Helms L.L. Potential Theory

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308 8 Newtonian Potential

where ξ

is a point on the line segment joining x to x+λ

.Sincetheﬁrstfac-

tor in the last integral is bounded uniformly for x, y ∈ Ω by Inequality (8.7)

and f is integrable on Ω,



f(x)=





k(x, y)ψ(|x − y|/)



f(y) dy.

It is clear from Equation (8.6) that the derivative in the latter integral is

continuous and bounded, and therefore K



f ∈ C

(Ω). Using the fact that

ψ(|x − y|/) = 1 whenever |x − y|/ ≥ 2,

g(x) − D



f(x)=





k(x, y)(1 − ψ(|x − y|/)



f(y) dy



Ω∩B

x,2



k(x, y)(1 − ψ(|x − y|/))



f(y) dy, x ∈ Ω.

By Inequality (8.2),





k(x, y)(1 − ψ(|x − y|/))





≤



|k(x, y)| +

|x − y|

n−1

,x,y∈ Ω.

It follows from Lemma 8.2.2 that

|g(x) − D



f(x)|≤



sup

|f|





Ω∩B

x,2





|k(x, y)| +

|x − y|

n−1



≤



sup

|f|









Ω∩B

x,2

|k(x, y)|dy +2Cσ



→ 0

uniformly for x ∈ Ω as  → 0. Since K

1/n

f converges uniformly to Kf

on Ω and D

1/n

f converges uniformly to g on Ω, D

Kf is deﬁned and

Kf = g on Ω by Theorem 0.2.2; that is,

Kf(x)=g(x)=



k(x, y)f(y) dy, x ∈ Ω.

Since it was shown above that g is continuous on Ω ∪ Σ, D

Kf has a con-

tinuous extension to Ω ∪Σ.Forx ∈ Ω ∪ Σ and >0,

Kf(x)|≤



Ω∩B

x,

k(x, y)||f(y)|dy +



Ω∼B

x,

k(x, y)||f(y)|dy

≤ sup

|f|



Ω∩B

x,

k(x, y)|dy +



n−1



|f(y)|dy.

The ﬁrst term on the right can be made less than 1 uniformly for x ∈ Ω ∪Σ

for some choice  = 

by Lemma 8.2.2 and the second term is then ﬁnite for

this choice of  by the integrability of f. Taking the supremum over x ∈ Ω ∪Σ

and then over 1 ≤ i ≤ n, it follows that [Kf]

1+0,Ω

< +∞.

8.2 Subnewtonian Kernels 309

To conclude that Kf is twice diﬀerentiable, hypotheses stronger than those

of the preceding theorem are required. The appropriate class of functions f

required for stronger results is the class of H¨older continuous functions.

Deﬁnition 8.2.8 (i) If f is a real-valued function deﬁned on a bounded

set Ω ⊂ R

∈ Ω,and0<α≤ 1, then f is H¨older continuous at x

with exponent α if there is an M(x

) > 0 such that

|f(x) − f(x

)|≤M(x

)|x − x

, for all x ∈ Ω.

(ii) f is uniformly H¨older continuous on Ω with exponent α if there is

an M(Ω) > 0 such that

|f(x) − f(y)|≤M(Ω)|x − y|

, for all x, y ∈ Ω.

(iii) f is locally H¨older continuous on Ω with exponent α, 0 <α≤ 1, if f

is uniformly H¨older continuous with exponent α on each compact subset

of Ω.

If α = 1 in (i), then f is customarily said to be Lipschitz continuous at x

The class of functions satisfying (ii) and (iii) are the H¨older spaces H

(Ω)

and H

α,loc

(Ω), respectively, deﬁned in Section 7.2.

Let Ω be a bounded open subset of R

with a boundary made up of a

ﬁnite number of smooth surfaces. Such a set will be said to have a piecewise

smooth boundary.If{

,...,

} is a vector base for R

, the inner product

of 

and n(x)atx ∈ ∂Ω will be denoted by γ

(x)=

·n(x) whenever n(x)

is deﬁned. The equation



u(y) dy =



∂Ω

u(y)γ

(y) dσ(y),u∈ C

(Ω

−

), (8.8)

is usually proved, as in [43], as a preliminary version of the Divergence

Theorem. In the course of the proof of the following theorem, reference will be

made to subsets of Ω of the form Ω

= {x ∈ Ω; d(x, ∼ Ω) >δ} where δ>0.

