In my homework problem, it asks to prove if $f \in L^1, g \in L^2, \widehat{f*g}=\hat{f} \hat{g}$ almost everywhere.
My attempts: First I showed $f*g \in L^2$ so $\widehat{f*g}=\lim_n S_n$ for some Schwarz functions $S_n$.
$\hat{f}\hat{g}=\hat{f} \lim_n \hat{g}_n$ where $g_n$ are in Schwartz Class. $\hat{f} \lim_n \hat{g}_n=\lim_n \hat{g}_n\hat{f}=\lim_n \widehat{g_n*f}$ this line is proved in the question before. And I wonder how to show $\lim_n \widehat{g_n*f}=\widehat{f*g}$. Is $g_n*f$ in the Schwartz Class? If so, why the result is almost everywhere? If not, how should I proceed?
We have covered no materials of $L^p$ space. So my guess is that the solution needs no $L^p$ theory.
Find a sequence $(g_{n})\subseteq S({\bf{R}}^{n})$ such that $g_{n}\rightarrow g$ in $L^{2}$, then \begin{align*} \|\widehat{f\ast g}-\widehat{f}\widehat{g}\|_{L^{2}}&\leq\|\widehat{f\ast g}-\widehat{f\ast g_{n}}\|_{L^{2}}+\|\widehat{f}\widehat{g_{n}}-\widehat{f}\widehat{g}\|_{L^{2}}\\ &=\|\widehat{f\ast g-f\ast g_{n}}\|_{L^{2}}+\|\widehat{f}(\widehat{g_{n}}-\widehat{g})\|_{L^{2}}\\ &=\|f\ast g-f\ast g_{n}\|_{L^{2}}+\|\widehat{f}(\widehat{g_{n}}-\widehat{g})\|_{L^{2}}\\ &\leq\|f\ast(g-g_{n})\|_{L^{2}}+\|\widehat{f}\|_{L^{\infty}}\|\widehat{g_{n}-g}\|_{L^{2}}\\ &\leq\|f\|_{L^{1}}\|g_{n}-g\|_{L^{2}}+\|f\|_{L^{1}}\|g_{n}-g\|_{L^{2}}\\ &\rightarrow 0, \end{align*} so $\widehat{f\ast g}=\widehat{f}\widehat{g}$ a.e.