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Poisson Problems and Interior Harmonic Estimates

1 · Prerequisites

2 · Summary

This page solves the Dirichlet problem on balls and half-spaces by explicit Poisson kernels and then develops the interior regularity estimates that follow from the Poisson representation. Inversion in a sphere converts the Laplace equation to itself, and reflection in the boundary hyperplane produces the half-space Green function; the same construction with a corrected pole gives the ball Green function, whose negative boundary normal derivative is the Poisson kernel. Positivity and unit mass of that kernel, together with the cap/complement estimate, prove boundary recovery for continuous data; the resulting ball Dirichlet theorem is the uniqueness and representation statement used throughout. The half-space kernel is treated separately, with its own bounded-data uniqueness proof by odd reflection.

Interior differentiation of the Poisson representation yields the derivative estimates for harmonic functions and their factorial Cauchy consequences; those estimates also give interior oscillation control for gradients. A separate Newtonian-potential argument proves the interior $C^{2, α}$ estimate for Poisson's equation and its gradient corollary. Real analyticity of harmonic functions and unique continuation follow from the coefficient bounds, and locally uniform limits of harmonic functions are shown to be smooth with all derivatives reconstructed. A Liouville corollary for sublinear entire growth and a remark transferring the two-dimensional disc theory to the cited complex-analysis page close the page. The sign convention is $- Δ Φ = δ_{0}$ with $Φ (x) = ∣ x ∣^{2 - n} / ((n - 2) ω_{n - 1})$ , every statement with integration or Green data assumes Countable Choice, and estimate constants depend only on the parameters indicated in their subscripts.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-10-02Open item page →

Local Hölder and scaled C-two-alpha norms on balls

Definition

Let $n \geq 1$ be an integer and $0 < α < 1$ , let $B_{r} (a) = {x \in R^{n} : ∣ x - a ∣ < r}$ be a Euclidean ball of radius $r > 0$ , and let $f : B_{r} (a) \to R$ or $C$ . Put $[f]_{0, α; B_{r} (a)} : = \sup {\frac{∣ f (x) - f (y) ∣}{∣ x - y ∣^{α}} : x, y \in B_{r} (a), x \neq y}, ∥ f ∥_{\infty; B_{r} (a)} : = \sup_{x \in B_{r} (a)} ∣ f (x) ∣,$ $∥ f ∥_{0, α; B_{r} (a)}^{*} : = ∥ f ∥_{\infty; B_{r} (a)} + r^{α} [f]_{0, α; B_{r} (a)} .$ Both displayed quantities take values in $[0, + \infty]$ , so a norm can be $+ \infty$ ; we say that $f$ is $α$ -Hölder on $B_{r} (a)$ when $[f]_{0, α; B_{r} (a)} < \infty$ .

For $u \in C^{2} (B_{r} (a))$ , a multi-index $γ$ , and $j = ∣ γ ∣ \leq 2$ , write $D^{γ} u$ for the partial derivative of $C^{k}$ maps and multi-index derivative notation in Euclidean space in its displayed canonical order, and set $∥ u ∥_{2, α; B_{r} (a)}^{*} : = \sum_{j = 0}^{2} r^{j} \max_{∣ γ ∣ = j} \sup_{x \in B_{r} (a)} ∣ D^{γ} u (x) ∣ + r^{2 + α} \max_{∣ γ ∣ = 2} [D^{γ} u]_{0, α; B_{r} (a)},$ where the inner maximum runs over the finitely many multi-indices with the stated order and the $j = 0$ term is $\sup_{B_{r} (a)} ∣ u ∣$ . We write $C^{2, α} (B_{r} (a))$ for the functions $u \in C^{2} (B_{r} (a))$ for which this quantity is finite.

Remarks

Scaling. If $r > 0$ , $a \in R^{n}$ , and $v (z) : = u (a + r z)$ on $B_{1} (0)$ , then $D^{γ} v (z) = r^{∣ γ ∣} D^{γ} u (a + r z)$ and therefore $∥ v ∥_{2, α; B_{1} (0)}^{*} = ∥ u ∥_{2, α; B_{r} (a)}^{*}, ∥ v ∥_{0, α; B_{1} (0)}^{*} = ∥ u ∥_{0, α; B_{r} (a)}^{*} .$ The factors $r^{j}$ and $r^{2 + α}$ are exactly what makes the two sides equal: the norm is computed from the radius of the ball it is taken over, while each derivative of $v$ carries the extra factor $r^{j}$ .
Local, not global. These are interior ball quantities. They are read off the open ball $B_{r} (a)$ alone and say nothing about the boundary; in particular no boundary Schauder seminorm, no global $C^{2, α}$ scale and no extension of $u$ beyond $B_{r} (a)$ are defined here.
Finiteness. $∥ u ∥_{2, α; B_{r} (a)}^{*} < \infty$ holds exactly when $u$ , its first derivative field and its second derivative field are bounded on $B_{r} (a)$ and every second partial derivative is $α$ -Hölder there. No third derivative is involved. A finite $∥ f ∥_{0, α; B_{r} (a)}^{*}$ makes $f$ bounded; whether a continuous extension to the closed ball exists is a separate question, not part of this definition.
The two seminorms with subscript $0, α$ are used for Hölder sources in the Poincaré-style interior estimate of this page, while $∥ \cdot ∥_{2, α}^{*}$ is the quantity estimated there.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Euclidean balls are bounded C-one domains with radial outward normal

Statement

Assume Countable Choice. In this item use one-based labels $e_{j} : = e_{j - 1}^{c a n}$ for $1 \leq j \leq n$ . Let $n \geq 2$ , $a \in R^{n}$ and $R > 0$ . The open ball $B_{R} (a) = {x \in R^{n} : ∣ x - a ∣ < R}$ is a bounded $C^{1}$ domain in the sense of Bounded C1 domains and their outward normals, and for every boundary point $y \in \partial B_{R} (a) = S_{R} (a)$ its outward unit normal is the radial vector $ν (y) = (y - a) / R$ .

Facts & Assumptions

Given: an integer $n \geq 2$ , a centre $a \in R^{n}$ and a radius $R > 0$ ; write $Ω : = B_{R} (a)$ .

[A1]

Countable Choice is assumed, as in the published surface-integration convention used in [F1] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A bounded $C^{1}$ domain is a nonempty bounded open set whose boundary is locally, after a rigid change of coordinates with orthogonal part $Q$ , the graph $t = h (y)$ of a $C^{1}$ function $h$ on an open ball $B \subseteq R^{n - 1}$ , with the domain locally exactly the subgraph $t < h (y)$ ; the outward normal in these coordinates is $(- D h (y), 1) / \sqrt{1 + ∣ D h (y) ∣^{2}}$ , transported by the orthogonal coordinate map (Bounded C1 domains and their outward normals).

[F2]

For $c \in R^{n}$ and $r > 0$ , ${\overline{B}}_{2} (c, r) = {x : ∣ x - c ∣ \leq r}$ and $S_{2} (c, r) = {x : ∣ x - c ∣ = r}$ are the Euclidean closed ball and the Euclidean sphere (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[F3]

For every subspace $W$ of a finite-dimensional inner product space $V$ , $\dim W + \dim W^{⊥} = \dim V$ (In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ ).

[F4]

Every finite-dimensional real or complex inner product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[F5]

If $(e_{0}, \dots, e_{r - 1})$ is an orthonormal basis of an inner product space, then $v = \sum_{i < r} ⟨ v, e_{i} ⟩ e_{i}$ and $∥ v ∥^{2} = \sum_{i < r} ∣ ⟨ v, e_{i} ⟩ ∣^{2}$ for every vector $v$ (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[F6]

For $α \in R$ the function $x \mapsto x^{α}$ is continuous and differentiable on $(0, \infty)$ with derivative $α x^{α - 1}$ (Continuity and derivatives of positive-base real powers).

[F7]

$D (g \circ f) (p) = D g (f (p)) \circ D f (p)$ when $f$ is totally differentiable at $p$ and $g$ is totally differentiable at $f (p)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Proof

technique · direct

1.1givenF2F3F4F5algebra

Fix a boundary point $y \in S_{R} (a)$ and put $u : = (y - a) / R$ , so that $∣ u ∣ = 1$ and $y = a + R u$ . By [F3] applied to $W = span (u)$ we have $\dim u^{⊥} = n - 1$ , so [F4] supplies an orthonormal basis $(v_{1}, \dots, v_{n - 1})$ of $u^{⊥}$ ; then $(v_{1}, \dots, v_{n - 1}, u)$ is an orthonormal basis of $R^{n}$ , because every $ξ$ equals $(ξ - ⟨ ξ, u ⟩ u) + ⟨ ξ, u ⟩ u$ with $ξ - ⟨ ξ, u ⟩ u \in u^{⊥}$ . Define the linear map $T (ξ) : = \sum_{i = 1}^{n - 1} ⟨ ξ, v_{i} ⟩ e_{i} + ⟨ ξ, u ⟩ e_{n}$ ; by [F5], $∣ T (ξ) ∣^{2} = \sum_{i < n} ∣ ⟨ ξ, v_{i} ⟩ ∣^{2} + ∣ ⟨ ξ, u ⟩ ∣^{2} = ∣ ξ ∣^{2}$ for all $ξ$ , so $T$ is orthogonal, and orthonormality gives $T (u) = e_{n}$ and $T (v_{i}) = e_{i}$ .

2.1givenstep 1.1

Define the rigid motion $Φ (p) : = T (p - a) - R e_{n}$ (orthogonal part $T$ , translation $- T (a) - R e_{n}$ ), the open cylinder $C : = B_{R / 2} (0) \times (- R / 2, R / 2)$ , the open set $W : = Φ^{- 1} (C)$ and the open ball $B : = B_{R / 2} (0) \subseteq R^{n - 1}$ ; also put $h (w) : = \sqrt{R^{2} - ∣ w ∣^{2}} - R$ for $w \in B$ . The point $y$ lies in $W$ , because $Φ (y) = T (R u) - R e_{n} = 0 \in C$ ; so $W$ is an open neighbourhood of $y$ . Moreover $T (Ω - a) = T (B_{R} (0)) = B_{R} (0)$ by orthogonality, so $Φ (Ω) = B_{R} (- R e_{n})$ .

3.1givenstep 2.1algebra

For $z = (w, t) \in C$ we have $Φ^{- 1} (z) = p$ and $T (p - a) = z + R e_{n}$ , so by orthogonality $∣ p - a ∣^{2} = ∣ z + R e_{n} ∣^{2} = ∣ w ∣^{2} + (t + R)^{2}$ . Thus $z \in Φ (Ω)$ exactly when $∣ w ∣^{2} + t^{2} + 2 R t < 0$ . For $∣ w ∣ < R / 2$ , put $s (w) : = \sqrt{R^{2} - ∣ w ∣^{2}}$ ; then $s (w) > \sqrt{3} R / 2$ , so the quadratic inequality is equivalent to $- R - s (w) < t < - R + s (w) = h (w)$ . Its lower root satisfies $- R - s (w) < - R / 2 < t$ , so it is automatic throughout $C$ . Also $- R / 2 < (\sqrt{3} / 2 - 1) R < h (w) \leq 0 < R / 2$ , so the graph lies inside the vertical interval of $C$ . Therefore the local set equations are $Φ (Ω \cap W) = Φ (Ω) \cap C = {(w, t) \in B \times (- R / 2, R / 2) : - R - s (w) < t < h (w)} = {(w, t) \in C : t < h (w)}$ .

3.2givenstep 2.1F6F7algebra

The polynomial $q (w) : = R^{2} - ∣ w ∣^{2}$ is positive on $B$ and $C^{1}$ there; by [F6] with $α = 1 / 2$ the map $s \mapsto s^{1 / 2}$ is differentiable on $(0, \infty)$ with derivative $\frac{1}{2} s^{- 1 / 2}$ ; the chain rule [F7] applied to $h = q^{1 / 2} - R$ therefore gives $D h (w) = - w / \sqrt{R^{2} - ∣ w ∣^{2}}$ on $B$ , a continuous expression, so $h \in C^{1} (B)$ .

4.1givenstep 2.1step 3.1step 3.2A1F1

By steps 2.1, 3.1 and 3.2 the arbitrary boundary point $y \in S_{R} (a)$ has a neighbourhood $W$ and a rigid motion $Φ$ for which $Φ (Ω \cap W) = {(w, t) \in C : t < h (w)}$ , where $C = B \times (- R / 2, R / 2)$ and $h \in C^{1} (B)$ ; thus the boundary is locally a $C^{1}$ graph and the domain is locally exactly its subgraph. The set $Ω$ is nonempty, bounded and open in $R^{n}$ with $n \geq 2$ . Applying the bounded-domain convention [F1] under [A1], $Ω = B_{R} (a)$ is a bounded $C^{1}$ domain.

5.1givenstep 1.1step 3.2step 4.1A1F1algebra∎

In the coordinates $z = (w, t)$ of step 3.1 the definition [F1] prescribes the outward normal $(- D h (w), 1) / \sqrt{1 + ∣ D h (w) ∣^{2}}$ on the graph $t = h (w)$ ; by step 3.2 this equals $(w / \sqrt{R^{2} - ∣ w ∣^{2}}, 1) \sqrt{R^{2} - ∣ w ∣^{2}} / R = (w, h (w) + R) / R$ , which at the graph point $z = (w, h (w))$ is exactly $(z + R e_{n}) / R$ ; transporting back by the orthogonal part $T$ gives the vector $T^{T} (z + R e_{n}) / R$ . At $y = a + R u$ we have $Φ (y) = 0$ and $z + R e_{n} = T (y - a)$ , so the transported normal is $T^{T} T (y - a) / R = (y - a) / R$ , a unit vector because $∣ y - a ∣ = R$ . Thus the normal prescribed by [F1] under [A1] is $ν (y) = (y - a) / R$ for every $y \in S_{R} (a)$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Kelvin inversion transforms harmonic functions

Statement

Use one-based coordinate labels $x_{j} : = x_{j - 1}^{c a n}$ and $z_{j} : = z_{j - 1}^{c a n}$ for $1 \leq j \leq n$ , including their derivatives. Let $n \geq 3$ , $R > 0$ and $a \in R^{n}$ . Write $I_{R} (x) = a + R^{2} (x - a) / ∣ x - a ∣^{2}$ for $x \neq a$ . If $u \in C^{2} (U)$ on an open set $U$ avoiding $a$ , define $K_{R} u (x) = (R / ∣ x - a ∣)^{n - 2} u (I_{R} (x))$ on $I_{R}^{- 1} (U)$ . Then $Δ (K_{R} u) (x) = (R / ∣ x - a ∣)^{n + 2} (Δ u) (I_{R} (x))$ . In particular inversion preserves harmonicity on the punctured domains on which both sides are defined.

Facts & Assumptions

Given: $n \geq 3$ , $R > 0$ , $a \in R^{n}$ , an open set $U$ with $a \notin U$ , and $u \in C^{2} (U)$ .

[F1]

The Laplacian is $Δ f = \sum_{i < n} \partial_{i} \partial_{i} f$ in the coordinate partial derivatives of Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , and a $C^{2}$ function with $Δ f = 0$ is called harmonic (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F2]

If $f$ is totally differentiable at $p$ and $g$ is totally differentiable at $f (p)$ , then $D (g \circ f) (p) = D g (f (p)) \circ D f (p)$ ; finite sums, products and compositions of $C^{2}$ Euclidean maps are $C^{2}$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[F3]

One-variable derivatives obey the product rule $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and for $q > 0$ the power $q^{α}$ has derivative $α q^{α - 1}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Continuity and derivatives of positive-base real powers).

Proof

technique · direct

1.1givenF2F3

Put $y : = x - a$ , $r : = ∣ y ∣$ and $z : = I_{R} (x) = a + R^{2} y / r^{2}$ . On the open set $V : = I_{R}^{- 1} (U)$ we have $x \neq a$ , $r > 0$ and $z \in U$ , and $x \mapsto z$ is smooth there, being built from the smooth coordinate functions $y_{i}$ and $r^{2}$ and the smooth factor $R^{2} / r^{2}$ ; hence $K_{R} u = R^{n - 2} r^{2 - n} (u \circ z)$ is $C^{2}$ on $V$ , and no value is taken at $r = 0$ .

2.1step 1.1F2F3algebra

Coordinate differentiation of $z$ gives, for all $1 \leq i, k \leq n$ , $\partial_{i} z_{k} = R^{2} (δ_{i k} / r^{2} - 2 y_{k} y_{i} / r^{4})$ and $Δ z_{k} = - 2 (n - 2) R^{2} y_{k} / r^{4}$ , together with the auxiliary identities $\sum_{i} \partial_{i} z_{k} \partial_{i} z_{l} = R^{4} δ_{k l} / r^{4}$ and $\sum_{i} y_{i} \partial_{i} z_{l} = - R^{2} y_{l} / r^{2}$ ; every occurrence of $r$ is positive on $V$ .

2.2step 1.1F2F3algebra

Put $w_{0} (x) : = r^{2 - n} u (z)$ , so that $K_{R} u = R^{n - 2} w_{0}$ . The product and chain rules give $\partial_{i} w_{0} = (2 - n) r^{- n} y_{i} u (z) + r^{2 - n} \sum_{k} \partial_{k} u (z) \partial_{i} z_{k}$ .

3.1step 2.2F1F2F3algebra

The chain and power rules give $\partial_{i} (r^{2 - n}) = (2 - n) r^{- n} y_{i}$ and $\partial_{i}^{2} (r^{2 - n}) = (2 - n) (r^{- n} - n r^{- n - 2} y_{i}^{2})$ ; summing and using $r^{2} = ∣ y ∣^{2}$ gives $Δ (r^{2 - n}) = (2 - n) (n r^{- n} - n r^{- n - 2} ∣ y ∣^{2}) = 0$ . The Laplacian product rule $Δ (f g) = f Δ g + 2 \nabla f \cdot \nabla g + g Δ f$ applied to $f = r^{2 - n}$ and $g = u \circ z$ therefore gives $Δ w_{0} = r^{2 - n} Δ (u \circ z) + 2 \nabla (r^{2 - n}) \cdot \nabla (u \circ z)$ .

3.2step 2.1step 2.2F2F3algebra

Two chain-rule evaluations. First, $Δ (u \circ z) = \sum_{k, l} \partial_{k} \partial_{l} u (z) \sum_{i} \partial_{i} z_{k} \partial_{i} z_{l} + \sum_{k} \partial_{k} u (z) Δ z_{k} = R^{4} Δ u (z) / r^{4} - 2 (n - 2) R^{2} y \cdot \nabla u (z) / r^{4}$ by step 2.1. Second, since $\nabla (r^{2 - n}) = (2 - n) r^{- n} y$ , step 2.1 gives $\nabla (r^{2 - n}) \cdot \nabla (u \circ z) = (2 - n) r^{- n} \sum_{l} \partial_{l} u (z) \sum_{i} y_{i} \partial_{i} z_{l} = (n - 2) R^{2} y \cdot \nabla u (z) / r^{n + 2}$ .

4.1step 3.1step 3.2algebra

Substituting step 3.2 into step 3.1, the first-order terms $- 2 (n - 2) R^{2} r^{- n - 2} y \cdot \nabla u (z)$ and $+ 2 (n - 2) R^{2} r^{- n - 2} y \cdot \nabla u (z)$ cancel, leaving $Δ w_{0} = R^{4} Δ u (z) / r^{n + 2}$ .

5.1step 2.2step 4.1algebra

Restoring the factor of step 2.2 gives $Δ (K_{R} u) (x) = R^{n - 2} Δ w_{0} (x) = R^{n + 2} r^{- n - 2} (Δ u) (I_{R} (x)) = (R / ∣ x - a ∣)^{n + 2} (Δ u) (I_{R} (x))$ for every $x \in I_{R}^{- 1} (U)$ .

