Helms L.L. Potential Theory

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108 3 Green Functions

Proof: It will be shown ﬁrst that G

is symmetric on B × B in the n ≥ 3

case, the n = 2 case being essentially the same. If x = y = z,thenG

(x, z)=

(z,x)=+∞.Ifx = y and z ∈ B ∼{x},then

(x, z)=

|z − x|

n−2

−

n−2

Since |y −z||y − z

∗

| = ρ

by deﬁnition of z

∗

(z,x)=

|x − z|

n−2

−

n−2

|z − y|

n−2

|x − z

∗

n−2

|x − z|

n−2

−

n−2

(|y − z||y − z

∗

n−2

|x − z|

n−2

−

n−2

(ρ

)

n−2

= G

(x, z).

Lastly, suppose that x = y, z = y,andz ∈ B ∼{x}.Then

(x, z)=

|z −x|

n−2

−

n−2

|x − y|

n−2

|z − x

∗

n−2

and it suﬃces to show that the second term on the right is a symmetric

function of x and z. To see this, let φ be the angle between the line segment

joining x to y and z to y. Then by deﬁnition of x

∗

|z − x

∗

= |z − y|

+ |y − x

∗

− 2|z − y||y −x

∗

|cos φ

= |z − y|

|y − x|

− 2|z − y|

|y −x|

cos φ

and so

|y − x|

|z − x

∗

= |z − y|

|y −x|

+ ρ

− 2ρ

|z − y||y − x|cos φ.

Similarly,

|y −z|

|x − z

∗

= |x − y|

|y −z|

+ ρ

− 2ρ

|x − y||y − z|cos φ

and it follows that

|y −x||z − x

∗

| = |y − z||x − z

∗

|. (3.1)

Thus, G

(x, z)=G

(z,x) whenever x = y, z = y,andz ∈ B ∼{x}.

It is clear from the deﬁnition that G

(x, ·) is superharmonic on B.Us-

ing Equation (1.5) and the deﬁnition of G

(x, z), it is easy to see that

lim

z→z

(x, z) = 0 for all z

∈ ∂B. It follows from Corollary 2.3.6 that

(x, ·) ≥ 0onB.IfG

(x, ·) attains its minimum value at a point of B,it

3.2 Green Functions 109

would have to be a constant function in contradiction to the fact that it is

arbitrarily large near x. Therefore, G

(x, ·) is strictly positive on B.

According to the following theorem, if a nonnegative function is harmonic

on a neighborhood of a point, then it either has an inﬁnite pole at the point

or has a harmonic extension to all of the neighborhood.

Theorem 3.2.2 (Bˆocher [3]) Let Ω be a neighborhood of x

∈ R

,andlet

u be deﬁned, harmonic, and nonnegative on Ω ∼{x

}. Then either

(i) u can be deﬁned at x

so as to be harmonic on Ω,or

(ii) there is a c>0 such that u = cu

+ v where v is harmonic on Ω.

Proof: Consider a closed ball B

−

= B

−

,δ

⊂ Ω. Deﬁne

w(x)=



+∞ if x = x

u(x)+u

(x)ifx ∈ Ω ∼{x

Clearly, w is superharmonic on Ω. By Theorem 2.5.2, there is a sequence of

superharmonic functions {w

} in C

∞

(Ω) such that w

↑ w on Ω,withthe

additional property that outside of a small neighborhood of x

, w

= w and

is harmonic for suﬃciently large j. The rest of the proof will be carried out

for the n ≥ 3case,then = 2 case only requiring that σ

(n − 2) be replaced

by 2π. By Theorem 1.5.3, for each x ∈ B

(x)=−

(n − 2)



∂B

(z)D

(x, z) dσ(z)

−

(n − 2)



(x, z)Δw

(z) dz.

