数理解析研究所講究録
1042 巻 1998 年 112-122

112

LAYER HEAT POTENTIALS FOR A BOUNDED CYLINDER
WITH FRACTAL LATERAL BOUNDARYHISAKO WATANABE (渡辺ピサ子)

Ochanomizu University
1. Introduction

Let $D$ be a bounded smooth domain in$\Omega_{D}=D\cross(0,T)$

The double layer heat potential(1.1)$\Phi f$$\mathrm{R}^{d}$andand set

$S_{D}=\partial D\cross[0,T]$.

of $f\in L^{p}(S_{D})$ is defined by$\Phi f(X)=-\int_{0}^{T}\int_{\partial D}\langle\nabla_{y}W(X-\mathrm{Y}), ny\rangle f(\mathrm{Y})d\sigma(y)ds$

is theis the inner product infor $X=(x,t)\in(R^{d}\backslash \partial D)\cross \mathrm{R}$, where$W$
andis the fundamental
is the surface measure on
unit outer normal toi.e.,operator,solution for the heat$\mathrm{R}^{d},$$\langle , \rangle$$\partial D,$$\sigma$$n_{y}$$\partial D$$W(X)=W(x,t)=$
The double layer heat potential is important not only physically but also mathematically. For example, R. M. Brown proved that the solution to the initial-Dirichletproblem in a Lipschitz cylinder for the heat operator can be written by a double
layer heat potential and the solution to the initial-Neumann problem in a Lipschitz,
cylinder for the heat operator is given by a single layer heat potential (cf.
$[B_{1}]$$[B_{2}])$.

and the surface measure
If $D$ is a bounded domain with fractal boundary, then$D$
can not be defined. But if has a smooth boundary and is a -function onwith compact support, then we see by the Green formula that for $X=(x, t)\in D\cross \mathrm{R}$$n_{y}$

$f$

(1.2)

$C^{1}$$\mathrm{R}^{d+1}$$\Phi f(X)=\int_{0}^{T}ds\int_{\mathrm{R}^{d}\backslash \overline{D}}\langle\nabla fy(\mathrm{Y}), \nabla W(y-x\mathrm{Y})\rangle dy$$+ \int_{0}^{T}d_{S}\int_{\mathrm{R}^{d}\backslash \overline{D}}f(\mathrm{Y})\triangle W(y-\mathrm{Y}x).dy$

Typeset by$Ay\mathrm{t}^{\mathrm{R}}?ffl$

113
and for$X=(x,t)\in(\mathrm{R}^{d}\backslash \overline{D})\cross \mathrm{R}$(1.3)$\Phi f(x)=-I_{0}^{\tau_{ds}}\int_{D}\langle\nabla_{y}f(\mathrm{Y}), \nabla W(x-\mathrm{Y})\rangle dyy$$- \int_{0}^{T}d_{\mathit{8}}\int_{D}f(\mathrm{Y})\triangle_{y}W(x-\mathrm{Y})dy$.

So we see that, if a function defined on
can be extended to be a functiononsuch that for each $t\in[0,T]$ the function
is a$(j=1,2,\cdots,d)$-function onand, for eachand eachthefunctionmeasurable,
isthen the right-hand sides of (1.2) and (1.3)
may be defined.
In this paper we assume that $D$ is a bounded domain inandis a-set satisfying $d-1\leq\beta<d$ . Here, according to [JW] we say that a closed set
is a -set if there exist a positive Radon measure on and positive real numbers$f$$S_{D}$$\mathrm{R}^{d}\cross[0,T]$$\mathcal{E}(f)$$C^{1}$$xrightarrow \mathcal{E}(f)(x,t)$$\mathrm{R}^{d}\backslash \partial D$$x\in \mathrm{R}^{d}\backslash \partial D$$j$$t \vdasharrow\frac{\partial \mathcal{E}\{f)}{\partial x_{j}}(x,t)$$\mathrm{R}^{d}(d\geq 2)$$\partial D$$F$$\beta$$\beta$$r_{0},$$b_{1},$$\mu$$b_{2}$$F$