6.1step 5.1F1∎

Since $R^{n + 2} r^{- n - 2} > 0$ on $V$ , step 5.1 shows that $Δ u$ vanishes at $z = I_{R} (x)$ exactly when $Δ (K_{R} u)$ vanishes at $x$ ; both sides are evaluated only at points with $x \neq a$ , and $I_{R}$ is an involution exchanging the two punctured domains, so inversion transfers harmonicity in both directions [F1]. No choice principle and no measure-theoretic input is used.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Reflection Green kernel for the half-space

Statement

Assume Countable Choice and $n \geq 3$ . Use one-based coordinate labels $x_{j} : = x_{j - 1}^{c a n}$ and $y_{j} : = y_{j - 1}^{c a n}$ for $1 \leq j \leq n$ . For $H = {x \in R^{n} : x_{n} > 0}$ , $y \in H$ and $y^{†} = (y^{'}, - y_{n})$ , define $G_{H} (x, y) = Φ (x - y) - Φ (x - y^{†}) (x \in H ∖ {y}),$ with $Φ$ the fundamental solution normalized by $- Δ Φ = δ_{0}$ . The kernel is symmetric off the diagonal and strictly positive for distinct $x, y \in H$ . For fixed $y$ , it is locally integrable on $H$ , smooth and harmonic in $x$ off $y$ , satisfies $- Δ_{x} G_{H} (\cdot, y) = δ_{y}$ distributionally on $H$ , and extends continuously to the boundary with zero trace. Its diagonal is the Green pole. It is a Green kernel for this unbounded half-space; the published bounded-domain definition is not being applied to $H$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a pole $y \in H = {x \in R^{n} : x_{n} > 0}$ and $y^{†} = (y^{'}, - y_{n})$ .

[F1]

With $ω_{n - 1} = ∣ S^{n - 1} ∣ > 0$ in the published chart/polar convention, the fundamental solution is $Φ (x) = ∣ x ∣^{2 - n} / ((n - 2) ω_{n - 1})$ for $x \neq 0$ and $n \geq 3$ , extended as a locally integrable function at the pole (Fundamental solution for the positive operator minus Laplacian).

[F2]

$Φ$ is smooth on $R^{n} ∖ {0}$ with $Δ Φ = 0$ there, and for every pole $z$ the translate $x \mapsto Φ (x - z)$ is harmonic on $R^{n} ∖ {z}$ (The Laplace fundamental solution is harmonic off its pole).

[F3]

The regular distribution $T_{Φ (\cdot - z)}$ of $x \mapsto Φ (x - z)$ satisfies $- Δ_{x} T_{Φ (\cdot - z)} = δ_{z}$ on $R^{n}$ for every $z$ (The negative Laplacian of the fundamental solution is the unit Dirac distribution).

[F4]

On an open set $Ω$ , distributions act on $C_{c}^{\infty} (Ω)$ , $(\partial_{i} T) (ϕ) = - T (\partial_{i} ϕ)$ and $Δ T = \sum_{i} \partial_{i}^{2} T$ , while $T_{f} (ϕ) = \int f ϕ$ is the regular distribution of $f \in L_{l o c}^{1} (Ω)$ ; and $δ_{a} (ϕ) = ϕ (a)$ for $a \in Ω$ (Distributional harmonicity and Poisson's equation on an open subset of Rn, Dirac delta and its derivatives).

[F5]

The kernel $Φ$ is locally integrable on $R^{n}$ (Local integrability of the Laplace fundamental kernel).

Proof

technique · direct

1.1givenF1F2F4F5

Since $y_{n} > 0$ , we have $y^{†} = (y^{'}, - y_{n}) \notin \overline{H}$ . Thus for $x \in H ∖ {y}$ both vectors $x - y$ and $x - y^{†}$ are nonzero, and the formula defines a real function smooth in $x$ off $y$ . By [F1] and [F5] the first term is locally integrable, while the second is continuous on all of $H$ because $∣ x - y^{†} ∣ \geq x_{n} + y_{n} > 0$ . Hence the difference is locally integrable on $H$ ; write $T$ for its regular distribution, which exists by [F4].

2.1givenstep 1.1F1algebra

Symmetry. For distinct $x, y \in H$ , the vectors $x - y^{†} = (x^{'} - y^{'}, x_{n} + y_{n})$ and $y - x^{†} = (y^{'} - x^{'}, y_{n} + x_{n})$ have equal Euclidean norms, because their first $n - 1$ coordinates differ only by a sign and their last coordinates agree; and $Φ$ is even, being a function of $∣ z ∣$ only. Hence $Φ (x - y^{†}) = Φ (y - x^{†})$ and $Φ (x - y) = Φ (y - x)$ , so $G_{H} (x, y) = G_{H} (y, x)$ .

2.2givenstep 1.1F1algebra

Strict positivity. For distinct $x, y \in H$ the $n - 1$ leading coordinates of $x - y^{†}$ and $x - y$ agree, so $∣ x - y^{†} ∣^{2} - ∣ x - y ∣^{2} = (x_{n} + y_{n})^{2} - (x_{n} - y_{n})^{2} = 4 x_{n} y_{n} > 0$ ; thus $∣ x - y^{†} ∣ > ∣ x - y ∣ \geq 0$ . Since $n \geq 3$ gives the negative exponent $2 - n < 0$ and $r \mapsto r^{2 - n}$ is strictly decreasing on $(0, \infty)$ (a quotient of positive powers, verified from $r^{2 - n} = 1 / r^{n - 2}$ ), and since the factor $1 / ((n - 2) ω_{n - 1})$ of [F1] is positive, we get $Φ (x - y^{†}) < Φ (x - y)$ , that is $G_{H} (x, y) > 0$ .

2.3givenstep 1.1F2

Harmonicity in $x$ off the pole. Fix $y \in H$ . By [F2] the translate $x \mapsto Φ (x - y)$ is smooth and harmonic on $R^{n} ∖ {y}$ , hence on $H ∖ {y}$ ; and $x \mapsto Φ (x - y^{†})$ is smooth and harmonic on all of $H$ , because $H$ is contained in $R^{n} ∖ {y^{†}}$ . A difference of harmonic smooth functions is smooth and harmonic, so $x \mapsto G_{H} (x, y)$ is smooth and harmonic on $H ∖ {y}$ .

2.4givenstep 1.1F1F2algebra

Zero boundary trace. Let $z \in \partial H = {x_{n} = 0}$ and let $x \to z$ with $x \in H$ . Then $x - y \to z - y = (z^{'} - y^{'}, - y_{n})$ and $x - y^{†} \to z - y^{†} = (z^{'} - y^{'}, y_{n})$ , and these two limit vectors have equal norms $∣ z^{'} - y^{'} ∣^{2} + y_{n}^{2}$ , a positive number because $y_{n} > 0$ ; in particular neither limit is the origin. By continuity of $Φ$ off the origin, $\lim_{x \to z, x \in H} G_{H} (x, y) = Φ (z - y) - Φ (z - y^{†}) = 0$ . As the formula is continuous on the closed set ${x : x_{n} \geq 0, x \neq y}$ , it extends $G_{H} (\cdot, y)$ continuously to $\overline{H} ∖ {y}$ with value $0$ on $\partial H$ .

2.5givenstep 1.1F3F4algebra

Distributional identity. Let $ϕ \in C_{c}^{\infty} (H)$ be a test function and let $ψ$ be its extension by zero to $R^{n}$ , which is smooth and compactly supported. By the derivative rules of [F4], $(Δ T) (ϕ) = T (Δ ϕ) = \int_{H} G_{H} (x, y) Δ ϕ (x) d x$ , hence $⟨ - Δ T, ϕ ⟩ = - \int_{H} G_{H} (x, y) Δ ϕ (x) d x = - \int_{R^{n}} Φ (x - y) Δ ψ (x) d x + \int_{R^{n}} Φ (x - y^{†}) Δ ψ (x) d x$ , because $ψ = ϕ$ on $H$ and $G_{H} (x, y) = Φ (x - y) - Φ (x - y^{†})$ there. By [F3] applied at the poles $y$ and $y^{†}$ , $- \int_{R^{n}} Φ (x - y) Δ ψ (x) d x = ψ (y) = ϕ (y)$ , while $+ \int_{R^{n}} Φ (x - y^{†}) Δ ψ (x) d x = - ψ (y^{†}) = 0$ ; the last equality holds because $supp ψ \subseteq H$ avoids $H^{c}$ . Therefore $⟨ - Δ T, ϕ ⟩ = ϕ (y) = δ_{y} (ϕ)$ for every test function, that is $- Δ_{x} T_{G_{H} (\cdot, y)} = δ_{y}$ on $H$ in the sense of [F4].

3.1givenstep 1.1step 2.1step 2.2step 2.3step 2.4step 2.5∎

Steps 2.1, 2.2, 2.3, 2.4 and 2.5 establish that the reflection kernel $G_{H} (\cdot, y)$ is symmetric, strictly positive at distinct points of $H$ , smooth and harmonic in $x$ off $y$ , has zero continuous boundary trace, and represents $- Δ_{x} G_{H} (\cdot, y) = δ_{y}$ distributionally on $H$ ; it therefore acts as the Green kernel of this unbounded half-space, and no bounded-domain Green definition is applied to $H$ anywhere above.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Dirichlet Green function of a Euclidean ball

Statement

Assume Countable Choice and $n \geq 3$ . For $B_{R} (a)$ , put $y^{*} = a + R^{2} (y - a) / ∣ y - a ∣^{2}$ when $y \neq a$ . With $- Δ Φ = δ_{0}$ , $G (x, y) = Φ (x - y) - (R / ∣ y - a ∣)^{n - 2} Φ (x - y^{*})$ for $y \neq a$ , and $G (x, a) = Φ (x - a) - Φ (R)$ . For $x, y \in B_{R} (a)$ , $x \neq y$ , this is the positive, symmetric Dirichlet Green function: it is harmonic in $x$ off $y$ , has the correct point singularity, vanishes continuously on the boundary, and its corrector is $C^{2}$ on the closed ball.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a centre $a \in R^{n}$ , a radius $R > 0$ and the ball $Ω : = B_{R} (a)$ .

[F1]

With $ω_{n - 1} = ∣ S^{n - 1} ∣ > 0$ , the fundamental solution is $Φ (x) = ∣ x ∣^{2 - n} / ((n - 2) ω_{n - 1})$ for $x \neq 0$ and $n \geq 3$ , extended as a locally integrable function at the pole (Fundamental solution for the positive operator minus Laplacian).

[F2]

[F3]

A Dirichlet Green function for $- Δ$ on a bounded domain $Ω$ is a map $G_{Ω}$ on ${(x, y) \in Ω \times Ω : x \neq y}$ such that for each pole $y$ there is a harmonic $H_{y} \in C^{2} (Ω) \cap C (\overline{Ω})$ with $H_{y} = Φ (\cdot - y)$ on $\partial Ω$ and $G_{Ω} (x, y) = Φ (x - y) - H_{y} (x)$ ; for fixed $y$ the function $G_{Ω} (\cdot, y)$ is harmonic away from $y$ , extends continuously to $\overline{Ω} ∖ {y}$ with zero boundary trace, and its locally integrable representative satisfies $- Δ_{x} T_{G_{Ω} (\cdot, y)} = δ_{y}$ in $D^{'} (Ω)$ (Dirichlet Green function for minus Laplacian).

[F4]

The regular distribution of $x \mapsto Φ (x - z)$ satisfies $- Δ_{x} T_{Φ (\cdot - z)} = δ_{z}$ on $R^{n}$ for every pole $z$ (The negative Laplacian of the fundamental solution is the unit Dirac distribution).

[F5]

For $x \neq a$ the inversion $I_{R} (x) = a + R^{2} (x - a) / ∣ x - a ∣^{2}$ is smooth, and it is an involution exchanging the punctured ball $B_{R} (a) ∖ {a}$ with the exterior ${x : ∣ x - a ∣ > R}$ (Kelvin inversion transforms harmonic functions).

[F6]

$B_{R} (a)$ is a bounded $C^{1}$ domain with outward unit normal $ν (y) = (y - a) / R$ at each $y \in \partial B_{R} (a)$ (Euclidean balls are bounded C-one domains with radial outward normal).

[F7]

If $Ω$ is a bounded $C^{1}$ domain carrying a Dirichlet Green function whose designated correctors satisfy $H_{y} \in C^{2} (\overline{Ω})$ for every $y$ , then $G_{Ω} (x, y) = G_{Ω} (y, x)$ for all distinct $x, y \in Ω$ (Symmetry of the Dirichlet Green function).

[F8]

On a bounded $C^{1}$ domain $Ω$ and real $u, v \in C^{2} (\overline{Ω})$ , $\int_{Ω} (v Δ u - u Δ v) d x = \int_{\partial Ω} (v \partial_{ν} u - u \partial_{ν} v) d S$ (Second Green identity).

[F9]

Countable Choice ${A C}_{ω}$ is the standing hypothesis under which the Green, distributional and surface-measure statements used here are formulated (The Axiom of Countable Choice ( ${A C}_{ω}$ )); the distributional vocabulary is that of Distributional harmonicity and Poisson's equation on an open subset of Rn with $δ_{y} (ϕ) = ϕ (y)$ for $y$ in the open set (Dirac delta and its derivatives).

Proof

technique · direct

1.1givenF1F5F9algebra

Work under the standing hypothesis [F9]. Let $n \geq 3$ , $a \in R^{n}$ , $R > 0$ and $Ω = B_{R} (a)$ ; let $Φ$ be the kernel of [F1]. For $y \in Ω$ with $y \neq a$ put $y^{*} : = I_{R} (y) = a + R^{2} (y - a) / ∣ y - a ∣^{2}$ and $κ_{y} : = (R / ∣ y - a ∣)^{n - 2}$ ; by [F5], $∣ y^{*} - a ∣ = R^{2} / ∣ y - a ∣ > R$ , so $y^{*} \notin \overline{Ω}$ . Define $G (x, y) : = Φ (x - y) - κ_{y} Φ (x - y^{*})$ for $x \in Ω ∖ {y}$ when $y \neq a$ , and $G (x, a) : = Φ (x - a) - Φ (R)$ for $x \in Ω ∖ {a}$ . Now put $u : = x - a$ and $v : = y - a \neq 0$ , so that $x - y^{*} = (x - a) - R^{2} v / ∣ v ∣^{2} = u - R^{2} v / ∣ v ∣^{2}$ and $x - y = u - v$ : expanding the square $∣ u - R^{2} v / ∣ v ∣^{2} ∣^{2} = ∣ u ∣^{2} - 2 R^{2} ⟨ u, v ⟩ / ∣ v ∣^{2} + R^{4} / ∣ v ∣^{2}$ and multiplying by $∣ v ∣^{2}$ gives $∣ y - a ∣^{2} ∣ x - y^{*} ∣^{2} = ∣ x - a ∣^{2} ∣ y - a ∣^{2} - 2 R^{2} ⟨ u, v ⟩ + R^{4}$ , while $R^{2} ∣ x - y ∣^{2} = R^{2} ∣ x - a ∣^{2} - 2 R^{2} ⟨ u, v ⟩ + R^{2} ∣ y - a ∣^{2}$ ; subtracting yields the first algebraic identity below, and the same expansion with the roles of $x$ and $y$ exchanged yields the second, since $∣ v ∣^{2} ∣ u ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2}$ , $∣ x^{*} - a ∣ = R^{2} / ∣ x - a ∣$ and $x^{*}, y^{*}$ are defined symmetrically. $∣ y - a ∣^{2} ∣ x - y^{*} ∣^{2} - R^{2} ∣ x - y ∣^{2} = (R^{2} - ∣ x - a ∣^{2}) (R^{2} - ∣ y - a ∣^{2}), ∣ y - a ∣^{2} ∣ x - y^{*} ∣^{2} = ∣ x - a ∣^{2} ∣ y - x^{*} ∣^{2} .$

2.1step 1.1F2F3

Correctors. Fix $y \in Ω$ with $y \neq a$ . Since $y^{*} \notin \overline{Ω}$ , the translate $x \mapsto Φ (x - y^{*})$ is smooth with vanishing Laplacian on a neighbourhood of the closed ball $\overline{Ω}$ by [F2], so $H_{y} : = κ_{y} Φ (\cdot - y^{*})$ lies in $C^{2} (\overline{Ω})$ and is harmonic on $Ω$ ; by construction $G (x, y) = Φ (x - y) - H_{y} (x)$ for $x \in Ω ∖ {y}$ . For the centre put $H_{a} : = Φ (R)$ , the constant corrector: it is $C^{2}$ on $\overline{Ω}$ , harmonic, and $G (x, a) = Φ (x - a) - H_{a} (x)$ by definition.

2.2step 1.1F1algebra

Boundary values of the correctors. If $y \neq a$ and $∣ x - a ∣ = R$ , the first identity of step 1.1 gives $∣ y - a ∣^{2} ∣ x - y^{*} ∣^{2} = R^{2} ∣ x - y ∣^{2}$ , hence $∣ x - y^{*} ∣ = (R / ∣ y - a ∣) ∣ x - y ∣$ ; with the formula of [F1] this yields $κ_{y} Φ (x - y^{*}) = R^{n - 2} (∣ y - a ∣ ∣ x - y^{*} ∣)^{2 - n} / ((n - 2) ω_{n - 1}) = R^{n - 2} (R ∣ x - y ∣)^{2 - n} / ((n - 2) ω_{n - 1}) = ∣ x - y ∣^{2 - n} / ((n - 2) ω_{n - 1}) = Φ (x - y)$ . For $y = a$ and $∣ x - a ∣ = R$ we have $H_{a} (x) = Φ (R) = Φ (x - a)$ by [F1]. So $H_{y} = Φ (\cdot - y)$ on $\partial Ω$ in both cases.

2.3step 1.1F1algebra

Positivity. Let $x, y \in Ω$ be distinct. If $y \neq a$ , then $∣ x - y ∣ > 0$ , and the first identity of step 1.1 together with $R^{2} - ∣ x - a ∣^{2} > 0$ , $R^{2} - ∣ y - a ∣^{2} > 0$ gives $∣ y - a ∣^{2} ∣ x - y^{*} ∣^{2} = R^{2} ∣ x - y ∣^{2} + (R^{2} - ∣ x - a ∣^{2}) (R^{2} - ∣ y - a ∣^{2}) > R^{2} ∣ x - y ∣^{2}$ , so $∣ y - a ∣ ∣ x - y^{*} ∣ > R ∣ x - y ∣ > 0$ ; because $n \geq 3$ makes the exponent $2 - n$ negative and $r \mapsto r^{2 - n}$ strictly decreasing, $κ_{y} Φ (x - y^{*}) = R^{n - 2} (∣ y - a ∣ ∣ x - y^{*} ∣)^{2 - n} / ((n - 2) ω_{n - 1}) < R^{n - 2} (R ∣ x - y ∣)^{2 - n} / ((n - 2) ω_{n - 1}) = Φ (x - y)$ , that is $G (x, y) > 0$ . If $y = a$ , then $0 < ∣ x - a ∣ < R$ and strict decrease of $r \mapsto r^{2 - n}$ gives $G (x, a) = Φ (x - a) - Φ (R) > 0$ .

2.4step 1.1F1algebra

Symmetry. Let $x, y \in Ω ∖ {a}$ . The second identity of step 1.1 gives $∣ y - a ∣ ∣ x - y^{*} ∣ = ∣ x - a ∣ ∣ y - x^{*} ∣$ , so $κ_{y} Φ (x - y^{*}) = R^{n - 2} (∣ y - a ∣ ∣ x - y^{*} ∣)^{2 - n} / ((n - 2) ω_{n - 1}) = R^{n - 2} (∣ x - a ∣ ∣ y - x^{*} ∣)^{2 - n} / ((n - 2) ω_{n - 1}) = κ_{x} Φ (y - x^{*})$ ; since $Φ$ depends only on the norm, $Φ (x - y) = Φ (y - x)$ , hence $G (x, y) = G (y, x)$ . For the case of the centre, $x \in Ω ∖ {a}$ : $G (a, x) = Φ (a - x) - κ_{x} Φ (a - x^{*})$ and $∣ x^{*} - a ∣ = R^{2} / ∣ x - a ∣ > R$ gives $κ_{x} Φ (a - x^{*}) = (R / ∣ x - a ∣)^{n - 2} (R^{2} / ∣ x - a ∣)^{2 - n} / ((n - 2) ω_{n - 1}) = R^{2 - n} / ((n - 2) ω_{n - 1}) = Φ (R)$ , whence $G (a, x) = Φ (a - x) - Φ (R) = Φ (x - a) - Φ (R) = G (x, a)$ .

