Let $f:\mathbb R^n\rightarrow \mathbb R$ be a smooth function such that $\|\nabla f(x)\|=1$ for all $x\in \mathbb R^n$ and $f(0)=0$.
I would like to prove that $f$ is linear.
I first looked at the solution of the O.D.E. $$\dfrac{d\gamma}{dt}(t)=\nabla f(\gamma(t))$$I noticed that there exists only one solution $\gamma_x$ passing through $x\in \mathbb R^n$ at $t=0$, and such a solution is defined over $\mathbb R$. I also proved that $$\gamma_x(t)=x+t\nabla f(x)$$ and $$f(\gamma_x(t))=f(x)+t.$$
Furthermore, the following hint is given :
Show that if $f(x)=f(y)$ then $\langle \nabla f(x),x-y\rangle=\langle \nabla f(y),x-y\rangle=0$.
I did it, but now, I don't understand why it follows that $f$ is linear.
Edit: I write here the proof of the hint;
Let $c:[0,1]\rightarrow \mathbb R^n$ be the (usual) parametrization of the segment $[x+t\nabla f(x),y]$. Since $f(x)=f(y)$, we get $$\begin{align*}|t|=|f(\gamma_x(t))-f(x)| & =|f(x+t\nabla f(x))-f(y)|\\ & =|\int_0^1 \langle \nabla f(c(s)),c'(s)\rangle ds| \\ & \leq \int_0^1 \|c'(s)\|ds \\ & =\|y-x-t\nabla f(x)\|\end{align*}$$From this inequality, which is true for any $t\in \mathbb R$ and using the fact that $\nabla f(x)$ is a unitary vector, it follows that :$$t^2\leq \|x-y\|^2 + t^2 +2t\langle x-y,\nabla f(x) \rangle$$Dividing by $t$ and taking limits to $\pm \infty$ gives the hint.