# Mass Renormalization in the Nelson Model

Authors: Fumio Hiroshima; Susumu Osawa
Journal: International Journal of Mathematics and Mathematical Sciences
 (2017)

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## Abstract

The asymptotic behavior of the effective massmeff(Λ) of the so-called Nelson model in quantum field theory is considered, where Λ is an ultraviolet cutoff parameter of the model. Let m be the bare mass of the model. It is shown that for sufficiently small coupling constant α of the model, meff(Λ)/m can be expanded as meff(Λ)/m=1+∑n=1∞an(Λ)α2n. A physical folklore is that an(Λ)=O(logΛ(n-1)) as Λ→∞. It is rigorously shown that 0<limΛ→∞a1(Λ)<C,  C1≤limΛ→∞a2(Λ)/log⁡Λ≤C2 with some constants C, C1, and C2.


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## Body

## 1. Introduction and Main Results

The model considered in this paper is the so-called Nelson model [1], which describes a nonrelativistic nucleon with bare mass m>0 interacting with a quantized scalar field with mass ν>0. The nucleon is governed by a Schrödinger operator. Let us first define the Nelson Hamiltonian. We use relativistic unit and employ the total momentum representation. Then the Hilbert space of states is the boson Fock space over L2(R3) which is given by (1)F=⨁n=0∞⁡⨂sn⁡L2R3,where ⨂sn denotes the n-fold symmetric tensor product and ⨂s0L2(R3)=C. Then Φ∈F can be written as Φ={Φ(0),Φ(1),Φ(2),…}, where Φ(n)∈⨂snL2(R3). The Fock vacuum Ω∈F is defined by Ω={1,0,0,…}. Let a(f),f∈L2(R3), be the annihilation operator and a(f)∗,f∈L2(R3), the creation operator on F, which are defined by (2)Daf∗=Ψ∈F∣∑n=0∞n+1Sn+1f⊗Ψn⊗nL2R32<∞,af∗Ψ0=0,af∗Ψn+1=n+1Sn+1f⊗Ψn,and a(f)=(a(f)∗)∗, where Sn is the symmetrizer, D(X) the domain of operator X, and ·K the norm on K. They satisfy canonical commutation relations as follows: (3)af,ag∗=f,g,af,ag=0,af∗,ag∗=0on a suitable dense domain, where [X,Y]=XY-YX and (·,·) is the inner product on K (linear in the second variable). Let T be a self-adjoint operator on L2(R3). Then we define the self-adjoint operator dΓ(T) on F by dΓ(T)=⨁n=0∞T(n), where(4)Tn=∑j=1nI⊗⋯⊗I⊗T˘jth⊗I⊗⋯⊗I⨂^nDT¯n≥1with T(0)=0. Here, for a closable operator T,T¯ denotes the closure of T. The operator dΓ(T) is called the second quantization of T. The free energy of the scalar field is given by Hf=dΓ(ω), where ω(k)=|k|2+ν2(k=(k1,k2,k3)∈R3,ν>0) is considered as a multiplication operator on L2(R3). Similarly the momentum of the scalar field is given by Pfμ=dΓ(kμ)(μ=1,2,3). The coupling of the nucleon and a scalar field is mediated through the Segal field operator Φs(g) defined by(5)Φsg=12ag+ag∗,where g is a cutoff function given by g(k)=ϕ^(k)/ω(k). Here ϕ^ is the form factor with infrared cutoff κ>0 and ultraviolet cutoff Λ>0, which are defined by (6)ϕ^k=0k<κ,2π-3/2κ≤k≤Λ,0k>Λ.The Nelson Hamiltonian with total momentum p∈R3 is given by a self-adjoint operator on F as follows:(7)Hp=12mp-Pf2+Hf+αΦsg,where α∈R is a coupling constant. Let E(p,α) be the energy-momentum relation (the infimum of the spectrum σ(H(p))) defined by(8)Ep,α=inf⁡σHp.Then the effective mass meff=meff(Λ) is defined by (9)1meff=13ΔpEp,α↾p=0.Here Δp denotes the three-dimensional Laplacian in the variable p. We are concerned with the asymptotic behavior of meff as the ultraviolet cutoff goes to infinity. It is however a subtle problem. Removal of the ultraviolet cutoff Λ through mass renormalization means finding sequences {m} and {Λ} such that m→0, Λ→∞, and meff converges. Since we can see that meff/m is a function of Λ/m, to achieve this, we want to find constants 0<γ<1 and 0<b0<∞ such that(10)limΛ→∞⁡meff/mΛ/mγ=b0.If we succeed in finding constants γ and b0 such as in (10), scaling the bare mass m as (11)m=1Λγ/1-γM,where M=(m∗/b0)1/(1-γ) with an arbitrary positive constant m∗, we have (12)limΛ→∞⁡meffΛ=m∗.The mass renormalization is, however, a subtle problem, and unfortunately, we cannot yet find constants γ and b0 such as in (10). For that reason we turn to perturbative renormalization, by which we try to guess the proper value of γ. Main results obtained in this paper are summarized as follows.Theorem 1.
Letκ>0. Then meff is an analytic function of α2 and can be expanded in the following power series for sufficiently small α: (13)meffm=1+∑n=1∞anΛα2n.Theorem 2.
There exists a strictly positive constantC such that (14)limΛ→∞⁡a1Λ=C.Theorem 3.
There exist some constantsC1 and C2 such that (15)C1≤limΛ→∞⁡a2Λlog⁡Λ≤C2.From Theorems2 and 3, if D=limΛ→∞a2(Λ)/log⁡Λ>0, it is suggested that γ=Dα2/C. So, the mass of the Nelson model is renormalizable for sufficiently small α.The effective mass and energy-momentum relation have been studied mainly in nonrelativistic electrodynamics. Spohn [2] investigates the upper and lower bound of the effective mass of the polaron model from a functional integral point of view. Hiroshima and Spohn [3] study a perturbative mass renormalization including fourth order in the coupling constant in the case of a spinless electron. Hiroshima and Ito [4, 5] study it in the case of an electron with spin 1/2. Bach et al. [6] show that the energy-momentum relation is equal to the infimum of the essential spectrum of the Hamiltonian for κ≥0. Fröhlich and Pizzo [7] investigate energy-momentum relation when infrared cutoff goes to 0.


## 2. Analytic Properties

In order to investigate the effective mass in a perturbation theory we have to check the analytic properties ofE(p,α).


### 2.1. Analytic Family in the Sense of Kato

Lemma 4.
H
(
p
) is an analytic family in the sense of Kato.Proof.
We proveH(p) is an analytic family of type (A). We see that(16)Hp=H0+∑μ=13pμ12mpμ-2Pfμ+αHI,where H0=(1/2m)Pf2+Hf and HI=Φs(g). Hence all we have to do is to prove the following facts. (a)
D(H0)⊂⋂μ=13D(Pfμ)∩D(HI).(b)
There exist real constantsaμ,bμ(μ=1,2,3), c, and d such that for any Ψ∈D(H0)(17)12mpμ-2PfμΨF≤aμH0ΨF+bμΨFμ=1,2,3,HIΨF≤cH0ΨF+dΨF.We prove (a) at first. Since ⋂μ=13D(Pfμ2)⊂D(Pfμ), we have D(H0)=⋂μ=13D(Pfμ2)∩D(Hf)⊂⋂μ=13D(Pfμ). Additionally, since ω-1/2gL2(R3)<∞, we have g∈D(ω-1/2). Furthermore, since ω is a nonnegative and injective self-adjoint operator on L2(R3), it follows that D(dΓ(ω)1/2)⊂D(a(g))∩D(a(g)∗)=D(HI). Hence we have D(H0)⊂D(dΓ(ω))⊂D(dΓ(ω)1/2). Together with them, (a) is proven. Next we prove (b). Let Ψ be an arbitrary vector in D(H0). Then we have (18)12mpμ-2PfμΨF≤pμ2mΨF+1mPfμΨF.Since PfμΨF2≤2mH01/2ΨF2, we have H01/2ΨF2≤(H0+1)ΨF2. Hence(19)12mpμ-2PfμΨF≤2mH0ΨF+pμ2m+2mΨF.Since D(H0)⊂D(dΓ(ω)1/2),(20)agΨF≤ω-1/2gL2R3Hf1/2ΨF,ag∗ΨF≤ω-1/2gL2R3Hf1/2ΨF+gL2R3ΨFhold. Hence(21)HIΨF≤2ω-1/2gL2R3Hf1/2ΨF+12gL2R3ΨF.From triangle inequality, we have Hf1/2ΨF≤HfΨF+ΨF. In addition,(22)H0ΨF2-HfΨF2=12mPf2ΨF2+1mRPf2Ψ,HfΨ.Since Pf2 and Hf are strongly commutative and nonnegative self-adjoint operators on F, (Pf2Ψ,HfΨ)≥0 holds. Hence HfΨF≤H0ΨF. Then we have(23)HIΨF≤2ω-1/2gL2R3H0ΨF+2ω-1/2gL2R3+12gL2R3ΨF.From (19) and (23), (b) is proven. Hence H(p) is an analytic family of type (A). Since every analytic family of type (A) is an analytic family of in the sense of Kato, it is an analytic family in the sense of Kato.We denote the ground state ofH(p) by ψg(p).Lemma 5.
(
1
)   E(p,α) is analytic in p and α if p and α are sufficiently small. (2)   ψg(p) is strongly analytic in p and α if p and α are sufficiently small.Proof.
From [8, Theorem XII.9], (1) follows, and from [8, Theorem XII.8], (2) follows.


### 2.2. Formula

In this section we expandm/meff with respect to α.Lemma 6.
The ratiom/meff can be expressed as(24)mmeff=1-23∑μ=13Pfμψg0,ψgμ′0ψg0,ψg0,where ψgμ′(0)=s-∂pμψg(p)↾p=0.Proof.
SinceE(p,α) is symmetry, E(p,α)=E(-p,α), we have(25)∂pμEp,α↾p=0=0,μ=1,2,3.Since H(p)ψg(p)=E(p,α)ψg(p), for any Ψ∈D(H(p)), (26)HpΨ,ψgp=Ep,αΨ,ψgpholds. Taking a derivative with respect to pμ on both sides above, we have(27)Hμ′pΨ,ψgp+HpΨ,ψgμ′p=Eμ′p,αΨ,ψgp+Ep,αΨ,ψgμ′p,Hμ′′pΨ,ψgp+2Hμ′pΨ,ψgμ′p+HpΨ,ψgμ′′p=Eμ′′p,αΨ,ψgp+2Eμ′p,αΨ,ψgμ′p+Ep,αΨ,ψgμ′′p.Here ′ denotes the derivative or strong derivative with respect to pμ, and Hμ′(p)=(1/m)(pμ-Pfμ),Hμ′′(p)=1/m. Setting Ψ=ψg(0) and p=0, we have(28)Eμ′′0,α=1mψg0,ψg0-2Pfμψg0,ψgμ′0ψg0,ψg0.This expression and the definition of the effective mass prove the lemma.


