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

Sketch:

Define $g: [-1,1]\to \mathbb R$ by $g(x) = \begin{cases} f(x) & \text{if } x \geq 0 \cr f(-x) & \text{if } x <0 \end{cases}$.

There is a sequence of even polynomial functions converging uniformly on $[-1,1]$ to $g$ (most proofs of Weierstrass's approximation theorem yield even (resp. odd) polynomials when one considers an even (resp. odd) function on $[-a,a]$).

Restrict to $[0,1]$.

Answer 2

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

Answer 3

Kahen basically has it, but here it is explicitly. There's a sequence of polynomials $\{p_n(x)\}$ converging to the even function $g(x),$ which is simply the natural extension of $f$ to $[-1, 1].$ But $\{p_n(-x)\}$ converges to $g(-x) = g(x).$ Add up $p_n(-x)$ and $p_n(x)$ and divide by $2.$