Order-weakly compact operators from vector-valued function spaces to Banach spaces

Let $E$ be an ideal of $L^{0}$ over a  $\sigma$-finite measure space $( \Omega, \Sigma,\mu)$, and let $E^{\sim}$ stand for the order dual of $E$. For a real Banach space $(X,\Vert\cdot\Vert _{_X})$ let $E(X)$ be a subspace of the space $L^{0}(X)$ of $\mu$-equivalence classes of strongly $\Sigma$-measurable functions $f : \Omega \longrightarrow X$ and consisting of all those $f \in L^{0}(X)$ for which the scalar function $\Vert f(\cdot)\Vert _{_X}$ belongs to $E$. For a real Banach space $(Y, \Vert\cdot\Vert _{_Y})$ a linear operator $T : E(X) \longrightarrow Y$ is said to be order-weakly compact whenever for each $u \in E^+$ the set $\,T(\{f\in E(X):\; \Vert f(\cdot)\Vert _{_X}\le u\})\,$ is relatively weakly compact in $Y$. In this paper we examine order-weakly compact operators $T : E(X) \longrightarrow Y$. We give a characterization of an order-weakly compact operator $T$ in terms of the continuity of the conjugate operator of $T$ with respect to some weak topologies. It is shown that if $(E,\Vert\cdot\Vert _{_E})$ is an order continuous Banach function space, $X$ is a Banach space containing no isomorphic copy of $l^{1}$ and $Y$ is a weakly sequentially complete Banach space, then every continuous linear operator $T : E(X) \longrightarrow Y$ is order-weakly compact. Moreover, it is proved that if $(E,\Vert\cdot\Vert _{_E})$ is a Banach function space, then for every Banach space $Y$ any continuous linear operator $T : E(X) \longrightarrow Y$ is order-weakly compact iff the norm $\Vert\cdot\Vert _{_E}$ is order continuous and $X$ is reflexive. In particular, for every Banach space $Y$ any continuous linear operator $T : L^{1}(X) \longrightarrow Y$ is order-weakly compact iff $X$ is reflexive.