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

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

dx1
=

0

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.
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