It can be seen from Equation (1.9) that −D

(x, z) ≥ 0onB.Bythe

Lebesgue monotone convergence theorem, for each x ∈ B

w(x)=−

(n − 2)



∂B

w(z)D

(x, z) dσ(z)

−

(n − 2)

lim

j→∞



(x, z)Δw

(z) dz. (3.2)

For each j ≥ 1andBorelsetM ⊂ B

−

, deﬁne

(M)=−

(n − 2)



Δw

(z) dz.

By Lemma 2.3.4, Δw

≤ 0onΩ and ω

is a measure on the Borel subsets

of B

−

.Moreover,ifM ⊂ B

−

∼ B

,ρ

for some ρ>0, then ω

(M)=0for

suﬃciently large j and lim

j→∞

(M) = 0. Now consider any y ∈ B,y = x

and a neighborhood Σ of x

whose closure Σ

−

does not contain y and Σ

−

⊂

B.Sincew(y) < +∞, it follows from Equation (3.2) that

110 3 Green Functions

lim

j→∞



(y,z) dω

(z) < +∞.

Since ω

−

∼ Σ

−

) = 0 for suﬃciently large j,

lim

j→+∞



−

(y,z) dω

(z) < +∞.

By Lemma 3.2.1, G

(y,·) > 0onΣ

−

. By the continuity of G

(y,·)onΣ

−

there is a constant m>0 such that G

(y,z) ≥ m for z ∈ Σ

−

. Therefore,



−

(y,z) dω

(z) ≥ mω

(Σ

−

and it follows that sup

j≥1

(Σ

−

) < +∞.Sinceω

 = ω

−

)=ω

(Σ

−

∼ Σ

−

)=ω

(Σ

−

), for suﬃciently large j there is a constant k>0

such that ω

≤k for all j ≥ 1. By Lemma 0.2.5, there is a measure ω and

a subsequence {ω

} such that ω

w∗

→ ω. It can be assumed that ω

w∗

→ ω by

replacing the sequence with the subsequence if necessary. Since ω is zero on

the complement of each small neighborhood of x

,ω is a nonnegative mass

concentrated on {x

}.Ifx = x

,thenG

(x, ·) is continuous at x

and

lim

j→∞



(x, z) dω

(z)=ω({x

})G

(x, x

)

and Equation (3.2) becomes

w(x)=PI(w : B)(x)+ω({x

})G

(x, x

It follows from the deﬁnition of w that

u(x)=PI(w : B)(x)+ω({x

})G

(x, x

) − u

for x ∈ B ∼{x

}.SinceG

(x, x

)=G

,x) by Lemma 3.2.1, G

(·,x

)

can be expressed as a sum of u

and a harmonic function. Therefore, for

x ∈ B ∼{x

u(x)=v(x)+(ω({x

}) − 1)u

where v is harmonic on B.Ifω({x

}) − 1=0,thenu can be deﬁned at x

to be harmonic on B by putting u(x

)=v(x

). It is not possible to have

ω({x

}) −1 < 0 for in this case it would be true that u(x) →−∞as x → x

contradicting the fact that u ≥ 0onΩ ∼{x

}.Ifω({x

}) − 1 > 0, then the

assertion is true with c = ω({x

}) − 1.

Historically, Green functions for general regions were constructed ﬁrst as-

suming smooth boundaries. It is possible to go directly to arbitrary regions.

Deﬁnition 3.2.3 A Green Function for an open set Ω ⊂ R

, if it exists, is

an extended real-valued function G

on Ω ×Ω with the following properties:

3.2 Green Functions 111

(i) G

(x, ·)=u

+ h

where for each x ∈ Ω,h

is harmonic on Ω.

(ii) G

≥ 0.

(iii) If for each x ∈ Ω, v

is a nonnegative superharmonic function on Ω and

is the sum of u

and a superharmonic function, then v

≥ G

(x, ·)onΩ.

In other words, for each x ∈ Ω, G

(x, ·) is minimal in the class of functions

described in (iii). The Green function for R

,n ≥ 3, will be denoted simply

by G; it will be seen later that this convention does not apply to the n =2

case since R

does not have a Green function. An open subset Ω of R

having

a Green function will be called a Greenian set.

