$\displaystyle\phi(\omega_{1})-\phi(\omega_{2})=l_{ac}\cdot(\omega_{1}-\omega_{2})-\frac{1}{\text{i}}\left((\tilde{r}_{\infty}(\omega_{1},j)-\tilde{r}_{\infty}(\omega_{1},k))-(\tilde{r}_{\infty}(\omega_{2},j)-\tilde{r}_{\infty}(\omega_{2},k))\right)$
(11.109)

As in the proof of Lemma 11.8, we denote

$\displaystyle l_{ac}^{\perp}$
$\displaystyle:=\left\\{\vec{w}\in\mathbb{R}^{\nu}:\vec{w}\cdot
l_{ac}=0,\text{ and }sl_{ac}+\vec{w}\in R_{ljk}(i_{n}),\text{ for some
$s\in\mathbb{R}$}\right\\},$ $\displaystyle I_{l_{ac}}(\vec{w})$
$\displaystyle:=\left\\{s\in\mathbb{R}:sl_{ac}+\vec{w}\in
R_{ljk}(i_{n})\right\\}$

For each $\vec{w}\in l_{ac}^{\perp}$ and $s_{1},s_{2}\in I_{l_{ac}}(\vec{w})$,
it follows from (11.109) that denoting $\Psi(s):=\phi(sl_{ac}+\vec{w})$,

$\displaystyle|\Psi(s_{1})-\Psi({s_{2}})|$
$\displaystyle\geq|\mathbb{C}_{j,k}^{T}\mathbb{A}^{-1}l||s_{1}-s_{2}|-|\left((\tilde{r}_{\infty}(\omega_{1},j)-\tilde{r}_{\infty}(\omega_{1},k))-(\tilde{r}_{\infty}(\omega_{2},j)-\tilde{r}_{\infty}(\omega_{2},k))\right)|$
$\displaystyle\overset{\eqref{rxcxcxxcqmcxc111},\eqref{rpishsnsdiwsxcsd},\eqref{oosdhwwdwd1ssd}}{\geq_{\mathtt{p_{e}}}}|s_{1}-s_{2}|,$

for sufficiently small $\varepsilon$, depending on $\mathtt{p_{e}}$. On the
other hand, that $\omega_{1},\omega_{2}\in R_{ljk}(i_{n})$ implies

$|\Psi(s_{1})-\Psi({s_{2}})|=|\phi(\omega_{1})-\phi(\omega_{2})|\overset{\eqref{define_rlkj}}{\leq_{\mathtt{p_{e}}}}\gamma\langle
l\rangle^{-\tau}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|\leq_{\mathtt{p_{e}}}\gamma\langle
l\rangle^{-\tau},$

where the last inequality follows from that $j,k<C_{\mathtt{H}2}$. This
implies $|I_{l_{ac}}(\vec{w})|\leq_{\mathtt{p_{e}}}\gamma\langle
l\rangle^{-\tau}$ for all $\vec{w}\in l^{\perp}_{ac}$. As in the proof of
Lemma 11.9, we have
$|l_{ac}^{\perp}|\leq_{\mathtt{p_{e}}}\varepsilon^{2(\nu-1)}$, therefore,

$|R_{ljk}(i_{n})|=\int_{l^{\perp}_{ac}}\int_{I_{l_{ac}}(\vec{w})}1dsd\vec{w}\leq_{\mathtt{p_{e}}}\varepsilon^{2(\nu-1)}\gamma\langle
l\rangle^{-\tau},$

which finishes the proof. ∎

##  12 Proof of Theorem 6.5

  * Proof of Theorem 6.5: Let us fix $\mathtt{p}$ as in (6.24) and fix $S_{0}^{+},\mathtt{M}$ and $S$ so that S.1,S.2 and the hypotheses $(\mathtt{H}1)$-$(\mathtt{H}3)$ hold. Let $\mathcal{C}_{\varepsilon}:=\cap_{n\geq 0}\mathcal{G}_{n}$. Then, Proposition 11.1 tells us that the sequence of embeddings $i_{n}$, constructed in (11.2), is well-defined for $\omega\in\mathcal{C}_{\varepsilon}$ for sufficiently small $\varepsilon$, depending on $\mathtt{p},b,\mathtt{M},S_{0}$ and satisfies that for each $\omega\in\mathcal{C}_{\varepsilon}$,

    1. (1)

$\mathcal{F}_{\omega}(i_{n}(\omega))\to 0$ as $n\to\infty$ in
$H^{s_{0}}_{\varphi,x}$, (see (11.7)).

    2. (2)

$i_{n}(\omega)=i_{triv}+\mathfrak{I}_{n}(\omega)$ is Cauchy sequence in
$H^{s_{0}+2\mu_{\mathtt{p}}}_{\varphi,x}$ (see (11.10) and (11.2)).

    3. (3)

$\rVert\mathfrak{I}_{n}(\omega)\rVert_{s_{0}+2\mu_{\mathtt{p}}}\leq_{\mathtt{p},\mathtt{M},S_{0},b}\varepsilon^{6-4b}$
(see (11.59)).

    4. (4)

$i_{n}(\omega)\in X_{i}\cap X_{i,\mathtt{M}}$, that is, $i_{n}(\omega)$ is a
reversible and $\frac{2\pi}{\mathtt{M}}$-translation invariant, (see (2) of
Proposition 11.1).

Therefore, for $\omega\in\mathcal{C}_{\varepsilon}$, we have that
$i_{\infty}(\omega):=\lim_{n\to\infty}i_{n}(\omega)$ is reversible and
$\frac{2\pi}{\mathtt{M}}$-translation invariant, and satisfies that
$\mathcal{F}_{\omega}(i_{\infty}(\omega))=0$ with the estimate

$\displaystyle\rVert\mathfrak{I}_{\infty}(\omega)\rVert_{s_{0}+2\mu_{\mathtt{p}}}\leq_{\mathtt{p},\mathtt{M},S_{0},b}\varepsilon^{6-4b},\text{
where $\mathfrak{I}_{\infty}(\omega):=i_{\infty}(\omega)-i_{triv}$.}$ (12.1)

From Proposition 11.7 and (11.59), we see that

$|\mathcal{C}_{\varepsilon}|\geq|\Omega_{\varepsilon}|-C_{\mathtt{p},\mathtt{M},S_{0},b}\varepsilon^{2(\nu-1)}\gamma,$

Furthermore (5.25) and (5.27) imply that $|\Omega_{\varepsilon}|\geq
c_{\mathtt{p},\mathtt{M},S_{0},b}\varepsilon^{2\nu}$. Therefore,

$\lim_{\varepsilon\to
0}\frac{|\mathcal{C}_{\varepsilon}|}{|\Omega_{\varepsilon}|}\geq
1-C_{\mathtt{p},\mathtt{M},S_{0},b}\lim_{\varepsilon\to
0}\gamma\varepsilon^{-2}\overset{\eqref{frequency_set2},\gamma=o(\varepsilon^{2})}{=}1.$

Since
$\mathcal{C}_{\varepsilon}\overset{\eqref{setoffreqences}}{\subset}\Omega_{\varepsilon}$,
we obtain (6.25).

