Helms L.L. Potential Theory

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268 7 Interpolation and Monotonicity

7.2 H¨older Spaces

As solutions of elliptic equations require at least that second-order partial

derivatives exist and be continuous, the approximation methods used in prov-

ing the existence of solutions must involve slightly stronger conditions. It is

not necessary to go so far as requiring the existence of third partials as some

kind of smoothness between continuous second partials and the existence of

third partials will suﬃce. Collections of such functions constitute the H¨older

spaces. The deﬁnitions of this section apply to any subset Ω of R

Deﬁnition 7.2.1 If k is a nonnegative integer, 0 <α≤ 1, and u ∈

(Ω), let

[u]

k+0,Ω

=sup

x∈Ω,|β|=k

u(x)|,

|u|

k+0,Ω

= u

k,Ω



j=0

[u]

j+0,Ω

[u]

k+α,Ω

=sup

x,y∈Ω,|β|=k

u(x) − D

u(y)|

|x − y|

|u|

k+α,Ω

= u

k,Ω

+[u]

k+α,Ω

For k ≥ 1, note that |u|

k+0,Ω

is not the same as |u|

k,Ω

= |u|

(k−1)+1,Ω

.The

collection of functions u on Ω for which |u|

k+α,Ω

< +∞ will be denoted by

k+α

(Ω). For u ∈ H

k+α

(Ω) it is clear from the deﬁnition of [u]

k+α,Ω

that

u, |β| = k, is uniformly continuous on Ω and has a continuous extension

to Ω

−

which will be denoted by the same symbol. It is easily seen that the

k+α

(Ω) spaces are normed linear spaces with norm |u|

k+α,Ω

. These spaces

are called H¨older spaces. In general, [u]

k+α,Ω

is only a seminorm. The

collection of functions u on Ω such that u ∈ H

k+α

(Γ ) for all compact subsets

Γ of Ω will be denoted by H

k+α,loc

(Ω).

In the remainder of this section, Ω will denote a bounded open subset

of R

.GrantedthatC

(Ω

−

) is a Banach space, it will be shown that C

(Ω

−

)

is a Banach space for every k ≥ 1.

Theorem 7.2.2 C

(Ω

−

),k ≥ 1, is a Banach space under the norm u

k,Ω

Proof: Let {u

}

∞

j=1

be a Cauchy sequence in C

(Ω

−

). Since

D

− D



0,Ω

≤u

− u



k,Ω

for each β with |β|≤k, the sequence {D

}

∞

j=1

is Cauchy in C

(Ω

−

), and

thus there is a v

∈ C

(Ω

−

) such that D

→ v

as j →∞uniformly on

7.2 H¨older Spaces 269

−

.Ifitcanbeshownthatv

= D

u for |β|≤k where u

→ u = v

j →∞uniformly on Ω

−

, it would then follow that

u

− u

k,Ω



|β|≤k

D

− D

u

0,Ω

→ 0asj →∞

and that u

→ u in C

(Ω

−

). This statement will be proved by a ﬁnite

induction argument on |β| using Theorem 0.2.2. Suppose 

is a multi-index

with all 0



sexceptfora1inthemth position, |β +

|≤k,andD

→ D

as j →∞uniformly on Ω

−

.SinceD



)=D

β+

→ v

β+

j →∞uniformly on Ω

−

β+

= D



u)=D

β+

u.Thus,u

→ u in

(Ω

−

Theorem 7.2.3 H

k+α

(Ω

−

) is a Banach space under the norm |u|

k+α,Ω

−

Proof: If {u

}

∞

j=1

is a Cauchy sequence in H

k+α

(Ω

−

), then it is a Cauchy

sequence in C

(Ω

−

)andthereisau ∈ C

(Ω

−

) such that u

−u

k,Ω

−

→ 0

as j →∞. Thus, it suﬃces to show that [D

− D

α,Ω

−

→ 0asj →∞

for each β with |β| = k.If|β| = k and >0, there is an N() ≥ 1 such that

− D

]

α,Ω

−

=sup

x,y∈Ω

−

x=y

|(D

(x) − D

(x)) − (D

(y) − D

(y))|

|x − y|

<

for i, j ≥ N(). Thus,

|(D

(x) − D

(x)) − (D

(y) − D

(y))|≤|x − y|

for x, y ∈ Ω

−

,i,j ≥ N(). Letting i →∞,

|(D

u(x) − D

(x)) − (D

u(y) − D

(y))|≤|x − y|

for x, y ∈ Ω

−

,j ≥ N(). Since the sequence {[D

]

α,Ω

} is bounded in R,say

by M>0, it follows from this inequality that [D

α,Ω

−

≤ [D

]

α,Ω

−

+  ≤

M + , j ≥ N(), that u ∈ H

k+α

(Ω

−

), and that

u − D

]

α,Ω

−

≤  for j ≥ N().

Thus, [D

u − D

]

α,Ω

−

→ 0asj →∞and u

→ u in H

k+α

(Ω

−

A useful tool for proving the existence of solutions of an equation is to

construct a sequence of functions in H

2+α

(Ω) and then extracting a conver-

gent subsequence to approximate the solution. The principle embodied in the

following theorem will be called the subsequence selection principle.The

subsequence selection principle also applies to sequences in H

k+α

(Ω),k ≥ 1.

The theorem will be stated and proved for k = 2. To simplify the notation,

D and D

will serve as generic symbols for ﬁrst- and second-order partial

derivatives, respectively.

270 7 Interpolation and Monotonicity

Theorem 7.2.4 If Ω is a convex, bounded, open subset of R

and {u

} is

asequenceinH

2+α

(Ω), 0 <α≤ 1 for which there is an M>0 such that

2+α,Ω

≤ M,j ≥ 1, then the u

,Du

,andD

have continuous extensions

to Ω

−

and there is a function u on Ω

−

and a subsequence {u

} of the

} sequence such that the sequences {u

}, {Du

},and{D

} converge

uniformly on Ω

−

to continuous functions u, Du,andD

u, respectively.

Proof: Note ﬁrst that the inequality

(x) − D

(y)|

|x − y|

≤ [u

]

2+α

≤ M, x,y ∈ Ω

implies that the D

are uniformly continuous on Ω and have continuous

extensions to Ω

−

which will be denoted by the same symbol. The inequality

also implies that the sequence { D

} is uniformly equicontinuous on Ω

−

Since |D

0,Ω

=[u

]

2+0,Ω

≤ M,j ≥ 1, the sequence {D

} is uniformly

bounded on Ω

−

. It follows from the Arzel´a-Ascoli Theorem, Theorem 0.2.3,

that there is a subsequence of the {D

} sequence, which can be assumed

to be the sequence itself by a change of notation if necessary, which converges

uniformly on Ω

−

to a continuous function v

(2)

on Ω

−

. By the mean value the-

orem of the calculus, there is a z

on the line segment joining x to y such that

|Du

(x) − Du

(y)|

|x − y|

= |∇Du

)|≤

√

n[u

]

2+0,Ω

≤

√

nM.

This shows that the Du

are uniformly continuous on Ω, have continuous ex-

tensions to Ω

−

which will be denoted by the same symbol, and that the {Du

}

are uniformly bounded and equicontinuous on Ω

−

. Again it follows from the

Arzel´a-Ascoli Theorem that there is a subsequence of the {Du

} sequence,

which can be assumed to be the sequence itself, which converges uniformly

on Ω

−

to a continuous function v

(1)

on Ω

−

. Applying the same argument to

the {u

} sequence, there is a subsequence of the {u

} sequence, which can be

assumed to be the sequence itself, which converges uniformly on Ω

−

to a con-

tinuous function u on Ω

−

. By Theorem 0.2.2, Du = v

(1)

= lim

j→∞

and

u = v

(2)

= lim

j→∞

on Ω.NotethatDu and D

u have continuous

extensions to Ω

−

The roles played by a function u on Ω and a function g on a portion Σ of

the boundary of Ω are distinctly diﬀerent and the norm of the latter will be

deﬁned diﬀerently based on the idea of considering g to be the restriction of a

function u to Σ as discussed in [25]. The following deﬁnition and subsequent

discussion is also applicable if stated with the 1+α replaced by k+α for k ≥ 1.

Deﬁnition 7.2.5 For g : Σ → R,let

|g|

1+α,Σ

=inf{|u|

1+α,Ω

; u ∈ H

1+α

(Ω),u|

= g}},

and let H

1+α

(Σ)bethesetofallg for which |g|

1+α,Σ

< +∞.

