1
$\begingroup$

Given $0 < a, b < 1$, consider the Hartogs figure $H$ given by \begin{equation*} H = \{ (z,w) \in \mathbb{D}\times \mathbb{D} \ \ | \ \ |z| > a \} \cup \{ (z,w) \in \mathbb{D} \times \mathbb{D} \ \ | \ \ |w| < b \}. \end{equation*} It is well known that $H$ is not a domain of holomorphy; any holomorphic function on $H$ is actually holomorphic on the whole of $\mathbb{D}\times\mathbb{D}$. Thus, by the well established equivalence between domains of holomorphy and holomorphically convex domains $H$ is not holomorphically convex. However, is it possible to prove that $H$ is not holomorphically convex straight from the definition without using any equivalent statements nor known facts about $H$?

Recall the definition of holomorphic convexity: a domain $U$ is said to be holomorphically convex if for every compact subset $K \subset U$, the holomorphic convex hull $\hat{K}_U$ is also compact in $U$. Here the holomorphic convex hull is \begin{equation*} \hat{K}_U = \{ z \in U \ \ | \ \ |f(z)| \leq \sup_{\zeta \in K}|f(\zeta)| \ \ \forall f \in \mathcal{O}(U) \}. \end{equation*}

$\endgroup$
2
$\begingroup$

In the end you will need to know something about the functions in $\mathcal{O}(H)$. Specifically we would need their behaviour near some point on the boundary $\partial H$, which would tell us that $H$ is not a d.o.h. However since we get the extension to $\mathbb{D}^2$ directly by the Cauchy integral formula, it is a lot easier using this.

If we know that $\mathcal{O}(H) = \mathcal{O}(\mathbb{D}^2)$ the result follows directly. Consider the Reinhardt domain $K= \{(z,w) \in \mathbb{C}^2 \; | \; |z|=r_1,\; |w|=r_2\}$ for real numbers $r_1,r_2$.

We know that holomorphic functions on a polydisc $\Delta_{r_1}\times \Delta_{r_2}$ have maximum modulus on the distinguished boundary $\Gamma = \partial \Delta_{r_1}\times \partial \Delta_{r_h}$ by the maximum principle. Thus the holomorphic hull is \begin{equation} \hat K = H \cap \{(z,w) \in \mathbb{C}^2 \; | \; |z| \le r_1,\; |w| \le r_2\}. \end{equation} In particular if we pick $a<r_1<1$, $b<r_2<1$ we get a holomorphic hull that is not relatively compact in $H$.

$\endgroup$
1
  • $\begingroup$ That is exactly the argument I used to solve the problem. Thanks for your help. $\endgroup$ Aug 11 '20 at 2:38

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.