# Behavior of harmonic functions comparing with a Green’s function

Theorem Let ${E \subset \subset B_1}$, ${u \in C^0 (\bar{B_1} \setminus E) \cap C^2 (B_1 \setminus E)}$ satisfies

$\displaystyle - \Delta u =0, \quad \text{in } B_1 \setminus E, \text{ with } u \ge 0 \text{ on } \partial B_1,$

If there exists ${G \in C^2(\bar{B_1} \setminus E)}$ such that

$\displaystyle - \Delta G =0, \quad \text{in } B_1 \setminus E \text{ and } \lim_{dist(x,E) \rightarrow 0} G(x) = +\infty.$

If ${\frac{u^-}{G(x)} \rightarrow 0}$ as ${dist(x,E) \rightarrow 0}$, then ${u \ge 0}$ in ${(B_1 \setminus E)}$.

Proof: Without loss of generality assume ${G \ge 0}$ on ${\partial B_1}$, or we can add a large constant on ${G}$ to make it positive on boundary. ${\forall \varepsilon >0}$, there exists ${0 < \delta(\epsilon) < \epsilon}$ such that

$\displaystyle \frac{u^-}{G(x)} \le \varepsilon \text{ whenever } 0 < dist(x,E) \le \delta.$

Since ${\frac{-u}{G} \le \frac{\max(0, -u)}{G} = \frac{u^-}{G} \le \varepsilon}$, we have

$\displaystyle u(x) \ge - \varepsilon G(x) \text{ for } 0 < dist(x,E) \le \delta.$

Also ${u \ge - \varepsilon G(x)}$ on ${\partial B_1}$ since ${G}$ is non-negative on boundary. Then maximum principle tells ${ u \ge - \varepsilon G}$ in ${B_1 \setminus E_\delta}$, where ${E_\delta = \{x\in B_1: dist(x,E) \le \delta \}}$. Then we can first take ${\delta \rightarrow 0^+}$ and then take ${\varepsilon \rightarrow 0^+}$ to get ${u \ge 0}$ in ${B_1 \setminus E}$. $\Box$

Remark The condition ${\frac{u^-}{G(x)} \rightarrow 0}$ is essential, otherwise one can take ${-G}$ plus a large constant as a counterexample.