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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". – Rudy the Reindeer May 6 '12 at 18:54
up vote 9 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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@Craig You are right, it's not clear at all, I only hint at how to show uniform continuity. It would have been much better to use Lebesgue integrability of the square root function as in definition (2) here‌​. I will edit the answer when I have time. Thanks a lot for pointing out this shortcoming. – Rudy the Reindeer Nov 23 '15 at 2:33
    
I think your argument just proves uniform continuity and not absolutely continuity. – Fardad Pouran Dec 3 '15 at 9:27
    
@FardadPouran Yes, as I point out in my comment above. – Rudy the Reindeer Dec 3 '15 at 10:48

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.
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The link is broken. – AJY Mar 21 at 17:18
    
Now I've removed the link, and just thrown up the everlasting citation. Thank you for letting me know. – mixedmath Mar 21 at 17:23

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."? – Rudy the Reindeer Sep 22 '12 at 5:46

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