# Simplex slicing: an asymptotically-sharp lower bound

Colin Tang111Carnegie Mellon University, United States of America,
<EMAIL_ADDRESS>
###### Abstract

We show that for the regular $n$-simplex, the $1$-codimensional central slice
that’s parallel to a facet will achieve the minimum area (up to a $1-o(1)$
factor) among all $1$-codimensional central slices, thus improving the
previous best known lower bound (Brzezinski 2013) by a factor of
$\frac{2\sqrt{3}}{e}\approx 1.27$. In addition to the standard technique of
interpreting geometric problems as problems about probability distributions
and standard Fourier-analytic techniques, we rely on a new idea, mainly
_changing the contour of integration_ of a meromorphic function.

##  1 Introduction

We are broadly interested in the following question: if $K$ is a given convex
body in $\mathbb{R}^{n}$, and $H$ is some hyperplane of codimension $1$, can
we obtain good upper and lower bounds on the area of the intersection $H\cap
K$? (“Area” meaning the $(n-1)$-dimensional volume.) We refer to $H\cap K$ as
a (hyperplane) section of $K$. If $H$ is constrained to pass through the
centroid of $K$, we call $H\cap K$ a central section.

Let us consider the case where $K$ is the hypercube
$Q_{n}\coloneqq[-\frac{1}{2},\frac{1}{2}]^{n}$ and $H$ is constrained to pass
through the origin. The first sharp result was discovered independently by
Hadwiger ([Had72]) and Hensley ([Hen79]), who showed that the area of $H\cap
Q_{n}$ is minimized by taking $H$ perpendicular to the vector
$(1,0,0,0,\dots,0)$. (In this case $H\cap Q_{n}$ is isometric to $Q_{n-1}$.)
Hensley’s proof relied on a technique which has become commonplace: converting
the original geometric problem into the language of probability distributions.
Later, Ball proved in [Bal86] that the area of $H\cap Q_{n}$ is maximized by
taking $H$ perpendicular to the vector $(1,1,0,0,0,0,\dots,0)$. (In this case
$H\cap Q_{n}$ is a rectangular prism with one side $\sqrt{2}$ and all other
sides of length $1$.) Ball used the aforementioned technique of recasting the
problem into the language of probability distributions, and then also used
another technique which has become commonplace: the Fourier inversion formula.
(See [KRZ04] for more exposition on the method of Fourier analysis in convex
geometry.) Incidentally, Ball’s result immediately implies that for $n\geq
10$, any central sections of $Q_{n}$ will have area less than that of any
central section of $B(\Gamma(\frac{n}{2}+1)^{1/n}/\sqrt{\pi})$, the $n$-ball
of volume $1$ (see [Bal88]). This gives a negative answer to the Busemann-
Petty problem for all $n\geq 10$.

Now let us turn our attention to the case where $K$ is the regular simplex
$\Delta_{n}$ (of side length $\sqrt{2}$). If $H$ is not constrained to pass
through the center of $\Delta_{n}$, Webb noted in the introduction of [Web96]
that some results of Ball (involving maximum-volume inscribed ellipsoids) can
be used to show that the area of $H\cap\Delta_{n}$ is maximized by letting $H$
be a facet of $\Delta_{n}$. Fradelizi (in [Fra99]) has given a different proof
of this same result. However, if $H$ is constrained to pass through the center
of $\Delta_{n}$, Webb showed in the same paper ([Web96]) that the area of
$H\cap\Delta_{n}$ is maximized by taking $H$ to pass through all but two
vertices of $\Delta_{n}$ (a quantitative version of Webb’s result is given in
[MTTT24]). Webb gave two proofs of this result, one of which used both the
technique of probability distributions and the technique of the Fourier
inversion formula. This leaves open the following question, which is the focus
of this paper:

######  Question 1.

If $H$ is constrained to pass through the center of $\Delta_{n}$, then what is
the minimum possible value of the area of $H\cap\Delta_{n}$? In other words,
what is the minimum central simplex section? Furthermore, which choices of $H$
attain this minimum value?

It has been conjectured that the minimum in Question 1 is attained by taking
$H$ to be $H_{\text{facet}}$, where $H_{\text{facet}}$ is any hyperplane
through the center of $\Delta_{n}$ and parallel to a facet (this is Conjecture
4 in [NT23]). Dirksen showed (Theorem 1.3 (i) in [Dir17]) that
$H_{\text{facet}}$ is a local minimizer; i.e. for all $H$ sufficiently close
to $H_{\text{facet}}$, the area of $H\cap\Delta_{n}$ is at least that of
$H_{\text{facet}}\cap\Delta_{n}$. In Theorem 1.1 of [Brz13], Brzezinski showed
a lower bound that is within a factor of $1.27$ of the conjectured minimizer;
i.e. Brzezinski proved that for any $H$, the area of $H\cap\Delta_{n}$ is at
least $\frac{e}{2\sqrt{3}}\approx\frac{1}{1.27}$ times that of
$H_{\text{facet}}\cap\Delta_{n}$. One reason why the question of minimizing
$H\cap\Delta_{n}$ is harder than the analogous question for $Q_{n}$ is that
Hensley was able to utilize the fact that $Q_{n}$ is a symmetric convex body,
so the probability distribution resulting from slicing $Q_{n}$ is a symmetric
probability distribution. The simplex is not a symmetric convex body, so the
resulting probability distribution is in general not symmetric.

In this paper we almost resolve the question of minimizing $H\cap\Delta_{n}$;
i.e. we prove a lower bound that is within a factor of $1-o(1)$ of the
conjectured minimizer (the notation $o(1)$ is with respect to taking
$n\nearrow\infty$).

######  Theorem 2.

For each hyperplane $H$ passing through the center of $\Delta_{n}$, we have

$\operatorname{vol}_{n-1}(H\cap\Delta_{n})\geq\frac{1}{e}\sqrt{\frac{n+1}{n}}\left(\frac{n+1}{n}\right)^{n-1}\operatorname{vol}_{n-1}(H_{\mathrm{facet}}\cap\Delta_{n}).$

Note that
$\frac{1}{e}\sqrt{\frac{n+1}{n}}\left(\frac{n+1}{n}\right)^{n-1}\nearrow 1$ as
$n\nearrow\infty$, so Theorem 2 is best possible (asymptotically speaking).

Our main new idea is to move the contour of integration. Following the
aforementioned techniques of probability distributions and of the Fourier
inversion formula, we rewrite Theorem 2 as a statement that the probability
density of a sum of independent exponential random variables takes a value of
at least $\frac{1}{e}$ at zero. Specifically, we show the following, which may
have independent interest:

######  Theorem 3.

Let $Y_{1},Y_{2},\dots,Y_{n+1}$ be i.i.d. standard exponential random
variables (mean $1$). For each unit vector $u\in\mathbb{R}^{n+1}$, we have
that the density at $0$ of the random variable $Z\coloneqq
u_{1}(Y_{1}-1)+u_{2}(Y_{2}-1)+\dots+u_{n+1}(Y_{n+1}-1)$ is at least
$\frac{1}{e}$. Equality holds if $u$ is of form $(1,0,0,0,\dots,0)$.

Using Fourier inversion, we rewrite Theorem 3 as a statement that the integral
of the characteristic function of $Z$ is at least $\frac{1}{e}$. However, $F$
is highly oscillatory along the real axis, and is difficult to bound. We
simply move the contour of integration to some new contour $\gamma$, so that
$F$ takes only positive real values along $\gamma$; now we can bound the
integral more easily.

Our proof appears to be highly dependent on the idiosyncracies of the regular
simplex, and do not generalize well to other convex bodies. In particular,
there is a curious inequality (Equations (17) and (18)), and it is not clear
if there is an analogue when the simplex is replaced with some other convex
body.

A failed approach that is worth mentioning is the following false statement,
strengthening Theorem 3:

######  Fake Theorem 4.

Let $f_{X}$ be the density function of a log-concave random variable $X$, such
that $X$ has mean $0$ and variance $1$. Then $f_{X}(0)\geq\frac{1}{e}$, with
equality holding if $X$ is equal to $Y-1$, where $Y$ is a standard exponential
random variable (mean $1$).

This is indeed a strengthening of Theorem 3, since the random variable $Z$
from Theorem 3 is the sum of independent log-concave random variables, and is
hence log-concave. However, the Fake Theorem 4 is indeed false, as can be seen
by taking $X$ to be the uniform distribution with mean $0$ and variance $1$;
we end up with $f_{X}(0)=\frac{1}{\sqrt{12}}<\frac{1}{e}$. (In fact,
$\frac{1}{\sqrt{12}}$ is precisely the minimum value of $f_{X}(0)$, as $X$
varies as described in the hypothesis of Fake Theorem 4.)

Finally, we will note in passing that, letting $K$ be a convex body and $H$ be
a hyperplane as before, people have considered constraining $H$, not so that
it passes through the centroid of $K$, but so that it is at a distance exactly
$t$ from the centroid of $K$, where $t>0$ is a fixed constant. In this case
$H\cap K$ is called a noncentral section. König showed (Theorem 1.1 of [Kö21])
that if $t$ is reasonably large, the area of $H\cap\Delta_{n}$ is maximized by
taking $H$ to be parallel to a facet of $\Delta_{n}$. Liu and Tkocz
investigated the cross-polytope ([LT20]), and Moody, Stone, Zach, and Zvavitch
investigated the hypercube ([MSZZ13]).

