Let $W \subseteq X \subseteq \C$ and $f\colon X\to\C$ . Then $f$ is Lipschitz on $W$ if there exists an $M\in\R$ such that, for all $x,y\in W$ , $x \neq y$ $$|f(x)-f(y)|\leq M|x-y|$$

If $a,b\in\R$ with $a<b$ and $f\colon [a,b]\to\R$ is Lipschitz on $(a,b)$ , then $f$ is absolutely continuous on $[a,b]$ .

Example: Is

$$f(x) = \frac{1}{\sqrt{x}},~~~x \in [0,1]$$

a Lipschitz function.

We need to estimate the constant $M$ .

$$|f(x) - f(y)| = \left|\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{y}} \right| = \left|\frac{\sqrt{x} - \sqrt{y}}{\sqrt{xy}} \right| = \left|\frac{x - y}{\sqrt{xy} (\sqrt{x} + \sqrt{y})} \right| = \frac{1}{|\sqrt{xy} (\sqrt{x} + \sqrt{y})|} |x-y|.$$

It follows that

$$ M = \frac{1}{|\sqrt{xy} (\sqrt{x} + \sqrt{y})|} $$

and $f(x)$ is not Lipschitz at $x=0$ .