Helms L.L. Potential Theory

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68 2 The Dirichlet Problem

This shows, among other things, that ρ

n−1

L(u : x, ρ)isintegrableon[0,δ

Since the indeﬁnite integral of an integrable function is absolutely contin-

uous, A(u : x, δ) is continuous on (0,δ

). The above equation also gives

arepresentationofA(u : x, δ) on any closed interval [a, b] ⊂ (0,δ

)asa

product of two bounded, absolutely continuous functions of δ on [a, b]and

it follows directly from the deﬁnition of such functions that A(u : x, δ)is

diﬀerentiable a.e. on (0,δ

). Recall also that the derivative of an indeﬁnite

integral exists and is equal to the integrand a.e. relative to Lebesgue measure.

Then for almost all δ ∈ (0,δ

A(u : x, δ)=

L(u : x, δ) −

nσ

n+1



n−1

L(u : x, ρ) dρ

[L(u : x, δ) − A(u : x, δ)],

since ν

= σ

/n.InviewofthefactthatL(u : x, ρ) is monotone decreasing

on (0,δ),

A(u : x, δ)=



n−1

L(u : x, ρ) dρ

≥





n−1

dρ



L(u : x, δ)

= L(u : x, δ), (2.3)

and D

A(u : x, δ) ≤ 0 for almost all δ ∈ (0,δ

). If 0 <a<δ,then

A(u : x, δ) − A(u : x, a)=



A(u : x, ρ) dρ,

and A(u : x, δ) is a monotone decreasing function of δ on (0,δ

). Since

A(u : x, 0) = u(x) ≥ A(u : x, δ) ≥ L(u : x, δ) for any δ ∈ (0,δ

), A(u : x, δ)

is monotone decreasing on [0,δ

] and right continuous at 0.

The fact that a superharmonic function is ﬁnite a.e. places a limitation on

the set of inﬁnities of a superharmonic function.

Theorem 2.4.5 If u and v are superharmonic on an open set Ω and c>0,

then cu, u + v, min (u, v) are superharmonic on Ω.

Proof: That cu is superharmonic on Ω is obvious from the deﬁnition. As

to u + v,notethatu + v cannot be identically +∞ on any component U

of Ω since each component has positive Lebesgue measure and both u and

v are ﬁnite a.e. on U.Moreover,u + v>−∞ on Ω and u + v is l.s.c. on

Ω since both have these properties. If B

−

x,δ

⊂ Ω,thenu(x) ≥ L(u : x, δ)

and v(x) ≥ L(v : x, δ)sothatu(x)+v(x) ≥ L(u + v : x, δ). Thus, u + v is

superharmonic on Ω. It will be shown that min (u, v) is super-mean-valued

2.4 Properties of Superharmonic Functions 69

on Ω, the other properties of a superharmonic function being easy to verify.

If B

−

x,δ

⊂ Ω,thenu(x) ≥ L(u : x, δ) ≥ L(min (u, v):x, δ)withthesame

holding for v.Thus,min(u, v)(x) ≥ L(min (u, v):x, δ).

Theorem 2.4.6 If u is both subharmonic and superharmonic on an open set

Ω,thenu is harmonic on Ω.

Theorem 2.4.7 If u is superharmonic on the open set Ω, x ∈ Ω,andΛ is

asubsetofΩ of Lebesgue measure zero, then u(x) = lim inf

y→x,y∈Λ∪{x}

u(y);

in particular, u(x) = lim inf

y→x,y=x

u(y).

Proof: By l.s.c.,

u(x) ≤ lim inf

y→x,y=x

u(y) ≤ lim inf

y→x,y∈Λ∪{x}

u(y).

If B

x,δ

is any ball with B

−

x,δ

⊂ Ω,then

u(x) ≥ A(u : x, δ) ≥ inf{u(y); y ∈ B

x,δ

,y ∈ Λ ∪{x}}.

Thus, u(x) ≥ lim inf

y→x,y∈Λ∪{x}

u(y) and the two are equal.

Convergence properties of monotone sequences of superharmonic functions

will be considered; more generally, right-directed families of such functions

will be considered.

Theorem 2.4.8 If { u

; i ∈ I} is a right-directed family of superharmonic

functions on the open set Ω, then on each component of Ω the function

u =sup

i∈I

is either superharmonic or identically +∞.

