LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-02

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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 $y \in S_{R} (a)$ .

Depends on

Used by

Cap and complement estimate for the ball Poisson integral Lemma
The ball Poisson kernel is positive and has unit mass Lemma
Continuous Dirichlet problem on a ball Theorem
Dirichlet Green function of a Euclidean ball Theorem
Poisson kernel of a Euclidean ball Theorem

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Thomas Schmidt, Partial Differential Equations I (2026) (standard reference, not scraped)