on universally optimal lattice ... - personal.math.ubc.ca

ON UNIVERSALLY OPTIMAL LATTICE PHASE TRANSITIONS AND

ENERGY MINIMIZERS OF COMPLETELY MONOTONE POTENTIALS

SENPING LUO, JUNCHENG WEI, AND WENMING ZOU

Abstract. We consider the minimizing problem for energy functionals with two types of com-peting 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 forsquare 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). Itturns 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 thispaper. Consider a set of N classical identical particles in Rd (d ≤ 3), interacting by pairs through apotential V depending on the distance between them. Let x1, x2, · · · , xN ∈ Rd and p1, p2, · · · , pN ∈Rd be the positions and momenta of these particles.

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

HN (x1, · · · , xN , p1, · · · , pN ) =

N∑i=1

|pi|2

2m+

∑1≤i<j≤N

V(|xi − xj |2). (1.1)

Here m is the mass of the particles and | · | is the Euclidean norm of Rd. When the temperatureis zero, the equilibrium states are the minima of the functional HN , which satisfy p1 = · · · = pN .If one is interested in these equilibrium states, then it suffices to consider the potential energy

E(xiNi=1) :=∑

1≤i 6=j≤N

V(|xi − xj |2), (1.2)

where V is referred as the potential between two particles.

Let Λ be a two-dimensional lattice generated by the two basis α1 and α2. Assume that thereare two types of competing particles A and B. Now we consider the dynamics of two dimensionalperiodical lattices alternated by type A particle and type B particle (see figure 1), where one islocated at the center of another. That is, we only consider a special type of (Ω1,Ω2), termed twospecies periodic assemblies of discs, denoted by (Ωα,1,Ωα,2), with

Ωα,1 =⋃λ∈Λ

B(ξ, r1) ∪B(ξ′, r1) : ξ =

3

4α1 +

1

4α2 + λ, ξ′ =

1

4α1 +

3

4α2 + λ

, (1.3)

Ωα,2 =⋃λ∈Λ

B(ξ, r2) ∪B(ξ′, r2) : ξ =

1

4α1 +

1

4α2 + λ, ξ′ =

3

4α1 +

3

4α2 + λ

. (1.4)

1

2 SENPING LUO, JUNCHENG WEI, AND WENMING ZOU

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

The A type particles are located on red spots, denoted by P4k+1, P4k+3; the B type particlesare located on yellow spots, denoted by P4k+2, P4k+4, where k ∈ N . Let V be the potential ofthe system. Then the energy (1.2) of the system consists of two parts, the self-interaction partof each type particles and the interaction between the two different types of particles, while thesetwo parts contribute to the total energy in distribution of κ and 1− κ, where κ ∈ (0, 1).

Since the system is an infinite particles system, we first consider a mix of N−particles (xiNi=1)of type A and N−particles (yiNi=1) of type B to approximate the infinite particles system. Forconvenience, we set N = M2. The energy of self interaction of the same type particles is

Ea :=∑

(k,j)∈SN×SN\(0,0)

1

2V(|kα1 + jα2|2)(M − |k|)(M − |j|),

where the set SN is defined by −(M−1),−(M−2), · · · , 0, · · · ,M−2,M−1. The limit per-energyof the self interaction of the same type particles is

Eper,1(xj∞j=1) = Eper,1(yj∞j=1) := limN→∞

EaN

= limM→∞

∑(k,j)∈SN×SN\(0,0)

V(|kα1 + jα2|2)(M − |k|)(M − |j|)

2M2.

Such a limit converges if the summation∑

(k,j)∈SN×SN\(0,0) V(|kα1 + jα2|2) is summable. One

has

Eper,1(xj∞j=1) = Eper,1(yj∞j=1) =1

2

∑(k,j) 6=(0,0)

V(|kα1 + jα2|2).

One can rewrite it in a concise form on the lattice Λ

Eper,1(xj∞j=1) = Eper,1(yj∞j=1) =1

2

∑P∈Λ\0

V(|P|2).

Next we consider the interaction between two different particles. The energy betweenN−particles(xiNi=1)of type A and N−particles(yiNi=1) of type B is

Eb :=∑

(k,j)∈SN×SN\(0,0)

V(|12

(α1 − α2) + kα1 + jα2|2)(M − |k|)(M − |j|).

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

Eper,2(xj∞j=1, yj∞j=1) : = limN→∞

EbN

=∑

(k,j)∈Z⊗

Z

V(|12

(α1 − α2) + kα1 + jα2|2).

MINIMUM PRINCIPLE 3

One can rewrite it as

Eper,2(xj∞j=1, yj∞j=1) =∑P∈Λ

V(|P + 1

2|2).

Therefore, the total energy of two dimensional periodical alternating system with contributionκ, 1− κ of same-type and different-types interaction is determined by

Eper(A,B) : = κ(Eper,1(xj∞j=1) + Eper,1(yj∞j=1)) + (1− κ)Eper,2(xj∞j=1, yj∞j=1)

= κ∑

P∈Λ\0

V(|P|2) + (1− κ)∑P∈Λ

V(|P + 1

2|2), (1.5)

where κ ∈ [0, 1] and we include the endpoint cases κ = 0, 1 for convenience. To find out whichshape of lattice Λ minimizes the lattice energy (1.5), we consider the minimum of the functionalEper(A,B) over Λ, 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, thecompletely monotone potential cases. A function f : (0,∞)→ R is completely monotone if it is ofclass C∞(0,∞) and

(−1)jf (j)(x) > 0, j = 0, 1, 2, · · ·∞.The Bernstein functions are non-negative functions whose derivative is completely monotone.

They are such that

f(x) > 0, (−1)j−1f (j)(x) > 0, j = 1, 2, · · ·∞.The Bernstein functions have rich connection to other fields, firstly in probability theory. Com-

plete Bernstein functions are used in complex analysis under the name Pick or Nevanlinna func-tions, while in matrix analysis and operator theory, the name operator monotone function is morecommon. When studying the positivity of solutions of Volterra integral equations, various typesof kernels appear which are related to Bernstein functions. See the monograph [20].

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

E(κ,Λ) =∑

P∈Λ\0

((1− κ)F(|P + 1

2|2) + κF(|P|2)

). (1.6)

The main aim of this paper is to study the optimal lattice shape which minimizes E(κ,Λ), whereκ ∈ [0, 1] is a parameter.

To write E(κ,Λ) in function form, we parametrize the lattice Λ with cell by Λ =√

πIm(z)

(Z⊕zZ

)where the only parameter is z which belongs to H := z = x+ iy ∈ C : y > 0. It follows that

F(|P|2) : =∑

(m,n)∈Z2\0

F(π

Im(z)|mz + n|2),

F(|P + 1

2|2) : =

∑(m,n)∈Z2\0

F(π

Im(w)|mw + n|2) |w= z+1

2.

Therefore, we can write

Eκ(z) := E(κ,Λ) =∑

(m,n)∈Z2\0

(1− κ)F(π

2 Im(z)|m(z + 1) + 2n|2) + κF(

π

Im(z)|mz + n|2). (1.7)

We introduce the special curve in H. Let the vertical interval be

Ωea := z : Re(z) = 0, 1 ≤ Im(z) ≤√

3,while the arc on the circle is denoted by

Ωeb := z : |z| = 1, 0 ≤ Re(z) ≤ 1

2.


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

Ωe := Ωea ∪ Ωeb. (1.8)

In PDE theory, the maximum/minimum principle states that a function satisfying some partialdifferential equation in some domain achieves its maximum/minimum on the boundary of thedomain. Now we state a minimum principle for Eκ(z) on H:

Theorem 1.1 (Minimum principle for sum of two completely monotone functions onthe lattice). Let F(x) be any completely monotone function, κ ∈ [0, 1] be the parameter and Eκ(z)be defined in (1.7). Then

minz∈HEκ(z) = min

z∈ΩeEκ(z).

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

E(a, b,Λ) :=∑

P∈Λ\0

(aF(|P + 1

2|2) + bF(|P|2)

).

After normalization, one can rewrite it as

E(a, b,Λ) = (a+ b)∑

P∈Λ\0

( a

a+ bF(|P + 1

2|2) +

b

a+ bF(|P|2)

)= (a+ b)E(

b

a+ b,Λ),

then it becomes the standard form in Theorem 1.1. Similar remarks are applied to the theorems1.2, 1.3 and 1.4 below.

In Theorem 1.1, we conclude that the minimizers of sum of two completely monotone functionsare located precisely on a special curve. As we will see later, this special curve is a partial boundaryof 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 e−br

r (a, b > 0),

and Born-Mayer potential ae−br(a, b > 0). There are also a large number of completely monotonefunctions. Examples include (a + b

xα )µ, here a ≥ 0, b ≥ 0, µ ≥ 0, 0 ≤ α ≤ 1; eax (a > 0),

1(a−be−x)µ , here a ≥ b > 0, µ > 0; 1

xa(Γ(1+ 1x ))x

, here a ≥ 1; Γ(x)Γ(x+a+b)Γ(x+a)Γ(x+b) (a, b ≥ 0); 2√

πx

∫√x0

e−t2

dt;

etc. There are rules to generate more completely monotone functions from known ones. Forexample, if f(x) is completely monotone, then f (2m)(x),−f (2m+1)(x), ef(x), f(1−e−x), and f(axα+b)(a ≥ 0, b ≥ 0, 0 ≤ α ≤ 1) are also completely monotone for m = 1, 2, 3, · · · . We refer to[12, 13, 2, 3, 4] and the references therein for more concrete examples and the rules to generatecompletely monotone functions. For the references on the physical applications of the completelymonotone functions, we refer to Betermin [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 C ⊂ Sn−1 is universally optimal if it (weakly)minimizes potential energy among all configurations of |C| points on Sn−1 for each completelymonotonic 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 z0) is universally optimal if it(weakly) minimizes potential energy among all configurations of lattice shapes for each completelymonotonic potential function.

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

MINIMUM PRINCIPLE 5

Theorem 1.2 (Universally optimal and an explicitly universal interval for square lat-tice). For any completely monotone potential function F , and for any κ ∈ [ 1

3 ,23 ],

minz∈HEκ(z)

is attained at i, corresponding to the square lattice by our setting and is unique up the group G2

defined in (2.6).

Remark 1.2. The explicitly universal interval [ 13 ,

23 ] 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 pa-per by Betermin-Knupfer-Faulhuber [7] for another competing system with Gaussian and Rieszinteractions.

Note that there are three factors in the minimization process of minz∈H Eκ(z): one is the function(there are a large class of completely monotone functions), one is parameter κ which representsthe competing strength of the two-component system, and the lattice parameter z. Theorem 1.2asserts that the minimizer of sum of two completely monotone functions is always located on i(corresponding to the square lattice) for the parameter κ in [ 1

3 ,23 ]. This seems rather surprising.

Recall that the minimizer of the lattice energy functional corresponds to the optimal latticeshape in the competing system. Therefore, combining Theorem 1.1 and Theorem 1.2, we havethe 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 largeclass of other systems.

Theorem 1.3 (Hexagonal-rhombic-square-rectangular lattice phases transition). Forany completely monotone potential function F , there exists κ1 < κ2 ∈ (0, 1) depending on thepotential function F such that

• Case A: κ ∈ [0, κ1], minz∈H Eκ(z) = minz∈Ωea Eκ(z); the minimizer corresponds to rectan-gular lattice in this case;

• Case B: κ ∈ [κ1, κ2], Minimizerz∈HEκ(z) = i; the minimizer corresponds to square latticein this case;

• Case C: κ ∈ [κ2, 1], minz∈H Eκ(z) = minz∈Ωeb Eκ(z); the minimizer corresponds to rhombiclattice in this case;

For the parameters κ1, κ2, for any potential function which is completely monotone, there has

• κ1 ≤ 13 ;

• κ2 ≥ 23 ;

• κ1 + κ2 = 1;

namely, [κ1, κ2] ⊇ [ 13 ,

23 ] in Case B. In particular, when κ = 1, Minimizerz∈HEκ(z) = 1

2 + i√

32 ,

which corresponds to hexagonal lattice; when κ = 0, Minimizerz∈HEκ(z) =√

3i, which corresponds

to√

3 : 1 rectangular lattice.

In physical and biological models, the functional Eκ(z) is referred to the energy of competingperiodical system and the parameter κ reflects the competing strength of the system. As theparameter κ varies continuously, we see from Theorem 1.3 clearly that, the optimal lattice shapegoes through hexagonal-rhombic-square-rectangular lattice continuously (as κ moves from 1 to0 continuously). Theorem 1.2 shows that under the completely monotone potentials and undercertain positive interval of interaction strength, the minimal energy lattice shape is the squarelattice shape. This phenomenon has appeared and conjectured in two component Bose-Einsteincondensates (see in Mueller-Ho [17] and Luo-Wei [15]). When κ = 1 the optimality of hexagonallattice 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 forgeneral completely monotone function F . To introduce the formula we define

Wκ(z;α) =∑

(m,n)∈Z2\0

(1−κ) exp(−α π

2 Im(z)|m(z+1)+2n|2)+κ exp(−α π

Im(z)|mz+n|2) (1.9)

which is the energy functional associated to the special Gaussian type completely monotone func-tion exp(−αx2) with a free parameter α > 0.

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

F (x) =

∫ ∞0

e−αxdλ(α).

As a consequence we obtain the following integral representation of Eκ(z): there is an non-negative finite Borel measure on [0,∞) with cumulative distribution function λ(α) such that

Eκ(z) =

∫ ∞0

Wκ(z;α)dλ(α). (1.10)

The formula (1.10) reduces the proof of Theorems 1.1-1.3 to the study of the universal propertiesWκ(z;α), which is the main contribution of this paper.

1.3. Competing systems with Riesz potential. One interesting and important competingpotential is the Riesz potential type. We discuss the results on competing systems with Rieszpotential here. Let F(x) := 1

xs , s > 1 be the Riesz potential which is a special case of completelymonotone functions. We consider the associated competing system energy functional

ER(z;κ, s) := ER(κ, s,Λ) =∑

P∈Λ\0

((1− κ)

1

|P+12 |2s

+ κ1

|P|2s), s > 1, (1.11)

with Λ =√

πIm(z)

(Z⊕ zZ

).

Observe that up to a constant (depending on s) the Riesz sum∑

P∈Λ\01|P|2s is the Eisenstein

series G(z, s) which is defined as

G(z, s) :=1

2

∑(m,n)∈Z2\0

(Im(z)

)s|mz + n|2s

. (1.12)

See [9].The lattice energy functional ER(z;κ, s) under Riesz potential (defined at (1.11)) can be written

up to a constant (depending on s) as

GR(z;κ, s) := (1− κ)G(z + 1

2, s) + κG(z, s), κ ∈ [0, 1], (1.13)

where G(z, s) := π−sΓ(s)G(z, s) is a rescaled Eisenstein function which also appears frequently innumber theory ([9]).

Theorems 1.1-1.3 can be applied to GR(z;κ, s). However an important observation, as shownin the appendix, is that although the s−deriveatives are not completely monotone functions, wecan still apply the techniques in proving Theorem 1.1-1.3 to the s−derivatives.

Theorem 1.4. Let GR(z;κ, s) be defined in (1.13).(a) For any κ ∈ [0, 1], s > 1, j = 1, 2, 3 · · · , it follows that

minz∈H

dj

dsjGR(z;κ, s) = min

z∈Ωe

dj

dsjGL(z;κ, s).

(b) There exists κ1 < κ2 ∈ (0, 1) depending on s such that [ 13 ,

23 ] ⊆ [κ1, κ2], κ1 + κ2 = 1 and

• Case A: κ ∈ [0, κ1], minz∈Hdj

dsj GR(z;κ, s) = minz∈Ωeadj

dsj GR(z;κ, s); the minimizer cor-responds to rectangular lattice in this case;

MINIMUM PRINCIPLE 7

• Case B: κ ∈ [κ1, κ2], Minimizerz∈Hdj

dsj GR(z;κ, s) = i; the minimizer corresponds tosquare lattice in this case;

• Case C: κ ∈ [κ2, 1], minz∈Hdj

dsj GR(z;κ, s) = minz∈Ωebdj

dsj GR(z;κ, s); the minimizer cor-responds to rhombic lattice in this case;

In particular, when κ = 1, Minimizerz∈Hdj

dsj G(z, s) = 12 + i

√3

2 , which corresponds to hexagonal

lattice; when κ = 0, Minimizerz∈Hdj

dsj G( z+12 , s) =

√3i, which corresponds to

√3 : 1 rectangular

lattice.

Remark 1.4. The Eisentein series G(z, s) is closely related to the Epstein zeta function ζQ(s),that is defined for Re(s) > 1 by

ζQ(s) :=1

2

∑(m,n)∈Z2\0

1

Q(m,n)s,

where Q(x, y) = ax2 + bxy + cy2 is a positive definite quadratic form with real coefficients(i.e.,a > 0, c > 0 and b2 − 4ac < 0).

Denote that z = −b2a +i

√4ac−b2

2a and D = b2−4ac. Then the Epstein zeta function and Eisensteinseries is connected by

ζQ(s) = (|D|4

)−s2G(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

d

ds|s=1 G(z, s) = −1

2log∣∣ Im (z)η(z)∣∣− log(2π). (1.15)

It follows that

d

ds|s=1 GR(z;κ, s), GR(z;κ, s) := (1− κ)G(

z + 1

2, s) + κG(z, s)

=− 1

2

(κ log

∣∣ Im (z)η(z)∣∣+ (1− κ) log∣∣ Im (z + 1

2

)η(z + 1

2

)∣∣)− log(2π),

(1.16)

where

η(z) = eπ3 zi

∞∏n=1

(1− e2πnzi

)4(1.17)

is the fourth power of the Dedekind eta function. The derivative formula (1.16) connects the latticeenergy functional with Riesz potential to lattice energy functional with Coulomb potential. Forconvenience, define

Nκ(z) := −(κ log

∣∣ Im (z)η(z)∣∣+ (1− κ) log∣∣ Im (z + 1

2

)η(z + 1

2

)∣∣). (1.18)

The functional Nκ(z) is of the form of Dedekind eta functions, is a type of lattice energy functionalwith Coulomb potential.

The reduced energy functional Nκ(z) arises in the study of Triblock Ohta-Kawasaki system.For κ = 1 it arises in the Diblock copolymer Ohta-Kawasaki system and also Ginzburg-Landautheory. 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 newresults here include these as a special case and we have proved these results without using any PDEtheory. This also confirms that the hexagonal-rhombic-square-rectangular lattice phasestransition in Theorem 1.3 is quite general in various periodical competing system.

In the numerical computation done in [14], we have for Nκ(z)

κ1 := 0.1867 . . . , κ2 := 0.8132 · · · .This confirms that the upper and lower bounds in Theorem 1.3, in which we proved that κ1 ≤13 , κ2 ≥ 2

3 .


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

2. Preliminaries: Invariances of Eκ(z) and estimates on theta functions

Let Eκ(z) be the lattice energy functional defined at (1.7). The end point cases of Eκ(z) are

E0(z) =∑

(m,n)∈Z2\0

F(π

2 Im(z)|m(z + 1) + 2n|2) (2.1)

andE1(z) =

∑(m,n)∈Z2\0

F(π

Im(z)|mz + n|2). (2.2)

Then Eκ can be written as interpolation of E0 and E1:

Eκ(z) = κE1(z) + (1− κ)E0(z). (2.3)

In this section, we collect some basic properties of the functional Eκ, including invariance andmonotone properties, and some key estimates of Wκ.

