[proofplan] We establish the equivalence between the two definitions of pseudoconvexity for a bounded domain $\Omega \subset \mathbb{C}^n$ with $C^2$ boundary: (i) existence of a continuous plurisubharmonic exhaustion function, and (ii) the Levi condition (positive semidefiniteness of the Levi form on the complex tangent space at every boundary point). For the direction from Levi pseudoconvexity to the exhaustion definition, we construct a global $C^2$ defining function $\rho$ for $\partial\Omega$ and show that $-\log(-\rho)$ is a psh exhaustion near the boundary by computing its complex Hessian in terms of the Levi form. For the reverse direction, we show that the sublevel sets of a psh exhaustion are Levi pseudoconvex at regular boundary points, and pass to the limit. [/proofplan] [step:Construct a global $C^2$ defining function $\rho$ for $\partial\Omega$] Since $\partial\Omega$ is compact and $C^2$, for each boundary point $p \in \partial\Omega$ there exists an open neighbourhood $V_p \subset \mathbb{C}^n$ and a $C^2$ function $\rho_p: V_p \to \mathbb{R}$ satisfying: - $\Omega \cap V_p = \{z \in V_p : \rho_p(z) < 0\}$, - $\rho_p = 0$ on $\partial\Omega \cap V_p$, - $\nabla\rho_p \neq 0$ on $\partial\Omega \cap V_p$. The collection $\{V_p\}_{p \in \partial\Omega}$ covers the compact set $\partial\Omega$, so extract a finite subcover $\{V_{p_1}, \dots, V_{p_N}\}$. Let $\{\chi_i\}_{i=1}^N$ be a $C^\infty$ [partition of unity](/page/Partition%20of%20Unity) subordinate to this cover. Define \begin{align*} \rho: \bigcup_{i=1}^N V_{p_i} &\to \mathbb{R} \\ z &\mapsto \sum_{i=1}^N \chi_i(z)\, \rho_{p_i}(z). \end{align*} On $\partial\Omega$, each $\rho_{p_i}$ vanishes, so $\rho = 0$. In $\Omega$ near $\partial\Omega$, each $\rho_{p_i} < 0$ and each $\chi_i \geq 0$, so $\rho < 0$. To verify $\nabla\rho \neq 0$ on $\partial\Omega$: at a point $z_0 \in \partial\Omega$, the gradient of $\rho$ can be computed using the product rule. Since all $\rho_{p_i}$ vanish on $\partial\Omega$ and have the same inward normal direction (they are all defining functions for the same hypersurface), the sum $\sum_i \chi_i \nabla\rho_{p_i}$ is a positive linear combination of outward normals, hence nonzero. Thus $\rho$ is a global $C^2$ defining function for $\partial\Omega$ in a neighbourhood $U$ of $\partial\Omega$. By extending $\rho$ smoothly to all of $\mathbb{C}^n$ (setting $\rho > 0$ outside $\overline{\Omega}$ and $\rho < 0$ in $\Omega$), we may assume $\rho$ is defined globally with $\Omega = \{\rho < 0\}$ and $\nabla\rho \neq 0$ on $\partial\Omega$. [guided] Why do we need a global defining function? Each local defining function $\rho_{p_i}$ characterises $\partial\Omega$ near a single boundary point, but the candidate exhaustion $-\log(-\rho)$ must be defined on all of $\Omega$ near $\partial\Omega$. A global $\rho$ ensures that $-\log(-\rho)$ is a single well-defined function, not a patchwork of local constructions. The strategy is to glue the local defining functions using a [partition of unity](/page/Partition%20of%20Unity). Since $\partial\Omega$ is compact and $C^2$, for each boundary point $p \in \partial\Omega$ there exists an open neighbourhood $V_p \subset \mathbb{C}^n$ and a $C^2$ function $\rho_p: V_p \to \mathbb{R}$ satisfying: - $\Omega \cap V_p = \{z \in V_p : \rho_p(z) < 0\}$, - $\rho_p = 0$ on $\partial\Omega \cap V_p$, - $\nabla\rho_p \neq 0$ on $\partial\Omega \cap V_p$. These three properties are the defining conditions for a local defining function of a $C^2$ hypersurface: the zero set is the boundary, the function is negative inside, and the gradient is nonvanishing (so the boundary is a regular level set by the [implicit function theorem](/page/Implicit%20Function%20Theorem)). The collection $\{V_p\}_{p \in \partial\Omega}$ covers the compact set $\partial\Omega$, so we can extract a finite subcover $\{V_{p_1}, \dots, V_{p_N}\}$. Why does compactness matter here? Because a [partition of unity](/page/Partition%20of%20Unity) subordinate to an infinite cover might not be $C^\infty$, and we need finite summation to preserve the $C^2$ regularity. Let $\{\chi_i\}_{i=1}^N$ be a $C^\infty$ [partition of unity](/page/Partition%20of%20Unity) subordinate to this finite cover, meaning $\chi_i \in C_c^\infty(V_{p_i})$, $\chi_i \geq 0$, and $\sum_{i=1}^N \chi_i = 1$ on a neighbourhood of $\partial\Omega$. Define \begin{align*} \rho: \bigcup_{i=1}^N V_{p_i} &\to \mathbb{R} \\ z &\mapsto \sum_{i=1}^N \chi_i(z)\, \rho_{p_i}(z). \end{align*} We must verify the three defining-function properties for $\rho$: **Zero set.** On $\partial\Omega$, each $\rho_{p_i}$ vanishes, so $\rho(z) = \sum_{i=1}^N \chi_i(z) \cdot 0 = 0$. Thus $\rho = 0$ on $\partial\Omega$. **Sign condition.** In $\Omega$ near $\partial\Omega$, each $\rho_{p_i}(z) < 0$ (by the local defining-function property) and each $\chi_i(z) \geq 0$, so every term $\chi_i(z)\,\rho_{p_i}(z) \leq 0$. Since $\sum_i \chi_i = 1$ and at least one $\chi_i(z) > 0$ with $\rho_{p_i}(z) < 0$, the sum $\rho(z) < 0$. **Non-vanishing gradient.** This is the most delicate verification. At a point $z_0 \in \partial\Omega$, we apply the product rule to each term $\chi_i \rho_{p_i}$: \begin{align*} \nabla\rho(z_0) = \sum_{i=1}^N \chi_i(z_0)\,\nabla\rho_{p_i}(z_0) + \sum_{i=1}^N \nabla\chi_i(z_0)\,\underbrace{\rho_{p_i}(z_0)}_{=0} = \sum_{i=1}^N \chi_i(z_0)\,\nabla\rho_{p_i}(z_0). \end{align*} The second sum vanishes because every $\rho_{p_i}(z_0) = 0$ on $\partial\Omega$. Why is the remaining sum nonzero? The partition-of-unity construction works because on the overlaps $V_{p_i} \cap V_{p_j} \cap \partial\Omega$, the local defining functions $\rho_{p_i}$ and $\rho_{p_j}$ both vanish and have gradients pointing in the same outward direction (since they define the same hypersurface with the same orientation). More precisely, each $\nabla\rho_{p_i}(z_0)$ is a positive scalar multiple of the unit outward normal to $\partial\Omega$ at $z_0$. Since $\sum_i \chi_i = 1$ and $\chi_i \geq 0$, the sum $\sum_{i=1}^N \chi_i(z_0)\,\nabla\rho_{p_i}(z_0)$ is a convex combination of vectors all pointing in the same direction, hence nonzero. Thus $\nabla\rho \neq 0$ on $\partial\Omega$. This confirms that $\rho$ is a global $C^2$ defining function for $\partial\Omega$ in a neighbourhood $U$ of $\partial\Omega$. By extending $\rho$ smoothly to all of $\mathbb{C}^n$ (setting $\rho > 0$ outside $\overline{\Omega}$ and $\rho < 0$ in $\Omega$), we