= 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, $E_{n-1}(f_n)\ge1$; the zero <polynomial> is feasible in the smaller space as well, giving the reverse inequality. Hence
$$
\boxed{f_n=T_{n+1},\qquad E_{n-1}(f_n)=E_n(f_n)=1.}
$$
A constant <function> would also give equality with both errors zero, but this example shows that equality can occur at a positive approximation error.
Back to article page
