Helms L.L. Potential Theory

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118 3 Green Functions

→ x ∈ Ω.SinceΩ

↑ Ω, x ∈ Ω

for some j

. Since only a ﬁnite number

of balls intersect Ω

−

,thereissomek

such that x

∈



j=1

∂B

for large k,

and it follows that x ∈ ∂B

for some j with 1 ≤ j ≤ k

.Thus,x ∈



∞

j=1

∂B

and



∞

j=1

∂B

is closed relative to Ω. It therefore can be assumed that there

is a sequence of balls satisfying

(i) Ω =



∞

j=1

(ii) B

−

⊂ Ω for all j ≥ 1,

(iii)



∞

j=1

∂B

is closed in Ω,and

(iv) each B

occurs inﬁnitely often in the sequence.

Theorem 3.3.9 If the open set Ω has a Green function, then G(x, y)=

G(y,x) for all x, y ∈ Ω;inparticular,G(·,y) is harmonic on Ω ∼{y} for

each y ∈ Ω.

Proof: Let {B

} be a sequence of balls satisfying the above four properties.

For each x ∈ Ω, G(x, ·)=u

+ h

where h

is harmonic on Ω. To simplify

the notation, u

(y)andh

(y) will be written u(x, y)andh(x, y), respectively.

For each x ∈ Ω,

(x, ·)=u

(x, ·)+h(x, ·)

where G

(x, ·)andu

(x, ·)denotethejth result of applying the sweeping

out method to functions G(x, ·)andu(x, ·), respectively. The second term on

the right does not depend upon j because harmonic functions are invariant

under the sweeping out process. As j →∞, G

(x, ·) approaches the greatest

harmonic minorant of G(x, ·), which is the zero function by Theorem 3.3.7;

likewise, u

(x, ·) → u

∞

(x, ·). Therefore,

0=u

∞

(x, ·)+h(x, ·)

and it follows that

G(x, ·)=u(x, ·) − u

∞

(x, ·). (3.3)

It will be shown next that u

∞

(·,y)isharmoniconΩ for each y ∈ Ω by

reexamining the sweeping out process deﬁning u

∞

(·,y). Consider any ﬁxed

y ∈ Ω.Ify ∈ B

,thenu

(x, y)=u(x, y) for all x ∈ Ω and u

(·,y)isharmonic

on Ω ∼{y}.Ify ∈ B

,thenu

(·,y)isharmoniconΩ ∼ B

−

since it agrees

with u(·,y)onΩ ∼ B

−

and is harmonic on B

since u

(·,y)=PI(u(·,y):B

)

on B

. Combining the two cases, u

(·,y)isharmoniconΩ ∼ ({y}∪∂B

Now consider u

(·,y). Since u

(·,y) agrees with u

(·,y)onΩ ∼ B

,itis

harmonic on (Ω ∼ ({y}∪∂B

)) ∼ B

−

.Sinceu

(·,y)isharmoniconB

and

therefore on (Ω ∼ ({y}∪∂B

)) ∩ B

, it follows that u

(·,y)isharmonicon

Ω ∼ ({y}∪∂B

∪ ∂B

). By an induction argument, u

(·,y)isharmonicon

Ω ∼ ({y}∪



i=1

∂B

). Since each ball occurs inﬁnitely often in the sequence

}, there is a subsequence {B

} such that y ∈ B

,k ≥ 1. Since u

(·,y)

is harmonic on Ω ∼



i=1

∂B

⊃ Ω ∼



∞

i=1

∂B

and the sequence {u

(·,y)}

decreases to u

∞

(·,y) on each component of the open set Ω ∼



∞

i=1

∂B

and

3.3 Symmetry of the Green Function 119

each u

(·,y) ≥ 0,u

∞

is harmonic on each component by Lemma 2.2.5 and

therefore is harmonic on Ω ∼



∞

i=1

∂B

.Sinceu

∞

(·,y) is independent of the

sequence of balls {B

}, u

∞

(·,y)isharmoniconΩ for each y ∈ Ω.Since

(x, y)=u

(y) − u

∞

(x, y)=u

(x) − u

∞

(x, y), for each y ∈ Ω

(i) u(·,y) − u

∞

(·,y) ≥ 0onΩ,and

(ii) u(·,y) − u

∞

(·,y)isthesumofu

and a harmonic function.

