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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.

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Search this document for "absolutely continuous function". –  Matt N. May 6 '12 at 18:54
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3 Answers

up vote 6 down vote accepted

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 $$ (\sqrt{x} - \sqrt{y})^2 \leq (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = x-y < \delta$$

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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)

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What do you mean "and these are the only absolutely continous functions."? –  Matt N. Sep 22 '12 at 5:46
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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 $x^a \sin (1/x)$, which shows that $x^{3/2} \sin(1/x)$ is AC but not Lipschitz on $[0,1]$.

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