# Fourier Transform Identity (Convolution & Symmetry rule)

I don't follow the part of the solution highlighted in green.

Use the symmetry rule to show that $${\cal F}\{f(x)g(x)\}=\dfrac1{2\pi}\left(\hat f(\omega)*\hat g(\omega)\right).$$

Convolution $\Rightarrow{\cal F}\{\hat f(\omega)*\hat g(\omega)\}={\cal F}\{\hat f(x)\}{\cal F}\{\hat g(x)\}$
= (symmetry formula) = $4\pi^2f(-\omega)g(-\omega)$.

$\color{green}{\boxed{\color{black}{\begin{array}{l} \text{Take RHS, change } \omega \text{ to } x \text{ and take transform again:} \\ {\cal F}\{4\pi^2f(-x)g(-x)\}=2\pi(\hat f(-\omega)*\hat g(-\omega))\text{ using the symmetry rule again.} \\ \text{Thus: }{\cal F}\{f(x)g(x)\}=(\hat f(\omega)*\hat g(\omega))/(2\pi),\text{ as required.} \end{array}}}}$

• Exactly what do you not understand? Jul 31, 2014 at 0:26
• I don't understand how to apply the symmetry rule for the second time nor the fact that you seem to be able to interchange $\omega$and $x$ willy nilly.
– user144895
Jul 31, 2014 at 9:25

Let us rewrite the result of the first part as $$4\pi^2f(-\omega)g(-\omega)=\mathcal{F}\,(\hat f(x)*\hat g(x)).$$ We change $x$ to $\omega$ and $\omega$ to $x$, to get $$4\pi^2f(-x)g(-x)=\mathcal{F}\,(\hat f(\omega)*\hat g(\omega)).$$ By taking the Fourier transform of both sides, we have $$\mathcal{F}\,(4\pi^2f(-x)g(-x))=\mathcal{F}\,\mathcal{F}\,(\hat f(\omega)*\hat g(\omega))=-2\pi\,\hat f(-\omega)*\hat g(-\omega),$$ where in the last step we have used the symmetry rule.