# Bounds for Periodic Energy and the Optimality of Two Periodic Point
Configurations in the Plane

Douglas P. Hardin  Department of Mathematics, Vanderbilt University  Nathaniel
Tenpas  Department of Mathematics, Vanderbilt University

(July 28, 2023)

###### Abstract

We develop lower bounds for the energy of configurations in $\mathbb{R}^{d}$
periodic with respect to a lattice. In certain cases, the construction of
sharp bounds can be formulated as a finite dimensional, multivariate
polynomial interpolation problem. We use this framework to show a scaling of
the equitriangular lattice $A_{2}$ is universally optimal among all
configurations of the form $\omega_{4}+A_{2}$ where $\omega_{4}$ is a 4-point
configuration in $\mathbb{R}^{2}$. Likewise, we show a scaling and rotation of
$A_{2}$ is universally optimal among all configurations of the form
$\omega_{6}+L$ where $\omega_{6}$ is a 6-point configuration in
$\mathbb{R}^{2}$ and $L=\mathbb{Z}\times\sqrt{3}\mathbb{Z}$.

###### Contents

  1. 1 Introduction and Overview of Results
  2. 2 Lattices and Linear Programming Bounds for Periodic Energy
    1. 2.1 Preliminaries: Lattices and Fourier Series
    2. 2.2 Lattice symmetry, symmetrized basis functions, and polynomial structure
    3. 2.3 Linear Programming Bounds for Periodic Energy
    4. 2.4 Moments for certain lattice configurations
    5. 2.5 Lattice theta functions
    6. 2.6 Polynomial interpolation and linear programming bounds for lattice configurations
  3. 3 The Linear Programming Framework for the families $\omega_{m^{2}},\omega_{2m^{2}},\omega_{3m^{2}},$ and $\omega_{6m^{2}}$
    1. 3.1 The Polynomials $P^{L}_{v}$ and $P^{A_{2}}_{v}$
    2. 3.2 Interpolation Nodes
    3. 3.3 Interpolation Problem for $\omega_{m^{2}}$
    4. 3.4 Interpolation Problem for $\omega^{*}_{2m^{2}}$
    5. 3.5 Interpolation Problems for $\omega^{*}_{3m^{2}}$
    6. 3.6 Interpolation Problems for $\omega^{*}_{6m^{2}}$
  4. 4 Universal Optimality of $\omega^{*}_{4}$
    1. 4.1 Constructing magic $\tilde{g}_{a}$
      1. 4.1.1 Small $a$
      2. 4.1.2 Large $a$
  5. 5 Universal Optimality of $\omega^{*}_{6}$
    1. 5.1 Verifying properties of $\tilde{g}$
    2. 5.2 Necessary Conditions Fulfilled and $\tilde{F}\geq\tilde{g}$ when $|t_{2}|\in[1/2,1]$
    3. 5.3 The critical region for small $a$ ($a<\pi^{2}$)
      1. 5.3.1 Proving the Linear Approximation Bound
    4. 5.4 The critical region for $a\geq 9.6$
  6. A Divided differences and univariate interpolation
  7. B Technical Estimates and Computations for $a<\pi^{2}$
    1. B.1 $\theta$ estimates
    2. B.2 $\tilde{F}\geq\tilde{g}$ for small $a$ and 4 points
    3. B.3 $\tilde{F}\geq\tilde{g}$ for small $a$ and 6 points
      1. B.3.1 Satisfying Necessary Conditions
      2. B.3.2 Computations for the Linear Approximation Bound
  8. C Technical Estimates and Computations for large $a$
    1. C.1 Basic Lemmas and Other Estimates
    2. C.2 Intermediate $a$ and 4 points
      1. C.2.1 calculations for Lemma 19
      2. C.2.2 Calculations for Lemma 17
    3. C.3 Large $a$ and 4 points
    4. C.4 Large $a$ and 6 points
      1. C.4.1 Coefficient Bounds
      2. C.4.2 Satisfying Necessary Conditions for large $a$
      3. C.4.3 Bounds for proofs of lemmas 31 and 34
      4. C.4.4 computations for proof of Lemma 32
      5. C.4.5 Positivity of $L_{1}$ for Lemma 33
      6. C.4.6 Log Derivative Estimates for the proof of Lemma 35
  9. D Equivalence of Different Notions of Universal Optimality

##  1 Introduction and Overview of Results

Let $\Lambda$ be a lattice in $\mathbb{R}^{d}$ (see subsection 2.1 for more
about lattices) and let $F:\mathbb{R}^{d}\rightarrow(-\infty,\infty]$ be
lower-semicontinuous and $\Lambda$-periodic (i.e., $F(\cdot+v)=F$ for all
$v\in\Lambda$). We refer to $\Lambda$ as the periodization lattice and $F$ as
the periodization kernel. For a finite multiset
$\omega_{n}=\\{x_{1},...,x_{n}\\}\subseteq\mathbb{R}^{d}$ of cardinality $n$,
we consider the $F$-energy of $\omega_{n}$ defined by

$E_{F}(\omega_{n}):=\sum_{i=1}^{n}\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{n}F(x_{i}-x_{j}).$

Without loss of generality, we may assume that $\omega_{n}$ lies in some
specified fundamental domain $\Omega_{\Lambda}:=\mathbb{R}^{d}/\Lambda$ since
replacing a point $x\in\omega_{n}$ with any point in $x+\Lambda$ does not
change $E_{F}(\omega_{n})$.

The minimal discrete $n$-point $F$-energy is defined as

$\mathcal{E}_{F}(n):=\inf\\{E_{F}(\omega_{n})\mid\omega_{n}\subseteq\mathbb{R}^{d},\,\lvert\omega_{n}\rvert=n\\},$
(1)

where $\lvert\omega\rvert$ denotes the cardinality of a multiset $\omega$. An
$n$-point configuration $\omega_{n}\subset\mathbb{R}^{d}$ satisfying
$E_{F}(\omega_{n})=\mathcal{E}_{F}(n)$ is called $F$-optimal. Note that the
lower-semicontinuity of $F$ and the compactness of $\Omega_{\Lambda}$ in the
torus topology implies the existence of at least one $F$-optimal
configuration.

More specifically, we consider periodic potentials generated by a function
$f:[0,\infty)\to[0,\infty]$ with $d$-rapid decay (i.e.
$f(r^{2})\in\mathcal{O}(r^{-s}),r\rightarrow\infty$, for some $s>d$) using

$F_{f,\Lambda}(x):=\sum_{v\in\Lambda}f(|x+v|^{2}).$ (2)

The potential $F_{f,\Lambda}$ has the following physical interpretation: if
$f(r^{2})$ represents the energy required to place a pair of unit charge
particles a distance $r$ from each other, then $F_{f,\Lambda}(x)$ is the
energy required to place such a particle at the point $x$ in the presence of
existing particles at points of $\Lambda$. We write the pair interaction in
terms of the distance squared since it will be more compatible with the notion
of universal optimality described below.

The periodization of Gaussian potentials $f_{a}(r^{2}):=\exp(-ar^{2})$ for
$a>0$ leads to a type of lattice theta function (cf. [BHS19, Chapter 10]) and
plays a central role in our analysis. For convenience, we write
$F_{a,\Lambda}:=F_{f_{a},\Lambda}$ or just $F_{a}$ when the choice of lattice
is unambiguous.

######  Definition 1.1.

Let $\Lambda$ be a lattice in $\mathbb{R}^{d}$.

  * •

We say that an $n$-point configuration $\omega_{n}\subset\mathbb{R}^{d}$ is
$\Lambda$-universally optimal if it is $F_{a,\Lambda}$-optimal for all $a>0$
(cf., [CK07]).

  * •

We say $\Lambda$ is universally optimal if for any sublattice
$\Phi\subseteq\Lambda$ of index $n$, the $n$-point configuration
$\Lambda\cap\Omega_{\Phi}$ is $\Phi$-universally optimal.

If $\omega_{n}$ is $\Lambda$-universally optimal, then it follows from a
theorem of Bernstein [Ber29] (see [CK07],[BHS19]) that $\omega_{n}$ is
$F_{f,\Lambda}$-optimal for any $f$ with $d$-rapid decay that is completely
monotone on $(0,\infty)$.111Recall that a function $g$ is completely monotone
on an interval $I$ if $(-1)^{n}g^{(n)}\geq 0$ on $I$ for all positive integers
$n$.

As we discuss in the Appendix, it follows from classical results of [Fis64]
that a lattice is universally optimal in the sense of Definition 1.1 if and
only if it is universally optimal in the more general sense of [CK07] or
[Coh+22], which we describe at the end of the section.

We further show that to establish the universal optimality of $\Lambda$, it is
sufficient to prove that there is some sublattice $\Phi\subseteq\Lambda$ such
that $\Lambda\cap\Omega_{m\Phi}$ is $m\Phi$-universally optimal for infinitely
many $m\in\mathbb{N}$. Observing that the notion of lattice universal
optimality in Definition 1.1 is invariant under linear similitudes, we find it
convenient to consider the $\Phi$-universal optimality of configurations of
the form

$\omega(\Phi,\Lambda,m)=\left(\frac{1}{m}\Lambda\right)\cap\Omega_{\Phi}$ (3)

for a sublattice $\Phi$ of a lattice $\Lambda$.

It was shown in [CK07] that $\mathbb{Z}$ is universally optimal in
$\mathbb{R}$ (see [Su15] for a simple, alternate proof). Recently it was shown
in [Coh+22] that the $E_{8}$ and Leech lattices are universally optimal in
dimensions 8 and 24, respectively. It was further conjectured that the
hexagonal $A_{2}$ lattice

$A_{2}:=\begin{bmatrix}1&1/2\\\ 0&\sqrt{3}/2\end{bmatrix}\mathbb{Z}^{2},$

is universally optimal in $\mathbb{R}^{2}$. Though $A_{2}$ has long been known
to be optimal for circle packing (see [FT40]) and was proved to be universally
optimal among lattices in [Mon88], surprisingly, its conjectured universal
optimality among all infinite configurations (of fixed density) remains open.

The proofs of universal optimality for $\mathbb{Z}$, $E_{8}$, and the Leech
lattice given in [CK07] and [Coh+22] are based on so-called “linear
programming bounds” originally developed in the context of coding theory for
point configurations on the $d$-dimensional sphere (e.g., see [DGS77],
[Lev98], [Yud92]) and extended to bounds for the energy and sphere-packing
density of point configurations in $\mathbb{R}^{d}$ in (e.g., see [CE03],
[CK07], and [Coh+22]). In fact, these linear programming bounds for point
configurations in $\mathbb{R}^{d}$ are first proved for general periodic
configurations, which we adapt for a fixed periodization lattice $\Lambda$ in
Proposition 7.

In Section 2, we formulate linear programming bounds for lattice periodic
configurations in $\mathbb{R}^{d}$ and find sufficient conditions on the
periodization lattice to permit a certain polynomial structure (see
Proposition 6). Furthermore, we develop conditions for the $F$-optimality of
configurations of the form (3) in terms of polynomial interpolation.

We apply this framework to the following four families of configurations
(arising from scalings of $A_{2}$) using the notation of (3).

  1. (a)

$\Phi=A_{2}$ and

$\omega^{*}_{m^{2}}:=\omega(\Phi,A_{2},m),$

conjectured to be $A_{2}$-universally optimal.

  2. (b)

$\Phi=L$ and

$\omega^{*}_{2m^{2}}:=\omega(\Phi,A_{2},m),$

conjectured to be $L$-universally optimal.

  3. (c)

$\Phi=\sqrt{3}R_{\pi/6}A_{2}$ (where $R_{\pi/6}$ denotes the $\pi/6$ rotation
matrix) and

$\omega^{*}_{3m^{2}}:=\frac{1}{\sqrt{3}}R_{-\pi/6}\omega(\Phi,A_{2},m),$

conjectured to be $A_{2}$-universally optimal (note
$\frac{1}{\sqrt{3}}R_{-\pi/6}\Phi=A_{2}$).

  4. (d)

$\Phi=\sqrt{3}R_{\pi/6}L$ and

$\omega^{*}_{6m^{2}}:=\frac{1}{\sqrt{3}}R_{-\pi/6}\omega(\Phi,A_{2},m),$

conjectured to be $L$-universally optimal (note
$\frac{1}{\sqrt{3}}R_{-\pi/6}\Phi=L$).

The universal optimality of the configurations $\omega_{\kappa m^{2}}^{*}$ for
$\kappa\in\\{1,2,3,6\\}$ is an immediate consequence of the conjectured
universal optimality of $A_{2}$ should this be true. Conversely, as discussed
above, the universal optimality of $A_{2}$ would follow if analogous results
are established for any of the four families
$\omega^{*}_{m^{2}},\omega^{*}_{2m^{2}},\omega^{*}_{3m^{2}},\omega^{*}_{6m^{2}}$
for infinitely many $m\in\mathbb{N}$.

The proofs of universal optimality of two of the base cases, $\omega_{2}^{*}$
and $\omega_{3}^{*}$, follow immediately from results on theta functions, some
classical and some from [Bae97] (cf. [Su15] or [FSZ23] for proofs in the
context of periodic energy). Our main results are proofs, utilizing the linear
programming bounds, of the universal optimality of the next two base cases,
$\omega_{4}^{*}$ and $\omega_{6}^{*}$:

######  Theorem 1.

The configurations $\omega_{4}^{*}$ and $\omega_{6}^{*}$ are $A_{2}$ and
$L$-universally optimal, respectively.

Figure 1: Points of $A_{2}/2$ are shown with the 4 larger points yielding an
$A_{2}$ universally optimal configuration which we denote $\omega^{*}_{4}$

Figure 2: The 6 larger points form an $L$-universally optimal configuration
which we denote $\omega^{*}_{6}$.

We can rephrase Theorem 1 in terms of the energies of infinite configurations,
for which we follow the notation of [Coh+22]. Let $B(x,r)$ be the ball of
radius $r>0$ centered at $x$. If $C$ is an infinite, multiset in
$\mathbb{R}^{d}$ such that every ball intersects finitely many points, we call
it an infinite configuration. Define $C_{r}:C\cap B(0,r)$ and the density of
$C$ as

$\lim_{r\rightarrow\infty}\frac{|C_{r}|}{{\rm Vol}(B(0,r))},$

assuming the limit exists and is finite. Similarly to above, for a lower semi-
continuous map $f:[0,\infty)\rightarrow[0,\infty]$ of $d$-rapid decay, we
define the $f$-energy of an $n$-point configuration
$\omega_{n}=\\{x_{1},\dots,x_{n}\\}\subseteq\mathbb{R}^{d}$ as

$E_{f}(\omega_{n}):=\sum_{\begin{subarray}{c}1\leq i,j\leq n\\\ i\neq
j\end{subarray}}f(|x_{i}-x_{j}|^{2}).$

and the infimum over all $n$-point configurations contained in some set $X$ as
$\mathcal{E}_{f}(n,X)$, where we extend the definition to $n\not\in\mathbb{N}$
via linear interpolation. Then for a configuration $C$ of density $\rho$, the
lower $f$-energy of $C$ is

$E^{l}_{f}(C):=\liminf_{r\rightarrow\infty}\frac{E_{f}(C_{r})}{|C_{r}|}.$

If the limit exists, we’ll write it as $E_{f}(C)$ and call it the $f$-energy
of $C$. A configuration $C^{\prime}$ of density $\rho$ is universally optimal
if its $f$-energy exists and satisfies

$E_{f}(C^{\prime})\leq E^{l}_{f}(C)$

for every other infinite configuration $C$ of density $\rho$ and
$f:x\rightarrow e^{-ax^{2}}$. Similarly, a configuration $C^{\prime}$ is
universally optimal among $S$ if we further restrict $C$ to elements of $S$.
We’ll also say an infinite configuration $C$ is an $N$-point
$\Lambda$-periodic configuration if

$C=\cup_{i=1}^{N}x_{i}+\Lambda$

for some set of representatives $\omega_{N}^{C}=\\{x_{1},\dots,x_{n}\\}$. Then
we have the following connection between the $f$-energy of $C$, and the
$F_{f,\Lambda}$ energy of $\omega_{N}^{C}$ (cf. [CK07, Lemma 9.1] or [BHS19,
Chapter 10]).

######  Proposition 2.

Let $C$ be an $N$-point $\Lambda$-periodic configuration with $\omega_{N}^{C}$
a set of representatives, and $f:x^{2}\rightarrow e^{-ax^{2}}$ for some $a>0$.
Recall $F_{a,\Lambda}$ is the $\Lambda$-periodization of $f$. Then $E_{f}(C)$
exists and

$E_{f}(C)=\frac{1}{N}\left(E_{F,\Lambda}(\omega_{N}^{X})+N\sum_{0\neq
v\in\Lambda}f(v)\right).$

Thus, Theorem 1 can be restated as follows: $A_{2}/2$ is universally optimal
among all 4-point $A_{2}$-periodic configurations, and a rotation and scaling
of $A_{2}$ is universal among all 6-point $L$-periodic configurations.

##  2 Lattices and Linear Programming Bounds for Periodic Energy

###  2.1 Preliminaries: Lattices and Fourier Series

We first gather some basic definitions and properties of lattices in
$\mathbb{R}^{d}$.

######  Definition 2.1.

Let $\Lambda\subset\mathbb{R}^{d}$.

  * •

$\Lambda$ is a lattice in $\mathbb{R}^{d}$ if
$\Lambda:=V\mathbb{Z}^{d}=\left\\{\sum_{i=1}^{d}a_{i}v_{i}\mid
a_{1},a_{2},\ldots,a_{d}\in\mathbb{Z}\right\\}$ for some nonsingular $d\times
d$ matrix $V$ with columns $v_{1},\ldots v_{d}$. We refer to $V$ as a
generator for $\Lambda$.

  * •

Once a choice of generator $V$ is specified, we let
$\Omega_{\Lambda}:=V\mathbb{[}0,1)^{d}$ denote the parallelepiped fundamental
domain for $\Lambda$. The co-volume of $\Lambda$ defined by
$\left|\Lambda\right|:=|\det V|$ is the volume of $\Omega_{\Lambda}$ which is,
in fact, the same for any Lebesgue measurable fundamental domain222A
fundamental domain for a group $G$ acting on a set $X$ is a subset of $X$
consisting of exactly one point from each $G$-orbit. Note that $X/G$ will be
used to denote both a fundamental domain and the set of $G$ orbits in $X$. for
$\mathbb{R}^{d}/\Lambda$ where $\Lambda$ acts on $\mathbb{R}^{d}$ by
translation.

  * •

The dual lattice $\Lambda^{*}$ of a lattice $\Lambda$ with generator $V$ is
the lattice generated by $V^{-T}=(V^{T})^{-1}$ or, equivalently,
$\Lambda^{*}:=\\{v\in\mathbb{R}^{d}\mid w\cdot v\in\mathbb{Z}\text{ for all
}w\in\Lambda\\}.$

  * •

We denote by $S_{\Lambda}$ the symmetry group of $\Lambda$ consisting of
isometries on $\mathbb{R}^{d}$ fixing $\Lambda$ and denote by $G_{\Lambda}$
the subgroup of $S_{\Lambda}$ fixing the origin (and thus can be considered as
elements of the orthogonal group $O(d)$). Note that
$G_{\Lambda}=S_{\Lambda}/\Lambda$ where we identify $v\in\Lambda$ with the
translation $\cdot+v$. Further, note that $G_{\Lambda^{*}}=G_{\Lambda}$ since
elements of $O(d)$ preserve inner products.

Let $\Lambda$ be a lattice in $\mathbb{R}^{d}$ with generator $V$ and
fundamental domain $\Omega_{\Lambda}$. We let $L^{2}(\Omega_{\Lambda})$ denote
the Hilbert space of complex-valued $\Lambda$-periodic functions on
$\mathbb{R}^{d}$ with inner product $\langle
f,g\rangle=\int_{\Omega_{\Lambda}}f(x)\overline{g(x)}\,dx$. Then $\\{e^{2\pi
iv\cdot x}\mid v\in\Lambda^{*}\\}$ forms an orthogonal basis of
$L^{2}(\Omega_{\Lambda})$ yielding the Fourier expansion of a function $g\in
L^{2}(\Omega_{\Lambda})$:

$g(x)=\sum_{v\in\Lambda^{*}}\hat{g}_{v}e^{2\pi iv\cdot x}$ (4)

with Fourier coefficients
$\hat{g}_{v}:=\frac{1}{|\Lambda|}\int_{\Omega_{\Lambda}}g(x)$ for
$v\in\Lambda^{*}$ where equality (and the implied unconditional limit on the
right hand side) holds in $L^{2}(\Omega_{\Lambda})$. Of course, elements of
$L^{2}(\Omega_{\Lambda})$ are actually equivalence classes of functions. If
$g\in L^{2}(\Omega_{\Lambda})$ contains an element of $C(\mathbb{R}^{d})$,
then we identify $g$ with its continuous representative and write $g\in
L^{2}(\Omega_{\Lambda})\cap C(\mathbb{R}^{d})$. As will be the case in our
applications, if $g\in L^{2}(\Omega_{\Lambda})$ is such that
$\sum_{v\in\Lambda^{*}}|\hat{g}_{v}|<\infty$, then the right-hand side of (4)
converges uniformly and unconditionally to $g$ and so $g\in
L^{2}(\Omega_{\Lambda})\cap C(\mathbb{R}^{d})$ and (4) holds pointwise for
every $x\in\mathbb{R}^{d}$.

###  2.2 Lattice symmetry, symmetrized basis functions, and polynomial
structure

Let $\Lambda$ be a lattice in $\mathbb{R}^{d}$,
$f:\mathbb{R}^{d}\to\mathbb{R}$ have rapid $d$-decay, and $\sigma\in
G_{\Lambda}$. Since $\sigma^{-1}\in G_{\Lambda}$ and $\sigma$ is an isometry,
we have

$F_{f,\Lambda}(\sigma x)=\sum_{v\in\sigma^{-1}\Lambda}f(\lvert\sigma x+\sigma
v\rvert^{2})=\sum_{v\in\Lambda}f(\lvert x+v\rvert^{2})=F_{f,\Lambda}(x).$

Then $F_{f,\Lambda}$ is also $\Lambda$-periodic and we obtain:

######  Proposition 3.

Suppose $f:[0,\infty)\to[0,\infty]$ has $d$-rapid decay and $\Lambda$ is a
lattice in $\mathbb{R}^{d}$. Then for all $\sigma\in G_{\Lambda}$,
$v\in\Lambda$ and $x\in\mathbb{R}^{2}$, we have $F_{f,\Lambda}(\sigma
x+v)=F_{f,\Lambda}(x)$ showing that $F_{f,\Lambda}$ is
$S_{\Lambda}$-invariant.

