Helms L.L. Potential Theory

Подождите немного. Документ загружается.

208 5 Dirichlet Problem for Unbounded Regions

∗

of a set Λ ⊂ Ω with the same properties; similarly, any function that is

continuous on the closure of Λ

∗

and harmonic on Λ

∗

is the Kelvin transform

∗

of a function h on Λ with the same properties. If u

∗

≥ h

∗

on ∂Λ

∗

for any

such function h

∗

,thenu ≥ h on ∂Λ and u ≥ h on Λ since u is superharmonic

on Ω.Thus,u

∗

≥ h

∗

on Λ

∗

and u

∗

is superharmonic on Ω

∗

by the alternative

deﬁnition of a superharmonic function discussed in Section 2.3.

Remark 5.3.12 Another facet of the Kelvin transformation is important

in the discussion of barriers and regular boundary points. Suppose Ω ⊂

∞

,x ∈ ∂

∞

Ω,u is a positive superharmonic function on Ω such that

lim

z→x,z∈Ω

u(z) = 0, and there is a closed ball B

−

y,ρ

⊂∼ Ω

−

.Ifx = ∞

and u

∗

is the Kelvin transform of u relative to ∂B

y,ρ

,then

lim

∗

→x

∗

∈Ω

∗

) = lim

z→x,z∈Ω

|z − y|

n−2

u(z)=0;

that is, a barrier u at x relative to Ω is transformed into a barrier u

∗

relative to Ω

∗

provided x = ∞.Ifx = ∞ and n ≥ 3, this result does

not necessarily hold because of the indeterminancy of (|z −y|

n−2

/ρ

n−2

)u(z)

as z →∞;if,however,x = ∞ and n = 2, then a positive superharmonic

function that approaches zero at ∞ has a Kelvin transform that is positive

and approaches zero at y. The same is true in the n =2caseifΩ ⊂ B

y,ρ

,y ∈

∂

∞

Ω,andu is a positive superharmonic function that approaches zero at y;

that is, the Kelvin transform of u is a positive superharmonic function on Ω

∗

that approaches zero at ∞.

An inversion relative to a sphere ∂B

y,ρ

can be extended to R

∞

by map-

ping y onto ∞ and ∞ onto y. This extension, however, may not preserve

harmonicity.

Lemma 5.3.13 If u is superharmonic on B

y,ρ

, harmonic on a neighborhood

of y,andu(y)=0, then the function u

∗

deﬁned on R

∞

∼ B

−

y,ρ

∗



n−2

∗

−y|

n−2

u(x),x

∗

∈ R

∼ B

−

y,ρ

0 x

∗

= ∞

is superharmonic on R

∞

∼ B

−

y,ρ

Proof: Consider the n ≥ 3 case ﬁrst. It is clear that u

∗

is l.s.c. on R

∞

∼ B

−

y,ρ

and from the above discussion that u

∗

is superharmonic on R

∼ B

−

y,ρ

.It

remains only to show that L(u

∗

: x, δ) ≤ 0=u

∗

(∞) whenever B

x,δ

⊃ B

−

y,ρ

Since u is harmonic on a neighborhood of y, 0=u(y)=L(u : y, r)for

all suﬃciently small r.SinceL(u : y, r)=(ρ

n−2

)L(u

∗

: y, ρ

/r)by

Equation (5.3), L(u

∗

: y, r) = 0 for all suﬃciently large r. By Theorem 2.5.3

and Green’s identity, Theorem 1.2.2, L(u

∗

: x, r) = 0 for suﬃciently large r.

Suppose r

and L(u

∗

: x, r

)=L(u

∗

: x, r

)=0.Foranyρ<r<r

5.3 PWB Method for Unbounded Regions 209

1/r

n−2

is between 1/r

n−2

and 1/r

n−2

which implies that there is a λ ∈ (0, 1)

such that 1/r

n−2

= λ(1/r

n−2

)+(1−λ)(1 −r

n−2

). By Theorem 2.5.3, there is

a concave function φ on (ρ, +∞) such that L(u

∗

: x, δ)=φ(1/δ

n−2

) whenever

δ>ρ. By the concavity of φ,

φ(1/r

n−2

) ≥ λφ(1/r

n−2

)+(1− λ)φ(1/r

n−2

That is,

0=L(u

∗

: x, r

) ≥ λL(u

∗

: x, r)+(1−λ)L(u

∗

: x, r

)=λL(u

∗

: x, r).

