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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27
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An isolated puncture is removable in complex dimension at least two

Statement

Let m≥2, let Ω⊆Cm be a domain, let a∈Ω, and let f:Ω∖{a}→C be holomorphic. Then there exists a unique holomorphic F:Ω→C such that F=f on Ω∖{a}.

Facts & Assumptions

Given: A domain Ω⊆Cm with m≥2, a point a=(a1,a′)∈Ω, and a holomorphic function f:Ω∖{a}→C.

[L1]

Every point of an open subset of Cm has a polydisc neighborhood inside that open set (Balls, polydiscs and the distinguished boundary in Cm).

[L2]

A contour integral of a jointly continuous integrand that is holomorphic in the parameter variable defines a holomorphic function of that parameter (A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic).

[L3]

The one-variable Cauchy integral formula recovers a holomorphic function on a disc from any interior circle (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

[L4]

A holomorphic function on a connected open set is determined by its values on a nonempty open subset (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L5]

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

Proof

technique · direct
1.1L1construct

By [L1], choose radii ρ1>ρ>0 and τ>0 with Δρ1(a1)×Δτ(a′)⊆Ω. For (z1,z′) with ∣z1−a1∣<ρ and ∣z′−a′∣<τ, define F(z1,z′):=12πi∫∣ζ−a1∣=ρf(ζ,z′)ζ−z1 dζ. When ∣ζ−a1∣=ρ and ∣z′−a′∣<τ, the point (ζ,z′) lies in Δρ1(a1)×Δτ(a′)∖{a}, so the integrand is well defined and continuous there.

2.1L2L5step 1.1algebra

Fix all variables except one. If the free variable is z1, the Cauchy kernel makes z1↦F(z1,z′) holomorphic on ∣z1−a1∣<ρ. If the free variable is one coordinate of z′, then [L2] applies to that single complex parameter. So F is separately holomorphic on Δρ(a1)×Δτ(a′). The same integral formula gives local bounds on compact subsets, so [L5] upgrades F to a jointly holomorphic function there.

2.2step 1.1L3

If z′≠a′, then the slice ζ↦f(ζ,z′) is holomorphic on the full disc ∣ζ−a1∣<ρ, because the deleted point a does not lie on that slice. Hence [L3] gives F(z1,z′)=f(z1,z′) for every ∣z1−a1∣<ρ.

3.1step 2.2L4

The set {(z1,z′)∈P:z′≠a′} is a nonempty open subset of P∖{a}, and step 2.2 shows that F and f agree there. Both are holomorphic on the connected punctured polydisc P∖{a}, so [L4] forces F=f on all of P∖{a}.

4.1step 2.1step 3.1L4∎

Thus F is a holomorphic extension of f across a on the neighborhood Δρ(a1)×Δτ(a′). Repeating the same construction at each puncture point gives a local extension, and uniqueness on overlaps again follows from [L4]. Therefore the local extensions glue to a unique holomorphic function on all of Ω.

Depends on

Used by

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Sources