= Solution
{c}
Choose the nodes as the $n+1$ roots of the <Chebyshev polynomial> $T_{n+1}$. Its leading coefficient is $2^n$, so the monic nodal product is $\omega=2^{-n}T_{n+1}$. Since $|T_{n+1}|\leq1$ on $[-1,1]$, part (b) bounds this particular interpolant by
$$
\|u-p_n\|_\infty\leq
\frac{\|u^{(n+1)}\|_\infty}{2^n(n+1)!}.
$$
The best error cannot exceed the error of a particular member of $P_n$. Therefore
$$
\boxed{E_n\leq
\frac{\|u^{(n+1)}\|_\infty}{2^n(n+1)!}.}
$$
The choice of Chebyshev roots controls the maximum of the nodal product, not the size of any individual <interpolation> coefficient.
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