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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.

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?

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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.

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