Helms L.L. Potential Theory
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288 7 Interpolation and Monotonicity
a function u on Ω such that u
(m)
m
→ u and Du
(m)
m
→ Du on Ω as n →∞.As
before, |u|
(1+b)
1+α,Ω
≤|g|
(1+b)
1+α,Σ
and u|Σ = g.Sinceu ∈ H
(1+b)
1+α
(Ω), |u|
(1+b)
1+α,Ω
=
|g|
(1+b)
1+α,Σ
.
7.6 Monotonicity
Geometrically, a spherical chip Ω is the smaller of two regions determined
by intersecting a solid ball with a hyperplane. It will be assumed in this
section that a coordinate system is chosen so that Ω = {x ∈ R
n
: |x − x
0
| <
ρ, x
n
> 0},wherex
0
=0orx
0
is a point on the negative x
n
-axis, and that
Σ is a relatively open subset of ∂Ω ∩{x : x
n
=0}.WhenΩ is a ball in
the statement of the theorem, Σ is the empty set in which case
*
d = d and
*
Ω
δ
= Ω
δ
.
Theorem 7.6.1 If Ω is a spherical chip with Σ a relatively open subset of
∂Ω ∩{x : x
n
=0} or a ball in R
n
with Σ = ∅, j, k =1, 2,b > −1, 0 ≤
α, β ≤ 1,j + α<k+ β,j + b + α ≥ 0,and>0, then there is a constant
C
0
(b, α, β, , d(Ω)) such that
[u]
(b)
j+α,Ω
≤ C
0
*
d
b
u
0,Ω
+ [u]
(b)
k+β,Ω
and
|u|
(b)
j+α,Ω
≤ C
0
*
d
b
u
0,Ω
+ [u]
(b)
k+β,Ω
.
Proof: Since δ<1,
|u|
j+α,
*
Ω
δ
= u
0,
*
Ω
δ
+
j
i=1
[u]
i+0,
*
Ω
δ
+[u]
j+α,
*
Ω
δ
,
*
d>δon
*
Ω
δ
,andδ
j−i+α
< 1for1≤ i ≤ j,
δ
j+b+α
|u|
j+α,
*
Ω
δ
≤δ
j+b+α
u
0,
*
Ω
δ
+
j
i=1
δ
j+b+α
[u]
i+0,
*
Ω
δ
+ δ
j+b+α
[u]
j+α,
*
Ω
δ
≤ d(Ω)
j+α
*
d
b
u
0,Ω
+
j
i=1
[u]
(b)
i+0,Ω
+[u]
(b)
j+α,Ω
(7.15)
and the second assertion follows from the first by repeated application of the
first to the terms on the right side. It also can be assumed that the right
members of the above inequalities are finite, for otherwise there is nothing to
prove. Consider the following four cases:
7.6 Monotonicity 289
(i) j =1,k =2,α= β =0
(ii) j ≤ k, α =0,β > 0
(iii) j ≤ k, α > 0,β =0
(iv) j ≤ k, α > 0,β >0.
It will be shown first that there is a constant C = C(b, α, β, ), not depending
on Ω, such that
[u]
(b)
2+0,Ω
≤ C[u]
(b)
1+0,Ω
+ [u]
(b)
2+β,Ω
. (7.16)
For this purpose, let x be a fixed point of Ω,letμ ≤ 1/2beapositive
number that will be specified later, let δ =
*
d
x
/2, ρ = μδ, B = B
x,ρ
,and
B
+
= B ∩{x : x
n
> 0}(B
+
= B if Ω is a ball). If y ∈ B
+
,then
*
d
y
≥
*
d
x
−|x − y|≥(3/4)
*
d
x
=(3/2)δ>δso that y ∈
*
Ω
δ
;thatis,B
+
⊂
*
Ω
δ
.
