the construction, $I(E_{n})$ is a single interval of length on order at most
$\delta_{n-2}^{5}\ll\|k_{n}\cdot\omega\|$ by Diophantine condition, thus
$\theta_{k_{n},-}$ and $\theta_{k_{n},-}+k_{n}\cdot\omega$ can not both belong
to it, a contradiction. ∎

######  Lemma 3.32.

The above $k_{n}$ satisfies $10l_{n}<\|k_{n}\|\leq 10l_{n+1}^{(1)}$.

###### Proof.

Suppose that $\|k_{n}\|_{1}\leq 10l_{n}$. By the previous lemma and
Proposition 3.28, $E_{n}$ is $C^{2}$-close to $E_{n-1}$ on order at most
$\delta_{n-1}^{10}$. Since $E_{n}$ belongs to Type 1, we have by Hypothesis 4,

$\pm E_{n}^{\prime}|_{I(E_{n})_{\pm}}\gtrsim\delta_{n-2}^{10}.$

Thus

$\pm E_{n-1}^{\prime}|_{I(E_{n})_{\pm}}\gtrsim\delta_{n-2}^{10}.$

By the two-monotonicity interval structure and mean value theorem, there
exists some point $\theta_{k_{n-1},-}$ with

$|\theta_{k_{n-1},-}-{\theta}_{k_{n},-}|\lesssim\delta_{n-1}^{10}\delta_{n-2}^{-10}<\delta_{n-1}^{9}$
(3.38)

such that

$E_{n-1}(\theta_{k_{n-1},-})=E_{n-1}(\theta_{k_{n-1},-}+k_{n-1}\cdot\omega).$

Since $\|k_{n}\|_{1}\leq 10l_{n}$, by the construction, $\theta_{k_{n-1},-}$
has a distance larger than $\delta_{n-1}^{9}$ from any simple resonant domain
of $E_{n-1}$. Thus by (3.38), ${\theta}_{k,-}$ does not belong to any simple
resonant domain of $E_{n-1}$, a contradiction with Lemma 3.31. ∎

By the previous lemma, we have $B_{n}\cap(B_{n}+k_{n})=\emptyset$. Since
$H_{B_{n}+k_{n}}(\theta)=H_{B_{n}}(\theta+k_{n}\cdot\omega)$,
$H_{B_{n}+k_{n}}(\theta)$ has an eigenvalue $E_{n}(\theta+k_{n}\cdot\omega)$.
For convenience, we denote

$B_{n,-}=B_{n},\ E_{n,-}(\theta)=E_{n}(\theta),$ (3.39)

and the eigenfunction of $E_{n,-}$ by $\psi_{n,-}$,
and denote

$B_{n,+}=B_{n}+k_{n},\ E_{n,+}(\theta)=E_{n}(\theta+k_{n}\cdot\omega),$ (3.40)

and the eigenfunction of $E_{n,+}$ by $\psi_{n,+}$.

######  Proposition 3.33.

If $\theta\in I^{(2)}_{k_{n},\cup}$, then for any
$\|x\|_{1}\leq(l_{n+1}^{(1)})^{4}$, $x\notin\\{o,k_{n}\\}$, we have

$\theta+x\cdot\omega\notin I(E_{n-1}),\ n\geq 2,$
$|v(\theta)-v(\theta+x\cdot\omega)|\geq 2\delta_{0}^{\frac{1}{400}},\ n=1.$

###### Proof.

For $n\geq 2$, $I(E_{n-1})$ has at most two connected component with length at
most $2\delta_{n-3}^{5}\ll(l_{n+1}^{(1)})^{-4\tau}$. Each of the component
contains at least one point of $\\{o,k_{n}\\}$. The proof follows from
Diophantine condition immediately. The case $n=1$ can be proved similarly as
Proposition 3.11. ∎

We choose a block a block $B_{n+1}^{(2)}$ satisfying

$\Lambda_{l_{n+1}^{(2)}}\subset
B_{n+1}^{(2)}\subset\Lambda_{l_{n+1}^{(2)}+50l_{n}}$

such that for all $\theta\in I_{k_{n},\cup}^{(2)}$, $B_{n+1}^{(2)}$ is
$n-1$-strongly regular relative to $(\theta,E)$ for all $E\in
B_{2\delta_{n-1}}(E_{n,-}(\theta))$.

######  Proposition 3.34.

The following holds for $\theta\in I^{(2)}_{k_{n},\cup}$:

  * (a).

$H_{B_{n+1}^{(2)}}(\theta)$ has exactly two eigenvalues
$E_{n+1}^{(2)}(\theta)$ and $\mathcal{E}_{n+1}^{(2)}(\theta)$ in
$[E_{n,-}(\theta)-10D\delta_{n-1}^{5},E_{n,-}(\theta)+10D\delta_{n-1}^{5}]$.
Moreover, any other $\hat{E}\in\sigma(H_{B_{n+1}^{(2)}}(\theta))$ must obey
$|\hat{E}-E_{n,-}(\theta)|\geq\delta_{n-1}$.

  * (b).

The corresponding eigenfunction of $E_{n+1}^{(2)}$(resp.
$\mathcal{E}_{n+1}^{(2)}$), $\psi_{n+1}$(resp. $\Psi_{n+1}$) decays
exponentially fast away from $o$ and $k_{n}$, i.e.,

$|\psi_{n+1}(x)|\leq
e^{-\frac{\gamma_{0}}{4}\|x\|_{1}}+e^{-\frac{\gamma_{0}}{4}\|x-k_{n}\|_{1}},$
$|\Psi_{n+1}(x)|\leq
e^{-\frac{\gamma_{0}}{4}\|x\|_{1}}+e^{-\frac{\gamma_{0}}{4}\|x-k_{n}\|_{1}}$

for $\operatorname{dist}(x,\\{o,k_{n}\\})\geq l_{n}^{6/7}.$ Moreover, the two
eigenfunctions can be expressed as

$\displaystyle\begin{split}\psi_{n+1}=A\psi_{n,-}+B\psi_{n,+}+O(\delta_{n}^{10}),\\\
\Psi_{n+1}=B\psi_{n,-}-A\psi_{n,+}+O(\delta_{n}^{10}),\end{split}$ (3.41)

where $A,B$ depend on $\theta$ satisfying $A^{2}+B^{2}=1$.

  * (c).

$\|G_{n+1}^{\perp\perp}(E_{n+1}^{(2)})\|,\|G_{n+1}^{\perp\perp}(\mathcal{E}_{n+1}^{(2)})\|\leq
2\delta_{n-1}^{-1}$, where $G_{n+1}^{\perp\perp}$ denotes the Green’s function
for $B_{n+1}^{(2)}$ on the orthogonal complement of the space spanned by
$\psi_{n+1}$ and $\Psi_{n+1}$.

###### Proof.

We first consider the case $\theta^{*}=\theta_{k_{n},-}$. In this case, since
$E_{n,-}(\theta^{*})=E_{n,+}(\theta^{*})$, by the decay of the eigenfunctions
(c.f. (3.28)), we have

$\|(H_{B_{n+1}^{(2)}}(\theta^{*})-E_{n,-}(\theta^{*}))\psi_{n,-}\|=\|\Gamma_{B_{n,-}}\psi_{n,-}\|\lesssim
e^{-\frac{\gamma_{0}}{4}l_{n}}\ll\delta_{n}^{10},$
$\|(H_{B_{n+1}^{(2)}}(\theta^{*})-E_{n,+}(\theta^{*}))\psi_{n,+}\|=\|\Gamma_{B_{n,+}}\psi_{n,+}\|\lesssim
e^{-\frac{\gamma_{0}}{4}l_{n}}\ll\delta_{n}^{10}.$

Since $B_{n,-}$ and $B_{n,+}$ are disjoint, $\psi_{n,-}$ and $\psi_{n,+}$
serve as two trial wave function for the operator
$H_{B_{n+1}^{(2)}}(\theta^{*})-E_{n,-}(\theta^{*})$. It follows that by Lemma
2.2, $H_{B_{n+1}^{(2)}}(\theta^{*})$ has two eigenvalues in
$[E_{n,-}(\theta^{*})-O(\delta_{n}^{10}),E_{n,-}(\theta^{*})+O(\delta_{n}^{10})]$.
The claim for $\theta\in I^{(2)}_{k_{n},\cup}$ follows from
$|I^{(2)}_{k_{n},\cup}|\leq 4\delta_{n-1}^{5}$ and the uniform bound
$|v^{\prime}|\leq D$.

To establish the decay of eigenfunctions, assume
$E\in\sigma(H_{B_{n+1}}^{(2)}(\theta))$ with
$|E-E_{n,-}(\theta)|\leq\delta_{n-1}$. We denote
$\Lambda=B^{(2)}_{n+1}\setminus(B_{n-1}\cup(B_{n-1}+k_{n}))$, which is an
$(n-1)$-regular set relative to $(\theta,E)$ and $E_{n-1}$ by the
construction. Moreover, by Proposition 3.33, $\Lambda$ is $(n-1)$-nonresonant,
thus $G_{\Lambda}(E)$ decays exponentially, which together with restricting
the eigenfunction equation to $\Lambda$ yields the exponential decay of
$\psi_{n+1}$.

To prove (3.41), it suffices to show $\psi_{n+1}$ and $\Psi_{n+1}$ are close
to a linear combination of $\psi_{n,-}$ and $\psi_{n,+}$ inside $B_{n,-}\cup
B_{n,+}$. We restrict the equation
$H_{B_{n+1}^{(2)}}(\theta)\psi_{n+1}=E_{n+1}^{(2)}(\theta)\psi_{n+1}$ on
$B_{n,-}$ to get

$\left(H_{B_{n,-}}-E_{n+1}^{(2)}\right)\psi_{n+1}=\Gamma_{B_{n,-}}\psi_{n+1}.$

Thus

$\|P_{n,-}^{\perp}\psi_{n+1}\|=\|G_{B_{n,-}}^{\perp}(E_{n+1}^{(2)})P_{n,-}^{\perp}\Gamma_{B_{n,-}}\psi_{n+1}\|=O(\delta_{n-1}^{-1}e^{-\frac{1}{4}\gamma_{0}l_{n}})\leq\delta_{n}^{10},$

where $P_{n,-}^{\perp}=I-\langle\psi_{n,-}|\psi_{n,-}\rangle$ is the
projection onto the orthogonal complement of $\psi_{n,-}$ and
$G_{B_{n,-}}^{\perp}(E_{n+1}^{(2)})$ is the Green’s function of $B_{n,-}$ on
$\operatorname{Image}P_{n,-}^{\perp}$ with upper bound $O(\delta_{n-1}^{-1})$
by $|E_{n,-}(\theta)-E_{n+1}^{(2)}(\theta)|\lesssim\delta_{n-1}^{5}$ together
with the eigenvalue separation result of simple resonance case (c.f.
Proposition 3.27 replacing $n$ by $n-1$).
Therefore inside $B_{n,-}$, we have

$P_{n,-}^{\perp}\psi_{n+1}=O(\delta_{n}^{10}).$

Thus

$\psi_{n+1}\chi_{B_{n,-}}=a\psi_{n,-}+O(\delta_{n}^{10}).$

where $a=\langle\psi_{n+1},\psi_{n,-}\rangle$.
Similarly,

$\psi_{n+1}\chi_{B_{n,+}}=b\psi_{n,+}+O(\delta_{n}^{10})$

with $b=\langle\psi_{n+1},\psi_{n,+}\rangle$.
Since the eigenfunction decays exponentially, we have

$\|\psi_{n+1}\chi_{B^{(2)}_{n+1}\setminus(B_{n,-}\cup
B_{n,+})}\|\leq\delta_{n}^{10}.$

Thus we can write

$\psi_{n+1}=a\psi_{n,-}+b\psi_{n,+}+O(\delta_{n}^{10}).$

Taking norm of the above equation, we obtain
$k:=a^{2}+b^{2}=1-O(\delta_{n}^{10})$. We set $A=a/k$ and $B=b/k$. Hence
$A^{2}+B^{2}=1$ and $|A-a|,|B-b|=O(\delta_{n}^{10})$, which gives the desired
expression of $\psi_{n+1}$. A similar argument gives
$\Psi_{n+1}=C\psi_{n,-}+D\psi_{n,+}+O(\delta_{n}^{10})$ with $C^{2}+D^{2}=1$.
For convenience, we write
$A=\cos\alpha,B=\sin\alpha,C=\sin\beta,D=-\cos\beta$. Since
$\langle\psi_{n+1},\Psi_{n+1}\rangle=0$, we get
$|\sin(\beta-\alpha)|=O(\delta_{n}^{10})$. We can choose $\beta$ satisfying
$|\beta-\alpha|=O(\delta_{n}^{10})$, thus
$|B-C|=|\sin\alpha-\sin\beta|=O(\delta_{n}^{10})$ and
$|A+D|=|\cos\alpha-\cos\beta|=O(\delta_{n}^{10})$, which gives the desired
expression $\Psi_{n+1}=B\psi_{n,-}-A\psi_{n,+}+O(\delta_{n}^{10})$. Assume
$\hat{E}\in\sigma(H_{B_{n+1}}^{(2)}(\theta))$ is a third eigenvalue in the
interval $|\hat{E}-E_{n,-}(\theta)|\leq\delta_{n-1}$. The above argument still
holds if we replace $E_{n+1}^{(2)}$ by $\hat{E}$. Thus the eigenfunction
$\hat{\psi}$ corresponding to $\hat{E}$ can also be expressed as

$\hat{\psi}=\hat{A}\psi_{n,-}+\hat{B}\psi_{n,+}+O(\delta_{n}^{10})$

with $\hat{A}^{2}+\hat{B}^{2}=1.$ By orthogonality, we have
$A\hat{A}+B\hat{B}=O(\delta_{n}^{10})$ and
$B\hat{A}-A\hat{B}=O(\delta_{n}^{10})$, a contradiction since
$(A\hat{A}+B\hat{B})^{2}+(B\hat{A}-A\hat{B})^{2}=1$. Hence any other third
eigenvalue must obey $|\hat{E}-E_{n,-}(\theta)|\geq\delta_{n-1}$. Finally, (c)
follows from (a) immediately. ∎

With the notation of the previous proposition, we denote

$E_{n+1,>}^{(2)}(\theta)=\max(E_{n+1}^{(2)}(\theta),\mathcal{E}_{n+1}^{(2)}(\theta)),\
E_{n+1,<}^{(2)}(\theta)=\min(E_{n+1}^{(2)}(\theta),\mathcal{E}_{n+1}^{(2)}(\theta)).$

Define

$P_{n,\pm}:=\langle\psi_{n,\pm}|\psi_{n,\pm}\rangle,\
Q_{\pm}:=\operatorname{Id}_{B_{n+1}^{(2)}}-P_{n,\pm},$

and consider the compressed operators

$H_{\pm}(\theta):=Q_{\pm}H_{B_{n+1}^{(2)}}(\theta)Q_{\pm}.$

######  Lemma 3.35.

For $\theta\in I^{(2)}_{k_{n},\cup}$, the operators $H_{\pm}(\theta)$ have
unique eigenvalues $\lambda_{\mp}(\theta)$ interlacing
$E_{n+1,>}^{(2)}(\theta)$ and $E_{n+1,<}^{(2)}(\theta)$:

$E_{n+1,<}^{(2)}(\theta)\leq\lambda_{\mp}(\theta)\leq
E_{n+1,>}^{(2)}(\theta).$ (3.42)

Moreover, $\lambda_{\pm}(\theta)$ are $C^{1}$-close to $E_{n,\pm}(\theta)$:

$|\frac{d^{s}}{d\theta^{s}}(\lambda_{\pm}(\theta)-E_{n,\pm}(\theta))|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}\delta_{n-1}^{-s}\ll\delta_{n}^{10},\ s=0,1.$
(3.43)

In particular, we have

$\pm\lambda_{\pm}^{\prime}(\theta)\gtrsim\delta_{n-2}^{10}\gg\delta_{n-1}.$
(3.44)

###### Proof.

We claim $H_{\pm}(\theta)$ have unique eigenvalues $\lambda_{\mp}(\theta)$ in
$[E_{n,-}(\theta)-\delta_{n-1},E_{n,-}(\theta)+\delta_{n-1}]$ satisfying

$|\lambda_{\pm}(\theta)-E_{n,\pm}(\theta)|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}.$

Since $Q_{\pm}\psi_{n,\mp}=\psi_{n,\mp}$,

$\|(H_{\pm}(\theta)-E_{n,\mp}(\theta))\psi_{n,\mp}\|=\|Q_{\pm}\Gamma_{B_{n,\mp}}\psi_{n,\mp}\|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}.$

The above estimate together with Lemma 2.2. yields the existence of
$\lambda_{\pm}$.

