Helms L.L. Potential Theory
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378 10 Apriori Bounds
Since D
(y
r
)
n
= −D
y
n
,r
1
= r
2
,andy
r
= y on B
0
2
,
u(x)ψ
ρ
(x)=−
B
0
2
(r
2
D
y
n
(uψ
ρ
) − uψ
ρ
D
y
n
r
2
) dσ(y)
+
B
+
2
r
1
Δ(uψ
ρ
) dy, x ∈ B
+
2
.
(10.6)
Adding Equations (10.5) and (10.6), for x ∈ B
+
1
u(x)=u(x)ψ
ρ
(x)=
1
2
B
+
2
(r
1
− r
2
)Δ(uψ
ρ
) dy.
Lemma 10.3.2 If Ω is a spherical chip with Σ = int(∂Ω ∩ R
n
0
) and u ∈
C
1
(Ω ∪Σ)∩C
2
(Ω) satisfies Poisson’s equation Δu = f on Ω for f ∈ H
α
(Ω),
then
|u; ρ|
2+α,B
+
1
≤ C(n, α))
u
0,B
+
2
+ ρ
2
|f; ρ|
α,B
+
2
.
Proof: By the preceding lemma,
u =
B
+
2
K
0
(·,y)(uΔψ
ρ
+ ψ
ρ
f +2∇u ·∇ψ
ρ
) dy on B
+
1
.
Using the fact that |x − y
r
|≥|x − y| for x, y ∈ B
+
2
, it is easily seen that
K
0
satisfies Inequalities (8.1) through (8.4) and is therefore a subnewtonian
kernel. After writing the above integral as a sum of three integrals, denote
the three terms by u
1
,u
2
,andu
3
, respectively. Since ψ
ρ
=1onB
+
5/4
,
u
1
=
B
+
2
∼B
+
5/4
K
0
(·,y)uΔψ
ρ
dy on B
+
1
.
For x ∈ B
+
1
and y ∈ B
+
2
∼ B
+
5/4
, |x − y|≥ρ/4, making it easy to ob-
tain bounds on the norms of the derivatives of u
1
using the bounds on the
D
β
ψ
ρ
0,B
+
2
with the result
|u
1
; ρ|
2+α,B
+
1
≤ C(n)u
0,B
+
2
. (10.7)
By Lemma 8.2.10,
[u
2
]
2+0,B
+
1
+ ρ
α
[u
2
]
2+α,B
+
1
≤ C(n, α)
ψ
ρ
f
0,B
+
2
+ ρ
α
[ψ
ρ
f]
α,B
+
2
.
Since [ψ
ρ
f]
α,B
+
2
≤ ρ
−α
f
0,B
+
2
+[f]
α,B
+
2
,
[u
2
]
2+0,B
+
1
+ ρ
α
[u
2
]
2+α,B
+
1
≤ C(n, α)
f
0,B
+
2
+ ρ
α
[f]
α,B
+
2
.
10.3 Mixed Boundary Conditions for Laplacian 379
It follows from the definition of u
2
that
u
2
0,B
+
1
≤ C(n)ρ
2
f
0,B
+
2
and
[u
2
]
1+0,B
+
1
= Du
2
0,B
+
1
≤ C(n)ρf
0B
+
2
.
Therefore,
|u
2
; ρ|
2+α,B
+
1
≤ C(n, α))ρ
2
f
0,B
+
2
+ ρ
α
[f]
α,B
+
2
= C(n, α))ρ
2
|f; ρ|
α,B
+
2
.
(10.8)
Consider now
u
3
=2
B
+
2
K
0
(·,y)∇u ·∇ψ
ρ
dy =
n
i=1
2
B
+
2
K
0
(·,y)
∂u
∂y
i
∂ψ
ρ
∂y
i
dy.
Letting v
i
denote the ith term of the sum, it suffices to estimate each
|v
i
; ρ|
2+α,B
+
1
.For1≤ i ≤ n,letτ
i
denote the projection map
τ
i
(y
1
,...,y
n
)=(y
1
,...,y
i−1
, 0,y
i+1
,...,y
n
).
