From b591904469ec93887138669222931f0ebd2e9bf3 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 16 Mar 2026 21:05:12 -0400 Subject: [PATCH] Added the dual of c0. --- src/fa/order/index.tex | 1 + src/fa/order/lattice.tex | 26 +++++++++++ src/fa/order/norm.tex | 84 +++++++++++++++++++++++++++++++++ src/measure/radon/c0.tex | 93 +++++++++++++++++++++++++++++++++++++ src/measure/radon/index.tex | 1 + src/topology/main/c0.tex | 2 +- 6 files changed, 206 insertions(+), 1 deletion(-) create mode 100644 src/fa/order/norm.tex create mode 100644 src/measure/radon/c0.tex diff --git a/src/fa/order/index.tex b/src/fa/order/index.tex index 1c6d5bd..dd4f0bc 100644 --- a/src/fa/order/index.tex +++ b/src/fa/order/index.tex @@ -2,3 +2,4 @@ \label{chap:order-structure} \input{./lattice.tex} +\input{./norm.tex} \ No newline at end of file diff --git a/src/fa/order/lattice.tex b/src/fa/order/lattice.tex index 9d8797d..99462bc 100644 --- a/src/fa/order/lattice.tex +++ b/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} diff --git a/src/fa/order/norm.tex b/src/fa/order/norm.tex new file mode 100644 index 0000000..970aa8b --- /dev/null +++ b/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} + + diff --git a/src/measure/radon/c0.tex b/src/measure/radon/c0.tex new file mode 100644 index 0000000..22461a5 --- /dev/null +++ b/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} + + + + diff --git a/src/measure/radon/index.tex b/src/measure/radon/index.tex index 2552f39..00883d4 100644 --- a/src/measure/radon/index.tex +++ b/src/measure/radon/index.tex @@ -3,4 +3,5 @@ \input{./radon.tex} \input{./riesz.tex} +\input{./c0.tex} diff --git a/src/topology/main/c0.tex b/src/topology/main/c0.tex index 74b6d9d..bc0419f 100644 --- a/src/topology/main/c0.tex +++ b/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 \[ (\phi f - f)(X) = \underbrace{(\phi f - f)(\bracs{f \not\in U})}_{0} + \underbrace{(\phi f - f)(\bracs{f \in U})}_{\in U} \in U \]