6 1. Selected Problems in One Complex Variable

Ai be the ball with the same center as Bi but with half the radius. Clearly,

UZAZ = U.

By Lemma 1.3.1, there are functions fi with 0 f%{x) 1 for all x,

fi(x) = 1 on 1^, and fi(x) — 0 on the complement of B{. We define fa — f\

and

^i = ( l - / i ) ( l - /

2

) . . . ( l - / i - i ) / i for i\.

Note that ^ = 0 on the complement of B{. Also, an induction argument

shows that

1 - (Pl + f2 + • • • + Pn) = (1 - / l ) ( l - /

2

) • • • (1 ~ /„)

and so

fa + fa + • • • + (fn = 1 on A\ U A2 U • • • U An.

Given a compact subset K of [/, there is an n so that K C A i U ^ U - • -UAn.

It follows that, for i n, fa vanishes identically on K. Thus, the infinite

sum J2 4i makes sense and is identically 1 on (7.

i

In what follows,

C°°(U)P

will denote the space of vectors of length p

with entries from C°°(U). We consider C°°([/)p a module over the algebra

C°°(U) through the usual coordinate-wise operations of addition and scalar

multiplication.

1.3.3 Theorem .

IfUcW2

is an open set, G is apxq matrix with entries

from C°°(U), and H e C°°(U)P, then the equation

GF = H

has a solution F G

C°°(U)q,

provided it has a solution in

C°°(V)q

for some

neighborhood V of each point of U.

Proof. We choose an open cover V of U by sets on which the equation has

a solution. We then choose a partition of unity {fa} subordinate to this

cover. Thus, each fa vanishes off a compact subset of a set Vi in V and there

is an F% e C°°(Vi)q such that GF{ = H on V-. If we define faF{ to be 0 in

the complement of V, then the result is a vector which is in

C°°(U)q,

as is

the sum F = ^2faF{. Furthermore,

i

GF = G^2ptFt =

Y;&GFi

= X # = H.

i i i

Thus, F is a global solution to our equation.