Helms L.L. Potential Theory

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238 5 Dirichlet Problem for Unbounded Regions

Let Ω be a Greenian subset of R

, possibly unbounded. Regularity of ﬁnite

boundary points has been characterized in terms of thinness of ∼ Ω at x,

and regularity of ∞ in the n ≥ 3 case has been established. It remains only

to characterize regularity of ∞ in the n =2case.

Theorem 5.7.18 Let Ω be an unbounded Greenian subset of R

.Then∞

is a regular boundary point for Ω if and only if ∼ Ω is not thin at ∞.

Proof: Consider any inversion y → y

∗

with respect to a sphere ∂B

x,δ

.By

Remark 5.4.11 and Theorem 5.4.2, ∞ is a regular boundary point for Ω if

and only if x is a regular boundary point for Ω

∗

. By the same results, the

latter is true if and only if R

∼ Ω

∗

is not thin at x, and this is true if and

only if R

∼ Ω is not thin at ∞.

The ﬁne topology on R

seems to be the natural topology for studying

the boundary behavior of superharmonic functions. In some cases, the ﬁne

limit of a superharmonic function can be identiﬁed.

Example 5.7.19 Let x ∈ R

,letO be a neighborhood of x,andletu be a

nonnegative superharmonic function on O ∼{x}.Then

f-lim

y→x,y∈O

u(y) = lim inf

y→x,y∈O

u(y).

This follows from the fact that u has a superharmonic extension to O, denoted

by u, and that both are then equal to u(x).

Example 5.7.20 If u is a nonnegative superharmonic function on Ω = R

∼

−

x,δ

,then

f-lim

y→∞,y∈Ω

u(y) = lim inf

y→x,y∈Ω

u(y)ifn =2

and

f-lim

y→∞,y∈Ω

u(y)

|y|

n−2

= lim inf

y→∞,y∈Ω

u(y)

|y|

n−2

if n ≥ 3.

Consider an inversion with respect to ∂B

x,δ

.Sinceu

∗

)=u(y)inthen =2

case,

f-lim

y→∞,y∈Ω

u(y)

= f-lim

∗

→x,y

∗

∈Ω

∗

) = lim inf

∗

→x,y

∗

∈Ω

∗

) = lim inf

y→∞,y∈Ω

u(y)

by the preceding example. The same argument can be applied to the second

equation.

Theorem 5.7.21 (Cartan) If x is a ﬁne limit point of the set Λ and g is

an extended real-valued function having a ﬁne limit λ at x, then there is a

ﬁnely open set V

of x such that lim

y→x,y∈Λ∩V

f g(y)=λ.

5.7 Thinness and Regularity 239

Proof: It can be assumed that λ is ﬁnite for otherwise g can be replaced by

arctan g. It also can be assumed that Λ is a subset of a ball B with center

x.If{I

} is a decreasing sequence of open intervals such that I

↓{λ},then

for each j ≥ 1 there is a ﬁne neighborhood U

of x such that g(y) ∈ I

for

all y ∈ (Λ ∼{x}) ∩ U

.Ifx is an interior point of inﬁnitely many of the

, then it suﬃces to take V

= B. It therefore can be assumed that x is an

interior point of at most ﬁnitely many of the U

and, in fact, that x is a limit

point of ∼ U

for every j ≥ 1. Since U

is a ﬁne neighborhood of x, ∼ U

thin at x and there is a positive superharmonic function u

such that

(x) < lim

y→x,y∈B∼U

(y)=+∞. (5.12)

Let {α

} be a sequence of positive numbers such that



(x) < +∞.

Then u =



is superharmonic and is ﬁnite at x.Nowlet

= B ∩ (∼ U

) ∩{y; α

(y) >j} ,

and let F = ∪

.Thenx ∈ F . It will now be shown that F is thin at x.