If $\hat{\lambda}$ is an eigenvalue of $H_{\pm}(\theta)$ satisfying
$|\hat{\lambda}-E_{n,-}(\theta)|\leq\delta_{n-1}$, then its eigenfunction
$\hat{\phi}\in\operatorname{Image}Q_{\pm}$ satisfies

$Q_{\pm}(H_{B_{n+1}^{(2)}}(\theta)-\hat{\lambda})\hat{\phi}=0.$

Thus

$(H_{B_{n+1}^{(2)}}(\theta)-\hat{\lambda})\hat{\phi}=a\psi_{n,\pm}.$

Employing the same argument as the proof of item (b) of Proposition 3.34
together with the exponential decay of $\psi_{n,\pm}$, one can prove that

$|\hat{\phi}(x)|\leq
e^{-\frac{\gamma_{0}}{5}\|x\|_{1}}+e^{-\frac{\gamma_{0}}{5}\|x-k_{n}\|_{1}}$

for $\operatorname{dist}(x,\\{o,k_{n}\\})\geq l_{n}^{7/8}$ and
$\hat{\phi}=\psi_{n,\mp}+O(e^{-\frac{1}{5}\gamma_{0}l_{n}}\delta_{n-1}^{-1})$
since $\hat{\phi}\in\operatorname{Image}Q_{\pm}$. Thus such an eigenvalue must
be unique or else it will violate the orthogonality. By Theorem 3.14 and item
(a) of Proposition 3.34, $\lambda_{\mp}(\theta)$ must lie between
$E_{n+1,<}^{(2)}(\theta)$ and $E_{n+1,>}^{(2)}(\theta)$. Let
$\phi_{\mp}\in\operatorname{Image}Q_{\pm}$ be the eigenfunctions of $H_{\pm}$
corresponding to $\lambda_{\mp}$. Since $\|\phi_{\mp}-\psi_{n,\mp}\|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}\delta_{n-1}^{-1}$, by the Feynman-Hellman
formula (1) from Lemma 2.4, we have

$\displaystyle|\lambda_{\mp}^{\prime}-E_{n,\mp}^{\prime}|$
$\displaystyle=|\left\langle\phi_{\mp},(H_{\pm})^{\prime}\phi_{\mp}\right\rangle-\left\langle\psi_{n,\mp},H^{\prime}\psi_{n,\mp}\right\rangle|$
$\displaystyle=|\left\langle\phi_{\mp},V^{\prime}\phi_{\mp}\right\rangle-\left\langle\psi_{n,\mp},V^{\prime}\psi_{n,\mp}\right\rangle-2\langle\phi_{\mp},(P_{n,\pm})^{\prime}H_{B_{n+1}^{(2)}}\phi_{\mp}\rangle|$
$\displaystyle\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}\delta_{n-1}^{-1}+|\langle\phi_{\mp},Q_{n,\pm}(P_{n,\pm})^{\prime}H_{B_{n+1}^{(2)}}Q_{n,\pm}\phi_{\mp}\rangle|.$

Thus it suffices to estimate
$Q_{n,\pm}P_{n,\pm}^{\prime}H_{B_{n+1}^{(2)}}Q_{n,\pm}$.
By differentiating the relation

$(H_{B_{n,\pm}}-E_{n,\pm})\psi_{n,\pm}=0,$

we get

$(H_{B_{n,\pm}}-E_{n,\pm})\psi_{n,\pm}^{\prime}=-(V^{\prime}-E_{n,\pm}^{\prime})\psi_{n,\pm}.$

Since $\langle\psi_{n,\pm}^{\prime},\psi_{n,\pm}\rangle=0$, we have

$\psi_{n,\pm}^{\prime}=-G_{B_{n,\pm}}^{\perp}(E_{n,\pm})V^{\prime}\psi_{n,\pm}.$

Thus

$\displaystyle P_{n,\pm}^{\prime}$
$\displaystyle=\langle\psi_{n,\pm}|\psi_{n,\pm}^{\prime}\rangle+\langle\psi_{n,\pm}^{\prime}|\psi_{n,\pm}\rangle$
$\displaystyle=-G_{B_{n,\pm}}^{\perp}(E_{n,\pm})V^{\prime}P_{n,\pm}-P_{n,\pm}V^{\prime}G_{B_{n,\pm}}^{\perp}(E_{n,\pm}).$

Since $P_{n,\pm}Q_{n,\pm}=Q_{n,\pm}P_{n,\pm}=0$, it follows that

$Q_{n,\pm}P_{n,\pm}^{\prime}H_{B_{n+1}^{(2)}}Q_{n,\pm}=-Q_{n,\pm}G_{B_{n,\pm}}^{\perp}(E_{n,\pm})V^{\prime}P_{n,\pm}H_{B_{n+1}^{(2)}}Q_{n,\pm}.$
(3.45)

We can employ the estimates

$\displaystyle\|P_{n,\pm}H_{B_{n+1}^{(2)}}Q_{n,\pm}\|$
$\displaystyle=\|P_{n,\pm}(H_{B_{n+1}^{(2)}}-E_{n,\pm})Q_{n,\pm}\|$
$\displaystyle\leq\|P_{n,\pm}(H_{B_{n+1}^{(2)}}-E_{n,\pm})\|$
$\displaystyle\leq\|\Gamma_{B_{n,\pm}}\psi_{n,\pm}\|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}$

and $\|G_{B_{n,\pm}}^{\perp}(E_{n,\pm})\|=O(\delta_{n-1}^{-1})$ to (3.45) and
complete the proof of (3.43).
Since $E_{n}$ belongs to Type 1, by Hypothesis 4, we have

$\pm E_{n,\pm}^{\prime}\gtrsim\delta_{n-2}^{10}.$

(3.44) follows from the above estimate and (3.43) immediately. ∎

######  Lemma 3.36.

There is a unique point $\theta_{s}\in I^{(2)}_{k_{n},\cup}$ satisfying
$|\theta_{s}-\theta_{k_{n},-}|<\delta_{n}^{10}$, such that

$\lambda_{+}(\theta_{s})=\lambda_{-}(\theta_{s}).$

Moreover, the Rellich children have quantitative separation away from
$\theta_{s}$:

$E_{n+1,>}^{(2)}(\theta)-E_{n+1,<}^{(2)}(\theta)\geq|\lambda_{+}(\theta)-\lambda_{-}(\theta)|\geq\delta_{n-1}|\theta-\theta_{s}|.$
(3.46)

As a corollary, if $E_{n+1}^{(2)}(\theta)=\mathcal{E}_{n+1}^{(2)}(\theta)$,
then $\theta=\theta_{s}$.

###### Proof.

We consider the difference function

$d(\theta):=\lambda_{+}(\theta)-\lambda_{-}(\theta).$

Since $E_{n,+}(\theta_{k_{n},-})=E_{n,-}(\theta_{k_{n},-})$, by (3.43) (the
version $s=0$), we have

$|d(\theta_{k_{n},-})|\leq|\lambda_{+}(\theta_{k_{n},-})-E_{n,+}(\theta_{k_{n},-})|+|\lambda_{-}(\theta_{k_{n},-})-E_{n,-}(\theta_{k_{n},-})|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}.$

By (3.44), we have

$d^{\prime}(\theta)\geq\delta_{n-1}.$

Thus there exists a unique point $\theta_{s}$ satisfying
$|\theta_{s}-\theta_{k_{n},-}|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{n}}\delta_{n-1}^{-1}<\delta_{n}^{10}$, such that
$d(\theta_{s})=0$, that is

$\lambda_{+}(\theta_{s})=\lambda_{-}(\theta_{s}).$

(3.46) follows from (3.42) and

$|\lambda_{+}(\theta)-\lambda_{-}(\theta)|=|d(\theta)|=|d(\theta)-d(\theta_{s})|\geq\delta_{n-1}|\theta-\theta_{s}|$

by mean value theorem. ∎

In the following, we assume that the Rellich functions are chosen so that
$E_{n+1}^{(2)}>\mathcal{E}_{n+1}^{(2)}$ on the right of $\theta_{s}$.

Suppose that
$|E_{n,+}^{\prime}(\theta_{k_{n},-})\geq|E_{n,-}^{\prime}(\theta_{k_{n},-})|$
(the argument is similar for the opposite case). Recalling the derivative
estimate for $E_{n}$ belong to Type 1 from Hypotheses 4, we define $1\leq
r\leq\delta_{n-1}^{-1}$ such that

$|E_{n,+}^{\prime}(\theta_{k_{n},-})|=r|E_{n,-}^{\prime}(\theta_{k_{n},-})|.$

######  Lemma 3.37.

For $\theta\in I_{k_{n},\cup}^{(2)}$, we have

$|(E_{n,+}^{\prime}+rE_{n,-}^{\prime})(\theta)|\leq\delta_{n-1}^{2}$

###### Proof.

Since $E_{n,+}^{\prime}$ and $E_{n,-}^{\prime}$ have opposite signs, we have

$E_{n,+}^{\prime}(\theta_{k_{n},-})+rE_{n,-}^{\prime}(\theta_{k_{n},-})=0.$

By the Feynman-Hellman formula (2) from Theorem 2.4, we have the bound

$|E_{n,\pm}^{\prime\prime}|\lesssim
D+\|G_{B_{n,\pm}}(E_{n,\pm})\|\lesssim\delta_{n-1}^{-1},$

where we employ the eigenvalue separation result of simple resonance case
(c.f. Proposition 3.27 replacing $n$ by $n-1$) to bound the term
$\|G_{B_{n,\pm}}(E_{n,\pm})\|$ in the above inequality.
Thus by mean value theorem, we have

$\displaystyle|(E_{n,+}^{\prime}+rE_{n,-}^{\prime})(\theta)|$
$\displaystyle\leq(1+r)\delta_{n-1}^{-1}|\theta-\theta_{k_{n},-}|$
$\displaystyle\lesssim r\delta_{n-1}^{-1}|I_{k_{n},\cup}^{(2)}|$
$\displaystyle\leq\delta_{n-1}^{2}.$

∎

######  Proposition 3.38.

For $\theta\in I_{k_{n},\cup}^{(2)}$, we have the following:

  * (a).

If $E_{n+1}^{(2)}(\theta)\neq\mathcal{E}_{n+1}^{(2)}(\theta)$, then

$\displaystyle(E_{n+1}^{(2)})^{\prime}$
$\displaystyle=(A^{2}-rB^{2})E_{n,-}^{\prime}+O(\delta_{n-1}^{2}),$ (3.47)
$\displaystyle(\mathcal{E}_{n+1}^{(2)})^{\prime}$
$\displaystyle=(B^{2}-rA^{2})E_{n,-}^{\prime}+O(\delta_{n-1}^{2})$ (3.48)

and

$\displaystyle(E_{n+1}^{(2)})^{\prime\prime}$
$\displaystyle=\frac{2\left\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\right\rangle^{2}}{E_{n+1}^{(2)}-\mathcal{E}_{n+1}^{(2)}}+O(\delta_{n-1}^{-1}),$
(3.49) $\displaystyle(\mathcal{E}_{n+1}^{(2)})^{\prime\prime}$
$\displaystyle=\frac{2\left\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\right\rangle^{2}}{\mathcal{E}_{n+1}^{(2)}-E_{n+1}^{(2)}}+O(\delta_{n-1}^{-1})$
(3.50)

with the notation from Proposition 3.34 and $r$ as above.

  * (b).

If $E_{n+1}^{(2)}(\theta)\neq\mathcal{E}_{n+1}^{(2)}(\theta)$ and
$|(E_{n+1}^{(2)})^{\prime}(\theta)|\leq\delta_{n-1}^{2}$, then
$|(E_{n+1}^{(2)})^{\prime\prime}(\theta)|\geq 2$. Moreover, the sign of
$(E_{n+1}^{(2)})^{\prime\prime}(\theta)$ is the same as that of
$E_{n+1}^{(2)}(\theta)-\mathcal{E}_{n+1}^{(2)}(\theta)$. The analogous
conclusion holds if we exchange $E_{n+1}^{(2)}(\theta)$ and
$\mathcal{E}_{n+1}^{(2)}(\theta)$.

  * (c).

If there is a level crossing (thus
$E_{n+1}^{(2)}(\theta_{s})=\mathcal{E}_{n+1}^{(2)}(\theta_{s})$ by Lemma
3.36), then for all $\theta\in I_{k_{n},\cup}^{(2)}$,

$(E_{n+1}^{(2)})^{\prime}(\theta)>\delta_{n-1}^{2},\
(\mathcal{E}_{n+1}^{(2)})^{\prime}(\theta)<-\delta_{n-1}^{2}.$ (3.51)

###### Proof.

We prove (3.47) (the proof of (3.48) is analogous). By (3.41) and Lemma 3.37,
we refer to Feynman-Hellman formula (1) from Lemma 2.4 to obtain

$\displaystyle(E_{n+1}^{(2)})^{\prime}$
$\displaystyle=\left\langle\psi_{n+1},V^{\prime}\psi_{n+1}\right\rangle=A^{2}\left\langle\psi_{n,-},V^{\prime}\psi_{n,-}\right\rangle+B^{2}\left\langle\psi_{n,+},V^{\prime}\psi_{n,+}\right\rangle+O(\delta_{n}^{10})$
$\displaystyle=A^{2}E_{n,-}^{\prime}+B^{2}E_{n,+}^{\prime}+O(\delta_{n}^{10})$
$\displaystyle=(A^{2}-rB^{2})E_{n,-}^{\prime}+B^{2}(E_{n,+}^{\prime}+rE_{n,-}^{\prime})+O(\delta_{n}^{10})$
$\displaystyle=(A^{2}-rB^{2})E_{n,-}^{\prime}+O(\delta_{n-1}^{2}).$

To prove (3.49) and (3.50), we use Feynman-Hellman formula (2) and (3) to
obtain

$(E_{n+1}^{(2)})^{\prime\prime}=\left\langle\psi_{n+1},V^{\prime\prime}\psi_{n+1}\right\rangle+2\frac{\left\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\right\rangle^{2}}{E_{n+1}^{(2)}-\mathcal{E}_{n+1}^{(2)}}-2\left\langle
V^{\prime}\psi_{n+1},G^{\perp\perp}_{n+1}(E_{n+1}^{(2)})V^{\prime}\psi_{n+1}\right\rangle.$

The first term is bounded by $D$ and the third term is bounded by

$2\|G^{\perp\perp}_{n+1}(E_{n+1}^{(2)})\|\|V^{\prime}\psi_{n+1}\|^{2}\lesssim\delta_{n-1}^{-1},$

where we use the estimate $\|G^{\perp\perp}_{n+1}(E_{n+1}^{(2)})\|\leq
2\delta_{n-1}^{-1}$ from item (c) of Proposition 3.34. Thus we finish the
proof of item (a).

Now we are going to prove (b). We will show the first term in (3.49) is large
if $|(E_{n+1}^{(2)})^{\prime}(\theta)|\leq\delta_{n-1}^{2}$.
Assume $|(E_{n+1}^{(2)})^{\prime}(\theta)|\leq\delta_{n-1}^{2}$, by (3.47), we
have

$|A^{2}-rB^{2}||E_{n,-}^{\prime}(\theta)|\leq|(E_{n+1}^{(2)})^{\prime}(\theta)|+O(\delta_{n-1}^{2})=O(\delta_{n-1}^{2}).$

Since

$|E_{n,-}^{\prime}(\theta)|\gtrsim\delta_{n-2}^{10}>\delta_{n-1},$

it follows that

$|A^{2}-rB^{2}|\lesssim\delta_{n-1}^{2}\delta_{n-1}^{-1}<\frac{1}{100}$

Since $A^{2}+B^{2}=1$ and $r\geq 1$, we obtain

$(1+r)B^{2}\geq A^{2}+B^{2}-|A^{2}-rB^{2}|\geq\frac{99}{100}$

and

$(1+\frac{1}{r})A^{2}\geq
A^{2}+B^{2}-\frac{1}{r}|A^{2}-rB^{2}|\geq\frac{99}{100}.$

Thus

$B^{2}\geq\frac{1}{4r},\ A^{2}\geq\frac{1}{4}.$

By (3.41), Lemma 3.37 and the previous lower bound for $A,B$, we have

$\displaystyle|\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\rangle|$
$\displaystyle=|AB(\langle\psi_{n,-},V^{\prime}\psi_{n,-}\rangle-\langle\psi_{n,+},V^{\prime}\psi_{n,+}\rangle)+O(\delta_{n}^{10})|$
$\displaystyle=|AB(E_{n,-}^{\prime}-E_{n,+}^{\prime})+O(\delta_{n}^{10})|$
$\displaystyle=|AB\left((1+r)E_{n,-}^{\prime}-(E_{n,+}^{\prime}+rE_{n,-}^{\prime})\right)+O(\delta_{n}^{10})|$
$\displaystyle\geq\frac{1+r}{4\sqrt{r}}(|E_{n,-}^{\prime}|-O(\delta_{n-1}^{2}))$
$\displaystyle\gtrsim\delta_{n-2}^{10}\gg\delta_{n-1}.$

By item (a) of Proposition 3.34, we obtain the estimate of the denominator
$|E_{n+1}^{(2)}-\mathcal{E}_{n+1}^{(2)}|\lesssim\delta_{n-1}^{5}$. Combining
the previous estimate of numerator, by (3.49), we obtain

$|(E_{n+1}^{(2)})^{\prime\prime}(\theta)|\gtrsim\delta_{n-1}^{2}\delta_{n-1}^{-5}-O(\delta_{n-1}^{-1})>2,$

whose sign is the same as that of
$E_{n+1}^{(2)}(\theta)-\mathcal{E}_{n+1}^{(2)}(\theta)$.

To prove (c), we first show (3.51) holds for $\theta=\theta_{s}$. Since
$E_{n+1}^{(2)}(\theta_{s})=\mathcal{E}_{n+1}^{(2)}(\theta_{s})$, by (3.42) we
have

$E_{n+1}^{(2)}(\theta_{s})=\mathcal{E}_{n+1}^{(2)}(\theta_{s})=\lambda_{+}(\theta_{s})=\lambda_{-}(\theta_{s}),$

and since $E_{n+1}^{(2)}>\mathcal{E}_{n+1}^{(2)}$ on the right of
$\theta_{s}$, we have for $\theta>\theta_{s}$,

$E_{n+1}^{(2)}(\theta)\geq\lambda_{+}(\theta)>\lambda_{-}(\theta)\geq\mathcal{E}_{n+1}^{(2)}(\theta).$

Thus by (3.44), we get

$(E_{n+1}^{(2)})^{\prime}(\theta_{s})\geq\lambda_{+}^{\prime}(\theta_{s})>\delta_{n-1},\
(\mathcal{E}_{n+1}^{(2)})^{\prime}(\theta_{s})\leq\lambda_{-}^{\prime}(\theta_{s})<-\delta_{n-1}.$

We next claim the inequalities hold for all $\theta\in I_{k_{n},\cup}^{(2)}$.
We prove the case that $(E_{n+1}^{(2)})^{\prime}(\theta)$ and
$\theta>\theta_{s}$ (the other cases are analogous). If it is not true, then
the set

$\\{\theta\in I_{k_{n},\cup}^{(2)}:\theta>\theta_{s},\
(E_{n+1}^{(2)})^{\prime}(\theta)\leq\delta_{n-1}^{2}\\}\neq\emptyset.$

Let $\theta^{*}$ be its infimum. Since $(E_{n+1}^{(2)})^{\prime}(\theta)$ is
continuous and $(E_{n+1}^{(2)})^{\prime}(\theta_{s})>\delta_{n-1}^{2}$, we
have $\theta^{*}>\theta_{s}$ and
$(E_{n+1}^{(2)})^{\prime}(\theta)>\delta_{n-1}^{2}\geq(E_{n+1}^{(2)})^{\prime}(\theta^{*})$
for $\theta\in[\theta_{s},\theta^{*})$, which implies
$(E_{n+1}^{(2)})^{\prime\prime}(\theta^{*})\leq 0$. However, by item (b) and
$E_{n+1}^{(2)}(\theta^{*})>\mathcal{E}_{n+1}^{(2)}(\theta^{*})$, we get
$(E_{n+1}^{(2)})^{\prime\prime}(\theta^{*})>2$, a contradiction. Thus we prove
our claim. ∎

Define

$E_{n,\vee}(\theta)=\max(E_{n,+}(\theta),E_{n,-}(\theta)),\
E_{n,\wedge}(\theta)=\min(E_{n,+}(\theta),E_{n,-}(\theta)).$

######  Proposition 3.39.

For $\theta\in I_{k_{n},\cup}^{(2)}$, we have

$|E_{n+1,>}^{(2)}(\theta)-E_{n,\vee}(\theta)|,|E_{n+1,<}^{(2)}(\theta)-E_{n,\wedge}(\theta)|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}}.$ (3.52)

###### Proof.

Fix $\theta\in I_{k_{n},\cup}^{(2)}$. We have

$\|(H_{B_{n+1}^{(2)}}-E_{n,\pm})\psi_{n,\pm}\|=\|\Gamma_{B_{n,\pm}}\psi_{n,\pm}\|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}}.$

By Lemma 2.2, we deduce $H_{B_{n+1}^{(2)}}$ must have an eigenvalue in
$[E_{n,\vee}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}}),E_{n,\vee}+O(e^{-\frac{1}{4}\gamma_{0}l_{n}})]$
and in
$[E_{n,\wedge}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}}),E_{n,\wedge}+O(e^{-\frac{1}{4}\gamma_{0}l_{n}})]$.
By item (a) of proposition 3.34, the eigenvalue must be $E_{n+1}^{(2)}$ or
$\mathcal{E}_{n+1}^{(2)}$. If the two intervals are disjoint, (3.52) must
hold. Otherwise, we have

$E_{n,\vee}-E_{n,\wedge}\lesssim e^{-\frac{1}{4}\gamma_{0}l_{n}}.$ (3.53)

Thus we have two trial function for $E_{n,\vee}$:

$\|(H_{B_{n+1}^{(2)}}-E_{n,\vee})\psi_{n,\pm}\|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}}.$

By Lemma 2.2, $H_{B_{n+1}^{(2)}}$ must have two eigenvalues in
$[E_{n,\vee}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}}),E_{n,\vee}+O(e^{-\frac{1}{4}\gamma_{0}l_{n}})].$
They are $E_{n+1}^{(2)}$ and $\mathcal{E}_{n+1}^{(2)}$. Combining (3.53), we
also obtain (3.52). ∎

The next proposition shows the two-monotonicity interval structure of
$E_{n+1,>}^{(2)}$ and $E_{n+1,<}^{(2)}$.