To finish the proof, it suffices to show that $i_{\infty}(\omega)$ is linearly
stable. Let $I(t):=(\dot{\theta}(t),\dot{y}(t),\dot{z}(t))$ be
$\frac{2\pi}{\mathtt{M}}$-translation invariant and solve the linearized
Hamiltonian system at $i_{\infty}(\omega)(\omega t)$, that is,

$\displaystyle\dot{I}(t)=d_{i}X_{H_{\zeta}}(i_{\infty}(\omega)(\omega
t))[I(t)].$ (12.2)

Since $H_{\zeta}$ is $\frac{2\pi}{\mathtt{M}}$-translation invariant
Hamiltonian, one can easily see that $z(t)$ is also
$\frac{2\pi}{\mathtt{M}}$-translation invariant, that is $z(t)\in
X_{\mathtt{M}}$. Since $i_{\infty}(\omega)$ solves
$\mathcal{F}_{\omega}(i_{\infty}(\omega))=0$, Lemma 7.17 tells us that

$\displaystyle A(t):=(\psi(t),\eta(t),w(t)):=DG_{\delta}(\bar{i}(\omega
t))^{-1}[I(t)]$ (12.3)

solves

$\displaystyle\begin{pmatrix}\dot{\psi}(t)\\\ \dot{\eta}(t)\\\
\dot{w}(t)\end{pmatrix}=\begin{pmatrix}K_{20}(\omega t)[\eta(t)]+K_{11}(\omega
t)^{T}[w(t)]\\\ 0\\\ \partial_{x}K_{02}(\omega t)[w(t)]+K_{11}(\omega
t)[\eta(t)]\end{pmatrix}.$ (12.4)

Clearly, $|\psi(t)|$ is bounded since $\psi(t)\in\mathbb{T}^{\nu}$, and
$\eta(t)$ does not evolve in time, hence $\eta(t)=\eta(0)$. Therefore, if we
prove that the Sobolev norm of $w$ is bounded uniformly in time, then the
estimate for $DG_{\delta}$ in Lemma 7.8 and (12.3) implies the linear system
(12.2) is stable (see Remark 6.6). Denoting $f(\omega t):=K_{11}(\omega
t)\eta(0)$, we see that the evolution of $w$ is described by

$\displaystyle\dot{w}(t)-\partial_{x}K_{02}(\omega t)[w(t)]=f(\omega t).$
(12.5)

Lemma 7.6 and (12.3) imply that $w(t)\in X_{\mathtt{M}}$ and Lemma 7.11
implies $f(\omega t)\in X_{\mathtt{M}}$. Indeed, in Section 10, (specially in
Subsection 10.7), we proved that there exist $\varphi$-dependent
transformations $\Phi_{1,\infty}(\varphi),\Phi_{2,\infty}(\varphi)$ between
the spaces of $\frac{2\pi}{\mathtt{M}}$-translation invariant functions in
$H_{S^{\perp}}$ such that

$\displaystyle\Phi_{1,\infty}(\varphi)\circ(\mathcal{D}_{\omega}-\partial_{x}K_{02}(\varphi))\circ\Phi_{2,\infty}(\varphi)=\mathcal{L}_{\infty}=\mathcal{D}_{\omega}-D_{\infty},\text{
see \eqref{dixcxc2sdcs},\eqref{liabssdw2sss2sd2}},$ (12.6)
$\displaystyle\text{ $D_{\infty}=\text{diag}_{j\in
S^{\perp}_{\mathtt{M}}}d_{\infty}(\omega,j)$ is real and reversible (see
\ref{kamredu13} of Proposition~{}\ref{kamresdsd})}.$ (12.7)

Using (12.6) and (12.5), and denoting

$\displaystyle v(t):=\left(\Phi_{2,\infty}(\omega t)\right)^{-1}w(t),\quad
g(\varphi):=\Phi_{1,\infty}(\varphi)[f(\varphi)],$ (12.8)

we see that $v$ satisfies

$\displaystyle\dot{v}(t)-D_{\infty}v(t)=g(\omega t),\quad
v(0)=v_{0}=\Phi_{2,\infty}(0)^{-1}w(0).$ (12.9)

whose solution is given by

$v(t)=v(t,x)=\sum_{j\in
S_{\mathtt{M}}^{\perp}}\widehat{v}_{j}(t)e^{\text{i}jx},\quad\widehat{v}_{j}(t):=(\widehat{v_{0}})_{j}e^{d_{\infty}(\omega,j)t}+\sum_{l\in\mathbb{Z}^{\nu}}\frac{\widehat{g}_{j}(l)}{\text{i}\omega\cdot
l-d_{\infty}(\omega,j)}e^{\text{i}\omega\cdot lt}.$

From (12.7), we have that $d_{\infty}(\omega,j)$ is purely imaginary. Hence,
the Sobolev norm of $v$ is uniformly bounded, therefore the Sobolev norm of
$w$ does not grow in time. $\Box$

##  Appendix A Conjugation with the operators
$\Lambda^{\alpha-1}_{a},\Upsilon^{\alpha-3}_{a}$

In this section, we compute the conjugation of the operators
$\Lambda^{\alpha}_{a},\Upsilon^{\alpha-3}$ with the transformation $\Psi_{1}$
in (10.8). We recall that $B(x):=x+\beta(x)$ and
$\Psi_{1}^{T}[h](x)=h(B^{-1}(x))$. We simply denote $\Psi:=\Psi_{1}$ and use
the notation from (10.11).