Moreover, the following fact will be used in the proof. If g is a diﬀerentiable

function on (0, ∞), then

g(|x − y|)=−D

g(|x − y|), (8.9)

Theorem 8.2.9 Let f be bounded and locally H¨older continuous with expo-

nent α, 0 <α≤ 1, on the bounded open set Ω,andletΩ

be any bounded open

set containing Ω

−

with a piecewise smooth boundary. If k(x, y)=k(|x − y|)

satisﬁes Inequalities (8.1) through (8.4)on Ω

−

× Ω

−

with Σ = ∂Ω

and

w(x)=Kf(x)=



k(|x − y|)f(y) dy,

310 8 Newtonian Potential

then w ∈ C

(Ω) and

w(x)=





k(|x − y|)



f(y) − f (x)



−f(x)



∂Ω

k(|x − y|)γ

(y) dσ(y),x∈ Ω, (8.10)

where f is extended to be 0 on Ω

∼ Ω.

Proof: It suﬃces to show that w ∈ C

(Ω

) for every δ>0withΩ

= ∅.

Fix δ>0 and consider any >0forwhich4<δ.Let|f(x)|≤M for all

x ∈ Ω and |f (y) − f (x)|≤M|x − y|

for all x, y ∈ Ω

δ/2

.Fixx ∈ Ω

.Then

x,2

⊂ Ω

δ/2

and by Inequality (8.3),







k(|x − y|)



f(y) − f (x)





≤



x,



k(|x − y|)



f(y) − f (x)





∼B

x,



k(|x − y|)



f(y) − f (x)



≤ CM



x,

|x − y|

+2CM



∼B

x,

|x − y|

= CM







|θ|=1

α−1

dθ dr +

2CM



(Ω

)

= CMσ



2CM



(Ω

) < +∞

Since x ∈ Ω

⊂ Ω

−

⊂ Ω

, by Inequality (8.2),





∂Ω

k(|x − y|)γ

(y) dσ(y)



≤ C



∂Ω

|x − y|

n−1

dσ(y)

≤

n−1

σ(∂Ω

) < +∞.

Deﬁne

i,j

(x)=







k(|x − y|)







f(y) − f (x)



dy (8.11)

−f(x)



∂Ω

k(|x − y|)γ

(y) dσ(y)

8.2 Subnewtonian Kernels 311

for x ∈ Ω

,i,j =1,...,n.Alsoletv = D

Kf on Ω

and let



(x)=





k(|x − y|)



ψ(|x − y|/)f(y) dy





k(|x − y|)



ψ(|x − y|/)f(y) dy,

where ψ is the function in the proof of Theorem 8.2.5. The last equation

holds since f =0onΩ

∼ Ω. Applying Theorem 8.2.7 to v,

|v(x) − v



(x)| =







k(|x − y|)



1 − ψ(|x − y|/)



f(y) dy





x,2



k(|x − y|)



1 − ψ(|x − y|/)



f(y) dy



≤ CM



x,2

|x − y|

n−1

= CMσ

2 → 0as → 0.

This shows that lim

→0



= v uniformly on Ω

. Since the kernel used to

deﬁne v



satisﬁes Inequalities (8.1) and (8.2), Theorem 8.2.7 is applicable by

virtue of Remark 8.2.6 so that v



∈ C

(Ω), and it follows from Theorem 8.2.7

that



(x)=







k(|x − y|)



ψ(|x − y|/)



f(y) dy (8.12)







k(|x − y|)



ψ(|x − y|/)



(f(y) − f (x)) dy

+f(x)







k(|x − y|)



ψ(|x − y|/)









k(|x − y|)



ψ(|x − y|/)



(f(y) − f (x)) dy

−f(x)



∂Ω



k(|x − y|)



ψ(|x − y|/)γ

(y) dσ(y),

the last equation holding by Equations (8.8) and (8.9). For y ∈ ∂Ω

|x − y| > 2 and ψ(|x − y|/) = 1 so that the last integral is equal to



∂Ω

k(|x − y|)γ

(y) dσ(y).

