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For a polynomial of degree at most $n$, this derivative estimate holds for $-1<x<1$. It controls the derivative away from the endpoints. Sampling it at Chebyshev polynomial roots supplies the hypotheses of the Chebyshev nodal derivative comparison, which controls the missing endpoint regions. The norm is taken on the entire interval, although the displayed pointwise bound concerns its interior.

Apply the Bernstein inequality for algebraic polynomials to the central interval $|x|\le \cos (\pi /(2n))$, using $\sin (\pi /(2n))\ge 1/n$. On the remaining regions apply the Chebyshev nodal derivative comparison to $p^{\prime }$. Since $T_n^{\prime }(\cos \theta )=n\sin (n\theta )/\sin \theta$, the identity expressing $\sin (n\theta )/\sin \theta$ as a sum of $n$ unit complex phases gives $|T_n^{\prime }|\le n^2$. The endpoint limit is $T_n^{\prime }(1)=n^2$, so the constant is sharp.

For $x=\cos \theta$ with $0<\theta <\pi$, differentiation of the Chebyshev polynomial givesBoth sides are polynomials in $x$, so the identity holds at the endpoints and, in fact, for every real $x$.

Order the roots of a polynomial here as $x_j=\cos {\theta }_j$, ${\theta }_j=(2j-1)\pi /(2n)$, so $x_1>\cdots >x_n$. At these simple roots,For a polynomial $q$ of degree at most $n-1$, the Lagrange interpolation polynomial isFor $x>x_1$, the Chebyshev polynomial is positive: all its roots are at most $x_1$ and its leading coefficient is positive. Hence ${\ell }_j(x)T_n^{\prime }(x_j)=T_n(x)/(x-x_j)>0$. The assumed nodal bounds and interpolation of the degree-$n-1$ polynomial $T_n^{\prime }$ now giveContinuity includes $x=x_1$. Apply this argument to $q(-x)$ and use the parity of the Chebyshev polynomial to obtain the negative side. Thus the Chebyshev nodal derivative comparison proves