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\begin{document}

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\section*{Stone-Weierstrass theorem}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{functional_analysis}{}\paragraph*{{Functional analysis}}\label{functional_analysis}

[[!include functional analysis - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{precise_statement}{Precise statement}\dotfill \pageref*{precise_statement} \linebreak
\noindent\hyperlink{outline_of_proof}{Outline of proof}\dotfill \pageref*{outline_of_proof} \linebreak
\noindent\hyperlink{variations}{Variations}\dotfill \pageref*{variations} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The \emph{Stone--Weierstrass theorem} says given a [[compact Hausdorff space]] $X$, one can uniformly approximate [[continuous functions]] $f: X \to \mathbb{R}$ by elements of any subalgebra that has enough elements to distinguish points. It is a far-reaching generalization of a classical theorem of Weierstrass, that real-valued continuous functions on a closed [[interval]] are uniformly approximable by [[polynomial]] functions.

\hypertarget{precise_statement}{}\subsection*{{Precise statement}}\label{precise_statement}

Let $X$ be a compact Hausdorff topological space; for a [[constructive mathematics|constructive]] version take $X$ to be a compact regular locale (see [[compactum]]). Recall that the algebra $C(X)$ of real-valued continuous functions $f: X \to \mathbb{R}$ is a commutative (real) [[Banach algebra]] with unit, under pointwise-defined addition and multiplication, and where the norm is the sup-norm

\begin{displaymath}
\|f\| := sup_{x \in X} |f(x)|
\end{displaymath}
A \textbf{subalgebra} of $C(X)$ is a vector subspace $A \subseteq C(X)$ that is closed under the unit and algebra multiplication operations on $C(X)$. A \textbf{Banach subalgebra} is a subalgebra $A \subseteq C(X)$ which is closed as a subspace of the metric space $C(X)$ under the sup-norm metric. We say that $A \subseteq C(X)$ \textbf{separates points} if, given distinct points $x, y \in X$, there exists $f \in A$ such that $f(x) \neq f(y)$.

\begin{uthm}
A subalgebra inclusion $A \subseteq C(X)$ is dense if and only if it separates points. Equivalently, a Banach subalgebra inclusion $A \subseteq C(X)$ is the identity if and only if it separates points.

\end{uthm}
\hypertarget{outline_of_proof}{}\subsection*{{Outline of proof}}\label{outline_of_proof}

\begin{itemize}%
\item The first step is the classical Weierstrass approximation, that polynomial functions $p(x): [a, b] \to \mathbb{R}$ are dense in $C([a, b])$; without loss of generality, take $a = -\frac1{4}$ and $b = \frac1{4}$. We use the method of constructing polynomials which [[approximation of the identity|approximate the identity]] of the convolution product (the [[distribution|Dirac distribution]]). Explicitly, consider the normalizations $K_n = s_n/\|s_n\|_1$ where $s_n(x) = (1 - x^2)^n$ over $[-1, 1]$ and is compactly supported on $[-1, 1]$, so that $K_n$ approximates the Dirac distribution concentrated at 0. Given $f: [-\frac1{4}, \frac1{4}] \to \mathbb{R}$, extend to a function compactly supported on $[-\frac1{2}, \frac1{2}]$. The convolution product\begin{displaymath}
(K_n * f)(x) = \int_{-1}^1 K_n(x-y) f(y) d y
\end{displaymath}
is polynomial (since by differentiating under the integral enough times, we eventually kill the convolving polynomial factor $K_n$), and one may verify that $(K_n * f)$ converges to $f$ in sup norm (i.e., uniformly) as $n \to \infty$, and in particular when restricted over the interval $[-\frac1{4}, \frac1{4}]$.



\end{itemize}
Now suppose given a Banach subalgebra $A \subseteq C(X)$.

\begin{itemize}%
\item Given $f \in A \subseteq C(X)$, the values of $f$ are contained in some interval $[a, b]$. If polynomials $p_n$ converge to the absolute value function $|\cdot |: [a, b] \to \mathbb{R}$ in sup norm, then $p_n(f) \in A$ converges to $|f| \in C(X)$ in sup norm. Since $A$ is closed in $C(X)$ with respect to the sup-norm, it follows that $|f| \in A$.


\item Next, $A$ is partially ordered by $f \leq g$ if $f(x) \leq g(x)$ for all $x \in X$, and we claim the poset $A$ is closed under binary meets and binary joins. For,

\begin{displaymath}
f \vee g = \frac1{2}(|f - g| + (f + g))
\end{displaymath}
and $f \wedge g = -((-f) \vee (-g))$, and we showed in the last step that $A$ is closed under the operation $f \mapsto |f|$.



