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✓ 13 results · all verified · 8 also independently AI-judged

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Green Functions, Harmonic Measure, and Conformal Invariance

1 · Prerequisites

2 · Summary

This page develops the canonical Green kernel of a proper plane domain as the least nonnegative logarithmic-pole candidate. For bounded domains, a Perron corrector constructs that kernel under Countable Choice; the resulting kernel is positive off its pole and tends to zero at regular boundary points, with no boundary value asserted at irregular points. The logarithmic-modulus lemma supplies the harmonic pole term; analytic-boundary exhaustion and a smooth-corrector lemma support symmetry and domain monotonicity. A Riemann map gives the simply connected formula, while conformal covariance transports kernels between domains.

On bounded regular plane domains, harmonic measure is the unique Radon probability measure representing continuous boundary data through the Perron solution. The page derives the disc Poisson density, conformal transport, and harmonicity and comparison for Borel boundary sets. The final representation theorem uses a bounded regular $C^{1}$ domain, Dependent Choice, a bounded Laplacian, and the stated boundary regularity; its normal-derivative density has the additional analytic-boundary or explicit smooth-corrector hypotheses. The coefficient and sign are fixed by the convention $- Δ g_{Ω} (\cdot, a) = 2 π δ_{a}$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-30Open item page →

The canonical Green kernel of a plane domain

Definition

Let $Ω \subset C$ be a proper plane domain, that is, a nonempty connected open set with $Ω \neq C$ (A complex domain is a nonempty connected open subset of $C$ ), and let $a \in Ω$ . Moduli are those of Real and imaginary parts, complex conjugation, and modulus, and harmonicity is that of Plane harmonic functions.

Logarithmic-pole candidates. A function $u : Ω ∖ {a} \to R$ is a logarithmic-pole candidate at $a$ when:

$u (z) \geq 0$ for every $z \in Ω ∖ {a}$ ;
$u$ is harmonic on $Ω ∖ {a}$ ;
$u + \log ∣ z - a ∣$ extends harmonically across $a$ : there is a harmonic function $h$ on $Ω$ with $h (z) = u (z) + \log ∣ z - a ∣ (z \in Ω ∖ {a}) .$

The function $h$ in clause 3 is the harmonic corrector of $u$ at $a$ ; it is unique when it exists, because two harmonic functions on the connected open set $Ω$ that agree on the nonempty open set $Ω ∖ {a}$ agree on $Ω$ .

Order. Candidates at $a$ are compared pointwise on $Ω ∖ {a}$ .

Canonical Green function. Suppose the family of candidates at $a$ is nonempty. Its canonical Green function is its pointwise least member, when such a member exists: a candidate $g$ with $g \leq u$ pointwise on $Ω ∖ {a}$ for every candidate $u$ . A pointwise least member is unique, and it is written $z \mapsto g_{Ω} (z, a)$ . If the family is empty, or if it is nonempty but has no pointwise least member, then $g_{Ω} (\cdot, a)$ is not defined by this definition. The domain $Ω$ is called Greenian when $g_{Ω} (\cdot, a)$ exists for every $a \in Ω$ .

The coefficient of $\log ∣ z - a ∣$ is fixed to equal exactly one, so for the canonical kernel the corrector $h_{a} : = g_{Ω} (\cdot, a) + \log ∣ \cdot - a ∣$ is harmonic on all of $Ω$ and finite at $a$ .

Remarks

Leastness is a genuine restriction. On the punctured disc $Ω = D ∖ {0}$ with $a \neq 0$ , if $g_{Ω} (\cdot, a)$ exists then for every $t > 0$ the function $z \mapsto g_{Ω} (z, a) + t \log (1 / ∣ z ∣)$ is another logarithmic-pole candidate at $a$ . The added term is nonnegative on $Ω$ and harmonic there by Logarithmic modulus is harmonic off its centre, so the corrector still extends harmonically across $a \neq 0$ . The new candidate is strictly larger on $Ω$ . Thus the candidate clauses alone do not designate a unique function on this domain; the pointwise least-member clause does.
Promised boundary behaviour. On a bounded domain the canonical kernel has zero limit at every regular boundary point, in the sense of Barriers and regular boundary points, and no pointwise limit is required or asserted at an irregular boundary point. Both statements are proved later on this page together with the existence theorem; they are not part of the definition and are not assumed here.
Normalization relative to the PDE kernel. The published kernel $Φ (w) = - (2 π)^{- 1} \log ∣ w ∣$ of Fundamental solution for the positive operator minus Laplacian satisfies $2 π Φ (w) = - \log ∣ w ∣$ , so twice- $π$ times a Dirichlet Green function of that PDE page is a logarithmic-pole candidate with logarithmic coefficient one whenever the PDE correctors exist. The identification of the two normalizations, and the distributional identity $- Δ_{z} g_{Ω} (z, a) = 2 π δ_{a}$ , are not part of this definition: they are proved later on this page from that supplier.
Properness. The setting is a proper domain, $Ω \neq C$ ; nothing below asserts the existence of a canonical kernel on the whole plane.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Logarithmic modulus is harmonic off its centre

Statement

For every $a \in C$ , the real-valued function $u_{a} (z) = \log ∣ z - a ∣$ is smooth and harmonic on $C ∖ {a}$ . No choice principle is required.

Facts & Assumptions

Given: $a \in C$ and the plane harmonicity and modulus conventions (Plane harmonic functions, Real and imaginary parts, complex conjugation, and modulus).

[F1]

The real logarithm has derivative $(\log t)^{'} = 1 / t$ on $(0, \infty)$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

Proof

technique · direct differentiation

1.1F1givenalgebra

Write $z = x + i y$ , $a = p + i q$ , $X = x - p$ , $Y = y - q$ and $R = X^{2} + Y^{2} > 0$ . Then $u_{a} = \frac{1}{2} \log R$ . Repeatedly differentiating [F1] gives $(d / d t)^{m} \log t = (- 1)^{m - 1} (m - 1)! t^{- m}$ for $m \geq 1$ ; composition with the polynomial $R$ therefore makes $u_{a}$ smooth on $R > 0$ . Its first derivatives are $u_{x} = X / R$ and $u_{y} = Y / R$ .

2.1step 1.1givenalgebra∎

A further differentiation gives $u_{x x} = (Y^{2} - X^{2}) / R^{2}$ and $u_{y y} = (X^{2} - Y^{2}) / R^{2}$ . Their sum is zero at every $z \neq a$ . Thus $u_{a}$ is harmonic on $C ∖ {a}$ by the plane harmonicity definition.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Conformal covariance of the canonical planar Green kernel

Statement

Let $Ω, Ω^{'} \subseteq C$ be proper plane domains that are Greenian in the sense of The canonical Green kernel of a plane domain, and let $f : Ω \to Ω^{'}$ be a biholomorphism (Biholomorphic maps between complex domains). Then for every $a \in Ω$ and every $z \in Ω ∖ {a}$ , $g_{Ω} (z, a) = g_{Ω^{'}} (f (z), f (a)) .$ No extension of $f$ to the Euclidean boundaries of the two domains is assumed: both inequalities are produced from the least-candidate characterization of the canonical Green kernel, not from boundary values.

Facts & Assumptions

Given: Proper plane domains $Ω \neq C$ and $Ω^{'} \neq C$ , a biholomorphism $f : Ω \to Ω^{'}$ , and a pole $a \in Ω$ . Moduli are those of Real and imaginary parts, complex conjugation, and modulus and harmonicity is that of Plane harmonic functions.

[F1]

For a proper plane domain $Ω$ and $a \in Ω$ the canonical Green function $g_{Ω} (\cdot, a)$ , when it exists, is the pointwise least nonnegative logarithmic-pole candidate at $a$ : a function nonnegative on $Ω ∖ {a}$ , harmonic on $Ω ∖ {a}$ , such that $u (z) + \log ∣ z - a ∣$ extends harmonically across $a$ ; a Greenian domain is one for which this least candidate exists for every pole (The canonical Green kernel of a plane domain).

[F2]

A biholomorphism is a holomorphic bijection with holomorphic inverse (Biholomorphic maps between complex domains). An injective holomorphic map on a complex domain has nowhere-zero derivative (An injective holomorphic map has no critical point and is biholomorphic onto its image), holomorphic functions are real analytic and smooth (Holomorphic functions are real analytic and smooth in their two real coordinates), and if $f$ is holomorphic on a neighbourhood of $a$ with ${ord}_{a} f = 1$ then on some neighbourhood of $a$ one has $f (z) = (z - a) q (z)$ with $q$ holomorphic and $q (a) \neq 0$ (The order of a zero is the exponent in its local holomorphic factorization).

[F3]

If $h$ is holomorphic and nowhere zero on a disc $D (a, r)$ , there is a holomorphic logarithm $L$ on that disc with $\exp L = h$ (A nonvanishing holomorphic function on a disc has a holomorphic logarithm); the real and imaginary parts of a holomorphic function with $C^{2}$ components are harmonic (The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

[F4]

If $u$ is harmonic on an open set $V$ and $ϕ$ is holomorphic on an open set $U$ with $ϕ (U) \subseteq V$ , then $u \circ ϕ$ is harmonic on $U$ , and sums and differences of harmonic functions are harmonic (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate); the chain rule gives $(f^{- 1})^{'} (f (a)) \cdot f^{'} (a) = 1$ for a biholomorphism (The chain rule for complex derivatives).

Proof

technique · direct

1.1F2F3algebra

Since $f$ is injective and holomorphic on the domain $Ω$ , [F2] gives $f^{'} (a) \neq 0$ ; thus the holomorphic function $w \mapsto f (w) - f (a)$ has a zero of order one at $a$ , so by [F2] there is a disc $D (a, ρ) \subseteq Ω$ and a holomorphic $q$ on $D (a, ρ)$ with $f (z) - f (a) = (z - a) q (z)$ and $q (a) = f^{'} (a) \neq 0$ . Making $ρ$ smaller, $q$ is nowhere zero on $D (a, ρ)$ , so [F3] provides a holomorphic $L$ on $D (a, ρ)$ with $\exp L = q$ ; then both components of $L$ are $C^{2}$ by [F2], so $Re L$ is harmonic on $D (a, ρ)$ by [F3], and for $0 < ∣ z - a ∣ < ρ$ $\log ∣ f (z) - f (a) ∣ = \log ∣ z - a ∣ + Re L (z),$ because $∣ f (z) - f (a) ∣ = ∣ z - a ∣ \cdot ∣ q (z) ∣ = ∣ z - a ∣ e^{Re L (z)}$ .

1.2F2F4step 1.1

The inverse $f^{- 1} : Ω^{'} \to Ω$ is holomorphic by [F2], and by [F4] its derivative satisfies $(f^{- 1})^{'} (f (a)) = 1 / f^{'} (a) \neq 0$ ; the same computation as in step 1.1, applied to $f^{- 1}$ at the pole $f (a)$ , therefore supplies a disc $D (f (a), ρ^{'}) \subseteq Ω^{'}$ and a harmonic function $R$ on it with $\log ∣ f^{- 1} (w) - a ∣ = \log ∣ w - f (a) ∣ + R (w) (0 < ∣ w - f (a) ∣ < ρ^{'}) .$

2.1F1F4step 1.1

Define $G (z) : = g_{Ω^{'}} (f (z), f (a))$ for $z \in Ω ∖ {a}$ . Then $G$ is a logarithmic-pole candidate at $a$ on $Ω$ : it is nonnegative because $g_{Ω^{'}}$ is, it is harmonic by [F4] because $g_{Ω^{'}} (\cdot, f (a))$ is harmonic on $Ω^{'} ∖ {f (a)}$ and $f$ is holomorphic on $Ω ∖ {a}$ with image $Ω^{'} ∖ {f (a)}$ , and $G (z) + \log ∣ z - a ∣ = [g_{Ω^{'}} (f (z), f (a)) + \log ∣ f (z) - f (a) ∣] - Re L (z)$ extends across $a$ to a harmonic function, because the bracket is the harmonic corrector of $g_{Ω^{'}}$ composed with $f$ and $Re L$ is harmonic by step 1.1.

3.1F1F4step 1.2

Symmetrically, $H (w) : = g_{Ω} (f^{- 1} (w), a)$ for $w \in Ω^{'} ∖ {f (a)}$ is a logarithmic-pole candidate at $f (a)$ on $Ω^{'}$ : nonnegativity, harmonicity and the logarithmic pole follow as in step 2.1 with the roles of $f$ and $f^{- 1}$ exchanged, the local harmonic remainder being supplied by step 1.2.

3.2F1step 2.1

By the least-candidate characterization in [F1] applied on $Ω$ , the candidate $G$ of step 2.1 dominates the canonical kernel: $g_{Ω} (z, a) \leq g_{Ω^{'}} (f (z), f (a))$ for all $z \in Ω ∖ {a}$ .

4.1F1step 3.1

Applying [F1] on $Ω^{'}$ to the candidate $H$ of step 3.1 gives $g_{Ω^{'}} (w, f (a)) \leq g_{Ω} (f^{- 1} (w), a)$ for all $w \in Ω^{'} ∖ {f (a)}$ ; substituting $w = f (z)$ for $z \in Ω ∖ {a}$ yields the reverse inequality $g_{Ω^{'}} (f (z), f (a)) \leq g_{Ω} (z, a)$ .

5.1F1step 3.2step 4.1∎

The two inequalities of steps 3.2 and 4.1 are opposite, so $g_{Ω} (z, a) = g_{Ω^{'}} (f (z), f (a))$ for every $z \in Ω ∖ {a}$ . Only the least-candidate order of [F1] and the local behaviour of the two biholomorphisms were used: neither $f$ nor $f^{- 1}$ was extended to a boundary point.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Analytic-boundary exhaustion of a plane domain

Statement

Every plane domain $Ω \subset C$ admits an increasing sequence $D_{1} \subseteq D_{2} \subseteq \dots$ of relatively compact connected open subsets of $Ω$ whose boundaries are real-analytic regular in the following one-sided sense: for every $j$ and every $ζ \in \partial D_{j}$ there are a neighbourhood $U$ of $ζ$ and a real-analytic function $g$ of one real variable, defined on an open interval, such that, after relabelling the two coordinate axes if necessary, $\partial D_{j} \cap U = {(x, y) \in U : y = g (x)},$ and $D_{j} \cap U$ is one of the two connected components of $U ∖ {(x, y) \in U : y = g (x)}$ ; such that every compact $K \subseteq Ω$ lies in $D_{j}$ for all sufficiently large $j$ . If $A \subseteq Ω$ is finite, the sequence may be chosen with $A \subseteq D_{1}$ from the outset.

Facts & Assumptions

Given: A plane domain $Ω \subset C$ , that is, a nonempty connected open set (A complex domain is a nonempty connected open subset of $C$ ), and a finite set $A \subseteq Ω$ . For $c \in C$ and $r > 0$ we write $D (c, r) : = {z \in C : ∣ z - c ∣ < r}$ ; open and closed sets, interior, closure and boundary are those of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, compactness is that of Open cover, subcover, compact metric space, and compact subset of a metric space, and $\overline{D (c, r)}$ is the closed disc. A boundary is called real-analytic regular when it has the one-sided local graph description fixed in the Statement; such a boundary is in particular locally the zero set of a real-analytic function with nonvanishing gradient.

[F1]

The rationals are countably infinite, a product of two at most countable sets is at most countable, $N \times N$ is countable, subsets of at most countable sets are at most countable, and a nonempty set presented by a surjection $s : N \to S$ has a least-index element $x \mapsto \min {k : s (k) = x}$ . Consequently the points of $Q [i]$ , the positive rational radii, finite tuples of points of $Q [i]$ , and the polynomials in two variables with rational coefficients all sit in fixed explicitly enumerated at most countable families. For any fixed endpoints, polygonal paths whose intermediate vertices lie in $Q [i]$ are indexed by such finite tuples. A nonempty subfamily of any of these enumerated families has a least-index member ( $Q$ is countably infinite, A product of two at most countable sets is at most countable, $N \times N \approx N$ , Every subset of an at most countable set is at most countable, A nonempty set is at most countable iff it is a surjective image of $N$ ).

[F2]

Between any two real numbers there is a rational number, and a point of $C$ is described by its real part, imaginary part and modulus, whose elementary order properties make $Q [i]$ dense: given $z = p + i q$ and $η > 0$ , choosing rationals $p^{'} \in (p - η / 2, p + η / 2)$ and $q^{'} \in (q - η / 2, q + η / 2)$ gives $∣ z - (p^{'} + i q^{'}) ∣ < η$ (ℚ is dense in every Archimedean ordered field, Real and imaginary parts, complex conjugation, and modulus).

[F3]

An open connected subset of $R^{n}$ is polygonally connected, so any two of its points are joined by a polygonal path inside it; every connected component of an open subset of $R^{n}$ is open and polygonally connected, and a component is the largest connected subset containing each of its points, so every connected subset of an open set that meets a component is contained in that component (For an open subset of $R^{n}$ , connectedness, path-connectedness and polygonal connectedness are equivalent, Every connected component of an open subset of $R^{n}$ is open and polygonally connected, Connected components, quasicomponents, and totally disconnected spaces).

[F4]

A convex subset of the plane is path connected by the straight segment $t \mapsto (1 - t) x + t y$ ; every open or closed Euclidean disc is convex by the triangle inequality. Every path-connected space is connected; a polygonal path is a continuous map of a compact interval with connected image; and the continuous image of a connected space is connected (Every path-connected space is connected, and every path component lies inside a component, A finite concatenation of straight segments in $R^{n}$ is a continuous path, A continuous image of a connected space is connected, and connectedness is a topological property).

[F5]

A subset of $R^{2}$ is compact exactly when it is closed and bounded, every open cover of a compact set has a finite subcover, the continuous image of a compact set is compact, a continuous real function on a nonempty compact metric space attains its infimum, and a finite union of compact sets is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open cover, subcover, compact metric space, and compact subset of a metric space, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F6]

Distance to a nonempty set $S$ is the infimum $d (x, S)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space); $d (\cdot, S)$ is one-Lipschitz and hence continuous, with $d (x, S) = d (x, \overline{S})$ and $d (x, S) = 0$ exactly when $x \in \overline{S}$ . If $S$ is nonempty compact, $w \mapsto ∣ x - w ∣$ attains its minimum on $S$ by the extreme-value theorem [F5], so the point-to-set distance is attained. Also $d (K, L) = \inf_{z \in K} d (z, L)$ for nonempty sets, and $d (z, S) \geq d (w, S) - ∣ z - w ∣$ for all $z, w$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F7]

A union of connected sets in which every member meets one fixed connected member is connected, and a union of connected sets with a common point is connected (A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member).

[F8]

On a nonempty compact metric space, a unital real subalgebra of the real continuous functions that separates points is uniformly dense (Real Stone--Weierstrass theorem for compact metric spaces). On a fixed closed disc $\overline{D (0, M)}$ with $M \geq 1$ , every real-coefficient polynomial $P (x, y) = \sum_{(k, l) \in I} a_{k l} x^{k} y^{l}$ can be uniformly approximated by a rational-coefficient polynomial: choose rationals $q_{k l}$ with $\sum_{(k, l) \in I} ∣ a_{k l} - q_{k l} ∣ M^{k + l} < ϵ$ , which is possible by density of $Q$ in $R$ and finiteness of $I$ ; then $∣ P (x, y) - \sum q_{k l} x^{k} y^{l} ∣ < ϵ$ throughout the disc. Thus the rational-coefficient polynomials are also uniformly dense there.

[F9]

For a $C^{\infty}$ map $U \to R$ on an open $U \subseteq R^{2}$ , the critical value set is null, a subset of a null set is null, and no nondegenerate interval is null (Morse-Sard for Euclidean maps, Measure zero and content zero in $R^{m}$ by countable and finite cube covers, A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero).

[F10]

Polynomials in the two real coordinates are real analytic and $C^{\infty}$ maps of the plane, finite sums and products of $C^{\infty}$ Euclidean maps are $C^{\infty}$ , and a real-analytic map is smooth at every point of its domain (Real-analytic maps between open subsets of the coordinate plane, $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[F11]

A real-analytic map with invertible derivative has a real-analytic local inverse. If a real-analytic function $P$ of two variables has $D_{y} P$ invertible at a point where $P = 0$ , then near that point its zero set is exactly the graph of a real-analytic function of the first variable (Real analytic inverse and implicit functions).

Proof

technique · direct

1.1F1F4F5F10F11construct

Fix $Ω$ and the finite set $A$ . If $Ω = C$ , put $R_{0} : = 1 + \sup {∣ a ∣ : a \in A}$ (and $R_{0} : = 1$ when $A = \emptyset$ ) and $D_{j} : = D (0, R_{0} + j)$ for $j \geq 1$ : then $A \subseteq D_{1}$ , the discs increase with $j$ , each boundary ${∣ z ∣ = r}$ is the zero set of the real-analytic polynomial $F (z) = ∣ z ∣^{2} - r^{2}$ whose gradient $2 z$ does not vanish there, so after relabelling the axes [F11] exhibits that zero set near each of its points as the graph of a real-analytic function $g$ with the disc $D (0, r) = {F < 0}$ equal to one of the two local sides; hence each $\partial D_{j}$ is real-analytic regular in the one-sided sense of the Statement, and every compact $K \subseteq C$ is bounded, hence lies in $D_{j}$ for all large $j$ . So assume from now on that $Ω \neq C$ , in which case $C ∖ Ω$ is a nonempty closed set.