### 2.3. Perturbative Expansions

We define operatorsA+ and A- by A+=(1/2)a(g)∗ and A-=(1/2)a(g). Then HI=A++A-. Moreover, let F(n)=⨂sn⁡L2(R3) and(29)ψg0=∑n=0∞αnn!φn.Since E(p,α) is symmetry E(p,-α)=E(p,α), we have(30)E0,α=∑n=0∞α2n2n!E2n.Since ker⁡H0≠{0}, H0 is not injective. However, we define the operator 1/H0 (for notational simplicity we write 1/H0 for H0-1 in what follows) on F as follows. (31)D1H0=Ψ=⨁n=0∞⁡Ψn∈F∣∑n=1∞βnΨn2<∞,1H0Ψ0=0,1H0Ψnk1,…,kn=βnk1,…,knΨnk1,…,knn≥1.Here(32)βn=βnk1,…,kn=11/2mk1+⋯+kn2+∑i=1nωki.We define the subspace Ffin of F as Ffin={{Ψ(n)}n=0∞∈F∣Ψ(l)=0forl≥qwithsomeq}.Lemma 7.
It holds thatFfin⊂D(1/H0).Proof.
LetΨ∈Ffin. Then (1/H0)Ψ2=∑n=1∞((1/H0)Ψ)(n)2<∞. Hence the lemma follows.Lemma 8.
Letψg(0)=∑n=0∞(αn/n!)φn. Then φ0=Ω,   φ1=-(1/H0)HIΩ, and the recurrence formulas(33)φ2l=1H0-2lHIφ2l-1+∑j=1l2l2jE2jφ2l-2jl≥1,(34)φ2l+1=1H0-2l+1HIφ2l+∑j=1l2l+12jE2jφ2l+1-2jl≥0follow, with(35)φ2l∈F2⊕F4⊕⋯⊕F2ll≥1,φ2l+1∈F1⊕F3⊕⋯⊕F2l+1l≥0,and E2l is given by(36)E2l=2lΩ,HIφ2l-1l≥1.Proof.
We haveE(0,0)=E0 by substituting α=0 in (30). Since E(0,0) is the ground state energy of H0, E(0,0)=0. Hence E0=0. Since φ0 is the ground state of H0, φ0 can be Ω. We can find that (φn,Ω)=δ0n for n=0,1,… holds in the same way as [3]. From now we set H=H(p),   ψg=ψg(0),   E=E(p,α), and ′ means (strong)derivative with respect to α.(37)HΨ,ψg=EΨ,ψgholds for Ψ∈D(H). Differentiating (37) with respect to α, we have (38)HIΨ,ψg+HΨ,ψg′=E′Ψ,ψg+EΨ,ψg′.Hence ψg′∈D(H) and we have(39)HIψg+Hψg′=E′ψg+Eψg′.Substituting p=0 and α=0 into (39) and taking into account (φn,Ω)=δ0n, we have φ1=-(1/H0)HIΩ. Differentiating (37) n times with respect to α, we also have (40)HΨ,ψgn+nΨ,HIψgn-1=∑j=1nnjEjΨ,ψgn-j.By the induction on n, we have ψg(n)∈D(H) and (41)Hψgn+nHIψgn-1=∑j=1nnjEjψgn-j.Substituting p=0 and α=0 into both sides above, we have (42)H0φ2l+2lHIφ2l-1=∑j=1l2l2jE2jφ2l-2jl≥1,H0φ2l+1+2l+1HIφ2l=∑j=1l2l+12jE2jφ2l+1-2jl≥0.From now on, we shall prove (43)φni=0,i>n,i=0,φ2l2i+1=0,l≥1,0≤i≤l-1,suppk∈R3·2i⁡φ2l2ik=S2ior∅,l≥1,1≤i≤l,φ2l+12i=0,l≥0,0≤i≤l,suppk∈R32i+1⁡φ2l+12i+1k=S2i+1or∅,l≥0,0≤i≤l,where we set φn={φn(i)}i=0∞ by induction for n≥1, and (44)Si=k1,…,ki∈R3i∣κ≤k1≤Λ,…,κ≤ki≤Λ.Since φ1=-(1/H0)HIΩ∈F(1), φ1(i)=0,   i>1,i=0. Moreover, since (45)φ11k1=-122π3/2ωk1Ek1χκ,Λk1,we have suppk1∈R3⁡φ1(1)(k1)=S1, where E(k)=|k|2/2m+ω(k). Assume that the assumption of the induction holds when n≤2l+1,(l≥0). Then(46)H0φ2l+2+2l+2HIφ2l+1=∑j=1l+12l+22jE2jφ2l+2-2j.It is derived that φ2l+2(i)=0,   i>2l+2,i=0, by (φn,Ω)=δ0n and (46). By the assumption of the induction, (HIφ2l+1)(2i+1)=0,   0≤i≤l, holds. When 1≤q≤l, it holds that(47)H0φ2l+22q=-2l+1HIφ2l+12q+∑j=1l+12l+22jE2jφ2l+2-2j2q=-2l+112q∑i=12q12π3/2χκ,Λki2ωkiφ2l+12q-1k1,…,ki^,…,k2q+2q+1∫12π3/2χκ,Λk2ωkφ2l+12q+1k,k1,…,k2qdk+∑j=1l+12l+22jE2jφ2l+1-2j2q,where ki^ means that ki is omitted. By the assumption of the induction, the supports of the functions (48)12π3/2χκ,Λki2ωkiφ2l+12q-1k1,…,k^i,…,k2q,∫12π3/2χκ,Λk2ωkφ2l+12q+1k,k1,…,k2qdkand φ2l+1-2j(2q) are S2q or ∅. Furthermore, (49)H0φ2l+22l+2k1,…,k2l+2=-2l+1A+φ2l+12l+2k1,…,k2l+2=-2l+1∑i=12l+212π3/2χκ,Λki2ωkiφ2l+12l+1k1,…,ki^,…,k2l+2holds. By the assumption of the induction, the support of the right hand side is S2l+2 or ∅. Hence we have suppk∈R3·2iφ2l+2(2i)(k)=S2i or ∅,   1≤i≤l+1. We can prove φ2l+3(i)=0,i>2l+3,i=0, φ2l+3(2i)=0,1≤i≤l+1, and suppk∈R3(2i+1)φ2l+3(2i+1)(k)=S2i+1 or ∅,   0≤i≤l+1, in a similar way. From the discussion so far, we have (50)-2lHIφ2l-1+∑j=1l2l2jE2jφ2l-2j∈Ffinl≥1,-2l+1HIφ2l+∑j=1l2l+12jE2jφ2l+1-2j∈Ffinl≥0.Hence we have (51)φ2l=1H0-2lHIφ2l-1+∑j=1l2l2jE2jφ2l-2j+b2lΩl≥1,φ2l+1=1H0-2l+1HIφl+∑j=1l2l+12jE2jφ2l+1-2j+b2l+1Ωl≥0,where b2l and b2l+1 are some constants. Since (φ2l,Ω)=0,  l≥1, and (φ2l+1,Ω)=0,l≥0, b2l=b2l+1=0. Hence (33) and (34) are proven. By the discussion so far, (35) are also proven. We can derive (36) by (33) and (φn,Ω)=δ0n.


## 2.1. Analytic Family in the Sense of Kato

Lemma 4.
H
(
p
) is an analytic family in the sense of Kato.Proof.
We proveH(p) is an analytic family of type (A). We see that(16)Hp=H0+∑μ=13pμ12mpμ-2Pfμ+αHI,where H0=(1/2m)Pf2+Hf and HI=Φs(g). Hence all we have to do is to prove the following facts. (a)
D(H0)⊂⋂μ=13D(Pfμ)∩D(HI).(b)
There exist real constantsaμ,bμ(μ=1,2,3), c, and d such that for any Ψ∈D(H0)(17)12mpμ-2PfμΨF≤aμH0ΨF+bμΨFμ=1,2,3,HIΨF≤cH0ΨF+dΨF.We prove (a) at first. Since ⋂μ=13D(Pfμ2)⊂D(Pfμ), we have D(H0)=⋂μ=13D(Pfμ2)∩D(Hf)⊂⋂μ=13D(Pfμ). Additionally, since ω-1/2gL2(R3)<∞, we have g∈D(ω-1/2). Furthermore, since ω is a nonnegative and injective self-adjoint operator on L2(R3), it follows that D(dΓ(ω)1/2)⊂D(a(g))∩D(a(g)∗)=D(HI). Hence we have D(H0)⊂D(dΓ(ω))⊂D(dΓ(ω)1/2). Together with them, (a) is proven. Next we prove (b). Let Ψ be an arbitrary vector in D(H0). Then we have (18)12mpμ-2PfμΨF≤pμ2mΨF+1mPfμΨF.Since PfμΨF2≤2mH01/2ΨF2, we have H01/2ΨF2≤(H0+1)ΨF2. Hence(19)12mpμ-2PfμΨF≤2mH0ΨF+pμ2m+2mΨF.Since D(H0)⊂D(dΓ(ω)1/2),(20)agΨF≤ω-1/2gL2R3Hf1/2ΨF,ag∗ΨF≤ω-1/2gL2R3Hf1/2ΨF+gL2R3ΨFhold. Hence(21)HIΨF≤2ω-1/2gL2R3Hf1/2ΨF+12gL2R3ΨF.From triangle inequality, we have Hf1/2ΨF≤HfΨF+ΨF. In addition,(22)H0ΨF2-HfΨF2=12mPf2ΨF2+1mRPf2Ψ,HfΨ.Since Pf2 and Hf are strongly commutative and nonnegative self-adjoint operators on F, (Pf2Ψ,HfΨ)≥0 holds. Hence HfΨF≤H0ΨF. Then we have(23)HIΨF≤2ω-1/2gL2R3H0ΨF+2ω-1/2gL2R3+12gL2R3ΨF.From (19) and (23), (b) is proven. Hence H(p) is an analytic family of type (A). Since every analytic family of type (A) is an analytic family of in the sense of Kato, it is an analytic family in the sense of Kato.We denote the ground state ofH(p) by ψg(p).Lemma 5.
(
1
)   E(p,α) is analytic in p and α if p and α are sufficiently small. (2)   ψg(p) is strongly analytic in p and α if p and α are sufficiently small.Proof.
From [8, Theorem XII.9], (1) follows, and from [8, Theorem XII.8], (2) follows.


## 2.2. Formula

In this section we expandm/meff with respect to α.Lemma 6.
The ratiom/meff can be expressed as(24)mmeff=1-23∑μ=13Pfμψg0,ψgμ′0ψg0,ψg0,where ψgμ′(0)=s-∂pμψg(p)↾p=0.Proof.
SinceE(p,α) is symmetry, E(p,α)=E(-p,α), we have(25)∂pμEp,α↾p=0=0,μ=1,2,3.Since H(p)ψg(p)=E(p,α)ψg(p), for any Ψ∈D(H(p)), (26)HpΨ,ψgp=Ep,αΨ,ψgpholds. Taking a derivative with respect to pμ on both sides above, we have(27)Hμ′pΨ,ψgp+HpΨ,ψgμ′p=Eμ′p,αΨ,ψgp+Ep,αΨ,ψgμ′p,Hμ′′pΨ,ψgp+2Hμ′pΨ,ψgμ′p+HpΨ,ψgμ′′p=Eμ′′p,αΨ,ψgp+2Eμ′p,αΨ,ψgμ′p+Ep,αΨ,ψgμ′′p.Here ′ denotes the derivative or strong derivative with respect to pμ, and Hμ′(p)=(1/m)(pμ-Pfμ),Hμ′′(p)=1/m. Setting Ψ=ψg(0) and p=0, we have(28)Eμ′′0,α=1mψg0,ψg0-2Pfμψg0,ψgμ′0ψg0,ψg0.This expression and the definition of the effective mass prove the lemma.


## 2.3. Perturbative Expansions

We define operatorsA+ and A- by A+=(1/2)a(g)∗ and A-=(1/2)a(g). Then HI=A++A-. Moreover, let F(n)=⨂sn⁡L2(R3) and(29)ψg0=∑n=0∞αnn!φn.Since E(p,α) is symmetry E(p,-α)=E(p,α), we have(30)E0,α=∑n=0∞α2n2n!E2n.Since ker⁡H0≠{0}, H0 is not injective. However, we define the operator 1/H0 (for notational simplicity we write 1/H0 for H0-1 in what follows) on F as follows. (31)D1H0=Ψ=⨁n=0∞⁡Ψn∈F∣∑n=1∞βnΨn2<∞,1H0Ψ0=0,1H0Ψnk1,…,kn=βnk1,…,knΨnk1,…,knn≥1.Here(32)βn=βnk1,…,kn=11/2mk1+⋯+kn2+∑i=1nωki.We define the subspace Ffin of F as Ffin={{Ψ(n)}n=0∞∈F∣Ψ(l)=0forl≥qwithsomeq}.Lemma 7.
It holds thatFfin⊂D(1/H0).Proof.
LetΨ∈Ffin. Then (1/H0)Ψ2=∑n=1∞((1/H0)Ψ)(n)2<∞. Hence the lemma follows.Lemma 8.
Letψg(0)=∑n=0∞(αn/n!)φn. Then φ0=Ω,   φ1=-(1/H0)HIΩ, and the recurrence formulas(33)φ2l=1H0-2lHIφ2l-1+∑j=1l2l2jE2jφ2l-2jl≥1,(34)φ2l+1=1H0-2l+1HIφ2l+∑j=1l2l+12jE2jφ2l+1-2jl≥0follow, with(35)φ2l∈F2⊕F4⊕⋯⊕F2ll≥1,φ2l+1∈F1⊕F3⊕⋯⊕F2l+1l≥0,and E2l is given by(36)E2l=2lΩ,HIφ2l-1l≥1.Proof.
We haveE(0,0)=E0 by substituting α=0 in (30). Since E(0,0) is the ground state energy of H0, E(0,0)=0. Hence E0=0. Since φ0 is the ground state of H0, φ0 can be Ω. We can find that (φn,Ω)=δ0n for n=0,1,… holds in the same way as [3]. From now we set H=H(p),   ψg=ψg(0),   E=E(p,α), and ′ means (strong)derivative with respect to α.(37)HΨ,ψg=EΨ,ψgholds for Ψ∈D(H). Differentiating (37) with respect to α, we have (38)HIΨ,ψg+HΨ,ψg′=E′Ψ,ψg+EΨ,ψg′.Hence ψg′∈D(H) and we have(39)HIψg+Hψg′=E′ψg+Eψg′.Substituting p=0 and α=0 into (39) and taking into account (φn,Ω)=δ0n, we have φ1=-(1/H0)HIΩ. Differentiating (37) n times with respect to α, we also have (40)HΨ,ψgn+nΨ,HIψgn-1=∑j=1nnjEjΨ,ψgn-j.By the induction on n, we have ψg(n)∈D(H) and (41)Hψgn+nHIψgn-1=∑j=1nnjEjψgn-j.Substituting p=0 and α=0 into both sides above, we have (42)H0φ2l+2lHIφ2l-1=∑j=1l2l2jE2jφ2l-2jl≥1,H0φ2l+1+2l+1HIφ2l=∑j=1l2l+12jE2jφ2l+1-2jl≥0.From now on, we shall prove (43)φni=0,i>n,i=0,φ2l2i+1=0,l≥1,0≤i≤l-1,suppk∈R3·2i⁡φ2l2ik=S2ior∅,l≥1,1≤i≤l,φ2l+12i=0,l≥0,0≤i≤l,suppk∈R32i+1⁡φ2l+12i+1k=S2i+1or∅,l≥0,0≤i≤l,where we set φn={φn(i)}i=0∞ by induction for n≥1, and (44)Si=k1,…,ki∈R3i∣κ≤k1≤Λ,…,κ≤ki≤Λ.Since φ1=-(1/H0)HIΩ∈F(1), φ1(i)=0,   i>1,i=0. Moreover, since (45)φ11k1=-122π3/2ωk1Ek1χκ,Λk1,we have suppk1∈R3⁡φ1(1)(k1)=S1, where E(k)=|k|2/2m+ω(k). Assume that the assumption of the induction holds when n≤2l+1,(l≥0). Then(46)H0φ2l+2+2l+2HIφ2l+1=∑j=1l+12l+22jE2jφ2l+2-2j.It is derived that φ2l+2(i)=0,   i>2l+2,i=0, by (φn,Ω)=δ0n and (46). By the assumption of the induction, (HIφ2l+1)(2i+1)=0,   0≤i≤l, holds. When 1≤q≤l, it holds that(47)H0φ2l+22q=-2l+1HIφ2l+12q+∑j=1l+12l+22jE2jφ2l+2-2j2q=-2l+112q∑i=12q12π3/2χκ,Λki2ωkiφ2l+12q-1k1,…,ki^,…,k2q+2q+1∫12π3/2χκ,Λk2ωkφ2l+12q+1k,k1,…,k2qdk+∑j=1l+12l+22jE2jφ2l+1-2j2q,where ki^ means that ki is omitted. By the assumption of the induction, the supports of the functions (48)12π3/2χκ,Λki2ωkiφ2l+12q-1k1,…,k^i,…,k2q,∫12π3/2χκ,Λk2ωkφ2l+12q+1k,k1,…,k2qdkand φ2l+1-2j(2q) are S2q or ∅. Furthermore, (49)H0φ2l+22l+2k1,…,k2l+2=-2l+1A+φ2l+12l+2k1,…,k2l+2=-2l+1∑i=12l+212π3/2χκ,Λki2ωkiφ2l+12l+1k1,…,ki^,…,k2l+2holds. By the assumption of the induction, the support of the right hand side is S2l+2 or ∅. Hence we have suppk∈R3·2iφ2l+2(2i)(k)=S2i or ∅,   1≤i≤l+1. We can prove φ2l+3(i)=0,i>2l+3,i=0, φ2l+3(2i)=0,1≤i≤l+1, and suppk∈R3(2i+1)φ2l+3(2i+1)(k)=S2i+1 or ∅,   0≤i≤l+1, in a similar way. From the discussion so far, we have (50)-2lHIφ2l-1+∑j=1l2l2jE2jφ2l-2j∈Ffinl≥1,-2l+1HIφ2l+∑j=1l2l+12jE2jφ2l+1-2j∈Ffinl≥0.Hence we have (51)φ2l=1H0-2lHIφ2l-1+∑j=1l2l2jE2jφ2l-2j+b2lΩl≥1,φ2l+1=1H0-2l+1HIφl+∑j=1l2l+12jE2jφ2l+1-2j+b2l+1Ωl≥0,where b2l and b2l+1 are some constants. Since (φ2l,Ω)=0,  l≥1, and (φ2l+1,Ω)=0,l≥0, b2l=b2l+1=0. Hence (33) and (34) are proven. By the discussion so far, (35) are also proven. We can derive (36) by (33) and (φn,Ω)=δ0n.


