📂Analysis

Proof of the Stone-Weierstrass Theorem

Theorem¹

Auxiliary Definitions

Let’s say $A \subset C(X)$ for $X$.

1. If for any distinct $x_{1}, x_{2} \in X$, there always exists $f \in A$ that satisfies $f(x_{1}) \ne f(x_{2})$, then we say $A$ separates the points of $X$.
2. If $X$ is a metric space and for all $\varepsilon > 0$ and $f \in C(X)$ there exists $g \in A$ that satisfies $| g - f | < \varepsilon$, then $A$ is said to be uniformly dense in $C(X)$.

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• $C \left( X \right)$ is a class of continuous function spaces with domain $X$ and codomain $\mathbb{R}$.

The Stone-Weierstrass Theorem

Suppose $X$ is a compact metric space. If $A$ is an algebra of $C(X)$ that includes constant functions and separates the points of $X$, then $A$ is uniformly dense in $C(X)$.

Explanation

The Stone-Weierstrass theorem assures that continuous functions can be approximated by other functions. However, this statement might feel overly abstract. The Stone-Weierstrass theorem for polynomials in $1$-dimension is written as follows:

Weierstrass Approximation Theorem: If $f$ is continuous on $[a,b]$, for given $\epsilon > 0$, there exists a polynomial function $p(x)$ satisfying $\displaystyle \max_{x \in [a,b]} | f(x) - p (x) | < \epsilon$.

While the plain statement that $p(x)$ exists holds its charm, reflecting on its significance now, it seems overly modest. Considering $\epsilon$ as a form of tolerance, it’s fair to say, “Given any continuous function, it can be represented by a polynomial function.”

Proof

Strategy: It’s by no means easy. For $F \in C(X)$, we specifically work with the closure $\overline{A}$ of $A$ to find a $G$ that makes $| F - G | < \varepsilon$ possible. To construct $G$, since $\overline{A}$ is a closed algebra, we must utilize good properties, and after pinpointing such a $G$, presenting just one sequence of $A$ that converges to it concludes the process.

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• Part 1. $a, b \in \mathbb{R} , x_{1} \ne x_{2} \implies \exists f \in A : \begin{cases} f(x_{1}) = a \\ f(x_{2}) = b \end{cases}$

  Since $A$ separates the points of $X$, for any distinct $x_{1} , x_{2}$, there exists $g \in A$ that satisfies $g(x_{1} ) \ne g (x_{2} )$.

  Algebra: A set $A$ in $C(X)$ is called an algebra if it satisfies the following three conditions:

  • (i): $\emptyset \ne A \subset C(X)$
  • (ii): $f,g \in A \implies (f+g) , fg \in A$
  • (iii): $f \in A , c \in \mathbb{R} \implies cf \in A$

  $$ f(t) := a {{ g(t) - g(x_{2} ) } \over { g(x_{1} ) - g( x_{2} ) }} + b {{ g(t) - g(x_{1} ) } \over { g(x_{2} ) - g( x_{1} ) }} $$

  Since $A$ is an algebra that includes constant functions, it also includes constant functions with value $g(x_{1}) , g(x_{2})$, and for $a, b \in \mathbb{R}$, defining $f$ as above, we have $f \in A$, and substituting $t=x_{1} , x_{2}$ yields $f(x_{1}) = a$ and $f(x_{2} ) = b$.

• Part 2. $f_{1} ,f_{2} \in \overline{A} \implies ( f_{1} \land f_{2} ), ( f_{1} \lor g_{2} ) \in \overline{A}$

  $\land$ and $\lor$ imply the following for $f,g \in C(X)$ and $x \in X$: $$ \begin{align*} (f \land g) (x) :=& \min \left\{ f(x) , g(x) \right\} \\ (f \lor g) (x) :=& \max \left\{ f(x) , g(x) \right\} \end{align*} $$

  Property of Uniform Closure: Let’s say $A \subset C(X)$ for the metric space $X$. If every sequence $\left\{ f_{n} \in A : n \in \mathbb{N} \right\}$ of $A$ converges to some $f \in A$ as $n \to \infty$, then $A$ is considered uniformly closed if $\displaystyle | f - f_{n} | \to 0$. If $X$ is a compact metric space and $A$ is a uniformly closed algebra of $C(X)$ that includes constant functions, the following holds: $$ f,g \in A \implies (f \land g), ( f \lor g ) \in A $$

  Considering the uniform closure $\displaystyle \overline{A} := \left\{ f \in C(X) : \lim_{n \to \infty} | f_{n} - f | = 0, f_{n} \in A \right\}$ of $A$, since $A$ is an algebra, $\overline{A}$ is also an algebra, and by the property of uniform closure,

  $$ f_{1} ,f_{2} \in \overline{A} \implies (f_{1} \land f_{2}), ( f_{1} \lor f_{2} ) \in \overline{A} $$

• Part 3. $\displaystyle | F - G | < {{\varepsilon} \over {2}}$

  Whenever $F \in C(X)$ and $\displaystyle {{\varepsilon} \over {2}} > 0$ are given, we aim to prove that $G \in \overline{A}$ satisfying $\displaystyle | F - G | < {{\varepsilon} \over {2}}$ exists.

