\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]
\label{definition:vector-lattice}
    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. 
\end{definition}


\begin{definition}[Absolute Value]
\label{definition:order-absolute-value}
    Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| = x \vee -x$ is the \textbf{absolute value} of $x$. 
\end{definition}

\begin{lemma}
\label{lemma:absolute-ge-0}
    Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| \ge 0$.
\end{lemma}
\begin{proof}
    For any $x \in E$,
    \[
    2|x| = 2(x \vee (-x)) \ge x + -x = 0
    \]
\end{proof}


\begin{definition}[Disjoint]
\label{definition:order-disjoint}
    Let $(E, \le)$ be a vector lattice and $x, y \in E$, then $x$ and $y$ are \textbf{disjoint}, denoted $x \perp y$, if $|x| \wedge |y| = 0$.
\end{definition}


\begin{proposition}[{{\cite[V.1.1]{SchaeferWolff}}}]
\label{proposition:lattice-properties}
    Let $(E, \le)$ be a vector lattice, then:
    \begin{enumerate}
        \item For any $x, y \in E$,
        \[
        x + y = x \vee y + x \wedge y
        \]

        \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 $x = x^+ - x^-$.
    \end{enumerate}

    For any $x, y \in E$ and $\lambda \in \real$, 
    \begin{enumerate}
        \item[(3)] $|\lambda x| = |\lambda| \cdot |x|$
        \item[(4)] $|x + y| \le |x| + |y|$.
    \end{enumerate}
    
    Finally, for any $x, y \in E$ with $x, y \ge 0$, 
    \begin{enumerate}
        \item[(5)] $[0, x] + [0, y] = [0, x + y]$.
    \end{enumerate}
    
\end{proposition}
\begin{proof}
    (1): By \autoref{proposition:ordered-vector-space-properties},
    \begin{align*}
        x \vee y + x \wedge y - x - y &= 0 \vee (y - x) + (x - y) \wedge 0 \\
        &= 0 \vee (y - x) - 0 \vee (y - x) = 0
    \end{align*}
    
    (2): By (1),
    \[
    x  = x + 0 = x \vee 0 + x \wedge 0 = x^+ - x^-
    \]

    By \autoref{proposition:ordered-vector-space-properties} and \autoref{lemma:absolute-ge-0},
    \[
        x^+ + x^- = x \vee 0 + (-x \vee 0) = x \vee -x \vee 0 = |x|
    \]
    
    and
    \[
    x^+ \vee x^- = x \vee 0 \vee (-x) \vee 0 = |x|
    \]

    Since $x^+, x^- \ge 0$, $|x^+| = x^+$ and $|x^-| = x^-$, so by (1),
    \[
    x^+ \wedge x^- = x^+ + x^- - x^+ \vee x^- = 0
    \]

    so $x^+ \perp x^-$. 

    Now, let $y, z \in E$ with $y, z \ge 0$ such that $x = y - z$, then
    \[
    x^+ = x \vee 0 = (y - z) \vee 0 \le y \vee 0 = y
    \]

    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 $z = x^-$. 

    (3): For any $\lambda > 0$, by (LO2),
    \[
    |\lambda x| = (\lambda x) \vee (-\lambda x) = \lambda (x \vee -x) = \lambda |x|
    \]

    (4): For any $x, y \in E$, by \autoref{proposition:ordered-vector-space-properties},
    \begin{align*}
    x^+ + y^+ &= (x \vee 0) + (y \vee 0) = x \vee y \vee (x + y) \vee 0 \\
    &\ge (x + y) \vee 0 = (x + y)^+
    \end{align*}
    
    Likewise, $x^- + y^- \ge (x + y)^-$. Thus
    \[
    |x+y| = (x+y)^+ + (x + y)^- \le x^+ + y^+ + x^- + y^- = |x| + |y|
    \]

