= Solution
Multiply $\Delta u+qu\leq0$ by $\zeta^2/u_\varepsilon$ and integrate. Since $\zeta$ is compactly supported, <integration by parts> and Young's inequality give
$$
\begin{aligned}
\int q\frac{u}{u_\varepsilon}\zeta^2
&\leq-\int\frac{\Delta u}{u_\varepsilon}\zeta^2\\
&=2\int\frac{\zeta Du\cdot D\zeta}{u_\varepsilon}
-\int\frac{\zeta^2|Du|^2}{u_\varepsilon^2}\\
&\leq\int|D\zeta|^2.
\end{aligned}
$$
As $\varepsilon\downarrow0$, $u/u_\varepsilon$ converges to the indicator of $\{u>0\}$. The <dominated convergence theorem> therefore gives
$$
\boxed{\int_{\{u>0\}}q\zeta^2\leq\int|D\zeta|^2.}
$$
Back to article page
