数理解析研究所講究録1102 巻 1999 年 9-34

9Saturation of the approximation by spectral decompositions
東北大理

1(Miho Tanigaki)

谷垣美保

Introduction.
Let$\Omega$be an open domain in theoperator $A=-\Delta$ inextension of . Let$A$$\{k_{\lambda}(t)\}$is the Laplacian. Denote by$\hat{A}$. Consider thea nonnegative selfadjoint$\kappa_{1},$$\kappa_{2}>0$such that$k_{\lambda}(t)\sqrt{t}^{n}/2-2\kappa 2+1/2\in L^{1}(0, \infty)$

are uniformly bounded in and $t\in[0, \infty)$ , and$\lambda$

converge to a nonzero constant as$\lambdaarrow\infty$$J_{\nu}$$(k_{\lambda}(t)-1)/\lambda^{-\kappa_{1}}t^{\kappa_{2}}$,is the Bessel function of order . We assume, furthermore,$\nu$

the following conditions(1.1)$\nu\int_{0}^{R}S^{2}-1d_{S}\hslash_{2}|\int_{s}^{R}r^{nj2-}I2\kappa_{2}+2(r)dr1\lambda=o(\lambda^{-\kappa_{1}})$(1.2)$| \int_{R}^{\infty}r^{\nu+1}I_{\lambda}(r)dr|=o(\lambda^{-\kappa_{1}})$

,

,

and
(1.3)

as

$\lambdaarrow\infty$,

for any $t\in[0, \infty)$ . Let$I_{\lambda}(r)=f_{0}^{\infty}k_{\lambda}(t^{2})J_{\nu}(rt)t\nu+1dt$

where $\nu=n/2-2\kappa_{2}+1$ and$R^{n}$

be a family of bounded piecewise smooth functions on $[0, \infty)$ .

Suppose we have two constants$(k_{\lambda}(t)-1)/\lambda^{-\hslash}1t^{\kappa}2$dimensional Euclidean spacewith the domain of definition $D(A)=C_{c}^{\infty}(\Omega)$ , where$L^{2}(\Omega)$$\Delta=\partial^{2}/\partial x_{1^{2}}+\cdots+\partial^{2}/\partial x_{n^{2}}$$n$$( \sum_{T=\sigma}^{\infty}T^{4}\kappa 2-3\max\tau\leq\epsilon\leq\tau+1|\int_{R}^{\infty}J_{\nu}(sr)I_{\lambda}(r)rdr|^{2})^{1/2}=o(\lambda^{-\kappa_{1}})$for any small $R>0$ .

10We shall consider the approximation operator$L_{1\circ \mathrm{c}}^{\infty}(\Omega)$

if for every compact set $K$ in$\Omega$$k_{\lambda}(\hat{A})$forthere is a constant$f\in L^{2}(\Omega)$. We say$\Delta f\in$such that$C_{K}$$| \int_{K}f(x)\triangle g(X)d_{X}|\leq C_{K}||g||_{L^{1}(}K)$

for any infinitely differentiable function$g$whose support is contained in $K$ . Let$\{\varphi_{\epsilon}\}$bean infinitely differentiable approximate identity with supports contained in $\{x;|x|<\epsilon\}$ .
For a functionas$\epsilonarrow 0^{+}$

$f$

on

$\Omega$and$x\in\Omega,$$f$is said to be$\mathrm{r}\mathrm{e}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}_{9}\mathrm{d}$

at

$x$

if

$f*\varphi_{\epsilon}(x)arrow f(x)$.

In 1970, Igari proved the following Theorem in [5].

Theorem A. Suppose that there exust a complete orthonormal system

functions in$L^{2}(\Omega)$

and a numerical sequence$f_{j}= \int_{\Omega}f(x)\overline{uj(X)}dx$$\{\bigwedge_{j}\}$for which,$f\in L^{2}(\Omega)$$\{u_{j}\}$$-\Delta u_{j}=\lambda_{j}u_{j}$of smooth
in

and
,

$f\in L^{2}(\Omega)$$s_{\lambda}^{\delta}f= \sum_{j\lambda\leq\lambda}(1-\frac{\lambda_{j}}{\lambda})^{\delta}f_{j}u_{j}$Let $\delta\geq(n+3)/2$ and$f\in L^{2}(\Omega)$be regulated in$\Omega$. Then the following hold.

(i) The following conditions are equivalent.

( a)$\mathrm{i}$$||s_{\lambda}^{\delta}f-f||_{L(K)}\infty=O(\lambda^{-1})$as$\lambdaarrow\infty$

for$(\mathrm{i}\mathrm{b})\Delta f\in L_{1_{\mathrm{o}\mathrm{c}}}^{\infty}(\Omega)$.

$P^{ver}y$compact set $K$ in$\Omega$.

.

$\Omega$

. Let

11
(ii) The following$condition\mathit{8}$are equivalent.

(ii a)$||s_{\lambda}^{\delta}f-f||_{L^{\infty}(K)}=o(\lambda^{-1})$

as(ii b)$\lambdaarrow\infty$$\Delta f$

for every compact set

vanishes in$\Omega$$K$

in$\Omega$.

.

Our aim is to give a generalization of Theorem A. Let$\{k_{\lambda}(t)\}$Borel functions on $[0, \infty)$ . We can define the bounded operator

be a family of bounded$k_{\lambda}(\hat{A})$Example 1. Suppose that there exist a complete orthonormal system

functions in$L^{2}(\Omega)$

and a sequence$\{\lambda_{j}\}$such that$f_{j}= \int_{\Omega}f(‘ X)\overline{uj(_{X)}}d_{X}$Let$\hat{A}$be the selfadjoint extension of$-\Delta$,$-\Delta u_{j}=\lambda_{j}u_{j}$$f\in L^{2}(\Omega)$.

defined by$D(\hat{A})=\{f\in L^{2}(\Omega)$;$\sum_{j=1}\lambda j\infty 2|f_{j}|^{2}<\infty\}$and$\hat{A}f=\sum_{j=1}\infty\lambda jfjuj$,$f\in D(\hat{A})$For any $f\in L^{2}(\Omega)$ the spectral decomposition of$f$.

is given by$E((- \infty, t])f=\sum_{j\lambda\leq t}f_{j}uj$

and$k_{\lambda}(\hat{A})$is defined by$k_{\lambda}( \hat{A})f=\sum_{=1}k_{\lambda(\lambda_{j}})j\infty f_{j}uj$’$f\in L^{2}(\Omega)$.

in$\Omega$in$L^{2}(\Omega)$$\{u_{j}\}$

. Let

.

of smooth12
Example 2. Let $\Omega=R^{n}$ . Let$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}^{n}}\int_{R^{n}}f(X)e-i\epsilon\cdot xdX$,

In this case, there is a unique nonnegative selfadjoint$D(\hat{A})=\{f\in L^{2}(R^{n})$;$f\in L^{2}(R^{n})$$\mathrm{e}\mathrm{X}\mathrm{t}\mathrm{s}^{\mathrm{n}\mathrm{S}\mathrm{i}\mathrm{o}\mathrm{n}}\hat{A}$.

of$-\Delta$

defined by$|\xi|^{2}\hat{f}(\xi)\in L^{2}(R^{n})\}$

and,$f\in D(\hat{A})$$\hat{A}f(x)=\frac{1}{\sim \mathit{2}\pi^{n}}\int_{R^{n}}|\xi|^{2}\hat{f}(\xi)e^{i\xi}x\cdot d\xi$.