It will now be shown that the Green function G

for a ball B as deﬁned in

Chapter 1 satisﬁes (i), (ii), and (iii) of this deﬁnition. It is apparent from the

deﬁnition of G

(x, ·) that (i) is satisﬁed. By Lemma 3.2.1, (ii) is satisﬁed.

Suppose now that for each x ∈ B,v

= u

+ w

where v

≥ 0andw

superharmonic on B.Moreover,letG

(x, ·)=u

+ h

where h

is harmonic

on B.Sincebothu

and h

have limits at points z

∈ ∂B, by Lemma 3.2.1

lim inf

z→z

,z∈B

(z) − h

(z))

= lim inf

z→z

,z∈B

(z)+w

(z) − (u

(z)+h

(z)))

= lim inf

z→z

,z∈B

(z)+w

(z)) − lim

z→z

,z∈B

(x, z)

= lim inf

z→z

,z∈B

(z)+w

(z))

= lim inf

z→z

.z∈B

(z) ≥ 0.

Since w

−h

is superharmonic on B, w

−h

≥ 0onB by Corollary 2.3.6;

that is, v

−G

(x, ·)=(u

)−(u

)=w

−h

≥ 0andG

(x, ·) ≤ v

on B.Thus,G

(x, ·) satisﬁes (iii) and can be justiﬁably called the Green

function for the ball B.

Theorem 3.2.4 The Green function for the open set Ω is unique if it exists.

Proof: Let g

be a second Green function for Ω according to the above

deﬁnition. By (iii) of the above deﬁnition, G

(x, ·) ≥ g

(x, ·)andg

(x, ·) ≥

(x, ·)foreachx ∈ Ω.Thus,G

(x, ·)=g

(x, ·)foreachx ∈ Ω.

Lemma 3.2.5 If the open set Ω has a Green function, then inf

y∈Ω

(x, y)=

0 for each x∈Ω.

Proof: Suppose c =inf

y∈Ω

(x, y) > 0forsomex ∈ Ω.ThenG

(x, ·)−c ≥

0onΩ.Moreover,G

(x, ·) − c is nonnegative superharmonic, and repre-

sentable as the sum of u

and a harmonic function. Since G

(x, ·) − c<

(x, ·), the minimality of G

(x, ·) is contradicted.

Lemma 3.2.6 If the open set Ω has a Green function and B is a ball with

x ∈ B ⊂ B

−

⊂ Ω,thenG

(x, ·) is bounded outside B.

112 3 Green Functions

Proof: G

(x, ·) is continuous and nonnegative on Ω ∼ B.Itmustbestrictly

positive on the component of Ω containing x,forifG

(x, y)=0forsome

point y in this component, then G

(x, ·) would be identically zero thereon

by the minimum principle, Corollary 2.3.6, in contradiction to the fact that

(x, x)=+∞.SinceB

−

⊂ Ω, B

−

is a subset of the component containing

x and G

(x, ·) is strictly positive on B

−

.Letm =sup

y∈∂B

(x, y) > 0.

Deﬁne

g(x, y)=



(x, y),y∈ B,

min (G

(x, y), 2m),y∈ Ω ∼ B.

Then g(x, ·) is superharmonic on B since it agrees with G

(x, ·)onB.Since

g(x, ·) is the minimum of two superharmonic functions on Ω ∼ B

−

,itis

superharmonic on Ω ∼ B

−

.SinceG

(x, ·) ≤ m<2m on ∂B and G

(x, ·)

is continuous at points of ∂B, g(x, ·) agrees with G

(x, ·) on a neighborhood

of ∂B and is superharmonic thereon. It follows that g(x, ·) is superharmonic

on Ω.Sinceg(x, ·) − u

= G

(x, ·) − u

on B ∼{x} and G

(x, ·) − u

harmonic on B, g(x, ·) − u

can be deﬁned at x to be superharmonic on Ω.

