Consider the converse of the [Fourier convolution theorem,](Fourier%20convolution%20theorem.md) [$\mathcal{F}[f(x)g(x)]_k = \frac{1}{2\pi}\big(\mathcal{F}[f(x)]_k*\mathcal{F}[g(x)]_k\big)(k).$](Fourier%20convolution%20theorem#^eec510) Notice that by [convolution property 2](Convolution#^c1e692) we may re-write the right-hand as $\frac{1}{2\pi}\big(\mathcal{F}[f(x)]_k*\mathcal{F}[g(x)]_k\big)(k)=\frac{1}{2\pi}\int dp\,\mathcal{F}[f(x)]_{k-p}\mathcal{F}[g(x)]_p$ ^80b837 If we take the Fourier transform of a product of two functions we obtain $\mathcal{F}[f(x)g(x)]_k = \int dx\, f(x)g(x)e^{-ikx}= \int dx\, \mathcal{F}^{-1}[\mathcal{F}[f(x)]_p]_x \mathcal{F}^{-1}[\mathcal{F}[g(x)]_q]_x e^{-ikx}$$=\frac{1}{2\pi}\int dp\,\mathcal{F}[f(x)]_{k-p}\mathcal{F}[g(x)]_p$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ ^e37591 #MathematicalFoundations/Analysis/FourierAnalysis/Integrals #Proofs