3.1step 2.1step 2.2F1F2algebra

Harmonicity, continuity and zero trace. If $y \neq a$ , [F2] makes $x \mapsto Φ (x - y)$ smooth and harmonic on $R^{n} ∖ {y} \supseteq Ω ∖ {y}$ , and $H_{y}$ is smooth harmonic on $Ω$ by step 2.1; hence $G (\cdot, y) = Φ (\cdot - y) - H_{y}$ is smooth and harmonic on $Ω ∖ {y}$ , and the same holds for $y = a$ with the constant $H_{a}$ . For the continuous extension: fix $y$ and let $x \to z \in \partial Ω$ with $x \in Ω$ . By [F1] and continuity of $Φ$ off the origin, $Φ (x - y) \to Φ (z - y)$ and $κ_{y} Φ (x - y^{*}) \to κ_{y} Φ (z - y^{*}) = Φ (z - y)$ by step 2.2 applied at the boundary point $z$ ; for $y = a$ , $Φ (x - a) \to Φ (z - a) = Φ (R)$ because $∣ z - a ∣ = R$ . Hence $G (x, y) \to 0$ for every $z \in \partial Ω$ , and $G (\cdot, y)$ extends continuously to $\overline{Ω} ∖ {y}$ with zero boundary trace.

3.2step 1.1step 2.1F1F4F8F9algebra

Distributional identity. Let $ϕ \in C_{c}^{\infty} (Ω)$ and let $ψ \in C_{c}^{\infty} (R^{n})$ be its extension by zero. By the definitions of [F9], $(Δ T_{G (\cdot, y)}) (ϕ) = T_{G (\cdot, y)} (Δ ϕ) = \int_{Ω} G (x, y) Δ ϕ (x) d x$ , so $⟨ - Δ T_{G (\cdot, y)}, ϕ ⟩ = - \int_{R^{n}} Φ (x - y) Δ ψ (x) d x + \int_{Ω} H_{y} (x) Δ ϕ (x) d x$ . The first term equals $ψ (y) = ϕ (y)$ by [F4], since $- Δ_{x} T_{Φ (\cdot - y)} = δ_{y}$ on $R^{n}$ . For the second term: $H_{y} \in C^{2} (\overline{Ω})$ is harmonic and $ϕ$ vanishes on a neighbourhood of $\partial Ω$ , so the second Green identity [F8] with $u = ϕ$ , $v = H_{y}$ gives $\int_{Ω} H_{y} Δ ϕ d x = \int_{Ω} ϕ Δ H_{y} d x + \int_{\partial Ω} (H_{y} \partial_{ν} ϕ - ϕ \partial_{ν} H_{y}) d S = 0$ , both boundary terms vanishing because $ϕ$ and its first derivatives are zero near $\partial Ω$ . Hence $⟨ - Δ T_{G (\cdot, y)}, ϕ ⟩ = ϕ (y) = δ_{y} (ϕ)$ for every test function $ϕ$ , that is $- Δ_{x} T_{G (\cdot, y)} = δ_{y}$ in $D^{'} (Ω)$ .

4.1step 1.1step 2.1step 2.2step 3.1step 2.3step 2.4step 3.2F3F6F7∎

Steps 2.1, 2.2 and 3.1 verify the corrector clause and the zero-trace clause of the Dirichlet Green definition [F3] for the ball $Ω = B_{R} (a)$ and the kernel $G$ of step 1.1, step 3.2 verifies its distributional clause, and step 2.3 gives strict positivity while step 2.4 gives symmetry; so $G$ is the positive symmetric Dirichlet Green function of $B_{R} (a)$ . Symmetry also follows independently from the published theorem [F7], whose hypotheses hold because $Ω$ is a bounded $C^{1}$ domain by [F6] and the correctors $H_{y}$ of step 2.1 lie in $C^{2} (\overline{Ω})$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Poisson kernel of a Euclidean ball

Statement

Assume Countable Choice and $n \geq 3$ . For $x \in B_{R} (a)$ , $y \in \partial B_{R} (a)$ , the negative outward boundary-slot normal derivative of the ball Green function is $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ . The formula defines a continuous function of $(x, y)$ on $B_{R} (a) \times \partial B_{R} (a)$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a centre $a \in R^{n}$ , a radius $R > 0$ , a point $x \in B_{R} (a)$ and a boundary point $y \in \partial B_{R} (a)$ .

[F1]

With $ω_{n - 1} = ∣ S^{n - 1} ∣ > 0$ , the fundamental solution is $Φ (z) = ∣ z ∣^{2 - n} / ((n - 2) ω_{n - 1})$ for $z \neq 0$ and $n \geq 3$ (Fundamental solution for the positive operator minus Laplacian).

[F2]

For a bounded $C^{1}$ domain $Ω$ carrying a Dirichlet Green function $G_{Ω}$ with correctors $H_{p} \in C^{2} (\overline{Ω})$ , the boundary-slot normal derivative at $x \in Ω$ , $y \in \partial Ω$ is $\partial_{ν_{y}} G_{Ω} (x, y) : = D_{z} (Φ (z - x) - H_{x} (z)) ∣_{z = y} \cdot ν_{Ω} (y)$ , and the Poisson kernel is $P_{Ω} (x, y) : = - \partial_{ν_{y}} G_{Ω} (x, y)$ (Poisson kernel from a Dirichlet Green function).

[F3]

For $B_{R} (a)$ with $n \geq 3$ the Dirichlet Green function is $G (x, z) = Φ (x - z) - (R / ∣ z - a ∣)^{n - 2} Φ (x - z^{*})$ for $z \neq a$ , where $z^{*} = a + R^{2} (z - a) / ∣ z - a ∣^{2}$ , and $G (x, a) = Φ (x - a) - Φ (R)$ ; the designated corrector for a pole $p$ is $H_{p} (z) = (R / ∣ p - a ∣)^{n - 2} Φ (z - p^{*})$ for $p \neq a$ and $H_{a} (z) = Φ (R)$ , and $G$ is symmetric (Dirichlet Green function of a Euclidean ball).

[F4]

$B_{R} (a)$ is a bounded $C^{1}$ domain with outward unit normal $ν (y) = (y - a) / R$ at every $y \in \partial B_{R} (a)$ (Euclidean balls are bounded C-one domains with radial outward normal).

[F5]

[F6]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F2F3F6

Work under [F6]. Put $u : = x - a$ and $w : = y - a$ , so that $∣ u ∣ < R$ and $∣ w ∣ = R$ ; for $x \neq a$ put $κ_{x} : = (R / ∣ u ∣)^{n - 2}$ and $x^{*} : = a + R^{2} u / ∣ u ∣^{2}$ , the inversion of $x$ , while for $x = a$ the corrector is the constant $H_{a} = Φ (R)$ by [F3]. By [F3] the corrector for the pole $x$ is $H_{x} (z) = κ_{x} Φ (z - x^{*})$ when $x \neq a$ ; its value at $y$ is well defined because $x^{*} \notin B_{R} (a)$ and $y \in \partial B_{R} (a)$ are different points, and $Φ$ is smooth there by [F1]. Also $D_{z} Φ (z - x) ∣_{z = y}$ is defined because $y \neq x$ .

2.1givenstep 1.1algebra

Magnitude identity. Let $x \neq a$ . From $x^{*} - a = R^{2} u / ∣ u ∣^{2}$ and $y - x^{*} = (y - a) - R^{2} u / ∣ u ∣^{2}$ we get $∣ y - x^{*} ∣^{2} = ∣ w ∣^{2} - 2 R^{2} ⟨ w, u ⟩ / ∣ u ∣^{2} + R^{4} / ∣ u ∣^{2}$ ; multiplying by $∣ u ∣^{2}$ and using $∣ w ∣^{2} = R^{2}$ gives $∣ u ∣^{2} ∣ y - x^{*} ∣^{2} = R^{2} ∣ u ∣^{2} - 2 R^{2} ⟨ u, w ⟩ + R^{4} = R^{2} ∣ x - y ∣^{2}$ , because $∣ x - y ∣^{2} = ∣ u - w ∣^{2} = ∣ u ∣^{2} - 2 ⟨ u, w ⟩ + R^{2}$ . Hence $∣ u ∣ ∣ y - x^{*} ∣ = R ∣ x - y ∣ > 0$ .

2.2givenstep 1.1algebra

Vector identity. Let $x \neq a$ . Adding and subtracting $u$ and using $x^{*} - a = R^{2} u / ∣ u ∣^{2}$ gives $(x - y) + \frac{∣ u ∣^{2}}{R^{2}} (y - x^{*}) = (u - w) + \frac{∣ u ∣^{2}}{R^{2}} (w - \frac{R^{2} u}{∣ u ∣^{2}}) = u - w + \frac{∣ u ∣^{2} w}{R^{2}} - u = - (1 - \frac{∣ u ∣^{2}}{R^{2}}) (y - a) = - \frac{R^{2} - ∣ x - a ∣^{2}}{R^{2}} (y - a) .$

3.1step 2.1F1F5algebra

The gradient of each term of [F2] at $z = y$ . By [F5] and [F1], the gradient of $Φ$ is $\nabla Φ (ζ) = - ∣ ζ ∣^{- n} ζ / ω_{n - 1}$ for $ζ \neq 0$ , since $\nabla ∣ ζ ∣^{2 - n} = (2 - n) ∣ ζ ∣^{- n} ζ$ and the prefactor is $1 / ((n - 2) ω_{n - 1})$ ; hence $D_{z} Φ (z - x) ∣_{z = y} = - (∣ y - x ∣^{- n} (y - x)) / ω_{n - 1} = ∣ x - y ∣^{- n} (x - y) / ω_{n - 1}$ , and $D_{z} [κ_{x} Φ (z - x^{*})] ∣_{z = y} = - κ_{x} ∣ y - x^{*} ∣^{- n} (y - x^{*}) / ω_{n - 1}$ . By step 2.1, $κ_{x} ∣ y - x^{*} ∣^{- n} = (R / ∣ u ∣)^{n - 2} (∣ u ∣ / R)^{n} ∣ x - y ∣^{- n} = (∣ u ∣^{2} / R^{2}) ∣ x - y ∣^{- n}$ .

4.1step 2.2step 3.1algebra

The boundary-slot derivative. Subtracting the two expressions of step 3.1 and using step 2.2, $D_{z} (Φ (z - x) - H_{x} (z)) ∣_{z = y} = \frac{1}{ω_{n - 1}} ∣ x - y ∣^{- n} [(x - y) + \frac{∣ u ∣^{2}}{R^{2}} (y - x^{*})] = - \frac{R^{2} - ∣ x - a ∣^{2}}{ω_{n - 1} R^{2} ∣ x - y ∣^{n}} (y - a)$ for $x \neq a$ .

4.2step 3.1F1F2F3F4algebra

The case of the centre. For $x = a$ the corrector is the constant $H_{a} = Φ (R)$ of [F3], so $D_{z} (Φ (z - a) - H_{a}) ∣_{z = y} = \nabla Φ (y - a) = - R^{- n} (y - a) / ω_{n - 1}$ by the gradient computation of step 3.1; dotting with $ν (y) = (y - a) / R$ gives $\partial_{ν_{y}} G (a, y) = - R^{- n} \cdot R / ω_{n - 1} = - R^{1 - n} / ω_{n - 1}$ and $P_{R, a} (a, y) = R^{1 - n} / ω_{n - 1}$ , which is exactly the formula $(R^{2} - ∣ a - a ∣^{2}) / (R ω_{n - 1} ∣ a - y ∣^{n}) = R^{2} / (R ω_{n - 1} R^{n})$ .

5.1step 4.1F2F4algebra

The Poisson kernel. Dotting step 4.1 with $ν (y) = (y - a) / R$ from [F4] gives $\partial_{ν_{y}} G (x, y) = - \frac{(R^{2} - ∣ x - a ∣^{2}) (y - a) \cdot (y - a)}{ω_{n - 1} R^{3} ∣ x - y ∣^{n}} = - \frac{R^{2} - ∣ x - a ∣^{2}}{R ω_{n - 1} ∣ x - y ∣^{n}}$ , because $(y - a) \cdot (y - a) = R^{2}$ ; hence by [F2], $P_{R, a} (x, y) = - \partial_{ν_{y}} G (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ for $x \neq a$ .

6.1step 5.1step 4.2F2algebra∎

Steps 5.1 and 4.2 give $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ for every $x \in B_{R} (a)$ and $y \in \partial B_{R} (a)$ . This explicit expression is continuous on $B_{R} (a) \times \partial B_{R} (a)$ : numerator and denominator are continuous there and the denominator $R ω_{n - 1} ∣ x - y ∣^{n}$ is nonzero at every point of the product because an interior point $x$ and a boundary point $y$ are never equal, so $∣ x - y ∣ > 0$ . Hence the negative boundary-slot normal derivative of the ball Green function is the continuous function displayed in the statement.

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-10-02Open item page →

The ball Poisson kernel is positive and has unit mass

Statement

Assume Countable Choice and $n \geq 3$ . For every ball $B_{R} (a)$ , interior point $x$ and boundary point $y$ , $P_{R, a} (x, y) > 0$ , and the kernel has total surface mass one: $\int_{\partial B_{R} (a)} P_{R, a} (x, y) d S_{y} = 1$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a centre $a \in R^{n}$ , a radius $R > 0$ , an interior point $x \in B_{R} (a)$ and a boundary point $y \in \partial B_{R} (a)$ .

[F1]

For $x \in B_{R} (a)$ and $y \in \partial B_{R} (a)$ the ball Poisson kernel is $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ , the negative outward boundary-slot normal derivative of the ball Green function, and it is a continuous function of $(x, y)$ on $B_{R} (a) \times \partial B_{R} (a)$ (Poisson kernel of a Euclidean ball).

[F2]

Let $Ω$ be a bounded $C^{1}$ domain carrying a Dirichlet Green function whose designated correctors satisfy $H_{y} \in C^{2} (\overline{Ω})$ ; let $P_{Ω} = - \partial_{ν_{y}} G_{Ω}$ . Then for every real $u \in C^{2} (\overline{Ω})$ and every $x \in Ω$ one has $u (x) = \int_{Ω} G_{Ω} (x, y) (- Δ u (y)) d y + \int_{\partial Ω} P_{Ω} (x, y) u (y) d S (y)$ , both integrals absolutely finite; moreover $P_{Ω} \geq 0$ on $Ω \times \partial Ω$ and $\int_{\partial Ω} P_{Ω} (x, y) d S (y) = 1$ for every $x \in Ω$ (Green representation for classical Poisson data).

[F3]

For $n \geq 3$ the ball $B_{R} (a)$ carries the Dirichlet Green function $G (x, z) = Φ (x - z) - (R / ∣ z - a ∣)^{n - 2} Φ (x - z^{*})$ (with the centre case $G (x, a) = Φ (x - a) - Φ (R)$ ), whose designated correctors all lie in $C^{2} (\overline{B_{R} (a)})$ , and whose negative outward boundary-slot normal derivative is the kernel of [F1] (Dirichlet Green function of a Euclidean ball, Poisson kernel of a Euclidean ball).

[F4]

$B_{R} (a)$ is a bounded $C^{1}$ domain (Euclidean balls are bounded C-one domains with radial outward normal).

[F5]

For $n \geq 1$ and $r > 0$ the sphere and ball measures are $∣ \partial B_{r} ∣ = ω_{n - 1} r^{n - 1}$ and $∣ B_{r} ∣ = ω_{n - 1} r^{n} / n$ , both finite and positive (Sphere and ball measures scale in Rn).

[F6]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F2F3F4F6

Work under [F6] and let $Ω : = B_{R} (a)$ . The hypotheses of [F2] are met: $Ω$ is a bounded $C^{1}$ domain by [F4], and by [F3] it carries a Dirichlet Green function whose designated correctors lie in $C^{2} (\overline{Ω})$ ; moreover the kernel $P_{R, a}$ of [F1] is by [F3] the negative boundary-slot normal derivative $P_{Ω}$ of that Green function, so the two notation systems denote the same function on $Ω \times \partial Ω$ .

2.1givenstep 1.1F1F5algebra

Strict positivity. By [F1], $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ . Since $x$ lies in the open ball, $0 \leq ∣ x - a ∣ < R$ and the numerator $R^{2} - ∣ x - a ∣^{2}$ is positive; by [F5] both $R > 0$ and $ω_{n - 1} > 0$ , and $∣ x - y ∣ > 0$ because an interior point and a boundary point of $B_{R} (a)$ cannot coincide. A quotient of positive numbers is positive, so $P_{R, a} (x, y) > 0$ .

2.2givenstep 1.1F2F3algebra

Unit mass. Apply the representation identity of [F2] on $Ω = B_{R} (a)$ to the constant function $u \equiv 1$ , which is real and lies in $C^{2} (\overline{Ω})$ with $Δ u = 0$ : for every $x \in Ω$ , $1 = u (x) = \int_{Ω} G_{Ω} (x, y) \cdot 0 d y + \int_{\partial Ω} P_{Ω} (x, y) \cdot 1 d S (y)$ . Step 1.1 identifies $P_{Ω}$ with $P_{R, a}$ , and [F2] guarantees that the second integral is absolutely finite, so $\int_{\partial B_{R} (a)} P_{R, a} (x, y) d S_{y} = 1$ .

3.1step 2.1step 2.2F1F2∎

Step 2.1 gives $P_{R, a} (x, y) > 0$ for every interior $x$ and boundary $y$ , and step 2.2 gives unit total surface mass $\int_{\partial B_{R} (a)} P_{R, a} (x, y) d S_{y} = 1$ for every $x \in B_{R} (a)$ ; this proves both assertions of the statement. The argument uses the Green representation formula rather than the ball Dirichlet theorem, so the boundary-convergence question is not presupposed.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Cap and complement estimate for the ball Poisson integral

Statement

Assume Countable Choice and $n \geq 3$ . Let $g \in C (\partial B_{R} (a); C)$ , $p \in \partial B_{R} (a)$ , $δ > 0$ , and $x \in B_{R} (a)$ with $∣ x - p ∣ < δ / 2$ . Write $U_{g} (x) = \int_{\partial B_{R} (a)} P_{R, a} (x, y) g (y) d S_{y}$ , an absolutely convergent integral under these hypotheses, and $ω_{g, p} (δ) = \sup {∣ g (y) - g (p) ∣ : y \in \partial B_{R} (a), ∣ y - p ∣ < δ}$ . Then $∣ U_{g} (x) - g (p) ∣ \leq ω_{g, p} (δ) + 2^{n + 1} R^{n - 2} δ^{- n} ∥ g ∥_{\infty} (R^{2} - ∣ x - a ∣^{2}) .$ In particular, $U_{g} (x) \to g (p)$ as $x \to p$ from inside the ball.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a centre $a \in R^{n}$ , a radius $R > 0$ , a datum $g \in C (\partial B_{R} (a); C)$ , a boundary point $p \in \partial B_{R} (a)$ , a number $δ > 0$ and an interior point $x \in B_{R} (a)$ with $∣ x - p ∣ < δ / 2$ .

[F1]

For $x \in B_{R} (a)$ and $y \in \partial B_{R} (a)$ the kernel is $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ , it is continuous on $B_{R} (a) \times \partial B_{R} (a)$ , it is strictly positive, and $\int_{\partial B_{R} (a)} P_{R, a} (x, y) d S_{y} = 1$ (Poisson kernel of a Euclidean ball, The ball Poisson kernel is positive and has unit mass).

[F2]

$B_{R} (a)$ is a bounded $C^{1}$ domain whose boundary is the sphere $\partial B_{R} (a)$ ; thus $\partial B_{R} (a)$ is a compact embedded $C^{1}$ hypersurface and the surface integral $\int_{\partial B_{R} (a)} f d S$ is defined for Borel $f$ with finite absolute integral, is additive over a Borel partition and obeys $∣ \int f d S ∣ \leq \int ∣ f ∣ d S$ (Euclidean balls are bounded C-one domains with radial outward normal, Surface integration on compact C1 hypersurfaces).

[F3]

$\partial B_{R} (a)$ is compact and nonempty, so a continuous real function on it is bounded and attains its extrema; hence $∥ g ∥_{\infty} : = \sup_{y \in \partial B_{R} (a)} ∣ g (y) ∣$ is finite, and the set defining $ω_{g, p} (δ)$ is nonempty because it contains $y = p$ (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F4]

For $n \geq 1$ and $r > 0$ one has $∣ \partial B_{r} ∣ = ω_{n - 1} r^{n - 1} > 0$ (Sphere and ball measures scale in Rn).

