LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Radial derivative of a spherical average

Statement

If $u \in C^{2} (Ω)$ , $B_{r} (x) ⋐ Ω$ , and $m (t) = M_{u} (x, t)$ , then $m^{'} (r) = \frac{r}{n} \frac{1}{∣ B_{r} (x) ∣} \int_{B_{r} (x)} Δ u (y) d y$ .

Proof

Given: $u \in C^{2} (Ω)$ and $B_{r} (x) ⋐ Ω$ .

1.1

Differentiation under the compact sphere integral gives $m^{'} (r) = ω_{n - 1}^{- 1} \int_{S^{n - 1}} \nabla u (x + r θ) \cdot θ d σ (θ)$ [given].

1.2

Integrating $\partial_{t} (t^{n - 1} \partial_{t} u (x + t θ))$ from $0$ to $r$ , then using polar coordinates, yields $r^{n - 1} ω_{n - 1} m^{'} (r) = \int_{B_{r} (x)} Δ u$ [given].

2.1

Divide by $∣ B_{r} ∣ = ω_{n - 1} r^{n} / n$ from Sphere and ball measures scale in Rn [step 1.2]. ∎

Depends on

Used by

Spherical mean-value property for harmonic functions Theorem

Dependency tree · two levels

13 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

PDE source treatment (standard reference, not scraped)