1.2F1F2F5F6choose

List as $Q_{1}, Q_{2}, \dots$ the closed discs $\overline{D (c, r)}$ with $c \in Q [i]$ , $r$ a positive rational and $\overline{D (c, r)} \subset Ω$ , in increasing order of the index of the datum $(c, r)$ in the enumeration of [F1]. Every compact $K \subseteq Ω$ is covered by finitely many of the $Q_{i}$ : for $z \in K$ openness gives $δ (z) : = d (z, C ∖ Ω) > 0$ by [F6], and [F1] together with [F2] supplies $c \in Q [i]$ and a positive rational $r$ with $∣ c - z ∣ < δ (z) / 4$ and $δ (z) / 4 < r < δ (z) / 2$ ; every $w$ with $∣ w - c ∣ \leq r$ then has $∣ w - z ∣ < 3 δ (z) / 4$ and therefore $d (w, C ∖ Ω) > δ (z) / 4 > 0$ by [F6], so $\overline{D (c, r)} \subset Ω$ while $z \in D (c, r)$ . The interiors of the listed discs thus cover the compact set $K$ , and compactness extracts a finite subcover, whose largest index we call $N$ .

1.3F5F6contradictiondischarge-contradiction

For every nonempty compact $K \subseteq Ω$ the number $δ (K) : = d (K, C ∖ Ω)$ is positive: by [F6] the continuous function $z \mapsto d (z, C ∖ Ω)$ attains over $K$ its minimum, which is $δ (K)$ , at some $z_{*} \in K$ , and $δ (K) = 0$ would put $z_{*}$ in $Ω \cap \overline{C ∖ Ω} = \emptyset$ . Moreover $d (z, C ∖ Ω) \geq δ (K) - d (z, K)$ for every $z \in C$ , by [F6] and the defining infimum of $d (z, K)$ .

1.4F1F2F3F4F5F6F7construct

Let $x_{0}$ be the least-indexed point of $Q [i]$ lying in $Ω$ , which exists by [F1] and [F2] because $Ω$ is nonempty and open, and let $c_{1}$ be the centre of $Q_{1}$ . For each fixed endpoint $p \in A \cup {c_{1}}$ , [F3] supplies a polygonal path in $Ω$ from $x_{0}$ to $p$ . Its compact image has positive distance from $C ∖ Ω$ : the distance function is continuous and positive on that compact subset of the open set $Ω$ , so it attains a positive minimum by [F5] and [F6]. Perturbing each non-endpoint vertex by less than half that distance keeps every segment in $Ω$ , because each corresponding point on a perturbed segment moves by at most the maximum endpoint perturbation. Density of $Q [i]$ therefore gives a path with the same endpoints and all intermediate vertices in $Q [i]$ . For each of these finitely many fixed endpoints, [F1] selects the least-indexed such path from the countable family of finite tuples of rational intermediate vertices; the endpoints, including arbitrary points of $A$ , remain fixed. Let $K_{1}$ be the union of these paths with $Q_{1}$ . Then $K_{1}$ is a nonempty compact connected subset of $Ω$ with $A \subseteq K_{1}$ : each path is a continuous image of a compact interval, hence compact and connected by [F4] and [F5], the disc $Q_{1}$ is compact and connected by [F4] and [F5], and every member of the union meets the fixed connected member that is the path from $x_{0}$ to $c_{1}$ , so [F7] applies.

2.1F6step 1.3constructalgebra

Fix an integer $j \geq 1$ and a nonempty compact connected set $K_{j} \subseteq Ω$ with $A \subseteq K_{j}$ ; the base case $K_{1}$ is supplied by step 1.4. Put $δ_{j} : = δ (K_{j}) > 0$ by step 1.3 and let $χ_{j} : [0, \infty) \to [- 1, 1]$ be the continuous function $χ_{j} (t) : = \max {- 1, \min {1, 1 - \frac{8}{δ_{j}} (t - \frac{δ_{j}}{4})}}, f_{j} (z) : = χ_{j} (d (z, K_{j})) .$ Then $χ_{j} = 1$ on $[0, δ_{j} / 4]$ and $χ_{j} = - 1$ on $[δ_{j} / 2, \infty)$ , so $f_{j}$ is continuous with $f_{j} = 1$ on ${z : d (z, K_{j}) < δ_{j} / 4}$ and $f_{j} = - 1$ on ${z : d (z, K_{j}) \geq δ_{j} / 2}$ .

3.1F1F5F6F8F10step 2.1construct

Let $M_{j} \geq 1$ be the least integer with ${z : d (z, K_{j}) \leq δ_{j} / 2} \subseteq D (0, M_{j} / 2)$ ; such an integer exists because that set is bounded, since compactness gives $B_{j}^{*} : = \max_{w \in K_{j}} ∣ w ∣ < \infty$ , and a nearest point $w \in K_{j}$ gives $∣ z ∣ \leq ∣ w ∣ + ∣ z - w ∣ \leq B_{j}^{*} + δ_{j} / 2$ by [F5] and [F6]. The real-coefficient polynomials in the two coordinates form a unital real subalgebra of $C (\overline{D (0, M_{j})}, R)$ containing both coordinate functions, hence separating points, so by [F8] some real-coefficient polynomial $P$ satisfies $\sup_{\overline{D (0, M_{j})}} ∣ P - f_{j} ∣ < 1 / 8$ . Since $M_{j} \geq 1$ , [F8] lets us approximate the finitely many coefficients of $P$ by rationals so that the resulting rational-coefficient polynomial differs from $P$ by less than $1 / 8$ uniformly on this disc. Thus some rational-coefficient polynomial $P_{j}$ satisfies $\sup_{\overline{D (0, M_{j})}} ∣ P_{j} - f_{j} ∣ < 1 / 4$ ; take the least-indexed one in the enumeration of [F1]. By [F10] the polynomial $P_{j}$ is real analytic and $C^{\infty}$ on the whole plane.

4.1F1F2F5F9step 3.1choose

Let $Z_{j} : = {z : ∣ z ∣ \leq M_{j}, \nabla P_{j} (z) = 0}$ , compact by [F5] and continuity of $\nabla P_{j}$ , and let $B_{j} : = P_{j} (Z_{j})$ , which is compact by [F5] and null by [F9] because it is a set of critical values of the $C^{\infty}$ function $P_{j}$ . The interval $(- \frac{1}{2}, \frac{1}{2})$ is nondegenerate, so it is not contained in $B_{j}$ by [F9]; being the complement of the closed set $B_{j}$ inside an interval, $(- \frac{1}{2}, \frac{1}{2}) ∖ B_{j}$ is open and nonempty, so by [F1] and [F2] it contains rationals, and we let $t_{j}$ be its least-indexed rational point. Consequently $\nabla P_{j} (z) \neq 0$ whenever $∣ z ∣ \leq M_{j}$ and $P_{j} (z) = t_{j}$ , since otherwise $t_{j} \in B_{j}$ .

5.1F3step 2.1step 3.1step 4.1construct

Let $D_{j}$ be the connected component of the open set ${z : P_{j} (z) > t_{j}}$ containing $K_{j}$ , which exists because $K_{j}$ is connected and $K_{j} \subseteq {P_{j} > t_{j}}$ : on $K_{j}$ we have $d (\cdot, K_{j}) = 0 < δ_{j} / 4$ , so $f_{j} = 1$ there by step 2.1, hence $P_{j} > 3 / 4 > t_{j}$ by steps 3.1 and 4.1. Thus $D_{j}$ is a nonempty open connected set with $K_{j} \subseteq D_{j}$ , the inclusion by maximality of components.

6.1F3step 2.1step 3.1step 4.1step 5.1

$D_{j} \subseteq D (0, M_{j})$ . Every point $w$ with $∣ w ∣ = M_{j}$ has $d (w, K_{j}) > δ_{j} / 2$ , because ${z : d (z, K_{j}) \leq δ_{j} / 2} \subseteq D (0, M_{j} / 2)$ by step 3.1; hence $f_{j} (w) = - 1$ by step 2.1 and $P_{j} (w) < - 3 / 4 < t_{j}$ by steps 3.1 and 4.1. So the circle ${∣ z ∣ = M_{j}}$ is disjoint from ${P_{j} > t_{j}}$ , and the connected set $D_{j}$ , which contains $K_{j} \subseteq D (0, M_{j} / 2)$ , lies in the component $D (0, M_{j})$ of the complement of that circle.

6.2F5F6step 1.3step 2.1step 3.1step 4.1step 5.1contradictiondischarge-contradiction

$D_{j} \subseteq Ω$ , and $\overline{D_{j}}$ is a compact subset of $Ω$ . Let $z \in D_{j}$ . If $d (z, K_{j}) \geq δ_{j} / 2$ then $f_{j} (z) = - 1$ by step 2.1 and $P_{j} (z) < - 3 / 4 < t_{j}$ by steps 3.1 and 4.1, contradicting $z \in D_{j}$ ; hence $d (z, K_{j}) < δ_{j} / 2$ , and by step 1.3 and [F6] $d (z, C ∖ Ω) \geq δ_{j} - d (z, K_{j}) > δ_{j} / 2 > 0$ , so $z \in Ω$ . Passing to closures, [F6] gives $\overline{D_{j}} \subseteq {z : d (z, K_{j}) \leq δ_{j} / 2}$ , and that set is closed, bounded by step 3.1 and contained in $Ω$ by the same distance inequality, hence compact by [F5].

7.1F3F4step 4.1step 6.1step 6.2contradictiondischarge-contradiction

$\partial D_{j} \subseteq {z : P_{j} (z) = t_{j}}$ , and $\nabla P_{j} \neq 0$ at every point of $\partial D_{j}$ . Let $z_{0} \in \partial D_{j}$ . Since $D_{j} \subseteq {P_{j} > t_{j}}$ and $P_{j}$ is continuous, while $z_{0}$ lies in the closure of $D_{j}$ , we get $P_{j} (z_{0}) \geq t_{j}$ ; and $z_{0} \in \overline{D_{j}} \subseteq D (0, M_{j})$ by steps 6.1 and 6.2. If $P_{j} (z_{0}) > t_{j}$ , then ${P_{j} > t_{j}}$ contains a disc $B$ around $z_{0}$ ; $B$ is connected by [F4] and meets $D_{j}$ because $z_{0}$ is a boundary point of $D_{j}$ , so $B \subseteq D_{j}$ by maximality of components, making $z_{0}$ an interior point of $D_{j}$ and contradicting $z_{0} \in \partial D_{j}$ . Hence $P_{j} (z_{0}) = t_{j}$ , and $\nabla P_{j} (z_{0}) \neq 0$ by step 4.1.

7.2F1F2F3F4F5F7step 1.4step 6.2construct

Construction of the next compact set. Let $y_{j}$ be the least-indexed point of $Q [i]$ lying in $D_{j}$ , which exists by [F1] and [F2] because $D_{j}$ is nonempty and open, let $c_{j + 1}$ be the centre of $Q_{j + 1}$ , and let $P_{j + 1}$ be the union of the two least-indexed polygonal paths whose intermediate vertices lie in $Q [i]$ , from $x_{0}$ to $y_{j}$ and from $x_{0}$ to $c_{j + 1}$ . These paths exist by the argument of step 1.4; their endpoints are rational as well. Then $K_{j + 1} : = \overline{D_{j}} \cup Q_{j + 1} \cup P_{j + 1}$ is a nonempty compact connected subset of $Ω$ with $A \subseteq K_{j + 1}$ , so that the construction of step 2.1 can be applied to it: compactness follows from [F5] and step 6.2, and connectedness from [F7], because $\overline{D_{j}}$ meets $P_{j + 1}$ at $y_{j}$ , while $Q_{j + 1}$ meets $P_{j + 1}$ at $c_{j + 1}$ , and each of the three members is connected by [F3], [F4] and step 6.2. Moreover $A \subseteq K_{j} \subseteq D_{j} \subseteq K_{j + 1}$ .

8.1F3F4F10F11step 7.1construct

The boundary $\partial D_{j}$ is real-analytic regular. Fix $z_{0} \in \partial D_{j}$ . By step 7.1, after relabelling axes, $D_{y} P_{j} (z_{0}) \neq 0$ . Apply the inverse assertion of [F11] to $H (x, y) = (x, P_{j} (x, y) - t_{j})$ , whose Jacobian determinant at $z_{0}$ is $D_{y} P_{j} (z_{0})$ . Restrict its analytic inverse $K$ to a rectangle $V = I \times (- ϵ, ϵ)$ about $H (z_{0})$ and put $U = K (V)$ . The first coordinate identity forces $K (x, s) = (x, k (x, s))$ . Thus the zero set in $U$ is the graph $y = g (x) : = k (x, 0)$ , and the positive and negative sides are respectively $K (I \times (0, ϵ))$ and $K (I \times (- ϵ, 0))$ . Both are connected, being continuous images of convex rectangles; they are the two components of the complement of the graph in $U$ . The positive side meets $D_{j}$ since $z_{0} \in \partial D_{j}$ , so maximality of the component $D_{j}$ puts that whole side in $D_{j}$ . Conversely $D_{j} \cap U$ lies in that side by its definition. Every graph point is approached by points of the positive side and belongs to neither open side, hence $\partial D_{j} \cap U$ is exactly the graph. This proves the required one-sided regularity without assuming the sign of $D_{y} P_{j}$ . The same local argument with the negative side applies to the discs in step 1.1.

8.2F3step 2.1step 3.1step 4.1step 5.1step 7.2

Applying the construction of steps 2.1 through 5.1 to the admissible set $K_{j + 1}$ of step 7.2 produces the connected component $D_{j + 1}$ of ${P_{j + 1} > t_{j + 1}}$ containing $K_{j + 1}$ , so $K_{j + 1} \subseteq D_{j + 1}$ . Since $D_{j} \subseteq \overline{D_{j}} \subseteq K_{j + 1}$ by step 7.2 and $D_{j + 1}$ is a component, maximality of components yields $D_{j} \subseteq D_{j + 1}$ .

9.1step 1.2step 1.4step 5.1step 7.2step 8.2cases

Every compact $K \subseteq Ω$ lies in $D_{m}$ for all sufficiently large $m$ . By step 1.2 there is $N$ with $K \subseteq Q_{1} \cup \dots \cup Q_{N}$ . For each $m \geq 1$ the set $Q_{m}$ satisfies $Q_{m} \subseteq K_{m}$ by steps 1.4 and 7.2, and $K_{m} \subseteq D_{m}$ by step 5.1 applied to $K_{m}$ , while $D_{m} \subseteq D_{m^{'}}$ for $m^{'} \geq m$ by step 8.2; hence $K \subseteq D_{N} \subseteq D_{m^{'}}$ for every $m^{'} \geq N$ .

10.1F1F2step 5.1step 6.2step 7.2step 8.1step 8.2step 9.1∎

The sequence $D_{1} \subseteq D_{2} \subseteq \dots$ obtained by applying steps 2.1 through 7.2 inductively, starting from $K_{1}$ of step 1.4, consists of nonempty relatively compact connected open subsets of $Ω$ with real-analytic regular boundary by steps 6.2 and 8.1, contains $A$ in $D_{1}$ because $A \subseteq K_{1} \subseteq D_{1}$ by steps 1.4 and 5.1, and exhausts $Ω$ in the required sense by step 9.1. Every selection made above is either a finite selection or a least-index selection in one of the fixed at most countable families of [F1], or the least-indexed rational point of a nonempty open set, whose existence is [F2]; no choice principle was used.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Green functions exist on all bounded plane domains

Statement

Assume Countable Choice. Let $Ω \subseteq C$ be a bounded complex domain and let $a \in Ω$ . Put $F_{a} (z) : = - \log ∣ z - a ∣$ , let $b_{a} : = F_{a} ∣_{\partial Ω}$ be the boundary datum it induces, and let $h_{a} : = H_{b_{a}}$ be the regularized Perron envelope of The Perron envelope and its regularization with datum $b_{a}$ . Then $g_{Ω} (z, a) : = F_{a} (z) - h_{a} (z) (z \in Ω ∖ {a})$ is the canonical positive Green kernel of The canonical Green kernel of a plane domain: it is harmonic on $Ω ∖ {a}$ , the function $g_{Ω} (\cdot, a) + \log ∣ \cdot - a ∣ = - h_{a}$ extends harmonically across $a$ , it is strictly positive off $a$ , it is bounded on ${z \in Ω : ∣ z - a ∣ \geq δ}$ for every $δ > 0$ , and $\lim_{\begin{array}{c} z \to ζ \\ z \in Ω \end{array}} g_{Ω} (z, a) = 0$ at every regular boundary point $ζ \in \partial Ω$ (Barriers and regular boundary points); no boundary value is prescribed at an irregular boundary point. Moreover $- Δ_{z} T_{g_{Ω} (\cdot, a)} = 2 π δ_{a}$ as distributions on $Ω$ . Countable Choice is used for the cited distributional Poisson identity; the cited Perron envelope theorem has a choice-free directed-supremum proof. The boundary values of $g_{Ω}$ at regular points are the only boundary information.

Facts & Assumptions

Given: A bounded complex domain $Ω \subseteq C$ (A complex domain is a nonempty connected open subset of $C$ ), a point $a \in Ω$ , and Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Perron families and envelopes are those of The Perron lower family for continuous boundary data and The Perron envelope and its regularization, the Perron datum is $b_{a} = F_{a} ∣_{\partial Ω}$ with $F_{a} (z) = - \log ∣ z - a ∣$ , harmonicity and subharmonicity are those of Plane harmonic functions and Subharmonic functions on plane domains, distributions are those of Distributional harmonicity and Poisson's equation on an open subset of Rn, and the kernel candidate for $- Δ$ is $Φ$ from Fundamental solution for the positive operator minus Laplacian.

[F1]

For a proper plane domain $D$ and $b \in D$ , the canonical Green function $g_{D} (\cdot, b)$ , when it exists, is the pointwise least nonnegative function that is harmonic on $D ∖ {b}$ and satisfies: $g_{D} (\cdot, b) + \log ∣ \cdot - b ∣$ extends harmonically across $b$ (The canonical Green kernel of a plane domain).

[F2]

For a continuous datum $φ$ on the boundary of a bounded complex domain, the Perron family $P (φ, Ω)$ is nonempty, every $v \in P (φ, Ω)$ satisfies $v \leq M : = \max_{\partial Ω} φ$ , the constant $m : = \min_{\partial Ω} φ$ lies in the family, and $m \leq U_{φ} \leq M$ (The Perron family is nonempty and uniformly bounded by the boundary data).

[F3]

The regularized Perron envelope $H_{φ}$ is harmonic on $Ω$ (The regularized Perron envelope is harmonic), and by definition $H_{φ} (z) = \lim_{ρ ↓ 0} \sup {U_{φ} (w) : w \in Ω, ∣ w - z ∣ < ρ}$ , so $U_{φ} \leq H_{φ}$ (The Perron envelope and its regularization).

[F4]

A harmonic function is $C^{2}$ with $Δ u = 0$ , a $C^{2}$ function with $Δ u \geq 0$ is subharmonic, and a sum of a subharmonic function and a harmonic function is subharmonic (Plane harmonic functions, A C^2 function is subharmonic exactly when its Laplacian is nonnegative, Positive linear combinations and finite maxima preserve subharmonicity, Subharmonic functions on plane domains).

[F5]

The function $\log ∣ \cdot ∣$ is harmonic on $C ∖ {0}$ (Logarithmic modulus is harmonic off its centre), and composition with translations and other holomorphic maps preserves harmonicity (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate).

[F6]

A nonnegative harmonic function on a domain in $R^{n}$ , $n \geq 2$ , is either identically zero or strictly positive everywhere (Nonnegative harmonic function with an interior zero vanishes).

[F7]

Assume Countable Choice. $Φ_{2} (x) = - \frac{1}{2 π} \log ∣ x ∣$ for $x \neq 0$ (Fundamental solution for the positive operator minus Laplacian); the associated regular distribution satisfies $- Δ T_{Φ (\cdot - y)} = δ_{y}$ in $D^{'} (R^{n})$ for every $y$ (The negative Laplacian of the fundamental solution is the unit Dirac distribution); the map $f \mapsto T_{f}$ from $L_{l o c}^{1}$ modulo almost-everywhere equality to distributions is linear (Locally integrable functions embed in distributions); and distributional differentiation extends classical differentiation of $C^{k}$ functions and is linear (Distributional differentiation is continuous and commutes).

[F8]

A regular boundary point $ζ$ of a bounded complex domain is one at which the regularized Perron envelope of every continuous datum has limit equal to the datum at $ζ$ (Barriers and regular boundary points).

[F9]

$\partial Ω$ is compact because $Ω$ is bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct

1.1F5F9given

Since $a$ is an interior point of the bounded domain $Ω$ , the distance $δ_{0} : = d (a, \partial Ω)$ is positive, and $∣ z - a ∣$ is bounded above on $Ω$ by the diameter of $Ω$ ; by [F9] the continuous function $b_{a}$ attains finite extrema $m : = \min_{\partial Ω} b_{a}$ and $M : = \max_{\partial Ω} b_{a}$ on $\partial Ω$ . Also $b_{a}$ is continuous: $z \mapsto ∣ z - a ∣$ is continuous and takes values bounded away from $0$ on $\partial Ω$ , and $t \mapsto - \log t$ is continuous and real on positive $t$ .

1.2F2F4F5given

Every Perron lower function is dominated by $F_{a}$ : let $v \in P (b_{a}, Ω)$ and put $w : = v - F_{a}$ on the bounded complex domain $Ω ∖ {a}$ . Then $w$ is subharmonic by [F4], since $v$ is subharmonic and $F_{a} = - \log ∣ \cdot - a ∣$ is harmonic on $Ω ∖ {a}$ by [F5]. At every boundary point of $Ω ∖ {a}$ the boundary limsup of $w$ is at most $0$ : at $η \in \partial Ω$ the function $F_{a}$ is continuous with value $b_{a} (η)$ , so ${lim sup}_{z \to η} w \leq {lim sup}_{z \to η} v - F_{a} (η) \leq b_{a} (η) - b_{a} (η) = 0$ ; at the puncture $z = a$ one has $v \leq M$ on $Ω$ by [F2] while $F_{a} (z) \to + \infty$ , so ${lim sup}_{z \to a} w \leq M - \infty < 0$ . Since $Ω ∖ {a}$ is a bounded complex domain and the datum $ψ \equiv 0$ is continuous on its boundary, $w \in P (0, Ω ∖ {a})$ , and [F2] applied to that domain gives $w \leq 0$ on $Ω ∖ {a}$ .

