Lipschitz continuous and matrix

Question

Suppose $f$ is Lipshitz continuous, that is $$\|f(t,u)-f(t,v)\|\le L^2\|u-v\|,$$ where $L^2$ is a Lipschitz constant. Why can we write this in the form: $$f^Tf\le L^2 x^Tx?$$

Answer 1

Consider a Lipschitz function where $f(t, 0) \neq 0$. For example $f(t,u) = \sqrt{u^2 + 5}$, with $u \in \mathbb{R}$. It has Lipschitz constant $L = 1$ (since $|\frac{\partial f}{\partial u}| = |\frac{u}{\sqrt{u^2+5}}| \leq 1$). However, $f^T f = u^2 + 5 \not\leq 1 u^2$