A special test function: can $\Delta \phi<0$ on the support of $\phi$?

Question

Is there a test function $\phi$ (smooth with compact support) with support in a ball of $\mathbb{R}^n$ with negative laplacian on its support:$\Delta \phi<0$ on its support?

Answer 1

Actually, this is impossible because the support of a function is a closed set. Thus if $\Delta \phi > 0$, by extreme value theorem, $\Delta \phi > c > 0$ on $\operatorname{supp} \phi$ for some constant $c$. But by continuity, this means $\Delta\phi>0$ on an open neighbourhood of the support, and therefore $\phi$ does not vanish outside its support, which is absurd.