[proofplan] We characterise plurisubharmonicity for $C^2$ functions by the positive semidefiniteness of the Levi form. The proof reduces the condition to a one-variable Laplacian computation: for each $z \in \Omega$ and $w \in \mathbb{C}^n$, we restrict $u$ to the complex line $\zeta \mapsto z + \zeta w$ and compute the Laplacian of the restriction at $\zeta = 0$ using the chain rule and Wirtinger calculus. The result is $4$ times the Levi form evaluated at $w$. Since a smooth function of one complex variable is subharmonic if and only if its Laplacian is non-negative, the equivalence follows. [/proofplan]

[step:Restrict $u$ to the complex line through $z$ in direction $w$ and compute the Laplacian at $\zeta = 0$]Fix $z \in \Omega$ and $w \in \mathbb{C}^n$, and define the restriction \begin{align*} \varphi: D &\to \mathbb{R} \\ \zeta &\mapsto u(z + \zeta w), \end{align*} where $D = \{\zeta \in \mathbb{C} : z + \zeta w \in \Omega\}$ is an open subset of $\mathbb{C}$. Since $u \in C^2(\Omega)$, the function $\varphi$ is $C^2$ on $D$. Write $\zeta = \xi + i\eta$ with $\xi, \eta \in \mathbb{R}$. By the chain rule applied to $\varphi(\zeta) = u(z_1 + \zeta w_1, \dots, z_n + \zeta w_n)$: \begin{align*} \frac{\partial \varphi}{\partial \zeta} = \sum_{j=1}^n \frac{\partial u}{\partial z_j}(z + \zeta w) \cdot w_j, \end{align*} where $\partial/\partial\zeta = \frac{1}{2}(\partial/\partial\xi - i\,\partial/\partial\eta)$ is the Wirtinger derivative. Taking the conjugate Wirtinger derivative $\partial/\partial\bar\zeta = \frac{1}{2}(\partial/\partial\xi + i\,\partial/\partial\eta)$: \begin{align*} \frac{\partial^2 \varphi}{\partial\bar\zeta\,\partial\zeta} = \sum_{j=1}^n \sum_{k=1}^n \frac{\partial^2 u}{\partial\bar{z}_k\,\partial z_j}(z + \zeta w)\, w_j\, \overline{w_k}. \end{align*} The Laplacian of $\varphi$ in the $\zeta$-variable is related to the mixed Wirtinger derivative by \begin{align*} \Delta_\zeta \varphi = \frac{\partial^2 \varphi}{\partial\xi^2} + \frac{\partial^2 \varphi}{\partial\eta^2} = 4\,\frac{\partial^2 \varphi}{\partial\bar\zeta\,\partial\zeta}. \end{align*} Evaluating at $\zeta = 0$: \begin{align*} \Delta_\zeta \varphi(0) = 4 \sum_{j,k=1}^n \frac{\partial^2 u}{\partial\bar{z}_k\,\partial z_j}(z)\, w_j\, \overline{w_k} = 4\,\mathcal{L}_u(z)(w, \bar{w}), \end{align*} where $\mathcal{L}_u(z)(w, \bar{w}) := \sum_{j,k=1}^n \frac{\partial^2 u}{\partial z_j\,\partial\bar{z}_k}(z)\, w_j\, \overline{w_k}$ is the Levi form of $u$ at $z$ evaluated on $w$.[/step]

[guided]The computation uses the Wirtinger calculus for functions of several complex variables. Write $z_j = x_j + iy_j$ and $w_j = a_j + ib_j$. The complex line $\zeta \mapsto z + \zeta w$ maps $\mathbb{C}$ into $\mathbb{C}^n$ by $\zeta \mapsto (z_1 + \zeta w_1, \dots, z_n + \zeta w_n)$. Each coordinate $z_j + \zeta w_j$ depends holomorphically on $\zeta$, so $\partial(z_j + \zeta w_j)/\partial\zeta = w_j$ and $\partial(z_j + \zeta w_j)/\partial\bar\zeta = 0$. Similarly, $\overline{z_j + \zeta w_j} = \bar{z}_j + \bar\zeta\,\bar{w}_j$ gives $\partial\bar{z}_j/\partial\bar\zeta = \bar{w}_j$ and $\partial\bar{z}_j/\partial\zeta = 0$. Applying the chain rule to $\varphi(\zeta) = u(z + \zeta w)$: \begin{align*} \frac{\partial\varphi}{\partial\zeta} &= \sum_{j=1}^n \frac{\partial u}{\partial z_j}\cdot w_j + \sum_{k=1}^n \frac{\partial u}{\partial\bar{z}_k}\cdot 0 = \sum_{j=1}^n \frac{\partial u}{\partial z_j}\cdot w_j. \end{align*} Now differentiate again with respect to $\bar\zeta$: \begin{align*} \frac{\partial^2\varphi}{\partial\bar\zeta\,\partial\zeta} &= \sum_{j=1}^n \left[\sum_{k=1}^n \frac{\partial^2 u}{\partial\bar{z}_k\,\partial z_j}\cdot \bar{w}_k \cdot w_j + \sum_{k=1}^n \frac{\partial^2 u}{\partial z_k\,\partial z_j}\cdot 0\right] = \sum_{j,k=1}^n \frac{\partial^2 u}{\partial\bar{z}_k\,\partial z_j}\, w_j\,\overline{w_k}. \end{align*} The identity $\Delta_\zeta = 4\,\partial_{\bar\zeta}\partial_\zeta$ is a standard result in one complex variable: writing $\zeta = \xi + i\eta$, one verifies $\partial_\zeta = \frac{1}{2}(\partial_\xi - i\partial_\eta)$ and $\partial_{\bar\zeta} = \frac{1}{2}(\partial_\xi + i\partial_\eta)$, so $4\partial_{\bar\zeta}\partial_\zeta = (\partial_\xi + i\partial_\eta)(\partial_\xi - i\partial_\eta) = \partial_\xi^2 + \partial_\eta^2 = \Delta_\zeta$. Therefore $\Delta_\zeta\varphi(0) = 4\sum_{j,k}\frac{\partial^2 u}{\partial z_j\,\partial\bar{z}_k}(z)\,w_j\,\overline{w_k} = 4\,\mathcal{L}_u(z)(w,\bar{w})$.[/guided]

