Real and Complex Analysis (Rudin)

ZEROS OF HOLOMORPHIC FUNCTIONS 313
the closed interval 1= [0, 1], with the supremum norm. In other words, the set of
all finite linear combinations of the functions
is dense in C(I). This is sometimes expressed by saying that the functions (1) span
C(I).
This suggests a question: If 0 < At < A2 < A3 < ... , under what conditions is
it true that the functions
span C(l)?
It turns out that this problem has a very natural connection with the
problem of the distribution of the zeros of a bounded holomorphic function in a
half plane (or in a disc; the two are conform ally equivalent). The surprisingly neat
answer is that the functions (2) span C(I) if and only if l:.1/ An = 00.
Actually, the proof gives an even more precise conclusion:
15.26 Theorem Suppose 0 < At < A2 < A3 < . .. qnd let X be the closure in
C(l) of the set of all finite linear combinations of the functions
(1)
(2)
(a) Ifl:.l/A n = 00, then X = C(I).
(b) If l:.1/ An < 00, and if A ¢ {An}, A #: 0, then X does not contain the function
e'.
PROOF It is a consequence of the Hahn-Banach theorem (Theorem 5.19) that
qJ E C(l) but qJ ¢ X if and only if there is a bounded linear functional on C(I)
which does not vanish at qJ but which vanishes on all of X. Since every
bounded linear functional on C(l) is given by integration with respect to a
complex Borel measure on I, (a) will be a consequence of the following proposition:
Ifl:.l/ An = 00 and if J.l is a complex Borel measure on I such that
Ie .. dJ.l(t) = 0 (n = 1, 2, 3, ... ), (1)
then also
(k = 1, 2, 3, ... ). (2)
For if this is proved, the preceding remark shows that X contains all
functions t'; since 1 E X, all polynomials are then in X, and the Weierstrass
theorem therefore implies that X = C(I).