter to reduced its
expected value and at the same time to take a convex combination with a fixed
equilibrium measure to satisfy the positivity conditions on the energy and log
moments. We now explain this in detail. If $\int xd\mu(x)\geq 2$, let
$\mu^{\prime}=\mu_{[0,4]}$ be the equilibrium measure on the interval $[0,4]$.
This is sufficient for the lemma since the potential is non-negative on the
complex plane. Otherwise, $E:=\int xd\mu(x)<2$. For $M>0$, let

$d\mu_{M}(x):=\begin{cases}\frac{d\mu(x)}{\mu([0,M])}&\text{ if $x<M$,}\\\
0&\text{ otherwise,}\end{cases}$

and

$E_{M}:=\int xd\mu_{M}(x).$

$\mu_{M}$ is the conditional probability of $\mu$ with condition $x<M$. It
follows that $E_{M}\leq E$ with equality if $\mu_{M}=\mu$. In fact, by writing
$E$ as a convex combination of conditional expectations on $x\leq M$ and
$x>M$, we have

$E=\mu([0,M])E_{M}+\mu([M,\infty])E_{M,\infty},$

where $E_{M,\infty}:=\frac{1}{\mu([M,\infty])}\int_{M}^{\infty}xd\mu(x)$. It
is clear that $E_{M}\leq M$ and $E_{M,\infty}>M$ unless $\mu([M,\infty])=0$.
For $N>4$ and $M>N$, define

$\mu_{M,N,\alpha}:=\alpha\mu_{M}+(1-\alpha)\mu_{[0,N]},$

where $\mu_{[0,N]}$ is the equilibrium measure on the interval $[0,N]$. We
have

$\displaystyle\int xd\mu_{M,N,\alpha}(x)=\alpha E_{M}+(1-\alpha)N/2.$

Next, we estimate the potential of $\mu_{M,N,\alpha}$. Our goal is to find
parameters $M,N,\alpha$ such that

(11) $\alpha E_{M}+(1-\alpha)N/2\leq E,$

and

(12) $U_{\mu_{M,N,\alpha}}(x)\geq\delta U_{\mu}(x)$

for every $x\in\mathbb{R}^{+}$ and some $1\geq\delta>0$. Inequality (11) is
equivalent to

$\alpha\geq\frac{N/2-E}{N/2-E_{M}}.$

We take $\alpha:=\frac{N/2-E}{N/2-E_{M}}$. So inequality (11) is equality with
this choice. By Theorem 2.2, it is enough to check inequality (12) for every
$0\leq x<M$ as $0\leq\delta\leq 1$. Suppose that $0\leq x<M$, we have

(13) $\displaystyle U_{\mu}(x)$
$\displaystyle=\int_{0}^{M}\log|x-z|d\mu(z)+\int_{M}^{\infty}\log|x-z|d\mu(z)$
$\displaystyle=\mu([0,M])U_{\mu_{M}}(x)+\int_{M}^{\infty}\log|x-z|d\mu(z)$
$\displaystyle\leq\mu([0,M])U_{\mu_{M}}(x)+\int_{M}^{\infty}\log|z|d\mu(z).$

Let $U_{\mu_{[0,N]}}(x)$ be the potential of $\mu_{[0,N]}$. We have

(14) $U_{\mu_{M,N,\alpha}}(x)=\alpha
U_{\mu_{M}}(x)+(1-\alpha)U_{\mu_{[0,N]}}(x)\geq\alpha
U_{\mu_{M}}(x)+(1-\alpha)\log(N/4).$

By (13) and (14), inequality (12) follows if

(15) $\alpha
U_{\mu_{M}}(x)+(1-\alpha)\log(N/4)\geq\delta\left(\mu([0,M])U_{\mu_{M}}(x)+\int_{M}^{\infty}\log|z|d\mu(z)\right).$

We take $\delta:=\frac{\alpha}{\mu([0,M])}$. We show that $0\leq\delta\leq 1$
which is equivalent to

$\mu([0,M])\geq\alpha=\frac{N/2-E}{N/2-E_{M}}.$

This is equivalent to

$N/2\leq\frac{1}{\mu([M,\infty])}\int_{M}^{\infty}xd\mu(x),$

which follows from $N/2\leq M$. By this choice of $\delta$, inequality (15),
is equivalent to

$(1-\alpha)\log(N/4)\geq\delta\int_{M}^{\infty}\log|z|d\mu(z),$

Substituting the value of $\alpha$ in the above, it remains to check

$\frac{E-E_{M}}{N/2-E_{M}}\log(N/4)\geq\delta\int_{M}^{\infty}\log|z|d\mu(z).$

By substituting the values of $E$ and $E_{M}$, This is equivalent to

$\int_{0}^{\infty}xd\mu(x)-\frac{1}{\mu([0,M])}\int_{0}^{M}xd\mu(x)\geq\int_{M}^{\infty}\delta\frac{N/2-E_{M}}{\log(N/4)}\log|z|d\mu(z),$

or

$\frac{1}{\mu([M,\infty])}\int_{M}^{\infty}z-\delta\frac{N/2-E_{M}}{\log(N/4)}\log|z|d\mu(z)\geq\frac{1}{\mu([0,M])}\int_{0}^{M}xd\mu(x)=E_{M}.$

Note that $E_{M}\leq E<2$. The above inequality follows if

$z-\delta\frac{N/2-E_{M}}{\log(N/4)}\log|z|\geq 2$

for every $z\geq M$. Let $N=4e$. Then, the above follows if

$z\geq 2e\log(z)+2.$

The above holds for every $z\geq 18$. This concludes the proof of our lemma. ∎

######  Lemma 2.7.

Recall the definition of $\Lambda_{A}$ from Proposition 2.1. There exists a
probability measure $\mu_{A}$ supported in $[0,18]$ satisfying all the above
constraints and $\int xd\mu_{A}(x)=\Lambda_{A}.$

###### Proof.

By definition of $\Lambda_{A}$ and Lemma 2.6, there exists a sequence of
probability measures $\\{\mu_{n}\\}$ that are supported on the interval
$[0,18]$ satisfying the constraints

$I(\mu_{n})\geq 0,$ $\int\log|Q(x)|d\mu_{n}(x)\geq 0$

for every $Q\in A$, and

$\lim_{n\to\infty}\int xd\mu_{n}(x)=\Lambda_{A}.$

By compactness of the space of probability measures supported on $[0,18]$,
there exists a sub-sequence $\mu_{a_{n}}$ that converges weakly to some
probability distribution $\mu_{0}$. Note that $\log|Q(z_{1},\dots,z_{m})|$ is
an upper semi-continuous function. Hence, by monotone convergence theorem, we
have

$\int\log|Q(z_{1},\dots,z_{m})|d\mu_{0}(z_{1})\dots
d\mu_{0}(z_{m})\geq\limsup_{n\to\infty}\int\log|Q(z_{1},\dots,z_{m})|d\mu_{a_{n}}(z_{1})\dots
d\mu_{a_{n}}(z_{m})\geq 0.$

This completes the proof of our lemma. ∎

###  2.1. Strong duality

In this section, we give a proof of Proposition 2.1. Recall the primal problem
and its dual from section 1.2 of the Introduction. Let $\mathcal{P}$ be the
space of Borel probability measures supported on $\mathbb{R}^{+}$ with finite
energy. Let $\mathcal{P}^{+}\subset\mathcal{P}$ be the subspace of the
probability measures with logarithmic energy greater or equal to zero on
$\mathbb{R}^{+}$. Let $A=\\{Q_{1},\dots,Q_{n}\\}$ be a finite subset of
integral polynomials. By Lemma 2.7, there exists a feasible solution
$\mu_{A}\in\mathcal{P}^{+}$ with expectation $\Lambda_{A}$ to the primal
problem. Next, we introduce the associated dual problem.

Define $\lambda_{A}$ to be the supremum over all $\lambda$ such that for every
positive real $x$ and some fixed elements $\mu_{j}\in\mathcal{P}^{+}$ and
constants $c_{i},d_{j}\geq 0$ then

$x\geq\lambda+\sum_{Q_{i}\in
A}c_{i}\log|Q_{i}(x)|+\sum_{\mu_{j}}d_{j}(2U_{\mu_{j}}(x)-I(\mu_{j})).$

By integrating the above inequality with respect to $\mu_{A}$ which satisfies
the constraints of the primal problem, it follows that

$\Lambda_{A}\geq\lambda_{A}.$

The above inequality is known as the weak duality, and if equality holds it is
known as the strong duality. Proposition 2.1 implies that
$\Lambda_{A}=\lambda_{A}$ and there exists a feasible optimal solution for the
dual problem. Finally, we give a proof of Proposition 2.1 which is similar to
the proof of strong duality from [BV04, section 5.3.2]

###### Proof of Proposition 2.1.

Let $\mathcal{P}_{1}^{+}\subset\mathcal{P}^{+}$ be the countable subset of
smooth probability measures with density function $p(x)dx$ on finite union of
intervals $\Sigma\subset\mathbb{R}^{+}$ and energy greater or equal to zero,
where $p(x)\in\mathbb{Q}[x]$ up to a scalar and the end points of $\Sigma$ are
rational numbers. Note that $\mathcal{P}_{1}^{+}$ is dense inside
$\mathcal{P}^{+}$ with respect to the weak topology. We note that
$\Lambda_{A}\leq 2$ since the equilibrium measure of the interval $[0,4]$
satisfies all the conditions of the primal problem and has expected value $2$.
We write

$\mathcal{P}_{1}^{+}:=\\{\mu_{i}:i\in\mathbb{N}\\}.$

For $N\geq 1$, we define $\psi_{N}:\mathcal{P}\to\mathbb{R}^{1+n+N}$ as
follows

$\psi_{N}(\mu):=\left(E_{\mu}(x),E_{\mu}(\log|Q_{1}|),\dots,E_{\mu}(\log|Q_{n}|),E_{\mu}(U_{\mu_{1}}(x)-I(\mu_{1})/2),\dots,E_{\mu}(U_{\mu_{N}}(x)-I(\mu_{N})/2)\right)$

where $E_{\mu}(f):=\int f(x)d\mu(x)$ is the expected value of $f$ with respect
to $\mu$. Let

$\mathcal{A}_{N}:=\psi_{N}(\mathcal{P})\subset\mathbb{R}^{1+n+N}$

be the image of $\psi_{N}$. For $\varepsilon>0$, let

$\mathcal{B}_{\varepsilon,N}:=\\{(x_{0},x_{1},\dots,x_{n+N}):x_{0}\leq\Lambda_{A}-\varepsilon,\text{
and }x_{i}\geq 0\text{ for }1\leq i\leq n+N\\}.$

Since $\mathcal{P}_{1}^{+}$ is dense inside $\mathcal{P}^{+}$, it follows from
the definition of $\Lambda_{A}$ that for every $\varepsilon>0$, there exists a
large enough $M$ such that

