Helms L.L. Potential Theory

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278 7 Interpolation and Monotonicity

If c is a positive integer, the expression [u]

in the statement of the following

theorem can be interpreted as [u]

c+0

or [u]

(c−1)+1

since [u]

(c−1)+1

≤ C[u]

c+0

If the theorem is proved for [u]

c+0

,thenitisalsotruefor[u]

(c−1)+1

Theorem 7.3.4 (H¨ormander [34]) If 0 ≤ a<c<b,0 <λ<1,and

c = λa +(1− λ)b,then

[u]

≤ C[u]

[u]

1−λ

(7.3)

and

|u|

≤ C|u|

|u|

1−λ

. (7.4)

where C = C(a, b, d(Ω)) is a constant.

Proof: If [u]

≤ [u]

, then by Theorem 7.3.2

[u]

≤ C[u]

= C[u]

[u]

1−λ

≤ C[u]

[u]

1−λ

where C = C(b, d(Ω)). It therefore can be assumed that [u]

< [u]

for the

remainder of the proof, which will be dealt with in several steps.

Step 1 Consider the case that there is no integer between a and b. Suppose

ﬁrst that 0 ≤ a<c<b≤ 1. Then for x, y ∈ Ω,

|u(x) − u(y)|

|x − y|



|u(x) − u(y)|

|x − y|





|u(x) − u(y)|

|x − y|



1−λ

so that [u]

≤ [u]

[u]

1−λ

when a>0and[u]

≤ 2

[u]

1−λ

when a =0.In

either case, [u]

≤ 2[u]

[u]

1−λ

≤ 2|u|

|u|

1−λ

. Suppose now that k ≤ a<c<

b ≤ k + 1 for some integer k ≥ 1. By the preceding argument,

[u]

=sup

|α|=k

c−k

≤

2sup

|α|=k



a−k

1−λ

b−k



≤ 2[u]

[u]

1−λ

≤ 2|u|

|u|

1−λ

Step 2 Suppose now that c is an integer k ≥ 1. It can be assumed that

|u|

> 0 for otherwise u = 0, a trivial case that can be excluded. For x ∈ Ω

and 0 <<1, let



(y)=u((1 − )x + y),y∈ Ω.

It is easily seen that [u



]

≤ 

[u]

,[u



]

≤ 

[u]

,andD



(x)=

u(x)if

|β| = k.Since0< [u]

< [u]

, 0 <<1 can be chosen so that 

[u]

= 

[u]

Since |u



/

[u]

≤ 1and|u



/

[u]

≤ 1, it follows from Lemma 7.3.3 that





(x)



[u]



≤ C(b, d(Ω)) if |β| = k,

7.3 Global Interpolation 279

where C(b, d(Ω)). Thus,



u(x)|=|D



(x)|≤C(b, d(Ω))

[u]

=C(b, d(Ω))(

[u]

)

(

[u]

)

1−λ

Since c = k = λa +(1− λ)b, |D

u(x)|≤C(b, d(Ω))[u]

[u]

1−λ

.Takingthe

supremum over x ∈ Ω and β satisfying |β| = k,

[u]

≤ C(b, d(Ω))[u]

[u]

1−λ

≤ C(b, d(Ω))|u|

|u|

1−λ

. (7.5)

Step 3 Let c(λ)=λa +(1− λ)b, 0 ≤ λ ≤ 1, and assume that the inequality

[u]

c(λ)

≤ C(a, b, d(Ω))|u|

|u|

1−λ

(7.6)

holds when λ = λ

and λ = λ

.If0<μ

< 1,μ

=1− μ

,andλ =

+ λ

,thenc(λ)=μ

c(λ

)+μ

c(λ

)=μ

c(λ

)+(1− μ

)c(λ

). By

Step 1, if there is no integer between c(λ

)andc(λ

), then

[u]

c(λ)

=[u]

c(λ

)+(1−μ

)c(λ

)

≤ 2[u]

c(λ

)

[u]

1−μ

c(λ

)

≤ C(a, b, d(Ω))



([u]

[u]

1−λ





[u]

1−λ



≤ C(a, b, d(Ω))|u|

|u|

1−λ

;

that is, Inequality (7.6) holds for c(λ) for all λ between λ

and λ

providing

there is no integer between c(λ

)andc(λ

). Note that the inequality holds

for λ =1andλ =0.

