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Let $\Omega \subseteq \mathbb{R}^n$ be a domain. Suppose $u$ is locally integrable (i.e. $u\in L_{loc}^1(\Omega)$) and has a locally integrable weak derivative $\partial_i u$.

Is there a way to find the weak derivative of $\lvert u \rvert$?

I tried to show that $\partial_i \lvert u \rvert = \chi_{u>0}\partial_i u - \chi_{u<0}\partial_i u$. For this, I picked an arbitrary text function $\varphi$ and (using the dominated convergence theorem) showed that $$ \int_{\Omega} \left(\chi_{u>0}\partial_i u - \chi_{u<0}\partial_i u\right)\phi = \lim_{\epsilon \searrow 0} \int_{\Omega} \left[\left(\chi_{u>0} - \chi_{u<0}\right)\phi\right]_\epsilon\cdot \partial_i u $$ where $\left[\left(\chi_{u>0} - \chi_{u<0}\right)\phi\right]_\epsilon$ denotes the mollified version of $\left(\chi_{u>0} - \chi_{u<0}\right)\phi$.

Then, since $\left[\left(\chi_{u>0} - \chi_{u<0}\right)\phi\right]_\epsilon$ is also a test function, there holds $$ \int_{\Omega} \left[\left(\chi_{u>0} - \chi_{u<0}\right)\phi\right]_\epsilon\cdot \partial_i u = -\int_{\Omega} \partial_i\left[\left(\chi_{u>0} - \chi_{u<0}\right)\phi\right]_\epsilon\cdot u $$

Unfortunetly, this doesn't seem to be going anywhere? Any input is appreciated!

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  • $\begingroup$ I remember seeing this problem solved somewhere on MathStackExchange. I will add a link if I find it, but you should try too, it's out there! $\endgroup$ – Michał Miśkiewicz Nov 10 '18 at 23:24
  • $\begingroup$ @MichałMiśkiewicz I tried looking for a similar question but unfortunately couldn't find it. Thank you! $\endgroup$ – Quoka Nov 10 '18 at 23:53
  • $\begingroup$ Here it is. The only differences is that you're just considering one partial derivative at a time, and $L^1$ instead of $L^2$. math.stackexchange.com/questions/2578760/… $\endgroup$ – Michał Miśkiewicz Nov 11 '18 at 8:43
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    $\begingroup$ Possible duplicate of If $u\in H^1(\Omega )$ then $|u|\in H^1(\Omega )$. $\endgroup$ – Quoka Nov 11 '18 at 21:44

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