such that(1.4)$b_{1}r^{\beta}\leq\mu(B(z, r)\cap F)\leq b_{2}r^{\beta}$, where $B(z, r)$ stands for the open ball infor all $z\in F$ and allwith centerand radius .
We note that, if $D$ is a bounded Lipschitz domain, thenis a $(d-1)$ -set and thesurface measure has the property (1.4) for $F=\partial D$ and $\beta=d-1.$ Krthermore
ifconsists of a finite number of self-similar sets, which satisfies the open set
condition, and whose similarity dimensions are , thenis a -set such that the-dimensional Hausdorff
restricted to
has the property (1.4) for$F=\partial D$(cf. [Hu]).
$0<\alpha\leq
1$ and $F$ be a closed set inLet. We denote bythe
Banach space of all continuous functions on $F\cross[0,T]$ such thatis -H\"older
continuous for every $t\in[0,T]$ with norm
$r\leq r_{0}$$\mathrm{R}^{d}$

$r$

$z$

$\partial D$$\mu$$\partial D$$\partial D$$\beta$measu.re$\beta$$\mathcal{H}^{\beta}$$\beta$$\partial D$$\mathrm{R}^{d}$$\Lambda_{\alpha}(F\cross[0,T])$$f(\cdot,t)$$f$$||f||_{\infty,\alpha}= \sup_{\mathrm{x}X\in F10},|f(x)|+\sup_{y\tau 1x,\in F,x\neq y,\iota\in 1^{0},\tau]}\frac{|f(x,t)-f(y,\iota)|}{|x-y|^{\alpha}}$FUrther let $0<\alpha,$$\lambda\leq 1$. We also denote by

all $f\in\Lambda_{\alpha}(F\cross[0,T])$ such thatvariable with norm$f$$\alpha$.

the Banach space ofis -H\"older continuous with respect to the time$\Lambda_{\alpha,\lambda}(F\cross[0, T])$$\lambda$$||f||_{\infty}, \alpha,\lambda=||f||_{\infty},\alpha+\sup_{s[0\tau],t\neq s},\frac{|f(x,t)-f(_{X},s)|}{|t-s|^{\lambda}}x\in F,t,\in$

We will prove the following lemma in \S 3.

.

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satisfying (1.4)
Lemma 1.1. Let $d-1\leq\beta<d$ and be a compac. -set into. Then there exists a bounded operator from
andwith the following properties:
for each $t\in[0,T]$ , and both of(i)is a -function on
and for eachare measurable for eachand$F$$\mathrm{R}^{d}$$t\beta$$F\subset B(\mathrm{O}, R/2)$$\Lambda_{\alpha}(F\cross[0,T])$$\mathcal{E}$$\Lambda_{\alpha}(\mathrm{R}^{d})$$C^{1}$$\mathcal{E}(f)(\cdot, t)$$\mathcal{E}(f)(x, \cdot)$$\mathrm{R}^{d}\backslash F$$x\in \mathrm{R}^{d}$$( \frac{\partial \mathcal{E}(f)}{\partial x_{j}})(x, \cdot)(j=1, \cdots, d)$$x\in \mathrm{R}^{d}\backslash F$

(ii)
(iii)

, respectively,on and$\mathcal{E}(f)=f$$F$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\mathcal{E}(f)(\cdot,t)\subset B(\mathrm{O}, 2R)$$| \frac{\partial \mathcal{E}(f)}{\partial y_{i}}(y, s)|\leq c||f||_{\infty,\alpha}\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(y, \partial D)^{\alpha-1}$

for every(iv) If$(y, s)\in(\mathrm{R}^{d}\backslash F)\cross[0, T]$$f\in\Lambda_{\alpha,\lambda}(F\cross[0,T])$$| \frac{\partial^{2}\mathcal{E}(f)}{\partial y_{i}\partial y_{k}}(y, s)|\leq c||f||_{\infty},\alpha \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(y, \partial D)^{\alpha-2}$.

, thenUsing Lemma 1.1 we define, for

(1.5)

,

for each $t\in[0,T]$ .

$\mathcal{E}(f)\in\Lambda_{\alpha,\lambda}(\mathrm{R}^{d}\cross[0,T])$$f\in\Lambda_{\alpha}(S_{D})$.

