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

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.

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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).

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I just found in Levin's book you mentioned that he also gives a result for the lower bound. – Vagabond Jul 21 '10 at 18:21
I take it that the $C$ in $f:G\to C$ is the complex numbers, but what's the $C$ in "constant $C$, $c>0$"? It doesn't appear anywhere else in the answer. Also, what happens if $f(0)=0$? – Gerry Myerson Jul 22 '10 at 0:27
Fixed. If $f(0) = 0$ apply the result to $f(x + t)$ for some $t \neq 0$ such that $f(t) \neq 0$. You need some way to exclude that $f \equiv 0$ having $f(0)$ enter the estimate is the most convenient one. – Helge Jul 22 '10 at 8:37
@Gerry: You can replace $\varepsilon = |f(0)|$ by $\varepsilon = \sup_{|z| < \frac{1}{2}} |f(z)|$ without changing the result, except for the value of the constants. It is then clear that one does not have a problem for non-zero $f$, but the constants get worse. – Helge Jul 22 '10 at 11:42
@Helge, thanks. – Gerry Myerson Jul 22 '10 at 11:51

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

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And what is $I$? – Mariano Suárez-Alvarez Jul 21 '10 at 17:21
sorry... $I$ is any interval of the real line... – Matheus Jul 21 '10 at 17:22
Thankyou Matheus, for the nice reference. – Vagabond Jul 21 '10 at 18:42