######  Lemma A.1.

Given $a=a(x,y)\in C^{\infty}(\mathbb{T}^{2})$, let us denote
$\tilde{a}(x):=a(x,x).$ We have that

$\displaystyle\Psi^{T}\Lambda^{\alpha-1}_{a}\Psi
h=a_{1}\Lambda^{\alpha-1}h+a_{2}h+\frac{1}{2}\partial_{x}(a_{1})\partial_{x}\Upsilon^{\alpha-3}h-\Upsilon^{\alpha-3}_{a_{3}}h,$
$\displaystyle\Psi^{T}[a\mathcal{A}[h]]=\Psi^{T}[aB^{\prime}]h,$
$\displaystyle\Psi^{T}\Upsilon^{\alpha-3}_{a}\Psi[h]=\Upsilon^{\alpha-3}_{a_{4}}h,$

where

$\displaystyle a_{1}=\Psi^{T}[(B^{\prime})^{\alpha}\tilde{a}],$ $\displaystyle
a_{2}=\Psi^{T}[\Lambda^{\alpha-1}_{a}B^{\prime}]+\int(2-2\cos(x-y))^{-\frac{\alpha}{2}}(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x))dy$
$\displaystyle
a_{3}=\frac{\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x)-\partial_{y}\mathfrak{a}_{1}(x,x)\sin(x-y)}{2-2\cos(x-y)},$
$\displaystyle\mathfrak{a}_{1}(x,y)=\left(\frac{2-2\cos(B^{-1}(x)-B^{-1}(y))}{2-2\cos(x-y)}\right)^{-\frac{\alpha}{2}}B_{2}^{-1}[a](x,y),$
$\displaystyle
a_{4}=\left(\frac{2-2\cos(B^{-1}(x)-B^{-1}(y))}{2-2\cos(x-y)}\right)^{1-\frac{\alpha}{2}}B_{2}^{-1}[a](x,y).$

###### Proof.

The conjugation $\Psi^{T}[a\Psi[h]]$ and
$\Psi^{T}\Upsilon^{\alpha-3}_{a}\Psi[h]$ follows in the same way as below.
Therefore, we compute $\Psi^{T}\Lambda^{\alpha-1}_{a}\Psi[h]$ only.

Since $\Psi^{T}\Lambda^{\alpha-1}_{a}\Psi[h](x)=\Lambda^{\alpha-1}_{a}(\Psi
h)(B^{-1}(x))$, direct computations show that

$\displaystyle\Lambda^{\alpha-1}_{a}$ $\displaystyle(\Psi h)(B^{-1}(x))$
$\displaystyle=\int(2-2\cos(B^{-1}(x)-B^{-1}(y)))^{-\frac{\alpha}{2}}B_{2}^{-1}[a](x,y)(B^{\prime}(B^{-1}(x))h(x)-B^{\prime}(B^{-1}(y))h(y))\frac{1}{B^{\prime}(B^{-1}(y))}dy$
$\displaystyle=\int(2-2\cos(B^{-1}(x)-B^{-1}(y)))^{-\frac{\alpha}{2}}B_{2}^{-1}[a](x,y)\left(\frac{B^{\prime}(B^{-1}(x))}{B^{\prime}(B^{-1}(y))}-1\right)dyh(x)$
$\displaystyle\
+\int(2-2\cos(B^{-1}(x)-B^{-1}(y)))^{-\frac{\alpha}{2}}B_{2}^{-1}[a](x,y)(h(x)-h(y))dy$
$\displaystyle=:A_{1}+A_{2}.$ (A.1)

For $A_{1}$, it follows from the change of variables $y\mapsto B(y)$ that

$\displaystyle A_{1}=\Psi^{T}[\Lambda^{\alpha-1}_{a}B^{\prime}]h.$ (A.2)

For $A_{2}$, we denote

$\displaystyle\mathfrak{a}_{1}(x,y):=\left(\frac{2-2\cos(B^{-1}(x)-B^{-1}(y))}{2-2\cos(x-y)}\right)^{-\frac{\alpha}{2}}B_{2}^{-1}[a](x,y),$
(A.3)

so that $A_{2}=\Lambda^{\alpha-1}_{\mathfrak{a}_{1}}h$. Writing

$\mathfrak{a}_{1}(x,y)=\mathfrak{a}_{1}(x,x)+\left(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x)\right),$

we have that

$\displaystyle A_{2}$
$\displaystyle=\mathfrak{a}_{1}(x,x)\Lambda^{\alpha-1}h+\int(2-2\cos(x-y))^{-\frac{\alpha}{2}}(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x))dyh(x)$
(A.4) $\displaystyle\
-\int(2-2\cos(x-y))^{-\frac{\alpha}{2}}(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x))h(y)dy.$

For the last integral, we write

$\displaystyle\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x)$
$\displaystyle=\partial_{y}\mathfrak{a}_{1}(x,x)\sin(x-y)+\left(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x)-\partial_{y}\mathfrak{a}_{1}(x,x)\sin(x-y)\right),$

and
$\mathfrak{a}_{2}(x,y):=\frac{\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x)-\partial_{y}\mathfrak{a}_{1}(x,x)\sin(x-y)}{2-2\cos(x-y)}$,
so that

$\displaystyle\int(2-2\cos(x-y))^{-\frac{\alpha}{2}}(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x))h(y)dy=\partial_{y}\mathfrak{a}_{1}(x,x)\frac{1}{2-\alpha}\partial_{x}\Upsilon^{\alpha-3}h+\Upsilon^{\alpha-3}_{\mathfrak{a}_{2}}h.$

Therefore, from (A), (A.2), (A.4), we have that

$\displaystyle\Psi^{T}\Lambda^{\alpha-1}_{a}\Psi h$
$\displaystyle=\mathfrak{a}_{1}(x,x)\Lambda^{\alpha-1}h+\left(\Psi^{T}[\Lambda^{\alpha-1}_{a}B^{\prime}]+\int(2-2\cos(x-y))^{-\frac{\alpha}{2}}(\mathfrak{a}_{1}(x,y)-\mathfrak{a}_{1}(x,x))dy\right)h$
(A.5) $\displaystyle\
-\partial_{y}\mathfrak{a}_{1}(x,x)\frac{1}{2-\alpha}\partial_{x}\Upsilon^{\alpha-3}h-\Upsilon^{\alpha-3}_{\mathfrak{a}_{2}}h.$

Direct computations from (A.3) shows that

$\displaystyle\mathfrak{a}_{1}(x,x)=\Psi^{T}[(B^{\prime})^{\alpha}\tilde{a}],\quad\text{
where $\tilde{a}(x):=a(x,x)$},$
$\displaystyle\partial_{y}\mathfrak{a}_{1}(x,x)=\frac{1}{2}\partial_{x}\tilde{\mathfrak{a}}_{1}(x),\quad\text{
where $\tilde{\mathfrak{a}}_{1}(x):=\mathfrak{a}_{1}(x,x)$}.$

