Added the dual of c0.

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}