f(x)=√|x| is a famous example of a function which is

not Lipschitz continuous but is uniformly continuous. This link shows detailed explanation of it.Here provides the figure of this function:

However, I am still confused about how to show √|x| is not Lipschitz continuous?

- Consider [−a,a], which is compact. f(y)−f(x)y−x should be bounded by L. This method is also used to prove f(x) is a uniformly continuous.
- On [a,∞) and (−∞,−a], f(x) has a bounded derivative.
So, based on 1 and 2, f(x)=√|x| is Lipschitz continuous.

I have no idea how to prove it is not a Lipschitz continuous; clearly speaking I do not know how to

distinguish the proof of Lipschitz continuity from uniformly continuity.

**Answer**

The derivative of √|x| is sgn(x)2√|x|. Let x0=14L2. The derivative at x0 is L, so, the derivative is unbounded.

**Attribution***Source : Link , Question Author : sleeve chen , Answer Author : lisyarus*