So, what I have to begin with is a so called complex scaled Dirac operator on $L^2(\mathbb{R}^3)$. It can be written in the form
$$D_\theta = -\frac{1}{a(x)}i\alpha \cdot \nabla + \beta +R (x,\partial _x) + V(x)$$
where $a$ is a complex-valued function such that $\operatorname{Re} a$ is uniformly bounded away from 0 and $\operatorname{Im} a$ is non-negative and supported in $|x|>2$. Next, $\alpha $ is a vector of so called Dirac matrices ($4\times 4$), $\beta $ is the constant matrix $\beta = \mathrm{diag}(I_2, -I_2)$ where $I_2$ is the 2x2 identity matrix.
Moreover $R (x,\partial _x)$ is a first order differential operator with smooth compactly supported coefficients supported in $2<|x|<3$ and $V(x)$ is Hermitian and supported in $|x|<1$.
Now, my wish is to obtain a bound of the form
$$\operatorname{Im} z \|u \|^2 \le C \|(D_\theta - z)u\| \quad \text{for }\operatorname{Im} z >0 \text{ and }\operatorname{Re}z > 1$$
where C is some positive constant. My plan of attack is to consider the quantity $\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle $. For instance the claim would readily follow if I can show
$$\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle \le -C\operatorname{Im}z\|u\|^2$$
Upon expanding the left hand side above and using the fact that the largest eigenvalue of $\beta $ is 1 I obtain
$$\operatorname{Im}\langle a(x)(D_\theta -z)u,u\rangle \le (1 - \operatorname{Re}z)\langle (\operatorname{Im}a(x))u,u\rangle + \operatorname{Im}\langle a(x)Ru,u\rangle -C\operatorname{Im}z\|u\|^2$$
Clearly the first term on the right is non-positive since $\operatorname{Re}z>1$. My problem is to handle the term $\operatorname{Im}\langle a(x)Ru,u\rangle$. There are essentially two ways to go about -- either one uses the term $(1 - \operatorname{Re}z)\langle (\operatorname{Im}a(x))u,u\rangle $ to absorb it, or one estimate it by $C\|(D_\theta - z)u\|$. The latter approach seems more likely, or at least a mixture of the two. I can get close, for if $\chi $ is a smooth bump function which equals 1 near $2<|x|<|3|$ and 0 for $|x|\le 1$, then I can show
$$\operatorname{Im}\langle a(x)Ru,u\rangle \le C\|(D_\theta - z)\chi u\| \|u\|.$$
But when I in turn want to estimate this from above by $C\|(D_\theta - z)u\|\|u\|$ the error is an annoying commutator term $\|[D_\theta , \chi ]u\|$. Actually I am working in the semi-classical setting and then this term is $\mathcal{O}(h)\|u\|_{\operatorname{supp}(\nabla \chi )}$, but this is not small enough to be absorbed by any other term since $\operatorname{Im}a(x)$ is zero on part of $\operatorname{supp}(\nabla \chi )$ and $\operatorname{Im}z$ is supposed to be even smaller in terms of $h$.
Any new ideas would be greatly appreciated.
