Helms L.L. Potential Theory

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258 6Energy

Similarly,

|μ − γ|

= |μ − β|

|α − β|

+[μ − β, β − α]

Therefore,

2|μ − γ|

= |μ − α|

+ |μ − β|

|α − β|

+[β − α, α − β]

= |μ − α|

+ |μ − β|

−

|α − β|

(6.4)

so that

|α − β|

=2|μ − α|

+2|μ − β|

− 4|μ − γ|

Let m =inf

ν∈F

|μ − ν|

.If{ν

} is any sequence in F,then

|ν

− ν

≤−4m

+2|μ − ν

Choosing a sequence {ν

} in F such that |μ − ν

↓ m, lim sup

j,k→∞

|ν

−

≤ 0 and it follows that the sequence {ν

} is strong Cauchy in F.Since

is complete and F is closed, there is an α ∈Fsuch that

|μ − α|

= m =inf

ν∈F

|μ − ν|

Uniqueness of a measure minimizing |μ − ν|

can be shown as follows. If α

and β are two measures satisfying |μ − α|

= |μ − β|

= m,then

≤ 2m

−

|α − β|

by Equation (6.4) and α = β.

Deﬁnition 6.4.2 If F is a nonempty, closed, convex subset of E

and μ ∈

, the measure μ

of the preceding lemma is called the projection of μ

onto F.

Lemma 6.4.3 Let F be a nonempty, closed, convex subset of E

and let

μ ∈E

.Thenμ

is the projection of μ on F if and only if [μ−μ

,ν−μ

]

≤ 0

for all ν ∈F.

Proof: If α ∈F,then|μ − α|

≤|μ − ν|

for all ν ∈Fif and only if

2[μ − α, ν − α]

≤|α − ν|

for all ν ∈F. It now will be shown that the latter inequality is true if and only

if 2[μ −α, ν −α]

≤ 0 for all ν ∈F. The suﬃciency of the latter statement is

trivial. If ν ∈Fand 0 <λ<1, then ν



=(1−λ)α+λν = α+λ(ν−α) ∈Fand

6.4 Projections of Measures 259

2[μ − α, λ(ν − α)]

≤|λ(ν − α)|

2[μ − α, ν − α]

≤ λ|ν − α|

Letting λ → 0, 2[μ −α, ν −α]

≤ 0 for all ν ∈F.Takingα = μ

, |μ −μ

≤

|μ − ν|

for all ν ∈Fif and only if [μ − μ

,ν − μ

] ≤ 0 for all ν ∈F.

Recall that a convex set F is a cone with vertex α if α + β ∈Fimplies

that α + λβ ∈Ffor all λ ≥ 0.

Lemma 6.4.4 Let F be a nonempty, closed, convex cone in E

with vertex

0 and let μ ∈E

.Thenμ

is the projection of μ onto F if and only if

(i) [μ − μ

,ν]

≤ 0 for all ν ∈Fand

(ii) [μ − μ

,μ

]

=0.

Proof: Assume that μ

is the projection of μ onto F. By the preceding

lemma,

[μ − μ

,ν− μ

]

≤ 0

for all ν ∈F. For any ν ∈F, (μ

+ ν)/2 ∈F,μ

+ ν ∈F,and[μ −μ

,ν]

≤ 0

from the preceding inequality. This proves (i). In particular, [μ−μ

,μ

]

≤ 0.

Putting ν = 0 in the above inequality, [μ − μ

,μ

]

≥ 0and(ii)isproved.

Conversely, assume that μ

satisﬁes (i)and(ii). Then

[μ − μ

,ν− μ

]

=[μ − μ

,ν]

− [μ − μ

,μ

]

≤ 0

for all ν ∈Fand μ

is the projection of μ onto F by the preceding lemma.

If Γ is a compact subset of Ω,F(Γ ) will denote the set of measures ν ∈E

with support in Γ . F(Γ ) is easily seen to be a convex cone in E

.ThatF(Γ )

is strongly closed can be seen as follows. Consider a sequence {μ

} in F(Γ )

that converges strongly to μ ∈E

. This implies that the sequence is norm

bounded and weakly convergent to μ. Since weak convergence and vague

convergence are equivalent for norm-bounded sequences, the sequence {μ

}

converges vaguely to μ.Sincetheμ

have support in Γ , the same is true of

μ;thatis,F(Γ ) is closed. It therefore makes sense to consider the projection

F(Γ )

of μ onto F(Γ ).

Theorem 6.4.5 If Γ is a compact subset of the Greenian set Ω and μ ∈E

then μ

∈F(Γ ) is the projection of μ onto F(Γ ) if and only if

(i) G

≤ G

μ on Ω and

(ii) G

= G

μ on Γ except possibly for a polar set.

260 6Energy

Proof: Suppose μ

is the projection of μ onto F(Γ ). By the preceding lemma,



)



μ − G

) dν ≤ 0 for all ν ∈F(Γ )and

(ii



)



μ − G

) dμ

=0.

Consider the set {x; G

μ(x) > G

(x)} and any ν ∈F(Γ ). Since the

restriction of ν to this set also belongs to F(Γ ),



μ>G

}

μ − G

) dν ≤ 0.

This implies that ν({x; G

μ(x) > G

(x)}) = 0 for all ν ∈F(Γ ). Taking

ν = μ

, G

μ ≤ G

a.e.(μ

). In conjunction with (ii



), this means that

μ = G

a.e.(μ

) on the support of μ

. By Theorem 6.3.5, G

≤

μ on Ω.Thus,μ

satisﬁes (i). By Theorem 6.2.3, G

μ = G

, except

possibly for a polar set since μ

∈E

. Conversely, suppose μ

satisﬁes (i)

and (ii). Then (ii) implies that G

= G

μa.e.(ν) for all ν ∈F(Γ ) ⊂E

by the same theorem. Properties (i



)and(ii



) follow immediately from the

preceding lemma.

Corollary 6.4.6 If Γ is a compact subset of the Greenian set Ω and μ ∈E

then G

F(Γ )

μ, ·).

Proof: Since G

μ ≥

μ, ·),

μ, ·) is the potential of a measure

∈E

by (ii) of Theorem 6.2.4. By Corollary 4.6.2, G

μ, ·)=

μ on Γ except possibly for a polar set. It follows from the preceding the-

orem that μ

= μ

F(Γ )

;thatis,

μ, ·)=G

F(Γ )

The Gauss-Frostman theorem, Theorem 6.2.9, can be generalized to po-

tentials G

μ, μ ∈E

, by deﬁning

(ν)=



νdν− 2



μdν, ν ∈E

Theorem 6.4.7 Let F be a nonempty, closed convex subset of E

and let

μ ∈E

.Thenμ

is the unique measure that minimizes Φ

over F.

Proof: |μ − ν|

= Φ

(ν)+|μ|

6.5 Wiener’s Test

Wiener’s original theorem characterized irregularity of a boundary point of

aregionΩ in terms of the capacities of parts of ∼ Ω in neighborhoods of

the boundary point. Since thinness of the complement and irregularity are

equivalent, it suﬃces to relate thinness of ∼ Ω to the capacities of certain

subsets of ∼ Ω.

6.5 Wiener’s Test 261

Let Ω be a Greenian subset of R

,letx ∈ Ω,andletΛ be any subset of Ω.

For α>1, let

= {y ∈ Ω; α

≤ u(x, y) ≤ α

j+1

let A

= ∪

∞

i=j

,letΛ

= Λ ∩ A

,j ≥ 1, and let Λ

= Λ ∩ A

.Choosek ≥ 1

large enough so that (A

)

−

⊂ Ω,diam(A

) < 1, and A

is a subset of the

component of Ω containing x. Several lemmas will be required to prove the

celebrated theorem of Wiener.

Theorem 6.5.1 (Wiener’s Test [64]) Λ is thin at x if and only if (a) the

series



∞

j=0

∗

(Λ

) converges or (b) the series



∞

j=0

(x) converges.

Several lemmas will be required to prove Wiener’s result. The proof that

the series



∞

j=0

(x) converges as a consequence of the thinness at x will

be accomplished by showing that the two series corresponding to odd num-

beredtermsandevennumberedtermsconverge. This amounts to considering

alternate annuli A

2j+1

and alternate annuli A

, respectively. To avoid cum-

bersome notation, only the annuli with even subscripts will be considered;

the treatment of odd subscripts being essentially the same.

Lemma 6.5.2 For each j ≥ k,letΓ

be a nonempty compact subset of Λ

and let Γ = ∪

∞

j=k

.Ifα>1, there is a constant m>0, depending only

upon α and k, such that

u(y, z)

u(y, x)

≤ mz∈ Γ

,y ∈ Γ ∼ Γ

,j ≥ k.