Step 4 Consider any c ∈ (a, b). If there is no integer in (a, b), then [u]

≤

2|u|

|u|

1−λ

by Step 1. If c is an integer in (a, b), then [u]

≤|u|

|u|

1−λ

by Step

2. Suppose c is not an integer and there are one or more integers k

,...,k

(a, b) ordered so that a<k

< ···<k

<b. By considering the three cases

c ∈ (a, k

),c ∈ (k

i−1

), and c ∈ (k

,b) and the fact that Inequality (7.5)

holds for each k

, Inequality (7.5) holds for each c ∈ (a, b). This completes

the proof of Inequality (7.3).

Step 5 Suppose a = j + α and c =  + γ,then

|u|



i=0

[u]



i=j+1

[u]

+[u]

For i =0,...,j,[u]

≤|u|

≤ C(a, b, d(Ω))|u|

|u|

1−λ

Theorem 7.3.2. For i = j +1,...,,a≤ i<bso that

[u]

≤ C(a, b, d(Ω))|u|

|u|

1−λ

by Step 2. Lastly, [u]

≤ C(a, b, d(Ω))|u|

|u|

1−λ

by the preceding step.

280 7 Interpolation and Monotonicity

In applying Step 3 to the interval (k

i−1

), C = C(k

i−1

,d(Ω)) can be

replaced by C(a, b, d(Ω)) since the k



s can be expressed as explicit functions

of a and b.

Theorem 7.3.5 If 0 ≤ a<c<band >0, there is a constant C =

C(a, b, c, , d(Ω)) such that

[u]

≤ C[u]

+ [u]

(7.7)

and

|u|

≤ C|u|

+ |u|

. (7.8)

Proof: Letting c = λa +(1−λ)b, 0 <λ<1, the result follows from Inequal-

ity (7.3) when [u]

= 0. It therefore can be assumed that [u]

= 0. It suﬃces

to prove the result when [u]

=1forthen

(

[u]

)

≤ C

(

[u]

)

+ 

implies that [u]

≤ C[u]

+ [u]

.Inthecase|u|

=1,[u]

≤ C[u]

Theorem 7.3.4, where C = C(a, b, d(Ω)). It is easily seen that there is a

linear function Kx + /C such that x

≤ Kx + /C for x ≥ 0whereK

depends upon , c,andC;infact,

K = λ





C(1 −λ)



(λ−1)/λ

where λ is determined from c.Thus,

[u]

≤ C[u]

≤ C(K[u]

+ /C)=CK[u]

+ ,

where CK may depend upon c through λ. The second assertion follows in

exactly the same way.

7.4 Interpolation of Weighted Norms

In this section, Ω can be any bounded open subset of R

.Convexityisnot

required. The section applies to any weight function φ>0 satisfying the

inequality |φ(x) − φ(y)|≤M|x − y|

,x,y ∈ Ω,forsomeﬁxedα ∈ (0, 1]. As

the results will be applied only to the weight function φ(x)=d(x)=d(x, ∂Ω),

the results of this section will be stated only for d. In the proof of the following

lemma, D

u will denote a generic symbol for a jth order partial derivative of

u. The lemma is a special case of a more general result that can be found in

[25] allowing arbitrary j, k ≥ 0 . As the consideration of many cases is required

in the proof, only a few cases will be dealt with to illustrate techniques.

