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Question Fourier Transform Smoothness/Compactness 
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#1
Feb411, 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. 


#2
Feb411, 04:52 PM

Sci Advisor
P: 6,066

ωF(ω) > 0 is a stronger statement than F(ω) > 0 for ω > ∞.



#3
Feb511, 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. 


#4
Feb511, 04:40 PM

Sci Advisor
P: 6,066

Question Fourier Transform Smoothness/Compactness



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