Re: Fourier Transform, Smooth Functions
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Sat, 25 Feb 2006 11:45:39 -0600
On 25 Feb 2006 08:22:58 -0800, "irchans" <infinitgames@xxxxxxxxx>
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)?
I'm not sure. My first guess was that p = 0 was the largest possible,
but a proof of that that I expected should work does not work.
It's easy to see that p = 1 is not possible: If p = 1 then it's
easy to see that f must extend to a function holomorphic in a
horizontal strip containing the real axis, and hence f (restricted
to the line) cannot have compact support, unless of course f = 0.
That's so easy that it seems like a stronger result must be
available with more work. I had a thought - perhaps the
Denjoy-Carleman theorem (see for example "quasi-analytic
classes" in Rudin "Real and Complex Analysis") would rule
out the case 0 < p < 1. But it doesn't, if I did the
calculations correctly: If you assume that g satisfies the
given bound that gives a bound on the size of the n-th
derivative of f. A change of variables shows that D^n f
is bounded by something like Gamma(n/p). But if p < 1 then
Stirling's formula shows that sum (1/Gamma(n/p)^(1/n) is
finite, so the class of f satisfying those bounds on the
derivatives is not quasi-analytic.
So I'm really not sure. I should know this... My guess
is that the best possible p is either 0 (that still seems
right to me, although I can't say why) or 1 - epsilon,
as the D-C theorem would tend to be indicating.
(You should check my work - if I made an error in the
calculations then the D-C theorem does show that p = 0
is the largest possible. On the other hand it seems
possible that I did the calculations correctly, because
what I did with the D-C theorem _does_ give a harder
proof that the case p = 1 is impossible...)
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
************************
David C. Ullrich
.
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