 Fourier product formula

For Schwartz functions $f,g$ and the Fourier transform convention $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$, the displayed formula exchanges a pointwise product for a convolution. Substitute Fourier inversion for $g$ into the Fourier transform of $fg$. Fubini's theorem, justified by $\Vert f{\Vert }_1\Vert \hat{g}{\Vert }_1<\infty$, givesUnlike the convolution theorem $\widehat{f\ast g}=\hat{f}\hat{g}$, this version cannot be asserted for arbitrary pairs in $L^1$ without additional hypotheses, since their pointwise product need not be an integrable function.