数理解析研究所講究録 1293 巻 2002 年 1-14

1
MARTIN BOUNDARY FOR UNION OF CONVEX SETS

Torbj\"om Lundh
平田賢太郎 (Kentaro Hirata)
島根大学総合理工学研究科 Department of Mathematics,
Chalmers University(Department of Mathematics ,of Technology, Sweden
Shimane University)

相川弘明 (Hiroaki Aikawa)
島根大学総合理工学部(Department of Mathematics ,Shimane University )1. Introduction, that can
We study Martin boundary points of aproper subdomain in , where
be represented as the union of open convex sets. Especially, we give acertain sufficientcondition for aboundary point to have exactly one (minimal) Martin boundary point.
, Ancona considered abounded domainthat can be represented as the
In theunion of open balls with the same radius. He assumed that(A) if two balls tangent to each other at aboundary point 4of , then there is atruncatedcircular cone, with vertex at and axis in the hyperplane tangent to such balls at 4,$\mathbb{R}^{n}$$\geq 2$$n$$\Omega$$1970’ \mathrm{s}$$\Omega$$\xi$

included in$\Omega$.

Under these assumption he showed that each boundary point has exactly one Martin boundary point and it is minimal ([4]).
However, this result is not applicable to domains with wedges. So we consider open
convex sets rather than open balls with the same radius. Obviously, we need adifferent
sufficient condition for aboundary point to have exactly one (minimal) Martin boundary

point.
for the closure and the boundary of aset , respectively. Let ,We write and$(x\neq y)$ and $r>0$ . We denote by $B(x, r)$ and $S(x, r)$ the open ball and the sphere of centerstand for the open circular cone of vertex ,and radius , respectively. For $\theta>0$ let
axis and aperture , i.e.,$E$$\partial E$$\overline{E}$$x$$y\in \mathbb{R}^{n}$

$x$

$x$

$\Gamma_{\theta}(x,y)$$r$$\theta$$\overline{xy}$$\Gamma_{\theta}(x,y):=\{z \in \mathbb{R}^{n} :Let$P\mathrm{o}>0$\angle zxy <\theta\}$and $A0\geq 1$ . We consider aproper subdomain $D$ in(I) $D$ is the union of afamily of open convex sets
$B(z_{\lambda}A_{0}p_{0})$(II) Let.

.

$\xi\in\partial D$. Then there are positive constants

such that the union of truncated circular cones
connected,$\mathrm{i}.\mathrm{e}.$$\{c_{\lambda}\}_{\lambda\in\Lambda}$$\mathbb{R}^{n}$such thatsuch that$\theta_{1}\leq\sin^{-1}(1/A\mathrm{o})$$\Gamma_{\theta_{1}}(\xi,y)\cap B(\xi,2p_{1})$$\Gamma_{\theta_{1}}(\xi,y)\cap B(\xi,2\rho_{1})\subset D$

andincluded in $D$ is$\mathrm{r}_{\theta_{1}}(\xi,y)\cap B(\xi,2p_{1})$,$y\in D\cup$$B(z_{\lambda},p\mathrm{o})\subset c_{\lambda}\subset$

is connected.

$\beta 1\leq p_{0}\cos\theta_{1}$2Remark 1. We note that the union in the condition (II) is non-empty (Lemma 3.2). The
condition (II) is the same as Ancona’s when $A0=1$ (Ancona’s setting).

Throughout this note, we simply write adomain instead of aproper subdomain inaGreenian domain we mean adomain with the Green function.
The main result is as follows.

Theorem. Let be a Greenian domain satisfying (I). If
exactly one Martin boundary point at 4and it is minimal.
$D$$\xi\in\partial D$$\mathbb{R}^{n}$. Bysatisfies (II), then there isRemark 2. We investigated in [3] that the number of minimal Martin boundary points at eachboundary point of aJohn domain is estimated by the John constant. Abounded domain
satisfying (I) is aJohn domain. As seen in Theorem, we obtain abetter result under thecondition (II).

Corollary. Suppose that is a bounded domain satisfying (I) and that eachsatisfies(II). Then the Martin boundary $ofD$ is homeomorphic to its Euclidean boundary. Moreover,each Martin boundary point is minimal.
$D$$\xi\in\partial D$

The following proposition implies the sharpness of bounds
in the condition (II).

$\theta_{1}\leq\sin^{-1}(1/A\mathrm{o})$and$p_{1}\leq$$p\circ\cos\theta_{1}$Proposition 1.1. Let $A0>1$ . Suppose either(i) $\theta_{1}>\sin^{-1}(1/A_{0})$ , or(ii) $0<\theta_{1}\leq\sin^{-1}(1/A\mathrm{o})$ and $p_{1}>p\circ\cos\theta_{1}$ .

Then there is a domain $D$ satisfying (I) andminimal Martin boundary points.

$\xi\in\partial D$

satisfies (II), and yet 4has multiple

This note is organized as follows. In Section 2, we shall show ageneral fact for the supportof the measure associated with a kernel function in the Martin representation. In Section 3,
we shall show geometrical properties. In Section 4, we shall prove aCarleson tyPe estimate
after showing the upper bound of anon-negative subharmonic function on abounded domain
and showing the integrability of the negative power of the distance function. In Section 5, weshall show a(uniform) boundary Harnack principle. In Section 6, we shall prove Theorem
and Corollary. In Section 7, we shall give examples for Proposition. In Section 8, we shallgive adomain satisfying (I) and (II) at each boundary point but not auniform domain.
By the symbolA we denote an absolute positive constant whose value is unimportant andmay change from line to line. If two positive functions $f$ and satisfy $A^{-1}f\leq g\leq Af$ forsome constant $A\geq 1$ , then we writeand callA the constant of comparison.
$g$$f\approx g$

2. General fact
In this section, we show general fact for the support of the measure of acorresponding toakemel function in the Martin representation. Let
and $x0\in D$ be fixed. Let denote
the Green function for $D$ . The Martin kernel (or the Martin boundary point) at , written, (:), is given as alimit function of the Martin kernels $K(\cdot,y_{j}):=G(\cdot,y_{j})/G(x_{0},y_{j})$ for
in $D$ converging to . We say that aproperty holds quasi-everywhere if
some sequence
it holds except apolar set. Afunction on $D$ is called akernel function at if is positive$G$$\xi\in\partial D$$\xi$$K(\cdot$$\xi$$\{\mathcal{Y}j\}$

$h$

$\xi$

$h$

3
and is bounded on, vanishes quasi-everywhere on
and harmonic on $D$ , satisfies
the subset
for each $r>0$ . We denote by the Martin boundary of $D$ , and byof all minimal elements in A. We also write A(4) for the set of all Martin boundary points. We say that is minimally thin
. Let $E\subset D$ andat 4, and let
denotes the regularized reduced function of anon-negative. Hereat ifsuperharmonic function relative to in $D$ .
The following lemma will be used in the proof of Theorem (Section 6).
$\partial D$$h(x\mathrm{o})=1$$\Delta_{1}$$\Delta$$D\backslash B(\xi, r)$$\hat{R}_{u}^{E}$$\hat{R}_{K(\cdot,y)}^{E}\neq K(\cdot,y)$

