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Added the Stone-Weierstrass Theorem.

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Bokuan Li
2026-05-25 20:38:20 -04:00
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13
spec.toml
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@@ -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
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src/topology/functions/index.tex
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@@ -4,3 +4,4 @@
\input{./ideal.tex}
\input{./set-systems.tex}
\input{./uniform.tex}
\input{./sw.tex}

143
src/topology/functions/sw.tex Normal file
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@@ -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}

48
src/topology/functions/uniform.tex
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@@ -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}