Similarly, if y ∈ B
+
,then
*
d
y
≤
*
d
x
+ |x − y|≤(5/4)
*
d
x
. Therefore,
3
4
*
d
x
≤
*
d
y
≤
5
4
*
d
x
,y∈ B
+
. (7.17)
If r is any real number, then
*
d
r
x
≤ A(r)
*
d
r
y
,y∈ B
+
(7.18)
where A(r) = max ((4/5)
r
, (4/3)
r
). For i =1,...,n−1, let x
i
,x
i
be the end
points of the diameter of B parallel to the x
i
-axis; for i = n,letx
n
= x and
x
n
= x + ρ
n
; i.e., the latter is the point of intersection in Ω of ∂B and a
line through x perpendicular to R
n
0
.Moreover,x will denote a point on the
line segment joining x
i
to x
i
resulting from an application of the mean value
theorem. (Note that convexity of Ω or
*
Ω
δ
is not required since the mean
value theorem is applied to the convex set B
+
.) Since
|D
j
u(x
i
) − D
j
u(x
i
)|
2ρ
≤
1
ρ
sup
y∈B
+
|D
j
u(y)|,i=1,...,n− 1
and
|D
j
u(x
i
) − D
j
u(x
i
)|
ρ
≤
2
ρ
sup
y∈B
+
|D
j
u(y)|,i= n,
for some point
x on the line segment joining x
i
to x
i
,
|D
ji
u(x)|≤
2
ρ
sup
y∈B
+
|D
j
u(y)|,i=1,...,n
by the mean value theorem. Since |x −
x|≤ρ = μδ,
|D
ji
u(x)|≤|D
ji
u(x)| +
|D
ji
u(x) − D
ji
u(x)|
|x − x|
β
|x − x|
β
≤
2
μδ
sup
y∈B
+
|D
j
u(y)|+ μ
β
δ
β
[u]
2+β,
*
Ω
δ
≤
2
μδ
[u]
1+0,
*
Ω
δ
+ μ
β
δ
β
[u]
2+β,
*
Ω
δ
.
290 7 Interpolation and Monotonicity
Multiplying both sides by δ
2+b
,
δ
2+b
|D
ji
u(x)|≤
2
μ
δ
1+b
[u]
1+0,
*
Ω
δ
+ μ
β
δ
2+b+β
[u]
2+β,
*
Ω
δ
≤
2
μ
[u]
(b)
1+0,Ω
+ μ
β
[u]
(b)
2+β.Ω
.
Since δ =
*
d(x)/2,
*
d(x)
2+b
|D
ij
u(x)|≤
2
3+b
μ
[u]
(b)
1+0,Ω
+2
2+b
μ
β
[u]
(b)
2+β.Ω
.
For x ∈
*
Ω
η
,
η
2+b
|D
ij
u(x)|≤
2
3+b
μ
[u]
(b)
1+0,Ω
+2
2+b
μ
β
[u]
(b)
2+β.Ω
.
Taking the supremum over x ∈
*
Ω
η
and then the supremum over 0 <η<1,
[u]
(b)
2+0,Ω
≤
2
3+b
μ
[u]
(b)
1+0,Ω
+2
2+b
μ
β
[u]
(b)
2+β,Ω
.
Choosing μ such that 2
2+b
μ
β
<,μ<1/2 and letting C =2
3+b
/μ, the proof
of Inequality (7.16) is complete.
Case (i) j =1,k =2,α = β =0.Forsomepoint
x on the line segment
joining x
i
to x
i
,
|D
i
u(x| =
|u(x
i
) − u(x
i
)|
2ρ
≤
1
ρ
sup
y∈B
+
|u(y)|,i=1,...,n−1
and
|D
i
u(x)| =
|u(x
i
) − u(x
i
)|
ρ
≤
2
ρ
sup
y∈B
+
|u(y)|,i= n
so that
|D
i
u(x)| =
D
i
u(x)+
x
x
D
ii
udx
i
≤
2
ρ
sup
y∈B
+
|u(y)|+ ρ sup
y∈B
+
|D
ii
u(y)|.
Multiplying both sides by δ
1+b
and using the fact that ρ = μδ ≤
*
d
x
/4,
δ
1+b
|D
i
u(x)|≤
2
1−b
μ
sup
y∈B
+
|
*
d
b
x
u(y)|+ μδ
2+b
[u]
2+0,
*
Ω
δ
≤
2
1−b
μ
A(b)
*
d
b
u
0,Ω
+ μ[u]
(b)
2+0,Ω
.
