Real and Complex Analysis (Rudin)

EXAMPLES OF BANACH SPACE TECHNIQUES 103
This gains in interest if we realize that E, as well as each Qf' is an uncountable
set:
5.13 Theorem In a complete metric space X which has no isolated points, no
countable dense set is a G.,.
PROOF Let x k be the points of a countable dense set E in X. Assume that E is
a G.,. Then E = n v.. , where each v.. is dense and open. Let
"
w" = v.. - U {x k }·
k=l
Then each w" is still a dense open set, but n w" = 0, in contradiction to
Baire's theorem.
IIII
Note: A slight change in the proof of Baire's theorem shows actually that
every dense G., contains a perfect set if X is as above.
Fourier Coefficients of I1-functions
5.14 As in Sec. 4.26, we associate to every IE LI(T) a function! on Z defined by
!(n) = -2 I(t)e-i"t dt
n -"
1 I"
(n E Z). (1)
It is easy to prove that!(n)--+ 0 as I nl--+ 00, for every IE E. For we know that
C(T) is dense in LI(T) (Theorem 3.14) and that the trigonometric polynomials are
dense in C(T) (Theorem 4.25). If E > 0 and IE LI(T), this says that there is a
g E C(T) and a trigonometric polynomial P such that III - gill < E and
Ilg - PII co < E. Since
if follows that III - Pill < 2E; and if I n I is large enough (depending on P), then
I !(n) I = I ;n f" {f(t) - P(t)}e-i"t dt I ~ III - Pill < 2E. (2)
Thus!(n)--+ 0 as n--+ ± 00. This is known as the Riemann-Lebesgue lemma.
The question we wish to raise is whether the converse is true. That is to say,
if {a,,} is a sequence of complex numbers such that a,,--+ 0 as n--+ ± 00, does it
follow that there is ani E LI(T) such that!(n) = a" for all n E Z? In other words,
is something like the Riesz-Fischer theorem true in this situation?
This can easily be answered (negatively) with the aid of the open mapping
theorem.