Then the spectral decomposition of $f\in L^{2}(R^{n})$ is given by$E \langle(-\infty, t])f(x)=\frac{1}{\sqrt{2\pi}^{n}}\int_{|\xi|^{2}\leq}t\hat{f}(\xi)edix\cdot\xi\xi$

and

is defined by$k_{\lambda}(\hat{A})$,$k_{\lambda}( \hat{A})f(x)=\frac{1}{\sqrt{2\pi}^{n}}\int_{R^{n}}k_{\lambda}(|\xi|^{2})\hat{f}(\xi)eix\cdot\epsilon_{d\xi}$For$\kappa_{2}>0$

open set$G$in

and $1<p\leq\infty$ , we say$\Omega$with the closure$\overline{G}$$(-\Delta)^{\kappa_{2}}f$belongs tocontained in$\Omega$$f\in L^{2}(R^{n})$$L_{1}^{p_{\mathrm{O}\mathbb{C}}}(\Omega)$.

if for every bounded, there is a constant$C_{G}$

such that$| \int_{\overline{G}}f(x)(-\Delta)\kappa_{2}g(_{X)dX}|\leq C_{G}||g||_{L()}p’\overline{G}$for any infinitely differentiable function$1/p’=1$ .

Our results are stated as follows.

$g$with support contained in$\overline{G}$, where $1/p+$

13
Main theorem. Let
tension $of-\Delta$ in$[0,\infty)$

and$\kappa_{1},$be an open domain in$\Omega$. Let$\Omega$such that$\kappa_{2}>0$

are uniformly bounded innonzero constant as
Suppose that$\{k_{\lambda}(t)\}$$\{k_{\lambda}(t)\}$a regulated function inand$\hat{A}$be a nonnegative selfadjoint ex-

be a family of bounded piecewise smooth functions on$k_{\lambda}(t)\sqrt{t}^{n}/2-2\kappa 2+1/2\in L^{1}(0,\infty),$

and $t\in[0, \infty)$ , and$\lambda$$\lambdaarrow\infty$$R^{n}$

for any$t\in[0, \infty)$$\lambda^{\kappa_{1}}t^{-\kappa_{2}}[k_{\lambda}(t)-1]$$\lambda^{\kappa_{1}}t^{-\kappa_{2}}[k_{\lambda}(t)-1]$.

satisfies the conditions (1.1), (1.2) and (1.3) as
Furthermore,$L^{2}(\Omega).$converge to a$\lambdaarrow\infty$that $1<p\leq\infty$ and$\mathit{8}uppoSe$. Let(i) The following two conditions are equivalent.

( a)$\mathrm{i}$$||k_{\lambda}(\hat{A})f-f||_{L^{\mathrm{p}}(K})=o(\lambda^{-\kappa_{1}})$for every compact$(\mathrm{i}\mathrm{b})(-\Delta)^{\kappa_{2}}f\in L_{1_{\mathrm{o}\mathrm{C}}}^{p}(\Omega)$(ii) Let$G\subset\Omega$$\mathit{8}etK$

in$\Omega$.

.

be any open set.

(ii a) Suppose that$(-\triangle)^{\kappa_{2}}f$vanishes in G. Then$||k_{\lambda}(\hat{A})f-f||_{L^{p}(K)}=o(\lambda^{-n_{1}})$as$(\mathrm{i}\mathrm{i}\mathrm{b})$$\lambdaarrow\infty$

for any compact set$K\subset G$.

If$||k_{\lambda}(\hat{A})f-f||_{L^{\mathrm{p}}(K)}=o(\lambda^{-\kappa_{1}})$as$\lambdaarrow\infty$

for any compact set$K\subset G$, then$(-\Delta)^{\kappa_{2}}f$be$f\in L_{1\mathrm{o}\mathrm{c}}^{p}(\Omega)$

Then the following hold.

$a\mathit{8}\lambdaarrow\infty$$f$vanishes in

$G$

.

.

14

If

and

$\delta>(n+3)/2$, then the conditions (1.1), (1.2) and$k_{\lambda}(t)=(1-t/\lambda^{2})_{+}^{\delta}$(1.3) are satisfied. Therefore we have the following:

Corolary 1. Let$exten\mathit{8}ionof-\Delta$

function inin$L^{2}(\Omega)$be an open domain in$\Omega$$\Omega$. Let$s_{\lambda}^{\delta}=(1-\hat{A}/\lambda^{2})_{+}^{\delta}$$R^{n}$

and$\hat{A}$be a nonnegative

and $\delta>(n+3)/2$ . Let. Suppose that $1<p\leq\infty$ and$f\in L_{1}^{\mathrm{p}_{\mathrm{O}\mathrm{C}}}(\Omega)$$f$$\mathit{8}elfadj_{oit}n$be a regulated. Then the followinghold.
(i) The following are equivalent.
( a)$\mathrm{i}$$||s_{\lambda}^{\delta}f-f||_{Lp(K)}=O(\lambda^{-2})$

as$\lambdaarrow\infty$

for every compact set$(\mathrm{i}\mathrm{b})\Delta f\in L_{1_{\mathrm{o}\mathrm{c}}}^{p}(\Omega)$(ii) Let$G\subset \mathrm{f}l$

$K$

in

$\Omega$.

.

be any open set.

(ii a) Suppose that$\Delta f$vanishes in G. Then$||s_{\lambda}^{\delta}f-f||_{L^{p}(K)}=o(\lambda^{-2})$as$(\mathrm{i}\mathrm{i}\mathrm{b})$$\lambdaarrow\infty$

for any compact set$K\subset G$.

If$||s_{\lambda}^{\delta}f-f||_{L^{p}(K)}=o(\lambda^{-2})$

as$\lambdaarrow\infty$

for any compact set$K\subset Q$, then$\Delta fvani\mathit{8}hes$

in

$G$

.

Our main theorem follows from Theorem 1 in \S 2 and Theorem 2 in \S 3. Corollary 1
is proved in \S 4.

15

Saturation of the approximation.

2

Let$\langle$

be an open domain in the$\Omega$

2.-1i$\mathrm{p}1$$\alpha=$$D^{\alpha}=(-i)^{|\alpha|}(\partial/\partial_{X_{1}})^{\alpha}1\ldots(\partial/\partial_{X_{n}})\alpha_{n}$$L^{2}(\Omega)$$R^{n}$. Let$(\alpha_{1} , \alpha_{2,:}..

and, \alpha_{n}),$$a_{\alpha}\in C^{\infty}(\Omega)$$|\alpha|=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$. We consider$A$bound , then$c$$\hat{A}$$\hat{A}$is a selfadjoint extension of$A$

is formallywith the same lower

can be represented in the form of$\hat{A}=\int_{c}^{\infty}tE(dt)$$\{k_{\lambda}(t)\}$$A$.

be a family of bounded Borel functions on $[c, \infty),$(2.2)$\psi_{\lambda}(t):=\frac{k_{\lambda}(t)-1}{\lambda^{-\kappa_{1}}t^{\kappa_{2}}}$$\kappa 1,$$\kappa_{2}>0$

and.

Suppose that(1)$\psi_{\lambda}(t)$are uniformly bounded in(2)$\psi_{\lambda}(t)$converge to a nonzero constantLemma 1. Ifas$\lambdaarrow\infty$.

$f\in L^{2}(\Omega)$and$\lambda$

and $t\in[c, \infty)$ , and$g\in D(\hat{A}^{\kappa_{2}})$

$C$

as

, then$\lambdaarrow\infty$,

as an operator

with the domain of definition $D(A)=C_{c}^{\infty}(\Omega)$ . Suppose that

selfadjoint and semibounded. If

LetEuclidean space$A= \sum_{\leq|\alpha|m}a(\alpha x)D^{\alpha}$

be a differential operator, wherein}$n-\mathrm{d}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{S}\mathrm{i}_{\mathrm{o}\mathrm{I}}$for any $t\in[c, \infty)$ .

