Question Fourier Transform Smoothness/Compactness

[mnb96]

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:

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

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

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

Thanks.

 

[mathman]

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

 

[mnb96]

Ok, but I thought that if you multiply F by \omega 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]

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:

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

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

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

Thanks.

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