数理解析研究所講究録 1255 巻 2002 年 103-123

103
Asymptotics of Green functions and Martin boundaries
for elliptic operators with periodic coefficients
東工大 . 理村田實 (Minoru Murata)
Department of Mathematics, Tokyo Institute of Technology
土田哲生 (Tetsuo Tsuchida)

名城大 . 理工 Department of Mathematics, Meijo University

1. INTRODUCTION

We consider asecond order elliptic operator o$\mathrm{R}^{d}$$\mathrm{n}$, $d\geq 2$ ,$L=- \sum_{i,j=1}^{d}\nabla_{i}a_{ij}(x)\nabla_{j}-\sum_{j=1}^{d}b_{j}(x)\nabla_{j}+c(x)=-\nabla\cdot a(x)\nabla-b(x)\cdot\nabla+c(x)$,. We assume that the coefficients have
and $x=$where.
periodicity, i.e. $a_{*j}.(x+z)=ajj(\#)$ , $b_{j}(x+z)=bj(x)$ and $c(x+z)=c(x)$ for any
and,Assume that the coefficients are real-valued, thatfor some
is symmetric and satisfies
and that the matrix$\gamma>0$ and all. In this paper we give asymptotics of the Green function $G(x, y)$,of as $|x-y|arrow\infty$ , and determine the Martin boundary for using the asymptotics.
Among many studies of elliptic operators with periodic coefficients let us note thefollowing. Agmon [A2] discussed positive solutions called exponential solutions to $(L-$. Developing his results, Pinsky [Pinsl] gave a$\lambda)u=0$ and the spectral properties forand the structure of the exponential solutions.
relation between the criticality ofFurther generalization to operators on manifolds with agroup action was achieved by, Schroeder
Lin and Pinchover [LP]. About asymptotics of the Green function as[S] gave an exponential decay rate by means of avariational quantity for Schr\"odinger
operators with periodic potentials. On p.87 in [Pinsl], aconjecture of the asymptotics
by Agmon was stated. In this paper we will give an asymptotics which is more precice
than his conjecture.
let $L(k)$ be an operator
We recall some results to state our theorems. For eachdefined byacting on functions on the -dimensional toru$\mathrm{Z}^{d}-$$(x_{1}, \ldots, x_{d})$$\nabla j=\partial/\partial xj$$z\in \mathrm{Z}^{d}$$a_{ij}$$c\in C^{\alpha}(\mathrm{R}^{d})$$\sum_{i,j=1}^{d}a_{ij}(x)\xi_{i}\xi j\geq\gamma|\xi|^{2}$$(a_{ij})$$x$$bj\in C^{1,\alpha}(\mathrm{R}^{d})$$\xi\in \mathrm{R}^{d}$

$L$

$L$

$L$

$L-\lambda$$|x|arrow\infty$$k\in \mathrm{C}^{d}$$d$$\mathrm{s}$$\mathrm{T}^{d}=\mathrm{R}^{d}/\mathrm{Z}^{d}$$L(k)=e^{-ik\cdot x}Le^{ik\cdot x}=-(\nabla+ik)\cdot a(x)(\nabla+ik)-b(x)\cdot(\nabla+ik)+c(x)$.

We regard $L(k)$ and $L(k)^{*}$ , the formal adjoint of $L(k)$ , as closed operators on, $L(i\beta)$ has an. By the Krein-Rutman theorem, for 46
with domainof multiplicity one such that the corresponding eigenspace iseigenvalue
is also an eigenvalue. Furthermore,generated by apositive function inof multiplicity one such that the corresponding eigenspace is generated byofthe(cf. Theorem 4.11.1 in [Pins2]). We callapositive function in$L(i\beta)$.
principal eigenvalue of$=0$ and$>0$ }.
be acone of positive solutions for $L:C_{L}=$ {Let
When apositive Green function exists for , is called subcritical. In this case$C^{\alpha}(\mathrm{T}^{d})$$\mathrm{R}^{d}$$C^{2,\alpha}(\mathrm{T}^{d})$$\Lambda(i\beta)\in \mathrm{R}$$\Lambda(i\beta)$$C^{2,\alpha}(\mathrm{T}^{d})$$L(i\beta)^{*}$$\Lambda(i\beta)$$C^{2,\alpha}(\mathrm{T}^{d})$$\psi$$C_{L}$$L$$L$$\in C^{2}(\mathrm{R}^{d});L\psi$$\psi$$C_{L}\neq\emptyset$

104(cf. [Pinsl]). When positive Green function does not exist for but, is called
critical. When
, is called supercritical. Putis{subcritical}.
It is known that $-\infty<\lambda_{c}<\infty$ , $L-\lambda$ is subcritical for, either subcritical or
critical for A
and supercritical for.
Suppose that is subcritical. For $R>0$ let
be the Dirichlet realization of in$L^{2}(B_{R})$ , whereis the ball $\{|x|<R\}$ . Then the resolvent
exists and the Greenfunctionis positive. Since is subcritical there exists the limit $G= \lim_{Rarrow\infty}G_{R}$which is called the minimal Green function.
Define;there exists
with
{
} and
$=e^{-\beta\cdotx}u>0$$(L-\lambda)\psi\geq0$;thereexistssuch that{with}.
Our arguments are based on results in [A2] and [Pinsl], so we extract them. Note thatthe relation between our function Aand afunction
in [Pinsl] is
.
$L$

$L$

$C_{L}=\emptyset$$\lambda_{c}=\sup$$L$$C_{L}\neq\emptyset$$\lambda;L-\lambda$$\lambda<\lambda_{c}$$=\lambda_{c}$$\lambda>\lambda_{c}$

$L$

$L$

$L_{R}$$B_{R}$$L_{R}^{-1}$$L$$G_{R}$$\Gamma_{\lambda}=$$K_{\lambda}=$$\beta\in \mathrm{R}^{d}$$\beta\in \mathrm{R}^{d}$$\psi$$=e^{-\beta\cdot x}u\in C_{L-\lambda}$$u\in C^{2}(\mathrm{T}^{d})$$\psi$$u\in C^{2}(\mathrm{T}^{d})$$\Lambda(i\beta)=-\lambda_{0}(-\beta)$$\lambda_{0}$.

Theorem(i) If, then. If A, thenwith sorne. If, thenis a -dimensional strictry convex compact set
with smooth boundary.
(ii) The function
is real analytic and strictly concave, and itsofHessianis negative definite.
and the supremum is attained uniquely atin (i), in particular,
and only.
and $Kx$(iv).
$\mathrm{A}\mathrm{P}$$\beta_{0}\in \mathrm{R}^{d}$$\lambda>\lambda_{c}$$\Gamma x$$\lambda<\lambda_{c}$$K_{\lambda}$$=K_{\lambda}=\emptyset$$=\lambda_{c}$$\Gamma_{\lambda}=K_{\lambda}=\{\beta_{0}\}$$d$$\Gamma_{\lambda}$$\Lambda(i\beta)$$\beta$$\in \mathrm{R}^{d}$$\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{\beta}\Lambda(i\beta)$$(ii_{l}.)\lambda_{c}=\sup_{\beta}\Lambda(i\beta)$$\nabla\rho\Lambda(:\beta)=0.\cdot f$$\mathrm{Y}_{\lambda}$$\beta_{0}$$\dot{l}f\beta=\beta_{0}$$=\{\beta\in \mathrm{R}^{d};\Lambda(i\beta)=\lambda\}$$=\{\beta\in \mathrm{R}^{d};\mathrm{A}(0)\geq\lambda\}$First assume that
. Then it follows ffom the above theorem that is
subcritical, and for each
there exists
uniquely such that the supremum
is attained at. For, choosesuch that$\{e_{s,1}, \cdots, e_{s,d-1}, s\}$ is an orthonormal basis of
. For
let
and
be positive-periodic solutions to (and$(L(:\beta)^{*}-\Lambda(\beta))v=0$ , respectively. Putforfunctionand . Our first main theorem is the following.
$L$$\sup_{\beta}\Lambda(:\beta)>0$$s\in \mathrm{S}^{d-1}$$\sup_{\beta\in\Gamma_{\mathrm{O}}}\beta\cdot$$\beta_{s}\in\Gamma_{0}$$\beta=\beta_{s}$$s$$s\in \mathrm{S}^{d-1}$$\{e_{s,j}\}_{j=1}^{d-1}\subset \mathrm{R}^{d}$$\mathrm{R}^{d}$$v\rho\in C^{2,\alpha}(\mathrm{R}^{d})$$\beta\in \mathrm{R}^{d}$$\mathrm{L}(\mathrm{i}(3)-\mathrm{A}(\mathrm{i}/3))\mathrm{u}=0$$(u, v)= \int_{\mathrm{T}^{d}}u(x)\overline{v}(x)dx$$u$$up\in C^{2,\alpha}(\mathrm{R}^{d})$$\mathrm{Z}^{d}$$L^{2}(\mathrm{T}^{d})$$v$

Theorem 1.1. Assume that
. Then the rninirnal Green function
has the following asymptotics as $|x-y|arrow\infty$ :$\sup_{\beta}\Lambda(i\beta)>0$

$G$

of

$L$

$G(x, y)= \frac{e^{-(x-y)\cdot\beta}}{(2\pi|x-y|)^{(d-1)/2}}..\frac{|\nabla\rho\Lambda(i\beta_{s})|^{(d-3)/2}}{(\det(-e_{s},{}_{\mathrm{j}}\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{\beta}\Lambda(i\beta_{s})e_{s,k})_{jk})^{1/2}}\frac{u\rho.(x)v\rho.(y)}{(u\rho.,v\rho.)}$$\cross(1+O(|x-y|^{-1}))$ ,

where

$s$

(1.1)

$=(x-y)/|x-y|$ and the tervn $O(|x-y|^{-1})$ satisfies that $|O(|x-y|^{-1})|\leq$
for $|x-y|>R$ with positive constants and independent of , .

$C|x-y|^{-1}$

$C$

$R$

$x$

$y$

In the next section, we shall reduce the proof of Theorem 1.1 to the folowing theorem,
where is regarded as aclosed operator on.
$L$$L^{2}(\mathrm{R}^{d})$Theorem 1.2. Assume $\Lambda(0)>0$ . Then the resolvent
has the same asymptotics as in Theorem 1.1.
of$L^{-1}$exists and the integral kernel$L^{-1}$$G$

Next assume that
. Then, by Theorem 2in [Pinsl], is critical if$d\geq3$
and subcritical if. Our second main theorem is the following.
$\sup_{\beta}\Lambda(i\beta)=0$$d\leq 2$

$L$

105
Theorem 1.3. Let $d\geq 3$ . Assume that
. Then the minimal Green function$|x-y|arrow\infty$ :. Put
has the following asymptotics as$\sup_{\beta}\Lambda(i\beta)=\Lambda(i\beta_{0})=0$$\Lambda(i\beta_{0})$

of

$G$

$L$

$H=-\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\beta$$G(x, y)= \frac{\Gamma(\frac{d-2}{2})}{2\pi^{d/2}(\det H)^{1/2}}\frac{e^{-(x-y)\cdot\beta_{0}}}{|H^{-1/2}(x-y)|^{d-2}}\frac{u_{\beta_{0}}(x)v_{\beta 0}(y)}{(u\rho_{0},v_{\beta_{0}})}(1+O(|x-y|^{-1}))$

,

(1.2)

where the term $O(|x-y|^{-1})$ satisfies that $|O(|x-y|^{-1})|\leq C|x-y|^{-1}$ for $|x-y|>R$
and $R$ independent of , .
with positive constants
$C$

$x$

$y$

Here, by applying Theorem 1.1, we explicitly determine the Martin boundary of. As for the definition and basic properties of Martin
for in the case
. Then the folowingi. Fix areference point
boundary, see [M] and1.1.
proposition is adirect consequence of Theorem$\mathrm{R}^{d}$$L$$\sup_{\beta}\Lambda(i\beta)>0$$\mathrm{R}^{d}$$x_{0}$$[\mathrm{P}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{l},2]$

Proposition 1.4. Assume thatsuch thatfor any sequence$\sup_{\beta}\Lambda(i\beta)>0$$\{y_{n}\}$$in$$\mathrm{R}^{d}$$\mathrm{n}$. Then the Green function satisfies thatand $y_{n}/|y_{n}|arrow\nu$ ,$|y_{n}|arrow\infty$,$\lim_{narrow\infty}\frac{G(x,y_{n})}{G(x_{0},y_{n})}=e^{-(x-x_{\mathrm{O}})\cdot\beta_{-\nu}}\frac{u_{\beta_{-\nu}}(x)}{u_{\beta_{-\nu}}(x_{0})}$$x\in \mathrm{R}^{d}$

.

