[guided]The goal of this step is to show that the convex domain $\Omega$ is pseudoconvex, meaning it admits a continuous plurisubharmonic exhaustion function. We appeal to the [Convex Domains Are Pseudoconvex](/theorems/3408) theorem, but the argument is worth seeing in full because it reveals the geometric mechanism connecting convexity to plurisubharmonicity.
**Why should convexity imply pseudoconvexity?** Pseudoconvexity asks for a plurisubharmonic function that "blows up" at the boundary. The natural candidate is $-\log d(\cdot, \partial\Omega)$, the negative logarithm of the distance to the boundary. The key insight is that convexity of $\Omega$ forces the distance function $d(\cdot, \partial\Omega)$ to be concave, and composing a concave function with $-\log$ (which is convex and decreasing) produces a convex function — and convex functions on $\mathbb{R}^{2n}$ are automatically plurisubharmonic.
**Step 1: Concavity of the distance function.** Define
\begin{align*}
d: \Omega &\to (0, \infty) \\
z &\mapsto d(z, \partial\Omega) := \inf_{w \in \partial\Omega} |z - w|.
\end{align*}
We claim that $d$ is concave on $\Omega$: for any $z, w \in \Omega$ and $t \in [0,1]$,
\begin{align*}
d((1-t)z + tw, \partial\Omega) \geq (1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega).
\end{align*}
Why does this hold? The open ball $B(z, d(z, \partial\Omega))$ is contained in $\Omega$, and so is $B(w, d(w, \partial\Omega))$. The Minkowski combination $(1-t)\,B(z, d(z, \partial\Omega)) + t\,B(w, d(w, \partial\Omega))$ equals the ball $B((1-t)z + tw,\; (1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega))$. Since $\Omega$ is convex, this Minkowski combination of two subsets of $\Omega$ remains in $\Omega$. Therefore the distance from $(1-t)z + tw$ to $\partial\Omega$ is at least the radius of this ball, which is $(1-t)\,d(z, \partial\Omega) + t\,d(w, \partial\Omega)$. This is exactly the concavity inequality.
**Step 2: Plurisubharmonicity of $-\log d$.** Define
\begin{align*}
\phi: \Omega &\to \mathbb{R} \\
z &\mapsto -\log d(z, \partial\Omega).
\end{align*}
Since $d(\cdot, \partial\Omega)$ is concave and positive on $\Omega$, and $-\log$ is a convex decreasing function on $(0, \infty)$, the composition $\phi = (-\log) \circ d(\cdot, \partial\Omega)$ is convex on $\Omega$ (this is the standard composition rule: convex decreasing composed with concave gives convex). A convex function on an open subset of $\mathbb{R}^{2n}$ is subharmonic along every affine two-dimensional slice. In particular, $\phi$ is subharmonic along every complex line $\{z_0 + \zeta v : \zeta \in \mathbb{C}\}$ for any $z_0 \in \Omega$ and direction $v \in \mathbb{C}^n$. This is precisely the definition of plurisubharmonicity. Hence $\phi$ is plurisubharmonic on $\Omega$.
**Step 3: The exhaustion problem and the $|z|^2$ correction.** A plurisubharmonic exhaustion function must have the property that all its sublevel sets $\{\phi < c\}$ are compactly contained in $\Omega$. The sublevel sets of $\phi$ are $\{\phi < c\} = \{d(\cdot, \partial\Omega) > e^{-c}\}$. When $\Omega$ is bounded, these sets are automatically bounded (they are subsets of $\Omega$) and stay away from $\partial\Omega$ (the distance exceeds $e^{-c}$), so they are compactly contained in $\Omega$. But when $\Omega$ is unbounded — for example, a half-space — the sublevel set $\{d(\cdot, \partial\Omega) > e^{-c}\}$ extends to infinity and is not compactly contained in $\Omega$. So $\phi$ alone is not an exhaustion in general.
The fix is to add a term that forces boundedness. Define the corrected exhaustion function
\begin{align*}
\psi: \Omega &\to \mathbb{R} \\
z &\mapsto -\log d(z, \partial\Omega) + |z|^2.
\end{align*}
Why $|z|^2$? First, $|z|^2$ is smooth and strictly plurisubharmonic: its complex Hessian with respect to $z$ is the $n \times n$ identity matrix $I_n$, since $\frac{\partial^2 |z|^2}{\partial z_j \partial \bar{z}_k} = \delta_{jk}$. So $\psi$ is strictly plurisubharmonic as the sum of a plurisubharmonic function ($\phi$) and a strictly plurisubharmonic function ($|z|^2$). Second, $\psi$ is a genuine exhaustion: as $z \to \partial\Omega$, the term $-\log d(z, \partial\Omega) \to +\infty$, and as $|z| \to \infty$ within $\Omega$, the term $|z|^2 \to +\infty$. Therefore $\psi(z) \to +\infty$ whenever $z$ leaves every compact subset of $\Omega$ (whether by approaching the boundary or by going to infinity). The sublevel sets $\{\psi < c\}$ are bounded (by the $|z|^2$ term) and bounded away from $\partial\Omega$ (by the $-\log d$ term), hence compactly contained in $\Omega$.
Thus $\psi$ is a strictly plurisubharmonic exhaustion function on $\Omega$, and $\Omega$ is pseudoconvex.[/guided]
