# How to construct a polynomial with minimum deviation from zero on the complex region?

I need to compute the analog of Chebyshev polynomials (which give the minimum deviation from zero on [-1,1]) on the given region $\Omega\subset \mathbb C$. More precisely: find $P_n$ such that $P_n(0)=1$ and $$\max_{z\in\Omega}|P_n(z)|\to \min.$$

Any help and references are strongly appreciated.

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The maximum modulus principle would help a great deal here... –  Guess who it is. Feb 8 '12 at 10:34
Hm... Clearly, we can replace maximum over domain $\Omega$ by maximum over boundary $\partial\Omega$. What next? –  faleichik Feb 8 '12 at 11:25

As J.M. noted, this is really a problem about a compact set ($\overline{\Omega}$ or $\partial \Omega$), which I will denote by $K$.

If $0\in K$, then of course $1$ is the best we can do, so this case is not interesting. The same happens when $K$ separates $0$ from $\infty$, due to the maximum principle. So the problem is interesting only when $0$ and $\infty$ are in the same connected component of $\overline{\mathbb C}\setminus K$.

It looks strange to me that you normalize the polynomials by $P(0)=1$ but call them "the analog of Chebyshev polynomials". The Chebyshev polynomials are normalized by their leading coefficient, not by their free term. The extremal problem $\max_K |P_n|\to \min$ (where $P_n$ is monic and $K\subset \mathbb C$) is a classical subject in complex function theory, and the key words here are Chebyshev constant, transfinite diameter, and Fekete points. This article gives a brief overview and some good references (of which I am partial to Goluzin's book).

If you insist on $P(0)=1$, then the best thing I can do is to introduce $Q_n(z) = z^nP_n(1/z)$, which is a monic polynomial, and minimize $\max_{\widetilde K}|Q_n|$ where $\widetilde K = \{1/z : z\in K\}$. This isn't a perfect reformulation because of the factor $z^n$, but at least you can get upper and lower bound on $\min_{P_n}\max_{ K}|P_n|$ in terms of the Chebyshev constant of $\widetilde K$.

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