7.4 The Stone-Weierstrass Theorem

Proof. Let $p_{1} = 0$. For each $n \in \natp$ and $x \in [0, 1]$, define

Assume inductively that $0 \le p_{n} \le \sqrt{t}$, then

and

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 $x \in [0, 1]$,

so $f(x) = \sqrt{x}$. Since $p_{n} \upto f$ pointwise, $p_{n} \upto f$ uniformly as $n \to \infty$ by Dini’s Theorem.$\square$

Proof. For each $f, g \in C(X; \real)$,

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 Lemma 7.4.1, 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, $|f| \in A$.$\square$

Proof, [Lemma 4.49, Fol99]. 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 $\norm{f - g}_{u} < \eps$.$\square$

Proof, [Theorem 4.45, Fol99]. For each $x, y \in X$ with $x \ne y$, the mapping

is an algebra homomorphism, and $\pi_{xy}(A)$ is a subalgebra of $\real^{2}$. By (S) and Lemma 7.4.4, 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:

1. (1)

   There exists $x_{0} \in X$ such that $f(x_{0}) = 0$ for all $f \in A$.

2. (2)

   For all $x, y \in X$ with $x \ne y$, $\pi_{xy}(A) = \real^{2}$.

If (1) holds, let $B = \bracs{f \in C(X; \real)| f(x_0) = 0}$. If (2) holds, let $B = C(X; \real)$. By Proposition 7.4.2, $A$ is a lattice. In both cases, $A$ satisfies condition (C) of Proposition 7.4.3 for all $f \in B$, therefore $A = B$.$\square$

Proof, [Theorem 4.51, Fol99]. 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 Stone-Weierstrass Theorem, so exactly one of the following holds:

1. (1)

   There exists $x_{0} \in X$ such that $A = \bracs{f \in C(X; \real)| f(x_0) = 0}$.

2. (2)

   $A = C(X; \real)$.

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 $A = B$.$\square$