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Question Fourier Transform Smoothness/Compactness

by mnb96
Tags: fourier, transform
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mnb96
#1
Feb4-11, 04:07 PM
P: 626
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.
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mathman
#2
Feb4-11, 04:52 PM
Sci Advisor
P: 6,106
ωF(ω) -> 0 is a stronger statement than F(ω) -> 0 for ω -> ∞.
mnb96
#3
Feb5-11, 01:25 AM
P: 626
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: 6,106
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.


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