$y$

$E$

$y\in\Delta_{1}$$\Delta_{1}(\xi):=\Delta(\xi)\cap\Delta_{1}$

$E$

$u$

. If is a kernel function at 4,
Lemma 2.1. Let be a Greenian domain and. Inthen the support of the measure associated with it in the Martin representation isis non-empty.
particular,$h$$\xi\in\partial D$$D$$\Delta_{1}(\xi)$$\Delta_{1}(\xi)$Proof By the Martin representation, there is aunique measure on$\mu$$h(x)= \int_{\Delta_{1}}K(x,y)d\mu(y)$$\Delta_{1}$such thatfor $x\in D$ .

and let $\{Ej\}$ be adecreasing sequence of compact
Let be acompact subset of A
in the Martin topology such that $(E1\cap D)\cap B(\xi,r1)=\emptyset$ for some $r1>0$
neighborhoods
and $\bigcap_{j}Ej=E$ . Then we have ([5, Corollary 9.1.4])$E$$\backslash \Delta(\xi)$$\mathrm{o}\mathrm{f}E$$\hat{R}_{h}^{E_{j}\cap D}(x)=\int_{\Delta_{1}}\hat{R}_{K(\cdot,y)}^{E_{j}\cap D}(x)$Noting that$\partial D$

since

$h$

$\lim_{jarrow\infty}\hat{R}_{h}^{E_{j}\cap D}$

for$d\mu(y)$$x$$\in D$.

is bounded and harmonic on $D$ and vanishes quasi-everywhere onis the kernel function at 4, we have

(2.1)

$0= \lim_{jarrow\infty}\hat{R}_{h}^{E_{j}\cap D}(x_{0})$by the monotone convergence. Leteach ([5, Lemma 9.1.4]) and so$j$$= \int_{\Delta_{1}}\lim_{jarrow\infty}\hat{R}_{K(\cdot,y)}^{E_{j}\cap D}(x_{0})$$y\in E\cap\Delta_{1}$$d\mu(y)$. Then $Ej\cap D$ is not minimally thin at for. Hence $\mu(E)=0$ by (2.1).
$y$$\lim_{jarrow\infty}\hat{R}_{K(\cdot,y)}^{E_{j}\cap D}(x_{0})=K(x0,y)=1$

Thus the lemma follows.

$\square$3. Geometrical properties
for dist(x,. We write
Let be aproper subdomain and ,and define the quasi-hyperbolic metric between and by
from to$\Omega$

$x$

$x$

an

$y\in\Omega$$\delta_{\Omega}(x)$$x$$k_{\Omega}$$(x,y):=. \mathrm{n}\mathrm{f}\int_{\gamma}\gamma\frac{ds}{\delta_{\Omega}(z)}$$\partial\Omega$), the distance$y$,where the infimum is taken over all rectifiable curves in connecting to .
Throughout this section we suppose that $D$ is adomain satisfying (I) and thatsatisfies (II). The main purpose of this section is to show the following lemma.
$\Omega$$x$$\gamma$. There is a positive constantLemma 3.1. Let
such that
For each $0<R<R\xi$ there is$\kappa=6/\sin\theta_{1}$$\mathcal{Y}R\in D\cap S(\xi,R)$$k_{D\cap B(\xi,\kappa R)}$$(x,y_{R}) \leq A_{\xi}\log\frac{R}{\delta_{D}(x)}+A_{\xi}$

where $A\xi\geq 1$ is independent ofx and R.

$R\xi$$\xi\in\partial D$with the following property.
and$\delta_{D}(y_{R})\geq A_{\xi}^{-1}R$

for x

$y$

$\in D\cap B(\xi,R)$ ,4Remark 3. In general, Lemma 3.1 does not hold for aJohn domain. We introduced in[3] ageometrical notion, asystem of local reference points of order $N$ . That is, for each
, , , in $D\cap S(\xi,R)$ such thatfor
$0<R<R\xi$ there are $N$ points, say$N$$i=1$ ,
, and$y_{R}^{1}$$\cdots$$y_{R}^{N}$$\delta_{D}(y_{R}^{i})\geq A_{\xi}^{-1}R$$\cdots$for$\min_{i=1,\cdots,N}\{k_{D\cap B(\xi,\kappa R)}(x,y_{R}^{i})\}\leq A\xi\log\frac{R}{\delta_{D}(x)}+A\xi$$x$$\in D\cap B(\xi,R)$.

Lemma 3.1 is the case N $=1$ .
$\xi$

In order to prove Lemma 3.1, in view of translation and dilation, we may suppose that$=0$ and $p_{1}=1$ for simplicity. We briefly write $\Gamma(x,y)$ for. Let$\Gamma_{\theta_{1}}(x,y)$$\Psi$$:=\{y\in S(0, 1):\Gamma(0,y)\cap B(0,2) \subset D\}$

Then the union in the condition (II) is
Lemma 3.1 after showing some lemmas.

$\bigcup_{y\in\Psi}\Gamma(0,y)\cap B(0,2)$.

, written

Lemma 3.2. There is a positive constant $R0<\kappa^{-1}$ such that if.
. In particular,$c_{\lambda}\cap\Psi$$\Psi$$\neq\emptyset$$\mathscr{C}(0)$. We prove, then$c_{\lambda}\cap B(0,R\mathrm{o})\neq\emptyset$$\neq\emptyset$

Proof. We show this by leading a contradiction. Suppose that there is asequencesuch that dist
. Let
and
subsequence if necessary, we may assume that converges, say to . Let
such that $xjarrow 0$ . Then, by continuity of the angle
XjZj and the distance$(0,c_{\lambda_{j}})arrow 0$$c_{\lambda_{j}}\cap\Psi$$=\emptyset$$B(\mathrm{Z}j,\beta 0)\subset c_{\lambda_{j}}\subset B(zj,A0p\mathrm{o})$$\mathrm{Z}\mathrm{q}$$\mathrm{Z}j$. Taking a$xj\in\partial c_{\lambda_{j}}$$|\cdot-xj|$$\angle$$\{c_{\lambda_{j}}\}$

be

,

.