## 3. Main Theorems

For notational simplicity we setϕ^j=ϕ^(kj) and ωj=ω(kj) for kj∈R3,j=1,2. Let (52)Ej=kj22m+ωj,j=1,2,E12=k1+k222m+ω1+ω2,ωr=r2+ν2,Fr=r22m+ωr.Theorem 9.
Letκ>0. Then meff is an analytic function of α2 and can be expanded in the following power series for sufficiently small |α|: (53)meffm=1+∑n=1∞anΛα2n.Proof.
By the power series (29), we have(54)ψg,ψg=∑n=0∞αnn!φn,∑m=0∞αmm!φm=∑n=0∞∑m=0∞αn+mn!m!φn,φm.By Lemma 8, (φn,φm)≠0 if and only if both n and m are even or odd. Then we have (55)ψg,ψg=1+∑n=1∞bnΛα2n.From the fact that both meff-1 and (ψg,ψg) are analytic functions of α2 and Lemma 6, we have the following power series: (56)-23∑μ=13Pfμψg,ψgμ′0=∑n=0∞cnΛα2n.Since ψgμ′(0) is an analytic function of α, we can write (57)ψgμ′0=∑n=0∞αnn!Φnμ.We note that (58)c0Λ=-23∑μ=13Pfμφ0,Φ0μ=-23∑μ=13dΓkμφ0,Φ0μ=0.Hence if |α| is sufficiently small, then we have the following power series:(59)meffm=ψg,ψgψg,ψg-2/3∑μ=13Pfμψg0,ψgμ′0=1+∑n=1∞bnΛα2n1+∑n=1∞bnΛ+cnΛα2n=1+∑n=1∞bnΛα2n∑n=0∞-∑l=1∞blΛ+clΛα2ln.This proves the theorem.Theorem 10.
There exists strictly positive constantC such that limΛ→∞⁡a1(Λ)=C.Proof.
From (59), we have (60)meffm=1+b1Λα2+Oα41-b1Λ+c1Λα2+Oα4=1-c1Λα2+Oα4.Therefore a1(Λ)=-c1(Λ). Since (61)∑n=1∞cnΛα2n=-23∑μ=13Pfμψg,ψgμ′0=-23∑μ=13Pfμ∑n=0∞αnn!φn,∑n=0∞αnn!Φnμ,we have(62)c1Λ=-23∑μ=1310!2!Pfμφ0,Φ2μ+11!1!Pfμφ1,Φ1μ+12!0!Pfμφ2,Φ0μ=-23∑μ=13Pfμφ1,Φ1μ+12Pfμφ2,Φ0μ.Substituting p=0 into (27) and using (25), we have(63)-1mPfμΨ,ψg+H0+αHIΨ,ψgμ′0=E0,αΨ,ψgμ′0.In addition, by setting α=0, we have -(Pfμ/m)φ0+H0Φ0μ=0. Since Pfμφ0=dΓ(kμ)Ω=0,H0Φ0μ=0 holds. Hence we have(64)Φ0μ=c0Ω,c0is  some  constant.Differentiating both sides of (63) with respect to α, we have (65)-1mPfμΨ,s-ddαψg+H0Ψ,s-ddαψgμ′0+HIΨ,ψgμ′0=E0,αΨ,s-ddαψgμ′0+ddαE0,αΨ,ψgμ′0.Substituting α=0 into both sides, we have (66)H0Ψ,Φ1μ=1mΨ,Pfμφ1-Ψ,HIΦ0μ.Therefore Φ1μ∈D(H0) and (67)H0Φ1μ=1mPfμφ1-HIΦ0μ=-1mPfμ1H0HIΩ-c0HIΩ.Since -(1/m)Pfμ(1/H0)HIΩ-c0HIΩ∈Ffin, we have(68)Φ1μ=-1m1H0Pfμ1H0HIΩ-c01H0HIΩ+c1Ω,where c1 is some constant. By a1(Λ)=-c1(Λ), (62), (64), and (68), we have (69)a1Λ=23∑μ=13Pfμφ1,Φ1μ=23m∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H0HIΩ-2c03∑μ=13Pfμ1H0HIΩ,1H0HIΩ-2c13∑μ=13Pfμ1H0HIΩ,Ω.It is also seen that (70)∑μ=13Pfμ1H0HIΩ,1H0HIΩ=∑μ=13Pfμ1H0HIΩ,Ω=0.Thus we have (71)a1Λ=23m∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H0HIΩ=23m∑μ=13Pfμ1H0A+Ω,1H0Pfμ1H0A+Ω=23m∫ϕ^k2ωkk2Ek3dk.Changing variables into polar coordinate, we have (72)23m∫ϕ^k2ωkk2Ek3dk=8π3m2π3∫κΛr4ωrFr3dr.Since r4/ω(r)F(r)3=O(r-3)(r→∞), the improper integral ∫κ∞(r4/ω(r)F(r)3)dr converges. It is trivial to see that limΛ→∞⁡a1(Λ)>0. Thus the theorem follows.Lemma 11.
It follows thatΦnμ∈Ffin for n∈N∪{0}.Proof.
By (64), we have Φ0μ∈Ffin. Assume that Φnμ∈Ffin holds when n≤k-1. Differentiating both sides of (63) k times with respect to α and substituting α=0, we have (73)-1mPfμΨ,φk+H0Ψ,Φkμ+kHIΨ,Φk-1μ=∑j=1kkjEjΨ,Φk-jμ;however, Ej=0 when j is odd. Since Φk-1μ∈Ffin, Φk-1μ∈D(HI) and (74)H0Ψ,Φkμ=1mΨ,Pfμφk-kΨ,HIΦk-1μ+Ψ,∑j=1kkjEjΦk-jμ.Thus Φkμ∈D(H0) and(75)H0Φkμ=1mPfμφk-kHIΦk-1μ+∑j=1kkjEjΦk-jμ.Since φk∈Ffin,  HIΦk-1μ∈Ffin, and Φk-jμ∈Ffin(j=1,…,k), by the assumption of induction, H0Φkμ∈Ffin. Hence Φnμ∈Ffin holds when n=k.Lemma 12.
It holds thatΦ0μ=c0Ω,   Φ1μ=-(1/m)(1/H0)Pfμ(1/H0)HIΩ-c0(1/H0)HIΩ+c1Ω, and the recurrence formulas(76)Φ2lμ=1H01mPfμφ2l-2lHIΦ2l-1μ+∑j=1l2l2jE2jΦ2l-2jμ+c2lΩl≥1,c2lis  some  constant.Φ2l+1μ=1H01mPfμφ2l+1-2l+1HIΦ2lμ+∑j=1l2l+12jE2jΦ2l+1-2jμ+c2l+1Ω,l≥0,c2l+1is  some  constant.Proof.
The first and second expressions are proven in Theorem10. From (75), it follows that (77)H0Φ2lμ=1mPfμφ2l-2lHIΦ2l-1μ+∑j=1l2l2jE2jΦ2l-2jμl≥1H0Φ2l+1μ=1mPfμφ2l+1-2l+1HIΦ2lμ+∑j=1l2l+12jE2jΦ2l+1-2jμl≥0.These prove the lemma.Lemma 13.
It is proven thata2(Λ) can be expanded as (78)a2Λ=23m∑j=18IjΛ+E2ΛmI9Λ-a1ΛI10Λ+a1Λ2,where Ij are given by (79)I1Λ=14∬ϕ^12ϕ^22ω1ω2k12E13+k22E231E1+1E21E12dk1dk2,I2Λ=18∬ϕ^12ϕ^22ω1ω2k12E14+k22E241E12dk1dk2,I3Λ=18∬ϕ^12ϕ^22ω1ω21E12+1E221E1+1E2k1,k2E122dk1dk2,I4Λ=14∬ϕ^12ϕ^22ω1ω2k12E12+k22E221E1+1E21E122dk1dk2,I5Λ=14∬ϕ^12ϕ^22ω1ω2E12E22k1,k2E12dk1dk2,I6Λ=18∬ϕ^12ϕ^22ω1ω21E1+1E22k12+k22E123dk1dk2,I7Λ=14∬ϕ^12ϕ^22ω1ω21E1+1E22k1,k2E123dk1dk2,I8Λ=14∬ϕ^12ϕ^22ω1ω21E1+1E2k1,k2E124dk1dk2,I9Λ=12∫ϕ^k2k2ωkEk4dk,I10Λ=12∫ϕ^k2ωkEk2dk.The proof of Lemma13 is given in the next section. The asymptotic behaviors of terms Ij(Λ) as Λ→∞ is given in the lemma below. Only two terms I1(Λ) and I2(Λ) logarithmically diverge, and other terms converge as Λ→∞.Lemma 14.
(1)–(3) follow the following:(1)
There exist some constantsC3 and C4 such that C3≤limΛ→∞⁡I1(Λ)/log⁡Λ≤C4.(2)
There exist some constantsC5 and C6 such that C5≤limΛ→∞⁡(I2(Λ)/log⁡Λ)≤C6.(3)
Forj=3,4,5,6,7,8limΛ→∞⁡IjΛ<∞.The proof of Lemma14 is technical and also given in the next section.Lemma 15.
It holds that(80)limΛ→∞⁡E2Λlog⁡Λ=-mπ2.Proof.
From (36), we have E2(Λ)=-(1/2π2)∫κΛr2/ω(r)F(r)dr and limr→∞⁡(r2/ωrFr)/(1/r)=2m. It implies (80).Now we are in the position to state the main theorem in this paper.Theorem 16.
There exist some constantsC1 and C2 such that (81)C1≤limΛ→∞⁡a2Λlog⁡Λ≤C2.Proof.
We haveI9(Λ)=(1/4π2)∫κΛr4/ω(r)F(r)4dr. Since r4/ω(r)F(r)4=O(r-5)(r→∞), we have(82)limΛ→∞⁡I9Λ<∞.We also have I10(Λ)=(1/4π2)∫κΛ(r2/ω(r)F(r)2)dr. Then r2/ω(r)F(r)2=O(r-3)(r→∞) and we also have(83)limΛ→∞⁡I10Λ<∞.By (82) and (83), Theorem 10, and Lemmas 13, 14, and 15 we can conclude the theorem.


## 4. Proof of Lemmas13 and 14

In this section we prove Lemmas13 and 14.


