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456
BRUCE K. DRIVER
†
24.
Hölder Spaces
Notation 24.1.
Let
Ω
be an open subset of
R
d
,BC
(
Ω
)
and
BC
(
¯
Ω
)
be the bounded
continuous functions on
Ω
and
¯
Ω
respectively. By identifying
f
∈
BC
(
¯
Ω
)
with
f

Ω
∈
BC
(
Ω
)
,
we will consider
BC
(
¯
Ω
)
as a subset of
BC
(
Ω
)
.
For
u
∈
BC
(
Ω
)
and
0
<β
≤
1
let
k
u
k
u
:= sup
x
∈
Ω

u
(
x
)

and
[
u
]
β
:= sup
x,y
∈
Ω
x
6
=
y
½

u
(
x
)
−
u
(
y
)


x
−
y

β
¾
.
If
[
u
]
β
<
∞
,
then
u
is
Hölder continuous
with holder exponent
43
β.
The collection
of
β
— Hölder continuous function on
Ω
will be denoted by
C
0
,β
(
Ω
):=
{
u
∈
BC
(
Ω
):[
u
]
β
<
∞
}
and for
u
∈
C
0
,β
(
Ω
)
let
(24.1)
k
u
k
C
0
,β
(
Ω
)
:=
k
u
k
u
+[
u
]
β
.
Remark
24.2
.
If
u
:
Ω
→
C
and
[
u
]
β
<
∞
for some
β>
1
,
then
u
is constant on
each connected component of
Ω
.
Indeed, if
x
∈
Ω
and
h
∈
R
d
then
¯
¯
¯
¯
u
(
x
+
th
)
−
u
(
x
)
t
¯
¯
¯
¯
≤
[
u
]
β
t
β
/t
→
0
as
t
→
0
which shows
∂
h
u
(
x
)=0
for all
x
∈
Ω
.
If
y
∈
Ω
is in the same connected component
as
x,
then by Exercise 17.5 there exists a smooth curve
σ
:[0
,
1]
→
Ω
such that
σ
(0) =
x
and
σ
(1) =
y.
So by the fundamental theorem of calculus and the chain
rule,
u
(
y
)
−
u
(
x
)=
Z
1
0
d
dt
u
(
σ
(
t
))
dt
=
Z
1
0
0
dt
=0
.
This is why we do not talk about Hölder spaces with Hölder exponents larger than
1
.
Lemma 24.3.
Suppose
u
∈
C
1
(
Ω
)
∩
BC
(
Ω
)
and
∂
i
u
∈
BC
(
Ω
)
for
i
=1
,
2
,...,d,
then
u
∈
C
0
,
1
(
Ω
)
,
i.e.
[
u
]
1
<
∞
.
The proof of this lemma is left to the reader as Exercise 24.1.
Theorem 24.4.
Let
Ω
be an open subset of
R
d
.
Then
(1)
Under the identi
f
cation of
u
∈
BC
¡
¯
Ω
¢
with
u

Ω
∈
BC
(
Ω
)
,BC
(
¯
Ω
)
is a
closed subspace of
BC
(
Ω
)
.
(2)
Every element
u
∈
C
0
,β
(
Ω
)
has a unique extension to a continuous func
tion (still denoted by
u
)
on
¯
Ω
.
Therefore we may identify
C
0
,β
(
Ω
)
with
C
0
,β
(
¯
Ω
)
⊂
BC
(
¯
Ω
)
.
(In particular we may consider
C
0
,β
(
Ω
)
and
C
0
,β
(
¯
Ω
)
to
bethesamewhen
β>
0
.
)
(3)
The function
u
∈
C
0
,β
(
Ω
)
→
k
u
k
C
0
,β
(
Ω
)
∈
[0
,
∞
)
is a norm on
C
0
,β
(
Ω
)
which make
C
0
,β
(
Ω
)
into a Banach space.
Proof. 1.
The
f
rst item is trivial since for
u
∈
BC
(
¯
Ω
)
,
the supnorm of
u
on
¯
Ω
agrees with the supnorm on
Ω
and
BC
(
¯
Ω
)
is complete in this norm.
43
If
β
=1
,u
is is said to be Lipschitz continuous.
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View Full DocumentANALYSIS TOOLS WITH APPLICATIONS
457
2.
Suppose that
[
u
]
β
<
∞
and
x
0
∈
∂
Ω
.
Let
{
x
n
}
∞
n
=1
⊂
Ω
be a sequence such
that
x
0
= lim
n
→∞
x
n
.
Then

u
(
x
n
)
−
u
(
x
m
)

≤
[
u
]
β

x
n
−
x
m

β
→
0
as
m, n
→∞
showing
{
u
(
x
n
)
}
∞
n
=1
is Cauchy so that
¯
u
(
x
0
) := lim
n
→∞
u
(
x
n
)
exists. If
{
y
n
}
∞
n
=1
⊂
Ω
is another sequence converging to
x
0
,
then

u
(
x
n
)
−
u
(
y
n
)

≤
[
u
]
β

x
n
−
y
n

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