if $ \forall K\in\mathbb{R},\exists x_{1},x_{2}\in\left(a,b\right):|f\left(x_{1}\right)-f\left(x_{2}\right)|>K|x_{1}-x_{2}| $ then

Question

Let $ f: (a,b)\to \mathbb{R} $ be a function that follows:

$ \forall K\in\mathbb{R},\exists x_{1},x_{2}\in\left(a,b\right):|f\left(x_{1}\right)-f\left(x_{2}\right)|>K|x_{1}-x_{2}|.$

Is it true that $ f $ is not uniformly continious in $ (a,b) $ ?

I know it would mean (obviously) that the function is not Lipschitz function, and it shouldn't mean that the function is not uniformly continious, but I can't find a counterexample.

Any ideas would help. Thanks in advance.

Answer 1

$f:(-1,1) \to \Bbb R$ defined by $$f(x)= \begin{cases} \sqrt{x} & x \ge 0 \\ - \sqrt{-x} & x < 0 \end{cases} $$

is uniformly coninuous, but it is not Lipschitz continuous.

That's because it can be continuously extended to $[-1;1]$, and in a compact set every continuous function is uniformly continuous.

However, it is not Lipschitz because it is differentiable everywhere except at $x=0$, where $$f'(0)= + \infty$$