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324 6. Introduction to Pseudodifferential Operators

(x ,ξ ) = e −ix ·ξ A Φ,0 e iξ ·• (x )
σ A Φ,0

= e −iΦ(x)·ξ A 0 e iξ ·Φ(·) (x)for x = Φ −1 (x )

after coordinate transform. With the substitution
∂Φ
Φ(z)= Φ(x)+ (x)(z − x)+ r(x, z)
∂x
we obtain

ξ )
∂Φ −n −i(x−z)·ξ

(x ,ξ ) = e (2π) e a(x, ξ) ×
−ix·( ∂x
σ A Φ,0
IR n Ω
∂Φ
ξ ) ir(x,z)·ξ
×e iz·( ∂x e dzdξ
= e −ix·ζ A ◦ Be i•·ζ
where A = a(x, D)and B is the multiplication operator defined by the symbol
σ B (z, ξ)= e ir(x,z)·ξ

with x and ξ fixed as constant parameters. With formula (6.1.29) we then
find
∂Φ

(x ,ξ ) ∼ σ A◦B (x, ζ) where x = Φ(x)and ζ = ξ .

σ A Φ,0
∂x
The symbol of A ◦ B is given by (6.1.37), which now implies
α α
1 ∂ ∂
ir(x,z)·ξ
(x ,ξ ) ∼ a(x, ξ) − i e
σ A Φ,0
α! ∂ξ ∂z
α≥0 | ξ=ζ | z=x
with constant ζ = ∂Φ ξ , which is the desired formula (6.1.50).

∂x
To show (6.1.52) we employ induction for α with |α|≥ 1. In particular
we see that
∂ iξ ·r(x,y) iξ ·r(x,z) ∂Φ ∂Φ

e = ie (z) − (x) · ξ =0
| z=x
∂z j ∂z j ∂x j
| z=x
and
∂ ∂ iξ ·r(x,z) 2 iξ ·r(x,z) ∂Φ ∂Φ

e = i e (z) − (x) · ξ ×
| z=x
∂z k ∂z j ∂z j ∂x j

∂Φ ∂Φ
(z) − (x) · ξ
∂z k ∂x k
2
∂ Φ
iξ ·r(x,z)

+ie (z) · ξ
∂z j ∂z k
| z=x
2
∂ Φ

= i (x) · ξ ,
∂x j ∂x k