may assume $\rho$ is defined globally with $\Omega = \{\rho < 0\}$ and $\nabla\rho \neq 0$ on $\partial\Omega$. Such an extension exists by standard smooth extension results: one can multiply by a smooth cutoff function that equals $1$ near $\partial\Omega$ and then add a term like $\operatorname{dist}(z, \partial\Omega)$ with appropriate sign outside $\overline{\Omega}$. [/guided] [/step] [step:Show that Levi pseudoconvexity implies $-\log(-\rho)$ is psh near $\partial\Omega$] Define the candidate exhaustion function \begin{align*} \phi: \Omega &\to \mathbb{R} \\ z &\mapsto -\log(-\rho(z)). \end{align*} Since $\rho < 0$ on $\Omega$ and $\rho(z) \to 0^-$ as $z \to \partial\Omega$, the function $\phi(z) \to +\infty$ as $z \to \partial\Omega$, so $\phi$ has the correct boundary blow-up for an exhaustion. We compute the complex Hessian of $\phi$. Set $h = -\rho > 0$ on $\Omega$, so $\phi = -\log h$. By the chain rule applied to $-\log$: \begin{align*} \frac{\partial^2 \phi}{\partial z_j\,\partial\bar{z}_k} = -\frac{1}{h}\frac{\partial^2 h}{\partial z_j\,\partial\bar{z}_k} + \frac{1}{h^2}\frac{\partial h}{\partial z_j}\frac{\partial h}{\partial\bar{z}_k}. \end{align*} Since $h = -\rho$, we have $\frac{\partial^2 h}{\partial z_j\,\partial\bar{z}_k} = -\frac{\partial^2 \rho}{\partial z_j\,\partial\bar{z}_k}$ and $\frac{\partial h}{\partial z_j} = -\frac{\partial \rho}{\partial z_j}$. Substituting: \begin{align*} \frac{\partial^2 \phi}{\partial z_j\,\partial\bar{z}_k} = \frac{1}{h}\frac{\partial^2 \rho}{\partial z_j\,\partial\bar{z}_k} + \frac{1}{h^2}\frac{\partial \rho}{\partial z_j}\frac{\partial \rho}{\partial\bar{z}_k}. \end{align*} The Levi form of $\phi$ evaluated on $w \in \mathbb{C}^n$ is therefore \begin{align*} \mathcal{L}_\phi(z)(w,\bar{w}) = \frac{1}{h(z)}\sum_{j,k=1}^n \frac{\partial^2\rho}{\partial z_j\,\partial\bar{z}_k}(z)\,w_j\,\overline{w_k} + \frac{1}{h(z)^2}\left|\sum_{j=1}^n \frac{\partial\rho}{\partial z_j}(z)\,w_j\right|^2. \end{align*} The second term is always $\geq 0$. For the first term, consider a point $z$ near $\partial\Omega$ and a vector $w$ in the complex tangent space to the level set $\{\rho = \rho(z)\}$, i.e., $\sum_j \frac{\partial\rho}{\partial z_j}(z)\,w_j = 0$. By the Levi pseudoconvexity hypothesis, the Levi form of $\rho$ is positive semidefinite on the complex tangent space at every boundary point. Since the level sets $\{\rho = c\}$ for $c$ close to $0$ are $C^2$-close to $\partial\Omega$, and the Levi form depends continuously on the base point, the Levi form of $\rho$ is positive semidefinite on the complex tangent space to $\{\rho = c\}$ for all $c$ in a neighbourhood of $0$. For a general vector $w \in \mathbb{C}^n$, decompose $w = w_\parallel + w_\perp$ where $w_\parallel$ is tangent to the level set (i.e., $\sum_j \frac{\partial\rho}{\partial z_j}\,w_{\parallel,j} = 0$) and $w_\perp$ is proportional to $\nabla\rho$. The Levi form of $\rho$ applied to $w_\parallel$ is $\geq 0$ by the Levi condition, and the second term $\frac{1}{h^2}|\sum_j \partial_{z_j}\rho\,w_j|^2$ handles the normal component. Both terms are non-negative, so $\mathcal{L}_\phi(z)(w,\bar{w}) \geq 0$ for all $w$, proving $\phi$ is psh near $\partial\Omega$. [guided] The