Plugging this into (A.5), the result follows. ∎

##  Appendix B Analysis of the non-resonance condition

In this appendix, we aim to prove Proposition 5.5. Recall from (4.3) and (4.4)
that

$\displaystyle m^{\circ}_{1,\alpha}(j)$
$\displaystyle=\int_{\mathbb{T}}(2-2\cos(\theta))^{-\frac{\alpha}{2}}(1-e^{-\text{i}j\theta})d\theta,$
(B.1) $\displaystyle T_{\alpha}$
$\displaystyle=\int_{\mathbb{T}}(2-2\cos(\theta))^{1-\frac{\alpha}{2}},$
$\displaystyle\lambda_{\alpha}(j)$ $\displaystyle=jm_{1,\alpha}(j)$
$\displaystyle m^{\circ}_{3,\alpha}(j,k)$
$\displaystyle=m^{\circ}_{4,\alpha}(j,-j,k,-k)=\int_{\mathbb{T}}(2-2\cos
x)^{-1-\frac{\alpha}{2}}(2-2\cos(jx))(2-2\cos(kx))dx,$
$\displaystyle\kappa_{j}$
$\displaystyle=-\frac{\pi}{2}\left(m^{\circ}_{1,\alpha}(j)-\frac{T_{\alpha}}{2}\right).$

We first derive expressions of $\mathbb{A}$ and $\vec{D}(\xi)$ in terms of
$\lambda_{\alpha},m_{1,\alpha}$ and $m_{3,\alpha}$ and analyze the dominating
terms in their expressions. Lastly, we will prove Proposition 5.5.

###  B.1 Expressions of $\mathbb{A}$ and $\vec{D}(\xi)$

#####  B.1.0.1 Expression of $\mathbb{A}$.

Plugging (B.1) into (4.3), we have

$\displaystyle H_{4,j,-j,k,-k}$
$\displaystyle=-\frac{\alpha(\alpha+4)\pi}{192}(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k))-\frac{\alpha^{2}\pi}{384}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(j+k)+m^{\circ}_{1,\alpha}(j-k))$
(B.2) $\displaystyle\
+\frac{\alpha\pi}{192}m^{\circ}_{3,\alpha}(j,k)+\frac{\alpha(\alpha+2)\pi
T_{\alpha}}{192}$
$\displaystyle=-\frac{\alpha\pi}{192}\left(({\alpha+4})(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k))\right.$
$\displaystyle\left.\
+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(j+k)+m^{\circ}_{1,\alpha}(j-k))-m^{\circ}_{3,\alpha}(j,k)-(\alpha+2)T_{\alpha}\right),$
$\displaystyle H_{3,j,k,-(j+k)}$
$\displaystyle=\frac{\alpha\pi}{24}(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j+k)-T_{\alpha}),$
$\displaystyle H_{3,j,-k,-(j-k)}$
$\displaystyle=\frac{\alpha\pi}{24}(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j-k)-T_{\alpha}).$

Therefore, plugging this into (4.16), we have that if $j=k$,

$\displaystyle H^{(3)}_{4,j,-j,j,-j}$
$\displaystyle=H_{4,j,-j,j,-j}-\frac{3}{4}\frac{j(H_{3,j,j,-2j})^{2}}{j\kappa_{2j}-j\kappa_{j}}$
$\displaystyle=-\frac{\alpha\pi}{192}\left(2(\alpha+4)m^{\circ}_{1,\alpha}(j)+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(2j)+2m^{\circ}_{1,\alpha}(0))-m^{\circ}_{3,\alpha}(j,j)-(\alpha+2)T_{\alpha}\right)+\frac{\alpha^{2}\pi}{2\cdot
192}\frac{(2m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(2j)-T_{\alpha})^{2}}{m^{\circ}_{1,\alpha}(2j)-m^{\circ}_{1,\alpha}(j)}$
$\displaystyle=\frac{\alpha\pi}{192}\left(\frac{\alpha}{2}\frac{(2m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(2j)-T_{\alpha})^{2}}{m^{\circ}_{1,\alpha}(2j)-m^{\circ}_{1,\alpha}(j)}-\left(2(\alpha+4)m^{\circ}_{1,\alpha}(j)+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(2j)+2m^{\circ}_{1,\alpha}(0))-m^{\circ}_{3,\alpha}(j,j)-(\alpha+2)T_{\alpha}\right)\right).$

If $j\neq k$, we have

$\displaystyle H^{(3)}_{4,j,-j,k,-k}$
$\displaystyle=H_{4,j,-j,k,-k}-\frac{3}{4}\left(\frac{(j+k)(H_{3,j,k,-(j+k)})^{2}}{(j+k)\kappa_{j+k}-j\kappa_{j}-k\kappa_{k}}+\frac{(j-k)(H_{3,j,-k,-(j-k)})^{2}}{(j-k)\kappa_{j-k}-j\kappa_{j}+k\kappa_{k}}\right)$
$\displaystyle=-\frac{\alpha\pi}{192}\left(({\alpha+4})(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k))+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(j+k)+m^{\circ}_{1,\alpha}(j-k))-m^{\circ}_{3,\alpha}(j,k)-(\alpha+2)T_{\alpha}\right)$
$\displaystyle\ +\frac{\alpha^{2}\pi}{2\cdot
192}\left(\frac{(j+k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j+k)-T_{\alpha})^{2}}{(j+k)m_{1,\alpha}(j+k)-jm^{\circ}_{1,\alpha}(j)-km^{\circ}_{1,\alpha}(j)}+\frac{(j-k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j-k)-T_{\alpha})^{2}}{(j-k)m^{\circ}_{1,\alpha}(j-k)-jm^{\circ}_{1,\alpha}(j)+km^{\circ}_{1,\alpha}(j)}\right)$
$\displaystyle=\frac{\alpha\pi}{192}\left(\frac{\alpha}{2}\left(\frac{(j+k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j+k)-T_{\alpha})^{2}}{(j+k)m^{\circ}_{1,\alpha}(j+k)-jm^{\circ}_{1,\alpha}(j)-km^{\circ}_{1,\alpha}(j)}+\frac{(j-k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j-k)-T_{\alpha})^{2}}{(j-k)m^{\circ}_{1,\alpha}(j-k)-jm^{\circ}_{1,\alpha}(j)+km^{\circ}_{1,\alpha}(j)}\right)\right.$
$\displaystyle\left.-\left(({\alpha+4})(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k))+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(j+k)+m^{\circ}_{1,\alpha}(j-k))-m^{\circ}_{3,\alpha}(j,k)-(\alpha+2)T_{\alpha}\right)\right)$

Therefore, it follows from (5.10) that for
$i,k\in\left\\{1,\ldots,\nu\right\\}$ and the tangential sites in (4.6),
$S^{+}=\left\\{0<j_{1}<\cdots<j_{\nu}\right\\}$,

$\displaystyle\mathbb{A}^{i}_{k}$
$\displaystyle=\frac{\alpha\pi}{192}A(j_{i},j_{k}),\text{ for
$i,k=1,\ldots,\nu$}$ (B.3)

where

$\displaystyle
A(j,j)=j^{2}\left(\frac{\alpha}{2}\frac{(2m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(2j)-T_{\alpha})^{2}}{m^{\circ}_{1,\alpha}(2j)-m^{\circ}_{1,\alpha}(j)}-\left(2(\alpha+4)m^{\circ}_{1,\alpha}(j)+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(2j)+2m^{\circ}_{1,\alpha}(0))-m^{\circ}_{3,\alpha}(j,j)-(\alpha+2)T_{\alpha}\right)\right),$

and for $j\neq k$,

$\displaystyle A(j,k)$
$\displaystyle=2jk\left(\frac{\alpha}{2}\left(\frac{(j+k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j+k)-T_{\alpha})^{2}}{(j+k)m^{\circ}_{1,\alpha}(j+k)-jm^{\circ}_{1,\alpha}(j)-km^{\circ}_{1,\alpha}(k)}+\frac{(j-k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j-k)-T_{\alpha})^{2}}{(j-k)m^{\circ}_{1,\alpha}(j-k)-jm^{\circ}_{1,\alpha}(j)+km^{\circ}_{1,\alpha}(k)}\right)\right.$
$\displaystyle\left.-\left(({\alpha+4})(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k))+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(j+k)+m^{\circ}_{1,\alpha}(j-k))-m^{\circ}_{3,\alpha}(j,k)-(\alpha+2)T_{\alpha}\right)\right).$

To simplify the notation, let us denote

$\displaystyle\mathtt{B}_{1}(j,k)$
$\displaystyle:=(\alpha+4)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k))+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(j+k)+m^{\circ}_{1,\alpha}(j-k))-(\alpha+2)T_{\alpha},$
(B.4) $\displaystyle\mathtt{B}_{2}(j,k)$
$\displaystyle:=\frac{(j+k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j+k)-T_{\alpha})^{2}}{(j+k)m^{\circ}_{1,\alpha}(j+k)-jm^{\circ}_{1,\alpha}(j)-km^{\circ}_{1,\alpha}(k)}-\frac{(j-k)(m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(k)+m^{\circ}_{1,\alpha}(j-k)-T_{\alpha})^{2}}{jm^{\circ}_{1,\alpha}(j)-(j-k)m^{\circ}_{1,\alpha}(j-k)-km^{\circ}_{1,\alpha}(k)},$
(B.5) $\displaystyle\mathtt{B}_{3}(j)$
$\displaystyle:=\frac{\alpha}{2}\frac{(2m^{\circ}_{1,\alpha}(j)+m^{\circ}_{1,\alpha}(2j)-T_{\alpha})^{2}}{m^{\circ}_{1,\alpha}(2j)-m^{\circ}_{1,\alpha}(j)},$
(B.6)

so that we can write $A$ as

$\displaystyle
A(j,k)=\begin{cases}j^{2}\mathtt{B}_{3}(j)-j^{2}\mathtt{B}_{1}(j,j)+j^{2}m^{\circ}_{3,\alpha}(j,j),&\text{
if $j=k$},\\\
2jk\left(\frac{\alpha}{2}\mathtt{B}_{2}(j,k)-\mathtt{B}_{1}(j,k)\right)+2jkm_{3,\alpha}^{\circ}(j,k),&\text{
if $j\neq k$.}\end{cases}$

Therefore, defining

$\displaystyle A_{1}(j,k):=\begin{cases}j^{2}m^{\circ}_{3,\alpha}(j,j),&\text{
if $j=k$,}\\\ 2jkm^{\circ}_{3,\alpha}(j,k)&\text{ if $j\neq
k$},\end{cases}\quad
A_{2}(j,k)=\begin{cases}j^{2}\mathtt{B}_{3}(j)-j^{2}\mathtt{B}_{1}(j,j)&\text{
if $j=k$},\\\
2jk\left(\frac{\alpha}{2}\mathtt{B}_{2}(j,k)-\mathtt{B}_{1}(j,k)\right),&\text{
if $j\neq k$,}\end{cases}$ (B.7)

we have $A(j,k)=A_{1}(j,k)+A_{2}(j,k)$. Therefore (B.3) gives that

$\displaystyle\mathbb{A}^{i}_{k}$
$\displaystyle=\frac{\alpha\pi}{192}A_{1}(j_{i},j_{k})+\frac{\alpha\pi}{192}A_{2}(j_{i},j_{k})$
(B.8)
$\displaystyle=:\frac{\alpha\pi}{192}\left(\mathbb{A}_{1}\right)^{i}_{k}+\frac{\alpha\pi}{192}\left(\mathbb{A}_{2}\right)^{i}_{k},\text{
for $i,k\in\left\\{1,\ldots,\nu\right\\}$ for
$S^{+}=\left\\{0<j_{1}<\cdots<j_{\nu}\right\\}$.}$

Note that using the evenness of $j\mapsto m^{\circ}_{1,\alpha}(j)$ and
$m_{3,\alpha}(j,k)=m_{3,\alpha}(k,j)$, which easily follow from (B.1), we see
that

$\displaystyle\mathbb{A}^{T}=\mathbb{A},\quad\mathbb{A}_{1}^{T}=\mathbb{A}_{1},\quad\mathbb{A}_{2}^{T}=\mathbb{A}_{2}.$
(B.9)

#####  B.1.0.2 Expression of $\vec{D}(\xi)$.

Using (B.2), we have that for $\xi\in S^{\perp}$ and $j_{k}\in
S^{+}=\left\\{0<j_{1}<\cdots<j_{\nu}\right\\}$ for $k=1,\ldots,\nu$, we have