We first study the invariance properties of Eκ. We use the following definition of fundamentaldomain which is slightly different from the classical definition (see [16]):

Definition 3 (page 108, [11]). The fundamental domain associated to group G is a connecteddomain D satisfying

• For any z ∈ H, there exists an element π ∈ G such that π(z) ∈ D;• Suppose z1, z2 ∈ D and π(z1) = z2 for some π ∈ G, then z1 = z2 and π = ±Id.

For example, let Γ be the group generated by the actions τ 7→ − 1τ , τ 7→ τ+1, then by Definition

3, the fundamental domain associated to modular group Γ is

DΓ := z ∈ H : |z| > 1, −1

2< x <

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 Eκ(z). The generators of these

groups are given by

G1 : the group generated by τ 7→ −1

τ, τ 7→ τ + 1, τ 7→ −τ , (2.5)

G2 : the group generated by τ 7→ −1

τ, τ 7→ τ + 2, τ 7→ −τ . (2.6)

It is easy to see that the fundamental domains associated to group Gj , j = 1, 2 denoted by DG1 ,DG2are

DG1 := z ∈ H : |z| > 1, 0 < x <1

2 (2.7)

DG2 := z ∈ H : |z| > 1, 0 < x < 1. (2.8)

Clearly we have that G2 ⊆ G1, DG1 ⊆ DG2 .The following two Lemmas can be checked directly from the definition.

Lemma 2.1. We have the following fundamental invariant properties of the functionals:

• E1(z) is invariant under the group G1, i.e., for any γ ∈ G1

E1(γ(z)) = E1(z).

• E0(z) is invariant under the group G2, i.e., for any π ∈ G2

E0(π(z)) = E0(z).

• Eκ(z)(κ ∈ (0, 1)) is invariant under the group G2, i.e., for any π ∈ G2

E0(π(z)) = E0(z).

MINIMUM PRINCIPLE 9

As a consequence, the fundamental domain of the functionals E1(z) and E0(z) is DG1 and DG2respectively, and the fundamental domain for the general function Eκ(z)(κ ∈ (0, 1)) is DG2 .

By the integral representation formula (1.10) and the monotonicity properties of W0 and W1

(see [15]), we have the following important monotonicity properties for the functionals E0(z), E1(z)in the corresponding domain:

Theorem 2.1. There holds

•∂

∂xE0(z) > 0, ∀ z ∈ DG2 ,

•∂

∂xE1(z) < 0, ∀ z ∈ ΩC1 .

Here

ΩC1 := z | 0 < x <1

2, y >

√x− x2.

Remark 2.1. Montgomery [16] proved that

∂

∂xE1(z) < 0, ∀ z ∈ ΩC0 := z ∈ H : y >

1

2, 0 < x <

1

2. (2.9)

Theorem 2.1 improves this result to a larger domain ΩC1 as ΩC0 ⊂ ΩC1 . Furthermore, ΩC1 containsa corner at z = 0, which makes the proof much more involved.

We state two corollaries related to the functional Eκ(z).

Corollary 2.1.∂

∂xE1(z) > 0, ∀z ∈ ΩC2 .

Here

ΩC2 := z | 1

2< x < 1, y >

√x− x2.

Proof. Since z 7→ 1− z ∈ G1, by Lemma 2.1, we have E1(1− z) = E1(z). Thus

∂

∂xE1(1− z) = − ∂

∂xE1(z). (2.10)

The result follows by Theorem 2.1.

By Theorem 2.1 and Corollary 2.1 we have

Theorem 2.2. For κ ∈ [0, 1],

∂

∂xEκ(z) > 0, ∀z ∈ RR, j = 1, 2.

Here

RR := ΩC2 ∩ DG2 = z | 1

2< Re(z) < 1, |z| > 1.

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

RL := z | 0 < Re(z) <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 firstrecall the following well-known Jacobi triple product formula:

∞∏m=1

(1− x2m)(1 + x2m−1y2)(1 +x2m−1

y2) =

∞∑n=−∞

xn2

y2n(2.12)


for complex numbers x, y with |x| < 1, y 6= 0.The Jacob theta function is defined as

ϑJ(z; τ) :=

∞∑n=−∞

eiπn2τ+2πinz,

and the classical one-dimensional theta function is given by

ϑ(X;Y ) := ϑJ(Y ; iX) =

∞∑n=−∞

e−πn2Xe2nπiY . (2.13)

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

ϑ(X;Y ) =

∞∏n=1

(1− e−2πnX)(1 + e−2(2n−1)πX + 2e−(2n−1)πX cos(2πY )). (2.14)

By Poisson Summation Formula, one has

ϑ(X;Y ) = X−12

∑n∈Z

e−π(n−Y )2

X . (2.15)

Thus the functional W1(z;α) can be written in terms of one-dimensional theta function asfollows:

W1(z;α) =∑

(m,n)∈Z2

e−απ1y |nz+m|

2

=∑n∈Z

e−απyn2 ∑m∈Z

e−απ(nx+m)2

y

=

√y

α

∑n∈Z

e−απyn2

ϑ(y

α;−nx) =

√y

α

∑n∈Z

e−απyn2

ϑ(y

s;nx)

= 2

√y

α

∞∑n=1

e−απyn2

ϑ(y

α;nx).

(2.16)

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

Lemma 2.2. Let Q(X;Y ) = − 1sin(2πY )

∂∂Y ϑ(X;Y ). Then there holds the following lower and upper

bounds of Q(X;Y )

ϑ(X) ≤ Q(X;Y ) ≤ ϑ(X), for any Y,

where

ϑ(t) =

1

maxβ∈[0,k] sin(2πβ) t− 3

2 (e−πk2

t (k − e−π(1−2k)

t (1− k))

+∑∞m=1 e

−π(k+m)2

t (k +m− e−π(2m+1)(1−2k)

t (m+ 1− k))) (0 < t ≤ k < 12 ),

(1− µ(t))4πe−πt (t > 0),

(2.17)

and

ϑ(t) =

t−

32

∑∞m=−∞ e−

πm2

t (1− 2πm2

t ) (0 < t ≤ k < π2 ),

(1 + µ(t))4πe−πt (t > 0).(2.18)

Here k and a are positive numbers and

µ(t) :=

∞∑n=2

n2e−π(n2−1)t. (2.19)

(For t small, we can use

Q(t; k) =1

sin(2πk)t−

32 (e−

πk2

t (k−e−π(1−2k)

t (1−k))+

∞∑m=1

e−π(k+m)2

t (k+m−e−π(2m+1)(1−2k)

t (m+1−k))).)

MINIMUM PRINCIPLE 11

3. Minimum principle: Proof of Theorem 1.1

In this section, we prove Theorem 1.1. First we observe that the functional Wκ(z;α) satisfiesthe following duality principle: For all b ∈ R,

Wb(w) =W1−b(z), z ∈ H and w =z − 1

z + 1∈ 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 κ ∈ [0, 1

2 ].Next by the transition formula

Wκ(z;α) = κW 12(z;α) + (1− 2κ)W0(z;α) (3.1)

we only need to study W 12

and W0. Furthermore since by Fourier transform

Wκ(z;α) = αWκ(z;1

α) (3.2)

we only need to consider α ∈ [1,+∞). By Theorem 2.2 we only need to consider z ∈ RL (definedat (2.11)). We also use the formula in (2.16),

W1(z;α) = 2

√y

α

∞∑n=1

e−απyn2

ϑ(y

α;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 = 12 and x2 + y2 = 1, we

have∂

∂xW 1

2(z;α) ≥ 0,∀z ∈ RL (3.3)

where RL is defined at (2.11). If the equality holds, then x = 0.

By Corollary 4.2 of [15], we also have ∂∂xW0(z;α) ≥ 0,∀z ∈ RL. 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 ∂∂xW 1

2(z;α) in different ranges of x.

Lemma 3.1. For x ∈ [ 14 ,

12 ] we have the following lower bound estimates for ∂

∂xW 12(z;α):

∂

∂xW 1

2(z;α)

≥√y

αe−

παy2 sin(πx)

( 1

2√

2ϑ(

y

2α)−√

2(1 + ε1(α))e−παy2 ϑ(

y

α)− (1 + ε2(α))e−

3παy2 ϑ(

y

2α)),

(3.4)

where ε1(α) < 3.18 · 10−9, ε2(α) < 2.782 · 10−6 and ϑ, ϑ are defined in (2.17)-(2.18).

Furthermore, for y ≥√

32 , α ≥ 1, there holds

1

2√

2ϑ(

y

2α)−√

2(1 + ε1(α))e−παy2 ϑ(

y

α)− (1 + ε2(α))e−

3παy2 ϑ(

y

2α) ≥ 0. (3.5)

Lemma 3.2. For x ∈ [0, 14 ] we have the following lower bound estimates for ∂

∂xW 12(z;α):

∂

∂xW 1

2(z;α)

≥√y

αe−

παy2 sin(πx)

( 1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)) (3.6)

where ε3(α) < 1.1 · 10−4 and ε4(α) < 5.7 · 10−16.


Furthermore, for y ≥√

154 , α ≥ 1, we have

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) > 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 toconsider the case α ≥ 1. Let us start with the computation

∂

∂xW 1

2(z;α) =

√y

α

( ∞∑n=1

1

2√

2ne−

παyn2

2∂

∂Yϑ(

y

2α;Y ) |Y=n x+1

2

+

∞∑n=1

ne−παyn2 ∂

∂Yϑ(y

α;Y ) |Y=nx

).

