[proofplan] The forward implication is immediate: the affine map $\zeta \mapsto a + \zeta v$ is holomorphic, so composition with $\phi$ inherits subharmonicity from the disc-invariance definition of plurisubharmonicity. The reverse implication splits into two parts. We first treat smooth $\phi$, where slice subharmonicity at each point in each direction is precisely non-negativity of the Levi form at that point in that direction, and the Levi Form Criterion converts this to plurisubharmonicity. For general upper semicontinuous $\phi$, we approximate by truncating from below at $-M$ and convolving with a smooth radial mollifier; the resulting approximants are smooth and inherit slice subharmonicity from $\phi$, hence are plurisubharmonic by the smooth case. Stability of plurisubharmonic functions under decreasing limits then transfers the property to $\phi$. [/proofplan]

[step:Forward direction via composition with affine holomorphic maps]Suppose $\phi$ is plurisubharmonic on $\Omega$. Fix $a \in \Omega$ and $v \in \mathbb{C}^n$. The map \begin{align*} \iota_{a,v}: S_{a,v} &\to \Omega \\ \zeta &\mapsto a + \zeta v \end{align*} is holomorphic on $S_{a,v}$ (its components are affine in $\zeta$). Let $V$ be any connected component of $S_{a,v}$. Then $\iota_{a,v}|_V$ is a holomorphic map from $V \subseteq \mathbb{C}$ into $\Omega$. By the definition of plurisubharmonicity, $\phi \circ \iota_{a,v}|_V = \phi_{a,v}|_V$ is subharmonic on $V$ (where, as standard, "subharmonic" allows the degenerate case of being identically $-\infty$ on $V$).[/step]

[guided]The disc-invariance definition of plurisubharmonicity says: $\phi$ is plurisubharmonic on $\Omega$ if and only if $\phi$ is upper semicontinuous and $\phi \circ g$ is subharmonic on $U$ for every open $U \subseteq \mathbb{C}$ and every holomorphic $g: U \to \Omega$. So the forward direction reduces to exhibiting affine slices as a special class of holomorphic maps. Fix $a$ and $v$, and let \begin{align*} \iota_{a,v}: S_{a,v} &\to \Omega \\ \zeta &\mapsto a + \zeta v. \end{align*} Each component $(\iota_{a,v})_j(\zeta) = a_j + \zeta v_j$ is a polynomial in $\zeta$ of degree $\le 1$, hence holomorphic. The domain $S_{a,v}$ is open because it is the preimage of $\Omega$ under the continuous map $\zeta \mapsto a + \zeta v$. On each connected component $V$ of $S_{a,v}$, the restriction $\iota_{a,v}|_V: V \to \Omega$ is holomorphic, so the definition applies and gives $\phi \circ \iota_{a,v}|_V = \phi_{a,v}|_V$ subharmonic on $V$.[/guided]

[step:Reverse direction, smooth case via the Levi form]Assume $\phi \in C^2(\Omega)$ and that $\phi$ has the slice subharmonicity property. Fix $a \in \Omega$ and $v \in \mathbb{C}^n$. Because $\phi \in C^2$, the slice $\phi_{a,v}: S_{a,v} \to \mathbb{R}$ is $C^2$ near $\zeta = 0$, and the chain rule for Wirtinger derivatives gives \begin{align*} \frac{\partial^2 \phi_{a,v}}{\partial \zeta \, \partial \bar\zeta}(0) &= \sum_{j,k=1}^n \frac{\partial^2 \phi}{\partial z_j \, \partial \bar z_k}(a)\, v_j\, \overline{v_k} = L_\phi(a)(v, v), \end{align*} where $L_\phi(a)$ denotes the Levi form of $\phi$ at $a$. Subharmonicity of $\phi_{a,v}$ on the component of $S_{a,v}$ containing $0$ implies $\Delta_\zeta \phi_{a,v}(0) \ge 0$, and $\Delta_\zeta = 4 \partial_\zeta \partial_{\bar\zeta}$, so \begin{align*} L_\phi(a)(v, v) = \tfrac{1}{4} \Delta_\zeta \phi_{a,v}(0) \ge 0. \end{align*} Since $a \in \Omega$ and $v \in \mathbb{C}^n$ were arbitrary, the Levi form of $\phi$ is non-negative on all of $\Omega$. By the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403), $\phi$ is plurisubharmonic on $\Omega$.[/step]

