Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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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.

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A C1 map is locally Lipschitz on compact coordinate subsets

Statement

Let F:MN be C1, let (U,φ) be a chart on M, let (V,ψ) be a chart on N with F(U)V, and let KU be compact. Then every point of K has an open neighbourhood WU with WU such that the coordinate representative ψFφ1 is Lipschitz on φ(W).

Facts & Assumptions

Given: A C1 map F:MN, charts (U,φ) and (V,ψ) with F(U)V, and a compact set KU.

[F1]

A C1 map has a C1 coordinate representative between Euclidean chart domains (Cr and smooth maps between smooth manifolds).

[L1]

A continuous real-valued function on a compact metric space attains a maximum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct
1.1

Let f:=ψFφ1. By [F1], f is C1 on the open set φ(U)Rm. For each xφ(K) choose an open Euclidean ball Bx with Bx compact and contained in φ(U).

F1givenchoose
2.1

The derivative norm Df is continuous on each compact ball Bx, so [L1] gives a finite bound Lx there. Since Bx is convex, [L2] makes f Lx-Lipschitz on Bx.

L1L2step 1.1
3.1

Put Wx:=φ1(Bx). Then Wx is an open neighbourhood of φ1(x) with WxU, and f is Lipschitz on φ(Wx)=Bx. Since x was arbitrary in φ(K), the claim follows.

step 2.1construct

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