Added the dual of c0.

src/dg/derivative/mvt.tex

@@ -113,9 +113,9 @@
 \begin{proof}
     Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \autoref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
 
-    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, $f(y) - f(x) = g(1) - g(0)$ is contained in
     \[
-    f(y) - f(x) = g(1) - g(0) \in \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
+    \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
 
 \end{proof}

src/fa/index.tex

@@ -7,3 +7,4 @@
 \input{./norm/index.tex}
 \input{./rs/index.tex}
 \input{./lp/index.tex}
+\input{./order/index.tex}

src/fa/lc/hahn-banach.tex

@@ -140,7 +140,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $|\phi| \le \rho_M \le \rho$.
+    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $|\phi| \le \rho_M \le \rho$.
 
     (2): By (1) applied to $M = \bracs{0}$.
 
@@ -149,7 +149,7 @@
 
 \begin{proposition}
 \label{proposition:seminorm-lsc}
-    Let $E$ be a locally convex space and $\rho: E \to [0, \infty)$ be a continuous seminorm, then $\rho: E_w \to [0, \infty)$ is lower semicontinuous and Borel measurable. 
+    Let $E$ be a locally convex space and $\rho: E \to [0, \infty)$ be a continuous seminorm, then $\rho: E_w \to [0, \infty)$ is lower semicontinuous and Borel measurable with respect to the weak topology.
 \end{proposition}
 \begin{proof}
     Let $x \in E$, then there exists $\phi_x \in E^*$ such that $\dpn{x, \phi_x}{E} = \rho(x)$ and $|\phi_x| \le \rho$. Thus

src/fa/order/index.tex

@@ -0,0 +1,5 @@
+\chapter{Order Structures}
+\label{chap:order-structure}
+
+\input{./lattice.tex}
+\input{./norm.tex}

src/fa/order/lattice.tex

@@ -0,0 +1,299 @@
+\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}
+
+
+
+
+

src/fa/order/norm.tex

@@ -0,0 +1,84 @@
+\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}
+
+

src/fa/tvs/spaces-of-linear.tex

@@ -188,7 +188,7 @@
 
     and
     \[
-    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - Tx
+    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
     \]
 
     are continuous with respect to the product topology. Since

src/measure/lebesgue-integral/complex.tex

@@ -28,7 +28,7 @@
 \label{definition:positive-negative-parts}
     Let $X$ be a set and $f: X \to \real$ be a function, then
     \[
-    f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
+    f^+ = f \vee 0 \quad f^- = -(f \wedge 0)
     \]
 
     are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
@@ -56,27 +56,14 @@
     Let $(X, \cm, \mu)$ be a measure space, then the integral is a linear functional on $\mathcal{L}^1(X)$ such that for any $f \in \mathcal{L}^1(X)$, $\abs{\int f d\mu} \le \int \abs{f}d\mu$.
 \end{proposition}
 \begin{proof}
-    Let $f, g \in \mathcal{L}^1(X)$ and $\lambda \in \complex$. First suppose that $f, g$ are $\real$-valued and $\lambda \in \real$. In which case,
+    By \autoref{lemma:positive-functional-extension}, the mapping
     \[
-    \int \lambda f d\mu = \int (\lambda f)^+ d\mu - \int (\lambda f)^- d\mu
-    = \begin{cases}
-        \lambda\int f^+ d\mu - \lambda\int f^- d\mu &\lambda \ge 0 \\
-        -\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
-    \end{cases}
+    I: \mathcal{L}^1(X; \real) \to \real \quad f \mapsto \int f^+ d\mu - \int f^- d\mu
     \]
 
-    by \autoref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
+    is a $\real$-linear functional on $\mathcal{L}^1(X; \real)$. 
 
