Helms L.L. Potential Theory

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348 9 Elliptic Operators

Proof: For the time being, ﬁx x ∈ Ω. Letting δ = d(x)/2,B = B

x,δ

,and

applying Theorem 9.4.3,

1+b

(x)|Du(x)| + d

2+b

(x)|D

u(x)|

≤ d

(x)|u; d|

2+α,B

≤ d

(x)C



sup

y∈B

|u(y)|+sup

y∈B

(y)|f(y)|

+sup





∈B

2+α





)

|f(y



) − f(y





− y





≤ C



sup

y∈B

(x)|u(y)|+sup

y∈B

(x)d

(y)|f(y)|

+sup





∈B

(x)d

2+α





)

|f(y



) − f (y





− y





Since d(y) <δ+ d(x)=(3/2)d(x)fory ∈ B, d(x) > (2/3)d(y)and

1+b

(x)|Du(x)| + d

2+b

(x)|D

u(x)|≤C



|u; d|

(b)

0,B

+ |f ; d|

(2+b)

α,B



≤ C



|u; d|

(b)

0,Ω

+ |f ; d|

(2+b)

α,Ω



Taking the supremum over x ∈ Ω,

|u; d|

(b)

2+0,Ω

≤ C



|u; d|

(b)

0,Ω

+ |f ; d|

(2+b)

α,Ω



. (9.24)

Consider now two points x, y ∈ Ω labeled so that d(x) ≤ d(y)withx = y

and deﬁne δ and B as before. As above, for |x − y|≤δ

2+α+b

(x, y)

u(x) − D

u(y)|

|x − y|

≤ d

(x)|u; d|

2+α,B

≤ C



|u; d|

(b)

0,Ω

+ |f ; d|

(2+b)

α,Ω



On the other hand, for |x − y| >δ

2+α+b

(x, y)

u(x) − D

u(y)|

|x − y|

≤ d

2+α+b

(x)

(|D

u(x)| + |D

u(y)|)

≤ 2

2+b

(x)|D

u(x)| + d

2+b

(y)|D

u(y)|)

≤ 4|u; d|

(b)

2+0,Ω

≤ 4C(|u; d|

(b)

0,Ω

+ |f ; d|

(2+b)

α,Ω

the latter inequality following from Inequality (9.24). Taking the supremum

over x, y ∈ Ω and combining the result with Inequality (9.24),

|u; d|

(b)

2+α,Ω

= |u; d|

(b)

2+0,Ω

+[u; d]

(b)

2+α,Ω

≤ C(|u; d|

(b)

0,Ω

+ |f ; d|

(2+b)

α,Ω

9.6 The Dirichlet Problem for a Ball 349

Lemma 9.6.2 Let L be a strictly elliptic operator with c ≤ 0 on a ball B =

,ρ

⊂ R

satisfying

|, |b

|, |c|≤K on B, i, j =1,...,n.

If u ∈ C

−

) ∩ C

(B),α ∈ (0, 1),b ∈ (−1, 0),f ∈ H

(2+b)

(B; d), Lu = f on

B,andu =0on ∂B,then|u; d|

(b)

0,B

≤ C|f; d|

(2+b)

0,B

where C = C(n, b, ρ, m, K).

Proof: Since f ∈ H

(2+b)

(B; d),

A =sup

x∈B

2+b

(x)|f(x)| < +∞.

The inequality is trivial if A =0,forthenf =0andu =0bytheweak

maximum principle, Corollary 9.5.3. It can be assumed that A>0. Let

(x)=(ρ

− r

)

−b

where r = |x − x

|.Ifx =(x

,...,x

)andx

,...,x

), then

(x)=



i=1

(2b)(ρ

− r

)

−b−1



i,j=1

4b(b +1)(ρ

− r

)

−b−2

− x

)(x

− x

)



i=1

(2b)(ρ

− r

)

−b−1

− x

)+c(x)(ρ

− r

)

−b

= b(ρ

− r

)

−b−2





i,j=1

4(b +1)a

− x

)(x

− x

)

+2b(ρ

− r

)

−b−1





i=1

+ b

− x

))



c(x)

(ρ

− r

)



≤ b(ρ

− r

)