In the remainder of this section we will sketch the proof of Theorem 2.

  * •

In Section 3, we first use a result of Webb (Equation (2)) to reduce Theorem 2
to a question about the distribution of a sum of independent centered
exponential random variables (Theorem 8, which is the same as Theorem 3). This
constitutes a use of the technique of probability distributions.

  * •

In Section 4, we take the Fourier transform, and the problem reduces to
showing that the integral of an explicit meromorphic function $F_{a}$ along
the real line is at least $\frac{1}{e}$ (Theorem 9). Here, $F_{a}$ is the
characteristic function of the probability distribution. This constitutes a
use of the technique of the Fourier inversion formula.

  * •

In Section 5, we change the contour of integration; this is our main new idea.
As noted previously, $F_{a}$ is too difficult to bound along the real axis, so
we instead change the contour of integration from the real axis to some path
$\gamma_{a}$, with the property that $F_{a}$ always attains positive real
values on $\gamma_{a}$. Since the integral of a meromorphic function does not
depend on the contour used222As long as one can move one contour to the other
contour without passing over any poles., the problem reduces to showing that
the integral of $F_{a}$ along $\gamma_{a}$ is at least $\frac{1}{e}$ (Theorem
12).

  * •

In Section 6, we give a lower bound for the values of $F_{a}$ along
$\gamma_{a}$. Essentially, we let $v$ denote the vector
$(1)\in\mathbb{R}^{1}$, and we show that $F_{a}(x)\geq F_{v}(x)$ for each $x$
(Theorem 13). We do this by showing that the derivative of $F_{a}$ is at least
the derivative of $F_{v}$ (Equation (12)). There is a curious inequality
involved (Equations (17) and (18)). We obtain the desired $\frac{1}{e}$ lower-
bound by evaluating the integral of $F_{v}$.

##  2 Definitions

Let $n\geq 2$ be a positive integer. We use $\Delta_{n}$ to denote the regular
$n$-simplex of side length $\sqrt{2}$, embedded in $\mathbb{R}^{n+1}$ as
follows: $\Delta_{n}$ is the convex hull of the points

$(1,0,0,\dots,0),(0,1,0,\dots,0),\dots,(0,0,0,\dots,1).$

Equivalently,

$\Delta_{n}=\left\\{(x_{1},x_{2},x_{3},\dots,x_{n+1})\in\mathbb{R}^{n+1}\bigm{|}x_{1}+x_{2}+x_{3}+\dots+x_{n+1}=1,\text{
and }x_{i}\geq 0\text{ for each }i\right\\}.$

For any set $Q$, we let $\operatorname{vol}_{k}(Q)$ denote the $k$-dimensional
Hausdorff measure of $Q$. Then the volume of $\Delta_{n}$ is given by

$\operatorname{vol}_{n}(\Delta_{n})=\frac{\sqrt{n+1}}{n!}.$

Let $\vec{\mathbf{1}}\in\mathbb{R}^{n+1}$ denote the unit vector
$\frac{1}{\sqrt{n+1}}(1,1,1,\dots,1)=\left(\frac{1}{\sqrt{n+1}},\frac{1}{\sqrt{n+1}},\frac{1}{\sqrt{n+1}},\dots,\frac{1}{\sqrt{n+1}}\right)$.
Observe that $\frac{1}{\sqrt{n+1}}\vec{\mathbf{1}}$ is the center of the
regular simplex $\Delta_{n}$. Furthermore, given any two vectors
$v,w\in\mathbb{R}^{n+1}$, we write $\langle v,w\rangle$ for their dot product,
and we say that $v\perp w$ if $\langle v,w\rangle=0$, i.e. if $v,w$ are
orthogonal.

Throughout this paper, let $a\in\mathbb{R}^{n+1}$ denote an arbitrary unit
vector satisfying $a\perp\vec{\mathbf{1}}$. Let $H(a)$ denote the hyperplane
perpendicular to $a$; i.e.

$H(a)=\left\\{x\in\mathbb{R}^{n+1}\bigm{|}x\perp a\right\\}.$

Note that since $a\perp\vec{\mathbf{1}}$, we have that $H(a)$ contains the
point $\frac{1}{\sqrt{n+1}}\vec{\mathbf{1}}$; i.e. $H(a)$ always passes
through the center of $\Delta_{n}$.

The intersection $H(a)\cap\Delta_{n}$ is an $(n-1)$-dimensional polytope
embedded in $\mathbb{R}^{n+1}$. We call this object the central section of
$\Delta_{n}$ in the direction of $a$, and denote it by $T(a)$. As in Question
1, we will be concerned with finding the minimum possible value of
$\operatorname{vol}_{n-1}(T(a))$ as $a$ varies over all unit vectors
perpendicular to $\vec{\mathbf{1}}$.

settings.outformat = ”pdf”; settings.prc = false; //settings.render = 8;
import three; import graph;

unitsize(3cm); currentprojection = perspective(6,4,2); real ax = 1.2; real sc
= 1.1; draw((0,0,0)–(ax,0,0),arrow=Arrow3()); label(”$x_{1}$”,sc*(ax,0,0));
draw((0,0,0)–(0,ax,0),arrow=Arrow3()); label(”$x_{2}$”,sc*(0,ax,0));
draw((0,0,0)–(0,0,ax),arrow=Arrow3()); label(”$x_{3}$”,sc*(0,0,ax)); path3
outline = (1,0,0)–(0,1,0)–(0,0,1)–cycle; draw(outline);
draw(surface(outline),gray(0.8)+opacity(0.6));
label(”$\Delta_{2}$”,sc*(-0.09,0.79,0.3)); real p1 = -10; real p2 = -3; real
p3 = 13;

triple a = (p3,p1,p2)/sqrt(p1*p1+p2*p2+p3*p3);
draw((0,0,0)–a,arrow=Arrow3(),dashed); label(”$a$”,sc*a); triple a1 =
(p2,0,-p3)/(p2-p3); triple a2 = (p1,-p3,0)/(p1-p3); path3 plane =
(0,0,0)–a1–a2–cycle; draw(plane,linewidth(1.5)); draw(surface(plane),black);

dot((1/3,1/3,1/3),linewidth(3.5));
label(”$\frac{1}{\sqrt{3}}\vec{\mathbf{1}}$”, sc*(0.2,0.4,0.4));

Figure 1: The gray triangle represents $\Delta_{2}$, and the black triangle
represents $H(a)$. The thick line at the intersection of the two triangles
represents the hyperplane section $T(a)$.

For convenience, with $a$ defined as above, we will write the entries of $a$
as

$a=(a_{1},a_{2},\dots,a_{n+1}).$ (1)

Observe that

$a_{1}+a_{2}+\dots+a_{n+1}=0$

and

$a_{1}^{2}+a_{2}^{2}+\dots+a_{n+1}^{2}=1.$

Define the unit vector
$a_{\text{facet}}\coloneqq\left(\frac{n}{\sqrt{n(n+1)}},-\frac{1}{\sqrt{n(n+1)}},-\frac{1}{\sqrt{n(n+1)}},-\frac{1}{\sqrt{n(n+1)}},\dots,-\frac{1}{\sqrt{n(n+1)}}\right)\in\mathbb{R}^{n+1}$.
Observe that $H(a_{\text{facet}})$ corresponds exactly to the hyperplane
$H_{\text{facet}}$ from Section 1, which corresponds to the conjectured
minimum central section (as in Question 1). We compute that
$T(a_{\text{facet}})$ is the convex hull of the $n$ vertices

$\begin{array}[]{*7{>{\displaystyle}c}}\bigg{(}\frac{1}{n+1},&\frac{n}{n+1},&0,&0,&0,&\dots,&0\bigg{)}\\\
\bigg{(}\frac{1}{n+1},&0,&\frac{n}{n+1},&0,&0,&\dots,&0\bigg{)}\\\
\bigg{(}\frac{1}{n+1},&0,&0,&\frac{n}{n+1},&0,&\dots,&0\bigg{)}\\\
\bigg{(}\frac{1}{n+1},&0,&0,&0,&\frac{n}{n+1},&\dots,&0\bigg{)}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
\bigg{(}\frac{1}{n+1},&0,&0,&0,&0,&\dots,&\frac{n}{n+1}\bigg{)}\\\
\end{array}$

and thus

$\operatorname{vol}_{n-1}(T(a_{\text{facet}}))=\operatorname{vol}_{n-1}\left(\frac{n}{n+1}\cdot\Delta_{n-1}\right)=\left(\frac{n}{n+1}\right)^{n-1}\operatorname{vol}_{n-1}(\Delta_{n-1})=\left(\frac{n}{n+1}\right)^{n-1}\frac{\sqrt{n}}{(n-1)!}.$

