[proofplan] The forward direction is a direct construction: when $\Omega \neq \mathbb{C}^n$, the function $\varphi(z) := |z|^2 - \log \delta_\Omega(z)$ is a sum of plurisubharmonic functions (the first strictly so), is continuous because $\delta_\Omega$ is continuous and positive, and its sublevel sets are bounded and bounded away from $\partial \Omega$, hence compact in $\Omega$. The reverse direction uses the Kontinuitätssatz for continuous families of analytic disks: a continuous plurisubharmonic exhaustion keeps every such family in $\Omega$ when the initial disk lies in $\Omega$ and the boundaries stay in a fixed compact subset of $\Omega$. Applying this disk-family property to the perturbed disks $\zeta \mapsto a + \zeta b + t e^{-(p(\zeta)+\varepsilon)} w$ gives the polynomial-domination inequality for $\delta_\Omega$ along complex lines; letting $\varepsilon \downarrow 0$ yields the subharmonicity of $-\log \delta_\Omega$ on every complex line and hence its plurisubharmonicity. [/proofplan] [step:Reduce the forward direction to the case $\Omega \neq \mathbb{C}^n$] Assume (1). If $\Omega = \mathbb{C}^n$, set \begin{align*} \varphi : \mathbb{C}^n &\to \mathbb{R} \\ z &\mapsto |z|^2. \end{align*} Then $\varphi$ is smooth and the Levi form $\partial \bar\partial \varphi = \sum_{j=1}^n dz_j \wedge d\bar z_j$ is positive definite, so $\varphi$ is (strictly) plurisubharmonic. For each $c \in \mathbb{R}$, the sublevel set $\{|z|^2 \le c\}$ is the closed Euclidean ball of radius $\sqrt{c}$ (or empty if $c < 0$), which is compact in $\mathbb{C}^n = \Omega$. Hence $\varphi$ is a continuous plurisubharmonic exhaustion, proving (2) in this case. From now on assume $\Omega \neq \mathbb{C}^n$, so $\partial \Omega \neq \varnothing$ and $\delta_\Omega(z) \in (0, \infty)$ for every $z \in \Omega$. [/step] [step:Construct $\varphi(z) = |z|^2 - \log \delta_\Omega(z)$ and verify plurisubharmonicity] Define \begin{align*} \varphi : \Omega &\to \mathbb{R} \\ z &\mapsto |z|^2 - \log \delta_\Omega(z). \end{align*} The function $z \mapsto \delta_\Omega(z)$ is continuous on $\Omega$ (it is $1$-Lipschitz: $|\delta_\Omega(z) - \delta_\Omega(z')| \le |z - z'|$ for all $z, z' \in \Omega$, by the triangle inequality applied to the infimum defining the distance) and strictly positive on $\Omega$. Hence $-\log \delta_\Omega$ is continuous on $\Omega$, and $\varphi$ is continuous. By hypothesis, $-\log \delta_\Omega$ is plurisubharmonic on $\Omega$. The function $z \mapsto |z|^2 = \sum_{j=1}^n |z_j|^2$ is smooth with $\partial \bar\partial |z|^2 = \sum_j dz_j \wedge d\bar z_j > 0$, hence is plurisubharmonic. By the [Stability Properties of PSH Functions](/theorems/3404), the sum of two plurisubharmonic functions is plurisubharmonic, so $\varphi$ is plurisubharmonic on $\Omega$. [/step] [step:Show $\varphi$ is an exhaustion of $\Omega$] Fix $c \in \mathbb{R}$ and set $K_c := \{z \in \Omega : \varphi(z) \le c\}$. We show $K_c$ is compact and contained in $\Omega$. Continuity of $\varphi$ already gives that $K_c$ is closed in $\Omega$; we must show it is also bounded in $\mathbb{C}^n$ and bounded away from $\partial \Omega$. **Bounded away from $\partial \Omega$.** For $z \in K_c$, \begin{align*} -\log \delta_\Omega(z) \;=\; \varphi(z) - |z|^2 \;\le\; c - |z|^2 \;\le\; c, \end{align*} so $\delta_\Omega(z) \ge e^{-c} > 0$. Hence $K_c \subseteq \{z \in \Omega : \delta_\Omega(z) \ge e^{-c}\}$, which is closed in $\mathbb{C}^n$ and contained in $\Omega$. **Bounded in $\mathbb{C}^n$.