−b−2



4(1 + b)mr

+2(ρ

− r

)(mn − K

√

nr)



Since the ﬁrst term within the parentheses has the limit 4(1 + b)mρ

r → ρ,thereis0<ρ

<ρsuch that Lw

(x) ≤−k(ρ

− r

)

−b−2

for some

constant k and ρ

≤ r<ρ.Since(ρ

− r

)

−b−2

=(ρ −r)

−b−2

(ρ + r)

−b−2

≥

(ρ − r)

−b−2

(2ρ)

−b−2

, there are positive constants c

and c

such that

(x) ≤



(ρ − r)

−b−2

0 ≤ r<ρ

−c

(ρ − r)

−b−2

≤ r<ρ.

(9.25)

Now let w

(x)=e

aρ

− e

a(x

−x

)

where a is a positive constant that will be

speciﬁed later. Then

(x)=−a

a(x

−x

)

− b

a(x

−x

)

+ c(x)(e

aρ

− e

a(x

−x

)

350 9 Elliptic Operators

Since c ≤ 0,e

aρ

− e

a(x

−x

)

≥ 0, and a

≥ m,

(x) ≤ e

a(x

−x

)

a(−ma + |b

= −me

a(x

−x

)



a −



Choosing a so that a ≥ 1+sup

(|b

|/m),



a −



≥



1+sup



1+sup

−



≥ 1

and therefore

(x) ≤−me

a(x

−x

)

≤−me

−aρ

on B.

Note that Lw

(x) ≤ 0onB.For0≤ r<ρ

, using the fact that b ∈ (−1, 0)

(x) ≤−me

−aρ

(ρ − r)

−b−2

(ρ − r)

2+b

≤−c

(ρ − r)

−b−2

where c

= me

−aρ

(ρ − ρ

)

2+b

. Therefore,

(x) ≤



−c

(ρ − r)

−b−2

if 0 ≤ r<ρ

0ifρ

≤ r<ρ.

(9.26)

Consider the constants η

=1/c

, η

=(1+

)/c

, and the function w =

+ η

. By Inequalities (9.25) and (9.26),

Lw(x) ≤−(ρ − r)

−b−2

= −d

−b−2

(x) ≤−

|f(x)|

Therefore, L(Aw ± u)(x) ≤−|f(x)|±f(x) ≤ 0. When r = ρ, w = η

aρ

−

a(x

−x

)

) ≥ 0. Since u =0on∂B by hypothesis, Aw ± u ≥ 0on∂B.By

the minimum principle, Corollary 9.5.3, Aw ±u ≥ 0onB;thatis,

|u(x)|≤Aw(x)onB.

An upper bound for the function w can be obtained as follows. First note that

(x)=(ρ

− r

)

−b

=(ρ − r)

−b

(ρ + r)

−b

≤ ρ

−b

(ρ − r)

−b

on B.

Since w

(x)=e

aρ

−e

a(x

−x

)

depends only upon the ﬁrst coordinate of x,it

canbeassumedthatx is on the x

-axis. When x

− x

= ρ, w

(x)=0.By

the mean value theorem, |w

(x)| = |∇

(x)

(z)|(ρ −r) ≤ ae

aρ

(ρ −r)

−b

1+b

(ρ−r)

−b

where z lies on the line segment joining x to x

+(ρ, 0,...,0). Thus,

|u(x)|≤AC(ρ − r)

−b

= ACd

−b

(x)

9.6 The Dirichlet Problem for a Ball 351

so that

|u; d|

(b)

0,B

=sup

x∈B

(x)|u(x)|≤AC = C sup

x∈B

2+b

(x)|f(x)| = C|f; d|

(2+b)

0,B

Theorem 9.6.3 If α ∈ (0, 1),b ∈ (−1, 0), B is a ball in R

,andL is a

strictly elliptic operator on Ω with coeﬃcients satisfying

α,B

, |b

α,B

, |c|

α,B

≤ K<+∞,i,j=1,...,n. (9.27)

then L is a bounded, linear map from H

(b)

2+α

(B; d) into H

(2+b)

(B; d).