$\Gamma(0,zo)$ $\cap B(0,2)$$\subset\bigcup_{j}(\Gamma(x_{j},z_{j})\cap B(x_{j},2))\subset\bigcup_{j}C_{\lambda_{j}}$Hence$\bigcup_{j}c_{\lambda_{j}}\cap\Psi\neq\emptyset$

Let us take$\mathcal{Y}1\in\Psi$, and this contradicts the assumption. Thus the lemma follows.

and fix. For $0<R<1$ we let

Lemma 3.3. There is a positive constant$k_{D\cap B(0,\kappa R)}$$\mathcal{Y}R:=Ry1$. Then$\delta_{D(\mathcal{Y}R}$)$\geq R\sin\theta 1$$\square$.

such that if $0<R<R0$ , then$(Ry,y_{R})\leq A$

for y$\in\Psi$.

is connected since the cone
is connected. We observe$0<r0\leq\sin\theta_{1}$that there is aclosed connected subset ofandsuch thatand dist. Then. In view of the compactness of , we cantake acurve injoining and such thatfor alland, where depends only on a covering constant of anddenotes the lengthof acurve . Let be the image of in $S(0,R)$ under dilation. Then we have

Proof Note that$\mathscr{C}(0)\cap S(0,1)$$\mathscr{C}(0)$$E$$\Psi$$\subset E$$(E,\partial \mathscr{C}(0))\geq r\mathit{0}$$\gamma$$\ell(\gamma)\leq Ar0$$\gamma$$\mathscr{C}(0)\cap S(0, 1)$$E$$\mathcal{Y},\mathcal{Y}1\in E$$\mathscr{C}(0)\cap S(0, 1)$$y$$\delta_{\mathscr{C}(0)}(z)$$\mathcal{Y}1$

$A$

$E$

.
$\square$

Let $[,y]$ denote the (open) line segment between
distance function 4is concave on , i.e. ,

$x$

and

$y$

. If is aconvex set, then the$C$$\overline{C}$whenever x, y$\in\gamma$$\gamma$$\mathit{7}R$

the lemma follows.

(3.1)

$z$

$\ell(\gamma)$$k_{D\cap B(0,\kappa R)}(Ry,y_{R}) \leq\int_{\gamma_{R}}\frac{ds}{\delta_{D}(z)}\leq\frac{Ar_{0}R}{r_{0}R/2}=2A$$\mathfrak{M}\mathrm{u}\mathrm{s}$$\geq r\mathrm{o}/2$&$(z) \geq\frac{|z-y|}{|x-y|}\ (x)+ \frac{|x-z|}{|x-y|}\ (y)$$\in\overline{C}(x\neq y)$

.

for

$z$

$\in[x,y]$,5Lemma 3.4. Let $0<R<R0$ . Ifsuch that$c_{\lambda}\cap B(0,R)\neq\emptyset$

and$y\in c_{\lambda}\cap\Psi$, then there exists $w\in$$C_{\lambda}\cap\Gamma(0,y)\cap B(0, 3R/\sin\theta_{1})$$\delta_{C_{\lambda}\cap\Gamma(0,y)}(w)\geq\frac{\sin\theta_{1}}{3}R$.

with $|w1|\leq R/\sin\theta_{1}$ . In fact, if, so that, then we may take $w1$ at which $[,y]$ intersects

Proof. We can take$w1\in c_{\lambda}\mathrm{n}\overline{\Gamma(0,y)}$$x\in c_{\lambda}\mathrm{n}B(0,R)\backslash$$\partial\Gamma(0,y)$$\overline{\Gamma(0,y)}$.

$|w_{1}|= \frac{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(w_{1},[0,y])}{\sin\theta_{1}}\leq\frac{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(x,[0,y])}{\sin\theta_{1}}\leq\frac{R}{\sin\theta_{1}}$Note that $|w_{1}-y|>5R/\sin\theta_{1}$ since $R<\kappa^{-1}=6^{-1}\sin\theta_{1}$ . Letbe such that $|w1-w2|=R/\sin\theta 1$ . Applying (3.1) to $C:=\Gamma(0,y)$ , we have$w_{2}\in[w_{1},y]\subset c_{\lambda}\mathrm{n}$ $\Gamma(0,y)$$\delta_{\Gamma(0,y)}(w_{2})\geq\frac{|w_{1}-w_{2}|}{|w_{1}-y|}\delta_{\Gamma(0,y)}(y)\geq\frac{R/\sin\theta_{1}}{1+R/\sin\theta_{1}}\sin\theta_{1}\geq\frac{2}{3}R$.

Noting that $|w2-z_{\lambda}|\geq P\mathrm{o}-2R/\sin\theta_{1}\geq 4R$ since $p0\geq 1\geq 6R/\sin\theta_{1}$ , we can take $w\in$yields thatwith $|w_{2}-w|=R/3$ . Then (3.1) with
$C:=c_{\lambda}$$[w_{2},z_{\lambda}]\subset C_{\lambda}$.

$\ _{\lambda}(w)\geq\frac{|w_{2}-w|}{|w_{2}-z_{\lambda}|}\ _{\lambda}(z_{\lambda}) \geq\frac{R/3}{A_{0}p_{0}}p0\geq\frac{\sin\theta_{1}}{3}R$

Hence we have$\delta_{\Gamma(0,y)\cap C_{\lambda}}(w)\geq\min\{\frac{2}{3}R-\frac{R}{3}$,$\frac{\sin\theta_{1}}{3}R\}=\frac{\sin\theta_{1}}{3}R$,

and$|w| \leq|w-w_{2}|+|w_{2}-w_{1}|+|w_{1}|\leq\frac{R}{3}+\frac{R}{\sin\theta_{1}}+\frac{R}{\sin\theta_{1}}<\frac{3R}{\sin\theta_{1}}$.

Thus the lemma follows.

Proofof Lemma 3.1. Let$\square$$x\in c_{\lambda}\cap B(0,R)$$w\in C_{\lambda}\cap\Gamma(0,y)\cap B(0,3R/\sin\theta_{1})$$\delta_{D}(z)$by (3.1) with

with

and

$y\in C_{\lambda}\cap\Psi$. By Lemma 3.4, we can take$\delta_{C_{\lambda}\cap\Gamma(0,y)}(w)\geq 3^{-1}R\sin\theta_{1}$$\geq\ _{\lambda}(z) \geq\frac{|x-z|}{|x-w|}\delta_{C_{\lambda}}(w)\geq\frac{\sin^{2}\theta_{1}}{12}|x-z|$$C:=c_{\lambda}$. Since$[x,w]\subset B(0, \kappa R/2)$. Then we havefor $z\in[x, w]$ ,, it follows that$k_{D\cap B(0,\kappa R)}(x,w) \leq\int[x,w]\frac{ds}{\delta_{D}(z)}\leq 1+\int_{\delta(x)}^{|x-w|}*\frac{12}{\sin^{2}\theta_{1}}\frac{dt}{t}\leq A\log\frac{R}{\delta_{D}(x)}+A$

where$(w,Ry)\leq A$ . In fact, sincedepends only on . We also have, it follows from (3.1) with $C:=\Gamma(0,y)$ that

