Helms L.L. Potential Theory

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138 3 Green Functions

according to Lemma 1.6.1

x,δ

(y)=



log

if |y − x|≤δ

log

|x−y|

if |y − x| >δ.

Putting τ

x,δ

= ν

x,δ/2

− ν

x,δ

, G

x,δ

is a continuous potential with

x,δ

(y)

⎧

⎨

⎩

≥ 0onΩ

=0 onΩ ∼ B

x,δ

≡ 0.

(3.6)

To see this, consider u

x,δ/2

− u

x,δ

.Then

(i) for y ∈ B

x,δ/2

(y) − u

x,δ

(y)=log

(δ/2)

− log

=log2> 0;

(ii) for y ∈ B

−

x,δ

∼ B

x,δ/2

(y) − u

x,δ

(y)=log

|x − y|

− log

= −log

|x − y|

≥ 0;

(iii) for y ∈ B

−

x,δ

x,δ/2

(y) − u

x,δ

(y)=log

|x − y|

− log

|x − y|

=0.

Since G

(y,·)=u

+ h

,whereh

is harmonic on Ω,

x,δ

(y)=



(z)+h

(z)) dν

x,δ/2

−



(z)+h

(z)) dν

x,δ

(z)

= u

x,δ/2

(y) − u

x.δ

(y)+



(z) dν

x,δ/2

(z) −



(z) dν

x,δ

(z).

Since the last two integrals are just averages of the harmonic function h

their diﬀerence is equal to h

(x) − h

(x)=0.Thus,

x,δ

(y)=u

x,δ/2

(y) − u

x,δ

(y),y∈ Ω

and G

x,δ

has the properties listed in (3.6). Since it is easily seen that a

translate of G

x,δ

is again a potential of the same kind, the collection K

of potentials {G

x,δ

; B

−

x,δ

⊂ Ω} is a total subset of K

Example 3.5.6 (n ≥ 3) Let Ω be a Greenian set, let B

−

x,δ

⊂ Ω, δ > 0,

and let ν

x,δ/2

and ν

x,δ

be deﬁned as in the preceding example. If τ

x,δ

x,δ/2

− ν

x,δ

, then the G

x,δ

have the same properties as in the preceding

example.

3.5 Riesz Decomposition 139

Theorem 3.5.7 (Cartan [10],[11],[12]) If Ω is a Greenian set, then the

collection K

of Green potentials {G

x,δ

; B

−

x,δ

⊂ Ω} is a total subset of

Proof: There is nothing to prove in view of the two preceding examples and

Theorem 3.5.3.

Theorem 3.5.8 Let μ and ν be measures on the Greenian set Ω such that

μ and G

ν are potentials. If G

μ = G

ν at points where both are

ﬁnite, then μ = ν. More generally, let Λ be an open subset of Ω such that

μ = G

ν + h on the subset of Λ where both are ﬁnite and where h is

harmonic on Λ.Thenμ and ν are identical on the Borel subsets of Λ.

Proof: Let Σ be the set of points in Λ where G

μ and G

ν are both ﬁnite.

For x ∈ Σ,G

μ(x)=G

ν(x)+h(x). Since G

μ = G

μ|

+ G

μ|

Ω∼Λ

and G

μ|

Ω∼Λ

is harmonic on Λ, G

μ(x)=G

μ|

(x)+h



(x),x ∈ Σ,

where h



is harmonic on Λ. Applying the same argument to G

ν, G

ν(x)=

ν|

(x)+h



(x),x ∈ Σ,whereh



is harmonic on Λ. Letting ˜μ = μ|

˜ν = ν|

, G

˜μ(x)=G

˜ν(x)+h

∗

(x),x ∈ Σ,whereh

∗

is harmonic on Λ.

Since Λ ∼ Σ has Lebesgue measure zero,

A(G

˜μ : x, δ)=A(G

˜ν : x, δ)+h

∗

(x)

whenever B

−

x,δ

⊂ Λ. Letting δ ↓ 0 and applying Lemma 2.4.4, G

˜μ(x)=

˜ν(x)+h

∗

(x) for all x ∈ Λ.SinceΛ is an open subset of Ω, it has a Green

function G

. Applying Theorem 3.4.8 to G

˜μ and G

˜ν, there is a harmonic

function h

∗

on Λ such that G

˜μ = G

˜ν + h

∗

on Λ.ForB

−

x,δ

⊂ Λ,

L(G

˜μ : x, δ)=L(G

˜ν : x, δ)+h

∗

(x).

