[proofplan] We prove four stability properties of plurisubharmonic (psh) functions: (1) the pointwise maximum of finitely many psh functions is psh, (2) a decreasing pointwise limit of psh functions is psh (or identically $-\infty$), (3) the composition $u \circ F$ with a holomorphic map $F$ is psh, and (4) the composition $\chi \circ u$ with a convex increasing function $\chi$ is psh. Each property is established by verifying the two defining conditions of plurisubharmonicity: upper semicontinuity and the sub-mean-value inequality along every complex line. [/proofplan]

[step:Prove that the pointwise maximum of finitely many psh functions is psh] Let $u_1, \dots, u_N: \Omega \to [-\infty, \infty)$ be plurisubharmonic. Define $u := \max(u_1, \dots, u_N)$. **Upper semicontinuity.** For each $c \in \mathbb{R}$, the sublevel set $\{u < c\} = \bigcap_{j=1}^N \{u_j < c\}$ is an intersection of open sets (since each $u_j$ is upper semicontinuous), hence open. Therefore $u$ is upper semicontinuous. **Sub-mean-value inequality along complex lines.** Fix $z \in \Omega$ and $w \in \mathbb{C}^n$. For any $\zeta_0 \in D := \{\zeta \in \mathbb{C} : z + \zeta w \in \Omega\}$ and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D$, define $\varphi_j(\zeta) := u_j(z + \zeta w)$. Each $\varphi_j$ is subharmonic on $D$, so \begin{align*} \varphi_j(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \varphi_j(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta) \leq \frac{1}{2\pi}\int_0^{2\pi} \max_{1 \leq k \leq N} \varphi_k(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta) \end{align*} for each $j$, where the [second inequality](/theorems/2136) uses $\varphi_j \leq \max_k \varphi_k$ pointwise. Let $j_0$ be the index achieving $\max_j \varphi_j(\zeta_0) = \varphi_{j_0}(\zeta_0)$. Then \begin{align*} \max_j \varphi_j(\zeta_0) = \varphi_{j_0}(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \max_k \varphi_k(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta), \end{align*} which is the sub-mean-value inequality for $\zeta \mapsto u(z + \zeta w)$. [/step]

[step:Prove that a decreasing pointwise limit of psh functions is psh] Let $(u_j)_{j \geq 1}$ be a sequence of psh functions on $\Omega$ with $u_1 \geq u_2 \geq \cdots$ pointwise, and define $u := \lim_{j \to \infty} u_j = \inf_j u_j$. **Upper semicontinuity.** For each $c \in \mathbb{R}$, $\{u < c\} = \bigcup_{j=1}^\infty \{u_j < c\}$ (since $u_j \searrow u$ implies $u(z) < c$ iff $u_j(z) < c$ for some $j$). Each $\{u_j < c\}$ is open by upper semicontinuity of $u_j$, so $\{u < c\}$ is open, and $u$ is upper semicontinuous. **Sub-mean-value inequality along complex lines.** Fix $z \in \Omega$, $w \in \mathbb{C}^n$, $\zeta_0 \in D$, and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D$. Define $\varphi_j(\zeta) := u_j(z + \zeta w)$ and $\varphi(\zeta) := u(z + \zeta w)$. Each $\varphi_j$ is subharmonic, so \begin{align*} \varphi_j(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \varphi_j(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} Since $\varphi_j \searrow \varphi$ pointwise, the sequence $(-\varphi_j)$ increases to $(-\varphi)$. Applying the [Monotone Convergence Theorem](/theorems/509) to the non-negative functions $\varphi_1(\zeta_0 + re^{i\theta}) - \varphi_j(\zeta_0 + re^{i\theta}) \nearrow \varphi_1(\zeta_0 + re^{i\theta}) - \varphi(\zeta_0 + re^{i\theta})$ (non-negative because $\varphi_j \leq \varphi_1$), with respect to the finite measure $\frac{1}{2\pi}d\mathcal{L}^1(\theta)$ on $[0, 2\pi]$: \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} \varphi_j(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta) \to \frac{1}{2\pi}\int_0^{2\pi} \varphi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta) \end{align*} as $j \to \infty$. Passing to the limit on both sides of the sub-mean-value inequality for $\varphi_j$: \begin{align*} \varphi(\zeta_0) = \lim_{j \to \infty} \varphi_j(\zeta_0) \leq \lim_{j \to \infty} \frac{1}{2\pi}\int_0^{2\pi} \varphi_j(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta) = \frac{1}{2\pi}\int_0^{2\pi} \varphi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} [/step]