$A$

$R\sin\theta_{1}$$\theta_{1}$$\delta_{D}(z)$$\geq\delta_{\Gamma(0,y)}(z)$$k_{D\cap B(0,\kappa R)}$$\geq\frac{|w-z|}{|w-Ry|}\delta_{\Gamma(0,y)}(Ry)\geq\frac{\sin^{2}\theta_{1}}{4}\downarrow w-z|$,$\delta_{\Gamma(0,y)}(Ry)\geq$

for $z\in[w,Ry]$ ,and so$k_{D\cap B(0,\kappa R)}(w,Ry) \leq\int_{[w,Ry]}\frac{ds}{\delta_{D}(z)}\leq 2+\int_{\delta)}^{\frac{\delta_{D}(Ry)}{+^{2}(w}}\frac{4}{\sin^{2}\theta_{1}}\frac{dt}{t}\leq A$

,

6
where

$A$

depends only on$k_{D\cap B(0,\kappa R)}$$\theta_{1}$. Hence we obtain from Lemma 3.3 that$(x,y_{R})\leq k_{D\cap B(0,\kappa R)}(x, w)+k_{D\cap B(0,\kappa R)}(w,Ry)+k_{D\cap B(0,\kappa R)}(Ry,y_{R})$$\leq A\log$ $\frac{R}{\delta_{D}(x)}+A$.

Thus Lemma 3.1 follows.

$\square$

4. Carleson tyPe estimate
In this section we show a Carleson type estimate. To this end, we prepare two lemmas.
One is refinement of Domar’s theorem ([6, Theorem 2]). Another is the integrability of
the negative power of the distance function. This is a local version of [1, Lemma 5].
We note first the following. Let be adomain and $x,y\in\Omega$ . We say that and are con, $Xj-1$
nected by aHarnack chain$N$
for $j=2$ , $\cdots,N$ and $xN=y$ . The number is called the length of the Harnack chain. We, then the shortest length of the Harnack chain connecting
observe that if. Therefore, the Harnack inequality yields that there is aand is comparable toand is apositive
constant $A\geq 1$ depending only on the dimension such that if ,,thenharmonic function on$\Omega$$x$$y$$\in B(xj, \frac{1}{2}\delta\Omega(xj))$$\{B(xj, \delta\Omega(xj))\}_{j=1}^{N}\mathrm{i}\mathrm{f}x\in B(x1, \frac{1}{2}\delta_{\Omega}(x1))$$x \not\in B(y, \frac{1}{2}\delta_{\Omega}(y))$$x$$k_{\Omega}(x,y)$$y$$y\in\Omega$

$x$

$h$

$\Omega$(4. 1)$\exp(-Ak_{\Omega}(x,y)-1)$ $\leq\frac{h(x)}{h(y)}\leq\exp(Ak_{\Omega}(x,y)+1)$

Lemma 4.1. Let
such that$\Omega$.

be a bounded domain. If is a non-negative subharmonicfunction on$u$$\Omega$for some$I:= \int_{\Omega}(\log^{+}u)^{n-1+\epsilon}dx<\infty$

then there is a positive constant
(4.2)

$A$

depending only on$\epsilon$,$\epsilon>0$

and the dimension such that$u(x) \leq\exp(2+A(\frac{I}{\delta_{\Omega}(x)^{n}})^{1/\epsilon})$.

We show first the following lemma. We write |E| for the volume of aset E.

Lemma 4.2. Let u be a subharmonicfixnction ont $>0$ and that
(4.3)$\Omega$$containing\overline{B(x,R)}$$R\geq L_{n}|\{y\in B(x,R):t/e<u(y)\leq et\}|^{1/n}$. Suppose that $u(x)\geq$,

where $L_{n}=(e^{2}/|B(0,1)|)^{1/n}$ . Then there exists $d$ $\in B(x,R)$ such that $u(d)>et$ .

Proof Suppose to the contrary that$u\leq et$on $B(x,R)$ . Noting that (4.3) is equivalent to,$\frac{|\{y\in B(x,R).t/e<u(y)\leq et\}|}{|B(x,R)|}.\leq\frac{1}{e^{2}}$we have$t \leq u(x)\leq\frac{1}{|B(x,R)|}\int_{B(x,R)}u(y)dy$$= \frac{1}{|B(x,R)|}(\int_{B(x,R)\cap\{u\leq t/e\}}u(y)dy+\int_{B(x,R)\cap\{t/e<u(y)\leq et\}}u(y)dy)$$\leq\frac{t}{e}+\frac{et}{e^{2}}<t$.

7This is acontradiction, and the lemma follows.

$\square$

Proof of Lemma 4.1. Since the right hand side of (4.2) is not less than$e^{2}$, it is sufficient toshow that
(4.4)

Let

$8\mathrm{Q}(\mathrm{x})\leq AI^{1/n}(\log u(x))^{-\epsilon/n}$$x_{1}\in\Omega$whenever $u(x)>e^{2}$ .

be such that $u(x1)>e^{2}$ , and let$R_{j}=L_{n}|\{y\in\Omega :.

e^{j-2}u(x_{1})<u(y)\leq e^{j}u(x_{1})\}|^{1/n}$as
Let us show (4.4) for $x=x1$ . We can choose a finite or infifinite sequence $\{xj\}$ in$>e^{j}u(x_{1})$$u(xj+1)$
follows. By Lemma 4.2, we can iteratively fifind $xj+1\in B(xj,Rj)$ with, then we stop this iteration, otherwise we continue.
. If
wheneverWe claim that$\Omega$$\delta_{\Omega}(xj)>Rj$$\delta_{\Omega}(xj)\leq R_{j}$(4.5)$\delta_{\Omega}(x_{1})\leq 2\sum_{j=1}^{\infty}R_{j}$Suppose first$\{xj\}$.

is finite. Noting that

(4.6)$\delta_{\Omega}(x_{1})\leq\sum_{j=1}^{N-1}|x_{j}-x_{j+1}|+\delta_{\Omega}(x_{N})$,is infifinite. Since $u(xj)\geq$, it follows from the local boundedness of asubharmonic function thatfor some $N$ , and (4.5) follows from (4.6).
goes to the boundary. Hence
To obtain (4.4) for $=x1$ , it is enough to show that

we obtain (4.5) by our choice of$\{xj\}$. Suppose next$\{xj\}$$Xj$$e^{j-1}u(\backslash x_{1})arrow\infty$$\delta_{\Omega}(xN)\leq\delta_{\Omega}(x1)/2$

$x$

(4.7)

Let

$j_{1}$$\sum_{j=1}^{\infty}R_{j}\leq AI^{1/n}(\log u(x_{1}))^{-\epsilon/n}$

be the integer such that $e^{j_{1}}<u(x1)\leq e^{j_{1}+1}$ . Then
$R_{j}\leq L_{n}|\{y\in\Omega :.