So

$\operatorname{vol}_{n-1}(H_{\text{facet}}\cap\Delta_{n})=\frac{\sqrt{n}}{(n-1)!}\left(\frac{n}{n+1}\right)^{n-1}.$

settings.outformat = ”pdf”; settings.prc = false; //settings.render = 8;
import three; import graph;

unitsize(3cm);

currentprojection = perspective(6,4,2); real ax = 1.2; real sc = 1.1;
draw((0,0,0)–(ax,0,0),arrow=Arrow3()); label(”$x_{1}$”,sc*(ax,0,0));
draw((0,0,0)–(0,ax,0),arrow=Arrow3()); label(”$x_{2}$”,sc*(0,ax,0));
draw((0,0,0)–(0,0,ax),arrow=Arrow3()); label(”$x_{3}$”,sc*(0,0,ax)); path3
outline = (1,0,0)–(0,1,0)–(0,0,1)–cycle; draw(outline);
draw(surface(outline),gray(0.8)+opacity(0.6));
label(”$\Delta_{2}$”,sc*(-0.09,0.79,0.3)); real p1 = -1; real p2 = -1; real p3
= 2;

triple a = (p3,p1,p2)/sqrt(p1*p1+p2*p2+p3*p3);
draw((0,0,0)–a,arrow=Arrow3(),dashed); label(”$a_{\mathrm{facet}}$”,sc*a);
triple a1 = (p2,0,-p3)/(p2-p3); triple a2 = (p1,-p3,0)/(p1-p3); path3 plane =
(0,0,0)–a1–a2–cycle; draw(plane,linewidth(1.5)); draw(surface(plane),black);

dot((1/3,1/3,1/3),linewidth(3.5));
label(”$\frac{1}{\sqrt{3}}\vec{\mathbf{1}}$”, sc*(0.2,0.4,0.4));

Figure 2: The gray triangle represents $\Delta_{2}$, and the black triangle
represents $H(a_{\text{facet}})$. The thick line at the intersection of the
two triangles represents the hyperplane section $T(a_{\text{facet}})$. Note
that $T(a_{\text{facet}})$ is parallel to one of the facets of $\Delta_{2}$.

Thus Theorem 2 is equivalent to the following:

######  Theorem 5.

For each unit vector $a\in\mathbb{R}^{n+1}$ satisfying
$a\perp\vec{\mathbf{1}}$, we have
$\operatorname{vol}_{n-1}(T(a))\geq\frac{\sqrt{n+1}}{(n-1)!}\frac{1}{e}$.

##  3 The technique of probability distributions

In Section 0 of [Web96] it is proven that if $Y_{1},Y_{2},\dots,Y_{n+1}$ are
i.i.d. standard exponential random variables (mean $1$), and $G_{a}(x)$
represents the probability density function of the sum
$a_{1}Y_{1}+a_{2}Y_{2}+\dots+a_{n+1}Y_{n+1}$, then the volume of $T(a)$ is
given by

$\operatorname{vol}_{n-1}(T(a))=\frac{\sqrt{n+1}}{(n-1)!}G_{a}(0).$ (2)

In other words, the volume of $T(a)$ is proportional to the density at $0$ of
the random variable $a_{1}Y_{1}+a_{2}Y_{2}+\dots+a_{n+1}Y_{n+1}$. So Question
1 and Theorem 5 may be converted to the following statements about probability
distributions:

######  Question 6.

What is the smallest possible value of $G_{a}(0)$, as $a$ varies over all unit
vectors in $\mathbb{R}^{n+1}$ which are perpendicular to $\vec{\mathbf{1}}$?

######  Theorem 7.

For each unit vector $a\in\mathbb{R}^{n+1}$ satisfying
$a\perp\vec{\mathbf{1}}$, we have $G_{a}(0)\geq\frac{1}{e}$.

Throughout this paper, let $u\in\mathbb{R}^{n+1}$ denote an arbitrary unit
vector (not necessarily satisfying $u\perp\vec{\mathbf{1}}$). Let $G_{u}(x)$
denote the probability density of the random variable
$u_{1}(Y_{1}-1)+u_{2}(Y_{2}-1)+\dots+u_{n+1}(Y_{n+1}-1)$ (sum of independent
centered exponential random variables), and observe that when
$u\perp\vec{\mathbf{1}}$ this agrees with the earlier definition of
$G_{a}(x)$. We will prove the following strengthening of Theorem 7, which may
be of independent interest:

######  Theorem 8.

For each unit vector $u\in\mathbb{R}^{n+1}$, not necessarily satisfying
$u\perp\vec{\mathbf{1}}$, we have $G_{u}(0)\geq\frac{1}{e}$. Equality occurs
if $u$ is of form $(1,0,0,0,\dots,0)$.

Note that this Theorem is precisely the same as Theorem 3.

Let us try to understand the probability distribution $G_{u}$. For each $1\leq
j\leq n+1$, define $f_{j}(x)$ to be the probability density function of the
random variable $u_{j}(Y_{j}-1)$. Note that the $u_{j}(Y_{j}-1)$ are mutually
independent. Since $G_{u}$ is defined to be the probability density function
of the sum of these independent random variables, we know that $G_{u}$ is the
convolution of the $f_{j}$; i.e. $G_{u}=f_{1}*f_{2}*\dots*f_{n+1}$.

Define the function $f:\mathbb{R}\rightarrow[0,+\infty)$ via

$f(x)=\begin{cases}e^{-x}&\text{ if }x\geq 0\\\ 0&\text{
otherwise}.\end{cases}$

Note that $f$ is the probability density function of a standard exponential
random variable (mean $1$), so each $Y_{j}$ has probability density function
given by $f$. It follows that $f_{j}(x)=\frac{1}{|u_{j}|}f(\frac{x}{u_{j}}+1)$
for each $j$, and hence we may write

$\begin{split}G_{u}(x)&=(f_{1}*f_{2}*\dots*f_{n+1})(x)\\\
&=\int_{x_{1}+x_{2}+\dots+x_{n+1}=x}f_{1}(x_{1})f_{2}(x_{2})\dots
f_{n+1}(x_{n+1})\,\mathrm{d}x_{1}\,\mathrm{d}x_{2}\,\cdots\,\mathrm{d}x_{n+1}\\\
&=\int_{x_{1}+x_{2}+\dots+x_{n+1}=x}\prod_{j=1}^{n+1}\frac{1}{|u_{j}|}f\left(\frac{x_{j}}{u_{j}}+1\right)\,\mathrm{d}x_{1}\,\mathrm{d}x_{2}\,\cdots\,\mathrm{d}x_{n+1}.\end{split}$

##  4 The technique of the Fourier inversion formula

We will now use the Fourier transform to convert the convolution into a
pointwise product. Defining the Fourier transform (Section 2 of [Web96])

$\widehat{G_{u}}(t)=\int_{-\infty}^{+\infty}G_{u}(x)e^{-itx}\,\mathrm{d}x$

we have

$\widehat{G_{u}}(t)=\prod_{j=1}^{n+1}\widehat{f_{j}}(t).$

From $\hat{f}(t)=\frac{1}{1+it}$, we calculate
$\widehat{f_{j}}(t)=\frac{e^{iu_{j}t}}{1+iu_{j}t}$ and thus

$\widehat{G_{u}}(t)=\prod_{j=1}^{n+1}\frac{e^{iu_{j}t}}{1+iu_{j}t}.$

Note that whenever $u$ has at least two nonzero entries, we have
$\widehat{G_{u}}\in L^{1}$ and hence the Fourier inversion formula is valid,
yielding

$G_{u}(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\widehat{G_{u}}(t)e^{ixt}\,\mathrm{d}t=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\left(\prod_{j=1}^{n+1}\frac{e^{iu_{j}t}}{1+iu_{j}t}\right)e^{ixt}\,\mathrm{d}t.$

If we want to show a lower bound on $G_{u}(0)$ (as in Theorem 8), it suffices
to prove the following inequality:

$\frac{1}{2\pi}\int_{-\infty}^{+\infty}\left(\prod_{j=1}^{n+1}\frac{e^{iu_{j}t}}{1+iu_{j}t}\right)\,\mathrm{d}t\geq\frac{1}{e}.$

Writing

$F_{u}(t)\coloneqq\prod_{j=1}^{n+1}\frac{e^{iu_{j}t}}{1+iu_{j}t},$ (3)

the inequality that we wish to show becomes the following:

######  Theorem 9.

For each unit vector $u\in\mathbb{R}^{n+1}$ such that $u$ has at least two
nonzero entries, we have

$\frac{1}{2\pi}\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t\geq\frac{1}{e}.$

##  5 Main idea: Changing the contour of integration

Figure 3: A plot of $F_{u}(t)$ on the Argand plane when
$u=(\sqrt{0.42},\sqrt{0.38},\sqrt{0.20})$. The horizontal axis denotes the
real part of $t$, and the vertical axis denotes the imaginary part of $t$. The
brightness represents the modulus of $F_{u}(t)$, while the hue represents the
argument of $F_{u}(t)$ (see the legends to the right of the plot). Note the
poles along the imaginary axis.
This image was produced using Wolfram Mathematica Student Edition.

In this section we will showcase our main new idea, which is to change the
contour of integration.

As noted in Section 1, it is difficult to bound the integral
$\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t$ as written, because $F_{u}$
oscillates on the real axis (see Figure 3). We will instead find a contour
$\gamma_{u}$ in the complex plane, along which $F_{u}$ takes positive real
values (Proposition 10); finally we will argue that the integral
$\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t$ equals the contour integral
$\int_{\gamma_{u}}F_{u}(t)\,\mathrm{d}t$ (see Equation (10)), allowing us to
change the integral in Theorem 9 to an easier integral.

The rest of this section will be dedicated to finding and defining
$\gamma_{u}$.

We want to find a function $y_{u}$ such that the graph of $y_{u}$ (i.e. the
set $\left\\{x+iy_{u}(x)\bigm{|}x\in\mathbb{R}\right\\}$) is exactly the
contour $\gamma_{u}$ that we seek. In order for this to be the case, we need

$\arg F_{u}(x+iy_{u}(x))\equiv 0\pmod{2\pi}$

for each $x$, where $\arg z$ denotes the argument of a nonzero complex number
$z$. Using the definition of $F_{u}(t)$ (Equation (3)), we can write

$\begin{split}\arg
F_{u}(x+iy)&=\arg\prod_{j=1}^{n+1}\frac{e^{iu_{j}(x+iy)}}{1+iu_{j}(x+iy)}\\\
&=x\left(\sum_{j=1}^{n+1}u_{j}\right)-\sum_{j=1}^{n+1}\arg(1+iu_{j}(x+iy))\\\
&=x\left(\sum_{j=1}^{n+1}u_{j}\right)-\sum_{j=1}^{n+1}\arg(1-u_{j}y+iu_{j}x)\\\
&\equiv
x\left(\sum_{j=1}^{n+1}u_{j}\right)-\sum_{j=1}^{n+1}\arctan\frac{u_{j}x}{1-u_{j}y}\pmod{2\pi}\end{split}$
(4)

so all we need to do is to choose a continuous function $y_{u}$ such that

$x\left(\sum_{j=1}^{n+1}u_{j}\right)-\sum_{j=1}^{n+1}\arctan\frac{u_{j}x}{1-u_{j}y_{u}(x)}\pmod{2\pi}\equiv
0\pmod{2\pi}$ (5)

holds for all $x$. For any fixed $x$, there may be multiple values of
$y_{u}(x)$ which satisfy Equation (5) (due to the fact that that equation is
modulo $2\pi$), but if we additionally stipulate that $y_{u}$ is continuous
and satisfies $y_{u}(0)=0$, then there is only one way to choose $y_{u}$. We
will describe how to choose $y_{u}$ below.

Let $x,y$ be arbitrary real numbers satisfying $x>0$. We will adopt the
convention that the $\operatorname{\mathrm{arccot}}$ function is a continuous
function from $\mathbb{R}$ to $\mathbb{R}$ that has range $(0,\pi)$. (Some
authors define $\operatorname{\mathrm{arccot}}\theta$ to have a jump
discontinuity at $\theta=0$; we will not use their convention here.) Assume
that all $u_{j}$ are nonzero (since we can just discard any zero entries).

For each $1\leq j\leq n+1$, we define
$\alpha_{j}=\alpha_{j}(x,y;u)\coloneqq\operatorname{\mathrm{arccot}}\frac{\frac{1}{u_{j}}-y}{x}$
if $u_{j}>0$, and we define
$\beta_{j}=\beta_{j}(x,y;u)\coloneqq\operatorname{\mathrm{arccot}}\frac{-(\frac{1}{u_{j}}-y)}{x}$
if $u_{j}<0$. We will not define $\alpha_{j}$ for $u_{j}<0$, nor will we
define $\beta_{j}$ for $u_{j}>0$. (We may suppress the dependence on $x,y,u$,
thus writing merely $\alpha_{j}$ for what should technically be written as
$\alpha_{j}(x,y;u)$, if it is clear from context.) We define the function

$\Phi_{u}(x,y)\coloneqq
x\sum_{j=1}^{n+1}u_{j}-\sum_{j:u_{j}>0}\alpha_{j}+\sum_{j:u_{j}<0}\beta_{j},$
(6)

and the computation in Equation (4) implies

$\Phi_{u}(x,y)\equiv\arg F_{u}(x+iy)$ (7)

for all $(x,y)$ satisfying $x>0$.

The advantage of defining $\Phi_{u}$ is that $\Phi_{u}$ takes values in
$\mathbb{R}$, whereas $\arg F_{u}(x+iy)$ is only defined up to multiples of
$2\pi$. So we can say that $\Phi_{u}(x,y)$ is a real-valued function that is a
_lifting_ of the function $\arg F_{u}(x+iy)$. In other words, $\Phi_{u}(x,y)$
is a real-valued function that always is equivalent to the argument of
$F_{u}(x+iy)$ modulo $2\pi$.

We can now say that the pair $(x,y)$ has good angle if $\Phi_{u}(x,y)=0$, this
equality holding in $\mathbb{R}$ (instead of just
$\mathbb{R}/2\pi\mathbb{Z}$).

unitsize(1.2cm); draw((0,0)–(3,0),arrow=Arrow());label(”Re”,(3.3,0));
draw((0,-2.5)–(0,2.5),arrow=Arrow());label(”Im”,(0,2.7)); pair P = (1.9,0.8);
dot(P); label(”$x+iy$”,P,E); real u1 = 1.5; real u2 = -0.8; real u3 = -2.2;
draw(P–(0,u1)); label(”$i\frac{1}{u_{1}}$”,(0,u1),W); draw(P–(0,u2));
label(”$i\frac{1}{u_{2}}$”,(0,u2),W); draw(P–(0,u3));
label(”$i\frac{1}{u_{3}}$”,(0,u3),W); real rad = 0.2;
draw(arc((0,u1),(0,u1-rad),point(circle((0,u1),rad),intersect(circle((0,u1),rad),P–(0,u1))[0])));
label(”$\alpha_{1}$”,(0,u1),2*SE+2*S);
draw(arc((0,u2),intersectionpoint(circle((0,u2),rad),P–(0,u2)),(0,u2+rad)));
label(”$\beta_{2}$”,(0,u2),2*NE+2*N);
draw(arc((0,u3),intersectionpoint(circle((0,u3),rad),P–(0,u3)),(0,u3+rad)));
label(”$\beta_{3}$”,(0,u3),2*NE+4*N);

Figure 4: For each $j$, we define
$\alpha_{j}\coloneqq\operatorname{\mathrm{arccot}}\frac{\frac{1}{u_{j}}-y}{x}$
if $u_{j}>0$, and we define
$\beta_{j}\coloneqq\operatorname{\mathrm{arccot}}\frac{-(\frac{1}{u_{j}}-y)}{x}$
if $u_{j}<0$.

Let $m_{+},m_{-}$ denote the number of indices $j$ such that $u_{j}$ is
positive (respectively, negative). So $m_{+}+m_{-}=n+1$. Since $\Phi_{u}$ is a
strictly decreasing function of $y$, we know that:

  1. 1.

If $x$ satisfies $-m_{-}\pi<x\sum_{j=1}^{n+1}u_{j}<m_{+}\pi$, then there
exists a unique $y$ such that $(x,y)$ has good angle. In this case we will set
$y_{u}(x)=y$.

  2. 2.

If $x$ does not satisfy $-m_{-}\pi<x\sum_{j=1}^{n+1}u_{j}<m_{+}\pi$, then
there does not exist $y$ such that $(x,y)$ has good angle.

Thus we have found a function $y_{u}$ defined on the domain

$D_{u}\coloneqq\left\\{x\biggm{|}x>0\text{ and
}-m_{-}\pi<x\sum_{j=1}^{n+1}u_{j}<m_{+}\pi\right\\}.$

(Observe that $D_{u}$ contains the interval $(0,\pi)$.) By definition,

$\Phi_{u}(x,y_{u}(x))=0$ (8)

for all $x\in D_{u}$. In other words, $y_{u}$ is contained in the zero locus
of $\Phi_{u}$.

We can obtain that $y_{u}$ is $\mathcal{C}^{\infty}$ (in fact, real analytic)
on $D_{u}$ using the Implicit Function Theorem ($\Phi_{u}$ is a real analytic
function of $(x,y)$ which is also strictly decreasing with respect to $y$).

