= Chebyshev interpolation represents external evaluation
{c}
{title2=$\|E_u\|=|T_n(u)|,\quad |u|>1$}
On real polynomials of degree at most $n\geq1$ with the <uniform norm> on $[-1,1]$, set $x_j=\cos(j\pi/n)$ and let $\ell_j$ be the <Lagrange interpolation> basis. If $u>1$, the denominator of $\ell_j(u)$ has exactly $j$ negative factors while its numerator is positive. Therefore $|\ell_j(u)|=(-1)^j\ell_j(u)$. Since $T_n(x_j)=(-1)^j$, interpolation gives $\sum_j|\ell_j(u)|=T_n(u)$. Hence $|P(u)|\leq\|P\|_\infty T_n(u)$, and $P=T_n$ attains equality. Reflection and Chebyshev parity give the case $u<-1$. Dividing the interpolation weights by $T_n(u)$ yields an exact signed evaluation representation of norm one on $n+1$ nodes. Constants give the immediate degree-zero case.
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