######  Proposition 3.40.

We have the following:

  * (a).

If there is no level crossing, then $E_{n+1}^{(2)}=E_{n+1,>}^{(2)}$ and
$\mathcal{E}_{n+1}^{(2)}=E_{n+1,<}^{(2)}$. Moreover, $E_{n+1}^{(2)}$ (resp.
$\mathcal{E}_{n+1}^{(2)}$) has the two-monotonicity interval structure with a
critical point $\theta_{>}$ (resp. $\theta_{<}$) satisfying
$|\theta_{>}-\theta_{k_{n},-}|<\delta_{n}^{10}$ (resp.
$|\theta_{<}-\theta_{k_{n},-}|<\delta_{n}^{10}$).

  * (b).

If there is a level crossing (thus
$E_{n+1}^{(2)}(\theta_{s})=\mathcal{E}_{n+1}^{(2)}(\theta_{s})$), then

$\displaystyle E_{n+1}^{(2)}(\theta)=\left\\{\begin{aligned}
&E_{n+1,<}^{(2)}(\theta)&\text{if $\theta<\theta_{s}$,}\\\
&E_{n+1,>}^{(2)}(\theta)&\text{if $\theta>\theta_{s}$,}\end{aligned}\right.$

and

$\displaystyle\mathcal{E}_{n+1}^{(2)}(\theta)=\left\\{\begin{aligned}
&E_{n+1,>}^{(2)}(\theta)&\text{if $\theta<\theta_{s}$,}\\\
&E_{n+1,<}^{(2)}(\theta)&\text{if $\theta>\theta_{s}$.}\end{aligned}\right.$

Moreover, $E_{n+1,>}^{(2)}$ and $E_{n+1,<}^{(2)}$ are piecewise $C^{1}$
functions (except at the point $\theta_{s}$) with two-monotonicity interval
structure.

Combining the previous Propositions of double resonance case, we deduce that
if the gap between the critical values of $E_{n+1,>}^{(2)}$ and
$E_{n+1,<}^{(2)}$ is larger than $3\delta_{n+1}$, then each of them belongs to
Type 2, otherwise, $E_{n+1,>}^{(2)}$ along with $E_{n+1,<}^{(2)}$ belongs to
Type 3.

###### Proof.

If there is no level crossing, then
$E_{n+1,>}^{(2)}(\theta)>E_{n+1,<}^{(2)}(\theta)$ for all $\theta\in
I_{k_{n},\cup}^{(2)}$. Since $E_{n+1}^{(2)}>\mathcal{E}_{n+1}^{(2)}$ on the
right of $\theta_{s}$, by continuity
$E_{n+1}^{(2)}(\theta)>\mathcal{E}_{n+1}^{(2)}(\theta)$ for all $\theta\in
I_{k_{n},\cup}^{(2)}$. Thus $E_{n+1}^{(2)}=E_{n+1,>}^{(2)}$ and
$\mathcal{E}_{n+1}^{(2)}=E_{n+1,<}^{(2)}$.
Now we show the existence of the critical point $\theta_{>}$ for
$E_{n+1}^{(2)}$ (the proof of $\theta_{<}$ and $\mathcal{E}_{n+1}^{(2)}$ is
analogous). Let $\theta_{-}:=\theta_{k_{n},-}-\delta_{n}^{10}$ and
$\theta_{+}:=\theta_{k_{n},-}+\delta_{n}^{10}$. Recalling $\pm
E_{n,\pm}^{\prime}\gtrsim\delta_{n-2}^{10}>\delta_{n}$ and (3.52), we have by
mean value theorem

$E_{n+1}^{(2)}(\theta_{+})\geq
E_{n,+}(\theta_{+})-O(e^{-\frac{1}{4}\gamma_{0}l_{n}})\geq
E_{n,+}(\theta_{k_{n},-})+\delta_{n}^{11}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}})>E_{n+1}^{(2)}(\theta_{k_{n},-})$

and

$E_{n+1}^{(2)}(\theta_{-})\geq
E_{n,-}(\theta_{-})-O(e^{-\frac{1}{4}\gamma_{0}l_{n}})\geq
E_{n,-}(\theta_{k_{n},-})+\delta_{n}^{11}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}})>E_{n+1}^{(2)}(\theta_{k_{n},-}).$

Thus $E_{n+1}^{(2)}$ must have a critical point
$\theta_{>}\in[\theta_{-},\theta_{+}]$ by mean value theorem. The two-
monotonicity interval structure follows from item (b) of Proposition 3.38 and
Lemma 2.3. Thus we finish the proof of (a).

If $E_{n+1}^{(2)}(\theta_{s})=\mathcal{E}_{n+1}^{(2)}(\theta_{s})$, by item
(c) of Proposition 3.38, $E_{n+1}^{(2)}$ is strictly increasing and
$\mathcal{E}_{n+1}^{(2)}$ is strictly decreasing. Thus
$E_{n+1}^{(2)}>\mathcal{E}_{n+1}^{(2)}$ on the right of $\theta_{s}$ and
$E_{n+1}<\mathcal{E}_{n+1}^{(2)}$ on the left of $\theta_{s}$. The proof of
(b) is obvious. ∎

####  3.5.3. Completion of the simple resonance case

In this part, we will complete the proof of Subcase B in the simple resonance
case. In this subcase, $E_{n}$ belongs to Type 3 and the simple resonant
interval we consider is

$J_{i_{n}}^{(1)}=[\sup
J(E_{n,<})-\frac{1}{2}\delta_{n-1}^{10}-3\delta_{n},\inf
J(E_{n,>})+\frac{1}{2}\delta_{n-1}^{10}+3\delta_{n}].$ (3.54)

Recall that if $E_{n}$ belong to Type 3, there exist two brother Rellich
functions with $0\leq\inf E_{n,>}(\theta)-\sup E_{n,<}(\theta)\leq
3\delta_{n}$.

gap$\leq 3\delta_{n}$$E_{n,>}$$E_{n,<}$$J_{i_{n}}^{(1)}$$\underbrace{\hskip
71.13188pt}$$\delta_{n-1}^{10}\sim$ Figure 9. A cartoon illustration of the
interval $J_{i_{n}}^{(1)}$ in Subcase B: $E_{n}$ belongs to Type 3, and the
gap between the Rellich brothers is smaller than $3\delta_{n}$.
$J_{i_{n}}^{(1)}$ with length $\sim\delta_{n-1}^{10}$ covers the nearby region
of the critical values of the two Rellich brothers.

######  Lemma 3.41.

$E_{n}$ must have some ancestor $E_{s}(1\leq s\leq n)$ generated from the
double resonant interval of its parent $E_{s-1}$.

###### Proof.

For the sake of contradiction, assume that
$E_{s}\in\mathcal{C}^{(1)}(E_{s-1})$ for all $1\leq s\leq n$. By the previous
analysis of the simple resonance case (c.f. Proposition 3.6, 3.10, 3.30),
$E_{n}$ must belong to Type 1 or Type 2, a contradiction. ∎

Let $m$ be the largest integral of $0\leq m\leq n-1$ such that $E_{m+1}$ is
generated from the double resonant interval of its parent $E_{m}$ among the
ancestors of $E_{n}$, that is, there exists some $k_{m}\in\mathbb{Z}^{d}$ with
$10l_{m}<\|k_{m}\|_{1}\leq 10l_{m+1}$ satisfying

$e_{k_{m}}:=E_{m}(\theta_{k_{m},-})=E_{m}(\theta_{k_{m},-}+k_{m}\cdot\omega)$

for some $\theta_{k_{m},-}\in I(E_{m})_{-}$ and
$E_{m+1,>},E_{m+1,<}\in\mathcal{C}^{(2)}(E_{m})$ are the Rellich children
generated from the double resonant interval
$\bar{B}_{\delta_{m-1}^{10}}(e_{k_{m}})$. Thus the results of the double
resonance case from the previous section hold true with $m$ replacing $n$.

######  Lemma 3.42.

For all $m+1\leq s\leq n-1$, $E_{s}$ belongs to Type 3 and $E_{s+1}$ is
generated from the interval $J_{i_{s}}^{(1)}$ ((3.54) replacing $n$ by $s$) of
$E_{s}$.

###### Proof.

By the definition of $m$, $E_{s+1}\in\mathcal{C}^{(1)}(E_{s})$ for all
$m+1\leq s\leq n-1$. If the lemma is not true, as the proof of Lemma 3.41,
$E_{n}$ must belong to Type 1 or Type 2, a contradiction. ∎

Define the domains

$I^{(1)}_{>}:=(E_{n,>})^{-1}\left(J_{i_{n}}^{(1)}\cap J(E_{n,>})\right),\
I^{(1)}_{<}:=(E_{n,<})^{-1}\left(J_{i_{n}}^{(1)}\cap J(E_{n,<})\right)$

and

$I^{(1)}_{\cap}:=I^{(1)}_{>}\cap I^{(1)}_{<},\ I^{(1)}_{\cup}:=I^{(1)}_{>}\cup
I^{(1)}_{<}.$

######  Lemma 3.43.

$I^{(1)}_{\cap}\neq\emptyset$. Thus $I^{(1)}_{\cup}$ is a single interval with
$\delta_{n-1}^{10}\lesssim|I^{(1)}_{\cup}|\lesssim\delta_{n-1}^{5}$. Moreover,
let $\theta_{n,>}$ (resp. $\theta_{n,<}$) be the critical point of $E_{n,>}$
(resp. $E_{n,<}$), then $\theta_{n,>},\theta_{n,<}\in I^{(1)}_{\cap}$.

###### Proof.

We assume that $\theta_{n,>}\geq\theta_{n,<}$. (The other case is analogous.)
Recalling Hypothesis 4, there exist two interlacing functions
$\lambda_{n,\pm}$ satisfying

$E_{n,<}(\theta)\leq\lambda_{n,\pm}(\theta)\leq E_{n,>}(\theta),$
$\pm\lambda_{n,\pm}^{\prime}(\theta)\geq\delta_{n-2}.$

Thus

$\displaystyle\delta_{n-2}(\theta_{n,>}-\theta_{n,<})$
$\displaystyle\leq\lambda_{n,+}(\theta_{n,>})-\lambda_{n,+}(\theta_{n,<})$
$\displaystyle\leq E_{n,>}(\theta_{n,>})-E_{n,<}(\theta_{n,<})\leq
3\delta_{n}.$

From the above inequality, we obtain

$|\theta_{n,>}-\theta_{n,<}|\leq
3\delta_{n}\delta_{n-2}^{-1}\ll\delta_{n-1}^{10}.$ (3.55)

Since $|J_{i_{n}}^{(1)}\cap J(E_{n,>})|\geq\frac{1}{2}\delta_{n-1}^{10}$, by
the uniform bound of $|v^{\prime}|$, we have

$B_{\frac{\delta_{n-1}^{10}}{2D}}(\theta_{n,>})\subset I^{(1)}_{>}.$ (3.56)

Similarly,

$B_{\frac{\delta_{n-1}^{10}}{2D}}(\theta_{n,<})\subset I^{(1)}_{<}.$ (3.57)

The lemma follows from (3.55), (3.56), (3.57) immediately. ∎

We choose a block a block $B_{n+1}^{(1)}$ satisfying

$\Lambda_{l_{n+1}^{(1)}}\subset
B_{n+1}^{(1)}\subset\Lambda_{l_{n+1}^{(1)}+50l_{n}}$

such that for all $\theta\in I^{(1)}_{\cup}$, $B_{n+1}^{(1)}$ is
$n-1$-strongly regular relative to $(\theta,E)$ for all $E\in
B_{2\delta_{n-1}}(E_{n,>}(\theta))$.

######  Proposition 3.44.

The following holds for $\theta\in I^{(1)}_{\cup}$:

  * (a).

$H_{B_{n+1}^{(1)}}(\theta)$ has two eigenvalues $E_{n+1,>}(\theta)$ and
$E_{n+1,<}(\theta)$ with $E_{n+1,>}(\theta)\geq E_{n+1,<}(\theta)$ satisfying

$\displaystyle\begin{split}|E_{n+1,>}(\theta)-E_{n,>}(\theta)|\lesssim
e^{-\frac{\gamma_{0}}{4}l_{n}}\ll\delta_{n}^{10},\\\
|E_{n+1,<}(\theta)-E_{n,<}(\theta)|\lesssim
e^{-\frac{\gamma_{0}}{4}l_{n}}\ll\delta_{n}^{10}.\end{split}$ (3.58)

Moreover, any other $\hat{E}\in\sigma(H_{B_{n+1}^{(1)}}(\theta))$ must obey
$|\hat{E}-E_{n,>}(\theta)|\geq\delta_{n-1}$.

  * (b).

The corresponding eigenfunction of $E_{n+1,>}$(resp. $E_{n+1,<}$),
$\psi_{n+1,>}$(resp. $\psi_{n+1,<}$) decays exponentially fast away from $o$
and $k_{m}$, i.e.,

$|\psi_{n+1,>}(x)|\leq
e^{-\frac{\gamma_{0}}{4}\|x\|_{1}}+e^{-\frac{\gamma_{0}}{4}\|x-k_{m}\|_{1}},$
$|\psi_{n+1,<}(x)|\leq
e^{-\frac{\gamma_{0}}{4}\|x\|_{1}}+e^{-\frac{\gamma_{0}}{4}\|x-k_{m}\|_{1}}$

for $\operatorname{dist}(x,\\{o,k_{m}\\})\geq l_{m}^{6/7}.$ Moreover, the two
eigenfunctions can be expressed as

$\displaystyle\begin{split}&\psi_{n+1,>}=A\psi_{m,-}+B\psi_{m,+}+O(\delta_{m}^{10}),\\\
&\psi_{n+1,<}=B\psi_{m,-}-A\psi_{m,+}+O(\delta_{m}^{10}),\end{split}$ (3.59)

where $A,B$ depend on $\theta$ satisfying $A^{2}+B^{2}=1$.

  * (c).

$\|G_{n+1}^{\perp\perp}(E_{n+1,>})\|,\|G_{n+1}^{\perp\perp}(E_{n+1,<})\|\leq
2\delta_{n-1}^{-1}$, where $G_{n+1}^{\perp\perp}$ denotes the Green’s function
for $B_{n+1}^{(1)}$ on the orthogonal complement of the space spanned by
$\psi_{n+1,>}$ and $\psi_{n+1,<}$.

###### Proof.

Fix $\theta\in I_{\cup}^{(1)}$. Denote by $\psi_{n,>}$ (resp. $\psi_{n,<}$)
the eigenfunction corresponding to $E_{n,>}$ (resp. $E_{n,<}$) of the
restriction operator $H_{B_{n}}$. By the exponential decay (3.28) of the
eigenfunctions from the inductive hypothesis, we have

$\|(H_{B_{n+1}^{(1)}}-E_{n,>})\psi_{n,>}\|=\|\Gamma_{B_{n}}\psi_{n,>}\|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}},$
$\|(H_{B_{n+1}^{(1)}}-E_{n,<})\psi_{n,<}\|=\|\Gamma_{B_{n}}\psi_{n,<}\|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}}.$

The above estimates together with Lemma 2.2 yield $H_{B_{n+1}^{(1)}}$ must
have an eigenvalue in
$[E_{n,>}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}}),E_{n,>}+O(e^{-\frac{1}{4}\gamma_{0}l_{n}})]$
and in
$[E_{n,<}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}}),E_{n,<}+O(e^{-\frac{1}{4}\gamma_{0}l_{n}})]$.
If the two intervals are disjoint, (3.58) must hold. Otherwise, we have

$E_{n,>}-E_{n,<}\lesssim e^{-\frac{1}{4}\gamma_{0}l_{n}}.$ (3.60)

Thus we have two trial function for $E_{n,>}$:

$\|(H_{B_{n+1}^{(1)}}-E_{n,>})\psi_{n,>}\|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}},$
$\|(H_{B_{n+1}^{(1)}}-E_{n,>})\psi_{n,<}\|\leq|E_{n,>}-E_{n,<}|+\|(H_{B_{n+1}^{(1)}}-E_{n,<})\psi_{n,<}\|\lesssim
e^{-\frac{1}{4}\gamma_{0}l_{n}}.$

By Lemma 2.2, $H_{B_{n+1}^{(1)}}$ must have two eigenvalues in
$[E_{n,>}-O(e^{-\frac{1}{4}\gamma_{0}l_{n}}),E_{n,>}+O(e^{-\frac{1}{4}\gamma_{0}l_{n}})].$
Combining (3.60), we also obtain (3.58). Thus we finish the existence part of
(a).

