PROBLEM SET 4, 18.155
BRIEF SOLUTIONS
Q1. Show that if u ∈ Cc−∞ (Rn ) there exists a sequence un ∈ Cc∞ (Rn )
such that un → u weakly, which just means that
(1)
un (f ) −→ u(f ) ∀ f ∈ C ∞ (Rn ).
Hint: You only have to ‘put things together’ to get this. We
know that there is a sequence φn ∈ Cc∞ (Rn ) such that φn ∗f → f
in S(Rn ) for any f ∈ S(Rn ) where the support of φn approaches
0. We defined un = φn ∗ u (even when u does not have compact
support) by (φn ∗u)(g) = u(φ̌n ∗g)) where φ̌n (x) = φn (−x) (you
can actually choose φn to be even). Then φ̌n ∗ g → g uniformly
with all derivatives on any compact set and this is enough to
prove (1).
Solution: Not much more than the hint
R above. Take φn (x) =
−n φ(x/) where 0 ≤ φ ∈ Cc∞ (Rn ) and φ = 1 then we know
that φn ∗ g → g in S(Rn ) if g ∈ S(Rn ). Since φn ∗ h = 0
in |x| ≥ R + C/n if h ∈ Cc∞ (Rn ) has support in |x| ≤ R, it
follows that if g ∈ C ∞ (Rn ) then un ∗ g → g uniformly with all
derivatives on compact sets, since for any given compact set
we may replace g by g 0 ∈ S(Rn ) with g 0 = g on a sufficiently
large ball. Now, if u ∈ Cc−∞ (Rn ) then from the definition of
convolution between a distribution and an element of Cc∞ (Rn ),
u ∗ φn (ψ) = u(φ̌n ∗ ψ) → u(ψ) ∀ ψ ∈ Cc∞ (Rn ),
since φ̌n ∗ ψ → ψ in Cc∞ (Rn ). We showed in class that u ∗ φn ∈
Cc∞ (Rn ) by rewriting the definition as
Z
u ∗ φn (ψ) = u(φ̌n ∗ ψ) = u(φ(· − x)ψ(x)dx.
You could also justify this by using the definition of the (Riemann) integral.
Q2. Using this, or otherwise, show that the Fourier transform of a
compactly supported distribution is a smooth function of slow
growth.
Hint: The density allows you so see that the Fourier transform of u ∈ Cc∞ (Rn ) is the function u(exp(−iξ · −) which makes
sense and is true when u ∈ Cc∞ (Rn ).
1
2
PROBLEM SET 4, 18.155 BRIEF SOLUTIONS
Solution: For u ∈ Cc∞ (Rn ) the Fourier transform can be written as the distributional pairing
(2)
û(ξ) = u(exp(−i · ξ).
The right side makes sense for u ∈ Cc−∞ (Rn ) and defines a
smooth function of ξ by the smoothness (use convergence of
difference quotients if needed) of exp(−ix · ξ) in ξ as an element
of C ∞ (Rn ) in x. Taking the approximation from Q1, as un → u
weakly, the RHS converges and does so in L1loc for instance by
Dominated Convergence. Thus the formula (2) continues to
hold for u ∈ Cc−∞ (Rn ) so the Fourier transform is a smooth
function. Slow growth comes from the bound by a seminorm –
a C k norm on some compact set so
|u(exp(−i · ξ)| ≤ C(1 + |ξ|)k .
Derivatives with respect to ξ satisfy the same bound for another
constant.
A more precise description of û(ξ) where u ∈ Cc−∞ (Rn ) is
the Paley-Weiner-Schwartz theorem which says it is an entire
function of ξ ∈ Cn which is ‘of exponential type’
|u(exp(−i · ξ)| ≤ C(1 + | Re ξ|)k exp(A| Im ξ|), ξ ∈ Cn .
Conversely, the is precisely the range of the Fourier transform.
We might get to this eventually.
Q3. Suppose γ1 and γ2 are disjoint closed cones in Rn \ 0. Show that
there is a constant c > 0 such that
(3)
|ξ − η| > c(|ξ| + |η|) if ξ ∈ γ1 , η ∈ γ2 .
Hint: Use homogeneity to pass to the sphere S2n−1 in R2n
and observe that then (ξ, η) ∈ S2n−1 with ξ ∈ γ1 , η ∈ γ2 lie in a
compact set on which |ξ − η| =
6 0.
Solution: Adding 0 to a closed cone γ ⊂ (Rn \ {0}) gives a
closed set in Rn . If γ1 ∩ γ2 = ∅ are two closed cones then
K = {(η, ξ) ∈ ((Γ1 ∪ {0}) × (Γ2 ∪ {0})×) ∩ S2n−1
is a closed subset of the sphere ({|η|2 +kξ|2 = 1} which does not
intersect η = ξ – since the cones are disjoint the only point of
intersection must have one term zero but then the other cannot
be zero. So the continuous function |η − ξ| does not vanish on
K. By homogeneity it follows that (3) holds with c > 0 since it
holds on K.
