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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.

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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}

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