We now prove (b). Assume $E\in\sigma(H_{B_{n+1}}^{(1)}(\theta))$ with
$|E-E_{n,>}(\theta)|\leq\delta_{n-1}$. Fix
$\operatorname{dist}(x,\\{o,k_{m}\\})\geq l_{m}^{6/7}$. Suppose $\|x\|_{1}\leq
l_{m+2}/2$, restricting the eigenfunction equation to the $(m-1)$-good set
$\Lambda:=B_{m+2}\setminus(B_{m-1}\cup(B_{m-1}+k_{m}))$ together with the
exponential decay of $G_{\Lambda}(E)$ yields the desired decay of
eigenfunctions. Suppose $l_{s}/2\leq\|x\|_{1}\leq l_{s+1}/2$ for some $m+2\leq
s\leq n-1$ or $l_{n}/2\leq\|x\|_{1}$ for $s=n$, restricting the eigenfunction
equation to the $(s-1)$-good set $\Lambda:=B_{s+1}\setminus B_{s-1}$ together
with the exponential decay of $G_{\Lambda}(E)$ yields the desired decay.
We explain that the above set $B_{s+1}\setminus B_{s-1}$ is
$(s-1)$-nonresonant for $m+2\leq s\leq n$ since $I(E_{s-1})$ is a single
interval satisfying $|I(E_{s-1})|\lesssim\delta_{s-3}^{5}$, which together
with Diophantine condition yields $\theta+x\cdot\omega\notin I(E_{s-1})$ for
all $x\in B_{s+1}\setminus B_{s-1}$. The $(s-1)$-regularity of
$B_{s+1}\setminus B_{s-1}$ is confirmed by the construction.

To prove (3.59), it suffices to show $\psi_{n+1,>}$ and $\psi_{n+1,<}$ are
close to a linear combination of $\psi_{m,-}$ and $\psi_{m,+}$ inside
$B_{m,-}\cup B_{m,+}$. We restrict the equation
$H_{B_{n+1}^{(1)}}(\theta)\psi_{n+1,>}=E_{n+1,>}(\theta)\psi_{n+1,>}$ on
$B_{m,-}$ to get

$\left(H_{B_{m,-}}-E_{n+1,>}\right)\psi_{n+1,>}=\Gamma_{B_{m,-}}\psi_{n+1,>}.$

Thus

$\|P_{m,-}^{\perp}\psi_{n+1,>}\|=\|G_{B_{m,-}}^{\perp}(E_{n+1,>})P_{m,-}^{\perp}\Gamma_{B_{m,-}}\psi_{n+1,>}\|=O(\delta_{m-1}^{-1}e^{-\frac{1}{4}\gamma_{0}l_{m}})\leq\delta_{m}^{10},$

where $P_{m,-}^{\perp}=I-\langle\psi_{m,-}|\psi_{m,-}\rangle$ is the
projection onto the orthogonal complement of $\psi_{m,-}$ and
$G_{B_{m,-}}^{\perp}(E_{n+1,>})$ is the Green’s function of $B_{m,-}$ on
$\operatorname{Image}P_{m,-}^{\perp}$ with upper bound $O(\delta_{m-1}^{-1})$
by
$|E_{n+1,>}(\theta)-E_{m,-}(\theta)|\lesssim|J_{i_{m+1}}^{(1)}|\lesssim\delta_{m-1}^{5}$
together with the eigenvalue separation result of simple resonance case (c.f.
Proposition 3.27 replacing $n$ by $m-1$).
Thus

$\psi_{n+1,>}\chi_{B_{m,-}}=a\psi_{m,-}+O(\delta_{m}^{10}).$

Similarly,

$\psi_{n+1,>}\chi_{B_{m,+}}=b\psi_{m,+}+O(\delta_{m}^{10}).$

The above two estimates together with the exponentially decay of the
eigenfunction and orthogonality of $\psi_{n+1,>}$ and $\psi_{n+1,<}$ yield
(3.59).

Since the eigenfunction corresponding to $E$ in the interval
$|E-E_{n,>}(\theta)|\leq\delta_{n-1}$ has the form (3.59), there are only two
such eigenvalues $E_{n+1,>}$ and $E_{n+1,<}$ or else it will violate the
orthogonality.

Finally, (c) follows from (a) and
$|E_{n+1,>}(\theta)-E_{n,>}(\theta)|,|E_{n+1,<}(\theta)-E_{n,>}(\theta)|\lesssim\delta_{n-1}^{5}$
immediately. ∎

Define

$P_{m,\pm}:=\langle\psi_{m,\pm}|\psi_{m,\pm}\rangle$

and

$Q_{s,\pm}:=\operatorname{Id}_{B_{s}}-P_{m,\pm},\ m+1\leq s\leq n+1.$

and consider the compressed operators

$H_{s,\pm}(\theta):=Q_{s,\pm}H_{B_{s}}(\theta)Q_{s,\pm}.$

######  Lemma 3.45.

For $m+1\leq s\leq n+1$ and $\theta\in I^{(1)}_{\cup}$, the operators
$H_{s,\pm}(\theta)$ have unique eigenvalues $\lambda_{s,\mp}(\theta)$
interlacing $E_{s,>}(\theta)$ and $E_{s,<}(\theta)$:

$E_{s,<}(\theta)\leq\lambda_{s,\mp}(\theta)\leq E_{s,>}(\theta),$ (3.61)

and the eigenfunctions $\phi_{s,\mp}$ corresponding to $\lambda_{s,\mp}$ decay
exponentially:

$|\phi_{s,\mp}(x)|\leq e^{-\frac{\gamma_{0}}{5}\|x\|_{1}},\ \|x\|_{1}\geq
l_{s}^{\frac{7}{8}}.$ (3.62)

Moreover, we have

$|\lambda_{s,\pm}^{\prime}(\theta)-E_{m,\pm}^{\prime}(\theta)|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{m}}\delta_{m-1}^{-1}\ll\delta_{m}^{10}.$ (3.63)

In particular,

$\pm\lambda_{s,\pm}^{\prime}(\theta)\gtrsim\delta_{m-2}^{10}\gg\delta_{m-1}.$
(3.64)

###### Proof.

We prove (3.61), (3.62) by induction. The case $s=m+1$ was proved in Lemma
3.35 replacing $n$ by $m$. Assuming that $\lambda_{s,\mp}$ have been
constructed, we are going to construct $\lambda_{s+1,\mp}$. First we claim
that $H_{s+1,\pm}(\theta)$ have unique eigenvalues $\lambda_{s+1,\mp}(\theta)$
in
$[\lambda_{s,\mp}(\theta)-\delta_{s-1},\lambda_{s,\mp}(\theta)+\delta_{s-1}]$
satisfying

$|\lambda_{s+1,\mp}(\theta)-\lambda_{s,\mp}(\theta)|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{s}}.$ (3.65)

Let $\phi_{s,\mp}\in\operatorname{Image}Q_{s,\pm}$ be the eigenfunctions of
$H_{s,\pm}(\theta)$ corresponding to $\lambda_{s,\mp}(\theta)$. By the
exponential decay (3.62) for $\phi_{s,\mp}$, we have

$\displaystyle\|(H_{s+1,\pm}(\theta)-\lambda_{s,\mp}(\theta))\phi_{s,\mp}\|$
$\displaystyle=\|Q_{s+1,\pm}(H_{B_{s+1}}(\theta)-\lambda_{s,\mp}(\theta))\phi_{s,\mp}\|$
$\displaystyle=\|Q_{s+1,\pm}\Gamma_{B_{s}}\phi_{s,\mp}\|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{s}}.$

The above estimate together with Lemma 2.2 yields the existence of
$\lambda_{s+1,\mp}$.

Next we show the uniqueness and they are the desired interlacing curves. Let
$\hat{\lambda}\in\sigma(H_{s+1,\pm})$ satisfying
$|\hat{\lambda}-\lambda_{s,\mp}|\leq\delta_{s-1}$ with corresponding
eigenfunction $\hat{\phi}\in\operatorname{Image}Q_{s+1,\pm}$. Then we have

$Q_{s+1,\pm}(H_{B_{s+1}}-\hat{\lambda})\hat{\phi}=0.$

Thus

$(H_{B_{s+1}}-\hat{\lambda})\hat{\phi}=a\psi_{m,\pm}.$

Employing the same argument as the proof of item (b) of Proposition 3.44,
together with the exponential decay of $\psi_{m,\pm}$, one can prove that

$|\hat{\phi}(x)|\leq
e^{-\frac{\gamma_{0}}{5}\|x\|_{1}}+e^{-\frac{\gamma_{0}}{5}\|x-k_{m}\|_{1}}$

for $\operatorname{dist}(x,\\{o,k_{m}\\})\geq l_{m}^{7/8}$ and
$\hat{\phi}=\psi_{m,\mp}+O(e^{-\frac{1}{5}\gamma_{0}l_{m}}\delta_{m-1}^{-1})$
since $\hat{\phi}\in\operatorname{Image}Q_{\pm}$. Thus such an eigenvalue must
be unique or else it will violate the orthogonality. Since
$E_{s,<}(\theta)\leq\lambda_{s,\mp}(\theta)\leq E_{s,>}(\theta)$ and
$E_{s+1,<},E_{s+1,>}$ are generated from the interval $J_{i_{s}}^{(1)}$ with
$|J_{i_{s}}^{(1)}|\ll\delta_{s-1}$, $\lambda_{s+1,\mp}(\theta)$ must lie
between $E_{s+1,<}(\theta)$ and $E_{s+1,>}(\theta)$ by Theorem 3.14. Thus we
finish the proof of induction.

By the Feynman-Hellman formula (1) in Lemma 2.4, we have

$\displaystyle|\lambda_{s+1,\mp}^{\prime}-E_{m,\mp}^{\prime}|$
$\displaystyle=|\left\langle\phi_{s+1,\mp},(H_{s+1,\pm})^{\prime}\phi_{s+1,\mp}\right\rangle-\left\langle\psi_{m,\mp},H^{\prime}\psi_{m,\mp}\right\rangle|$
$\displaystyle=|\left\langle\phi_{s+1,\mp},V^{\prime}\phi_{s+1,\mp}\right\rangle-\left\langle\psi_{m,\mp},V^{\prime}\psi_{m,\mp}\right\rangle-2\langle\phi_{s+1,\mp},(P_{m,\pm})^{\prime}H_{B_{s+1}}\phi_{s+1,\mp}\rangle|$
$\displaystyle\leq|\left\langle\phi_{s+1,\mp},V^{\prime}\phi_{s+1,\mp}\right\rangle-\left\langle\psi_{m,\mp},V^{\prime}\psi_{m,\mp}\right\rangle|+2|\langle\phi_{s+1,\mp},(P_{m,\pm})^{\prime}H_{B_{s+1}}\phi_{s+1,\mp}\rangle|$
$\displaystyle\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{m}}\delta_{m-1}^{-1}+|\langle\phi_{s+1,\mp},Q_{s+1,\pm}(P_{m,\pm})^{\prime}H_{B_{s+1}}Q_{s+1,\pm}\phi_{s+1,\mp}\rangle|,$

where we use the estimate $\|\phi_{s+1,\mp}-\psi_{m,\mp}\|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{m}}\delta_{m-1}^{-1}$ to bound the first term on
the third line of the inequality.

Thus it suffices to estimate
$Q_{s+1,\pm}P_{m,\pm}^{\prime}H_{B_{s+1}}Q_{s+1,\pm}$.
Recall that

$\displaystyle P_{m,\pm}^{\prime}$
$\displaystyle=\langle\psi_{m,\pm}|\psi_{m,\pm}^{\prime}\rangle+\langle\psi_{m,\pm}^{\prime}|\psi_{m,\pm}\rangle$
$\displaystyle=-G_{B_{m,\pm}}^{\perp}(E_{m,\pm})V^{\prime}P_{m,\pm}-P_{m,\pm}V^{\prime}G_{B_{m,\pm}}^{\perp}(E_{m,\pm}).$

Since $Q_{s+1,\pm}P_{m,\pm}=P_{m,\pm}Q_{s+1,\pm}=0$, it follows that

$Q_{s+1,\pm}P_{m,\pm}^{\prime}H_{B_{s+1}}Q_{s+1,\pm}=-Q_{s+1,\pm}G_{B_{m,\pm}}^{\perp}(E_{m,\pm})V^{\prime}P_{m,\pm}H_{B_{s+1}}Q_{s+1,\pm}.$
(3.66)

We can employ the estimates

$\displaystyle\|P_{m,\pm}H_{B_{s+1}}Q_{s+1,\pm}\|$
$\displaystyle=\|P_{m,\pm}(H_{B_{s+1}}-E_{m,\pm})Q_{s+1,\pm}\|$
$\displaystyle\leq\|P_{m,\pm}(H_{B_{s+1}}-E_{m,\pm})\|$
$\displaystyle\leq\|\Gamma_{B_{m,\pm}}\psi_{m,\pm}\|\lesssim
e^{-\frac{1}{5}\gamma_{0}l_{m}}$

and $\|G_{B_{m,\pm}}^{\perp}(E_{m,\pm})\|=O(\delta_{m-1}^{-1})$ to (3.66) and
complete the proof of (3.63).
(3.64) follows from $\pm E_{m,\pm}^{\prime}\gtrsim\delta_{m-2}^{10}$ and
(3.63) immediately. ∎

######  Lemma 3.46.

With the notation from Lemma 3.43, there is a unique point $\bar{\theta}\in
I^{(1)}_{\cup}$ satisfying
$|\bar{\theta}-\theta_{n,>}|\lesssim\delta_{n}\delta_{n-2}^{-2}$, such that

$\lambda_{n+1,+}(\bar{\theta})=\lambda_{n+1,-}(\bar{\theta}).$

Moreover, the Rellich children have quantitative separation away from
$\bar{\theta}$:

$E_{n+1,>}(\theta)-E_{n+1,<}(\theta)\geq|\lambda_{n+1,+}(\theta)-\lambda_{n+1,-}(\theta)|\geq\delta_{m-1}|\theta-\bar{\theta}|.$
(3.67)

As a corollary, if $E_{n+1,>}(\theta)=E_{n+1,<}(\theta)$, then
$\theta=\bar{\theta}$.

###### Proof.

We consider the difference function

$d(\theta):=\lambda_{n+1,+}(\theta)-\lambda_{n+1,-}(\theta).$

By (3.64), we have

$d^{\prime}(\theta)\geq 2\delta_{m-1}.$ (3.68)

Recalling (3.55), it follows that

$\displaystyle|\lambda_{n,+}(\theta_{n,>})-\lambda_{n,-}(\theta_{n,>})|$
$\displaystyle\leq E_{n,>}(\theta_{n,>})-E_{n,<}(\theta_{n,>})$
$\displaystyle\leq
E_{n,>}(\theta_{n,>})-E_{n,<}(\theta_{n,<})+|E_{n,<}(\theta_{n,<})-E_{n,<}(\theta_{n,>})|$
$\displaystyle\leq 3\delta_{n}+O(\delta_{n}\delta_{n-2}^{-1}).$

By the above estimate and approximation (3.65), we obtain

$|d(\theta_{n,>})|=|\lambda_{n+1,+}(\theta_{n,>})-\lambda_{n+1,-}(\theta_{n,>})|\lesssim\delta_{n}\delta_{n-2}^{-1}.$

It follows that by mean value theorem there exists a unique point
$\bar{\theta}$ satisfying
$|\bar{\theta}-\theta_{n,>}|\lesssim\delta_{n}\delta_{n-2}^{-1}\delta_{m-1}^{-1}\leq\delta_{n}\delta_{n-2}^{-2}$,
such that $d(\bar{\theta})=0$, that is

$\lambda_{n+1,+}(\bar{\theta})=\lambda_{n+1,-}(\bar{\theta}).$

(3.67) follows from (3.61), (3.68) and

$|\lambda_{n+1,+}(\theta)-\lambda_{n+1,-}(\theta)|=|d(\theta)|=|d(\theta)-d(\bar{\theta})|\geq\delta_{m-1}|\theta-\bar{\theta}|$

by mean value theorem. ∎

Suppose that
$|E_{m,+}^{\prime}(\theta_{k_{m},-})\geq|E_{m,-}^{\prime}(\theta_{k_{m},-})|$
(the argument is similar for the opposite case), and define $1\leq
r\lesssim\delta_{m-1}^{-1}$ such that

$|E_{m,+}^{\prime}(\theta_{k_{m},-})|=r|E_{n,-}^{\prime}(\theta_{k_{m},-})|.$

Recalling Lemma 3.37, for $\theta\in I_{\cup}^{(1)}$,

$|(E_{m,+}^{\prime}+rE_{m,-}^{\prime})(\theta)|\leq\delta_{m-1}^{2}$ (3.69)

######  Proposition 3.47.

For $\theta\in I_{\cup}^{(1)}$, we have the following:

  * (a).

If $E_{n+1,>}(\theta)\neq E_{n+1,<}(\theta)$, then

$\displaystyle(E_{n+1,>})^{\prime}$
$\displaystyle=(A^{2}-rB^{2})E_{m,-}^{\prime}+O(\delta_{m-1}^{2}),$ (3.70)
$\displaystyle(E_{n+1,<})^{\prime}$
$\displaystyle=(B^{2}-rA^{2})E_{m,-}^{\prime}+O(\delta_{m-1}^{2})$ (3.71)

and

$\displaystyle(E_{n+1,>})^{\prime\prime}$
$\displaystyle=\frac{2\left\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\right\rangle^{2}}{E_{n+1,>}-E_{n+1,<}}+O(\delta_{n-1}^{-1}),$
(3.72) $\displaystyle(E_{n+1,<})^{\prime\prime}$
$\displaystyle=\frac{2\left\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\right\rangle^{2}}{E_{n+1,<}-E_{n+1,>}}+O(\delta_{n-1}^{-1})$
(3.73)

with the notation from Proposition 3.44 and $r$ as above.

  * (b).

Suppose $E_{n+1,>}(\theta)\neq E_{n+1,<}(\theta)$. If
$|(E_{n+1,>})^{\prime}(\theta)|\leq\delta_{m-1}^{2}$, then
$(E_{n+1,>})^{\prime\prime}(\theta)\geq 2$. Similarly, if
$|(E_{n+1,<})^{\prime}(\theta)|\leq\delta_{m-1}^{2}$, then
$(E_{n+1,<})^{\prime\prime}(\theta)\leq-2$.

  * (c).