Proof: Consider the n = 2 case ﬁrst. If u(y, x) ≤ α

2j−1

and α

≤ u(x, z) ≤

2j+1

,then

u(y, z)

u(y, x)

−log |y − z|

u(y, x)

≤

−log (|y − x|−|x − z|)

u(y, x)

≤ 1 − α

−2k+1

log



1 −

|x − z|

|x − y|



Since

|x − z|

|x − y|

−u(x,z)

−u(x,y)

≤ e

−α

2j−1

(α−1)

and the last expression has the limit 0 as j →∞uniformly in y and z,it

follows that u(y, z)/u(y, x) is bounded uniformly in j and y, z as described

above. Suppose now that u(x, y) ≥ α

2j+2

and α

≤ u(x, z) ≤ α

2j+1

.Inthis

case

u(y, z)

u(y, x)

≤ 1 −

log



|x−z|

|x−y|

− 1



u(y, x)

262 6Energy

Since

|x − z|

|x − y|

≥ e

2j+1

(α−1)

u(y, z)

u(y, x)

≤ 1 −

log



2j+1

(α−1)

− 1



2j+2

where the last term has the limit 1+(1−α)/α as j →∞, and again it follows

that u(y, z)/u(y,x) is bounded uniformly in j and y, z as described above.

This proves the assertion in the n = 2 case. Consider now the n ≥ 3case.

Similarly, if u(x, y) ≤ α

2j−1

and α

≤ u(x, z) ≤ α

2j+1

,then

u(y, z)

u(y, x)

≤



1/(n−2)

− 1



n−2

and u(y,z)/u(y, x) is uniformly bounded in j and y,z as described above; if

u(x, y) ≥ α

2j+2

and α

≤ u(x, z) ≤ α

2j+1

,then

u(y, z)

u(y, x)

≤



1/(n−2)

− 1



n−2

and u(y, z)/u(y,x) is uniformly bounded in j and y, z as described above.

Lemma 6.5.3 There is a constant c>0, depending only on k, such that

(x, ·) ≤ u(x, ·) ≤ cG

(x, ·) on (A

)

−

Proof: Since G

(x, ·)=u(x, ·) − ghm

u(x, ·)onΩ,G

(x, ·) ≤ u(x, ·)on

)

−

.NotethatA

= B

x,δ

∼{x} where δ = α

k/(n−2)

if n ≥ 3andδ = e

−α

in the n = 2 case. Since G

(x, ·)andu(x, ·) are positive and continuous on

∂B

x,δ

, there is a constant c>1 such that cG

(x, ·) −u(x, ·) ≥ 0on∂B

x,δ

that

lim inf

y→z,y∈B

x,δ

(cG

(x, y) − u(x, y)) ≥ 0

for z ∈ ∂B

x,δ

.Since

lim inf

y→x

(cG

(x, y) − u(x, y)) = lim inf

y→x

((c − 1)u(x, y) − cghm

u(x, y)) = +∞,

lim inf

y→x,y∈B

x,δ

∼{x}

(cG

(x, y) − u(x, y)) ≥ 0on∂(B

x,δ

∼{x}). By Corol-

lary 2.3.6, cG

(x, ·) − u(x, y) ≥ 0onB

x,δ

∼{x} = A

.Thus,u(x, ·) ≤

(x, ·)on(A

)

−

.Sincec>1, (1/c)G

(x, ·) ≤ G

(x, ·) ≤ u(x, ·)on(A

)

−

Lemma 6.5.4 If Γ is a compact subset of the Greenian set Ω,then

lim

|x−y|→0,x,y∈Γ

(x, y)

u(x, y)

=1.

6.5 Wiener’s Test 263

Proof: Since Γ is covered by ﬁnitely many components of Ω, it suﬃces to

prove the assertion for the part of Γ in each of the ﬁnitely many components.

Thus, it can be assumed that Ω is connected. Since G

(x, y)=u(x, y)−h

(y),

where h

is a nonnegative harmonic function on Ω, G

(x, y)/u(x, y)=1−

(y)/u(x, y) whenever |x − y| is suﬃciently small. It suﬃces to show that

there is a constant m>0 such that h

(y) ≤ m for all x, y ∈ Γ .Bythe

minimum principle, each h

=0orh

> 0onΩ.Considerx ∈ Γ for which

the latter holds, any y ∈ Γ , and ﬁxed y

∈ Γ . By Harnack’s inequality,

Theorem 2.2.2, there is a constant k>0 such that h

(y) ≤ kh

). Since

and u are symmetric functions, h

)=h

(x)andh

(y) ≤ h

(x).

