LECTURE 8, 18.155, 4 OCTOBER, 2011
This week: Sobolev Embedding and Schwartz Structure Theorems.
We now know that the Fourier transform ‘extends’ to an isomorphism
of S 0 (Rn ). This was done by using the formula, for a pair of functions
in S(Rn )
Z
Z
φ̂ψ = φψ̂
to define the Fourier transform by
(1)
û(φ) = u(φ̂), u ∈ S 0 (Rn ), φ ∈ S(Rn )
since we know that S(Rn ) 3 φ 7−→ φ̂ ∈ S(Rn ) is continuous (in fact it
is a topological isomorphism).
Then we showed that the Fourier transform restricts from S 0 (Rn ) (on
which it is a bijection, but we have not given any continuity statement)
to a bijection on L2 (Rn ) which is essentially an isometric isomorphism,
i.e. (2π)−n/2 F is unitary on L2 (Rn ) – this follows from by the version
of (1)
Z
Z
−n
φψ = (2π)
φ̂ψ̂ ∀ φ, ψ ∈ S(Rn ).
Next we showed that the L2 based Sobolev spaces are readily characterized in terms of the Fourier transforms of their elements:
u ∈ H k (Rn ) ⇐⇒ hξik û ∈ L2 .
Here k ∈ N but we can use this as a definition of H k (Rn ) for all k ∈ R
where I usually replace the integer-looking k by s or m. Then it follows
that these spaces decrease with increasing order
0
H s (Rn ) ⊂ H s (Rn ) ⇐⇒ s ≥ s0
which may be unfortunate but that is the way things are.
From this we easily deduce a form of ‘weak=strong’ for derivatives.
Namely, u ∈ L2 (Rn ) is said to have a strong derivative, v = ∂j u if there
is a sequence un ∈ S(Rn ) (or Cc∞ (Rn ) it is the same in the end) such
that uk → u in L2 and ∂j uk → v in L2 (Rn ). Such a strong derivative
(of any order) is automatically a weak derivative since we can integrate
by parts before taking the limit so
u(−∂j φ) = lim uk (−∂j φ) = lim (∂j uk )(φ) = v(φ).
k→∞
k→∞
1
2
LECTURE 8, 18.155, 4 OCTOBER, 2011
In particular if a strong derivative exists then it is unique since weak
derivatives are already unique.
Proposition 1 (weak=strong). If k ∈ N then S(Rn ) ⊂ H k (Rn ) is
dense (in the norm on H k ) and each u ∈ H k (Rn ) has strong derivatives
or all orders |α| ≤ k.
Proof. Take u ∈ H k (Rn ), so û ∈ L2 (Rn ) and û(ξ) = hξi−k w, w ∈
L2 (Rn ) by the result above. Since S(Rn ) is dense in L2 we can choose
a sequence wl → w in L2 . Then choose ul ∈ S(Rn ) by ûl = hξi−k wk
where we use the fact that h·i−k is a multiplier on S and that the Fourier
transform is an isomorphism. The norm on u ∈ H k is equivalent to
kwkL2 when u and w are related as above, so wl → w in L2 implies
ul → u in H k (Rn ) and we have the density.
Since ξ α ûl = ξ α hξh−k wl , the same argument shows that ∂ α ul converges in L2 , so strong derivatives exist for |α| ≤ k.
Of course if s ≥ k then u ∈ H s (Rn ) has strong derivatives of any
order up to k – but what about the ‘extra’ s − k ‘derivatives’ ? We will
get to this later.
From this discussion the basic result relating ‘L2 -derivatives’ to ordinary derivatives also follows.
Theorem 1 (Sobolev embedding). If k ∈ N then
n
H s (Rn ) ⊂ C0k (Rn ) if s > + k.
2
This result is sharp. The spaces on the right are the spaces of functions
with k continuous, bounded derivatives each vanishing at infinity.
Proof. The main point here is that hξi−s L2 (Rn ) ⊂ L1 (Rn ) if (and only
if) s > n2 . By this I mean the space of those v ∈ S 0 (Rn ) such that
v = h·i−s w, w ∈ L2 . Indeed, this is just the fact that h·i−p ∈ L1 (Rn ) iff
p > n. So, if we apply the Cauchy-Schwarz inequality,
Z
Z
|v| = hξi−s |w| ≤ ki·i−2s kL2 kwkL2
the result follows. Well, we really need to check that the integral makes
sense first! This however is the usual argument by continuity – take
an approximating sequence wn → w with wn ∈ Cc0 (Rn ) for which the
computation does make sense, and hence deduce that vn = h·i−s wn → v
a.e. is an approximating sequence in L1 , so v ∈ L1 and the estimate
holds.
Now, we apply this to v = û, where by assumption u ∈ H s (Rn )
means that w(ξ) = hξis û ∈ L2 (Rn ). Thus, we see that v ∈ L1 and so,
LECTURE 8, 18.155, 4 OCTOBER, 2011
3
as you showed in the last problem set,
n
.
2
For higher derivatives, we just need to use density an iteration as
usual. Take a sequence as in the preceding result, ul → u in H s (Rn )
with ul ∈ S(Rn ). Then, ∂ α ul is a convergent sequence in H s−|α| (Rn ), so
if s − |α| > n2 it follows that ∂ α ul → ∂ α u ∈ C00 (Rn ) where the notation
is for the weak (or strong) derivative of u. That is, we conclude that
the weak derivatives up to order k are continuous. In fact we have been
through these arguments before, the fact that the sequence converges
with respect to the C k norm (supremum of derivatives up to order k)
means the limit is in C0k (Rn ) – one just needs to quote the completeness
of C0k (Rn ).
u = G û ∈ C00 (Rn ), if u ∈ H s (Rn ), s >
Now, we ‘know’ that the Fourier transform exchanges regularity and
growth. Let’s look at another (extreme) example of this.
