Localization for monotone quasi-periodic Schrödinger operators]Localization for  Lipschitz  monotone quasi-periodic Schrödinger operators on $\Z^d$ via Rellich functions analysis
Cao]Hongyi Cao
[H. Cao] School of Mathematical Sciences,
Peking University,
Beijing 100871,
Shi]Yunfeng Shi
[Y. Shi] School of Mathematics,
Sichuan University,
Chengdu 610064,

Zhang]Zhifei Zhang
[Z. Zhang] School of Mathematical Sciences,
Peking University,
Beijing 100871,

We establish the Anderson localization and exponential dynamical localization for  a class of quasi-periodic Schrödinger operators on $\Z^d$ with bounded or unbounded Lipschitz  monotone potentials via multi-scale analysis based on Rellich function analysis in the perturbative regime. We show that at each scale, the resonant  Rellich function uniformly  inherits the Lipschitz monotonicity property of the potential via a novel Schur complement argument.

§ INTRODUCTION  AND MAIN RESULTS

In this paper, we consider a class of quasi-periodic Schrödinger operators on $\Z^d$
\begin{align}\label{model}
H(\theta)=\varepsilon \Delta+v(\theta+ x\cdot{\omega})\delta_{x, y},\ x,y\in\Z^d,
\end{align}

where $\varepsilon\geq0$ and  the discrete Laplacian $\Delta$ is defined as

\begin{align*}
\Delta(x, y)=\delta_{{\|x- y\|_{1}, 1}},\ \| x\|_{1}:=\sum_{i=1}^{d}\left|x_{i}\right|.
\end{align*}

For the diagonal part of (<ref>),  we call $\theta$ the phase  and $\omega$ the frequency.  We  let $\theta\in\R$ and $\omega$ be Diophantine  (denoted by $\omega\in 	{\rm DC}_{\tau, \gamma}$ for some $\tau>d,\gamma>0$), where
\begin{equation}\label{DC}
{\rm DC}_{\tau, \gamma}=\left\{\omega\in[0,1]^d:\ \|x\cdot\omega\|_\T=\inf_{l\in\mathbb{Z}}|l- x\cdot\omega|\geq \frac{\gamma}{\| x\|_1^{\tau}},\ \forall\  x\in\mathbb{Z}^d\setminus\{o\}\right\}
\end{equation}

with $x\cdot\omega=\sum\limits_{i=1}^dx_i \omega_i$  and $o$ being the origin of $\Z^d$. We   assume the  function $v(\theta)$ belongs to the bounded Lipschitz  monotone (BLM) class  or the unbounded  Lipschitz  monotone  (UBLM) class, defined as follows:

(BLM)   $v$ is a 1-periodic, real-valued function defined on $\R$, continuous on $[0,1)$, satisfying
for some	constant $L>0$,

\begin{equation}\label{LC}
v(\theta_2)-v(\theta_1)\geq L(\theta_2-\theta_1), \ \  0\leq\theta_1\leq \theta_2<1,
\end{equation}

$$v(0)=0, \ \ v(1-0)=1,$$
where $``-0"$ denotes the  left limit.

(UBLM) $v$ is a 1-periodic, real-valued function defined on $\R\setminus\Z$, continuous on $(0,1)$, satisfying for some $L>0$,

\begin{equation*}\label{LC'}
v(\theta_2)-v(\theta_1)\geq L(\theta_2-\theta_1), \ \  0<\theta_1\leq \theta_2<1,
\end{equation*}

$$v(0+0)=-\infty, \ \ v(1-0)=+\infty,$$
where $``+0"$ denotes  the right  limit.

The class BLM (resp. UBLM) was first introduced in  [20]  (resp. [24]) as a natural generation of the famous Maryland potential $v(\theta)=\tan\pi\theta$ (resp. $v(\theta)=\theta\mod 1$ originated from [4])  dropping  the meromorphy assumption.  Indeed, the Anderson localization (i.e., pure point spectrum with exponentially decaying eigenfunctions) of quasi-periodic Schrödinger operator with  tangent  potential (called the Maryland model)  plays a central role in the study of quantum suppression of classical chaos concerning the quantum kicked rotor (c.f. [8]).  The monotonicity property of Maryland type potentials results in the absence of resonances. As a consequence, the perturbative KAM diagonalization method can be employed to prove the Anderson localization [4, 2, 26]. More importantly,  it turns out that the exact Maryland model is solvable (c.f. [14])  and all couplings Anderson localization holds true [10, 28].  The perturbative  localization results proved in  the 1980s  require Diophantine type frequencies, and the uniform in  phase   Anderson localization[Anderson localization holds true for all $\theta$ with the coupling perturbation  $\varepsilon$ being independent of $\theta$.]  via the KAM method  [2] even hinges  on the meromorphy property  of monotone potentials.    Recently, many efforts have been made  to investigate  localization for quasi-periodic Schrödinger operators with monotone potentials mainly going  beyond the above restrictions.  For the one-dimensional case,  Jitomirskaya-Liu [22] gave an almost complete description of the spectral types for the standard Maryland model  for all parameters   (c.f. [23, 15] for more results).   Jitomirskaya and Kachkovskiy [20, 24]  established  all couplings Anderson localization  for  one-dimensional quasi-periodic Schrödinger operators with BLM or UBLM potentials   via the non-perturbative method, which was  first developed for  almost Mathieu operators in the breakthrough works of Jitomirskaya [16, 18].   In [21], they even proved the universality of sharp arithmetic localization for all one-dimensional
quasi-periodic Schrödinger operators with anti-Lipschitz monotone potentials.  For the multi-dimensional case, Shi [27] developed a Nash-Moser iteration type diagonalization  method to prove power-law localization, thus extended some results of Aizenman-Molchanov [1]  to the quasi-periodic operators case.  The uniform in phase Anderson localization for a   class of  multi-dimensional  quasi-periodic Schrödinger   operators with unbounded monotone potentials was established by Kachkovskiy-Parnovski-Shterenberg  [25].   Their  method, without the need of multiple KAM-type steps in the perturbative setting, is based on the analysis of   the convergence of perturbation series for  eigenvalues and eigenfunctions, and also applies to  some non-strictly monotone potentials with  a flat segment under some additional assumptions on the length of the segment and the frequencies.  However, the full localization  of [25] relies  on the  unboundedness  of $v$ and does   not cover the  BLM potentials, such as the one given by  $v(\theta)=\theta \mod 1$.

In this paper, we prove the uniform in phase Anderson localization and exponential dynamical localization for (<ref>) with bounded or unbounded Lipschitz  monotone $v$ in the  perturbative regime by  extending  the multi-scale analysis scheme of [12, 13, 5, 6] to the monotone potentials one. We construct a sequence of increasing blocks  $B_n\subset\Z^d$ and  associates each $B_n$ with  a unique   Rellich function (parameterized eigenvalue) $E_n(\theta)$, the one   resonant with the previous scale one $E_{n-1}(\theta)$. And then we verify  that $E_n(\theta)$ uniformly  inherits the Lipschitz monotonicity property of $v(\theta)$.   The construction and verification  are  based on a  novel  Schur complement type estimate  concerning the energy parameters. As a consequence, resonant points  of each scale are uniformly separated, which yields   localization for all phases.  Specifically, our main results are:

§.§ Main results

Assume that $\omega\in 	{\rm DC}_{\tau, \gamma}$ and $v$ belongs to the BLM class with some Lipschitz parameter $L>0$.	Then there exists some $\varepsilon_0=\varepsilon_0(L,d,\tau,\gamma)>0$ such that for all $0\leq \varepsilon\leq \varepsilon_0$, we have

* For any $\theta\in \R$,	 $H(\theta)$ satisfies the  Anderson localization.

* For any $\theta\in \R$,	$H(\theta)$ satisfies the  exponential dynamical localization,  that is, there exists some $C(\varepsilon)$ depending on $\varepsilon$, such that for any $x,y\in \Z^d$,

\begin{equation}\label{1536}
\sup_{t\in \mathbb{R}}|\langle e^{itH(\theta)}{\bm e}_x, {\bm e}_y\rangle|\leq C(\varepsilon)e^{-\frac{1}{800}|\ln\varepsilon|\cdot\|x-y\|_1},
\end{equation}

where $\{\bm e_x\}_{x\in\Z^d}$ is  the standard basis of $\ell^2(\Z^d)$.

This uniform in phase  localization theorem is new even in the special case that $v(\theta)=\theta\mod1$ in the multi-dimensional case. One should mention that  the Anderson localization of this model for a fixed $\theta$ ($\theta=0$) and small $\varepsilon$ was  established in a recent work of  Facchi-Ligabò [9] via the KAM method.

Assume that $\omega\in 	{\rm DC}_{\tau, \gamma}$ and $v$ belongs to the UBLM class with some Lipschitz parameter $L>0$.	Then there exists some $\varepsilon_0=\varepsilon_0(L,d,\tau,\gamma)>0$ such that for all $0\leq \varepsilon\leq \varepsilon_0$, we have

* For any $\theta\in \R\setminus(\Z+\Z^d\cdot\omega)$,	 $H(\theta)$ satisfies the  Anderson localization.

* For any $\theta\in \R\setminus(\Z+\Z^d\cdot\omega)$,	$H(\theta)$ satisfies the  exponential dynamical localization with the same estimate as in (<ref>).

Our approach  applies to piecewise continuous   Lipschitz  monotone  functions $v$, that is, $v$ is continuous on $(0,1)$ except on a finite set of points $0=\vartheta_1<\vartheta_2<\cdots<\vartheta_K<\vartheta_{K+1}=1$ with
$$v(\vartheta_k-0)<v(\vartheta_k+0), \ \ 2\leq k\leq K,$$

$$ v(\theta_2)-v(\theta_1)\geq L(\theta_2-\theta_1), \ \ \vartheta_k<\theta_1< \theta_2<\vartheta_{k+1},$$

and also applies to exponentially decaying  long-range operators, namely, the discrete Laplacian $\Delta$ in (<ref>)  can be replaced by a Toeplitz operator $S$ satisfying

$$|S(x,y)|\leq e^{-\rho\|x-y\|_1}, \ \ \rho>0.$$

For the special case of meromorphic monotone quasi-periodic potentials, we also give an alternative proof of the classical localization result of Bellissard et al. [2], in which the uniform in phase Anderson localization was established by taking account of both the phase  and space directions  in  a KAM diagonalization scheme.

The Anderson localization results in Theorem <ref> and <ref> are also uniform in the sense of  Jitomirskaya  [17]  that every normalized  eigenfunction   $\psi_{\theta,s}$ satisfies
$$|\psi_{\theta,s}(x)|\leq C(\varepsilon) |\psi_{\theta,s}(x_{\theta,s})|e^{-\frac{1}{400}|\ln\varepsilon|\cdot\|x-x_{\theta,s}\|_1}$$

with   $x_{\theta,s}\in \Z^d$  such that $|\psi_{\theta,s}(x_{\theta,s})|=  \|\psi_{\theta,s}\|_{\ell^\infty(\Z^d)}$ (c.f. Corollary <ref>).

The above exponential decay rates “$\frac{1}{400}"$ and   “$\frac{1}{800}"$  are not optimal. In fact, it can be seen from the proof that the exponential decay rate can be $c(\varepsilon)|\ln\varepsilon|$ with $c(\varepsilon)\to 1$ as $\varepsilon\to 0$.

To our best knowledge, this is the first application of multi-scale analysis method to monotone potentials setting.

§.§ The main new ingredients of the proof

Our  proof relies on Green's function estimates  via the multi-scale analysis  in the perturbative regime originated from [11]. Of particular importance includes the analysis of Rellich functions in the spirit of  [12, 13, 5, 6] together with the adaptation  of  the Schur complement argument of Bourgain [3]  (c.f. [7] for recent refinement).  Although the general scheme of our proof seems expected, almost all of the known elements are not available and have to be replaced by new ingredients or a more complicated setup.  The details will be explained as follows.

§.§.§ Schur complement argument
In [3], Bourgain  proved  the  $(\frac{1}{2}-)$-Hölder continuity  of the integrated density  of states (IDS)  for   almost Mathieu operator  via  introducing  the  Schur complement  argument and Weierstrass preparation theorem.  The key point is to analyze the roots of the determinants of  the finite Dirichlet restriction of $H(\theta)$   for a fixed energy $E$.  Resonances at the  $n$-scale induction  there  depend on the pairs of the roots $\{\theta_{n,1}(E),\theta_{n,2}(E)\}$. This celebrated  method was later largely extended by the authors  [7] to prove arithmetic Anderson localization for multi-dimensional quasi-periodic Schrödinger operators with  cosine potential.  Definitely, all these  estimates of [3, 7] were obtained by fixing  the energy parameter $E$ and varying the phase  parameters $\theta.$ As we will see below, in the present work, we employ  some type of Schur complement argument via  varying   energy parameters.

§.§.§ Rellich functions analysis

In [13], Forman and VandenBoom developed a novel  inductive procedure by constructing the Rellich functions   tree to handle the resonances depending on energy,
thus removed  the crucial  symmetrical assumption in [12].  This scheme was recently extended by the authors [6] to  the multi-dimensional case via overcoming the level crossing issue.
More precisely,  for a finite subset $\Lambda\subset\Z^d$,   denote by $H_\Lambda(\theta)$ the Dirichlet restriction of $H(\theta)$ to $\Lambda$. It follows from the perturbation theory of  one-parameter self-adjoint operator families that   $H_\Lambda(\theta)$ exhibits $|\Lambda|$ branches of  Rellich functions. The idea is then based on carefully choosing  increasing  scales of finite blocks and   constructing each generation of  the Rellich function, which is resonant with the previous generation.
The main issue is to verify that the  Rellich  functions maintain the  crucial properties of  $v$.

In the previous works [13, 6], this issue was settled by eigenvector perturbation and eigenvalue variation arguments from [12]. Concretely,  restricting  $H$ to  a fixed block  $B$, one can try to fix an eigenvalue and  its eigenvector and  extend it by a small perturbation to a new eigenvalue and the corresponding  eigenvector  of a bigger block. Then one can use  Feynman-Hellman eigenvalue variation formulas to compute the derivatives of the new  eigenvalue parameterized by $\theta$.  Together  with the Cauchy  interlacing  idea originated from [13], using the above two  arguments   can  successfully yield the two-monotonicity  Morse structure of  Rellich functions for  quasi-periodic Schrödinger operators with $C^2$-cosine type  potentials (c.f. [6] for details).

In the monotone  potentials  setting, the lack of resonances leads to obvious conveniences:  One can easily establish the local Lipschitz  monotone property of all branches of   Rellich functions (c.f. (<ref>) and (<ref>) below), and establish the  Lipschitz continuity of the integrated density of states by the argument of [24] (c.f.  Theorem 4.1 of  [24]). However, there comes the key obstacle for proving the localization,  namely, the  presence of jump discontinuities  of  $v$  at integer points, that will destroy the continuity of the  Rellich function,  making it extraordinary for the   monotonicity  property of the  Rellich function  to hold at these  jump discontinuities. This becomes a very delicate issue for following reasons:  Since the size of  the Dirichlet block  will tend  to infinity, one may eventually lose the separation and  stability  of  eigenvalues, making the jump discontinuities unclear.  Moreover, the  Rellich function of a given block  can be very closed  to other eigenvalues, in which case a rank one perturbation will totally change  the order of eigenvalues, making it hard to compare the size of the  Rellich function near these discontinuities (c.f. Remark <ref>),  where the previous eigenvector perturbation and eigenvalue variation ideas do not seem to work.

To analyze  these  jump discontinuities of the Rellich function  in the present paper, we construct the the Rellich function in a   more quantitative way.   We  deal with  the root of the  characteristic polynomial of a certain scale Dirichlet restriction in a small neighborhood of the value of  the Rellich function from the previous scale, and utilize  Schur complement argument and  Rouché theorem to obtain a unique eigenvalue, which is  chosen to be the value of the  present scale Rellich function (c.f. Proposition <ref>).  This  idea is motivated by the work of Bourgain [3], which made use of   the Schur complement as a perturbation of quadratic polynomial concerning   the phase parameters  $\theta$. The  new aspect   is that we regard the Schur complement as a perturbation of linear function concerning the energy parameters  $E$ (c.f. Lemma <ref>). Now the issue of jump discontinuities of the Rellich function can be handled by comparing the roots of the two Schur complements with a disparity of a rank one perturbation due to the jump discontinuity  of $v$.   By carefully estimating the difference of the two Schur complements via the multi-scale analysis argument (c.f. Proposition <ref> and Lemma <ref>), we confirm the sign of this difference and ultimately prove that all the jump discontinuities except those at integer points  of the Rellich function  are non-negative.  Hence each generation of the Rellich function maintains the uniform Lipschitz monotonicity property, which enables the whole induction.

§.§ Outline of the inductive scheme

The main motivation of  multi-scale analysis is to establish “good”  Green's function estimates for certain subsets of $\Z^d$, which will be specified by “good” later.  Results of this variety are well-established in the literature (c.f. e.g. [11, 13, 6]).  We sketch this scheme relative to our specific application for the reader's convenience.

First, we fix a sufficiently small  $\varepsilon_0$ depending on $L,d,\tau,\gamma$ and let $\varepsilon$ be such that  $0\leq \varepsilon\leq \varepsilon_0$.  We choose  $\delta_0:=\varepsilon_0^\frac{1}{20}\gg\varepsilon$ to be the $0$-scale  resonance parameter. For fixed  $\theta$ and $E$, the $0$-resonant points set will be defined as

$$S_0(\theta,E):=\left\{p\in \Z^d:\ |v (\theta+p\cdot \omega)-E|<\delta_0\right\}.$$

If a finite set $\Lambda\subset\Z^d$ satisfies
\begin{equation}\label{1418}
\Lambda\cap S_0(\theta,E)=\emptyset,
\end{equation}
then one can obtain  the “good” estimates (i.e.,  one over the resonance bound  of the operator norm and off-diagonal exponential decay of  elements) of the Green's function $G_\Lambda^{\theta,E}:=\left ( H_\Lambda (\theta)-E\right) ^{-1}$ by the Neumann series argument. If (<ref>) is not fulfilled,  one can still try to estimate the Green's function via the by-now classical resolvent identity iteration method, which roughly means that if for every $x\in \Lambda$, there exists  a “good” subset $U(x)\subset \Lambda$  with $x\in U(x)$ such that $G_{U(x)}^{\theta,E}$ has “good” estimates, then one can obtain  “good” estimates of   $G_\Lambda^{\theta,E}$, with a small loss on the exponential decay rate, by iterating the resolvent identity. Hence to employ the  resolvent identity method, one needs to associate a “good”  block $U(x)$ with  every $x\in 	\Lambda\cap S_0(\theta,E)$. For this purpose, we will choose  $l_1:=[|\ln\delta_0|^4]$  ($[\cdot]$ denotes the integer part) to be the first scale length and consider the block

$$B_1(x):=\{y\in \Z^d:  \  \|y-x\|_1\leq l_1\}$$
for each $x\in 	\Lambda\cap S_0(\theta,E)$.
Since the Dirichlet restrictions of the   original operator satisfy the following translation property

\begin{equation}\label{1538}
\end{equation}

with $B_1=\{x\in \Z^d:\ \|x\|\leq l_1\}$, it suffices to consider  $H_{B_1}(\theta)$ for all $\theta$.
Since by self-adjointness, we have

\begin{equation}\label{1540}
\|G_{B_1}^{\theta,E}\|=\frac{1}{\operatorname{dist}(\operatorname{Spec}(H_{B_1}(\theta), E))},
\end{equation}

and it follows that the issue of  operator norm bound   turns into  the one  of the difference between the eigenvalue of $H_{B_1}(\theta)$ and $E$. Since   $x\in 	\Lambda\cap S_0(\theta,E)$, it follows that   $|v (\theta+x\cdot \omega)-E|<\delta_0$. Thus by (<ref>) and (<ref>),   it suffices to consider  the  eigenvalue of $H_{B_1}(\theta+x\cdot\omega)$ that  is closed  to  $E$ (hence is  also closed  to   $v(\theta+x\cdot\omega)$).  Since Diophantine property of $\omega$ and Lipschitz monotonicity property of $v$ provide sufficiently large separation of  resonant points, we can  utilize the Schur complement argument and the  smallness of $\varepsilon$ to construct a unique Rellich function $E_1(\theta)$ of $H_{B_1}(\theta)$ in a certain neighborhood of $v(\theta)$.  Moreover, we can verify that $E_1(\theta)$ also satisfies  the  Lipschitz monotonicity property. Now the resonance depends on the difference $|E_1(\theta+x\cdot\omega)-E|$ and we classify the blocks $B_1(x)$ as “good”  or “bad”  depending on the difference  bigger or smaller than the first scale resonance parameter $\delta_1:=e^{-l_1^{2/3}}$. Moreover, by the special structure of $B_1(x)$, one can derive the off-diagonal exponential  decay  of the  Green's function from the sub-exponential operator norm  bound via the resolvent identity argument. Hence we can establish “good”  estimates of $G_\Lambda^{\theta,E}$ provided all the blocks $B_1(x)\subset\Lambda$ with $x\in 	\Lambda\cap S_0(\theta,E)$ being “good”, completing   the first inductive step.

