Remez inequality

In mathematics, the Remez inequality, discovered by the Soviet mathematician Evgeny Yakovlevich Remez, gives a bound on the sup norms of certain polynomials, the bound being attained by the Chebyshev polynomials.

The inequality
Let σ be an arbitrary fixed positive number. Define the class of polynomials πn(σ) to be those polynomials p of the nth degree for which


 * $$|p(x)| \le 1$$

on some set of measure ≥ 2 contained in the closed interval [−1, 1+σ]. Then the Remez inequality states that


 * $$\sup_{p \in \pi_n(\sigma)} \left\|p\right\|_\infty = \left\|T_n\right\|_\infty$$

where Tn(x) is the Chebyshev polynomial of degree n, and the supremum norm is taken over the interval [−1, 1+σ].

Observe that Tn is increasing on $$[1, +\infty]$$, hence
 * $$ \|T_n\|_\infty = T_n(1+\sigma). $$

The R.i., combined with an estimate on Chebyshev polynomials, implies the following corollary: If J ⊂ R is a finite interval, and E ⊂ J is an arbitrary measurable set, then

for any polynomial p of degree n.

Extensions: Nazarov–Turán lemma
Inequalities similar to ($$) have been proved for different classes of functions, and are known as Remez-type inequalities. One important example is Nazarov's inequality for exponential sums :


 * Nazarov's inequality. Let
 * $$ p(x) = \sum_{k = 1}^n a_k e^{\lambda_k x} $$
 * be an exponential sum (with arbitrary λk ∈C), and let J ⊂ R be a finite interval, E ⊂ J—an arbitrary measurable set. Then
 * $$ \max_{x \in J} |p(x)| \leq e^{\max_k |\Re \lambda_k| \, \mathrm{mes} J} \left( \frac{C \,\, \textrm{mes} J}{\textrm{mes} E} \right)^{n-1} \sup_{x \in E} |p(x)|~, $$
 * where C > 0 is a numerical constant.

In the special case when λk are pure imaginary and integer, and the subset E is itself an interval, the inequality was proved by Pál Turán and is known as Turán's lemma.

This inequality also extends to $$L^p(\mathbb{T}),\ 0\leq p\leq2$$ in the following way


 * $$ \|p\|_{L^p(\mathbb{T})} \leq e^{A(n - 1) \textrm{mes }(\mathbb{T}\setminus E)}\|p\|_{L^p(E)} $$

for some A>0 independent of p, E, and n. When


 * $$\mathrm{mes} E <1-\frac{\log n}{n}$$

a similar inequality holds for p > 2. For p=∞ there is an extension to multidimensional polynomials.

Proof: Applying Nazarov's lemma to $$E=E_\lambda=\{x\,:\ |p(x)|\leq\lambda\},\ \lambda>0$$ leads to


 * $$\max_{x \in J} |p(x)| \leq e^{\max_k |\Re \lambda_k| \, \mathrm{mes} J} \left( \frac{C \,\, \textrm{mes} J}{\textrm{mes} E_\lambda} \right)^{n-1} \sup_{x \in E_\lambda} |p(x)| \leq e^{\max_k |\Re \lambda_k| \, \mathrm{mes} J} \left( \frac{C \,\, \textrm{mes} J}{\textrm{mes} E_\lambda} \right)^{n-1} \lambda $$

thus


 * $$\textrm{mes} E_\lambda\leq C \,\, \textrm{mes} J\left(\frac{\lambda e^{\max_k |\Re \lambda_k| \, \mathrm{mes} J}}{\max_{x \in J} |p(x)|} \right )^{\frac{1}{n-1}}$$

Now fix a set $$E$$ and choose $$\lambda$$ such that $$\textrm{mes} E_\lambda\leq\tfrac{1}{2}\textrm{mes} E$$, that is


 * $$\lambda =\left(\frac{\textrm{mes} E}{2C \mathrm{mes} J}\right)^{n-1}e^{-\max_k |\Re \lambda_k| \, \mathrm{mes} J}\max_{x \in J} |p(x)| $$

Note that this implies:
 * 1) $$ \textrm{mes}E\setminus E_{\lambda}\ge \tfrac{1}{2} \textrm{mes}E .$$
 * 2) $$ \forall x \in E \setminus E_{\lambda} : |p(x)| > \lambda .$$

Now


 * $$\begin{align}

\int_{x\in E}|p(x)|^p\,\mbox{d}x &\geq \int_{x\in E \setminus E_\lambda}|p(x)|^p\,\mbox{d}x \\[6pt] &\geq \lambda^p\frac{1}{2}\textrm{mes} E \\[6pt] &= \frac{1}{2}\textrm{mes} E \left(\frac{\textrm{mes} E}{2C \mathrm{mes} J}\right)^{p(n-1)}e^{-p\max_k |\Re \lambda_k| \, \mathrm{mes} J}\max_{x \in J} |p(x)|^p \\[6pt] &\geq \frac{1}{2} \frac{\textrm{mes} E}{\textrm{mes} J}\left(\frac{\textrm{mes} E}{2C \mathrm{mes} J}\right)^{p(n-1)}e^{-p\max_k |\Re \lambda_k| \, \mathrm{mes} J}\int_{x \in J} |p(x)|^p\,\mbox{d}x, \end{align}$$

which completes the proof.

Pólya inequality
One of the corollaries of the R.i. is the Pólya inequality, which was proved by George Pólya, and states that the Lebesgue measure of a sub-level set of a polynomial p of degree n is bounded in terms of the leading coefficient LC(p) as follows:


 * $$ \textrm{mes} \left\{ x \in \R : \left|P(x)\right| \leq a \right\} \leq 4 \left(\frac{a}{2 \mathrm{LC}(p)}\right)^{1/n}, \quad a > 0~.$$