I am trying to prove that every convex domain in $\mathbb{C^n}$ is a weak domain of Holomorphy.

Let $G$ be a convex domain. I pick a point $p \in \partial{G}$. Then By Hahn-Banach Separation theorem, there exists a continuous complex-linear Functional $l: G \to \mathbb{C}$ such that $Re(l(z)) \lt Re(l(p))$ for all $z \in G$

Now I consider the function $$f_{p}(z)=\frac{1}{l(z)-l(p)}$$

I need to show that $f_{p}$ is completely singular at $p$. It is intuitively clear that $f_{p}(z)$ is so but I am unable to show it mathematically.

  • $\begingroup$ What is your definition of "completely singular at $p$"? It is clear that $f_p$ can't be extended across $z=p$ since $f_p$ is unbounded near $p$. $\endgroup$
    – mrf
    Nov 23, 2015 at 20:43
  • $\begingroup$ @mrf By "completely singular" at a point $p$, for every neighborhood $U(p)$ of $p$ , and every connected component of $C$ of $U(p) \cap G$, there doesnot exist any holomorphic function $g$ such that $g|_{C}=f|_{C}$ $\endgroup$ Nov 27, 2015 at 2:04

1 Answer 1


We can get a stronger result than weak holomorphy. It is in fact a domain of holomorphy.

If your domain $D \subseteq \mathbb{C}^n$ is convex, in the sense of affine mappings, then it is holomorphically convex. That is $\widehat{D} : = \{ z \in D : \left| f(z) \right| \leq \sup_{z \in K} \left| f(z) \right| \}$ is contained in $D$, where $\widehat{D}$ denotes the holomorphically convex hull. This is easy to see since the class of affine functions embeds into the class of holomorphic functions.

Now, we know that the domains of holomorphy in $\mathbb{C}^n$ are the domains which are holomorphically convex. For the proof of this result, see the end of chapter 3 of Shabat's Introduction to Complex Analysis -- Part II.


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