$j_{1}\geq 2$ande^{j_{1}+j-2}<u(y)\leq e^{j_{1}+j+1}\}|^{1/n}$.

Since the family of intervals $\{(e^{j_{1}+j-2}, e^{j_{1}+j+1}]\}j$ overlaps at most three times, it follows
from H\"older’s inequality that$\sum_{j=1}^{\infty}R_{j}\leq 3L_{n}\sum_{j=j_{1}}^{\infty}|\{y\in\Omega :e^{j-1}<u(y)\leq e^{j}\}|^{1/n}$$\leq 3L_{n}(\sum_{j=j_{1}}^{\infty}\frac{1}{j^{(n-1+\epsilon)/(n-1)}})^{(n-1)/n}(\sum_{j=j_{1}}^{\infty}j^{n-1+\epsilon}|\{y\in\Omega :e^{j-1}<u(y)\leq e^{j}\}|)^{1/n}$$\leq Aj_{1}^{-\epsilon/n}(\int_{\Omega}(\log^{+}u(y))^{n-1+\epsilon}dy)^{1/n}$$\leq A(\log u(x_{1}))^{-\epsilon/n}I^{1/n}$

where

$A$

depends only on$\epsilon$

,

and

$n$

. Thus (4.7) follows and Lemma 4.1 is proved.

$\square$8
Lemma 4.3. Let D be a domain satisfying (I) and. If $0<R<Po$ , then there arepositive constants and A depending only on A0 and the dimension such that$\xi\in\partial D$$\tau$$\int_{D\cap B(\xi,R)}(\frac{R}{\delta_{D}(x)})^{\tau}dx\leq AR^{n}$

Proof. For each j

Let
Let ,
that

$x \in\bigcup_{j=k+1j}^{\infty}V$$y$$y’\in[x,z\lambda]$

we put$\in \mathrm{N}\cup\{0\}$$V_{j}:=$

{x

.

:

$\in D\cap B(\xi,R+\frac{A_{0}+1}{2^{j-1}}R)$. Then there is$c_{\lambda}$

so that$x\in C_{\lambda}$$\frac{R}{2^{j+1}}\leq\delta_{D}(x)<\frac{R}{2^{j}}\}$, and let.

$B(z_{\lambda},p_{0})\subset C_{\lambda}\subset B(z_{\lambda},A_{0}\rho 0)$.

be such that $\delta_{D}(y)=R/2^{k}$ and $\delta_{D}(y’)=(R/2^{k+1}+R/2^{k})/2$ . Then we seeby (3.1), and that $B(y’,R/2^{k+2})\subset Vk\cap B(y,A\circ R/2^{k})$ . Hence we obtain$x\in\overline{B(y,A0R/2^{k})}$(4.8)$|B(y,$ $\frac{5A_{0}R}{2^{k}})|\leq A_{1}|V_{k}\cap B(y,\frac{A_{0}R}{2^{k}})|$,

where depends only on $A0$ and the dimension. We also have,where is the point associated with as above. Hence the covering lemma yields that thereissuch thatandare mutually disjoint.
Then we obtain from (4.8) that
$A_{1}$$\bigcup_{j=k+1}^{\infty}V_{j}\subset\bigcup_{y}\overline{B(y,A_{0}R/2^{k})}$

$x$

$y$

$\bigcup_{j=k+1j}^{\infty}V\subset\bigcup_{j}\overline{B(\mathcal{Y}j,5A0R/2^{k})}$$\{\mathcal{Y}j\}$$\{B(\mathcal{Y}j,A0R/2^{k})\}$$\infty$$\sum$$|V_{j}$$|$$=$$|j$ $=k\cup\infty$$V_{j}$$|$$|B$ $(y_{j})$ $\frac{5A\mathrm{o}R}{2^{k}})$$\leq$$\sum_{j}$$j=k$ $+$

$1$

$|$

$\leq$$A1$$|$$V_{k}$ $\cap$$B$ $(y_{j})$$\frac{A\mathrm{o}R}{2^{k}})$$|$$\leq$$\sum_{j}$$A1$ $|V_{k}$

$|$

.

$+1$

Let t $=1+1/2A0$ . Then
$A_{1}\Sigma t^{k+1}|V_{k}|\geq\Sigma\Sigma t^{k+1}|V_{j}|=\Sigma\Sigma t^{k+1}|V_{j}|\geq\Sigma^{J}\Sigma t^{k+1}|V_{j}|NNN+11j-1N-1$$k=0j=k+1$$k=0$$j=1k=0$$j=1k=0$$= \sum_{j=1}^{N}\frac{t^{j+1}-t}{t-1}|V_{j}|=\frac{1}{t-1}\sum_{j=0}^{N}t^{j+1}|V_{j}|-\frac{t}{t-1}\sum_{j=0}^{N}|V_{j}|$,and so$\sum_{j=0}^{N}t^{j+1}|V_{j}|\leq\frac{t}{1-(t-1)A_{1}}\sum_{j=0}^{N}|V_{j}|$Letting$Narrow\infty$.

, we have$\sum_{j=0}^{\infty}t^{j+1}|V_{j}|\leq\frac{t}{1-(t-1)A_{1}}\sum_{j=0}^{\infty}|V_{j}|\leq A|B(\xi,R+2(A_{0}+1)R)|\leq AR^{n}$Since$t^{j}<(R/\delta_{D}(x))^{\tau}\leq t^{j+1}$

for $x\in Vj$ with $\tau=\log t/\log 2>0$, we obtain$\int_{D\cap B(\xi,R)}(\frac{R}{\delta_{D}(x)})^{\tau}dx\leq\sum_{j=0}^{\infty}t^{j+1}|V_{j}|\leq AR^{n}$Thus the lemma follows.

.

.
$\square$9
Lemma 4.4 (Carleson type estimate). Suppose that is a domain satisfying (I) and thatsatisfies (II). Let $0<R<R_{\xi}$ . If is a positive bounded harmonic function on, thenvanishing quasi-everywhere on$D$$/;\in\partial D$$h$$D\cap B(\xi, \kappa R)$$\partial D\cap B(\xi, \kappa R)$$h(x)\leq Ah(\mathcal{Y}R)$

where

$A$

is independent of ,
$X$

$R$

for

$x\in D\cap\overline{B(\xi,\kappa^{-1}R)}$

,

and .
$h$

Proof. By (4.1) and Lemma 3.1 we have
(4.9)$\frac{h(x)}{h(y_{R})}\leq A2(\frac{R}{\delta_{D}(x)})^{\alpha}$

for $x\in D\cap B(\xi,R)$ ,and are positive constants depending only on and the dimension. We note thathas anon-negative subharinonic extension to $B(\xi,R)$ with zero values on([5, Theorem 5.2.1]). Let $u=h^{*}/A_{2}h(yR)$ . Using the inequality$\alpha$$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}A2$$A\xi$$h$$h^{*}$$B(\xi,R)\backslash \mathrm{Z}\mathrm{t}$$[ \log(\frac{R}{\delta_{D}(x)})]^{n}\leq(\frac{n}{\tau})^{n}(\frac{R}{\delta_{D}(x)})^{\tau}$

for $x\in D\cap B(\xi,R)$ ,where $\tau>0$ is as in Lemma 4.3, we obtain from (4.9) and Lemma 4.3 that
$I= \int_{B(\xi,R)}(\log^{+}u)^{n}dx\leq A\int_{D\cap B(\xi,R)}(\frac{R}{\delta_{D}(x)})^{\tau}dx\leq AR^{n}$Hence it follows from Lemma 4.1 that $u\leq A$ on
yields thatfor
$h(x)\leq Ah(\mathcal{Y}R)$$S(\xi, \kappa^{-1}R)$.

