Helms L.L. Potential Theory

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88 2 The Dirichlet Problem

Since the two extreme limits are the same,

lim sup

z→x,z∈Ω

(z) = lim sup

z→x,z∈Ω

u(z)=0.

Remark 2.6.30 It is worth noting an interesting fact in the preceding proof.

Namely, the fact that lim

z→y,z∈Ω

(αv(z) − u(z)) ≥ 0 for all y ∈ ∂Ω except

possibly for a single point was enough to prove that αv − u ≥ 0onΩ,as

though a single point is negligible. The proof was accomplished by exploiting

the fact that u

=+∞ at x. The same proof would have worked for more

than one point if a superharmonic function u could be found that takes on

the value +∞ on the set of points where lim

z→y,z∈Ω

(αv(z) − u(z)) ≥ 0 fails

to hold. This suggests that sets of points where a superharmonic function

takes on the value +∞ are negligible. Such sets are called polar sets and

will be taken up later.

Corollary 2.6.31 If Ω is an open subset of R

,n ≥ 2, then there is an

increasing sequence {Ω

} of regular bounded open sets with closures in Ω

such that Ω = ∪Ω

Proof: Let {Γ

} be an increasing sequence of compact sets such that Ω =

∪Γ

.ConsiderΓ

and a ﬁnite covering of Γ

by balls B

(1)

,...,B

(1)

having

closures in Ω. By slightly increasing the radii of some of the balls, it can

be assumed that each boundary point of Ω

= ∪

j=1

(1)

is the vertex of a

truncated closed solid cone lying outside Ω

; that is, it can be assumed that

is a regular region containing Γ

. Apply the same procedure to Ω

−

∪Γ

etc.

2.7 The Radial Limit Theorem

If h = PI(μ : B) on the ball B and μ is the indeﬁnite integral of a continuous

boundary function f relative to surface area σ on ∂B, then lim

z→x,z∈B

h(z)=

f(x) for all x ∈ ∂B. Since the Radon-Nikodym derivative of μ with respect to

σ is equal to fa.e.(σ), the boundary behavior of h is related to the derivative

dμ/dσ. Since there are several ways of deﬁning a derivate, diﬀerent boundary

limit theorems may be obtained depending upon the deﬁnition of the derivate

used. Generally speaking, a deﬁnition that places stringent conditions on μ

for the existence of the derivate will yield the strongest results. In this section,

a weak derivate will be used ﬁrst to describe the behavior of h along radial

approaches to the boundary of B.

As a matter of notational convenience, let B = B

0,1

. It is not essential

the 0 is the center of the ball or that the radius is equal to 1. If z ∈ ∂B and

0 ≤ α ≤ π/2,C

z,α

will denote a closed cone with vertex 0, axis coincident

with the vector z, and half-angle α;thatis,

2.7 The Radial Limit Theorem 89

z,α

= {y;cosα ≤

(y,z)

|y||z|

≤ 1} .

Each such cone determines a closed polar cap K

z,α

= C

z,α

∩∂B having center

z and spherical angle α.

Let μ be a signed measure of bounded variation on the Borel subsets of

∂B.

Deﬁnition 2.7.1 If lim

σ(K)→0

μ(K)/σ(K) exists, where K is any closed po-

lar cap centered at x, the limit is called the symmetric derivate of μ with

respect to σ at x and is denoted by D

μ(x).

The symmetric derivate is a weak derivate and results in the following

weak boundary limit theorem. Although the theorem is stated for harmonic

functions, versions of the theorem are true for superharmonic functions.

Littlewood [41] proved an a.e. radial limit theorem for potentials G

μ in

the n = 2 case, F. Riesz [53] extended Littlewood’s theorem to positive su-

perharmonic functions in the n = 2 case, and Privalov [51, 52] extended the

latter to the n ≥ 3case.

Theorem 2.7.2 Let μ be a signed measure of bounded variation on the Borel

subsets of ∂B and let h = PI(μ : B).IfD

μ exists at x ∈ ∂B,then

lim

λ→1

− h(λx)=D

μ(x).

