[proofplan] All three conditions are pointwise statements at each $z \in \Omega$, so it suffices to verify the equivalences locally. The equivalence (i) $\Leftrightarrow$ (ii) is the smooth-function criterion for plurisubharmonicity, which translates the subharmonicity of complex-line restrictions into the positivity of the Hermitian quadratic form $(\phi_{j\bar k})$. The equivalence (ii) $\Leftrightarrow$ (iii) is purely linear-algebraic: expanding $i\partial\bar\partial\phi$ in coordinates shows that its Hermitian coefficient matrix is exactly $(\phi_{j\bar k})$, and a real $(1,1)$-form is semipositive iff its coefficient matrix is positive semidefinite — that is, iff its associated Hermitian form is nonnegative. [/proofplan]

[step:Reduce the three conditions to pointwise statements at each $z \in \Omega$] Each of the three conditions in the theorem is local: condition (i) means that for every $z \in \Omega$ and every $\xi \in \mathbb{C}^n$ the slice \begin{align*} \phi_{z,\xi} : D_{z,\xi} &\to \mathbb{R},\\ \zeta &\mapsto \phi(z + \zeta \xi), \end{align*} defined on the [open set](/page/Open%20Set) $D_{z,\xi} := \{\zeta \in \mathbb{C} : z + \zeta\xi \in \Omega\} \subseteq \mathbb{C}$, is subharmonic; condition (ii) is the pointwise inequality $\mathcal{L}_\phi(z;\xi) \geq 0$; condition (iii) is the pointwise condition that $(i\partial\bar\partial\phi)(z)$ be a semipositive Hermitian form on $T_z^{1,0}\Omega \cong \mathbb{C}^n$. Hence it suffices to fix $z \in \Omega$ and verify the three pointwise statements are equivalent. [/step]

[step:Apply the Levi form criterion to obtain (i) $\Leftrightarrow$ (ii)]Since $\Omega \subseteq \mathbb{C}^n$ is open and $\phi \in C^2(\Omega;\mathbb{R})$, the hypotheses of the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403) are satisfied: complex-line slices are defined on open subsets of $\mathbb{C}$, and the second complex partial derivatives of $\phi$ exist and are continuous. Therefore $\phi$ is plurisubharmonic on $\Omega$ if and only if its Levi form is positive semidefinite at every point, i.e.\ \begin{align*} \mathcal{L}_\phi(z;\xi) \;=\; \sum_{j,k=1}^n \phi_{j\bar k}(z)\, \xi_j\, \bar\xi_k \;\geq\; 0 \qquad \text{for all } z \in \Omega,\ \xi \in \mathbb{C}^n. \end{align*} This is exactly the equivalence (i) $\Leftrightarrow$ (ii).[/step]

[guided]The openness of $\Omega$ matters here: for each $z \in \Omega$ and $\xi \in \mathbb{C}^n$, the set $D_{z,\xi} := \{\zeta \in \mathbb{C} : z + \zeta\xi \in \Omega\}$ is open in $\mathbb{C}$, so the complex-line restriction $\zeta \mapsto \phi(z + \zeta\xi)$ is defined on a planar domain. Plurisubharmonicity is defined by subharmonicity of all such complex-line restrictions. To get an infinitesimal criterion we differentiate twice in $\zeta$ at $\zeta = 0$: a direct calculation gives \begin{align*} \frac{1}{4}\Delta_\zeta \bigl[ \phi(z + \zeta\xi) \bigr]\Big|_{\zeta = 0} \;=\; \frac{\partial^2}{\partial \zeta\, \partial \bar\zeta}\,\phi(z + \zeta \xi)\Big|_{\zeta = 0} \;=\; \sum_{j,k=1}^n \phi_{j\bar k}(z)\, \xi_j\, \bar\xi_k \;=\; \mathcal{L}_\phi(z;\xi), \end{align*} where we used $\partial_\zeta = \sum_j \xi_j \partial_{z_j}$, $\partial_{\bar\zeta} = \sum_k \bar\xi_k\, \partial_{\bar z_k}$ on the slice. For a $C^2$ function on a planar domain, subharmonicity is equivalent to nonnegativity of the Laplacian. Combining these — and quantifying over $\xi$ — gives the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403): $\phi$ is psh iff $\mathcal{L}_\phi(z;\xi) \geq 0$ for all $(z,\xi)$. This is the content of (i) $\Leftrightarrow$ (ii). The smoothness hypothesis $\phi \in C^2$ is exactly the hypothesis needed by the cited criterion; no regularization is required.[/guided]

