# Where does polynomials come from in Stone Weierstrass?

It might be a silly question to ask but most versions of Stone Weierstrass theorem define polynomials inductively and shows that it converges to a particular function. To make the previous sentence meaningful, consider the following theorem

Define a sequence $$(P_n)_{n=0}^\infty$$ of polynomials recursively by the relations $$P_0(t)=0$$ and $$P_{n+1}(t)=P_n(t)+1/2 [ t^2-P_n(t)^2]$$. Then $$P_n\to \mid t \mid$$ uniformly on $$[-1,1]$$.

Now my question is where do these polynomials come from? Thanks for any help!

• You might be interested in "Bernstein polynomials". en.wikipedia.org/wiki/Bernstein_polynomial – awkward Oct 3 '18 at 12:55
• "most versions of Stone Weierstrass theorem define polynomials inductively"??? I don't know a version of the theorem that even mentions polynomials. Weierstass' theorem on polynomial approximation is a special case of S-W, not a "version of" S-W. – David C. Ullrich Oct 3 '18 at 14:03
• @DavidC.Ullrich All versions of Stone Weierstrass Theorem rely on (function) subalgebra. Along the way you need to show that uniform closure is closed for $\wedge$ and $\vee$ operations. For this fact you need a theorem like the one I put in my question. So every version implicitly relies on polynomials defined inductively!!!! – user64066 Oct 3 '18 at 15:45
• I gave an answer to your question, but now I am no longer sure what you are really asking. Is it why polynomials are used at all to approximate functions or why, precisely, the above $P_n$ are used to approximate $\lvert x \rvert$? – Paul Frost Oct 3 '18 at 22:44
• If you mean the latter, I shall delete my answer. – Paul Frost Oct 3 '18 at 23:00

In its most general version the Stone-Weierstrass theorem is a result about generating dense subalgebras of $$C(X)$$ = unitary algebra of continuous functions $$f : X \to \mathbb{R}$$ endowed with the compact-open topology (which is the topology of uniform convergence on all compact subsets of $$X$$). Note that if $$X$$ is compact, then $$C(X)$$ has the topology of uniform convergence.

Any subset $$D \subset C(X)$$ generates a smallest unitary subalgebra $$A(D) \subset C(X)$$. It is easy to see that $$A(D)$$ is the set of functions having the form $$p(f_1,\dots,f_n)$$, where $$f_i \in D$$ and $$p$$ ranges over all polynomials in $$n \ge 1$$ variables (all $$n$$ allowed).

This is the true origin of the polynomials in the Stone-Weierstrass theorem which says the following:

If $$D$$ separates points (which means that for any two distinct $$x,x' \in X$$ there exists $$f \in D$$ such that $$f(x) \ne f(x')$$), then $$A(D)$$ is dense in $$C(X)$$.

Now let $$X = [-1,1]$$. Then $$D = \{ id \}$$, $$id(x) = x$$, trivially separates points and $$P(X) = A(D)$$ is the set of all polynomials in the variable $$x$$. That $$P(X)$$ is dense in $$C(X)$$ is the "classical" Stone-Weierstrass theorem. One can certainly give a direct proof by constructing inductively a sequence of polynomials converging to a given $$f \in C([-1,1])$$, but in my opinion this is an unusual (though valid) approach.

Edited:

A key ingredient in the proof of the Stone-Weierstrass Theorem is the following result:

Let $$A \subset C(X)$$ be a unitary subalgebra. If $$f \in A$$, then $$\lvert f \rvert \in \overline{A}$$.

This is proved as follows (here we only consider compact $$X$$, but the argument can be generalized to arbitrary $$X$$).

For $$f : X \to \mathbb{R}$$ let $$R = \sup_{x \in X} \lvert f(x) \rvert$$. Then $$f/R \in C(X)$$ and $$(f/R)(X) \subset [-1,1]$$. If we can find a sequence of polynomials $$p_n(t)$$ such that $$p_n(t) \to \lvert t \rvert$$ uniformly on $$[-1,1]$$, then $$p_n(f(x)/R) \to \lvert f(x)/R \rvert$$ uniformly on $$X$$. Defining $$f^*_n(x) = Rp_n(f(x)/R)$$ we see that $$f^*_n(x) \to \lvert f(x) \rvert$$ uniformly on $$X$$. But we have $$f^*_n \in A$$ because $$f/R \in A$$ and $$p_n$$ is a polynomial (recall that $$A$$ is a unitary algebra). Note that to conclude $$p_n(f/R) \in A$$ it is essential that $$p_n$$ is a polynomial.

So how to find these $$p_n$$? There are certainly many approaches. Not all of of them construct the $$p_n$$ recursively as in your question. An alternative is to construct polynomials $$q_n(x)$$ such that $$q_n(x) \to \sqrt{x}$$ uniformly on $$[0,1]$$. Then $$p_n(x) = q_n(x^2)$$ is a solution. But now it is known that

$$1 - \sqrt{1-y} = \sum_{i=1}^\infty \left\lvert \binom{\frac{1}{2}}{i} \right\rvert y^i$$

uniformly on $$[0,1]$$. Therefore $$q_n(x) = 1 - \sum_{i=1}^n \left\lvert \binom{\frac{1}{2}}{i} \right\rvert (1-x)^i$$ will do.

Conclusion:

Polynomials are essential to prove the Stone-Weierstrass theorem. The reason is that $$A$$ being a unitary subalgebra of $$C(X)$$ is equivalent to the following:

For all polynomials $$p$$ in $$n \ge 1$$ variables and all $$f_1,\dots,f_n \in A$$ one has $$p(f_1,\dots,f_n) \in A$$.

However, in the general formulation of the Stone-Weierstrass theorem this fact is not explicitly mentioned. But it is implicit in the assertion that $$A(D)$$ is dense in $$C(X)$$ because $$A(D)$$ is generated by polynomials.