(1.3)

(1.3) was conjectured by Pinchover, as was mentioned in p.90 of [Pinsl]. Denote by$K(x, \nu)$ the right hand side of (1.3). Then I\’u
, $K(x_{0}, \nu)=1$ , and, is minimal,. Furthermore, it is well-known that for any
ifthen $=CK(\cdot$ , for some positiveon, i.e., if
satisfies
in
$C$ .
as
for
Hence we can explicitly determine the Martin boundary of
constant
follows$K(\cdot, \nu)\neq$$(\cdot, \nu)\in C_{L}$$\nu\in \mathrm{S}^{d-1}$$\nu\neq\mu$$K(\cdot, \mu)$$\psi\in C_{L}$$C_{L}$$\psi(\cdot)\leq \mathrm{L}(\mathrm{k})\nu)$$\mathrm{R}^{d}$$K(\cdot$$\nu)$$\nu)$$\psi$$\mathrm{R}^{d}$$L$. Then the Martin boundary and the miniTheorem 1.5. Suppose thatat infinity which ismal Martin boundary of
for are both equal to the surface;and the Martin
is equal tohomeomorphic to
the Martin kernel at
equipped withcompactification offor is equal to
the standard topology.
and $d\geq 3$ , we obtain the following propsition andIn the case where
theorem. These results, however, are also simple consequences of the known results
thatis one dimensional in this case.
$\sup_{\beta}\Lambda(i\beta)>0$$\mathrm{R}^{d}$$\mathrm{S}^{d-1}$$L$$\nu\in \mathrm{S}^{d-1}$$\Gamma_{0i}$$\mathrm{R}^{d}$$K(\cdot, \nu)$$\{x\in \mathrm{R}^{d}; |x|<1\}\cup[1, \infty]\cross \mathrm{S}^{d-1}$$L$$\sup_{\beta}\Lambda(i\beta)=0$$C_{L}$

Proposition 1.6. Let $d\geq 3$ . Assume that$in$such thatsequence$\{y_{n}\}$$\mathrm{R}^{d}$$\sup_{\beta}\Lambda(i\beta)=\Lambda(i\beta_{0})=0$. Then for any$|y_{n}|arrow\infty_{f}$$\lim_{narrow\infty}\frac{G(x,y_{n})}{G(x_{0},y_{n})}=e^{-(x-x_{\mathrm{O}})\cdot\beta 0}\frac{u_{\beta 0}(x)}{u\rho_{0}(x_{0})}$,$x\in \mathrm{R}^{d}$

.

(1.4)

and $d\geq 3$ . Then the Martin boundary andTheorem 1.7. Suppose that
the Martin
the minimal Martin boundary of
for are both equal to one pointcompactification
kernel at oo is equal to the right hand side of (1.4); and the Martinof.
compactification
point
one
is
to
the
equal
of
for
$\sup_{\beta}\Lambda(i\beta)=0$$\mathrm{R}^{d}$$\mathrm{R}^{d}$$L$$\infty,\cdot$$\mathrm{R}^{d}\cup\{\infty\}$$L$$\mathrm{R}^{d}$In the rest of the paper we prove Theorems 1.1, 1.2 and 1.3. In \S 2, we study thespectra of $L(k)$ and , and give an integral expression of the resolvent of in term
$L$

$L$

106
of the resolvent of $L(k)$ . At the end of the section, we prove Theorem 1.1 under theassumption that Theorem 1.2 is true. In \S 3, we analyse the set of zeros of Aandan asymptotics of $L(k)^{-1}$ near the zero set. Furthermore, we present asaddle point
method, which is abasic tool in obtaining the asymptotics of the Green function. In\S 4, using results in \S 2 and \S 3, we show Theorem 1.2. Finally, Theorem 1.3 is proved in\S 5.

2. JNTEGRAL EXPRESSION

In the following, $L(k)$ and are regarded as closed operators on
andwith domains, respectively. For an operator , we denote byandandthe spectrum and the resolvent set of , respectively. We first study thespectrum of $L(k)$ .
$L$$H^{2}(\mathrm{T}^{d})$$L^{2}(\mathrm{T}^{d})$$\rho(T)$$\sigma(T)$$T$

Proposition 2.1. Let ,In particular, for any k$\alpha$$\beta\in \mathrm{R}^{d}$$\in \mathrm{R}^{d}$

, {A

and A

$\in \mathrm{C}$with$\mathrm{A}(\mathrm{i}\mathrm{p})>{\rm Re}$$\in \mathrm{C};{\rm Re}\lambda<\Lambda(0)$Proof. We have only to show that if$u\equiv 0$$L^{2}(\mathrm{R}^{d})$$T$$H^{2}(\mathrm{R}^{d})$}$\subset\rho(L(k))$$u\in H^{2}(\mathrm{T}^{d})$. Using Kato’s inequality$\nabla\cdot$

A. Thensatisfies$\lambda\in\rho(L(\alpha+i\beta))$.

.

$\mathrm{L}(\mathrm{a}+\mathrm{i}\mathrm{p})\mathrm{u}=\lambda u,\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$$a(x)\nabla|u|\geq{\rm Re}[(\mathrm{s}\mathrm{g}\mathrm{n}\overline{u})(\nabla+i\alpha)\cdot \mathrm{a}(\mathrm{x})(\mathrm{V}+\mathrm{i}\mathrm{a})\mathrm{u}]$in the sense of distributions (see Lemma Ain [Ka]), we have$L(\dot{\iota}\beta)|u|=[-\nabla\cdot a(x)\nabla+\nabla\cdot a(x)\beta+\beta\cdot a(x)\nabla-\beta\cdot a(x)\beta-b(x)\cdot(\nabla-\beta)+c(x)]|u|$$\leq{\rm Re}[-(\mathrm{s}\mathrm{g}\mathrm{n}\overline{u})(\nabla+i\alpha)\cdot \mathrm{a}(\mathrm{x})(\mathrm{V}+\mathrm{i}\mathrm{a})\mathrm{u}]$$+[\nabla\cdot a(x)\beta+\beta\cdot a(x)\nabla-\beta\cdot a(x)\beta-b(x)\cdot(\nabla-\beta)+c(x)]|u|$$={\rm Re}[(\mathrm{s}\mathrm{g}\mathrm{n}\overline{u})[-(\nabla+i\alpha)\cdot a(x)\beta-\beta\cdot a(x)(\nabla+i\alpha)+\beta\cdot a(x)\beta$$+b(x)\cdot(\nabla+i(\alpha+i\beta))-c(x)+\lambda]u]$$+[\nabla\cdot a(x)\beta+\beta\cdot a(x)\nabla-\beta\cdot a(x)\beta-b(x)\cdot(\nabla-\beta)+c(x)]|u|$$={\rm Re}\lambda|u|$Let$\psi$.

(2.1)$>0$ be an eigenfunction to $L(\beta)^{*}\psi=\Lambda(:\beta)\psi$. Then by (2.1), we have${\rm Re} \lambda\int_{\mathrm{T}^{d}}|u|\psi\geq\int_{\mathrm{T}^{d}}L(i\beta)|u|\psi=\Lambda(i\beta)\int_{\mathrm{T}^{d}}|u|\psi$

This shows$u\equiv 0$

by the assumption

Proposition 2.2. Let$\Lambda(i\beta)>{\rm Re}$$\alpha\in \mathrm{R}^{d}\backslash (2\pi \mathrm{Z})^{d}$

Proof. We have only to show that if

and

A.

$\beta\in \mathrm{R}^{d}$.

$\square$. Then$\Lambda(i\beta)\in\rho(L(\alpha+i\beta))$.

satisfies $L(\alpha+i\beta)u=\Lambda(i\beta)u$ , then. First we show that $L(i\beta)|u|=\Lambda(i\beta)|u|$ . As in the proof of Proposition 2.1,
by Kato’s inequality, we havefor any.
$u\in H^{2}(\mathrm{T}^{d})$$u\equiv 0$$\int_{\mathrm{T}^{d}}(L(i\beta)-\Lambda(i\beta))|u|\varphi\leq 0$$0\leq\varphi\in C_{0}^{\infty}(\mathrm{T}^{d})$107
Suppose that there exists
an eigenfunction to$\varphi_{0}\geq 0$such thatand take$\int_{\mathrm{T}^{d}}(L(i\beta)-\Lambda(i\beta))|u|\varphi_{0}<0$$L(i\beta)^{*}\psi=\Lambda(i\beta)\psi$$\epsilon$$>0$ such that. Let

$0\leq\epsilon\varphi_{0}<\psi$$>0$ be$\psi$. Then$\Lambda(i\beta)\int_{\mathrm{T}^{d}}|u|\psi=\int_{\mathrm{T}^{d}}L(i\beta)|u|\psi=\int_{\mathrm{T}^{d}}L(i\beta)|u|\epsilon\varphi_{0}+\int_{\mathrm{T}^{d}}L(i\beta)|u|(\psi-\epsilon\varphi_{0})$$< \Lambda(i\beta)\int_{\mathrm{T}^{d}}|u|\psi$.

. Therefore
for any
This is acontradiction. Hence,$L(i\beta)|u|=\Lambda(i\beta)|u|$ . This implies that either $|u|>0$ or.
shows that
calculation. Suppose that $|u|>0$ . Then adirect
Next we show$\varphi\geq 0$$\int_{\mathrm{T}^{d}}(L(i\beta)-\Lambda(i\beta))|u|\varphi=0$$u\equiv 0$$u\equiv 0$$(L(i \beta)-\Lambda(i\beta))|u|=-|u|(\frac{{\rm Im}(\overline{u}\nabla u)}{|u|^{2}}+\alpha)\cdot a(x)(\frac{{\rm Im}(\overline{u}\nabla u)}{|u|^{2}}+\alpha)$(cf. the proof of Theorem 3.1 in [Pinsl]). Since $L(i\beta)|u|=\Lambda(i\beta)|u|$ ,,. Since. Put $v=u/|u|$ . Then we have;;and so $\nabla v+iv\alpha=0$ . This implies that
. Thus,$u=c|u|e^{-i\alpha\cdot x}$ . But since,and sofor some constant . Henceis not periodic. This is acontradiction.
Next we study the spectrum of , and give an integral expression of the resolvent of-valued functions on. Let 7{ be an -space of. Letwith measure $(2\pi)^{-d}dk$ :$|u|^{-2}{\rm Im}(\overline{u}\nabla u)+\alpha=$$v\overline{v}=1$${\rm Im}(\overline{v}\nabla v)=|u|^{-2}{\rm Im}(\overline{u}\nabla u)=-\alpha$$0$$\nabla(ve^{i\alpha\cdot x})=0$$\mathrm{v}\mathrm{V}\mathrm{v}=-i\alpha$${\rm Re}(\overline{v}\nabla v)=0$$ve^{i\alpha\cdot x}=c$$\alpha\in \mathrm{R}^{d}\backslash (2\pi \mathrm{Z})^{d}$$\mathrm{c}$$\square$

$u$

$L$

$L$

$L^{2}$$2\pi \mathrm{T}^{d}=\mathrm{R}^{d}/(2\pi \mathrm{Z})^{d}$$2\pi \mathrm{T}^{d}$$L^{2}(\mathrm{T}^{d})$$\mathcal{H}=L^{2}(2\pi \mathrm{T}^{d}, \frac{dk}{(2\pi)^{d}};L^{2}(\mathrm{T}^{d}))=\int_{2\pi \mathrm{T}^{d}}^{\oplus}L^{2}(\mathrm{T}^{d})\frac{dk}{(2\pi)^{d}}$Define an operator$\mathcal{F}$from$L^{2}(\mathrm{R}^{d})$(Ff)(k, )to 7{ by

$x$

Then

$\mathcal{F}$$= \sum_{l\in \mathrm{Z}^{d}}f(x-l)e^{-i(x-l)\cdot k}$

is aunitary operator, and the adjoint$\mathcal{F}^{*}$.

is given by, for,$( \mathcal{F}^{*}g)(x-l)=\int_{2\pi \mathrm{T}^{d}}\frac{dk}{(2\pi)^{d}}e^{i(x-l)\cdot k}g(k, x)$$x\in \mathrm{T}^{d}$(see Lemma on p.289 of [RS] or Theorem 2.2.5 in [Ku]). For$(\nabla_{x}+ik)\mathcal{F}f=\mathcal{F}(\nabla f)$Letdomain$\tilde{L}=\int_{2\pi \mathrm{T}^{d}}^{\oplus}L(k)\frac{dk}{(2\pi)^{d}}$$D(\tilde{L})=${$g\in \mathcal{H}$,,$l\in \mathrm{Z}^{d}$$f\in H^{1}(\mathrm{R}^{d})$, we have

.

