Let $f\_w:\mathbb C \to \mathbb C$ be an entire function such that $(0,1) \ni w \mapsto f\_w$ is real-analytic.


Assuming that there is a dense subset $D \subset (0,1)$ such that for $w \in D$ the function $f\_w$ has infinitely many zeros. What does this imply for the number of zeros of $f\_w$ with $w\in(0,1) \setminus D?$


In a previous question I already learned that I cannot expect $f\_w$ to have always infinitely many zeros as well.


[This happened in this question.](https://mathoverflow.net/questions/423377/zeros-of-entire-functions-with-parameter)


I wonder whether $f\_w$ has to have infinitely many zeros


1.) for some $w \in (0,1) \setminus D?$


2.) for almost all $w \in (0,1)$ in the Lebesgue sense?


3.) for generic $w \in (0,1)$ in the Baire sense?


4.) for all $w\in (0,1)$ aside from finitely many?



Zeros of entire functions
The answer for 3.) is **True**.


Lemma. If $f\_w$ has at least $n$ roots, then there exists a neighbourhood $W$ of $w$ such that all functions $f\_z$ $(z \in W)$ has at least $n$ roots.


Let $L\_n$ denote the subset of $(0,1)$ for which $w \in L\_n$ iff $f\_w$ has at least $n$ zeros. By the lemma above, $L\_n$ is an open set. As $D$ is dense and contained in $L\_n$, $L\_n$ is dense. It turns out that $(0,1)\backslash L\_n$ is a nowhere dense set.


The set of $w$ for which $f\_w$ has finitely many zeros is $\underset{n=1}{\overset{\infty}{\bigcup}}(0,1)\backslash L\_n$, and is (Baire) first category because the sets $(0,1)\backslash L\_n$ are nowhere dense.


