[proofplan] Set $M := \phi(a)$. We first establish a *local* propagation: at any point where $\phi$ has a local maximum with value $M$, the function is identically $M$ on a whole Euclidean ball. The key tool is that plurisubharmonicity is precisely the requirement that the restriction of $\phi$ to every complex line is subharmonic; applying the one-variable strong maximum principle along each complex line through the maximum point sweeps out a full Euclidean ball of constancy. We then promote this local statement to a global one by a connectedness argument: the set $E$ of points around which $\phi \equiv M$ on a neighbourhood is open by construction, nonempty by the local step, and closed in $\Omega$ because near any limit point $b$ one finds a nearby point of $E$, fits a complex disc through it that reaches $b$, and reapplies the one-variable strong maximum principle to conclude $\phi \equiv M$ on a whole ball around $b$. Connectedness of $\Omega$ forces $E = \Omega$. [/proofplan] [step:Fix notation and isolate a ball on which $\phi$ is bounded above by $\phi(a)$] Write $M := \phi(a) \in \mathbb{R}$. By hypothesis, there is an open neighbourhood $U \subseteq \Omega$ of $a$ with $\phi \le M$ on $U$. Since $U$ is open in $\mathbb{C}^n \cong \mathbb{R}^{2n}$, choose $r_0 > 0$ such that the Euclidean ball \begin{align*} B(a, r_0) := \{z \in \mathbb{C}^n : |z - a| < r_0\} \end{align*} satisfies $\overline{B(a, r_0)} \subset U \subseteq \Omega$. In particular $\phi \le M$ on $B(a, r_0)$ and $\phi(a) = M$, so $\phi$ attains its supremum over $B(a, r_0)$ at the centre $a$, with value $M$. [/step] [step:Prove that $\phi \equiv M$ on the ball $B(a, r_0)$ by sweeping complex lines through $a$] Fix a unit vector $v \in \mathbb{C}^n$ (i.e.\ $|v| = 1$) and define the one-variable slice \begin{align*} \psi_v: \{\zeta \in \mathbb{C} : |\zeta| < r_0\} &\to [-\infty, +\infty), \\ \zeta &\mapsto \phi(a + \zeta v). \end{align*} For $|\zeta| < r_0$ we have $|a + \zeta v - a| = |\zeta| < r_0$, so $a + \zeta v \in B(a, r_0) \subset \Omega$ and $\psi_v$ is well-defined. Because $\phi$ is plurisubharmonic on $\Omega$, by definition its restriction to every complex affine line, intersected with $\Omega$, is subharmonic on the corresponding open set in $\mathbb{C}$ (see the [Stability Properties of PSH Functions](/theorems/3404)). Applied to the line $\{a + \zeta v : \zeta \in \mathbb{C}\}$ intersected with $\Omega$, this gives that $\psi_v$ is subharmonic on the open disc $\{|\zeta| < r_0\}$. On this disc, $\psi_v(\zeta) = \phi(a + \zeta v) \le M$ for all $\zeta$, and $\psi_v(0) = \phi(a) = M$. Hence $\psi_v$ attains its supremum $M$ over the connected [open set](/page/Open%20Set) $\{|\zeta| < r_0\}$ at the interior point $\zeta = 0$. By the strong maximum principle for subharmonic functions (the subharmonic analogue of the [Strong Maximum Principles](/theorems/32) for harmonic functions, which has an identical proof via the sub-mean-value inequality), $\psi_v$ is constant on $\{|\zeta| < r_0\}$, equal to $M$. Now let $z \in B(a, r_0)$ with $z \ne a$. Set $v := (z - a)/|z - a|$, a unit vector, and $\zeta_0 := |z - a| \in (0, r_0)$. Then $z = a + \zeta_0 v$, so $\phi(z) = \psi_v(\zeta_0) = M$. Combined with $\phi(a) = M$, we obtain $\phi \equiv M$ on $B(a, r_0)$. [guided] We need to extract local constancy of $\phi$ at $a$ in *all* complex directions simultaneously. The definition of plurisubharmonic is tailored for exactly this: $\phi$ is plurisubharmonic on $\Omega$ if and only if it is upper semicontinuous and its restriction to every complex line, intersected with $\Omega$, is subharmonic on the resulting open subset of $\mathbb{C}$ (the equivalent characterisation underpinning the [Stability Properties of PSH Functions](/theorems/3404)). So fix any unit vector $v \in \mathbb{C}^n$ and slice: \begin{align*} \psi_v: \{|\zeta| < r_0\} \to [-\infty, +\infty), \quad \zeta \mapsto \phi(a + \zeta v). \end{align*} The domain check is immediate: $|a + \zeta v - a| = |\zeta| |v| = |\zeta| < r_0$, so $a + \zeta v \in B(a, r_0) \subseteq \Omega$. By the PSH characterisation, $\psi_v$ is subharmonic on the disc $\{|\zeta| < r_0\}$. Why