How to prove $\langle x^{2n+1}: n\in \mathbb{N}\rangle$ is dense in $\{ f\in C([0,1]): f(0)=0\}$ I'm trying to prove that $\langle x^{2n+1}: n\in \mathbb{N}_0\rangle$ is dense in $\{ f\in C([0,1]): f(0)=0\}$ without the use of the Müntz–Szász theorem.
I know how to prove this for even exponents by using the Stone-Weierstrass theorem, however $\langle x^{2n+1}:n\in \mathbb{N}\rangle$ isn't an algebra so the same proof won't work. I then thought to show there is a sufficiently good approximation for an even exponent polynomial by odd exponent ones but that didn't seem to go anywhere.
I've been given a hint to prove it first for functions with the added condition of being differentiable at $0$ but I'm not sure how that makes the problem any easier. I'm quite stuck so any help would be really appreciated.
 A: *

*Consider the continuous odd functions in the interval $C[-1,1].$ By the Weierstrass theorem every such function $g(x)$ is the uniform limit of polynomials, say $p_n(x).$
Then the polynomials $$q_n(x)={1\over 2}[p_n(x)-p_n(-x)]$$ are uniformly convergent to $$g(x)={1\over 2}[g(x)-g(-x)]$$ The polynomials $q_n(x)$ are odd functions, hence only odd powers $x^{2k+1}$ are involved. The space $\{f\in C[0,1]\,: \,f(0)=0\}$ can be identified with odd continuous functions on $[-1,1],$ by the correspondence $g(x)=f(x) $ for $x\ge 0$ and $g(x)=-f(-x)$ for $x<0.$


*Another solution could be as follows. Consider $h(x)=g(\sqrt{x}),$ where $g\in C[0,1].$ Then by the Weierstrass theorem there exist polynomials $p_n(x)$ uniformly convergent to $h(x).$  Thus the polynomials $r_n(x)=xp_n(x^2)$ converge uniformly to $xg(x).$ Every function $f\in C[0,1],$ $f(0)=0,$  can be approximated uniformly by functions of the form $xg(x),$ where $g\in C[0,1].$
Indeed, let $f\in C[0,1]$ and $f(0)=0.$ Define
$$f_n(x)=\begin{cases} nxf(1/n) & 0\le x\le 1/n\\
f(x) & 1/n \le x\le 1\end{cases}$$
Then $f_n\rightrightarrows f.$ Moreover $g_n(x)=f_n(x)/x\in C[0,1]$ and $f_n(x)=xg_n(x).$ By the way, the functions $f_n$ are differentiable at $x=0.$
Remark The Weierstrass theorem is sufficient for proving the claim. Also the fact that the linear span of $x^{2n}$ is dense in $C[0,1]$ does not require the Stone-Weierstrass theorem. It suffices to approximate $f(\sqrt{x})$ by polynomials $p_n(x).$ Then the polynomials $p_n(x^2)$ approximate $f(x).$