We next recall that $g\in L^{2}(\Omega_{\Lambda})$ is $\sigma$-invariant for
$\sigma\in G_{\Lambda}$ if and only if the Fourier coefficients of $g$ are
$\sigma$-invariant, as described in the next proposition.

######  Proposition 4.

Suppose $g\in L^{2}(\Omega_{\Lambda})$ and $\sigma\in G_{\Lambda}$. Then
$g(\sigma x)=g(x)$ for a.e. $x\in\mathbb{R}^{d}$ if and only if
$\hat{g}_{\sigma v}=\hat{g}_{v}$ for all $v\in\Lambda^{*}$.

###### Proof.

Since $\sigma^{-1}\in G_{\Lambda^{*}}=G_{\Lambda}$, we have

$g(\sigma x)=\sum_{v\in\Lambda^{*}}\hat{g}_{v}e^{2\pi iv\cdot(\sigma
x)}=\sum_{v\in\sigma^{-1}\Lambda^{*}}\hat{g}_{v}e^{2\pi i(\sigma
v)\cdot(\sigma x)}=\sum_{v\in\Lambda^{*}}\hat{g}_{\sigma v}e^{2\pi iv\cdot
x}.$

The proposition then follows from uniqueness properties of the Fourier
expansion. ∎

Let $\Gamma$ be a subgroup of $G_{\Lambda}$. For $v\in\Lambda^{*}$, let
$C_{v}^{\Gamma}$ be the $\Lambda$-periodic function defined by

$C_{v}^{\Gamma}(x):=\frac{1}{|\Gamma|}\sum_{\sigma\in\Gamma}e^{2\pi i(\sigma
v)\cdot x}=\frac{1}{|\Gamma(v)|}\sum_{v^{\prime}\in\Gamma(v)}e^{2\pi
iv^{\prime}\cdot x},\quad x\in\mathbb{R}^{d}.$ (5)

where $\Gamma(v)$ denotes the orbit $\Gamma(v)=\\{\sigma
v\mid\sigma\in\Gamma\\}.$ We write $C_{v}$ for $C_{v}^{\Gamma}$ when $\Gamma$
is unambiguous. If $g\in L^{2}(\Omega_{\Lambda})$ and $g$ is
$G_{\Lambda}$-invariant (i.e., if $g(\sigma\cdot)=g$ for all $\sigma\in
G_{\Lambda}$), then we may rewrite (4) as

$g(x)=\sum_{v\in\Lambda^{*}/\Gamma}|\Gamma(v)|\,\hat{g}_{v}\,C_{v}^{\Gamma}(x).$
(6)

We next consider the case of a rectangular lattice by which we mean a lattice
of the form $\Lambda_{R}=(a_{1}\mathbb{Z})\times\cdots(a_{d}\mathbb{Z})$ with
$a_{1},\ldots,a_{d}>0$. The symmetry group of a rectangular lattice in
$\mathbb{R}^{d}$ contains the subgroup $H$ of order $2^{d}$ generated by the
coordinate reflections

$R_{j}(x_{1},\dots,x_{j},\ldots,x_{d})=(x_{1},\dots,-x_{j},\ldots,x_{d}),\qquad
j=1,2,\ldots,d.$ (7)

Let $v\in\Lambda^{*}=(1/a_{1})\mathbb{Z}\times\cdots(1/a_{d})\mathbb{Z}$, and
note that $v=(k_{1}/a_{1},k_{2}/a_{2},\dots,k_{d}/a_{d})$ for some
$k_{1},\dots,k_{d}\in\mathbb{Z}$. A straightforward induction on $d$ gives

$C_{v}^{H}(x)=\prod_{i=1}^{d}\cos(2\pi
k_{i}x_{i}/a_{i}))=\prod_{i=1}^{d}T_{|k_{i}|}(\cos(2\pi x_{i}/a_{i})).$ (8)

Recall the $\ell$th Chebyshev polynomial of the first kind defined by
$\cos(\ell t)=T_{\ell}(\cos t)$ for $\ell=0,1,2,\ldots$. We then have the
following proposition.

######  Proposition 5.

Let $\Lambda_{R}=(a_{1}\mathbb{Z})\times\cdots(a_{d}\mathbb{Z})$ with
$a_{1},\ldots,a_{d}>0$. If $v\in\Lambda_{R}^{*}$, then
$v=(k_{1}/a_{1},k_{2}/a_{2},\dots,k_{d}/a_{d})$ for some
$k_{1},\dots,k_{d}\in\mathbb{Z}$ and

$C_{v}^{H}(x)=\prod_{i=1}^{d}T_{|k_{i}|}(t_{i}),$ (9)

where $t_{i}=\cos(2\pi x_{i}/a_{i})\in[-1,1]$ for $i=1,2,\ldots,d$.

We next deduce a polynomial structure for $C_{v}^{\Lambda}$ for lattices
$\Lambda$ that are invariant under the coordinate reflections $R_{j}$; i.e.,
such that $H\subseteq G_{\Lambda}$.

######  Proposition 6.

Let $\Lambda\subseteq\mathbb{R}^{d}$ be a lattice such that $H\subseteq
G_{\Lambda}$. Then $\Lambda$ contains a rectangular lattice
$\Lambda_{R}=(a_{1}\mathbb{Z})\times\cdots\times(a_{d}\mathbb{Z})$ and the
function $C^{G_{\Lambda}}_{v}(x)$ is a polynomial in the variables
$t_{j}=\cos(2\pi x_{j}/a_{j})$ for $j=1,2,\ldots,d$ and any $v\in\Lambda^{*}$.

###### Proof.

We first show that $\Lambda$ must contain some rectangular sublattice (i.e.,
of the form
$\Lambda_{R}=(a_{1}\mathbb{Z})\times\cdots\times(a_{d}\mathbb{Z})$). Since
$\Lambda$ is full-rank, for each $j=1,2,\ldots,d$, there is some
$w^{j}\in\Lambda$ such that $a_{j}:=2w^{j}\cdot e^{j}\neq 0$ where $e^{j}$
denotes the $j$-th coordinate unit vector. Then
$a_{j}e^{j}=w^{j}-R_{j}w^{j}\in\Lambda$, and so the rectangular lattice
$(a_{1}\mathbb{Z})\times\cdots\times(a_{d}\mathbb{Z})$ is a sublattice of
$\Lambda$.

Let $v\in\Lambda^{*}$. Since $\Lambda_{R}\subseteq\Lambda$,
$\Lambda^{*}\subseteq\Lambda_{R}^{*}$, so $v\in\Lambda_{R}^{*}$. Let
$C=\\{\sigma_{1},\dots,\sigma_{[G_{\lambda}:H]}\\}$ be a set of right coset
representatives of $H$ in $G_{\Lambda}$, so that $|C||H|=|G|$. Then we have

$\begin{split}C^{G_{\Lambda}}_{v}&=\frac{1}{|G_{\Lambda}|}\sum_{g\in
G_{\Lambda}}e^{2\pi igv\cdot x}\\\ &=\frac{1}{|C|}\sum_{\sigma\in
C}\frac{1}{|H|}\sum_{h\in H}e^{2\pi ih\sigma v\cdot x}\\\
&=\frac{1}{|C|}\sum_{\sigma\in C}C^{H}_{\sigma v}.\end{split}$ (10)

Proposition 5 implies $C^{H}_{\sigma v}$ is polynomial in the variables
$t_{j}=\cos(2\pi x_{j}/a_{j})$ and thus so is $C^{G_{\Lambda}}_{v}$. ∎

Proposition 6 motivates the change of variables

$t_{i}:=\cos(2\pi x_{i}/a_{i}),\qquad i=1,...,d.$ (11)

We then let $T_{a_{1},...,a_{d}}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}$ be
defined by

$T_{a_{1},\dots,a_{d}}(x_{1},...,x_{d}):=(t_{1},\dots,t_{d}).$ (12)

For any function $h$ defined on a subset of
$X\subseteq[0,a_{1}/2]\times\cdots[0,a_{d}/2]$, $\tilde{h}$ will refer to the
function defined on $\tilde{X}\subseteq[-1,1]^{d}$ by the formula

$\tilde{h}(t)=h\left(\frac{a_{1}\arccos t_{1}}{2\pi},\dots,\frac{a_{d}\arccos
t_{2}}{2\pi}\right),$

which ensures $\tilde{h}(t)=h(x)$.

It follows by Proposition 6 that the maps

$P^{\Phi}_{v}:=\tilde{C}^{G_{\Phi}}_{v},\qquad v\in\Phi^{*},$ (13)

are polynomials in the variables $t_{i},\dots,t_{d}$. We shall also write
$P_{v}$ when the choice of $\Phi$ is clear. Similarly, the
$T_{a_{1},\dots,a_{d}}$ image of any subset
$D\subseteq[0,1/2]\times[0,\sqrt{3}/2]$ will be denoted $\tilde{D}$. In any
case where we do so, the choice of rectangular lattice (and hence the choice
of $a_{i}$’s will be clear).

We also remark that the collection of polynomials $\\{P_{v}^{G_{\Phi}}\mid
v\in\Phi^{*}/G_{\Phi}\\}$ is orthogonal with respect to the measure
$(1-t_{1}^{2})^{-1/2}\cdots(1-t_{d}^{2})^{-1/2}dt_{1}\,\cdots\,dt_{d}$ on
$[-1,1]^{d}$.

###  2.3 Linear Programming Bounds for Periodic Energy

We say that $g\in L^{2}(\Omega_{\Lambda})$ is positive semi-definite (PSD) if
the Fourier coefficients $\hat{g}_{v}\geq 0$ for all
$v\in\Lambda^{*}\setminus\\{0\\}$ and
$\sum_{v\in\Lambda^{*}}\hat{g}_{v}<\infty$. (This is a slightly nonstandard
choice of notation since we are not requiring $\hat{g}_{0}\geq 0$.) If $g\in
L^{2}(\Omega_{\Lambda})$ is PSD, then its Fourier series converges uniformly
and so, without loss of generality, we may assume that a PSD $g\in
L^{2}(\Omega_{\Lambda})$ is continuous and equal to its Fourier series at
every $x\in\mathbb{R}^{d}$.

If $g\in L^{2}(\Omega_{\Lambda})$ is PSD and $\omega_{n}$ is an arbitrary
$n$-point configuration in $\mathbb{R}^{d}$, then the following fundamental
lower bound holds:

$\begin{split}E_{g}(\omega_{n})&=\sum_{x\neq
y\in\omega_{n}}g(x-y)=-ng(0)+\sum_{x,y\in\omega_{n}}g(x-y)\\\
&=-ng(0)+\sum_{v\in\Lambda^{*}}\hat{g}_{v}\sum_{x,y\in\omega_{n}}e^{2\pi
iv\cdot x}e^{-2\pi iv\cdot y}\\\
&=-ng(0)+\sum_{v\in\Lambda^{*}}\hat{g}_{v}\left\lvert\sum_{x\in\omega_{n}}e^{2\pi
iv\cdot x}\right\rvert^{2}\\\ &\geq n^{2}\hat{g}_{0}-ng(0).\end{split}$ (14)

For $v\in\mathbb{R}^{d}$ we refer to

$M_{v}(\omega_{n}):=\sum_{x\in\omega_{n}}e^{2\pi iv\cdot x},$

as the $v$-moment of $\omega_{n}$. Note that equality holds in (14) if and
only if

$\hat{g}_{v}M_{v}(\omega_{n})=0,\qquad v\in\Lambda^{*}\setminus\\{0\\}.$ (15)

The next proposition follows immediately from (14) and the condition (15) for
equality in (14). This proposition is essentially contained in the proof of
the linear programming bounds of [CK07, Proposition 9.3] and is closely
related to Delsarte-Yudin energy bounds for spherical codes (cf. [BHS19,
Chapters 5.5 and 10.4]).

######  Proposition 7.

Let $F:\mathbb{R}^{d}\to[0,\infty]$ be $\Lambda$-periodic, and suppose $g\in
L^{2}(\Omega_{\Lambda})$ is PSD such that $g\leq F$. Then for any $n$-point
configuration $\omega_{n}$, we have

$E_{F}(\omega_{n})\geq E_{g}(\omega_{n})\geq n^{2}\hat{g}_{0}-ng(0)$ (16)

with equality holding throughout (16) if and only if the following two
conditions hold:

  1. (a)

$g(x-y)=F(x-y)$ for all $x\neq y\in\omega_{n}$,

  2. (b)

$\hat{g}_{v}M_{v}(\omega_{n})=0$, for all $v\in\Lambda^{*}\setminus\\{0\\}$.

If (a) and (b) hold, then $E_{F}(\omega_{n})=\mathcal{E}_{F}(n)$.

######  Remark.

If $F:\mathbb{R}^{d}\to[0,\infty]$ is $\Lambda$-periodic and
$G_{\Lambda}$-invariant and $g\in L^{2}(\Omega_{\Lambda})$ is PSD such that
$g\leq F$, then the $S_{\Lambda}$-invariant function

$g^{\rm sym}(x):=\frac{1}{|G_{\Lambda}|}\sum_{\sigma\in G_{\Lambda}}g((\sigma
v)\cdot x),\qquad x\in\mathbb{R}^{d}$

is also PSD and satisfies $g^{\rm sym}\leq F$. Thus, we may restrict our
search for functions $g$ to use in Proposition 7 to those of the form given in
(6) in which case we only need verify the condition that $g\leq F$ on the
fundamental domain of the action of $S_{\Lambda}$ on $\mathbb{R}^{d}$. In
particular, when $\Lambda=A_{2}$, we have the representative set

$\Delta_{A_{2}}:=\\{(x_{1},x_{2})\mid 0\leq x_{1}\leq\frac{1}{2},0\leq
x_{2}\leq x_{1}/\sqrt{3}\\}$

and when $\Lambda=L$, we’ll consider the representative set
$[0,1/2]\times[0,\sqrt{3}/2]$.

###  2.4 Moments for certain lattice configurations

We consider moments of configurations obtained by restricting scalings of a
lattice $\Lambda$ to the fundamental domain of a sublattice $\Phi$.

######  Theorem 8.

Suppose $\Phi$ is a sublattice of a lattice $\Lambda$ in $\mathbb{R}^{d}$. Let
$\omega_{\Phi}:=\Lambda\cap\Omega_{\Phi}$ and $\kappa:=|\omega_{\Phi}|$ denote
the index of $\Phi$ in $\Lambda$. For $m\in\mathbb{N}$, let

$\mu_{\kappa m^{d}}:=(\frac{1}{m}\Lambda)\cap\Omega_{\Phi}.$ (17)

Then for $v\in\Phi^{*}$, we have

$M_{v}(\mu_{\kappa m^{d}})=\begin{cases}\kappa m^{d},&v\in m\Lambda^{*},\\\
0,&\text{otherwise.}\end{cases}$ (18)

Furthermore, if $G_{\Phi}\subset G_{\Lambda}$, then for any $v\in\Phi^{*}$ and
$\sigma\in G_{\Phi}$, we have $M_{\sigma v}(\mu_{\kappa
m^{d}})=M_{v}(\mu_{\kappa m^{d}})$.

###### Proof.

Let $\Lambda=V\mathbb{Z}^{d}$; i.e., $V$ is a generator for $\Lambda$. Since
$\Phi$ is a sublattice of $\Lambda$, there is some integer $d\times d$ matrix
$W$ such that $VW$ is a generator for $\Phi$. Then $W$ is an that can be
written in Smith Normal Form as $W=SDT$ where $S$ and $T$ are integer matrices
with determinant $\pm 1$ (equivalently, their inverses are also integer
matrices) and $D$ is a diagonal matrix with positive integer diagonal entries
$\lambda_{1},\ldots,\lambda_{d}$. Then $\widetilde{V}=VS$ is a generator for
$\Lambda$ and $U=\widetilde{V}D$ is a generator for $\Phi$. Choosing the
fundamental domains $\Omega_{\Lambda}=\widetilde{V}[0,1)^{d}$ and
$\Omega_{\Phi}=U[0,1)^{d}$ we may write

$\mu_{\kappa m^{d}}=\\{\frac{1}{m}\widetilde{V}j\mid j\in
I_{m\lambda_{1}}\times\cdots\times I_{m\lambda_{d}}\\},$

where $I_{p}:=\\{0,1,2,\ldots,p-1\\}$. Let $v\in\Phi^{*}$ so that
$v=U^{-T}k={\widetilde{V}}^{-T}D^{-1}k$ for some
$k=(k_{1},k_{2},\dots,k_{d})\in\mathbb{Z}^{d}.$ Then
$v\cdot(\frac{1}{m}\widetilde{V}j)=\frac{1}{m}j\cdot(D^{-1}k)$ and so

$\displaystyle M_{v}(\mu_{\kappa m^{d}})$ $\displaystyle=\sum_{j\in
I_{m\lambda_{1}}\times\cdots\times I_{m\lambda_{d}}}e^{2\pi i\frac{1}{m}j\cdot
D^{-1}k}=\prod_{\ell=1}^{d}\left(\sum_{j_{\ell}=0}^{m\lambda_{\ell}-1}e^{2\pi
i\frac{j_{\ell}k_{\ell}}{m\lambda_{\ell}}}\right)$
$\displaystyle=\begin{cases}m^{d}\lambda_{1}\cdots\lambda_{d},&k\in
mD\mathbb{Z}^{d},\\\ 0,&\text{otherwise},\end{cases}$

where we used the finite geometric sum formula in the last equality. Noting
that $\kappa=\lambda_{1}\cdots\lambda_{d}$ and that $v\in m\Lambda^{*}$ if and
only if $k\in mD\mathbb{Z}^{d}$ establishes (18).

Finally, if $\sigma\in G_{\Phi}$ and $G_{\Phi}\subset G_{\Lambda}$, then
$\sigma v\in m\Lambda^{*}$ if and only if $v\in m\Lambda^{*}$ which completes
the proof. ∎

We define the index of a configuration $\omega_{n}\subset\mathbb{R}^{d}$ with
respect to a lattice $\Phi\subset\mathbb{R}^{d}$ by

$\mathcal{I}_{\Phi}(\omega_{n})=\\{v\in\Phi^{*}\mid M_{v}(\omega_{n})=0\\}.$
(19)

It then follows from Theorem 8 that $\mathcal{I}_{\Phi}(\mu_{\kappa
m^{d}})=\Phi^{*}\setminus(m\Lambda^{*})$.

###  2.5 Lattice theta functions

For $c>0$, the classical Jacobi theta function of the third type, is defined
by

$\theta(c;x):=\sum^{\infty}_{k=-\infty}e^{-\pi k^{2}c}e^{2\pi ikx},\qquad
x\in\mathbb{R}.$ (20)

Via Poisson Summation on the integers, we have

$\theta(c;x)=c^{-1/2}\sum^{\infty}_{k=-\infty}e^{-\frac{\pi(k+x)^{2}}{c}},$
(21)

and so

$F_{a,\mathbb{Z}}(x)=(a/\pi)^{1/2}\theta(\frac{\pi}{a};x).$

We note that $\theta(c;x)$ is $\mathbb{Z}$-periodic and
$\theta(c;x)=\theta(c;-x)$ for any $x$. In terms of our earlier language for
periodizing gaussians by lattices, we have

$F_{a,\mathbb{Z}}(x)=\sqrt{\frac{a}{\pi}}\theta(\frac{\pi}{a};x).$

We’ll also use

$\tilde{\theta}(c;t):=\theta\left(c,\frac{\arccos t}{2\pi}\right),\quad
t\in[-1,1].$

It follows from the symmetries of $\theta(c,x)$ that for all $x\in\mathbb{R}$,

$\tilde{\theta}(c;\cos 2\pi x)=\theta(c;x),$

and moreover, as shown below, $\tilde{\theta}$ is absolutely monotone on
$[-1,1]$. First, we recall the Jacobi triple product formula.

######  Theorem 9.

Jacobi Triple Product Formula Let $z,q\in\mathbb{C}$ with $\lvert q\rvert<1$
and $z\neq 0$. Then

$\prod^{\infty}_{r=1}(1-q^{2r})(1+q^{2r-1}z^{2})(1+q^{2r-1}z^{-2})=\sum^{\infty}_{k=-\infty}q^{k^{2}}z^{2k}.$

######  Proposition 10.

For any $c>0$, $\tilde{\theta}(c;\cdot):[-1,1]\rightarrow(0,\infty)$ is
strictly absolutely monotone on $[-1,1]$.

###### Proof.

Applying the Jacobi triple product with $q=e^{-\pi c}$ and $z=e^{\pi ix}$, we
observe

$\displaystyle\theta(c;x)$ $\displaystyle=\prod^{\infty}_{r=1}(1-e^{-2\pi
rc})(1+e^{-(2r-1)\pi c}e^{2\pi ix})(1+e^{-2\pi rc})(1+e^{-(2r-1)\pi c}e^{-2\pi
ix})$ (22) $\displaystyle=\prod^{\infty}_{r=1}(1-e^{-2\pi rc})(1+2e^{-2\pi
rc}\cos(2\pi x)+e^{-2(2r-1)\pi c}).$ (23)

So

$\tilde{\theta}(c;t)=\prod^{\infty}_{r=1}(1-e^{-2\pi rc})(1+2e^{-2\pi
rc}t+e^{-2(2r-1)\pi c}).$

It’s elementary to verify that $\tilde{\theta}(c;\cdot)$ is entire and
repeated applications of the product rule show the strict absolute
monotonicity of $\tilde{\theta}$ on $[-1,1]$. ∎

We also have

######  Proposition 11.

For all $n\in\mathbb{N}\cup\\{0\\}$ and $t\in[-1,1]$, $\tilde{\theta}$
satisfies

$(-1)^{n}\left(\frac{\tilde{\theta}^{\prime}}{\tilde{\theta}}\right)^{(n)}(t)>0.$

In other words, $\frac{\tilde{\theta}^{\prime}}{\tilde{\theta}}$ is strictly
completely monotone.