Recalling that G

(x, y)=u

(y)+h

(y), the symmetry of the latter two

functions, and applying Tonelli’s theorem

L(G

˜μ : x, δ)=



L((u

(y)+h

(y)) : x, δ) d˜μ(y)



L((u

(·)+h

(·)) : x, δ) d˜μ(y)



x,δ

(y)+h

(y)) d˜μ(y)



x,δ

(y) d˜μ(y)+



(y) d˜μ(y).

Applying the same argument to G

˜ν,



x,δ

(y) d˜μ(y)+



(y) d˜μ(y)=



x,δ

(y) d˜ν(y)+



(y) d˜ν(y)+h

∗

(x).

140 3 Green Functions

Replacing δ by δ/2 and subtracting,



x,δ

(y) − u

x,δ/2

(y)) d˜μ(y)=



x,δ

(y) − u

x,δ/2

(y)) ˜ν(y).

Since the functions u

x,δ

− u

x,δ/2

form a total subset of K

(Λ), ˜μ =˜ν by

Theorem 3.5.4. It follows that μ and ν agree on the Borel subsets of Λ.

Lemma 3.5.9 Let u be superharmonic on the open set Ω,andletB be a ball

with B

−

⊂ Ω. Then there is a unique measure μ on B such that u = G

μ+h

where h is the greatest harmonic minorant of u on B.

Proof: The proof will be carried out only in the n ≥ 3case,then =2diﬀer-

ing only by certain constants appearing before integrals. Since the conclusion

concerns only the representation of u on B, it can be assumed that Ω is a ball

on which u is bounded below. It also can be assumed that u is harmonic on

Ω ∼ B

−

for the following reason. Let u

∞

be the reduction of u over Ω ∼ B

−

Since u is bounded below on Ω, u

∞

is harmonic on Ω ∼ B

−

and u

∞

= u on

B by Theorem 3.3.2. The function u

∞

need not be superharmonic on Ω but

the lower regularization v of u

∞

is superharmonic on Ω by Theorem 2.4.9

since u

∞

is the limit of a decreasing sequence of superharmonic functions.

The lower regularization v diﬀers from u

∞

only on ∂B. Therefore, v is su-

perharmonic on Ω, v = u on B,andv is harmonic on Ω ∼ B

−

. In so far as

the representation of u over B is concerned, it can be assumed that u has

thesamepropertiesasv;thatis,u is harmonic on Ω ∼ B

−

.LetB

be a

ball such that B

−

⊂ B

−

⊂ Ω and let Λ be a neighborhood of B

−

such

that Λ

−

⊂ B

. By Theorem 2.5.2 there is a sequence {u

} of superharmonic

functions in C

∞

(Ω) such that

(i) u

↑ u as j →∞,

(ii) u

= u on Ω ∼ Λ

−

for all suﬃciently large j,and

(iii) u

is harmonic on Ω ∼ Λ

−

for all suﬃciently large j.

By discarding a ﬁnite number of the u

’s, it can be assumed that (ii) and

(iii) hold for all j. By Theorem 1.5.3, for x ∈ B

(x)=−

(n − 2)



∂B

(z)D

(x, z) dσ(z)

−

(n − 2)



(x, z)Δu

(z) dz.

Since u

= u on ∂B

, the ﬁrst integral does not depend upon j and is, in fact,

PI(u : B

) which will be denoted by h

.IfM is a Borel subset of B

−

,let

(M)=−

(n − 2)



Δu

(z) dz.

