 Solution

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.