Helms L.L. Potential Theory

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128 3 Green Functions

therefore that U

is superharmonic on R

whenever μ has compact support.

The potentials U

and G

μ can be compared in a sense.

Theorem 3.4.10 If Ω ⊂ R

,n ≥ 2,isGreenianandμ is a measure with

compact support, then U

is superharmonic and diﬀers from G

μ on Ω by

a harmonic function.

Proof: Letting G

(x, ·)=u

+ h

μ(x)=



(y) dμ(y)+



(y) dμ(y),x∈ R

Since h

is bounded on the support of μ,



(y) dμ(y) < +∞ and the second

term of the above equation is harmonic on Ω by Theorem 1.7.13. Thus,

μ = U

+ h with h harmonic on Ω.

As an application of the results of this section, a region Ω ⊂ R

having an

irregular boundary point will be constructed using a potential function.

Example 3.4.11 (Lebesgue Spine[37, 38]) Consider a measure concen-

trated on the line segment {(x, 0, 0); 0 ≤ x ≤ 1} andhavingadensity

f(x, 0, 0) = x, 0 ≤ x ≤ 1. The Newtonian potential u(x, y, z)ofthismea-

sure at a point not on the x-axis is given by

u(x, y, z)=



((ξ − x)

+ y

+ z

)

1/2

dξ



(ξ − x)

((ξ − x)

+ y

+ z

)

1/2

dξ + x



((ξ − x)

+ y

+ z

)

1/2

dξ.

Using elementary calculus, the integrals can be evaluated to obtain

u(x, y, z) = ((1 − x)

+ y

+ z

)

1/2

− (x

+ y

+ z

)

1/2

+x log (1 − x + ((1 − x)

+ y

+ z

)

1/2

)

−x log (−x +(x

+ y

+ z

)

1/2

)

= ((1 − x)

+ y

+ z

)

1/2

− (x

+ y

+ z

)

1/2

+x log (1 − x + ((1 − x)

+ y

+ z

)

1/2

)

+x log (x +(x

+ y

+ z

)

1/2

) − x log (y

+ z

the last term arising as a result of multiplying and dividing the expression

−x +(x

+ y

+ z

)

by x +(x

+ y

+ z

)

. It is easily seen that u has the

limit 1 as (x, y, z) → (0, 0, 0) subject to the condition that y

+ z

> 0, has

the limit 0 as |(x, y, z)|→+∞, and has the limit +∞ as (x, y, z) approaches

any point on the line segment {(x, 0, 0); 0 <x<1}. Consider the open set

Ω = {(x, y, z); <u(x, y, z) < 1+c

3.4 Green Potentials 129

where 0 <<1andc

> 1. The set {(x, y, z); u(x, y, z) >} is a neighbor-

hood of (0, 0, 0) and the set {(x, y, z); u(x, y, z) < 1+c

} is an open set which

excludes the line segment {(x, 0, 0); 0 <x<1}. It is easily seen that

lim

|(x,y,z)|→0

u(x, y, z)=1− lim

|(x,y,z)|→0

x log ρ

where ρ

= y

+ z

.If(x, y, z) → (0, 0, 0) along a path for which x>0,y

=0andx

= ρ

for some positive integer k,thenu(x, y, z) → 1; note

also that such points will be in Ω if suﬃciently close to (0, 0, 0). If, however,

(x, y, z) → (0, 0, 0) along a path for which x>0,y

+ z

=0andρ =

exp (−c/2x)withc>c

,thenu(x, y, z) → 1+c; note also that such points

will be in ∼ Ω if suﬃciently close to (0, 0, 0). It follows that (0, 0, 0) is a

boundary point for Ω. It remains only to show that (0, 0, 0) is an irregular

boundary point for the region Ω. The function u, as the Newtonian potential

of a measure with compact support, is harmonic on the open set Ω. As such,

it is the solution of the Dirichlet problem for the boundary function which is

equal to  on the boundary of the region u>and equal to 1 + c

on the

boundary of the region u<1+c

.Forx<0,

u(x, 0, 0) =



ξ − x

dξ =1+x log (1 − x) −x log |x|

so that lim

x→0−

u(x, 0, 0) = 1 < 1+c

In general, little can be said about the continuity of a Green potential

μ unless the measure μ is absolutely continuous with respect to Lebesgue

measure. If Ω is a Greenian set and f is a measurable function on Ω,let

f(x)=



(x, y)f(y) dy

provided the integral is deﬁned for all x ∈ Ω.

