Real and Complex Analysis (Rudin)

HOLOMORPHIC FOURIER TRANSFORMS 379
Conversely, supposefis defined on the real axis andfe C{n!}. In other
words,
We claim that the representation
(n = 0, 1, 2, ... ). (2)
f(x) = f (Dnf)(a) (x - a)n
n=O n!
(3)
is valid for all a e R 1 if a -
formula
B- 1 < X < a + B- 1. This follows from Taylor's
f(x) = ~f (D~~(a) (x - aY + ( ~ 1)' rx(x - tt- 1(Dnf)(t) dt, (4)
}=o J. n. Ja
which one obtains by repeated integrations by part. By (2) the last term in (4)
(the" remainder") is dominated by
nPBnlix(X - t)n-1 dtl = PIB(x - a) In. (5)
If I B(x - a) I < 1, this tends to 0 as n- 00, and (3) follows.
We can now replace x in (3) by any complex number z such that
I z - a I < liB. This defines a holomorphic function Fa in the disc with center
at a and radius liB, and Fix) = f(x) if x is real and I x - a I < liB. The
various functions Fa are therefore analytic continuations of each other; they
form a holomorphic extension F off in the strip I y I < II B.
If 0 < J < liB and z = a + iy, I y I < J, then
I F(z) I = I Fiz) I = I Jo (D~~(a) (iyt I ~ Pnt (BJt = 1 ! BJ'
This shows that F is bounded in the strip I y I < J, and the proof is complete.
IIII
19.10 Theorem The class C{Mn} is quasi-analytic if and only if C{Mn} contains
no nontrivial function with compact support.
PROOF If C{Mn} is quasi-analytic, iff e C{Mn}, and iffhas compact support,
then evidently f and all its derivatives vanish at some point, hence f(x) = 0
for all x.
Suppose C{Mn} is not quasi-analytic. Then there exists an fe C{Mn}
such that (D'1)(O) = 0 for n = 0, 1,2, ..., butf(xo) =I- 0 for some Xo' We may
assume Xo > O. If g(x) = f(x) for x ~ 0 and g(x) = 0 for x < 0, then g e
C{Mn}. Put h(x) = g(x)g(2xo - x). By Theorem 19.7, he C{Mn}. Also,
h(x) = 0 if x < 0 and if x > 2xo. But h(xo) = P(xo) =I- O. Thus h is a nontrivial
member of C{Mn} with compact support.
IIII