7.6 Monotonicity 291
Therefore, for all x ∈
*
Ω
η
, 0 <η<1
*
d
1+b
x
|D
i
u(x)|≤
4A(b)
μ
*
d
b
u
0,Ω
+ μ2
1+b
[u]
(b)
2+0,Ω
.
Taking the supremum over x ∈
*
Ω
η
and then the supremum over 0 <η<1,
choosing μ such that μ2
1+b
<, and taking C =4A(b)/μ,
[u]
(b)
1+0,Ω
≤ C
*
d
b
u
0,Ω
+ [u]
(b)
2+0,Ω
. (7.19)
Case (ii) j ≤ k, α =0,β > 0. Suppose first that j =1,k =2.ByInequal-
ity (7.16), there is a constant C
0
such that
[u]
(b)
2+0,Ω
≤ C
0
[u]
(b)
1+0,Ω
+ [u]
(b)
2+β,Ω
.
Choose
1
> 0 such that
1
C
0
< 1/2and
1
< 1/2. Applying Inequality (7.19)
with this choice of
1
< 1, there is a constant C
1
such that
[u]
(b)
1+0,Ω
≤ C
1
*
d
b
u
0,Ω
+
1
[u]
(b)
2+0,Ω
≤ C
1
*
d
b
u
0,Ω
+
1
C
0
[u]
(b)
1+0,Ω
+
1
[u]
(b)
2+β,Ω
.
Since
1
C
0
< 1/2, the second term on the right can be transposed to the left
resulting in
[u]
(b)
1+0,Ω
≤ 2C
*
d
b
u
0,Ω
+ [u]
(b)
2+β,Ω
(7.20)
where C =2C
1
. Suppose now that j =1andk =1.Inthiscase,
|D
i
u(x)|≤|D
i
u(x)| +
|D
i
u(x) − D
i
u(x)|
|x − x|
β
|x − x|
β
≤
2
ρ
sup
y∈B
+
|u(y)|+ μ
β
δ
β
[u]
1+β,
*
Ω
δ
.
Multiplying both sides by δ
1+b
=(
*
d
x
/2)
1+b
,
*
d
1+b
x
2
1+b
|D
i
u(x)|≤
2
1−b
μ
sup
y∈B
+
|
*
d
b
x
u(y)| + μ
β
[u]
(b)
1+β,Ω
≤
2
1−b
A(b)
μ
*
d
b
u
0,Ω
+ μ
β
[u]
(b)
1+β,Ω
.
Thus, for all x ∈
*
Ω
η
*
d
1+b
x
|D
i
u(x)|≤
4A(b)
μ
*
d
b
u
0,Ω
+2
1+b
μ
β
[u]
(b)
1+β,Ω
.
292 7 Interpolation and Monotonicity
Taking the supremum over x ∈
*
Ω
η
and then the supremum over 0 <η<1,
choosing μ such that 2
1+b
μ
β
<,μ<1/2, and taking C =4A(b)/μ,
[u]
(b)
1+0,Ω
≤ C
*
d
b
u
0,Ω
+ [u]
(b)
1+β,Ω
. (7.21)
Suppose now that j =2andk = 2. By Inequality (7.16), there is a constant
C
0
such that
[u]
(b)
2+0,Ω
≤ C
0
[u]
(b)
1+0,Ω
+
2
[u]
(b)
2+β,Ω
.
Choose
1
> 0 such that
1
C
0
</2. By Inequality (7.20), there is a constant
C
1
such that
[u]
(b)
1+0,Ω
≤ C
1
*
d
b
u
0,Ω
+
1
[u]
(b)
2+β,Ω
.
Thus,
[u]
(b)
2+0,Ω
≤ C
0
C
1
*
d
b
u
0,Ω
+ C
0
1
[u]
(b)
2+β,Ω
+
2
[u]
(b)
2+β,Ω
≤ C
*
d
b
u
0,Ω
+ [u]
(b)
2+β,Ω
(7.22)
where C = C
0
C
1
, completing the proof of Case (ii).