$\lambda^{\kappa_{1}}(k_{\lambda}(\hat{A})f-f,g)arrow C(f,\hat{A}^{\kappa}2g)$

16
Proof. By the definition of$k_{\lambda}(\hat{A})$, we have$\lambda^{\kappa_{1}}(k_{\lambda}(\hat{A})f-f,g)=\lambda^{\kappa_{1}}l^{\infty}[k_{\lambda}(\mathrm{t})-1](E(dt)f,g)$$= \int_{c}^{\infty}\frac{k_{\lambda}(t)-1}{\lambda^{-\kappa_{1}}}(f, E(dt)g)$$=..f_{\mathrm{c}}^{\infty} \frac{k_{\lambda}(t)-1}{\lambda^{-\kappa_{1}}t^{\kappa_{2}}}t^{\kappa_{2}}(f, E(dt)g)$$= \int_{\mathrm{c}}^{\infty}\psi_{\lambda}(t)t^{\kappa_{2}}(f, E(dt)g)$$=(f,\overline{\psi_{\lambda}}(\hat{A})\hat{A}^{\kappa_{2}}g)$$=l^{\infty}\psi_{\lambda}(t)(f,E(dt)\hat{A}\kappa 2g)$

Let $\rho=(f,$$E(\cdot)\hat{A}^{\kappa}2g)$and$|\rho|$be the total variation of$\rho$.

. Then.

$\int_{c}^{\infty}|\rho|(dt)\leq||f||_{L^{2}}(\Omega)||\hat{A}\kappa 2|g|L^{2}(\Omega)<\infty$

Therefore, by Lebesgue’s dominated convergence theorem, it follows that$\lim_{\lambdaarrow\infty}\lambda\kappa 1(k_{\lambda}(\hat{A})f-f,g)=\lim_{\lambdaarrow\infty}Ie)\psi_{\lambda}(t\rho\infty(dt)$$= \int_{c}^{\infty}\lim_{\lambdaarrow\infty}\psi_{\lambda}(t)\rho(dt)=C\int_{c}^{\infty}\rho(dt)=C(f,\hat{A}^{\kappa_{2}}g)$.

Thus Lemma 1 is proved.

Let

$G$

be an open subset in$\Omega$with compact closure$\overline{G}$and $1<p\leq\infty$ . We say

if$A^{\kappa_{2}}f\in L^{p}(\overline{G})$$||A^{\kappa}2f||_{Lp}( \overline{G})^{:=\mathrm{s}}||g|\mathrm{I}_{L},\mathrm{u}p(^{\frac{\mathrm{p}}{G})}=1|\int_{\Omega}f(x)\overline{\hat{A}^{\kappa}2g(x)}dX|<\infty$

where $1/p+1/p’=1$ andcontailled in$\overline{G}$

.

$g$

,

is an infinitely differentiable function whose support is17
Theorem 1. Let

be an open domain in$\Omega$$R^{n}$

bounded differential operator with coefficients in$i_{\mathit{8}}$a selfadjoint extension of$A$of bounded Borel functions onof Borel functions on$1<p\leq\infty$

and

$G$

$\mathrm{f}^{c},$$\infty$

and $A$ be a formally selfadjoint semi$C^{\infty}(\Omega)$

given by (2.1). Suppose thatwith the same lower bound . Let$c$$[c, \infty)$

and$\kappa_{1},$$\kappa_{2}>0$$\{k_{\lambda}(t)\}$be any open$\mathit{8}et$inwith compact closure$\Omega$

be a familysuqh that the sequence) given by (2.2) satisfies (1) and (2). Let$\overline{G}$$\hat{A}$$\{\psi_{\lambda}(t)\}$$f\in L^{2}(\Omega)$,. Then the followinghold.
(i) If$||k_{\lambda}(\hat{A})f-f||_{L^{p}}(\overline{G})=O(\lambda^{-\kappa_{1)}}$as$\lambdaarrow\infty$, then$A^{\kappa_{2}}f\in L^{p}(\overline{G})$.

(ii) If$||k_{\lambda}(\hat{A})f-f||Lp(\overline{G})^{=}o(\lambda-\kappa 1)$

as$\lambdaarrow\infty$

Proof. Let

Suppose that
(2.3)

, then$g$$A^{\kappa_{2}}f$vanishes in$\overline{G}$.

be an infinitely differentiable function and$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}g\subset\overline{G}$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}g$

be the support of. Then by Lemma 1$\lambda^{\kappa_{1}}(k_{\lambda}(\hat{A})f-f,g)arrow C(f,\hat{A}^{\kappa_{2}}g)$as$\lambdaarrow\infty$.

On the other hand, we have
(2.4)$|\lambda^{\kappa_{1}}(k_{\lambda}(\hat{A})f-f,g)|\leq\lambda^{\kappa_{1}}||k_{\lambda}(\hat{A})f-f||_{Lp}(\overline{G})g||||Lp’(\overline{G})$

.

$g$

.

18

If

$||k_{\lambda}(\hat{A})f-f||_{L^{\mathrm{p}()}}\overline{G}=O(\lambda^{-\kappa_{1}})$as$\lambdaarrow\infty$, then by (2.4) for any$\lambda$$|\lambda^{\kappa_{1}}(k_{\lambda}(\hat{A})f-f,g)|\leq C’||g||L^{p}(’\overline{G})$with$\mathrm{s}o\mathrm{m}\mathrm{e}$constant $C’>0$ . Therefore, by (2.3), we have$| \int_{\Omega}f(x)\overline{\hat{A}\kappa_{2}(Xg)}dx|=|(f,\hat{A}^{\kappa_{2}}g)|\leq C^{-1}C’||g||_{L^{p’}}(\overline{c})$for any
If$g$. Thus (i) is proved.

$||k_{\lambda}(\hat{A})f-f||_{L^{p()}}\overline{G}=o(\lambda^{-\kappa_{1}})$as$\lambdaarrow\infty$, then in the same way as in (i), (ii) is

proved.

Examples. (1) Riesz summation: For $\kappa>0$ and $\delta>0$ , the Riesz summation is. In this case,given by the multiplier$k_{\lambda}(t)=[1-(t/\lambda^{2})^{\kappa}]_{+}^{\delta}$

uniformly bounded inand $t\in[c, \infty)$ with a constant $c>0$ and$\lambda$$(\lambda^{2}/t)^{\kappa}[k_{\lambda}(t)-1]$

are$\lim_{\lambdaarrow\infty}\frac{kx(t)-1}{(\lambda^{-2}t)^{\kappa}}=\lim_{\mathrm{q}arrow+0}\frac{(1-s^{\kappa})^{\delta}-1}{s^{\kappa}}=-\lim_{sarrow+0}\delta(1-s)\delta-1=-\delta$

for any $t\in[c, \infty)$ . Thus$\kappa_{1}=2\kappa,$$\kappa_{2}=\kappa$

and $C=-\delta,$ where $C$ is a constant in (2).

(2) Fej\’er-Korovkin summation is defined by$(1- \frac{t}{\lambda^{2}})\mathrm{c}o\mathrm{s}\frac{\pi t}{\lambda^{2-}}+\frac{1}{\lambda^{\mathit{3}}}\cot\frac{\pi}{\lambda^{2}}\sin\frac{\pi t}{\lambda^{2}}$$t<\lambda^{2}$,$0$$t\geq\lambda^{2}$.

$k_{\lambda}(t)=$In this case,$(\lambda^{2}/t)^{2}[k_{\lambda}(t)-1]$

are uniformly bounded in$\lambda$

and $t\in[c, \infty)$ and$\lim_{\lambdaarrow\infty}\frac{k_{\lambda}\langle t)-1}{(\lambda^{-2}t)^{2}}=\lim_{sarrow+0}\frac{\cos\pi\neq-1}{s^{2}}=\lim\underline{\cos^{2}\pi s-1}=-\lim_{sarrow+0}\frac{\sin^{2}\pi s}{s^{2}(\cos\pi s+1)}=-\frac{\pi^{2}}{2}$$sarrow+0S^{2}(\cos\pi \mathit{8}+1)$

for any $t\in[c, \infty)$ . Thus$\kappa_{1}=4,$$\kappa_{2}=2$

and $C=-\pi^{2}/2$ .

19(3) Rogosinski summation is given by$\cos\frac{\pi t}{2\lambda^{2}}$$t<\lambda^{2}$,$0$$t\geq\lambda^{2}$.