(3.8)

To refine the estimates, we divide the proof into two parts by x ∈ [0, 14 ] or [ 1

4 ,12 ]. We first prove

(3.4) of Lemma 3.1.Proof of (3.4) of Lemma 3.1: Let x ∈ [ 1

4 ,12 ]. We rewrite

∂

∂xW 1

2(z;α) =

√y

αe−

παy2

( 1

2√

2

∂

∂Yϑ(

y

2α;Y ) |Y= x+1

2+e−

παy2

∂

∂Yϑ(y

α;Y ) |Y=x

+1√2e−

3παy2

∂

∂Yϑ(

y

2α;Y ) |Y=x+1

)+

√y

α

(2e−4παy ∂

∂Yϑ(y

α;Y ) |Y=2x +

∞∑n=3

1

2√

2ne−

παyn2

2∂

∂Yϑ(

y

2α;Y ) |Y=n x+1

2

+

∞∑n=3

ne−παyn2 ∂

∂Yϑ(y

α;Y ) |Y=nx

):=

√y

αe−

παy2 · J1 + (

y

α)

12 · J2,

(3.9)where J1 and J2 are defined respectively at the last equality.

By the lower and upper bounds of ϑ in Lemma 2.2, we have

J1 ≥1

2√

2ϑ(

y

2α) sin(πx)− e−

παy2 ϑ(

y

α) sin(2πx)− 1√

2e−

3παy2 ϑ(

y

2α) sin(2πx)

= sin(πx)( 1

2√

2ϑ(

y

2α)− e−

παy2 ϑ(

y

α)2 cos(πx)− 1√

2e−

3παy2 ϑ(

y

2α)2 cos(πx)

)≥ sin(πx)

( 1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)),

(3.10)

where ϑ and ϑ are defined at (2.17) and (2.18) respectively.

Here we have used the fact that sin(π(x+ 1)) ≤ 0, cos(πx) ≤√

22 for x ∈ [ 1

4 ,12 ].

Next, we are going to estimate J2. By (3.9), using | sin(ny)| ≤ n| sin(y)|, y ∈ R and Lemma 2.2,we have

J2 ≥ −1

4√

2ϑ(

y

2α) sin(2πx)

∞∑n=3

n2e−παyn2

2 − ϑ(y

α) sin(2πx)

∞∑n=3

n2e−παyn2

. (3.11)


Combining (3.10) and (3.11) we have

J1 + eπαy2 J2 ≥( 1

2√

2ϑ(

y

2α) sin(πx)− (1 + ε1(α))e−

παy2 ϑ(

y

α) sin(2πx)− (1 + ε2(α))

1√2e−

3παy2 ϑ(

y

2α) sin(2πx)

)≥ sin(πx)

( 1

2√

2ϑ(

y

2α)− (1 + ε1(α))

√2e−

παy2 ϑ(

y

α)− (1 + ε2(α))e−

3παy2 ϑ(

y

2α)), for x ∈ [

1

4,

1

2],

(3.12)where the constants ε1(α) and ε2(α) are defined by

ε1(α) :=

∞∑n=3

n2e−παy(n2−1) ≤∞∑n=3

n2e−π√

3(n2−1)2 (3.13)

and

ε2(α) :=1

4

∞∑n=3

n2e−παy(n2−4)

2 ≤ 1

4

∞∑n=3

n2e−π√

3(n2−4)2 (3.14)

It is easy to see that ε1 < 3.18 · 10−9, ε2 < 2.782 · 10−6.The inequality (3.4) 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 ∈ [ 14 ,

12 ] (equivalently, y ≥

√3

2 ), (3.5) holds by assuming ε1 = ε2 = 0. The proof can beeasily modified when two small constants are added. We split the proof into three subcases asfollows

A :y

α,y

2α∈ (0, k]; B :

y

α,y

2α∈ (k,+∞); C :

y

2α∈ (0, k],

y

α∈ (k,+∞),

where the parameter k ∈ (0, 12 ] is to be chosen later.

Case A : yα ,y

2α ∈ (0, k]. By Lemma 2.2, we have

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)

≥ 1

2√

2(y

2α)−

32

1

sin(2πk)

(e−πk

2 2αy k − e−π(1−k)2 2α

y (1− k)

+ e−π(1+k)2 2αy (1 + k − e−3π(1−2k) 2α

y ))−√

2e−παy2 (

y

α)−

32

(1 + (2− 16π

α

y)e−4π αy

)− e−

3παy2 (

y

2α)−

32

(1 + (2− 16π

2α

y)e−4π 2α

y ))

≥ (y

2α)−

32

( 1

2√

2

1

sin(2πk)e−πk

2 2αy (k − εk)− 2e−

παy2 − e−

3παy2

)≥ (

y

2α)−

32

( 1

2√

2e−πk

2 2αy (k − εk)− 2e−

παy2 − e−

3παy2

).

(3.15)

Here εk is determined by

εk := maxyα∈[0,k]

e−π(1−k)2 2αy (1− k)

e−πk2 2αy k

≤ 1− kk

e−π(1−2k) 2k .


Let k ∈ [0, 0.3]. Then it is easy to see that εk ≤ ε0.3 = 0.000536 · · · < 35000 which yields that

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)

≥(y

2α)−

32

( 1

2√

2e−πk

2 2αy (k − 3

5000)− 2e−

παy2 − e−

3παy2

)=e−πk

2 2αy (

y

2α)−

32

(k − 35000

2√

2− 2e−πα

y2−4k2

2y − e−(πα y2−4k2

2y +παy))

≥e−πk2 2αy (

y

2α)−

32

(k − 35000

2√

2− 2(1 +

1

5000)e−πα

y2−4k2

2y

)≥e−πk

2 2αy (

y

2α)−

32

(k − 35000

2√

2− 2(1 +

1

5000)e−π

y2−4k2

2k

), since α ≥ y

k

≥e−πk2 2αy (

y

2α)−

32

(k − 35000

2√

2− 2(1 +

1

5000)e−π( 3

8k−2k)), since y2 ≥ 1− x2 =

3

4.

(3.16)

Here we have used the fact that

maxyα∈[0,k],k∈[0,0.3]

e−παy

2≤ e−π

10.3y

2

2≤ e−π

10.3

34

2= 0.00019 · · · < 1

5000, since y2 ≥ 1− x2 ≥ 3

4.

An elementary computation shows that the function

f(k) :=k − 3

5000

2√

2− 2(1 +

1

5000)e−π( 3

8k−2k) > 0 (3.17)

for k ∈ [0, kf ] where kf = 0.25094 · · · .Therefore, by (3.18),

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α) ≥ e−πk

2 2αy (

y

2α)−

32 f(k) > 0 (3.18)

if k ∈ (0, kf ], where kf ≈ 0.25094 · · · . This gives the first range of the free parameter k.

Case B : yα ,y

2α ∈ (k,∞). By Lemma 2.2 again, we have

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)

≥ 1

2√

2(1− µ(k))4πe−π

y2α −

√2e−

παy2 (1 + µ(k))4πe−π

yα − e−

3παy2 (1 + µ(k))4πe−π

y2α

=4πe−πy2α

( 1

2√

2(1− µ(k))−

√2(1 + µ(k))e−

πy2 (α+ 1

α ) − (1 + µ(k))e−3παy

2

)≥4πe−π

y2α

( 1

2√

2(1− µ(k))− (

√2 + δ)(1 + µ(k))e−

πy2 (α+ 1

α ))

≥4πe−πy2α

( 1

2√

2(1− µ(k))− (

√2 + e−

π√

34 )(1 + µ(k))e−πy

), since

1

2(α+

1

α) ≥ 1, α > 0

≥4πe−πy2α

( 1

2√

2(1− µ(k))− (

√2 + e−

π√

34 )(1 + µ(k))e−

√3π2

), since y ≥

√1− x2 ≥

√3

2.

(3.19)

Here δ is defined by

δ = maxα≥1,y≥

√1−x2,x∈[ 14 ,

12 ]

e−3παy

2

e−πy2 (α+ 1

α ).

Then

δ = maxα≥1,y≥

√1−x2,x∈[ 14 ,

12 ]e−πy(α− 1

2α ) ≤ e−πy(1− 12 ) ≤ e−

π√

34 .


Let

g(k) : =1

2√

2(1− µ(k))− (

√2 + e−

π√

34 )e−

√3π2 (1 + µ(k)), k ∈ [0,

1

2]

= (1

2√

2+ (√

2 + e−π√

34 )e−

√3π2 )(

12√

2− (√

2 + e−π√

34 )e−

√3π2

12√

2+ (√

2 + e−π√

34 )e−

√3π2

− µ(k)).

(3.20)

Then g′(k) > 0 for any k > 0 by µ′(k) > 0 and g(k) > 0 if and only if k > kg. Numerically,kg = 0.22247 · · · .

Therefore by (3.10), (3.19) and (3.20), one has

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α) ≥ 4πe−π

y2α g(k) > 0 if k > kg = 0.22247 · · · .

This gives the second range of k to be selected.

Case C : y2α ∈ (0, k], yα ∈ (k,∞). By Lemma 2.2

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)

≥ 1

2√

2(y

2α)−

32

1

maxx∈[0,k] sin(2πx)

(e−πk

2 2αy k − e−π(1−k)2 2α

y (1− k)

+ e−π(1+k)2 2αy (1 + k − e−3π(1−2k) 2α

y ))

−√

2e−παy2 (1 + µ(k))4πe−π

yα − e−

3παy2 (

y

2α)−

32 (1 + (2− 4π

2α

y)e−π

2αy )

≥ 1

2√

2(y

2α)−

32 e−πk

2 2αy (k − εk)−

√2e−

παy2 (1 + µ(k))4πe−π

yα − e−

3παy2 (

y

2α)−

32 ,

(3.21)

where εk is defined by

εk :=e−π(1−k)2 2α

y (1− k)

e−πk2 2αy k

. (3.22)

Since the function εk will also appear in the proof of the next lemma, we single out the elementarybut useful property here:

∂

∂kεk = (1− k)(π − k

1− k)e−

π(1−2k)k

k3. (3.23)

Hence εk is decreasing with respect to k if k ∈ (0, π1+π ].