Case (iii) j ≤ k, α > 0,β = 0. The hypothesis that j + α<k+ β requires
in this case that j =1,k =2,andα ∈ (0, 1). Fix δ>0, let x, y be distinct
points in
*
Ω
2δ
,letμ<1/2 be a constant to be specified later, let ρ = μδ,and
let B = B
x,ρ
.Ifz ∈ B
+
,then
*
d
z
≥
*
d
x
−|x − z| > (2 − μ)δ ≥ 3δ/2 >δ,
showing that B
+
⊂
*
Ω
δ
.Forsomepointx on the line segment joining x to y,
ρ
1+b+α
|D
i
u(x) − D
i
u(y)|
|x − y|
α
= ρ
1+b+α
|D
i
u(x) − D
i
u(y)|
|x − y|
|x − y|
1−α
χ
|x−y|<ρ
+ ρ
1+b+α
|D
i
u(x)| + |D
i
u(y)|
ρ
α
χ
|x−y|≥ρ
≤ ρ
2+b
|∇D
i
u(x)| +2ρ
1+b
[u]
1+0,
*
Ω
2δ
.
Therefore,
δ
1+b+α
|D
i
u(x) − D
i
u(y)|
|x − y|
α
≤ μ
1−α
√
nδ
2+b
[u]
2+0,
*
Ω
δ
+2μ
−α
δ
1+b
[u]
1+0,
*
Ω
2δ
≤
√
nμ
1−α
[u]
(b)
2+0,Ω
+2
−b
μ
−α
[u]
(b)
1+0,Ω
.
Taking the supremum over x, y ∈
*
Ω
2δ
and then the supremum over 0 <δ<1,
[u]
(b)
1+α,Ω
≤
√
nμ
1−α
[u]
(b)
2+0,Ω
+2
−b
μ
−α
[u]
(b)
1+0,Ω
.
Applying Inequality (7.19) with = μ, there is a constant C
0
such that
[u]
(b)
1+0,Ω
≤ C
0
*
d
b
u
0,Ω
+ μ[u]
(b)
2+0,Ω
.
7.6 Monotonicity 293
Therefore,
[u]
(b)
1+α,Ω
≤
√
nμ
1−α
[u]
(b)
2+0,Ω
+2
−b
C
0
μ
−α
*
d
b
u
0,Ω
+2
−b
μ
1−α
[u]
(b)
2+0,Ω
.
Choosing μ such that (
√
n +2
−b
)μ
1−α
<,μ<1/2, and letting C =
2
−b
C
0
μ
−α
,
[u]
(b)
1+α,Ω
≤ C
*
d
b
u
0,Ω
+ [u]
(b)
2+0,Ω
, (7.23)
completing the proof of Case (iii).
Case (iv) j ≤ k, α > 0,β > 0. The j =1,k = 2 case follows by applying
Inequality (7.23) and then Inequality (7.22). Only the j = k =1, 2 cases need
to be discussed in detail. Suppose first that j = k =1.Fixδ>0, let x, y be
distinct points in
*
Ω
2δ
,andletμ, ρ, B be defined relative to x as above. Then
ρ
1+b+α
|D
i
u(x) − D
i
u(y)|
|x − y|
α
= ρ
1+b+α
|D
i
u(x) − D
i
u(y)|
|x − y|
β
|x − y|
β−α
χ
|x−y|<ρ
+ ρ
1+b
(|D
i
u(x)| + |D
i
u(y)|)
so that
μ
1+b+α
[u]
(b)
1+α,Ω
≤ μ
1+b+β
[u]
(b)
1+β,Ω
+2
−b
μ
1+b
[u]
(b)
1+0,Ω
.
Therefore,
[u]
(b)
1+α,Ω
≤ μ
β−α
[u]
(b)
1+β,Ω
+2
−b
μ
−α
[u]
(b)
1+0,Ω
.
Applying Inequality (7.21) with = μ
β
, there is a constant C
0
such that
[u]
(b)
1+α,Ω
≤ μ
β−α
[u]
(b)
1+β,Ω
+2
−b
C
0
μ
−α
*
d
b
u
0,Ω
+2
−b
μ
β−α
[u]
(b)
1+β,Ω
.