can we apply the strong maximum principle? We need the standard hypotheses: a subharmonic function on a connected open subset of $\mathbb{C}$ attaining its supremum at an interior point. Connectedness of $\{|\zeta| < r_0\}$ is obvious. The supremum is attained: $\psi_v \le M$ on the whole disc because $\phi \le M$ on $B(a, r_0)$ (Step 1), and $\psi_v(0) = M$. So $\sup_{|\zeta| < r_0} \psi_v = M$, attained at the interior point $\zeta = 0$. The strong maximum principle for subharmonic functions—the direct analogue of the [Strong Maximum Principles](/theorems/32) for harmonic functions, proved by the same submean-value plus open/closed argument—forces $\psi_v \equiv M$. The geometric upshot: $\phi(a + \zeta v) = M$ for every $\zeta \in \mathbb{C}$ with $|\zeta| < r_0$ and every unit $v$. The union of all such points $a + \zeta v$ as $v$ ranges over the unit sphere of $\mathbb{C}^n$ and $|\zeta| < r_0$ is precisely $B(a, r_0)$: any $z \in B(a, r_0) \setminus \{a\}$ has the unique representation $z = a + |z-a| \cdot \frac{z-a}{|z-a|}$. Therefore $\phi \equiv M$ on $B(a, r_0)$. [/guided] [/step] [step:Define the propagation set $E$ and verify it is open and nonempty] Define \begin{align*} E := \{z \in \Omega : \text{there exists } s > 0 \text{ with } B(z, s) \subseteq \Omega \text{ and } \phi \equiv M \text{ on } B(z, s)\}. \end{align*} $E$ is open in $\Omega$: if $z \in E$ with witness $s > 0$ and if $y \in B(z, s)$, define \begin{align*} \varepsilon_y := s - |y - z| > 0. \end{align*} For any $w \in B(y, \varepsilon_y)$, the triangle inequality gives \begin{align*} |w - z| \le |w - y| + |y - z| < \varepsilon_y + |y - z| = s, \end{align*} so $B(y, \varepsilon_y) \subseteq B(z, s) \subseteq \Omega$. Since $\phi \equiv M$ on $B(z, s)$, we have $\phi \equiv M$ on $B(y, \varepsilon_y)$, and therefore $y \in E$. Thus $B(z, s) \subseteq E$, proving that $E$ is open in $\Omega$. Step 2 shows that $a \in E$ with witness $r_0$, so $E \ne \varnothing$. [/step] [step:Show $E$ is closed in $\Omega$ by reapplying the one-variable maximum principle along discs reaching limit points] Let $b \in \Omega$ be a limit point of $E$ inside $\Omega$. We prove $b \in E$. Since $\Omega$ is open, choose $\rho > 0$ with $\overline{B(b, 2\rho)} \subset \Omega$. As $b$ is a limit point of $E$, there exists \begin{align*} c \in E \cap B(b, \rho/2). \end{align*} Pick a witness $s > 0$ for $c \in E$: that is, $B(c, s) \subseteq \Omega$ and $\phi \equiv M$ on $B(c, s)$. By replacing $s$ with $\min(s, \rho/2)$, we may assume $0 < s \le \rho/2$. We claim that $\phi \equiv M$ on the ball $B(c, \rho)$. Granting this, $|b - c| < \rho/2 < \rho$ gives $b \in B(c, \rho)$, and the open ball $B(b, \rho/2) \subseteq B(c, \rho)$ (by the triangle inequality: $|w - b| < \rho/2$ implies $|w - c| \le |w - b| + |b - c| < \rho/2 + \rho/2 = \rho$) serves as a witness placing $b \in E$. It remains to prove the claim. First, $B(c, \rho) \subseteq \Omega$, since for $w \in B(c, \rho)$, \begin{align*} |w - b| \le |w - c| + |c - b| < \rho + \rho/2 = 3\rho/2 < 2\rho, \end{align*} so $w \in B(b, 2\rho) \subset \Omega$. Fix any $z \in B(c, \rho)$ with $z \ne c$, and set \begin{align*} v := \frac{z - c}{|z - c|} \in \mathbb{C}^n \text{ (unit vector)}, \qquad \zeta_1 := |z - c| \in (0, \rho). \end{align*} Define the slice \begin{align*} \tilde\psi: \{\zeta \in \mathbb{C} : |\zeta| < \rho\} &\to [-\infty, +\infty), \\ \zeta &\mapsto \phi(c + \zeta v). \end{align*} For $|\zeta| < \rho$ we have $|c + \zeta v - c| = |\zeta| < \rho$, so $c + \zeta v \in B(c, \rho) \subset \Omega$, and $\tilde\psi$ is well-defined. By plurisubharmonicity of $\phi$ (the [Stability Properties of PSH Functions](/theorems/3404) characterisation via complex-line restrictions), $\tilde\psi$ is subharmonic on the disc $\{|\zeta| < \rho\}$. Moreover, $\tilde\psi \equiv M$ on the smaller disc $\{|\zeta| < s\}$: for $|\zeta| < s$, the point $c + \zeta v$ lies in $B(c, s)$, where $\phi \equiv M$. In particular, $\tilde\psi(0) = M$, and on the open neighbourhood $\{|\zeta| < s\}$ of $0$ we have $\tilde\psi(\zeta) = M = \tilde\psi(0)$, so $0$ is a local maximum of $\tilde\psi$ in the disc $\{|\zeta| < \rho\}$. We invoke the local-maximum form of the strong maximum principle for subharmonic functions: *if $u$ is subharmonic on a connected [open set](/page/Open%20Set) $D \subset \mathbb{C}$ and $u$ has a local maximum at some point of $D$, then $u$ is constant on $D$.