1.3F1F4givenalgebra

Leastness among all candidates: let $k$ be any nonnegative logarithmic-pole candidate at $a$ on $Ω$ . Near $a$ the function $k + \log ∣ \cdot - a ∣$ agrees with a harmonic function $ϕ$ on some disc $B (a, r) \subseteq Ω$ by [F1], so on $B (a, r) ∖ {a}$ one has $F_{a} - k = (F_{a} + \log ∣ \cdot - a ∣) - (k + \log ∣ \cdot - a ∣) = - (k + \log ∣ \cdot - a ∣) = - ϕ$ , since $F_{a} + \log ∣ \cdot - a ∣ = 0$ ; gluing the harmonic functions $F_{a} - k$ on $Ω ∖ {a}$ and $- ϕ$ on $B (a, r)$ along their agreement on the connected set $B (a, r) ∖ {a}$ produces a harmonic extension $\tilde{V}$ of $F_{a} - k$ to all of $Ω$ .

1.4F8given

Boundary behaviour: at a regular boundary point $ζ \in \partial Ω$ one has $\lim_{z \to ζ} h_{a} (z) = b_{a} (ζ)$ by [F8], while $F_{a}$ is continuous at $ζ$ with $F_{a} (ζ) = b_{a} (ζ)$ ; hence $\lim_{z \to ζ} g_{Ω} (z, a) = b_{a} (ζ) - b_{a} (ζ) = 0$ . At an irregular boundary point no limit is asserted, and none was used: the construction of $g_{Ω}$ involved only $F_{a}$ and the Perron envelope of $b_{a}$ .

2.1F2F3step 1.1

Let $h_{a} : = H_{b_{a}}$ . By [F3] the function $h_{a}$ is harmonic on $Ω$ , and since $m \leq U_{b_{a}} \leq M$ while $H_{b_{a}}$ is the limit of suprema of values of $U_{b_{a}}$ over shrinking discs, $m \leq h_{a} \leq M$ on $Ω$ ; so $h_{a}$ is bounded.

2.2F2F3step 1.2

Consequently $U_{b_{a}} (z) = \sup {v (z) : v \in P (b_{a}, Ω)} \leq F_{a} (z)$ for every $z \in Ω ∖ {a}$ by [F2] and step 1.2, and then, since $F_{a}$ is continuous at every $z \neq a$ , [F3] gives $h_{a} (z) = \lim_{ρ ↓ 0} \sup {U_{b_{a}} (w) : ∣ w - z ∣ < ρ} \leq \lim_{ρ ↓ 0} \sup {F_{a} (w) : ∣ w - z ∣ < ρ} = F_{a} (z) .$ Hence $g_{Ω} (z, a) : = F_{a} (z) - h_{a} (z) \geq 0$ for $z \in Ω ∖ {a}$ .

2.3F1F2F3F4step 1.3

The extension $\tilde{V}$ of step 1.3 belongs to $P (b_{a}, Ω)$ : it is harmonic, hence subharmonic, on $Ω$ by [F4], and at each $η \in \partial Ω$ its boundary limsup is $lim sup (F_{a} - k) \leq b_{a} (η) - lim inf k \leq b_{a} (η)$ , because $k \geq 0$ . Therefore $U_{b_{a}} \geq \tilde{V}$ on $Ω$ by [F2] and [F3], so $h_{a} \geq U_{b_{a}} \geq \tilde{V}$ ; restricting to $Ω ∖ {a}$ , where $\tilde{V} = F_{a} - k$ , gives $F_{a} - h_{a} \leq k$ , that is $g_{Ω} (z, a) \leq k (z)$ .

3.1F1F4step 2.1step 2.2

The function $g_{Ω} (\cdot, a)$ is harmonic on $Ω ∖ {a}$ , being the difference of the harmonic functions $F_{a}$ and $h_{a}$ there by [F4] and steps 2.1, 2.2. Moreover $g_{Ω} (z, a) + \log ∣ z - a ∣ = F_{a} (z) + \log ∣ z - a ∣ - h_{a} (z) = - h_{a} (z) (z \in Ω ∖ {a}),$ and the right-hand side is harmonic on all of $Ω$ by step 2.1; so $g_{Ω} (\cdot, a) + \log ∣ \cdot - a ∣$ extends harmonically across $a$ and $g_{Ω} (\cdot, a)$ is a nonnegative logarithmic-pole candidate at $a$ in the sense of [F1].

3.2step 2.1step 2.2givenalgebra

Boundedness away from the pole: fix $δ > 0$ . On the set ${z \in Ω : ∣ z - a ∣ \geq δ}$ the function $F_{a}$ satisfies $∣ F_{a} (z) ∣ \leq \max {∣ \log δ ∣, ∣ \log R ∣}$ where $R$ is the diameter of $Ω$ , and $∣ h_{a} ∣ \leq \max {∣ m ∣, ∣ M ∣}$ by step 2.1; hence $∣ g_{Ω} (z, a) ∣ \leq ∣ F_{a} (z) ∣ + ∣ h_{a} (z) ∣$ is bounded there.

4.1F6step 2.1step 2.2step 3.1

The candidate is strictly positive off the pole: if $g_{Ω} (w, a) = 0$ for some $w \neq a$ , then the nonnegative harmonic function $g_{Ω} (\cdot, a)$ on the complex domain $Ω ∖ {a}$ would be identically zero by [F6]; but $g_{Ω} (z, a) \geq F_{a} (z) - M \to + \infty$ as $z \to a$ by steps 2.1 and 2.2, so $g_{Ω} (\cdot, a)$ is unbounded and not identically zero. Hence $g_{Ω} (z, a) > 0$ for every $z \in Ω ∖ {a}$ .

4.2F4F7step 2.1step 3.1

Distributional normalization: on $Ω ∖ {a}$ one has $F_{a} = 2 π Φ_{2} (\cdot - a)$ by the two-dimensional branch of [F7], and $h_{a} \in C^{2} (Ω)$ with $Δ h_{a} = 0$ ; extend $g_{Ω} (\cdot, a)$ arbitrarily at the single point $a$ . By the linearity of the embedding $f \mapsto T_{f}$ in [F7], $T_{g_{Ω} (\cdot, a)} = T_{F_{a}} - T_{h_{a}} = 2 π T_{Φ_{2} (\cdot - a)} - T_{h_{a}}$ , and by the linearity of distributional differentiation and its agreement with classical differentiation on $C^{2}$ functions, $- Δ T_{g_{Ω} (\cdot, a)} = - 2 π Δ T_{Φ_{2} (\cdot - a)} + Δ T_{h_{a}} = 2 π δ_{a} - T_{Δ h_{a}} = 2 π δ_{a},$ because $- Δ T_{Φ_{2} (\cdot - a)} = δ_{a}$ by [F7] and $Δ T_{h_{a}} = T_{Δ h_{a}} = T_{0} = 0$ by [F7] and [F4]. This is the sense in which $- Δ_{z} g_{Ω} (z, a) = 2 π δ_{a}$ on $Ω$ .

5.1F1step 3.1step 4.1step 2.3

Since $k$ was an arbitrary nonnegative logarithmic-pole candidate, steps 3.1, 4.1 and 2.3 show that $g_{Ω} (\cdot, a)$ is the pointwise least such candidate and is strictly positive; by [F1] it is the canonical Green kernel $g_{Ω} (\cdot, a)$ of $Ω$ at $a$ .

6.1F3F7given∎

The stated Countable Choice is used exactly in the cited distributional identity and classical-differentiation comparison of [F7]. The cited Perron envelope theorem [F3] now uses a choice-free directed-supremum argument, and the construction of $b_{a}$ , the comparison of Perron lower functions and the boundary limits at regular points require no additional choice principle.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

A planar barrier forces the regularized Perron envelope to have the prescribed boundary limit

Statement

Let $Ω \subseteq C$ be a bounded complex domain, let $ζ \in \partial Ω$ , and let $b$ be a barrier at $ζ$ in the sense of the published definition (Barriers and regular boundary points). Then $ζ$ is a regular boundary point: for every continuous boundary datum $φ : \partial Ω \to R$ the regularized Perron envelope satisfies $\lim_{\begin{array}{c} z \to ζ \\ z \in Ω \end{array}} H_{φ} (z) = φ (ζ) .$ The limit is produced from the lower Perron family by squeezing the envelope between the barrier bounds; no boundary limit of $H_{φ}$ at any other boundary point and no converse implication is used.

Facts & Assumptions

Given: A bounded complex domain $Ω \subseteq C$ (A complex domain is a nonempty connected open subset of $C$ ), a point $ζ \in \partial Ω$ , a barrier $b : Ω \to [- \infty, 0)$ at $ζ$ , a continuous datum $φ : \partial Ω \to R$ , and $ε > 0$ . Here $\partial Ω$ is the topological boundary (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space), boundedness is boundedness of the diameter (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and compactness is that of Open cover, subcover, compact metric space, and compact subset of a metric space. A barrier at $ζ$ is a subharmonic $b < 0$ on $Ω$ with $b (z) \to 0$ as $z \to ζ$ inside $Ω$ and with the property that for every neighbourhood $V$ of $ζ$ there is $c_{V} < 0$ such that ${lim sup}_{z \to η, z \in Ω} b (z) \leq c_{V}$ for every $η \in \partial Ω ∖ V$ .

[F1]

A complex domain is a nonempty connected open subset of $C$ , and the Perron lower family $P (φ, Ω)$ consists of the subharmonic $v : Ω \to [- \infty, \infty)$ with ${lim sup}_{z \to η, z \in Ω} v (z) \leq φ (η)$ at every $η \in \partial Ω$ ; the Perron envelope is the pointwise supremum $U_{φ} = \sup {v : v \in P (φ, Ω)}$ and its upper semicontinuous regularization is $H_{φ} (z) = \lim_{ρ ↓ 0} \sup {U_{φ} (w) : w \in Ω, ∣ w - z ∣ < ρ}$ (A complex domain is a nonempty connected open subset of $C$ , The Perron lower family for continuous boundary data, The Perron envelope and its regularization).

[F2]

A barrier at $ζ$ is a subharmonic function $b$ on $Ω$ with $b < 0$ , with $b (z) \to 0$ as $z \to ζ$ inside $Ω$ , and with the stated family of negative constants $c_{V}$ (Barriers and regular boundary points).

[F3]

Nonnegative linear combinations of subharmonic functions are subharmonic, and every harmonic function is subharmonic because a $C^{2}$ function is subharmonic exactly when $Δ u \geq 0$ (Positive linear combinations and finite maxima preserve subharmonicity, A C^2 function is subharmonic exactly when its Laplacian is nonnegative, Plane harmonic functions).

[F4]

Every member $v$ of $P (ψ, Ω)$ for a continuous datum $ψ$ satisfies $v \leq \max_{\partial Ω} ψ$ on $Ω$ (The Perron family is nonempty and uniformly bounded by the boundary data).

[F5]

A subset of $C$ is compact exactly when it is closed and bounded, a closed subset of a compact set is compact, and a continuous real function on a nonempty compact set attains its maximum and minimum (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open cover, subcover, compact metric space, and compact subset of a metric space, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct

1.1F2F5choosealgebra

The boundary $\partial Ω$ is closed, and it is bounded because $Ω$ is bounded, so $\partial Ω$ is compact by [F5]. Hence there is a neighbourhood $V$ of $ζ$ with $∣ φ (η) - φ (ζ) ∣ < ε$ for every $η \in \partial Ω \cap V$ , namely a disc around $ζ$ whose intersection with the boundary lies inside the open set $φ^{- 1} (φ (ζ) - ε, φ (ζ) + ε)$ . By the barrier property [F2] there is $c_{V} < 0$ with ${lim sup}_{z \to η, z \in Ω} b (z) \leq c_{V}$ for every $η \in \partial Ω ∖ V$ . The set $S : = \partial Ω ∖ V$ is a closed subset of the compact set $\partial Ω$ , hence compact by [F5]. Put $D = 0$ if $S = \emptyset$ ; otherwise, since the two displayed functions are continuous on the nonempty compact set $S$ , put $D : = \max (0, \max_{η \in S} \max {φ (ζ) - ε - φ (η), φ (η) - φ (ζ) - ε}) .$ This finite nonnegative number bounds both deviations that will be needed. Let $A$ be the least positive integer with $A \cdot (- c_{V}) > D$ , which exists because $(- c_{V}) > 0$ . Then for every $η \in S$ both $φ (ζ) - ε + A c_{V} \leq φ (η)$ and $φ (η) + A c_{V} \leq φ (ζ) + ε$ , since $A (- c_{V}) > D$ bounds respectively $φ (ζ) - ε - φ (η)$ and $φ (η) - φ (ζ) - ε$ .

2.1F1F2F3step 1.1cases

The function $ℓ (z) : = φ (ζ) - ε + A b (z)$ is subharmonic on $Ω$ by [F3], since $b$ is subharmonic, $A \geq 0$ and constants are harmonic. It belongs to $P (φ, Ω)$ : at a boundary point $η \in \partial Ω \cap V$ its limsup is at most $φ (ζ) - ε + 0 < φ (η)$ by the choice of $V$ and $b < 0$ , and at $η \in \partial Ω ∖ V$ it is at most $φ (ζ) - ε + A c_{V} \leq φ (η)$ by step 1.1. Therefore $U_{φ} \geq ℓ$ on $Ω$ by [F1], that is $U_{φ} (z) \geq φ (ζ) - ε + A b (z) (z \in Ω) .$

2.2F1F3F4step 1.1cases

Let $v \in P (φ, Ω)$ and put $w : = v + A b$ . Then $w$ is subharmonic on $Ω$ by [F3], and at every boundary point its limsup is at most $φ (ζ) + ε$ : at $η \in \partial Ω \cap V$ we have $lim sup w \leq φ (η) + 0 < φ (ζ) + ε$ by step 1.1 and $b < 0$ , while at $η \in \partial Ω ∖ V$ we have $lim sup w \leq φ (η) + A c_{V} \leq φ (ζ) + ε$ by the upper-deviation bound in step 1.1. So $w \in P (φ (ζ) + ε, Ω)$ for the constant datum $φ (ζ) + ε$ , and [F4] gives the pointwise bound $v (z) \leq φ (ζ) + ε - A b (z) (z \in Ω) .$

3.1F1F2step 2.1step 2.2

Fix $δ > 0$ . Since $b (z) \to 0$ as $z \to ζ$ inside $Ω$ [F2], there is a neighbourhood $N$ of $ζ$ with $- δ \leq b \leq 0$ on $N \cap Ω$ . Steps 2.1 and 2.2 apply to every $z \in N \cap Ω$ and give, after taking suprema in $v$ and using that $b \leq 0$ only strengthens the upper bound, $φ (ζ) - ε - A δ \leq ℓ (z) \leq U_{φ} (z) \leq φ (ζ) + ε + A δ (z \in N \cap Ω) .$

4.1F1step 3.1

The regularization inherits the two bounds on a smaller neighbourhood. Indeed $H_{φ} \geq U_{φ}$ by the defining limit in [F1], so $H_{φ} \geq φ (ζ) - ε - A δ$ on $N \cap Ω$ ; and if $w \in N^{'} \cap Ω$ for a neighbourhood $N^{'}$ of $ζ$ with ${∣ w - y ∣ < ρ_{0}} \subseteq N$ for some $ρ_{0} > 0$ , then every $y \in Ω$ with $∣ y - w ∣ < ρ_{0}$ lies in $N \cap Ω$ , so the supremum defining $H_{φ} (w)$ is at most $φ (ζ) + ε + A δ$ and hence so is its limit $H_{φ} (w)$ $∣ H_{φ} (w) - φ (ζ) ∣ \leq ε + A δ (w \in N^{'} \cap Ω) .$

5.1F1F2F5step 1.1step 3.1step 4.1∎

Given $η > 0$ , apply step 1.1 with $ε : = η / 2$ and step 3.1 with $δ : = η / (2 A)$ , where $A \geq 1$ is the positive integer of step 1.1; then step 4.1 yields a neighbourhood of $ζ$ on which $∣ H_{φ} - φ (ζ) ∣ \leq η$ . Hence the limit exists and equals $φ (ζ)$ , that is, $ζ$ is regular in the sense of [F2]; the argument used only the barrier $b$ at $ζ$ , the continuity of $φ$ at $ζ$ and the compactness of $\partial Ω$ , and it made no use of boundary behaviour of $H_{φ}$ at any other point.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Green correctors are smooth at analytic boundaries

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $D \subseteq C$ be a bounded complex domain whose boundary is a compact real-analytic curve: for every $ζ \in \partial D$ there are an open interval $(- ε, ε)$ , a real-analytic parametrization $γ : (- ε, ε) \to C$ with $γ (0) = ζ$ and $γ^{'} (0) \neq 0$ , and a neighbourhood $U$ of $ζ$ with $\partial D \cap U = γ ((- ε, ε))$ for which $D \cap U$ is one of the two components of $U ∖ γ ((- ε, ε))$ . Let $a \in D$ and let $g_{D} (\cdot, a)$ be the canonical Green kernel of Green functions exist on all bounded plane domains, with Perron corrector $h_{a} : = - \log ∣ \cdot - a ∣ - g_{D} (\cdot, a)$ . Then every boundary point of $D$ is regular (Barriers and regular boundary points), the corrector $h_{a}$ extends to a function of class $C^{2}$ on the closure $\overline{D}$ , and consequently $g_{D} (\cdot, a) = - \log ∣ \cdot - a ∣ - h_{a}$ extends to a $C^{2}$ function on $\overline{D} ∖ {a}$ whose trace on $\partial D$ is identically zero. The extension is obtained locally from a holomorphic chart and the odd harmonic reflection across the analytic arc.

Facts & Assumptions

Given: Countable Choice and a bounded complex domain $D \subseteq C$ with the real-analytic boundary parametrizations of the statement, a point $a \in D$ , and $ζ \in \partial D$ with its parametrization $γ : (- ε, ε) \to C$ and neighbourhood $U$ (A real-analytic function on an open subset of $R$ is locally represented by a convergent real power series). Green kernels and Perron correctors are those of Green functions exist on all bounded plane domains.

[F1]

For the bounded domain $D$ and $a \in D$ , the canonical Green kernel exists and equals $g_{D} (z, a) = F_{a} (z) - h_{a} (z)$ with $F_{a} = - \log ∣ \cdot - a ∣$ and $h_{a} = H_{b_{a}}$ the regularized Perron envelope of $b_{a} = F_{a} ∣_{\partial D}$ ; $h_{a}$ is harmonic on $D$ and bounded, $g_{D} (\cdot, a)$ is positive and harmonic on $D ∖ {a}$ , and $g_{D} (z, a) \to 0$ as $z \to η \in \partial D$ through $D$ at every regular boundary point $η$ (Green functions exist on all bounded plane domains).

[F2]

Suppose there are a neighbourhood $U_{0}$ of a boundary point $η$ and a subharmonic $q$ on $D \cap U_{0}$ with: $q < 0$ on $D \cap U_{0}$ ; $q (z) \to 0$ as $z \to η$ ; and $\sup {q (z) : z \in D \cap \partial W} < 0$ for some smaller neighbourhood $W ⋐ U_{0}$ of $η$ . Then $D$ has a global barrier at $η$ , hence $η$ is regular (A local strict subharmonic peak function globalizes, A planar barrier forces the regularized Perron envelope to have the prescribed boundary limit).

[F3]

If a function $u$ is harmonic on $D^{+} = {∣ z ∣ < 1, Im z > 0}$ , continuous on its closure and vanishes on $(- 1, 1)$ , then its odd reflection $U (z) = u (z)$ for $Im z \geq 0$ and $U (z) = - u (\overline{z})$ for $Im z < 0$ is harmonic on the full unit disc (Harmonic and holomorphic Schwarz reflection across the real axis); every harmonic function is smooth, indeed real-analytic (Plane harmonic functions are smooth and real analytic).

[F4]

If $f$ is nonconstant and holomorphic on a complex domain and $f^{'} (a) \neq 0$ , then $f$ is biholomorphic between neighbourhoods of $a$ and $f (a)$ (Holomorphic inverse function theorem and local-degree criterion).

[F5]

A real-analytic $γ$ equals its convergent power series $γ (t) = \sum_{n \geq 0} c_{n} t^{n}$ near $0$ (A real-analytic function on an open subset of $R$ is locally represented by a convergent real power series); the same series with complex coefficients converges on a disc in $C$ and defines a holomorphic function there (Complex series, absolute convergence, complex power series, and radius of convergence, The sum of a complex power series is analytic throughout its open disc of convergence), whose derivative at $0$ is the coefficient $c_{1}$ (A power-series sum is infinitely differentiable inside its radius and satisfies $a_{n} = f^{(n)} (c) / ι (n!)$ at its centre).

[F6]

The function $\log ∣ \cdot ∣$ is harmonic on $C ∖ {0}$ , composition with a holomorphic map preserves harmonicity, holomorphic functions have smooth real and imaginary components (Holomorphic functions are real analytic and smooth in their two real coordinates), and the $C^{2}$ real and imaginary components of a holomorphic function are harmonic (Logarithmic modulus is harmonic off its centre, Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair); harmonic functions are subharmonic by the $C^{2}$ Laplacian criterion (A C^2 function is subharmonic exactly when its Laplacian is nonnegative). Thus $F_{a} = - \log ∣ \cdot - a ∣$ is harmonic, and smooth, on $D ∖ {a}$ .

[F7]

Countable Choice supplies a choice function for every countable family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The Green-kernel existence and regular-boundary clause of [F1] inherit this hypothesis; [F1] is used at steps 5.1, 6.1 and 7.1.

