# Algebraic Proof of Stone-Weierstrass

I have seen a few proofs of the Stone-Weierstrass theorem today for the first time. While some were completely analytic, the proof given in Rudin's Principles of Mathematical Analysis hinted at an algebraic approach. He defines a sequence of normalized polynomials which he convolutes with the arbitrary continuous function $f(x)$ to obtain polynomial approximations. I am aware that convolution may be thought of as an inner product, hence was curious if there is a proof which uses algebraic language and minimizes the analysis.

Of course, I do not expect the analysis to disappear, but perhaps it can be marginalized. For example, while the easiest proofs of the fundamental theorem of algebra are entirely analytic, there are mostly algebraic proofs which only use analysis to assert that odd degree polynomials have a real root.

I seek such a proof because 1) I think it would be interesting, and 2) My intuition works best with algebra.

• Note that convolution isn't an "inner product", since it's not scalar valued. (The convolution of two functions is a function.) Instead, it is the product operation for a group algebra – Jim Belk Jun 26 '13 at 16:31
• That's fair. If we fix an $x_0$ at which to evaluate the convolution, then we get an inner product, no? Perhaps this is not a useful observation. I do not have enough experience to know. – tghyde Jun 27 '13 at 0:45

The norm in $C(X)$ (the supremum norm) does not come from an inner product. More strongly, the topology in $C(X)$ is not induced by an inner product. The use of the convolution in Rudin's argument allows him to produce uniform approximations.