By the minimality of the Green function G

(y,x) ≤ u(x, y) − u

∞

(x, y)=

(x, y). Interchanging x and y, G

(x, y)=G

(y,x).

Corollary 3.3.10 Let Ω be a Greenian set, and let Ω

be a component of Ω.

If x ∈ Ω

,thenG

(x, y)=0for y ∈ Ω ∼ Ω

and G

(x, ·) > 0 on Ω

Moreover, the Green function for Ω

is given by G

= G

×Ω

Proof: Consideraﬁxedx ∈ Ω

.SinceG

(x, ·)=u

+ h

≥ 0, u

has a

harmonic minorant on Ω,namely,−h

. It therefore has a greatest harmonic

minorant h

on Ω

. Letting

∗



on Ω

on Ω ∼ Ω

∗

is harmonic on Ω and u

≥ h

∗

on Ω. By Equation (3.3), −h

is the

greatest harmonic minorant of u

and therefore −h

≥ h

∗

.Thus,G

(x, ·)=

+ h

≤ u

−h

∗

=0onΩ ∼ Ω

. Since it is always true that G

(x, ·) ≥ 0,

=0onΩ ∼ Ω

.IfG

(x, y)=0forsomepointy ∈ Ω

,thenG

(x, ·)=0

on Ω

by the minimum principle, contradicting the fact that G

(x, x)=+∞.

As to the second assertion, let {B

} be a subsequence of the sequence {B

}

of balls deﬁning the reduction of u

(·)=u(x, ·)overΩ such that B

−

⊂ Ω

Using such a sequence, it is easy to see that

∞,Ω

(x, ·)=u

∞

(x, ·)onΩ

By Equation (3.3), for x ∈ Ω

(x, ·)=u(x, ·) − u

∞,Ω

(x, ·)=u(x, ·) − u

∞

(x, ·)=G

(x, ·)

on Ω

. Hence, G

= G

×Ω

Theorem 3.3.11 If {Ω

} is an increasing sequence of Greenian sets with

Ω =



, then either

(i) Ω is Greenian and lim

j→∞

= G

on Ω × Ω,or

(ii) Ω is not Greenian and lim

j→∞

=+∞ on Ω × Ω.

Proof: Since G

(x, ·)andG

(x, ·), if deﬁned, vanish on components of Ω

not containing x, it can be assumed that Ω is connected. For j ≥ 1and

x ∈ Ω

, by Equation (3.3) G

(x, ·)=u

− h

(j)

where h

(j)

is the greatest

120 3 Green Functions

harmonic minorant of u

on Ω

.SincetheG

(x, ·) are eventually deﬁned and

increase at each point of Ω by Theorem 3.2.12, the h

(j)

are eventually deﬁned

and decrease at each point of Ω,andh

= lim

j→∞

(j)

is deﬁned on Ω.By

Lemma 2.2.7, h

is either identically −∞ or harmonic on Ω.Ifh

is harmonic

on Ω for some x,thenh

is harmonic on Ω for all y ∈ Ω since h

(j)

(x)=h

(j)

(y)

by the symmetry of the Green function, and so h

(x)=h

(y); in this case,

= u

− h

∈ B

for each x ∈ Ω and the Green function G

exists by

Lemma 3.2.9. If h

∗

is any harmonic minorant of u

on Ω,thenh

∗

≤ h

(j)

, and therefore h

∗

≤ h

on Ω;thatis,h

is the greatest harmonic minorant

of u

on Ω, and therefore G

(x, ·)=u

− h

= lim

j→∞

(x, ·)onΩ.On

the other hand, if h

is identically −∞ for some x ∈ Ω,thenh

is identically

−∞ for all x ∈ Ω;ifΩ were Greenian, the fact that G

(x, ·) ≤ G

(x, ·)

would imply that G

(x, ·)=+∞ on Ω, a contradiction.