Proof

technique · direct

1.1F5F4given

Complexifying the chart: by [F5] the parametrization satisfies $γ (t) = \sum_{n \geq 0} c_{n} t^{n}$ for $∣ t ∣$ small, with $c_{0} = ζ$ and $c_{1} = γ^{'} (0) \neq 0$ . The complex power series $Γ (w) : = \sum_{n \geq 0} c_{n} w^{n}$ converges on a disc $D (0, r_{0})$ and defines a holomorphic function there with $Γ (t) = γ (t)$ for real $∣ t ∣ < r_{0}$ and $Γ^{'} (0) = c_{1} \neq 0$ . By [F4] the map $Γ$ restricts to a biholomorphism from some disc $D (0, r) \subseteq D (0, r_{0})$ onto an open neighbourhood $V$ of $ζ$ , and the two components of $D (0, r) ∖ (- r, r)$ map onto the two components of $V ∖ γ ((- r, r))$ , the latter being an arc of $\partial D$ when $r \leq ε$ . Replacing $γ$ by $t \mapsto γ (- t)$ if necessary, we may assume $Γ$ maps the upper half-disc $D_{r}^{+} : = {w : ∣ w ∣ < r, Im w > 0}$ onto $D \cap V$ .

2.1F6F4step 1.1algebra

A local peak on the domain: define, for $w \in D_{r}^{+}$ , $q (w) : = - Re \sqrt{- i w},$ with the principal square root. Since $w \mapsto \sqrt{- i w}$ is holomorphic on the half-disc --- there $- i w$ has positive real part, so it avoids $(- \infty, 0]$ --- both components of that holomorphic square root are smooth by [F6], so the $C^{2}$ components theorem makes $q$ harmonic and the Laplacian criterion makes it subharmonic on $D_{r}^{+}$ . Writing $w = ρ e^{i φ}$ with $0 < φ < π$ gives $- i w = ρ e^{i (φ - π / 2)}$ with $φ - π / 2 \in (- π / 2, π / 2)$ , so $\sqrt{- i w} = \sqrt{ρ} e^{i (φ - π / 2) / 2}$ and $q (w) = - \sqrt{ρ} \cos (\frac{φ}{2} - \frac{π}{4}) \leq - \frac{\sqrt{ρ}}{\sqrt{2}} < 0,$ because $∣ φ / 2 - π / 4 ∣ < π / 4$ ; moreover $q (w) \to 0$ as $w \to 0$ . Hence $q \circ Γ^{- 1}$ is subharmonic (by [F6], applied to the holomorphic $Γ^{- 1}$ ) and negative on $D \cap V$ , and it tends to $0$ at $ζ$ .

3.1step 1.1step 2.1algebra

The peak is bounded away from $0$ on the boundary of a smaller neighbourhood. Let $W : = Γ (D (0, r / 2))$ . Then $W$ is a neighbourhood of $ζ$ with $W ⋐ V$ , and $D \cap \partial W = Γ (D_{r}^{+} \cap {∣ w ∣ = r / 2})$ by step 1.1; on that set $ρ = r / 2$ , so step 2.1 gives $\sup {q (Γ^{- 1} (z)) : z \in D \cap \partial W} \leq - (r / 2)^{1 / 2} / \sqrt{2} < 0$ .

4.1F2step 2.1step 3.1

Conclusion of regularity: step 3.1 verifies hypothesis 3 of [F2] for the subharmonic function $q \circ Γ^{- 1}$ of step 2.1, whose hypotheses 1 and 2 were also verified there; hence $D$ has a global barrier at $ζ$ and $ζ$ is a regular boundary point by [F2]. As $ζ \in \partial D$ was arbitrary, every boundary point of $D$ is regular.

5.1F1F7F3F6step 1.1step 4.1

Smoothness near the arc by reflection: Countable Choice [F7] licenses the Green kernel and its regular-boundary limits from [F1]. Fix $ζ$ and keep the chart $Γ$ of step 1.1. Choose $0 < R < r$ small enough that $\overline{D (0, R)}$ remains in the chart and $Γ (\overline{D_{R}^{+}}) \subseteq \overline{D} ∖ {a}$ ; this is possible since $Γ (0) = ζ \neq a$ , while step 1.1 puts the open upper half-disc in $D$ and its real diameter on $\partial D$ . Define $G (w) : = g_{D} (Γ (w), a)$ for $w \in D_{R}^{+}$ . Then $G$ is harmonic there by [F6], since $g_{D} (\cdot, a)$ is harmonic on $D ∖ {a}$ by [F1] and $Γ$ is holomorphic. At a real $t \in (- R, R)$ define the boundary value $G (t) : = 0$ . This is a continuous extension across the diameter: whenever $w_{n}$ approaches $t$ from the upper half-disc, $Γ (w_{n}) \in D$ approaches the regular boundary point $Γ (t)$ , so [F1] and step 4.1 give $G (w_{n}) \to 0$ . On the upper semicircle of radius $R$ , the image lies in $D$ except at its two real endpoints; the same interior continuity and regular-boundary limits give continuity on the whole closed half-disc. Rescale by $w \mapsto w / R$ and apply [F3]; the odd reflection is harmonic on $D (0, R)$ and smooth there. Choose $0 < r_{1} < R$ ; its restriction to $\overline{D_{r_{1}}^{+}}$ is therefore $C^{2}$ , including the real diameter near $0$ .

6.1F1F6step 5.1algebra

The corrector near the arc is $C^{2}$ : on the smaller closed half-disc $\overline{D_{r_{1}}^{+}}$ of step 5.1 one has $h_{a} (Γ (w)) = F_{a} (Γ (w)) - G (w)$ for interior $w$ , and this equality extends continuously to its real diameter using the regular boundary values. The reflected extension of $G$ is $C^{2}$ by step 5.1. Also $F_{a} \circ Γ$ is smooth on a neighbourhood of this closed half-disc: its compact image lies in $\overline{D} ∖ {a}$ , hence at positive distance from $a$ , and $F_{a} = - \log ∣ \cdot - a ∣$ is smooth on the ambient open set $C ∖ {a}$ . Thus $F_{a} \circ Γ - G$ gives a $C^{2}$ extension of $h_{a} \circ Γ$ across the real diameter. Transport through the local biholomorphism $Γ$ gives a $C^{2}$ extension of $h_{a}$ to an ambient neighbourhood of the boundary arc near $ζ$ .

7.1F1F6step 4.1step 6.1∎

Since $ζ$ was arbitrary, step 6.1 supplies a $C^{2}$ ambient extension near every boundary point; interior harmonicity supplies smoothness at every interior point. On overlaps the restrictions of these extensions to $D$ equal the same $h_{a}$ , and continuity makes their boundary values and one-sided derivatives agree on $\overline{D}$ . The extensions need not agree outside $D$ ; their local existence is exactly the asserted $C^{2}$ regularity on the closure. Finally $g_{D} (z, a) = - \log ∣ z - a ∣ - h_{a} (z)$ for $z \in D ∖ {a}$ ; since $- \log ∣ \cdot - a ∣$ is smooth away from $a$ , $g_{D} (\cdot, a)$ is $C^{2}$ on $\overline{D} ∖ {a}$ in the same local-extension sense, and its trace on $\partial D$ vanishes by the boundary limits of [F1] in step 4.1.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Canonical Green kernels are unique, symmetric and domain monotone

Statement

Assume Countable Choice. Let $Ω$ be a Greenian plane domain (The canonical Green kernel of a plane domain). Then:

a canonical Green kernel is unique: if a pointwise least logarithmic-pole candidate at $a \in Ω$ exists, it is unique, so the notation $g_{Ω} (\cdot, a)$ is unambiguous;
symmetry: $g_{Ω} (z, a) = g_{Ω} (a, z)$ for all distinct $a, z \in Ω$ ;
domain monotonicity: if $Ω_{1} \subseteq Ω_{2}$ are Greenian plane domains and $a, z \in Ω_{1}$ are distinct, then $g_{Ω_{1}} (z, a) \leq g_{Ω_{2}} (z, a)$ . The inequality is in this direction: enlarging the domain increases the Green kernel.

Countable Choice is used only through the cited bounded-domain existence theorem and the cited PDE Green symmetry theorem.

Facts & Assumptions

Given: Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )); a Greenian plane domain $Ω$ (The canonical Green kernel of a plane domain), so $Ω$ is a nonempty connected open set with $Ω \neq C$ (A complex domain is a nonempty connected open subset of $C$ ); harmonicity in the sense of Plane harmonic functions; and distinct points $x, y \in Ω$ in the symmetry part.

[F1]

A logarithmic-pole candidate at $a$ on a proper plane domain is a nonnegative function on $Ω ∖ {a}$ that is harmonic there and whose sum with $\log ∣ z - a ∣$ extends harmonically across $a$ ; the canonical Green function is the pointwise least candidate, when such a member exists, a pointwise least member is unique, and $Ω$ is Greenian when $g_{Ω} (\cdot, a)$ exists for every $a \in Ω$ (The canonical Green kernel of a plane domain).

[F2]

Assume Countable Choice. If $D$ is a bounded complex domain and $p \in D$ , then with $F_{p} (z) : = - \log ∣ z - p ∣$ , $b_{p} : = F_{p} ∣_{\partial D}$ and $h_{p} : = H_{b_{p}}$ one has $g_{D} (z, p) = F_{p} (z) - h_{p} (z)$ is the canonical positive Green kernel of $D$ at $p$ , and $- Δ_{z} T_{g_{D} (\cdot, p)} = 2 π δ_{p}$ as distributions on $D$ (Green functions exist on all bounded plane domains).

[F3]

Every plane domain admits an increasing sequence $D_{1} \subseteq D_{2} \subseteq \dots$ of relatively compact connected open subsets whose boundaries are real-analytic regular in the one-sided sense: for every $ζ \in \partial D_{j}$ there are a neighbourhood $U$ of $ζ$ and a real-analytic function $g$ of one real variable with, after relabelling the two coordinate axes if necessary, $\partial D_{j} \cap U = {(x, y) \in U : y = g (x)}$ and $D_{j} \cap U$ one of the two connected components of $U ∖ {(x, y) \in U : y = g (x)}$ . Every compact $K \subseteq Ω$ lies in $D_{j}$ for all sufficiently large $j$ , and if $A \subseteq Ω$ is finite the sequence may be chosen with $A \subseteq D_{1}$ (Analytic-boundary exhaustion of a plane domain).

[F4]

Let $D$ be a bounded complex domain whose boundary is a compact real-analytic curve, locally parametrized by a real-analytic $γ$ with $γ^{'} (0) \neq 0$ and $D$ on one side. Then for $a \in D$ the Perron corrector $h_{a} : = - \log ∣ \cdot - a ∣ - g_{D} (\cdot, a)$ extends to a function of class $C^{2}$ on $\overline{D}$ ; consequently $g_{D} (\cdot, a) = - \log ∣ \cdot - a ∣ - h_{a}$ extends to a $C^{2}$ function on $\overline{D} ∖ {a}$ whose trace on $\partial D$ is identically zero (Green correctors are smooth at analytic boundaries).

[F5]

Assume Countable Choice. Let $n \geq 2$ and let $Ω \subset R^{n}$ be a bounded $C^{1}$ domain carrying a Dirichlet Green function $G_{Ω}$ for $- Δ$ whose designated harmonic correctors satisfy $H_{y} \in C^{2} (\overline{Ω})$ ; then $G_{Ω} (x, y) = G_{Ω} (y, x)$ for all distinct $x, y$ (Symmetry of the Dirichlet Green function).

[F6]

Assume Countable Choice. A Dirichlet Green function for $- Δ$ on $Ω$ is a function $G_{Ω}$ on pairs of distinct points such that for each pole $y$ there is a harmonic $H_{y}$ with $H_{y} = Φ (\cdot - y)$ on $\partial Ω$ and $G_{Ω} (x, y) = Φ (x - y) - H_{y} (x)$ , such that $G_{Ω} (\cdot, y)$ is harmonic off $y$ with zero boundary trace, and such that $- Δ T_{G_{Ω} (\cdot, y)} = δ_{y}$ in $D^{'} (Ω)$ (Dirichlet Green function for minus Laplacian).

[F7]

A bounded $C^{1}$ domain is a nonempty bounded open set whose boundary is locally, after a rigid change of coordinates, the graph of a $C^{1}$ function with the set locally exactly the corresponding subgraph; connectedness is not required (Bounded C1 domains and their outward normals).

[F8]

Assume Countable Choice and $n \geq 2$ . The fundamental solution is $Φ (x) = ∣ x ∣^{2 - n} / ((n - 2) ω_{n - 1})$ for $n \geq 3$ and $Φ (x) = - (2 π)^{- 1} \log ∣ x ∣$ for $n = 2$ , $x \neq 0$ (Fundamental solution for the positive operator minus Laplacian).

[F9]

An increasing sequence of harmonic functions on a complex domain either tends to $+ \infty$ at every point or converges locally uniformly to a harmonic limit (An increasing harmonic sequence converges locally uniformly to a harmonic limit or diverges to +infinity).

[F10]

A harmonic function on a punctured disc that is bounded on that punctured disc extends harmonically across the puncture (A bounded harmonic function near an isolated puncture extends harmonically).

[F11]

Countable Choice: every family $(X_{n})$ of nonempty sets indexed by $N$ has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F12]

The function $\log ∣ \cdot ∣$ is harmonic on $C ∖ {0}$ (Logarithmic modulus is harmonic off its centre), and precomposition of a harmonic function with a holomorphic map is harmonic (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate); hence $z \mapsto \log ∣ z - p ∣$ is harmonic on $C ∖ {p}$ .

[F13]

A harmonic function is $C^{2}$ with $Δ u = 0$ (Plane harmonic functions), and finite sums of $C^{2}$ functions are $C^{2}$ with linear Laplacian ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition); hence sums and differences of harmonic functions are harmonic.

[F14]

Assume Countable Choice. The map $f \mapsto T_{f}$ is complex-linear from $L_{l o c}^{1}$ into distributions, distributional differentiation is linear and continuous on $D^{'} (Ω)$ , and it extends classical smooth differentiation (Locally integrable functions embed in distributions, Distributional differentiation is continuous and commutes).

[F15]

A real-analytic function of one real variable is locally the sum of a convergent power series (A real-analytic function on an open subset of $R$ is locally represented by a convergent real power series), and such sums have derivatives of every order (A power-series sum is infinitely differentiable inside its radius and satisfies $a_{n} = f^{(n)} (c) / ι (n!)$ at its centre); hence a real-analytic function is $C^{1}$ .

Proof

technique · direct

1.1F1

By [F1] the canonical Green function on a Greenian $Ω$ is defined as the pointwise least member of the family of logarithmic-pole candidates at $a$ , and the definition records that a pointwise least member is unique; hence whenever it exists it is unique, as claimed in clause 1.

1.2F1

Domain monotonicity. Let $Ω_{1} \subseteq Ω_{2}$ be Greenian plane domains and let $a, z \in Ω_{1}$ be distinct. The restriction of $g_{Ω_{2}} (\cdot, a)$ to $Ω_{1} ∖ {a}$ is a logarithmic-pole candidate at $a$ on $Ω_{1}$ : it is nonnegative, it is harmonic on $Ω_{1} ∖ {a} \subseteq Ω_{2} ∖ {a}$ , and the corrector $g_{Ω_{2}} (\cdot, a) + \log ∣ \cdot - a ∣$ agrees with a harmonic function on $Ω_{2}$ by [F1] whose restriction to $Ω_{1}$ is harmonic, so the sum extends harmonically across $a$ inside $Ω_{1}$ . Leastness of $g_{Ω_{1}} (\cdot, a)$ gives $g_{Ω_{1}} (z, a) \leq g_{Ω_{2}} (z, a)$ , which is clause 3.

1.3F3F11

Symmetry setup. Assume Countable Choice and fix distinct $x, y \in Ω$ . Apply [F3] with the finite set $A = {x, y}$ : there are relatively compact connected open subsets $D_{1} \subseteq D_{2} \subseteq \dots$ of $Ω$ with ${x, y} \subseteq D_{1}$ , real-analytic regular one-sided boundaries, and every compact subset of $Ω$ contained in $D_{j}$ for all large $j$ .

2.1F3F4F7F15step 1.3

For each $j$ , $D_{j}$ is a bounded complex domain, its boundary is a compact real-analytic curve in the sense of [F4], and $D_{j}$ is a bounded $C^{1}$ domain in the sense of [F7]. Indeed $D_{j}$ is nonempty, open, connected and relatively compact, hence bounded; for $ζ \in \partial D_{j}$ the regularity of [F3] provides $U$ and a real-analytic $g$ with $\partial D_{j} \cap U = {(x, y) \in U : y = g (x)}$ and $D_{j} \cap U$ equal to one of the two components of $U ∖ {(x, y) \in U : y = g (x)}$ . Parametrizing that graph, after translating the parameter, by $γ (t) : = (t, g (t))$ gives a real-analytic curve with $γ^{'} (t) = (1, g^{'} (t)) \neq 0$ for which $D_{j} \cap U$ is one of the two components of $U ∖ γ$ , so [F4] applies; and since $g$ is $C^{1}$ by [F15], after relabelling the axes and if necessary reflecting one of them the boundary is locally a $C^{1}$ graph with $D_{j}$ locally the corresponding subgraph, which is the structure required by [F7].

2.2F1F3step 1.2step 1.3

Fix a pole $p \in {x, y}$ and $z \in Ω ∖ {p}$ . Since the sequence exhausts $Ω$ and $z$ is an interior point, $z \in D_{j}$ for all large $j$ , and ${p} \subseteq D_{1}$ ; step 1.2 applied to the Greenian domains $D_{j} \subseteq D_{j + 1}$ shows $g_{D_{j}} (z, p) \leq g_{D_{j + 1}} (z, p)$ , and applied to $D_{j} \subseteq Ω$ it shows $g_{D_{j}} (z, p) \leq g_{Ω} (z, p)$ , a finite bound by [F1]. Hence the limit $u_{p} (z) : = \lim_{j \to \infty} g_{D_{j}} (z, p)$ exists and lies in $[0, g_{Ω} (z, p)]$ .

3.1F2F4step 2.1

For each $j$ and each pole $p \in {x, y}$ the canonical kernel $g_{D_{j}} (\cdot, p)$ exists by [F2] because $D_{j}$ is a bounded complex domain, and [F4] applied to $D = D_{j}$ shows that the Perron corrector $h_{p}^{(j)} : = - \log ∣ \cdot - p ∣ - g_{D_{j}} (\cdot, p)$ is harmonic on $D_{j}$ , extends to $C^{2} (\overline{D_{j}})$ , and that $g_{D_{j}} (\cdot, p) = - \log ∣ \cdot - p ∣ - h_{p}^{(j)}$ extends to a $C^{2}$ function on $\overline{D_{j}} ∖ {p}$ with trace identically zero on $\partial D_{j}$ .

3.2F1F9F3step 2.2

The limit $u_{p}$ is harmonic on $Ω ∖ {p}$ . Let $W \subseteq Ω ∖ {p}$ be nonempty, open and relatively compact; by [F3] there is $j_{0}$ with $\overline{W} \subseteq D_{j_{0}}$ , so $(g_{D_{j_{0} + n}} (\cdot, p) ∣_{W})_{n \geq 0}$ is an increasing sequence of harmonic functions on $W$ bounded above by $g_{Ω} (\cdot, p)$ , which is finite by [F1]. The first alternative of [F9] is therefore impossible and the second applies: the limit is harmonic on $W$ and the convergence is locally uniform there. As $W$ is arbitrary, $u_{p}$ is harmonic on $Ω ∖ {p}$ .

4.1F2F6F8F14step 2.1step 3.1

For each $j$ , $G_{j} : = g_{D_{j}} / (2 π)$ with correctors $H_{p}^{(j)} : = h_{p}^{(j)} / (2 π)$ is a Dirichlet Green function for $- Δ$ on $D_{j}$ in the sense of [F6] with designated $C^{2} (\overline{D_{j}})$ correctors. Correctors: for $z \in D_{j} ∖ {p}$ , $Φ (z - p) - H_{p}^{(j)} (z) = - \frac{1}{2 π} \log ∣ z - p ∣ - \frac{1}{2 π} h_{p}^{(j)} (z) = \frac{1}{2 π} (- \log ∣ z - p ∣ - h_{p}^{(j)} (z)) = \frac{1}{2 π} g_{D_{j}} (z, p) = G_{j} (z, p)$ by [F8] and step 3.1, and $H_{p}^{(j)}$ is harmonic with $H_{p}^{(j)} = Φ (\cdot - p)$ on $\partial D_{j}$ because $g_{D_{j}} (\cdot, p)$ has zero boundary trace; harmonicity off the pole and the zero trace of $G_{j} (\cdot, p)$ are step 3.1. Dirac identity: by [F2] and step 2.1, $- Δ T_{g_{D_{j}} (\cdot, p)} = 2 π δ_{p}$ in $D^{'} (D_{j})$ , and linearity of the embedding and of distributional differentiation [F14] gives $- Δ T_{G_{j} (\cdot, p)} = \frac{1}{2 π} (- Δ T_{g_{D_{j}} (\cdot, p)}) = δ_{p}$ .

4.2F1F10F12F13step 3.1step 2.2step 3.2

The limit $u_{p}$ is a logarithmic-pole candidate at $p$ on $Ω$ : it is nonnegative by step 2.2, harmonic by step 3.2, and $Φ_{p} : = u_{p} + \log ∣ \cdot - p ∣$ is harmonic on $Ω ∖ {p}$ by [F12] and [F13]. Near $p$ the function $Φ_{p}$ is bounded: below, $u_{p} \geq g_{D_{1}} (\cdot, p)$ by step 2.2, so $Φ_{p} \geq g_{D_{1}} (z, p) + \log ∣ z - p ∣ = - h_{p}^{(1)} (z)$ by step 3.1, and $h_{p}^{(1)} \in C^{2} (\overline{D_{1}})$ is bounded on a neighbourhood of $p$ ; above, $u_{p} \leq g_{Ω} (\cdot, p)$ by step 2.2, so $Φ_{p} \leq g_{Ω} (z, p) + \log ∣ z - p ∣$ , and the right-hand side is harmonic on $Ω$ , hence bounded on a neighbourhood of $p$ by [F13]. Thus $Φ_{p}$ is harmonic and bounded on a punctured disc about $p$ , and [F10] extends it harmonically across $p$ ; so $u_{p} + \log ∣ \cdot - p ∣$ extends harmonically to $Ω$ and $u_{p}$ is a candidate in the sense of [F1].

