Stone-Weierstrass for cones 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.
 A: 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)).  
A: 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.
