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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.

1 Answers1

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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

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

David C. Ullrich
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  • Deducing that $m \in L^\infty$ from $| m | _E | _2 ^2 \le c |E|$ seems to be the crux of the problem, but it is precisely this step that is quickly skipped over and left unproved. – Alex M. Oct 09 '21 at 16:09
  • @AlexM. Seems fairly obvious to me; in any case it's clear from the Lebesgue Differentiation Theorem – David C. Ullrich Oct 09 '21 at 17:00
  • The Lebesgue Differentiation Theorem seems overkill to me. From $\int |m|^2 1_E \le c |E|$ one could deduce that $\int |m|^2 \sum \alpha_i 1_{E_i} \le c \int \sum \alpha_i 1_{E_i}$ for every simple function with $\alpha_i \ge 0$, whence $\int |m|^2 f \le c \int f$ for every integrable $f \ge 0$, whence $\left| \int |m|^2 f \right| \le \int |m|^2 |f| \le c \int |f|$, so the left hand side is a continuous linear functional on $L^1$, therefore $|m|^2 \in L^\infty$, whence $m \in L^\infty$. (I've written it in full detail for potential future readers.) – Alex M. Oct 09 '21 at 17:19
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    @AlexM. In fact it's much simpler than that. Say $\lambda> c$ and let $E\subset E_\lambda={|m|^2>\lambda}$.with $|E|<\infty$. Then the inequality implies $|E|=0$. Hence $|E_\lambda|=0$, which says $||m^2||\infty<\lambda$ _by definition of $||\cdot||\infty$. So $||m^2||\infty\le c$. Imo that's simple enough to go without saying. LDT is overkill but it is faster to type and will convince the reader that yes it really is clear. – David C. Ullrich Oct 09 '21 at 21:37