Let $f(x):\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a globally Lipschitz function. Does this mean that there exists a matrix $M(x,y)$ such that:
$$f(x)-f(y)=M(x,y)[x-y]$$
I do not want to use the Jacobian matrix because the function is not differentiable. If this is true, can I write $M(x,y)$ as follows
$$M(x,y)=\left\{\begin{array}{cc} \big[f(x)-f(y)\big]\frac{(x-y)^T}{\|x-y\|^2} & x\neq y\\ 0 & x=y \end{array}\right.$$