Sequences, Series, and Special Functions 523Proof We know (Theorem 4.19) that the series converges uniformly to F(z)on lzl ::; r1 where 0::; r1 < r, hence on every compact subset of lzl < r.
Since the terms fn(z) = anzn are analytic functions on lzl ::; r1, it follows
from Corollary 8.2 that the function F(z) defined by the power series is
analytic in lzl < r.
Theorem 8.2 (Weierstrass). Suppose that:
1. The functions Fn(z) (n = 1, 2, ... ) are analytic in some open set A.
2. Fn(z) ~ F(z) for all z EA as z ~ oo, the convergence being uniform
on every compact subset B of A.Then:- F( z) is analytic in A.
- F~k)(z) ~ p(k)(z) as n ~ oo for every z E A, the convergence being
uniform on compact subsets of A.
Proof Let z EA. Since A is open, there exists a neighborhood N(z) CA,
and also a closed disk D(z), such that D(z) C N(z) CA (Fig. 8.1). Since
D(z) is compact, we have Fn(() ~ F(() for all (ED. Hence we have also
Fn(() ~ F(() for all (in the open disk D(z), which is a simply connected
region. Thus Corollary 8.3 applies, so that F( () is analytic in D( z) and, in
particular, at z. Since z is arbitrary in A, it follows that F(z) is analytic
in A.
Let C be the boundary of D(z). On C we have also Fn(() ~ F((). By
formula (7.21-2) for the derivatives, we have
C;AFig. 8.1p~k)(z) = kl J
27l'i
c+(8.1-2)