5.1F5step 2.1step 4.1

Symmetry on the exhaustion domains. [F5] applies to the bounded $C^{1}$ domain $D_{j}$ of step 2.1, to the Dirichlet Green function $G_{j}$ of step 4.1 and to its correctors $H_{p}^{(j)} \in C^{2} (\overline{D_{j}})$ : hence $G_{j} (u, v) = G_{j} (v, u)$ for all distinct $u, v \in D_{j}$ , and multiplying by $2 π$ , $g_{D_{j}} (u, v) = g_{D_{j}} (v, u)$ .

5.2F1step 2.2step 4.2

Identification of the limit. Leastness of $g_{Ω} (\cdot, p)$ among the candidates on the Greenian domain $Ω$ [F1] gives $g_{Ω} (\cdot, p) \leq u_{p}$ , while step 2.2 gives $u_{p} (z) \leq g_{Ω} (z, p)$ for every $z \in Ω ∖ {p}$ ; hence $u_{p} = g_{Ω} (\cdot, p)$ , that is $\lim_{j} g_{D_{j}} (z, p) = g_{Ω} (z, p)$ for every $z \in Ω ∖ {p}$ .

6.1step 5.1step 5.2

Symmetry. Step 5.1 gives $g_{D_{j}} (x, y) = g_{D_{j}} (y, x)$ for every $j$ . Taking $j \to \infty$ and applying step 5.2 with $p = y$ on the left and with $p = x$ on the right yields $g_{Ω} (x, y) = g_{Ω} (y, x)$ , which is clause 2 for the given pair; as $x, y$ were arbitrary distinct points of $Ω$ , symmetry holds throughout.

7.1F2F5F11step 1.1step 1.2step 2.2step 3.1step 3.2step 4.2step 5.1step 5.2step 6.1∎

Choice accounting and scope. Countable Choice is used exactly through the bounded-domain existence theorem [F2], applied to each $D_{j}$ in step 3.1, and through the PDE Green symmetry theorem [F5] in step 5.1; the exhaustion [F3], the monotone bound of step 2.2, the Harnack limit of step 3.2, the removable-singularity step 4.2 and the comparison steps 1.1-1.2 and 5.2 use no choice principle. Steps 1.1-1.2, 6.1 establish the three clauses: 1.1 the uniqueness, 1.2 the domain monotonicity for arbitrary Greenian pairs, and 6.1 the symmetry for the Greenian $Ω$ fixed in step 1.3.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Green kernel of a simply connected plane domain from a Riemann map

Statement

Assume the Axiom of Choice for the existence of the Riemann map (Every proper homologically simply connected plane domain is conformally equivalent to the unit disc). Let $Ω ⊊ C$ be a homologically simply connected complex domain (Homologically simply connected complex domains) with $Ω \neq C$ , let $a \in Ω$ , and suppose $φ : Ω \to D$ is a biholomorphism with $φ (a) = 0$ . Then the canonical Green kernel of The canonical Green kernel of a plane domain is $g_{Ω} (z, a) = - \log ∣ φ (z) ∣ (z \in Ω ∖ {a}),$ and this value is independent of the biholomorphism chosen: any other biholomorphism $ψ : Ω \to D$ with $ψ (a) = 0$ gives the same function. Once $φ$ is supplied, the identity uses no choice principle; the Axiom of Choice is used only by the cited existence theorem.

Facts & Assumptions

Given: A homologically simply connected complex domain $Ω ⊊ C$ , a point $a \in Ω$ , and a biholomorphism $φ : Ω \to D$ onto the unit disc (The unit disc, the upper half-plane, and Blaschke factors, Biholomorphic maps between complex domains) with $φ (a) = 0$ ; moduli are those of Real and imaginary parts, complex conjugation, and modulus and harmonicity is that of Plane harmonic functions.

[F1]

For a proper plane domain $Ω$ and $a \in Ω$ the canonical Green function is the pointwise least nonnegative logarithmic-pole candidate at $a$ : a function nonnegative on $Ω ∖ {a}$ , harmonic on $Ω ∖ {a}$ , with $u + \log ∣ z - a ∣$ extending harmonically across $a$ (The canonical Green kernel of a plane domain).

[F2]

A biholomorphism is a holomorphic bijection with holomorphic inverse; an injective holomorphic map on a complex domain has nowhere-zero derivative; a holomorphic function with a zero of order one at $a$ factors as $f (z) = (z - a) q (z)$ with $q$ holomorphic and $q (a) \neq 0$ near $a$ (Biholomorphic maps between complex domains, An injective holomorphic map has no critical point and is biholomorphic onto its image, The order of a zero is the exponent in its local holomorphic factorization).

[F3]

$\log ∣ \cdot ∣$ is harmonic on $C ∖ {0}$ (Logarithmic modulus is harmonic off its centre); composition with a holomorphic map preserves harmonicity, and sums and differences of harmonic functions are harmonic (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate); a nowhere-zero holomorphic function on a disc has a holomorphic logarithm there, whose real part equals $\log ∣ q ∣$ when its exponential is $q$ , and is harmonic by the preceding logarithmic-modulus and composition facts (A nonvanishing holomorphic function on a disc has a holomorphic logarithm).

[F4]

A harmonic function on a bounded domain that extends continuously to the closure attains its minimum on the boundary (Maximum and minimum principles for plane harmonic functions); a holomorphic self-map of $D$ fixing $0$ that attains equality in $∣ f (z) ∣ \leq ∣ z ∣$ is a rotation (Schwarz lemma with the equality cases).

[F5]

Assume the Axiom of Choice: every homologically simply connected $Ω ⊊ C$ and $z_{0} \in Ω$ admit a biholomorphism $Ω \to D$ with $F (z_{0}) = 0$ (Every proper homologically simply connected plane domain is conformally equivalent to the unit disc, The Axiom of Choice).

Proof

technique · direct

1.1F2F3algebra

Because $φ$ is an injective holomorphic map on the domain $Ω$ , [F2] gives $φ^{'} (a) \neq 0$ ; hence $w \mapsto φ (w)$ has a zero of order one at $a$ , and [F2] provides a disc $D (a, ρ) \subseteq Ω$ with $φ (z) = (z - a) q (z)$ for a holomorphic $q$ that is nowhere zero on $D (a, ρ)$ . By [F3] there is a holomorphic $L$ on $D (a, ρ)$ with $\exp L = q$ , and $Re L$ is harmonic on $D (a, ρ)$ $- \log ∣ φ (z) ∣ + \log ∣ z - a ∣ = - Re L (z) (0 < ∣ z - a ∣ < ρ),$ because $∣ φ (z) ∣ = ∣ z - a ∣ ∣ q (z) ∣ = ∣ z - a ∣ e^{Re L (z)}$ . The inverse $φ^{- 1} : D \to Ω$ is holomorphic by [F2].

1.2F1F3F4

On the unit disc the least logarithmic-pole candidate at $0$ is $- \log ∣ w ∣$ . Indeed $- \log ∣ w ∣$ is positive on $D ∖ {0}$ , harmonic there by [F3], and $- \log ∣ w ∣ + \log ∣ w ∣ = 0$ extends harmonically across $0$ , so it is a candidate. If $U$ is any candidate at $0$ , then $D : = U + \log ∣ w ∣$ agrees on $D ∖ {0}$ with a function harmonic on $D$ , hence is harmonic on $D$ by [F3]; on the circle $∣ w ∣ = r$ it satisfies $D \geq \log r$ because $U \geq 0$ , so the minimum principle [F4] applied on ${∣ w ∣ \leq r}$ gives $D \geq \log r$ on that disc, and letting $r ↑ 1$ yields $D \geq 0$ , that is $U \geq - \log ∣ w ∣$ .

1.3F2F4algebra

If $ψ : Ω \to D$ is another biholomorphism with $ψ (a) = 0$ , then $h : = ψ \circ φ^{- 1}$ is a biholomorphic self-map of $D$ fixing $0$ , so $∣ h (w) ∣ \leq ∣ w ∣$ for all $w$ ; the same bound applied to $h^{- 1}$ gives $∣ h (w) ∣ = ∣ w ∣$ , so $h$ is a rotation by [F4] and therefore $∣ ψ (z) ∣ = ∣ h (φ (z)) ∣ = ∣ φ (z) ∣$ for every $z \in Ω$ . Hence $- \log ∣ ψ (z) ∣ = - \log ∣ φ (z) ∣$ on $Ω ∖ {a}$ .

2.1F1F3step 1.1

The function $z \mapsto - \log ∣ φ (z) ∣$ is a logarithmic-pole candidate at $a$ on $Ω$ : it is positive because $∣ φ (z) ∣ < 1$ on $Ω$ , it is harmonic on $Ω ∖ {a}$ because it is the composite of the harmonic function $- \log ∣ \cdot ∣$ on $C ∖ {0}$ with the holomorphic $φ$ by [F3], and its corrector across $a$ is the harmonic function $- Re L$ of step 1.1.

2.2F1F3step 1.1

Let $u$ be an arbitrary logarithmic-pole candidate at $a$ on $Ω$ and put $U (w) : = u (φ^{- 1} (w))$ for $w \in D ∖ {0}$ . Then $U$ is a candidate at $0$ on $D$ : it is nonnegative, harmonic by [F3] because $φ^{- 1}$ is holomorphic by step 1.1, and $U (w) + \log ∣ w ∣ = u (φ^{- 1} (w)) + \log ∣ φ^{- 1} (w) - a ∣ - \log \frac{∣ φ^{- 1} (w) - a ∣}{∣ w ∣}$ extends harmonically across $0$ , because the first two terms are the harmonic corrector of $u$ composed with $φ^{- 1}$ and the last term is $- \log ∣ g (w) ∣$ for the holomorphic function $g (w) : = (φ^{- 1} (w) - a) / w$ , which satisfies $g (0) = (φ^{- 1})^{'} (0) \neq 0$ , so that it has a holomorphic logarithm near $0$ by [F3].

3.1F1F5step 1.2step 1.3step 2.1step 2.2∎

By disc leastness, step 1.2 applied to the candidate $U$ of step 2.2 gives $u (φ^{- 1} (w)) \geq - \log ∣ w ∣$ for every $w \in D ∖ {0}$ ; writing $w = φ (z)$ yields $u (z) \geq - \log ∣ φ (z) ∣$ on $Ω ∖ {a}$ . So $- \log ∣ φ ∣$ is the pointwise least candidate and hence $g_{Ω} (z, a) = - \log ∣ φ (z) ∣$ by [F1]; by step 1.3 the same formula holds for every biholomorphism sending $a$ to $0$ . The supplied biholomorphism is the only place where a choice principle could enter, and by [F5] its existence is exactly what the Axiom of Choice is assumed for.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Harmonic measure on a bounded regular plane domain

Definition

Let $Ω \subseteq C$ be a bounded complex domain (A complex domain is a nonempty connected open subset of $C$ ) every boundary point of which is regular in the sense of Barriers and regular boundary points, and let $z \in Ω$ . The Euclidean boundary $\partial Ω$ is closed and, $Ω$ being bounded, also bounded, hence compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and it carries the Borel $σ$ -algebra (The Borel sigma-algebra of a topological space).

For every real continuous function $φ : \partial Ω \to R$ , let $H_{φ}$ denote the regularized Perron envelope of the bounded plane Dirichlet problem with boundary datum $φ$ (The Perron envelope and its regularization).

A harmonic measure for $Ω$ at $z$ is a Radon Borel probability measure $ω_{Ω}^{z}$ on $\partial Ω$ , in the sense of Radon measure on an LCH space, such that

$H_{φ} (z) = \int_{\partial Ω} φ d ω_{Ω}^{z}$ for every real continuous $φ : \partial Ω \to R$ .

The measure is written in the superscript slot $z$ because it is a measure attached to the point $z$ ; for a fixed Borel set $E \subseteq \partial Ω$ the assignment $z \mapsto ω_{Ω}^{z} (E)$ is a scalar function on $Ω$ , a distinct object from the measure itself.

Remarks

Existence and uniqueness are not part of this definition. A harmonic measure for $Ω$ at $z$ is a Radon probability measure satisfying the displayed identity for all continuous data. Existence and uniqueness are proved later on this page, for every bounded regular plane domain and every $z \in Ω$ ; this item only fixes the object and its test identity.
No probabilistic interpretation is used. This library defines no Brownian motion and no hitting distribution, and none is invoked: the defining property above is the totality of what "harmonic measure" means here.
The test identity is linear and normalized. Taking $φ \equiv 1$ in the defining identity and using that the constant function $1$ solves its own Dirichlet problem gives $\int 1 d ω_{Ω}^{z} = 1$ for every candidate measure, which is why probability measures rather than arbitrary finite measures are used.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Existence and uniqueness of harmonic measure on a bounded regular plane domain

Statement

Assume Dependent Choice, as required by the published positive $C_{0}$ Riesz-Markov representation theorem (Positive C_0(X) functionals have finite regular representing measures, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Then for every bounded regular plane domain $Ω$ , in the sense of Harmonic measure on a bounded regular plane domain, and every $z \in Ω$ there is exactly one Radon Borel probability measure $ω_{Ω}^{z}$ on $\partial Ω$ with $H_{φ} (z) = \int_{\partial Ω} φ d ω_{Ω}^{z}$ for every real continuous $φ : \partial Ω \to R$ . Moreover, for each such $φ$ the function $z \mapsto \int_{\partial Ω} φ d ω_{Ω}^{z} = H_{φ} (z)$ is the unique continuous extension to $\overline{Ω}$ that is harmonic on $Ω$ and agrees with $φ$ on $\partial Ω$ .

Facts & Assumptions

Given: A bounded complex domain $Ω$ every boundary point of which is regular (A complex domain is a nonempty connected open subset of $C$ , Barriers and regular boundary points, Harmonic measure on a bounded regular plane domain) and a point $z \in Ω$ . Harmonicity is that of Plane harmonic functions; the Perron family and envelope are those of The Perron lower family for continuous boundary data and The Perron envelope and its regularization; Radon measures are as in Radon measure on an LCH space.

[F1]

The boundary $\partial Ω$ is closed, hence compact because $Ω$ is bounded, and carries the Borel sigma-algebra (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, The Borel sigma-algebra of a topological space); the regularized Perron envelope $H_{φ}$ of every continuous datum is harmonic on $Ω$ (The regularized Perron envelope is harmonic), and at every regular boundary point $H_{φ} (w) \to φ (ζ)$ as $w \to ζ$ inside $Ω$ (Barriers and regular boundary points).

[F2]

Two functions continuous on $\overline{Ω}$ and harmonic on $Ω$ with equal boundary values are equal (The bounded plane Dirichlet problem has at most one continuous harmonic solution); the constant $0$ belongs to the Perron lower family of any datum $φ \geq 0$ , the envelope satisfies $U_{φ} \leq \max_{\partial Ω} φ$ , and $U_{φ} \leq H_{φ}$ (The Perron family is nonempty and uniformly bounded by the boundary data, The Perron envelope and its regularization).

[F3]

Assume Dependent Choice. For a locally compact Hausdorff space $X$ and a bounded positive linear $L : C_{0} (X; R) \to R$ there is a unique finite regular Borel measure $μ$ with $L (f) = \int f d μ$ and $μ (X) = ∥ L ∥$ (Positive C_0(X) functionals have finite regular representing measures).

Proof

technique · direct

1.1F1F2

For a continuous datum $φ$ define $L (φ) : = H_{φ} (z)$ . Each $H_{φ}$ is harmonic on $Ω$ and has the boundary limit $φ$ at every boundary point by [F1], so the function equal to $H_{φ}$ on $Ω$ and to $φ$ on $\partial Ω$ is continuous on $\overline{Ω}$ ; by [F2] it is the unique continuous harmonic extension of $φ$ .

2.1F1step 1.1algebra

The map $L$ is linear: for real $α, β$ and continuous $φ, ψ$ the function $α H_{φ} + β H_{ψ}$ is harmonic on $Ω$ and extends continuously to the boundary with values $α φ + β ψ$ , so it equals $H_{α φ + β ψ}$ by the uniqueness in step 1.1, and evaluating at $z$ gives $L (α φ + β ψ) = α L (φ) + β L (ψ)$ .

2.2F2step 1.1algebra

The map $L$ is positive and normalized: if $φ \geq 0$ then the constant $0$ lies in the Perron family of $φ$ by [F2], so $U_{φ} \geq 0$ and hence $H_{φ} \geq U_{φ} \geq 0$ ; and $H_{1} = 1$ because the constant function $1$ is a continuous harmonic extension of the boundary datum $1$ , so it equals $H_{1}$ by step 1.1. Consequently $∣ L (φ) ∣ \leq \max_{\partial Ω} ∣ φ ∣$ for every continuous $φ$ , by applying positivity to $\max ∣ φ ∣ - φ$ and $\max ∣ φ ∣ + φ$ , and $∥ L ∥ = 1$ .

3.1F1F3step 2.2

The boundary $\partial Ω$ is compact by [F1], hence a locally compact Hausdorff space on which every continuous function has compact support, so $C (\partial Ω) = C_{0} (\partial Ω)$ ; by [F3] and $D C$ there is a unique finite regular Borel measure $ω$ on $\partial Ω$ with $L (φ) = \int_{\partial Ω} φ d ω$ for all continuous $φ$ and $ω (\partial Ω) = ∥ L ∥ = 1$ .

4.1F3step 3.1given

The measure $ω$ of step 3.1 is a Radon Borel probability measure representing every continuous boundary datum at $z$ , so it is a harmonic measure for $Ω$ at $z$ in the sense of the definition. If $ω^{'}$ were another one, then $\int φ d ω^{'} = H_{φ} (z) = L (φ)$ for every continuous $φ$ , so $ω^{'} = ω$ by the uniqueness in [F3]; hence the harmonic measure is unique.

5.1F1F3step 1.1step 4.1∎

Finally, for fixed continuous $φ$ the function $z \mapsto \int_{\partial Ω} φ d ω_{Ω}^{z}$ coincides with $H_{φ}$ by the defining identity, so it is harmonic on $Ω$ and has the boundary values $φ$ ; by step 1.1 it is the unique continuous harmonic extension. This is the only place where $D C$ is used, through the representation theorem [F3]; the Perron input [F1] was used as a completed theorem.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Poisson density of harmonic measure on a disc

Statement

Assume Dependent Choice for the general representing-measure interface. Let $c \in C$ , $R > 0$ , and let $ω_{D (c, R)}^{z}$ be the harmonic measure of the disc $D (c, R)$ at $z \in D (c, R)$ , in the sense of Harmonic measure on a bounded regular plane domain. Writing $ξ = c + R e^{i t}$ for the boundary point of angle $t$ , one has, for every Borel subset $E \subseteq \partial D (c, R)$ and with $s$ the arclength parameter on the circle, $ω_{D (c, R)}^{z} (E) = \int_{t \in [0, 2 π) : c + R e^{i t} \in E} \frac{R^{2} - ∣ z - c ∣^{2}}{∣ ξ - z ∣^{2}} \frac{d t}{2 π} = \int_{E} \frac{R^{2} - ∣ z - c ∣^{2}}{2 π R ∣ ξ - z ∣^{2}} d s (ξ) .$ The explicit Poisson kernel identity itself is a choice-free calculation; Dependent Choice enters only through the uniqueness theorem for harmonic measure.

Facts & Assumptions

Given: A centre $c \in C$ , a radius $R > 0$ , a point $z \in D (c, R)$ , and Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain) for the uniqueness theorem. The Poisson kernel of the disc is as in The Poisson kernel on the unit disc, harmonic measure as in Harmonic measure on a bounded regular plane domain, and regular boundary points as in Barriers and regular boundary points.

[F1]

For continuous $ψ : \partial D \to R$ the Poisson integral $P [ψ]$ is harmonic on the unit disc, continuous on its closure, and equal to $ψ$ on the boundary, and it is the unique such function (The Poisson integral gives the unique continuous harmonic extension on the closed unit disc).

[F2]

Under Dependent Choice, every bounded regular plane domain has exactly one harmonic measure at each interior point (Existence and uniqueness of harmonic measure on a bounded regular plane domain); the functions continuous on $\overline{Ω}$ and harmonic on $Ω$ with equal boundary values coincide (The bounded plane Dirichlet problem has at most one continuous harmonic solution).

[F3]

If $b$ is a barrier at a boundary point $ζ$ of a bounded complex domain $Ω$ , then $ζ$ is regular: for every continuous boundary datum the regularized Perron envelope has limit $φ (ζ)$ at $ζ$ (A planar barrier forces the regularized Perron envelope to have the prescribed boundary limit, Barriers and regular boundary points); a barrier at $ζ$ is a subharmonic $b < 0$ on $Ω$ with $b (w) \to 0$ as $w \to ζ$ inside $Ω$ and with each boundary point outside a neighbourhood of $ζ$ kept away from $0$ uniformly.

[F4]

The function $\log ∣ \cdot ∣$ is harmonic on $C ∖ {0}$ (Logarithmic modulus is harmonic off its centre), and harmonicity is preserved by composition with holomorphic maps (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate).