(2.2)

be an operator on $H$ defined by $(\tilde{L}g)(k)=L(k)g(k)$ with.

$g\in \mathcal{H};g(k)\in D(L(k))=H^{2}(\mathrm{T}^{d})\mathrm{a}.\mathrm{e}$$k$

and $L(k)g(k)\in \mathcal{H}$ }.

Since $L(k)$ is closed, is closed. Clearly,. Then we see thatshow the opposite inclusion. Let$\tilde{L}$.

. Let us$D(\tilde{L})\supset L^{2}(2\pi \mathrm{T}^{d}, (2\pi)^{-d}dk;H^{2}(\mathrm{T}^{d}))$$g\in D(\tilde{L})$$g$is ameasurable squar

108
integrable-valued function. In fact, the measurablity follows from, andthe square integrabhty folows from $||g(k)||_{H^{2}}\leq c(||L(k)g(k)||_{L^{2}}+||g(k)||_{L^{2}})$ . By (2.2),
we have.
(2.3)
Let. By Proposition 2.1, we see that $(L(k)-\lambda)^{-1}$ is areal analytic
function fromto the Banach space of bounded operators on. Thus,by Theorem XIII.83 in [RS], we can define abounded operator $M$ on
by $M=$$H^{2}(\mathrm{T}^{d})$$g\in \mathcal{H}$$\mathcal{F}L=\tilde{L}\mathcal{F}$${\rm Re}\lambda<\Lambda(0)$$2\pi \mathrm{T}^{d}$$L^{2}(\mathrm{T}^{d})$$\mathcal{H}$$\int_{2\pi \mathrm{T}^{d}}^{\oplus}(L(k)-\lambda)^{-1}\frac{dk}{(2\pi)^{d}}$.

Proposition 2.3. Let Rel
and f
for any x$\lambda<\Lambda(0)$$\in \mathrm{T}^{d},$$\in \mathrm{Z}^{d}$. Then$\in L^{2}(\mathrm{R}^{d})$$\lambda\in\rho(L)$,and (L$(L- \lambda)^{-1}f(x-l)=\int_{2\pi \mathrm{T}^{d}}F(k)\frac{dk}{(2\pi)^{d}}$, :. e.,$-\lambda)^{-1}=\mathcal{F}^{*}M\mathcal{F}$

,

(2.4)

where

.

(2.5)

$F(k)=e- \iota)\cdot k(:\mathrm{t}xL(k)-\lambda)^{-1}(\sum_{m\in \mathrm{Z}^{d}}f(\cdot-m)e^{-:(\cdot-m)\cdot k})(x)$

Proof. For any, put $g=Mf\in ll$ . Then $g(k)=(L(k)-\lambda)^{-1}f(k)$ for.
$(\tilde{L}-\lambda)g=f$; hence. This implies
that $M$ is aright inverse of –A. For any $g\in D(\tilde{L})$ , put $f=(\tilde{L}-\lambda)g$ . Then$f(k)=(L(k)-\lambda)g(k)$ for. . Thus $(L(k)-\lambda)^{-1}f(k)=g(k)$ and,,$Mf=g$ . This implies that $M$ is aleft inverse of -A. Hence $(\tilde{L}-\lambda)^{-1}=M$ . By theunitary equivalence (2.3) of and , we have that
and
.
$k$

$f\in \mathcal{H}$$\mathrm{a}.\mathrm{e}$. Thus $(L(k)-\lambda)g(k)=f(k)$ and$g(k)\in H^{2}(\mathrm{T}^{d})$$\tilde{L}$$\mathrm{a}.\mathrm{e}$$k$$f(k)\in \mathcal{H}$$\mathrm{i}.\mathrm{e}.$$\tilde{L}$$L$$\tilde{L}$$\lambda\in\rho(L)$$(L-\lambda)^{-1}=\mathcal{F}^{*}M\mathcal{F}$$\square$Lemma 2.4. The spectr um of $L(k)$ and $L(k+2\pi z)$ coincide for eacha $nd$. If $(L(k)-\lambda)^{-1}$ exists forand, the $F(k)=F(k+2\pi z)$ for any$k\in \mathrm{C}^{d}$$z\in \mathrm{Z}^{d}$$z\in \mathrm{Z}^{d}$$\lambda\in \mathrm{C}$$k\in \mathrm{C}^{d}$$n$.

Proof The first claim clearly holds. Let us show that the second. Note$e^{i2\pi z\cdot x}(L(k+2\pi z)-\lambda)^{-1}=(L(k)-\lambda)_{C}^{-1:2\pi z\cdot x}$.

Then we have$F(k+2 \pi z)=e-l)\cdot ke(:(x\dot{\iota}2\pi z\cdot xL(k+2\pi z)-\lambda)^{-1}(\sum_{m\in \mathrm{Z}^{d}}f(\cdot-m)e^{-:(\cdot-m)\cdot(k+2\pi z)})(x)$$=e-l) \cdot k(:(xL(k)-\lambda)^{-1}(e\sum_{m\in \mathrm{Z}^{d}}:2\pi z\cdot(\cdot)f(\cdot-m)e^{-:(\cdot-m)\cdot(k+2\pi z)})(x)=F(k)$.

Cl

We close this section by showing that Theorem 1.1 follows from Theorem 1.2.
Proof of Theorem 1.1. Suppose that Theorem 1.2 holds. Assume that the operator
satisfies $\sup_{\beta}\Lambda(i\beta)>0$ . Choosesthat $\Lambda(i\beta_{0})>0$ , and consider
the operator. Then the principal eigenvalueof $L_{1}(i\beta)=$
is equal to, and so $\Lambda_{1}(0)>0$ . By Proposition 2.3,
$L$$\beta_{0}\in \mathrm{R}^{d}$$L_{1}=e^{\beta_{0}\cdot x}Le^{-\beta_{0}\cdot x}$$e^{\beta\cdot x}L_{1}e^{-\beta\cdot x}$$\Lambda(i\beta+i\beta_{0})$$\mathrm{u}\mathrm{c}\mathrm{h}$$\Lambda_{1}(i\beta)$$\inf{\rm Re} \mathrm{a}(\mathrm{L}\mathrm{i})\geq$109
is the integralof. Thus $\inf{\rm Re}\sigma(L_{1})>0$ . Since the minimal Green function
has the(cf. Theorem 2.3 in [M]), the Green function
kernel of the resolvent
same asymptotics as in Theorem 1.1. On the other hand, the minimal Green function”. Thus we obtain the asymptotics of inof satisfies
Theorem 1.1.
$L_{1}$$G_{1}$$\Lambda_{1}(0)$$G_{1}$$L_{1}^{-1}$

$G$

$G$

$G_{1}(x, y)=e^{\beta_{0}\cdot x}G(x, y)e^{-\beta_{0}}$$L$$\square$

3. ANALYSIS OF$\Lambda(k)$AND $L(k)^{-1}$

be the vector defined, let
In this section we assume $\Lambda(0)>0$ . For-valued smooth. We see that is smooth in . Choos
in \S 1. Put$\{e_{s,1},\ldots, e_{s,d-1}, s\}$$e_{s}=(e_{s,1},\ldots,e_{s,d-1})$,such that for any
onfunction
is nondegenerate,. Since the principal eigenvalue
is an orthonormal basis of
has an analytic continuation
the analytic perturbation theory shows that$N$
, which is also anondegenerate eigenvalue of $L(k)$ for anyof
to aneighborhood$k\in N$ (cf. Theorem XII.8 in [RS]). We introduce new coordinates $(w, z)$ near
such
that
$s$

$\eta_{s}=\beta_{s}/|\beta_{s}|$$\in \mathrm{R}^{d}$$\beta_{s}\in\Gamma_{0}$$\mathrm{R}^{d(d-1)}$$s$$\eta_{s}$$\mathrm{e}$$s\in \mathrm{S}^{d-1}$$\mathrm{S}^{d-1}$$\mathrm{R}^{d}$$\Lambda(i\beta)$$\Lambda(k)$$\Lambda(i\beta)$$i\mathrm{R}_{\beta}^{d}$$i\beta_{s}$$k=w \eta_{s}+z\cdot e_{s}=w\eta_{s}+\sum_{j=1}^{d-1}z_{j}e_{s,j}$,$w\in \mathrm{C}$,$z=(z_{1}, \ldots, z_{d-1})\in \mathrm{C}^{d-1}$.

We write $\Lambda_{s}(w, z)=\Lambda(w\eta_{s}+z\cdot e_{s})$ .
Lemma 3.1. There exist $R>0$ and afunction $w(s, z)$ of $(s, z)\in D=\mathrm{S}^{d-1}\cross\{z\in$; $|z|<R$ } such that $w(s, z)\eta_{s}+z\cdot e_{s}\in N$ for $(s, z)\in D$ , $w(s, 0)=i|\beta_{s}|$ andis holomorphic in $z\in\{z\in$, $z)=0$ on D. For each
; $|z|<R$ }.
$C^{\infty}$$\mathrm{C}^{d-1}$$s\in \mathrm{S}_{f}^{d-1}\mathrm{w}(\mathrm{s}, z)$$\Lambda_{s}(w(s, z)$$\mathrm{C}^{d-1}$

Proof. Note that$\Lambda_{s}(i|\beta_{s}|, 0)=\Lambda(i\beta_{s})=0$

(3.1)

.

It follows from the assumption $\Lambda(0)>0$ that $s\cdot\beta_{s}>0$ andsome $c>0$ . Thus$\nabla_{\beta}\Lambda(i\beta)|_{\beta=\beta_{*}}=-cs$

for. By the implicit function theorem,onand aunique smooth function
there exist
for each$z)=0$
andsuch thatsuch, we can choose afinite number of. By the compactness ofonholomorphic
in
is
.
Since
. Put
that
$(w, z)$ , it follows from the implicit function theorem for holomorhic functions thaton $(\{|s-sj|<R_{s_{\mathrm{j}}}\}\cap\{|s-$are holomorphic on $\{|z|<R\}$ . Thus$s_{k}|<R_{s_{k}}\})\cross\{|z|<R\}$ .
So we obtain adesired function $w(s, z)$ on $D$ by taking$w(s, z)=w_{s_{\mathrm{j}}}(s, z)$ on. The last claim has been shown already.
$s_{0}\in \mathrm{S}^{d-1}$$i \frac{\partial\Lambda_{s}}{\partial w}(i|\beta_{s}|, 0)=\nabla_{\beta}\Lambda(i\beta_{s})\cdot\eta_{s}<0$$w_{s_{\mathrm{O}}}(s, z)$$R_{s_{\mathrm{O}}}>0$$\{|s-s_{0}|<R_{s_{\mathrm{O}}}\}\cross\{|z|<R_{s_{\mathrm{O}}}\}$$w_{s_{\mathrm{O}}}(s_{0},0)=i|\beta_{s_{\mathrm{O}}}|$$\{sj\}$$\mathrm{S}^{d-1}$$D_{s\mathrm{o}}$$\mathrm{S}^{d-1}\cross\{z=0\}\subset\cup jDs_{j}$$D_{s_{\mathrm{O}}}=$$\Lambda_{s}(w_{s_{\mathrm{O}}}(s, z),$$\Lambda_{s}(w, z)$$R=\mathrm{m}\mathrm{i}\mathrm{n}jRs_{j}$$w_{s_{j}}$$w_{s_{\mathrm{j}}}(s, z)=w_{s_{k}}(s, z)$$\square$$D\cap D_{s_{j}}$Write $w_{s}(z)=w(s, z)$ . Since$\frac{\partial\Lambda_{s}}{\partial z_{j}}(i|\beta_{s}|, 0)=0,1\leq j\leq d-1$$\frac{\partial w_{s}}{\partial z_{j}}(0)=0$,$1\leq j\leq d-1$

, we see that

,

and
,

$\frac{\partial^{2}w_{s}}{\partial z_{j}\partial z_{k}}(0)=-(\frac{\partial\Lambda_{s}}{\partial w}(i|\beta_{s}|, 0))^{-1}e_{s,j}\cdot(\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{k}\Lambda)(i\beta_{s})e_{s,k}$$1\leq j$, $k\leq d-1$ ,110where HessjtA. Note that$=( \frac{\partial^{2}\Lambda}{\partial k_{m}\partial k_{n}})_{1\leq m,n\leq d}$,$\frac{\partial\Lambda_{s}}{\partial w}(i|\beta_{s}|,0)=\eta_{s}\cdot(\nabla_{k}\Lambda)(i\beta_{s})=(-i)\eta_{s}\cdot\nabla_{\beta}\Lambda(i\beta)|_{\beta=\beta}.$$\eta_{s}\cdot\nabla\rho\Lambda(:\beta)|_{\beta=\beta}$.

