LECTURE 7, 18.155, 29 SEPTEMBER 2011
(1) Last time I defined the Sobolev spaces of integral order in terms
of weak derivatives. To understand the L2 based spaces,
H k (Rn ) = {u ∈ L2 (Rn ); ∂ α u ∈ L2 (Rn ) ∀ |α| ≤ k}
we will use the Fourier transform. Just as with derivatives last
time, this will be defined ‘weakly’ (so the definition above says
– u in L2 ,→ S 0 (Rn ) and its derivatives, ∂ α u ∈ S 0 (Rn ) happen
to be in (the image of) L2 ⊂ S 0 (Rn ). Let me first remind you
of the process.
(2) We showed earlier (I think you did all the work) that multiplication by functions of slow growth gives a map on S(Rn )
– if f ∈ C ∞ (Rn ) and for each k there exists N such that
|∂ α | ≤ CN (1 + |x|)N for all |α| ≤ k then
×f : S(Rn ) −→ S(Rn ).
We can see that this has a ‘weak definition’ as well. Namely, if
we regard ψ ∈ S(Rn ) as a distribution, denoted for the moment
Z
Uψ (φ) = ψφ
then we get the identity
Z
Uf ψ (φ) = f ψφ = Uψ (f φ).
We want to extend multiplication by f to S 0 (Rn ), consistent
with its action on S(Rn ), so we want f Uψ = Uf ψ and then we
see how to do it in general, just define
f u(φ) = u(f φ) ∀ u ∈ S 0 (Rn ), φ ∈ S(Rn ).
This makes sense since f u ∈ S 0 (Rn ), because of the continuity
on S of multiplication by f. Clearly this gives a linear map from
S 0 (Rn ) to itself.
All trivial, but we do the same thing for the Fourier transform
(3) So we need to get a ‘weak’ formulation of the Fourier transform
on S(Rn ). What is ψ̂ as a distribution? For one last time lets
denote it Uψ̂ and compute away:
Z
Z
Uψ̂ (φ) = ψ̂φ = ψ̂(ξ)φ(ξ).
1
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LECTURE 7, 18.155, 29 SEPTEMBER 2011
Now, we can substitute the definition of ψ̂ and not have to
worry too much about convergence, since everything is in S :
Z Z
Z
Z
−ix·ξ
e−ix·ξ ψ(x)φ(ξ)dxdξ.
ψ̂(ξ) = e
ψ(x)dx =⇒ ψ̂φ =
We can exchange the order of integration without much trouble
(do a limiting argument if you are not sure) and conclude that
Z
Z
Z
Z
−ix·ξ
ψ̂φ = ψ(x) e
φ(ξ)dξdx = ψ φ̂.
This is the weak formulation of the Fourier transform we want,
it says that
Uψ̂ (φ) = Uψ (φ̂).
Thus, we can extend the Fourier transform to all of S 0 (Rn ) by
defining
(1)
û(φ) = u(φ̂) ∀ φ ∈ S(Rn ).
Proposition 1. The Fourier transform of tempered distributions, defined by (1) is a bijection on S 0 (Rn ).
Proof. First observe that if u ∈ S 0 (Rn ) then (1) does indeed
define an element û ∈ S 0 (Rn ). In fact û = u◦F : S(Rn ) −→ C is
the composite of continuous linear maps, so is itself continuous
an linear hence an element of S 0 (Rn ). The fact that F is a
bijection on S(Rn ) means that Fu = û defines a bijection –
since û = 0 means u(φ̂) = 0 for all φ and hence u(ψ) = 0 for
all ψ ∈ S(Rn ). So it is an injective linear map and its inverse is
given by the inverse of the Fourier transform on S(Rn ),
v = û =⇒ u = Gv, Gv(φ) = v(Gφ) ∀ v ∈ S 0 (Rn ), φ ∈ S(Rn ).
This follows from the identity
c = u(φ).
G û(φ) = û(Gφ) = u(Gφ)
(4) For the moment I will not discuss continuity of the Fourier
transform, but it is in fact continuous in the natural topologies
on S 0 (Rn ).
We do want to observe that the relationship of derivation
and Fourier transform, which we found on S(Rn ) carries over
to S 0 (Rn ) :
d
α u = i|α| ξ α û, x
β u = (−i)|β| ∂ β û.
∂d
LECTURE 7, 18.155, 29 SEPTEMBER 2011
3
These follow from the weak definition; for instance
α u(φ) = (∂ α u)(φ̂) = u((−1)|α| ∂ α φ̂)
∂d
α u) = i|α| û(ξ α φ) = i|α| ξ α û(φ)
= (−1)|α| u((−i)α ξd
using the identities on S in the middle.
(5) Next we consider the restriction of F from S 0 (Rn ) to L2 (Rn ); in
fact we will discuss this by extension from S(Rn ). Going back
to the identity (2), or (3), we can set φ = µ̂ where µ ∈ S(Rn )
so
Z
Z
ψ̂φ = ψ̂ µ̂.
Now, µ is the inverse Fourier transform of its Fourier transform:Z
−n
µ(x) = (2π)
eix·ξ µ̂(ξ)dξ.