Proof: The assertion follows if it can be shown that there is a constant C

such that

u; d|

(2+b)

α,B

, |b

u; d|

(2+b)

α,B

, |cu; d|

(2+b)

α,B

≤ C|u; d|

(b)

2+α,B

for u ∈ H

(b)

2+α

(B; d). By Inequality (9.10),

u; d|

(2+b)

α,B

≤|a

; d|

(0)

α,B

u; d|

(2+b)

α,B

≤ max (1,d(B)

)K|D

u; d|

(2+b)

α,B

u; d|

(2+b)

α,B

≤|b

; d|

(0)

α,B

u; d|

(2+b)

α,B

≤ max (1,d(B)

)K|D

u; d|

(2+b)

α,B

|cu; d|

(2+b)

α,B

≤|c; d|

(0)

α,B

|u; d|

(2+b)

α,B

≤ max (1,d(B)

)K|u; d|

(2+b)

α,B

so that it suﬃces to show that there is a constant C such that |D

u; d|

(2+b)

α,B

u; d|

(2+b)

α,B

, |u; d|

(2+b)

α,B

≤ C|u; d|

(b)

2+α

. An upper bound for the second-order

term follows from the inequality

u; d|

(2+b)

α,B

≤ [u; d]

(b)

2+0,B

+[u; d]

(b)

2+α,B

≤|u; d|

(b)

2+α,B

Consider two points x, y ∈ B so labeled that d(x) ≤ d(y). A simple convexity

argument shows that if z is a point on the line segment joining x to y,then

d(z) ≥ d(x). Applying the mean value theorem,

u; d|

2+b)

α,B

=sup

x∈B

2+b

(x)|D

u(x)| +sup

x,y∈B

2+b+α

(x, y)

u(x) − D

u(y)|

|x − y|

≤ d(B)[u; d]

(b)

1+0,B

+ d(B)

√

n[u; d]

(b)

2+0,B

≤ d(B)(1 +

√

n)|u; d|

(b)

2+α,B

Another application of the mean value theorem results in the inequality

|u; d|

(2+b)

α,B

≤ d

(B)[u; d]

(b)

0,B

+ d

(B)

√

n[u; d]

(b)

1+0,B

≤ d

(B)(1 +

√

n)|u; d|

(b)

2+α,B

352 9 Elliptic Operators

This shows that

|Lu; d|

(2+b)

α,B

≤ C|u; d|

(b)

2+α,B

for all u ∈ H

(b)

2+α

(B; d).

Lemma 9.6.4 If α ∈ (0, 1),b ∈ (−1, 0), B = B

,ρ

is a ball in R

,and

f ∈ H

(2+b)

(B; d), then there is a unique u ∈ C

−

) ∩H

(b)

2+α

(B; d) such that

Δu = f on B and u =0on ∂B.

Proof: It is easily seen that f ∈ H

α,loc

(B). By Theorem 8.4.4, there is a

unique u ∈ C

−

) ∩ C

(B) such that Δu = f on B and u =0on∂B.By

Lemma 9.3.1,

|u; d|

2+α,B

≤ C(u

0,B

+ |f ; d|

(2)

α,B

) ≤ C(u

0,B

+ ρ

−b

|f; d|

(2+b)

α,B

) < +∞

so that u ∈ H

2+α

(B; d). By Theorem 8.4.4,

[u; d]

(b)

0,B

≤ [f ; d]

(2+b)

0,B

≤|f ; d|

2+b)

α,B

< +∞.

The hypotheses of Lemma 9.6.1 are satisﬁed and so u ∈ H

(b)

2+α

(B; d).

Theorem 9.6.5 Let B be a ball in R

,letα ∈ (0, 1),b ∈ (−1, 0),andletL

be a strictly elliptic operator on B with c ≤ 0 satisfying Inequality (9.27). If

|f; d|

(2+b)

α,B

< +∞, then there is a unique u ∈ C

−

) ∩H

(b)

2+α

(B; d) satisfying

Lu = f on B, u =0on ∂B,and

|u; d|

(b)

2+α,B

≤ C(|u; d|

(b)

0,B

+ |f; d|

(2+b)

α,B

) < +∞ (9.28)

where C = C(n, α, b, m, K).

