The question is to proof: If the harmonic function on the unbounded region in $\mathbb R^3$ tends to $0$ at infinity, then the order in which the function tends to $0$ is at least $O(\frac 1r)$.
I have learnt the Kelvin transformation and the singularity removal theorem, but I didn't find how to use them...
Thanks in advance!
Let $\mathcal{U}(0;1)$ denote the set of points $\boldsymbol{x}=(x_1,x_2,x_3)\in\mathbb{R}^3$ such that $\|\boldsymbol{x}\|^2=x_1^2+x_2^2+x_3^2<1$, that is the open unit ball. Suppose $f : \mathbb{R}^3 \setminus \mathcal{U}(0;1)$ is continuous and satisfies that it is harmonic in $\{\boldsymbol{x}\, :\, \|\boldsymbol{x}\|^2>1\}$. Then let $g : \overline{\mathcal{U}}(0;1) \to \mathbb{R}$ be the function $g(\boldsymbol{x}) = \frac{1}{\|\boldsymbol{x}\|} f\left(\frac{\boldsymbol{x}}{\|\boldsymbol{x}\|^2}\right)$ where $\overline{\mathcal{U}}(0;1)$ is the closure of $\mathcal{U}(0;1)$, namely $\{\boldsymbol{x}\, :\, \|\boldsymbol{x}\|^2\leq 1\}$. Then $g$ is harmonic. This requires a calculation, but it is stated on Wikipedia here https://en.wikipedia.org/wiki/Method_of_image_charges#Method_of_inversion.