# Fourier Transform Properties - Proving

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

1. $\hat{f'(x)}=2\pi iξ\hat{f}(ξ)$
2. $\hat{-2\pi ixf(x)}=d\hat{f}(ξ)/dξ$
• Can you provide your definition of the Fourier transfrom, the constants are sometimes different. – mvw Mar 26 '15 at 14:17
• Do you know how to differentiate an integral (sometimes called the Fundamental Theorem of Calculus)? That is, if $$f(x) = \int_{-\infty}^\infty \hat{f}(t)\exp(i2\pi xt)\,\mathrm dt,$$ do you know how to find $f^\prime(x) = \frac{\mathrm d}{\mathrm dx}f(x)$ by differentiating the integral on the right with respect to $x$? – Dilip Sarwate Mar 26 '15 at 14:20

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.

• how can i formulate an analog of the above two when I replace $f '(x)$ by $f^{n}(x)$ and replacing $\hat{f} (ξ)$ by $\hat{f} ^{n}(ξ)$? @kobe – samsonite Mar 26 '15 at 15:07
• Just apply statement 1. $n$ times and statement 2. $n$ times. – kobe Mar 26 '15 at 15:10

It will be roughly like this:

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