Helms L.L. Potential Theory

Подождите немного. Документ загружается.

78 2 The Dirichlet Problem

for all x ∈ ∂Ω,andu

is bounded below by the same constant bounding

u;thatis,u

is a superharmonic member of U

. Therefore, the family of

superharmonic functions in U

is a saturated family.

Deﬁnition 2.6.8 If H

= H

and both are harmonic, then f is called a

resolutive boundary function ;inthiscase,H

= H

is called the

Dirichlet solution for f .

The solution of the Dirichlet problem need not be unique. In fact, given a

continuous boundary function f on the half-space R

= {(x

,...,x

; x

0}⊂R

, according to Theorem 1.9.1 there is a one-parameter family of

Dirichlet solutions corresponding to f. The diﬃculty here is related to the

fact that the region R

is unbounded. This diﬃculty will be avoided by

assuming that Ω is bounded for the remainder of this section. It is also

possible that the Perron-Wiener method will result in the same harmonic

function for two diﬀerent, but not too diﬀerent, boundary functions. Let

Ω = B

0,1

∼{0}⊂R

,letB = B

0,1

,andlet

f =



1at0

0on∂B.

Consider the function u(x)=1/|x|,x∈ Ω, the restriction of the fundamental

harmonic function u

to Ω. For every j ≥ 1,u/j ∈ U

,sothatH

on Ω. Since the zero function belongs to L

, 0 ≤ H

≤ H

=0andso

= H

=0onΩ. On the other hand, if g =0on∂Ω,then

=0.Thus,f and g determine the same harmonic function although

0=g(0) = f(0) = 1. In this case, the set of points at which f and g diﬀer

does not aﬀect the Dirichlet solution. Such sets will be studied in greater

detail later.

What follows in this section is dependent upon Corollary 2.3.6 in which it

is assumed that Ω is bounded. Since the conclusions of the corollary will be

extended to unbounded Ω in Theorem 5.2.5, the results of the remainder of

this section are valid for unbounded sets.

Lemma 2.6.9 Let Ω be a bounded open subset of R

.Ifu ∈ L

and v ∈ U

then u ≤ v and H

≤ H

on Ω.

Proof: ItcanbeassumedthatΩ is connected. If either v is not superhar-

monic or u is not subharmonic, then u ≤ v trivially. It suﬃces to consider

the case where v − u is superharmonic. If x ∈ ∂Ω and f(x) is ﬁnite, then

lim inf

y→x,y∈Ω

(v −u)(y) ≥ lim inf

y→x,y∈Ω

v(y) − lim sup

y→x,y∈Ω

u(y) ≥ f (x) − f (x)=0;

if f (x)=+∞, then lim inf

y→x,y∈Ω

v(y)=+∞ and lim inf

y→x,y∈Ω

(v −

u)(y) ≥ 0 by virtue of the fact that u ∈ L

is bounded above on Ω and,

similarly, if f(x)=−∞. Therefore, lim inf

y→x,y∈Ω

(v − u)(y) ≥ 0 for all

2.6 Perron-Wiener Method 79

x ∈ ∂Ω,andv ≥ u on Ω by Corollary 2.3.6. Using the fact that v ≥ u for all

v ∈ U

and each u ∈ L

, H

≥ u for all u ∈ L

and therefore H

≥ H

If there is any merit to the Perron-Wiener method, it should give solutions

that are consistent with certain natural solutions.

Theorem 2.6.10 Let Ω be a bounded open subset of R

.Iff is bounded on

∂Ω and there is a harmonic function h on Ω such that lim

y→x,y∈Ω

h(y)=

f(x),x∈ ∂Ω,thenf is resolutive and H

= h.

Proof: As was noted in Section 0.1, h has a continuous extension to Ω

−

and is

bounded on Ω. Therefore, h belongs to both L

and U

and H

≤ h ≤ H

Thus, H

= H

= h, and since h is a harmonic function, f is a resolutive

boundary function and h = H

In particular, if h is harmonic on a neighborhood of a compact set Ω and

the Perron-Wiener method is applied to h|

∂Ω

, the solution is just what it

should be.

Eventually necessary and suﬃcient conditions will be determined in or-

der that a boundary function be resolutive. For the time being, it is more

convenient to study the class of resolutive functions. Each of the assertions

of the following lemma has been proved already or is easily proved from the

deﬁnitions.