Proof: For 0 ≤ t ≤ 1, let

u = tLu +(1− t)Δu.

Since L and Δ are bounded linear operators from the Banach space B

(b)

2+α

(B; d) into the Banach space B

= H

(2+b)

(B; d), the same is true of each

with the m in Inequality (9.2) replaced by m

=min(1,m)andtheK in

Inequality (9.27) replaced by K

=max(1,K). Consider the Dirichlet prob-

lem L

u = f on B and u =0on∂B. Suppose there is a u ∈ H

(b)

2+α

(B; d)sat-

isfying the equation L

u = f.Sinced(x)

(b)

|u(x)|≤[u; d]

(b)

0,B

≤|u; d|

(b)

2+α,B

+∞, the function u has a continuous extension, denoted by the same symbol,

to B

−

with u =0on∂B. By Lemma 9.6.2,

|u; d|

(b)

0,B

≤ C|f; d|

(2+b)

0,B

≤ C|f; d|

(2+b)

α,B

< +∞.

9.7 Dirichlet Problem for Bounded Domains 353

Using the fact that d(B)

≤ d(x)

for x ∈ B, it is easily seen that u ∈

(b)

2+α

(B; d) implies u ∈ H

2+α

(B; d) and it follows from Lemma 9.6.1 that

|u; d|

(b)

2+α,B

≤ C

(|u; d|

(b)

0,B

+ |f ; d|

(2+b)

α,B

) ≤ C|f; d|

(2+b)

α,B

or, equivalently,

u

≤ CL

u

where C is independent of t. Since the Laplacian L

= Δ is an invertible map

from B

onto B

by Lemma 9.6.4, it follows from Theorem 9.2.2 that L is

invertible. Thus, given f ∈ H

(2+b)

(B; d)thereisau ∈ H

(b)

2+α

(B; d) such that

Lu = f on B and u =0on∂B. Uniqueness follows from Theorem 9.5.4.

Theorem 9.6.6 Let B be a ball in R

,letα ∈ (0, 1),b ∈ (−1, 0),andlet

L be a strictly elliptic operator with c ≤ 0 and coeﬃcients satisfying In-

equalities (9.22). If |f; d|

(2+b)

α,B

< +∞ and g ∈ C

(∂B),thereisaunique

u ∈ C

−

) ∩ H

2+α,loc

(B) that solves the problem Lu = f on B and u = g

on ∂B.

Proof: Consider any g ∈ C

−

) and the Dirichlet problem Lv = f −Lg on

B and v =0on∂B.Sincef − Lg ∈ H

(2+b)

(B; d), by the preceding theorem

there is a unique v ∈ C

−

) ∩ H

(b)

2+α

(B; d) that solves this problem and

satisﬁes Inequality (9.28). Letting u = v + g, Lu = f on B, u = g on ∂B,and

u ∈ C

−

)∩H

(b)

2+α

(B; d). Consider any g ∈ C

(∂B). It can be assumed that

g ∈ C

−

) by the Tietze Extension Theorem. Now let {g

} be a sequence

in C

−

) that converges uniformly to g on B

−

.Foreachj ≥ 1, let v

the solution of the problem Lv

= f − Lg

on B and v

= g

on ∂B as in

the preceding step. Letting u

= v

+ g

, Lu

= f on B, u

= g

on ∂B,

and u

∈ C

−

) ∩ H

(b)

2+α

(B; d). Since L(u

− u

)=0fori, j ≥ 1, by the

weak maximum principle, Corollary 9.5.3, u

−u



0,B

−

≤g

−g



0,∂B

,the

sequence {u

} is Cauchy on B

−

, and therefore uniformly bounded on B

−

.By

Corollary 9.4.4, there is a subsequence that converges uniformly on compact

subsets of B to a function u ∈ H

2+α,loc

(B)forwhichLu = f on B and u = g

on ∂B. Since the sequence {u

} converges uniformly on B

−

, u also belongs

to C

−

9.7 Dirichlet Problem for Bounded Domains

Throughout this section, Ω will be a bounded open subset of R

, L will be

a strictly elliptic operator with c ≤ 0 and coeﬃcients a

,c,i,j =1,...n,

in H

α,loc

(Ω). Unless otherwise noted, f will be a ﬁxed function in C

(Ω) ∩

(2+b)

(Ω; d),α ∈ (0, 1),b∈ (−1, 0).