It follows from Equations (8.11) and (8.12) that

i,j

(x) − D



(x)|









k(|x − y|)



1 − ψ(|x − y|/)





(f(y) − f (x)) dy



312 8 Newtonian Potential

Since ψ(|x − y|/)=1for|x − y| > 2,

i,j

− D



(x)|





x,2





k(|x − y|)



1 − ψ(|x − y|/)





(f(y) − f(x)) dy



≤ M



x,2







k(|x − y|)



1 − ψ(|x − y|/)







|x − y|

≤ M



x,2



(1 − ψ(|x − y|/))D

k(|x − y|)



k(|x − y|)





(|x − y|/)



− y

)

|x − y|



|x − y|

dy (8.13)

≤ M



x,2



k(|x − y|)| +



k(|x − y|)|



|x − y|

≤ CM



x,2



|x − y|



|x − y|

n−1



|x − y|

≤

CMσ

(2)

+ CM



(2)

α+1

α +1

→ 0as → 0

uniformly for x ∈ Ω

.Sincethev



converge uniformly to v = D

Kf on Ω

and the D



converge uniformly to u

i,j

on Ω

, by Theorem 0.2.2 D

w =

Kf = u

i,j

on Ω

, i, j =1,...,n. Returning to Equation (8.12),



(x)=



Ω∼B

x,





k(|x − y|)



ψ(|x − y|/)



f(y) dy.

It is easily seen that the integrand is dominated by an integrable function,

namely, a constant, and that D



is continuous on Ω

.SincetheD



converge uniformly to u

i,j

= D

w on Ω

, the latter function is continuous

on Ω

. Since this is true for each δ>0,w ∈ C

(Ω).

Recall that the symbol C is used as a generic symbol for a constant

that may very well change from line to line or even within a line.

Whenever possible, the dependence of C on certain parameters will be in-

dicated by C = C(n, a, b), etc. It will be assumed in the remainder of this

section that k(x, y) is a function of |x − y|;thatis,k(x, y)=k(|x − y|). If

x =(x

,...,x

) ∈ R

,letx =(x



)wherex



=(x

,...,x

n−1

), and let

= {(x



) ∈ R

; x

> 0}

= {(x



) ∈ R

; x

=0}

−

= {(x



) ∈ R

; x

< 0} ;

8.2 Subnewtonian Kernels 313

moreover, if A ⊂ R

,letA

= A ∩ R

and A

= A ∩ R

.LetΩ be a

bounded open subset of R

∈ Ω,B

= B

,ρ

and B

= B

,2ρ

with

⊂ Ω. The balls B

and B

may or may not have a nonempty intersection

with R

. The function f on B

of the following lemma is understood to be

zero outside B

. Recall that  D

u 

0,Ω

stands for the supremum of norms

of derivatives of u of order k.

Lemma 8.2.10 If B

= B

,ρ

and B

= B

,2ρ

are concentric balls with

center x

in Ω,f ∈ H

),α ∈ (0, 1),andw = Kf on Ω,thenw ∈

2+α

) and

[w]

2+0,B

+ ρ

[w]

2+α,B

≤ C(n, α)(f 

0,B

+ ρ

[f]

α,B

Proof: Note that f is bounded and uniformly H¨older continuous on B

Suppose i = n or j = n, the latter, for example. If y ∈ ∂B

∩ R

,then

(y) = 0, and it follows from Equation (8.10) that

w(x)=



k(|x − z|)(f(z) − f(x)) dz

−f(x)



(∂B

)

k(|x − z|)γ

(z) dσ(z)

for x ∈ B

.Ifi = n instead, interchange i and j in this equation. Using

Inequalities (8.2) and (8.3) and the fact that B

⊂ B

,3ρ

w(x)|≤



k(|x − z|)||f(z) − f (x)|dz

+|f(x)|



(∂B

)

k(|x − y|)|dσ(z)

≤ [f]

α,B



x,3ρ

C(n)

|x − z|

|z − x|

+f

0,B



(∂B

)

C(n)

|x − z|

n−1

dσ(z)

≤

C(n)σ

[f]

α,B

+ C(n)f

0,B

(2ρ)

n−1

(2ρ)

n−1

Thus, there is a constant C(n, α) such that

w(x)|≤C(n, α)(ρ

[f]

