TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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An interior local maximum of the modulus forces a scalar holomorphic function to be constant

Statement

Let $m \geq 1$ , let $U \subseteq C^{m}$ be a nonempty connected open set, and let $f : U \to C$ be holomorphic. Suppose there are $a \in U$ and an open ball $B \subseteq U$ centred at $a$ such that

$∣ f (z) ∣ \leq ∣ f (a) ∣ (z \in B) .$

Then $f$ is constant on $U$ .

Facts & Assumptions

Given: A nonempty connected open set $U \subseteq C^{m}$ , a holomorphic function $f : U \to C$ , a point $a \in U$ , and an open ball $B \subseteq U$ centred at $a$ such that $∣ f (z) ∣ \leq ∣ f (a) ∣$ for every $z \in B$ .

[L1]

The composite of holomorphic maps is holomorphic and its complex Jacobian is the product (The composite of holomorphic maps is holomorphic and its complex Jacobian is the product).

[L2]

If the modulus of a holomorphic function on a complex domain has an interior local maximum, then the function is constant (Local maximum modulus principle).

[L3]

A holomorphic function vanishing on a nonempty open subset of a connected open set in $C^{m}$ vanishes identically (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L4]

Balls in $C^{m}$ are the Euclidean balls of Balls, polydiscs and the distinguished boundary in $C^{m}$ .

[L5]

Sums, products and nonvanishing quotients of holomorphic functions are holomorphic (Sums, products and nonvanishing quotients of holomorphic functions are holomorphic).

Proof

technique · direct

1.1

Because $B$ is an open ball centred at $a$ , there is $r > 0$ such that ${z \in C^{m} : ∥ z - a ∥ < r} \subseteq B$ in the Euclidean norm of [L4].

givenL4

2.1

Fix a vector $v \in C^{m}$ with $∥ v ∥ = 1$ , and define $g_{v} (ξ) : = f (a + ξ v)$ on the disc $D (0, r)$ . The map $ξ \mapsto a + ξ v$ is holomorphic, so [L1] makes $g_{v}$ holomorphic on $D (0, r)$ ; and for every $ξ \in D (0, r)$ one has $a + ξ v \in B$ , hence $∣ g_{v} (ξ) ∣ = ∣ f (a + ξ v) ∣ \leq ∣ f (a) ∣ = ∣ g_{v} (0) ∣$ . Therefore [L2] makes $g_{v}$ constant on all of $D (0, r)$ .

step 1.1L1L2

3.1

Let $z \in B$ be arbitrary. If $z = a$ there is nothing to prove. Otherwise put $v : = (z - a) / ∥ z - a ∥$ and $ξ : = ∥ z - a ∥$ ; then $∥ v ∥ = 1$ , $∣ ξ ∣ < r$ , and $z = a + ξ v$ , so step 2.1 gives $f (z) = g_{v} (ξ) = g_{v} (0) = f (a)$ . Thus $f$ is constant on the nonempty open set $B$ , and [L5] makes $f - f (a)$ holomorphic on $U$ ; applying [L3] to $f - f (a)$ and the open set $B$ yields $f \equiv f (a)$ on $U$ .

step 2.1L3L5∎

Remarks

The argument gives constancy on the whole local ball. The slice theorem is applied on the entire disc cut out by the ball, not merely near the origin, so the proof first shows that $f$ is constant on all of $B$ and only then extends that constancy to $U$ by the several-variable identity theorem.

Depends on

Used by

A function whose modulus attains its maximum only on the distinguished boundary of a bidisc Example

Dependency tree · two levels

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Sources

J. Lebl, Tasty Bits of Several Complex Variables, v4.4, Thm. 1.2.8 (standard reference, not scraped)
M. Jabbari, Notes for Analysis and Geometry of Several Complex Variables, Thm. 22(8) (standard reference, not scraped)