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I've been trying to find a rigorous proof of the dual convolution / multiplication, but I found nothing, can you give me a hand with this?

\begin{align} f(t) * g(t) &\overset{\mathcal F}{\iff} F(j\omega)G(j\omega)\\ f(t)g(t) &\overset{\mathcal F}{\iff}\frac1{2\pi} F(j\omega) * G(j\omega)\\ \end{align}

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closed as unclear what you're asking by Matt L., Stanley Pawlukiewicz, lennon310, A_A, Peter K. Dec 12 '18 at 16:13

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    $\begingroup$ Have you checked Wikipedia? $\endgroup$ – Tendero Dec 5 '18 at 15:09
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Just do the double integration:

$$\begin{align*}\mathscr{F}\left\{f(t) * g(t)\right\} &= \mathscr{F}\left\{\int_{-\infty}^\infty f(\tau)g(t-\tau)d\tau\right\} \\ \\ &= \int_{-\infty}^\infty\left[\int_{-\infty}^\infty f(\tau)g(t-\tau)d\tau\right]e^{-j\omega t}dt\\ \\ &= \int_{-\infty}^\infty f(\tau)\left[\int_{-\infty}^\infty g(t-\tau)e^{-j\omega t}dt\right]d\tau\\ \\ &= \int_{-\infty}^\infty f(\tau)e^{-j\omega \tau}G(\omega)d\tau\\ \\ &= F(\omega)G(\omega)\\ \end{align*}$$

The above derivation used Fubini's Theorem to switch the order of integration and the Fourier Transform Shift Theorem.

The proof for convolution in the frequency domain is analogous to the one above.

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