[Math] For every $f\in C[0,1]$ there is a sequence of even polynomials which converges uniformly on $[0,1]$ to f

Question

For every $f\in C[0,1]$ there is a sequence of even polynomials which converges uniformly on $[0,1]$ to f ?

What I have tried:

f is continuous on $D:=[0,1]$, let $(x_k)_{k\in \mathbb{N}} \in D$ converge to $y \in D$, then it must hold that (sequence definition of continuity): $$\lim _{k \rightarrow \infty} x_k=y\Rightarrow \lim_{n \rightarrow \infty} f(x_k)=f(y) $$

a sequence of even polynomials: $a_k= \sum_{k=0}^{x_k}a_kx^{2k} $

don't see anything more…

how does one show this ? Doesn't one need to know that every f is analytic to show this ?

edit: the theorem by stone-weierstrass was proven already at this point…

Answer 1

Let $g(x)=f(\sqrt x)$, find a sequence of polynomials with $p_n\to g$. Then $p_n(x^2)\to g(x^2)=f(x)$.