0
$\begingroup$

Let $U$ be an open subset of $\mathbb{C}$ and $\{f_n\}$ be a sequence of holomorphic functions. Suppose that $f_n$ converges uniformly to a function $f$ on compact subsets of $U$ and that $f$ is not identically zero in $U$ and $f(w)=0$ for some $w \in U$.

Prove that there exists $N \in \mathbb{N}$ and a sequence $\{z_n\}$ such that $f_n(z_n) = 0$ for all $n \geq N$ and $\lim_{n \rightarrow \infty}z_n = w$.

How can I prove the existence of $z_n$ and $f_n(z_n)$ is exactly 0? Any idea?

$\endgroup$
  • $\begingroup$ Feels like an $\epsilon/3$ argument to me. $\endgroup$ – Adrian Keister Aug 7 '18 at 19:17
  • $\begingroup$ Hint: try with Hurwitz's theorem. $\endgroup$ – Bob Aug 7 '18 at 19:27
1
$\begingroup$

Hint: Let $\Gamma$ be a circle around $w$ such that $\Gamma$ and its interior are in $U$, and $f_n$ is nonzero on $\Gamma$. Then $$\dfrac{1}{2\pi i} \oint_\Gamma \dfrac{f_n'(z)\; dz}{f_n(z)}$$ is the number of zeros of $f_n$ (counted by multiplicity) inside $\Gamma$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.