If $f$, $g$ and $f*g$ are in $L^{2}(\mathbb{R}^{n})$, does it follow that $\mathcal{F}(f*g)=\mathcal{F}(f)\mathcal{F}(g)$ ? Here $\mathcal{F}$ denotes the Fourier transform on $L^{2}(\mathbb{R}^{n})$.
I know that if we assume that $f\in L^{1}(\mathbb{R}^{n})\cap L^{2}(\mathbb{R}^{n})$ and $g\in L^{2}(\mathbb{R}^{n})$ then $f*g\in L^{2}(\mathbb{R}^{n})$ and the equality above holds true.
However, if I have only that $f\in L^{2}(\mathbb{R}^{n})$, I was thinking of taking a sequence of functions in $L^{1}(\mathbb{R}^{n})\cap L^{2}(\mathbb{R}^{n})$ converging to $f$ in $L^{2}(\mathbb{R}^{n})$ and apply the result above but I can't pass to the limit.
This makes me think that maybe there are counterexamples.