Theorem 3.4.12 If Ω is a Greenian set and f is a bounded Lebesgue inte-

grable function on Ω,thenG

f is continuous on Ω.

Proof: Since the positive and negative parts f

and f

−

areintegrableover

Ω and G

f = G

− G

−

, it can be assumed that 0 ≤ f ≤ m for some

constant m. By Theorem 3.4.4, G

f is a potential. Let B = B

y,ρ

be a ball

with B

−

⊂ Ω of radius less that 1/2. Then G

f = G

fχ

+ G

fχ

Ω∼B

Since the density fχ

Ω∼B

assigns zero measure to B,G

fχ

Ω∼B

is harmonic

on B by Theorem 3.4.7 and therefore continuous on B.ConsidernowG

fχ

If it can be shown that G

fχ

is continuous on B ∼{y}, it would follow that

f is continuous on B ∼{y}; but since B = B

y,ρ

is an arbitrary ball with

closure in Ω, this would imply that G

f is continuous on Ω.SinceG

fχ

diﬀers from G

f by a function harmonic on B by Theorem 3.4.8, it suﬃces

to prove that G

f is continuous on B ∼{y}. It is necessary to consider the

130 3 Green Functions

n =2andn ≥ 3 cases separately for the rest of the proof. Suppose n =2

and x

∈ B ∼{y} and consider any ball B

with closure in B.Since

(x, z)=log

|y − x|

|z − x

∗

|z − x|

for z ∈ B ∼{x},wherex

∗

is the inverse of x relative to ∂B,

f(x) − G

f(x

≤





B∼B

log

|y −x|

|y − x

|z − x

∗

|z − x|

|z − x

|z −x

∗

f(z) dz





(x, z)f(z) dz +



,z)f(z) dz.

Recall that G

(x, z) is minimal in the class of nonnegative functions of the

form −log |x − z| + w

(z)wherew

is harmonic on B.Since|x − z| < 1for

x, z ∈ B, −log |x − z| is positive and majorizes G

(x, z). Thus for any x ∈ B,



(x, z)f(z) dz ≤ m



−log |x − z|dz.

The integral on the right becomes larger if taken over B

x,r

rather than B

Thus,



(x, z)f(z) dz ≤ m



x,r

−log |x − z|dz

=2πm



−ρ log ρdρ.

Since the latter integral approaches zero as r → 0, given >0, r

can be

chosen so that r

< |x

− y| and



(x, z)f(z) dz +



,z)f(z) dz < 

for any x ∈ B. It follows that

f(x) − G

f(x

≤





B∼B

log

|y −x|

|y −x

|z − x

∗

|z − x|

|z − x

∗

f(z) dz



+ .

Consider only those x ∈ B

. Since the logarithmic factor in the in-

tegral approaches zero boundedly as x → x

, by the Lebesgue dominated

convergence theorem,

3.4 Green Potentials 131

lim sup

x→x

,x∈B

f(x) − G

f(x

)|≤.

Since  is arbitrary, lim

x→x

,x∈B

f(x)=G

f(x

)andG

f is continuous

on B ∼{y}. The proof of the n ≥ 3 case is just the same except for the fact

that G

(x, z) ≤|x − z|

−n+2

and the estimate



(x, z)f(z) dz ≤ σ



ρdρ.

A digression into approximation theory is necessary in order to simplify

later proofs.

