Lemma 38.8.5.label Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^{\infty}(X; \complex)\to L^{\infty}(Y; \complex)$ such that:
- (a)
$T$ is an isometric isomorphism.
- (b)
For each $f, g \in L^{\infty}(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$.
then:
- (1)
$T^{*}(L^{1}(Y; \complex)) \subset L^{1}(X; \complex)$.
- (2)
$T$ is weak*-continuous.
Proof. Let $\phi \in L^{1}(Y; \complex)$ with $\phi \ge 0$, then $T^{*}\phi \in L^{\infty}(X; \complex)^{*}$. For each $\seq{B_n}\subset \cm$ and $B \in \cm$ with $B_{n} \upto B$, $\one_{B} = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^{\infty}(X; \complex)$. Thus (b) and Theorem 38.8.3 imply that
Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the Radon-Nikodym Theorem, there exists $f \in L^{1}(X; \complex)$ with $f \ge 0$ such that $\int_{B} f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$.
Let $g \in L^{\infty}(X; [0, 1])$. By Lemma 26.6.4, there exists simple functions $\seq{g_n}\subset \Sigma(X; [0, 1])$ such that $g_{n} \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_{n}$ as an element of $L^{\infty}(X; \real)$, so (b) and Theorem 38.8.3 imply that,
By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)}= \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^{\infty}(X; \complex)$. Therefore $T^{*}(L^{1}(Y; \complex)) \subset L^{1}(X; \complex)$, and $T$ is weak*-continuous.$\square$
Post a Comment