For the general inductive step,  suppose  that for $1\leq k\leq n$, we have inductively constructed increasing block $B_k\subset\Z^d$ centered at origin  and the   corresponding Rellich function $E_k(\theta)$ satisfying the  Lipschitz monotonicity property. Suppose further that we  have  established “good”   estimates of $G_\Lambda^{\theta,E}$ for  those  $\Lambda$ which do not contain any  $k$-scale “bad”   block $B_k(x)$ with $|E_k(\theta+x\cdot\omega)-E|$ smaller than the $k$-scale resonance  parameter $\delta_k=e^{-l_k^{2/3}}$. Then,  since resonant points  are uniformly separated by Diophantine property of $\omega$ and Lipschitz monotonicity property of $E_k$,  one can properly choose  a bigger block $B_{n+1}$, and utilize the Schur complement argument together with the smallness of the perturbation term  due to the  off-diagonal exponential decay of Green's function from the  inductive hypothesis to construct a unique Rellich function $E_{n+1}(\theta)$ of $H_{B_{n+1}}(\theta)$ in a certain neighborhood of $E_n(\theta)$. Moreover, one  can   verify that $E_{n+1}(\theta)$ also   satisfies  the Lipschitz monotonicity property. It remains  to  use resolvent identity to  establish “good”   estimates of $G_\Lambda^{\theta,E}$ for  those  $\Lambda$ which do not contain any  $(n+1)$-scale “bad”   block $B_{n+1}(x)$ with $|E_{n+1}(\theta+x\cdot\omega)-E|<\delta_{n+1}$.  Since $\varepsilon_0$ is small, the loss of  exponential decay rate in each scale is super-exponentially small, providing a uniformly positive decay rate.

Finally, the localization results  follow  from the  Green's function estimates  for “good”  sets, together with the uniform separation of resonant points due to Diophantine property of $\omega$ and Lipschitz monotonicity property of Rellich functions  by  the standard Schnol's  type argument.

§.§ Organization of the paper

Since most of   arguments in the unbounded  potentials case  are similar  to the bounded one  except the analysis of the jump discontinuities of the Rellich function (Proposition <ref> v.s. Proposition <ref>), we will mainly focus the multi-scale analysis  on  the  bounded case.   In <ref>,  we introduce  some basic notation throughout the paper.  In <ref>, we present  the center arguments of this paper, constructing the Rellich functions
and establishing Green’s function estimates via  multi-scale  analysis in the bounded case. In <ref>, we complete  the analysis of the  unbounded potentials case. In <ref>,  we employ the results from <ref> and <ref> to finish the proof of Theorems  <ref> and <ref>.

§ SOME NOTATION

In this section, we explain the meanings of some notation appearing  in this paper.

* For $L>0$, we denote by $Q_L$ the $L$-size lattice cube centered at the  origin:

$$Q_L:=\{x\in \Z^d:\ \|x\|_1\leq L\}.$$

* For $a\in \C,R>0$, we denote

$$D(a,R):=\{z\in \C :\ |z-a|<R\}.$$

* For two sets $X,Y\subset\Z^d$, we  denote
$$\operatorname{dist}_1(X,Y)=\min_{\substack{x\in X, y\in Y} }\|x-y\|_1,$$

$$\partial_Y^-X=\{x\in X:\ \text{there exists }y\in Y\setminus X\text{ with }\|y-x\|_1=1\},$$

$$\partial_Y^+X=\{y\in Y\setminus X:\ \text{there exists }x\in X\text{ with }\|y-x\|_1=1\},$$

$$\partial_YX=\{(x,y):\ x\in X,\  y\in Y\setminus X \text{ with }\|y-x\|_1=1 \}.$$

* For $\Lambda\subset\Z^d, p\in \Z^d$, we  denote
$$\Lambda+p:=\{x+p:\ x\in \Lambda\}.$$

* We denote by $\|c\|$ the $\ell^2$-norm of a vector $c$ and denote by $\|A\|$ the operator ($\ell^2$) norm of a matrix $A$.

* We denote by $A^*$ the adjugate matrix of a matrix $A$.

§ MULTI-SCALE ANALYSIS

Let $ v$ belong to the BLM class. Let  $\varepsilon_0>0$ be sufficiently small depending on $L,d,\tau ,\gamma$. Let $0\leq \varepsilon\leq \varepsilon_0$.  Define $ \delta_{0}:=\varepsilon_0^{\frac{1}{20}}$.

§.§ The initial scale

§.§.§ Definition of $E_0(\theta)$ and $S_0(\theta,E)$

The $0$-generation  Rellich function $E_0(\theta)$ is exactly  $v(\theta)$. Fix $\theta\in \R,E\in \C$. The $0$-resonant points set $S_0(\theta,E)$ is defined as

$$S_0(\theta,E):=\left\{p\in \Z^d:\ |v (\theta+p\cdot \omega)-E|<\delta_0\right\}.$$

§.§.§ Green's function estimates for $0$-nonresonant sets

Fix $\theta,E$. we say that a finite set $\Lambda\subset \Z^d$ is $0$-nonresonant related to  the pair $(\theta,E)$ if $\Lambda\cap S_0(\theta,E)=\emptyset$.

Let $\gamma_0=\frac{1}{2}|\ln\varepsilon|$.	If $\Lambda\cap S_0(\theta,E)=\emptyset$, then  for any $E^*$ such that  $|E-E^*|<\delta_0/5$, we have

\begin{align*}
\|G_\Lambda^{\theta,E^*}\|&\leq 10\delta_0^{-1},\\
|G_\Lambda^{\theta,E^*}(x,y)|&\leq e^{-\gamma_0\|x-y\|_1}, \ \|x-y\|_1\geq 1.
\end{align*}

Since $\Lambda\cap S_0(\theta,E)=\emptyset$, it follows that 	for any $x\in \Lambda$ and  $|E-E^*|<\delta_0/5$, we have

$$|v (\theta+x\cdot \omega) -E^*|\geq |v (\theta+x\cdot \omega) -E|-|E-E^*|\geq \delta_0-\delta_0/5>\delta_0/2. $$
Thus $V_\Lambda(\theta)$,  the diagonal part of $H_\Lambda (\theta)$, satisfies   $\| (V_\Lambda (\theta)-E^*)^{-1}\|\leq2\delta_0^{-1}$.  Since $\|\Delta\|\leq2d$, it follows that

$$\varepsilon\|\left (V_\Lambda(\theta)-E^*\right)^{-1} \Delta\| \leq 4d\varepsilon\delta_0^{-1}\leq\frac{1}{2}.$$
Hence  by Neumann series, we have

\begin{align*}
\left(H_\Lambda (\theta)-E^*\right)^{-1} & =\left (\varepsilon \Delta+V_\Lambda(\theta)-E^*\right)^{-1} \\
& =\sum_{l=0}^{\infty} (-1)^l \varepsilon^l\left[\left (V_\Lambda(\theta)-E^*\right)^{-1} \Delta\right]^l\left (V_\Lambda(\theta)-E^*\right)^{-1}.
\end{align*}

\begin{align*}
\|\left(H_\Lambda (\theta)-E^*\right)^{-1}\|&\leq \sum_{l=0}^{\infty}\left (\frac{4 d\varepsilon}{\delta_0}\right)^l \|\left (V_\Lambda(\theta)-E^*\right)^{-1}\| \\ &\leq2\|\left (V_\Lambda(\theta)-E^*\right)^{-1}\|\\&\leq4\delta_0^{-1}.
\end{align*}

Since $\Delta$ has
only nearest neighbor matrix elements, provided  $\|x-y\|_1\geq1$, we have

\begin{align*}
\left|\left(H_\Lambda (\theta)-E^*\right)^{-1} (x, y)\right|  &\leq \sum_{l=\|x-y\|_1}^{\infty}\left (\frac{4 d\varepsilon}{\delta_0}\right)^l \|\left (V_\Lambda(\theta)-E^*\right)^{-1}\| \\&\leq \frac{4}{\delta_0}\left (\frac{4 d\varepsilon}{\delta_0}\right)^{\|x-y\|_1}\\
&\leq e^{-\gamma_0(\frac{3}{2}\|x-y\|_1-\frac{1}{2})}\\
&\leq e^{-\gamma_0\|x-y\|_1}.
\end{align*}

§.§ The first inductive scale

Let $l_1:=[|\ln\delta_0|^4]$ be the first inductive length and $B_1=Q_{l_1}$ be the first scale block.
Recall that $H_{B_1}(\theta)$ is the Dirichlet restriction of $H(\theta)$ on $\ell^2(B_1)$.    Listing the   eigenvalues in non-decreasing order,  $H_{B_1}(\theta)$ has  $|B_1|$ branches of  Rellich functions  $\lambda_i(\theta)$ $(1\leq i\leq |B_1|)$. By the perturbation theory (Min-Max argument) for self-adjoint operators, the Rellich  functions $\lambda_i(\theta)$ are $1$-periodic and continuous on $[0,1)$ except on the finite set of points $0=\alpha_1<\alpha_2<\cdots<\alpha_{|B_1|}<\alpha_{|B_1|+1}=1$, where

$$\{\alpha_1,\alpha_2,\cdots,\alpha_{|B_1|}\}=\left\{\{-x\cdot\omega\}\right\}_{x\in B_1}$$

with $\{-x\cdot\omega\}$ being the decimal part of $ -x\cdot\omega$.

Let us denote $\{-x\cdot\omega\}$ by $\beta_x$.
From the below equation

$$H_{B_1}(\beta_x)-H_{B_1}(\beta_x-0)=-{\bm e}_x{\bm e}_x^{\rm T},$$
it follows that   all  discontinuities  of $\lambda_i(\theta)$ are caused by  negative rank one perturbations due to the $x$-site diagonal element of  the matrix jumping from  $1$ to $0$, as  the jumps at finite energies will affect all eigenvalues.
Lipschitz monotonicity property (<ref>) implies the operator inequality

\begin{equation}\label{mno}
H_{B_1}(\theta_2)-H_{B_1}(\theta_1)\geq L(\theta_2-\theta_1)\operatorname{Id}_{B_1}, \ \ \alpha_k\leq \theta_1\leq \theta_2<\alpha_{k+1},
\end{equation}
where $\operatorname{Id}_{B_1}$ is the identity operator on $\ell^2(B_1)$.
Hence it follows from the  Min-Max argument that  the Rellich functions $\lambda_i(\theta)$ also satisfy the local Lipschitz monotonicity property

\begin{equation}\label{mne}
\lambda_i(\theta_2)-\lambda_i(\theta_1)\geq L(\theta_2-\theta_1), \ \ \alpha_k\leq \theta_1\leq \theta_2<\alpha_{k+1}.
\end{equation}

The goal of this section is to construct the first generation Rellich function $E_1(\theta)$, which is a  $1$-periodic, single-valued, real-valued Rellich function of $H_{B_1}(\theta)$ essentially defined on $[0,1)$. The construction is done by locally choosing  the  unique branch of $\lambda_i(\theta)$ which is resonant with the previous scale Rellich function $E_0(\theta)$. We will prove that in our construction, on each small interval $[\alpha_k,\alpha_{k+1})$,  $E_1(\theta)$ coincides with one branch of $\lambda_i(\theta)$, and more importantly, $E_1(\theta)$ has  non-negative jumps at $\alpha_k$ except $\alpha_1(=0)$, that is,

$$E_1(\alpha_k)\geq E_1(\alpha_k-0), \ \ 2\leq k\leq |B_1|. $$

It may seem strange at the first sight that the Rellich function $E_1(\theta)$ has   non-negative jumps since  negative rank one perturbations will cause all the eigenvalues to drop down. The reason of these non-negative jumps is that $E_1(\theta)$ coincides with different branches of $\lambda_i(\theta)$ before and after $\alpha_k$. Hence $E_1(\theta)$ maintains the Lipschitz monotonicity property from $v(\theta)$ on the whole interval $[0,1)$. After constructing $E_1(\theta)$ and verifying the properties of it, for fixed $\theta$ and $E$, we relate the first scale resonant points  $p\in S_1(\theta,E)$ to the difference  $|E_1(\theta+p\cdot\omega)-E|$ and prove good Green's function estimates for $1$-good  set.

§.§.§ Construction of $E_1(\theta)$

For any $\theta\in \R$ and  $x\in B_1$ such that $\|x\|_1\neq 0$, we have
$$|v(\theta+x\cdot\omega)-v(\theta)|\geq 20\delta_0.$$

By (<ref>), (<ref>) and $l_1^{-\tau}\geq  |\ln\delta_0|^{-4\tau}$, provided $\delta_0$ sufficiently small, we have
$$|v(\theta+x\cdot\omega)-v(\theta)|\geq L\|x\cdot\omega\|_\T\geq L\gamma l_1^{-\tau}\geq  L\gamma  |\ln\delta_0|^{-4\tau} \geq 20\delta_0.$$

For any $\theta\in \R$, $H_{B_1}(\theta)$ has a unique eigenvalue $E_1(\theta)$ such that $|E_1(\theta)-v(\theta)|\leq \varepsilon$. Moreover, any other eigenvalues of  $H_{B_1}(\theta)$, $\hat E$ except $E_1(\theta)$  satisfy $|\hat E-v(\theta)|>10\delta_0$.

Fix $\theta\in \R$. Consider $d_\theta(z):=\operatorname{det}[z\operatorname{Id}_{B_1}-H_{B_1}(\theta)]$, which is the characteristic polynomial of $H_{B_1}(\theta)$. In the following we restrict $z$ in the complex neighborhood

It suffices to prove that $d_\theta(z)$ has a unique zero $E_1(\theta)$ in $\bar{D}$ satisfying $|E_1(\theta)-v(\theta)|\leq \varepsilon$.
Let us  denote $B_1^o$ by $B_1\setminus \{o\}$. We can write

$$	z\operatorname{Id}_{B_1}-H_{B_1}(\theta)=\begin{pmatrix}
z-v(\theta) & c^{\text{T}} \\
c & z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\theta)
\end{pmatrix}.$$

By Lemma <ref>, for $x\in B_1^o$ and $z\in \bar{D}$, we have

$$|z-v(\theta+x\cdot\omega)|\geq |v(\theta)-v(\theta+x\cdot\omega)|-|z-v(\theta)|\geq20\delta_0-10\delta_0=10\delta_0. $$

Thus $B_1^o\cap S_0(\theta,z)=\emptyset$ and hence by Proposition <ref>, we have

\begin{equation}\label{gb}
\|G_{B_1^o}^{\theta,z}\|\leq 10\delta_0^{-1}.
\end{equation}

Denote the Schur complement by

\begin{align}\label{923}
s_\theta(z):&=z-v(\theta) -c^{\text{T}}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\theta)\right)^{-1}c\nonumber\\
\end{align}

It follows from Schur complement formula that  $d_\theta(z)=0$ if and only if $s_\theta(z)=0$. By (<ref>) and $\|c\|\leq 2d\varepsilon$, we have
\begin{equation}\label{556}
|r_\theta(z)|\leq 40d^2\varepsilon^2\delta_0^{-1}\leq \varepsilon.
\end{equation}

$$|r_\theta(z)|<|z-v(\theta)|=10\delta_0, \ \ z\in \partial D.$$

It follows that by Rouché theorem, $s_\theta(z)$ has the same number of zero(s) as $z-v(\theta)$ in $\bar{D}$. Denote by $E_1(\theta)$ the unique zero of $s_\theta(z)$ in $\bar{D}$. Since   $s_\theta(E_1(\theta))=E_1(\theta)-v(\theta)-r_\theta(E_1(\theta))=0$ and (<ref>), we have

\begin{equation}\label{841}
|E_1(\theta)-v(\theta)|=|r_\theta(E_1(\theta))|\leq \varepsilon.
\end{equation}

Since $v(\theta)$ is  $1$-periodic, $H_{B_1}(\theta)$ is $1$-periodic and self-adjoint, it follows that $E_1(\theta)$ is a $1$-periodic,   real-valued function.

With the notation in the proof of Proposition <ref>, we have the following approximation, for  $z\in D$,
\begin{equation}\label{app}
\end{equation}

Moreover, for  $z\in D':=D(v(\theta),5\delta_0)$,
\begin{equation}\label{De}
s_\theta'(z)=1+O(\delta_0) 	.
\end{equation}

The above  $``O(\delta_0)"$ are understood by  analytic functions of $z$ bounded by  $\delta_0$.

Since $E_1(\theta)$ is the unique zero of $s_\theta(z)$ in $\bar{D}$, the function

is analytic in $D$. Moreover, by (<ref>) and (<ref>), one has  for $z\in \partial D$,

\begin{align*}
|h_\theta(z)|=\left|\frac{E_1(\theta)-v(\theta)-r_\theta(z)}{z-E_1(\theta)}\right|&\leq\left|\frac{E_1(\theta)-v(\theta)}{z-E_1(\theta)}\right| +\left|\frac{r_\theta(z)}{z-E_1(\theta)}\right|\\&\leq \frac{2\varepsilon}{10\delta_0-\varepsilon}\\
\end{align*}

Hence by maximum principle, we have

$$\sup_{z\in D}|h_\theta(z)|\leq \sup_{z\in \partial D}|h_\theta(z)|<\delta_0$$
and (<ref>) follows. By (<ref>), we have  $s_\theta'(z)=1-r_\theta'(z)$. By   (<ref>) and gradient estimate, it follows that

$$\sup_{z\in D'} |r_\theta'(z)|\leq \delta_0^{-1}\sup_{z\in D} |r_\theta(z)|\leq \delta_0^{-1}\varepsilon\leq \delta_0.$$
Thus we finish the proof of (<ref>).

Analogues of Proposition <ref> and Lemma <ref> hold true for the left limit operator $H_{B_1}(\theta-0)$. We state them below:

For any $\theta\in \R$, $H_{B_1}(\theta-0)$ has a unique eigenvalue $E_1(\theta-0)$ such that $|E_1(\theta-0)-v(\theta-0)|\leq \varepsilon$. Moreover, since $H_{B_1}(\theta)=H_{B_1}(\theta-0)$ for $\theta\notin \left\{\{-x\cdot\omega\}\right\}_{x\in B_1}$, it follows that $E_1(\theta)=E_1(\theta-0)$ for $\theta\notin \left\{\{-x\cdot\omega\}\right\}_{x\in B_1}$.

For $z\in  D(v(\theta-0),10\delta_0)$, the Schur complement

\begin{align}\label{923.}
s_{\theta-0}(z):&=z-v(\theta-0) -c^{\text{T}}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\theta-0)\right)^{-1}c\nonumber	\\
\end{align}

exists and 	satisfies

\begin{align*}
s_{\theta-0}(z)&=(z-E_1(\theta-0))(1+O(\delta_0)) ,  & z\in  D(v(\theta-0),10\delta_0),\\
s_{\theta-0}'(z)&=1+O(\delta_0) , & z\in  D(v(\theta-0),5\delta_0).
\end{align*}

§.§.§ Verification of Lipschitz monotonicity property for $E_1(\theta)$

In this part, we prove that $E_1(\theta)$ inherits the Lipschitz monotonicity property of $v(\theta)$.

Let $E_1(\theta)$ be the eigenvalue of $H_{B_1}(\theta)$ constructed in Proposition <ref>.  Then $E_1(\theta)$ satisfies the following two properties:

* On each small interval  $[\alpha_k,\alpha_{k+1})$, $E_1(\theta)$ coincides with exactly one branch of $\lambda_i(\theta)$, and hence is continuous and satisfies the local Lipschitz monotonicity property  on $[\alpha_k,\alpha_{k+1})$ by (<ref>):

\begin{equation*}
E_{1}(\theta_2)-E_{1}(\theta_1)\geq L(\theta_2-\theta_1), \ \ \alpha_k\leq \theta_1\leq \theta_2<\alpha_{k+1}.
\end{equation*}

* At  $\alpha_k$ $(2\leq k\leq |B_1|)$, the   discontinuities of $H_{B_1}(\theta)$,  $E_1(\theta)$ has  non-negative “jumps”:

$$E_1(\alpha_k)\geq E_1(\alpha_k-0). $$

As a consequence, $E_1(\theta)$ satisfies the uniform Lipschitz monotonicity property:
$$E_1(\theta_2)-E_1(\theta_1)\geq L(\theta_2-\theta_1), \ \ 0\leq\theta_1\leq \theta_2<1.$$

To prove the first item, we assume that $E_1(\alpha_k)=\lambda_i(\alpha_k)$ for some branch $\lambda_i(\theta)$. Define the set $X=\{\theta\in[\alpha_k,\alpha_{k+1}) :\ E_1(\theta)=\lambda_i(\theta) \}$. We prove  $X=[\alpha_k,\alpha_{k+1})$ below. By the  definition,  $\alpha_k\in X$, thus $X\neq\emptyset$. Assume there is  a sequence  $\{\theta_j\}_{j=1}^{\infty}\subset X$ such that $\theta_j\to\theta_\infty\in [\alpha_k,\alpha_{k+1})$. Hence by Proposition <ref>, one has

\begin{equation}\label{721}
|\lambda_i(\theta_j)-v(\theta_j)|=|E_1(\theta_j)-v(\theta_j)|\leq \varepsilon.
\end{equation}