Thus, L(u

∗

: x, r) ≤ 0=u

∗

(∞) whenever B

x,r

⊃ B

−

y,ρ

. This shows that u

∗

superharmonic on R

∞

∼ B

−

y,ρ

. The proof of the n = 2 case is basically the

same using the fact that L(u

∗

: x, r) is a concave function of −log r.

Lemma 5.3.14 Let Ω be an open subset of R

∞

such that ∼ Ω

−

= ∅.If

f ∈ C

(∂Ω) and >0, then there are continuous superharmonic functions

u and v deﬁned on a neighborhood of Ω

−

such that sup

x∈∂Ω

|f(x) − (u(x) −

v(x))| <.

Proof: If ∞ ∈ Ω

−

, the assertion is just that of Lemma 2.6.15. The two

cases ∞∈Ω and ∞∈∂

∞

Ω will be considered separately. Suppose ﬁrst that

∞∈Ω.Since∼ Ω

−

= ∅,thereisaballB

y,ρ

with B

−

y,ρ

⊂∼ Ω

−

.Aninversion

relative to ∂B

y,ρ

maps Ω into Ω

∗

⊂ B

y,ρ

with y an interior point of Ω

∗

and the corresponding Kelvin transformation induces a continuous function

∗

on ∂Ω

∗

. Applying Lemma 2.6.15 to the compact set ∂Ω

∗

∪{y} and the

continuous function that agrees with f

∗

on ∂Ω

∗

and is equal to zero at y,

there are continuous superharmonic functions u

∗

and v

∗

on B

y,ρ

such that

sup

∗

∈∂Ω

∗

) − (u

∗

) − v

∗

))| <



and

∗

(y) − v

∗

(y)| <



Since L(u

∗

: y, r)increasestou

∗

(y)asr↓0andL(u

∗

: y, r)=PI(u

∗

y,r

)(y),

PI(u

∗

y,r

)(y)increasestou

∗

(y)asr ↓ 0. Similarly for v

∗

.Thus,r can be

chosen small enough so that the continuous superharmonic functions

∗



∗

on B

y,ρ

∼ B

y,r

PI(u

∗

y,r

)onB

y,r

∗



∗

on B

y,ρ

∼ B

y,r

PI(v

∗

y,r

)onB

y,r

satisfy

sup

∗

∈∂Ω

∗

) − (u

∗

) − v

∗

))| <



210 5 Dirichlet Problem for Unbounded Regions

and

∗

(y) − v

∗

(y)| <



The functions u

∗

and v

∗

are then harmonic on a neighborhood of y.Let

∗

= u

∗

−u

∗

(y)andv

∗

= v

∗

− v

∗

(y). Then u

∗

and v

∗

are superharmonic on

y,ρ

, are harmonic on a neighborhood of y, and vanish at y.Moreover,

sup

∗

∈∂Ω

∗

) − (u

∗

) − v

∗

))|≤ sup

∗

∈∂Ω

∗

) − (u

∗

) − v

∗

))|

+ |u

∗

(y) − v

∗

(y)|

<.

By the preceding lemma, the Kelvin transforms u and v of u

∗

and v

∗

,re-

spectively, are superharmonic on Ω, provided both are deﬁned to be zero

at ∞. It is clear that u and v are continuous superharmonic functions on a

neighborhood of Ω

−

.Since

sup

x∈∂Ω

|f(x) − (u(x) − v(x))| =sup

∗

∈∂Ω

∗

− y|

n−2

∗

) − (u

∗

) − v

∗

))|

≤ sup

∗

∈∂Ω

∗

) − (u

∗

) − v

∗

)| <,

the assertion is proved in the ∞∈Ω case. Suppose now that ∞∈∂

∞

and that f is continuous on ∂

∞

Ω.Sincef can be approximated uniformly

by a function that is constant on a neighborhood of ∞, it suﬃces to prove

the result for such functions. It also suﬃces to prove the result for functions

that are zero in a neighborhood of ∞, for then an appropriate constant can

be added to u or v. It therefore can be assumed that f is continuous on

∂

∞

Ω and equal to zero on a neighborhood of ∞. The Kelvin transform f

∗

of f relative to a ball B

y,ρ

⊂∼ Ω

−

is then continuous on ∂Ω

∗

and zero on

a neighborhood of y ∈ ∂Ω

∗

. The remainder of the proof is the same as the

proof of the preceding case.