,$\Phi f(X)=\int_{0}^{T}d_{S}\int_{\mathrm{R}^{d}\backslash \overline{D}}\langle\nabla \mathcal{E}(f)(y, S), \nabla_{y}W(yX-\mathrm{Y})\rangle dy$$+ \int^{\tau_{ds}}\int_{\mathrm{R}^{d}\backslash \overline{D}}\mathcal{E}(f)(y, s)\triangle_{y}W(X-\mathrm{Y})dy$

for $X=(x,t)\in D\cross \mathrm{R}$ and(1.6)$\Phi f(X)=-\int_{0}^{T}d_{S}\int_{D}\langle \mathrm{v}_{y}\mathcal{E}(f)(y, S), \nabla W(yX-\mathrm{Y})\rangle dy$$- \int_{0}^{T}d_{S}\int_{D}\mathcal{E}(f)(y, s)\triangle_{y}W(X-\mathrm{Y})dy$for$X=(x,t)\in(\mathrm{R}^{d}\backslash \overline{D})\cross \mathrm{R}$.

Furthermore we also define the operator $K$ by(1.7)$Kf(Z)= \frac{1}{2}(I_{1}(Z)+I_{2}(Z))$

,

where
$I_{1}(Z)= \int_{0}^{\tau_{d_{\mathit{8}}}}\int_{\mathrm{R}^{d}\backslash \overline{D}}\langle\nabla \mathcal{E}(f)(y, s), \nabla_{y}W(z-\mathrm{Y})\rangle dy$$+ \int_{0}^{\tau_{d_{\mathit{8}}}}\int_{\mathrm{R}^{d}\backslash \overline{D}}(\mathcal{E}(f)(\mathrm{Y})-f(z))\triangle W(Zy-\mathrm{Y})dy$$+f(Z) \int_{(\mathrm{R}^{d}}\backslash \overline{D})\mathrm{x}\{0\}W(z-\mathrm{Y})dy$

.

115
and
$I_{2}(Z)=- \int_{0}^{T}d_{S}\int_{D}\langle\nabla \mathcal{E}(f)(y, s)"\nabla yW(Z-\mathrm{Y})\rangle dy$$- \int_{0}^{T}d_{S}\int_{D}(\mathcal{E}(f)(\mathrm{Y})-f(z))\triangle W(y-\mathrm{Y})zdy-f(Z)\int_{D\mathrm{x}\{0\}}W(z-\mathrm{Y})dy$

Under these notations we will prove the following theorem in \S 3.
Theorem. Assume that $D$ is a bounded domain insuch thatis a -set.
$Z\in\partial D\cross[0,T]$ ,$0\leq\beta-(d-1)<\alpha<1$
,
and
then,
each
If
for
$\mathrm{R}^{d}$$\partial D$$\beta$$f\in\Lambda_{\alpha,\alpha/2}(S_{D})$(1.8)$\lim_{Xarrow z,x\epsilon D\mathrm{x}\mathrm{t}0,\tau)}\Phi f(x)=Kf(Z)+\frac{1}{2}f(Z)$

and

(1.9)

$\Phi f(X)=Kf(Z)-\frac{1}{2}f(z)$$\lim$.

$Xarrow Z,\mathrm{x}\in(\mathrm{R}^{d}\backslash \overline{D}\rangle \mathrm{x}\langle 0,\tau)$Thus we see that our double layer heat potentials have the same boundary behavior as the usual ones for a bounded cylinder with smooth lateral boundary.
Remark. In this paper we shall treat the double layer heat potentials of H\"older
continuous functions on. But under a similar consideration we can also the
double layer heat potentials of functions in a Besov space on
and prove thatthey have the parabolically non-tangential limit at. $Z\in S_{D}$ .
$S_{D}$$S_{D}$$\mathrm{a}.\mathrm{e}$2. Properties of $W$
In this section we recall and study properties of the function $W$ . To do so, we
use the parabolic metric defined by$\delta$for $X=(x,t)$ and $\mathrm{Y}=(y, s)$ .

$\delta(X, \mathrm{Y})=(|x-y|^{2}+|t-S|)^{1/2}$

Lemma 2.1. (i) $W(X)\leq c\delta(X, 0)^{-d}$ ,(ii) $|\nabla_{x}W(X)|\leq c\delta(X, \mathrm{o})-d-1$ if $X\neq 0$ ,(iii)$| \frac{\partial^{2}}{\partial x.\partial x_{j}}.W(X)|\leq c\delta(X,\mathrm{o})-d-2,$$| \frac{\partial}{\partial t}W(X)|\leq c\delta(X, 0)-d-2$

