I'm trying to find proofs for the following identities:
- $\quad F({f \cdot g}) = F({f}) \ast F({g})$
- $\quad F({f(\alpha x)})=\frac{1}{|\alpha|} \cdot F(f(\frac{k}{\alpha}))$
where $F$ denotes the Fourier transform.
I'm aware that 1) is a form of the convolution theorem, but I struggle to find a proof of it and instead I always just find the proof for the form $F(f \ast g)=F(f) \cdot F(g)$. I can't really find a way to proof this form since I don't know how to express $F({f}) \ast F({g})$ in integrals.
For 2) I think I know how to start, but I can't go on from here: $$F(f(\alpha x))= \int _{-\infty}^\infty f(\alpha x) \exp(-2\pi ikx) \, dx = \int_{-\infty}^\infty f(u) \exp\left(-2\pi i\frac{k}{\alpha}u\right) \, du$$
Any help or just a link would be greatly appreciated.
\begin{align} \mathcal{F}(f)(\alpha t) & := \int f(x) e^{-2\pi i x \cdot \alpha t} \, dx \\ & = \int f(x) e^{-2\pi i (\alpha x) \cdot t} \, dx \end{align}
Taking the change of variables $y = \alpha x$, this is
$$\int f\left(\frac{y}{\alpha}\right) e^{-2 \pi i y t} \frac{dy}{\alpha}, \text{ if } \alpha > 0.$$
If $\alpha < 0$ then the order of the integral gets reversed and you'll end up with an extra $-$ out front, giving you
$$\frac{1}{|\alpha|} \int f\left(\frac{y}{\alpha}\right) e^{-2\pi i y t} \, dy,$$ as claimed.