Lemma 5.3.15 If Ω is an open subset of R

∞

with ∼ Ω

−

= ∅ and f ∈

(∂

∞

Ω),thenf is resolutive.

Proof: Since f can be approximated on ∂

∞

Ω by the diﬀerence of two con-

tinuous superharmonic functions deﬁned on a neighborhood of ∂Ω by the

preceding lemma and the restriction of such functions to ∂

∞

Ω are resolutive

by Lemma 5.3.10, the assertion follows from Lemma 5.3.9.

5.4 Boundary Behavior

In order to replace the condition that ∼ Ω

−

= ∅ in the last result of the

preceding section by the condition that ∼ Ω is not polar, it is necessary to

extend the concepts of regular boundary point and barrier.

5.4 Boundary Behavior 211

Let Ω be any open subset of R

∞

.Apointx ∈ ∂

∞

Ω is a regular boundary

point if lim

z→x

(z)=f(x) for all f ∈ C

(∂

∞

Ω); a function u is a local

barrier at x ∈ ∂

∞

Ω if u is a positive superharmonic function deﬁned on

Λ ∩ Ω for some neighborhood Λ of x with lim

z→x,z∈Λ∩Ω

u(z)=0;if,in

addition, Λ = Ω,thenu is called a barrier at x ∈ ∂

∞

Ω.

Lemma 5.4.1 Let Ω be an open subset of R

∞

such that ∼ Ω

−

= ∅.Ifthere

is a local barrier at x ∈ ∂

∞

Ω, then there is a barrier w at x deﬁned on Ω with

the additional property that inf{w(z); z ∈ Ω ∼ Λ

} > 0 for all neighborhoods

of x.

Proof: If ∞ ∈ Ω

−

, the assertion is the same as that of Theorem 2.6.20.

It therefore can be assumed that ∞∈Ω

−

. Suppose ﬁrst that ∞∈∂

∞

Ω.

If x = ∞ and n ≥ 3, there is always a barrier at x with the additional

property, namely, the fundamental harmonic function |z −y|

−n+2

with ﬁxed

pole y ∈∼ Ω

−

. It therefore can be assumed that x ∈ ∂Ω.LetB

−

y,ρ

⊂∼ Ω

−

let Ω

∗

be the inversion of Ω relative to ∂B

y,ρ

,andletv be a local barrier at x.

Then the Kelvin transform

∗

|z − y|

n−2

v(z),z∈ Ω

is a local barrier at x

∗

∈ ∂Ω

∗

. By Theorem 2.6.20 there is a barrier w

∗

at x

∗

deﬁned on Ω

∗

such that inf{w

∗

); z

∗

∈ Ω

∗

∼ V

∗

} > 0 for all

neighborhoods V

∗

of x

∗

. It is easy to see that the Kelvin transform w of

∗

is a barrier at x deﬁned on Ω with the additional property. Consider

now the case ∞∈Ω.Inthiscase,∂Ω is a compact subset of R

. Suppose

there is a local barrier at x ∈ ∂Ω. The boundary function m(z)=|z − x|

is then continuous on ∂Ω and resolutive by Lemma 5.3.15. The proof that

lim

z→x,z∈Ω

(z) = 0 is precisely the same as in the proof of Theorem 2.6.20.

