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6.2 Elliptic Pseudodifferential Operators on Ω ⊂ IR n 327

m
Definition 6.2.2. A parametrix Q 0 for the operator A ∈L (Ω) is a prop-
−∞
erly supported operator which is a two–sided inverse for A modulo L (Ω):
A ◦ Q 0 − I = C 1 ∈L −∞ (Ω)
and (6.2.6)
Q 0 ◦ A − I = C 2 ∈L −∞ (Ω)
Theorem 6.2.2. (H¨ormander[129], Taylor [302, Chap. III, Theorem 1.3])
m
A ∈L (Ω) is elliptic of order m if and only if there exists a parametrix
−m
Q 0 ∈L (Ω) satisfying (6.2.6).
m
Proof: (i) Let A be a given operator in L (Ω) which is elliptic of order
m. Then to the complete symbol class of A there is a representative a(x, ξ)
satisfying (6.2.2). If Q would be known then (6.1.37) with (6.2.6) would imply

α
1 ∂
σ Q◦A =1 ∼ q −m−j (x, ξ)a(x, ξ)+ q −m− (x, ξ) ×
α! ∂ξ
0≤j 1≤|α| 0≤
∂
α
× − i a(x, ξ) . (6.2.7)
∂x
n
So, to a we choose a function χ ∈ C (Ω ×IR ) with χ(x, ξ) = 1 for all x ∈ Ω
∞
and |ξ|≥ C ≥ 1 where C is chosen appropriately large, and χ = 0 in some
neighbourhood of the zeros of a. Then define
q −m (x, ξ):= χ(x, ξ)a(x, ξ) −1 (6.2.8)
and, in view of (6.2.7), recursively for j =1, 2,...,

q −m−j (x, ξ) := (6.2.9)
1 ∂ α ∂ α
− q −m−j+|α| (x, ξ) − i a(x, ξ) q −m (x, ξ) .
α! ∂ξ ∂x
1≤|α|≤j
n
From this construction one can easily show that q −m−j ∈ S S S −m−j (Ω × IR ).
The sequence {q −m−j } can be used to define an operator Q ∈ OPS −m (Ω ×
n
IR ) with ∞ q −m−j as the asymptotic expansion of σ Q , see Theorem 6.1.3.
j=0
With Q available, we use the decomposition of Q = Q 0 + R, Theorem 6.1.9,
n
where Q 0 ∈ OPS −m (Ω × IR ) is properly supported and still has the same
0
symbol as Q. Then we have with Theorem 6.1.14 that Q 0 ◦A ∈L (Ω)and by
−∞
construction in view of (6.1.37) σ Q 0 ◦A ∼ 1. Hence, Q 0 ◦A−I = R 2 ∈L (Ω).
−∞
In the next step it remains to show that A ◦ Q 0 − I ∈L (Ω), too.
The choice of q −m also shows that Q 0 is an elliptic operator of order
(−1) m
−m. Therefore, to Q 0 there exists an operator Q 0 ∈ OPS which is also
(−1)
properly supported and satisfies Q 0 ◦ Q 0 − I = R 1 ∈L −∞ (Ω), where R 1
is properly supported, due to Proposition 6.1.6. Then