Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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Every rectifiable path factors through its arc-length function as a unit-speed path on [0,L]

Statement

Let γ:[a,b]→Rn be rectifiable, put L=L(γ), and let s=sγ. There is a unique map γˉ:[0,L]→Rn such that

γ=γˉ∘s.

It is 1-Lipschitz and, for every 0≤r≤q≤L,

L[r,q](γˉ∣[r,q])=q−r.

Thus γˉ has metric unit speed. If L=0, its domain is a singleton and the formula reads 0=0.

Facts & Assumptions

Given: The rectifiable path, length L, and arc-length function s.

[L2]

Every chord is at most the length of the corresponding subpath; in particular, a path of zero length is constant (Every endpoint chord is no longer than the arc: ∥γ(b)−γ(a)∥2≤L(γ)).

[L3]

Length is invariant under a continuous surjective monotone reparametrization (Arc length is invariant under every continuous surjective monotone reparametrization, including pauses and reversal).

Proof

technique · factorization
1.1

By continuity and the endpoint values in [L1], s([a,b])=[0,L]. If s(u)=s(v) with u≤v, [L1] makes the intervening length zero and [L2] gives γ(u)=γ(v).

givenL1L2
2.1

For r∈[0,L], define γˉ(r) to be the unique common value γ(t) of all t with s(t)=r. Existence follows from surjectivity and well-definedness from step 1.1. This definition immediately gives γ=γˉ∘s and uniqueness.

step 1.1construct
3.1

For r<q, take u≤v with s(u)=r and s(v)=q. The chord bound and [L1] give ∥γˉ(q)−γˉ(r)∥2≤L[u,v](γ)=q−r. Hence γˉ is 1-Lipschitz and continuous.

step 2.1L1L2
4.1

The restriction s∣[u,v] is a continuous surjective nondecreasing map onto [r,q], and γ∣[u,v]=γˉ∣[r,q]∘s∣[u,v]. By [L3], L(γˉ∣[r,q])=L(γ∣[u,v])=q−r.

step 2.1step 3.1L1L3
5.1

If L=0, [L2] makes γ constant, s has singleton image, and the construction gives the unique constant map on [0,0]; the subinterval formula is 0=0.

givenL1L2∎

Depends on

Used by

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Dependency tree · two levels

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Sources