This time, however, it is not true that the function |z − x| is in the lower

class L

since it is not subharmonic at ∞∈Ω.Ifitcanbeshownthatthere

is a number α>0 such that H

(z) ≥ α|z − x| on Ω in a neighborhood

of x, then the minimum principle can be used to show that H

has the

required additional property. By considering the Dirichlet problem for Ω

∗

and the Kelvin transform m

∗

of m, it can be seen that H

is strictly positive

on Ω sinceitistrueofH

∗

. Consider any ball B

x,ρ

⊃ ∂Ω,andletk =

inf{H

(z); z ∈ ∂B

x,ρ

} > 0. Choosing α<1 such that αρ

<k/4,α|z −x| <

k/4forz ∈ ∂B

x,ρ

. Moreover, there is a function u ∈ L

such that u(z) >k/2

for z ∈ ∂B

x,ρ

. The function

w(z)=



max (u(z),α|z − x|) z ∈ Ω ∩ B

−

x,ρ

u(z) z ∈ Ω ∼ B

x,ρ

is a member of L

. This shows that H

(z) ≥ α|x − z| for all z ∈ Ω ∩B

x,ρ

212 5 Dirichlet Problem for Unbounded Regions

The condition that ∼ Ω

−

= ∅ in the following theorem can be replaced by

the less restrictive condition that ∼ Ω is nonpolar. See Remark 5.4.11.

Theorem 5.4.2 Let Ω be an open subset of R

∞

such that ∼ Ω

−

= ∅.If

the function f is bounded above on ∂

∞

Ω and there is a barrier at x ∈ ∂Ω,

then lim sup

z→x,z∈Ω

(z) ≤ lim sup

z→x,z∈∂

∞

f(z). A ﬁnite point x ∈ ∂Ω

is regular if and only if there is a barrier at x.Thepoint∞ is always a regular

boundary point if n ≥ 3. A ﬁnite point x ∈ ∂

∞

Ω is a regular boundary point if

and only if x

∗

is a regular boundary point for Ω

∗

, the inversion of Ω relative

to ∂B

y,ρ

⊂ B

−

y,ρ

⊂∼ Ω

−

;if∞∈∂Ω and n =2,then∞ is a regular boundary

point if and only if y is a regular boundary point for Ω

∗

Proof: The proof of the ﬁrst assertion is basically the same as the proof of

Lemma 2.6.21. It follows from the ﬁrst assertion that a boundary point x is

regular if there is a barrier at x. As to the converse, suppose x is a ﬁnite

regular boundary point for Ω.Ifg(z)=min(|z −x|, 1),z ∈ ∂Ω,thenH

a barrier at x. If x = ∞ and n ≥ 3, there is always a barrier at x,namely,

the fundamental harmonic function with a pole in ∼ Ω

−

. Since a boundary

point x is regular if and only there is a barrier at x, the last two statements

follow from Remark 5.3.12.

Lemma 5.4.3 If Ω is an open subset of R

∞

such that ∼ Ω

−

= ∅, then the

set I of irregular boundary points of Ω is a polar set.

Proof: Letting B

−

y,ρ

⊂ B

y,δ

⊂ B

−

y,δ

⊂∼ Ω

−

= ∅,I ⊂ R

∞

∼ B

−

y,ρ

.The

inversion I

∗

of the set I relative to ∂B

y,ρ

iscontainedinthesetofirregular

boundary points of Ω

∗

by the preceding theorem. (Note that y may be an

irregular boundary point for Ω

∗

,whereasy

∗

= ∞ is not an irregular boundary

point for Ω in the n ≥ 3 case.) Since I

∗

is a polar subset of B

y,ρ

,thereis

a superharmonic function v

∗

on B

y,ρ

such that v

∗

=+∞ on I

∗

.Consider

the Kelvin transform v of v

∗

, deﬁned only on R

∼ B

−

y,ρ

.Sincev =+∞ on

I ∼{∞}, I ∼{∞}is a polar set. If n =2and∞∈I,thenI is a polar

set since {∞} is a polar subset of R

∞

by Theorem 5.3.8; if n ≥ 3, then ∞

cannot belong to I since ∞ is always a regular boundary point when n ≥ 3,

and therefore I is a polar subset of R

∞

,n≥ 3.

Lemma 5.4.4 If Ω is an open subset of R

∞

with ∼ Ω

−

= ∅ and Z is a polar

subset of ∂

∞

Ω, then there is a positive superharmonic function v deﬁned on

a neighborhood of Ω

−

such that lim

z→x,z∈Ω

v(z)=+∞ for all x ∈ Z.

