We assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$ and we further assume that $g \in L^1(\mathbb{R})$ and $f * g \in L^1(\mathbb{R})$. Admitting that, for all real $\xi$, $$\int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t) \,\mathrm{d}t\right) \mathrm{d}x \quad \text{and} \quad \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t) \,\mathrm{d}x\right) \mathrm{d}t$$ exist and are equal, show that $\widehat{f * g} = \hat{f}\hat{g}$.
We assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$ and we further assume that $g \in L^1(\mathbb{R})$ and $f * g \in L^1(\mathbb{R})$. Admitting that, for all real $\xi$,
$$\int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t) \,\mathrm{d}t\right) \mathrm{d}x \quad \text{and} \quad \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \mathrm{e}^{-\mathrm{i}x\xi} f(t) g(x-t) \,\mathrm{d}x\right) \mathrm{d}t$$
exist and are equal, show that $\widehat{f * g} = \hat{f}\hat{g}$.