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I'm trying to find proofs for the following identities:

  1. $\quad F({f \cdot g}) = F({f}) \ast F({g})$
  2. $\quad F({f(\alpha x)})=\frac{1}{|\alpha|} \cdot F(f(\frac{k}{\alpha}))$
    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.

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    $\begingroup$ See math.stackexchange.com/questions/1709352/… $\endgroup$
    – Kenny Wong
    Feb 25, 2017 at 14:36
  • $\begingroup$ @KennyWong Thanks! do you have any hint on the second one, as well? $\endgroup$
    – Max
    Feb 25, 2017 at 14:40
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    $\begingroup$ 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 $dx = du/|\alpha |$. $\endgroup$
    – Kenny Wong
    Feb 25, 2017 at 14:54
  • $\begingroup$ @KennyWong You're totally right. Thanks a lot. $\endgroup$
    – Max
    Feb 25, 2017 at 14:56

2 Answers 2

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\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.

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\begin{align} \int _{-\infty}^\infty f(\alpha x) \exp(-2\pi ikx) \, dx \ne \int_{-\infty}^\infty f(u) \exp\left(-2\pi i\frac{k}{\alpha}u\right) \, & du \\[10pt] & \updownarrow \\[10pt] \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) \, & \frac{du} {|\alpha|} \end{align} Two points:

  • That denominator needs to be there.
  • Why the absolute value sign? If $\alpha>0$ then that's not needed. If $\alpha<0,$ then as $x$ goes from $-\infty$ to $+\infty,$ $u$ goes from $+\infty$ to $-\infty,$ so omitting the absolute value and interchanging $\pm\infty$ would be correct. Then interchanging those two multiplies the whole thing by $-1,$ and putting the absolute value sign there also multiplies the whole thing by $-1.$
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