Proof: Letting B

−

y,ρ

⊂ B

y,δ

⊂ B

−

y,δ

⊂∼ Ω

−

= ∅,Z ⊂ R

∞

∼ B

−

y,δ

.Ifx ∈

Z ∼{∞}, there is a neighborhood W ⊂ R

∼ B

−

y,δ

of x and a superharmonic

function w =+∞ on W ∩Z. Then the Kelvin transform of w relative to ∂B

y,ρ

is +∞ on W

∗

∩Z

∗

. It follows that the image of Z ∼{∞}is a polar subset of

y,ρ

. Since adjoining a single point of R

to a polar set results in a polar set,

∗

is a polar subset of B

y,ρ

.Nowletj

be the smallest integer greater than δ,

5.4 Boundary Behavior 213

let Z

= Z ∩(B

−

y,j

∼ B

y,δ

), and let Z

= Z ∩(B

−

y,j

∼ B

y,j−1

),j ≥ j

.Then

Z ∼{∞}= ∪

j≥j

and each Z

∗

is a polar subset of B

y,ρ

. It follows that

there is a superharmonic function v

∗

on B

y,ρ

such that v

∗

=+∞ on Z

∗

.By

replacing v

∗

on B

y,ρ

by its Poisson integral as in Lemma 2.4.11, it can be

assumed that v

∗

is harmonic on a neighborhood of y without aﬀecting its

values on Z

∗

. By adding a constant, it can be assumed that v

∗

(y)=0.Then

the Kelvin transform v

of v

∗

,withv

(∞) deﬁned to be zero, is superharmonic

on R

∞

∼ B

−

y,δ

by Lemma 5.3.13 and v

=+∞ on Z

.Sincev

is l.s.c. on the

compact set R

∞

∼ B

y,δ

, a constant can be added to v

to make it positive

thereon without aﬀecting its inﬁnities. Now let {α

}

j≥j

be a sequence of

positive numbers such that



j≥j

(∞) < +∞ on Z ∼{∞}.Since∞

cannot be in the polar set Z in the n ≥ 3 case, the proof is complete in this

case; if ∞∈Z in the n = 2 case, replace v(z)byv(z)+log(|z − y|/δ).

Lemma 5.4.5 Let Ω be an open subset of R

∞

with ∼ Ω

−

= ∅.Iff and g

are functions on ∂

∞

Ω such that f = g on ∂

∞

Ω except possibly for a polar

subset of ∂

∞

Ω,thenH

= H

and H

= H

; if, in addition, f is resolutive,

then g is also and H

= H

Proof: Using the preceding lemma, the result follows as in the proofs of

Theorem 4.2.19 and Corollary 4.2.20.

The method used to prove the following lemma is due to Schwarz [57] and

is known as the alternating method. The method was originally used to

construct the Dirichlet solution on a region that is the union of two regions

known to be regular (c.f. Kellogg [35]).

Lemma 5.4.6 If Ω is an open subset of R

∞

such that ∼ Ω is not polar and

Z is a polar subset of ∂

∞

Ω, then there is a positive superharmonic function

v on Ω such that lim

y→z,y∈Ω

v(y)=+∞ for all z ∈ Z.

Proof: If ∼ Ω

−

= ∅, the assertion is just that of the preceding lemma.

