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Every rectifiable path factors through its arc-length function as a unit-speed path on $[0, L]$

Statement

Let $γ : [a, b] \to R^{n}$ be rectifiable, put $L = L (γ)$ , and let $s = s_{γ}$ . There is a unique map $\overset{ˉ}{γ} : [0, L] \to R^{n}$ such that

$γ = \overset{ˉ}{γ} \circ s .$

It is $1$ -Lipschitz and, for every $0 \leq r \leq q \leq L$ ,

$L_{[r, q]} (\overset{ˉ}{γ} ∣_{[r, q]}) = q - r .$

Thus $\overset{ˉ}{γ}$ 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 $s$ .

[L1]

The function $s$ is continuous and nondecreasing, maps $a$ to $0$ and $b$ to $L$ , and $s (v) - s (u)$ is the length on $[u, v]$ (The arc-length function $s_{γ} (t) = L (γ ∣_{[a, t]})$ of a rectifiable path, The arc-length function is continuous and nondecreasing, with increments equal to subpath lengths; it is strictly increasing exactly when no nondegenerate subpath is constant).

[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: $∥ γ (b) - γ (a) ∥_{2} \leq L (γ)$ ).

[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 $u \leq v$ , [L1] makes the intervening length zero and [L2] gives $γ (u) = γ (v)$ .

givenL1L2

2.1

For $r \in [0, L]$ , define $\overset{ˉ}{γ} (r)$ 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 $γ = \overset{ˉ}{γ} \circ s$ and uniqueness.

step 1.1construct

3.1

For $r < q$ , take $u \leq v$ with $s (u) = r$ and $s (v) = q$ . The chord bound and [L1] give $∥ \overset{ˉ}{γ} (q) - \overset{ˉ}{γ} (r) ∥_{2} \leq L_{[u, v]} (γ) = q - r$ . Hence $\overset{ˉ}{γ}$ 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 $γ ∣_{[u, v]} = \overset{ˉ}{γ} ∣_{[r, q]} \circ s ∣_{[u, v]}$ . By [L3], $L (\overset{ˉ}{γ} ∣_{[r, q]}) = L (γ ∣_{[u, v]}) = q - r$ .

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∎

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Sources

U. Lang, Differential Geometry I, Lemma 1.1 (standard reference, not scraped)

Every rectifiable path factors through its arc-length function as a unit-speed path on [0,L][0,L]