How large (small) can be the measure of a set where a polynomial takes small values ? - MathOverflow

How large (small) can be the measure of a set where a polynomial takes small values ?

2010-07-21T16:57:53Z

A $n$-th degree polynomial has precisely $n$ roots. So it is natural to ask the question how large ( and small) can be the measure of a set where a polynomial takes small values ?

This, and other interesting variation of this must have been studied in depth.

I would really appreciate any reference to the relevant literature.

Also, if there are some interesting variation of this problem I would like to know.

Thank you.

Answer by Helge for How large (small) can be the measure of a set where a polynomial takes small values ?

2010-07-21T17:10:31Z

There is first Polya's estimate that if $f$ is a monic polynomial, then $$ |\{x\in \mathbb{R}:\quad |f(x)|\leq 2\}| \leq 4. $$ A proof can be found in the book "Proofs from the book". One can obtain inequalities for non-monic polynomials by rescaling.

Second there is Cartan's lemma or estimate. It can for example be found in Levin's book on entire functions. The estimate even holds for analytic functions.

The basic statement is:

Let $f: G \to \mathbb{C}$ be analytic and assume that $f$ is bounded by $1$ on a disc of radius $2$. Then there are constant $C, c > 0$ such that $$ |\{z\in \mathbb{C}:\quad |z| < 1, | f(z)| \leq e^{-s}\}| \leq C \exp\left( - \frac{c}{\log(\varepsilon^{-1})} s \right) $$ where $\varepsilon = |f(0)|$. In fact, this is sharper, since it provides some information on how the set looks. For a polynomial it's just the union of its degree many disks. (For analytic functions countably many).

Answer by Matheus for How large (small) can be the measure of a set where a polynomial takes small values ?

2010-07-21T17:19:04Z

Dear Vagabond,

I think a useful reference containing a lemma in the direction of your question is:

http://www.ams.org/mathscinet-getitem?mr=1652916 (see in particular proposition 3.2)

In this paper, Kleinbock and Margulis show the following result:

$\lambda({x\in I: |f(x)|<\varepsilon})\leq 2k (k+1)^{1/k} (\varepsilon/\|f\|_I)^{1/k} \lambda(I)$

Here $\lambda$ is the Lebesgue measure, $f$ is a polynomial of degree $k$ and $\|f\|_I$ is the supremum of $f$ on $I$.

Best,

Matheus