PROBLEM SET 2, 18.155
BRIEF SOLUTIONS
The norms we use on Schwartz space here are
kukk =
sup
x∈Rn ,|α|≤k
|(1 + |x|2 )k/2 |Dxα u(x)|, k ∈ N0 = {0, 1, . . . , }.
(1) Say a few words in support of the statement that if U : S(Rn ) −→
C is linear then
U ∈ S 0 (Rn ) ⇐⇒ ∃ k ∈ N0 , C > 0 s.t. |U (u)| ≤ Ckukk ∀ u ∈ S(Rn ).
(1)
Solution: We did this in class and you do not need to say this
much! The metric on S(Rn ) is
X
ku − vkk
d(u, v) =
2−k
.
1
+
ku
−
vk
k
k
If d(u, v) < 2−k δ where d < 1/4 then ku − vkk < 2δ shows
the continuity of k · kk and hence the second condition in (1)
implies the continuity of U. Conversely if U is continuous then
U −1 ({|z| < 1}) contains a ball d(u, 0) < . Taking N such
that 2−N +1 < /2 and noting that the norms increase with k it
follows that U −1 ({|z| < 1}) contains the norm ball kukN < /10
and the bound (1) follows.
(2) Show that for some constant C 0 > 0
(2) k(1 + |x|2 )−k/2 ukk ≤ C 0 kukC k ,
kukC k =
sup
|Dα u(x)| ∀ u ∈ S(Rn ).
x∈Rn ,|α|≤k
Solution: We showed in class that for any s and α there is a
constant C such that
|Dα (1 + |x|2 )s/2 | ≤ C(1 + |x|2 )s−|α|/2 =⇒
(3)
sup |(1 + |x|2 )−s/2 Dα (1 + |x|2 )s/2 | < ∞.
x
This in turn follows from the fact, proved by induction, that
(4)
Dα (1 + |x|2 )s/2 = pα,s (x)(1 + |x|2 )s/2−|α|
where pα,s is a polynomial of degree (at most) |α|.
1
2
PROBLEM SET 2, 18.155 BRIEF SOLUTIONS
(5)
Applying Leibniz formula gives, for |α| ≤ k,
X α
α
2 −k/2
D [(1 + |x| )
u] =
[Dα−β (1 + |x|2 )−k/2 ]Dβ u =⇒
β
β≤α
X
(1 + |x|2 )k/2 Dα [(1 + |x|2 )−k/2 u] ≤ C
sup |Dβ u|
β≤α
Summing this over α gives (2).
Remark: The same argument works to show that for any
l ≥ k there exists a constant such that
(6)
k(1 + |x|2 )−l/2 ukk ≤ Cl0 kukC k .
(3) Conclude that if U ∈ S 0 (Rn ) then, for some k and c, V =
(1 + |x|2 )−k/2 U satisfies
|V (u)| ≤ ckukC k ∀ u ∈ S(Rn ).
Solution: If V = V = (1 + |x|2 )−k/2 U then for u ∈ S(Rn ),
|V (u)| = |U (V (1 + |x|2 )−k/2 u)| ≤ Ck(1 + |x|2 )−k/2 ukk
≤ ckukC k ∀ u ∈ S(Rn ).
(4) ‘Recall’ the Sobolev embedding theorem in the form that
kukC k ≤ C 00 kukH N , N > k + n/2, u ∈ S(Rn ).
Solution: I recalled it, now what?
(5) Conclude that if V is as above then for some N ∈ N
|V (u)| ≤ CN kukH N ∀ u ∈ S(Rn ).
Solution: Right, so from (3) and (4) we get (5).
(6) Conclude that if u ∈ S 0 (Rn ) then there exist f ∈ L2 (Rn ) and
N ∈ N such that
n
X
2 N
N
U = (1 + |x| ) (1 + ∆) f, ∆ =
Di2 .
i=1
Solution: Now, we can recall from class that ‘with respect to
the extension of the distributional pairing’ H −N (Rn ) is the dual
of H N (Rn ) for any N ∈ R. This is really just the fact that
(7)
u ∈ H N (Rn ) ⇐⇒ (1 + |ξ|2 )N/2 û ∈ L2 (Rn )
so v pairs with u iff (1+|ξ|2 )−N/2 v̂ ∈ L2 (Rn ) so that the product
with (1 + |ξ|2 )N/2 û is integrable.
So, (5) means that (1 + |ξ|2 )−N/2 V̂ ∈ L2 (Rn ) where N ≥ 0
is a non-negative integer and can be increased. So fˆ = (1 +
PROBLEM SET 2, 18.155
BRIEF SOLUTIONS
3
|ξ|2 )−N V̂ ∈ L2 (Rn ) and undoing the Fourier transform this
means
V = (1 + ∆)N f.
(8)
Now, as remarked above we could have defined V = (1 +
|x|2 )−N U by replacing k/2 by the larger N and still have arrived at (3) for the same k. Doing this gives
U = (1 + |x|2 )N (1 + ∆)N f, f ∈ L2 (Rn )
(9)
as desired.
(7) The Dirac delta ‘function’ δ ∈ S 0 (Rn ) defined by
δ(φ) = φ(0) ∀ φ ∈ S(Rn )
is amongst the most important distributions (it is a measure).
A) Find explicit formulae for the derivatives ∂ α δ evaluated on
test functions
B) Compute the Fourier transform of ∂ α δ.
C) Show that
∂ α δ ∈ H −|α|−n/2− (Rn )
for > 0 but not for = 0.
Solution:
A) By definition of the action of differentiation on distributions
∂ α U (u) = U ((−1)|α| ∂ α u)
so for Dirac’s distributions
∂ α δ(u) = (−1)|α| ∂ α u(0)
(10)
evaluates the corresponding derivative at the origin.
B) Again, for a general tempered distribution
Û (u) = U (û)
so
(11)
|α| α
α δ(u) = (−1) ∂ û(0) = (i)
∂d
|α|
Z
α δ = (iξ)α
xα u =⇒ ∂d
(as an equality of functions of slow growth in S 0 (Rn )).
C) This is equivalent to the statement that
(12)
(1 + |ξ|2 )−|α|/2−n/4−/2 ξ α ∈ L2 (Rn ) > 0
4
(13)
PROBLEM SET 2, 18.155 BRIEF SOLUTIONS
but not for = 0 and follows both ways from the boundedness (for > 0 and unboundedness (for = 0) of the
Riemann integral
Z
(1 + |ξ|2 )−|α|−n/2− |ξ α |2 as R → ∞.
|x|<R