Denoting the extension by [g(x, ·) − u

0 ≤ g(x, ·)=u

+[g(x, ·) − u

where the bracket function is superharmonic on Ω and G

(x, ·) ≤ g(x, ·)

by the deﬁnition of the Green function. But since g(x, ·) ≤ 2m on Ω ∼

B,G

(x, ·) ≤ 2m on Ω ∼ B.

The existence of a Green function for an open set Ω can be established

by showing that Ω supports suﬃciently many nonnegative superharmonic

functions.

Deﬁnition 3.2.7 If Ω is an open subset of R

and x ∈ Ω,let

= {v

; v

≥ 0,v

= u

+ w

∈S(Ω)}.

The class B

will generally depend upon Ω and x.ItmaywellbethatB

= ∅

for some x ∈ Ω.

Lemma 3.2.8 B

is a saturated family of superharmonic functions on Ω ∼

{x}.

Proof: Since there is nothing to prove if B

= ∅, it can be assumed that

= ∅.Ifv

∈B

,thenv

≥ 0andv

= u

+ w

,i =1, 2, where w

and w

are superharmonic on Ω.Sincemin(v

)=u

+min(w

)and

min (w

) is superharmonic on Ω,min(v

) is nonnegative and super-

harmonic on Ω; that is, the minimum of two elements of B

belongs to B

Suppose now that v = u

+ w ∈B

where w is superharmonic on Ω.LetB

be any ball with B

−

⊂ Ω ∼{x}. By Lemma 2.4.11,



PI(v : B)onB

v on Ω ∼ B

3.2 Green Functions 113

is superharmonic on Ω.Sinceu

is harmonic on a neighborhood of B

−

, PI(v :

B)=PI(u

+w : B)=u

+PI(w : B). Thus, v

= u

is superharmonic

on Ω and v

∈B

. This completes the proof that B

is saturated on Ω ∼{x}.

Lemma 3.2.9 If B

= ∅ for each x ∈ Ω,thenΩ has a Green function

(x, ·)=inf{v

; v

∈B

Proof: Consider a ﬁxed x ∈ Ω and let w

=inf{v

; v

∈B

}.SinceB

a saturated family of superharmonic functions on Ω ∼{x} and w

≥ 0, w

is harmonic on Ω ∼{x} by Theorem 2.6.2. According to Bˆocher’s theorem,

Theorem 3.2.2, there is a constant c ≥ 0 such that w

= cu

on Ω ∼{x},

where h

is harmonic on Ω. It will be shown now that c = 1. Consider a ball

B = B

x,δ

with B

−

⊂ Ω and the Green function G

(x, ·). It follows from (iii)

of Deﬁnition 3.2.3 and the persuant discussion that v

≥ G

(x, ·)onB for

all v

∈B

, and therefore that w

≥ G

(x, ·)onB and that c =0.Since

+ h

= w

≥ G

(x, ·)=u

+ h

∗

on B where h

∗

is harmonic on B,

∗

− h

) ≤ (c − 1)u

on B ∼{x}.

If c<1, then the harmonic function h

∗

− h

would have a limit −∞ at x,

a contradiction. Therefore, c ≥ 1andw

/c = u

+ H

on Ω where H

harmonic on Ω. Therefore, w

/c ∈B

and w

≤ w

/c ≤ w

so that c =1.

Therefore, G

(x, ·)=w

(·) is the Green function for Ω.

Theorem 3.2.10 Let Ω be an open subset of R

.If(i)n =2and Ω is

bounded, (ii) n =2and Ω is a subset of a set having a Green function, or

(iii) n ≥ 3,thenΩ has a Green function.

Proof:

(i) If n =2andΩ is bounded, then for each x ∈ Ω there is a constant c

such that u

+ c

∈B

(ii) If n =2andΩ ⊂ Σ,andΣ has a Green function G

,thenG

(x, ·)|

∈

for all x ∈ Ω.

(iii) If n ≥ 3,u

∈B

for all x ∈ Ω.

