Added the dual of c0.

src/fa/order/norm.tex

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+\section{Banach Lattices}
+\label{section:banach-lattice}
+
+\begin{definition}[Banach Lattice]
+\label{definition:banach-lattice}
+    Let $(E, \normn{\cdot}_E)$ be a Banach space over $\real$ and $\le$ be a partial order on $E$, then $(E, \normn{\cdot}_E, \le)$ is a \textbf{Banach lattice} if for any $x, y \in E$, $|x| \le |y|$ implies that $\normn{x}_E \le \normn{x}_E$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:banach-lattice-properties}
+    Let $(E, \normn{\cdot}_E, \le)$ be a Banach lattice, then:
+    \begin{enumerate}
+        \item For any $x \in E$, $\normn{\ |x|\ }_E = \normn{x}_E$.
+        \item For any $x \in E$, $\normn{x^+}_E, \normn{x^-}_E \le \normn{x}_E$.
+        \item For any $x \in E$ and positive linear functional $\phi \in E^*$, 
+        \[
+        \norm{\phi}_{E^*} = \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x \ge 0, \norm{x}_E = 1}
+        \]
+    \end{enumerate}
+
+    % More to be added.
+
+\end{proposition}
+\begin{proof}
+    (1): Let $x \in E$, then $|\ |x|\ | = |x|$, so $\normn{\ |x|\ }_E = \normn{x}_E$.
+
+    (2): Let $x \in E$, then $|x| = x^+ + x^-$, so
+    \[
+    \normn{x}_E = \normn{x^+ + x^-}_E \ge \normn{x^+}_E, \normn{x^-}_E
+    \]
+
+    (3): Since $|x| = x \vee (-x)$,
+    \[
+    \norm{\phi}_{E^*} = \sup_{\substack{x \in E \\ \norm{x}_E = 1}}\dpn{x, \phi}{E} \le \sup_{\substack{x \in E \\ \norm{x}_E = 1}}\dpn{|x|, \phi}{E} \le \sup_{\substack{x \in E \\ x \ge 0 \\ \norm{x}_E = 1}}\dpn{x, \phi}{E} \le \norm{\phi}_{E^*}
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:banach-lattice-dual}
+    Let $(E, \normn{\cdot}_E, \le)$ be a Banach lattice and $(E^*, \normn{\cdot}_{E^*}, \le)$ be its dual, equipped with the canonical ordering, then $E^*$ is a Banach lattice. 
+\end{proposition}
+\begin{proof}
+    Let $x, y \in E$ and $z \in [x, y]$, then
+    \[
+    |z| = z \vee (-z) \le y \vee (-x)
+    \]
+
+    so $[x, y]$ is bounded by $\norm{x}_E \vee \norm{y}_E$. Thus $E^* \subset E^b$. 
+
+    Let $\phi \in E^*$, then by \autoref{proposition:order-vector-dual}, $0 \vee \phi$ exists in $E^b$. For any $x \in E$ with $x \ge 0$, 
+    \[
+    (0 \vee \phi)(x) = \sup(\phi([0, x])) \le \norm{\phi}_{E^*} \cdot \norm{x}_E
+    \]
+
+    so for arbitrary $x \in E$,
+    \begin{align*}
+    (0 \vee \phi)(x) &= (0 \vee \phi)(x^+) - (0 \vee \phi)(x^-) \\
+    &\le \norm{\phi}_{E^*}(\normn{x^+}_E \vee \normn{x^-}_E) \le \norm{\phi}_{E^*} \cdot \norm{x}_E
+    \end{align*}
+    
+
+    and $\norm{0 \vee \phi}_{E^*} \le \norm{\phi}_{E^*}$. Therefore $E^*$ is a vector lattice. 
+
+    Now, let $x \in E$ with $\norm{x}_E = 1$, then
+    \begin{align*}
+        \dpn{x, \phi}{E^*} &= \dpn{x, \phi^+}{E} + \dpn{-x, \phi^-}{E} \\
+        &\le \dpn{|x|, \phi^+}{E} + \dpn{|x|, \phi^-}{E} \\
+        &= \dpn{|x|, \phi^+ + \phi^-}{E} \le \norm{\ |\phi|\ }_{E^*}
+    \end{align*}
+
+    On the other hand, by \autoref{proposition:order-vector-dual}, for any $x \in E$ with $x \ge 0$ and $\norm{x}_E = 1$,
+    \[
+        \dpn{x, |\phi|}{E^*} = \sup\bracsn{\dpn{y, \phi}{E}|y \in E, |y| \le x} \le \norm{\phi}_{E^*}
+    \]
+
+    so $\norm{\ |\phi|\ }_{E^*} \le \norm{\phi}_{E^*}$. Therefore for any $\phi, \psi \in E$ with $|\phi| \le |\psi|$, 
+    \begin{align*}
+        \norm{\phi}_{E^*} &= \norm{\ |\phi|\ }_{E^*} = \sup\bracsn{\dpn{x, |\phi|}{E}|x \in E, x \ge 0, \norm{x}_E = 1} \\
+        &\le \sup\bracsn{\dpn{x, |\psi|}{E}|x \in E, x \ge 0, \norm{x}_E = 1} \\
+        &= \norm{\ |\psi|\ }_{E^*} \norm{\psi}_E
+    \end{align*}
+\end{proof}
+
+