Letting dy
i
denote integration with respect to y
1
,...,y
i−1
,y
i+1
,...,y
n
,
B
+
2
K
0
(·,y)
∂u
∂y
i
∂ψ
ρ
∂y
i
dy =
τ
i
B
+
2
ζ
+
i
ζ
−
i
K
0
(·,y)
∂u
∂y
i
∂ψ
ρ
∂y
i
dy
i
dy
i
where ζ
−
i
and ζ
+
i
are the ith components of point of intersection of the τ
i
y-
section of B
+
2
. Integrating by parts,
ζ
+
i
ζ
−
i
K
0
(·,y)
∂u
∂y
i
∂ψ
ρ
∂y
i
dy= K
0
(·,y)u
∂ψ
ρ
∂y
i
ζ
+
i
ζ
−
i
−
ζ
+
i
ζ
−
i
u
∂
∂y
i
K
0
(·,y)
∂ψ
ρ
∂y
i
dy
i
.
For 1 ≤ i ≤ n−1, (y
1
,...,y
i−1
,ζ
±
i
,y
i+1
,...,y
n
)isapointof∂B
+
2
∼ B
0
2
where
∂ψ
ρ
/∂y
i
isequalto0.Wheni = n, the same is true of (y
1
,...,y
n−1
,ζ
+
n
); at
the point (y
1
,...,y
n−1
,ζ
−
n
),K
0
(·,y) = 0. The first term on the right is thus
equal to 0. Thus,
B
+
2
K
0
(·,y)
∂u
∂y
i
∂ψ
ρ
∂y
i
dy = −
B
+
2
u
∂
∂y
i
K
0
(·,y)
∂ψ
ρ
∂y
i
dy.
Since ∂ψ
ρ
/∂y
i
=0onB
+
5/4
,
v
i
= −2
B
+
2
∼B
+
5/4
u
K
0
(·,y)
∂
2
ψ
ρ
∂y
2
i
+
∂
∂y
i
K
0
(·,y)
∂ψ
ρ
∂y
i
dy.
380 10 Apriori Bounds
As was the case for u
1
,forx ∈ B
+
1
and y ∈ B
+
2
∼ B
+
5/4
, |x − y|≥ρ/4
making it easy to show that |v
i
; ρ|
2+α,B
+
1
≤ C(n)u
0,B
+
2
. It follows that
|u
3
; ρ|
2+α,B
+
1
≤ C(n)u
0,B
+
2
. Combining this result with Inequalities (10.7)
and (10.8) completes the proof.
Before stating the next result, it is necessary to define a weighted norm of
a boundary function. Let Ω be a bounded open subset of R
n
and let Σ be a
nonempty, relatively open subset of ∂Ω.Ifd>0isaconstantandg : Σ → R,
let
|g; d|
1+α,Σ
=inf{|u; d|
1+α,Ω
; u ∈ H
1+α
(Ω; d),u|
Σ
= g}.
In using this definition in a proof, g should be replaced by a u ∈ H
1+α
(Ω; d)
and then take the infimum of the |u; d|
1+α,Ω
. For all practical purposes, this
amounts to considering g as a function in H
1+α
(Ω; d).
The following lemma applies to both spherical chips with Σ = int(∂Ω ∩
R
n
0
) and balls with Σ = ∅. The latter case is dealt with in Theorem 8.4.2
and is a special case of the following lemma with B
+
i
= B
i
and B
0
6
= ∅.
Although the condition on Ω,namely,ρ
0
>ρ/
√
2, in Lemma 8.5.6 is not
required for the following lemma, the choice of the balls B
t
ensures that the
argument of g in the definition of ψ below is a point in Σ.Theρ
0
>ρ
√
2
condition was imposed to achieve the same result. It follows that the estimates
for ψ
0,B
+
2
, Dψ
0,B
+
2
, D
2
ψ
0,B
+
2
,and[D
2
ψ]
α,B
+
2
are valid in the present
context.
Lemma 10.3.3 If B
+
6
= B
+
x
0
,6ρ
⊂ Ω, u ∈ C
1
(Ω ∪ Σ) ∩ C
2
(Ω),Δu = f on
Ω for f ∈ H
α
(Ω),α∈ (0, 1),andM
0
u = g on Σ for g ∈ H
1+α
(Σ),then
|u; ρ|
2+α,B
+
1
≤ C(u
0,B
+
2
+ ρ|g; ρ|
1+α,B
0
6
+ ρ
2
|f; ρ|
α,B
+
2
)
where C = C(α, β
n
).