Proof: (B = B

0,1

)Itwillbeshownﬁrstthata = D

μ(x),x ﬁxed, can be

assumed to be zero. Consider the signed measure μ − aσ which is also of

bounded variation. For this measure, D

(μ−aσ)(x)=D

μ(x)−aD

σ(x)=0

and PI(μ−aσ : B)=PI(μ : B)−aPI(σ : B)=h−a.Ifitcanbeshownthat

lim

λ→1

−

(h(λx) −a) = 0, it would follow that lim

λ→1

−

h(λx)=a. Therefore,

by replacing μ by μ − aσ, it can be assumed that a =0.Forx ∈ ∂B,

h(λx)=



∂B

1 −|λx|

|z − λx|

dμ(z)



∂B

1 − λ

|z − λx|

dμ(z).

If γ denotes the angle between λx and the point z ∈ ∂B,then

h(λx)=



∂B

1 − λ

(1 + λ

− 2λ cos γ)

n/2

dμ(z)



∂B

1 − λ

((1 − λ)

+4λ sin

(γ/2))

n/2

dμ(z).

With λ ﬁxed, the integrand depends only upon γ and a “zonal region” method

can be used to evaluate the integral. Letting α(γ)=μ(K

x,γ

), 0 ≤ γ ≤ π, α is

of bounded variation on [0,π]and

90 2 The Dirichlet Problem

h(λx)=



P (λ, γ) dα(γ)

where

P (λ, γ)=

1 − λ

((1 − λ)

+4λ sin

(γ/2))

n/2

, 0 <λ<1, 0 ≤ γ ≤ π.

Since lim

γ→0

μ(K

x,γ

)/σ(K

x,γ

)=0,α(0) = μ(K

x,0

) = 0. For each λ ∈

(0, 1),P(λ, ·) is positive, monotone decreasing on [0,π], and has a continuous

derivative P

(λ, ·)on[0,π]. Consider also the function β(γ)=σ(K

x,γ

), 0 ≤

γ ≤ π,forwhich

lim

γ→0

|α(γ)|

β(γ)

= lim

γ→0



μ(K

x,γ

σ(K

x,γ

)



= |D

μ(x)| =0.

Expressing the surface area of K

x,γ

as an integral and transforming to spher-

ical coordinates (r, θ),

β(γ)=σ(K

x,γ

)=σ

n−1



sin γ

n−2

(1 − r

)

1/2

dr ≥

n−1

n − 1

sin

n−1

and so

lim sup

γ→0

n−1

β(γ)

≤ lim sup

γ→0

n − 1

n−1



sin γ



n−1

n − 1

n−1

< +∞.

Choose >0 such that

lim sup

γ→0

n−1

β(γ)



and then choose φ ∈ (0,π) such that |α(γ)|≤β(γ)andγ

n−1

≤ β(γ)for

all γ ∈ [0,φ]. Integrating by parts and noting that P

(λ, γ) ≤ 0,



P (λ, γ) dα(γ)=α(φ)P(λ, φ) −



α(γ)P

(λ, γ) dγ

≤ α(φ)P(λ, φ) −



|α(γ)|P

(λ, γ) dγ

≤ α(φ)P(λ, φ) −



β(γ)P

(λ, γ) dγ

= P (λ, φ)(α(φ) −β(φ)) + 



P (λ, γ) dβ(γ)

≤ P (λ, φ)(|α(φ)|−β(φ)) + 



P (λ, γ) dβ(γ)

≤ 



P (λ, γ) dβ(γ).

2.7 The Radial Limit Theorem 91

Since

1=PI(σ : B)(λx)=



∂B

P (λ, γ) dσ(z)=



P (λ, γ) dβ(γ),

integrating by parts

1 ≥



P (λ, γ) dβ(γ)

= β(φ)P (λ, φ) −



β(γ)P

(λ, γ) dγ

≥ β(φ)P (λ, φ) −



n−1

(λ, γ) dγ

= P (λ, φ)(β(φ) −φ

n−1

)+(n − 1)



P (λ, γ)γ

n−2

dγ

≥ 



P (λ, γ)γ

n−2

dγ.