α,B

+ f 

0,B

),x∈ B

. (8.14)

Now let y be any other point in B

,t = |x − y|, ξ =

(x + y), and B

ξ,t

∩ R

.Then

314 8 Newtonian Potential

w(y) − D

w(x)=

f(x)



(∂B

)

k(|x − z|) − D

k(|y − z|)γ

(z) dσ(z)

+(f(x) − f(y))



(∂B

)

k(|y − z|)γ

(z) dσ(z)



ξ,t

∩B

k(|x − z|)(f(x) − f(z)) dz



ξ,t

∩B

k(|y − z|)(f(z) − f (y)) dz

+(f(x) − f(y))



∼B

ξ,t

k(|x − z|) dz



∼B

ξ,t

k(|x − z|) − D

k(|y − z|)(f (y) − f (z)) dz.

Denote the terms on the right by I

,...,I

, respectively. For some point ˆx

on the line segment joining x to y,

|≤f

0,B

|x − y|



(∂B

)

|∇

(x)

k(|ˆx − z|)|dσ(z).

Using Inequality (8.3) and the fact that |ˆx − z|≥ρ for z ∈ (∂B

)

|≤f

0,B

|x − y|



(∂B

)

C(n)

|ˆx − z|

dσ(z)

≤f 

0,B

|x − y|C(n)

(2ρ)

n−1

Since |x − y|≤|x − y|

(2ρ)

1−α

, there is a constant C(n, α) such that

|≤C(n, α)f

0,B

|x − y|

Similarly, |I

|≤C(n)[f]

α,B

|x−y|

. Using Inequality (8.3) and the fact that

ξ,t

⊂ B

x,3t/2

|≤



ξ,t

∩B

k(|x − z|)(f(x) − f(z))|dz

≤ C(n)[f]

α,B



x,3t/2

|x − z|

α−n

≤ C(n, α)[f ]

α,B

|x − y|

8.2 Subnewtonian Kernels 315

Similarly,

|≤C(n, α)[f ]

α,B

|x − y|

By Equation (8.8),

|≤[f]

α,B

|x − y|





∼B

ξ,t

k(|x − z|) dz



=[f]

α,B

|x − y|





∂(B

∼B

ξ,t

)

k(|x − z|)γ

(z) dσ(z)



Since |x − z| >ρfor z ∈ ∂B

and |x − z| >t/2forz ∈ ∂B

ξ,t





∂(B

∼B

ξ,t

)

k(|x − z|)γ

(z) dσ(z)



≤



∂B

C(n)

|x − z|

n−1

dσ(z)



∂B

ξ,t

C(n)

|x − z|

n−1

dσ(z)

≤ C(n)2

n−1

+ C(n)2

n−1

= C(n)2

Therefore,

|≤[f]

α,B

|x − y|

C(n)2

Lastly, for some ˆx on the line segment joining x to y,

|≤|x − y|



∼B

ξ,t

|∇

(x)

k(|ˆx − z|)||f(y) − f (z)|dz.

By Inequality (8.4),

|≤C(n)|x − y|[f]

α,B



∼B

ξ,t

|y −z|

|ˆx − z|

n+1

dz.

Since |y − z|≤

|ξ − z|≤3|ˆx − z| for z ∈ B

∼ B

ξ,t

, t = |x − y|,and

∼ B

ξ,t

⊂∼ B

ξ,t

|≤C(n, α)|x − y|[f ]

α,B

α−1

= C(n, α)[f]

α,B

|x − y|

Thus, there is a constant C(n, α) such that

w(y) − D

w(x)|≤C(n, α)|x − y|



f

0,B

+[f]

α,B



316 8 Newtonian Potential

and it follows that

α,B

≤ C(n, α)



f

0,B

+[f]

α,B



. (8.15)

Taking the supremum over pairs (i, j) in Inequalities (8.14) and (8.15) and

then combining,

[w]

2+0,B

+ ρ

[w]

2+α,B

≤ C(n, α)



f

0,B

+ ρ

[f]

α,B



Since f is bounded and integrable on Ω, |w|

1+0,B

< +∞ by Theorem 8.2.5

and Theorem 8.2.7. This fact combined with the preceding inequality estab-

lishes that w ∈ H

2+α

This proof fails in the α = 1 case because the bound on |I

| involves a

divergent integral.