Definition 1. An element u ∈ S 0 (Rn ) is said to be of compact support
(or have compact support, or be compactly supported) if there exists
R > 0 such that
u(φ) = 0 ∀ φ ∈ S(Rn ) s.t. φ(x) = 0 in |x| < R.
We will discuss the notion of the support of a distribution soon.
Another way of looking at this property is to say that there exists χ ∈
Cc∞ (Rn ) such that χu = u. Indeed, take χ ∈ Cc∞ (Rn ) to be identically
equal to 1 in say |x| < R + 1 and to vanish identically in |x| > R + 2.
Then if φ ∈ S(Rn ) it follows that (1−χ)φ = 0 in |x| < R and (1−χ)φ ∈
S(Rn ) so u((1 − χ)φ) = 0 for all φ ∈ S(Rn ) which is u(φ) = (χu)(φ)
for all φ ∈ S(Rn ). Conversely, if χu = u with χ ∈ Cc∞ (Rn ) it follows
χ(x) = 0 in |x| > R − 1 for some R and then if φ ∈ S(Rn ), and φ = 0
in |x| < R it follows that χφ ≡ 0 and hence χu(φ) = u(χφ) = 0 which
is the the definition.
One of the Assignment Questions is: Discuss the Paley-WienerSchwartz Theorem. Here is an extremely watered-down version (the
true version characterizes the Fourier transforms of compactly supported distributions in terms of entire functions of several complex
variables).
Lemma 1. If u ∈ S 0 (Rn ) has compact support then û ∈ C ∞ (Rn ) is a
function of slow growth.
Proof. The important properties of a distribution of compact support
are (duh!) it is a distribution and it has compact support. The first
4
LECTURE 8, 18.155, 4 OCTOBER, 2011
means that there are constants C and N such that
X
|u(φ)| ≤ CkφkN = C
sup |xα ∂ β φ|.
|α|+|β|≤N
The second means u = χu so u(φ) = u(χφ) and we see what happens
to the right side when φ is replaced by χφ :
X
sup |xα ∂ β (χφ)| ≤ C
sup |∂ β φ|
|β|≤N
|x|≤T
where T is such that χ = 0 in |x| ≥ T. So, for compactly supported
distributions we get a better estimate in terms of the C N seminorm on
a fixed set |x| ≤ T
|u(φ)| ≤ CkφkC N ({|x|≤T }) .
So, what is the Fourier transform of u, well, it is defined by
û(φ) = u(φ̂)
Let’s consider the function involved in the definition of the Fourier
transform, exp(ix · ξ) as a function of x. It is smooth, but of course not
in S(Rn ). However, if χ ∈ Cc∞ (Rn ) then χ(x) exp(ix · ξ) ∈ S(Rn ). So
we can define a function of ξ by
F (ξ) = u(χ(x) exp(ix · ξ) ∈ C ∞ (Rn ).
First, it is continuous, since as ξ varies, χ(x) exp(−ix · ξ) varies continuously in S(Rn ) and u is continuous, being a distribution. Moreover,
we can take the difference quotient and see that
h−1 (χ(x) exp(−ix · (ξ + hej )) − χ(x) exp(−ix · ξ)) → −ixχ exp(ix · ξ)
converges in S(Rn ). Iterating this argument gives the smoothness of
F (ξ).
Finally then, we claim that is that F (ξ) is the Fourier transform of
u. It has slow growth, since the norm estimate above shows |F (ξ)| ≤
ChξiN so it is a tempered distribution. Moreover, if we integrate
against χ(ξ)φ̂ where φ ∈ S(Rn ),
Z
F (ξ)φ(ξ)dξ = u(χφ̂) = u(φ̂) = û(φ)
showing that F (ξ) = û in S 0 (Rn ). The first step is the crucial one.
Think about computing u(χφ̂) in terms of φ. Let’s assume that φ itself has compact support. Then the integral for φ̂ is a Riemann integral
and so is the limit of a sum
Z
X
φ̂(x) = exp(−iξ · x)φ(ξ) = lim 2−nj
exp(−iξk,j · x)φ(ξk,j )
j→∞
k
LECTURE 8, 18.155, 4 OCTOBER, 2011
5
Here ξk,j is an ordering of the finite number of points in 2−j Zn in a big
ball containing the support of φ and the constant is the volume of the
corresponding cube(s).
As a function of x in a compact set this approximation to the Riemann integral actually converges uniformly with all its derivatives –
just differentiate to see this. So
X
χ(x)2−nj
exp(−iξk,j · x)φ(ξk,j ) → χ(x)φ̂(x) in S(Rn ).
k
By the continuity of u this implies
X
exp(−iξk,j · x)φ(ξk,j )
u(χφ̂) = lim u(χ(x)2−nj
j→∞
k
→ χ(x)φ̂(x) in S(Rn )
Z
X
=⇒ lim
F (ξkj )φ = F (ξ)φ(ξ)
j→∞
k
again by the properties of Riemann integrals.
So, for χ of compact support, û(φ) = F (φ) but both sides are tempered distributions so by density of Cc∞ (Rn ) in S(Rn ) û = F.
I said we would check that the extension of the Fourier transform
(etc) to S 0 (Rn ) was forced by continuity here is the first version. We
can say that a sequence uj ∈ S 0 (Rn ) converges weakly to u ∈ S 0 (Rn )
if uj (φ) → u(φ) in C for all φ ∈ S(Rn ). The limit is certainly unique
and we can use the Fourier transform on S(Rn ) to prove that S(Rn ) ⊂
S 0 (Rn ) is weakly dense.
I did not finish the proof of this last result.