From 65fdc1cc8d11be6a360fce764896f22132c1d048 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 25 May 2026 20:38:20 -0400 Subject: [PATCH] Added the Stone-Weierstrass Theorem. --- spec.toml | 13 ++- src/topology/functions/index.tex | 1 + src/topology/functions/sw.tex | 143 +++++++++++++++++++++++++++++ src/topology/functions/uniform.tex | 48 ++++++++++ 4 files changed, 203 insertions(+), 2 deletions(-) create mode 100644 src/topology/functions/sw.tex diff --git a/spec.toml b/spec.toml index 556bcf7..5f3eecd 100644 --- a/spec.toml +++ b/spec.toml @@ -17,8 +17,17 @@ primaryColour = "violet" neutralColour = "grey" searchLimit = 16 maxSearchPages = 48 -recentChanges = 0 -tableOfContentsDepth = 1 + +recentChanges = 10 +tableOfContentsDepth = 2 +tableOfContentsUnfoldDepth = 1 +tableOfContentsShortDepth = 1 + hoverPreview = false + copyLabelButton = false + advertiseSpec = true + +displayComments = true +allowComments = true diff --git a/src/topology/functions/index.tex b/src/topology/functions/index.tex index 1464d22..359d69b 100644 --- a/src/topology/functions/index.tex +++ b/src/topology/functions/index.tex @@ -4,3 +4,4 @@ \input{./ideal.tex} \input{./set-systems.tex} \input{./uniform.tex} +\input{./sw.tex} diff --git a/src/topology/functions/sw.tex b/src/topology/functions/sw.tex new file mode 100644 index 0000000..9134d3a --- /dev/null +++ b/src/topology/functions/sw.tex @@ -0,0 +1,143 @@ +\section{The Stone-Weierstrass Theorem} +\label{section:stone-weierstrass} + +\begin{lemma} +\label{lemma:sqrt-polynomial} + There exists a sequence of polynomials $\seq{p_n} \subset \real[x]$ such that $p_n(x) \upto \sqrt{x}$ uniformly on $[0, 1]$ as $n \to \infty$. +\end{lemma} +\begin{proof} + Let $p_1 = 0$. For each $n \in \natp$ and $x \in [0, 1]$, define + \[ + p_{n+1}(x) = p_n(x) + \frac{x - p_n(x)^2}{2} + \] + + Assume inductively that $0 \le p_n \le \sqrt{t}$, then + \[ + p_{n+1}(x) = p_n(x) + \frac{x - p_n(x)^2}{2} \ge p_n(x) \ge 0 + \] + + and + + \begin{align*} + \sqrt{x} - p_{n+1}(x) &= \sqrt{x} - p_n(x) - \frac{x - p_n(x)^2}{2} \\ + &= \sqrt{x} - p_n(x) - \frac{1}{2}(\sqrt{x} - p_n(x))(\sqrt{x} + p_n(x)) \\ + &= (\sqrt{x} - p_n(x))\paren{1 - \frac{\sqrt{x} + p_n(x)}{2}} \ge 0 + \end{align*} + + for any $x \in [0, 1]$. + + Since $\seq{p_n} \subset \real[x]$ is increasing on $[0, 1]$ with $p_n(x) \le \sqrt{x}$ for all $x \in [0, 1]$ and $n \in \natp$, there exists $f: [0, 1] \to [0, 1]$ such that $p_n \to f$ pointwise as $n \to \infty$. For each $x \in [0, 1]$, + \[ + f(x) = \limv{n}p_{n+1}(x) = \limv{n}p_n(x) + \frac{x - p_n(x)^2}{2} = f(x) + \frac{x - f(x)^2}{2} + \] + + so $f(x) = \sqrt{x}$. Since $p_n \upto f$ pointwise, $p_n \upto f$ uniformly as $n \to \infty$ by \hyperref[Dini's Theorem]{theorem:dini}. +\end{proof} + +\begin{proposition} +\label{proposition:algebra-lattice} + Let $X$ be a compact Hausdorff space and $A \subset C(X; \real)$ be a closed subalgebra, then $A$ is a lattice. +\end{proposition} +\begin{proof} + For each $f, g \in C(X; \real)$, + \[ + f \vee g = \frac{f + g + |f - g|}{2} \quad f \wedge g = \frac{f + g - |f - g|}{2} + \] + + so it is sufficient to show that for each $f \in A$, $|f| \in A$. + + Let $f \in A$ and assume without loss of generality that $\norm{f}_u \le 1$. By \autoref{lemma:sqrt-polynomial}, there exists $\seq{p_n} \subset \real[x]$ such that $p_n(x) \to \sqrt{x}$ uniformly on $[0, 1]$ as $n \to \infty$. In which case, $p_n(f^2) \in A$ and $p_n(f^2) \to \sqrt{f^2} = |f|$ uniformly as $n \to \infty$. Since $A$ is closed, $|f| \in A$. +\end{proof} + +\begin{proposition} +\label{proposition:lattice-copy} + Let $X$ be a compact Hausdorff space, $A \subset C(X; \real)$ be a closed lattice, and $f \in C(X; \real)$. If + \begin{enumerate} + \item[(C)] For any $x, y \in X$, there exists $g_{xy} \in A$ such that $g_{xy}(x) = f(x)$ and $g_{xy}(y) = f(y)$. + \end{enumerate} + + then $f \in A$. +\end{proposition} +\begin{proof}[Proof, {{\cite[Lemma 4.49]{Folland}}}. ] + Fix $\eps > 0$. Let $x \in X$, then for each $y \in Y$, $\bracs{g_{xy} > f - \eps} \in \cn_X(y)$. By compactness of $X$, there exists $Y_x \subset Y$ finite such that $X = \bigcup_{y \in Y_x}\bracs{g_{xy} > f - \eps}$. In which case, let $g_x = \bigvee_{y \in Y_x}g_{xy}$, then $g \in A$ and $g_x(x) = f(x)$ and $g > f - \eps$. + + Similarly, for each $x \in X$, $\bracs{g_x < f + \eps} \in \cn_X(x)$. By compactness of $X$, there exists $X_0 \subset X$ finite such that $X = \bigcup_{x \in X_0}\bracs{g_x < f + \eps}$. Let $g = \bigwedge_{x \in X_0}g_x$, then $g \in A$ and $\norm{f - g}_u < \eps$. +\end{proof} + +\begin{lemma} +\label{lemma:algebra-r2} + Let $A \subset \real^2$ be an algebra. If + \begin{enumerate} + \item[(S)] There exists $(x, y) \in A$ such that $x \ne y$. + \end{enumerate} + + then $A$ is one of the following: + \begin{enumerate} + \item $\real^2$. + \item $\text{span}\bracs{(0, 1)}$. + \item $\text{span}\bracs{(1, 0)}$. + \end{enumerate} + + +\end{lemma} +% Proof omitted because it is obvious. + + +\begin{theorem}[Stone-Weierstrass] +\label{theorem:stone-weierstrass} + Let $X$ be a compact Hausdorff space and $A \subset C(X; \real)$ be a closed subalgebra. If + \begin{enumerate} + \item[(S)] For each $x, y \in X$, there exists $f \in A$ such that $f(x) \ne f(y)$. + \end{enumerate} + + + then exactly one of the following holds: + \begin{enumerate} + \item There exists $x_0 \in X$ such that $A = \bracs{f \in C(X; \real)| f(x_0) = 0}$. + \item $A = C(X; \real)$. + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem 4.45]{Folland}}}. ] + For each $x, y \in X$ with $x \ne y$, the mapping + \[ + \pi_{xy}: A \to \real^2 \quad f \mapsto (f(x), f(y)) + \] + + is an algebra homomorphism, and $\pi_{xy}(A)$ is a subalgebra of $\real^2$. By (S) and \autoref{lemma:algebra-r2}, there exists no $x, y \in X$ with $x \ne y$ such that $\pi_{xy}(A) = \bracs{0}$, so exactly one of the following holds: + \begin{enumerate} + \item There exists $x_0 \in X$ such that $f(x_0) = 0$ for all $f \in A$. + \item For all $x, y \in X$ with $x \ne y$, $\pi_{xy}(A) = \real^2$. + \end{enumerate} + + If (1) holds, let $B = \bracs{f \in C(X; \real)| f(x_0) = 0}$. If (2) holds, let $B = C(X; \real)$. By \autoref{proposition:algebra-lattice}, $A$ is a lattice. In both cases, $A$ satisfies condition (C) of \autoref{proposition:lattice-copy} for all $f \in B$, therefore $A = B$. + +\end{proof} + +\begin{theorem}[Stone-Weierstrass (Complex)] +\label{theorem:complex-stone-weierstrass} + Let $X$ be a compact Hausdorff space and $A \subset C(X; \complex)$ be a closed subalgebra. If + \begin{enumerate} + \item[(S)] For each $x, y \in X$, there exists $f \in A$ such that $f(x) \ne f(y)$. + \item[(*)] For each $f \in A$, the complex conjugate $\ol f$ is also in $A$. + \end{enumerate} + + then exactly one of the following holds: + \begin{enumerate} + \item There exists $x_0 \in X$ such that $A = \bracs{f \in C(X; \complex)| f(x_0) = 0}$. + \item $A = C(X; \complex)$. + \end{enumerate} + +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem 4.51]{Folland}}}. ] + By (*), for each $f \in A$, $\text{Re}(f), \text{Im}(f) \in A$ as well. Let $A_{sa} = A \cap C(X; \real)$, then $A_{sa} \subset C(X; \real)$ satisfies (S) of the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, so exactly one of the following holds: + \begin{enumerate} + \item There exists $x_0 \in X$ such that $A = \bracs{f \in C(X; \real)| f(x_0) = 0}$. + \item $A = C(X; \real)$. + \end{enumerate} + + If (1) holds, let $B = \bracs{f \in C(X; \real)| f(x_0) = 0}$. If (2) holds, let $B = C(X; \real)$. For any $f \in B$, $f = \text{Re}(f) + \text{Im}(f)$, where $\text{Re}(f), \text{Im}(f) \in A_{sa} \subset A$. Therefore $f \in A$, and $A = B$. +\end{proof} + + + + diff --git a/src/topology/functions/uniform.tex b/src/topology/functions/uniform.tex index 5c83877..1591f92 100644 --- a/src/topology/functions/uniform.tex +++ b/src/topology/functions/uniform.tex @@ -59,3 +59,51 @@ Let $f \in \overline{C(X; Y)} \subset Y^X$ with respect to the $\sigma$-uniformity. By \autoref{proposition:uniform-limit-continuous}, $f \in C(S; Y)$ for all $S \in \sigma$, so $f \in C(X; Y)$ by (3) of \autoref{definition:final-topology}. \end{proof} +\begin{proposition} +\label{proposition:compact-continuous-space} + Let $X$ be a compact topological space and $E$ be a TVS over $K \in \RC$, then: + \begin{enumerate} + \item $C(X; E)$ equipped with the uniform topology is a topological vector space over $K$. + \item If $E$ is locally convex and $\rho: E \to [0, \infty)$ is a continuous seminorm, then + \[ + [\cdot]_{u, \rho}: C(X; E) \to [0, \infty) \quad f \mapsto \sup_{x \in X}\rho(f(x)) + \] + + is a continuous seminorm on $C(X; E)$ with respect to the uniform topology. + + The uniform topology on $C(X; E)$ is locally convex, and induced by seminorms of the form $[\cdot]_{u, \rho}$, where $\rho$ ranges over all continuous seminorms on $E$. + \item If $E$ is normed, then + \[ + [\cdot]_u: C(X; E) \quad f \mapsto \sup_{x \in X}\norm{f(x)}_E + \] + + is a norm on $C(X; E)$ that induces the uniform topology. + \item If $E$ is complete, then so is $C(X; E)$. + \end{enumerate} + + As such, the uniform topology is the canonical topology on $C(X; E)$. +\end{proposition} +\begin{proof} + (1): Since $X$ is compact, for any $f \in C(X; E)$, $f(X) \in \mathfrak{B}(E)$, so (1) holds by \autoref{proposition:tvs-set-uniformity}. + + (2): By \autoref{proposition:lc-spaces-linear-map}. + + (4): By \autoref{proposition:uniform-limit-continuous}. +\end{proof} + + + +\begin{theorem}[Dini] +\label{theorem:dini} + Let $X$ be a compact topological space, $\angles{f_\alpha}_{\alpha \in A} \subset C(X; \real)$, and $f \in C(X; \real)$ such that $f_\alpha \upto f$ pointwise, then $f_\alpha \upto f$ uniformly. +\end{theorem} +\begin{proof} + Let $\eps > 0$, then for each $\alpha \in A$, $\bracs{f - f_\alpha < \eps} \subset X$ is open, and $X = \bigcup_{\alpha \in A}\bracs{f - f_\alpha < \eps}$. By compactness, there exists $B \subset A$ finite such that $X = \bigcup_{\beta \in B}\bracs{f - f_\beta < \eps}$. Let $\alpha_0 \in A$ with $\alpha_0 \ge \beta$ for all $\beta \in B$, then since $f_\alpha \upto f$ pointwise, + \[ + \bracs{f - f_\alpha < \eps} = \bigcup_{\beta \in B}\bracs{f - f_\beta < \eps} = X + \] + + for any $\alpha \ge \alpha_0$. +\end{proof} + +