key computation is the complex Hessian of $-\log(-\rho)$. The formula produces two terms: (1) $\frac{1}{h}\mathcal{L}_\rho$, which is the Levi form of $\rho$ scaled by $1/h$, and (2) $\frac{1}{h^2}|\partial\rho \cdot w|^2$, which is a non-negative rank-one correction. The Levi condition guarantees that term (1) is non-negative on vectors $w$ satisfying $\sum_j \partial_{z_j}\rho\,w_j = 0$, and term (2) is non-negative on all vectors. Together, they give $\mathcal{L}_\phi \geq 0$ on all of $\mathbb{C}^n$. Why does the argument work near $\partial\Omega$ but potentially fail deep inside $\Omega$? Because the Levi condition is a statement about boundary points, and the level sets $\{\rho = c\}$ approximate $\partial\Omega$ only for $c \approx 0$. Deep inside $\Omega$, the level sets of $\rho$ may have nothing to do with $\partial\Omega$, and the Levi form of $\rho$ need not be positive semidefinite on their tangent spaces. This is why the exhaustion $\phi$ is only guaranteed to be psh in a neighbourhood of $\partial\Omega$. To obtain a global psh exhaustion on all of $\Omega$, one can modify $\phi$ in the interior. Choose a compact set $K \subset \Omega$ such that $\phi$ is psh on $\Omega \setminus K$. Since $K$ is compact and contained in $\Omega$, one can find a smooth function $\psi$ that is psh on $\Omega$ and satisfies $\psi > \phi$ on $K$ (for instance, a large constant). The function $\max(\phi, \psi)$, after smoothing, gives a global psh exhaustion. [/guided] [/step] [step:Extend $\phi$ to a global psh exhaustion on $\Omega$] The function $\phi = -\log(-\rho)$ is psh on a neighbourhood $V$ of $\partial\Omega$ within $\Omega$, and $\phi(z) \to +\infty$ as $z \to \partial\Omega$. Choose $c_0 \in \mathbb{R}$ large enough that $\{z \in \Omega : \phi(z) \geq c_0\} \subset V$. On the compact set $K = \{z \in \Omega : \phi(z) \leq c_0 + 1\}$, the function $\phi$ is bounded. Choose a smooth convex increasing function $\chi: \mathbb{R} \to \mathbb{R}$ with $\chi(t) = t$ for $t \geq c_0$ and $\chi$ convex everywhere. Then $\chi \circ \phi$ is psh wherever $\phi$ is psh (since $\chi$ is convex and increasing, and the composition of a convex increasing function with a psh function is psh). On the interior region $\{\phi < c_0\}$, which is compactly contained in $\Omega$, set $\Phi = \max(\chi \circ \phi, C)$ for a suitable constant $C$, and smooth the maximum. The result is a smooth psh exhaustion function on all of $\Omega$. The sublevel sets $\{\Phi < c\}$ are compactly contained in $\Omega$ for every $c \in \mathbb{R}$: for $c$ large, $\{\Phi < c\}$ avoids $\partial\Omega$ because $\Phi \to +\infty$ there; for $c$ small, $\{\Phi < c\}$ is contained in the compact set $K$. Therefore $\Omega$ is pseudoconvex in the exhaustion sense. [/step] [step:Show that the existence of a psh exhaustion implies the Levi condition at boundary points] Suppose $\Omega$ admits a $C^2$ psh exhaustion $\phi$. For each regular value $c$ of $\phi$, the sublevel set $\Omega_c = \{\phi < c\}$ has $C^2$ boundary $\partial\Omega_c = \{\phi = c\}$ with defining function $\phi - c$. At a boundary point $z_0 \in \partial\Omega_c$, the Levi form of $\phi$ is positive semidefinite on all of $\mathbb{C}^n$ (since $\phi$ is psh). In particular, it is positive semidefinite on the complex tangent space $T_{z_0}^{\mathbb{C}}\partial\Omega_c = \{w \in \mathbb{C}^n : \sum_j \frac{\partial\phi}{\partial z_j}(z_0)\,w_j = 0\}$. This means $\Omega_c$ is Levi pseudoconvex. Now take an increasing sequence of regular values $c_m \nearrow \sup_\Omega \phi$ with $\Omega_{c_m} \nearrow \Omega$. Since $\Omega$ has $C^2$ boundary, for each boundary point $p \in \partial\Omega$ and each complex tangent vector $w \in T_p^{\mathbb{C}}\partial\Omega$, there exist sequences $z_m \in \partial\Omega_{c_m}$ converging to $p$ and vectors $w_m \to w$ with $w_m \in T_{z_m}^{\mathbb{C}}\partial\Omega_{c_m}$. The Levi form of $\rho$ at the level set $\{\rho = \rho(z_m)\}$ can be related to the Levi form of $\phi$ by the chain rule. Since $\partial\Omega_{c_m}$ is a level set of $\phi$ and $\partial\Omega$ is a level set of $\rho$, and both $\phi$ and $\rho$ are $C^2$, the Levi forms are related by a continuous transformation. The Levi pseudoconvexity of each $\Omega_{c_m}$ gives $\mathcal{L}_\rho(z_m)(w_m, \bar{w}_m) \geq 0$, and passing to the limit $m \to \infty$ by continuity of the Levi form in both the base point and the vector yields \begin{align*} \mathcal{L}_\rho(p)(w, \bar{w}) \geq 0 \end{align*} for all $p \in \partial\Omega$ and $w \in T_p^{\mathbb{C}}\partial\Omega$. This is the Levi condition. [guided] The subtlety in this direction is that the psh exhaustion $\phi$ need not equal $-\log(-\rho)$ or have any particular relationship to the defining function $\rho$. Nevertheless, plurisubharmonicity of $\phi$ is a pointwise condition ($\mathcal{L}_\phi \succeq 0$), and this condition restricts the Levi form on each level set. The key logical chain is: 1. $\phi$ is psh, so $\mathcal{L}_\phi(z)(w, \bar{w}) \geq 0$ for all $z \in \Omega$ and all $w \in \mathbb{C}^n$. 2. At a regular level set $\{\phi = c\}$, the Levi form of the defining function $\phi - c$ restricted to the complex tangent space is exactly $\mathcal{L}_\phi$ restricted to that subspace, which is $\geq 0$. 3. The exhaustion property $\phi \to +\infty$ as $z \to \partial\Omega$ ensures that the level sets $\{\phi = c\}$ approach $\partial\Omega$. 4. The $C^2$ regularity of $\partial\Omega$ allows us to transfer the Levi condition from the level sets of $\phi$ to $\partial\Omega$ by a continuity argument. What could go wrong without $C^2$ boundary? If $\partial\Omega$ had corners or cusps, the Levi form of $\rho$ would not be defined at the singular boundary points, and the limiting argument in step 4 would fail. The exhaustion definition would still hold, but one could not express it as a pointwise Levi condition. [/guided] [/step] [step:Conclude the equivalence] Combining the two directions: if $\Omega$ is Levi pseudoconvex, then by the construction in the previous steps, $\phi = -\log(-\rho)$ extends to a psh exhaustion of $\Omega$, so $\Omega$ is pseudoconvex in the exhaustion sense. Conversely, if $\Omega$ admits a psh exhaustion, the Levi condition holds at every boundary point by the limiting argument. Therefore, for bounded domains with $C^2$ boundary, the two definitions of pseudoconvexity are equivalent. [/step]