By cases A and B, we will select the parameter by k ∈ (kg, kf ) = (0.22247 · · · , 0.25094 · · · ).

Let

k ∈ (kg, kf ).

We further take k ∈ (ka, kb], where ka = kg and kb = minkf , 14. The number 1

4 is taken by thebound to control that

1

maxx∈[0,k] sin(2πx)≥ 1

provided that k ∈ (0, kb]. We keep the symbols ka, kb to make the estimates clearer, and keep inmind that ka = 0.22247 · · · , kb = 1

4 .

On the other hand it is easy to see that εk < εkf = 0.005843 · · · < 3500 . Note that y

2α ∈ (0, k]implies that

α ≥ y

2k. (3.24)

Our method here works for all the k belongs to the range (ka, kb]. Here for simplicity, we set

k =1

4.


In this case, by (3.24),

α ≥√

3. (3.25)

Here we have used that y ≥√

32 implied by x ∈ [ 1

4 ,12 ] since z = (x, y) belongs to the domain RL.

Now to deal with the last inequality in (3.21), we rewrite

αy = α2 y

α≥ 3

y

α. (3.26)

The bounds of yα in this case implies that 1

k ≤2αy ≤

2k . Fixing all the parameters except k in

the expression e−πk2 2αy (k − εk), then we have

∂

∂k

(e−πk

2 2αy (k − εk)

)= −e−πk

2 2αy(2α

yπ(k − εk)− 1

)(3.27)

which is positive since trivially

2α

yπ(k − εk)− 1 ≥ π(1− εk

k)− 1 >

π

2− 1 > 0.

To simplify the notations, we set γ = yα , hence γ ∈ (k, 2k] = ( 1

4 ,12 ].

In view of (3.21), by the monotonicity in (3.27),

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)

≥γ− 32 e−

π2

1γ (

1

4− εk)−

√2e−

π2 α

2γ(1 + µ(1

4))4πe−πγ − e− 3π

2 α2γ(

γ

2)−

32 .

(3.28)

Using the elementary but useful bound for α in (3.25), we further get the lower bound with theonly variable γ, i.e.,

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α)

≥γ− 32 e−

π2

1γ (

1

4− εk)−

√2e−

3π2 γ(1 + µ(

1

4))4πe−πγ − e− 9π

2 γ(γ

2)−

32 .

(3.29)

Now the only variable in above is the parameter γ, to make the structure clear, let

h(γ) := C1γ− 3

2 e−π2

1γ − C2e

− 5π2 γ − C3e

− 9π2 γ(

γ

2)−

32 (3.30)

with the constants determined by (3.29)

C1 =1

4− 3

500= 0.244

C2 =√

2(1 + µ(1

4))4π = 24.80995446 · · ·

C3 = 1,

(3.31)

where µ(X) is defined in (2.19) and µ( 14 ) = 0.39605042 · · · .

By (3.29),

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α) ≥ h(γ), γ =

y

α. (3.32)

It remains to show that

h(γ) > 0, for γ ∈ (1

4,

1

2]. (3.33)

Simple calculation shows that (γ−

32 e−

π2

1γ)′

= γ−72 e−

π2

1γ (π

2− 3

2γ2), (3.34)


while γ ∈ ( 14 ,

12 ] in (3.33), one has π

2 −32γ

2 > 0. Hence, we infer that h(γ) is monotonically

increasing for γ ∈ ( 14 ,

12 ] and it follows that

h(γ) > h(1

4) = 0.003734 · · · > 0, γ ∈ (

1

4,

1

2]. (3.35)

Therefore, by (3.29) and (3.30), we have

1

2√

2ϑ(

y

2α)−√

2e−παy2 ϑ(

y

α)− e−

3παy2 ϑ(

y

2α) > 0. (3.36)

By the same proof, we can take the parameter k to be 14 and then all the estimates work. In

fact, using the same method above, for any k ∈ (kg,14 ], we can show that the inequality (3.36)

holds.Notice that the constants εj(α), j = 1, 2 are very close to zero. With almost the same compu-

tations in the lower bound estimates (3.36), we have that for x ∈ [ 14 ,

12 ] and y ≥

√3

2

1

2√

2ϑ(

y

2α)− (1 + ε1(α))

√2e−

παy2 ϑ(

y

α)− (1 + ε2(α))e−

3παy2 ϑ(

y

2α) > 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 haveProof of (3.6) of Lemma 3.2: Let x ∈ [0, 1

4 ]. Similar to the proof of Lemma 3.1, we divide

∂

∂xW 1

2(z;α)

=e−παy2

√y

α

( 1

2√

2

∂

∂Yϑ(

y

2α;Y ) |Y= x+1

2+e−

παy2

∂

∂Yϑ(y

α;Y ) |Y=x

+1√2e−

3παy2

∂

∂Yϑ(

y

2α;Y ) |Y=x+1 +2e−

7παy2

∂

∂Yϑ(y

α;Y ) |Y=2x

)+

√y

α

( 3

2√

2e−

9παy2

∂

∂Yϑ(

y

2α;Y ) |Y=3 x+1

2+

∞∑n=4

1

2√

2ne−

παyn2

2∂

∂Yϑ(

y

2α;Y ) |Y=n x+1

2

+

∞∑n=3

ne−παyn2 ∂

∂Yϑ(y

α;Y ) |Y=nx

)=

√y

α

(e−

παy2 J3 + J4

).

(3.37)

Here J3 and J4 are defined at the last equality.For the lower bound of J3, since sin(4πx) ≥ 0, by Lemma 2.2

J3 ≥1

2√

2ϑ(

y


παy2 ϑ(

y

α) sin(2πx)

− 1√2e−

3παy2 ϑ(

y

2α) sin(2πx)− 2e−

7παy2 ϑ(

y

α) sin(4πx).

(3.38)

Observe that for x ∈ [0, 14 ]

∂

∂Yϑ(

y

2α;Y ) |Y=3 x+1

2≥ ϑ(

y

2α) sin(3πx) ≥ 0.

To compare the terms in J3, using | sin(ny)| ≤ n| sin(y)|, y ∈ R and Lemma 2.2, we estimate that

J4 ≥ −1

4√

2

∞∑n=4

n2e−παyn2

2 ϑ(y

2α) sin(2πx)−

∞∑n=3

n2e−παyn2

ϑ(y

α) sin(2πx). (3.39)


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

J3 + eπαy2 J4 ≥

1

2√

2ϑ(

y


παy2 ϑ(

y

α) sin(2πx)

(1 + 4e−3παy +

∑n=3

n2e−παy(n2−1))

− 1√2e−

3παy2 ϑ(

y

2α) sin(2πx)

(1 +

1

4

∞∑n=4

n2e−παy(n2−4)

2

)≥ sin(πx)

( 1

2√

2ϑ(

y

2α)− 2e−

παy2 ϑ(

y

α)(1 + ε3(α))−

√2e−

3παy2 ϑ(

y

2α)(1 + ε4(α))

).

(3.40)The constants εj(α), j = 3, 4 are defined by

ε3(α) = 4e−3παy +∑n=3

n2e−παy(n2−1, ε4(α) =1

4

∞∑n=4

n2e−παy(n2−4)

2 .

Since x ∈ [0, 14 ], y ≥

√1− x2 ≥

√154 . A direct calculation shows that ε3(α) < 1.1 · 10−4, ε4(α) <

5.7 · 10−16.Combining (3.37) and (3.40), we obtain (3.6).

Proof of (3.7) of Lemma 3.2: It suffices to prove (3.7) for y ≥√

1− x2 ≥√

154 .

The proof is similar to that of (3.4) of Lemma 3.1. We first divide the proof into three subcases

A :y

α,y

2α∈ (0, k]; B :

y

α,y

2α∈ (0, k] ∈ (k,+∞); C :

y

α∈ (0, k],

y

2α∈ (k,+∞)

where k will be chosen appropriately later.

Case A : yα ,y

2α ∈ (0, k]. By Lemma 2.2

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)

≥ 1

2√

2(y

2α)−

32

1

sin(2πk)

(e−πk

2 2αy k − e−π(1−k)2 2α

y (1− k))

− 2(1 + ε3(α))e−παy2 (

y

α)−

32 −√

2(1 + ε4(α))e−3παy

2 (y

2α)−

32

≥(y

2α)−

32 e−πk

2 2αy

( 1

2√

2(k − εk)− e−

πα(y2−4k2)2y 2

√2(1 + ε3(α))(1 +

√2(1 + η2(α))e−παy

2√

2(1 + ε3(α)))).

Here εk is determined by

εk := maxyα∈[0,k]

e−π(1−k)2 2αy (1− k)

e−πk2 2αy k

.

Let

c(α) := 2√

2(1 + ε3(α))(1 +

√2(1 + ε4(α))e−παy

2√

2(1 + η1(α)))

and

F0(k) :=1

2√

2(k − εk)− c(α)e−

πα(y2−4k2)2y .

Then1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) ≥ (

y

2α)−

32 e−πk

2 2αy F0(k).

(3.41)

Let k ∈ [0, 0.3]. Since yα ∈ [0, k] and y ≥

√154 , one has

εk <3

5000, c(α) < 2.8287


and

F0(k) ≥ F (k) :=1

2√

2(k − εk)− c(α)e−π( 15

32k−2k). (3.42)

Hence, by (3.41), one has

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) ≥ (

y

2α)−

32 e−πk

2 2αy F (k).

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

k ∈ (0, kF ), kF = 0.28689 · · · .Therefore, by (3.43), one has

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) ≥ (

y

2α)−

32 e−πk

2 2αy F (k) > 0

provided that k ∈ (0, kF ) with kF = 0.28689 · · · .This gives the first range of k to be selected in completing our proof.