[step:Prove that the composition with a holomorphic map is psh]Let $u: \Omega \to [-\infty, \infty)$ be psh and $F: \Omega' \to \Omega$ holomorphic, where $\Omega' \subset \mathbb{C}^m$ is open. Define $v := u \circ F: \Omega' \to [-\infty, \infty)$. **Upper semicontinuity.** Since $F$ is holomorphic, it is continuous. For each $c \in \mathbb{R}$, $\{v < c\} = F^{-1}(\{u < c\})$. The set $\{u < c\}$ is open (by upper semicontinuity of $u$), and $F^{-1}$ preserves openness (by continuity of $F$), so $\{v < c\}$ is open. Therefore $v$ is upper semicontinuous. **Sub-mean-value inequality along complex lines.** Fix $z' \in \Omega'$ and $w' \in \mathbb{C}^m$. Define the holomorphic map of one complex variable \begin{align*} h: D' &\to \Omega \\ \zeta &\mapsto F(z' + \zeta w'), \end{align*} where $D' = \{\zeta \in \mathbb{C} : z' + \zeta w' \in \Omega'\}$, and set $\psi := u \circ h: D' \to [-\infty, \infty)$. The map $h$ is holomorphic as the composition of $\zeta \mapsto z' + \zeta w'$ (affine, hence holomorphic) with $F$ (holomorphic by hypothesis). We must show $\psi$ is subharmonic on $D'$. **Step (a): Approximate $u$ from above by smooth psh functions.** For each $j \geq 1$, define $\Omega_j := \{z \in \Omega : \operatorname{dist}(z, \partial\Omega) > 1/j\}$ and $u_j := u * \eta_{1/j}$ on $\Omega_j$, where $\eta_{1/j}$ is the standard mollifier. Each $u_j$ is smooth and psh on $\Omega_j$ (the sub-mean-value property passes through convolution with a non-negative radially symmetric kernel), and $u_j \searrow u$ pointwise as $j \to \infty$. **Step (b): Compute $\Delta_\zeta(u_j \circ h) \geq 0$.** Fix $\zeta_0 \in D'$ and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D'$. For $j$ large enough that $h(\overline{B}(\zeta_0, r)) \subset \Omega_j$, the composition $u_j \circ h$ is $C^2$ on a neighbourhood of $\overline{B}(\zeta_0, r)$. Writing $h = (h_1, \dots, h_n)$ in components, each $h_a: D' \to \mathbb{C}$ holomorphic, the chain rule gives \begin{align*} \frac{\partial^2(u_j \circ h)}{\partial\zeta\,\partial\bar{\zeta}}(\zeta) = \sum_{a,b=1}^n \frac{\partial^2 u_j}{\partial z_a\,\partial\bar{z}_b}(h(\zeta))\,h_a'(\zeta)\,\overline{h_b'(\zeta)}, \end{align*} using $\frac{\partial h_a}{\partial\bar{\zeta}} = 0$ and $\frac{\partial\bar{h}_b}{\partial\zeta} = 0$ (holomorphicity of $h$). Therefore \begin{align*} \Delta_\zeta(u_j \circ h)(\zeta) = 4\sum_{a,b=1}^n \frac{\partial^2 u_j}{\partial z_a\,\partial\bar{z}_b}(h(\zeta))\,h_a'(\zeta)\,\overline{h_b'(\zeta)} = 4\,\mathcal{L}_{u_j}(h(\zeta))(h'(\zeta), \overline{h'(\zeta)}) \geq 0, \end{align*} where the inequality holds because $u_j$ is smooth and psh, so by the [Levi Form Criterion](/theorems/3403) its Levi form $\mathcal{L}_{u_j} \succeq 0$ everywhere. Since $\Delta_\zeta(u_j \circ h) \geq 0$, the function $u_j \circ h$ is subharmonic on its domain. **Step (c): The sub-mean-value inequality for $u_j \circ h$.** Since $u_j \circ h$ is subharmonic: \begin{align*} u_j(h(\zeta_0)) \leq \frac{1}{2\pi}\int_0^{2\pi} u_j(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta). \end{align*} **Step (d): Pass to the limit using property (2).** On the left, $u_j(h(\zeta_0)) \searrow u(h(\zeta_0))$. On the right, $u_j(h(\zeta_0 + re^{i\theta})) \searrow u(h(\zeta_0 + re^{i\theta}))$ for each $\theta$. By the [Monotone Convergence Theorem](/theorems/509) applied to the non-negative functions $u_1(h(\zeta_0 + re^{i\theta})) - u_j(h(\zeta_0 + re^{i\theta})) \nearrow u_1(h(\zeta_0 + re^{i\theta})) - u(h(\zeta_0 + re^{i\theta}))$ with respect to the finite measure $\frac{1}{2\pi}d\mathcal{L}^1(\theta)$ on $[0, 2\pi]$: \begin{align*} \psi(\zeta_0) = u(h(\zeta_0)) = \lim_{j \to \infty} u_j(h(\zeta_0)) \leq \lim_{j \to \infty} \frac{1}{2\pi}\int_0^{2\pi} u_j(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta) = \frac{1}{2\pi}\int_0^{2\pi} \psi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} Therefore $\psi = u \circ h$ is subharmonic on $D'$, confirming that $v = u \circ F$ is psh on $\Omega'$.[/step]

