# Absolutely continuous function Sobolev space

Today we had a $$L^1((0,A);\mathbb{R})$$-function $$G: (0,A)\rightarrow \mathbb{R}$$ with $$G(a)=\int_0^ag(s)ds$$ for all $$a\in (0,A)$$ which is absolutely continuous, $$A>0$$. Why is $$G\in W^{1,1}(0,A;\mathbb{R})$$ then? They only said proof with approximation and that's it. How do I do so?

I know that $$C_c^{\infty}$$ is dense in $$L^1$$ so if $$G\in L^1((0,A))$$ then $$G_{\epsilon}=\eta_{\epsilon}\ast G$$ is a smooth function with $$G_{\epsilon} \rightarrow G$$ when $$\epsilon\rightarrow 0$$. To show $$G\in W^{1,1}(0,A;\mathbb{R})$$, I need to show $$G,G'\in L^1((0,A))$$, right? Can I use $$G'=g\in L^1((0,A))$$?

Thanks for any help!

1. Yes, you need to show precisely that. Assuming the interval is finite and that $$g$$ is $$L^1$$ notice that $$|G(t)| \le \|g\|_1$$ for all $$t$$ and so $$\int_0^A |G|\le \int_0^A(\int_0^x |g|) dx\le \int_0^A (\int_0^A|g(t)|dt)dx= A \|g \|_1$$ so there you have the first step.
• Thank you for your help and the recommentation of the book! In the first step you proved that the integral of $G$ over $(0,A)$ is bounded, therefore $G \in L^1((0,A))$, right? And in the second step I need to show $G'\in L^1$?