Helms L.L. Potential Theory

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198 5 Dirichlet Problem for Unbounded Regions

the Euclidean boundary of Ω will be denoted by ∂Ω, whereas the boundary

of Ω relative to the one-pont compactiﬁcation R

∞

will be denoted by ∂

∞

Ω.

The following deﬁnition is consistent with the earlier deﬁnition.

Deﬁnition 5.2.1 The extended real-valued function u on the open set Ω is

superharmonic on Ω if it is superharmonic on Ω ∼{∞}in the usual sense,

and if ∞∈Ω,then

(i) u is l.s.c. at ∞ and

(ii) u(∞) ≥ L(u : x, δ) for all x ∈ R

and δ>0 such that ∼ B

x,δ

⊂ Ω;

the function u is subharmonic on Ω if −u is superharmonic on Ω;ifboth

u and −u are superharmonic on Ω,thenu is harmonic on Ω.

The constant functions are obviously harmonic on R

∞

. Consider the func-

tion u

(y)=|x − y|

−n+2

deﬁned on R

∞

,n ≥ 3, by putting u

(∞)=0;

although this function is superharmonic on R

, it is not superharmonic on

∞

since 0 = u

(∞) < L(u : x, δ) for any x ∈ R

and δ>0. It is, however,

subharmonic on R

∞

∼{x}. It will be shown shortly that the only functions

that are superharmonic on R

∞

are the constant functions.

A Poisson integral type representation of functions harmonic on a neigh-

borhood of ∞ is possible.

Lemma 5.2.2 If u is nonnegative and harmonic on R

∼ B

−

y,ρ

, continuous

on R

∼ B

y,ρ

,andα = lim

x→∞

u(x) exists in R,then

(i) for x ∈ R

∼ B

−

y,ρ

u(x)=

2πρ



∂B

y,ρ

|y −x|

− ρ

|z −x|

u(z) dσ(z)

and α = L(u : y, r) for all r ≥ ρ;

(ii) for x ∈ R

∼ B

−

y,ρ

,n≥ 3,

u(x)=



∂B

y,ρ

|y −x|

− ρ

|z − x|

u(z) dσ(z) −

αρ

n−2

|x − y|

n−2

+ α;

moreover, if α = L(u : y, ρ),thenL(u : y, r)=α for all r>ρ.

Proof: If x ∈ R

∼ B

y,ρ

,letx

∗

denote the inverse of x relative to ∂B

y,ρ

deﬁned in Equation (1.11). Under this inversion map, the region R

∼ B

y,ρ

maps onto the region B

−

y,ρ

∼{y} and induces a function u

∗

on B

−

y,ρ

∼{y}

by means of the equation

∗

− y|

n−2

∗

)=u(x),x

∗

∈ B

−

y,ρ

∼{y} (5.1)

∗

)=u(x)inthen = 2 case). The function u

∗

is harmonic on B

y,ρ

∼{y}

and continuous on B

−

y,ρ

∼{y} by Theorem 1.8.1. By Theorem 3.2.2, there is

5.2 Exterior Dirichlet Problem 199

a constant c ≥ 0 and a harmonic function v on B

y,ρ

such that u

∗

= cu

+ v

on B

y,ρ

∼{y}.Notethatv has a continuous extension to B

−

y,ρ

so that the

preceding equation holds on B

−

y,ρ

∼{y}.Then =2andn ≥ 3 cases will be

treated separately for part of the proof.

(i)(n =2)Usingthefactthat|x − y||x

∗

− y| = ρ

u(x)=u

∗

)=−c log |x

∗

− y| + v(x

∗

)=−c log

|x − y|

+ v(x

∗

Since α = lim

x→∞

u(x)existsinR, c =0,andu(x)=u

∗

)=v(x

∗

Applying the Poisson integral formula to v, using Equation (1.5), the fact

that |x

∗

− y||x − y| = ρ

, and the fact that u = v on ∂B

y,ρ

u(x)=

2πρ



∂B

y,ρ

−|x

∗

− y|

|z −x

∗

v(z) dσ(z)

2πρ



∂B

y,ρ

|y −x|

− ρ

|z −x|

u(z) dσ(z).

