10
$\begingroup$

A version of the Stone-Weierstrass Theorem asserts: If A is a linear subspace of C(K), the set of continuous functions on a compact space, and if A is a subalgebra that contains the constant functions and separates points, then A is dense in C(K) relative to the uniform (or sup-norm) topology. I am looking for a version for cones along the lines: if A is a subcone of X, itself a cone in C(K), if A is closed with respect to products, and if it contains constants and separates points in K, then A is ``dense'' in X. An example, would be the statement that the set of nondecreasing polynomials on [0,1] is dense in the set of nondecreasing continuous functions on [0,1]. (Is this true?)

I would appreciate references to such results, or to counterexamples.

Improve this question
$\endgroup$
5
  • $\begingroup$ Your example statement is certainly true: Given nondecreasing $f\in C[0,1]$, approximate $f$ by a nondecreasing smooth function $g$. Then approximate $g'$ by a polynomial $p$ which is nonnegative on $[0,1]$ (approximate to within $\epsilon$, then add constant $\epsilon$ to get nonnegative approximand within $2\epsilon$), then integrate $p$. $\endgroup$
    – Harald Hanche-Olsen
    Commented Jul 6, 2010 at 19:46
  • $\begingroup$ Any time "Stone-Weierstrass" and some flavour of positivity get mentioned, I'm reminded of Korovkin's theorem (eom.springer.de/k/k110130.htm ) which is not quite what is being asked for but might feed into the general picture. $\endgroup$
    – Yemon Choi
    Commented Jul 6, 2010 at 20:08
  • $\begingroup$ Seems like you need some restrictions on the size of X. For instance, X = C(K) is trivially a cone, but A is generally not going to be dense in C(K) (e.g. your example). $\endgroup$
    – Nate Eldredge
    Commented Jul 6, 2010 at 20:10
  • $\begingroup$ One case where this does work is if $K$ is a compact subset of $\mathbb{R}^n$, $X$ is the set of nonnegative functions on $K$, and $A$ is the set of [sums of] squares of polynomial functions on $K$. The way to prove it is to take any function $f\in X$, approximate $\sqrt{f}$ with $g\in \mathbb{R}[x_1,\ldots, x_n]$ using Stone-Weierstrass, and note that $g^2$ approximates $f$. $\endgroup$
    – Noah Stein
    Commented Jul 6, 2010 at 21:43
  • 7
    $\begingroup$ One small request: I'm guessing such results already exist, but if they do not and you find a nice one which you end up writing a paper about, could you title it "The Cone Weierstrass Theorem"? $\endgroup$
    – Noah Stein
    Commented Jul 6, 2010 at 22:08

2 Answers 2

Reset to default
3
$\begingroup$

A simple counterexample to the general "Cone Stone-Weierstrass" is the cone $A$ of all $C^\infty$ functions with all derivatives (including the function itself) non-negative (that one is clearly closed under multiplication, contains constants and separates points) in the cone $X$ of all non-negative non-decreasing functions in $C([-1,1]$. The reason is that every $f\in A$ can be extended to the unit disk as an analytic function and, moreover, the maximum of the absolute value of the extension will be controlled by the maximum of the absolute value of $f$ on $[-1,1]$ (just look at the Taylor series at $0$). So, the uniform closure of $A$ will also consist of traces of analytic functions (actually, it'll just be $A$ itself).

I believe one can create some result in positive direction here that is closer to what you asked than robin's quote but it seems easier just to ask what you really need.

Improve this answer
$\endgroup$
2
$\begingroup$

Your example is true because $A=X\cap H$ with $H$ dense in $C(K)$ and $X$ closed.

Anyway, there is a "cone Weierstrass theorem" which is:

If $S$ is a cone in $C(K)$ that contains constants and separates the points in $K$ then it is total in $C(K)$.

(it results from the fact that the linear span of $S$ is an algebra see here (it is in french unfortunatly)).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .