 Chebyshev nodal derivative comparison

Let $x_j=\cos ((2j-1)\pi /(2n))$ be the roots of the Chebyshev polynomial, and let $q$ have degree at most $n-1$ with $|q(x_j)|\le |T_n^{\prime }(x_j)|$. Its Lagrange interpolation polynomial uses basis ${\ell }_j(x)=T_n(x)/[(x-x_j)T_n^{\prime }(x_j)]$. For $x>x_1$, $T_n(x)>0$ and $x-x_j>0$, so ${\ell }_j(x)T_n^{\prime }(x_j)>0$. Therefore $|q(x)|\le {\sum }_j|{\ell }_j(x)T_n^{\prime }(x_j)|=T_n^{\prime }(x)$, the last equality being interpolation of the derivative. Continuity includes $x=x_1$ and reflection proves the left-hand region.