, and the maximum principle$x\in D\cap\overline{B(\xi,\kappa^{-1}R)}$.

Thus the lemma follows.

$\square$

5. Boundary Harnack principle
The purpose of this section is to show a(uniform) boundary Harnack principle, which isuseful to obtain properties of Martin kernels. The proofs in this section are based on [2] for
aunifom domain.
For $r>0$ we let$U(r):=\{x\in D:\delta_{D}(x)<r\}$.

We denote by 03 $(x,E, U)$ the harmonic measure of aset for an open set evaluated at .
We writefor the volume of aset . Let us start with an estimate of aharmonic measure.
$U$

$E$

$x$

$E$

$|E|$

Lemma 5.1. Let be a domain satisfying (I). Then there are constants
such that if $0<r<P\mathrm{o}/2$, then$D$

} $<1$ and$0<\mathrm{q}$$A_{3}\geq 1$$\omega(x, U(r)\cap S(x,A3r),$ $U(r)\cap B(x,A3r))\leq\epsilon 0$$forx\in U(r)$.

.
, and let
Proof. Let $x\in U(r)$ . Then there is so that
Take
with
. Then we have $|x-w|\leq 2A0r$ by (3.1), and so$B(x,3A0r)$
and the dimension such
. Hence there is) $<1$ depending only on$c_{\lambda}$$w\in[x,z\lambda]$$\backslash U(r)$$B(z\lambda,p\mathrm{o})\subset c_{\lambda}\subset B(z\lambda,A0p\mathrm{o})$$x\in c_{\lambda}$$B(w,r)\subset$$\delta_{D}(w)=2r$$A0$$0<\mathrm{q}$that$\frac{|U(r)\cap B(x,3A_{0}r)|}{|B(x,3A_{0}r)|}\leq \mathrm{a}$.

10
Let $A_{3}:=3A0+1$ . We note that 0)( , $U(r)\cap S(x,A3^{\gamma)}$ , $U(r)\cap B(x,A3^{\gamma))}$ has asubharmonic( , Theorem 5.2.1]). Henceextension to $B(x,A3r)$ with zero values on$\cdot$$B(x,A3r)\backslash \overline{U(r)}$$\omega$$[5$$\omega(x)\leq\frac{1}{|B(x,3A_{0}r)|}\int_{B(x,3A_{0}r)}\omega(y)dy\leq\epsilon_{0}$.

Thus the lemma follows.

$[]$

Lemma 5.2. Let D be a domain satisfying (I) andbe as in Lemma 5.1. Then there ispositive constantsuch that if r $>0$ and R $>0$, then$A_{3}$$a$$A_{4}\leq 1$(5.1)$\omega(x,U(r)\cap S(x,R),U(r)\cap B(x,R))\leq\exp(A3-A4\frac{R}{r})$for $x\in U(r)$ .

Proof. Note that if $R\leq A3r$ , then (5.1) clearly holds since the right hand side of (5.1) is notless than 1. Letbe such that $kA3^{\Gamma}<R\leq(k+1)A3r$ . We claim that$k\in \mathrm{N}$(5.2)$\omega(\cdot,U(r)\cap S(x,R),U(r)\cap B(x,R))\leq\epsilon_{0}^{j}$$\sup$$U(r)\cap B(x,R-jA_{3}r)$, , where } is as in Lemma 5.1. We show this by induction.
for $j=0$ ,, then (5.2)
clearly holds. We assume that (5.2) holds for $j-1$ , and show (5.2) for . Let $y\in U(r)\cap$$S(x,R-jA3r)$ . since $S(y,A3r)\subset\overline{B(x,R-(j-1)A3r)}$ , it follows from the assumption, themaximum principle and Lemma 5.1 that$\cdots$$k$$\mathrm{I}\mathrm{f}j=0$$\mathrm{q}$$j$$\omega(y,U(r)\cap S(x,R),U(r)\cap B(x,R))\leq\epsilon_{0}^{j-1}\omega(y,U(r)\cap S(y,A3r),U(r)\cap B(y,A3r))$$\leq\epsilon_{0}^{j}$.

Since is an arbitrary point in $U(r)\cap S(x,R-jA3r)$ , the maximum principle yields (5.2) for. Finally, noting that $R/A_{3}r\leq 2k$ , we obtain from (5.2) with $j:=k$ that
$y$

$j$

$\omega(x,U(r)\cap S(x,R),U(r)\cap B(x,R))\leq\exp((\mathrm{q})-1)k)\leq\exp(\frac{\mathrm{q}_{1}-1}{2A_{3}}\frac{R}{r})$.

Thus the lemma follows.

$\square$Lemma 5.3. Suppose that is a domain satisfying (I) and that
$0<R<R\xi$ . If is a positive bounded harmonicfirnction on, theneverywhere on
$D$

$h$

$\xi\in\partial D$$D\cap B(\xi, \kappa R)$

satisfies (II). Let
vanishing quasi-$\partial D\cap B(\xi, \kappa R)$$\omega(x,D\cap S(\xi, \kappa^{-1}R),D\cap B(\xi, \kappa^{-1}R))\leq A\frac{h(x)}{h(y_{R})}$

where

$A$

is independent $ofx$,$R$

Proof. By Lemma 4.4, we have$e^{-1}$

that

and

$h$

$forx\in D\cap B(\xi, \kappa^{-2}R)$

,

.
on

.

be suchon $D\cap B(\xi, \kappa^{-1}R)$ , and put $u:=A5h/h(y_{R})$ . Then it follows from (4.1) and Lemma 3.1(5.3)$h\leq Ah(y_{R})$$u(x) \geq A(\frac{\delta_{D}(x)}{R})^{\alpha}$$D\cap B(\xi, \kappa^{-1}R)$$\mathrm{L}\mathrm{e}\mathrm{t}A5$$\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}A_{\mathit{5}}h/h(y_{R})\leq$

for $x\in D\cap B(\xi, \kappa^{-1}R)$ .