[step:Expand $i\partial\bar\partial\phi$ in holomorphic coordinates] We compute $i\partial\bar\partial\phi$ in the global coordinates $(z_1,\dots,z_n)$ on $\Omega \subseteq \mathbb{C}^n$. The Dolbeault operators are \begin{align*} \partial \phi \;=\; \sum_{j=1}^n \frac{\partial \phi}{\partial z_j}\, dz_j, \qquad \bar\partial \phi \;=\; \sum_{k=1}^n \frac{\partial \phi}{\partial \bar z_k}\, d\bar z_k. \end{align*} Applying $\partial$ to $\bar\partial \phi$ and using $\partial(d\bar z_k) = 0$ gives \begin{align*} \partial\bar\partial \phi \;=\; \sum_{j,k=1}^n \frac{\partial^2 \phi}{\partial z_j\, \partial \bar z_k}\, dz_j \wedge d\bar z_k \;=\; \sum_{j,k=1}^n \phi_{j\bar k}\, dz_j \wedge d\bar z_k. \end{align*} Hence \begin{align*} i\partial\bar\partial \phi \;=\; i \sum_{j,k=1}^n \phi_{j\bar k}\, dz_j \wedge d\bar z_k. \end{align*} The coefficient matrix $H(z) := (\phi_{j\bar k}(z))_{j,k=1}^n$ is Hermitian: since $\phi$ is real-valued, the equality of mixed partials gives \begin{align*} \overline{\phi_{j\bar k}(z)} \;=\; \overline{\frac{\partial^2 \phi}{\partial z_j\, \partial \bar z_k}(z)} \;=\; \frac{\partial^2 \phi}{\partial \bar z_j\, \partial z_k}(z) \;=\; \frac{\partial^2 \phi}{\partial z_k\, \partial \bar z_j}(z) \;=\; \phi_{k\bar j}(z), \end{align*} so $H(z)^* = H(z)$. In particular $i\partial\bar\partial\phi$ is a real $(1,1)$-form. [/step]

[step:Identify semipositivity of $i\partial\bar\partial\phi$ with positive semidefiniteness of $(\phi_{j\bar k})$]A real $(1,1)$-form on $\Omega$ written in coordinates as $\omega = i \sum_{j,k} h_{j\bar k}\, dz_j \wedge d\bar z_k$ with $(h_{j\bar k})$ Hermitian is, by definition, **semipositive at $z$** iff the associated Hermitian quadratic form $\xi \mapsto \sum_{j,k} h_{j\bar k}(z)\, \xi_j \bar\xi_k$ on $\mathbb{C}^n$ is nonnegative. We verify that this definition agrees with evaluation on $(1,0)$-tangent vectors. For $\xi \in \mathbb{C}^n$ set \begin{align*} X_\xi \;:=\; \sum_{j=1}^n \xi_j\, \frac{\partial}{\partial z_j} \;\in\; T_z^{1,0}\Omega, \qquad \overline{X_\xi} \;:=\; \sum_{k=1}^n \bar\xi_k\, \frac{\partial}{\partial \bar z_k} \;\in\; T_z^{0,1}\Omega. \end{align*} Here $\overline{X_\xi}$ denotes the conjugate $(0,1)$-tangent vector determined by $X_\xi$. Using $dz_j(\partial_{z_l}) = \delta_{jl}$, $d\bar z_k(\partial_{\bar z_l}) = \delta_{kl}$, $dz_j(\partial_{\bar z_l}) = 0$, and the antisymmetry of the wedge product: \begin{align*} (dz_j \wedge d\bar z_k)(X_\xi, \overline{X_\xi}) \;=\; dz_j(X_\xi)\, d\bar z_k(\overline{X_\xi}) - dz_j(\overline{X_\xi})\, d\bar z_k(X_\xi) \;=\; \xi_j\, \bar\xi_k - 0 \;=\; \xi_j\, \bar\xi_k. \end{align*} Therefore \begin{align*} -i\, \omega(z)(X_\xi, \overline{X_\xi}) \;=\; \sum_{j,k=1}^n h_{j\bar k}(z)\, \xi_j\, \bar\xi_k. \end{align*} The factor $-i$ converts the form-value (an imaginary number times $i$) into the real Hermitian form, and confirms that **semipositivity of $\omega$ is equivalent to positive semidefiniteness of the matrix $(h_{j\bar k})$**. Specialising to $\omega = i\partial\bar\partial\phi$ — for which $h_{j\bar k} = \phi_{j\bar k}$ by the expansion above — we obtain \begin{align*} -i\, (i\partial\bar\partial\phi)(z)(X_\xi, \overline{X_\xi}) \;=\; \sum_{j,k=1}^n \phi_{j\bar k}(z)\, \xi_j\, \bar\xi_k \;=\; \mathcal{L}_\phi(z;\xi). \end{align*} Hence $i\partial\bar\partial\phi$ is semipositive at $z$ if and only if $\mathcal{L}_\phi(z;\xi) \geq 0$ for every $\xi \in \mathbb{C}^n$. Quantifying over $z$, this is the equivalence (ii) $\Leftrightarrow$ (iii).[/step]