Lemma 3.4.13 If φ is a ﬁnite-valued, continuous function on the Greenian

set Ω withcompactsupportinaclosedballB

−

⊂ Ω and B

is a ball with

−

⊂ B

−

⊂ Ω, then there is a sequence of diﬀerences {u

− v

} of

ﬁnite-valued, continuous potentials on Ω supported by measures in B

such

that u

= v

on Ω ∼ B

and lim

j→∞

− v

)=φ uniformly on Ω.

Proof: (n ≥ 3) Put φ(x)=0forx ∈ R

∼ Ω, B = B

x,δ

and B

= B

x,δ

Consider any j

≥ 1forwhich1/j

< (δ

− δ)/4andtheballB

x,δ

where

δ =(δ

+ δ)/2. For j ≥ j

, deﬁne J

1/j

φ to be zero on Ω ∼ B

.Then

1/j

φ}

j≥j

is a sequence in C

∞

) which converges uniformly to φ on Ω.

By Theorem 1.5.3, for x ∈ B

1/j

φ(x)=−

(n − 2)



(x, z)(ΔJ

1/j

φ)(z) dz

(n − 2)



(x, z)(ΔJ

1/j

φ)

−

(z) dz

−

(n − 2)



(x, z)(ΔJ

1/j

φ)

(z) dz

(n − 2)

(ΔJ

1/j

φ)

−

(n − 2)

(ΔJ

1/j

φ)

The functions on the right are potentials of measures having supports in B

x,δ

and are continuous by Theorem 3.4.12. By Theorem 3.4.8,

1/j

φ =

(n − 2)

(ΔJ

1/j

φ)

−

(n − 2)

(ΔJ

1/j

φ)

+ h

(3.5)

where h

is harmonic on B

.SinceJ

1/j

φ =0onΩ ∼ B

x,δ

and G

(ΔJ

1/j

φ)

are harmonic on Ω ∼ B

x,δ

has a harmonic extension to Ω. Letting

(n − 2)

(ΔJ

1/j

φ)

−

and v

(n − 2)

(ΔJ

1/j

φ)

132 3 Green Functions

1/j

φ = u

−v

+ h

.SinceJ

1/j

φ =0onΩ ∼ B

x,δ

−u

= h

on Ω ∼ B

x,δ

Thus, v

= u

≥ h

on Ω ∼ B

x,δ

and, in particular, v

≥ h

on ∂B

x,δ

.By

the minimum principle, v

≥ h

on B

x,δ

and therefore v

≥ h

on Ω.Since

the greatest harmonic minorant of v

on Ω is the zero function, h

≤ 0onΩ.

A similar argument applied to u

shows that h

≥ 0. Thus, h

=0onΩ and

1/j

φ = u

− v

. Since the sequence {J

1/j

φ} converges uniformly to φ,the

same is true of the sequence {u

−v

}. Finally, it follows from Equation (3.5)

that u

= v

on Ω ∼ B

since J

1/j

φ =0onΩ ∼ B

and h

=0onΩ.

The result of the following theorem is known as a partition of unity.

The proof is taken from [48].

Theorem 3.4.14 (L. Schwartz [55]) Let Γ be a compact subset of R

and

let {U

,...,U

} be an open covering of Γ .Forj =1,...,m,thereisa

∈ C

∞

) with 0 ≤ φ

≤ 1 such that

x ∈ Γ ⇒



j=1

(x)=1.

Proof: For j =1,...,m,letV

be an open subset of U

with compact closure

−

⊂ U

such that {V

,...,V

} is an open covering of Γ .Then

dist(Γ, ∼∪

j=1

) > 0,

and for each j =1,...,m,dist(V

−

,∂U

) > 0. Let V

m+1

=∼∪

j=1

−

and

choose δ>0 such that

dist(Γ, ∼∪

j=1

) >δ

dist(V

−

,∂U

) >δ, j=1,...m.

Each V

,j =1,...,m+1, can be covered by balls with centers y

∈ V

−

,i≥ 1,

of radius δ so that at most ﬁnitely many of the balls intersect any compact

set. Choosing the constant c of Section 2.5 so that m(0) = 1, let

(x)=





x − y



,j=1,...,m+1.

