Fourier transform of $L^1$ function is the convolution of an $L^2$ function with itself

Question

I am interested in showing that if $g \in L^1(\mathbb{R})$, then the Fourier transform $\mathcal{F}(g)$ is equal to the convolution $f \ast f$ for some $f \in L^2(\mathbb{R})$. If it happens that $\sqrt{g} \in L^1(\mathbb{R})$ also, then by the convolution theorem we have $$\mathcal{F}(g) = \mathcal{F}(\sqrt{g} \cdot \sqrt{g}) = \mathcal{F}(\sqrt{g}) \ast \mathcal{F}(\sqrt{g}),$$ so we take $f = \mathcal{F}(\sqrt{g})$. This suggests we want to approximate $g$ by a sequence of functions $g_n \in L^1$ such that $\sqrt{g_n} \in L^1$. However, is it true that $g_n \to g$ in $L^1$ implies $\sqrt{g_n} \to \sqrt{g}$ in $L^2$?