# If $\|f'(x)(v)\|=\|v\|,\forall v\in \mathbb{R}^m,\forall x \in \mathbb{R}^m\Longrightarrow f$ is an isometry

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?
Thanks Tomás for your solution, but I have problems with $(f\circ\alpha)(1)=f(x+\frac{y-x}{||y-x||})$ and definition $l(f\circ\alpha)[0,1]=\sum_{i=1}^{n}||(f\circ \alpha)(t_i)-(f\circ \alpha)(t_{i-1}))||$ –  felipeuni Oct 23 '12 at 1:17
Sorry, it is not $[0,1]$, it is $[0,\|y-x\|]$. I will edit it –  Tomás Oct 23 '12 at 11:37