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I'm trying to find proofs for the following identities:
1) $F({f \cdot g}) = F({f}) \ast F({g})$
2) $F({f(\alpha x)})=\frac{1}{|\alpha|} \cdot F(f(\frac{k}{\alpha}))$
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: $F(f(\alpha x))= \int _{-\infty}^\infty f(\alpha x) exp(-2\pi ikx)dx=\int _{-\infty}^\infty f(u) exp(-2\pi i\frac{k}{\alpha}u)du$

Any help or just a link would be greatly appreciated.

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    See http://math.stackexchange.com/questions/1709352/fourier-transform-proof-mathcal-ffxgx-frac12-pifsgs?rq=1 – 2017-02-25
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    @KennyWong Thanks! do you have any hint on the second one, as well? – 2017-02-25
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    Yes. In fact your method for the second one is fine. Your $\int_{-\infty}^\infty f(u) \exp(-2\pi i \frac k \alpha u) du $ is precisely $F(f)(\frac k \alpha)$. But I think you accidentally missed the factor of $1 / | \alpha |$, coming from $dx = du/|\alpha |$. – 2017-02-25
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    @KennyWong You're totally right. Thanks a lot. – 2017-02-25

1 Answers 1

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$\mathcal{F}(f)(\alpha t) := \int f(x) e^{-2\pi i x \cdot \alpha t} dx$

$= \int f(x) e^{-2\pi i (\alpha x) \cdot t} dx$

Taking the change of variables $y = \alpha x$, this is

$\int f(\frac{y}{\alpha}) e^{-2 \pi i y t} \frac{dy}{\alpha}$, 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

$\frac{1}{|\alpha|} \int f(\frac{y}{\alpha}) e^{-2\pi i y t} dy$, as claimed.