$\mathcal{A}_{N}\cap\mathcal{B}_{\varepsilon,N}=\emptyset$

for every $N\geq M.$ Otherwise, there exists a sequence of probability
measures $\mu_{N}\in\mathcal{P}$ for any $N\geq 0$ such that

$E_{\mu_{N}}(x)\leq\Lambda_{A}-\varepsilon<2,$ $E_{\mu_{N}}(\log|Q_{i}|)\geq
0,$

for every $1\leq i\leq n$ and

$E_{\mu_{N}}(U_{\mu_{j}}(x))\geq I(\mu_{j})/2\geq 0$

for every $j\leq N$. Since $E_{\mu_{N}}(x)<2$, $\\{\mu_{N}\\}$ is a tight
family of probability measures on $\mathbb{R}^{+}$. By Prokhorov’s theorem,
there exists a convergent subsequence $\\{N_{a}\\}$. Suppose that

$\mu_{0}=\lim_{a\to\infty}\mu_{N_{a}}.$

It follows that

$E_{\mu_{0}}(x)\leq\Lambda_{A}-\varepsilon,$ $E_{\mu_{0}}(\log|Q_{i}|)\geq 0,$

for every $1\leq i\leq n$, and

$E_{\mu_{0}}(U_{\mu_{j}}(x))\geq I(\mu_{j})/2\geq 0$

for every $\mu_{j}\in\mathcal{P}_{1}^{+}$. Next, we show that $I(\mu_{0})\geq
0$, which gives a contradiction as $\mu_{0}$ is a feasible solution with
expected value less than $\Lambda_{A}$. By Lemma 2.4, it is enough to show
that for every probability measure $\eta$ supported on $\mathbb{R}^{+}$ with
$I(\eta)=0$, we have $I(\mu_{0},\eta)\geq 0$. By density of
$\mathcal{P}_{1}^{+}$ inside $\mathcal{P}^{+}$, there exists a sequence
$\mu_{a_{n}}\in\mathcal{P}_{1}^{+}$ such that

$\lim_{n\to\infty}\mu_{a_{n}}=\eta.$

This implies that

$U_{\eta}(x)\geq\limsup_{n\to\infty}U_{\mu_{a_{n}}}(x)$

for every $x\in\mathbb{C}$. This implies that

$I(\eta,\mu_{0})=E_{\mu_{0}}(U_{\eta}(x))\geq\limsup_{n\to\infty}E_{\mu_{0}}(U_{\mu_{a_{n}}}(x))\geq
0.$

Hence, there exists a large enough $M$ such that

$\mathcal{A}_{N}\cap\mathcal{B}_{\varepsilon,N}=\emptyset$

for every $N\geq M.$ Suppose that $N\geq M$. Note that
$\mathcal{A}_{N},\mathcal{B}_{\varepsilon,N}\subset\mathbb{R}^{1+n+N}$ are
both convex regions with

$\mathcal{A}_{N}\cap\mathcal{B}_{\varepsilon,N}=\emptyset.$

By the separating hyperplane theorem, it follows that there exists some
coefficients $(c_{0},\dots,c_{n},d_{1},\dots,d_{N})$ such that

(16) $\sum_{i=0}^{n}c_{i}x_{i}+\sum_{j=1}^{N}d_{j}x_{j+n}\geq
c_{0}E_{\mu}(x)+\sum_{i=1}^{n}c_{i}E_{\mu}(\log|Q_{i}|)+\sum_{j=1}^{N}d_{j}E_{\mu}(U_{\mu_{j}}(x)-I(\mu_{j})/2)$

for every $(x_{0},x_{1},\dots,x_{n+N})\in\mathcal{B}_{\varepsilon,N}$ and
$\mu\in\mathcal{P}.$ Since $x_{i}\geq 0$ could be arbitrarily large for $i\geq
1$, so we must have $c_{i},d_{j}\geq 0$ for every $i,j\geq 1$. Similarly,
$x_{0}\leq\Lambda_{A}-\varepsilon$ can be arbitrarily small. So we must have
$c_{0}\leq 0$. Next, we show that $c_{0}\neq 0.$ Let $\mu_{eq}$ be the the
equilibrium measure of $[0,18]$. We note that

$E_{\mu_{eq}}(x)=9,$ $E_{\mu_{eq}}(\log|Q_{i}|)\geq\deg(Q_{i})\log(9/2)$

and by inequality(6), we have

$E_{\mu_{eq}}(U_{\mu_{j}}(x)-I(\mu_{j})/2)\geq I(\mu_{eq})/2=\log(9/2)/2.$

By substituting
$(x_{0},x_{1},\dots,x_{n+N})=(\Lambda_{A}-\varepsilon,0,\dots,0)$ and
$\mu=\mu_{eq}$ in (16) and the above identities, we have

$c_{0}(\Lambda_{A}-\varepsilon-9)\geq\sum_{i=1}^{n}c_{i}\deg(Q_{i})\log(9/2)+\sum_{j=1}^{N}d_{j}\log(9/2)/2.$

Since $\Lambda_{A}\leq 2$, the above inequality implies that $c_{0}\neq 0$ and
$c_{0}<0$. We divide both side of (16), by $c_{0}$ and obtain

$\Lambda_{A}-\varepsilon\leq
E_{\mu}(x)+\sum_{i=1}^{n}c_{i}^{\prime}E_{\mu}(\log|Q_{i}|)+\sum_{j=1}^{N}d_{j}^{\prime}E_{\mu}(U_{\mu_{j}}(x)-I(\mu_{j})/2)$

for every $\mu\in\mathcal{P}$, where $c_{i}^{\prime}:=\frac{c_{i}}{c_{0}}\leq
0$ and $d_{j}^{\prime}:=\frac{d_{j}}{c_{0}}\leq 0$. Fix $x\in\mathbb{R}$ and
let $\delta_{x}$ be the Dirac mass at $x$. Since there exists a sequence
$\\{\mu_{k}\\}\subset\mathcal{P}$ such that
$\lim_{k\to\infty}\mu_{k}=\delta_{x}$, it follows that

(17) $\Lambda_{A}-\varepsilon\leq
x+\sum_{i=1}^{n}c_{i}^{\prime}\log|Q_{i}|(x)+\sum_{j=1}^{N}d_{j}^{\prime}\left(U_{\mu_{j}}(x)-I(\mu_{j})/2\right)$

for every $x\in\mathbb{R}^{+}$, and some coefficients
$c_{i}^{\prime},d_{i}^{\prime}\leq 0$. Let

$\mu_{0}:=\frac{1}{d^{\prime}}\sum_{j}d_{j}^{\prime}\mu_{j}$

where $d^{\prime}=\sum_{j}d_{j}^{\prime}$. By linearity of the logarithmic
potential and Lemma 2.3, we have

$\sum_{j=1}^{N}d_{j}^{\prime}\left(U_{\mu_{j}}(x)-I(\mu_{j})/2\right)\leq
d^{\prime}\left(U_{\mu_{0}}(x)-I(\mu_{0})/2\right).$

By substituting the above in (17), we have

(18) $\Lambda_{A}-\varepsilon\leq
x+\sum_{i=1}^{n}c_{i}^{\prime}\log|Q_{i}|(x)+d^{\prime}\left(U_{\mu_{0}}(x)-I(\mu_{0})/2\right)$

for every $x\in\mathbb{R}^{+}$ and some $c_{i}^{\prime}\leq 0$ and
$d^{\prime}\leq 0$. By integrating the above inequality with respect to
$\mu_{eq}$, we obtain

$\Lambda_{A}-10\leq\Lambda_{A}-\varepsilon-9\leq\sum_{i=1}^{n}c_{i}^{\prime}\deg(Q_{i})\log(9/2)+d^{\prime}\log(9/2)/2.$

Define the compact subset $\Omega\subset\mathbb{R}^{n+1}$ as follows:

$\Omega:=\\{(x_{0},\dots,x_{n}):\Lambda_{A}-10\leq
x_{0}\log(9/2)/2+\sum_{i=1}^{n}x_{i}\log(9/2)\deg(Q_{i}),x_{i}\leq 0\text{ for
every }i\\}$

Note that coefficients $c_{1}^{\prime},\dots,c_{n}^{\prime}$ and $d^{\prime}$
depend on $\varepsilon>0$ and $\Omega$ is independent of $\varepsilon.$ By
definition of $\Omega$, we have
$(d^{\prime},c_{1}^{\prime},\dots,c_{n}^{\prime})\in\Omega$. Now we can take
$\varepsilon_{m}=\frac{1}{m}$ and obtain a sequence of points
$p_{m}:=(d_{m}^{\prime},c_{1,m}^{\prime},\dots,c_{n,m}^{\prime})\in\Omega.$
Let $(D,C_{1},\dots,C_{n})$ be a limiting point of a sub-sequence. By (17), we
obtain

(19) $\Lambda_{A}-\frac{1}{m}\leq
x+\sum_{i=1}^{n}C_{i}\log|Q_{i}|(x)+D\left(U_{\mu_{m}}(x)-I(\mu_{m})/2\right)$

for some sequence of probability measures $\mu_{m}$. By Lemma 2.7, there
exists a feasible solution $\mu\in\mathcal{P}$ with optimal trace
$\Lambda_{A}$ to the primal problem. We integrate the above inequality with
respect to $\mu$ and obtain

(20)
$\Lambda_{A}-\frac{1}{m}\leq\Lambda_{A}+\sum_{i=1}^{n}C_{i}E_{\mu}(\log|Q_{i}|(x))+DE_{\mu}\left(U_{\mu_{m}}(x)-I(\mu_{m})/2\right)$

for every $m\geq 0$. Note that $E_{\mu}(\log|Q_{i}|(x))\geq 0$, $C_{i}\leq
0,D\leq 0$, and by [Ahl10, Lemma 2.1],

$E_{\mu}\left(U_{\mu_{m}}(x)-I(\mu_{m})/2\right)=-I(\mu-\mu_{m})/2+I(\mu)/2,$

where

$-I(\mu-\mu_{m})=\lim_{R\to\infty}\int\int_{|z|<R}\left[\int\frac{d\mu(x)-d\mu_{m}(x)}{|z-x|}\right]^{2}d\textup{vol}(z)\geq
0.$

By taking $m\to\infty$ in inequality (20), we have

(21)
$0\leq\sum_{i=1}^{n}C_{i}E_{\mu}(\log|Q_{i}|(x))+D\left(I(\mu)/2+\limsup_{m\to\infty}-I(\mu-\mu_{m})/2\right).$

We note that the right-hand side is the sum of non-positive terms. It follows
that

$I(\mu)=0,$ $E_{\mu}(\log|Q_{i}|(x))=0$

if $C_{i}\neq 0$, and

$\limsup_{m\to\infty}-I(\mu-\mu_{m})=0$

which implies $\lim_{m\to\infty}\mu_{m}=\mu$. This implies that

$0=I(\mu)\geq\limsup_{m\to\infty}I(\mu_{m}).$

On the other hand, by our assumptions $I(\mu_{m})\geq 0.$ So the following
limit holds

$\lim_{m\to\infty}I(\mu_{m})=0.$

This implies that

$\lim_{m\to\infty}U_{\mu_{m}}(x)=U_{\mu}(x)$

for every $x$ in the support of $\mu$. From the above, (19), and (21), it
follows that

$\Lambda_{A}\leq x+\sum_{i=1}^{n}C_{i}\log|Q_{i}|(x)+D\left(U_{\mu}(x)\right)$

with equality for every $x$ inside the support of $\mu$. ∎

Next, we show that the support of $\mu_{A}$ is a finite union of intervals.
This proves part of Theorem 1.1.

