From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Complex analysis
Date: 17 Oct 1999 20:58:05 GMT
Newsgroups: sci.math
Keywords: series of analytic functions

[deletia --djr]

The answer to Eric Brunet's question about asymptotic series is no.
That is, there are functions F(z,c) which are entire functions
of z (i.e. analytic on the complex plane C) for all c > 0, have 
asymptotic series sum_n a_n(z) c^n as c -> 0+, but a_n are not all
entire.   
Here are two examples.

1) Let a_0(z) = 1 for Im z > 0, 0 for Im z <= 0,
and a_n(z) = 0 for n >= 1.
Using Runge's Theorem, you can show that there
is a sequence of polynomials p_j(z) such that
for each z, |p_j(z) - a_0(z)| < 2^(-j) for j 
sufficiently large (of course, what is "sufficiently
large" depends on z).  Let F(z,c) = 
sum_{j=1}^infinity exp(-(j-1/c)^2) p_j(z) /
 sum_{j=1}^infinity exp(-(j-1/c)^2).
Then F(z,c) is analytic in a neighbourhood of 
C x (0,infinity), and has the asymptotic series
sum_{n=0}^infinity a_n(z) c^n as c -> 0+ since
|F(z,c) - a_0(z)| = O(2^(-1/c)) = O(c^n)  as c -> 0+
for all n. 

  
2) The entire function 
g(z) = int_0^infinity exp(z t)/t^t dt
has the property that |g(z)| <= 1/(|Im z| - pi/2)
for |Im z| > pi/2: see
D.J. Newman, Amer. Math. Monthly 83 (1976) 192-193.
Take F(z,c) = g(i z^2 exp(2/c) + z exp(1/c)), 
which is analytic on C x (C \ {0}).
For z <> 0 we have |F(z,c)| = O(exp(-1/c)) = O(c^n)
as c -> 0+, while F(0,c) = g(0) > 0.  Thus
F(z,c) has the asymptotic series 
sum_{n=0}^infinity a_n(z) c^n as c -> 0+, where
a_0(0) = g(0), a_n(z) = 0 otherwise.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2