$\displaystyle\frac{12}{\pi}H_{4,j_{k},-j_{k},\xi,-\xi}$
$\displaystyle=-\frac{\alpha}{16}\left(({\alpha+4})(m^{\circ}_{1,\alpha}(\xi)+m^{\circ}_{1,\alpha}(j_{k}))+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(\xi+j_{k})+m^{\circ}_{1,\alpha}(\xi-
j_{k}))-m^{\circ}_{3,\alpha}(\xi,j_{k})-(\alpha+2)T_{\alpha}\right),$

and using $\lambda_{\alpha}(\xi)=\xi m^{\circ}_{1,\alpha}(\xi)$ for all
$\xi\in\mathbb{Z}\backslash\left\\{0\right\\}$ (see (2.98) and (2.97)),

$\displaystyle\frac{18}{\pi^{2}}$
$\displaystyle\frac{(\xi+j_{k})(H_{3,j_{k},\xi,-(\xi+j_{k})})^{2}}{\lambda_{\alpha}(\xi+j_{k})-\lambda_{\alpha}(\xi)-\lambda_{\alpha}(j_{k})}=\frac{\alpha^{2}}{32}\frac{(\xi+j_{k})(m^{\circ}_{1,\alpha}(\xi)+m^{\circ}_{1,\alpha}(j_{k})+m^{\circ}_{1,\alpha}(\xi+j_{k})-T_{\alpha})^{2}}{(\xi+j_{k})m_{1,\alpha}^{\circ}-\xi
m_{1,\alpha}^{\circ}(\xi)-j_{k}m^{\circ}_{1,\alpha}(j_{k})},$
$\displaystyle\frac{18}{\pi^{2}}$ $\displaystyle\frac{(\xi-
j_{k})(H_{3,j_{k},\xi,-(\xi-
j_{k})})^{2}}{\lambda_{\alpha}(\xi)-\lambda_{\alpha}(\xi-
j_{k})-\lambda_{\alpha}(j_{k})}=\frac{\alpha^{2}}{32}\frac{(\xi-
j_{k})(m^{\circ}_{1,\alpha}(\xi)+m^{\circ}_{1,\alpha}(j_{k})+m^{\circ}_{1,\alpha}(\xi-
j_{k})-T_{\alpha})^{2}}{\xi m^{\circ}_{1,\alpha}(\xi)-(\xi-
j_{k})m^{\circ}_{1,\alpha}(\xi-j_{k})-j_{k}m^{\circ}_{1,\alpha}(j_{k})}.$

Plugging this into (5.1.0.2), we have that for $k=1,\ldots,\nu$,
$S^{+}=\left\\{0<j_{1}<\cdots<j_{\nu}\right\\}$ and $\xi\in S^{\perp}$,

$\displaystyle(\vec{D}(\xi))_{k}$
$\displaystyle=\frac{\alpha}{16}m^{\circ}_{3,\alpha}(\xi,j_{k})j_{k},$
$\displaystyle-
j_{k}\frac{\alpha}{16}\left(({\alpha+4})(m^{\circ}_{1,\alpha}(\xi)+m^{\circ}_{1,\alpha}(j_{k}))+\frac{\alpha}{2}(m^{\circ}_{1,\alpha}(0)+m^{\circ}_{1,\alpha}(\xi+j_{k})+m^{\circ}_{1,\alpha}(\xi-
j_{k}))-(\alpha+2)T_{\alpha}\right)$
$\displaystyle+\frac{\alpha^{2}}{32}j_{k}\left(\frac{(\xi+j_{k})(m^{\circ}_{1,\alpha}(\xi)+m_{1,\alpha}(j_{k})+m^{\circ}_{1,\alpha}(\xi+j_{k})-T_{\alpha})^{2}}{(\xi+j_{k})m_{1,\alpha}^{\circ}-\xi
m_{1,\alpha}^{\circ}(\xi)-j_{k}m^{\circ}_{1,\alpha}(j_{k})}-\frac{(\xi-
j_{k})(m^{\circ}_{1,\alpha}(\xi)+m^{\circ}_{1,\alpha}(j_{k})+m^{\circ}_{1,\alpha}(\xi-
j_{k})-T_{\alpha})^{2}}{\xi m^{\circ}_{1,\alpha}(\xi)-(\xi-
j_{k})m^{\circ}_{1,\alpha}(\xi-
j_{k})-j_{k}m^{\circ}_{1,\alpha}(j_{k})}\right).$

Hence, using the notation in (B.4) and (B.5), we write $\vec{D}(\xi)_{k}$ as

$\displaystyle(\vec{D}(\xi))_{k}=\begin{cases}\frac{\alpha}{16}m^{\circ}_{3,\alpha}(\xi,j_{k})j_{k}-\frac{\alpha}{16}j_{k}\mathtt{B}_{1}(\xi,j_{k})+\frac{\alpha^{2}}{32}j_{k}\mathtt{B}_{2}(\xi,j_{k}),&\text{
if $\xi\in S^{\perp}$,}\\\ 0,&\text{ if $\xi=0$},\end{cases}$ (B.10)

for $k=1,\ldots,\nu$, $S^{+}=\left\\{0<j_{1}<\cdots<j_{\nu}\right\\}$.

Note that using the evenness of $j\mapsto m^{\circ}_{1,\alpha}(j)$ and
$m_{3,\alpha}(j,k)=m_{3,\alpha}(k,j)$, which easily follow from (B.1), we see
that

$\displaystyle\vec{D}(-\xi)=\vec{D}(\xi).$ (B.11)

###  B.2 Estimates for $\mathtt{B}_{1},\mathtt{B}_{2},\mathtt{B}_{3}$ and
$m^{\circ}_{3,\alpha}$

From (B.8) and (B.10) we see that $\mathbb{A}$ and $\vec{D}(\xi)$ are
completely determined by the choice of the tangential sites $S^{+}$ in (4.6),
that is, by the choice of $S_{0}^{+}$ and $\mathtt{M}$. In order to reduce the
complexity of the computation, we will take $\mathtt{M}$ large so that the
nature of $\mathbb{A}$ and $\vec{D}(\xi)$ is determined by the asymptotic
behavior of $m^{\circ}_{1,\alpha}$ and $m_{3,\alpha}^{\circ}$. To do so, we
first prove the necessary estimates for $\mathtt{B}_{1},\mathtt{B}_{1}$ and
$m^{\circ}_{3,\alpha}$ in this subsection.