### 4.1. Proof of Lemma13

From (59) and a1(Λ)=-c1(Λ), we have a2(Λ)=-c2(Λ)-b1(Λ)a1(Λ)+a1(Λ)2. Here(84)b1Λ=φ1,φ1=12∫ϕ^k2ωkEk2dk,(85)c2Λ=-2310!4!∑μ=13Pfμφ0,Φ4μ+11!3!∑μ=13Pfμφ1,Φ3μ+12!2!∑μ=13Pfμφ2,Φ2μ+13!1!∑μ=13Pfμφ3,Φ1μ+14!0!∑μ=13Pfμφ4,Φ0μ=-19∑μ=13Pfμφ1,Φ3μ-16∑μ=13Pfμφ2,Φ2μ-19∑μ=13Pfμφ3,Φ1μ.Using recurrence formulas (33), (34), and (76), we have (86)φ2=21H0HI2Ω,φ3=-61H0HI3Ω-3E21H02HIΩ,Φ2μ=2m1H0Pfμ1H0HI2Ω+2m1H0HI1H0Pfμ1H0HIΩ+2c01H0HI2Ω-2c11H0HIΩ+c2Ω,Φ3μ=-6m1H0Pfμ1H0HI3Ω-6m1H0HI1H0Pfμ1H0HI2Ω-6m1H0HI21H0Pfμ1H0HIΩ-3mE21H0Pfμ1H02HIΩ-3mE21H02Pfμ1H0HIΩ-6c01H0HI3Ω+6c11H0HI2Ω-3c21H0HIΩ-3c0E21H02HIΩ+c3Ω.Substituting them into (85), we have (87)c2Λ=-23m∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H0HI3Ω+∑μ=13Pfμ1H0HIΩ,1H0HI1H0Pfμ1H0HI2Ω+∑μ=13Pfμ1H0HIΩ,1H0HI21H0Pfμ1H0HIΩ+∑μ=13Pfμ1H0HI2Ω,1H0Pfμ1H0HI2Ω+∑μ=13Pfμ1H0HI2Ω,1H0HI1H0Pfμ1H0HIΩ+∑μ=13Pfμ1H0HI3Ω,1H0Pfμ1H0HIΩ-E23m∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H02HIΩ+∑μ=13Pfμ1H0HIΩ,1H02Pfμ1H0HIΩ+∑μ=13Pfμ1H02HIΩ,1H0Pfμ1H0HIΩ-2c03∑μ=13Pfμ1H0HIΩ,1H0HI3Ω+2c13∑μ=13Pfμ1H0HIΩ,1H0HI2Ω-c23∑μ=13Pfμ1H0HIΩ,1H0HIΩ-c0E23∑μ=13Pfμ1H0HIΩ,1H02HIΩ+c39∑μ=13Pfμ1H0HIΩ,Ω-2c03∑μ=13Pfμ1H0HI2Ω,1H0HI2Ω+2c13∑μ=13Pfμ1H0HI2Ω,1H0HIΩ-c23∑μ=13Pfμ1H0HI2Ω,Ω-2c03∑μ=13Pfμ1H0HI3Ω,1H0HIΩ+2c13∑μ=13Pfμ1H0HI3Ω,Ω-c0E23∑μ=13Pfμ1H02HIΩ,1H0HIΩ+c1E23∑μ=13Pfμ1H02HIΩ,Ω=∑j=121j.We estimate 21 terms (1)–(21) above. We can however directly see that 0=(10)=⋯=(21) as follows: (88)∑μ=13Pfμ1H0HIΩ,1H0HI3Ω=∑μ=13Pfμ1H0HIΩ,1H0HI2Ω=∑μ=13Pfμ1H0HIΩ,1H0HIΩ=∑μ=13Pfμ1H0HIΩ,1H02HIΩ=∑μ=13Pfμ1H0HIΩ,Ω=∑μ=13Pfμ1H0HI2Ω,1H0HI2Ω=∑μ=13Pfμ1H0HI2Ω,1H0HIΩ=∑μ=13Pfμ1H0HI2Ω,Ω=∑μ=13Pfμ1H0HI3Ω,1H0HIΩ=∑μ=13Pfμ1H0HI3Ω,Ω=∑μ=13Pfμ1H02HIΩ,1H0HIΩ=∑μ=13Pfμ1H02HIΩ,Ω=0.We can compute remaining terms (1)–(9) as (89)1∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H0HI3Ω=∑μ=13A+1H0Pfμ1H0Pfμ1H0A+Ω,1H0A+2Ω=18∬ϕ^12ϕ^22ω1ω2k12E13+k22E231E1+1E21E12dk1dk2,2∑μ=13Pfμ1H0HIΩ,1H0HI1H0Pfμ1H0HI2Ω=∑μ=13A+1H0Pfμ1H0A+Ω,1H0Pfμ1H0A+2Ω=18∬ϕ^12ϕ^22ω1ω2k12E12+k22E221E1+1E21E122dk1dk2+18∬ϕ^12ϕ^22ω1ω21E12+1E221E1+1E2k1,k2E122dk1dk2,3∑μ=13Pfμ1H0HIΩ,1H0HI21H0Pfμ1H0HIΩ=∑μ=13A+1H0Pfμ1H0A+Ω,1H0A+1H0Pfμ1H0A+Ω=18∬ϕ^12ϕ^22ω1ω2k12E14+k22E241E12dk1dk2+14∬ϕ^12ϕ^22ω1ω2E12E22k1,k2E12dk1dk2,4∑μ=13Pfμ1H0HI2Ω,1H0Pfμ1H0HI2Ω=∑μ=13Pfμ1H0A+2Ω,1H0Pfμ1H0A+2Ω=18∬ϕ^12ϕ^22ω1ω21E1+1E22k12+k22E123dk1dk2+14∬ϕ^12ϕ^22ω1ω21E1+1E22k1,k2E123dk1dk2,5∑μ=13Pfμ1H0HI2Ω,1H0HI1H0Pfμ1H0HIΩ=∑μ=13Pfμ1H0A+2Ω,1H0A+1H0Pfμ1H0A+Ω=18∬ϕ^12ϕ^22ω1ω2k12E12+k22E221E1+1E21E122dk1dk2+14∬ϕ^12ϕ^22ω1ω21E1+1E2k1,k2E124dk1dk2,6∑μ=13Pfμ1H0HI3Ω,1H0Pfμ1H0HIΩ=∑μ=131H0A+2Ω,A+1H0Pfμ1H0Pfμ1H0A+Ω=18∬ϕ^12ϕ^22ω1ω2k12E13+k22E231E1+1E21E12dk1dk2,7∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H02HIΩ=12∫ϕ^k2k2ωkEk4dk,8∑μ=13Pfμ1H0HIΩ,1H02Pfμ1H0HIΩ=12∫ϕ^k2k2ωkEk4dk,9∑μ=13Pfμ1H02HIΩ,1H0Pfμ1H0HIΩ=12∫ϕ^k2k2ωkEk4dk.Thus the lemma follows.