Lemma 2.6.11 Let Ω be a bounded open subset of R

,letf and g be ex-

tended real-valued functions on ∂Ω,andletc be any real number.

(i) If f = c on ∂Ω,thenf is resolutive and H

= c on Ω.

(ii)

f+c

= H

+ c and H

f+c

= H

+ c.Iff is resolutive, then f + c is

resolutive and H

f+c

= H

+ c.

(iii) If c>0,then

= cH

and H

= cH

.Iff is resolutive, then cf is

resolutive and H

= cH

for c ≥ 0.

(iv) If f ≤ g,then

≤ H

and H

≤ H

(v)

−f

= −H

.Iff is resolutive, then −f is resolutive with H

−f

= −H

(vi)

f+g

≤ H

+ H

and H

f+g

≥ H

+ H

whenever the sums are deﬁned.

If f and g are resolutive and f + g is deﬁned, then f + g is resolutive with

f+g

= H

+ H

Lemma 2.6.12 Let Ω be a bounded open subset of R

. The resolutive bound-

ary functions form a linear class. Moreover, if { f

} is a sequence of such

functions that converges uniformly to f ,thenf is resolutive and the sequence

} converges uniformly to H

Proof: If |f

− f| <,thenf

− <f<f

+ , H

≤ H

+

= H

+ ,

and

≥ H

−

= H

− . Therefore, |H

− H

| = |H

− H

| <,

and the sequence {H

} converges uniformly to H

.Inthesameway,the

sequence {H

} converges uniformly to H

.Thus,H

= H

and since both

are ﬁnite by the above inequalities, f is resolutive.

80 2 The Dirichlet Problem

Lemma 2.6.13 Let Ω be a bounded open subset of R

.If{f

} is an increas-

ing sequence of boundary functions, f = lim

j→∞

,andH

> −∞ on Ω,

then

= lim

j→∞

; if, in addition, the f

are resolutive, then H

= H

and f is resolutive if at least one of H

or H

is ﬁnite on each component

of Ω.

Proof: ItcanbeassumedthatΩ is connected. Since

≤ H

, there is noth-

ing to prove if lim

j→∞

=+∞ on Ω. Therefore, it can be assumed that

lim

j→∞

) < +∞ for some x

∈ Ω.Then−∞ < H

) ≤ H

) ≤

lim

j→∞

) < +∞ for all j ≥ 1 and each of the functions H

is har-

monic by Lemma 2.6.7. Since the limit of an increasing sequence of harmonic

functions is either harmonic or identically +∞ and lim

j→∞

) < +∞,

lim

j→∞

is harmonic on Ω. Suppose >0. For each j ≥ 1, choose v

∈ U

such that

) < H

)+2

−j

and deﬁne

v = lim

i→∞

∞



j=1

− H

For each j ≥ 1,v

− H

is superharmonic on Ω, nonnegative, and less than

2

−j

at x

. From the inequality

v ≥ lim

i→∞

+(v

− H

) = ( lim

i→∞

− H

)+v

and the fact that lim

i→∞

≥ H

,v ≥ v

for each j ≥ 1. Since v

bounded below, the same is true of v. By Theorem 2.4.8, v is superharmonic

on Ω. Moreover, for each x ∈ ∂Ω and each j ≥ 1,

lim inf

y→x,y∈Ω

v(y) ≥ lim inf

y→x,y∈Ω

(y) ≥ f

(x).

Therefore, lim inf

y→x,y∈Ω

v(y) ≥ f(x) for all x ∈ ∂Ω and consequently v ∈

. This means that v(x

) ≥ H

). Since

lim

j→∞

) ≤ H

) ≤ v(x

) ≤ lim

j→∞

)+

and  is arbitrary,

) = lim

j→∞

). Since the latter quantity is

ﬁnite at x

by assumption, H

) is ﬁnite and H

is therefore harmonic.

Now

−lim

j→∞

is a nonnegative harmonic function that vanishes at x

and therefore must be the constant zero function by the minimum principle.