7.4 Interpolation of Weighted Norms 281

Lemma 7.4.1 If j + α<2+β where j =0, 1, 2, 0 ≤ α, β ≤ 1,andu ∈

2+β

(Ω; d), then for any >0 there is a constant C = C(, j, α, β) such that

[u; d]

j+α,Ω

≤ Cu

0,Ω

+ [u; d]

2+β,Ω

(7.9)

|u; d|

j+α,Ω

≤ Cu

0,Ω

+ [u; d]

2+β,Ω

(7.10)

Proof: Since [u; d]

0+0,Ω

= u

0,Ω

, Inequality (7.9) is trivially satisﬁed in any

case with j =0,α=0.Thecases(j =1,α =1,β =0), (j =2, 0 ≤ α<1,β =

0), and (j =2,α = 1) are excluded by the requirement that j + α<2+β.

Let x, y be two points of Ω labeled so that d(x) <d(y), let 0 <μ≤

let δ = μd(x)andB = B

x,δ

. For any y ∈ B, d(y) ≥ d(x)/2. Consider the

j =1,α=0,β =0case.Foraﬁxedi =1,...,n,letx



and x



be the points

where a diameter of B parallel to the x

-axis intersects ∂B.Forsomepoint

x =(x

,...,x

i−1

, x

i+1

,...,x

) on the diameter,

u(x)| =

|u(x



) − u(x



2δ

≤

u

0,Ω

and

u(x)|≤|D

u(x| +





u(x

,...,x

i−1

i+1

,...,x

) dy



≤

u

0,Ω

+ δ sup

y∈B

u(y)|

≤

u

0,Ω

+ δ sup

y∈B

−2

(y)d

(y)|D

u(y)|.

Thus,

d(x)|D

u(x)|≤

d(x)

u

0,Ω

4δd(x)

(x)

sup

y∈B

(y)|D

u(y)|

≤ μ

−1

u

0,Ω

+4μ[u; d]

2+0,Ω

Thus,

[u; d]

1+0,Ω

≤ μ

−1

u

0,Ω

+4μ[u; d]

2+0,Ω

Given >0, choose μ<min (1/2,/4). Then

[u; d]

1+0,Ω

≤ C()u

0,Ω

+ [u; d]

2+0,Ω

(7.11)

where C()=1/μ. This proves Inequality (7.9) in the j =1,α= β =0case.

Suppose now that j =0, 1,α∈ (0, 1),β =0.Forsomepointz on the line

segment joining x to y ∈ B,

282 7 Interpolation and Monotonicity

j+α

(x, y)

u(x) − D

u(y)|

|x − y|

= d

j+α

(x, y)|x − y|

1−α

|∇(D

u)(z)|

≤ d

j+α

(x)δ

1−α

−j−1

(z)d

j+1

(z)|∇(D

u)(z)|

≤ 2

j+1

α−1

(x)δ

1−α

√

n[u; d]

j+1+0,Ω

j+1

√

nμ

1−α

[u; d]

j+1+0,Ω

On the other hand, if y ∈ B,then

j+α

(x, y)

u(x) − D

u(y)|

|x − y|

≤

j+α

(x, y)

(|D

u(x)| + |D

u(y)|)

≤ μ

−α

(x)|D

u(x)| + d

u(y)|D

u(y)|)

≤ 2μ

−α

[u; d]

j+0,Ω

Combining these two cases and taking the supremum over x, y ∈ Ω,

[u; d]

j+α,Ω

≤ 2μ

−α

[u; d]

j+0,Ω

j+1

√

nμ

1−α

[u; d]

j+1+0,Ω

Given >0, choose μ such that 2

j+1

√

nμ

1−α

<.Thenforj =0, 1,α∈ (0, 1)

and β =0,

[u; d]

j+α,Ω

≤ C()[u; d]

j+0,Ω

+ [u; d]

j+1+0,Ω

(7.12)

where C()=C(n, α, j, )=2μ

−α

.Considernowthej =1or2,α =0,and

0 <β≤ 1 case. Deﬁning x





, x as above,

u(x)=

j−1

u(x



) − D

j−1

u(x



2δ

≤

sup

y∈B

j−1

u(y)|.