Note that α = lim

x→∞

u(x)=L(u : y, ρ). Since u

∗

= v on B

−

y,ρ

∼{y},for

r ≥ ρ,

L(u : y, r)=

2π



|θ|=1

u(y + rθ) dσ(θ)

2π



|θ|=1

∗

(y +(ρ

/r)θ) dσ(θ)

2π



|θ|=1

v(y +(ρ

/r)θ) dσ(θ)

= v(y).

Therefore, α = L(u : y, ρ)=L(u : y, r) for all r>ρ.

(ii)(n ≥ 3) Since

u(x)=

∗

− y|

n−2

∗

− y|

n−2



∗

− y|

n−2

+ v(x

∗

)



and v is harmonic on B

y,ρ

α = lim

x→∞

u(x)=

n−2

;

that is, u

∗

= αρ

n−2

+ v on B

y,ρ

∼{y}.Sinceu

= ρ

−n+2

on ∂B

y,ρ

,u =

∗

= v + α on ∂B

y,ρ

and v = u − α on ∂B

y,ρ

. Returning to the equation

∗

= αρ

n−2

+ v and applying the Poisson integral representation to v

200 5 Dirichlet Problem for Unbounded Regions

∗

αρ

n−2

∗

− y|

n−2

− α +



∂B

y,ρ

−|x

∗

− y|

∗

− z|

u(z) dσ(z) (5.2)

for x

∗

∈ B

y,ρ

∼{y}. Using the fact that |x − y||x

∗

− y| = ρ

and

Equation (1.5),



∂B

y,ρ

−|x

∗

− y|

|z − x

∗

u(z) dσ(z)

|x − y|

n−2



∂B

y,ρ

|x − y|

− ρ

|z − x|

u(z) dσ(z).

By Equation (5.2),

u(x)=

∗

− y|

n−2

∗

)

= α −

αρ

n−2

|x − y|

n−2



∂B

y,ρ

|x − y|

− ρ

|z −x|

u(z) dσ(z)

for x ∈ R

∼ B

−

y,ρ

. To prove the second assertion of (ii), a relation between

the spherical averages of u and u

∗

will be established. If δ ≥ ρ,then

L(u : y, δ)=



|θ|=1

u(y + δθ) dσ(θ)

n−2



|θ|=1

|δθ|

n−2

u(y + δθ) dσ(θ).

Since the integrand is just u

∗

(y +(ρ

/δ)θ),

L(u : y, δ)=

n−2



|θ|=1

∗

(y +

θ) dσ(θ)

n−2

L(u

∗

: y, ρ

/δ). (5.3)

Returning to the equation u

∗

= αρ

n−2

+ v,

L(u

∗

: y, ρ

/δ)=αρ

n−2

L(u

: y, ρ

/δ)+L(v : y, ρ

/δ).

Since u

= δ

n−2

/ρ

2(n−2)

on ∂B

y,ρ

/δ

L(u

∗

: y, ρ

/δ)=α

n−2

+ v(y).

In particular, L(u

∗

: y,ρ)=α + v(y). Using the fact that u = u

∗

on ∂B

y,ρ

α = L(u : y, ρ)=L(u

∗

: y, ρ)=α + v(y)sothatv(y)=0andthat

5.2 Exterior Dirichlet Problem 201

L(u : y, ρ)=

n−2

L(u

∗

: y, ρ)=α

whenever δ ≥ ρ.

The requirement that α = L(u : y, ρ)in(ii) of the above lemma need not

be satisﬁed. For example, if u

=1/|x − y|

n−2

,n ≥ 3,α = lim

y→∞

(y)=

0 < L(u

: x, ρ)=ρ

−n+2

Note that the equation in (ii) of the lemma can be put into a simpler form

under the additional hypothesis that

α = L(u : y, ρ)=



∂B

y,ρ

u(z)

n−2

dσ(z).

