Let $f$ be absolutely continuous on $[a,b]$, then is it Lipschitz on all compact subsets of $(a,b)$?
I am not able to prove or disprove it.
My attempt to disprove: The favorite example $f(x)=\sqrt x$ on $[0,1]$ doesn't seem to work here since it is Lipschitz on compact subsets of $(0,1)$. (bounded derivatives on compact subset)
My attempt to prove: \begin{align*} |f(y)-f(x)|&=\left|\int_x^y f'(t)\,dt\right|\\ &\leq\int_x^y |f'(t)|\,dt \end{align*}
I am unable to proceed since $f'$ is not necessarily bounded.
Thanks for any help!
Hint: $\sqrt{x}$ will work, but consider it in an interval larger than $(0,1)$.
Edit: So, we consider $$f(x)=\left\{\begin{array}{c l} 0, &-1\leq x\leq 0 \\ \sqrt{x}, & 0<x\leq 2\end{array}\right..$$ Then $f$ is differentiable on $(-1,2)\setminus\{0\}$, with $$f'(x)=\left\{\begin{array}{c l} 0, &-1<x<0 \\ \frac{1}{2\sqrt{x}}, & 0<x<2\end{array}\right.,$$ which is in $L^1\left([-1,2]\right)$, and $$f(x)=f(-1)+\int_{-1}^xf'(t)\,dt$$ for all $x\in[-1,2]$. Therefore $f$ is absolutely continuous on $[-1,2]$. However, $f$ is not Lipschitz on $[0,1]$: if it was, then, for some $M>0$ and all $x\in[0,1]$, $$\sqrt{x}=|f(x)-f(0)|\leq M|x-0|=Mx\Rightarrow M\sqrt{x}\geq 1,$$ which is a contradiction.