# Abstract Nonsense

## Stone-Weierstrass Theorem (Real deal)

In this section we extend the idea of the last post to arbitrary compact Hausdorff spaces. Admittedly a lot of information will be assumed known. Including facts about lattices and subalgebras of $\mathcal{C}[X,\mathbb{R}]$.

We begin by recalling that if $f,g:X\to\mathbb{R}$ are continuous then so is $f\vee g=\min\{f,g\}$ and $f\wedge g=\max\{f,g\}$

It thus makes sense to talk about sublattices of $\mathcal{C}[X,\mathbb{R}]$.

Our first theorem is a potent one, it gives sufficient conditions for a closed sublattice of $\mathcal{C}[X,\mathbb{R}]$ to be the full space.

Theorem: Let $X$ be compact Hausdorff with more than one point and  $\mathcal{L}$ be a closed sublattice of $\mathcal{C}[X,\mathbb{R}]$ with the property that given any distinct $x,y\in X$ and any real numbers $a,b$ there exists some $g\in\mathcal{L}$ such that $g(x)=a,g(y)=b$. Then, $\mathcal{L}=\mathcal{C}[X,\mathbb{R}]$

Proof: Since $\mathcal{L}$ is closed it suffices to show that given any $f\in\mathcal{C}[X,\mathbb{R}]$ there exists some $g\in\mathcal{L}$ such that $f(x)-\varepsilon for any $\varepsilon>0$.

We start by considering a fixed but arbitrary $x\in X$. For each $y\in X-\{x\}$ there exists some $\varphi_y\in\mathcal{L}$ such that $\varphi_y(x)=f(x)$ and $\varphi_y(y)=f(y)$. Define $G_y=\left\{z:\varphi_y(z). Clearly this is equivalent to $(\varphi_y-f)^{-1}(-\infty,\varepsilon)$ and since $\varphi_y-f$ is continuous it follows that $G_y$ is open. Also, since $\varphi_y(y)=f(y) and $\varphi_y(x)=f(x) and so $x,y\in G_y$. Thus, $\left\{G_y\right\}_{y\in X-\{x\}}$ is an open cover for $X$ and thus by it’s compactness it admits a finite subcover $\left\{G_{y_1},\cdots,G_{y_n}\right\}$. Let $\varphi_{y_1},\cdots,\varphi_{y_n}$ be the associated functions. Clearly then since $\varphi_{y_k}\in\mathcal{L}$ we have (by virtue of the fact that it’s a lattice) that $g_x=\varphi_{y_1}\vee\cdots\vee\varphi_{y_n}\in\mathcal{L}$. Also, given any $z\in X$ we have that $z\in G_{y_k}$ for some $k$ and so $g_x(z)\leqslant \varphi_{y_k}(z). Since $z$ was arbitrary it follows that $g_x(z) for every $z\in X$.

Now, considering the same procedure as above for each $x\in X$ we consider the sets $H_x=\left\{z\in X:f(z)-\varepsilon for each $x\in X$. Since

$g_x(x)=\min\{\varphi_{y_1}(x),\cdots,\varphi_{y_n}(x)\}=\min\{f(x),\cdots,f(x)\}=f(x)>f(x)-\varepsilon$

we see that $x\in H_x$. Noticing once again that $H_x=(g_x-f)^{-1}(-\varepsilon,\infty)$ we see that $H_x$ is open for each $x\in X$. Thus, combining these two comments we see that $\left\{H_x\right\}_{x\in X}$ is an open cover for $X$. Appealing to compactness again we see that it has a finite subcover $\left\{H_{x_1},\cdots,H_{x_n}\right\}$. Thus, let $\{g_{x_1},\cdots,g_{x_n}\}$ be the associated functions. Since each are in $\mathcal{L}$ we have that $g=g_{x_1}\wedge\cdots\wedge g_{x_n}\in\mathcal{L}$. Now, let $z\in X$ be arbitrary. Then, $z\in H_{x_k}$ for some $k$ and so $g(z)\geqslant g_{x_k}(z)>f(z)-\varepsilon$. Since $z$ was arbitrary it follows that $g(x)>f(x)-\varepsilon$ for all $x\in X$. Also noting though that

$g_{x_1}(z),\cdots,g_{x_n}(z)

for every $z\in X$ We see that

$f(z)-\varepsilon.

Thus, by previous comment the conclusion follows. $\blacksquare$

Now, it is a problem in elementary analysis to prove that

$\displaystyle f\vee g=\frac{f+g-|f-g|}{2}$

and

$\displaystyle f\wedge g=\frac{f+g+|f-g|}{2}$

It follows that every linear subspace of $\mathcal{C}[X,\mathbb{R}]$ which contains $|f|$ is a sublattice. From this we come to the following theorem

Theorem: Let $\mathcal{A}$ be a closed subalgebra of $\mathcal{C}[X,\mathbb{R}]$ then it is also a closed sublattice.

Proof: It follows from prior comment that we must merely show that $f\in\mathcal{A}\implies |f|\in\mathcal{A}$. Since $\mathcal{A}$ is closed we must merely show that $|f|$ is an adherent point of $\mathcal{A}$. Thus, let $\varepsilon>0$ be given.

