From f5ebcd7979b32236794d75fe3d5f127a42990be4 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 7 Jul 2026 14:40:27 -0400 Subject: [PATCH] Reworked the order chapter. --- src/fa/order/index.tex | 2 ++ src/fa/order/lattice.tex | 61 ++------------------------------------- src/fa/order/order.tex | 56 +++++++++++++++++++++++++++++++++++ src/fa/order/positive.tex | 34 ++++++++++++++++++++++ src/fa/tvs/complexify.tex | 44 ++++++++++++++++++++++++---- 5 files changed, 132 insertions(+), 65 deletions(-) create mode 100644 src/fa/order/order.tex create mode 100644 src/fa/order/positive.tex diff --git a/src/fa/order/index.tex b/src/fa/order/index.tex index dd4f0bc..90c8495 100644 --- a/src/fa/order/index.tex +++ b/src/fa/order/index.tex @@ -1,5 +1,7 @@ \chapter{Order Structures} \label{chap:order-structure} +\input{./order.tex} +\input{./positive.tex} \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 8a11b6e..3076dc6 100644 --- a/src/fa/order/lattice.tex +++ b/src/fa/order/lattice.tex @@ -1,63 +1,6 @@ \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] @@ -249,7 +192,7 @@ \end{proof} -\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}] +\begin{proposition} \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} @@ -264,7 +207,7 @@ \end{proposition} -\begin{proof} +\begin{proof}[Proof, {{\cite[V.1.4]{SchaeferWolff}}}. ] (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])) diff --git a/src/fa/order/order.tex b/src/fa/order/order.tex new file mode 100644 index 0000000..6eff7e8 --- /dev/null +++ b/src/fa/order/order.tex @@ -0,0 +1,56 @@ +\section{Ordered Vector Spaces} +\label{section:ovs} + +\begin{definition}[Ordered Vector Space] +\label{definition:ordered-vector-space} + Let $E$ be a vector space over $K \in \RC$ 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} + + The set $C = \bracs{x \in E|x \ge 0}$ is the \textbf{positive cone} of $E$. +\end{definition} + +\begin{definition}[Ordered Topological Vector Space] +\label{definition:ordered-tvs} + Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an \textbf{ordered topological vector space} if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed. +\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} diff --git a/src/fa/order/positive.tex b/src/fa/order/positive.tex new file mode 100644 index 0000000..7b0f553 --- /dev/null +++ b/src/fa/order/positive.tex @@ -0,0 +1,34 @@ +\section{The Order Dual} +\label{section:order-dual} + + +\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 over $K \in \RC$ and $\Phi^+ \in \hom(E; K)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\text{Re}\dpn{x, \Phi^+}{E} \ge 0$. The subspace $E^+ \subset \hom(E; K)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. +\end{definition} + + +\begin{theorem}[Bauer-Namioka] +\label{theorem:bauer-namioka} + Let $E$ be an ordered vector space over $K \in \RC$ with positive cone $C$, $\topo$ be a vector space topology on $E$\footnote{The order and the topology need not to be compatible. }, $F \subset E$ be a subspace, and $\phi \in (F, \topo)^*$, then the following are equivalent: + \begin{enumerate} + \item There exists a continuous positive linear functional $\Phi \in (E, \topo)^*$ such that $\Phi|_F = \phi$. + \item There exists $U \in \cn_\topo(0)$ convex such that + \[ + \sup\bracs{\text{Re}\dpn{x, \phi}{F}|x \in F \cap (U - C)} < \infty + \] + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[V.5.4]{SchaeferWolff}}}. ] + (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_\topo(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E} \le \text{Re}\dpn{x, \Phi}{E} < 1$. + + (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \autoref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. + + After rescaling, assume without loss of generality that $\bracs{\phi = \alpha} \subset \bracs{\Phi = \alpha}$, then $\Phi \in E^*$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E} < \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E} \ge 0$, so $\Phi$ is the desired extension. +\end{proof} diff --git a/src/fa/tvs/complexify.tex b/src/fa/tvs/complexify.tex index 4fd7013..6881b47 100644 --- a/src/fa/tvs/complexify.tex +++ b/src/fa/tvs/complexify.tex @@ -119,8 +119,7 @@ The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and \begin{enumerate}[start=4] \item If $E$ is locally convex, then so is $\complex(E)$. - \item If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric. - \item[(F)] For any vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes: + \item[(F)] For any topological vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes: \[ \xymatrix{ \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\ @@ -132,6 +131,8 @@ \[ \complex(T)(x + iy) = Tx + iTy \] + + Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. \end{enumerate} \end{definition} @@ -141,25 +142,56 @@ (U): By (U) of the \hyperref[complexification]{definition:complexification}, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}. (4): By \autoref{proposition:finite-lc-product}, the direct sum and product of finitely many locally convex spaces coincide. By \autoref{proposition:lc-projective-topology}, this topology is locally convex. + - (5): Let $\norm{\cdot}_E: E \to [0, \infty)$ be the norm of $E$, and define + (F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$. + + For the isometry, + \begin{align*} + \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \norm{\complex(T)} + \end{align*} + +\end{proof} + +\begin{definition}[Complexification of Normed Spaces] +\label{definition:complexification-of-normed-spaces} + Let $E$ be a normed vector space over $\real$, then \[ \norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E \] - then for any $\phi \in [0, 2\pi]$ and $x, y \in E$, + is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric. + + Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. +\end{definition} +\begin{proof} + For any $\phi \in [0, 2\pi]$ and $x, y \in E$, \begin{align*} \normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\ &= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\ &= \norm{(x, y)}_{\complex(E)} \end{align*} - so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$, + so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$, \[ \norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\ \] Therefore $\iota: E \to \complex(E)$ is isometric. - (F): By (U) applied to $\iota \circ T$. + Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then + \begin{align*} + \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1} \\ + &\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1} = \norm{T}_{L(E; F)} + \end{align*} + + On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that + \begin{align*} + \normn{\complex(T)(x, y)}_{\complex(F)} &= \normn{(Tx, Ty)}_{\complex(F)} = \norm{\cos(\theta)Tx + \sin(\theta)Ty}_F \\ + &\le \norm{T}_{L(E; F)} \cdot \norm{\cos(\theta)x + \sin(\theta)y}_E \\ + &\le \norm{T}_{L(E; F)} \cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_E \\ + &= \norm{T}_{L(E; F)} \cdot \norm{(x, y)}_E + \end{align*} + + \end{proof}