** Fix any $z_0 \in \partial \Omega$ (which exists since $\Omega \neq \mathbb{C}^n$). For $z \in \Omega$ we have $\delta_\Omega(z) \le |z - z_0| \le |z| + |z_0|$, so \begin{align*} \log \delta_\Omega(z) \;\le\; \log(|z| + |z_0|). \end{align*} Therefore for $z \in K_c$, \begin{align*} |z|^2 \;=\; \varphi(z) + \log \delta_\Omega(z) \;\le\; c + \log(|z| + |z_0|). \end{align*} Since $|z|^2 - \log(|z| + |z_0|) \to +\infty$ as $|z| \to \infty$, there exists $R_c < \infty$ depending on $c$ and $z_0$ such that $|z| \le R_c$ for all $z \in K_c$. Combining, $K_c$ is a closed and bounded subset of $\mathbb{C}^n$ contained in the [open set](/page/Open%20Set) $\{\delta_\Omega \ge e^{-c}\} \subseteq \Omega$. By the [Heine–Borel theorem](/theorems/313), $K_c$ is compact in $\mathbb{C}^n$, and since $K_c \subseteq \Omega$, it is compact in $\Omega$. Thus $\varphi$ is an exhaustion, completing the proof of (1) $\Rightarrow$ (2). [/step] [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] [step:Translate the disk property into subharmonicity of $-\log \delta_\Omega$ along complex lines] To show $-\log \delta_\Omega$ is plurisubharmonic on $\Omega$, we use the following characterization (a definition of plurisubharmonicity, see [Stability Properties of PSH Functions](/theorems/3404)): an upper semicontinuous function $u : \Omega \to [-\infty, \infty)$ is plurisubharmonic iff for every $a \in \Omega$ and every $b \in \mathbb{C}^n$, the slice \begin{align*} \zeta \;\longmapsto\; u(a + \zeta b) \end{align*} is subharmonic on the open subset $\{\zeta \in \mathbb{C} : a + \zeta b \in \Omega\}$ of $\mathbb{C}$. The function $-\log \delta_\Omega$ is continuous on $\Omega$ (Step 2), hence upper semicontinuous. We verify the slice condition. Fix $a \in \Omega$ and $b \in \mathbb{C}^n \setminus \{0\}$ (the case $b = 0$ is trivial: the slice is the constant $-\log \delta_\Omega(a)$). Pick $r > 0$ with $\overline{D_r} := \{a + \zeta b : |\zeta| \le r\} \subseteq \Omega$. Define the parameter disk and its closure by \begin{align*} \mathbb{D}_r &:= \{\zeta \in \mathbb{C} : |\zeta| < r\}, & \overline{\mathbb{D}_r} &:= \{\zeta \in \mathbb{C} : |\zeta| \le r\}. \end{align*} A continuous function $u : \overline{\mathbb{D}_r} \to \mathbb{R}$ is subharmonic on $\mathbb{D}_r$ iff for every polynomial $p \in \mathbb{C}[\zeta]$ with $u(\zeta) \le \operatorname{Re} p(\zeta)$ on $|\zeta| = r$, the same inequality holds on $|\zeta| \le r$. Indeed, the forward implication follows by applying the maximum principle to the subharmonic function $u - \operatorname{Re} p$. Conversely, real parts of complex polynomials give the harmonic extensions of trigonometric polynomials on $\partial\mathbb{D}_r$, and trigonometric polynomials are uniformly dense in continuous real-valued boundary data; therefore the polynomial majorant condition implies the usual harmonic-majorant characterization of subharmonicity. So fix a polynomial $p \in \mathbb{C}[\zeta]$ satisfying \begin{align*} -\log \delta_\Omega(a + \zeta b) \;\le\; \operatorname{Re} p(\zeta) \qquad \text{for all } |\zeta| = r, \end{align*} i.e. $\delta_\Omega(a + \zeta b) \ge e^{-\operatorname{Re} p(\zeta)} = |e^{-p(\zeta)}|$ on $|\zeta| = r$. We must show \begin{align*} \delta_\Omega(a + \zeta b) \;\ge\; |e^{-p(\zeta)}| \qquad \text{for all } |\zeta| \le r. \tag{$\ast$} \end{align*} Fix $\varepsilon > 0$ and define the shifted polynomial $p_\varepsilon \in \mathbb{C}[\zeta]$ by $p_\varepsilon(\zeta) := p(\zeta) + \varepsilon$. For each unit vector $w \in \mathbb{C}^n$ with $|w| = 1$ and each $t \in [0,1]$, define \begin{align*} \Delta_{w,\varepsilon,t} : \overline{\mathbb{D}_r} &\to \mathbb{C}^n \\ \zeta &\mapsto a + \zeta b + t e^{-p_\varepsilon(\zeta)} w. \end{align*} The