If there is a level crossing (thus
$E_{n+1,>}(\bar{\theta})=E_{n+1,<}(\bar{\theta})$ by Lemma 3.46), then
$E_{n+1,>},E_{n+1,<}$ are piecewise differentiable (except the point
$\bar{\theta}$), and for $\theta>\bar{\theta}$,

$(E_{n+1,>})^{\prime}(\theta)>\delta_{m-1}^{2},\
(E_{n+1,<})^{\prime}(\theta)<-\delta_{m-1}^{2},$

for $\theta<\bar{\theta}$,

$(E_{n+1,>})^{\prime}(\theta)<-\delta_{m-1}^{2},\
(E_{n+1,<})^{\prime}(\theta)>\delta_{m-1}^{2}.$

###### Proof.

We are going to prove (3.70). (The proof of (3.71) is analogous.) By (3.59)
and (3.69), we refer to Feynman-Hellman formula (1) Lemma 2.4 to obtain

$\displaystyle(E_{n+1,>})^{\prime}$
$\displaystyle=\left\langle\psi_{n+1},V^{\prime}\psi_{n+1}\right\rangle=A^{2}\left\langle\psi_{m,-},V^{\prime}\psi_{m,-}\right\rangle+B^{2}\left\langle\psi_{m,+},V^{\prime}\psi_{m,+}\right\rangle+O(\delta_{m}^{10})$
$\displaystyle=A^{2}E_{m,-}^{\prime}+B^{2}E_{m,+}^{\prime}+O(\delta_{m}^{10})$
$\displaystyle=(A^{2}-rB^{2})E_{m,-}^{\prime}+B^{2}(E_{m,+}^{\prime}+rE_{m,-}^{\prime})+O(\delta_{m}^{10})$
$\displaystyle=(A^{2}-rB^{2})E_{m,-}^{\prime}+O(\delta_{m-1}^{2}).$

To prove (3.72) and (3.73), we use Feynman-Hellman formula (2) and (3) to
obtain

$(E_{n+1,>})^{\prime\prime}=\left\langle\psi_{n+1},V^{\prime\prime}\psi_{n+1}\right\rangle+2\frac{\left\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\right\rangle^{2}}{E_{n+1,>}-E_{n+1,<}}-2\left\langle
V^{\prime}\psi_{n+1},G^{\perp\perp}_{n+1}(E_{n+1,>})V^{\prime}\psi_{n+1}\right\rangle.$

The first term is bounded by $D$ and the third term is bounded by

$2\|G^{\perp\perp}_{n+1}(E_{n+1,>})\|\|V^{\prime}\psi_{n+1}\|^{2}\lesssim\delta_{n-1}^{-1},$

where we use the estimate $\|G^{\perp\perp}_{n+1}(E_{n+1,>})\|\leq
2\delta_{n-1}^{-1}$ from item (c) of Proposition 3.44. Thus we finish the
proof of item (a).

Now we are going to prove (b). We will show the first term in (3.72) is large
if $|(E_{n+1,>})^{\prime}(\theta)|\leq\delta_{m-1}^{2}$.
Assume $|(E_{n+1,>})^{\prime}(\theta)|\leq\delta_{m-1}^{2}$, by (3.70), we
have

$|A^{2}-rB^{2}||E_{m,-}^{\prime}(\theta)|\leq|(E_{n+1,>})^{\prime}(\theta)|+O(\delta_{m-1}^{2})\lesssim\delta_{m-1}^{2}.$

Since

$|E_{m,-}^{\prime}(\theta)|\gtrsim\delta_{m-2}^{10},$

it follows that

$|A^{2}-rB^{2}|\lesssim\delta_{m-1}^{2}\delta_{m-2}^{-10}<\frac{1}{100}$

Since $A^{2}+B^{2}=1$ and $r\geq 1$, we obtain

$(1+r)B^{2}\geq A^{2}+B^{2}-|A^{2}-rB^{2}|\geq\frac{99}{100}$

and

$(1+\frac{1}{r})A^{2}\geq
A^{2}+B^{2}-\frac{1}{r}|A^{2}-rB^{2}|\geq\frac{99}{100}.$

Thus

$B^{2}\geq\frac{1}{4r},\ A^{2}\geq\frac{1}{4}.$

By (3.59), (3.69) and the previous lower bound for $A,B$, we have

$\displaystyle|\langle\psi_{n+1},V^{\prime}\Psi_{n+1}\rangle|$
$\displaystyle=|AB(\langle\psi_{m,-},V^{\prime}\psi_{m,-}\rangle-\langle\psi_{m,+},V^{\prime}\psi_{m,+}\rangle)+O(\delta_{m}^{10})|$
$\displaystyle=|AB(E_{m,-}^{\prime}-E_{m,+}^{\prime})+O(\delta_{m}^{10})|$
$\displaystyle=|AB\left((1+r)E_{m,-}^{\prime}-(E_{m,+}^{\prime}+rE_{m,-}^{\prime})\right)+O(\delta_{m}^{10})|$
$\displaystyle\geq\frac{1+r}{4\sqrt{r}}(|E_{m,-}^{\prime}|-O(\delta_{m-1}^{2}))$
$\displaystyle\gtrsim\delta_{m-2}^{10}\gg\delta_{m-1}.$

The denominator $0<E_{n+1,>}-E_{n+1,<}\lesssim\delta_{n-1}^{5}$ since
$|J_{i_{n}}^{(1)}|\lesssim\delta_{n-1}^{10}$ and
$|I_{\cup}^{(1)}|\lesssim\delta_{n-1}^{5}$. Combining the previous estimate of
numerator, by (3.72), we obtain

$(E_{n+1,>})^{\prime\prime}(\theta)\gtrsim\delta_{m-1}^{2}\delta_{n-1}^{-5}-O(\delta_{n-1}^{-1})>2,$

and

$-(E_{n+1,<})^{\prime\prime}(\theta)\gtrsim\delta_{m-1}^{2}\delta_{n-1}^{-5}-O(\delta_{n-1}^{-1})>2.$

Suppose there is a level crossing, then

$E_{n+1,>}(\bar{\theta})=\lambda_{n+1,+}(\bar{\theta})=\lambda_{n+1,-}(\bar{\theta})=E_{n+1,<}(\bar{\theta}).$

Thus for $\theta>\bar{\theta}$,

$E_{n+1,>}(\theta)\geq\lambda_{n+1,+}(\theta)>\lambda_{n+1,-}(\theta)\geq
E_{n+1,<}(\theta).$

The above inequality together with (3.64) yields the left derivative of
$E_{n+1,>}$ and $E_{n+1,<}$ satisfy we get

$(E_{n+1,>})^{\prime}_{\operatorname{Left}}(\bar{\theta})\geq\lambda_{n+1,+}^{\prime}(\bar{\theta})>\delta_{m-1},\
(E_{n+1,<})^{\prime}_{\operatorname{Left}}(\bar{\theta})\leq\lambda_{n+1,-}^{\prime}(\bar{\theta})<-\delta_{m-1}.$

We next claim the inequalities hold for all $\theta>\bar{\theta}$. If it is
not true, then the set

$\\{\theta\in I_{\cup}^{(1)}:\theta>\bar{\theta},\ \
(E_{n+1,>})^{\prime}(\theta)\leq\delta_{m-1}^{2}\\}\neq\emptyset.$

Let $\theta^{*}$ be its infimum. Since $E_{n+1,>}^{\prime}(\theta)$ is
continuous for $\theta\geq\bar{\theta}$ and
$(E_{n+1,>})^{\prime}_{\operatorname{Left}}(\bar{\theta})>\delta_{m-1}^{2}$,
we have $\theta^{*}>\bar{\theta}$ and
$(E_{n+1,>})^{\prime}(\theta)>\delta_{m-1}^{2}\geq(E_{n+1,>})^{\prime}(\theta^{*})$
for $\theta\in[\bar{\theta},\theta^{*})$, which implies
$(E_{n+1,>})^{\prime\prime}(\theta^{*})\leq 0$. However, by item (b), we get a
contradiction that $(E_{n+1,>})^{\prime\prime}(\theta^{*})>2$. Thus we prove
our claim. The case of $\theta<\bar{\theta}$ is analogous. ∎

######  Remark 3.7.

We can deduce from the above proposition that $E_{n+1,>}$ and $E_{n+1,<}$ have
the two-monotonicity interval structure. Moreover, if the gap between the
critical values of $E_{n+1,>}$ and $E_{n+1,<}$ is larger than $3\delta_{n+1}$,
then each of them belongs to Type 2, otherwise, $E_{n+1,>}$ along with
$E_{n+1,<}$ belongs to Type 3. We finish the proof of Subcase B.

####  3.5.4. A sketch of the domain adjustment

At the end of the construction, we employ the argument in Section 3.2.3 to
adjust the domains of Rellich children by $O(\delta_{n}^{10})$ so that every
Rellich child $E_{n+1}$ has the same image on each of its monotonicity
interval. After the domain adjustment, Hypothesis 5 still holds for $m=n+1$
because the overlap of the adjacent codomains ($\simeq 3\delta_{n}$) is much
larger than the codomain contraction in the adjustment. We henceforth denote
by $I(E_{n+1})$ (resp. $J(E_{n+1})$) the modified domain (codomain) of
$E_{n+1}$.

####  3.5.5. Green’s function estimates

In this part, we show the Green’s function estimates for $(n+1)$-good sets by
a multi-scale analysis argument and finish the proof of induction.

Recall that $\tilde{J}(E_{n+1})$ is defined as:
For $E_{n+1}$ belonging to Type 1 or 2,
(1). if $E_{n+1}$ has no critical point in $I(E_{n+1})$, then

$\tilde{J}(E_{n+1}):=[\inf J(E_{n+1})+\frac{9}{8}\delta_{n},\sup
J(E_{n+1})-\frac{9}{8}\delta_{n}],$

(2). if $E_{n+1}$ attains its minimum at a critical point in $I(E_{n+1})$,
then

$\tilde{J}(E_{n+1}):=[\inf J(E_{n+1})-2\delta_{n+1},\sup
J(E_{n+1})-\frac{9}{8}\delta_{n}],$

(3). if $E_{n+1}$ attains its maximum at a critical point in $I(E_{n+1})$,
then

$\tilde{J}(E_{n+1}):=[\inf J(E_{n+1})+\frac{9}{8}\delta_{n},\sup
J(E_{n+1})+2\delta_{n+1}],$

and for $E_{n+1}$ belonging to Type 3,

$\tilde{J}(E_{n+1}):=[\inf J(E_{n+1,<})+\frac{9}{8}\delta_{n},\sup
J(E_{n+1,>})-\frac{9}{8}\delta_{n}].$

Fix $\theta^{*},E^{*}$. Let $E_{n+1}\in\mathcal{C}_{n+1}$ be the Rellich
function such that $E^{*}\in\tilde{J}(E_{n+1})$ (if exists). The set of
$(n+1)$-resonant points $S_{n+1}(\theta^{*},E^{*})$ (relative to
$\theta^{*},E^{*}$ and $E_{n+1}$) is defined as:
For $E_{n+1}$ belonging to Type 1 or 2,

$S_{n+1}(\theta^{*},E^{*}):=\\{x\in\mathbb{Z}^{d}:\ \theta^{*}+x\cdot\omega\in
I(E_{n+1}),\ |E_{n+1}(\theta^{*}+x\cdot\omega)-E^{*}|<\delta_{n+1}\\},$

and for $E_{n+1}$ belonging to Type 3,

$S_{n+1}(\theta^{*},E^{*}):=\\{x\in\mathbb{Z}^{d}:\ \theta^{*}+x\cdot\omega\in
I(E_{n+1,<})\cup I(E_{n+1,>}),\
\min_{\bullet\in\\{>,<\\}}|E_{n+1,\bullet}(\theta^{*}+x\cdot\omega)-E^{*}|<\delta_{n+1}\\}.$

We say that a set $\Lambda\subset\mathbb{Z}^{d}$ is $(n+1)$-nonresonant
(relative to $(\theta^{*},E^{*})$ and $E_{n+1}$) if $\Lambda\cap
S_{n+1}(\theta^{*},E^{*})=\emptyset$ and is $(n+1)$-regular if
$(\Lambda_{2l_{i+1}}+x)\subset\Lambda$ for any $x\in
S_{i}(\theta^{*},E^{*})\cap\Lambda$ relative to each ancestor $E_{i}$ $(0\leq
i\leq n)$ of $E_{n+1}$. We say that a set is $(n+1)$-good if it is both
$(n+1)$-nonresonant and $(n+1)$-regular. We are going to prove the Green’s
function estimates for $(n+1)$-good sets:

######  Theorem 3.48.

Fix $\theta^{*},E^{*}$ and a finite set $\Lambda\subset\mathbb{Z}^{d}$. Let
$E_{n+1}\in\mathcal{C}_{n+1}$ such that $E^{*}\in\tilde{J}(E_{n+1})$ (if
exists). If $\Lambda$ is $(n+1)$-good, then for
$|\theta-\theta^{*}|<\delta_{n+1}/(10D),|E-E^{*}|<\delta_{n+1}/5$,

$\|G_{\Lambda}(\theta,E)\|\leq 10\delta_{n+1}^{-1},$ $\
|G_{\Lambda}(\theta,E;x,y)|\leq e^{-\gamma_{n+1}\|x-y\|_{1}},\ \|x-y\|_{1}\geq
l_{n+1}^{\frac{5}{6}},$

where
$\gamma_{n+1}=(1-O(l_{n+1}^{-\frac{1}{30}}))\gamma_{n}\geq\frac{1}{2}\gamma_{0}$.
The above estimates also hold if $\Lambda$ is $(n+1)$-regular and
$E^{*}\notin\tilde{J}(E_{n+1})$ for any $E_{n+1}\in\mathcal{C}_{n+1}$.

###### Proof.

If $E^{*}\notin\tilde{J}(E_{n})$ for any $E_{n}\in\mathcal{C}_{n}$, then the
theorem follows from the inductive hypothesis. By the overlap of two adjacent
modified codomains $\tilde{J}$ where the Rellich functions do not attain their
critical values (c.f. Hypothesis 5), it follows that
$\cup_{E_{n}\in\mathcal{C}_{n}}\tilde{J}(E_{n})=\cup_{E_{n}\in\mathcal{C}_{n}}\tilde{\tilde{J}}(E_{n})$.
We may thus assume $E^{*}\in\tilde{\tilde{J}}(E_{n})$ for some
$E_{n}\in\mathcal{C}_{n}$.

First we assume that $E^{*}\in\tilde{J}(E_{n+1})\subset\tilde{J}(E_{n})$ for
some $E_{n+1}\in\mathcal{C}^{(j)}(E_{n})$. We define the set
$S_{n}^{\Lambda}:=S_{n}(\theta^{*},E^{*})\cap\Lambda$. If
$S_{n}^{\Lambda}=\emptyset$, then $\Lambda$ is $n$-nonresonant and the theorem
follows from the inductive hypothesis. We thus assume $S_{n}^{\Lambda}$ is
nonempty. We wish to find a finite family
$\\{B_{n+1}^{(j)}+p\\}_{p\in\bar{S}_{n}^{\Lambda}}$ of translates of
$B_{n+1}^{(j)}$ such that each $x\in S_{n}^{\Lambda}$ is near the center of
$B_{n+1}^{(j)}+p$ for some $p\in\bar{S}_{n}^{\Lambda}$ and
$\|G_{B_{n+1}^{(j)}+p}(\theta^{*},E^{*})\|\leq\delta_{n+1}^{-1}$ for all
$p\in\bar{S}_{n}^{\Lambda}$. We do so by case analysis:
(1). $j=1$: We claim that for any $x\in S_{n}^{\Lambda}$, if $E_{n+1}$ is
generated from the interval in Subcase A, then $\theta^{*}+x\cdot\omega\in
I(E_{n+1})$, and if $E_{n+1}$ is generated from the interval (3.54) from
Subcase B, then $\theta^{*}+x\cdot\omega\in I(E_{n+1,<})\cup I(E_{n+1,>})$,
where $E_{n+1,<}$ and $E_{n+1,>}$ are the two Rellich children generated from
the interval (3.54). The proof proceeds also by case analysis:
Subcase A (a). $E_{n+1}$ has no critical point.
Since $E^{*}\in\tilde{J}(E_{n+1})$ and $x\in S_{n}^{\Lambda}$, by the
definition of $\tilde{J}(E_{n+1})$, we obtain

$E_{n}(\theta^{*}+x\cdot\omega)\in[\inf J(E_{n+1})+\frac{1}{8}\delta_{n},\sup
J(E_{n+1})-\frac{1}{8}\delta_{n}].$

Since $j=1$, by approximation (3.34), we have

$[\inf J(E_{n+1})+\frac{1}{8}\delta_{n},\sup
J(E_{n+1})-\frac{1}{8}\delta_{n}]\subset(E_{n})_{\pm}(I(E_{n+1})_{\pm}).$

Thus $E_{n}(\theta^{*}+x\cdot\omega)\in(E_{n})_{\pm}(I(E_{n+1})_{\pm})$ and so
$\theta^{*}+x\cdot\omega\in I(E_{n+1})$.
Subcase A (b). $E_{n+1}$ has a critical point.
Suppose that this critical point is a minimum (the other case is analogous),
then $E_{n}$ likewise achieves its minimum at a critical point in
$I(E_{n+1})$. As above, we have

$E_{n}(\theta^{*}+x\cdot\omega)\in[\inf J(E_{n}),\sup
J(E_{n+1})-\frac{1}{8}\delta_{n}].$

and

$[\inf J(E_{n}),\sup
J(E_{n+1})-\frac{1}{8}\delta_{n}]\subset(E_{n})_{\pm}(I(E_{n+1})_{\pm}).$

Thus $\theta^{*}+x\cdot\omega\in I(E_{n+1})$.
Subcase B. $E_{n}$ belongs to Type 3 and $E_{n+1,<}$ and $E_{n+1,>}$ are the
two Rellich children generated from the interval (3.54) . Since
$E^{*}\in\tilde{J}(E_{n+1})$ and $x\in S_{n}^{\Lambda}$, by the definition of
$\tilde{J}(E_{n+1})$, we obtain

$E_{n,\bullet}(\theta^{*}+x\cdot\omega)\in[\inf
J(E_{n+1,<})+\frac{1}{8}\delta_{n},\sup J(E_{n+1,>})-\frac{1}{8}\delta_{n}]$
(3.74)

for some $\bullet\in\\{>,<\\}$. If (3.74) holds for $\bullet=<$, by the
approximation (3.58), it follows that

$E_{n,<}(\theta^{*}+x\cdot\omega)\geq\inf
J(E_{n+1,<})+\frac{1}{8}\delta_{n}\geq\inf(E_{n,<})_{\pm}(I(E_{n+1,<})),$

and so $\theta^{*}+x\cdot\omega\in I(E_{n+1,<}).$
Likewise, if (3.74) holds for $\bullet=>$, then $\theta^{*}+x\cdot\omega\in
I(E_{n+1,>}).$ Thus we finish the proof of the claim.

