6 1. Selected Problems in One Complex Variable
Ai be the ball with the same center as Bi but with half the radius. Clearly,
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\
^i = ( l - / i ) ( l - /
) . . . ( 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 - /
) (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.
In what follows,
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
1.3.3 Theorem .
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
provided it has a solution in
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
as is
the sum F = ^2faF{. Furthermore,
GF = G^2ptFt =
= X # = H.
i i i
Thus, F is a global solution to our equation.
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