Commutant of $L^{\infty}(X)$ inside $\mathcal{B}(L^2(X))$

Question

Represent $L^{\infty}(X)$ in $\mathcal{B}(L^{2}(X)$ by the multiplication operator given by $f \to M_f$ where $M_f(g)=fg$. I want to prove that the commutant of $L^{\infty}(X)$ when regarded as a subalgebra of $\mathcal{B}(L^{2}(X)$ is $L^{\infty}(X)$. Do I need to make any assumptions about the measure to hold these results? Is this representation faithful?

Answer 1

Ah, one of those things I know I know how to prove but I can never remember right away how to prove it.

New news: It's not true for an arbitrary measure space. First we give the proof for $\sigma$-finite spaces, then the counterexample.

First, if $0<\mu(E)<\infty$ then $f\mapsto\int_ETf$ is a bounded linear functional on $L^2$, so there exists $m_E\in L^2$ with $$\int_E Tf=\int m_Ef\quad(f\in L^2).$$

Now if $f=0$ on $E$ then $$\int m_Ef=\int_ETf=\int\chi_ETf=\int T(\chi_Ef)=\int T0=0.$$On a $\sigma$-finite (or just semi-finite) measure space this implies that

$m_E=0$ a.e. on $X\setminus E$

(If not then there exists $F\subset X\setminus E$ with $0<\mu(F)<\infty$ and $|m|>0$ on $F$, which makes it easy to concoct $f\in L^2$ supported on $F$, hence vanishing on $E$, with $\int m_Ef\ne0$.)

Next,

If $E'\subset E$ then $m_{E'}=\chi_{E'}m_E$

Because for every $f\in L^2$ we have $$\int m_{E'}f=\int_{E'}Tf=\int_E\chi_{E'}Tf =\int_E T(\chi_{E'}f)=\int m_E(\chi_{E'}f)=\int(\chi_{E'}m_E)f.$$

So $m_{E'}-\chi_{E'}m_E\in L^2$ is orthogonal to every $f\in L^2$.

And so

Theorem If $\mu(X)<\infty$ there exists $m\in L^\infty$ such that $Tf=mf$.

Indeed, let $$m=m_X.$$The above shows that $m_E=m\chi_E$, so for every $f\in L^2$ we have $$\int_ETf=\int m_Ef=\int_Emf.$$ So $Tf=mf$, and now if $f\mapsto mf$ is bounded on $L^2$ it follows that $m\in L^\infty$.

You can easily patch things to get the result for a $\sigma$-finite measure space. (Start by writing $X$ as the union of a sequence of disjoint sets of finite measure...)

Counterexample

Say $X$ is an uncountable set. Let $M$ be the $\sigma$-algebra of $E$ such that $E$ or $X\setminus E$ is countable, and consider the measure space $(X, M,\mu)$, where $\mu$ is counting measure restricted to $M$.

Fix $E_0\subset X$ with $E_0\notin M$, and set $$m_0=\chi_{E_0}.$$

Now, even though $m_0$ is not measurable, we have

Curious Fact If $f\in L^2$ then $m_0f$ is measurable.

Proof: $\{f(x)\ne0\}$ is $\sigma$-finite, hence countable.

So we can define a bounded linear operator on $L^2$ by $$Tf=m_0f.$$

It's clear that $T(gf)=gTf$ for $g\in L^\infty$. And if $m$ is a bounded function such that $Tf=mf$ for all $f\in L^2$ then it's clear that $m=m_0$ (because $\chi_{\{x\}}$ is a non-zero $L^2$ function), hence $m\notin L^\infty$.

Note that $\mu$ is semi-finite.

The next section is now obsolete; I'm leaving it here in case anyone wants to verify that this counterexample is consistent with my ruminations about how the problem is equivalent to the "patching problem":

The General Case

Just in case someone wants to try to fix it, I might explain what the problem is for a general measure space. If $f\in L^2$ then $\{f(x)\ne0\}$ is $\sigma$-finite; one could define $m_E$ for $\sigma$-finite $E$ and try to patch them together.

Def. Suppose $S$ is a collection of $\sigma$-finite measurable sets and for every $E\in S$ we have a measurable function $m_E:E\to\Bbb C$. Suppose $m_E=m_{E'}$ on $E\cap E'$. We will say $S$ can be patched (or $S$ is patchable) (or, better, $(m_E)_{E\in S}$ can be patched) if there exists a measurable function $m:\bigcup_{E\in S}E\to\Bbb C$ such that $m=m_E$ almost everywhere on $E$.

<p>If <span class="math-container">$S$</span> is countable then <span class="math-container">$S$</span> can be patched.</p>

Proof: Suppose $S=\{E_1,E_S,\dots\}$. For $x\in\bigcup S$ define $n(x)=\min\{k:x\in E_k\}$ and $m(x)=m_{E_{n(x)}}(x)$. It's not hard to show that $m=m_E$ almost everywhere on $E$.

If $S$ is uncountable one could well-order $S$ and then give an analogous definition of $n(x)$ and $m(x)$. But showing that $m=m_E$ almost everywhere on $E$ seems to involve uncountable unions of null sets...(for that matter, the idea that $m$ has to be measurable seems implausible for large $S$).

Edit: I just realized that the patching problem for the class of $\sigma$-finite measurable sets in an arbitrary measure space is equivalent to the original question for arbitrary measures. If as I suspect the answer to the no, then it seems this could be a way to look for a counterexample.

Suppose then that $S$ is the collection of $\sigma$-finite measurable sets and $(m_E)_{E\in S}$ cannot be patched. Note that since $\Bbb C$ is homeomorphic to the unit disk we can assume that $||m_E||_\infty\le 1$.

For $f\in L^2$ define $$E_f= \{f(x)\ne0\}$$and then $$Tf=\int m_{E_f}f.$$Then you can show that $T$ is a bounded linear operator and that if there exists $m\in L^\infty$ with $Tf=mf$ then $(m_E)_{E|in S}$ can be patched.