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For any polynomial $P_{n}(x)$ with degree $n$ that all its zeros lie in$[-1,1]$, then $\|P_{n}'(x)\|>C \sqrt{n} \|P_{n}(x)\|$ where $C$ is an absolute constant.

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    $\begingroup$ The standard reference is Borwein-Erdelyi. I understand you may not have access to it (neither do I at the moment), but the reference seems worth mentioning. $\endgroup$ – user53153 Jan 7 '13 at 7:28
  • $\begingroup$ You are right, that is mentioned on Page434 in that book, and provide some steps to prove this statement. I'm trying now. $\endgroup$ – Larry Eppes Jan 7 '13 at 9:25
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    $\begingroup$ I think you should consult the original article by Turan. It is very nice - goo.gl/4J92t [pdf] $\endgroup$ – ivan Jan 10 '13 at 7:15
  • $\begingroup$ good, thank you, ivan. I've scanned Turan's article, and I should say the reference book ,5PM had mentioned, shares the same idea as Turan's article. and I had gave a prove by 5PM's reference $\endgroup$ – Larry Eppes Jan 19 '13 at 14:04
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What you have seems to be incorrect. For instance, consider $$P_2(x) = (x-a)(x-b)$$ Then $P_2'(x) = 2x - (a+b)$. Consider $x^* = \dfrac{a+b}2$. Then we have $$P_2'(x^*) = 0$$ whereas $$P_2(x^*) = \left(\dfrac{b-a}2\right)\left(\dfrac{a-b}2\right) = - \dfrac{(a-b)^2}4$$ which clearly violates the inequality.

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  • $\begingroup$ I think the inequality involves supremum norms of P and of its derivative. $\endgroup$ – user53153 Jan 7 '13 at 6:42
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    $\begingroup$ Notation: $\|P_{n}(x)\|=max_{x\in [-1,1]}|P_{n}(x)|$, and$\|P_{n}'(x)\|=max_{x\in [-1,1]}|P_{n}'(x)|$. so the $x^{*}$ for $P_{n}'(x)$ and $P_{n}(x)$ may not the same one $\endgroup$ – Larry Eppes Jan 7 '13 at 7:23
  • $\begingroup$ @Larry In Marvis's defense, this is a really bad notation: to write $P_n(x)$ when you mean $P_n$ is to invite misinterpretation. $\endgroup$ – user53153 Jan 7 '13 at 7:30

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