# Reconciling two definitions of domains of holomorphy

I have seen the following two definitions of a domain of holomorphy; I was wondering if they are actually equivalent:

1) A domain $\Omega$ is a domain of holomorphy if there exists a holomorphic function on $\Omega$ that does not extend to a larger domain $\Omega' \supset \Omega$;

2) A domain $\Omega$ is a domain of holomorphy if for every domain $\Omega'$ which intersects the boundary of $\Omega$, and for every connected component $U$ of $\Omega \cap \Omega'$, there exists a holomorphic function $\varphi$ on $\Omega$ such that $\varphi|_U$ does not extend to $\Omega'$.

I think that at least the second condition implies the first; but does the first imply the second?

The first is not a good definition. Define $\Omega \subset {\mathbb C}^2$ in the following way: $$\Omega = \{ (z,w) \mid z e^{i|w|^2} \text{is not a nonnegative real number} \}$$ For each fixed $w$, the set is simply a branch cut of the plane. It is not difficult to show that for the complex line given by $z = -r$ for a negative real number $-r$ hits the boundary of $\Omega$ at a single point. That is the boundary of omega except for the set $z=0$ is actually a strictly pseudoconvex hypersurface. But it is $\Omega$ on both sides of it.
Take any $f$ holomorphic on $\Omega$. You can see that according to the second (correct) definition, $\Omega$ is not a domain of holomorphy. The $f$ will extend from the pseudoconcave side. You should also note how the "every connected component" plays into this definition.
On the other hand, there clearly is not a large domain $\Omega'$ for the first definition. You can just take a branch of $\log z$ so that it is holomorphic in $\Omega$. That will be discontinuous at the branch cuts and so cannot extend.
• Why does the complex line $z = -r$ intersecting the boundary of $\Omega$ at a single point imply that $\partial\Omega$ is a strictly pseudoconvex hypersurface? Jul 1 at 16:16