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.

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  • $\begingroup$ Thanks. But any lower than 0 will lead to square root of a negative number, which is undefined right? E.g. $\sqrt x$ on $[-\epsilon, 2]$? $\endgroup$ – yoyostein Nov 15 '16 at 8:34
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    $\begingroup$ Try defining it differently for $x<0$. $\endgroup$ – detnvvp Nov 15 '16 at 8:36
  • $\begingroup$ Nice. I am trying $f(x)=\sqrt x$ on $[0,2]$, $f(x)=0$ on $[-1,0]$. $\endgroup$ – yoyostein Nov 15 '16 at 8:40
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    $\begingroup$ Right; this will be continuous on $[-1,2]$, and not Lipschitz on $[0,1]$. $\endgroup$ – detnvvp Nov 15 '16 at 8:41
  • $\begingroup$ Since your answer does not really answer the original question, perhaps you should rewrite to reflect the comments. $\endgroup$ – zhw. Nov 15 '16 at 18:17

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