###### Proof.

Our proposition is equivalent to showing that $h^{\prime}$ is strictly
completely monotone, where $h:=\log\tilde{\theta}$. Recall that
$\tilde{\theta}$ can be expressed via the Jacobi triple product as

$\tilde{\theta}(c;t)=\prod^{\infty}_{r=1}(1-e^{-2\pi rc})(1+2e^{-2\pi
rc}t+e^{-2(2r-1)\pi c}).$

so

$h=\sum_{r=1}^{\infty}\log\left[(1-e^{-2\pi rc})(1+2e^{-2\pi
rc}t+e^{-2(2r-1)\pi c})\right].$

Let $h_{r}$ be the $r$th term in this sum. It suffices to show that each
$h_{r}^{\prime}$ is strictly completely monotone. Indeed, we have

$h_{r}^{\prime}=\frac{2e^{-2\pi rc}}{1+2e^{-2\pi rc}t+e^{-2(2r-1)\pi c}}$

so

$[h_{r}^{\prime}]^{(n)}=\frac{(-1)^{n}[2e^{-2\pi rc}]^{n+1}}{1+2e^{-2\pi
rc}t+e^{-2(2r-1)\pi c}}$

from which the claim follows since $1+2e^{-2\pi rc}t+e^{-2(2r-1)\pi c}\geq 0$
for all $r,c>0$ and $t\geq-1$. ∎

If $\Lambda_{R}=(a_{1}\mathbb{Z})\times\cdots\times(a_{d}\mathbb{Z})$ is a
rectangular lattice, then $F_{a,\Lambda_{R}}(x)$ is a tensor product of such
functions:

$\begin{split}F_{a,\Phi}(x)&=\sum_{v\in\Phi}e^{-a\|x+v\|^{2}}=\sum_{k\in\mathbb{Z}^{d}}\prod_{i=1}^{d}e^{-aa_{i}(x_{i}/a_{i}+k_{i})^{2}}\\\
&=\prod_{i=1}^{d}\left(\sum_{k_{i}\in\mathbb{Z}}e^{-aa_{i}(\frac{x_{i}}{a_{i}}+k_{i})^{2}}\right)=\prod_{i=1}^{d}F_{aa_{i},\mathbb{Z}}(x_{i}/a_{i}).\end{split}$
(24)

If $\Lambda$ contains a rectangular sublattice $\Lambda_{R}$, then we may
write $F_{\Lambda}$ as a sum of such tensor products.

######  Proposition 12.

Suppose $\Lambda$ is a lattice in $\mathbb{R}^{d}$ that contains a rectangular
sublattice $\Lambda_{R}=(a_{1}\mathbb{Z})\times\cdots\times(a_{d}\mathbb{Z})$
and let $\omega_{\Phi}=\Lambda\cap\Omega_{\Lambda_{R}}$. Then

$F_{a,\Lambda}(x)=\sum_{y\in\omega_{\Lambda_{R}}}F_{a,\Lambda_{R}}(x+y).$ (25)

###### Proof.

The formula follows immediately from $\Lambda=\omega_{\Phi}+\Phi$. ∎

###  2.6 Polynomial interpolation and linear programming bounds for lattice
configurations

Combining the previous results in this section, we obtain the following
general polynomial interpolation framework for linear programming bounds. For
convenience, we shall write

$W_{\Phi}:=\Phi^{*}/G_{\Phi}\text{ and
}\Delta_{\Phi}:=\mathbb{R}^{d}/S_{\Phi},$ (26)

to denote some choice of the respective fundamental domains for a lattice
$\Phi\subset\mathbb{R}^{d}$.

######  Theorem 13.

Let $\Phi\subset\mathbb{R}^{d}$ be such that $H\subseteq G_{\Phi}$ where $H$
is the coordinate symmetry group (see Sec. 2.2) and suppose
$F:\mathbb{R}^{d}\to(-\infty,\infty]$ is $S_{\Phi}$ invariant. By Proposition
6, $\Phi$ contains a rectangular sublattice

$a_{1}\mathbb{Z}\times\cdots\times a_{d}\mathbb{Z},$

which induces the change of variables $T:=T_{a_{1},\ldots,a_{d}}$ defined in
(12) and associated polynomials $P^{G_{\Phi}}_{v}$ defined in (13). Suppose
$a>0$ and $(c_{v})_{v\in W_{\Phi}}$ is such that (a) $c_{v}\geq 0$, for all
nonzero $v\in W_{\Phi}$ , (b) $\sum_{v\in W_{\Phi}}c_{v}<\infty$, and (c) the
continuous function

$\tilde{g}_{a}:=c_{0}+\sum_{v\in W\\_Phi}c_{v}P^{G_{\Phi}}_{v}.$

satisfies $\tilde{g}_{a}\leq\tilde{F}$ on $\tilde{\Delta}_{\Phi}$.

Then for any $n$-point configuration
$\omega_{n}=\\{x_{1},\ldots,x_{n}\\}\subset\mathbb{R}^{d}$, we have

$E_{F_{a,\Phi}}(\omega_{n})\geq
E_{g_{a}}(\omega_{n})=n^{2}c_{0}-n\tilde{g}_{a}(1,\ldots,1),$ (27)

where equality holds if and only if

  1. 1.

$\tilde{g}_{a}(t)=\tilde{F}_{a,\Phi}(t)$ for all $t\in T(\\{x_{i}-x_{j}\mid
i\neq j\in\\{1,\ldots,n\\}\\})$ and

  2. 2.

$c_{v}M_{\sigma v}(\omega_{n})=0$ for all $v\in W_{\Phi}$ and $\sigma\in
G_{\Phi}$.

We now consider sufficient conditions to for the $F$-optimality of
configurations of the form $\mu_{\kappa
m^{d}}=\frac{1}{m}\Lambda\cap\Omega_{\Phi}$ as in (17). For such a
configuration we define

$\tau_{\kappa m^{d}}:=(\frac{1}{m}\Lambda)\cap\Delta_{\Phi}),$ (28)

which equals $\mu_{\kappa m^{d}}\cap\Delta_{\Phi}$ if we choose
$\Delta_{\Phi}\subset\Omega_{\Phi}.$

######  Corollary 14.

Suppose $\Phi$, $T:=T_{a_{1},\ldots,a_{d}}$, $\tilde{g}_{a}$, and $F$ are as
in Theorem 13 and that $\Phi\subseteq\Lambda$, and $G_{\Phi}\subseteq
G_{\Lambda}$ for some lattice $\Lambda\subset\mathbb{R}^{d}$. Then the
configuration $\mu_{\kappa m^{d}}=\frac{1}{m}\Lambda\cap\Omega_{\Phi}$ defined
in Prop. 8 is $F$-optimal if

  1. 1.

$c_{v}=0$ for all $v\in\left(m\Lambda^{*}\right)\cap W_{\Phi}$, and

  2. 2.

$\tilde{g}_{a}(t)=\tilde{F}_{a,\Phi}(t)$ for all $t\in\tilde{\tau}_{\kappa
m^{d}}\setminus\\{\mathbf{1}\\}$ where
$\mathbf{1}=(1,1,\ldots,1)\in\mathbb{R}^{d}$.

If such a $\tilde{g}_{a}$ exists, we refer to it as a ‘magic’ interpolant.

##  3 The Linear Programming Framework for the families
$\omega_{m^{2}},\omega_{2m^{2}},\omega_{3m^{2}},$ and $\omega_{6m^{2}}$

We explicitly apply Corollary 14 linear programming framework for the four
families of periodic problems described in the introduction to obtain
bivariate polynomial interpolation problems whose solutions would verify the
$A_{2}$-universal optimality of $\omega_{m^{2}}$ and $\omega_{3m^{2}}$, and
the $L$-universal optimality of $\omega_{2m^{2}}$ and $\omega_{6m^{2}}$.
Observe that our four families of point configurations are of the form
$\mu_{\kappa m^{2}}$ with the following choices of rectangular lattice
$\Phi\subseteq\Lambda$:

  1. 1.

$\omega^{*}_{m^{2}}$: $\Phi=A_{2},\Lambda=A_{2}$

  2. 2.

$\omega^{*}_{2m^{2}}$: $\Phi=L,\Lambda=A_{2}$

  3. 3.

$\omega^{*}_{3m^{2}}$: $\Phi=A_{2},\Lambda=\frac{1}{\sqrt{3}}A_{2}^{\pi/6}$

  4. 4.

$\omega^{*}_{3m^{2}}$: $\Phi=L,\Lambda=\frac{1}{\sqrt{3}}A_{2}^{\pi/6}$.

It is straightforward to check that in all cases, $\Phi$ and $\Lambda$ satisfy
the conditions of Theorem 13. Since both choices of $\Phi$, $A_{2}$ and $L$,
contain $L$ as a rectangular sublattice, we will work with the following
change of variables to induce our polynomial structure, as described in
Proposition 6:

$\displaystyle(t_{1},t_{2}):=\left(\cos(2\pi x_{1}),\cos\left(\frac{2\pi
x_{2}}{\sqrt{3}}\right)\right).$ (29)

We will use $T$ to denote the change of variables
$T(x)=(t_{1}(x_{1}),t_{2}(x_{2}))$.

Figure 3: The region $\tilde{\Delta}_{A_{2}}$, pictured above, is our region
of interpolation for the families $\omega^{*}_{m^{2}}$ and
$\omega^{*}_{3m^{2}}$ (see sec. 3.4).

Importantly, the maps $F_{a,\Phi}$ are also well-behaved under this change of
variables, as seen through decomposing $F_{a,\Phi}$ into $\theta$ functions as
described in Proposition 25.

When $\Phi=L$, we apply Prop. 5 to obtain

$\displaystyle
F_{a,L}(x)=\frac{\pi}{\sqrt{3}a}\theta(\frac{\pi}{a};x_{1})\theta(\frac{\pi}{3a};\frac{x_{2}}{\sqrt{3}}).$
(30)

As a result,

$\displaystyle\tilde{F}_{a,L}(t_{1},t_{2})$
$\displaystyle:=F(\frac{\arccos(t_{1})}{2\pi},\frac{\sqrt{3}\arccos(t_{2})}{2\pi})$
$\displaystyle=\frac{\pi}{\sqrt{3}a}\theta(\frac{\pi}{a};\frac{\arccos(t_{1})}{2\pi})\theta(\frac{\pi}{3a};\frac{\arccos(t_{2})}{2\pi})$
$\displaystyle=\frac{\pi}{\sqrt{3}a}\tilde{\theta}(\frac{\pi}{a};t_{1})\tilde{\theta}(\frac{\pi}{3a};t_{2})$

for $t_{1},t_{2}\in[-1,1]$. Thus, for fixed $t_{1}\in[-1,1]$,
$\tilde{F}_{a,L}$ is strictly absolutely monotone as a function of $t_{2}$ and
vice versa. We will use the absolute monotonicity in $t_{1}$ and $t_{2}$
repeatedly, and the symmetry of this formula for $\tilde{F}$ is one of the
main motivations for considering the sublattice $L$.

On the other hand, when $\Phi=A_{2}$, we arrive at the following formula, a
consequence of Proposition 25, which also appears in [Bae97] and [Su15]:

$F(x):=F_{a,A_{2}}(x)=\frac{\pi}{\sqrt{3}a}\left(\theta(\frac{\pi}{a};x_{1})\theta(\frac{\pi}{3a};\frac{x_{2}}{\sqrt{3}})+\theta(\frac{\pi}{3a};\frac{x_{2}}{\sqrt{3}}+\frac{1}{2})\theta(\frac{\pi}{a};x_{1}+\frac{1}{2})\right),$
(31)

which gives

$\begin{split}\tilde{F}(t)&:=F(\frac{\arccos
t_{1}}{2\pi},\frac{\sqrt{3}\arccos t_{2}}{2\pi})\\\
&=\frac{\pi}{\sqrt{3}a}(\tilde{\theta}(\frac{\pi}{a};t_{1})\tilde{\theta}(\frac{\pi}{3a};t_{2})+\tilde{\theta}(\frac{\pi}{a};-t_{1})\tilde{\theta}(\frac{\pi}{3a};-t_{2})).\end{split}$
(32)

The next corollary follows immediately from the absolute monotonicity of
$\tilde{\theta}$ (see Proposition 10).

######  Corollary 15.

For any nonnegative integers $l_{1}$ and $l_{2}$ satisfying $l_{1}+l_{2}\equiv
0\pmod{2}$,

$\frac{\partial^{l_{1}+l_{2}}\tilde{F}}{\partial^{l_{1}}t_{1}\partial^{l_{2}}t_{2}}>0,\quad\forall
t_{1},t_{2}\in[-1,1].$

Finally, we have the following lemma from [Bae97] (also see [Su15]).

######  Lemma 16.

$\displaystyle\frac{\partial\tilde{F}}{\partial t_{1}}(t_{1},t_{2})$
$\displaystyle>0\quad\forall t_{1}\in[-1,1],\forall t_{2}\in[-1,1]$ (33)
$\displaystyle\frac{\partial\tilde{F}}{\partial t_{2}}(t_{1},t_{2})$
$\displaystyle\geq 0\quad\forall t_{1}\in[-1,1],\forall t_{2}\in[-1,1]$ (34)

where the equality holds if and only if $t_{1}=-1$, $t_{2}=\frac{1}{2}$. In
particular, these inequalities hold in the entire $t_{1},t_{2}$ region.

###### Proof.

Since even partial deriviatives of $\tilde{F}$ are positive and every point
$(t_{1},t_{2})\in\tilde{\Delta}_{A_{2}}$ satisfies $t_{1}\geq-1$ and
$t_{2}\geq\frac{1}{2}$, it suffices to verify the inequalities

$\frac{\partial\tilde{F}}{\partial
t_{1}}(t_{1},t_{2})>0,\frac{\partial\tilde{F}}{\partial
t_{2}}(t_{1},t_{2})\geq 0$

at $(-1,\frac{1}{2})$. See [Bae97] or [Su15]. ∎

As observed in [Su15] (also see [BHS19, Chapter 10] and [FSZ23]), this Lemma
16 suffices to proves the $A_{2}$-universal optimality of the 2 and 3-point
configurations discussed in the introduction (see section 3.5 for more
detail).

We will also make the following choices of fundamental domains
$\mathbb{R}^{2}/S_{\Phi}$ and $\Phi^{*}/G_{\Phi}$. When $\Phi=L$, we take as a
choice for $\Phi^{*}/G_{L}$ the set

$W_{L}:=\left\\{\begin{bmatrix}k_{1}\\\ k_{2}/\sqrt{3}\end{bmatrix}\mid
k_{1},k_{2}\in\mathbb{Z}\text{ and }k_{1},k_{2}\geq 0\right\\}$

and $[0,1/2]\times[0,\sqrt{3}/2]$ for $\mathbb{R}^{2}/S_{L}$. Likewise for
$A_{2}$, we take the sets

$W_{A_{2}}:=\left\\{\begin{bmatrix}k_{1}\\\ k_{2}/\sqrt{3}\end{bmatrix}\in
L^{*}\mid 0\leq k_{2}\leq k_{1}\text{ and }k_{1}\equiv
k_{2}\pmod{2}\,\right\\}$

and $\Delta_{A_{2}}$ (see Sec. 2.3) for $\Phi^{*}/G_{A_{2}}$ and
$\mathbb{R}^{2}/S_{A_{2}}$, respectively.

Finally, the following characterizations of the dual lattices $\Lambda$ will
be useful for determining which degree polynomials are available to us for
interpolation. First, $L^{*}=\\{[k_{1},k_{2}/\sqrt{3}]^{T}\mid
k_{1},k_{2}\in\mathbb{Z}\\}$. Then $A_{2}^{*}=\\{v\in L^{*}\mid v\cdot
e_{1}\equiv v\cdot\sqrt{3}e_{2}\pmod{2}\\}$, and
$\left(\frac{1}{\sqrt{3}}A_{2}^{\pi/6}\right)^{*}=\\{v\in A_{2}^{*}\mid
v\cdot\sqrt{3}e_{2}\equiv 0\pmod{3}\\}$. Thus, the set of all $v$ for which
$c_{v}$ may be non-zero in the construction of an interpolant $\tilde{g}_{a}$
are expressed by the index sets defined

$\displaystyle\mathcal{I}_{m^{2}}:=\mathcal{I}_{A_{2}}(\omega_{m^{2}}^{*})\cap
W_{A_{2}}$ $\displaystyle=W_{A_{2}}/(mA_{2}^{*})$ (35)
$\displaystyle=\\{[k_{1},k_{2}/\sqrt{3}]\mid k_{1},k_{2}\geq 0,k_{1}\equiv
k_{2}\bmod 2,[k_{1},k_{2}]\neq m[j_{1},j_{2}]\\}$ (36)
$\displaystyle\mathcal{I}_{2m^{2}}:=\mathcal{I}_{L}(\omega_{2m^{2}}^{*})\cap
W_{L}$ $\displaystyle=W_{L}/(mA_{2}^{*})$ (37)
$\displaystyle=\\{[k_{1},k_{2}/\sqrt{3}]\mid k_{1},k_{2}\geq
0,[k_{1},k_{2}]\neq m[j_{1},j_{2}]\text{ for some }j_{1}\equiv j_{2}(\bmod
2)\\}$ (38)
$\displaystyle\mathcal{I}_{3m^{2}}:=\mathcal{I}_{A_{2}}(\omega_{3m^{2}}^{*})\cap
W_{A_{2}}$ $\displaystyle=W_{A_{2}}/(mA_{2}^{*})$ (39)
$\displaystyle=\\{[k_{1},k_{2}/\sqrt{3}]\mid k_{1},k_{2}\geq 0,k_{1}\equiv
k_{2}\bmod 2,[k_{1},k_{2}]\neq m[j_{1},j_{2}]\text{ for some }j_{2}\equiv
0\bmod 3\\}$ (40)
$\displaystyle\mathcal{I}_{6m^{2}}:=\mathcal{I}_{L}(\omega_{6m^{2}}^{*})\cap
W_{L}$ $\displaystyle=W_{L}/(mA_{2}^{*})$ (41)
$\displaystyle=\\{[k_{1},k_{2}/\sqrt{3}]\mid k_{1},k_{2}\geq
0,[k_{1},k_{2}]\neq m[j_{1},j_{2}]\text{ for some }j_{1}\equiv j_{2}(\bmod
2),j_{2}\equiv 0\bmod 3\\}.$ (42)

###  3.1 The Polynomials $P^{L}_{v}$ and $P^{A_{2}}_{v}$

When $\Phi=L$, we have already shown that the functions are $P^{L}_{v}$ are
tensors of Chebyshev Polynomials

$P^{L}_{v}=T_{k_{1}}(t_{1})T_{k_{2}}(t_{2})$

where $v=[k_{1},k_{2}/\sqrt{3}]^{T}$, $k_{1},k_{2}\geq 0$ is an arbitrary
element of $W_{L}$. (see Proposition 5).
What can be said in the case when $\Phi=A_{2}$? These polynomials have been
studied extensively (see [LSX08], [LSX12], and references therein). Of
particular importance to $P^{A_{2}}_{v}$ are the polynomials $P_{v^{\prime}}$
and $P_{v^{\prime\prime}}$, where $v^{\prime}=[1,1/\sqrt{3}]^{T}$ is the
shortest non-zero vector in $W_{A_{2}}$ and we’ll let
$v^{\prime\prime}=[2,0]^{T}$ be the next shortest vector. We have

$\displaystyle P_{v^{\prime}}$
$\displaystyle=\frac{1}{3}(-1+2t_{2}(t_{1}+t_{2})),$ (43) $\displaystyle
P_{v^{\prime\prime}}$
$\displaystyle=\frac{1}{3}(-1+2t_{1}(t_{1}-3t_{2}+4t_{2}^{3})).$ (44)

It turns out that every other $P_{v}$ can be expressed as a bivariate
polynomial in $P_{v^{\prime}}$ and $P_{v^{\prime\prime}}$, i.e. for any $v\in
A_{2}^{*}$, there exist coefficients $c_{i,j}$ (with only finitely many
nonzero) such that

$P_{v}=\sum_{i,j\geq
0}c_{i,j}(-1+2t_{2}(t_{1}+t_{2}))^{i}(-1+2t_{1}(t_{1}-3t_{2}+4t_{2}^{3}))^{j}$

First, note that since $P_{v^{\prime}}$ and $P_{v^{\prime\prime}}$ contain
only monomials of even total degree, the same is true of arbitrary $P_{v}$ To
further understand these bivariate polynomials, we set
$\alpha=P_{v^{\prime}}$, $\beta=P_{v^{\prime\prime}}$ and introduce a notion
of degree on monomials introduced in [PM10]:

######  Definition 3.1.

The $A_{2}$-degree of a bivariate monomial $\alpha^{k_{0}}\beta^{k_{1}}$ is
$2k_{0}+3k_{1}$.

If $v\in W_{A_{2}}$, then $v=k_{0}v^{\prime}+k_{1}v^{*}$ for some unique
$k_{0},k_{1}\geq 0$, and so we can likewise introduce the notion of the
$A_{2}$ degree of $v\in W_{A_{2}}$ as $2k_{0}+3k_{1}$. We will denote the
degree function as $\mathcal{D}$ for both monomials and elements of
$W_{A_{2}}$. Now we can introduce an ordering on monomials by $m$-degree and
break ties via the power of $\alpha$. Then the leading term (by
$A_{2}$-degree) of $P_{v}$ is $\alpha^{k_{0}}\beta^{k_{1}}$. Certainly, this
is true for our first polynomials, $P_{0}=1$, $P_{v^{\prime}}=\alpha$, and
$P_{v^{\prime\prime}}=\beta$, and then an examination of the recursion
generating the polynomials shows that the claim holds inductively (cf.
[LSX12]).