Proof: Since each u

is l.s.c. and the supremum of any collection of l.s.c.

functions is again l.s.c., u is l.s.c. on Ω. For any i

∈ I,u ≥ u

> −∞ on Ω.

If B

−

x,δ

⊂ Ω, then by Lemma 2.2.10

u(x)=sup

i∈I

(x) ≥ sup

i∈I

L(u

: x, δ)=L(sup

i∈I

: x, δ)=L(u : x, δ).

Thus, u is super-mean-valued and is superharmonic on each component of Ω

provided it is not +∞ thereon.

The preceding theorem applies to monotone increasing sequences of su-

perharmonic functions, but the analogous theorem for monotone decreasing

sequences is not true. Consider, for example, the superharmonic function u

on R

with pole y.Forj ≥ 1, deﬁne u

= u

/j.Then{u

} is a monotone

decreasing sequence of superharmonic functions. Letting u =inf

j≥1

,u=0

except at y where it is equal to +∞. Clearly, u is not superharmonic since

it is not l.s.c. at y. Not only is the lack of semicontinuity of the limit a diﬃ-

culty, but the requirement that the limit not take on the value −∞ is another

problem.

70 2 The Dirichlet Problem

Theorem 2.4.9 If {u

; i ∈ I} is a left-directed family of superharmonic func-

tions on the open set Ω which is locally bounded below, then the lower regu-

larization ˆu of u =inf

i∈I

is superharmonic on Ω.

Proof: The fact that the family is locally bounded below implies that

ˆu cannot take on the value −∞. For any i ∈ I, u

≥ u and u

(x)=

lim inf

y→x,y=x

(y) ≥ lim inf

y→x,y=x

u(y)=ˆu(x),x ∈ Ω, by Theorem 2.4.7

and ˆu is not identically +∞ on any component of Ω. It will be shown that

ˆu is superharmonic using the deﬁnition of superharmonic function following

Theorem 2.3.8. Let U be an open subset of Ω with compact closure U

−

⊂ Ω.

Let h be harmonic on U, continuous on U

−

,andˆu ≥ h on ∂U.Thenu

≥ h

on ∂U for each i ∈ I. Since each u

is superharmonic on Ω, u

≥ h on U

for each i ∈ I.Thenu ≥ h on U and ˆu ≥

h = h on U. Therefore, ˆu is

superharmonic on Ω.

It is apparent that the maximum of a convex function and a linear function

is again a convex function that is partially linear. There is an analogous

operation on superharmonic functions deﬁned as follows.

Deﬁnition 2.4.10 If u is superharmonic on the open set Ω and B is a ball

with B

−

⊂ Ω,let



PI(u : B)onB

u on Ω ∼ B.

If u is subharmonic on Ω, u

is deﬁned similarly. The function u

)is

called the lowering (lifting)ofu over B.

Lemma 2.4.11 If u is superharmonic on the open set Ω and B is a ball

with B

−

⊂ Ω,thenu

≤ u on Ω, harmonic on B, and superharmonic on

Ω. Moreover, if v is a superharmonic function on B with v ≥ u

on B,then

the function

w =



min (u, v) on B

u on Ω ∼ B

is superharmonic on Ω.

Proof: By Theorem 2.4.3, u

is deﬁned, harmonic on B,andu

≤ u on Ω.

The fact that u

is superharmonic is a special case of the second assertion

by taking v = u

on B,inwhichcasew = u

.Letv be any superharmonic

function on B satisfying v ≥ u

on B.Thenw ≥ u

on Ω with w = u

= u

on Ω ∼ B. Clearly, w is superharmonic on Ω ∼ ∂B so that it suﬃces to show

that w is l.s.c. and super-mean-valued at points of ∂B.Consideranyx ∈ ∂B.

Then

lim inf

y→x,y∈∼B

w(y) = lim inf

y→x,y∈∼B

u(y) ≥ lim inf

y→x,y∈Ω

u(y) ≥ u(x)=w(x).

