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PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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A smooth map with pointwise operator norm at most c is c lipschitz for riemannian distance

Statement

Let M,N be connected Riemannian manifolds. If smooth F:MN satisfies dFpvhcvg for a finite c0 and all p,v, then dh(Fp,Fq)cdg(p,q).

Facts & Assumptions

Given: The stated differential bound and two source points.

[F1]

Riemannian distance on a connected manifold: On a connected Riemannian manifold define dg(p,q)=inf{Lg(γ):γ is piecewise C1 from p to q}. Lengths are those of def-riemannian-speed-and-length. For each pair p,q, lem-any-two-points-in-a-connected-smooth-manifold-can-be-joined-by-a-piecewise-c-one-curve supplies a curve, so the set of lengths is nonempty, contains a finite real number and is bounded below by zero. Applying the least-upper-bound property cor-cauchy-reals-lub-complete to the negatives gives a finite nonnegative infimum. On the empty connected manifold this defines the empty distance function; there are no pairs to evaluate. No minimizing curve is part of this definition.

[F2]

Riemannian speed and length: The Riemannian speed on a C1 piece is γ˙(t)g=gγ(t)(γ˙(t),γ˙(t)). Its length is Lg(γ)=jtj1tjγ˙(t)gdt. The curve convention is def-piecewise-c-one-curve-on-a-manifold and the norm is def-pointwise-norm-and-angle-from-a-riemannian-metric. Each integrand is continuous on its closed piece with the one-sided endpoint derivative, hence Riemann integrable and nonnegative. Values chosen at the finitely many corners do not change its integral. For a singleton interval the empty sum is zero; a constant curve also has zero length. Partition independence is established next.

[F3]

Any two points in a connected smooth manifold can be joined by a piecewise c one curve: Any two points in a nonempty connected smooth manifold can be joined by a finite piecewise C1 curve.

Proof

technique · direct
1.1

Every source competitor γ maps to a target competitor Fγ. The chain rule gives (Fγ)h=dFγ˙hcγ˙g on each piece, and integration gives Lh(Fγ)cLg(γ). Hence dh(Fp,Fq)cLg(γ).

F1F2F3given
2.1

If c=0, any such path yields dh(Fp,Fq)=0. If c>0, for every ε>0 choose a competitor of length less than dg(p,q)+ε/c. The preceding bound gives dh(Fp,Fq)<cdg(p,q)+ε; letting ε decrease to zero gives the required inequality. Empty source has no point pairs.

F1step 1.1

Source locator

Lee, Chapter 13, pp.337–340, Proposition 13.25, Lemma 13.28 and Theorem 13.29; finite piecewise C1 refinements and pauses are treated explicitly here.

Depends on

Used by

Dependency tree · two levels

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Sources