Abstract Nonsense

Crushing one theorem at a time

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<g(x)<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)<f(z)+\varepsilon\right\}. 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)<f(y)+\varepsilon and \varphi_y(x)=f(x)<f(x)+\varepsilon 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)<f(z)+\varepsilon. Since z was arbitrary it follows that g_x(z)<f(z)+\varepsilon 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<g_x(z)\right\} for each x\in X. Since


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)<f(z)+\varepsilon\implies g(z)<f(z)+\varepsilon

for every z\in X We see that

f(z)-\varepsilon<g(z)<f(z)+\varepsilon\implies \|f-g\|_{\infty}<\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}


\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


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 - Posted by | General Topology, Topology, Uncategorized | , ,


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