# On universally optimal lattice phase transitions and energy minimizers of
completely monotone potentials

Senping Luo  ,  Juncheng Wei  and  Wenming Zou  School of Mathematics and
statistics, Jiangxi Normal University, Nanchang, 330022, China  Department of
Mathematics, University of British Columbia, Vancouver, B.C., Canada, V6T 1Z2
Department of Mathematical Sciences, Tsinghua University, Beijing 100084,
China<EMAIL_ADDRESS>or<EMAIL_ADDRESS><EMAIL_ADDRESS>zou-
<EMAIL_ADDRESS>
###### Abstract.

We consider the minimizing problem for energy functionals with two types of
competing particles and completely monotone potential on a lattice. We prove
that the minima of sum of two completely monotone functions among lattices is
located exactly on a special curve which is part of the boundary of the
fundamental region. We also establish a universal result for square lattice
being the optimal in certain interval, which is surprising. Our result
establishes the hexagonal-rhombic-square-rectangular transition lattice shapes
in many physical and biological system (such as Bose-Einstein condensates and
two-component Ginzburg-Landau systems). It turns out, our results also apply
to locating the minimizers of sum of two Eisenstein series, which is new in
number theory.

##  1\. Introduction and main results

###  1.1. Lattice energy functional

We first introduce the energy functional considered in this paper. Consider a
set of $N$ classical identical particles in ${\mathbb{R}}^{d}$ ($d\leq 3$),
interacting by pairs through a potential $\mathcal{V}$ depending on the
distance between them. Let $x_{1},x_{2},\cdots,x_{N}\in{\mathbb{R}}^{d}$ and
$p_{1},p_{2},\cdots,p_{N}\in{\mathbb{R}}^{d}$ be the positions and momenta of
these particles.

The Hamiltonian dynamics of the model, based on the energy, is given by

$\mathcal{H}_{N}(x_{1},\cdots,x_{N},p_{1},\cdots,p_{N})=\sum_{i=1}^{N}\frac{|p_{i}|^{2}}{2m}+\sum_{1\leq
i<j\leq N}\mathcal{V}(|x_{i}-x_{j}|^{2}).$ (1.1)

Here $m$ is the mass of the particles and $|\cdot|$ is the Euclidean norm of
${\mathbb{R}}^{d}$. When the temperature is zero, the equilibrium states are
the minima of the functional $\mathcal{H}_{N}$, which satisfy
$p_{1}=\cdots=p_{N}$. If one is interested in these equilibrium states, then
it suffices to consider the potential energy

$\mathcal{E}(\\{x_{i}\\}_{i=1}^{N}):=\sum_{1\leq i\neq j\leq
N}\mathcal{V}(|x_{i}-x_{j}|^{2}),$ (1.2)

where $\mathcal{V}$ is referred as the potential between two particles.

Let $\Lambda$ be a two-dimensional lattice generated by the two basis
$\alpha_{1}$ and $\alpha_{2}$. Assume that there are two types of competing
particles $A$ and $B$. Now we consider the dynamics of two dimensional
periodical lattices alternated by type A particle and type B particle (see
figure 1), where one is located at the center of another. That is, we only
consider a special type of $(\Omega_{1},\Omega_{2})$, termed two species
periodic assemblies of discs, denoted by
$(\Omega_{\alpha,1},\Omega_{\alpha,2})$, with

$\displaystyle\Omega_{\alpha,1}$
$\displaystyle=\bigcup_{\lambda\in\Lambda}\Big{\\{}B(\xi,r_{1})\cup
B(\xi^{\prime},r_{1}):\xi=\frac{3}{4}\alpha_{1}+\frac{1}{4}\alpha_{2}+\lambda,\
\xi^{\prime}=\frac{1}{4}\alpha_{1}+\frac{3}{4}\alpha_{2}+\lambda\Big{\\}},$
(1.3) $\displaystyle\Omega_{\alpha,2}$
$\displaystyle=\bigcup_{\lambda\in\Lambda}\Big{\\{}B(\xi,r_{2})\cup
B(\xi^{\prime},r_{2}):\xi=\frac{1}{4}\alpha_{1}+\frac{1}{4}\alpha_{2}+\lambda,\
\xi^{\prime}=\frac{3}{4}\alpha_{1}+\frac{3}{4}\alpha_{2}+\lambda\Big{\\}}.$
(1.4)

The A type particles are located on red spots, denoted by $P_{4k+1},P_{4k+3}$;
the B type particles are located on yellow spots, denoted by
$P_{4k+2},P_{4k+4}$, where $k\in N$. Let $\mathcal{V}$ be the potential of the
system. Then the energy (1.2) of the system consists of two parts, the self-
interaction part of each type particles and the interaction between the two
different types of particles, while these two parts contribute to the total
energy in distribution of $\kappa$ and $1-\kappa$, where $\kappa\in(0,1)$.

Figure 1. Two lattices with centers at the lattice points and the half lattice
points.

Since the system is an infinite particles system, we first consider a mix of
$N-$particles $(\\{x_{i}\\}_{i=1}^{N})$ of type A and $N-$particles
$(\\{y_{i}\\}_{i=1}^{N})$ of type B to approximate the infinite particles
system. For convenience, we set $N=M^{2}$. The energy of self interaction of
the same type particles is

$\displaystyle\mathcal{E}_{a}:=\sum_{(k,j)\in S_{N}\times
S_{N}\setminus\\{(0,0)\\}}\frac{1}{2}\mathcal{V}(|k\alpha_{1}+j\alpha_{2}|^{2}){(M-|k|)(M-|j|)},$

where the set $S_{N}$ is defined by
$\\{-(M-1),-(M-2),\cdots,0,\cdots,M-2,M-1\\}$. The limit per-energy of the
self interaction of the same type particles is

$\displaystyle\mathcal{E}_{per,1}({\\{x_{j}\\}_{j=1}^{\infty}})=\mathcal{E}_{per,1}({\\{y_{j}\\}_{j=1}^{\infty}}):=\lim_{N\rightarrow\infty}\frac{\mathcal{E}_{a}}{N}$
$\displaystyle=$ $\displaystyle\lim_{M\rightarrow\infty}\sum_{(k,j)\in
S_{N}\times
S_{N}\setminus\\{(0,0)\\}}\mathcal{V}(|k\alpha_{1}+j\alpha_{2}|^{2})\frac{(M-|k|)(M-|j|)}{2M^{2}}.$

Such a limit converges if the summation $\sum_{(k,j)\in S_{N}\times
S_{N}\setminus\\{(0,0)\\}}\mathcal{V}(|k\alpha_{1}+j\alpha_{2}|^{2})$ is
summable. One has

$\displaystyle\mathcal{E}_{per,1}({\\{x_{j}\\}_{j=1}^{\infty}})=\mathcal{E}_{per,1}({\\{y_{j}\\}_{j=1}^{\infty}})=\frac{1}{2}\sum_{(k,j)\neq\\{(0,0)\\}}\mathcal{V}(|k\alpha_{1}+j\alpha_{2}|^{2}).$

One can rewrite it in a concise form on the lattice $\Lambda$

$\mathcal{E}_{per,1}({\\{x_{j}\\}_{j=1}^{\infty}})=\mathcal{E}_{per,1}({\\{y_{j}\\}_{j=1}^{\infty}})=\frac{1}{2}\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\mathcal{V}(|\mathbb{P}|^{2}).$

Next we consider the interaction between two different particles. The energy
between $N-$particles$(\\{x_{i}\\}_{i=1}^{N})$ of type A and
$N-$particles$(\\{y_{i}\\}_{i=1}^{N})$ of type B is

$\displaystyle\mathcal{E}_{b}:=\sum_{(k,j)\in S_{N}\times
S_{N}\setminus\\{(0,0)\\}}\mathcal{V}(|\frac{1}{2}(\alpha_{1}-\alpha_{2})+k\alpha_{1}+j\alpha_{2}|^{2}){(M-|k|)(M-|j|)}.$

The limit per-energy of the interaction between two different type particles
$A,B$ is determined by

$\displaystyle\mathcal{E}_{per,2}({\\{x_{j}\\}_{j=1}^{\infty}},{\\{y_{j}\\}_{j=1}^{\infty}}):$
$\displaystyle=\lim_{N\rightarrow\infty}\frac{\mathcal{E}_{b}}{N}$
$\displaystyle=\sum_{(k,j)\in\mathbb{Z}\bigotimes\mathbb{Z}}\mathcal{V}(|\frac{1}{2}(\alpha_{1}-\alpha_{2})+k\alpha_{1}+j\alpha_{2}|^{2}).$

One can rewrite it as

$\displaystyle\mathcal{E}_{per,2}({\\{x_{j}\\}_{j=1}^{\infty}},{\\{y_{j}\\}_{j=1}^{\infty}})=\sum_{\mathbb{P}\in\Lambda}\mathcal{V}(|\frac{\mathbb{P}+1}{2}|^{2}).$

Therefore, the total energy of two dimensional periodical alternating system
with contribution $\kappa,1-\kappa$ of same-type and different-types
interaction is determined by

$\displaystyle\mathcal{E}_{per}(A,B):$
$\displaystyle=\kappa(\mathcal{E}_{per,1}({\\{x_{j}\\}_{j=1}^{\infty}})+\mathcal{E}_{per,1}({\\{y_{j}\\}_{j=1}^{\infty}}))+(1-\kappa)\mathcal{E}_{per,2}({\\{x_{j}\\}_{j=1}^{\infty}},{\\{y_{j}\\}_{j=1}^{\infty}})$
(1.5)
$\displaystyle=\kappa\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\mathcal{V}(|\mathbb{P}|^{2})+(1-\kappa)\sum_{\mathbb{P}\in\Lambda}\mathcal{V}(|\frac{\mathbb{P}+1}{2}|^{2}),$

where $\kappa\in[0,1]$ and we include the endpoint cases $\kappa=0,1$ for
convenience. To find out which shape of lattice $\Lambda$ minimizes the
lattice energy (1.5), we consider the minimum of the functional
$\mathcal{E}_{per}(A,B)$ over $\Lambda$, which is the main contribution of
this paper.

###  1.2. Sum of two completely monotone functions on the lattice and their
minimums

We now consider the potential in the lattice energy functional with special
form, namely, the completely monotone potential cases. A function
$f:(0,\infty)\rightarrow{\mathbb{R}}$ is completely monotone if it is of class
$C^{\infty}(0,\infty)$ and

$(-1)^{j}f^{(j)}(x)>0,j=0,1,2,\cdots\infty.$

The Bernstein functions are non-negative functions whose derivative is
completely monotone. They are such that

$f(x)>0,(-1)^{j-1}f^{(j)}(x)>0,j=1,2,\cdots\infty.$

The Bernstein functions have rich connection to other fields, firstly in
probability theory. Complete Bernstein functions are used in complex analysis
under the name Pick or Nevanlinna functions, while in matrix analysis and
operator theory, the name operator monotone function is more common. When
studying the positivity of solutions of Volterra integral equations, various
types of kernels appear which are related to Bernstein functions. See the
monograph [20].

Let $\mathcal{F}(x)$ be completely monotone function and $\Lambda$ be a two
dimensional lattice with fixed area. In (1.5) we choose the potential
$\mathcal{V}=\mathcal{F}$. Now we consider the lattice summation

$\mathcal{E}(\kappa,\Lambda)=\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\Big{(}(1-\kappa)\mathcal{F}(|\frac{\mathbb{P}+1}{2}|^{2})+\kappa\mathcal{F}(|\mathbb{P}|^{2})\Big{)}.$
(1.6)

The main aim of this paper is to study the optimal lattice shape which
minimizes $\mathcal{E}(\kappa,\Lambda)$, where $\kappa\in[0,1]$ is a
parameter.

To write $\mathcal{E}(\kappa,\Lambda)$ in function form, we parametrize the
lattice $\Lambda$ with cell by
$\Lambda=\sqrt{\frac{\pi}{\operatorname{Im}(z)}}\Big{(}{\mathbb{Z}}\oplus
z{\mathbb{Z}}\Big{)}$ where the only parameter is $z$ which belongs to
$\mathbb{H}:=\\{z=x+iy\in\mathbb{C}:y>0\\}$. It follows that

$\displaystyle\mathcal{F}(|\mathbb{P}|^{2}):$
$\displaystyle=\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}\mathcal{F}(\frac{\pi}{\operatorname{Im}(z)}|mz+n|^{2}),$
$\displaystyle\mathcal{F}(|\frac{\mathbb{P}+1}{2}|^{2}):$
$\displaystyle=\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}\mathcal{F}(\frac{\pi}{\operatorname{Im}(w)}|mw+n|^{2})\mid_{w=\frac{z+1}{2}}.$

Therefore, we can write

$\mathcal{E}_{\kappa}(z):=\mathcal{E}(\kappa,\Lambda)=\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}(1-\kappa)\mathcal{F}(\frac{\pi}{2\operatorname{Im}(z)}|m(z+1)+2n|^{2})+\kappa\mathcal{F}(\frac{\pi}{\operatorname{Im}(z)}|mz+n|^{2}).$
(1.7)

We introduce the special curve in $\mathbb{H}$. Let the vertical interval be

$\displaystyle\Omega_{ea}:=\\{z:\operatorname{Re}(z)=0,1\leq\operatorname{Im}(z)\leq\sqrt{3}\\},$

while the arc on the circle is denoted by

$\displaystyle\Omega_{eb}:=\\{z:|z|=1,0\leq\operatorname{Re}(z)\leq\frac{1}{2}\\}.$

The union of the vertical interval and the arc is denoted as

$\displaystyle\Omega_{e}:=\Omega_{ea}\cup\Omega_{eb}.$ (1.8)

In PDE theory, the maximum/minimum principle states that a function satisfying
some partial differential equation in some domain achieves its maximum/minimum
on the boundary of the domain. Now we state a minimum principle for
$\mathcal{E}_{\kappa}(z)$ on $\mathbb{H}$:

######  Theorem 1.1 (Minimum principle for sum of two completely monotone
functions on the lattice).

Let $\mathcal{F}(x)$ be any completely monotone function, $\kappa\in[0,1]$ be
the parameter and $\mathcal{E}_{\kappa}(z)$ be defined in (1.7). Then

$\min_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=\min_{z\in\Omega_{e}}\mathcal{E}_{\kappa}(z).$

######  Remark 1.1.

Theorem 1.1 holds for any two nonnegative combination of two completely
monotone functions on the lattice. In fact, consider the nonnegative
combination,

$\mathcal{E}(a,b,\Lambda):=\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\Big{(}a\mathcal{F}(|\frac{\mathbb{P}+1}{2}|^{2})+b\mathcal{F}(|\mathbb{P}|^{2})\Big{)}.$

After normalization, one can rewrite it as

$\displaystyle\mathcal{E}(a,b,\Lambda)$
$\displaystyle=(a+b)\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\Big{(}\frac{a}{a+b}\mathcal{F}(|\frac{\mathbb{P}+1}{2}|^{2})+\frac{b}{a+b}\mathcal{F}(|\mathbb{P}|^{2})\Big{)}$
$\displaystyle=(a+b)\mathcal{E}(\frac{b}{a+b},\Lambda),$

then it becomes the standard form in Theorem 1.1. Similar remarks are applied
to the theorems 1.2, 1.3 and 1.4 below.

In Theorem 1.1, we conclude that the minimizers of sum of two completely
monotone functions are located precisely on a special curve. As we will see
later, this special curve is a partial boundary of the fundamental region
under the group defined in (2.6).

Many of the physical meaningful potentials are completely monotone functions,
including Riesz potentials $r^{-a}(a>0),$ screened Coulomb potential (also
called Yukawa potential) $a\frac{e^{-br}}{r}(a,b>0)$, and Born-Mayer potential
$ae^{-br}(a,b>0)$. There are also a large number of completely monotone
functions. Examples include
$(a+\frac{b}{x^{\alpha}})^{\mu},\;\hbox{here}\;a\geq 0,b\geq 0,\mu\geq
0,0\leq\alpha\leq 1$; $e^{\frac{a}{x}}(a>0)$,
$\frac{1}{(a-be^{-x})^{\mu}},\;\hbox{here}\;a\geq b>0,\mu>0$;
$\frac{1}{x^{a}(\Gamma(1+\frac{1}{x}))^{x}},\;\hbox{here}\;a\geq 1$;
$\frac{\Gamma(x)\Gamma(x+a+b)}{\Gamma(x+a)\Gamma(x+b)}(a,b\geq 0)$;
$\frac{2}{\sqrt{\pi x}}\int_{0}^{\sqrt{x}}e^{-t^{2}}dt$; etc. There are rules
to generate more completely monotone functions from known ones. For example,
if $f(x)$ is completely monotone, then $f^{(2m)}(x),-f^{(2m+1)}(x),e^{f(x)}$,
$f(1-e^{-x})$, and $f(ax^{\alpha}+b)$($a\geq 0,b\geq 0,0\leq\alpha\leq 1$) are
also completely monotone for $m=1,2,3,\cdots$. We refer to [12, 13, 2, 3, 4]
and the references therein for more concrete examples and the rules to
generate completely monotone functions. For the references on the physical
applications of the completely monotone functions, we refer to Bétermin [6]
and the references therein.