Then each ψ

∈ C

∞

)and

x ∈ Γ ⇒



j=1

(x) =0,ψ

m+1

(x)=0.

Letting

(x)=

(x)



j=1

(x)

,x∈ R

,j =1,...,m,

the functions φ

,...,φ

satisfy the above requirements.

3.4 Green Potentials 133

Theorem 3.4.15 If φ is a ﬁnite-valued, continuous function on R

having

compact support Γ in the Greenian set Ω and O is a neighborhood of Γ with

compact closure O

−

⊂ Ω, then there is a sequence of diﬀerences {u

− v

}

of ﬁnite-valued, continuous potentials of measures having support in O such

that u

= v

on Ω ∼ O

−

and lim

j→∞

− v

)=φ uniformly on Ω.

Proof: Let B

,...,B

be a ﬁnite collection of balls covering Γ with ∪

j=1

−

⊂ O. Consider a partition of unity ψ

,...,ψ

where each ψ

∈ C

∞

)has

compact support in B

and



j=1

(x)=1forx ∈ Γ .Thenφ =



j=1

φψ

where each φψ

is ﬁnite-valued and continuous with compact support in B

For j =1,...,m, there is a sequence of diﬀerences {u

−v

} of ﬁnite-valued,

continuous potentials such that u

= v

on Ω ∼ B

⊃ Ω ∼∪

j=1

⊃ Ω ∼

−

, and lim

i→∞

− v

)=φψ

uniformly on Ω.Thus,

lim

i→∞

⎛

⎝



j=1

−



j=1

⎞

⎠

= φ

uniformly on Ω and



j=1



j=1

on Ω ∼ O

−

Generally speaking, the boundary behavior of a potential on a region hav-

ing a Green function reﬂects the boundary behavior of the Green function.

Theorem 3.4.16 If μ is a measure with compact support Γ in an open ball

B,thenlim

x→x

,x∈B

μ(x)=0for x

∈ ∂B.

Proof: Let B = B

y,ρ

and let z

be a ﬁxed point of Γ .AlsoletΛ be a con-

nected neighborhood of Γ such that Λ

−

⊂ B.Forx ∈ B ∼ Λ

−

, the functions

(x, ·) are strictly positive harmonic functions on Λ. By Harnack’s inequal-

ity, Theorem 2.2.2, there is a constant k (depending upon Λ and Γ ) such that

(x, z) ≤ kG

(x, z

) for all x ∈ B ∼ Λ

−

,z ∈ Γ .SinceG

(x, z

) → 0as

x → x

∈ ∂B by Lemma 3.2.1, G

(x, z) → 0asx → x

∈ ∂B uniformly for

z ∈ Γ . Since the integrand in the equation G

μ(x)=



(x, z) dμ(z)can

be made uniformly small as x → x

, lim

x→x

μ(x)=0.

The requirement that the support of μ be compact can be relaxed provided

it has a density.

Lemma 3.4.17 Let Ω be a Greenian connected set with ﬁnite Lebesgue mea-

sure if n ≥ 3 and bounded if n =2,andletx

∈ ∂Ω.Iff is a bounded

measurable function on Ω and lim

x→x

,x∈Ω

(x, z

)=0for some z

∈ Ω,

then lim

x→x

,x∈Ω

f(x)=0.

Proof: Suppose n ≥ 3. By consideration of the positive and negative parts

of f , it can be assumed that 0 ≤ f ≤ m.Sincef is integrable on Ω, G

is a potential by Theorem 3.4.4. Now let {Γ

} be an increasing sequence of

compact subsets of Ω such that Ω = ∪

and let

(x)=



(x, z)f(z) dz.

134 3 Green Functions

Then φ

↑ G

f by the Lebesgue monotone convergence theorem. Given

>0, choose j

such that the Lebesgue measure λ

of Ω ∼ Γ

is less than 

and let α denote the Lebesgue measure of Ω.SinceG

(x, z) is majorized by

|x − z|

−n+2

|φ

(x) − φ

(x)|≤



∼Γ

|x − z|

n−2

f(z) dz

≤ m



Ω∼Γ

|x − z|

n−2

dz.

