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 $x\in [a,b]$.

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