* This is the standard subharmonic counterpart of the [Strong Maximum Principles](/theorems/32) for harmonic functions, proved by the same sub-mean-value plus open/closed argument. Applied to $\tilde\psi$ on the connected disc $\{|\zeta| < \rho\}$ with local maximum at $\zeta = 0$, it yields $\tilde\psi \equiv M$ on $\{|\zeta| < \rho\}$. Evaluating at $\zeta_1 \in (0, \rho)$, \begin{align*} \phi(z) = \phi(c + \zeta_1 v) = \tilde\psi(\zeta_1) = M. \end{align*} Since $z \in B(c, \rho) \setminus \{c\}$ was arbitrary and $\phi(c) = M$, the claim $\phi \equiv M$ on $B(c, \rho)$ holds. As noted, this places $b \in E$, proving $E$ is closed in $\Omega$. [guided] This is the heart of the proof: bootstrapping from local constancy at $c$ to local constancy at the nearby point $b$. The geometric question is: how do we get $\phi$ to "carry" the value $M$ from the small ball $B(c, s)$ around $c$ across the gap to $b$ and beyond? The answer is to fit a complex disc through $c$ that reaches all the way out to radius $\rho$ in every direction. We chose $\rho$ small enough that $\overline{B(b, 2\rho)} \subset \Omega$, and the geometry was arranged so that the entire ball $B(c, \rho)$ stays inside $\Omega$: any $w \in B(c, \rho)$ satisfies $|w - b| \le |w - c| + |c - b| < \rho + \rho/2 = 3\rho/2 < 2\rho$, hence $w \in B(b, 2\rho) \subset \Omega$. For a target point $z \in B(c, \rho)$, we slice $\phi$ along the complex line through $c$ in direction $v = (z - c)/|z - c|$, getting a subharmonic function $\tilde\psi$ on the disc $\{|\zeta| < \rho\}$. The key feature: $\tilde\psi \equiv M$ on the smaller disc $\{|\zeta| < s\}$ because that small disc lies in $B(c, s)$ where $\phi \equiv M$ by hypothesis. Now, *crucially*, we cannot directly say "$\tilde\psi \le M$" on the full disc $\{|\zeta| < \rho\}$—we don't have any a priori upper bound on $\phi$ on $B(c, \rho)$, since $b$ might not even be a local max yet. So the "$\sup$ attained" version of the maximum principle does not apply directly. Instead, we invoke the stronger local-maximum form: a subharmonic function on a connected open subset of $\mathbb{C}$ that has a local maximum at *any* point must be constant. This is the standard strong maximum principle for one-variable subharmonic functions (the subharmonic counterpart of [Strong Maximum Principles](/theorems/32)). The two formulations are equivalent: from "local max attained" one shrinks to a small disc where the local max becomes a global max, applies the sup-attained form to conclude constancy on that small disc, and then propagates by the same open/closed argument we are running here in $\mathbb{C}^n$. Does $\tilde\psi$ have a local maximum at some point of $\{|\zeta| < \rho\}$? Yes: at $\zeta = 0$, because $\tilde\psi \equiv M$ on the open neighbourhood $\{|\zeta| < s\}$ of $0$, which immediately gives $\tilde\psi(\zeta) \le \tilde\psi(0) = M$ for $\zeta$ in this neighbourhood. So the strong maximum principle applies and yields $\tilde\psi \equiv M$ on the entire disc $\{|\zeta| < \rho\}$. Plugging in $\zeta_1 := |z - c| \in (0, \rho)$ recovers $\phi(z) = \tilde\psi(\zeta_1) = M$. Since $z \in B(c, \rho)$ was arbitrary, $\phi \equiv M$ on the entire ball $B(c, \rho)$. This ball contains $b$ (as $|b - c| < \rho/2 < \rho$) and even a neighbourhood of $b$ (since $B(b, \rho/2) \subset B(c, \rho)$ by the triangle inequality), placing $b \in E$ with witness $\rho/2$. [/guided] [/step] [step:Conclude $E = \Omega$ by connectedness and read off the result] The set $E$ is open in $\Omega$ (Step 3), nonempty (Step 3, as $a \in E$), and closed in $\Omega$ (Step 4). Since $\Omega$ is connected, \begin{align*} E = \Omega. \end{align*} By the definition of $E$, every $z \in \Omega$ has a neighbourhood on which $\phi \equiv M$; in particular $\phi(z) = M$ for every $z \in \Omega$. Hence $\phi \equiv M = \phi(a)$ on $\Omega$, completing the proof. [/step]