Variational calculus - inequality

Question

Let $u \in L^1_{loc}(\Omega)$, and suppose that $\int _{\Omega} u(x)\eta (x)\; dx \geq 0$, $\forall \eta \in C ^{\infty}_{0} (\Omega)$ . Then $u(x) \geq 0$ , a. e. $x \in\Omega$

Answer 1

Suppose $u < 0$ on some positive-measure set. Then we can pick $\delta>0$ such that $u < - \delta$ on a bounded set $A$. Let $\epsilon>0$ and pick $K_{\epsilon} \subset A$ compact such that $|A \setminus K_{\epsilon}| < \epsilon/2$. Then pick $U_{\epsilon}$ containing $K_{\epsilon}$ such that $|U_{\epsilon} \setminus K_{\epsilon}| < \epsilon/2$ and $U_{\epsilon}$ is a finite collection of intervals. Then there exists a $C_{c}^{\infty}$ function $\eta_{\epsilon}$ that is 1 on $K_{\epsilon}$ and 0 outside of $U_{\epsilon}$, and $0 \leq \eta_{\epsilon} \leq 1$ on $U_{\epsilon} \setminus K_{\epsilon}$. Thus we calculate \begin{align*} \int_{\Omega} u\eta_{\epsilon} &= \int_{K_{\epsilon}}u\eta_{\epsilon} + \int_{U_{\epsilon} \setminus K_{\epsilon}} u\eta_{\epsilon} \\ &\leq \int_{K_{\epsilon}}-\delta+\int_{U_{\epsilon} \setminus K_{\epsilon}} u \\ & = -\delta|K_{\epsilon}|+\int_{U_{\epsilon} \setminus K_{\epsilon}} u. \end{align*} As we pick $\epsilon$ smaller, $K_{\epsilon}$ gains robust measure because $|A \setminus K_{\epsilon}| < \epsilon$. Also, the fact that $u \in L_{loc}^{1}$ implies that the last integral on the right tends to 0 by refining our choice of $\epsilon$. But then, for some small enough $\epsilon$, the right hand side becomes negative, a contradiction.