# Sign of Laplacian Green's function in 3D

I am trying to prove that on a "nice" domain $\Omega$ in $\mathbb{R}^{3}$, the Green's function $G$ of $\bigtriangleup$ (the Laplacian) on $\Omega$ is always negative. I would like to use the Maximum Principle for harmonic functions, since $G(\mathbf{x}, \mathbf{x}_{0})=0$ on $\partial \Omega$, so naturally I want to show that $G(\mathbf{x}, \mathbf{x}_{0})$ is harmonic on $\Omega$. Here's what I have so far:

Writing $G(\mathbf{x}, \mathbf{x}_{0})=H(\mathbf{x}, \mathbf{x}_{0})+\mathbf{\Phi}$ (where $H$ is harmonic on $\Omega$ and is such that $G=0$ on $\partial \Omega$, $\mathbf{\Phi}=-\frac{1}{4\pi |\mathbf{x}-\mathbf{x}_{0}|}$ is the fundamental solution in 3D), $G$ must be harmonic on $\Omega^{'}=\Omega -B_{\epsilon}(\mathbf{x}_{0})$, where $B_{\epsilon}(\mathbf{x}_{0})$ is the ball centered at $\mathbf{x}_{0}$ of radius $\epsilon$.

I want to show that this is equivalent to $G$ being harmonic on $\Omega$ as $\epsilon \rightarrow 0$ so that I may then evoke the aforementioned Maximum Principle to show that $G$ is bounded above by $0$, but I'm not sure of how to do this.

If you apply maximum principle on $\Omega \setminus B_{\epsilon}(\mathbf{x}_0)$, there are two boundaries. As you have said Green function takes zero on $\partial \Omega$.
For another boundary, note that $H$, being harmonic, is a twice continuously differentiable function in $\Omega$, so $|H(\mathbf{x}_0)|<\infty$, which implies that $\mathbf{x}_0$ is a blow-up singularity for G (where $\Phi$ kills $H$), $G(\mathbf{x}_0,\mathbf{x}_0)$ is $-\infty$.
You may argue that $\forall M>0$, $\exists\epsilon>0$, such that $G(\mathbf{x},\mathbf{x}_0)<-M$ for $\mathbf{x}\in \partial B$; then by maximum priciple you are done with $\Omega \setminus B_{\epsilon}(\mathbf{x}_0)$, finally just take $\epsilon \to 0$.