Proposition 40.8.3.label Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^{\infty}(X; \complex)$ act on $L^{2}(X; \complex)$ by multiplication, then:
- (1)
The weak* topology on $L^{\infty}(X; \complex)$ is equal to the weak operator topology of $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$.
- (2)
$L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is a von Neumann algebra.
- (3)
$L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is maximal abelian.
Proof. (2): Let $A \subset B(L^{2}(X; \complex))$ be the von Neumann algebra generated by $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$. By Theorem 16.3.4, $L^{\infty}(X; \complex)$ is the dual of $L^{1}(X; \complex)$, so (1) and the Banach-Alaoglu Theorem imply that the closed unit ball of $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is weak-operator closed. By the Kaplansky Density Theorem, $\ol{B_{L^\infty(X; \complex)}(0, 1)}= \ol{B_A(0, 1)}$. Therefore $L^{\infty}(X; \complex) = A$.
(3): Using Proposition 39.4.2, it is sufficient to show that $L^{\infty}(X; \complex)' = L^{\infty}(X; \complex)$. For each $g \in L^{\infty}(X; \complex)$, denote $M_{g} \in B(L^{2}(X; \complex))$ as the multiplication map by $g$.
First suppose that $\mu(X) < \infty$ and let $f = T\one_{X}$, then for each $g \in L^{\infty}(X; \complex)$,
By Lemma 40.8.2, $f \in L^{\infty}(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))}\le \norm{T}_{B(L^2(X; \complex))}$. By Proposition 16.1.10, $\Sigma(X; \complex) \subset L^{\infty}(X; \complex)$ is dense in $L^{2}(X; \complex)$, so $M_{f}g = Tg$ for all $g \in L^{2}(X; \complex)$.
Now suppose that $\mu$ is an arbitrary localisable measure. For each $A \in \cm$ with $\mu(A) < \infty$, the finite case shows that there exists $f_{A} \in L^{\infty}(X; \complex)$ such that
- (i)
$f_{A}|_{X \setminus A}= 0$ almost everywhere.
- (ii)
$Tg = M_{f_A}g$ for all $g \in L^{2}(X; \complex)$ with $f_{A}|_{X \setminus A}= 0$ almost everywhere.
By (ii), $f_{A}|_{A \cap B}= f_{B}|_{A \cap B}$ for all $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$. By the gluing lemma for meaasurable functions, there exists $f \in L^{\infty}(X; \complex)$ such that $f|_{A} = f_{A}|_{A}$ for all $A \in \cm$ with $\mu(A) < \infty$.
For each $g \in L^{2}(X; \complex)$, $\bracsn{g \ne 0}$ is $\sigma$-finite. As such, there exists $\seq{A_n}\subset \cm$ such that $\bracsn{g \ne 0}= \bigsqcup_{n \in \natp}A_{n}$ and $\mu(A_{n}) < \infty$ for all $n \in \natp$. In which case,
By Lemma 40.8.2, $f \in L^{\infty}(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))}\le \norm{T}_{B(L^2(X; \complex))}$. Therefore $T = M_{f}$, and $L^{\infty}(X; \complex)$ is maximal abelian.$\square$
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