###  3.2 Interpolation Nodes

Our final step is to calculate for each family the nodes at which a
hypothetical interpolant $\tilde{g}_{a}$ must interpolate $\tilde{F}_{a,\Phi}$
to meet the conditions of our linear programming bounds (including $0$ to
simplify formulas). We end up with

$\displaystyle\tau_{\kappa m^{2}}$ $\displaystyle:=\omega_{\kappa
m^{2}}\cap\Delta_{A_{2}},\hskip 28.45274pt\kappa=1,3$ (45)
$\displaystyle\tau_{\kappa m^{2}}$ $\displaystyle:=\omega_{\kappa
m^{2}}\cap([0,1/2]\times[0,\sqrt{3}/2]),\hskip 28.45274pt\kappa=2,6.$ (46)

Going straight from the definitions of $\omega_{\kappa m^{2}}$ and our
$t_{1},t_{2}$ change of variables, we then compute

$\displaystyle\tilde{\tau}_{m^{2}}$
$\displaystyle=\left\\{\left(\cos(\frac{\pi k_{1}}{m}),\cos(\frac{\pi
k_{2}}{m})\right)\mid 0\leq 3k_{2}\leq k_{1}\leq m,k_{1}\equiv k_{2}(\bmod
2)\right\\}$ (47) $\displaystyle\tilde{\tau}_{2m^{2}}$
$\displaystyle=\left\\{\left(\cos(\frac{\pi k_{1}}{m}),\cos(\frac{\pi
k_{2}}{m})\right)\mid 0\leq k_{1},k_{2}\leq m,k_{1}\equiv k_{2}(\bmod
2)\right\\}$ (48) $\displaystyle\tilde{\tau}_{3m^{2}}$
$\displaystyle=\left\\{\left(\cos(\frac{\pi k_{1}}{m}),\cos(\frac{\pi
k_{2}}{3m})\right)\mid 0\leq k_{2}\leq k_{1}\leq m,k_{1}\equiv k_{2}(\bmod
2)\right\\}$ (49) $\displaystyle\tilde{\tau}_{6m^{2}}$
$\displaystyle=\left\\{\left(\cos(\frac{\pi k_{1}}{m}),\cos(\frac{\pi
k_{2}}{3m})\right)\mid 0\leq 3k_{1},k_{2}\leq 3m,k_{1}\equiv k_{2}(\bmod
2)\right\\}.$ (50)

###  3.3 Interpolation Problem for $\omega_{m^{2}}$

With all the machinery now set up, we address the family $\omega_{m^{2}}^{*}$
and its base case, $\omega_{4}^{*}$. Recall $v^{\prime}:=[1,1/\sqrt{3}]$ is
the shortest vector in $W_{A_{2}}$ and
$P_{v^{\prime}}=\frac{1}{3}(-1+2t_{2}(t_{1}+t_{2}))$. In Section 4, we prove
the $A_{2}$-universal optimality of the base case $\omega_{4}^{*}$ by
constructing for each $a>0$ a polynomial of the form
$g_{a}(t_{1},t_{2}):=c_{0}+c_{1}P_{v^{\prime}}(t_{1},t_{2})$ with $c_{1}\geq
0$ such that $g_{a}\leq F_{a,A_{2}}$.

Figure 4: $\tilde{g}_{a}$ must stay below $\tilde{F}_{a}$ on
$\tilde{\Delta}_{A_{2}}$ with equality at the corner point $(-1,1)$.

For general $m$, recalling the background on $G_{2}$ polynomials in Sec. 3.1,
we note that

$\\{v\in W_{A_{2}}\mid\mathcal{D}(v)<2m\\}\subset\mathcal{I}_{m^{2}}.$

The containment holds because if $v=k_{0}v^{\prime}+k_{1}v^{\prime\prime}$ and
$\mathcal{D}(v)<2m$, then $k_{0}<m$, and so $v\not\in mA_{2}^{*}$ (i.e.
$v\in\mathcal{I}_{m^{2}}$). For the $m=2$ base case already discussed, our
interpolant $\tilde{g}_{a}$ satisfies

$\tilde{g}_{a}\in\text{span}\\{P_{v}:v\in\Delta_{A_{2}},\mathcal{D}(v)<2m\\}.$

###  3.4 Interpolation Problem for $\omega^{*}_{2m^{2}}$

Now to the case of $\omega^{*}_{2m^{2}}$ with base case $\omega^{*}_{2}$.

Figure 5: $\omega^{*}_{2}$, pictured above is $L$-universally optimal, and
analogous results hold for any rectangular lattice.

The universal optimality of $\omega^{*}_{2}$ follows from

$F_{a,L}(x)=\frac{\pi}{\sqrt{3}a}\theta(\frac{\pi}{a};x_{1})\theta(\frac{\pi}{3a};\frac{x_{2}}{\sqrt{3}}),$

which takes its minimum at $(1/2,\sqrt{3}/2)$ for all $a$, as $\theta(c,x)$
takes its minimum at $x=1/2$ for all $c>0$ (see Proposition 10). Since the
$F_{a,L}$ energy of a two-point configuration is determined only by the
difference of the two points in the configuration, the universal optimality of
$\omega^{*}_{2}$ immediately follows, and the same argument can be used to
show for any rectangular lattice (cf. [FSZ23]) that a point at the origin and
a point at the centroid of a rectangular fundamental domain yield a 2-point
universally optimal configuration.

For the general case,

$\\{[k_{1},k_{2}/\sqrt{3}]^{T}:0\leq k_{1},k_{2}\text{ and
}k_{1}+k_{2}<2m\\}\subseteq\mathcal{I}_{2m^{2}},$

and then

$\text{span}\\{P_{v}\mid v=[k_{1},k_{2}/\sqrt{3}]^{T}:0\leq k_{1},k_{2}\text{
and }k_{1}+k_{2}<2m\\}=\mathcal{P}_{2m-1}(t_{1},t_{2}),$

where $\mathcal{P}_{n}(t_{1},t_{2})$ is the set of bivariate polynomials of
total degree at most $n$. For the first non-trivial case, $\omega^{*}_{8}$, we
have numerical evidence that for each $a>0$, an interpolant $\tilde{g}_{a}$
exists in $P_{3}$ and satisfies the conditions of Corollary 14.

Figure 6: The 8 larger points comprise $\omega^{*}_{8}$

Figure 7: After applying the linear programming framework, it suffices to find
for each $a>0$ an interpolant $\tilde{g}_{a}\in\mathcal{C}_{8}$ such that
$\tilde{g}_{a}\leq\tilde{F}_{a}$ on $[-1,1]$ with equality at the three points
shown.

###  3.5 Interpolation Problems for $\omega^{*}_{3m^{2}}$

Now to the case of $\omega^{*}_{3m^{2}}$ with base case $\omega^{*}_{3}$.

Figure 8: The 3 larger points comprise $\omega^{*}_{3}$

Figure 9: Almost the exact same analysis needed for the $A_{2}$-universal
optimality of $\omega^{*}_{3}$ yields $A_{2}$-universal optimality of the two
point honeycomb configurations, one of which is pictured above.

The universal optimality of $\omega^{*}_{3}$ (cf. [Su15] and [FSZ23]) follows
from Lemma 16, which is used to show a global minimum of $F_{a,A_{2}}$ occurs
at $(1/2,\sqrt{3}/6)$ for all $a$. This point, $(1/2,\sqrt{3}/6)$, is also the
only difference $x-y$ (up to $S_{A_{2}}$ action) for $x,y\in\omega^{*}_{6}$.
Thus for an arbitrary 6-point configuration $\omega_{6}$, we have

$E_{F}(\omega_{6})\geq 6F(1/2,\sqrt{3}/6)=E_{F}(\omega^{*}_{6})$

and so $\omega^{*}_{6}$ is $A_{2}$-universally optimal. This same line of
argument is also used in [Su15] to show that the 2-point honeycomb
configurations pictured below are $A_{2}$-universally optimal.

More generally, we suggest invoking $A_{2}$-degree as in the $m^{2}$ case to
find a nice subset of $\mathcal{I}_{3m^{2}}$. We have the containment

$\\{v\in W_{A_{2}}\mid\mathcal{D}(v)<3m\\}\subset\mathcal{I}_{3m^{2}}.$

The containment holds because if $v=k_{0}v^{\prime}+k_{1}v^{\prime\prime}$ and
$\mathcal{D}(v)<3m$, then either $v\not\in mA_{2}^{*}$ or
$v=mv^{\prime}\not\in(\frac{1}{\sqrt{3}}A_{2}^{\pi/6})^{*}$.

###  3.6 Interpolation Problems for $\omega^{*}_{6m^{2}}$

It remains to consider the family $\omega_{6}^{*}$ and its base case
$\omega_{6}^{*}$, whose universal optimality involves our most complex
application of the linear programming bounds. In section 5, we prove the
$L$-universal optimality of $\omega^{*}_{6}$ by constructing for each $a>0$ an
interpolant of the form

$g_{a}(t_{1},t_{2})=b_{0,0}+b_{1,0}t_{1}+b_{0,1}t_{2}+b_{1,1}t_{1}t_{2}+b_{0,2}t^{2}_{2}$

where $b_{i,j}\geq 0$ for $(i,j)\neq 0$. In that section, we will explain in
greater detail why such a $\tilde{g}_{a}$ satisfies the conditions of
Corollary 14.

Figure 10: $\tilde{g}_{a}$ must stay below $\tilde{F}_{a}$ on $[-1,1]^{2}$
with equality at the three points shown

Finally, for arbitrary $m$, we propose a few nice subsets of
$\mathcal{I}_{6m^{2}}$. First, we have the set

$\\{[k_{1},k_{2}/\sqrt{3}]^{T}:0\leq k_{1}<2m,0\leq
k_{2}<3m\\}\subseteq\mathcal{I}_{6m^{2}},$

and

$\text{span}\\{P_{v}\mid v=[k_{1},k_{2}/\sqrt{3}]^{T}:0\leq k_{1}<2m0\leq
k_{2}<3m\\}=\mathcal{P}_{2m-1}(t_{1})\times\mathcal{P}_{3m-1}(t_{2}).$

Working with such a tensor space of polynomials is natural due to the tensor
product nature of

$\tilde{F}_{a,L}(x)=\frac{\pi}{\sqrt{3}a}\tilde{\theta}(\frac{\pi}{a};t_{1})\tilde{\theta}(\frac{\pi}{3a};t_{2}).$

Notably, our interpolant, $\tilde{g}_{a}$, for $\omega^{*}_{6}$ satisfies
$\tilde{g}_{a}\in\mathcal{P}_{1}(t_{1})\times\mathcal{P}_{2}(t_{2})$.

##  4 Universal Optimality of $\omega^{*}_{4}$

To prove $\omega^{*}_{4}$ is $A_{2}$-universally optimal, it remains to show
for each $a>0$ that there are $c_{0},c_{1}\in\mathbb{R}$ with $c_{1}\geq 0$
such that the resulting interpolant
$\tilde{g}_{a}(t_{1},t_{2}):=c_{0}+c_{1}P_{v^{\prime}}=c_{0}+c_{1}/3(-1+t_{2}(t_{1}+t_{2}))$
satisfies $\tilde{g}_{a}\leq\tilde{F}_{a}$ on $\tilde{\Delta}_{A_{2}}$ with
equality at $(-1,1)$ or, equivalently, finding such an interpolant of the form

$\ \tilde{g}_{a}(t_{1},t_{2}):=\tilde{F}_{a}(-1,1)+b_{1}t_{2}(t_{1}+t_{2})$
(51)

for $b_{1}\geq 0$.
Our formulas for $\tilde{g}_{a}$ are defined piecewise333We suspect that
$b_{1}$ need not be defined piecewise. In fact the choice
$b_{1}=\frac{\partial\tilde{F}}{\partial t_{2}}\bigg{|}_{-1,1}$ numerically
appears to lead to $\tilde{g}_{a}\leq\tilde{F}_{a}$ for all $a>0$. But our
most simple proofs come from this piecewise definition of $b_{1}$. in $a$. We
set

$b_{1}=\begin{cases}2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}&\text{if }0<a<21\\\ \frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1}&\text{if }a\geq 21.\end{cases}$ (52)

Due to the convergence properties of different formulas for $\theta$ (see (20)
and (21)), we find it convenient to rescale $\tilde{F}$ by a factor of
$\sqrt{3}\pi/a$ for small $a$ case. Defining

$\begin{split}\tilde{f_{1}}(t_{1})&:=\begin{cases}\tilde{\theta}(\frac{\pi}{a};t_{1}),&0<a\leq\pi^{2}\\\
\sqrt{\frac{\pi}{a}}\tilde{\theta}(\frac{\pi}{a};t_{1})&a>\pi^{2},\end{cases}\\\
\tilde{f_{2}}(t_{2})&:=\begin{cases}\tilde{\theta}(\frac{\pi}{a};t_{2}),&0<a\leq\pi^{2}\\\
\sqrt{\frac{\pi}{3a}}\tilde{\theta}(\frac{\pi}{a};t_{2})&a>\pi^{2}.\end{cases}\end{split}$
(53)

With this rescaling convention, it follows from (31) that

$\tilde{F}(t_{1},t_{2})=\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t_{2})+\tilde{f_{1}}(-t_{1})\tilde{f_{2}}(-t_{2}).$
(54)

###  4.1 Constructing magic $\tilde{g}_{a}$

For all $a>0$, we will establish

######  Lemma 17.

For all points $(t_{1}^{\prime},t_{2}^{\prime})\in\tilde{\Delta}_{A_{2}}$,
$\displaystyle\frac{\partial^{3}\tilde{F}}{\partial t_{1}\partial
t_{2}^{2}}\bigg{|}_{t_{1}^{\prime},t_{2}^{\prime}}\geq 0.$

###### Proof.

Since even partial derivatives of $\tilde{F}$ are positive, it suffices to
check the inequality at the minimal $t_{1}$ and $t_{2}$ values, when
$t_{1}=-1$ and $t_{2}^{\prime}=\frac{1}{2}$. This check is handled in in the
appendix with large $a$ and small $a$ cases handled separately. ∎

Likewise, we have

######  Lemma 18.

Let $\tilde{h}$ be of the form $\tilde{F}(-1,1)+c_{1}t_{2}(t_{1}+t_{2})$ such
that $\tilde{h}(-1,1/2)\leq\tilde{F}(-1,1/2)$ and
$\frac{\partial\tilde{F}-\tilde{h}}{\partial t_{2}}\bigg{|}_{-1,1}\leq 0$.
Then for all $t_{2}\in[1/2,1]$, $\tilde{F}(-1,t_{2})\geq\tilde{h}(-1,t_{2})$.

###### Proof.

We abuse notation here and use $\tilde{F},\tilde{h}$ to refer to the one
variable functions in $t_{2}$ obtained by fixing $t_{1}=-1$. By assumption on
the form of $\tilde{h}$, Lemma 16, and the two assumed inequalities, we have
$\tilde{F}(1/2)\geq\tilde{h}(1/2)$,
$\tilde{F}^{\prime}(1/2)=\tilde{h}^{\prime}(1/2)=0$
$\tilde{F}(1)=\tilde{h}(1)$, and
$\tilde{F}^{\prime}(1)\leq\tilde{h}^{\prime}(1)$. It follows that there are
exists some point in $[1/2,1]$ at which
$\tilde{F}^{\prime\prime}\leq\tilde{h}^{\prime\prime}$. Let
$t_{2}^{\prime}\leq t_{2}^{\prime\prime}$ be such that $t_{2}^{\prime}$ and
$t_{2}^{\prime\prime}$ respectively are the minimal and maximal points in
$[1/2,1]$ at which $\tilde{F}^{\prime\prime}\leq\tilde{h}^{\prime\prime}$. For
$t_{2}\geq t_{2}^{\prime\prime}$ we have
$\tilde{F}^{\prime\prime}\geq\tilde{h}^{\prime\prime}$ and so we get
$\tilde{F}\geq\tilde{h}$ by bounding $\tilde{F}-\tilde{h}$ below with a
tangent line of $\tilde{F}-\tilde{h}$ at $1$. Similarly, for $t_{2}\leq
t_{2}^{\prime}$, we get the desired inequality with tangent approximation from
$\frac{1}{2}$. For $t_{2}\in[t_{2}^{\prime\prime},t_{2}^{\prime}]$, we note
that $\tilde{F}^{\prime\prime}=\tilde{h}^{\prime\prime}$ at the endpoints of
the interval and $(\tilde{F}-\tilde{h}){{}^{\prime\prime}}$ is convex (since
$\tilde{F}^{(4)}>0$), so $(\tilde{F}-\tilde{h})^{\prime\prime}\leq 0$ for the
whole interval. Thus, $\tilde{F}(t_{2})\geq\tilde{h}(t_{2})$ by bounding the
difference below with its secant line (since we’ve already established
$\tilde{F}\geq\tilde{h}$ at the endpoints). ∎

####  4.1.1 Small $a$

Let $0<a\leq 21$. We will refer to $\tilde{g}_{a}$ as simply $\tilde{g}$.

Next, we claim that for this range of $a$, the following inequalities hold:

$\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1},4(\tilde{F}(-1,1)-\tilde{F}(-1,1/2))\leq
2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}\leq\frac{\partial^{2}(\tilde{F}-\tilde{g})}{\partial
t_{1}\partial t_{2}}\bigg{|}_{-1,1/2}.$

Moreover, we claim that these 3 inequalities444Though in the small $a$ case,
we have set $b_{1}=2\frac{\partial\tilde{F}}{\partial t_{1}}\bigg{|}_{-1,1/2}$
for simplicity, in fact, we could set $b_{1}$ to be any element of the (non-
empty) interval $\left[\max\left\\{\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1},4(\tilde{F}(-1,1)-\tilde{F}(-1,1/2))\right\\},2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}\right]$ and the exact same proof would work. suffice
to show $\tilde{F}\leq\tilde{g}$.

Let’s first address why these 3 inequalities are sufficient. Assume they hold,
then first off we certainly have $b_{1}\geq 0$ since
$b_{1}=2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}\geq\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1}\geq 0$ where the first inequality holds by assumption
and the next by Lemma 16. Next, we have

$(\tilde{F}-\tilde{g})(-1,1/2)=\tilde{F}(-1,1/2)-\tilde{F}(-1,1)+\frac{1}{4}b_{1}\geq
0$

and likewise

$\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{2}}\bigg{|}_{-1,1}=\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1}-b_{1}\leq 0.$

Applying Lemma 18 and the previous two inequalities to $\tilde{g}$, we obtain
$\tilde{F}\geq\tilde{g}$ for all points $(-1,t_{2})$ with $t_{2}\in[1/2,1]$,
and since $\tilde{F}-\tilde{g}$ is convex in $t_{1}$, it remains to show that
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\geq 0$ for all points of
the form $(-1,t_{2})$, $t_{2}\in[1/2,1]$ (recall the picture of
$\tilde{\Delta}_{A_{2}}$, Figure 4). By Lemma 17,
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}$ is convex in $t_{2}$ in
$\tilde{\Delta}_{A_{2}}$, so we just need to show

$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{-1,1/2}\geq
0,\frac{\partial^{2}(\tilde{F}-\tilde{g})}{\partial t_{1}\partial
t_{2}}\bigg{|}_{-1,1/2}\geq 0.$

But these follow directly from our assumptions on $b_{1}$. Indeed,

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,1/2}$ $\displaystyle=\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}-\frac{b_{1}}{2}=0$ (55)
$\displaystyle\frac{\partial^{2}(\tilde{F}-\tilde{g})}{\partial t_{1}\partial
t_{2}}\bigg{|}_{-1,1/2}$ $\displaystyle=\frac{\partial^{2}\tilde{F}}{\partial
t_{1}\partial t_{2}}\bigg{|}_{-1,1/2}-b_{1}\geq 0.$ (56)

Thus, it remains to establish the following555The reason we don’t use this
approach for all $a>9.6$ is Lemma 19 fails at roughly $a=22$. Namely, the
terms $\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1},4(\tilde{F}(-1,1)-\tilde{F}(-1,1/2))$ both have lead
exponential terms on the order of $e^{-a/4}$, while
$\frac{\partial\tilde{F}}{\partial t_{1}}\bigg{|}_{-1,1/2}$ is on the order of
$e^{-a/3}$. is the same as for $0<a<\pi^{2}$, but we use estimates on
$\tilde{\theta}$ derived from the ‘large $a$’ formula for $\tilde{\theta}$. So
for this range it suffices to show in the appendix that Lemma 19 holds for
$9.6\leq a\leq 21$. in the appendix:

######  Lemma 19.

For $0<a<\pi^{2}$, we have

$\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1},4(\tilde{F}(-1,1)-\tilde{F}(-1,1/2))\leq
2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}\leq\frac{\partial^{2}(\tilde{F}-\tilde{g})}{\partial
t_{1}\partial t_{2}}\bigg{|}_{-1,1/2}$

We handle the proof piecewise, splitting into 2 cases, $0<a\leq\pi^{2}$ and
$\pi^{2}<a\leq 21$ depending on which formulas we use for $\tilde{f_{1}}$ and
$\tilde{f_{2}}$.

####  4.1.2 Large $a$

Throughout, we assume $a\geq 21$ and refer to $\tilde{g}_{a}$ as $\tilde{g}$.

Figure 11: The figure depicts our strategy for showing
$\tilde{F}\geq\tilde{g}$ in the large $a$ case. We show
$\tilde{F}\leq\tilde{g}$ on the rectangular region A (which includes points
outside of $\tilde{\Delta}_{A_{2}}$) in Lemma 22. The remaining points of
$\tilde{\Delta}_{A_{2}}$, in region B, are handled by Lemma 23

.

First, we show

######  Lemma 20.

$\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{(-1,1)}\geq\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{(-1,1)}$

The proof is contained in the appendix, using bounds also contained there.

Now recall for this range of $a$ that

$b_{1}=\frac{\partial\tilde{F}}{\partial t_{2}}\bigg{|}_{-1,1}.$

We first show that this choice of $b_{1}$ ensures
$(\tilde{F}-\tilde{g})(-1,1/2)\geq 0$:

######  Lemma 21.

For $a\geq 21$, $(\tilde{F}-\tilde{g})(-1,1/2)\geq 0$.

The proof is contained in the appendix using our bounds on $\theta$. Along
with our selection of $b_{1}$, we can then apply Lemma 18 to conclude
$\tilde{F}\geq\tilde{g}$ whenever $t_{1}=-1$ in $\tilde{\Delta}_{A_{2}}$.
Similarly, the inequality $\tilde{F}\geq\tilde{g}$ for all points $(t_{1},1)$
with $t_{1}\in[-1,1]$ is immediate from Lemma 20, our choice of $b_{1}$, and
the convexity of the $\tilde{F}-\tilde{g}$ in $t_{1}$.

