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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!

Mittens
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Uhmm
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  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.

  2. That is a true fact, but the proof takes a few steps. See the chapter 3 of the book Real Analysis (Princeton Lectures in Analysis, book 3) by E. Stein and R. Shakarchi for a proof.

Marko Karbevski
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    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 $G'\in L^1$? – Uhmm Apr 12 '22 at 18:05
  • All your statements above are correct with a minor modification: in the first step i showed that G is bounded and hence integrable, as we are looking at a finite interval. There is no approximation used though. An approach through distributions may be more "approximative" and closer to what your teacher had in mind. See proposition 1.4.5 here: http://www.math.u-psud.fr/~pgerard/Distributions2019_Chap1,2,3.pdf – Marko Karbevski Apr 12 '22 at 18:15
  • @Uhmm see the link above for a solution closer to approximations – Marko Karbevski Apr 13 '22 at 00:19