Show that $\| f(x) - f(y) \| \leq 2 \| x - y \| $

Question

Let $V$ be a vector space with a given norm $\| \cdot \|$. Define a function $f:V \to V$ in the following way: $$ f(x) = \begin{cases} x, & \|x\| \leq 1 \\ x/ \|x\|, & \|x\| > 1 \end{cases} $$

Prove, that $\| f(x) - f(y) \| \leq 2 \|x-y\|$ for all $x,y \in E$.

I tried to consider all $3$ cases, namely $\|x\|,\|y\| \lessgtr1$ and $\|x\| > 1, \|y\| \leq 1$, but it did not help much.

Answer 1

We will consider two cases.

First case: WLOG, assume $\| x\| \geq 1$ and $ \|y \| \geq 1$, then we see that \begin{align} \| f(x) -f(y)\| \leq&\ \Bigg\| \frac{x}{\|x\|}-\frac{y}{\|y\|}\Bigg\| \leq \frac{1}{\|x\| \|y\|}\Big\| x\|y\|-y\|x\|\Big\|\\ \leq&\ \frac{1}{\|y\|} \Big\| x\|y\|-y\|y\|+y\|y\|-y\|x\|\Big\|\\ \leq&\ \|x-y\|+ \Big|\|x\|-\|y\| \Big|\\ \leq&\ 2\|x-y\|. \end{align}

Second case: WLOG, assume $\|x\| \leq 1$ and $\|y\| >1$, then we see that \begin{align} \|f(x)-f(y)\| \leq&\ \Bigg\| x-\frac{y}{\|y\|}\Bigg\| \leq \frac{1}{\|y\|}\Big\|x\|y\|-y\Big\|\\ \leq&\ \frac{1}{\|y\|} \Big\| x\|y\| -y\|y\|+y\|y\|-y\Big\|\\ \leq&\ \|x-y\|+\Bigg\|y-\frac{y}{\| y\|} \Bigg\|\\ \leq&\ \|x-y\|+\|y-x \| =2\|x-y\| \end{align} where the last inequality comes from the fact that $y/\|y\|$ on the unit ball is distance minimizing from the point $y$.