Real and Complex Analysis (Rudin)

368 REAL AND COMPLEX ANALYSIS
and/(ei~ =1= O/or every real lJ. Then
(2)
PROOF We let A be the space of all complex functions / on the unit circle
which satisfy (1), with the norm
-00
(3)
It is clear that A is a Banach space. In fact, A is isometrically isomorphic to
tl, the space of all complex functions on the integers which are integrable
with respect to the counting measure. But A is also a commutative Banach
algebra, under pointwise multiplication. For if g e A and g(e~ = r.b ll eiIl8,
then
(4)
and hence
II/gil = L I L CII-kbk I ~ L Ibkl L ICII-kl = II/II . IlglI· (5)
II k k II
Also, the function 1 is the unit of A, and 11111 = 1.
Put/o(ei~ = ei8, as before. Then/o e A, 1//0 e A, and II/all = 1 for n = 0,
± 1, ± 2, .... If h is any complex homomorphism of A and h(/o) = A, the fact
that IIhll :s; 1 implies that
(n = 0, ± 1, ± 2, ... ). (6)
Hence 1,1.1 = 1. In other words, to each h corresponds a point e ilZ e T such
that h(/o) = e ilZ , so
(n = 0, ± 1, ± 2, ... ). (7)
If/is given by (1), then/= r.clI/o. This series converges in A; and since
h is a continuous linear functional on A, we conclude from (7) that
(fe A). (8)
Our hypothesis that / vanishes at no point of T says therefore that / is
not in the kernel of any complex homomorphism of A, and now Theorem
18.17 implies that/is invertible in A. But this is precisely what the theorem
asserts. / / / /