[proofplan] We show that $\log|f|$ is plurisubharmonic on $\Omega$ for any holomorphic $f: \Omega \to \mathbb{C}$, $f \not\equiv 0$. The argument has two parts: verifying upper semicontinuity (which follows from continuity of $|f|$ and the convention $\log 0 = -\infty$), and verifying the sub-mean-value inequality along every complex line (which reduces to the classical fact that $\log|g|$ is subharmonic for a [holomorphic function](/page/Holomorphic%20Function) $g$ of one complex variable). [/proofplan]

[step:Verify that $\log|f|$ is upper semicontinuous on $\Omega$] Since $f \in \mathcal{O}(\Omega)$, the modulus $|f|: \Omega \to [0, \infty)$ is continuous. The function $\log: (0, \infty) \to \mathbb{R}$ is continuous, and we adopt the convention $\log 0 = -\infty$. With this convention, the extended function \begin{align*} \log|\cdot|: [0, \infty) &\to [-\infty, \infty) \\ t &\mapsto \begin{cases} \log t & \text{if } t > 0, \\ -\infty & \text{if } t = 0 \end{cases} \end{align*} is upper semicontinuous on $[0, \infty)$: for each $c \in \mathbb{R}$, the set $\{t \geq 0 : \log|t| < c\} = [0, e^c)$ is open in $[0, \infty)$. Since the composition of a continuous function with an upper semicontinuous function is upper semicontinuous, $\log|f| = (\log|\cdot|) \circ |f|$ is upper semicontinuous on $\Omega$. [/step]

[step:Reduce the sub-mean-value inequality to the one-variable case by restricting to complex lines] Fix $z \in \Omega$ and $w \in \mathbb{C}^n$. Define \begin{align*} g: D &\to \mathbb{C} \\ \zeta &\mapsto f(z + \zeta w), \end{align*} where $D = \{\zeta \in \mathbb{C} : z + \zeta w \in \Omega\}$ is open in $\mathbb{C}$. Since $f$ is holomorphic on $\Omega$ and the map $\zeta \mapsto z + \zeta w$ is holomorphic (in fact, affine), $g$ is holomorphic on $D$. The restriction satisfies $\log|f(z + \zeta w)| = \log|g(\zeta)|$, so it suffices to show that $\log|g|$ is subharmonic on $D$ for a [holomorphic function](/page/Holomorphic%20Function) $g: D \to \mathbb{C}$ of one variable. [/step]

[step:Prove that $\log|g|$ is subharmonic for a holomorphic function $g$ of one complex variable]Let $g: D \to \mathbb{C}$ be holomorphic with $g \not\equiv 0$ on any connected component of $D$. We verify the sub-mean-value inequality: for every $\zeta_0 \in D$ and $r > 0$ with $\overline{B}(\zeta_0, r) \subset D$, \begin{align*} \log|g(\zeta_0)| \leq \frac{1}{2\pi}\int_0^{2\pi} \log|g(\zeta_0 + re^{i\theta})|\,d\mathcal{L}^1(\theta). \end{align*} **Case 1: $g(\zeta_0) = 0$.** Then $\log|g(\zeta_0)| = -\infty$, and the right-hand side is either finite or $-\infty$. If $g \not\equiv 0$, then the zero set of $g$ is discrete (by the [Identity Principle](/theorems/3357) in one variable), so $g$ is not identically zero on the circle $|\zeta - \zeta_0| = r$ (for all but countably many values of $r$). The function $\log|g|$ is locally integrable near an isolated zero: if $g$ has a zero of order $m$ at $\zeta_0$, then $g(\zeta) = (\zeta - \zeta_0)^m h(\zeta)$ with $h(\zeta_0) \neq 0$, so $\log|g(\zeta)| = m\log|\zeta - \zeta_0| + \log|h(\zeta)|$ near $\zeta_0$, and $\log|\zeta - \zeta_0|$ is locally integrable. Therefore the right-hand side is finite (for generic $r$), hence $\geq -\infty = \log|g(\zeta_0)|$. **Case 2: $g(\zeta_0) \neq 0$.** Since the zeros of $g$ are isolated, we may choose $r > 0$ small enough that $g$ has no zeros on $\overline{B}(\zeta_0, r)$ (or if zeros exist on the disc but not at $\zeta_0$, we argue as follows). If $g$ has no zeros in $B(\zeta_0, r)$, then $\log g$ has a holomorphic branch on $B(\zeta_0, r)$ (since $B(\zeta_0, r)$ is simply connected and $g \neq 0$ there), and $\log|g| = \operatorname{Re}(\log g)$ is harmonic on $B(\zeta_0, r)$. Harmonic functions satisfy the mean-value equality, so the sub-mean-value inequality holds (with equality). If $g$ has zeros in $B(\zeta_0, r)$ but not at $\zeta_0$, let $\zeta_1, \dots, \zeta_N$ be the zeros of $g$ in $B(\zeta_0, r)$ counted with multiplicity. Define the Blaschke factor \begin{align*} b(\zeta) = \prod_{j=1}^N \frac{\zeta - \zeta_j}{r - \overline{(\zeta_j - \zeta_0)}(\zeta - \zeta_0)/r}, \end{align*} which satisfies $|b(\zeta)| = 1$ on $|\zeta - \zeta_0| = r$, and $g/b$ is holomorphic and nonvanishing on $B(\zeta_0, r)$. Then $\log|g/b| = \operatorname{Re}(\log(g/b))$ is harmonic on $B(\zeta_0, r)$, and the mean-value property gives \begin{align*} \log|g(\zeta_0)/b(\zeta_0)| = \frac{1}{2\pi}\int_0^{2\pi} \log|g(\zeta_0 + re^{i\theta})/b(\zeta_0 + re^{i\theta})|\,d\mathcal{L}^1(\theta). \end{align*} Since $|b(\zeta_0 + re^{i\theta})| = 1$ on the circle, the right-hand side equals $\frac{1}{2\pi}\int_0^{2\pi}\log|g(\zeta_0 + re^{i\theta})|\,d\mathcal{L}^1(\theta)$. Since $|b(\zeta_0)| = \prod_j |\zeta_0 - \zeta_j|/r \cdot |r/\overline{(\zeta_j - \zeta_0)}| \leq 1$ (each factor has modulus $< 1$ because $|\zeta_j - \zeta_0| < r$), we have $\log|b(\zeta_0)| \leq 0$, so \begin{align*} \log|g(\zeta_0)| = \log|g(\zeta_0)/b(\zeta_0)| + \log|b(\zeta_0)| \leq \log|g(\zeta_0)/b(\zeta_0)| = \frac{1}{2\pi}\int_0^{2\pi}\log|g(\zeta_0 + re^{i\theta})|\,d\mathcal{L}^1(\theta). \end{align*}[/step]