3.5 Riesz Decomposition 141

Since u

is superharmonic on Ω, Δu

≤ 0andν

is a measure on the Borel

subsets of B

−

with ν

(∂B

) = 0. The above representation of u

on B

then

can be written u

= G

on B

.Sinceu

↑ u on Ω,w = lim

j→∞

is deﬁned on B

. It will now be shown that w is the potential of a measure

on B

.Considerν

 = μ

). By Green’s identity, Theorem 1.2.2,

)=−

(n − 2)



Δu

(z) dz = −

(n − 2)



∂B

(z) dσ(z);

but since u

= u outside Λ

−

(z)=D

u(z)on∂B

and the latter in-

tegral is independent of j. Therefore, ν

 is independent of j and there is

a subsequence of the sequence {ν

} that converges to a measure ν on B

−

the w

∗

-topology by Theorem 0.2.5. It can be assumed that the subsequence

is the sequence itself. Letting j →∞in the equation

(x)=



(x, z) dν

(z)+h

(z),x∈ B

requires justiﬁcation since the integrand G

(x, z) is not a bounded contin-

uous function on B

−

. Suppose B

−

x,δ

⊂ B

. By Tonelli’s theorem,

L(G

: x, δ)=



L(G

(·,z):x, δ) dν

(z).

By Theorem 3.3.12, L(G

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

and,

in fact, has a continuous extension to B

−

since L(G

(·,z):x, δ)=G

(x, z)

for z ∈ B

∼ B

−

x,δ

.Since

L(u

: x, δ)=



L(G

(·,z):x, δ) dν

(z)+h

(x),

∗

→ ν,andu

↑ u,

L(u : x, δ)=



L(G

(·,z):x, δ) dν(z)+h

(x).

Since u and G

(·,z) are superharmonic functions, the surface averages in-

crease to u(x)andG

(x, z), respectively, as δ ↓ 0andu = G

ν + h

.SinceG

ν = G

ν|

+ G

ν|

∼B

and G

ν|

∼B

is harmonic on

B, u = G

ν|

+ h

on B where h

is harmonic on B. By Theorem 3.4.8,

ν|

diﬀers from G

ν|

by a function that is harmonic on B. Letting

μ = ν|

, it follows that u = G

μ + h where h is harmonic on B.Sincethe

greatest harmonic minorant of G

μ on B is the zero function, the greatest

harmonic minorant of u on B is h by Lemma 3.3.6. Let u = G

˜μ + h

∗

be a

second such representation. Since the greatest harmonic minorant of u on B

142 3 Green Functions

is unique, h = h

∗

and so G

μ = G

˜μ. It follows from the preceding theorem

that μ =˜μ and that the representation of u is unique.

Lemma 3.5.10 Let u be superharmonic on the open set Ω,andletΛ be

an open subset of Ω with compact closure Λ

−

⊂ Ω. Then there is a unique

measure μ on Λ such that u = G

μ + h,whereh is the greatest harmonic

minorant of u on Λ.

Proof: Suppose ﬁrst that Λ is a subset of a ball B with B

−

⊂ Ω.Bythe

preceding lemma, there is a measure ν on B such that u = G

ν+h

,whereh

is the greatest harmonic minorant of u on B.SinceG

ν = G

ν|

B∼Λ

and the latter term is harmonic on Λ, u = Gν|

+ h,whereh is harmonic

on Λ. Letting μ = ν|

, the assertion is true for this particular case. Returning

to the general case, suppose only that Λ

−

is a compact subset of Ω.Then

Λ = ∪

i=1

,whereeachΛ

is an open subset of a ball with closure in Ω.

It was just shown that the assertion is true if p = 1. Since the proof is by

induction on p, it suﬃces to consider the case Λ = Λ

∪ Λ

,whereΛ

and

are open sets with compact closures in Ω for which the assertion is true.

Then there are meausres μ

on Λ

and harmonic functions h

on Λ

,i=1, 2,

such that

u = G

+ h

on Λ

u = G

+ h

on Λ

(3.7)

Let Λ

= Λ

∩Λ

. Comparing potentials, there are harmonic functions h

∗

such that u = G

+ h

∗

,i=1, 2. It follows from Theorem 3.5.8 that

= μ

.Thus,thereisameasureμ on the Borel subsets of Λ = Λ

∪Λ

such that μ

= μ|

,i=1, 2. Since G

μ = G

μ|

Λ∼Λ

and the latter

is harmonic on Λ

, G

μ = G

+ h

∗

on Λ

with h

∗

harmonic on Λ

,i=1, 2.

