Helms L.L. Potential Theory

Подождите немного. Документ загружается.

98 2 The Dirichlet Problem

Suppose x = λz and λ ↑ 1. Then

lim

λ↑1

h(λz)=

2πρ

lim

λ↑1

− λ

(1 − λ)

2πρ

lim

λ↑1

1+λ

1 − λ

=+∞.

Thus, if x ∈ B approaches z along a radial line to z,thenh(x) → +∞.

Suppose now that x → z in the following manner. Taking the radial line

through z as a polar axis, let (r, θ) be the polar coordinates of x where r = |x|

and θ is the angle between the x and z vectors. Now let x → z subject to the

condition r = ρ cos θ.Inthiscase,

lim

x→z,|x|=ρ cos θ

h(x)=

2πρ

lim

θ→0

(1 − cos

θ)

(1 − cos

θ)

2πρ

The values +∞ and 1/2πρ are not the only possible limiting values; for

example, if x → z subject to the condition r = ρ cos

θ, then the limiting

value is 1/πρ. In the latter two cases, x approaches z along a path that is

internally tangent to ∂B. If anything is to be said about the limiting behavior

of h at z, then such tangential approaches to z must be excluded.

For the remainder of this section, B will denote a ﬁxed ball with center at

0ofradiusρ.

Deﬁnition 2.8.9 (i)AStolz domain with vertex z ∈ ∂B and half-angle

θ, denoted by S

z,θ

, is deﬁned to be an open circular cone with vertex z,with

the radial line to z as axis, and half-angle θ<π/2. (ii) A function u on B is

said to have nontangential limit  at z ∈ ∂B, denoted by θ − lim

x→z

u(x),

if lim

x→z,x∈S

z,θ

∩B

u(x)= for all Stolz domains S

z,θ

Lemma 2.8.10 Let z ∈ ∂B,S

z,θ

a Stolz domain, 0 <δ<1,andB

δz,r

aball

with center δz which is internally tangent to ∂S

z,θ

. Then there is a constant

m, depending only upon θ, such that if x, x



∈ B

δz,r

and h is any positive

harmonic function on B,thenh(x)/h(x



) ≤ m; the constant m is given by

m =

cos



1+sinθ

1 − sin θ



Proof: Consider any >0 for that θ + <π/2andaballB

δz,r



that is inter-

nally tangent to C

z,θ+

. The ball B

δz,r



has radius r



=(r/ sin θ)sin(θ + ).

Applying Theorem 2.2.1 to B

δz,r

⊂ B

δz,r



h(x)

h(x



)

≤

(r/ sin θ)

sin

(θ + )

(r/ sin θ)

sin

(θ + ) − r



(r/ sin θ)sin(θ + )+r

(r/sinθ)sin(θ + ) − r



for all positive harmonic functions h on B and all x, x



∈ B

δz,r

. The constant

m is obtained by letting θ +  → π/2.

2.8 Nontangential Boundary Limit Theorem 99

Theorem 2.8.11 (Fatou [20]) Let μ be a signed measure of bounded vari-

ation on the Borel subsets of the boundary of B,andleth = PI(μ : B) on

B.Thenθ − lim

x→z

h(x) exists and is equal to Dμ(z) a.e.(σ).

Proof: It will be shown ﬁrst that it suﬃces to prove that

ν is a measure on ∂B,Dν(z)=0⇒ θ − lim

x→z

PI(ν : B)(x)=0a.e.(σ). (2.9)

Assume that this is the case and consider the decomposition μ = μ

+ μ

μ relative to σ,whereμ

and μ

are the absolutely continuous and singular

parts of μ, respectively. Since μ

= μ

−μ

−

,whereμ

and μ

−

are the positive

and negative parts of μ

, respectively, Dμ

= D μ

−Dμ

−

=0a.e.(σ)by(ii)

of Theorem 2.8.7 and it follows that

θ − lim

x→z

PI(μ

: B)(x)=0a.e.(σ). (2.10)

Now consider μ

which can be written

(M)=



fdσ

for each Borel set M ⊂ ∂B and some integrable function f on ∂B.By(i)of

Theorem 2.8.7, D μ

= fa.e.(σ). For each Borel set M ⊂ ∂B and z ∈ ∂B,

(μ

− f (z)σ)(M)=



(f(y) − f (z)) dσ(y)

and

|μ

− f (z)σ|(M)=



|f(y) − f (z)|dσ(y).