Choosing μ so that (1 + 2
−b
)μ
β−α
<and taking C =2
−b
C
0
μ
−α
,
[u]
(b)
1+α,Ω
≤ C
*
d
b
u
0,Ω
+ [u]
(b)
1+β,Ω
.
The j = k = 2 case is essentially the same.
Corollary 7.6.2 If the conditions of the preceding theorem are satisfied and
j =1, 2,b>−1, 0 ≤ α ≤ 1, then there is a constant C
0
= C
0
(b, α, d(Ω)) such
that
|u|
(b)
j+α,Ω
≤ C
0
*
d
b
u
0,Ω
+[u]
(b)
j+α,Ω
.
Proof: Apply Theorem 7.6.1 to the terms [u]
(b)
i+0,Ω
in Inequality (7.15).
It will be assumed in the remainder of this section that Ω and the
*
Ω
δ
are
bounded, convex, open subsets of R
n
.
294 7 Interpolation and Monotonicity
Theorem 7.6.3 If 0 ≤ a
1
<a
2
,b
1
≤ b
2
,a
1
+ b
1
≥ 0, 0 <λ<1,and
(a, b)=λ(a
1
,b
1
)+(1− λ)(a
2
,b
2
),then
|u|
(b)
a,Ω
≤ C
|u|
(b
1
)
a
1
,Ω
λ
|u|
(b
2
)
a
2
,Ω
1−λ
where C = C(a
1
,a
2
,b
1
,b
2
,d(Ω)).
Proof: By Theorem 7.3.4,
δ
a+b
|u|
a,
*
Ω
δ
≤ Cδ
a+b
|u|
λ
a
1
,
*
Ω
δ
|u|
1−λ
a
2
,
*
Ω
δ
= C
δ
a
1
+b
1
|u|
a
1
,
*
Ω
δ
λ
δ
a
2
+b
2
|u|
a
2
,
*
Ω
δ
1−λ
≤ C
|u|
(b
1
)
a
1
,Ω
λ
|u|
(b
2
)
a
2
,Ω
1−λ
.
The result follows by taking the supremum over δ ∈ (0, 1).
In the course of the proof of the following and subsequent proofs, the
symbol C will be used as a generic symbol for a constant that may
very well change from line to line or even within a line.
Lemma 7.6.4 If a>1 and a + b ≥ 0, then there is a constant C =
C(a, d(Ω)) such that
|u|
(b)
a,Ω
≤ C
|α|≤1
|D
α
u|
(b+1)
a−1,Ω
and
|α|≤1
|D
α
u|
(b+1)
a−1,Ω
≤ C|u|
(b)
a,Ω
. (7.24)
Proof: Suppose 1 ≤ k ≤ a<k+1. Then
|u|
a,
*
Ω
δ
≤u
0,
*
Ω
δ
+
1≤|α|≤k
D
α
u
0,
*
Ω
δ
+
|α|=k
[D
α
u]
a−k,
*
Ω
δ
.
Recalling that
j
is a unit vector with 1 in the jth place, for 1 ≤|α|≤k, α =
α
+
j
for some j with |α
|≤k −1. Thus,
|u|
a,
*
Ω
δ
≤u
0,
*
Ω
δ
+
|α
|≤k−1
n
j=1
D
α
(D
j
u)
0,
*
Ω
δ
+
|α
|=k−1
n
j=1
[D
α
(D
j
u)]
(a−1)−(k−1),
*
Ω
δ
≤u
0,
*
Ω
δ
+ C
n
j=1
|D
j
u|
a−1,
*
Ω
δ
.
Since u
0,
*
Ω
δ
≤|u|
a−1,
*
Ω
δ
,
|u|
a,
*
Ω
δ
≤|u|
a−1,
*
Ω
δ
+ C
n
j=1
|D
j
u|
a−1,
*
Ω
δ
≤ C
|α|≤1
|D
α
u|
a−1,
*
Ω
δ
.