\end{itemize}
Finally, suppose the Banach subalgebra $A$ separates points. Given $g \in C(X)$ and $\varepsilon \gt 0$, the last step is to show there exists $f \in A$ such that $\|f - g\| \leq \varepsilon$.

\begin{itemize}%
\item \begin{ulemma}
Given $x \in X$, there exists $f_x \in A$ such that $f_x(x) = g(x)$ and $g(y) \leq f_x(y) + \varepsilon$ for all $y \in X$.

\end{ulemma}
\begin{proof}
For each $y \in X$, $y \neq x$ we can choose $s \in A$ such that $s(x) \neq s(y)$, since $A$ separates points. Thus, given any $x, y$, there exist $s \in A$ and scalars $a$ and $b$ such that

\begin{displaymath}
a + b \cdot s(x) = g(x), \qquad a + b \cdot s(y) = g(y)
\end{displaymath}
Denote $a \cdot 1 + b s$ by $f_{x, y}$ (to indicate dependence on $x$ and $y$). For each $y$, choose a neighborhood $V_y$ so that $g(z) \leq f_{x, y}(z) + \varepsilon$ for all $z \in V_y$. Finitely many such neighborhoods $V_{y_1}, \ldots, V_{y_n}$ cover $X$; let $f_x$ be the join of $f_{x, y_1}, \ldots, f_{x, y_n}$. Then $g(y) \leq f_x(y) + \varepsilon$ for all $y \in X$.

\end{proof}

\item Given a choice of $f_x$ for each $x \in X$, as in the preceding lemma, we may choose a neighborhood $V_x$ such that $f_x(z) - \varepsilon \leq g(z)$ for all $z \in V_x$. Finitely many such neighborhoods $V_{x_1}, \ldots, V_{x_n}$ cover $X$; let $f$ be the meet of $f_{x_1}, \ldots, f_{x_n}$. Then

\begin{displaymath}
f(y) - \varepsilon \leq g(y) \leq f(y) + \varepsilon
\end{displaymath}
for all $y \in X$, as was to be shown.



\end{itemize}
\hypertarget{variations}{}\subsection*{{Variations}}\label{variations}

There is a complex-valued version of Stone--Weierstrass. Let $C(X, \mathbb{C})$ denote the commutative $C^*$-[[C-star algebra|algebra]] of complex-valued functions $f: X \to \mathbb{C}$, where the star operation is pointwise-defined conjugation. A $C^*$-\textbf{subalgebra} is a subalgebra $A \subseteq C(X, \mathbb{C})$ which is closed under the star operation.

\begin{uthm}
A $C^*$-subalgebra $A \subseteq C(X, \mathbb{C})$ is dense if and only if it separates points.

\end{uthm}
There is also a [[locally compact space|locally compact]] version. Let $X$ be a locally compact Hausdorff space and let $C_0(X)$ be the space of (say real-valued) functions $f$ which ``vanish at infinity'': for every $\varepsilon \gt 0$ there exists a compact set $K \subseteq X$ such that $|f(x)| \lt \varepsilon$ for all $x$ outside $K$. ($C_0(X)$ is no longer a Banach space, but it is [[locally convex topological vector space|locally convex]] and complete in its [[uniform space|uniformity]], and a Fr\'e{}chet space if $X$ is second countable.) Under pointwise multiplication, $C_0(X)$ is a commutative algebra \emph{without} unit. As before, we have a notion of subalgebra $A \subseteq C_0(X)$.

\begin{uthm}
$A \subseteq C_0(X)$ is dense if and only if it separates points and for no $x \in X$ is it true that every $f \in A$ vanishes at $x$.

\end{uthm}
\hypertarget{references}{}\subsection*{{References}}\label{references}

Named after [[Karl Weierstraß]] and [[Marshall Stone]].

See

\begin{itemize}%
\item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem}{Stone-Weierstrass theorem}}

\end{itemize}
Discussion in [[constructive mathematics]] is in

\begin{itemize}%
\item [[B. Banaschewski]], [[C. J. Mulvey]], \emph{A constructive proof of the Stone-Weierstrass theorem}, J. Pure Appl. Algebra \textbf{116} (1997), pp. 25--40, 

\end{itemize}
[[!redirects Stone–Weierstrass theorem]] [[!redirects Stone--Weierstrass theorem]]



\end{document}