Therefore,

u(x)|≤|D

u(x)| + |D

u(x) − D

u(x)|

≤

sup

y∈B

−j+1

(y)d

j−1

(y)|D

j−1

u(y)|

+ δ

sup

y∈B

−j−β

(x, y)d

j+β

(x, y)

u(x) − D

u(y)|

|x − y|

≤

j−1

δd

j−1

(x)

[u; d]

j−1+0,Ω

j+β

(x)

[u; d]

j+β,Ω

so that

u(x)|D

u(x)|≤2

j−1

−1

[u; d]

j−1+0,Ω

j+β

[u; d]

j+β

If >0, by taking the supremum over x ∈ Ω and choosing μ such that

j+β

<, there is a constant C()=C(j, , β) such that

[u; d]

j+0,Ω

≤ C()[u; d]

j−1+0,Ω

+ [u; d]

j+β,Ω

(7.13)

7.4 Interpolation of Weighted Norms 283

whenever j =1or2,α =0,and0<β≤ 1. Inequalities (7.11), (7.12), and

(7.13) are not quite enough to prove Inequality (7.9). Consider the j =2,α=

0, 0 <β≤ 1case.Given>0, choose 

> 0 such that 2

<.Thereisthen

a constant C

(

) satisfying Inequality (7.13). Choosing 

> 0 such that

1 − C

(

)

≥ 1/2, there is a constant C

(

) satisfying Inequality (7.11).

Thus,

[u; d]

2+0,Ω

≤ C

(

)[u; d]

1+0,Ω

+ 

[u; d]

2+β,Ω

≤ C

(

)(C

(

)u

0,Ω

+ 

[u; d]

2+0,Ω

)+

[u; d]

2+β,Ω

from which it follows that

[u; d]

2+0,Ω

≤

(

)

1 − C

(

)

u

0,Ω



1 − C(

)

[u; d]

2+β,Ω

≤ C()u

0,Ω

+ [u; d]

2+β,Ω

(7.14)

where C() is the coeﬃcient of u

in the ﬁrst line. By successive applica-

tions of Inequalities (7.11), (7.12), (7.13), and (7.14), the following cases of

Inequality (7.9) can be established:

(i) j =0, 1, 2andα =0

(ii) j =0, 1andα ∈ (0, 1).

Cases (j =0,α =1), (j =1,α =1, 0 <β≤ 1), and (j =2,α ∈ (0, 1), 0 <

β ≤ 1) require separate arguments. Consider ﬁrst the j =0,α =1case.As

in the argument leading to Inequality (7.12),

[u; d]

0+1,Ω

≤ 2μ

−1

u

0,Ω

√

n[u; d]

1+0,Ω

Now apply Inequality (7.11) to the second term and then Inequality (7.14).

Consider now the j =1,α =1, 0 <β≤ 1 case. Using the fact that d(x, y)=

d(x) ≤ 2d(z



) for any z



∈ B,forsomepointz on the line segment joining x

and y,

d(x, y)

|Du(x) − Du(y)|

|x − y|

≤ d(x, y)

|Du(x) − Du(y)|

|x − y|

(y)

+ d(x, y)

|Du(x)|+ |Du(y)|

∼B

(y)

≤ d(x, y)

|∇Du(z)|χ

(y)+

2d(x)

[u]

1+0,Ω

≤ 4[u; d]

2+0,Ω

[u]

1+0,Ω

Choose μ =

and apply Inequalities (7.11) and (7.14) to the latter two

terms. Lastly, consider the j =2,α∈ (0, 1) case. If y ∈ B,then

284 7 Interpolation and Monotonicity

2+α

(x, y)

u(x) − D

u(y)|

|x − y|

= d

2+α

(x)|x − y|

β−α

−2−β

(x, y)d

2+β

(x, y)

u(x) − D

u(y)|

|x − y|

≤ d

α−β

(x)δ

β−α

[u; d]

2+β,Ω

= μ

β−α

[u; d]

