Showing that if an operator commutes with all multiplication operator, it must be a multiplication operator

Question

Here comes the context: I'm looking at von Neumann algebras and in particular here, we are considering Hilbert space $\mathcal{H} = L^2(\mathbb{R})$ and von Neumann algebra, $\mathcal{M}$ to be the set of multiplication operators that multiply by essentially bounded functions, that is: $$\mathcal{M} = \{ A\in \mathcal{B}(\mathcal{H}) \ | \ (A\psi)(x) = a(x)\psi(x), a\in L^{\infty}(\mathbb{R}), \forall \psi\in L^2(\mathbb{R})\},$$

where $\mathcal{B}(\mathcal{H})$ denotes the set of all bounded linear operators on the Hilbert space. Also for extra context (as it may be helpful) to take the cyclic and separating vector $\Omega \in \mathcal{M}$ to be $\Omega \psi = \Omega(x) \psi(x),$ where $\|\Omega\| = 1$, $\Omega(x) \neq 0$ for all $x$.

I wish to prove that $\mathcal{M} = \mathcal{M}'$, that is, the von Neumann algebra in question is equal to its own commutant defined as: $$\mathcal{M}' = \{ A\in \mathcal{B}(\mathcal{H}) | AB=BA \ \forall B\in \mathcal{M} \}.$$

The direction $\mathcal{M} \subset \mathcal{M}'$ is fairly trivial since all multiplication operator commute with each other. The other direction however is causing me trouble. I recognise that the idea here is to show that if $A\in \mathcal{M}'$, meaning that $A$ commutes with all operators in $\mathcal{M}$ then it must also be a multiplication operator. However, I'm struggling to argue that mathematically without being handwavey.

Answer 1

To save typing: $E$ and $E_j$ will always denote a set of positive finite measure.

Say $T$ commutes with every multiplication operator: $$T(\phi f)=\phi T(f) \quad (f\in L^2,\phi\in L^\infty).$$

Suppose $E=E_1\cap E_2$. Then $$T(\chi_E)=T(\chi_{E_1}\chi_{E2}) =\chi_{E_1}T(\chi_{E_2})=\chi_{E_2}T(\chi_{E_1}).$$Hence

$T(\chi_{E_1})=T(\chi_{E_2})$ on $E_1\cap E_2$.

So there is a well-defined $m:\mathbb R\to\mathbb C$ such that $$m(x)=T(\chi_E)(x),\quad(x\in E),$$or in other words,

$T(\chi_E)=m$ on $E$.

Now $T(\chi_E)=T(\chi_E^2)=\chi_ET(\chi_E)$, so $T(\chi_E)$ is supported on $E$. With the above this shows that

$T(\chi_E)=m\chi_E$.

Since $T$ is bounded, $$\int_E |m|^2=||T(\chi_E)||_2^2\le c||\chi_E||_2^2=c|E|,$$which implies (by the Lebesgue Differentiation Theorem, for example) that

$m\in L^\infty$.

And now since $T$ is linear and bounded, as is the operator of multiplication by $m$, the fact that $T(\chi_E)=m\chi_E$ shows that

$T(f)=mf\quad(f\in L^2)$.