combined map $(t,\zeta) \mapsto \Delta_{w,\varepsilon,t}(\zeta)$ is continuous on $[0,1] \times \overline{\mathbb{D}_r}$, and for each fixed $t$ the map $\Delta_{w,\varepsilon,t}$ is holomorphic on $\mathbb{D}_r$. On the boundary $|\zeta| = r$, \begin{align*} |\Delta_{w,\varepsilon,t}(\zeta) - (a + \zeta b)| &= t |e^{-p_\varepsilon(\zeta)}| |w| \\ &\le e^{-\varepsilon} |e^{-p(\zeta)}| \\ &\le e^{-\varepsilon} \delta_\Omega(a + \zeta b) \\ &< \delta_\Omega(a + \zeta b). \end{align*} Thus $\Delta_{w,\varepsilon,t}(\zeta)$ lies in the open ball centered at $a + \zeta b$ with radius $\delta_\Omega(a + \zeta b)$, which is contained in $\Omega$ by the definition of the boundary distance. Hence the boundary image \begin{align*} \{\Delta_{w,\varepsilon,t}(\zeta) : t \in [0,1],\ |\zeta| = r\} \end{align*} is a compact subset of $\Omega$. Also $\Delta_{w,\varepsilon,0}(\overline{\mathbb{D}_r}) = \overline{D_r} \subseteq \Omega$. After the affine change of variables $\zeta = r\tau$, define \begin{align*} A_{w,\varepsilon} : [0,1] \times \overline{\mathbb{D}} &\to \mathbb{C}^n \\ (t,\tau) &\mapsto a + r\tau b + t e^{-p_\varepsilon(r\tau)} w. \end{align*} The preceding paragraph verifies the hypotheses of the Kontinuitätssatz for continuous disk families proved in Step 4. Therefore \begin{align*} a + \zeta b + t e^{-p_\varepsilon(\zeta)} w \in \Omega \end{align*} for every $|\zeta| \le r$, every $t \in [0,1]$, and every unit vector $w \in \mathbb{C}^n$. Fix $|\zeta| \le r$. Since every vector $v \in \mathbb{C}^n$ with $|v| \le |e^{-p_\varepsilon(\zeta)}|$ can be written as $v = t e^{-p_\varepsilon(\zeta)}w$ for some $t \in [0,1]$ and some unit vector $w \in \mathbb{C}^n$ (with $t=0$ when $v=0$), the closed ball of radius $|e^{-p_\varepsilon(\zeta)}|$ around $a + \zeta b$ is contained in $\Omega$. Hence \begin{align*} \delta_\Omega(a + \zeta b) \ge |e^{-p_\varepsilon(\zeta)}| = e^{-\varepsilon}|e^{-p(\zeta)}|. \end{align*} Letting $\varepsilon \downarrow 0$ gives ($\ast$). [/step] [step:Conclude plurisubharmonicity of $-\log \delta_\Omega$ and finish the equivalence] By Step 5, for every $a \in \Omega$, every $b \in \mathbb{C}^n$, and every $r > 0$ with $\{a + \zeta b : |\zeta| \le r\} \subseteq \Omega$, the function \begin{align*} \zeta \;\longmapsto\; -\log \delta_\Omega(a + \zeta b) \end{align*} satisfies the polynomial-harmonic-majorant criterion on $\{|\zeta| \le r\}$ and is therefore subharmonic on $\{|\zeta| < r\}$. Define the slice domain \begin{align*} S_{a,b} := \{\zeta \in \mathbb{C} : a + \zeta b \in \Omega\}. \end{align*} The affine map $\zeta \mapsto a + \zeta b$ is continuous and $\Omega$ is open, so $S_{a,b}$ is open in $\mathbb{C}$. For any $\zeta_0 \in S_{a,b}$, choose $\rho > 0$ such that $\{\zeta_0 + \eta : |\eta| \le \rho\} \subseteq S_{a,b}$. Applying Step 5 with base point $a_0 := a + \zeta_0 b$, direction $b$, and parameter variable $\eta$ shows that $\eta \mapsto -\log \delta_\Omega(a_0 + \eta b)$ is subharmonic on $\{|\eta| < \rho\}$. Equivalently, $\zeta \mapsto -\log \delta_\Omega(a + \zeta b)$ is subharmonic in a neighborhood of $\zeta_0$. Since $\zeta_0$ was arbitrary, the slice is subharmonic on its full domain. Combined with the continuity (hence upper semicontinuity) of $-\log \delta_\Omega$ on $\Omega$ (Step 2), the slice characterization of plurisubharmonicity invoked in Step 5 gives that $-\log \delta_\Omega$ is plurisubharmonic on $\Omega$. Hence $\Omega$ is pseudoconvex, establishing (2) $\Rightarrow$ (1). The two implications together yield the equivalence (1) $\Leftrightarrow$ (2), completing the proof. $\blacksquare$ [/step]