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Continuity principle for domains of holomorphy

Statement

Let ΩCm be a domain of holomorphy, and let Φ:[0,1]×DCm be a continuous family of analytic discs. Assume that there is a compact set KΩ with Φt(D)K for every t, and that Φ0(D)Ω. Then

Φt(D)Ω(t[0,1]).

Facts & Assumptions

Given: A domain of holomorphy Ω and a continuous family of analytic discs Φ satisfying the boundary and initial-disc hypotheses.

[L1]

The family and its boundary hypotheses are those of Continuous families of analytic discs.

[L2]

A domain of holomorphy is holomorphically convex, so the holomorphic hull of a compact subset is compactly contained in the domain (Cartan-Thullen theorem).

[L3]

A function holomorphic on a disc and continuous on its closure is bounded there by its boundary maximum (Boundary maximum modulus principle on a bounded domain).

Proof

technique · direct
1.1

Put H:=K^Ω. By [L2], HΩ. Let S:={t[0,1]:Φt(D)Ω}. The initial-disc hypothesis gives 0S. If tS, then the compact set Φt(D) lies in the open set Ω; uniform continuity of Φ on the compact parameter product shows that the same containment holds for all parameters sufficiently close to t. Thus S is open in [0,1].

L1L2given
2.1

If tS and fO(Ω), then fΦt is holomorphic on D and continuous on its closure. By [L3] and the boundary hypothesis, f(Φt(λ))supDfΦtsupKf(λD). Since this holds for every f, one has Φt(D)H.

L1L3step 1.1
3.1

Let tnS and tnt. For every λD, step 2.1 gives Φtn(λ)H. The compact set H is closed in Cm, so continuity of Φ yields Φt(λ)HΩ. Hence tS, and S is closed. Since [0,1] is connected and S is nonempty, open, and closed, S=[0,1].

step 1.1step 2.1L2

Depends on

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