Question Fourier Transform Smoothness/Compactness

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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.
 

Answers and Replies

  • #2
mathman
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ωF(ω) -> 0 is a stronger statement than F(ω) -> 0 for ω -> ∞.
 
  • #3
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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.
 
Last edited:
  • #4
mathman
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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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