  • Part 3-1. $\displaystyle g_{x_{0}} ( x ) < F(x) + {{\varepsilon} \over {2}}$

    Fixing $x_{0} \in X$ and setting $y \ne x_{0}$, according to Part 1, there exists a continuous function $f_{y} \in A \subset \overline{A} \subset C(X)$ satisfying

    $$ \begin{align*} f_{y} (x_{0}) =& F ( x_{0} ) \\ f_{y} ( y ) =& F ( y ) \end{align*} $$ Since $f_{y}$ and $F$ are continuous functions, $$ V_{y} := \left\{ x \in X : f_{y} (x) < F(x) + {{ \varepsilon } \over { 2 }} \right\} $$ is an open set, and $$ X = \bigcup_{y \ne x_{0}} V_{y} $$ Thus, since $X$ is a compact set, there exist finite elements $y_{1} , \cdots , y_{N_{1}} \in X$ satisfying $$ X = \bigcup_{i=1}^{N_{1}} V_{i} $$ Now, for $i = 1 , \cdots , N_{1}$, if we define it as $$ \begin{align*} f_{i} :=& f_{y_{i}} \\ g_{y_{0}} :=& f_{1} \land \cdots \land f_{N_{1}} \end{align*} $$ then by Part 2, we have $g_{x_{0}} \in \overline{A}$. Substituting $x = x_{0}$, we get $$ \begin{align*} g_{x_{0}} ( x_{0} ) =& f_{1} ( x_{0} ) \land \cdots \land f_{N_{1}} ( x_{0} ) \\ =& F ( x_{0} ) \land \cdots \land F ( x_{0} ) \\ =& F ( x_{0} ) \end{align*} $$ If $x \in X$, it means that $x$ belongs to at least one of $V_{y_{1}} , \cdots , V_{y_{N_{1}}}$, so for at least one $1 \le k \le N_{1}$, $$ f_{k} (x) < F(x) + {{\varepsilon} \over {2}} $$ holds, and by the definition of $g_{x_{0}}$, for all $i = 1, \cdots , N_{1}$, $g_{x_{0}} ( x ) \le f_{i} (x)$, thus we obtain $$ g_{x_{0}} ( x ) < F(x) + {{\varepsilon} \over {2}} $$

  • Part 3-2. $\displaystyle F(x) - {{\varepsilon} \over {2}} < G(x) < F(x) + {{\varepsilon} \over {2}}$

    Similarly to $\left\{ V_{y_{i}} \right\}_{i=1}^{N_{1}}$, let’s define a finite collection of open sets $\left\{ W_{x_{i}} \right\}_{i=1}^{N_{2}}$ that cover $X$ as follows: $$ W_{x_{i}} := \left\{ x \in X : g_{x_{i}} (x) > F(x) - {{ \varepsilon } \over { 2 }} \right\} $$ Just like in Part 3-1, for each $x_{1}, \cdots , x_{N_{2}}$, the respective $g_{x_{i}}$ satisfies $$ g_{x_{i}} ( x ) > F(x) - {{\varepsilon} \over {2}} $$ for $x_{i} \in X$. Now, let’s define the following functions for $i = 1, \cdots , N_{2}$: $$ \begin{align*} g_{i} :=& g_{x_{i}} \\ G :=& g_{1} \lor \cdots \lor g_{N_{2}} \end{align*} $$ Then, for all $x \in X$, we have $\displaystyle G(x) > F(x) - {{\varepsilon} \over {2}}$. Meanwhile, if $x \in X$, it signifies that $x$ belongs to at least one of $W_{x_{1}} , \cdots , W_{x_{N_{2}}}$, hence for at least one $j$, $$ F(x) - {{\varepsilon} \over {2}} < g_{j} \le G(x) $$

  In summary, for all $x \in X$, since $\displaystyle F(x) - {{\varepsilon} \over {2}} < G(x) < F(x) + {{\varepsilon} \over {2}}$, we derive $$ | F(x) - G(x) | < {{\varepsilon} \over {2}} $$

• Part 4. $A$ is Uniformly Dense

  Since $\overline{A}$ is the uniform closure of $A$, there exists a sequence $\left\{ G_{n} \right\}_{n \in \mathbb{N}}$ of $A$ that converges to $G \in \overline{A}$. In other words, for all $\displaystyle {{\varepsilon} \over {2}} >0$, there exists $N \in \mathbb{N}$ satisfying $$ n \ge N \implies | G_{n} - G | < {{\varepsilon} \over {2}} $$ For all $\varepsilon > 0$ and given function $F \in C(X)$, with $G \in \overline{A}$ satisfying $$ | F(x) - G(x) | < {{\varepsilon} \over {2}} $$ and $N$ satisfying $$ | G_{N} - G | < {{\varepsilon} \over {2}} $$ we always have $$ \begin{align*} | F - G_{N} | \le & | F - G | + | G - G_{N} | \\ =& {{\varepsilon} \over {2}} + {{\varepsilon} \over {2}} \\ =& \varepsilon \end{align*} $$ That is, for all $\varepsilon > 0$ and given function $F \in C(X)$, there always exists $G_{N} \in A$ that satisfies $| F - G_{N} | < \varepsilon$, hence $A$ is uniformly dense in $C(X)$.

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