In this case,

u(x)=



∂B

y,ρ



|y −x|

− ρ

|z − x|

−

|x − y|

n−2



u(z) dσ(z)

for x ∈ R

∼ B

−

y,ρ

Deﬁnition 5.2.3 If f is a Borel measurable function on ∂B

y,ρ

that is in-

tegrable relative to surface area, the Poisson Integral of f relative to

∞

∼ B

y,ρ

is deﬁned for x ∈ R

∞

∼ B

−

y,ρ

PI(f :∼ B

y,ρ

)(x)



∂B

y,ρ



|y − x|

− ρ

|z − x|

n−2

−

|x − y|

n−2



f(z) dσ(z)

when n ≥ 3andby

PI(f :∼ B

y,ρ

)(x)=

2πρ



∂B

y,ρ

|y −x|

− ρ

|z − x|

f(z) dσ(z)

when n = 2. The integrand in the n ≥ 3caseisdeﬁnedtobeρ

−n+2

when

x = ∞.

Note that the Poisson integral has been deﬁned so that PI(f :∼ B

y,ρ

)is

harmonic on ∼ B

y,ρ

Theorem 5.2.4 (Brelot [8]) If f is a Borel measurable function on ∂B

y,ρ

that is integrable relative to surface area, then u = PI(f :∼ B

y,ρ

) is har-

monic on R

∞

∼ B

−

y,ρ

and lim sup

z→x,z∈∼B

−

y,ρ

u(z) ≤ lim sup

z→x,z∈∂B

y,ρ

f(z)

for all x ∈ ∂B

y,ρ

; if, in addition, f is continuous at x ∈ ∂B

y,ρ

,then

lim

z→x,z∈∼B

−

y,ρ

u(z)=f(x).

202 5 Dirichlet Problem for Unbounded Regions

Proof: The n ≥ 3 case will be proved with only minor modiﬁcations required

for the n = 2 case. Applying the Laplacian Δ

(x)

to both sides of the equations

deﬁning the Poisson integral, it is easy to justify taking the Laplacian under

the integral; since the integrand is harmonic, u is harmonic on R

∼ B

−

y,ρ

.It

is also easily seen that

lim

x→∞

u(x)=

n−1



∂B

y,ρ

f(z) dσ(z)=u(∞);

that is, u is continuous at ∞ and u(∞)=L(f : y, ρ). It remains only to show

that u(∞)=L(u : x, δ) whenever ∼ B

x,δ

⊂ R

∼ B

−

y,ρ

. This will be done

in two steps, the ﬁrst of which speciﬁes that x = y and δ>ρ. Applying the

inversion relative to ∂B

y,ρ

∗

n−2

∗

− y|

n−2

L(f : y, ρ) − L(f : y, ρ)



∂B

y,ρ

−|x

∗

− y|

|z −x

∗

f(z) dσ(z). (5.4)

Since the last term on the right is PI(f : B

y,ρ

)andPI(f : B

y,ρ

)(y)=

L(f : y, ρ),

L(u

∗

: y, ρ

/δ)=L(f : y,ρ)

n−2

− L(f : y, ρ)+L(f : y, ρ)

= L(f : y, ρ)

n−2

whenever δ>ρ. From the proof of the preceding lemma,

L(u : y, δ)=

n−2

L(u

∗

: y, ρ

/δ),

so that L(u : y, δ)=L(f : y, ρ)=u(∞) whenever δ>ρ.Considernow

any closed ball B

−

,δ

with ∼ B

,δ

⊂ R

∼ B

−

y,ρ

.Chooseδ

>ρsuch that

−

y,δ

⊂ B

,δ

.ThenB

−

,δ

∼ B

y,δ

⊂ R

∼ B

−

y,ρ

.Sinceu is harmonic on

the latter region, u is harmonic on a neighborhood of B

−

,δ

∼ B

y,δ

and by

Green’s identity, Theorem 1.2.2,



∂B

y,δ

u(z) dσ(z)=



∂B

,δ

u(z) dσ(z).

Since L(u : y, δ)=L(u : y, δ

) for all δ>δ

as was just shown, the integral on

the left is zero by Theorem 2.5.3 and so both integrals are zero. By the same

theorem, L(u : x

,δ



)isaconstantforδ



>δ.Sinceu is continuous at ∞,its

spherical average over ∂B

,δ



converges to u(∞)=L(f : y,ρ)asδ



→ +∞.

