Can a disk that is limit of disks in a pseudoconvex domain be partially contained in the boundary?

Question

A holomorphic disk is the image of an injective holomorphic map $f:\mathbb D \to \mathbb C^n$ from the unit disk $\mathbb D \subset \mathbb C$ to $\mathbb C^n$.

Let $\Omega$ be a pseudoconvex domain in $\mathbb C^n$. Let $D=f(\mathbb D)\subset \overline\Omega$ be a holomorphic disk in the closure of $\Omega$ that is the uniform limit of holomorphic disks in $\Omega$. The Continuity Theorem implies, that if the boundary $\partial D$ is contained in $\Omega$, then so is $D$.

What if an inner point $p\in D$ of the disk lies in the boundary? Can another inner point of the disk lie in $\Omega$?

Preliminary results:

By the continuity theorem, an inner point of $D$ in the boundary of $\Omega$ immediately implies that there is also at least one point of $\partial D$ in $\partial \Omega$. Points of $\partial D$ in $\partial \Omega$ are not enough to force $D$ into the boundary, as $\mathbb D \times{0} \subset \mathbb D \times \mathbb D$ shows.

If both $D$ and $\partial \Omega$ are smooth (at $p$), then at $p$ we have a complex tangent vector to both $D$ and $\partial \Omega$, so by Levi-pseudoconvexity a real tangent direction in which $D$ enters inside $\Omega$ necessitates a corresponding real tangent direction in which $D$ exits $\overline \Omega$. So in this case the answer is no.

Passing to smaller subdisks around the point in $\partial \Omega \cap D$ shows that such a point is never isolated, by the continuity theorem.

Since a singularity of $D$ is isolated, we can always find a regular point of $D$ in $\partial \Omega$ in our setting.

Bonus question:

What if $D$ is not a limit of inner disks?

Answer 1

Here is an example with a "negative" result: take a plurisubharmonic function $u:\Bbb C^n\to[0,\infty)$, not identically zero, with $u(\lambda z)=|\lambda|u(z)$, $\lambda\in\Bbb C, z\in\Bbb C^n$, and its zeroes are dense in $\Bbb C^n$. Such a function exists according the work of J. Siciak. Put $h(z):=u(z)+\|z\|$ and $D:=\{z\in\Bbb C^n: h(z)<1\}$. Then $D$ is pseudoconvex sitting in the unit ball. Take points $a, a_j$ in the boundary of the unit ball with $u(a_j)=0$ and $u(a)> 0$ such that $a_j\to a$. Then the discs $\lambda\to\lambda a_j$ are sitting in $D$ and converge uniformly to the disk $\phi(\lambda):=\lambda a$. Note that the limit disc is sitting partially in $D$, partially on the boundary of $D$.

Answer 2

Take a bounded pseudoconvex domain $D\subset\Bbb C^n$ which allows a continuous function $u:\overline D\to(-\infty,0]$, plurisubharmonic on $D$, not identically zero, and zero on the boundary. Let $\phi_j:\Bbb D\to D$ be holomorphic and let $\phi_j\to\phi:\Bbb D\to\overline D$ uniformly. Assume there is a point $\lambda_0\in\Bbb D$ with $\phi(\lambda_0)\in\partial D$. Since $u$ is uniformly continuous on $\overline D$, we get that $u\circ\phi_j$ converges locally uniformly to $u\circ\phi$. Since the functions $u\circ\phi_j$ are subharmonic, the limit function is also. Then by the maximum principle $u\circ\phi$ vanishes identically, implying that the limit disk lies totally in $\partial D$. Such a function $u$ exists if $D$ is hyperconvex according a result of Blocki. Moreover, if $D$ has a H"older-continuous boundary, then it is hyperconvex according to a recent result by Bo-Yong Chen.

Answer 3

Take the so-called Hartog's triangle $\{(z,w)\in C^2:|w|<|z|<1\}$ and the disk $\lambda\to (\lambda,0)$.