From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Convergence (Fourier Transform Related)
Date: 12 Feb 2001 19:41:52 GMT
Newsgroups: sci.math.research
Summary: Fourier transform need not preserve uniform convergence

In article <969726$55js$1@rook.le.ac.uk>,
Rob Brownlee <rab14@leicester.ac.uk> wrote:
>I have a sequence w_k of real valued functions that I have shown converges
>uniformly on compact sets to a real valued function w. In fact, I know that
>w_k = 1/((f_k)^) and w=1/(f^) for some functions f_k and f. Here, ^ denotes
>the Fourier transform of f. Can I conlcude that (f_k)^ converges to f^
>uniformly on compact sets? Does it now follow that f_k converges to f
>uniformly on compact sets?

If (f_k)^ and f^ are continuous (which is the case if f_k and f are in 
L^1), then (f_k)^ -> f^ uniformly on compact sets.  I'm assuming that by
saying w = 1/f^ you imply that f^ is never 0.  Then on any compact set K,
f^ is bounded above and below (say 0 < m <= |f^| <= M), and then since 
1/m >= |w| >= 1/M,  for k sufficiently large 2/m >= |w_k| >= 1/(2M).
Since the function t -> 1/t is a homeomorphism of [1/(2M), 2/m] onto
[m/2, 2M], uniform convergence of w_k to w on K implies uniform 
convergence of (f_k)^ to f^ there.
  
However, an L^1 function is not even defined pointwise, so no similar
conclusion can be drawn about the convergence of f_k to f.  In fact, 
it is not even true that f_k -> f in L^1.  For example, take
f(x) = 2 exp(-pi x^2) and f_k(x) = exp(-pi x^2)(2 + cos(pi k^2 x^2)).
Then f_k does not converge to f in L^1 (or pointwise, or in any L^p
space), but f^ - f_k^ is the convolution f^ * phi_k^ where
phi_k(x) = (1/2) cos(pi k^2 x^2), and  
phi_k^(t) = (const/k) (sin(pi k^2 t^2) + cos(pi k^2 t^2)) -> 0 uniformly
as k -> infinity, so |f^ - f_k^| -> 0 uniformly.

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