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Infinity-pole Green function recovered from a circular conductor

Statement

Assume the Axiom of Choice. Let a∈C, r>0, let K:=D(a,r)‾ be the closed disc, and let Ω:={z∈C:∣z−a∣>r} be its exterior. Then the normalized infinity-pole Green function of Ω is

gΩ(z,∞)=g(z):=log⁡∣z−a∣r(z∈Ω),

so that gΩ(z,∞)=VK−UμK(z) on Ω. Moreover g has boundary limit 0 at every point of the boundary circle {z:∣z−a∣=r}, with no exceptional set, and g(z)−log⁡∣z∣→−log⁡r as ∣z∣→∞.

The Axiom of Choice is spent through the equilibrium-measure input (Capacity of a disc and its circular equilibrium measure and Green function at infinity from the equilibrium potential); the explicit radial computations for g are choice-free.

Facts & Assumptions

Given: a∈C, r>0, the closed disc K:=D(a,r)‾, its exterior Ω:={z:∣z−a∣>r}, the Axiom of Choice, and the conventions of Logarithmic potential and energy of a positive compactly supported measure, Robin constant and logarithmic capacity of a compact set, Capacity-polar sets, quasi-everywhere, and subharmonic polar sets and Green function with a pole at infinity.

[F1]

Assume the Axiom of Choice. With K:=D(a,r)‾ and μ the normalized arclength measure dμ=dt/(2π) on the circle w=a+reit, μ is the unique equilibrium measure of K, Uμ(z)=log⁡1r(∣z−a∣≤r),Uμ(z)=log⁡1∣z−a∣(∣z−a∣≥r), and cap⁡(K)=r with Robin constant VK=log⁡(1/r); the same potential, capacity and equilibrium measure hold for the boundary circle (Capacity of a disc and its circular equilibrium measure).

[F2]

A Green function of Ω with pole at infinity and Robin constant VK is a function g:Ω→R that is positive and harmonic on Ω, satisfies g(z)−log⁡∣z∣→VK as ∣z∣→∞, is locally bounded near every point of ∂Ω, and has boundary limit 0 outside a Borel capacity-polar subset of ∂Ω; if existence and uniqueness hold the function is written gΩ(⋅,∞) (Green function with a pole at infinity).

[F3]

A compact set K with cap⁡(K)>0 is nonempty, Ω:=Ω(K) denotes the unbounded connected component of C∖K and is a complex domain with compact boundary ∂Ω⊆K, and VK=inf⁡μ∈P(K)I(μ) is a real number (Green function with a pole at infinity, Robin constant and logarithmic capacity of a compact set).

[F4]

Assume the Axiom of Choice. For K⊆C compact with cap⁡(K)>0, the unbounded component Ω of C∖K, the equilibrium measure μK and VK=log⁡1cap⁡(K), the function g(z)=VK−UμK(z) satisfies properties 1-4 of [F2], every function satisfying properties 1-4 equals it, and gΩ(z,∞)=VK−UμK(z) on Ω; hence K has exactly one Green function with pole at infinity (Green function at infinity from the equilibrium potential).

[F5]

For every a∈C the function ua(z)=log⁡∣z−a∣ is smooth and harmonic on C∖{a}; no choice principle is required (Logarithmic modulus is harmonic off its centre, Plane harmonic functions).

[F6]

If K⊆C is compact, the complement C∖K has exactly one unbounded connected component and every other component is bounded; for K=D(a,r)‾ that component is {z:∣z−a∣>r}, which is therefore a complex domain (The complement of a compact plane set has exactly one unbounded connected component, A complex domain is a nonempty connected open subset of C).

[F7]

A set is capacity-polar when every compact subset of it has capacity zero; ∅ is capacity-polar, and a subset of a capacity-polar set is capacity-polar (Capacity-polar sets, quasi-everywhere, and subharmonic polar sets).

[F8]

The Axiom of Choice implies Dependent Choice, which implies Countable Choice (AC implies DC implies countable choice).

