Is $\frac{u}{|u|}$ 1- Lipschitz? I want to prove that the application  $\{s\in \mathbb{R}^3:|s|\geq 1\} \to S^2 ,u\mapsto \frac{u}{|u|}$ is Lipschitz with constant $1$. If I add ($\frac{u|u|-u|u}{|u||v|}|$),  I find the constant equals $2$ ( the norm used is the euclidien norm).
Is it not $1$-Lipschitz ?
Thank you
 A: It is $1-$ Lipschitz. $|\frac u {|u|} -\frac v {|v|}|^{2}=1+1-2\frac {u.v} {|u||v|}$ and $|u-v|^{2}=|u|^{2}+|v|^{2}-2u.v$ or $2(u.v) (1-\frac 1  {|u||v|}) \leq |u|^{2}+|v|^{2} -2$. Applying C-S inequality the Lipschitz property reduces to  $2ab(1-\frac 1 {ab}) \leq a^{2}+b^{2}-2$ where $a=|u|$ and $b=|b|$ and this reduces to $(a-b)^{2} \geq 0$ which is true.
A: A proof using projection on convex sets
$B_3$ the closed unit ball of $\mathbb R^3$ is a closed convex subset. The requested result consists in proving that the projection on $S^2$ of points in $\mathbb R^3 \setminus B_3$ is contracting.
A proof can be found here.
You can mimic that proof in your special case if you wish.
A: Geometric proof using intercept theorem
Suppose without loss of generality that $1 \le \vert u \vert \le \vert v \vert$ and denote $v^\prime = v \frac{\vert u \vert}{\vert v \vert}$. Notice that $\vert v^\prime \vert = \vert u \vert$.
Using intercept theorem for the triangles $0 \ \frac{u}{\vert u \vert }\ \frac{v}{\vert v \vert }$ and $0 \ u \ v^\prime$ you get
$$\begin{aligned}
\vert u - v \vert^2&= \vert u - v^\prime\vert^2 + \vert v^\prime - v\vert^2 + 2 \langle u - v^\prime, v^\prime -v \rangle\\
 &\ge \vert u - v^\prime \vert^2\\&= \vert u \vert^2 \left\vert\frac u {|u|} -\frac v {|v|}\right\vert^2\\
&\ge \left\vert\frac u {|u|} -\frac v {|v|}\right\vert^2
\end{aligned}$$
$\langle u - v^\prime, v^\prime -v \rangle \ge 0$ because $\angle u v^\prime v \ge \frac{\pi}{2}$.
