Real and Complex Analysis (Rudin)

104 REAL AND COMPLEX ANALYSIS
Let Co be the space of all complex functions (() on Z such that (()(n)-4 0 as
n-4 ± 00, with the supremum norm
11(()1100 = sup {I (()(n) I: n E Z}. (3)
Then Co is easily seen to be a Banach space. In fact, if we declare every subset of
Z to be open, then Z is a locally compact Hausdorff space, and Co is nothing but
Co(Z)·
The following theorem contains the answer to our question:
5.15 Theorem The mapping f -41 is a one-to-one bounded linear transformation
of E(T) into (but not onto) co.
PROOF Define A by Af =! It is clear that A is linear. We have just proved
that A maps E(T) into Co, and formula 5.14(1) shows that I!(n) I ~ Ilflll' so
that IIAII ~ 1. (Actually, IIAII = 1; to see this, take f = 1.) Let us now prove
that A is one-to-one. Supposef E E(T) and!(n) = 0 for every n E Z. Then
f/(t)g(t) dt = 0 (1)
if g is any trigonometric polynomial. By Theorem 4.25 and the dominated
convergence theorem, (1) holds for every g E C(T). Apply the dominated convergence
theorem once more, in conjunction with the Corollary to Lusin's
theorem, to conclude that (1) holds if g is the characteristic function of any
measurable set in T. Now Theorem 1.39(b) shows thatf = 0 a.e.
lf the range of A were all of co, Theorem 5.10 would imply the existence
of a 0 > 0 such that
1111100 ~ ollflll (2)
for every f E E(T). But if D.(t) is defined as in Sec. 5.11, then D. E E(T),
1115.1100 = 1 for n = 1,2,3, ... , and IID.lll-4 00 as n-4 00. Hence there is no
o > 0 such that the inequalities
(3)
hold for every n.
This completes the proof.
IIII
The Hahn-Banach Theorem
5.16 Theorem If M is a subspace of a normed linear space X and iff is a
bounded linear functional on M, then f can be extended to a bounded linear
functional F on X so that II F II = II f II.
Note that M need not be closed.