Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 62 / 3

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

Assume $n\ge 1$ and $\Vert p{\Vert }_{[-1,1]}\le 1$. The given Bernstein inequality for algebraic polynomials shows that $q=p^{\prime }$ satisfies the nodal hypotheses of (a). Consequently, in the two endpoint regions $\cos (\pi /(2n))\le |x|\le 1$, $|p^{\prime }(x)|\le |T_n^{\prime }(x)|$.

For $0<\theta <\pi$, the finite geometric identitybounds its absolute value by $n$. It follows that $|T_n^{\prime }(\cos \theta )|\le n^2$, and the endpoint limit gives $T_n^{\prime }(1)=n^2$. In the central interval, the Bernstein inequality for algebraic polynomials givessince $\sin t\ge 2t/\pi$ on $[0,\pi /2]$. Combining the regions and rescaling proves the Markov inequality for polynomial derivatives:The Chebyshev polynomial itself attains equality, so this constant is sharp. Constant polynomials have zero derivative.