To introduce next Theorem, we follow the terminology of ”universally optimal”
from Cohn-Kumar [10] as defined by

######  Definition 1 (Cohn-Kumar [10]).

A finite subset $\mathcal{C}\subset S^{n-1}$ is universally optimal if it
(weakly) minimizes potential energy among all configurations of
$|\mathcal{C}|$ points on $S^{n-1}$ for each completely monotonic potential
function.

In parallel, we shall state our definition of universally optimal among all
lattice shapes.

######  Definition 2.

A lattice shape(determined by a complex variable $z_{0}$) is universally
optimal if it (weakly) minimizes potential energy among all configurations of
lattice shapes for each completely monotonic potential function.

The weak minimizer(s) here means that the minimizer is not necessarily unique.
Now we state our universally optimal result for the parameter locating on some
interval with positive measure:

######  Theorem 1.2 (Universally optimal and an explicitly universal interval
for square lattice).

For any completely monotone potential function $\mathcal{F}$, and for any
$\kappa\in[\frac{1}{3},\frac{2}{3}]$,

$\min_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)$

is attained at $i$, corresponding to the square lattice by our setting and is
unique up the group $\mathcal{G}_{2}$ defined in (2.6).

######  Remark 1.2.

The explicitly universal interval $[\frac{1}{3},\frac{2}{3}]$ is not sharp and
can be improved to be optimal by our method through refining the estimates.
See Section 4 below.

######  Remark 1.3.

The universality of optimal square lattice has also been observed in a recent
paper by Betermin-Knupfer-Faulhuber [7] for another competing system with
Gaussian and Riesz interactions.

Note that there are three factors in the minimization process of
$\min_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)$: one is the function (there are
a large class of completely monotone functions), one is parameter $\kappa$
which represents the competing strength of the two-component system, and the
lattice parameter $z$. Theorem 1.2 asserts that the minimizer of sum of two
completely monotone functions is always located on $i$ (corresponding to the
square lattice) for the parameter $\kappa$ in $[\frac{1}{3},\frac{2}{3}]$.
This seems rather surprising.

Recall that the minimizer of the lattice energy functional corresponds to the
optimal lattice shape in the competing system. Therefore, combining Theorem
1.1 and Theorem 1.2, we have the following phases transition result which has
appeared in many competing systems(like Bose-Einstein system, see in Mueller-
Ho [17] and Luo-Wei [15]). Nevertheless this result applies to large class of
other systems.

######  Theorem 1.3 (Hexagonal-rhombic-square-rectangular lattice phases
transition).

For any completely monotone potential function $\mathcal{F}$, there exists
$\kappa_{1}<\kappa_{2}\in(0,1)$ depending on the potential function
$\mathcal{F}$ such that

  * •

Case A: $\kappa\in[0,\kappa_{1}]$,
$\min_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=\min_{z\in\Omega_{ea}}\mathcal{E}_{\kappa}(z)$;
the minimizer corresponds to rectangular lattice in this case;

  * •

Case B: $\kappa\in[\kappa_{1},\kappa_{2}]$,
$Minimizer_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=i$; the minimizer
corresponds to square lattice in this case;

  * •

Case C: $\kappa\in[\kappa_{2},1]$,
$\min_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=\min_{z\in\Omega_{eb}}\mathcal{E}_{\kappa}(z)$;
the minimizer corresponds to rhombic lattice in this case;

For the parameters $\kappa_{1},\kappa_{2}$, for any potential function which
is completely monotone, there has

  * •

$\kappa_{1}\leq\frac{1}{3}$;

  * •

$\kappa_{2}\geq\frac{2}{3}$;

  * •

$\kappa_{1}+\kappa_{2}=1$;

namely, $[\kappa_{1},\kappa_{2}]\supseteq[\frac{1}{3},\frac{2}{3}]$ in Case B.
In particular, when $\kappa=1$,
$Minimizer_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=\frac{1}{2}+i\frac{\sqrt{3}}{2}$,
which corresponds to hexagonal lattice; when $\kappa=0$,
$Minimizer_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=\sqrt{3}i$, which
corresponds to $\sqrt{3}:1$ rectangular lattice.

In physical and biological models, the functional $\mathcal{E}_{\kappa}(z)$ is
referred to the energy of competing periodical system and the parameter
$\kappa$ reflects the competing strength of the system. As the parameter
$\kappa$ varies continuously, we see from Theorem 1.3 clearly that, the
optimal lattice shape goes through hexagonal-rhombic-square-rectangular
lattice continuously (as $\kappa$ moves from $1$ to $0$ continuously). Theorem
1.2 shows that under the completely monotone potentials and under certain
positive interval of interaction strength, the minimal energy lattice shape is
the square lattice shape. This phenomenon has appeared and conjectured in two
component Bose-Einstein condensates (see in Mueller-Ho [17] and Luo-Wei [15]).
When $\kappa=1$ the optimality of hexagonal lattice for sum of Gaussian
lattice has been investigated recently in [8].

The key tool in proving Theorems 1.1-1.3 is the so-called Bernstein
representation formula for general completely monotone function $\mathcal{F}$.
To introduce the formula we define

$\mathcal{W}_{\kappa}(z;\alpha)=\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}(1-\kappa)\exp(-\alpha\frac{\pi}{2\operatorname{Im}(z)}|m(z+1)+2n|^{2})+\kappa\exp(-\alpha\frac{\pi}{\operatorname{Im}(z)}|mz+n|^{2})$
(1.9)

which is the energy functional associated to the special Gaussian type
completely monotone function $\exp(-\alpha x^{2})$ with a free parameter
$\alpha>0$.

The celebrated Bernstein theorem ([5]) states that for any completely monotone
function $F$, there is an non-negative finite Borel measure on $[0,\infty)$
with cumulative distribution function $\lambda(\alpha)$ such that

$F(x)=\int_{0}^{\infty}e^{-\alpha x}d\lambda(\alpha).$

As a consequence we obtain the following integral representation of
$\mathcal{E}_{\kappa}(z)$: there is an non-negative finite Borel measure on
$[0,\infty)$ with cumulative distribution function $\lambda(\alpha)$ such that

$\mathcal{E}_{\kappa}(z)=\int_{0}^{\infty}\mathcal{W}_{\kappa}(z;\alpha)d\lambda(\alpha).$
(1.10)

The formula (1.10) reduces the proof of Theorems 1.1-1.3 to the study of the
universal properties $\mathcal{W}_{\kappa}(z;\alpha)$, which is the main
contribution of this paper.

###  1.3. Competing systems with Riesz potential

One interesting and important competing potential is the Riesz potential type.
We discuss the results on competing systems with Riesz potential here. Let
$\mathcal{F}(x):=\frac{1}{x^{s}},s>1$ be the Riesz potential which is a
special case of completely monotone functions. We consider the associated
competing system energy functional

$\mathcal{E}_{R}(z;\kappa,s):=\mathcal{E}_{R}(\kappa,s,\Lambda)=\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\Big{(}(1-\kappa)\frac{1}{|\frac{\mathbb{P}+1}{2}|^{2s}}+\kappa\frac{1}{|\mathbb{P}|^{2s}}\Big{)},s>1,$
(1.11)

with $\Lambda=\sqrt{\frac{\pi}{\operatorname{Im}(z)}}\Big{(}{\mathbb{Z}}\oplus
z{\mathbb{Z}}\Big{)}$.

Observe that up to a constant (depending on $s$) the Riesz sum
$\sum_{\mathbb{P}\in\Lambda\backslash\\{0\\}}\frac{1}{|\mathbb{P}|^{2s}}$ is
the Eisenstein series $G(z,s)$ which is defined as

$G(z,s):=\frac{1}{2}\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}\frac{\big{(}\operatorname{Im}(z)\big{)}^{s}}{|mz+n|^{2s}}.$
(1.12)

See [9].

The lattice energy functional $\mathcal{E}_{R}(z;\kappa,s)$ under Riesz
potential (defined at (1.11)) can be written up to a constant (depending on
$s$) as

$\mathbf{G}_{R}(z;\kappa,s):=(1-\kappa)\mathbf{G}(\frac{z+1}{2},s)+\kappa\mathbf{G}(z,s),\kappa\in[0,1],$
(1.13)

where $\mathbf{G}(z,s):=\pi^{-s}\Gamma(s)G(z,s)$ is a rescaled Eisenstein
function which also appears frequently in number theory ([9]).

Theorems 1.1-1.3 can be applied to $\mathbf{G}_{R}(z;\kappa,s)$. However an
important observation, as shown in the appendix, is that although the
$s-$deriveatives are not completely monotone functions, we can still apply the
techniques in proving Theorem 1.1-1.3 to the $s-$derivatives.

######  Theorem 1.4.

Let $\mathbf{G}_{R}(z;\kappa,s)$ be defined in (1.13).

(a) For any $\kappa\in[0,1],s>1,j=1,2,3\cdots$, it follows that

$\min_{z\in\mathbb{H}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)=\min_{z\in\Omega_{e}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{L}(z;\kappa,s).$

(b) There exists $\kappa_{1}<\kappa_{2}\in(0,1)$ depending on $s$ such that
$[\frac{1}{3},\frac{2}{3}]\subseteq[\kappa_{1},\kappa_{2}],\kappa_{1}+\kappa_{2}=1$
and

  * •

Case A: $\kappa\in[0,\kappa_{1}]$,
$\min_{z\in\mathbb{H}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)=\min_{z\in\Omega_{ea}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)$;
the minimizer corresponds to rectangular lattice in this case;

  * •

Case B: $\kappa\in[\kappa_{1},\kappa_{2}]$,
$Minimizer_{z\in\mathbb{H}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)=i$;
the minimizer corresponds to square lattice in this case;

  * •

Case C: $\kappa\in[\kappa_{2},1]$,
$\min_{z\in\mathbb{H}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)=\min_{z\in\Omega_{eb}}\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)$;
the minimizer corresponds to rhombic lattice in this case;

In particular, when $\kappa=1$,
$Minimizer_{z\in\mathbb{H}}\frac{d^{j}}{ds^{j}}\mathbf{G}(z,s)=\frac{1}{2}+i\frac{\sqrt{3}}{2}$,
which corresponds to hexagonal lattice; when $\kappa=0$,
$Minimizer_{z\in\mathbb{H}}\frac{d^{j}}{ds^{j}}\mathbf{G}(\frac{z+1}{2},s)=\sqrt{3}i$,
which corresponds to $\sqrt{3}:1$ rectangular lattice.

######  Remark 1.4.

The Eisentein series $G(z,s)$ is closely related to the Epstein zeta function
$\zeta_{Q}(s)$, that is defined for $\operatorname{Re}(s)>1$ by

$\zeta_{Q}(s):=\frac{1}{2}\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}\frac{1}{Q(m,n)^{s}},$

where $Q(x,y)=ax^{2}+bxy+cy^{2}$ is a positive definite quadratic form with
real coefficients$($i.e., $a>0$, $c>0$ and $b^{2}-4ac<0$$)$.

Denote that $z=\frac{-b}{2a}+i\frac{\sqrt{4ac-b^{2}}}{2a}$ and $D=b^{2}-4ac$.
Then the Epstein zeta function and Eisenstein series is connected by

$\zeta_{Q}(s)=(\frac{|D|}{4})^{-\frac{s}{2}}G(z,s).$ (1.14)

See [9]. We also have an analogue theorem for Epstein zeta function.

By the Kronecker limit formula(See for example [19] and [18] for the details),
one has

$\displaystyle\frac{d}{ds}\mid_{s=1}G(z,s)=-\frac{1}{2}\log\big{|}\operatorname{Im}\big{(}z\big{)}\eta\big{(}z\big{)}\big{|}-\log(2\pi).$
(1.15)

It follows that

$\displaystyle\frac{d}{ds}\mid_{s=1}G_{R}(z;\kappa,s),{G}_{R}(z;\kappa,s):=(1-\kappa){G}(\frac{z+1}{2},s)+\kappa{G}(z,s)$
(1.16) $\displaystyle=$
$\displaystyle-\frac{1}{2}\Big{(}\kappa\log\big{|}\operatorname{Im}\big{(}z\big{)}\eta\big{(}z\big{)}\big{|}+(1-\kappa)\log\big{|}\operatorname{Im}\big{(}\frac{z+1}{2}\big{)}\eta\big{(}\frac{z+1}{2}\big{)}\big{|}\Big{)}-\log(2\pi),$

where

$\eta(z)=e^{\frac{\pi}{3}zi}\prod_{n=1}^{\infty}\big{(}1-e^{2\pi
nzi}\big{)}^{4}$ (1.17)

is the fourth power of the Dedekind eta function. The derivative formula
(LABEL:YYY) connects the lattice energy functional with Riesz potential to
lattice energy functional with Coulomb potential. For convenience, define

$\mathcal{N}_{\kappa}(z):=-\Big{(}\kappa\log\big{|}\operatorname{Im}\big{(}z\big{)}\eta\big{(}z\big{)}\big{|}+(1-\kappa)\log\big{|}\operatorname{Im}\big{(}\frac{z+1}{2}\big{)}\eta\big{(}\frac{z+1}{2}\big{)}\big{|}\Big{)}.$
(1.18)

The functional $\mathcal{N}_{\kappa}(z)$ is of the form of Dedekind eta
functions, is a type of lattice energy functional with Coulomb potential.

The reduced energy functional $\mathcal{N}_{\kappa}(z)$ arises in the study of
Triblock Ohta-Kawasaki system. For $\kappa=1$ it arises in the Diblock
copolymer Ohta-Kawasaki system and also Ginzburg-Landau theory. See [14] and
[18]. Applying Theorem 1.4 we then recover the results in our previous paper
[14], in which we proved these results by a different method. We state them
here since our new results here include these as a special case and we have
proved these results without using any PDE theory. This also confirms that the
hexagonal-rhombic-square-rectangular lattice phases transition in Theorem 1.3
is quite general in various periodical competing system.

In the numerical computation done in [14], we have for
$\mathcal{N}_{\kappa}(z)$

$\kappa_{1}:=0.1867\dots,\;\kappa_{2}:=0.8132\cdots.$

This confirms that the upper and lower bounds in Theorem 1.3, in which we
proved that $\kappa_{1}\leq\frac{1}{3},\;\kappa_{2}\geq\frac{2}{3}.$

The paper is organized as follows: In section 2, we aim to give some basic
properties of the lattice energy functional and provide some basic tools in
estimates. We prove Theorem 1.1 and Theorem 1.2, Theorem 1.3 in Section 3 and
Section 4 respectively.

##  2\. Preliminaries: Invariances of $\mathcal{E}_{\kappa}(z)$ and estimates
on theta functions

Let $\mathcal{E}_{\kappa}(z)$ be the lattice energy functional defined at
(1.7). The end point cases of $\mathcal{E}_{\kappa}(z)$ are

$\mathcal{E}_{0}(z)=\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}\mathcal{F}(\frac{\pi}{2\operatorname{Im}(z)}|m(z+1)+2n|^{2})$
(2.1)

and

$\mathcal{E}_{1}(z)=\sum_{(m,n)\in\mathbb{Z}^{2}\backslash\\{0\\}}\mathcal{F}(\frac{\pi}{\operatorname{Im}(z)}|mz+n|^{2}).$
(2.2)

Then $\mathcal{E}_{\kappa}$ can be written as interpolation of
$\mathcal{E}_{0}$ and $\mathcal{E}_{1}$:

$\mathcal{E}_{\kappa}(z)=\kappa\mathcal{E}_{1}(z)+(1-\kappa)\mathcal{E}_{0}(z).$
(2.3)

In this section, we collect some basic properties of the functional
$\mathcal{E}_{\kappa}$, including invariance and monotone properties, and some
key estimates of $\mathcal{W}_{\kappa}$.