2.5 Approximation of Superharmonic Functions 71

Since u is l.s.c. on ∂B,

lim inf

y→x,y∈B

w(y) ≥ lim inf

y→x,y∈B

(y) ≥ lim inf

y→x,y∈∂B

u(y) ≥ u(x)=w(x)

by Lemma 1.7.5. This shows that w is l.s.c. at points of ∂B and therefore

l.s.c. on Ω. Consider again x ∈ ∂B.Sincew ≤ u on Ω, for any closed ball

−

x,δ

⊂ Ω

w(x)=u(x) ≥ L(u : x, δ) ≥ L(w : x, δ),

which shows that w is super-mean-valued at points of ∂B and therefore locally

super-mean-valued on Ω.

2.5 Approximation of Superharmonic Functions

Consider an extended real-valued function u on an open set Ω ⊂ R

.In

order to smoothen u, at the expense of reducing its domain, a molliﬁer

m : R

→ R deﬁned by

m(x)=



−

1−|x|

if |x|≤1

0if|x| > 1,

(2.4)

will be employed, where c is a constant chosen so that m is a probability

density. It is known that m ∈ C

∞

Let u be a locally integrable function on Ω and let h>0. For x ∈ Ω

{y ∈ Ω; d(y, ∂Ω) >h}, deﬁne

u(x)=





y − x



u(y) dy =



x,h



y − x



u(y) dy. (2.5)

It is easily seen that the operator J

has the following properties:

(i) u ≥ 0onΩ implies that J

u ≥ 0onΩ

(ii) J

u ∈ C

∞

(Ω

(iii) J

1=1onΩ

(iv) u linear on Ω implies that J

u = u on Ω

The limit lim

h→0

u(x),x ∈ Ω, is understood to be over those h for which

x ∈ Ω

Lemma 2.5.1 If u is superharmonic on the open set Ω and h>0,thenJ

is a C

∞

(Ω

) superharmonic function on Ω

, J

u ≤ J

u ≤ u on Ω

whenever

0 <k≤ h,andlim

h→0

u(x)=u(x) for each x ∈ Ω; if, in addition, u is

harmonic on Ω,thenJ

u = u on Ω

72 2 The Dirichlet Problem

Proof: That J

u ∈ C

∞

(Ω

) follows from the fact that m ∈ C

∞

). It will

be shown now that u is locally super-mean-valued on Ω

. Consider any ball

x,δ

⊂ B

−

x,δ

⊂ Ω

.Then

A(J

u : x, δ)=



x,δ



|z|≤1

m(z)u(y + hz) dz dy



|z|≤1

m(z)





x,δ

u(y + hz) dy



≤



|z|≤1

m(z)u(x + hz) dz

= J

u(x),

and J

u is locally super-mean-valued on Ω

. Using spherical coordinates

(ρ, θ) relative to the pole x ∈ Ω

,thefactthatm(ρθ/h) is independent of θ,

and the superharmonicity of u,

u(x)=



|θ|=1



ρθ



u(x + ρθ)ρ

n−1

dθ dρ





ρθ



n−1





|θ|=1

u(x + ρθ) dθ



dρ

≤





ρθ



n−1

u(x) dρ

= u(x)J

1(x)

= u(x).

Thus, J

u ≤ u on Ω

.Consideranyx ∈ Ω.Sinceu(x) = lim inf

y→x,y=x

u(y),

given >0thereisaneighborhoodN of x such that inf

y∈(N∩Ω)∼{x}

u(y) >

u(x) − .Consideranyh>0forwhichx ∈ Ω

and B

x,h

⊂ N.Then

u(x)=



x,h



y − x



u(y) dy

≥ (u(x) − )J

1(x)

= u(x) − .

Thus for each x ∈ Ω, lim

h→0

u(x)=u(x). If u is harmonic on Ω, then the

above argument is applicable to both u and −u so that J

u ≤ u and J

u ≥ u

on Ω

and the two are equal. It remains only to show that 0 <k<himplies

u ≤ J

u on Ω

.Consideranyx ∈ Ω

and k<h. Letting ρθ =(y − x)/h

where θ is a point on the unit sphere with center at 0,

u(x)=



x,h



y −x



u(y) dy

2.5 Approximation of Superharmonic Functions 73



|θ|=1

m(ρθ)u(x + ρhθ)ρ

n−1

dθ dρ.

