Added vector lattices.
This commit is contained in:
Bokuan Li
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\input{./norm/index.tex}
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\input{./rs/index.tex}
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\input{./lp/index.tex}
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\input{./order/index.tex}
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src/fa/order/index.tex
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src/fa/order/index.tex
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\chapter{Order Structures}
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\label{chap:order-structure}
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\input{./lattice.tex}
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src/fa/order/lattice.tex
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src/fa/order/lattice.tex
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\section{Vector Lattices}
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\label{section:vector-lattice}
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\begin{definition}[Ordered Vector Space]
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\label{definition:ordered-vector-space}
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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
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\begin{enumerate}
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\item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
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\item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
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\end{enumerate}
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\end{definition}
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\begin{proposition}
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\label{proposition:ordered-vector-space-properties}
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Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then
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\begin{enumerate}
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\item $\sup(A + B) = \sup(A) + \sup(B)$.
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\item $\sup(A) = -\inf (-A)$
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(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 $\sup(A) + \sup(B) = \sup(A + B)$.
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(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 $\sup(A) \le -\inf(-A)$.
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\end{proof}
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\begin{definition}[Interval]
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\label{definition:ordered-vector-space-interval}
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Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
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\[
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[x, y] = \bracs{z \in E| x \le z \le y}
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\]
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is the \textbf{order interval} with endpoints $x$ and $y$.
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\end{definition}
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\begin{definition}[Order Bounded]
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\label{definition:ordered-vector-space-bounded}
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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 $A \subset [x, y]$.
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\end{definition}
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\begin{definition}[Order Complete]
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\label{definition:order-vector-complete}
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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.
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\end{definition}
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\begin{definition}[Order Bounded Dual]
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\label{definition:order-bounded-dual}
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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 $E$.
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\end{definition}
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\begin{definition}[Order Dual]
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\label{definition:order-dual}
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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 $E$.
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\end{definition}
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\begin{definition}[Vector Lattice]
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\label{definition:vector-lattice}
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Let $(E, \le)$ be an ordered vector space, then $E$ is a \textbf{vector lattice} if for any $x, y \in E$, $x \vee y = \sup\bracs{x, y}$ and $x \wedge y = \inf\bracs{x, y}$ exist.
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\end{definition}
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\begin{definition}[Absolute Value]
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\label{definition:order-absolute-value}
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Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| = x \vee -x$ is the \textbf{absolute value} of $x$.
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\end{definition}
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\begin{lemma}
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\label{lemma:absolute-ge-0}
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Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| \ge 0$.
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\end{lemma}
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\begin{proof}
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For any $x \in E$,
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\[
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2|x| = 2(x \vee (-x)) \ge x + -x = 0
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\]
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\end{proof}
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\begin{definition}[Disjoint]
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\label{definition:order-disjoint}
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Let $(E, \le)$ be a vector lattice and $x, y \in E$, then $x$ and $y$ are \textbf{disjoint}, denoted $x \perp y$, if $|x| \wedge |y| = 0$.
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\end{definition}
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\begin{proposition}[{{\cite[V.1.1]{SchaeferWolff}}}]
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\label{proposition:lattice-properties}
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Let $(E, \le)$ be a vector lattice, then:
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\begin{enumerate}
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\item For any $x, y \in E$,
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\[
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x + y = x \vee y + x \wedge y
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\]
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\item Let $x \in E$, $x^+ = x \vee 0 \ge 0$, and $x^- = -(x \wedge 0) \ge 0$, then $x = x^+ - x^-$ and $|x| = x^+ + x^-$. Moreover, $(x^+, x^-)$ are the unique disjoint non-negative elements of $E$ such that $x = x^+ - x^-$.
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\end{enumerate}
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For any $x, y \in E$ and $\lambda \in \real$,
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\begin{enumerate}
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\item[(3)] $|\lambda x| = |\lambda| \cdot |x|$
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\item[(4)] $|x + y| \le |x| + |y|$.
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\end{enumerate}
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Finally, for any $x, y \in E$ with $x, y \ge 0$,
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\begin{enumerate}
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\item[(5)] $[0, x] + [0, y] = [0, x + y]$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): By \autoref{proposition:ordered-vector-space-properties},
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\begin{align*}
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x \vee y + x \wedge y - x - y &= 0 \vee (y - x) + (x - y) \wedge 0 \\
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&= 0 \vee (y - x) - 0 \vee (y - x) = 0
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\end{align*}
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(2): By (1),
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\[
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x = x + 0 = x \vee 0 + x \wedge 0 = x^+ - x^-
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\]
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By \autoref{proposition:ordered-vector-space-properties} and \autoref{lemma:absolute-ge-0},
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\[
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x^+ + x^- = x \vee 0 + (-x \vee 0) = x \vee -x \vee 0 = |x|
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\]
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and
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\[
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x^+ \vee x^- = x \vee 0 \vee (-x) \vee 0 = |x|
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\]
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Since $x^+, x^- \ge 0$, $|x^+| = x^+$ and $|x^-| = x^-$, so by (1),
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\[
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x^+ \wedge x^- = x^+ + x^- - x^+ \vee x^- = 0
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\]
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so $x^+ \perp x^-$.
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Now, let $y, z \in E$ with $y, z \ge 0$ such that $x = y - z$, then
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\[
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x^+ = x \vee 0 = (y - z) \vee 0 \le y \vee 0 = y
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\]
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and $x^- \le z$. If $y \perp z$, then $y - x^+ \perp z - x^-$. Since $y - x^+ = z - x^-$, $y - x^+ = z - x^- = 0$, so $y = x^+$ and $z = x^-$.