We first study the invariance properties of $\mathcal{E}_{\kappa}$. We use the
following definition of fundamental domain which is slightly different from
the classical definition (see [16]):

######  Definition 3 (page 108, [11]).

The fundamental domain associated to group $G$ is a connected domain
$\mathcal{D}$ satisfying

  * •

For any $z\in\mathbb{H}$, there exists an element $\pi\in G$ such that
$\pi(z)\in\overline{\mathcal{D}}$;

  * •

Suppose $z_{1},z_{2}\in\mathcal{D}$ and $\pi(z_{1})=z_{2}$ for some $\pi\in
G$, then $z_{1}=z_{2}$ and $\pi=\pm Id$.

For example, let $\Gamma$ be the group generated by the actions
$\tau\mapsto-\frac{1}{\tau},\;\;\tau\mapsto\tau+1$, then by Definition 3, the
fundamental domain associated to modular group $\Gamma$ is

$\displaystyle\mathcal{D}_{\Gamma}:=\\{z\in\mathbb{H}:|z|>1,\;-\frac{1}{2}<x<\frac{1}{2}\\}.$
(2.4)

which is open. Note that the fundamental domain can be open. (See [page 30,
[1]].)

Next we introduce another two groups related to the functional
$\mathcal{E}_{\kappa}(z)$. The generators of these groups are given by

$\displaystyle\mathcal{G}_{1}:\hbox{the group generated
by}\;\;\tau\mapsto-\frac{1}{\tau},\;\;\tau\mapsto\tau+1,\;\;\tau\mapsto-\overline{\tau},$
(2.5) $\displaystyle\mathcal{G}_{2}:\hbox{the group generated
by}\;\;\tau\mapsto-\frac{1}{\tau},\;\;\tau\mapsto\tau+2,\;\;\tau\mapsto-\overline{\tau}.$
(2.6)

It is easy to see that the fundamental domains associated to group
$\mathcal{G}_{j},j=1,2$ denoted by
$\mathcal{D}_{\mathcal{G}_{1}},\mathcal{D}_{\mathcal{G}_{2}}$ are

$\displaystyle\mathcal{D}_{\mathcal{G}_{1}}:=\\{z\in\mathbb{H}:|z|>1,\;0<x<\frac{1}{2}\\}$
(2.7)
$\displaystyle\mathcal{D}_{\mathcal{G}_{2}}:=\\{z\in\mathbb{H}:|z|>1,\;0<x<1\\}.$
(2.8)

Clearly we have that
$\mathcal{G}_{2}\subseteq\mathcal{G}_{1},\;\;\mathcal{D}_{\mathcal{G}_{1}}\subseteq\mathcal{D}_{\mathcal{G}_{2}}$.

The following two Lemmas can be checked directly from the definition.

######  Lemma 2.1.

We have the following fundamental invariant properties of the functionals:

  * •

$\mathcal{E}_{1}(z)$ is invariant under the group $\mathcal{G}_{1}$, i.e., for
any $\gamma\in\mathcal{G}_{1}$

$\mathcal{E}_{1}(\gamma(z))=\mathcal{E}_{1}(z).$

  * •

$\mathcal{E}_{0}(z)$ is invariant under the group $\mathcal{G}_{2}$, i.e., for
any $\pi\in\mathcal{G}_{2}$

$\mathcal{E}_{0}(\pi(z))=\mathcal{E}_{0}(z).$

  * •

$\mathcal{E}_{\kappa}(z)$$(\kappa\in(0,1))$ is invariant under the group
$\mathcal{G}_{2}$, i.e., for any $\pi\in\mathcal{G}_{2}$

$\mathcal{E}_{0}(\pi(z))=\mathcal{E}_{0}(z).$

As a consequence, the fundamental domain of the functionals
$\mathcal{E}_{1}(z)$ and $\mathcal{E}_{0}(z)$ is
$\mathcal{D}_{\mathcal{G}_{1}}$ and $\mathcal{D}_{\mathcal{G}_{2}}$
respectively, and the fundamental domain for the general function
$\mathcal{E}_{\kappa}(z)$$(\kappa\in(0,1))$ is
$\mathcal{D}_{\mathcal{G}_{2}}$.

By the integral representation formula (1.10) and the monotonicity properties
of $\mathcal{W}_{0}$ and $\mathcal{W}_{1}$ (see [15]), we have the following
important monotonicity properties for the functionals
$\mathcal{E}_{0}(z),\mathcal{E}_{1}(z)$ in the corresponding domain:

######  Theorem 2.1.

There holds

  * •

$\displaystyle\frac{\partial}{\partial
x}\mathcal{E}_{0}(z)>0,\;\;\forall\;z\in\mathcal{D}_{\mathcal{G}_{2}},$

  * •

$\frac{\partial}{\partial
x}\mathcal{E}_{1}(z)<0,\;\;\forall\;z\in\Omega_{\mathcal{C}_{1}}.$

Here

$\displaystyle\Omega_{\mathcal{C}_{1}}:=\\{z\mid
0<x<\frac{1}{2},y>\sqrt{x-x^{2}}\\}.$

######  Remark 2.1.

Montgomery [16] proved that

$\displaystyle\frac{\partial}{\partial
x}\mathcal{E}_{1}(z)<0,\;\;\forall\;z\in\Omega_{\mathcal{C}_{0}}:=\\{z\in\mathbb{H}:y>\frac{1}{2},\;0<x<\frac{1}{2}\\}.$
(2.9)

Theorem 2.1 improves this result to a larger domain $\Omega_{\mathcal{C}_{1}}$
as $\Omega_{\mathcal{C}_{0}}\subset\Omega_{\mathcal{C}_{1}}$. Furthermore,
$\Omega_{\mathcal{C}_{1}}$ contains a corner at $z=0$, which makes the proof
much more involved.

We state two corollaries related to the functional $\mathcal{E}_{\kappa}(z)$.

######  Corollary 2.1.

$\frac{\partial}{\partial x}\mathcal{E}_{1}(z)>0,\;\forall
z\in\Omega_{\mathcal{C}_{2}}.$

Here

$\displaystyle\Omega_{\mathcal{C}_{2}}:=\\{z\mid\frac{1}{2}<x<1,y>\sqrt{x-x^{2}}\\}.$

###### Proof.

Since $z\mapsto 1-\overline{z}\in\mathcal{G}_{1}$, by Lemma 2.1, we have
$\mathcal{E}_{1}(1-\overline{z})=\mathcal{E}_{1}(z)$. Thus

$\displaystyle\frac{\partial}{\partial
x}\mathcal{E}_{1}(1-\overline{z})=-\frac{\partial}{\partial
x}\mathcal{E}_{1}(z).$ (2.10)

The result follows by Theorem 2.1.

∎

By Theorem 2.1 and Corollary 2.1 we have

######  Theorem 2.2.

For $\kappa\in[0,1]$,

$\frac{\partial}{\partial x}\mathcal{E}_{\kappa}(z)>0,\;\forall
z\in{\mathcal{R}_{R}},j=1,2.$

Here

${\mathcal{R}_{R}}:=\Omega_{\mathcal{C}_{2}}\cap\mathcal{D}_{\mathcal{G}_{2}}=\\{z\mid\frac{1}{2}<\operatorname{Re}(z)<1,|z|>1\\}.$

By Theorem 2.2, it remains to consider functional on the domain

${\mathcal{R}_{L}}:=\\{z\mid 0<\operatorname{Re}(z)<\frac{1}{2},|z|>1\\}.$
(2.11)

We shall solve this problem in Section 3.

Before going to the proof, we also need some auxiliary functions from number
theory. We first recall the following well-known Jacobi triple product
formula:

$\displaystyle\prod_{m=1}^{\infty}(1-x^{2m})(1+x^{2m-1}y^{2})(1+\frac{x^{2m-1}}{y^{2}})=\sum_{n=-\infty}^{\infty}x^{n^{2}}y^{2n}$
(2.12)

for complex numbers $x,y$ with $|x|<1$, $y\neq 0$.

The Jacob theta function is defined as

$\displaystyle\vartheta_{J}(z;\tau):=\sum_{n=-\infty}^{\infty}e^{i\pi
n^{2}\tau+2\pi inz},$

and the classical one-dimensional theta function is given by

$\displaystyle\vartheta(X;Y):=\vartheta_{J}(Y;iX)=\sum_{n=-\infty}^{\infty}e^{-\pi
n^{2}X}e^{2n\pi iY}.$ (2.13)

Hence by the Jacobi triple product formula (2.12), it holds

$\displaystyle\vartheta(X;Y)=\prod_{n=1}^{\infty}(1-e^{-2\pi
nX})(1+e^{-2(2n-1)\pi X}+2e^{-(2n-1)\pi X}\cos(2\pi Y)).$ (2.14)

By Poisson Summation Formula, one has

$\displaystyle\vartheta(X;Y)=X^{-\frac{1}{2}}\sum_{n\in\mathbb{Z}}e^{-\frac{\pi(n-Y)^{2}}{X}}.$
(2.15)

Thus the functional $\mathcal{W}_{1}(z;\alpha)$ can be written in terms of
one-dimensional theta function as follows:

$\displaystyle\mathcal{W}_{1}(z;\alpha)$
$\displaystyle=\sum_{(m,n)\in\mathbb{Z}^{2}}e^{-\alpha\pi\frac{1}{y}|nz+m|^{2}}=\sum_{n\in\mathbb{Z}}e^{-\alpha\pi
yn^{2}}\sum_{m\in\mathbb{Z}}e^{-\frac{\alpha\pi(nx+m)^{2}}{y}}$ (2.16)
$\displaystyle=\sqrt{\frac{y}{\alpha}}\sum_{n\in\mathbb{Z}}e^{-\alpha\pi
yn^{2}}\vartheta(\frac{y}{\alpha};-nx)=\sqrt{\frac{y}{\alpha}}\sum_{n\in\mathbb{Z}}e^{-\alpha\pi
yn^{2}}\vartheta(\frac{y}{s};nx)$
$\displaystyle=2\sqrt{\frac{y}{\alpha}}\sum_{n=1}^{\infty}e^{-\alpha\pi
yn^{2}}\vartheta(\frac{y}{\alpha};nx).$

We state the following very important properties of the theta function whose
proof will be proved in the appendix. We use these properties in estimates of
the proof of Theorem 1.1.

######  Lemma 2.2.

Let $Q(X;Y)=-\frac{1}{\sin(2\pi Y)}\frac{\partial}{\partial Y}\vartheta(X;Y)$.
Then there holds the following lower and upper bounds of $Q(X;Y)$

$\underline{\vartheta}(X)\leq Q(X;Y)\leq\overline{\vartheta}(X),\;\hbox{for
any}\;Y,$

where

$\underline{\vartheta}(t)=\begin{cases}\frac{1}{\max_{\beta\in[0,k]}{\sin(2\pi\beta)}}t^{-\frac{3}{2}}(e^{-\frac{\pi
k^{2}}{t}}(k-e^{-\frac{\pi(1-2k)}{t}}(1-k))\\\
+\sum_{m=1}^{\infty}e^{-\frac{\pi(k+m)^{2}}{t}}(k+m-e^{-\frac{\pi(2m+1)(1-2k)}{t}}(m+1-k)))\;(0<t\leq
k<\frac{1}{2}),\\\ (1-\mu(t))4\pi e^{-\pi t}\;(t>0),\end{cases}$ (2.17)

and

$\overline{\vartheta}(t)=\begin{cases}t^{-\frac{3}{2}}\sum_{m=-\infty}^{\infty}e^{-\frac{\pi
m^{2}}{t}}(1-\frac{2\pi m^{2}}{t})\;(0<t\leq k<\frac{\pi}{2}),\\\
(1+\mu(t))4\pi e^{-\pi t}\;(t>0).\end{cases}$ (2.18)

Here $k$ and $a$ are positive numbers and

$\mu(t):=\sum_{n=2}^{\infty}n^{2}e^{-\pi(n^{2}-1)t}.$ (2.19)

(For $t$ small, we can use

$Q(t;k)=\frac{1}{\sin(2\pi k)}t^{-\frac{3}{2}}(e^{-\frac{\pi
k^{2}}{t}}(k-e^{-\frac{\pi(1-2k)}{t}}(1-k))+\sum_{m=1}^{\infty}e^{-\frac{\pi(k+m)^{2}}{t}}(k+m-e^{-\frac{\pi(2m+1)(1-2k)}{t}}(m+1-k))).)$

##  3\. Minimum principle: Proof of Theorem 1.1

In this section, we prove Theorem 1.1. First we observe that the functional
$\mathcal{W}_{\kappa}(z;\alpha)$ satisfies the following duality principle:
For all $b\in\mathbb{R}$,

$\mathcal{W}_{b}(w)=\mathcal{W}_{1-b}(z),\ \ z\in\mathbb{H}\ \ \mbox{and}\ \
w=\frac{z-1}{z+1}\in\mathbb{H}.$

By the duality principle(see more details in Lemma 4.1) and the Bernstein
representation formula (1.10), we only need to consider the case
$\kappa\in[0,\frac{1}{2}]$.

Next by the transition formula

$\displaystyle\mathcal{W}_{\kappa}(z;\alpha)=\kappa\mathcal{W}_{\frac{1}{2}}(z;\alpha)+(1-2\kappa)\mathcal{W}_{0}(z;\alpha)$
(3.1)

we only need to study $\mathcal{W}_{\frac{1}{2}}$ and $\mathcal{W}_{0}$.
Furthermore since by Fourier transform

$\displaystyle\mathcal{W}_{\kappa}(z;\alpha)=\alpha\mathcal{W}_{\kappa}(z;\frac{1}{\alpha})$
(3.2)

we only need to consider $\alpha\in[1,+\infty)$. By Theorem 2.2 we only need
to consider $z\in{\mathcal{R}_{L}}$ (defined at (2.11)). We also use the
formula in (2.16),

$\displaystyle\mathcal{W}_{1}(z;\alpha)=2\sqrt{\frac{y}{\alpha}}\sum_{n=1}^{\infty}e^{-\alpha\pi
yn^{2}}\vartheta(\frac{y}{\alpha};nx).$

Then we use the bounds in Lemma 2.2 to prove the estimates.

The following is the main result we shall prove in this section:

######  Theorem 3.1.

On the half fundamental domain bounded by $x=0,x=\frac{1}{2}$ and
$x^{2}+y^{2}=1$, we have

$\displaystyle\frac{\partial}{\partial
x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)\geq 0,\forall
z\in\overline{\mathcal{R}_{L}}$ (3.3)

where $\mathcal{R}_{L}$ is defined at (2.11). If the equality holds, then
$x=0$.

By Corollary 4.2 of [15], we also have $\frac{\partial}{\partial
x}\mathcal{W}_{0}(z;\alpha)\geq 0,\forall z\in\overline{\mathcal{R}_{L}}$.
Theorem 1.1 then follows from (3.3) and Theorem 2.2.

The proof of Theorem 3.1 follows from the following two lemmas which establish
the lower bound of $\frac{\partial}{\partial
x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)$ in different ranges of $x$.

######  Lemma 3.1.

For $x\in[\frac{1}{4},\frac{1}{2}]$ we have the following lower bound
estimates for $\frac{\partial}{\partial
x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)$:

$\displaystyle\frac{\partial}{\partial x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)$
(3.4) $\displaystyle\geq$
$\displaystyle\sqrt{\frac{y}{\alpha}}e^{-\frac{\pi\alpha y}{2}}\sin(\pi
x)\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}(1+\epsilon_{1}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-(1+\epsilon_{2}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\Big{)},$

where $\epsilon_{1}(\alpha)<3.18\cdot
10^{-9},\;\epsilon_{2}(\alpha)<2.782\cdot 10^{-6}$ and
$\overline{\vartheta},\underline{\vartheta}$ are defined in (2.17)-(2.18).

Furthermore, for $y\geq\frac{\sqrt{3}}{2},\alpha\geq 1$, there holds

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}(1+\epsilon_{1}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-(1+\epsilon_{2}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq 0.$ (3.5)

######  Lemma 3.2.

For $x\in[0,\frac{1}{4}]$ we have the following lower bound estimates for
$\frac{\partial}{\partial x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)$:

$\displaystyle\frac{\partial}{\partial x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)$
(3.6) $\displaystyle\geq$
$\displaystyle\sqrt{\frac{y}{\alpha}}e^{-\frac{\pi\alpha y}{2}}\sin(\pi
x)\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\Big{)}$

where $\epsilon_{3}(\alpha)<1.1\cdot 10^{-4}$ and
$\epsilon_{4}(\alpha)<5.7\cdot 10^{-16}.$

Furthermore, for $y\geq\frac{\sqrt{15}}{4},\alpha\geq 1$, we have

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})>0.$ (3.7)

In the rest of this section we shall prove Lemma 3.1 and Lemma 3.2. By (3.2),
we only need to consider the case $\alpha\geq 1$. Let us start with the
computation

$\displaystyle\frac{\partial}{\partial
x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)=\sqrt{\frac{y}{\alpha}}\Big{(}$
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2\sqrt{2}}ne^{-\frac{\pi\alpha
yn^{2}}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=n\frac{x+1}{2}}$ (3.8)
$\displaystyle+\sum_{n=1}^{\infty}ne^{-\pi\alpha
yn^{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=nx}\Big{)}.$

To refine the estimates, we divide the proof into two parts by
$x\in[0,\frac{1}{4}]$ or $[\frac{1}{4},\frac{1}{2}]$. We first prove
(LABEL:HHHAA) of Lemma 3.1.