### 4.2. Proof of Lemma14

Proof ofC3≤limΛ→∞(I1Λ/log⁡Λ) and C5≤limΛ→∞(I2Λ/log⁡Λ).
Changing variables to polar coordinates, we have(90)I1Λ=2π22π6∫-11∫κΛ∫κΛr12r22ωr1ωr2r12Fr13+r22Fr231Fr1+1Fr21Lr1,r2,zdzdr1dr2,where (91)Lr1,r2,z=r12+r22+2r1r2z2m+ωr1+ωr2.We define h(r1,r2),   h1(r1,r2),   h2(r1,r2),   S(Λ),   S1(Λ) and S2(Λ) as (92)hr1,r2=r12r22ωr1ωr2r12Fr13+r22Fr231Fr1+1Fr21Lr1,r2,1,h1r1,r2=r12r24ωr1ωr2Fr24Lr1,r2,1,h2r1,r2=hr1,r2-h1r1,r2,SΛ=∫κΛ∫κΛhr1,r2dr1dr2,S1Λ=∫κκ+1∫κ+1+ν+mΛh1r1,r2dr2dr1,S2Λ=∫κκ+1∫κκ+1+ν+mh1r1,r2dr2dr1+∫κ+1Λ∫κΛh1r1,r2dr1dr2+∫κΛ∫κΛh2r1,r2dr1dr2.Then(93)4π22π6SΛ≤I1Λ.In addition, S(Λ)=S1(Λ)+S2(Λ) follows. Since h1(r1,r2)>0 and h2(r1,r2)>0, S2(Λ)>0. Hence(94)SΛ>S1Λ.Let r2 satisfy κ≤r2≤κ+1. Suppose that κ+1+ν+m≤r1≤Λ. Since ν<r1, r12+ν2<2r12 holds. Therefore we have ω(r1)<2r1. Since r2<r1, we have r1r2<r12 and r22<r12. Thus L(r1,r2,1)<2(1/m+2)r12. So, (95)∫κ+1+ν+mΛr12ωr1Lr1,r2,1dr1>1221/m+2∫κ+1+ν+mΛr1-1dr1=1221/m+2log⁡Λ-log⁡κ+1+ν+mfollows. When κ≤r2≤κ+1, we have (96)ωr2≤κ+12+ν2,Fr2≤κ+122m+κ+12+ν2.Then(97)S1Λ=∫κκ+1r24ωr2Fr24dr2∫κ+1+ν+mΛr12ωr1Lr1,r2,1dr1>K221/m+2log⁡Λ-log⁡κ+1+ν+m∫κκ+1r24dr2=Kκ+15-κ51021/m+2log⁡Λ-log⁡κ+1+ν+m,where (98)K=1κ+12+ν2κ+12/2m+κ+12+ν24.
From (93), (94), and (97), C3≤limΛ→∞(I1Λ/log⁡Λ) follows.
The proof ofC5≤limΛ→∞(I2Λ/log⁡Λ) is similar to that of C3≤limΛ→∞(I1Λ/log⁡Λ). Then we omit it.Proof oflimΛ→∞(I1Λ/logΛ)≤C4.
We redefineh(r1,r2),   h1(r1,r2),   h2(r1,r2),   S1(Λ),   S2(Λ), and S(Λ) as(99)hr1,r2=r12r22ωr1ωr2r12Fr13+r22Fr231Fr1+1Fr21Lr1,r2,-1,h1r1,r2=r12r24ωr1ωr2Fr24Lr1,r2,-1,h2r1,r2=r12r24ωr1ωr2Fr1Fr23Lr1,r2,-1,S1Λ=∫κΛ∫κΛh1r1,r2dr1dr2,S2Λ=∫κΛ∫κΛh2r1,r2dr1dr2,SΛ=∫κΛ∫κΛhr1,r2dr1dr2.Then we have(100)I1Λ≤4π22π6SΛ.Since h(r1,r2)=h1(r1,r2)+h1(r2,r1)+h2(r1,r2)+h2(r2,r1), we have(101)SΛ=2S1Λ+S2Λ.Let B be B=((κ+1)3-κ3)/6ν2. Since ν<ω(r1) and 2ν<L(r1,r2,-1),(102)∫κκ+1r12ωr1Lr1,r2,-1dr1<Bfollows. Let Y be Y=2r2+κ+1. Since r1<ω(r1) and r1<L(r1,r2,-1),(103)∫κ+1Yr12ωr1Lr1,r2,-1dr1<2r2holds. When Y≤r1, since 2r2<r1, we have r1-r2>r1/2. Then L(r1,r2,-1)>(r1-r2)2/2m>r12/8m. So,(104)∫YΛr12ωr1Lr1,r2,-1dr1<8m∫YΛdr1r1<8mlog⁡Λfollows. From (102), (103), and (104), we have (105)∫κΛr12ωr1Lr1,r2,-1dr1<B+2r2+8mlog⁡Λ.Using this, we see that S1(Λ)<∫κΛ(B+2r2+8mlog⁡Λ)(r24/ω(r2)F(r2)4)dr2. Since r2<ω(r2) and r22/2m<F(r2), we have(106)S1Λ<∫κΛ2r2+B+8mlog⁡Λ16m4r25dr2=32m431κ3-1Λ3+4m4B+8mlog⁡Λ1κ4-1Λ4<32m5κ4log⁡Λ+32m43κ3+4m4Bκ4.Since r1<L(r1,r2,-1), (107)∫κΛr12ωr1Fr1Lr1,r2,-1dr1<2m∫κΛdr1r12<2mκ.Then we have(108)S2Λ=∫κΛr24ωr2Fr23dr2∫κΛr12ωr1Fr1Lr1,r2,-1dr1<2mκ∫κΛr24ωr2Fr23dr2<16m4κ∫κΛdr2r23<8m4κ3.From (101), (106), and (108), it follows that(109)SΛ<64m5κ4log⁡Λ+112m43κ3+8m2Bκ4.From (100) and (109), the lemma follows.Proof oflimΛ→∞(I2Λ/log⁡Λ)≤C6.
We redefineh(r1,r2),   h1(r1,r2),   S(Λ), and S1(Λ) as(110)hr1,r2=r12r22ωr1ωr2r12Fr14+r22Fr241Lr1,r2,-1,h1r1,r2=r12r24ωr1ωr2Fr24Lr1,r2,-1,SΛ=∫κΛ∫κΛhr1,r2dr1dr2,S1Λ=∫κΛ∫κΛh1r1,r2dr1dr2.We have (111)I2Λ=π22π6∫-11∫κΛ∫κΛr12r22ωr1ωr2r12Fr14+r22Fr241Lr1,r2,zdzdr1dr2.Then(112)I2Λ≤2π22π6SΛholds. Since h(r1,r2)=h1(r1,r2)+h1(r2,r1), we have(113)SΛ=2S1Λ.We have (114)∫κΛr12ωr1Lr1,r2,-1dr1<B+2r2+8mlog⁡Λin the same way as the proof of limΛ→∞(I1Λ/log⁡Λ)≤C4. Since r2<ω(r2) and r22/2m<F(r2), we have(115)S1Λ<∫κΛB+2r2+8mlog⁡Λr24ωr2Fr24dr2<32m4∫κΛdr2r24+16m4B+8mlog⁡Λ∫κΛdr2r25<32m5κ4log⁡Λ+32m43κ3+4m4Bκ4.From (112), (113), and (115), the lemma follows.Proof oflimΛ→∞(I3Λ/log⁡Λ)=0.
We defineh(r1,r2,z) as (116)hr1,r2,z=zr13r23ωr1ωr21Fr12+1Fr221Fr1+1Fr21Lr1,r2,z2and redefine S(Λ) as (117)SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr2dr1.Then we have I3(Λ)=(π2/(2π)6)S(Λ). We divide S(Λ) in the following way.(118)SΛ=∫-10∫κΛ∫κΛhr1,r2,zdzdr2dr1+∫01∫κΛ∫κΛhr1,r2,zdzdr2dr1=-∫10∫κΛ∫κΛhr1,r2,-zdzdr2dr1+∫01∫κΛ∫κΛhr1,r2,zdzdr2dr1=∫01∫κΛ∫κΛhr1,r2,z+hr1,r2,-zdzdr2dr1=∫01∫κΛ∫κΛgr1,r2,zdzdr2dr1,where(119)gr1,r2,z=hr1,r2,z+hr1,r2,-z=-2z2r14r24mωr1ωr21Fr12+1Fr221Fr1+1Fr2r12+r22/m+2ωr1+2ωr2Lr1,r2,z2Lr1,r2,-z2.Since g(r1,r2,z)≤0, S(Λ) is decreasing in Λ.(120)SΛ=∫01∫κΛ∫r2Λgr1,r2,zdzdr2dr1+∫01∫κΛ∫r1Λgr1,r2,zdzdr1dr2=2∫01∫κΛ∫r2Λgr1,r2,zdzdr2dr1=2∫01∫κΛ∫r22r2gr1,r2,zdzdr2dr1+2∫01∫κΛ∫2r2Λgr1,r2,zdzdr2dr1.Since κ≤r1, we have 1≤r1/κ. Hence r12+ν2≤(κ2+ν2/κ2)r12. Therefore we have ω(r1)≤(κ2+ν2/κ)r1, and similarly ω(r2)≤(κ2+ν2/κ)r2. When 0≤z≤1, we have(121)Lr1,r2,z>r122m.Then(122)Lr1,r2,-z=r1-r22+2r1r21-z2m+ωr1+ωr2>ωr1>r1.When r2≤r1, we have r22/m≤r12/m. Then it holds that (123)2ωr1≤2r12κ2κ2+ν2,2ωr2≤2r2κκ2+ν2≤2r12κ2κ2+ν2.Thus we have(124)r12m+r22m+2ωr1+2ωr2≤21m+2κ2+ν2κ2r12.When r2≤r1≤2r2, we have(125)1Fr1+1Fr2<10mr12,1Fr12+1Fr22<68m2r14.From (119), (121), (122), (124), and (125), it follows that (126)-gr1,r2,z≤2mz2r14r24r1r268m2r1410mr1221m+2κ2+ν2κ2r122mr1221r12=68·10·16m41m+2κ2+ν2κ2z2r23r17.Hence(127)-2∫01∫κΛ∫r22r2gr1,r2,zdzdr2dr1≤1190m4κ21m+2κ2+ν2κ2.When 2r2≤r1, we have(128)1Fr1+1Fr2<5m2r22,1Fr12+1Fr22<17m24r24.Since r2≤r1/2, we have r1/2≤r1-r2. Then we have(129)Lr1,r2,±z=r1-r22+2r1r21±z2m+ωr1+ωr2>r1-r222m≥r128m.From (119), (124), (128), and (129), it follows that (130)-gr1,r2,z≤2mz2r14r24r1r217m24r245m2r2221m+2κ2+ν2κ2r128mr124=211·5·17m61m+2κ2+ν2κ2z2r13r23.Hence(131)-2∫01∫κΛ∫2r2Λgr1,r2,zdzdr2dr1≤27·5·17m6κ41m+2κ2+ν2κ2.Then by (120), (127), and (131), we have (132)-SΛ≤1190m4κ21m+2κ2+ν2κ2+27·5·17m6κ41m+2κ2+ν2κ2.Since S(Λ) is decreasing and bounded below, it converges as Λ→∞. This fact proves the lemma.Proof oflimΛ→∞(I4Λ/log⁡Λ)=0.
We redefineh(r1,r2,z) as (133)hr1,r2,z=r12r22ωr1ωr2r12Fr12+r22Fr221Fr1+1Fr21Lr1,r2,z2.Then we have I4(Λ)=(2π2/(2π)6)∫-11∫κΛ∫κΛh(r1,r2,z)dzdr1dr2. We define J(Λ) as (134)JΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr1dr2.
Step 1. We define S(Λ,z) as(135)SΛ,z=∫κΛ∫κΛhr1,r2,zdr1dr2.Our first task is to prove that limΛ→∞S(Λ,z) exists for all z∈I=[-1,1]. Since h(r1,r2,z)>0,   S(Λ,z) is increasing in Λ. Let (136)S1Λ,z=2∫κΛ∫r22r2hr1,r2,zdr2dr1,S2Λ,z=2∫κΛ∫2r2Λhr1,r2,zdr2dr1.Then(137)SΛ,z=∫κΛ∫r2Λhr1,r2,zdr2dr1+∫κΛ∫r1Λhr1,r2,zdr1dr2=2∫κΛ∫r2Λhr1,r2,zdr2dr1=2∫κΛ∫r22r2hr1,r2,zdr2dr1+2∫κΛ∫2r2Λhr1,r2,zdr2dr1=S1Λ,z+S2Λ,zholds. We have(138)1Fr1+1Fr2<2m1r12+1r22,r12Fr12+r22Fr22<4m21r12+1r22,(139)r1<Lr1,r2,z.Let r2≤r1≤2r2. Since 1/r22≤4/r12, it holds that(140)1r12+1r22≤5r12.Then from (138) and (140), it follows that(141)1Fr1+1Fr2<10mr12,r12Fr12+r22Fr22<20m2r12.Hence from (139) and (141), it follows that h(r1,r2,z)<200m3r2/r15. Therefore we have(142)S1Λ,z<400m3∫κΛr2dr2∫r22r2dr1r15=375m381κ2-1Λ2<375m38κ2.Let 2r2≤r1. Since r1/2≤r1-r2, we have(143)r128m≤r1-r222m<Lr1,r2,z.Since r22≤r12, we have(144)1Fr1+1Fr2<4mr22,r12Fr12+r22Fr22<8m2r22.Hence from (143) and (144), it follows that h(r1,r2,z)<2048m5/r13r23. Therefore we have(145)S2Λ,z<4096m5∫κΛdr2r23∫2r2Λdr1r13<512m5∫κΛdr2r25<128m5κ4.From (137), (142), and (145), it follows that (146)SΛ,z<375m38κ2+128m5κ4.Since S(Λ,z) is increasing in Λ and bounded above for all z∈I, it converges as Λ goes to infinity.
Step 2. Our second task is to prove that J(Λ) converges when Λ goes to infinity. Let M(r1,r2) be (147)Mr1,r2=r12r22ωr1ωr2r12Fr12+r22Fr221Fr1+1Fr21Lr1,r2,-12.hr1,r2,z≤M(r1,r2) holds for all (r1,r2,z)∈[κ,Λ]2×I, and by Step 1 there exists (148)limΛ→∞⁡∫κΛ∫κΛMr1,r2dr1dr2.Since(149)MΛ=∫κΛ∫κΛMr1,r2dr1dr2⟶M∞=limΛ→∞⁡∫κΛ∫κΛMr1,r2dr1dr2,from Cauchy convergence condition, for any ϵ>0, there exists Λ0∈[κ,∞) such that if Λ0<Λ1≤Λ2,   MΛ2-MΛ1<ϵ. Then for Λ0<Λ1≤Λ2 and all z∈I, (150)SΛ2,z-SΛ1,z=∫κΛ1∫Λ1Λ2hr1,r2,zdr1dr2+∫κΛ1∫Λ1Λ2hr1,r2,zdr2dr1+∫Λ1Λ2∫Λ1Λ2hr1,r2,zdr1dr2≤∫κΛ1∫Λ1Λ2Mr1,r2dr1dr2+∫κΛ1∫Λ1Λ2Mr1,r2dr2dr1+∫Λ1Λ2∫Λ1Λ2Mr1,r2dr1dr2=MΛ2-MΛ1<ϵ.Therefore supz∈ISΛ2,z-SΛ1,z≤MΛ2-MΛ1<ϵ holds. Since family of functions (S(Λ,·))Λ∈[κ,∞) on I satisfies uniform Cauchy conditions, it converges uniformly on I. Since [κ,Λ]2 is a Jordan measurable bounded closed set of R2, the function S(Λ,z) is continuous on I. Hence (151)S∞,z=limΛ→∞⁡∫κΛ∫κΛhr1,r2,zdr1dr2is continuous on I. Since both S(Λ,z) and S(∞,z) are integrable on Jordan measurable set I, by uniform convergence theorem, we have (152)limΛ→∞⁡∫-11SΛ,zdz=∫-11S∞,zdz.It implies that J(Λ) converges as Λ→∞.Proof oflimΛ→∞I5(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   g(r1,r2,z), and S(Λ) as (153)hr1,r2,z=zr13r23ωr1ωr2Fr12Fr22Lr1,r2,z,gr1,r2,z=hr1,r2,z+hr1,r2,-z,SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr1dr2.Then I5(Λ)=2π2/(2π)6S(Λ). We have (154)SΛ=∫01∫κΛ∫κΛgr1,r2,zdzdr2dr1in the same way as (118). Since (155)gr1,r2,z=-2z2r14r24m2ωr1ωr2Fr12Fr22Lr1,r2,zLr1,r2,-z≤0,S(Λ) is decreasing in Λ. Since r1<L(r1,r2,z), we have (156)r14ωr1Fr12Lr1,r2,z<4m2r12.Similarly, we have (157)r24ωr2Fr22Lr1,r2,-z<4m2r22.Hence (158)-SΛ=2m2∫01z2dz∫κΛr24ωr2Fr22Lr1,r2,-zdr2∫κΛr14ωr1Fr12Lr1,r2,zdr1<23m2∫κΛ4m2r22dr2∫κΛ4m2r12dr1<32m23κ2.Since S(Λ) is decreasing and bounded below, it converges as Λ→∞.Proof oflimΛ→∞I6(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   J(Λ),   S(Λ,z),   S1(Λ,z), and S2(Λ,z) as (159)hr1,r2,z=r12r22ωr1ωr21Fr1+1Fr22r12+r22Lr1,r2,z3,JΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr2dr1,SΛ,z=∫κΛ∫κΛhr1,r2,zdr1dr2,S1Λ,z=2∫κΛ∫r22r2hr1,r2,zdr2dr1,S2Λ,z=2∫κΛ∫2r2Λhr1,r2,zdr2dr1.We have I6(Λ)=π2/(2π)6J(Λ).
Step 1. Our first task is to prove that limΛ→∞S(Λ,z) exists for all z∈I. Since h(r1,r2,z)>0,   S(Λ,z) is increasing in Λ. We have(160)SΛ,z=S1Λ,z+S2Λ,zin the same way as (137). When r2≤r1, it holds that(161)r12+r22≤2r12,1Fr1+1Fr2<4mr22.When r2≤r1≤2r2, it also holds that (162)1Fr1+1Fr2<10mr12.Then we have(163)∫r22r2hr1,r2,zdr1<∫r22r2r12r22r1r210mr1222r12r13dr1=175m23r22.Hence(164)S1Λ,z<350m23∫κΛdr2r22<350m23κ.Let 2r2≤r1. Since r1/2≤r1-r2, we have(165)r128m≤r1-r222m<Lr1,r2,z.Then(166)∫2r2Λhr1,r2,dr1<∫2r2Λr12r22r1r24mr2228mr1232r12dr1=16384m5r23∫2r2Λdr1r13<2048m5r25.Therefore(167)S2Λ,z<4096m5∫κΛdr2r25<1024m5κ4.From (160), (164), and (167), it follows that S(Λ,z)<350m2/3κ+1024m5/κ4. Since S(Λ,z) is increasing in Λ and bounded above, it converges as Λ goes to infinity.
Step 2. Our second task is to prove J(Λ) converges as Λ goes to infinity. This step is the same as that of limΛ→∞I4(Λ)/log⁡Λ=0.Proof oflimΛ→∞I7(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   g(r1,r2,z), and S(Λ) as(168)hr1,r2,z=zr13r23ωr1ωr21Fr1+1Fr221Lr1,r2,z3,gr1,r2,z=hr1,r2,z+hr1,r2,-z,SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr1dr2.Then we have I7(Λ)=2π2/(2π)6S(Λ), and(169)SΛ=∫01∫κΛ∫κΛgr1,r2,zdzdr1dr2in the same way as limΛ→∞I3(Λ)/log⁡Λ=0. We define g1(r1,r2,z) and g2(r1,r2,z) as(170)g1r1,r2,z=-6z2r14r24mωr1ωr21Fr1+1Fr22r12+r22/2m+ωr1+ωr22Lr1,r2,z3Lr1,r2,-z3,g2r1,r2,z=-2z4r16r26m3ωr1ωr21Fr1+1Fr221Lr1,r2,z3Lr1,r2,-z3.Then we have g(r1,r2,z)=g1(r1,r2,z)+g2(r1,r2,z). We redefine S1(Λ) and S2(Λ) by(171)S1Λ=∫01∫κΛ∫κΛg1r1,r2,zdzdr1dr2,S2Λ=∫01∫κΛ∫κΛg2r1,r2,zdzdr1dr2.Then(172)SΛ=S1Λ+S2Λ.Since g1(r1,r2,z)≤0,   S1(Λ) is decreasing in Λ. We divide S1(Λ) in the following way:(173)S1Λ=∫01∫κΛ∫r1Λg1r1,r2,zdzdr1dr2+∫01∫κΛ∫r2Λg1r1,r2,zdzdr2dr1=∫01∫κΛ∫r1Λg1r1,r2,zdzdr1dr2+∫01∫κΛ∫r1Λg1r2,r1,zdzdr1dr2=2∫01∫κΛ∫r1Λg1r1,r2,zdzdr1dr2.Let κ≤r1≤r2. Then we have(174)r12+r222m+ωr1+ωr2≤1m+2κ2+ν2κ2r22in the same way as (124). Let r1≤r2≤2r1. Then we also have(175)1Fr1+1Fr2<10mr22.Therefore(176)-g1r1,r2,z≤6mz2r14r24r1r210mr2221m+2κ2+ν2κ2r2222mr2231r23=4800m41m+2κ2+ν2κ22z2r13r26,-∫r12r1g1r1,r2,zdr2≤4800m41m+2κ2+ν2κ22z2r13∫r12r1dr2r26=930m41m+2κ2+ν2κ22z2r12.Hence(177)-2∫01∫κΛ∫r12r1g1r1,r2,zdzdr1dr2≤1860m41m+2κ2+ν2κ22∫01z2dz∫κΛdr2r22<620m4κ1m+2κ2+ν2κ22.Let 2r1≤r2. Then we have(178)1Fr1+1Fr2<5m2r12.In addition, since r2/2≤r2-r1, we can see that (179)r228m≤r2-r122m<Lr1,r2,-z.Therefore(180)-g1r1,r2,z≤6mz2r14r24r1r25m22r1221m+2κ2+ν2κ2r2222mr2238mr223=153600m91m+2κ2+ν2κ22z2r1r25.Then we have(181)-∫2r1Λg1r1,r2,zdr2≤153600m9z2r11m+2κ2+ν2κ22∫2r1Λdr2r25≤2400m91m+2κ2+ν2κ22z2r15.Hence(182)-2∫01∫κΛ∫2r1Λg1r1,r2,zdzdr1dr2≤4800m91m+2κ2+ν2κ22∫01z2dz∫κΛdr1r15<400m9κ41m+2κ2+ν2κ22.Then we have(183)-S1Λ<620m4κ1m+2κ2+ν2κ22+400m9κ41m+2κ2+ν2κ22.Since S1(Λ) is decreasing and bounded below, it converges as Λ→∞. Since g2(r1,r2,z)≤0,   S2(Λ) is also decreasing in Λ. Let r1≤r2. Then (184)1Fr1+1Fr2<4mr12.Therefore(185)-g2r1,r2,z≤2m3z4r16r26r1r24mr1228m3r261r23=256m2z4r1r24.Then(186)-∫r1Λg2r1,r2,zdr2≤256m2z4r1∫r1Λdr2r24≤256m2z43r12.Hence(187)-S2Λ≤512m23∫01z4dz∫κΛdr1r12<512m215κ.Since S2(Λ) is decreasing in Λ and bounded below, it converges. Since both S1(Λ) and S2(Λ) converge, S(Λ) converges.Proof oflimΛ→∞I8(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   g(r1,r2,z),   g1(r1,r2,z),   g2(r1,r2,z),   S(Λ),   S1(Λ), and S2(Λ) as(188)hr1,r2,z=zr13r23ωr1ωr21Fr1+1Fr21Lr1,r2,z4,gr1,r2,z=hr1,r2,z+hr1,r2,-z,(189)g1r1,r2,z=-2z2r14r24mωr1ωr21Fr1+1Fr21Lr1,r2,zLr1,r2,-z4,(190)g2r1,r2,z=-2z2r14r24mωr1ωr21Fr1+1Fr2Gr1,r2,z,(191)SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr2dr1,S1Λ=∫κΛ∫r2Λ∫01-1/r11/4r21/2g1r1,r2,zdr2dr1dz,S2Λ=∫κΛ∫r2Λ∫1-1/r11/4r21/21g1r1,r2,zdr2dr1dz,where(192)Gr1,r2,z=1Lr1,r2,z2Lr1,r2,-z3+1Lr1,r2,z3Lr1,r2,-z2+1Lr1,r2,z4Lr1,r2,-z.Furthermore, we define S3(Λ) as(193)S3Λ=∫01∫κΛ∫r2Λg2r1,r2,zdzdr2dr1.Then we have I8(Λ)=2π2/(2π)6S(Λ), and(194)SΛ=2∫01∫κΛ∫r2Λgr1,r2,zdzdr2dr1in the same way as the proof of limΛ→∞(I3Λ/log⁡Λ)=0. Since g(r1,r2,z)=g1(r1,r2,z)+g2(r1,r2,z), it holds that(195)SΛ=2S1Λ+2S2Λ+2S3Λ.Since g1(r1,r2,z)≤0 and g2(r1,r2,z)≤0,Si(Λ)(i=1,2,3) are decreasing in Λ. Let r2≤r1. Then(196)1Fr1+1Fr2<4mr22.Let 0≤z≤1-1/r11/4r21/2. Then we have(197)11-z4≤r1r22.We have(198)Lr1,r2,-z=r1-r22+2r1r21-z2m+ωr1+ωr2>r1r21-zm.Using (197) and (198), we have(199)1Lr1,r2,-z4<m4r14r241-z4≤m4r13r22.From (189), (196), and (199), it follows that (200)-g1r1,r2,z≤2mz2r14r24r1r24mr222mr12m4r13r22=16m5r12r2.Hence we have (201)-S1Λ≤16m5∫κΛdr2r2∫r2Λ1r121-1r11/4r21/2dr1<16m5∫κΛdr2r2∫r2Λdr1r12<16m5κ.Since S1(Λ) is decreasing in Λ and bounded below, it converges. When r2≤r1 and 1-1/r11/4r21/2≤z≤1, from (189) and (196), it holds that (202)-g1r1,r2,z<2mr14r24r1r24mr222mr121r14=16mr2r13.Hence we have (203)-S2Λ<16m∫κΛ∫r2Λ∫1-1/r11/4r21/21r2r13dr2dr1dz=16m∫κΛ∫r2Λr2r13r1-1/4r2-1/2dr2dr1=16m∫κΛr21/2dr2∫r2Λr1-13/4dr1=64m9∫κΛr21/2r2-9/4-Λ-9/4dr2<64m9∫κΛr2-7/4dr2=256m27κ-3/4-Λ-3/4<256m27κ3/4.Since S2(Λ) is decreasing in Λ and bounded below, it converges. We have(204)Gr1,r2,z<4m2r17+8m3r18+16m4r19.From (190), (196), and (204), we have (205)-g2r1,r2,z≤2mz2r14r24r1r24mr224m2r17+8m3r18+16m4r19<8r13r24m2r17+8m3r18+16m4r19=32m2r2r14+64m3r2r15+128m4r2r16.Hence (206)-S3Λ<32m2∫κΛr2dr2∫r2Λdr1r14+64m3∫κΛr2dr2∫r2Λdr1r15+128m4∫κΛr2dr2∫r2Λdr1r16<32m23∫κΛ1r22dr2+16m3∫κΛ1r23dr2+128m45∫κΛ1r24dr2<32m23κ+8m3κ2+128m415κ3.Since S3(Λ) is decreasing in Λ and bounded below, it converges. Since Si(Λ)(i=1,2,3) converge, I8(Λ) converges by (195).