######  Proposition 2.8.

Let $\mu_{A}$ be the optimal solution to the primal problem in Proposition 2.1
with support $\Sigma\subset[0,18]$ and $A$ being a finite subset of integer
polynomials with only real roots. Then
$\Sigma=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$ and every gap interval
$(a_{2i+1},a_{2i+2})$ contains a root of some $Q\in A$.

###### Proof.

By Proposition 2.1 and Lemma 2.7, there exists a unique probability measure
$\mu$ satisfying all the conditions of Theorem 1.1, but to check the support
of $\mu$ is a finite union of intervals separated by the roots of $Q\in A$.
Let $\Sigma$ be the support of $\mu$. Suppose that $a,b\in\Sigma$ and $Q(x)$
does not have any root inside the interval $[a,b]$ for every $Q\in A$. We
claim that $[a,b]\subset\Sigma.$ Now suppose the contrary that there is an
open interval gap inside $\Sigma\cap[a,b]$ say $(c,d)$ where $c$ and $d$ are
maximal, i.e. $c,d\in\Sigma$ and $(c,d)\cap\Sigma=\emptyset$. Our argument is
based on the following perturbation. Let $\delta>0$ be a small enough
parameter that we choose later. Define $\mu_{1}:=\mu|(c-\delta,c)$ and
$\mu_{2}:=\mu|(d,d+\delta)$ to be the restriction of $\mu$ to two small
intervals around $c$ and $d$. Let

$0<m<\min(\mu_{1}((c-\delta,c)),\mu_{2}((d,d+\delta)))$

be smaller than the minimum of the two masses which by our assumption is
strictly positive that we choose later. Let $\mu_{3}$ be the equilibrium
probability measure of the interval $[c-\delta,d]$. Let $\mu^{\prime}$ be the
probability measure obtained by taking $m/2$ mass of $\mu_{1}$ and $\mu_{2}$
and replacing it with the $m\mu_{3}$, namely

$\mu^{\prime}:=\mu-\lambda_{1}-\lambda_{2}+\lambda_{3}$

where $\lambda_{1}:=\frac{m}{2\mu_{1}((c-\delta,c))}\mu_{1}$,
$\lambda_{2}:=\frac{m}{2\mu_{2}((d,d+\delta))}\mu_{2}$ and
$\lambda_{3}:=m\mu_{3}$. We have

$E_{\mu^{\prime}}(x)-E_{\mu}(x)=-E_{\lambda_{1}}(x)-E_{\lambda_{2}}(x)+E_{\lambda_{3}}(x)<m\left(\frac{-c+\delta}{2}-d/2+\frac{d+c-\delta}{2}\right)=0$

Suppose that $y<c<d$. Since $\lambda_{1}$ is supported on $[c-\delta,c]$, we
have

$U_{\lambda_{1}}(y)\leq m\max(\log(c-y),\log(\delta))/2,$

and similarly,

$U_{\lambda_{2}}(y)\leq m\log(d-y+\delta)/2=m(\log(d-y)/2+O(\delta)),$

where the implicit coefficient in $O(\delta)$ depends only on $d-c$. Hence, we
have

(22)
$\begin{split}U_{\mu^{\prime}}(y)-U_{\mu}(y)&=-U_{\lambda_{1}}(y)-U_{\lambda_{2}}(y)+U_{\lambda_{3}}(y)\\\
&\geq
m\left(\log\left(\frac{c-y+d-y+2\sqrt{(c-y)(d-y)}}{4}\right)-\log(c-y)/2-\log(d-y)/2+O(\delta)\right)\\\
&=m\left(\log\left(\frac{1}{2}+\frac{c-y+d-y}{4\sqrt{(c-y)(d-y)}}\right)+O(\delta)\right)>0,\end{split}$

where we used the explicit value of the potential of the equilibrium measure
$\mu_{3}$ and the fact that $\delta$ is small comparing to $d-c$. Similarly,
we have

$U_{\mu^{\prime}}(y)-U_{\mu}(y)>0$

for $c<d<y.$ Since every $Q\in A$ does not vanish inside $[a,b]$, this implies
that

$\sum_{Q(y)=0}U_{\mu^{\prime}}(y)>\sum_{Q(y)=0}U_{\mu}(y)\geq 0$

for small enough $\delta$. Finally, we note that

$\begin{split}I(\mu^{\prime})-I(\mu)=2I(\mu,-\lambda_{1}-\lambda_{2}+\lambda_{3})+I(-\lambda_{1}-\lambda_{2}+\lambda_{3},-\lambda_{1}-\lambda_{2}+\lambda_{3})\end{split}$

By (22) and the fact that $\Sigma\cap(c,d)=\emptyset$, we have

$I(\mu,-\lambda_{1}-\lambda_{2}+\lambda_{3})=\int-
U_{\lambda_{1}}(x)-U_{\lambda_{2}}(x)+U_{\lambda_{3}}(x)d\mu(x)=km>0$

for some constant $k>0$. We also note that

$I(-\lambda_{1}-\lambda_{2}+\lambda_{3},-\lambda_{1}-\lambda_{2}+\lambda_{3})=O(m^{2}).$

Therefore, by choosing $m$ small enough

$I(\mu^{\prime})>I(\mu).$

We constructed $\mu^{\prime}$ with expected value less than the optimal value
$\Lambda_{A}$ and $\mu^{\prime}$ satisfies all the constraints of the primal
problem. This contradicts with the fact that the optimal expectation is
$\Lambda_{A}$. This completes the proof of our main theorem. ∎

##  3\. Explicit computation of logarithmic potential

Fix a compact set $\Sigma:=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$ which is a
finite union of closed intervals. In this section, we construct measures with
explicit potentials that we need to express the optimal solution to Theorem
1.1.

###  3.1. Measures with locally constant potential

It is well-known that the equilibrium measure of $\Sigma$ has a constant
logarithmic potential on $\Sigma$. Its density function is

$f(x)dx=\frac{|p(x)|}{\pi\sqrt{|H(x)|}}dx$

where $H(x)=\prod_{i=0}^{2l+1}(x-a_{i})$ and
$p(x)=\prod_{i=0}^{l-1}(x-\sigma_{i})$ where
$\sigma_{i}\in[a_{2i+1},a_{2i+2}]$ are the critical points of the Green’s
function of $\Sigma$ uniquely determined by the following equations:

(23) $\int_{a_{2i+1}}^{a_{2i+2}}\frac{p(x)}{\sqrt{|H(x)|}}dx=0$

for every $0\leq i\leq l-1$ [ST97]. Generalizing this, we state the following
lemma which expresses the density function of measures with locally constant
potentials.

######  Lemma 3.1.

Suppose that $\mu$ is a measure with density function

$f(x)dx=\frac{|p(x)|}{\pi\sqrt{|H(x)|}}dx$

where $H(x)=\prod_{i=0}^{2l+1}(x-a_{i})$ and
$p(x)=\prod_{i=0}^{l-1}(x-\sigma_{i})$ where
$\sigma_{i}\in[a_{2i+1},a_{2i+2}]$. Then the logarithmic potential of $\mu$ is
locally constant and the difference of the value between consecutive intervals
is

$U_{\mu}(a_{2i+2})-U_{\mu}(a_{2i+1})=\int_{a_{2i+1}}^{a_{2i+2}}\frac{p(x)}{\pi\sqrt{|H(x)|}}dx.$

###### Proof.

It follows from the Schwarz-Christoffel map. ∎

###  3.2. Measures with logarithmic potentials

Suppose that $\alpha\notin\Sigma$ is given. The following proposition gives
the measure with potential $\log(x-\alpha)$ on $\Sigma$. Let
$\psi_{\alpha}(z):=\frac{1}{z-\alpha}$ which sends $\alpha$ to $\infty$ and
let $\mu_{\alpha}$ be the equilibrium measure of
$\Sigma_{\alpha}:=\psi_{\alpha}(\Sigma)$. Let $\nu_{\alpha}$ be the
probability measure supported on $\Sigma$ which is the push-forward of
$\mu_{\alpha}$ with map $\psi_{\alpha}^{-1}(z):=\frac{1}{z}+\alpha$.

######  Lemma 3.2.

Let $\nu_{c}$ be as above. We have

$U_{\nu_{c}}(x)=\log(x-\alpha)-U_{\mu_{\alpha}}(0)+C_{\Sigma_{\alpha}}.$

###### Proof.

We have

$\displaystyle U_{\nu_{\alpha}}(x)=\int\log|x-y|d\nu_{\alpha}(y)$
$\displaystyle=\int\log\left|\alpha+\frac{1}{s}-\left(\alpha+\frac{1}{t}\right)\right|d\nu_{\alpha}\left(\alpha+\frac{1}{t}\right)$
$\displaystyle=\int\log\left|\frac{t-s}{st}\right|d\mu_{\alpha}(t)=\log\left(\frac{1}{s}\right)-U_{\mu_{\alpha}}(0)+C_{\Sigma_{\alpha}}.$

where we write $x=\alpha+\frac{1}{s}$ and $y=\alpha+\frac{1}{t}$ for some
$t,s\in\psi(\Sigma)$. This completes the proof of our lemma. ∎

###  3.3. Measures with linear potentials

######  Lemma 3.3.

Let $f(x)dx$ be a probability density on $\Sigma$ with a locally constant
potential. Let $\eta$ be the signed measure with density $-xf(x)dx$ on
$\Sigma$. Then the potential of $\eta$ is $x$ plus a locally constant function
on $\Sigma$.

###### Proof.

Suppose that $y\in\Sigma$. There exists some $i$ such that
$y\in[a_{i},b_{i}]$. Define

$I_{y}:=[a_{i},y]\subset\Sigma.$

By the assumption, we have

$C_{i}=\int_{\Sigma}\log|x-t|f(x)dx$

for every $t\in I_{y}$ and some constant $C_{i}$. We integrate the above
identity over $I_{y}$ and obtain

(24) $C_{i}(y-a_{i})=\int_{\Sigma}\int_{a_{i}}^{y}\log|x-t|dtf(x)dx.$

We integrate the inner integral and obtain

$\int_{a_{i}}^{y}\log|x-t|dt=(y-x)\log|y-x|-(y-x)+(x-a_{i})\log|x-a_{i}|-(x-a_{i}).$

We substitute the above in (24) and obtain

$\displaystyle C_{i}(y-a_{i})$
$\displaystyle=\int_{\Sigma}\left[(y-x)\log|y-x|-(y-x)+(x-a_{i})\log|x-a_{i}|-(x-a_{i})\right]f(x)dx.$

By simplifying the above identity, we deduce that

$y=\int_{\Sigma}-x\log|x-y|f(x)dx+c_{i}$

where

$c_{i}:=a_{i}+\int_{\Sigma}x\log|x-a_{i}|f(x)dx$

is a constant that only depends on $i$. ∎

In order to make constant $c_{i}$ independent of $i$, we use Lemma 3.1 to find
a measure $\mu$ such that

$U_{\mu}(a_{2i+2})-U_{\mu}(a_{2i+1})=\beta(c_{i+1}-c_{i})$

for some small enough $\beta>0$. By adding this measure to $\mu$, we construct
a measure with linear potential.

