= Solution
For $\eta>0$, use the <weak maximum principle for elliptic operators> perturbation $\phi_\eta(x)=\phi(x)+\eta|x|^2$. Its <Laplacian> is at least $4\eta$, so the preceding argument puts its maximum on the boundary. If $R^2=\max_{\bar\Omega}|x|^2$, then, for any $x\in\bar\Omega$,
$$
\phi(x)\leq\max_{\partial\Omega}\phi+\eta R^2.
$$
Letting $\eta\downarrow0$ gives $\phi(x)\leq\max_{\partial\Omega}\phi$. The boundary is compact, so this maximum is attained. This proves the weak <weak maximum principle for elliptic operators> even when the <Laplacian> may vanish.
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