Theorem 3.3.12 If the open set Ω is Greenian, then G

is continuous on

Ω × Ω, in the extended sense at points of the diagonal of Ω × Ω. Moreover,

if B

−

x,δ

⊂ Ω,thenL(G

(z,·):x, δ) is a continuous function of z ∈ Ω.

Proof: G

is obviously continuous in the extended sense at points of the

diagonal of Ω ×Ω.If(x

) ∈ Ω ×Ω, x

= y

,letB

and B

be balls with

centers x

and y

, respectively, with disjoint closures in Ω.Ifx

and y

are

in diﬀerent components of Ω,thenG

=0onB

×B

and G

is trivially

continuous at (x

). It therefore can be assumed that x

and y

are in the

same component of Ω. By Lemma 3.2.6, there is a constant m>0 such that

, ·) ≤ m on B

.SinceG

(·,y) is positive and harmonic on B

for

each y ∈ B

, there is a constant k>0 such that G

(x, y)/G

,y) ≤

k for all x ∈ B

and y ∈ B

by Theorem 2.2.2. Therefore, G

(x, y)=

(x, y)/G

,y))G

,y) ≤ km for all x ∈ B

and y ∈ B

.This

shows that the family {G

(x, ·); x ∈ B

} is uniformly bounded on B

Since G

(x, y)=u

(y)+h

(y)andu

(y) is bounded on B

× B

,the

family of harmonic functions {h

; x ∈ B

} is uniformly bounded on B

and

equicontinuous on any compact subset of B

according to Theorem 2.2.4.

Since h

(y) is a symmetric function,

(x, y) − G

)|≤|u

(y) − u

)| + |h

(y) − h

≤|u

(y) − u

)| + |h

(y) − h

+ |h

(x) − h

)|.

The ﬁrst term on the right can be made arbitrarily small by the joint continu-

ity of u

(y)at(x

), the second term can be made arbitrarily small by the

equicontinuity of the family {h

; x ∈ B

}, and the last term can be made ar-

bitrarily small by the continuity of h

at x

.Thus,G

(x, y) is continuous at

). Now let B

−

x,δ

⊂ Ω,andletν

x,δ

be a unit mass uniformly distributed

on ∂B

x,δ

.Consideranyz

∈ Ω ∼ ∂B

x,δ

.Ifz is not in the component of Ω con-

taining B

x,δ

,thenG

(z,·)=0on∂B

x,δ

for all z near z

and L(G

(z,·):x, δ)

3.3 Symmetry of the Green Function 121

is obviously continuous at z

. Suppose now that z

and B

x,δ

belong to the

same component Ω

of Ω.Chooseρ>0 such that B

−

,ρ

∩ ∂B

x,δ

= ∅.By

the preceding corollary, the functions G

(z,·),z ∈ B

,ρ

, are harmonic and

positive on Ω

∼ B

−

,ρ

. By Harnack’s inequality, Theorem 2.2.2, there is a

constant K such that G

(z,y) ≤ KG

,y),y ∈ ∂B

x,δ

and z ∈ B

,ρ

.Since

, ·) is bounded on ∂B

x,δ

, the functions G

(z,·) are dominated on ∂B

x,δ

by the integrable function G

, ·). By the Lebesgue dominated convergence

theorem,

L(G

(z,·):x, δ)=



(z,y) dν

x,δ

is continuous at z

. Finally, suppose that z

∈ ∂B

x,δ

. Since the func-

tions L(G

(z,·):x, δ + r),r > 0 are continuous at z

and increase to

L(G

(z,·):x, δ)asr ↓ 0, L(G

(z,·):x, δ) is l.s.c. at z

; likewise,

the functions L(G

(z,·):x, δ − r) are continuous at z

and decrease to

L(G

(z·):x, δ)asr ↓ 0, so that L(G

(z·):x, δ) is u.s.c. at z

.Thus,

L(G

(z,·):x, δ) is continuous at z

According to Theorem 3.2.10, every open subset Ω of R

,n ≥ 3, has a

Green function. The following theorem will help to narrow the class of open

subsets of R

without Green functions.

Theorem 3.3.13 If Ω ⊂ R

is a nonconnected open set, then Ω is Greenian.

Proof: Let {Ω

} be the components of Ω, of which there are at least two.

