Hartogs's Theorem Let $f$ be a holomorphic function on a set $G \setminus K$, where $G$ is an open subset of $\mathbb{C}^n$ ($n \ge 2$) and $K$ is a compact subset of $G$. If the complement $G\setminus K$ is connected, then $f$ can be extended to a unique to a unique holomorphic function on $G$.

This theorem can be used to show the following result about the zeros of analytic functions of several variables.

Suppose that $f$ is an analytic function on some open set $U$ and that $f$ is not identically zero on $U \subset \mathbb{C}^n$ with $n \ge 2$. Then, the set of zeros of $f$ (i.e. $\Lambda(f)=\{ z: f(z)=0\}$) is not compact.

Since $\Lambda(f)$ is not compact we can have the following three possibilities:

  1. $\Lambda(f)$ is closed but is not bound
  2. $\Lambda(f)$ is not closed but bounded
  3. $\Lambda(f)$ is not closed and not bounded

My question is the following: Can we come up with examples of $f$ for each of the three cases?

Here is an example of the function that satisfies the first case. Let $f_1(z_1,z_2)=z_1 \cdot z_2$ then \begin{align} \Lambda(f_1)= \{ (z_1,z_2) : z_1=0 \} \cup \{ (z_1,z_2) : z_2=0 \}. \end{align} where $\Lambda(f_1)$ is closed but not bounded.


The cases 2 and 3 cannot exist by continuity. Since $f\colon U \rightarrow \mathbb{C}$ is holomorphic, it is continuous. Hence $\{0\}$ being a closed subset implies that $f^{-1}(0) = \Lambda(f)$ is closed in $U$.

  • $\begingroup$ Ok. Thanks. So a more accurate statement about the set of zeros is that it is closed and unbounded, right? $\endgroup$ – Boby Aug 15 '18 at 20:20
  • $\begingroup$ Closed and unbounded in $U$. $\endgroup$ – Alan Muniz Aug 15 '18 at 23:25
  • $\begingroup$ Can you remind me what unbounded in $U$ means? There exists no ball $B$ that contains $\Lambda(f)$ such that $B \subset U$? $\endgroup$ – Boby Aug 16 '18 at 0:01
  • $\begingroup$ It means that the set $\Lambda(f)$ is not contained in any compact subset of $U$. I want to emphasise that because $U$ may be bounded in $\mathbb{C}^n$. For example, let $U = B(0,1)$ the unitary ball centered at the origin. Then being unbounded in $U$ means that the closure of $\Lambda(f)$ needs to meet the boundary of $U$. However $\Lambda(f)$ is bounded as a subset of $\mathbb{C}^n$. $\endgroup$ – Alan Muniz Aug 16 '18 at 0:09
  • 1
    $\begingroup$ I'd recommend "From holomorphic functions to complex manifolds" by Fritzsche and Grauert. It is an introductory book but worth to be read. $\endgroup$ – Alan Muniz Aug 16 '18 at 1:12

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.