Proof

technique · direct

1.1F3F4

Every boundary point of a disc is regular. Fix $ξ \in \partial D (c, R)$ and put $ζ_{*} : = c + 2 (ξ - c)$ , so that $∣ ζ_{*} - c ∣ = 2 R$ and $∣ ζ_{*} - ξ ∣ = R$ . Define $b (w) : = \log R - \log ∣ w - ζ_{*} ∣$ for $w \in D (c, R)$ . Then $b$ is harmonic on $D (c, R)$ by [F4], since $w \mapsto \log ∣ w - ζ_{*} ∣$ is the composition of $\log ∣ \cdot ∣$ with the translation $w \mapsto w - ζ_{*}$ , which is holomorphic and nowhere zero on the disc. Moreover $∣ w - ζ_{*} ∣ > R$ for $w \in D (c, R)$ , so $b < 0$ there; $b (ξ) = \log R - \log R = 0$ , so $b \to 0$ at $ξ$ by continuity of $b$ at $ξ$ . For every neighbourhood $V$ of $ξ$ , choose an open neighbourhood $W$ with $ξ \in W \subseteq V$ . If $\partial D (c, R) ∖ W$ is empty, the uniform separation condition in [F3] is vacuous and any $c_{V} < 0$ works. Otherwise the continuous boundary extension of $b$ attains a strictly negative maximum on the nonempty compact set $\partial D (c, R) ∖ W$ ; choosing this maximum as $c_{V}$ gives the required bound at every point of $\partial D (c, R) ∖ V$ . Hence $b$ is a barrier at $ξ$ and $ξ$ is regular by [F3]; since $ξ$ was arbitrary, $D (c, R)$ is a bounded regular plane domain.

1.2F1algebra

For a continuous $φ : \partial D (c, R) \to R$ put $ψ (t) : = φ (c + R e^{i t})$ and $u (ζ) : = P [ψ] ((ζ - c) / R)$ for $ζ \in D (c, R)$ . By [F1] applied to the unit disc, $u$ is harmonic on $D (c, R)$ and continuous on its closure with boundary values $φ$ . For $ζ = z$ the definition of the Poisson kernel gives, with $ξ = c + R e^{i t}$ , $u (z) = \frac{1}{2 π} \int_{0}^{2 π} φ (ξ) \frac{R^{2} - ∣ z - c ∣^{2}}{∣ ξ - z ∣^{2}} d t,$ because $1 - ∣ (z - c) / R ∣^{2} = (R^{2} - ∣ z - c ∣^{2}) / R^{2}$ and $∣ e^{i t} - (z - c) / R ∣ = ∣ ξ - z ∣ / R$ .

2.1F2step 1.1step 1.2

Since all boundary points of $D (c, R)$ are regular by step 1.1, the regularized Perron envelope $H_{φ}$ of a continuous datum $φ$ is harmonic on $D (c, R)$ and has the boundary limit $φ$ at every boundary point; it is therefore a continuous harmonic extension of $φ$ to the closure, and so is $u$ by step 1.2. Uniqueness [F2] gives $H_{φ} = u$ , hence by step 1.2 $H_{φ} (z) = \frac{1}{2 π} \int_{0}^{2 π} φ (c + R e^{i t}) \frac{R^{2} - ∣ z - c ∣^{2}}{∣ c + R e^{i t} - z ∣^{2}} d t .$

3.1F2step 2.1algebra

Define the measure $ν$ on $\partial D (c, R)$ by $ν (E) : = \frac{1}{2 π} \int_{{t \in [0, 2 π) : c + R e^{i t} \in E}} \frac{R^{2} - ∣ z - c ∣^{2}}{∣ c + R e^{i t} - z ∣^{2}} d t$ . The integrand is continuous and positive, so $ν$ is a finite Borel measure on the compact circle, and step 2.1 says exactly that $\int φ d ν = H_{φ} (z)$ for every continuous $φ$ ; taking $φ \equiv 1$ and using the representation theorem of [F2], $ν$ is a probability measure. Hence $ν$ is a harmonic measure for $D (c, R)$ at $z$ , and by the uniqueness in [F2], $ν = ω_{D (c, R)}^{z}$ . Finally, the parametrization $t \mapsto c + R e^{i t}$ is arclength measured in units $d s = R d t$ , so the density of $ω_{D (c, R)}^{z}$ with respect to $s$ is $(R^{2} - ∣ z - c ∣^{2}) / (2 π R ∣ ξ - z ∣^{2})$ , which is the displayed second form.

4.1F2step 2.1step 3.1∎

Consequently the harmonic measure of the disc $D (c, R)$ at $z$ has the Poisson density of the statement, in both its angle form and its arclength form, and it is a probability measure on the boundary circle. The kernel computation of steps 1.2 and 3.1 is choice-free; $D C$ was used only in step 2.1 through the uniqueness theorem [F2] and in the identification of $ν$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Conformal invariance of harmonic measure

Statement

Assume Dependent Choice. Let $Ω, Ω^{'} \subseteq C$ be bounded regular plane domains, in the sense of Harmonic measure on a bounded regular plane domain, and let $F : Ω \to Ω^{'}$ be a biholomorphism (Biholomorphic maps between complex domains) that extends to a homeomorphism $\overline{F} : \overline{Ω} \to \overline{Ω^{'}}$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological). Then for every $z \in Ω$ the pushforward boundary measure satisfies ${\overline{F}}_{*} ω_{Ω}^{z} = ω_{Ω^{'}}^{F (z)}$ on all Borel subsets of $\partial Ω^{'}$ . No pushforward of a boundary measure is asserted without the closure homeomorphism: the transport is proved by equality of continuous harmonic extensions and not by a boundary correspondence alone.

Facts & Assumptions

Given: Bounded plane domains $Ω, Ω^{'}$ all of whose boundary points are regular (A complex domain is a nonempty connected open subset of $C$ , Harmonic measure on a bounded regular plane domain), a biholomorphism $F : Ω \to Ω^{'}$ , and a homeomorphism $\overline{F} : \overline{Ω} \to \overline{Ω^{'}}$ extending $F$ ; also Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[F1]

Under Dependent Choice the harmonic measures $ω_{Ω}^{z}$ and $ω_{Ω^{'}}^{F (z)}$ exist and are the unique Radon Borel probability measures on the compact boundaries representing their Perron envelopes (Existence and uniqueness of harmonic measure on a bounded regular plane domain): for continuous data $ψ$ on $\partial Ω$ and $φ$ on $\partial Ω^{'}$ , $H_{Ω, ψ} (z) = \int ψ d ω_{Ω}^{z}$ and $H_{Ω^{'}, φ} (F (z)) = \int φ d ω_{Ω^{'}}^{F (z)}$ .

[F2]

Regularity of every boundary point means $H_{Ω, ψ} (w) \to ψ (ζ)$ as $w \to ζ$ inside $Ω$ , for every $ζ \in \partial Ω$ and every continuous $ψ$ ; hence $H_{Ω, ψ}$ is continuous on $\overline{Ω}$ when set equal to $ψ$ on $\partial Ω$ , and analogously for $Ω^{'}$ . Two continuous functions on $\overline{Ω}$ , harmonic on $Ω$ , with equal boundary values coincide (Harmonic measure on a bounded regular plane domain, The bounded plane Dirichlet problem has at most one continuous harmonic solution).

[F3]

Composition with a holomorphic map preserves harmonicity (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate), and a homeomorphism between the closures restricting to a bijection $Ω \to Ω^{'}$ carries $\partial Ω$ onto $\partial Ω^{'}$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, Continuity of a map of topological spaces at a point and globally).

[F4]

Two finite regular Borel measures on a compact space that agree on all continuous functions coincide (Positive C_0(X) functionals have finite regular representing measures, Radon measure on an LCH space).

Proof

technique · direct

1.1F3given

The map $\overline{F}$ is a bijection of the compact sets $\overline{Ω}$ and $\overline{Ω^{'}}$ restricting to the bijection $F : Ω \to Ω^{'}$ ; therefore it maps $\partial Ω = \overline{Ω} ∖ Ω$ onto $\overline{Ω^{'}} ∖ Ω^{'} = \partial Ω^{'}$ , and it is a homeomorphism between the two boundaries. Consequently, for continuous $φ : \partial Ω^{'} \to R$ the pullback $φ \circ \overline{F}$ is continuous on $\partial Ω$ , and the pushforward $({\overline{F}}_{*} ω_{Ω}^{z}) (E) : = ω_{Ω}^{z} ({\overline{F}}^{- 1} (E))$ is well defined on Borel subsets of $\partial Ω^{'}$ .

2.1F2F3step 1.1

For continuous $φ$ on $\partial Ω^{'}$ and $ψ : = φ \circ \overline{F}$ on $\partial Ω$ , the function $u : = H_{Ω^{'}, φ} \circ F$ is harmonic on $Ω$ by [F3], since $H_{Ω^{'}, φ}$ is harmonic on $Ω^{'}$ , and it extends continuously to $\partial Ω$ with boundary values $ψ$ , because $H_{Ω^{'}, φ}$ extends continuously to $\overline{Ω^{'}}$ with values $φ$ by [F2] and $\overline{F}$ maps $\partial Ω$ onto $\partial Ω^{'}$ by step 1.1.

2.2F1step 1.1

The pushforward of step 1.1 represents the same value: by the defining property of $ω_{Ω}^{z}$ in [F1] and the change of variables defining the pushforward, $\int_{\partial Ω^{'}} φ d ({\overline{F}}_{*} ω_{Ω}^{z}) = \int_{\partial Ω} (φ \circ \overline{F}) d ω_{Ω}^{z} = H_{Ω, ψ} (z) .$

3.1F2step 2.1

The continuous harmonic extensions $u = H_{Ω^{'}, φ} \circ F$ and $H_{Ω, ψ}$ of step 2.1 have the same boundary values $ψ$ on $\partial Ω$ , so they coincide on $Ω$ by [F2]; at $z$ this is $H_{Ω, ψ} (z) = H_{Ω^{'}, φ} (F (z)) .$

4.1F1F4step 3.1step 2.2

Combining steps 3.1 and 2.2 with the defining property of $ω_{Ω^{'}}^{F (z)}$ in [F1] gives, for every continuous $φ : \partial Ω^{'} \to R$ , $\int_{\partial Ω^{'}} φ d ({\overline{F}}_{*} ω_{Ω}^{z}) = H_{Ω^{'}, φ} (F (z)) = \int_{\partial Ω^{'}} φ d ω_{Ω^{'}}^{F (z)} .$ Both sides are finite regular Borel measures on the compact boundary $\partial Ω^{'}$ , so by [F4] they coincide as measures, and in particular on every Borel subset of $\partial Ω^{'}$ .

5.1F1step 4.1∎

Therefore ${\overline{F}}_{*} ω_{Ω}^{z} = ω_{Ω^{'}}^{F (z)}$ on all Borel boundary sets. Dependent Choice was used only through the existence and uniqueness theorem [F1]; the transport itself is the identification of two continuous harmonic extensions with common boundary data, and no boundary behaviour of $F$ beyond the given closure homeomorphism was assumed.

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Borel harmonicity and comparison of harmonic measure

Statement

Assume Dependent Choice. Let $Ω$ be a bounded regular plane domain, in the sense of Harmonic measure on a bounded regular plane domain. Then for every Borel set $E \subseteq \partial Ω$ the function $z \mapsto ω_{Ω}^{z} (E)$ is harmonic on $Ω$ (Plane harmonic functions) and takes values in $[0, 1]$ , and for every fixed $z \in Ω$ the assignment $E \mapsto ω_{Ω}^{z} (E)$ is countably additive. If $Ω_{1} \subseteq Ω_{2}$ are bounded regular domains and $E \subseteq \partial Ω_{1} \cap \partial Ω_{2}$ is Borel, then $ω_{Ω_{1}}^{z} (E) \leq ω_{Ω_{2}}^{z} (E) (z \in Ω_{1}) .$ The inequality is in this direction: enlarging the domain does not decrease the harmonic mass of a common boundary piece.

Facts & Assumptions

Given: Bounded regular plane domains in the sense of Harmonic measure on a bounded regular plane domain and A complex domain is a nonempty connected open subset of $C$ , a point $z$ of the relevant domain, and Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Harmonicity is that of Plane harmonic functions, distances to subsets are as in Distance from a point to a subset, and Radon measures are as in Radon measure on an LCH space.

[F1]

Under Dependent Choice, for every bounded regular plane domain $Ω$ and every $z \in Ω$ the harmonic measure $ω_{Ω}^{z}$ exists and is the unique Radon Borel probability measure on $\partial Ω$ with $H_{φ} (z) = \int_{\partial Ω} φ d ω_{Ω}^{z}$ for every real continuous $φ$ ; moreover $z \mapsto H_{φ} (z)$ is the unique continuous extension to $\overline{Ω}$ that is harmonic on $Ω$ and agrees with $φ$ on $\partial Ω$ (Existence and uniqueness of harmonic measure on a bounded regular plane domain, Harmonic measure on a bounded regular plane domain).

[F2]

A Radon measure $μ$ on a locally compact Hausdorff space satisfies $μ (E) = \inf_{E \subseteq V open} μ (V)$ for every Borel $E$ , $μ (U) = \sup_{K \subseteq U compact} μ (K)$ for every open $U$ , and $μ (K) < \infty$ for every compact $K$ (Radon measure on an LCH space).

[F3]

If $(u_{m})$ is an increasing sequence of harmonic functions on a complex domain, then either $u_{m} \to + \infty$ pointwise everywhere or $u_{m}$ converges locally uniformly to a harmonic function (An increasing harmonic sequence converges locally uniformly to a harmonic limit or diverges to +infinity).

[F4]

If $0 \leq f_{1} \leq f_{2} \leq \dots$ are measurable with $f_{m} ↑ f$ pointwise, then $\int f_{m} d μ ↑ \int f d μ$ (Monotone convergence for the integral).

[F5]

For a bounded complex domain $Ω$ and $u$ continuous on $\overline{Ω}$ and harmonic on $Ω$ , $\inf_{\overline{Ω}} u = \inf_{\partial Ω} u$ and $\sup_{\overline{Ω}} u = \sup_{\partial Ω} u$ (Maximum and minimum principles for plane harmonic functions).

[F6]

For nonempty $A \subseteq X$ in a metric space, $x \mapsto d (x, A)$ is $1$ -Lipschitz, hence continuous, and the set ${x : d (x, A) \geq c}$ is closed for every $c$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, Distance from a point to a subset).

[F7]

If $K$ is nonempty compact, $C$ nonempty closed and $K \cap C = \emptyset$ in a real or complex normed space, then there is $δ > 0$ with $∥ k - c ∥ \geq δ$ for all $k \in K$ , $c \in C$ (A compact set and a disjoint closed set have a positive norm-distance gap).

[F8]

A closed subset of a compact metric space is a compact subset of it (A closed subset of a compact metric space is compact).

[F9]

Compactness of a subset is intrinsic: a set that is a compact subset of one ambient space is a compact subset of every ambient space inducing the same topology on it (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[F10]

The Borel sigma-algebra on a topological space is the sigma-algebra generated by its open sets; a family of subsets that contains every open set and is closed under complements and countable unions contains every Borel set (The Borel sigma-algebra of a topological space).

[F11]

A lambda-system contains the whole space, is closed under differences $B ∖ A$ when $A \subseteq B$ are members, and is closed under increasing countable unions (Lambda-systems, or Dynkin systems). The family of open subsets of a topological space is a pi-system, and Dynkin's pi-lambda theorem says that every lambda-system containing a pi-system contains the sigma-algebra it generates (Pi-systems, Dynkin's pi-lambda theorem).

[F12]

Finite linear combinations of harmonic functions are harmonic, since harmonic functions are $C^{2}$ with Laplacian zero and the Laplacian is linear (Plane harmonic functions).

Proof

technique · direct

1.1F6F7given

Fix a compact nonempty $K \subseteq \partial Ω$ and for $m \geq 1$ define $φ_{m} (x) = \max {0, 1 - m d (x, K)}$ on $\partial Ω$ . Each $φ_{m}$ is continuous and $0 \leq φ_{m + 1} \leq φ_{m} \leq 1$ ; and $φ_{m} = 1$ on $K$ , while for $x \notin K$ the compact set $K$ and the closed set ${x}$ are disjoint, so $d (x, K) > 0$ by [F7] and $φ_{m} (x) = 0$ for all $m \geq 1 / d (x, K)$ . Thus $φ_{m} ↓ 1_{K}$ pointwise on $\partial Ω$ .

1.2F1F6F7F8given

Let $U \subseteq \partial Ω$ be open. The cases $U = \partial Ω$ and $U = \emptyset$ give the constant functions $1$ and $0$ , which are harmonic; assume $U \notin {\emptyset, \partial Ω}$ . For $m \geq 1$ put $K_{m} = {x \in \partial Ω : d (x, \partial Ω ∖ U) \geq 1 / m}$ . Since $\partial Ω ∖ U$ is a nonempty closed set and $d (\cdot, \partial Ω ∖ U)$ is continuous, each $K_{m}$ is closed in $\partial Ω$ , hence compact because $\partial Ω$ is compact; clearly $K_{m} \subseteq K_{m + 1}$ . If $x \in K_{m}$ then $d (x, \partial Ω ∖ U) \geq 1 / m > 0$ , so $x \notin \partial Ω ∖ U$ (that set being closed, a point of it would have distance zero), hence $x \in U$ and $K_{m} \subseteq U$ . Conversely every $x \in U$ lies outside the nonempty closed set $\partial Ω ∖ U$ , so by [F7] applied to the compact singleton ${x}$ we have $d (x, \partial Ω ∖ U) > 0$ , hence $x \in K_{m}$ for all large $m$ ; therefore $⋃_{m} K_{m} = U$ .

1.3F1F5F9given

Now let $Ω_{1} \subseteq Ω_{2}$ be bounded regular domains, let $z \in Ω_{1}$ , and let $K \subseteq \partial Ω_{1} \cap \partial Ω_{2}$ be compact. Such a $K$ is a compact subset of both $\partial Ω_{1}$ and $\partial Ω_{2}$ by [F9]. Let $φ_{2} : \partial Ω_{2} \to R$ be continuous with $φ_{2} \geq 1_{K}$ , and let $u_{2}$ be the continuous harmonic extension of $φ_{2}$ on $Ω_{2}$ provided by [F1]. Evaluating the representation of [F1] for $Ω_{2}$ at $z$ gives $\int_{\partial Ω_{2}} φ_{2} d ω_{Ω_{2}}^{z} = u_{2} (z)$ . Since $Ω_{1} \subseteq Ω_{2}$ we have $\overline{Ω_{1}} \subseteq \overline{Ω_{2}}$ , and $u_{2} \geq 0$ on $\overline{Ω_{2}}$ by [F5] and $φ_{2} \geq 0$ ; restricted to $\overline{Ω_{1}}$ it is continuous, harmonic on $Ω_{1}$ , and hence is the unique continuous harmonic extension of its trace $u_{2} ∣_{\partial Ω_{1}}$ . Applying the representation of [F1] to $Ω_{1}$ with that trace gives $\int_{\partial Ω_{1}} u_{2} d ω_{Ω_{1}}^{z} = u_{2} (z)$ . Finally $u_{2} \geq 1_{K}$ at every point of $K \subseteq \partial Ω_{1} \cap \partial Ω_{2}$ , because $u_{2} = φ_{2}$ on $\partial Ω_{2}$ , and $u_{2} \geq 0$ everywhere on $\partial Ω_{1}$ , so $\int_{\partial Ω_{1}} u_{2} d ω_{Ω_{1}}^{z} \geq ω_{Ω_{1}}^{z} (K)$ . Chaining the three displays, $ω_{Ω_{1}}^{z} (K) \leq \int_{\partial Ω_{2}} φ_{2} d ω_{Ω_{2}}^{z}$ .

1.4F2F6F7given

The compact set $K \subseteq \partial Ω_{2}$ satisfies $ω_{Ω_{2}}^{z} (K) = \inf {\int_{\partial Ω_{2}} φ d ω_{Ω_{2}}^{z} : φ \in C (\partial Ω_{2}), φ \geq 1_{K}}$ . The inequality " $\geq$ " is monotonicity of the integral against the positive measure $ω_{Ω_{2}}^{z}$ ; for " $\leq$ " fix $ε > 0$ and use outer regularity [F2] to choose open $V \supseteq K$ with $ω_{Ω_{2}}^{z} (V) < ω_{Ω_{2}}^{z} (K) + ε$ . If $\partial Ω_{2} ∖ V = \emptyset$ the constant function $1$ is admissible and $\int 1 d ω_{Ω_{2}}^{z} = 1 = ω_{Ω_{2}}^{z} (V) < ω_{Ω_{2}}^{z} (K) + ε$ . Otherwise $\partial Ω_{2} ∖ V$ is nonempty closed and disjoint from $K$ , so $δ : = d (K, \partial Ω_{2} ∖ V) > 0$ by [F7]; the function $φ (x) = \max {0, 1 - d (x, K) / δ}$ is continuous by [F6], satisfies $1_{K} \leq φ \leq 1_{V}$ , and hence $\int φ d ω_{Ω_{2}}^{z} \leq ω_{Ω_{2}}^{z} (V) < ω_{Ω_{2}}^{z} (K) + ε$ . Letting $ε ↓ 0$ proves the displayed infimum.

1.5F2F8given

Let $E \subseteq \partial Ω_{1} \cap \partial Ω_{2}$ be Borel and let $ε > 0$ . The measure $ω_{Ω_{1}}^{z}$ has total mass one and is outer regular on Borel sets by [F2]; applied to the Borel set $\partial Ω_{1} ∖ E$ it yields an open $V \supseteq \partial Ω_{1} ∖ E$ with $ω_{Ω_{1}}^{z} (V) < ω_{Ω_{1}}^{z} (\partial Ω_{1} ∖ E) + ε$ . Then $C : = \partial Ω_{1} ∖ V$ is closed in $\partial Ω_{1}$ , hence compact by [F8] and $\partial Ω_{1}$ being compact, satisfies $C \subseteq E$ , and $ω_{Ω_{1}}^{z} (C) = 1 - ω_{Ω_{1}}^{z} (V) > 1 - ω_{Ω_{1}}^{z} (\partial Ω_{1} ∖ E) - ε = ω_{Ω_{1}}^{z} (E) - ε$ .