if $X\neq 0$ ,if $X\neq 0$ ,(iv)(v) $|W(X-\mathrm{Y})-W(z-\mathrm{Y})|\leq c\delta(X, Z)\epsilon\{\delta(X,\mathrm{Y})^{-}d-\epsilon+\delta(Z,\mathrm{Y})-d-\epsilon\}$,andif$|\nabla_{y}W(X-\mathrm{Y})-\nabla
Wy(Z-\mathrm{Y})|\leq
c\delta(X, Z)\epsilon\{\delta(x,\mathrm{Y})^{-d1-}-\epsilon+\delta(Z,\mathrm{Y})-d-1-\epsilon\}$
(vi)
and
.
if
$| \frac{\partial^{3}}{\partial x:\partial x_{\mathrm{j}}\partial x_{k}}W(X)|\leq c\delta(X,\mathrm{o})-d-3,$$0\leq\epsilon\leq 1$$X\neq \mathrm{Y},$$Z\neq \mathrm{Y}$$0\leq\epsilon\leq 1$$X\neq \mathrm{Y},$$Z\neq \mathrm{Y}$$| \frac{\partial^{2}}{\partial x_{j}\partial t}W(X)|\leq c\delta(X, \mathrm{o})-d-3$

116
Proof. The assertions (i), (ii), (iii) and (iv) are well known (cf. [B2, p.5]). The

assertions (v) and
2.3 in [W2].
Let
tions on$D_{0}$$\langle$$\mathrm{v}\mathrm{i})$will be shown by the same method as in the proof of Lemmabe a bounded piecewise smooth domain in
and $u,$. Using the divergence theorem, we obtain$\mathrm{R}^{d}$$v$

be smooth func-$\overline{D_{0}}\cross[0, \rho]$(2.1)$\int_{0}^{\rho}\int_{D_{0}}(uL^{*}v-vLu)dxdt$

uvdx,$= \int_{0}^{\rho}dt\int_{\partial D_{\mathrm{O}}}\langle u\nabla xv-v\nabla xu, n_{x}\rangle d\sigma(x)-\int_{D\mathrm{x}\mathrm{t}^{t}=0\}}uvdx+\int_{D\cross \mathrm{t}}t=\rho\}$where.

and$L= \triangle-\frac{\partial}{\partial t}$

If $Lu=L^{*}v=0$ in$D_{0}\cross(0,\rho)$$L^{*}= \triangle+\frac{\partial}{\partial t}$.

, then (2.1) implies$(2.2) \int_{0}^{\rho}dt\int_{\partial D_{\mathrm{O}}}\langle u\nabla_{x}v-v\nabla_{xx}u, n\rangle d\sigma(X)-\int_{D_{\mathrm{O}}\mathrm{x}\mathrm{t}^{t=}0}\}Xuvd+\int_{D_{\mathrm{O}}\mathrm{x}\{t}=\rho\}$$uvdx=0$ .

Let $X=(x,t)(0\leq t\leq T)$ be an exterior point of $D_{0}\cross(0,T)$ . Then, setting$u=1$ and $v(\mathrm{Y})=W(X-\mathrm{Y})$ and noting that $W(X-\mathrm{Y})=0$ for $\mathrm{Y}=(y,T)$ , wededuce from (2.2)
(2.3)

.

$\int_{0}^{T}d_{S}\int_{\partial D_{0}}\langle\nabla_{y}W(x-\mathrm{Y}),n\rangle yd\sigma(y)-\int_{D_{\mathrm{O}}\mathrm{x}\{}S=0\}W(x-\mathrm{Y})dy=0$

Hereafter we assume that $D$ is a bounded domain in$\mathrm{R}^{d}$.

such that$\partial D$

is a -set$\beta$$\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{S}}\mathrm{f}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{g}\overline{D}\subset B(O, R/2)$

Let us use the Whitney decomposition to approximate $D$ andp.167]). Let
be the Whitney decomposition of $D$ and define$\mathrm{R}^{d}\backslash \overline{D}$(cf.

$[\mathrm{S}$,$\mathcal{V}(D)$$A_{n}= \bigcup_{k}^{n}=k\mathrm{o}Q\cup\in \mathcal{V}k\mathrm{t}D)Q$

where{ $Q\in \mathcal{V}(D);Q$ is athat.
Similarly we also define$\mathcal{V}_{k}(D)=$$k$-cube} and$k_{0}$,

is the smallest integer$\mathcal{V}_{k}(D)\neq\emptyset$$B_{n}=( \bigcup_{k\backslash \overline{D}}^{n}\bigcup_{Q}=-\infty\epsilon v_{k}\mathrm{t}\mathrm{R}^{d})Q)$

Then we have the following lemma.

