# proof of Stone-Weierstrass theorem

Let $\overline{\mathcal{A}}$ denote the closure^{} of $\mathcal{A}$ in ${C}^{0}(X,\mathbb{R})$ according to the uniform convergence topology^{}. We want to show that, if conditions 1 and 2 are satisfied, then $\overline{\mathcal{A}}={C}^{0}(X,\mathbb{R})$.

First, we shall show that, if $f\in \overline{\mathcal{A}}$, then $|f|\in \overline{\mathcal{A}}$. Since $f$ is a continuous function^{} on a compact space $f$ must be bounded^{} – there exists constants $a$ and $b$ such that $a\le f\le b$. By the Weierstrass approximation theorem^{}, for every $\u03f5>0$, there exists a polynomial such that $$ when $x\in [a,b]$. (By the way, one does not need the full-blown Weierstrass approximation theorem to show that $P$ exists – see the entry “proof of Weierstrass approximation theorem” for an elementary construction of $P$) Define $g:X\to \mathbb{R}$ by $g(x)=P(f(x))$. Since $\overline{\mathcal{A}}$ is an algebra, $g\in \overline{\mathcal{A}}$. For all $x\in X$, $$. Since $\overline{\mathcal{A}}$ is closed under the uniform convergence topology, this implies that $|f|\in \overline{\mathcal{A}}$.

A corollary of the fact just proven is that if $f,g\in \overline{\mathcal{A}}$, then $\mathrm{max}(f,g)\in \overline{\mathcal{A}}$ and $\mathrm{min}(f,g)\in \overline{\mathcal{A}}$. The reason for this is that one can write

$$\mathrm{max}(a,b)=\frac{1}{2}\left(a+b+|a-b|\right)$$ |

$$\mathrm{min}(a,b)=\frac{1}{2}\left(a+b-|a-b|\right)$$ |

Second, we shall show that, for every $f\in {C}^{0}(X,\mathbb{R})$, every $x\in X$, and every $\u03f5>0$, there exists ${g}_{x}\in \overline{\mathcal{A}}$ such that ${g}_{x}\le f+\u03f5$ and ${g}_{x}>f$. By condition 1, if $y\ne x$, there exists a function ${\stackrel{~}{h}}_{xy}\in \mathcal{A}$ such that ${\stackrel{~}{h}}_{xy}(x)\ne {\stackrel{~}{h}}_{xy}(y)$. Define ${h}_{xy}$ by ${h}_{xy}(z)=p{\stackrel{~}{h}}_{xy}(z)+q$, where the constants $p$ and $q$ have been chosen so that

$${h}_{xy}(x)=f(x)+\frac{\u03f5}{2}$$ |

$${h}_{xy}(y)=f(y)-\frac{\u03f5}{2}$$ |

By condition 2, ${h}_{xy}\in \mathcal{A}$. (Note: This is the only place in the proof where condition 2 is used, but it is crucial since, otherwise, it might not be possible to construct a function which takes arbitrary preassigned values at two distinct points of $X$. The necessity of condition 2 can be shown by a simple example: Suppose that $\mathcal{A}$ is the algebra of all continuous functions on $f$ which vanish at a point $O\in X$. It is easy to see that this algebra satisfies all the hypotheses of the theorem except condition 2 and the conclusion^{} of the theorem
does not hold in this case.)

For every $y\ne x$, define the set ${U}_{xy}$ as

$$ |

Since $f$ and ${h}_{xy}$ are continuous, ${U}_{xy}$ is an open set. Because $x\in {U}_{xy}$ and $y\in {U}_{xy}$, $\{{U}_{xy}\mid y\in X\setminus \{x\}\}$ is an open cover of $X$. By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset $\{{y}_{1},{y}_{2},\mathrm{\dots},{y}_{n}\}\subset X$ such that $X={\bigcup}_{m=0}^{n}{U}_{x{y}_{m}}$. Define

$${g}_{x}=\mathrm{min}({h}_{x{y}_{1}},{h}_{x{y}_{2}},\mathrm{\dots},{h}_{x{y}_{n}}).$$ |

By the corollary of the first part of the proof, ${g}_{x}\in \overline{\mathcal{A}}$. By construction, ${g}_{x}(x)=f(x)+\u03f5/2$ and $$.

Third, we shall show that, for every $f\in {C}^{0}(X,\mathbb{R})$ and every $\u03f5>0$, there exists a function $g\in \overline{\mathcal{A}}$ such that $$. This will complete^{} the proof becauase it implies that $\overline{\mathcal{A}}={C}^{0}(X,\mathbb{R})$. For every $x\in X$, define the set ${V}_{x}$ as

$${V}_{x}=\{z\in X\mid {g}_{x}(z)>f(x)\}$$ |

where ${g}_{x}$ is defined as before. Since $f$ and ${g}_{x}$ are continuous, ${V}_{x}$ is an open set. Because ${g}_{x}(x)=f(x)+\u03f5/2>f(x)$, $x\in {V}_{x}$. Hence $\{{V}_{x}\mid x\}$ is an open cover of $X$. By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset $\{{x}_{1},{x}_{2},\mathrm{\dots}{x}_{n}\}\subset X$ such that $X={\bigcup}_{m=0}^{n}{V}_{{x}_{n}}$. Define $g$ as

$$g(z)=\mathrm{max}\{{g}_{{x}_{1}}(z),{g}_{{x}_{2}}(z),\mathrm{\dots},{g}_{{x}_{n}}(z)\}$$ |

By the corollary of the first part of the proof, $g\in \overline{\mathcal{A}}$. By construction, $g>f$. Since $$ for every $x\in X$, $$.

Title | proof of Stone-Weierstrass theorem |
---|---|

Canonical name | ProofOfStoneWeierstrassTheorem |

Date of creation | 2013-03-22 14:35:01 |

Last modified on | 2013-03-22 14:35:01 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 18 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 46E15 |