##  4\. Proof of Theorem 1.1

Finally, we give a proof of Theorem 1.1, assuming the results of this section.

###### Proof of Theorem 1.1.

By Proposition 2.1, there exists a unique measure $\mu_{A}$ which is supported
on $\Sigma\subset[0,18]$ such that

$\begin{array}[]{cc}\int xd\mu_{A}(x)=\lambda_{A},\\\ I(\mu_{A})=0,\\\
x\geq\lambda_{A}+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}U_{\mu_{A}}(x),\text{ and}\\\
\int\log|Q(x)|d\mu_{A}=0\text{ if $\lambda_{Q}\neq 0$,}\par\end{array}$

where in the third line the inequality holds for non-negative $x$ and equality
holds for every $x\in\Sigma$ with some scalars $\lambda_{Q}\geq 0$ and
$\lambda_{0}>0$. By Proposition 2.8,
$\Sigma=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$ is a finite union of intervals.
Furthermore, every gap interval $(a_{2i+1},a_{2i+2})$ contains a root of some
$Q\in A$. Since $\lambda_{0}\neq 0$, we have

(25) $U_{\mu_{A}}(x)=\frac{1}{\lambda_{0}}\left(x-\lambda_{A}-\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|\right)$

for every $x\in\Sigma$. Let

$H(x)=\prod_{i=0}^{l}(x-a_{2i})(x-a_{2i+1}).$

By Lemma 3.3 and Lemma 3.1, there exists a unique measure $\mu_{lin}$ with
density function

$\frac{f_{lin}(x)}{\sqrt{|H(x)|}}$

for some polynomial $f_{lin}(x)\in\mathbb{R}[x]$ with $\deg(f)=l+1$ such that
$U_{\mu_{lin}}(x)=x$ for every $x\in\Sigma$. Furthermore, by Lemma 3.2 and
Lemma 3.1, for every $Q\in A$, there exists probability measure $\mu_{Q}$ with
density function

$\frac{f_{Q}(x)}{Q(x)\sqrt{|H(x)|}}$

for some polynomial $f_{Q}(x)\in\mathbb{R}[x]$ with $\deg(f)=l+\deg(Q)$ such
that $U_{\mu_{Q}}(x)=\frac{\log|Q(x)|}{\deg(Q)}+C_{Q}$ for every $x\in\Sigma$
and some constant $C_{Q}$. By (25), we have

$\mu_{A}=\frac{1}{\lambda_{0}}\mu_{lin}-\sum_{Q\in
A}\frac{\lambda_{Q}\deg(Q)}{\lambda_{0}}\mu_{Q}+c\mu_{eq},$

where $c$ is a constant and $\mu_{eq}$ is the equilibrium measure of $\Sigma.$
This implies that the density function of $\mu_{A}$ is

$\frac{|f(x)|}{\sqrt{|H(x)|}\prod_{\lambda_{Q}\neq 0}|Q(x)|},$

where $f(x)\in\mathbb{R}[x]$ with $\deg(f)=\sum_{Q\in A}\deg(Q)+l+1$. By Lemma
4.1, $f(x)=H(x)p(x)$ for some polynomials $p(x)$ with $\deg(p)=\sum_{Q\in
A}\deg(Q)-l-1$ then $\mu_{A}$ has the following density function up to a
scalar:

$\frac{|p(x)|\sqrt{|H(x)|}}{\prod_{\lambda_{Q}\neq 0}|Q(x)|}.$

This concludes the proof of Theorem 1.1. ∎

###  4.1. Continuity of the Density Functions

In this section, we show that the density function of the optimal probability
measure to the primal problem is a continuous function over the positive real
line. By the previous section, it is enough to show that the density function
vanishes at the positive end points of the intervals in $\Sigma$.

######  Proposition 4.1.

Let $\mu$ be a solution to a primal problem. The density function of $\mu$ as
a function on $(0,\infty)$ is continuous. In particular, it vanishes on the
positive boundary points of $\Sigma$.

###### Proof.

By the proof of Theorem 1.1 in the previous page, the optimal measure $\mu$
will be supported on a finite union of intervals
$\Sigma:=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$, and its potential function
satisfies the following on $\Sigma$:

(26) $x=\lambda_{A}+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}U_{\mu_{A}}(x),$

where $\lambda_{Q}\geq 0$ and $\lambda_{0}>0$ for all $Q\in A$. Suppose to the
contrary that the density function does not vanish at an end point $a_{i}$ for
some $i$. We show this implies that the following inequality

(27) $x\geq\lambda_{A}+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}U_{\mu_{A}}(x),$

fails to hold inside a neighborhood of $a_{i}$. This is a contradiction with
Proposition 2.1. Suppose that $a_{i}=a_{2k}$ for some $k$ where $a_{i}\neq 0$.
Let $\delta>0$ be a small parameter that we choose later. By inequality (27)
and identity (26), we have

$\frac{U(a_{2k}-\delta)-U(a_{2k})}{\delta}\leq\frac{1}{\lambda_{0}}\left(-1+\sum_{Q\in
A}\lambda_{Q}\frac{\log|Q(a_{2k})|-\log|Q(a_{2k}-\delta)|}{\delta}\right).$

This implies that

(28)
$\frac{U(a_{2k}-\delta)-U(a_{2k})}{\delta}\leq\frac{1}{\lambda_{0}}\left(-1+\sum_{Q\in
A}\lambda_{Q}\frac{Q^{\prime}(a_{2k})}{Q(a_{2k})}\right)+O(\delta),$

where the coefficients of $O$ depends only on $A$ and $\Sigma$. We note that
if the density function of $\mu$ does not vanish at $a_{2k}$, then the density
function goes to infinity near $a_{2k}$ with the following asymptotic
behaviour

$d\mu(x)\gg(x-a_{2k})^{-1/2}dx$

for $a_{2k}<x<a_{2k+1}$. This implies that

$\begin{split}\frac{U(a_{2k}-\delta)-U(a_{2k})}{\delta}&\gg\frac{1}{\delta}\int_{a_{2k}}^{a_{2k}+c}\left(\log(x-a_{2k}+\delta)-\log(x-a_{2k})\right)(x-a_{2k})^{-1/2}dx+O(1)\\\
&=\frac{1}{\delta}\int_{0}^{c}\log\left(1+\frac{\delta}{t}\right)t^{-1/2}dt+O(1)\\\
&=\delta^{-1/2}\int_{0}^{c/\delta}\log\left(1+\frac{1}{u}\right)u^{-1/2}du+O(1)\\\
&\gg\delta^{-1/2}+O(1).\end{split}$

where $t=\delta u$ and $a_{2k+1}-a_{2k}>c>0$. The above is contradiction with
(28). The case where $a_{i}=a_{2k+1}$ for some $k$ is analogous. This
completes the proof of our theorem. ∎

Given the explicit density function proven in Theorem 1.1, we can actually
prove differentiability of the potential of these pareto optimal measures.

######  Lemma 4.2.

Let $\mu$ be a probability measure supported on
$\Sigma:=\bigcup_{i=1}^{n}[a_{2i},a_{2i+1}]$ with density function

$f(x):=\frac{|(x-a_{j})P(x)|}{|Q(x)\sqrt{|H(x)|}|},$

where $a_{j}$ is any boundary point of $\Sigma,$
$H(x)=\prod_{i=1}^{n}(x-a_{2i})(x-a_{2i+1}),$ and $P(x),Q(x)$ are some real
coefficients polynomials with roots outside $\Sigma$. Then $\mu$ has a
differentiable potential function at boundary point $a_{j}$.

###### Proof.

The argument is the same for each end point. For simplicity, we prove it at
$a_{0}$. We know that $U_{\mu}$ is right-differentiable at $a_{0}$ by our
explicit form for $U_{\mu}^{\prime}(x)$ in $\Sigma$. So it suffices to show
that

$\lim_{h\to
0^{+}}\int_{\Sigma}\frac{\log|a_{0}+h-x|+\log|a_{0}-h-x|-2\log|a_{0}-x|}{h}f(x)dx=0.$

The left-hand side of this is equal to

$L:=\lim_{h\to
0^{+}}\int_{\Sigma}\frac{\log\left|1-\frac{h^{2}}{(a_{0}-x)^{2}}\right|}{h}f(x)dx.$

Doing the substitution $z=1-\frac{h^{2}}{(a_{0}-x)^{2}}$ and noticing that
$f(x)\leq C\sqrt{|x-a_{0}|}$ for some positive constant $C>0$, we see that
$|L|$ is at most a constant multiple of

$\lim_{h\to
0^{+}}\sqrt{h}\int_{-\infty}^{1}\frac{|\log|z||}{\sqrt{|1-z|^{3}}}dx=0.$

This completes the proof. ∎

##  5\. Proving Special Case Bounds

In this section, we use the integrals computed in the Appendix A to prove
Corollaries 1.3 and 1.4. To do this, we compute the measures $\mu_{\emptyset}$
and $\mu_{\\{x\\}}$ that satisfy the requirements of Theorem 1.1 explicitly.
Let $A\in\\{\emptyset,\\{x\\}\\}$. We need the following:

(29) $\displaystyle\lambda_{A}$ $\displaystyle=\int xd\mu,$ (30)
$\displaystyle I({\mu_{A}})$ $\displaystyle=0,\text{ and }$ (31)
$\displaystyle\int\log xd\mu_{A}$ $\displaystyle=0$ $\displaystyle\text{for
}A=\\{x\\}.$

To find this measure, we use (9), (39), (41), and (42). To use these formulas,
we need to make a choice of $\alpha,\beta$ and $\gamma$. Notice that a measure
of the form given in section A has density function
$\frac{\alpha+\frac{\beta\sqrt{ab}}{x}+\gamma(m-x)}{\pi\sqrt{(b-x)(x-a)}}$.
Since this density must be positive, we have that
$\alpha+\frac{\beta\sqrt{ab}}{x}+\gamma(m-x)\geq 0$ which rearranges to

$\gamma x^{2}-(\gamma m+\alpha)x-\beta\sqrt{ab}\geq 0.$

This must hold for all $x\in[a,b]$. We make the unique choice of
$\alpha,\beta,$ and $\gamma$ that makes the density 0 on the boundary, i.e. at
$a$ and $b$, subject to the constraint $\beta=1-\alpha$. The equality case of
this quadratic inequality gives

$x=\frac{\gamma m+\alpha\pm\sqrt{(\gamma
m+\alpha)^{2}+4\gamma\beta\sqrt{ab}}}{2\gamma}.$

Thus we must have that $\frac{\gamma m+\alpha}{2\gamma}=m$ and
$\frac{\sqrt{(\gamma
m+\alpha)^{2}+4\gamma\beta\sqrt{ab}}}{2\gamma}=\frac{b-a}{2}$.

