Best polynomial approximation, show that $E_{n+1}(f) < E_{n}(f)$.

Question

Given $f \in C[a, b]$ such that $|f^{(k)}(x)|>0$ for all $x$ and $k = 1, 2, \dots$ and $E_n = \inf_{w_{n} \in \Pi_{n}}\lVert f-w_n\rVert $ (supremum norm), I want to show, that for $n \geq 0$: $$E_{n+1}(f) < E_{n}(f)$$ Now, There is an obvious inequality $E_{n+1}(f) \leq E_{n}(f)$, because if $w \in \Pi_{n}$, then also $w \in \Pi_{n+1}$ so approximations of $f$ in $\Pi_{n+1}$ are at least as good as those in $\Pi_{n}$. This means, I need to show that $E_{n+1}(f) = E_{n}(f)$ is impossible. Now, this equality would mean that $w_n = w_{n+1}$, so showing that there exists $w_{n+1}$ such that $\deg(w_{n+1}) = n+1$ which, by Chebyshew alternation theorem would be equivalent to finding polynomial $w_{n+1} \in \Pi_{n+1}$such that $|f(x_i)-w_{n+1}(x_i)| = E_{n+1}$ for $n+3$ points, would be enough. I think I am heading in good direction, but I don't know how find such a polynomial, and make use of a fact that derivatives are positve. I hope someone can shed some light on correct approach.

Answer 1

So i figured out the solution: Assume $E_n$ = $E_{n+1}$. This means that $w_n = w_{n+1}$ and so $deg(w_{n+1}) \leq n$, since the best approximation in supremum norm is unique. Define $e_{n+1} = f-w_{n+1}$. By Chebyshev alternation theorem, there are $n+3$ points such that $|e(x_i)| = E_n$ and $e(x_i) = -e(x_{i+1})$ for $i = 1, 2, \dots, n+2$. In particular, this means that $e_n$ changes sign $n+2$ times and thus has $n+2$ roots. From this follows, that $e_n'$ has $n+1$ roots, $e_n^{(2)}$ has $n$ roots, ... , $e_n^{(n+1)}$ has $1$ root. $$e_n^{(n+1)} = f^{(n+1)}-w_{n+1}^{(n+1)}$$ But as estabilished earlier, $deg(w_{n+1}) \leq n$, so its' $n+1$st derivative vanishes, meaning $e_n^{(n+1)} = f^{(n+1)}$. This gives contradiction, since by assumpion $|f^{n+1}(x)| > 0$ for all $x$, so it cannot have any roots.