Stone Weierstrass, All Continuous Functions are Accessible

The Closure of a Subring is a Subring

If a subring of c(S) does not contain the constant functions, (it might not even contain 1), we will assume it can be scaled by R. It is thus an R vector space. If S is compact, the closure of a subring is another subring.

Let W be such a subring and let W′ be the closure of W in c(S), the set of functions uniformly approximated by functions in W. If g is in W′, approximated by a sequence f₁ f₂ f₃ etc, scale the sequence by a constant, and the limit g scales accordingly. Also, add two sequences together and add their limits. Thus the functions of W′ form an R vector space.

Product is a little trickier, and that's where we need S to be compact, so that all functions are bounded in R. Let the functions d₁ d₂ d₃ etc approach e, and let f₁ f₂ f₃ etc approach g. Let b be the larger of |e| and |g|, then add 1 for good measure. Build a new sequence of functions h_i = d_i×f_i. The product of continuous functions is continuous, hence each h_i is in c(S). Also, W is a subring, hence each h_i is in W. If j = e×g, then j(x) is the limit of h(x). This because the product of two limits is the limit of the product sequence in real space. The tricky part is uniformity.

Go far enough out in the sequences d and f, so that for every x, d_n is within ε of e, and f_n is within ε of g. The product is h_n, and when this is subtracted from j_n, the error term is no worse than 2bε+ε². This approaches 0 for small ε, hence h_n approaches j uniformly, j is in W′, and W′ is a subring of c(S).

All Continuous Functions are Accessible

Let W be a subring of c(S) that contains the constant functions, and separates any pair of points in S. We wish to show W′ is a lattice. In other words, W′ includes the min and max of any two functions in W′.

Since W′ is a ring containing the constant functions, f in W′ implies p(f) in W′ for any polynomial p. If |f| ≤ 1, Compose f with p(x) from the previous lemma, and we can approximate the function that is the absolute value of f. (I'm going to call this function |f|, which is poor notation, because |f| also means the norm of f. Sorry about that.) Since W′ is closed, it contains |f|.

If f is any function, divide f by b, the bound on f, take the absolute value, and multiply by b. Thus f ∈ W′ implies |f| ∈ W′.

Note that (f+|f|)/2 extracts the positive portions of f, and leaves the negative portions on the cuttingroom floor. Similarly, (f-|f|)/2 retains the negative portions. Given f and g, f plus the positive portions of g-f yields the max, while f plus the negative portions of g-f yields the min. Therefore W′ is a lattice, it separates points, and it is closed under scaling and translation. This makes W′ dense in c(S).

Remember that W′ is closed, and dense in c(S), hence W′ = c(S). Every function in c(S) is approximated by the functions of W.

Polynomials in the Coordinates of S

As a corollary, let S be any closed bounded set in Rⁿ, hence compact, and let W be the ring of polynomials built from the n coordinate projections of Rⁿ onto R. The constant polynomials become the constant functions from S into R. If x and y difffer in the i^th coordinate, then the i^th projection maps x and y to different values of R. In other words, W separates the points of S. It follows that W is dense in c(S), and any continuous function from S into R can be uniformly approximated by a polynomial in the coordinates of Rⁿ.