$V\geq 0$, set $$\label{eq:B1}
B_{1,\pm, V}(i\sigma ) = (I\pm \vhalf R_0(i\sigma)\vhalf)^{-1/2} \vhalf
 T(i\sigma) \vhalf (I\pm \vhalf R_0(i\sigma)\vhalf)^{-1/2}$$ and notice that if $\sigma >0$ (for the “$+$” sign) or $\sigma > 2(\| V\|_{\infty} +1)$ (for the “$-$” sign), $B_{1,\pm V}(i\sigma )$ is a self-adjoint trace class operator. Comparing (\[eq:Fmrewrite\]), we see that $$F_{m,\pm V}(i\sigma)= \det(I \pm i m B_{1,\pm, V}(i\sigma)).$$ Hence for sufficiently large $\sigma$ $$\begin{aligned}
\label{eq:detprod}
|F_{m,\pm V}(i\sigma)| &  =  \left|\prod (I+ im \lambda_j(B_{1,\pm, V}(i\sigma))) \right| \nonumber 
 \\ & = \prod \left|(I+ im \lambda_j(B_{1,\pm, V}(i\sigma)))\right|
\nonumber \\ &  = \prod 
\sqrt{1+ m^2  \lambda_j^2(B_{1,\pm, V}(i\sigma))}\end{aligned}$$ where $\lambda_j(B_{1,\pm, V})$ are the nonzero eigenvalues of $B_{1,\pm, V}$, repeated according to multiplicity and arranged in decreasing order of magnitude: $|\lambda_1(B_{1,\pm, V})|
\geq |\lambda_2(B_{1,\pm, V})| \geq ...$.

Now we turn to $V_1$ and $V_2$, and $\sigma$ as in the statement of the lemma. Note that $$\begin{gathered}
B_{1,\pm, V_1}(i\sigma)=  
(I \pm \vhalf_1 R_0(i\sigma) \vhalf_1)^{-1/2} \frac{\vhalf_1}{\vhalf_2}
(I \pm \vhalf_2 R_0(i\sigma) \vhalf_2)^{1/2}  B_{1, \pm , V_2}(i\sigma)  \\
\times 
(I \pm \vhalf_2 R_0(i\sigma) \vhalf_2)^{1/2}\frac{\vhalf_1}{\vhalf_2}(I \pm \vhalf_1 R_0(i\sigma) \vhalf_1)^{-1/2}.\end{gathered}$$ Again we use the convention that $\vhalf_1/\vhalf_2$ is $0$ outside the support of $V_1$. The lemma now follows from (\[eq:detprod\]) and Lemmas \[l:normbd\] and \[l:eigenvaluebd\].

In order to obtain the lower 