3Function spaces
III Analysis of Partial Differential Equations
3.5 Sobolev inequalities
Before we can move on to PDE’s, we have to prove some Sobolev inequalities.
These are inequalities that compare different norms, and allows us to “trade”
different desirable properties. One particularly important thing we can do is
to trade differentiability for continuity. So we will know that if
u ∈ W
k,p
(
U
)
for some large
k
, then in fact
u ∈ C
m
(
U
) for some (small)
m
. The utility of
these results is that we would like to construct our solutions in
W
k,p
spaces,
since these are easier to work with, but ultimately, we want an actual, smooth
solution to our equation. Sobolev inequalities let us do so, since if
u ∈ W
k,p
(
U
)
for all k, then it must be in C
m
as well.
To see why we should be expected to be able to do that, consider the space
H
1
0
([0
,
1]). A priori, if
u ∈ H
1
0
([0
,
1]), then we only know it exists as some
measurable function, and there is no canonical representative of this function.
However, we can simply assign
u(x) =
Z
x
0
u
0
(t) dt,
since we know
u
0
is an honest integrable function. This gives a well-defined
representative of the function
u
, and even better, we can bound its supremum
using ku
0
k
L
2
([0,1])
.
Before we start proving our Sobolev inequalities, we first prove the following
lemma:
Lemma. Let n ≥ 2 and f
1
, . . . , f
n
∈ L
n−1
(R
n−1
). For 1 ≤ i ≤ n, denote
˜x
i
= (x
1
, . . . , x
i−1
, x
i+1
, . . . , x
n
),
and set
f(x) = f
1
(˜x
1
) ···f
n
(˜x
n
).
Then f ∈ L
1
(R
n
) with
kfk
L
1
(R
n
) ≤
n
Y
i=1
kf
i
k
L
n−1
(R
n−1
)
.
Proof. We proceed by induction on n.
If n = 2, then this is easy, since
f(x
1
, x
2
) = f
1
(x
2
)f
2
(x
1
).
So
Z
R
2
|f(x
1
, x
2
)| dx =
Z
|f
1
(x
2
)| dx
2
Z
|f
2
(x
1
)| dx
1
= kf
1
k
L
1
(R
1
)
kf
2
k
L
1
(R
1
)
.
Suppose that the result is true for n ≥ 2, and consider the n + 1 case. Write
f(x) = f
n+1
(˜x
n+1
)F (x),
where F (x) = f
1
(˜x
1
) ···f
n
(˜x
n
). Then by H¨older’s inequality, we have
Z
x
1
,...,x
n
|f( ·, x
n+1
)| dx ≤ kf
n+1
k
L
n
(R
n
)
kF ( ·, x
n+1
)k
L
n/(n−1)
(R
n
)
.
We now use the induction hypothesis to
f
n/(n−1)
1
( ·, x
n+1
)f
n/(n−1)
2
( ·, x
n+1
) ···f
n/(n−1)
n
( ·, x
n+1
).
So
Z
x
1
,...,x
n
|f( ·, x
n+1
)| dx ≤ kf
n+1
k
L
n
(R
n
)
n
Y
i=1
kf
n
n−1
i
( ·, x
n
)k
L
n−1
(R
n−1
)
!
n−1
n
= kf
n+1
k
L
n
(R
n
)
n
Y
i=1
kf
i
( ·, x
m
)k
L
n
(R
n−1
)
.
Now integrate over x
n+1
. We get
kfk
L
1
(R
n+1
)
≤ kf
n+1
k
L
n
(R
n
)
Z
x
n+1
n
Y
i=1
kf
i
( ·, x
n+1
)k
L
n
(R
n−1
)
dx
n
.
≤ kf
n+1
k
L
n
(R
n+1
)
n
Y
i=1
Z
x
n+1
kf
i
( ·, x
n+1
)k
n
L
n
(R
n−1
)
dx
n+1
!
1/n
= kf
n+1
k
L
n
(R
n
)
n
Y
i=1
kf
i
k
L
n
(R
n
)
.
Theorem
(Gagliardo–Nirenberg–Sobolev inequality)
.
Assume
n > p
. Then we
have
W
1,p
(R
n
) ⊆ L
p
∗
(R
n
),
where
p
∗
=
np
n −p
> p,
and there exists c > 0 depending on n, p such that
kuk
L
p
∗
(R
n
)
≤ ckuk
W
1,p
(R
n
)
.
