# Lipschitz-continuity of $x\mapsto\frac{x}{||x||}$ in a general Banach space

Let $(X,||.||)$ be a Banach space. Assume we have constants $0<C_1<C_2<\infty$. Define the set

$A:=\{x\in X\text{ }|\text{ } C_1\le ||x||\le C_2\}$.

Is the map

$f\colon A\rightarrow X$, $x\mapsto\frac{x}{||x||}$,

Lipschitz-continuous? (I.e., does there exist a $C\ge0$ s.t. $\|f(x)-f(y)\|\le C||x-y||$ for all $x,y\in A$.)

If not, is $f$ Lipschitz-continuous in the case where $X$ is a Hilbert space?

• It does not seem to be a map to $\mathbb{R}$ – Aman May 26 '16 at 12:06

Yes, it is Lipschitz from $A$ to $X$. You have \begin{align*} \frac{x}{\|x\|} - \frac{y}{\|y\|} &= \frac{\|y\| \, x - \|y\|\,y + \|y\|\,y - \|x\|\,y}{\|x\|\,\|y\|} \\ &= \frac{ x - y}{\|x\|} + \frac{\|y\|- \|x\|}{\|x\|\,\|y\|} \, y. \end{align*} Hence, \begin{align*} \Bigg\| \frac{x}{\|x\|} - \frac{y}{\|y\|} \Bigg\| \le \frac{ \| x - y \|}{\|x\|} + \frac{\big|\|y\|- \|x\|\big|}{\|x\|} \le 2 \, \frac{ \| x - y \|}{\|x\|} \le \frac{2}{C_1} \, \| x - y \|. \end{align*} Thus, you can even choose $C_2 = \infty$.