Theorem 3.2.11 G(x, y)=u

(y) is the Green function for R

,n≥ 3.

Proof: By the preceding theorem, the Green function G for R

,n ≥ 3,

exists. Fix x ∈ R

.ThenG(x, ·)=u

+ h

where h

is harmonic on

.Sinceu

∈B

+ h

≤ u

and h

≤ 0. By Picard’s theorem,

Theorem 1.5.7, h

= c with c ≤ 0. Suppose c<0. Since u

(y) → 0as

|y|→+∞,G(x, y) < 0forlarge|y|. But this contradicts the fact that

inf

y∈R

G(x, y) = 0 by Lemma 3.2.5. Therefore, c =0andh

=0.

Theorem 3.2.12 If Ω

,Ω

are Greenian subsets of R

with Ω

⊂ Ω

,then

(x, ·) ≤ G

(x, ·) for each x ∈ Ω

114 3 Green Functions

Proof: Let B

(Ω

)betheB

class relative to Ω

.Thus,G

(x, ·)|

∈

(Ω

)andG

(x, ·) ≤ G

(x, ·)foreachx ∈ Ω

Theorem 3.2.13 If the open set Ω is not everywhere dense in R

,thenit

is Greenian.

Proof: By hypothesis, there is a point x ∈ R

and a closed ball B

−

x,δ

⊂∼ Ω.

Suppose, for example, that Ω ∩B

−

0,1

= ∅.ThepointsofR

will be interpreted

as complex numbers in what follows. For any x, y ∈ Ω, x

−1

∈ B

0,1

and



−



< 2.

Then

log



−



=log

2|xy|

|x − y|

> 0.

Consider

(y)=log



−



=log2|x| +log|y|−log |x − y|.

It has just been shown that v

≥ 0onΩ for each x ∈ Ω.Forﬁxedx ∈ Ω,the

ﬁrst term on the right is a constant on Ω, the second term is the fundamental

harmonic function with pole at 0 and is harmonic on Ω, and the third term

is the fundamental harmonic function with pole at x. Hence

(y)=h

(y)+log

|x − y|

on Ω

where h

is harmonic on Ω. It follows that B

= ∅ for each x ∈ Ω and the

Green function for Ω exists by Theorem 3.2.9

Theorem 3.2.14 R

does not have a Green function.

Proof: Assume that R

does have a Green function

G(x, y)=−log |x − y|+ h

(y)

where h

is harmonic on R

for each x ∈ R

. Consider a ﬁxed x.Since

−log |x − y|→−∞as |y|→+∞,h

(y) → +∞ as |y|→+∞.Ifm is any

positive integer, then there is a ball with its center at 0 such that h

≥ m

on the boundary of the ball. By the minimum principle, Corollary 1.5.10,

≥ m on the ball. Therefore, h

(0) = +∞, contradicting the fact that a

harmonic function is real-valued. Thus, R

does not have a Green function.

3.3 Symmetry of the Green Function 115

Theorem 3.2.15 If Λ ⊂ R

is a nonempty closed set consisting only of

isolated points, then Ω = R

∼ Λ does not have a Green function.

Proof: Consider any x ∈ Ω.SinceΩ is open, there is a neighborhood U ⊂ Ω

of x outside of which G

(x, ·) is bounded by Lemma 3.2.6. Consider any

y ∈ Ω. There is a neighborhood V of y that does not contain x or any of the

other points of Λ such that G

(x, ·) is bounded and harmonic on V ∼{y}.By

Theorem 3.2.2, G

(x, ·) can be deﬁned at y so as to be harmonic on V .Since

this is true of each y ∈ Ω, G

(x, ·) can be extended so that it is harmonic

on Ω ∼{x}.WritingG

(x, ·)=u

+ h

,whereh

is harmonic on Ω,this

means that h

has a harmonic extension to all of R

.TheclassB

)

relative to R

is then nonempty. The same can be seen to be true of B

)

for any x

∈ R

by means of a translation taking x to x

, since harmonic

functions map into harmonic functions and u

goes into u

under such a

translation. It follows that R

has a Green function, contrary to the preceding

theorem.