Hence,

h(λx)=



P (λ, γ) dα(γ)+



P (λ, γ) dα(γ)

≤





P (λ, γ) β(γ)+



P (λ, γ) dα(γ)



P (λ, γ) dβ(γ)

≤  +



P (λ, γ) dα(γ)





P (λ, γ)γ

n−2

dγ

Letting V

denote the total variation of α,

|h(λx)|≤ +



P (λ, γ) dV

(γ)





P (λ, γ)γ

n−2

dγ

≤  +

P (λ, φ)



(γ)





P (λ, γ)γ

n−2

dγ

≤  +

|μ|





P (λ,γ)γ

n−2

P (λ,φ)

dγ

92 2 The Dirichlet Problem

By Fatou’s Lemma, applied sequentially,

lim inf

λ→1

−



P (λ, γ)γ

n−2

P (λ, φ)

dγ ≥



lim inf

λ→1

−

P (λ, γ)γ

n−2

P (λ, φ)

dγ



sin

(φ/2)

n−2

sin

(γ/2)

dγ

=+∞,

and therefore lim sup

λ→1

−

|h(λx)|≤.Since can be taken arbitrarily small,

lim

λ→1

−

h(λx)=0.

Example 2.7.3 Consider B = B

0,1

and the function

f(x, y)=



1(x, y) ∈ ∂B,y ≥ 0

0(x, y) ∈ ∂B,y < 0.

Let μ(M)=



fdσfor each Borel set M ⊂ ∂B.ThenD

μ exists for (x, y) ∈

∂B and is equal to 1, 0, or 1/2accordingtoy>0,y <0, or y = 0, respectively.

Consider the harmonic function h = PI(μ : B). Then lim

λ→1

−

h(λx, λy)=

1, 0, or 1/2for(x, y) ∈ ∂B according to y>0,y < 0, or y = 0, respectively.

2.8 Nontangential Boundary Limit Theorem

Obtaining stronger results concerning the boundary behavior of harmonic

functions necessitates replacement of the symmetric derivate by a more re-

strictive derviate.

Deﬁnition 2.8.1 If lim

σ(K)→0

μ(K)/σ(K) exists, where K is any closed po-

lar cap containing x, the limit is called the derivate of μ with respect to σ

at x and is denoted by Dμ(x).

Proof of the existence of a derivate requires a discussion of the Vitali

covering theorem. Let X be a metric space with metric d.IfF ⊂ X,the

diameter of F is deﬁned to be d(F )=sup{d(x, y); x, y ∈ F }.IfF ⊂ X and

x ∈ X, the distance from F to x is denoted by d(F, x)=inf{d(y, x); y ∈ F }.

An -neighborhood of F ⊂ X is denoted by N(F, )={x; d(F, x) <}.

Deﬁnition 2.8.2 Let μ be a measure deﬁned on the Borel subsets of a com-

pact metric space X.AsetA ⊂ X is said to be covered in the sense of

Vitali by a family F of closed sets if each F ∈Fhas positive μ measure and

there is a constant α such that each point of A is contained in elements F ∈F

of arbitrarily small positive diameter for which μ(N(F, 3d(F )))/μ(F ) ≤ α.

Note that if A is covered in the sense of Vitali by F, then any subset of A

is also. A proof of the following theorem can be found in [18].

2.8 Nontangential Boundary Limit Theorem 93

Theorem 2.8.3 (Vitali Covering Theorem) Let μ be the completion of

a ﬁnite measure on the σ-algebra of Borel subsets of a compact metric space

X.IfA ⊂ X is covered in the sense of Vitali by a family F of closed sets,

then there is a sequence {F

} of disjoint sets in F such that μ(A ∼∪F

)=0.

The Vitali covering theorem will be applied in the following context. For X

take ∂B where B = B

0,ρ

⊂ R

. The family F will consist of all nondegenerate

closed polar caps K ⊂ ∂B. To show that F covers ∂B in the sense of Vitali,

it is necessary to estimate the surface area of polar caps.

Let K = K

x,α

, 0 <α≤ π/2, be a closed polar cap on ∂B.Makingfree

use of the fact that surface area is invariant under rotations of the sphere by

assuming that x =(0, 0....,0,ρ),

σ(K)=



···



+···+y

n−1

≤ρ

sin

(ρ

− y

−···−y

n−1

)

1/2

...dy

n−1

Introducing the spherical coordinates (r, θ)for(y

,...,y

n−1

σ(K)=ρσ

n−1



ρ sin α

n−2

(ρ

− r

)

1/2

dr.

