Hi I have encountered the following statement:

Let $\displaystyle f: S^{n-1}\rightarrow \mathbb{R}$ be a function with Lipschitz constant L, $\displaystyle |f(x)-f(y)|\leq L|x-y|$. If we define $\displaystyle g=|x|f(\frac{x}{|x|})$, "it is easy to show" that $\displaystyle g$ has Lipschitz constant at most $\displaystyle 3L$.

First of all it I showed that $\displaystyle |f(\frac{x}{|x|})-f(\frac{y}{|y|})|\leq \left\{\begin{array}{cc} \frac{2L}{|x|}|x-y| \\ \frac{2L}{|y|}|x-y| \end{array}\right.$.

I used this to show

$\displaystyle ||x|f(\frac{x}{|x|})-|y| f(\frac{y}{|y|}) |\leq |x| |f(\frac{x}{|x|})-f(\frac{y}{|y|})| + |f(\frac{y}{|y|})| ||x|-|y| | $

$\displaystyle =2L|x-y|+|f(\frac{y}{|y|})| ||x|-|y|| $

I have tried to do further manipulations with the last term to get another $\displaystyle L|x-y|$ but was not successful.

Any ideas or references to a similar problem would be great.

Thanks