One of the most important consequences of the Weierstrass theorem (Thm 7.26 of Rudin) is the following (obvious) corollary:

Corollary 1Let be a positive real number. Then there is a sequence of polynomials which converges uniformly to on the interval , i.e.,

Exercise 23 in Rudin shows how you can prove this result directly without using the Weierstrass theorem. The reason that this particular corrollary is so important is that it provides a key step in the proof of the *Stone-Weierstrass* theorem:

Theorem 2Let be a compact metric space and let denote the set of real valued continuous functions on . Let be analgebraof functions with the following properties:

separates pointsof :given with there is a function such thatvanishes at no point of: for every there is a function such that .Then given any there exists a sequence such that converges to uniformly on .

Remark 3By saying that is analgebrawe mean that

- is a vector sub-space of , i.e., if and then the function .
- If then the function .
Here, and in 1 and 2 are the following:

Thus the Stone-Weierstrass theorem says that *an algebra of continuous functions on a compact set that is nowhere vanishing and separates points is dense in (in the uniform topology).*

You can find the full proof of the Stone-Weierstrass theorem in Rudin (Theorem 7.32). The main steps of the proof are:

- Let be the closure of in the uniform metric (). We must show that
- Using Corollary 1 we prove that if then . To see this let and let be a sequence that converges to uniformly. We may assume that for all and . (Why?) Now let be a sequence of polynomials that converge uniformly to on the interval . Then is a sequence in that converges uniformly to . (Why are ? Notice how we need the fact that is an algebra!)
- From 2, we conclude that
are in whenever and are. Repeatedly applying these identities, we can prove that and are in whenever are.

- Now let and let . For each we can find such that and (Why? This uses the fact that separates points and is no-where vanishing.) By continuity and compactness for each we can find points and an open cover such that
Thus

satisfies

Now we find points and an open cover such that

so

satsifies

Note how every hypothesis of the theorem was used in an essential way.

As an example of an important corrollary we have the following result regarding trigonometric polynomials. A *trigonometric polynomial* is a linear combination of the form

The numbers are called the *coefficients* and are called the *frequencies of the polynomial. *(Note that using trig identities we can reduce and to expressions of this form. Thus the name *trigonometric polynomial* makes sense.)

Theorem 4Let . Then there exists a sequence of trigonometric polynomials that converge uniformly to on . Indeed there exist a sequence of trigonometric polynomials with frequencies restricted to the set that converge uniformly to on .

*Proof:* Let be the set of all trigonometric polynomials restricted to . Since

and

it is easy to see that is an algebra. It is not identically zero since for every either or . Likewise if and then for some and and . It follows that separates points. Thus the first claim follows from Stone-Weierstrass.

To see that we may restrict the frequencies as in the second claim is left as an exercise. You must show that the result set is an algebra and separtes points. That it is an algebra follows from the angle sum identities and the facts that and . To see that it separates points note that the functions and already do the job.