3Function spaces
III Analysis of Partial Differential Equations
3.2 Sobolev spaces
The properties of H¨older spaces are not difficult to understand, but on the other
hand they are not too useful. This is not too surprising, perhaps, because the
supremum norm only sees the maximum of the function, and ignores the rest.
In contrast, the
L
p
norm takes into account the values at all points. This gives
rise to the notion of Sobolev spaces.
Definition
(
L
p
space)
.
Let
U ⊆ R
n
be open, and suppose 1
≤ p ≤ ∞
. We
define the space L
p
(U) by
L
p
(U) = {u : U → R measurable | kuk
L
p
(U)
< ∞}/{equality a.e.}.
where, if p < ∞, we define
kuk
L
p
(U)
=
Z
U
|u(x)|
p
dx
1/p
,
and
kuk
L
∞
(U)
= inf{C ≥ 0 | |u(x)| ≤ C almost everywhere}.
Theorem. L
P
(U) is a Banach space with the L
p
norm.
We can also define local versions of
L
p
spaces by saying
u ∈ L
p
loc
(
U
) if
u ∈ L
p
(
V
) for every
V b U
, i.e.
¯
V ⊆ U
and
¯
V
is compact. This is read as “
V
is compactly contained in
U
”. By working with
L
p
loc
(
U
), we ignore any possible
blowing up at the boundary. Note that
L
p
loc
(
U
) is not Banach, but is a Fr´echet
space.
What we want to do is to define differentiability for these things. If we try to
define them via limits, then we run into difficulties since the value of an element
in
L
p
(
U
) at a point is not well-defined. To proceed, we use the notion of a weak
derivative.
Definition
(Weak derivative)
.
Suppose
u, v ∈ L
1
loc
(
U
) and
α
is a multi-index.
We say that v is the αth weak derivative of u if
Z
U
uD
α
φ dx = (−1)
|α|
Z
U
vφ dx
for all
φ ∈ C
∞
c
(
U
), i.e. for all smooth, compactly supported function on
U
. We
write v = D
α
u.
Note that if
u
is a genuine smooth function, then D
α
u
is the
α
th weak
derivative of u, as integration by parts tells us.
For those who have seen distributions, this is the same as the definition of a
distributional derivative, except here we require that the derivative is an
L
1
loc
function.
Lemma.
Suppose
v, ˜v ∈ L
1
loc
(
U
) are both
α
th weak derivatives of
u ∈ L
1
loc
(
U
),
then v = ˜v almost everywhere.
Proof. For any φ ∈ C
∞
c
(U), we have
Z
U
(v − ˜v)φ dx = (−1)
|α|
Z
U
(u − u)D
α
φ dx = 0.
Therefore v − ˜v = 0 almost everywhere.
Now that we have weak derivatives, we can define the Sobolev spaces.
Definition
(Sobolev space)
.
We say that
u ∈ L
1
loc
(
U
) belongs to the Sobolev
space W
k,p
(U) if u ∈ L
p
(U) and D
α
u exists and is in L
p
(U) for all |α| ≤ k.
If p = 2, we write H
k
(U) = W
k,2
(U), which will be a Hilbert space.
If p < ∞, we define the W
k,p
(U) norm by
kuk
W
k,p
(U)
=
X
|α|≤k
Z
U
|D
α
u|
p
dx
1/p
.
If p = ∞, we define
kuk
W
k,∞
(U)
=
X
|α|≤k
kD
α
uk
L
∞
(U)
.
We denote by
W
k,p
0
(
U
) the completion of
C
∞
c
(
U
) in this norm (and again
H
k
0
(U) = W
k,2
0
(U)).
To see that these things are somehow interesting, it would be nice to find
some functions that belong to these spaces but not the C
k
spaces.
Example. Let u = B
1
(0) be the unit ball in R
n
, and set
u(x) = |x|
−α
when x ∈ U, x 6= 0. Then for x 6= 0, we have
D
i
u =
−αx
i
|x|
α+1
.
By considering
φ ∈ C
∞
c
(
B
1
(0)
\{
0
}
), it is clear that if
u
is weakly differentiable,
then it must be given by
D
i
u =
−αx
i
|a|
α+1
(∗)
We can check that u ∈ L
1
loc
(U) iff α < n, and
x
i
|x|
α+1
∈ L
1
loc
(U) if α < n − 1.
So if we want
u ∈ W
1,p
(
U
), then we must take
α < n −
1. To check (
∗
) is
indeed the weak derivative, suppose
φ ∈ C
∞
c
(
U
). Then integrating by parts, we
get
−
Z
U−B
ε
(0)
uφ
x
i
dx =
Z
U−B
ε
(0)
D
i
uφ dx −
Z
∂B
ε
(0)
uφν
i
dS,
where ν = (ν
1
, . . . , ν
n
) is the inwards normal. We can estimate
Z
∂B
ε
(0)
uφν
i
dS
≤ kφk
L
∞
· ε
−α
· Cε
n−1
≤
˜
Cε
n−1−α
→ 0 as ε → 0
for some constants
C
and
˜
C
. So the second term vanishes. So by, say, dominated
convergence, it follows that (∗) is indeed the weak derivative.
Finally, note that D
i
u ∈ L
p
(
U
) iff
p
(
α
+ 1)
< n
. Thus, if
α <
n−p
p
, then
u ∈ W
1,p
(
U
). Note that if
p > n
, then the condition becomes
α <
0, and
u
is
continuous.
Note also that if α >
n
p
, then u 6∈ W
1,p
(U).
Theorem.
For each
k
= 0
,
1
, . . .
and 1
≤ p ≤ ∞
, the space
W
k,p
(
U
) is a Banach
space.
Proof.
Homogeneity and positivity for the Sobolev norm are clear. The triangle
inequality follows from the Minkowski inequality.
For completeness, note that
kD
α
uk
L
p
(U)
≤ kuk
W
k,p
(U)
for |α| ≤ k.
So if (
u
i
)
∞
i=1
is Cauchy in
W
k,p
(
U
), then (D
α
u
i
)
∞
i=1
is Cauchy in
L
p
(
U
) for
|α| ≤ k. So by completeness of L
p
(U), we have
D
α
u
i
→ u
α
∈ L
p
(U)
for some
u
α
. It remains to show that
u
α
= D
α
u
, where
u
=
u
(0,0,...,0)
. Let
φ ∈ C
∞
c
(U). Then we have
(−1)
|α|
Z
U
u
j
D
α
φ dx =
Z
U
D
α
u
j
φ dx
for all j. We send j → ∞. Then using D
α
u
j
→ u
α
in L
p
(U), we have
(−1)
|α|
Z
U
uD
α
φ dx =
Z
U
u
α
φ dx.
So D
α
u = u
α
∈ L
p
(U) and we are done.