Can a Lipshitz Continuous function on an interval with a Uniformly continuous derivative have a nonexistant second derivative.

Question

if $f:\left[a,b\right]\mapsto\mathbb R$ Is Lipschitz Continuous

And

$f':\left[a,b\right]\mapsto\mathbb R$ exists for all $x \in [a,b]$ and is is Uniformly Continuous

then Does $f'':\left[a,b\right]\mapsto\mathbb R$ exist on all $x \in \mathbb [a,b]$

Edit This is part of a much larger question that is also on this site. Also I have completely reworded the question.

Answer 1

$$f(x) = |x|.$$

Lipschitz Continuous on $[-1,1]$, does not even have a derivative in $0$, therefore $f'$ is not uniformly continuous on $[-1,1]$.

As the answer to your edit: You cannot prove the converse. $f'$ being uniformly continuous imples that $f$ is Lipschitz continuous, but the converse is not true.

Edit: You completely reworded your question, so here is the answer to your current question:

Take $$f(x) = \begin{cases}x^2&\text{ if } x\geq0\\ -x^2&\text{ if } x\leq 0\end{cases}$$ defined on $[-1,1]$.

$f'(x) = |x|$ is then uniformly continuous and $f$ is Lipschitz continuous. $f''$ does not exist for $x=0$