Thus, L(u : x

,δ



)=u(∞)=L(f : y, ρ) for all δ



>δ. Since this is true

5.2 Exterior Dirichlet Problem 203

for all δ for which ∼ B

,δ

⊂ R

∼ B

−

y,ρ

,italsoholdswhenδ



= δ.Thus,

u(∞)=L(u : x

,δ) whenever ∼ B

x,δ

⊂ R

∼ B

−

y,ρ

. Now consider any point

x ∈ ∂B

y,ρ

. By Equation (5.1), Lemma 1.7.5, and Equation (5.4),

lim sup

z→x,z∈∼B

−

y,ρ

u(z) = lim sup

∗

→x,z

∗

∈B

y,ρ

∗

− y|

n−2

∗

)

= lim sup

∗

→x,z

∗

∈B

y,ρ

PI(f : B

y,ρ

)(z

∗

)

≤ lim sup

z→x,z∈∂B

y,ρ

f(z).

The last assertion follows from Lemma 1.7.6.

Theorem 5.2.5 If u is superharmonic on an open connected set Ω ⊂ R

∞

then u satisﬁes the minimum principle on Ω.IfΩ is any open subset of R

∞

is superharmonic on Ω,andlim inf

z→x,z∈Ω

u(z) ≥ 0 for all x ∈ ∂

∞

Ω,then

u ≥ 0 on Ω.

Proof: If ∞ ∈ Ω, the ﬁrst assertion is just Corollary 2.3.6. If ∞∈Ω,let

m =inf{u(y); y ∈ Ω} and let M = {y ∈ Ω; u(y)=m}.ThenM is relatively

closed by the l.s.c. of u.Ify ∈ M ∩ R

,thenu must be equal to m on a

neighborhood of y. Suppose ∞∈M .Since∞∈Ω,thereisanx

and δ>0

such that ∂B

,ρ

⊂ Ω whenever ρ ≥ δ.Since∞∈M, u(∞)=L(u : x

,ρ)

for all ρ ≥ δ and therefore u = m on ∼ B

,δ

. This shows that u = m on

a neighborhood of ∞. Hence, M is both open and relatively closed. Since Ω

is connected, M = Ω or M = ∅; in the ﬁrst case, u is constant on Ω and in

the second case u does not attain its minimum on Ω. To prove the second

assertion, deﬁne a function ˆu on Ω

−

as follows:

ˆu(y)=



u(y)ify ∈ Ω

lim inf

z→y,z∈Ω

u(z)ify ∈ ∂

∞

Ω.

Then ˆu is l.s.c. on Ω

−

and ˆu ≥ 0on∂

∞

Ω. Consider any component Λ of Ω.

Then ∂

∞

Λ ⊂ ∂

∞

Ω and ˆu ≥ 0on∂

∞

Λ.SinceR

∞

is compact, Λ

−

is compact

and ˆu attains its minimum on Λ

−

.Ifˆu attains a negative minimum at an

interior point of Λ, then the minimum principle is contradicted. Thus, ˆu ≥ 0

on each component Λ of Ω and therefore ˆu ≥ 0onΩ.

Corollary 5.2.6 If u is superharmonic on R

∞

,thenu is a constant function.

Proof: Since R

∞

is compact and u is l.s.c. on R

∞

, u attains its minimum

on R

∞

and is therefore constant by the minimum principle.

Example 5.2.7 (Green Function for a Half-space) Let

Ω = R

= {(x

,...,x

) ∈ R

; x

> 0},n≥ 2,

204 5 Dirichlet Problem for Unbounded Regions

and consider u(x, y)=u

(y), the fundamental harmonic function with

pole at x ∈ Ω.Forx =(x

,...,x

) ∈ R

,letx =(x



), where



=(x

,...,x

n−1

) ∈ R

n−1

,andletx

=(x

,...,x

n−1

, −x

) be the reﬂec-

tion of x across the hyperplane x

=0.Notethatu((x



, 0),y)=u((x



, 0),y

)

since |(x



, 0) −y| = |(x



, 0) − y

|. It now will be shown that u(x, y) −u(x, y

)

is the Green function for the half-space Ω by showing that the diﬀer-

ence satisﬁes (i), (ii), and (iii) of Deﬁnition 3.2.3. Using the fact that

|x −y|≤|x −y

|,u(x, y) −u(x, y

) ≥ 0forx, y ∈ Ω.Sinceu(x, y

)=u(x

,y)

for x, y ∈ Ω, it is clear that u(x, y

) is a harmonic function of y so that

u(x, y) −u(x, y

) has the form of a sum of u

and a harmonic function on Ω.

Fix x ∈ Ω and let v

be a nonnegative superharmonic function on Ω that is

the sum of u

and a superharmonic function on Ω. Consider the function

(y) − u(x, y)+u(x, y

)=v

(y) − u(x, y)+u(x

,y)

and any y

∈ R

= {(y

,...,y

) ∈ R

; y

=0}.Then

lim inf

y→y

,y∈Ω

(y) − u(x, y)+u(x

,y)) = lim inf

y→y

,y∈Ω

(y) ≥ 0

since v

≥ 0onΩ.Moreover,

lim inf

|y|→+∞,y∈Ω

(y) − u(x, y)+u(x

,y)) ≥ lim inf

|y|→+∞,y∈Ω

(y)

+ lim inf

|y|→+∞,y∈Ω

(u(x

,y) − u(x, y)).

The ﬁrst term on the right is nonnegative since v

≥ 0onΩ, and it is easily

seen that lim

|y|→+∞,y∈Ω

(u(x

,y) − u(x, y)) = 0. Thus,

lim inf

y→y

,y∈Ω

(y) − u(x, y)+u(x, y

)) ≥ 0 for all y

∈ ∂

∞

Ω.

By the preceding theorem, v

(y) ≥ u(x, y) − u(x, y

)onΩ and (iii) of Deﬁ-

nition 3.2.3 is satisﬁed. Thus, G

(x, y)=u(x, y) − u(x, y

5.3 PWB Method for Unbounded Regions

The ﬁrst few results of this section have been studied before and do not war-

rant elaboration. The following theorem follows easily from the deﬁnitions.

Theorem 5.3.1 If u and v are superharmonic on the open set Ω ⊂ R

∞

and

c>0,then

(i) cu is superharmonic on Ω,

(ii) u + v is superharmonic on Ω,and

(iii) min (u, v) is superharmonic on Ω.

5.3 PWB Method for Unbounded Regions 205

Lemma 5.3.2 Let u be superharmonic on the open set Ω ⊂ R

∞

.IfΛ is an

open subset of Ω with Λ

−

⊂ Ω, h is continuous on Λ

−

and harmonic on Λ,

and u ≥ h on ∂

∞

Λ,thenu ≥ h on Λ.

Proof: Since lim inf

z→x,z∈Λ

(u − h)(z) ≥ 0on∂

∞

Λ, the result follows from

Theorem 5.2.5.

As in Lemma 2.4.11, the part of a superharmonic function over the com-

plement of a ball can be lowered to obtain a harmonic patch.

Lemma 5.3.3 If Ω is an open subset of R

∞

with R

∞

∼ B ⊂ Ω for some

ball B = B

x,δ

and u is superharmonic on Ω, the function

∼B



u on B

−

∩ Ω

PI(u :∼ B) on R

∞

∼ B

−

is superharmonic on Ω, harmonic on R

∞

∼ B

−

,andu ≥ u

∼B

on Ω.

Proof: Let {φ

} be an increasing sequence of continuous functions on ∂

∞

(∼

B) such that φ

↑ u and let



on ∂

∞

(∼ B)

PI(φ

:∼ B)onΩ ∼ B

−

By Theorem 5.2.4, each h

is continuous on R

∞

∼ B and harmonic on

∞

∼ B

−

.Sinceu ≥ h

on ∂

∞

(∼ B),u ≥ h

on R

∞

∼ B by the preceding

lemma; that is, u ≥ PI(φ

:∼ B)onR

∞

∼ B

−

.Sinceφ

↑ u on ∂

∞

(∼ B),

u ≥ PI(u :∼ B)=u

∼B

on R

∞

∼ B

−

Obviously, u ≥ u

∼B

on Ω. The proof that u

∼B

is superharmonic on Ω is the

same as in the proof of Lemma 2.4.11.