Now, the function $\eta:\left[\gamma,\|f\|_{\infty}\right]\to\mathbb{R}$ where $\displaystyle \gamma=\min\left\{\inf_{x\in X}f(x),-\|f\|_{\infty}\right\}$ given by $x\mapsto |x|$ is continuous. And from the previous post there exists some $a_0+\cdots+a_n x^n-p\in\mathbb{P}[\gamma,-\|f\|_{\infty}]$ such that $\left|\eta-p\right|_{\infty}<\frac{\varepsilon}{2}$. In particular, $\left|\eta(0)-p(0)\right|=|a_0|<\frac{\varepsilon}{2}$. Thus, $q(x)=p(x)-a_0$ is a polynomial without a constant such that

$\left\|\eta-q\right\|_{\infty}\leqslant \left\|\eta-p\right\|+|a_0|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$.

Now, since $f\in\mathcal{A}$ and $\mathcal{A}$ is a subalgebra we have that $q(f)\in\mathcal{A}$. From there it is fairly easy to see that $\left\||f|-q(f)\right\|_{\infty}<\varepsilon$. By previous comment the conclusion follows. $\blacksquare$

We are now prepared to prove the main theorem in this post

Theorem: Let $X$ be a compact Hausdorff space and $\mathcal{A}$ a closed subalgebra of $\mathcal{C}[X,\mathbb{R}]$ which contains a non-zero constant function and separates points. Then, $\mathcal{A}=\mathcal{C}[X,\mathbb{R}]$.

Proof: If $X$ has one point then $\mathcal{C}[X,\mathbb{R}]$ consists of all the constant functions on $X$ and since $\mathcal{A}$ contains a non-zero constant function it follows that $\mathcal{A}=\mathcal{C}[X,\mathbb{R}]$.

Thus, we may assume WLOG that $\text{card }X\geqslant 2$. Thus, it follows from our first theorem that given any distinct $x,y\in\mathcal{A}$ and $a,b\in\mathbb{R}$ there exists some $\varphi\in\mathcal{A}$ such that $\varphi(x)=a,\varphi(y)=b$. Now, since $\mathcal{A}$ separates points there exists some $f\in\mathcal{A}$ such that $f(x)\ne f(y)$. So define

$\displaystyle \varphi(z)=a\frac{f(z)-f(y)}{f(x)-f(y)}+b\frac{f(x)-f(z)}{f(x)-f(y)}$

Clearly $\varphi$ is a linear combination of elements of $\mathcal{A}$ and so $\varphi\in\mathcal{A}$. Also, $\varphi(x)=a,\varphi(y)=b$. The conclusion follows. $\blacksquare$.

There is an analgous theorem dealing with $\mathcal{C}[X,\mathbb{C}]$ be it is omitted here.

We discuss only one application of the above theorem. But before it we need a theorem.

Theorem: Let $\mathcal{A}$ be a subalgebra of $\mathcal{C}[X,\mathbb{R}]$ then $\overline{\mathcal{A}}$ is also a subalgebra.

Proof: We prove that it is closed under vector addition, for scalar and vector multiplication follow similarly.

Let $f,g\in\overline{A}$ then there exists $f_0,g_0\in\mathcal{A}$ such that $\|f-f_0\|_{\infty},\|g-g_0\|_{\infty}<\frac{\varepsilon}{2}$. It follows from the fact that $\mathcal{A}$ is a subalgebra that $f_0+g_0\in\mathcal{A}$ and

$\left\|f+g-(f_0+g_0)\right|_{\infty}=\left\|f-f_0+g-g_0\right\|_{\infty}\leqslant\left\|f-f_0\right\|_{\infty}+\left\|g-g_0\right\|_{\infty}<\varepsilon$.

It follows that $f+g$ is an adherent point of $\mathcal{A}$ and so $f+g\in\overline{\mathcal{A}}$.

The other follows similarly. $\blacksquare$

Theorem: Let $X=[a_1,b_1]\times\cdots\times[a_n,b_n]$ then $\mathbb{P}[X]$ is dense in $\mathcal{C}[X,\mathbb{R}]$.

Proof: Clearly $1(x)\in\mathbb{P}[X]$. Also, let $(x_1,\cdots,x_n),(y_1,\cdots,y_n)\in X$ be distinct, in other words $x_k\ne y_k$ for some $k$. Define $p:X\to\mathbb{R}$ by

$\displaystyle (z_1,\cdots,z_k,\cdots,z_n)\to z_k-x_k$

Clearly $p(x_1,\cdots,x_n)=0$ but $p(y_1,\cdots,y_n)\ne 0$. It follows that $\mathbb{P}[X]$ separates points. Thus, $\overline{\mathbb{P}[X]}$ is a closed subalgebra of $\mathcal{C}[X,\mathbb{R}]$ which contains a non-zero constant function and separates points. The conclusion follows from the previous theorem since $X$ is a compact Hausdorff space. $\blacksquare$.

March 28, 2010 -