We may also show that $\lim_{x\searrow 0}y_{u}(x)=0$. To do this, let
$\epsilon\in(0,1)$ be arbitrary and fixed. If $j$ is such that $u_{j}>0$, we
note that $\frac{1}{u_{j}}-\epsilon>\frac{1}{1}-1=0$, so
$(\operatorname{\mathrm{arccot}}\frac{\frac{1}{u_{j}}-\epsilon}{x})/(\frac{x}{\frac{1}{u_{j}}-\epsilon})$
goes to $1$ as $x\searrow 0$. Similarly, if $j$ is such that $u_{j}<0$, we
have
$(\operatorname{\mathrm{arccot}}\frac{-(\frac{1}{u_{j}}-\epsilon)}{x})/(\frac{x}{-(\frac{1}{u_{j}}-\epsilon)})$
goes to $1$ as $x\searrow 0$. So

$\frac{1}{x}\left(\sum_{j:u_{j}>0}\operatorname{\mathrm{arccot}}\frac{\frac{1}{u_{j}}-\epsilon}{x}-\sum_{j:u_{j}<0}\operatorname{\mathrm{arccot}}\frac{-(\frac{1}{u_{j}}-\epsilon)}{x}\right)\rightarrow\sum_{j:u_{j}>0}\frac{1}{\frac{1}{u_{j}}-\epsilon}-\sum_{j:u_{j}<0}\frac{1}{-\left(\frac{1}{u_{j}}-\epsilon\right)}=\sum_{j=1}^{n+1}\frac{1}{\frac{1}{u_{j}}-\epsilon}$

as $x\searrow 0$. In other words,

$\lim_{x\searrow
0}\frac{1}{x}\left(\sum_{j:u_{j}>0}\alpha_{j}(x,\epsilon;u)-\sum_{j:u_{j}<0}\beta_{j}(x,\epsilon;u)\right)=\sum_{j=1}^{n+1}\frac{1}{\frac{1}{u_{j}}-\epsilon}.$

Since
$\sum_{j=1}^{n+1}\frac{1}{\frac{1}{u_{j}}-\epsilon}>\sum_{j=1}^{n+1}u_{j}$, we
see that when $x$ is sufficiently close to $0$, we have
$\Phi_{u}(x,\epsilon)<0$. Using the fact that $\Phi_{u}$ is strictly
decreasing with respect to $y$, we obtain $y_{u}(x)<\epsilon$ when $x$ is
sufficiently close to $0$. Similarly, we can show that $y_{u}(x)>-\epsilon$
when $x$ is sufficiently close to $0$. Since $\epsilon$ was arbitrary, this
proves $\lim_{x\searrow 0}y_{u}(x)=0$, as desired.

Thus $y_{u}$ extends continuously to the point $x=0$, by setting
$y_{u}(0)\coloneqq 0$.

Let us extend $y_{u}$ by reflection about the $y$-axis. Explicitly, we define
an expanded domain

$E_{u}\coloneqq\left\\{x\bigm{|}x=0\text{ or }x\in D_{u}\text{ or }-x\in
D_{u}\right\\}$

and extend $y_{u}$ to this expanded domain by setting $y_{u}(x)=y_{u}(-x)$
whenever $-x\in D_{u}$. So $y_{u}$ will be a continuous function on this
expanded domain, and is in fact $\mathcal{C}^{\infty}$ away from the point
$x=0$.

We will now show that $y_{u}$ is $\mathcal{C}^{\infty}$ on the entirety of
$E_{u}$. Observe that for $x>0$ and $y\in(-1,1)$, we have

$\begin{split}\frac{1}{x}\Phi_{u}(x,y)&=\sum_{j=1}^{n+1}u_{j}-\sum_{j:u_{j}>0}\frac{1}{x}\operatorname{\mathrm{arccot}}\frac{\frac{1}{u_{j}}-y}{x}+\sum_{j:u_{j}<0}\frac{1}{x}\operatorname{\mathrm{arccot}}\frac{-(\frac{1}{u_{j}}-y)}{x}\\\
&=\sum_{j=1}^{n+1}u_{j}-\sum_{j:u_{j}>0}\frac{1}{x}\arctan\frac{x}{\frac{1}{u_{j}}-y}+\sum_{j:u_{j}<0}\frac{1}{x}\arctan\frac{x}{-(\frac{1}{u_{j}}-y)}\end{split}$

since $\operatorname{\mathrm{arccot}}\theta=\arctan(1/\theta)$ whenever
$\theta>0$. Now we let $\phi$ denote the real analytic function given by
$\phi(\theta)\coloneqq(\arctan\theta)/\theta$, so we have

$\begin{split}\frac{1}{x}\Phi_{u}(x,y)&=\sum_{j=1}^{n+1}u_{j}-\sum_{j:u_{j}>0}\frac{1}{\frac{1}{u_{j}}-y}\phi\left(\frac{x}{\frac{1}{u_{j}}-y}\right)-\sum_{j:u_{j}<0}\frac{1}{\frac{1}{u_{j}}-y}\phi\left(\frac{x}{-(\frac{1}{u_{j}}-y)}\right)\\\
\end{split}$

and since $\phi$ is an even function, we can write

$\begin{split}\frac{1}{x}\Phi_{u}(x,y)&=\sum_{j=1}^{n+1}u_{j}-\sum_{j=1}^{n+1}\frac{1}{\frac{1}{u_{j}}-y}\phi\left(\frac{x}{\frac{1}{u_{j}}-y}\right).\\\
\end{split}$

The right-hand-side can be extended to a real analytic function
$\psi_{u}(x,y)$ on all of $\mathbb{R}\times(-1,1)$. Moreover, we compute

$\left(\frac{\partial}{\partial
y}\psi_{u}\right)(0,0)=-\sum_{j=1}^{n+1}\frac{1}{\left(\frac{1}{u_{j}}\right)^{2}}-\sum_{j=1}^{n+1}\frac{1}{\left(\frac{1}{u_{j}}\right)}\cdot
0=-\sum_{j=1}^{n+1}u_{j}^{2}=-1<0,$

so by the Implicit Function Theorem we know that near $(0,0)$, the zero locus
of $\psi_{u}$ is the graph of a real analytic function. But this zero locus is
exactly the graph of $y_{u}$, so we obtain that $y_{u}$ is a real analytic
function of $x$ at the point $x=0$, as desired.

Figure 5: A plot of $y_{u}(x)$ when
$u=(\sqrt{0.62},-\sqrt{0.19},-\sqrt{0.19})$, for $-20<x<20$.
This image was produced using Wolfram Mathematica Student Edition.

######  Remark.

What happens for other probability distributions (other than the exponential
distribution that we have been considering)? Are we still able to define a
smooth $y_{u}$ on which $F_{u}$ takes positive real values?

We have our function $y_{u}$ which is $\mathcal{C}^{\infty}$ on $E_{u}$. By
construction, we have that $y_{u}$ satisfies Equation (5) for all $x\in
E_{u}$. Now if we define

$\gamma_{u}\coloneqq\left\\{x+iy_{u}(x)\bigm{|}x\in E_{u}\right\\}$

to be the graph of $y_{u}$, then $\gamma_{u}$ is exactly the contour that we
seek. Indeed, the key property of $\gamma_{u}$ is the following:

######  Proposition 10.

For each $t\in\gamma_{u}$, we have that $F_{u}(t)$ is a positive real number.

Now that we have our contour $\gamma_{u}$, we would like to say the following
(in order to change the integral in Theorem 9 to an easier integral):

######  Theorem 11.

For each unit vector $u\in\mathbb{R}^{n+1}$ such that $u$ has at least two
nonzero entries, we have that the integral

$\frac{1}{2\pi}\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t$

equals the integral

$\frac{1}{2\pi}\int_{\gamma_{u}}F_{u}(t)\,\operatorname{\mathrm{Re}}(\mathrm{d}t)=\frac{1}{2\pi}\int_{s\in
E_{u}}F_{u}(s+iy_{u}(s))\,\mathrm{d}s.$

Note that the latter integral exists because the integrand is always a
positive real number on the path of integration.

To show Theorem 11, we will use the fact that _the integral of a meromorphic
function does not depend on the contour, as long as one contour can be moved
to the other without passing over any poles_. This is the main idea of our
paper: moving the contour of integration. We will approximate both integrals
in Theorem 11 by integrals over bounded contours, and then we will use
Cauchy’s Integral Theorem to shift the contour of integration. We must use an
upper bound on the magnitude of $F_{u}(t)$; this is because when shifting the
contour, we will incur a small error at each of the fringes, and we need to
know that this error goes to zero as we take the fringes to be further and
further away from the origin.

###### Proof of Theorem 11.

To upper bound the magnitude of $F_{u}(t)$, we now use the fact that $u$ has
at least two nonzero entries. Hence there exists a constant $C>0$, depending
only on $u$, such that for all pairs $(x,y)\in\mathbb{R}^{2}$ with $x\not=0$,
we have

$|F_{u}(x+iy)|\leq
C\exp\left(-y\sum_{j=1}^{n+1}u_{j}\right)\left/\left(x^{2}+\min_{j=1}^{n+1}\left(\frac{1}{u_{j}}-y\right)^{2}\right)\right..$
(9)

(In particular, the integral
$\frac{1}{2\pi}\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t$ in Theorem 9 is
well-defined.)

We split into cases based on the sign of $\sum_{j=1}^{n+1}u_{j}$.