For $j=1$, we define $\bar{S}_{n}^{\Lambda}=S_{n}^{\Lambda}$.
For Subcase A, since $x\in S_{n}^{\Lambda}$,

$|E_{n}(\theta^{*}+x\cdot\omega)-E^{*}|\leq\delta_{n},$

and since $\Lambda$ is $(n+1)$-nonresonant,

$\delta_{n+1}\leq|E_{n+1}(\theta^{*}+x\cdot\omega)-E^{*}|.$

For Subcase B, since $x\in S_{n}^{\Lambda}$,

$\min_{\bullet\in\\{>,<\\}}|E_{n,\bullet}(\theta^{*}+x\cdot\omega)-E^{*}|\leq\delta_{n},$

and since $\Lambda$ is $(n+1)$-nonresonant,

$\delta_{n+1}\leq\min_{\bullet\in\\{>,<\\}}|E_{n+1,\bullet}(\theta^{*}+x\cdot\omega)-E^{*}|.$

By the eigenvalue separation estimates from Proposition 3.27, 3.44, it follows
that $|\hat{E}-E^{*}|\geq\delta_{n+1}$ for any eigenvalue $\hat{E}$ of
$H_{B_{n+1}^{(1)}+x}(\theta^{*})$, and so

$\|G_{B_{n+1}^{(1)}+x}(\theta^{*},E^{*})\|\leq\delta_{n+1}^{-1}.$

(2). $j=2$: Recalling that in double resonance case, $E_{n+1}$ has a brother,
we denote by $E_{n+1,>}$ the larger one and $E_{n+1,<}$ the smaller one. We
prove that for any $x\in S_{n}^{\Lambda}$, $\theta^{*}+p\cdot\omega\in
I(E_{n+1,<})\cup I(E_{n+1,>})$ for some $p\in\\{x,x-k_{n}\\}$, where the
definition of $k_{n}$ is from (3.37).
Since $E^{*}\in\tilde{J}(E_{n+1})$ and $x\in S_{n}^{\Lambda}$, by the
definition of $\tilde{J}(E_{n+1})$, we obtain

$E_{n}(\theta^{*}+x\cdot\omega)\in[\inf
J(E_{n+1,<})+\frac{1}{8}\delta_{n},\sup J(E_{n+1,>})-\frac{1}{8}\delta_{n}].$

Since $j=2$, by approximation (3.52), we have

$[\inf J(E_{n+1,<})+\frac{1}{8}\delta_{n},\sup
J(E_{n+1,>})-\frac{1}{8}\delta_{n}]\subset(E_{n})_{-}\left([\inf
I(E_{n+1,>}),\sup I(E_{n+1,<})]\right)$

and

$[\inf J(E_{n+1,<})+\frac{1}{8}\delta_{n},\sup
J(E_{n+1,>})-\frac{1}{8}\delta_{n}]\subset(E_{n})_{+}\left([\inf
I(E_{n+1,<}),\sup I(E_{n+1,>})]+k_{n}\cdot\omega\right).$

Thus

$\theta^{*}+x\cdot\omega\in[\inf I(E_{n+1,>}),\sup I(E_{n+1,<})]\cup[\inf
I(E_{n+1,<}),\sup I(E_{n+1,>})]+k_{n}\cdot\omega.$

It follows that $\theta^{*}+x\cdot\omega\in[\inf I(E_{n+1,>}),\sup
I(E_{n+1,<})]$ or $\theta^{*}+(x-k_{n})\cdot\omega\in[\inf I(E_{n+1,<}),\sup
I(E_{n+1,>})]$. Thus

$\displaystyle\theta^{*}+p\cdot\omega$ $\displaystyle\in[\inf
I(E_{n+1,>}),\sup I(E_{n+1,<})]\cup[\inf I(E_{n+1,<}),\sup I(E_{n+1,>})]$
$\displaystyle=I(E_{n+1,<})\cup I(E_{n+1,>})$

for some $p\in\\{x,x-k_{n}\\}$.
For $j=2$, we define $\bar{S}_{n}^{\Lambda}$ to be the set of all point $p$
coming from $x\in S_{n}^{\Lambda}$ as above. Since $\Lambda$ is
$(n+1)$-regular and $\|k_{n}\|_{1}\leq 10l_{n+1}^{(1)}<2l_{n+1}^{(2)}$, it
follows that $\bar{S}_{n}^{\Lambda}\subset\Lambda$. Since $\Lambda$ is
$(n+1)$-nonresonant, by the eigenvalue separation estimates from Proposition
3.34, it follows that $|\hat{E}-E^{*}|\geq\delta_{n+1}$ for any eigenvalue
$\hat{E}$ of $H_{B_{n+1}^{(2)}+p}(\theta^{*})$, and so

$\|G_{B_{n+1}^{(2)}+p}(\theta^{*},E^{*})\|\leq\delta_{n+1}^{-1}.$

Now we assume that $E^{*}\notin\tilde{J}(E_{n+1})$ for all
$E_{n+1}\in\mathcal{C}(E_{n})$. Recall that $\tilde{\tilde{J}}(E_{n})$ is
almost covered by $\cup_{E_{n+1}\in\mathcal{C}(E_{n})}\tilde{J}(E_{n+1})$
except the regions near a critical value of some
$E_{n+1}\in\mathcal{C}(E_{n})$ (c.f. Hypothesis 5 for $m=n+1$). Thus the

$\theta\in\bigcup_{E_{n+1}\in\mathcal{C}(E_{n})}I(E_{n+1})$

minimizing

$\operatorname{dist}(E^{*},E_{n+1}(\theta)),\ E_{n+1}\in\mathcal{C}(E_{n})$

must attain at a critical point of some $E_{n+1}\in\mathcal{C}^{(j)}(E_{n})$.
By the same argument as above, it follows that for this $E_{n+1}$,
$\theta^{*}+p\cdot\omega\in I(E_{n+1})$ for some $p\in\\{x,x-k_{n}\\}$. By the
definition of $\tilde{J}(E_{n+1})$, we have

$\delta_{n+1}\leq|E_{n+1}(\theta^{*}+p\cdot\omega)-E^{*}|.$

Thus we also have

$\|G_{B_{n+1}^{(j)}+p}(\theta^{*},E^{*})\|\leq\delta_{n+1}^{-1}.$

For both $j=1,2$, by the $(n+1)$-regularity of $\Lambda$,
$(B_{n+1}+p)\subset\Lambda$ for all $p\in\bar{S}_{n}^{\Lambda}$. Moreover, by
Diophantine condition, the separation of $p\in\bar{S}_{n}^{\Lambda}$ is larger
than the size of $B_{n+1}$, thus the blocks
$\\{B_{n+1}+p\\}_{p\in\bar{S}_{n}^{\Lambda}}$ are non-overlapping.

First we prove the case that $\Lambda=B_{n+1}+p$ is a single
$(n+1)$-nonresonant block. By the previous discussion, we have

$\|G_{\Lambda}(\theta^{*},E^{*})\|\leq\delta_{n+1}^{-1}.$

It follows that by Neumann series, for
$|\theta-\theta^{*}|<\delta_{n+1}/(10D)$ and
$|E-E^{*}|<\frac{2}{5}\delta_{n+1}$,

$\|G_{\Lambda}(\theta,E)\|\leq 2\delta_{n+1}^{-1}.$

Let $x,y\in\Lambda$ satisfying $\|x-y\|_{1}\geq l_{n+1}^{\frac{4}{5}}$. Since
$G_{\Lambda}$ is self-adjoint, we may assume $\|x-p\|_{1}\geq
l_{n+1}^{\frac{3}{4}}$. Let $\Omega_{n+1,p}$ be an
$l_{n+1}^{\frac{2}{3}}$-size block centered at $p$ such that
$\Lambda\setminus\Omega_{n+1,p}$ is $n$-good. Recalling the estimates for
$n$-good sets from Hypothesis 6, by resolvent identity, we obtain

$\displaystyle|G_{\Lambda}(\theta,E;x,y)|$
$\displaystyle=|G_{\Lambda\setminus\Omega_{n+1,p}}(x,y)\chi(y)+\sum_{z,z^{\prime}}G_{\Lambda\setminus\Omega_{n+1,p}}(x,z)\Gamma_{z,z^{\prime}}G_{\Lambda}(z^{\prime},y)|$
$\displaystyle\leq
e^{-\gamma_{n}\|x-y\|_{1}}+C(d)\sup_{z,z^{\prime}}e^{-\gamma_{n}\|x-z\|_{1}}|G_{\Lambda}(z^{\prime},y)|$
$\displaystyle\leq
e^{-\gamma_{n}\|x-y\|_{1}}+C(d)\sup_{z,z^{\prime}}e^{-\gamma_{n}\|x-z\|_{1}}e^{-\gamma_{n}(\|z^{\prime}-y\|_{1}-l_{n+1}^{\frac{3}{4}})}\delta_{n+1}^{-1}$
$\displaystyle\leq e^{-\gamma^{\prime}_{n}\|x-y\|_{1}}$

with $\gamma^{\prime}_{n}=(1-O(l_{n+1}^{-\frac{1}{30}}))\gamma_{n}$, where we
used if $\|z^{\prime}-y\|\leq l_{n+1}^{\frac{3}{4}}$, then

$|G_{\Lambda}(z^{\prime},y)|\leq\|G_{\Lambda}\|\leq 2\delta_{n+1}^{-1}\leq
2e^{-\gamma_{n}(\|z^{\prime}-y\|_{1}-l_{n+1}^{\frac{3}{4}})}\delta_{n+1}^{-1},$

if $\|z^{\prime}-y\|\geq l_{n+1}^{\frac{3}{4}}$, then

$\displaystyle|G_{\Lambda}(z^{\prime},y)|=|G_{\Lambda}(y,z^{\prime})|$
$\displaystyle\leq\sum_{w,w^{\prime}}|G_{\Lambda\setminus\Omega_{n+1,p}}(y,w)\Gamma_{w,w^{\prime}}G_{\Lambda}(w^{\prime},z^{\prime})|$
$\displaystyle\leq C(d)e^{-\gamma_{n}\|y-w\|_{1}}\|G_{\Lambda}\|$
$\displaystyle\leq
C(d)e^{-\gamma_{n}(\|y-z^{\prime}\|_{1}-l_{n+1}^{\frac{3}{4}})}\delta_{n+1}^{-1}$

and $\delta_{n+1}^{-1}=e^{l_{n+1}^{\frac{2}{3}}}\ll e^{\gamma_{n}\|x-y\|_{1}}$
to bound the second term. Thus we finish the case that $\Lambda$ is a single
$(n+1)$-nonresonant block.
Now assume $\Lambda$ is an arbitrary $(n+1)$-good set. We must show that
$G_{\Lambda}$ exist. By Schur’s test Lemma, it suffices to prove

$\sup_{x}\sum_{y}|G_{\Lambda}(\theta,E+io;x,y)|<C<\infty.$ (3.75)

Define

$\Lambda^{\prime}:=\Lambda\setminus\bigcup_{p\in\bar{S}_{n}^{\Lambda}}\Omega_{n+1,p}.$

Then $\Lambda^{\prime}$ is $n$-nonresonant since
$S_{n}^{\Lambda}\subset\cup_{p\in\bar{S}_{n}^{\Lambda}}\Omega_{n+1,p}$ and is
$n$-regular since $\Lambda$ is $(n+1)$-regular and each
$B_{n+1}+p\setminus\Omega_{n+1,p}$ is $n$-regular. Let
$\tilde{\Omega}_{n+1,p}$ be a $2l_{n+1}^{\frac{2}{3}}$-size cube centered at
$p$. For
$x\in\Lambda\setminus\cup_{p\in\bar{S}_{n}^{\Lambda}}\tilde{\Omega}_{n+1,p}$,
by resolvent identity, we have

$\displaystyle\sum_{y}|G_{\Lambda}(x,y)|$
$\displaystyle\leq\sum_{y}|G_{\Lambda^{\prime}}(x,y)|+\sum_{z,z^{\prime},y}|G_{\Lambda^{\prime}}(x,z)\Gamma_{z,z^{\prime}}G_{\Lambda}(z^{\prime},y)|$
$\displaystyle\leq
C(d)\delta_{n}^{-2}+C(d)e^{-l_{n+1}^{\frac{2}{3}}}\sup_{z^{\prime}}\sum_{y}|G_{\Lambda}(z^{\prime},y)|.$

For $x\in\tilde{\Omega}_{n+1,p}$, by resolvent identity, we have

$\displaystyle\sum_{y}|G_{\Lambda}(x,y)|$
$\displaystyle\leq\sum_{y}|G_{B_{n+1}+p}(x,y)|+\sum_{z,z^{\prime},y}|G_{B_{n+1}+p}(x,z)\Gamma_{z,z^{\prime}}G_{\Lambda}(z^{\prime},y)|$
$\displaystyle\leq\delta_{n+1}^{-2}+C(d)e^{-\frac{1}{2}l_{n+1}}\sup_{z^{\prime}}\sum_{y}|G_{\Lambda}(z^{\prime},y)|.$

Taking supremum for $x$ on the left hand side of the above two inequalities,
we get

$\sup_{x}\sum_{y}|G_{\Lambda}(x,y)|\leq\delta_{n+1}^{-2}+\frac{1}{2}\sup_{x}\sum_{y}|G_{\Lambda}(x,y)|,$

thus

$\sup_{x}\sum_{y}|G_{\Lambda}(x,y)|\leq 2\delta_{n+1}^{-2},$

which gives (3.75). Since for $|\theta-\theta^{*}|<\delta_{n+1}/(10D)$ and
$|E-E^{*}|<\frac{2}{5}\delta_{n+1}$, $G_{\Lambda}(\theta,E)$ exists, we obtain
$\operatorname{dist}(\sigma(H_{\Lambda}(\theta)),E^{*})\geq\frac{2}{5}\delta_{n+1}$
and hence
$\operatorname{dist}(\sigma(H_{\Lambda}(\theta)),E)\geq\frac{1}{5}\delta_{n+1}$
for all $|E-E^{*}|<\frac{1}{5}\delta_{n+1}$. Thus we get the desired operator
norm

$\|G_{\Lambda}(\theta,E)\|=\|\left(H_{\Lambda}(\theta)-E\right)^{-1}\|=\frac{1}{\operatorname{dist}(\sigma(H_{\Lambda}(\theta)),E)}\leq
10\delta_{n+1}^{-1}.$

It remains to prove the off-diagonal decay of $G_{\Lambda}$. Let
$x,y\in\Lambda$ such that $\|x-y\|_{1}\geq l_{n+1}^{\frac{5}{6}}$. We define

$B_{x}:=\left\\{\begin{aligned}
&\Lambda_{\lambda_{1}^{\frac{1}{2}}}(x)\cap\Lambda\quad\text{if
}x\in\Lambda\setminus\bigcup_{p\in\bar{S}_{0}^{\Lambda}}\tilde{\Omega}_{1,p},\\\
&B_{m}+p_{m}\ \quad\text{if
}x\in\tilde{\Omega}_{m,p_{m}},\end{aligned}\right.$

where $p_{m}\in\bar{S}_{m-1}^{\Lambda}$ and $m$ satisfying $1\leq m\leq n+1$
is the smallest integral such that $p_{m}\notin S_{m}(\theta^{*},E^{*})$.

The set $B_{x}$ has the following two properties: (1). $B_{x}$ is $m$-good for
some $0\leq m\leq n+1$; (2). $x$ is close to the center of $B_{x}$ and away
from its relative boundary $\partial_{\Lambda}B_{x}$. We can iterate the
resolvent identity to obtain

$\displaystyle|G_{\Lambda}(x,y)|$
$\displaystyle\leq\prod_{s=0}^{L-1}(C(d)l_{m_{s}}^{d}e^{-\gamma_{m_{s}-1}^{\prime}\|x_{s}-x_{s+1}\|_{1}})|G_{\Lambda}(x_{L},y)|$
$\displaystyle\leq
e^{-\gamma_{n}^{\prime\prime}\|x-x_{L}\|_{1}}|G_{\Lambda}(x_{L},y)|,$ (3.76)

where $x_{0}:=x$, $x_{s+1}\in\partial B_{x_{s}}$ with
$\|x_{s}-x_{s+1}\|_{1}\geq\frac{1}{2}l_{m_{s}}\gg l_{m_{s}}^{\frac{5}{6}}$. We
stop the iteration until $y\in B_{x_{L}}$. By resolvent identity again, we get

$\displaystyle|G_{\Lambda}(x_{L},y)|$
$\displaystyle\leq|G_{B_{x_{L}}}(x_{L},y)|+\sum_{z,z^{\prime}}|G_{B_{x_{L}}}(x_{L},z)\Gamma_{z,z^{\prime}}G_{\Lambda}(z^{\prime},y)|$
$\displaystyle\leq
C(d)e^{-\gamma_{n}^{\prime}(\|x_{L}-y\|_{1}-l_{n+1}^{\frac{4}{5}})}\delta_{n+1}^{-1},$
(3.77)

where we have used the off-diagonal exponential decay of $G_{B_{x_{L}}}$ and
the operator norm estimate $\|G_{\Lambda}\|\leq 10\delta_{n+1}^{-1}$. Since
$\|x-y\|_{1}\geq l_{n+1}^{\frac{5}{6}}$, (3.76) and (3.77) give the desired
off-diagonal estimate

$|G_{\Lambda}(x,y)|\leq e^{-\gamma_{n+1}\|x-y\|_{1}}$

with $\gamma_{n+1}=(1-O(l_{n+1}^{-\frac{1}{30}}))\gamma_{n}$. Thus we finish
the proof. ∎

##  4\. Proof of main Theorem

In this section, we will relate the Rellich functions we have constructed in
the previous section to spectrum information about the limiting operator $H$
and give the proof of Theorem 1.1.