######  Lemma B.1.

For $\alpha\in(1,2)$ and
$j,j_{1},j_{2},k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that
$j_{1}\neq j_{2},\ |j|,|j_{1}|,|j_{2}|\neq|k|$, we have that

$\displaystyle|m_{1,\alpha}(j+k)-m_{1,\alpha}(j)|\leq_{\alpha}|k||j|^{\alpha-2},$
(B.12)
$\displaystyle|2\lambda_{\alpha}(j)-\lambda_{\alpha}(j+k)-\lambda_{\alpha}(j-k)|$
$\displaystyle\leq_{\alpha}|k|^{2}|j|^{\alpha-2},$ (B.13)

and

$\displaystyle|m_{1,\alpha}(j_{1}+k)-m_{1,\alpha}(j_{2}+k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2},$ (B.14)
$\displaystyle|(m_{1,\alpha}(j_{1}+k)-m_{1,\alpha}(j_{1}))$
$\displaystyle-(m_{1,\alpha}(j_{2}+k)-m_{1,\alpha}(j_{2}))|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-3}|k|,$
(B.15)
$\displaystyle|(2\lambda_{\alpha}(j_{1})-\lambda_{\alpha}(j_{1}+k)-\lambda_{\alpha}(j_{1}-k))$
$\displaystyle-(2\lambda_{\alpha}(j_{2})-\lambda_{\alpha}(j_{2}+k)-\lambda_{\alpha}(j_{2}-k))|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-3}|k|^{2}.$
(B.16)

###### Proof.

Since the proof for each estimate is similar, we will prove (B.16) only. Let
us denote

$T:=|(2\lambda_{\alpha}(j_{1})-\lambda_{\alpha}(j_{1}+k)-\lambda_{\alpha}(j_{1}-k))-(2\lambda_{\alpha}(j_{2})-\lambda_{\alpha}(j_{2}+k)-\lambda_{\alpha}(j_{2}-k))|.$

We claim that

$\displaystyle
T\leq_{\alpha}|j_{1}-j_{2}|(|j_{1}|^{\alpha-3}+|j_{2}|^{\alpha-3})|k|^{2}.$
(B.17)

Let us assume the claim for a moment and derive the estimate (B.16). If
$|j_{1}|\geq 2|j_{2}|$, then it follows from (B.13) that

$T\leq_{\alpha}|k|^{2}|j_{1}|^{\alpha-2},$

while,

$|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-3}|k|^{2}\geq_{\alpha}|j_{1}|^{\alpha-2}|k|^{2},$

therefore, we obtain (B.16). Similarly, we have (B.16) in case
$|j_{1}|\leq\frac{1}{2}|j_{2}|$. If $\frac{1}{2}|j_{2}|\leq|j_{1}|\leq
2|j_{2}|$, then clearly the estimate (B.17) is equivalent to (B.16).
Therefore, it suffices to prove (B.17) to obtain (B.16).

Towards (B.17), we see from Proposition 2.34 that
$\xi\mapsto\lambda_{\alpha}(\xi)$ is odd. Therefore without loss of
generality, we can assume that

$\displaystyle j_{1}>0,\ |j_{1}|>|j_{2}|,\ k>0,$ (B.18)

since the estimate for this case can cover the all possible cases of
$j_{1},j_{2},k$ in the estimate (B.13). We further decompose into more cases:
$j_{1}>0>j_{2}$ and $j_{1}>j_{2}>0$:

Case $j_{1}>0>j_{2}$: In this case, (B.13) and the triangular inequality yield
that

$T\leq|k|^{2}(|j_{1}|^{\alpha-2}+|j_{2}|^{\alpha-2})\overset{\eqref{amspsdjxx1}}{\leq}2|j_{1}|^{\alpha-2}|k|^{2}\overset{j_{1}>0>j_{2}}{\leq}_{\alpha}|j_{1}-j_{2}||j_{1}|^{\alpha-3}|k|^{2},$

which gives (B.17).

For the case $j_{1}>j_{2}>0$, we divide it into three cases:
$0<j_{2}<j_{1}\leq 2k,\ 0<j_{2}\leq 2k<j_{1},\ 0<2k<j_{2}<j_{1}$.

Case $0<j_{2}<j_{1}\leq 2k$: In this case, we have

$T=\left|\int_{j_{2}}^{j_{1}}2\partial_{\xi}\lambda_{\alpha}(\xi)-\partial_{\xi}\lambda_{\alpha}(\xi+k)-\partial_{\xi}\lambda_{\alpha}(\xi-k)d\xi\right|,$

while each term of the integrand is bounded by
$C_{\alpha}(|\xi|^{\alpha-1}+|k|^{\alpha-1})\leq_{\alpha}|k|^{\alpha-1}$.
Therefore, we have $T\leq|j_{1}-j_{2}||k|^{\alpha-1}$. Using that
$|k|^{\alpha-1}=|k|^{\alpha-3}|k|\leq_{\alpha}|j_{1}|^{\alpha-3}|k|$, the
estimate (B.17) follows.

Case $0<j_{2}\leq 2k<j_{1}$: We compute

$\displaystyle T$
$\displaystyle=\left|\int_{j_{2}}^{j_{1}}2\partial_{\xi}\lambda_{\alpha}(\xi)-\partial_{\xi}\lambda_{\alpha}(\xi+k)-\partial_{\xi}\lambda_{\alpha}(\xi-k)d\xi\right|$
$\displaystyle\leq\left|\int_{j_{2}}^{2k}2\partial_{\xi}\lambda_{\alpha}(\xi)-\partial_{\xi}\lambda_{\alpha}(\xi+k)-\partial_{\xi}\lambda_{\alpha}(\xi-k)d\xi\right|+\left|\int_{2k}^{j_{1}}2\partial_{\xi}\lambda_{\alpha}(\xi)-\partial_{\xi}\lambda_{\alpha}(\xi+k)-\partial_{\xi}\lambda_{\alpha}(\xi-k)d\xi\right|$
$\displaystyle\leq_{\alpha}|2k-j_{2}||k|^{\alpha-1}+\left|\int_{2k}^{j_{1}}\int_{0}^{k}\int_{-t}^{t}\partial_{\xi\xi\xi}\lambda_{\alpha}(\xi+u)dudtd\xi\right|$
$\displaystyle\leq_{\alpha}|2k-j_{2}||k|^{\alpha-1}+\left|\int_{2k}^{j_{1}}\int_{0}^{k}\int_{-t}^{t}|\xi|^{\alpha-3}dudtd\xi\right|$
$\displaystyle\leq_{\alpha}|2k-j_{2}||k|^{\alpha-1}+|j_{1}-2k||k|^{\alpha-1}$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}||k|^{\alpha-1}.$

Using that $0<j_{2}<2k$, we have
$|j_{1}-j_{2}||k|^{\alpha-1}\leq_{\alpha}|j_{1}-j_{2}||j_{2}|^{\alpha-3}|k|^{2}$,
therefore (B.17) follows.