[guided]We want to show that $v := u \circ F: \Omega' \to [-\infty, \infty)$ is plurisubharmonic, where $u: \Omega \to [-\infty, \infty)$ is psh and $F: \Omega' \to \Omega$ is holomorphic with $\Omega' \subset \mathbb{C}^m$ open. The essential point is that a holomorphic map $h: D' \to \Omega$ from a domain in $\mathbb{C}$ to $\mathbb{C}^n$ pulls back psh functions to subharmonic functions. Why does this work? The deep reason is that subharmonicity in one variable is characterised by the sub-mean-value inequality, and holomorphic maps preserve the relevant structure because their Jacobians interact well with the Levi form. **Upper semicontinuity.** Since $F$ is holomorphic, it is continuous. For each $c \in \mathbb{R}$, $\{v < c\} = F^{-1}(\{u < c\})$. The set $\{u < c\}$ is open by upper semicontinuity of $u$, and the preimage $F^{-1}(\{u < c\})$ is open by continuity of $F$. Therefore $\{v < c\}$ is open for every $c \in \mathbb{R}$, which means $v$ is upper semicontinuous. Note that this argument uses only the continuity of $F$; the holomorphicity is consumed in the sub-mean-value step below. **Sub-mean-value inequality along complex lines.** Fix $z' \in \Omega'$ and $w' \in \mathbb{C}^m$. We need to show that the one-variable function \begin{align*} \psi: D' &\to [-\infty, \infty) \\ \zeta &\mapsto u(F(z' + \zeta w')) \end{align*} is subharmonic on the open set $D' = \{\zeta \in \mathbb{C} : z' + \zeta w' \in \Omega'\}$. Define the holomorphic map of one complex variable \begin{align*} h: D' &\to \Omega \\ \zeta &\mapsto F(z' + \zeta w'), \end{align*} so that $\psi = u \circ h$. The map $h$ is holomorphic because it is the composition of $\zeta \mapsto z' + \zeta w'$ (an affine map $\mathbb{C} \to \mathbb{C}^m$, hence holomorphic) with $F: \Omega' \to \Omega$ (holomorphic by hypothesis). Fix $\zeta_0 \in D'$ and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D'$. Set $z_0 = h(\zeta_0) = F(z' + \zeta_0 w') \in \Omega$. We want to establish \begin{align*} \psi(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \psi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta), \end{align*} i.e., $u(h(\zeta_0)) \leq \frac{1}{2\pi}\int_0^{2\pi} u(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta)$. One natural attempt is to use the mean-value property of $h$ itself. Since $h$ is holomorphic, its real and imaginary parts are harmonic, so \begin{align*} h(\zeta_0) = \frac{1}{2\pi}\int_0^{2\pi} h(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} But this tells us about the mean of $h$, not the mean of $u \circ h$. Since $u$ is not linear, we cannot simply apply $u$ to both sides and preserve the equality. And since $h$ is not generally linear (it is holomorphic but may have nonzero higher-order terms), we cannot reduce $u \circ h$ to a restriction of $u$ along a single complex line. So a direct argument from the definition requires more care. The clean resolution is a **smooth approximation argument**. Approximate $u$ from above by a decreasing sequence of smooth psh functions $u_j \searrow u$. This is possible by convolution with a standard mollifier: if $u$ is psh on $\Omega$, define $u_j := u * \eta_{1/j}$ on the set $\Omega_j := \{z \in \Omega : \operatorname{dist}(z, \partial\Omega) > 1/j\}$. The convolution $u * \eta_{1/j}$ is smooth and psh (the sub-mean-value property passes through