5. Regularizations of nonintegrable functions 19

Then (5.18) can be rewritten in the form

(5.21)

1

|x|2 − k2 ± i0

= p.v.

1

|x|2 − k2

∓

iπδ(|x|2

−

k2).

Example 5.5 (Distributions

1

S±i0

, p.v.

1

S

, δ(S), where S(x) = 0 is a smooth

surface). A surface S(x) = 0 is called smooth if S(x) ∈

C∞(Rn),

S(x) is

real-valued, and ∇S(x0) = (

∂S(x0)

∂x1

, . . . ,

∂S(x0)

∂xn

) = 0 when S(x0) = 0. We

have

1

S ± iε

(ϕ) =

Rn

ϕ(x)

S(x) ± iε

dx.

Let {αj(x)} be a partition of unity in

Rn,

i.e., αj(x) ∈ C0

∞(Rn)

and

∑

j

αj(x)

≡ 1. We choose a partition of unity such that if supp αj ∩ {S(x) = 0} = ∅,

then

∂S

∂xnj

= 0 on supp αj for some nj, and the change of variables

(5.22) S(x) = t, xnj = xnj

has a

C∞

inverse on supp αj. Here xnj are n−1 remaining coordinates after

xnj is deleted.

Denote by

(5.23) xnj = h(t, xnj ), xnj = xnj

the inverse map to (5.22) . Making the change of variables (5.23), we get

Rn

αj(x)ϕ(x)dx

S(x) ± iε

=

Rn

αj(h(t, xnj ),xnj )ϕ(h(t, xnj ),xnj )

∂h(t,xnj )

∂t

t ± iε

dtdxnj .

Taking the limit as ε → 0 we obtain, as in (5.18):

(5.24) lim

ε→0

Rn

αj(x)ϕ(x)dx

S(x) ± iε

= p.v.

Rn

αjϕ

∂h(t,xnj )

∂t

t

dtdxnj

∓ iπ

Rn−1

αj(h(0,xnj ),xnj )ϕ(h(0,xnj ),xnj )

∂h(0,xnj )

∂t

dxnj .

Denote by dσ the surface area element of the surface S(x) = 0. On supp αj

we have:

(5.25) dσ = 1 + |∇xnj h(0,xnj )|2dxnj .

Since S(xnj , h(t, xnj )) ≡ t, we have:

(5.26)

∂S

∂xnj

∂h

∂t

= 1,

∂S

∂xk

+

∂S

∂xnj

∂h

∂xk

= 0 for k = nj.

Therefore

(5.27) dσ = 1 +

∂S

∂xnj

−2

|∇xnj S|2dxnj = |∇xS|

∂h

∂t

dxnj .