The intersection of domains of holomorphy is again a domain of holomorphy

Question

The original problem is this:

$\{\Omega_j\}_{j\in J}$ is a family of domains of holomorphy. Prove that $\mathrm{int}\left(\bigcap_{j\in J} \Omega_j \right)$ is a domain of holomorphy.

I used $\textbf{Cartan-Thullen}$ first, which implies that $U$ is a domain of holomorphy if and only if it is holomorphically convex.

Now I can prove the theorem if $|J|$ is finite by mean of the Lemma which reads:

$$K\subset U\subset V \Longrightarrow\widehat{K}_U \subset \widehat{K}_V.$$

However, I cannot prove the theorem if $|J|$ is not finite. Specifically, I have no idea what the interior of the domains means here. I want to prove that the convex hull does not touch the boundary, but when interior operator is given, many points goes away. Then how can I prove that those points are not in the convex hull?

Thanks in advance for your help!

Answer 1

This might not be the exact answer that you were expecting to get, but my suggestion is to try using the following equivalent characterization of domains of holomorphy, which is likely to be an easier approach to get a simple solution:

A domain $\Omega\subset\mathbb{C}^n$ is a domain of holomorphy if and only if there does not exist a domain $\omega\subset\mathbb{C}^n$ such that (1) $\omega$ intersects both $\Omega$ and $\Omega^C$, and (2) there exists a connected component $\omega_0$ of $\omega\cap\Omega$ such that for every holomorphic function $f$ on $\Omega$, there exists $f_0\in \mathcal{O}(\omega)$ such that $f=f_0$ on $\omega_0$.

Let $\Omega=\mathrm{int}(\bigcap_j \Omega_j)$ and suppose that it is not a domain of holomorphy. Then there exists $\omega$ satisfying (1) and (2). Deduce from (1) that there exists $j_0\in J$ satisfying $\omega\cap\Omega_{j_0}\ne\emptyset$ and $\omega\cap\Omega_{j_0}^C\ne\emptyset$. (Otherwise, $\omega$ would be an open subset of $\bigcap_j\Omega_j$ and then it is contained in the interior of $\bigcap_j\Omega_j$.) Then let $\omega_0'$ be the connected component of $\omega\cap\Omega_{j_0}$ containing $\omega_0$ and check that $\omega_0'$ also satisfies (2), with $\Omega$ replaced by $\Omega_{j_0}$. This implies that $\Omega_{j_0}$ is not a domain of holomorphy, so we get a contradiction.

Answer 2

I guess the question is where you want to start at. My first instinct would be to look at the distance function to the boundary $r(z)$ and show that $- \log r(z)$ is plurisubharmonic. You can set up the intersection to be a decreasing sequence. Then take these functions for the domains in this sequence and you'll have an increasing sequence of plurisubharmonic functions. Therefore the limit is plurisubharmonic. That shows that the intersection is Hartogs pseudoconvex. Hence it is a domain of holomorphy by a solution of the Levi-problem. The advantage of this approach is that it is immediately obvious what goes wrong at the points that aren't in the interior. The sequence of functions goes to infinity.

Answer 3

In fact, an open set $\Omega\subset\mathbb{C}^n$ is a domain of holomorphy iff for any $p\in\partial\Omega$, there exists a function $f\in \mathcal{O}(\Omega)$ such that $\limsup_{z\rightarrow p}|f(z)|=+\infty$.
Note that $\partial(\cap_j\Omega_j) \subset \cup_j \partial\Omega_j$, we can easily get the result.