This proves the ﬁrst assertion that

= lim

j→∞

. Suppose now that the

are resolutive. Then H

= H

and

= lim

j→∞

= lim

j→∞

= lim

j→∞

≤ H

;

2.6 Perron-Wiener Method 81

it follows that H

= H

and that f is resolutive if at least one of H

and

is ﬁnite on each component of Ω.

Lemma 2.6.14 If u is a bounded superharmonic function on the bounded

open set Ω such that f(x) = lim

y→x,y∈Ω

u(y) exists for all x ∈ ∂Ω,thenf is

a resolutive boundary function.

Proof: Note that f,

,andH

are all bounded and that the latter two are

harmonic functions. Clearly, u ∈ U

and u ≥ H

. Then lim sup

y→x,y∈Ω

≤

lim sup

y→x,y∈Ω

u(y)=f(x) for all x ∈ ∂Ω and therefore H

∈ L

. It follows

that

≤ H

; since the opposite inequality always holds, the two are equal

and f is resolutive.

Lemma 2.6.15 Let K be a compact subset of R

,andletf be a continuous

function on K. Given >0 and a ball B ⊃ K, there is a function u that is

the diﬀerence of two continuous superharmonic functions deﬁned on B such

that sup

x∈K

|u(x) − f (x)| <.

Proof: By the Stone-Weierstrass theorem, there is a polynomial function u of

the n coordinate variables such that sup

x∈K

|f(x) −u(x)| <.LetB be any

ball containing K,andletv(y)=−|y|

.ThenΔv = −2n, v is superharmonic

by Lemma 2.3.4, and λv is a continuous superharmonic function on B for

λ ≥ 0. Choose λ

such that Δ(u + λ

v) ≤ 0onB.Thenu =(u + λ

v) −λ

with u + λ

v and λ

v continuous superharmonic functions on B.

The existence of a harmonic function associated with each continuous

boundary function will be established now.

Theorem 2.6.16 (Wiener) If f is a continuous function on the boundary

∂Ω of a bounded open set Ω,thenf is resolutive.

Proof: Since ∂Ω is compact, the preceding lemma is applicable. Let B be a

ball containing ∂Ω. Then there is a function of the form u = v − w,where

v and w are continuous superharmonic functions on B, which approximates

f uniformly on ∂Ω.Sincev|

∂Ω

and w|

∂Ω

are resolutive by Lemma 2.6.14,

∂Ω

= v|

∂Ω

− w|

∂Ω

is resolutive. Since u|

∂Ω

approximates f uniformly on

∂Ω, f is resolutive by Lemma 2.6.12.

Although Wiener’s theorem associates a harmonic function with each

continuous boundary function, it leaves open the connection between the

boundary behavior of the harmonic function and the boundary function.

Deﬁnition 2.6.17 Apointx ∈ ∂Ω is a regular boundary point if

lim

y→x,y∈Ω

(y)=f(x)

for all f ∈ C(∂Ω). Otherwise, x is an irregular boundary point. Ω is

a regular region or Dirichlet region if every point of ∂Ω is a regular

boundary point.

82 2 The Dirichlet Problem

Example 2.6.18 Let Ω = B

0,1

∼{0}⊂R

. Then 0 is an irregular bound-

ary point. This follows from the fact that H

=0onΩ for the continuous

boundary function

f(x)=



1ifx =0

0if|x| =1.

In this case, lim

y→0,y∈Ω

(y)=0=1=f(0).

The irregular boundary point of this example is an isolated point of the

boundary. It will be seen later that this is true of all isolated boundary points.

Necessary and suﬃcient conditions for a boundary point to be a regular

boundary point will be taken up now.

Deﬁnition 2.6.19 A function w is a local barrier at x ∈ ∂Ω if w is deﬁned

on Λ ∩Ω for some neighborhood Λ of x and (i) w is superharmonic on Λ ∩Ω,

(ii) w>0onΛ ∩ Ω, and (iii) lim

y→x,y∈Λ∩Ω

w(y) = 0. The function w is a

barrier at x ∈ ∂Ω on Ω if (i), (ii), and (iii) are satisﬁed with Λ = Ω,and

a strong barrier at x ∈ ∂Ω if, in addition, inf {w(z); z ∈ Ω ∼ Λ} > 0for

each neighborhood Λ of x.