2.1F1step 1.1

For each $m$ let $u_{m} : = H_{φ_{m}}$ be the corresponding envelope for $Ω$ ; by [F1] each $u_{m}$ is harmonic on $Ω$ , continuous on $\overline{Ω}$ , and satisfies $u_{m} (w) = \int_{\partial Ω} φ_{m} d ω_{Ω}^{w}$ for every $w \in Ω$ . Since $φ_{m + 1} \leq φ_{m}$ and $ω_{Ω}^{w}$ is a positive measure, $u_{m + 1} (w) = \int φ_{m + 1} d ω_{Ω}^{w} \leq \int φ_{m} d ω_{Ω}^{w} = u_{m} (w)$ and $u_{m} (w) \geq 0$ for every $w$ .

2.2step 1.3step 1.4

Combining steps 1.3 and 1.4, for every compact $K \subseteq \partial Ω_{1} \cap \partial Ω_{2}$ we have $ω_{Ω_{1}}^{z} (K) \leq \inf_{φ_{2} \geq 1_{K}} \int_{\partial Ω_{2}} φ_{2} d ω_{Ω_{2}}^{z} = ω_{Ω_{2}}^{z} (K)$ ; for $K = \emptyset$ both sides are $0$ .

3.1F1F4step 1.1step 2.1

Fix $w \in Ω$ . The functions $φ_{1} - φ_{m}$ are nonnegative, measurable, and increase to $φ_{1} - 1_{K}$ pointwise by step 1.1, so monotone convergence [F4] gives $\int (φ_{1} - φ_{m}) d ω_{Ω}^{w} ↑ \int (φ_{1} - 1_{K}) d ω_{Ω}^{w}$ . All these integrals are finite because $ω_{Ω}^{w}$ is a probability measure and $0 \leq φ_{1} - φ_{m} \leq 1$ ; subtracting the common finite value $\int φ_{1} d ω_{Ω}^{w}$ gives $u_{m} (w) = \int φ_{m} d ω_{Ω}^{w} ↓ ω_{Ω}^{w} (K)$ .

4.1F1F3step 2.1step 3.1

The sequence $- u_{m}$ is increasing by step 2.1 and bounded above by $0$ , so the divergent alternative of [F3] is excluded and $- u_{m}$ converges locally uniformly on $Ω$ to a harmonic function; equivalently $u_{m}$ converges locally uniformly to a harmonic function $u$ , and by step 3.1 $u (w) = \lim_{m} u_{m} (w) = ω_{Ω}^{w} (K)$ for every $w \in Ω$ . Hence $w \mapsto ω_{Ω}^{w} (K)$ is harmonic on $Ω$ for every nonempty compact $K \subseteq \partial Ω$ ; for $K = \emptyset$ it is the constant $0$ , which is harmonic.

5.1F1step 4.1

Since every $ω_{Ω}^{w}$ is a probability measure, $0 \leq ω_{Ω}^{w} (K) \leq 1$ for every $w$ and every compact $K$ .

6.1F3step 4.1step 5.1step 1.2

For each $w$ , continuity from below for the measure $ω_{Ω}^{w}$ gives $ω_{Ω}^{w} (U) = \lim_{m} ω_{Ω}^{w} (K_{m})$ by step 1.2. The functions $w \mapsto ω_{Ω}^{w} (K_{m})$ are harmonic by step 4.1, the sequence is increasing in $m$ and takes values in $[0, 1]$ by step 5.1, so its limit $w \mapsto ω_{Ω}^{w} (U)$ is harmonic on $Ω$ by [F3].

7.1F1F9step 2.2step 1.5given∎

Let $A$ be the family of Borel sets $E \subseteq \partial Ω$ for which $w \mapsto ω_{Ω}^{w} (E)$ is harmonic on $Ω$ . It contains $\partial Ω$ because the harmonic measure is a probability, and it contains every open set by steps 1.2 and 6.1 and the two trivial open cases there. If $A, B \in A$ with $A \subseteq B$ , then for every $w$ the measure identity gives $ω_{Ω}^{w} (B ∖ A) = ω_{Ω}^{w} (B) - ω_{Ω}^{w} (A)$ ; the right side is a difference of harmonic functions, hence harmonic by [F12], so $B ∖ A \in A$ . If $E_{1} \subseteq E_{2} \subseteq \dots$ are in $A$ , then continuity from below for each measure gives $ω_{Ω}^{w} (⋃_{n} E_{n}) = \lim_{n} ω_{Ω}^{w} (E_{n})$ . This is an increasing sequence of harmonic functions bounded above by $1$ , so its limit is harmonic by [F3]. Thus $A$ is a lambda-system by [F11]. The open subsets of $\partial Ω$ form a pi-system that generates its Borel sigma-algebra, so Dynkin's pi-lambda theorem [F11] implies that $A$ contains every Borel set. Hence $w \mapsto ω_{Ω}^{w} (E)$ is harmonic for every Borel $E$ , its values lie in $[0, 1]$ because each $ω_{Ω}^{w}$ is a probability measure, and countable additivity in $E$ at fixed $w$ is the measure property of $ω_{Ω}^{w}$ . [F1, F3, F10, F11, F12, step 5.1, step 6.1, step 1.2] 8.1 The compact set $C$ of step 1.5 is a compact subset of $\partial Ω_{1} \cap \partial Ω_{2}$ and hence of $\partial Ω_{2}$ by [F9]; if $C \neq \emptyset$ step 2.2 gives $ω_{Ω_{1}}^{z} (C) \leq ω_{Ω_{2}}^{z} (C)$ , while for $C = \emptyset$ this inequality is trivial. Since $C \subseteq E \subseteq \partial Ω_{2}$ , monotonicity of $ω_{Ω_{2}}^{z}$ gives $ω_{Ω_{2}}^{z} (C) \leq ω_{Ω_{2}}^{z} (E)$ . Therefore $ω_{Ω_{1}}^{z} (E) < ω_{Ω_{1}}^{z} (C) + ε \leq ω_{Ω_{2}}^{z} (E) + ε$ for every $ε > 0$ , so $ω_{Ω_{1}}^{z} (E) \leq ω_{Ω_{2}}^{z} (E)$ . Dependent Choice was used only through the existence and uniqueness theorem [F1]; the compact and Borel approximation arguments and the comparison itself are choice-free.

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Green and harmonic-measure representation with the $2 π$ sign

Statement

Assume Dependent Choice, hence Countable Choice (Dependent choice implies countable choice). Let $Ω \subseteq C$ be a bounded $C^{1}$ regular plane domain: a bounded $C^{1}$ domain in the sense of Bounded C1 domains and their outward normals which is a complex domain (A complex domain is a nonempty connected open subset of $C$ ) every boundary point of which is regular (Barriers and regular boundary points). Let $u \in C^{2} (Ω) \cap C^{1} (\overline{Ω})$ be real-valued with $Δ u$ bounded on $Ω$ , and write $d A$ for two-dimensional Lebesgue measure. Then for every $z \in Ω$ $u (z) = \int_{\partial Ω} u (ξ) d ω_{Ω}^{z} (ξ) - \frac{1}{2 π} \int_{Ω} g_{Ω} (z, y) Δ u (y) d A (y),$ and both integrals are absolutely finite.

If in addition $\partial Ω$ is real analytic, by which is meant the parametrization hypothesis of Green correctors are smooth at analytic boundaries: for every $ζ \in \partial Ω$ there are $ε > 0$ , a real-analytic $γ : (- ε, ε) \to C$ with $γ (0) = ζ$ and $γ^{'} (0) \neq 0$ , and a neighbourhood $U$ of $ζ$ with $\partial Ω \cap U = γ ((- ε, ε))$ for which $Ω \cap U$ is one of the two components of $U ∖ γ ((- ε, ε))$ ; then $d ω_{Ω}^{z} (ξ) = - \frac{1}{2 π} \partial_{ν_{ξ}} g_{Ω} (z, ξ) d s (ξ),$ where $ν$ is the outward unit normal of $\partial Ω$ and $d s$ its arclength element; explicitly $ω_{Ω}^{z} (E) = - \frac{1}{2 π} \int_{E} \partial_{ν_{ξ}} g_{Ω} (z, ξ) d s (ξ)$ for every Borel set $E \subseteq \partial Ω$ . The normal derivative in the boundary slot is the classical one (Classical normal derivative) of the trace $ξ \mapsto g_{Ω} (z, ξ) = g_{Ω} (ξ, z)$ , which symmetry (Canonical Green kernels are unique, symmetric and domain monotone) identifies with the trace of $ξ \mapsto g_{Ω} (ξ, z)$ , a function of class $C^{2}$ near $\partial Ω$ under the regularity hypothesis used below. The same conclusion holds if instead there is a uniformly dense set of continuous real boundary data, each admitting a harmonic extension of class $C^{2} (\overline{Ω})$ , and the correctors $H_{y}$ of $G_{Ω} : = g_{Ω} / (2 π)$ are of class $C^{2} (\overline{Ω})$ for every pole $y$ , so that the hypotheses of Green representation for classical Poisson data are met.

Neither a pointwise Poisson density for arbitrary continuous boundary data, nor the representation identity under the weaker hypothesis $u \in C^{2} (Ω) \cap C^{1} (\overline{Ω})$ with $Δ u$ merely finite, is asserted.

Facts & Assumptions

Given: Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); the bounded $C^{1}$ regular plane domain $Ω$ with its boundary normal and surface measure; the real function $u \in C^{2} (Ω) \cap C^{1} (\overline{Ω})$ with $Δ u$ bounded; a point $z \in Ω$ ; and, for the density clause, either the real-analytic boundary hypothesis or the dense-class hypothesis stated above.

[A1]

Dependent Choice is The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain; it implies Countable Choice (Dependent choice implies countable choice), and Countable Choice says that every at most countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For a bounded complex domain $Ω$ and $a \in Ω$ the canonical kernel exists and equals $g_{Ω} (z, a) = - \log ∣ z - a ∣ - h_{a} (z)$ , where $h_{a} = H_{b_{a}}$ is the regularized Perron envelope of the continuous datum $b_{a} = - \log ∣ \cdot - a ∣$ on $\partial Ω$ ; the kernel is harmonic and strictly positive on $Ω ∖ {a}$ , its corrector $- h_{a}$ is harmonic on $Ω$ , it tends to $0$ at every regular boundary point, and $- Δ_{z} T_{g_{Ω} (\cdot, a)} = 2 π δ_{a}$ (Green functions exist on all bounded plane domains, The canonical Green kernel of a plane domain).

[F2]

On a Greenian plane domain the canonical kernel is symmetric, $g_{Ω} (z, a) = g_{Ω} (a, z)$ , and it is monotone under domain enlargement: $g_{Ω_{1}} (z, a) \leq g_{Ω_{2}} (z, a)$ for Greenian $Ω_{1} \subseteq Ω_{2}$ and distinct $z, a \in Ω_{1}$ (Canonical Green kernels are unique, symmetric and domain monotone).

[F3]

For a bounded regular plane domain and each interior point there is exactly one Radon Borel probability measure $ω_{Ω}^{z}$ on $\partial Ω$ with $H_{φ} (z) = \int_{\partial Ω} φ d ω_{Ω}^{z}$ for every real continuous $φ$ , and $H_{φ}$ is the unique continuous extension to $\overline{Ω}$ that is harmonic on $Ω$ and agrees with $φ$ on $\partial Ω$ (Existence and uniqueness of harmonic measure on a bounded regular plane domain, Harmonic measure on a bounded regular plane domain).

[F4]

For a continuous datum $φ$ on the boundary of a bounded complex domain with $m = \min_{\partial Ω} φ$ and $M = \max_{\partial Ω} φ$ , the Perron family is nonempty and $m \leq U_{φ} \leq M$ , and the regularized envelope satisfies $m \leq H_{φ} \leq M$ (The Perron family is nonempty and uniformly bounded by the boundary data, The Perron envelope and its regularization, The Perron lower family for continuous boundary data).

[F5]

A harmonic function on a bounded complex domain that extends continuously to the closure has its supremum and infimum on the boundary, and two functions continuous on $\overline{Ω}$ and harmonic on $Ω$ with equal boundary values coincide (Maximum and minimum principles for plane harmonic functions, The bounded plane Dirichlet problem has at most one continuous harmonic solution).

[F6]

The normalized kernel $Φ = - (2 π)^{- 1} \log ∣ \cdot ∣$ of Fundamental solution for the positive operator minus Laplacian is locally integrable on $R^{2}$ , with $\int_{B_{R}} ∣ Φ ∣ d A = \int_{0}^{R} r ∣ \log r ∣ d r$ finite for every $R > 0$ , and Lebesgue measure on $R^{2}$ is translation invariant (Local integrability of the Laplace fundamental kernel, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F7]

Distributions on an open set are continuous linear functionals on the real test functions $C_{c}^{\infty}$ , with $(\partial_{i} T) (ϕ) = - T (\partial_{i} ϕ)$ and $Δ T = \sum_{i} \partial_{i}^{2} T$ ; for $u \in C^{2} (Ω)$ one has $Δ T_{u} = T_{Δ u}$ , because distributional differentiation extends classical differentiation; the map $f \mapsto T_{f}$ is linear and injective modulo almost-everywhere equality (Distributional harmonicity and Poisson's equation on an open subset of Rn, Distributional differentiation is continuous and commutes, Regular distribution from a locally integrable function, Locally integrable functions embed in distributions).

[F8]

Under Countable Choice, a distribution with $Δ T = 0$ on an open set is $T_{h}$ for a unique smooth harmonic $h$ (Weyl's lemma for the Laplacian).

[F9]

Fubini's theorem computes a double integral of an $L^{1}$ function on a sigma-finite product as an iterated integral, and dominated convergence applies to measurable functions converging almost everywhere under one integrable majorant (Fubini's theorem for L^1 functions on a sigma-finite product, Dominated convergence).

[F10]

A bounded $C^{1}$ domain is locally, after a rigid change of coordinates, the subgraph of a $C^{1}$ function, and $F \in C^{2} (\overline{Ω})$ means that $F$ and its derivatives through order two extend continuously to the closure; on the boundary of such a domain the chart integral defines a finite Borel measure $d S$ with a continuous outward unit normal that agrees on chart overlaps, and the classical normal derivative of $F \in C^{1} (\overline{Ω})$ is $D F \cdot ν$ on $\partial Ω$ (Bounded C1 domains and their outward normals, Chart and partition independence of surface measure, Surface integration on compact C1 hypersurfaces, Classical normal derivative).

[F11]

Assume Countable Choice. Let $Ω$ be a bounded $C^{1}$ domain carrying a Dirichlet Green function $G_{Ω}$ for $- Δ$ whose correctors satisfy $H_{y} \in C^{2} (\overline{Ω})$ , and let $P_{Ω} (x, y) = - \partial_{ν_{y}} G_{Ω} (x, y)$ be its Poisson kernel, where the boundary-slot normal derivative is the trace of $D_{z} (Φ (z - x) - H_{x} (z)) \cdot ν_{Ω} (y)$ at $z = y$ from inside. Then for every real $U \in C^{2} (\overline{Ω})$ and $x \in Ω$ , $U (x) = \int_{Ω} G_{Ω} (x, y) (- Δ U (y)) d y + \int_{\partial Ω} P_{Ω} (x, y) U (y) d S (y),$ both integrals absolutely finite, and $P_{Ω} \geq 0$ with $\int_{\partial Ω} P_{Ω} (x, y) d S (y) = 1$ (Green representation for classical Poisson data, Poisson kernel from a Dirichlet Green function, Dirichlet Green function for minus Laplacian).

[F12]

If a bounded complex domain $D$ has a compact real-analytic boundary curve in the sense of the parametrization hypothesis, then every boundary point of $D$ is regular and for each $a \in D$ the Perron corrector $- \log ∣ \cdot - a ∣ - g_{D} (\cdot, a)$ extends to a function of class $C^{2} (\overline{D})$ with trace $- \log ∣ ξ - a ∣$ on $\partial D$ ; the Green kernel itself extends in class $C^{2}$ away from the pole and has zero boundary trace (Green correctors are smooth at analytic boundaries).

[F13]

A real-analytic parametrization is $C^{1}$ , sums, products and compositions of real-analytic functions are real analytic, a real-analytic function equals its power series near the centre, the same coefficients define a holomorphic function on a disc, a $C^{1}$ map with invertible derivative is a local diffeomorphism, a holomorphic map with nonzero derivative is a local biholomorphism, holomorphic functions have smooth real and imaginary components (Holomorphic functions are real analytic and smooth in their two real coordinates), and the real part of a holomorphic function with $C^{2}$ components is harmonic (A real-analytic function on an open subset of $R$ is locally represented by a convergent real power series, Real-analytic functions are closed under sums, products and compositions, and under quotients where the denominator is nonzero, Complex series, absolute convergence, complex power series, and radius of convergence, The sum of a complex power series is analytic throughout its open disc of convergence, The Euclidean inverse function theorem, Holomorphic inverse function theorem and local-degree criterion, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

[F14]

A harmonic function on a half-disc that is continuous on the closure and vanishes on the straight edge has a harmonic odd reflection to the full disc; plane harmonic functions are smooth; and harmonicity is preserved by composition with a holomorphic map (Harmonic and holomorphic Schwarz reflection across the real axis, Plane harmonic functions are smooth and real analytic, Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate).

[F15]

A unital subalgebra of $C (K, R)$ that separates points of a nonempty compact metric space $K$ is uniformly dense (Real Stone--Weierstrass theorem for compact metric spaces).

[F16]

Under Countable Choice every Borel measure finite on compact sets on a second-countable locally compact Hausdorff space is regular, hence Radon (Locally finite Borel measures on second-countable LCH spaces are regular, Radon measure on an LCH space, Second countability: an at most countable basis for the topology); the rational open boxes are a countable basis of $R^{2}$ ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis).

[F17]

A bounded subset of $R^{2}$ has compact closure, compact subsets of Euclidean space are closed and bounded, and a continuous real function on a nonempty compact Euclidean set is bounded and attains its bounds (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A compact subset of a metric space is closed and bounded, For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[F18]

Continuous maps pull Borel sets back to Borel sets, sums, products and absolute values of measurable functions are measurable, and every Borel subset of $R^{n}$ is Lebesgue measurable (A continuous map has Borel preimages of Borel sets, Arithmetic and lattice operations preserve measurability whenever they are defined, Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable).

[F19]

If $0 \leq f_{1} \leq f_{2} \leq \dots$ are measurable and increase pointwise to $f$ , then $\int f_{n} ↑ \int f$ (Monotone convergence for the integral).

[F20]

The nonnegative integral is monotone and additive on nonnegative Borel functions, the absolute value of an integral is at most the integral of the absolute value, and the Lebesgue integral is linear on $L^{1}$ (Monotonicity and nonnegative homogeneity of the nonnegative integral, The modulus of an integral is bounded by the integral of the modulus, The Lebesgue integral is linear on $L^{1} (μ)$ , Integrable real and complex functions, and their integrals).

[F21]

The mean value theorem and $\log^{'} (x) = 1 / x$ give $∣ \log a - \log b ∣ \leq ∣ a - b ∣ / d$ for real $a, b \geq d > 0$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ , The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

Proof

technique · direct

1.1givenF3F10F17F18

$Ω$ is nonempty, bounded, open and connected, so $\overline{Ω}$ and $\partial Ω$ are compact, and $u$ extends continuously to $\overline{Ω}$ with boundary values $u ∣_{\partial Ω}$ . The Laplacian $Δ u$ is continuous on $Ω$ , hence Borel and bounded there, and $f : = - Δ u$ lies in $L^{\infty} (Ω)$ . Every boundary point of $Ω$ is regular, so $Ω$ is a bounded regular plane domain in the sense of [F3].

1.2F6F20algebra

Uniform integrability of the kernel. Let $D : = diam (Ω)$ . For every $z \in Ω$ the domain is contained in the ball $B (z, D)$ , so translation invariance, polar coordinates and [F6] give $\int_{Ω} ∣ \log ∣ z - y ∣ ∣ d A (y) \leq \int_{B (0, D)} ∣ \log ∣ w ∣ ∣ d A (w) = 2 π \int_{0}^{D} r ∣ \log r ∣ d r = : J_{0} < \infty,$ and this bound is uniform over $z \in Ω$ .

2.1F1step 1.1

For every pole $a \in Ω$ the canonical kernel exists. Writing $b_{a} : = - \log ∣ \cdot - a ∣ \in C (\partial Ω)$ and $h_{a} : = H_{b_{a}}$ for its regularized Perron envelope, one has $g_{Ω} (z, a) = - \log ∣ z - a ∣ - h_{a} (z)$ for $z \in Ω ∖ {a}$ , the kernel is harmonic and strictly positive off $a$ , its corrector $- h_{a}$ is harmonic on $Ω$ , $g_{Ω} (z, a) \to 0$ as $z \to ζ$ at every boundary point $ζ$ , and $- Δ_{z} T_{g_{Ω} (\cdot, a)} = 2 π δ_{a}$ . In particular $Ω$ is Greenian.

3.1F1F2F4F17step 2.1algebra

A uniform logarithmic bound. Fix $c \in Ω$ and put $R : = 1 + \max_{y \in \overline{Ω}} ∣ y - c ∣$ , so that $\overline{Ω} \subset B : = D (c, R)$ , where $B$ is a bounded complex domain. By monotonicity $0 \leq g_{Ω} (z, y) \leq g_{B} (z, y)$ for distinct $z, y \in Ω$ , and by [F1] applied on $B$ , $g_{B} (z, y) = - \log ∣ z - y ∣ - H_{y}^{(B)} (z)$ with $H_{y}^{(B)}$ the Perron envelope of the datum $- \log ∣ \cdot - y ∣$ on $\partial B$ . By [F4], $H_{y}^{(B)} (z) \geq \min_{\partial B} (- \log ∣ \cdot - y ∣) = - \log (R + ∣ y - c ∣)$ , so $0 \leq g_{Ω} (z, y) \leq - \log ∣ z - y ∣ + \log (R + ∣ y - c ∣) \leq ∣ \log ∣ z - y ∣ ∣ + C_{*},$ with $C_{*} : = \log (2 R)$ , because $R + ∣ y - c ∣ < 2 R$ . The constant $C_{*}$ is independent of $z$ and $y$ .

