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Suppose $f:[-1, 1]\rightarrow\mathbb{R}$ is a polynomial. I am curious what the minimal degree of $f$ can be such that for $0<a<b<1$, $f$ satisfies the following two properties:

1) $\forall x\in[-a, a], ~~~f(x)\in[1-\varepsilon, 1+\varepsilon]$

2) $\forall x\in[-1, -b]\cup[b,1]$, $~~~|f(x)|\leq\varepsilon$

How does the minimal degree depend on $a, b$ and $\varepsilon$?

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    $\begingroup$ By using symmetry, it's not hard to show that this is equivalent to finding (double) the minimal degree of $g: [0, 1] \rightarrow \mathbb{R}$ such that $\forall x \in [0, \sqrt{a}], |g(x) - 1| < \epsilon, \forall x \in [\sqrt{b}, 1], |g(x)| < \epsilon$. $\endgroup$
    – user44191
    Commented Oct 11, 2019 at 1:26

1 Answer 1

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By the remark of user44191, this is equivalent to the question of the best uniform approximation of piecewise constant function on two intervals by polynomials of fixed degree. (Minimize $\epsilon$ for fixed degree instead of minimizing the degree for fixed $\epsilon$). This last problem has been solved in the paper:

Polynomials of the best uniform approximation to sgn(x) on two intervals, J. d'Analyse math., 114 (2011) 285-315, arXiv:1008.3765

It gives an asymptotics of minimal $\epsilon$ as a function of degree $d$ when $d\to\infty$, and also a description of extremal polynomials.

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  • $\begingroup$ Thanks a lot! This is exactly what I was looking for. BTW, is there an analog in multi-dimension (indicator function of $l_\infty$ ball in \mathbb{R}^d? It seems the optimal way is to take a product of $d$ such approximations....but I'm not sure how to formalize it. $\endgroup$
    – ignescent
    Commented Oct 17, 2019 at 0:10
  • $\begingroup$ I am not aware on any multi-dimensional result. $\endgroup$
    – Alexandre Eremenko
    Commented Oct 17, 2019 at 2:06

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