Suppose $(\Omega, \mathcal{A},\mu)$ is a $\sigma$-finite measure space and that $\phi:\Omega\to\mathbb{R}$ is an $\mathcal{A}$-measurable function such that $$f\in L^2(\mu)\implies \phi f\in L^2(\mu).$$($L^2$ is the Hilbert space of square integrable functions on $\Omega$).
I am trying to show that the linear map $T:L^2(\mu)\to L^2(\mu)$ given by $Tf=\phi f$ is bounded. However, I am not getting anywhere. Has anyone any ideas?
This is a classical application of the closed graph theorem: Assume that $f_n\to f$ in $L^2$ and $\phi f_n\to g$ in $L^2$. Then $f_n\to f$ in measure (or a subsequence converges a.e. if you don't know convergence in measure) and thus also $\phi f_n\to \phi f$ in measure (or a.e. along a subsequence). Hence $g=\phi f$ in $L^2$. By the closed graph theorem, $T$ is bounded.
As a side remark, this proof also works for noncommutative $L^p$ spaces over semi-finite von Neumann algebras.