Injections, Embeddings and the Trace Theorem
We want to derive a number of properties for the spaces H k U for U a bounded open set in
R n . It turns out that the most efficient way to do this is to consider first the much simpler
case where U = R n .
1. Fourier Characterization of the Hilbert-Sobolev Spaces
For φ ∈ C ∞c R n  let the Fourier transform be defined by,
T F φ = 2π −n/2 ∫ n φxe −ix⋅z dx = φ̂ z
R
We have showed previously that the Fourier transform is an isometric isomorphism on
L 2 = C ∞c R n  ∩ L 2 R n . In particular, we have
a
b
∀u ∈ L 2
|| u|| 0 = || û|| 0
Parseval relation
u, v 0 = û, v̂  0
∀u, v ∈ L 2 ,
Plancherel relation
The results (a) and (b) hold for every test function, and since the test functions are dense in
L 2 R n , they extend to L 2 R n  by continuity. Then the Fourier transform can be extended to
L 2 R n  by continuity as well. In addition,
i if u, D α u ∈ L 2
then
ii if u, x α ux ∈ L 2
T F D α u = iz α ûz ∈ L 2
T F x α ux = iD z  α ûz ∈ L 2 .
then
i.e.,
T F D α u = ∫ n D α ux e −ix⋅z dx̂ = −1 |α| ∫ n ux D αx e −ix⋅z dx̂
R
R
= −1 |α| ∫ n ux −iz α e −ix⋅z dx̂ = iz α ûz;
R
T F x α ux = i α ∫ n ux −ix α e −ix⋅z dx̂ = i α ∫ n ux D αz  e −ix⋅z  dx̂ = iD z  α ûz.
R
R
The result (i) asserts that when u is smooth, û decays rapidly at infinity and result (ii) asserts
the converse. This suggested the definition,
for s ≥ 0,
Hs =
s/2
u ∈ L 2 : 1 + | z| 2  ûz ∈ L 2
with
||u|| 2s = ∫R n 1 + | z| 2  ûz 2 dz,
and
u, v s = ∫R n 1 + | z| 2  ûz v̂ z dz.
s
s
Then
H 0 = L 2 , and s ≥ t ≥ 0 implies H s ⊂ H t ⊂ H 0 .
To see how partial differential operators act on these spaces, let
P m z = ∑ |α|≤m c α z α
and define
Then
PD x ux = T −1
F P m i z ûz
for u ∈ H m R n  ⊂ H 0 = L 2 .
PD x ux = ∫ n e ix⋅z ∑ |α|≤m c α iz α ûz dẑ
R
1
= ∑ |α|≤m c α ∫ n e ix⋅z iz α ûz dẑ
R
= ∑ |α|≤m c α D αx ux
where the last step follows from i. Then PD x  is a differential operator of order m. For
s > m we have
||PD x ux|| s−m = ∫R n 1 + | z| 2 
= ∫ n 1 + | z| 2 
R
s−m
s
P m i z 2 ûz 2 dz,
P m i z 2
m
1 + | z| 2 
ûz 2 dz ≤ C m ||u|| 2s ,
where
C m = max z
|P m i z| 2
m < ∞.
1 + | z| 2 
This implies that any differential operator of order m is a bounded linear mapping from
H s into H s−m for s > m.
Recall that the definition of the norm in H m R n  = u ∈ L 2 : D α u ∈ L 2 , |α| ≤ m was
defined as
|| u|| 2m = ∑ |α|≤m ||D α u|| 20
But
||D α u|| 20 = || z α ûz|| 20 = ∫R n | z| 2α ûz 2 dz
hence
|| u|| 2m = ∑ |α|≤m ∫R n | z| 2α ûz 2 dz
= ∫ n ∑ |α|≤m
R
m
| z| 2α
1 + | z| 2  ûz 2 dz ≤ C m || u|| 2m
2 m
1 + | z| 
where
1 + | z| 2  ≤ ∑ |α|≤m | z| 2α ≤ C m 1 + | z| 2 
m
m
This shows that the Fourier norm is equivalent to the original Sobolev norm on H m , and the
two definitions of the space are therefore also equivalent. We will now prove the first of the
important embedding results for these spaces.
2. Inclusions and Embeddings
We prove first a result that says that if u ∈ H s for s sufficiently large, then u is equivalent
(i.e., equal almost everywhere to) a function that is continuous.
