I am trying to prove the converse of convolution theorem: $$ \mathscr{F}[f(x)g(x)]=\frac{1}{\sqrt{2\pi}}\,\widetilde{f}(\omega)*\widetilde{g}(\omega)$$
I try to apply the definition of convolution and Fourier tranform: \begin{align*} \frac{1}{\sqrt{2\pi}}\,\widetilde{f}(\omega)*\widetilde{g}(\omega) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\widetilde{f}(\omega)\widetilde{g}(z-\omega)\,d\omega \\ &= \left(\frac{1}{\sqrt{2\pi}}\right)^3\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}f(x)e^{-i\omega x}\,dx\int_{-\infty}^{\infty}g(y)e^{-i(z-\omega) y}\,dy\right]\,d\omega \end{align*} obviously my aim is to obtain $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)g(x)e^{-i\omega x}\,dx$, but I don't have any idea on how to proceed. Could anyone give me some suggestions? Thank you.
I found the solution here: http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html