Theorem 2.6.20 (Bouligand) Let Ω be a bounded open subset of R

.If

there is a local barrier at x ∈ ∂Ω, then there is a harmonic strong barrier h

at x on Ω satisfying

(i) lim inf

y→x



,y∈Ω

h(y) > 0 for all x



∈ ∂Ω ∼{x},and

(ii) inf {h(y); y ∈ Ω ∼ Λ

} > 0 for each neighborhood Λ

of x.

Proof: Let w be a local barrier at x ∈ ∂Ω. Deﬁne a function m on ∂Ω by

putting m(y)=|y − x|,y ∈ ∂Ω. It will be shown that the function h = H

has the required properties. Since the function |y − x| on Ω is subharmonic

by Lemma 2.3.4 and belongs to the lower class L

, |y − x|≤H

(y)onΩ.

Moreover,

lim inf

y→x



,y∈Ω

(y) ≥ lim inf

y→x



,y∈Ω

|y −x| = |x



− x| > 0

for all x



∈ ∂Ω,x



= x. It remains only to show that lim

y→x,y∈Ω

(y)=0.

Since w is a local barrier at x,thereisaballB

x,r

such that w>0onB

x,r

∩Ω,

superharmonic on B

x,r

∩Ω and lim

y→x,y∈B

x,r

∩Ω

w(y) = 0. Consider any ρ for

which 0 <ρ<rand let B = B

x,ρ

. Two cases will be considered depending

upon whether or not ∂B ∩ Ω is empty. Assume ﬁrst that ∂B ∩ Ω = ∅,in

which case ∂(B ∩Ω) ⊂ B

−

∩∂Ω.Sincem ≥ 0on∂Ω, it can be assumed that

=sup{u; u ∈ L

,u≥ 0}. For any such u and y ∈ ∂(B ∩Ω) ⊂ B

−

∩∂Ω,

lim sup

z→y,z∈B∩Ω

u(z) ≤ lim sup

z→y,z∈Ω

u(z) ≤ m(y) ≤ ρ.

By Corollary 2.3.6, u ≤ ρ on B ∩ Ω, and since this is true of all such

u ∈ L

≤ ρ on B ∩ Ω. Now assume that ∂B ∩ Ω = ∅,andchoose

2.6 Perron-Wiener Method 83

aclosedsetF ⊂ ∂B ∩ Ω such that σ((∂B ∩ Ω) ∼ F ) <ρ/M where M =

sup {|y −x|; y ∈ ∂Ω}.Notethatm ≤ M on ∂Ω and let k =inf{w(y); y ∈ F }

> 0. Deﬁne a function f on ∂B by putting

f(y)=



M if y ∈ (∂B ∩ Ω) ∼ F

0otherwise.

For u ∈ L

,u≥ 0, consider the function

v = u − ρ − (M/k)w − PI(f : B)

which is subharmonic on B ∩Ω. It will be shown that lim sup

z→y,z∈B∩Ω

v(z)

≤ 0 for all y ∈ ∂(B ∩ Ω). Note ﬁrst the following facts pertaining to such y:

(i) Since w is l.s.c. at points of F ,

− lim inf

z→y,z∈B∩Ω

w(z) ≤−k for y ∈ F ;

otherwise,

− lim inf

z→y,z∈B∩Ω

w(z) ≤ 0.

(ii) Since m ≤ M and lim sup

z→y,z∈B∩Ω

u(z) ≤ M for y ∈ ∂Ω,u ≤ M on Ω

by Corollary 2.3.6 and

lim sup

z→y,z∈B∩Ω

u(z) ≤ M for y ∈ ∂B ∩ Ω;

also,

lim sup

z→y,z∈B∩Ω

u(z) ≤ m(y) ≤ ρ for y ∈ B ∩ ∂Ω.

(iii) On the set (∂B ∩ Ω) ∼ F, f is a continuous function and

lim

z→y,z∈B∩Ω

PI(f : B)=f (y)=M for y ∈ (∂B ∩ Ω) ∼ F ;

otherwise,

− lim inf

z→y,z∈B∩Ω

PI(f : B) ≤ 0.