This is trivially true if x ∈ F

−

and it therefore can be assumed that x is

a limit point of F .Ifp is a positive integer, α

(y) >pfor y ∈ F

,j ≥ p

and u(y)=



(y) ≥ p.Foreachj<p, there is a neighborhood V

of x

such that α

(y) ≥ p for y ∈ F

∩V

by Inequality (5.12). Then y ∈∩

p−1

j=1

implies that u(y)=



(y) ≥ p.Sincep is an arbitrary positive integer,

lim

y→x,y∈F

u(y)=+∞ and that F is thin at x. Letting V

be the ﬁne interior

of ∼ F, V

is then a ﬁne neighborhood of x.IfW

= B ∩{y : α

(y) >j},

then W

∩ (∼ U

)=F

⊂ W

∩ F and

⊃ W

∩ U

⊃ W

∩ (∼ F ) ⊃ W

∩ V

If y ∈ W

∩ V

∩ (Λ ∼{x}) ⊂ U

∩ (Λ ∼{x}), then g(y) ∈ I

.Thisproves

that lim

y→x,y∈V

∩Λ

g(y)=λ.

Theorem 5.7.22 (Brelot) If F is a closed subset of R

that is thin at x

and v is a positive superharmonic function on U ∼ F for some neighborhood

U of x,thenv has a ﬁne limit at x.

Proof: If x is not a limit point of F , then there is nothing to prove since

in this case v is ﬁnely continuous at x. It therefore can be assumed that x

is a limit point of F . There is then a superharmonic function u such that

u(x) < lim

y→x,y∈F

u(y)=+∞.Sinceu is ﬁnely continuous at x,

u(x) = f-lim

y→x,y∈∼F

u(y) < +∞.

240 5 Dirichlet Problem for Unbounded Regions

If lim inf

y→x,y∈∼F

(v+u)(y)=+∞, then lim

y→x,y∈∼F

(v+u)(y) exists and, in

particular, f-lim

y→x,y∈∼F

(v + u)(y)=+∞. It follows that f-lim

y→x,y∈∼F

v(y)

exists and is equal to +∞. Suppose now that λ = lim inf

y→x,y∈∼F

(v+u)(y) <

+∞ and let p be a positive integer with p>λ. Since lim

y→x,y∈F

u(y)=

+∞, there is a neighborhood O of x having compact closure in U such that

u(y) >pfor y ∈ O

−

∩ (F ∼{x}) ⊃ O ∩ (F ∼{x}). Since F is closed,

lim inf

z→y,z∈∼F

(v + u)(z) ≥ lim inf

z→y,z∈∼F

u(z) ≥ u(y) >pfor y ∈ O ∩

(∂F ∼{x}). Let

w =



min (u + v),p)on(O ∼ F ) ∼{x}

p on (O ∩F ) ∼{x}.

At each point y ∈ O ∩ (∂F ∼{x}), w is equal to p on a neighborhood of y.

Thus, w is superharmonic on O ∼{x} and since O has compact closure in

U, it is bounded below on O. By Theorem 4.2.15, w has a superharmonic

extension to O. Noting that lim inf

y→x,y∈∼F

(v + u)(y)=λ<p,wagrees

with v + u on ∼ F in a neighborhood of x.Sincew is ﬁnely continuous

at x, f-lim

y→x,y∈∼F

(v + u)(y) exists. Lastly, f-lim

y→x,y∈∼F

v(y)existssince

f-lim

y→x,y∈∼F

u(y) exists.

Chapter 6

Energy

6.1 Introduction

In 1828, Green used a physical argument to introduce a function, which he

called a potential function, to calculate charge distributions on conducting

bodies. The lack of mathematical rigor led Gauss in a 1840 paper to pro-

pose a procedure for ﬁnding equilibrium distributions based on the fact that

such a distribution should have minimal potential energy. This led to the

study of the functional



(G − 2f)σdS where σ is a density function and dS

denotes integration with respect to surface area on the boundary of a con-

ducting body. Gauss assumed the existence of a distribution minimizing the

functional. Frostman proved the existence of such a minimizing distribution

in a 1935 paper. After deﬁning the energy of a measure, properties of en-

ergy will be related to capacity and equilibrium distributions as developed in

Section 4.4. The chapter will conclude with Wiener’s necessary and suﬃcient

condition for regularity of a boundary point for the Dirichlet problem.

