2 I. Theory of Distributions

Laplace equations the traditional approach is simpler, and we offer this

approach as problems (with some hints).

1. Spaces of infinitely differentiable functions

Denote by C0

∞(Rn)

the space of infinitely differentiable complex-valued func-

tions in

Rn

with compact supports, i.e., ϕ(x) ∈ C0

∞(Rn)

if ϕ(x) has contin-

uous partial derivatives of every order and ϕ(x) = 0 when |x| R for some

R depending of ϕ(x). For example,

(1.1) χ(x) =

e

−

1

1−|x|2 if |x| 1,

0 if |x| ≥ 1, |x| = x1

2

+ · · · + xn,2

is a C0

∞

function.

Definition 1.1. Let f(x) be a continuous function in

Rn.

The support of

f(x) is the closure of the set where f(x) = 0.

We denote the support of f(x) by supp f.

Example 1.1.

a) The support of χ(x) is the closed ball of radius 1.

b) The support of f(x1) = x1

2

− 1 is

R1.

Definition 1.2. Let ϕ(x) be a measurable bounded and ψ(x) a Lebesgue

integrable functions in

Rn,

ψ

L1

=

Rn

|ψ(x)|dx +∞. Then the convo-

lution of ϕ(x) and ψ(x) (denoted by (ϕ ∗ ψ)(x) ) is the following integral:

(1.2) (φ ∗ ψ)(x) =

Rn

φ(x − y)ψ(y)dy.

Proposition 1.1. If ϕ(x) ∈ L1 and ψ(x) ∈ L1, then the integral (1.2) exists

for almost all x ∈

Rn,

ϕ ∗ ψ ∈

L1(Rn)

and

(1.3) ϕ ∗ ψ

L1

≤ ϕ

L1

ψ

L1

.

The proof of Proposition 1.1 is given at the end of this section.

1.1. Properties of the convolution.

a) ϕ ∗ ψ = ψ ∗ ϕ.

Proof: We have

(1.4) (ϕ ∗ ψ)(x) =

Rn

ϕ(x − y)ψ(y)dy.

Changing the variables x − y = t, we get

Rn

ϕ(x − y)ψ(y)dy =

Rn

ϕ(t)ψ(x − t)dt = ψ ∗ ϕ.