[F5]

For $f, h \in L^{1}$ the integral is additive, $\int (f + h) = \int f + \int h$ , and additive over a Borel partition of the domain (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F6]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F2F3F4F6

Work under [F6] and set $κ : = R^{2} - ∣ x - a ∣^{2}$ , $C : = {y \in \partial B_{R} (a) : ∣ y - p ∣ < δ}$ and $D : = \partial B_{R} (a) ∖ C$ . Since $x$ is interior, $∣ x - a ∣ < R$ and $κ > 0$ . By [F3] the numbers $∥ g ∥_{\infty}$ and $ω_{g, p} (δ)$ are finite, and the integrand $y \mapsto P_{R, a} (x, y) g (y)$ is Borel with $∣ P_{R, a} (x, y) g (y) ∣ \leq ∥ g ∥_{\infty} \sup_{z \in \partial B_{R} (a)} P_{R, a} (x, z)$ , a finite bound by [F1] and [F3]; the sphere has finite surface measure by [F4], so $U_{g} (x)$ is absolutely convergent.

2.1givenstep 1.1algebra

For $y \in D$ one has $∣ y - p ∣ \geq δ$ , hence $∣ x - y ∣ \geq ∣ y - p ∣ - ∣ x - p ∣ > δ - δ / 2 = δ / 2$ , so $1 / ∣ x - y ∣^{n} \leq 2^{n} δ^{- n}$ .

2.2step 1.1F1F2algebra

The cap carries mass at most one: $C$ is open in $\partial B_{R} (a)$ , hence Borel, $0 \leq P_{R, a} (x, \cdot) 1_{C} \leq P_{R, a} (x, \cdot)$ pointwise by [F1], and the surface integral is monotone by [F2]; therefore $\int_{C} P_{R, a} (x, \cdot) d S \leq \int_{\partial B_{R} (a)} P_{R, a} (x, \cdot) d S = 1$ by the unit-mass clause of [F1].

2.3step 1.1F3

The modulus vanishes at small scales: $g$ is continuous at $p$ on the sphere, so for every $η > 0$ there is $δ_{0} > 0$ with $∣ g (y) - g (p) ∣ < η$ whenever $y \in \partial B_{R} (a)$ and $∣ y - p ∣ < δ_{0}$ ; the set over which the supremum in $ω_{g, p} (δ_{0})$ is taken is nonempty by [F3], so $0 \leq ω_{g, p} (δ_{0}) \leq η$ .

3.1step 1.1step 2.1F1F3algebra

Consequently, for every $y \in D$ , [F1] and step 2.1 give $P_{R, a} (x, y) = κ / (R ω_{n - 1} ∣ x - y ∣^{n}) \leq κ 2^{n} / (R ω_{n - 1} δ^{n})$ , and also $∣ g (y) - g (p) ∣ \leq ∣ g (y) ∣ + ∣ g (p) ∣ \leq 2 ∥ g ∥_{\infty}$ by [F3].

4.1step 3.1F2F4algebra

The complement carries little mass: by [F2], [F4] and step 3.1, $\int_{D} P_{R, a} (x, \cdot) d S \leq \frac{κ 2^{n}}{R ω_{n - 1} δ^{n}} ∣ \partial B_{R} (a) ∣ = \frac{κ 2^{n}}{R ω_{n - 1} δ^{n}} \cdot ω_{n - 1} R^{n - 1} = 2^{n} R^{n - 2} δ^{- n} κ$ .

5.1step 3.1step 4.1step 2.2F1F2F5algebra

Splitting by [F2] and [F5] and bounding each piece, $∣ U_{g} (x) - g (p) ∣ \leq \int_{C} P_{R, a} (x, \cdot) ∣ g - g (p) ∣ d S + \int_{D} P_{R, a} (x, \cdot) ∣ g - g (p) ∣ d S \leq ω_{g, p} (δ) \int_{C} P_{R, a} (x, \cdot) d S + 2 ∥ g ∥_{\infty} \int_{D} P_{R, a} (x, \cdot) d S \leq ω_{g, p} (δ) + 2^{n + 1} R^{n - 2} δ^{- n} ∥ g ∥_{\infty} κ$ , which is the displayed estimate.

6.1step 5.1step 2.3algebra

Therefore $U_{g} (x) \to g (p)$ as $x \to p$ from inside: given $η > 0$ , choose $δ_{0}$ as in step 2.3, keep it fixed and let $x \to p$ with $∣ x - p ∣ < δ_{0} / 2$ ; step 5.1 gives $∣ U_{g} (x) - g (p) ∣ \leq η + 2^{n + 1} R^{n - 2} δ_{0}^{- n} ∥ g ∥_{\infty} (R^{2} - ∣ x - a ∣^{2})$ , and $R^{2} - ∣ x - a ∣^{2} \to R^{2} - ∣ p - a ∣^{2} = 0$ because $∣ p - a ∣ = R$ , so ${lim sup}_{x \to p} ∣ U_{g} (x) - g (p) ∣ \leq η$ .

7.1step 5.1step 6.1∎

Since $η > 0$ was arbitrary, the limsup in step 6.1 is zero; thus the displayed estimate holds for all admissible $x, δ$ and the integral tends to $g (p)$ as $x \to p$ from inside the ball, which proves both assertions of the statement. The argument uses the kernel formula, its positivity and its unit mass, but never the ball Dirichlet solution theorem, so no circularity arises with the later boundary-trace theorems.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-10-02Open item page →

Continuous Dirichlet problem on a ball

Statement

Assume Countable Choice and $n \geq 3$ . For every real or complex $g \in C (\partial B_{R} (a))$ the integral $U_{g} (x) = \int_{\partial B_{R} (a)} P_{R, a} (x, y) g (y) d S_{y}$ is absolutely convergent, smooth and harmonic on $B_{R} (a)$ , and $U_{g}$ extends continuously to $\overline{B_{R} (a)}$ with boundary trace $g$ . It is the unique function in $C^{2} (B_{R} (a)) \cap C (\overline{B_{R} (a)})$ that is harmonic on $B_{R} (a)$ and equals $g$ on $\partial B_{R} (a)$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a centre $a \in R^{n}$ , a radius $R > 0$ , and a complex-valued datum $g \in C (\partial B_{R} (a))$ .

[F1]

For $x \in B_{R} (a)$ , $y \in \partial B_{R} (a)$ the kernel is $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ , positive, continuous on $B_{R} (a) \times \partial B_{R} (a)$ , with $\int_{\partial B_{R} (a)} P_{R, a} (x, y) d S_{y} = 1$ (Poisson kernel of a Euclidean ball, The ball Poisson kernel is positive and has unit mass).

[F2]

Under $∣ x - p ∣ < δ / 2$ one has $∣ U_{g} (x) - g (p) ∣ \leq ω_{g, p} (δ) + 2^{n + 1} R^{n - 2} δ^{- n} ∥ g ∥_{\infty} (R^{2} - ∣ x - a ∣^{2})$ , and $U_{g} (x) \to g (p)$ as $x \to p$ from inside the ball (Cap and complement estimate for the ball Poisson integral).

[F3]

If $Ω$ is bounded, nonempty and open and real $u \in C^{2} (Ω) \cap C (\overline{Ω})$ has $Δ u \geq 0$ , then $\max_{\overline{Ω}} u = \max_{\partial Ω} u$ (Weak maximum principle for the laplacian); $B_{R} (a)$ is bounded, open and nonempty (Euclidean balls are bounded C-one domains with radial outward normal).

[F4]

On a measure space $(X, μ)$ and an open interval $I$ , suppose $f : X \times I \to C$ has integrable $x$ -slices for every $t \in I$ , is differentiable in $t$ outside a fixed measurable null set, has measurable derivative slices (extended by zero where undefined), and satisfies $∣ \partial_{t} f (x, t) ∣ \leq G (x)$ for all $t$ outside a fixed null set, with $G \geq 0$ measurable and $\int G d μ < \infty$ . Then $\frac{d}{d t} \int f (x, t) d μ (x) = \int \partial_{t} f (x, t) d μ (x)$ (Differentiation under the integral sign).

[F5]

The surface integral on the compact sphere is defined by chart integration, is additive over Borel partitions and monotone, bounded Borel integrands over finite measure have finite integrals, and dominated convergence applies to a pointwise convergent dominated family (Surface integration on compact C1 hypersurfaces, Sphere and ball measures scale in Rn, Dominated convergence).

[F6]

Calculus interface: $C^{k}$ maps on Euclidean domains are closed under sums, products and composition; $(t^{α})^{'} = α t^{α - 1}$ for $t > 0$ ; the chain rule and product rule hold; the Laplacian is $Δ f = \sum_{i} \partial_{i} \partial_{i} f$ , and multi-indices, $D^{α}$ and $C^{k}$ are as fixed in the notation item ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, Continuity and derivatives of positive-base real powers, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , $C^{k}$ maps and multi-index derivative notation in Euclidean space, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F7]

Compact subsets of Euclidean space are closed and bounded, closed bounded Euclidean subsets are compact, and continuous real-valued functions on nonempty compact metric spaces attain their extrema. Hence for any nonempty compact $K \subset B_{R} (a)$ the product $K \times \partial B_{R} (a)$ is closed and bounded in $R^{2 n}$ and therefore compact; the continuous functions $(x, y) \mapsto ∣ x - y ∣$ and $(x, y) \mapsto ∣ D_{x}^{α} P (x, y) ∣$ attain their extrema there. Also $\partial B_{R} (a)$ is compact and nonempty (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, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

[F8]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F5F7F8

Work under [F8]. Fix $x_{0} \in B_{R} (a)$ . By [F1] and [F7], $y \mapsto P_{R, a} (x_{0}, y)$ is continuous on the compact sphere, hence bounded, and $∥ g ∥_{\infty} < + \infty$ ; the sphere has finite surface measure by [F5], so $∣ P_{R, a} (x_{0}, \cdot) g ∣ \leq ∥ g ∥_{\infty} \sup_{y} P_{R, a} (x_{0}, y)$ is integrable and $U_{g} (x_{0})$ is absolutely convergent. Moreover $∥ U_{g} ∥_{\infty} \leq ∥ g ∥_{\infty}$ on $B_{R} (a)$ by [F1] and [F5].

1.2givenF6F7

Smoothness of the kernel and of the parametrised integrals. The map $(x, y) \mapsto ∣ x - y ∣^{2} = \sum_{i} (x_{i} - y_{i})^{2}$ is a polynomial, hence $C^{\infty}$ on $R^{n} \times R^{n}$ , and it is strictly positive on the set where $x \neq y$ ; composing with $t \mapsto t^{- n / 2}$ , which is $C^{\infty}$ on $(0, \infty)$ by [F6], and multiplying by the polynomial $R^{2} - ∣ x - a ∣^{2}$ shows that $P$ is $C^{\infty}$ on its domain by [F6]. Consequently for every multi-index $α$ the function $(x, y) \mapsto D_{x}^{α} P (x, y)$ is continuous there, and for every nonempty compact $K \subset B_{R} (a)$ the distance $d_{K} : = \min {∣ x - y ∣ : x \in K, y \in \partial B_{R} (a)}$ is positive and $M_{α, K} : = \sup_{K \times \partial B_{R} (a)} ∣ D_{x}^{α} P ∣ < + \infty$ , by [F7] and [F6] applied to the continuous function $(x, y) \mapsto ∣ x - y ∣$ on the compact set $K \times \partial B_{R} (a)$ .

1.3givenF1F6algebra

The kernel is harmonic in the interior variable. Fix $y \in \partial B_{R} (a)$ and $x \in B_{R} (a)$ , put $m (x) : = R^{2} - ∣ x - a ∣^{2}$ and $ρ (x) : = ∣ x - y ∣ > 0$ . Direct differentiation gives $\partial_{i} m = - 2 (x_{i} - a_{i})$ , $Δ m = - 2 n$ , $\partial_{i} ρ = (x_{i} - y_{i}) / ρ$ , and for a $C^{2}$ radial profile $q$ the formulas $\partial_{i} (q \circ ρ) = q^{'} (ρ) (x_{i} - y_{i}) / ρ$ and $Δ (q \circ ρ) = q^{''} (ρ) + (n - 1) q^{'} (ρ) / ρ$ ; with $q (ρ) = ρ^{- n}$ this gives $\nabla (ρ^{- n}) = - n ρ^{- n - 2} (x - y)$ and $Δ (ρ^{- n}) = 2 n ρ^{- n - 2}$ by [F6]. Hence $Δ (m ρ^{- n}) = Δ m \cdot ρ^{- n} + 2 \nabla m \cdot \nabla (ρ^{- n}) + m Δ (ρ^{- n}) = - 2 n ρ^{- n} + 4 n (x - a) \cdot (x - y) ρ^{- n - 2} + 2 n (R^{2} - ∣ x - a ∣^{2}) ρ^{- n - 2}$ , and the identity $∣ x - y ∣^{2} = ∣ x - a ∣^{2} - 2 (x - a) \cdot (y - a) + R^{2}$ shows that the last two terms equal $2 n ρ^{- n}$ , so $Δ (m ρ^{- n}) = 0$ . Since $P_{R, a} (x, y) = m (x) ρ (x)^{- n} / (R ω_{n - 1})$ , we get $Δ_{x} P_{R, a} (x, y) = 0$ for all $x \in B_{R} (a)$ , $y \in \partial B_{R} (a)$ .

2.1step 1.2F4F5F6induction

Higher derivatives under the integral. Induct on the length of an ordered word of coordinate derivatives. The empty word gives the defining integral for $U_{g}$ . Suppose a word gives $V (x) = \int Q (x, y) g (y) d S_{y}$ , where $Q$ is the same ordered derivative of $P$ . Fix $x_{0}$ and a closed ball $K$ with $x_{0} \in int K$ and $K \subset B_{R} (a)$ . By step 1.2, both $Q$ and $\partial_{i} Q$ are continuous and bounded on $K \times \partial B_{R} (a)$ . For $x = x_{0} + t e_{i}$ on a sufficiently small open interval, each slice $Q (x, \cdot) g$ is Borel and integrable, and its $t$ -derivative is Borel and bounded by $\sup_{K \times \partial B_{R} (a)} ∣ \partial_{i} Q ∣ ∥ g ∥_{\infty}$ , an integrable constant by [F5]. Thus all hypotheses of [F4] hold, with empty exceptional set, and $\partial_{i} V (x_{0}) = \int \partial_{i} Q (x_{0}, y) g (y) d S_{y}$ . The integral expressions for both $V$ and this derivative are continuous near $x_{0}$ by dominated convergence [F5], using the respective bounded continuous kernels on $K$ . This proves existence and continuity for every ordered derivative, hence $U_{g} \in C^{\infty}$ under [F6]; choosing the canonical word for a multi-index gives $D^{α} U_{g} (x) = \int D_{x}^{α} P (x, y) g (y) d S_{y}$ . No interchange of derivative order is required.

3.1step 1.3step 2.1F5F6algebra

$U_{g}$ is smooth and harmonic. Step 2.1 with $α = 0$ gives $C^{\infty}$ , and for $∣ α ∣ = 2$ it gives $Δ U_{g} (x) = \sum_{i} \int \partial_{i} \partial_{i} P_{R, a} (x, y) g (y) d S_{y} = \int Δ_{x} P_{R, a} (x, y) g (y) d S_{y}$ by [F5] and the Laplacian definition of [F6]; step 1.3 makes every value of $Δ_{x} P_{R, a}$ vanish, so $Δ U_{g} = 0$ on $B_{R} (a)$ and $U_{g}$ is smooth harmonic.

3.2step 2.1F2

Boundary trace and continuity on the closed ball. Interior continuity holds by step 2.1 with $α = 0$ . Define $\tilde{U} : = U_{g}$ on $B_{R} (a)$ and $\tilde{U} : = g$ on $\partial B_{R} (a)$ . At a boundary point $p$ , [F2] gives $U_{g} (x) \to g (p) = \tilde{U} (p)$ along every interior approach, and $g$ is continuous on the sphere by hypothesis; hence $\tilde{U}$ is continuous at every point of $\overline{B_{R} (a)}$ and $U_{g}$ extends continuously to the closed ball with trace $g$ .

4.1step 3.1step 3.2F3cases

Uniqueness. Let $v \in C^{2} (B_{R} (a)) \cap C (\overline{B_{R} (a)})$ be harmonic on $B_{R} (a)$ with $v = g$ on $\partial B_{R} (a)$ , and put $w : = v - \tilde{U}$ , which is continuous on the closure, $C^{2}$ inside and harmonic inside by step 3.1. Apply [F3] to $R e w$ and to $- R e w$ , and to $I m w$ and $- I m w$ : on the boundary all four functions vanish, so their maxima over $\overline{B_{R} (a)}$ are zero. Hence $w \equiv 0$ and $v = U_{g}$ .

5.1step 1.1step 3.1step 3.2step 4.1∎

Steps 1.1, 3.1 and 3.2 show that $U_{g}$ is absolutely convergent, smooth harmonic and continuously extendible with trace $g$ , and step 4.1 shows that every such classical solution equals $U_{g}$ ; this is exactly the assertion. The boundary convergence was obtained from the cap/complement estimate [F2], which depends only on the kernel formula, its positivity and its unit mass, so the later uniform-radial corollary is not presupposed.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Ball Poisson integrals converge uniformly along radial boundary approaches

Statement

Assume Countable Choice and $n \geq 3$ . For $g \in C (\partial B_{R} (a); C)$ let $U_{g}$ be its ball Poisson integral. Then $\sup_{θ \in S^{n - 1}} ∣ U_{g} (a + r θ) - g (a + R θ) ∣ ⟶ 0 (r ↑ R) .$

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , a centre $a \in R^{n}$ , a radius $R > 0$ , and a datum $g \in C (\partial B_{R} (a); C)$ .

[F1]

With $ω_{g, p} (δ) = \sup {∣ g (y) - g (p) ∣ : y \in \partial B_{R} (a), ∣ y - p ∣ < δ}$ and $∣ x - p ∣ < δ / 2$ , $∣ U_{g} (x) - g (p) ∣ \leq ω_{g, p} (δ) + 2^{n + 1} R^{n - 2} δ^{- n} ∥ g ∥_{\infty} (R^{2} - ∣ x - a ∣^{2})$ (Cap and complement estimate for the ball Poisson integral).

[F2]

The sphere $\partial B_{R} (a)$ is compact and nonempty, so continuous real functions on it are bounded and a continuous $g$ on it is uniformly continuous (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[F3]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F2F3

Work under [F3]. By [F2] the quantity $∥ g ∥_{\infty} = \sup_{\partial B_{R} (a)} ∣ g ∣$ is finite and $g$ is uniformly continuous on the sphere: for every $ε > 0$ there is $δ > 0$ with $∣ g (y) - g (z) ∣ \leq ε$ whenever $y, z \in \partial B_{R} (a)$ and $∣ y - z ∣ < δ$ . In particular, for every $p \in \partial B_{R} (a)$ and every cap radius $δ$ with this property, $ω_{g, p} (δ) = \sup {∣ g (y) - g (p) ∣ : ∣ y - p ∣ < δ} \leq ε$ by [F1].

2.1step 1.1F1algebra

Fix such an $ε$ and an associated $δ > 0$ , and let $r < R$ with $R - r < δ / 2$ . For $p = a + R θ$ and $x = a + r θ$ with $θ \in S^{n - 1}$ we have $∣ x - p ∣ = (R - r) ∣ θ ∣ = R - r < δ / 2$ , so [F1] applies and gives $∣ U_{g} (x) - g (p) ∣ \leq ε + 2^{n + 1} R^{n - 2} δ^{- n} ∥ g ∥_{\infty} (R^{2} - r^{2})$ .

3.1step 1.1step 2.1F2algebra

Choose $r$ additionally so close to $R$ that $2^{n + 1} R^{n - 2} δ^{- n} ∥ g ∥_{\infty} (R^{2} - r^{2}) \leq ε$ ; this is possible because $R^{2} - r^{2} \to 0$ as $r ↑ R$ . Then step 2.1 gives $∣ U_{g} (a + r θ) - g (a + R θ) ∣ \leq 2 ε$ for every $θ \in S^{n - 1}$ simultaneously, since neither the bound $ε$ from [F2] nor the factor $R^{2} - r^{2}$ depends on $θ$ .

