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Minus log boundary distance is plurisubharmonic on a half-space

Example

For the half-space

H={z∈Cm:Re⁡z1>0},

one has

δH(z)=Re⁡z1,−log⁡δH(z)=−log⁡(Re⁡z1),

and this function is plurisubharmonic on H.

Facts & Assumptions

Given: The half-space H={Re⁡z1>0}.

[L1]

Hartogs pseudoconvexity is defined through the function −log⁡δH (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
1.1L1givenalgebra

The equal-radius polydisc about z stays in the half-space exactly while its first-coordinate radius is smaller than Re⁡z1, so δH(z)=Re⁡z1. The function u(z)=−log⁡(Re⁡z1) is C2 on H and satisfies ∂2u∂z1∂z‾1(z)=14(Re⁡z1)2,∂2u∂zj∂z‾k(z)=0 for (j,k)≠(1,1).

2.1L2step 1.1∎

Hence the Levi form of u is Lu(z;v)=∣v1∣24(Re⁡z1)2≥0. By [L2], u=−log⁡δH is plurisubharmonic on H.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources