# Question Fourier Transform Smoothness/Compactness

by mnb96
Tags: fourier, transform
 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: $$F(\omega) = \ldots = \frac{1}{i\omega}\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx$$ 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.
 P: 626 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.
 Quote by 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.
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.