$k_{\lambda}(t)=\{$In this case,$(\lambda^{2}/t.)^{2}\{k_{\lambda}-\langle t$) $-1$ ] are uniformly bounded in$\lambda$

and $t\in[c, \infty)$ and$\lim_{\lambdaarrow\infty}..\frac{k_{\lambda}(t)-1}{\langle\lambda^{--2}t)^{2}}=Sarrow+\lim\frac{\mathrm{c}\mathrm{o}s\frac{\pi}{2}\mathit{8}-1}{\mathscr{S}}=-0sarrow\lim\frac{\mathrm{s}^{A}\ln\frac{\pi}{2}2s}{s^{2}(\cos\frac{\pi}{2}s+1)}=-(+0)\frac{\pi}{2}2$

for any$t\in \mathrm{f}^{c},$$\infty$). Thus$\kappa_{1}=4,$$\kappa_{2}=2$.

$\frac{1}{2}=-\frac{\pi^{2}}{8}$and $C=-\pi^{2}/8$ .

(4) Jackson summation is given by$k_{\lambda}(t)=\{\vee$

In this case,,$1- \frac{3}{2}(\frac{t}{\lambda^{2}})^{2}+\frac{3}{4}(\frac{t}{\lambda^{2}})^{3}$$t<\lambda^{2}$$\frac{1}{4}(2-\frac{t}{\lambda^{2}})^{\mathrm{a}}$$\lambda^{2}\leq t<2\lambda^{2}$$0$$t\geq 2\lambda^{2}$

are umiformly bounded in$(\lambda^{2}/t)^{2}[k_{\lambda}(t)-11$$\lim_{\lambdaarrow\infty}(\lambda^{2}/t)^{2}[k_{\lambda}(t)-1]=-3/2$. Thus$\kappa_{1}=4,$$\kappa_{2}=2$$\lambda$,.
and $t\in[c, \infty)$ andand $C=-3/2$ .

(5) Gauss-Weierstrass summation: We consider the multiplier $k_{\lambda}^{W}(t)=\exp(-t/\lambda)$ .
The function of $t(\lambda/t)[k_{\lambda}(t)-1]$ is bounded uniformly in$\lambda$, and we have$\lim_{\lambdaarrow\infty}\frac{k_{\lambda}(t)-1}{\lambda^{-1}t}=\lim_{arrow s+0}\frac{e^{-s}-1}{s}=-\lim_{0sarrow+}e-\mathit{8}=-1$

Thus$\kappa_{1}=\kappa_{2}=1$$\exp(-\sqrt{t}/\lambda)$3

and$C=-1_{\lrcorner}$, and we have

Estimates of.

Poisson summation is given by the function$\kappa_{1}=1$

and $\kappa_{2}=1/2$ .

$k_{\lambda}(\hat{A})f-f$.

The aim of this section is to prove the following theorem.

$k_{\lambda}^{P}(t)=$20Theorem 2. Let
extension $of-\Delta$ in

of

$K$

be an open domain in$\Omega$$\Omega$$0,$ $\infty$

and$\kappa_{1}>0$$\hat{A}$$\{k_{\lambda}(X)\}$

be a nonnegative$\Omega$$\mathit{8}elfadj_{oit}n$and $K’$ is a closed subset

be a family of bounded piecewise smooth) such that $k_{\lambda}(t)\sqrt{t}^{n/\kappa}2-22+1/2\in L^{1}(0, \infty)$ with a constantand $k_{\lambda}(0)=1$ for any

Suppose thatand. Suppose that $K$ is a compact set inwith dist $(K’, K^{c})>0_{-}$ Let

functions on I

$R^{n}$$\lambda$$\kappa_{2}>0$.

$\{k_{\lambda}(t)\}$satisfies the conditions (1.1), (1.2) and (1.3) with a constant$0<R<\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(K’,K^{c})>0$

. Let

$f$

be a regulated function in$L^{2}(\Omega)$.

Suppose that $1<p\leq\infty$ and $f\in L^{p}(K)$ . Then the following hold.

(i) If$(-\Delta)^{\kappa_{2}}f\in L^{p}(K)$, then$||k_{\lambda}(\hat{A})f-f||_{Lp(K’})=O(\lambda^{-\kappa_{1}})$(ii) If$(-\Delta)^{\hslash_{2}}f$$\lambdaarrow\infty$as$\lambdaarrow\infty$.

vanishes in $K$ , then$||k_{\lambda}(\hat{A})f-f||_{L^{p}(K’})=o(\lambda^{-\kappa_{1}})$

3.1

as

.

Generalized eigenfunction system.

In order to prove Theorem 2, we shall use the generalized eigenfunction system cor-

responding to an ordered representation of$L^{2}(\Omega)$

associated with the Laplace operator.

21
We shall begin with several defimitions. We consider $A=-\Delta$ as an operator in$L^{2}(\Omega)$with the domain of

extension of

$A$

. Let

corresponding to$H(h)$$\wedge.=$$=$$\hat{A}$$\mathfrak{B}$$\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\dot{\mathrm{m}}\mathrm{t}\mathrm{i}_{0}\mathrm{n}D(A)=C_{c}^{\infty}(\Omega)$. Let$\hat{A}$be a nonnegative selfadjoint

be the Borel field on $R$ and $E$ be the unique spectral measure. For$h\in L^{2}(\Omega)$$\{F(\hat{A})h;F$, we define the following closed subspace of$L^{2}(\Omega)$:

is a Borel function on $R$ and $h\in D(F(\hat{A}))\}$$\{F(\hat{A})h;F\in L^{2}(R, \mathfrak{B}, (E(\cdot)h, h))\}$.

If $f\in H(h)$ , then we can write uniquely $f=F(\hat{A})h$ , where $F\in L^{2}(R, \mathfrak{B}, (E(\cdot)h, h))$and$||f||_{L^{2}(\Omega)}=(I_{R}^{|F(}t)|2(E(dt)h, h))1/2$

Therefore we can define an isomorphism$U_{h}f:=F$from $H(h)$ onto $L^{2}(R, \mathfrak{B}, (E(\cdot)h, h))$ by, which preserves inner products.

There exist a sequence of functions$\mathfrak{B}$$U_{h}$$\{h_{j}\}\subset L^{2}(\Omega)$

and a sequence of sets$\{e_{j}\}\subset$, called the set of multiplicity, with the following properties (see [3, XII.3.16] or [4,Chap.14]):(I)$L^{2}( \Omega)=\bigoplus_{j}H(h_{j})$That is, $H(h_{j})$ are mutually orthogonal and span.

$L^{2}(\Omega)$

.

.

(II)

$R=e_{1}\supseteq e_{2}\supseteq\cdots$(III)$(E(e)h_{j}, h_{j})=(E(e\cap e_{j})h_{1}, h_{1})$for any$e\in$B.

22
By (I), for $f\in L^{2}(\Omega)$ we can write uniquely$f= \sum_{j}F_{j()h}\hat{A}j$

where$F_{j}\in L^{2}(R, \mathfrak{B}, (E(\cdot)h_{j}, h_{j}))$,and$( \sum_{j}\int_{R}|F_{j}(t)|^{2}(E(dt)hj, h_{j})11/2=(\sum_{j}||Fj(\hat{A})hj||_{L}^{2}2(\Omega))^{1/}2|=|f||_{L()}2\Omega<\infty$

Therefore we can define an isometry

from

$U$

$L^{2}(\Omega)$

onto

.

$\bigoplus_{j}L^{2}(R, \mathfrak{B}, (E(\cdot)h_{j}, h_{j}))$,which is equivalent to say$L^{2}(\Omega)rightarrow\{\{F_{j}\};F_{j}\in L^{2}(R, \mathfrak{B}, (E(\cdot)h_{j}, h_{j}))$and $\sum_{j}\int_{R}|F_{j}(t)|^{2}(E(dt)h_{j}, h_{j})<\infty\}$ ,and the correspondence is given by $Uf:=\{F_{j}\}$ . We denote $F_{j}=:(Uf)_{j}$ .

