Bernstein's theorem (polynomials)

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

$\underset{|z|=1}{\max }|P^{\prime }(z)|\le n\underset{|z|=1}{\max }|P(z)|$.

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

$$\begin{gathered} P(z)=\alpha z^n, \\ |\alpha |=\underset{|z|=1}{\max }|P(z)| \end{gathered}$$.

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 $|z|\ge 1$. We show first that if $|P(z)|<|Q(z)|$ on $|z|=1$, then $|P^{\prime }(z)|<|Q^{\prime }(z)|$ on $|z|\ge 1$. By Rouché's theorem, $$\begin{gathered} P(z)+\epsilon \\ Q(z) \end{gathered}$$ with $|\epsilon |\ge 1$ has all its zeros in $|z|<1$. By virtue of the Gauss–Lucas theorem, $$\begin{gathered} P^{\prime }(z)+\epsilon \\ {Q}^{\prime }(z) \end{gathered}$$ has all its zeros in $|z|<1$ as well. It follows that $|P^{\prime }(z)|<|Q^{\prime }(z)|$ on $|z|\ge 1$, otherwise we could choose an $\epsilon$ with $|\epsilon |\ge 1$ such that $P^{\prime }(z)+\epsilon Q^{\prime }(z)$ has a zero in $|z|\ge 1$. For an arbitrary polynomial $P(z)$ of degree $n$, we obtain Bernstein's Theorem by applying the above result to the polynomials $Q(z)=C{z}^n$, where $C$ is an arbitrary constant exceeding $\underset{|z|=1}{\max }|P(z)|$.

Bernstein's inequality

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,

$\underset{|z|\le 1}{\max }(|P^{(k)}(z)|)\le \frac{n!}{(n-k)!}\cdot \underset{|z|\le 1}{\max }(|P(z)|).$

Similar results

Paul Erdős conjectured that if $P(z)$ has no zeros in $|z|<1$, then $\underset{|z|=1}{\max }|P^{\prime }(z)|\le \frac{n}{2}\underset{|z|=1}{\max }|P(z)|$. This was proved by Peter Lax. M. A. Malik showed that if $P(z)$ has no zeros in $|z|<k$ for a given $k\ge 1$, then $\underset{|z|=1}{\max }|P^{\prime }(z)|\le \frac{n}{1+k}\underset{|z|=1}{\max }|P(z)|$.

See also

• Markov brothers' inequality
• Remez inequality

Further reading

• Frappier, Clément (2004). "Note on Bernstein's inequality for the third derivative of a polynomial" (PDF). J. Inequal. Pure Appl. Math. 5 (1). Paper No. 7. ISSN 1443-5756. Zbl 1060.30003.
• Natanson, I.P. (1964). Constructive function theory. Volume I: Uniform approximation. Translated by Alexis N. Obolensky. New York: Frederick Ungar. MR 0196340. Zbl 0133.31101.
• 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.