Proof of (LABEL:HHHAA) of Lemma 3.1: Let $x\in[\frac{1}{4},\frac{1}{2}]$. We
rewrite

$\displaystyle\frac{\partial}{\partial x}\mathcal{W}_{\frac{1}{2}}(z;\alpha)=$
$\displaystyle\sqrt{\frac{y}{\alpha}}e^{-\frac{\pi\alpha
y}{2}}\Big{(}\frac{1}{2\sqrt{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=\frac{x+1}{2}}+e^{-\frac{\pi\alpha
y}{2}}\frac{\partial}{\partial Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=x}$
(3.9)
$\displaystyle\;\;\;\;\;\;\;\;\;\;+\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=x+1}\Big{)}$
$\displaystyle+\sqrt{\frac{y}{\alpha}}\Big{(}2e^{-4\pi\alpha
y}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=2x}+\sum_{n=3}^{\infty}\frac{1}{2\sqrt{2}}ne^{-\frac{\pi\alpha
yn^{2}}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=n\frac{x+1}{2}}$
$\displaystyle\;\;\;\;\;\;\;\;\;\;+\sum_{n=3}^{\infty}ne^{-\pi\alpha
yn^{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=nx}\Big{)}$ $\displaystyle:=$
$\displaystyle\sqrt{\frac{y}{\alpha}}e^{-\frac{\pi\alpha y}{2}}\cdot
J_{1}+(\frac{y}{\alpha})^{\frac{1}{2}}\cdot J_{2},$

where $J_{1}$ and $J_{2}$ are defined respectively at the last equality.

By the lower and upper bounds of $\vartheta$ in Lemma 2.2, we have

$\displaystyle J_{1}$
$\displaystyle\geq\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})\sin(\pi
x)-e^{-\frac{\pi\alpha y}{2}}\overline{\vartheta}(\frac{y}{\alpha})\sin(2\pi
x)-\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\sin(2\pi x)$ (3.10)
$\displaystyle=\sin(\pi
x)\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})2\cos(\pi
x)-\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})2\cos(\pi x)\Big{)}$
$\displaystyle\geq\sin(\pi
x)\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\Big{)},$

where $\underline{\vartheta}$ and $\overline{\vartheta}$ are defined at (2.17)
and (2.18) respectively.

Here we have used the fact that $\sin(\pi(x+1))\leq 0,\cos(\pi
x)\leq\frac{\sqrt{2}}{2}\;\hbox{for}\;x\in[\frac{1}{4},\frac{1}{2}]$.

Next, we are going to estimate $J_{2}$. By (3.9), using $|\sin(ny)|\leq
n|\sin(y)|,y\in{\mathbb{R}}$ and Lemma 2.2, we have

$\displaystyle
J_{2}\geq-\frac{1}{4\sqrt{2}}\overline{\vartheta}(\frac{y}{2\alpha})\sin(2\pi
x)\sum_{n=3}^{\infty}n^{2}e^{-\frac{\pi\alpha
yn^{2}}{2}}-\overline{\vartheta}(\frac{y}{\alpha})\sin(2\pi
x)\sum_{n=3}^{\infty}n^{2}e^{-\pi\alpha yn^{2}}.$ (3.11)

Combining (3.10) and (3.11) we have

$\displaystyle J_{1}+e^{\frac{\pi\alpha y}{2}}J_{2}\geq$ (3.12)
$\displaystyle\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})\sin(\pi
x)-(1+\epsilon_{1}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})\sin(2\pi
x)-(1+\epsilon_{2}(\alpha))\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\sin(2\pi x)\Big{)}$
$\displaystyle\geq\sin(\pi
x)\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-(1+\epsilon_{1}(\alpha))\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-(1+\epsilon_{2}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\Big{)},\;\hbox{for}\;x\in[\frac{1}{4},\frac{1}{2}],$

where the constants $\epsilon_{1}(\alpha)$ and $\epsilon_{2}(\alpha)$ are
defined by

$\epsilon_{1}(\alpha):=\sum_{n=3}^{\infty}n^{2}e^{-\pi\alpha
y(n^{2}-1)}\leq\sum_{n=3}^{\infty}n^{2}e^{\frac{-\pi\sqrt{3}(n^{2}-1)}{2}}$
(3.13)

and

$\epsilon_{2}(\alpha):=\frac{1}{4}\sum_{n=3}^{\infty}n^{2}e^{-\frac{\pi\alpha
y(n^{2}-4)}{2}}\leq\frac{1}{4}\sum_{n=3}^{\infty}n^{2}e^{-\frac{\pi\sqrt{3}(n^{2}-4)}{2}}$
(3.14)

It is easy to see that $\epsilon_{1}<3.18\cdot 10^{-9},\epsilon_{2}<2.782\cdot
10^{-6}$.

The inequality (LABEL:HHHAA) then follows from (3.9) and (3.12).

∎

Proof of (3.5) of Lemma 3.1: In the following we prove that on the half
fundamental domain with $x\in[\frac{1}{4},\frac{1}{2}]$ (equivalently,
$y\geq\frac{\sqrt{3}}{2}$), (3.5) holds by assuming
$\epsilon_{1}=\epsilon_{2}=0$. The proof can be easily modified when two small
constants are added. We split the proof into three subcases as follows

$\mathbf{A}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(0,k];\mathbf{B}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(k,+\infty);\mathbf{C}:\frac{y}{2\alpha}\in(0,k],\frac{y}{\alpha}\in(k,+\infty),$

where the parameter $k\in(0,\frac{1}{2}]$ is to be chosen later.

Case $\mathbf{A}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(0,k]$. By Lemma 2.2, we
have

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.15) $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\frac{1}{\sin(2\pi
k)}\Big{(}e^{-\pi
k^{2}\frac{2\alpha}{y}}k-e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)$
$\displaystyle+e^{-\pi(1+k)^{2}\frac{2\alpha}{y}}(1+k-e^{-3\pi(1-2k)\frac{2\alpha}{y}})\Big{)}-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}(\frac{y}{\alpha})^{-\frac{3}{2}}\Big{(}1+(2-16\pi\frac{\alpha}{y})e^{-4\pi\frac{\alpha}{y}}\Big{)}$
$\displaystyle-e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}1+(2-16\pi\frac{2\alpha}{y})e^{-4\pi\frac{2\alpha}{y}})\Big{)}$
$\displaystyle\geq(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{1}{2\sqrt{2}}\frac{1}{\sin(2\pi
k)}e^{-\pi k^{2}\frac{2\alpha}{y}}(k-\varepsilon_{k})-2e^{-\frac{\pi\alpha
y}{2}}-e^{-\frac{3\pi\alpha y}{2}}\Big{)}$
$\displaystyle\geq(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{1}{2\sqrt{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}(k-\varepsilon_{k})-2e^{-\frac{\pi\alpha
y}{2}}-e^{-\frac{3\pi\alpha y}{2}}\Big{)}.$

Here $\varepsilon_{k}$ is determined by

$\displaystyle\varepsilon_{k}:=\max_{\frac{y}{\alpha}\in[0,k]}\frac{e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)}{e^{-\pi
k^{2}\frac{2\alpha}{y}}k}\leq\frac{1-k}{k}e^{-\pi(1-2k)\frac{2}{k}}.$

Let $k\in[0,0.3]$. Then it is easy to see that
$\varepsilon_{k}\leq\varepsilon_{0.3}=0.000536\cdots<\frac{3}{5000}$ which
yields that

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.16) $\displaystyle\geq$
$\displaystyle(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{1}{2\sqrt{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}(k-\frac{3}{5000})-2e^{-\frac{\pi\alpha
y}{2}}-e^{-\frac{3\pi\alpha y}{2}}\Big{)}$ $\displaystyle=$ $\displaystyle
e^{-\pi
k^{2}\frac{2\alpha}{y}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{k-\frac{3}{5000}}{2\sqrt{2}}-2e^{-\pi\alpha\frac{y^{2}-4k^{2}}{2y}}-e^{-(\pi\alpha\frac{y^{2}-4k^{2}}{2y}+\pi\alpha
y)}\Big{)}$ $\displaystyle\geq$ $\displaystyle e^{-\pi
k^{2}\frac{2\alpha}{y}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{k-\frac{3}{5000}}{2\sqrt{2}}-2(1+\frac{1}{5000})e^{-\pi\alpha\frac{y^{2}-4k^{2}}{2y}}\Big{)}$
$\displaystyle\geq$ $\displaystyle e^{-\pi
k^{2}\frac{2\alpha}{y}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{k-\frac{3}{5000}}{2\sqrt{2}}-2(1+\frac{1}{5000})e^{-\pi\frac{y^{2}-4k^{2}}{2k}}\Big{)},\hbox{since}\;\alpha\geq\frac{y}{k}$
$\displaystyle\geq$ $\displaystyle e^{-\pi
k^{2}\frac{2\alpha}{y}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\Big{(}\frac{k-\frac{3}{5000}}{2\sqrt{2}}-2(1+\frac{1}{5000})e^{-\pi(\frac{3}{8k}-2k)}\Big{)},\hbox{since}\;y^{2}\geq
1-x^{2}=\frac{3}{4}.$

Here we have used the fact that

$\max_{\frac{y}{\alpha}\in[0,k],k\in[0,0.3]}\frac{e^{-\pi\alpha
y}}{2}\leq\frac{e^{-\pi\frac{1}{0.3}y^{2}}}{2}\leq\frac{e^{-\pi\frac{1}{0.3}\frac{3}{4}}}{2}=0.00019\cdots<\frac{1}{5000},\hbox{since}\;y^{2}\geq
1-x^{2}\geq\frac{3}{4}.$

An elementary computation shows that the function

$f(k):=\frac{k-\frac{3}{5000}}{2\sqrt{2}}-2(1+\frac{1}{5000})e^{-\pi(\frac{3}{8k}-2k)}>0$
(3.17)

for $k\in[0,k_{f}]$ where $k_{f}=0.25094\cdots$.

Therefore, by (3.18),

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq e^{-\pi
k^{2}\frac{2\alpha}{y}}(\frac{y}{2\alpha})^{-\frac{3}{2}}f(k)>0$ (3.18)

if $k\in(0,k_{f}]$, where $k_{f}\approx 0.25094\cdots$. This gives the first
range of the free parameter $k$.

Case $\mathbf{B}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(k,\infty)$. By Lemma
2.2 again, we have

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.19) $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(1-\mu(k))4\pi
e^{-\pi\frac{y}{2\alpha}}-\sqrt{2}e^{-\frac{\pi\alpha y}{2}}(1+\mu(k))4\pi
e^{-\pi\frac{y}{\alpha}}-e^{-\frac{3\pi\alpha y}{2}}(1+\mu(k))4\pi
e^{-\pi\frac{y}{2\alpha}}$ $\displaystyle=$ $\displaystyle 4\pi
e^{-\pi\frac{y}{2\alpha}}\Big{(}\frac{1}{2\sqrt{2}}(1-\mu(k))-\sqrt{2}(1+\mu(k))e^{-\frac{\pi
y}{2}(\alpha+\frac{1}{\alpha})}-(1+\mu(k))e^{-\frac{3\pi\alpha y}{2}}\Big{)}$
$\displaystyle\geq$ $\displaystyle 4\pi
e^{-\pi\frac{y}{2\alpha}}\Big{(}\frac{1}{2\sqrt{2}}(1-\mu(k))-(\sqrt{2}+\delta)(1+\mu(k))e^{-\frac{\pi
y}{2}(\alpha+\frac{1}{\alpha})}\Big{)}$ $\displaystyle\geq$ $\displaystyle
4\pi
e^{-\pi\frac{y}{2\alpha}}\Big{(}\frac{1}{2\sqrt{2}}(1-\mu(k))-(\sqrt{2}+e^{-\frac{\pi\sqrt{3}}{4}})(1+\mu(k))e^{-{\pi
y}}\Big{)},\;\hbox{since}\;\frac{1}{2}(\alpha+\frac{1}{\alpha})\geq
1,\alpha>0$ $\displaystyle\geq$ $\displaystyle 4\pi
e^{-\pi\frac{y}{2\alpha}}\Big{(}\frac{1}{2\sqrt{2}}(1-\mu(k))-(\sqrt{2}+e^{-\frac{\pi\sqrt{3}}{4}})(1+\mu(k))e^{-\frac{\sqrt{3}\pi}{2}}\Big{)},\;\hbox{since}\;y\geq\sqrt{1-x^{2}}\geq\frac{\sqrt{3}}{2}.$

Here $\delta$ is defined by

$\delta=\max_{\alpha\geq
1,y\geq\sqrt{1-x^{2}},x\in[\frac{1}{4},\frac{1}{2}]}\frac{e^{-\frac{3\pi\alpha
y}{2}}}{e^{-\frac{\pi y}{2}(\alpha+\frac{1}{\alpha})}}.$

Then

$\delta=\max_{\alpha\geq
1,y\geq\sqrt{1-x^{2}},x\in[\frac{1}{4},\frac{1}{2}]}e^{-\pi
y(\alpha-\frac{1}{2\alpha})}\leq e^{-\pi y(1-\frac{1}{2})}\leq
e^{-\frac{\pi\sqrt{3}}{4}}.$

Let

$\displaystyle g(k):$
$\displaystyle=\frac{1}{2\sqrt{2}}(1-\mu(k))-(\sqrt{2}+e^{-\frac{\pi\sqrt{3}}{4}})e^{-\frac{\sqrt{3}\pi}{2}}(1+\mu(k)),\;k\in[0,\frac{1}{2}]$
(3.20)
$\displaystyle=(\frac{1}{2\sqrt{2}}+(\sqrt{2}+e^{-\frac{\pi\sqrt{3}}{4}})e^{-\frac{\sqrt{3}\pi}{2}})(\frac{\frac{1}{2\sqrt{2}}-(\sqrt{2}+e^{-\frac{\pi\sqrt{3}}{4}})e^{-\frac{\sqrt{3}\pi}{2}}}{\frac{1}{2\sqrt{2}}+(\sqrt{2}+e^{-\frac{\pi\sqrt{3}}{4}})e^{-\frac{\sqrt{3}\pi}{2}}}-\mu(k)).$

Then $g^{\prime}(k)>0$ for any $k>0$ by $\mu^{\prime}(k)>0$ and $g(k)>0$ if
and only if $k>k_{g}$. Numerically, $k_{g}=0.22247\cdots$.

Therefore by (3.10), (LABEL:J1B) and (3.20), one has

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq 4\pi
e^{-\pi\frac{y}{2\alpha}}g(k)>0\;\hbox{if}\;k>k_{g}=0.22247\cdots.$

This gives the second range of $k$ to be selected.

Case $\mathbf{C}:\frac{y}{2\alpha}\in(0,k],\frac{y}{\alpha}\in(k,\infty)$. By
Lemma 2.2

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.21) $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\frac{1}{\max_{x\in[0,k]}{\sin(2\pi
x)}}\Big{(}e^{-\pi
k^{2}\frac{2\alpha}{y}}k-e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)$
$\displaystyle\;\;\;\;+e^{-\pi(1+k)^{2}\frac{2\alpha}{y}}(1+k-e^{-3\pi(1-2k)\frac{2\alpha}{y}})\Big{)}$
$\displaystyle\;\;\;\;-\sqrt{2}e^{-\frac{\pi\alpha y}{2}}(1+\mu(k))4\pi
e^{-\pi\frac{y}{\alpha}}-e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}(1+(2-4\pi\frac{2\alpha}{y})e^{-\pi\frac{2\alpha}{y}})$
$\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}(k-\epsilon_{k})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}(1+\mu(k))4\pi e^{-\pi\frac{y}{\alpha}}-e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}},$

where $\epsilon_{k}$ is defined by

$\epsilon_{k}:=\frac{e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)}{e^{-\pi
k^{2}\frac{2\alpha}{y}}k}.$ (3.22)

Since the function $\epsilon_{k}$ will also appear in the proof of the next
lemma, we single out the elementary but useful property here:

$\frac{\partial}{\partial
k}\epsilon_{k}=(1-k)(\pi-\frac{k}{1-k})\frac{e^{-\frac{\pi(1-2k)}{k}}}{k^{3}}.$
(3.23)

Hence $\epsilon_{k}$ is decreasing with respect to $k$ if
$k\in(0,\frac{\pi}{1+\pi}]$.

