# Zeros of entire functions

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?

• But how are $f_w$ for different $w$ related? May 27 at 10:30
• There is a paper math.purdue.edu/~eremenko/dvi/sokal7.pdf which addresses this question when $(z,w)\mapsto f_w(z)$ is complex analytic. May 27 at 15:07

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.