Comparing G

and G

, there is a harmonic function h

∗

on Λ

such that

μ = G

+ h

∗

on Λ

,i=1, 2. It follows from Equations (3.7) that

u = G

μ + H

on Λ

u = G

μ + H

on Λ

where H

is harmonic on Λ

,i =1, 2. It follows that H

= H

on Λ

∩ Λ

and a harmonic function h canbedeﬁnedonΛ = Λ

∪ Λ

such that h|

,i =1, 2. The above equations then become simply u = G

μ + h.Asin

the preceding proof, h is the greatest harmonic minorant of u on Λ and μ is

unique.

The measure ν of the following theorem is known as the Riesz measure

associated with the superharmonic function u and the region Ω.Theresult

will be referred to as the Riesz Decomposition Theorem to distinguish

it from the Riesz representation theorem concerning integral representations

of linear functionals on spaces of continuous functions.

3.5 Riesz Decomposition 143

Theorem 3.5.11 (F. Riesz [53]) Let Ω be a Greenian set and let u be su-

perharmonic on Ω. Then there is a unique measure ν on Ω such that if Λ is

any open subset of Ω with compact closure Λ

−

⊂ Ω,thenu = G

ν|

+ h

where h

is the greatest harmonic minorant of u on Λ; if, in addition, u ≥ 0

on Ω,thenu = G

ν+h where h is the greatest harmonic minorant of u on Ω.

Proof: Let {Ω

} be an increasing sequence of open sets with compact closures

in Ω such that Ω = ∪

.Letν

and h

be the measure and harmonic

function on Ω

, respectively, of the preceding lemma. Then u = G

+ h

on Ω

. By Theorem 3.5.8, ν

and ν

j+k

must agree on the Borel subsets of

for every k ≥ 1. If Λ is any Borel subset of Ω,then{ν

(Ω

∩ Λ)} is a

nondecreasing sequence and ν(Λ) = lim

j→∞

(Ω

∩ Λ) is deﬁned. If ν

extended to be zero on Ω ∼ Ω

,thenν is the limit of an increasing sequence

of measures and therefore is a measure. In particular, if Λ is a Borel subset

of Ω

,thenν

(Ω

∩Λ)=ν

(Λ)=ν

j+k

(Λ)=ν

j+k

(Ω

j+k

∩Λ) for all k ≥ 1so

that ν

(Λ) = lim

k→∞

j+k

(Ω

j+k

∩ Λ)=ν(Λ); that is, ν

= ν|

.NowletΛ

be any open set with compact closure Λ

−

⊂ Ω and let u = G

+ h

the decompostion of the preceding lemma. Choose j

such that Λ ⊂ Ω

for

all j ≥ j

and consider only such j.Sinceu = G

+ h

on Ω

with h

harmonic on Ω

and G

= G

+ G

∼Λ

with the latter term

harmonic on Λ, u = G

+ h

∗

with h

∗

harmonic on Λ.SinceG

and G

diﬀer by a harmonic function on Λ, u = G

+ h

∗

where h

∗

harmonic on Λ. It follows from the uniqueness of the representation of u on Λ

that ν

= ν

= ν|

. This shows that u = G

ν|

+ h

∗

on Λ. It follows from

this equation that h

∗

is independent of j on Λ, and therefore u = G

+ h

where h

is harmonic on Λ. Suppose u ≥ 0onΩ.Thenu = G

ν|

+ h

where h

is harmonic on Ω

.SinceG

≤ G

j+1

on Ω

×Ω

,forx ∈ Ω

ν|

(x)=



(x, y) dν(y)

≤



j+1

(x, y) dν(y)

≤



j+1

(x, y) dν(y)

= G

j+1

ν|

j+1

(x),

showing that the sequence {G

j+k

ν|

j+k

}

k≥1

increases on Ω

and that the

sequence {h

j+k

}

k≥1

decreases on Ω

.Infact,forx ∈ Ω

J+k

ν|

j+k

(x)=



j+k

(x, y) dν(y)

⏐



G(x, y) dν(y)=G

ν(x)

by Theorem 3.3.11. Since 0 ≤ u = G

ν|

+ h

on Ω

, it follows from

Theorem 3.3.6 that h

≥ 0foreachj ≥ 1. Thus, the sequence {h

j+k

}

k≥1

a decreasing sequence of nonnegative harmonic functions on Ω

which has a

144 3 Green Functions

limit that is harmonic on Ω

by Theorem 2.2.7. Thus, a harmonic function

h = lim

k→∞

j+k

can be deﬁned on Ω

which does not depend upon j.It

follows that u = G

ν + h on Ω. Using Theorem 3.3.6 and the fact that the

greatest harmonic function of G

ν is the zero function, h is the greatest

harmonic minorant of u.

