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.
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