# Positive subharmonic test function

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?

• @Bernard: Thank you. – M. Rahmat Oct 1 '20 at 23:06

## 1 Answer

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.