Next, we show that if $t_{1}\leq\cos(2\pi\frac{\sqrt{3}}{4})$, we have
$\tilde{F}\geq\tilde{g}$ in $\tilde{\Delta}_{A_{2}}$, and in fact we’ll show
the stronger claim that $\tilde{F}\geq\tilde{g}$ in the region
$[-1,\cos(2\pi\frac{\sqrt{3}}{4})]\times[1/2,1]$ (corresponding to rectangle A
in Figure 11).
To do so, we’ll use that the log derivative of $\tilde{\theta}$ is strictly
completely monotone (cf. Proposition 11. One consequence of this property is
that if $H_{\tilde{F}}$ denotes the Hessian matrix of $\tilde{F}$, we have for
all $t_{1},t_{2}\in[-1,1]$ that

$\det(H_{\tilde{F}})<0.$

Indeed, recalling the formulas,

$\displaystyle
f_{1}(x_{1})=\sqrt{\frac{a}{\pi}}\theta(\frac{\pi}{a},x)=\sum^{\infty}_{k=-\infty}e^{-a(k+x)^{2}}$
(57) $\displaystyle
f_{2}(x_{2})=\sqrt{\frac{3a}{\pi}}\theta(\frac{\pi}{3a},x)=\sum^{\infty}_{k=-\infty}e^{-3a(k+x)^{2}},$
(58)

we obtain in the $A_{2}$ case,

$\tilde{F}(t_{1},t_{2})=\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t_{2})+\tilde{f_{1}}(-t_{1})\tilde{f_{2}}(-t_{2}).$

By Proposition 11, for $i\in\\{1,2\\}$,

$0>\left(\frac{\tilde{f}_{i}^{\prime}}{\tilde{f}_{i}}\right)^{\prime}=\frac{\tilde{f}_{i}^{\prime\prime}\tilde{f}_{i}-(\tilde{f}_{i}^{\prime})^{2}}{\tilde{f}_{i}^{2}}$

and so

$\tilde{f}_{i}^{\prime\prime}\tilde{f}_{i}<(\tilde{f}_{i}^{\prime})^{2}$

Thus, we compute

$\displaystyle\det(H_{\tilde{F}})$
$\displaystyle=\tilde{f_{1}}(t_{1})^{\prime\prime}\tilde{f_{2}}(t_{2})\tilde{f_{2}}^{\prime\prime}(t_{2})\tilde{f_{1}}(t_{1})-(\tilde{f_{1}}^{\prime}(t_{1})\tilde{f_{2}}^{\prime}(t_{2}))^{2}+$
(59)
$\displaystyle\tilde{f_{1}}(-t_{1})^{\prime\prime}\tilde{f_{2}}(-t_{2})\tilde{f_{2}}^{\prime\prime}(-t_{2})\tilde{f_{1}}(-t_{1})-(\tilde{f_{1}}^{\prime}(-t_{1})\tilde{f_{2}}^{\prime}(-t_{2}))^{2}$
(60)
$\displaystyle=(\tilde{f_{1}}^{\prime\prime}(t_{1})\tilde{f_{1}}(t_{1}))(\tilde{f_{2}}^{\prime\prime}(t_{2})\tilde{f_{2}}(t_{2}))-\tilde{f_{1}}^{\prime}(t_{1})^{2}\tilde{f_{2}}^{\prime}(t_{2})^{2}+$
(61)
$\displaystyle(\tilde{f_{1}}^{\prime\prime}(-t_{1})\tilde{f_{1}}(-t_{1}))(\tilde{f_{2}}^{\prime\prime}(-t_{2})\tilde{f_{2}}(-t_{2}))-\tilde{f_{1}}^{\prime}(-t_{1})^{2}\tilde{f_{2}}^{\prime}(-t_{2})^{2}$
(62) $\displaystyle<0$ (63)

as desired.

######  Definition 4.1.

Fix $b,c\in[-1.1]$. Then we define

$\tilde{g}_{b,c}:=\tilde{g}-b_{1}t_{1}t_{2}+b_{1}(bt_{2}+ct_{1}-bc).$

Note that if $t_{1}\geq b$, $t_{2}\leq c$, then

$(t_{1}-b)(t_{2}-c)\leq 0$

and so $t_{1}t_{2}\leq bt_{2}+ct_{1}-bc$. Thus, if $t_{1}\geq b$, $t_{2}\leq
c$, then $\tilde{g}\leq\tilde{g}_{b,c}$ with equality if $t_{1}=b$ or
$t_{2}=c$.

The useful thing about $\tilde{g}_{b,c}$ for any choice of $b,c$ is that it
satisfies

$\det(H_{\tilde{F}-\tilde{g}_{b,c}})=\det(H_{\tilde{F}})-2b_{1}\tilde{f_{1}}^{\prime\prime}(t_{1})\tilde{f_{2}}(t_{2})<\det(H_{\tilde{F}})<0.$

Thus, to check $\tilde{F}\geq\tilde{g}_{a,b}$ on a rectangular region, it
suffices to consider the boundary by the second derivative test. If that
rectangular region is chosen such that $g_{b,c}\geq\tilde{g}$, then we have a
means of showing $\tilde{F}\geq\tilde{g}$ without needing to consider a full
2-dimensional region.

######  Lemma 22.

For $t_{1},t_{2}\in[-1,\cos(2\pi\frac{\sqrt{3}}{4})]\times[\frac{1}{2},1]$,
$\tilde{F}\geq\tilde{g}$.

###### Proof.

First, we show the inequality for the rectangle
$[-1,\cos(2\pi\frac{\sqrt{3}}{4})]\times[.7,1]$, As explained above, it
suffices to prove that $\tilde{F}\geq\tilde{g}_{-1,1}$ on the boundary of the
rectangle. Since $\tilde{g}_{-1,1}=\tilde{g}\leq\tilde{F}$ when $t_{1}=-1$ or
$t_{2}=1$, it suffices to prove the inequality for the other two sides.
There, we use our approximations of $\theta$ and $b_{1}$ to provide a simple
upper bound, call it $\tilde{g}_{-1,1}^{\prime}$ for $\tilde{g}_{-1,1}$. The
key feature of this upper bound is that $e^{a/4}\tilde{g}_{-1,1}^{\prime}$ is
linear in $a$. Meanwhile, as a lower bound for $\tilde{F}$, we use the
function

$\tilde{F}_{T}:=(e^{-ax^{2}}+e^{-a(x-1)^{2}})e^{-3au^{2}}+e^{-a((\frac{1}{2}-x)^{2}+3(\frac{1}{2}-u)^{2})}$

where

$x=\frac{\arccos(t_{1})}{2\pi},\hskip
14.22636ptu=\frac{\arccos(t_{2})}{2\pi}.$

We know $\tilde{F}_{T}\leq\tilde{F}$ because we have simply truncated positive
terms from the formulas for $f_{1}$ and $f_{2}$ to obtain $\tilde{F}_{T}$.
It’s also straightforward to see that at any point, $e^{a/4}\tilde{F}_{T}$ is
convex in $a$. Thus, the difference
$e^{a/4}(\tilde{F}_{T}-\tilde{g}_{-1,1}^{\prime})$ is convex in $a$, and so to
show that $\tilde{F}_{T}\geq\tilde{g}_{-1,1}$ at some
$t_{1}^{\prime},t_{2}^{\prime}$ for all $a\geq 21$, it suffices to show
$\tilde{F}_{T}(t_{1}^{\prime},t_{2}^{\prime})\geq\tilde{g}_{-1,1}^{\prime}(t_{1}^{\prime},t_{2}^{\prime})$
when $a=21$ and also that

$\frac{\partial\left[e^{a/4}(\tilde{F}_{T}-\tilde{g}_{-1,1}^{\prime})\right]}{\partial
a}\bigg{|}_{a=21,t_{1}=t_{1}^{\prime},t_{2}=t_{2}^{\prime}}\geq 0.$

These computations are carried out in the appendix.
For the remaining region $[-1,\cos(2\pi\frac{\sqrt{3}}{4})]\times[.5,.7]$, we
carry out the same procedure with $\tilde{g}_{-1,.7}$ functioning as our upper
bound for $\tilde{g}$ when $t_{2}\in[.6,.7]$ and $\tilde{g}_{-1,.6}$
functioning as our upper bound when $t_{2}\in[.5,.6]$. For the former case,
when $t_{2}=.7$ or $t_{1}=-1$, $\tilde{g}_{-1,.7}=\tilde{g}$, which we have
just shown stays below $\tilde{F}$ on those segments. So it remains to show
that the inequality holds on the sides where
$t_{1}=\cos(2\pi\frac{\sqrt{3}}{4})$ and $t_{2}=.6$. Likewise, we only need to
show the inequality $\tilde{g}_{-1,.6}\leq\tilde{F}_{T}$ when
$t_{1}=\cos(2\pi\frac{\sqrt{3}}{4})$, $t_{2}\in[1/2,.6]$ and when $t_{2}=1/2$,
$t_{1}\in[-1,\cos(2\pi\frac{\sqrt{3}}{4})]$. These computations are carried
out as in the previous case in the appendix. ∎

Finally, we show that $\tilde{F}-\tilde{g}$ increases in $t_{1}$ for every
point in $\tilde{\Delta}_{A_{2}}$ with
$t_{1}\geq\cos(2\pi\frac{\sqrt{3}}{4})$.

######  Lemma 23.

For all $a\geq 21$ and every $p=(t_{1},t_{2})\in\tilde{\Delta}_{A_{2}}$ with
$t_{1}\geq\cos(2\pi\frac{\sqrt{3}}{4})$,
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{p}\geq 0$.

###### Proof.

Because of the convexity of the difference in $t_{1}$ and Lemma 17, it
suffices to show that at
$P=(\cos(2\pi\frac{\sqrt{3}}{4}),\cos(2\pi\frac{\sqrt{3}}{12})):=(t_{1}^{\prime},t_{2}^{\prime})$,

$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{P}\geq 0,\hskip
28.45274pt\frac{\partial^{2}(\tilde{F}-\tilde{g})}{\partial t_{1}\partial
t_{2}}\bigg{|}_{P}\geq 0$

which is handled in the appendix. ∎

##  5 Universal Optimality of $\omega^{*}_{6}$

We consider the $m=1$ case of the interpolation problem from Section 3.6. In
this case we have interpolation conditions at the nodes
$\tilde{\tau}_{6}=\\{(-1,-1),(1,-\frac{1}{2}),(-1,\frac{1}{2})\\}$. Using the
same rescaling convention as in the previous section, we have

$\tilde{F}(t_{1},t_{2})=\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t_{2}),$

where $\tilde{f_{1}},\tilde{f_{2}}$ are as in (53). For $m=1$, we may choose
an interpolant
$\tilde{g}=\tilde{g}_{a}\in\mathcal{P}_{1}(t_{1})\times\mathcal{P}_{2}(t_{2})$;
i.e $\tilde{g}$ of the form

$\tilde{g}(t_{1},t_{2})=\sum_{i=0}^{1}\sum_{j=0}^{2}b_{i,j}t_{1}^{i}t_{2}^{j}.$

We require that $\tilde{g}$ and $\tilde{F}$ agree at the three points in
$\tilde{\tau}_{6}$ and remark that the condition $\tilde{g}\leq\tilde{F}$
further requires $\partial\tilde{g}/\partial t_{2}=\partial\tilde{F}/\partial
t_{2}$ at the points $(-1,1/2)$ and $(1,-1/2)$ giving a total of 5 linearly
independent conditions on
$\mathcal{P}_{1}(t_{1})\times\mathcal{P}_{2}(t_{2})$.

Figure 12: $\tilde{g}$ has 5 necessary equality interpolation conditions in
order to provide a sharp bound, the 3 value conditions from
$\tilde{\tau}_{6}$, plus two derivative conditions.

Noting that $q(t_{1},t_{2})=(1+t_{1})(t_{2}+1/2)^{2}$ vanishes on
$\tilde{\tau}_{6}$ and that $\frac{\partial q}{\partial t_{2}}$ vanishes on
$\\{(-1,1/2),(1,-1/2)\\}$ shows that $\tilde{g}$ can be written as

$\begin{split}\tilde{g}(t_{1},t_{2})&=\frac{(1-t_{1})}{2}\tilde{f_{1}}(-1)H_{\\{-1,\frac{1}{2},\frac{1}{2}\\}}(\tilde{f_{2}})(t_{2})+\frac{(1+t_{1})}{2}\tilde{f_{1}}(1)H_{\\{-\frac{1}{2},-\frac{1}{2}\\}}(\tilde{f_{2}})(t_{2})+cq(t_{1},t_{2}),\end{split}$
(64)

where $H_{T}(f)$ is the Hermite interpolant to $f$ on the node set $T$ which
can be expressed in terms of divided differences (see Appendix A). In
particular,

$H_{\\{-1,\frac{1}{2},\frac{1}{2}\\}}(\tilde{f_{2}})(t_{2})=\tilde{f_{2}}(-1)+\tilde{f_{2}}[-1,\frac{1}{2}](t_{2}+1)+\tilde{f_{2}}[-1,\frac{1}{2},\frac{1}{2}](t_{2}+1)(t_{2}-\frac{1}{2}),$

and
$H_{\\{-\frac{1}{2},-\frac{1}{2}\\}}(\tilde{f_{2}})(t_{2})=\tilde{f_{2}}(-\frac{1}{2})+\tilde{f_{2}}^{\prime}(-\frac{1}{2})(t_{2}+\frac{1}{2})^{2}$.
Since $T_{1}(t)=t$ and $T_{2}(t)=2t^{2}-1$, it easily follows that $\tilde{g}$
is PSD if and only if $b_{i,j}\geq 0$ for $(i,j)\neq 0$. From (64), it follows
that
$b_{1,2}=-\frac{1}{2}\tilde{f_{1}}(-1)\tilde{f_{2}}[-1,\frac{1}{2},\frac{1}{2}]+c$.
Observing that $q\geq 0$ on $[-1,1]^{2}$, we choose
$c=\frac{1}{2}\tilde{f_{1}}(-1)\tilde{f_{2}}[-1,\frac{1}{2},\frac{1}{2}]$ as
small as possible in which case $b_{1,2}=0$.

In addition, the following derivative equality,

$\tilde{f_{1}}(-1)\tilde{f_{2}}^{\prime}(-1/2)=\tilde{f_{1}}(1)\tilde{f_{2}}^{\prime}(1/2).$
(65)

proved in [Bae97] (also see [Su15]) implies that $b_{1,1}=b_{0,2}$. Hence we
may express $\tilde{g}$ in the form

$\tilde{g}(t_{1},t_{2})=a_{0,0}+a_{1,0}t_{1}+a_{0,1}t_{2}+a_{0,2}(t_{1}t_{2}+t^{2}_{2}+1/4),$
(66)

where $a_{0,0}=b_{0,0}-b_{0,2}/4$ and $a_{i,j}=b_{i,j}$ otherwise.

From (64), we then compute

$\displaystyle a_{0,1}$
$\displaystyle=\tilde{f_{1}}(-1)\tilde{f_{2}}^{\prime}(-1/2)$ (67)
$\displaystyle a_{0,0}$
$\displaystyle=\frac{\tilde{f_{1}}(1)\tilde{f_{2}}(-1/2)+\tilde{f_{1}}(-1)\tilde{f_{2}}(1/2)}{2}$
(68) $\displaystyle a_{1,0}$
$\displaystyle=\frac{\tilde{f_{1}}(1)\tilde{f_{2}}(-1/2)-\tilde{f_{1}}(-1)\tilde{f_{2}}(1/2)}{2}+\frac{a_{0,1}}{2}$
(69) $\displaystyle a_{0,2}$
$\displaystyle=\tilde{f_{1}}(-1)\tilde{f_{2}}[-1,\frac{1}{2},\frac{1}{2}]=\frac{4}{9}(\tilde{f_{1}}(-1)\tilde{f_{2}}(-1)+a_{0,1}+a_{1,0}-a_{0,0}).$
(70)

###  5.1 Verifying properties of $\tilde{g}$

It remains to verify that the $a_{i,j}$’s are nonnegative and
$\tilde{g}\leq\tilde{F}$.

######  Lemma 24.

$\tilde{F}(-1,t_{2})\geq\tilde{g}(-1,t_{2})$ for all $t_{2}\in[-1,1]$.

###### Proof.

Note that $\tilde{F}(-1,t_{2})=\tilde{f_{1}}(-1)\tilde{f_{2}}(t_{2})$ is
strictly absolutely monotone as a function of $t_{2}$ on [-1,1] and so the
inequality follows from the error formula (82) with strict inequality if
$t_{2}\neq-1,1/2$. ∎

######  Proposition 25.

$a_{0,0},a_{1,0},a_{0,1},a_{0,2}\geq 0.$

###### Proof.

That $a_{0,1}$, $a_{0,0}$, $a_{0,2}$ are nonnegative follows immediately from
the absolute monotonicity of $\tilde{f_{1}}$ and $\tilde{f_{2}}$.

Finally, by lemma 24, $\tilde{g}(-1,-1/2)<\tilde{F}(-1,-1/2)$. Moreover, by
definition, $\tilde{g}(1,-1/2)=\tilde{F}(1,-1/2)$. Since
$(\tilde{F}-\tilde{g})(t_{1},-1/2)$ is convex in $t_{1}$, we must have

$a_{1,0}-\frac{1}{2}a_{0,2}=\frac{\partial\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,-1/2}\geq\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,-1/2}>0.$

So $a_{1,0}-\frac{1}{2}a_{0,2}>0$ which implies $a_{1,0}>0$ since we have
already established $a_{0,2}\geq 0$. ∎

Thus, $\tilde{g}$ has a nonnegative Chebyshev expansion as desired.

###  5.2 Necessary Conditions Fulfilled and $\tilde{F}\geq\tilde{g}$ when
$|t_{2}|\in[1/2,1]$

Finally, we must show $\tilde{g}\leq\tilde{F}$ on $[-1,1]^{2}$, which is by
far our most technical task. Our equality conditions on $\tilde{g}$ yield
necessary inequality conditions (for $\tilde{g}\leq\tilde{F}$) on the
derivatives of $\tilde{g}$. First, we have the necessary conditions

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,-1}\geq 0$ (71)
$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,\frac{1}{2}}\geq 0$ (72)

Likewise, we have

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{1,-\frac{1}{2}}\leq 0$ (73)

each of which we show is satisfied in the appendix. These necessary conditions
are actually sufficient to obtain $\tilde{F}\geq\tilde{g}$ when
$|t_{2}|\in[1/2,1]$ without any more calculations.

######  Proposition 26.

If $\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{(-1,t^{\prime}_{2})}\geq 0$ and
$\tilde{f_{1}}(-1)\tilde{f_{2}}(t^{\prime}_{2})\geq\tilde{g}(-1,t^{\prime}_{2})$
for some $t^{\prime}_{2}\in[-1,1]$, then

$\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t^{\prime}_{2})\geq\tilde{g}(t_{1},t^{\prime}_{2})$

for all $t_{1}\in[-1,1]$ with strict inequality if $t_{1}\neq-1$. The same
inequality holds if
$\frac{\partial(\tilde{f_{1}}\tilde{f_{2}}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{(1,t_{2})}\leq 0$ and
$\tilde{f_{1}}(1)\tilde{f_{2}}(t_{2})\geq\tilde{g}(1,t_{2})$, with strict
inequality if $t_{1}\neq 1$.

###### Proof.

Note that for fixed $t_{2}$, $\tilde{g}(t_{1},t_{2})$ is linear in $t_{1}$.
Thus by the absolute monotonicity of $\tilde{f_{1}}$, $\tilde{F}-\tilde{g}$ is
convex in $t_{1}$. It follows that if
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{(-1,t^{\prime}_{2})}\geq 0$ and
$\tilde{f_{1}}(-1)\tilde{f_{2}}(t^{\prime}_{2})=\tilde{g}(-1,t^{\prime}_{2})$,
then

$\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t^{\prime}_{2})\geq\tilde{g}(t_{1},t^{\prime}_{2})$

for all $t_{1}\in[-1,1]$ with strict inequality for $t_{1}>-1$. Again invoking
the convexity of $\tilde{F}-\tilde{g}$ in $t_{1}$, we obtain the same
conclusion if $\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{(1,t^{\prime}_{2})}\leq 0$ and
$\tilde{f_{1}}(1)\tilde{f_{2}}(t^{\prime}_{2})\geq\tilde{g}(1,t_{2})$ with
strict inequality if $t_{1}\neq 1$. ∎

######  Lemma 27.

$\tilde{F}(t_{1},t_{2})\geq\tilde{g}(t_{1},t_{2})$ in $\tilde{\square}_{L}$
whenever $t_{2}\geq 1/2$, with equality only at $(-1,1/2)$.

###### Proof.

From our necessary conditions, we have

$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{-1,1/2}\geq
0,\hskip 14.22636pt\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{1,-1/2}\leq 0.$

Since $\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{t_{1},-1/2}$ is increasing in $t_{1}$,
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{1,-1/2}\leq 0$
implies $\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,-1/2}\leq 0$. We also know that
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{-1,t_{2}}$ is
convex in $t_{2}$ by the absolute monotonicity of $\tilde{f_{1}}$ and
$\tilde{f_{2}}$, from which we obtain
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{-1,t_{2}}\geq
0$ for all $t_{2}\geq 1/2$. Indeed, fixing $t_{1}=-1$, we can bound
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}$ below for $t_{2}\geq
1/2$ by its tangent approximation, which is positive at $t_{2}=1/2$ and has
positive slope since the average slope from $t_{2}=-1/2$ to $t_{2}=1/2$ is
positive. Finally, lemma 24 yields
$\tilde{F}(-1,t_{2})\geq\tilde{g}(-1,t_{2})$ for all $t_{2}\in[1/2,1]$.
Combining these two facts, we invoke lemma 26 to assert
$\tilde{F}-\tilde{g}\geq 0$ whenever $t_{2}\geq 1/2$. ∎

Next, we have $\tilde{F}\geq\tilde{g}$ when $t_{1}=1$.