Let $Dj:=\{x\in D:\exp(-2^{j+1})\leq u(x)<\exp(-2^{j})\}$ and $Uj:=\{x\in D:u(x)<\exp(-2^{j})\}$ .
Then, by (5.3), we have
$Uj \cap B(\xi, \kappa^{-1}R)\subset Vj:=\{x\in D:\delta_{D}(x)\leq A6R\exp(-\frac{2^{j}}{\alpha})\}$.

11
Let $\{Rj\}$ be asequence defined by $R0:=\kappa^{-1}R$ and.

$R_{j}:=( \kappa^{-1}-\frac{6(\kappa^{-1}-\kappa^{-2})}{\pi^{2}}\sum_{k=1}^{j}\frac{1}{k^{2}})R$Then. We briefly write$Rj\downarrow\kappa^{-2}R$$\omega$)$:=\omega(\cdot,D\cap S(\xi, \kappa^{-1}R),D\cap B(\xi, \kappa^{-1}R))$if$D_{j}\cap B(\xi,R_{j})\neq\emptyset$

0if$D_{j}\cap B(\xi,R_{j})=\emptyset$$-\alpha)$$\sup$, and put,$d_{j}:=\{$ $D_{j}\cap B(\xi,R_{j})u$.

is bounded by aconstant independent
It suffices to show thatand $x\in Uj\cap B(\xi,Rj)$ . Then the maximum principle yields that$\sup_{j\geq 0j}d$

(5.4)

and

$\mathrm{o}\mathrm{f}R$$\infty(x)\leq\omega(x, U_{j}\cap S(\xi,R_{j-1}),U_{j}\cap B(\xi,R_{j-1}))+d_{j-1}\iota\ell(x)$$u$. Let $j>0$.

Since $B(x,R_{j-1}-R_{j})\subset B(\xi,R_{j-1})$ , the first term of the right hand side of (5.4) is not greaterthan$\omega(x,Vj\cap S(x,Rj-1-Rj),Vj\cap B(x,Rj-1-Rj))\leq\exp(A3-A4\frac{R_{j-1}-R_{j}}{A_{6}R\exp(-2j/\alpha)})$by Lemma 5.2. Let us divide the both sides of (5.4) by $u(x)$ and take the supremum over$Dj\cap B(\xi,Rj)$ . Then we have$d_{j} \leq\exp(2^{j+1}+A_{3}-A_{4}\frac{6(\kappa^{-1}-\kappa^{-2})}{\pi^{2}}\frac{\exp(2^{j}/\alpha)}{A_{6}j^{2}})+d_{j-1}$

Since$d_{0}\leq e^{2}$.

, we obtain$d_{j} \leq\sum_{j=1}^{\infty}\exp(2^{j+1}+A_{3}-A_{4}\frac{6(\kappa^{-1}-\kappa^{-2})}{\pi^{2}}\frac{\exp(2^{j}/\alpha)}{A_{6}j^{2}})+d_{0}<\infty$.

Thus the lemma follows.

$\square$Lemma 5.4 (Boundary Harnack principle). Suppose that is a domain satisfying (I) andthat
satisfies (II). Let $0<R<R_{\xi}$ . $Ifu$ and are positive bounded hamonicfunctions, thenvanishing quasi-everywhere on
on$D$$\xi\in\partial D$$v$$\partial D\cap B(\xi, \kappa R)$$D\cap B(\xi, \kappa R)$$\frac{u(y)}{v(y)}\approx\frac{u(y’)}{v(\oint)}$for$y, \oint$$\in D\cap B(\xi, \kappa^{-2}R)$

where the constant of comparison is independent of ,
$y$

$y’$

, $R$,

$u$

,

and .
$v$

Proof By Lemma 4.4 the maximum principle and Lemma 5.3, we have$u(y) \leq Au(y_{R})\omega(y,D\cap S(\xi, \kappa^{-1}R),D\cap B(\xi, \kappa^{-1}R))\leq Au(y_{R})\frac{v(y)}{v(y_{R})}$

for $y\in D\cap B(\xi, \kappa^{-2}R)$ . Changing the roles of and , we have$u$$v(y’) \leq Av(\mathcal{Y}R)\frac{u(\sqrt)}{u(y_{R})}$

for

Hence two inequalities above yield the lemma.

$v$$\sqrt$$\in D\cap B(\xi, \kappa^{-2}R)$.
$\square$12
Remark 4. We note that the constant of comparison andi Lemma 5.4 depends on 4. For
abounded uniform domain, these constants could be taken uniformly for 4([2, Theorem1]). Using this fact, the first author showed the uniqueness of akernel function at 4([2,Lemma 4and Proof of Theorem 3]). However, in view of Lemma 2.1, we need not takethose constants uniformly in order to prove Theorem. We also note that there is aboundeddomain satistying (I) and each boundary point satisfies (II) but not auniform domain (seeexample in Section 8).
$R\xi$$\mathrm{n}$6. Proof of Theorem and Corollary
Suppose that $D$ is adomain satisfying (I) and thatsatisfies (II). We note first thatevery Martin kernel at is akemel function at . In fact, let $R>0$ be small enough and. Applying Lemma 5.4 to $u:=G(x$ , and $v:=G(0$ , , we see that eachMartin kernel at 4is bounded on, and is kernel function at 4.
$\xi\in\partial D$

$\xi$

$\xi$

$x\in D\backslash B(\xi, \kappa R)$$\cdot$$)$$\cdot$$)$$D\backslash B(\xi, \kappa R)$

Proof of Theorem. Let ,and $R>0$ be small enough. Then, by definition of
the Martin kernel at 4, there are sequences
andin $D$ converging to such that
$K(\cdot,yj)arrow u$ and, respectively. Since $K(x,y_{j}) \approx K(x,\oint_{j})$ forby Lemma 5.4 if is sufficiently large, we have $u(x)\approx v(x)$ for. Sincethe constant of comparison is independent of , it follows from the minimality of andand $u(x\mathrm{o})=1=v(x\mathrm{o})$ that
. Henceis a singleton. Furthemore, it follows from
Lemma 2.1 that A(4). Theorem is proved.
$u$$v\in\Delta_{1}(\xi)$$\{\mathcal{Y}j\}$$\{f_{j}\}$$\xi$$K( \cdot,\oint_{j})arrow v$$x$ $\in D\backslash B(\xi, \kappa R)$$j$$x\in D\backslash B(\xi, \kappa R)$$R$$u\equiv v$$u$$\Delta_{1}(\xi)$$=\Delta 1(\xi)$$\square$

Proofof Corollary. Let $x\in D$ . By$\overline{D}\backslash \{x\mathrm{o}\}$if$\xi_{1}$,$v$, we see that $K(x$ ,extends continuously to. Moreover, it follows from the first paragraph of this section thatare distinct. Thus Corollary follows.
$\mathfrak{M}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{m}$$\cdot$$)$$K(\cdot,\xi_{1})\neq K(\cdot,\xi_{2})$$\xi_{2}\in\partial D$$\square$

7. Remark for bounds in condition (II)Let $x=$ (I)$\cdots$,$x_{n}$)$\in \mathbb{R}^{n}$and let$\mathbb{H}+:=\{x\in \mathbb{R}^{n} :x_{n}>0\}$and$\mathbb{H}_{-}:=\{x\in \mathbb{R}^{n} :x_{n}<0\}$.

