Lineally convex Hartogs domains (Q1373065)

A set \(A\subset \mathbb{C}^n\) is said to be lineally convex if \(\mathbb{C}^n \setminus A\) is a union of complex hyperplanes. Lineal convexity is a kind of complex convexity intermediate between usual convexity and pseudoconvexity. However, as opposed to the latter two notions, lineal convexity is not a local property. The main results of the paper read as follows. Theorem 1. Let \(\Omega\) be a bounded complete Hartogs domain in \(\mathbb{C}^2\) (i.e. \(\Omega= \{(z,t)\in \mathbb{C} \times \mathbb{C}\); \(|t|<R(z)\}\), where \(R\) is a real-valued function on \(\mathbb{C}^n)\) with boundary of class \({\mathcal C}^2\). Let \(H\) and \(L\) denote, respectively, the Hessian form and the Levi form at a boundary point \(a\) of \(\Omega\). If \[ H(s)\leq L(s) \quad \text{for all} \quad s\in T_C(a) \tag{*} \] at all boundary points, then \(\Omega\) is lineally convex. Thus, for such domains, the property of being lineally convex is a local property. Theorem 2. Let \(\omega\) be a bounded open set in \(\mathbb{C}\) such that \(\overline\omega\) is not a disk. Then one can find a Hartogs domain \(\Omega\) over \(\omega\) (i.e. such that \(R(z)>0\) for \(z\in \omega)\) and two open sets \(\omega_0\) and \(\omega_1\) such that the Hartogs domains \(\Omega_j\) over \(\omega_j\) are lineally convex, \(j=1,2\), but \(\Omega= \Omega_0 \cup \Omega_1\) is not. If on the other hand \(\omega\) is a disk, and \(\Omega\) is a Hartogs domain over \(\omega\) satisfying (*) at all boundary points, then \(\Omega\) is lineally convex.