## 4.1. Proof of Lemma13

From (59) and a1(Λ)=-c1(Λ), we have a2(Λ)=-c2(Λ)-b1(Λ)a1(Λ)+a1(Λ)2. Here(84)b1Λ=φ1,φ1=12∫ϕ^k2ωkEk2dk,(85)c2Λ=-2310!4!∑μ=13Pfμφ0,Φ4μ+11!3!∑μ=13Pfμφ1,Φ3μ+12!2!∑μ=13Pfμφ2,Φ2μ+13!1!∑μ=13Pfμφ3,Φ1μ+14!0!∑μ=13Pfμφ4,Φ0μ=-19∑μ=13Pfμφ1,Φ3μ-16∑μ=13Pfμφ2,Φ2μ-19∑μ=13Pfμφ3,Φ1μ.Using recurrence formulas (33), (34), and (76), we have (86)φ2=21H0HI2Ω,φ3=-61H0HI3Ω-3E21H02HIΩ,Φ2μ=2m1H0Pfμ1H0HI2Ω+2m1H0HI1H0Pfμ1H0HIΩ+2c01H0HI2Ω-2c11H0HIΩ+c2Ω,Φ3μ=-6m1H0Pfμ1H0HI3Ω-6m1H0HI1H0Pfμ1H0HI2Ω-6m1H0HI21H0Pfμ1H0HIΩ-3mE21H0Pfμ1H02HIΩ-3mE21H02Pfμ1H0HIΩ-6c01H0HI3Ω+6c11H0HI2Ω-3c21H0HIΩ-3c0E21H02HIΩ+c3Ω.Substituting them into (85), we have (87)c2Λ=-23m∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H0HI3Ω+∑μ=13Pfμ1H0HIΩ,1H0HI1H0Pfμ1H0HI2Ω+∑μ=13Pfμ1H0HIΩ,1H0HI21H0Pfμ1H0HIΩ+∑μ=13Pfμ1H0HI2Ω,1H0Pfμ1H0HI2Ω+∑μ=13Pfμ1H0HI2Ω,1H0HI1H0Pfμ1H0HIΩ+∑μ=13Pfμ1H0HI3Ω,1H0Pfμ1H0HIΩ-E23m∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H02HIΩ+∑μ=13Pfμ1H0HIΩ,1H02Pfμ1H0HIΩ+∑μ=13Pfμ1H02HIΩ,1H0Pfμ1H0HIΩ-2c03∑μ=13Pfμ1H0HIΩ,1H0HI3Ω+2c13∑μ=13Pfμ1H0HIΩ,1H0HI2Ω-c23∑μ=13Pfμ1H0HIΩ,1H0HIΩ-c0E23∑μ=13Pfμ1H0HIΩ,1H02HIΩ+c39∑μ=13Pfμ1H0HIΩ,Ω-2c03∑μ=13Pfμ1H0HI2Ω,1H0HI2Ω+2c13∑μ=13Pfμ1H0HI2Ω,1H0HIΩ-c23∑μ=13Pfμ1H0HI2Ω,Ω-2c03∑μ=13Pfμ1H0HI3Ω,1H0HIΩ+2c13∑μ=13Pfμ1H0HI3Ω,Ω-c0E23∑μ=13Pfμ1H02HIΩ,1H0HIΩ+c1E23∑μ=13Pfμ1H02HIΩ,Ω=∑j=121j.We estimate 21 terms (1)–(21) above. We can however directly see that 0=(10)=⋯=(21) as follows: (88)∑μ=13Pfμ1H0HIΩ,1H0HI3Ω=∑μ=13Pfμ1H0HIΩ,1H0HI2Ω=∑μ=13Pfμ1H0HIΩ,1H0HIΩ=∑μ=13Pfμ1H0HIΩ,1H02HIΩ=∑μ=13Pfμ1H0HIΩ,Ω=∑μ=13Pfμ1H0HI2Ω,1H0HI2Ω=∑μ=13Pfμ1H0HI2Ω,1H0HIΩ=∑μ=13Pfμ1H0HI2Ω,Ω=∑μ=13Pfμ1H0HI3Ω,1H0HIΩ=∑μ=13Pfμ1H0HI3Ω,Ω=∑μ=13Pfμ1H02HIΩ,1H0HIΩ=∑μ=13Pfμ1H02HIΩ,Ω=0.We can compute remaining terms (1)–(9) as (89)1∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H0HI3Ω=∑μ=13A+1H0Pfμ1H0Pfμ1H0A+Ω,1H0A+2Ω=18∬ϕ^12ϕ^22ω1ω2k12E13+k22E231E1+1E21E12dk1dk2,2∑μ=13Pfμ1H0HIΩ,1H0HI1H0Pfμ1H0HI2Ω=∑μ=13A+1H0Pfμ1H0A+Ω,1H0Pfμ1H0A+2Ω=18∬ϕ^12ϕ^22ω1ω2k12E12+k22E221E1+1E21E122dk1dk2+18∬ϕ^12ϕ^22ω1ω21E12+1E221E1+1E2k1,k2E122dk1dk2,3∑μ=13Pfμ1H0HIΩ,1H0HI21H0Pfμ1H0HIΩ=∑μ=13A+1H0Pfμ1H0A+Ω,1H0A+1H0Pfμ1H0A+Ω=18∬ϕ^12ϕ^22ω1ω2k12E14+k22E241E12dk1dk2+14∬ϕ^12ϕ^22ω1ω2E12E22k1,k2E12dk1dk2,4∑μ=13Pfμ1H0HI2Ω,1H0Pfμ1H0HI2Ω=∑μ=13Pfμ1H0A+2Ω,1H0Pfμ1H0A+2Ω=18∬ϕ^12ϕ^22ω1ω21E1+1E22k12+k22E123dk1dk2+14∬ϕ^12ϕ^22ω1ω21E1+1E22k1,k2E123dk1dk2,5∑μ=13Pfμ1H0HI2Ω,1H0HI1H0Pfμ1H0HIΩ=∑μ=13Pfμ1H0A+2Ω,1H0A+1H0Pfμ1H0A+Ω=18∬ϕ^12ϕ^22ω1ω2k12E12+k22E221E1+1E21E122dk1dk2+14∬ϕ^12ϕ^22ω1ω21E1+1E2k1,k2E124dk1dk2,6∑μ=13Pfμ1H0HI3Ω,1H0Pfμ1H0HIΩ=∑μ=131H0A+2Ω,A+1H0Pfμ1H0Pfμ1H0A+Ω=18∬ϕ^12ϕ^22ω1ω2k12E13+k22E231E1+1E21E12dk1dk2,7∑μ=13Pfμ1H0HIΩ,1H0Pfμ1H02HIΩ=12∫ϕ^k2k2ωkEk4dk,8∑μ=13Pfμ1H0HIΩ,1H02Pfμ1H0HIΩ=12∫ϕ^k2k2ωkEk4dk,9∑μ=13Pfμ1H02HIΩ,1H0Pfμ1H0HIΩ=12∫ϕ^k2k2ωkEk4dk.Thus the lemma follows.


