Let $f_w:\mathbb C \to \mathbb C$ be an entire function with $f_w(0)=1$ and at least one root for any choice of $w \in (0,1)$. Assume further that for a dense set of $w$ the function $f_w$ has infinitely many distinct roots and that $w\mapsto f_w$ depends real-analytically on $w.$ Does $f_w$ necessarily have infinitely many distinct roots for all $w \in (0,1)$?
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- 2$\begingroup$ $f(z)=1-(w-1/2)\sin z$ is a counterexample. $\endgroup$Christian Remling– Christian Remling2022-05-26 19:40:49 +00:00Commented May 26, 2022 at 19:40
- 1$\begingroup$ @ChristianRemling : This function has no zeroes for $w=1/2$. $\endgroup$Iosif Pinelis– Iosif Pinelis2022-05-26 19:59:38 +00:00Commented May 26, 2022 at 19:59
- 2$\begingroup$ @IosifPinelis: This is trivial to fix, of course (for example, consider $(1+z)f$). $\endgroup$Christian Remling– Christian Remling2022-05-26 20:08:22 +00:00Commented May 26, 2022 at 20:08
- $\begingroup$ @ChristianRemling : You are right, this is a simper example. $\endgroup$Iosif Pinelis– Iosif Pinelis2022-05-26 20:38:06 +00:00Commented May 26, 2022 at 20:38
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1 Answer
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$$f_w(z)=(1-z)e^z+(2w-1)(e^z-1)$$ is a counterexample.
More specifically, the only root of $f_{1/2}$ is $1$, whereas for each $w\in(0,1)\setminus\{1/2\}$ the function $f_w$ has infinitely many roots, of the form $2w+W_k\left(e^{-2 w}(1-2 w)\right)$, where $k$ is any integer and $W_k$ is the $k$th branch of the Lambert $W$ function.
answered May 26, 2022 at 19:58