Added the dual of c0.

src/fa/order/lattice.tex

@@ -214,6 +214,10 @@
         \item $E^b = E^+$.
         \item The order bound dual $E^b$ equipped with its canonical ordering is a vector lattice.
         \item $E^b$ is order complete. 
+        \item For any $\phi \in E^b$ and $x \in E$ with $x \ge 0$,
+        \[
+        |\phi|(x) = \sup\bracs{\phi(y)|y \in E, |y| \le x}
+        \]
     \end{enumerate}
     
 \end{proposition}
@@ -265,6 +269,28 @@
     \]
 
     is a positive linear functional on $E$. By definition $\Phi \ge f$ for all $f \in A$. If $\psi \in \hom(E; \real)$ is a linear functional such that $\psi \ge f$ for all $f \in A$, then for any $x \in C$, $\psi(x) \ge \sup_{f \in A}f(x)$. Therefore $\Phi = \sup(A)$, and $E^b$ is order complete. 
+
+    (3): Let $\phi \in E^b$ and $x \in E$ with $x \ge 0$, then by (1), 
+    \begin{align*}
+    |\phi|(x) &= \sup(\phi([0, x])) + \sup(-\phi([0, x])) \\
+    &= \sup\bracs{\phi(u) - \phi(v)| u, v \in [0, x]}
+    \end{align*}
+
+    For any $u, v \in [0, x]$, $|u - v| \le x$, so
+    \[
+    |\phi|(x) \le \sup\bracs{\phi(y)|y \in E, |y| \le x}
+    \]
+
+    On the other hand, for any $y \in E$ with $|y| \le x$, $y^+, y^- \le x$, so
+    \[
+    \phi(y) = \phi(y^+) - \phi(y^-) \le \sup\bracs{\phi(u) - \phi(v)| u, v \in [0, x]} = |\phi|(x)
+    \]
+
+    Therefore
+    \[
+    |\phi|(x) = \sup\bracs{\phi(y)|y \in E, |y| \le x}
+    \]
+    
 \end{proof}