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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) = $4\pi^2f(-\omega)g(-\omega)$.

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

Hakim
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  • Exactly what do you not understand? – timur Jul 31 '14 at 00:26
  • 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. –  Jul 31 '14 at 09:25

1 Answers1

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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.

timur
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