$<0$ ,. is positive definite.

$(\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{k}\Lambda)(i\beta_{s})=-\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\rho\Lambda(i\beta)|\rho_{=}\rho$

Hence we have the folowing.
Lemma 3.2. For every $1\leq j$ , $k\leq d-1$ ,. (3.2)$\frac{\partial^{2}w_{s}}{\partial z_{j}\partial z_{k}}(0)=.\frac{\partial^{2}{\rm Im} w_{s}}{\partial z_{j}\partial z_{k}}(0)=i(\eta_{s}\cdot\nabla_{\beta}\Lambda(i\beta)|_{\beta=\beta}.)^{-1}e_{s},\cdot {}_{j}\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\rho \mathrm{A}(:\beta)|_{\beta=\beta}.e_{s,k}$

Furthe rmore, the matrix
and there exist$\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}{\rm Im} w_{s}(0)$$p$$\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}{\rm Im} w_{s}(0)=(\frac{\partial^{2}{\rm Im} w_{s}}{\partial z_{j}\partial z_{k}}(0))_{1\leq j,k\leq d-1}$, $p’>0$ independent ofsatisfies$p\leq\lambda_{s}\leq p’$$s\in \mathrm{S}^{d-1}$.

is positive definite,such that any eigenvalue$\lambda_{s}$ofProof. We have only to note that the upper and lower estimates of the eigenvalues

follow from the positivity and the continuity ofin.
In the folowing lemma we take afamily of solutions to $L(w_{s}(z)\eta_{s}+z\cdot e_{s})u=0$
depending on parameters $(s, z)$ .
$\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}{\rm Im} w_{s}(0)$$s\in \mathrm{S}^{d-1}$$\square$Lemma 3.3. There exists $>0$ such that $u_{s,z}(x)$ in (8.3), $(s, z,x)\in \mathrm{S}^{d-1}\cross\{z\in$; $|z|<r$ }, is a non-zero-solution to $L(w_{s}(z)\eta_{s}+z\cdot e_{s})u=0$ . For
thermore, it is continuous in $(s, z)\in \mathrm{S}^{d-1}\cross\{z\in \mathrm{C}^{d-1}; |z|<r\}$ and holomorphic in$z\in\{z\in \mathrm{C}^{d-1} ; |z|<r\}$as a-valued function. In particular,
for fixed
it follows that for any multiindex ,with a constant $C_{\gamma}>0$$r$$\mathrm{C}^{d-1}$$\cross \mathrm{T}^{d}$$C^{2,\alpha}$$s\in \mathrm{S}^{d-1}$$\gamma$independent of$s\in \mathrm{S}^{d-1}$

$r\}$

and

$\overline{v_{s,z}}$$||\partial_{z}^{\gamma}u_{s,z}||_{C^{\alpha}(\mathrm{T}^{d})}\leq C_{\gamma}$.

The proof is ommited.
$L(w_{s}(z)\eta_{s}+z\cdot es)^{*}v$$C^{\alpha}$

Similarly, we can take anon-zero$C^{2,\alpha}$-solution$v_{s,z}$to$=0$ such that $v_{s,z}$ is continuous in $(s, z)\in \mathrm{S}^{d-1}\cross\{z\in \mathrm{C}^{d-1}$ ; $|z|<$is holomorphic in$z$

for fixed

$s$

as a

$C^{\alpha}$-valued function.

Proposition 3.4. There exists $>0$ such that for eachwith $|\alpha|<r$ the inverse
has a simple poleand has the following asymptotics at the pole$s\in \mathrm{S}^{d-1}$$r$$L(w\eta_{s}+\alpha\cdot e_{s})^{-1}$$L(w \eta_{s}+\alpha\cdot e_{s})^{-1}\sim\frac{A_{s,\alpha}}{w-w_{s}(\alpha)}$$w_{s}(\alpha)$

and eachas a function of ,$\alpha\in \mathrm{R}^{d-1}$

$w$

,

where
$i(\cdot, v_{s,\alpha})u_{s,\alpha}$$A_{s,\alpha}=\overline{(}$,$\eta_{s}\cdot[2a(\nabla+i(w_{s}(\alpha)\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha}$$v_{s,\alpha})$.

111
In $particular_{f}$ for $\alpha=0$ we have$A_{s,0}= \frac{i(\cdot,v_{s,0})u_{s,0}}{\eta_{s}\cdot\nabla_{\beta}\Lambda(i\beta)|_{\beta=\beta_{s}}(u_{s,0},v_{s,0})}$

.

(3.4)

. We write $L(w\eta_{s}+\alpha\cdot e_{s})^{-1}=(1-K(w))^{-1}L(0)^{-1}$ with aShatten-von$Pro\mathrm{o}/$Neumann class holomorphic operator$K(w)=L$ (0)$(L(0)-L(w\eta_{s}+\alpha\cdot e_{s}))$.

Here the existence of $L(0)^{-1}$ follows from the assumption $\Lambda(0)>0$ and Proposition 2.1.
Put $w_{0}=w_{s}(\alpha)$ . By the Fredholm theory (cf. Theorem VI.14 in [RS]), we can assume
that $(1-K(w))^{-1}$ has the following form
(3.5)$(1-K(w))^{-1}= \frac{A_{n}}{(w-w_{0})^{n}}+\cdots+\frac{A_{1}}{(w-w_{0})^{1}}+r(w)$with some $n\geq 1$ , finite rank operator Aj, $1\leq j\leq n$ , and holomorphic $r(w)$ . From a
relation$1=(1- \mathrm{K}(\mathrm{w}))" 1-K(w))^{-1}=(1-K(w))[\frac{A_{n}}{(w-w_{0})^{n}}+\cdots+\frac{A_{1}}{(w-w_{0})^{1}}+r(w)]$,

we have $(w-w_{0})^{n}=(1-K(w))A_{n}+O((w-\mathrm{w}\mathrm{o})\mathrm{n}$ , hence$(1-K(w_{0}))A_{n}=0$ .

(3.6)

Similarly, $A_{n}(1-K(w_{0}))=0$ . These imply that$L$(

$w_{0}\eta_{s}+\alpha\cdot$es)An $=0$ ,From these, since the kernels ofmust be of the form:.

$L(w_{0}\eta_{s}+\alpha\cdot e_{s})^{*}(L(0)^{*})^{-1}A_{n}^{*}=0$$L(w_{0}\eta_{s}+\alpha\cdot e_{s})$and$L(w_{0}\eta_{s}+\alpha\cdot e_{s})^{*}$

are one dimensional,$A_{n}$(3.7)$A_{n}=c(\cdot, L(0)^{*}v_{s,\alpha})u_{s,\alpha}$with some constant . Here note that$c$

$L(0)^{*}v_{s,\alpha}\neq 0$. Furthermore, we have by (3.6)$A_{n}+(1-K(w))^{-1}(K(w)-K(w_{0}))A_{n}=0$

and by the definition of $K(w)$$K(w)-K(w_{0})=(w-w_{0})L(0)^{-1}$$\cross[i(\eta_{s}\cdot a(\nabla+i\alpha\cdot e_{s})+(\nabla+i\alpha\cdot e_{s})\cdot a\eta_{s}+b\cdot\eta_{s})-(w+w_{0})\eta_{s}\cdot$$a\eta_{s}]$.

From these and (3.5), it follows that
$A_{n}L(0)^{-1}i\eta_{s}\cdot$$[2a(\nabla+i(w_{0}\eta_{s}+\alpha\cdot \mathrm{e},))+\nabla\cdot a+b]A_{n}=0$,$n\geq 2$$A_{n}+A_{n}L(0)^{-1}i\eta_{s}\cdot$ $[2a(\nabla+i(w_{0}\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]A_{n}=0$,,$n=1$ .

(3.8)

112
First consider the case $n\geq 2$ . By (3.7),$c^{2}(\cdot,L(0)^{*}v_{s,\alpha})(\eta_{s}\cdot[2a(\nabla+i(w_{0}\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha},v_{s,\alpha})u_{s,\alpha}=0$

Let us show the factor
for smal. When $\alpha=0$ , since $w_{0}=\mathrm{A}(0)=i|\beta_{s}|$ , we have$(\eta_{s}\cdot[2a(\nabla+i(w_{0}\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha}, v_{s,\alpha})$.

is non-vanishing$\alpha$$(\eta_{s}\cdot[2a(\nabla+i(w_{0}\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha}, v_{s,\alpha})$$=\eta_{s}\cdot([2a(\nabla-\beta_{s})+\nabla\cdot a+b]u_{s,0},$ $v_{s,0})$.

$=\eta_{s}\cdot\nabla_{\beta}\Lambda(i\beta)|\rho=\rho.(u_{s,0}, v_{s,0})\neq 0$

Here, in the second equality, we have used Theorem
in [Pinsl]. Hence because of
the continuity in of the quantity, the conclusion holds. Thus the constant in (3.7)
must be zero if $n\geq 2$ , so we have $n=1$ in (3.5). By (3.7) and (3.9) it folows that$5(\mathrm{i}\mathrm{i})$$\alpha$$c$$c=i(\eta_{s}\cdot[2a(\nabla+:(w_{0}\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha},$ $v_{s,\alpha})^{-1}$Hence we have by (3.7)$i(\cdot, v_{s,\alpha})u_{s,\alpha}$$A_{s,\alpha}=A_{1}L(0)^{-1}=\overline{(}$,$\eta_{s}\cdot[2a(\nabla+i(w_{0}\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha}$.

$v_{s,\alpha})$Thus we have shown the proposition.
We describe asaddle point method which we shall use in proving Theorem 1.2.
$\square$Proposition 3.5. Letwith $c>0$ , here
on a neighborhood of
$U$

$U$

$B_{c}$$\overline{U}$$\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\varphi(0)=\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}{\rm Re}\varphi(0)$such that
Then the as ymptotics

be an open neighborhood of the origin $in$
satisfying
is the ball $\{|x|<c\}$ . Let
and $a(x)$ be-functions
satisfying
and $||a||_{C^{6}(U)}\leq b_{2}$ . Assume thatand it is positive definite and satisfies that there exists $p>0$
a $nd$ ${\rm Re}(\varphi(x)-\varphi(0))\geq p|x|^{2}/4$ for $x\in U$ .
for$\mathrm{R}^{d}$$c\infty$$\varphi(x)$$||\varphi||_{C^{9}(U)}\leq b_{1}$$p|x|^{2}\leq x\cdot \mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\varphi(0)x$$x\in \mathrm{R}^{d}$$\int_{U}e^{-\lambda\varphi(x)}a(x)dx=(\frac{2\pi}{\lambda})^{d/2}\frac{e^{-\lambda\varphi(\mathrm{O})}}{(\det \mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\varphi(0))^{1/2}}(a(0)+O(\lambda^{-1}))$

holds, where the $tem$
dependent only on ,$O(\lambda^{-1})$

$C$

$c$

$b_{1}$$\overline{B_{c}}\subset$,$b_{2}$

satisfies
,

$p$

and

$|O(\lambda^{-1})|\leq C\lambda^{-1},$ $\lambda>1$
$d$

.

as

$\lambdaarrow\infty$, with a positive constant

The proof is omitted.
4. Proof OF THEOREM 1.2

By Proposition 2.3, the resolvent
exists. It remains to show (1.1) under theassumption $\Lambda(0)>0$ . Put $F_{0}(L)=\{k\in \mathrm{C}^{d};L(k)u=0$ , for some non-zero $u\in$which is called the Fermi variety. For
and $>0$ let
be an$L^{-1}$$H^{2}(\mathrm{T}^{d})\}$$s\in \mathrm{S}^{d-1}$$\delta$$U_{s,\delta}$