4.1step 3.1F1∎

Taking the supremum over $θ$ and letting $ε ↓ 0$ shows $\sup_{θ} ∣ U_{g} (a + r θ) - g (a + R θ) ∣ \to 0$ as $r ↑ R$ , which is the assertion. The estimate used is the pointwise cap/complement bound; the ball Dirichlet solution theorem is not needed for this uniformity statement, and no structure of $U_{g}$ beyond the integral formula is used.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Poisson kernel and bounded Dirichlet problem on a half-space

Statement

Assume Countable Choice and $n \geq 3$ . Use one-based labels $x_{j} : = x_{j - 1}^{c a n}$ and $e_{j} : = e_{j - 1}^{c a n}$ , $1 \leq j \leq n$ . For $x = (x^{'}, t) \in H = {x_{n} > 0}$ and $z \in R^{n - 1}$ , $P_{H} ((x^{'}, t), z) = \frac{2 t}{ω_{n - 1} (∣ x^{'} - z ∣^{2} + t^{2})^{n / 2}}$ is the negative outward boundary derivative of the reflected Green kernel $G_{H}$ of the half-space, is positive, and satisfies $\int_{R^{n - 1}} P_{H} ((x^{'}, t), z) d z = 1$ . For bounded continuous real or complex $g$ on $\partial H = R^{n - 1}$ , the function $U_{g} (x) = \int_{R^{n - 1}} P_{H} (x, z) g (z) d z$ is bounded, smooth and harmonic on $H$ , and $U_{g} (x) \to g (z_{0})$ as $x \to (z_{0}, 0)$ from inside $H$ . It is the unique bounded harmonic function on $H$ , continuous on $\overline{H}$ , with trace $g$ ; boundedness is the growth condition at infinity that makes the solution unique.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 3$ , the upper half-space $H = {x = (x^{'}, x_{n}) : x_{n} > 0}$ , the outward normal $ν = - e_{n}$ of its boundary plane $\partial H = {x_{n} = 0} = R^{n - 1}$ , and a bounded continuous $g : \partial H \to C$ , $∥ g ∥_{\infty} = \sup_{\partial H} ∣ g ∣ < + \infty$ .

[F1]

The reflected kernel $G_{H} (x, y) = Φ (x - y) - Φ (x - y^{†})$ , $y^{†} = (y^{'}, - y_{n})$ , is symmetric off the diagonal and strictly positive for distinct $x, y \in H$ , smooth and harmonic in $x$ off $y$ , has $- Δ_{x} G_{H} (\cdot, y) = δ_{y}$ distributionally in $H$ and has zero continuous boundary trace (Reflection Green kernel for the half-space).

[F2]

For $n \geq 3$ , $Φ (w) = ∣ w ∣^{2 - n} / ((n - 2) ω_{n - 1})$ is smooth and harmonic on $R^{n} ∖ {0}$ (Fundamental solution for the positive operator minus Laplacian, The Laplace fundamental solution is harmonic off its pole).

[F3]

The continuous Dirichlet problem on a ball is uniquely solvable by the Poisson integral, for real and complex data (Continuous Dirichlet problem on a ball).

[F4]

On a bounded nonempty open set, a $C^{2} \cap C (\overline{Ω})$ function with $Δ u \geq 0$ attains its maximum on the boundary (Weak maximum principle for the laplacian).

[F5]

A bounded harmonic function on all of $R^{n}$ is constant (Liouville theorem for bounded harmonic functions).

[F6]

Toolkit for the normalisation: polar coordinates in $R^{n - 1}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma), the $C^{1}$ change-of-variables formula for nonnegative measurable functions (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions), $B (p, q) = \int_{0}^{1} t^{p - 1} (1 - t)^{q - 1} d t$ (Euler's real Beta integral), $B (p, q) = Γ (p) Γ (q) / Γ (p + q)$ (The real Beta--Gamma identity), $Γ (1 / 2) = \sqrt{π}$ ( $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral), $Γ (s + 1) = s Γ (s)$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ), $V_{n} (1) = π^{n / 2} / Γ (n / 2 + 1)$ (The closed form for the volume of the unit $n$ -ball), $∣ \partial B_{r} ∣ = ω_{n - 1} r^{n - 1}$ and $∣ B_{r} ∣ = ω_{n - 1} r^{n} / n$ (Sphere and ball measures scale in Rn), and $ω_{n - 1} = ∣ S^{n - 1} ∣$ for the polar surface measure (Agreement with the existing polar sphere measure).

[F7]

Differentiation under the integral sign over a general measure space, and dominated convergence (Differentiation under the integral sign, Dominated convergence).

[F8]

Calculus interface: chain rule, product rule, real-power derivatives, closure of $C^{k}$ maps under algebra and composition, and $Δ = \sum_{i} \partial_{i} \partial_{i}$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Continuity and derivatives of positive-base real powers, $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F9]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF9

Work under [F9] and put $R (x, z) : = ∣ x - (z, 0) ∣ = (∣ x^{'} - z ∣^{2} + t^{2})^{1 / 2} > 0$ for $x = (x^{'}, t) \in H$ . Write $y = (z, 0)$ .

1.2givenF2F8algebra

Derivative of the fundamental kernel: since $Φ (w) = ∣ w ∣^{2 - n} / ((n - 2) ω_{n - 1})$ on $w \neq 0$ , the chain rule and real-power rule [F8] give $\nabla Φ (w) = ∣ w ∣^{- n} w \cdot (2 - n) / ((n - 2) ω_{n - 1}) = - w / (ω_{n - 1} ∣ w ∣^{n})$ for every $w \neq 0$ .

1.3givenF6algebra

Normalisation, first reduction. By [F6] applied to the nonnegative measurable function $z \mapsto P_{H} (x, z)$ and the $C^{1}$ diffeomorphism $z = x^{'} + t w$ with Jacobian $t^{n - 1}$ , $\int_{R^{n - 1}} P_{H} (x, z) d z = \frac{2 t}{ω_{n - 1}} \int_{R^{n - 1}} \frac{d z}{(∣ x^{'} - z ∣^{2} + t^{2})^{n / 2}} = \frac{2 t t^{n - 1}}{ω_{n - 1} t^{n}} \int_{R^{n - 1}} \frac{d w}{(1 + ∣ w ∣^{2})^{n / 2}} = \frac{2}{ω_{n - 1}} I, I : = \int_{R^{n - 1}} \frac{d w}{(1 + ∣ w ∣^{2})^{n / 2}} .$

2.1step 1.1step 1.2F1algebra

The boundary derivative. Fix $x = (x^{'}, t) \in H$ and a boundary coordinate $z \in R^{n - 1}$ . For $0 < s < t$ , the formula in [F1] gives $G_{H} (x, (z, s)) = Φ (x^{'} - z, t - s) - Φ (x^{'} - z, t + s)$ . Both arguments stay nonzero through $s = 0$ , so this explicit expression extends smoothly to the boundary pole $(z, 0)$ . By step 1.2, differentiating in $s$ at $0$ gives $\partial_{s} G_{H} (x, (z, s)) ∣_{s = 0} = 2 t / (ω_{n - 1} (∣ x^{'} - z ∣^{2} + t^{2})^{n / 2})$ . Since the outward normal is $ν = - e_{n}$ , the negative outward derivative is $- \partial_{ν} G_{H} = + \partial_{s} G_{H}$ , which is the displayed positive kernel. This calculation uses the explicit reflected formula and its smooth boundary extension; it does not apply the interior-pole statement of [F1] at a boundary pole.

2.2step 2.1F6algebra

Polar evaluation of $I$ . With $m = n - 1 \geq 2$ and $w = r θ$ , [F6] gives $I = ω_{n - 2} \int_{0}^{\infty} \frac{r^{n - 2} d r}{(1 + r^{2})^{n / 2}}$ ; substituting $s = r^{2}$ , $r^{n - 2} d r = \frac{1}{2} s^{(n - 1) / 2 - 1} d s$ , this is $I = \frac{ω_{n - 2}}{2} \int_{0}^{\infty} s^{(n - 1) / 2 - 1} (1 + s)^{- n / 2} d s$ . The further substitution $t = \frac{s}{1 + s}$ turns the last integral into $\int_{0}^{1} t^{(n - 1) / 2 - 1} (1 - t)^{1 / 2 - 1} d t = B (\frac{n - 1}{2}, \frac{1}{2})$ by [F6]; hence $I = \frac{ω_{n - 2}}{2} B (\frac{n - 1}{2}, \frac{1}{2}) < + \infty$ . [step 1.3, F6, algebra] 3.1 Positivity: for $x \in H$ we have $t > 0$ and $ω_{n - 1} > 0$ by [F6], while the denominator is a positive real number; hence $P_{H} (x, z) > 0$ for every $z \in R^{n - 1}$ .

3.2step 1.3step 2.2F6algebra

Evaluation of the constants. By [F6], $∣ B_{1}^{n} ∣ = π^{n / 2} / Γ (n / 2 + 1)$ and $∣ B_{1}^{n} ∣ = ω_{n - 1} / n$ , so $ω_{n - 1} = n π^{n / 2} / Γ (n / 2 + 1) = 2 π^{n / 2} / Γ (n / 2)$ ; replacing $n$ by $n - 1$ gives $ω_{n - 2} = 2 π^{(n - 1) / 2} / Γ ((n - 1) / 2)$ . By [F6] again, $B (\frac{n - 1}{2}, \frac{1}{2}) = Γ (\frac{n - 1}{2}) Γ (\frac{1}{2}) / Γ (\frac{n}{2}) = Γ (\frac{n - 1}{2}) \sqrt{π} / Γ (\frac{n}{2})$ . Substituting into steps 1.3 and 2.2, $\int_{R^{n - 1}} P_{H} (x, z) d z = \frac{2}{ω_{n - 1}} \cdot \frac{ω_{n - 2}}{2} \cdot \frac{Γ (\frac{n - 1}{2}) \sqrt{π}}{Γ (\frac{n}{2})} = \frac{2 π^{(n - 1) / 2}}{2 π^{n / 2}} \cdot \sqrt{π} = 1$ .

3.3step 1.3step 2.2F7algebra

Boundary convergence. Fix $z_{0} \in \partial H$ and $η > 0$ ; by continuity of $g$ at $z_{0}$ choose $δ > 0$ with $∣ g (z) - g (z_{0}) ∣ < η$ for $∣ z - z_{0} ∣ < δ$ . Let $C = {z : ∣ z - z_{0} ∣ < δ}$ and $D = R^{n - 1} ∖ C$ . For $z \in C$ , $∣ g (z) - g (z_{0}) ∣ \leq η$ ; for $z \in D$ we use $∣ g (z) - g (z_{0}) ∣ \leq 2 ∥ g ∥_{\infty}$ , and the mass of $D$ is small: if $∣ x^{'} - z_{0} ∣ < δ / 2$ and $z \in D$ , then $∣ x^{'} - z ∣ \geq ∣ z - z_{0} ∣ - ∣ x^{'} - z_{0} ∣ > δ / 2$ , so by step 1.3 $\int_{D} P_{H} (x, z) d z \leq \frac{2}{ω_{n - 1}} \int_{∣ w ∣ > δ / (2 t)} \frac{d w}{(1 + ∣ w ∣^{2})^{n / 2}}$ , and this tail tends to $0$ as $t ↓ 0$ by [F7] and the finiteness in step 2.2, since the integrands are dominated by the integrable function $(1 + ∣ w ∣^{2})^{- n / 2}$ and vanish pointwise on the shrinking domain. Hence $∣ U_{g} (x) - g (z_{0}) ∣ \leq η + 2 ∥ g ∥_{\infty} \int_{D} P_{H} (x, z) d z$ , so ${lim sup}_{x \to (z_{0}, 0)} ∣ U_{g} (x) - g (z_{0}) ∣ \leq η + 2 ∥ g ∥_{\infty} \cdot 0 = η$ , and $η > 0$ was arbitrary; so $U_{g} (x) \to g (z_{0})$ as $x \to (z_{0}, 0)$ from inside $H$ .

4.1step 2.1step 3.2F8algebra

Derivative bounds and integrability. Every partial derivative $D_{x}^{α} P_{H} (x, z)$ is continuous on $H \times R^{n - 1}$ and, on each compact $K \subset H$ , satisfies $∣ D_{x}^{α} P_{H} (x, z) ∣ \leq C_{α, K} (1 + ∣ z ∣)^{- n}$ . Indeed, writing $λ = ∣ x' - z ∣^{2} + t^{2}$ , on $K$ the height $t$ is bounded away from $0$ and both $t$ and $∣ x' ∣$ are bounded above; for large $∣ z ∣$ , $λ$ is comparable to $∣ z ∣^{2}$ . Each horizontal derivative of $λ^{- n / 2}$ contributes a factor $O (∣ z ∣)$ and one extra factor $λ^{- 1}$ , gaining decay; each vertical derivative either differentiates the numerator $t$ , leaving the base decay $O (∣ z ∣^{- n})$ , or differentiates a denominator factor and gains decay with bounded factors of $t$ . Repeating these rules shows that no derivative decays more slowly than $∣ z ∣^{- n}$ ; bounded $z$ are covered by compactness and smoothness on $K$ . Since $n > n - 1$ , this majorant is integrable over $R^{n - 1}$ . Also $∣ U_{g} (x) ∣ \leq ∥ g ∥_{\infty} \int_{R^{n - 1}} P_{H} (x, z) d z = ∥ g ∥_{\infty}$ by steps 3.1 and 3.2; in particular $U_{g}$ is absolutely convergent and bounded on $H$ .

4.2step 3.3F3F4F5cases

Uniqueness. Let $w$ be bounded and harmonic on $H$ , continuous on $\overline{H}$ , with $w = 0$ on $\partial H$ ; it suffices to show $w \equiv 0$ . If $w$ is complex-valued, apply the argument below separately to its real and imaginary parts, so assume $w$ is real-valued. Fix $p \in \partial H$ and a ball $B : = B_{ρ} (p)$ with $ρ > 0$ . Define $g_{ρ}$ on $\partial B$ by $g_{ρ} (y) = w (y)$ for $y_{n} \geq 0$ and $g_{ρ} (y) = - w (y^{'}, - y_{n})$ for $y_{n} < 0$ ; this is continuous on $\partial B$ because $w$ is continuous on $\overline{H}$ and $w = 0$ on the plane, where the two clauses agree. By [F3] let $W$ be the harmonic function on $B$ with trace $g_{ρ}$ ; since $g_{ρ}$ is odd under the reflection $σ (y) = (y^{'}, - y_{n})$ , the function $y \mapsto - W (σ (y))$ is harmonic on $B$ with the same trace $g_{ρ}$ (because $g_{ρ} \circ σ = - g_{ρ}$ ), so [F3] gives $W (σ (y)) = - W (y)$ : $W$ is odd. In particular $W = 0$ on the flat part $\partial B \cap \partial H$ , and on the upper half ball $B^{+} : = B \cap H$ both $W$ and $w$ are harmonic, continuous on the closure of $B^{+}$ , and agree on its boundary (the upper hemisphere carries $g_{ρ} = w$ , and the flat part carries $w = 0 = W$ ); the weak maximum principle [F4] applied to $W - w$ and to $w - W$ gives $W = w$ on $B^{+}$ . Therefore the odd extension $\tilde{w}$ of $w$ (namely $\tilde{w} (y) = w (y)$ for $y_{n} > 0$ and $\tilde{w} (y) = - w (y^{'}, - y_{n})$ for $y_{n} < 0$ ) coincides with the harmonic function $W$ on $B$ , hence is harmonic on a neighbourhood of $p$ ; as $p$ was arbitrary and $\tilde{w}$ is harmonic off the plane, $\tilde{w}$ is harmonic on all of $R^{n}$ . It is bounded by $∥ w ∥_{\infty}$ , so [F5] makes it constant, and its value at the plane is $0$ ; hence $\tilde{w} \equiv 0$ and $w \equiv 0$ .

5.1step 1.2step 4.1F7F8algebra

Smoothness and harmonicity. By step 4.1 the domination hypothesis of [F7] holds on every compact $K \subset H$ and all admissible derivatives, so induction over the coordinate directions as in [F7] gives $U_{g} \in C^{\infty} (H)$ with $D^{α} U_{g} (x) = \int_{R^{n - 1}} D_{x}^{α} P_{H} (x, z) g (z) d z$ . Moreover $Δ_{x} (w_{n} ∣ w ∣^{- n}) = 0$ for $w \neq 0$ : by [F8] and step 1.2, $Δ w_{n} = 0$ , $\nabla (w_{n}) = e_{n}$ , $\nabla ∣ w ∣^{- n} = - n ∣ w ∣^{- n - 2} w$ and $Δ ∣ w ∣^{- n} = 2 n ∣ w ∣^{- n - 2}$ , so the product rule gives $Δ (w_{n} ∣ w ∣^{- n}) = 2 e_{n} \cdot (- n ∣ w ∣^{- n - 2} w) + w_{n} \cdot 2 n ∣ w ∣^{- n - 2} = 0$ . Since $P_{H} (x, z) = \frac{2}{ω_{n - 1}} φ (x - (z, 0))$ with $φ (w) = w_{n} ∣ w ∣^{- n}$ and $x - (z, 0)$ never vanishes for $x \in H$ , the chain rule gives $Δ_{x} P_{H} (x, z) = 0$ for all $x \in H$ , $z \in R^{n - 1}$ , and therefore $Δ U_{g} (x) = \int_{R^{n - 1}} Δ_{x} P_{H} (x, z) g (z) d z = 0$ .

6.1step 2.1step 3.1step 3.2step 4.1step 5.1step 3.3step 4.2cases∎

If $v$ is any bounded harmonic function on $H$ , continuous on $\overline{H}$ , with trace $g$ , then $w : = v - U_{g}$ is bounded, harmonic by step 5.1, continuous on $\overline{H}$ and zero on the plane by step 3.3, so step 4.2 gives $w \equiv 0$ and $v = U_{g}$ ; for complex data both $U_{g}$ and the difference are complex, and the maximum-principle and Liouville steps were applied to the real and imaginary parts. Together with steps 2.1, 3.1, 3.2, 4.1, 5.1 and 3.3 this proves every clause of the statement.

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Interior derivative estimates for harmonic functions

Statement

Assume Countable Choice and $n \geq 2$ . Let $Ω \subseteq R^{n}$ be open, let $u$ be real or complex harmonic on $Ω$ , let $B_{r} (x) ⋐ Ω$ with $r > 0$ , and let $α$ be a multi-index. Then $∣ D^{α} u (x) ∣ \leq C_{n, α} r^{- n - ∣ α ∣} \int_{B_{r} (x)} ∣ u (y) ∣ d y,$ where the constant $C_{n, α}$ depends only on $n$ and $α$ , not on $u$ , $x$ , $r$ or $Ω$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , an open set $Ω \subseteq R^{n}$ , a harmonic $u$ on $Ω$ , a point $x \in Ω$ and a radius $r > 0$ with $\overline{B_{r} (x)} \subset Ω$ , and a multi-index $α$ .

[F1]

If $u \in C^{2} (Ω)$ and $Δ u = 0$ , then $u (x) = M_{u} (x, r)$ for every $B_{r} (x) ⋐ Ω$ , and consequently the ball mean value property $u (y) = \frac{1}{∣ B_{ρ} (y) ∣} \int_{B_{ρ} (y)} u$ holds whenever $B_{ρ} (y) ⋐ Ω$ (Spherical mean-value property for harmonic functions, Ball mean-value property for harmonic functions under Countable Choice).

[F2]

A continuous function on an open set with the ball mean value property lies in $C^{\infty}$ and is harmonic (Continuous ball-mean-value functions are harmonic).

[F3]

For $n \geq 3$ and continuous data $g$ on a sphere, the Poisson integral is the unique $C^{2} \cap C ({\overline{B}}_{R} (a))$ harmonic function on $B_{R} (a)$ with trace $g$ ; its kernel is $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ (Continuous Dirichlet problem on a ball, Poisson kernel of a Euclidean ball).

[F4]

For $n \geq 2$ and real smooth data $g \in C^{\infty} (\partial B_{R} (a))$ the unique $C^{\infty} \cap C ({\overline{B}}_{R} (a))$ harmonic function with trace $g$ is the Poisson integral with the same kernel formula (Smooth sphere data have a harmonic replacement under Countable Choice).

