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Let $\gamma:[0,1]\to\mathbb{R}^d$ be a rectifiable curve and define the length of the curve as $$\operatorname{Length}(\gamma)=\sup\left\{\sum\limits_{k=0}^{n-1}\|\gamma(t_{k+1})-\gamma(t_k)\|,n\ge 1, 0=t_0<t_1<\ldots<t_n=1 \right\}.$$ Is there an example where $\operatorname{Length}(\gamma)\neq \int_0^1\|\dot{\gamma}(t)\|\,dt$ (assuming this quantity is well defined)?

RobPratt
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edamondo
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  • I suspect the graph of the Cantor function is such a case. – Michael Hardy May 31 '21 at 00:14
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    If $\phi$ is continuously differentiable or piece wise continuously differentiable you have equality in your expressions. – Mittens May 31 '21 at 00:16
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    @OliverDiaz I am not sure if the definition of the length I gave is the same as the "arc length" but in this discussion if I understood correctly they say the arc length of the Cantor function is 2, so it would be a valid example? – edamondo May 31 '21 at 01:01
  • @edamondo: yes it is. Anyway, I did not calculate the length of the Cantor map (just did a mental deduction and seems I am wrong). In any event, now you have tow example of rectifiable curves where the integral representation does not hold. In both there are pathologies. – Mittens May 31 '21 at 01:17
  • @OliverDiaz, thank you. I am not sure I got to the bottom of the other example (not the Cantor one, the other). I understand that there we have $\operatorname{Length}(\gamma)=\infty$. But isn't $\int_0^1 |\dot{\gamma}(t)|dt=\infty$ as well? – edamondo May 31 '21 at 13:34
  • @edamondo: I think you are right, I should not stay on MSE past my bedtime. [Here] (https://math.stackexchange.com/q/889066/121671) is another link you mat found useful for your problem. – Mittens May 31 '21 at 18:08
  • @OliverDiaz, anyway your replies are still always useful. I checked the linked post on the length of the Cantor curve and dropped upvotes everywhere – edamondo May 31 '21 at 22:22

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