354 9 Elliptic Operators

Deﬁnition 9.7.1 u ∈ C

(Ω

−

)isasuperfunction if for every ball B with

closure in Ω and v ∈ C

−

) ∩ C

(B),

Lv = f on B,u ≥ v on ∂B implies u ≥ v on B.

Reversing the inequalities of this deﬁnition results in the deﬁnition of a sub-

function.Thefactthatu is a superfunction does not mean that −u is a

subfunction. Otherwise, superfunctions and subfunctions share most of the

properties of superharmonic and subharmonic functions.

Deﬁnition 9.7.2 If B = B

x,δ

,α ∈ (0, 1),b ∈ (−1, 0), |f; d|

(2+b)

α,B

< +∞,and

g ∈ C

−

), LS(x, B, g) will denote the unique solution of the problem

Lv = f on B, v = g on ∂B.

Lemma 9.7.3 u ∈ C

(Ω

−

) is a superfunction if and only if u ≥ LS(x, B, u)

on B for every ball B = B

x,δ

with closure in Ω.

Proof: Suppose ﬁrst that u is a superfunction on Ω and let v = LS(x, B, u)

on B.ThenLv = f on B and v = u on ∂B.Thus,u ≥ v = LS(x, B, u)

on B. On the other hand, suppose u ≥ LS(x, B, u)onB.Considerany

v ∈ C

−

) ∩ C

(B) satisfying

Lv = f on B and u ≥ v on ∂B.

Since L(LS(x, B, u) − LS(x, B, v)) = 0 on B and LS(x, B, u) ≥ LS(x, B, v)

on ∂B, by Theorem 9.5.4

u ≥ LS(x, B, u) ≥ LS(x, B, v)=v on B,

showing that u is a superfunction on Ω.

Lemma 9.7.4 u ∈ C

(Ω

−

)∩C

(Ω) is a superfunction if and only if Lu ≤ f

on Ω; moreover, if u ∈ C

(Ω

−

) ∩ C

(Ω) is a superfunction and k ≥ 0,then

u + k is a superfunction.

Proof: Suppose ﬁrst that Lu ≤ f on Ω.LetB be a ball with closure in

Ω and let v ∈ C

−

) ∩ C

(B)satisfyLv = f on B and u ≥ v on ∂B.

By Theorem 9.5.4, u ≥ v on B and u is a superfunction. Now let u be

a superfunction and assume that Lu(x

) >f(x

)forsomepointx

∈ Ω.

Then there is a ball B = B

,δ

such that Lu>f on B. By Theorem 9.6.6,

there is a function u

on B

−

such that Lu

= f on B and u

= u on

∂B.SinceL(u − u

) >f− f =0onB, by the weak maximum principle,

Corollary 9.5.3, u − u

≤ sup

∂B

(u − u

)

=0onB so that u ≤ u

on B.

Since u is a superfunction, u ≥ u

on B, and therefore u = u

on B so that

Lu = Lu

on B.Thus,f(x

)=Lu

)=Lu(x

) >f(x

), a contradiction,

and it follows that Lu ≤ f on Ω. The second assertion follows easily from

the ﬁrst and the fact that Lk = ck ≤ 0.

9.7 Dirichlet Problem for Bounded Domains 355

Lemma 9.7.5 If v ∈ C

(Ω

−

) is a superfunction, u ∈ C

(Ω

−

) is a subfunc-

tion, and v ≥ u on ∂Ω,thenv ≥ u on Ω.

Proof: Let m =inf

−

(v − u)andletΓ = {y ∈ Ω; v(y) − u(y)=m}.If

Γ = ∅,thenv − u attains its minimum value on ∂Ω, and therefore v −u ≥ 0

on Ω

−

since v − u ≥ 0on∂Ω. Assume Γ = ∅ and let z be a point of Γ such

that d(z)=d(Γ, ∂Ω) > 0. Let B = B

z,δ

be a ball with closure in Ω.Note

that B ∩ (Ω ∼ Γ ) = ∅ since z ∈ ∂Γ. By Theorem 9.6.6, there is then a v

∈

−

) ∩C

(B) satisfying Lv

= f on B and v

= v on ∂B.Thus,v

≤ v

on B since v is a superfunction. Likewise, there is a u

∈ C

−

) ∩ C

(B)

satisfying Lu

= f on B, u

= u on ∂B,andu

≥ u on B.Notethat

L(v

− u

)=0onB and that m = v(z) − u(z) ≥ v

(z) − u

(z). Since

− u

= v − u ≥ m on ∂B, v

− u

≥ m on B by the weak maximum

principle, Corollary 9.5.3. It follows that v

(z) −u

(z)=m so that v

−u

attains its minimum value at an interior point of B;ifm ≤ 0, then v

− u

would be constant on B by the strong maximum principle, Theorem 9.5.6.

But this contradicts the fact that B ∩ (Ω ∼ Γ ) = ∅ where v

− u

>m.

Thus, m ≥ 0andv ≥ u on Ω.

Lemma 9.7.6 If u ∈ C

(Ω

−

) is a superfunction, B is a ball with closure in

Ω,Lv = f on B, v = u on ∂B,then



v on B

u on Ω ∼ B

is a superfunction. If u and v are superfunctions on Ω and u ≥ v on Ω,then

≥ v

on Ω.

Proof: Let B

be an arbitrary ball with closure in Ω,andletw

∈ C

−

)∩

)satisfyLw

= f on B

= u

on ∂B

.Sinceu is a superfunction,

u ≥ u

on Ω and, in particular, u ≥ w

on ∂B

. Using the fact that u is a

superfunction again, u ≥ w

on B

.Sinceu

= u on Ω ∼ B,u

= u ≥ w

on B

∼ B.NotingthatLu

= Lv = f on B, L(u

− w

)=0onB ∩ B

Since u

= u ≥ w

on ∂B ∩ B

and u

≥ w

on ∂B

∩ B, u

− w

≥ 0on

(∂B∩B

)∪(∂B

∩B). By continuity, u

−w

≥ 0on∂(B∩B

) ⊂ (∂B∩B

−

)∪

(∂B

∩ B

−

). It follows from the weak maximum principle, Corollary 9.5.3,

that u

≥ w

on B ∩ B

and therefore on B

. This proves that u

is a

superfunction. Let u and v be superfunctions on Ω with u ≥ v on Ω.Since

L(u

− v

)=0onB and u

− v

= u − v ≥ 0on∂B, u

− v

≥ 0bythe

weak maximum principle.

The function u

of the preceding lemma will be called the lowering of

u over B; the corresponding operation for subfunctions will be called the

lifting of u over B and will be denoted by u

356 9 Elliptic Operators

Lemma 9.7.7 If u

∈ C

(Ω

−

) are superfunctions, then min (u

) is a

superfunction.

Proof: Let u =min(u

) and consider any ball B with closure in Ω.

Suppose Lw = f on B and u ≥ w on ∂B.Thenu

≥ w on ∂B so that u

≥ w

on B, i =1, 2. Therefore, u =min(u

) ≥ w on B and u is a superfunction.

Deﬁnition 9.7.8 A function v ∈ C

(Ω

−

)isasupersolution relative to

the bounded function g on ∂Ω if v is a superfunction and v ≥ g on ∂Ω.

The set of supersolutions relative to g will be denoted by U(f,g). There is

a corresponding deﬁnition of subsolution relative to the bounded function

g on ∂Ω; the set of such functions will be denoted by L(f,g). According to

Lemma 9.7.5, each subsolution relative to g is less than or equal to each super-

solution relative to g. The hypothesis of the following lemma is an exception

to the requirement that f be a ﬁxed function.

Lemma 9.7.9 If F⊂C

(Ω) ∩ H

(2+b)

(Ω; d) with sup

f∈F

f

0,Ω

≤ M<

+∞, G is a collection of bounded functions on ∂Ω with sup

g∈G

g

0,∂Ω

≤

N<+∞,then

f∈F,g∈G



U(f,g) ∩ C

(Ω

−

)



= ∅ and

f∈F,g∈G



L(f,g) ∩C

(Ω

−

)



= ∅;

moreover, the set



f∈F,g∈G

U(f,g)





f∈F,g∈G

L(f,g)



is uniformly bounded

below (above) by a constant.

