Talk:Remez inequality

Is this the right inequality?
I've done a bit of digging around to try to nail down the exact form in which this inequality was originally given, without much success. I don't have access to the JSTOR articles, for example.

I did find this useful web page, constructed by professor Tamás Erdélyi, who has written several papers about Remez-type inequalities. In The Remez Inequality for Linear Combinations of Shifted Gaussians (page 3) Erdélyi says


 * "The classical Remez inequality states that if p is a polynomial of degree at most n, s &isin; (0, 2), and



m(\lbrace x \in [-1, 1] : |p(x)| \le 1 \rbrace) \ge 2-s, $$


 * then



\lVert p(x) \rVert_{[-1,1]} \le T_n \left( \frac{2+s}{2-s} \right), $$


 * where Tn(x) = cos(n&thinsp;arccos&thinsp;x) is the Chebyshev polynomial of degree n."

In other words, the supremum of |p(x)| on the interval [&minus;1, 1] is bounded by a value of Tn(y), y on the interval (1, &infin;), the exact point y in that interval depending on the measure of a set within which |p(x)| &le; 1.

I can't square this up with the way the inequality is stated in this article, so I'm confused. Can anybody clear this up for me? DavidCBryant 22:29, 20 July 2007 (UTC)


 * Hi,
 * both versions are equivalent. Indeed, let $$ A_\sigma $$ be a linear function that maps $$[-1, 1]$$ onto $$[-1, 1 + \sigma]$$. If $$P$$ is a polynomial on $$[-1, 1+\sigma]$$ such that
 * $$\mathrm{mes} \left\{ x \in [-1, 1+\sigma] \, \big| \, |P| \leq 1 \right\} \geq 2 \quad (1)$$,
 * then $$Q = P \circ A_\sigma$$ satisfies
 * $$\mathrm{mes} \left\{ x \in [-1, 1] \, \big| \, |P| \leq 1 \right\} \geq \frac{4}{2+\sigma} \quad (2)$$,
 * and vice versa. Of course, $$\|P\|_\infty = \|Q\|_\infty$$ (with some abuse of notation, since the first norm is on $$[-1, 1+\sigma]$$, whereas the second one is on $$[-1, 1]$$.)
 * The inequality in the Wiki-article states that
 * $$ \|P\|_\infty \leq \|T_n\|_\infty = T_n(1+\sigma) \quad (3) $$,
 * whereas Erdélyi's version is
 * $$ \|Q\|_\infty \leq T_n \left( \frac{2+s}{2-s} \right) \quad (4)$$.
 * To make the (equivalent) assumptions (1), (2) identical to those of Erdélyi, you should take
 * $$s = \frac{2\sigma}{2+\sigma} $$,
 * and then the conclusions (3), (4) are also identical.
 * Best, Sasha  —Preceding unsigned comment added by Sodin (talk • contribs) 16:38, 19 September 2008 (UTC)