Since m(ρθ) does not depend upon θ, by Theorem 2.4.4

u(x)=



m(ρθ)ρ

n−1

L(u : x, ρh) dρ

≤



m(ρθ)ρ

n−1

L(u : x, ρk) dρ

= J

u(x)

and J

u ≤ J

u ≤ u on Ω

whenever 0 ≤ k<h.

Theorem 2.5.2 If u is superharmonic on the open set Ω ⊂ R

and har-

monic on Ω ∼ Γ where Γ is a compact subset of Ω, then there is an increasing

sequence of superharmonic functions { u

} on Ω such that

(i) u

∈ C

∞

(Ω) for all j ≥ 1,

(ii) lim

j→∞

= u on Ω,and

(iii) if U is any neighborhood of Γ with compact closure U

−

⊂ Ω,thenu

= u

on Ω ∼ U

−

for suﬃciently large j.

Proof: Let U be a neighborhood of Γ with compact closure U

−

⊂ Ω.Since

{Ω

; h>0} is an open covering of the compact set U

−

,thereisanh

> 0

such that Γ ⊂ U ⊂ U

−

⊂ Ω

and h

<d(Γ,∼ U). Let {h

} be a sequence

in (0,h

) that decreases to 0 and let

(x)=



u(x)ifx ∈ Ω

u if x ∈ Ω ∼ Ω

Since the Ω

increase to Ω, the sequence {u

} increases to u on Ω.Since

is superharmonic on Ω

⊃ U

−

is superharmonic on U ;moreover,

∈ C

∞

(Ω

) implies that u

∈ C

∞

(U). It will be shown now that u

= u

on a neighborhood of Ω ∼ U, thus implying that u

∈ C

∞

(Ω) and is harmonic

on Ω ∼ U

−

. By deﬁnition, u

= u on Ω ∼ Ω

.Consideranyx ∈ Ω

∼ U.

Since x ∈ Ω

and h

<d(Γ, ∼ U), δ

> 0 can be chosen so that B

x,δ

⊂ Ω

and h

+ δ

<d(Γ, ∼ U ). For y ∈ B

x,δ

,d(y, Γ) ≥ d(x, Γ ) − d(x, y) >d(Γ, ∼

U) − δ

so that B

y,δ

⊂ Ω ∼ Γ .Thus,u

(y)=J

u(y)=u(y) for all

y ∈ B

x,δ

. This proves that u

= u on a neighborhood of Ω ∼ U.

It was shown in Lemma 2.4.4 that L(u : x, δ) is a decreasing function of δ

if u is superharmonic. Much more is true. Recall that a function φ is concave

on an interval (a, b)if−φ is convex on the interval; that is,

φ(λx +(1− λy) ≥ λφ(x)+(1− λ)φ(y)

whenever a<x<y<band 0 <λ<1.

74 2 The Dirichlet Problem

Theorem 2.5.3 Let u be superharmonic on the open set Ω and let B

x,ρ

aballwith∂B

x,ρ

⊂ Ω for r

≤ ρ ≤ r

.ThenL(u : x, ρ) is a concave function

of −log ρ if n =2and a concave function of ρ

−n+2

if n ≥ 3 on (r

);in

particular, L(u : x, ρ) is a continuous function of ρ on (r

). If, in addition,

u is harmonic on Ω,then

L(u : x, ρ)=



2π



∂B

x,r

u(z) dσ(z)



log ρ + const. (2.6)

on (r

) if n =2,and

L(u : x, ρ)=



−

(n − 2)



∂B

x,r

u(z) dσ(z)



−n+2

+ const. (2.7)

on (r

) if n ≥ 3.

Proof: Assume ﬁrst that u ∈ C

(Ω). Letting A

denote the closed annulus

−

x,ρ

∼ B

x,r

<ρ≤ r

,u has a bounded normal derivative on ∂A

.Since

u is superharmonic on Ω,Δu ≤ 0onΩ by Lemma 2.3.4. Applying Green’s

theorem to the region A

0 ≥



Δu dz =



∂B

x,ρ

u(z) dσ(z) −



∂B

x,r

u(z) dσ(z)

for r

<ρ≤ r

. Using the fact that D

u is bounded on ∂A



∂B

x,ρ

u(z) dσ(z)=



|θ|=1

n−1

(x + rθ)|

r=ρ

,dσ(θ)

= ρ

n−1



|θ|=1

lim

η→0

u(x +(ρ + η)θ) − u(x + ρθ)

dσ(θ)

= ρ

n−1

lim

η→0





|θ|=1

u(x +(ρ + η)θ) dσ(θ)

−



|θ|=1

u(x + ρθ) dσ(θ)



= σ

n−1

L(u : x, ρ).