###  4.1. Anderson localization

We first construct the set where Anderson localization occurs.
For each function $E_{n}\in\mathcal{C}_{n}(n\geq 1)$, noticing that
$\delta_{n-2}^{10}\ll(l_{n+2})^{-2\tau}$, one can employ the argument from the
proof of Lemma 3.24 to show that there exists at most one $\tilde{k}_{n}$ with
$0\neq\|{\tilde{k}_{n}}\|_{1}\leq 10l_{n+2}$, such that there is some
$\theta_{{\tilde{k}_{n}},-}\in I(E_{n})_{-}$ satisfying

$\theta_{{\tilde{k}_{n}},-}+{\tilde{k}_{n}}\cdot\omega\in I(E_{n})_{+},\
e_{\tilde{k}_{n}}:=E_{n}(\theta_{{\tilde{k}_{n}},-})=E_{n}(\theta_{{\tilde{k}_{n}},-}+{\tilde{k}_{n}}\cdot\omega).$

Moreover, if such $\tilde{k}_{n}$ exists, then $E_{n}$ belongs to Type 1.
If the above $\tilde{k}_{n}$ exists, we define

$J^{DR}(E_{n}):=B_{\delta_{n-1}^{10}}(e_{\tilde{k}_{n}}).$ (4.1)

We now define the sets

$K_{n}:=\bigcup_{E_{n}\in\mathcal{C}_{n}}\bigcup_{\|k\|_{1}\leq
l_{n+1}}(E_{n}^{-1}(J^{DR}(E_{n}))+k\cdot\omega)$ (4.2)

and

$K:=\bigcap_{s\geq 1}\bigcup_{n\geq s}K_{n}$ (4.3)

######  Lemma 4.1.

$|K|=0.$

###### Proof.

By construction (c.f. Proposition 3.25(5)), for any $E_{s}\in\mathcal{C}_{s}$,
$|\mathcal{C}(E_{s})|\leq 2\delta_{s-1}^{-10}$. Thus for $n\geq 1$,

$|\mathcal{C}_{n}|\leq\prod_{s=0}^{n-1}(2\delta_{s-1}^{-10})\leq\delta_{n-2}^{-11}.$
(4.4)

Fix $E_{n}\in\mathcal{C}_{n}$. Since each $|J^{DR}(E_{n})|=2\delta_{n-1}^{10}$
and $E_{n}$ satisfies the Morse condition with a two-monotonicity interval
structure, we can employ Lemma 2.3 to obtain

$|E_{n}^{-1}(J^{DR}(E_{n}))|\leq 2\sqrt{2|J^{DR}(E_{n})|}\leq
4\delta_{n-1}^{5}.$ (4.5)

Combining the estimates (4.4) and (4.5), we get

$|K_{n}|\lesssim\delta_{n-2}^{-11}(l_{n+1})^{d}\delta_{n-1}^{5}\leq\delta_{n-1}$

and so

$\sum_{n=1}^{+\infty}|K_{n}|<+\infty.$

Thus by Borel-Cantelli Lemma, $|K|=0$. ∎

Thus $\Theta:=\mathbb{T}\setminus K$ is a set of full measure. For the
remainder of this section, we fix $\theta\in\Theta$. Thus there exists some
$N_{1}$ such that $\theta\notin K_{n}$ for all $n\geq N_{1}$.

To show that $H(\theta)$ satisfies Anderson localization, it suffices by
Schnol’s Lemma to show that every generalized eigenfunction $\psi$ of
$H(\theta)$ that grows at most polynomial bound
$(|\psi(x)|\leq(1+\|x\|_{1})^{d})$ in fact decays exponentially. For the
remainder of this section, we fix a generalized eigenvalue $E$ and
corresponding generalized eigenfunction $\psi$.

######  Lemma 4.2.

There exists some $N_{2}$ such that for all $n\geq N_{2}$,
$E\in\tilde{J}(E_{n})$ for some $E_{n}\in\mathcal{C}_{n}$, moreover, there
exists an $x_{n}\in\mathbb{Z}^{d}$ satisfying $\|x_{n}\|_{1}\leq 100l_{n}$
such that $x_{n}\in S_{n}(\theta,E)$ relative to this $E_{n}$.

###### Proof.

If not, there exists an increasing integer sequence
$\\{n_{i}\\}_{i=0}^{\infty}$ such that $\Lambda_{100l_{n_{i}}}$ is
$n_{i}$-nonresonant and a set sequence $\\{U_{i}\\}_{i=0}^{\infty}$ such that
$U_{i}$ is $n_{i}$-regular and satisfies

$\Lambda_{50l_{n_{i}}}\subset U_{i}\subset\Lambda_{100l_{n_{i}}}.$

Thus $U_{i}$ satisfies the condition of Hypothesis 6 at scale $n_{i}$. It
follows that $G_{U_{i}}(\theta,E;x,y)$ decays exponentially for
$\|x-y\|_{1}\geq l_{n_{i}}^{\frac{5}{6}}$.
For any $\|x\|_{1}\leq\frac{1}{3}n_{i}$, $\operatorname{dist}(x,\partial
U_{i})\geq 4l_{n_{i}}>l_{n_{i}}^{\frac{5}{6}}$; thus the Poisson formula gives

$|\psi(x)|\leq\sum_{z,z^{\prime}}|G_{U_{i}}(x,z)\Gamma_{z,z^{\prime}}\psi(z^{\prime})|\lesssim
l_{n_{i}}^{2d}e^{-2\gamma_{0}l_{n_{i}}}\leq e^{-\gamma_{0}l_{n_{i}}}.$

Fixing $x\in\mathbb{Z}^{d}$, we have $\|x\|_{1}\leq\frac{1}{3}n_{i}$, as
$i\to\infty$, and the right-hand side of the above inequality tends to $0$.
Thus we must have $\psi(x)=0$. But $\psi$ can not be identity zero. ∎

###### Proof of Anderson localization.

First we note that fixing $E_{n}\in\mathcal{C}_{n}$, if $E_{n}$ belongs to
Type 3, then for any $\theta\in I(E_{n,>})\cup I(E_{n,<}),0<\|y\|\leq
10l_{n+2}$, we have $\theta+y\cdot\omega\notin I(E_{n,>})\cup I(E_{n,<})$
since $I(E_{n,>})\cup I(E_{n,<})$ is a single interval with length $\ll
l_{n+2}^{-\tau}$, and if $\theta\in I(E_{n})\setminus
E_{n}^{-1}(J^{DR}(E_{n}))$, then $\theta$ is in the simple resonant interval.
An analogous argument in the proof of Proposition 3.26 yields that for
$0<\|y\|\leq 10l_{n+2}$ with $\theta+y\cdot\omega\in I(E_{n})$,

$|E_{n}(\theta)-E_{n}(\theta+y\cdot\omega)|\geq 3\delta_{n}.$ (4.6)

Let $n\geq\max(N_{1},N_{2}):=N$. By Lemma 4.2 and the definition of
$S_{n}(\theta,E)$, there exists some $x_{n}$ satisfying $\|x_{n}\|_{1}\leq
100l_{n}\leq l_{n+1}$, such that $\theta+x_{n}\cdot\omega\in I(E_{n})$ and

$|E_{n}(\theta+x_{n}\cdot\omega)-E|\leq\delta_{n}.$

Since $\theta\notin K_{n}$, by the definition of $K_{n}$, we have
$\theta+x_{n}\cdot\omega\notin E_{n}^{-1}(J^{DR}(E_{n}))$. Thus by (4.6) (with
$\theta+x_{n}\cdot\omega$ in the role of $\theta$), for any $x$ with
$0<\|x-x_{n}\|_{1}\leq 10l_{n+2}$ and $\theta+x\cdot\omega\in I(E_{n})$,

$|E_{n}(\theta+x\cdot\omega)-E|\geq|E_{n}(\theta+x\cdot\omega)-E_{n}(\theta+x_{n}\cdot\omega)|-|E_{n}(\theta+x_{n}\cdot\omega)-E|\geq\delta_{n}.$

It follows that the annual $\Lambda_{9l_{n+2}}\setminus\Lambda_{l_{n+1}}$ is
$n$-nonresonant. Thus there exists an $n$-good set $\Lambda$, which is a
deformation of $\Lambda_{8l_{n+2}}\setminus\Lambda_{2l_{n+1}}$ satisfying

$(\Lambda_{8l_{n+2}}\setminus\Lambda_{2l_{n+1}})\subset\Lambda\subset(\Lambda_{9l_{n+2}}\setminus\Lambda_{l_{n+1}}).$

Let $x\in\Lambda_{6l_{n+2}}\setminus\Lambda_{4l_{n+1}}.$ Then

$\operatorname{dist}(x,\partial\Lambda)\geq\frac{1}{5}\|x\|_{1}>l_{n}^{\frac{5}{6}}.$

Thus by Poisson formula,

$\displaystyle|\psi(x)|$
$\displaystyle\leq\sum_{z,z^{\prime}}|G_{\Lambda}(x,z)\Gamma_{z,z^{\prime}}\psi(z^{\prime})|$
$\displaystyle\lesssim(l_{n+2})^{2d}e^{-\frac{1}{10}\gamma_{0}\|x\|_{1}}.$
$\displaystyle\leq e^{-\frac{1}{20}\gamma_{0}\|x\|_{1}}.$

Since

$\left(\mathbb{Z}^{d}\setminus\bigcup_{n\geq
N}(\Lambda_{6l_{n+2}}\setminus\Lambda_{4l_{n+1}})\right)\subset\Lambda_{4l_{N+1}},$

the above estimate shows the exponential decay of $\psi(x)$ for
$\|x\|_{1}>4l_{N+1}$ and so we finish the proof of Anderson localization for
$\theta\in\Theta$. ∎

###  4.2. Dynamical localization

We first construct the set where Dynamical localization occurs.
Fix $E_{n}\in\mathcal{C}_{n}(n\geq 1)$. Recalling the definition of
$J^{DR}(E_{n})$ in the previous section (c.f. (4.1)), we define analogously
the set $J^{DR}_{\rho}$ for $\rho>0$,

$J^{DR}_{\rho}(E_{n}):=B_{\rho\delta_{n-1}^{9}}(\tilde{k}_{n}),$

and define the set

$K_{\rho}:=\bigcup_{n\geq
1}\bigcup_{E_{n}\in\mathcal{C}_{n}}\bigcup_{\|k\|_{1}\leq
l_{n+1}}(E_{n}^{-1}(J^{DR}_{\rho}(E_{n}))+k\cdot\omega).$

As the proof of Lemma 4.1, one can prove $|K_{\rho}|\leq\rho^{\frac{1}{2}}.$
Recalling the set $K$ (c.f. (4.3)), we have $K\subset K_{\rho}$ for any
$\rho>0$. Thus Anderson localization holds for $\theta\in\mathbb{T}\setminus
K_{\rho}$ by the previous theorem. We will show Dynamical localization also
occurs on this set.

For the reminder of this section, we fix $\theta\in\mathbb{T}\setminus
K_{\rho}$ and denote by
$\\{\lambda_{\alpha},\varphi_{\alpha}\\}_{\alpha\in\mathbb{N}}$ the complete
set of eigenvalues and corresponding eigenfunctions of $H(\theta)$. Then

${\bm{e}}_{0}=\sum_{\alpha}\varphi_{\alpha}(0)\varphi_{\alpha},$

where $\varphi_{\alpha}(0)=\langle\varphi_{\alpha},{\bm{e}}_{0}\rangle.$ Hence

$e^{itH}{\bm{e}}_{0}=\sum_{\alpha}e^{it\lambda_{\alpha}}\varphi_{\alpha}(0)\varphi_{\alpha}.$

Thus, to bound

$\sum_{x\in\mathbb{Z}^{d}}(1+\|x\|_{1})^{q}|\langle
e^{itH(\theta)}{\bm{e}}_{0},{\bm{e}}_{x}\rangle|,$

it suffices to estimate

$\sum_{\alpha}\left(\sum_{x}(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|\right)|\varphi_{\alpha}(0)|.$

Let $A_{0}=\emptyset$ and $A_{j}=\\{\alpha:\
|\varphi_{\alpha}(0)|>e^{-\gamma_{0}l_{j}}\\}\ (j\geq 1).$ Then we can rewrite
the above summation as

$\sum_{j=1}^{+\infty}\sum_{\alpha\in A_{j}\setminus
A_{j-1}}\left(\sum_{x}(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|\right)|\varphi_{\alpha}(0)|.$
(4.7)

######  Lemma 4.3.

Let $\alpha\in A_{j}$. Then for all $n\geq j$, there exists some
$E_{n}\in\mathcal{C}_{n}$ such that $\lambda_{\alpha}\in\tilde{J}(E_{n})$.
Moreover, there exists an $x_{n}\in\mathbb{Z}^{d}$ satisfying
$\|x_{n}\|_{1}\leq 100l_{n}$ such that $x_{n}\in
S_{n}(\theta,\lambda_{\alpha})$ relative to this $E_{n}$.

###### Proof.

The proof is analogous to Lemma 4.2. ∎

Let $m\geq 1$ be the smallest integer such that $\rho\geq\delta_{m-1}$. From
the above lemma, as the proof of the exponential decay for generalized
eigenfunctions in Anderson localization (via the Poisson formula argument),
one can prove that for $\alpha\in A_{j},x\in\mathbb{Z}^{d}$ satisfying
$\|x\|_{1}>4\max(l_{m+1},l_{j+1}):=L_{\rho,j}$,

$|\varphi_{\alpha}(x)|\leq e^{-\frac{1}{20}\gamma_{0}\|x\|_{1}}.$

By the exponential decay of the eigenfunction, we employ the Hilbert-Schmidt
argument to obtain

$\displaystyle L_{\rho,j}^{d}$ $\displaystyle\gtrsim\sum_{\|x\|_{1}\leq
L_{\rho,j}}\sum_{\alpha}|\varphi_{\alpha}(x)|^{2}$
$\displaystyle\geq\sum_{\alpha\in A_{j}}\sum_{\|x\|_{1}\leq
L_{\rho,j}}|\varphi_{\alpha}(x)|^{2}$ $\displaystyle=\sum_{\alpha\in
A_{j}}(1-\sum_{\|x\|_{1}>L_{\rho,j}}|\varphi_{\alpha}(x)|^{2})$
$\displaystyle\geq\frac{1}{2}|A_{j}|.$

Thus $|A_{j}|\lesssim L_{\rho,j}^{d}$.

Using the estimate $|\varphi_{\alpha}(x)|\leq
e^{-\frac{1}{20}\gamma_{0}\|x\|_{1}}$ for $\alpha\in A_{m}$ and
$\|x\|_{1}>L_{\rho,m}=4l_{m+1}$, we get

$\displaystyle\sum_{j=1}^{m}$ $\displaystyle\sum_{\alpha\in A_{j}\setminus
A_{j-1}}\left(\sum_{x}(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|\right)|\varphi_{\alpha}(0)|\leq\sum_{\alpha\in
A_{m}}\left(\sum_{x}(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|\right)$
$\displaystyle\leq|A_{m}|\sup_{\alpha\in A_{m}}\left(\sum_{\|x\|_{1}\leq
4l_{m+1}}+\sum_{\|x\|_{1}>4l_{m+1}}\right)(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|$
$\displaystyle\leq C({q,d})l_{m+1}^{q+2d}.$ (4.8)

Since $|\varphi_{\alpha}(0)|\leq e^{-\gamma_{0}l_{j-1}}$ for $\alpha\notin
A_{j-1}$, using the estimate $|\varphi_{\alpha}(x)|\leq
e^{-\frac{1}{20}\gamma_{0}\|x\|_{1}}$ for $\alpha\in A_{j}(j>m)$ and
$\|x\|_{1}>L_{\rho,j}=4l_{j+1}$, we get

$\displaystyle\ \ \ \sum_{\alpha\in A_{j}\setminus
A_{j-1}}\left(\sum_{x}(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|\right)|\varphi_{\alpha}(0)|$
$\displaystyle\leq|A_{j}|e^{-\gamma_{0}l_{j-1}}\sup_{\alpha\in
A_{j}}\left(\sum_{\|x\|_{1}\leq
4l_{j+1}}+\sum_{\|x\|_{1}>4l_{j+1}}\right)(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|$
$\displaystyle\leq C(q,d)l_{j+1}^{q+2d}e^{-\gamma_{0}l_{j-1}},$

Summing up the above estimates for $j$ from $m+1$ to $+\infty$, we get

$\displaystyle\sum_{j=m+1}^{+\infty}\sum_{\alpha\in A_{j}\setminus
A_{j-1}}\left(\sum_{x}(1+\|x\|_{1})^{q}|\varphi_{\alpha}(x)|\right)|\varphi_{\alpha}(0)|\leq
C(q,d)e^{-\frac{1}{2}\gamma_{0}l_{m}}$ (4.9)

From (4.2) and (4.9), we obtain

$\displaystyle\eqref{kuai}$ $\displaystyle\leq C({q,d})l_{m+1}^{q+2d}$ (4.10)
$\displaystyle\leq C({q,d})\max(|\log\delta_{m-2}|^{100(q+2d)},l_{2}^{q+2d})$
$\displaystyle\leq
C({q,d})\max(|\log\rho|,|\log\varepsilon_{0}|)^{100(q+2d)}<+\infty,$

where on the third line, we use $\rho\leq\delta_{m-2}$ for $m\geq 2$ by the
definition of $m$.
Hence we finish proving $H(\theta)$ satisfies Dynamical localization on the
set $\mathbb{T}\setminus K_{\rho}$ for all $\rho>0$. Since
$|K_{\rho}|\leq\rho^{\frac{1}{2}}$, it follows that Dynamical localization
holds for Lebesgue almost $\theta\in\mathbb{T}$.

Next we prove the expectation version of Dynamical localization. For this
purpose, we employ the estimate from (4.10) to get

$\displaystyle\ \ \ \left(\int_{\mathbb{T}\setminus
K_{\delta_{1}}}+\sum_{n=1}^{+\infty}\int_{K_{\delta_{n}}\setminus
K_{\delta_{n+1}}}\right)\sup_{t\in\mathbb{R}}\sum_{x\in\mathbb{Z}^{d}}(1+\|x\|_{1})^{q}|\langle
e^{itH(\theta)}{\bm{e}}_{0},{\bm{e}}_{x}\rangle|d\theta$ $\displaystyle\leq
C({q,d})\left(|\log\delta_{1}|^{100(q+2d)}+\sum_{n=1}^{+\infty}|\log\delta_{n+1}|^{100(q+2d)}|K_{\delta_{n}}|\right)$
$\displaystyle<+\infty.$

###  4.3. $(\frac{1}{2}-)$-Hölder regularity of IDS

In this section, we fix $\theta\in\mathbb{T},E\in\mathbb{R}$ and $\eta>0$. We
are going to estimate the number of eigenvalues of $H_{\Lambda_{N}}(\theta)$
belonging to $[E-\eta,E+\eta]$ as $N\to+\infty$. For this purpose, we first
prove a lemma relating the operator norm of Green’s function to the number of
eigenvalues inside a certain interval $[E-\eta,E+\eta]$.

######  Lemma 4.4.

Let $H$ be a self-adjoint operator on $\mathbb{Z}^{d}$ and
$\Lambda\subset\mathbb{Z}^{d}$ be a finite set. Fix $E\in\mathbb{R}$. Assume
there exists a set $\Lambda^{\prime}\subset\mathbb{Z}^{d}$ such that
$|\Lambda\Delta\Lambda^{\prime}|\leq M$ and
$\|G_{\Lambda^{\prime}}(E)\|\leq(2\eta)^{-1}$, where
$G_{\Lambda^{\prime}}(E)=(H_{\Lambda^{\prime}}-E)^{-1}$. Then the number of
eigenvalues of $H_{\Lambda}$ inside $[E-\eta,E+\eta]$ dose not exceed $3M$.