2+β,Ω

On the other hand, if y ∈ B,then

2+α

(x, y)

u(x) − D

u(y)|

|x − y|

≤

2+α

(x, y)

−2

(x, y)(d

(x)|D

u(x)| + d

(y)|D

u(y)|) ≤ 2μ

−α

[u; d]

2+0,Ω

Applying Inequality (7.14) to [u; d]

2+0,Ω

with  = μ

2α

[u; d]

2+α,Ω

≤ μ

β−α

[u; d]

2+β,Ω

+2μ

−α

(μ

2α

)u

+ μ

2α

[u; d]

2+β,Ω

)

=2C

(μ

2α

)μ

−α

u

0,Ω

+(μ

β−α

+2μ

)[u; d]

2+β,Ω

Given >0, choose μ so that μ

β−α

+ μ

<to obtain Inequality (7.9).

Lastly, Inequality (7.10) follows by applying (7.9) to each of the terms of

|u; d|

j+α,Ω



i=0

[u; d]

i+0,Ω

+[u; d]

2+α,Ω

7.5 Inner Norms

If Ω is a bounded open subset of R

and Σ is a relatively open subset of ∂Ω

with ∂Ω ∼ Σ = ∅,

d(x) is deﬁned by the equation

d(x)=dist(x, ∂Ω ∼ Σ),x∈ Ω.

Since ∂Ω ∼ Σ is closed,

d is positive and continuous on Ω;infact,|

d(x) −

d(y)|≤|x − y| for x, y ∈ Ω. Recalling that a ∧ b denotes the minimum of

the two real numbers a and b,let

d(x, y)=

∧

for x, y ∈ Ω.For

δ>0, let

= {x ∈ Ω;

d(x) >δ}, if nonempty. Thus,

is deﬁned for

0 <δ<sup {η>0;

= ∅} andisanopenset.IfΩ is a convex set, it does

not necessarily follow that

is convex; but if Σ = ∅,then

= Ω

and is

convex. It will be assumed at the end of this section that Ω and the

are

bounded, convex, open subsets of R

.Inaddition,itwillbeassumed

also that Ω is stratiﬁable, a condition assuring that the annulus between

and

2δ

is of uniform width. Since the ﬁrst theorem does not require this

assumption, the deﬁnition of stratiﬁable appears after this theorem. All of

these assumptions are fulﬁlled for a particular case of interest here; namely,

Ω is a spherical chip and Σ is the interior of the ﬂat part of the boundary.

7.5 Inner Norms 285

Deﬁnition 7.5.1 If Ω is a bounded open subset of R

, Σ is a relatively

open subset of ∂Ω,0 ≤ α ≤ 1, and k + b + α ≥ 0, let

|u|

(b)

k+α,Ω

=sup{δ

k+α+b

|u|

k+α,

;0<δ<1,

= ∅},

[u]

(b)

k+α,Ω

=sup{δ

k+α+b

[u]

k+α,

;0<δ<1,

= ∅},

and let H

(b)

k+α

(Ω)bethesetofu such that |u|

(b)

k+α,Ω

< +∞.

There is no restriction that Σ be a proper subset of ∂Ω in this deﬁnition. In

the Σ = ∅ or Σ = ∂Ω cases,

and Ω

are taken as the same.

Remark 7.5.2 The requirement that δ ∈ (0, 1)) in the deﬁnition is based

on the fact that the behavior of |u|

is of interest only for small δ.Any

number δ

∈ (0, 1) for which

= ∅ for all δ<δ

could serve as well at the

expense of increasing the bound by a factor that does not depend upon u;

that is, if sup

0<δ≤δ

a+b

|u|

≤ M,then

sup {δ

a+b

|u|

;0<δ<1,

= ∅} ≤





a+b

To see this, suppose the ﬁrst inequality is true. Consider those δ for which

≤ δ<1and

= ∅ and let q be the smallest integer such that 1 ∧d(Ω) ≤

qδ

.Sinceδ/q < (1)/q ≤ δ

and

⊂

δ/q





a+b

|u|

≤





a+b

|u|

δ/q

≤ M

so that

a+b

|u|

≤ q

a+b

M, δ

≤ δ<1.