Case I: $\sum_{j=1}^{n+1}u_{j}>0$.
In this case,
$E_{u}=(-m_{+}\pi/\sum_{j=1}^{n+1}u_{j},m_{+}\pi/\sum_{j=1}^{n+1}u_{j})$ is a
finite interval. Now let us choose $x_{1}\in D_{u}$ and $x_{2}>x_{1}$. Let
$K_{x_{1},x_{2}}$ denote the path consisting of the line segment from
$x_{1}+iy_{u}(x_{1})$ to $x_{2}+iy_{u}(x_{1})$, concatenated with the line
segment from $x_{2}+iy_{u}(x_{1})$ to $x_{2}$; we will use $K_{x_{1},x_{2}}$
as a fringe to relate the two contours ($\gamma_{u}$ and the real axis).
Equation (9) implies that $|F_{u}(x+iy)|\leq
C\exp\left(-y\sum_{j=1}^{n+1}u_{j}\right)/x^{2}$ for all $x\not=0$. By
breaking $K_{x_{1},x_{2}}$ into its constituent line segments, we obtain

$\left|\int_{K_{x_{1},x_{2}}}F_{u}(t)\,\mathrm{d}t\right|\leq
C\exp\left(-y_{u}(x_{1})\sum_{j=1}^{n+1}u_{j}\right)\big{/}x_{1}+(C/x_{2}^{2})\left|\int_{0}^{y_{u}(x_{1})}\exp\left(-t\sum_{j=1}^{n+1}u_{j}\right)\,\mathrm{d}t\right|.$

Now note that $\lim_{x\nearrow
m_{+}\pi/\sum_{j=1}^{n+1}u_{j}}y_{u}(x)=+\infty$. If $x_{1}$ is chosen
sufficiently close to the right endpoint of $D_{u}$, we will have
$y_{u}(x_{1})>0$, and then we can write

$\left|\int_{K_{x_{1},x_{2}}}F_{u}(t)\,\mathrm{d}t\right|\leq
C\exp\left(-y_{u}(x_{1})\sum_{j=1}^{n+1}u_{j}\right)\big{/}x_{1}+(C/x_{2}^{2})\int_{0}^{+\infty}\exp\left(-t\sum_{j=1}^{n+1}u_{j}\right)\,\mathrm{d}t.$

Thus

$\lim_{\begin{subarray}{c}x_{1}\nearrow m_{+}\pi/\sum_{j=1}^{n+1}u_{j}\\\
x_{2}\nearrow+\infty\end{subarray}}\left|\int_{K_{x_{1},x_{2}}}F_{u}(t)\,\mathrm{d}t\right|=0;$

i.e. the integral along the fringe $K_{x_{1},x_{2}}$ goes to zero, from which
we conclude

$\lim_{x\nearrow
m_{+}\pi/\sum_{j=1}^{n+1}u_{j}}\int_{-x}^{x}F_{u}(s+iy_{u}(s))\,\left(\mathrm{d}s+i\frac{\mathrm{d}y_{u}(s)}{\mathrm{d}s}\,\mathrm{d}s\right)=\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t$

by Cauchy’s Integral Theorem (this step is where we have shifted the contour
from the real axis to $\gamma_{u}$).

We are not done yet, since the integrand on the left-hand-side of the above
equation still contains an extra term of $F_{u}(s+iy_{u}(s))\cdot
i\frac{\mathrm{d}y_{u}(s)}{\mathrm{d}s}\,\mathrm{d}s$. Since $y_{u}(x)$ is an
even function of $x$, we know that $\frac{\mathrm{d}y_{u}(x)}{\mathrm{d}x}$ is
an odd function of $x$. Additionally, $F_{u}(x+iy_{u}(x))$ is an even function
of $x$, so
$\int_{-x}^{x}F_{u}(s+iy_{u}(s))\frac{\mathrm{d}y_{u}(s)}{\mathrm{d}s}\,\mathrm{d}s=0$
and we may delete the extra term from the integrand:

$\lim_{x\nearrow
m_{+}\pi/\sum_{j=1}^{n+1}u_{j}}\int_{-x}^{x}F_{u}(s+iy_{u}(s))\,\mathrm{d}s=\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t,$

from which it follows that

$\int_{s\in
E_{u}}F_{u}(s+iy_{u}(s))\,\mathrm{d}s=\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t.$
(10)

This proves Theorem 11, as desired. This concludes Case I.

Case II: $\sum_{j=1}^{n+1}u_{j}<0$.
This case is similar to Case I.

Case III: $\sum_{j=1}^{n+1}u_{j}=0$.
In this case, we have that $E_{u}$ is the entire real line $\mathbb{R}$. Then
Equation (9) becomes

$|F_{u}(x+iy)|\leq
C\left/\left(x^{2}+\min_{j=1}^{n+1}\left(\frac{1}{u_{j}}-y\right)^{2}\right)\right.$

for all $x\not=0$. In particular, we have $|F_{u}(x+iy)|\leq C/x^{2}$. For
convenience, let $M>0$ be large enough so that $1/u_{j}\in(-M,M)$ for all $j$,
so that we have $|F_{u}(x+iy)|\leq C/(x^{2}+(|y|-M)^{2})$ if $|y|\geq M$. For
any $x>0$, we may define the path $K_{x}$ to be the line segment from
$x+iy_{u}(x)$ to $x$. As before, we will use $K_{x}$ as a fringe to relate the
two contours ($\gamma_{u}$ and the real axis). We obtain

$\begin{split}\left|\int_{K_{x}}F_{u}(t)\,\mathrm{d}t\right|&\leq\int_{x-i\infty}^{x+i\infty}|F_{u}(t)|\,|\mathrm{d}t|\\\
&=\int_{x-i(M+1)}^{x+i(M+1)}|F_{u}(t)|\,|\mathrm{d}t|+\int_{x-i\infty}^{x-i(M+1)}|F_{u}(t)|\,|\mathrm{d}t|+\int_{x+i(M+1)}^{x+i\infty}|F_{u}(t)|\,|\mathrm{d}t|\\\
&\leq(C/x^{2})(2\cdot(M+1))+\int_{1}^{+\infty}C/(x^{2}+s^{2})\,\mathrm{d}s+\int_{1}^{+\infty}C/(x^{2}+s^{2})\,\mathrm{d}s\end{split}$

where the first term comes from the bound $|F_{u}(x+iy)|\leq C/x^{2}$ for
$|y|\leq M+1$, and the other two terms come from the bound $|F_{u}(x+iy)|\leq
C/(x^{2}+(|y|-M)^{2})$ for $|y|\geq M+1$. Thus

$\lim_{x\nearrow+\infty}\left|\int_{K_{x}}F_{u}(t)\,\mathrm{d}t\right|=0;$

i.e. the integral along the fringe $K_{x}$ goes to zero, from which we
conclude

$\lim_{x\nearrow+\infty}\int_{-x}^{x}F_{u}(s+iy_{u}(s))\,\left(\mathrm{d}s+i\frac{\mathrm{d}y_{u}(s)}{\mathrm{d}s}\,\mathrm{d}s\right)=\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t$

by Cauchy’s Integral Theorem (this step is where we have shifted the contour
from the real axis to $\gamma_{u}$). Arguing as in Case I, we note that
$\int_{-x}^{x}F_{u}(s+iy_{u}(s))\frac{\mathrm{d}y_{u}(s)}{\mathrm{d}s}\,\mathrm{d}s=0$,
so we may delete the $F_{u}(s+iy_{u}(s))\cdot
i\frac{\mathrm{d}y_{u}(s)}{\mathrm{d}s}\,\mathrm{d}s$ term from the integrand,
resulting in

$\lim_{x\nearrow+\infty}\int_{-x}^{x}F_{u}(s+iy_{u}(s))\,\mathrm{d}s=\int_{-\infty}^{+\infty}F_{u}(t)\,\mathrm{d}t,$

which proves Theorem 11, as desired. This concludes Case III.

∎

We have proven that Theorem 11 holds in all cases. This allows us to change
the integral in Theorem 9. We have successfully applied our main idea:
changing the contour of integration.

It remains to show that $\frac{1}{2\pi}\int_{s\in
E_{u}}F_{u}(s+iy_{u}(s))\,\mathrm{d}s\geq\frac{1}{e}$ for each $u$. Defining

$\tilde{F}_{u}(x)\coloneqq F_{u}(x+iy_{u}(x)),$

we wish to show the following:

######  Theorem 12.

For each unit vector $u\in\mathbb{R}^{n+1}$, we have

$\frac{1}{2\pi}\int_{s\in E_{u}}\tilde{F}_{u}(s)\,\mathrm{d}s\geq\frac{1}{e}.$

Equality occurs if $u$ is of form $(1,0,0,0,\dots,0)$.

##  6 A pointwise bound via differential equations

Let us first prove the equality case of Theorem 12. For convenience, define
$v\coloneqq(1)\in\mathbb{R}^{1}$, and note that $E_{v}=(-\pi,\pi)$. We wish to
show that

$\frac{1}{2\pi}\int_{-\pi}^{\pi}F_{v}(s+iy_{v}(s))\,\mathrm{d}s=\frac{1}{e}.$
(11)

We calculate explicitly $F_{v}(t)=e^{it}/(1+it)$ and $y_{v}(x)=1-x\cot x$ when
$x\in(-\pi,\pi)$. It follows that
$|F_{v}(x+iy)|=e^{-y}/\sqrt{(1-y)^{2}+x^{2}}$ whenever $(x,y)\not=(0,1)$.

