6. Supports of distributions 21
Proof: If (6.3) does not hold for any m and C, then there exists a
sequence {ϕp}, ϕp C0
∞(BR),
such that
(6.4) |f(ϕp)| p|[ϕp]|p,BR , p = 1, 2,....
Let
(6.5) ψp =
ϕp
p|[ϕp]|p,BR
.
Then ψp C0
∞(BR),
|[ψp]|p,Bk =
1
p
0, and |f(ψp)| 1. Note that
ψp 0 in D since for any k, |[ψp]|k,Bk |[ψp]|p,Bk =
1
p
if k p and therefore
limp→∞ |[ψp]|k,BR = 0 for ∀k, i.e., ψp 0 in D. However, |f(ψp)| 1, and
this is a contradiction.
Theorem 6.2. Any distribution f with support at 0 has the form
(6.6) f =
m
|k|=0
ck
∂kδ
∂xk
.
Proof: Denote by β(x) a C0

function such that β(x) = 1 for |x|
1
2
,
β(x) = 0 for |x| 1.
For any ε 0 we have that 1 β(
x
ε
) = 0 for |x|
1
2
ε and therefore
f
(
(1 β(
x
ε
))ϕ
)
= 0 for any ϕ D since supp f = {0}. Hence
(6.7) f(ϕ) = f β
x
ε
ϕ for any ε 0.
Applying Proposition 6.1 to f and the ball B1, we get
(6.8) |f(ϕ)| C|[ϕ]|m,B1 for some C and m.
By the Taylor formula,
(6.9) ϕ(x) =
m
|k|=0
1
k!
∂kϕ(0)
∂xk
xk
+ Rm(x),
where Rm(x)
C∞
and |Rm(x)|
C|x|m+1.
Therefore
(6.10) f(ϕ) =
m
k=0
dk
∂kϕ(0)
∂xk
+ f β
x
ε
Rm(x) ,
where
(6.11) dk = f
xk
k!
β
x
ε
.
Note that dk does not depend on ε since supp f = {0}. Applying the
estimate (6.8) to β(
x
ε
)Rm(x), we get
f β
x
ε
Rm(x) C β
x
ε
Rm(x)
m,B1
.
Previous Page Next Page