## 4.2. Proof of Lemma14

Proof ofC3≤limΛ→∞(I1Λ/log⁡Λ) and C5≤limΛ→∞(I2Λ/log⁡Λ).
Changing variables to polar coordinates, we have(90)I1Λ=2π22π6∫-11∫κΛ∫κΛr12r22ωr1ωr2r12Fr13+r22Fr231Fr1+1Fr21Lr1,r2,zdzdr1dr2,where (91)Lr1,r2,z=r12+r22+2r1r2z2m+ωr1+ωr2.We define h(r1,r2),   h1(r1,r2),   h2(r1,r2),   S(Λ),   S1(Λ) and S2(Λ) as (92)hr1,r2=r12r22ωr1ωr2r12Fr13+r22Fr231Fr1+1Fr21Lr1,r2,1,h1r1,r2=r12r24ωr1ωr2Fr24Lr1,r2,1,h2r1,r2=hr1,r2-h1r1,r2,SΛ=∫κΛ∫κΛhr1,r2dr1dr2,S1Λ=∫κκ+1∫κ+1+ν+mΛh1r1,r2dr2dr1,S2Λ=∫κκ+1∫κκ+1+ν+mh1r1,r2dr2dr1+∫κ+1Λ∫κΛh1r1,r2dr1dr2+∫κΛ∫κΛh2r1,r2dr1dr2.Then(93)4π22π6SΛ≤I1Λ.In addition, S(Λ)=S1(Λ)+S2(Λ) follows. Since h1(r1,r2)>0 and h2(r1,r2)>0, S2(Λ)>0. Hence(94)SΛ>S1Λ.Let r2 satisfy κ≤r2≤κ+1. Suppose that κ+1+ν+m≤r1≤Λ. Since ν<r1, r12+ν2<2r12 holds. Therefore we have ω(r1)<2r1. Since r2<r1, we have r1r2<r12 and r22<r12. Thus L(r1,r2,1)<2(1/m+2)r12. So, (95)∫κ+1+ν+mΛr12ωr1Lr1,r2,1dr1>1221/m+2∫κ+1+ν+mΛr1-1dr1=1221/m+2log⁡Λ-log⁡κ+1+ν+mfollows. When κ≤r2≤κ+1, we have (96)ωr2≤κ+12+ν2,Fr2≤κ+122m+κ+12+ν2.Then(97)S1Λ=∫κκ+1r24ωr2Fr24dr2∫κ+1+ν+mΛr12ωr1Lr1,r2,1dr1>K221/m+2log⁡Λ-log⁡κ+1+ν+m∫κκ+1r24dr2=Kκ+15-κ51021/m+2log⁡Λ-log⁡κ+1+ν+m,where (98)K=1κ+12+ν2κ+12/2m+κ+12+ν24.
From (93), (94), and (97), C3≤limΛ→∞(I1Λ/log⁡Λ) follows.
The proof ofC5≤limΛ→∞(I2Λ/log⁡Λ) is similar to that of C3≤limΛ→∞(I1Λ/log⁡Λ). Then we omit it.Proof oflimΛ→∞(I1Λ/logΛ)≤C4.
We redefineh(r1,r2),   h1(r1,r2),   h2(r1,r2),   S1(Λ),   S2(Λ), and S(Λ) as(99)hr1,r2=r12r22ωr1ωr2r12Fr13+r22Fr231Fr1+1Fr21Lr1,r2,-1,h1r1,r2=r12r24ωr1ωr2Fr24Lr1,r2,-1,h2r1,r2=r12r24ωr1ωr2Fr1Fr23Lr1,r2,-1,S1Λ=∫κΛ∫κΛh1r1,r2dr1dr2,S2Λ=∫κΛ∫κΛh2r1,r2dr1dr2,SΛ=∫κΛ∫κΛhr1,r2dr1dr2.Then we have(100)I1Λ≤4π22π6SΛ.Since h(r1,r2)=h1(r1,r2)+h1(r2,r1)+h2(r1,r2)+h2(r2,r1), we have(101)SΛ=2S1Λ+S2Λ.Let B be B=((κ+1)3-κ3)/6ν2. Since ν<ω(r1) and 2ν<L(r1,r2,-1),(102)∫κκ+1r12ωr1Lr1,r2,-1dr1<Bfollows. Let Y be Y=2r2+κ+1. Since r1<ω(r1) and r1<L(r1,r2,-1),(103)∫κ+1Yr12ωr1Lr1,r2,-1dr1<2r2holds. When Y≤r1, since 2r2<r1, we have r1-r2>r1/2. Then L(r1,r2,-1)>(r1-r2)2/2m>r12/8m. So,(104)∫YΛr12ωr1Lr1,r2,-1dr1<8m∫YΛdr1r1<8mlog⁡Λfollows. From (102), (103), and (104), we have (105)∫κΛr12ωr1Lr1,r2,-1dr1<B+2r2+8mlog⁡Λ.Using this, we see that S1(Λ)<∫κΛ(B+2r2+8mlog⁡Λ)(r24/ω(r2)F(r2)4)dr2. Since r2<ω(r2) and r22/2m<F(r2), we have(106)S1Λ<∫κΛ2r2+B+8mlog⁡Λ16m4r25dr2=32m431κ3-1Λ3+4m4B+8mlog⁡Λ1κ4-1Λ4<32m5κ4log⁡Λ+32m43κ3+4m4Bκ4.Since r1<L(r1,r2,-1), (107)∫κΛr12ωr1Fr1Lr1,r2,-1dr1<2m∫κΛdr1r12<2mκ.Then we have(108)S2Λ=∫κΛr24ωr2Fr23dr2∫κΛr12ωr1Fr1Lr1,r2,-1dr1<2mκ∫κΛr24ωr2Fr23dr2<16m4κ∫κΛdr2r23<8m4κ3.From (101), (106), and (108), it follows that(109)SΛ<64m5κ4log⁡Λ+112m43κ3+8m2Bκ4.From (100) and (109), the lemma follows.Proof oflimΛ→∞(I2Λ/log⁡Λ)≤C6.
We redefineh(r1,r2),   h1(r1,r2),   S(Λ), and S1(Λ) as(110)hr1,r2=r12r22ωr1ωr2r12Fr14+r22Fr241Lr1,r2,-1,h1r1,r2=r12r24ωr1ωr2Fr24Lr1,r2,-1,SΛ=∫κΛ∫κΛhr1,r2dr1dr2,S1Λ=∫κΛ∫κΛh1r1,r2dr1dr2.We have (111)I2Λ=π22π6∫-11∫κΛ∫κΛr12r22ωr1ωr2r12Fr14+r22Fr241Lr1,r2,zdzdr1dr2.Then(112)I2Λ≤2π22π6SΛholds. Since h(r1,r2)=h1(r1,r2)+h1(r2,r1), we have(113)SΛ=2S1Λ.We have (114)∫κΛr12ωr1Lr1,r2,-1dr1<B+2r2+8mlog⁡Λin the same way as the proof of limΛ→∞(I1Λ/log⁡Λ)≤C4. Since r2<ω(r2) and r22/2m<F(r2), we have(115)S1Λ<∫κΛB+2r2+8mlog⁡Λr24ωr2Fr24dr2<32m4∫κΛdr2r24+16m4B+8mlog⁡Λ∫κΛdr2r25<32m5κ4log⁡Λ+32m43κ3+4m4Bκ4.From (112), (113), and (115), the lemma follows.Proof oflimΛ→∞(I3Λ/log⁡Λ)=0.
We defineh(r1,r2,z) as (116)hr1,r2,z=zr13r23ωr1ωr21Fr12+1Fr221Fr1+1Fr21Lr1,r2,z2and redefine S(Λ) as (117)SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr2dr1.Then we have I3(Λ)=(π2/(2π)6)S(Λ). We divide S(Λ) in the following way.(118)SΛ=∫-10∫κΛ∫κΛhr1,r2,zdzdr2dr1+∫01∫κΛ∫κΛhr1,r2,zdzdr2dr1=-∫10∫κΛ∫κΛhr1,r2,-zdzdr2dr1+∫01∫κΛ∫κΛhr1,r2,zdzdr2dr1=∫01∫κΛ∫κΛhr1,r2,z+hr1,r2,-zdzdr2dr1=∫01∫κΛ∫κΛgr1,r2,zdzdr2dr1,where(119)gr1,r2,z=hr1,r2,z+hr1,r2,-z=-2z2r14r24mωr1ωr21Fr12+1Fr221Fr1+1Fr2r12+r22/m+2ωr1+2ωr2Lr1,r2,z2Lr1,r2,-z2.Since g(r1,r2,z)≤0, S(Λ) is decreasing in Λ.(120)SΛ=∫01∫κΛ∫r2Λgr1,r2,zdzdr2dr1+∫01∫κΛ∫r1Λgr1,r2,zdzdr1dr2=2∫01∫κΛ∫r2Λgr1,r2,zdzdr2dr1=2∫01∫κΛ∫r22r2gr1,r2,zdzdr2dr1+2∫01∫κΛ∫2r2Λgr1,r2,zdzdr2dr1.Since κ≤r1, we have 1≤r1/κ. Hence r12+ν2≤(κ2+ν2/κ2)r12. Therefore we have ω(r1)≤(κ2+ν2/κ)r1, and similarly ω(r2)≤(κ2+ν2/κ)r2. When 0≤z≤1, we have(121)Lr1,r2,z>r122m.Then(122)Lr1,r2,-z=r1-r22+2r1r21-z2m+ωr1+ωr2>ωr1>r1.When r2≤r1, we have r22/m≤r12/m. Then it holds that (123)2ωr1≤2r12κ2κ2+ν2,2ωr2≤2r2κκ2+ν2≤2r12κ2κ2+ν2.Thus we have(124)r12m+r22m+2ωr1+2ωr2≤21m+2κ2+ν2κ2r12.When r2≤r1≤2r2, we have(125)1Fr1+1Fr2<10mr12,1Fr12+1Fr22<68m2r14.From (119), (121), (122), (124), and (125), it follows that (126)-gr1,r2,z≤2mz2r14r24r1r268m2r1410mr1221m+2κ2+ν2κ2r122mr1221r12=68·10·16m41m+2κ2+ν2κ2z2r23r17.Hence(127)-2∫01∫κΛ∫r22r2gr1,r2,zdzdr2dr1≤1190m4κ21m+2κ2+ν2κ2.When 2r2≤r1, we have(128)1Fr1+1Fr2<5m2r22,1Fr12+1Fr22<17m24r24.Since r2≤r1/2, we have r1/2≤r1-r2. Then we have(129)Lr1,r2,±z=r1-r22+2r1r21±z2m+ωr1+ωr2>r1-r222m≥r128m.From (119), (124), (128), and (129), it follows that (130)-gr1,r2,z≤2mz2r14r24r1r217m24r245m2r2221m+2κ2+ν2κ2r128mr124=211·5·17m61m+2κ2+ν2κ2z2r13r23.Hence(131)-2∫01∫κΛ∫2r2Λgr1,r2,zdzdr2dr1≤27·5·17m6κ41m+2κ2+ν2κ2.Then by (120), (127), and (131), we have (132)-SΛ≤1190m4κ21m+2κ2+ν2κ2+27·5·17m6κ41m+2κ2+ν2κ2.Since S(Λ) is decreasing and bounded below, it converges as Λ→∞. This fact proves the lemma.Proof oflimΛ→∞(I4Λ/log⁡Λ)=0.
We redefineh(r1,r2,z) as (133)hr1,r2,z=r12r22ωr1ωr2r12Fr12+r22Fr221Fr1+1Fr21Lr1,r2,z2.Then we have I4(Λ)=(2π2/(2π)6)∫-11∫κΛ∫κΛh(r1,r2,z)dzdr1dr2. We define J(Λ) as (134)JΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr1dr2.
Step 1. We define S(Λ,z) as(135)SΛ,z=∫κΛ∫κΛhr1,r2,zdr1dr2.Our first task is to prove that limΛ→∞S(Λ,z) exists for all z∈I=[-1,1]. Since h(r1,r2,z)>0,   S(Λ,z) is increasing in Λ. Let (136)S1Λ,z=2∫κΛ∫r22r2hr1,r2,zdr2dr1,S2Λ,z=2∫κΛ∫2r2Λhr1,r2,zdr2dr1.Then(137)SΛ,z=∫κΛ∫r2Λhr1,r2,zdr2dr1+∫κΛ∫r1Λhr1,r2,zdr1dr2=2∫κΛ∫r2Λhr1,r2,zdr2dr1=2∫κΛ∫r22r2hr1,r2,zdr2dr1+2∫κΛ∫2r2Λhr1,r2,zdr2dr1=S1Λ,z+S2Λ,zholds. We have(138)1Fr1+1Fr2<2m1r12+1r22,r12Fr12+r22Fr22<4m21r12+1r22,(139)r1<Lr1,r2,z.Let r2≤r1≤2r2. Since 1/r22≤4/r12, it holds that(140)1r12+1r22≤5r12.Then from (138) and (140), it follows that(141)1Fr1+1Fr2<10mr12,r12Fr12+r22Fr22<20m2r12.Hence from (139) and (141), it follows that h(r1,r2,z)<200m3r2/r15. Therefore we have(142)S1Λ,z<400m3∫κΛr2dr2∫r22r2dr1r15=375m381κ2-1Λ2<375m38κ2.Let 2r2≤r1. Since r1/2≤r1-r2, we have(143)r128m≤r1-r222m<Lr1,r2,z.Since r22≤r12, we have(144)1Fr1+1Fr2<4mr22,r12Fr12+r22Fr22<8m2r22.Hence from (143) and (144), it follows that h(r1,r2,z)<2048m5/r13r23. Therefore we have(145)S2Λ,z<4096m5∫κΛdr2r23∫2r2Λdr1r13<512m5∫κΛdr2r25<128m5κ4.From (137), (142), and (145), it follows that (146)SΛ,z<375m38κ2+128m5κ4.Since S(Λ,z) is increasing in Λ and bounded above for all z∈I, it converges as Λ goes to infinity.
Step 2. Our second task is to prove that J(Λ) converges when Λ goes to infinity. Let M(r1,r2) be (147)Mr1,r2=r12r22ωr1ωr2r12Fr12+r22Fr221Fr1+1Fr21Lr1,r2,-12.hr1,r2,z≤M(r1,r2) holds for all (r1,r2,z)∈[κ,Λ]2×I, and by Step 1 there exists (148)limΛ→∞⁡∫κΛ∫κΛMr1,r2dr1dr2.Since(149)MΛ=∫κΛ∫κΛMr1,r2dr1dr2⟶M∞=limΛ→∞⁡∫κΛ∫κΛMr1,r2dr1dr2,from Cauchy convergence condition, for any ϵ>0, there exists Λ0∈[κ,∞) such that if Λ0<Λ1≤Λ2,   MΛ2-MΛ1<ϵ. Then for Λ0<Λ1≤Λ2 and all z∈I, (150)SΛ2,z-SΛ1,z=∫κΛ1∫Λ1Λ2hr1,r2,zdr1dr2+∫κΛ1∫Λ1Λ2hr1,r2,zdr2dr1+∫Λ1Λ2∫Λ1Λ2hr1,r2,zdr1dr2≤∫κΛ1∫Λ1Λ2Mr1,r2dr1dr2+∫κΛ1∫Λ1Λ2Mr1,r2dr2dr1+∫Λ1Λ2∫Λ1Λ2Mr1,r2dr1dr2=MΛ2-MΛ1<ϵ.Therefore supz∈ISΛ2,z-SΛ1,z≤MΛ2-MΛ1<ϵ holds. Since family of functions (S(Λ,·))Λ∈[κ,∞) on I satisfies uniform Cauchy conditions, it converges uniformly on I. Since [κ,Λ]2 is a Jordan measurable bounded closed set of R2, the function S(Λ,z) is continuous on I. Hence (151)S∞,z=limΛ→∞⁡∫κΛ∫κΛhr1,r2,zdr1dr2is continuous on I. Since both S(Λ,z) and S(∞,z) are integrable on Jordan measurable set I, by uniform convergence theorem, we have (152)limΛ→∞⁡∫-11SΛ,zdz=∫-11S∞,zdz.It implies that J(Λ) converges as Λ→∞.Proof oflimΛ→∞I5(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   g(r1,r2,z), and S(Λ) as (153)hr1,r2,z=zr13r23ωr1ωr2Fr12Fr22Lr1,r2,z,gr1,r2,z=hr1,r2,z+hr1,r2,-z,SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr1dr2.Then I5(Λ)=2π2/(2π)6S(Λ). We have (154)SΛ=∫01∫κΛ∫κΛgr1,r2,zdzdr2dr1in the same way as (118). Since (155)gr1,r2,z=-2z2r14r24m2ωr1ωr2Fr12Fr22Lr1,r2,zLr1,r2,-z≤0,S(Λ) is decreasing in Λ. Since r1<L(r1,r2,z), we have (156)r14ωr1Fr12Lr1,r2,z<4m2r12.Similarly, we have (157)r24ωr2Fr22Lr1,r2,-z<4m2r22.Hence (158)-SΛ=2m2∫01z2dz∫κΛr24ωr2Fr22Lr1,r2,-zdr2∫κΛr14ωr1Fr12Lr1,r2,zdr1<23m2∫κΛ4m2r22dr2∫κΛ4m2r12dr1<32m23κ2.Since S(Λ) is decreasing and bounded below, it converges as Λ→∞.Proof oflimΛ→∞I6(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   J(Λ),   S(Λ,z),   S1(Λ,z), and S2(Λ,z) as (159)hr1,r2,z=r12r22ωr1ωr21Fr1+1Fr22r12+r22Lr1,r2,z3,JΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr2dr1,SΛ,z=∫κΛ∫κΛhr1,r2,zdr1dr2,S1Λ,z=2∫κΛ∫r22r2hr1,r2,zdr2dr1,S2Λ,z=2∫κΛ∫2r2Λhr1,r2,zdr2dr1.We have I6(Λ)=π2/(2π)6J(Λ).
Step 1. Our first task is to prove that limΛ→∞S(Λ,z) exists for all z∈I. Since h(r1,r2,z)>0,   S(Λ,z) is increasing in Λ. We have(160)SΛ,z=S1Λ,z+S2Λ,zin the same way as (137). When r2≤r1, it holds that(161)r12+r22≤2r12,1Fr1+1Fr2<4mr22.When r2≤r1≤2r2, it also holds that (162)1Fr1+1Fr2<10mr12.Then we have(163)∫r22r2hr1,r2,zdr1<∫r22r2r12r22r1r210mr1222r12r13dr1=175m23r22.Hence(164)S1Λ,z<350m23∫κΛdr2r22<350m23κ.Let 2r2≤r1. Since r1/2≤r1-r2, we have(165)r128m≤r1-r222m<Lr1,r2,z.Then(166)∫2r2Λhr1,r2,dr1<∫2r2Λr12r22r1r24mr2228mr1232r12dr1=16384m5r23∫2r2Λdr1r13<2048m5r25.Therefore(167)S2Λ,z<4096m5∫κΛdr2r25<1024m5κ4.From (160), (164), and (167), it follows that S(Λ,z)<350m2/3κ+1024m5/κ4. Since S(Λ,z) is increasing in Λ and bounded above, it converges as Λ goes to infinity.
Step 2. Our second task is to prove J(Λ) converges as Λ goes to infinity. This step is the same as that of limΛ→∞I4(Λ)/log⁡Λ=0.Proof oflimΛ→∞I7(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   g(r1,r2,z), and S(Λ) as(168)hr1,r2,z=zr13r23ωr1ωr21Fr1+1Fr221Lr1,r2,z3,gr1,r2,z=hr1,r2,z+hr1,r2,-z,SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr1dr2.Then we have I7(Λ)=2π2/(2π)6S(Λ), and(169)SΛ=∫01∫κΛ∫κΛgr1,r2,zdzdr1dr2in the same way as limΛ→∞I3(Λ)/log⁡Λ=0. We define g1(r1,r2,z) and g2(r1,r2,z) as(170)g1r1,r2,z=-6z2r14r24mωr1ωr21Fr1+1Fr22r12+r22/2m+ωr1+ωr22Lr1,r2,z3Lr1,r2,-z3,g2r1,r2,z=-2z4r16r26m3ωr1ωr21Fr1+1Fr221Lr1,r2,z3Lr1,r2,-z3.Then we have g(r1,r2,z)=g1(r1,r2,z)+g2(r1,r2,z). We redefine S1(Λ) and S2(Λ) by(171)S1Λ=∫01∫κΛ∫κΛg1r1,r2,zdzdr1dr2,S2Λ=∫01∫κΛ∫κΛg2r1,r2,zdzdr1dr2.Then(172)SΛ=S1Λ+S2Λ.Since g1(r1,r2,z)≤0,   S1(Λ) is decreasing in Λ. We divide S1(Λ) in the following way:(173)S1Λ=∫01∫κΛ∫r1Λg1r1,r2,zdzdr1dr2+∫01∫κΛ∫r2Λg1r1,r2,zdzdr2dr1=∫01∫κΛ∫r1Λg1r1,r2,zdzdr1dr2+∫01∫κΛ∫r1Λg1r2,r1,zdzdr1dr2=2∫01∫κΛ∫r1Λg1r1,r2,zdzdr1dr2.Let κ≤r1≤r2. Then we have(174)r12+r222m+ωr1+ωr2≤1m+2κ2+ν2κ2r22in the same way as (124). Let r1≤r2≤2r1. Then we also have(175)1Fr1+1Fr2<10mr22.Therefore(176)-g1r1,r2,z≤6mz2r14r24r1r210mr2221m+2κ2+ν2κ2r2222mr2231r23=4800m41m+2κ2+ν2κ22z2r13r26,-∫r12r1g1r1,r2,zdr2≤4800m41m+2κ2+ν2κ22z2r13∫r12r1dr2r26=930m41m+2κ2+ν2κ22z2r12.Hence(177)-2∫01∫κΛ∫r12r1g1r1,r2,zdzdr1dr2≤1860m41m+2κ2+ν2κ22∫01z2dz∫κΛdr2r22<620m4κ1m+2κ2+ν2κ22.Let 2r1≤r2. Then we have(178)1Fr1+1Fr2<5m2r12.In addition, since r2/2≤r2-r1, we can see that (179)r228m≤r2-r122m<Lr1,r2,-z.Therefore(180)-g1r1,r2,z≤6mz2r14r24r1r25m22r1221m+2κ2+ν2κ2r2222mr2238mr223=153600m91m+2κ2+ν2κ22z2r1r25.Then we have(181)-∫2r1Λg1r1,r2,zdr2≤153600m9z2r11m+2κ2+ν2κ22∫2r1Λdr2r25≤2400m91m+2κ2+ν2κ22z2r15.Hence(182)-2∫01∫κΛ∫2r1Λg1r1,r2,zdzdr1dr2≤4800m91m+2κ2+ν2κ22∫01z2dz∫κΛdr1r15<400m9κ41m+2κ2+ν2κ22.Then we have(183)-S1Λ<620m4κ1m+2κ2+ν2κ22+400m9κ41m+2κ2+ν2κ22.Since S1(Λ) is decreasing and bounded below, it converges as Λ→∞. Since g2(r1,r2,z)≤0,   S2(Λ) is also decreasing in Λ. Let r1≤r2. Then (184)1Fr1+1Fr2<4mr12.Therefore(185)-g2r1,r2,z≤2m3z4r16r26r1r24mr1228m3r261r23=256m2z4r1r24.Then(186)-∫r1Λg2r1,r2,zdr2≤256m2z4r1∫r1Λdr2r24≤256m2z43r12.Hence(187)-S2Λ≤512m23∫01z4dz∫κΛdr1r12<512m215κ.Since S2(Λ) is decreasing in Λ and bounded below, it converges. Since both S1(Λ) and S2(Λ) converge, S(Λ) converges.Proof oflimΛ→∞I8(Λ)/log⁡Λ=0.
We redefineh(r1,r2,z),   g(r1,r2,z),   g1(r1,r2,z),   g2(r1,r2,z),   S(Λ),   S1(Λ), and S2(Λ) as(188)hr1,r2,z=zr13r23ωr1ωr21Fr1+1Fr21Lr1,r2,z4,gr1,r2,z=hr1,r2,z+hr1,r2,-z,(189)g1r1,r2,z=-2z2r14r24mωr1ωr21Fr1+1Fr21Lr1,r2,zLr1,r2,-z4,(190)g2r1,r2,z=-2z2r14r24mωr1ωr21Fr1+1Fr2Gr1,r2,z,(191)SΛ=∫-11∫κΛ∫κΛhr1,r2,zdzdr2dr1,S1Λ=∫κΛ∫r2Λ∫01-1/r11/4r21/2g1r1,r2,zdr2dr1dz,S2Λ=∫κΛ∫r2Λ∫1-1/r11/4r21/21g1r1,r2,zdr2dr1dz,where(192)Gr1,r2,z=1Lr1,r2,z2Lr1,r2,-z3+1Lr1,r2,z3Lr1,r2,-z2+1Lr1,r2,z4Lr1,r2,-z.Furthermore, we define S3(Λ) as(193)S3Λ=∫01∫κΛ∫r2Λg2r1,r2,zdzdr2dr1.Then we have I8(Λ)=2π2/(2π)6S(Λ), and(194)SΛ=2∫01∫κΛ∫r2Λgr1,r2,zdzdr2dr1in the same way as the proof of limΛ→∞(I3Λ/log⁡Λ)=0. Since g(r1,r2,z)=g1(r1,r2,z)+g2(r1,r2,z), it holds that(195)SΛ=2S1Λ+2S2Λ+2S3Λ.Since g1(r1,r2,z)≤0 and g2(r1,r2,z)≤0,Si(Λ)(i=1,2,3) are decreasing in Λ. Let r2≤r1. Then(196)1Fr1+1Fr2<4mr22.Let 0≤z≤1-1/r11/4r21/2. Then we have(197)11-z4≤r1r22.We have(198)Lr1,r2,-z=r1-r22+2r1r21-z2m+ωr1+ωr2>r1r21-zm.Using (197) and (198), we have(199)1Lr1,r2,-z4<m4r14r241-z4≤m4r13r22.From (189), (196), and (199), it follows that (200)-g1r1,r2,z≤2mz2r14r24r1r24mr222mr12m4r13r22=16m5r12r2.Hence we have (201)-S1Λ≤16m5∫κΛdr2r2∫r2Λ1r121-1r11/4r21/2dr1<16m5∫κΛdr2r2∫r2Λdr1r12<16m5κ.Since S1(Λ) is decreasing in Λ and bounded below, it converges. When r2≤r1 and 1-1/r11/4r21/2≤z≤1, from (189) and (196), it holds that (202)-g1r1,r2,z<2mr14r24r1r24mr222mr121r14=16mr2r13.Hence we have (203)-S2Λ<16m∫κΛ∫r2Λ∫1-1/r11/4r21/21r2r13dr2dr1dz=16m∫κΛ∫r2Λr2r13r1-1/4r2-1/2dr2dr1=16m∫κΛr21/2dr2∫r2Λr1-13/4dr1=64m9∫κΛr21/2r2-9/4-Λ-9/4dr2<64m9∫κΛr2-7/4dr2=256m27κ-3/4-Λ-3/4<256m27κ3/4.Since S2(Λ) is decreasing in Λ and bounded below, it converges. We have(204)Gr1,r2,z<4m2r17+8m3r18+16m4r19.From (190), (196), and (204), we have (205)-g2r1,r2,z≤2mz2r14r24r1r24mr224m2r17+8m3r18+16m4r19<8r13r24m2r17+8m3r18+16m4r19=32m2r2r14+64m3r2r15+128m4r2r16.Hence (206)-S3Λ<32m2∫κΛr2dr2∫r2Λdr1r14+64m3∫κΛr2dr2∫r2Λdr1r15+128m4∫κΛr2dr2∫r2Λdr1r16<32m23∫κΛ1r22dr2+16m3∫κΛ1r23dr2+128m45∫κΛ1r24dr2<32m23κ+8m3κ2+128m415κ3.Since S3(Λ) is decreasing in Λ and bounded below, it converges. Since Si(Λ)(i=1,2,3) converge, I8(Λ) converges by (195).