[guided]For a $C^2$ function the Levi form \begin{align*} L_\phi(a)(v, w) = \sum_{j,k=1}^n \frac{\partial^2 \phi}{\partial z_j \, \partial \bar z_k}(a)\, v_j\, \overline{w_k} \end{align*} is a Hermitian form on $\mathbb{C}^n$, and the Levi Form Criterion (theorem 3403) says: a $C^2$ function on $\Omega$ is plurisubharmonic if and only if $L_\phi(a)(v, v) \ge 0$ for all $a \in \Omega$, $v \in \mathbb{C}^n$. So we just need to deduce non-negativity of the Levi form from slice subharmonicity. The link is the Wirtinger chain rule. For $f(\zeta) = a + \zeta v$, we have $\partial f_j / \partial \zeta = v_j$ and $\partial f_j / \partial \bar\zeta = 0$, so for any $C^2$ function $\phi$ on $\Omega$: \begin{align*} \frac{\partial}{\partial \zeta}\bigl[\phi(a + \zeta v)\bigr] &= \sum_j \frac{\partial \phi}{\partial z_j}(a + \zeta v)\, v_j, \\ \frac{\partial^2}{\partial \bar\zeta \, \partial \zeta}\bigl[\phi(a + \zeta v)\bigr] &= \sum_{j,k} \frac{\partial^2 \phi}{\partial \bar z_k \, \partial z_j}(a + \zeta v)\, v_j\, \overline{v_k} = L_\phi(a + \zeta v)(v, v). \end{align*} Evaluating at $\zeta = 0$: \begin{align*} \tfrac{1}{4}\Delta_\zeta \phi_{a,v}(0) = \frac{\partial^2 \phi_{a,v}}{\partial \bar\zeta \, \partial \zeta}(0) = L_\phi(a)(v, v). \end{align*} Now subharmonicity of $\phi_{a,v}$ at $0$ — concretely, the local sub-mean-value inequality — forces $\Delta_\zeta \phi_{a,v}(0) \ge 0$ because $\phi_{a,v}$ is $C^2$ (a $C^2$ subharmonic function has non-negative Laplacian, since otherwise a Taylor expansion would produce a strict super-mean-value at a small radius). Now subharmonicity of $\phi_{a,v}$ at $0$ — concretely, the local sub-mean-value inequality — forces $\Delta_\zeta \phi_{a,v}(0) \ge 0$. Indeed, since $\phi_{a,v}$ is $C^2$ near $0$, Taylor expansion in polar coordinates gives \begin{align*} \frac{1}{2\pi}\int_0^{2\pi}\phi_{a,v}(re^{i\theta})\,d\mathcal{L}^1(\theta) = \phi_{a,v}(0) + \frac{r^2}{4}\Delta_\zeta\phi_{a,v}(0) + o(r^2) \quad \text{as } r \to 0^+, \end{align*} since the linear terms vanish upon angular averaging and the second-order terms collect as $\frac{r^2}{4}\Delta_\zeta\phi_{a,v}(0)$. If $\Delta_\zeta\phi_{a,v}(0) < 0$, the right-hand side would be strictly less than $\phi_{a,v}(0)$ for all sufficiently small $r > 0$, contradicting the sub-mean-value inequality. Hence $L_\phi(a)(v, v) \ge 0$. As $a, v$ are arbitrary, the Levi form is non-negative semidefinite at every point. Theorem 3403 now closes the loop: a $C^2$ function with everywhere non-negative semidefinite Levi form is plurisubharmonic.[/guided]

[step:Truncate from below to control the polar set]Now treat general upper semicontinuous $\phi$ satisfying the slice subharmonicity hypothesis. For each $M > 0$, define \begin{align*} \phi^M: \Omega &\to \mathbb{R} \\ z &\mapsto \max\{\phi(z), -M\}. \end{align*} Then $\phi^M$ is upper semicontinuous (the maximum of two upper semicontinuous functions is upper semicontinuous), takes values in $[-M, \infty)$, and is therefore locally bounded. For any $a \in \Omega$, $v \in \mathbb{C}^n$, and any component $V$ of $S_{a,v}$: - If $\phi_{a,v}|_V$ is subharmonic, then $\max\{\phi_{a,v}|_V, -M\}$ is subharmonic on $V$, since the maximum of a subharmonic function and a constant is subharmonic. - If $\phi_{a,v}|_V \equiv -\infty$, then $\max\{\phi_{a,v}|_V, -M\} \equiv -M$, which is subharmonic because every real-valued constant function satisfies the sub-mean-value inequality with equality. In either case, $(\phi^M)_{a,v}|_V$ is subharmonic. Thus $\phi^M$ is upper semicontinuous, locally bounded, and slice-subharmonic in the strong sense (subharmonic on every component, no degenerate cases).[/step]

