Bernstein's theorem (polynomials)

Bernstein's theorem is an inequality relating the maximum modulus of a complex polynomial function on the unit disk with the maximum modulus of its derivative on the unit disk. It was proven by Sergei Bernstein while he was working on approximation theory.

Statement

Let $\max _{|z|=1}|f(z)|$ denote the maximum modulus of an arbitrary function f(z) on |z| = 1, and let f′(z) denote its derivative. Then for every polynomial P(z) of degree n we have

$\max _{|z|=1}|P'(z)|\leq n\max _{|z|=1}|P(z)|.$

The inequality is best possible with equality holding if and only if

$P(z)=\alpha z^{n},\ |\alpha |=\max _{|z|=1}|P(z)|.$

Proof

Let P(z) be a polynomial of degree $n$. and let Q(z) be another polynomial of the same degree with no zeros in $|z|\geq 1$. We show that if $|P(z)|<|Q(z)|\ on\ |z|=1$, then $|P'(z)|\leq |Q'(z)|\ on\ |z|\geq 1$.

From Rouché's theorem we have $P(z)+\varepsilon \ Q(z)$, with $|\varepsilon |>1$ has all it's zeros in |z| < 1. By virtue of Gauss–Lucas theorem, $P'(z)+\varepsilon \ Q'(z)$ has all its zeros in $|z|\leq 1$. It follow's that $|P'(z)|\leq |Q'(z)|$ on $|z|>1$, otherwise it would have been possible to choose $\varepsilon$ such that $P(z)+\varepsilon \ Q(z)$ had a zero in |z| > 1, with $|\varepsilon |>1$. Since P′(z) and Q′(z) are both continuous, we have $|P'(z)|\leq |Q'(z)|{\text{ on }}|z|\geq 1$.

For an arbitrary polynomial P(z) of degree n, choosing $Q(z)=(\max _{|z|=1}|P(z)|)z^{n}$ proves Bernstein's Theorem.

Similar results

Paul Erdős had conjectured that if P(z) has no zeros in |z| < 1, then $\max _{|z|=1}|P'(z)|\leq {\frac {n}{2}}\max _{|z|=1}|P(z)|$, and was proven by Peter Lax.

Prof M.A. Malik showed that if P(z) has no zeros in $|z|\geq k,\ k\geq 1$, then $\max _{|z|=1}|P'(z)|\leq {\frac {n}{1+k^{n}}}\max _{|z|=1}|P(z)|$