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Question. If $f(z)\in\mathbb{C}[z]$ is a monic polynomial of degree $n$, is it true that $$\max\{\,\vert f(x)\vert: \, -1\leq x\leq 1\}\geq 2^{1-n} \,\, ?$$

Context. This came up while working on some convexity problems.

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Yes. Proved by Chebyshev. The extreme case ($\max = 2^{1-n}$) is given by $2^{n-1}\cdot f(x)$ being a Chebyshev polynomial of first kind. https://en.wikipedia.org/wiki/Chebyshev_polynomials

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  • $\begingroup$ Okay. Any reference? $\endgroup$ – T. Amdeberhan Feb 15 '17 at 12:57
  • $\begingroup$ Check the lead section of Wikipedia. $\endgroup$ – Max Alekseyev Feb 15 '17 at 12:58
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    $\begingroup$ Cool. Maybe one adds a line of argument to complex polynomials. But, that is okay. $\endgroup$ – T. Amdeberhan Feb 15 '17 at 13:12
  • $\begingroup$ Interpolate your polynomial in the extrema points of $T_n$ and look at the leading coefficient. $\endgroup$ – Fedor Petrov Feb 15 '17 at 17:58
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You are essentially asking to approximate $x^n$ on $[-1,1]$ by a polynomial of degree less than $n$. The Remez algorithm will find the best possible aproximation. In Maple you can enter

numapprox[minimax](x^n,x=-1..1,[n-1,0]);

If you take $n=10$ or so and work to 100 digits you will quickly convince yourself that the maximum error is precisely $2^{1-n}$. It also looks as though the optimal approximation $f(x)$ has $2^{1-n}f(x)\in\mathbb{Z}[x]$. One could probably find a formula for the coefficients experimentally with the help of OEIS and then work back to a proper proof. But sadly I need to go and do less interesting things now.

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