Proof: Choose d>0sothatΩ ⊂{x; −d ≤ x

≤ d}.Let

β =sup

x∈Ω,1≤i≤n

(|b

(x)|/m),

let ρ>β+1,and let L

= L − c.Sinceβ<ρ− 1,

ρ(d+x

)

= a

ρ(d+x

)

+ b

ρe

ρ(d+x

)

≥ mρ(ρ − β)e

ρ(d+x

)

≥ m.

Letting

= N +

2ρd

− e

ρ(d+x

)

)M,

= L

+ cv

= L

(−e

ρ(d+x

)

)M/m + cv

≤−M ≤−f 

≤ f

for all f ∈F.Sincev

∈ C

(Ω

−

) ∩ C

(Ω)andv

≥ N ≥ g on ∂Ω,v

is a

supersolution relative to g for all g ∈Gand therefore

f∈F,g∈G



U(f,g) ∩ C

(Ω

−

)



= ∅.

9.7 Dirichlet Problem for Bounded Domains 357

Now let

−

= −N −

2ρd

− e

ρ(d+x

)

)M.

Then

−

= L

−

+ cv

−

≥ L

ρ(d+x

)

)M/m ≥ M ≥f

≥ f.

for all f ∈F.Sincev

−

∈ C

(Ω

−

) ∩C

(Ω)andv

−

≤−N ≤ g on ∂Ω,v

−

is a

subsolution relative to g for all g ∈Gand it follows that

f∈F,g∈G



L(f,g) ∩ C

(Ω

−

)



= ∅.

Consider now a ﬁxed f and g. Since each subsolution is less than or

equal to each supersolution, U(f, g) is bounded below by each element of

L(f,g)andL(f, g) is bounded above by each element of U(f,g). Since

≤ N + e

2ρd

M/m, the elements of L(f, g) are uniformly bounded above by

the constant N + e

2ρd

M/m. Similarly, the elements of U(f, g) are uniformly

bounded below by the constant −N − e

2ρd

M/m.

Deﬁnition 9.7.10 The Perron supersolution for the bounded function g

on ∂Ω is the function H

f,g

=inf{v; v ∈ U (f, g)} and the Perron subsolu-

tion for g is the function H

−

f,g

=sup{v; v ∈ L(f,g)}.

It can be assumed that U(f,g) is uniformly bounded on Ω by considering

only those u for which u ≤ u

for some u

∈ U(f,g)since

f,g

=inf{u ∧ u

; u ∈ U(f, g)}.

It follows that H

f,g

is bounded on Ω as is H

−

f,g

. Clearly, H

−

f,g

≤ u for all

u ∈ U(f,g) and consequently H

−

f,g

≤H

f,g

. The same assumption can be

made for L(f, g).

Deﬁnition 9.7.11 If H

−

f,g

= H

f,g

on Ω,theng is said to be L-resolutive

and H

f,g

is deﬁned to be the common value.

Lemma 9.7.12 If g is a bounded function on ∂Ω,thenH

f,g

∈ H

2+α

(B; d)

for every closed ball B with B

−

⊂ Ω and satisﬁes Poisson’s equation LH

f,g

f on Ω;inparticular,H

f,g

∈ C

(Ω).

Proof: By Lemma 2.2.8, there is a countable set I

⊂ U(f,g) such that for

every l.s.c. function w on Ω, w ≤H

f,g

whenever w ≤ inf {v; v ∈ I

}.By

Lemma 9.7.7, it can be assumed that I

is a decreasing sequence {u

} of

uniformly bounded functions on Ω.Letu = lim

j→∞

and let B be a ﬁxed

ball with B

−

⊂ Ω. By replacing each u

by (u

)

, if necessary, it can be

assumed that each u

satisﬁes the equation Lu

= f on B. By Theorem 9.4.4,

there is a subsequence {u

} that converges uniformly on compact subsets

of B to a function w ∈ H

2+α

(B; d)forwhichLw = f on B.Sincew = u