-    Let $h = f + g$, then $h = h^+ - h^- = f^+ + g^+ - f^- - g^-$, so
-    \begin{align*}
-        h^+ + f^- + g^- &= h^- + f^+ + g^+ \\
-        \int h^+d\mu + \int f^- d\mu + \int g^-d\mu &= \int h^- d\mu + \int f^+ d\mu + \int g^+ d\mu \\
-        \int h^+ d\mu - \int h^- d\mu &= \int f^+ d\mu - \int f^- d\mu + \int g^+ d\mu - \int g^- d\mu \\
-        &= \int f d\mu + \int g d\mu
-    \end{align*}
-    by \autoref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
-
-    Now suppose that $f, g$ are $\complex$-valued and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
+    Let $f, g \in L^1(X; \complex)$ and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
     \begin{align*}
         \int (\alpha f)d\mu &= \int \text{Re}(\lambda f)d\mu + i\int \text{Im}(\lambda f)d\mu \\
         &= \int \alpha \text{Re}(f) - \beta \text{Im}(f)d\mu + i\int \beta\text{Re}(f) + \alpha \text{Im}(f)d\mu \\

src/measure/radon/c0.tex

@@ -0,0 +1,93 @@
+\section{Dual of $C_0$}
+\label{section:dual-of-c0}
+
+\begin{lemma}[Jordan Decomposition]
+\label{lemma:positive-linear-jordan}
+    Let $X$ be a topological space and $I \in C_0(X; \real)^*$, then there exists positive linear functionals $I^+, I^- \in C_0(X; \real)^*$ such that:
+    \begin{enumerate}
+        \item $I = I^+ - I^-$.
+        \item $I^+ \perp I^-$. 
+        \item $\norm{I^+}_{C_0(X; \real)^*}, \norm{I^-}_{C_0(X; \real)^*} \le \norm{I}_{C_0(X; \real)^*}$.
+    \end{enumerate}
+    
+\end{lemma}
+\begin{proof}
+    (1), (2): Since $C_0(X; \real)$ is a Banach lattice, $C_0(X; \real)^*$ is also a Banach lattice by \autoref{proposition:banach-lattice-dual}. Therefore there exists positive linear functionals $I^+, I^- \in C(X; \real)^*$ such that $I = I^+ - I^-$ and $I^+ \perp I^-$. 
+
+    (3): By \autoref{proposition:banach-lattice-properties}.
+\end{proof}
+
+\begin{definition}[Radon Measure]
+\label{definition:radon-measure-extended}
+    Let $X$ be a LCH space and $\mu$ be a Borel signed/vector measure on $X$, then $\mu$ is \textbf{Radon} if $|\mu|$ is Radon. 
+\end{definition}
+
+
+
+\begin{definition}[Space of Finite Radon Measures]
+\label{definition:space-radon-measures}
+    Let $X$ be a LCH space and $E$ be a normed space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
+\end{definition}
+\begin{proof}
+    Let $\mu, \nu \in M_R(X; E)$, then for any $A \in \cb_X$, $|\mu + \nu|(A) \le |\mu|(A) + |\nu|(A)$. Let $\eps > 0$, then by outer regularity and \autoref{proposition:radon-measurable-description}, there exists $K \subset A$ compact and $U \in \cn^o(A)$ such that $(|\mu| + |\nu|)(A \setminus K), (|\mu| + |\nu|)(U \setminus A) < \eps$. Therefore $|\mu + \nu|$ is regular on all Borel sets, and hence Radon. 
+\end{proof}
+
+
+\begin{theorem}[Riesz Representation Theorem]
+\label{theorem:riesz-radon-c0}
+    Let $X$ be an LCH space. For each $\mu \in M_R(X; \complex)$, let
+    \[
+    I_\mu: C_0(X; \complex) \to \complex \quad \dpn{f, I_\mu}{C_0(X; \complex)} = \int f d\mu
+    \]
+    
+    then the map
+    \[
+    M_R(X; \complex) \to C_0(X; \complex) \quad \mu \mapsto I_\mu
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    Let $f \in C_0(X; \complex)$, then
+    \[
+    \abs{\int f d\mu} \le \int \norm{f}_u d\mu \le \norm{f}_u \cdot |\mu|(X)
+    \]
+
+    so $\norm{I_\mu}_{C_0(X; \complex)} \le \norm{\mu}_{\text{var}}$. 
+    
+    On the other hand, let $\seqf{A_j} \subset \cb_X$ such that $X = \bigsqcup_{j = 1}^nA_j$. Let $\eps > 0$, then by \autoref{proposition:radon-measurable-description} of $|\mu|$, there exists compact sets $\seqf{K_j}$ such that for each $1 \le j \le n$, $K_j \subset A_j$ and $|\mu(K_j) - \mu(A_j)| < \eps$. By outer regularity and \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{\phi_j} \subset C_c(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < \eps/|\mu(K_j)|$. Let $\phi = \sum_{j = 1}^n \ol{\sgn(\mu(K_j))}\phi_j$, then $\norm{\phi}_u \le 1$ and
+    \[
+        \abs{\sum_{j = 1}^n |\mu(A_j)| - \int \phi d\mu} \le \sum_{j = 1}^n |\mu(A_j) - \mu(K_j)| + \sum_{j = 1}^n \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < 2n\eps
+    \]
+
+    so
+    \[
+    \abs{\int \phi d\mu} \le \sum_{j = 1}^n |\mu(A_j)| + 2n\eps
+    \]
+
+    As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)} \ge \sum_{j = 1}^n |\mu(A_j)|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)} \ge |\mu|(X)$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
+
+    Finally, let $I \in C_0(X; \complex)^*$, then there exists bounded linear functionals $I_r, I_i \in C_0(X; \real)^*$ such that for any $f \in C_0(X; \real)$, 
+    \[
+    \dpn{f, I}{C_0(X; \real)^*} = \dpn{f, I_r}{C_0(X; \real)} + i\dpn{f, I_i}{C_0(X; \real)}
+    \]
+
+    By \autoref{lemma:positive-linear-jordan}, there exists bounded positive linear functionals $I_r^+, I_r^-, I_i^+, I_i^-$ such that for any $f \in C_0(X; \real)$, 
+    \begin{align*}
+    \dpn{f, I}{C_0(X; \real)} &= \dpn{f, I_r^+}{C_0(X; \real)} - \dpn{f, I_r^-}{C_0(X; \real)} \\
+    &+ i\dpn{f, I_i^-}{C_0(X; \real)} - i\dpn{f, I_i^-}{C_0(X; \real)}
+    \end{align*}
+
+    Thus by the Riesz representation theorem, there exists finite Radon measures $\mu_r^+, \mu_r^-, \mu_i^+, \mu_i^- \in M_R(X; \complex)$ such that for any $f \in C_0(X; \real)$, 
+    \[
+    \dpn{f, I}{C_0(X; \real)} = \int f d\mu_r^+ - \int f d\mu_r^- + i\int f d\mu_i^+ - i\int f d\mu_i^-
+    \]
+
+    Let $\mu = \mu_r^+ - \mu_r^- + i\mu_i^+ - i\mu_i^-$, then $I = I_\mu$, and the map $\mu \mapsto I_\mu$ is surjective. 
+    
+    
+\end{proof}
+
+
+
+