[guided]The mollification step ahead requires the integrand to be locally integrable. The raw function $\phi$ may take the value $-\infty$ on a large set in principle (the slice hypothesis controls $\{\phi = -\infty\}$ only along complex lines), and $\int \phi(z - w) \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w)$ need not be well-defined a priori. Truncation from below sidesteps this issue cleanly: $\phi^M$ is bounded below by $-M$ and bounded above locally by upper semicontinuity, so it is locally bounded, hence in $L^\infty_{\mathrm{loc}}(\Omega)$. We will mollify $\phi^M$, send $\varepsilon \to 0$ to get plurisubharmonicity of $\phi^M$, then send $M \to \infty$ to recover plurisubharmonicity of $\phi$. The check that $\phi^M$ inherits slice subharmonicity uses two stability facts for subharmonic functions of one complex variable: (i) the maximum of a subharmonic function and a constant is subharmonic (because the constant is itself subharmonic and the maximum of two subharmonic functions is subharmonic), and (ii) constant functions are subharmonic. The first handles components where $\phi_{a,v}$ is genuinely subharmonic; the second handles components where $\phi_{a,v} \equiv -\infty$ (the slice hypothesis allows this exceptional case, and on such a component $\max\{-\infty, -M\} = -M$ is constant).[/guided]

[step:Mollify $\phi^M$ and inherit slice subharmonicity]Choose a non-negative radial mollifier $\eta \in C_c^\infty(B(0,1))$ on $\mathbb{C}^n \cong \mathbb{R}^{2n}$ with $\int_{\mathbb{C}^n} \eta\, d\mathcal{L}^{2n} = 1$, and set $\eta_\varepsilon(w) = \varepsilon^{-2n} \eta(w/\varepsilon)$. Let \begin{align*} \Omega_\varepsilon &:= \{z \in \Omega : \operatorname{dist}(z, \partial \Omega) > \varepsilon\} \end{align*} and define \begin{align*} \phi^M_\varepsilon: \Omega_\varepsilon &\to \mathbb{R} \\ z &\mapsto \int_{B(0,\varepsilon)} \phi^M(z - w)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w). \end{align*} Because $\phi^M$ is locally bounded, the integral converges absolutely and $\phi^M_\varepsilon \in C^\infty(\Omega_\varepsilon)$ by the standard properties of convolution with a $C_c^\infty$ kernel. [claim:$\phi^M_\varepsilon$ is slice-subharmonic on $\Omega_\varepsilon$] For every $a \in \Omega_\varepsilon$ and $v \in \mathbb{C}^n$, the affine slice $\zeta \mapsto \phi^M_\varepsilon(a + \zeta v)$ is subharmonic on every connected component of $\{\zeta \in \mathbb{C} : a + \zeta v \in \Omega_\varepsilon\}$. [/claim] [proof] Fix $a \in \Omega_\varepsilon$, $v \in \mathbb{C}^n$, and a component $V$ of $\{\zeta : a + \zeta v \in \Omega_\varepsilon\}$. For each $w \in B(0,\varepsilon)$, define \begin{align*} \psi_w: V &\to \mathbb{R} \\ \zeta &\mapsto \phi^M(a - w + \zeta v). \end{align*} For $\zeta \in V$ we have $a + \zeta v \in \Omega_\varepsilon$ and $|w| < \varepsilon$, so $a - w + \zeta v \in \Omega$; thus $\psi_w$ is well-defined on $V$. Moreover $\psi_w$ is the restriction of the affine slice $(\phi^M)_{a - w, v}$ to the [open set](/page/Open%20Set) $V \subseteq S_{a-w, v}$. Since $V$ is connected and contained in $S_{a-w, v}$, it lies in a single component of $S_{a-w, v}$, on which $(\phi^M)_{a-w, v}$ is subharmonic by Step 3. Restriction of a subharmonic function to an open subset is subharmonic, so $\psi_w$ is subharmonic on $V$. By a translation of integration variables ($w \leftrightarrow$ same $w$, no change), for $\zeta \in V$, \begin{align*} \phi^M_\varepsilon(a + \zeta v) &= \int_{B(0,\varepsilon)} \phi^M\bigl((a + \zeta v) - w\bigr)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w) = \int_{B(0,\varepsilon)} \psi_w(\zeta)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w). \end{align*} Fix a closed disc $\overline{B}(\zeta_0, r) \subset V$. For each $w$, the sub-mean-value inequality for the subharmonic function $\psi_w$ on $V$ gives \begin{align*} \psi_w(\zeta_0) \le \frac{1}{2\pi} \int_0^{2\pi} \psi_w(\zeta_0 + r e^{i\theta})\, d\mathcal{L}^1(\theta). \end{align*} Multiplying by $\eta_\varepsilon(w) \ge 0$, integrating over $w \in B(0,\varepsilon)$, and applying [Fubini's theorem](/theorems/2961) (upper semicontinuity of $\phi^M$ implies Borel measurability, so $(w, \theta) \mapsto \psi_w(\zeta_0 + re^{i\theta})\, \eta_\varepsilon(w)$ is Borel-measurable on $B(0,\varepsilon) \times [0, 2\pi]$; it is bounded above by $\|\eta_\varepsilon\|_\infty\sup_K \phi^M$ and below by $-M\|\eta_\varepsilon\|_\infty$, where $K := a + \overline{B}(\zeta_0,r)v - \overline{B}(0,\varepsilon) \subset \Omega$ is compact, so it is absolutely integrable): \begin{align*} \phi^M_\varepsilon(a + \zeta_0 v) &= \int_{B(0,\varepsilon)} \psi_w(\zeta_0)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w) \\ &\le \int_{B(0,\varepsilon)} \frac{1}{2\pi} \int_0^{2\pi} \psi_w(\zeta_0 + r e^{i\theta})\, d\mathcal{L}^1(\theta)\, \eta_\varepsilon(w)\, d\mathcal{L}^{2n}(w) \\ &= \frac{1}{2\pi} \int_0^{2\pi} \phi^M_\varepsilon(a + (\zeta_0 + r e^{i\theta}) v)\, d\mathcal{L}^1(\theta). \end{align*} This is the sub-mean-value inequality for $\zeta \mapsto \phi^M_\varepsilon(a + \zeta v)$ at $\zeta_0$ on radius $r$. Combined with continuity of $\phi^M_\varepsilon$ (hence upper semicontinuity) and that $\phi^M_\varepsilon$ is not identically $-\infty$ (it is real-valued), this proves $(\phi^M_\varepsilon)_{a,v}|_V$ is subharmonic. [/proof] By the smooth case (Step 2), $\phi^M_\varepsilon$ is plurisubharmonic on $\Omega_\varepsilon$.[/step]

[guided]First we verify that the convolution is a legitimate smooth function. The truncated function \begin{align*} \phi^M: \Omega &\to \mathbb{R} \\ z &\mapsto \max\{\phi(z),-M\} \end{align*} is locally bounded by the preceding step. The mollifier is the smooth compactly supported function \begin{align*} \eta_\varepsilon: \mathbb{C}^n &\to [0,\infty) \\ w &\mapsto \varepsilon^{-2n}\eta(w/\varepsilon), \end{align*} with support contained in $B(0,\varepsilon)$ and total integral $1$ with respect to $\mathcal{L}^{2n}$. Therefore the map \begin{align*} \phi^M_\varepsilon: \Omega_\varepsilon &\to \mathbb{R} \\ z &\mapsto \int_{B(0,\varepsilon)} \phi^M(z-w)\,\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w) \end{align*} is well-defined and real-valued. Standard convolution regularity applies because the kernel is $C_c^\infty$ and the integrand is locally bounded on every compact subset of $\Omega$; hence $\phi^M_\varepsilon \in C^\infty(\Omega_\varepsilon)$. Now fix $a \in \Omega_\varepsilon$, $v \in \mathbb{C}^n$, and a connected component $V$ of \begin{align*} \{\zeta \in \mathbb{C}: a+\zeta v \in \Omega_\varepsilon\}. \end{align*} For each $w \in B(0,\varepsilon)$ define the translated slice \begin{align*} \psi_w: V &\to \mathbb{R} \\ \zeta &\mapsto \phi^M(a-w+\zeta v). \end{align*} This is well-defined because if $\zeta \in V$, then $a+\zeta v \in \Omega_\varepsilon$, so $\operatorname{dist}(a+\zeta v,\partial\Omega)>\varepsilon$; since $|w|<\varepsilon$, the point $a-w+\zeta v$ lies in $\Omega$. Moreover $V \subseteq S_{a-w,v}$, because every $\zeta \in V$ satisfies $a-w+\zeta v\in\Omega$. Since $V$ is connected, it is contained in one connected component of $S_{a-w,v}$. The preceding truncation step says that $(\phi^M)_{a-w,v}$ is subharmonic on that component, and restriction of a subharmonic function to an open subset remains subharmonic. Thus $\psi_w$ is subharmonic on $V$ for every $w \in B(0,\varepsilon)$. For each $\zeta \in