By (III) we have$\bigoplus_{j}L^{2}(R, (E(\cdot)h_{j}.’Let $p:=(E(\cdot)h_{1}, h_{1})$ . Then$U$h_{j}))=\bigoplus_{j}L^{2}(e_{j}, (E(\cdot)h1, h1))$

is an isomorphism from$L^{2}(\Omega)$

onto

.

$\oplus_{j}L^{2}(e_{j}, \rho)$

whichpreserves imer products, that is, for any $f,g\in L^{2}(\Omega)$ it holds that
(3.1)
$U$

$(f,g)_{L^{2}(\Omega)}= \sum_{j}\int_{e_{j}}(Uf)_{j}(t)\overline{(Ug)j(t)}_{P(d}t)$

is called an ordered representation of$L^{2}(\Omega)$with respect to.

$\hat{A}$.

With these understood, there exists a sequence of functions $\{u_{j}(x, t)\}$ defined on the
product space of$\Omega\cross R$

such that the following conditions are satisfied (see [3, XII.3 andXIV.6] or [4, Chap.15] :
$)$

(i)

The functions $u_{j}(x,t)$ are $dx\cross d\rho(t)$ -measurable and vanish outsidewhere $dx$ is the Lebesgue measure.

$\Omega\cross e_{j}$

,

23
(ii)

For any fixed $t\in R$ , each
(3.2)(\"ui)$u_{j}(x, t)$belongs$C^{\infty}(\Omega)$$-\triangle u_{j}(X, t)=tuj(X, t)$

For each compact subset $K$ of$\Omega$,and satisfies$x\in\Omega$and each bounded Borel set$\mathrm{e}\mathrm{s}\mathrm{s}\sup_{x\epsilon K}\int_{e}|u_{j}(x, t)|2(\rho dt)<\infty$(iv)

For each$(Uf)_{j(t)=} \int_{\Omega}f(X)\overline{u_{j(}x,t)}dX$

where the integral exists in the sense of

For each$f\in L^{2}(\Omega)$

(3.4)

for

(3.5)

and each$L^{2}(e_{j,\rho})$,.

$e\in \mathfrak{B}$$s\mathrm{t}\mathrm{s}$,and the series converges in the sense ofis called the generalized eigenfunction system of$f\in L^{2}(\Omega)$$\hat{A}$$L^{2}(\Omega)$corresponding to

we have$f(X)= \sum j\int_{R}(Uf)_{j(}t)u_{j}(x,t)\rho(dt)$

and
(3.6)

in $R$.

$E(e)f(_{X)\sum}=j \int_{e}(Uf)_{j(}t)uj(x, t)\rho(dt)$where the integral exi$\{u_{j}\}$$e$$f\in L^{2}(\Omega)$

(3.3)

(v)

.

$k_{\lambda}( \hat{A})f(x)=\sum_{j}\int_{R}k_{\lambda}(t)(Uf)_{j(t)}u_{j}(x,\mathrm{t})p(dt)$

.

.
$U$

. By (v),24

Decomposition of

3.2

Throughout what follows,$s$elfadjoint extension ofrespect to

with

$U$

$\hat{A},$$-\Delta$$\Omega$$k_{\lambda}(\hat{A})f-f$

denotes an open domain in

. Let

$U$

$R^{n}$

and$\hat{A}$$p$$\Gamma$$\omega_{n}$. LetLemma 2. Let$\kappa_{2}$$S^{n-1}$by$\sigma$$\mathrm{Z}^{2}(\Omega)$withthe measure associated. We denote the gamma function by , the unit sphere inbyis a nonnegative

denote an ordered representation of

the generalized eigenfunction system and$\{u_{j}\}$Lebesgue measure on the unit sphere$S^{n-1}$.

and the surface area$R^{n}$

by$S^{n-1}$, the$2\sqrt{\pi}^{n}/\Gamma(n/2)$

ofbe a constant in (1.1), (1.2) and (1.3), and $\nu=n/2-2\kappa_{2}+1$ .

$f\in L^{2}(\Omega)_{f}x\in\Omega$

and $R>0$ . Then$k_{\lambda}(\hat{A})f(x)-f(X)$$=- \sum_{j}f_{0}^{\infty}t(I7f)j(t)u_{j}(x, t)\rho(dt)f0)I\lambda(rr^{\nu+}dr\int_{0}^{r}R\frac{J_{\nu+1(\sqrt{t}}S)}{(\sqrt{t}s)\mathcal{V}+1}1Sd_{S}$$+ \sum_{\mathrm{i}}\int_{0}^{\infty}\frac{(Uf\rangle_{j}\not\in t\rangle\ j\mathrm{t}x,l)}{\sqrt{t}^{\nu}} \rho(dt)\int_{R}^{\infty}I_{\lambda()}rJ\nu(\sqrt{t}r)rdr$$-f(x) \mathrm{X}\frac{1}{2^{\nu}\Gamma(\nu+1)}\int_{R}^{\infty}I_{\lambda}(r)rd\nu+1r$

,

where
$I_{\lambda}(r)= \int_{0}^{\infty}k_{\lambda}(s^{2})J_{\nu}(rs)s^{\nu+1}dS$

Proof. First observe that the function$k_{\lambda}(t)\sqrt{t}^{\nu-1}$$k_{\lambda}(t)$.

is piecewise smooth on$[0, \infty)$

andis integrable on $(0, \infty)$ . By Hankel’s integral formula $([2, \mathrm{p}.73,(60)])$ , wehave

$k_{\lambda}(t)$$=$$\frac{1}{\sqrt{t}^{\nu}}\int_{0}^{\infty}J_{\nu}(\sqrt{t}r)rdr\int_{0}^{\infty}k_{\lambda}(s^{2})J_{\nu}(rs)s^{\nu+1}d_{S}$$=$$\frac{1}{\sqrt{t}^{\nu}}\int_{0}^{\infty}I_{\lambda}(r)J_{\nu}(\sqrt{t}r)r$

dr.

25
Then, by (3.5), (3.6) and the fact that$k_{\lambda}(\mathrm{O})=1$, we have$k_{\lambda}(\hat{A})f(x)-f(x)$$= \sum_{j}\int_{0}^{\infty}\{k_{\lambda}(t)-k_{\lambda}(\mathrm{o}\rangle\}(Uf)_{j}(t)uj(x, t)\rho(dt)$$= \sum_{j}f_{\alpha}^{\infty}(Uf)_{j}(t)u_{j}(_{X}, t)\rho(dt)f_{\mathrm{t}}^{\infty}\}\dagger\frac{\sqrt{\nu}(\sqrt\overline{t}r)}{\sqrt{t}^{\nu}}-\frac{r^{\nu}}{2^{\nu}\Gamma(\nu+1)}\}I_{\lambda}(r)rdr$$= \sum_{j};_{0}^{\infty}(Uf)j(t)u\mathrm{j}(X-,P)p(dt)f0R\dagger\frac{J_{\nu}(\sqrt{t}r)}{\sqrt{t}^{\nu}}-\frac{r^{\nu}}{2^{\nu}\Gamma(\nu+1)}\mathrm{I}I\lambda(r)rdr$$+ \sum_{j}\int_{0}^{\infty}(Uf)j(t\rangle uj(X, t)\rho(dt)fR\infty\{\frac{J_{\nu}(\sqrt{t}r)}{\sqrt{t}^{\nu}}-\frac{r^{\nu}}{2^{\nu}\Gamma(\nu+1)}\}I\lambda(r)r$dr.

Now apply the formula ( , p.45])$\mathrm{I}7$$\frac{J_{\nu}(\sqrt{t}r)}{\sqrt{t}^{\nu}}-\frac{r^{\nu}}{2^{\nu}\Gamma(\nu+1)}=-tr^{\nu}\int 0\frac{J_{\nu+1}(\sqrt{t}\mathit{8})}{(\sqrt{t}s)^{\nu}+1}Sd_{S}r$.