[guided]The point of this step is purely linear-algebraic: a real $(1,1)$-form is a section of $\Lambda^{1,1}T^*\Omega \cap \Lambda^2_{\mathbb{R}}T^*\Omega$, and at each point such a form is determined by an $n \times n$ Hermitian matrix of coefficients via $\omega = i\sum h_{j\bar k}\, dz_j \wedge d\bar z_k$. The convention "$\omega$ is semipositive" should match the convention that the Kähler form $i\sum dz_j \wedge d\bar z_j$ (which has $h_{j\bar k} = \delta_{jk}$) is positive — and indeed for $n = 1$, \begin{align*} i\, dz \wedge d\bar z \;=\; i\,(dx + i\,dy) \wedge (dx - i\,dy) \;=\; i\, ({-2i})\, dx\wedge dy \;=\; 2\, dx \wedge dy, \end{align*} which is the positive standard volume form on $\mathbb{C}$. Now fix $\xi \in \mathbb{C}^n$ and define the corresponding tangent vectors by \begin{align*} X_\xi \;:=\; \sum_{j=1}^n \xi_j\, \frac{\partial}{\partial z_j} \;\in\; T_z^{1,0}\Omega, \qquad \overline{X_\xi} \;:=\; \sum_{k=1}^n \bar\xi_k\, \frac{\partial}{\partial \bar z_k} \;\in\; T_z^{0,1}\Omega. \end{align*} Here $\overline{X_\xi}$ denotes the conjugate $(0,1)$-tangent vector determined by $X_\xi$. Using $dz_j(\partial_{z_l}) = \delta_{jl}$, $d\bar z_k(\partial_{\bar z_l}) = \delta_{kl}$, $dz_j(\partial_{\bar z_l}) = 0$, and $d\bar z_k(\partial_{z_l}) = 0$, the wedge-product evaluation is \begin{align*} (dz_j \wedge d\bar z_k)(X_\xi, \overline{X_\xi}) &= dz_j(X_\xi)\, d\bar z_k(\overline{X_\xi}) - dz_j(\overline{X_\xi})\, d\bar z_k(X_\xi)\\ &= \xi_j\, \bar\xi_k - 0\\ &= \xi_j\, \bar\xi_k. \end{align*} Therefore \begin{align*} -i\,\omega(z)(X_\xi, \overline{X_\xi}) = \sum_{j,k=1}^n h_{j\bar k}(z)\,\xi_j\,\bar\xi_k. \end{align*} Thus evaluating $\omega$ on $(X_\xi, \overline{X_\xi})$ and multiplying by $-i$ recovers the Hermitian form attached to $(h_{j\bar k})$. Hence the geometric condition, semipositivity of $\omega$ in complex-tangent evaluation, and the algebraic condition, positive semidefiniteness of $(h_{j\bar k})$, are the same. For $\omega = i\partial\bar\partial\phi$, Step 3 identified the coefficient matrix as $(\phi_{j\bar k})$. The Hermitian quadratic form built from this matrix is, by inspection, the Levi form $\mathcal{L}_\phi(z;\xi)$. So semipositivity of $i\partial\bar\partial\phi$ at $z$ is literally the same statement as $\mathcal{L}_\phi(z;\xi) \geq 0$ for all $\xi$ — there is no further analytic content to extract.[/guided]

[step:Combine the two equivalences] Step 2 gives (i) $\Leftrightarrow$ (ii) and Step 4 gives (ii) $\Leftrightarrow$ (iii). Composing them yields the three-way equivalence (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii), which is the statement of the theorem. $\blacksquare$ [/step]