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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 $x\in [a,b]$.

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    The current title is very uninformative. Guys, can you come up with something with something more informative; better titles help future better search this website & benefit from the answers (I don't have a suggestion though). – 2012-08-15

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

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    I see so now I have the following |f(x)-g(x)|=|f(x)-f(a) + \int_{a}^{x}h(t)dt|=\int_{a}^{x}|f'(t)-h(t)|dt<\epsilon. But since $h(x)=\chi_{E_m}f$ I have that $f'$ is bounded on $E_M$ and $f$ is absolutely continuous so $h(x)$ is Lipschitz. – 2012-08-16
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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.

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    I asked the professor today about this question and he said that g should be uniform Lipschitz(global) not just local – 2012-08-16