Since h

is bounded on Γ , there is a constant m such that h

(y) ≤ m

whenever x, y ∈ Γ with h

> 0onΩ.Ifh

=0onΩ, this result is trivially

true.

Lemma 6.5.5 If Γ is a compact subset of the Greenian set Ω,thereisa

constant c>0 such that G

(x, y) ≤ cu(x, y),x,y ∈ Γ .

Proof: If x and y belong to diﬀerent components of Ω, G

(x, y)=0and

the result is trivially true for any choice of c. Only the case with x, y in the

same component need be considered. Since Γ is covered by ﬁnitely many

components of Ω, it suﬃces to prove the result for the part of Γ in each of

the components and then take c to be the maximum of a ﬁnite collection

of constants. It therefore can be assumed that Ω is connected. The result

follows from a simple compactness argument and the preceding lemma.

Lemma 6.5.6 The series



∞

j=0

∗

(Λ

) converges if and only if the series



∞

j=0

(x) converges.

Proof: Since Λ

−

is compact,

is a potential by (vii) of Theorem 4.3.5.

Letting

= G

, the measure μ

is supported by cl

⊂ Λ

−

by Theo-

rem 5.7.13. By Theorem 6.2.8, C

∗

(Λ

)=μ

(Ω)and

∗

(Λ

)=α

(Ω)=α



1 dμ

≤



u(x, y) μ

(dy) ≤ α

j+1

(Ω)=α

j+1

∗

(Λ

Thus,

∗

(Λ

) ≤ U

(x) ≤ α

j+1

∗

(Λ

Since

(x)=G

(x) ≤ U

(x), the convergence of



∞

j=0

∗

(Λ

)im-

plies the convergence of



∞

j=0

(x). By Lemma 6.5.3, there is a constant

c>0 such that u(x, ·) ≤ cG

(x, ·)on(A

)

−

.Thus,

∗

(Λ

) ≤ U

(x) ≤ cG

(x)=c

(x)

264 6Energy

so that convergence of the series



∞

j=0

(x) implies convergence of the

series



∞

j=0

∗

(Λ

The following proof of Wiener’s theorem simpliﬁes early versions and can

be found in [17].

Proof of Theorem 6.5.1: Suppose ﬁrst that



∞

j=0

(x) < +∞.By

Theorem 4.7.6, for j ≥ k

∪

i=j

(x) ≤



i=j

(x) ≤

∞



i=j

(x).

Since the term on the left increases to

(x)asm →∞by Theorem 4.6.5,

(x) ≤

∞



i=j

(x).

Since the latter term has the limit 0 as j →∞, lim

j→∞

(x) = 0. Since

(x)=

∪{x}

(x) by Corollary 4.6.4, Λ

∪{x} is a neighborhood of x,

and

∪{x}

(x) < 1forlargej, Λ is thin at x according to Theorem 5.7.11.

Assume now that Λ is thin at x. It can be assumed that x ∈ ∂Λ ∼ Λ for

the following reasons. Since Λ is thin at x, x is not a ﬁne limit point of

Λ and there is a ﬁne neighborhood O

of x such that O

∩ Λ = ∅,which

means that x ∈ Λ;ifitweretruethatx ∈∼ Λ

−

, it would then follow that

= ∅ for large j so that C

∗

(Λ

)=0forlargej and the series



∞

j=0

∗

(Λ

equivalently



∞

j=0

(x), would converge trivially. For each j,let{Γ

} be

an increasing sequence of compact sets such that Γ

↑ Λ

.Since

↑

by Theorem 4.6.5, for each j ≥ k there is a compact set Γ

⊂ Λ

such that

(x) >

(x) − (1/2

). To show that the series



∞

j=0

(x)converges,

it suﬃces to show that the two series



∞

j=k

(x)and



∞

j=k

2j+1

(x)

converge. As indicated at the beginning of this section, only the ﬁrst series

will be shown to be convergent, the argument for the second series being the

same. Let Γ = ∪

∞

j=k

. If the intersection of Γ with some neighborhood of

x is polar, then at most a ﬁnite number of terms of the series



∞

j=0

(x)

are nonzero and the series converges. It therefore can be assumed that the

intersection of Γ with each neighborhood of x is nonpolar. Since Γ ⊂ Λ and

Λ is thin at x, Γ is thin at x. By Theorem 5.6.4, there is a potential u such

that u(x) < +∞ and lim

y→x,y∈Γ ∼{x}

u(y)=+∞.Since

= R

= u quasi

everywhere on Γ ,

has the limit +∞ at x on Γ except possibly for a polar

set. Since

(x) ≤ u(x) < +∞,

is ﬁnite at x.Since

≤ u and u is a

potential,

is a potential with

= G

μ for some measure μ supported

by the compact set Γ ∪{x}.Forj ≥ k,let

6.5 Wiener’s Test 265

= μ|

,μ



= μ|

Ω∼Γ

so that μ = μ

+ μ



. By Lemmas 6.5.2, 6.5.3, and 6.5.5 there is a constant

c>0, not depending upon j, such that

(z,y) ≤ cG

(x, y),y∈ Γ ∼ Γ

,z ∈ Γ

,j ≥ k.