3.3 Symmetry of the Green Function

The principal result of this section is that the Green function is a sym-

metric function of its arguments. This result is achieved by introducing

the concept of harmonic minorant and a smoothing operation known

as balayage or sweeping out, the latter having its origin in the work of

Poincar´e[50].

Let Ω be an open subset of R

,letΛ be an open subset of Ω,andletu be

a superharmonic function on Ω.Let{B

} be a sequence of balls such that

(i) B

−

⊂ Λ for all j ≥ 1,

(ii) Λ = ∪

∞

j=1

,and

(iii) for each j ≥ 1, B

occurs inﬁnitely often in the sequence.

Deﬁne

= u



PI(u : B

)onB

u on Ω ∼ B

and, inductively, for j ≥ 2

=(u

j−1

)



PI(u

j−1

: B

)onB

j−1

on Ω ∼ B

By Theorem 2.4.3, u

≤ u

j−1

on Ω, u

is harmonic on B

,andu

is su-

perharmonic on Ω. Since the sequence {u

} is monotone decreasing, u

∞,Λ

lim

j→∞

is deﬁned on Ω.

116 3 Green Functions

Deﬁnition 3.3.1 u

∞,Λ

is called the reduction of u over Λ relative to Ω.

The second subscript Λ will be omitted when Λ = Ω. It will be shown shortly

that u

∞,Λ

is independent of the sequence {B

} whenever u is bounded below

by a harmonic function.

Theorem 3.3.2 If u is superharmonic on the open set Ω and Λ is an open

subset of Ω,thenu

∞,Λ

is either harmonic or identically −∞ on each com-

ponent of Λ and u

∞,Λ

= u on Ω ∼ Λ.

Proof: By considering components, it can be assumed that Λ is connected.

Let {B

} be the sequence of balls and {u

} the corresponding sequence of

superharmonic functions deﬁning u

∞,Λ

. Consider a ﬁxed j ≥ 1. Since B

occurs inﬁnitely often in the sequence of balls, there is a subsequence {B

}

such that B

= B

for all k ≥ 1. On B

∞,Λ

= lim

k→∞

.Sinceu

harmonic on B

= B

∞,Λ

is either harmonic or identically −∞ on B

Theorem 2.2.5. It is easy to see using a connectedness argument that u

∞,Λ

is either harmonic or identically −∞ on Λ.Sinceu

= u on Ω ∼ Λ for each

j ≥ 1, u

∞,Λ

= u on Ω ∼ Λ.

Even if u is not identically −∞ on any component of Λ, it cannot be

concluded that u

∞,Λ

is superharmonic on Ω, since the limit of a decreasing

sequence of l.s.c. functions need not be l.s.c.

Lemma 3.3.3 If u is superharmonic on the open set Ω,Λ is an open subset

of Ω, v is subharmonic on Λ,andv ≤ u on Λ,thenv ≤ u

∞,Λ

≤ u on Λ.

Proof: Let {B

} be the sequence of balls deﬁning u

∞,Λ

.OnB

,v ≤ PI(v :

) ≤ PI(u : B

)=u

.Moreover,v ≤ u = u

on Λ ∼ B

.Thus,v ≤ u

Λ and, by induction, v ≤ u

on Λ for all j ≥ 1.

Deﬁnition 3.3.4 If u is superharmonic on the open set Ω, h is harmonic

on Ω,andh ≤ u on Ω,thenh is called a harmonic minorant of u.The

function h is the greatest harmonic minorant of u, denoted by ghm

h,if

h is a harmonic minorant and v ≤ h whenever v is a harmonic minorant of u.