6.2 Energy Principle

Throughout this section, Ω will denote a connected Greenian subset of R

with Green function G

Deﬁnition 6.2.1 If μ, ν are Borel measures on Ω,themutual energy

[μ, ν]

of μ and ν is deﬁned by

[μ, ν]



μdν

and the energy |μ|

is deﬁned by |μ|

=[μ, μ]

. The set of Borel measures μ

of ﬁnite energy |μ|

will be denoted by E

L.L. Helms, Potential Theory, Universitext, 241

 Springer-Verlag London Limited 2009

242 6Energy

Note that the integral deﬁning [μ, ν] is always deﬁned, even though it may

be inﬁnite, and that

0 ≤ [μ, ν]



μdν =



νdμ=[ν, μ]

≤ +∞,

by the reciprocity theorem, Theorem 3.5.1. It follows that [μ, ν]

is a sym-

metric, bilinear form. If G

μ is bounded by m and μ has compact support

Γ ⊂ Ω,thenμ has ﬁnite energy and

|μ|



μdμ ≤ mμ(Γ ).

An example will be given later to show that G

μ cantakeonthevalue+∞

even though μ has ﬁnite energy.

Example 6.2.2 Let Ω = B

x,ρ

⊂ R

,letB = B

x,δ

where 0 <δ<ρ,andlet

μ = δ

∂B

(x, ·). By deﬁnition of δ

∂B

(x, ·),

|μ|



μdμ



(y,z) δ

∂B

(x, dz) δ

∂B

(x, dy)



∂B

(x, ·))(y) δ

∂B

(x, dy).

In the n =2case,G

(x, z)=log(ρ/|x − z|)=log(ρ/δ)forz ∈ ∂B,

andinthen ≥ 3case,G

(x, z)=(1/|x − z|

n−2

) − (1/ρ

n−2

). Since

∂B

(x, ·),z)=G

(x, z)forz ∈ ∂B,δ

∂B

(x, ·) has support in ∂B,and

∂B

(x, ∂B)=1,

|μ|



log

if n =2

n−2

−

n−2

if n ≥ 3

It is easily seen that δ

∂B

(x, ·) is a unit uniform measure on ∂B.

Theorem 6.2.3 An analytic set Z is polar if and only if μ(Z)=0for all

μ ∈E

with support in Z.

Proof: It is easily seen that the assertion is true if it is true for any analytic

subset of B

0,n

,n ≥ 1. It therefore can be assumed that Z is a subset of a

Greenian set Ω. Assume that Z is an analytic nonpolar set. Then Z has

positive capacity relative to Ω by Theorem 4.5.2. It follows that there is a

compact set Γ ⊂ Z with C(Γ ) > 0. Consider the capacitary distribution

for which 0 < C(Γ )=μ

(Γ ) < +∞ and G

≤ 1. Thus,

∈E

and μ

(Z) > 0. This proves the suﬃciency. Suppose now that Z is

an analytic polar set and that μ ∈E

has support in Z. By Theorem 4.3.11,

6.2 Energy Principle 243

there is a nonnegative superharmonic function w on Ω such that w =+∞

on Z.Sinceμ has ﬁnite energy, G

μ<+∞a.e.(μ). Let Λ = {x; w(x)=+∞}

and Λ

= Λ ∩{x; G

μ(x) ≤ j},j ≥ 1. Since Λ ∼∪

= Λ ∩{x; G

μ(x)=

+∞},μ(Λ ∼∪

)=0.Fixj ≥ 1, let Γ be any compact subset of Λ

,and

let ν = μ|

.SinceG

ν ≤ G

μ, G

ν ≤ j on Γ .Sinceν has support in Γ

and w =+∞ on Γ, G

ν ≤ w on the support of ν for every >0. By the

Maria-Frostman domination principle, Theorem 4.4.8, G

ν ≤ w on Ω for

every >0 and it follows that G

ν =0.Thus,ν is the zero measure and

ν(Γ )=μ(Γ ) = 0 for all compact Γ ⊂ Λ

.Since

μ(Λ

)= sup

Γ ⊂Λ

,Γ ∈K(Ω)

μ(Γ ),

μ(Λ

) = 0 for all j ≥ 1. Since ∪Λ

⊂ Λ and μ(Λ ∼∪Λ

)=0,μ(Λ)=0and

μ(Z)=0.

