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 and BR1 + 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 ∞.
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