Hi folks! I'm trying to answer this one exercise:
Let $f:\mathbb R^m\to\mathbb R^m$ be a $C^1$ function such that, for all $x\in\mathbb R^m$ $|f'(x)\cdot v|=||v||$ (where $||\cdot||$ is the euclidean norm). Show that $||f(x)-f(y)||=||x-y||$.
$\bullet$ Using the Mean Value Inequality, it's easy to see that $||f(x)-f(y)||\leq||x-y||$.
$\bullet$ This exercise was on the Inverse Function Theorem section.
I'm stuck trying to prove that $||f(x)-f(y)||\geq||x-y||$.
Here is a sketch - feel free to leave comments or ask further questions.
Since you want to apply the inverse function theorem, you could perhaps begin by verifying that at every $x \in \mathbb R^m$, the linear map $f'(x)$ is invertible. (It should be possible to get injectivity from the fact that $|| f'(x) v || = || v ||$ for all $v \in \mathbb R^m$; surjectivity then follows on dimensional grounds.)
Having done this, the inverse function theorem tells us that around every $x \in \mathbb R^m$, there exists an open neighbourhood $U$ such that $f : U \to f(U)$ is bijective. Furthermore, we learn that the local inverse function $g : f(U) \to U$ is also continuously differentiable, and its derivative at a given point $y \in f(U)$ is $ g'(y) = (f'(x))^{-1}$, where $x \in U$ is the point such that $f(x) = y$.
Having found this expression for the derivative of $g$, you can now show that $g$ obeys a similar condition to $f$, i.e. $|| g'(y) w || = || w ||$ for every $y \in f(U)$ and for every vector $w \in \mathbb R^m$. Applying your mean value theorem argument to $g$ instead of $f$, you should be able to prove that $||f(x_1) - f(x_2) || \geq || x_1 - x_2 ||$ for all $x_1, x_2 \in U$, which is very close to what you asked for!
Unfortunately, we've only shown that $f$ is locally an isometry, i.e. we've only shown that we can cover $\mathbb R^m$ with open sets $U_\alpha$ such that for each $U_\alpha$, we have $ ||f(x_1) - f(x_2) || = || x_1 - x_2 || $ for all $x_1, x_2 \in U_\alpha$. We need to turn this into a global statement, valid for all $x_1, x_2 \in \mathbb R^m$. One strategy is to observe that, since $f$ is locally an isometry, $f$ maps short line segments to line segments of the same length - but then, $f$ also maps long line segments to line segments of the same length, because long line segments can always be broken down into short line segments. See this answer for more details.