It therefore can be assumed that ∼ Ω = ∂

∞

Ω and that ∂

∞

Ω is not po-

lar. Fix B

−

y,ρ

⊂ Ω.ThenZ is a polar subset of the boundary of Ω ∼ B

−

y,ρ

Consider ∂

∞

(Ω ∼ B

−

y,ρ

)=∂

∞

Ω ∪ ∂B

y,ρ

.Sincethepointsof∂B

y,ρ

are reg-

ular boundary points for Ω ∼ B

−

y,ρ

and ∂

∞

Ω is not polar, some point of

∂

∞

Ω must be a regular boundary point for Ω ∼ B

−

y,ρ

by Lemma 5.4.3. By

Lemma 5.4.4, there is a positive superharmonic function w on Ω ∼ B

−

y,ρ

such

that lim

y→z,y∈Ω

w(y)=+∞ for all z ∈ Z ⊂ ∂

∞

Ω.Moreover,ifB is a ball

containing B

−

y,ρ

with B

−

⊂ Ω, then it can be assumed that w is harmonic

on B ∼ B

−

y,ρ

; for if this is not the case, w can be replaced by

Ω∼B

−

rel-

ative to Ω ∼ B

−

y,ρ

without changing its boundary behavior at points of ∂Ω

according to Lemma 4.3.2. Choose ρ

>ρ

>ρsuch that B

−

y,ρ

⊂ B.Thenw

is harmonic on a neighborhood of B

−

y,ρ

∼ B

y,ρ

. Two sequences of functions

} and {v

} on B

y,ρ

and Ω ∼ B

y,ρ

, respectively, will be constructed as

follows. Deﬁning

214 5 Dirichlet Problem for Unbounded Regions



PI(w : B

y,ρ

)onB

y,ρ

w on ∂B

y,ρ

is continuous on B

y,ρ

and harmonic on the interior. Let



0on∂

∞

− w on ∂B

y,ρ

and deﬁne



on Ω ∼ B

−

y,ρ

− w on ∂B

y,ρ

where H

is the Dirichlet solution corresponding to f

and the region Ω ∼

−

y,ρ

. Since all points of ∂B

y,ρ

are regular boundary points for Ω ∼ B

−

y,ρ

is continuous on Ω ∼ B

y,ρ

and harmonic on the interior. Now let



PI(w + v

: B

y,ρ

)onB

y,ρ

w + v

on ∂B

y,ρ

Then u

is continuous on B

−

y,ρ

and harmonic on the interior. Again letting



0on∂

∞

− w on ∂B

y,ρ

and deﬁning



on Ω ∼ B

−

y,ρ

− w on ∂B

y,ρ

is continuous on Ω ∼ B

y,ρ

and harmonic on the interior. The sequences

} and {v

} are so deﬁned inductively. On ∂B

y,ρ

= u

− w and so

j+1

− v

= u

j+1

− u

on ∂B

y,ρ

for j ≥ 1; whereas on ∂B

y,ρ

j+1

= w + v

and so

j+1

− u

= v

− v

j−1

on ∂B

y,ρ

for j ≥ 2. Let

=sup

z∈∂B

y,ρ

j+1

(z) − v

(z)| =sup

z∈∂B

y,ρ

j+1

(z) − u

(z)|

and let

=sup

z∈∂B

y,ρ

j+1

(z) − u

(z)| =sup

z∈∂B

y,ρ

(z) − v

j−1

(z)|

for j ≥ 2. Consider the boundary function g thatisequalto1on∂B

y,ρ

and

0on∂

∞

Ω and the associated Dirichlet solution H

.Sinceallpointsof∂B

y,ρ

5.4 Boundary Behavior 215

and some points of ∂

∞

Ω are regular for Ω ∼ B

−

y,ρ

, 0 <H

< 1onΩ ∼ B

−

y,ρ

It follows that there is a constant q<1 such that H

<qon ∂B

y,ρ

.If˜g

is any continuous function on the boundary of Ω ∼ B

−

y,ρ

such that |˜g|≤g,

then |H

˜g

| <qon ∂B

y,ρ

.Since(v

j+1

− v

)/α

= H

j+1

−f

)/α

on Ω ∼

−

y,ρ

, |(f

j+1

−f

)/α

| = |(u

j+1

−u

)/α

|≤1on∂B

y,ρ

,and|f

j+1

−f

| =0

on ∂Ω,

|(v

j+1

− v

)/α

| = |H

j+1

−f

)/α

| <q on ∂B

y,ρ

Thus,

j+1

− v

|≤qα

on ∂B

y,ρ

Therefore, β

j+1

≤ qα

,j ≥ 1. Since u

j+1

− u

is harmonic on B

y,ρ

j+1

− u

|≤ sup

z∈∂B

y,ρ

j+1

(z) − u

(z)| = β

on ∂B

y,ρ

and it follows that α

≤ β

,j ≥ 1. Hence, α

j+1

≤ β

j+1

≤ qα

for

j ≥ 1, and the positive series



is dominated by a convergent geometric

series. This shows that the series

∞



j=1

j+1

− v

)

