User:Rjones122/Sandbox

Theorem
Let $f$ and $g$ be two functions in $L^1\mathbb{R}^n$ with convolution $f\ast g$. Denote with $\mathcal{F}\{f\}$ the Fourier transform of $f$, given by $\int_{\mathbb{R}^n} f(t)e^{-2\pi it\cdot\nu}\, dt$. Then $\mathcal{F}\{f\ast g\} = \mathcal{F}\{f\}\mathcal{F}\{g\}$ up to the normalization constant of the Fourier transform.

Proof
By Fubini's theorem, if $h = f\ast g$ then $h$ is in $L^1\mathbb{R}^n$. The Fourier transform of $h$ is then given as:
 * $\displaystyle \mathcal{F}\{h\} = \int_{\mathbb{R}^n} \int_{\mathbb{R}^n}f(\tau) g(t-\tau) \,d\tau\, e^{-2\pi it\cdot\nu}\, dt$
 * $\displaystyle \mathcal{F}\{h\} = \int_{\mathbb{R}^n} f(\tau) \left( \int_{\mathbb{R}^b}g(t-\tau) e^{-2\pi it\cdot\nu}\,dt \right) \,d\tau$

Let $x=t-\tau$. We now have:
 * $\displaystyle \mathcal{F}\{h\} = \int_{\mathbb{R}^n} f(\tau) \left( \int_{\mathbb{R}^b}g(x) e^{-2\pi i(x+\tau)\cdot\nu}\,dx \right) \,d\tau$

which simplifies to:
 * $\displaystyle \mathcal{F}\{h\} = \left(\int_{\mathbb{R}^n} f(\tau)e^{-2\pi i\tau\cdot\nu}\,d\tau\right)\left(\int_{\mathbb{R}^n} g(x)e^{-2\pi ix\cdot\nu}\,dx\right)$

which is $\mathcal{F}\{f\}\mathcal{F}\{g\}$.