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I was looking for a simple application of the Stone-Weierstrass theorem.

First I thought that if $X$ is any compact measure space then Stone-Weierstrass implies that $C_c(X)$ is dense in $L^p$.

But I have to assume that $X$ is compact otherwise I don't have $1$ in $C_c(X)$. That of course makes it a boring example since then $C_c(X) = C(X)$. Can someone show me a slightly more interesting but still simple example? Thank you.

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Well, one standard application of the theorem is that the trigonometric polynomials are dense in $C([0,2\pi])$, say. Note that this is not immediate from Fourier Analyis (only if you consider Cesaro summation). –  user20266 Jul 13 '12 at 18:48
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I don't see how Stone-Weierstrass implies the density of $C_c(X)$ in $L^p$, since this theorem only gives density with respect to the supremum norm. –  Davide Giraudo Jul 13 '12 at 19:02
    
@DavideGiraudo Doh! facepalm Thank you! –  Rudy the Reindeer Jul 13 '12 at 19:03
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@MattN. The Stone-Weierstrass theorem implies that any commutative $C^*$-algebra is equivalent to $C_0(X)$ for some locally compact $X$. –  azarel Jul 13 '12 at 19:18

3 Answers 3

This is an example I learned only recently from one of my professors (I don't know much about probability, so I'm not sure whether this is an old hat):

We can use Stone-Weierstrass to prove the Kolmogorov extension theorem (or a version thereof). Namely, given a collection $X = \{\mu_\alpha\}_{\alpha}$ of probability measures on $[0,1]$, there exists a probability measure $\mu$ on $\prod_{\alpha} [0,1]$ such that $$\mu(\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) ) = \prod_{i=1}^n \mu_{\alpha_i}(A_i)$$ for all $\alpha_1, \dots, \alpha_n$, where $\pi_\alpha: X \to [0,1]$ denotes projection.

Sketch of proof: The set $Q \subset C(X)$ of continuous maps, which depend only on "finitely many components", i.e. those $f\in C(X)$ which can be written in the form $$f(x) = g(\pi_{\alpha_1}(x), \dots, \pi_{\alpha_n}(x)) \qquad \text{ for some }\alpha_1, \dots, \alpha_n,\, n\in \mathbb N \text{ and } g\in C([0,1]^n)$$ is a unital algebra which separates points. By Tychonoff's theorem we know that $X$ is compact, so $Q$ is dense by Stone-Weierstrass.

We can now define a continuous linear functional $I: C(X) \to \mathbb R$ as follows: Given $f(x) = g(\pi_{\alpha_1}(x), \dots, \pi_{\alpha_n}(x)) \in Q$ as above, we define $$I(f) = \int_{[0,1]^n} g(x_1, \dots, x_n) \, dx_1 \dots dx_n$$ Then $I$ is well-defined and is a bounded linear functional on the dense subspace $Q\subset C(X)$. Therefore it can be extended uniquely to a bounded linear functional on all of $C(X)$.

Finally, Riesz' representation theorem for compact Hausdorff spaces shows that to every such functional $I$ there corresponds a unique Radon measure $\mu$ such that $I(f) = \int_X f\, d\mu$.

It is now not hard to show that this $\mu$ satisfies $\mu(\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) ) = \prod_{i=1}^n \mu_{\alpha_i}(A_i)$ by finding a suitable sequence of continuous functions approximating the indicator function of $\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) $. $\square$

I had seen other proofs of this extension theorem (which I didn't like too much, because I couldn't easily see what's going on), so I was really amazed when I first heard of the above argument. I especially like how various big theorems suddenly pop up and fit together so well! =)

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Some examples:

  • if $(K,d)$ is a compact metric space, then $C(K)$ endowed with the supremum norm is separable. To see this, take $\{x_n\}$ a countable dense subset, and consider the algebra generated by the maps $f_{m,n}:=\max\{1/n,d(x,x_m)\}$.
  • the set of trigonometric polynomials is dense in $C[0,2\pi]$ endowed with the supremum norm.
  • Consider two compact metric spaces $(K_1,d_1)$, $(K_2,d_2)$. The set of maps of the form $$f(x,y):=\sum_{k=1}^nf_k(x)g_k(y),$$ $n\in\Bbb N$, $f_k\in C(K_1), g_k\in C(K_2)$, $1\leq k\leq n$, is dense in $C(K_1\times K_2)$.
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  • Given that $C[0,1]$ is $\|.\|_{L^2}$-dense in $L^2[0,1]$, Stone-Weierstrass easily implies that the trigonometric polynomials are as well.
  • Look at this question (but not the answer).
  • Take a functional analysis book (e.g. Conway) and look at the exercises following Stone-Weierstrass.
  • Try to formulate and deduce a similar statement about $C_c(X)$ for $X$ not necessarily compact. As you noticed, it need not contain $1$, so think about how to modify this condition. (If you get stuck, consult wikipedia or again a functional analysis book.)
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