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I'm interested in polynomials in several variables $p(x_1,\ldots,x_n)$, with complex coefficients, such that the maximum modulus of $p$ on the unit complex $n$-ball $$ \max \{ |p(z_1,\ldots,z_n)| : \left|z_1\right|^2 + \cdots + \left|z_n\right|^2 \leq 1\} $$ is known exactly. I can think of some, but these are mostly constructed from a 1-variable polynomial in the manner of $p(x,y) = q((x+y)/2)$ or such like. I'm particularly interested in the case $n = 2,3,4$ and polynomials of order less than a hundred or so.

Any takers?

Thanks in advance!

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up vote 1 down vote accepted

Here is a different family of examples. Let $p(z)=Az_1+Bz_2^b+Cz_3^c+Dz_4^d$ where $b,c,d\ge 2$ and $|A|\ge 2\max(|B|,|C|,|D|)$. I claim that $\max_{|z|\le 1} |p|=|A|$. Clearly, at $(1,0,0,0)$ this value is attained. On the other hand, for all $z$ with $|z|\le 1$ we have $$|Bz_2^b+Cz_3^c+Dz_4^d|\le \max(|B|,|C|,|D|)(|z_2|^2+|z_3|^2+|z_4|^2)\le \frac{|A|}{2}(1-|z_1|^2)\le |A|(1-|z_1|)$$ Hence $|p(z)|\le |A||z_1|+|A|(1-|z_1|)=|A|$.

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Thanks Pavel, exactly the sort of thing I was looking for. – n00b Dec 30 '12 at 12:05
As a postscript to this, there is a slight error in the proof (since $1-x^2 \geq 1-x$ for $x\in(0,1]$) but this can be repaired if we insist that $|A| \geq 2\max(|B|,|C|,|D|)$ and apply a little calculus. – n00b Dec 31 '12 at 18:03
@noob Thanks, I fixed the error. – user53153 Dec 31 '12 at 18:14

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