By Theorem 2.8.8, D|μ

−f (z)σ|(z) = 0 for almost all (σ)pointsz ∈ ∂B.Let

be the exceptional set of σ-measure zero. For z ∈ Z

, D|μ

−f(z)σ|(z)=0

and by (2.9)

θ − lim

x→z

PI(|μ

− f (z)σ| : B)(x)=0.

Since

|PI(μ

− f (z)σ : B)(x)|≤PI(|μ

− f (z)σ| : B)(x),

θ − lim

x→z

(PI(μ

: B)(x) − f(z)) = 0

for z ∈ Z

. Include in the set Z

the null set {z ∈ ∂B; Dμ

(z) = f(z)}.Then

for z ∈ Z

θ − lim

x→z

PI(μ

: B)(x)=f(z)=Dμ

(z).

In other words, θ − lim

x→z

PI(μ

: B)(x)=Dμ

(z) a.e.(σ). It follows from

this result and Equation (2.10) that

θ − lim

x→z

h(x)=θ − lim

x→z

PI(μ : B)(x)=θ − lim

x→z

(PI(μ

: B)(x)=Dμ

= Dμ

100 2 The Dirichlet Problem

for almost all (σ)pointsz ∈ ∂B. This shows that it suﬃces to prove (2.9).

Suppose ν is a measure on ∂B and Dν(z) = 0. Letting h = PI(ν : B)on

B,lim

λ→1

−

h(λz) = 0 by Theorem 2.7.2. If S

z,θ

is a Stolz domain, by the

preceding lemma there is a constant m, depending only upon θ, such that

h(x)/h(x



) ≤ m whenever x is suﬃciently close to z and x



is the projection

of x onto the radial line through z.Then

|h(x



) − h(x)| = h(x



)



1 −

h(x)

h(x



)



≤ h(x



)(1 + m).

Since h(x



) → 0asx → z, h(x) → 0asx → z in S

z,θ

2.9 Harmonic Measure

Wiener’s theorem settles the problem of showing that each continuous func-

tion on the boundary of a bounded open set Ω ⊂ R

determines a harmonic

function. The same problem can be considered for more general boundary

functions. By Wiener’s theorem, Theorem 2.6.16, each f ∈ C

(∂Ω)andx ∈ Ω

determines a real number H

(x). Consider the map L

: C

(∂Ω) → R de-

ﬁned by L

(f)=H

(x). Using Lemma 2.6.11, it is easily seen that L

positive linear functional on C

(∂Ω). It follows from the Riesz representa-

tion theorem that there is a unique unit measure μ

on the Borel subsets of

∂Ω such that H

(x)=L

(f)=



fdμ

for all x ∈ Ω, f ∈ C

(∂Ω).

Showing that two measures are equal can be a tedious process involving

several steps, the ﬁrst of which is to show that the two measures agree on

certain elementary sets. The following digression into measure theory will

simplify this process.

Let F

be a collection of subsets of ∂Ω that contains ∂Ω and is closed

under ﬁnite intersections. A system G of subsets of ∂Ω is called a Dynkin

system or d-system if

(i) ∂Ω ∈G,

(ii) E,F ∈Gwith E ⊂ F implies that F ∼ E ∈G,and

(iii) if {E

} is an increasing sequence in G,thenE = ∪E

∈G.

