PROBLEM SET 3, 18.155
BRIEF SOLUTIONS
(1) Suppose Ω ⊂ Rn is open, consider
(1)
(2)
C ∞ (Ω) = {φ : Ω −→ C; φ infinitely differentiable}.
Take an exhaustion of Ω by an increasing sequence Kj of compact subsets (meaning every compact subset of Ω is contained
in one of the Kj ) and show that C ∞ (Ω) is complete as a metric
space where
X
kφ − ψk(k)
d(φ, ψ) =
2−k
,
1 + kφ − ψk(k)
k
kφk(k) =
sup
|Dα φ(x)|.
|α|≤k, x∈Kk
[Note that these are only seminorms, but I think we did enough
before to prove the triangle inequality for d].
Solution: The seminorms in (2) increase with k. Since the Kj
exhaust Ω, if kφk(k) = 0 for all k then φ = 0 in Ω. Thus d is
a norm – since d(φ, ψ) = 0 implies φ = ψ and we showed the
triangle inequality earlier.
Completeness follows from the fact that a Cauchy sequence
φn is uniformly Cauchy on each Kj so converges, uniformly on
compact subsets, to a continuous function φ. The same is true of
all derivatives Dα φn → φα . In a small ball around each point in
Ω the fundamental
theorem of calculus shows that φn (x+sej ) =
Rs
φn (x)+ 0 ∂j φn (x+tej )dt and passing to the limit, the Riemann
integral converges and it follows that ∂j φ = φej for each j.
Thus φ is continuously differentiable and by induction infinitely
differentiable with derivatives the φα . It follows that φn → φ
with respect to each of the seminorms and hence with respect
to the metric, since to make d(φn , φ) < 2−k+1 it suffices to make
kφn − φk(k) < 2−k−2 . Thus C ∞ (Ω) is complete.
It is okay to quote some theorem about uniform convergence
of derivatives.
(2) Conclude that an element of the dual space denote Cc−∞ (Ω) (this
is just apparently weird notation for the continuous linear maps
u : C ∞ (Ω) −→ C) is precisely a linear map such that for some
1
2
PROBLEM SET 3, 18.155 BRIEF SOLUTIONS
C and k
(3)
|u(φ)| ≤ Ckφk(k) ∀ φ ∈ C ∞ (Ω).
Solution: Each of the seminorms is continuous with respect
to the metric, so (3) implies that u is continous on C ∞ (Ω) and
conversely, if u is continous then u−1 ({|z| < 1}) contains an
open ball around 0 of radius > 0 and hence a seminorm ball
kφk(k) < 2−k−2 for k so large that 2−k < /4.
Probably enough to quite a general result about Fréchet spaces.
(3) Observe that there is a restriction map S(Rn ) −→ C ∞ (Ω) and
use this to show that
(4)
u0 (f ) = u(f ) ∀ f ∈ S(Rn ) defines u0 ∈ S 0 (Rn ).
Solution: The restriction map S(Rn ) −→ C ∞ (Ω) is continuous, since the seminorm kφk(k) is bounded by some global norm
(5)
(1 + |x|2 )k/2 |Dα φ|
sup
x∈Rn ,|α|≤k
which is continuous on S(Rn ). Thus u0 as defined is an element
of S 0 (Rn ).
(4) Show that the image of the the resulting map Cc−∞ (Ω) −→
S 0 (Rn ) consists of the tempered distributions with compact support contained in Ω. [You might want to use approproate cut-off
functions to prove this.]
Solution: Each element u0 ∈ S 0 (Rn ) obtained by restriction
of u ∈ Cc−∞ (Ω) vanishes on the complement of the compact set
Kj for which the estimate (3) holds, since the seminorm, and
hence u(φ), vanishes on any φ ∈ Cc∞ (Rn ) with support disjoint
from Kj .
