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A regular $C^{1}$ path has a $C^{1}$ arc-length reparametrization with derivative of Euclidean norm one

Statement

Let $n \geq 1$ and $a < b$ . Suppose $γ : [a, b] \to R^{n}$ is continuous, differentiable on $(a, b)$ , and its derivative extends continuously to $v : [a, b] \to R^{n}$ with $∥ v (t) ∥_{2} > 0$ for every $t$ . Put

$s (t) = \int_{a}^{t} ∥ v (u) ∥_{2} d u, L = s (b) .$

Then $s$ is a continuously differentiable increasing bijection from $[a, b]$ onto $[0, L]$ . Its inverse $τ$ is continuously differentiable, and

$σ (r) : = γ (τ (r))$

is a $C^{1}$ reparametrization with $∥ σ^{'} (r) ∥_{2} = 1$ for every $r \in [0, L]$ , using relative derivatives at the endpoints.

Facts & Assumptions

Given: The regular $C^{1}$ path and its continuous velocity extension $v$ .

[L1]

The path has length $\int_{a}^{b} ∥ v ∥_{2}$ , and its arc-length function is the corresponding partial integral (If $γ : [a, b] \to R^{n}$ is continuous, differentiable on $(a, b)$ , and $γ^{'}$ extends continuously to $[a, b]$ , then $L (γ) = \int_{a}^{b} ∥ γ^{'} (t) ∥_{2} d t$ , The arc-length function $s_{γ} (t) = L (γ ∣_{[a, t]})$ of a rectifiable path).

[L2]

The first fundamental theorem gives $s^{'} (t) = ∥ v (t) ∥_{2}$ at every point, including relative endpoints (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive).

[L3]

A continuous injective function on an interval has an inverse, and if its derivative is nonzero then the inverse derivative is its reciprocal (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ ).

[L4]

The chain rule gives $(f \circ g)^{'} = (f^{'} \circ g) g^{'}$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ); vector differentiation and continuity are componentwise (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

[L5]

A continuous function with positive derivative on an interval is increasing by the mean value theorem (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · inverse-function

1.1

By [L1]--[L2], $s^{'} (t) = ∥ v (t) ∥_{2} > 0$ . By [L5], $s$ is increasing, and continuity plus its endpoint values makes it a bijection onto $[0, L]$ ; in particular $L > 0$ .

givenL1L2L5

2.1

By [L3], the inverse $τ : [0, L] \to [a, b]$ is differentiable and $τ^{'} (r) = 1 / ∥ v (τ (r)) ∥_{2}$ . This derivative is continuous because $v$ , the norm, $τ$ , and reciprocal on positive reals are continuous.

step 1.1L3

3.1

Apply [L4] componentwise to $σ = γ \circ τ$ to get $σ^{'} (r) = v (τ (r)) / ∥ v (τ (r)) ∥_{2}$ .

step 2.1L4

4.1

Absolute homogeneity of the Euclidean norm gives $∥ σ^{'} (r) ∥_{2} = 1$ , and the formula is continuous in $r$ , so $σ$ is $C^{1}$ . The relative endpoint derivatives follow from the relative forms in [L2] and [L3].

step 3.1algebra∎

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Sources

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

A regular C1C1 path has a C1C1 arc-length reparametrization with derivative of Euclidean norm one