[F5]

On a measure space and an open parameter interval, differentiation under the integral sign holds when every integrand slice is integrable, the parameter derivative exists off a fixed measurable null set, its slices are measurable (with zero extension), and its modulus has one nonnegative measurable integrable majorant for all parameters off a fixed null set. Bounded continuous integrands on the compact sphere have finite surface integrals, and dominated convergence applies to measurable pointwise convergent families with an integrable majorant (Differentiation under the integral sign, Surface integration on compact C1 hypersurfaces, Dominated convergence).

[F6]

Calculus interface: sums, products and compositions of $C^{k}$ maps are $C^{k}$ ; $(t^{β})^{'} = β t^{β - 1}$ for $t > 0$ ; the chain rule and the product rule hold; $D^{α}$ is the iterated coordinate derivative of the multi-index notation, and $Δ = \sum_{i} \partial_{i} \partial_{i}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, Continuity and derivatives of positive-base real powers, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , $C^{k}$ maps and multi-index derivative notation in Euclidean space, The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F7]

For $n \geq 1$ and $s > 0$ , the closed ball and sphere are compact, the sphere is nonempty, and $∣ \partial B_{s} (x) ∣ = ω_{n - 1} s^{n - 1}$ and $∣ B_{s} (x) ∣ = ω_{n - 1} s^{n} / n$ . Compact Euclidean sets are closed and bounded, so the product of the closed ball ${∣ ζ ∣ \leq 1 / 2}$ and unit sphere, viewed in $R^{2 n}$ , is closed and bounded and hence compact; continuous functions on nonempty compact metric spaces attain extrema (Sphere and ball measures scale in Rn, For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, 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, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F8]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF8

Work under [F8] and suppose first that $u$ is real-valued. Put $ρ : = r / 4 > 0$ , so that $\overline{B_{2 ρ} (x)} \subset B_{r} (x) \subset Ω$ and $u$ is harmonic, hence $C^{2}$ , on a neighbourhood of $\overline{B_{2 ρ} (x)}$ .

1.2givenF1F7algebra

Mean-value bound on the inner sphere. For $y \in \partial B_{ρ} (x)$ and $w \in B_{ρ} (y)$ we have $∣ w - x ∣ \leq ∣ w - y ∣ + ∣ y - x ∣ < 2 ρ = r / 2 < r$ , so $B_{ρ} (y) \subset B_{r} (x)$ ; by [F1] and [F7], $∣ u (y) ∣ = ∣ \frac{1}{∣ B_{ρ} (y) ∣} \int_{B_{ρ} (y)} u ∣ \leq \frac{1}{∣ B_{ρ} ∣} \int_{B_{r} (x)} ∣ u ∣ = \frac{n}{ω_{n - 1} ρ^{n}} \int_{B_{r} (x)} ∣ u ∣$ .

2.1step 1.1F1F2F3F4cases

Representation on the inner ball. For $n \geq 3$ put $g : = u ∣_{\partial B_{ρ} (x)} \in C (\partial B_{ρ} (x))$ and note that $u \in C^{2} (B_{ρ} (x)) \cap C (\overline{B_{ρ} (x)})$ is harmonic with trace $g$ ; the uniqueness clause of [F3] gives $u (z) = U_{g} (z) = \int_{\partial B_{ρ} (x)} P_{ρ, x} (z, y) g (y) d S_{y}$ for $z \in B_{ρ} (x)$ . For $n = 2$ : $u$ is continuous on $Ω$ and has the ball mean value property by [F1], so [F2] makes it $C^{\infty}$ ; its restriction $g$ to the sphere is then real and $C^{\infty}$ , and $u$ is a $C^{\infty} \cap C (\overline{B_{ρ} (x)})$ harmonic function with trace $g$ , so uniqueness in [F4] gives $u (z) = \int_{\partial B_{ρ} (x)} \frac{ρ^{2} - ∣ z - x ∣^{2}}{ρ ω_{n - 1} ∣ z - y ∣^{n}} g (y) d S_{y}$ for $z \in B_{ρ} (x)$ . Thus in both dimensions $u$ on $B_{ρ} (x)$ is the Poisson integral of $g$ with the same kernel.

2.2step 1.1F6F7algebra

Kernel derivative bound. Write $z = x + ρ ζ$ and $y = x + ρ η$ with $∣ η ∣ = 1$ ; the kernel is $P_{ρ, x} (z, y) = ρ^{1 - n} (1 - ∣ ζ ∣^{2}) / (ω_{n - 1} ∣ ζ - η ∣^{n})$ , whose denominator is bounded below on the compact set ${∣ ζ ∣ \leq 1 / 2} \times {∣ η ∣ = 1}$ by $2^{- n}$ . For every multi-index $α$ , the partial derivatives $D_{ζ}^{α} [(1 - ∣ ζ ∣^{2}) ∣ ζ - η ∣^{- n}]$ are continuous by [F6] on that compact set and hence bounded in modulus by a constant $c_{n, α}$ by [F7]; rescaling gives, for $∣ ζ ∣ \leq 1 / 2$ , $∣ D_{z}^{α} P_{ρ, x} (z, y) ∣ = ρ^{1 - n - ∣ α ∣} ∣ D_{ζ}^{α} [(1 - ∣ ζ ∣^{2}) ∣ ζ - η ∣^{- n}] ∣ / ω_{n - 1} \leq C_{n, α} ρ^{1 - n - ∣ α ∣}$ .

3.1step 2.1step 2.2F5F6F7induction

Derivatives of $u$ . By step 2.1, $u (z) = \int_{\partial B_{ρ} (x)} P_{ρ, x} (z, y) g (y) d S_{y}$ . On $\overline{B_{ρ / 2} (x)} \times \partial B_{ρ} (x)$ every ordered $z$ -derivative of the smooth kernel is continuous and bounded, by the compactness argument of step 2.2. Multiplying by the bounded continuous $g$ gives Borel integrable slices; the next coordinate derivative has an integrable constant majorant on the finite sphere. Thus [F5] applies on each sufficiently small open coordinate interval, with no exceptional points. Induction over ordered coordinate derivatives, with dominated convergence for their continuity, gives $D^{α} u (z) = \int_{\partial B_{ρ} (x)} D_{z}^{α} P_{ρ, x} (z, y) g (y) d S_{y}$ for $z \in B_{ρ / 2} (x)$ , using the canonical order for $D^{α}$ . At $z = x$ step 2.2 then yields $∣ D^{α} u (x) ∣ \leq C_{n, α} ρ^{1 - n - ∣ α ∣} \int_{\partial B_{ρ} (x)} ∣ u (y) ∣ d S_{y}$ .

4.1step 3.1F7algebra

Bounding the boundary integral by the sphere area, step 3.1 and [F7] give $∣ D^{α} u (x) ∣ \leq C_{n, α} ρ^{1 - n - ∣ α ∣} \cdot ω_{n - 1} ρ^{n - 1} \cdot \sup_{y \in \partial B_{ρ} (x)} ∣ u (y) ∣ = C_{n, α} ω_{n - 1} ρ^{- ∣ α ∣} \sup_{\partial B_{ρ} (x)} ∣ u ∣$ .

5.1step 1.2step 4.1F7algebra

Substituting the mean-value bound of step 1.2 into step 4.1 yields $∣ D^{α} u (x) ∣ \leq C_{n, α} ω_{n - 1} ρ^{- ∣ α ∣} \cdot \frac{n}{ω_{n - 1} ρ^{n}} \int_{B_{r} (x)} ∣ u ∣ = C_{n, α} n ρ^{- n - ∣ α ∣} \int_{B_{r} (x)} ∣ u ∣ = C_{n, α} n 4^{n + ∣ α ∣} r^{- n - ∣ α ∣} \int_{B_{r} (x)} ∣ u ∣$ , and absorbing $n 4^{n + ∣ α ∣}$ into the constant gives the displayed estimate with a constant depending only on $n$ and $α$ .

6.1step 5.1casesalgebra

For complex $u$ , apply steps 1.1–5.1 to $R e u$ and to $I m u$ , which are real harmonic functions on $Ω$ with $\overline{B_{r} (x)} \subset Ω$ : $∣ D^{α} u (x) ∣ \leq ∣ D^{α} R e u (x) ∣ + ∣ D^{α} I m u (x) ∣ \leq C_{n, α} r^{- n - ∣ α ∣} \int_{B_{r} (x)} (∣ R e u ∣ + ∣ I m u ∣) \leq 2 C_{n, α} r^{- n - ∣ α ∣} \int_{B_{r} (x)} ∣ u ∣$ , and $2 C_{n, α}$ again depends only on $n$ and $α$ .

7.1step 5.1step 6.1algebra∎

Steps 5.1 and 6.1 give the estimate for real and complex $u$ with a constant independent of $u, x, r, Ω$ ; the value $r > 0$ is unavoidable because the estimate divides by $r$ , and the hypothesis $\overline{B_{r} (x)} \subset Ω$ was used only to place $B_{2 ρ} (x)$ and the mean-value balls inside $Ω$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Harmonic Cauchy estimates in supremum norm

Statement

Assume Countable Choice and $n \geq 2$ . Let $u$ be real or complex harmonic on an open $Ω \subseteq R^{n}$ with $B_{r} (x) ⋐ Ω$ , $r > 0$ , and let $α$ be a multi-index. Then $∣ D^{α} u (x) ∣ \leq C_{n, α}^{'} r^{- ∣ α ∣} \sup_{B_{r} (x)} ∣ u ∣,$ with $C_{n, α}^{'}$ depending only on $n$ and $α$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , an open set $Ω$ , a harmonic $u$ on $Ω$ , $x \in Ω$ , $r > 0$ with $\overline{B_{r} (x)} \subset Ω$ , and a multi-index $α$ .

[F1]

Under these hypotheses, $∣ D^{α} u (x) ∣ \leq C_{n, α} r^{- n - ∣ α ∣} \int_{B_{r} (x)} ∣ u (y) ∣ d y$ with $C_{n, α}$ independent of $u, x, r, Ω$ (Interior derivative estimates for harmonic functions).

[F2]

For $n \geq 1$ and $r > 0$ , $∣ B_{r} ∣ = ω_{n - 1} r^{n} / n$ , finite and positive (Sphere and ball measures scale in Rn).

[F3]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF3

Work under [F3] and set $M : = \sup_{B_{r} (x)} ∣ u ∣ \in [0, + \infty]$ . Since $∣ u ∣$ is continuous and $B_{r} (x)$ is bounded, the integral $\int_{B_{r} (x)} ∣ u ∣$ is defined in $[0, + \infty]$ .

2.1step 1.1F2algebra

If $M < + \infty$ , then $∣ u (y) ∣ \leq M$ for every $y \in B_{r} (x)$ , so by monotonicity of the integral $\int_{B_{r} (x)} ∣ u ∣ \leq M ∣ B_{r} (x) ∣ = M ω_{n - 1} r^{n} / n$ by [F2].

3.1step 2.1F1F2algebra

Substituting step 2.1 into [F1] gives $∣ D^{α} u (x) ∣ \leq C_{n, α} r^{- n - ∣ α ∣} \cdot M ω_{n - 1} r^{n} / n = (C_{n, α} ω_{n - 1} / n) r^{- ∣ α ∣} M$ , so the stated estimate holds with $C_{n, α}^{'} : = C_{n, α} ω_{n - 1} / n$ , a constant depending only on $n$ and $α$ .

4.1step 1.1step 3.1cases∎

If $M = + \infty$ the right-hand side of the stated inequality is $+ \infty$ while $∣ D^{α} u (x) ∣$ is a finite real number, so the inequality holds trivially; for the local applications of this estimate one always takes a compactly contained ball on which $∣ u ∣$ , being continuous, is bounded, so the case $M = + \infty$ never carries mathematical content.

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Interior estimate for the Poisson equation with Hölder data

Statement

Assume Countable Choice, $n \geq 2$ and $0 < α < 1$ . Let $u \in C^{2} (B_{r} (a)) \cap L^{\infty} (B_{r} (a))$ and let $f \in C^{0, α} (B_{r} (a))$ have finite Hölder seminorm, with $- Δ u = f$ pointwise on $B_{r} (a)$ . Then $u \in C^{2, α} (B_{r / 2} (a))$ and $∥ u ∥_{2, α; B_{r / 2} (a)}^{*} \leq C_{n, α} (∥ u ∥_{\infty; B_{r} (a)} + r^{2} ∥ f ∥_{\infty; B_{r} (a)} + r^{2 + α} [f]_{0, α; B_{r} (a)}),$ with $C_{n, α}$ independent of $r$ , $a$ , $u$ and $f$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , $0 < α < 1$ , a centre $a \in R^{n}$ , a radius $r > 0$ , a function $u \in C^{2} (B_{r} (a)) \cap L^{\infty} (B_{r} (a))$ and $f \in C^{0, α} (B_{r} (a))$ with $- Δ u = f$ pointwise and $∥ f ∥_{0, α; B_{r} (a)}^{*} < \infty$ .

[F1]

The local Hölder and scaled $C^{2, α}$ quantities are $[f]_{0, α; B} : = \sup_{x \neq y} ∣ f (x) - f (y) ∣ / ∣ x - y ∣^{α}$ and $∥ w ∥_{2, α; B}^{*} = \sum_{j = 0}^{2} ρ^{j} \max_{∣ γ ∣ = j} \sup_{B} ∣ D^{γ} w ∣ + ρ^{2 + α} \max_{∣ γ ∣ = 2} [D^{γ} w]_{0, α; B}$ on a ball $B$ of radius $ρ$ ; under the scaling $v (z) = w (a + ρ z)$ one has $∥ v ∥_{2, α; B_{1} (0)}^{*} = ∥ w ∥_{2, α; B_{ρ} (a)}^{*}$ (Local Hölder and scaled C-two-alpha norms on balls).

[F2]

The Newtonian potential is $N G (x) = \int_{R^{n}} Φ (x - y) G (y) d y$ (Newtonian potential of compactly supported data).

[F3]

For $0 < α < 1$ and $G \in C_{c}^{0, α} (R^{n}; C)$ with finite global seminorm, the Newtonian potential $w = N G$ is $C^{2}$ with $- Δ w = G$ , its second derivatives are locally $α$ -Hölder, and for every compact $K$ the $C^{2, α}$ size of $w$ on $K$ is bounded by a constant times $∥ G ∥_{C^{0, α}} = \sup ∣ G ∣ + [G]_{α; R^{n}}$ (Hölder data give a classical Newtonian solution).

[F4]

For $0 < ρ < R$ there is a smooth $η : R^{n} \to [0, 1]$ with $η = 1$ on ${\overline{B}}_{ρ} (0)$ and $supp η \subseteq B_{R} (0)$ (A smooth bump between concentric Euclidean balls).

[F5]

If $u$ is harmonic on an open set containing $\overline{B_{R} (y)}$ , then $∣ D^{γ} u (y) ∣ \leq C_{n, γ} R^{- n - ∣ γ ∣} \int_{B_{R} (y)} ∣ u ∣$ for every multi-index $γ$ (Interior derivative estimates for harmonic functions).

[F6]

$∣ B_{ρ} ∣ = ω_{n - 1} ρ^{n} / n$ (Sphere and ball measures scale in Rn).

[F7]

The Laplacian is $Δ = \sum_{i} \partial_{i} \partial_{i}$ (Fundamental solution for the positive operator minus Laplacian).

[F8]

The chain rule computes derivatives of compositions (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F9]

The real mean value theorem applies to a real-valued function continuous on a segment and differentiable in its interior (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 $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F10]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F7F8algebra

Work under [F10]. Rescale to the unit ball: put $v (z) : = u (a + r z)$ and $F (z) : = r^{2} f (a + r z)$ for $z \in B_{1} (0)$ . Differentiating twice with the chain rule [F8] and Laplacian convention [F7] gives $Δ v (z) = r^{2} (Δ u) (a + r z) = - r^{2} f (a + r z) = - F (z)$ , so $- Δ v = F$ pointwise on $B_{1} (0)$ ; moreover $∥ v ∥_{\infty; B_{1} (0)} = ∥ u ∥_{\infty; B_{r} (a)}$ , $∥ F ∥_{\infty; B_{1} (0)} = r^{2} ∥ f ∥_{\infty; B_{r} (a)}$ and $[F]_{0, α; B_{1} (0)} = r^{2 + α} [f]_{0, α; B_{r} (a)}$ by [F1].

2.1step 1.1F4algebra

Cutoff. By [F4] fix a smooth $η : R^{n} \to [0, 1]$ with $η = 1$ on ${\overline{B}}_{3 / 4} (0)$ and $supp η \subseteq B_{7 / 8} (0)$ , and let $G : = η F$ on $B_{7 / 8} (0)$ , extended by $0$ to all of $R^{n}$ . Then $G$ is continuous and compactly supported, and its global Hölder seminorm satisfies $[G]_{α; R^{n}} \leq C_{n, α} (∥ F ∥_{\infty; B_{1} (0)} + [F]_{0, α; B_{1} (0)})$ : for $x, y$ in the support one uses $∣ G (x) - G (y) ∣ \leq ∣ η (x) ∣ ∣ F (x) - F (y) ∣ + ∣ η (x) - η (y) ∣ ∣ F (y) ∣$ and the smoothness of the fixed cutoff, while if one point lies outside the support the estimate follows from $∣ G ∣ \leq ∥ F ∥_{\infty}$ , the vanishing of $η$ at the support boundary and $∣ x - y ∣^{α} \geq ∣ x - y ∣$ for $∣ x - y ∣ \leq 1$ ; the constant depends only on the fixed cutoff, hence only on $n$ and $α$ .

3.1step 2.1F2F3algebra

The Newtonian potential. Put $w : = N G$ using [F2]; by [F3] the potential is $C^{2}$ on $R^{n}$ with $- Δ w = G$ , and on the compact set $K : = {\overline{B}}_{3 / 4} (0)$ its size is controlled: $\sum_{j = 0}^{2} \max_{∣ γ ∣ = j} \sup_{K} ∣ D^{γ} w ∣ + \max_{∣ γ ∣ = 2} [D^{γ} w]_{0, α; K} \leq C_{n, α} (∥ G ∥_{\infty} + [G]_{α; R^{n}}) \leq C_{n, α}^{'} (∥ F ∥_{\infty; B_{1} (0)} + [F]_{0, α; B_{1} (0)})$ by step 2.1 and the size bounds on $η$ .

4.1step 1.1step 3.1F7algebra

The remainder is harmonic. Since $η = 1$ on ${\overline{B}}_{3 / 4} (0)$ , we have $G = F$ on $B_{3 / 4} (0)$ , so $- Δ (v - w) = - F + G = 0$ there by steps 1.1 and 3.1; thus $h : = v - w$ is harmonic on $B_{3 / 4} (0)$ . Moreover $∥ h ∥_{\infty; B_{3 / 4} (0)} \leq ∥ v ∥_{\infty; B_{1} (0)} + ∥ w ∥_{\infty; B_{3 / 4} (0)} \leq ∥ u ∥_{\infty; B_{r} (a)} + C_{n, α}^{'} (∥ F ∥_{\infty; B_{1}} + [F]_{0, α; B_{1}})$ by step 3.1.

5.1step 4.1F5F6F8F9algebra

Estimates for the harmonic part. For every $y \in B_{1 / 2} (0)$ , the closed ball ${\overline{B}}_{1 / 8} (y)$ lies in $B_{5 / 8} (0) \subset B_{3 / 4} (0)$ , where $h$ is harmonic. Applying [F5] with radius $1 / 8$ gives, for every multi-index $γ$ with $∣ γ ∣ \leq 3$ , $∣ D^{γ} h (y) ∣ \leq C_{n, γ} 8^{n + ∣ γ ∣} \int_{B_{1 / 8} (y)} ∣ h ∣ \leq C_{n}^{''} ∥ h ∥_{\infty; B_{3 / 4}}$ , using [F6] to bound the ball's volume. If $∣ γ ∣ = 2$ and $x, y \in B_{1 / 2} (0)$ , their segment stays in $B_{1 / 2} (0)$ . Apply the real mean value theorem [F9] separately to the real and imaginary parts of $t \mapsto D^{γ} h (x + t (y - x))$ on $[0, 1]$ (only the real part is needed when $h$ is real); the chain rule [F8] and the bounds just obtained for derivatives of order three then give $∣ D^{γ} h (x) - D^{γ} h (y) ∣ \leq C_{n}^{''} ∣ x - y ∣ ∥ h ∥_{\infty; B_{3 / 4}}$ . Since $0 < ∣ x - y ∣ < 1$ implies $∣ x - y ∣ \leq ∣ x - y ∣^{α}$ for $0 < α < 1$ , this bounds $[D^{γ} h]_{0, α; B_{1 / 2}}$ by $C_{n}^{''} ∥ h ∥_{\infty; B_{3 / 4}}$ ; the radius factors for the scaled norm on $B_{1 / 2}$ only change the constant.