Theorem 2.1 (Sobolev embedding theorem, primitive version) If s > m +
continuously injected into C m
n
2
then H s is
−s/2
| v̂ z| dz
Proof- We will first prove the result for m = 0. If u ∈ H s then
s
v̂ z = 1 + | z| 2  ûz ∈ H 0 = L 2 . Then note that
|| ûz|| L 1 = ∫R n | ûz| dz = ∫R n 1 + | z| 2 
−s/2
1 + | z| 2  | ûz| dz = ∫R n 1 + | z| 2 
s/2
2
and
|| ûz|| L 1 ≤
1 + | z| 2 
Now
and
∞
1 + | z| 2 
−s/2
2
0
−s/2
0
|| v̂ || 0 .
∞
−s
−s
= ∫ n 1 + | z| 2  dz = ∫ ∫ 1 + r 2  r n−1 dr dω,
Ω 0
R
∞
2 −s n−1
∫ 0 1 + r  r dr ~ ∫ 1 r n−1−2s dr ~ r n−2s | ∞1 .
Then the integral is finite if n − 2s < 0; i. e. , if s > n2 . We have proven that u ∈ H s , s > n2
implies û ∈ L 1 . But û ∈ L 1 implies, in turn, that ũ = T −1
F û is in the space, C 0 , of continuous
functions which tend to zero at infinity. That is to say, after possibly modifying u on a set of
measure zero, it is a continuous function.
For m > 0, and u ∈ H s for s > m + n2 , we have, from the result about differential
operators of order m, that D α u ∈ H s−m for | α| ≤ m, and hence, if s − m > n2 then
D α u ∈ C 0 for | α| ≤ m. ■
We interpret this result to mean that for s > m + n2 , every u ∈ H s can be identified with
a unique function ũ ∈ C m and that this injection is continuous. More precisely, if u ∈ H s then
there exists a ũ ∈ C m such that ||u − ũ|| H s = 0 and ||ũ|| C m ≤ ||u|| H s .
The next result we are going to prove has to do with the embedding of H 10 U into
H U. For U a bounded open set in R n , let u m  ∈ H 10 U be such that || u m || 1 ≤ M, ∀m.
Since H 10 U is a Hilbert space, it follows from a classical property of Hilbert spaces that
u m  must contain a weakly convergent subsequence. Moreover, since H 10 U is
continuously injected into H 0 U, the subsequence must also converge weakly in
H 0 U. The somewhat surprising fact is that the subsequence is actually strongly
convergent in H 0 U; i.e.,, the continuous injection of H 10 U into H 0 U is a compact
embedding.
To prove the compact embedding result, we first define, for u ∈ H 10 U,
0
Zux =
ux if x ∈ U
0
extension by zero
if x ∉ U
Since the support of u is compactly contained in U, it follows that
|| Zu|| H 1 R n  = || u|| H 1 U
and
Zu ∈ H 1 R n .
More generally,
u ∈ H m0 U = completion of C ∞c U in the norm ||⋅|| m implies
Zu ∈ H s
s ≤ m.
For u ∈ H m U, it is not generally the case that the extension by zero, Zu, belongs to
H m R n  since extending by zero introduces a sharp change in u at the boundary of the set
U. For example, ux = 1, x ∈ 0, 1 belongs to H 1 0, 1 but Zux is just the indicator
function of 0, 1. Since this is a discontinuous function, it is not in H 1 R.
Theorem 2.2 (Rellich’s lemma, primitive version) Suppose U is open and bounded in
R n . Then the natural injection
i : H 10 U  H 0 U
is a compact mapping.
i.e., every sequence that is bounded in H 10 U contains a subsequence that is strongly
3
convergent in H 0 U.
Proof- Suppose u m  ∈ H 10 U is such that || u m || 1 ≤ M, ∀m. Since u m  ∈ H 10 U it is
evident that Zu m  ∈ H 1 ⊂ L 2 and Zu m has a Fourier transform, û m ∈ L 2 . Then
ix⋅z
Zu m x = T −1
dz = ∫
F û m  = ∫ n û m z e
R
|z|<A
û m z e ix⋅z dz + ∫
|z|>A
û m z e ix⋅z dz,
−1
= T −1
F û m zI A z + T F û m z1 − I A z.
Now û m z I A z has compact support which implies that T −1
F û m z I A z is smooth.
Moreover, we will use the hypothesis || u m || 1 ≤ M, ∀m, to show that the second term here
can be made arbitrarily small by choosing A large. In fact,
||T −1
F û m z1 − I A z|| 0 = ∫
2
|z|>A
û m z 2 dz = ∫
−1
|z|>A
1 + | z| 2  1 + | z| 2 û m z 2 dz,
2
−1
≤ max |z|>A 1 + | z| 2  || u m || 21 ≤ M2 , ∀m.
A
−1
Then, for any  > 0, choose A > M
 , to obtain ||T F û m z1 − I A z|| 0 ≤ .