By cases $\mathbf{A}$ and $\mathbf{B}$, we will select the parameter by
$k\in(k_{g},k_{f})=(0.22247\cdots,0.25094\cdots)$. Let

$k\in(k_{g},k_{f}).$

We further take $k\in(k_{a},k_{b}]$, where $k_{a}=k_{g}$ and
$k_{b}=\min\\{k_{f},\frac{1}{4}\\}$. The number $\frac{1}{4}$ is taken by the
bound to control that

$\displaystyle\frac{1}{\max_{x\in[0,k]}{\sin(2\pi x)}}\geq 1$

provided that $k\in(0,k_{b}]$. We keep the symbols $k_{a},k_{b}$ to make the
estimates clearer, and keep in mind that
$k_{a}=0.22247\cdots,k_{b}=\frac{1}{4}$.

On the other hand it is easy to see that
$\epsilon_{k}<\epsilon_{k_{f}}=0.005843\cdots<\frac{3}{500}$. Note that
$\frac{y}{2\alpha}\in(0,k]$ implies that

$\displaystyle\alpha\geq\frac{y}{2k}.$ (3.24)

Our method here works for all the $k$ belongs to the range $(k_{a},k_{b}]$.
Here for simplicity, we set

$\displaystyle k=\frac{1}{4}.$

In this case, by (3.24),

$\displaystyle\alpha\geq\sqrt{3}.$ (3.25)

Here we have used that $y\geq\frac{\sqrt{3}}{2}$ implied by
$x\in[\frac{1}{4},\frac{1}{2}]$ since $z=(x,y)$ belongs to the domain
$\overline{{\mathcal{R}_{L}}}$. Now to deal with the last inequality in
(LABEL:J1C), we rewrite

$\displaystyle\alpha y=\alpha^{2}\frac{y}{\alpha}\geq 3\frac{y}{\alpha}.$
(3.26)

The bounds of $\frac{y}{\alpha}$ in this case implies that
$\frac{1}{k}\leq\frac{2\alpha}{y}\leq\frac{2}{k}$. Fixing all the parameters
except $k$ in the expression $e^{-\pi
k^{2}\frac{2\alpha}{y}}(k-\epsilon_{k})$, then we have

$\displaystyle\frac{\partial}{\partial k}\big{(}e^{-\pi
k^{2}\frac{2\alpha}{y}}(k-\epsilon_{k})\big{)}=-e^{-\pi
k^{2}\frac{2\alpha}{y}}\big{(}\frac{2\alpha}{y}\pi(k-\epsilon_{k})-1\big{)}$
(3.27)

which is positive since trivially

$\displaystyle\frac{2\alpha}{y}\pi(k-\epsilon_{k})-1\geq\pi(1-\frac{\epsilon_{k}}{k})-1>\frac{\pi}{2}-1>0.$

To simplify the notations, we set $\gamma=\frac{y}{\alpha}$, hence
$\gamma\in(k,2k]=(\frac{1}{4},\frac{1}{2}].$

In view of (LABEL:J1C), by the monotonicity in (3.27),

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.28) $\displaystyle\geq$
$\displaystyle\gamma^{-\frac{3}{2}}e^{-\frac{\pi}{2}\frac{1}{\gamma}}(\frac{1}{4}-\epsilon_{k})-\sqrt{2}e^{-\frac{\pi}{2}\alpha^{2}\gamma}(1+\mu(\frac{1}{4}))4\pi
e^{-\pi\gamma}-e^{-\frac{3\pi}{2}\alpha^{2}\gamma}(\frac{\gamma}{2})^{-\frac{3}{2}}.$

Using the elementary but useful bound for $\alpha$ in (3.25), we further get
the lower bound with the only variable $\gamma$, i.e.,

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.29) $\displaystyle\geq$
$\displaystyle\gamma^{-\frac{3}{2}}e^{-\frac{\pi}{2}\frac{1}{\gamma}}(\frac{1}{4}-\epsilon_{k})-\sqrt{2}e^{-\frac{3\pi}{2}\gamma}(1+\mu(\frac{1}{4}))4\pi
e^{-\pi\gamma}-e^{-\frac{9\pi}{2}\gamma}(\frac{\gamma}{2})^{-\frac{3}{2}}.$

Now the only variable in above is the parameter $\gamma$, to make the
structure clear, let

$\displaystyle
h(\gamma):=C_{1}\gamma^{-\frac{3}{2}}e^{-\frac{\pi}{2}\frac{1}{\gamma}}-C_{2}e^{-\frac{5\pi}{2}\gamma}-C_{3}e^{-\frac{9\pi}{2}\gamma}(\frac{\gamma}{2})^{-\frac{3}{2}}$
(3.30)

with the constants determined by (LABEL:J1C1new)

$\displaystyle C_{1}$ $\displaystyle=\frac{1}{4}-\frac{3}{500}=0.244$ (3.31)
$\displaystyle C_{2}$
$\displaystyle=\sqrt{2}(1+\mu(\frac{1}{4}))4\pi=24.80995446\cdots$
$\displaystyle C_{3}$ $\displaystyle=1,$

where $\mu(X)$ is defined in (2.19) and $\mu(\frac{1}{4})=0.39605042\cdots$.

By (LABEL:J1C1new),

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq
h(\gamma),\gamma=\frac{y}{\alpha}.$ (3.32)

It remains to show that

$\displaystyle h(\gamma)>0,\hbox{for}\;\gamma\in(\frac{1}{4},\frac{1}{2}].$
(3.33)

Simple calculation shows that

$\displaystyle\big{(}\gamma^{-\frac{3}{2}}e^{-\frac{\pi}{2}\frac{1}{\gamma}}\big{)}^{\prime}=\gamma^{-\frac{7}{2}}e^{-\frac{\pi}{2}\frac{1}{\gamma}}(\frac{\pi}{2}-\frac{3}{2}\gamma^{2}),$
(3.34)

while $\gamma\in(\frac{1}{4},\frac{1}{2}]$ in (3.33), one has
$\frac{\pi}{2}-\frac{3}{2}\gamma^{2}>0$. Hence, we infer that $h(\gamma)$ is
monotonically increasing for $\gamma\in(\frac{1}{4},\frac{1}{2}]$ and it
follows that

$\displaystyle
h(\gamma)>h(\frac{1}{4})=0.003734\cdots>0,\;\;\gamma\in(\frac{1}{4},\frac{1}{2}].$
(3.35)

Therefore, by (LABEL:J1C1new) and (3.30), we have

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})>0.$ (3.36)

By the same proof, we can take the parameter $k$ to be $\frac{1}{4}$ and then
all the estimates work. In fact, using the same method above, for any
$k\in(k_{g},\frac{1}{4}]$, we can show that the inequality (3.36) holds.

Notice that the constants $\epsilon_{j}(\alpha),j=1,2$ are very close to zero.
With almost the same computations in the lower bound estimates (3.36), we have
that for $x\in[\frac{1}{4},\frac{1}{2}]$ and $y\geq\frac{\sqrt{3}}{2}$

$\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-(1+\epsilon_{1}(\alpha))\sqrt{2}e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-(1+\epsilon_{2}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})>0.$

The proof of (3.5) is thus complete.

∎

Finally we prove Lemma 3.2. The proof is similar to that of Lemma 3.1. We
first have

Proof of (LABEL:HHHBB) of Lemma 3.2: Let $x\in[0,\frac{1}{4}].$ Similar to the
proof of Lemma 3.1, we divide

$\displaystyle\frac{\partial}{\partial x}$
$\displaystyle\mathcal{W}_{\frac{1}{2}}(z;\alpha)$ (3.37) $\displaystyle=$
$\displaystyle e^{-\frac{\pi\alpha
y}{2}}\sqrt{\frac{y}{\alpha}}\Big{(}\frac{1}{2\sqrt{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=\frac{x+1}{2}}+e^{-\frac{\pi\alpha
y}{2}}\frac{\partial}{\partial Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=x}$
$\displaystyle\;\;\;+\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=x+1}+2e^{-\frac{7\pi\alpha
y}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=2x}\Big{)}$
$\displaystyle\;\;\;+\sqrt{\frac{y}{\alpha}}\Big{(}\frac{3}{2\sqrt{2}}e^{-\frac{9\pi\alpha
y}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=3\frac{x+1}{2}}+\sum_{n=4}^{\infty}\frac{1}{2\sqrt{2}}ne^{-\frac{\pi\alpha
yn^{2}}{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=n\frac{x+1}{2}}$
$\displaystyle\;\;\;+\sum_{n=3}^{\infty}ne^{-\pi\alpha
yn^{2}}\frac{\partial}{\partial
Y}\vartheta(\frac{y}{\alpha};Y)\mid_{Y=nx}\Big{)}$
$\displaystyle=\sqrt{\frac{y}{\alpha}}\Big{(}e^{-\frac{\pi\alpha
y}{2}}J_{3}+J_{4}\Big{)}.$

Here $J_{3}$ and $J_{4}$ are defined at the last equality.

For the lower bound of $J_{3}$, since $\sin(4\pi x)\geq 0$, by Lemma 2.2

$\displaystyle J_{3}$
$\displaystyle\geq\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})\sin(\pi
x)-e^{-\frac{\pi\alpha y}{2}}\overline{\vartheta}(\frac{y}{\alpha})\sin(2\pi
x)$ (3.38) $\displaystyle\;\;\;-\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\sin(2\pi
x)-2e^{-\frac{7\pi\alpha y}{2}}\overline{\vartheta}(\frac{y}{\alpha})\sin(4\pi
x).$

Observe that for $x\in[0,\frac{1}{4}]$

$\frac{\partial}{\partial
Y}\vartheta(\frac{y}{2\alpha};Y)\mid_{Y=3\frac{x+1}{2}}\geq\underline{\vartheta}(\frac{y}{2\alpha})\sin(3\pi
x)\geq 0.$

To compare the terms in $J_{3}$, using $|\sin(ny)|\leq
n|\sin(y)|,y\in{\mathbb{R}}$ and Lemma 2.2, we estimate that

$\displaystyle J_{4}$
$\displaystyle\geq-\frac{1}{4\sqrt{2}}\sum_{n=4}^{\infty}n^{2}e^{-\frac{\pi\alpha
yn^{2}}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\sin(2\pi
x)-\sum_{n=3}^{\infty}n^{2}e^{-\pi\alpha
yn^{2}}\overline{\vartheta}(\frac{y}{\alpha})\sin(2\pi x).$ (3.39)

In fact, combining (3.38) and (3.39), by collecting the terms, one has

$\displaystyle J_{3}+e^{\frac{\pi\alpha y}{2}}J_{4}$
$\displaystyle\geq\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})\sin(\pi
x)-e^{-\frac{\pi\alpha y}{2}}\overline{\vartheta}(\frac{y}{\alpha})\sin(2\pi
x)\Big{(}1+4e^{-3\pi\alpha y}+\sum_{n=3}n^{2}e^{-\pi\alpha y(n^{2}-1)}\Big{)}$
(3.40) $\displaystyle\;\;\;-\frac{1}{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\sin(2\pi
x)\Big{(}1+\frac{1}{4}\sum_{n=4}^{\infty}n^{2}e^{-\frac{\pi\alpha
y(n^{2}-4)}{2}}\Big{)}$ $\displaystyle\geq\sin(\pi
x)\Big{(}\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})(1+\epsilon_{3}(\alpha))-{\sqrt{2}}e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})(1+\epsilon_{4}(\alpha))\Big{)}.$

The constants $\epsilon_{j}(\alpha),j=3,4$ are defined by

$\epsilon_{3}(\alpha)=4e^{-3\pi\alpha y}+\sum_{n=3}n^{2}e^{-\pi\alpha
y(n^{2}-1},\
\epsilon_{4}(\alpha)=\frac{1}{4}\sum_{n=4}^{\infty}n^{2}e^{-\frac{\pi\alpha
y(n^{2}-4)}{2}}.$

Since $x\in[0,\frac{1}{4}]$, $y\geq\sqrt{1-x^{2}}\geq\frac{\sqrt{15}}{4}$. A
direct calculation shows that $\epsilon_{3}(\alpha)<1.1\cdot
10^{-4},\epsilon_{4}(\alpha)<5.7\cdot 10^{-16}.$

Combining (3.37) and (3.40), we obtain (LABEL:HHHBB).

∎

Proof of (3.7) of Lemma 3.2: It suffices to prove (3.7) for
$y\geq\sqrt{1-x^{2}}\geq\frac{\sqrt{15}}{4}$.

The proof is similar to that of (LABEL:HHHAA) of Lemma 3.1. We first divide
the proof into three subcases

$\mathbf{A}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(0,k];\mathbf{B}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(0,k]\in(k,+\infty);\mathbf{C}:\frac{y}{\alpha}\in(0,k],\frac{y}{2\alpha}\in(k,+\infty)$

where $k$ will be chosen appropriately later.

Case $\mathbf{A}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(0,k]$. By Lemma 2.2

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\frac{1}{\sin(2\pi
k)}\big{(}e^{-\pi
k^{2}\frac{2\alpha}{y}}k-e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)\big{)}$
$\displaystyle-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}(\frac{y}{\alpha})^{-\frac{3}{2}}-\sqrt{2}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}$ $\displaystyle\geq$
$\displaystyle(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}\Big{(}\frac{1}{2\sqrt{2}}(k-\varepsilon_{k})-e^{-\frac{\pi\alpha(y^{2}-4k^{2})}{2y}}2\sqrt{2}(1+\epsilon_{3}(\alpha))(1+\frac{\sqrt{2}(1+\eta_{2}(\alpha))e^{-\pi\alpha
y}}{2\sqrt{2}(1+\epsilon_{3}(\alpha))})\Big{)}.$

Here $\varepsilon_{k}$ is determined by

$\varepsilon_{k}:=\max_{\frac{y}{\alpha}\in[0,k]}\frac{e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)}{e^{-\pi
k^{2}\frac{2\alpha}{y}}k}.$

Let

$c(\alpha):=2\sqrt{2}(1+\epsilon_{3}(\alpha))(1+\frac{\sqrt{2}(1+\epsilon_{4}(\alpha))e^{-\pi\alpha
y}}{2\sqrt{2}(1+\eta_{1}(\alpha))})$

and

$F_{0}(k):=\frac{1}{2\sqrt{2}}(k-\varepsilon_{k})-c(\alpha)e^{-\frac{\pi\alpha(y^{2}-4k^{2})}{2y}}.$

Then

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}F_{0}(k).$ (3.41)

Let $k\in[0,0.3]$. Since $\frac{y}{\alpha}\in[0,k]$ and
$y\geq\frac{\sqrt{15}}{4}$, one has

$\varepsilon_{k}<\frac{3}{5000},\;c(\alpha)<2.8287$

and

$\displaystyle F_{0}(k)\geq
F(k):=\frac{1}{2\sqrt{2}}(k-\varepsilon_{k})-c(\alpha)e^{-\pi(\frac{15}{32k}-2k)}.$
(3.42)

Hence, by (3.41), one has

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}F(k).$ (3.43)

Now $F(k)$ is an elementary function and is decreasing on $[0,0.3]$, and
computation shows that $F(k)>0$ equivalents to

$k\in(0,k_{F}),k_{F}=0.28689\cdots.$

Therefore, by (3.43), one has

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}F(k)>0$

provided that $k\in(0,k_{F})$ with $k_{F}=0.28689\cdots$.

This gives the first range of $k$ to be selected in completing our proof.