Using the fact that this inequality is trivially true when y = x,forj ≥ k



≤ cG



(x) ≤ cG

μ(x)=c

(x) ≤ cu(x) < +∞ on Γ

Since u =

= G

μ = G

+ G



quasi everywhere on Γ, G

≥

u − cu(x) quasi everywhere on Γ, G

≥ 1 quasi everywhere on Γ

for

suﬃciently large j ≥ k.Thus,

≤ 1 ≤ G

quasi everywhere on Γ

for suﬃciently large j ≥ k.Since

is a potential

of a measure with support in Γ

≤ G

on Ω by Theorem 6.3.5.

Thus,



j≥i

(x) ≤



j≥i

(x) ≤ G

μ(x) ≤

(x) ≤ u(x) < +∞

for suﬃciently large i, proving that the series



∞

j=k

(x)converges.

Example 6.5.7 (Zaremba’s Cone Condition) Let Λ be a subset of R

n ≥ 3, let x ∈ ∂Λ,andletK be a closed cone with vertex at x,nonempty

interior, and K ⊂ Λ. Zaremba’s cone condition in Theorem 2.6.29 implies

that x is a regular boundary point of ∼ Λ. Wiener’s theorem can be used to

prove that Λ is not thin at x as follows. Fix α>1 and consider the series



∞

j=0

C(Λ

). Since Λ

= Λ∩A

⊃ A

∩K, it suﬃces to show that the series



∞

j=0

C(A

∩K) diverges. Consider any orthogonal transformation φ of R

about x. By symmetry,

φ(A

∩K)

(y)=

∩K

(φ

−1

y) and using the equation

for calculating C(φ(A

∩ K)) in Example 4.4.2, C(φ(A

∩ K)) = C(A

∩ K).

By the compactness of A

, there a ﬁnite number of such transformations

(j)

,...,φ

(j)

such that A

⊂∪

i=1

(j)

∩ K). By Lemma 4.4.20, C(A

) ≤



i=1

C(φ

(j)

∩ K)) = mC(A

∩ K). As calculated in Example 4.4.4,

C(A

)=α

−

n−2

Therefore,

∞



j=0

C(A

∩K) ≥

∞



j=0

C(A

∞



j=0

n−3

n−2

=+∞.

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Chapter 7

Interpolation and Monotonicity

7.1 Introduction

In order to take advantage of known facts about the Dirichlet and Neumann

problems, it is necessary to study transformations from one region onto an-

other and their eﬀects on smoothness properties of transformed functions. In

later chapters, it will be shown that the solution of the Neumann problem

for a spherical chip with a speciﬁed normal derivative on the ﬂat part of

the boundary can be morphed into a solution of an elliptic equation with an

oblique derivative boundary condition. The method for accomplishing this

is called the continuity method and requires showing that the mapping

from known solutions to potential solutions is a bounded transformation.

The essential problem here is to establish inequalities known as apriori in-

equalities whicharebasedontheassumptionthat there are solutions. The

passage from spherical chips to regions Ω with curved boundaries requires

relating the norms of functions and their transforms. The ﬁnal step involves

the adaptation of the Perron-Wiener method and the extraction of convergent

subsequences in certain Banach spaces called H¨older spaces.

Eventually, very strong conditions will have to be imposed on Ω. So strong,

in fact, that the reader might reasonably conclude that only a spherical chip,

or some topological equivalent, will satisfy all the conditions. As the ultimate

application will involve spherical chips, the reader might beneﬁt from assum-

ing that Ω is a spherical chip, at least up to the point of passing to regions

with curved boundaries . But as some of the inequalities require less stringent

conditions on Ω, the conditions on Ω will be spelled out at the beginning of

each section. Since these conditions will not be repeated in the formal state-

ments, it is important to refer back to the beginning of each section for a

description of Ω.

L.L. Helms, Potential Theory, Universitext, 267

 Springer-Verlag London Limited 2009