Now choose ρ such that the volume of B

x,ρ

is λ

;thatis,λ

= ν

ρ =(λ

/ν

)

1/n

.Then

|φ

(x) − φ

(x)|≤m



Ω∼Γ

|x − z|

n−2

≤ m



x,ρ

|x − z|

n−2

mρ

)

2/n

≤



)

2/n

uniformly with respect to x ∈ Ω and j ≥ 1. As in the preceding proof,

lim

x→x

,x∈Ω

(x, z) = 0 uniformly with respect to z ∈ Γ

and there is a

δ>0 such that G

(x, z) <for all z ∈ Γ

and all x satisfying |x −x

| <δ;

for such x,

|φ

(x)|≤



(x, z)f(z) dz ≤ mα.

Thus, if |x − x

| <δand j ≥ 1, then

|φ

(x)|≤|φ

(x) − φ

(x)| + |φ

(x)|

≤



)

2/n

+ mα.

Letting j →∞,

f(x)|≤



)

2/n

+ mα;

but since >0 is arbitrary and the constants on the right do not depend upon

 and x, G

f(x) can be made arbitrarily small by taking x suﬃciently close

to x

. The proof in the n = 2 case is precisely the same except that G

(x, z)

need not be majorized by −log |x − z|. Assuming that Ω is bounded, there is

a constant c such that log (c/|x − z|) is positive on Ω and therefore majorizes

(x, z).

3.5 Riesz Decomposition 135

3.5 Riesz Decomposition

It will be shown in this section that a nonnegative superharmonic function can

be represented as the sum of a potential and a harmonic function. A potential

theory version of Tonelli’s theorem will be proved ﬁrst.

Theorem 3.5.1 (Reciprocity Theorem) Let Ω be a Greenian subset of

,andletμ, ν be two measures on the Borel subsets of Ω.Then



Gμdν =



Gνdμ.

Proof: By Tonelli’s theorem and the symmetry of the Green function,



Gμ(x) dν(x)=







G(x, y) dμ(y)



dν(x)







G(x, y) dν(x)



dμ(y)







G(y,x) dν(x)



dμ(y)



Gν(y) dμ(y).

Note that the reciprocity theorem holds even if Gμ is not integrable relative

to ν,inwhichcasethesameistrueofGν relative to μ.

A digression is necessary in order to develop some tools for showing that

two measures are equal if they produce the same potential function. K

will

denote the subset of C

(Ω) consisting of nonnegative, continuous functions

with compact support.

Deﬁnition 3.5.2 A subset K

⊂ K

is called total if for each v ∈ K

with compact support Γ , each neighborhood Λ of Γ ,andeach>0there

corresponds a ﬁnite linear combination f =



j=1

of elements of K

with α

> 0,f

∈ K

,andf

having compact support in Λ for each j such

that

v −f =sup

x∈Ω

|v(x) − f(x)| <.

If it is necessary to consider more than one region, the notation K

(Ω) will

be used.

Theorem 3.5.3 (Cartan [10]) Let K

be a subset of K

with the follow-

ing properties:

(i) If f ∈ K

, then every translate of f is in K

136 3 Green Functions

(ii) If x ∈ Ω and δ>0, then there is an f ∈ K

that vanishes outside B

x,δ

but does not vanish identically.

