The [[Fourier transform]] of the [[Convolution]] of two functions is given by the product of their respective Fourier transforms: $\mathcal{F}[(f*g)(x)]_k = \mathcal{F}[f(x)]_p\,\mathcal{F}[g(x)]_q.$ ^ee6b1b The converse of the [Fourier convolution theorem](Fourier%20convolution%20theorem.md) is also true: $\mathcal{F}[f(x)g(x)]_k = \frac{1}{2\pi}\big(\mathcal{F}[f(x)]_k*\mathcal{F}[g(x)]_k\big)(k).$ This means that the [Fourier transform](Fourier%20transform.md) of the product $f(x)g(x)$ is the [convolution](Convolution.md) of their individual Fourier transforms. %%Go to pg 51 of fft book for more elementary and graphical based discussions of convultion. where the steps to getting a convolution are broken down in terms of folding, displacement, multiplication, and integration.%% ^6d0a2e --- # Proofs and examples ## Proof of the Fourier convolution theorem    ## Proof of the converse of the Fourier convolution theorem   #MathematicalFoundations/Analysis/FourierAnalysis/Integrals