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I'm not sure how to go about proving this theorem:

Let $f:\mathbb{R}^m \longrightarrow \mathbb{R}^m,f\in C^1(\mathbb{R}^m)$ such that:

$\|f'(x)(v)\|=\|v\|,\forall v\in \mathbb{R}^m,\forall x \in \mathbb{R}^m$

Prove that $\|f(x)-f(y)\|=\|x-y\|,\forall x,y\in \mathbb{R}^m$.

Any hints would be appreciated.

By the Theorem of the Mean Value Inequality, we have $\|f(x)-f(y)\|\le\|x-y\|,\forall x,y\in \mathbb{R}^m$.

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Let $\alpha(t)=x+t\frac{(y-x)}{\|y-x\|}$ with $t\in[0,\|y-x\|]$ and $x,y\in\mathbb{R}^{n}$. So $\alpha$ is parametrized byt arc length, therefore, $\|(f\circ\alpha)'(t)\|=\|f'(\alpha(t))\alpha'(t)\|=\|\alpha'(t)\|=1$, hence the curve $(f\circ\alpha)(t)$ is parametrized by arc length. By using these facts and the definition of arc length, can you finish?

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    Sorry, it is not $[0,1]$, it is $[0,\|y-x\|]$. I will edit it2012-10-23