Re: Fourier Transform, Smooth Functions
- From: "david petry" <david_lawrence_petry@xxxxxxxxx>
- Date: 26 Feb 2006 17:13:31 -0800
irchans wrote:
I have two questions about the relationship between the smoothness of a
function and its Fourier transform. I know that if a function is very
smooth, then its Fourier transform can decay faster than 1/s^p for any
p>0. But, can its Fourier transform decay faster than that if the
function is zero outside of [-1,1]? More precisely,
1) Does there exist a function f(x) such that f is continuous, f(x) is
zero when abs(x) > 1, and its Fourier transform, g(s), is order exp( -
abs(s)^p) where p > 1/2?
2) What is the largest value of p such that there exists a function f
such that f is continuous, f(x) is zero when abs(x) > 1, and its
Fourier transform, g(s), is order exp( - abs(s)^p)?
Here's a way to produce functions f(x) such that g(s) is especially
rapidly decreasing.
Let {a_k} be an infinite sequence of positive reals such that
sum(a_k) < oo, and let it be such that the sum converges very very
slowly. Then look at the function g(s) = product sin(a_k s)/(a_k s).
Let f(x) be the Fourier transform of g(s).
Then f(x) will have compact support, and g(s) will decay faster than
1/s^p for all p. Now you want to find something that decays even
faster. I contend, without giving a proof, that if the series {a_k} is
well chosen and the sum converges slowly enough, then we can ensure
that g(s) will decay more rapidly than exp( -s^m ) for any particular m
< 1.
Unfortunately, I'm not absolutely certain I'm right. I remember looking
at this problem years ago, and the above is how I remember solving it.
.
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