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A Hartogs domain with a strictly plurisubharmonic exhaustion

Example

Assume the Axiom of Choice (AC). Put s(z,w):=∣w∣2e2∣z∣2 and Ω:={(z,w)∈C2:s(z,w)<1}={(z,w):∣w∣<e−∣z∣2}. Then Ω is a domain, the function ψ(z,w):=∣z∣2+∣w∣2+(1−s(z,w))−1 is a C∞ strictly plurisubharmonic exhaustion of Ω, and Ω is Levi pseudoconvex with strict positivity on complex tangents: Ls(p;ξ)>0 for every nonzero complex tangent vector ξ at every boundary point p of Ω. In particular Ω carries the continuous plurisubharmonic exhaustion function ψ.

Facts & Assumptions

Given: The Axiom of Choice; the functions s(z,w)=∣w∣2e2∣z∣2 and ψ=∣z∣2+∣w∣2+(1−s)−1; and the domain Ω={(z,w):s(z,w)<1}={(z,w):∣w∣<e−∣z∣2}.

[F1]

The Levi form of a C2 function u is Lu(a;v):=∑j,k∂2u∂zj∂zˉk(a)vjvk‾, and u is strictly plurisubharmonic when Lu(a;v)>0 for every a and every v≠0 (The Levi form and strict plurisubharmonicity).

[F2]

A domain Ω with C2 boundary is Levi pseudoconvex when every boundary point p has a neighbourhood U and a C2 function ρ with Ω∩U={ρ<0}, dρ(p)≠0, and Lρ(p;v)≥0 for every complex tangent vector v with ∑j∂ρ∂zj(p)vj=0 (Levi pseudoconvex domains).

[F3]

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

[F4]

A continuous plurisubharmonic exhaustion of Ω is a continuous plurisubharmonic u with {u≤c} compact in Ω for every real c (Plurisubharmonic exhaustions and Hartogs pseudoconvexity).

[F5]

The Wirtinger operators are ∂zk=12(∂xk−i∂yk) and ∂zˉk=12(∂xk+i∂yk) (Wirtinger operators in Cm).

[F6]

A path in a set A from x to y is a continuous γ:[0,1]→A with γ(0)=x, γ(1)=y, and A is path-connected when every two of its points are joined by a path (Paths, path-connected spaces and path components); every path-connected space is connected (Every path-connected space is connected, and every path component lies inside a component).

[F8]

Through Φ(z,w)=(Re⁡z,Im⁡z,Re⁡w,Im⁡w), the metric, the balls, the open sets and the compact sets of C2 are verbatim those of R4 (Complex m-space and its real coordinate dictionary).

[F9]

AC states that every family of nonempty sets has a choice function (The Axiom of Choice).

Choice use. AC is the ambient hypothesis recorded in the Example, and [F9] is cited as that hypothesis. The proof selects nothing: the star-shaped paths, the function ψ, the open sublevel bounds and the halving radius arguments are explicit formulas.

Verification

technique · direct
1.1F5F9givenalgebra

Put E:=e2∣z∣2, so that s=∣w∣2E is C∞ on C2; the Wirtinger operators of [F5] give ∂zs=2zˉs, ∂zˉs=2zs, ∂ws=wˉE, ∂wˉs=wE, and differentiating once more gives ∂2s/∂z∂zˉ=2s(1+2∣z∣2), ∂2s/∂w∂wˉ=E, ∂2s/∂z∂wˉ=2zˉwE, ∂2s/∂w∂zˉ=2zwˉE.

1.2F6givenalgebra

Ω={s<1} is open and nonempty (s(0,0)=0<1), and it is star-shaped about the origin: if (z,w)∈Ω and 0<t≤1, then ∣w∣<e−∣z∣2 gives ∣tw∣=t∣w∣<te−∣z∣2≤e−t2∣z∣2, since t≤1 and t2∣z∣2≤∣z∣2, so (tz,tw)∈Ω; at t=0 the point is (0,0)∈Ω. Thus each radial segment t↦(tz,tw) lies in Ω and joins (z,w) to the origin, so Ω is path-connected and, by [F6], connected. Hence Ω is a domain.

1.3givenalgebra

The function h(t):=(1−t)−1 is C∞ and satisfies h′(t)=(1−t)−2>0 and h′′(t)=2(1−t)−3>0 on (−∞,1), so h is strictly increasing and convex there; since s<1 on Ω, the function ψ=∣z∣2+∣w∣2+h(s) is C∞ and real-valued on Ω.