113
open neighborhood of the origin given by. We
can take $\delta>0$ so small that $F_{0}(L)\cap\{(-\pi, \pi)^{d}+\{i\eta_{s}t;0\leq t<|\beta_{s}|+2\delta\}\}$ consists
only of. In fact, suppose that for each integer $n\geq 1$$e_{s_{n}}\in F_{0}(L)\cap$there exist,
andsuch that$\{(-\pi, \pi)^{d}+\{i\eta_{s_{n}}t;0\leq t<|\beta_{s_{n}}|+1/n\}\}$ and. Then by Proposition 2.1,
we can take asubsequence of (an, $sn$ ,for) such that
$F_{0}(L)$some. Note thatis closed. Hence it follows that$e_{s}\in F_{0}(L)$ . So by Proposition 2.2, $x=0$ and $\alpha=0$ hold. But thiscontradicts to that $w=w_{s}(z)$ is the unique solution to $\Lambda(w\eta_{s}+z\cdot e_{s})=0$ near $s=s_{0}$ ,$z=0$ and. Furthermore, using Lemma 3.2, if necessary choose $>0$ sosmall that there exists $c>0$ independent of, wheresuch thatisthe ball $B_{c}=\{|\alpha|<c\}$ , andonwith some $p>0$ .
$P$Let be aprojection along onto the plane spanned by, i.e.
$t_{1}(\alpha)=\min\{t;t\eta_{s}+\alpha\cdot
e_{s}\in[-\pi, \pi]^{d}\}$$Q=P[-\pi,\pi]^{d}$, and letput
. For each
and $t_{2}( \alpha)=\max\{t;t\eta_{s}+\alpha\cdot e_{s}\in[-\pi, \pi]^{d}\}$ . We can write
as $[-\pi, \pi]^{d}=$$\{t\eta_{s}+\alpha\cdot e_{s};\alpha\in Q, t_{1}(\alpha)\leq t\leq t_{2}(\alpha)\}$ .
Let
andbe $(d-1)$ -dimensional
cubes given by $M_{j}=$and$\{(k_{1}, \ldots, k_{j-1}, -\pi, kj+1, \ldots, k_{d});-\pi\leq k_{i}\leq\pi, i\neq j\}$ for $1\leq j\leq d$ .
Take $Nj\in$$1\leq
j\leq
d$
$Q=P(
\bigcup_{j=1}^{d}Nj)$,, such that
and.
Then putting, we haveand$U_{s,\delta}=\{\alpha\in \mathrm{R}^{d-1} ; \mathrm{w}3(\mathrm{a})<|\beta_{s}|+\delta\}$$\{w_{s}(\alpha)\eta_{s}+\alpha\cdot e_{s} ; \alpha\in U_{s,2\delta}\}$$\alpha_{n}\in \mathrm{R}^{d-1}$$s_{n}\in \mathrm{S}^{d-1}$$w_{n}\in \mathrm{C}$$w_{n}\eta_{s_{n}}+\alpha_{n}\cdot$$w_{n}\neq w_{s_{n}}(\alpha_{n})$$(\alpha_{n}, s_{n}, w_{n})arrow(\alpha, s_{0}, x+i|\beta_{s}|)$$w_{n}$$(\alpha, s_{0}, x)\in \mathrm{R}^{d-1}\cross \mathrm{S}^{d-1}\cross \mathrm{R}$$(x+i|\beta_{s}|)\eta_{s}+\alpha\cdot$$\delta$$w=i|\beta_{s_{\mathrm{O}}}|$$s\in \mathrm{S}^{d-1}$$\overline{B}_{c}\subset \mathrm{U}\mathrm{s},\mathrm{s}$${\rm Im}(w_{s}(\alpha)-\mathrm{w}\mathrm{s}(0))\geq p|\alpha|^{2}$$U_{s,\delta}$$P:t\eta_{s}+\alpha\cdot e_{s}arrow$$\{e_{s}\}$$\eta_{s}$$\alpha\cdot$$B_{c}$$\alpha\in Q$$e_{s}$$[-\pi, \pi]^{d}$$M_{j}$$\tilde{M}_{j}$$\{(k_{1}, \ldots, \mathrm{R}\mathrm{d}-1, \pi, k_{j+1}, \ldots, \mathrm{k}\mathrm{d});-\pi\leq k_{i}\leq\pi, i\neq j\}$$\{Mj,\tilde{M}j\}$$\tilde{M}_{j}=$$\bigcup_{j=1}^{d}Nj=\{t_{1}(\alpha)\eta_{s}+\alpha\cdot e_{s};\alpha\in Q\}$$\tilde{N}j=(M_{j}\cup\tilde{M}_{j})\backslash N_{j}$$Q=P( \bigcup_{j=1}^{d}\tilde{N}_{j})$$\bigcup_{j=1}^{d}\tilde{N}j=\{t_{2}(\alpha)\eta_{s}+\alpha\cdot e_{s}; \alpha\in Q\}$.

Recalling the integral expression (2.4), we have by Lemma 2.4 that for any,
and$x\in \mathrm{T}^{d}$$l\in \mathrm{Z}^{d}$$(L^{-1}f)(x-l)=(2 \pi)^{-d}\int_{[-\pi,\pi]^{d}}F(k)dk$

with $F(k)$ in (2.5). We change the integral variables fromthat $k=t\eta_{s}+\alpha\cdot$ . By Fubini’s theorem, we have

$k$

to

$(t, \alpha)\in \mathrm{R}\cross \mathrm{R}^{d-1}$such$e_{s}$$(L^{-1}f)(x-l)= \frac{|D_{s}|}{(2\pi)^{d}}\int_{Q}$da$\int_{t_{1}(\alpha)}^{t_{2}(\alpha)}dtF(t\eta_{s}+\alpha\cdot e_{s})$

,

(4.1)

where $D_{s}=\det(\eta_{s}, e_{s,1}, \cdots, e_{s,d-1})$ . For each
let $C=C_{1}\cup C_{2}\cup C_{3}\cup C_{4}$ andbe closed contours in given by$\alpha\in Q$$\tilde{C}=C_{1}\cup\tilde{C}_{2}\cup\tilde{C}_{3}\cup\tilde{C}_{4}$$\mathrm{C}$, $C_{2}=\{t_{2}(\alpha)+it;t : 0arrow|\beta_{s}|+2\delta\}$ ,$C_{3}=$}, $C_{4}=\{t_{1}(\alpha)+it;t : |\beta_{s}|+2\deltaarrow 0\}$ ,{ $t+i(|\beta_{s}|+2\delta);t$ : ta (o)$C_{1}=\{t :t_{1}(\alpha)arrow t_{2}(\alpha)\}$$arrow t_{1}(\alpha)$$\tilde{C}_{2}=\{t_{2}(\alpha)+it;t :0arrow|\beta_{s}|+\delta/2\},\tilde{C}_{3}=$$\tilde{C}_{4}=\{t_{1}(\alpha)+it;t :|\beta_{s}|+\delta/2arrow 0\}$

By the argument above, fornear and inside , and for$C$$\alpha\in U_{s,\delta}${ $t+i(|\beta_{s}|+\delta/2);t$ : ta (o)$arrow t_{1}(\alpha)$},.

the integrand in (4.1) has only asimple pole
the integrand in (4.1) is holomorphic near and$\alpha\in Q\backslash \mathrm{U}\mathrm{s},\mathrm{s}$$w_{s}(\alpha)$

114
inside$\tilde{C}$. Hence, it follows from the residue theorem that$(L^{-1}f)(x-l)=I_{1}f+I_{2}f$ ,$I_{1}f= \frac{2\pi i|D_{s}|}{(2\pi)^{d}}\int_{U}\cdot,\delta$

da $\exp[i(x-l)\cdot(w_{s}(\alpha)\eta_{s}+\alpha\cdot e_{s})]$$\cross-$()$:( \sum_{m}f(\cdot-m)e^{-:\mathrm{t}\cdot-m)\cdot(w.(\alpha\rangle\eta.+\alpha\cdot e.)}, v_{s,\alpha})u_{s,\alpha}(x)$$\eta_{s}\cdot[2a(\nabla+:(w_{s}(\alpha)\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha}$,’$v_{s,\alpha}$$I_{2}f= \frac{|D_{s}|}{(2\pi)^{d}}(\int_{U}\cdot.\delta d\alpha\int_{C_{2}\cup C_{3}\cup C_{4}}dw+\int_{Q\backslash U..s}d\alpha\int_{\tilde{C}_{2}\cup\tilde{C}_{3}\cup\tilde{C}_{4}}dw)F(w\eta_{s}+\alpha\cdot e_{s})$By Fubini’s theorem, the integral kernel $I_{1}(x-l, y)$ ,$y\in \mathrm{R}^{d}$, of$I_{1}$.

is$I_{1}(x-l, y)=$,$\frac{-|D_{s}|}{(2\pi)^{d-1}}\int_{U..s}d\alpha\frac{\exp[i(x-l-y)\cdot(w_{s}(\alpha)\eta_{s}+\alpha\cdot e_{s})]\overline{v_{s,\alpha}(y)}u_{s,\alpha}(x)}{(\eta_{s}\cdot[2a(\nabla+(w_{s}(\alpha)\eta_{s}+\alpha\cdot e_{s}))+\nabla\cdot a+b]u_{s,\alpha},v_{s,\alpha})}$is regarded as a -periodic function in. Take $=(x-l-$$y)/|x-l-y|$ . Note that $(x-l-y)\cdot$
and $(x-l-y)\cdot$. In view of Lemma
3.2, we apply the saddle point method (Proposition 3.5) to obtain that $I_{1}(x-l, y)$ has
the asymptotics
where$\mathrm{Z}^{d}$$v_{s,\alpha}(y)$$C^{2,\alpha}(\mathrm{R}^{d})$$\eta_{s}>0$$s$$(\alpha\cdot e_{s})=0$.

$I_{1}(x-l, y)= \frac{-|D_{s}|}{(2\pi)^{d-1}}(\frac{2\pi}{(x-l-y)\cdot\eta_{s}})^{(d-1)/2}\frac{e^{-(x-l-y)\cdot\beta}}{(\det \mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}{\rm Im} w_{s}(0))^{1/2}}$$\cross(\frac{u_{s,0}(x)v_{s,\mathrm{O}}(y)}{\eta_{s}\cdot\nabla_{\beta}\Lambda(i\beta_{s})(u_{s,\mathrm{O}},v_{s,\mathrm{O}})}+O(|x-l-y|^{-1}))$$= \frac{|D_{s}|}{(2\pi)^{(d-1)/2}}\frac{(-\eta_{S}\cdot\nabla_{\beta}\Lambda(i\beta_{s}))^{(d-3)/2}}{(\det(-e_{s},{}_{j}\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{\beta}\Lambda(i\beta_{s})e_{s,k}))^{1/2}}.\frac{e^{-(x-l-y)\cdot\beta}\cdot u_{s,\mathrm{O}}(x)v_{s,0}(y)}{((x-l-y)\cdot\eta_{s})^{(d-1)/2}(u_{s,\mathrm{O}},v_{s,0})}$$\cross(1+O(|x-l-y|^{-1}))$ ,where the term $O(|x-l-y|^{-1})$ satisfies $|O(|x-l-y|^{-1})|\leq C|x-l-y|^{-1}$ with aconstant $C>0$ independent of,a. We have used (3.2) inthe second equality. Noting that $x-l-y$ md
have the same direction and, we have$x\in \mathrm{T}^{d}$$y\in \mathrm{R}^{d}$$\mathrm{d}$$\mathit{1}\in \mathrm{Z}^{d}$$-\nabla\rho\Lambda(i\beta_{s})$$|D_{s}|=-\eta_{s}\cdot$$\nabla\rho\Lambda(i\beta_{s})/|\nabla\rho\Lambda(i\beta_{s})|$$I_{1}(x-l,y)= \frac{e^{-(x-l-y)\cdot\beta}}{(2\pi|x-l-y|)^{(d-1)/2}}..\frac{|\nabla\rho\Lambda(i\beta_{s})|^{(d-3)/2}}{\det(-e_{s},{}_{j}\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}\rho\Lambda(\beta_{s})e_{s,k})^{1/2}}\frac{u\rho.(x)v\rho.(y)}{(u\rho.,v\rho.)}$$\cross$$(1 +O(|x-l-y|^{-1}))$ .

