$${\left\lbrace z: {\operatorname{Re}}z \geq 2 N_{*} \right\rbrace}  \setminus \bigcup_{k = N_*}^{\infty} \mathcal{D}_k \label{eq:resdom}$$ outside of the disks $\mathcal{D}_k$, $k \geq N_*$, if $N_*$ is large enough.

It follows from the Neumann-Riesz decomposition $$\begin{aligned}
\begin{aligned}
R & = R^0 + R^0 W R^0 + R^0 W R^0 W R^0 + \dotsb\\
& = R^0 + \sum_{j = 1}^{\infty} U_j,
\end{aligned} \label{eq:rdecomp} \\
\intertext{where}
\begin{aligned}
U_0 &= R^0 = (z - L^0)^{-1},\\
U_j & = R^0 W U_{j - 1} = U_{j - 1} W R^0, \quad j \geq 1.
\end{aligned} \label{eq:udef}\end{aligned}$$ Of course, the convergence of the series should be explained at least in . This is done in [@AM12], [@MiSi]; now I’ll remind only the estimates of $N_*$ because it will be important later (see Theorem \[thm:eigendistr\], and Corollary XXX) in accounting for points of the spectrum $\sigma(L)$ outside of the real line.

Define a diagonal operator $K$, $$K e_j = \frac{1}{\sqrt{z - z_j}} e_j,\, \,  j = 0, 1, \dotsc, \, \, {\operatorname{Im}}z \neq 0 \label{eq:kdef}$$ with understanding that $$\sqrt{\xi} = r^{1/2} e^{i \varphi / 2} \text{ if } \xi = r e^{i \varphi}, \quad - \pi < \varphi \leq \pi. \label{eq:sqrootdef}$$ Then $K^2 = R^0$, $z \in {\mathbb{C}}\setminus 