Conversely, suppose v ∈ S 0 (Rn ) has support K b Ω. Choose
χ ∈ Cc∞ (Ω) with χ = 1 in a neighbourhood of K. Then χv = v
and we may define u(φ) = v(χφ) for all φ ∈ C ∞ (Ω). This is
clearly a linear map and the continuity of v means that
(6)
|u(φ)| = |v(χφ)| ≤ C sup (1 + |x|2 )k/2 |Dα (χφ)| ≤ C 0 kφ|(j)
x,|α|≤k
for j ≥ k large enough that supp χ ⊂ Kj . Thus u is continous
and u0 (φ) = v(χφ) = v(φ) for φ ∈ S(Rn ). Thus each such
distribution arises this way.
(5) Let ∆ = D12 + · · · + Dn2 be the (positive) Laplacian on Rn . Show
that if u ∈ S 0 (Rn ) and ∆u = 0 then u is a polynomial.
Solution: If ∆u = 0 and u ∈ S 0 (Rn ) then |ξ|2 û = 0. This
implies that supp û ⊂ {0} and we know that this means that
PROBLEM SET 3, 18.155
BRIEF SOLUTIONS
3
û is a finite linear combination of the derivatives of the delta
function at 0. Thus u itself is a polynomial – called an harmonic
polynomial.
(6) For n ≥ 3 compute ∆|x|2−n in |x| > 0 and show that it vanishes; show that |x|2−n is integrable near 0 and hence defines an
element U ∈ S 0 (Rn ) (by integration of course). Conclude that
∆U has support at 0 and cannot vanish identically.
Solution: In x 6= 0, |x|2−n is a smooth function and we can
differentiate directly to see
∂j |x|2−n = (2 − n)|x|1−n
(7)
xj
.
|x|
Differentiating again,
(8)
−∂j2 |x|2−n
−n
= −(2−n)(1−n)|x|
x2
x2j
1−n 1
1−n j
−(2−n)|x|
+(2−n)|x|
.
|x|2
|x|
|x|2
Combining the first and last term and then summing over j
gives 0.
Now, |x|s is locally integrable across 0 if s > −n so |x|2−n
is a well-defined tempered distribution – outside a compact set
it is equal to a symbol. So ∆U is well-defined as a tempered
distribution and we have shown it has support contained in the
origin. It cannot vanish since U is not a polynomial.
(7) Complete the discussion above by finding the constant c so that
∆|x|2−n = cδ0 .
Solution: Well, it has to be a multiple of δ0 by considering
homogeneity.
(8) Consider Lojasieciwz Theorem (where the j is wrong) that for
every non-trivial (i.e. not identically zero) polynomail P in n
variables, has a tempered fundamental solution, U ∈ S 0 (Rn )
such that P (D)U = δ0 . Show that to do this it suffices to show
there is F ∈ S 0 (Rn ) such that |P (ξ)|2 f = 1 on Rn . As far
as I am aware there are three proofs of this and none are easy.
There are estimates by Hörmander, simplifying earlier estimates
by Lojasieciwz. There is a clever proof by Atiyah using the
Hironaka resolution theorem from algebraic geoemtry and there
is a proof showing the meromorphy of |P (ξ)|z . Fortunately it is
not important for us but you might like to look at one of them.
(9) Prove this theorem for n = 1 – it is still non-trivial but the
results from Thursday should allow you to do it.
4
PROBLEM SET 3, 18.155 BRIEF SOLUTIONS
(10) In Problem Set 2 you showed that
[
(9)
S 0 (Rn ) = (1 + |x|2 )N H −2N (Rn )
N
where the space on the right is a Hilbert space – the elemenst
are tempered distributions of the form u = (1 + |x|2 )N U where
U ∈ H −2N (Rn ) and the norm on u is the H −2N (Rn ) norm on U.
Give S 0 (Rn ) the corresponding inductive limit topology – sets
are open if and only if they meet these Hilbert spaces in open
sets (you might want to observe that as N increases the spaces
get bigger and the smaller ones are continuously included in the
larger ones). Show that the dual of S 0 (Rn ) with respect to this
topology is S(Rn ).