I'm trying to understand the following construction:
Let $E$ a Banach lattice. Basically we want to construct for each $\nu \in E^*$ (A positive linear functional $\nu:E \rightarrow \mathbb{C})$, a real Banach lattice $F_\nu$.
So in $E$ we consider the (continuous)seminorm $p_\nu(x):= \nu(|x|) \, \, \forall x \in E$ (Why?). Let now $I_\nu := \{x \in E \, : \, p_\nu(x)=0\}$, this is a lattice ideal of $E$ (why?).
It is known (not to me) that the quotient space $E/I_\nu$, obtained in the usual way, is a vector lattice, and $p_\nu$ induces in $E/I_\nu$ a lattice norm $\tilde{p}_\nu(\hat{x}):=p_\nu(x) \,\,(x \in \hat{x}) \quad \forall \hat{x} \in E/I_\nu$. This comes from the fact that the canonical map $x \mapsto \hat{x}$ of $E$ into $E/I_\nu$ is a lattice homomorphism. Now let $F_\nu$ be the completion of vector lattice $E/I_\nu$ with the norm $\tilde{p}_\nu$. Then $F_\nu$ is a real Banach lattice and also a $AL-$ space, cause the norm is additive on the cone of positive elements. By the Kakutani representation theorem, $F_\nu$ as Banach lattice is isomorphic to some $L^1(\mu)$.
Really grateful to anyone giving me some background to understand this. (Even just the first part).
EDIT: As requested, I'm adding more informations to the question, adding the definition of Lattice Ideal:
A subset $S$ of a vector lattice $E$ is called solid if $x \in S$, $|y| \le |x|$ implies $y \in S$. A solid linear subspace of a vector lattice is called ideal. One could verify that a subspace $I$ of a Banach lattice $E$ is an ideal iif $$x \in I \Rightarrow x \in I \quad and \quad 0\le y \le x \in I \Rightarrow y \in I$$
