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Question Fourier Transform Smoothness/Compactness

  1. Feb 4, 2011 #1
    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] ?

  2. jcsd
  3. Feb 4, 2011 #2


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    ωF(ω) -> 0 is a stronger statement than F(ω) -> 0 for ω -> ∞.
  4. Feb 5, 2011 #3
    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.

    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: Feb 5, 2011
  5. Feb 5, 2011 #4


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    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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