Kontinuitätssatz

Remarks

Let $f$ be holomorphic on $D$. By the maximum principle applied to $f \circ \varphi_j$, the values of $\abs{f}$ on $\varphi_j(\bar\Delta)$ are at most $\max_K \abs{f}$. So every disc lies in the holomorphically convex hull $\hat K_D$, which is compact by the Cartan–Thullen theorem. Take $L = \hat K_D$.

Example: the Hartogs figure $H = \set{(z_1, z_2) : \abs{z_1} < 1, \abs{z_2} < 1, \ \abs{z_1} < r \text{ or } \abs{z_2} > s}$ fails the conclusion. Fix $\rho$ with $s < \rho < 1$, choose $c_j$ with $\abs{c_j} < r$ and $c_j \to c$ with $\abs{c} = r$, and let $\varphi_j(\zeta) = (c_j, \rho\zeta)$. These discs lie in $H$, and their boundaries lie in the compact set $\set{\abs{z_1} \le r, \abs{z_2} = \rho} \subseteq H$. But the centres $(c_j, 0)$ tend to $(c, 0) \notin H$, so no compact $L \subseteq H$ contains all the discs. Hence $H$ is not a domain of holomorphy.

The converse also holds: an open set with the disc property of the statement is pseudoconvex, and hence a domain of holomorphy by the solution of the Levi problem.

Going deeper

For a student

The continuity principle is one of the oldest forms of pseudoconvexity. It is a convenient practical test. A single family of discs whose rims stay inside but whose centres leave the domain shows that the domain is not a domain of holomorphy. The proof above shows how it follows from holomorphic convexity and the maximum principle.