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

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

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

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

^[2]

Proof edit

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

By Rouché's theorem, ${\displaystyle P(z)+\varepsilon \ Q(z)}$  with ${\displaystyle |\varepsilon |\geq 1}$  has all its zeros in ${\displaystyle |z|<1}$ . By virtue of the Gauss–Lucas theorem, ${\displaystyle P'(z)+\varepsilon \ Q'(z)}$  has all its zeros in ${\displaystyle |z|<1}$  as well. It follows that ${\displaystyle |P'(z)|<|Q'(z)|}$  on ${\displaystyle |z|\geq 1}$ , otherwise we could choose an ${\displaystyle \varepsilon }$  with ${\displaystyle |\varepsilon |\geq 1}$  such that ${\displaystyle P'(z)+\varepsilon Q'(z)}$  has a zero in ${\displaystyle |z|\geq 1}$ .

For an arbitrary polynomial ${\displaystyle P(z)}$  of degree ${\displaystyle n}$ , we obtain Bernstein's Theorem by applying the above result to the polynomials ${\displaystyle Q(z)=Cz^{n}}$ , where ${\displaystyle C}$  is an arbitrary constant exceeding ${\displaystyle \max _{|z|=1}|P(z)|}$ .

In mathematical analysis, Bernstein's inequality states that on the complex plane, within the disk of radius 1, the degree of a polynomial times the maximum value of a polynomial is an upper bound for the similar maximum of its derivative. Taking the k-th derivative of the theorem,

${\displaystyle \max _{|z|\leq 1}(|P^{(k)}(z)|)\leq {\frac {n!}{(n-k)!}}\cdot \max _{|z|\leq 1}(|P(z)|).}$ 

Paul Erdős conjectured that if ${\displaystyle P(z)}$  has no zeros in ${\displaystyle |z|<1}$ , then ${\displaystyle \max _{|z|=1}|P'(z)|\leq {\frac {n}{2}}\max _{|z|=1}|P(z)|}$ . This was proved by Peter Lax.^[3]

M. A. Malik showed that if ${\displaystyle P(z)}$  has no zeros in ${\displaystyle |z|<k}$  for a given ${\displaystyle k\geq 1}$ , then ${\displaystyle \max _{|z|=1}|P'(z)|\leq {\frac {n}{1+k}}\max _{|z|=1}|P(z)|}$ .^[4]

• Markov brothers' inequality
• Remez inequality

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• Rahman, Q. I.; Schmeisser, G. (2002). Analytic theory of polynomials. London Mathematical Society Monographs. New Series. Vol. 26. Oxford: Oxford University Press. ISBN 0-19-853493-0. Zbl 1072.30006.