 Solution

The trigonometric Chebyshev alternation theorem says that $p_n^{\ast }\in {\mathcal{T}}_n$ is best exactly when its error has at least $2n+2$ cyclically ordered extrema of equal magnitude and alternating sign. Let $5^m\le n<5^{m+1}$ and set $x_j=j\pi /5^{m+1}$. For every $k\ge m+1$,ThusThere are $2\cdot 5^{m+1}\ge 2n+2$ such extrema, while the triangle inequality bounds the tail by their common magnitude. Hence