From: israel@math.ubc.ca (Robert Israel) Subject: Re: A simple analysis question Date: 10 May 2000 23:00:54 GMT Newsgroups: sci.math Summary: [missing] In article , David Petry wrote: >Given a sequence a_0, a_1, a_2 ... of complex numbers >such that sum | a_k | diverges, is it necessarily true that >the function f(x) = sum a_k*x^k is unbounded on the >unit disk? No, it isn't. Example: f(x) = exp((x-1)/(x+1)). Note that if sum |a_k| converged, f would have to be continuous on the unit circle (but of course it isn't continuous near x=-1). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: ullrich@math.okstate.edu Subject: Re: A simple analysis question Date: Thu, 11 May 2000 19:19:07 GMT Newsgroups: sci.math In article , "David Petry" wrote: > > ullrich@math.okstate.edu wrote in message <8felkr$93b$1@nnrp1.deja.com>... > >In article , > > "David Petry" wrote: > >> > >> Given a sequence a_0, a_1, a_2 ... of complex numbers > >> such that sum | a_k | diverges, is it necessarily true that > >> the function f(x) = sum a_k*x^k is unbounded on the > >> unit disk? > > > There are also > >counterexamples continuous on the closed disk, and in fact > >examples where the given series converges _uniformly_ on > >the closed disk, even though sum |a_k| diverges. > > I'd like to see explicitly such a counterexample. Ah, a challenge. I don't know if there's a one-liner, unless of course you use Clifford algebras. According to Zygmund (Trigonometric Series, Chapter 5 Section 4) the series sum(a_n e^(int)) is uniformly convergent if a_n = e^(i*c*n*log(n)) * n^-(alpha + 1/2) with alpha > 0. If alpha < 1/2 then the sum of |a_n| is infinite. There's your explicit counterexample. The proof is in Zygmund. There's a simpler example you could try to verify yourself, although it's not _quite_ what you asked for, being a "sine series" instead of an "analytic" series: If e_n -> 0 then the series sum((e_n/n) * sin(nt)) converges uniformly. You show this by showing that the sum for N < n < M tends to zero uniformly as N, M -> infinity; this you can do by summation by parts. (You write it as e_n * (sin(nt)/n), you sum the sin(nt)/n and "differentiate" the e_n. It's well-known that the partial sums of sin(nt)/n are uniformly bounded, which makes it all work. It's clear that the partial sums of e^(int)/n are not uniformly bounded, so you can't use exactly the same argument in exactly the situation you asked about.) Sent via Deja.com http://www.deja.com/ Before you buy.