Let d(F

) denote the smallest d-system containing F

;thatis,d(F

)isthe

intersection of all d-systems containing F

and is itself a d-system. Then

d(F

) ⊃F

. Also, let σ(F

) denote the smallest σ-algebra containing F

Since σ(F

) is a d-system containing F

,d(F

) ⊂ σ(F

). It will be shown

now that d(F

)=σ(F

). Since a d-system is closed under complementation,

it is closed under unions if closed under intersections. It will be shown ﬁrst

that d(F

) is closed under ﬁnite intersections. Let

= {E ∈ d(F

); E ∩ F ∈ d(F

) for all F ∈F

}⊂d(F

2.9 Harmonic Measure 101

Since F

is closed under ﬁnite intersections, F

⊂H

. It is easily checked

that H

is a d-system, and therefore that d(F

) ⊂H

.SinceH

⊂ d(F

the two are equal. Now let

= {E ∈ d(F

); E ∩ F ∈ d(F

) for all F ∈ d(F

)}⊂d(F

Again, H

is a d-system. If F ∈F

,thenE ∩ F ∈ d(F

) for all E ∈H

d(F

) by deﬁnition of H

and consequently F

⊂H

. Therefore, d(F

) ⊂

and the two are equal. But this shows that d(F

) is closed under ﬁnite

intersections, and therefore under ﬁnite unions. Since d(F

) is closed under

monotone increasing limits and ﬁnite unions, d(F

) is closed under countable

unions; that is, d(F

)isaσ-algebra and d(F

)=σ(F

). The advantage of

using d-systems will be apparent in the proof of the following theorem.

Theorem 2.9.1 If f is a lower bounded (or upper bounded) Borel measurable

function on ∂Ω,thenH

(x)=H

(x)=



fdμ

for all x ∈ Ω; if, in addition,

f is bounded, then f is resolutive and H

(x)=



fdμ

for all x ∈ Ω.

Proof: Let F

be the collection of compact (=closed) subsets of ∂Ω.IfC ∈

, there is a sequence {f

} in C

(∂Ω) such that f

↓ χ

, the indicator

function of C,with0≤ f

≤ 1 for all j ≥ 1. It follows from Lemma 2.6.13

that χ

is a resolutive boundary function and that H

(x)=



dμ

,x∈

Ω.LetG be the collection of sets E ⊂ ∂Ω such that χ

is resolutive and

(x)=



dμ

,x ∈ Ω.ThenF

⊂G.Notethat∂Ω ∈Gsince χ

∂Ω

and the constant functions are resolutive. If E, F ∈Gwith E ⊂ F ,then

F ∼E

= χ

− χ

.Sinceχ

and χ

are resolutive, χ

F ∼E

is resolutive by

Lemma 2.6.11 and

F ∼E

(x)=H

(x) − H

(x)=



dμ

−



dμ



F ∼E

dμ

for all x ∈ Ω. This shows that G is closed under relative complements. Sup-

pose {E

} is an increasing sequence in G and E = ∪E

. By Lemma 2.6.13,

is resolutive and

(x) = lim

j→∞

(x) = lim

j→∞



dμ



dμ

,x∈ Ω.

This shows that G is closed under unions of increasing sequences and there-

fore that G is a d-system containing F

.Thus,G = d(G)=σ(G) ⊃ σ(F

)

so that G contains all Borel subsets of ∂Ω; that is, the indicator function of

any Borel subset of ∂Ω is resolutive. It follows that any simple Borel mea-

surable function χ on ∂Ω is resolutive with H

(x)=



χdμ

,x∈ Ω.Iff is a

nonnegative Borel measurable function on ∂Ω, then there is a sequence {f

}

of simple Borel measurable functions such that 0 ≤ f

≤ f and f

↑ f.By

Lemma 2.6.13,

102 2 The Dirichlet Problem

(x)=H

(x) = lim

j→∞

(x) = lim

j→∞



dμ



fdμ

,x∈ Ω,

by the Lebesgue monotone convergence theorem. If f is only bounded below

by α on ∂Ω, the result can be applied to f − α.Iff is bounded, then

and H

are ﬁnite on Ω by Lemma 2.6.11 and harmonic by Lemma 2.6.7.

Theorem 2.9.2 If Λ is a component of the bounded open set Ω, the collec-

tion of Borel subsets of ∂Ω of μ

measure zero is independent of x ∈ Λ.

Proof: If E is a Borel subset of ∂Ω,thenχ

is resolutive and H

(x)=



dμ

= μ

(E),x ∈ Ω.Ifμ

(E)=0forsomex

∈ Λ, then the nonneg-

ative harmonic function H

= μ



(E) must be identically zero on Λ by the

minimum principle, Corollary 1.5.10.

Let f be a bounded Borel measurable function on ∂Ω and let Λ be a

component of Ω. Using the notation of Lemma 2.6.6, H

= H

∂Λ

on Λ,so

that the value of H

= H

at a point x ∈ Λ is independent of the values of

f on ∂Ω ∼ ∂Λ. Therefore, μ

(∂Ω ∼ ∂Λ) = 0 for all x ∈ Λ;thatis,eachμ

has support on the boundary of the component containing x. This argument

shows that μ

is just the restriction of μ

to ∂Λ.

If E denotes any Borel subset of ∂Ω and Z denotes any subset of a Borel

subset of ∂Ω of μ

measure zero, the class F

of sets of the form (E ∼

Z) ∪ (Z ∼ E)isaσ-algebra of subsets of ∂Ω. Letting F = ∩

x∈Ω

, F is a

σ-algebra of subsets of ∂Ω which includes the Borel subsets of ∂Ω. Since each

can be uniquely extended to F

, the same is true of F. The extension of

to F will be denoted by the same symbol.

Deﬁnition 2.9.3 Sets in F are called μ



-measurable sets and extended

real-valued functions measurable relative to F are called μ



-measurable.

The measure μ

on F is called the harmonic measure relative to Ω and x.

A μ



-measurable function f on ∂Ω is μ



-integrable if it is integrable relative

to each μ

,x∈ Ω.

Lemma 2.9.4 If f is a nonnegative μ



-measurable function on ∂Ω,then

(x)=H

(x)=



fdμ

for all x ∈ Ω;iff is any μ



-measurable function

on ∂Ω which is μ

-integrable for some y in each component of Ω,thenf is



-integrable, resolutive, and H

(x)=



fdμ

,x∈ Ω.

Proof: Consider any F ∈F.Foreachx ∈ Ω,F ∈F

,andF has the form

(E ∼ Z) ∪ (Z ∼ E)whereE is a Borel subset of ∂Ω and Z is a subset of a

Borel set A ⊂ ∂Ω with μ

(A)=0(E and Z may depend upon x). Note that

= χ

+χ

−2χ

E∩Z

and that χ

is resolutive by Theorem 2.9.1. Since 0 ≤

(x) ≤ H

(x)andH

(x)=μ

(A) = 0 by Theorem 2.9.1,

the nonnegative functions H

and H

are harmonic on the component

of Ω containing x.SinceH

(x)=H

(x)=0,H

= H

=0onthe

2.9 Harmonic Measure 103

component of Ω containing x by the minimum principle, Corollary 1.5.10.

Likewise, H

E∩Z

= H

E∩Z

= 0 on the component of Ω containing x.By

Lemma 2.6.11,

0 ≤ H

≤ H

+ H

− 2H

E∩Z

= H

on the component of Ω containing x. Likewise,

≥ H

+ H

− 2H

E∩Z

= H

on the component of Ω containing x so that H

= H

.Thus,χ

is resolutive and

(x)=



dμ

,,x∈ Ω,

by Theorem 2.9.1. Suppose now that f is a simple μ



-measurable function.

Then f is resolutive by Lemma 2.6.11 and

(x)=H

(x)=



fdμ

,x∈ Ω,

by the preceding step. Now let f be a nonnegative μ



-measurable function

on ∂Ω. Then there is a sequence {f

} of μ



-measurable simple functions such

that 0 ≤ f

≤ f and f

↑ f on ∂Ω. By Lemma 2.6.13,

(x)=H

(x) = lim

j→∞

(x) = lim

j→∞



dμ



fdμ

,x∈ Ω.

If, in addition, f is μ

-integrable for some y in each component of Ω, then this

equation shows that H

and H

are ﬁnite at some point of each component

of Ω and therefore are harmonic on Ω. It follows that f is resolutive and

(x)=



fdμ

,x∈ Ω.

Lastly, if f is any μ



-measurable function on ∂Ω that is μ

-integrable for

some y in each component of Ω, then the above argument can be applied to

and f

−

, the positive and negative parts of f.

Theorem 2.9.5 (Brelot [6]) Let Ω be a bounded open set. A function f

on ∂Ω is resolutive if and only if it is μ



-integrable; in which case, H

(x)=



fdμ

for all x ∈ Ω.

Proof: The suﬃciency is just the preceding lemma. Assume that f is reso-

lutive. If it can be shown that f is measurable relative to each F

,x ∈ Ω,

then f is measurable relative to F = ∩

x∈Ω

.Fixx ∈ Ω and let Λ be the

component of Ω containing x. In order to show that f is F

-measurable,

it suﬃces to show that f|

∂Λ

is equal a.e. (μ

) to a Borel measurable func-

tion on ∂Λ since μ(∂Ω ∼ ∂Λ) = 0 implies that every subset of ∂Ω ∼ ∂Λ

104 2 The Dirichlet Problem

belongs to F

. Note also that the ﬁniteness of



fdμ

is independent of the

values of f on ∂Ω ∼ ∂Λ. Therefore it can be assumed that Ω is connected.

Let x

be a ﬁxed point of Ω and let {v

} be a sequence in U

such that

lim

j→∞

)=H

). It can be assumed that the sequence {v

} is de-

creasing. Consider a ﬁxed j ≥ 1. Since v

∈ U

is bounded below by some

as is the Borel measurable function f

(x) = lim inf

y→x,y∈Ω

(y). Since

∈ U

≥ H

≥ α

. It follows that the functions H

and H

are

harmonic on Ω. By Theorem 2.9.1, each f

is resolutive. Since {v

} is a de-

creasing sequence, the sequence {f

} is also decreasing and f

∗

= lim

j→∞

is deﬁned. Using the fact that v

∈ U

(x) = lim inf

y→x,y∈Ω

(y) ≥ f (x)

and f

∗

≥ f.Sincethef

’s are Borel measurable, the same is true of f

∗

.Since

≥ H

∗

≥ H

∗

≥ H

and lim

j→∞

)=H

), H

∗

)=H

). It follows that H

∗

and H

∗

are harmonic and equal

to H

;thatis,f

∗

is resolutive with H

∗

= H

. Similarly, there is an

increasing sequence {g

} of Borel measurable functions on ∂Ω such that

lim

j→∞

= f

∗

≤ f and H

∗

= H

.Sincef

≥ f

∗

≥ f ≥ f

∗

≥ g

for each j

and f

are μ

-integrable for each x ∈ Ω, f

∗

and f

∗

are μ

-integrable for

each x ∈ Ω.InviewofthefactthatH

∗

(x) − H

∗

(x)=



∗

− f

∗

) dμ

and

∗

− f

∗

≥ 0,f

∗

= f

∗

a.e.(μ

). Therefore, f

∗

= f

∗

= fa.e.(μ

)foreachx ∈ Ω

and f is F

-measurable for each x ∈ Ω.Thus,f is F-measurable. Since f

∗

-integrable for each x ∈ Ω, the same is true of f and so f is μ



- integrable.

Lastly,

(x)=H

∗

(x)=



∗

dμ



fdμ

,x∈ Ω.

Harmonic measures will now be examined from the viewpoint of the classic

spaces, p ≥ 1. Let Ω be a connected, bounded, open subset of R

. Accord-

ing to Lemma 2.9.4, the class of μ

-integrable boundary functions is indepen-

dent of x ∈ Ω. This class constitutes the Banach space L

(μ

)consistingof

-integrable functions f on ∂Ω with norm deﬁned by f

1,μ



|f|dμ

under the usual proviso that functions that are equal a.e. relative to μ

are

identiﬁed. Since the class of Borel subsets of ∂Ω of μ

measure zero is inde-

pendent of x, L

(μ

) is independent of x ∈ Ω; but it need not be true that

·

1,μ

is independent of x. For any p>1,L

(μ

) consists of equivalence

classes of μ

-measurable boundary functions f for which

f

p,μ





|f|

dμ



1/p

< +∞.

Since f

p,μ

= H

|f|

(x) and the latter function is harmonic, the Banach

space L

(μ

) is independent of x ∈ Ω; but again the norm f 

p,μ

may

depend upon x.Consideranyp ≥ 1. There are two notions of convergence

in L

(μ

), weak and strong convergence, which could possibly depend upon

x. Consider any two points x, y ∈ Ω.Iff

p,μ

=0,thenH

|f|

(x)=0

and H

|f|

=0onΩ;inparticular,f

p,μ

= H

|f|

(y)=0.Thisshows

2.9 Harmonic Measure 105

that f

p,μ

=0ifandonlyiff

p,μ

=0.LetK = {x, y}.ByHarnack’s

inequality, Theorem 2.2.2, there is a constant k such that h(x



)/h(x



) ≤ k for





∈ K whenever h is a positive harmonic function on Ω.Iff

p,μ

> 0,

then f

p,μ



> 0onΩ and

f

p,μ

f

p,μ

|f|

(x)

|f|

(y)

≤ k;

that is, f 

p,μ

≤ k

1/p

f

p,μ

, with this inequality obviously holding if

f

p,μ

= 0. This shows that the norms ·

p,μ

and ·

p,μ

are equivalent

norms on L

(μ



)=L

(μ

)=L

(μ

). Thus, strong convergence in L

(μ

)is

independent of x ∈ Ω.Sincethenorms·

p,μ

and ·

p,μ

are equivalent,

the adjoint space L

∗

(μ

)=L

(μ

), (1/p +1/q =1ifp>1; q = ∞ if p =1)

is independent of x and the weak topology on L

(μ

) is independent of x.

Thus, weak convergence in L

(μ

) is independent of x ∈ Ω.Moreover,if

; α ∈ A} is a convergent net in Ω having the limit x ∈ Ω,thefactthat

is harmonic whenever f ∈ C

(∂Ω) implies that



fdμ

→



fdμ

= H

(x),

and therefore that μ

∗

→ μ

. Lastly, note that the measures {μ

; x ∈ Ω} are

mutually absolutely continuous; if μ

(A)=0forsomeμ



-measurable A ∈ ∂Ω,

then μ

(A) = 0 for any y ∈ Ω since the class of zero sets is independent of x.

This page intentionally left blank

Chapter 3

Green Functions

3.1 Introduction

In 1828, Green introduced a function, which he called a potential, for cal-

culating the distribution of a charge on a surface bounding a region in R

in the presence of external electromagnetic forces. The argument that led to

the function G

introduced in Section 1.5 is precisely the argument made

by Green in [26]. The function introduced by Green is now called the Green

function. After Green, the most general existence theorem for Green functions

was proven by Osgood in 1900 for simply connected regions in the plane (see

[49]). In this chapter it will be shown that a Green function can be deﬁned

for some sets, but not all, which serve the same purpose as G

.Thesets

for which this is true are called Greenian sets. This chapter will culminate

in an important theorem of F. Riesz which states that a nonnegative super-

harmonic on a Greenian set has a decomposition into the sum of a Green

potential and a harmonic function.

3.2 Green Functions

In Chapter 1, the Green function for a ball led to a solution of the Dirichlet

problem for a measurable boundary function by way of the Poisson integral

formula. The Green function for the ball B = B

y,ρ

will be reexamined to

establish certain properties.

Lemma 3.2.1 G

is symmetric on B×B; moreover, for each x ∈ B, G

(x, ·)

is stricly positive on B, superharmonic on B,andlim

z→z

(x, z)=0for

all z

∈ ∂B.

L.L. Helms, Potential Theory, Universitext, 107

 Springer-Verlag London Limited 2009