7.2 H¨older Spaces 271

According to the following theorem, it can be assumed that a boundary

function g in H

1+α

(Σ) has an extension ˜g on Ω of the same norm.

Theorem 7.2.6 If Ω is a convex, bounded, open subset of R

,Σ is a rela-

tively open subset of ∂Ω,andg : Σ → R with |g|

1+α,Σ

< +∞, then there is

a u ∈ H

1+α

(Ω) such that u|

= g and |u|

1+α,Ω

= |g|

1+α,Σ

Proof: Let {u

} be a sequence in H

1+α

(Ω)forwhich|u

1+α,Ω

↓|g|

1+α,Σ

and u

= g. Since the sequence {|u

1+α,Ω

} is bounded, by the subsequence

selection principle there is a subsequence {u

} and a function u on Ω

−

such

that the sequences {u

} and {Du

} converge uniformly on Ω

−

to u and Du,

respectively, as j →∞.Since|u

0,Ω

−

→|u|

0,Ω

−

, |Du

0,Ω

−

→|Du|

0Ω

−

and lim inf

j→∞

|Du

α,Ω

−

≤|g|

1+α,Σ

−|u|

0,Ω

−

−|Du|

0,Ω

−

, for any x, y ∈ Ω

−

lim inf

j→∞

|Du

(x) − Du

(y)|

|x − y|

≤ lim inf

j→∞

[Du

]

α,Ω

−

≤|g|

1+α,Σ

−|u|

0,Ω

−

−|Du|

0,Ω

−

Since Du

(x) → Du(x)andDu

(y) → Du(y)asj →∞, [Du]

α,Ω

−

≤

|g|

1+α,Σ

−|u|

0,Ω

−

−|Du|

0,Ω

−

so that |u|

1+α,Ω

−

≤|g|

1+α,Σ

.Sinceu belongs

to the class deﬁning |g|

1+α,Σ

, the two are equal.

The proof of the following theorem is an easy exercise.

Theorem 7.2.7 H

1+α

(Σ) is a normed linear space under the norm |g|

1+α,Σ

The norms |u|

k+α,Ω

on H

k+α

(Ω) are global in that the points of Ω are

treated equally. An alternative is to deemphasize points of Ω near the bound-

ary by using a weight function.

Deﬁnition 7.2.8 If φ(x) is a positive, real-valued function on Ω, k is a non-

negative integer, φ(x, y)=min(φ(x),φ(y)),u ∈ C

(Ω

−

), 0 <α≤ 1,b ∈ R,

and k + α + b ≥ 0, let

[u; φ]

(b)

k+0,Ω

=sup

x∈Ω,|β|=k

k+b

(x)|D

u(x)|

|u; φ|

k+0,Ω

=  u; φ 

(b)

k,Ω



j=0

[u; φ]

(b)

j+0,Ω

[u; φ]

(b)

k+α,Ω

=sup

x,y∈Ω,|β|=k

k+α+b

(x, y)

u(x) − D

u(y)|

|x − y|

|u; φ|

(b)

k+α,Ω

= | u; φ |

(b)

k+0,Ω

+[u; φ]

(b)

k+α,Ω

|u; φ|

k+α,Ω

= |u; φ|

(0)

k+α,Ω

272 7 Interpolation and Monotonicity

H¨older classes H

(b)

(Ω; φ),H

(b)

(Ω; φ),H

(b)

k+0

(Ω; φ), and H

(b)

k+α

(Ω; φ)arede-

ﬁned as above and are easily seen to be normed linear spaces. These classes

will be of particular interest for

(i) φ(x)=c,aconstant,

(ii) φ(x)=d

= d(x)=dist(x, ∂Ω).

When φ(x)=1, [u;1]

(b)

k+0,Ω

, u;1

(b)

k,Ω

, [u;1]

(b)

k+α,Ω

,and|u;1|

(b)

k+α,Ω

are the

same as [u]

k+0,Ω

, u

k,Ω

, [u]

k+α,Ω

,and|u|

k+α,Ω

, respectively, as deﬁned

above. The superscript (b) is dropped when b =0;thatis,|u; φ|

(0)

k+α,Ω

written simply |u; φ|

k+α

. Whenever the notation H

(b)

2+α

(Ω; φ) is used, it is

understood that 2 + b + α ≥ 0.

