Asymptotic behavior of fundamental solution to Laplace's equation.

Question

Let $K \subset \mathbb{R}^{3}$ be compact. I want to prove that there exist $a,b > 0$ such that \begin{align} \left \vert \frac{1}{\lvert x -y \rvert} - \frac{1}{\lvert x \rvert} - \frac{x \cdot y}{\lvert x \rvert^{3}} \right \vert \le \frac{a}{\lvert x \rvert^{3}} \end{align} for all $x$ such that $\lvert x \rvert \ge b$ and all $y \in K$. I'm not entirely sure how to go about this. I was thinking to perform a Taylor series expansion of \begin{align} f(s) = \frac{s \lvert x \rvert }{\lvert x - s \lvert x \rvert y \rvert} \end{align} around $s = 0$, and then set $s = 1/\lvert x \rvert$. But perhaps there is a much simpler way?

Answer 1

Take a look at Taylor expansion of Inverse Distance. The meaning of $\mathcal{O}\left(\frac{|y|^2}{|x|^3}\right)$ is that there is a constant $A>0$ and a $\delta>0$ such that for all $(x,y)$ satisfying $\frac{|y|^2}{|x|^3}\leq \delta$, we have \begin{align} \left|\frac{1}{|x-y|}-\frac{1}{|x|}-\frac{x\cdot y}{|x|^3}\right|\leq A\frac{|y|^2}{|x|^3}\tag{$*$} \end{align} To get the uniform estimate you have, we only need $K$ to be bounded, compactness is not necessary. So, suppose $K$ lies inside a ball of radius $M>0$. Then, provided $|x|\geq b:=\left(\frac{M^2}{\delta}\right)^{1/3}$, we have for each $y\in K$ that \begin{align} \frac{|y|^2}{|x|^3}\leq M^2\cdot \frac{\delta}{M^2}=\delta \end{align} Hence, by the estimate $(*)$, \begin{align} \left|\frac{1}{|x-y|}-\frac{1}{|x|}-\frac{x\cdot y}{|x|^3}\right|\leq A\frac{|y|^2}{|x|^3}\leq \frac{AM^2}{|x|^3}. \end{align} In other words, we have $a=AM^2$.