the convolution), and $u_j \searrow u$ pointwise as $j \to \infty$ (since psh functions are upper semicontinuous, the mollification decreases to the function). For each smooth $u_j$, the composition $u_j \circ h: D' \to \mathbb{R}$ is $C^2$ (since $u_j$ is smooth and $h$ is holomorphic, hence smooth). We can compute the Laplacian of $u_j \circ h$ directly by the chain rule. Writing $h = (h_1, \dots, h_n)$ in components, where each $h_a: D' \to \mathbb{C}$ is holomorphic: \begin{align*} \Delta_\zeta(u_j \circ h)(\zeta) = 4\frac{\partial^2(u_j \circ h)}{\partial\zeta\,\partial\bar{\zeta}}(\zeta) = 4\sum_{a,b=1}^n \frac{\partial^2 u_j}{\partial z_a\,\partial\bar{z}_b}(h(\zeta))\,h_a'(\zeta)\,\overline{h_b'(\zeta)}. \end{align*} Why does this formula hold? By the chain rule, $\frac{\partial}{\partial\zeta}(u_j \circ h) = \sum_a \frac{\partial u_j}{\partial z_a}(h(\zeta))\,h_a'(\zeta)$, using the fact that $h$ is holomorphic so $\frac{\partial h_a}{\partial\bar{\zeta}} = 0$ and $\frac{\partial\bar{h}_b}{\partial\zeta} = 0$. Differentiating again with respect to $\bar{\zeta}$: \begin{align*} \frac{\partial^2(u_j \circ h)}{\partial\zeta\,\partial\bar{\zeta}}(\zeta) = \sum_{a,b=1}^n \frac{\partial^2 u_j}{\partial z_a\,\partial\bar{z}_b}(h(\zeta))\,h_a'(\zeta)\,\overline{h_b'(\zeta)}. \end{align*} The right-hand side is exactly the Levi form of $u_j$ evaluated on the tangent vector $h'(\zeta) = (h_1'(\zeta), \dots, h_n'(\zeta)) \in \mathbb{C}^n$: \begin{align*} \Delta_\zeta(u_j \circ h)(\zeta) = 4\,\mathcal{L}_{u_j}(h(\zeta))(h'(\zeta), \overline{h'(\zeta)}) \geq 0, \end{align*} where the inequality uses the [Levi Form Criterion](/theorems/3403): since $u_j$ is smooth and psh, its Levi form $\mathcal{L}_{u_j}(z)(\xi, \bar{\xi}) = \sum_{a,b} \frac{\partial^2 u_j}{\partial z_a\,\partial\bar{z}_b}(z)\,\xi_a\,\bar{\xi}_b$ is positive semidefinite at every point $z$, and evaluating it on the vector $\xi = h'(\zeta)$ gives a non-negative result. The fact that $\Delta_\zeta(u_j \circ h) \geq 0$ means $u_j \circ h$ is subharmonic on $D'$ for each $j$. Since $u_j \circ h$ is subharmonic, the sub-mean-value inequality holds: \begin{align*} u_j(h(\zeta_0)) \leq \frac{1}{2\pi}\int_0^{2\pi} u_j(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta). \end{align*} Now we pass to the limit $j \to \infty$. On the left, $u_j(h(\zeta_0)) \searrow u(h(\zeta_0))$. On the right, $u_j(h(\zeta_0 + re^{i\theta})) \searrow u(h(\zeta_0 + re^{i\theta}))$ for each $\theta$. By the [Monotone Convergence Theorem](/theorems/509) applied to the non-negative functions $u_1(h(\zeta_0 + re^{i\theta})) - u_j(h(\zeta_0 + re^{i\theta})) \nearrow u_1(h(\zeta_0 + re^{i\theta})) - u(h(\zeta_0 + re^{i\theta}))$ with respect to the finite measure $\frac{1}{2\pi}d\mathcal{L}^1(\theta)$ on $[0, 2\pi]$, we obtain \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} u_j(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta) \to \frac{1}{2\pi}\int_0^{2\pi} u(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta). \end{align*} Passing to the limit in the sub-mean-value inequality for $u_j \circ h$: \begin{align*} \psi(\zeta_0) = u(h(\zeta_0)) = \lim_{j \to \infty} u_j(h(\zeta_0)) \leq \lim_{j \to \infty} \frac{1}{2\pi}\int_0^{2\pi} u_j(h(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta) = \frac{1}{2\pi}\int_0^{2\pi} \psi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} This confirms that $\psi = u \circ h$ is subharmonic on $D'$, and therefore $v = u \circ F$ is psh on $\Omega'$. This smooth approximation argument is the cleanest route because it avoids the need for a direct proof of the sub-mean-value inequality for the composition $u \circ h$ when $u$ is merely upper semicontinuous. The Levi form computation reveals the geometric reason: the holomorphic map $h$ transforms the one-dimensional Laplacian into a non-negative expression precisely because the Levi form of a psh function is positive semidefinite.[/guided]