Since $\lambda_i(\theta)$ and $v(\theta)$ are continuous on $[\alpha_k,\alpha_{k+1})$, taking a limit of (<ref>) yields $|\lambda_i(\theta_\infty)-v(\theta_\infty)|\leq \varepsilon$. By the definition of $E_1(\theta_\infty)$ and the uniqueness of eigenvalue of $H_{B_1}(\theta_\infty)$ in the energy interval $[v(\theta_\infty)-10\delta_0,v(\theta_\infty)+10\delta_0]$, we get $E_1(\theta_\infty)=\lambda_i(\theta_\infty)$ and hence $X$ is closed in $[\alpha_k,\alpha_{k+1})$.
Assume that $\theta_0\in X$. Then by Proposition <ref>,  for any $j\neq i$, one has

\begin{equation}\label{858}
\end{equation}

By the continuity of $\lambda_j(\theta)$ and $v(\theta)$ in $[\alpha_k,\alpha_{k+1})$, the inequality (<ref>) can be extended to  a neighborhood   (denoted by $U$) of $\theta_0$ in $[\alpha_k,\alpha_{k+1})$. Hence by the definition of $E_1(\theta)$, $\lambda_j(\theta)\neq E_1(\theta)$ for any $\theta\in U$ and  $j\neq i$. So it must be that $\lambda_i(\theta)= E_1(\theta)$ in $U$. Hence $X$ is open in $[\alpha_k,\alpha_{k+1})$. Thus we finish the proof of the first item.
For the second item, 	since $\alpha_k\neq 0$, there exists some $x\in B_1$ with  $x\neq o$ such   that $\alpha_k=\beta_x(=\{-x\cdot\omega\})$. Since $\beta_x\neq 0$, we have  $v(\beta_x)=v(\beta_x-0)$.
It suffices to prove
\begin{equation}\label{1052}
E_1(\beta_x)\geq E_1(\beta_x-0).
\end{equation}
Recalling the notation $s_\theta(z)$, $r_\theta(z)$, $s_{\theta-0}(z)$, $r_{\theta-0}(z)$ (c.f. (<ref>), (<ref>)),
we will restrict $z$ in the interval $I:=(v(\beta_x)-5\delta_0,v(\beta_x)+5\delta_0)$ in the following discussion  so that (<ref>) holds and  all the functions concerning $z$ will be real-valued.
If we can  prove
\begin{equation}\label{1055}
s_{\beta_x}(z)\leq s_{\beta_x-0}(z),
\end{equation}

$$s_{\beta_x}(E_1(\beta_x-0))\leq s_{\beta_x-0}(E_1(\beta_x-0))=0=s_{\beta_x}(E_1(\beta_x)).$$
(<ref>) will follow from the above inequality and the monotonicity of  $s_{\beta_x}(z)$ (since by (<ref>), $s_{\beta_x}'(z)\approx1$). The remainder of the proof  aims at verifying (<ref>). Recalling (<ref>), (<ref>), since $v(\beta_x)=v(\beta_x-0)$,  we compute

\begin{align}
&s_{\beta_x}(z)-s_{\beta_x-0}(z)\nonumber\\ =&r_{\beta_x-0}(z)-r_{\beta_x}(z)\nonumber\\
=&c^{\text{T}}\left[ \left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x-0)\right)^{-1}- \left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x)\right)^{-1}\right]c\nonumber\\
=&c^{\text{T}}\left[ \left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x-0)\right)^{-1}{\bm e}_x {\bm e}_x^{\rm T}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x)\right)^{-1}\right]c. \label{1210}
\end{align}

On the last line of the above equation  we use the resolvent identity and the below equation
$$H_{B_1^o}(\beta_x-0)-H_{B_1^o}(\beta_x)={\bm e}_x {\bm e}_x^{\rm T}.$$

By Cramer's rule, it follows that

A simple computation shows that

\begin{equation}\label{1437}
\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x-0)\right)^{*}{\bm e}_x=\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x)\right)^{*}{\bm e}_x.
\end{equation}

We denote by $b(z)$ the vector in (<ref>).

\begin{equation}\label{1528}
\eqref{1210}=\frac{1}{\operatorname{det}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x-0)\right)}\frac{1}{\operatorname{det}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x)\right)}\left(c^{\rm T} b(z)\right)^2.
\end{equation}

From (<ref>), it follows that  the sign of $s_{\beta_x}(z)-s_{\beta_x-0}(z)$ is the same as the sign of  the product

Denote by $B_1^1$ the set $B_1^o\setminus \{x\}$.
$$\langle H_{B_1^o}(\beta_x-0){\bm e}_x,{\bm e}_x\rangle=v(1-0)=1, \  \ \langle H_{B_1^o}(\beta_x){\bm e}_x,{\bm e}_x\rangle=v(0)=0$$
and  $H_{B_1^1}(\beta_x -0)=H_{B_1^1}(\beta_x )$, we  can write

z-1 & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)
\end{pmatrix}

z-0 & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)
\end{pmatrix}.

By Lemma <ref> and $|z-v(\beta_x)|<5\delta_0$, it follows that  $B_1^1\cap S_0(\beta_x,z)=\emptyset$. By Proposition <ref>, we get the resolvent bound

\begin{equation}\label{1623}
\left \|\left( z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)  \right)^{-1}\right\|\leq 10\delta_0^{-1}.
\end{equation}

Hence by Schur complement formula,

\begin{align}\label{351}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x-0)\right)\\
\left [z-1-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)\right).
\end{align}

By Lemma <ref>, we have
$$|v(\beta_x)-v(1-0)|=|v(\beta_x)-1|\geq 20\delta_0.$$
Since $v(\beta_x)\leq v(1-0)=1$, it follows that $v(\beta_x)-1\leq -20\delta_0$.
By (<ref>), $\|\tilde{c}\|\leq 2d\varepsilon$, $|z-v(\beta_x)|\leq 5\delta_0$, we have

\begin{equation}\label{352}
z-1-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)\right)^{-1}\tilde{c}\leq v(\beta_x) -1+10\delta_0\leq -20\delta_0+10\delta_0<0.
\end{equation}

\begin{align}\label{353}
\nonumber &	 \operatorname{det}\left(z\operatorname{Id}_{B_1^o}-H_{B_1^o}(\beta_x)\right)\\
=&\left [z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}                                                           _{B_1^1}-H_{B_1^1}(\beta_x)\right).
\end{align}

By Lemma <ref>, we have
$$|v(\beta_x)-v(0)|=|v(\beta_x)-0|\geq 20\delta_0.$$

Since $v(\beta_x)\geq v(0)=0$, it follows that $v(\beta_x)-0\geq 20\delta_0$. By (<ref>), $\|\tilde{c}\|\leq 2d\varepsilon$, $|z-v(\beta_x)|\leq 5\delta_0$, we have

\begin{equation}\label{354}
z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_1^1}-H_{B_1^1}(\beta_x)\right)^{-1}\tilde{c}\geq  v(\beta_x) -0-10\delta_0\geq 20\delta_0-10\delta_0>0.
\end{equation}

Combing (<ref>), (<ref>), (<ref>) and (<ref>), we get

and hence
(<ref>) holds true  by (<ref>). Thus we finish the proof of (<ref>) and the second item.

Another  proof of Proposition <ref> depends on eigenvalue perturbation theory. The separation property of the  diagonal  elements
\begin{equation}\label{842}
\min_{\substack{x,y\in B_1\\ x\neq y}}|v(\theta+x\cdot\omega)-v(\theta+y\cdot\omega)| \geq L\gamma (2dl_1)^{-\tau}>10d\varepsilon
\end{equation}
yields  the separation of the Rellich functions $\lambda_i(\theta)$ after adding an $\varepsilon$-perturbative Laplacian.
Thus one can verify that $E_1(\theta)$ coincides with $\lambda_k(\theta)$ (the $k$-branch of Rellich functions in non-decreasing order) on $[\alpha_k,\alpha_{k+1})$. The non-negative “jumps” at $\alpha_k$ follow from a min-max principle argument for rank one perturbation.
However, this easier idea seems difficult to   be  extended to large scale case since (<ref>) will be violated as the size of the block increases.

§.§.§ Green's function estimates for $1$-good sets

In this part, we relate the resonances to $E_1(\theta)$ and establish Green's function estimates for $1$-good sets.
For $p\in \Z^d$, we denote

Fix $\theta\in \R$, $E\in \C$. Let $\delta_1:=e^{-l_1^\frac{2}{3}}$. Define the first scale resonant points set (related to $(\theta,E)$) as
$$S_1(\theta,E):=\left\{p\in \Z^d:\ |E_1 (\theta+p\cdot \omega)-E|<\delta_1\right\}.$$

Let $\Lambda\subset\Z^d$ be a finite set. Related to $(\theta,E)$,  we say

* $\Lambda$ is $1$-nonresonant  if $\Lambda\cap S_1(\theta,E)=\emptyset$.

* $\Lambda$ is $1$-regular if $p\in \Lambda\cap S_0(\theta,E)$ $\Rightarrow$ $B_1(p)\subset\Lambda$.

* $\Lambda$ is $1$-good if it is both $1$-nonresonant and $1$-regular.

Fix $\theta\in \R$, $E\in \C$. If $p\in S_0(\theta,E)$ and $p \notin S_1(\theta,E)$, then for any $E^*$ such that $|E-E^*|<\delta_1/2$, we have

\begin{align}\label{1l2}
\|G_{B_1(p)} ^{\theta,E^*}\|&\leq10\delta_1^{-1}, \\
\label{1ex}
|G_{B_1(p)}^{\theta,E^*}(x,y)|&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|x-y\|_1}, \ \|x-y\|_1\geq l_1^{\frac{4}{5}}.
\end{align}

Since $p\in S_0(\theta,E)$, by definition we have $|v(\theta+p\cdot \omega )-E|\leq \delta_0$. Thus
$$ |v(\theta+p\cdot \omega )-E^*|\leq |v(\theta+p\cdot \omega )-E| +|E-E^*|\leq \delta_0+\delta_1/2.$$
By Proposition <ref> and translation property, $E_1(\theta+p\cdot \theta )$ is the unique eigenvalue of $H_{B_1(p)}(\theta)$ in $[v(\theta+p\cdot \omega )-10\delta_0,v(\theta+p\cdot \omega )+10\delta_0]$ and moreover

\begin{align*}\
|E_1(\theta+p\cdot \omega )-E^*|&\leq  |v(\theta+p\cdot \omega )-E^*|+|E_1(\theta+p\cdot \omega )-v(\theta+p\cdot \omega )|\\&\leq \delta_0+\delta_1/2+\varepsilon\\&<2\delta_0.
\end{align*}

Hence we get

$$\operatorname{dist}\left (\operatorname{Spec}(H_{B_1(p)}(\theta )),E^* \right )=|E_1(\theta+p\cdot \omega )-E^*|.$$

Since $p \notin S_1(\theta,E)$, we have
$$|E_1(\theta+p\cdot \omega )-E^*|\geq |E_1(\theta+p\cdot \omega )-E|-|E-E^*|\geq \delta_1-\delta_1/2\geq \delta_1/2.$$

Since $H_{B_1(p)}(\theta )$ is self-adjoint,  we get
$$\|G_{B_1(p)} ^{\theta,E^*}\|=\frac{1}{\operatorname{dist}\left (\operatorname{Spec}(H_{B_1(p)}(\theta )),E^* \right )}=\frac{1}{|E_1(\theta+p\cdot \omega )-E^*|}\leq 2\delta_1^{-1}.$$
Thus we finish the proof of (<ref>).
Denote  $B_1^o(p):=B_1(p)\setminus\{p\}$. By Lemma <ref>, it follows that for any $x\in B_1^o(p)$, we have
\begin{align*}
|v(\theta+x\cdot\omega)-E|&\geq|v(\theta+x\cdot\omega)-v(\theta+p\cdot\omega)| -|v(\theta+p\cdot\omega)-E|\\&\geq 20\delta_0-\delta_0\\&>\delta_0.
\end{align*}
Thus $B_1^o(p)\cap S_0(\theta,E)=\emptyset$. Hence by Proposition <ref>, we have

\begin{align}
\|G_{B_1^o(p)} ^{\theta,E^*}\|&\leq10\delta_0^{-1}, \nonumber\\
\label{1200}
|G_{B_1^o(p)}^{\theta,E^*}(x,y)|&\leq e^{-\gamma_0\|x-y\|_1}, \ \|x-y\|_1\geq 1.
\end{align}
Let $x,y\in B_1(p)$ such that $\|x-y\|_1\geq l_1^{\frac{4}{5}}$. By self-adjointness, we may assume $\|x-p\|_1\geq \frac{1}{2}l_1^\frac{4}{5}$.
By the resolvent identity, one has

\begin{align}\label{1156}
G_{B_1(p)}^{\theta,E^*}(x,y)=\chi_{B_1^o(p)}(y)G_{B_1^o(p)}^{\theta,E^*}(x,y)-\varepsilon\sum_{(w,w')\in \partial_{B_1(p)}B_1^o(p)} G_{B_1^o(p)}^{\theta,E^*}(x,w)G_{B_1(p)}^{\theta,E^*}(w',y).
\end{align}

Since $w \in  \partial^-_{B_1(p)}B_1^o(p)$, $w' \in  \partial^+_{B_1(p)}B_1^o(p)=\{p\}$, one has

$$\|x-w\|_1\geq \|x-p\|_1-\|w-p\|_1\geq \|x-p\|_1-1.$$
Hence by  (<ref>) and  (<ref>), we have

\begin{equation}\label{1205}
|G_{B_1(p)}^{\theta,E^*}(x,y)|\leq e^{-\gamma_0\|x-y\|_1}+e^{-\gamma_0(\|x-p\|_1-1)}|G_{B_1(p)}^{\theta,E^*}(p,y)|.
\end{equation}

* If $\|y-p\|_1\leq l_1^\frac{3}{4}$, then
\begin{align*}
\|x-p\|_1-1&\geq \|x-y\|_1-\|y-p\|_1-1\\
&\geq \left (1-(\l_1^\frac{3}{4}+1) l_1^{-\frac{4}{5}}\right) \|x-y\|_1\\
&\geq \left (1- 2l_1^{-\frac{1}{20}}\right) \|x-y\|_1.
\end{align*}

Bounding the term $|G_{B_1(p)}^{\theta,E^*}(p,y)|$ in (<ref>) by $\|G_{B_1(p)}^{\theta,E^*}\|\leq 10\delta_1^{-1}=10e^{l_1^\frac{2}{3}}$, we get
\begin{align*}
\eqref{1205}&\leq e^{-\gamma_0\|x-y\|_1}+10e^{-\gamma_0(1- 2l_1^{-\frac{1}{20}})\|x-y\|_1}e^{l_1^\frac{2}{3}}  \\
&\leq e^{-\gamma_0(1- 2l_1^{-\frac{1}{20}}-(l_1^\frac{2}{3}+10) l_1^{-\frac{4}{5}})\|x-y\|_1}\\
&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|x-y\|_1}.
\end{align*}

* If $\|y-p\|_1> l_1^\frac{3}{4}$, we can use  (<ref>), (<ref>) and the resolvent identity  to bound the term   $|G_{B_1(p)}^{\theta,E^*}(p,y)|$ in (<ref>) by

\begin{align*}
|G_{B_1(p)}^{\theta,E^*}(p,y)|=|G_{B_1(p)}^{\theta,E^*}(y,p)|&\leq \varepsilon\sum_{(w,w')\in \partial_{B_1(p)}B_1^o(p)} |G_{B_1^o(p)}^{\theta,E^*}(y,w)G_{B_1(p)}^{\theta,E^*}(w',p)|\\
&\leq e^{-\gamma_0(\|y-p\|_1-1)} \delta_1^{-1}.
\end{align*}

\begin{align*}
\eqref{1205} &\leq e^{-\gamma_0\|x-y\|_1}+e^{-\gamma_0(\|x-p\|_1-1)}e^{-\gamma_0(\|y-p\|_1-1)} \delta_1^{-1}\\&\leq e^{-\gamma_0(1-(l_1^\frac{2}{3} +10) l_1^{-\frac{4}{5}})\|x-y\|_1}\\
&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|x-y\|_1}.
\end{align*}

Thus we finish the proof of (<ref>).

Fix $\theta\in \R$, $E\in \C$. Assume that  a finite set $\Lambda$ is $1$-good related to $(\theta,E)$,  then for any $E^*$ such that $|E-E^*|<\delta_1/5$, we have

\begin{align}\label{1l}
\|G_{\Lambda} ^{\theta,E^*}\|&\leq10\delta_1^{-1},
\\
|G_{\Lambda} ^{\theta,E^*}(x,y)|&\leq e^{-\gamma_1\|x-y\|_1}, \ \|x-y\|_1\geq l_1^{\frac{5}{6}},\label{1e}
\end{align}

where $\gamma_1=\gamma_0(1-l_1^{-\frac{1}{80}})$.

Define the set $S_0^\Lambda:=S_0(\theta,E)\cap\Lambda$. If $S_0^\Lambda=\emptyset$, then (<ref>) and (<ref>) follows from  Proposition <ref> immediately. Thus we assume that $S_0^\Lambda\neq \emptyset$.
To prove (<ref>), by self-adjointness it suffices to prove

\begin{equation}\label{sd}
\operatorname{dist}\left (\operatorname{Spec}(H_{\Lambda}(\theta )),E \right )\geq  \delta_1 /2.
\end{equation}

Fix $E'$ such that $|E'-E|<\delta_1/2$ and $E'\notin\operatorname{Spec}(H_{\Lambda}(\theta ))$.
For any $p\in S_0^\Lambda$, since $\Lambda$ is $1$-regular, it follows that  $B_1(p)\subset\Lambda$. Since $\Lambda$ is $1$-nonresonant, it follows that  $p\notin S_1(\theta,E)$. Hence Lemma <ref> holds true for such $p$.
Denote $\Lambda_0:=\Lambda\setminus S_0^\Lambda$. Then $\Lambda_0\cap S_0(\theta,E)=\emptyset$ and hence Proposition <ref> holds true for $\Lambda_0$.

* Let $x\in \Lambda$ such that $\operatorname{dist}_1(x,S_0^\Lambda)\geq l_1^\frac{2}{3}$. By the resolvent identity, one has

$$G_\Lambda^{\theta,E'}(x,y)=\chi_{\Lambda_0}(y)G_{\Lambda_0}^{\theta,E'}(x,y)-\varepsilon\sum_{(w,w')\in \partial_{\Lambda}\Lambda_0} G_{\Lambda_0}^{\theta,E'}(x,w)G_{\Lambda}^{\theta,E'}(w',y).$$

\begin{align}\label{1835}
\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|&\leq \sum_{y\in \Lambda_0} |G_{\Lambda_0}^{\theta,E'}(x,y)|+\varepsilon\sum_{\substack{(w,w')\in \partial_{\Lambda}\Lambda_0\\y\in \Lambda}} |G_{\Lambda_0}^{\theta,E'}(x,w)||G_{\Lambda}^{\theta,E'}(w',y)|.
\end{align}

Since  $|E-E'|<\delta_1/2$, by Proposition <ref>, the first summation on the right-hand side of (<ref>) satisfies

\begin{align*}
\sum_{y\in \Lambda_0} |G_{\Lambda_0}^{\theta,E'}(x,y)|&\leq \|G_{\Lambda_0}^{\theta,E'}\|+\sum_{y\in\Lambda_0:\|x-y\|_1\geq1 }|G_{\Lambda_0}^{\theta,E'}(x,y)|\\
&\leq 10\delta_0^{-1}+\sum_{y:\|x-y\|_1\geq1 } e^{-\gamma_0\|x-y\|_1}\\
&\leq 20\delta_0^{-1}.
\end{align*}

In the second summation on the right-hand side of (<ref>), since $w\in \partial_{\Lambda}^-\Lambda_0,$ we have

$$\|x-w\|_1\geq\operatorname{dist}_1(x,S_0^\Lambda)-\operatorname{dist}_1(w,S_0^\Lambda)\geq l_1^\frac{2}{3}-1.$$
It follows that

\begin{align*}
&\varepsilon\sum_{\substack{(w,w')\in \partial_{\Lambda}\Lambda_0\\y\in \Lambda}} |G_{\Lambda_0}^{\theta,E'}(x,w)||G_{\Lambda}^{\theta,E'}(w',y)|\\ \leq&2d\varepsilon \sum_{w\in \partial_{\Lambda}^-\Lambda_0} e^{-\gamma_0\|x-w\|_1}	\sup_{w'\in \partial_{\Lambda}^+\Lambda_0}\sum_y|G_\Lambda^{\theta,E'}(w',y)|\\
\leq & 2d\varepsilon \sum_{{w:\|x-w\|_1\geq l_1^\frac{2}{3}-1}} e^{-\gamma_0\|x-w\|_1}	\sup_{w'\in \partial_{\Lambda}^+\Lambda_0}\sum_y|G_\Lambda^{\theta,E'}(w',y)|
\\ \leq &\frac{1}{4}	\sup_{w'\in \Lambda}\sum_y|G_\Lambda^{\theta,E'}(w',y)|.
\end{align*}

\begin{equation}\label{1906}
\sup_{x: \operatorname{dist}_1(x,S_0^\Lambda)\geq l_1^\frac{2}{3}}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|\leq 20\delta_0^{-1}+ \frac{1}{4}	\sup_{w'\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(w',y)|.
\end{equation}

* Let $x\in \Lambda$ such that $\operatorname{dist}_1(x,S_0^\Lambda)< l_1^\frac{2}{3}$. Then $\|x-p\|_1<l_1^\frac23$ and  $x\in B_1(p)$ for some $p\in S_0^\Lambda$.
By the  resolvent identity, one has

\begin{equation*}
G_\Lambda^{\theta,E'}(x,y)=\chi_{B_1(p)}(y)G_{B_1(p)}^{\theta,E'}(x,y)-\varepsilon\sum_{(w,w')\in \partial_\Lambda B_1(p)} G_{B_1(p)}^{\theta,E'}(x,w)G_{\Lambda}^{\theta,E'}(w',y).
\end{equation*}

\begin{align}\label{1911}
\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|\leq \sum_{y\in B_1(p)} |G_{B_1(p)}^{\theta,E'}(x,y)|+\varepsilon\sum_{\substack{(w,w')\in \partial_\Lambda B_1(p)\\ y\in \Lambda}} |G_{B_1(p)}^{\theta,E'}(x,w)||G_{\Lambda}^{\theta,E'}(w',y)|.
\end{align}

By Lemma <ref>,  the first summation on the right-hand side of (<ref>) satisfies

\begin{align*}
\sum_{y\in B_1(p)} |G_{B_1(p)}^{\theta,E'}(x,y)|&\leq (3l_1)^d\|G_{B_1(p)}^{\theta,E'}\|\leq \delta_1^{-2}.
\end{align*}

In the second summation on the right-hand side of (<ref>), since $w\in \partial_\Lambda^-B_1(p)$, we have  $\|w-p\|_1\geq l_1$. It follows that
$$\|x-w\|_1\geq \|w-p\|_1- \|x-p\|_1\geq  l_1-l_1^\frac{2}{3}\geq l_1^\frac{4}{5}.$$
Since $|E'-E|<\delta_1/2$, by Lemma <ref>, one has

\begin{align*}
&\varepsilon\sum_{\substack{(w,w')\in \partial_\Lambda B_1(p)\\ y\in \Lambda}} |G_{B_1(p)}^{\theta,E'}(x,w)||G_{\Lambda}^{\theta,E'}(w',y)|\\	\leq& 2d\varepsilon \sum_{w\in \partial_\Lambda^-B_1(p)} |G_{B_1(p)}^{\theta,E'}(x,w)|	\sup_{w'\in \partial_\Lambda^+B_1(p)}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(w',y)|\\ \leq &
2d\varepsilon \sum_{w:\|x-w\|_1\geq l_1^\frac{4}{5}} e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|x-w\|_1}	\sup_{w'\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(w',y)|\\ \leq& \frac{1}{4}	\sup_{w'\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(w',y)|.
\end{align*}

\begin{equation}\label{1940}
\sup_{x: \operatorname{dist}_1(x,S_0^\Lambda)< l_1^\frac{2}{3}}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|\leq \delta_1^{-2}+ \frac{1}{4}	\sup_{w'\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(w',y)|.
\end{equation}

By (<ref>) and (<ref>), we get

$$	\sup_{x\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|\leq \delta_1^{-2}+ \frac{1}{4}	\sup_{w'\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(w',y)|$$
and hence

$$	\sup_{x\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|\leq 2\delta_1^{-2}.$$

By Schur's test and self-adjointness, we get

\begin{equation}\label{1947}
\| G_\Lambda^{\theta,E'}\|\leq	\sup_{x\in \Lambda}\sum_{y\in \Lambda}|G_\Lambda^{\theta,E'}(x,y)|\leq 2\delta_1^{-2} .
\end{equation}

Since the uniform bound (<ref>) holds true for all $E'$ in  $|E'-E|<\delta_1/2$  except a finite subset of $\operatorname{Spec}(H_{\Lambda}(\theta ))$,   (<ref>) must  hold true for all $E'$ in  $|E'-E|<\delta_1/2$ as the resolvent is unbounded near the spectrum. Thus we finish the proof of (<ref>). To prove (<ref>), we let $x,y\in \Lambda$ such that $\|x-y\|_1\geq l_1^{\frac{5}{6}}$. For any $q\in \Lambda$, we introduce a subset $U(q)\subset\Lambda$ with $q\in U(q)$ to iterate the resolvent identity.

