Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 75 / 3 / a / Solution

A nontrivial example is $f_n=T_{n+1}$, the Chebyshev polynomial characterized by $T_{n+1}(\cos \theta )=\cos ((n+1)\theta )$. Its supremum norm is one. At the $n+2$ points $\cos (j\pi /(n+1))$, $0\le j\le n+1$, its values alternate between $1$ and $-1$; reversing their order gives increasing nodes with the same alternation.

The zero polynomial therefore satisfies the Chebyshev alternation theorem for degree at most $n$, so $E_n(f_n)=1$. Since the smaller polynomial space is contained in the larger one, $E_{n-1}(f_n)\ge 1$; the zero polynomial is feasible in the smaller space as well, giving the reverse inequality. HenceA constant function would also give equality with both errors zero, but this example shows that equality can occur at a positive approximation error.