524 Chapter 8
Since the integrand in (8.1-2) is a continuous function of ( along C, by
applying Theorem 8.1 we get
lim F(k)(z) = ~ J limn-+oo Fn(() d(
n-+O n 27ri (( - z)k+^1
c+
= ~ J F(() d( = F(k)(z) (8.1-3)
27ri (( - z)k+I
c+
To prove that the convergence in (8.1-3) is uniform on compact subsets
of A, let B C A be compact, and let p = d(B, 8A). For every z E B
consider the open disk D(z) = {(: IC-zl <r}, where r = p/3 if pis
finite. Otherwise, choose for r any positive real number (e.g., r = 1).
Then G = {D(z)lzeB is an open covering of B and, by the definition of
compactness, there exists a finite subcovering G1 = {D(zi)}:, 1.
The convergence of F~k\z) to F(k)(z) is uniform on each open disk
D(zi)· In fact, letting Ci: ( - Zi = 2reit, 0 :::; t :::; 271", for any given€> 0
choose 0 < €^1 < rkE/2k!. Then there is a positive number Ni(E') such that
n > N;( E^1 ) implies
!Fn(() -F(()! < €^1
for all (EC; (since Ci is compact and contained in A.) Hence if z E D(z;)
we have, for n > N;,
k! €^1 2k!€^1
:::; -2 k+1 (47rr) = -k-< €
7r r r
(8.1-4)
for all z E D(zi)· Now by taking N = max(N1,N2, ... ,Nm) we see that
the inequality (8.1-4) holds for all z EB provided that n > N.
Corollary 8.5 Suppose that:
1. The functions fn(z) are analytic in some open set A (n = 1, 2, ... ).
2. L::::'=l f n(z) = F(z) for all z E A, the convergence being uniform on
every compact subset of A.
Then:
- F( z) is analytic in A.
- F(k)(z) = L::::'=i Jik)(z) for z E A, the convergence being uniform on
compact subsets of A.
