If the weak derivative $\nabla u$ of $u\in L^2(\Omega)$ exists, then $\int_\Omega|\nabla u|^2=\int_\Omega|\nabla u^+|^2+\int_\Omega|\nabla u^-|^2$

Question

Let

• $\Omega\subseteq\mathbb{R}^n$ be bounded
• $u\in \mathcal{L}^2(\Omega)$ be weakly differentiable, i.e. $$\int_\Omega u\nabla\psi\;d\lambda^n=-\int\psi\nabla u\;d\lambda^n\;\;\;\text{for all }\psi\in C_0^\infty(\Omega)$$ (where we carefully denote the weak derivative of $u$ by $\nabla u$)
• $x^+:=\max(x,0)$ and $x^-:=\max(-x,0)$ for $x\in\mathbb{R}$

How can we show, that $$\int_\Omega\left|\nabla u\right|^2\;d\lambda^n=\int_\Omega\left|\nabla u^+\right|^2\;d\lambda^n+\int_\Omega\left|\nabla u^-\right|^2\;d\lambda^n\tag{1}$$ and what can we derive for $\nabla u^\pm$? (e.g. $\nabla u^\pm=\nabla u$ almost everywhere in $\left\{u^\pm >0\right\}$ or even $\nabla u=\nabla u^+-\nabla u^-$ almost everywhere).

Answer 1

I think theorem 4 (iii) on page 131 in this book is good enough to answer your question.