3.2F3F4F5F21step 2.1algebra

Continuity off the diagonal. For fixed $y \in Ω$ the function $z \mapsto h_{y} (z) = H_{b_{y}} (z)$ is harmonic on $Ω$ and extends continuously to $\overline{Ω}$ with boundary values $b_{y}$ by [F3]; hence $(z, y) \mapsto g_{Ω} (z, y) = - \log ∣ z - y ∣ - h_{y} (z)$ is continuous on ${(z, y) \in Ω \times Ω : z \neq y}$ once $y \mapsto h_{y}$ is controlled uniformly on compact sets. Indeed, for $y, y^{'} \in K$ with $K \subset Ω$ compact and every $z \in Ω$ , $∣ h_{y} (z) - h_{y^{'}} (z) ∣ \leq \max_{\partial Ω} ∣ b_{y} - b_{y^{'}} ∣ \leq \frac{∣ y - y^{'} ∣}{dist (K, \partial Ω)},$ the first inequality because $φ \mapsto H_{φ}$ is additive and $∣ H_{τ} ∣ \leq \max_{\partial Ω} ∣ τ ∣$ by [F5] and [F4], and the second by [F21]. Given $(z, y)$ with $z \neq y$ , choose a compact $K \subset Ω$ containing $y$ and avoiding a neighbourhood of $z$ ; both estimates together with continuity of $z \mapsto h_{y} (z)$ give joint continuity at $(z, y)$ . Hence the kernel is Borel measurable on that open set by [F18].

3.3F1F10F12F13step 2.1algebra

The analytic case: structure and correctors. Assume now that $\partial Ω$ is real analytic in the stated sense. Fix $ζ \in \partial Ω$ with its parametrization $γ$ and neighbourhood $U$ . Since $γ^{'} (0) \neq 0$ , after relabelling the two coordinates one has $γ_{1}^{'} (0) \neq 0$ , and the inverse function theorem applied to the $C^{1}$ map $t \mapsto γ_{1} (t)$ gives a $C^{1}$ inverse $x \mapsto t (x)$ near $x_{0} = γ_{1} (0)$ with $γ_{1} (t (x)) = x$ . Hence near $ζ$ the boundary is the graph of the $C^{1}$ function $x \mapsto γ_{2} (t (x))$ and $Ω \cap U$ is locally one of the two components of the complement of that graph; reflecting the second coordinate if necessary, a rigid change of coordinates, makes $Ω$ locally the subgraph. Therefore $Ω$ is a bounded $C^{1}$ domain, so $\partial Ω$ carries the finite surface measure $d s$ and the continuous outward unit normal $ν$ of [F10]. Moreover [F12] applies with $D = Ω$ : every boundary point of $Ω$ is regular and the Perron corrector $h_{a} = - \log ∣ \cdot - a ∣ - g_{Ω} (\cdot, a)$ is of class $C^{2} (\overline{Ω})$ for every $a \in Ω$ . Consequently $G_{Ω} : = g_{Ω} / (2 π)$ is a Dirichlet Green function for $- Δ$ on $Ω$ whose correctors $H_{a} = h_{a} / (2 π)$ satisfy $H_{a} \in C^{2} (\overline{Ω})$ , since $Φ (ξ - a) = - (2 π)^{- 1} \log ∣ ξ - a ∣ = H_{a} (ξ)$ for $ξ \in \partial Ω$ .

4.1F18F20step 3.1step 3.2step 1.2given

The volume potential. With $f = - Δ u \in L^{\infty} (Ω)$ define $V (z) : = \frac{1}{2 π} \int_{Ω} g_{Ω} (z, y) f (y) d A (y) (z \in Ω) .$ For fixed $z$ the integrand is Borel measurable in $y$ by step 3.2, and step 3.1 together with step 1.2 bounds its integral by $\frac{1}{2 π} ∥ f ∥_{\infty} (C_{*} ∣ Ω ∣ + J_{0}) < \infty$ ; hence $V (z)$ is a well-defined finite real number for every $z \in Ω$ .

4.2F6F9F20step 3.1step 3.2step 2.1algebra

$V$ is continuous on $Ω$ . Let $z_{n} \to z$ in $Ω$ , all terms in a compact $K \subset Ω$ , and let $ε > 0$ . Put $J (δ) : = \int_{B (0, 2 δ)} ∣ \log ∣ w ∣ ∣ d A (w)$ ; by [F6] the function $∣ \log ∣ \cdot ∣ ∣$ is integrable on $B (0, 1)$ , and the integrands $1_{B (0, 2 δ)} ∣ \log ∣ w ∣ ∣$ decrease to $0$ off the origin as $δ ↓ 0$ , so dominated convergence [F9] gives $J (δ) \to 0$ . Fix $δ \in (0, 1)$ so small that $∥ f ∥_{\infty} (2 C_{*} π δ^{2} + 2 J (δ)) < ε$ . On $B (z, δ)$ the estimates of step 3.1 bound $∣ g_{Ω} (z_{n}, y) - g_{Ω} (z, y) ∣$ by $2 C_{*} + ∣ \log ∣ z_{n} - y ∣ ∣ + ∣ \log ∣ z - y ∣ ∣$ , whose integral over $B (z, δ)$ is at most $2 C_{*} π δ^{2} + 2 J (δ)$ by translation invariance, so the contribution of $B (z, δ)$ to $∣ V (z_{n}) - V (z) ∣$ is less than $ε / (2 π)$ . On $Ω ∖ B (z, δ)$ one has $∣ z_{n} - y ∣ \geq δ / 2$ for large $n$ , so the bound of step 3.1 gives $∣ g_{Ω} (z_{n}, y) f (y) ∣ \leq (C_{*} + \max {\log D, \log (2 / δ)}) ∥ f ∥_{\infty}$ , an integrable majorant on the bounded set $Ω$ ; the integrands converge pointwise to $g_{Ω} (z, y) f (y)$ off the null set ${z}$ by step 3.2, so dominated convergence makes this contribution tend to $0$ . Hence $V (z_{n}) \to V (z)$ .

4.3F1F9F20step 2.1step 3.1step 1.2algebra

$V$ vanishes at the boundary. Fix $ζ \in \partial Ω$ and $δ \in (0, 1)$ . For $z \in Ω \cap B (ζ, δ)$ , $\int_{Ω \cap B (ζ, 2 δ)} g_{Ω} (z, y) d A (y) \leq \int_{B (ζ, 2 δ)} (C_{*} + ∣ \log ∣ z - y ∣ ∣) d A (y) \leq C_{*} π (2 δ)^{2} + J (2 δ)$ by steps 3.1 and 1.2, a bound independent of $z$ that tends to $0$ with $δ$ . On the complement $Ω ∖ B (ζ, 2 δ)$ the kernel obeys $g_{Ω} (z, y) \leq C_{*} + \max {\log D, \log (1 / δ)}$ for $z \in Ω \cap B (ζ, δ)$ , and for each fixed $y \in Ω ∖ B (ζ, 2 δ)$ one has $g_{Ω} (z, y) \to 0$ as $z \to ζ$ by the boundary limit of step 2.1 at the regular point $ζ$ . Given $ε > 0$ , choose $δ$ first and then $z$ close enough to $ζ$ ; dominated convergence on the finite-measure set $Ω ∖ B (ζ, 2 δ)$ makes the second contribution small, so $\lim_{z \to ζ} V (z) = 0$ .

4.4F7F9F20step 2.1step 3.1step 1.2given

Distributional Laplacian of $V$ . Let $ϕ \in C_{c}^{\infty} (Ω)$ be a real test function with compact support $K$ . The double integral $\int_{Ω} \int_{Ω} ∣ g_{Ω} (z, y) f (y) Δ ϕ (z) ∣ d A (y) d A (z)$ is finite because for $z \in K$ the inner integral is at most $∥ f ∥_{\infty} (C_{*} ∣ Ω ∣ + J_{0})$ by steps 3.1 and 1.2. Fubini's theorem and the definition of the distributional Laplacian therefore give $⟨ Δ T_{V}, ϕ ⟩ = \int_{Ω} V Δ ϕ = \frac{1}{2 π} \int_{Ω} f (y) [\int_{Ω} g_{Ω} (z, y) Δ ϕ (z) d A (z)] d A (y) .$ The inner bracket is $⟨ T_{g_{Ω} (\cdot, y)}, Δ ϕ ⟩ = ⟨ Δ T_{g_{Ω} (\cdot, y)}, ϕ ⟩ = - 2 π ϕ (y)$ by the distributional identity of step 2.1, so $⟨ Δ T_{V}, ϕ ⟩ = - \int_{Ω} f ϕ = - ⟨ T_{f}, ϕ ⟩$ : that is $Δ T_{V} = - T_{f}$ .

4.5F3F13F14F15step 2.1step 3.3algebra

The analytic case: a dense class with $C^{2}$ harmonic extensions. Let $A \subseteq C (\partial Ω, R)$ be the set of restrictions to $\partial Ω$ of polynomials in the two real coordinates. Then $A$ contains the constants, is closed under sums and products, and separates points of the compact metric space $\partial Ω$ ; by [F15] it is uniformly dense in $C (\partial Ω, R)$ . Fix $ϕ = p ∣_{\partial Ω} \in A$ . Complexifying the power series of $γ$ at $0$ gives a holomorphic $Γ$ on a disc $D (0, r)$ with $Γ (0) = ζ$ , $Γ = γ$ on the real interval and $Γ^{'} (0) \neq 0$ ; by the holomorphic inverse function theorem, after shrinking $r$ , $Γ$ is a biholomorphism onto a neighbourhood $V$ of $ζ$ that maps the upper half-disc onto $Ω \cap V$ (replacing $γ$ by $t \mapsto γ (- t)$ if necessary). Put $w : = H_{ϕ} \circ Γ$ ; by [F14] and [F3] the function $w$ is harmonic on the half-disc and continuous on its closure with $w (t) = ϕ (γ (t))$ for $t \in (- r, r)$ . Here $t \mapsto ϕ (γ (t)) = p (γ (t))$ is real analytic by [F13], so it equals $\sum_{n} a_{n} t^{n}$ on some interval $(- r^{'}, r^{'})$ ; the sum $F (w) : = \sum_{n} a_{n} w^{n}$ is holomorphic on $∣ w ∣ < r^{'}$ , both components of $F$ are smooth by [F13], so the real part $u_{l o c} : = Re F$ is harmonic there by the $C^{2}$ components theorem with $u_{l o c} (t) = ϕ (γ (t))$ on the edge, and $v : = w - u_{l o c}$ is harmonic on the half-disc, continuous on its closure and zero on the edge. By [F14] the odd reflection of (a rescaled) $v$ is harmonic on the full disc, so $v$ is $C^{2}$ up to the edge and $w = u_{l o c} + v$ is $C^{2}$ on the closed half-disc; transferring through the biholomorphism $Γ$ shows that $H_{ϕ}$ agrees near the arc with a $C^{2}$ function on a neighbourhood of the boundary. As $ζ$ was arbitrary, $H_{ϕ} \in C^{2} (\overline{Ω})$ , and $H_{ϕ}$ is harmonic with $Δ H_{ϕ} = 0$ .

4.6F2F10F11step 3.3algebra

The kernel in terms of the Green function. By symmetry [F2], $g_{Ω} (x, y) = g_{Ω} (y, x)$ for distinct $x, y$ , and by step 3.3 the function $y \mapsto g_{Ω} (y, x) = - \log ∣ y - x ∣ - h_{x} (y)$ is of class $C^{2}$ near $\partial Ω$ ; hence the trace $y \mapsto g_{Ω} (x, y)$ has a classical normal derivative $\partial_{ν_{y}} g_{Ω} (x, y)$ there. The boundary-slot derivative of [F11] is the trace of $D_{z} (Φ (z - x) - H_{x} (z)) \cdot ν (y)$ , and $Φ (z - x) - H_{x} (z) = g_{Ω} (z, x) / (2 π)$ ; by symmetry, $\partial_{ν_{y}} g_{Ω} (z, x) ∣_{z = y} = \partial_{ν_{y}} g_{Ω} (x, y)$ . Therefore $P_{Ω} (x, y) = - \frac{1}{2 π} \partial_{ν_{y}} g_{Ω} (x, y)$ for every $y \in \partial Ω$ .

5.1F7step 4.4given

$u - V$ is distributionally harmonic. Since $u \in C^{2} (Ω)$ , [F7] gives $Δ T_{u} = T_{Δ u} = T_{- f}$ ; subtracting the identity of step 4.4 and using linearity of the embedding and of distributional differentiation, $Δ T_{u - V} = T_{- f} + T_{f} = 0$ as distributions on $Ω$ .

5.2F3F11step 3.3step 4.5given

Applying the PDE representation formula. In the analytic case, steps 3.3 and 4.5 provide: the bounded $C^{1}$ domain $Ω$ ; the Dirichlet Green function $G_{Ω} = g_{Ω} / (2 π)$ with $C^{2} (\overline{Ω})$ correctors; and the uniformly dense class $A$ of continuous data each of which has a harmonic $C^{2} (\overline{Ω})$ extension, namely $H_{ϕ}$ . In the alternative hypothesis of the statement the corresponding dense class and $C^{2}$ harmonic extensions, together with the $C^{2}$ corrector condition, are assumed, and the assumed extension of $ϕ$ coincides with $H_{ϕ}$ by the uniqueness in [F3]. In both cases [F11] applies with $U : = H_{ϕ}$ for $ϕ \in A$ , and $Δ H_{ϕ} = 0$ , so $H_{ϕ} (x) = \int_{\partial Ω} P_{Ω} (x, y) ϕ (y) d S (y), P_{Ω} (x, y) = - \partial_{ν_{y}} G_{Ω} (x, y) \geq 0, \int_{\partial Ω} P_{Ω} (x, y) d S (y) = 1$ for every $x \in Ω$ ; and by [F3], $\int_{\partial Ω} ϕ d ω_{Ω}^{x} = H_{ϕ} (x)$ .

6.1A1F6F8F18step 4.2step 5.1given

$u - V$ is harmonic. By [A1] Countable Choice holds, so [F8] applies and there is a unique smooth harmonic $h$ on $Ω$ with $T_{u - V} = T_{h}$ ; injectivity of the embedding modulo almost-everywhere equality gives $u - V = h$ almost everywhere. Both $u - V$ (by step 4.2 and continuity of $u$ ) and $h$ are continuous on $Ω$ , so the set where they differ is open and null, hence empty: a nonempty open set contains a ball of radius $r > 0$ , whose area is $π r^{2} > 0$ by the polar and translation formulas in [F6]; therefore $u - V = h$ everywhere on $Ω$ .

7.1F3step 1.1step 4.3step 6.1

The boundary values and harmonic measure. By step 4.3, $V (z) \to 0$ as $z \to ζ$ for every $ζ \in \partial Ω$ , while $u (z) \to u (ζ)$ by continuity; hence $h (z) = u (z) - V (z) \to u (ζ)$ . Thus $h$ extends continuously to $\overline{Ω}$ with boundary values $u ∣_{\partial Ω}$ , and the uniqueness of the continuous harmonic extension in [F3] gives $h (z) = H_{u ∣_{\partial Ω}} (z) = \int_{\partial Ω} u d ω_{Ω}^{z}$ for every $z \in Ω$ .

8.1A1F3F8F9F20step 3.1step 1.2step 7.1given

The representation formula. For $z \in Ω$ , $u (z) = h (z) + V (z) = \int_{\partial Ω} u d ω_{Ω}^{z} + \frac{1}{2 π} \int_{Ω} g_{Ω} (z, y) f (y) d A (y) = \int_{\partial Ω} u d ω_{Ω}^{z} - \frac{1}{2 π} \int_{Ω} g_{Ω} (z, y) Δ u (y) d A (y) .$ The boundary integral is absolutely finite because $ω_{Ω}^{z}$ is a probability measure, and the volume integral because $\frac{1}{2 π} \int_{Ω} ∣ g_{Ω} (z, y) Δ u (y) ∣ d A (y) \leq \frac{1}{2 π} ∥ Δ u ∥_{\infty} (C_{*} ∣ Ω ∣ + J_{0})$ by steps 3.1 and 1.2. Dependent Choice supplies harmonic measure through [F3] and implies the Countable Choice used in the kernel's distributional normalization [F1], Weyl's lemma [F8], and the measure and integration interfaces [F6], [F9], [F19] and [F20]; the density clause also uses it through [F11], [F12] and [F16]. No stronger choice principle is used. The statement is formulated for real $u$ ; a complex-valued $u$ is handled by applying the result to its real and imaginary parts. This proves clause 1.

9.1

Passage to all continuous data and identification of the measure. For $ϕ \in A$ and $x \in Ω$ , steps 5.2 and 4.6 give $\int_{\partial Ω} ϕ d ω_{Ω}^{x} = \int_{\partial Ω} ϕ (y) P_{Ω} (x, y) d S (y)$ . Define $ν (E) : = \int_{E} P_{Ω} (x, y) d S (y)$ for Borel $E \subseteq \partial Ω$ , the surface integral of the nonnegative Borel function $P_{Ω} (x, \cdot) 1_{E}$ as in [F10]. Countable additivity of $ν$ follows from the finite chart sum defining $d S$ and additivity of the Lebesgue integral over countable families of nonnegative functions [F19]; and $ν (\partial Ω) = \int_{\partial Ω} P_{Ω} (x, y) d S (y) = 1$ by step 5.2. The boundary $\partial Ω$ is a compact metric subspace of $R^{2}$ , and the intersections with $\partial Ω$ of the rational open boxes of $R^{2}$ form a countable basis of its topology, so $\partial Ω$ is a second-countable locally compact Hausdorff space and [F16] makes $ν$ Radon. For $ϕ \in A$ the two probability integrals agree, and if $ψ \in C (\partial Ω, R)$ is arbitrary then uniform density of $A$ and the bound $∥ ψ - ϕ ∥_{\infty}$ for both probability measures extend the identity to $ψ$ ; hence $\int_{\partial Ω} ψ d ν = H_{ψ} (x)$ for every continuous $ψ$ , that is, $ν$ is a harmonic measure for $Ω$ at $x$ . By the uniqueness in [F3], $ν = ω_{Ω}^{x}$ , and step 4.6 converts this into $ω_{Ω}^{x} (E) = - \frac{1}{2 π} \int_{E} \partial_{ν_{y}} g_{Ω} (x, y) d s (y)$ for every Borel set $E \subseteq \partial Ω$ , which is the density clause. In the analytic case this used [F12] and the polynomial class; in the alternative case it used the assumed dense class and correctors. No pointwise Poisson density for arbitrary continuous data and no $C^{2} (Ω) \cap C^{1} (\overline{Ω})$ representation without the bounded-Laplacian hypothesis is claimed. ∎

Source notes

Lyubich §§10.8-10.9, printed pp. 171-172, defines harmonic measure as the measure representing evaluation of the Dirichlet solution at an interior point and defines the Green function by the Dirichlet zero boundary condition with a logarithmic pole; the present item combines those two objects and fixes the $2 π$ normalization used throughout this page. Axler-Bourdon-Ramey Chapter 11, printed pp. 223-237, treats the bounded-domain Dirichlet problem and boundary behavior; the present proof uses only the Perron envelope, the maximum principle and the analytic-boundary reflection argument, which are developed in this library's own items. Saff §3, printed pp. 186-189, records the Green function with a finite pole, Green's formula, and the identification of the equilibrium measure with $(2 π)^{- 1} \partial g / \partial n d s$ in the outer normal direction; the sign convention here is the opposite one, because the normal is the outward normal of $Ω$ and the coefficient is $- 1 / (2 π)$ , and it is derived from the PDE Poisson kernel rather than quoted. The dominated-convergence and Fubini arguments controlling the singular integrand, the boundary-limit estimate, and the a.e.-to-everywhere upgrade through Weyl's lemma are proved here and are not attributed to a source.

5 · Examples, counterexamples and false statements

None yet.

Green Functions, Harmonic Measure, and Conformal Invariance

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The canonical Green kernel of a plane domain

Definition

Remarks

Logarithmic modulus is harmonic off its centre

Statement

Facts & Assumptions

Proof

Conformal covariance of the canonical planar Green kernel

Statement

Facts & Assumptions

Proof

Analytic-boundary exhaustion of a plane domain

Statement

Facts & Assumptions

Proof

Green functions exist on all bounded plane domains

Statement

Facts & Assumptions

Proof

A planar barrier forces the regularized Perron envelope to have the prescribed boundary limit

Statement

Facts & Assumptions

Proof

Green correctors are smooth at analytic boundaries

Statement

Facts & Assumptions

Proof

Canonical Green kernels are unique, symmetric and domain monotone

Statement

Facts & Assumptions

Proof

Green kernel of a simply connected plane domain from a Riemann map

Statement

Facts & Assumptions

Proof

Harmonic measure on a bounded regular plane domain

Definition

Remarks

Existence and uniqueness of harmonic measure on a bounded regular plane domain

Statement

Facts & Assumptions

Proof

Poisson density of harmonic measure on a disc

Statement

Facts & Assumptions

Proof

Conformal invariance of harmonic measure

Statement

Facts & Assumptions

Proof

Borel harmonicity and comparison of harmonic measure

Statement

Facts & Assumptions

Proof

Green and harmonic-measure representation with the 2π2π sign

Statement

Facts & Assumptions

Proof

Source notes

5 · Examples, counterexamples and false statements

Sources

Green and harmonic-measure representation with the $2 π$ sign