Theorem 6.2.4 Let Ω be a Greenian subset of R

and let {μ

}, {ν

} be

sequences of Borel measures on Ω.

(i) If G

≤ G

,j =1, 2,then[μ

,μ

]

≤ [ν

,ν

]

(ii) If μ, ν are Borel measures on Ω with G

μ ≤ G

ν,then|μ|

≤|ν|

;

moreover, if |μ|

= |ν|

< +∞,thenμ = ν.

(iii) If {G

} and {G

} are increasing sequences of potentials on Ω with

potentials G

μ and G

ν as limits, respectively, then

lim

j→∞

[μ

,ν

]

=[μ, ν]

(iv) If {G

} and {G

} are decreasing sequences of potentials, then the

lower regularizations of their limits are potentials G

μ and G

ν, respec-

tively, and

lim

j→∞

[μ

,ν

]

=[μ, ν]

provided |μ

, |ν

,and[μ

,ν

]

are ﬁnite.

Proof: (i) By the reciprocity theorem, Theorem 3.5.1,

[μ

,μ

]



dμ

≤



dμ



dν

≤



dν

=[ν

,ν

]

(ii)Takeμ

= μ, ν

= ν, i =1, 2, in (i)toobtain|μ|

≤|ν|

;if

|μ|

= |ν|

< +∞, then the above inequalities are equalities so that G

μ =

νa.e.(μ). Since G

ν<+∞a.e.(ν), G

μ ≤ G

ν on Ω by the Maria-

Frostman domination principle, Theorem 4.4.8. Similarly, G

ν ≤ G

μ on

Ω.Thus,G

μ = G

ν and μ = ν by Theorem 3.5.8.

244 6Energy

(iii)By(i), {[μ

,ν

]

} is an increasing sequence with lim

j→∞

[μ

,ν

]

≤

[μ, ν]

.SinceG

≤ G

for j ≥ i, by the Lebesgue monotone convergence

theorem and the reciprocity theorem

lim

j→∞

[μ

,ν

]

= lim

j→∞



dν

≥ lim inf

j→∞



dν

= lim

j→∞



dμ



νdμ



dν.

Since the latter sequence of integrals increases to [μ, ν]

, lim

j→∞

[μ

,ν

]

≥

[μ, ν]

and the two are equal.

(iv)By(i), [μ

,ν

]

≤ [μ

,ν

]

for all j ≥ 1 and lim

j→∞

[μ

,ν

]

is de-

ﬁned and ﬁnite; moreover, each G

is ν

- integrable and each G

-integrable. Note also that lim

j→∞

= G

μ quasi everywhere and

lim

j→∞

= G

ν quasi everywhere by Theorem 4.4.10. Since polar sets

are null sets for measures of ﬁnite energy by the above theorem,

lim

j→∞

[μ

,ν

]

= lim

j→∞



dν

≥ lim

j→∞



( lim

i→∞

) dν

= lim

j→∞



μdν



dμ

≥



( lim

i→∞

) dμ =



νdμ=[μ, ν]

On the other hand, for each i ≥ 1,

lim

j→∞

[μ

,ν

]

≤ lim inf

j→∞



dν

= lim inf

j→∞



dμ

Since the sequence {G

}

j≥i

is dominated by the μ

-integrable function

lim

j→∞

[μ

,ν

]

≤



( lim

j→∞

) dμ



νdμ



dν ≥



( lim

i→∞

) dν



μdν =[μ, ν]

Thus, lim

j→∞

[μ

,ν

]

=[μ, ν]

It was shown in Theorem 4.3.5 that for a nonnegative superharmonic func-

tion u on Ω and a compact set Γ ⊂ Ω,

is a potential.