converges uniformly on Ω ∼ B

−

y,ρ

. Similarly, β

j+1

≤ qβ

, and the positive se-

ries



is dominated by a convergent geometric series, thereby establishing

that

u = lim

j→∞

= u

∞



j=1

j+1

− u

)

deﬁnes a function that is harmonic on B

y,ρ

.Sincev

= u

−w on ∂B

y,ρ

,v =

u − w on ∂B

y,ρ

.Also,u

j+1

= w + v

on ∂B

y,ρ

implies that w = u − v on

∂B

y,ρ

;thatis,u = v + w on both ∂B

y,ρ

and ∂B

y,ρ

, but since both u and

v + w are harmonic on B

y,ρ

∼ B

−

y,ρ

,u= v + w on B

y,ρ

∼ B

−

y,ρ

. It follows

that the function

˜w =



u on B

y,ρ

v + w on Ω ∼ B

−

y,ρ

has the desired properties.

Corollary 5.4.7 (Bouligand [4]) If Ω is an open subset of R

∞

such that

∼ Ω is not polar and h is a bounded harmonic function on Ω such that

lim

y→z,y∈Ω

h(y)=0for all z ∈ ∂

∞

Ω except possibly for a polar set, then

h =0on Ω.

Proof: Let Z be the exceptional subset of ∂

∞

Ω. By the preceding lemma,

there is a positive superharmonic function w on Ω such that lim

y→z,y∈Ω

w(y)

=+∞ for all z ∈ Z. Then for every >0,h+ w is superharmonic on Ω and

lim inf

y→z,y∈Ω

(h+w)(y) ≥ 0 for all z ∈ ∂

∞

Ω. By Theorem 5.2.5, h+w ≥ 0

216 5 Dirichlet Problem for Unbounded Regions

on Ω.Sincew is ﬁnite a.e. on Ω ∩ R

,h ≥ 0 a.e.on Ω ∩ R

, and it follows

from the continuity of h on Ω that h ≥ 0onΩ. Applying the same argument

to −h, h =0onΩ.

Lemma 5.4.8 Let Ω be an open subset of R

∞

such that ∼ Ω is not polar

and let B

−

y,ρ

⊂ Ω ∩ R

.Iff is a bounded function on ∂Ω,then

=inf{u; u hyperharmonic on Ω, u bounded below onΩ, and

lim inf

y→z,y∈Ω

u(y) ≥ f (z) for all z ∈ ∂Ω that are regular boundary

points for Ω ∼ B

−

y,ρ

Proof: Since the right side of the equation is less than or equal to

,it

suﬃces to show that u ≥

for every u in the indicated set. This inequality

is obviously true on any component of Ω on which u is identically +∞.It

therefore can be assumed that u is superharmonic on Ω. Note ﬁrst that the

set Z of irregular boundary points of Ω ∼ B

−

y,ρ

is a polar subset of ∂

∞

by Lemma 5.4.3. By the preceding lemma, there is a positive superharmonic

function w on Ω such that lim

y→z,y∈Ω

w(y)=+∞ for all z ∈ Z. For every

>0,u + w ∈ U

and therefore u + w ≥ H

.Sincew is ﬁnite a.e. on

Ω ∩ R

,u ≥ H

a.e. on Ω ∩ R

.Thus,forx ∈ Ω ∩R

,u(x) ≥ lim

δ→0

A(u :

x, δ) ≥ lim

δ→0

A(H

: x, δ)=H

(x). This shows that u ≥ H

on Ω ∩ R

;

if ∞∈Ω, it follows from Deﬁnition 5.2.1 that for any x ∈ R

and δ>0for

which ∼ B

x,δ

⊂ Ω,u(∞) ≥ L(u : x, δ) ≥ L(H

: x, δ)=H

(∞), showing

that u ≥

on Ω.

Theorem 5.4.9 Let Ω be an open subset of R

∞

such that ∼ Ω is not polar.

Then each f ∈ C

(∂

∞

Ω) is resolutive.

