# Absolute Continuity, Lipschitz

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 $x\in [a,b]$.

-
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

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 $g(x) = f(a) + \int_a^x h(t)dt$?

-
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.

-
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