 Solution

Approximate $g\in L^2$ by Schwartz $g_n$. Young's inequality gives
$\Vert f\ast (g_n-g){\Vert }_2\le \Vert f{\Vert }_1\Vert g_n-g{\Vert }_2$, while multiplication by bounded $\hat{f}$ is continuous in $L^2$. Pass to the limit in
$\widehat{f\ast g_n}=\hat{f}{\hat{g}}_n$.