Proof: Since f is bounded on the compact set ∂

∞

Ω, the same is true of H

and H

and both are harmonic. Let m =sup

z∈∂Ω

|f(z)| and let B

−

y,ρ

⊂ Ω.

The Dirichlet solution H

corresponding to Ω ∼ B

−

y,ρ

and the boundary

function

g =



m +1 on ∂B

y,ρ

f on ∂

∞

is harmonic on Ω ∼ B

−

y,ρ

and lim

z→x,z∈Ω∼B

−

y,ρ

(z)=f(x) whenever x ∈

∂

∞

Ω is a regular boundary point for Ω ∼ B

−

y,ρ

. Moreover, the function

u =



m +1 onB

−

y,ρ

on Ω ∼ B

−

y,ρ

is superharmonic on Ω and lim

z→x,z∈Ω

u(z)=f(x) whenever x ∈ ∂

∞

Ω is

a regular boundary point for Ω ∼ B

−

y,ρ

. By the preceding lemma, u ≥ H

Similarly, there is a subharmonic function v on Ω such that v ≤ H

and

lim

z→x,z∈Ω

v(z)=f(x) whenever x ∈ ∂

∞

Ω is a regular boundary point for

5.4 Boundary Behavior 217

Ω ∼ B

−

y,ρ

. It follows that lim

z→x,z∈Ω

−H

)(z) = 0 for all x ∈ ∂

∞

Ω that

are regular boundary points for Ω ∼ B

−

y,ρ

; but since the irregular boundary

points of Ω ∼ B

−

y,ρ

constitute a polar set by Lemma 5.4.3, H

− H

=0by

the preceding corollary.

By the preceding theorem, each f ∈ C

(∂

∞

Ω) determines a harmonic

function H

on Ω.Foreachx ∈ Ω,themapx → H

(x) is a linear functional

on C

(∂

∞

Ω) to which the Riesz representation theorem may be applied to

obtain the integral representation

(x)=



fdμ

where μ

is a Borel measure on ∂

∞

Ω with μ

(∂

∞

Ω)=1sinceH

(x)=1.If

} is an increasing sequence of boundary functions such that H

> −∞ on

Ω and f = lim

j→∞

, then the proof of Lemma 2.6.13 is valid in the present

context and results in

= lim

j→∞

;ifthef

are also resolutive, then

(x)=H

(x)=



fdμ

(5.5)

on Ω, where the three quantities are all +∞ if any one is +∞. This result

can be used to show that the indicator functions of compact subsets of ∂

∞

are resolutive. Using the minimum principle of Theorem 5.2.5 as in The-

orem 2.9.2, the class of μ

null sets is independent of x on any component

of Ω.Ifforeachx ∈ Ω, F

is the completed Borel σ-algebra of subsets of ∂

∞

relative to μ

,thenF = ∩

x∈Ω

is called the σ-algebra of μ



-measurable

sets; those F-measurable functions f on ∂

∞

Ω for which



|f|dμ

< +∞

for some x in each component of Ω are called μ



-integrable functions. As

before, μ

, extended to F, will be called harmonic measure relative to x

and Ω. If necessary, the dependence of harmonic measure on the set Ω will

be noted by using the notation μ

For the remainder of this section, Ω is an open subset of R

having a

Green function. The point ∞∈R

∞

is allowed to be a boundary point of Ω

but not an interior point. As a subset of R

∞

, ∞∈∼Ω and ∼ Ω is not polar

in the n ≥ 3case;inthen =2case,R

∼ Ω is not polar by Theorem 4.6.1 so

that R

∞

∼ Ω is not polar. The results of this chapter are therefore applicable

to Ω as a subset of R

∞

.IfF is a relatively closed subset of Ω, H

Ω∼F

will

denote the Dirichlet solution corresponding to f and Ω ∼ F , regarded as a

subset of R

∞

Theorem 5.4.10 Let Ω be a Greenian subset of R

such that ∼ Ω is nonpo-

lar. If Z is a polar subset of ∂

∞

Ω,thenμ

(Z)=0for all x ∈ Ω. Moreover,

if f is a resolutive function on ∂

∞

Ω and g = f q.e. on ∂

∞

Ω,theng is

resolutive on ∂

∞

Ω and H

= H

on Ω.