# Differentiability of the arc length function : $\sigma (t)=l(\alpha)[a,t]$

Let $\alpha:[a,b]\longrightarrow \mathbb{R}^n$ be a rectifiable path (not necessarily continuous)

The function arc length $\sigma:[a,b]\longrightarrow \mathbb{R}$ is defined as:$$\sigma(t)=\mathrm{length}(\alpha)[a,t]=\mathrm{length}(\alpha |[a,t])$$

Is the following statement true?

$\hspace{4.8cm}\sigma$ is differentiable $\Longleftrightarrow$ $\alpha$ is differentiable

We say that the path $\alpha:[a,b]\longrightarrow \mathbb{R}^n$ is rectifiable if $:$

$\mathrm{length}(\alpha)[a,b]=\underset{P}{\text{Sup}}\;l(\alpha,P)<\infty$ $\;:$ path length at $[a,b]$

$P=\{a=t_0<t_1<\cdots<t_k=b\}$ $\;:$ partition of $[a,b]$

$\mathrm{length}(\alpha,P)=\sum_{i=1}^{k}$ $||\alpha(t_i)-\alpha(t_{i-1})||$ $\;:$ polygonal path length

Any hints would be appreciated.

1. Consider $\alpha(t)=(t,|t|)$, a path in $\mathbb R^2$. What is $\sigma$ here?
2. If $\alpha$ is $C^1$, then so is $\sigma$, with $\sigma' = |\alpha'|$.
3. But the mere differentiability of $\alpha$ does not imply the differentiability of $\sigma$. Consider $\alpha(t)=(t,0)$ for $t\le 0$, $\alpha(t)=(t,t^2\sin(1/t))$ for $t>0$. Since $|\alpha'|$ is bounded below by a positive constant on the set $\{t\in (0,1/10):|\cos (1/t)|>1/2\}$, the function $\sigma$ does not satisfy $\sigma(t)/t\to 1$ as $t\to 0^+$.
• Thanks for your solution, $1.$ If $\alpha:[-1,1]\longrightarrow \mathbb{R}^2$ then $\sigma (t)=\sqrt{2}(t+1)$ differentiable. – felipeuni Dec 31 '13 at 8:21