$h$ is strictly monotone iff polynomials in $h$ are uniformly dense Suppose $h$ is continuous on $[0,1]$. Then, I have to prove the following:

$h$ is strictly monotone iff every continuous function on $[0,1]$ can be uniformly approximated on $[0,1]$ by a polynomial in $h$.

I know we get to use here the Stone—Weierstrass theorem. I proved one direction, that is if $h$ is continuous on $[0,1]$ and $h$ is strictly monotone then every function on $[0,1]$ can be uniformly approximated by a polynomial in $h$. But I am having a hard time doing the converse part. 
 A: Here is a hint: if $h$ is continuous on $[0,1]$, it is strictly monotone if and only if it is one-to-one. 
A: By hypothesis, we have that $h$ is continuous. Assume, by contrapositive, that $h$ is not strictly monotone: in particular, since $h$ is continuous, there exist two points $x<y$ such that $h(x)=h(y)$ (can you see why?).
Now, consider $f$ to be the identity function on $[0,1]$. We can show that no family of polynomials $(P_n\circ h)_n$ in $h$ can uniformly approximate $f$. Indeed, take any family of polynomials $(P_n)_n$, and let $\varepsilon \stackrel{\rm def}{=} \frac{y-x}{3}$.
Since $h(x)=h(y)$, for every $n\geq 0$ we have $P_n\circ h(x)=P_n\circ h(y)$, and therefore by the triangle inequality 
$$
\lvert P_n\circ h(x) - f(x)\rvert + \lvert P_n\circ h(y) - f(y)\rvert
= \geq \lvert x-y\rvert = 3\varepsilon
$$
so, for every $n\geq 0$, $\lVert P_n\circ h - f\rVert_\infty \geq \max(\lvert P_n\circ h(x) - f(x)\rvert,\lvert P_n\circ h(y) - f(y)\rvert) \geq \frac{3}{2}\varepsilon$.
Therefore, $h$ not strictly monotone implies that there exists a function on $[0,1]$ which cannot be uniformly approximated by a family of polynomials in $h$. $\square$
A: First part: if $h$ is strictly monotone, polynomials in $h$ are dense in $C^0[0,1]$ with respect to the supremum norm. Assume that $f$ is the function we want to uniformly approximate. Due to Weierstrass approximation theorem $f\circ h^{-1}$ is a continuous function that can be uniformly approximated by a sequence of polynomials $\{p_n(x)\}_{n\geq 1}$, hence the sequence $\{(p_n\circ h)(x)\}_{n\geq 1}$ provides a uniform approximation of $f$ as wanted.
Second part: if $h$ is not strictly monotone, there is some $f\in C^0[0,1]$ that cannot be uniformly approximated by any sequence of the form $\{(p_n\circ h)(x)\}_{n\geq 1}$. We may assume that $h$ has a stationary point at some $\xi\in(0,1)$, and so does any $(p_n\circ h)$. If we take some $f\in C^1[0,1]$ that is strictly monotone and has a very large derivative at $\xi$, the uniform convergence of $\{(p_n\circ h)(x)\}_{n\geq 1}$ is doomed to fail for any sequence of polynomials $\{p_n(x)\}_{n\geq 1}$.
There is a little of handwaving in the second part: I leave to you the mission to remove it$^{(*)}$ and turn my second part into a full proof, I just wanted to sketch some ideas.
$(*)$ This is an interesting fix: there are $\xi_1\neq \xi_2$ in $[0,1]$ such that $h(\xi_1)=h(\xi_2)$; the same applies to any $p_n\circ h$, hence some continuous function such that $f(\xi_1)\neq f(\xi_2)$ cannot be uniformly approximated by a $\{(p_n\circ h)(x)\}_{n\geq 1}$ sequence.
