# Weak derivative of absolute value of function

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!

• 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! – Michał Miśkiewicz Nov 10 '18 at 23:24
• @MichałMiśkiewicz I tried looking for a similar question but unfortunately couldn't find it. Thank you! – Quoka Nov 10 '18 at 23:53
• 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/… – Michał Miśkiewicz Nov 11 '18 at 8:43
• Possible duplicate of If $u\in H^1(\Omega )$ then $|u|\in H^1(\Omega )$. – Quoka Nov 11 '18 at 21:44