This Wikipedia page on the convolution theorem outlines several versions of the convolution theorem for functions $f$ and $g$. I'm particularly interested in these two:
$$\displaystyle \mathcal{F}\{f \ast g\} = \mathcal{F}\{f\} \ \cdot \ \mathcal{F}\{g\}, \\ \mathcal{F}\{f \cdot g\} = \mathcal{F}\{f\} \ast \mathcal{F}\{g\}.$$
Suppose $f, g : \mathbb{R}^n \rightarrow \mathbb{R}^n$ are generic functions. What are the precise properties that $f, g$ need to have for either of these relations to hold? Is it sufficient to have $f, g \in L^{1}(\mathbb{R}^n) \cap L^{2}(\mathbb{R}^n)$? Does $f$ need to be smooth, continuous? In the Wikipedia article, the first formula insists that we must have $f,g \in L^{1}(\mathbb{R}^n).$ This seems reasonable given the proof of the first identity in that article. Is that sufficient for the second formula to hold, also? Or must additional restrictions be imposed on $f,g$?
If you assume $f,g\in L^1$, then $f * g \in L^1$ and $\mathcal{F}(f *g)=\mathcal{F}(f)\mathcal{F}(g)$. That works out without further assumptions, and their proof works just fine because all functions are absolutely integrable, which allows the interchange of integrals.
But it seems to me that the identity $\mathcal{F}(fg)=\mathcal{F}(f)*\mathcal{F}(g)$ requires $\mathcal{F}(f),\mathcal{F}(g)\in L^1$ in order for the same method of proof to work (which they seem to be claiming is true.) $f,g \in L^1\cap L^2$ is not going to lead to $\mathcal{F}(f),\mathcal{F}(g)\in L^1$.