This gives the main term of the asymptotics (1.1)

(4.2)

115
Next we estimate the integral kernel ofhave$I_{2}$. We abbreviate$\eta_{s}$and$e_{s}$

to$\eta$

and . We$e$

$I_{2}f= \frac{|D_{s}|}{(2\pi)^{d}}(\int_{Q}d\alpha\int_{\overline{C}_{2}\cup\tilde{C}_{4}}dw+\int_{U_{s,\delta}}d\alpha\int_{C_{2}\backslash \tilde{C}_{2}\cup C_{3}\cup C_{4}\backslash \overline{C}_{4}}dw$$+ \int_{Q\backslash U_{*,\delta}}d\alpha\int_{\tilde{C}_{3}}dw)F(w\eta+\alpha\cdot e)$

Let us show that the first term vanishes. By Lemma 2.4 andhave$|D_{s}| \int_{Q}$

da

(4.3)

.

$N_{j}\equiv\tilde{N}_{j}\mathrm{m}\mathrm{o}\mathrm{d} 2\pi \mathrm{Z}^{d}$, we$\int_{\tilde{C}_{2}\cup\tilde{C}_{4}}\cdot dwF(w\eta+\alpha\cdot e)$$=|D_{s}|( \int_{\bigcup_{j=1}^{d}PN_{\mathrm{j}}}d\alpha\int_{\tilde{C}_{2}}dw+\int_{\bigcup_{j=1}^{d}P\tilde{N}_{\mathrm{j}}}d\alpha\int_{\tilde{C}_{4}}dw)F(w\eta+\alpha\cdot e)$$=i|D_{s}| \sum_{j=1}^{d}(\int_{PN_{j}}$da$\int_{0}^{|\beta_{\iota}|+\delta/2}dtF((t_{2}(\alpha)+it)\eta+\alpha\cdot e)$$- \int_{P\tilde{N}_{\mathrm{j}}}d\alpha\int_{0}^{|\beta_{\iota}|+\delta/2}dtF((t_{1}(\alpha)+it)\eta+\alpha\cdot e))$$=i \sum_{1<j<d}|\eta_{j}|(\int_{N_{j}}dk’\int_{0}^{|\beta_{*}|+\delta/2}dtF(k’+it\eta)-\int_{\tilde{N}_{j}}dk’\int_{0}^{|\beta_{*}|+\delta/2}dtF(k’+it\eta))$$|P\overline{N}_{j}\overline{|}\neq 0$

$=0$

,

.
if
where is the -th component of and $dk’=dk_{1}\cdots$ $dkj-1dkj+1\cdots$
be functions from $[0, 1]$andDenote the kernel of $L(k)^{-1}$ by $E_{k}(x, y)$ . Let , ,and , respectively. For $n\geq 0$,,
to , which parametrize contours
, put
and ,
integer, ,$k’\in N_{j}\cup\tilde{N}_{j}$$dk_{d}$$j$$\eta j$$\eta$$\varphi_{1}$$C_{2}\backslash \tilde{C}_{2}$$C_{3}$$\varphi_{2}$$\varphi_{3}$$C_{4}\backslash \tilde{C}_{4}$$\varphi_{4}$$\tilde{C}_{3}$$\mathrm{C}$$x$$y\in \mathrm{T}^{d}$$l$$m\in \mathrm{Z}^{d}$$H_{n}(k)= \exp[i(x-l-y+m)\cdot k]\sum_{j=0}^{n}$$(\begin{array}{l}nj\end{array})$$(i(x-y)\cdot\eta)^{j}(\eta\cdot\partial_{k})^{n-j}E_{k}(x, y)$

By Fubini’s theorem, the integral kernel $I_{2}(x-l, y-m)$ of I2 is written,, ,$\mathrm{T}^{d}$$l$

as, for.
$x$, $y\in$$m\in \mathrm{Z}^{d}$$I_{2}(x-l, y-m)= \frac{|D_{s}|}{(2\pi)^{d}}\int_{U_{\iota,\delta}}d\alpha\sum_{j=1}^{3}\int_{0}^{1}dt\dot{\varphi}_{j}(t)H_{0}(\varphi_{j}(t)\eta+\alpha\cdot e)$$+ \frac{|D_{s}|}{(2\pi)^{d}}\int_{Q\backslash U_{e,\delta}}$

where
equality$\dot{\varphi}j(t)=\frac{d}{dt}\varphi j(t)$da$\int_{0}^{1}dt\dot{\varphi}_{4}(t)H_{0}(\varphi_{4}(t)\eta+\alpha\cdot e)$

,

(4.4)

. Note $(m-l)\cdot\eta\neq 0$ for $m-l$ sufficiently large. Using the$e \dot{\varphi}_{j}=:(m-l)\cdot(\varphi_{j}(t)\eta+\alpha\cdot \mathrm{e})\frac{1}{i(m-l)\cdot\eta}\partial_{t}e^{\dot{\iota}(m-l)\cdot(\varphi j(t)\eta+\alpha\cdot e)}$,116
we integrate by parts for in each integral in (4.4) toobtain$t$$\int_{0}^{1}dt\dot{\varphi}_{j}(t)H_{0}(\varphi_{j}(t)\eta+\alpha\cdot e)=\frac{1}{i(m-l)\cdot\eta}[H_{0}(\varphi_{j}(1)\eta+\alpha\cdot e)-H_{0}(\varphi_{j}(0)\eta+\alpha\cdot e)]$$- \frac{1}{i(m-l)\cdot\eta}\int_{0}^{1}dt\dot{\varphi}_{j}(t)H_{1}(\varphi_{j}(t)\eta+\alpha\cdot e)$

By$\varphi_{1}(1)=\varphi_{2}(0)\mathrm{m}\mathrm{d}$$\varphi_{2}(1)=\varphi s$.

(0), we have$I_{2}(x-l,y-m)$$= \frac{1}{i(m-l)\cdot\eta}\frac{|D_{s}|}{(2\pi)^{d}}[\int_{U..s}d\alpha[H_{0}(\varphi_{3}(1)\eta+\alpha\cdot e)-H_{0}(\varphi_{1}(0)\eta+\alpha\cdot e)]$$+ \int_{Q\backslash U..s}d\alpha[H_{0}(\varphi_{4}(1)\eta+\alpha\cdot e)-H_{\mathrm{O}}(\varphi_{4}(0)\eta+\alpha\cdot e)]$$-( \int_{U.,s}d\alpha\int_{C_{2}\backslash \tilde{C}_{2}\cup c_{\epsilon}\cup C_{4}\backslash \tilde{c}_{4}}dw+\int_{Q\backslash U..s}d\alpha\int_{\tilde{C}_{\theta}}dw)H_{1}(w\eta+\alpha\cdot e)]$

.
(4.5)

We claim that the sum of the ffist md the second teminorder to show the claim, we need alemma.

Lemma 4.1. Suppose thatany z$\in \mathrm{Z}^{d}$, and$E_{k}(x,$y) exists for k$\in \mathrm{C}^{d}$$[\cdots]$of (4.5) vanishes. In. Then $E_{k+2\pi z}(x,$ y) exists for$\exp(i(x-y+l)\cdot k)(\eta\cdot\partial_{k})^{n}E_{k}(x, y)=\exp(i(x-y+l)\cdot(k+2\pi z))(\eta\cdot\partial_{k})^{n}E_{k+2\pi z}(x, y)$for any ,$z$$l\in \mathrm{Z}^{d}$,$\eta\in \mathrm{S}^{d-1}$

Proof. Note thatand $n\geq 0$ integer. In particular $H_{n}(k)=H_{n}(k+2\pi z)$ .

$(\eta\cdot\partial_{k})^{n}E_{k}(x, y)$is of the form:$( \eta\cdot\partial_{k})^{n}E_{k}=\sum_{m}C_{m}E_{k}*(\eta\cdot\partial_{k})^{j_{1}}L(k)E_{k}*\cdots*(\eta\cdot\partial_{k})^{j_{m}}L(k)E_{k}$where $E*F(x,y)= \int_{\mathrm{T}^{d}}E(x, z)F(z,y)dz$ for two functions
and$j_{1}$, (4.6)$E\mathrm{m}\mathrm{d}$, $\ldots,j_{m}=1,2$ . Hence to see (4.6) we have only to notice$eE_{k+2\pi z}(x,y)=E_{k}(x, y)e:2\pi z\cdot x:2\pi z\cdot y$

$F$

,

,

(4.7)

$\mathrm{m}\mathrm{d}\sum_{s=1}^{m}j_{s}=n$

that

,

$e(:2\pi z\cdot x\eta\cdot\partial_{k})^{\mathrm{j}}L(k+2\pi z)=(\eta\cdot\partial_{k})^{j}L(k)e:2\pi z\cdot x$,$j=1,2$ .

$\square$Note that $\varphi_{1}(0)=t_{2}(\alpha)+i(|\beta|+\delta/2)$ , $\varphi_{3}(1)=t_{1}(\alpha)+i(|\beta|+\delta/2)$,$i(|\beta|+\delta/2)$
and $\varphi_{4}(1)=t_{1}(\alpha)+i(|\beta|+\delta/2)$ . The sum of theand the second term$\varphi_{4}(0)=t_{2}(\alpha)+$$\mathrm{f}_{\mathrm{i}}\mathrm{r}\mathrm{s}\mathrm{t}$117$\mathrm{n}[\cdots]$in (4.5) vanishes since we have by Lemma 4.1 and$N_{j}\equiv\tilde{N}j\mathrm{m}\mathrm{o}\mathrm{d} 2\pi \mathrm{Z}^{d}$$|D_{s}|( \int_{U_{\epsilon,\delta}}d\alpha[H_{0}(\varphi_{3}(1)\eta+\alpha\cdot e)-H_{0}(\varphi_{1}(0)\eta+\alpha\cdot e)]$$+ \int_{Q\backslash U_{s,\delta}}d\alpha[H_{0}(\varphi_{4}(1)\eta+\alpha\cdot e)-H_{0}(\varphi_{4}(0)\eta+\alpha\cdot e)])$$=|D_{s}|( \int_{\bigcup_{\mathrm{j}=1}^{d}PN_{j}}d\alpha H_{0}([t_{1}(\alpha)+i(|\beta|+\delta/2)]\eta+\alpha\cdot e)$$- \int_{\bigcup_{j=1}^{d}P\tilde{N}_{\mathrm{j}}}d\alpha H_{0}([t_{2}(\alpha)+i(|\beta|+\delta/2)]\eta+\alpha\cdot e))$$= \sum_{1<j<d}|\eta_{j}|(\int_{N_{\mathrm{j}}}dk’H_{0}(k’+i(|\beta|+\delta/2)\eta)-\int_{\tilde{N}_{\mathrm{j}}}dk’H_{0}(k’+i(|\beta|+\delta/2)\eta))$$|P\overline{N}_{\mathrm{j}}\overline{|}\neq 0$

$=0$

,

.
if
where i the -th component of and $dk’=dk_{1}\cdots$ $dkj-1dkj+1\ldots$thein
We repeat this integration by parts for , $(d-1)$ -times. By Lemma 4.1 we havesame way as above$dk_{d}$$j$$\mathrm{s}$$\eta j$$k’\in N_{j}\cup\tilde{N}_{j}$$\eta$$t$$I_{2}(x-l, y-m)= \frac{|D_{s}|}{(2\pi)^{d}}(\frac{i}{(m-l)\cdot\eta})^{d-1}$$\cross$($\int_{U_{*,\delta}}$doda$\int_{C_{2}\backslash \tilde{C}_{2}\cup C_{3}\cup C_{4}\backslash \tilde{C}_{4}}dw+\int_{Q\backslash U_{s,\delta}}$$\int_{\overline{C}_{3}}dw$)$H_{d-1}(w\eta+\alpha\cdot e)$

.
(4.8)

on the integral domain in (4-8)
Lemma 4.2. The absolute value of the integrandindependentis majorized by $C\exp[-(|\beta_{s}|+\delta/2)(x-l-y+m)\cdot\eta]$ with a constant $C>0$$\mathrm{H}\mathrm{d}-\mathrm{i}$

.
,
of ,
or the second term
Proof Note that if belongs to the integral domain of the firstintegral
domain such$x$$y\in \mathrm{T}^{d},$$l$$m\in \mathrm{Z}^{d}$$k$in (4.8), there exists aconstant $M_{d}>0$ independent ofthat$|E_{k}(x, y)| \leq M_{2}(1+\log\frac{1}{|x-y|})$$|\partial_{x}E_{k}(x, y)|\leq M_{d}|x-y|^{1-d}$

By the definition of$\mathrm{H}\mathrm{d}-\mathrm{i},$, $d=2$ ,$k$in the$|E_{k}(x, y)|\leq M_{d}|x-y|^{2-d}$, $d\geq 3$ ,
(4.9)

.