######  Proposition 28.

$\tilde{f_{1}}(1)\tilde{f_{1}}(t_{2})\geq\tilde{g}(1,t_{2})$ for all
$t_{2}\in[-1,1]$ with equality only at $t_{2}=-1,1/2$.

###### Proof.

Suppose by way of contradiction that there exists $t_{2}^{\prime}\in[-1,1]$
such that $t_{2}^{\prime}\neq-\frac{1}{2}$ and
$\tilde{f_{1}}(1)\tilde{f_{2}}(t_{2}^{\prime})\leq\tilde{g}(1,t_{2}^{\prime})$.
Then there must be some point $-1/2\neq p\in[-1,1]$ such that
$\tilde{f_{1}}(1)\tilde{f_{2}}(p)\leq\tilde{g}(1,p)$. Indeed, either
$t_{2}^{\prime}$ is such a point, or
$\tilde{f_{1}}(1)\tilde{f_{2}}(t_{2}^{\prime})<\tilde{g}(1,t_{2}^{\prime})$.
We have from our necessary derivative conditions and lemma 27 that
$\tilde{f_{1}}(1)\tilde{f_{2}}(\pm 1)>\tilde{g}(1,\pm 1)$, which yield two
cases for $t_{2}^{\prime}$. If $t_{2}^{\prime}<-1/2$, then there exists
$p\in(-1,t_{2}^{\prime})$ such that
$\tilde{f_{1}}(1)\tilde{f_{2}}(p)=\tilde{g}(1,p)$ by the intermediate value
theorem. If $t_{2}^{\prime}>-1/2$, instead apply the intermediate value
theorem to $t_{2}=t_{2}^{\prime},1$ to get our desired $p\in[1/2,1]$.
Similarly to the proof that $\tilde{F}\geq\tilde{g}$ for $t_{2}=-1$,
$\tilde{g}(1,t_{2})$ is a polynomial of degree at most 2, and interpolates
$\tilde{f_{1}}(1)\tilde{f_{1}}(t_{2})$ for $T=\\{p,-1/2,-1/2\\}$. By the
uniqueness of hermite interpolants, $\tilde{g}$ satisfies the error formula,
that is,

$\tilde{f_{1}}(1)\tilde{f_{2}}(t_{2})-\tilde{g}(1,t_{2})=\tilde{f_{1}}(1)\tilde{f_{2}}^{(3)}(\xi)(t_{2}-p)(t_{2}+1/2)^{2}$

for some $\xi\in[-1,1]$ . By absolute monotonicity of $\tilde{f_{2}}$,
$\tilde{f_{1}}(1)\tilde{f_{2}}^{(3)}(\xi)\geq 0$. But for $t_{2}=-1$,
$(t_{2}-p)(t_{2}+1/2)^{2}<0$, implying
$\tilde{f_{1}}(1)\tilde{f_{2}}(-1)<\tilde{g}(1,-1)$ a contradiction. We
conclude $\tilde{f_{1}}(1)\tilde{f_{1}}(t_{2})\geq\tilde{g}(1,t_{2})$ for all
$t_{2}\in[-1,1]$ as desired. ∎

######  Proposition 29.

For all $(t_{1},t_{2})\in[-1,1]\times[-1,-1/2]$
$\tilde{f_{1}}(t_{1})\tilde{f_{1}}(t_{2})\geq\tilde{g}(t_{1},t_{2})$.

###### Proof.

From lemma 24, we have $\tilde{F}\geq\tilde{g}$ for $t_{1}=-1$. By the
necessary derivative conditions and lemma 26, we have the same inequality when
$t_{2}=-1/2,-1$. Finally, by lemma 28, we have the inequality for $t_{1}=1$.
All of these inequalities are strict in $[-1,1]\times[-1,1/2]$ except for at
$(-1,-1)$ and $(1,-1/2)$. Thus, it remains to show the inequality on the
interior. For an arbitrary point $p=(p_{1},p_{2})$ on the interior, the line
$l$ through $p$ and $(1,-1/2)$ has positive slope. Let its direction vector in
the positive $t_{1}$ direction be $u_{l}$.
Let $\tilde{F}^{l}(r)$ and $\tilde{g}^{l}$ be the one-variable functions
obtained by parametrizing $\tilde{F}$ and $\tilde{g}$ along $l$ with constant
speed so that $\tilde{F}^{l}(1)=\tilde{F}(1,-1/2)$ and
$\tilde{F}^{l}(0)=\tilde{F}(p^{\prime})$ for some $p^{\prime}$ on the boundary
of $[-1,1]\times[-1,1/2]$ where $t_{1}=-1$ or $t_{2}=-1$ . Since $l$ has
positive slope, and $\tilde{f_{1}}$, $\tilde{f_{2}}$ are both strictly
absolutely monotone, $\tilde{F}^{l}$ is also strictly absolutely monotone on
$[0,1]$. Likewise, $\tilde{g}^{l}$ is a polynomial of total degree at most 2
since $\tilde{g}$ is a polynomial of total degree 2.
We know for all $\epsilon>0$ sufficiently small that
$(\tilde{F}^{l}-\tilde{g}^{l})(\epsilon)>0$. Indeed, if $l$ does not go
through $(-1,-1)$, then $(\tilde{F}^{l}-\tilde{g}^{l})(0)>0$ and the result
follows by continuity, and if $l$ does go through $(-1,1)$, then we can use
the necessary derivative conditions to see

$\nabla(\tilde{F}-\tilde{g})\bigg{|}_{-1,-1}\cdot u_{l}>0,$

from which the inequality holds for all desired $\epsilon$. Similarly, the
necessary derivative inequality and equality conditions at $(1,-1/2)$ imply
$\nabla(\tilde{F}-\tilde{g})\bigg{|}_{-1,1/2}\cdot u_{l}<0$. Together with the
fact that $(\tilde{F}^{l}-\tilde{g}^{l})(1)=0$, we get
$(\tilde{F}^{l}-\tilde{g}^{l})(1-\epsilon)>0$ for all $\epsilon$ sufficiently
small.
Now supposing for a contradiction that $\tilde{F}^{L}(r^{\prime})<0$ for some
$r^{\prime}\in[0,1]$, we apply the intermediate value theorem twice to get
some points $0<r_{1}<r^{\prime}<r_{2}<1$ such that
$(\tilde{F}^{l}-\tilde{g}^{l})(r_{1})=(\tilde{F}^{l}-\tilde{g}^{l})(r_{2})=0$.
But now as with the previous lemmas, $\tilde{g}^{l}$ is a polynomial of degree
at most 2 which interpolates $\tilde{F}^{l}$ for $T={r_{1},r_{2},1}$, leading
to a contradiction with the error formula applied at $0$ to
$\tilde{F}^{l}-\tilde{g}^{l}$. We conclude that
$\tilde{F}^{l}\geq\tilde{g}^{l}$ on $[0,1]$ as desired, and in fact, a similar
argument shows we have strict inequality on $(0,1)$. ∎

Thus, we have proved $\tilde{F}\geq\tilde{g}$ on $[-1,1]^{2}$ whenever
$t_{2}\geq 1/2$ or $t_{2}\leq-1/2$. Our proof of the inequality when $-1/2\leq
t_{2}\leq 1/2$, our so-called ‘critical region,’ requires much more
calculation.

###  5.3 The critical region for small $a$ ($a<\pi^{2}$)

For $a<\pi^{2}$, we take a linear approximation approach. Recall that for
fixed $t_{2}^{\prime}$, $(\tilde{F}-\tilde{g})(t_{1},t_{2}^{\prime})$ is
convex in $t_{1}$. Thus, for $-1\leq t_{1}^{\prime}\leq 0$,

$(\tilde{F}-\tilde{g})(t_{1}^{\prime},t_{2}^{\prime})\geq(\tilde{F}-\tilde{g})(-1,t_{2}^{\prime})+(t_{1}+1)\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}=\tilde{F}(-1,t_{2}^{\prime})+(t+1)\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}-\tilde{g}(t_{1}^{\prime},t_{2}^{\prime})$

since $\tilde{g}$ is linear in $t_{1}$. Since we already have
$(\tilde{F}-\tilde{g})(-1,t_{2}^{\prime})\geq 0$, then if
$(\tilde{F}-\tilde{g})(-1,t_{2}^{\prime})+(t_{1}+1)\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}\geq 0$ for $t_{1}=0$, also
$(\tilde{F}-\tilde{g})(-1,t_{2}^{\prime})+(t_{1}+1)\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}$ for all $t_{1}\leq 0$. Thus, to show
$(\tilde{F}-\tilde{g})(t_{1}^{\prime},t_{2}^{\prime})\geq 0$ for all
$t_{1}^{\prime}\leq 0,\frac{-1}{2}\leq t_{2}\leq\frac{1}{2}$, we need only
show for all $t_{2}\in[\frac{-1}{2},\frac{1}{2}]$ that

$(\tilde{F}-\tilde{g})(-1,t_{2}^{\prime})+\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}=\tilde{F}(-1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}-\tilde{g}(0,t_{2}^{\prime})\geq 0.$

Likewise, when $0\leq t_{1}^{\prime}\leq 1$,

$(\tilde{F}-\tilde{g})(t_{1}^{\prime},t_{2}^{\prime})\geq(\tilde{F}-\tilde{g})(1,t_{2}^{\prime})+(t_{1}-1)\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}=\tilde{F}(1,t_{2}^{\prime})+(t_{1}-1)\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}-\tilde{g}(t_{1}^{\prime},t_{2}^{\prime}).$

Since we already have $(\tilde{F}-\tilde{g})(1,t_{2}^{\prime})\geq 0$, then if
$(\tilde{F}-\tilde{g})(1,t_{2}^{\prime})+(t_{1}-1)\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}\geq 0$ for $t_{1}=0$, also
$(\tilde{F}-\tilde{g})(-1,t_{2}^{\prime})+(t_{1}-1)\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}$ for all $t_{1}\geq 0$. Thus, to show
$(\tilde{F}-\tilde{g})(t_{1}^{\prime},t_{2}^{\prime})\geq 0$ for all
$t_{1}^{\prime}\geq 0,\frac{-1}{2}\leq t_{2}\leq\frac{1}{2}$, we need only
show for all $t_{2}\in[\frac{-1}{2},\frac{1}{2}]$ that

$(\tilde{F}-\tilde{g})(1,t_{2}^{\prime})+\frac{\partial\tilde{F}-\tilde{g}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}=\tilde{F}(1,t_{2}^{\prime})-\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}-\tilde{g}(0,t_{2}^{\prime})\geq 0.$

Now we claim for all $t_{2}^{\prime}$ that

$\tilde{F}(-1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}-\tilde{g}(0,t_{2}^{\prime})\geq\tilde{F}(1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}-\tilde{g}(0,t_{2}^{\prime}),$

which would imply that showing
$\tilde{F}(1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}-\tilde{g}(0,t_{2}^{\prime})\geq 0$ suffices
to prove $\tilde{F}\geq\tilde{g}$ in the entire critical region. Now 5.3 holds
because $\tilde{F}(-1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}$ and
$\tilde{F}(1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}-\tilde{g}(0,t_{2}^{\prime})$ are both
tangent approximations of $\tilde{F}(0,t_{2}^{\prime})$ from $t_{1}=-1,1$
respectively. Using the Lagrange Remainder of the Taylor Polynomial, we have

$\tilde{F}(-1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,t_{2}^{\prime}}=\tilde{F}(0,t_{2}^{\prime})-\frac{\tilde{f_{1}}^{\prime\prime}(\chi_{1})}{2!}\geq\tilde{F}(0,t_{2}^{\prime})-\frac{\tilde{f_{1}}^{\prime\prime}(\chi_{2})}{2!}=\tilde{F}(1,t_{2}^{\prime})+\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}^{\prime}}$

where $\xi_{1}\in[-1,0],\xi_{2}\in[0,1]$ and using absolute monotonicity of
$\tilde{f_{1}}$.

####  5.3.1 Proving the Linear Approximation Bound

So it suffices to show
$\phi(t_{2}):=\tilde{F}(1,t_{2})-\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{1,t_{2}}-\tilde{g}(0,t_{2})\geq 0$ for all
$t_{2}\in[-1/2,1/2]$. We can express $\phi$ as

$\phi(t_{2})=(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1))\tilde{f_{2}}(t_{2})-b_{0,0}-b_{0,1}t_{2}-b_{0,2}t_{2}^{2}.$

Using our technical bounds on $\tilde{\theta}$, we show in the appendix that
for $a<\pi^{2}$, we have:

######  Lemma 30.

$\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\geq 0$.

Thus, $\phi^{(3)}\geq 0$, and so the 2nd degree Taylor polynomial at
$t_{2}=-\frac{1}{2}$ yields a lower bound on $\phi$ for $t_{2}\geq-1/2$:

$\phi(t_{2})\geq A+B(t_{2}+1/2)+\frac{C}{2}(t_{2}+1/2)^{2}$

where

$\displaystyle A=\phi(-1/2)$
$\displaystyle=\left(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}(-1/2)-b_{0,0}+\frac{b_{0,1}}{2}-\frac{b_{0,2}}{4}$
(74)
$\displaystyle=\left(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}(-1/2)-a_{0,0}+\frac{b_{0,1}}{2}-\frac{b_{0,2}}{2}$
(75)
$\displaystyle=\left(\frac{1}{2}\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}(-1/2)+\frac{1}{6}\tilde{f_{1}}(-1)\tilde{f_{2}}^{\prime}(1/2)-\tilde{f_{1}}(-1)\left(\frac{2}{9}\tilde{f_{2}}(-1)+\frac{5}{18}\tilde{f_{2}}(1/2)\right)$
(76) $\displaystyle B=\phi^{\prime}(-1/2)$
$\displaystyle=(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1))\tilde{f_{2}}^{\prime}(-1/2)-b_{0,1}+b_{0,2}$
(77)
$\displaystyle=b_{0,2}-\tilde{f_{1}}^{\prime}(1)\tilde{f_{2}}^{\prime}(-1/2)$
(78)
$\displaystyle=\frac{4}{9}\tilde{f_{1}}(-1)\left(\tilde{f_{2}}(-1)-\tilde{f_{2}}(1/2)+\frac{3}{2}\tilde{f_{2}}^{\prime}(1/2)\right)-\tilde{f_{1}}^{\prime}(1)\tilde{f_{2}}^{\prime}(1/2)$
(79) $\displaystyle C=\phi^{\prime\prime}(-1/2)$
$\displaystyle=\left(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}^{\prime\prime}(-1/2)-2b_{0,2}$
(80)
$\displaystyle=\left(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}^{\prime\prime}(-1/2)-\frac{8}{9}\tilde{f_{1}}(-1)\left(\tilde{f_{2}}(-1)-\tilde{f_{2}}(1/2)+\frac{3}{2}\tilde{f_{2}}^{\prime}(1/2)\right).$
(81)

With these representations, we’ll produce lower bounds on $A$ and $C$ that
show $A,C>0$. So then it suffices to show that the quadratic has no zeroes,
i.e. if $B^{2}-2AC<0$. Moreover, if $B>0$, there’s nothing to show (as $\phi$
would definitely be positive), so to bound $B^{2}$ above, we can bound $B$
below and then square that bound. Thus, an upper bound on $B^{2}-2AC$ is
obtained by using our lower bound for $B$ and lower bounds for $A,C$, and we
show this upper bound for $B^{2}-2AC$ is negative, completing the proof when
$a<\pi^{2}$.

###  5.4 The critical region for $a\geq 9.6$

Figure 13: The figure depicts our proof strategy for showing
$\tilde{F}\geq\tilde{g}$ on the critical region when $a$ is large. The points
$p_{1}$ and $p_{2}$ are located at $(-\sqrt{2}/2,0),(0,-1/5)$, respectively.

The inequality $\tilde{F}\geq\tilde{g}$ for Regions A,B,C,D, and E from figure
13 is handled in lemmas 31, 32, 33, 34, and 35, respectively. We begin by
showing $\tilde{F}\leq\tilde{g}$ using the same linearization approach as in
the large $a$ case of the 4-point problem. Similarly as before, we define for
$b,c\in[-1,1]$

$\tilde{g}_{b,c}:=\tilde{g}-b_{0,2}t_{1}t_{2}+b_{0,2}(bt_{2}+ct_{2}-bc),$

with the important resulting property that
$\det(H_{\tilde{F}-\tilde{g}_{b,c}})<0$ everywhere.

######  Lemma 31.

For all $(t_{1},t_{2})$ such that $-1\leq t_{1}\leq-\sqrt{2}/2$, $0\leq
t_{2}\leq\frac{1}{2}$, $\tilde{F}\geq\tilde{g}$.

###### Proof.

First, we show the inequality for $t_{1}\leq-\sqrt{2}/2$, $\frac{1}{4}\leq
t_{2}\leq\frac{1}{2}$. In this region, we have
$\tilde{g}_{-1,.5}\geq\tilde{g}$ with equality when $t_{2}=1/2$ or $t_{1}=-1$.
As in the 4-point case, it suffices to show that
$\tilde{g}_{-1,.5}\leq\tilde{F}$ on the region, and it further suffices to
show this inequality holds on the boundary.
For the sections of the boundary where $t_{1}=-1$, or $t_{2}=1/2$, we have
$\tilde{g}_{-1,1/2}=\tilde{g}$ and so $\tilde{g}\leq\tilde{F}$ by lemmas 24
and 26. For the other two sections of the boundary, $t_{1}=-\sqrt{2}/2$ and
$t_{2}=\frac{1}{4}$, we proceed analogously to the 4-point case. Using
approximations of the $a_{i,j}$ coefficients (listed in the appendix) yields a
relatively simple upper bound, again call it $\tilde{g}_{-1,1/2}^{\prime}$ for
$\tilde{g}_{-1,1/2}$, which is linear in $a$, up to a factor of $e^{-a/3}$. As
a lower bound for $\tilde{F}$, we use the similar function

$\tilde{F}_{T}:=(e^{-ax^{2}}+e^{-a(x-1)^{2}})e^{-3au^{2}}$

where as usual

$x=\frac{\arccos(t_{1})}{2\pi},\hskip
14.22636ptu=\frac{\arccos(t_{2})}{2\pi}.$

and as before, $e^{a/3}\tilde{F}_{T}$ is convex in $a$. making the difference
$e^{a/3}(\tilde{F}_{T}-\tilde{g}_{-1,1/2}^{\prime})$ convex in $a$. So it
suffices to show
$\tilde{F}_{T}(t_{1}^{\prime},t_{2}^{\prime})\geq\tilde{g}_{t_{1}^{\prime},t_{2}^{\prime}}$
when $a=9.6$ and also that

$\frac{\partial\left[e^{a/3}(\tilde{F}_{T}-\tilde{g}_{-1,1/2}^{\prime})\right]}{\partial
a}\bigg{|}_{a=9.6,t_{1}=t_{1}^{\prime},t_{2}=t_{2}^{\prime}}\geq 0.$

In the appendix, we carry out these two calculations in a computer assisted
manner for the boundary portions $\\{(-\sqrt{2}/2,t_{2})\mid\frac{1}{4}\leq
t_{2}\leq\frac{1}{2}\\}$ and $\\{(t_{1},\frac{1}{4})\mid-1\leq
t_{1}\leq-\sqrt{2}/2\\}$, which complete the proof that
$\tilde{F}\geq\tilde{g}$ for
$(t_{1},t_{2})\in[-1,-\sqrt{2}/2]\times[\frac{1}{4},\frac{1}{2}]$.

To obtain $\tilde{F}\geq\tilde{g}$ for
$(t_{1},t_{2})\in[-1,-\sqrt{2}/2]\times[0,\frac{1}{4}]$, we conduct the same
procedure with $\tilde{g}_{-1,\frac{1}{4}}$ instead of
$\tilde{g}_{-1,\frac{1}{2}}$. Two sections of the boundary are immediate: when
$t_{1}=-1$ or $t_{2}=\frac{1}{4}$, $\tilde{g}=\tilde{g}_{-1,\frac{1}{4}}$, and
we have just established $\tilde{F}\geq\tilde{g}$ on those segments. The
remaining two segments of the boundary are handled exactly as in the first
section of the proof in the appendix. ∎

######  Lemma 32.

For all $(t_{1},t_{2})$ such that $-\sqrt{2}/2\leq t_{1}\leq 1$ and $0\leq
t_{2}\leq\frac{1}{2}$, $\tilde{F}\geq\tilde{g}$.

###### Proof.

By the convexity of $\tilde{F}-\tilde{g}$ in $t_{1}$, it satisfies to show the
following:

  1. 1.

$\tilde{F}(-\sqrt{2}/2,t_{2})\geq\tilde{g}(-\sqrt{2}/2,t_{2})$ for all
$t_{2}\in[0,1/2]$

  2. 2.

$\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-\sqrt{2}/2,t_{2}}\geq 0$ for $t_{2}\in[0,1/2]$.

The first of these follows from lemma 31. To prove the second, it actually
suffices to just show that $\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-\sqrt{2}/2,0}\geq 0$, which is handled in the appendix. This
sufficiency is because $\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\geq 0$ is convex in $t_{2}$ and satisfies
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-\sqrt{2}/2,-1/2}<0$ (due to the necessary condition
$\frac{\partial(\tilde{F}-\tilde{g})}{\partial t_{1}}\bigg{|}_{1,-1/2}<0$). ∎

######  Lemma 33.

For all $(t_{1},t_{2})$ such that $t_{1}\geq 0$ and $t_{2}\leq 0$,
$\tilde{F}\geq\tilde{g}$.

