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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02
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Smooth strictly plurisubharmonic exhaustion of a pseudoconvex domain

Statement

Assume the Axiom of Choice (AC). Let Ω⊆Cn be a domain, n≥1, that is Hartogs pseudoconvex (Plurisubharmonic exhaustions and Hartogs pseudoconvexity). Then there exist a function S∈C∞(Ω) that is strictly plurisubharmonic on Ω (The Levi form and strict plurisubharmonicity) and a strictly increasing sequence c1<c2<⋯ with ck→+∞ such that, writing Ωk:={z∈Ω:S(z)<ck}:

  1. every ck is a regular value of S, each ∂Ωk={z∈Ω:S(z)=ck} is a nonempty C∞ hypersurface of Ω, and Ωk‾⊆Ωk+1 with ⋃k≥1Ωk=Ω, so every Ωk‾ is a compact subset of Ω;
  2. each sublevel is strongly pseudoconvex along its boundary: for every k, every p∈∂Ωk and every v∈Cn∖{0} with ∑j<n(∂S/∂zj)(p) vj=0 one has LS(p;v)>0.

Facts & Assumptions

Given: The Axiom of Choice; a domain Ω⊆Cn with n≥1 that is Hartogs pseudoconvex.

[F1]

A function u:Ω→R is a continuous plurisubharmonic exhaustion when u is continuous, plurisubharmonic, and every sublevel set {z∈Ω:u(z)≤c} is compact in Ω for every real number c (Plurisubharmonic exhaustions and Hartogs pseudoconvexity).

[F2]

The domain Ω is Hartogs pseudoconvex when −log⁡δΩ is plurisubharmonic on Ω, and the whole space is Hartogs pseudoconvex by the empty-complement convention (Plurisubharmonic exhaustions and Hartogs pseudoconvexity).

[F3]

If a domain Ω⊆Cm is Hartogs pseudoconvex, then it admits a continuous plurisubharmonic exhaustion function (Hartogs pseudoconvexity yields a continuous plurisubharmonic exhaustion).

[F4]

Assume AC and ACω; for every continuous plurisubharmonic exhaustion u on a domain there are S∈C∞(Ω) strictly plurisubharmonic and a strictly increasing sequence ck→+∞ such that each ck is a regular value of S, each ∂Ωk with Ωk={S<ck} is a nonempty C∞ hypersurface, Ωk‾⊆Ωk+1 and ⋃kΩk=Ω, and LS(p;v)>0 for all p∈∂Ωk and all v≠0 with ∑j(∂S/∂zj)(p)vj=0 (Smooth strict plurisubharmonic regularization of a psh exhaustion).

[F5]

AC⇒DC⇒ACω in ZF (AC implies DC implies countable choice).

[F6]

The Axiom of Countable Choice selects from every at most countable family of nonempty sets (The Axiom of Countable Choice (ACω)).

[F7]

The Axiom of Choice supplies a choice function for every family of nonempty sets (The Axiom of Choice).

Choice use. AC is the ambient hypothesis recorded in the Statement and cited as [F7]; the regularization lemma [F4] is stated under AC and ACω, and the countable instance [F6] is obtained from the ambient AC by the exact implication [F5] in step 2.1. The proof selects no family of nonempty sets.

Proof

technique · direct
1.1F1F2F3given

By the defining property [F2] the Hartogs pseudoconvexity of Ω says that −log⁡δΩ is plurisubharmonic on Ω, so the equivalence theorem [F3] supplies a continuous plurisubharmonic exhaustion u:Ω→R, that is, u is continuous, plurisubharmonic, and every sublevel set {u≤c} is compact in Ω by [F1].

2.1F4F5F6step 1.1given

The regularization lemma [F4], whose hypotheses are assumed AC together with ACω here, applies to the continuous plurisubharmonic exhaustion u produced in step 1.1 and yields S∈C∞(Ω) strictly plurisubharmonic together with a strictly increasing sequence ck→+∞ such that each ck is a regular value of S, each ∂Ωk is a nonempty C∞ hypersurface of Ω, Ωk‾⊆Ωk+1 and ⋃kΩk=Ω, and LS(p;v)>0 whenever p∈∂Ωk and v≠0 satisfies ∑j(∂S/∂zj)(p)vj=0; the countable instance required by that lemma is supplied from the ambient AC by the implication [F5] and its content [F6].

3.1F7step 2.1∎

The function S and the sequence ck produced in step 2.1 have exactly the properties listed as claims 1 and 2 of the Statement: strictly plurisubharmonic and smooth on Ω, increasing regular values tending to infinity, sublevels with nonempty smooth boundary, increasing relatively compact closures exhausting Ω, and strong pseudoconvexity along each boundary. The ambient hypothesis is the AC cited as [F7].

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