It is easy to see that d(K)=2ρ sin α. There is also an α

with 0 <α

<π/2

such that for α ≤ α

σ(N(K, 3d(K)) ≤ ρσ

n−1



7ρ sin α

n−2

(ρ

− r

)

1/2

= ρσ

n−1



ρ sin α

n−2

(ρ

− 49r

)

1/2

dr.

Therefore,

σ(N(K, 3d(K))

σ(K)

≤

n−1



ρ sin α

n−2

/(ρ

− 49r

)

1/2

) dr



ρ sin α

n−2

/(ρ

− r

)

1/2

) dr

Considering the integrals as functions of their upper limits and applying the

second mean value theorem of the diﬀerential calculus, there is a point ξ with

0 <ξ<ρsin α such that the quotient on the right is equal to

n−1

(ξ

n−2

/(ρ

− 49ξ

)

1/2

)

n−2

/(ρ

− ξ

)

1/2

Therefore,

σ(N(K, 3d(K)))

σ(K)

≤ 7

n−1

(ρ

− ξ

)

1/2

(ρ

− 49ξ

)

1/2

94 2 The Dirichlet Problem

As α → 0,ξ → 0 and the right member has the limit 7

n−1

. This shows that

there is a constant α



with 0 <α



<π/2 such that

σ(N(k, 3d(K)))

σ(K)

≤ constant

whenever K = K

x,α

is a closed polar cap with spherical angle α<α



.There-

fore, the family F of closed polar caps covers ∂B in the sense of Vitali. Unless

stated otherwise, K will denote a closed polar cap in ∂B where B = B

0,ρ

Lemma 2.8.4 Let μ be a measure on the Borel subsets of ∂B,andlet

0 <r<+∞.

(i) If for each x in a set A ⊂ ∂B

lim inf

σ(K)→0,x∈K

μ(K)

σ(K)

<r,

then each neighborhood of A contains an open set O such that σ(A ∼ O)=

0 and μ(O) <rσ(O).

(ii) If for each x in a set A ⊂ ∂B

lim sup

σ(K)→0,x∈K

μ(K)

σ(K)

>r,

then each neighborhood of A contains a Borel set A

such that σ(A ∼

)=0and μ(A

) >rσ(A

Proof: To prove (i), let U be any open set with A ⊂ U ⊂ ∂B,andletF be

the family of all closed polar caps K ⊂ U satisfying μ(K) <rσ(K). Then

F covers A in the sense of Vitali. By the Vitali covering theorem, there is a

sequence {K

} of disjoint closed polar caps in F such that σ(A ∼∪K

)=0.

Let K

◦

be the interior, relative to ∂B,ofK

.Sinceσ(K

∼ K

◦

)=0,σ(A ∼



◦

)=0and

μ(K

◦

) ≤ μ(K

) <rσ(K

)=rσ(K

◦

Letting O =



◦

, O is open, O ⊂ U,andσ(A ∼ O)=0.Moreover,

μ(O)=

∞



j=1

μ(K

◦

) <r

∞



j=1

σ(K

◦

)=rσ(O).

To prove (ii), let U be an open set such that A ⊂ U ⊂ ∂B.Byhypothesis,

the family F of closed polar caps K ⊂ U satisfying μ(K) >rσ(K)covers

A in the sense of Vitali. By the Vitali covering theorem, there is a sequence

} of disjoint closed polar caps in F such that σ(A ∼



)=0.Let



.Since

2.8 Nontangential Boundary Limit Theorem 95

μ(A

∞



j=1

μ(K

) >r

∞



j=1

σ(K

)=rσ(A

satisﬁes the requirements of (ii).

It is assumed throughout the remainder of this section that σ is a complete

measure.

Lemma 2.8.5 Let μ be a measure on the Borel subsets of ∂B such that

Dμ(x) = lim

σ(K)→0,x∈K

μ(K)/σ(K) exists for almost all (σ) points x ∈ ∂B,

and deﬁne Dμ arbitrarily at points where the limit does not exist. Then Dμ

is Lebesgue measurable on ∂B.

