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6.1 Basic Theory of Pseudodifferential Operators 313

equivalent to (6.1.22). Consequently, (6.1.21) and (6.1.22) are satisfied if A
is properly supported.
Conversely, if (6.1.21) and (6.1.22) hold, then i) in Lemma 6.1.4 is already
satisfied. In order to show ii) let us choose any compact subset K x of Ω and
∞
let K y be the corresponding compact subset in (6.1.22). Let v ∈ C (Ω \K y ).
0
Then Av ∈ C (Ω) due to (6.1.21). Now, let ψ ∈ C (Ω) be any function
∞
∞
0
0
with supp ψ ⊂ K x . Then supp A ψ ⊂ K y because of (6.1.22). Hence,

Av, ψ = v, A ψ =0
for any such ψ. Therefore, supp(Av)∩K x = ∅ which is the second proposition
ii) of Lemma 6.1.4.
As an obvious consequence of Corollary 6.1.5, the following proposition
is valid.
∞
Proposition 6.1.6. If A and B : C (Ω) → C (Ω) are properly supported,
∞
0
then the composition A ◦ B is properly supported, too.
n
m
Theorem 6.1.7. If A ∈ OPS (Ω × IR ) is properly supported then
a(x, ξ)= e −ix·ξ (Ae iξ• )(x) (6.1.23)

is the symbol (see Taylor [302, Chap. II, Theorem 3.8]).
n
m
To express the transposed operator A of A ∈ OPS (Ω × IR ) with the
n
m
given symbol a(x, ξ) ,a ∈ S S S (Ω × IR ) we use the distributional relation

Au, v = u, A v = Au(x)v(x)dx = u(x)A v(x)dx for u, v ∈ C (Ω) .
∞
0
Ω Ω
Then from the definition (6.1.8) of A, we find

A v(x)=(2π) −n e i(x−y)·ξ a(y, −ξ)v(y)dydξ . (6.1.24)

IR n Ω
From this representation it is not transparent that A is a standard
pseudodifferential operator since it does not have the standard form (6.1.8).
In fact, the following theorem is valid.
m
Theorem 6.1.8. (Taylor [302, Chap. II, Theorem 4.2]) If A ∈ OPS (Ω ×
n
n
m
IR ) is properly supported, then A ∈ OPS (Ω × IR ).

n
m
If, however, A ∈ OPS (Ω × IR )is not properly supported, then A
belongs to a slightly larger class of operators. Note that if we let

a(x, y, ξ):= a(y, −ξ)