= Chebyshev nodal derivative comparison
{c}
{title2=$|q(x)|\le|T_n'(x)|\quad\text{for }|x|\ge\cos(\pi/(2n))$}
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 $|q(x_j)|\le |T_n'(x_j)|$. Its <Lagrange interpolation polynomial> uses basis $\ell_j(x)=T_n(x)/[(x-x_j)T_n'(x_j)]$. For $x>x_1$, $T_n(x)>0$ and $x-x_j>0$, so $\ell_j(x)T_n'(x_j)>0$. Therefore $|q(x)|\le\sum_j|\ell_j(x)T_n'(x_j)|=T_n'(x)$, the last equality being interpolation of the derivative. Continuity includes $x=x_1$ and reflection proves the left-hand region.
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