Case $\mathbf{B}:\frac{y}{\alpha},\frac{y}{2\alpha}\in(k,\infty)$: By Lemma
2.2

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(1-\mu(k))4\pi
e^{-\pi\frac{y}{2\alpha}}-2(1+\epsilon_{3}(\alpha))(1+\mu(k))4\pi
e^{-\pi\frac{y}{\alpha}}e^{-\frac{\pi\alpha y}{2}}$
$\displaystyle-\sqrt{2}(1+\epsilon_{4}(\alpha))(1+\mu(k))4\pi
e^{-\pi\frac{y}{2\alpha}}e^{-\frac{3\pi\alpha y}{2}}$ $\displaystyle=$
$\displaystyle 4\pi
e^{-\frac{y}{2\alpha}}\Big{(}\frac{1}{2\sqrt{2}}(1-\mu(k))-(2(1+\epsilon_{3}(\alpha)+\sqrt{2}(1+\epsilon_{3}(\alpha))e^{-\pi
y(\alpha-\frac{1}{2\alpha})}))e^{-\frac{\pi
y}{2}(\alpha+\frac{1}{\alpha})}(1+\mu(k))\Big{)}.$

Let

$d(\alpha):=(2(1+\epsilon_{3}(\alpha)+\sqrt{2}(1+\epsilon_{3}(\alpha))e^{-\pi
y(\alpha-\frac{1}{2\alpha})}))e^{-\frac{\pi y}{2}(\alpha+\frac{1}{\alpha})}$

and

$\displaystyle
G(k):=\frac{\frac{1}{2\sqrt{2}}-d(\alpha)}{\frac{1}{2\sqrt{2}}+d(\alpha)}-\mu(k).$
(3.44)

Then

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq\frac{4\pi
e^{-\pi\frac{y}{2\alpha}}}{\frac{1}{2\sqrt{2}}+d(\alpha)}G(k).$

Since $y\geq\frac{\sqrt{15}}{4}$, then $d(\alpha)\leq 0.11025(\alpha\geq 1)$
and $G^{\prime}(k)=-\mu^{\prime}(k)>0$. Hence numerically

$G(k)>0\;\hbox{if}\;k>k_{G}:=0.22263\cdots.$

Therefore

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ $\displaystyle\geq$
$\displaystyle\frac{4\pi
e^{-\pi\frac{y}{2\alpha}}}{\frac{1}{2\sqrt{2}}+d(\alpha)}G(k)>0\;\hbox{if}\;k>k_{G}.$

This gives the second range of $k$ to be selected, i.e.,
$k>k_{G}(=0.22263\cdots)$.

Case $\mathbf{C}:\frac{y}{2\alpha}\in(0,k],\frac{y}{\alpha}\in(k,\infty)$. In
this case, we first set $k\in(k_{G},k_{F})=(0.22263\cdots,0.28689\cdots)$ by
the computations in subcases D and E above. In using the comparison inequality
later, we further set that

$\displaystyle k\in(k_{A},k_{B}]$

where

$\displaystyle
k_{A}:=\max\\{k_{g},k_{G}\\},\;\;k_{B}:=\min\\{k_{f},k_{F},\frac{1}{4}\\}.$

Then, $k\in(k_{1},k_{2}]=(0.22263\cdots,\frac{1}{4}]$. On the other hand,
since $\frac{y}{2\alpha}\in(0,k]$, then $\alpha\geq\frac{y}{2k}$. Again the
bound $\frac{1}{4}$ is taken for controlling

$\displaystyle\frac{1}{\max_{x\in[0,k]}{\sin(2\pi x)}}\geq 1$

provided that $k\in(0,k_{B}]$.

By Lemma 2.2

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}1(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.45) $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}\frac{1}{\max_{x\in[0,k]}{\sin(2\pi
x)}}\big{(}e^{-\pi
k^{2}\frac{2\alpha}{y}}k-e^{-\pi(1-k)^{2}\frac{2\alpha}{y}}(1-k)\big{)}$
$\displaystyle\;\;\;\;-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}(1+\mu(k))4\pi
e^{-\pi\frac{y}{\alpha}}-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2\sqrt{2}}(k-\epsilon_{k})(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}-2(1+\epsilon_{3}(\alpha))(1+\mu(k))4\pi
e^{-\frac{\pi\alpha y}{2}}e^{-\pi\frac{y}{\alpha}}$
$\displaystyle\;\;\;\;-\sqrt{2}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}$ $\displaystyle=$
$\displaystyle(k-\epsilon_{k})(\frac{y}{\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}-2(1+\epsilon_{3}(\alpha))(1+\mu(k))4\pi
e^{-\frac{\pi\alpha y}{2}}e^{-\pi\frac{y}{\alpha}}$
$\displaystyle\;\;\;\;-\sqrt{2}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}(\frac{y}{2\alpha})^{-\frac{3}{2}}.$

Our method works for all $k\in(k_{A},k_{B}]$, to make the structure clear, we
set that

$\displaystyle k=k_{B}=\frac{1}{4}.$

Since $\frac{y}{2\alpha}\in(0,k]$ and $y\geq\frac{\sqrt{15}}{4}$, one has

$\displaystyle\alpha\geq\frac{y}{2k}=2y\geq\frac{\sqrt{15}}{2}.$

To simplify the lower bounds in (LABEL:Feee1), we use the following

$\displaystyle\alpha
y=\alpha^{2}\frac{y}{\alpha}\geq\frac{15}{4}\frac{y}{\alpha}.$

Continuing from (LABEL:Feee1), one has

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.46) $\displaystyle\geq$
$\displaystyle(\frac{1}{4}-\frac{3}{500})(\frac{y}{\alpha})^{-\frac{3}{2}}e^{-\frac{\pi}{2}\frac{\alpha}{y}}-2(1+\epsilon_{3}(\alpha))(1+\mu(\frac{1}{4}))4\pi
e^{-\frac{23}{8}\pi\frac{y}{\alpha}}$
$\displaystyle\;\;\;\;-\sqrt{2}(1+\epsilon_{4}(\alpha))e^{-\frac{45\pi\alpha
y}{8}}(\frac{y}{2\alpha})^{-\frac{3}{2}}.$

Let

$\displaystyle\gamma=\frac{y}{\alpha}.$

Then in this case

$\displaystyle\gamma=\frac{y}{\alpha}\in(k,2k]=(\frac{1}{4},\frac{1}{2}].$

Let

$\displaystyle
H(\gamma):=\mathcal{C}_{1}\gamma^{-\frac{3}{2}}e^{-\frac{\pi}{2}\frac{1}{\gamma}}-\mathcal{C}_{2}e^{-\frac{23}{8}\pi\gamma}-\mathcal{C}_{3}e^{-\frac{45\pi}{8}\gamma}(\frac{\gamma}{2})^{-\frac{3}{2}},$
(3.47)

here

$\displaystyle\mathcal{C}_{1}$ $\displaystyle=\frac{1}{4}-\frac{3}{500}$
$\displaystyle\mathcal{C}_{2}$
$\displaystyle=2(1+\epsilon_{3}(\alpha))(1+\mu(\frac{1}{4}))4\pi$
$\displaystyle\mathcal{C}_{3}$
$\displaystyle=\sqrt{2}(1+\epsilon_{4}(\alpha)).$

By (LABEL:Feee2),

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})\geq
H(\gamma),\gamma=\frac{y}{\alpha}.$

Then, it suffices to prove that

$\displaystyle H(\gamma)>0,\;\;\gamma\in(\frac{1}{4},\frac{1}{2}].$

The first part of the function $H(\gamma)$ was studied in (3.35), the second
and third parts of the function are decreasing, thus $H(\gamma)$ is increasing
on $(\frac{1}{4},\frac{1}{2}]$, hence

$\displaystyle
H(\gamma)>H(\frac{1}{4}),\;\;\gamma\in(\frac{1}{4},\frac{1}{2}].$

The numerical result shows that

$\displaystyle H(\frac{1}{4})>0.$

Thus we complete the proof of case C.

In summary of cases A, B and C, we have

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})$ (3.48) $\displaystyle\geq$
$\displaystyle(\frac{y}{2\alpha})^{-\frac{3}{2}}e^{-\pi
k^{2}\frac{2\alpha}{y}}F(k)>0,\hbox{if}\;\frac{y}{\alpha}\in(0,k]\;,k\in(0,k_{F}),k_{F}=0.28689\cdots$
$\displaystyle\geq$ $\displaystyle\frac{4\pi
e^{-\pi\frac{y}{2\alpha}}}{\frac{1}{2\sqrt{2}}+d(\alpha)}G(k)>0\;\hbox{if}\;k>k_{G},k_{G}=0.22263\cdots$
$\displaystyle\geq$ $\displaystyle
H(\gamma)>0,\gamma=\frac{y}{\alpha},\hbox{if}\;\frac{y}{2\alpha}\in(0,k],\frac{y}{\alpha}\in(k,\infty),k=\frac{1}{4},$

see the functions $F(k),G(k),H(\gamma)$ in (3.42), (3.44) and (3.47)
respectively. Therefore, by selecting $k=\frac{1}{4}$ in (LABEL:KeyF), one has

$\displaystyle\frac{1}{2\sqrt{2}}\underline{\vartheta}(\frac{y}{2\alpha})-2(1+\epsilon_{3}(\alpha))e^{-\frac{\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{\alpha})-{\sqrt{2}}(1+\epsilon_{4}(\alpha))e^{-\frac{3\pi\alpha
y}{2}}\overline{\vartheta}(\frac{y}{2\alpha})>0.$

The proof of (3.7) is complete.

∎

##  4\. proof of Theorem 1.2 and Theorem 1.3

In this section, we study the monotonicity properties of
$\mathcal{W}_{\kappa}(z;\alpha)$ on the imaginary axis $z=iy,y\geq 1$ and
prove Theorems 1.2 and 1.3.

Recall that from the definition of $\mathcal{W}_{\kappa}(z;\alpha)$ we have

$\displaystyle\mathcal{W}_{0}(iy;\alpha)$
$\displaystyle=\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}+\sum_{n,m}e^{-2\pi\alpha(n^{2}y+\frac{m^{2}}{y})}$
$\displaystyle\mathcal{W}_{1}(iy;\alpha)$
$\displaystyle=\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}.$

Here for convenience, throughout this section, we denote
$\sum_{n,m}=\sum_{n=-\infty}^{n=\infty}\sum_{m=-\infty}^{m=\infty}.$

In this section, we aim to establish that

######  Theorem 4.1.

For $\sigma\in[0,2]$, the function
$\mathcal{W}_{1}(iy;\alpha)+\sigma\mathcal{W}_{0}(iy;\alpha),y\in(0,\infty)$
achieves its unique minimum at $y=1$.

As a consequence, by the Bernstein Theorem [5], we have

######  Theorem 4.2.

For $\sigma\in[0,2]$, the function
$\mathcal{E}_{1}(iy;\alpha)+\sigma\mathcal{E}_{0}(iy;\alpha),y\in(0,\infty)$
achieves its unique minimum at $y=1$.

We first recall the duality property of $\mathcal{W}_{\kappa}$, which can be
proved by Lemma 2.1 and we omit the details here.

######  Lemma 4.1.

Under the transform $z\rightarrow w=\frac{z-1}{z+1}$ of $\mathbb{H}$,

$\mathcal{W}_{1}(z)=\mathcal{W}_{0}(w)\ \ \mbox{and}\ \
\mathcal{W}_{0}(z)=\mathcal{W}_{1}(w),\ \ z\in\mathbb{H}\ \ \mbox{and}\ \
w=\frac{z-1}{z+1}\in\mathbb{H}.$

Consequently, for all $b\in\mathbb{R}$,

$\mathcal{W}_{b}(w)=\mathcal{W}_{1-b}(z),\ \ z\in\mathbb{H}\ \ \mbox{and}\ \
w=\frac{z-1}{z+1}\in\mathbb{H}.$

More generally, if $h:z^{\prime}\rightarrow
w^{\prime}=\frac{z^{\prime}-1}{z^{\prime}+1}$ and $g_{1}:z\rightarrow
z^{\prime}$, $g_{2}:w^{\prime}\rightarrow w$ are transforms in
${\mathcal{G}_{2}}$, then

$\mathcal{W}_{b}(w)=\mathcal{W}_{1-b}(z),\ \ z\in\mathbb{H}\ \ \mbox{and}\ \
w=g_{2}\circ h\circ g_{1}(z)\in\mathbb{H}.$

Let us write $z=x+yi$ henceforth, and set

$X_{b}(z)=\frac{\partial\mathcal{W}_{b}(z)}{\partial
x}=bX_{1}(z)+(1-b)X_{0}(z),\ \
Y_{b}(z)=\frac{\partial\mathcal{W}_{b}(z)}{\partial
y}=bY_{1}(z)+(1-b)Y_{0}(z).$

The following two formulas relate $f_{b}$ on the upper half of the unit circle
to $f_{1-b}$ on the upper half of the imaginary axis.

######  Lemma 4.2.

Let the upper half of the unit circle be parametrized by $u+i\sqrt{1-u^{2}}$,
$u\in(-1,1)$. Then $\frac{\sqrt{1-u^{2}}}{1-u}i$ parametrizes the upper half
of the imaginary axies, and

$\displaystyle X_{b}(u+i\sqrt{1-u^{2}})$
$\displaystyle=\frac{\sqrt{1-u^{2}}}{1-u}Y_{1-b}\Big{(}\frac{\sqrt{1-u^{2}}}{1-u}i\Big{)}$
(4.1) $\displaystyle Y_{b}(u+i\sqrt{1-u^{2}})$
$\displaystyle=\frac{-u}{1-u}Y_{1-b}\Big{(}\frac{\sqrt{1-u^{2}}}{1-u}i\Big{)}$
(4.2)

hold for $u\in(-1,1)$.

###### Proof.

Consider the transform in Lemma 4.1, $z\rightarrow w=\frac{z-1}{z+1}$. With
$z=x+yi$ and $w=u+vi$,

$u=\frac{x^{2}+y^{2}-1}{(x+1)^{2}+y^{2}},\ \ v=\frac{2y}{(x+1)^{2}+y^{2}}.$
(4.3)

Conversely,

$x=\frac{1-u^{2}-v^{2}}{(1-u)^{2}+v^{2}},\ \ y=\frac{2v}{(1-u)^{2}+v^{2}}.$
(4.4)

Differentiate $f_{b}(w)=f_{1-b}(z)$ with respect to $u$ and $v$ to find

$\displaystyle X_{b}(w)=X_{1-b}(z)\frac{\partial x}{\partial
u}+Y_{1-b}(z)\frac{\partial y}{\partial u}$ $\displaystyle
Y_{b}(w)=X_{1-b}(z)\frac{\partial x}{\partial v}+Y_{1-b}(z)\frac{\partial
y}{\partial v}.$

When $w$ is on the unit circle, $z$ is on the imaginary axis. Since
$\mathcal{W}_{1-b}$ is invariant under the reflection about the imaginary
axis, $X_{1-b}(z)=0$ on the imaginary axis. Also

$\frac{\partial y}{\partial u}\Big{|}_{|w|=1}=\frac{\sqrt{1-u^{2}}}{1-u},\ \
\frac{\partial y}{\partial v}\Big{|}_{|w|=1}=\frac{-u}{1-u}$ (4.5)

from which the lemma follows. ∎

By Lemma 2.1, one has

$\displaystyle\mathcal{W}_{1}(i\frac{1}{y};\alpha)+\sigma\mathcal{W}_{0}(i\frac{1}{y};\alpha)=\mathcal{W}_{1}(iy;\alpha)+\sigma\mathcal{W}_{0}(iy;\alpha).$
(4.6)

Then to prove Theorem 4.1, we will show that

######  Lemma 4.3.

Assume that $\sigma\in[0,2]$. Then

$\displaystyle\frac{d}{dy}\Big{(}\mathcal{W}_{1}(iy;\alpha)+\sigma\mathcal{W}_{0}(iy;\alpha)\Big{)}\geq
0,\;\;\hbox{for}\;\;y\geq 1,\alpha\geq 1.$ (4.7)

The proof of (4.7) is through the operator $\mathcal{L}$

$\mathcal{L}Y:=Y^{\prime\prime}+\frac{2}{y}Y^{\prime}=y^{-2}(y^{2}Y^{{}^{\prime}})^{{}^{\prime}},\
y>0$ (4.8)

Observe that if $\mathcal{L}f(y)\geq 0$ for $y\geq 1>0$ and $f^{\prime}(1)=0$,
then $f^{\prime}(y)>0\;\;\hbox{for}\;\;y>1.$ From (4.6) we see that
$\frac{d}{dy}(\mathcal{W}_{1}(iy;\alpha)+\sigma\mathcal{W}_{0}(iy;\alpha))\bigl{|}_{y=1}=0$.
In view of this fact, to prove Lemma 4.3, it suffices to prove that

######  Lemma 4.4.

Assume that $\sigma\in[0,2]$. Then

$\displaystyle\mathcal{L}\Big{(}\mathcal{W}_{1}(iy;\alpha)+\sigma\mathcal{W}_{0}(iy;\alpha)\Big{)}\geq
0,\;\;\hbox{for}\;\;y\geq 1,\alpha\geq 1.$

Note that $\mathcal{L}\mathcal{W}_{1}(iy;\alpha)\geq$ for $\alpha>0,y\geq 1$
is proved by Montgomery [16] and stated in Lemma 4.6. To prove Lemma 4.4, it
suffices to prove that it holds for $\sigma=2$.

