22
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ @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. $\endgroup$ – Rudy the Reindeer Nov 23 '15 at 2:33
  • 2
    $\begingroup$ I think your argument just proves uniform continuity and not absolutely continuity. $\endgroup$ – Fardad Pouran Dec 3 '15 at 9:27
  • 1
    $\begingroup$ @FardadPouran Yes, as I point out in my comment above. $\endgroup$ – Rudy the Reindeer Dec 3 '15 at 10:48
11
$\begingroup$

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.
$\endgroup$
1
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ What do you mean "and these are the only absolutely continous functions."? $\endgroup$ – Rudy the Reindeer Sep 22 '12 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.