Projection on unit vector lipschitz condition if i have a vector $ v \in C^{n}$, where we do not want to look at vectors with norm $||v||\le 1$ with an arbitrary norm of this space and I am asked whether the map $ f(x)=\frac{x}{||x||}$ satisfies the lipschitz boundary condition $|| f(x)-f(y)|| \le L ||x-y||$. how do i proof this? Somehow it is clear, that lipschitz constant 1 will be sufficient for this, but I have troubles to show it. maybe one needs a clever inequality. 
 A: The statement is in general false taking the constant to be equal to $1$. The best possible constant in general is $2$.
It easy to prove that
\begin{equation}
\left\| \frac{x}{\| x \|} - \frac{y}{\| y \|}  \right\|= \left\| \frac{(x-y) \| y \| + y(\| y \| - \| x \|)}{\|x \| \| y \|} \right\| \leqslant \frac{2 \|x-y\| \| y \|}{\|x \| \| y \|} \leqslant 2 \| x-y \|.
\end{equation}
To show that this is, indeed, the best possible constant consider $\mathbb{R}^2$ with the $L^1$ norm given by
\begin{equation}
\| (x_1, x_2 ) \| = |x_1| + |x_2|
\end{equation}
Let $\mathbf{x} = (1,0)$ and $\mathbf{y} = (1,a)$ with $a >0$. Then it is easy to see that the following hold
\begin{align}
\| \mathbf{x} \| &= 1, \quad \| \mathbf{y} \| = 1+a\\
f(\mathbf{x}) &= (1,0), \quad f(\mathbf{y}) = \left(  \frac{1}{a+1}, \frac{a}{a+1} \right)\\
\mathbf{x} - \mathbf{y} &= (0, -a), \| \mathbf{x} - \mathbf{y} \| = a\\
f(\mathbf{x}) - f(\mathbf{y}) &= \left( \frac{a}{a+1}, -\frac{a}{a+1}  \right),\\
\| f(\mathbf{x}) - f(\mathbf{y}) \| &= \frac{2a}{a+1}.
\end{align}
Finally, we have
\begin{equation}
\lim_{a \to 0} \frac{\| f(\mathbf{x}) - f(\mathbf{y}) \|}{\| \mathbf{x} - \mathbf{y} \|} = 2.
\end{equation}
