Square root x is not Lipschitz on [0,1]

$f(x)=\sqrt x$ is not Lipschitz on $[0,1]$:

Suppose there exists $C\geq 0$ such that for all $x,y\in [0,1]$, $x\neq y$, $\displaystyle \frac{|f(x)-f(y)|}{|x-y|}\leq C.$

By Mean Value Theorem, this means that $|f'(\xi)|\leq C$ for some $\xi$ between $x$ and $y$.

However, $f'(x)=\frac{1}{2}x^{-1/2}$ is unbounded on $[0,1]$, a contradiction.

Note however, that $\sqrt x$ is absolutely continuous on [0,1]. So Lipschitz is a stronger condition than absolutely continuous.