* If $\operatorname{dist}_1(q,S_0^\Lambda)\geq l_1^\frac{2}{3}$, we define

\begin{equation}\label{sf1}
U(q)=\{u\in \Lambda:\ \|q-u\|_1\leq l_1^\frac{1}{2}\}.
\end{equation}

By the resolvent identity,  one has

$$G_\Lambda^{\theta,E^*}(q,y)=\chi_{U(q)}(y)G_{U(q)}^{\theta,E^*}(q,y)-\varepsilon\sum_{(w,w')\in \partial_\Lambda U(q)} G_{U(q)}^{\theta,E^*}(q,w)G_{\Lambda}^{\theta,E^*}(w',y).$$

Since $\operatorname{dist}_1(q,S_0^\Lambda)\geq l_1^\frac{2}{3}>l_1^\frac{1}{2}$, it follows that $U(q)\cap S_0(\theta,E)=\emptyset.$ By Proposition <ref> and $\|q-w\|_1\geq l_1^\frac{1}{2}$ for $w\in \partial_\Lambda^- U(q)$, we have

\begin{align}\label{fenzi}
\nonumber	|G_\Lambda^{\theta,E^*}(q,y)|&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+\varepsilon\sum_{(w,w')\in \partial_\Lambda U(q)} |G_{U(q)}^{\theta,E^*}(q,w)||G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber	&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+2dl_1^d\varepsilon\sup_{(w,w')\in \partial_\Lambda U(q)} |G_{U(q)}^{\theta,E^*}(q,w)||G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber	&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+l_1^d\sup_{(w,w')\in \partial_\Lambda U(q)} e^{-\gamma_0\|q-w\|_1}|G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber		&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+\sup_{(w,w')\in \partial_\Lambda U(q)} e^{-\gamma_0(\|q-w\|_1-d\ln l_1)}|G_{\Lambda}^{\theta,E^*}(w',y)|\\
&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+\sup_{w'\in \partial_\Lambda^+ U(q)} e^{-\gamma_0(1-l_1^{-\frac13})\|q-w'\|_1}|G_{\Lambda}^{\theta,E^*}(w',y)|.
\end{align}

For $y\in U(q)$, since by Proposition <ref>,

$$|G_{U(q)}^{\theta,E^*}(q,y)|\leq  e^{-\gamma_0\|q-y\|_1},\ \ \|q-y\|_1\geq 1,$$

$$|G_{U(q)}^{\theta,E^*}(q,y)|\leq\|G_{U(q)}^{\theta,E^*}\| \leq 10\delta_0^{-1} ,\ \ \|q-y\|_1< 1,$$

we have

\begin{equation}\label{f1}
|G_{U(q)}^{\theta,E^*}(q,y)|\leq10\delta_0^{-1} e^{-\gamma_0(\|q-y\|_1-1)}.
\end{equation}

* If $\operatorname{dist}_1(q,S_0^\Lambda)< l_1^\frac{2}{3}$, then $\|q-p\|_1\leq l_1^{\frac{2}{3}}$ and $q\in B_1(p)$ for some $p\in S_0^\Lambda$. We define

\begin{equation}\label{sf2}
U(q)=\{B_1(p):\ q\in B_1(p)\text{ with }p\in S_0^\Lambda\}.
\end{equation}

By the resolvent identity,  one has

$$G_\Lambda^{\theta,E^*}(q,y)=\chi_{U(q)}(y)G_{U(q)}^{\theta,E^*}(q,y)-\varepsilon\sum_{(w,w')\in \partial_\Lambda U(q)} G_{U(q)}^{\theta,E^*}(q,w)G_{\Lambda}^{\theta,E^*}(w',y).$$

Since  $w\in \partial_\Lambda^- U(q)=\partial_\Lambda^- B_1(p)$ and $\|q-p\|_1\leq l_1^{\frac{2}{3}}$, it follows that

\begin{equation}\label{yya}
\|q-w\|_1\geq \|w-p\|_1-\|p-q\|_1\geq l_1-l_1^{\frac{2}{3}}\geq l_1^{\frac45}.
\end{equation}

Since $p\notin S_1(\theta,E)$, by Lemma <ref> and (<ref>), one can get a similar estimate as (<ref>):

\begin{align}\label{fenzi2}
\nonumber|G_\Lambda^{\theta,E^*}(q,y)|&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+\varepsilon\sum_{(w,w')\in \partial_\Lambda U(q)} |G_{U(q)}^{\theta,E^*}(q,w)||G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber	&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+2dl_1^d\varepsilon\sup_{(w,w')\in \partial_\Lambda U(q)} |G_{U(q)}^{\theta,E^*}(q,w)||G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber	&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+l_1^d\sup_{(w,w')\in \partial_\Lambda U(q)} e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|q-w\|_1}|G_{\Lambda}^{\theta,E^*}(w',y)|\\
&\leq\chi_{U(q)}(y)|G_{U(q)}^{\theta,E^*}(q,y)|+\sup_{w'\in \partial_\Lambda^+ U(q)} e^{-\gamma_0(1-l_1^{-\frac{1}{60}})\|q-w'\|_1}|G_{\Lambda}^{\theta,E^*}(w',y)|.
\end{align}

For $y\in U(q)$, since by Lemma <ref>,

$$|G_{U(q)}^{\theta,E^*}(q,y)|\leq  e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|q-y\|_1},\ \ \|q-y\|_1\geq l_1^\frac{4}{5},$$

$$|G_{U(q)}^{\theta,E^*}(q,y)|\leq\|G_{U(q)}^{\theta,E^*}\| \leq 10\delta_1^{-1} ,\ \ \|q-y\|_1< l_1^\frac{4}{5},$$

we have

\begin{equation}\label{f2}
|G_{U(q)}^{\theta,E^*}(q,y)|\leq10\delta_1^{-1} e^{-\gamma_0(1-l_1^{-\frac{1}{50}})(\|q-y\|_1-l_1^\frac{4}{5})}.
\end{equation}

Let $q_0=x$. Applying the estimates (<ref>) and (<ref>) $K$ times yields

\begin{align}\label{1217}
|G_\Lambda^{\theta,E^*}(x,y)|\leq \sum_{k=0}^{K-1}\chi_{U(q_k)}(y)|G_{U(q_k)}^{\theta,E^*}(q_k,y)|+
\prod_{k=0}^{K-1}e^{-\gamma_0(1-l_1^{-\frac{1}{60}})\|q_k-q_{k+1}\|_1}|G_{\Lambda}^{\theta,E^*}(q_K,y)|,
\end{align}

where $q_{k+1} \in \partial_\Lambda^+U(q_k)$ takes the supremum in the last line of (<ref>) or (<ref>).
We choose $K$ such that $y\notin U(q_k)$ for $k\leq K-1$ and $y\in U(q_K)$, thus the first summation in (<ref>) vanishes. Hence,

\begin{align}\label{2221}
\prod_{k=0}^{K-1}e^{-\gamma_0(1-l_1^{-\frac{1}{60}})\|q_k-q_{k+1}\|_1}|G_{\Lambda}^{\theta,E^*}(q_K,y)|.
\end{align}

By (<ref>) and    (<ref>), we have

\begin{align}\label{2204}
|G_{\Lambda}^{\theta,E^*}(q_K,y)|\leq |G_{U(q_K)}^{\theta,E^*}(q_K,y)|+\sup_{w'\in \partial_\Lambda^+ U(q_K)} e^{-\gamma_0(1-l_1^{-\frac{1}{60}})\|q_K-w'\|_1}|G_{\Lambda}^{\theta,E^*}(w',y)|.
\end{align}

\begin{align*}
e^{-\gamma_0\|q_K-y\|}_1&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{50}})\|q_K-y\|_1},\\
e^{\gamma_0}&\leq e^{\frac{1}{2}\gamma_0l_1^\frac{4}{5}} \leq e^{\gamma_0(1-l_1^{-\frac{1}{50}})l_1^\frac{4}{5}}
\end{align*}

it follows from (<ref>) and (<ref>) that the first term on the right-hand side of (<ref>) satisfies

\begin{equation}\label{25}
|G_{U(q_K)}^{\theta,E^*}(q_K,y)|\leq10\delta_1^{-1} e^{-\gamma_0(1-l_1^{-\frac{1}{50}})(\|q_K-y\|_1-l_1^\frac{4}{5})}.
\end{equation}

For $y\in U(q_K)$, $ w'\in \partial_\Lambda^+ U(q_K)$, we have:

* If $U(q_K)$ is in the  form of (<ref>), then
$$\|q_K-w'\|_1\geq \|q_K-y\|_1.$$

* If $U(q_K)$ is in the  form of (<ref>) with $U(q_K)=B_1(p)$, then

\begin{align*}
\|q_K-w'\|_1&\geq \|p-w'\|_1-\|p-q_K\|_1\\&\geq \|p-y\|_1-\|p-q_K\|_1\\&\geq \|q_K-y\|_1-\|p-q_K\|-\|p-q_K\|_1\\
&\geq \|q_K-y\|_1-2l_1^\frac{2}{3} .
\end{align*}

It follows that the second term on the right-hand side of (<ref>) satisfies

\begin{align}\label{2223}
\nonumber&	\sup_{w'\in \partial_\Lambda^+ U(q_K)} e^{-\gamma_0(1-l_1^{-\frac{1}{60}})\|q_K-w'\|_1}|G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber\leq &	\sup_{w'\in \partial_\Lambda^+ U(q_K)} e^{-\gamma_0(1-l_1^{-\frac{1}{60}})(\|q_K-y\|_1-2l_1^\frac{2}{3})}|G_{\Lambda}^{\theta,E^*}(w',y)|\\
\nonumber \leq & e^{-\gamma_0(1-l_1^{-\frac{1}{60}})(\|q_K-y\|_1-2l_1^\frac{2}{3})}\|G_{\Lambda}^{\theta,E^*}\|\\
\leq& 10\delta_1^{-1}e^{-\gamma_0(1-l_1^{-\frac{1}{60}})(\|q_K-y\|_1-2l_1^\frac23)}.
\end{align}

By (<ref>), (<ref>), (<ref>), (<ref>) and $\delta_1^{-1}=e^{l_1^\frac{2}{3}}$, provided $\|x-y\|\geq l_1^\frac56 $, we have

\begin{align*}
20\delta_1^{-1}	e^{-\gamma_0(1-l_1^{-\frac{1}{60}})(\sum\limits_{k=0}^{K-1}\|q_k-q_{k+1}\|_1+\|q_K-y\|_1-3l_1^\frac45)}\\
&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{60}})(\|x-y\|_1-3l_1^\frac{4}{5}-2l_1^\frac{2}{3})}\\
&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{60}})(1-(3l_1^\frac{4}{5}+2l_1^\frac{2}{3})l_1^{-\frac{5}{6}})\|x-y\|_1}\\
&\leq e^{-\gamma_0(1-l_1^{-\frac{1}{80}})\|x-y\|_1}.
\end{align*}

Replacing $\theta$ by $\theta-0$, we can estimate the Green's function $G_{\Lambda} ^{\theta-0,E^*}$ of the left limit operator  $H_\Lambda(\theta-0)$       for  $1$-good set $\Lambda$ related to $(\theta-0,E)$ with   $|E-E^*|<\delta_1/5$, by analogous definitions
\begin{align*}
S_0(\theta-0,E)&:=\left\{p\in \Z^d:\ |v (\theta+p\cdot \omega-0)-E|<\delta_0\right\},\\
S_1(\theta-0,E)&:=\left\{p\in \Z^d:\ |E_1 (\theta+p\cdot \omega-0)-E|<\delta_1\right\}.
\end{align*}

§.§ Inductive hypotheses

In this section, we state the inductive hypotheses of each scale. First, we introduce some inductive definitions.

Recalling that $l_1=[|\ln\delta_0|^4],\delta_0=\varepsilon_0^{\frac{1}{20}},\gamma_0=\frac{1}{2}|\ln\varepsilon|$ defined in the previous section, for $n\geq 1$, we define the inductive parameters:

* $l_n:=(l_1)^{2^{n-1}}$  (the $n$-scale length)

* $\delta_n:=e^{-l_n^\frac{2}{3}}$(the $n$-scale resonance parameter)

* $\gamma_n:=\prod\limits_{k=1}^{n}(1-l_k^{-\frac{1}{80}})\gamma_0$ (the Green's function decay rate)

We let $\varepsilon$ be sufficiently small so that
\begin{equation}\label{rate}
\gamma_n\geq \gamma_\infty=\prod\limits_{k=1}^{\infty}(1-l_k^{-\frac{1}{80}})\gamma_0\geq \frac{1}{2}\gamma_0\geq 10.
\end{equation}

Inductive hypotheses (scale $n$). The following inductive  hypotheses hold true for $0\leq m\leq n $:

There exists a block $B_m$ satisfying
\begin{equation}\label{h11}
Q_{l_m}\subset B_m\subset Q_{l_m+50l_{m-1}}.
\end{equation}

The Dirichlet restriction  $H_{B_m}(\theta)$ has a $1$-periodic, single-valued, real-valued  Rellich function $E_m(\theta)$ satisfying the following properties:

* For $m=0$, $B_0=\{o\}$  and $E_0(\theta)=v(\theta)$.

* For $m\geq1$, $E_m(\theta)$ is well-approached by the previous generation Rellich function $E_{m-1}(\theta)$:

\begin{equation}\label{h12}
|E_m(\theta)- E_{m-1}(\theta)|\leq e^{-l_{m-1}},
\end{equation}
where $e^{-l_0}$ is understood by $\varepsilon$.

* $E_m(\theta)$ is the unique eigenvalue of $H_{B_m}(\theta)$ in $D(E_{m-1}(\theta),10\delta_{m-1})$.

* Denote by  $0=\alpha_1<\alpha_2<\cdots<\alpha_{|B_m|}<\alpha_{|B_m|+1}=1$ the discontinuities of $H_{B_m}(\theta)$, where

$$\{\alpha_1,\alpha_2,\cdots,\alpha_{|B_m|}\}=\left\{\{-x\cdot\omega\}\right\}_{x\in B_m}.$$

Then $E_m(\theta)$ is continuous on each interval $[\alpha_k,\alpha_{k+1})$  with  the local Lipschitz monotonicity property:
\begin{equation}\label{1639}
E_m(\theta_2)-E_m(\theta_1)\geq L(\theta_2-\theta_1), \ \ \alpha_k\leq \theta_1\leq \theta_2<\alpha_{k+1},
\end{equation}

and moreover, $E_m(\theta)$ has  non-negative jumps at $\alpha_k$  ($2\leq k\leq |B_m|$):

\begin{equation}\label{1640}
E_m(\alpha_k)\geq E_m(\alpha_k-0).
\end{equation}

By (<ref>) and (<ref>), $E_m(\theta)$ in fact satisfies the uniform Lipschitz monotonicity property:

\begin{equation}\label{Lc}
E_m(\theta_2)-E_m(\theta_1)\geq L(\theta_2-\theta_1), \ \ 0\leq \theta_1 \leq \theta_2<1.
\end{equation}

To simplify the notation, for an $m$-generation  Rellich function $E_m(\theta)$ describing in Hypothesis  <ref>,  $\theta\in \R\setminus\Z$ and  $R>0$, we denote the region

\begin{equation}\label{AR}
A_m(\theta,R):=\bigcup_{a\in [E_m(\theta-0),E_m(\theta)]}D(a,R).
\end{equation}

For the case $E_m(\theta-0)=E_m(\theta)$, the above  $[E_m(\theta-0),E_m(\theta)]$ is understood by the single point  $E_m(\theta)$ and thus $A_m(\theta,R)=D(E_m(\theta),R)$.

[blue!20!white] (2,0) circle (2);	[blue!20!white] (5,0) circle (2); 	[blue!20!white] (2,2) rectangle (5,-2);
(2,2) arc (90:270:2);
(5,2) arc (90:-90:2);
(2,2)– (5,2);		 	(2,-2)– (5,-2);
(2,-0.2)node[below]$E_m(\theta-0)$;  	(5,-0.2)node[below]$E_m(\theta)$;	(2,0)circle(1pt);	(5,0)circle(1pt);
[->](2,0)–(2,2);  	(2,1)node[right]$R$;
[blue!20!white] (11.5,0) circle (2);	(11.5,-0.2)node[below]$E_m(\theta)$;(11.5,0)circle(1pt); (13.5,0) arc (0:360:2);	(11.5,1)node[right]$R$;	[->](11.5,0)–(11.5,2);  (3.5,-2.5)node[below]$E_m(\theta-0)<E_m(\theta)$; (11.5,-2.5)node[below]$E_m(\theta-0)=E_m(\theta)$;

A picture for $A_m(\theta,R)$

Denote  $B_m^o:=B_m\setminus\{o\}$.
$$	z\operatorname{Id}_{B_m}-H_{B_m}(\theta)=\begin{pmatrix}
z-v(\theta) & c^{\text{T}} \\
c & z\operatorname{Id}_{B_m^o}-H_{B_m^o}(\theta)
\end{pmatrix}.$$

If  the resolvent $(z\operatorname{Id}_{B_m^o}-H_{B_m^o}(\theta))^{-1}$ exists, we 	denote the Schur complement by
$$s_{\theta,m}(z):=z-v(\theta) -c^{\text{T}}\left(z\operatorname{Id}_{B_m^o}-H_{B_m^o}(\theta)\right)^{-1}c.$$

* For any $\theta\in \R\setminus\Z$, $z\in A_{m-1}(\theta,10\delta_{m-1})$ and  $\xi\in \{\theta,\theta-0\}$, the resolvent $(z\operatorname{Id}_{B_m^o}-H_{B_m^o}(\xi))^{-1}$ exists and  we have approximations for the Schur complement:

\begin{align}\label{sa1}
s_{\xi,m}(z)&=(z-E_m(\xi))\left (1+O(\sum_{k=0}^{m-1}\delta_k) \right),& z\in A_{m-1}(\theta,10\delta_{m-1}),\\
\label{sa2}
s_{\xi,m}'(z)&=1+O(\sum_{k=0}^{m-1}\delta_k), & z\in  A_{m-1}(\theta,5\delta_{m-1}).
\end{align}

* For any $z\in D(E_{m-1}(0),10\delta_{m-1})$,   the resolvent $(z\operatorname{Id}_{B_m^o}-H_{B_m^o}(0))^{-1}$ exists  and

\begin{align}\label{sa10}
s_{0,m}(z)&=(z-E_m(0))\left (1+O(\sum_{k=0}^{m-1}\delta_k) \right),&  z\in D(E_{m-1}(0),10\delta_{m-1}),\\
\label{sa20}
s_{0,m}'(z)&=1+O(\sum_{k=0}^{m-1}\delta_k), & z\in  D(E_{m-1}(0),5\delta_{m-1}).
\end{align}

* For any $z\in D(E_{m-1}(1-0),10\delta_{m-1})$, the resolvent $(z\operatorname{Id}_{B_m^o}-H_{B_m^o}(1-0))^{-1}$ exists  and

\begin{align*}
s_{1-0,m}(z)&=(z-E_m(1-0))\left (1+O(\sum_{k=0}^{m-1}\delta_k) \right), &  z\in D(E_{m-1}(1-0),10\delta_{m-1}),\\
s_{1-0,m}'(z)&=1+O(\sum_{k=0}^{m-1}\delta_k),& z\in  D(E_{m-1}(1-0),5\delta_{m-1}).
\end{align*}

The above  $``O(\sum_{k=0}^{m-1}\delta_k)"$ are understood by analytic functions of $z$ bounded by  $\sum_{k=0}^{m-1}\delta_k$.

