[step:State and prove the analytic disk lemma for the reverse direction]
We now prove (2) $\Rightarrow$ (1). Assume there exists a continuous plurisubharmonic exhaustion $\varphi : \Omega \to \mathbb{R}$. If $\Omega = \mathbb{C}^n$, condition (1) holds by definition; assume $\Omega \neq \mathbb{C}^n$ for the remainder. Write $\mathbb{D} := \{\zeta \in \mathbb{C} : |\zeta| < 1\}$ for the open unit disk and $\overline{\mathbb{D}}$ for its closure.
[claim:Kontinuitätssatz for continuous disk families]
Let
\begin{align*}
A : [0,1] \times \overline{\mathbb{D}} &\to \mathbb{C}^n
\end{align*}
be continuous. For each $t \in [0,1]$, define
\begin{align*}
A_t : \overline{\mathbb{D}} &\to \mathbb{C}^n \\
\zeta &\mapsto A(t,\zeta).
\end{align*}
Assume $A_t$ is holomorphic on $\mathbb{D}$ for each $t \in [0,1]$, $A_0(\overline{\mathbb{D}}) \subseteq \Omega$, and $A([0,1] \times \partial\mathbb{D})$ is contained in a compact subset of $\Omega$. Then $A_t(\overline{\mathbb{D}}) \subseteq \Omega$ for every $t \in [0,1]$.
[/claim]
[proof]
Set
\begin{align*}
K := A([0,1] \times \partial\mathbb{D}).
\end{align*}
By hypothesis $K$ is contained in a compact subset of $\Omega$, so
\begin{align*}
M := \sup_{z \in K} \varphi(z)
\end{align*}
is finite. Define the sublevel set
\begin{align*}
L := \{z \in \Omega : \varphi(z) \le M\}.
\end{align*}
Since $\varphi$ is an exhaustion, $L$ is compact in $\Omega$.
Let
\begin{align*}
S := \{t \in [0,1] : A_t(\overline{\mathbb{D}}) \subseteq \Omega\}.
\end{align*}
The set $S$ is nonempty because $0 \in S$ by hypothesis. If $t \in S$, define
\begin{align*}
u_t : \overline{\mathbb{D}} &\to \mathbb{R} \\
\zeta &\mapsto \varphi(A_t(\zeta)).
\end{align*}
The map $u_t$ is continuous on $\overline{\mathbb{D}}$. Since $A_t$ is holomorphic on $\mathbb{D}$ and $\varphi$ is plurisubharmonic on $\Omega$, the pullback $u_t$ is subharmonic on $\mathbb{D}$ by the holomorphic-pullback stability of plurisubharmonic functions, as recorded in [Stability Properties of PSH Functions](/theorems/3404). For $\zeta \in \partial\mathbb{D}$ we have $A_t(\zeta) \in K$, so $u_t(\zeta) \le M$. The maximum principle for subharmonic functions gives $u_t(\zeta) \le M$ for every $\zeta \in \overline{\mathbb{D}}$. Hence
\begin{align*}
A_t(\overline{\mathbb{D}}) \subseteq L \qquad \text{for every } t \in S. \tag{1}
\end{align*}
We prove that $S$ is closed in $[0,1]$. Let $(t_k)_{k \ge 1} \subseteq S$ be a sequence with $t_k \to t \in [0,1]$. For each fixed $\zeta \in \overline{\mathbb{D}}$, relation (1) gives $A(t_k,\zeta) \in L$ for every $k$. Since $L$ is compact, hence closed in $\mathbb{C}^n$, continuity of $A$ gives $A(t,\zeta) \in L$. Thus $A_t(\overline{\mathbb{D}}) \subseteq L \subseteq \Omega$, so $t \in S$.
We prove that $S$ is open in $[0,1]$. Let $t \in S$. By relation (1), $A_t(\overline{\mathbb{D}}) \subseteq L$. Since $L$ is a compact subset of the [open set](/page/Open%20Set) $\Omega$, define
\begin{align*}
\eta := \operatorname{dist}(L, \mathbb{C}^n \setminus \Omega) > 0.
\end{align*}
The map $A$ is uniformly continuous on the compact set $[0,1] \times \overline{\mathbb{D}}$. Therefore there exists $\rho > 0$ such that, whenever $s \in [0,1]$ and $|s-t| < \rho$,
\begin{align*}
|A(s,\zeta) - A(t,\zeta)| < \frac{\eta}{2} \qquad \text{for every } \zeta \in \overline{\mathbb{D}}.
\end{align*}
For such $s$ and every $\zeta \in \overline{\mathbb{D}}$, the point $A(s,\zeta)$ has distance less than $\eta/2$ from $L$, and hence belongs to $\Omega$ by the definition of $\eta$. Thus $s \in S$, proving that $S$ is open.
The interval $[0,1]$ is connected, and $S \subseteq [0,1]$ is nonempty, open, and closed. Hence $S = [0,1]$, which proves $A_t(\overline{\mathbb{D}}) \subseteq \Omega$ for every $t \in [0,1]$.
[/proof]
[/step]