.

$k$

such

117
Lemma 2.2. Set
and$g_{n}(X)= \int_{0}^{T}\int_{A_{n}}\triangle_{y}W(X-\mathrm{Y})dy$Then$\lim_{narrow\infty^{g_{n}}}(X)$and$h_{n}(X)= \int_{0}^{T}\int_{B_{n}}\triangle_{y}W(X-\mathrm{Y})dy$

enist on$\lim_{narrow\infty^{h_{n}}}(X)$$\mathrm{R}^{d}\cross[0, T]$

and for.

$X\in \mathrm{R}^{d}\cross(0,T]$$\lim_{narrow\infty}g_{n}(X)=\int_{D\cross\{\}}0-W(x-\mathrm{Y})dy\chi D(X)$and$\lim_{narrow\infty}h_{n}(x)=\int_{\langle\backslash \overline{D})}\mathrm{R}^{d}\mathrm{x}$Proof. Let$X=(x, t)\in \mathrm{R}^{d}\cross(0,T]$o$W(X-\mathrm{Y})dy-\chi_{\mathrm{R}}d\overline{D}(X)$$\backslash$and $t>\rho>0$ . Applying (2.2) to$A_{n}\cross(0, \rho)$,

we have$\int_{0}^{\rho}ds\int_{\partial A_{n}}\langle\nabla_{y}W(x-\mathrm{Y}), n\rangle yd\sigma(y)-\int_{A_{n}\mathrm{x}\mathrm{t}\}}0)W(X-\mathrm{Y}dy$$+ \int_{A_{\mathfrak{n}}\cross\{\rho\}}W(X-\mathrm{Y})dy=0$

Using the divergence theorem for$A_{n}$

in.
$\mathrm{R}^{d}$, we have$\int_{0}^{\rho}ds\int_{A_{n}}\triangle_{y}W(X-\mathrm{Y})dy-\int_{A_{n}\cross \mathrm{t}0}\}W(X-\mathrm{Y})dy$$+ \int_{A_{n}\cross\{\rho\}}W(X-\mathrm{Y})dy=0$As$\rhoarrow t$.

and $narrow\infty$ , we obtain,$\lim_{narrow\infty}g_{n}(X)=\int_{D\mathrm{x}\mathrm{t}^{0}\}}W(x-\mathrm{Y})dy-\chi D(X)$

On the other hand $g_{n}(X)=0$ for $t=0$ . Hence$X\in \mathrm{R}^{d_{\mathrm{X}}}[0,T]$.

Similarly we can also prove the conclusion for$h_{n}$$\lim_{narrow\infty}g_{n}(X)$.

3. Double layer heat potentials
In this section we first prove Lemma 1.1 in \S 1.

.
exists for each118
Proof of Lemma 1.1 We use the extension$C^{\infty}$-function$\phi_{0}$in [ , p.172] and choose a$\mathrm{S}$$\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}.\mathcal{E}0$such thaton$\phi_{0}=1$$B(\mathrm{O}, R)$,$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\phi_{0}\subset B(0,2R)$

and$0\leq\phi_{0}\leq 1$.

We define$\mathcal{E}(f)(_{X},t)=\mathcal{E}0(f(\cdot,t))(x)\phi \mathrm{o}(x)$for$f\in\Lambda_{\alpha}(F)$and$(x,t)\in(\mathrm{R}^{d}\backslash F)\cross[0,T]$$\mathcal{E}(f)(x,t)=f(x,t)$

and

on $(x,t)\in F\cross[0,T]$ .

Then properties (i), (ii), (iii) follow from the definition and (13) on p.174 in [S]. Since
is linear, positive and maps the constant function 1 to 1, (iv) is also
the operator
valid.
$\mathcal{E}_{0}$In [W1] we gave the following lemma.

$0$

Lemma A. Let. Then$\delta,$$k$

be non-negative numbers satisfying $d-\beta>\delta$ and $d-\delta-k>$$\partial D-\delta|y-Z|-kdy\leq Crd-\delta-k$$\int_{B\langle z,\Gamma}))\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(y,$for every

and $r>0$ .

