Is $\frac{u}{|u|}$ 1- Lipschitz?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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 $\angle u v^\prime v \ge \frac{\pi}{2}$.