Therefore,

n−1

L(u : x, ρ)=



Δu(z) dz +



∂B

x,r

u(z) dσ(z).

2.6 Perron-Wiener Method 75

If u is harmonic on Ω, then the ﬁrst term on the right is zero, the second term

is a constant, and an integration with respect to ρ results in Equations (2.6)

and (2.7). Suppose now that n ≥ 3. Letting ζ =1/ρ

n−2

,ρ(ζ)theinverse

function, and k the second term on the right of the last equation,

−σ

(n − 2)D

L(u : x, ρ(ζ)) =



ρ(ζ)

Δu(z) dz + k.

As ρ increases,



Δu(z) dz decreases; thus, D

L(u : x, ρ(ζ)) increases as

ζ decreases on (ζ

,ζ

)whereζ

=1/r

n−2

,ζ

=1/r

n−2

. Therefore, L(u :

x, ρ(ζ)) is a concave function of ζ =1/ρ

n−2

;thatis,L(u : x, ρ)isaconcave

function of 1/ρ

n−2

. The proof of the n = 2 case is accomplished in the same

way by letting ζ = −log ρ. Consider now an arbitrary superharmonic function

u and the n = 3 case. By Lemma 2.5.1, there is a C

∞

(Ω) sequence { u

} of

superharmonic functions on a neighborhood of B

−

x,r

∼ B

x,r

with u

↑ u.

Note that the sequence {L(u

: x, ρ)} is increasing and uniformly bounded

on (r

)since

L(u

: x, ρ) ≤ L(u

: x, r

) ≤ L(u : x, r

) < +∞,

the latter inequality holding because u is bounded on ∂B

x,r

.Foreachj ≥ 1,

there is a concave function ψ

deﬁned on (ζ

,ζ

) such that L(u

: x, ρ)=

(ρ

−n+2

)=ψ

(ζ). Note that the sequence {ψ

} is increasing and uniformly

bounded on (ζ

,ζ

). It is easily seen that ψ = lim

j→∞

is also a concave

function and that L(u : x, ρ)=ψ(ρ

−n+2

)forρ ∈ (r

). The same argument

applies to the n =2case.ContinuityofL(u : x, ρ)on(r

) follows from

the fact that a concave function on an open interval is continuous.

2.6 Perron-Wiener Method

Let Ω be an open subset of R

with nonempty boundary, and let f be an

extended real-valued function on ∂Ω. The generalized Dirichlet problem is

that of ﬁnding a harmonic function h corresponding to the boundary function

f. Having shown that such a function exists, there is then the problem of

relating the limiting behavior of h at boundary points of Ω to the values

of f at such points. A standard method of proving the existence of an h is

to approximate from above and below, and then show that the diﬀerence

between the two can be made arbitrarily small.

Deﬁnition 2.6.1 A family F of superharmonic functions on an open set Ω

is saturated if

(i) u, v ∈Fimplies min (u, v) ∈F

(ii) u ∈Fand B aballwithB

−

⊂ Ω implies u

∈Fwhere

76 2 The Dirichlet Problem



PI(u : B)onB

u on Ω ∼ B.

Theorem 2.6.2 If F is a saturated family of superharmonic functions on

Ω,theninf

u∈F

u is either identically −∞ or harmonic on each component

of Ω.

Proof: Let w =inf

u∈F

u and consider any closed ball B

−

⊂ Ω.Sincethe

assertion of the theorem pertains to the components of Ω, it can be assumed

that Ω is connected. For u ∈F,u

≤ u,harmoniconB, and superharmonic

on Ω by Lemma 2.4.11. Moreover, w =inf{u

; u ∈F}.Consideranyu, v ∈F

and the corresponding u

.SinceF is left-directed, min (u, v) ∈Fso

that (min (u, v)

≤ min (u

). It follows that the family {u

; u ∈F}

is left-directed. By Theorem 2.4.8, w =inf{u

; u ∈F}is identically −∞

or superharmonic on Ω. Since the functions u

are harmonic on B, w is

identically −∞ or harmonic on B by Theorem 2.2.7. If w is not identically

−∞ on Ω, then it is superharmonic on Ω and therefore harmonic on each

ball B since superharmonic functions are ﬁnite a.e. on compact subsets of Ω

by Theorem 2.4.2.

