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Hartogs pseudoconvexity yields a continuous plurisubharmonic exhaustion

Statement

Let ΩCm be a domain. If Ω is Hartogs pseudoconvex, that is, if logδΩ is plurisubharmonic on Ω, then Ω admits a continuous plurisubharmonic exhaustion function.

Facts & Assumptions

Given: A domain ΩCm.

[L1]

Hartogs pseudoconvexity means exactly that logδΩ is plurisubharmonic, while a continuous plurisubharmonic exhaustion is defined by compact sublevel sets (Plurisubharmonic exhaustions and Hartogs pseudoconvexity).

[L2]

Finite maxima preserve plurisubharmonicity (Basic stability operations for plurisubharmonic functions).

[L3]

The squared norm has positive Levi form, so it is plurisubharmonic (The Levi form and strict plurisubharmonicity, The C^2 Levi criterion for plurisubharmonicity).

Proof

technique · direct
1.1

Assume that logδΩ is plurisubharmonic. The function q(z)=z12++zm2 is plurisubharmonic by [L3]. Since δΩ is a distance-to-the-complement function for the sup norm, it is continuous on Ω, so logδΩ is continuous as well. Therefore [L2] makes u(z):=max{logδΩ(z),q(z)} a continuous plurisubharmonic function.

L2L3given
2.1

As z approaches Ω, the term logδΩ(z) tends to +, and as z inside Ω, the term q(z) tends to +. Hence the sublevel sets of u are compact in Ω, so u is a continuous plurisubharmonic exhaustion.

L1step 1.1given

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