Bernstein inequality for algebraic polynomials ($|p^{\prime }(x)|\le n\Vert p{\Vert }_{[-1,1]}/\sqrt{1-x^2}$)

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.

If a real degree-at-most-$n$ polynomial has absolute value at most one at every extremum $\cos (j\pi /n)$ of the Chebyshev polynomial $T_n$, its $k$th derivative on $[-1,1]$ is bounded by $T_n^{(k)}(1)$, for $1\le k\le n$. This strengthens the Markov inequality for polynomial derivatives by replacing a bound on the whole interval with these finitely many samples. The node set is rigid, as quantified by uniqueness of Chebyshev derivative-norming nodes.

For $n\ge 1$, any distinct node set in $[-1,1]$ different from the $n+1$ Chebyshev polynomial extrema admits one degree-$n$ polynomial $q$, bounded by one at all those nodes, whose every nonconstant derivative violates the corresponding sharp endpoint bound. In decreasing order set $q(t_i)=(-1{)}^i$. The endpoint derivative signs of Lagrange cardinal polynomials give $q^{(k)}(1)={\sum }_i|L_i^{(k)}(1)|$. Interpolating $T_n$ on the same nodes showsEach term is nonnegative and at least one is positive, because the other node set includes a nonextremal point. Uniqueness is a claim about node sets: permuting the same nodes does not change their norming property.