Functions for which $\mathcal{F}g = f \ast f$ Suppose one is given $f \in L^{2}(\mathbb{R})$, my question is whether or not there exists a $g \in L^{1}(\mathbb{R})$ such that $f \ast f = \mathcal{F}g$ where $\mathcal{F}$ is the Fourier transform.
Conversely, suppose one is given a $g \in L^{1}(\mathbb{R})$, then does there exist an $f \in L^{2}(\mathbb{R})$ such that $\mathcal{F}g = f \ast f$?
Formally, the answer for the first question should be $g := \mathcal{F}^{-1}(f \ast f)$. But of course all I know is that $f \ast f \in L^{\infty}(\mathbb{R})$.

As you have observed that, $g$ should be, $(f\ast f)^{\vee}.$ Next, observe that, $g= f^{\vee}\cdot f^{\vee}= (f^{\vee})^{2}.$ Since $f\in L^{2}(\mathbb R),$ we have $f^{\vee}\in L^{2}(\mathbb R)$ (by Planchrel) and  therefore $(f^{\vee})^{2}\in L^{1}(\mathbb R);$ and so $g\in L^{1}(\mathbb R).$
On the other hand,  Given $g\in L^{1}(\mathbb R),$ to find $f\in L^{2}(\mathbb R)$ so that $f\ast f= \hat{g}.$
Observer that, $(f^{\vee})^{2}= g,$ then $f^{\vee} =(g)^{\frac{1}{2}};$ (Why ?, this step may not be true) and since, $g\in L^{1},$ we have $f^{\vee}\in L^{2},$ therefore by Planchrel, $f\in L^{2}.$