Corollary 3.5.12 A positive superharmonic function on a Greenian set is

the potential of a measure if and only if its greatest harmonic minorant is the

zero function.

Proof: The necessity is just Theorem 3.3.7. The suﬃciency follows from the

preceding theorem.

3.6 Properties of Potentials

In general, little can be said about the continuity properties of potentials

μ without assuming that the measure μ has a density. The discontinuities

of a potential of a measure can be removed by giving up a small part of the

measure. This is the point of the second theorem below, but the ﬁrst theorem

is of interest in its own right.

Theorem 3.6.1 (Evans [19] and Vasilesco [60]) Let Ω be a Greenian

set, let μ be a measure with compact support Γ ⊂ Ω,andletu = G

μ.

If u|

is continuous at x

∈ Γ ,thenu is continuous at x

Proof: Only the n ≥ 3 case will be proved. If u(x

)=+∞, the result is

trivially true by the l.s.c. of u. It therefore can be assumed that u(x

) < ∞.

According to Theorem 3.4.2, the Newtonian potential U

= Gμ of the mea-

sure μ is superharmonic. By Theorem 3.4.10, u and U

diﬀer by a harmonic

function on Ω.Sinceu|

is continuous at x

, the same is true of U

.If

is continuous at x

, then the same will be true of u so that it suﬃces to

show that U

is continuous at x

.AmapfromΩ to Γ is deﬁned as follows.

If x ∈ Ω,chooseapointz

in Γ nearest x.Sincex

∈ Γ ,

− x

|≤|z

− x| + |x − x

|≤2|x − x

for any x ∈ Ω so that z

→ x

as x → x

.Fory ∈ Γ , |x − z

|≤|x − y| and

− y|≤|z

− x| + |x − y|≤2|x − y|.

Therefore,

|x − y|

n−2

≤ 2

n−2

− y|

n−2

3.6 Properties of Potentials 145

If ν is any measure with support in Γ,byintegrating

(x)=



|x − y|

n−2

dν(y) ≤ 2

n−2



− y|

n−2

dν(y)=2

n−2

Since U

) < +∞,given>0thereisaballB ⊂ Ω with center at x

such that



− y|

n−2

dμ(y) <

by absolute continuity of the integral with respect to μ.Letμ



= μ|

and



= μ|

Ω∼B

. By Theorem 3.4.7, U



is harmonic on B and, in particular,

continuous at x

.SinceU

= U



+ U



(x) − U

)|≤|U



(x) − U



)|+ |U



(x) − U



≤|U



(x) − U



)|+2

n−2



)+.

Since U

is continuous at x

and U



is continuous at x

, U



continuous at x

.Sincez

∈ Γ and z

→ x

as x → x

, lim

x→x



) <.Thus,forx suﬃciently close to x

(x) − U

)|≤|U



(x) − U



)| +2

n−2

 + .

By continuity of U



at x

, the ﬁrst term can be made arbitrarily small by

taking x suﬃciently close to x

;thatis,U

is continuous at x

.SinceU

diﬀers from u by a harmonic function on Ω, u is continuous at x

Theorem 3.6.2 Let Ω be a Greenian set, and let μ be a measure with com-

pact support Γ ⊂ Ω such that G

μ<+∞ on Γ . Then given any >0,there

is a compact set Γ

⊂ Γ such that (i) μ(Γ ∼ Γ

) <, (ii) G

(μ|

) < +∞

on Γ

,and(iii) G

(μ|

) is continuous on Ω.

