TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Separate holomorphy implies joint holomorphy in finite dimensions

Statement

Let $m \geq 1$ , let $U \subseteq C^{m}$ be open, and let $f : U \to C$ be separately holomorphic. Then $f$ is holomorphic on $U$ .

Facts & Assumptions

Given: An open set $U \subseteq C^{m}$ and a separately holomorphic function $f : U \to C$ .

[L1]

Separate holomorphy means one-variable holomorphy on every coordinate slice (Separately holomorphic functions).

[L2]

A separately holomorphic function is locally bounded on every smaller polydisc (Separate holomorphy forces local boundedness on smaller polydiscs).

[L3]

A separately holomorphic function that is locally bounded is jointly holomorphic (Locally bounded and separately holomorphic implies holomorphic).

Proof

technique · direct

1.1

Fix $p \in U$ . Because $U$ is open, there is a polydisc neighborhood $P \subseteq U$ of $p$ . The restriction $f ∣_{P}$ is still separately holomorphic by [L1], so [L2] makes it locally bounded near $p$ .

givenL1L2

2.1

Applying [L3] to that same restriction shows that $f$ is holomorphic on a neighborhood of $p$ . As $p$ was arbitrary, $f$ is holomorphic on all of $U$ .

step 1.1L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Paul Garrett, Hartogs' Theorem: separate analyticity implies joint (standard reference, not scraped)
J. Lebl, Tasty Bits of Several Complex Variables, §1.2 (standard reference, not scraped)