MA54200 HOMEWORK

ASSIGNMENT 1: SOLUTIONS

1.3. Show that the principal value integral

Z Z − Z ∞ 
φ(x) φ(x) φ(x)
p.v. dx = lim dx + dx
x →0+ −∞ x  x
exists for all φ ∈ Cc∞ (R), and is a distribution. What is its order?
Solution. Note that we can write
Z ∞
φ(x) − φ(−x)
Z
φ(x)
p.v. dx = dx.
x 0 x
The latter integral exist, since
φ(x) − φ(−x)
≤ 2 sup |φ0 |.
x
Moreover, if supp φ ⊂ [−a, a], then
 Z 
p.v. φ(x) dx ≤ 2a sup |φ0 |,
 
 x 
which implies that the p.v. of 1/x is a distribution of order at most 1. Finally, the
order cannot be 0. Indeed, if 0 ≤ φ ≤ 1 is such that supp φ ⊂ [, 4] and φ = 1
on [2, 3] then Z
φ(x) 1
p.v. ≥ sup |φ |
x 4

1.5. Show that
∞
X
hu, φi = ∂ k φ(1/k)
k=1
is a distribution on (0, ∞), but that there is no v ∈ D0 (R) whose restriction to
(0, ∞) is equal to u.
Solution. Let φ be such that supp φ ⊂ [1/N, N ]. Then
N
X N
X
k
hu, φi = ∂ φ(1/k) ≤ sup |∂ k φ|.
k=1 k=1 [1/N,N ]
Since the compacts [1/N, N ] exhaust (0, ∞), it follows that u is a distribution on
(0, ∞). 
Assume now u = v (0,∞) for v ∈ D0 (R). Then there must exist N0 and C0 such
that
N0
X
|hv, φi| ≤ C0 sup |∂ k φ|, supp φ ⊂ [−1, 1].
k=1
So, if we take N > N0 , we will have
1
2 ASSIGNMENT 1: SOLUTIONS

N0
X
|∂ N φ(1/N )| = |hu, φi| = |hu, φi| ≤ C0 sup |∂ k φ|
k=1

if supp φ ⊂ ( N1+1 , N 1−1 ). N

This would imply that ∂ δ1/N is of order at most N0 < N
and consequently that ∂ N δ is of order at most N0 on a small interval (−, ). We
claim that this is impossible. Indeed, let ψ ∈ Cc∞ ((−, )) be such that ∂ N ψ(0) 6= 0.
Consider then the test functions
ψλ (x) = λN ψ(x/λ)
for small λ > 0. We have supp ψλ ⊂ (−λ, λ). Moreover,
∂ N ψλ (0) = ∂ N ψ(0)
and
∂ k ψλ = λN −k ∂ k ψ.
Thus, we would have an estimate
N0
X
|∂ N ψ(0)| ≤ C0 λN −N0 sup |∂ k ψ|
k=1

for any λ > 0. This is clearly a contradiction for small λ. 

1.6. Let u ∈ D0 (Rn ) have the property that hu, φi ≥ 0 for all real valued nonnega-
tive φ ∈ Cc∞ (Rn ). Show that u is of order 0.
Proof. Let K ⊂⊂ Rn and ψK ∈ Cc∞ (Rn ) be a nonnegative cut-off function such
that ψK = 1 on K. Then for real-valued test functions φ with supp φ ⊂ K we have


sup |φ| ψK (x) − φ(x) ≥ 0.
K

Hence


hu, sup |φ| ψK (x) − φ(x)i ≥ 0.
K
This implies
hu, φ(x)i ≤ hu, ψK i(sup |φ|).
K
For complex valued φ we obtain
|hu, φ(x)i| ≤ 2hu, ψK i(sup |φ|)
K

by considering the real and imaginary parts of φ. 

1.9. Let (ck )k∈Z be complex numbers which satisfy

|ck | ≤ C(1 + |k|)m , k ∈ Z,
for some constants C ≥ 0 and m. Show that
X
u= ck eikx
k∈Z
0
converges in D (R).
 MA54200 HOMEWORK 3

Proof. Let φ ∈ Cc∞ (R). Then

(−1)m+2
Z Z
ikx
e φ(x)dx = eikx ∂ m+2 φ(x)dx, k 6= 0.
R (ik)m+2 R
Hence  Z 
C
e φ(x)dx ≤ 2 sup |∂ m+2 φ|, k 6= 0.
ikx
 
ck

R k
thus, the series hu, φi converges and defines a distribution of order at most m + 2.