Fourier Transform Properties - Proving

Question

How do I go about proving the following properties of Fourier transforms? I do not have a textbook (professor didn't issue one) so it is very hard for me to understand these concepts.

1. $\hat{f'(x)}=2\pi iξ\hat{f}(ξ)$
2. $\hat{-2\pi ixf(x)}=d\hat{f}(ξ)/dξ$

Answer 1

Judging from your identities, I assume your definition of the Fourier transform of a function $g$ is

$$\hat{g}(\xi) = \int_{-\infty}^\infty g(x)e^{-2\pi ix\xi}\, dx.$$

I'll also assume that $f$ is Schwartz. We find

$$\hat{f'}(\xi) = \int_{-\infty}^\infty f'(x)e^{-2\pi ix\xi}\, dx \underset{(*)}{=} \int_{-\infty}^\infty -f(x) \frac{d}{dx}(e^{-2\pi ix\xi})\, dx = 2\pi i\xi \int_{-\infty}^\infty f(x)e^{-2\pi ix\xi}\, dx.$$

Equality $(*)$ follows from integration by parts. Since

$$2\pi i\xi \int_{-\infty}^\infty f(x)e^{-2\pi ix\xi}\, dx = 2\pi i\xi \hat{f}(\xi),$$

we deduce statement $1.$

To prove statement $2.$, we compute

$$\frac{d\hat{f}(\xi)}{d\xi} = \int_{-\infty}^\infty \frac{\partial}{\partial \xi} f(x)e^{-2\pi ix\xi}\, dx = \int_{-\infty}^\infty f(x)(-2\pi ix)e^{-2\pi ix\xi}\, dx$$ $$ = \int_{-\infty}^\infty -2\pi i xf(x)e^{-2\pi ix\xi}\, dx.$$

The last expression is $\mathcal{F}(-2\pi i xf)(\xi)$ (here $\mathcal{F}$ denotes the Fourier transform). So statement $2.$ holds.

Answer 2

It will be roughly like this:

Using integration by parts gives $$ \hat{f'}(\xi) = \int f'(x) e^{-2\pi i\xi x} dx = \left[f(x)(-2\pi i \xi)e^{-2\pi i \xi x}\right]_{x \to -\infty}^{x \to +\infty} - \int f(x) (-2\pi i \xi) e^{-2\pi i\xi x} dx = 2\pi i \xi \hat{f}(\xi) $$