1. Spaces of infinitely differentiable functions 5

b) ϕm(x) =

1

m

χ(

x

m

) does not converge to 0 in D: although |

∂kϕm

∂xk

| ≤

ck

m|k|+1

→ 0 as m → ∞, since supp ϕm = supp χ(

x

m

) = Bm spreads as

m → ∞.

1.3. Proof of Proposition 1.1.

Since the repeated integral

Rn Rn

|ϕ(x − y)| |ψ(y)|dx dy =

Rn Rn

|ϕ(t)|dt |ψ(y)|dy

=

Rn

|ϕ(t)|dt

Rn

|ψ(y)|dy

is finite, Fubini’s theorem implies that the repeated integral

Rn Rn

|ϕ(x − y)||ψ(y)|dy dx

and the double integral

Rn Rn

|ϕ(x − y)||ψ(y)|dxdy

are finite and all three integrals are equal. Moreover, Fubini’s theorem (see,

for example, [R]) implies that the double integral

Rn×Rn

ϕ(x − y)ψ(y)dydx

exists and for almost all x ∈

Rn

the integral

Rn

ϕ(x−y)ψ(y)dy exists. Thus

ϕ ∗ ψ

L1

=

Rn Rn

ϕ(x − y)ψ(y)dy dx

≤

Rn Rn

|ϕ(x − y)||ψ(y)|dydx = ϕ

L1

ψ

L1

.

1.4. Proof of property b) of the convolution.

We shall need the following theorem [R]:

Theorem 1.4 (Lebesgue convergence theorem). If fm(x) → f(x) almost

everywhere as m → ∞ and |fm(x)| ≤ g(x), where

Rn

g(x)dx +∞, then

Rn

fm(x)dx →

Rn

f(x)dx as n → ∞.

Since ϕ(x) is continuous and bounded, we have that |ϕ(x)| ≤ M and

ϕ(xm −y) → ϕ(x0 −y) as xm → x0. Therefore, by the Lebesgue convergence

theorem,

Rn

ϕ(xm − y)ψ(y)dy →

Rn

ϕ(x0 − y)ψ(y)dy,

since |ϕ(xm − y)ψ(y)| ≤ M|ψ(y)| and ψ ∈ L1.

Hence ϕ ∗ ψ is a continuous function, and |(ϕ ∗ ψ)(x)| ≤

Rn

M|ψ(y)|dy.