4. Convergence of distributions 11
Example 4.1. If fn(x) f(x) a.e. and |fn(x)| g(x) for all n, where g(x)
is a locally integrable function, then, by the Lebesgue convergence theorem,
fn(ϕ) =
Rn
fn(x)ϕ(x)dx
Rn
f(x)ϕ(x)dx = f(ϕ), ∀ϕ D.
In particular, consider
(4.2) ln(x1 + iε) = ln |x1 + iε| + i arg(x1 + iε),
where ε 0 and 0 arg(x1 + iε) π for all x1
R1.
Denote
(4.3) ln(x1 + i0) = lim
ε→+0
ln(x1 + iε) = ln |x1| + iπθ(−x1),
where θ(−x1) = 0 for x1 0 and θ(−x1) = 1 for x1 0. Since | ln(x1+iε)|
π + 1 + | ln |x1|| for |x1| 1, and | ln(x1 + iε)| π + ln ||x1| + 1| for |x1|
1, 0 ε 1, we get by the Lebesgue convergence theorem that the regular
functionals, corresponding to ln(x1 + iε), converge as ε 0 to the regular
functional, corresponding to ln(x1 + i0).
Example 4.2. Denote fn(x1) = n if 0 x1
1
n
, and fn(x1) = 0 if x1 does
not belong to [0,
1
n
]. Then

−∞
fn(x1)dx1 = 1 and

−∞
fn(x1)ϕ(x1)dx1 ϕ(0) = n
1
n
0
(ϕ(x1) ϕ(0))dx1
max
0≤x1≤
1
n
|ϕ(x1) ϕ(0)| 0
as n ∞. Therefore
fn(ϕ) δ(ϕ)
as n ∞.
Theorem 4.1. If fm f in D , then for any k = (k1,...,kn),
∂kfm
∂xk

∂kf
∂xk
in D .
Proof: We have, by the definition of the derivative of a distribution:
∂kfm
∂xk
(ϕ) =
(−1)|k|fm(
∂kϕ
∂xk
)
(−1)|k|f(
∂kϕ
∂xk
) =
∂kf
∂xk
(ϕ).
Example 4.3. Let λ be a complex number. Consider
(x1+iε)λ
=
ln(x1+iε),
where ε 0 and ln(x1 + iε) is defined as in Example 4.1. Thus
(x1 +
iε)λ
= |x1 +
iε|λeiλ arg(x1+iε),
0 arg(x1 + iε) π.
Denote
(4.4) (x1 +
i0)λ
= lim
ε→+0
(x1 +
iε)λ
=
|x1|λeiλπθ(−x1).
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