Bernstein's inequality for polynomials

Remarks

Proof idea: let $M = \max_{\abs{z} = 1} \abs{P(z)} > 0$ and let $\lambda \in \C$ with $\abs{\lambda} > 1$. On $\abs{z} = 1$ one has $\abs{\lambda M z^n} > \abs{P(z)}$, so by Rouché's theorem all $n$ roots of $P(z) - \lambda M z^n$ lie in $\abs{z} < 1$. By the Gauss–Lucas theorem, so do the roots of $P'(z) - \lambda n M z^{n-1}$. If $\abs{P'(z_0)} > n M$ for some $\abs{z_0} = 1$, then $\lambda = P'(z_0) / (n M z_0^{n-1})$ has $\abs{\lambda} > 1$ and makes $z_0$ such a root, a contradiction.

The inequality has an equivalent trigonometric form: a real or complex trigonometric polynomial $T$ of degree at most $n$ satisfies $\max \abs{T'} \le n \max \abs{T}$. With $x = \cos \theta$, the trigonometric form gives, for a real polynomial $P$ of degree at most $n$ and $-1 < x < 1$, the bound $\abs{P'(x)} \le \frac{n}{\sqrt{1 - x^2}} \max_{[-1, 1]} \abs{P}$.

Example: $P(z) = z^n$ gives $\abs{P'(z)} = n$ on the unit circle, so equality holds.

Going deeper

For a student

Bernstein's inequality is the basic tool for inverse theorems in approximation theory: good approximation by polynomials of low degree forces smoothness. Its form on $[-1, 1]$ is sharp in the interior, while near the endpoints the Markov brothers' inequality gives the correct bound $n^2$. The proof through the Gauss–Lucas theorem shows how geometric facts about roots give analytic estimates.