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

.