In other words, W
1,p
(R
n
) is continuously embedded in L
p
∗
(R
n
).
Proof.
Assume
u ∈ C
∞
c
(
R
n
), and consider
p
= 1. Since the support is compact,
u(x) =
Z
x
i
−∞
u
x
i
(x
1
, . . . , x
i−1
, y
i
, x
i+1
, . . . , x
n
) dy
i
.
So we know that
|u(x)| ≤
Z
∞
−∞
|Du(x
1
, . . . , x
i−1
, y
i
, x
i+1
, . . . , x
n
)| dy
i
≡ f
i
(˜x
i
).
Thus, applying this once in each direction, we obtain
|u(x)|
n/(n−1)
≤
n
Y
i=1
f
i
(˜x
i
)
1/(n−1)
.
If we integrate and then use the lemma, we see that
kuk
L
n/(n−1)
(R
n
)
n/(n−1)
≤ C
n
Y
i=1
kf
1/(n−1)
i
k
L
n−1
(R
n−1
)
= kDuk
n/(n−1)
L
1
(R
n
)
.
So
kuk
L
n/(n−1)
(R
n
)
≤ CkDuk
L
1
(R
n
)
.
Since C
∞
c
(R
n
) is dense in W
1,1
(R
n
), the result for p = 1 follows.
Now suppose p > 1. We apply the p = 1 case to
v = |u|
γ
for some γ > 1, which we choose later. Then we have
Dv = γ sgn u · |u|
γ−1
Du.
So
Z
R
n
|u|
γn
n−1
dx
n−1
n
≤ γ
Z
R
n
|u|
γ−1
|Du| dx
≤ γ
Z
R
n
|u|
(γ−1)
p
p−1
dx
p−1
p
Z
R
n
|Du|
p
dx
1
p
.
We choose γ such that
γn
n −1
=
(γ − 1)p
p −1
.
So we should pick
γ =
p(n −1)
n −p
> 1.
Then we have
γn
n −1
=
np
n −p
= p
∗
.
So
Z
R
n
|u|
p
∗
dx
n−1
n
≤
p(n −1)
n −p
Z
R
n
|u|
p
∗
dx
p−1
p
kDuk
L
p
(R
n
)
.
So
Z
R
n
|u|
p
∗
dx
1/p
∗
≤
p(n −1)
n −p
kDuk
L
p
(R
n
)
.
This argument is valid for
u ∈ C
∞
c
(
R
n
), and by approximation, we can extend
to W
1,p
(R
n
).
We can deduce some corollaries of this result:
Corollary.
Suppose
U ⊆ R
n
is open and bounded with
C
1
-boundary, and
1 ≤ p < n. Then if p
∗
=
np
n−p
, we have
W
1,p
(U) ⊆ L
p
∗
(U),
and there exists C = C(U, p, n) such that
kuk
L
p
∗
(U)
≤ Ckuk
W
1,p
(U)
.
Proof.
By the extension theorem, we can find
¯u ∈ W
1,p
(
R
n
) with
¯u
=
u
almost
everywhere on U and
k¯uk
W
1,p
(R
n
)
≤ Ckuk
W
1,p
(U)
.
Then we have
kuk
L
p
∗
(U)
≤ k¯uk
L
p
∗
(R
n
)
≤ ck¯uk
W
1,p
(R
n
)
≤
˜
Ckuk
W
1,p
(U)
.
Corollary.
Suppose
U
is open and bounded, and suppose
u ∈ W
1,p
0
(
U
). For
some 1 ≤ p < n, then we have the estimates
kuk
L
q
(U)
≤ CkDuk
L
p
(U)
for any q ∈ [1, p
∗
]. In particular,
kuk
L
p
(U)
≤ CkDuk
L
p
(U)
.
Proof.
Since
u ∈ W
1,p
0
(
U
), there exists
u
0
∈ C
∞
c
(
U
) converging to
u
in
W
1,p
(
U
).
Extending u
m
to vanish on U
c
, we have
u
m
∈ C
∞
c
(R
n
).
Applying Gagliardo–Nirenberg–Sobolev, we find that
ku
m
k
L
p
∗
(R
n
)
≤ CkDu
m
k
L
p
(R
n
)
.