For $p\in \Z^d$, we denote
Fix $\theta\in \R$, $E\in \C$.  Define the $m$-scale resonant points set as
\begin{equation}\label{sm}
S_m(\theta,E):=\left\{p\in \Z^d:\ |E_m (\theta+p\cdot \omega)-E|<\delta_m\right\}.
\end{equation}

Let $\Lambda\subset\Z^d$ be a finite set. Related to $(\theta,E)$,  we say

* $\Lambda$ is $m$-nonresonant  if $\Lambda\cap S_m(\theta,E)=\emptyset$.

* $\Lambda$ is $m$-regular if $p\in \Lambda\cap S_k(\theta,E)$ $\Rightarrow$  $B_{k+1}(p)\subset\Lambda$ ($0\leq k\leq m-1$).

* $\Lambda$ is $m$-good if it is both $m$-nonresonant and $m$-regular.

Fix $\theta\in \R$, $E\in \C$. Assume that  a finite set $\Lambda$ is $m$-good related to $(\theta,E)$,  then for any $E^*$ such that $|E-E^*|<\delta_m/5$, we have

\begin{align*}
\|G_{\Lambda} ^{\theta,E^*}\|&\leq10\delta_m^{-1}, \\
|G_{\Lambda} ^{\theta,E^*}(x,y)|&\leq e^{-\gamma_m\|x-y\|_1}, \ \|x-y\|_1\geq l_m^{\frac{5}{6}}.
\end{align*}

Moreover, as mentioned in Remark <ref>,	analogous estimates of  $G_{\Lambda} ^{\theta-0,E^*}$ hold true  for $m$-good set  $\Lambda $ related to $(\theta-0,E)$ with  $|E-E^*|<\delta_m/5$.

$B_m$ satisfies  the following properties:

* For any $E\in D(E_{m-1}(0),10\delta_{m-1})$, the set  $B_m$ is
$(m-1)$-regular related to $(0,E)$.

* For any $E\in D(E_{m-1}(1-0),10\delta_{m-1})$, the set  $B_m$ is
$(m-1)$-regular related to $(1-0,E)$.

* For any $\theta\in\R\setminus\Z$, $E\in A_{m-1}(\theta,10\delta_{m-1})$, the set  $B_m$ is
$(m-1)$-regular related to $(\theta,E)$ and $(\theta-0,E)$.

§.§ Verification of the hypotheses for $n=1$

By the analysis in section <ref>, we can prove
There exists $\varepsilon_0=\varepsilon_0(L,d,\tau,\gamma)>0$ such that for all $0\leq \varepsilon\leq \varepsilon_0$, the inductive hypotheses hold true at the scale $n=1$.

We check the hypotheses one by one.

* The $1$-periodic, single-valued, real-valued  properties of $E_1(\theta)$ follows  from the construction (c.f. Remark <ref>).  Item (2) and  (3)  follow from Proposition <ref>.
(<ref>) and (<ref>) follow from  Proposition <ref>.

Lemma <ref> and <ref> yield Hypothesis <ref>.

* The good Green's function estimates for $0$-nonresonant and $1$-good sets are verified in  Proposition <ref> and <ref>.

* There is nothing to check with  Hypothesis <ref> for $n=1$ since we make a convention  that any set is

§.§ Verification of inductive hypotheses for the general inductive scale
In this section, we will prove the following theorem:

There exists $\varepsilon_0=\varepsilon_0(L,d,\tau,\gamma)>0$ such that for all $0\leq \varepsilon\leq \varepsilon_0$ and $n\geq 1$, if the  inductive hypotheses hold true at the scale $n$, then they  remain valid  at the  scale $n+1$.

The proof of Theorem <ref> is divided into the following  several parts. First,  we prove two lemmas  that will be used  later.

For $0\leq k\leq m\leq n$, with the notation from Definition <ref>,  we have the inclusion relations:

* for  $\theta\in \R,$

$$D(E_m(\theta),10\delta_m)\subset D(E_k(\theta),10\delta_k),$$

$$D(E_m(\theta-0),10\delta_m)\subset D(E_k(\theta-0),10\delta_k).$$

* for $\theta\in \R\setminus\T,$

$$A_m(\theta,10\delta_m)\subset A_k(\theta,10\delta_k). $$

Moreover, they remain valid if all the $``10"$ above replaced by $``5"$.

By inductive hypothesis (<ref>), for $\theta\in \R$,  we have
$$|E_k(\theta)-E_m(\theta)|\leq \sum_{j=k}^{m-1}e^{-l_j}\leq \delta_k-\delta_m.$$

Then the lemma follows immediately.

For $0\leq k\leq m\leq n$, with the notation from Definition <ref>, for any $\theta\in \R\setminus\Z$ and $\tilde\theta\in \R$, we have

\begin{align*}
\min(|E_m(\tilde\theta)-E_m(\theta)|,|E_m(\tilde\theta)-E_m(\theta-0)|)& \leq  \operatorname{dist}(E_m(\tilde\theta),A_m(\theta,R))+R,\\
\min(|E_m(\tilde\theta-0)-E_m(\theta)|,|E_m(\tilde\theta-0)-E_m(\theta-0)|)& \leq  \operatorname{dist}(E_m(\tilde\theta-0),A_m(\theta,R))+R.
\end{align*}

We give the proof of  the first inequality since the second is analogous.
Since the function $E_m$ is real-valued   and does not take value in the interval $(E_m(\theta-0),E_m(\theta))$ (c.f. Hypothesis <ref>, item ($4$)), we have $E(\tilde\theta)\in\R\setminus (E_m(\theta-0),E_m(\theta))$. It follows that for  $z\in  A_m(\theta,R)$ with $\Re z \in (E_m(\theta-0),E_m(\theta))$, then

\begin{align*}
|z-E_m(\tilde\theta)|&\geq |E_m(\theta)-E_m(\tilde\theta)|,  &\text{if} \ \ E_m(\tilde\theta)\geq E_m(\theta) ,\\
|z-E_m(\tilde\theta)|&\geq |E_m(\theta-0)-E_m(\tilde\theta)|, &\text{if} \ \  E_m(\tilde\theta)\leq E_m(\theta-0).
\end{align*}

For $z\in  A_m(\theta,R)$ with $\Re z \notin (E_m(\theta-0),E_m(\theta))$, we have if $\Re z\geq E_m(\theta)$, then

$$|z-E_m(\tilde\theta)|+R\geq  |z-E_m(\tilde\theta)|+|z-E_m(\theta)|\geq  |E_m(\theta)-E_m(\tilde\theta)|,$$
and if $\Re z\leq E_m(\theta-0)$, then

$$|z-E_m(\tilde\theta)|+R\geq  |z-E_m(\tilde\theta)|+|z-E_m(\theta-0)|\geq  |E_m(\theta-0)-E_m(\tilde\theta)|.$$

§.§.§ Construction of the block $B_{n+1}$

In this section, we construct the  block $B_{n+1}$ satisfying hypothesis (<ref>) and Hypothesis <ref>.

Let $0\leq m\leq n$.
Define the sets
$$X_m^{(1)}=\{ x\in \Z^d : \text{ there exists  $E\in D(E_{n}(0),10\delta_{n})$  with $x\in S_m(0,E)$}\},$$

$$X_m^{(2)}=\{ x\in \Z^d : \text{ there exists  $E\in D(E_{n}(1-0),10\delta_{n})$  with $x\in S_m(1-0,E)$}\},$$

$$X_m^{(3)}=\{ x\in \Z^d : \text{ there exist    $\theta\in \R\setminus\Z$, $E\in A_{n}(\theta,10\delta_{n})$  with $x\in S_m(\theta,E)\cup S_m(\theta-0,E)$}\}$$

$$X_m=X_m^{(1)}\cup X_m^{(2)}\cup X_m^{(3)}.$$
Then we have

$$\inf_{\substack{x,y\in X_m\\x\neq y}}\|x-y\|_1\geq 100l_{m+1}.$$

We analyze by cases:

* Assume $x\in X_m^{(1)}$. Then  by the definition of $S_m(\theta,E)$ (c.f. (<ref>)) and $ X_m^{(1)}$, there exists $E\in D(E_{n}(0),10\delta_{n})\subset D(E_m(0),10\delta_m) $ (c.f. Lemma <ref>) such that

$$|E_m(0+x\cdot \omega )-E|<\delta_m.$$
It follows that
$$|E_m(0+x\cdot \omega )-E_{m}(0)|\leq |E_m(0+x\cdot \omega )-E|+|E-E_m(0)|\leq 11\delta_m.$$

Thus by the Lipschitz monotonicity property of $E_m(\theta)$ (c.f. (<ref>)), we have

\begin{equation}\label{713}
L\|x\cdot \omega \|_\T	\leq |E_m (0+x\cdot \omega)-E_m (0)|\leq 11\delta_m.
\end{equation}

* Assume $x\in X_m^{(2)}$. Then  by the definition of $S_m(\theta-0,E)$ and  $ X_m^{(2)}$, there exists $E\in D(E_{n}(1-0),10\delta_{n})\subset D(E_m(1-0),10\delta_m) $ such that

$$|E_m(1+x\cdot \omega -0)-E|<\delta_m.$$
It follows that
$$|E_m(1+x\cdot \omega-0 )-E_{m}(1-0)|\leq |E_m(1+x\cdot \omega -0)-E|+|E-E_m(1-0)|\leq 11\delta_m.$$

Thus by the Lipschitz monotonicity property of $E_m(\theta)$, we have

\begin{equation}\label{713,}
L\|x\cdot \omega \|_\T	\leq |E_m (1+x\cdot \omega -0)-E_m (1-0)|\leq 11\delta_m.
\end{equation}

* Assume $x\in X_m^{(3)}$. Then  by the definition of $S_m(\theta,E)$, $S_m(\theta-0,E)$ and  $ X_m^{(3)}$, there exist $\theta\in \R\setminus\Z$, $\xi_1\in \{\theta,\theta-0\}$,  $E\in A_{n}(\theta,10\delta_{n})\subset  A_{m}(\theta,10\delta_{m})$   such that

$$|E_m(\xi_1+x\cdot \omega)-E|<\delta_m  .$$
It follows from Lemma <ref> that

\begin{align*}
&	\min \left(|E_m(\theta)-E_m(\xi_1+x\cdot \omega)|,|E_m(\theta-0)-E_m(\xi_1+x\cdot \omega)|\right)\\
\leq& \operatorname{dist}\left(A_{m}(\theta,10\delta_{m}),E_m(\xi_1+x\cdot \omega)\right)+10\delta_m\\
\leq &|E_m(\xi_1+x\cdot \omega)-E|+10\delta_m\\\leq& 11\delta_m .
\end{align*}

Hence there exists  some  $\xi_2\in \{\theta,\theta-0\},$   such that

$$|E_m(\xi_1+x\cdot \omega)-E_m(\xi_2)|\leq 11\delta_m.$$

Thus by the Lipschitz monotonicity property of $E_m(\theta)$, we have

\begin{equation}\label{713,,}
L\|x\cdot \omega \|_\T	\leq |E_m(\xi_1+x\cdot \omega)-E_m(\xi_2)|\leq 11\delta_m.
\end{equation}

By (<ref>), (<ref>), (<ref>), for any $x\in X_m$, we have $	L\|x\cdot \omega \|_\T\leq 11\delta_m.$ Let $y\in  X_m$ with $x\neq y$. Similar, we have  $	L\|y\cdot \omega \|_\T\leq 11\delta_m.$
By  (<ref>) and $x\neq y$, it follows that

$$L\gamma \|x-y\|_1^{-\tau}\leq  L\|(x-y)\cdot \omega \|_\T\leq L\|x\cdot \omega \|_\T+L\|y\cdot \omega \|_\T\leq 22\delta_m.$$

Since $\delta_m= e^{-l_m^\frac{2}{3}}$, it follows from  the above inequality that

$$100l_{m+1}=100l_{m}^2\leq e^{\frac{1}{\tau}l_m^\frac{2}{3}}(\frac{L\gamma}{22})^{\frac{1}{\tau}}=(\frac{L\gamma}{22\delta_m})^{\frac{1}{\tau}}\leq \|x-y\|_1$$

provided $\varepsilon_0$ sufficiently small.

Let $L\geq 1$ and  $X\subset\Z^d$ satisfy the below condition:

\begin{equation}\label{844}
\inf_{\substack{x,y\in X\\x\neq y}}\|x-y\|_1\geq 10L.
\end{equation}

Then for any set $B\subset \Z^d$, there exists another set $\tilde{B}$  satisfying the following two  properties:

* $B\subset \tilde{B}\subset \{x\in \Z^d: \ \operatorname{dist}_1(x,B)\leq 2 L\}$.

* $(Q_L+x)\subset \tilde{B}$ for any $x\in X$ such that  $(Q_L+x)\cap \tilde{B}\neq \emptyset$.

We define

\begin{equation}\label{839}
\tilde{B}:=B\bigcup\left(\bigcup_{x\in X:(Q_L+x)\cap B\neq \emptyset }(Q_L+x)\right)
\end{equation}

and claim that $\tilde{B}$ satisfies the desired properties. For any $y\in \tilde{B}$, it follows that  $y\in B$ or $ y\in (Q_L+x)$ for some $x\in X$ with  $(Q_L+x)\cap B\neq \emptyset$. Hence,

$$\operatorname{dist}_1(y,B)\leq \|y-x\|_1+\operatorname{dist}_1(x,B)\leq L+L\leq 2L.$$
Thus we finish the proof of the first property. For the second,
assume  there exists $x\in X$ such that $(Q_L+x)\cap \tilde{B}\neq \emptyset$. Then  by (<ref>), if
\begin{equation}\label{849}
(Q_L+x)\cap B= \emptyset,
\end{equation}
then there exists  $y\in X$ with
\begin{equation}\label{851}
(Q_L+y)\cap B\neq \emptyset
\end{equation}
such that  $(Q_L+x)\cap (Q_L+y)\neq \emptyset$. Thus $\|x-y\|_1\leq 2L$. By (<ref>), it follows that $x=y$, which contradicts with (<ref>) and (<ref>). Thus (<ref>) does not hold true. Then by (<ref>), it follows that  $(Q_L+x)\subset \tilde{B}$.

There exists a block $B_{n+1}$ satisfying the following  properties:

* $	Q_{l_{n+1}}\subset B_{n+1}\subset Q_{l_{n+1}+50l_{n}}$.

* $x\in B_{n+1}\cap S_k(0,E)$  for some    $E\in D(E_n(0) ,10\delta_n)$ $\Rightarrow$ $B_{k+1}(x)\subset B_{n+1}$  ($0\leq k\leq n-1$).

* $x\in B_{n+1}\cap S_k(1-0,E)$  for some    $E\in D(E_n(1-0) ,10\delta_n)$ $\Rightarrow$ $B_{k+1}(x)\subset B_{n+1}$  ($0\leq k\leq n-1$).

* $x\in B_{n+1}\cap (S_k(\theta,E)\cup S_k(\theta-0,E) )$  for some $\theta\in \R\setminus\Z$,   $E\in A_n(\theta ,10\delta_n)$ $\Rightarrow$ $B_{k+1}(x)\subset B_{n+1}$  ($0\leq k\leq n-1$).

Define the sets
$$X_k^{(1)}=\{ x\in \Z^d : \text{ there exists  $E\in D(E_{n}(0),10\delta_{n})$  with $x\in S_k(0,E)$}\},$$

$$X_k^{(2)}=\{ x\in \Z^d : \text{ there exists  $E\in D(E_{n}(1-0),10\delta_{n})$  with $x\in S_k(1-0,E)$}\},$$

$$X_k^{(3)}=\{ x\in \Z^d : \text{ there exist    $\theta\in \R\setminus\Z$, $E\in A_{n}(\theta,10\delta_{n})$  with $x\in S_k(\theta,E)\cup S_k(\theta-0,E)$}\}$$

$$X_k=X_k^{(1)}\cup X_k^{(2)}\cup X_k^{(3)}.$$

By  Lemma <ref>, we have

$$\inf_{\substack{x,y\in X_k\\x\neq y}}\|x-y\|_1\geq 100l_{k+1}.$$

We start with $B^{(0)}:=Q_{l_{n+1}}$.	Setting $L=10l_{n}$, $X=X_{n-1}$ in  Lemma <ref> yields  that there exists $B^{(1)}$ satisfying

* $B^{(0)}\subset B^{(1)}\subset \{x\in \Z^d: \ \operatorname{dist}_1(x,B^{(0)})\leq 20l_n\} .$

* $(Q_{10l_n}+x)\subset B^{(1)}$ for any $x\in X_{n-1}$ such that  $(Q_{10l_n}+x)\cap B^{(1)}\neq \emptyset$.

By setting $L=10l_{n-k}$, $X=X_{n-k-1}$ in Lemma <ref>,  we inductively	define  $B^{(k+1)}$ ($1\leq k\leq n-1$), such that

\begin{equation}\label{1321}
B^{(k)}\subset B^{(k+1)}\subset \{x\in \Z^d: \ \operatorname{dist}_1(x,B^{(k)})\leq 20l_{n-k}\},
\end{equation}

\begin{equation}\label{1916}
\text{ 	 $(Q_{10l_{n-k}}+x)\subset B^{(k+1)}$ for $x\in X_{n-k-1}$ with $(Q_{10l_{n-k}}+x)\cap B^{(k+1)}\neq \emptyset$.}
\end{equation}

We claim that $B^{(n)}$ is the desired block. Since
$$\sum_{k=m}^{n-1}20l_{n-k}\leq 30l_{n-m},$$
it follows from (<ref>)  that

\begin{equation}\label{1316}
B^{(n)}\subset \{x\in \Z^d :\ \operatorname{dist}_1(x,B^{(m)})\leq 30l_{n-m}\}.
\end{equation}

$$B^{(n)}\subset \{x\in \Z^d :\ \operatorname{dist}_1(x,B^{(0)})\leq  30l_n\}\subset  Q_{l_{n+1}+50l_{n}}.$$

Let $x$ be such that
\begin{equation}\label{1429}
x\in B^{(n)}\cap X_k
\end{equation}
for some $0\leq k\leq n-1$.  By (<ref>) with $m=n-k$, we have

\begin{equation}\label{1858}
B^{(n)}\subset \{x\in \Z^d :\ \operatorname{dist}_1(x,B^{(n-k)})\leq 30l_{k}\}.
\end{equation}

By (<ref>), (<ref>) and $30l_k\leq l_{k+1}$, it follows that

$$(Q_{10l_{k+1}}+x)\cap B^{(n-k)}\neq \emptyset.$$

By (<ref>), it follows that $(Q_{10l_{k+1}}+x)\subset  B^{(n-k)}.$
Hence, by the inductive hypothesis  (<ref>) and $l_{k+1}+50l_k\leq 10l_{k+1}$, it follows that

$$B_{k+1}(x)\subset (Q_{10l_{k+1}}+x)\subset  B^{(n-k)}\subset B^{(n)}.$$

Hence the set $B_{n+1}:=B^{(n)}$,  with $	Q_{l_{n+1}}\subset B_{n+1}\subset Q_{l_{n+1}+50l_{n}}$ satisfies

* For any $E\in D(E_{n}(0),10\delta_{n})$, the set    $B_{n+1}$ is
$n$-regular related to $(0,E)$.

* For any $E\in D(E_{n}(1-0),10\delta_{n})$, the set   $B_{n+1}$ is
$n$-regular related to $(1-0,E)$.

* For any $\theta\in\R\setminus\Z$, the set  $E\in A_{n}(\theta,10\delta_{n})$,  $B_{n+1}$ is
$n$-regular related to $(\theta,E)$ and $(\theta-0,E)$.

§.§.§ Construction of $E_{n+1}(\theta)$

Having constructed the $(n+1)$-scale  block $B_{n+1}$ in the previous section,  we construct the $(n+1)$-scale Rellich function $E_{n+1}(\theta )$ of the Dirichlet restriction $H_{B_{n+1}}(\theta)$ satisfying
$$|E_{n+1}(\theta)-E_{n}(\theta)|\leq e^{-l_n}$$
in this section. To this end, we utilize the Schur complement perturbation argument to show the existence and uniqueness of such $E_{n+1}(\theta)$.

Let $m$ be an integer such that  $0\leq m\leq n$. For any $\theta\in\R$,  $\xi_1,\xi_2\in \{\theta,\theta-0\}$  and $x\in B_{m+1}$ such that $x\neq o$, we have
$$|E_{m}(\xi_1+x\cdot \omega )-E_m(\xi_2)|\geq 20\delta_m. $$

Let $x\in B_{m+1}$ with $x\neq o$. Since  $B_{m+1}\subset Q_{l_{m+1}+50l_m}\subset Q_{2l_{m+1}}$, it follows that  $\|x\|_1\leq 2l_{m+1}$. Thus by (<ref>), inductive hypothesis (<ref>) and $l_{m+1}^{-\tau}=l_m^{-2\tau}=|\ln\delta_m|^{-3\tau}$, it follows that

$$|E_{m}(\xi_1+x\cdot\omega)-E_m(\xi_2)|\geq L\|x\cdot\omega\|_\T \geq L\gamma( 2l_{m+1})^{-\tau}\geq L\gamma2^{-\tau}|\ln\delta_m|^{-3\tau}\geq 20\delta_m.$$

Let $m$ be an integer of  $0\leq m\leq n$. Denote  the set  $B_{m+1}^o:=B_{m+1}\setminus \{o\}$. We  have the followings:

* For any $E\in D(E_m(0),10\delta_m)$, the set  $B_{m+1}^o$ is $m$-good related to $(0,E)$.