6.1step 3.1step 4.1step 5.1algebra

Combining on the half ball. By step 3.1 the derivatives of $w$ through order two are bounded on ${\overline{B}}_{1 / 2} (0) \subset K$ by $C_{n, α}^{'} (∥ F ∥_{\infty} + [F]_{0, α})$ , and its second derivatives have $α$ -Hölder seminorm on ${\overline{B}}_{1 / 2} (0)$ bounded by the same quantity; step 5.1 gives the corresponding bounds for $h$ by $C_{n}^{''} ∥ h ∥_{\infty; B_{3 / 4}}$ , which step 4.1 bounds by $∥ u ∥_{\infty; B_{r} (a)} + C_{n, α}^{'} (∥ F ∥_{\infty} + [F]_{0, α})$ ; summing, $∥ v ∥_{2, α; B_{1 / 2} (0)}^{*} \leq C_{n, α} (∥ u ∥_{\infty; B_{r} (a)} + ∥ F ∥_{\infty; B_{1} (0)} + [F]_{0, α; B_{1} (0)})$ .

7.1step 1.1step 6.1F1algebra

Undoing the scaling. The scaling identity of [F1] applied to the sub-ball of radius $1 / 2$ gives $∥ u ∥_{2, α; B_{r / 2} (a)}^{*} = ∥ v ∥_{2, α; B_{1 / 2} (0)}^{*}$ , and step 1.1 converts $∥ F ∥_{\infty} + [F]_{0, α}$ into $r^{2} ∥ f ∥_{\infty; B_{r} (a)} + r^{2 + α} [f]_{0, α; B_{r} (a)}$ ; hence step 6.1 gives exactly the displayed estimate with a constant depending only on $n$ and $α$ . In particular $∥ u ∥_{2, α; B_{r / 2} (a)}^{*} < \infty$ , so $u \in C^{2, α} (B_{r / 2} (a))$ .

8.1step 2.1step 3.1step 4.1F3F5∎

The quantitative estimate for the potential and the identity $- Δ w = G$ come from [F3], and the estimate for the harmonic remainder comes from [F5]. Although [F3] also gives a cancellation formula for the singular Hessian, the proof uses its stated $C^{2, α}$ bound and does not differentiate $F$ ; neither the weak maximum principle nor a ball Dirichlet theorem is needed.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Interior oscillation controls the harmonic gradient

Statement

Assume Countable Choice and $n \geq 2$ . Let $u$ be real or complex harmonic on an open set $Ω \subseteq R^{n}$ with $\overline{B_{r} (x)} \subset Ω$ , $r > 0$ . Then $∣ D u (x) ∣ \leq C_{n} r^{- 1} {osc}_{B_{r} (x)} u, {osc}_{B_{r} (x)} u : = \sup_{y, z \in B_{r} (x)} ∣ u (y) - u (z) ∣ .$ The constant $C_{n}$ depends only on $n$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , an open set $Ω$ , a harmonic $u$ on $Ω$ , a point $x \in Ω$ and $r > 0$ with $\overline{B_{r} (x)} \subset Ω$ .

[F1]

For harmonic $v$ on an open set containing $\overline{B_{r} (x)}$ and every multi-index $α$ , $∣ D^{α} v (x) ∣ \leq C_{n, α}^{'} r^{- ∣ α ∣} \sup_{B_{r} (x)} ∣ v ∣$ (Harmonic Cauchy estimates in supremum norm).

[F2]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF2algebra

Work under [F2] and put $v : = u - u (x)$ , which is harmonic on $Ω$ with the same derivatives as $u$ , in particular $D v (x) = D u (x)$ ; moreover $∣ v (y) ∣ = ∣ u (y) - u (x) ∣ \leq {osc}_{B_{r} (x)} u$ for every $y \in B_{r} (x)$ , so $\sup_{B_{r} (x)} ∣ v ∣ \leq {osc}_{B_{r} (x)} u$ .

2.1step 1.1F1algebra

Apply [F1] with $α = e_{i}$ to the harmonic function $v$ on $B_{r} (x)$ : $∣ \partial_{i} u (x) ∣ = ∣ \partial_{i} v (x) ∣ \leq C_{n, e_{i}}^{'} r^{- 1} \sup_{B_{r} (x)} ∣ v ∣ \leq C_{n, e_{i}}^{'} r^{- 1} {osc}_{B_{r} (x)} u$ , with a constant depending only on $n$ and the coordinate; taking $C_{n} : = \max_{i} C_{n, e_{i}}^{'}$ gives $∣ \partial_{i} u (x) ∣ \leq C_{n} r^{- 1} {osc}_{B_{r} (x)} u$ for every $i$ .

3.1step 2.1algebra∎

Summing the coordinate bounds, $∣ D u (x) ∣ = (\sum_{i} ∣ \partial_{i} u (x) ∣^{2})^{1 / 2} \leq \sqrt{n} \max_{i} ∣ \partial_{i} u (x) ∣ \leq \sqrt{n} C_{n} r^{- 1} {osc}_{B_{r} (x)} u$ ; absorbing $\sqrt{n}$ into the constant gives the assertion with a constant depending only on $n$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Entire harmonic functions of sublinear growth are constant

Statement

Assume Countable Choice and $n \geq 2$ . Use one-based basis labels $e_{i} : = e_{i - 1}^{c a n}$ for $1 \leq i \leq n$ . Let $u : R^{n} \to R$ or $C$ be harmonic. If $\lim_{R \to \infty} R^{- 1} \sup_{B_{R} (0)} ∣ u ∣ = 0,$ then $u$ is constant.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , a harmonic $u$ on all of $R^{n}$ with $\lim_{R \to \infty} R^{- 1} \sup_{B_{R} (0)} ∣ u ∣ = 0$ .

[F1]

Under the compact-ball hypotheses, $∣ D^{α} u (x) ∣ \leq C_{n, α}^{'} r^{- ∣ α ∣} \sup_{B_{r} (x)} ∣ u ∣$ (Harmonic Cauchy estimates in supremum norm).

[F2]

A continuous function on an interval whose derivative vanishes at every interior point is constant there (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant); the chain rule computes the derivative of a restriction to a line (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F3]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF3algebra

Work under [F3] and fix $x \in R^{n}$ . For every $R > ∣ x ∣$ put $ρ_{R} : = (R - ∣ x ∣) / 2 > 0$ . If $y \in {\overline{B}}_{ρ_{R}} (x)$ , then $∣ y ∣ \leq ∣ y - x ∣ + ∣ x ∣ \leq ρ_{R} + ∣ x ∣ = (R + ∣ x ∣) / 2 < R$ , so ${\overline{B}}_{ρ_{R}} (x) \subset B_{R} (0)$ .

2.1step 1.1F1algebra

Apply [F1] with radius $ρ_{R}$ and coordinate multi-index $α = e_{i}$ for each $i = 1, \dots, n$ . With $K_{n} : = 2 \sqrt{n} \max_{i} C_{n, e_{i}}^{'}$ , step 1.1 gives $∣ D u (x) ∣ \leq \sqrt{n} \max_{i} ∣ \partial_{i} u (x) ∣ \leq \sqrt{n} \max_{i} C_{n, e_{i}}^{'} ρ_{R}^{- 1} \sup_{B_{ρ_{R}} (x)} ∣ u ∣ \leq K_{n} \frac{R}{R - ∣ x ∣} R^{- 1} \sup_{B_{R} (0)} ∣ u ∣$ for every $R > ∣ x ∣$ .

3.1step 2.1algebra

Letting $R \to \infty$ in step 2.1, the factor $R / (R - ∣ x ∣) \to 1$ and $R^{- 1} \sup_{B_{R} (0)} ∣ u ∣ \to 0$ by hypothesis, so $∣ D u (x) ∣ \leq \lim_{R \to \infty} K_{n} \frac{R}{R - ∣ x ∣} R^{- 1} \sup_{B_{R} (0)} ∣ u ∣ = 0$ ; hence $D u (x) = 0$ for every $x \in R^{n}$ .

4.1step 3.1F2∎

Therefore $u$ is constant: for fixed $x \in R^{n}$ and each coordinate $i$ , if $u$ is real-valued then $t \mapsto u (x + t e_{i})$ has zero derivative for every real $t$ by step 3.1, so it is constant on $R$ by [F2]; if $u$ is complex-valued, apply [F2] separately to the real and imaginary parts of this line restriction, whose derivatives also vanish by step 3.1. Thus each coordinate line restriction is constant, and changing the coordinates one at a time connects any two points of $R^{n}$ , so $u$ has the same value everywhere. The chain rule identifies each line derivative with the corresponding partial derivative. No bounded-Liouville theorem is invoked; the sublinear growth hypothesis is used exactly in step 3.1.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Locally uniform limits of harmonic functions are smooth, with all derivatives converging

Statement

Assume Countable Choice and $n \geq 2$ . Let $Ω \subseteq R^{n}$ be open and let $u_{j} : Ω \to R$ or $C$ be harmonic with $u_{j} \to u$ locally uniformly on $Ω$ . Then $u$ is smooth and harmonic, and for every compact $K \subset Ω$ and every multi-index $α$ , $\sup_{x \in K} ∣ D^{α} u_{j} (x) - D^{α} u (x) ∣ ⟶ 0.$

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , an open set $Ω \subseteq R^{n}$ , harmonic functions $u_{j}$ on $Ω$ converging locally uniformly to $u$ , a compact set $K \subset Ω$ and a multi-index $α$ .

[F1]

A locally uniform limit of harmonic functions is harmonic (Locally uniform limits of harmonic functions are harmonic).

[F2]

Every harmonic function is real analytic and hence $C^{\infty}$ , so all derivatives $D^{α} u$ exist (Harmonic functions are real analytic).

[F3]

Supremum Cauchy estimates: for harmonic $v$ on an open set containing $\overline{B_{ρ} (x)}$ , $∣ D^{α} v (x) ∣ \leq C_{n, α}^{'} ρ^{- ∣ α ∣} \sup_{B_{ρ} (x)} ∣ v ∣$ (Harmonic Cauchy estimates in supremum norm).

[F4]

A compact set and a disjoint closed set in a normed space keep a positive distance (A compact set and a disjoint closed set have a positive norm-distance gap), and a closed bounded subset of $R^{n}$ 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).

[F5]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F2F5

Work under [F5]. By [F1] the limit $u$ is harmonic, and hence $C^{\infty}$ with all derivatives existing, by [F2].

2.1step 1.1F4cases

If $K = \emptyset$ , the uniform-convergence assertion on $K$ is vacuous, so assume $K \neq \emptyset$ . If $Ω \neq R^{n}$ , its complement is nonempty, closed and disjoint from $K$ , so [F4] gives $δ : = dist (K, R^{n} ∖ Ω) > 0$ ; put $ρ : = δ / 2$ . If $Ω = R^{n}$ , put $ρ : = 1$ . In either case let $K_{ρ} : = {x \in R^{n} : dist (x, K) \leq ρ}$ . Since $K$ is compact, it is bounded; the distance function is continuous, so $K_{ρ}$ is closed and bounded and hence compact by [F4]. In the first case $K_{ρ} \subseteq Ω$ , since every point of the complement has distance at least $δ$ from $K$ ; in the second case this inclusion is automatic. By local uniform convergence, $ε_{j} : = \sup_{K_{ρ}} ∣ u_{j} - u ∣ \to 0$ .

3.1step 1.1step 2.1F2F3

The difference $u_{j} - u$ is harmonic on $Ω \supseteq \overline{B_{ρ} (x)}$ for every $x \in K$ , so [F2] and [F3] give $∣ D^{α} u_{j} (x) - D^{α} u (x) ∣ \leq C_{n, α}^{'} ρ^{- ∣ α ∣} \sup_{B_{ρ} (x)} ∣ u_{j} - u ∣ \leq C_{n, α}^{'} ρ^{- ∣ α ∣} ε_{j}$ , a bound independent of $x \in K$ .

4.1step 3.1algebra

Taking the supremum over $x \in K$ in step 3.1 gives $\sup_{K} ∣ D^{α} u_{j} - D^{α} u ∣ \leq C_{n, α}^{'} ρ^{- ∣ α ∣} ε_{j} \to 0$ as $j \to \infty$ , for the arbitrary compact $K$ and multi-index $α$ ; this proves the derivative convergence.

5.1step 1.1step 4.1cases∎

Complex-valued $u_{j}$ are handled by applying the argument to real and imaginary parts, whose differences are harmonic and whose absolute values control $∣ D^{α} u_{j} - D^{α} u ∣ \leq ∣ D^{α} R e (u_{j} - u) ∣ + ∣ D^{α} I m (u_{j} - u) ∣$ ; the constant is doubled. Together with step 1.1 this proves that $u$ is smooth harmonic and that every derivative converges uniformly on compacta.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Harmonic functions are real analytic

Statement

Assume Countable Choice and $n \geq 2$ . Every real or complex harmonic $u$ on an open set $Ω \subseteq R^{n}$ is real analytic: for every $a \in Ω$ there is $ρ > 0$ such that $u (a + h) = \sum_{α \in N^{n}} \frac{D^{α} u (a)}{α!} h^{α}$ with absolute convergence whenever $∣ h ∣ < ρ$ . In particular, if $\overline{B_{2 r} (a)} \subset Ω$ and $M = \sup_{B_{2 r} (a)} ∣ u ∣$ , then $∣ D^{α} u (a) ∣ \leq M C_{n}^{∣ α ∣} ∣ α ∣! r^{- ∣ α ∣}$ for every multi-index $α$ , with $C_{n}$ depending only on $n$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , an open set $Ω \subseteq R^{n}$ , a real or complex harmonic $u$ on $Ω$ , and a point $a \in Ω$ .

[F1]

For $n \geq 3$ the Poisson kernel of $B_{R} (a)$ is $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ , positive with unit mass (Poisson kernel of a Euclidean ball, The ball Poisson kernel is positive and has unit mass); the continuous Dirichlet problem on a ball is uniquely solved by the Poisson integral, for real and complex data (Continuous Dirichlet problem on a ball).

[F2]

Differentiation under the integral sign and dominated convergence for integrals over the compact sphere (Differentiation under the integral sign, Surface integration on compact C1 hypersurfaces, Dominated convergence).

[F3]

The multivariable Taylor formula with Lagrange remainder: for $f \in C^{k + 1} (U)$ on an open convex $U ∋ a, a + h$ there is $θ \in (0, 1)$ with $f (a + h) = T_{k} f (a; h) + \sum_{∣ α ∣ = k + 1} D^{α} f (a + θ h) h^{α} / α!$ (Multivariable Taylor formula with a Lagrange remainder along a line segment), and the multinomial theorem gives $\sum_{∣ α ∣ = k} 1 / α! = n^{k} / k!$ by evaluating the expansion of $(x_{1} + \dots + x_{n})^{k}$ at $x_{i} = 1$ (The multinomial coefficient equals $n! / \prod_{i < m} k_{i}!$ , and $(x_{0} + \dots + x_{m - 1})^{n} = \sum ι ⁣ (\binom{n}{k}) \prod_{i < m} x_{i}^{k_{i}}$ in $R$ ).

[F4]

Real analyticity means representation by an absolutely convergent multi-indexed power series $f (x) = \sum_{α} c_{α} (x - a)^{α}$ with $c_{α} = D^{α} f (a) / α!$ on a polydisc (Real analytic germs in several variables, Multi-indexed power series in $C^{m}$ and their absolute convergence).

[F5]

Sphere and ball measures: $∣ \partial B_{R} ∣ = ω_{n - 1} R^{n - 1}$ ; multi-index notation $D^{α}$ , $α!$ , $h^{α}$ (Sphere and ball measures scale in Rn, $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F6]

Calculus interface for the kernel computation: chain rule, product rule, real-power derivatives and closure of $C^{k}$ maps under algebra and composition (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Continuity and derivatives of positive-base real powers, $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[F7]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F8]

Closed Euclidean balls and spheres of positive radius are compact; compact Euclidean subsets are closed and bounded and closed bounded subsets are compact; continuous real-valued functions on nonempty compact metric spaces attain their extrema. Thus the closed ball used in step 1.1 is compact, its continuous $u$ is bounded there, and the compact product of the closed interior ball with the boundary sphere in step 2.1 supports the uniform derivative bounds (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, 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, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F9]

Under Countable Choice, a classical harmonic function has the ball mean-value property, and a continuous function with that property is $C^{\infty}$ (Ball mean-value property for harmonic functions under Countable Choice, Continuous ball-mean-value functions are harmonic).

[F10]

For $n \geq 2$ , real $C^{\infty}$ data on a sphere have a unique smooth harmonic replacement on the ball, given by the explicit Poisson kernel formula (Smooth sphere data have a harmonic replacement under Countable Choice).

Proof

technique · direct

1.1givenF7F8

Work under [F7] and suppose first that $u$ is real. Since $Ω$ is open and $a \in Ω$ , choose $r > 0$ with $\overline{B_{2 r} (a)} \subset Ω$ . Then $u$ is continuous on the compact set $\overline{B_{2 r} (a)}$ , so $M : = \sup_{B_{2 r} (a)} ∣ u ∣$ is a finite nonnegative number.

2.1step 1.1F1F2F8F9F10

Poisson representation and derivative bounds from the kernel. For $n \geq 3$ , $u$ equals the Poisson integral of its trace $g : = u ∣_{\partial B_{2 r} (a)}$ on $B_{2 r} (a)$ by [F1], since both functions are $C^{2} \cap C (\overline{B_{2 r} (a)})$ harmonic with trace $g$ . For $n = 2$ , [F9] makes $u$ smooth on a neighbourhood of $\overline{B_{2 r} (a)}$ , so $g$ is smooth; [F10] then gives the same Poisson representation and uniqueness. In both cases the kernel is $P_{2 r, a}$ . Differentiating the representation through the integral by [F2] (for $x$ in the compact ball $\overline{B_{r} (a)}$ the sphere is separated from $x$ , and all kernel derivatives are bounded there), we get $D^{α} u (x) = \int_{\partial B_{2 r} (a)} D_{x}^{α} P_{2 r, a} (x, y) u (y) d S_{y}$ for every multi-index $α$ and every $x \in B_{r} (a)$ .

3.1step 2.1F6algebra

Kernel derivative bound. Write $x = a + 2 r ζ$ , $y = a + 2 r η$ , with $∣ ζ ∣ \leq 1 / 2$ and $∣ η ∣ = 1$ . Then $P_{2 r, a} (x, y) = (2 r)^{1 - n} g (ζ, η) / ω_{n - 1}$ , where $g = (1 - ∣ ζ ∣^{2}) ∣ ζ - η ∣^{- n}$ . Fix $ζ_{0}$ with $∣ ζ_{0} ∣ \leq 1 / 2$ , put $d = ζ_{0} - η$ and $A = ∣ d ∣^{2} \geq 1 / 4$ , and write $ζ = ζ_{0} + u$ . Then $∣ ζ - η ∣^{- n} = A^{- n / 2} (1 + P_{0} (u))^{- n / 2}, P_{0} (u) = A^{- 1} (2 d \cdot u + ∣ u ∣^{2}) .$ For $∣ u ∣ \leq 1 / 4$ , $∣ P_{0} (u) ∣ \leq 16 ∣ u ∣$ , so the binomial series for $(1 + P_{0})^{- n / 2}$ converges near $u = 0$ , for example when $∣ u ∣ < 1 / 32$ . Its coefficients satisfy $∣ (\binom{- n / 2}{m}) ∣ = \prod_{j = 1}^{m} (1 + (n / 2 - 1) / j) \leq (n / 2)^{m}$ . The coefficients of the linear and quadratic terms of $P_{0}$ are bounded by $12$ and $4$ , respectively, and it has at most $2 n$ monomials. For total degree $k \geq 1$ , only powers $m \leq k$ contribute; counting at most $(2 n)^{m}$ products in $P_{0}^{m}$ , then multiplying by the degree-two polynomial $1 - ∣ ζ_{0} + u ∣^{2}$ and by $A^{- n / 2} \leq 2^{n}$ , bounds each Taylor coefficient of $g$ of total degree $k$ by $C_{2}^{k}$ for a constant $C_{2} (n)$ . Since $D^{α} g (ζ_{0}, η) = α!$ times its $u^{α}$ coefficient and $α! \leq ∣ α ∣!$ , this gives $∣ D_{ζ}^{α} g ∣ \leq C_{2}^{∣ α ∣} ∣ α ∣!$ for $∣ α ∣ \geq 1$ , uniformly in $ζ_{0}, η$ . Thus for $k = ∣ α ∣ \geq 1$ , $∣ D_{x}^{α} P_{2 r, a} (x, y) ∣ \leq (2 r)^{1 - n - k} ω_{n - 1}^{- 1} C_{2}^{k} k!$ .

