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Homeproof of Stone-Weierstrass theorem

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# proof of Stone-Weierstrass theorem

Let $\bar{\mathcal{A}}$ 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 $\bar{\mathcal{A}}=C^{0}(X,\mathbb{R})$.

First, we shall show that, if $f\in\bar{\mathcal{A}}$, then $|f|\in\bar{\mathcal{A}}$. Since $f$ is a continuous function on a compact space $f$ must be bounded – there exists constants $a$ and $b$ such that $a\leq f\leq b$. By the Weierstrass approximation theorem, for every $\epsilon>0$, there exists a polynomial such that $|P(x)-|x||<\epsilon$ 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 $\bar{\mathcal{A}}$ is an algebra, $g\in\bar{\mathcal{A}}$. For all $x\in X$, $|g(x)-|f(x)||<\epsilon$. Since $\bar{\mathcal{A}}$ is closed under the uniform convergence topology, this implies that $|f|\in\bar{\mathcal{A}}$.

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

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

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

Second, we shall show that, for every $f\in C^{0}(X,\mathbb{R})$, every $x\in X$, and every $\epsilon>0$, there exists $g_{x}\in\bar{\mathcal{A}}$ such that $g_{x}\leq f+\epsilon$ and $g_{x}>f$. By condition 1, if $y\neq x$, there exists a function ${\tilde{h}}_{{xy}}\in\mathcal{A}$ such that ${\tilde{h}}_{{xy}}(x)\neq{\tilde{h}}_{{xy}}(y)$. Define $h_{{xy}}$ by $h_{{xy}}(z)=p{\tilde{h}}_{{xy}}(z)+q$, where the constants $p$ and $q$ have been chosen so that

$h_{{xy}}(x)=f(x)+\frac{\epsilon}{2}$ |

$h_{{xy}}(y)=f(y)-\frac{\epsilon}{2}$ |

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 $y\neq x$, define the set $U_{{xy}}$ as

$U_{{xy}}=\{z\in X\mid h_{{xy}}(z)<f(z)+\epsilon\}$ |

Since $f$ and $h_{{xy}}$ are continuous, $U_{{xy}}$ is an open set. Because $x\in U_{{xy}}$ and $y\in U_{{xy}}$, $\left\{U_{{xy}}\mid y\in X\setminus\{x\}\right\}$ 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 $\{y_{1},y_{2},\ldots,y_{n}\}\subset X$ such that $X=\bigcup_{{m=0}}^{n}U_{{xy_{m}}}$. Define

$g_{x}=\min(h_{{xy_{1}}},h_{{xy_{2}}},\ldots,h_{{xy_{n}}}).$ |

By the corollary of the first part of the proof, $g_{x}\in\bar{\mathcal{A}}$. By construction, $g_{x}(x)=f(x)+\epsilon/2$ and $g_{x}<f+\epsilon$.

Third, we shall show that, for every $f\in C^{0}(X,\mathbb{R})$ and every $\epsilon>0$, there exists a function $g\in\bar{\mathcal{A}}$ such that $f\leq g<f+\epsilon$. This will complete the proof becauase it implies that $\bar{\mathcal{A}}=C^{0}(X,\mathbb{R})$. 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 $g_{x}(x)=f(x)+\epsilon/2>f(x)$, $x\in V_{x}$. Hence $\left\{V_{x}\mid x\right\}$ 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 $\{x_{1},x_{2},\ldots x_{n}\}\subset X$ such that $X=\bigcup_{{m=0}}^{n}V_{{x_{n}}}$. Define $g$ as

$g(z)=\max\{g_{{x_{1}}}(z),g_{{x_{2}}}(z),\ldots,g_{{x_{n}}}(z)\}$ |

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

## Mathematics Subject Classification

46E15*no label found*

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