To prove Lemma 4.4 we compute first

######  Lemma 4.5.

One first has the following two identities by direct computation

$\displaystyle\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}$
$\displaystyle=I_{a}(y)-I_{b}(y),$ (4.9)
$\displaystyle\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
$\displaystyle=J_{a}(y)-J_{b}(y),$

where

$\displaystyle I_{a}(y):$
$\displaystyle=\sum_{n,m}\pi^{2}\alpha^{2}(\frac{m^{2}}{y^{2}}-n^{2})^{2}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})},$
(4.10) $\displaystyle I_{b}(y):$ $\displaystyle=\sum_{n,m}\frac{2\pi\alpha
n^{2}}{y}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})},$ $\displaystyle J_{a}(y):$
$\displaystyle=\sum_{n,m}\frac{\pi^{2}\alpha^{2}}{4}\big{(}\frac{(2m+1)^{2}}{y^{2}}-(2n+1)^{2}\big{)}^{2}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}},$
$\displaystyle J_{b}(y):$
$\displaystyle=\frac{\pi\alpha}{y}\sum_{n,m}(2n+1)^{2}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}.$

The following Lemma is due to Montgomery [16].

######  Lemma 4.6.

  * •

$\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}\geq
0$ for $y\in[1,\infty)$ with equality holds only when $y=1$;

  * •

$\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}+\sum_{n,m}e^{-2\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}\geq
0$ for $y\in[\sqrt{3},\infty)$ with equality holds only when $y=\sqrt{3}$.

The
$\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
will change sign on the interval $[1,\sqrt{3}]$, however, we establish the
following

######  Lemma 4.7.

At least for $\sigma=2$, it holds that

$\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}+\sigma\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\geq
0,\hbox{for}\;\;y\in[1,\sqrt{3}].$ (4.11)

As a consequence, (4.11) holds for $\sigma\in[0,2]$ as well.

The proof of Lemma 4.7 is quite technique and we postpone it to the end of
this section. Define $\sigma_{M}>0$ as

$\displaystyle\sigma_{M}:=\max_{\sigma}\\{\sigma\mid:\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}+\sigma\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\geq
0,\hbox{for}\;\;y\in[1,\sqrt{3}],\alpha\geq 1\\}.$

Here, as the invariance for the parameter $\alpha$ in (4.6), one only needs to
consider the case $\alpha\geq 1$.

We omit the proof of the following lemma that can be proved by the
monotonicity property and Lemma 4.7.

######  Lemma 4.8.

$\sigma_{M}$ satisfies

$\sigma_{M}\leq-\frac{\mathcal{L}^{\prime}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}\mid_{y=1,\alpha=1}}{\mathcal{L}^{\prime}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\mid_{y=1,\alpha=1}}.$

######  Remark 4.1.

The expression in Lemma 4.8 is well-defined since one notes that

$\displaystyle{\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}\mid_{y=1}}={\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\mid_{y=1}}=0$
$\displaystyle{\mathcal{L}^{\prime}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}\mid_{y=1}}>0,{\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\mid_{y=1}}<0.$

######  Theorem 4.3.

For any $\kappa\in[\frac{1}{1+\sigma_{M}},\frac{\sigma_{M}}{1+\sigma_{M}}]$,

$Minimizer_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=i,$

which is the purely imaginary number $i$, corresponding to the square lattice
by our setting and is unique up the group $\mathcal{G}_{2}$ defined in (2.6).
In particular, via Lemma 4.7,

$[\frac{1}{1+\sigma_{M}},\frac{\sigma_{M}}{1+\sigma_{M}}]\supseteq[\frac{1}{3},\frac{2}{3}].$

###### Proof.

Note that

$\displaystyle\mathcal{E}_{1}(z;\alpha)+\sigma\mathcal{E}_{0}(z;\alpha)$
(4.12) $\displaystyle=$
$\displaystyle(1+\sigma)\Big{(}\frac{1}{1+\sigma}\mathcal{E}_{1}(z;\alpha)+\frac{\sigma}{1+\sigma}\mathcal{E}_{0}(z;\alpha)\Big{)}$
$\displaystyle=$ $\displaystyle(1+\sigma)\mathcal{E}_{\frac{1}{1+\sigma}}(z).$

By minimum principle in Theorem 1.1, one has

$\min_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=\min_{z\in\Omega_{e}}\mathcal{E}_{\kappa}(z),$
(4.13)

the minimizer may happen on $\mathbb{H}\backslash\Omega_{e}$, while it must be
modular to some minimizer on $\Omega_{e}$ via the group $\mathcal{G}_{2}$. By
Theorem 4.2 and (4.12), for $\kappa\in[\frac{1}{3},1]$ (or more generally
$\kappa\in[\frac{1}{1+\sigma_{M}},1]$),

$Minimizer_{z\in\Omega_{ea}}\mathcal{E}_{\kappa}(z)=i.$ (4.14)

By Lemma 4.2 and (4.14), for $\kappa\in[0,\frac{2}{3}]$ (or more generally
$\kappa\in[0,\frac{\sigma_{M}}{1+\sigma_{M}}]$),

$Minimizer_{z\in\Omega_{eb}}\mathcal{E}_{\kappa}(z)=i.$ (4.15)

Therefore, it follows by (4.13), (4.14) and (4.15), for
$\kappa\in[\frac{1}{1+\sigma_{M}},\frac{\sigma_{M}}{1+\sigma_{M}}]$,

$Minimizer_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=i.$ (4.16)

The proof is thus complete.

∎

Next, we turn to the proof of Theorem 1.3.

###### Proof.

By definition

$\mathcal{W}_{\kappa}(z;\alpha)=\kappa\mathcal{W}_{1}(z;\alpha)+(1-\kappa)\mathcal{W}_{0}(z;\alpha).$

To comparison, by deformation one has, for any $\kappa,\kappa_{0}(\neq 0)$

$\mathcal{W}_{\kappa}(z;\alpha)=\frac{\kappa}{\kappa_{0}}\mathcal{W}_{\kappa_{0}}(z;\alpha)+\frac{\kappa_{0}-\kappa}{\kappa_{0}}\mathcal{W}_{0}(z;\alpha).$
(4.17)

Define the least and largest number such that

$Minimizer_{z\in\mathbb{H}}\mathcal{E}_{\kappa}(z)=i$

is $\kappa_{1}$ and $\kappa_{2}$ respectively. It follows by duality Lemma 4.1
that

$\kappa_{1}+\kappa_{2}=1.$

Theorem 4.3 asserts that

$\kappa_{1}\leq\frac{1}{3},\;\kappa_{2}\geq\frac{2}{3}.$

We also define that

$\kappa_{1}^{\prime}:=\frac{1}{1+\sigma_{M}},\;\kappa_{2}^{\prime}:=\frac{\sigma_{M}}{1+\sigma_{M}}.$

It follows by the definitions that

$\kappa_{1}^{\prime}\geq\kappa_{1},\;\kappa_{2}^{\prime}\leq\kappa_{2}.$

Consider $\kappa\in[0,\kappa_{1}^{\prime}]$. By the definition of
$\sigma_{M}$, one has

$Minimizer_{z\in\Omega_{eb}}\mathcal{W}_{\kappa_{1}^{\prime}}(z;\alpha)=i,$
(4.18)

and the minimizer is unique by the monotonicity induced in Lemma 4.2, Lemma
4.1 and Lemma 4.6. On the other hand, via Lemma 4.6 and 4.2, one has the
functional $\mathcal{W}_{0}(z)$ is monotonically increasing on the circle
$\Omega_{eb}$, i.e,

$Minimizer_{z\in\Omega_{eb}}\mathcal{W}_{0}(z;\alpha)=i,$ (4.19)

and the minimizer is uniquely achieved.

From the comparison formula (4.17), we have

$\mathcal{W}_{\kappa}(z;\alpha)=\frac{\kappa}{\kappa_{1}^{\prime}}\mathcal{W}_{\kappa_{1}^{\prime}}(z;\alpha)+\frac{\kappa_{1}^{\prime}-\kappa}{\kappa_{1}^{\prime}}\mathcal{W}_{0}(z;\alpha).$
(4.20)

Combining (4.18) and (4.19), one has, for $\kappa\in[0,\kappa_{1}^{\prime}]$,

$Minimizer_{z\in\Omega_{eb}}\mathcal{W}_{\kappa}(z;\alpha)=i.$ (4.21)

The minimum principle in Theorem 1.1 gives that

$\min_{z\in\mathbb{H}}\mathcal{W}_{\kappa}(z;\alpha)=\min_{z\in\Omega_{e}}\mathcal{W}_{\kappa}(z;\alpha)$
(4.22)

Therefore, for $\kappa\in[0,\kappa_{1}^{\prime}]$, the location identity
(4.21) and minimum principle (4.22)(recall that
$\Omega_{e}=\Omega_{ea}\cup\Omega_{eb}$) conclude that

$\min_{z\in\mathbb{H}}\mathcal{W}_{\kappa}(z;\alpha)=\min_{z\in\Omega_{ea}}\mathcal{W}_{\kappa}(z;\alpha).$
(4.23)

The minimizer is unique up to the group $\mathcal{G}_{2}$.

For $\kappa\in[\kappa_{1}^{\prime},\kappa_{2}^{\prime}]$, by Theorem 4.3,

$Minimizer_{z\in\mathbb{H}}\mathcal{W}_{\kappa}(z;\alpha)=i,$ (4.24)

which is unique up the group $\mathcal{G}_{2}$. For
$\kappa\in[\kappa_{2}^{\prime},1]$, by the duality Lemma 4.1 and (4.23), one
has

$\min_{z\in\mathbb{H}}\mathcal{W}_{\kappa}(z;\alpha)=\min_{z\in\Omega_{eb}}\mathcal{W}_{\kappa}(z;\alpha).$

Therefore, the proof of Theorem 1.3 is complete.

∎

Finally, by Lemma 4.7, one has $\sigma_{M}\geq 2$, hence
$[\frac{1}{1+\sigma_{M}},\frac{\sigma_{M}}{1+\sigma_{M}}]\supseteq[\frac{1}{3},\frac{2}{3}]$.
Theorem 1.2 then follows easily from Theorem 4.3.

It remains to prove Lemma 4.7. We postpone its proof to the end of this
section. Before this, we introduce some auxiliary functions and lemmas to the
proof. From Lemma 4.5 and (4.10), we note that each term in
$I_{a}(y),J_{a}(y)$ is non-negative. To control $I_{b}(y),J_{b}(y)$, we
introduce the auxiliary functions as follows

$\displaystyle\mu_{1}(y,k):$
$\displaystyle=\sum_{n=k}^{\infty}e^{-\pi(n^{2}-1)y}$ (4.25)
$\displaystyle\mu_{2}(y,k):$
$\displaystyle=\sum_{n=k}^{\infty}n^{2}e^{-\pi(n^{2}-1)y}$
$\displaystyle\nu_{1}(y,k):$ $\displaystyle=\sum_{n=k}^{\infty}e^{-2\pi
n(n+1)y}$ $\displaystyle\nu_{2}(y,k):$
$\displaystyle=\sum_{n=k}^{\infty}(2n+1)^{2}e^{-2\pi n(n+1)y}$

for $k\geq 2$, these terms are small since $k\geq 2$, and they are used to
express the error terms.

Within these auxiliary functions, from Lemma 4.5, one can have the lower bound
estimates

######  Lemma 4.9.

We obtain the first lower bound estimate

$\displaystyle\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}\geq\sum_{n=-1}^{1}\sum_{m=-1}^{1}\pi^{2}\alpha^{2}(\frac{m^{2}}{y^{2}}-n^{2})^{2}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}$
(4.26) $\displaystyle-\frac{4\pi\alpha}{y}e^{-\pi\alpha
y}\Big{(}1+2e^{-\pi\alpha\frac{1}{y}}+Error_{1}(\alpha,y)\Big{)},$

where

$Error_{1}(\alpha,y):=2\mu_{2}(\alpha
y,2)e^{-\pi\alpha\frac{1}{y}}+2\mu_{2}(\alpha
y,2)\mu_{1}(\alpha\frac{1}{y},2),$ (4.27)

and the second lower bound estimate

$\displaystyle\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
(4.28)
$\displaystyle\geq\sum_{n=-1}^{1}\sum_{m=-1}^{1}\frac{\pi^{2}\alpha^{2}}{4}\big{(}\frac{(2m+1)^{2}}{y^{2}}-(2n+1)^{2}\big{)}^{2}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}$
$\displaystyle-\frac{4\pi\alpha}{y}e^{-\frac{\pi\alpha}{2}(y+\frac{1}{y})}(1+Error_{2}(\alpha,y)),$

where

$Error_{2}(\alpha,y):=\nu_{1}(\alpha,1)+\nu_{2}(\alpha,1)+\nu_{1}(\alpha,1)\nu_{2}(\alpha,1).$
(4.29)

Before going to the precise estimates, we use the main order analysis to show
the sketch of the proof. Notice that here

$y\in[1,\sqrt{3}].$

Here $y$ is variable however it is well controlled by the lower and upper
bounds.

Notice that in the lower bound estimates in (4.26) and (LABEL:Lformb), the
main order terms only have three possibilities as follows

$\displaystyle e^{-\pi\alpha
y},e^{-\pi\alpha\frac{1}{y}},\;e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}.$

To illustrate the main idea, we only capture the main order terms in Lemma
4.9.

By Lemma 4.9, the leading order terms of
$\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}$
are given by

$\displaystyle\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}$
(4.30) $\displaystyle=$ $\displaystyle
2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+\big{(}2\pi^{2}\alpha^{2}-\frac{4\pi\alpha}{y}\big{)}e^{-\pi\alpha
y}$

and the leading order terms of
$\mathcal{L}\Big{(}\sum_{n,m}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}\Big{)}$
are given by

$\displaystyle\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
(4.31)
$\displaystyle=\pi^{2}\alpha^{2}(\frac{1}{y^{2}}-1)^{2}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}-\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}.$

The term
$\pi^{2}\alpha^{2}(\frac{1}{y^{2}}-1)^{2}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}$
will be dropped in estimates since it is nonnegative and it is zero as $y=1$.
Now by (LABEL:VVV1) and (LABEL:VVV2),

$\displaystyle\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}+\sigma\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
(4.32) $\displaystyle\geq$ $\displaystyle
2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+\big{(}2\pi^{2}\alpha^{2}-\frac{4\pi\alpha}{y}\big{)}e^{-\pi\alpha
y}-\sigma\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}.$

To simplify the notations, let

$\displaystyle\mathcal{P}(\alpha,y;\sigma):=2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+\big{(}2\pi^{2}\alpha^{2}-\frac{4\pi\alpha}{y}\big{)}e^{-\pi\alpha
y}-\sigma\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}.$ (4.33)

Thus

$\displaystyle\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}+\sigma\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\geq\mathcal{P}(\alpha,y;\sigma).$
(4.34)

We shall prove that $\mathcal{P}(\alpha,y;\sigma)>0$ for some positive
$\sigma$.

Since $y\geq 1,\alpha\geq 1$, one has
$\frac{4\pi\alpha}{y}\leq\frac{2}{\pi}2\pi^{2}\alpha^{2}$, thus

$\displaystyle
2\pi^{2}\alpha^{2}-\frac{4\pi\alpha}{y}\geq(1-\frac{2}{\pi})2\pi^{2}\alpha^{2}.$

Then

$\displaystyle\mathcal{P}(\alpha,y;\sigma)\geq
2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+(1-\frac{2}{\pi})2\pi^{2}\alpha^{2}e^{-\pi\alpha
y}-\sigma\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}.$ (4.35)

By the basic mean value inequality, one has, for any positive
$\mathcal{X},\mathcal{Y}$, there has

$\displaystyle\mathcal{X}e^{-\pi\alpha
y}+\mathcal{Y}e^{-\pi\alpha\frac{1}{y}}\geq
2\sqrt{\mathcal{X}\mathcal{Y}}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})},$

with equality holds only if $y=1$ and $\mathcal{X}=\mathcal{Y}$. It follows
that, at least for

$\displaystyle\sigma=\sqrt{1-\frac{2}{\pi}}\pi,$ (4.36)

there holds that

$\displaystyle
2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+(1-\frac{2}{\pi})2\pi^{2}\alpha^{2}e^{-\pi\alpha
y}>\sigma\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})},$

which implies that

$\displaystyle\mathcal{P}(\alpha,y;\sigma)>0.$

In view of (4.35), for the value of $\sigma$ in (4.36), one has

$\displaystyle\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}+\sigma\mathcal{L}_{m}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}\geq
0.$ (4.37)

By the main order analysis, we give the heuristic proof of Lemma 4.7. In the
next, we give full proof of Lemma 4.7 and provide the scheme of finding the
optimal value of $\sigma$.

