Remez inequality

Remarks

The bound is sharp: $p = T_n$ and $E = [-1, 1]$ give equality, since $T_n$ is increasing on $[1, \infty)$. Proof idea: compare $p$ with a rescaled Chebyshev polynomial whose set where it is at most $1$ in absolute value has measure $2$. A count of sign changes, as in the minimal deviation property of $T_n$, shows that $p$ cannot be larger.

Example: for $n = 1$, a line with $\abs{p} \le 1$ on a set of measure $2$ in $[-1, 1 + \sigma]$ is at most $1 + \sigma = T_1(1 + \sigma)$ in absolute value on the whole interval.

By an affine change of variable, for an interval $I$ and a measurable $E \subseteq I$ of positive measure, $$\sup_{I} \abs{p} \le T_n\left(\frac{2 \abs{I}}{\abs{E}} - 1\right) \sup_{E} \abs{p} \le \left(\frac{4 \abs{I}}{\abs{E}}\right)^n \sup_{E} \abs{p}.$$ Without a lower bound on the measure of $E$ there is no bound: $p(x) = c x$ is small on a tiny interval around $0$ for every $c$.

Going deeper

For a student

The inequality is a quantitative form of the fact that a polynomial is determined by its values on any set of positive measure. It extends Chebyshev's extremal property, which concerns the growth of $T_n$ outside $[-1, 1]$, to arbitrary measurable sets. It is used to bound the size of sublevel sets of polynomials and has versions for several variables and for exponential sums.