[proofplan]
We show that every domain of holomorphy is pseudoconvex by exhibiting a continuous plurisubharmonic exhaustion function. The strategy is to invoke the [Cartan--Thullen Theorem](/theorems/3385) to obtain holomorphic convexity, then use the result that the [log-distance function is PSH on domains of holomorphy](/theorems/3406) to produce the exhaustion. The sublevel sets of the log-distance function are shown to be compactly contained in $\Omega$ by a direct geometric argument.
[/proofplan]
[step:Establish that $\Omega$ is holomorphically convex via the Cartan--Thullen theorem]
Since $\Omega$ is a domain of holomorphy, the [Cartan--Thullen Theorem](/theorems/3385) implies that $\Omega$ is holomorphically convex: for every compact $K \subset \Omega$, the holomorphic hull
\begin{align*}
\hat{K}_\Omega := \{z \in \Omega : |f(z)| \leq \sup_K |f| \text{ for all } f \in \mathcal{O}(\Omega)\}
\end{align*}
is again compact in $\Omega$.
[/step]
[step:Invoke the plurisubharmonicity of the log-distance function]
Define the log-distance function
\begin{align*}
\phi: \Omega &\to \mathbb{R} \\
z &\mapsto -\log d(z, \partial\Omega),
\end{align*}
where $d(z, \partial\Omega) = \inf_{w \in \partial\Omega} |z - w|$ is the Euclidean distance from $z$ to the boundary. The function $\phi$ is continuous on $\Omega$ since $d(\cdot, \partial\Omega)$ is continuous and positive on $\Omega$.
By the theorem that the [log-distance is PSH on domains of holomorphy](/theorems/3406), the function $\phi$ is plurisubharmonic on $\Omega$. This result uses the holomorphic convexity established above: the [distance characterisation of the holomorphic hull](/theorems/3402) shows that $d(\hat{K}_\Omega, \partial\Omega) = d(K, \partial\Omega)$ for every compact $K \subset \Omega$, and this distance equality is equivalent to the sub-mean-value property of $\phi$ along complex lines.
[guided]
The plurisubharmonicity of $\phi = -\log d(\cdot, \partial\Omega)$ is not a general fact about arbitrary domains -- it relies on the holomorphic convexity of $\Omega$. For a domain that is not holomorphically convex, the holomorphic hull $\hat{K}_\Omega$ can escape to $\partial\Omega$, meaning $d(\hat{K}_\Omega, \partial\Omega) < d(K, \partial\Omega)$. In that case, the sub-mean-value inequality for $\phi$ along complex lines would fail, and $\phi$ would not be psh.
The proof of the [log-distance PSH theorem](/theorems/3406) works as follows: fix $z_0 \in \Omega$, $w \in \mathbb{C}^n$ with $|w| \leq 1$, and consider $\psi(\zeta) = -\log d(z_0 + \zeta w, \partial\Omega)$ on the set $D = \{\zeta \in \mathbb{C} : z_0 + \zeta w \in \Omega\}$. For $\zeta_0 \in D$ and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D$, let $K = \{z_0 + \zeta w : |\zeta - \zeta_0| \leq r\}$. The [maximum modulus principle](/page/Maximum%20Modulus%20Principle) places $z_0 + \zeta_0 w$ in $\hat{K}_\Omega$, and the distance characterisation gives $d(z_0 + \zeta_0 w, \partial\Omega) \geq d(K, \partial\Omega) = \min_{|\zeta - \zeta_0| = r} d(z_0 + \zeta w, \partial\Omega)$. Applying $-\log$ reverses the inequality, yielding $\psi(\zeta_0) \leq \max_{|\zeta - \zeta_0| = r} \psi(\zeta)$, which implies the sub-mean-value inequality.
[/guided]
[/step]
[step:Verify that the sublevel sets of $\phi$ are compactly contained in $\Omega$]
For each $c \in \mathbb{R}$, the sublevel set is
\begin{align*}
\{\phi < c\} = \{z \in \Omega : -\log d(z, \partial\Omega) < c\} = \{z \in \Omega : d(z, \partial\Omega) > e^{-c}\}.
\end{align*}
We must show this set is compactly contained in $\Omega$, i.e., its closure in $\mathbb{C}^n$ is a compact subset of $\Omega$.
First, the closure of $\{\phi < c\}$ avoids $\partial\Omega$: every point in $\overline{\{\phi < c\}}$ satisfies $d(z, \partial\Omega) \geq e^{-c} > 0$, so $\overline{\{\phi < c\}} \cap \partial\Omega = \varnothing$.
Second, $\overline{\{\phi < c\}}$ is bounded. If $\Omega$ is bounded, this is immediate since $\overline{\{\phi < c\}} \subset \overline{\Omega}$. If $\Omega$ is unbounded, then $\Omega \neq \mathbb{C}^n$ (since $\mathbb{C}^n$ has no boundary and is not a proper domain of holomorphy). In the unbounded case, note that $\Omega$ being a domain of holomorphy with $\partial\Omega \neq \varnothing$ ensures that $d(z, \partial\Omega)$ remains bounded as $|z| \to \infty$ within $\Omega$ (otherwise $\Omega$ would contain arbitrarily large balls and, by a normal families argument, every [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ would extend to $\mathbb{C}^n$). Thus $\{d(z, \partial\Omega) > e^{-c}\}$ is bounded.
Since $\overline{\{\phi < c\}}$ is closed, bounded, and contained in $\Omega$, it is a compact subset of $\Omega$. Therefore $\{\phi < c\} \subset\subset \Omega$ for every $c$.
[/step]
[step:Conclude that $\Omega$ is pseudoconvex]
The function $\phi = -\log d(\cdot, \partial\Omega)$ is continuous, plurisubharmonic on $\Omega$, and satisfies $\{\phi < c\} \subset\subset \Omega$ for every $c \in \mathbb{R}$. By definition, $\phi$ is a continuous psh exhaustion of $\Omega$, so $\Omega$ is pseudoconvex.
[/step]
