Lipschitz function

Let $W\subseteq X\subseteq ℂ$ and $f:X\to ℂ$. Then $f$ is on $W$ if there exists an $M\in ℝ$ such that, for all $x,y\in W$, $x\ne y$

$$|f(x)-f(y)|\le M|x-y|$$

If $a,b\in ℝ$ with and $f:[a,b]\to ℝ$ 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 $x=0$.