So we know that
ku
m
k
L
p
∗
(U)
≤ CkDu
m
k
L
p
(U)
.
Sending m → ∞, we obtain
kuk
L
p
∗
(U)
≤ CkDuk
L
p
(U)
.
Since U is bounded, by H¨older, we have
Z
U
|u|
q
dx
1/q
≤
Z
U
1 dx
1/rq
Z
U
|u|
qs
ds
1/sq
≤ Ckuk
L
p
∗
(U)
provided
q ≤ p
∗
, where we choose
s
such that
qs
=
p
∗
, and
r
such that
1
r
+
1
s
= 1.
The previous results were about the case
n > p
. If
n < p < ∞
, then we
might hope that if u ∈ W
1,p
(R
n
), then u is “better than L
∞
”.
Theorem
(Morrey’s inequality)
.
Suppose
n < p < ∞
. Then there exists a
constant C depending only on p and n such that
kuk
C
0,γ
(R
n
)
≤ Ckuk
W
1,p
(R
n
)
for all u ∈ C
∞
c
(R
n
) where C = C(p, n) and γ = 1 −
n
p
< 1.
Proof. We first prove the H¨older part of the estimate.
Let Q be an open cube of side length r > 0 and containing 0. Define
¯u =
1
|Q|
Z
Q
u(x) dx.
Then
|¯u −u(0)| =
1
|Q|
Z
Q
[u(x) −u(0)] dx
≤
1
|Q|
Z
Q
|u(x) −u(0)| dx.
Note that
u(x) −u(0) =
Z
1
0
d
dt
u(tx) dt =
X
i
Z
1
0
x
i
∂u
∂x
i
(tx) dt.
So
|u(x) −u(0)| ≤ r
Z
1
0
X
i
∂u
∂x
i
(tx)
dt.
So we have
|¯u −u(0)| ≤
r
|Q|
Z
Q
Z
1
0
X
i
∂u
∂x
i
(tx)
dt dx
=
r
|Q|
Z
1
0
t
−n
Z
tQ
X
i
∂u
∂x
i
(y)
dy
!
dt
≤
r
|Q|
Z
1
0
t
−n
n
X
i=1
∂u
∂x
i
L
p
(tQ)
|tQ|
1/p
0
!
dt.
where
1
p
+
1
p
0
= 1.
Using that |Q| = r
n
, we obtain
|¯u −u(0)| ≤ cr
1−n+
n
p
0
kDuk
L
p
(R
n
)
Z
1
0
t
−n+
n
p
0
dt
≤
c
1 −n/p
r
1−n/p
kDuk
L
p
(R
n
)
.
Note that the right hand side is decreasing in
r
. So when we take
r
to be very
small, we see that u(0) is close to the average value of u around 0.
Indeed, suppose
x, y ∈ R
n
with
|x −y|
=
r
2
. Pick a box containing
x
and
y
of side length
r
. Applying the above result, shifted so that
x
,
y
play the role of
0, we can estimate
|u(x) −u(y)| ≤ |u(x) − ¯u| + |u(y) − ¯u| ≤
˜
Cr
1−n/p
kDuk
L
p
(R
n
)
.
Since r < kx − yk, it follows that
|u(x) −u(y)|
|x −y|
1−n/p
≤ C · 2
1−n/p
kDuk
L
p
(R
n
)
.
So we conclude that [u]
C
0,γ
(R
n
)
≤ CkDuk
L
p
(R
n
)
.
Finally, to see that
u
is bounded, any
x ∈ R
n
belongs to some cube
Q
of side
length 1. So we have
|u(x)| ≤ |u(x) − ¯u + ¯u| ≤ |¯u| + CkDuk
L
p
(R
n
)
.
But also
|¯u| ≤
Z
Q
|u(x)| dx ≤ kuk
L
p
(R
n
)
k1k
L
p
(Q)
= kuk
L
p
(R
n
)
.
So we are done.
Corollary.
Suppose
u ∈ W
1,p
(
U
) for
U
open, bounded with
C
1
boundary.
Then there exists
u
∗
∈ C
0,γ
(
U
) such that
u
=
u
∗
almost everywhere and
ku
∗
k
C
0,γ
(U)
≤ Ckuk
W
1,p
(U)
.
By applying these results iteratively, we can establish higher order versions
W
k,p
⊆ L
q
(U)
with some appropriate q.