Problem #7.2, MA598R, Summer 2009, Weigel
Define the Fourier transform of $ f \in L^1(\mathbb{R}) $ by
$ \widehat{f}(x) = \int_{-\infty}^{\infty} f(t) e^{-ixt}dt $
If $ f $, $ g \in L(\mathbb{R}) $, show
$ \widehat{f * g}(x) = \widehat{f}(x)\widehat{g}(x) $
Proof
Applying the definitions of Fourier transform and convolution, followed by Fubini (since $ f, g \in L(\mathbb{R}) $) we have:
$ \begin{align}\widehat{f * g}(x) &= \int_{\mathbb{R}}(f * g)(t)e^{-ixt}dt\\ &= \int_{\mathbb{R}}\left(\int_{\mathbb{R}}f(t-y)g(y)dy\right)e^{-ixt}dt\\ &= \int_{\mathbb{R}}g(y)\left(\int_{\mathbb{R}}f(t-y)e^{-ixt}dt\right)dy\\ &= \int_{\mathbb{R}}g(y)e^{-ixy}\left(\int_{\mathbb{R}}f(t-y)e^{-ix(t-y)}dt\right)dy\\ &= \int_{\mathbb{R}}g(y)e^{-ixy}\widehat{f}(x)dy\\ &= \widehat{f}(x)\widehat{g}(x) \end{align} $
