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.
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 $\mathbb{R}^+$.