[proofplan] We prove that every convex domain $\Omega \subset \mathbb{C}^n$ is a domain of holomorphy by combining two results: the [Convex Domains Are Pseudoconvex](/theorems/3408) theorem, which establishes that convexity implies pseudoconvexity via a plurisubharmonic exhaustion function, and the [Solution of the Levi Problem](/theorems/3416), which establishes that pseudoconvexity is equivalent to the domain-of-holomorphy property. The key analytic ingredient is that the distance function $d(\cdot, \partial\Omega)$ is concave on a convex domain, so $-\log d(\cdot, \partial\Omega) + |z|^2$ furnishes a strictly plurisubharmonic exhaustion function. [/proofplan] [step:Establish pseudoconvexity of $\Omega$ via the convexity hypothesis] By the [Convex Domains Are Pseudoconvex](/theorems/3408) theorem, every convex domain $\Omega \subset \mathbb{C}^n$ admits a continuous plurisubharmonic exhaustion function. Specifically, the function \begin{align*} \phi: \Omega &\to \mathbb{R} \\ z &\mapsto -\log d(z, \partial\Omega) \end{align*} is plurisubharmonic on $\Omega$. The proof of this uses the concavity of $d(\cdot, \partial\Omega)$ on convex domains: for $z, w \in \Omega$ 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*} which holds because convexity of $\Omega$ ensures that the Minkowski combination of two balls contained in $\Omega$ remains in $\Omega$. Composing with $-\log$ (a convex decreasing function) yields convexity of $-\log d(\cdot, \partial\Omega)$, and a convex function on $\mathbb{R}^{2n}$ is subharmonic along every affine two-dimensional slice, in particular along every complex line through any point. Hence $-\log d(\cdot, \partial\Omega)$ is plurisubharmonic. When $\Omega$ is bounded, the sublevel sets $\{\phi < c\} = \{d(\cdot, \partial\Omega) > e^{-c}\}$ are compactly contained in $\Omega$, so $\phi$ is already an exhaustion. When $\Omega$ is unbounded, these sublevel sets need not be bounded (e.g., a half-space), so $\phi$ alone is not an exhaustion. To handle both cases uniformly, define the corrected exhaustion function \begin{align*} \psi: \Omega &\to \mathbb{R} \\ z &\mapsto -\log d(z, \partial\Omega) + |z|^2. \end{align*} The term $|z|^2$ is smooth and strictly plurisubharmonic (its complex Hessian is the identity matrix), so $\psi$ is strictly plurisubharmonic as the sum of a plurisubharmonic function and a strictly plurisubharmonic function. For exhaustion, observe that $\psi(z) \to +\infty$ both as $z \to \partial\Omega$ (since $-\log d(z, \partial\Omega) \to +\infty$) and as $|z| \to \infty$ within $\Omega$ (since $|z|^2 \to +\infty$). Therefore 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, and $\Omega$ is pseudoconvex. [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] [/step] [step:Apply the Levi problem equivalence to conclude $\Omega$ is a domain of holomorphy] Since $\Omega$ is pseudoconvex (admits a strictly plurisubharmonic exhaustion $\psi$), the [Solution of the Levi Problem](/theorems/3416) gives the equivalence: $\Omega$ is pseudoconvex if and only if $\Omega$ is a domain of holomorphy. Therefore $\Omega$ is a domain of holomorphy. [guided] We established in the previous step that $\Omega$ admits the strictly plurisubharmonic exhaustion function $\psi(z) = -\log d(z, \partial\Omega) + |z|^2$, so $\Omega$ is pseudoconvex. The question is: why does pseudoconvexity imply the domain-of-holomorphy property? Recall what it means for $\Omega$ to be a domain of holomorphy: there is no pair of open sets $U_1, U_2 \subset \mathbb{C}^n$ with $U_1 \subset \Omega \cap U_2$, $U_2 \not\subset \Omega$, and $U_1$ connected, such that every $f \in \mathcal{O}(\Omega)$ extends to $U_2$. Equivalently, one cannot "push" the domain outward and still extend all holomorphic functions. The [Solution of the Levi Problem](/theorems/3416) establishes the equivalence: an [open set](/page/Open%20Set) $\Omega \subset \mathbb{C}^n$ is pseudoconvex if and only if $\Omega$ is a domain of holomorphy. The forward direction (domain of holomorphy implies pseudoconvex) is elementary — the function $-\log d(z, \partial\Omega)$ is automatically plurisubharmonic on any domain of holomorphy. The deep direction is the converse: pseudoconvex implies domain of holomorphy. This was proved by Oka (for $n = 2$), Bremermann, and Norguet, and later streamlined by Hörmander using $L^2$ estimates for the $\bar\partial$-operator. We apply this theorem to $\Omega$. The hypothesis requires that $\Omega$ be a pseudoconvex open subset of $\mathbb{C}^n$. We verified this: $\psi = -\log d(\cdot, \partial\Omega) + |z|^2$ is a strictly plurisubharmonic exhaustion function on $\Omega$, which is one of the equivalent definitions of pseudoconvexity. The conclusion is that $\Omega$ is a domain of holomorphy. Why is the $|z|^2$ correction necessary? Consider the half-space $\Omega = \{z \in \mathbb{C}^n : \operatorname{Re}(z_1) > 0\}$, which is convex. Here $d(z, \partial\Omega) = \operatorname{Re}(z_1)$, so $-\log d(z, \partial\Omega) = -\log \operatorname{Re}(z_1)$. The sublevel set $\{-\log \operatorname{Re}(z_1) < c\} = \{\operatorname{Re}(z_1) > e^{-c}\}$ is unbounded in every direction except $\operatorname{Re}(z_1)$, so it is not compactly contained in $\Omega$. Adding $|z|^2$ ensures the sublevel sets are bounded, giving a genuine exhaustion. The plurisubharmonicity of $\psi$ is unaffected because $|z|^2$ is strictly plurisubharmonic. [/guided] [/step]