###### Proof.

We claim that for this portion of the critical region, it suffices to show at
each point that

$L_{1}(t_{1},t_{2}):=\frac{\tilde{f_{2}}^{\prime}\tilde{f_{1}}}{\tilde{f_{1}}^{\prime}\tilde{f_{2}}}-\frac{\frac{\partial\tilde{g}}{\partial
t_{2}}}{\frac{\partial\tilde{g}}{\partial t_{1}}}>0.$

Indeed, this would imply that $\tilde{g}$ increases along the level curves of
$\tilde{F}$ as $t_{1}$ increases. First, we note that if we take a point
$(t_{1}^{\prime},t_{2}^{\prime})$ on the boundary of the region where
$t_{2}=-\frac{1}{2}$ or $t_{1}=1$, then by the implicit function theorem, the
equation
$\tilde{F}=\tilde{f_{1}}(t_{1}^{\prime})\tilde{f_{2}}(t_{2}^{\prime})$
implicitly defines a function $t_{2}=p(t_{1})$ with negative slope from
$(t_{1}^{\prime},t_{2}^{\prime})$ to a point on the portion of the boundary
with $t_{1}=0$ or $t_{2}=0$. At any point on the curve, the vector
$[\frac{\tilde{f_{2}}^{\prime}}{\tilde{f_{2}}},-\frac{\tilde{f_{1}}^{\prime}}{\tilde{f_{1}}}]^{T}$
is parallel and in the positive $t_{1}$ direction of $p$, so $\tilde{g}$
increases along $p$ if

$\frac{\tilde{f_{2}}^{\prime}}{\tilde{f_{2}}}\frac{\partial\tilde{g}}{\partial
t_{1}}-\frac{\partial\tilde{g}}{\partial
t_{2}}\frac{\tilde{f_{1}}^{\prime}}{\tilde{f_{1}}}>0$

as needed. Thus, if the previous inequality holds, then $\tilde{F}-\tilde{g}$
is minimized along the right and bottom boundaries of the region, where we
have already showed it to be nonnegative in the previous section. We show
$L_{1}(t_{1},t_{2})>0$ in the appendix. ∎

######  Lemma 34.

For all $(t_{1},t_{2})$ such that $-\sqrt{2}/2\leq t_{1}\leq 0$,
$-\frac{1}{5}\leq t_{2}\leq 0$, $\tilde{F}\geq\tilde{g}$.

###### Proof.

The proceeds very similarly to that of lemma 31. We use $\tilde{F}_{T}$ in the
same fashion, and when $-\sqrt{2}/2\leq t_{1}\leq 0$ and $-.1\leq t_{2}\leq
0$, we use $\tilde{g}_{-\sqrt{2}/2,0}$. For $-\sqrt{2}/2\leq t_{1}\leq 0$ and
$-.2\leq t_{2}\leq-.1$, we use $\tilde{g}_{-\sqrt{2}/2,.1}$ The precise
calculations are carried out in the appendix. ∎

######  Lemma 35.

For all $(t_{1},t_{2})$ such that $t_{1}\leq 0$ and $t_{2}\leq-.2$,
$\tilde{F}\geq\tilde{g}$. Similarly, for all $(t_{1},t_{2})$ such that
$t_{1}\leq-\sqrt{2}/2$ and $t_{2}\leq 0$, $\tilde{F}\geq\tilde{g}$.

###### Proof.

Let $S$ be the set described in the lemma set. Fix $-.5\leq t_{2}^{\prime}\leq
0$. We need to show $\tilde{F}\geq\tilde{g}$ on
$S_{t_{2}^{\prime}}:=\\{t_{1}\mid(t_{1},t_{2}^{\prime})\in S\text{ and
}g(t_{1},t_{2}^{\prime})\geq\tilde{F}(-1,t_{2}^{\prime})\\}$. Since
$\tilde{g}$ is increasing in $t_{1}$ for $t_{2}\geq-1/2$, we have
$S_{t_{2}^{\prime}}$ is a closed interval. Either $S_{t_{2}^{\prime}}$ is
empty and there’s nothing to show, or we have $\tilde{F}-\tilde{g}\geq 0$ at
the right endpoint from our previous lemmas, namely 33 and 34. Equivalently,
at this right endpoint,

$0\leq\log(\frac{\tilde{F}}{\tilde{g}})=\log(\tilde{f_{1}})+\log(\tilde{f_{2}})-\log(\tilde{g}).$

Thus, to prove $\tilde{F}\geq\tilde{g}$ for all $t_{1}\in S_{t_{2}^{\prime}}$,
it suffices to show

$\frac{\partial\log(\frac{\tilde{F}}{\tilde{g}})}{\partial
t_{1}}=\frac{\tilde{f_{1}}^{\prime}}{\tilde{f_{1}}}-\frac{\frac{\partial\tilde{g}}{\partial
t_{1}}}{\tilde{g}}\leq 0$

throughout $S$. Now
$\frac{\tilde{f_{1}}^{\prime}}{\tilde{f_{1}}}-\frac{\frac{\partial\tilde{g}}{\partial
t_{1}}}{\tilde{g}}\leq 0$ is equivalent to

$\frac{\tilde{f_{1}}}{\tilde{f_{1}}^{\prime}}-\frac{\tilde{g}}{\frac{\partial\tilde{g}}{\partial
t_{1}}}=\frac{\tilde{f_{1}}}{\tilde{f_{1}}^{\prime}}-t_{1}-\frac{\tilde{g}(0,t_{2})}{a_{1,0}+a_{0,2}t_{2}}\geq
0,$

since each of
$\tilde{f_{1}},\tilde{f_{1}}^{\prime},\tilde{g},\frac{\partial\tilde{g}}{\partial
t_{1}}\geq 0.$ Notably, $\frac{\tilde{f_{1}}}{\tilde{f_{1}}^{\prime}}-t_{1}$
is a function only in $t_{1}$, while
$\frac{\tilde{g}(0,t_{2})}{a_{1,0}+a_{0,2}t_{2}}$ depends only on $t_{2}$. Let

$L_{2}(t_{1},t_{2}):=\frac{\tilde{f_{1}}}{\tilde{f_{1}}^{\prime}}-t_{1}-\frac{\tilde{g}(0,t_{2})}{a_{1,0}+a_{0,2}t_{2}}.$

If we can show that $L_{2}$ is decreasing in $t_{1}$ and $t_{2}$, then to show
$L_{2}\geq 0$ on all of $S$, we only need to show
$L_{2}(-\sqrt{2}/2,0),L_{2}(0,-.2)\geq 0$, which we handle in the appendix.
So it remains to show $L_{2}$ is decreasing in both $t_{1}$ and $t_{2}$.

$\frac{\partial
L_{2}}{t_{1}}=(\frac{\tilde{f_{1}}}{\tilde{f_{1}}^{\prime}})^{\prime}-1=\frac{(\tilde{f_{1}}^{\prime})^{2}-\tilde{f_{1}}^{\prime\prime}\tilde{f_{1}}}{\tilde{f_{1}}^{\prime
2}}-1=1-\frac{\tilde{f_{1}}^{\prime\prime}\tilde{f_{1}}}{(\tilde{f_{1}}^{\prime})^{2}}-1<0$

as needed. Similarly,

$\frac{\partial
L_{2}}{t_{2}}=\frac{b_{0,0}b_{0,2}-b_{0,1}b_{1,0}-b_{0,2}t_{2}(2b_{1,0}+b_{0,2}t_{2})}{(b_{1,0}+b_{0,2}t_{2})^{2}}$

whose sign depends only on
$N_{2}(t_{2}):=b_{0,0}b_{0,2}-b_{0,1}b_{1,0}-b_{0,2}t_{2}(2b_{1,0}+b_{0,2}t_{2})$.
Now

$N_{2}^{\prime}(t_{2})=-2b_{0,2}(b_{1,0}+b_{0,2}t_{2})=-2b_{0,2}\left(\frac{\partial\tilde{g}}{\partial
t_{1}}\right)\leq 0$

for $t_{2}\geq\frac{-1}{2}$. So the negativity of $N_{2}(t_{2})$ and (thus
$\frac{\partial L_{2}}{t_{2}}$) follows from $N_{2}(-1/2)<0$, which we check
in the appendix using our bounds on the $b_{i,j}$’s. ∎

Acknowledgements. The authors thank Henry Cohn, Denali Relles, Larry Rolen, Ed
Saff, and Yujian Su for helpful discussions at various stages of this project.

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##  Appendix A Divided differences and univariate interpolation

We review some basic results concerning one-dimensional polynomial
interpolation (e.g., see [BHS19, Section 5.6.2]). Let $f\in C^{m}[a,b]$ for
some $a,b$ be given along with some multiset

$T=\\{t_{0},t_{1},...,t_{m}\\}\subseteq[a,b].$

Then there then exists a unique polynomial $H_{T}(f)(t)$ of degree at most $m$
(called a Hermite interpolant of $f$) such that for each $\alpha\in T$, we
have $H^{(\ell)}_{T}(f)(\alpha)=f^{(\ell)}(\alpha)$ for $0\leq\ell<k_{\alpha}$
where $k_{\alpha}$ denotes the multiplicity of $\alpha$ in $T$. Let
$f[t_{0},...,t_{m}]$ denote the coefficient of $t^{m}$ in $H_{T}(f)(t)$. This
coefficient is called the m-th divided difference of $f$ for $T$. Then
$H_{T}(f)$, may be expressed as

$H_{T}(f)(t)=\sum_{k=0}^{m}f[t_{0},t_{1},...,t_{k}]p_{k}(T;t)$

where the partial products $p_{k}$ are defined by

$p_{0}(T;t):=1\text{ and }p_{j}(T;t):=\prod_{i<j}(t-t_{i}),\hskip
5.69046ptj=1,2,...,m.$

Then a generalization of the mean value theorem implies that there is some
$\xi\in[a,b]$ such that

$\frac{f^{(m)}(\xi)}{m!}=f[t_{0},t_{1},...,t_{m}].$

Putting these together, we arrive at the classical Hermite error formula:

$f(t)-H_{T}(f)(t)=f[t_{0},t_{1},...,t_{m},t]\prod_{i=0}^{m}(t-t_{i})=\frac{f^{(m+1)}(\xi)}{(m+1)!}\prod_{i=0}^{m}(t-t_{i}).$
(82)

In the case that $f$ is absolutely monotone on $[a,b]$, such as with
$\tilde{f_{1}}$ and $\tilde{f_{2}}$, then the sign of $f(t)-H_{T}(f)(t)$
equals the sign of $\prod^{m}_{i=0}(t-t_{i})$.

##  Appendix B Technical Estimates and Computations for $a<\pi^{2}$

###  B.1 $\theta$ estimates

Recall that for $a<\pi^{2}$, we define
$\tilde{f}_{1}(t_{1})=\tilde{\theta}(\frac{\pi}{a};t_{1})$ and
$\tilde{f}_{2}(t_{2})=\tilde{\theta}(\frac{\pi}{3a};t_{2})$. We then aim to
find an interpolant for $\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t_{2})$ in the $L$
case and
$\tilde{f_{1}}(t_{1})\tilde{f_{2}}(t_{2})+\tilde{f_{1}}(-t_{1})\tilde{f_{2}}(-t_{2})$
in the $A_{2}$ case.

For $0<a<\pi^{2}$, we will be using truncations of the formula

$\theta(\frac{\pi}{a};x):=\sum^{\infty}_{k=-\infty}e^{-dk^{2}}e^{2\pi
ikx}=1+\sum_{k\geq 1}2e^{-dk^{2}}\cos(2\pi kx).$

where $d:=\frac{\pi^{2}}{a}>1$, to obtain bounds on $\theta$. Thus, we will
define

$\displaystyle\tilde{f_{1}}(t_{1},j)=\tilde{f_{1}}(x_{1},j)$
$\displaystyle:=1+\sum_{k=1}^{j}2e^{-dk^{2}}\cos(2\pi kx_{1})$ (83)
$\displaystyle\tilde{f_{2}}(t_{2},j)=f_{2}(\frac{x_{2}}{\sqrt{3}},j)$
$\displaystyle:=1+\sum_{k=1}^{j}2e^{\frac{-dk^{2}}{3}}\cos(\frac{2\pi
kx_{2}}{\sqrt{3}})$ (84)

Here are the relevant technical bounds on $\tilde{f_{1}}$, $\tilde{f_{2}}$,
and their derivatives. Let’s first bound the tails of these series:

$\displaystyle\left\lvert\sum_{k\geq 3}2e^{-dk^{2}}\cos(2\pi kx)\right\rvert$
$\displaystyle\leq\sum_{k\geq 3}2e^{-dk^{2}}$ (85) $\displaystyle=2\sum_{k\geq
0}2e^{-d(k+3)^{2}}$ (86) $\displaystyle\leq 2e^{-4d}e^{-5}\sum_{k\geq
0}e^{-(k^{2}+6k)}$ (87) $\displaystyle\leq 2e^{-4d}e^{-5}\sum_{k\geq
0}e^{-(7k)}$ (88)
$\displaystyle=e^{-4d}\frac{2}{e^{5}(1-e^{-7})}\leq\frac{e^{-4d}}{50}.$ (89)

Similarly, we have

$\displaystyle\left\lvert\sum_{k\geq 5}2e^{-d/3k^{2}}\cos(2\pi
kx)\right\rvert$ $\displaystyle\leq\sum_{k\geq 5}2e^{-d/3k^{2}}$ (90)
$\displaystyle=2\sum_{k\geq 0}2e^{-d/3(k+5)^{2}}$ (91) $\displaystyle\leq
2e^{-16d/3}e^{-3}\sum_{k\geq 0}e^{-(k^{2}+10k)/3}$ (92) $\displaystyle\leq
2e^{-16d/3}e^{-3}\sum_{k\geq 0}e^{-11k/3}$ (93)
$\displaystyle=e^{-16d/3}\frac{2}{e^{3}(1-e^{-11/3})}\leq\frac{e^{-16d/3}}{5}.$
(94)

The following bounds on $\tilde{f_{1}}$ and $\tilde{f_{2}}$ immediately
follow:

######  Proposition 36.

For any $t_{1},t_{2}\in[-1,1]$,

$\displaystyle\tilde{f_{1}}(t_{1},2)-\frac{e^{-4d}}{50}$
$\displaystyle\leq\tilde{f_{1}}(t_{1})\leq\tilde{f_{1}}(t_{1},2)+\frac{e^{-4d}}{50}$
(95) $\displaystyle\tilde{f_{2}}(t_{2},4)-\frac{e^{-16d/3}}{5}$
$\displaystyle\leq\tilde{f_{2}}(t_{2})\leq\tilde{f_{2}}(t_{2},4)+\frac{e^{-16d/3}}{5}$
(96)

We also need bounds on the derivatives of $\tilde{f_{1}}$ and $\tilde{f_{2}}$.
With $x=\frac{\arccos(t)}{2\pi}$, then $\tilde{\theta}(c,t)=\theta(c,x)$, by
the chain rule, we have

$\frac{\partial\tilde{\theta}}{\partial t}=\frac{\partial\theta}{\partial
x}\frac{\partial x}{\partial t}.$

So

$\frac{\partial\tilde{\theta}}{\partial t}(\frac{\pi}{a},t)=\left(-\sum_{k\geq
1}2e^{-dk^{2}}(2\pi k)\sin(2\pi
kx)\right)\frac{-1}{2\pi\sqrt{1-t^{2}}}=\frac{\sum_{k\geq
1}2ke^{-dk^{2}}\sin(2\pi kx)}{\sin(2\pi x)}.$

For $t=\pm 1$ ($x=0,\frac{1}{2}$), we need to use L’Hopital’s rule. For
example, when $t=1$, so $x=0$,

$\frac{\partial\tilde{\theta}}{\partial t}(\frac{\pi}{a},t)=\frac{\sum_{k\geq
1}2ke^{-dk^{2}}(2\pi k)\cos(2\pi kx^{\prime})}{(2\pi)\cos(2\pi
x^{\prime})}=\sum_{k\geq 1}2k^{2}e^{-dk^{2}}$

and likewise, when $t=-1$

$\frac{\partial\tilde{\theta}}{\partial t}(\frac{\pi}{a},t)=\sum_{k\geq
1}(-1)^{n+1}2k^{2}e^{-dk^{2}}.$

Then we again bound the tails by comparison with geometric series. For
example,

$\displaystyle\sum_{k\geq 3}2k^{2}e^{-dk^{2}}$ $\displaystyle=\sum_{k\geq
0}2(k+3)^{2}e^{-d(k+3)^{2}}$ (97) $\displaystyle=e^{-9d}2\sum_{k\geq
0}(k+3)^{2}e^{-d(k^{2}+6k)}$ (98) $\displaystyle\leq e^{-4d}e^{-5}2\sum_{k\geq
0}9e^{k}e^{-(k^{2}+6k)}$ (99) $\displaystyle\leq e^{-4d}18e^{-5}\sum_{k\geq
0}e^{-6k}$ (100) $\displaystyle=\frac{18e^{-4d}}{e^{5}(1-e^{-6})}$ (101)
$\displaystyle\leq\frac{e^{-4d}}{8}$ (102)

To see this, we just used that $d>1$ and for all $k\geq 0$, $(k+3)^{2}\leq
9e^{k}$. In the $d/3$ case, since $(k+6)^{2}\leq 36e^{k/3}$, we analogously
have

$\displaystyle\sum_{k\geq 6}2k^{2}e^{-d/3k^{2}}$ $\displaystyle=\sum_{k\geq
0}2(k+6)^{2}e^{-d/3(k+6)^{2}}$ (103)
$\displaystyle=e^{-25d/3}2e^{-11/3}\sum_{k\geq 0}(k+6)^{2}e^{-d/3(k^{2}+12k)}$
(104) $\displaystyle\leq e^{-25d/3}2e^{-11/3}\sum_{k\geq
0}36e^{k/3}e^{-1/3(13k)}$ (105) $\displaystyle\leq
e^{-25d/3}72e^{-11/3}\sum_{k\geq 0}e^{-4k}$ (106)
$\displaystyle=e^{-25d/3}\frac{72}{e^{11/3}(1-e^{-4})}$ (107)
$\displaystyle\leq 2e^{-25d/3}.$ (108)

When $t=\pm\frac{1}{2}$,

$\frac{\sum_{k>j}2ke^{-dk^{2}}\sin(2\pi kx)}{\sin(2\pi
x)}\leq\frac{\sum_{k>j}\left\lvert 2ke^{-dk^{2}}\sin(2\pi
kx)\right\rvert}{\lvert\sin(2\pi
x)\rvert}\leq\sum_{k>j}4ke^{-dk^{2}}\leq\sum_{k>j}4k^{2}e^{-dk^{2}}$

and we can apply the previous bounds for $\sum_{k\geq 6}2k^{2}e^{-dk^{2}/3}$
Thus, we have for $d>1$ the following bounds, where again
$\tilde{f_{1}}^{\prime}(t,j),\tilde{f_{2}}^{\prime}(t,j)$ indicate the
truncation of the sum involved in
$\tilde{f_{1}}^{\prime}(t),\tilde{f_{2}}^{\prime}(t)$ after $j$ terms:

######  Proposition 37.

For $t_{1}=\pm 1$, $t_{2}\in\\{-1,-\frac{1}{2},\frac{1}{2},1\\}$

$\displaystyle\tilde{f_{1}}^{\prime}(t_{1},2)-\frac{e^{-4d}}{8}$
$\displaystyle\leq\tilde{f_{1}}^{\prime}(t_{1})\leq\tilde{f_{1}}^{\prime}(t_{1},2)+\frac{e^{-4d}}{8}$
(109) $\displaystyle\tilde{f_{2}}^{\prime}(t_{2},5)-4e^{-25d/3}$
$\displaystyle\leq\tilde{f_{2}}^{\prime}(t_{2})\leq\tilde{f_{2}}^{\prime}(t_{2},5)+4e^{-25d/3}$
(110)

Finally, we need bounds for $\tilde{f_{2}}(\pm\frac{1}{2})$. Again, using the
chain rule, we obtain

$\displaystyle\tilde{\theta}^{\prime\prime}(d/3,-1/2)$
$\displaystyle=\left(\sum_{k\geq 2}2ke^{-\frac{d}{3}k^{2}}\frac{\cot(2\pi
x)\sin(2\pi kx)-k\cos(2\pi kx)}{\sin^{2}(2\pi x)}\right)\bigg{|}_{x=-1/2}$
(111) $\displaystyle=\sum_{k\geq 2}-8/3ke^{-\frac{d}{3}k^{2}}\left(k\cos(2\pi
k/3)+\frac{1}{\sqrt{3}}\sin(2\pi k/3)\right).$ (112)

Likewise,

$\tilde{\theta}^{\prime\prime}(d/3,1/2)=\sum_{k\geq
2}-8/3ke^{-\frac{d}{3}k^{2}}\left(k\cos(2\pi k/3)-\frac{1}{\sqrt{3}}\sin(2\pi
k/3)\right)$

Note

$\left\lvert-8/3ke^{-\frac{d}{3}k^{2}}\left(k\cos(2\pi
k/3)+\frac{1}{\sqrt{3}}\sin(2\pi
k/3)\right)\right\rvert\leq\frac{8}{3}k(k+1)e^{-\frac{d}{3}k^{2}},$

so using the fact that $(k+6)(k+7)\leq 42e^{k/3}$ for $k\geq 0$ yields

$\displaystyle\left\lvert\sum_{k\geq
6}-8/3ke^{-\frac{d}{3}k^{2}}\left(k\cos(2\pi k/3)+\frac{1}{\sqrt{3}}\sin(2\pi
k/3)\right)\right\rvert$ $\displaystyle\leq\sum_{k\geq
6}\frac{8}{3}k(k+1)e^{-\frac{d}{3}k^{2}}$ (113)
$\displaystyle=\frac{8}{3}\sum_{k\geq 0}(k+6)(k+7)e^{-\frac{d}{3}(k+6)^{2}}$
(114) $\displaystyle\leq e^{-25d/3}\frac{8}{3}e^{-11/3}\sum_{k\geq
0}(k+6)(k+7)e^{-\frac{d}{3}(k^{2}+12k)}$ (115) $\displaystyle\leq
e^{-25d/3}\frac{8}{3}e^{-11/3}\sum_{k\geq 0}42e^{k/3}e^{-(k^{2}+12k)/3}$ (116)
$\displaystyle\leq e^{-25d/3}\frac{8}{3}e^{-11/3}\sum_{k\geq 0}42e^{-4k}$
(117) $\displaystyle=\frac{112e^{-25d/3}}{e^{11/3}(1-e^{-4})}\leq
5e^{-25d/3}.$ (118)

Thus, we obtain our final bounds

######  Proposition 38.

$\displaystyle\tilde{f_{2}}^{\prime\prime}(\pm\frac{1}{2},5)-5e^{-25d/3}\leq\tilde{f_{2}}^{\prime\prime}(\pm\frac{1}{2})\leq\tilde{f_{2}}^{\prime\prime}(\pm\frac{1}{2},5)+5e^{-25d/3}$
(119)

As a final remark, it is straightforward to check that each of these lower
bounds is at least 0 for all $d>1$.