Note also that ∂(B ∩Ω) ⊂ ((∂B ∩Ω) ∼ F ) ∪F ∪(∂Ω ∩B

−

). By considering

the three cases y ∈ ((∂B ∩ Ω) ∼ F ),y ∈ F ,andy ∈ (∂Ω ∩ B

−

), it is easily

seen that lim sup

z→y,z∈B∩Ω

v(z) ≤ 0fory ∈ ∂(B ∩Ω)sothatv ≤ 0onB ∩Ω

by Corollary 2.3.6. Thus,

u ≤ ρ +(M/k)w + PI(f : B)onB ∩ Ω.

But since u is an arbitrary element of L

≤ ρ +(M/k)w + PI(f : B)onB ∩ Ω.

84 2 The Dirichlet Problem

Since the average value of f is ρ/M and PI(f : B)(x)=L(f : x, B) ≤

M(ρ/M )=ρ and PI(f : B) is continuous at x, lim sup

y→x,y∈B∩Ω

PI(f :

B)(y) ≤ ρ. Since lim

y→x,y∈Ω

w(y)=0,

0 ≤ lim sup

y→x,y∈Ω

(y) = lim sup

y→x,y∈B∩Ω

(y) ≤ 2ρ

in the ∂B ∩ Ω = ∅ case. In either case, lim sup

y→x,y∈Ω

(y)=0and(i)is

satisﬁed. The function h = H

clearly satisﬁes (ii).

Simply put, if there is a local barrier at x ∈ ∂Ω for the bounded open set

Ω, then the function H

|y−x|

is a strong barrier at x. It should also be noted

that the existence of a barrier at a boundary point is a local property of Ω;

that is, the existence of a barrier at x ∈ Ω depends only upon the relationship

between x and the part of Ω near x. Since the assumption that there is a

local barrier at a boundary point can be replaced by the assumption of a

barrier or a strong barrier, the shorter term will be used in the statement of

theorems.

Lemma 2.6.21 Let Ω be a bounded open set. If f is bounded above on ∂Ω

and there is a barrier at x ∈ ∂Ω,then

lim sup

y→x,y∈Ω

(y) ≤ lim sup

y→x,y∈∂Ω

f(y);

or if f is bounded below on ∂Ω and there is a barrier at x ∈ ∂Ω,then

lim inf

y→x,y∈Ω

(y) ≥ lim inf

y→x,y∈∂Ω

f(y).

Proof: By the preceding theorem, there is a barrier w at x deﬁned on Ω

such that inf {w(y); y ∈ Ω ∼ Λ

} > 0 for each neighborhood Λ

of x.IfL =

lim sup

y→x,y∈∂Ω

f(y)and>0, then there is a neighborhood V of x such

that f (y) <L+  for all y ∈ ∂Ω ∩ V .Choosec>0 such that

L +  + c



inf

y∈Ω∼V

w(y)



> sup

y∈∂Ω

f(y)

and let u = L++cw.Thenu is superharmonic on Ω and bounded below since

w>0onΩ. Moreover, lim inf

y→x



,y∈Ω

u(y) ≥ L +  + c lim inf

y→x



,y∈Ω

w(y)

for x



∈ ∂Ω.Ifx



∈ ∂Ω ∼ V , then lim inf

y→x



,y∈Ω

w(y) ≥ inf

y∈Ω∼V

w(y)and

lim inf

y→x



,y∈Ω

u(y) ≥ L +  + c



inf

y∈Ω∼V

w(y)



> sup

y∈∂Ω

f(y) ≥ f(x



On the other hand, if x



∈ ∂Ω ∩ V ,thenL + >f(x



)and

lim inf

y→x



,y∈Ω

u(y) ≥ L + >f(x



2.6 Perron-Wiener Method 85

This shows that lim inf

y→x



,y∈Ω

u(y) ≥ f (x



) for all x



∈ ∂Ω and that u ∈ U

Therefore,

≤ u on Ω and

lim sup

y→x,y∈Ω

(y) ≤ lim sup

y→x,y∈Ω

u(y) ≤ L +  + c lim sup

y→x,y∈Ω

w(y)=L + .

Since  is arbitrary, lim sup

y→x,y∈Ω

(y) ≤ L = lim sup

y→x,y∈∂Ω

f(y).

Corollary 2.6.22 Let Ω be a bounded open set with a barrier at x ∈ ∂Ω.If

f is bounded on ∂Ω and continuous at x,then

lim

y→x,y∈Ω

(y) = lim

y→x,y∈Ω

(y)=f(x).