Theorem 7.2.9 H

(b)

k+α

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

(b)

k+α,Ω

Proof: If δ ∈ (0, 1), let Ω

= {y ∈ Ω; d(y) >δ}. There is then a constant C

such that

C|u|

k+α,Ω

≤|u; d|

(b)

k+α,Ω

(7.1)

where

C =min{d

j+b

(Ω),d

k+α+b

(Ω),δ

j+b

,δ

k+α+b

,j =0,...,k},

where d(Ω) is the diameter of Ω.Let{u

} be a Cauchy sequence in

(b)

k+α

(Ω; d). By the above inequality, the sequence of restrictions of the u

−

is Cauchy in H

k+α

(Ω

) and has a limit u. The function u has an obvious

extension to Ω which is independent of δ. Since the sequence {|u

; d|

(b)

k+α,Ω

}

is Cauchy in R, there is a constant M such that |u

; d|

(b)

k+α,Ω

≤ M for all

j ≥ 1. If β is a multi-index with |β| =  ≤ k,then

+b

(x)|D

(x)|≤M, x ∈ Ω,j ≥ 1.

Letting j →∞,d

+b

(x)|D

u(x)|≤M so that [u; d]

(b)

+0,Ω

≤ M . Similarly,

[u; d]

(b)

k+α,Ω

≤ M ,provingthatu ∈ H

(b)

k+α

(Ω; d). If >0and|β| =  ≤ k,

there is an m such that

+b

(x)|D

(x) − D

(x)|≤|(u

− u

); d|

(b)

k+α,Ω



k +1

,i,j≥ m.

Letting j →∞and taking the supremum over x ∈ Ω and β with |β| = ,

− u; d|

(b)

+0,Ω



k +1

,i≥ m;

Similarly, [u

−u; d]

(b)

k+α,Ω

</(k + 1) and it follows that |u

−u; d|

(b)

k+α,Ω

<

for i ≥ m; that is, the sequence {u

} converges to u in H

(b)

k+α

(Ω; d).

7.3 Global Interpolation 273

7.3 Global Interpolation

Given norms |·|

, |·|

,and|·|

on a normed linear space X with s<u<t,

interpolation inequalities compare |·|

to some function of |·|

and |·|

usually a linear function of the latter of the form

|x|

≤ a|x|

+ b|x|

,x∈ X.

Such inequalities can be derived from certain convexity properties of norms

due to H¨ormander. It will be assumed throughout this section that Ω is a

bounded, convex, open subset of R

. Since only norms of functions on Ω

will be considered in this section, the subscript Ω will be omitted from

the norm notation; that is, |u|

k+α,Ω

will appear as |u|

k+α

Lemma 7.3.1 If k is a positive integer and k<b= m + σ, 0 <σ≤ 1,then

there is a constant C = C(b, d(Ω)) = n

m/2

(max (d(Ω)

,d(Ω), 1))

such that

|u|

≤ C|u|

Proof: Let y

be a ﬁxed point of Ω,let

v(y)=



|τ |=m

τ!

u(y

and let w = u −v. Consider any multi-index β with |β| = m.ThenD

v(y)=

u(y

),y ∈ Ω,andD

w(y)=D

u(y)−D

v(y)=D

u(y)−D

u(y

). Note

that D

w(y

)=0andthat[w]

=[u]

since

[w]

=sup

|β|=m

=sup

|β|=m

u − D

u(y

)]

=sup

|β|=m

=[u]

(7.2)

Now consider any multi-index β with |β| = m − 1. For 1 ≤ j ≤ n,let

the vector having 1 as its jth component and 0 as its other components. By

the mean value theorem, there is a point z on the line segment joining x to

y such that

w(x) − D

w(y)|

|x − y|

= |∇D

w(z)|

= |(D

β+

w(z),...,D

β+

w(z)|





β+

w(z) − D

β+

w(y

),...,D

β+

w(z) − D

β+

w(y

)









i=1



β+

w(z) − D

β+

w(y

|z − y



|z −y

2σ



≤

√

nd(Ω)

[w]

m+σ

√

nd(Ω)

[w]

274 7 Interpolation and Monotonicity

This shows that [w]

=[w]

(m−1)+1

≤

√

nd(Ω)

[w]

≤

√

nd(Ω)

[u]

Equation (7.2). Consider [u]

=[w + v]

≤ [w]

+[v]

and the last term in

particular. By the mean value theorem,

[v]

=sup

|τ |=m−1

sup

x,y∈Ω

x=y

v(x) − D

v(y)|

|x − y|

=sup

|τ |=m−1

sup

x,y∈Ω

x=y

|∇D

v(z)|

where z is a point on the line segment joining x to y.Since

|∇D

v(z)| = |(D

τ +

v(z),...,D

τ +

v(z)|

= |(D

τ +

u(y

),...,D

τ +

u(y

)|,

|∇D

v(z)|≤

√

n[u]

m+0

so that

[v]

≤

√

n[u]

m+0

Therefore,

[u]

≤ [w]

+[v]

≤

√

nd(Ω)

[u]

√

n[u]

m+0

≤

√

n max (d(Ω)

, 1)([u]

m+0

+[u]

)

where

√

n max (d(Ω)

, 1) ≥ 1. Thus,

|u|

= |u|

(m−1)+1

= u

m−1

+[u]

≤

m−1



j=0

[u]

j+0

√

n max (d(Ω)

, 1)([u]

m+0

+[u]

)

≤

√

n max (d(Ω)

, 1)|u|

Replacing b by m and m by m − 1,

|u|

m−1

≤

√

n max (d(Ω), 1)|u|

≤ n(max (d(Ω)

,d(Ω), 1))

|u|

Repeating the last step m − k +1times,

|u|

≤ n

(m−k+1)/2



max (d(Ω)

,d(Ω), 1)



m−k+1

|u|

The conclusion follows with

C = C(b, d(Ω)) = n

m/2

(max (d(Ω)

,d(Ω), 1))

The growth of |u|

as a function of a will be considered now.

7.3 Global Interpolation 275

Theorem 7.3.2 If 0 ≤ a<b= m + σ, 0 <σ≤ 1,then|u|

≤ C|u|

with

C = C(b, d(Ω)) = n

m/2

max (1,d(Ω)

,d(Ω)).

Proof: Three cases will be considered according to a>m,a= m, a < m.

Case (i) (a>m)Inthiscase,a = m + ρ where 0 <ρ<1,ρ < σ.Since

≤ d(Ω)

σ−ρ

whenever |β| = m,[u]

m+ρ

≤ d(Ω)

σ−ρ

[u]

m+σ

and

|u|



j=0

[u]

j+0

+[u]

m+ρ

≤



j=0

[u]

j+0

+ d(Ω)

σ−ρ

[u]

m+σ

≤ max (1,d(Ω)

σ−ρ

)|u|

Since d(Ω)

σ−ρ

≤ 1ifd(Ω) ≤ 1andd(Ω)

σ−ρ

≤ d(Ω)

if d(Ω) > 1, |u|

≤

max (1,d(Ω)

)|u|

. In this case, the constant in the assertion can be taken as

max (1,d(Ω))

Case (ii) (a = m) This case is covered by the preceding lemma with the

constant n

m/2

(max (1,d(Ω)

,d(Ω))

Case (iii) (a<m) Suppose a =  + ρ<m<b= m + σ where 0 <ρ≤ 1.

Consider any multi-index β with |β| = .Since

=sup

x,y∈Ω

x=y

u(x) − D

u(y)|

|x − y|

1−ρ

≤ d(Ω)

1−ρ

sup

x,y∈Ω

x=y

|∇D

u(z)|

≤

√

nd(Ω)

1−ρ

[u]

+1+0

where z is a point on the line segment joining x to y,

[u]

+ρ

≤

√

nd(Ω)

1−ρ

[u]

+1+0

and

|u|



j=0

[u]

j+0

+[u]

+ρ

≤



j=0

[u]

j+0

√

nd(Ω)

1−ρ

[u]

+1+0

≤

√

n max (1,d(Ω)

1−ρ

)

+1



j=0

[u]

j+0

276 7 Interpolation and Monotonicity

Since  +1 ≤ m, the latter sum is less than or equal to |u|

.Asabove,

d(Ω)

1−ρ

≤ max (1,d(Ω)) and the constant of the assertion can be taken as

√

n max (1,d(Ω)). The constant C(b, d(Ω)) = n

m/2

max (1,d(Ω)

,d(Ω)) will

suﬃce for all three cases.