* For any $E\in D(E_m(1-0),10\delta_m)$, the set  $B_{m+1}^o$ is $m$-good related to $(1-0,E)$.

* For any $\theta\in \R\setminus\Z $ and $E\in A_m(\theta ,10\delta_m)$, the set  $B_{m+1}^o$ is $m$-good related to $(\theta,E)$ and $(\theta-0,E)$.

We give the proof of  the third item since the others are analogous. Fix $\theta\in \R\setminus\Z$ and $E\in A_m(\theta ,10\delta_m)$. 	First, we prove that  $B_{m+1}^o$ is $m$-nonresonant related to $(\theta,E)$ and $(\theta-0,E)$. It follows from Lemma <ref> and Lemma <ref>  that  for  $x\in B_{m+1}^o$, $\xi\in\{\theta,\theta-0\}$ we have
\begin{align*}
|E_m(\xi+x\cdot\omega ) -E|&\geq \operatorname{dist}(E_m(\xi+x\cdot\omega ),A_m(\theta,10\delta_m))\\&\geq \min_{\xi'\in\{\theta,\theta-0\}} |E_m(\xi+x\cdot\omega ) -E_m(\xi')|-10\delta_m\\
&\geq 20\delta_m-10\delta_m\\
&\geq 10\delta_m.
\end{align*}

Thus $B_{m+1}^o\cap( S_m(\theta,E)\cup S_m(\theta-0,E))=\emptyset$.  It remains to prove that  $B_{m+1}^o$ is $m$-regular related to $(\theta,E)$ and  $(\theta-0,E)$. By Hypothesis <ref> and  Proposition <ref>, it follows that  $B_{m+1}$ is $m$-regular related to $(\theta,E)$ and $(\theta-0,E)$.  Let $0\leq k\leq m-1$ and $x\in B_{m+1}^o\cap( S_{k}(\theta,E) \cup S_{k}(\theta-0,E))$. Since $B_{m+1}$ is $m$-regular and $x\in B_{m+1}^o\cap( S_{k}(\theta,E) \cup S_{k}(\theta-0,E))\subset  B_{m+1}\cap( S_{k}(\theta,E) \cup S_{k}(\theta-0,E))$, it follows that
\begin{equation}\label{1545}
B_{k+1}(x)\subset B_{m+1}.
\end{equation}

By the definition of  $S_{k}(\theta,E)$ and $S_{k}(\theta-0,E)$, there exists some  $\xi_1\in \{\theta,\theta-0\}$ such that $|E_k(\xi_1+x\cdot\omega)-E|<\delta_k.$ Since $E\in A_m(\theta ,10\delta_m)\subset A_k(\theta ,10\delta_k)$ (c.f. Lemma <ref>),  it follows from Lemma <ref>  that

\begin{align*}
&	\min \left(|E_k(\theta)-E_k(\xi_1+x\cdot \omega)|,|E_k(\theta-0)-E_k(\xi_1+x\cdot \omega)|\right)\\
\leq& \operatorname{dist}\left(A_{k}(\theta,10\delta_{k}),E_k(\xi_1+x\cdot \omega)\right)+10\delta_k\\
\leq &|E_k(\xi_1+x\cdot \omega)-E|+10\delta_k\\
\leq &11\delta_k .
\end{align*}

Hence there exists  some  $\xi_2\in \{\theta,\theta-0\},$   such that

$$|E_k(\xi_1+x\cdot \omega)-E_k(\xi_2)|\leq 11\delta_k.$$

Thus by $x\neq o$, (<ref>) and  the inductive hypothesis (<ref>),

\begin{align*}
L\gamma\|x\|_1^{-\tau}\leq L \|x\cdot \omega\|_\T	\leq |E_k(\xi_1+x\cdot\omega)-E_k(\xi_2)|
&\leq 11\delta_k.
\end{align*}

It follows that

\begin{equation}\label{1606}
\|x\|_1\geq (L\gamma/10)^{\frac{1}{\tau}}\delta_k^{-\frac{1}{\tau}}= (L\gamma/10)^{\frac{1}{\tau}}e^{\frac{1}{\tau}l_k^{\frac{2}{3}}}\geq 10l_{k+1}.
\end{equation}

By (<ref>) and $B_{k+1}(x)\subset   (Q_{l_{k+1}+50l_{k}}+x)$, it follows that

\begin{equation}\label{1610}
o\notin B_{k+1}(x).
\end{equation}

By (<ref>) and (<ref>), we get $B_{k+1}(x)\subset B_{m+1}^o$.

By translation argument,  Lemma <ref> yields the following corollary
Let $m$ be an integer of  $0\leq m\leq n$ and $x\in \Z^d$. Denote $B_{m+1}^o(x):=B^o_{m+1}+x$. Then we have the followings:

* For any $E\in D(E_m(0),10\delta_m)$, the set  $B_{m+1}^o(x)$ is $m$-good related to $(\{-x\cdot\omega\},E)$.

* For any $E\in D(E_m(1-0),10\delta_m)$, the set  $B_{m+1}^o(x)$ is $m$-good related to $(\{-x\cdot\omega\}-0,E)$.

* For any  $\theta\in \R$ with $\theta+x\cdot\omega\notin \Z$ and   $E\in A_m(\theta+x\cdot\omega ,10\delta_m)$, then the set  $B^o_{m+1}(x)$ is $m$-good related to $(\theta,E)$ and $(\theta-0,E)$.

For any $\theta\in \R$, $H_{B_{n+1}}(\theta)$ has a unique eigenvalue $E_{n+1}(\theta)$ such that $|E_{n+1}(\theta)-E_{n}(\theta)|\leq e^{-l_n}.$ Moreover, any other eigenvalues of  $H_{B_{n+1}}(\theta)$, $\hat E$ except $E_{n+1}(\theta)$  satisfy $|\hat E- E_n(\theta)|>10\delta_n$.

Fix $\theta\in \R$. Consider $d_{\theta,n+1}(z):=\operatorname{det}[z\operatorname{Id}_{B_{n+1}}-H_{B_{n+1}}(\theta)]$, which is the characteristic polynomial of $H_{B_{n+1}}(\theta)$. In the following, we restrict $z$ in the complex neighborhood
D(E_n(0),10\delta_n),\ \ &\theta\in \Z,\\
A_n(\theta,10\delta_n),\ \ &\theta\notin\Z.
\end{cases}$$

It suffices to prove that $d_{\theta,n+1}(z)$ has a unique zero $E_{n+1}(\theta)$ in $\bar{D}$ satisfying $|E_{n+1}(\theta)-E_n(\theta)|\leq e^{-l_n}$. Denote $B_{n+1}^o:=B_{n+1}\setminus \{o\}$.
We can write

$$	z\operatorname{Id}_{B_{n+1}}-H_{B_{n+1}}(\theta)=\begin{pmatrix}
z-v(\theta) & c^{\text{T}} \\
c & z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\theta)
\end{pmatrix}.$$

By Lemma <ref>  and  Lemma <ref>, $B_{n+1}^o$ (resp. $B_n^o$) is $n$-good (resp. $(n-1)$-good) related to $(\theta,z)$. Hence by Hypothesis <ref>, we have

\begin{align}\label{bn+1}
\|G_{B_{n+1}^o}^{\theta,z}\|&\leq10\delta_n^{-1},  \\
\label{bn+}
|G_{B_{n}^o}^{\theta,z}(x,y)|&\leq e^{-\gamma_{n-1}\|x-y\|_1}, \ \|x-y\|_1\geq l_{n-1}^{\frac{5}{6}}.
\end{align}

Denote the Schur complement by
\begin{align}\label{923+}
s_{\theta,n+1}(z):&=z-v(\theta) -c^{\text{T}}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\theta)\right)^{-1}c \nonumber \\
\end{align}

It follows from Schur complement formula, $d_{\theta,n+1}(z)=0$ if and only if $s_{\theta,n+1}(z)=0$.
Since the vector $c$ has only nonzero elements on the nearest neighbor of $o$, by the  resolvent identity, it follows that

\begin{align}
\nonumber s_{\theta,n+1}(z)&=s_{\theta,n}(z)-\varepsilon \sum_{\substack{\|x\|_1,\|y\|_1=1\\(w,w')\in \partial_{B_{n+1}^o}B_{n}^o}}c(x)G_{B_{n}^o}^{\theta,z}(x,w)G_{B_{n+1}^o}^{\theta,z}(w',y)c(y)\\
&=s_{\theta,n}(z)-g_{\theta,n+1}(z), \label{1945}
\end{align}

$$s_{\theta,n}(z)= z-v(\theta) -\tilde{c}^{\text{T}}\left(z\operatorname{Id}_{B_{n}^o}-H_{B_{n}^o}(\theta)\right)^{-1}\tilde{c}$$

with $\tilde{c}$ being  the restriction of $c$ on $B_{n}^o$.
Since $	Q_{l_n}\subset B_n$, it follows that  for $\|x\|_1=1$ and  $w\in \partial^-_{B_{n+1}^o}B_{n}^o$, we have

$$\|x-w\|_1\geq l_n-1\geq l_{n-1}^\frac{5}{6}.$$

By (<ref>) and (<ref>) and $\gamma_{n-1}\geq 10$ (c.f. (<ref>)), it follows that

\begin{align}\label{res}
|	g_{\theta,n+1}(z)|\leq l_n^d e^{-\gamma_{n-1}(l_n-1)}\delta_n^{-1}\leq e^{-\gamma_{n-1}(l_n-2l_n^{\frac{2}{3}})}\leq e^{-3l_n}.
\end{align}

By the inductive  hypotheses (<ref>), (<ref>) and Lemma <ref>, we have

\begin{align}
s_{\theta,n+1}(z)&=	(z-E_n(\theta))\left (1+O(\sum_{k=0}^{n-1}\delta_k)\right )-	g_{\theta,n+1}(z)\label{1924}\\
&=\left (1+O(\sum_{k=0}^{n-1}\delta_k)\right )\left(z-E_n(\theta)-\tilde{g}_{\theta,n+1}(z)\right ) \label{1845}
\end{align}

\begin{equation}\label{1853}
|\tilde{g}_{\theta,n+1}(z)|\leq 2|g_{\theta,n+1}(z)|\leq  e^{-2l_n}.
\end{equation}

\begin{equation}\label{15}
|\tilde{g}_{\theta,n+1}(z)|<10\delta_n\leq |z-E_n(\theta)|, \ \ z\in \partial D.
\end{equation}

By (<ref>), it follows that  $s_{\theta,n+1}(z)$ and  $z-E_n(\theta)-\tilde{g}_{\theta,n+1}(z)$ have the same zero(s) in $\bar{D}$. And by (<ref>)  together with Rouché theorem, $z-E_n(\theta)-\tilde{g}_{\theta,n+1}(z)$ has the same number of zero(s) as $z-E_{n}(\theta)$ in $\bar{D}$. Denote by $E_{n+1}(\theta)$ the unique zero of $s_{\theta,n+1}(z)$ in $\bar{D}$. Since   $s_{\theta,n+1}(E_{n+1}(\theta))=0$, by (<ref>), it follows that $E_{n+1}(\theta)-E_n(\theta)-\tilde{g}_{\theta,n+1}(E_{n+1}(\theta))=0$. Hence by  (<ref>), we have

\begin{equation}\label{1844}
|E_{n+1}(\theta)-E_n(\theta)|=|\tilde{g}_{\theta,n+1}(E_{n+1}(\theta))|\leq e^{-2l_n}.
\end{equation}

Since $E_n(\theta)$ is  $1$-periodic  and   $H_{B_{n+1}}(\theta)$ is  $1$-periodic and self-adjoint, it follows that $E_{n+1}(\theta)$ is a $1$-periodic,   real-valued function.

With the notation in the proof of Proposition <ref>, we have the following approximation, for  $z\in D$,
\begin{equation}\label{appn}
s_{\theta,n+1}(z)=(z-E_{n+1}(\theta))\left (1+O(\sum_{k=0}^{n}\delta_k)\right ).
\end{equation}

Moreover, for $z\in D'$ with
$$ D':=\begin{cases}
D(E_n(0),5\delta_n),\ \ &\theta\in \Z,\\
A_n(\theta,5\delta_n),\ \ &\theta\notin\Z,
\end{cases}$$

we have
\begin{equation}\label{Den}
s_{\theta,n+1}'(z)=1+O(\sum_{k=0}^{n}\delta_k)	.
\end{equation}

The above  $``O(\sum_{k=0}^{n}\delta_k)"$ are understood by analytic functions of $z$ bounded by  $\sum_{k=0}^{n}\delta_k$.

Since $E_{n+1}(\theta)$ is the unique zero of $s_{\theta,n+1}(z)$ in $\bar{D}$, the function

is analytic in $D$. Moreover, by (<ref>), (<ref>) and (<ref>), one has for $z\in \partial D$

\begin{align*}
&\leq\left|\frac{E_{n+1}(\theta)-E_n(\theta)}{z-E_{n+1}(\theta)}\right| +\left|\frac{g_{\theta,n+1}(z)}{z-E_{n+1}(\theta)}\right| +\left|\frac{z-E_{n}(\theta)}{z-E_{n+1}(\theta)}\right|\sum_{k=0}^{n-1}\delta_k\\
&\leq \frac{2e^{-2l_n}}{10\delta_n-e^{-2l_n}}+\frac{10\delta_n}{10\delta_n-e^{-2l_n}}\sum_{k=0}^{n-1}\delta_k\\
\end{align*}

Hence by maximum principle, we have

$$\sup_{z\in D}|h_\theta(z)|\leq \sup_{z\in\partial  D}|h_\theta(z)|<\sum_{k=0}^{n}\delta_k$$
and (<ref>) follows. By (<ref>), we have  $s_{\theta,n+1}'(z)=s_{\theta,n}'(z)-g_{\theta,n+1}'(z)$. By   (<ref>) and gradient estimate, it follows that

\begin{equation}\label{1955}
\sup_{z\in D'} |g_{\theta,n+1}'(z)|\leq \delta_n^{-1}\sup_{z\in D} |g_{\theta,n+1}(z)|\leq \delta_n^{-1} e^{-2l_n}\leq \delta_n
\end{equation}

By the inductive hypotheses (<ref>), (<ref>) and Lemma <ref>, we have
\begin{equation}\label{1954}
\end{equation}
Then (<ref>) follows from   (<ref>) and (<ref>).

Analogues of Proposition <ref> and Lemma <ref> hold true for the left limit operator $H_{B_{n+1}}(\theta-0)$. We state them below:

For any $\theta\in \R$, $H_{B_{n+1}}(\theta-0)$ has a unique eigenvalue $E_{n+1}(\theta-0)$ such that $|E_{n+1}(\theta-0)-E_n(\theta-0)|\leq e^{-l_n}$. Moreover, since $H_{B_{n+1}}(\theta)=H_{B_{n+1}}(\theta-0)$ and $E_n(\theta )=E_n(\theta-0)$ for $\theta\notin \left\{\{-x\cdot\omega\}\right\}_{x\in B_{n+1}}$, it follows that $E_{n+1}(\theta)=E_{n+1}(\theta-0)$ for $\theta\notin \left\{\{-x\cdot\omega\}\right\}_{x\in B_{n+1}}$.

$$\tilde D:=\begin{cases}
D(E_n(1-0),10\delta_n),\ \ &\theta\in \Z,\\
A_n(\theta,10\delta_n),\ \ &\theta\notin\Z,
\end{cases}$$

$$\tilde D':=\begin{cases}
D(E_n(1-0),5\delta_n),\ \ &\theta\in \Z,\\
A_n(\theta,5\delta_n),\ \ &\theta\notin\Z.
\end{cases}$$

For  $z\in\tilde  D$, the Schur complement

\begin{align}\label{923.*}
s_{\theta-0,n+1}(z):&=z-v(\theta-0) -c^{\text{T}}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\theta-0)\right)^{-1}c\nonumber	\\
\end{align}

exists and 	satisfies

\begin{align*}
s_{\theta-0,n+1}(z)&=(z-E_{n+1}(\theta-0))\left(1+O(\sum_{k=0}^{n}\delta_k)\right) , & z\in  \tilde D,\\
s_{\theta-0,n+1}'(z)&=1+O(\sum_{k=0}^{n}\delta_k) ,& z\in  \tilde D'.
\end{align*}

§.§.§ Verification of the Lipschitz monotonicity property for $E_{n+1}(\theta)$

Denote $\{-x\cdot\omega\}$ by $\beta_x$ and denote by  $0=\alpha_1<\alpha_2<\cdots<\alpha_{|B_{n+1}|}<\alpha_{|B_{n+1}|+1}=1$ the discontinuities of $H_{B_{n+1}}(\theta)$, where

$$\{\alpha_1,\alpha_2,\cdots,\alpha_{|B_{n+1}|}\}=\left\{\beta_x\right\}_{x\in B_{n+1}}.$$

Also denote  the Rellich functions of $H_{B_{n+1}}(\theta)$ in non-decreasing order by $\lambda_i(\theta)$ ($1\leq i\leq |B_{n+1}|$) which are $1$-periodic and continuous on $[0,1)$ except on the set $\{\alpha_1,\alpha_2,\cdots,\alpha_{|B_{n+1}|}\}$ with the  local Lipschitz monotonicity property as  mentioned at the beginning of section <ref>.

Let $E_{n+1}(\theta)$ be the eigenvalue of $H_{B_{n+1}}(\theta)$ constructed in Proposition <ref>. Then $E_{n+1}(\theta)$ satisfies the following two properties:

* On each small interval $[\alpha_k,\alpha_{k+1})$, $E_{n+1}(\theta)$ coincides with exactly one branch of $\lambda_i(\theta)$, and hence is continuous and satisfies the local Lipschitz monotonicity property on $[\alpha_k,\alpha_{k+1})$:

\begin{equation*}\label{400}
E_{n+1}(\theta_2)-E_{n+1}(\theta_1)\geq L(\theta_2-\theta_1), \ \ \alpha_k\leq \theta_1\leq \theta_2<\alpha_{k+1}.
\end{equation*}

* $E_1(\theta)$ has  non-negative “jumps” at  $\alpha_k$ $(2\leq k\leq |B_{n+1}|)$:

$$E_{n+1}(\alpha_k)\geq E_{n+1}(\alpha_k-0). $$

To prove the first property, we assume that $E_{n+1}(\alpha_k)=\lambda_i(\alpha_k)$ for some branch $\lambda_i(\theta)$. Define the set $X=\{\theta\in[\alpha_k,\alpha_{k+1}) :\ E_{n+1}(\theta)=\lambda_i(\theta) \}$. We prove   $X=[\alpha_k,\alpha_{k+1})$ below. By the definition,  $\alpha_k\in X$, thus $X\neq\emptyset$. Assume there is  a sequence  $\{\theta_j\}_{j=1}^{\infty}\subset X$ such that $\theta_j\to\theta_\infty\in [\alpha_k,\alpha_{k+1})$. Hence by Proposition <ref>, one has

\begin{equation}\label{721n}
|\lambda_i(\theta_j)-E_n(\theta_j)|=|E_{n+1}(\theta_j)-E_n(\theta_j)|\leq e^{-l_n}.
\end{equation}

Since the discontinuities of $E_n(\theta)$ are contained in the set
$$\{\beta_x\}_{x\in B_n}\subset \{\alpha_1,\alpha_2,\cdots,\alpha_{|B_{n+1}|}\} $$
by the item (4) of Hypothesis <ref>, it follows that $E_n(\theta)$ is continuous on $[\alpha_k,\alpha_{k+1})$.
Since $\lambda_i(\theta)$ and $E_n(\theta)$ are continuous on $[\alpha_k,\alpha_{k+1})$, taking a limit of (<ref>) yields $|\lambda_i(\theta_\infty)-E_n(\theta_\infty)|\leq e^{-l_n}$. By the definition of $E_{n+1}(\theta_\infty)$ and the uniqueness of eigenvalue of $H_{B_{n+1}}(\theta_\infty)$ in the energy interval $[E_n(\theta_\infty)-10\delta_n,E_n(\theta_\infty)+10\delta_n]$, we get $E_{n+1}(\theta_\infty)=\lambda_i(\theta_\infty)$ and hence $X$ is closed in $[\alpha_k,\alpha_{k+1})$.
Assume that $\theta_0\in X$. Then by Proposition <ref>,  for any $j\neq i$, one has

\begin{equation}\label{858n}
\end{equation}

By the continuity of $\lambda_j(\theta)$ and $E_n(\theta)$ in $[\alpha_k,\alpha_{k+1})$, the inequality (<ref>) can be extended to  a neighborhood   (denoted by $U$) of $\theta_0$ in $[\alpha_k,\alpha_{k+1})$. Hence by the definition of $E_{n+1}(\theta)$, $\lambda_j(\theta)\neq E_{n+1}(\theta)$ for any $\theta\in U$ and  $j\neq i$. So it must be that $\lambda_i(\theta)= E_{n+1}(\theta)$ in $U$. Hence $X$ is open in $[\alpha_k,\alpha_{k+1})$. Thus we finish the proof of the first property.
For the second property, 	since $\alpha_k\neq 0$, there exists some $x\in B_{n+1}$ with  $x_1\neq o$ such that  $\alpha_k=\beta_x$.
Recall the notation $s_{\theta,n+1}(z)$, $r_{\theta,n+1}(z)$, $s_{\theta-0,n+1}(z)$, $r_{\theta-0,n+1}(z)$  (c.f. (<ref>), (<ref>)).
We will restrict $z$ in the interval
\begin{equation*}
\end{equation*}
in the following discussion  so that all the functions concerning $z$ will be real-valued. Since $\beta_x\in \R\setminus\Z$ and $I\subset A_n(\beta_x,5\delta_n)$,  it follows that  (<ref>) holds true.
If we can  prove
\begin{equation}\label{1055+}
s_{\beta_x,n+1}(z)\leq s_{\beta_x-0,n+1}(z),\ \ z\in I,
\end{equation}

\begin{equation}\label{ky}
s_{\beta_x,n+1}(E_{n+1}(\beta_x-0))\leq s_{\beta_x-0,n+1}(E_{n+1}(\beta_x-0))=0=s_{\beta_x,n+1}(E_{n+1}(\beta_x)).
\end{equation}
The desired conclusion
\begin{equation}\label{key}
E_{n+1}(\beta_x)\geq  E_{n+1}(\beta_x-0)
\end{equation}
will follow from (<ref>) and the monotonicity of  $s_{\beta_x,n+1}(z)$ (since by (<ref>), $s_{\beta_x,n+1}'(z)\approx1$). The remainder of the proof  aims at verifying (<ref>). Recalling (<ref>), we compute

\begin{align}
&s_{\beta_x,n+1}(z)-s_{\beta_x-0,n+1}(z)\nonumber\\ =&r_{\beta_x-0,n+1}(z)-r_{\beta_x,n+1}(z)\nonumber\\
=&c^{\text{T}}\left[ \left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)^{-1}- \left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)^{-1}\right]c\nonumber\\
=&c^{\text{T}}\left[ \left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)^{-1}{\bm e}_x {\bm e}_x^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)^{-1}\right]c. \label{1210+}
\end{align}

On the last line of the above equation  we use the resolvent identity and the below equation
$$H_{B_{n+1}^o}(\beta_x-0)-H_{B_{n+1}^o}(\beta_x)={\bm e}_x {\bm e}_x^{\rm T}.$$

By Cramer's rule, it follows that

A simple computation shows that

\begin{equation}\label{1437+}
\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)^{*}{\bm e}_x=\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)^{*}{\bm e}_x.
\end{equation}

We denote by $b(z)$ the vector in (<ref>).

\begin{equation}\label{1528+}
\eqref{1210+}=\frac{1}{\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)}\frac{1}{\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)}\left(c^{\rm T} b(z)\right)^2.
\end{equation}

From (<ref>), it follows  that the sign of $s_{\beta_x,n+1}(z)-s_{\beta_x-0,n+1}(z)$ is the same as the sign of  the product
\begin{equation}\label{532}
\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right).
\end{equation}

The following technical Lemma <ref> shows that the sign of (<ref>) is negative and hence by  (<ref>) and  (<ref>), it follows that (<ref>) holds true. Thus we finish the proof of (<ref>) and the second property.

