Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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  • 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:M→N be C1, let (U,φ) be a chart on M, let (V,ψ) be a chart on N with F(U)⊆V, and let K⊆U be compact. Then every point of K has an open neighbourhood W⊆U with W‾⊆U such that the coordinate representative ψ∘F∘φ−1 is Lipschitz on φ(W‾).

Facts & Assumptions

Given: A C1 map F:M→N, charts (U,φ) and (V,ψ) with F(U)⊆V, and a compact set K⊆U.

[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
1.1F1givenchoose

Let f:=ψ∘F∘φ−1. By [F1], f is C1 on the open set φ(U)⊆Rm. For each x∈φ(K) choose an open Euclidean ball Bx with Bx‾ compact and contained in φ(U).

2.1L1L2step 1.1

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‾.

3.1step 2.1construct∎

Put Wx:=φ−1(Bx). Then Wx is an open neighbourhood of φ−1(x) with Wx‾⊆U, and f is Lipschitz on φ(Wx‾)=Bx‾. Since x was arbitrary in φ(K), the claim follows.

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Sources