If $f,g\in L^1(\mathbb{R}^n)$ and
$$\mathcal{F}(f):=(2\pi)^{-n/2}\int e^{i\langle x,y\rangle}f(y)dy$$
is the Fourriertransform.
Then we have $\mathcal{F}(f\cdot g)=(2\pi)^{-n/2}\mathcal{F}(f)*\mathcal{F}(g)$.
But what if only $f\in \mathcal{S}(\mathbb{R}^n)$ and $g\in L^2(\mathbb{R}^n)$ such that $fg \in L^{1}\cap L^2$?
If $g\in L^2$ we define $\mathcal{F}g:=L^2\text{-}\lim\limits_{R\rightarrow \infty} \mathcal{F}\chi_{B_R(0)} g$.