# Fourier transformation and test function -- Question in the derivation

1. Dec 31, 2015

### Small bugs

How does it work? (The derivative rules of FT)
We look at $$F[x(t)]=\hat{x}(f)$$
$$\mathcal{l} \text{ is a distribution, with}\tilde{x}=tx(t)$$
$$\mathcal{F}[Dl(x)]=\mathcal{F}l'(x)=2\pi il(\mathcal{F}\tilde{x})=2\pi i \mathcal{F}l(\tilde{x})$$
Till here I fully understand. But next step:
$$=2\pi i fFl(x)$$
How do they drag out f and make x~=tx(t) to x?

**Well, of course this is just a proof of derivative rule of FT, which is not hard
$$\mathcal{F}x'=2\pi i f \hat{x}$$

2. Jan 5, 2016