In order to prove the equality case of Theorem 12, we will use Cauchy’s
Residue Theorem. For $x\in(0,\pi)$, let $P_{x}$ denote the path which is the
line segment from $-x+iy_{v}(x)$ to $x+iy_{v}(x)$. We have

$\left|\int_{P_{x}}F_{v}(t)\,\mathrm{d}t\right|\leq 2x\cdot
e^{-y_{v}(x)}/(y_{v}(x)-1)\leq 2\pi\cdot e^{-y_{v}(x)}/(y_{v}(x)-1)$

whenever $x$ is such that $y_{v}(x)>1$. Since
$\lim_{x\nearrow\pi}y_{v}(x)=+\infty$, we have

$\lim_{x\nearrow\pi}\left|\int_{P_{x}}F_{v}(t)\,\mathrm{d}t\right|=0.$

Thus we can write

$\begin{split}\frac{1}{2\pi}\int_{-\pi}^{\pi}F_{v}(s+iy_{v}(s))\,\mathrm{d}s&=\lim_{x\nearrow\pi}\frac{1}{2\pi}\int_{-x}^{x}F_{v}(s+iy_{v}(s))\,\mathrm{d}s\\\
&=\lim_{x\nearrow\pi}\frac{1}{2\pi}\int_{-x}^{x}F_{v}(s+iy_{v}(s))\,\left(\mathrm{d}s+i\frac{\mathrm{d}y_{v}(s)}{\mathrm{d}s}\,\mathrm{d}s\right)\\\
&=\lim_{x\nearrow\pi}\frac{1}{2\pi}\int_{\text{follow }\gamma_{v}\text{ from
}-x+iy_{v}(-x)\text{ to }x+iy_{v}(x)}F_{v}(t)\,\mathrm{d}t\\\
&=\lim_{x\nearrow\pi}\frac{1}{2\pi}\int_{\begin{subarray}{c}\text{follow
}\gamma_{v}\text{ from }-x+iy_{v}(-x)\text{ to }x+iy_{v}(x)\\\ \text{then
follow }P_{x}\text{ from }x+iy_{v}(x)\text{ back to
}-x+iy_{v}(-x)\end{subarray}}F_{v}(t)\,\mathrm{d}t.\\\ \end{split}$

Since $F_{v}(t)$ has a simple pole at $t=i$, we can use the Cauchy Residue
Theorem to obtain

$\frac{1}{2\pi}\int_{-\pi}^{\pi}F_{v}(s+iy_{v}(s))\,\mathrm{d}s=\lim_{x\nearrow\pi}\frac{1}{2\pi}2\pi
i\cdot\operatorname{\mathrm{Res}}(F_{v},i)=\lim_{x\nearrow\pi}e^{-1}=e^{-1},$

proving Equation (11), and thus proving the equality case of Theorem 12.

Now that we know that $v$ achieves the equality case in Theorem 12, we can
prove Theorem 12 just by showing $\int_{s\in
E_{u}}\tilde{F}_{u}(s)\,\mathrm{d}s\geq\int_{s\in
E_{v}}\tilde{F}_{v}(s)\,\mathrm{d}s$ for each $u$. Note that for any $u$, we
have $E_{v}=(-\pi,\pi)\subseteq E_{u}$. Thus we can write

$\int_{s\in E_{u}}\tilde{F}_{u}(s)\,\mathrm{d}s\geq\int_{s\in
E_{v}}\tilde{F}_{u}(s)\,\mathrm{d}s$

since the integrand is positive. It follows that it suffices to prove
$\int_{s\in E_{v}}\tilde{F}_{u}(s)\,\mathrm{d}s\geq\int_{s\in
E_{v}}\tilde{F}_{v}(s)\,\mathrm{d}s$, which would be implied by the following
pointwise bound:

######  Theorem 13.

For each unit vector $u\in\mathbb{R}^{n+1}$, and for any $x\in(-\pi,\pi)$, we
have $\tilde{F}_{u}(x)\geq\tilde{F}_{v}(x)$.

The remainder of this section will be devoted to proving Theorem 13.

We first note that we only have to prove Theorem 13 when $x\in(0,\pi)$ (since
$\tilde{F}_{u}(0)=\tilde{F}_{v}(0)=1$, and since $\tilde{F}_{u}(x)$,
$\tilde{F}_{v}(x)$ are both even functions of $x$).

It suffices to show that $\log\tilde{F}_{u}(x)\geq\log\tilde{F}_{v}(x)$ for
all $x\in(0,\pi)$. To do this, we will show that

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log\tilde{F}_{u}(x)\right)\geq\frac{\mathrm{d}}{\mathrm{d}x}\left(\log\tilde{F}_{v}(x)\right)$
(12)

holds for every $x\in(0,\pi)$; this suffices because
$\tilde{F}_{u}(0)=\tilde{F}_{v}(0)=1$ as noted above.

From now on, we suppress the variable $x$ when writing $y_{u}$, so we write
simply $y_{u}$ and $y_{v}$ instead of $y_{u}(x)$ and $y_{v}(x)$. Now we will
give a differential equation that $y_{u}$ must satisfy. For all $x\in D_{u}$,
we have $\Phi_{u}(x,y_{u})=0$, which implies

$\sum_{j:u_{j}>0}\operatorname{\mathrm{arccot}}\frac{\frac{1}{u_{j}}-y_{u}}{x}-\sum_{j:u_{j}<0}\operatorname{\mathrm{arccot}}\frac{-(\frac{1}{u_{j}}-y_{u})}{x}=x\sum_{j=1}^{n+1}u_{j}.$

Differentiating with respect to $x$, we obtain

$\sum_{j=1}^{n+1}\left(-\frac{-\frac{\frac{1}{u_{j}}-y_{u}}{x^{2}}+\frac{-y_{u}^{\prime}}{x}}{1+\left(\frac{\frac{1}{u_{j}}-y_{u}}{x}\right)^{2}}\right)=\sum_{j=1}^{n+1}u_{j}$

where $y_{u}^{\prime}$ denotes $\frac{\mathrm{d}}{\mathrm{d}x}y_{u}$. We can
further simplify to

$\sum_{j=1}^{n+1}\frac{\frac{1}{u_{j}}-y_{u}+xy_{u}^{\prime}}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}=\sum_{j=1}^{n+1}u_{j}.$

Rearranging terms yields

$\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\cdot
y_{u}^{\prime}=\sum_{j=1}^{n+1}\frac{-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}$
(13)

which is our desired differential equation, valid for all $x\in D_{u}$. In
particular, by plugging in $v$ for $u$, we obtain

$y_{v}^{\prime}=\frac{-y_{v}+(x^{2}+y_{v}^{2})}{x},$

valid for all $x\in(0,\pi)$.

From Equation (13) and the fact that each $u_{j}\in[-1,1]$, we obtain

$\begin{split}\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\cdot
y_{u}^{\prime}&\leq\sum_{j=1}^{n+1}\frac{-y_{u}+1\cdot(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\\\
&=\sum_{j=1}^{n+1}\frac{-y_{u}+(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\\\
\end{split}$

from which it follows that

$y_{u}^{\prime}\leq\frac{-y_{u}+(x^{2}+y_{u}^{2})}{x}$ (14)

for all $x\in D_{u}$.

We must first prove an auxiliary lemma relating $y_{u}$ and $y_{v}$:

######  Lemma 14.

For each $x\in(-\pi,\pi)$, we have $-y_{v}\leq y_{u}\leq y_{v}$.

###### Proof.

It suffices to prove that $y_{u}\leq y_{v}$ for all $x\in(0,\pi)$ (because we
can obtain the other cases by replacing $x$ with $-x$ and by replacing $u$
with $-u$).

Recall the $\Phi$ function from Equations (6) and (7). (As a reminder,
$\Phi_{u}(x,y)$ is a function taking values in $\mathbb{R}$, defined so that
$\Phi_{u}(x,y)$ always agrees with the argument of $F_{u}(x+iy)$ modulo
$2\pi$.) Note that for each $u$, we have that $\Phi_{u}$ is a
$\mathcal{C}^{\infty}$ function of $(x,y)$ (in fact, it is a real analytic
function).

If we plug in $v$ for $u$ in Equation (6), we obtain

$\Phi_{v}(x,y)=x-\operatorname{\mathrm{arccot}}\frac{1-y}{x}$

for all $(x,y)\in\mathbb{R}^{2}$ satisfying $x>0$. Furthermore, we note that
by the definition of $y_{v}$, we have that $\Phi_{v}$ vanishes along
$\gamma_{v}$ (see Equation (8)).

Our general proof strategy will be to show that $\Phi_{v}$ is nondecreasing
along $\gamma_{u}$. Combined with the fact that $\Phi_{v}(x,y)$ is a strictly
decreasing function of $y$, we can conclude that $y_{u}\leq y_{v}$ for all
$x\in(0,\pi)$, as desired.

