R_0(i\sigma) \vhalf_j$ is a positive operator, $$\begin{aligned}
& (I\pm V_j R_0(i\sigma) )\vhalf_j (I\pm \vhalf_j R_0(i\sigma)\vhalf_j)^{-1}\\
& 
 = \vhalf_j (I\pm \vhalf_j R_0(i\sigma)\vhalf_j)
(I\pm \vhalf_j R_0(i\sigma)\vhalf_j)^{-1}
\\& = \vhalf_j.\end{aligned}$$ Thus $$\vhalf_j(I\pm \vhalf_j R_0(i\sigma) \vhalf_j)^{-1} 
\vhalf_j = (I \pm V_j R_0(i\sigma))^{-1}V_j,\; j=1,2$$ for $\sigma>0 $ sufficiently large. Applying [@vasy Lemma 2.2], and using that $V_2\geq V_1$, we get $$(I + V_2 R_0(i\sigma))^{-1}V_2\geq (I + V_1 R_0(i\sigma))^{-1}V_1.$$ When we take the “$-$” sign, again applying [@vasy Lemma 2.2], $$(I - V_2 R_0(i\sigma))^{-1}V_2\geq (I - V_1 R_0(i\sigma))^{-1}V_1$$ when $\sigma>2( \|V_2\|_\infty +1)$. Here we note our convention differs somewhat from [@vasy], in that we take $V_j\geq 0$. Summarizing, $$\vhalf_2(I \pm \vhalf_2 R_0(i\sigma) \vhalf_2)^{-1} \vhalf_2
\geq \vhalf_1(I \pm \vhalf_1 R_0(i\sigma) \vhalf_1)^{-1} \vhalf_1$$ when $\sigma>0$ (for the “$+$” sign) or $\sigma>2(\|V\|+1)$ (for the “$-$” sign). For the remainder of the proof, we shall assume $\sigma>0$ satisfies these requirements and suppress the argument $i\sigma$.

Now let $\chi_{V_2}$ be the characteristic function of the support of $V_2$ and recall $\vhalf_1 \chi_{V_2}=\vhalf_1$ and note that $\chi_{V_2}(I\pm \vhalf_2 R_0\vhalf_2)= (I\pm \vhalf_2 R_0\vhalf_2)\chi_{V_2}$. Then $$\chi_{V_2} (I \pm \vhalf_2 R_0 \vhalf_2)^{-1} \chi_{V_2} 
\geq  \frac{ \vhalf_1}{\vhalf_2}(I \pm \vhalf_1 R_0 \vhalf_1)^{-1} \frac{\vhalf_1}
{\vhalf_2}.$$ This implies $$\chi_{V_2} \geq (I \pm \vhalf_2 R_0 \vhalf_2)^{1/2}
\frac{ \vhalf_1}{\vhalf_2}(I \pm \vhalf_1 