Quadratic form in $L^{2}$ is closed

Question

Let $V\in L_{\text{loc}}^{1}(\mathbb{R}^{n})$ and consider the quadratic form $q(u):=\langle Vu, u\rangle_{L^{2}(dx)}$. I want to show that this form is closed in $L^{2}$ with $\operatorname{dom}(q)={L^{2}(dx)}\cap {L^{2}(Vdx)}$. My approach is to show that if $\{f_{n}\}\subset {L^{2}(dx)}\cap {L^{2}(Vdx)}$ with $$\|f_{n}-f\|_{L^{2}(dx)}\to 0, ~q(f_{n}-f_{m})\to 0,$$ then $f\in L^{2}(dx)\cap L^{2}(Vdx)$ and $q(f_{n}-f)\to 0$. Any ideas?

Answer 1

This follows from the fact that $({L^{2}(dx)}\cap {L^{2}(Vdx)}, ||.||_{q})$ is a Hilbert space where $$||u||_{q}^{2}=q(u)+||u||_{L^{2}}^{2}.$$