convexity of the range I'm currently trying to prove this one statement (IDK if it's true yet):
Let $f:\mathbb R^m \to \mathbb R^m$ be a $C^1$ function such that, for all $x,v\in \mathbb R^m$, we have 
$\langle f'(x)\cdot v,f'(x)\cdot v\rangle = \langle v, v\rangle$ $\big($  i.e. $|f'(x)\cdot v|=|v|\big)$.
Show that the line segment $[f(x);f(y)]\subset f(\mathbb R^m)$ for all $x,v\in \mathbb R^m$ (in others words, $f(\mathbb R^m)$ is a convex subset of $\mathbb R^m$).
[My results]
$\bullet$ $f$ is a local diffeomorphism (i.e. for all $a\in\mathbb R^m$, there is opens sets $a\in V_a$ and $f(a)\in V_{f(a)}$ such that $f|_{V_a}:V_a\to V_{f(a)}$ is a diffeomorphism)
$\bullet$ $f|_{V_a}:V_a\to V_{f(a)}$ also preserve the distance (i.e. $|f(x)-f(y)|=|x-y|$ for all $x,y\in V_a$)
Any help would be really appreciated!
 A: First, note that if $f'(x)$ is an isometry, by the polarization identity, it has to belong to $O(m)$, the orthogonal group. Let $x, x_0 \in \mathbb{R}^m$, and define the path joining $x_0$ with $x$:
$$p(s) = s x + (1 - s) x_0,$$
with $0 \leq s \leq 1$. Define $\int_0^1 (f\circ p)'(s)\, ds$ integrating element-wise, then the fundamental theorem of calculus implies
$$\int_0^1 (f\circ p)'(s)\, ds = f(x) - f(x_0).$$
Applying the chain rule to the left hand side of the previous equation, it turns out that 
$$\int_0^1 f'\circ p (s)\, (x - x_0) \, ds = f(x) - f(x_0).$$
Since we are integrating element-wise, the constant vector $x - x_0$ can be factored out of the previous equation, so that 
$$\left(\int_0^1 f'\circ p (s) \, ds\right)\, (x - x_0) = f(x) - f(x_0).$$
Let $A = \int_0^1 f'\circ p (s) \, ds$, again, element-wise integration implies that $A \in O(m)$, and therefore
$$f(x) = A(x - x_0) + f(x_0).$$
i.e. $f(x)$ is an affine transformation. In particular, $f(\mathbb{R}^m) = \mathbb{R}^m$.
