4 I. Theory of Distributions

Proof: Let cn =

Rn

χ(x)dx, where χ(x) is the same as in (1.1). Set

χδ(x) =

1

cnδn

χ

x

δ

.

Then χδ(x) ≥ 0, and changing the variables y =

x

δ

, we get

Rn

χδ(x)dx =

1

cnδn

Rn

χ

x

δ

dx =

1

cn

Rn

χ(y)dy = 1.

Given a continuous function ϕ(x) with support in BR, define ϕδ(x) = χδ ∗ϕ.

Since supp ϕ is a closed set and BR is open, there exists R1 R such that

supp ϕ ⊂ BR1 . If 0 δ R − R1, then, by Proposition 1.2, supp ϕδ ⊂ BR,

since supp χδ ⊂ Bδ and BR1 + Bδ ⊂ BR.

By property b) of the convolution ϕδ(x) ∈

C∞.

Therefore it remains to

proof that ϕδ(x) converges to ϕ(x) uniformly as δ → 0. We have

|ϕδ(x) − ϕ(x)| =

Rn

ϕ(x − y)χδ(y)dy − ϕ(x)

Rn

χδ(y)dy

≤

|y|≤δ

χδ(y)|ϕ(x − y) − ϕ(x)|dx.

Since ϕ(x) is uniformly continuous, for any ε 0 there exists δ0 0 such

that |ϕ(x − y) − ϕ(x)| ε for all x ∈

Rn

and all |y| δ0. Thus

|ϕδ(x) − ϕ(x)| ε

|y|≤δ

χδ(y)dy = ε

when 0 δ δ0.

Finally, we introduce a sequential topology in the space C0

∞(Rn).

Definition 1.3. We call D the space of all C0

∞

functions with the follow-

ing notion of convergence: a sequence ϕm ∈ C0

∞(Rn)

converges to ϕ(x) ∈

C0

∞(Rn)

if

a) there exists R 0 such that supp ϕm ⊂ BR for all m = 1, 2,... ;

b) maxx∈Rn |ϕm(x) − ϕ(x)| → 0 and maxx∈Rn |

∂kϕm(x)

∂xk

−

∂kϕ(x)

∂xk

| → 0 as

m → ∞ for all k = (k1,...,kn).

Remark 1.1. In Definition 1.3 we have defined the convergence of sequences

in D, i.e., the sequential topology in D. We will not define open sets in D,

i.e., we do not describe the topological structure of D, since we will not use

it in this book.

Example 1.2.

a) ϕm(x) =

e−mχ(x)

sin mx → 0 in D, since supp ϕm ⊂ supp χ ⊂ B1

and |

∂kϕm

∂xk

| ≤

Cke−mm|k|

→ 0 as m → ∞.