The Stone-Weierstrass Theorem guarantees that every continuous function defined on a closed interval [a,b] can be approximated as closely as desired by a polynomial.

I'm curious in polynomials of the form $P(x) = a_0 + a_1 f(x)^1 + a_2 f(x)^2 + a_3 f(x)^3$ where $f(x)$ is some continuous function f(x) could equal $e^x$ for example. What can be said for polynomials of this form? Does something similar to the Stone-Weierstrass Theorem hold?

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    $\begingroup$ Well, functions in the form $p\circ f$ for $p$ polynomial and $f$ a fixed continuous function $f$ will approximate uniformly functions in the form $g\circ f$ for some continuous function $g$. $\endgroup$ – Saucy O'Path Jun 4 '19 at 20:49
  • $\begingroup$ I don't quite know what to think about continuous functions in the form $g\circ f$ where $g$ is not continuous, though. Of course, the limit must be in the form $g\circ f$ for some function $g$. $\endgroup$ – Saucy O'Path Jun 4 '19 at 20:50
  • $\begingroup$ You need some restriction on the definition of $f(x)$. For example $f(x)=c$ won't work. $\endgroup$ – herb steinberg Jun 4 '19 at 20:53
  • $\begingroup$ Depends - if you have $f$ one-to-one on $[a,b]$ then the range of $f$ is of the form $[c,d]$ for some $c\leq d.$ If $f$ is not constant, and $g$ is continuous on $[a,b]$ we can write $g(x)=g\circ f^{-1}\circ f(x)$ and $g\circ f^{-1}$ is a continuous function on $[c,d]$ and thus can be approximated by a polynomial $p(x).$ Then $p\circ f$ is an approximation for $g,$ sinc $f$ is continuous and hence uniformly continuous. $\endgroup$ – Thomas Andrews Jun 4 '19 at 20:54
  • $\begingroup$ More generally, you can approximate $g$ by $p(f(x))$ if and only if, for all $x,y\in[a,b],$ with $f(x)=f(y),$ you have $g(x)=g(y).$ $\endgroup$ – Thomas Andrews Jun 4 '19 at 20:57

If $f$ is injective and continuous, then the Stone-Weierstrass allows us to deduce that the functions of the form $\sum_{k=0}^na_kf^k$ are dense in $C\bigl([a,b]\bigr)$. Actually, assuming that $f$ is continuous, the set that I described is dense if and only if $f$ is injective.

  • $\begingroup$ I'm having a bit of trouble following your explanation. What is meant by 'dense' in this context? $\endgroup$ – user2944352 Jun 4 '19 at 22:25
  • $\begingroup$ It means that for every $g\in C\bigl([a,b]\bigr)$ and for every $\varepsilon>0$, there are numbers $a_0,a_1,\ldots,a_n$ such that$$\sup_{x\in[0,1]}\left\lvert g(x)-\sum_{k=0}^na_kf^k(x)\right\rvert<\varepsilon.$$ $\endgroup$ – José Carlos Santos Jun 4 '19 at 22:37
  • $\begingroup$ That's exactly what I wanted to know. Thanks! $\endgroup$ – user2944352 Jun 4 '19 at 22:40

You seem to be a bit confused about what the Stone-Weierstrass theorem is. The Weierstrass approximation theorem (not the Stone-Weierstrass theorem!) says that any continuous function on a closed interval is a uniform limit of polynomials. The Stone-Weierstrass theorem is a vast generalization of the Weierstrass approximation theorem, and (one version of it) says that for any compact Hausdorff space $X$, any unital subalgebra of $C(X,\mathbb{R})$ that separates points is dense. In particular, if $f$ is any injective function, the unital subalgebra it generates (i.e., the set of polynomials in $f$) is dense.

In other words, when $f$ is injective on $[a,b]$, the Stone-Weierstrass theorem itself says that every continuous function on $[a,b]$ is a uniform limit of polynomials in $f$. (Note that the assumption that $f$ is injective is obviously necessary, since if $f(c)=f(d)$ then the same will be true of any uniform limit of polynomials in $f$).

  • $\begingroup$ I think you're right here. I have mistakenly conflated the two. $\endgroup$ – user2944352 Jun 4 '19 at 22:26

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