For each $g\in L^\infty$, let $M_g:L^2\rightarrow L^2$ be the multiplication operator defined by $M_g(f)=fg$. Show that the range of $M_g$ is closed if and only if $g$ is a characteristic function.
I'm not sure how to approach this. For the backwards direction I tried to assume that the range of $M_{1_A}$ is closed, by showing $M_{1_A}^{-1}:range(M_{1_A})\rightarrow L^2$ is continuous. But I couldn't get anywhere in trying to prove to $M_{1_A}^{-1}$ maps convergent sequences to convergent sequences. The other thought I had is that we use the Hilbert space properties of $L^2$, like the existence of an orthonormal basis But I'm not sure how to use it here.
$g \equiv 2$ is not a characteristic function but $M_g$ has closed range (namley whole of $L^{2}$) in this case.