Hartogs figure not holomorphically convex 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*}
 A: 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 $H$.