Proof: By Lusin’s theorem, given >0thereisacompactsetΓ

⊂ Γ

such that μ(Γ ∼ Γ

) <and G

μ is continuous on Γ

. Restrict all func-

tions to Γ

for the time being. Since G

(μ|

)=G

μ − G

(μ|

Γ ∼Γ

)on

, G

μ is continuous on Γ

,andG

(μ|

Γ ∼Γ

) is l.s.c. on Γ

, G

(μ|

)is

u.s.c. on Γ

. Since the function G

(μ|

) is also l.s.c. on Γ

, G

(μ|

continuous on Γ

. By the preceding theorem, G

(μ|

) is continuous on Γ

Since G

(μ|

)isharmoniconΩ ∼ Γ

, it is continuous on Ω.

The proof of the preceding theorem can be modiﬁed to obtain an approx-

imation of a potential by continuous potentials.

Theorem 3.6.3 Let Ω be a Greenian set, and let μ be a measure on Ω

with the property that there is a Borel set Σ ⊂ Ω with μ(Ω ∼ Σ)=0and

μ<+∞ on Σ. Then there is a sequence of measures {μ

} with disjoint

compact supports in Σ such that μ =



∞

n=1

, G

μ =



∞

n=1

,and

for each n ≥ 1, G

is ﬁnite-valued and continuous on Ω.

146 3 Green Functions

Proof: Assume ﬁrst that μ is a ﬁnite measure. Applying Lusin’s theorem

to the set Σ, there is a disjoint sequence of compact sets {Γ

} in Σ such

that μ(Ω ∼∪

∞

n=1

)=0andG

μ|

is ﬁnite-valued and continuous on Γ

Let μ

be the restriction of μ to Γ

. As in the preceding proof, G

ﬁnite-valued and continuous on Ω. The general case can be reduced to the

ﬁnite measure case using the fact that μ is σ-ﬁnite.

There is not much to be said about convergence properties of sequences of

potentials.

Theorem 3.6.4 Let {μ

} be a sequence of measures on the Greenian set Ω

that converges in the w

∗

-topology to the measure μ.Then

μ ≤ lim inf

j→∞

Proof: Consider any x ∈ Ω.ThenG

(x, ·) is nonnegative and l.s.c. on Ω,

and there is an increasing sequence {φ

} in C

(Ω) such that φ

↑ G

(x, ·).

Thus,



dμ = lim

j→∞



dμ

≤ lim inf

j→∞



(x, ·)dμ

Letting i →∞, G

μ(x) ≤ lim inf

j→∞

(x).

More stringent conditions are needed to improve on this result.

Theorem 3.6.5 If Ω is a Greenian subset of R

, {μ

} is a sequence of mea-

sureswithsupportsinthecompactsetΓ ⊂ Ω,μ

w∗

→ μ,and{G

} is an

increasing sequence on Ω,thenG

μ = lim

j→∞

and is superharmonic

on Ω.

Proof: Since μ

(Γ ) → μ(Γ ), the latter is ﬁnite and G

μ is a potential by

Theorem 3.4.4. By Theorem 2.4.8, the function u = lim

j→∞

is either

superharmonic or identically +∞ on each component of Ω. It can be assumed

that Ω is connected. Suppose x ∈ Ω ∼ Γ and B

x,ρ

⊂ Ω ∼ Γ .SinceG

(x, ·)

is bounded outside B

x,ρ

by a constant M

x,ρ

(x)=



(x, y) dμ

(y)=



∼B

x,ρ

(x, y) dμ

(y) ≤ M

x,ρ

(Γ ).

Thus, u(x) = lim

j→∞

(x) ≤ M

x,ρ

μ(Γ ) < +∞ and it follows that u is

superharmonic on Ω.Consideranyx ∈ Ω and any ball B

x,ρ

with B

−

x,ρ

⊂ Ω.

According to Theorem 3.3.12, L(G

(y,·):x, r) is a continuous function of

y ∈ Ω. By Tonelli’s theorem,

L(u : x, r) = lim

j→∞

L(G

: x, r)

3.6 Properties of Potentials 147

= lim

j→∞



L(G

(·,z):x, r) dμ

(z)



L(G

(·,z):x, r) dμ(z)

= L(G

μ : x, r).

Letting r ↓ 0,u(x)=G

μ(x) by Lemma 2.4.4.