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 $t\to 0^+$.

  • $\begingroup$ Thanks for your solution, $1.$ If $\alpha:[-1,1]\longrightarrow \mathbb{R}^2$ then $\sigma (t)=\sqrt{2}(t+1)$ differentiable. $\endgroup$ – felipeuni Dec 31 '13 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.