Proof: For each positive integer m,let{k

,...,k

} be a ﬁnite collection

of closed polar caps which covers ∂B and σ(K

) < 1/m, j =1,...,j

Deﬁne a sequence of functions {f

} on ∂B as follows. Let K

= ∅.If

x ∈ K

∼∪

j−1

i=1

,j =1,...,j

,letf

(x)=μ(K

)/σ(K

). Then

each f

is measurable and the sequence {f

} converges a.e.(σ)toDμ.

Theorem 2.8.6 (Lebesgue) Let μ be a signed measure of bounded varia-

tion on the Borel subsets of ∂B.ThenDμ(x) exists for almost all (σ) points

x ∈ ∂B and is Lebesgue measurable.

Proof: By decomposing μ into its positive and negative variations, it can be

assumed that μ is a measure. Let A be the set of points x ∈ ∂B for which

lim sup

σ(K)→0,x∈K

μ(K)

σ(K)

> lim inf

σ(K)→0,x∈K

μ(K)

σ(K)

If i is a nonnegative integer and j is a positive integer, let A

be the set of

x ∈ ∂B where

lim sup

σ(K)→0,x∈K

μ(K)

σ(K)

i +1

> lim inf

σ(K)→0,x∈K

μ(K)

σ(K)

Suppose σ

∗

(A) > 0whereσ

∗

is the outer measure on ∂B induced by σ.Since

A =



,σ

∗

) > 0forsomei and j.Given>0, there is an open set U

such that A

⊂ U and σ(U ) <σ

∗

)+. Applying (i) of Lemma 2.8.4 with

r replaced by i/j, there is an open set O such that O ⊂ U, σ(A

∼ O)=0,

and

μ(O) <

σ(O) <

(σ

∗

)+).

Applying (ii) of Lemma 2.8.4 to the set A

∩O,thereisaBorelsetM ⊂ O

such that σ((A

∩ O) ∼ M)=0andμ(M) > ((i +1)/j)σ(M). Since

⊂ M ∪ (A

∼ O) ∪ ((A

∩ O) ∼ M )

96 2 The Dirichlet Problem

and

σ(A

∼ O)=0=σ((A

∩O) ∼ M ),

σ(M) ≥ σ

∗

). Thus,

(σ

∗

)+) >μ(O) ≥ μ(M ) >

i +1

σ(M) ≥

i +1

∗

Letting  → 0, (i/j)σ

∗

) ≥ ((i +1)/j)σ

∗

), a contradiction. Therefore,

σ(A)=0andDμ(x) = lim

σ(K)→0,x∈K

μ(K)/σ(K) exists as a ﬁnite or inﬁnite

limit for all x ∈ A.LetZ = {x ∈∼ A; Dμ(x)=+∞}.By(ii) of Lemma 2.8.4,

for each positive integer j there is a Borel set Z

⊂ ∂B such that σ(Z ∼

)=0andμ(Z

) >jσ(Z

). Since σ

∗

(Z) ≤ σ

∗

(Z ∩ Z

)+σ

∗

(Z ∼ Z

∗

(Z ∩ Z

) ≤ σ

∗

+∞ >μ(∂B) ≥ μ(Z

) >jσ(Z

) ≥ jσ

∗

(Z)

for all positive integers j. Therefore, σ

∗

(Z)=0andDμ<+∞a.e.(σ)on

∼ A. The measurability of Dμ follows from the preceding lemma.

The proof of the principal boundary limit theorem requires the exclusion

of null sets of boundary points in addition to those where Dμ is not deﬁned.

Theorem 2.8.7 (Lebesgue) Let μ be a signed measure of bounded varia-

tion on the Borel subsets of ∂B.

(i) If f is an integrable function on ∂B and μ(M)=



fdμfor each Borel

set M ⊂ ∂B,thenDμ = fa.e.(σ).

(ii) If μ is singular relative to σ,thenDμ =0a.e.(σ).

Proof: Suppose μ(M)=



fdσwith f integrable. It can be assumed that

f ≥ 0 by decomposing f into its positive and negative parts, if necessary.