Suppose x ∈ Ω

. By Theorem 3.2.13, Ω

has a Green function; that is, there

is a nonnegative superharmonic function v

on Ω

of the form v

= u

+ h

where h

is harmonic on Ω

.Notethatu

is harmonic on the remaining

components of Ω.Extendh

to Ω by putting h

= −u

on Ω ∼ Ω

.Then

+ h

∈ B

and B

= ∅. Since this can be done for each component of Ω,

= ∅ for all x ∈ Ω and Ω has a Green function.

Combining this theorem with Theorem 3.2.13, only the connected dense

subsets of R

present a problem in so far as the existence of a Green function

is concerned.

The boundary behavior of the Green function can be related to regular

boundary points for the Dirichlet problem.

Lemma 3.3.14 If Ω is a bounded open set, then G

(x, ·)=u

− H

Proof: Since u

is in the upper class deﬁning H

− H

≥ 0onΩ.

By Lemma 3.2.9, u

− H

≥ G

(x, ·)onΩ. Also from the deﬁnition of

the Green function, G

(x, ·)=u

+ h

,whereh

is harmonic on Ω.Since

(x, ·) ≥ 0,u

≥−h

on Ω and −h

belongs to the lower class deﬁning H

;

that is, −h

≤ H

.Moreover,u

−H

≥ u

+ h

implies that − h

≥ H

This shows that −h

= H

and that G

(x·)=u

− H

Theorem 3.3.15 Let Ω be a bounded open set. Then x ∈ ∂Ω is a regular

boundary point for Ω if and only if lim

y→x,y∈Ω

(z,y)=0for all z ∈ Ω.

122 3 Green Functions

Proof: Suppose x is a regular boundary point for the Dirichlet problem. Con-

sider any z ∈ Ω.Sinceu

is continuous at x, lim

y→x,y∈Ω

(z,y)=u

(x) −

lim

y→x,y∈Ω

= u

(x)−u

(x) = 0. Now suppose that lim

y→x,y∈Ω

(z,y)

=0forallz ∈ Ω.Let{Ω

} be the set of components of Ω and let z

be a

ﬁxed point of Ω

,j ≥ 1. It is easy to see that the function w =



, ·)

is a barrier at x.

3.4 Green Potentials

Physically, a potential function represents the potential energy of a unit mass

(charge) at a point of R

due to a mass (charge) distribution on R

. As such,

potential functions are of interest in their own right.

Deﬁnition 3.4.1 If μ is a signed measure on the Greenian set Ω,let

μ(x)=



(x, y) dμ(y),

if deﬁned for all x ∈ Ω.Ifμ is a measure on Ω and G

μ is superharmonic on

Ω,thenG

μ is called the potential of μ.Ifμ = μ

−μ

−

is a signed measure

and both G

and G

−

are superharmonic functions, then G

μ is called

a Green potential.

Under suitable conditions on μ, G

μ is deﬁned and superharmonic on Ω or

a diﬀerence of two such functions.

Lemma 3.4.2 If μ is a Borel measure on the Greenian set Ω,thenG

μ is

either superharmonic or identically +∞ on each component of Ω.

Proof: Let{ Ω

} be an increasing sequence of open subsets of Ω with compact

closures Ω

−

⊂ Ω such that Ω =



∞

j=1

.Foreachx ∈ Ω, deﬁne m

(x, ·)=

min (G

(x, ·),j)and

(x)=



(x, y) dμ(y).

Since μ is a Borel measure, μ(Ω

) ≤ μ(Ω

−

) < +∞ for j ≥ 1. Since G

continuous in the extended sense on Ω ×Ω, G

,y) → G

(x, y)asx

→ x

and the same is true of m

.Sincem

is bounded by j,



,y) dμ(y) →



(x, y) dμ(y)=u

(x)

as x

→ x by the Lebesgue dominated convergence theorem. Thus, each u

continuous on Ω.SinceΩ

↑ Ω and m

(x, ·) ↑ G

(x, ·)asj →∞,u

↑ G

3.4 Green Potentials 123

as j →∞by the Lebesgue monotone convergence theorem. As the limit of

an increasing sequence of continuous functions, G

μ is l.s.c. on Ω. For any

−

x,δ

⊂ Ω,

L(G

μ : x, δ)=

n−1



∂B

x,δ





(z,y) dμ(y)



dσ(z).