Proof: By the preceding lemma, continuity of f at x, and the fact that

≤ H

lim sup

y→x,y∈Ω

(y) ≤ f(x) ≤ lim inf

y→x,y∈Ω

(y) ≤ lim inf

y→x,y∈Ω

(y).

Thus, lim

y→x,y∈Ω

(y)=f(x). Similarly, lim

y→x,y∈Ω

(y)=f(x).

Theorem 2.6.23 Let Ω be a bounded open set. A point x ∈ ∂Ω is a regular

boundary point if and only if there is a barrier at x.

Proof: The suﬃciency is immediate from the deﬁnition of a regular boundary

point, the preceding corollary and Theorem 2.6.16. As to the necessity, assume

that x is a regular boundary point, and let m(y)=|y −x|,y ∈ ∂Ω.Asinthe

proof of Theorem 2.6.20, H

(y) ≥|y − x| > 0,y ∈ Ω.Sincex is a regular

boundary point and m is continuous on ∂Ω,lim

y→x,y∈Ω

(y)=m(x)=0.

This shows that H

is a barrier at x.

Theorem 2.6.24 If x is a regular boundary point for the bounded open set

Ω and Ω

is a subset of Ω for which x ∈ ∂Ω

,thenx is regular for Ω

Proof: A barrier at x relative to Ω is a barrier at x relative to Ω

when

restricted to the latter.

Theorem 2.6.25 (Poincar´e) If Ω is a bounded open set, if x ∈ ∂Ω,and

there is a ball B with B ∩Ω = ∅ and x ∈ ∂B,thenx is a regular boundary

point for Ω.

Proof: By taking a ball internally tangent to B at x, it can be assumed that

∂B ∩ ∂Ω = {x}.IfB = B

y,ρ

,aconstantc can be added to the fundamental

harmonic function u

having pole y so that u

+ c vanishes on ∂B.The

function w = u

+ c, restricted to Ω, is then a barrier at x for Ω. It follows

that x is regular by Theorem 2.6.23.

86 2 The Dirichlet Problem

Example 2.6.26 The boundary points of a disk, square, and annulus in

are regular boundary points. The boundary points of a bounded convex

subset of R

are regular boundary points.

In the one example of an irregular boundary point that has been consid-

ered, namely the point 0 of the boundary of Ω = B

0,1

∼{0}⊂R

, the region

Ω is not simply connected. Generally speaking, irregular boundary points are

more diﬃcult to illustrate in R

than in R

,n≥ 3.

Theorem 2.6.27 If x is a boundary point of the bounded open set Ω ⊂ R

having the property that there is a line segment I ⊂∼ Ω with x as an end

point, then x is a regular boundary point.

Proof: Points of R

will be regarded as complex numbers. Let B = B

x,δ

be a

ball such that I ∩∂B = ∅ and δ<1. The function log (z −x) can be deﬁned

to be continuous on B ∼ I by taking the branch cut along the ray starting

at x andinthedirectionI. Letting

w(z)=−Re



log (z − x)



= −

log |z − x|

|log (z − x)|

w is harmonic on Ω ∩B since it is the real part of a function that is analytic

on Ω ∩B.Moreover,w>0onΩ ∩B since |z −x| < 1onΩ ∩B. It is easily

seen that w is a barrier at x.

More is true in the n = 2 case than is stated in this theorem. Recall that

a continuum is a set containing more than one point which is closed and

connected. For the sake of completeness, the following theorem is included

here but the proof is deferred until Section 5.7.

Theorem 2.6.28 (Lebesgue) If x is a boundary point of the bounded open

set Ω ⊂ R

having the property that there is a continuum  ⊂∼ Ω with x ∈ ,

then x is a regular boundary point.

The phrase “right circular cone” in the statement of the following theorem

excludes a cone with zero half-angle.

Theorem 2.6.29 (Zaremba) If x is a boundary point of the bounded open

set Ω ⊂ R

,n ≥ 2, and there is a truncated right circular cone in ∼ Ω with

vertex at x,thenx is a regular boundary point.

