Re: Fourier Transform, Smooth Functions

On 26 Feb 2006 17:13:31 -0800, "david petry"
<david_lawrence_petry@xxxxxxxxx> wrote:

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 f will be infinitely differentiable.

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.

I believe this may very well be right. Here's a sloppy version of
a proof (various statements of the form a ~ b would need to be
converted to explicit inequalities to make an actual proof):

First note that there exists c > 0 such that if |t| < 1 then

(i) sin(t)/t <= exp(-c t^2).

(Note that "c" below will denote various constants,
differing from line to line.)

Now suppose that m < 1. Choose eps > 0 so that

(ii) 2 - m > (1 + 2 eps)/(1 + eps)

(such an eps > 0 exists since 2 - m > 1).

Let a_k = 1/k^(1+eps).

(Note that the argument must use the fact that m < 1, since
m = 1 is impossible, as I showed in another post. Here we
see where m < 1 is used; we need sum a_k < infinity to give
f compact support, and that requires eps > 0, hence requires
m < 1. The fact that we can see where m < 1 is used seems to
me to lend some credibility...)

Suppose that s is large. Choose N so that s ~ 1/a_N. Now
(i) shows that

(iii) g(s) <= exp(-c sum_{k=N}^infinity a_k^2 s^2).

Our choice of a_k shows that

sum_{k=N}^infinity a_k^2 ~ 1/N^(1 + 2eps)

= a_N ^ ((1 + 2 eps)/(1 + eps))

So (ii) gives

sum_{k=N}^infinity a_k^2 > c a_N ^ (2 - m) ~ s^(m-2),

so

sum_{k=N}^infinity a_k^2 s^2 > c s^m,

hence (iii) shows

g(s) <= exp(-c s^m).

Keen.

************************

David C. Ullrich
.

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