Best approximation (infinity norm) of even functions

Question

Let $f\in C([-1,1])$ be an even function, i.e., $f(-x)=f(x)$ for all $x$. In fact, I've shown that its best approximation $p^{*}_{n}$ is even as well (Here the best approximation is in the sense of infinity norm, i.e., $\lVert f-p^{*}_{n}\rVert$). I want to show that for any $n\geq 0$, $p^{*}_{2n}=p^{*}_{2n+1}$ and if $f$ is not even, then for sufficiently large value of $n$, its best approximation $p^{*}_{n}$ is not even.

Answer 1

Note that the best approximating polynomial is unique.

Let $(Rf)(x) = f(-x)$ and let $p$ be the best approximation to $f$.

We have $\|f -p\| = \|Rf -Rp\| = \|f-Rp\|$. In particular, since the norm is convex, $p^*={1 \over 2} (p + Rp)$ satisfies $\|f-p^*\| \le \|f-p\|$ and so $p^* = p$.

Since $p^*$ is even, we obtain the desired result.

Suppose that $f$ is not even. Then there is some $x^*$ such that $f(x^*) \neq f(-x^*)$.Choose $\epsilon < {1 \over 4} |f(x^*)-f(-x^*)|$ and $p$ such that $\|f-p\| < \epsilon$.

Then $|p(x^*)-p(-x^*)| \ge |f(x^*)-f(-x^*)| - | p(-x^*)-f(-x^*)| -|p(x^*)-f(x^*)| > {1 \over 2} |f(x^*)-f(-x^*)|$ and so $p$ is not even.