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Reworked the order chapter.

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Bokuan Li
2026-07-07 14:40:27 -04:00
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src/fa/order/index.tex
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\chapter{Order Structures}
\label{chap:order-structure}
\input{./order.tex}
\input{./positive.tex}
\input{./lattice.tex}
\input{./norm.tex}
src/fa/order/lattice.tex
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\section{Vector Lattices}
\label{section:vector-lattice}
\begin{definition}[Ordered Vector Space]
\label{definition:ordered-vector-space}
Let $E$ be a vector space over $\real$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
\begin{enumerate}
\item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
\item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
\end{enumerate}
\end{definition}
\begin{proposition}
\label{proposition:ordered-vector-space-properties}
Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then
\begin{enumerate}
\item $\sup(A + B) = \sup(A) + \sup(B)$.
\item $\sup(A) = -\inf (-A)$
\end{enumerate}
\end{proposition}
\begin{proof}
(1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
(2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$.
\end{proof}
\begin{definition}[Interval]
\label{definition:ordered-vector-space-interval}
Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
\[
[x, y] = \bracs{z \in E| x \le z \le y}
\]
is the \textbf{order interval} with endpoints $x$ and $y$.
\end{definition}
\begin{definition}[Order Bounded]
\label{definition:ordered-vector-space-bounded}
Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
\end{definition}
\begin{definition}[Order Complete]
\label{definition:order-vector-complete}
Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
\end{definition}
\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 and $\Phi^+ \in \hom(E; \real)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\Phi^+(x) \ge 0$. The subspace $E^+ \subset \hom(E; \real)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$.
\end{definition}
\begin{definition}[Vector Lattice]
@@ -249,7 +192,7 @@
\end{proof}
\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
\begin{proposition}
\label{proposition:order-vector-dual}
Let $(E, \le)$ be an ordered vector space. If for any $x, y \in E$ with $x, y \ge 0$, $[0, x] + [0, y] = [0, x + y]$, then:
\begin{enumerate}
@@ -264,7 +207,7 @@
\end{proposition}
\begin{proof}
\begin{proof}[Proof, {{\cite[V.1.4]{SchaeferWolff}}}. ]
(1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
\[
\Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))
src/fa/order/order.tex
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\section{Ordered Vector Spaces}
\label{section:ovs}
\begin{definition}[Ordered Vector Space]
\label{definition:ordered-vector-space}
Let $E$ be a vector space over $K \in \RC$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
\begin{enumerate}
\item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
\item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
\end{enumerate}
The set $C = \bracs{x \in E|x \ge 0}$ is the \textbf{positive cone} of $E$.
\end{definition}
\begin{definition}[Ordered Topological Vector Space]
\label{definition:ordered-tvs}
Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an \textbf{ordered topological vector space} if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed.
\end{definition}
\begin{proposition}
\label{proposition:ordered-vector-space-properties}
Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then
\begin{enumerate}
\item $\sup(A + B) = \sup(A) + \sup(B)$.
\item $\sup(A) = -\inf (-A)$
\end{enumerate}
\end{proposition}
\begin{proof}
(1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
(2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$.
\end{proof}
\begin{definition}[Interval]
\label{definition:ordered-vector-space-interval}
Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
\[
[x, y] = \bracs{z \in E| x \le z \le y}
\]
is the \textbf{order interval} with endpoints $x$ and $y$.
\end{definition}
\begin{definition}[Order Bounded]
\label{definition:ordered-vector-space-bounded}
Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
\end{definition}
\begin{definition}[Order Complete]
\label{definition:order-vector-complete}
Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
\end{definition}
src/fa/order/positive.tex
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\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}