[step:Prove that the composition with a convex increasing function is psh]Let $u: \Omega \to [-\infty, \infty)$ be psh and $\chi: \mathbb{R} \to \mathbb{R}$ be convex and increasing (we extend $\chi$ to $[-\infty, \infty)$ by $\chi(-\infty) := \lim_{t \to -\infty}\chi(t)$, which exists in $[-\infty, \infty)$ since $\chi$ is increasing). Define $v := \chi \circ u: \Omega \to [-\infty, \infty)$. **Upper semicontinuity.** Since $\chi$ is convex on $\mathbb{R}$, it is continuous on $\mathbb{R}$. Combined with the convention $\chi(-\infty) = \lim_{t \to -\infty}\chi(t)$, the function $\chi$ is upper semicontinuous on $[-\infty, \infty)$. The composition of an upper semicontinuous function with an increasing upper semicontinuous function is upper semicontinuous: for each $c \in \mathbb{R}$, $\{v < c\} = \{\chi(u) < c\} = \{u < \chi^{-1}(c)\}$ (interpreting $\chi^{-1}$ appropriately as a generalised inverse), which is open. **Sub-mean-value inequality along complex lines.** Fix $z \in \Omega$ and $w \in \mathbb{C}^n$. Define $\varphi(\zeta) := u(z + \zeta w)$, which is subharmonic on $D$. The sub-mean-value inequality for $\varphi$ gives \begin{align*} \varphi(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} \varphi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} Since $\chi$ is increasing: \begin{align*} \chi(\varphi(\zeta_0)) \leq \chi\!\left(\frac{1}{2\pi}\int_0^{2\pi} \varphi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta)\right). \end{align*} Since $\chi$ is convex, by [Jensen's inequality](/theorems/1977) applied to the probability measure $\frac{1}{2\pi}d\mathcal{L}^1(\theta)$ on $[0, 2\pi]$: \begin{align*} \chi\!\left(\frac{1}{2\pi}\int_0^{2\pi} \varphi(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta)\right) \leq \frac{1}{2\pi}\int_0^{2\pi} \chi(\varphi(\zeta_0 + re^{i\theta}))\,d\mathcal{L}^1(\theta). \end{align*} Combining: \begin{align*} (\chi \circ \varphi)(\zeta_0) \leq \frac{1}{2\pi}\int_0^{2\pi} (\chi \circ \varphi)(\zeta_0 + re^{i\theta})\,d\mathcal{L}^1(\theta), \end{align*} which is the sub-mean-value inequality for $\chi \circ \varphi(\zeta) = \chi(u(z + \zeta w)) = v(z + \zeta w)$.[/step]

[guided]The two ingredients are the monotonicity of $\chi$ (to pass from $\varphi(\zeta_0) \leq \text{mean}$ to $\chi(\varphi(\zeta_0)) \leq \chi(\text{mean})$) and the convexity of $\chi$ (to apply [Jensen's inequality](/theorems/9) and push $\chi$ inside the integral). Why do we need $\chi$ to be increasing, not just convex? Because the sub-mean-value inequality $\varphi(\zeta_0) \leq \frac{1}{2\pi}\int\varphi$ goes in the direction $\leq$. Applying a decreasing function would reverse this inequality, and we would get $\chi(\varphi(\zeta_0)) \geq \chi(\text{mean})$, which is the wrong direction. Why do we need $\chi$ to be convex, not just increasing? Because we need to interchange $\chi$ with the integral. An increasing function that is not convex would give \begin{align*} \chi(\varphi(\zeta_0)) \leq \chi\!\left(\frac{1}{2\pi}\int\varphi\right), \end{align*} but without convexity we have no way to bound $\chi(\frac{1}{2\pi}\int\varphi)$ by $\frac{1}{2\pi}\int\chi(\varphi)$. [Jensen's inequality](/theorems/1977) is exactly the tool that provides this bound, and it requires convexity. The application of [Jensen's inequality](/theorems/9) requires that $\varphi$ be integrable with respect to the probability measure $\frac{1}{2\pi}d\mathcal{L}^1$ on $[0, 2\pi]$. This holds because $\varphi$ is upper semicontinuous on the compact circle $\{|\zeta - \zeta_0| = r\}$, hence bounded above, and the integral $\int\varphi$ is well-defined in $[-\infty, \infty)$ (subharmonic functions are integrable over circles whenever they are not identically $-\infty$).[/guided]