Taking complex conjutates we see that
Z
−n
b = (2π)−n φ̂.
µ(x) = (2π)
e−ix·ξ µ̂(ξ)dξ =⇒ µ(x) = (2π)−n µ
(2)
Substituting this in (3) gives the Plancherel/Parseval identity
(I am hopelessly confused about which is which)
Z
Z
n
φ̂µ̂ = (2π)
ψµ ∀ φ, µ ∈ S(Rn ).
Now, from this it follows that
Proposition 2. The Fourier transform extends by continuity
(using the density of S(Rn ) in L2 (Rn ) to an essentially isometric isomorphism F : L2 (Rn ) −→ L2 (Rn );
(3)
kFukL2 = (2π)−n/2 kukL2 .
Proof. The norm identity follows from the discussion above
when u ∈ S(Rn ). Then if S(Rn ) 3 uk → u in L2 (Rn ) it follows, applying the identity to the norm of differences, that the
sequence Fuk is Cauchy, and hence convergent, in L2 (Rn ) and
(3) holds for the limit – from which it follows that F : L2 −→ L2
is well-defined and satisfies this identity. It is necessarily injective (since Fuk = 0 implies u = 0) and has inverse given by G
to which the same extension discussion applies.
(6) So, now we have a simple ‘characterization’ of the image under
the Fourier transform of L2 (Rn ) ⊂ S 0 (Rn ), namely it is L2 (Rn )
again. Thus if you believe that F ‘interchanges growth and
regularity’ then L2 has the same growth and regularity! We can
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LECTURE 7, 18.155, 29 SEPTEMBER 2011
use this basic result to get a rather useful characterization of
the Fourier transforms of the L2 -based Sobolev spaces H k (Rn )
for k ∈ N.
Before doing so, let’s recall some simple estimates. The usual
notation |x| for x ∈ Rn is the Euclidean norm, |x|2 = x21 + · · · +
x2n . We could also consider the ‘l1 ’ norm |x1 | + · · · + |xn |. As
norms on finite dimensional vector spaces these are equivalent,
namely
(4)
1
(|x1 | + · · · + |xn |) ≤ |x| ≤ |x1 | + · · · + |xn |.
n
I also introduced the notatio
1
hxi = (1 + |x|2 ) 2
which has the virtue of being a smooth function. In fact this,
and its powers, satisfy symbol estimates
(5)
∂xα hxhm ≤ Cm,α hxim−|α| .
The point to note is that the three functions
(6)
(1 + |x|), (1 + |x1 | + |x2 | + · · · + |xn |), hxh
are all the ‘same size’ on Rn – each is bounded by a positive
multiple of the other two, and hence above an below and the
same is true for any fixed power m (the same for all three of
course). You can check this from (4) and
1
(1 + |x|)2 ≤ (1 + |x|2 ) ≤ (1 + |x|)2 .
2
Since the quotients are bounded continuous functions for u ∈
2
L and m > 0 the three conditions
(7)
(1 + |x1 | + · · · + |xn |)m u ∈ L2 , (1 + |x|)m u ∈ L2 , hxim u ∈ L2
are all equivalent.
As already noted the advantage of the last one is that the
‘multiplier’ is smooth and of slow growth so hxim u ∈ S 0 (Rn ) for
any u ∈ S 0 (Rn ) makes sense, an observation which will be used
below.
(7) Now for the characterization of L2 -based Sobolev spaces
Proposition 3. If k ∈ N and u ∈ S 0 (Rn ) then u ∈ H k (Rn ) if
and only if u ∈ L2 (Rn ) and
(8)
hξik û ∈ L2 (Rn ).
LECTURE 7, 18.155, 29 SEPTEMBER 2011
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Proof. Suppose first that u ∈ H k (Rn ), for some fixed k ∈ N.
Then by definition u ∈ L2 (Rn ) and the (weak) derivatives
∂ α u ∈ L2 (Rn ) for all |α| ≤ k. From the discussion of the Fourier
tranform we conclude that
α u = iξ α û ∈ L2 (Rn ).
∂d
From this it follows that |ξ α |û ∈ L2 (Rn ) and hence that
X
(
|ξ α |)û ∈ L2 (Rn ).
|α|≤k
From the inequalities above this in turn implies (8).
Conversely, (8) implies that ξ α û ∈ L2 (Rn ) for |α| ≤ k since
then xiα hξi−k is a bounded continuous function, so a multiplier
on L2 (Rn ). Reversing the argument above, ∂xα u(x) ∈ L2 (Rn )
and hence u ∈ H k (Rn ).
(8) In fact the condition (8) alone on u ∈ L2 (Rn ) implies that
u ∈ H k (Rn ) as follows from the argument above. However,
this makes sense with k replaced by any m ∈ R, positive or
negative and not necessarily integral. So we may simply define,
consistently with the earlier definition,
H m (Rn ) = {u ∈ S 0 (Rn ); hξim û ∈ L2 (Rn ).
For m < 0 this does not imply that u ∈ L2 (Rn ) and in general
the elements of negative Sobolev spaces are not functions.