Real and Complex Analysis (Rudin)

312 REAL AND COMPLEX ANALYSIS
Our assumption that fEN is equivalent to the existence of a constant
C < 00 which exceeds the right side of (3) for all r, 0 < r < 1. It follows that
k
n I IX" I ~ C- 1 1
n=1
The inequality persists, for every k, as r -4 1. Hence
f(O) I ri. (4)
00
n I IX" I ~ C- 1 I f(O) I > O. (5)
"=1
By Theorem 15.5, (5) implies (1).
Corollary Iff E H OO (or even iff EN), if 1Xl> 1X2' 1X3' ••• are the zeros off in U,
and if I;(1 - I IX" I) = 00, then f(z) = 0 for all z E U.
For instance, no nonconstant bounded holomorphic function in U can have
a zero at each of the points (n - 1)/n (n = 1, 2, 3, ... ).
We conclude this section with a theorem which describes the behavior of a
Blaschke product near the boundary of U. Recall that as a member of H oo , B has
radial limits B*(ei~ at almost all points of T.
IIII
15.24 Theorem If B is a Blaschke product, then I B*(ei~ I = 1 a.e. and
lim -2 1 I" log I B(rei~ I ~O = O.
r-l 1l -x
(1)
PROOF The existence of the limit is a consequence of the fact that the integral
is a monotonic function of r. Suppose B(z) is as in Theorem 15.21, and put
BN(Z) = Ii IX" ~ Z • ~
"=N 1 -IX"Z IXn
Since log (I BIBN I) is continuous in an open set containing T, the limit (1) is
unchanged if B is replaced by BN. If we apply Theorem 15.19 to BN we
therefore obtain
log I BJO) I ~ !~n: 2n 1 I" _"tog I B(rei~ I dO ~ 2n 1 I" _"tog I B*(e I8 ) I dO ~ O. (3)
As N -400, the first term in (3) tends to O. This gives (1), and shows that
J log I B* 1= O. Since log I B* I ~ 0 a.e., Theorem l.39(a) now implies that
log I B* I = 0 a.e.
IIII
The Miintz-Szasz Theorem
15.25 A classical theorem of Weierstrass ([26], Theorem 7.26) states that the
polynomials are dense in C(I), the space of all continuous complex functions on
(2)