, it suffices to show that$|x-y|^{d-1-j}|(\eta\cdot\partial_{k})^{j}E_{k}(x, y)|\leq C$

for

$k$

,

$0\leq j\leq d-1$

(4.10),

in the integral domain. By (4.7), this follows from$|x-y|^{d-1-j}|E_{k}*(\eta\cdot\partial_{k})^{j_{1}}L(k)E_{k}*\cdots*(\eta\cdot\partial_{k})^{j_{m}}L(k)E_{k}|\leq C$

,

118
where

and

$\sum_{s=1}^{m}j_{s}=j$$j_{1}$,

for $d=2$$\ldots$, $j_{m}=1,2$ . To see this, by (4.9) we have only to note$\int_{\mathrm{T}^{2}}(1+\log\frac{1}{|x-x_{1}|})|x_{1}-y|^{-1}dx_{1}\leq C$

that

,

md for $d\geq 3$$\int_{\mathrm{T}^{d}}\cdots\int_{\mathrm{T}^{d}}|x-x_{1}|^{2-d}|x_{1}-x_{2}|^{j_{1}-d}\cdots|x_{m}-y|^{j_{m}-d}dx_{1}\cdots dx_{m}$$C|x-y|^{2+j-d}$$\leq\{$$C(C’+ \log\frac{1}{|x-y|})$$C$$2+j-d<0$ ,$2+j-d=0$ ,$2+j-d>0$ .

$\square$From this lemma, it follows that$|I_{2}(x-l, y-m)|\leq C|l-m|^{1-d}\exp[-(|\beta_{s}|+\delta/2)(x-l-y+m)\cdot\eta]$with aconstmt $C>0$ independent of ,shows (1.1).
$x$5.

$y\in \mathrm{T}^{d}$$\mathrm{p}_{\mathrm{R}\mathrm{O}\mathrm{O}\mathrm{F}\mathrm{O}\mathrm{F}}$, ,$m\in \mathrm{Z}^{d}$$l$

THEOREM 1.3

Then $\nabla\Lambda(0)=0$ . By Proposition 2.2, $L(k)^{-1}$ exists if$\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{k}$A(0)$=-\mathrm{H}\mathrm{e}\mathrm{s}\mathrm{s}_{\beta}$. This together with (4.2).
$k\in \mathrm{R}^{d}\backslash 2\pi \mathrm{Z}^{d}$A(0).

Proposition 5.1. There exists
the forrn$\delta$$>0$ such thatfor$k\in \mathrm{R}^{d}$, $0<|k|<\delta$, $L(k)^{-1}\dot{l}S$ of,$L(k)^{-1}= \frac{2(\cdot,v_{0})u_{0}}{k\cdot Hk(u_{0},v_{0})}+\frac{A(\omega)}{|k|}+B(k)+Q(k)$

where
and
Fuhhermore,$u_{0}$$v_{0}$. Put $H=$

(5.1)

is a positive $solut$ : to $L(0)u_{\mathrm{O}}=0$ and $L(0)*v_{\mathrm{O}}=0$ , $respect\cdot.vely$.
is a finite rank operator-valued funciion$=k/|k|$ and the integral$A(\omega)$$of\omega$$0<|k|<\delta andcontinuouS\dot{l}n(x,y)andallde\Gamma l.vat\dot{l}vesofB_{k}(x, y)inkarebo.undedonrankoperator- valuedfunctionofkandtheintegralkernelB_{k}(x,y)ofB(k)_{lS}C^{\infty}onkemelA_{\omega}(x,y)ofA(\omega)isC^{\infty}in\omega\in \mathrm{S}^{d-1}andcontinuou\mathit{8}in(x,y).B(k)isafinite$$\{0<|k|<\delta\}\cross \mathrm{T}^{d}\cross \mathrm{T}^{d}$. $Q(k)$ is a real analytic funciion on

kernel $Q_{k}(x, y)$ of $Q(k)$ satisfies,$|x-y|^{j}|(\eta\cdot\partial_{k})^{l}Q_{k}(x, y)|\leq C$

for some constant$C$

independent $of|k|<\delta$,$j$$|k|<\delta$, $l\geq 0$ , $j+l=d-1$ ,$\eta\in \mathrm{S}^{d-1}$

and the integral(5.2)

and $x,y$ .

Proof. By the regular perturbation theory, since $\Lambda(0)=0$ is nondegenerate, there
exist , $\delta’>0$ such that if $|k|<\delta$ the eigenfunctionof $L(k)$ is the ony point of the$\delta$$\mathrm{A}(\mathrm{k})$119( is analytic) eigenfunction of $L(k)^{*}$ corresponding to. Since
Using these, we have $P(k)=(u_{k}, v_{k})^{-1}$ ,$\overline{\Lambda}(k):(L(k)^{*}-\overline{\Lambda}(k))v_{k}=0$$\overline{v_{k}}$$v_{k})u_{k}$$(\cdot$$\Lambda(k)$.

is nondegenerate, theequality, where$L(k)^{-1}=\Lambda(k)^{-1}P(k)+\mathrm{L}(\mathrm{k})$$Q(k)= \frac{1}{2\pi i}\oint_{|\zeta|=\delta}$,$\zeta^{-1}(L(k)-\zeta)^{-1}d\zeta$,by the expansions $u_{k}=u0+u_{1}\cdot$ $k+$andholds. Expressing the functions ,with someand $\Lambda(k)=k$ .
$O(k^{2})$ ,, we obtain that the integral kernel of $\Lambda(k)^{-1}P(k)$ equals
and,$u_{k}$$\Lambda(k)$$v_{k}$$Hk/2+ \sum_{|\alpha|=3}\tilde{H}_{\alpha}k^{\alpha}+O(k^{4})$$\overline{v_{k}}=v_{0}+\overline{v_{1}}\cdot k+O(k^{2})$$\tilde{H}_{\alpha}$$u_{1}$$v_{1}$$+B_{k}(x, y)$$= \frac{2u_{0}(x)v_{0}(y)}{k\cdot Hk(u_{0},v_{0})}+\frac{A_{\omega}(x,y)}{|k|}+B_{k}(x, y)$.

the proposition except
The each term of the right hand of this has the property stated in(4.10) with $E_{k}(x, y)$for (5.2). The same argument as in the proof of Lemma 4.2 shows
replaced by the integral kernel of $(L(k)-\zeta)^{-1}$ , which implies (5.2).
in $L^{2}(B_{R})$ , where
be the Dirichlet realization of
and $R>0$ let
For[A1], since
has apositive solution, the$\{|x|<R\}$ . By Theorem 3.1 inis theis positive. By Theorem
exists and the Green function
resolventexists when $d\geq 3$ . Since
2 in [Pinsl], the, we, $0<R\leq R’$ , andHence.
of satisfies
can see that the minimal Green function$\square$$L+\epsilon$$(L+\epsilon)_{R}$$\epsilon\geq 0$$L+\epsilon$$B_{R}$$\mathrm{b}\mathrm{a}\mathbb{I}$$G_{R,\epsilon}(x, y)$$(L+\epsilon)_{R}^{-1}$$G_{R,\epsilon}\leq G_{R,0}\leq$$\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}R\lim_{arrow\infty}G_{R,\epsilon}=G_{\infty,\epsilon}$$G_{R,\epsilon}\leq G_{\infty,\epsilon}\leq G_{\infty,\epsilon’}\leq G_{\infty,0},0\leq\epsilon’\leq\epsilon$$G_{R’,0}\leq G_{\infty,0},0\leq\epsilon$$L$$G_{\infty,0}$by the integral expression for$(L+\epsilon)^{-1}$,$\epsilon$$G_{\infty,0}= \lim_{\epsilon\downarrow 0}G_{\infty,\epsilon}$$>0$ , we have with $G=G_{\infty},0$$G(x-l, y-m)= \lim_{e\downarrow 0}\int_{[-\pi,\pi]^{d}}e^{i(x-l-y+m)\cdot k}E_{k}^{\epsilon}(x, y)\frac{dk}{(2\pi)^{d}}$

,

$x$

,

$y\in \mathrm{T}^{d}$, , $m,$$l$$\in \mathrm{Z}^{d}$,$E_{k}(x, y)$ beis the integral kemel of the resolvent $(L(k)+\epsilon)^{-1}$ . Let
where. We can see that for
the integral kernel of the resolvent $L(k)^{-1}$ foris. Furthermore,asand $x\neq y$ ,$x\neq y$ . In fact, choosebounded by some integrable function of $k\in[-\pi, \pi]^{d}$ for fixed
and $|k|\leq\delta/2$ . Since we haveforso small that$E_{k}^{\epsilon}(x, y)$$k\in[-\pi, \pi]^{d}\backslash 0$$E_{k}^{\epsilon}(x, y)arrow E_{k}(x, y)$$k\in[-\pi, \pi]^{d}\backslash 0$$\epsilon$$|E_{k}^{\epsilon}(x, y)|$$\downarrow 0$$\epsilon_{0}$$|\Lambda(k)+\epsilon|<\delta’$$0\leq\epsilon$$\leq\epsilon_{0}$$(L(k)+\epsilon)^{-1}=(\Lambda(k)+\epsilon)^{-1}P(k)+Q(k)$where $P(k)$ and $Q(k)$ is given in the
an integrable function$\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}^{)}\mathrm{o}\mathrm{f}$,$|k|\leq\delta/2$

Proposition 5.1,,

$0\leq\in$,is bounded byby Proposition 5.1$|E_{k}^{\epsilon}(x, y)|$$\frac{2u_{\mathrm{O}}(x)v\mathrm{o}(y)}{\overline{k}\cdot Hk(\mathrm{u}_{\mathrm{O}},v_{\mathrm{O}})}+\frac{|A\omega(x,y)|}{|k|}+|B_{k}(x, y)|+|Q_{k}(x, y)|$$\leq\epsilon_{0}$120
If $|k|>\delta/2$ md, then we have $|E_{k}^{e}(x, y)|\leq C|x-y|^{2-d}$ withsome constantC $>0$ independent of kmd . Thus by theLebesgue’s convergence theorem, the Greenfunction of L is expressed by$0\leq\epsilon$$\leq\epsilon_{0}$$\epsilon$$G(x-l,y-m)= \int_{[-\pi,\pi]^{d}}e^{i(x-l-y+m)\cdot k}E_{k}(x,y)\frac{dk}{(2\pi)^{d}}$Let.

be the least eigenvalue of $H$ . Takesuch thaton
andon. By Proposition 5.1, divide
the Green function into four parts, where each is given by$h_{\mathrm{O}}>0$$C^{\infty}(0, \infty)- \mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\chi(\mathrm{r})$$\chi(\mathrm{r})=1$$0<’\leq\sqrt{h_{\mathrm{O}}}\delta/3$$\chi(\mathrm{r})=0$$2\sqrt{h_{0}}\delta/3\leq \mathrm{r}$$G= \sum_{j=1}^{4}I_{j}$$I_{j}$,$I_{1}(_{X-l,y-m)=\int_{[-\pi,\pi]^{d}}\chi(|\sqrt{H}k|)e^{i(x-l-y+m)\cdot k_{\frac{2u_{0}(x)v_{0}(y)}{k\cdot Hk(u_{0},v_{0})}\frac{dk}{(2\pi)^{d}}}}}$$I_{2}(x-l,y-m)= \int_{[-\pi,\pi]^{d}}\chi(|\sqrt{H}k|)_{C}^{:(x-l-y+m)\cdot k_{\frac{A.(x,y)}{|k|}\frac{dk}{(2\pi)^{d}}}}$$I_{3}(x-l,y-m)= \int_{[-\pi,\pi]^{d}}\chi(|\sqrt{H}k|)eB_{k}(x,y)\frac{dk}{(2\pi)^{d}}:(x-l-y+m)\cdot k$,,$I_{4}(x-l,y-m)$$= \int_{[-\pi,\pi]^{d}}e-l-y+m)\cdot k[(x\chi(|\sqrt{H}:k|)Q_{k}(x,y)+(1-\chi(|\sqrt{H}k|))E_{k}(x,y)]\frac{dk}{(2\pi)^{d}}$

for

, ,

,

.
Lemma 5.2. The following asymptotics holds$x,y\in \mathrm{T}^{d}$$l$$m\in \mathrm{Z}^{d}$$I_{1}(x-l, y-m)= \frac{\Gamma(\frac{d-2}{2})}{2\pi^{d/2}}\frac{(\det H)^{-1/2}}{|H^{-1/2}(x-l-y+m)|^{d-2}}\frac{u_{\mathrm{O}}(x)v_{\mathrm{O}}(y)}{(u_{0},v_{0})}(1+O(|x-l-y+m|^{-1}))$,$where$ $O(|x-l-y+m|^{-1})sat_{\dot{l}S}fies$

positive constant $C$ independent$|O(|x-l-y+m|^{-1})|\leq C|x-l-y+m|^{-1}$ with

of ,
$x$

$y\in \mathrm{T}^{d}$This gives the main term in (1.2) with
Pmof. It suffices to show that for

, ,
$l$

$m\in \mathrm{Z}^{d}$$\beta_{0}=0$

.