Now the smooth functions v m x = T −1
F û m zI A z can be shown to be a uniformly
bounded and equicontinuous family of continuous functions. Postponing the proof of this
fact for the moment, we note that this implies the existence (via Ascoli-Arzela) of a
subsequence v m ′ x which converges uniformly on U to a limit v ∈ CŪ. Since uniform
convergence on U implies convergence in L 2 U, it follows that this subsequence converges
strongly to v in L 2 U. In particular then, it must be true of the subsequence that
for m ′ , k ′ > N 
|| v m ′ − v k ′ || H 0 U < 
and
|| u m ′ − u k ′ || H 0 U ≤ || v m ′ − v k ′ || 0 + ||T −1
F û m ′ z − û k ′ z1 − I A z|| 0
for m ′ , k ′ > N  .
≤  + 2
Repeat this argument now for the subsequence v m ′ = T −1
F û m ′ zI A z replacing  by /2.
Iterate the procedure in order to generate a sequence of subsequences to which we can
apply a diagonal procedure to get, finally, a subsequence u m"  of u m  that is Cauchy in
H 0 U.
To see that v m x = T −1
F û m zI A z is uniformly bounded and equicontinuous, note
first that for any x and every m,
| v m x| = ∫
|z|<A
û m z e ix⋅z dz ≤ ∫
|z|<A
| û m z | dz ≤ C A || û m || 0 = C M || u m || 0 ≤ C M ||u m || 1 ≤ M. C M .
Thus the family is uniformly bounded. Next,
| v m x − v m y| = ∫
|z|<A
û m z e ix⋅z − e iy⋅z dz ≤ ∫
|z|<A
| û m z e ix⋅z 1 − e iy−x⋅z | dz
≤ sup |z|≤A |1 − e iy−x⋅z | C A M → 0 as y → x uniformly in m.
Since
sup |z|≤A |1 − e iy−x⋅z | → 0 as y → x uniformly in m, it follows that the family is
equicontinuous.■
4
Corollary- For U open and bounded in R n , and m > 1, H m0 U is compactly embedded in
H m−1 U.
Proof- Suppose u m  ⊂ H m0 U with || u m || m ≤ M ∀m. Then for each α, |α| ≤ m − 1, we
have || D α u m || 1 ≤ M and it follows from the theorem that there is a subsequence
u m ′  ⊂ u m  such that D α u m ′  is Cauchy in H 0 U. Then a diagonal argument implies the
existence of a single subsequence u m"  ⊂ u m  such that D α u m"  is Cauchy in H 0 U for
every α, |α| ≤ m − 1. Then this subsequence is Cauchy in H m−1 U. ■
We will think of this result in the following way. Any sequence
u m  ⊂ H m0 U with || u m || m ≤ M ∀m contains a weakly convergent (in H m0 U)
subsequence just by virtue of the fact that H m0 U is a Hilbert space. Then we use the
embedding lemma to conclude that this subsequence is, in fact, strongly convergent in
H m−1 U. So, a bounded sequence in H m0 U contains a subsequence that is strongly
convergent in H m−1 U.
3. The Trace Theorem
Since functions in H 0 U, are, in fact, equivalence classes of functions they are defined only
up to sets of measure zero. Then restriction of such functions to the boundary of U (a set of
measure zero) has no meaning. On the other hand, for u ∈ H m U for m sufficiently large
relative to n, it will be possible to give meaning to the restrictions to ∂U not just for the
functions, but for some derivatives of the functions as well. We will begin with the simplest
possible case of a set with a boundary, the set
In this case,
U = R n+ = x ′ , x n  : x ′ ∈ R n−1 , 0 < x n < ∞ = R n−1 × R + .
∂U = R n−1 = x ′ , x n  : x ′ ∈ R n−1 , x n = 0 .
For u ∈ C ∞ R n+  define
T j ux ′  = lim x n →0+ ∂ jn ux ′ , x n ,
0 ≤ j ≤ m − 1.
We will show that this operator, called the trace operator of order j, can be extended to
H m R n+ .
Theorem 3.1 (Trace theorem, primitive version)
1.
2.
3.
||T j u|| H m−j−1/2 R n−1 ≤ C || u|| H m R n+  ∀u ∈ C ∞ R n+ 
linear mapping
T j : H m R n+   H m−j−1/2 R n−1 
hence T j extends as a bounded
T j maps H m R n+  onto H m−j−1/2 R n−1 
T j u = 0 if and only if u ∈ H m0 R n+ 
idea of the proof- In order to use the Fourier transform techniques, the functions must
belong to H m R n  rather than H m R n+ . Thus for u ∈ H m R n+  we define
Zux =
ux ′ , x n  if
xn > 0
0
xn < 0
if
.
