Re: Fourier Transform, Smooth Functions
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Sat, 25 Feb 2006 16:06:41 -0600
On 25 Feb 2006 13:04:50 -0800, Han.deBruijn@xxxxxxxxxxxxxx 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]? [ ... snip ... ]
I can only guess what you really want,
That's bcause you didn't read what he wrote - he stated _very_
clearly exactly what he wanted. He wants to know the largest
p such that there exists a (non-zero) function f with compact
support, such that the fourier transform g(s) is bounded
by exp(-|s|^p).
but here is my two cents worth.
A Fourier transform can even decay infinitely fast. For example, if the
original function is a simple constant, then the Fourier transform is a
delta function. You can see this if you determine the Fourier transform
of a rectangle: f(x) = 0 for x < -a/2, f(x) = 1 for -a/2 < x < +a/2 and
f(x) = 0 for x > +a/2. Then F(s) = a/pi.sinc(a/2.s) [sinc(x) = sin(x)/x
for x <> 0 and sinc(0) = 1] The integral F(s).ds from -oo to +oo equals
= 1 independent of a . Consider the case where a becomes large -> oo.
Then the rectangle f(x) becomes a constant 1 and the Fourier transform
F(s) becomes the delta function (iff I made no mistakes). The reverse
is also true. Therefore a "wide" function in the original domain causes
a "narrow" function in the Fourier domein and vice versa.
Restricting your original f(x) to an interval [-1,+1] makes its Fourier
transform F(s) less decaying. There is a general theorem in the theory
which is very much alike (in fact it _is_) the Heisenberg uncertainity
principle. It sounds like the following. Form the variance dx^2 of the
original function as the integral (-oo,+oo) x^2 f(x) dx , the variance
ds^2 of the Fourier transform as integral (-oo,+oo) s^2 F(s) ds / (2Pi)
apart from norming constants. Then there is the theorem: dt.ds > 1/2 ,
It's very easy to see that this is false (even if we add absolute
value signs to your definitions, integral (-oo,+oo) x^2 |f(x)| dx
and similarly for the other one, so that we're at least
talking about _positive_ quantities. For example, say
dx dt = H(f). If g(x) = f(2x) then H(g) = H(f)/sqrt(2).
Repeat as needed until you get a function with H < 1/2.
Why not look up the correct statement before posting it?
Which is equivalent to Heisenberg's uncertainity principle in quantum
mechanics (!) Here the limiting value dt.ds = 1/2 is reached if we take
f(x) = exp(-(x/sigma)^2/2)/(sigma.sqrt(2.pi)), F(s) = exp(-(s.sigma)^2)
a combination which is "optimal" in the sense that dx.ds is minimal.
Hope this helps.
Han de Bruijn
************************
David C. Ullrich
.
- Follow-Ups:
- Re: Fourier Transform, Smooth Functions
- From: Han . deBruijn
- Re: Fourier Transform, Smooth Functions
- References:
- Fourier Transform, Smooth Functions
- From: irchans
- Re: Fourier Transform, Smooth Functions
- From: Han . deBruijn
- Fourier Transform, Smooth Functions
- Prev by Date: Re: Peano's Axioms from the Field Axioms alone?
- Next by Date: Steiner tree (2)
- Previous by thread: Re: Fourier Transform, Smooth Functions
- Next by thread: Re: Fourier Transform, Smooth Functions
- Index(es):
Relevant Pages
|