PROBLEM SET 4, 18.155
BRIEF SOLUTIONS
3
Q4. Suppose u and v are continuous functions on Rn , which are
both rapidly decreasing, in the sense that for each N ∈ N there
exists CN such that
|u(ξ)|, |v(ξ)| ≤ CN (1 + |ξ|)−N
(4)
R
show that u(η − ξ)v(ξ)dξ is rapidly decreasing in the same
sense.
Hint: (1 + |η − ξ|)(1 + |ξ|) ≥ 21 (1 + |η| + |ξ|) ≥ 21 min(1 +
|η|, 1 + |ξ|).
Solution: By the triangle inequality
1
|η − ξ| + |ξ| ≥ (|η| + |ξ|)) =⇒
2
1
(1 + |η − ξ|)(1 + |ξ|) ≥ (1 + |ξ| + η|)
2
Thus
(1 + |η − ξ|)−1 (1 + |ξ|)−1 ≤ C(1 + |ξ| + η|)−1 .
For any N it follows that the integrand has a bound
|u(η − ξ)v(ξ)| ≤ CN (1 + |ξ|)−N +n (1 + |η|)−n
which integrates to show rapid decrease of the integral.
Q5. Suppose u and v are continuous on Rn and u is rapidly decreasing everywhere and v is polynomially bounded on Rn and
rapidly decreasingRin an open cone Γ ⊂ Rn \ 0 (see Lecture 8
notes), show that u(η − ξ)v(ξ)dξ is rapidly decreasing in the
same cone, Γ.
Hint: This means rapid decay on each closed subcone of Γ.
For each closed subcone γ ⊂ Γ use a ‘conic cutoff’ φ(ξ/|ξ|) to
divide v into a part which is zero on a closed cone containing
γ in it interior and a globally rapidly decreasing part. Apply
the preceding result to the second part and use the estimate
(3) for η in the smaller closed cone and ξ in the closure of the
complement of the bigger one to show that the second term is
rapidly decreasing in the smaller closed cone.
Solution: Take any closed cone γ ⊂ γ and second closed cone
0
γ ⊂ Γ which contains γ in its interior. Use a conic cutoff to
write v as the sum v = v 0 + v 00 of term v 0 with support in γ 0
which is rapidly decreasing and a term v 00 with support in a
closed cone, plus the origin, not meeting γ. Then the integral
u∗v 0 is rapidly decreasing by Q4 and u∗v 00 is rapidly decreasing
in γ by applying the same argument using Q3.
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PROBLEM SET 4, 18.155 BRIEF SOLUTIONS
Q6. Conclude that if u ∈ Cc−∞ (Rn ) has (p, ξ) ∈
/ WF(u) according
to the definition in Lecture 8, then (p, ξ) ∈
/ WF(φu) for any
∞
n
φ ∈ C (R ).
Solution: I did this in lecture.
Q7. Consider the distribution δ(x1 ) ∈ C −∞ (Rn ), n > 1, defined by
Z
(5)
δ(x1 )(φ) = φ(0, x0 )dx0 , φ ∈ Cc∞ (Rn ).
(6)
Show that WF(δ(x1 )) = {(0, x0 ; ξ1 , 0) ∈ Rn × (Rn \ 0)}.
Solution: Multiplying by ψ ∈ Cc∞ (Rn ) gives
Z
u(φ) = ψδ(x1 )(φ) = ψ(0, x0 )φ(0, x0 )dx0 , φ ∈ Cc∞ (Rn ).
So the Fourier transform Zis
û(φ) = u(φ̂) = ψ(0, ξ 0 )φ̂(0, ξ 0 )dξ 0 .
From the Fourier inversion formula this means that û = F 0 ψ(0, ·)
is constant in ξ and equal to the Fourier transform of in the last
n − 1 variables. It follows that û is rapidly decreasing in any
opne cone which does not contain the points (±1, 0) on the
unit sphere. Since the support of the distribution is contained
in x1 = 0 the result follows.
Q8. Show that the ‘extended wave front set’
g
(7)
WF(u)
= WF ∪(singsupp(u) × {0}) ⊂ Rn × Rn
is a closed set.
Solution: This is just the fact that π1 (WF) = singsupp .
Q9. Given any (relatively of course) closed conic subset in Rn ×(Rn ×
{0}) show that there is a distribution with this as wavefront set.
The usual argument is called ‘condensation of singularities’ if
you want to look it up.
Q10. Work out the propagation of singularities theorem for the wave
equation in two dimensions, P = Dt2 −Dx2 = (Dt −Dx )(Dt +Dx ).
Every solution is of the form U (t, x) = u(t − x) + v(t + x) for
distributions in one variable u and v (show that this makes sense
¯ ∈ WF(U )
at least). Use this to show that if a point (t̄, x̄, τ̄ , ξ)
¯
for a solution then τ̄ = ±ξ (non-zero of necessity) and the whole
line
¯ τ̄ , ξ)
¯ ∈ WF(U ), ∀ s ∈ R.
(8)
(t̄ + sτ̄ , x̄ − sξ,
This is a form of Huyghen’s Principle.