The first equality gives us $\alpha=m\gamma$. The latter inequality gives us

$(\gamma m+\alpha)^{2}+4\gamma\beta\sqrt{ab}=(b-a)^{2}\gamma^{2}.$

Using $\alpha=\gamma m$ and $\beta=1-\alpha$, we get

$4\gamma(1-\gamma m)\sqrt{ab}=-4ab\gamma^{2}.$

Dividing by $4\gamma\sqrt{ab}$ and rearranging, we finally have
$\gamma=\frac{1}{m-\sqrt{ab}}$. Just as we use $m$ to represent the arithmetic
mean of $a$ and $b$, we shall use $g$ to represent the geometric mean of $a$
and $b$ so that

(32) $\alpha=\frac{m}{m-g},\beta=\frac{-g}{m-g},\text{ and
}\gamma=\frac{1}{m-g}.$

This change of variables from $(a,b)$ to $(m,g)$ will prove useful in reducing
the complexity of (39), (41), and (42) to prove (29), (30), and (31). As such,
we will rewrite (39), (41), and (42) in terms of only $m$ and $g$. Rewriting
(39), we get

$E:=\int_{a}^{b}xd\mu=\frac{m+g}{2}.$

Rewriting (41), we get

$L:=\int_{a}^{b}\log xd\nu=\frac{m+g}{m-g}\log\frac{m+g}{2}-\frac{2g}{m-g}\log
g-1.$

Rewriting (42), we get

$I:=\frac{1}{2}\log\frac{m^{2}-g^{2}}{4}-\frac{2g^{2}}{(m-g)^{2}}\log\frac{m+g}{2g}+\frac{3g-m}{2(m-g)}.$

The last equality we need to hold is (9). Looking at the constant terms of
this expression, equation (38) gives us that

$\lambda_{A}=\frac{-(\alpha\log(c/2)-\beta\log\frac{a+b+2\sqrt{ab}}{b-a}-\gamma
m)}{\gamma}$

assuming $I=0$. Using the change of variables, we have

$\lambda_{A}=m-\frac{m-g}{2}\log\frac{m-g}{2}-\frac{m+g}{2}\log\frac{m+g}{2}.$

Lastly, we rewrite the equalities $L=0,I=0,\lambda_{A}=E$ in terms of $E$ and
$g$ We see that $L=0$ is equivalent to

(33) $E\log E-E=g\log g-g.$

Similarly, $\lambda_{A}=E$ is equivalent to

$E\log E-E=-(E-g)\log(E-g)-g.$

Consequently, this equation and (33) gives

(34) $g\log g=(g-E)\log(E-g).$

We now show that (33) and (34) imply $I=0$.

$\displaystyle I$
$\displaystyle=\frac{1}{2}\log\frac{m^{2}-g^{2}}{4}-\frac{2g^{2}}{(m-g)^{2}}\log\frac{m+g}{2g}+\frac{3g-m}{2(m-g)}$
$\displaystyle=\frac{1}{2}\log(E-g)+\left(\frac{1}{2}-\frac{2g^{2}}{(m-g)^{2}}\right)\log
E+\frac{2g^{2}}{(m-g)^{2}}\log g+\frac{3g-m}{2(m-g)}$
$\displaystyle=\left(\frac{1}{2}-\frac{g}{m-g}\right)\log(E-g)+\left(\frac{1}{2}-\frac{2g^{2}}{(m-g)^{2}}\right)\log
E+\frac{3g-m}{2(m-g)}$

by (34). We note that $\frac{1}{2}(1-\frac{2g}{m-g})$ is not 0 because $m>g$.
Dividing the last line above by this quantity gives

$\log(E-g)+\left(1+\frac{2g}{m-g}\right)\log E-1=0$

which follows from (33) and (34).

Therefore in the case $A=\\{x\\}$, if we have our requirements that
$\lambda_{A}=E,I=0$, and $L=0$ then (33) and (34) hold simultaneously with no
other required relations. We claim that this system of equations
$\begin{cases}g\log g-g=E\log E-E\\\ g\log g=(g-E)\log(E-g)\end{cases}$ has a
unique solution.

######  Proposition 5.1.

The system of equations

$\begin{cases}x\log x-x=y\log y-y\\\ x\log x=(x-y)\log(y-x)\end{cases}$

has at most one real solution.

###### Proof.

Note that the top equation is equivalent to
$\left(\frac{x}{e}\right)^{\frac{x}{e}}=\left(\frac{y}{e}\right)^{\frac{y}{e}}$
and that from the second equation, we must have $y>x$. Furthermore,
$f(x)=x^{x}$ has a strictly increasing derivative. So $x^{x}=c$ has at most
two solutions for a given $c$. Since $f^{\prime}(x)=0$ has the unique solution
$x=\frac{1}{e}$ and $x<y$, any solution to the system must satisfy $x<1<y$.
This means that for a given $0<x<1$, there is a unique $y>1$ such that the
first equation holds. Call this number $y_{1}(x)$. Taking $\frac{d}{dx}$ of
both sides of the top equation, we see that $\frac{dy_{1}}{dx}=\frac{\log
x}{\log y_{1}}$. Since $0<x<1<y_{1}$, we thus have that $y_{1}(x)$ is
decreasing.

Similarly, for a fixed $0<x<1$, the second equation has exactly one possible
corresponding $y$. This is because we have $(y-x)\log(y-x)=-x\log x$ has
exactly one solution due to $-x\log x$ being positive and the fact that $z\log
z>0$ precisely when $z>1$. Furthermore, $z\log z$ has strictly positive
derivative for $z>1$. So as in the first equation, for each $0<x<1$, there is
a unique $y$ satisfying the second equation. Call this $y_{2}(x)$. Taking
$\frac{d}{dx}$ of both sides of the bottom equation, we get
$\frac{dy_{2}}{dx}=1-\frac{\log(x)+1}{\log(y_{2}-x)+1}$. We saw that $z=y-x>1$
and so $y>x+1$. Hence $\frac{dy_{2}}{dx}>-\log(x)>0$. Hence $y_{2}(x)$ is
increasing. Since $y_{1}(x)$ is decreasing and defined only on $x\in(0,1)$ and
$y_{2}(x)$ is increasing on $(0,1)$, it follows that the curves intersect at
most once. ∎

###### Corollary 1.4.

From (5), we have that $(1+\nu)\log(1+\nu^{-1})+\frac{\log\nu}{1+\nu}=1$ has a
unique solution and Siegel derived the lowerbound $T:=e(1+\nu^{-1})^{-\nu}$.
We wish to show that $T=\lambda_{\\{x\\}^{+}}$. Choose
$T^{\prime}=e(1+\nu^{-1})^{-(1+\nu)}$. Then $(T^{\prime},T)$ is a solution to
the system of equations in Proposition 5.1 and hence the only solution.
Indeed, the first equation in the system is a simple identity, and the second
equation is equivalent to (5). From our discussion above, it follows that
$T=E=\lambda_{\\{x\\}^{+}}$. ∎

###### Corollary 1.3.

We can re-use the computation done for Corollary 1.4, except now we require
that $\lambda_{\emptyset}=E,I=0$, and $\beta=0$ instead of $L=0$. Since
$\beta=0$, by (32) we have that $g=0$. Since $g=\sqrt{ab}=0$ and $a<b$, it
follows that $a=0$. Note: In Section A, we assume $a>0$ to avoid $g(x)\equiv
0$; however all other computations are identical without this assumption. From
this, we derive that $\alpha=1$ and $\gamma=\frac{1}{m}=\frac{2}{b}.$ Now
using (42), we have $\log\left(\frac{b}{4}\right)=\frac{1}{2}$ since $I=0$.
Hence $b=4\sqrt{e}$. Thus $\lambda_{\emptyset}=E$ implies
$\lambda_{\emptyset}=\frac{m}{2}=\sqrt{e}$ as desired. ∎

##  6\. Numerical results

In this section, we give numerical lower bounds on the Schur-Siegel-Smyth
trace problem. First, we compute explicitly the optimal measure $\mu_{A}$ in
terms of its support $\Sigma_{A}$. Next, we explain our gradient descent
algorithm for approximating $\Sigma_{A}$.

###  6.1. Expressing $\mu_{A}$

Recall our notations from Theorem 1.1. Fix $A=\\{Q_{1},\dots,Q_{n}\\}$ and let
$\Sigma_{A}:=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$ be the support of the optimal
measure $\mu_{A}$ in Theorem 1.1. Recall from the proof of Theorem 1.1 in
section 4 that $\mu_{eq}$, $\mu_{lin}$, $\mu_{Q}$ are the equilibrium measure,
the measure with linear potential $U_{\mu_{lin}}(t)=t$, and the probability
measure with potential $U_{\mu_{Q}}(t)=\frac{\log(|Q(t)|)}{\deg(Q)}+C_{Q}$ for
some constant $C_{Q}$ supported on $\Sigma_{A}$ respectively. By Lemma 3.2,
the density function of $\mu_{Q}$ is given by the following explicit
expression:

$d\mu_{Q}(x)=\frac{1}{\deg(Q)}\sum_{Q(\alpha)=0}\frac{|P_{\alpha}(x-\alpha)|\sqrt{H(\alpha)}}{|x-\alpha|\pi\sqrt{H(x)}},$

where $P(x):=x^{\deg(P_{\alpha})}p_{\alpha}(1/x)$ is the polynomial that
appears in the numerator of the equilibrium measure of the image of $\Sigma$
under the map $z\to\frac{1}{z-\alpha}$. Suppose that we have exactly one root
of $Q\in A$ inside every gap, then $\deg(p)=0$ in Theorem 1.1 and up to a
scalar $\mu_{A}$ has the following density function

$\frac{\sqrt{|H(x)|}}{\prod_{\lambda_{Q}\neq 0}|Q(x)|}.$

By the proof of Theorem 1.1 in section 4, we have the following identity

$\begin{split}X_{eq}\frac{|p_{eq}(x)|}{\pi\sqrt{H(x)}}+X_{lin}\frac{-x|p_{eq}(x)|+|P_{gap}(x)|}{\pi\sqrt{H(x)}}+\sum_{Q}\frac{X_{Q}}{\deg(Q)}\sum_{Q(\alpha)=0}\frac{|P_{\alpha}(x-\alpha)|\sqrt{H(\alpha)}}{|x-\alpha|\pi\sqrt{H(x)}}\\\
=\frac{c|H(x)|}{\sqrt{H(x)}\prod_{Q\in\tilde{A}}|Q(x)|},\end{split}$

where $X_{eq},X_{lin}>0,X_{Q}<0$ are some coefficients, and $c$ is normalized
such that the right-hand side of the equation is a probability measure on
$\Sigma$ and $\tilde{A}=\\{Q\in A:X_{Q}\neq 0\\}$. It follows that the
expressions inside the absolute values in the above have the same signs, so we
may take the absolute values outside the sums. By comparing the top
coefficient of both side, we have

