Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying for a qualifying exam and can't seem to solve this problem. Any suggestions would be appreciated!

Let $f:[a,b] \rightarrow \mathbb R$ be absolutely continuous. Show, for each $\epsilon>0$, that there is a uniformly Lipschitz function(global) $g:[a,b] \rightarrow \mathbb R$ such that $|f(x)-g(x)|<\epsilon$ for all $x\in [a,b]$.

share|cite|improve this question
By $\,\Re\,,\,Re\,$ , did you mean $\,\Bbb R\,$ = the real numbers? – DonAntonio Aug 15 '12 at 2:33
@DonAntonio I would be very surprised if he didn't. – Pedro Tamaroff Aug 15 '12 at 2:36
What do you mean by uniformly Lipschitz? If it means something similar to what I call Lipschitz, then, hint: the Weierstrass approximation theorem. (And if so, then absolutely continuous is a red herring). – Nate Eldredge Aug 15 '12 at 2:47
So would I, @PeterTamaroff – DonAntonio Aug 15 '12 at 2:48
Sorry I did mean the real numbers. Yes Lipschitz and uniformly Lipschitz are the same for this instructor apparently:p – Dave Aug 15 '12 at 3:11
up vote 2 down vote accepted

To expand on my comment:

One approach is just to invoke the Weierstrass approximation theorem. This works even if $f$ is merely continuous, and it gives a $g$ which is a polynomial, which is drastically stronger than just being Lipschitz or even $C^\infty$.

You could also give a more direct proof. An absolutely continuous function has a derivative which is $L^1$; a Lipschitz function has a derivative which is bounded. You can approximate $L^1$ functions by bounded functions. Now, to get from the derivative back to the function, what could you do...?

Indeed, integrate. So find a bounded measurable function $h$ which is close to $f'$ in $L^1$ norm. What can you say about the difference between the integrals (from $a$ to $x$) of $f'$ and $h$?

So we can get $\int_a^x f'(t) dt - \int_a^x h(t)dt$ to be small, right? Or in other words, we can get $f(x) - f(a) - \int_a^x h(t)dt$ to be small. So what if we set $g(x) = f(a) + \int_a^x h(t)dt$?

share|cite|improve this answer
Integrate? Your second approach is what I tried to figure out before but I can't seem to get anywhere. – Dave Aug 15 '12 at 3:49
That's what I tried doing and I get that $f(x)-h(x) = \int_{a}^{x}f'(t) + f(a) - \int_{a}^{x}h'(t) - h(a)$. But I don't quite see how or why $f(a)$ and $h(a)$ are related? – Dave Aug 15 '12 at 4:11
@Dave: Careful, I am choosing $h$ to be close to $f'$, not to $f$. My $h$ is merely bounded measurable and I don't expect it to be differentiable at all. We want to be comparing $f$ with $\int_a^x h(t)dt$. See edit. – Nate Eldredge Aug 15 '12 at 13:19
So do I want something like $h(x)=\chi_{E_M}f'(x)$ where $E_{M}={x:|f'(x)\leq M|}$ is the set such that $f'(x)$ is bounded? How do I know that $h(x)$ is close enough? – Dave Aug 16 '12 at 1:01
@Dave: For any $\delta > 0$, you can find $M$ so large that the corresponding $h$ has $\int |f'(x) - h(x)| < \delta$. Now see what kind of bound on $|f-g|$ you get in terms of $\delta$. – Nate Eldredge Aug 16 '12 at 1:07

Theorem: Let M be a metric space. Then any continuous function f:M→R can be uniformly approximated by a locally Lipschitz functions. See here.

Remember that if a function is absolutely continuous on $[a,b]$ then it is continuous on $[a,b]$. The converse is not true.

share|cite|improve this answer
Yes that's what I've been trying to use but without success. – Dave Aug 15 '12 at 4:01
@Dave:Check the editing. – Mhenni Benghorbal Aug 15 '12 at 5:12
I asked the professor today about this question and he said that g should be uniform Lipschitz(global) not just local – Dave Aug 16 '12 at 1:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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