Added the dual of c0.

src/fa/order/index.tex

@@ -2,3 +2,4 @@
 \label{chap:order-structure}
 
 \input{./lattice.tex}
+\input{./norm.tex}

src/fa/order/lattice.tex

@@ -214,6 +214,10 @@
         \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}
@@ -265,6 +269,28 @@
     \]
 
     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/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
     \[
     (\phi f - f)(X) = \underbrace{(\phi f - f)(\bracs{f \not\in U})}_{0} + \underbrace{(\phi f - f)(\bracs{f \in U})}_{\in U} \in U
     \]