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Is there any complex polynomial $p$ of one variable having no zeros within the unit square: $-1 < \Re(z) , \Im(z) < 1$ such that $\left|p(0)\right|$ is strictly smaller than $\left|p(z)\right|$ whenever $z$ is either on the corner of the square or on the middle of one the square's sides. That is, when $z \in$ {1, 1+i, i, -1+i, -1, -1-i, -i, 1-i}.

This problem may be quite arbitrary, so any theorems are welcome where some finite number of inequalities (where on each side of each inequality the only relevant parameters are evaluations of the polynomial in a finite number of points) implies a strong bound on some zero of the polynomial.

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Try polynomial $z^8+2$. –  Gerald Edgar May 12 '13 at 16:48
    
Gerald's answer is much better and simpler than my. I was considering deleting my answer, but decided to leave it. –  Johan Andersson May 12 '13 at 17:14
    
Ok, but the zeros from $z^8+2$ are still "very close" to the unit square. What if I ask for a polynomial satisfying the same inequalities, but having no zero within a square with a (large) side $l$ and center 0. Is there some bound on $l$ for which we can find an example? –  Thibaut Demaerel May 12 '13 at 17:30
    
$p(z)=z^8+k$, for $k$ large, has no zeros inside a large square. But still $|p(0)|$ smaller than the specified points on the unit square. –  Gerald Edgar May 12 '13 at 22:37
    
Thanks, with hindsight this problem had a pretty easy resolution. –  Thibaut Demaerel May 13 '13 at 14:21

4 Answers 4

Edit: I had the feeling my original answer below was much too complicated. Here is a better answer.

New answer: Given $n$ complex numbers $z_k$ and $a_k$, $k=1,\ldots,n$ we can use standard interpolation arguments to find a polynomial $q(z)$ of degree $n$ such that $q(z_k)=a_k$. For example by choosing $a_k=k$ we will get $|q(z_k)|<|q(z_j)|$ iff $ k < j$. For any bounded set $D$, we can now add a large positive constant $K$ such that $p(z)=q(z)+K$ is zero free on $D$. In your example we can choose $z_1=0$ and $z_2,\ldots,z_9$ the points on the boundary of the unit square, and we can choose the constant $K$ sufficiently large so that $p(z)$ is zero free on the unit square and we get that $|p(0)|<|p(z_k)|$ for these points

Old answer: Sure, there is. In fact given any finitely number of points on the boundary, say $z_1,\ldots,z_n$, there exists a polynomial such that $|p(0)|<|p(z_k)|$ that is zero free on the unit disc. An observation is that it follows by looking at the meromorphic function $$ f_{\epsilon}(z)=\prod_{k=1}^n z_k(z-z_k(1+\epsilon))^{-1} $$ By the construction $\lim_{\epsilon \to 0^+} f_\epsilon(0)=1$ and $\lim_{\epsilon \to 0^+} |f_\epsilon(z_k)|=\infty$. Thus we can choose an $\epsilon>0$ such that $|f_\epsilon(0)|<3/2<2<|f_\epsilon(z_k)|$. Since $f_\epsilon(z)$ is continuous and zero free on the closed unit square and analytic in the open unit square, a variant of Mergelyan's theorem of mine, http://arxiv.org/abs/1010.0850 shows that we can approximate the function arbitrarily closely (in sup norm) on the unit square by a polynomial without zeros (this is where my variant is needed) in the unit square. If we find such a polynomial $p(z)$ that approximates the function $f_\epsilon(z)$ with an error less that $1/4$ then the inequality $|p(0)|<|p(z_k)|$ holds.

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haha, nice there! –  Per Alexandersson May 12 '13 at 17:14
    
"unit disc" is replaceable by "unit square" in the previous post? –  Thibaut Demaerel May 12 '13 at 17:19
    
unit disc was a mistake that I just corrected. Thanks, I meant the unit square. The same result holds however for the unit disc, for example for any closed Jordan domain with the inequality holding for finitely many boundary points. –  Johan Andersson May 12 '13 at 17:33

Since $f$ is non-zero in a neighbourhood of the unit square, it follows that $1/f$ is holomorhic in the unit square. By the maximum modulus principle, $|1/f|$ attains its maximum on the boundary. If follows that the minimum of $|f|$ is attained at the boundary, of the unit square. This sort of hints that it should be very hard to find such a polynomial.

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Use Lagrange Interpolation Formula: Specify values at $1, i, 1+i$ etc to be numbers of very high absolute value and value at 0 to be of much smaller absolute value. By this formula we will get a polynomial. The problem could be it can have roots of small absolute value. To avoid this and meet your requirement, change the polynomial $f(z)$ obtained to $f(cz)$ with the scaling constant chosen to push out the zeros far away from 0. Note that $f(cz)$ has the same constant term as $f(z)$ and so your condition will be preserved.

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This does not seem to work since we do not longer know that $|f(c(1+i))|>|f(0)|$ after scaling with the constant $c$. By using the lagrange interpolation formula in a differnt way (as in my new answer above) to be say $1$ for $1,i,1+i$ etc and $0$ for $z=0$ and adding a sufficiently large positive constant $k$, we are sure that $0<p(0)<p(z)$ (and of course the same inequality holds when taking absolute values) when $z=1,i,1+i$, etc and also that the polynomial is zero-free. –  Johan Andersson May 14 '13 at 15:45
    
I see the snag. Can't we, with hindsight, specify values to be even higher at 1, $1+i$ so that scaling constant can't bring down. Of course this depends on the least absolute value of the zero. I'll try to rework again. Thanks, Johan! –  P Vanchinathan May 15 '13 at 0:07
  1. You can construct a polynomial $p$ with prescribed values at all your points.

  2. Take $\exp(p)$, and then approximate this by sufficiently long partial sum of the Taylor series, so that the result does not have zeros in your square, and the values at the points of interest satisfy the desired inequalities.

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