$z\in\partial D$

We next show that the double layer heat potential defined by (1.5) and (1.6)

converges.
Lemma 3.1. Let $0\leq\beta-(d-1)<\alpha<1$ andin$(\mathrm{R}^{d}\backslash \partial D)\cross \mathrm{R}$Proof. Set, for(3.1)$f\in\Lambda_{\alpha}(S_{D})$.

$X=(x, t)\in D\cross \mathrm{R}$. Then$\Phi f$is$calor\dot{\tau}c$,$J_{1}(X)= \int_{0}^{T}d_{S}\int_{\mathrm{R}^{d}\backslash \overline{D}}\langle\nabla_{y}\mathcal{E}(f)(y, s), \nabla_{y}W(X-\mathrm{Y})\rangle dy$. If $X=(x,t)\in$
and let $X_{0}=(x_{0},t_{0})\in D$ . Choose $\rho>0$ satisfying, then we deduce from Lemmas 2.1 and 1.1 and Lemma A$\overline{B(x0,2\rho)}\subset D$$B(x_{0}, \rho)\cross \mathrm{R}$,$|J_{1}(X)| \leq\int_{0}^{T}d_{S}\int_{\mathrm{R}^{d}\backslash \overline{D}}\mathrm{d}\mathrm{i}\mathrm{S}\mathrm{t}(y, \partial D)\alpha-1\delta(x,\mathrm{Y})^{-1d}-dy\leq c_{1}\rho^{-1-}d||f||_{\infty,\alpha}$converges locally uniformly in $D$ . We denote by
whenceright-hand side on (3.1). Since$J_{1}$$| \nabla_{y}\frac{\partial^{2}}{\partial x_{i}\partial_{X_{j}}}W(\mathrm{x}-\mathrm{Y})|\leq c_{2}\delta(X,\mathrm{Y})^{-}d-3$and$g_{1}$

the integrand of the$| \nabla_{y}\frac{\partial}{\partial t}W(X-\mathrm{Y})|\leq c_{3}\delta(X,\mathrm{Y})^{-}d-3$,

119
we see that the integral of
on $D$ . Therefore
Next, setsatisfie.s$J_{1}$

over
the heat equation in$Lg_{1}$$(\mathrm{R}^{d}\backslash \overline{D})\cross[0,T]$

also converges locally uniformlyR.

D.

$\mathrm{x}$.

$J_{2}(X)= \int_{0}^{T}d_{S}\int_{\mathrm{R}^{d}\backslash \overline{D}}\mathcal{E}(f)(y, s)\triangle W(X-\mathrm{Y})dy$also converges
Using Lemma 1.1, (iii), we can show by the above method that
$D$
we
that
equation.
Thus
heatsatisfies
theandlocally uniformly in$\Phi f=J_{1}+J_{2}$ has the same properties inalso has the
R. We can show that.
same properties in$J_{2}$$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{l}^{\mathrm{c}}\mathrm{u}\mathrm{d}\mathrm{e}$$D\cross$$\Phi f$$(\mathrm{R}^{d}\backslash \overline{D})\cross R$

Using Lemma 2.1, (iv), (v) and Lemma , we can prove the following lemma bya similar method to that in the proof of [Wl, Lenma 3.3].
$\mathrm{A}$

Lemma 3.2. Let $0\leq\beta-(d-1)<\alpha<1$ andthe. function defined by (3.1) and the function defined by$f\in\Lambda_{\alpha,\alpha/2(s_{D})}$. Then both of$J_{3}$$J_{1}$$J_{3}(X)= \int_{\mathrm{o}\mathrm{R}^{d}\backslash \overline{D}}^{\tau_{dSI(}}(\mathcal{E}(f)(\mathrm{Y})-\mathcal{E}(f)(x))\triangle WX-\mathrm{Y})dy$

are are continuous onby replacing$\mathrm{R}^{d}\cross[0,\tau]$$\mathrm{R}^{d}\backslash \overline{D}$.

