TITLE: Applying the Stone-Weierstrass Theorem to approximate even functions
QUESTION [5 upvotes]: Let $f:[-1,1] \rightarrow \mathbb{R}$ be any even continuous function on $[-1,1]$ (i.e. $f(-x)=f(x)$ $\forall x \in [-1,1]$). Let $\epsilon>0$. Prove that there exists an even polynomial $p$ such that $$|f(x)-p(x)|< \epsilon$$ $$\forall x \in [-1,1]$$

Here, "even polynomial" means that $p(-x)=p(x)$, not simply that it has even degree.
I think I should use the Stone-Weierstrass theorem to show that the subalgebra of even polynomials, call it $\mathcal{A}$, over this interval is dense, from which the result follows immediately.
For this to work I require that $\mathcal{A}$ contains the constants (obviously true) and separates points...which is not true, unfortunately. Anyone have any hints? I would prefer hints only, rather than solutions.
Oh yes, and I should mention that the version of the Stone-Weierstrass theorem that I can use says that if a subalgebra of $C(\mathbb{R})$ contains the constants and separates points, then it is dense in $C(\mathbb{R})$.

REPLY [5 votes]: Just for reference, here is another trick: Since $x\mapsto f(\sqrt{x})$ is continuous on $[0, 1]$, for $\epsilon > 0$ there exists a polynomial $q(x)$ such that
$$ \sup_{x\in[0,1]} |f(\sqrt{x}) - q(x)| < \epsilon. \tag{*}$$
Now let $p(x) = q(x^2)$ and notice that $p$ is even and
$$ \sup_{x\in[-1,1]}|f(x) - p(x)| = \sup_{x\in[0,1]} |f(\sqrt{x}) - q(x)| \stackrel{\text{by (*)}}{<} \epsilon $$