Since q<

+1,

a+b

|u|

≤





a+b

M, δ

≤ δ<1.

As this inequality is trivially true for 0 <δ<δ

sup {δ

a+b

|u|

;0<δ<1,

= ∅} ≤





a+b

It is easily seen that for 0 <δ

< 1, there is a constant C = C(δ

) such that

|u|

(b)

a,Ω

≤ sup {δ

a+b

|u|

;0<δ<δ

= ∅} ≤ C|u|

(b)

a,Ω

286 7 Interpolation and Monotonicity

Theorem 7.5.3 If k is a nonnegative integer, 0 <α≤ 1,andk + b + α ≥ 0,

then H

(b)

k+α

(Ω) is a Banach space under the norm |u|

(b)

k+α,Ω

Proof: It is easily checked that H

(b)

k+α

(Ω) is a normed linear space. Let

} be a Cauchy sequence in H

(b)

k+α

(Ω). The sequence {|u

(b)

k+α,Ω

} is then

bounded by K,say.Let{δ(j)} be a sequence in (0, 1) decreasing to zero.

Since δ(j)

k+b+α

k+α,

δ(j)

≤|u

(b)

k+α,Ω

≤ K, the sequence {|u

k+α,

δ(j)

}

is bounded for each j ≥ 1. Using the subsequence selection principle, an in-

duction argument, and letting u

(0)

= u

, the functions u

,Du

,...,D

have continuous extensions to

−

δ(j)

, there is a function v

−

δ(j)

,anda

subsequence {u

(j)

} of the {u

(j−1)

} sequence such that u

(j)

→ v

,Du

(j)

→

,...,D

(j)

→ D

−

δ(j)

with v

= v

j−1

−

δ(j−1)

.Thus,there

is a function v on Ω that agrees with v

−

δ(j)

,j ≥ 1. Consider the di-

agonal sequence {u

(j)

} that is a subsequence of all of the above sequences

so that u

(j)

→ v, Du

(j)

→ Dv,...,D

(j)

, → D

v on Ω as j →∞. It will

be shown now that v ∈ H

(b)

k+α

(Ω). Note ﬁrst that |u

(j)

(b)

k+α,Ω

≤ K for all

j ≥ 1. Consider any δ ∈ (0, 1) and x, y ∈

.Sinceδ

k+b+α

(j)

(x)|

≤

(j)

(b)

k+α,Ω

≤ K<+∞, letting j →∞, taking the supremum over x ∈

and then taking the supremum over δ, |v|

(b)

0,Ω

≤ K<+∞. Similar arguments

apply to |Du

0,Ω

,...,|D

(j)

0,Ω

,and[D

(j)

]

α,Ω

.Thus,|v|

(b)

k+α

< +∞ and

v ∈ H

(b)

k+α

(Ω). To show that u

(j)

→ v in H

(b)

k+α

(Ω), consider any δ ∈ (0, 1)

and x, y ∈

. Using the fact that the sequence is Cauchy, given >0there

is an N () ≥ 1 such that |u

(i)

− u

(j)

(b)

k+α,Ω

</(k + 1) for all i, j ≥ N().

For any δ ∈ (0, 1),δ

k+b+α

(i)

− u

(j)

k+α,

</(k + 1). Expanding the

norm according to its deﬁnition, the last term δ

k+b+α

(i)

− D

(j)

]

dominates

k+b+α

(i)

(x) − D

(j)

(x) − D

(i)

(y) − D

(j)

(y)|

|x − y|

≤



k +1

Letting j →∞and taking the supremum over x, y ∈

,δ

k+b+α

(i)

−

α,

</(k + 1). Similar arguments can be applied to the other

terms, resulting in δ

k+b+α

(i)

− v|

k+α,

<. Taking the supremum over

δ ∈ (0, 1), |u

(i)

− v|

(b)

k+α,Ω

<for all i ≥ N(). This proves that the sequence

(i)

} converges to v ∈ H

(b)

k+α

(Ω). Since the sequence {u

} is Cauchy and a

subsequence converges to v ∈ H

()b)

k+α

(Ω), the sequence {u

} converges to v in

()b)

k+α

(Ω).

