"Convex" polynomials Let me define "convex" polynomials, as the smallest class $\mathcal{C}$ of functions $p:\mathbb{R}\rightarrow \mathbb{R}$ defined (inductively) as:
UPDATED (case 0 was missing): 
0) $p(x)=x$, i.e., the identity, is in $\mathcal{C}$.
1) $p(x)= 1$ is in $\mathcal{C}$ 
2) $p(x)= 1- q(x)$, for $q\in \mathcal{C}$, is in $\mathcal{C}$.
3) $p(x) = c \cdot q(x) + (1-c) r(x)$, for some $c\in [0,1]$ and $q,r\in\mathcal{C}$, is in $\mathcal{C}$.   
4) $p(x) = q(x) \cdot r(x)$, with $q,r\in\mathcal{C}$, is in $\mathcal{C}$.
So basically $\mathcal{C}$ include all constant $c\in[0,1]$ functions, and is closed under the operations of convex combination, $1-$, and multiplication.
Clearly for every $p\in \mathcal{C}$ and $x\in [0,1]$, $p(x)\in [0,1]$ too.
Question: Is this class of functions known? Can anything interesting be said about this class? For example I think that every continuous function $f:[0,1]\rightarrow[0,1]$ can be approximated by functions in $\mathcal{C}$. Would this be some (trivial?) consequence of the Stone–Weierstrass theorem?
Thanks!

(Too long for a comment)
Lemma 1. If $f\in\mathcal C$ is not constant then $f^{-1}(\{0,1\})\subseteq \{0,1\}$.
Proof. The claim clearly holds for the functions from rules 0 and 1.
If in rule 2 it holds for $q$, then also for $p$ because $0$ and $1$ are simply swapped. If in rule 3 it holds for $q$ and $r$, it also holds for $p$ because a convex combination within $[0,1]$ can yield a boundary value only if at least one input is this boundary value. If in rule 4 it holds for $q$ and $r$, it also holds for $p$ because a product i s$0$ iff one factor is $0$ and the product is $1$ if both factos are $1$. 
Hence the claim follows by induction. $_\square$
As a corollary, Alex Ravsky's tentative conjecture is busted: $\mathcal C$ does not contain all polynomials that map $[0,1]\to[0,1]$. For example, $p(x)=4x^2-4x$ is missing.
(I thought I had more, but had to scratch it again)