Lemma 5.3.4 If {u

; i ∈ I} is a right-directed family of harmonic functions

on the open set Ω ⊂ R

∞

,thenu =sup

i∈I

is either identically +∞ or

harmonic on each component of Ω.

Proof: Since the assertion concerns components, it can be assumed that Ω

is connected; it might as well be assumed that ∞∈Ω, for otherwise the

assertion is just Lemma 2.2.7. By the same lemma, u is either identically

+∞ or harmonic on Ω ∼{∞}.If∼ B

x,δ

⊂ Ω,thenu

= PI(u

:∼ B

x,δ

)

on R

∞

∼ B

−

x,δ

for each i ∈ I and u = PI(u :∼ B

x.δ

)onR

∞

∼ B

−

x,δ

Lemma 2.2.10. If u is harmonic on Ω ∼{∞},itisintegrableon∂B

x,δ

and

therefore harmonic on R

∞

∼ B

−

x,δ

.Ifu is identically +∞ on Ω ∼{∞},then

u(∞)=L(u : x, δ)=+∞ and u is identically +∞ on Ω.

Deﬁnition 5.3.5 A family F of superharmonic functions on the open set

Ω ⊂ R

∞

is saturated over Ω if it is saturated over Ω ∼{∞}in the usual

sense and if ∞∈Ω,then

206 5 Dirichlet Problem for Unbounded Regions

(i) u, v ∈Fimplies min (u, v) ∈F,and

(ii) for every ball B ⊂ R

with B

−

⊂ Ω and u ∈F, the function

∼B



u on B

−

PI(u :∼ B)onΩ ∼ B

−

belongs to F.

Theorem 5.3.6 If F is a saturated family of superharmonic functions on

the open set Ω ⊂ R

∞

, then the function v =inf

u∈F

u is either identically

−∞ or harmonic on each component of Ω.

Proof: It can be assumed that Ω is connected and that ∞∈Ω.By

Theorem 2.6.2, v is identically −∞ or harmonic on Ω ∼{∞}.Consider

any ball B ⊂ R

with B

−

⊂ Ω.Ifu ∈F,thenu

∼B

≤ u, u

∼B

is harmonic on

Ω ∼ B

−

,andv =inf

u∈F

∼B

on Ω ∼ B

−

. As in the proof of Theorem 2.6.2,

the family {u

∼B

: u ∈F}is left-directed. By Lemma 5.3.4, v is identically

−∞ or harmonic on Ω ∼ B

−

.SinceΩ ∼ B

−

and Ω ∼{∞}overlap, v is

identically −∞ or harmonic on Ω.

The concept of polar set has an extension to the one point compactiﬁcation

of R

Deﬁnition 5.3.7 AsetZ ⊂ R

∞

is a polar set if to each x ∈ Z there

corresponds a neighborhood Λ of x and a superharmonic function u on Λ

such that u =+∞ on Z ∩ Λ.

Theorem 5.3.8 {∞} is a polar subset of R

∞

but not of any R

∞

,n≥ 3.

Proof: First consider R

∞

. The function u(x)=log|x| is harmonic on R

∼

−

0,1

and has the limit +∞ as |x|→+∞. Deﬁning u(∞)=+∞,u is easily

seen to be superharmonic on R

∞

∼ B

−

0,1

. This shows that {∞} is a polar

subset of R

∞

.NowletΛ be any neighborhood of ∞ in R

∞

,n≥ 3, and let u

be superharmonic on Λ.Then∼ B

x,δ

⊂ Λ for some x and δ>0. Let δ

>δ.

By Theorem 2.5.3, L(u : x, ρ) is a concave function of ρ

−n+2

for ρ ≥ δ

.This

implies that there are constants a and b such that L(u : x, ρ) ≤ aρ

−n+2

+ b

for ρ ≥ δ

. Therefore, lim sup

ρ→+∞

L(u : x, ρ) ≤ b<+∞.Ifu(∞)=+∞,

then L(u : x, ρ) would have to approach +∞ as ρ → +∞ by the l.s.c. of u

at ∞. Therefore, u(∞) < +∞ for any such u and ∞ is not a polar subset of

∞

,n≥ 3.

A polar subset of R

∞

,n ≥ 3, is just a polar subset of R

as previously

deﬁned; but a polar subset of R

∞

may include the point at inﬁnity. Note

that Theorems 4.2.9 and 4.2.10 cannot be generalized in the context of R

∞

in view of Corollary 5.2.6.

As in Chapter 2, a function u on Ω ⊂ R

∞

is said to be hyperharmonic

(hypoharmonic) on Ω if on each component of Ω the function u is su-

perharmonic (subharmonic) or identically +∞ (−∞). If f is an extended

real-valued function on ∂

∞

Ω, the upper and lower classes associated with Ω

and f are deﬁned by

5.3 PWB Method for Unbounded Regions 207

= {u; lim inf

z→x,z∈Ω

u(z) ≥ f (x) for all x ∈ ∂

∞

Ω,u hyperharmonic

and bounded below on Ω},

= {u; lim sup

z→x,z∈Ω

u(z) ≤ f (x) for all x ∈ ∂

∞

Ω,u hypoharmonic

and bounded above on Ω},

respectively; the corresponding upper and lower solutions are deﬁned as

=inf{u; u ∈ U

} and H

=sup{u; u ∈ L

}. Noting that the assertions of

Lemma 2.6.11 follow from Corollary 2.3.6 and the order properties of func-

tions, the same assertions are valid in the present context with Corollary 2.3.6

replaced by Theorem 5.2.5. The boundary function f is said to be resolutive

for Ω if

= H

and both are harmonic functions. Since the collection of

functions in U

, which are superharmonic on a given component of Ω,forms

a suaturated family over that component, on each component of Ω the func-

tion

is either identically +∞, identically −∞, or harmonic. The same is

true of H

Several of the following results will require that the complement of an open

subset Ω of R

∞

be nonpolar. In the n = 2 case, this means that R

∼ Ω must

be nonpolar so that Ω ∩ R

is Greenian; but Ω may or may not contain ∞.

This requirement is of no consequence in the n ≥ 3 case for the following

reason. Suppose Ω ⊂ R

∞

,n≥ 3, and ∼ Ω is polar. In this case, ∞∈Ω and

∂

∞

Ω ⊂ R

.Ifh were any bounded harmonic function on Ω,thenitcould

be extended to a harmonic function on R

∞

by Corollary 4.2.16; but such

functions are constant functions by Corollary 5.2.6. Thus, the PWB method

applied to a bounded Borel measurable boundary function will yield only

constant functions.

Lemma 5.3.9 If Ω is an open subset of R

∞

with ∼ Ω nonpolar and {f

}

is a sequence of resolutive boundary functions that converge uniformly to f,

then f is resolutive and the sequence {H

} converges uniformly to H

Proof: The proof is the same as the proof of Lemma 2.6.12.

Lemma 5.3.10 Let Ω be an open subset of R

∞

with ∼ Ω nonpolar. If u is

a bounded superharmonic function on Ω such that f(x) = lim

z→x,z∈Ω

u(z)

exists for all x ∈ ∂Ω,thenf is a resolutive boundary function.

Proof: The proof is the same as the proof of Lemma 2.6.14.

Remark 5.3.11 Recalling that the Kelvin transformation preserves positiv-

ity and harmonicity, it will now be shown that it preserves superharmonicity

as well. Suppose Ω ⊂ R

∼{y} and u is superharmonic on Ω.LetΩ

∗

⊂ R

be the inversion of Ω relative to ∂B

y,ρ

,andletu

∗

be the Kelvin transform

of u. Note that the Kelvin transformation preserves the order properties of

functions. An open subset of Ω

∗

with compact closure in Ω

∗

is the inversion