Using Tonelli’s theorem and the fact that G

(·,y)=G

(y,·), is superhar-

monic on Ω,

L(G

μ : x, δ)=



n−1





∂B

x,δ

(z,y) dσ(z)



dμ(y)



L(G

(·,y):x, δ) dμ(y)

≤



(x, y) dμ(y)=G

μ(x).

It follows that G

μ is nonnegative on Ω, l.s.c. on Ω, super-mean-valued on Ω,

and therefore superharmonic on any component on which it is not identically

+∞.

Lemma 3.4.3 If μ is a measure on the Greenian set Ω such that μ(Ω) <

+∞ and μ(B)=0for some closed ball B

−

⊂ Ω,thenG

μ is superharmonic

on the component of Ω containing B.

Proof: If B = B

x,δ

,thenG

(x, ·) is bounded outside B by Lemma 3.2.6.

Since μ(B)=0,

μ(x)=



(x, y) dμ(y)=



Ω∼B

(x, y) dμ(y) < +∞.

Therefore, G

μ is not identically +∞ on the component of Ω containing B

and is superharmonic on that component by the preceding lemma.

Theorem 3.4.4 If μ is a measure on the Greenian set Ω and μ(Ω) < +∞,

then G

μ is a potential.

Proof: Assume ﬁrst that Ω is connected. Consider any x ∈ Ω and a closed

ball B

−

= B

−

x,δ

⊂ Ω.Thenμ = μ|

+ μ|

Ω∼B

where μ|

and μ|

Ω∼B

are measures vanishing on Ω ∼ B and B, respectively. Clearly, G

μ =

μ|

Ω∼B

.Sinceμ|

Ω∼B

(B)=0andμ|

Ω∼B

(Ω) < +∞, G

μ|

Ω∼B

superharmonic on Ω by the preceding lemma. Since B

−

⊂ Ω, there is a closed

ball B

−

⊂ Ω ∼ B

−

such that μ|

) = 0. Since μ|

(Ω) < +∞, G

μ|

superharmonic on Ω by the preceding lemma. It follows from Theorem 2.4.5

that G

μ = G

μ|

+ G

μ|

Ω∼B

is superharmonic on Ω, assuming that Ω

is connected. Suppose now that Ω is not connected, and let Ω =



124 3 Green Functions

where {Ω

} is the countable collection of components of Ω.Letμ

= μ|

Then μ

(Ω

) < +∞. By Corollary 3.3.10, G

= G

×Ω

.SinceG

(x, ·)

vanishes outside of the component of Ω containing x by the same corollary

μ(x)=



(x, y) dμ(y)=



(x, y) dμ(y)=G

(x)

for all x ∈ Ω

. By the ﬁrst part of the proof, G

is superharmonic on Ω

and therefore G

μ is superharmonic on Ω.

Theorem 3.4.5 Let Ω be a Greenian set and let G

μ be the potential of a

measure μ. Then the greatest harmonic minorant of G

μ is the zero function.

Proof: Let u = G

μ and let {B

} be a sequence of balls satisfying the re-

quirements for deﬁning the reduction u

∞

of the superharmonic function u

over Ω. By Theorem 3.3.5, u

∞

is the greatest harmonic minorant of u so that

the assertion requires proving that u

∞

=0onΩ.Let{u

} and {G

Ω,j

(·,y)}

be the sequences of functions deﬁning u

∞

and G

Ω,∞

(·,y), respectively. By

Theorem 3.3.7, the sequence {G

Ω,j

(·,y)} decreases to zero on Ω.Therestof

the proof is based on the following result which will be proved using mathe-

matical induction.