## 5. Concluding Remarks

(1) The Nelson model is defined as the self-adjoint operator(207)HV=-12Δ+V⊗1+1⊗Hf+α∫R3⊕ϕxdx,acting in the Hilbert space L2(R)⊗F≅∫R3⊕Fdx. Here V:R3→R is an external potential and (208)ϕx=12∫a†ke-ikxφ^kωk+akeikxφ^-kωkdk.In the case of V=0, HV=0 is translation invariant and the relationship between HV and H(p) is given by (209)HV=0=∫R3⊕Hpdp.Furthermore the ground state energy ofH(p=0) coincides with that of HV=0.(2) We show thatmeff(Λ)/m=1+∑n=1∞an(Λ)α2n and limΛ→∞a2(Λ)=±∞. It is also expected that limΛ→∞an(Λ) diverges and the signatures are alternatively changed. Hence limΛ→∞meff(Λ)/m may converge but it is not trivial to see it directly.(3) The relativistic Nelson model is defined by replacing-(1/2)Δ+V with the semirelativistic Schrödinger operator -Δ+1+V in (207); that is, (210)HVrel⁡=-Δ+1+V⊗1+1⊗Hf+∫R3⊕ϕxdx.Then it follows that (211)HV=0rel⁡=∫R3⊕Hrel⁡pdp,whereHrel⁡(p)=(p-Pf)2+1+Hf+ϕ(0). Then the effective mass meff(Λ) of Hrel⁡(p) is defined in the same way as that of H(p). We are also interested in seeing the asymptotic behavior of meff(Λ) as Λ→∞. However (p-Pf)2+1 is a nonlocal operator and then estimates are rather complicated.Another interesting nonlocal model is the so-called semirelativistic Pauli-Fierz model defined by(212)HVPF=-i∇⊗-α∫R3⊕Axdx2+1+V⊗1+1⊗Hf,where A(x) is a quantized radiation field. See [9] for the detail. Then it follows that (213)HV=0PF=∫R3⊕HPFpdp,where HPF(p)=p-Pf-αA(0)2+1+Hf. It is also interesting to investigate the asymptotic behavior of the effective mass of the semirelativistic Pauli-Fierz model.



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*Source: 4760105-2017-06-06.xml*