## Stone-Weierstrass Theorems

This post is the first in a line of posts which deal with theorems of approximation due to Stone and/or Weierstrass. The simplest and first to be discussed is that the set of all polynomials defined on (denoted ) is dense in . From there we will make a generalization to where is compact Hausdorff. Finally, after a brief discussion of locally compact Hausdorff spaces we shall make our final statement regarding on such spaces. But, we begin with baby steps.

**Theorem:** The subspace is dense in .

**Proof:** To do this we must merely show that given any and there exists some such that . We take a constructionist approach.

We first note that it suffices to prove the case when .

So, define to be the th *Bernstein Polynomial* given by

We begin by noting some identities. By the binomial theorem

Differentiating with respect to and doing some rearrangement gives

Differentiating with respect to again gives us

Splitting the sum over the addition and apply the first identity to the first sum gives

Lastly, multiplying by gives

Now, using the first identity we see that

And so

And so

Since is a continuous map on a compact metric space it follows that is uniformly continuous and so we may find a such that

From here we break our sum into two terms

Clearly the first sum is less than (just replace the term with by , yank it out, and apply the first identity). So, we must merely show the second sum can be made less than independent of . We know appealing to ‘s continuity again that it is bounded and so for all . It clearly follows that

Thus, we merely need to make the right hand sum less than . Note that the last identity in the above shows that

and thus

But, it is easily verified that for and so

Taking larger than implies that

and so

.

It follows that

The conclusion follows. .

With this theorem we may prove some very cool stuff. For example:

**Theorem:** is separable.

**Proof:** Let be the set of all polynomials with rational coefficients. We prove that is dense in . To do this let and be arbitrary. Since is dense in there exists some such that .

Now, the mapping given by is continuous and so it attains a maximum on . So, choose numbers such that . Then, clearly and

It follows that and so

Thus, is dense in . It remains to show that it is countable. To see this define by . Clearly is surjective and since is the countable union of countable sets it’s countable. The conclusion follows. .

The next problem is a particular favorite of mine, for no particular reason.

**Theorem:** Define a *moment* of by . If have the same moments for then .

**Proof:** We prove (mainly for fun) a lemma from real analysis.

**Lemma:** Let be continuous. Then, only if .

**Proof:** Suppose that then for some , call that point . Since we clearly have for some . So, by ‘s continuity we see that there exists some such that . Taking the concentric closed ball of radius we see that for every . Thus, by ‘s continuity and the compactness of we have that for some . It follows that

Which of course contradicts the assumption that the integral is zero. The conclusion follows. .

Using this lemma we may assume that otherwise we would have that

and so we’d be done.

So, under this assumption define the functional by

We claim that is continuous. To see this and be given. Then, choosing such that

we see that

‘s continuity follows. We next note that that is linear and so

The last part gotten by noticing that

It follows that and since the right hand side is closed we see that

It follows that for every . In particular

The conclusion follows. .

We now prove a nice little extension of the Weierstrass theorem.

**Theorem:** Let be closed and bounded. Prove that is dense in .

**Proof: **Note that since is closed and bounded it is a subset of for some . Now, given any there exists by Tietze’s extension theorem an extension such that . Now, for every there exists some such that . Let . Furthermore, let

and finally let . Clearly, . So, it remains to show that given any and there exists some such that . To see this let and be as above and let . Then, clearly and

The conclusion follows.

**Corollary:** Using the exact same idea except taking one may prove that is separable for every closed and bounded .

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