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}


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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Have you checked Wikipedia? $\endgroup$ – Tendero Dec 5 '18 at 15:09

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.