Real and Complex Analysis (Rudin)

ZEROS OF HOLOMORPHIC FUNCTIONS 315
interval from - 1 + iR to - 1 - iR. The integral over the semicircle tends to
o as R --400, so we are left with
But
f(z) = - ~ foo
f( -1 .+ is) ds
2n _ 00 - 1 + IS - Z
1 = (1 tz-is dt
1 + z - is Jo
Hence (6) can be rewritten in the form
(Re z > -1).
(Re z > -1).
f(z) = (\z{~ foo f( -1 + is)e-iS log r dS} dt.
Jo 2n - 00
The interchange in the order of integration was legitimate: If the integrand in
(8) is replaced by its absolute value, a finite integral results.
Put g(s) = f( -1 + is). Then the inner integral in (8) is.g(log t), where 9 is
the Fourier transform of g. This is a bounded continuous function on (0, 1],
and if we set dJl(t) = g(log t) dt we obtain a measure which represents f in the
desired form (3).
This completes the proof.
15.27 Remark The theorem implies that whenever {I, tAl, tA2, ... } spans C(J),
then some infinite subcollection of the tAi can be removed without altering
the span. In particular, C(J) contains no minimal spanning sets of this type.
This is in marked contrast to the behavior of orthonormal sets in a Hilbert
space: if any element is removed from an orthonormal set, its span is diminished.
Likewise, if {I, tAl, t A2 ... } does not span C(J), removal of any of its
elements will diminish the span; this follows from Theorem 15.26(b).
(6)
(7)
(8)
IIII
Exercises
I Suppose {a.} and {b.} are sequences of complex numbers such that l: I a. - b.1 < 00. On what sets
will the product
fI z - an
.=1 z-b.
converge uniformly? Where will it define a holomorphic function?
2 Suppose/is entire, A is a positive number, and the inequality
I/(z)l < exp (lzIA)
holds for all large enough I z I. (Such functions / are said to be of finite order. The greatest lower
bound of all A for which the above condition holds is the order off,) If/(z) = l:a.z·, prove that the
inequality
(d)'IA
la.l::;; - n