###### Proof.

Define the operator $T:=H-E$. Let $\\{\xi_{l}\\}_{l=1}^{L}$ be the collection
of orthonormal eigenfunctions of $H_{\Lambda}$ with corresponding eigenvalues
belonging to $[E-\eta,E+\eta]$. Then for every
$\xi\in\\{\xi_{l}\\}_{l=1}^{L}$, $\|T_{\Lambda}\xi\|\leq\eta$ and so

$\displaystyle\eta\geq\|R_{\Lambda\cap\Lambda^{\prime}}TR_{\Lambda}\xi\|$
$\displaystyle=\|(R_{\Lambda^{\prime}}-R_{\Lambda^{\prime}\setminus\Lambda})TR_{\Lambda}\xi\|$
$\displaystyle=\|R_{\Lambda^{\prime}}TR_{\Lambda^{\prime}}\xi+R_{\Lambda^{\prime}}TR_{\Lambda\setminus\Lambda^{\prime}}\xi-
R_{\Lambda^{\prime}\setminus\Lambda}TR_{\Lambda}\xi\|.$

The above estimate together with $\|G_{\Lambda^{\prime}}(E)\|\leq(2\eta)^{-1}$
yields

$\|R_{\Lambda^{\prime}}\xi+G_{\Lambda^{\prime}}(E)(R_{\Lambda^{\prime}}TR_{\Lambda\setminus\Lambda^{\prime}}-R_{\Lambda^{\prime}\setminus\Lambda}TR_{\Lambda})\xi\|\leq
1/2.$ (4.11)

Define the subspace
$\mathcal{H}:=\operatorname{Image}G_{\Lambda^{\prime}}(E)(R_{\Lambda^{\prime}}TR_{\Lambda\setminus\Lambda^{\prime}}-R_{\Lambda^{\prime}\setminus\Lambda}TR_{\Lambda})$.
Thus

$\operatorname{dim}\mathcal{H}\leq\operatorname{Rank}R_{\Lambda\setminus\Lambda^{\prime}}+\operatorname{Rank}R_{\Lambda^{\prime}\setminus\Lambda}\leq
M.$ (4.12)

From (4.11), we deduce

$\|R_{\Lambda^{\prime}}\xi\|^{2}-\|P_{\mathcal{H}}R_{\Lambda^{\prime}}\xi\|^{2}=\|P_{\mathcal{H}}^{\perp}R_{\Lambda^{\prime}}\xi\|^{2}\leq
1/4.$ (4.13)

(4.12) and (4.13) yield

$\displaystyle L=\sum_{l=1}^{L}\|\xi_{l}\|^{2}$
$\displaystyle=\sum_{l=1}^{L}\|R_{\Lambda^{\prime}}\xi_{l}\|^{2}+\sum_{l=1}^{L}\|R_{\Lambda\setminus\Lambda^{\prime}}\xi_{l}\|^{2}$
$\displaystyle\leq
L/4+\sum_{l=1}^{L}\|P_{\mathcal{H}}R_{\Lambda^{\prime}}\xi_{l}\|^{2}+\sum_{l=1}^{L}\|R_{\Lambda\setminus\Lambda^{\prime}}\xi_{l}\|^{2}$
$\displaystyle\leq
L/4+\operatorname{dim}\mathcal{H}+|\Lambda\setminus\Lambda^{\prime}|$
$\displaystyle\leq L/4+2M.$

It follows that $L\leq\frac{8}{3}M$ and so we finish the proof. ∎

Let $N$ be sufficiently large depending on $\eta$. The proof proceeds by case
analysis:
(1). $\varepsilon_{0}=\delta_{0}^{20}<\eta$.
We define the set

$S_{\eta}:=\\{x\in\Lambda_{N}:\ |v(\theta+x\cdot\omega)-E|\leq(2d+2)\eta\\}.$

Since $v$ has a two-monotonicity interval structure and satisfies the Morse
condition,

$|v^{-1}\left([E-(2d+2)\eta,E+(2d+2)\eta]\right)|\leq C(d)\eta^{\frac{1}{2}}.$

From the uniform distribution of $\\{x\cdot\omega\\}_{x\in\mathbb{Z}^{d}}$, we
deduce that $|S_{\eta}|\leq C(d)\eta^{\frac{1}{2}}|\Lambda_{N}|$ for
sufficiently large $N$. Let $\Lambda^{\prime}=\Lambda_{N}\setminus S_{\eta}$.
Then by Neumann series,

$\|G_{\Lambda^{\prime}}(E;\theta)\|\leq((2d+2)\eta-\varepsilon\|\Delta\|)^{-1}\leq((2d+2)\eta-2d\varepsilon)^{-1}\leq(2\eta)^{-1}.$

By Lemma 4.4, $H_{\Lambda_{N}}(\theta)$ has at most
$C(d)\eta^{\frac{1}{2}}|\Lambda_{N}|$ eigenvalues in $[E-\eta,E+\eta]$. Thus

$\mathcal{N}_{\Lambda_{N}}(E+\eta;\theta)-\mathcal{N}_{\Lambda_{N}}(E-\eta;\theta)\leq
C(d)\eta^{\frac{1}{2}}.$

(2). $\eta\in(\delta_{n+1}^{20},\delta_{n}^{20}]$ for some $n\geq 0$.
Replacing $\delta_{n+1}$ with $\eta$ in the proof of Theorem 3.48 where we
essentially use the relation $|\log\delta_{n}|\ll|\log\delta_{n+1}|\ll
l_{n+1}$, one can obtain the following lemma:

######  Lemma 4.5.

Let $E_{n+1}\in\mathcal{C}_{n+1}$, such that $E\in\tilde{J}(E_{n+1})$ or
$E_{n+1}$ minimizes $\operatorname{dist}(\tilde{J}(\textbf{E}),E)$ among
$\textbf{E}\in\mathcal{C}_{n+1}$. Let $\Lambda\subset\mathbb{Z}^{d}$ be an
$(n+1)$-regular set satisfying that

$|E_{n+1}(\theta+x\cdot\omega)-E|\geq 20\eta$

for any $x\in\Lambda$ with $\theta+x\cdot\omega\in I(E_{n+1})$.
Then

$\|G_{\Lambda}(\theta,E)\|\leq\frac{1}{2}\eta^{-1}.$

We find an $(n+1)$-regular set $\Lambda$ such that

$\Lambda_{N}\subset\Lambda\subset\Lambda_{N+50l_{n+1}}$

and define the set

$S_{\eta}:=\\{x\in\Lambda:\ \theta+x\cdot\omega\in I(E_{n+1}),\
|E_{n+1}(\theta+x\cdot\omega)-E|<20\eta.\\}$

Then the set $\Lambda^{\prime}:=\Lambda\setminus(\bigcup_{x\in
S_{\eta}}B_{n+1}+x)$ satisfies the assumption of Lemma 4.5 and so

$\|G_{\Lambda^{\prime}}(\theta,E)\|\leq\frac{1}{2}\eta^{-1}.$ (4.14)

As above we have for sufficiently large $N$,

$|S_{\eta}|\leq C(d)\eta^{\frac{1}{2}}|\Lambda_{N+50l_{n+1}}|\leq
C(d)\eta^{\frac{1}{2}}|\Lambda_{N}|.$

If follows that

$\displaystyle|\Lambda_{N}\setminus\Lambda^{\prime}|+|\Lambda^{\prime}\setminus\Lambda_{N}|$
$\displaystyle\leq|\Lambda\setminus\Lambda^{\prime}|+|\Lambda\setminus\Lambda_{N}|$
(4.15) $\displaystyle\leq C(d)(l_{n+1}^{d}|S_{\eta}|+l_{n+1}N^{d-1})$
$\displaystyle\leq
C(d)(l_{n+1}^{d}\eta^{\frac{1}{2}}|\Lambda_{N}|+l_{n+1}N^{d-1}).$

Combining (4.14), (4.15), we employ Lemma 4.4 to obtain

$\displaystyle\mathcal{N}_{\Lambda_{N}}(E+\eta;\theta)-\mathcal{N}_{\Lambda_{N}}(E-\eta;\theta)$
$\displaystyle\leq C(d)(l_{n+1}^{d}\eta^{\frac{1}{2}}+l_{n+1}/N)$
$\displaystyle\leq C(d)\eta^{\frac{1}{2}}|\log\eta|^{16d}$

provided $N$ sufficiently large, where we use the estimate

$l_{n+1}\leq l_{n}^{8}\leq|\log\delta_{n}|^{16}\leq|\log\eta|^{16}$

on the second line.

We finish the proof by combining the above two cases and letting
$N\to+\infty$.

##  Appendix A Proof of Lemma 2.1

###### Proof.

Without loss of generation, we may assume $E^{*}=0$. It suffices to show the
smallest $m$ eigenvalues of the positive semidefinite operator $H^{2}$,
$0\leq\lambda_{1}\leq\cdots\leq\lambda_{m}$ satisfy

$\sum_{k=1}^{m}\lambda_{k}\leq m\delta^{2}.$

Denote by $P$ the orthogonal projection on the space spanned by
$\psi_{k}(1\leq k\leq m)$. Thus the compressed operator $PH^{2}P$ has $m$
eigenvalues $0\leq\mu_{1}\leq\cdots\leq\mu_{m}$ satisfying
$\lambda_{k}\leq\mu_{k}$ by min-max principle. Thus we obtain

$\displaystyle\sum_{k=1}^{m}\lambda_{k}\leq\sum_{k=1}^{m}\mu_{k}$
$\displaystyle=\operatorname{Trace}(PH^{2}P)$
$\displaystyle\leq\sum_{k=1}^{m}\langle\psi_{k},PH^{2}P\psi_{k}\rangle$
$\displaystyle=\sum_{k=1}^{m}\|H\psi_{k}\|^{2}$ $\displaystyle\leq
m\delta^{2}$

and finish the proof. ∎

##  Appendix B Proof of Lemma 2.3

###### Proof.

Without loss of generality, we may consider the case
$\left|E^{\prime}\right|\leq\delta$ implies $E^{\prime\prime}\geq 2$ and
$\theta_{s}\leq\theta_{1}\leq\theta_{2}$. Since
$E^{\prime}\left(\theta_{s}\right)=0$, by assumption
$E^{\prime\prime}\left(\theta_{s}\right)\geq 2$. Let $\theta_{d}$ be the
largest point with the following property:

$E^{\prime\prime}(\theta)\geq 2\text{ for
}\theta_{s}\leq\theta\leq\theta_{d}.$

This implies that $E(\theta)$ is an increasing function near the right of
$\theta_{s}$. By definition of $\theta_{d}$, we have
$E^{\prime\prime}\left(\theta_{d}+\Delta\theta_{n}\right)<2$ for a sequence
$\Delta\theta_{n}\to 0^{+}$; therefore
$E^{\prime}\left(\theta_{d}+\Delta\theta_{n}\right)>\delta$. This inequality
must hold for every $\theta>\theta_{d}$ or else we would have a point
$\theta^{*}>\theta_{d}$ where
$E^{\prime}(\theta^{*})=\delta,E^{\prime}(\theta^{*}-\Delta\theta)>\delta$ for
small $\Delta\theta>0$ but $E^{\prime\prime}(\theta^{*})\geq 2>0$ by
$|E^{\prime}(\theta^{*})|\leq\delta$ and the assumption of $E$. This is
impossible. Therefore

$E^{\prime}(\theta)\geq\delta\text{ for }\theta>\theta_{d}.$

There are three cases.
Case 1: $\theta_{s}\leq\theta_{1}<\theta_{2}\leq\theta_{d}$.

$\displaystyle
E(\theta_{2})-E(\theta_{1})=E^{\prime}(\theta_{1})(\theta_{2}-\theta_{1})+\frac{1}{2}E^{\prime\prime}(\xi)(\theta_{2}-\theta_{1})^{2}\geq(\theta_{2}-\theta_{1})^{2}.$

Case 2: $\theta_{d}\leq\theta_{1}<\theta_{2}$.

$\displaystyle
E(\theta_{2})-E(\theta_{1})=E^{\prime}(\xi)(\theta_{2}-\theta_{1})\geq\delta(\theta_{2}-\theta_{1})\geq(\theta_{2}-\theta_{1})^{2}.$

Case 3: $\theta_{s}\leq\theta_{1}\leq\theta_{d}\leq\theta_{2}$.

$\displaystyle E(\theta_{2})-E(\theta_{1})$
$\displaystyle=E(\theta_{2})-E(\theta_{d})+E(\theta_{d})-E(\theta_{1})$
$\displaystyle\geq(\theta_{2}-\theta_{d})^{2}+(\theta_{d}-\theta_{1})^{2}$
$\displaystyle\geq\frac{1}{2}(\theta_{2}-\theta_{1})^{2}.$

To prove the second inequality, we assume $\theta_{s}\leq\theta$. (The case
$\theta\leq\theta_{s}$ is analogous.) We need to consider two cases.
Case 1: $\theta_{s}\leq\theta\leq\theta_{d}$.

$E^{\prime}(\theta)=E^{\prime}(\theta_{s})+E^{\prime\prime}(\xi)(\theta-\theta_{s})\geq\theta-\theta_{s}.$

Case 2: $\theta_{d}\leq\theta$. In this case we have
$E^{\prime}(\theta)\geq\delta$.
Hence we finish the proof. ∎

##  Appendix C Proof of Lemma 2.4

###### Proof.

The locally $C^{2}$ smoothness of $E(\theta)$ is inherited from the operators
since $E(\theta^{*})$ is a simple eigenvalue. Noting that
$\langle\psi,\psi\rangle\equiv 1$, we have
$\langle\psi,\psi^{\prime}\rangle=0$. Differentiating the equation
$E=\langle\psi,H\psi\rangle$ yields

$E^{\prime}=\langle\psi,H^{\prime}\psi\rangle+2\langle\psi^{\prime},H\psi\rangle=\langle\psi,H^{\prime}\psi\rangle,$
(C.1)

where we used $H\psi=E\psi$ and $\langle\psi,\psi^{\prime}\rangle=0$.
Now we prove (2). Differentiate (C.1) gives
$E^{\prime\prime}=\langle\psi,H^{\prime\prime}\psi\rangle+2\langle\psi^{\prime},H^{\prime}\psi\rangle$.
Thus it suffices to show

$\langle\psi^{\prime},H^{\prime}\psi\rangle=-\langle
H^{\prime}\psi,G^{\perp}(E)H^{\prime}\psi\rangle.$

Since $\psi^{\prime}$ is orthogonal to $\psi$, we have

$\displaystyle\langle\psi^{\prime},H^{\prime}\psi\rangle=\langle
P^{\perp}\psi^{\prime},H^{\prime}\psi\rangle=\langle
G^{\perp}(E)(H-E)\psi^{\prime},H^{\prime}\psi\rangle=-\langle
H^{\prime}\psi,G^{\perp}(E)H^{\prime}\psi\rangle,$

where we used $G^{\perp}(E)\psi=0$ and
$(H^{\prime}-E^{\prime})\psi=-(H-E)\psi^{\prime}$ since $(H-E)\psi\equiv 0$.
(3) follows from

$G^{\perp}(E)=((H-E)^{\perp})^{-1}=\sum_{E^{\prime}\neq
E}\frac{1}{E^{\prime}-E}P_{E^{\prime}}$

and

$G^{\perp\perp}(E)=((H-E)^{\perp\perp})^{-1}=\sum_{E^{\prime}\neq
E,\mathcal{E}}\frac{1}{E^{\prime}-E}P_{E^{\prime}}$

immediately. ∎

##  Appendix D Lemmas for constructing the desired regular deformation

The desired deformation can be made by the following two lemmas:

######  Lemma D.1.

Let $\theta^{*}\in\mathbb{T},E^{*}\in\mathbb{R},E_{n}\in\mathcal{C}_{n}$. Then
any three different points $x_{1},x_{2},x_{3}\in S_{n}(\theta^{*},E^{*})$
relative to $E_{n}$ satisfy

$\max(\|x_{1}-x_{2}\|_{1},\|x_{2}-x_{3}\|_{1},\|x_{1}-x_{3}\|_{1})\gtrsim\delta_{n}^{-\frac{1}{2\tau}}\gg
l_{n}.$

The above separation also holds uniformly for $\theta$ and $E$ belonging to a
small neighbourhood (on order $\delta_{n}$) of $(\theta^{*},E^{*})$.

###### Proof.

By the two-monotonicity interval structure and Morse condition of $E_{n}$, it
follows that the preimage $E_{n}^{-1}(B_{\delta_{n}}(E^{*}))$ has at most two
connected components with length $\lesssim\delta_{n}^{\frac{1}{2}}$. The lemma
follows from the Diophantine condition and pigeonhole principle immediately. ∎

######  Lemma D.2.

Let $L>1$. Assume a set $S\subset\mathbb{Z}^{d}$ satisfies the following
condition:
Any three different points $x_{1},x_{2},x_{3}\in S$ satisfy

$\max(\|x_{1}-x_{2}\|_{1},\|x_{2}-x_{3}\|_{1},\|x_{1}-x_{3}\|_{1})>100L.$

Then for any set $B\subset\mathbb{Z}^{d}$, there exists a deformation $B^{*}$
satisfying

$B\subset B^{*}\subset(B+\Lambda_{50L})=\\{x\in\mathbb{Z}^{d}:\
\operatorname{dist}(x,B)\leq 50L\\},$

such that $(\Lambda_{L}+x)\subset B^{*}$ for any $x\in S$ with
$(\Lambda_{L}+x)\cap B^{*}\neq\emptyset$.

###### Proof.

We inductively define $B_{0}:=B$,

$B_{0}\subsetneqq B_{1}\subsetneqq\cdots\subsetneqq B_{N}$ (D.1)

satisfying

$B_{n+1}=B_{n}\bigcup\left(\bigcup_{\\{x\in S:\ (\Lambda_{L}+x)\cap
B_{n}\neq\emptyset\\}}\Lambda_{L}+x\right)$

for $n\geq 0$, where $N$ is the largest integer such that the “$\subsetneqq$”
relationship holds in (D.1). We claim that $N\leq 2$. Otherwise, there exist
three different point $x_{1},x_{2},x_{3}\in S$ such that

$(\Lambda_{L}+x_{1})\cap B_{0}\neq\emptyset,\ \
(\Lambda_{L}+x_{2})\cap(\Lambda_{L}+x_{1})\neq\emptyset,\ \
(\Lambda_{L}+x_{3})\cap(\Lambda_{L}+x_{2})\neq\emptyset.$

Thus $\max(\|x_{1}-x_{2}\|_{1},\|x_{2}-x_{3}\|_{1},\|x_{1}-x_{3}\|_{1})\leq
4L,$ a contradiction. Then $B^{*}:=B_{N}$ is the desired deformation. ∎

## Acknowledgments

Y. Shi was partially supported by NSFC (12271380). Z. Zhang was partially
supported by NSFC (12171010, 12288101).

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