From 69c4e5e0303dbd28c14df69f03fed637362b171a Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 16 Mar 2026 16:01:07 -0400 Subject: [PATCH] Added vector lattices. --- src/fa/index.tex | 1 + src/fa/order/index.tex | 4 + src/fa/order/lattice.tex | 273 ++++++++++++++++++++++ src/measure/lebesgue-integral/complex.tex | 23 +- 4 files changed, 283 insertions(+), 18 deletions(-) create mode 100644 src/fa/order/index.tex create mode 100644 src/fa/order/lattice.tex diff --git a/src/fa/index.tex b/src/fa/index.tex index d99d2a4..4e4b2e1 100644 --- a/src/fa/index.tex +++ b/src/fa/index.tex @@ -7,3 +7,4 @@ \input{./norm/index.tex} \input{./rs/index.tex} \input{./lp/index.tex} +\input{./order/index.tex} diff --git a/src/fa/order/index.tex b/src/fa/order/index.tex new file mode 100644 index 0000000..1c6d5bd --- /dev/null +++ b/src/fa/order/index.tex @@ -0,0 +1,4 @@ +\chapter{Order Structures} +\label{chap:order-structure} + +\input{./lattice.tex} diff --git a/src/fa/order/lattice.tex b/src/fa/order/lattice.tex new file mode 100644 index 0000000..9d8797d --- /dev/null +++ b/src/fa/order/lattice.tex @@ -0,0 +1,273 @@ +\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. + \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. +\end{proof} + + + + + diff --git a/src/measure/lebesgue-integral/complex.tex b/src/measure/lebesgue-integral/complex.tex index deaef8f..6f024fe 100644 --- a/src/measure/lebesgue-integral/complex.tex +++ b/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 \\