 Chebyshev interpolation represents external evaluation

On real polynomials of degree at most $n\ge 1$ 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 $|P(u)|\le \Vert P{\Vert }_{\infty }{T}_n(u)$, 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.