Fourier Transform, Smooth Functions
- From: "irchans" <infinitgames@xxxxxxxxx>
- Date: 25 Feb 2006 08:22:58 -0800
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)?
If you understand the questions, then you can stop reading here and
reply with your thoughts. For everyone else, I will try to define the
terms in those questions.
Given a function f(x) that maps reals to reals. Assume that f(x) has
the following properties:
1) f(x) is continuous,
2) f(x) is 0 when x<-1 or x>1, and
3) 0<= f(x) <= 1 for all x.
Define the Fourier transform of f, to be the function g that maps reals
into complex numbers with the formula
g(s) = Integral[ f(x)* exp( - 2 * pi * i * x * s), {x, -Infinity,
Infinity}].
Define the "set of functions of order h(s)", denoted O( h(s) ), to be
the set of all complex valued functions w(s) with one real argument
such that there exists a real constant c obeying
abs( w(s) ) < c h(s) for all s.
How quickly can g(s) decay?
Or more precisely, does there exist a function f(x) obeying the
assumptions 1 - 3, such that its Fourier transform g(s) is contained in
the set O( exp( - abs(s)^p ) ) for p = 1/4? How about for p=1/2, 1, or
2?
I don't know the answer, but I suspect that the answer is yes for p<0.5
and no for p>1.
Cheers,
Irchans
.
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