3. Operations with distributions 9

∀ϕ ∈ D(0, +∞). One can prove (cf. Problem 8.12) that this functional is

not a restriction to (0, +∞) of any f ∈ D

(R1).

3. Operations with distributions

3.1. Derivative of a distribution.

Let f(x1) be an absolutely continuous function in

R1.

Then the formula of

integration by parts holds:

(3.1)

∞

−∞

df(x1)ϕ(x1)dx1

dx1

= −

∞

−∞

f(x1)

dϕ(x1)dx1,

dx1

∀ϕ ∈ D.

Thus, for the regular functionals corresponding to

df

dx1

and f(x1) we get

(3.2)

df

dx1

(ϕ) = −f

dϕ

dx1

.

Formula (3.2) leads to the following definition of a derivative of a distribu-

tion:

Definition 3.1. For any f ∈ D and any k = (k1,...,kn),

(3.3)

∂kf

∂xk

(ϕ) =

(−1)|k|f

∂kϕ

∂xk

.

The right hand side of (3.3) is well defined since

∂kϕ

∂xk

∈ D and ϕm → ϕ

in D implies that

∂kϕm

∂xk

→

∂kϕ

∂xk

in D. Therefore

(−1)|k|f(

∂kϕ

∂xk

) is a linear

continuous functional on D and this functional, by definition, is

∂kf

∂xk

(ϕ).

Example 3.1.

i) Consider the regular functional θ corresponding to the function

θ(x1) = 1 for x1 0, θ(x1) = 0 for x1 0. Then

d

dx1

θ(ϕ) = −θ

dϕ

dx1

= −

∞

0

dϕ

dx1

(x1)dx1 = ϕ(0) = δ(ϕ).

ii)

dkδ

dx1

k

(ϕ) =

(−1)|k|δ

dkϕ

dx1

k

=

(−1)|k|

dkϕ(0)

dx1k

.

3.2. Multiplication of a distribution by a

C∞-function.

Again, if f(ϕ) is a regular functional corresponding to f(x) and if a(x) ∈

C∞(Rn),

then, obviously,

(3.4)

Rn

(a(x)f(x))ϕ(x)dx =

Rn

f(x)(a(x)ϕ(x))dx.

This identity leads to the following definition.