7.5 Inner Norms 287

The above deﬁnition of inner norm cannot be applied to a boundary func-

tion if Σ = ∂Ω, as an open set in its own right, because

d(x)andΣ

are

not deﬁned in this case. As in Deﬁnition 7.2.5, the inner norm of a boundary

function can be based on the idea that g is the restriction of a function u on

−

to Σ.

Deﬁnition 7.5.4 For g : Σ → R,let

|g|

(1+b)

1+α,Σ

=inf{|u|

(1+b)

1+α,Ω

; u ∈ H

(1+b)

1+α

(Ω),u|

= g}

and let H

(1+b)

1+α

(Σ)bethesetofallg for which |g|

(1+b)

1+α,Σ

< +∞.

Theorem 7.5.5 If Σ = ∅, 0 <α≤ 1, 2+b + α ≥ 0, H

(1+b)

1+α

(Σ) is a normed

linear space under the norm |g|

(1+b)

1+α,Σ

Proof: As it is a routine exercise to show that |g|

(1+b)

1+α,Σ

is a seminorm, only the

fact that |g|

(1+b)

1+α,Σ

= 0 implies that g =0onΣ will be proved. Assuming that

|g|

(1+b)

1+α,Σ

= 0, there is a sequence {u

} in H

(1+b)

1+α

(Ω) such that |u

(1+b)

1+α,Ω

↓ 0

as n →∞. Since the sequence of norms is bounded, |u

(1+b)

1+α,Ω

≤ K, n ≥ 1,

for some K>0. For each δ ∈ (0, 1),δ

2+b+α

1+α,

≤ K, and therefore

each sequence {|u

1+α,

} is bounded. Let { δ(m)} be a sequence in (0, 1)

that decreases to 0. For each m ≥ 1, {|u

1+α,

δ(m)

} is a bounded sequence

in H

1+α,

δ(m)

. As in the proof of Theorem 7.5.3 for each m ≥ 1, there is a

subsequence {u

(m)

} of the {u

(m−1)

} sequence and a function v

δ(m)

such that u

(m)

→ v

and Du

(m)

→ Dv

uniformly on

−

δ(m)

and v

= v

m−1

δ(m− 1)

. Consider the diagonal sequence {u

(m)

} which is a subsequence

of all of the above sequences. It follows that there is a function v on Ω such

that u

(m)

→ v and Du

(m)

→ Dv on Ω as m →∞.Given>0, there is

an N ≥ 1 such that |u

(1+b)

1+α,Ω

<for all n ≥ N.Bylookingattheterms

of |u

(m)

(1+b)

1+α,Ω

pointwise and taking the limit as m →∞, |v|

(1+b)

1+α,Ω

<for

every >0sothat|v|

(1+b)

1+α,Ω

= 0 and, in particular, that [v]

=0forevery

δ ∈ (0, 1). Thus, v =0onΩ.Sincetheu

(m)

= g on Σ,v = g on Σ,and

therefore g =0onΣ.

Theorem 7.5.6 If g : Σ → R and |g|

(1+b)

1+α,Σ

< +∞, then there is a u ∈

(1+b)

1+α

(Ω) such that u|

= g and |u|

(1+b)

1+α,Ω

= |g|

(1+b)

1+α,Σ

Proof: Let {u

} be a sequence in H

(1+b)

1+α

(Ω)forwhich|u

(1+b)

1+α,Ω

↓|g|

(1+b)

1+α,Σ

and u

= g. As in the preceding proof there is a subsequence {u

(m)

} and