A polynomial approximation is needed for the proof of the next lemma.

Given a function f on a neighborhood U of the origin in R

and a ﬁnite set

Ξ ⊂ U , the polynomial p in n variables of degree k is said to interpolate f

on Ξ if p = f on Ξ. The number of terms of such a polynomial is



n + k





j=0



n + j − 1



the jth term of the sum being the number of terms having degree j (c.f., Feller

[21], pp. 36, 62). It is shown in [14] that there is a subset Ξ = {ξ

,...,ξ

}

of U having M =



n+k



points such that the polynomial

p(y)=



j=0

(y)f(ξ

)

interpolates f on Ξ. The polynomial p is unique, the p

areofdegreeless

than or equal to k and are products of linear factors, and the p

depend only

upon n, k, Ξ and not on any f.

Lemma 7.3.3 (H¨omander [34]) Let 0 <a≤ k<bwhere k is an integer.

If [u]

≤ 1, [u]

≤ 1,andα is a multi-index with |α| = k, then there is a

constant C = C(a, b)) and a polynomial function P of d(Ω) that increases

with d(Ω) such that D

u

≤ C max (d(Ω)

,d(Ω)

)P (d(Ω)).

Proof: Suppose the result is true when a ≤ 1. Consider any multi-index β

with |β| = j.Ifa = j + σ, 0 <σ≤ 1, then 0 <a− j ≤ k − j<b− j.

Since [D

a−j

≤ [u]

≤ 1, [D

b−j

≤ [u]

≤ 1, and a − j ≤ 1, for any

multi-index γ with |γ| = k − j, D

u)

≤ C.Thus,D

u

≤ C for

any α with |α| = k. It therefore can be assumed that a ≤ 1. Since neither

the hypothesis nor the conclusion is aﬀected by adding a constant to u,it

canbeassumedthatthereisapointx

∈ Ω such that u(x

) = 0. Since

[u]

≤ 1, |u(y)| = |u(y) − u(x

|≤|y − x

≤ d(Ω)

and u

≤ d(Ω)

.Let

b = m + γ,0 <γ≤ 1. For each x ∈ Ω, consider the Taylor formula of order

m − 1 with remainder

u(y)=

m−1



j=0



|α|=j

u(x)

α!

(y − x)



|α|=m

u(z)

α!

(y − x)

,y∈ Ω,

where z is a point on the line segment joining x to y. Letting

(y)=



j=0



|α|=j

u(x)

α!

(y −x)

,y∈ Ω,

7.3 Global Interpolation 277

Taylor’s formula can be written

u(y)=P

(y)+



|α|=m

u(z) − D

u(x)

α!

(y −x)

Therefore,

|u(y) − P

(y)|≤



|α|=m

u(x) − D

u(z)|

α!

|y −x|

|α|

≤ [u]



|α|=m

α!

|y −x|

m+γ

≤ C(b)d(Ω)

Combining this inequality with the bound for u

, there is a constant C =

C(a, b)) such that |P

(y)|≤C max (d(Ω)

,d(Ω)

),y ∈ Ω.Letp be the unique

polynomial of degree m that interpolates the function



j=0



|α|=j

u(x)

α!

,z∈ R

on the set Ξ = {ξ

,...,ξ

} discussed above. Since this function interpolates

itself on Ξ and p is unique,

(y)=



j=0

(y − x)P

(ξ

),y∈ Ω.

If α is any multi-index with |α|≤m,then

(y)=



j=0

(y − x)P

(ξ

)

so that

(y)|≤



j=0

(y −x)||P

(ξ

≤ C max (d(Ω)

,d(Ω)

)



j=0

(y − x)|,y∈ Ω.

Note that the sum is bounded by a polynomial function P of d(Ω)that

increases with d(Ω). Thus, there is a constant C = C(a, b) such that

(y)|≤C max (d(Ω)

,d(Ω)

)P (d(Ω)) for |α|≤m, y ∈ Ω.Since

(x)=D

u(x), the conclusion follows from the preceding inequality.