$x\in \Real$, $|x|\geq 1$, $$\begin{aligned}
\frac{h'(x)}{h(x)}& =\frac{f'(x)}{f(x)}+ 
\frac{(g_1/g_2)'(x)}{g_1(x)/g_2(x)}.\end{aligned}$$ By applying our assumptions on $f$ and Lemma \[l:cproductreals\], we find that for $x>1$, $|\int_1^x h'(t)/h(t) dt| = O(x^\rho)$, and likewise for $x<-1$, $|\int _{x}^{-1} h'(t)/h(t) dt|=O(|x|^{\rho})$. By Lemma \[l:p-lapp\], there is a constant $C$ so that $$\label{eq:hbd}
\log |h(z)|\leq C(1+ |z|^\rho),\;  \text{when}\; z\in \Omega.$$

Now we write $f(z) = g_2(z)h(z)/g_1(z)$, holomorphic in a neighborhood of $\Omega$. Given $R>1$ and $\delta$ satisfying $0<\delta<1$, as above we choose $r_1\in [R,R(1-\delta)^{-1}]$ so that (\[eq:g2lb\]) holds for $g_1$. Using in addition (\[eq:gjbound\]) and (\[eq:hbd\]), we find there is a constant so that $$\log |f(z)|\leq C(1+(R(1-\delta )^{-1})^{\rho})\; \text{for}\; |z|=r_1,
\; {\operatorname{\rm Im}\nolimits}z\geq 0.$$ As in the proof of the bound on $h$, since $|h(x)|=|f(x)|$ for $x\in \Real \cap \Omega_R$, we find then that there is a constant $C$ so that $$\max_{z\in \Omega_R}\log |f(z)|\leq C(1+R^{\rho}).$$

A scalar function having zeros at the poles of the resolvent {#s:defF}
============================================================

We recall the derivation of some identities commonly used in the study of resonances for Schrödinger operators. Let $V\in L^{\infty}_{\comp}(\Real^d)$ and let $d\geq 2$ be even. There is no need to make an assumption on the sign of $V$ here. We recall the notation $R_V(\lambda)=(-\Delta +V-\lambda^2)^{-1}$ when $\lambda \in 