3Function spaces
III Analysis of Partial Differential Equations
3.4 Extensions and traces
If
U ⊆ R
n
is open and bounded, then there is of course a restriction map
W
1,p
(
R
n
)
→ W
1,p
(
U
). It turns out under mild conditions, there is an extension
map going in the other direction as well.
Theorem
(Extension of
W
1.p
functions)
.
Suppose
U
is open, bounded and
∂U
is
C
1
. Pick a bounded
V
such that
U b V
. Then there exists a bounded linear
operator
E : W
1,p
(U) → W
1.p
(R
n
)
for 1 ≤ p < ∞ such that for any u ∈ W
1,p
(U),
(i) Eu = u almost everywhere in U
(ii) Eu has support in V
(iii) kEuk
W
1,p
(R
n
)
≤ Ckuk
W
1,p
(U)
, where the constant
C
depends on
U, V, p
but not u.
Proof.
First note that
C
1
(
¯
U
) is dense in
W
1,p
(
U
). So it suffices to show that
the above theorem holds with
W
1,p
(
U
) replaced with
C
1
(
¯
U
), and then extend
by continuity.
We first show that we can do this locally, and then glue them together using
partitions of unity.
Suppose
x
0
∈ ∂U
is such that
∂U
near
x
0
lies in the plane
{x
n
= 0
}
. In
other words, there exists r > 0 such that
B
+
= B
r
(x
0
) ∩ {x
n
≥ 0} ⊆
¯
U
B
−
= B
r
(x
0
) ∩ {x
n
≤ 0} ⊆ R
n
\ U.
The idea is that we want to reflect
u|
B
+
across the
x
n
= 0 boundary to get a
function on
B
−
, but the derivative will not be continuous if we do this. So we
define a “higher order reflection” by
¯u(x) =
(
u(x) x ∈ B
+
−3u(x
0
, −x
n
) + 4
ux
0
, −
x
n
2
x ∈ B
−
x
n
u
−x
−
x
2
x
We see that this is a continuous function. Moreover, by explicitly computing the
partial derivatives, we see that they are continuous across the boundary. So we
know ¯u ∈ C
1
(B
r
(x
0
)).
We can then easily check that we have
k¯uk
W
1,p
(B
r
(x
0
))
≤ Ckuk
W
1,p
(B
+
)
for some constant C.
If
∂U
is not necessarily flat near
x
0
∈ ∂U
, then we can use a
C
1
diffeomor-
phism to straighten it out. Indeed, we can pick
r >
0 and
γ ∈ C
1
(
R
n−1
) such
that
U ∩ B
r
(p) = {(x
0
, x
n
) ∈ B
r
(p) | x
n
> γ(x
0
)}.
We can then use the C
1
-diffeomorphism Φ : R
n
→ R
n
given by
Φ(x)
i
= x
i
i = 1, . . . , n − 1
Φ(x)
n
= x
n
− γ(x
1
, . . . , x
n
)
Then since
C
1
diffeomorphisms induce bounded isomorphisms between
W
1,p
,
this gives a local extension.
Since
∂U
is compact, we can take a finite number of points
x
0
i
∈ ∂W
, sets
W
i
and extensions u
i
∈ C
1
(W
i
) extending u such that
∂U ⊆
N
[
i=1
W
i
.
Further pick
W
0
b U
so that
U ⊆
S
N
i=0
W
i
. Let
{ζ
i
}
N
i=0
be a partition of unity
subordinate to {W
i
}. Write
¯u =
N
X
i=0
ζ
i
¯u
i
where ¯u
0
= u. Then ¯u ∈ C
1
(R
n
), ¯u = u on U , and we have
k¯uk
W
1,p
(R
n
)
≤ Ckuk
W
1,p
(U)
.
By multiplying ¯u by a cut-off, we may assume supp ¯u ⊆ V for some V c U.
Now notice that the whole construction is linear in
u
. So we have constructed
a bounded linear operator from a dense subset of
W
1,p
(
U
) to
W
1,p
(
V
), and there
is a unique extension to the whole of
W
1,p
(
U
) by the completeness of
W
1,p
(
V
).
We can see that the desired properties are preserved by this extension.
Trace theorems
A lot of the PDE problems we are interested in are boundary value problems,
namely we want to solve a PDE subject to the function taking some prescribed
values on the boundary. However, a function
u ∈ L
p
(
U
) is only defined up to
sets of measure zero, and
∂U
is typically a set of measure zero. So naively, we
can’t naively define
u|
∂U
. We would hope that if we require
u
to have more
regularity, then perhaps it now makes sense to define the value at the boundary.
This is true, and is given by the trace theorem
Theorem
(Trace theorem)
.
Assume
U
is bounded and has
C
1
boundary. Then
there exists a bounded linear operator
T
:
W
1,p
(
U
)
→ L
p
(
∂U
) for 1
≤ p < ∞
such that T u = u|
∂U
if u ∈ W
1,p
(U) ∩ C(
¯
U).
We say T u is the trace of u.
Proof.
It suffices to show that the restriction map defined on
C
∞
functions is a
bounded linear operator, and then we have a unique extension to
W
1,p
(
U
). The
gist of the argument is that Stokes’ theorem allows us to express the integral of
a function over the boundary as an integral over the whole of
U
. In fact, the
proof is indeed just the proof of Stokes’ theorem.
By a general partition of unity argument, it suffices to show this in the case
where U = {x
n
> 0} and u ∈ C
∞
¯
U with supp u ⊆ B
R
(0) ∩
¯
U. Then
Z
R
n−1
|u(x
0
, 0)|
p
dx
0
=
Z
R
n−1
Z
∞
0
∂
∂x
n
|u(x
0
, x
n
)|
p
dx
n
dx
0
=
Z
U
p|u|
p−1
u
x
n
sgn u dx
n
dx
0
.
We estimate this using Young’s inequality to get
Z
R
n−1
|u(x
0
, 0)|
p
dx
0
≤ C
p
Z
U
|u|
p
+ |u
x
n
|
p
dU ≤ C
p
kuk
p
W
1,p
(U)
.
So we are done.
We can apply this to each derivative to define trace maps
W
k,p
(
U
)
→
W
k−1,p
(U).
In general, this trace map is not surjective. So in some sense, we don’t
actually need to use up a whole unit of differentiability. In the example sheet,
we see that in the case p = 2, we only lose “half” a derivative.
Note that
C
∞
c
(
U
) is dense in
W
1,p
0
(
U
), and the trace vanishes on
C
∞
c
(
U
).
So
T
vanishes on
W
1,p
0
(
U
). In fact, the converse is true — if
T u
= 0, then
u ∈ W
1,p
0
(U).