Proof: By Theorem 7.2.6, it can be assumed that g ∈ H
1+α
(Ω)andthat
|g|
1+α,Σ
= |g|
1+α,Ω
.IfB
2
⊂ R
n
+
, the inequality follows from Theorem 8.4.2
without the ρ|g; ρ|
1+α,B
0
6
term. It can be assumed therefore that B
2
∩R
n
0
= ∅
so that x
0n
≤ 2ρ where x
0
=(x
0
,x
0n
). Consider the function ψ,modifiedby
a constant factor 1/β
n
, defined by Equation (8.30) using the extension ˜g of g
to R
n
0
and the function η ∈ C
2
(R
n
0
) having compact support in B
0,1
defined
in the proof of Lemma 8.5.6. For (x
,x
n
) ∈ B
+
2
and y
∈ B
0,1
⊂ R
n
0
,
|x
0
− (x
− x
n
y
, 0)|
≤|x
0
− (x
0
, 0)| + |(x
0
, 0) − (x
, 0)| + |(x
, 0) − (x
− x
n
y
, 0)|
< 6ρ
showing that (x
− x
n
y
, 0) ∈ B
x
0
,6ρ
and therefore that x
− x
n
y
∈ B
0
6
;that
is, for x ∈ B
+
2
, the function ψ is defined by
ψ(x)=ψ(x
,x
n
)=
x
n
β
n
R
n
0
g(x
− x
n
y
, 0)η(y
) dy
.
10.3 Mixed Boundary Conditions for Laplacian 381
By Equations (8.31) to (8.35) and Inequality (8.36), ψ
0,B
+
2
≤Cρ[g]
0,B
0
6
,
Dψ
0,B
+
2
≤C[g]
0,B
0
6
, D
2
ψ
0,B
+
2
≤C[g]
1+0,B
0
6
,and[D
2
ψ]
α,B
+
2
≤ C[g]
1+α,B
0
6
.
It follows that ψ ∈ H
2+α
(B
+
2
)andthatΔψ ∈ H
α
(B
+
2
). Combining these
inequalities
|ψ; ρ|
2+α,B
+
2
≤ Cρ|g; ρ|
1+α,B
0
6
(10.9)
where C = C(n, β
n
). Note that D
x
i
ψ =0for1≤ i ≤ n − 1whenx
n
=0
and D
x
n
ψ = g when x
n
=0sothatM
0
ψ = g on Σ. Letting v = u − ψ on
B
+
2
,Δv = f − Δψ ∈ H
α
(B
+
2
)andM
0
v = M
0
u − M
0
ψ =0onB
0
2
.Bythe
preceding lemma, Lemma 10.3.2,
|v; ρ|
2+α,B
+
1
≤ C(v
0,B
+
2
+ ρ
2
|f − Δψ; ρ|
α,B
+
2
)
≤ C(u
0,B
+
2
+ ψ
0,B
+
2
+ ρ
2
|f; ρ|
α,B
+
2
+ ρ
2
|Δψ; ρ|
α,B
+
2
)
where C = C(n, α). Since ψ
0,B
+
2
≤C(n, β
n
)ρ[g]
0,B
0
6
≤ C(n, β
n
)ρ|g; ρ|
1+α,B
0
6
and
ρ
2
|Δψ; ρ|
α,B
+
2
≤ nρ
2
|D
2
ψ; ρ|
α,B
+
2
≤ Cρ
2
(D
2
ψ
0,B
+
2
+ ρ
α
[D
2
ψ]
α,B
+
2
)
≤ Cρ
2
([ψ]
2+0,B
+
2
+ ρ
α
[ψ]
2+α,B
+
2
)
≤ Cρ
2
([g]
1+0,B
0
6
+ ρ
α
[g]
1+α,B
0
6
)
≤ Cρ|g; ρ|
1+α,B
0
6
.
|v; ρ|
2+α,B
+
1
≤ C(u
0,B
+
2
+ ρ|g; ρ|
1+α,B
0
6
+ ρ
2
|f; ρ|
α,B
+
2
).