Note that for the second term, we have$\sum_{j}\int_{0}^{\infty}(Uf)j\langle t)u\mathrm{j}(X, t)\rho(dt)\int^{\infty}R\{\frac{J_{\nu}(\sqrt{t}r)}{\sqrt{t}^{\nu}}-\frac{r^{\nu}}{2^{\nu}\Gamma(\nu+1)}\}I_{\lambda}(r)rdr$$= \sum_{j}\int_{0}\infty\frac{(Uf)_{j}(X)u_{j}(x,t)}{\sqrt{t}^{\nu}}\rho(dt)\int_{R}\infty\nu Ix(r)J(\sqrt{t}r)rdr$$-f(x) \cross\frac{1}{2^{\nu}\Gamma(\nu+1)}\int_{R}^{\infty}I_{\lambda}(r)r^{\nu+_{d}}r1$.

Thus we get Lemma 2.

3.3

Proof of Theorem 2.

Let

$f$

be a regulated function in$L^{2}(\Omega)$. Let $K$ be a compact set in$\Omega$and $K’$ bea closed set in $K$ with dist $(K’,K^{c})>0$ . We choose $0<R<$ dist $(K’, KC)$ . Letand$\kappa_{2}$$\kappa_{1}$be constants in (1.1), (1.2) and (1.3). Let $\nu=n/2-2\kappa_{2}+1$ and $1<p\leq\infty$ .

26

Suppose that $f\in L^{p}(K)$ and$(-\Delta)^{\kappa_{2}}f\in L^{p}(K)$. By Lemma 2, we have$||k_{\lambda}( \hat{A})f-f||_{Lp}(K’)\leq\}|f\}|L^{p}(K’)\cross\frac{1}{2^{\nu}\Gamma(\nu+1)}|\int_{R}^{\infty}I\lambda(r)r\nu+1dr|$$+||I_{\theta}^{I_{\lambda}(r}R \int_{\alpha}^{r})rdrsds\sum_{j}\int^{\infty}\nu+1t(Uf)j(t)uj(\cdot, t)\frac{J_{\nu+1}(\sqrt{t}s)}{(\sqrt{t}s)^{\nu}+1}0\rho(dt)||_{L^{\mathrm{p}}(K)}$

,

(3.7)

$+|| \sum_{j}\int_{0}^{\infty}\frac{(Uf)_{j}(t)u_{j(\cdot,t)}}{\sqrt{t}^{\nu}}\rho(dt)\int_{R}\infty\nu I_{\lambda(r)J}(\sqrt{t}r)rdr||_{L^{\infty}(K)},\cdot$.
Lemma 3. We have$|| \int_{0}RI_{\lambda}(r)rd\nu+1\int r\int \mathrm{o}t(Uf)j(t)u_{j}(\cdot,t)\frac{J_{\nu+1(\sqrt{t})}S}{(\sqrt{t}s)^{\nu}+1}\infty 0r_{\mathit{8}d_{S}\sum_{j}p(dt)}||_{L^{\mathrm{p}}(K’})$$\leq C\lambda^{-\kappa 1}||(-\triangle)\kappa_{2}f||L\mathrm{p}(K)$.

Proof. Let $x\in K’$ and $0<s<R$ . Put,$g_{s}(y)$$= \frac{1}{s^{\nu+1}|y|^{n}/2-1}\int_{0}^{\infty}J_{\nu+1}(Sr)Jn/2-1(|y|r)dr$$g_{s}^{x}(y)$$=g_{s}(x-y)$.

If $|y|>s$ , then $g_{s}(y)=0([7, \mathrm{p}.404,(6)])$ . Therefore$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}g_{s}^{x}\subset K\subset\Omega$. Then, by (3.3),

we have$(Ug_{s}^{x})_{j}(t)$$=$$\int_{\Omega}g_{s}^{x}(y)\overline{uj(y,t)}dy$$=$$\int_{\Omega}gs(y)\overline{uj(x-y,t)}dy$$– \frac{1}{s^{\nu+1}}I^{\overline{u_{j}(x-y}},t)dy\frac{1}{1y\downarrow n/2-1}\int_{0}\infty(J\nu+1(sr)J_{n/2}-1|y|r)dr$$=$$\frac{1}{s^{\nu+1}}\int_{0}^{\infty}q^{n/2}dq\int_{S^{n-1}}\overline{uj(x-qw,t)}\sigma(dw)\int_{0}^{\infty}J_{\nu}+1(sr)J_{n}/2-1(qr)$

dr.

27
On the other hand, by (3.2),
for$y\in\Omega$$u_{j}(y, t)\in C^{\infty}(\Omega)$, and we have $-\Delta u_{j}(y, t)=tuj(y, t)$. Therefore, by the mean-value formula, we have.

$\int_{S^{n-1}}u_{j}(X-qw,t)\sigma(dw)=\sqrt{2\pi}^{n}\frac{\sqrt n/2-1(\sqrt{t}q)}{(\sqrt{t}q)^{n/}2-1}u_{j(Xt)},$Thus, by Hankel’s formula, we have$(Ug_{s}^{x})_{j}(t)$$=$$\frac{\sqrt{2\pi}^{n}}{\sqrt{t}^{n/2-1}S\nu+1}\overline{u_{j}(x,t)}\int 0-(\sqrt{t}q)\infty J_{n}/21qdq\int_{0}\infty J_{\nu+1}(Sr)J_{n/}2-1(qr)dr$$=$$\frac{\sqrt{2\pi}^{n}J_{\nu+1}(\sqrt{t}s)}{\sqrt{t}^{n/2}s\nu+1}\overline{u_{j}(x,t)}$We can assume that$f\in C_{c}^{\infty}\text{ノ}(\Omega)$.

by approximation. Then, by (3.1), we have$\sum_{j}\int_{0}^{\infty}t(Uf)j(t)uj(x, t)\frac{J_{\nu+1}(\sqrt{t}s)}{(\sqrt{t}s)^{\nu}+1}\rho(dt)$$= \frac{1}{\sqrt{2\pi}^{n}}\sum_{j}\int_{e}t^{\hslash_{2}}(Uf)_{j(}t)\overline{(Ugxs)_{j}}j(t)\rho(dt)$$= \frac{1}{\sqrt{2\pi}^{n}}\int_{\Omega}[(-\Delta)^{\kappa}2f(y)]g_{s}(xy)dy$.

$= \frac{1}{\sqrt{2\pi}^{n}}\int_{\Omega}[(-\triangle)^{\kappa_{2}}f(y)]g_{s}(X-y)dy$

Therefore we have$\int_{0}^{R}I\lambda(r)r\nu+1dr\int^{r_{Sd}}0\sum SI0\frac{J_{\nu+1}(\sqrt{t}s)}{(\sqrt{t}s)^{\nu}+1}t(Uf)j(t)uj(X,t)\rho j\infty(dt)$$= \frac{1}{\sqrt{2\pi}^{n}}\int_{0}^{R}I_{\lambda}(r)rd\nu+1r\int_{0}^{r}sdS\int|y|<sy[(-\Delta)^{\kappa}2f(X-)]g_{s}(y)dy$$= \frac{1}{\sqrt{2\pi}^{n}}\int_{0}^{R}sdS\int s\nu I_{\lambda(}r)r+1Rrd\int_{|y|<\mathit{8}}[(-\Delta)^{\kappa_{2}}f(x-y)]g_{\theta}(y)dy$.