We are now in a position to prove Lemma 4.7.

Proof of Lemma 4.7.

###### Proof.

We only need to consider the cases $\alpha\geq 1$ and $y\geq 1$ by Lemma 4.1
and identity (4.6) respectively.

By Lemma 4.9, one has

$\displaystyle\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}+\sigma\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
(4.38) $\displaystyle\geq$
$\displaystyle\sum_{n=-1}^{1}\sum_{m=-1}^{1}\pi^{2}\alpha^{2}(\frac{m^{2}}{y^{2}}-n^{2})^{2}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}-\frac{4\pi\alpha}{y}e^{-\pi\alpha
y}\Big{(}1+2e^{-\pi\alpha\frac{1}{y}}\Big{)}$ $\displaystyle+$
$\displaystyle\sigma\Big{(}\sum_{n=-1}^{1}\sum_{m=-1}^{1}\frac{\pi^{2}\alpha^{2}}{4}\big{(}\frac{(2m+1)^{2}}{y^{2}}-(2n+1)^{2}\big{)}^{2}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}$
$\displaystyle-\frac{4\pi\alpha}{y}e^{-\frac{\pi\alpha}{2}(y+\frac{1}{y})})\Big{)}$
$\displaystyle-$ $\displaystyle\mathcal{R}_{1}(\alpha,y).$

Here

$\displaystyle\mathcal{R}_{1}(\alpha,y)=\frac{4\pi\alpha}{y}e^{-\pi\alpha
y}Error_{1}(\alpha,y)+\frac{4\pi\alpha}{y}e^{-\frac{\pi\alpha}{2}(y+\frac{1}{y})}Error_{2}(\alpha,y),$

where $Error_{1}(\alpha,y)$ and $Error_{2}(\alpha,y)$ are defined in (4.27)
and (4.29) respectively. Now we define

$\displaystyle\mathcal{A}(\alpha,y;\sigma):=$
$\displaystyle\sum_{n=-1}^{1}\sum_{m=-1}^{1}\pi^{2}\alpha^{2}(\frac{m^{2}}{y^{2}}-n^{2})^{2}e^{-\pi\alpha(n^{2}y+\frac{m^{2}}{y})}-\frac{4\pi\alpha}{y}e^{-\pi\alpha
y}\Big{(}1+2e^{-\pi\alpha\frac{1}{y}}\Big{)}$ (4.39) $\displaystyle+$
$\displaystyle\sigma\Big{(}\sum_{n=-1}^{1}\sum_{m=-1}^{1}\frac{\pi^{2}\alpha^{2}}{4}\big{(}\frac{(2m+1)^{2}}{y^{2}}-(2n+1)^{2}\big{)}^{2}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}$
$\displaystyle-\frac{4\pi\alpha}{y}e^{-\frac{\pi\alpha}{2}(y+\frac{1}{y})})\Big{)}.$

Then

$\displaystyle\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}+\sigma\mathcal{L}\Big{(}\sum_{n,m}e^{-\frac{\pi\alpha}{2}\big{(}(2n+1)^{2}y+\frac{(2m+1)^{2}}{y}\big{)}}\Big{)}$
(4.40) $\displaystyle\geq$
$\displaystyle\mathcal{A}(\alpha,y;\sigma)-\mathcal{R}_{1}(\alpha,y).$

The remainder term $\mathcal{R}_{1}(\alpha,y)$ is very small comparing the
major term $\mathcal{A}(\alpha,y;\sigma)$ for $\sigma\geq 0$. We note that the
estimate in $\mathcal{A}(\alpha,y;\sigma)$ is much refined than that in the
heuristic estimate. In fact, after dropping the nonnegative terms in
$\mathcal{A}(\alpha,y;\sigma)$, one has

$\displaystyle\mathcal{A}(\alpha,y;\sigma)\geq$ $\displaystyle
2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+\big{(}2\pi^{2}\alpha^{2}-\frac{4\pi\alpha}{y}(1+2e^{-\pi\alpha\frac{1}{y}})\big{)}e^{-\pi\alpha
y}-\sigma\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}$ (4.41)
$\displaystyle=$
$\displaystyle\mathcal{P}(\alpha,y;\sigma)-\frac{8\pi\alpha}{y}e^{-\pi\alpha(y+\frac{1}{y})}.$

The lower bound is quite similar to the one in
$\mathcal{P}(\alpha,y;\sigma)$((4.35)). Same as the estimates in (4.35),
similar to calculation of the value of $\sigma$ in (4.36), i.e., for

$\displaystyle\sigma=\sqrt{1-\frac{2}{\pi}}\pi-2e^{-\pi},$

one has

$\displaystyle
2\pi^{2}\alpha^{2}\frac{1}{y^{4}}e^{-\pi\alpha\frac{1}{y}}+\big{(}2\pi^{2}\alpha^{2}-\frac{4\pi\alpha}{y}(1+2e^{-\pi\alpha\frac{1}{y}})\big{)}e^{-\pi\alpha
y}-\sigma\frac{4\pi\alpha}{y}e^{-\pi\alpha\frac{1}{2}(y+\frac{1}{y})}>0$
(4.42)

for $\alpha\geq 1$ and $y\geq 1$. Hence by (4.41) and (4.42),

$\displaystyle\mathcal{A}(\alpha,y;\sqrt{1-\frac{2}{\pi}}\pi-2e^{-\pi})>0\;\;\hbox{for}\;\;\alpha\geq
1,y\in[1,\sqrt{3}].$

The value of $\sigma$ such that $\mathcal{A}(\alpha,y;\sigma)$ is positive on
the strip $\alpha\geq 1$ and $y\in[1,\sqrt{3}]$ can be improved by the
auxiliary function $\mathcal{A}(\alpha,y;\sigma)$.

In the next, we show that

$\displaystyle\mathcal{A}(\alpha,y;2)>0\;\;\hbox{for}\;\;\alpha\geq
1,y\in[1,\sqrt{3}].$ (4.43)

To use the monotonicity, we set

$\displaystyle\mathcal{A}(\alpha,y;\sigma)=e^{-\pi\alpha
y}y^{-4}\mathcal{B}(\alpha,y;\sigma),$ (4.44)

where

$\displaystyle\mathcal{B}(\alpha,y;\sigma)=$ $\displaystyle
2\pi^{2}\alpha^{2}y^{4}+2\pi^{2}\alpha^{2}e^{\pi\alpha(y-\frac{1}{y})}+4\pi^{2}\alpha^{2}(y^{2}-1)^{2}e^{-\pi\alpha\frac{1}{y}}$
$\displaystyle-4\pi\alpha y^{3}-8\pi\alpha
y^{3}e^{-\pi\alpha\frac{1}{y}}-\sigma\Big{(}\pi^{2}\alpha^{2}(y^{2}-1)^{2}e^{\frac{\pi\alpha}{2}(y-\frac{1}{y})}-4\pi\alpha
y^{3}e^{\frac{\pi\alpha}{2}(y-\frac{1}{y})}\Big{)}.$

By an elementary but complicated argument, one has

$\displaystyle\frac{\partial}{\partial\alpha}\mathcal{B}(\alpha,y;\sigma)>0,\frac{\partial}{\partial
y}\mathcal{B}(\alpha,y;\sigma)>0\;\;\hbox{for}\;\;y\in[1,\sqrt{3}],\alpha\geq
1,\sigma\in[0,2].$

It follows that

$\displaystyle\min_{\alpha\geq
1,y\in[1,\sqrt{3}]}\mathcal{B}(\alpha,y;2)=\mathcal{B}(1,1;2).$

Now the numerical result shows that

$\displaystyle\mathcal{B}(1,1;2)>0.$

Therefore by (4.44), (4.43) is proved.

Since the remainder term $\mathcal{R}_{1}(\alpha,y)$ is so small comparing the
major term $\mathcal{A}(\alpha,y;\sigma)$, one has

$\displaystyle\mathcal{A}(\alpha,y;2)-\mathcal{R}_{1}(\alpha,y)>0\;\;\hbox{for}\;\;\alpha\geq
1,y\in[1,\sqrt{3}].$

By (4.39), the proof is complete.

∎

Remark on the Proof of Lemma 4.7.

The bound $\sigma=2$ is not sharp as we can see in the proof. However we can
improve this bound by some auxiliary functions. By taking more terms in the
identity of Lemma 4.5, one can get more accurate upper bound.

Acknowledgment: The research of S. Luo was partially supported by double
thousand plan of Jiangxi(jxsq2019101048) and NSFC(No. 12001253). The research
of J. Wei was partially supported by NSERC of Canada. The research of W. Zou
was partially supported by NSFC(Nos. 11801581,11025106, 11371212, 11271386).
S. Luo is grateful to Prof.H.J. Zhao (Wuhan University) for his constant
support and encouragement.

##  5\.  Appendix: The proof of Lemma 2.2

We separate the proof into two parts. On the one hand, by the Jacobi tri-
product formula, we have

$\displaystyle\vartheta(X;Y)$ $\displaystyle=\prod_{n=1}^{\infty}(1-e^{-2\pi
nX})(1+e^{-(2n-1)\pi X}e^{2\pi iY})(1+e^{-(2n-1)\pi X}e^{-2\pi iY})$ (5.1)
$\displaystyle=\prod_{n=1}^{\infty}(1-e^{-2\pi nX})(1+e^{-2(2n-1)\pi
X}+2e^{-(2n-1)\pi X}\cos(2\pi Y)).$

Taking logarithmic on both sides of (5.1) and differentiating with respect to
$Y$, one gets

$\displaystyle-\frac{\frac{\partial}{\partial Y}\vartheta(X;Y)}{\sin(2\pi Y)}$
$\displaystyle=4\pi\sum_{n=1}^{\infty}e^{-(2n-1)\pi
X}\frac{\vartheta(X;Y)}{1+e^{-2(2n-1)\pi X}+2e^{-(2n-1)\pi X}\cos(2\pi Y)}$
(5.2) $\displaystyle=4\pi\sum_{n=1}^{\infty}e^{-(2n-1)\pi X}\prod_{m\neq
n,m=1}^{\infty}(1-e^{-2\pi mX})(1+e^{-2(2m-1)\pi X}+2e^{-(2m-1)\pi X}\cos(2\pi
Y)).$

One sees from (5.2) that the function $-\frac{\frac{\partial}{\partial
Y}\vartheta(X;Y)}{\sin(2\pi Y)}$ has a period $1$, is decreasing on
$[0,\frac{1}{2}]$ and a even function for $Y$, just like $\cos(2\pi Y)$.

Then

$\displaystyle\lim_{Y\rightarrow\frac{1}{2}}-\frac{\frac{\partial}{\partial
Y}\vartheta(X;Y)}{\sin(2\pi Y)}$
$\displaystyle\leq-\frac{\frac{\partial}{\partial Y}\vartheta(X;Y)}{\sin(2\pi
Y)}\leq\lim_{Y\rightarrow 0}-\frac{\frac{\partial}{\partial
Y}\vartheta(X;Y)}{\sin(2\pi Y)}.$ (5.3)

It follows that

$\displaystyle\frac{1}{2\pi}\frac{\partial^{2}}{\partial
Y^{2}}\vartheta(X;Y)\mid_{Y=\frac{1}{2}}\leq-\frac{\frac{\partial}{\partial
Y}\vartheta(X;Y)}{\sin(2\pi Y)}\leq-\frac{1}{2\pi}\frac{\partial^{2}}{\partial
Y^{2}}\vartheta(X;Y)\mid_{Y=0}$ (5.4)

by L’Hospital rule.

By (2.13), one has,

$\displaystyle\frac{\partial^{2}}{\partial Y^{2}}\vartheta(X;Y)\mid_{Y=0}$
$\displaystyle=4\pi e^{-\pi X}(1+\sum_{n=2}^{\infty}n^{2}e^{-\pi(n^{2}-1)X})$
(5.5) $\displaystyle\frac{1}{2\pi}\frac{\partial^{2}}{\partial
Y^{2}}\vartheta(X;Y)\mid_{Y=\frac{1}{2}}$
$\displaystyle=4\pi\sum_{n=1}^{\infty}(-1)^{n-1}n^{2}e^{-n^{2}\pi X}\geq 4\pi
e^{-\pi X}(1-\sum_{n=2}^{\infty}n^{2}e^{-\pi(n^{2}-1)X}).$

Combining (5.3), (5.4) and (5.5), one gets the result for lower and upper for
$t>a>0$. For more details see [15].

On the other hand, for t is small and closes to 0, by Poisson Summation
Formula, one has

$\displaystyle\vartheta(X;Y)=X^{-\frac{1}{2}}\sum_{n\in\mathbb{Z}}e^{-\frac{\pi(n-Y)^{2}}{X}}.$

We have

$\displaystyle\frac{\partial}{\partial Y}\vartheta(t;Y)$ $\displaystyle=-2\pi
t^{-\frac{3}{2}}\sum_{m=-\infty}^{\infty}e^{-\frac{\pi(m-Y)^{2}}{t}}(Y-m)$
$\displaystyle=-2\pi
t^{-\frac{3}{2}}(\sum_{m=0}^{\infty}e^{-\frac{\pi(k+m)^{2}}{t}}(k+m-e^{-\frac{\pi(2m+1)(1-2k)}{t}}(m+1-k)))$

Notice that for $Y\leq 1,t\leq 2\pi$

$\displaystyle\frac{d}{dY}\big{(}e^{-\frac{\pi(2m+1)(1-2Y)}{t}(m+1-Y)}\big{)}=\frac{e^{-\frac{\pi(2m+1)(1-2Y)}{t}}}{t}(4m^{2}\pi+6m\pi+2\pi-4Ym\pi-2Y\pi-t)>0.$

Then by this, one gets the lower and upper bounds for
$0<t<k\leq\frac{\pi}{2}$. The proof of Lemma 2.2 is thus complete.

##  6\. Appendix: the proof of Theorem 1.4

We start with the scaling (by parameter a) of the Gamma function

$\frac{\Gamma(s)}{\pi^{s}}\frac{1}{a^{s}}=\int_{0}^{\infty}e^{-\pi
at}t^{s}\frac{dt}{t}$ (6.1)

which can be rewritten in convergent form

$\displaystyle\frac{\Gamma(s)}{\pi^{s}}\frac{1}{a^{s}}=(\frac{1}{s-1}-\frac{1}{s})+\int_{1}^{\infty}(e^{-\pi
at}-1)(t^{s}+t^{1-s})\frac{dt}{t}.$

It follows that

$\displaystyle\mathbf{G}_{R}(z;\kappa,s)=(\frac{1}{s-1}-\frac{1}{s})+\int_{1}^{\infty}(\mathcal{W}_{\kappa}(z;\alpha)-1)(t^{s}+t^{1-s})\frac{dt}{t}.$
(6.2)

See $\mathbf{G}_{R}(z;\kappa,s)$ and $\mathcal{W}_{\kappa}(z;\alpha)$ in
(1.13) and (1.9) respectively. The identity (6.2) can be viewed as an analogue
of the Bernstein Theorem in Theorem (1.10). Taking derivative with respect to
$s$, one has

$\displaystyle\frac{d^{j}}{ds^{j}}\mathbf{G}_{R}(z;\kappa,s)=\frac{d^{j}}{ds^{j}}(\frac{1}{s-1}-\frac{1}{s})+\int_{1}^{\infty}(\mathcal{W}_{\kappa}(z;\alpha)-1)\frac{d^{j}}{ds^{j}}(t^{s}+t^{1-s})\frac{dt}{t}.$
(6.3)

Notice that

$\displaystyle\frac{d^{j}}{ds^{j}}(t^{s}+t^{1-s})=(\log t)^{j}t^{s}+(-\log
t)^{j}t^{1-s}$

is nonnegative for any $j\geq 1,t\geq 1,s\geq\frac{1}{2}$.

By the bridge identity (6.3), Theorem 1.4 can be proved by following similar
proofs of Theorem 1.1 and Theorem 1.2. We omit the details.

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