By Inequality (10.9),
|u; ρ|
2+α,B
+
1
≤|v; ρ|
2+α,B
+
1
+ |ψ; ρ|
2+α,B
+
1
≤|v; ρ|
2+α,B
+
1
+ Cρ|g; ρ|
1+α,B
0
6
≤ C(u
0,B
+
2
+ ρ|g; ρ|
1+α,B
0
6
+ ρ
2
|f; ρ|
α,B
+
2
)
where C = C(n, α, β
n
).
Under the conditions of the preceding lemma, if b>−1, x
0
∈ Ω,and
ρ =
*
d
x
0
/8, then
ρ
1+b
|g; ρ|
1+α,B
0
6
≤ 3|g|
(1+b)
1+α,Σ
(10.10)
and
ρ
2+b
|f; ρ|
α,B
+
2
≤ 2|f |
(2+b)
α,Ω
. (10.11)
To prove the first inequality, prove it first with Σ replaced by Ω and then
use the fact that |g|
(1+b)
1+α,Ω
= |g|
(1+b)
1+α,Σ
. The second follows in a similar way.
382 10 Apriori Bounds
Theorem 10.3.4 If Ω is a spherical chip with Σ = int(∂Ω ∩ R
n
0
) or a ball
in R
n
with Σ = ∅,α∈ (0, 1),b>−1,u∈ C
1
(Ω ∪Σ)∩C
2
(Ω) satisfies Δu = f
on Ω for f ∈ H
(2+b)
α
(Ω),andM
0
u = g on Σ for g ∈ H
(1+b)
1+α
(Σ), then there
is a constant C
0
= C
0
(b, α, β
n
,d(Ω)) such that
|u|
(b)
2+α,Ω
≤ C
0
*
d
bu
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
. (10.12)
Proof: Since |u|
(b)
2+α,Ω
≤|u|
(b)
2+0,Ω
+[u]
(b)
2+α,Ω
, Theorem 7.6.1 can be applied
to the first term to show that there is a constant C
0
= C
0
(b, α, d(Ω)) such
that
|u|
(b)
2+0,Ω
≤ C
0
*
d
bu
0,Ω
+[u]
(b)
2+α,Ω
.
It therefore suffices to prove the inequality for [u]
(b)
2+α,Ω
.Consideranyx ∈ Ω.
Letting ρ =
*
d
x
/8andB
t
= B
x,tρ
,
*
d
x
=8ρ>7ρ so that x ∈
*
Ω
7ρ
and B
+
6
=
B
+
x,6ρ
⊂
*
Ω
ρ
.Ify ∈ B
+
2
and b ≥ 0, then
*
d
y
≥
*
d
x
−|y −x| >
*
d
x
−2ρ =(3/4)
*
d
x
so that
*
d
b
x
≤ (4/3)
b
*
d
b
y
;ify ∈ B
+
2
and b<0, then
*
d
y
≤
*
d
x
+ |y − x| < 2
*
d
x
so
that
*
d
b
x
<
*
d
b
y
/2
b
. In either case, there is a constant C = C(b) such that
*
d
b
x
≤ C
*
d
b
y
,y∈ B
+
2
. (10.13)
Noting that g ∈ H
(1+b)
1+α
(Σ) implies g ∈ H
1+α
(B
0
6
) by Inequality (10.10) and
applying Lemma 10.3.3,
*
d
2+b
x
|D
2
u(x)|≤
*
d
b
x
(8ρ)
2
D
2
u
0,B
+
1
≤ C
*
d
b
x
u
0,B
+
2
+ ρ|g; ρ|
1+α,B
0
6
+ ρ
2
|f; ρ|
α,B
+
2
.
