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318 6. Introduction to Pseudodifferential Operators

and we have the asymptotic expansion for the complete symbol class:

1 ∂ α ∂ α
σ A◦B ∼ σ A (x, ξ) − i σ B (x, ξ) . (6.1.37)
α! ∂ξ ∂x
α≥0
Here, σ A (x, ξ)and σ B (x, ξ) denote one of the respective representatives in
the corresponding equivalence classes of the complete symbol classes.
For the proof, we remark that with Theorem 6.1.9 we have either
A = A 0 + R and B = B 0 or A = A 0 and B = B 0 + Q. Then

A ◦ B = A 0 ◦ B 0 + R ◦ B 0
or
A ◦ B = A 0 ◦ B 0 + A 0 ◦ Q.

Since A 0 and B 0 are properly supported, R ◦ B 0 or A 0 ◦ Q are continuous

mappings from E (Ω)to C (Ω) and, hence, are smoothing operators. (Note
∞
that for A and B both not properly supported, the composition generates
the term R ◦ Q, the composition of two regularizers, which is not defined in
general.)
For the remaining products A 0 ◦ B 0 , we can use that without loss of
generality, A 0 = A 0 (x, D) ∈ OPS m 1 (Ω)and B 0 = B 0 (x, D) ∈ OPS m 2 (Ω).
Hence, one obtains for C := A 0 ◦ B 0 a representation in the form (6.1.26)
with the amplitude function

c(x, z, η)= a(x, ξ)e i(x−y)·ξ b(y, η)e i(y−z)·η dydξ
IR n Ω
n
for which one needs to show c ∈ S m 1 +m 2 (Ω × Ω × IR ) and, with the Leibniz
rule, to evaluate the asymptotic expansion (6.1.28) for c. For details see e.g.
[306, Chap. I, Theorem 3.2].
As an immediate consequence of Theorem 6.1.14, the following corollary
holds.

Corollary 6.1.15. Let A ∈L m 1 (Ω) ,B ∈L m 2 (Ω) and one of them be
properly supported. Then the commutator satisfies
m 1 +m 2 −1
[A, B]:= A ◦ B − B ◦ A ∈L (Ω) . (6.1.38)
Proof: For A ◦ B and B ◦ A we have Theorem 6.1.14 and the commutator’s
symbol has the asymptotic expansion
1 ∂ α ∂ α

σ [A,B] ∼ σ A (x, ξ) − i σ B (x, ξ)
α! ∂ξ ∂x
|α|≥1
∂ ∂
α α
− σ B (x, ξ) − i σ A (x, ξ) .
∂ξ ∂x
Hence, the order of [A, B] equals m 1 − 1+ m 2 .