V$, the definition of convolution gives the identity \begin{align*} \phi^M_\varepsilon(a+\zeta v) &= \int_{B(0,\varepsilon)} \phi^M((a+\zeta v)-w)\,\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w) \\ &= \int_{B(0,\varepsilon)} \psi_w(\zeta)\,\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w). \end{align*} So the slice of the mollified function is a non-negative weighted average of the subharmonic functions $\psi_w$. To prove subharmonicity directly, take a closed disc $\overline{B}(\zeta_0,r)\subset V$. For each fixed $w\in B(0,\varepsilon)$, the sub-mean-value inequality for $\psi_w$ gives \begin{align*} \psi_w(\zeta_0) \le \frac{1}{2\pi}\int_0^{2\pi}\psi_w(\zeta_0+re^{i\theta})\,d\mathcal{L}^1(\theta). \end{align*} Multiplying by the non-negative weight $\eta_\varepsilon(w)$ and integrating over $B(0,\varepsilon)$ yields \begin{align*} \int_{B(0,\varepsilon)}\psi_w(\zeta_0)\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w) \le \int_{B(0,\varepsilon)}\frac{1}{2\pi}\int_0^{2\pi}\psi_w(\zeta_0+re^{i\theta})\,d\mathcal{L}^1(\theta)\,\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w). \end{align*} We may apply [Fubini's theorem](/theorems/2961) to the right-hand side. Indeed, upper semicontinuity of $\phi^M$ implies Borel measurability, so the map \begin{align*} (w,\theta) \mapsto \psi_w(\zeta_0+re^{i\theta})\eta_\varepsilon(w) \end{align*} is Borel-measurable on $B(0,\varepsilon)\times[0,2\pi]$. It is bounded below by $-M\|\eta_\varepsilon\|_\infty$. It is bounded above by $\|\eta_\varepsilon\|_\infty\sup_K\phi^M$, where \begin{align*} K := a+\overline{B}(\zeta_0,r)v-\overline{B}(0,\varepsilon) \subset \Omega \end{align*} is compact and $\sup_K\phi^M<\infty$ by upper semicontinuity on a compact set. Hence the integrand is absolutely integrable with respect to $\mathcal{L}^{2n}\otimes\mathcal{L}^1$. Fubini gives \begin{align*} \phi^M_\varepsilon(a+\zeta_0v) &= \int_{B(0,\varepsilon)}\psi_w(\zeta_0)\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w) \\ &\le \frac{1}{2\pi}\int_0^{2\pi}\int_{B(0,\varepsilon)}\psi_w(\zeta_0+re^{i\theta})\eta_\varepsilon(w)\,d\mathcal{L}^{2n}(w)\,d\mathcal{L}^1(\theta) \\ &= \frac{1}{2\pi}\int_0^{2\pi}\phi^M_\varepsilon(a+(\zeta_0+re^{i\theta})v)\,d\mathcal{L}^1(\theta). \end{align*} This is exactly the sub-mean-value inequality for the slice $\zeta\mapsto\phi^M_\varepsilon(a+\zeta v)$ on $V$. Since $\phi^M_\varepsilon$ is smooth, the slice is continuous, hence upper semicontinuous, and it is real-valued, so it is not identically $-\infty$. Therefore the slice is subharmonic on $V$. We have proved that $\phi^M_\varepsilon$ is smooth and slice-subharmonic on $\Omega_\varepsilon$. Applying the smooth case already proved, equivalently the [Levi Form Criterion for Smooth PSH Functions](/theorems/3403) through Step 2, gives that $\phi^M_\varepsilon$ is plurisubharmonic on $\Omega_\varepsilon$.[/guided]

[step:Pass to the limit $\varepsilon \to 0$ to obtain plurisubharmonicity of $\phi^M$]Fix $M > 0$. We claim $\phi^M_\varepsilon \searrow \phi^M$ pointwise on $\Omega$ as $\varepsilon \searrow 0$ (where for each fixed $z \in \Omega$ the statement is interpreted for $\varepsilon < \operatorname{dist}(z, \partial \Omega)$). For a radial mollifier $\eta_\varepsilon$, write $\eta(w) = h(|w|)$ for some $h \in C_c^\infty([0,1))$ with $h \ge 0$. Switching to polar coordinates on $\mathbb{R}^{2n}$ with $w = r u$, $r \ge 0$, $u \in S^{2n-1}$, the polar formula $d\mathcal{L}^{2n}(w) = r^{2n-1}\, d\mathcal{H}^{2n-1}(u)\, d\mathcal{L}^1(r)$ gives \begin{align*} \phi^M_\varepsilon(z) = \int_0^\varepsilon \varepsilon^{-2n} h(r/\varepsilon)\, r^{2n-1} \biggl( \int_{S^{2n-1}} \phi^M(z - r