In view of dilation, we give examples for $p0=1$ .

Example of (i)$B(0, 1)\cup\{0\}$$(\theta 1>\sin^{-1}(1/A\mathrm{o}))$. Let$w0=(0, \cdots,0,A\mathrm{o})$. We consider the domain

and let

$V_{1}$$D:=(B(0,A_{0}+1)\backslash (\overline{B(0,A_{0}-1)\cap \mathbb{H}_{+}}))\cup V_{1}$be the convex hull of.

Then $D$ satisfies (I) and the unionin the condition (II) at 0is, that is,
the origin satisfies (II). But there are two minimal Martin boundary points at the origin.
$B(0,2p_{1})\cap \mathbb{H}_{-}$$\mathscr{C}(0)$Example of (ii) for $1<A0\leq 2$ ( $0<\theta_{1}\leq\sin^{-1}(1/A_{0})$ and$w1=$ $( 0, \cdots,0, 1)$ ,$p_{1}>p_{0}\cos\theta_{1}$$w2=(\sqrt{1-(2-A\mathrm{o})^{2}},0, \cdots,0, -1)$

$w3=$ $(\sqrt{1-(2-A\mathrm{o})^{2}},0, \cdots,0,A0 -1)$.

). Letand13 13Let$V_{2}$

be the convex hull of $B(w2, 1)\cup\{w3\}$ . We consider the domain$D:=(B(0, 5)\backslash (\overline{B(0,3)\cap \mathbb{H}_{+}}))\cup B(w_{1}, 1)\cup V_{2}$Then $D$ satisfies (I) andpoints at the origin.

$\mathscr{C}(0)=B(0, 2p_{1})\cap \mathrm{I}\mathrm{H}\mathrm{I}_{-}$.

. But there are two minimal Martin boundary

Example of (ii) for $A0>2$ (and$(1, 0, \cdots,0, 1-A\mathrm{o})$ and $w_{3}’=(1,0, \cdots,0, 1)$ . Let. We consider the domain$0<\theta_{1}\leq\sin^{-1}(1/A\mathrm{o})$$P1>p\circ\cos\theta_{1}$$\sqrt{2}=$$V_{3}$). Let $\sqrt{1}=(0, \cdots,0, 1)$ ,be the convex hull of$B(\sqrt{2},1)\cup$$\{w_{3}’\}$$D:=(B(0,5)\backslash (\overline{B(0,3)\cap \mathbb{H}_{+}}))\cup B$

Then $D$ satisfies (I) andpoints at the origin.

$\mathscr{C}(0)=B(0,2p_{1})\cap \mathbb{H}_{-}$( I) $1)\cup V3$ .

. But there are two minimal Martin boundaryand
It is easy to check, in each case, that $D$ is represented as the union of ballswith the same
, and that includes ball of radius 1and is included in aball of radius$(0,y)\cap
B(0,2p_{1})$is not included
center. We also observe that any truncated circular cone
, so thatin. Moreover, we observe that one limit function
and another limit function
is bounded on
obtained by approaching from,so that the origin has two
is bounded on
obtained by approaching fromminimal Martin boundary points.
$B(z\lambda, 1)$$A\circ$$V_{i}$$V_{i}$$\mathrm{r}_{\theta_{1}}$$\mathscr{C}(0)=B(0,2p_{1})\cap \mathbb{H}_{-}$$D\cap \mathbb{H}_{+}$$D\cap \mathbb{H}_{+}$$D\cap \mathbb{H}_{-}$$D\cap \mathbb{H}_{-}$$D\cap \mathbb{H}+$8. Example of adomain satisfying (I) and (II) but not auniform domain
is called auniform if there exists apositive constant with the following
Adomainproperty. For each pair of points , $x2\in\Omega$ there is arectifiable curve in joining and$x2$ such that$A$$\Omega$$\Omega$$x1$$\gamma$$x_{1}$(i) $\ell(\gamma)\leq A|x_{1}-x_{2}|$ ,(ii) $\min\{\ell(\gamma(x1,z)),\ell(\gamma(z,x2))\}\leq A\delta_{\Omega}(z)$,for all
are the length of and the subarc of between and , respectively.
where
and
For simplicity, we give an example when $n=2$ .
$\ell(\gamma)$$\gamma(z,w)$$z$$\in\gamma$$z$$\gamma$$w$

Example. Let $a=(0,2)$ , $b=(0, -2)$ and $c=(-2,0)$ . Suppose$\Omega:=B(a,2)\cup B(b,2)$ $\cup B(c,2)$

Then$\Omega$.

satisfies (I) and each boundary point satisfies (II) but not auniform domain.

In fact, let $p=$ $(0, 1)$ and $w=(x,y)$ be apoint in $S(p, 1)$ such that $>0$ and $0<y<1$ ,. Then $y=1-(1-x^{2})^{1/2}$ . Let be an arbitrary rectifiable curve inand letjoining and . Then must hit -axis $\{x=0\}$ , and we have$x$$\overline{w}=(x, -y)$$w$$\Omega$$\gamma_{w}$$\overline{w}$$\ell(\gamma_{w})\geq \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(w,$$\gamma_{w}$

{x

$y$

$=0 \})=x=\frac{x}{1-(1-x^{2})^{1/2}}y=\frac{1}{2}\frac{x}{1-(1-x^{2})^{1/2}}|w-\overline{w}|$This inequality shows that aconstant$A$

satisfying (i) does not exist$\lim\underline{x}=+\infty$$xarrow 0+1-(1-x^{2})^{1/2}$.

$\sin \mathrm{c}$.

14
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H. AIKAWA; Boundary Harnack principle and Martin boundary for a uniform domain, J. Math. Soc.
Japan 53, no. 1, 119-145, 2001.
[3] H. AIKAWA, K. HIRATA AND T. LUNDH; Martin boundary points ofJohn domains and unions of convex[2]

sets, preprint.

A. ANCONA; Une propriiti de la compactification de Martin d’un domaine euclidien, Ann. Inst.
Fourier (Grenoble) 29, n0.4, , 71-90, 1979.
[51 D. H. ARMITAGE AND S. J. GARDINER; Classical potential theory, Springer-Verlag London Ltd., Lon[4]$\mathrm{i}\mathrm{x}$don, 2001.
[6] Y. DOMAR; On the existence of a largest subharmonic minorant of a givenfunction, Ark. Mat. 3(1957),

429-440