By Inequalities (10.10), (10.11), and (10.13) and the facts that B
0
6
⊂ Σ
ρ
,
B
+
2
⊂
*
Ω
ρ
, for any x ∈ Ω
*
d
2+b
x
|D
2
u(x)|≤C
*
d
b
x
u
0,B
+
2
+ ρ
1+b
|g; ρ|
1+α,B
0
6
+ ρ
2+b
|f; ρ|
α,B
+
2
≤ C
*
d
b
u
0,B
+
2
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
≤ C
*
d
b
u
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
. (10.14)
Consider now any x, y ∈ Ω, labeled so that
*
d
x
≤
*
d
y
. Define ρ and B
i
relative
to x as above. Applying Inequality (10.14) to
*
d
2+b
x
|D
2
u(x)| and
*
d
2+b
y
|D
2
u(y)|
and applying Lemma 10.3.3 to [D
2
u]
α,B
+
1
,
10.3 Mixed Boundary Conditions for Laplacian 383
*
d
2+b+α
xy
|D
2
u(x) − D
2
u(y)|
|x − y|
α
≤
*
d
2+b+α
x
|D
2
u(x) − D
2
u(y)|
|x − y|
α
χ
|x−y|<ρ
+
*
d
2+b+α
x
|D
2
u(x)| + |D
2
u(y)|
ρ
α
χ
|x−y|≥ρ
≤ 8
2+b+α
ρ
2+b+α
[D
2
u]
α,B
+
1
+8
α
*
d
2+b
x
|D
2
u(x)| +
*
d
2+b
y
|D
2
u(y)|
≤ C
ρ
b
(u
0,B
+
2
+ ρ|g; ρ|
1+α,B
0
6
+ ρ
2
|f; ρ|
α,B
+
2
)
+
*
d
b
u
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
≤ C
*
d
b
u
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
.
Taking the supremum over x, y ∈
*
Ω
η
, 0 <η<1 and using the fact that
*
d
xy
>ηfor x, y ∈
*
Ω
η
,
η
2+b+α
[u]
2+α,
*
Ω
η
≤ C
*
d
b
u
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
.
Taking the supremum over η,
[u]
(b)
2+α,Ω
≤ C
*
d
b
u
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
where c = C(α, b, β
n
).
The preceding results apply only to Poisson’s equation u = f. The first
step in passing to elliptic operators on general domains is to extend these
results to elliptic operators L
0
with constant coefficients. Let T : R
n
→ R
n
be the nonsingular transformation defined by the matrix of coefficients {a
ij
}
as in Section 9.3. If Ω ⊂ R
n
and Σ ⊂ ∂Ω,letΩ
∗
= T (Ω)andΣ
∗
=
T (Σ). Since T is one-to-one and onto, T preserves the set operations so that
T (∂Ω ∼ Σ)=T (∂Ω) ∼ T (Σ)=∂Ω
∗
∼ Σ
∗
, (∂Ω)
∗
= ∂(Ω
∗
),Σ
∗
⊂ ∂Ω
∗
,and
T (∂Ω)=∂Ω
∗
. Recalling that
*
d(x)=d(x, ∂Ω ∼ Σ), the
*
d notation will be
augmented by adding the subscript Ω in order to deal with the two sets Ω
and Ω
∗
simultaneously. Using the fact that T is one-to-one and onto, it is
easily checked that there is a constant c = c(a
ij
;1≤ i, j ≤ n) such that
1
c
*
d
Ω
(x) ≤
*
d
Ω
∗
(y) ≤ c
*
d
Ω
(x),y= Tx,x∈ Ω.
Moreover, there are also constants c = c(a
ij
;1≤ i, j ≤ n)andk = k(a
ij
;1≤
i, j ≤ n) such that
T (
*
Ω
δ
) ⊂
(Ω
∗
)
cδ
(10.15)
and
(Ω
∗
)
δ
⊂ T (
*
Ω
kδ
).
384 10 Apriori Bounds
If u is any function on Ω,letu
∗
(y)=u(x)fory = T (x),x ∈ Ω. Clearly,
u
0,Ω
= u
∗
0,Ω
∗
.Inparticular,u
0,
*
Ω
δ
= u
∗
0,T (
*
Ω
δ
)
. By Relation (10.15),
there is a constant c = c(a
ij
;1≤ i, j ≤ n) such that
u
0,
*
Ω
δ
= u
∗
0,T (
*
Ω
δ
)
≤u
∗
0,
(Ω
∗
)
cδ
.
As was shown in Section 9.3,
∂u
∂x
j
=
n
i=1
∂u
∗
∂y
i
1
√
λ
i
o
ij
and
1
√
λ
i
o
ij
≤
n
i=1
1
√
λ
i
|o
ij
|≤c
√
n
Therefore,
D
x
j
u
0,
*
Ω
δ
=sup{|D
x
j
u(x)|; x ∈
*
Ω
δ
}
≤ C sup {
n
i=1
|D
y
i
u
∗
(y)|; T
−1
y ∈
*
Ω
δ
}
= C sup {
n
i=1
|D
y
i
u
∗
(y)|; y ∈ T (
*
Ω
δ
)}
≤ C sup {
n
i=1
|D
y
i
u
∗
(y)|; y ∈
(Ω
∗
)
cδ
}
≤ CDu
∗
0,
(Ω
∗
)
cδ
Similar results hold for D
2
u
∗
0,
(Ω
∗
)
δ
and [D
2
u
∗
]
α,
(Ω
∗
)
δ
.Thus,thereare
constants C and c, not depending on Ω, such that
1
C
|u|
2+α,
˜
Ω
δ/c
≤|u
∗
|
2+α,
(Ω
∗
)
δ
≤ C|u|
2+α,
˜
Ω
cδ
.
Multiplying both sides by δ
2+α+b
, adjusting constants, and taking the supre-
mum over δ,
1
C
|u|
(b)
2+α,Ω
≤|u
∗
|
(b)
2+α,Ω
∗
≤ C|u|
(b)
2+α,Ω
(10.16)
where C = C(a
ij
,α,b).
If u ∈ C
2
(Ω)oru ∈ H
α
(Ω), then u
∗
∈ C
2
(Ω
∗
)oru
∗
∈ H
α
(Ω
∗
), respec-
tively; moreover, if g ∈ H
(1+b)
1+α
(Σ), then g
∗
∈ H
(1+b)
1+α
(Σ
∗
). If u ∈ C
2
(Ω),f ∈
H
(2+b)
α
(Ω), and g ∈ H
(1+b)
1+α
(Σ), as in Section 9.3 the oblique derivative bound-
ary value problem
L
0
u = f on Ω, M
0
u = g on Σ
10.4 Nonconstant Coefficients 385
is transformed by the transformation T into the problem
Δu
∗
= f
∗
on Ω
∗
, M
1
u
∗
= g
∗
on Σ
∗
where
M
1
u
∗
(y)=
n
j=1
β
∗
j
D
y
j
u
∗
(y)+γu
∗
(y)
with
β
∗
j
=
n
i=1
β
i
o
ji
λ
j
,j=1,...,n.
The following lemma is an immediate consequence of Theorem 10.3.4 and
Inequality (10.16).
Lemma 10.3.5 If Ω is a spherical chip with Σ = int(∂Ω ∩ R
n
0
) or a ball
in R
n
with Σ = ∅,α ∈ (0, 1),b > −1,andu ∈ C
1
(Ω ∪ Σ) ∩ C
2
(Ω) satisfies
L
0
u = f on Ω for f ∈ H
(2+b)
α
(Ω),andM
0
u = g on Σ for g ∈ H
(1+b)
1+α
(Σ),
then there is a constant C
0
= C
0
(a
ij
,b,α,β
i
,γ,d(Ω)) such that
|u|
(b)
2+α,Ω
≤ C
0
*
d
b
u
0,Ω
+ |g|
(1+b)
1+α,Σ
+ |f |
(2+b)
α,Ω
.
10.4 Nonconstant Coefficients
As in the preceding section, Ω will denote a spherical chip with Σ = int(∂Ω∩
R
n
0
) or a ball in R
n
with Σ = ∅. In this section, the general elliptic operator L
as defined in Section 9.1 will be assumed to satisfy Inequalities (9.2). Consider
the oblique derivative problem Lu = f on Ω for f ∈ H
(2+b)
α
(Ω) subject to the
boundary condition Mu = g on Σ for g ∈ H
(1+b)
1+α
(Σ). If x
0
is a fixed point
of Ω, define operators L
0
and M
0
with constant coefficients by the following
equations:
L
0
u(x)=
n
i,j=1
a
ij
(x
0
)
∂
2
u
∂x
i
∂x
j
(x),x∈ Ω (10.17)
and
M
0
u(x
, 0) =
n
i=1
β
i
(x
0
)
∂u
∂x
i
(x
, 0) + γ(x
0
)u(x
, 0),x
∈ Σ. (10.18)
These operators will be used as crude approximations to L and M.The
following inequalities, with appropriate restrictions on the parameters α, b, b
,
along with repeated applications of Inequalities (7.25) and (7.26) will be used
to justify these approximations.