28
Applying successively Minkowski’s inequality for integral, we have$|| \int_{0}^{R}I_{\lambda}(r)rd\nu+1r\int^{r}0SdS\sum_{j}\int^{\infty}0t(Uf)_{j}(t)uj(\cdot,t)\frac{j_{\nu+1}(\sqrt{t}s)}{(\sqrt{t}s)^{\nu}+1}\rho(dt)||_{L^{\mathrm{p}}(K)}$,$\leq\frac{1}{\sqrt{2\pi}^{n}}\int_{0}^{R}sds||\int_{s}^{R}I_{\lambda}(r)r^{\nu}d+1rI_{1\mathrm{I}}y<s)[(-\Delta\kappa_{2}f(\cdot-y)]gs(y)dy||_{L^{\mathrm{p}}(K’)}$$= \frac{1}{\sqrt{2\pi}^{n}}\int_{0}^{R}sd_{S}|\int_{s}^{R}I_{\lambda}(r)r\nu+1dr|||\int_{|y|<s}[.(-\Delta)\kappa 2f(\cdot-y)]g_{\beta}(y)dy||_{L^{\mathrm{p}}(K’\rangle}$$\leq\frac{1}{\sqrt{2\pi}^{n}}\int_{0}^{R}sds|\int_{s}^{R}I_{\lambda}(r)rd\nu+1r|\int_{|y|<s}||(-\Delta)^{\kappa 2}f(\cdot-y)||Lp(K^{l})|g_{s}(y)|dy$$\leq\frac{1}{\sqrt{2\pi}^{n}}||(-\triangle)\kappa_{2}f||L\mathrm{p}(K)\int_{0}^{R}sdS|I_{s}^{R}I_{\lambda}(r)r^{\nu+}1dr|\int_{|y|<}s|g_{s}(y)|dy$.

On the other hand, we have$\int_{|y|<S}|g\mathit{8}(y)|dy$$=$$\frac{1}{s^{\nu+1}}\int_{|y|<}s\frac{1}{|y|^{n/21}-}dy|\int_{0}^{\infty}J_{\nu+}1(sr)J_{n/}2-1(|y|r)dr|$$=$$\frac{\omega_{n}}{s^{\nu+1}}\int_{0}^{s}q^{n/2}dq|\int_{0}^{\infty}J_{\nu+1}(sr)jn/2-1(qr)dr|$$=$$\frac{\omega_{n}\Gamma((2\nu+n+2)/4)}{\Gamma(n/2)\mathrm{r}((2_{\mathcal{U}}-n+6)/4)s\nu+n/2+1}$$\cross\int_{0}^{s}|_{2}F_{1}((2\nu+n+2)/4,$ $-(2\nu-n+2)/4;n/2;q2/S^{2})|q^{n-1}dq$$\leq$

where$2F1(\alpha,\beta;\gamma;Z)$$\frac{C_{\kappa_{2}}}{s^{\nu-n/2+1}’}$is Gauss’ hypergeometric function.

Therefore the last term isbounded by$C_{\kappa_{2}}||(- \triangle)\kappa 2f||_{L^{p}(K)}\int^{R}0s^{2\kappa 2}-1d_{S}|\int_{s}^{R}I_{\lambda}(r)r^{\nu+1}dr|$

By the condition (1.1), we get the boundLemma 3 is proved.

$C\lambda^{-\kappa_{1}}||(-\Delta)\kappa_{2}f||_{L^{p}()}K$.

for the last term. Thus

29
We shall use the folowing lemma ([1, p.655]).

Lemma 4. Under the assumptions above, if$Ki\mathit{8}$a compact set contained in$( \sum_{j}\int_{\tau\leq\sqrt{t}T}\leq+1|u_{j(}X,t)|2\rho(dt))^{1/}2n\leq cK(T+1)^{(}-1)/2$

where$\Omega$, then,is a constant independent of $T\geq 0$ and $x\in K$ .

$C_{K}$

Lemma 5. We have$||j \sum\int_{0}^{\infty}\frac{(Uf)_{j()}tu_{j(\cdot,t)}}{\sqrt{t}^{\nu}}\rho(dt)\int_{R}\infty\lambda I(r)J\nu(\sqrt{t}r)rdr||_{L(K)}\infty=o(\lambda^{-\kappa_{1}})$as$\lambdaarrow\infty$.

Proof. We have, by Schwarz’s inequality,$| \sum_{j}\int_{0}^{\infty}\frac{(Uf)_{j()u()}tjx,t}{\sqrt{t}^{\nu}}\rho(dt)\int RI\infty\lambda(r)J_{\nu}(\sqrt{t}r)rdr|$$\leq(\sum_{j}\int_{e_{j}}|(Uf)_{j}(t)|^{2}p(dt))1/2$$\cross(\sum_{j}\int_{0}^{\infty}\frac{|u_{j}(X,t)|^{2}}{t^{\nu}}\rho(dt)|\int_{R}^{\infty}I_{\lambda}(r)J_{\nu}(\sqrt{t}r)rdr|^{2})^{1/2}$

Now, by (3.1), we have$( \sum_{j}\int_{e_{j}}|(Uf)_{j}(t)|^{2}\rho(dt))^{1/}2f=||||_{L^{2}(}\Omega)$.

30
By Lemma 4, there exists a constant$C_{K}$

such that$( \sum_{j}\int_{0}\infty\frac{|u_{j(,)1^{2}}Xt}{t^{\nu}}\rho(dt)|\int R\sqrt{t}\infty I\lambda(r)J\nu(r)rdr|2)^{1/2}$$\leq C_{K}(\sum_{T=0}^{\infty}\tau^{4\kappa_{2}-}3\max_{sT\leq\leq T+1}|\int_{R}^{\infty}I_{\lambda(r)(Sr)r}J\nu dr|^{2})^{1/2}$

umiformly in $x\in K$ . Therefore, by (1.3), we have$| \sum_{j}\int_{0}^{\infty}\frac{(Uf)_{j()(x,t)}tu_{j}}{\sqrt{t}^{\nu}}p(dt)\int_{R}\infty\lambda I(r)J\nu(\sqrt{t}r)rdr|=o(\lambda^{-\kappa_{1}})$

uniformly in $x\in K$ as

We remark that$\lambdaarrow\infty$. Thus Lemma 5 is proved.

$| \int_{R}^{\infty}I_{\lambda}(r)r^{\nu}d+1r|=o(\lambda^{-\kappa_{1}})$By (3.7) together with Lemmas 3 and 5,

If$(-\Delta)^{\kappa_{2}}f$by the assumption (1.2).

$||k_{\lambda}(\hat{A})f-f||_{L^{\mathrm{p}}(K},$$=O(\lambda^{-\kappa_{1}})$)as$\lambdaarrow\infty$

vanishes in $K$ , then by Lemma 3$=0$$|| \int_{0}^{R}I\lambda(r)r^{\nu+1}dr\int^{r}0Sd_{S}\sum_{j}I_{0}^{\infty}t(Uf)_{j()u}tj(\cdot, t)\frac{J_{\nu+1(\sqrt{t}}S)}{(\sqrt{t}s)^{\nu}+1}\rho(dt)$.

$L^{p}(K’)$

Therefore, by (3.7) and Lemma 5, we have

Consequently, Theorem 2 is proved.

$||k_{\lambda}(\hat{A})f-f||_{L^{p}(K},$$=o(\lambda^{-\kappa_{1}})$)as$\lambdaarrow\infty$

.

.

31

4

Applications of main theorem.

4.1

Proof of Corollary 1.

Let$k_{\lambda}(t)=(1-t/\lambda 2)^{\delta}+\cdot$

Then we have the fomula (see [2,$\mathrm{p}.92,(34)]$),$k_{\lambda}(t)= \frac{2^{\delta}\Gamma(\delta+1)}{\lambda^{\delta-n}/2\sqrt{t}n/2-1}\int^{\infty}0\frac{J_{n/2+s}(\lambda r)J_{n/-}21(\sqrt{t}r)}{r^{\delta}}dr$and can take$\kappa_{2}=1$. We have.

