On my notes I found an unproven statement:
"If $f \in \mathcal L^1(\mathbb R)$ then $fx^n \in \mathcal S'(\mathbb{R})$"
Why is it so? I know that $\mathcal S(\mathbb{R})$ is contained in $\mathcal L^1(\mathbb{R})$, here it seems to suggest the opposite. Then a continuous functional on $\mathcal S(\mathbb{R})$ would of course be a tempered distribution...
We want to prove that $fx^n$ (assuming $n\geq 0$) is a tempered distribution, that is, the functional \begin{align*}\mathcal{S}&\to \mathbb{R}\\ g&\mapsto \int_{\mathbb{R}} fx^n g \end{align*} is continuous. By Holder's inequality, and since $g\in \mathcal{S}\Rightarrow x^ng\in \mathcal{S}\subset L^{\infty}$, $$\int_{\mathbb{R}} |f(x)x^n g(x)|dx\leq \|f\|_{L^1}\|x^n g\|_{L^\infty}<\infty$$ Moreover, the following implications hold: $$g_k\to^{\mathcal{S}} 0\Rightarrow x^ng_k\to^{\mathcal{S}}0 \Rightarrow \|x^ng_k\|_{L^{\infty}}\to 0 $$ Thus if $g_k\to^{\mathcal{S}} 0$ then by the previous inequality $\int_{\mathbb{R}}fx^ng_k \to 0$, and the functional is continuous.