(35) $X_{lin}=c\pi$

Moreover, by computing residues at $x-\alpha$, we deduce that

$\frac{-X_{Q}}{\deg(Q)\pi}=\frac{c\sqrt{|H(\alpha)|}}{|Q^{\prime}_{\alpha}(\alpha)|\prod_{\stackrel{{\scriptstyle
Q\in\tilde{A}}}{{Q(\alpha)\neq 0}}}|Q(\alpha)|}$

where $Q_{\alpha}\in\tilde{A}$ is such that $Q_{\alpha}(\alpha)=0$. Fix
$Q\in\tilde{A}$ and let ${\alpha_{1},\dots,\alpha_{\deg(Q)}}$ be roots of $Q$.
The above identity implies that

$\frac{H(\alpha_{1})}{\left(Q^{\prime}_{\alpha}(\alpha_{1})\prod_{\stackrel{{\scriptstyle
Q\in\tilde{A}|Q(\alpha_{1})|}}{{Q(\alpha_{1})\neq
0}}}|Q(\alpha_{1})|\right)^{2}}=\dots=\frac{H(\alpha_{\deg(Q)})}{\left(Q^{\prime}_{\alpha_{\deg(Q)}}(\alpha_{\deg(Q)})\prod_{\stackrel{{\scriptstyle
Q\in\tilde{A}|Q(\alpha_{\deg(Q)})|}}{{Q(\alpha_{\deg(Q)})\neq
0}}}|Q(\alpha_{\deg(Q)})|\right)^{2}}.$

We may write the above as

(36)
$X_{Q}=-\frac{\deg(Q)c\pi\sqrt{|\text{Res}(H,Q)^{1/\deg(Q)}|}}{|\text{Disc}(Q)|^{1/\deg(Q)}|\prod_{Q_{i}\neq
Q}|\text{Res}(Q,Q_{i})|^{1/\deg(Q)}|}.$

###  6.2. Gradient descent algorithm

To compute the appropriate $\Sigma$ for a given set $A$, the authors
implemented a gradient descent algorithm. Given
$\Sigma=\bigcup_{i}[a_{2i},a_{2i+1}]$, we are able to compute $X_{lin}$,
$X_{eq}$ and $X_{Q}$ for each $Q\in A$ as in section 6.1. We then compute
$\mu_{eq,\Sigma}$, $\mu_{lin,\Sigma}$, and $\mu_{Q,\Sigma}$ for each $Q\in A$
as shown in section 3. Define

$\mu_{\Sigma}:=X_{eq}\mu_{eq,\Sigma}+X_{lin,\Sigma}\mu_{lin,\Sigma}+\sum_{Q\in
A}X_{Q}\mu_{Q,\Sigma}.$

With this density function, we can compute all of the required quantities of a
particular measure. So the high-level algorithm is to perform gradient descent
on $\Sigma$ numerically (where $a_{i}$’s are the parameters) minimizing the
sum of squares of the energy, the value of the density function at the
$a_{i}$’s, and the values of $\int\log|Q(x)|d\mu_{\Sigma}$ for $Q\in A$,
because these values are 0 from Theorem 1.1.

In order to compute $d\mu_{eq,\Sigma}$, we use the equalities in (23). To
achieve these equalities, we can compute
$\int_{a_{2i}}^{a_{2i+1}}\frac{x^{j}}{\sqrt{H(x)}}dx$ for each $i,j$ and solve
a linear system to get the coefficients of the polynomial in the numerator.
Similar systems can be solved to compute $d\mu_{lin,\Sigma}$ and
$d\mu_{Q,\Sigma}$ since they can be derived from equilibrium measures. To save
computation time, the authors combine the systems at each iteration so only
one system needs solved per density function.

Lastly, to decrease the time taken by each iteration, the authors parallelized
the computation of the gradient. They did this using multiprocessing. By using
a different process for numerically computing the partial derivative of the
objective function on each endpoint of $\Sigma$, we can compute the gradient
much faster. For this, we use Pool from the multiprocessing module in Python.

One issue in the implementation of this algorithm is the numerical instability
of integrating functions of the form $\frac{|p(x)|}{\sqrt{H(x)}}$ near the
endpoints of the intervals. To fix this problem, when integrating on
$[a_{i},a_{i+1}]$, we perform the change of variables
$u=\frac{a_{i+1}-a_{i}}{2}\cos(x)+\frac{a_{i}+a_{i+1}}{2}$. This removes the
factor of $\sqrt{(a_{i+1}-x)(x-a_{i})}$ from the denominator of the integrand
and hence removes the singularities. This allows for more stable numerical
integration and consequently greatly increases the speed of the algorithm.

The above change of variables fixed the issue with the speed of each iteration
in the gradient descent, but the descent itself was still quite slow. We found
that instead of squaring the density of the measure at the boundary, we should
square the density at the boundary divided by the density of the equilibrium
measure at the boundary. This has given a great empirical improvement in the
rate of the descent.

###  6.3. Validity of Results

Suppose there is a finite subset $A\subset\mathbb{Z}[x]$, a finite union of
intervals $\Sigma=\bigcup_{i=0}^{l}[a_{2i},a_{2i+1}]$, a Borel probability
measure $\mu$ with support $\Sigma$, and $\lambda>0,\lambda_{0}>0,$ and
$\lambda_{Q}>0$ for all $Q\in A$ such that $x=\lambda+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}\left(U_{\mu}(x)-\frac{I(\mu)}{2}\right)$
for all $x\in\Sigma$. A priori, this alone is not sufficient in order to
indicate whether $\lambda$ is a lower bound to the Schur-Siegel-Smyth trace
problem. It is an additional requirement that

(37) $x\geq\lambda+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}\left(U_{\mu}(x)-\frac{I(\mu)}{2}\right)\text{
for all }x\in\mathbb{R}^{+}.$

We prove that equality on $\Sigma$ is indeed sufficient for special classes of
$A$.

######  Proposition 6.1.

Let $A,\Sigma,\lambda,\lambda_{0},$ and $\lambda_{Q}$ for all $Q\in A$ be
defined as above. Suppose $\Sigma\subset[0,18]$, and $\mu_{eq}$ is the
equilibrium of $\Sigma$. Define $\mathcal{R}$ to be the set
$\\{\alpha\in\mathbb{R}:\exists Q\in A,Q(\alpha)=0\\}$. Suppose that
$\mathcal{R}\cap(a_{2i+1},a_{2(i+1)})=\\{r_{i}\\}$ for each $i=0,\dots,l-1$
and $\mathcal{R}\cap\big{(}[0,a_{0})\cup(a_{2l+1},\infty)\big{)}=\\{0\\}$.
Moreover, suppose that for some $\delta>0$ and any boundary point $a_{i}$ of
$\Sigma$, we have

$\delta_{i}:=\lim_{x\to
a_{i},x\in\Sigma}\frac{d\mu(x)}{d\mu_{eq}(x)}\leq\delta,$

for any $x\in(a_{2i+1},r_{i})$,

$\sum_{Q\in
A}\sum_{Q(\alpha)=0}\frac{\lambda_{Q}}{(x-\alpha)^{2}}+\lambda_{0}\int\frac{f_{2i+1}(y)}{(x-y)^{2}}dy>0$

where $f_{i}(y)$ is the density function of $\mu-\delta_{i}\mu_{eq}$, and
similarly for any $x\in(r_{i},a_{2(i+1)})$

$\sum_{Q\in
A}\sum_{Q(\alpha)=0}\frac{\lambda_{Q}}{(x-\alpha)^{2}}+\lambda_{0}\int\frac{f_{2i+1}(y)}{(x-y)^{2}}dy>0.$

Then $\lambda-\delta\lambda_{0}\log(18)$ is a lower bound on $\lambda_{A}$.

###### Proof.

Suppose that $\mu_{A}$ is the optimal measure in Theorem 1.1 with support
$\Sigma_{A}\subset[0,18]$, and

$\int xd\mu_{A}(x)=\lambda_{A}.$

It suffices to prove

$x\geq\lambda-\delta\lambda_{0}\log(18)+\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|+\lambda_{0}\left(U_{\mu}(x)-\frac{I(\mu)}{2}\right)\text{
for all }x\in[0,18].$

Since by taking the average of both side with respect to $d\mu_{A}(x)$, it
follows that

$\lambda_{A}\geq\lambda-\delta\lambda_{0}\log(18).$

Let

$g(x):=x-\lambda+\delta\lambda_{0}\log(18)-\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|-\lambda_{0}\left(U_{\mu}(x)-\frac{I(\mu)}{2}\right).$

It remains to show that $g(x)\geq 0$ for $0\leq x\leq 18$. By assumption,
$g(x)=\delta\lambda_{0}\log(18)>0$ on $\Sigma$. Let
$\\{r_{i}\\}=\mathcal{R}\cap(a_{2i+1},a_{2(i+1)})$. We show that $g(x)\geq 0$
for any $x\in(a_{2i+1},r_{i})$. The argument for other intervals is similar.
By our assumption, there exists $\delta_{2i+1}<\delta$ such that the density
function of $\mu-\delta_{2i+1}\mu_{eq}$ vanishes at $a_{2i+1}$. Let

$f(x):=x-\lambda-\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|-\lambda_{0}\left(U_{\mu-\delta_{2i+1}\mu_{eq}}(x)-\frac{I(\mu)}{2}\right).$

Note that $f(x)=\delta_{2i+1}C_{\Sigma}\lambda_{0}\geq 0$ on $\Sigma$, where
$C_{\Sigma}>0$ is the logarithmic capacity of $\Sigma$ and

$g(x)-f(x)=\delta\lambda_{0}\log(18)-\delta_{2i+1}\lambda_{0}U_{\mu_{eq}}(x)\geq
0$

for $0\leq x\leq 18$. It is enough to show that $f(x)\geq 0$ for any
$x\in(a_{2i+1},r_{i}).$ The key reason that $f(x)\geq 0$ for
$x\in\mathbb{R}^{+}\setminus\Sigma$ is that

$f^{\prime\prime}(x)=\sum_{Q\in
A}\sum_{Q(\alpha)=0}\frac{\lambda_{Q}}{(x-\alpha)^{2}}+\lambda_{0}\int\frac{f_{2i+1}(y)}{(x-y)^{2}}dy>0,$

where $f_{2i+1}(y)$ is the density function of $\mu-\delta_{2i+1}\mu_{eq}$,
which is positive by our assumption. We can differentiate under the integral
by Liebnitz rule since the complement of the support of $f$ is open. Thus
$f(x)$ has strictly increasing derivative outside $\Sigma$ on intervals of
continuity. Furthermore, by Lemma 4.2, $f^{\prime}(x)=0$ at $a_{2i+1}$. By our
assumption on $\mathcal{R}$, this implies $g(x)\geq 0$ for all $x\geq 0$. ∎

Thus to prove the corollaries in section 1.3, by Proposition 6.1 and the
explicit construction of measures with prescribed potential, it suffices to
show that the resulting density functions we give are indeed positive. The
numerators of the density functions are polynomial combinations of absolute
values of polynomials. Thus we simply need to verify positivity of the
numerators between the roots of these polynomials via polynomial inequalities.
This can be done for each corollary given proving the bounds claimed in
section 1.3.

