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!

  • $\begingroup$ You might be interested in "Bernstein polynomials". en.wikipedia.org/wiki/Bernstein_polynomial $\endgroup$ – awkward Oct 3 '18 at 12:55
  • $\begingroup$ "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. $\endgroup$ – David C. Ullrich Oct 3 '18 at 14:03
  • $\begingroup$ @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!!!! $\endgroup$ – user64066 Oct 3 '18 at 15:45
  • 1
    $\begingroup$ 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 $\lvert x \rvert$? $\endgroup$ – Paul Frost Oct 3 '18 at 22:44
  • $\begingroup$ If you mean the latter, I shall delete my answer. $\endgroup$ – 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.


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.