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312 6. Introduction to Pseudodifferential Operators

Next, we show the sufficiency of i) and ii).Let K x be any compact subset
of Ω and K y the corresponding subset in ii) such that supp v∩K y = ∅ implies
supp(Av) ∩ K x = ∅.Wewanttoshowthatsupp K A ∩ (K x × Ω) is compact.
∞
To this end, we consider ψ ∈ C (Ω) with supp ψ ⊂ K x . Then
0

Av, ψ = K A (x, y)v(y)ψ(x)dydx
Ω Ω

= K A (x, y)v(y)dyψ(x)dx =0

Ω Ω\K y
∞
for every v ∈ C (Ω) with supp v ∩K y = ∅ due to ii). This implies supp K A ∩
0
K x × (Ω \ K y ) = ∅;so,
supp K A ∩ (K x × Ω) = supp K A ∩ (K x × K y ) ⊂ K x × K y .

Hence, supp K A ∩(K x ×Ω) is compact because K x ×K y is compact. To show
that for any chosen compact K y Ω, the set supp K A ∩(Ω ×K y ) is compact
we invoke i) and take K x to be the corresponding compact set in Ω. Then
for any v ∈ C (Ω) with supp v ⊂ K y , we have supp(Av) ⊂ K x . Hence, for
∞
0
∞
every ψ ∈ C (Ω \ K x ),
0

K A (x, y)v(y)dyψ(x)dx = Av, ψ =0 .
Ω Ω
As a consequence, we find supp K A ∩ (Ω × K y ) = supp K A ∩ (K x × K y )
⊂ K x × K y ; so, supp K A ∩ (Ω × K y ) is compact. Thus, K A is properly
supported.

Lemma 6.1.4 implies the following corollary.
Corollary 6.1.5. The operator A : C (Ω) → C ∞ is properly supported if
∞
0
and only if the following two conditions hold:
i) For any compact subset K y Ω there exists a compact K x Ω
such that
∞
∞
A : C (K y ) → C (K x ) is continuous. (6.1.21)
0 0
ii) For any compact subset K x Ω there exists a compact K y Ω
such that
∞
A : C (K x ) → C (K y ) is continuous. (6.1.22)
∞
0 0
∞
Proof: Since A : C (Ω) → C (Ω) is continuous, i) in Lemma 6.1.4 is
∞
0
equivalent to (6.1.21).
Now, if A is properly supported then K (x, y)= K A (y, x)isthe

A
Schwartz kernel of A and, hence, also properly supported. Therefore i)
in Lemma 6.1.4 is valid for K A (x, y)= K (y, x) which, for this case, is

A