Prove that $u \circ f $ is plurisubharmonic on $\Omega_1$ I'm trying to show that the theorem in my book:

Let $f: \Omega_1 \to \Omega_2$ be a holomorphic map between open sub - sets $\Omega_1, \Omega_2$ of $\Bbb C^n$. If $u$ is plurisubharmonic on $\Omega_2$ ($u \in PSH(\Omega_2)$) then $u \circ f $ is plurisubharmonic on $\Omega_1$ ($u \in PSH(\Omega_1)$).

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Assume that $u \in C^2(\Omega_2)$. I tried to show that $\forall z \in \Omega_1, \omega \in \Bbb C^n$ then $$\sum_{j,k=1}^n \dfrac{\partial^2 (u\circ f)}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k} \ge 0 \tag I$$


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*Since $u \in PSH(\Omega_2)$ we have $$\sum_{j,k=1}^n \dfrac{\partial^2 u}{\partial z_j \partial \overline{z_k}}(z)\cdot \omega_j \overline{\omega_k} \ge 0 \tag 1$$


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How can we show that $(I)$?
Any help is always appreciated! Thanks!
 A: For $u$ twice differentiable, it's a simple consequence of the chain rule. Since the $f_m$ are holomorphic, the general
$$\frac{\partial (u\circ f)}{\partial z_j} = \sum_{m}\left(\frac{\partial u}{\partial w_m}\circ f\right)\cdot \frac{\partial f_m}{\partial z_j} + \sum_m \left(\frac{\partial u}{\partial \overline{w_m}}\circ f\right)\cdot \frac{\partial \overline{f_m}}{\partial z_j}$$
simplifies to
$$\frac{\partial (u\circ f)}{\partial z_j} = \sum_{m} \left(\frac{\partial u}{\partial w_m}\circ f\right)\cdot \frac{\partial f_m}{\partial z_j}.$$
No $\frac{\partial u}{\partial \overline{w}_m}$ occur since the components $f_m$ are holomorphic, so $\frac{\partial \overline{f_m}}{\partial z_j} = 0$. Then differentiating with respect to $\overline{z_k}$ yields
$$\frac{\partial^2(u\circ f)}{\partial z_j\partial \overline{z_k}} = \sum_{m,n} \left(\frac{\partial^2 u}{\partial w_m \partial \overline{w_n}}\circ f\right)\cdot \frac{\partial f_m}{\partial z_j}\cdot \overline{\frac{\partial f_n}{\partial z_k}},$$
where the holomorphy of $f$ causes all terms $\frac{\partial^2 u}{\partial z_j\partial z_k}$ to be annihilated, and for the matrix of the Levi form, you get
$$L_{u\circ f} = J_f^\ast\cdot  L_u\cdot J_f$$
(or with $\overline{J_f}$ instead of $J_f$, depends on which is the row and which the column index). If $L_u$ is positive semidefinite, any $A^\ast\cdot L_u\cdot A$ is also positive semidefinite.