386 10 Apriori Bounds
|fg|
(b+b
)
α,Ω
≤ C|f |
(b)
α,Ω
|g|
(b
)
α,Ω
, (10.19)
|fg|
(b+b
)
1+α,Ω
≤ C|f |
(b)
1+α,Ω
|g|
(b
)
1+α,Ω
. (10.20)
The proof of Inequality (10.19) is similar, but easier, than the proof of
Inequality (10.20) and only the latter will be proved. Since
|fg|
1+α,
*
Ω
δ
=[fg]
0,
*
Ω
δ
+[fg]
1+0,
*
Ω
δ
+[fg]
1+α,
*
Ω
δ
≤ [f ]
0,
*
Ω
δ
[g]
0,
*
Ω
δ
+[f]
0,
*
Ω
δ
[g]
1+0,
*
Ω
δ
+[f]
1+0,
*
Ω
δ
[g]
0,
*
Ω
δ
+sup
x,y∈
*
Ω
δ
|f(x)Dg(x)+g(x)Df(x) − f(y)Dg(y) − g(y)Df(y)|
|x − y|
α
≤ [f ]
0,
*
Ω
δ
[g]
0,
*
Ω
δ
+[f]
0,
*
Ω
δ
[g]
1+0,
*
Ω
δ
+[f]
1+0,
*
Ω
δ
[g]
0,
*
Ω
δ
+[f]
0,
*
Ω
δ
[g]
1+α,
*
Ω
δ
+[f]
α,
*
Ω
δ
[g]
1+0,
*
Ω
δ
+[f]
1+α,
*
Ω
δ
[g]
0,
*
Ω
δ
+[f ]
1+0,
*
Ω
δ
[g]
α,
*
Ω
δ
.
Multiplying both sides by δ
b+b
+1+α
, using the fact that only δ<1 need be
considered, and using Inequality (7.33),
δ
b+b
+1+α
|fg|
1+α,
*
Ω
δ
≤ δ
b
[f]
0,
*
Ω
δ
δ
b
[g]
0,
*
Ω
δ
+ δ
b
[f]
0,
*
Ω
δ
δ
1+b
[g]
1+0,
*
Ω
δ
+ δ
1+b
[f]
1+0,
*
Ω
δ
δ
b
[g]
0,
*
Ω
δ
+ δ
b
[f]
0,
*
Ω
δ
δ
1+b
+α
[g]
1+α,
*
Ω
δ
+ δ
b+α
[f]
α,
*
Ω
δ
δ
1+b
[g]
1+0,
*
Ω
δ
+ δ
1+α+b
[f]
1+α,
*
Ω
δ
δ
b
[g]
0,
*
Ω
δ
+ δ
1+b
[f]
1+0,
*
Ω
δ
δ
b
+α
[g]
α,
*
Ω
δ
≤ +[f]
(b)
0,Ω
[g]
(b
)
0,Ω
+[f]
(b)
0,Ω
[g]
(b
)
1+0,Ω
+[f]
(b)
1+0,Ω
[g]
(b
)
0,Ω
+[f]
(b)
0,Ω
[g]
(b
)
1+α,Ω
+[f]
(b)
α,Ω
[g]
(b
)
1+0,Ω
+[f]
(b)
1+α,Ω
[g]
(b
)
0,Ω
+[f]
(b)
1+0,Ω
[g]
(b
)
α,Ω
≤ C|f |
(b)
1+α,Ω
|g|
(b
)
1+α,Ω
.
Inequality (10.20) follows by taking the supremum over δ>0.
It will be assumed throughout this section that the coefficients of L and
M satisfy the inequality
A =sup
1≤i,j≤n
a
i,j
0,Ω
+ |a
ij
|
(0)
α,Ω
+ |b
i
|
(1)
α,Ω
+ |c|
(2)
α,Ω
+ β
i
0,Σ
+ |β
i
|
(0)
1+α,Σ
+ γ
0,Σ
+ |γ|
(1)
1+α,Σ
< +∞.