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I'm in the process of reading a paper and I believe there's a mistake, but it could also be me not noticing something.

Here's the deal:

Let $\bar\Pi_k$ be the set of all polynomials of degree at most $k$ that satisfy $p(1) = 1$.

Next, let $\mu_i, i = 1...n$ be a finite set of numbers in the open interval $(-1,1)$.

In one of the proofs in the paper, they appear to make the following assumption: If $p \in \bar\Pi$, then $$ \max_i\, p^2(\mu_i) \le p^2(\max_i\, | \mu_i |)$$

However, I find absolutely no reason for this to be true. Note: The coefficients of the polynomials are not guaranteed to be non-negative, otherwise like Jensen's inequality might have done the trick.

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Your title says that the polynomials are monotone but the body doesn't. – Qiaochu Yuan Jan 16 '11 at 18:05
Well, if the polynomials 'were' monotone, then the statement would be true, wouldn't it? So that's what I was wondering. – Lagerbaer Jan 16 '11 at 19:15
up vote 3 down vote accepted

Counterexample: $p(X)=2-X^2$ then $p(0)\geq p(x)$ for all $x\in(-1,1)$. The same property is true for $p^2$. One can check that $p(1)=1$. Take $\mu_0=0$ and $\mu_1$ whatever value different from zero to get a couterexample.

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Do you mean $\mu_1$ and $\mu_2$? – PEV Jan 16 '11 at 17:38
@trevor. $0$ is the maximum of $p^2$, any finite set with $0$ (i.e. $\{0,\mu_1,\mu_2,\ldots\}$) is a couterexample – Nabyl Bod Jan 16 '11 at 17:41
I really should start looking for examples first, before trying to prove something :D – Lagerbaer Jan 16 '11 at 17:59
@Lagerbaer. ;) Remark that the result is true if the coefficients are non-negative. In this case $p^2(x)$ has non-negative coefficients and a triangular inequality gives you the result. – Nabyl Bod Jan 16 '11 at 18:05

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