Case B : yα ,y

2α ∈ (k,∞): By Lemma 2.2

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)

≥ 1

2√

2(1− µ(k))4πe−π

y2α − 2(1 + ε3(α))(1 + µ(k))4πe−π

yα e−

παy2

−√

2(1 + ε4(α))(1 + µ(k))4πe−πy2α e−

3παy2

=4πe−y2α

( 1

2√

2(1− µ(k))− (2(1 + ε3(α) +

√2(1 + ε3(α))e−πy(α− 1

2α )))e−πy2 (α+ 1

α )(1 + µ(k))).

Let

d(α) := (2(1 + ε3(α) +√

2(1 + ε3(α))e−πy(α− 12α )))e−

πy2 (α+ 1

α )

and

G(k) :=

12√

2− d(α)

12√

2+ d(α)

− µ(k). (3.44)

Then

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) ≥ 4πe−π

y2α

12√

2+ d(α)

G(k).

Since y ≥√

154 , then d(α) ≤ 0.11025(α ≥ 1) and G′(k) = −µ′(k) > 0. Hence numerically

G(k) > 0 if k > kG := 0.22263 · · · .Therefore

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)

≥ 4πe−πy2α

12√

2+ d(α)

G(k) > 0 if k > kG.

This gives the second range of k to be selected, i.e., k > kG(= 0.22263 · · · ).

Case C : y2α ∈ (0, k], yα ∈ (k,∞). In this case, we first set k ∈ (kG, kF ) = (0.22263 · · · , 0.28689 · · · )

by the computations in subcases D and E above. In using the comparison inequality later, wefurther set that

k ∈ (kA, kB ]


where

kA := maxkg, kG, kB := minkf , kF ,1

4.

Then, k ∈ (k1, k2] = (0.22263 · · · , 14 ]. On the other hand, since y

2α ∈ (0, k], then α ≥ y2k . Again

the bound 14 is taken for controlling

1

maxx∈[0,k] sin(2πx)≥ 1

provided that k ∈ (0, kB ].By Lemma 2.2

1

2√

2ϑ(

y

2α)− 2(1 + ε31(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)

≥ 1

2√

2(y

2α)−

32

1

maxx∈[0,k] sin(2πx)

(e−πk

2 2αy k − e−π(1−k)2 2α

y (1− k))

− 2(1 + ε3(α))e−παy2 (1 + µ(k))4πe−π

yα −√

2(1 + ε4(α))e−3παy

2 (y

2α)−

32

≥ 1

2√

2(k − εk)(

y

2α)−

32 e−πk

2 2αy − 2(1 + ε3(α))(1 + µ(k))4πe−

παy2 e−π

yα

−√

2(1 + ε4(α))e−3παy

2 (y

2α)−

32

=(k − εk)(y

α)−

32 e−πk

2 2αy − 2(1 + ε3(α))(1 + µ(k))4πe−

παy2 e−π

yα

−√

2(1 + ε4(α))e−3παy

2 (y

2α)−

32 .

(3.45)

Our method works for all k ∈ (kA, kB ], to make the structure clear, we set that

k = kB =1

4.

Since y2α ∈ (0, k] and y ≥

√154 , one has

α ≥ y

2k= 2y ≥

√15

2.

To simplify the lower bounds in (3.45), we use the following

αy = α2 y

α≥ 15

4

y

α.

Continuing from (3.45), one has

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)

≥(1

4− 3

500)(y

α)−

32 e−

π2αy − 2(1 + ε3(α))(1 + µ(

1

4))4πe−

238 π

yα

−√

2(1 + ε4(α))e−45παy

8 (y

2α)−

32 .

(3.46)

Let

γ =y

α.

Then in this case

γ =y

α∈ (k, 2k] = (

1

4,

1

2].

Let

H(γ) := C1γ−32 e−

π2

1γ − C2e−

238 πγ − C3e−

45π8 γ(

γ

2)−

32 , (3.47)


here

C1 =1

4− 3

500

C2 = 2(1 + ε3(α))(1 + µ(1

4))4π

C3 =√

2(1 + ε4(α)).

By (3.46),

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) ≥ H(γ), γ =

y

α.

Then, it suffices to prove that

H(γ) > 0, γ ∈ (1

4,

1

2].

The first part of the function H(γ) was studied in (3.35), the second and third parts of the functionare decreasing, thus H(γ) is increasing on ( 1

4 ,12 ], hence

H(γ) > H(1

4), γ ∈ (

1

4,

1

2].

The numerical result shows that

H(1

4) > 0.

Thus we complete the proof of case C.In summary of cases A, B and C, we have

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α)

≥(y

2α)−

32 e−πk

2 2αy F (k) > 0, if

y

α∈ (0, k] , k ∈ (0, kF ), kF = 0.28689 · · ·

≥ 4πe−πy2α

12√

2+ d(α)

G(k) > 0 if k > kG, kG = 0.22263 · · ·

≥H(γ) > 0, γ =y

α, if

y

2α∈ (0, k],

y

α∈ (k,∞), k =

1

4,

(3.48)

see the functions F (k), G(k), H(γ) in (3.42), (3.44) and (3.47) respectively. Therefore, by selectingk = 1

4 in (3.48), one has

1

2√

2ϑ(

y

2α)− 2(1 + ε3(α))e−

παy2 ϑ(

y

α)−√

2(1 + ε4(α))e−3παy

2 ϑ(y

2α) > 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 Wκ(z;α) on the imaginary axis z =iy, y ≥ 1 and prove Theorems 1.2 and 1.3.

Recall that from the definition of Wκ(z;α) we have

W0(iy;α) =∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

)+∑n,m

e−2πα(n2y+m2

y )

W1(iy;α) =∑n,m

e−πα(n2y+m2

y ).

Here for convenience, throughout this section, we denote∑n,m =

∑n=∞n=−∞

∑m=∞m=−∞ .

In this section, we aim to establish that


Theorem 4.1. For σ ∈ [0, 2], the function W1(iy;α) + σW0(iy;α), y ∈ (0,∞) achieves its uniqueminimum at y = 1.

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

Theorem 4.2. For σ ∈ [0, 2], the function E1(iy;α) + σE0(iy;α), y ∈ (0,∞) achieves its uniqueminimum at y = 1.

We first recall the duality property of Wκ, which can be proved by Lemma 2.1 and we omit thedetails here.

Lemma 4.1. Under the transform z → w = z−1z+1 of H,

W1(z) =W0(w) and W0(z) =W1(w), z ∈ H and w =z − 1

z + 1∈ H.

Consequently, for all b ∈ R,

Wb(w) =W1−b(z), z ∈ H and w =z − 1

z + 1∈ H.

More generally, if h : z′ → w′ = z′−1z′+1 and g1 : z → z′, g2 : w′ → w are transforms in G2, then

Wb(w) =W1−b(z), z ∈ H and w = g2 h g1(z) ∈ H.

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

Xb(z) =∂Wb(z)

∂x= bX1(z) + (1− b)X0(z), Yb(z) =

∂Wb(z)

∂y= bY1(z) + (1− b)Y0(z).

The following two formulas relate fb on the upper half of the unit circle to f1−b on the upperhalf of the imaginary axis.

Lemma 4.2. Let the upper half of the unit circle be parametrized by u + i√

1− u2, u ∈ (−1, 1).

Then√

1−u2

1−u i parametrizes the upper half of the imaginary axies, and

Xb(u+ i√

1− u2) =

√1− u2

1− uY1−b

(√1− u2

1− ui)

(4.1)

Yb(u+ i√

1− u2) =−u

1− uY1−b

(√1− u2

1− ui)

(4.2)

hold for u ∈ (−1, 1).

Proof. Consider the transform in Lemma 4.1, z → w = z−1z+1 . With z = x+ yi and w = u+ vi,

u =x2 + y2 − 1

(x+ 1)2 + y2, v =

2y

(x+ 1)2 + y2. (4.3)

Conversely,

x =1− u2 − v2

(1− u)2 + v2, y =

2v

(1− u)2 + v2. (4.4)

Differentiate fb(w) = f1−b(z) with respect to u and v to find

Xb(w) = X1−b(z)∂x

∂u+ Y1−b(z)

∂y

∂u

Yb(w) = X1−b(z)∂x

∂v+ Y1−b(z)

∂y

∂v.

When w is on the unit circle, z is on the imaginary axis. Since W1−b is invariant under thereflection about the imaginary axis, X1−b(z) = 0 on the imaginary axis. Also

∂y

∂u

∣∣∣|w|=1

=

√1− u2

1− u,∂y

∂v

∣∣∣|w|=1

=−u

1− u(4.5)


from which the lemma follows.

By Lemma 2.1, one has

W1(i1

y;α) + σW0(i

1

y;α) =W1(iy;α) + σW0(iy;α). (4.6)

Then to prove Theorem 4.1, we will show that

Lemma 4.3. Assume that σ ∈ [0, 2]. Then

d

dy

(W1(iy;α) + σW0(iy;α)

)≥ 0, for y ≥ 1, α ≥ 1. (4.7)

The proof of (4.7) is through the operator L

LY := Y ′′ +2

yY ′ = y−2(y2Y

′)′, y > 0 (4.8)

Observe that if Lf(y) ≥ 0 for y ≥ 1 > 0 and f ′(1) = 0, then f ′(y) > 0 for y > 1. From (4.6) wesee that d

dy (W1(iy;α) + σW0(iy;α))∣∣y=1

= 0. In view of this fact, to prove Lemma 4.3, it suffices

to prove that

Lemma 4.4. Assume that σ ∈ [0, 2]. Then

L(W1(iy;α) + σW0(iy;α)

)≥ 0, for y ≥ 1, α ≥ 1.

Note that LW1(iy;α) ≥ for α > 0, y ≥ 1 is proved by Montgomery [16] and stated in Lemma4.6. To prove Lemma 4.4, it suffices to prove that it holds for σ = 2.