## References

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##  Appendix A Ideal Measures on an Interval

This section serve as a reference to potential and energy computations of the
set of measures considered in this paper on $[a,b]$ where $0<a<b$. Use
$m:=\frac{a+b}{2}$ to denote the midpoint of the interval and
$c:=\frac{b-a}{2}$ to denote half the length of the interval. Set
$f(x):=\frac{1}{\pi\sqrt{(b-x)(x-a)}}$, $g(x)=\frac{\sqrt{ab}}{\pi
x\sqrt{(b-x)(x-a)}}$ and $h(x)=\frac{m-x}{\pi\sqrt{(b-x)(x-a)}}$.

###  A.1. Integrals

This section simply verifies that $f$ and $g$ are density functions
(corresponding to $\mu_{[a,b]}$ and $\nu_{[a,b]}$ respectively) and that $\int
hd\eta=0$ where $\eta$ is the Lebesgue measure.

  * •

$\int fd\eta=1$.
This holds because $fd\eta$ is the density of the equilibrium measure on
$[a,b]$.

  * •

$\int gd\eta=1.$
This holds because $\nu_{[a,b]}$ is the pull-back measure of the equilibrium
measure on $[b^{-1},a^{-1}]$ under the reciprocol map.

  * •

$\int hd\eta=0.$
This holds because
$\int_{a}^{b}h(x)dx=\int_{-c}^{c}\frac{-u}{\pi\sqrt{c^{2}-u^{2}}}du$ where
$u=x-m$. The latter integrand is odd and therefore the integral is 0.

We have that if $\alpha,\beta,\gamma\in\mathbb{R}$, then $\tilde{f}:=\alpha
f+\beta g+\gamma h$ has integral 1 so long as $\alpha+\beta=1$. For
$\tilde{f}$ to be a density function, it remains to consider only
$\alpha,\beta,$ and $\gamma$ for which $\tilde{f}\geq 0$.

###  A.2. Potentials

Now we compute the potential of such $\tilde{f}$. To do so, we integrate
$f,g$, and $h$ against logarithm. For this section, we assume $x,y\in[a,b]$.

  * •

$U_{\mu_{[a,b]}}(x)=\int_{a}^{b}\log|x-y|f(y)dy=\log(c/2)$ by the definition
of the equilibrium measure.

  * •

$U_{\nu_{[a,b]}}(x)=\int_{a}^{b}\log|x-y|g(y)dy=\log
x-\log\frac{a+b+2\sqrt{ab}}{b-a}$ by Lemma 3.2.

  * •

Lastly, $\int\log|x-y|h(y)dy=x-m$. The computation for $a=-c$ and $b=c$ is
computed in Lemma 3.3.

Therefore for a valid choice of $\alpha,\beta,$ and $\gamma$, we have that

(38) $U_{\tilde{\mu}}(x)=\int\log|x-y|\tilde{f}(y)dy=\gamma x+\beta\log
x+\left(\alpha\log(c/2)-\beta\log\frac{a+b+2\sqrt{ab}}{b-a}-\gamma m\right)$

where $\tilde{\mu}$ is the measure with density function $\tilde{f}$ for
$a\leq x\leq b$.

###  A.3. Expectations

We compute the expectation of $\tilde{\mu}$ by using the integrals of
$xf(x),xg(x)$, and $xh(x)$. These integrals will be helpful in computing the
energies.

  * •

$\int xf(x)dx=m$.
We can see this as follows:

$\displaystyle\int_{a}^{b}xf(x)dx$
$\displaystyle=\int_{a}^{b}\frac{x}{\pi\sqrt{(b-x)(x-a)}}dx$
$\displaystyle=\int_{-c}^{c}\frac{u+m}{\pi\sqrt{c^{2}-u^{2}}}dx$
$\displaystyle\text{where }u=x-m.\text{ So}$
$\displaystyle=\int_{-c}^{c}\frac{m}{\pi\sqrt{c^{2}-u^{2}}}dx$
$\displaystyle\because\frac{u}{\sqrt{c^{2}-u^{2}}}\text{ is odd. Thus}$
$\displaystyle=m.$

The last equality holds, because we already computed that
$\int_{-c}^{c}\frac{du}{\pi\sqrt{c^{2}-u^{2}}}=1.$

  * •

$\int xg(x)dx=\sqrt{ab}.$
This one is simpler because $xg(x)=\sqrt{ab}f(x)$. Thus
$\int_{a}^{b}xg(x)=\sqrt{ab}$ as desired.

  * •

$\int xh(x)dx=\frac{-c^{2}}{2}.$
This computation is slightly more involved. First, we start as usual.

$\int_{a}^{b}xh(x)dx=\int_{a}^{b}\frac{mx-x^{2}}{\pi\sqrt{(b-x)(x-a)}}dx=-\int_{-c}^{c}\frac{u^{2}+um}{\pi\sqrt{c^{2}-u^{2}}}du$

where $u=x-m$. Recall that $\frac{u}{\pi\sqrt{c^{2}-u^{2}}}$ is odd, so it
suffices to show that
$\int_{-c}^{c}\frac{u^{2}du}{\pi\sqrt{c^{2}-u^{2}}}=\frac{c^{2}}{2}$. Since
$\frac{u^{2}}{\sqrt{c^{2}-u^{2}}}$ is even, we can write

$\int_{-c}^{c}\frac{u^{2}}{\pi\sqrt{c^{2}-u^{2}}}du=\frac{2}{\pi}\int_{0}^{c}\frac{u^{2}}{\sqrt{c^{2}-u^{2}}}du=\frac{2}{\pi}\left(\int_{0}^{c}\frac{c^{2}}{\sqrt{c^{2}-u^{2}}}du-\int_{0}^{c}\sqrt{c^{2}-u^{2}}du\right).$

It remains to notice that

$\int_{0}^{c}\frac{c^{2}}{\sqrt{c^{2}-u^{2}}}du=c^{2}\arcsin(u/c)|_{u=0}^{c}=\frac{1}{2}\pi
c^{2}$

and that

$\int_{0}^{c}\sqrt{c^{2}-u^{2}}du=\frac{1}{4}\pi c^{2}$

since the latter represents the area of a quarter circle of radius $c$.
Putting everything together, we get

$\int_{a}^{b}xh(x)dx=-\int_{-c}^{c}\frac{u^{2}}{\pi\sqrt{c^{2}-u^{2}}}du=\frac{-2}{\pi}\left(\frac{1}{4}\pi
c^{2}\right)=\frac{-c^{2}}{2}.$

An interesting observation is that the expectation of $\mu_{[a,b]}$ is the
arithmetic mean of $a$ and $b$ where as that of $\nu_{[a,b]}$ is the geometric
mean of $a$ and $b$. In particular, decreasing $\alpha$ gives a non-increasing
expectation. Furthermore, since $\int xh(x)dx<0$, increasing $\gamma$
decreases expectation. Finally, the expectation of $\tilde{\mu}$ is

(39) $\int_{a}^{b}xd\tilde{\mu}=\alpha m+\beta\sqrt{ab}-\frac{\gamma
c^{2}}{2}=\sqrt{ab}+\alpha(m-\sqrt{ab})-\frac{\gamma c^{2}}{2}.$

###  A.4. Logarithmic Integrals

We now compute when $\int\log|x|d\tilde{\mu}\geq 0$. These results are also
useful in computing the energies.

  * •

$\int\log xd\mu_{[a,b]}=U_{\mu_{[a,b]}}(0)=\log\frac{a+b+2\sqrt{ab}}{4}.$
This is the value in Smith’s notation [Smi21] which he takes from Serre in
[AP08].

  * •

$\int\log xd\nu_{[a,b]}=U_{\nu_{[a,b]}}(0)=\log\frac{4ab}{a+b+2\sqrt{ab}}.$
This can be derived from $U_{\mu_{[a,b]}}(0)$ using the $z\mapsto 1/z$ map.

  * •

$\int_{a}^{b}\frac{(m-x)\log x}{\pi\sqrt{(b-x)(x-a)}}dx=\sqrt{ab}-m.$
We know that $\int_{a}^{b}\frac{m\log
xdx}{\pi\sqrt{(b-x)(x-a)}}=mU_{\mu_{[a,b]}}(0)=m\log\frac{a+b+2\sqrt{ab}}{4}$.
We compute $\int_{a}^{b}\frac{x\log x}{\pi\sqrt{(b-x)(x-a)}}dx$ similarly to
how we computed $\int xf(x)d\mu$. First, we see that

(40)
$\int_{a}^{b}\frac{\log|y-x|dx}{\pi\sqrt{(b-x)(x-a)}}=2\log\frac{\sqrt{a-y}+\sqrt{b-y}}{2}$

for $y\leq a$. Integrating the left-hand side of (40) with respect to $y$
gives (up to a constant)

$\int_{a}^{b}\frac{(y-x)\log|y-x|-y}{\pi\sqrt{(b-x)(x-a)}}dx=y\left(2\log\left(\frac{\sqrt{a-y}+\sqrt{b-y}}{2}\right)-1\right)-\int_{a}^{b}\frac{x\log|y-x|dx}{\pi\sqrt{(b-x)(x-a)}}.$

Integrating the right-hand side of (40) with respect to $y$ gives (up to a
constant)

$\sqrt{a-y}\sqrt{b-y}-(a+b-2y)\log\left(\frac{\sqrt{a-y}+\sqrt{b-y}}{2}\right).$

Therefore, we have that

$\int_{a}^{b}\frac{x\log|y-x|dx}{\pi\sqrt{(b-x)(x-a)}}=(a+b)\log\left(\frac{\sqrt{a-y}+\sqrt{b-y}}{2}\right)-\sqrt{a-y}\sqrt{b-y}-y+C.$

Finally,

$\int_{a}^{b}\frac{(m-x)\log|y-x|dx}{\pi\sqrt{(b-x)(x-a)}}=\sqrt{a-y}\sqrt{b-y}+y-m$

where the constant is $-m$ because we know that at $y=a$ the integral should
be $-c$. In particular, when $y=0$, this is $\sqrt{ab}-m$.

