Weak completeness properties of the $L^1 $-space of a spectral measure
Volume 519 / 2016
Abstract
Given a Banach-space-valued vector measure $m$, its associated space $L^1 (m)$ is weakly complete iff it is finite-dimensional, weakly quasi-complete iff $L^1 (m)$ is reflexive, and weakly sequentially complete iff $L^1 (m)$ contains no copy of $c_0$. If $m$ takes its values in a locally convex Hausdorff space $Y$, then the situation is more complicated. The first question is the completeness of $L^1 (m)$ for its given $L^1$-topology.
It is shown that $L^1 (m)$ is complete iff it is quasi-complete and, if $Y$ is sequentially complete, that $L^1 (m)$ is complete iff $\Sigma (m) := \{f \in L^1 (m): f $ is $\{0,1\}$-valued$\}$ is relatively weakly compact in $L^1 (m)$. Weak completeness properties of $L^1 (m)$ have no reason\-able characterization, and little is known about the dual space $L^1 (m)^*$. However, if $m := P $ is a spectral measure acting in a Banach space (with $m$ strong operator $\sigma$-additive), then more can be said. Two remarkable features arise: $f \in L^1 (P)$ iff $f \in L^\infty (P)$ and a concrete description of $L^1 (P)^*$ is available. Some sample consequences are: $L^1 (P)$ is complete iff it is quasi-complete, iff it is weakly quasi-complete, iff $\Sigma (P)$ is relatively weakly compact in $L^1(P)$. If $\Sigma (P)$ is not relatively weakly compact, then $L^1 (P)$ may fail to be weakly sequentially complete, but never if $P$ is atomic.
