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 [ilmath]A[/ilmath], that is A[ilmath]\implies[/ilmath]B but B (does not imply) A

• TODO: Find LaTeX not implies symbol [ilmath]\not{\implies} [/ilmath] is awful, [ilmath]\nRightarrow[/ilmath] is too short

Forms

A

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

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

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

B

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

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

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

Proof of claim

Proof that [ilmath]A\implies B[/ilmath]

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 [ilmath]\mathbb{K}:\eq\mathbb{R} [/ilmath] and [ilmath]\mathbb{K}:\eq\mathbb{C} [/ilmath] - we have this covered by the first case. We require [ilmath]\mathcal{A} [/ilmath] 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 [ilmath]\mathbb{R} [/ilmath] 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 [ilmath]B[/ilmath] doesn't imply [ilmath]A[/ilmath]

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 [ilmath]\mathbb{K}:\eq\mathbb{R} [/ilmath] that:
   • [ilmath]\forall x\in K[g(t)>f(t)-\epsilon][/ilmath] [ilmath]\implies[/ilmath] [ilmath]\forall x\in K[ f(t)-g(t) <\epsilon][/ilmath] [ilmath]\implies[/ilmath] [ilmath]\Vert f-g\Vert_\infty<\epsilon[/ilmath].
     • If this can be shown for other norms, there should not be a problem extending it.