[guided]The core of this step is the classical fact that $\log|g|$ is subharmonic whenever $g$ is holomorphic and not identically zero. Let us explain the two cases in detail. **When $g(\zeta_0) = 0$:** The inequality $\log|g(\zeta_0)| \leq \frac{1}{2\pi}\int\log|g|$ reduces to $-\infty \leq \frac{1}{2\pi}\int\log|g|$, which is automatic as long as the integral is well-defined (possibly $= -\infty$). The integral is well-defined because $\log|g|$ is bounded above on compact sets (since $|g|$ is continuous) and its only singularities are logarithmic at the zeros of $g$, which are integrable in one dimension: near a zero of order $m$, $\log|g(\zeta)| \sim m\log|\zeta - \zeta_0| + \text{bounded}$, and $\int_0^\delta \log t\,d\mathcal{L}^1(t) = \delta\log\delta - \delta > -\infty$. **When $g(\zeta_0) \neq 0$:** The idea is to remove the zeros by dividing by a Blaschke product. If $g$ has no zeros in $B(\zeta_0, r)$, then $g$ has a holomorphic logarithm on the simply connected domain $B(\zeta_0, r)$ (since $g \neq 0$ there and $B(\zeta_0, r)$ is simply connected, the [winding number](/page/Winding%20Number) of $g$ around any closed path in $B(\zeta_0, r)$ is zero, so a branch of $\log g$ exists). Then $\log|g| = \operatorname{Re}(\log g)$ is harmonic, and the mean-value equality holds. If $g$ has finitely many zeros $\zeta_1, \dots, \zeta_N$ in $B(\zeta_0, r)$, we factor them out using the Blaschke product $b(\zeta) = \prod_{j=1}^N \frac{\zeta - \zeta_j}{r - \overline{(\zeta_j - \zeta_0)}(\zeta - \zeta_0)/r}$. This product has the property that $|b| = 1$ on the circle $|\zeta - \zeta_0| = r$ (each factor is a Mobius transformation of $B(\zeta_0, r)$ and has unit modulus on the boundary). Inside the disc, $|b(\zeta)| < 1$ (since each factor maps the disc to the disc). Dividing out, $g/b$ is holomorphic and nonvanishing on $B(\zeta_0, r)$, so $\log|g/b|$ is harmonic. The mean-value property applied to $\log|g/b|$, combined with $|b| = 1$ on the circle and $|b(\zeta_0)| \leq 1$, gives the desired inequality for $\log|g|$.[/guided]

[step:Conclude that $\log|f|$ is plurisubharmonic on $\Omega$] Combining the three preceding steps: $\log|f|$ is upper semicontinuous on $\Omega$ (first step), and for every $z \in \Omega$ and $w \in \mathbb{C}^n$, the restriction $\zeta \mapsto \log|f(z + \zeta w)| = \log|g(\zeta)|$ is subharmonic on $D$ (second and third steps). Therefore $\log|f|$ is plurisubharmonic on $\Omega$. [/step]