. Furthermore the function (resp. ) obtained) is also continuous on(resp.
in the definition of$\mathrm{R}^{d}\cross[0,T]$

with

$D$

$J_{1}’$$J_{3}’$$J_{3}$$J_{1}$

such that
Lemma 3.3. Let $0\leq\beta-(d-1)<\alpha<1$ andis bounded, suppfor every. Let $X=(x,t)\in \mathrm{R}^{d}\cross(0,T]$ and set, for $0<\rho\leq T$ ,for every$g\in\Lambda_{\alpha,\alpha/2(\mathrm{R}^{d}}\cross[0,T])$$g(\cdot,t)\in C^{1}(\mathrm{R}^{d})$$g(\cdot, s)\subset B(\mathrm{O},r_{0})$$t\in[0,T]and_{\overline{\partial}x_{j}}\partial s_{-}(X, \cdot)$$x\in \mathrm{R}^{d}$$A_{\rho}g(X)= \int_{0}^{\rho}ds\int_{\mathrm{R}^{d}\backslash \overline{D}}\langle\nabla g(\mathrm{Y}), \nabla_{y}W(X-\mathrm{Y})\rangle dy$$+ \int^{\rho}\mathrm{o}ds\int \mathrm{R}^{d}\backslash \overline{D}((g(\mathrm{Y})-gX))\triangle_{y}W(x-\mathrm{Y})dy$$+g(X) \int_{(\mathrm{R}^{d}\backslash )\mathrm{X}\{\}}\overline{D}0)W(x-\mathrm{Y}dy$and$B_{\rho}g(X)=- \int_{0}^{\rho}ds\int_{D}\langle\nabla g(\mathrm{Y}), \nabla_{y}W(X-\mathrm{Y})\rangle dy$$- \int_{\mathrm{o}D}^{\rho}dSI^{(g(\mathrm{Y}})-g(x))\triangle yW(X-\mathrm{Y})dy$$-g(X) \int_{D\mathrm{x}\mathrm{t}}0\})W(x-\mathrm{Y}dy$

.

120

Then
(3.2)

$A\tau g(X)=B\tau g(X)+g(X)$for$X\in \mathrm{R}^{d}\cross(0,T]$

instead of $A_{\rho}g(X)$ and$B_{\rho}g(X)$ , respectively. We first show (3.2) in case $D=D_{0}$ is a bounded piecewise.
smooth domain. Let $X=(x, t)$ and set, for $0<\rho<t$ ,Proof. To simplify the notation, we use$A_{\rho}(x)$and$B_{\rho}(X)$$I_{\rho}(X)=- \int_{0}^{\rho}ds\int_{\partial D_{0}}g(\mathrm{Y})\langle\nabla_{y}W(X-\mathrm{Y}), n\rangle y(y)d\sigma$

The Green formula for(3.3)$D_{0}$.

yields$I_{\rho}(X)=- \int_{0}^{\rho}ds\int_{D_{\mathrm{O}}}\{\nabla g(\mathrm{Y}),$$\nabla_{y}W(\mathrm{x}-\mathrm{Y})\rangle dy$$- \int_{0}^{\rho}ds\int_{D_{\mathrm{O}}}(g(\mathrm{Y})-g(X))\triangle_{y}W(x-\mathrm{Y})dy$$-g(X) \int_{0}^{\rho}ds\int_{D_{\mathrm{O}}}\triangle_{y}W(X-\mathrm{Y})dy$

From (2.2) we deduce$\int_{0}^{\rho}ds\int_{D_{\mathrm{O}}}\triangle_{y}W(X-\mathrm{Y})dy$,$= \int_{D_{\mathrm{O}}\mathrm{x}\mathrm{t}}0\}W(X-\mathrm{Y})dy-\int_{D_{0}\cross\{}\rho\})W(x-\mathrm{Y}dy$whence$\int_{0}^{t}ds\int_{D_{0}}\triangle_{y}W(X-\mathrm{Y})dy=\int_{D_{\mathrm{O}\cross}\mathrm{t}0\}}W\langle X-\mathrm{Y})dy-\chi D\mathrm{O}(x)$.

This and (3.3) imply(3.4)$I_{t}(X)=B_{t}(X)+g(x)x_{D0}(X)$ for $X\in(\mathrm{R}^{d}\backslash \partial D_{0})\cross(0,T]$

Similarly, using the Green formula for$B(\mathrm{O},r)\backslash \overline{D}_{0}$and$rarrow\infty$, we obtain$I_{t}(X)=A_{t}(X)-g(X)x_{\mathrm{R}\backslash \overline{D}_{\mathrm{O}}}d(X)$for$X\in(\mathrm{R}^{d}\backslash \partial D_{0})\cross(0, T]$.

