# On Lipschitz continuity of $\sqrt{x}$

The fact that the function $x\mapsto\sqrt{x}$ is not Lipschitz on the interval $[0,b]$ with $b>0$ has already been addressed in the forum; see here. My question is related to the following definition of Lipschitz continuity.

Definition. A function $f:[a,b]\rightarrow \mathbb{R}$, where $[a,b] \subset \mathbb{R}$, is Lipschitz on $[a,b]$ if and only if all of the following hold (cf. here or wikipedia article, here):

• $f$ is absolutely continuous (a.c.) on $[a,b]$
• there exists a real-valued constant $M$ such that $\vert f'(x) \vert \leq M$ for almost every (a.e.) $x \in [a,b]$.

Now, consider the function $g(x)=\sqrt{x}$ with $x \in [0,b]$, where $b>0$. It is known that $g$ is a.c. but not Lipschitz on $[0,b]$ because $\lim_{x \downarrow 0}g'(x)=\infty$; see wikipedia.

My problem / question: Clearly, the derivative $\vert g'(x)\vert$ is not bounded away from infinity everywhere on $x \in [0,b]$. But why can't we say that $\vert g'(x)\vert \leq M < \infty$ almost everywhere (except on the null set $\{0\}$) and deduce from this that $g$ is Lipschitz? Or can we say that $g$ is Lipschitz a.e. on $[0,b]$?

• You will not find such an $M$. Can you give me one? EDIT: Downvoter: Show yourself! – Friedrich Philipp Mar 10 '16 at 22:10

It is not true that $\lvert g'(x)\rvert \leq M$ for all $x > 0$. You need to remove an open neighborhood containing $0$, as otherwise there is a sequence of points $x_i$ in the domain with $x_i \to 0$ and therefore $g'(x_i) \to \infty$.
So your $g$ is not Lipschitz, and it is not Lipschitz a.e. on $[0,b]$.