Corollary 8.2.11 If Ω is a bounded open subset of R

∈ Ω, B

= B

,ρ

= B

,2ρ

with B

⊂ Ω, f ∈ H

),α ∈ (0, 1),andw = Kf on Ω,then

w ∈ H

2+α

) and

[w]

2+0,B

+ ρ

[w]

2+α,B

≤ C(n, α)



f

0,B

+ ρ

[f]

α,B



Proof: If Ω ⊂ R

,thenB

= B

and B

= B

.IfΩ is any subset of R

the preceding case can be applied by translating Ω onto R

Theorem 8.2.12 Let Ω be a bounded open subset of R

and let Σ be a

relatively open subset of ∂Ω ∩ R

.Ifα ∈ (0, 1),f ∈ H

(Ω),andw = Kf on

Ω,thenw ∈ C

(Ω) ∩ C

(Ω ∪ Σ).

Proof: It follows from the preceding corollary that w ∈ C

(Ω). The assertion

that w ∈ C

(Ω ∪ Σ) follows from Theorem 8.2.7.

8.3 Poisson’s Equation

Taking the kernel of the preceding section to be the Newtonian kernel u(|x −

y|), the results therein can be used to show that Poisson’s equation Δw = f

has a solution and that w inherits smoothness properties of f . Recall that

the notations u(x, y),u

(y),u(|x − y|),u(r) for the fundamental harmonic

function are used interchangeably.

If f : R

→ R is Lebesgue integrable and has compact support, the New-

tonian potential of f is the function Uf : R

→ R deﬁned by

Uf(x)=



u(|x − y|)f(y) dy.

8.3 Poisson’s Equation 317

The kernel u(x, y) is continuous in the extended sense on R

× R

and con-

tinuous oﬀ the diagonal of R

×R

. It will be shown now that u(x, y)satisﬁes

Inequalities (8.1), (8.2), (8.3), and (8.4). Letting δ

= 0 or 1 according to

j = i or j = i, respectively, the following derivatives and inequalities are

easily veriﬁed:

u(x, y)=

⎧

⎨

⎩

−

−y

|x−y|

if n =2

−(n − 2)

−y

|x−y|

if n ≥ 3,

(8.16)

u(x, y)=

⎧

⎪

⎨

⎪

⎩

−

|x−y|

−y

)(x

−y

)

|x−y|

if n =2

(n − 2)



−

|x−y|

+ n

−y

)(x

−y

)

|x−y|

n+2



if n ≥ 3,

(8.17)

u(x, y)|≤

⎧

⎨

⎩

|x−y|

if n =2

n−2

|x−y|

n−1

if n ≥ 3,

(8.18)

u(x, y)|≤

⎧

⎨

⎩

|x−y|

if n =2

(n−2)(n+1)

|x−y|

if n ≥ 3

(8.19)

where i, j =1, 2,...,n. More generally, there is a constant C(n, β) such that

u|≤C(n, β)

|x − y|

n−2+k

if |β| = k ≥ 1. (8.20)

All of the results of the preceding section are applicable to u(x, y)onΩ×Ω for

any bounded open subset Ω of R

. The statement that u ∈ C

(Ω

−

) ∩C

(Ω)

means that the restriction of u to Ω

−

is in C

(Ω

−

) and that the restriction

of u to Ω is in C

(Ω).

Theorem 8.3.1 If the function f is bounded and locally H¨older continuous

with exponent α ∈ (0, 1], on the bounded open set Ω, then the function

u =

(2 − n)σ

belongs to C

(Ω

−

) ∩ C

(Ω) and satisﬁes Δu = f on Ω; if, in addition, g is

continuous on ∂Ω, then there is a unique u in C

(Ω

−

) ∩ C

(Ω) satisfying

Δu = f on Ω and lim

y→x

u(y)=g(x) for each regular boundary point x ∈

∂Ω.

Proof: (n ≥ 3) Extending f to be 0 outside Ω, the function v = Uf is con-

tinuous on R

by Theorem 8.2.5 and belongs to C

(Ω) by Theorem 8.2.9. If

is any bounded open set containing Ω with a piecewise smooth boundary,

the second partials of v have the representation