. This and (3.4) lead to$A_{t}(X)=B_{t}(X)+g(X)$ for $X\in(\mathrm{R}^{d}\backslash \partial D_{0})\cross(0, T]$ .

121Noting that $A_{t}(X)=A_{T}(X)$ and $B_{t}(X)=B_{T}(X)$ , we obtain (3.2) forby Lemma 3.2, (3.2)
are continuous onand. Sinceholds for a bounded piecewise smooth domain $D=D_{0}$ .
is a -set. We use (3.2)
We next show (3.2) for a bounded domain such thatfor $D_{0}=A_{n}$ . Since$X\in(\mathrm{R}^{d}\backslash$$A_{T}$$\partial D)\cross(\mathrm{O},T]$$\mathrm{R}^{d}\cross(0,T]$$B_{T}$$\partial D$$\int_{0}^{T}\int_{\mathrm{R}^{d}}[\mathrm{v}_{g(}\mathrm{Y})||\nabla_{y}W(x-\mathrm{Y})|dydS<\infty$$\beta$,$\int_{0}^{\tau}ds\int_{\mathrm{R}^{d}}|g(\mathrm{Y})-g(x)||\triangle W(\mathrm{x}-\mathrm{Y})|yy<\infty d$and,$\int_{\mathrm{R}^{d}\mathrm{x}\{0}\}W(X-\mathrm{Y})dy<\infty$

we see that (3.2) holds for the domain
Lemma 3.4.

for$g=\mathcal{E}(f)$

.

$D$

as

$narrow\infty$

Let $0\leq\beta-(d-1)<\alpha$ and

Sketch of Proof. Let. We define, for$f\in\Lambda_{a,\alpha/2}(S_{D})$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}v_{m}\subset B(0,1/m)$

and

.

口

$f\in\Lambda_{\alpha,\alpha/2(s_{D})}$$\{v_{m}\}$. Then (3.2) holds

be a mollifier on$\mathrm{Y}=(y, s)\in \mathrm{R}^{d}\cross[0,T]$$g_{m}(\mathrm{Y})=(\mathcal{E}(f)(\cdot, s)*vm)(y)$$\mathrm{R}^{d}$such that,.

Lemma 3.3 yields$A\tau g_{m}(X)=B\tau g_{m}(X)+g_{m}(X)$for$X\in \mathrm{R}^{d}\cross(0,T]$ :Using $g_{m}(X)arrow \mathcal{E}(f)(X)$ uniformly as $marrow\infty$ and Lemmas
show that$A\tau g_{m}(X)arrow A_{T}\mathcal{E}(f)(X)$$\mathrm{A},$$1.1$

and 2.1, we can.

and$B\tau g_{m}(X)arrow B_{T}\mathcal{E}(f)(X)$

for$X\in \mathrm{R}^{d}\cross[0, T]$

as $marrow\infty$ .
口We can also show the following lemma.

Lemma 3.5. Let $0\leq\beta-(d-1)<\alpha<\dot{1}$ . Then the operator $K$ defined by.
to(1.7) is a bounded operator$f_{\Gamma om}\Lambda\alpha,\alpha/2(sD)$Let us prove our thorem.

$\Lambda_{\alpha,\alpha/2}(sD)$122
Proof of Theorem. Let $X\in D\cross(0,T]$ . Using Lemma 2.2, we have $\Phi f(X)=$$A\tau f(X)$. Since$A_{T}f$

is continuous on$\mathrm{R}^{d}\cross(0, T]$

by Lemma 3.2, we have$\Phi f(X)=ATf(Z)$ .

$\lim$$Xarrow Z,X\in D\mathrm{x}\mathrm{t}0,\tau)$

On the other hand Lemma 3.4 yields$Kf(Z)= \frac{1}{2}(A\tau f(z)+BTf(Z))=A_{T}f(z)-\frac{1}{2}f(z)$.

Therefore we have (1.8). Similarly we can show (1.9).

口References[B1]

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J. E. Hutchinson, Fractals and selfsimilarity, Indiana Univ. Math. J. 30(1981), 713-747.
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E. M. Stein, Singular integrals and differentiability properties of funct.ions,
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$?\mathrm{k}\mathrm{a}\mathrm{n}\mathrm{S}$$L_{p}$$\mathrm{t}\mathrm{h}\dot{\mathrm{e}}$