(x)=



Ω,j

(x, y) dμ(y),x∈ Ω, j ≥ 1. (3.4)

The ﬁrst step in the induction proof will be omitted because of its similarity

with the general case. By deﬁnition,

(x)=



PI(u

j−1

: B

)(x)ifx ∈ B

j−1

(x)ifx ∈ Ω ∼ B

with a similar deﬁnition of G

Ω,j

(x, y)forﬁxedy. Assuming that Equa-

tion (3.4) holds for j − 1, two cases will be considered. Suppose ﬁrst that

x ∈ B

.Thenu

(x)=u

j−1

(x),G

Ω,j

(x, y)=G

Ω,j−1

(x, y), and

(x)=u

j−1

(x)=



Ω,j−1

(x, y) dμ(y)=



Ω,j

(x, y) dμ(y)

for x ∈ B

,j ≥ 1. Now suppose that x ∈ B

. By Tonelli’s theorem,

(x)=PI(u

j−1

: B

)(x)

= PI





Ω,j−1

(·,y) dμ(y):B



(x)



PI(G

Ω,j−1

(·,y):B

)(x) dμ(y)



Ω,j

(x, y) dμ(y),x∈ B

,j ≥ 1.

3.4 Green Potentials 125

Thus, Equation (3.4) is true for all j ≥ 1. Since G

μ is superharmonic on Ω,

it is ﬁnite a.e. on Ω.Chooseanypointx

∈ Ω such that G

μ(x

) < +∞.

Then

μ(x



,y) dμ(y) < +∞;

that is, G

, ·) is integrable relative to μ.Since0≤ G

Ω,j

, ·) ≤ G

, ·)

and lim

j→∞

Ω,j

, ·)=0,

∞

) = lim

j→∞

) = lim

j→∞



Ω,j

,y) dμ(y)=0

by the Lebesgue dominated convergence theorem. Since u

∞

) ≥ 0andis

harmonic on Ω, u

∞

=0onΩ by the minimum principle.

Corollary 3.4.6 If Ω is a Greenian set, μ is a measure on Ω,andG

μ is

harmonic on Ω,thenμ is the zero measure.

Proof: Let h = G

μ be harmonic on Ω.Sinceh is a harmonic mi-

norant of itself, h = 0 by the preceding theorem. Thus, for any x ∈

Ω,



(x, y) dμ(y) = 0. Since G

(x, ·) > 0 on the component of Ω con-

taining x, each component of Ω has μ-measure zero.

It is possible for a potential to be harmonic on a proper subset of Ω.

Theorem 3.4.7 If μ is a measure on the Greenian set Ω such that G

μ is

a potential on Ω,thenG

μ is harmonic on any open set of μ-measure zero.

Proof: Let Λ be a nonempty open subset of Ω with μ(Λ)=0.Foranyx ∈ Ω,

μ(x)=



Ω∼Λ

(x, y) dμ(y).

Suppose B

−

x,δ

⊂ Λ. By Tonelli’s theorem,

A(G

μ : x, δ)=A





Ω∼Λ

(·,y) dμ(y):x, δ





Ω∼Λ

A(G

(·,y):x, δ) dμ(y).

For y ∈ Ω ∼ Λ, G

(·,y)isharmoniconΛ.SinceB

−

x,δ

⊂ Λ, A(G

(·,y):

x, δ)=G

(x, y)and

A(G

μ : x, δ)=



Ω∼Λ

A(G

(·,y):x, δ) dμ(y)



Ω∼Λ

(x, y) dμ(y)

= G

μ(x).

126 3 Green Functions

Since G

μ is superharmonic on Ω, it is locally integrable on Ω by

Theorem 2.4.2 and, as in the proof of Theorem 1.7.11, continuous on Ω.

It follows from Theorem 2.4.1 that G

μ is harmonic on Λ.

It is possible to compare the Green potential for a region with that of

another region, especially when one is a subset of the other.

Theorem 3.4.8 Let Ω be a Greenian set, let Λ be an open subset of Ω,and

let μ be a measure on Ω such that μ(Ω ∼ Λ)=0and G

μ is a potential. Then

there is a nonnegative harmonic function h on Λ such that G

μ = G

μ|

on Λ.

Proof: Since Λ ⊂ Ω, G

≤ G

on Λ by Theorem 3.2.12. Thus, for each x ∈ Λ,

μ|

(x)=



(x, y) dμ(y) ≤



(x, y) dμ(y)=G

μ(x).