u)\, d\mathcal{H}^{2n-1}(u) \biggr) d\mathcal{L}^1(r). \end{align*} Set $\sigma_{2n-1} := \mathcal{H}^{2n-1}(S^{2n-1})$ and define the spherical mean \begin{align*} \mathcal{M}(\phi^M, \cdot, \cdot): \{(z, r) : z \in \Omega,\ 0 < r < \operatorname{dist}(z, \partial\Omega)\} &\to \mathbb{R} \\ (z, r) &\mapsto \frac{1}{\sigma_{2n-1}} \int_{S^{2n-1}} \phi^M(z + r u)\, d\mathcal{H}^{2n-1}(u). \end{align*} By radial symmetry of $\eta$ (substituting $u \mapsto -u$), \begin{align*} \phi^M_\varepsilon(z) = \int_0^\varepsilon \varepsilon^{-2n} h(r/\varepsilon)\, r^{2n-1}\, \sigma_{2n-1}\, \mathcal{M}(\phi^M, z, r)\, d\mathcal{L}^1(r). \end{align*} The normalisation $\int_{\mathbb{C}^n} \eta_\varepsilon\, d\mathcal{L}^{2n} = 1$ rewrites as $\int_0^\varepsilon \varepsilon^{-2n} h(r/\varepsilon)\, r^{2n-1}\, \sigma_{2n-1}\, d\mathcal{L}^1(r) = 1$, so $\phi^M_\varepsilon(z)$ is a weighted average of $\mathcal{M}(\phi^M, z, \cdot)$ over $r \in (0, \varepsilon)$. [claim:Monotonicity of spherical means] For each $z \in \Omega$, the map $r \mapsto \mathcal{M}(\phi^M, z, r)$ is non-decreasing on $(0, \operatorname{dist}(z, \partial \Omega))$. [/claim] [proof] Identify $S^{2n-1} \subset \mathbb{C}^n$ as the disjoint union of circles $\{e^{i\theta} u : \theta \in [0, 2\pi)\}$ as $u$ ranges over the base of the Hopf fibration $S^{2n-1} \to \mathbb{CP}^{n-1}$ given by the $S^1$-action $\theta \cdot u = e^{i\theta} u$. Let $\nu$ denote the unique $S^1$-invariant Borel probability measure on $\mathbb{CP}^{n-1}$. Disintegration of $\mathcal{H}^{2n-1}$ along this fibration (each fibre carries arc-length measure on a circle of radius $1$, total mass $2\pi$) gives the equality \begin{align*} \mathcal{M}(\phi^M, z, r) = \int_{\mathbb{CP}^{n-1}} \biggl( \frac{1}{2\pi} \int_0^{2\pi} \phi^M(z + r e^{i\theta} u)\, d\mathcal{L}^1(\theta) \biggr) d\nu(u), \end{align*} where the normalising constants on the two sides match because $\nu$ is a probability measure and the inner integral is normalised by $1/(2\pi)$. For each fixed $u \in S^{2n-1}$, the inner integral \begin{align*} A_u(r) := \frac{1}{2\pi} \int_0^{2\pi} \phi^M(z + r e^{i\theta} u)\, d\mathcal{L}^1(\theta) \end{align*} is the circular mean at radius $r$ of the slice $\zeta \mapsto \phi^M(z + \zeta u)$ along the complex line through $z$ in direction $u$. By Step 3 (slice subharmonicity of $\phi^M$), this slice is subharmonic on the open disc $\{\zeta \in \mathbb{C} : |\zeta| < \operatorname{dist}(z, \partial \Omega)\}$ (which is contained in $S_{z, u}$); circular means of a subharmonic function are non-decreasing in the radius. Thus $r \mapsto A_u(r)$ is non-decreasing for $r < \operatorname{dist}(z, \partial \Omega)$. Integrating over $u$ against the non-negative probability measure $\nu$ preserves the monotonicity, so $r \mapsto \mathcal{M}(\phi^M, z, r)$ is non-decreasing. [/proof] For fixed $z$, the weighting in $\phi^M_\varepsilon(z)$ concentrates near $r = 0$ as $\varepsilon \searrow 0$, and the integrand $\mathcal{M}(\phi^M, z, r)$ is non-decreasing in $r$; hence $\varepsilon \mapsto \phi^M_\varepsilon(z)$ is non-decreasing in $\varepsilon$ (equivalently, non-increasing as $\varepsilon \searrow 0$). Indeed, for $0 < \varepsilon_1 < \varepsilon_2 < \operatorname{dist}(z, \partial \Omega)$, applying the substitution $r = \varepsilon_i s$ (so $d\mathcal{L}^1(r) = \varepsilon_i\, d\mathcal{L}^1(s)$, and the domain $r \in (0, \varepsilon_i)$ maps to $s \in (0, 1)$) in each integral gives: \begin{align*} \phi^M_{\varepsilon_i}(z) = \int_0^1 h(s)\, s^{2n-1}\, \sigma_{2n-1}\, \mathcal{M}(\phi^M, z, \varepsilon_i s)\, d\mathcal{L}^1(s), \end{align*} and since $\mathcal{M}(\phi^M, z, \varepsilon_1 