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Given a polynomial with real coefficients, that satisfies $\forall x \in \left[ -1,1 \right]: \ |p(x)|\leqslant 1$, I have to show, that its leading coefficient, $a_m$ satisfies $a_m \leqslant 2^{m-1}$.

I have obtained some easy partial results: for example under the given conditions the sum of all even coefficients has to be in $\left[ -1,1 \right]$ (and the sum of the odd in $\left[ -2,2 \right]$), but that didn't help me much...

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up vote 2 down vote accepted

This is directly related to Chebyshev polynomials.

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But isn't there also a more "low-level" proof ? The argument using the Chebyshev polynomials isn't really giving an explanation, why this holds - I mean, it is giving an explanation, but just using some other "high-level" argument. It would interest me to also find a proof that uses just elementary methods. – resu May 18 '11 at 8:33

This is a property of Chebyshev polynomials. See the section "Minimal $\infty$-norm" there, where there's also a proof.

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