Putting these together, we have that

(41) $\int_{a}^{b}\log
xd\tilde{\mu}=\alpha\log\frac{a+b+2\sqrt{ab}}{4}+\beta\log\frac{4ab}{a+b+2\sqrt{ab}}+\gamma\left(\sqrt{ab}-m\right).$

###  A.5. Energies

We wish to compute the energy of $\tilde{\mu}$. We do this by integrating the
potentials of $\mu_{[a,b]}$ and $\nu_{[a,b]}$ and also by computing
$\int\int\log|x-y|h(x)h(y)dxdy$.

  * •

$I({\mu_{[a,b]}})=\log(c/2)$.
This holds because
$\int_{a}^{b}U_{\mu_{[a,b]}}(x)d\mu_{[a,b]}(x)=\int_{a}^{b}\log(c/2)d\mu_{[a,b]}(x)=\log(c/2)$.

  * •

$I({\nu_{[a,b]}})=U_{\nu_{[a,b]}}(0)-\log\frac{a+2\sqrt{ab}+b}{b-a}.$
To compute this, we simply use the potentials we computed.

$\displaystyle I({\nu_{[a,b]}})$
$\displaystyle=\int_{a}^{b}U_{\nu_{[a,b]}}(x)d\nu_{[a,b]}$
$\displaystyle=\int_{a}^{b}\frac{(\log
x-\log\frac{a+2\sqrt{ab}+b}{b-a})\sqrt{ab}}{\pi x\sqrt{(b-x)(x-a)}}dx$
$\displaystyle=\int_{a}^{b}\frac{\sqrt{ab}\log x}{\pi
x\sqrt{(b-x)(x-a)}}dx-\log\frac{a+2\sqrt{ab}+b}{b-a}\int_{a}^{b}\frac{\sqrt{ab}}{\pi
x\sqrt{(b-x)(x-a)}}dx$
$\displaystyle=U_{\nu_{[a,b]}}(0)-\log\frac{a+2\sqrt{ab}+b}{b-a}$

as desired.

  * •

$\int\int\log|x-y|h(x)h(y)dxdy=\frac{-c^{2}}{2}.$
To do this, we recall that

$\int_{a}^{b}\log|x-y|h(y)dy=x-m.$

Since $\int_{a}^{b}h(x)dx=0$, we just need to show that

$\int_{a}^{b}(x-m)h(x)dx=\int_{a}^{b}xh(x)dx=\frac{-c^{2}}{2}.$

This is shown in section A.3.

Finally, we have that

$\displaystyle I({\tilde{\mu}})$ $\displaystyle=\int_{a}^{b}\gamma x+\beta\log
x+\left(\alpha\log(c/2)-\beta\log\frac{a+2\sqrt{ab}+b}{b-a}-\gamma
m\right)d\tilde{\mu}$
$\displaystyle=\gamma\int_{a}^{b}xd\tilde{\mu}+\int_{a}^{b}\beta\log x(\alpha
f(x)+\beta g(x)+\gamma
h(x))dx+\left(\alpha\log(c/2)-\beta\log\frac{a+b+2\sqrt{ab}}{b-a}-\gamma
m\right)$ (42)
$\displaystyle=\log(c/2)-2\beta^{2}\log\frac{m+\sqrt{ab}}{2\sqrt{ab}}-2\beta\gamma(m-\sqrt{ab})-\frac{\gamma^{2}c^{2}}{2}.$

##  Appendix B Details on Computed Measures

This appendix gives details of each of the measures suggested in Corollaries
1.3 to 1.7. First we discuss Corollaries 1.3 and 1.4. Their exact density
functions are already given in the statement of these corollaries and these
bounds are proven in Section 5.

(a) Density function of $\mu_{\emptyset}$

(b) Density function of $\mu_{\\{x\\}}$

Figure 2. Schur and Siegel Density Plots

The density functions for the cases of Schur’s bound and Siegel’s bound are
given in figures 2(a) and 2(b), respectively.

###  B.1. Numerical results

In this section, we give our numerical results when
$A=\\{Q_{0},Q_{1},Q_{2},Q_{3}\\}$ where
$Q_{0}:=x,Q_{1}:=x-1,Q_{2}:=x^{2}-3x+1,$ and $Q_{3}:=x^{3}-5x^{2}+6x-1$ that
gives a proof of Corollary 1.7. The proof of Corollaries 1.5 and 1.6 is
similar, and we just present the values of the support of the measure which is
enough to reproduce their proof.

###### Proof of Corollary 1.7.

Let $\Sigma:=\bigcup_{k=0}^{6}[a_{2k},a_{2k+1}],$ where

$\begin{split}[a_{0},\dots,a_{13}]=[0.04299491807925009,0.1859914049100783,0.2104756929603335,\\\
0.33831784130520615,0.42840285408953455,0.804598700230365,\\\
1.2208782797696687,1.5167041310779525,1.5933467096187162,2.474901146984523,\\\
2.7690575129408894,3.17915998503281,3.316051282538935,5.451640604980446]\end{split}$

First, we compute the normalizing coefficient $c$ using the following
equation:

$\int_{\Sigma}\frac{c\sqrt{H(x)}}{\prod_{Q\in\tilde{A}}|Q(x)|}dx=1.$

By (35), we let $X_{lin}$ to be

$X_{lin}=0.71081641\approx c\pi.$

It is easy to check that the pairwise resultant of the above polynomials is
$\pm 1$ (for distinct polynomials). By (36) and $\text{Disc}(Q_{2})=5$ and
$\text{Disc}(Q_{3})=7^{2}$, we let the coefficients of the density functions
as

$X_{Q_{0}}=-0.33947579\approx-c\pi|\text{Res}(Q_{0},H)|^{1/2},$
$X_{Q_{1}}=-0.30194521\approx-c\pi|\text{Res}(Q_{1},H)|^{1/2},$
$X_{Q_{2}}=-0.21076402\approx-\frac{2c\pi|\text{Res}(Q_{2},H)|^{1/4}}{\sqrt{5}},$
$X_{Q_{3}}=-0.11411836\approx-\frac{3c\pi|\text{Res}(Q_{3},H)|^{1/6}}{7^{2/3}}.$

Finally, we let

$X_{eq}=3.92688835,$

and define

$\mu:=\frac{1}{X_{eq}+X_{lin}\mu_{lin}(\Sigma)+\sum_{i=0}^{3}X_{Q}}\left(X_{eq}\mu_{eq}+\sum_{i=0}^{3}X_{Q}\mu_{Q}+X_{lin}\mu_{lin}\right).$

One can check that $\mu$ is a probability measure on $\Sigma$ and the density
function of $\mu$ is given by

$\frac{|P(x)|}{\pi\sqrt{|H(x)|}\prod_{Q\in A}|Q(x)|}$

where

$\begin{split}P(x)\approx
0.71081641(0.2280881534672178-9.942352366966269\textperiodcentered x^{1}\\\
+147.10493680314798\textperiodcentered
x^{2}-1094.9380145113385\textperiodcentered
x^{3}+4752.050538316919\textperiodcentered x^{4}\\\
-12957.61855762317\textperiodcentered
x^{5}+23205.561916616443\textperiodcentered
x^{6}-28090.29239337309\textperiodcentered x^{7}\\\
+23381.53369081627\textperiodcentered
x^{8}-13464.42922996917\textperiodcentered
x^{9}+5330.017929691111\textperiodcentered x^{10}\\\
-1417.4672356750882\textperiodcentered
x^{11}+240.90364774955518\textperiodcentered
x^{12}-23.532521064518708\textperiodcentered x^{13}+1.0\textperiodcentered
x^{14})\end{split}$

The equilibrium measure is given by

$d\mu_{eq}=\frac{|P_{eq}(x)|}{\pi\sqrt{|H(x)|}}$

where

$\begin{split}P_{eq}(x)\approx-1.0055206637281864+10.050734776080317\textperiodcentered
x^{1}-33.139572351202375\textperiodcentered x^{2}+\\\
46.13961639295653\textperiodcentered
x^{3}-30.057211222358188\textperiodcentered
x^{4}+9.008044896183025\textperiodcentered x^{5}-x^{6}\end{split}$

One can check that for $x\in\Sigma$,

$x=\lambda-\sum_{Q\in
A}\lambda_{Q}\log|Q(x)|-\lambda_{0}\left(U_{\mu}(x)-\frac{I(\mu)}{2}\right)$

for every $x\in\Sigma$ where

$\lambda\geq 1.79824993.$

Next, we check the conditions of Proposition 6.1. Let

$\delta_{i}:=\lim_{x\to
a_{i},x\in\Sigma}\frac{d\mu(x)}{d\mu_{eq}(x)}=\frac{P(a_{i})}{P_{eq}(a_{i})\prod_{Q\in
A}Q(a_{i})}.$

One can check that $\delta_{i}\leq 10^{-7}$. Moreover, the inequality

$\int\frac{f_{2i}(y)}{(x-y)^{2}}dy>0$

for $x$ in a neighborhood of $a_{2i}$ is equivalent to

$\frac{d}{dx}\frac{|P(x)|}{\prod_{Q}|Q(x)|}|_{x=a_{2i}}\geq\frac{d}{dx}\big{|}P_{eq}(x)|_{x=a_{2i}}.$

One can check the following numerical values

$\begin{split}\left[\frac{d}{dx}\frac{|P(x)|}{\prod_{Q}|Q(x)|}|_{x=a_{2i}}:0\leq
i\leq 6\right]=[54.17,7.09,5.17,4.32,7.42,21.92,71.51]\\\
\left[\frac{d}{dx}|P_{eq}(x)|_{x=a_{2i}}:0\leq i\leq
6\right]=[-7.44,1.19,0.96,0.45,1.80,6.39,31.36]\end{split}$

that verifies the required inequality for $x$ in a neighbourhood of $a_{2i}$.
Similarly, the inequality

$\int\frac{f_{2i+1}(y)}{(x-y)^{2}}dy>0$

in a neighborhood of $a_{2i+1}$ is equivalent to

$\frac{d}{dx}\frac{|P(x)|}{\prod_{Q}|Q(x)|}|_{x=a_{2i+1}}\leq\frac{d}{dx}|P_{eq}(x)|_{x=a_{2i+1}}$

One can check the following numerical values

$\begin{split}\left[\frac{d}{dx}\frac{|P(x)|}{\prod_{Q}|Q(x)|}|_{x=a_{2i+1}}:0\leq
i\leq 6\right]=[-8.80,-4.55,-5.54,-6.22,-20.33,-47.74,-2003.52]\\\
\left[\frac{d}{dx}|P_{eq}(x)|_{x=a_{2i+1}}:0\leq i\leq
6\right]=[-1.79,-0.62,-0.56,-1.35,-3.96,-12.42,4826.03]\end{split}$

For other values of $x$ away from the boundary points, one can check the
required inequalities numerically. By Proposition 6.1, it follows that

$\lambda_{A}\geq 1.7982494.$

On the other hand, it is easy to check that $\mu$ satisfies all the
constraints of Theorem 1.1 and

$\int xd\mu\leq 1.79825002.$

The above implies that

$\lambda_{A}\leq 1.79825002.$

This completes the proof of Corollary 1.7. ∎