Theorem 3.3.5 If u is superharmonic on the open set Ω and u has a har-

monic minorant on Ω,thenu has a unique greatest harmonic minorant,

namely u

∞

Proof: Let h be a harmonic minorant of u. By the preceding lemma, h ≤

∞

≤ u on Ω.Sinceh is harmonic on Ω, u

∞

cannot take on the value −∞

on Ω and is therefore harmonic on Ω by Theorem 3.3.2. Since h can be any

harmonic minorant of u, it follows that u

∞

is the greatest harmonic minorant

of u. Uniqueness is obvious.

In deﬁning u

∞,Λ

, it was not shown that u

∞,Λ

is independent of the se-

quence of balls {B

}.Ifu has a harmonic minorant on Λ,thenu

∞,Λ

is inde-

pendent of the sequence used, since it is the greatest harmonic minorant of u

3.3 Symmetry of the Green Function 117

over Λ, which is unique. Since u

∞,Λ

will be considered only when a harmonic

minorant exists, further discussion is not warranted.

Lemma 3.3.6 If u

are superharmonic on the open set Ω and have har-

monic minorants on Ω, then the greatest harmonic minorant of u

exists

and is the sum of the greatest harmonic minorants of u

and u

Proof: Let {B

} be a sequence of balls and let {u

} be the corresponding

sequence of superharmonic functions deﬁning u

∞

. Since the Poisson integral

is linear in the boundary function, (u

+ u

)

=(u

)

+(u

)

. Therefore,

+ u

)

∞

=(u

)

∞

+(u

)

∞

and the lemma follows from Theorem 3.3.5.

Theorem 3.3.7 If the open set Ω is Greenian, then the greatest harmonic

minorant of G

(x, ·),x∈ Ω, is the zero function.

Proof: It is known that G

(x, ·) ≥ 0onΩ and that G

(x, ·) is superharmonic

on Ω. By Theorem 3.3.5, G

(x, ·) has a greatest harmonic minorant, since

the zero function is a harmonic function. Let h

be a harmonic minorant of

(x, ·). Then G

(x, ·) − h

≥ 0 and it can be written as the sum of u

and a harmonic function; that is, G

(x, ·) − h

∈B

. Therefore, G

(x, ·) ≤

(x, ·)−h

by the minimality of G

(x, ·). This shows that h

≤ 0andthat

the zero function is the greatest harmonic minorant of G

(x, ·).

The following theorem is a classic version of the domination principle

(c.f.Theorem 4.4.8).

Theorem 3.3.8 Let Ω be a Greenian subset of R

.Ifx ∈ Ω, Λ is a neigh-

borhood of x with compact closure in Ω,andu is a positive superharmonic

function deﬁned on a neighborhood of Ω ∼ Λ with u ≥ G

(x, ·) on a neigh-

borhood of ∂Λ,thenu ≥ G

(x, ·) on Ω ∼ Λ.

Proof: Letting

w =



(x, ·)onΛ

min (G

(x, ·),u)onΩ ∼ Λ,

w is superharmonic on Ω.IfG

(x, ·) − w is deﬁned to be 0 at x,then

it is a subharmonic minorant of G

(x, ·)onΩ. By Lemma 3.3.3 and the

fact that the greatest harmonic minorant of G

(x, ·) is the zero function,

(x, ·) − w ≤ 0sothatu ≥ G

(x, ·)onΩ ∼ Λ.

The method used to deﬁne u

∞,Λ

will be used to prove that the Green

function is a symmetric function of its arguments. The proof will require a

modiﬁcation of the sequence of balls {B

} so that



∞

j=1

∂B

is closed relative

to Ω. Such a sequence can be constructed as follows. Let {Ω

} be a sequence

of open sets with compact closures such that Ω

↑ Ω and Ω

−

⊂ Ω, k ≥ 1. It

is easily seen using compactness arguments that there is a sequence of balls

} such that only a ﬁnite number of balls intersect any Ω

−

. If the sequence

is modiﬁed so that each ball occurs inﬁnitely often in the sequence, then this

will still be true. Suppose now that {x

} is a sequence in



∞

j=1

∂B

with