We will have to take the derivative
$\frac{\mathrm{d}}{\mathrm{d}x}(\Phi_{v}(x,y_{u}(x)))$ and show that it is
nonnegative.

Note that $\frac{\partial}{\partial
x}\Phi_{v}(x,y)=1-\frac{1-y}{x^{2}+(1-y)^{2}}$ and $\frac{\partial}{\partial
y}\Phi_{v}(x,y)=\frac{-x}{x^{2}+(1-y)^{2}}$. Now, for $x\in D_{u}$, we compute

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}x}(\Phi_{v}(x,y_{u}(x)))&=\left(\frac{\partial}{\partial
x}\Phi_{v}\right)(x,y_{u})+y_{u}^{\prime}\cdot\left(\frac{\partial}{\partial
y}\Phi_{v}\right)(x,y_{u})\\\
&=1-\frac{1-y_{u}}{x^{2}+(1-y_{u})^{2}}+y_{u}^{\prime}\cdot\frac{-x}{x^{2}+(1-y_{u})^{2}}\end{split}$

and since $\frac{-x}{x^{2}+(1-y_{u})^{2}}$ is negative, we may use Equation
(14) to write

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}x}(\Phi_{v}(x,y_{u}(x)))&\geq
1-\frac{1-y_{u}}{x^{2}+(1-y_{u})^{2}}+\frac{-y_{u}+(x^{2}+y_{u}^{2})}{x}\cdot\frac{-x}{x^{2}+(1-y_{u})^{2}}\\\
&=1-\frac{1-y_{u}}{x^{2}+(1-y_{u})^{2}}+\frac{y_{u}-x^{2}-y_{u}^{2}}{x^{2}+(1-y_{u})^{2}}\\\
&=1+\frac{-1+2y_{u}-y_{u}^{2}-x^{2}}{x^{2}+(1-y_{u})^{2}}\\\ &=0.\end{split}$

We conclude that when $x\in D_{u}$, the function $\Phi_{v}(x,y_{u}(x))$ is a
nondecreasing function of $x$. Since
$\lim_{x\searrow 0}y_{u}(x)=0$, we obtain $\lim_{x\searrow
0}\Phi_{v}(x,y_{u}(x))=0$, from which it follows that

$\Phi_{v}(x,y_{u}(x))\geq 0$

for all $x\in D_{u}$.

As observed above, we have $\Phi_{v}(x,y_{v}(x))=0$ for all $x\in(0,\pi)$. So

$\Phi_{v}(x,y_{u}(x))\geq\Phi_{v}(x,y_{v}(x))$

for all $x\in(0,\pi)$. Since $\Phi_{v}(x,y)$ is a strictly decreasing function
of $y$, we must have

$y_{u}(x)\leq y_{v}(x),$

as desired. ∎

We turn our attention to Equation (12). For $x\in D_{u}$, we compute

$\begin{split}&\qquad\frac{\mathrm{d}}{\mathrm{d}x}\tilde{F}_{u}(x)\\\
&=\tilde{F}_{u}(x)\cdot\left(\sum_{j=1}^{n+1}iu_{j}\left(1+iy_{u}^{\prime}\right)-\frac{iu_{j}(1+iy_{u}^{\prime})}{1+iu_{j}(x+iy_{u})}\right)\\\
&=\tilde{F}_{u}(x)\cdot\sum_{j=1}^{n+1}\frac{-u_{j}^{2}(x+iy_{u})(1+iy_{u}^{\prime})}{1+iu_{j}(x+iy_{u})}\\\
&=\tilde{F}_{u}(x)\cdot\left(\sum_{j=1}^{n+1}\frac{-u_{j}^{2}(x+iy_{u})}{1-u_{j}y_{u}+iu_{j}x}+y_{u}^{\prime}\cdot\sum_{j=1}^{n+1}\frac{-iu_{j}^{2}(x+iy_{u})}{1-u_{j}y_{u}+iu_{j}x}\right)\\\
&=\tilde{F}_{u}(x)\cdot\left(\sum_{j=1}^{n+1}\frac{-x+i(-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2}))}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}+y_{u}^{\prime}\cdot\sum_{j=1}^{n+1}\frac{-(-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2}))-ix}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)\\\
&=\tilde{F}_{u}(x)\cdot\left(\sum_{j=1}^{n+1}\frac{-x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}+y_{u}^{\prime}\cdot\sum_{j=1}^{n+1}\frac{-(-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2}))}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)\\\
\end{split}$

where the last step follows from Equation (13) (namely, we cancelled the
imaginary part). Using Equation (13) again, we may write

$\frac{\mathrm{d}}{\mathrm{d}x}\tilde{F}_{u}(x)=\tilde{F}_{u}(x)\cdot(-1)\cdot\frac{\left(\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}+\left(\sum_{j=1}^{n+1}\frac{-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}}{\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}}$

and thus

$\frac{\mathrm{d}}{\mathrm{d}x}\log\tilde{F}_{u}(x)=-\frac{\left(\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}+\left(\sum_{j=1}^{n+1}\frac{-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}}{\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}}$
(15)

holds for all $x\in D_{u}$.

In particular, by substituting $u=v$ we obtain

$\frac{\mathrm{d}}{\mathrm{d}x}\log\tilde{F}_{v}(x)=-\frac{x^{2}+y_{v}^{2}}{x}$
(16)

for all $x\in(0,\pi)$.

We are ready to use Equations (15) and (16) to prove Equation (12). The crux
is to use the Cauchy-Schwarz inequality to obtain the following curious
inequalities:

$\begin{split}\left(\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}&=\left(\sum_{j=1}^{n+1}\frac{(x/u_{j})\cdot
u_{j}}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}\\\
&\leq\left(\sum_{j=1}^{n+1}\frac{(x/u_{j})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}\right)\left(\sum_{j=1}^{n+1}u_{j}^{2}\right)\\\
&=\sum_{j=1}^{n+1}\frac{(x/u_{j})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}\end{split}$
(17)

and

$\begin{split}\left(\sum_{j=1}^{n+1}\frac{-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}&=\left(\sum_{j=1}^{n+1}\frac{(-y_{u}/u_{j}+x^{2}+y_{u}^{2})\cdot
u_{j}}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}\\\
&\leq\left(\sum_{j=1}^{n+1}\frac{(-y_{u}/u_{j}+x^{2}+y_{u}^{2})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}\right)\left(\sum_{j=1}^{n+1}u_{j}^{2}\right)\\\
&=\sum_{j=1}^{n+1}\frac{(-y_{u}/u_{j}+x^{2}+y_{u}^{2})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}.\end{split}$
(18)

Thus

$\begin{split}&\left(\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}+\left(\sum_{j=1}^{n+1}\frac{-y_{u}+u_{j}\cdot(x^{2}+y_{u}^{2})}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}\right)^{2}\\\
&\leq\sum_{j=1}^{n+1}\frac{(x/u_{j})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}+\sum_{j=1}^{n+1}\frac{(-y_{u}/u_{j}+x^{2}+y_{u}^{2})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}\\\
&=\sum_{j=1}^{n+1}\frac{(x/u_{j})^{2}+(-y_{u}/u_{j}+x^{2}+y_{u}^{2})^{2}}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}\\\
&=\sum_{j=1}^{n+1}\frac{(x^{2}+y_{u}^{2})\cdot\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)}{\left(x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}\right)^{2}}\\\
&=\sum_{j=1}^{n+1}\frac{x^{2}+y_{u}^{2}}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}.\end{split}$

Plugging this into Equation (15) yields

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}x}\log\tilde{F}_{u}(x)&\geq-\frac{\sum_{j=1}^{n+1}\frac{x^{2}+y_{u}^{2}}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}}{\sum_{j=1}^{n+1}\frac{x}{x^{2}+\left(\frac{1}{u_{j}}-y_{u}\right)^{2}}}\\\
&=-\frac{x^{2}+y_{u}^{2}}{x}\\\ &\geq-\frac{x^{2}+y_{v}^{2}}{x}\\\
\end{split}$

for all $x\in(0,\pi)$ because $-y_{v}\leq y_{u}\leq y_{v}$ holds for all
$x\in(0,\pi)$ (from Lemma (14)). Combining this with Equation (16) yields

$\frac{\mathrm{d}}{\mathrm{d}x}\log\tilde{F}_{u}(x)\geq\frac{\mathrm{d}}{\mathrm{d}x}\log\tilde{F}_{v}(x)$

which is exactly Equation (12), as desired. This proves Theorem 13.

##  7 A final question

We showed that $H_{\text{facet}}$ is the smallest central slice, but only up
to a factor of $1-o(1)$.

######  Question 15.

Can we remove the factor of $1-o(1)$, so that we can say that
$\operatorname{vol}_{n-1}(H_{\mathrm{facet}}\cap\Delta_{n})$ is exactly the
minimum value of a central section?

##  8 Acknowledgements

I thank Tomasz Tkocz (Carnegie Mellon University) for introducing me to this
problem, and for many stimulating discussions.

I was supported by the ARCS Foundation Scholar Award.

I relied on Wolfram Mathematica Student Edition to test various hypotheses,
and also to create some of the figures in this paper.

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