Notes:Stone-Weierstrass theorem

Overview

I have come up with something that is not logically equivalent to the form in the lecture notes. I document them here in the hope of showing it to someone who can tell me what form to commit to.

I claim B is strictly corollary to $A$, that is A$\implies$B but B (does not imply) A

• TODO: Find LaTeX not implies symbol $\not{\implies}$ is awful, $\nRightarrow$ is too short

Forms

A

Let $(K,\mathcal{ J })$ be a compact topological space and let $\mathcal{A}\in\mathcal{P}($$C(K,\mathbb{K})$$)$ be given^Important: such that:

1. $\mathcal{A}$ is a sub-algebra of $C(K,\mathbb{K})$
2. $\forall x\in K\exists f\in\mathcal{A}[f(x)\neq 0]$ - $\mathcal{A}$ vanishes at no points in $K$
3. $\forall x_1,x_2\in K\big[x_1\neq x_2\implies \exists f\in \mathcal{A}[f(x_1)\neq f(x_2)]\big]$ - $\mathcal{A}$ separates points in $K$
4. $\forall f\in\mathcal{A}\exists g\in\mathcal{A}\forall x\in K[f(x)\eq (g(x))^\ast]$ - $\mathcal{A}$ is closed under complex conjugate

Then $\mathcal{A}$ is dense in $C(K,\mathbb{K})$ where $C(K,\mathbb{K})$ is considered with the uniform norm (AKA: infinity norm, sup norm), $$\Vert\cdot\Vert_\infty:f\mapsto\mathop{\text{Sup} }_{x\in K}(\vert f(x)\vert)$$ .

B

Let $(K,d)$ be a compact metric space, Let $\mathcal{A}\in\mathcal{P}($$C(K,\mathbb{R})$$)$ such that:

1. $\mathcal{A}$ is a sub-algebra of $C(K,\mathbb{R})$
2. $\exists c\in\mathbb{R}\exists f\in\mathcal{A}\forall x\in K[c\neq 0\wedge f(x)\eq c]$ - there exists a non-zero constant function in $\mathcal{A}$
3. $\forall x_1,x_2\in K\big[x_1\neq x_2\implies \exists f\in \mathcal{A}[f(x_1)\neq f(x_2)]\big]$ - $\mathcal{A}$ separates points in $K$

Then $\mathcal{A}$ is dense in $C(K,\mathbb{R})$ - where $C(K,\mathbb{R})$ is considered with the uniform norm (AKA: infinity norm, sup norm), $$\Vert\cdot\Vert_\infty:f\mapsto\mathop{\text{Sup} }_{x\in K}(\vert f(x)\vert)$$

Proof of claim

Proof that $A\implies B$

Let the conditions of B be given. I.e. a compact metric space and so forth. All metric spaces are topological spaces (see topology induced by a metric) so that is met. Claim A covers both $\mathbb{K}:\eq\mathbb{R}$ and $\mathbb{K}:\eq\mathbb{C}$ - we have this covered by the first case. We require $\mathcal{A}$ to be a sub-algebra, the existence of a constant non-zero function implies the vanishes nowhere condition of A. They both require the separation of points, because B deals with $\mathbb{R}$ the complex conjugate thing is satisfied as the complex conjugate of a real valued function is just itself.

Thus by A we have B. As required.

Proof that $B$ doesn't imply $A$

Suppose we have a non-metricisable topological space, we cannot invoke A to speak of its truth or falseness.

Comments

A

1. Note: I only used this once in the proof. I showed that in the case of $\mathbb{K}:\eq\mathbb{R}$ that:
   • $\forall x\in K[g(t)>f(t)-\epsilon]$ $\implies$ $\forall x\in K[ f(t)-g(t) <\epsilon]$ $\implies$ $\Vert f-g\Vert_\infty<\epsilon$.
     • If this can be shown for other norms, there should not be a problem extending it.