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 ∈ (0, 2), and

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

then

$\lVert p(x)\rVert _{[-1,1]}\leq T_{n}\left({\frac {2+s}{2-s}}\right),$

where T_n(x) = cos(n arccos x) is the Chebyshev polynomial of degree n."

In other words, the supremum of |p(x)| on the interval [−1, 1] is bounded by a value of T_n(y), y on the interval (1, ∞), the exact point y in that interval depending on the measure of a set within which |p(x)| ≤ 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)