Let $\beta_x=\{-x\cdot\omega\}$ with $x\neq o$. Then for
\begin{equation}\label{I}
z\in I:=(E_n(\beta_x-0)-5\delta_n,E_n(\beta_x)+5\delta_n),
\end{equation}
we have

\begin{equation}\label{key1}
\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)<0.
\end{equation}

Fix $z\in I$. For $0\leq m\leq n$, we define the set

$$S_m:=S_m(\beta_x,z)\cup S_m(\beta_x-0,z).$$

First, we prove (<ref>) in  the case that
\begin{equation}\label{1509}
\text{$B_{n+1}^o\cap S_m=\emptyset$ ($0\leq m\leq n$). }
\end{equation}

Denote $B_{n+1}^1:=B_{n+1}^o\setminus\{x\}$.
$$\langle H_{B_{n+1}^o}(\beta_x-0){\bm e}_x,{\bm e}_x\rangle=v(1-0)=1, \  \ \langle H_{B_{n+1}^o}(\beta_x){\bm e}_x,{\bm e}_x\rangle=v(0)=0$$
and  $H_{B_{n+1}^1}(\beta_x -0)=H_{B_{n+1}^1}(\beta_x )$, we  can write

z-1 & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)
\end{pmatrix}

z-0 & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)
\end{pmatrix}.
By (<ref>), it follows  that $B_{n+1}^1\cap S_0(\beta_x,z)=\emptyset$.
Hence by Proposition <ref>, we get the resolvent bound

\begin{equation}\label{1623+}
\left \|\left( z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)  \right)^{-1}\right\|\leq 10\delta_0^{-1}.
\end{equation}

Using Schur complement formula yields

\begin{align}\label{351+}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)\\=&\left [z-1-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right).
\end{align}

By (<ref>), we have $x\notin S_0(\beta_x-0,z)$. Thus
\begin{equation}\label{1539}
|1-z|=|v(1-0)-z|=	|v(\beta_x+x\cdot\omega-0)-z|\geq  \delta_0.
\end{equation}

Since  $z\in I$ yields  $z \leq E_n(\beta_x)+5\delta_n$, it follows  from  inductive hypothesis (<ref>) that

\begin{align}\label{1544}
\nonumber	z-1\leq z-v(\beta_x)&\leq z-E_n(\beta_x)+\sum_{k=0}^{n-1}|E_{k}(\beta_x)-E_{k+1}(\beta_x)|\\ \nonumber
&\leq 5\delta_n+\varepsilon+\sum_{k=1}^{n-1}e^{-l_k}\\
\end{align}

By (<ref>) and (<ref>), we have $z-1\leq -\delta_0$.
Together with  (<ref>), $\|\tilde{c}\|\leq 2d\varepsilon$, we obtain

\begin{equation}\label{352+}
z-1-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\leq -\delta_0+\frac{40d^2\varepsilon^2}{\delta_0}<0.
\end{equation}

\begin{align}\label{353+}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)\\=&\left [z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}                                                           _{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right) .
\end{align}

And since $x\notin S_0(\beta_x,z)$ we have
\begin{equation}\label{1539.}
|z|=|v(0)-z|=	|v(\beta_x+x\cdot\omega)-z|\geq  \delta_0.
\end{equation}

Since  $z\in I$ yields  $z \geq E_n(\beta_x-0)-5\delta_n$, it follows  from  inductive hypothesis (<ref>) that

\begin{align}\label{1544.}
\nonumber	z-0\geq z-v(\beta_x-0)&\geq z-E_n(\beta_x-0)-\sum_{k=0}^{n-1}|E_{k}(\beta_x-0)-E_{k+1}(\beta_x-0)|\\ \nonumber
&\geq -5\delta_n-\varepsilon-\sum_{k=1}^{n-1}e^{-l_k}\\
\end{align}

By (<ref>) and (<ref>), we have $z-0\geq \delta_0$.
Together with  (<ref>), $\|\tilde{c}\|\leq 2d\varepsilon$, we obtain

\begin{equation}\label{354+}
z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\geq \delta_0-\frac{40d^2\varepsilon^2}{\delta_0}>0.
\end{equation}

Then 	(<ref>) follows from  (<ref>), (<ref>), (<ref>) and (<ref>).

Next we consider the  negation case of (<ref>).
Let  $m$ be the  largest integer of $0\leq m\leq n$ such that $B_{n+1}^o\cap S_m\neq \emptyset$ and let  $x_1\in S_m\cap B_{n+1}^o$.	Without loss of generality,  we assume $x_1\in S_m(\beta_x,z)\cap B_{n+1}^o$ since the case $x_1\in S_m(\beta_x-0,z)\cap B_{n+1}^o$ is analogous. Since $z\in I$, it follows from Lemma <ref> that

$$z\in I\subset A_n(\beta_x,5\delta_n)\subset A_m(\beta_x,5\delta_m).$$

And since $x_1\in S_m(\beta_x,z)\cap B_{n+1}^o$, we have $0<\|x_1\|_1\leq 2l_{n+1}$ and $|E_m(\beta_x+x_1\cdot\omega)-z|\leq\delta_m$.
Then  by (<ref>) and Lemma <ref> , it follows that

\begin{align*}
L\|x_1\cdot \omega\|_\T	\leq &\min (|E_m(\beta_x+x_1\cdot \omega )-E_m(\beta_x)|,|E_m(\beta_x+x_1\cdot \omega )-E_m(\beta_x-0)|)
\\ \leq &\operatorname{dist}(E_m(\beta_x+x_1\cdot\omega),A_m(\beta_x,5\delta_m))+5\delta_m\\
\leq & |E_m(\beta_x+x_1\cdot\omega)-z|+5\delta_m\\
\leq& 6\delta_m.
\end{align*}

Hence by (<ref>) and $0<\|x_1\|_1\leq 2l_{n+1}$, we have

$$L\gamma(2l_{n+1})^{-\tau}\leq L\|x_1\cdot \omega\|_\T 	\leq 6 \delta_m=6e^{-l_m^{\frac{2}{3}}}\leq  L\gamma(2l_m^{16})^{-\tau}.$$

It follows from the above inequality that $l_{n+1}\geq l_m^{16}$ and hence $m\leq n-3$.

The proof proceeds by case analysis:

Case 1: Assume that

\begin{equation}\label{case1}
x_1\in S_m(\beta_x,z)\cap B_{n+1}^o,\ \  x_1=x.
\end{equation}

We denote  $B_{n+1}^1:=B_{n+1}^o\setminus\{x\}$ as before. Then $H_{B_{n+1}^1}(\beta_x)=H_{B_{n+1}^1}(\beta_x-0)$. Let  $p\in S_k(\beta_x,z)\cap B_{n+1}^1$.  (By the maximal property of $m$, it follows that $k\leq m$.) Then $B_{k+1}(p)\subset B_{n+1}^o$ since $B_{n+1}^o$ is $n$-regular related to $(\beta_x,z)$ (c.f. Lemma <ref>).

Since   $x_1\in S_m(\beta_x,z)$ and $p\in S_k(\beta_x,z)$, it follows that

$$|E_m(\beta_x+x\cdot\omega)-z|\leq \delta_m, \ \ |E_k(\beta_x+p\cdot\omega)-z|\leq \delta_k. $$

Since $p\neq x$, estimating as before by (<ref>), (<ref>) and (<ref>), we have
\begin{align*}
L\gamma\|p-x\|_1^{-\tau} &\leq  L\gamma\|(p-x)\cdot\omega\|_\T\\ &\leq |E_k(\beta_x+p\cdot\omega)-E_k(\beta_x+x\cdot \omega )|\\&\leq |E_k(\beta_x+p\cdot\omega)-z|+|z-E_m(\beta_x+x\cdot\omega)|\\ &+|E_m(\beta_x+x\cdot\omega)-E_k(\beta_x+x\cdot\omega)|
\\&\leq \delta_k+\delta_m +\sum_{j=k}^{m-1}e^{-l_j}\leq 10\delta_k,
\end{align*}
yielding $\|p-x\|_1\geq (L\gamma/10)^\frac{1}{\tau}\delta_k^{-\tau}\geq 10l_{k+1}$. Since $B_{k+1}\subset Q_{2l_{k+1}}$,  it follows that  $x\notin B_{k+1}(p)$ and hence $B_{k+1}(p)\subset B_{n+1}^1$. Thus $B_{n+1}^1$ is $(m+1)$-regular related to $(\beta_x,z)$. Since  $S_{m+1}(\beta_x,z)\cap B_{n+1}^1=\emptyset$ by the maximal property of $m$, it follows that $B_{n+1}^1$ is $(m+1)$-good related to $(\beta_x,z)$.
Hence by Hypothesis <ref>, we get the resolvent bound

\begin{equation}\label{1623++}
\left \|\left( z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)  \right)^{-1}\right\|\leq 10\delta_{m+1}^{-1}.
\end{equation}

$$\langle H_{B_{n+1}^o}(\beta_x-0){\bm e}_x,{\bm e}_x\rangle=v(1-0)=1, \  \ \langle H_{B_{n+1}^o}(\beta_x){\bm e}_x,{\bm e}_x\rangle=v(0)=0$$
and  $H_{B_{n+1}^1}(\beta_x -0)=H_{B_{n+1}^1}(\beta_x )$,
we can write

z-1 & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)
\end{pmatrix}

z-0 & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)
\end{pmatrix}.

Hence by Schur complement formula,

\begin{align}\label{351++}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)\\=&\left [z-1-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)
\end{align}

\begin{align}\label{353++}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)\\=&\left [z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}                                                           _{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right) .
\end{align}

Denote $B_{m+1}^o(x):=B_{m+1}(x)\setminus\{x\}$. Since $x\in S_m(\beta_x,z)\cap B_{n+1}^o$ and $B_{n+1}^o$ is $n$-regular related to $(\beta_x,z)$ (c.f. Lemma  <ref>), it follows that $B_{m+1}(x)\subset B_{n+1}^o$ and hence $B_{m+1}^o(x)\subset B_{n+1}^1 $.   Since $\beta_x+x\cdot\omega\in \Z $ and  $x\in S_m(\beta_x,z)$, it follows that $|E_m(\beta_x+x\cdot\omega)-z|=|E_m(0)-z|\leq \delta_m$, yielding $z\in D(E_m(0),10\delta_m)$. It  follows from Corollary <ref>  that $B_{m+1}^o(x)$ is $m$-good related to $(\beta_x,z)$. Thus

\begin{equation}\label{bm+}
|G_{B_{m+1}^o(x)}^{\beta_x,z}(\tilde y,y)|\leq e^{-\gamma_{m}\|\tilde y-y\|_1}, \ \|\tilde y-y\|_1\geq l_{m}^{\frac{5}{6}}.
\end{equation}

By the resolvent identity, (<ref>), (<ref>) and (<ref>), estimating as in (<ref>), we have

\begin{align}
\nonumber	&z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\\
\nonumber	=&z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{m+1}^o(x)}-H_{B_{m+1}^o(x)}(\beta_x)\right)^{-1}\tilde{c}\\-&\nonumber\varepsilon \sum_{\substack{\|\tilde{y}-x\|_1,\|y-x\|_1=1\\(w,w')\in \partial_{B_{n+1}^1}B_{m+1}^o(x)}}\tilde c(\tilde y)G_{B_{m+1}^o(x)}^{\beta_x,z}(\tilde y,w)G_{B_{n+1}^1}^{\beta_x,z}(w',y)\tilde c(y)\\
\nonumber=&z-0-c^{\rm T}\left(z\operatorname{Id}_{B_{m+1}^o}-H_{B_{m+1}^o}(\beta_x+x\cdot\omega)\right)^{-1}c-O(e^{-3l_{m+1}})\\
\nonumber	=&\left(z-E_{m+1}(\beta_x+x\cdot \omega) -O(e^{-2l_{m+1}})\right)\left(1+O(\sum_{k=0}^m\delta_k)\right)\\
=&\left(z-E_{m+1}(0) -O(e^{-2l_{m+1}})\right)\left(1+O(\sum_{k=0}^m\delta_k)\right).\label{2036}
\end{align}

The above $O(\star)$ are understood by  numbers bounded by $\star$.
Since  $x\notin S_{m+1}(\beta_x,z)$ by the maximal property of $m$, it follows that

\begin{equation}\label{2038}
|z-E_{m+1}(\beta_x+x\cdot\omega)|=|z-E_{m+1}(0)|\geq \delta_{m+1}.
\end{equation}

Since $z\geq E_n(\beta_x-0)-5\delta_n$,     it follows from   inductive hypotheses (<ref>), (<ref>) that

\begin{align}\label{2}
\nonumber	z-E_{m+1}(0)&\geq z-E_{m+1}(\beta_x-0)\\
\nonumber	&\geq z-E_{n}(\beta_x-0)-|E_n(\beta_x-0)-E_{m+1}(\beta_x-0)|\\
\nonumber	&\geq -5\delta_n-\sum_{k=m+1}^{n-1}e^{-l_k}\\
\end{align}

By (<ref>) and (<ref>), we have 	$z-E_{m+1}(0)\geq \delta_{m+1}>e^{-2l_{m+1}}$. Hence by (<ref>), it follows that

\begin{equation}\label{0}
z-0-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}>0.
\end{equation}

Since $x\in S_m(\beta_x,z)$, we have $|z-E_m(0)|<\delta_m$. Hence by (<ref>) and  (<ref>), it follows that

\begin{align}
\nonumber	&z-1-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^1}-H_{B_{n+1}^1}(\beta_x)\right)^{-1}\tilde{c}\\
\nonumber	=&\left(z-E_{m+1}(0) -O(e^{-2l_{m+1}})\right)\left(1+O(\sum_{k=0}^m\delta_k)\right)-1\\
\nonumber	\leq &2(|z-E_m(0)|+|E_m(0)-E_{m+1}(0)|+e^{-2l_{m+1}})-1\\
\leq &2(\delta_m+\delta_m)-1<0.\label{1}
\end{align}

Then (<ref>) follows from (<ref>), (<ref>), (<ref>) and (<ref>).

Case 2: Assume that
\begin{equation}\label{case2}
x_1\in S_m(\beta_x,z)\cap B_{n+1}^o,\ \  x_1\neq x.
\end{equation}

We denote $B_{n+1}^{(1)}=B_{n+1}^o\setminus \{x_1\}$.
Let  $p\in S_k(\beta_x,z)\cap B_{n+1}^{(1)}$.  (By the maximal property of $m$, it follows that $k\leq m$.) Then $B_{k+1}(p)\subset B_{n+1}^o$ since $B_{n+1}^o$ is $n$-regular related to $(\beta_x,z)$ (c.f. Lemma  <ref>).
Since $p\neq x_1$, estimating as before  by (<ref>), (<ref>) and (<ref>), we have

\begin{align*}
L\gamma\|p-x_1\|_1^{-\tau} &\leq  L\gamma\|(x_1-p)\cdot\omega\|_\T\\ &\leq |E_k(\beta_x+p\cdot\omega)-E_k(\beta_x+x_1\cdot \omega )|\\&\leq |E_k(\beta_x+p\cdot\omega)-z|+|z-E_m(\beta_x+x_1\cdot\omega)|\\ &+|E_m(\beta_x+x_1\cdot\omega)-E_k(\beta_x+x_1\cdot\omega)|
\\&\leq \delta_k+\delta_m +\sum_{j=k}^{m-1}e^{-l_j}\leq 10\delta_k,
\end{align*}
yielding $\|p-x_1\|_1\geq (L\gamma/10)^\frac{1}{\tau}\delta_k^{-\tau}\geq 10l_{k+1}$. It follows that  $x_1\notin  B_{k+1}(p)$ and hence $B_{k+1}(p)\subset B_{n+1}^{(1)}$. Thus $B_{n+1}^{(1)}$ is $(m+1)$-regular related to $(\beta_x,z)$. Since  $S_{m+1}(\beta_x,z)\cap B_{n+1}^{(1)}=\emptyset$ by the maximal property of $m$, it follows that $B_{n+1}^{(1)}$ is $(m+1)$-good related to $(\beta_x,z)$.
Hence by Hypothesis <ref>, we get the resolvent bound

\begin{equation}\label{1623,}
\left \|\left( z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x)  \right)^{-1}\right\|\leq 10\delta_{m+1}^{-1}.
\end{equation}

Let  $p\in S_k(\beta_x-0,z)\cap B_{n+1}^{(1)}$.  (By the maximal property of $m$, it follows that $k\leq m$.) Then $B_{k+1}(p)\subset B_{n+1}^o$ since $B_{n+1}^o$ is $n$-regular related to $(\beta_x-0,z)$ (c.f. Lemma  <ref>).
Since $p\neq x_1$, estimating as before  by (<ref>), (<ref>) and (<ref>), we have

\begin{align*}
L\gamma\|p-x_1\|_1^{-\tau} &\leq  L\gamma\|(x_1-p)\cdot\omega\|_\T\\ &\leq |E_k(\beta_x+p\cdot\omega-0)-E_k(\beta_x+x_1\cdot \omega )|\\&\leq |E_k(\beta_x+p\cdot\omega-0)-z|+|z-E_m(\beta_x+x_1\cdot\omega)|\\ &+|E_m(\beta_x+x_1\cdot\omega)-E_k(\beta_x+x_1\cdot\omega)|
\\&\leq \delta_k+\delta_m +\sum_{j=k}^{m-1}e^{-l_j}\leq 10\delta_k,
\end{align*}
yielding $\|p-x_1\|_1\geq (L\gamma/10)^\frac{1}{\tau}\delta_k^{-\tau}\geq 10l_{k+1}$. It follows that  $x_1\notin B_{k+1}(p)$ and hence $B_{k+1}(p)\subset B_{n+1}^{(1)}$. Thus $B_{n+1}^{(1)}$ is $(m+1)$-regular related to $(\beta_x-0,z)$. Since  $S_{m+1}(\beta_x-0,z)\cap B_{n+1}^{(1)}=\emptyset$ by the maximal property of $m$, it follows that $B_{n+1}^{(1)}$ is $(m+1)$-good related to $(\beta_x-0,z)$.
Hence by Hypothesis <ref>, we get the resolvent bound

\begin{equation}\label{1623,,}
\left \|\left( z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x-0)  \right)^{-1}\right\|\leq 10\delta_{m+1}^{-1}.
\end{equation}

Since $x_1\neq x$, it follows that $v(\beta_x-0+x_1\cdot\omega)=v(\beta_x+x_1\cdot\omega)$ and hence
$$\langle H_{B_{n+1}^o}(\beta_x-0){\bm e}_{x_1},{\bm e}_{x_1}\rangle =\langle H_{B_{n+1}^o}(\beta_x){\bm e}_{x_1},{\bm e}_{x_1}\rangle=v(\beta_x+x_1\cdot\omega).$$
Thus we  can write  write

z-v(\beta_x+x_1\cdot\omega) & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x-0)
\end{pmatrix}

z-v(\beta_x+x_1\cdot\omega) & \tilde{c}^{\text{T}} \\
\tilde{c} & z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x)
\end{pmatrix}.

