Recall that $f:[a,b]\to\mathbb{R}$ is absolutely continuous if for all $\epsilon$ there exists $\delta$ such that for any $(a_i,b_i)$ pairwise disjoint intervals we have $\sum(b_i-a_i)<\delta\implies\sum|f(b_i)-f(a_i)|<\epsilon$, or equivalently $f$ is a.e. differentiable and:
$$f(x)-f(y)=\int_y^xf'(t)\mathrm{d}t,$$
for all $x,y\in[a,b]$, and it is Lipschitz continuous (or just Lipschitz) if there exists a constant $C$ such that, for $x,y\in[a,b]$:
$$|f(x)-f(y)|\leq C|x-y|.$$
Choosing $\delta=\frac{\epsilon}{C}$, one verifies that the definition of absolute continuity holds for Lipschitz functions.
The space of absolutely continuous functions $\operatorname{AC}([a,b])$ can be normed with:
$$\|f\|:=|f(a)|+\|f'\|_{L^1([a,b])}.$$
It turns out (see 1 and 2) that this norm makes $\operatorname{AC}$ into a Banach space.
$\operatorname{AC}$ is also the same as $W^{1,1}$, and we know that Lipschitz functions are dense there, since smooth functions are, and they are Lipschitz.
So I was wondering: are Lipschitz functions dense in the other norm as well? Mollifying doesn't really seem to help, because I do get convergence in $L^1$ of the derivatives, but the convergence at $a$ is troublesome. I mean, convergence in that norm equates to $f_n'\to f'$ in $L^1$ and $f_n(a)\to f(a)$, so I was trying to mollify and prove those two convergences. Any hints?
Perhaps I could show that norming $\operatorname{AC}$ with $|f(x)|$ instead of $|f(a)|$, where $x\in(a,b)$, results in an equivalent norm, and then mollify and use that, for continuous functions on $(a,b)$, the mollifications converge uniformly in any compact subset, hence pointwise?
Or maybe I can extend uniform convergence of mollifications to any compact subset including the whole $[a,b]$? I'd have a problem there, I guess, because how do I define the mollification at the endpoints?