src/measure/radon/index.tex

@@ -3,4 +3,5 @@
 
 \input{./radon.tex}
 \input{./riesz.tex}
+\input{./c0.tex}
 

src/topology/main/c0.tex

@@ -31,7 +31,7 @@
 
     If $E$ is complete, then $BC(X; E)$ is complete by \autoref{definition:bounded-continuous-function-space}. Since $C_0(X; E)$ is a closed subspace, it is complete by \autoref{proposition:complete-closed}.
 
-    (3): Let $f \in C_0(X; E)$ and $U \in \cn_E^o(0)$ be balanced. By \hyperref[Urysohn's Lemma]{lemma:lch-urysohn}, there exists $\phi \in C_c(X; [0, 1])$ such that $\phi|_{\bracs{f \not\in U}} = 1$. In which case, $\phi f \in C_c(X; E)$ with
+    (3): Let $f \in C_0(X; E)$ and $U \in \cn_E^o(0)$ be balanced. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\phi \in C_c(X; [0, 1])$ such that $\phi|_{\bracs{f \not\in U}} = 1$. In which case, $\phi f \in C_c(X; E)$ with
     \[
     (\phi f - f)(X) = \underbrace{(\phi f - f)(\bracs{f \not\in U})}_{0} + \underbrace{(\phi f - f)(\bracs{f \in U})}_{\in U} \in U
     \]