Deﬁnition 6.2.5 A subset Λ of the Greenian set Ω is energizable if

the potential of a measure μ

6.2 Energy Principle 245

It is clear that a subset of an energizable set is again energizable, that compact

subsets of Ω are energizable, and that polar subsets of Ω are energizable. Even

though μ

is not deﬁned when Λ is not energizable, the formal expression

|μ

=+∞ will be used in this case so that |μ

is deﬁned for all subsets

of Ω.

Lemma 6.2.6 The set E

of energizable subsets of Ω has the following prop-

erties:

(i) If Λ ∈E

, then there is an open set O such that Λ ⊂ O ∈E

(ii) If Λ ∈E

,then|μ

= μ

(cl

Λ)=μ

(Ω).

(iii) |μ

is an increasing function of Λ ∈E

Proof: Consider the open set U = {x ∈ Ω;

> 1/2}.Since

= R

quasi everywhere on Λ, U covers Λ except possibly for a polar subset of Λ.By

Theorem 4.3.11, there is a potential u that is identically +∞ on the polar set

Λ ∼ U . Letting O = U ∪{u>1}, O is an open set containing Λ.Since2

is a potential with 1 ≤ e

+ u on O,

≤ R

≤ 2

+ u. Since the latter

is a potential,

is a potential so that O ∈E

and O is an energizable open

set containing Λ.(ii)Since

=1oncl

Λ by Theorem 5.7.11 and μ

supported by cl

Λ by Theorem 5.7.13,

|μ



dμ



dμ

= μ

(cl

Λ)=μ

(Ω), (6.1)

proving (ii). Consider Λ

,Λ

∈E

with Λ

⊂ Λ

⊂ Ω,thenG

≤

= G

and |μ



dμ

≤



dμ



dμ

≤



dμ

= |μ

by the reciprocity theorem,

Theorem 3.5.1.

Theorem 6.2.7 If Λ is an energizable subset of the Greenian set Ω,then

|μ

=inf{|μ

; Λ ⊂ O ∈O(Ω)} .

Proof: Since there is nothing to prove if |μ

=+∞, it can be assumed

that μ

has ﬁnite energy. By the preceding lemma, there is an energizable

open set U ⊃ Λ so that it suﬃces to consider only those open sets O with

Λ ⊂ O ⊂ U.ForsuchO,

is a potential. By Lemma 4.6.12,

=inf{R

; O ∈O(Ω),Λ⊂ O ⊂ U}.

By Choquet’s Lemma, Lemma 2.2.8, there is a decreasing sequence {O

} of

open sets with Λ ⊂ O

⊂ U such that

is equal to the lower regulariza-

tion of

lim

j→∞

= lim

j→∞

By part (iv) of Theorem 6.2.4, lim

j→∞

|μ

= |μ

246 6Energy

Theorem 6.2.8 If Λ is an energizable subset of the the Greenian set Ω,then

∗

(Λ) ≤ μ

(Ω)=μ

(cl

Λ)=C

∗

(Λ);

if, in addition, Λ is capacitable, then μ

(Ω)=|μ

= C(Λ).

Proof: If Γ is any compact subset of Λ,thenC(Γ )=μ

(Γ )=μ

(Ω) ≤

(Ω) by Lemma 6.2.6. Taking the supremum over such compact sets

Γ, C

∗

(Λ) ≤ μ

(Ω). Now let O be any open set containing Λ and let {Γ

}

be an increasing sequence of compact subsets of O such that Γ

↑ O.

Then |μ

∩Λ

= μ

∩Λ

(Ω) ≤ μ

(Ω)=C(Γ

) ≤C(O) for all j ≥ 1. By

Theorem 4.6.5, G

∩Λ

↑

= G

on Ω and it follows from

Theorem 6.2.4 that |μ

∩Λ

↑|μ

. Therefore, μ

(Ω)=|μ

≤C(O) for all

open sets O ⊃ Λ. Taking the inﬁmum over such sets, μ

(Ω) ≤C

∗

(Λ). Thus,

∗

(Λ) ≤|μ

= μ

(Ω) ≤C

∗

(Λ). It remains only to show that μ

(Ω)=

∗

(Λ). Since open sets O are capacitable, |μ

= μ

(Ω)=C(O)=C

∗

(O).

