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

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 $y$. 
