Reworked the order chapter.

src/fa/order/positive.tex

@@ -0,0 +1,34 @@
+\section{The Order Dual}
+\label{section:order-dual}
+
+
+\begin{definition}[Order Bounded Dual]
+\label{definition:order-bounded-dual}
+    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
+\end{definition}
+
+
+\begin{definition}[Order Dual]
+\label{definition:order-dual}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$ and $\Phi^+ \in \hom(E; K)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\text{Re}\dpn{x, \Phi^+}{E} \ge 0$. The subspace $E^+ \subset \hom(E; K)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
+\end{definition}
+
+
+\begin{theorem}[Bauer-Namioka]
+\label{theorem:bauer-namioka}
+    Let $E$ be an ordered vector space over $K \in \RC$ with positive cone $C$, $\topo$ be a vector space topology on $E$\footnote{The order and the topology need not to be compatible. }, $F \subset E$ be a subspace, and $\phi \in (F, \topo)^*$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a continuous positive linear functional $\Phi \in (E, \topo)^*$ such that $\Phi|_F = \phi$. 
+        \item There exists $U \in \cn_\topo(0)$ convex such that
+        \[
+        \sup\bracs{\text{Re}\dpn{x, \phi}{F}|x \in F \cap (U - C)} < \infty
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[V.5.4]{SchaeferWolff}}}. ]
+    (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_\topo(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E} \le \text{Re}\dpn{x, \Phi}{E} < 1$. 
+
+    (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \autoref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. 
+
+    After rescaling, assume without loss of generality that $\bracs{\phi = \alpha} \subset \bracs{\Phi = \alpha}$, then $\Phi \in E^*$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E} < \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E} \ge 0$, so $\Phi$ is the desired extension.
+\end{proof}