A weak minimum principle.

I am not able to solve the following question that I came across: Let $\Omega \subset R^n$ be a bounded set. A function $u$ is called a weak solution of the differential inequality $$\begin{cases} -\Delta u \ge 0 &\text{in} \ \Omega \\ u = 0 &\text{in} \ \partial\Omega, \end{cases}$$ if $u\in H_0^1(\Omega)$ and $\int_\Omega \nabla u \cdot \nabla \phi \ dx \ge 0$ for all $\phi \in H_0^1$ such that $\phi \ge 0$ a.e

What I want to show is that any such weak solution $u$ satisfies weak minimum principle that $u\ge 0$ a.e in $\Omega$.

Thank you for your help .

Hint: Try taking $\phi = u^-$, where $u^-(x) = \max\{-u(x), 0\}$.
• Given any $v \in H^1_0$, show that $$v^- \in H^1_0 \quad \text{and} \quad \nabla v^- = -1_{\{v < 0\}} \nabla v. \quad (*)$$ (This is a good exercise in properties of Sobolev space. Start by considering $v \in C^\infty_0(\Omega)$. Find functions $\psi_n \in C^\infty(\mathbb{R})$ with $\psi_n(s) \to s^-$ in some appropriate sense. Show that $\psi_n \circ v \to v^-$ and $\nabla(\psi_n \circ v) \to -1_{\{v < 0\}} \nabla v$ in $L^2$. This shows (*) holds for $v \in C^\infty_0(\Omega)$. Now given arbitrary $v \in H^1_0$, approximate by $v_n \in C^\infty_0$.)
• It follows from (*) that if $u$ is a weak solution, $$\int_\Omega |\nabla u^-|^2 = -\int_\Omega \nabla u \cdot \nabla u^- \le 0.$$ This means $\nabla u^- = 0$ a.e. and it follows that $u^- = 0$ a.e.
• Sir , what does $1_{v<0}\nabla v$ mean ? I am trying to work on with the proof that u have sketched . – Theorem Jul 29 '12 at 7:20
• $1_{\{v < 0\}}$ is the function which is 1 on the set where $v < 0$, and 0 otherwise. In other notation, $$1_{\{v < 0\}}(x) = \begin{cases} 1, & v(x) < 0 \\ 0, & v(x) \ge 0. \end{cases}$$ And $1_{\{v < 0\}} \nabla v$ is just this function multiplied by the gradient of $v$. – Nate Eldredge Jul 29 '12 at 14:17