 Solution

For real continuous $f$, the Chebyshev alternation theorem says that $p\in {\mathcal{P}}_n$ is the unique best uniform approximation to $f$ precisely when there are $n+2$ ordered points $0\le x_0<\cdots <x_{n+1}\le 1$ and a sign $\sigma \in \{-1,1\}$ such thatHere ${\mathcal{P}}_n$ means degree at most $n$. Best approximants exist: a minimizing sequence is bounded in the finite-dimensional polynomial space, and finite-dimensional equivalence of norms gives a convergent coefficient subsequence. When $f=p$, the zero error is treated separately; the derivative-sign hypothesis below excludes that case.

Let $p_n$ be a best degree-at-most-$n$ approximant and suppose for contradiction that $E_n(f)=E_{n+1}(f)$. Since $f^{(n+1)}>0$, $f$ is not a degree-at-most-$n$ polynomial and its error is nonzero. The same $p_n$ would be best in ${\mathcal{P}}_{n+1}$, so the Chebyshev alternation theorem at that degree supplies $n+3$ alternating error extrema. The intermediate value theorem gives at least $n+2$ distinct zeros of $e=f-p_n$, one between each consecutive pair of extrema.

Repeated Rolle's theorem then forces a zero of $e^{(n+1)}$. But $p_n^{(n+1)}=0$, whereas $e^{(n+1)}=f^{(n+1)}>0$ everywhere. This contradiction provesThe argument proves strict decrease of best polynomial approximation under a derivative sign; an everywhere negative $(n+1)$st derivative gives the same strict conclusion after replacing $f$ by $-f$.