Proposition 38.8.2.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.
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 $L^{\infty}(X; \complex) = A$.$\square$
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