Fourier transform of a product with one integrable transform

F1 and the bound F2 applied to $\widehat{f}$ give $\mathrm{\mid }{f}_c(x)\mathrm{\mid }\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1$, hence $\mathrm{\parallel }{f}_{c}g{\mathrm{\parallel }}_1\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1<\mathrm{\infty }$. Changing either input on a null set changes its product only on the union of those two sets, so the integrable product class is well-defined. Also the convolution integral in the conclusion is absolutely convergent at every frequency, bounded by $\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$ using F2.

Insert the inverse integral for $f_c$ into $\int f_c(x)g(x)e^{-2\pi ix\cdot \xi }dx$. The product integrand has modulus $\mathrm{\mid }\widehat{f}(\eta )\mathrm{\mid }\mathrm{\mid }g(x)\mathrm{\mid }$ with double integral $\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$; measurability follows from coordinate pullbacks and the continuous exponential. F3 exchanges the integrals, giving $\int \widehat{f}(\eta )[\int g(x)e^{-2\pi ix\cdot (\xi -\eta )}dx]d\eta$, the required expression. Exchanging the roles of f and g proves the symmetric variant under its stated hypothesis. Countable choice is inherited from F1. No assertion that arbitrary products of two integrable functions are integrable is used.