7.6 Monotonicity 295
Multiplying both sides of this inequality by δ
a+b
and taking the supremum
over δ ∈ (0, 1), the first inequality of (7.24) is obtained. On the other hand,
|α|≤1
|D
α
u|
a−1,
*
Ω
δ
= |u|
a−1,
*
Ω
δ
+
|α|=1
|D
α
u|
a−1,
*
Ω
δ
≤|u|
a−1,
*
Ω
δ
+
|α|=1
|β|≤k−1
D
β
(D
α
u)
0,
*
Ω
δ
+
|β|=k−1
[D
β
(D
α
u)]
(a−1)−(k−1),
*
Ω
δ
= |u|
a−1,
*
Ω
δ
+
1≤|β|≤k
D
β
u
0,
*
Ω
δ
+
|β|=k
[D
β
u]
a−k,
*
Ω
δ
.
Replacing |u|
a−1,
*
Ω
δ
by its definition,
|α|≤1
|D
α
u|
a−1,
*
Ω
δ
≤
|β|≤k−1
D
β
u
0,
*
Ω
δ
+
|β|=k−1
[D
β
u]
(a−1)−(k−1),
*
Ω
δ
+
1≤|β|≤k
D
β
u
0,
*
Ω
δ
+
|β|=k
[D
β
u]
a−k,
*
Ω
δ
≤ 2
|β|≤k
D
β
u
0,
*
Ω
δ
+
|β|=k−1
sup
x,y∈
*
Ω
δ
|D
β
u(x) − D
β
u(y)|
|x − y|
a−k
+
|β|=k
[D
β
u]
a−k,
*
Ω
δ
.
By the mean value theorem for derivatives and the fact that 1 − a + k ≥ 0,
there is a constant K = K(a, d(Ω)) such that
|α|≤1
|D
α
u|
a−1,
*
Ω
δ
≤ 2
|β|≤k
D
β
u
0,
*
Ω
δ
+
|β|=k−1
sup
x,y∈
*
Ω
δ
∇D
β
u
0,
*
Ω
δ
|x − y|
|x − y|
a−k
+
|β|=k
[D
β
u]
a−k,
*
Ω
δ
≤ 2
|β|≤k
D
β
u
0,
*
Ω
δ
+ K
|β|=k
D
β
u
0,
*
Ω
δ
+
|β|=k
[D
β
u]
a−k,
*
Ω
δ
≤ C
|β|≤k
D
β
u
0,
*
Ω
δ
+
|β|=k
[D
β
u]
a−k,
*
Ω
δ
≤ C|u|
a,
*
Ω
δ
.
296 7 Interpolation and Monotonicity
For each α with |α|≤1, |D
α
u|
a−1,
*
Ω
δ
≤ C|u|
a,
*
Ω
δ
. Multiplying both sides by
δ
a+b
and taking the supremum over δ ∈ (0, 1), |D
α
u|
(b+1)
a−1,Ω
≤ C|u|
(b)
a,Ω
.By
summing over |α|≤1, the second inequality of (7.24) is obtained.
Definition 7.6.5 If O is a bounded open subset of R
n
with
*
O
2δ
0
= ∅ for
some δ
0
> 0, then O is stratifiable if there is a constant K>0withthe
property that for 0 <δ≤ δ
0
and x ∈
*
O
δ
,thereisay ∈
*
O
2δ
satisfying
|x − y|≤Kδ.
Example 7.6.6 Let ρ
0
> 0, let y be any point in R
n
with y
n
< 0and
|y
n
| <ρ
0
,letΩ = B
y,ρ
0
∩ R
n
+
,andletΣ = B
y,ρ
0
∩ R
n
0
. The latter is the
flat face of a spherical chip. Taking δ
0
=(ρ
0
+ y
n
)/4,
*
Ω
δ
= B
y,ρ
0
−δ
∩R
n
+
for
0 <δ≤ δ
0
. Note that each
*
Ω
δ
is a bounded, convex, open subset of Ω.For
each x ∈
*
Ω
δ
, 0 <δ<δ
0
,thereisay ∈
*
Ω
2δ
with |x − y|≤Kδ where
K =
ρ
0
((ρ
0
− 2δ
0
)
2
− y
2
n
)
1
2
;
thus, spherical chips are stratifiable.
