Register to reply

Question Fourier Transform Smoothness/Compactness

by mnb96
Tags: fourier, transform
Share this thread:
mnb96
#1
Feb4-11, 04:07 PM
P: 625
Hello,
my question arises from reading the section on Smoothness/Compactness from Bracewell's "The Fourier Transform and Its Applications" page 162.

I don't quite understand the following reasoning:

[tex]F(\omega) = \ldots = \frac{1}{i\omega}\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx[/tex]

and at this point the author says that when [itex]\omega\to\infty[/itex] then [itex]\omega F(\omega) \to 0[/itex].

But why [itex]\omega F(\omega)[/itex] is supposed to tend to zero, and not just [itex]F(\omega) \to 0[/itex] ?

Thanks.
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
mathman
#2
Feb4-11, 04:52 PM
Sci Advisor
P: 6,039
ωF(ω) -> 0 is a stronger statement than F(ω) -> 0 for ω -> ∞.
mnb96
#3
Feb5-11, 01:25 AM
P: 625
Ok, but I thought that if you multiply F by [itex]\omega[/itex] that quantity does not tend to zero anymore.

In a similar way 1/x -> 0 but x(1/x) clearly tends to 1.

*EDIT:
On the other hand, I was thinking that if F(w) itself didnt tend to zero, its inverse FT, that is f(x), would not exist.

mathman
#4
Feb5-11, 04:40 PM
Sci Advisor
P: 6,039
Question Fourier Transform Smoothness/Compactness

Quote Quote by mnb96 View Post
Hello,
my question arises from reading the section on Smoothness/Compactness from Bracewell's "The Fourier Transform and Its Applications" page 162.

I don't quite understand the following reasoning:

[tex]F(\omega) = \ldots = \frac{1}{i\omega}\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx[/tex]

and at this point the author says that when [itex]\omega\to\infty[/itex] then [itex]\omega F(\omega) \to 0[/itex].

But why [itex]\omega F(\omega)[/itex] is supposed to tend to zero, and not just [itex]F(\omega) \to 0[/itex] ?

Thanks.
The author is simply saying that [itex]|\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx| [/itex] tends to 0. This makes sense unless f'(x) is completely pathological.


Register to reply

Related Discussions
Fourier transform smoothness General Math 7
Fourier Transform Question Engineering, Comp Sci, & Technology Homework 1
Fourier Transform question Calculus 3
Fourier Series / Fourier Transform Question Electrical Engineering 6
The difference between Fourier Series, Fourier Transform and Laplace Transform General Physics 1