By Theorem 6.2.7, μ

(Ω)=|μ

=inf{C

∗

(O); Λ ⊂ O ∈O(Ω)} = C

∗

(Λ).

The following fact will be used in the course of the next proof. Let { μ

} be a

sequence of measures on a compact Hausdorf space X which converges in the

∗

-topology to the measure μ. The sequence of product measures {μ

×μ

}

on X × X then converges in the w

∗

-topology to the product measure μ × μ.

For continuous fucntions of the form f (x)g(y)onX ×X where f, g ∈ C

(X),



f(x)g(y) dμ

(x) dμ

(y)=





f(x) dμ

(x)





g(y) dμ

(y)



→





f(x) dμ(x)





g(y) dμ(y)





f(x)g(y) dμ(x) dμ(y)

as j →∞. This results extends to ﬁnite sums of such products. Since the

class A of such functions is a subalgebra of C

(X ×X) which separates points

and contains the constant functions, A is a dense subset of C

(X × X)by

the Stone-Weierstrass theorem. The above convergence can be extended to

any function in C

(X × X) by a simple approximation.

The following theorem was proposed, but not justiﬁed, as a procedure for

solving the Dirichlet problem by Gauss in 1840.

Theorem 6.2.9 (Gauss-Frostman) Let Γ be a compact subset of the

Greenian set Ω,letf ∈ C

(Γ ),andforeachmeasureμ with support in

Γ let

6.2 Energy Principle 247

(μ)=



μ − 2f) dμ.

Then there is a measure μ ∈E

with support in Γ that minimizes Φ

.More-

over, G

μ ≤ f on the support of μ and G

μ ≥ f on Γ except possibly for a

polar set.

Proof: By (vii) of Theorem 4.3.5, there is a measure ν with support in Γ such

that 1 ≥

= G

ν on Ω; for this measure, |Φ

(ν)|≤ν(Γ )(1 + 2f

0,Γ

) <

+∞. Replacing ν by ν for 0 <<1,Φ

(ν)=



Gνdν − 



fdν ≤

Φ

(ν). If Φ

(ν) ≤ 0, then inf

(μ) ≤ 0; if Φ

(ν) ≥ 0, then there are

measures μ with Φ

(μ) arbitrarily close to 0 so that inf

(μ) ≤ 0. Let

α =infΦ

(μ) ≤ 0,β =inf{G

(x, y); x, y ∈ Γ },andγ =sup{f (y); y ∈ Γ }.

Then

(μ)=



(x, y) dμ(x) dμ(y) − 2



f(x) dμ(x) ≥ βμ

(Γ ) − 2γμ(Γ ).

Since βm

−2γm has a minimum value of −γ

/β, α ≥−γ

/β > −∞. Letting

{μ

} be a sequence of measures with supports in Γ such that Φ

(μ

) → α as

j →∞,

(μ

) ≥ βμ

(Γ ) − 2γμ

(Γ )=μ

(Γ )(βμ

(Γ ) − 2γ)

and it follows that the sequence {μ

(Γ )} is bounded. There is then a subse-

quence of the sequence { μ

}, which can be assumed to be the sequence itself,

and a measure μ with support in Γ such that μ

∗

→ μ.Letk be any positive

integer. Since μ

×μ

∗

→ μ ×μ and min (k, G

(x, y)) is a bounded continuous

function on Γ × Γ ,



(x, y) dμ(x) dμ(y) = lim

k→∞



min (k, G

(x, y)) dμ(x) dμ(y)

= lim

k→∞

lim

j→∞



min (k, G

(x, y)) dμ

(x) dμ

(y)

≤ lim inf

j→∞



(x, y) dμ

(x) dμ

(y).

Therefore,

α ≤ Φ

(μ)=



(x, y) dμ(x) dμ(y) − 2



f(x) dμ(x)

≤ lim inf

j→∞



(x, y) dμ

(x) dμ

(y) − 2 lim

j→∞



f(x) dμ

(x)

= lim

j→∞

(μ

)=α.

This shows that μ minimizes Φ

and that μ ∈E

. It will be shown now

that G

μ ≤ f on the support of μ. Suppose x is in the support of μ and