How to calculate the Lipschitz constant for the normalization function $f(x)=\frac{x}{\|x\|}$?

Question

I want to calculate the Lipschitz constant for the normalization function $f(x)=\frac{x}{\|x\|}$. Any idea how I can solve this problem?

Just as indicated @Yiorgos S. Smyrlis, it is it self not Lipschitz. But I really appreciate that when we assume $\|x\|$ is bounded below, we can get the Lipschitz constant.

Answer 1

The function $$ f(x)=\frac{x}{\|x\|}, \quad x\in X\setminus\{0\}, $$ where $X$ is a normed space IS NOT Lipschitz continue. Not even uniformly continuous. Let $e\in X$, with $\|e\|=1$. Then $$ \|f(e/n)-f(-e/n)\|= \left\|\frac{e/m}{\|e/m\|}-\frac{-e/n}{\|-e/n\|}\right\| = \left\|\frac{e}{\|e\|}+\frac{e}{\|e\|}\right\|=2, $$ while $\|e/n-(-e/n)\|\to 0$.

However, $f$ is indeed Lipschitz continuous in $X\setminus B_r(0)$, where $B_r(0)$ is the open ball, centered at 0, with radius $r>0$. $$ \|f(u)-f(v)\|= \left\|\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\| =\frac{1}{\|u\|\|v\|}\big\|\|v\|u-\|u\|v\big\| \\ = \frac{1}{\|u\|\|v\|}\big\|\|v\|(u-v)-(\|u\|-\|v\|)v\big\| \\\le \frac{1}{\|u\|\|v\|}\big\|\|v\|(u-v)\big\|+ \frac{1}{\|u\|\|v\|}\big\|\|(\|u\|-\|v\|)v\big\| \\ \le \frac{1}{\|u\|}\big\|u-v\big\|+ + \frac{1}{\|u\|}\big|\|u\|-\|v\|\big| \le \frac{2}{r}\big\|u-v\big\|. $$

Answer 2

As already remarked in comments, the map is not Lipschitz on the whole space (without 0). I assume therefore that you actually mean the radial retraction, that is, the map $$f(x)=\begin{cases}\frac x{\lVert x\rVert}&\text{if $\lVert x\lVert \ge1$,}\\x&\text{if $\lVert x\rVert\le1$.}\end{cases}$$ The Lipschitz constant of this map depends actually on the norm. Of course, the Lipschitz constant is at most $2$, but in many cases the constant is smaller.

In fact, if the norm comes from an inner product, it can be shown that this map has Lipschitz constant $1$. On the other hand, this property characterizes inner product spaces in a sense, see e.g. https://www.mat.univie.ac.at/~esiprpr/esi1787.pdf