To prove Lemma 4.4 we compute first

Lemma 4.5. One first has the following two identities by direct computation

L(∑n,m

e−πα(n2y+m2

y ))

= Ia(y)− Ib(y),

L(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))= Ja(y)− Jb(y),

(4.9)

where

Ia(y) : =∑n,m

π2α2(m2

y2− n2)2e−πα(n2y+m2

y ),

Ib(y) : =∑n,m

2παn2

ye−πα(n2y+m2

y ),

Ja(y) : =∑n,m

π2α2

4

( (2m+ 1)2

y2− (2n+ 1)2

)2e−

πα2

((2n+1)2y+

(2m+1)2

y

),

Jb(y) : =πα

y

∑n,m

(2n+ 1)2e−πα2

((2n+1)2y+

(2m+1)2

y

).

(4.10)

The following Lemma is due to Montgomery [16].

Lemma 4.6. • L(∑

n,m e−πα(n2y+m2

y ))≥ 0 for y ∈ [1,∞) with equality holds only when

y = 1;

• L(∑

n,m e−πα2

((2n+1)2y+

(2m+1)2

y

)+∑n,m e

−2πα(n2y+m2

y ))≥ 0 for y ∈ [

√3,∞) with equality

holds only when y =√

3.

The L(∑

n,m e−πα2

((2n+1)2y+

(2m+1)2

y

))will change sign on the interval [1,

√3], however, we

establish the following


Lemma 4.7. At least for σ = 2, it holds that

L(∑n,m

e−πα(n2y+m2

y ) + σ∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))≥ 0, for y ∈ [1,

√3]. (4.11)

As a consequence, (4.11) holds for σ ∈ [0, 2] as well.

The proof of Lemma 4.7 is quite technique and we postpone it to the end of this section. DefineσM > 0 as

σM := maxσσ |: L

(∑n,m

e−πα(n2y+m2

y ) + σ∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))≥ 0, for y ∈ [1,

√3], α ≥ 1.

Here, as the invariance for the parameter α in (4.6), one only needs to consider the case α ≥ 1.We omit the proof of the following lemma that can be proved by the monotonicity property and

Lemma 4.7.

Lemma 4.8. σM satisfies

σM ≤ −L′(∑

n,m e−πα(n2y+m2

y ))|y=1,α=1

L′(∑

n,m e−πα2

((2n+1)2y+

(2m+1)2

y

))|y=1,α=1

.

Remark 4.1. The expression in Lemma 4.8 is well-defined since one notes that

L(∑n,m

e−πα(n2y+m2

y ))|y=1 = L

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))|y=1 = 0

L′(∑n,m

e−πα(n2y+m2

y ))|y=1 > 0,L

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))|y=1 < 0.

Theorem 4.3. For any κ ∈ [ 11+σM

, σM1+σM

],

Minimizerz∈HEκ(z) = i,

which is the purely imaginary number i, corresponding to the square lattice by our setting and isunique up the group G2 defined in (2.6). In particular, via Lemma 4.7,

[1

1 + σM,

σM1 + σM

] ⊇ [1

3,

2

3].

Proof. Note thatE1(z;α) + σE0(z;α)

=(1 + σ)( 1

1 + σE1(z;α) +

σ

1 + σE0(z;α)

)=(1 + σ)E 1

1+σ(z).

(4.12)

By minimum principle in Theorem 1.1, one has

minz∈HEκ(z) = min

z∈ΩeEκ(z), (4.13)

the minimizer may happen on H\Ωe, while it must be modular to some minimizer on Ωe via thegroup G2. By Theorem 4.2 and (4.12), for κ ∈ [ 1

3 , 1] (or more generally κ ∈ [ 11+σM

, 1]),

Minimizerz∈ΩeaEκ(z) = i. (4.14)

By Lemma 4.2 and (4.14), for κ ∈ [0, 23 ] (or more generally κ ∈ [0, σM

1+σM]),

Minimizerz∈ΩebEκ(z) = i. (4.15)

Therefore, it follows by (4.13), (4.14) and (4.15), for κ ∈ [ 11+σM

, σM1+σM

],

Minimizerz∈HEκ(z) = i. (4.16)


The proof is thus complete.

Next, we turn to the proof of Theorem 1.3.

Proof. By definitionWκ(z;α) = κW1(z;α) + (1− κ)W0(z;α).

To comparison, by deformation one has, for any κ, κ0(6= 0)

Wκ(z;α) =κ

κ0Wκ0(z;α) +

κ0 − κκ0

W0(z;α). (4.17)

Define the least and largest number such that

Minimizerz∈HEκ(z) = i

is κ1 and κ2 respectively. It follows by duality Lemma 4.1 that

κ1 + κ2 = 1.

Theorem 4.3 asserts that

κ1 ≤1

3, κ2 ≥

2

3.

We also define that

κ′1 :=1

1 + σM, κ′2 :=

σM1 + σM

.

It follows by the definitions thatκ′1 ≥ κ1, κ

′2 ≤ κ2.

Consider κ ∈ [0, κ′1]. By the definition of σM , one has

Minimizerz∈ΩebWκ′1(z;α) = i, (4.18)

and the minimizer is unique by the monotonicity induced in Lemma 4.2, Lemma 4.1 and Lemma4.6. On the other hand, via Lemma 4.6 and 4.2, one has the functional W0(z) is monotonicallyincreasing on the circle Ωeb, i.e,

Minimizerz∈ΩebW0(z;α) = i, (4.19)

and the minimizer is uniquely achieved.From the comparison formula (4.17), we have

Wκ(z;α) =κ

κ′1Wκ′1

(z;α) +κ′1 − κκ′1

W0(z;α). (4.20)

Combining (4.18) and (4.19), one has, for κ ∈ [0, κ′1],

Minimizerz∈ΩebWκ(z;α) = i. (4.21)

The minimum principle in Theorem 1.1 gives that

minz∈HWκ(z;α) = min

z∈ΩeWκ(z;α) (4.22)

Therefore, for κ ∈ [0, κ′1], the location identity (4.21) and minimum principle (4.22)(recall thatΩe = Ωea ∪ Ωeb) conclude that


z∈ΩeaWκ(z;α). (4.23)

The minimizer is unique up to the group G2.For κ ∈ [κ′1, κ

′2], by Theorem 4.3,

Minimizerz∈HWκ(z;α) = i, (4.24)

which is unique up the group G2. For κ ∈ [κ′2, 1], by the duality Lemma 4.1 and (4.23), one has


z∈ΩebWκ(z;α).

Therefore, the proof of Theorem 1.3 is complete.


Finally, by Lemma 4.7, one has σM ≥ 2, hence [ 11+σM

, σM1+σM

] ⊇ [ 13 ,

23 ]. 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, weintroduce some auxiliary functions and lemmas to the proof. From Lemma 4.5 and (4.10), we notethat each term in Ia(y), Ja(y) is non-negative. To control Ib(y), Jb(y), we introduce the auxiliaryfunctions as follows

µ1(y, k) : =

∞∑n=k

e−π(n2−1)y

µ2(y, k) : =

∞∑n=k

n2e−π(n2−1)y

ν1(y, k) : =∞∑n=k

e−2πn(n+1)y

ν2(y, k) : =

∞∑n=k

(2n+ 1)2e−2πn(n+1)y

(4.25)

for k ≥ 2, these terms are small since k ≥ 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

L(∑n,m

e−πα(n2y+m2

y ))≥

1∑n=−1

1∑m=−1

π2α2(m2


y )

−4πα

ye−παy

(1 + 2e−πα

1y + Error1(α, y)

),

(4.26)

where

Error1(α, y) := 2µ2(αy, 2)e−πα1y + 2µ2(αy, 2)µ1(α

1

y, 2), (4.27)

and the second lower bound estimate

L(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))

≥1∑

n=−1

1∑m=−1

π2α2

4

( (2m+ 1)2

y2− (2n+ 1)2

)2e−

πα2

((2n+1)2y+

(2m+1)2

y

)− 4πα

ye−

πα2 (y+ 1

y )(1 + Error2(α, y)),

(4.28)

where

Error2(α, y) := ν1(α, 1) + ν2(α, 1) + ν1(α, 1)ν2(α, 1). (4.29)

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

y ∈ [1,√

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 (4.28), the main order terms only have

three possibilities as follows

e−παy, e−πα1y , e−πα

12 (y+ 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 L(∑

n,m e−πα(n2y+m2

y ))

are given by

Lm(∑n,m

e−πα(n2y+m2

y ))

=2π2α2 1

y4e−πα

1y +

(2π2α2 − 4πα

y

)e−παy

(4.30)

and the leading order terms of L(∑

n,m e−πα(n2y+m2

y ))

are given by

Lm(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))= π2α2(

1

y2− 1)2e−πα

12 (y+ 1

y ) − 4πα

ye−πα

12 (y+ 1

y ).

(4.31)

The term π2α2( 1y2 − 1)2e−πα

12 (y+ 1

y ) will be dropped in estimates since it is nonnegative and it is

zero as y = 1. Now by (4.30) and (4.31),

Lm(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))+ σLm

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))≥2π2α2 1

y4e−πα

1y +

(2π2α2 − 4πα

y

)e−παy − σ 4πα

ye−πα

12 (y+ 1

y ).

(4.32)

To simplify the notations, let

P(α, y;σ) := 2π2α2 1

y4e−πα

1y +

(2π2α2 − 4πα

y

)e−παy − σ 4πα

ye−πα

12 (y+ 1

y ). (4.33)

Thus

Lm(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))+ σLm

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))≥ P(α, y;σ). (4.34)

We shall prove that P(α, y;σ) > 0 for some positive σ.Since y ≥ 1, α ≥ 1, one has 4πα

y ≤2π2π2α2, thus

2π2α2 − 4πα

y≥ (1− 2

π)2π2α2.

