Suppose $B$ is a ball in $\mathbb{R}^n$ and $n>1$. Is there a non-zero test function $\phi$ in $B$ ($C^\infty$-smooth function with compact support in $B$) that is subharmonic everywhere, i.e., the laplacian $\Delta\phi\geq0$ everywhere?

  • $\begingroup$ @Bernard: Thank you. $\endgroup$
    – M. Rahmat
    Oct 1, 2020 at 23:06

1 Answer 1


I don't believe so. Let $\phi$ be such a function, and let $K = \text{supp} \phi$ be the (compact, contained in $B$) support of $\phi$. Since $\phi$ is subharmonic, it obeys a maximum principle: $$ \max_{\bar{U}} \phi = \max_{\partial U}\phi. $$ Therefore, since $\phi = 0$ on $\partial B \subset K^c$, we have that $\phi \leq 0$ in all of $U$. But we also have a mean-value property for subharmonic functions: $$ \phi(x) \leq \frac{1}{| B(x, r)|}\int_{B(x, r)} \phi(y)\, dy $$ whenever $B(x, r) \subset B$. Letting $x \in K^c$ be closer to $K$ than to $\partial B$ and letting $r > 0$ such that $B(x, r) \subset B$ intersects a large enough region where $\phi \leq 0$ then yields a contradiction in the above formula.


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