$a$

.

$z\in \mathrm{R}^{d}$$\int_{\mathrm{R}^{d}}\chi(|\sqrt{H}k|)\frac{ez\cdot k}{k\cdot Hk}.\cdot dk=\frac{(2\pi)^{d/2}2^{\nu-1}\Gamma(\nu)}{(\det H)^{1/2}|H^{-1/2_{Z|^{d-2}}}}(1+O(|z|^{-1}))$

here $\nu=(d-2)/2$ . By a chmge of variables $k’=\sqrt{H}k$ ,as$|z|arrow\infty$

the left hand side of this isequal to.

$\int_{\mathrm{R}^{d}}\chi(|k’|)\frac{\exp(iz\cdot H^{-1/2}k’)}{\det H^{1/2}|k|^{2}},dk’$Use the polar coordinates$k’=\mathrm{r}\omega$,$f$$\geq 0$,$\omega\in \mathrm{S}^{d-1}$$\frac{1}{\det H^{1/2}}\int_{0}^{\infty}\chi(t)t^{d-3}d\mathrm{r}$,, to obtain that this equal$\int\exp(iz\cdot H^{-1/2}\mathrm{r}\omega)h$$= \frac{1}{\det H^{1/2}}\int_{0}^{\infty}\chi(\mathrm{r})_{\Gamma}^{d-3}(2\pi)^{d/2}\frac{J_{\nu}(t|H^{-1/2}z|)}{(\mathrm{r}|H^{-1/2_{Z|)^{\nu}}}}d\mathrm{r}$

,

121
where
that

$J_{\nu}(r)$is the Bessel function of order . Put A $=|H^{-1/2}z|$ . We have only to show$\nu$$\int_{0}^{\infty}\chi(r)r^{d-3}\frac{J_{\nu}(\lambda r)}{(\lambda r)^{\nu}}dr=\lambda^{2-d}\Gamma(\nu)2^{\nu-1}(1+O(\lambda^{-1}))$,equivalently, we show that$\int_{0}^{\infty}\chi(r/\lambda)r^{\nu-1}J_{\nu}(r)dr=\Gamma(\nu)2^{\nu-1}+O(\lambda^{-1})$

Let us prove this by induction on . When $\nu=1/2$ , bysee that$\nu$,$\nu=(d-2)/2$ .

$J_{1/2}(r)= \sqrt{\frac{2}{\pi}}\frac{\sin r}{\sqrt{f}}$(5.3), it is easy to$\int_{0}^{\infty}\chi(r/\lambda)r^{-1/2}J_{1/2}(r)dr=\int_{0}^{\infty}\chi(r/\lambda)\sqrt{\frac{2}{\pi}}\frac{\sin r}{r}dr=\sqrt{\frac{\pi}{2}}+O(\lambda^{-1})$$\mathrm{B}\mathrm{y}-\frac{d}{rdr}(r^{-\nu+1}J_{\nu-1}(r))=r^{-\nu}J_{\nu}(r)$.

, integration by parts yields$\int_{0}^{\infty}\chi(r/\lambda)r^{\nu-1}J_{\nu}(r)dr=\chi(r/\lambda)r^{\nu-1}J_{\nu-1}(r)|_{r=0}$$+(2 \nu-2)\int_{0}^{\infty}\chi(r/\lambda)r^{\nu-2}J_{\nu-1}(r)dr+\int_{0}^{\infty}\lambda^{-1}\chi’(r/\lambda)r^{\nu-1}J_{\nu-1}(r)$

dr.
(5.4)

For the moment, suppose that the following estimate holds: for any integer $N\geq 1$ there
exists aconstant $C_{N,\nu}>0$ such that$| \int_{0}^{\infty}\lambda^{-1}\chi’(r/\lambda)r^{\nu}J_{\nu}(r)dr|\leq C_{N,\nu}\lambda^{-N}$

.

(5.5)

The proof of (5.5) is given at the end of the proof of the lemma. When $\nu=1$ , since$J\mathrm{o}(0)=1$ , (5.3) follows from (5.4) and(5.5). Suppose that (5.3) holds for $1/2\leq\nu\leq\nu_{0}$ .
From (5.4) for $\nu=\nu_{0}+1$ , we have (5.3) for $\nu=\nu_{0}+1$ by the induction hypothesis and(5.5).
It remains to prove (5.5). Similarly as in (5.4), integration by parts yields$\int_{0}^{\infty}\lambda^{-1}\chi’(r/\lambda)r^{\nu}J_{\nu}(r)dr$$= \int_{0}^{\infty}\lambda^{-2}\chi’(r/\lambda)r^{\nu}J_{\nu-1}(r)+(2\nu-1)\lambda^{-1}\chi’(r/\lambda)r^{\nu-1}J_{\nu-1}(r)$dr.

Repeating this $N$ -times for each term, we have$\int_{0}^{\infty}\lambda^{-1}\chi’(r/\lambda)r^{\nu}J_{\nu}(r)dr=\int_{0}^{\infty}(\sum_{j=1}^{N+1}C_{N,j}\lambda^{-j}\chi^{(j)}(r/\lambda)r^{\nu+j-N-1})J_{\nu-N}(r)dr$with some constants. Since the support ofis in $\{c\lambda<r<c’\lambda\}$ and$J_{\nu-N}(r)$ is bounded forlarge, the absolute value of the right hand side of this isestimated by. Thus we have proved (5.5).
$C_{N,j}$$\chi^{(j)}(r/\lambda)$$r$$C_{N}\lambda^{\nu-N}$$\square$122
Lemma 5.3. The following estimates hold$|I_{2}(x-l, y-m)|\leq C|x-l-y+m|^{1-d}$ ,(5.6)$|I_{3}(x-l,y-m)|\leq C|x-l-y+m|^{1-d}$ ,(5.7)$|I_{4}(x-l,y-m)|\leq C|x-l-y+m|^{1-d}$ ,(5.8)

with a positive constant $C$ independent of ,$x$$y\in \mathrm{T}^{d}$

, ,
$l$

$m\in \mathrm{Z}^{d}$.

Theorem 1.3 follows from Lemmas 5.2 and 5.3.
Proof of (5.6). Put $\lambda=|x-l-y+m|$ and change the integral variable as $k’=\lambda k$in the integral of . Then we have with $s=(x-l-y+m)/|x-l-y+m|$$I_{2}$.

$I_{2}(x-l, y-m)= \int_{\mathrm{R}^{d}}\chi(|\sqrt{H}k’|/\lambda)e\frac{\lambda A.(x,y)}{|k|}:s\cdot k’,\frac{dk’}{(2\pi)^{d}\lambda^{d}}$Hence we have only to show that
$J( \lambda, s)=\int_{\mathrm{R}^{d}}\chi(|\sqrt{H}k|/\lambda)e\frac{A.(x,y)}{|k|}:s\cdot kdk$

is abounded function of$\lambda\geq 1$

and$s\in \mathrm{S}^{d-1}$. Since$\frac{\partial}{\partial\lambda}J(\lambda, s)=\frac{-1}{\lambda^{2}}\int_{\mathrm{R}^{d}}|\sqrt{H}k|\chi’(|\sqrt{H}k|/\lambda)e\frac{A_{1d}(x,y)}{|k|}:s\cdot kdk:=\frac{-1}{\lambda^{2}}\tilde{J}(\lambda, s)$it suffices to show that$\tilde{J}(\lambda, s)$is bounded for$\lambda\geq 1$

and$s\in \mathrm{S}^{d-1}$,. In fact,$\tilde{J}(\lambda, s)=\lambda^{d}\int_{\mathrm{R}^{d}}|\sqrt{H}k|\chi’(|\sqrt{H}k|)e\frac{A_{\omega}(x,y)}{|k|}:\lambda s\cdot kdk$

function Thusand the integral is the Fourier transform of afor Alarge.
Proof of (5.7). It suffices to estimate aquantity$C_{0}^{\infty}$$\tilde{J}(\lambda, s)$decays rapidly$\int_{\mathrm{R}^{d}}\chi(|\sqrt{H}k|)eB_{k}:\lambda s\cdot k(x, y)dk$

for Alarge and$s\in \mathrm{S}^{d-1}$. Divide this into two parts$\int_{\mathrm{R}^{d}}\chi(|\sqrt{H}k|)\chi(\lambda^{e}|k|)eB_{k}:xs\cdot k(x,y)dk+\int_{\mathrm{R}^{d}}\chi(|\sqrt{H}k|)(1-\chi(\lambda^{e}|k|))eB_{k}:\lambda s\cdot k(x, y)dk$

since$=1-1/d$. It is easy to see that the first term is majorized by$B_{k}(x, y)$ is bounded. For the second, by using, it suffices torepeat the integration by parts $N$ times with $N\geq d(d-1)$ since derivatives of $B_{k}(x, y)$are bounded.

here$C\lambda^{1-d}$$\epsilon$$(i\lambda)^{-1}s\cdot$$\partial_{k}e:\lambda s\cdot k=e:\lambda s\cdot k$123Proof of (5.8). Put $\eta=(m-l)/|m-l|$ . By using$-i|m-l|^{-1}\eta\cdot\partial_{k}e^{i(m-l)\cdot k}=e^{i(m-l)\cdot k}$

and periodicity (4.6), the (d $-1)$ -times integration by parts yields$\frac{(d-1)!}{\alpha!\beta!\gamma!}(i\eta\cdot(x-y))^{\alpha}$$I_{4}(x-l, y-m)= \int_{[-\pi,\pi]^{d}}\frac{dk}{(2\pi)^{d}}\frac{i^{d-1}e^{i(x-l-y+m)\cdot k}}{|m-l|^{d-1}}\alpha+$$\alpha,\beta,\gamma\geq 0\sum_{\beta+\gamma=d-1},$$\cross[(\eta\cdot\partial_{k})^{\beta}Q_{k}(x, y)(\eta\cdot\partial_{k})^{\gamma}\chi(|\sqrt{H}k|)+(\eta\cdot\partial_{k})^{\beta}E_{k}(x, y)(\eta\cdot\partial_{k})^{\gamma}(1-\chi(|\sqrt{H}k|))]$.

It suffices to show that the each term of the summation in the integral is abounded
function of $(k, x, y)$ . Consider the terms of the case $\gamma=0$ in the summation. By (4.10)
and (5.2), they are bounded. For the terms of the case $\gamma>0$ in the summation, i.e.
$(i\eta\cdot(x-y))^{\alpha}(\eta\cdot\partial_{k})^{\beta}(Q_{k}(x, y)-E_{k}(x, y))(\eta\cdot\partial_{k})^{\gamma}\chi(|\sqrt{H}k|)$we can see that,is abounded function on the support ofby Proposition 5.1. Hence they are bounded.

$(\eta\cdot\partial_{k})^{\beta}(Q_{k}(x, y)-E_{k}(x, y))$$(\eta\cdot\partial_{k})^{\gamma}\chi(|\sqrt{H}k|)$$\square$

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