Then

P(α, y;σ) ≥ 2π2α2 1

y4e−πα

1y + (1− 2

π)2π2α2e−παy − σ 4πα

ye−πα

12 (y+ 1

y ). (4.35)

By the basic mean value inequality, one has, for any positive X ,Y, there has

X e−παy + Ye−πα1y ≥ 2

√XYe−πα

12 (y+ 1

y ),

with equality holds only if y = 1 and X = Y. It follows that, at least for

σ =

√1− 2

ππ, (4.36)

there holds that

2π2α2 1

y4e−πα

1y + (1− 2

π)2π2α2e−παy > σ

4πα

ye−πα

12 (y+ 1

y ),

which implies thatP(α, y;σ) > 0.

In view of (4.35), for the value of σ in (4.36), one has

Lm(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))+ σLm

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))≥ 0. (4.37)


By the main order analysis, we give the heuristic proof of Lemma 4.7. In the next, we give fullproof of Lemma 4.7 and provide the scheme of finding the optimal value of σ.

We are now in a position to prove Lemma 4.7.

Proof of Lemma 4.7.

Proof. We only need to consider the cases α ≥ 1 and y ≥ 1 by Lemma 4.1 and identity (4.6)respectively.

By Lemma 4.9, one has

L(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))+ σL

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))

≥1∑

n=−1

1∑m=−1

π2α2(m2


y ) − 4πα

ye−παy

(1 + 2e−πα

1y

)+σ( 1∑n=−1

1∑m=−1

π2α2

4

( (2m+ 1)2

y2− (2n+ 1)2

)2e−

πα2

((2n+1)2y+

(2m+1)2

y

)− 4πα

ye−

πα2 (y+ 1

y )))

−R1(α, y).

(4.38)

Here

R1(α, y) =4πα

ye−παyError1(α, y) +

4πα

ye−

πα2 (y+ 1

y )Error2(α, y),

where Error1(α, y) and Error2(α, y) are defined in (4.27) and (4.29) respectively. Now we define

A(α, y;σ) :=

1∑n=−1

1∑m=−1

π2α2(m2


y ) − 4πα

ye−παy

(1 + 2e−πα

1y

)+σ( 1∑n=−1

1∑m=−1

π2α2

4

( (2m+ 1)2

y2− (2n+ 1)2

)2e−

πα2

((2n+1)2y+

(2m+1)2

y

)− 4πα

ye−

πα2 (y+ 1

y ))).

(4.39)

Then

L(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))+ σL

(∑n,m

e−πα2

((2n+1)2y+

(2m+1)2

y

))≥A(α, y;σ)−R1(α, y).

(4.40)

The remainder term R1(α, y) is very small comparing the major term A(α, y;σ) for σ ≥ 0. Wenote that the estimate in A(α, y;σ) is much refined than that in the heuristic estimate. In fact,after dropping the nonnegative terms in A(α, y;σ), one has

A(α, y;σ) ≥2π2α2 1

y4e−πα

1y +

(2π2α2 − 4πα

y(1 + 2e−πα

1y ))e−παy − σ 4πα

ye−πα

12 (y+ 1

y )

=P(α, y;σ)− 8πα

ye−πα(y+ 1

y ).

(4.41)

The lower bound is quite similar to the one in P(α, y;σ)((4.35)). Same as the estimates in (4.35),similar to calculation of the value of σ in (4.36), i.e., for

σ =

√1− 2

ππ − 2e−π,


one has

2π2α2 1

y4e−πα

1y +

(2π2α2 − 4πα

y(1 + 2e−πα

1y ))e−παy − σ 4πα

ye−πα

12 (y+ 1

y ) > 0 (4.42)

for α ≥ 1 and y ≥ 1. Hence by (4.41) and (4.42),

A(α, y;

√1− 2

ππ − 2e−π) > 0 for α ≥ 1, y ∈ [1,

√3].

The value of σ such that A(α, y;σ) is positive on the strip α ≥ 1 and y ∈ [1,√

3] can be improvedby the auxiliary function A(α, y;σ).

In the next, we show that

A(α, y; 2) > 0 for α ≥ 1, y ∈ [1,√

3]. (4.43)

To use the monotonicity, we set

A(α, y;σ) = e−παyy−4B(α, y;σ), (4.44)

where

B(α, y;σ) =2π2α2y4 + 2π2α2eπα(y− 1y ) + 4π2α2(y2 − 1)2e−πα

1y

− 4παy3 − 8παy3e−πα1y − σ

(π2α2(y2 − 1)2e

πα2 (y− 1

y ) − 4παy3eπα2 (y− 1

y )).

By an elementary but complicated argument, one has

∂

∂αB(α, y;σ) > 0,

∂

∂yB(α, y;σ) > 0 for y ∈ [1,

√3], α ≥ 1, σ ∈ [0, 2].

It follows that

minα≥1,y∈[1,

√3]B(α, y; 2) = B(1, 1; 2).

Now the numerical result shows that

B(1, 1; 2) > 0.

Therefore by (4.44), (4.43) is proved.Since the remainder term R1(α, y) is so small comparing the major term A(α, y;σ), one has

A(α, y; 2)−R1(α, y) > 0 for α ≥ 1, y ∈ [1,√

3].

By (4.40), the proof is complete.

Remark on the Proof of Lemma 4.7.The bound σ = 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 moreaccurate upper bound.

Acknowledgment: The research of S. Luo was partially supported by double thousand plan ofJiangxi(jxsq2019101048) and NSFC(No. 12001253). The research of J. Wei was partially sup-ported 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, wehave

ϑ(X;Y ) =

∞∏n=1

(1− e−2πnX)(1 + e−(2n−1)πXe2πiY )(1 + e−(2n−1)πXe−2πiY )

=

∞∏n=1

(1− e−2πnX)(1 + e−2(2n−1)πX + 2e−(2n−1)πX cos(2πY )).

(5.1)

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

−∂∂Y ϑ(X;Y )

sin(2πY )= 4π

∞∑n=1

e−(2n−1)πX ϑ(X;Y )

1 + e−2(2n−1)πX + 2e−(2n−1)πX cos(2πY )

= 4π

∞∑n=1

e−(2n−1)πX∞∏

m6=n,m=1

(1− e−2πmX)(1 + e−2(2m−1)πX + 2e−(2m−1)πX cos(2πY )).

(5.2)

One sees from (5.2) that the function −∂∂Y ϑ(X;Y )

sin(2πY ) has a period 1, is decreasing on [0, 12 ] and a even

function for Y , just like cos(2πY ).Then

limY→ 1

2

−∂∂Y ϑ(X;Y )

sin(2πY )≤ −

∂∂Y ϑ(X;Y )

sin(2πY )≤ limY→0

−∂∂Y ϑ(X;Y )

sin(2πY ). (5.3)

It follows that

1

2π

∂2

∂Y 2ϑ(X;Y ) |Y= 1

2≤ −

∂∂Y ϑ(X;Y )

sin(2πY )≤ − 1

2π

∂2

∂Y 2ϑ(X;Y ) |Y=0 (5.4)

by L’Hospital rule.By (2.13), one has,

∂2

∂Y 2ϑ(X;Y ) |Y=0 = 4πe−πX(1 +

∞∑n=2

n2e−π(n2−1)X)

1

2π

∂2

∂Y 2ϑ(X;Y ) |Y= 1

2= 4π

∞∑n=1

(−1)n−1n2e−n2πX ≥ 4πe−πX(1−

∞∑n=2

n2e−π(n2−1)X).

(5.5)

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

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

ϑ(X;Y ) = X−12

∑n∈Z

e−π(n−Y )2

X .

We have

∂

∂Yϑ(t;Y ) = −2πt−

32

∞∑m=−∞

e−π(m−Y )2

t (Y −m)

= −2πt−32 (

∞∑m=0

e−π(k+m)2

t (k +m− e−π(2m+1)(1−2k)

t (m+ 1− k)))

Notice that for Y ≤ 1, t ≤ 2π

d

dY

(e−

π(2m+1)(1−2Y )t (m+1−Y )

)=e−

π(2m+1)(1−2Y )t

t(4m2π + 6mπ + 2π − 4Y mπ − 2Y π − t) > 0.

Then by this, one gets the lower and upper bounds for 0 < t < k ≤ π2 . 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

Γ(s)

πs1

as=

∫ ∞0

e−πattsdt

t(6.1)

which can be rewritten in convergent form

Γ(s)

πs1

as= (

1

s− 1− 1

s) +

∫ ∞1

(e−πat − 1)(ts + t1−s)dt

t.

It follows that

GR(z;κ, s) = (1

s− 1− 1

s) +

∫ ∞1

(Wκ(z;α)− 1)(ts + t1−s)dt

t. (6.2)

See GR(z;κ, s) and Wκ(z;α) in (1.13) and (1.9) respectively. The identity (6.2) can be viewed asan analogue of the Bernstein Theorem in Theorem (1.10). Taking derivative with respect to s, onehas

dj

dsjGR(z;κ, s) =

dj

dsj(

1

s− 1− 1

s) +

∫ ∞1

(Wκ(z;α)− 1)dj

dsj(ts + t1−s)

dt

t. (6.3)

Notice thatdj

dsj(ts + t1−s) = (log t)jts + (− log t)jt1−s

is nonnegative for any j ≥ 1, t ≥ 1, s ≥ 12 .

By the bridge identity (6.3), Theorem 1.4 can be proved by following similar proofs of Theorem1.1 and Theorem 1.2. We omit the details.

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(S. Luo) School of Mathematics and statistics, Jiangxi Normal University, Nanchang, 330022, China

(J. Wei) Department of Mathematics, University of British Columbia, Vancouver, B.C., Canada, V6T

1Z2

(W. Zou) Department of Mathematical Sciences, Tsinghua University, Beijing 100084, ChinaE-mail address, S. Luo: [email protected] or [email protected]

E-mail address, J. Wei: [email protected]

E-mail address, W. Zou: [email protected]

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