Examples of absolutely continuous functions that are not Lipschitz. I have just solved an exercise, which asked to show that function $f$ is Lipschitz implies that $f$ is absolutely continuous. However, I'm wondering if the converse is true. I can't seem to think of any counterexamples at the moment. I think I'm brain dead or something, so I could use some help.
 A: See we know that indefinite integral of an integrable function is absolutely continous and these are the only absolutely continous functions.
So we want to find such f which is integrable but unbounded then its indefinite integral F(say) will be absolutely continous but derivative of F which is precisely f a.e. is unbounded and hence F is not lipschitz function.
we can define $f:(0,1)\to \mathbb R$ as
$f(x)=(x)^\frac{-1}{2} = \frac{1}{\sqrt{x}}$
then clearly $f$ is unbounded 
but it is an integrable function as its integral is $2$ on $(0,1)$ which is finite.
So its indefinite integral which is $2\sqrt(x)$ is our required absolutely continous function which is not lipschitz(as its derivative $f$ is unbounded)
A: Take any unbounded integrable function $f(x)$. Then its antiderivative $F(x) = \int_0^x f(t)$ is absolutely continuous.
But as $f$ is unbounded, F is not Lipschitz.
There is also the enjoyable article In Praise of $x^a \sin (1/x)$ [citation below], which shows that $x^{3/2} \sin(1/x)$ is AC but not Lipschitz on $[0,1]$.


*

*Kaptanoğlu, H. Turgay. "In Praise of $y= x^\alpha \sin \left(\frac{1}{x}\right)$." The American Mathematical Monthly 108.2 (2001): 144-150.

A: Consider $f(x) = \sqrt{x}$ on $[0,c]$. Then $f^\prime (x) = \frac{1}{2 \sqrt{x}}$ is not bounded and hence $f$ is not Lipschitz.
But $f(x) = \sqrt{x}$ is absolutely continuous on $[0,c]$. To see this observe
(i) $(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = x - y$
(ii) Since $- \sqrt{y} \leq \sqrt{y}$ you have $(\sqrt{x} - \sqrt{y}) \leq (\sqrt{x} + \sqrt{y})$
Now let $\varepsilon > 0$. Choose $\delta := \varepsilon^2$ then
$$ (\sqrt{x} - \sqrt{y})^2 \leq (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = x-y  < \delta$$
