Theorem 38.8.3 (Order Structure of $L^{\infty}$).label Let $(X, \cm, \mu)$ be a localisable measure space, then
- (1)
$L^{\infty}(X; \real)$ is order complete.
- (2)
For each $\phi \in L^{1}(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^{\infty}(X; \real)$,
\[\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)}= \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}\]
Proof. By Proposition 38.8.2, $L^{\infty}(X; \complex)$ acting on $L^{2}(X; \complex)$ is a von Neumann algebra.
(1): Since $L^{\infty}(X; \real)$ is a lattice, it is order complete by Theorem 37.3.1.
(2): By Theorem 37.3.1, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of Proposition 38.8.2, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)}= \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.$\square$
Post a Comment