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An isolated puncture is removable in complex dimension at least two

Statement

Let m2, let ΩCm be a domain, let aΩ, and let f:Ω{a}C be holomorphic. Then there exists a unique holomorphic F:ΩC such that F=f on Ω{a}.

Facts & Assumptions

Given: A domain ΩCm with m2, 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
1.1

By [L1], choose radii ρ1>ρ>0 and τ>0 with Δρ1(a1)×Δτ(a)Ω. For (z1,z) with z1a1<ρ and za<τ, define F(z1,z):=12πiζa1=ρf(ζ,z)ζz1dζ. When ζa1=ρ and za<τ, the point (ζ,z) lies in Δρ1(a1)×Δτ(a){a}, so the integrand is well defined and continuous there.

L1construct
2.1

Fix all variables except one. If the free variable is z1, the Cauchy kernel makes z1F(z1,z) holomorphic on z1a1<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.

L2L5step 1.1algebra
2.2

If za, then the slice ζf(ζ,z) is holomorphic on the full disc ζa1<ρ, because the deleted point a does not lie on that slice. Hence [L3] gives F(z1,z)=f(z1,z) for every z1a1<ρ.

step 1.1L3
3.1

The set {(z1,z)P:za} 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}.

step 2.2L4
4.1

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.

step 2.1step 3.1L4

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