Verification

technique · direct
1.1F1F3F6given

Setup. With K=D(a,r)‾ and Ω={z:∣z−a∣>r}, [F6] identifies Ω with the unbounded connected component of C∖K and makes it a complex domain with ∂Ω={z:∣z−a∣=r}; by [F1] cap⁡(K)=r>0, VK=log⁡1r, and the equilibrium measure μK has potential UμK(z)=log⁡1∣z−a∣ for ∣z−a∣≥r and UμK(z)=log⁡1r for ∣z−a∣≤r.

1.2F5algebragiven

Positivity and harmonicity. Define g(z):=log⁡∣z−a∣r for z∈Ω. Then g(z)>0 on Ω, since ∣z−a∣>r; and g is harmonic on Ω, because log⁡∣z−a∣ is harmonic on the open set C∖{a}⊇Ω by [F5] and subtracting the constant log⁡r leaves its Laplacian zero.

1.3givenalgebra

Boundary values, local boundedness and the infinity normalization. If ξ∈∂Ω, that is ∣ξ−a∣=r, then for z∈Ω, ∣g(z)−0∣=log⁡∣z−a∣r→log⁡rr=0 as z→ξ, and for every δ>0 one has sup⁡{g(z):z∈Ω, ∣z−ξ∣<δ}≤log⁡r+δr<+∞; moreover g(z)−log⁡∣z∣=log⁡∣z−a∣∣z∣−log⁡r→−log⁡r as ∣z∣→∞, because ∣z−a∣∣z∣→1 and log⁡ is continuous at 1.

2.1step 1.1step 1.2F1

Identification with the equilibrium potential. On Ω one has ∣z−a∣>r, so by [F1] UμK(z)=log⁡1∣z−a∣ there and VK−UμK(z)=log⁡1r−log⁡1∣z−a∣=log⁡∣z−a∣r=g(z).

2.2step 1.2step 1.3F2F7

The explicit function is a Green function. Steps 1.2 and 1.3 give properties 1, 2 and 3 of [F2] for g with Robin constant VK=log⁡1r; property 4 holds with the exceptional set E:=∅, because step 1.3 gives the boundary limit 0 at every point of ∂Ω, and ∅ is Borel and capacity-polar by [F7]. Hence g is a Green function of Ω with pole at infinity and Robin constant VK in the sense of [F2].

3.1step 1.1step 2.1step 2.2F4

Uniqueness and the notation. By step 1.1, K is compact with cap⁡(K)=r>0, so [F4] applies and gives: the Green function of Ω with pole at infinity exists, every function satisfying properties 1-4 of [F2] equals VK−UμK, and the notation gΩ(⋅,∞) is licensed with gΩ(z,∞)=VK−UμK(z) on Ω. By step 2.2 the function g satisfies properties 1-4, so g=VK−UμK=gΩ(⋅,∞) on Ω by step 2.1, and the boundary and normalization assertions are step 1.3.

4.1step 2.1step 3.1F1F4F8∎

Assembly and choice. Assertions of the Statement are exactly steps 2.1 and 3.1 (identification, notation, boundary limit on the entire circle and the infinity normalization); the boundary set is empty, so no exceptional set is needed. The Axiom of Choice is used only through [F1] and [F4], which by [F8] also supply Dependent and Countable Choice to their equilibrium-measure and Frostman inputs; the radial computations of steps 1.2, 1.3 and 2.1 are choice-free.

Remarks

Direct radial computation, not the general quasi-everywhere machinery. The properties of g are verified here by direct radial computation: positivity and harmonicity came from log⁡∣z−a∣ being harmonic off its centre, the boundary limit 0 holds at every boundary point because g=log⁡(∣z−a∣/r) extends continuously to the closed exterior with value 0 on the circle, and the normalization at infinity is the elementary limit log⁡(∣z−a∣/∣z∣)→0. In particular the exceptional set in property 4 of Green function with a pole at infinity may be taken empty here; the general quasi-everywhere uniqueness theorem (Green function at infinity from the equilibrium potential) is invoked only for the uniqueness clause, where the ordinary maximum principle on the exterior alone would not suffice for candidates whose boundary limit is assumed only quasi-everywhere.

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