Proof: By reducing the half-angle of the cone, it can be assumed that the

truncated cone intersects ∂Ω only in {x}.LetC

be a closed right circular

cone with vertex x such that C

∩ B

x,ρ

∩ Ω = ∅ for some ρ>0, and let

= Ω ∪ (B

x,ρ

∼ C

). Then x is a boundary point for Ω

.Ifitcanbe

shown that x is a regular boundary point for Ω

, then it would follow from

Theorem 2.6.24 that x is a regular boundary point for Ω. Since a barrier at

x on B

x,ρ

∼ C

is a local barrier at x on Ω ∩ (B

x,ρ

∼ C

), it is enough to

2.6 Perron-Wiener Method 87

show that there is a barrier at x on B

x,ρ

∼ C

; that is, there is a positive

superharmonic function w on B

x,ρ

∼ C

for which lim

y→x,y∈B

x,ρ

∼C

w(y)=

0. In other words, it suﬃces to prove that x is a regular boundary point for

x,ρ

∼ C

. Henceforth, it will be assumed that Ω = B

x,ρ

∼ C

. According to

Theorem 2.6.25, all boundary points of Ω, except possibly for x, are regular

boundary points. Let Ω

be a magniﬁcation of Ω by a factor of 2; that is,

= {x +2(y − x):y ∈ Ω}.Alsoletr(y)=|y − x|,y ∈ ∂Ω

. Consider the

Dirichlet solution H

on Ω

corresponding to the boundary function r on ∂Ω

Since H

is harmonic on Ω

and, as in the proof of Theorem 2.6.20, H

(z) ≥

|z − x| > 0forz ∈ Ω

, it remains only to show that lim

z→x,z∈ Ω

(z)=0.

Deﬁne a function u on Ω

−

∼{x} by putting

u(y)=



(y) y ∈ Ω

r(y) y ∈ ∂Ω

∼{x}

and a second function v on Ω

−

∼{x} by putting v(y)=u(x +2(y −x)). The

functions u and v are continuous on Ω

−

∼{x} and Ω

−

∼{x}, respectively. It

will be shown next that u<von ∂Ω∩∂B

x,ρ

.Sincer ≤ 2ρ on ∂Ω

, 2ρ ∈ U

that u = H

≤ 2ρ on Ω

.Sinceu is harmonic on Ω

, it satisﬁes the maximum

principle on Ω

by Corollary 1.5.10, and since it is clearly not constant on Ω

it cannot attain its maximum value at a point of Ω

.Thus,u<2ρ on Ω

and

on Ω

∩∂Ω ∩∂B

x,ρ

in particular. But since v =2ρ on Ω

∩∂Ω ∩∂B

x,ρ

,u<v

on the same set. Since points y ∈ ∂Ω

∩ ∂Ω are regular boundary points

for Ω

, lim

z→y,z∈∂Ω∩Ω

u(z)=ρ;forsuchpoints,u(y)=ρ<2ρ = v(y).

Thus, u<von ∂Ω ∩ ∂B

x,ρ

. Since both functions are continuous on the

compact set ∂Ω ∩ ∂B

x,ρ

, there is a constant α with 1/2 <α<1such

that u<αvon ∂Ω ∩ ∂B

x,ρ

. On the remainder of ∂Ω ∼{x},u =(1/2)v.

Therefore, u<αvon ∂Ω ∼{x}.Sinceu and v are continuous on Ω

−

∼

{x}, lim

z→y,z∈Ω

(αv(z) − u(z)) ≥ 0fory ∈ ∂Ω ∼{x}. Consider the function

αv −u + (u

+ c)where>0andc has been chosen so that u

+ c ≥ 0on

−

.Then

lim inf

z→y,z∈Ω

(αv −u + (u

+ c)) ≥



0ify ∈ ∂Ω ∼{x}

+∞ if y = x.

By Theorem 2.3.6, αv −u +(u

+ c) ≥ 0onΩ. For any closed ball B

−

y,δ

⊂ Ω,

αA(v : y,δ) − A(u : y, δ)+(A(u

: y, δ)+c) ≥ 0.

Fixing δ>0 and letting  → 0, and then letting δ → 0,αv(y) − u(y) ≥ 0for

arbitrary y ∈ Ω. Therefore,

lim sup

z→x,z∈Ω

u(z) ≤ α lim sup

z→x,z∈Ω

v(z)=α lim sup

z→x,z∈Ω

u(x +2(z − x)).