Schwartz convolution and product laws

By [F1]–[F3], $h={\mathcal{F}}^{-1}(\widehat{f}\widehat{g})$ is Schwartz and integrable. Also the convolution integral exists for every $x$, bounded absolutely by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_1$. It is continuous: for any $x_k\to x$, its integrands converge pointwise by continuity of $f$ and are dominated by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }g\mathrm{\mid }$; [F7] gives convergence of the integrals. The sequential continuity criterion is valid under countable choice. By [F4] the integrable convolution class has transform $\widehat{f}\widehat{g}$, so [F6] identifies it with $h$ almost everywhere. Two continuous functions equal almost everywhere are equal everywhere, since a nonzero difference persists on a ball containing a box of positive measure. Hence the actual convolution is $h\in \mathcal{S}$.

The product $fg$ is Schwartz by [F2]. By [F1] and [F3], $f,g,\widehat{f}$ are integrable, so [F5] applies. Its continuous inverse representative of $f$ is $f$ itself by [F1]. Thus its identity gives the second displayed formula everywhere; step 1.1 and [F4] give the first.