Given a boundary function f, classes of functions that approximate the

solution of the Dirichlet problem from above and from below will be consid-

ered.

Deﬁnition 2.6.3 An extended real-valued function u on Ω is hyperhar-

monic (hypoharmonic) if it is superharmonic (subharmonic) or identically

+∞ (−∞) on each component of Ω.

Deﬁnition 2.6.4 The upper class U

of functions determined by the func-

tion f on ∂Ω is given by

= {u; u hyperharmonic on Ω, lim inf

y→x,y∈Ω

u(y) ≥ f (x)

for all x ∈ ∂Ω,u bounded below on Ω}.

The lower class L

of functions determined by the function f on ∂Ω is given

= {u; u hypoharmonic on Ω, lim sup

y→x,y∈Ω

u(y) ≤ f (x)

for all x ∈ ∂Ω,u bounded above on Ω}.

Note that U

contains the function that is identically +∞ on Ω,andL

contains the function that is identically −∞ on Ω.AlsonotethatU

)

may contain only the identically +∞ (−∞) function, and also that the re-

quirement that elements of U

be bounded below may not be fulﬁlled by any

nontrivial hyperharmonic function.

2.6 Perron-Wiener Method 77

Deﬁnition 2.6.5 The Perron upper solution for the boundary function

f is given by

=inf{u; u ∈ U

}.ThePerron lower solution is given by

=sup{u; u ∈ L

Whenever it is necessary to compare upper classes and upper solutions

for two diﬀerent regions Ω and Σ, the upper classes will be denoted by U

and U

and the upper solutions by H

and H

, respectively, with a similar

convention for lower classes and lower solutions. As a result of the following

lemma, it can be assumed usually that Ω is connected.

Lemma 2.6.6 If f is a function on ∂Ω and Λ is a component of Ω,then

f|∂Λ

= H

and H

f|∂Λ

= H

Proof: Let U

f|∂Λ

be the upper class for Λ and f|

∂Λ

. It will be shown ﬁrst

that U

f|∂Λ

⊂ U

= {u|

; u ∈ U

}.Ifu ∈ U

f|∂Λ

,letu

∗

= u on Λ and

∗

=+∞ on Ω ∼ Λ.Thenu

∗

∈ U

and U

f|∂Λ

⊂ U

. Suppose now that

u ∈ U

and x ∈ ∂Λ.Thenx ∈ ∂Ω and

lim inf

y→x,y∈Λ

u(y) ≥ lim inf

y→x,y∈Ω

u(y) ≥ f (x),x∈ ∂Λ.

Therefore, u|

∈ U

f|∂Λ

and U

⊂ U

f|Λ

.Thus,U

f|∂Λ

= U

and H

f|∂Λ

Lemma 2.6.7 H

) is identically −∞ (+∞) or harmonic on each com-

ponent of Ω.

Proof: ItcanbeassumedthatΩ is connected. If U

contains only the iden-

tically +∞ function, then

=+∞ and the assertion is true. Suppose U

contains a hyperharmonic function that is not identically +∞ on Ω.Then

=inf{u; u ∈ U

,u superharmonic on Ω}. If it can be shown that the

family of superharmonic functions in U

is a saturated family, it would then

follow from Theorem 2.6.2 that

is either identically −∞ or harmonic

on Ω. Accordingly, let u

and u

be two superharmonic members of U

.If

x ∈ ∂Ω, then lim inf

y→x,y∈Ω

(y) ≥ f(x) and it is easy to show that

lim inf

y→x,y∈Ω

min (u

)(y) ≥ f(x).

Since u

and u

are each bounded below on Ω, the same is true of the super-

harmonic function min (u

) and it has been shown that min (u

) ∈ U

If u is a superharmonic function in U

and B is a ball with B

−

⊂ Ω,then

is superharmonic on Ω by Lemma 2.4.11,

lim inf

y→x,y∈Ω

(y) = lim inf

y→x,y∈Ω

u(y) ≥ f (x)