$I_{\lambda}(r)= \int_{0}^{\infty}k_{\lambda}(t^{2})J_{n/2-1}(rt)tn/2dt=2^{\delta}\mathrm{r}(\delta+1)\lambda^{n}/2-\delta J/2+\delta(n\lambda r)r-\delta-1$

To check the conditions (1.1), (1.2) and (1.3), let $R>0$ and $\delta>(n-3)/2$ . Then wehave$| \int_{R}^{\infty}I_{\lambda(}r)rdr\iota/2|r=2\delta\Gamma(\delta+1)\lambda^{n}/2-\delta|\int_{R}^{\infty}\frac{J_{n/2+\delta}(\lambda r)}{r^{\delta-n/2}+1}dr|\leq C_{\delta,R}\lambda^{(}n-3)/2-\delta$.

On the other hand, we have$\int_{0}^{R}sds$$| \int_{s}^{R}I_{\lambda}(r)r/n2dr|=2^{s}\Gamma(\delta+1)\lambda^{n}/2-\delta\int^{R}\mathrm{o}Sd\mathit{8}|\int_{s}^{R}\frac{J_{n/2+\delta(\lambda}r)}{r^{\delta-n/2+}1}dr|$$C_{\delta}\lambda^{(n-}3)/2-\delta$$\leq\{$

if $(n-3)/2<\delta<(n+1)/2$ ,$c_{s^{\lambda^{(n})/\delta}\log\lambda}-32-$

if$\delta=(n+1)/2$ ,$C_{\delta}\lambda^{-2}$if$\delta>(n+1)/2$.

We now apply the estimates (see [6, p.202, Lemma 18.10$\mathrm{a}]$)$| \int_{R}^{\infty}\frac{J_{n/2+\delta}(\lambda r)Jn/2-1(_{S}r)}{r^{\delta}}dr|$$C_{\delta,R}\lambda-1/2_{S^{-}}1/2$$\leq\{$$C_{\delta,R^{\frac{\lambda^{-3/21}\mathit{8}/2}{\lambda-s}+}}c_{s,R}\lambda-3/2-S1/2$$C_{\delta,R} \frac{\lambda^{1/2-3}S\mathit{1}2}{s-\lambda}+c_{\delta},R\lambda-1/23/2s^{-}$,

if$s,$ $\lambda>0$if$0<s<\lambda$,

if$0<\lambda<s$.

32Then we have$( \sum_{T=0}^{\infty}\tau_{\tau\leq s}\max\leq T+1|\int_{R}^{\infty}I_{\lambda}(r)J_{n}/2-1(sr)rdr|^{2})^{1/2}$$=2^{\delta} \Gamma(\delta+1)\lambda^{n}/2-\delta(\sum_{T=0}^{\infty}T_{\tau\leq}\max|s\leq T+1I_{R}^{\infty}\frac{J_{n/2+\delta}(\lambda r)\sqrt n/2-1(\mathit{8}r)}{r^{\delta}}dr|^{2})^{1}/2$$\leq C_{\delta,R}\lambda^{(}n-1)/2-\delta$.

If $\delta>(n+3)/2$ , then the last term is$o(\lambda^{-2})$. Thus Corollary 1 follows from Main

theorem.

The Gauss-Weierstrass summation.

4.2
Let

$k_{\lambda}^{W}(t)=e^{-t/\lambda}(\lambdaarrow\infty)$. We then have$(4.1) \int_{0}^{\infty}k_{\lambda}W(t^{2})J_{\nu}(rt)t^{\nu+1}dt=\int_{0}^{\infty}e^{-t^{2}/\lambda}J_{\nu}(rt)t^{\nu+1}d\mathrm{t}=\frac{\lambda^{\nu+1}r^{\nu}}{2^{I’+1}}\exp(-\frac{\lambda r^{2}}{4})$(cf. [2, 7.7.3]). Let

extension of$-\Delta$

Corollary 2. Let$f\in L_{loc}^{p}(\Omega)$inbe an open domain in$\Omega$$\Omega$$f$$R^{n}$

and$\hat{A}$be a nonnegative selfadjoint.

be a regulated function in$L^{2}(\Omega)$.

$Suppo\mathit{8}e$. Then the following hold.

(i) The following are equivalent.

( a)$\mathrm{i}$$||k_{\lambda}^{W}(\hat{A})f-f||_{L^{p(K)}}=O(\lambda^{-1})$as$\lambdaarrow\infty$

for every compact set$(\mathrm{i}\mathrm{b})\Delta f\in L_{1}^{p}(\mathrm{o}\mathrm{c}\Omega)$

.

$K$

in

$\Omega$.

that$1<p\leq\infty$

and

33
(ii) Let$G\subset\Omega$be any open set.

(iia) Suppose that$\Delta f$vanishes in G. Then$||k_{\lambda}^{W}(\hat{A})f-f||_{L^{p}(K)}=o(\lambda^{-1})$as$(\mathrm{i}\mathrm{i}\mathrm{b})$$\lambdaarrow\infty$

for any compact set$K\subset G$.

If$||k_{\lambda}^{W}(\hat{A})f-f||_{L^{p}()}K=o(\lambda^{-1})$as$\lambdaarrow\infty$

for any compact set$K\subset G$, then$\Delta f$vanishes in

Proof. For the Gauss-Weierstrass summation method we take$G$$\kappa_{2}=1$

.

. Let

$R$

be a

small positive number. By (4.1), we have$\int^{\infty}Rr^{n/2}drl^{\infty}k_{\lambda}^{W}(t^{2})J_{\nu}(rt)t\nu+1dt=(\frac{\lambda}{2})^{n/2}\int_{R}^{\infty}r-1n\mathrm{p}\mathrm{e}\mathrm{x}(-\frac{\lambda r^{2}}{4}\mathrm{I}^{d}r=O(\lambda^{-}1)$$\int_{0}^{R}sdS|\int_{s}^{R}r^{n/2}dr\int_{0}\infty k\lambda W(t^{2})J_{\nu}(rt)t^{\nu+1}dt|$$=( \frac{\lambda}{2})^{n/2}\int_{0}RdSs\int sr^{n}-1\exp R(-\frac{\lambda r^{2}}{4})dr=O(\lambda^{-1})$

and$( \sum_{T=0}^{\infty}T\max\tau\leq s\leq\tau+1|\int_{R}^{\infty}J_{n/-}21(Sr)rdr\int_{0}\infty k^{W}\lambda(t^{2})J_{\nu}(rt)t\nu+1dt|^{2})^{1/2}$$=( \frac{\lambda}{2})^{n/2}(\int_{R}^{\infty}r^{n}-1\exp(-\frac{\lambda r^{2}}{2})dr)1/2(=\mathit{0}\lambda^{-1})$

Thus Corollary 2 follows from Main theorem.

.

,

34

参考文献
[1] S. A. Alimov and V. A.

$\mathrm{I}\mathrm{l}’ \mathrm{i}\mathrm{n}$, Conditions for the convergence of spectral expansions

corresponding to selfadjoint extensions of elliptic operators II (selfadjoint extensionsof Laplace’s operator with arbitrary spectra), Differential Equations 7 (1971), 651-

670.
[2] H. Bateman, Higher transcendental functions, Volume II,$\mathrm{M}\mathrm{c}\mathrm{G}\mathrm{r}\mathrm{a}\mathrm{w}$-Hill Book Com-pany, Inc. New York, 1953.
[3] N. Dunford and J. T. Schwartz, Linear operators, Part II (Spectral theory), Pureand Appl. Math. VII, Interscience Publishers, New York, 1963.
[4] H. Fujita, S.-T. Kuroda and S. It\^o, Fhnctional Analysis (Japanese), Iwanami Shoten,
T\^oky\^o, 1991.

[5] S. Igari, Saturation of the approximation by eigenfunction expansions associated with

the Laplace operator, T\^ohoku Math. J. 22 (1970), 231-239.
[6] E. C. Titchmarsh, Eigenfunction expansions associated with second order differential

equations, Part II, Oxford Univ. Press, Oxford, 1958.
[7] G. N. Watson, A treatise on the theory of Bessel functions, Cambridge Univ. Press,

Cambridge, 1944.

-mail

$E$

$addre\mathit{8}s$: tanigaki@math.tohoku.ac.jp