Real and Complex Analysis (Rudin)

ANALYTIC CONTINUATION 321
integers. These points form a dense subset of T; and since the set of all
singular points offis closed,fhas T as its natural boundary.
That this example is a power series with large gaps (i.e., with many zero
coefficients) is no accident. The example is merely a special case of Theorem
16.6, due to Hadamard, which we shall derive from the following theore,tU of
Ostrowski:
'
16.5 Theorem Suppose A., Pk' and qk are positive integers,
and
Suppose
PI < P2 < P3 < ... ,
(k = 1, 2, 3, ... ). (1)
co
f(z) = L anzn (2)
"=0
has radius of convergence 1, and an = 0 whenever Pk < n < qk for some k. If
sp(z) is the pth partial sum of (2), and if P is a regular point off on T, then the
sequence {Spk(Z)} converges in some neighborhood of p.
Note that the full sequence {sp(z)} cannot converge at any point outside U.
The gap condition (1) ensures the existence of a subsequence which converges in
a neighborhood of p, hence at some points outside U. This phenomenon is called
overconvergence.
PROOF If g(z) = f(pz), then g also satisfies the gap condition. Hence we may
assume, without loss of generality, that P = 1. Then f has a hoi om orphic
extension to a region Q which contains U u {I}. Put
and define F(w) = f(cp(w» for all w such that cp(w) E Q. If I wi:$; 1 but w :F 1,
then I cp(w) I < 1, since 11 + wi < 2. Also, cp(l) = 1. It follows that there exists
an E > 0 such that cp(D(O; 1 + E» c: Q. Note that the region cp(D(O; 1 + E»
contains the point 1. The series
converges if I wi < 1 + E.
(3)
co
F(w) = L bmwm (4)
m=O