Cartan-Thullen theorem

Property 1 implies property 2 by [L1].

Assume property 3. If $\mathrm{\Omega }={\mathbb{C}}^m$, property 1 holds directly from [L3]. Otherwise fix $p\in \mathrm{\partial }\mathrm{\Omega }$ and a connected open set $C\subseteq \mathrm{\Omega }$ with $p\in \bar{C}$. Let $(D_n)$ be an increasing compact exhaustion of $\mathrm{\Omega }$. Construct increasing holomorphically convex compact sets $K_n$ and points $p_n\to p$ recursively. Start with $K_1:={\widehat{D_1}}_{\mathrm{\Omega }}$. Once $K_n⋐\mathrm{\Omega }$ is chosen, choose $p_n\in C\setminus K_n$ with $\mathrm{\mid }{p}_n-p\mathrm{\mid }<1\mathrm{/}n$, and put $K_{n+1}:={\widehat{K_n\cup D_{n+1}\cup \{p_n\}}}_{\mathrm{\Omega }}\mathrm{.}$ Property 3 and idempotence of hulls make each $K_n$ compactly contained in $\mathrm{\Omega }$, and $p_n\in K_j$ whenever $j>n$.

Since $p_n\mathrm{\notin }{\widehat{K}}_n{}_{\mathrm{\Omega }}=K_n$, choose $g_n\in \mathcal{O}(\mathrm{\Omega })$ with $\mathrm{\mid }{g}_n(p_n)\mathrm{\mid }>{\sup }_{K_n}\mathrm{\mid }{g}_n\mathrm{\mid }$. Taking a sufficiently large positive power makes the ratio of these two quantities arbitrarily large; scaling that power then gives $f_n\in \mathcal{O}(\mathrm{\Omega })$ such that ${\sup }_{K_n}\mathrm{\mid }{f}_n\mathrm{\mid }<2^{-n}\text{and}\mathrm{\mid }{f}_n(p_n)\mathrm{\mid }>n+1+{\sum }_{j<n}\mathrm{\mid }{f}_j(p_n)\mathrm{\mid }\mathrm{.}$ Every compact subset of $\mathrm{\Omega }$ lies in some $K_N$, so ${\sum }_n{f}_n$ converges uniformly on compact subsets to a holomorphic function $f$. Moreover, $p_n\in K_j$ for $j>n$, so ${\sum }_{j>n}\mathrm{\mid }{f}_j(p_n)\mathrm{\mid }<{\sum }_{j>n}{2}^{-j}<1.$ The reverse triangle inequality and the second displayed bound give $\mathrm{\mid }f(p_n)\mathrm{\mid }>n$.

Assume property 2, and let $K⋐\mathrm{\Omega }$ be compact. By [L2], the hull ${\widehat{K}}_{\mathrm{\Omega }}$ is closed in $\mathrm{\Omega }$ and coordinate-bounded. Property 2 gives ${\delta }_{\mathrm{\Omega }}({\widehat{K}}_{\mathrm{\Omega }})={\delta }_{\mathrm{\Omega }}(K)>0$, so every point of ${\widehat{K}}_{\mathrm{\Omega }}$ carries a positive-radius polydisc contained in $\mathrm{\Omega }$. Hence every Euclidean limit point of ${\widehat{K}}_{\mathrm{\Omega }}$ still lies in $\mathrm{\Omega }$, and the closedness from [L2] makes ${\widehat{K}}_{\mathrm{\Omega }}$ closed in ${\mathbb{C}}^m$. Being closed and bounded in finite-dimensional Euclidean space, it is compact; the positive boundary-radius lower bound keeps it away from $\mathrm{\partial }\mathrm{\Omega }$. Thus ${\widehat{K}}_{\mathrm{\Omega }}⋐\mathrm{\Omega }$, so property 3 holds.

Suppose toward a contradiction that domains $U_1,U_2$ witness failure of property 1 as in [L3]. Choose $a\in U_1$ and $b\in U_2\setminus \mathrm{\Omega }$. By [L4], join them by a path in $U_2$, and let $p$ be its first point on $\mathrm{\partial }\mathrm{\Omega }$. The part of the path before $p$ lies in the connected component $C$ of $\mathrm{\Omega }\cap U_2$ containing $U_1$, so $p\in \bar{C}$. Apply steps 1.2 and 2.1 with this $p$ and $C$, obtaining $f\in \mathcal{O}(\mathrm{\Omega })$ and $p_n\in C$ with $p_n\to p$ and $\mathrm{\mid }f(p_n)\mathrm{\mid }>n$. By the assumed simultaneous-extension property, $f$ has an extension $F\in \mathcal{O}(U_2)$ agreeing with $f$ on $U_1$. The identity theorem [L5] gives $F=f$ on $C$, but continuity makes $F$ bounded near $p\in U_2$, contradicting $\mathrm{\mid }F(p_n)\mathrm{\mid }>n$. Thus property 3 implies property 1, and the three properties are equivalent.