s) \le \mathcal{M}(\phi^M, z, \varepsilon_2 s)$ pointwise by monotonicity, $\phi^M_{\varepsilon_1}(z) \le \phi^M_{\varepsilon_2}(z)$. To identify the limit, upper semicontinuity of $\phi^M$ gives $\limsup_{r \to 0^+} \mathcal{M}(\phi^M, z, r) \le \phi^M(z)$. We apply the reverse [Fatou lemma](/theorems/510) to the family $u \mapsto \phi^M(z + r u)$ on $(S^{2n-1}, \mathcal{H}^{2n-1})$ as $r \to 0^+$. The hypothesis of reverse Fatou is a uniform integrable upper bound: by local boundedness of $\phi^M$, there exist $\rho > 0$ and $C \in \mathbb{R}$ such that $\phi^M(y) \le C$ for every $y \in \overline{B}(z, \rho)$; the constant function $C$ is $\mathcal{H}^{2n-1}$-integrable on $S^{2n-1}$ (which has finite measure $\sigma_{2n-1}$); and for $r < \rho$ we have $\phi^M(z + r u) \le C$ for every $u \in S^{2n-1}$. Reverse Fatou then gives \begin{align*} \limsup_{r \to 0^+} \frac{1}{\sigma_{2n-1}} \int_{S^{2n-1}} \phi^M(z + r u)\, d\mathcal{H}^{2n-1}(u) \le \frac{1}{\sigma_{2n-1}} \int_{S^{2n-1}} \limsup_{r \to 0^+} \phi^M(z + r u)\, d\mathcal{H}^{2n-1}(u) \le \phi^M(z), \end{align*} where the last inequality uses $\limsup_{r \to 0^+} \phi^M(z + r u) \le \phi^M(z)$ for every $u$ by upper semicontinuity of $\phi^M$ at $z$. Combined with the sub-mean-value inequality $\phi^M(z) \le \mathcal{M}(\phi^M, z, r)$ (which holds for subharmonic functions of one complex variable applied to slices, then averaged as above), we get $\lim_{r \to 0^+} \mathcal{M}(\phi^M, z, r) = \phi^M(z)$. Hence $\phi^M_\varepsilon(z) \to \phi^M(z)$ as $\varepsilon \to 0^+$. Combined with monotonicity, $\phi^M_\varepsilon \searrow \phi^M$. By [Stability Properties of PSH Functions](/theorems/3404), the decreasing limit of plurisubharmonic functions on a fixed [open set](/page/Open%20Set) is either plurisubharmonic on that set or identically $-\infty$ on each connected component of that set. We apply this fact one $\varepsilon$-shell at a time. Fix $\varepsilon_0 > 0$. For every $\varepsilon \in (0, \varepsilon_0]$, $\Omega_{\varepsilon_0} \subseteq \Omega_\varepsilon$, and $\phi^M_\varepsilon$ is plurisubharmonic on $\Omega_\varepsilon$ by Step 4, hence its restriction is plurisubharmonic on $\Omega_{\varepsilon_0}$. The sequence $(\phi^M_\varepsilon|_{\Omega_{\varepsilon_0}})_{\varepsilon \le \varepsilon_0}$ is decreasing with pointwise limit $\phi^M|_{\Omega_{\varepsilon_0}}$. Since $\phi^M \ge -M > -\infty$, the identically $-\infty$ alternative is excluded on every component, so $\phi^M$ is plurisubharmonic on $\Omega_{\varepsilon_0}$. As $\varepsilon_0 > 0$ was arbitrary and $\Omega = \bigcup_{\varepsilon_0 > 0} \Omega_{\varepsilon_0}$, plurisubharmonicity at every point of $\Omega$ follows from the local nature of plurisubharmonicity (each $z \in \Omega$ has a neighbourhood contained in some $\Omega_{\varepsilon_0}$). Thus $\phi^M$ is plurisubharmonic on $\Omega$.[/step]

[guided]We need two ingredients to push the smooth approximation $\phi^M_\varepsilon$ down to the original $\phi^M$: (i) monotone convergence $\phi^M_\varepsilon \searrow \phi^M$, and (ii) closure of plurisubharmonicity under decreasing limits (theorem 3404). For (i), the key is the [polar decomposition](/theorems/3074). Writing the mollifier as a function of $r = |w|$ via $\eta(w) = h(r)$, the convolution $\phi^M_\varepsilon(z)$ becomes a weighted integral of the spherical mean $\mathcal{M}(\phi^M, z, r)$ over $r \in (0, \varepsilon)$. If $r \mapsto \mathcal{M}(\phi^M, z, r)$ is non-decreasing, then concentrating the weight near $r = 0$ (by shrinking $\varepsilon$) gives a decreasing function of $\varepsilon$. The one-line computation that makes this precise is the change of variables $r = \varepsilon s$, $d\mathcal{L}^1(r) = \varepsilon\, d\mathcal{L}^1(s)$, which transforms \begin{align*} \phi^M_\varepsilon(z) = \int_0^\varepsilon \varepsilon^{-2n} h(r/\varepsilon)\, r^{2n-1}\, \sigma_{2n-1}\, \mathcal{M}(\phi^M, z, r)\, d\mathcal{L}^1(r) \end{align*} into \begin{align*} \phi^M_\varepsilon(z) = \int_0^1 h(s)\, s^{2n-1}\, \sigma_{2n-1}\, \mathcal{M}(\phi^M, z, \varepsilon s)\, d\mathcal{L}^1(s). \end{align*} The weight $h(s) s^{2n-1} \sigma_{2n-1}$ is now independent of $\varepsilon$ and non-negative, while the only dependence on $\varepsilon$ sits inside $\mathcal{M}(\phi^M, z, \varepsilon s)$, which is non-decreasing in $\varepsilon$ for each fixed $s$ by the monotonicity of spherical means. Hence $\varepsilon \mapsto \phi^M_\varepsilon(z)$ is non-decreasing in $\varepsilon$, equivalently non-increasing as $\varepsilon \searrow 0$. The monotonicity of spherical means in $\mathbb{R}^{2n}$ from slice subharmonicity in $\mathbb{C}^n$ is the heart of the argument. The sphere $S^{2n-1} \subset \mathbb{C}^n$ has a natural $S^1$-action (multiplication by $e^{i\theta}$ in $\mathbb{C}^n$). The orbits are circles $\{e^{i\theta} u : \theta \in [0, 2\pi)\}$ — each orbit lies in a complex line through the origin, namely $\mathbb{C} \cdot u$. Disintegrating the spherical measure along this fibration writes $\mathcal{M}(\phi^M, z, r)$ as a $\nu$-average over $\mathbb{CP}^{n-1}$ of circular means $A_u(r)$ of $\phi^M$ along complex lines through $z$. Each $A_u(r)$ is the circular mean of the subharmonic slice $\phi^M_{z,u}|_{\{|\zeta| < r\}}$ at radius $r$, which is non-decreasing in $r$ by the standard property of subharmonic functions of one complex variable. Averaging over $u$ preserves monotonicity. For the identification of the limit: USC gives $\limsup_{r \to 0^+} \mathcal{M}(\phi^M, z, r) \le \phi^M(z)$ via reverse Fatou (the integrand is uniformly bounded above near $z$ because $\phi^M$ is locally bounded, and USC says $\limsup_{y \to z}\phi^M(y) \le \phi^M(z)$ which integrates against the surface measure on small spheres). The reverse inequality $\phi^M(z) \le \mathcal{M}(\phi^M, z, r)$ comes from the sub-mean-value inequality on each slice: $\phi^M(z) = \phi^M_{z, u}(0) \le A_u(r)$, then average over $u$. So $\mathcal{M}(\phi^M, z, r) \to \phi^M(z)$ as $r \to 0^+$, and consequently $\phi^M_\varepsilon(z) \to \phi^M(z)$ as $\varepsilon \to 0^+$. For (ii), theorem 3404 packages the standard fact that if $(u_k)$ is a decreasing sequence of plurisubharmonic functions on $\Omega$, then $u := \lim u_k = \inf u_k$ is either plurisubharmonic or identically $-\infty$ on each connected component. Here $\phi^M \ge -M$, so the identically-$-\infty$ alternative is excluded, and we conclude $\phi^M$ is plurisubharmonic on each $\Omega_\varepsilon$ — hence on all of $\Omega$, since plurisubharmonicity is local and $\Omega = \bigcup_\varepsilon \Omega_\varepsilon$.[/guided]

[step:Pass to the limit $M \to \infty$ to recover plurisubharmonicity of $\phi$] As $M \to \infty$, $\phi^M = \max\{\phi, -M\}$ is a decreasing sequence of plurisubharmonic functions on $\Omega$ (each plurisubharmonic by Step 5) with pointwise limit \begin{align*} \lim_{M \to \infty} \phi^M(z) = \phi(z) \quad \text{for every } z \in \Omega \end{align*} (if $\phi(z) > -\infty$ then $\phi^M(z) = \phi(z)$ for all $M > -\phi(z)$; if $\phi(z) = -\infty$ then $\phi^M(z) = -M \to -\infty$). By hypothesis, $\phi$ is not identically $-\infty$ on any connected component of $\Omega$. By [Stability Properties of PSH Functions](/theorems/3404), the decreasing limit $\phi$ is plurisubharmonic on $\Omega$. This completes the reverse direction and the proof. [/step]