4.1step 2.1step 3.1F5algebra

Factorial derivative bound on the inner ball. For $k : = ∣ α ∣ \geq 1$ , combining steps 2.1 and 3.1 with [F5] gives, for $x \in B_{r} (a)$ , $∣ D^{α} u (x) ∣ \leq C_{2}^{k} k! (2 r)^{1 - n - k} ω_{n - 1}^{- 1} \int_{\partial B_{2 r} (a)} ∣ u ∣ d S \leq M C_{2}^{k} k! (2 r)^{- k} .$ For $k = 0$ , the bound $∣ u (x) ∣ \leq M$ follows directly from the definition of $M$ .

5.1step 2.1step 4.1F3algebra

Taylor remainder. Let $k \geq 0$ and let $h$ satisfy $∣ h ∣ < r$ . The ball $B_{r} (a)$ is convex and open, contains $a$ and $a + h$ , and $u$ is $C^{k + 1}$ on it by step 2.1, so [F3] gives some $θ \in (0, 1)$ with $u (a + h) - T_{k} u (a; h) = \sum_{∣ α ∣ = k + 1} D^{α} u (a + θ h) h^{α} / α!$ . Since $a + θ h \in B_{r} (a)$ , step 4.1 bounds each term by $M (2 r)^{- (k + 1)} C_{2}^{k + 1} (k + 1)! ∣ h^{α} ∣ / α!$ , and $\sum_{∣ α ∣ = k + 1} ∣ h^{α} ∣ / α! \leq \sum_{∣ α ∣ = k + 1} ∣ h ∣^{k + 1} / α! = n^{k + 1} ∣ h ∣^{k + 1} / (k + 1)!$ by [F3]; the two $(k + 1)!$ factors cancel and the remainder is at most $M (n C_{2} ∣ h ∣ / (2 r))^{k + 1}$ .

5.2step 4.1algebra

The factorial bound. If $\overline{B_{2 r} (a)} \subset Ω$ and $M = \sup_{B_{2 r} (a)} ∣ u ∣$ , step 4.1 gives the claimed estimate for $∣ α ∣ \geq 1$ with $C_{n} : = C_{2} / 2$ ; for $∣ α ∣ = 0$ it is $∣ u (a) ∣ \leq M$ . The constant depends only on $n$ .

6.1step 4.1step 5.1F3F4F5algebra

Convergence and analyticity. Choose $0 < ρ < 2 r / (n C_{2})$ . For $∣ h ∣ < ρ$ the Taylor remainder bound in step 5.1 tends to zero, so the Taylor polynomials converge to $u (a + h)$ . The degree-zero term is at most $M$ , while for each $k \geq 1$ step 4.1 and the multinomial bound in [F3] give $\sum_{∣ α ∣ = k} ∣ D^{α} u (a) h^{α} ∣ / α! \leq M (n C_{2} ∣ h ∣ / (2 r))^{k}$ . The geometric series converges, so the Taylor series converges absolutely and equals $u (a + h)$ ; the ball $∣ h ∣ < ρ$ contains the polydisc $∣ h_{i} ∣ < ρ / \sqrt{n}$ , hence $u$ is real analytic at $a$ in the sense of [F4].

7.1step 6.1step 5.2cases∎

Complex $u$ : apply steps 1.1 through 6.1 to $R e u$ and $I m u$ , which are real harmonic; the Taylor coefficients of $u$ are the sums of the corresponding coefficients, and the two real series give an absolutely convergent complex series. For $∣ α ∣ = 0$ , $∣ u (a) ∣ \leq M$ directly. For $k : = ∣ α ∣ \geq 1$ , the real estimates give $∣ D^{α} u (a) ∣ \leq 2 M C_{n}^{k} k! r^{- k} \leq M (2 C_{n})^{k} k! r^{- k}$ . Thus the stated estimate, including order zero, holds with constant $2 C_{n}$ in place of $C_{n}$ . Since $a \in Ω$ was arbitrary, every harmonic function on $Ω$ is real analytic.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Interior gradient bound for Poisson solutions

Statement

Assume Countable Choice, $n \geq 2$ and $0 < α < 1$ . Let $u \in C^{2} (B_{r} (a)) \cap L^{\infty} (B_{r} (a))$ and let $f \in C^{0, α} (B_{r} (a))$ have finite Hölder seminorm, with $- Δ u = f$ pointwise. Then $∥ D u ∥_{\infty; B_{r / 2} (a)} \leq C_{n} (r^{- 1} ∥ u ∥_{\infty; B_{r} (a)} + r ∥ f ∥_{\infty; B_{r} (a)}) .$ No Hölder seminorm of $f$ occurs on the right-hand side.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , $0 < α < 1$ , a centre $a \in R^{n}$ , a radius $r > 0$ , $u \in C^{2} (B_{r} (a)) \cap L^{\infty} (B_{r} (a))$ and $f \in C^{0, α} (B_{r} (a))$ with finite Hölder seminorm and $- Δ u = f$ pointwise.

[F1]

The normalized kernel is $Φ (z) = ∣ z ∣^{2 - n} / ((n - 2) ω_{n - 1})$ for $n \geq 3$ and $Φ (z) = - (2 π)^{- 1} \log ∣ z ∣$ for $n = 2$ , with $\nabla Φ (z) = - z / (ω_{n - 1} ∣ z ∣^{n})$ for $n \geq 3$ and $\nabla Φ (z) = - z / (2 π ∣ z ∣^{2})$ for $n = 2$ (Fundamental solution for the positive operator minus Laplacian, Continuity and derivatives of positive-base real powers, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ); $Φ$ is locally integrable (Local integrability of the Laplace fundamental kernel).

[F2]

For nonnegative Borel $F$ , polar coordinates give $\int_{B_{s}} F (∣ z ∣) d z = ω_{n - 1} \int_{0}^{s} F (ρ) ρ^{n - 1} d ρ$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Sphere and ball measures scale in Rn).

[F3]

The Newtonian potential of $G$ is $N G (x) = \int_{R^{n}} Φ (x - y) G (y) d y$ (Newtonian potential of compactly supported data).

[F4]

For compactly supported $α$ -Hölder $G$ , its Newtonian potential is $C^{2}$ and satisfies $- Δ N G = G$ pointwise (Hölder data give a classical Newtonian solution).

[F5]

Dominated convergence for Lebesgue integrals on $R^{n}$ (Dominated convergence).

[F6]

For $0 < ρ < R$ there is a smooth $η$ with $η = 1$ on ${\overline{B}}_{ρ} (0)$ and $supp η \subseteq B_{R} (0)$ (A smooth bump between concentric Euclidean balls); rescaled and translated, such cutoffs exist between any two concentric Euclidean balls.

[F7]

[F8]

For harmonic $v$ on an open set containing $\overline{B_{ρ} (x)}$ : $∣ D^{α} v (x) ∣ \leq C_{n, α}^{'} ρ^{- ∣ α ∣} \sup_{B_{ρ} (x)} ∣ v ∣$ (Harmonic Cauchy estimates in supremum norm).

[F9]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF9

Work under [F9], let $x_{0} \in B_{r / 2} (a)$ be arbitrary, and put $ρ : = r / 4$ , so that $\overline{B_{ρ} (x_{0})} \subset B_{r} (a)$ ; write $M_{u} : = ∥ u ∥_{\infty; B_{r} (a)}$ and $M_{f} : = ∥ f ∥_{\infty; B_{r} (a)}$ . Both are finite: $M_{u} < \infty$ is given, and the finite Hölder seminorm bounds $∣ f (x) ∣ \leq ∣ f (a) ∣ + [f]_{0, α; B_{r} (a)} (2 r)^{α}$ for every $x \in B_{r} (a)$ .

2.1givenF1F2algebra

Kernel integrals. By [F1] and [F2], $∣ D Φ (z) ∣ \leq C_{n} ∣ z ∣^{1 - n}$ and $\int_{B_{s}} ∣ D Φ (z) ∣ d z \leq C_{n} s$ for every $s > 0$ . For the fixed scale $ρ$ of step 1.1, a change of variables $z = ρ ζ$ in the power-kernel case $n \geq 3$ , and the identity $Φ (ρ ζ) - Φ (ρ) = - (2 π)^{- 1} \log ∣ ζ ∣$ when $n = 2$ , give $\int_{B_{3 ρ / 2}} ∣ Φ (z) - Φ (ρ) ∣ d z \leq C_{n} ρ^{2} .$ The scaled integral is finite in every dimension by polar coordinates; constants depend only on $n$ .

2.2step 1.1F3F4F6

Cutoff and normalized potential at $x_{0}$ . By [F6] fix a smooth cutoff $η$ with $η = 1$ on ${\overline{B}}_{3 ρ / 4} (x_{0})$ and $supp η \subseteq B_{ρ} (x_{0})$ , and put $G : = η f$ on $B_{ρ} (x_{0})$ , extended by $0$ to $R^{n}$ . Then $G$ is continuous, compactly supported and has finite $α$ -Hölder seminorm. Define $w (x) : = \int_{R^{n}} (Φ (x - y) - Φ (ρ)) G (y) d y = N G (x) - Φ (ρ) \int_{R^{n}} G (y) d y .$ By [F3] and [F4], $w \in C^{2} (R^{n})$ and $- Δ w = G$ pointwise; the subtracted term is constant in $x$ .

3.1step 2.2algebra

The remainder $h : = u - w$ is harmonic on $B_{3 ρ / 4} (x_{0})$ : there $η = 1$ , so $G = f$ and $- Δ h = - Δ u + Δ w = f - G = 0$ by the hypothesis and step 2.2.

3.2step 2.1step 2.2F1F2F5F7algebra

Explicit form and bound for $D w$ . Fix $x \in B_{3 ρ / 4} (x_{0})$ and a coordinate $k$ . For $y$ with $∣ x - y ∣ > 2 ∣ t ∣$ , the real mean value theorem [F7] and $∣ D Φ (z) ∣ \leq C_{n} ∣ z ∣^{1 - n}$ give $∣ \frac{Φ (x + t e_{k} - y) - Φ (x - y)}{t} ∣ \leq C_{n} ∣ x - y ∣^{1 - n},$ since every point on the segment between $x - y$ and $x + t e_{k} - y$ has norm at least $∣ x - y ∣ / 2$ . The right side is integrable on the bounded support of $G$ . On this far region the quotients converge pointwise for $y \neq x$ to $\partial_{k} Φ (x - y)$ , so dominated convergence [F5], with the indicator of $∣ x - y ∣ > 2 ∣ t ∣$ , gives convergence of the far-region integrals to $\int \partial_{k} Φ (x - y) G (y) d y$ . On the near region $∣ x - y ∣ \leq 2 ∣ t ∣$ , the quotient integral is bounded by $\frac{M_{f}}{∣ t ∣} (\int_{B_{2 ∣ t ∣} (t e_{k})} ∣ Φ (z) ∣ d z + \int_{B_{2 ∣ t ∣} (0)} ∣ Φ (z) ∣ d z),$ which tends to zero: it is $O (∣ t ∣)$ for $n \geq 3$ and $O (∣ t ∣ (1 + ∣ \log ∣ t ∣ ∣))$ for $n = 2$ , by polar coordinates [F1, F2]. The integral of $∣ D Φ (x - y) G (y) ∣$ over that near region is $O (M_{f} ∣ t ∣)$ by step 2.1. Hence $\partial_{k} w (x) = \int \partial_{k} Φ (x - y) G (y) d y$ . At $x = x_{0}$ , this yields $∣ D w (x_{0}) ∣ \leq M_{f} \int_{B_{ρ} (x_{0})} ∣ D Φ (x_{0} - y) ∣ d y \leq C_{n} M_{f} ρ$ . For $x \in B_{ρ / 2} (x_{0})$ , the normalized kernel and the inclusion $B_{ρ} (x_{0}) \subset B_{3 ρ / 2} (x)$ give $∣ w (x) ∣ \leq M_{f} \int_{B_{3 ρ / 2}} ∣ Φ (z) - Φ (ρ) ∣ d z \leq C_{n} M_{f} ρ^{2}$ by step 2.1.

4.1step 3.1step 3.2F8algebra

Harmonic gradient bound. Since $h$ is harmonic on $B_{3 ρ / 4} (x_{0}) \supseteq \overline{B_{ρ / 2} (x_{0})}$ , apply [F8] separately to each coordinate derivative $\partial_{i} h (x_{0})$ , $0 \leq i < n$ . The vector norm satisfies $∣ D h (x_{0}) ∣ \leq \sqrt{n} \max_{i} ∣ \partial_{i} h (x_{0}) ∣$ , so, absorbing $\sqrt{n}$ into $C_{n}^{'}$ , $∣ D h (x_{0}) ∣ \leq C_{n}^{'} (2 / ρ) \sup_{B_{ρ / 2} (x_{0})} ∣ h ∣ \leq C_{n}^{'} (2 / ρ) (M_{u} + C_{n} M_{f} ρ^{2})$ by steps 3.1 and 3.2.

5.1step 3.2step 4.1algebra

Combining steps 3.2 and 4.1 at the point $x_{0}$ , $∣ D u (x_{0}) ∣ \leq ∣ D w (x_{0}) ∣ + ∣ D h (x_{0}) ∣ \leq C_{n} M_{f} ρ + C_{n}^{'} (2 / ρ) (M_{u} + C_{n} M_{f} ρ^{2}) \leq C_{n}^{''} (ρ^{- 1} M_{u} + ρ M_{f}) = C_{n}^{''} (4 r^{- 1} M_{u} + \frac{1}{4} r M_{f})$ , and absorbing the numerical factors into $C_{n}^{''}$ gives $∣ D u (x_{0}) ∣ \leq C_{n} (r^{- 1} M_{u} + r M_{f})$ .

6.1step 5.1cases∎

Since $x_{0} \in B_{r / 2} (a)$ was arbitrary, taking the supremum over $x_{0}$ gives $∥ D u ∥_{\infty; B_{r / 2} (a)} \leq C_{n} (r^{- 1} ∥ u ∥_{\infty; B_{r} (a)} + r ∥ f ∥_{\infty; B_{r} (a)})$ , the displayed estimate; the constants encountered in steps 2.1, 3.2 and 4.1 depend only on $n$ , and the Hölder seminorm of $f$ entered only through the qualitative $C^{2}$ clause of [F4] used to define $w$ and $h$ , never through a quantitative bound. The argument covers complex-valued $u$ and $f$ by applying the real case to real and imaginary parts.

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Dimension split and the separate Poisson-disc theory

Remark

This page splits its statements by dimension, and this remark records where the split sits so that no item silently overclaims.

The $n \geq 3$ construction. The Kelvin/image construction (Dirichlet Green function of a Euclidean ball), the explicit ball kernel (Poisson kernel of a Euclidean ball) and the half-space kernel with its bounded Dirichlet problem (Poisson kernel and bounded Dirichlet problem on a half-space) are stated for $n \geq 3$ . The reflection formula for the half-space kernel is the reflection of the fundamental solution, whose profile is $∣ z ∣^{2 - n}$ precisely for $n \geq 3$ ; in the plane the corresponding profile is logarithmic and the kernel constants change. The bounded uniqueness argument requires its own planar proof. The statements above do not claim to cover the planar case.

The full planar Dirichlet theory is cited; a smooth-data lemma is used. The continuous-data disc Dirichlet theorem (The Poisson integral gives the unique continuous harmonic extension on the closed unit disc) and the full disc Poisson theory are developed on their own page. The $n = 2$ branches of the interior derivative estimates and real-analyticity theorem also use Smooth sphere data have a harmonic replacement under Countable Choice, whose planar case gives the Poisson representation for smooth circle data after harmonic regularity is established. This restricted smooth-data result does not reprove the full continuous-data theorem or its boundary-convergence theorem.

Results including $n = 2$ . The interior derivative estimates (Interior derivative estimates for harmonic functions) and the real-analyticity theorem (Harmonic functions are real analytic) cover $n \geq 2$ ; their planar arguments use mean-value regularity and the smooth-data sphere lemma above. The interior $C^{2, α}$ Poisson estimate and its gradient corollary (Interior estimate for the Poisson equation with Hölder data) also cover $n \geq 2$ and handle the planar case with the logarithmic Newtonian potential. These arguments use separate formulas in dimension two and dimensions at least three, without extending the image construction to the plane.

What this remark does not say. This is a statement about the scope of the items on this page, not a mathematical claim that the $n = 2$ kernels fail to exist. The disc and half-plane kernels exist; they are simply developed elsewhere and referenced here.

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Unique continuation for harmonic functions

Statement

Assume Countable Choice and $n \geq 2$ . Let $Ω \subseteq R^{n}$ be connected and open, and let $u$ be real or complex harmonic on $Ω$ . If $u$ vanishes on a nonempty open subset of $Ω$ , then $u$ vanishes identically on $Ω$ .

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 2$ , a connected open set $Ω \subseteq R^{n}$ , a harmonic $u$ on $Ω$ , and a nonempty open set $V \subseteq Ω$ with $u = 0$ on $V$ .

[F1]

Every harmonic function on $Ω$ is real analytic: for each $a \in Ω$ there is $ρ > 0$ with $u (a + h) = \sum_{α} D^{α} u (a) h^{α} / α!$ absolutely convergent for $∣ h ∣ < ρ$ ; in particular $u$ is $C^{\infty}$ (Harmonic functions are real analytic).

[F2]

A topological space is connected exactly when its only clopen (simultaneously open and closed) subsets are the whole space and the empty set (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[F3]

Multi-index notation $D^{α}$ , with $D^{0} u = u$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F4]

Countable Choice ${A C}_{ω}$ is the standing hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF3F4

Work under [F4] and let $F : = {x \in Ω : D^{α} u (x) = 0 for every multi-index α}$ , the set of points where all derivatives vanish. Since $V$ is open and $u = 0$ identically on $V$ , every derivative of $u$ vanishes on $V$ (a derivative of the zero function), so $V \subseteq F$ and $F \neq \emptyset$ .

2.1step 1.1F1

$F$ contains $V$ , and $F$ is closed in $Ω$ : by [F1] each function $D^{α} u$ is continuous on $Ω$ , and $F = ⋂_{α} {x \in Ω : D^{α} u (x) = 0}$ is an intersection of closed subsets of $Ω$ .

2.2step 1.1F1F3algebra

$F$ is open in $Ω$ : let $x \in F$ . By [F1] choose $ρ > 0$ such that $u (x + h) = \sum_{α} D^{α} u (x) h^{α} / α!$ with absolute convergence for $∣ h ∣ < ρ$ . Since $x \in F$ , every coefficient $D^{α} u (x)$ vanishes, so the series is identically zero and $u = 0$ on the ball $B_{ρ} (x) \subseteq Ω$ ; that ball is open, so every derivative of $u$ vanishes on it and $B_{ρ} (x) \subseteq F$ . Hence $F$ is open in $Ω$ .

3.1step 2.1step 2.2F2F3∎

Therefore $F$ is clopen in $Ω$ and nonempty, while $Ω$ is connected; by [F2] the only clopen subsets of $Ω$ are $\emptyset$ and $Ω$ , so $F = Ω$ . Hence all derivatives of $u$ vanish everywhere and, in particular, $u (x) = D^{0} u (x) = 0$ for every $x \in Ω$ by [F3]; that is, $u$ vanishes identically on $Ω$ . The argument applies to real and complex $u$ alike because the Taylor representation and continuity are available in both cases.

5 · Examples, counterexamples and false statements

None yet.