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ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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Minus log boundary distance is plurisubharmonic on a half-space

Example

For the half-space

H={zCm:Rez1>0},

one has

δH(z)=Rez1,logδH(z)=log(Rez1),

and this function is plurisubharmonic on H.

Facts & Assumptions

Given: The half-space H={Rez1>0}.

[L1]

Hartogs pseudoconvexity is defined through the function logδH (Plurisubharmonic exhaustions and Hartogs pseudoconvexity).

[L2]

A C2 function is plurisubharmonic exactly when its Levi form is semipositive (The C^2 Levi criterion for plurisubharmonicity).

Verification

technique · direct
1.1

The equal-radius polydisc about z stays in the half-space exactly while its first-coordinate radius is smaller than Rez1, so δH(z)=Rez1. The function u(z)=log(Rez1) is C2 on H and satisfies 2uz1z1(z)=14(Rez1)2,2uzjzk(z)=0 for (j,k)(1,1).

L1givenalgebra
2.1

Hence the Levi form of u is Lu(z;v)=v124(Rez1)20. By [L2], u=logδH is plurisubharmonic on H.

L2step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources