 Solution

For real-valued $f\in C[-1,1]$, there is a unique best uniform approximation $p_{\ast }\in {\mathcal{P}}_n$, where ${\mathcal{P}}_n$ means degree at most $n$. If $E=\Vert f-p_{\ast }{\Vert }_{\infty }>0$, the Chebyshev alternation theorem characterizes it by the existence of $n+2$ ordered pointsand a sign $\epsilon \in \{-1,1\}$ such thatConversely, a polynomial in ${\mathcal{P}}_n$ whose error has this alternation is the best one. If the best error is zero, $f$ itself is the exact polynomial approximant. The real-valued qualification matters: this sign-alternation formulation is not the corresponding criterion for arbitrary complex-valued approximation.

The obstruction to a strict improvement is concrete. If $q$ had smaller error at every alternation point, $q-p_{\ast }$ would have alternating strict signs there. The intermediate value theorem would give at least $n+1$ distinct zeros, impossible for a nonzero polynomial of degree at most $n$.