\in \Omega_k;\quad |y-\eta |<|y|/2 \quad \}\\
\Sigma_3&=&\Omega_k\setminus (\Sigma_1\cup \Sigma_2).\end{aligned}$$ Then in addition for $|y|>2$, $$\label{1020e5}
|G(y,\eta)-G(0,\eta)|\le \left\{\begin{array}{ll}
C(\log |y|+|\log |\eta ||),\quad \eta\in \Sigma_1,\\
C(\log |y|+|\log |y-\eta ||),\quad \eta\in \Sigma_2,\\
C|y|/|\eta |,\quad \eta \in \Sigma_3.
\end{array}
\right.$$

[**Proof:**]{} The expression for $G(y,\eta)$ is $$G(y,\eta)=-\frac{1}{2\pi}\log |y-\eta |+\frac 1{2\pi}\log (\frac{|y|}{\epsilon_k^{-1}}|\frac{\epsilon_k^{-2}y}{|y|^2}-\eta |), \quad y,\eta\in \Omega_k.$$ In particular $$G(0,\eta)=-\frac{1}{2\pi}\log |\eta |+\frac 1{2\pi}\log \epsilon_k^{-1},\quad \eta\in \Omega_k.$$ Therefore we write $G(y,\eta)-G(0,\eta)$ as $$\label{1020e3}
G(y,\eta)-G(0,\eta)=\frac 1{2\pi}\log \frac{|\eta |}{|y-\eta |}
+\frac 1{2\pi}\log |\frac{y}{|y|}-\frac{|y|\eta}{\epsilon_k^{-2}}|.$$

The proof of (\[1020e5\]) for $\eta\in \Sigma_1$ is obvious. For $\eta \in \Sigma_2$, (\[1020e5\]) also obviously holds if either $|y|$ or $|\eta |$ is less than $\frac 78 \epsilon_k^{-1}$ because in this case $$\bigg |\log \big | \frac{y}{|y|}-\epsilon_k^2|y|\eta \big |\bigg |\le C.$$ Consequently $$\begin{aligned}
|G(y,\eta)-G(0,\eta)|&\le &C(\log |\eta |+\big |\log |y-\eta |\big |+C)\\
&\le &C(\log |\eta  |+\big |\log |y-\eta |\big |).\end{aligned}$$ Therefore for $\eta\in \Sigma_2$ we only need to consider the case when $|y|,|\eta|>\frac 78\epsilon_k^{-1}$. In this case it is immediate to observe that $$\bigg |\log |\frac{y}{|y|}-\epsilon_k^2|y|\eta |\bigg |<C, \quad \mbox{ if }\,\, \angle (\frac{y}{|y}, \frac{\eta}{|\eta |})>\frac{\pi}8$$ where $\angle(\cdot,\cdot)$ is the angle between two unit vectors. Thus for $\eta\in \Sigma_2$ we only consider the situation when $|y|,|\eta|>\frac 78\epsilon_k^{-1}$, $\angle(\frac{y}{|y|},\frac{\eta}{|\eta |})<\frac{\pi}8$. For this case we estimate $G(y,\eta)-G(0,\eta)$ as 