1.4F1F5algebra

For a real-valued C2 function u and a C2 function ϕ of one real variable one has ∂∂ˉ(ϕ∘u)=ϕ′(u) ∂∂ˉu+ϕ′′(u) ∂u∧∂ˉu, hence at every point Lϕ∘u(a;ξ)=ϕ′(u(a))Lu(a;ξ)+ϕ′′(u(a))∣∑j∂zju(a)ξj∣2.

2.1F3step 1.1algebra

The Hermitian matrix of the coefficients of step 1.1 is M=(2s(1+2∣z∣2)2zˉwE2zwˉEE): its diagonal entries are nonnegative, its determinant is 2s(1+2∣z∣2)E−4∣z∣2∣w∣2E2=2∣w∣2E2≥0, and for w=0 it is diag⁡(0,E) with E>0; hence Ls(a;ξ)=∑j,kξjMjkξˉk≥0 for all a,ξ, so [F3] makes s plurisubharmonic on C2, and at every point with w≠0 the matrix M is even positive definite (trace at least E>0, determinant 2∣w∣2E2>0).

2.2step 1.1step 1.2algebra

The boundary of Ω is {s=1}: a point with s(p)<1 lies in the open set Ω and a point with s(p)>1 has a neighbourhood disjoint from Ω, so neither is a boundary point; and if s(p)=1 then w≠0, so ∂wˉs(p)=wE≠0 and ds(p)≠0, and the points p−t∇s(p) for small t>0 satisfy s(p−t∇s(p))=1−t∣∇s(p)∣2+O(t2)<1, hence lie in Ω and converge to p, while p∉Ω; therefore p∈∂Ω.

3.1F1step 2.1step 1.3step 1.4algebra

Applying step 1.4 to ϕ=h and u=s, and adding the strictly plurisubharmonic term ∣z∣2+∣w∣2 with L∣z∣2+∣w∣2(a;ξ)=∣ξ1∣2+∣ξ2∣2, gives at every a∈Ω and every ξ≠0 the bound Lψ(a;ξ)=∣ξ1∣2+∣ξ2∣2+h′(s(a))Ls(a;ξ)+h′′(s(a))∣∂s(a;ξ)∣2≥∣ξ∣2>0, because h′>0, h′′>0 and Ls≥0 by step 2.1; hence ψ is strictly plurisubharmonic on Ω by [F1].

3.2F2step 2.1step 2.2

On {s=1} the matrix M of step 2.1 is positive definite, because w≠0 there; hence with the global defining function ρ:=s−1, the neighbourhood U=C2 and Lρ=Ls one has Ω={ρ<0}, dρ(p)≠0 and Lρ(p;ξ)>0 for every nonzero complex tangent vector ξ at every boundary point p; in particular Ω is Levi pseudoconvex in the sense of [F2].

4.1F4F7F8step 1.2step 1.3step 3.1step 3.2algebra∎

For c≤0 the sublevel set {ψ≤c} is empty; for c>0 it is contained in {∣z∣2+∣w∣2≤c}∩{s≤1−c−1}, on which ψ is continuous, so {ψ≤c} is a closed subset of C2; it is bounded, and it lies in Ω because s≤1−c−1<1; by [F8] it is a closed and bounded subset of R4, hence compact by [F7], and a compact subset of C2 contained in Ω is compact in Ω. Thus every sublevel set of ψ is compact, and with steps 1.2, 1.3 and 3.1 the function ψ is a continuous strictly plurisubharmonic exhaustion of Ω in the sense of [F4].

Remarks

  • Relation to the boundary-distance formulation. The library defines Hartogs pseudoconvexity by plurisubharmonicity of −log⁡δΩ (Plurisubharmonic exhaustions and Hartogs pseudoconvexity), and the direction Hartogs pseudoconvexity implies the existence of a continuous plurisubharmonic exhaustion is Hartogs pseudoconvexity yields a continuous plurisubharmonic exhaustion. The converse direction, which would upgrade the exhaustion ψ constructed here to plurisubharmonicity of −log⁡δΩ, is not part of the published statement of that theorem. This example therefore establishes the exhaustion and the strict Levi boundary condition, and records the identification with Hartogs pseudoconvexity as an obligation rather than assuming it.

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