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(3): For any $\lambda > 0$, by (LO2),
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\[
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|\lambda x| = (\lambda x) \vee (-\lambda x) = \lambda (x \vee -x) = \lambda |x|
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\]
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(4): For any $x, y \in E$, by \autoref{proposition:ordered-vector-space-properties},
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\begin{align*}
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x^+ + y^+ &= (x \vee 0) + (y \vee 0) = x \vee y \vee (x + y) \vee 0 \\
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&\ge (x + y) \vee 0 = (x + y)^+
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\end{align*}
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Likewise, $x^- + y^- \ge (x + y)^-$. Thus
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\[
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|x+y| = (x+y)^+ + (x + y)^- \le x^+ + y^+ + x^- + y^- = |x| + |y|
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\]
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(5): Let $x, y \in E$ with $x, y \ge 0$, then $[0, x] + [0, y] \subset [0, x + y]$. For any $z \in [0, x + y]$, let $u = z \wedge x$ and $v = z - u$, then
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\[
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v = z - z \wedge x = z + (-z \vee -x) = (0 \vee z - x) \le z - x \le y
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\]
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\end{proof}
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\begin{lemma}
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\label{lemma:positive-functional-extension}
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Let $(E, \le)$ be a vector lattice, $C = \bracs{x \in E|x \ge 0}$ and $\phi: C \to [0, \infty)$ such that:
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\begin{enumerate}
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\item For any $x \in C$ and $\lambda \in \real$ with $\lambda \ge 0$, $\phi(\lambda x) = \lambda \phi(x)$.
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\item For any $x, y \in C$, $\phi(x + y) = \phi(x) + \phi(y)$.
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\end{enumerate}
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then the mapping
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\[
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\Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
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\]
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is a positive linear functional on $E$.
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\end{lemma}
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\begin{proof}
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For any $\lambda \in \real$ with $\lambda \ge 0$,
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\begin{align*}
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\Phi(\lambda x) &= \phi((\lambda x)^+) - \phi((\lambda x)^-) \\
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&= \lambda\phi(x^+) - \lambda\phi(x^-) = \lambda\Phi(x)
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\end{align*}
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Likewise, if $\lambda < 0$, then
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\begin{align*}
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\Phi(\lambda x) &= \phi((\lambda x)^+) - \phi((\lambda x)^-) \\
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&= -\lambda\phi(x^-) + \lambda\phi(x^+) = \lambda\Phi(x)
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\end{align*}
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For any $x, y \in E$, let $z = x + y$, then $z = z^+ - z^- = x^+ + y^+ - x^- - y^-$. Thus
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\begin{align*}
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z^+ + x^- + y^- &= z^- + x^+ + y^+ \\
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\phi(z^+) + \phi(x^-) + \phi(y^-) &= \phi(z^-) + \phi(x^+) + \phi(y^+) \\
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\phi(z^+) - \phi(z^-) &= \phi(x^+) - \phi(x^-) + \phi(y^+) - \phi(y^-) \\
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\Phi(z) &= \Phi(x) + \Phi(y)
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\end{align*}
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\end{proof}
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\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
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\label{proposition:order-vector-dual}
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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:
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\begin{enumerate}
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\item $E^b = E^+$.
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\item The order bound dual $E^b$ equipped with its canonical ordering is a vector lattice.
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\item $E^b$ is order complete.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
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\[
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\Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))
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\]
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then for any $x, y \in C$ and $\lambda \in \real$ with $\lambda \ge 0$,
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\begin{align*}
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\Phi^+(\lambda x + y) &= \sup(f([0, \lambda x + y])) = \sup(f(\lambda [0,x]) + f([0, y])) \\
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&= \lambda \sup(f([0, x])) + \sup(f([0, y])) = \lambda \Phi^+(x) + \Phi^+(y)
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\end{align*}
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Let
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\[
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\Phi: E \to \real \quad x \mapsto \Phi^+(x^+) - \Phi^+(x^-)
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\]
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then $\Phi$ is a positive linear functional by \autoref{lemma:positive-functional-extension}.
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For any $x \in C$, $\Phi(x) \ge \phi(x)$, so $\Phi \ge \phi$. On the other hand, let $\psi \in \hom(E; \real)$ such that $\psi \ge 0$ and $\psi \ge \phi$, then for each $x \in C$, $\psi(x) \ge \sup(\phi([0, x]))$. Therefore $\Phi = 0 \vee \phi$.
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Since $0 \wedge \phi = -(0 \vee -\phi)$, $\phi = (0 \vee \phi) - (0 \vee -\phi)$ is a sum of two positive linear functionals, so $E^b = E^+$.
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(2): For any $x, y \in E^b$, $x \wedge y = (x - y) \wedge 0 + y$ and $x \vee y = (x - y) \vee 0 + y$, so $E^b$ is a lattice.
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(3): Let $A \subset E^b$ be order bounded, then since $E^b$ is a lattice, assume without loss of generality that $A$ is upward-directed under the canonical ordering. Let
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\[
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\phi: C \to [0, \infty) \quad x \mapsto \sup_{f \in A}f(x)
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\]
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then for any $\lambda \ge 0$ and $x \in C$, $\phi(\lambda x) = \lambda \phi(x)$. Let $x, y \in C$, then for any $f \in A$,
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\[
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f(x + y) = f(x) + f(y) \le \phi(x) + \phi(y)
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\]
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As this holds for all $f \in A$, $\phi(x + y) \le \phi(x) + \phi(y)$. On the other hand, for any $f, g \in A$, there exists $h \in A$ such that
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\[
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h(x + y) = h(x) + h(y) \ge f(x) + g(y)
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\]
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Since such an $h \in A$ exists for all $f, g \in A$, $\phi(x + y) \ge \phi(x) + \phi(y)$.
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By \autoref{lemma:positive-functional-extension}, the mapping
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\[
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\Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
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\]
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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.
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\end{proof}
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