Hence by Schur complement formula,

\begin{align}\label{351,}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)\\=&\left [z-v(\beta_x+x_1\cdot\omega)-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x-0)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x-0)\right)
\end{align}

\begin{align}\label{353,}
&\nonumber	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)\\=&\left [z-v(\beta_x+x_1\cdot\omega)-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x)\right)^{-1}\tilde{c}\right ]\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x)\right).
\end{align}

Denote $B_{m+1}^o(x_1):=B_{m+1}(x_1)\setminus\{x_1\}$.  Since $x_1\in S_m(\beta_x,z)\cap B_{n+1}^o$ and $B_{n+1}^o$ is $n$-regular related to $(\beta_x,z)$ (c.f. Lemma  <ref>), it follows that $B_{m+1}(x_1)\subset B_{n+1}^o$ and hence $B_{m+1}^o(x_1)\subset B_{n+1}^1 $.
Since $x_1\in S_m(\beta_x,z)$ yields $|E_m(\beta_x+x_1\cdot\omega)-z|\leq \delta_m$, it follows that    $z\in A_m(\beta_x+x_1\cdot\omega,10\delta_m)$ (by $x_1\neq x$, we have $\beta_x+x_1\cdot\omega\notin\Z$).   It follows from Corollary <ref> that $B_{m+1}^o(x_1)$ is $m$-good related to $(\beta_x,z)$ and $(\beta_x-0,z)$. Thus  by Hypothesis <ref>, we have

\begin{equation}\label{bm,-}
|G_{B_{m+1}^o(x_1)}^{\beta_x-0,z}(\tilde y,y)|\leq e^{-\gamma_{m}\|\tilde y-y\|_1}, \ \|\tilde y-y\|_1\geq l_{m}^{\frac{5}{6}},
\end{equation}

\begin{equation}\label{bm,}
|G_{B_{m+1}^o(x_1)}^{\beta_x,z}(\tilde y,y)|\leq e^{-\gamma_{m}\|\tilde y-y\|_1}, \ \|\tilde y-y\|_1\geq l_{m}^{\frac{5}{6}}.
\end{equation}

By the resolvent identity, (<ref>), (<ref>), (<ref>), (<ref>) and (<ref>), estimating as in (<ref>), we have

\begin{align}
\nonumber	&z-v(\beta_x+x_1\cdot\omega)-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x-0)\right)^{-1}\tilde{c}\\
\nonumber	=&z-v(\beta_x+x_1\cdot\omega)-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{m+1}^o(x_1)}-H_{B_{m+1}^o(x_1)}(\beta_x-0)\right)^{-1}\tilde{c}\\-&\nonumber\varepsilon \sum_{\substack{\|\tilde{y}-x_1\|_1,\|y-x_1\|_1=1\\(w,w')\in \partial_{B_{n+1}^{(1)}}B_{m+1}^o(x_1)}}\tilde c(\tilde y)G_{B_{m+1}^o(x_1)}^{\beta_x-0,z}(\tilde y,w)G_{B_{n+1}^{(1)}}^{\beta_x-0,z}(w',y)\tilde c(y)\\
\nonumber	=&z-v(\beta_x+x_1\cdot\omega)-c^{\rm T}\left(z\operatorname{Id}_{B_{m+1}^o}-H_{B_{m+1}^o}(\beta_x+x_1\cdot \omega -0)\right)^{-1}c-O(e^{-3l_{m+1}})\\
=&\left(z-E_{m+1}(\beta_x+x_1\cdot \omega-0) -O(e^{-2l_{m+1}})\right)\left(1+O(\sum_{k=0}^m\delta_k)\right) \label{2036,-}
\end{align}

\begin{align}
\nonumber	&z-v(\beta_x+x_1\cdot\omega)-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x)\right)^{-1}\tilde{c}\\
\nonumber	=&z-v(\beta_x+x_1\cdot\omega)-\tilde{c}^{\rm T}\left(z\operatorname{Id}_{B_{m+1}^o(x_1)}-H_{B_{m+1}^o(x_1)}(\beta_x)\right)^{-1}\tilde{c}\\-&\nonumber\varepsilon \sum_{\substack{\|\tilde{y}-x_1\|_1,\|y-x_1\|_1=1\\(w,w')\in \partial_{B_{n+1}^{(1)}}B_{m+1}^o(x_1)}}\tilde c(\tilde y)G_{B_{m+1}^o(x_1)}^{\beta_x,z}(\tilde y,w)G_{B_{n+1}^{(1)}}^{\beta_x,z}(w',y)\tilde c(y)\\
\nonumber	=&z-v(\beta_x+x_1\cdot\omega)-c^{\rm T}\left(z\operatorname{Id}_{B_{m+1}^o}-H_{B_{m+1}^o}(\beta_x+x_1\cdot\omega)\right)^{-1} c-O(e^{-3l_{m+1}})\\
=&\left(z-E_{m+1}(\beta_x+x_1\cdot \omega) -O(e^{-2l_{m+1}})\right)\left(1+O(\sum_{k=0}^m\delta_k)\right) \label{2036,}.
\end{align}

The product of (<ref>) and (<ref>) is positive.

Since $x_1\neq x$, then $\{\beta_x+x_1\cdot \omega \}\neq 0$. Hence by the inductive hypothesis (<ref>), we have $E_{m+1}(\beta_x+x_1\cdot \omega)\geq E_{m+1}(\beta_x+x_1\cdot \omega-0)$. And since $x_1\notin S_{m+1}$ (by the maximum property of $m$), it follows that
\begin{equation}\label{845}
|E_{m+1}(\beta_x+x_1\cdot \omega)-z|\geq \delta_{m+1}\end{equation}

\begin{equation}\label{846}
|E_{m+1}(\beta_x+x_1\cdot \omega-0)-z|\geq \delta_{m+1}.
\end{equation}

Since $x_1\neq o$, we have $\beta_x\neq \{\beta_x+x_1\cdot\omega\}$.
If $\beta_x<\{\beta_x+x_1\cdot\omega\}$, then by (<ref>), (<ref>) and  $z\leq E_n(\beta_x)+5\delta_n$, we have

\begin{align}\label{>}
\nonumber	&E_{m+1}(\beta_x+x_1\cdot \omega-0)-z\\ \nonumber\geq& E_{m+1}(\beta_x+x_1\cdot \omega-0)-E_{m+1}(\beta_x) -|E_{m+1}(\beta_x)-E_n(\beta_x)|-5\delta_n \\
\geq &0-\sum_{k=m+1}^{n-1}e^{-l_k}-5\delta_n\nonumber\\
\end{align}

Then by (<ref>), (<ref>), we have

$$E_{m+1}(\beta_x+x_1\cdot \omega)-z\geq E_{m+1}(\beta_x+x_1\cdot \omega-0)-z\geq \delta_{m+1} >e^{-2l_{m+1}},$$
yielding the first factors of (<ref>) and (<ref>) are both negative.

If $\beta_x>\{\beta_x+x_1\cdot\omega\}$, then by (<ref>), (<ref>) and  $z\geq E_n(\beta_x-0)- 5\delta_n$, we have

\begin{align}\label{<}
\nonumber	&E_{m+1}(\beta_x+x_1\cdot \omega)-z\\ \nonumber\leq& E_{m+1}(\beta_x+x_1\cdot \omega)-	E_{m+1}(\beta_x-0)+|E_{m+1}(\beta_x-0)-E_n(\beta_x-0)| +5\delta_n\\
\leq &0+\sum_{k=m+1}^{n-1}e^{-l_k}+5\delta_n\nonumber\\
\end{align}

Then by (<ref>), (<ref>), we have

$$E_{m+1}(\beta_x+x_1\cdot \omega-0)-z\leq E_{m+1}(\beta_x+x_1\cdot \omega)-z\leq -\delta_{m+1} <-e^{-2l_{m+1}},$$
yielding the first factors of (<ref>) and (<ref>) are both positive.  Hence the claim is proved.

By (<ref>), (<ref>),   (<ref>), (<ref>) and Claim <ref>,  we get the following proposition:
Let  $m$ be the  largest integer of $0\leq m\leq n$ such that $B_{n+1}^o\cap S_m\neq \emptyset$ and let  $x_1\in S_m\cap B_{n+1}^o$ with $x_1\neq x$.  Then the sign of

\begin{equation}\label{956}
\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^o}-H_{B_{n+1}^o}(\beta_x)\right)
\end{equation}

is the same as the sign of
$$	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(1)}}-H_{B_{n+1}^{(1)}}(\beta_x)\right)$$

with $B_{n+1}^{(1)}:=B_{n+1}^o\setminus\{x_1\}$.

Proposition <ref> means that deleting the largest order resonant point $x_1(\neq x)$ of $B_{n+1}^o$ does not change the sign of the  determinant product (<ref>). Moreover,  we   can iterate it into the following form:
Let $\{B_{n+1}^{(j)}\}_{j=0}^J$ be a set sequence such that $B_{n+1}^{(0)}=B_{n+1}^o$ and $B_{n+1}^{(j)}=B_{n+1}^{(j-1)}\setminus\{x_j\}$ where $x_j(\neq x)$ is the largest order resonant point in $B_{n+1}^{(j-1)}$ (that is, $x_j\in S_{m_j}\cap B_{n+1}^{(j-1)}$ with $m_j$ being the largest integer such that $B_{n+1}^{(j-1)}\cap S_{m_j}\neq \emptyset$).
Then the sign of

\begin{equation*}
\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(j-1)}}-H_{B_{n+1}^{(j-1)}}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(j-1)}}-H_{B_{n+1}^{(j-1)}}(\beta_x)\right)
\end{equation*}

is the same as the sign of
$$	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(j)}}-H_{B_{n+1}^{(j)}}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(j)}}-H_{B_{n+1}^{(j)}}(\beta_x)\right).$$

The proof of Proposition <ref> is the same as in Case 2 (c.f. (<ref>)), with $B_{n+1}^o$ (resp. $B_{n+1}^{(1)}$, $x_1$) replaced by $B_{n+1}^{(j-1)}$ (resp. $B_{n+1}^{(j)}$, $x_j$). One issue is to verify that   $B_{n+1}^{(j-1)}$ is $(m_j+1)$-regular. This  follows from the $n$-regular property of $B_{n+1}^o$ and the sufficiently large separation of the   resonant points from the reason below.

Let  $p_1\in S_k$ with $k\leq m_j$.  Then  for $p_2\in \{x_1,\cdots,x_{j-1}\}$, there exist   $\xi_1,\xi_2\in \{\beta_x,\beta_x-0\}$ such that

\begin{align*}
&L\gamma \|p_1-p_2\|^{-\tau}\\ \leq& L\gamma \|(p_1-p_2)\cdot\omega\|_\T\\
\leq& |E_k(\xi_1+p_1\cdot\omega)-E_k(\xi_2+p_2\cdot\omega)|\\
\leq & |E_k(\xi_1+p_1\cdot\omega)-z|+|z-E_{m_j}(\xi_2+p_2\cdot\omega)|+|E_{m_j}(\xi_2+p_2\cdot\omega)-E_k(\xi_2+p_2\cdot\omega)|\\
\leq&  \delta_k+2\delta_{m_j}+\sum_{s=k}^{m_j-1}e^{-l_s}\leq 10\delta_k,
\end{align*}

yielding   $\|p_1-p_2\|_1\geq 10l_{k+1}$ and hence  $p_2\notin B_{k+1}(p_1)$. Thus the whole blocks $B_{k+1}(p)$ ($p\in S_k\cap B_{n+1}^{(j-1)} $, $k\leq m_j$) are contained in $B_{n+1}^{(j-1)}$.

By Proposition <ref>, we can find a set $B_{n+1}^{(J)}=B_{n+1}^o\setminus\{x_1,\cdots,x_J\}$ with $x_j\neq x$ such that

* either  $B_{n+1}^{(J)}\cap S_m=\emptyset$ ($0\leq m\leq n$) or $x\in B_{n+1}^{(J)}$ is the largest order resonant point in $B_{n+1}^{(J)}$.

* The sign of
is the same as the sign of

$$	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(J)}}-H_{B_{n+1}^{(J)}}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(J)}}-H_{B_{n+1}^{(J)}}(\beta_x)\right).$$

Replacing $B_{n+1}^o$ by $B_{n+1}^{(J)}$ in  (<ref>) or Case 1  (c.f. <ref>)  yields

$$	\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(J)}}-H_{B_{n+1}^{(J)}}(\beta_x-0)\right)\operatorname{det}\left(z\operatorname{Id}_{B_{n+1}^{(J)}}-H_{B_{n+1}^{(J)}}(\beta_x)\right)<0$$

and hence (<ref>).

§.§.§ Green's function estimates for $(n+1)$-good sets

In this part, we relate the resonances to $E_{n+1}(\theta)$ and establish Green's function estimates for $(n+1)$-good sets.
For $p\in \Z^d$, we denote

Fix $\theta\in \R$, $E\in \C$. Let $\delta_{n+1}:=e^{-l_{n+1}^\frac{2}{3}}$. Define the $(n+1)$-scale resonant points set (related to $(\theta,E)$) as
$$S_{n+1}(\theta,E):=\left\{p\in \Z^d:\ |E_{n+1} (\theta+p\cdot \omega)-E|<\delta_{n+1}\right\}.$$

Let $\Lambda\subset\Z^d$ be a finite set. Related to $(\theta,E)$,  we say

* $\Lambda$ is $(n+1)$-nonresonant  if $\Lambda\cap S_{n+1}(\theta,E)=\emptyset$.

* $\Lambda$ is $(n+1)$-regular if $p\in \Lambda\cap S_k(\theta,E)$ $\Rightarrow$  $B_{k+1}(p)\subset\Lambda$ ($0\leq k\leq n$).

* $\Lambda$ is $(n+1)$-good if it is both $(n+1)$-nonresonant and $(n+1)$-regular.

Let $m$ be an integer of $0\leq m\leq n$.	Fix $\theta\in \R$, $E\in \C$. If $p\in S_m(\theta,E)$ and $p \notin S_{m+1}(\theta,E)$, then for any $E^*$ such that $|E-E^*|<\delta_{m+1}/2$, we have

\begin{align}\label{1l2`}
\|G_{B_{m+1}(p)} ^{\theta,E^*}\|&\leq10\delta_{m+1}^{-1},
\\
\label{1ex`}
|G_{B_{m+1}(p)}^{\theta,E^*}(x,y)|&\leq e^{-\gamma_m(1-l_{m+1}^{-\frac{1}{50}})\|x-y\|_1}, \ \|x-y\|_1\geq l_{m+1}^{\frac{4}{5}}.
\end{align}

Since $p\in S_m(\theta,E)$, by definition we have $|E_m(\theta+p\cdot \omega )-E|\leq \delta_m$. Thus
$$ |E_m(\theta+p\cdot \omega )-E^*|\leq |E_m(\theta+p\cdot \omega )-E| +|E-E^*|\leq \delta_m+\delta_{m+1}/2.$$
By item ($3$) of Hypothesis <ref>,  Proposition <ref> and translation property, $E_{m+1}(\theta+p\cdot \theta )$ is the unique eigenvalue of $H_{B_{m+1}(p)}(\theta)$ in $[E_m(\theta+p\cdot \omega )-10\delta_m,E_m(\theta+p\cdot \omega )+10\delta_m]$ and moreover

\begin{align*}\
|E_{m+1}(\theta+p\cdot \omega )-E^*|&\leq  |E_m(\theta+p\cdot \omega )-E^*|+|E_{m+1}(\theta+p\cdot \omega )-E_m(\theta+p\cdot \omega )|\\&\leq \delta_m+\delta_{m+1}/2+e^{-l_m}\\&<2\delta_m.
\end{align*}

Hence we get

$$\operatorname{dist}\left (\operatorname{Spec}(H_{B_{m+1}(p)}(\theta )),E^* \right )=|E_{m+1}(\theta+p\cdot \omega )-E^*|.$$

Since $p \notin S_{m+1}(\theta,E)$, we have
$$|E_{m+1}(\theta+p\cdot \omega )-E^*|\geq |E_{m+1}(\theta+p\cdot \omega )-E|-|E-E^*|\geq \delta_{m+1}-\delta_{m+1}/2\geq \delta_{m+1}/2.$$

Since $H_{B_{m+1}(p)}(\theta )$ is self-adjoint,  we get
$$\|G_{B_{m+1}(p)} ^{\theta,E^*}\|=\frac{1}{\operatorname{dist}\left (\operatorname{Spec}(H_{B_{m+1}(p)}(\theta )),E^* \right )}=\frac{1}{|E_{m+1}(\theta+p\cdot \omega )-E^*|}\leq 2\delta_{m+1}^{-1}.$$
Thus we finish the proof of (<ref>).
Denote  $B_{m+1}^o(p):=B_{m+1}(p)\setminus\{p\}$. Since $|E_m(\theta+p\cdot \omega )-E|\leq \delta_m$ it follows that $E\in D(E_m(\theta+p\cdot\omega),\delta_m)$ and hence by Corollary <ref>,   $B_{m+1}^o(p)$ is $(m+1)$-good related to $(\theta,E)$. Hence by Hypothesis <ref>, we have

\begin{align}\label{1200`}
\|G_{B_{m+1}^o(p)} ^{\theta,E^*}\|&\leq10\delta_m^{-1},\nonumber\\
|G_{B_{m+1}^o(p)}^{\theta,E^*}(x,y)|&\leq e^{-\gamma_m\|x-y\|_1}, \ \|x-y\|_1\geq l_m^\frac{5}{6}.
\end{align}
Let $x,y\in B_{m+1}(p)$ such that $\|x-y\|_1\geq l_{m+1}^{\frac{4}{5}}$. By self-adjointness, we may assume $\|x-p\|_1\geq \frac{1}{2}l_{m+1}^\frac{4}{5}$.
By the resolvent identity, one has

\begin{align}
\nonumber		G_{B_{m+1}(p)}^{\theta,E^*}(x,y)&=\chi_{B_{m+1}^o(p)}(y)G_{B_{m+1}^o(p)}^{\theta,E^*}(x,y)\\
-&\varepsilon\sum_{(w,w')\in \partial_{B_{m+1}(p)}B_{m+1}^o(p)} G_{B_{m+1}^o(p)}^{\theta,E^*}(x,w)G_{B_{m+1}(p)}^{\theta,E^*}(w',y),\label{1156`}
\end{align}

Since $w \in  \partial^-_{B_{m+1}(p)}B_{m+1}^o(p)$, $w' \in  \partial^+_{B_{m+1}(p)}B_{m+1}^o(p)=\{p\}$, one has

$$\|x-w\|_1\geq \|x-p\|_1-\|w-p\|_1\geq \|x-p\|_1-1>l_m^\frac{5}{6}.$$
Hence by  (<ref>) and  (<ref>), we have

\begin{equation}\label{1205`}
|G_{B_{m+1}(p)}^{\theta,E^*}(x,y)|\leq e^{-\gamma_m\|x-y\|_1}+e^{-\gamma_m(\|x-p\|_1-1)}|G_{B_{m+1}(p)}^{\theta,E^*}(p,y)|.
\end{equation}

* If $\|y-p\|_1\leq l_{m+1}^\frac{3}{4}$, then
\begin{align*}
\|x-p\|_1-1&\geq \|x-y\|_1-\|y-p\|_1-1\\
&\geq \left (1-(l_{m+1}^\frac{3}{4}+1) l_{m+1}^{-\frac{4}{5}}\right) \|x-y\|_1\\
&\geq \left (1- 2l_{m+1}^{-\frac{1}{20}}\right) \|x-y\|_1.
\end{align*}

Bounding the term $|G_{B_{m+1}(p)}^{\theta,E^*}(p,y)|$ in (<ref>) by $\|G_{B_{m+1}(p)}^{\theta,E^*}\|\leq 10\delta_{m+1}^{-1}=10e^{l_{m+1}^\frac{2}{3}}$, we get
\begin{align*}
\eqref{1205`}&\leq e^{-\gamma_m\|x-y\|_1}+10e^{-\gamma_m(1- 2l_{m+1}^{-\frac{1}{20}})\|x-y\|_1}e^{l_{m+1}^\frac{2}{3}}  \\
&\leq e^{-\gamma_m(1- 2l_{m+1}^{-\frac{1}{20}}-(l_{m+1}^\frac{2}{3}+10) l_{m+1}^{-\frac{4}{5}})\|x-y\|_1}\\
&\leq e^{-\gamma_m(1-l_{m+1}^{-\frac{1}{50}})\|x-y\|_1}.
\end{align*}

* If $\|y-p\|_1> l_{m+1}^\frac{3}{4}$, we can use  (<ref>), (<ref>) and the resolvent identity  to bound the term   $|G_{B_{m+1}(p)}^{\theta,E^*}(p,y)|$ in (<ref>) by

\begin{align*}
|G_{B_{m+1}(p)}^{\theta,E^*}(p,y)|&=|G_{B_{m+1}(p)}^{\theta,E^*}(y,p)|\\&\leq \varepsilon\sum_{(w,w')\in \partial_{B_{m+1}(p)}B_{m+1}^o(p)} |G_{B_{m+1}^o(p)}^{\theta,E^*}(y,w)G_{B_{m+1}(p)}^{\theta,E^*}(w',p)|\\
&\leq e^{-\gamma_m(\|y-p\|_1-1)} \delta_{m+1}^{-1}.
\end{align*}

\begin{align*}
\eqref{1205} &\leq e^{-\gamma_m\|x-y\|_1}+e^{-\gamma_m(\|x-p\|_1-1)}e^{-\gamma_m(\|y-p\|_1-1)} \delta_{m+1}^{-1}\\&\leq e^{-\gamma_m(1-(l_{m+1}^\frac{2}{3} +10) l_{m+1}^{-\frac{4}{5}})\|x-y\|_1}\\
&\leq e^{-\gamma_m(1-l_{m+1}^{-\frac{1}{50}})\|x-y\|_1}.
\end{align*}

Thus we finish the proof of (<ref>).