Let A = {x ∈ ∂B; f(x) > D μ(x)} and assume that σ(A) > 0. There are then

positive numbers α and β such that A

αβ

= {x ∈ ∂B; f(x) >α>β>Dμ(x)}

and σ(A

αβ

) > 0. Since σ is a regular Borel measure, there is a compact set

Γ ⊂ A

αβ

with σ(Γ ) > 0. Thus,

μ(Γ )=



fdσ≥ ασ(Γ ). (2.8)

Since Dμ(x) = lim

σ(K)→0,x∈K

μ(K)/σ(K) <βfor all x ∈ Γ , each neighbor-

hood of Γ contains an open set O such that μ(O) <βσ(O)andσ(Γ ∼ O)=0

by part (i) of Lemma 2.8.4. Since σ(Γ ∼ O)=0andμ is the indeﬁnite

integral of an integrable function, a small open set can be adjoined to O

so that Γ ⊂ O and μ(O) <βσ(O). Since Γ is the intersection of a se-

quence of such open sets O, μ(Γ ) ≤ βσ(Γ ) <ασ(Γ ) ≤ μ(Γ ), the latter

inequality holding by Inequality (2.8), a contradiction. Therefore, σ(A)=0.

In a similar manner, using part (ii) of Lemma 2.8.4, it can be shown that

σ({x ∈ ∂B; f(x) < Dμ(x)}) = 0. Therefore, Dμ = fa.e.(σ). Suppose now

2.8 Nontangential Boundary Limit Theorem 97

that μ is singular relative to σ.Sinceμ = μ

− μ

−

and the positive and

negative variations of μ are singular relative to σ, it can be assumed that μ

is a measure and Dμ ≥ 0. There is then a measurable set M ⊂ ∂B such that

σ(M)=0andμ(∂B ∼ M )=0.LetP = {x ∈ ∂B; Dμ(x) > 0} and assume

that σ(P) > 0. Since Dμ is measurable and σ(M)=0,thereisan>0and

acompactsetΓ ⊂ P ∼ M such that σ(Γ) > 0andDμ(x) >for x ∈ Γ .

By (ii) of Lemma 2.8.4, each neighborhood Γ has μ measure greater than

σ(Γ). Since Γ is the intersection of such neighborhoods, μ(Γ ) ≥ σ(Γ ) > 0;

but this contradicts the fact that 0 ≤ μ(Γ ) ≤ μ(P ∼ M) ≤ μ(∂B ∼ M)=0.

Therefore, σ(P )=0andDμ =0a.e.(σ).

Points x ∈ ∂B satisfying the equation of the following theorem are called

Lebesgue points.

Theorem 2.8.8 (Lebesgue) Let f be a measurable function on ∂B which is

integrable relative to σ,andletμ(M)=



fdσfor each Borel set M ⊂ ∂B.

Then

lim

σ(K)→0,x∈K

σ(K)



|f(y) − f (x)|dσ(y)=0

a.e.(σ) on ∂B.

Proof: Let {r

} be an enumeration of the rationals. By the preceding the-

orem, for each i ≥ 1thereisasetZ

⊂ ∂B with σ(Z

) = 0 such that for

x ∈ Z

lim

σ(K)→0.x∈K

σ(K)



|f(y) − r

|dσ(y)=|f(x) − r

Then Z =



has surface area zero. Let x ∈ ∂B ∼ Z,let>0, and choose

such that |f(x) − r

| </2. Then

lim

σ(K)→0,x∈K

σ(K)



|f(y) − f (x)|dσ(y)

≤ lim sup

σ(K)→0,x∈K

σ(K)



(|f(y) − r

|+ |r

− f (x)|) dσ(y)

≤ lim sup

σ(K)→0,x∈K

σ(K)



|f(y) − r

|dσ(y)+



= |f(x) − r

| +



<.

It is not diﬃcult to construct examples of harmonic functions h on ∂B ⊂

that behave badly in a neighborhood of a point z ∈ ∂B. In fact, suppose

h = PI(δ

: B)onB = B

0,ρ

,whereδ

is a unit measure supported by the

singleton {z}.Forx ∈ B,

h(x)=

2πρ

−|x|

|z −x|