[step:Characterise subharmonicity in one variable by non-negativity of the Laplacian for $C^2$ functions] A $C^2$ function $\varphi: D \to \mathbb{R}$ on an [open set](/page/Open%20Set) $D \subset \mathbb{C}$ is subharmonic on $D$ if and only if $\Delta_\zeta\varphi(\zeta) \geq 0$ for all $\zeta \in D$. This is the classical characterisation: the sub-mean-value property $\varphi(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \varphi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta)$ for all small $r > 0$ is equivalent to $\Delta_\zeta\varphi \geq 0$ pointwise when $\varphi$ is twice continuously differentiable. [/step]

[step:Conclude the equivalence between plurisubharmonicity and positive semidefiniteness of the Levi form]By definition, $u$ is plurisubharmonic on $\Omega$ if and only if $\varphi(\zeta) = u(z + \zeta w)$ is subharmonic in $\zeta$ for every $z \in \Omega$ and every $w \in \mathbb{C}^n$. Since $u \in C^2$, the restriction $\varphi$ is $C^2$, so by the previous step, $\varphi$ is subharmonic if and only if $\Delta_\zeta\varphi(\zeta) \geq 0$ for all $\zeta \in D$. In particular, evaluating at $\zeta = 0$ (which is the point corresponding to $z$ itself): \begin{align*} \varphi \text{ is subharmonic near } \zeta = 0 \iff \Delta_\zeta\varphi(0) \geq 0 \iff 4\,\mathcal{L}_u(z)(w, \bar{w}) \geq 0 \iff \mathcal{L}_u(z)(w, \bar{w}) \geq 0. \end{align*} Since this must hold for every $z \in \Omega$ and every $w \in \mathbb{C}^n$, and the Laplacian condition at general $\zeta$ reduces to the same Levi form condition at the point $z + \zeta w$ (by translation), we conclude: \begin{align*} u \text{ is plurisubharmonic on } \Omega \iff \mathcal{L}_u(z) \succeq 0 \text{ for all } z \in \Omega, \end{align*} where $\mathcal{L}_u(z) \succeq 0$ means $\sum_{j,k=1}^n \frac{\partial^2 u}{\partial z_j\,\partial\bar{z}_k}(z)\,w_j\,\overline{w_k} \geq 0$ for all $w \in \mathbb{C}^n$.[/step]

[guided]The final logical step is to ensure that the pointwise condition at $\zeta = 0$ is sufficient for subharmonicity of $\varphi$ on all of $D$, not just at $\zeta = 0$. This is immediate: for any $\zeta_0 \in D$, we can reparametrise by setting $z' = z + \zeta_0 w$, so $\varphi(\zeta_0 + \zeta) = u(z' + \zeta w)$. The Laplacian at $\zeta_0$ is then $4\,\mathcal{L}_u(z')(w, \bar{w})$, which is non-negative if and only if the Levi form at $z'$ is positive semidefinite on $w$. Since $z' = z + \zeta_0 w$ ranges over all of $\Omega$ as $z$ and $\zeta_0$ vary, the condition $\mathcal{L}_u \succeq 0$ everywhere on $\Omega$ guarantees $\Delta_\zeta\varphi \geq 0$ everywhere on $D$, hence $\varphi$ is subharmonic on all of $D$. Conversely, if $u$ is plurisubharmonic, then for every $z \in \Omega$ and $w \in \mathbb{C}^n$, $\varphi$ is subharmonic, hence $\Delta_\zeta\varphi(0) \geq 0$, hence $\mathcal{L}_u(z)(w,\bar{w}) \geq 0$. Since $z$ and $w$ are arbitrary, $\mathcal{L}_u(z) \succeq 0$ for all $z \in \Omega$.[/guided]