# Closable Multiplication Operator

I have the operator $M:Dom (M)\subset L^2(\mathbb{R}^N)\rightarrow L^2(\mathbb{R}^N)$, $Mf(x)=m(x)f(x)$, where $m$ is a continuous function and $Dom(M)=\{f\in L^2(\mathbb{R}^N)| mf\in L^2(\mathbb{R}^N) \}$ and $M_0$ the 'same' operator but with domain $Dom(M_0)=C^{\infty}_c(\mathbb{R}^N)$. It's clear thar $M_0$ is closable operator and $M$ is a closed operator. I have prove that a closure of $M_0$ is $M$, the problem is show that exists a sequence $(f_n)\subset C^{\infty}_c(\mathbb{R}^N)$ such that $mf_n$ converge in $L^2(\mathbb{R}^N)$, that is possible only assuming that $m$ is continuous?