Show that $L^{\infty}(\Omega,\mu)$ is a maximal abelian subalgebra of $B(L^{2}(\Omega,\mu))$

Question

Let $\Omega $ be a compact Hausdorff space and let $\mu$ be a positive regular Borel measure on $\Omega$ .Let $\phi \in L^{\infty}(\Omega,\mu)$.

Define $M_{\phi} \in B(L^2(\Omega, \mu))$ by $M_{\phi}(f) = \phi f, \ f\in L^2(\Omega, \mu)$ .

I have shown that $\phi \to M_{\phi}$ is an isometric isomorphism of $L^{\infty}(\Omega,\mu)$ onto a $C^*$ subalgebra $B$ of $ B(L^2(\Omega, \mu))$.

But now the next step is to prove that $B' = B$, i.e. $B$ is same as its commutant.

I require some hints to proceed with the problem.

Answer 1

$\def\one{\mathbf 1}$ Hint: Let $T \in B'$. We need to find an $L^\infty$ function $\varphi$ such that $T = M_\varphi$. There is a good candidate for $\varphi$, namely $\varphi = T(\one)$, where $\one$ denotes the constant function with value $1$ on $\Omega$. Note $\one \in L^\infty \subset L^2$, so $\varphi = T(\one)$ at least makes sense. There is quite a bit to check to see that this works.