###  B.2 $\tilde{F}\geq\tilde{g}$ for small $a$ and 4 points

First, we prove that $\frac{\partial^{3}\tilde{F}}{\partial t_{1}\partial
t_{2}^{2}}\bigg{|}_{-1,\frac{1}{2}}\geq 0$ to complete the proof of Lemma 17.
We’ll use the notation $\tilde{f_{1}}^{u}(t),\tilde{f_{1}}^{l}(t)$ to denote
the upper and lower bounds given in the previous section and likewise for
$\tilde{f_{2}}$. We have

$\displaystyle\frac{\partial^{3}\tilde{F}}{\partial t_{1}\partial
t_{2}^{2}}\bigg{|}_{-1,\frac{1}{2}}$
$\displaystyle\geq\prescript{}{l}{\tilde{f_{1}}}^{\prime}(-1)\prescript{}{l}{\tilde{f_{2}}}^{\prime\prime}(1/2)-\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)\prescript{}{u}{\tilde{f_{2}}}^{\prime\prime}(-1/2)$
(120)

In the accompanying mathematica notebook, we expand the products and simplify,
resulting in an analytic expression whose positivity is easy to verify. Next,
we prove Lemma 19

###### Proof.

$2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}\leq\frac{\partial^{2}\tilde{F}-\tilde{g}}{\partial
t_{1}\partial t_{2}}\bigg{|}_{-1,1/2}.$

Using the bounds from the previous section,,

$\displaystyle 2\frac{\partial\tilde{F}}{\partial t_{1}}\bigg{|}_{-1,1/2}$
$\displaystyle\leq
2\left[\prescript{}{u}{\tilde{f_{1}}}^{\prime}(-1)\prescript{}{u}{\tilde{f_{2}}}(1/2)-\prescript{}{l}{\tilde{f_{1}}}^{\prime}(1)\prescript{}{l}{\tilde{f_{2}}}(-1/2)\right],$
(121) $\displaystyle\frac{\partial^{2}\tilde{F}}{\partial t_{1}\partial
t_{2}}\bigg{|}_{-1,1/2}$
$\displaystyle=\tilde{f_{1}}^{\prime}(-1)\tilde{f_{2}}^{\prime}(1/2)+\tilde{f_{1}}^{\prime}(1)\tilde{f_{2}}^{\prime}(-1/2)$
(122)
$\displaystyle\geq\prescript{}{l}{\tilde{f_{1}}}^{\prime}(-1)^{\prime}\prescript{}{l}{\tilde{f_{2}}}(1/2)+\prescript{}{l}{\tilde{f_{1}}}^{\prime}(1)\prescript{}{l}{\tilde{f_{2}}}(-1/2).$
(123)

We show the desired inequality in a similar fashion as with
$\frac{\partial^{3}\tilde{F}}{\partial t_{1}\partial
t_{2}^{2}}\bigg{|}_{-1,\frac{1}{2}}$ in the Mathematica file.

It remains to show

$4(\tilde{F}(-1,1)-\tilde{F}(-1,1/2)),\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1}\leq 2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}.$

Again with our bounds we have:

$\displaystyle 4(\tilde{F}(-1,1)-\tilde{F}(-1,1/2))$
$\displaystyle=4\left[\tilde{f_{1}}(-1)(\tilde{f_{2}}(1)-\tilde{f_{2}}(-1/2))+\tilde{f_{1}}(1)(\tilde{f_{2}}(-1)-\tilde{f_{2}}(-1/2))\right]$
(124) $\displaystyle\leq
4\left[\prescript{}{u}{\tilde{f_{1}}}(-1)(\prescript{}{u}{\tilde{f_{2}}}(1)-\prescript{}{l}{\tilde{f_{2}}}(-1/2)+\prescript{}{u}{\tilde{f_{1}}}(1)(\prescript{}{u}{\tilde{f_{2}}}(-1)-\prescript{}{l}{\tilde{f_{2}}}(-1/2))\right]$
(125) $\displaystyle 2\frac{\partial\tilde{F}}{\partial
t_{1}}\bigg{|}_{-1,1/2}$ $\displaystyle\geq
2\left[\prescript{}{l}{\tilde{f_{1}}}^{\prime}(-1)\prescript{}{l}{\tilde{f_{2}}}(1/2)-\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)\prescript{}{u}{\tilde{f_{2}}}(-1/2)\right].$
(126)

The proof that the corresponding difference is nonnegative is in the notebook.

Finally,

$\displaystyle\frac{\partial\tilde{F}}{\partial
t_{2}}\bigg{|}_{-1,1}\leq\prescript{}{u}{\tilde{f_{1}}}(-1)\prescript{}{u}{\tilde{f_{2}}}^{\prime}(1)-\prescript{}{l}{\tilde{f_{1}}}(1)\prescript{}{l}{\tilde{f_{2}}}^{\prime}(-1)$
(127)

which we compare in mathematica with the upper bound for
$2\frac{\partial\tilde{F}}{\partial t_{1}}\bigg{|}_{-1,1/2}$ given above. ∎

###  B.3 $\tilde{F}\geq\tilde{g}$ for small $a$ and 6 points

First, we handle the necessary derivative conditions:

####  B.3.1 Satisfying Necessary Conditions

Recall we aim to show

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,-1}\geq 0$ (128)
$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,\frac{1}{2}}\geq 0$ (129)
$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{1,-\frac{1}{2}}\leq 0.$ (130)

###### Proof.

First, we’ll compute bounds for $b_{1,0},b_{0,2}$. Recall

$\displaystyle 2b_{1,0}$
$\displaystyle=\tilde{f_{1}}(1)\tilde{f_{2}}(-1/2)-\tilde{f_{1}}(-1)\tilde{f_{2}}(1/2)+\tilde{f_{1}}(-1)\tilde{f_{2}}^{\prime}(1/2)$
(131) $\displaystyle\frac{9}{4}b_{0,2}$
$\displaystyle=\tilde{f_{1}}(-1)(\tilde{f_{2}}(-1)-\tilde{f_{2}}(\frac{1}{2})+\frac{3}{2}\tilde{f_{2}}^{\prime}(\frac{1}{2})).$
(132)

Thus, using our bounds,

$\displaystyle\prescript{}{l}{\tilde{f_{1}}}(1)\prescript{}{l}{\tilde{f_{2}}}(-1/2)-\prescript{}{u}{\tilde{f_{1}}}(-1)\prescript{}{u}{\tilde{f_{2}}}(1/2)+\prescript{}{l}{\tilde{f_{1}}}(-1)\prescript{}{l}{\tilde{f_{2}}}^{\prime}(1/2)\leq
2a_{1,0}$ (133) $\displaystyle
2a_{1,0}\leq\prescript{}{u}{\tilde{f_{1}}}(1)\prescript{}{u}{\tilde{f_{2}}}(-1/2)-\prescript{}{l}{\tilde{f_{1}}}(-1)\prescript{}{l}{\tilde{f_{2}}}(1/2)+\prescript{}{u}{\tilde{f_{1}}}(-1)\prescript{}{u}{\tilde{f_{2}}}^{\prime}(1/2)$
(134)
$\displaystyle\prescript{}{l}{\tilde{f_{1}}}(-1)(\prescript{}{l}{\tilde{f_{2}}}(-1)-\prescript{}{u}{\tilde{f_{2}}}(1/2)+\frac{3}{2}\prescript{}{l}{\tilde{f_{2}}}^{\prime}(1/2))\leq\frac{9}{4}b_{0,2}\leq\prescript{}{u}{\tilde{f_{1}}}(-1)(\prescript{}{u}{\tilde{f_{2}}}(-1)-\prescript{}{l}{\tilde{f_{2}}}(1/2)+\frac{3}{2}\prescript{}{u}{\tilde{f_{2}}}^{\prime}(1/2)).$
(135)

Call these upper and lower bounds $a^{u}_{i,j},a^{l}_{i,j}$ respectively. Now
we compute

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,-1}$
$\displaystyle=\tilde{f_{1}}^{\prime}(d,-1)\tilde{f_{2}}(d,-1,5)-(a_{1,0}-a_{0,2})$
(136)
$\displaystyle\geq\prescript{}{l}{\tilde{f_{1}}}^{\prime}(-1)\prescript{}{l}{\tilde{f_{2}}}(-1)-a^{u}_{1,0}+a^{l}_{0,2}$
(137)

which we show positivity of in the notebook.

The other conditions are similar. We must check the positivity of the lower
bound

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{-1,\frac{1}{2}}\geq\prescript{}{l}{\tilde{f_{1}}}^{\prime}(-1)\prescript{}{l}{\tilde{f_{2}}}(\frac{1}{2})-a^{u}_{1,0}-\frac{1}{2}a^{u}_{0,2}$
(138)

and the upper bound

$\displaystyle\frac{\partial(\tilde{F}-\tilde{g})}{\partial
t_{1}}\bigg{|}_{1,-\frac{1}{2}}\leq\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)\prescript{}{u}{\tilde{f_{2}}}(-\frac{1}{2})-a^{l}_{1,0}+\frac{1}{2}a^{u}_{0,2}$
(139)

which completes the proof of the necessary conditions in the small $a$ case. ∎

####  B.3.2 Computations for the Linear Approximation Bound

It remains to show $\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)>0$ and produce
lower bounds on $A,B,C$ such that $A>0,C>0$ and $B^{2}-2AC<0$. Equivalently,
with those same bounds, $\frac{B}{C}-\frac{2A}{B}>0$ (as we’ll assume $B<0$
else there is nothing to show).

######  Lemma 39.

The following inequalities hold:

  1. 1.

$\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\geq 0$

  2. 2.

$A,C>0$

###### Proof.

We’ll first show the stronger statement that
$\tilde{f_{1}}(1)-2\tilde{f_{1}}^{\prime}(1)\geq 0$. from the aforementioned
bounds,

$\displaystyle\tilde{f_{1}}(1)-2\tilde{f_{1}}^{\prime}(1)$
$\displaystyle\geq\prescript{}{l}{\tilde{f_{1}}}(1)-\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)$
(140)

whose nonnegativity is in the mathematica notebook. Next we want a lower bound
for $A$. Recall

$\displaystyle 2A$
$\displaystyle=\left(\tilde{f_{1}}(1)-2\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}(-1/2)+\frac{1}{3}\tilde{f_{1}}(-1)\tilde{f_{2}}^{\prime}(1/2)-\tilde{f_{1}}(-1)\left(\frac{4}{9}\tilde{f_{2}}(-1)+\frac{5}{9}\tilde{f_{2}}(1/2)\right)$
(141)
$\displaystyle\geq\left(\prescript{}{l}{\tilde{f_{1}}}(1)-2\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)\right)\prescript{}{l}{\tilde{f_{2}}}(-1/2)+\frac{1}{3}\prescript{}{l}{\tilde{f_{1}}}(-1)\prescript{}{l}{\tilde{f_{2}}}^{\prime}(1/2)-\prescript{}{u}{\tilde{f_{1}}}(-1)\left(\frac{4}{9}\prescript{}{u}{\tilde{f_{2}}}(-1)+\frac{5}{9}\prescript{}{u}{\tilde{f_{2}}}(1/2)\right).$
(142)

Finally, we generate a lower bound for $C$. Again recall

$\displaystyle C$
$\displaystyle=\left(\tilde{f_{1}}(1)-\tilde{f_{1}}^{\prime}(1)\right)\tilde{f_{2}}^{\prime\prime}(-1/2)-2b_{0,2}$
(144)
$\displaystyle\leq\left(\prescript{}{l}{\tilde{f_{1}}}(1)-\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)\right)\prescript{}{u}{\tilde{f_{2}}}^{\prime\prime}(-1/2)-2b^{l}_{0,2}$
(145)

We check the positivity of both bounds in the notbeook. ∎

It remains to show We also need to provide a lower bound for $B$. Recall

$\displaystyle B$
$\displaystyle=b_{0,2}-\tilde{f_{1}}^{\prime}(1)\tilde{f_{2}}^{\prime}(-1/2)$
(146) $\displaystyle\geq
b^{l}_{0,2}-\prescript{}{u}{\tilde{f_{1}}}^{\prime}(1)\prescript{}{u}{\tilde{f_{2}}}^{\prime}(-1/2).$
(147)

Utilizing the computations in the mathematica notebook we’ve found lower
bounds for $A,B,C$, with the $A$ and $C$ bounds shown to be positive:

$\displaystyle A_{L}$
$\displaystyle:=2951/600e^{-16d/3}+4879/600e^{-13d/3}-2429/200e^{-4d}+2e^{-3d}$
(148) $\displaystyle B_{L}$
$\displaystyle:=14309/450e^{-16d/3}-65/4e^{-13d/3}$ (149) $\displaystyle
C_{L}$
$\displaystyle:=3687/25e^{-7d}-688/45e^{-19d/3}-9377/225e^{-16d/3}-24e^{-3d}+16e^{-7d/3},$
(150)

and either $B\geq 0$ in which case the taylor polynomial is positive for
$t_{2}\geq-\frac{1}{2}$ or $B<0$ and so $B^{2}-2AC<B_{L}^{2}-2A_{L}C_{L}$,
where $B_{l}$ is our lower bound for $B$. Thus, we’ll assume $B<0$ and show
$B_{L}^{2}-2A_{L}C_{L}<0$, which is equivalent to
$B_{L}/C_{L}-2A_{L}/B_{L}>0$. We plug in $d=1$ and see it holds there. Then,
we claim $B_{L}/C_{L}$ is increasing while $2A_{L}/B_{L}$ is decreasing, which
would complete the proof. This final claim is proved in the mathematica
notebook. For example, the sign of the derivative of
$B_{L}/C_{L}=\frac{e^{13d/3}}{e^{13d/3}C_{l}}$ depends only on the sign of
$(B_{l}e^{13d/3})^{\prime}C_{l}-B_{l}(e^{13d/3}C_{l})^{\prime}$, which we
check to be positive for $d\geq 1$. The one proposition we use repeatedly for
showing the positivity of various bounds is:

######  Proposition 40.

Let $h(d)=a_{1}e^{b_{1}d}+\dots+a_{n}e^{b_{n}d}$, where the $b_{i}$’s are
increasing and there is some $j$ such that $a_{1},\dots,a_{j-1}\leq 0\leq
a_{j},\dots,a_{n}$. Then $h(d)\geq 0$ for all $d\geq 1$ if and only if
$h(1)\geq 0$.

###### Proof.

Note $h(d)\geq 0$ if and only if $h(d)e^{-b_{j}d}\geq 0$.

$h(d)e^{-b_{j}d}=a_{1}e^{(b_{1}-b_{j})d}+\dots+a_{j-1}e^{(b_{j-1}-b_{j})d}a_{j}+\dots+a_{n}e^{b_{n}d}.$

By our assumptions on the $a_{i}$’s and $b_{i}$’s, for $i<j$, $a_{i}$ and
$b_{i}-b_{j}$ are both negative, so $a_{i}e^{(b_{i}-b_{j})}$ is nondereasing.
For $i\geq j$, both $a_{i}$ and $b_{i}-b_{j}$ are nonnegative, and so again
$a_{i}e^{(b_{i}-b_{j})}$ is nondecreasing. Thus, $h(d)e^{-b_{j}d}\geq 0$ is
nondecreasing, which suffices. ∎

##  Appendix C Technical Estimates and Computations for large $a$

Throughout, assume that $a\geq 9.6$. We’ll set $\epsilon=\frac{1}{1000}$ so
that for all $a\geq 9.6$:

$\displaystyle\epsilon\geq 2\sum_{n\geq 1}e^{-an}\text{ and }\epsilon\geq
5e^{-a}.$ (151)

Then set $\epsilon_{2}:=\frac{1}{100}\geq 4(1+\epsilon)^{2}\sum_{n\geq
1}e^{-2(9.6)n/3}$.

In the large $a$ case, it is preferable to use the following formula for
$\theta$ because of its rapid convergence:

$\theta(c;x)=c^{-1/2}\sum^{\infty}_{k=-\infty}e^{-\frac{\pi(k+x)^{2}}{c}}.$

Thus, we’ll use the formulas

$\displaystyle\tilde{f_{1}}(x_{1})=\sqrt{\frac{a}{\pi}}\theta(\frac{\pi}{a},x)=\sum^{\infty}_{k=-\infty}e^{-a(k+x)^{2}}$
(152) $\displaystyle
f_{2}(x_{2})=\sqrt{\frac{3a}{\pi}}\theta(\frac{\pi}{3a},x)=\sum^{\infty}_{k=-\infty}e^{-3a(k+x)^{2}},$
(153)

yielding in the $A_{2}$ case

$F(x)=\tilde{f_{1}}(x_{1})f_{2}(x_{2})+\tilde{f_{1}}(\frac{1}{2}-x_{1})f_{2}(\frac{1}{2}-x_{2}),$

and in the $L$ case,

$F(x)=\tilde{f_{1}}(x_{1})f_{2}(x_{2}).$

###  C.1 Basic Lemmas and Other Estimates

We first establish a couple basic workhorse lemmas bounding the tails of
$\theta$ and $\tilde{\theta}^{\prime}$ which we’ll use repeatedly.

######  Lemma 41.

For $0\leq x\leq\frac{1}{2}$,

$\displaystyle
e^{-ax^{2}}+e^{-a(x-1)^{2}}\leq\theta(\frac{\pi}{a},x)\leq(1+\epsilon)(e^{-ax^{2}}+e^{-a(x-1)^{2}})$
$\displaystyle=(1+\epsilon)e^{-ax^{2}}(1+e^{-a(1-2x)})\leq
2(1+\epsilon)e^{-ax^{2}}$ (154) $\displaystyle
e^{-ax^{2}}\leq\theta(\frac{\pi}{a},x)\leq e^{-ax^{2}}(1+2\sum_{n\geq
1}e^{-a(n^{2}-2nx)}).$ (155)

In particular, for $u=\frac{\arccos(t_{2})}{2\pi}$, $t_{2}\geq-\frac{1}{2}$,

$\displaystyle e^{-3au^{2}}$
$\displaystyle\leq\tilde{f_{2}}(t_{2})\leq(1+\epsilon)e^{-3au^{2}}$ (156)
$\displaystyle 1$ $\displaystyle\leq\tilde{f_{1}}(1),\tilde{f_{2}}(1)\leq
1+\epsilon$ (157) $\displaystyle 2e^{-a/4}$
$\displaystyle\leq\tilde{f_{1}}(-1)\leq 2(1+\epsilon)e^{-a/4}$ (158)
$\displaystyle 2e^{-3a/4}$ $\displaystyle\leq\tilde{f_{2}}(-1)\leq
2(1+\epsilon)e^{-3a/4}.$ (159)

###### Proof.

Recall $\theta(\frac{\pi}{a},x)=\sum_{n\in\mathbb{Z}}e^{-a(n+x)^{2}}$. The
lower bounds follow from simply truncating the series. To obtain the first
upper bound, observe

$\displaystyle\sum_{n\in\mathbb{Z}}e^{-a(n+x)^{2}}$ $\displaystyle=\sum_{n\geq
0}e^{-a(n+x)^{2}}+\sum_{n\leq-1}e^{-a(n+x)^{2}}$ (160)
$\displaystyle=e^{-ax^{2}}\sum_{n\geq
0}e^{-a(n^{2}+2nx)}+e^{-a(x-1)^{2}}\sum_{n\geq 0}e^{-a(n^{2}+2(1-x)n)}$ (161)
$\displaystyle\leq e^{-ax^{2}}\sum_{n\geq
0}e^{-an^{2}}+e^{-a(x-1)^{2}}\sum_{n\geq 0}e^{-a(n^{2})}$ (162)
$\displaystyle\leq(e^{-ax^{2}}+e^{-a(1-x)^{2}})\sum_{n\geq 0}e^{-an}$ (163)
$\displaystyle\leq(1+\epsilon)(e^{-ax^{2}}+e^{-a(1-x)^{2}}).$ (164)

The second upper bound follows in a similar fashion using the fact that for
$n\geq 1$, the $n$th term is larger than the $-(n+1)$th term. The particular
bounds all follow immediately except the first, where we use that if
$t_{2}\geq-\frac{1}{2}$, $u\in[0,1/3]$ and so

$2\sum_{n\geq 1}e^{-3a(n^{2}-2nx)}\leq 2\sum_{n\geq
1}e^{-3a(n-2n(1/3))}=2\sum_{n\geq 1}e^{-a}\leq\epsilon.$

∎

######  Lemma 42.

With $x=\frac{\arccos(t)}{2\pi}$,

$\displaystyle\frac{ae^{-ax^{2}}(x-(1-x)e^{-a(1-2x)})}{\pi\sin(2\pi
x)}\leq\frac{\partial\tilde{\theta}(\frac{\pi}{a},x)}{\partial
t}\leq\frac{axe^{-ax^{2}}}{\pi\sin(2\pi x)}$ (165)

###### Proof.

Using $x=\frac{\arccos(t)}{2\pi}$, we have

$\frac{\partial\tilde{\theta}(\frac{\pi}{a},x)}{\partial
t}=\frac{\sum_{n\in\mathbb{Z}}-2a(n+x)e^{-a(n+x)^{2}}}{-2\pi\sin(2\pi
x)}=\frac{a\sum_{n\in\mathbb{Z}}(n+x)e^{-a(n+x)^{2}}}{\pi\sin(2\pi x)}.$

Let $s_{n}=(n+x)e^{-a(n+x)^{2}}$. Now to obtain the lower bound, we simply
show check that for $n\geq 1$, $s_{n}+s_{-n-1}\geq 0$. Thus, $s_{0}+s_{-1}$
yields a lower bound. Similarly, we check $s_{n}+s_{-n}\leq 0$, so $s_{0}$
yields an upper bound. It will be independently useful that $s_{0}\geq s_{-1}$
for $\frac{1}{4}\leq x\leq\frac{1}{2}$. Indeed, in this case, taking
$v=\frac{1}{2}-x$,

$ae^{-ax^{2}}(x-(1-x)e^{-a(1-2x)}=ae^{-ax^{2}}(1/2-v-(1/2+v)e^{-2av})$

and $(1/2-v-(1/2+v)e^{-2av})$ is concave in $v$ for all $a\geq 9.6$,