Monotonicity of |u|
(b)
a,Ω
with respect to each of the parameters a and b will
be considered now. It is easy to see that
|u|
(b)
a,Ω
≤|u|
(b
)
a,Ω
whenever b>b
. (7.25)
The following theorem and its proof are adapted from the paper cited
below that deals with the Σ = ∅ case.
Theorem 7.6.7 (Gilbarg and H¨ormander[24]) If Ω is a stratifiable,
bounded, convex open subset of R
n
0
,each
*
Ω
δ
is convex, 0 ≤ a
<a,a
+ b ≥ 0
and b is not a negative integer or 0, then there is a constant C = C(a, b, d(Ω))
such that
|u|
(b)
a
,Ω
≤ C|u|
(b)
a,Ω
. (7.26)
Proof: Consider the a ≤ 1 case first. Suppose it has been shown that
|u|
(b)
0,Ω
≤ C|u|
(b)
a,Ω
if b>0 (7.27)
and
|u|
(b)
−b,Ω
≤ C|u|
(b)
a,Ω
if b<0,a
+ b ≥ 0. (7.28)
In the first case, a
= λ · 0+(1− λ)a for some λ ∈ (0, 1] and it follows from
Theorem 7.6.3 that
|u|
(b)
a
,Ω
≤ C
|u|
(b)
0,Ω
λ
|u|
(b)
a,Ω
1−λ
≤ C|u|
(b)
a,Ω
.
7.6 Monotonicity 297
In the second case, a
= λ(−b)+(1− λ)a for some λ ∈ (0, 1] and it follows
from Theorem 7.6.3 that
|u|
(b)
a
,Ω
≤ C
0
|u|
(b)
−b,Ω
λ
|u|
(b)
a,Ω
1−λ
≤ C|u|
(b)
a,Ω
.
It therefore suffices to prove Inequalities (7.27) and (7.28) for the a ≤ 1case.
To prove the first, assume that b>0andletδ
0
and K be the constants
specified in the stratifiable definition. It can be assumed that |u|
(b)
a,Ω
is finite,
for otherwise there is nothing to prove. For any δ ≤ δ
0
,
δ
b+a
sup
x,y∈
*
Ω
δ
|u(x) − u(y)|
|x − y|
a
≤|u|
(b)
a,Ω
.
Consider only those δ for which δ<δ
0
/2. For any x ∈
*
Ω
δ
,choosek ≥ 1such
that δ
0
/2 < 2
k
δ ≤ δ
0
. By the preceding lemma, there is a x
1
∈
*
Ω
2δ
such that
|x − x
1
|≤Kδ.Thus,
δ
b+a
|u(x) − u(x
1
)|
(Kδ)
a
≤|u|
(b)
a,Ω
or
|u(x) − u(x
1
)|≤K
a
δ
−b
|u|
(b)
a,Ω
.
Thus,
|u(x)|≤|u(x
1
)| + |u(x) − u(x
1
)|
≤|u(x
1
)| + K
a
δ
−b
|u|
(b)
a,Ω
.
Applying the same argument to x
1
∈
*
Ω
2δ
,thereisax
2
∈
*
Ω
4δ
such that
|u(x)|≤|u(x
2
)|+ K
a
(2δ)
−b
|u|
(b)
a,Ω
+ K
a
δ
−b
|u|
(b)
a,Ω
.
Continuing in this way, there is a finite sequence x
1
....,x
k
with x
k
∈
*
Ω
2
k
δ
⊂
*
Ω
δ
0
/2
such that
|u(x)|≤|u(x
k
)| + K
a
k−1
j=0
(2
j
δ)
−b
|u|
(b)
a,Ω
.
Since (δ
0
/2)
a+b
|u|
a,
*
Ω
δ
0
/2
≤|u|
(b)
a,Ω
and x
k
∈
*
Ω
δ
0
/2
,
|u(x
k
)|≤|u|
a,
*
Ω
δ
0
/2
≤
2
a+b
δ
a+b
0
|u|
(b)
a,Ω
.