This proves that G

μ|

is a potential. Since G

μ is superharmonic, it is

ﬁnite-valued a.e. on Ω. Except possibly for x in a subset of Λ of Lebesgue

measure zero,

μ(x) − G

μ|

(x)=



(x, y) − G

(x, y)) dμ(y).

Since G

(x, y)=u

(y)+h

(y)andG

(x, y)=u

(y)+h

∗

(y), where h

and h

∗

are harmonic on Ω and Λ, respectively, G

(x, y) −G

(x, y)=H

(x)

where H

(x) is the symmetric function h

(y) − h

∗

(y)onΛ × Λ.Foreach

y ∈ Λ, H

is harmonic on Λ,andforeachx ∈ Λ, H

(x)isharmoniconΛ.

Suppose B

−

x,δ

⊂ Λ. By Tonelli’s theorem, for x ∈ Λ as described above

+∞ > A(G

μ − G

μ|

: x, δ)=



A(G

(·,y) − G

(·,y)) : x, δ) dμ(y)



A(H

: x, δ) dμ(y)



(x) dμ(y).

The last integral is harmonic on Λ by Theorem 1.7.13. Writing the last equa-

tion as

A(G

μ : x, δ)=A(G

μ|

: x, δ)+



(x) dμ(y)

and noting that the solid ball averages increase to G

μ(x)andG

μ|

(x),

respectively, as δ ↓ 0 by Lemma 2.4.4,

μ(x)=G

μ|

(x)+



(x) dμ(y)

on Λ.

The above theorem can be used to prove a converse to Theorem 3.4.7.

3.4 Green Potentials 127

Theorem 3.4.9 Let Ω be a Greenian set. If Gμ is the potential of a measure

μ and is harmonic on the open set Λ ⊂ Ω,thenμ(Λ)=0.

Proof: Since G

μ = G

μ|

+ G

μ|

Ω∼Λ

and G

μ|

Ω∼Λ

is harmonic on

Λ by Theorem 3.4.7, G

μ|

is harmonic on Λ. By the preceding theorem,

μ|

= G

μ|

+ h where h is harmonic on Λ. This implies that G

μ|

harmonic on Λ, and by Corollary 3.4.6, μ|

(Λ)=0=μ(Λ).

Historically, the Green potential was preceded by the logarithmic potential

in the n = 2 case and the Newtonian potential in the n ≥ 3case.The

logarithmic potential of a measure μ on R

having compact support is

deﬁned by

(x)=−



log |x − y|dμ(y),x∈ R

;

the Newtonian potential U

of a measure on R

,n≥ 3, is deﬁned by

(x)=



|x − y|

n−2

dμ(y),x∈ R

Since the Newtonian potential U

of a measure is just the Green potential

Gμ,whereG is the Green function for R

, the preceding results apply di-

rectly to U

in the n ≥ 3case;forexample,ifμ(R

) < +∞,thenU

superharmonic on R

. A similar argument for the logarithmic potential can-

not be made since R

does not have a Green function. In order to show that

is superharmonic in the n = 2 case when μ has compact support, let

,δ

be any ball and let C be the support of μ.Since−log |x − y| is con-

tinuous in the extended sense on B

−

,δ

× C, there is a constant c such that

−log |x − y| + c ≥ 0forx ∈ B

−

,δ

,y ∈ C.Forx ∈ B

−

,δ

(x)+cμ(C)=



(−log |x − y| + c) dμ(y).

As in the proof of Lemma 3.4.2, U

is l.s.c. on R

. Calculating spherical

averages and using Tonelli’s theorem,

L(U

: x

,δ)+cμ(C)=



L(−log |x − y| + c : x

,δ) dμ(y)

≤



(−log |x

− y| + c) dμ(y)

= U

)+cμ(C).

Therefore, U

) ≥ L(U

: x

,δ) for any ball B

,δ

and U

is super-mean-

valued. Using the compactness of the support of μ, it can be seen that U

cannot take on the value −∞.Ifx

∈ C,then−log |x

− y| is bounded on

C and U

) < +∞. This shows that U

is not identically +∞ on R

and