5
In general, u ∈ H m R n+  does not imply Zu ∈ H m R n . Thus we define an operator,
E : H m R n+  → H m R n ,
ux ′ , x n 
Eux =
if
xn > 0
∑ a k ux ′ , −kx n  if
xn < 0
m
k=1
Here the a ′k s are required to satisfy
m
∑ a k −k j = 1 for j = 0, 1, . . . , m − 1.
k=1
This (unique) choice of the a ′k s ensures
lim x n →0− ∂ jn Eux ′ , x n  = lim x n →0+ ∂ jn ux ′ , x n 
for j = 0, 1, . . . , m − 1.
Then u ∈ H m R n+  implies Eu ∈ H m R n  and, moreover
|| Eu|| H m R n  ≤ C 0 || u|| H m R n+ 
For future reference we also define
ux ′ , x n 
E 1 ux ′ , x n  =
if
′
ax n  Eux , x n  if
xn > 0
xn < 0
where
ax n  ∈ C ∞ R 1 ,
ax =
1 if x x > 0
0 if x n < −1
= a smooth cutoff function
Then E 1 u ∈ H m R n−1  for all u ∈ H m R n+  where R n−1 = x ′ , x n  : x ′ ∈ R n−1 , − 1 < x n < ∞ .
This modified extension operator smoothly extends the function u ∈ H m R n+  to a
neighborhood of the boundary of R n+ .
For v ∈ H m R n  let
v̂ z = v̂ z ′ , z n  = ∫ n e −ix⋅z vx dx̂
R
=∫
=∫
∫ e −ix ′ ⋅z ′ e −ix n ⋅z n vx ′ , x n  dx̂ ′ dx̂ n
R n−1 R
′
R n−1
′
e −ix ⋅z v̂ x ′ , z n  dx̂ ′ .
where
v̂ x ′ , z n  denotes the Fourier transform of vx ′ , x n  with respect to the variable
x n only; i.e., v̂ x ′ , z n  := T F vx ′ , x n  n
This splitting off of the x n to z n part of the Fourier transform now allows us to define the
action of the mapping T j in a convenient way for purposes of telling to which space
H s R n−1  the function T j v belongs. We write,
vx ′ , 0 = ∫ e −ix n ⋅z n | x n =0 v̂ x ′ , z n  dẑ n = ∫ v̂ x ′ , z n  dẑ n
R
R
and
T j vx ′  = ∂ jn vx ′ , 0 = ∫ iz n  j v̂ x ′ , z n  dẑ n ,
R
j = 0, 1, . . . , m − 1.
6
This leads to
||T j v|| 2H s
and since
R n−1
=∫
R n−1
1 + | z′ |2
s
T F T j vz ′  2n−1 dẑ ′ ,
T F T j vz ′  n−1 = transform in x ′ only
T F T j vz ′  n−1 = T F ∫ iz n  j v̂ x ′ , z n  dẑ n
R
n−1
= ∫ iz n  j T F v̂ x ′ , z n  n−1 dẑ n ,
R
we have
||T j v|| 2H s
R n−1
=∫
1 + | z′ |2
R n−1
s
2
∫R iz n  j v̂ z
dẑ n dẑ ′
Then, using several clever tricks, we can show
||T j v|| 2H s
R n−1
≤ C 1 || v|| H2 m R n  for v ∈ H m R n , s ≤ m − j − 1/2.
and
E : H m R n+  → H m R n 
T j : H m R n  → H m−j−1/2 R n−1 
so the composition
T j ∘ E : H m R n+  → H m−j−1/2 R n−1  is linear and bounded.
To show that T j ∘ E is onto we will construct a continuous right inverse for T j . Define,
for v ∈ H s R n−1 
−x n
Kvx ′ , x n  = T −1
F e
Then
1+ z ′
2
1/2
v̂ z ′ 
=∫
e ix ⋅z e −x n
n−1
′
R
′
1+ z ′
2
1/2
v̂ z ′ dẑ ′ .
Kvx ′ , 0 = vx ′  which implies T 0 Kv = v ∀v ∈ H s R n−1 
Note that this definition of K implies that Kvx satisfies
1 − ∇ 2 Kvx = 0 in R n+
T 0 Kv = v on ∂R n+
Then for every v ∈ H s R n−1  this boundary value problem is uniquely solvable; i.e., there
exists a u = Kv and we can show that
|| Kv|| H m R n+  ≤ C || v|| H m−1/2
R n−1
This is the beginning of the idea of the proof that each T j has a continuous right inverse;
i.e., T j ∘ K = id, and T j maps H m R n+  onto H m−j−1/2 R n−1  continuously. The choice of the
operator K is not unique.
Finally, it is easy to see that T j u = 0 for all u ∈ H m R n+ . That the converse is also true
is more difficult to show and we will omit this proof.■
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