Then K

is a total subset of K

Proof: Without loss of generality, it can be assumed that 0 ∈ Ω.Letv be an

element of K

with compact support Γ .Ifv is identically zero and >0, then

a scalar multiple of any function satisfying (ii) will approximate v uniformly

on Ω within . It therefore can be assumed that v is not identically zero. Let

Λ be a neighborhood of Γ with Λ

−

⊂ Ω,letλ = d(Γ, Ω ∼ Λ) > 0, and let

>0. Since v is uniformly continuous on Ω,thereisaδ<λ/2 such that

|v(x) − v(y)| </2 whenever |x − y| <δ.Thend(Γ, Ω ∼ Λ) > 2δ.By(ii),

there is an f

∈ K

that vanishes outside B

0,δ

, does not vanish identically,

and



(z) dz =1.Sincef

is uniformly continuous, there is a ρ>0such

that

(x) − f

(y)| <





v(z) dz

whenever |x − y| <ρ. It can be assumed that δ + ρ<λ. Moreover, there

is a sequence {M

} of disjoint measurable sets such that R

= ∪

,diam

<ρ, and thus δ+diamM

<δ+ ρ<λ. It can be assumed that

only ﬁnitely many of the M

have a nonempty intersection with Γ . Such a

sequence of M

’s can be constructed by covering R

by a sequence of balls

each of diameter less than ρ with only a ﬁnite number intersecting Γ and

then forming a disjoint sequence of measurable sets from the sequence of

balls. Deﬁne the real-valued function h by

h(x)=



(x − y)v(y) dy, x ∈ R

Then

|h(x) − v(x)|≤



x,δ

(y)|v(y) − v(x)|dy ≤





x,δ

(y) dy =



for all x ∈ R

;thatis,h approximates v uniformly within /2. It will now

be shown that h can be approximated uniformly within /2 by a ﬁnite linear

combination of translates of f

.Foreachj ≥ 1, choose y

∈ M

.Thenfor

any x ∈ R



h(x) −



(x − y

)



v(y) dy





(x − y)v(y) dy −



(x − y

)



v(y) dy







(x − y) − f

(x − y

)v(y) dy



3.5 Riesz Decomposition 137

Since diam M

<ρ,|(x − y) − (x − y

)| = |y

− y| <ρfor y ∈ M

and the

ﬁrst factor in the integral is bounded in absolute value by /(2



v(y) dy).

Therefore,



h(x) −



(x − y

)



v(y) dy



≤





v(y) dy





v(y) dy =



This shows that h is approximated uniformly within /2 by the function

f(x)=



(x − y

)



v(y) dy

so that f approximates v uniformly within . Note that the sum in the last

equation is a ﬁnite sum since only a ﬁnite number of the M

have a nonempty

intersection with the support of v. It remains only to show that each function

(x)=f

(x − y

)



v(y) dy

has support in Λ provided it is not identically zero. Suppose M

∩Γ = ∅ and

(x−y

) =0.Then|x−y

| <δand d(x, Γ ) ≤ d(x, y

)+d(y

,Γ) <δ+ρ<λ.

Since λ = d(Γ, Ω ∼ Λ),x ∈ Ω ∼ Λ. This means that the support of those g

that are not identically zero is contained in Λ.

Theorem 3.5.4 If Ω is an open subset of R

,μ

and μ

are measures on

Ω,K

is a total subset of K

,and



fdμ



fdμ

for all f ∈ K

,then

= μ

Proof: Since any g ∈ K

can be approximated uniformly by a ﬁnite linear

combination of elements of K

, the above equation holds for all f ∈ K

.Let

Γ be a compact subset of Ω and let χ

be its indicator function. Then there

is a sequence {φ

} in K

such that φ

↓ χ

with 0 ≤ φ

≤ 1; for example,

let φ

(x)=max(2e

−jd(x,Γ)

− 1, 0). Since



dμ



dμ

,j ≥ 1,

(Γ )=



dμ



dμ

= μ

(Γ )

by the Lebesgue dominated convergence theorem. Hence, μ

= μ

since both

are regular Borel measures.

The following examples will be needed to construct total sets of continuous

functions.

Example 3.5.5 (n=2) Let Ω be a Greenian set, let B

−

x,δ

⊂ Ω, δ > 0, and

let ν

x,δ

be a unit measure uniformly distributed on ∂B

x.δ

. Letting

x,δ



∂B

x,δ

log

|y −z|

dν

x,δ

(z),y∈ R