    (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
    \[
        v = z - z \wedge x = z + (-z \vee -x) = (0 \vee z - x) \le z - x \le y
    \]
\end{proof}

\begin{lemma}
\label{lemma:positive-functional-extension}
    Let $(E, \le)$ be a vector lattice, $C = \bracs{x \in E|x \ge 0}$ and $\phi: C \to [0, \infty)$ such that:
    \begin{enumerate}
        \item For any $x \in C$ and $\lambda \in \real$ with $\lambda \ge 0$, $\phi(\lambda x) = \lambda \phi(x)$.
        \item For any $x, y \in C$, $\phi(x + y) = \phi(x) + \phi(y)$.
    \end{enumerate}
    
    then the mapping
    \[
    \Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
    \]

    is a positive linear functional on $E$. 
\end{lemma}
\begin{proof}
    For any $\lambda \in \real$ with $\lambda \ge 0$, 
    \begin{align*}
    \Phi(\lambda x) &= \phi((\lambda x)^+) - \phi((\lambda x)^-) \\
    &= \lambda\phi(x^+) - \lambda\phi(x^-) = \lambda\Phi(x)
    \end{align*}
    
    Likewise, if $\lambda < 0$, then 
    \begin{align*}
    \Phi(\lambda x) &= \phi((\lambda x)^+) - \phi((\lambda x)^-) \\
    &= -\lambda\phi(x^-) + \lambda\phi(x^+) = \lambda\Phi(x)
    \end{align*}

    For any $x, y \in E$, let $z = x + y$, then $z = z^+ - z^- = x^+ + y^+ - x^- - y^-$. Thus
    \begin{align*}
        z^+ + x^- + y^- &= z^- + x^+ + y^+ \\
        \phi(z^+) + \phi(x^-) + \phi(y^-) &= \phi(z^-) + \phi(x^+) + \phi(y^+) \\
        \phi(z^+) - \phi(z^-) &= \phi(x^+) - \phi(x^-) + \phi(y^+) - \phi(y^-) \\
        \Phi(z) &= \Phi(x) + \Phi(y)
    \end{align*}
\end{proof}


\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
\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}
        \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}
\begin{proof}
    (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
    \[
    \Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))
    \]

    then for any $x, y \in C$ and $\lambda \in \real$ with $\lambda \ge 0$,
    \begin{align*}
    \Phi^+(\lambda x + y) &= \sup(f([0, \lambda x + y])) = \sup(f(\lambda [0,x]) + f([0, y])) \\
    &= \lambda \sup(f([0, x])) + \sup(f([0, y])) = \lambda \Phi^+(x) + \Phi^+(y)
    \end{align*}

    Let
    \[
    \Phi: E \to \real \quad x \mapsto \Phi^+(x^+) - \Phi^+(x^-)
    \]

    then $\Phi$ is a positive linear functional by \autoref{lemma:positive-functional-extension}.

    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 $\Phi = 0 \vee \phi$. 
    
    Since $0 \wedge \phi = -(0 \vee -\phi)$, $\phi = (0 \vee \phi) - (0 \vee -\phi)$ is a sum of two positive linear functionals, so $E^b = E^+$. 
    
    (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. 

    (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
    \[
    \phi: C \to [0, \infty) \quad x \mapsto \sup_{f \in A}f(x)
    \]

    then for any $\lambda \ge 0$ and $x \in C$, $\phi(\lambda x) = \lambda \phi(x)$. Let $x, y \in C$, then for any $f \in A$, 
    \[
    f(x + y) = f(x) + f(y) \le \phi(x) + \phi(y)
    \]

    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
    \[
    h(x + y) = h(x) + h(y) \ge f(x) + g(y)
    \]

    Since such an $h \in A$ exists for all $f, g \in A$, $\phi(x + y) \ge \phi(x) + \phi(y)$. 

    By \autoref{lemma:positive-functional-extension}, the mapping 
    \[
    \Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
    \]

    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}





