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For $n\in\mathbb{N}$ let $p_n$ be a polynomial of degree $n$. Suppose there exists $c>0$ such that

$\bullet$ if $z\in\mathbb{C}$ is a zero of a $p_n$, then $|z^2+c|\leq c$ (note that in particular the $p_n$'s have no real positive zeros)

$\bullet$ $p_n(x)\xrightarrow[n\to\infty]{}f(x)$ for all $x>0$, where $f$ is an analytic function

Can we conclude that $f$ is never zero on $\mathbb{R^+}$?

I think Vitali's theorem should play a role, but I don't see precisely how.

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  • $\begingroup$ Is $c$ dependent on $n$? $\endgroup$ – 23rd Apr 11 '13 at 18:02
  • $\begingroup$ yes $c$ is the same for every $n$ $\endgroup$ – qwertyuio Apr 11 '13 at 18:25
  • $\begingroup$ If so, I think $f\equiv 0$. $\endgroup$ – 23rd Apr 11 '13 at 18:28
  • $\begingroup$ why? $p_n$ is never zero on $\mathbb{R}^+$ $\endgroup$ – qwertyuio Apr 11 '13 at 18:37
  • $\begingroup$ Would you mind if I post an answer for proving $f\equiv 0$? $\endgroup$ – 23rd Apr 11 '13 at 18:41
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Let $R=\sqrt{2c}>0$. Write $p_n$ as $p_n(z)=a_n\cdot\prod_{i=1}^n(z-z_{n,i})$. Then for $i=1,\dots, n$, from $|z_{n,i}^2+c|\le c$ we know that $|z_{n,i}|\le R$. Fix an integer $k\ge 1$. It follows that when $|z|\le kR$, $|p_n(z)|\le |a_n|\cdot[(k+1)R]^n$; when $|z|\ge (2k+3)R$, $|p_n(z)|\ge |a_n|\cdot[2(k+1)R]^n$. In particular, if $0<x\le kR<(2k+3)R\le y$, then $|p_n(x)|\le \frac{|p_n(y)|}{2^n}$. Letting $n\to\infty$, it follows that $f(x)=0$ for every $0<x\le kR$. Since $k$ is arbitrary, $f\equiv 0$ on $\mathbb{R}^+$.

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  • $\begingroup$ I read your argument and I don't find an error, but I'have a counter-example! $\endgroup$ – qwertyuio Apr 11 '13 at 19:48
  • $\begingroup$ @qwertyuio: Could you please show me your example? $\endgroup$ – 23rd Apr 11 '13 at 19:57
  • $\begingroup$ It's a bit complicated because it descends from a theorem and so it would require some effort to put it in an explicit form. So I prefer to read more carefully your proof $\endgroup$ – qwertyuio Apr 11 '13 at 20:01
  • $\begingroup$ @qwertyuio: OK. $\endgroup$ – 23rd Apr 11 '13 at 20:06
  • $\begingroup$ I'm terribly sorry. Your proof is correct. My functions $p_n$ are not polynomials but rational functions.. $\endgroup$ – qwertyuio Apr 11 '13 at 20:07

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