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In a previews question I asked here I used the following definition of path length:$\gamma=(x(t),y(t),z(t))$ : $L(\gamma)=\intop_{a}^{b}\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}$.

In the answer another definition was used:

$$ l(f)= \sup_{P}\sum_{i=0}^k |f(t_{i+1})-f(t_i)| $$ where $f:[0,1]\to \mathbb{R}^3$, $f(0)=a$, $f(1)=b$ and the $\sup$ is taken over all partitions $P$ of $[0,1]$.

How can I show they are equivalent ?

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Is $|\cdot|$ the euclidian norm? – Davide Giraudo May 14 '12 at 18:45… – Blah May 14 '12 at 18:48
up vote 2 down vote accepted

Since $|\gamma(t_{i+1}) - \gamma(t_i)| = \sum_j \sqrt{(\gamma_j(t_{i+1}) - \gamma_j(t_i))^2}$, the mean value theorem gives: $\sup_P\sum_{i = 0}^k |\gamma(t_{i+1}) - \gamma(t_i))| = \sup_P\sum_{i = 0}^k \Delta_i \sqrt{\sum_j \gamma_j'(\tilde{t}_i)^2}$, where $\Delta_i = |t_{i + 1} - t_i|$ and $\tilde{t_i}$ is some point between $t_{i + 1}$ and $t_i$. Taking the supremum over all the partitions gives you the integral that you want (this is definition of Riemann integration, where the integral exists when the map is continuous, which I assume it is--say $\gamma$ has continuous derivatives).

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