Case $0<2k<j_{2}<j_{1}$: In this case, we have

$T=\left|\int_{j_{2}}^{j_{1}}\int_{0}^{k}\int_{-t}^{t}\partial_{\xi\xi\xi}\lambda\alpha(\xi+u)dudtd\xi\right|\leq_{\alpha}\int_{j_{2}}^{j_{1}}\int_{0}^{k}\int_{-t}^{t}\langle\xi+u\rangle^{\alpha-3}dudtd\xi\leq_{\alpha}|j_{1}-j_{2}||j_{2}|^{\alpha-3}|k|^{2},$

therefore we obtain (B.17). ∎

######  Lemma B.2.

For $\alpha\in(1,2)$ and
$j,j_{1},j_{2},k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that
$j_{1}\neq j_{2},\ |j|,|j_{1}|,|j_{2}|\neq|k|$, we have that

$\displaystyle|\mathtt{B}_{1}(j,k)|$
$\displaystyle\leq_{\alpha}(|j|^{\alpha-1}+|k|^{\alpha-1}),\text{ for all
$j,k\in\mathbb{Z},k\neq 0$}$ (B.19)
$\displaystyle|\mathtt{B}_{1}(j_{1},k)-\mathtt{B}_{1}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}.$ (B.20)

###### Proof.

In the expression of $\mathtt{B}_{1}$ in (B.4), we can replace
$m^{\circ}_{1,\alpha}(j,k)$ by $m_{1,\alpha}$ since $j,k\neq 0$. Then the
estimates (B.19) and (B.20) follow immediately from the fact that $\xi\mapsto
m_{1,\alpha}(\xi)\in\mathcal{S}^{\alpha-1}$ and the estimate (B.14). ∎

Now, we estimate $\mathtt{B}_{2}$ We further decompose $\mathtt{B}_{2}$ as

$\displaystyle\mathtt{B}_{2}(j,k)$
$\displaystyle=\mathtt{B}_{21}(j,k)+\mathtt{B}_{22}(j,k),$ (B.21)
$\displaystyle\mathtt{B}_{21}(j,k)$
$\displaystyle=\frac{(j+k)(m_{1,\alpha}(j)+m_{1,\alpha}(k)+m_{1,\alpha}(j+k)-T_{\alpha})^{2}-(j-k)(m_{1,\alpha}(j)+m_{1,\alpha}(k)+m_{1,\alpha}(j-k)-T_{\alpha})^{2}}{\lambda^{\circ}_{\alpha}(j+k)-\lambda^{\circ}_{\alpha}(j)-\lambda^{\circ}_{\alpha}(k)},$
$\displaystyle=:\frac{\mathtt{B}_{21}^{N}(j,k)}{\mathtt{B}^{D}_{21}(j,k)}$
(B.22) $\displaystyle\mathtt{B}_{22}(j,k)$
$\displaystyle=\frac{(j-k)(m_{1,\alpha}(j)+m_{1,\alpha}(k)+m_{1,\alpha}(j-k)-T_{\alpha})^{2}(2\lambda_{\alpha}^{\circ}(j)-\lambda_{\alpha}^{\circ}(j-k)-\lambda_{\alpha}^{\circ}(j+k))}{(\lambda^{\circ}_{\alpha}(j+k)-\lambda^{\circ}_{\alpha}(j)-\lambda^{\circ}_{\alpha}(k))(\lambda^{\circ}_{\alpha}(j)-\lambda^{\circ}_{\alpha}(j-k)-\lambda^{\circ}_{\alpha}(k))}$
$\displaystyle=:\frac{\mathtt{B}^{N}_{22}(j,k)}{\mathtt{B}^{D}_{22}(j,k)}.$
(B.23)

######  Lemma B.3.

For $\alpha\in(1,2)$ and
$j,j_{1},j_{2},k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that
$|j|,|j_{1}|,|j_{2}|\neq|k|$, it holds that

$\displaystyle|\mathtt{B}^{N}_{21}(j,k)|$
$\displaystyle\leq_{\alpha}|k|(|j|^{2\alpha-2}+|k|^{2\alpha-2}),$ (B.24)
$\displaystyle|\mathtt{B}^{N}_{21}(j_{1},k)-\mathtt{B}^{N}_{21}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|k||j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}(|j_{1}|^{\alpha-1}+|j_{2}|^{\alpha-1}+|k|^{\alpha-1}),$
(B.25) $\displaystyle|\mathtt{B}^{N}_{22}(j,k)|$
$\displaystyle\leq_{\alpha}|k|^{2}|j|^{\alpha-2}(|j|^{2\alpha-1}+|k|^{2\alpha-1}),$
(B.26)
$\displaystyle|\mathtt{B}^{N}_{22}(j_{1},k)-\mathtt{B}^{N}_{22}(j_{1},k)|$
$\displaystyle\leq_{\alpha}|k|^{2}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|+|k|)^{2\alpha-1}(|j_{1}|+|j_{2}|)^{\alpha-3},$
(B.27) $\displaystyle|\mathtt{B}^{D}_{21}(j,k)|$
$\displaystyle\leq_{\alpha}|k|(|j|^{\alpha-1}+|k|^{\alpha-1}),$ (B.28)
$\displaystyle|\mathtt{B}^{D}_{21}(j_{1},k)-\mathtt{B}^{D}_{21}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}||k|(|j_{1}|+|j_{2}|)^{\alpha-2},$
(B.29) $\displaystyle|\mathtt{B}^{D}_{22}(j,k)|$
$\displaystyle\leq_{\alpha}|k|^{2}(|j|+|k|)^{2\alpha-2},$ (B.30)
$\displaystyle|\mathtt{B}^{D}_{22}(j_{1},k)-\mathtt{B}^{D}_{22}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}||k|^{2}(|j_{1}|+|j_{2}|)^{\alpha-2}(|j_{1}|+|j_{2}|+|k|)^{\alpha-1}.$
(B.31)

###### Proof.

Proof of (B.24), (B.25). Applying (B.12) and (B.15) to

$\displaystyle
m_{1,\alpha}(j+k)-m_{1,\alpha}(j-k)=(m_{1,\alpha}(j+k)-m_{1,\alpha}(j))+(m_{1,\alpha}(j)-m_{1,\alpha}(j-k)),$

we see that

$\displaystyle|m_{1,\alpha}(j+k)-m_{1,\alpha}(j-k)|$
$\displaystyle\leq_{\alpha}|k||j|^{\alpha-2},$ (B.32)
$\displaystyle|(m_{1,\alpha}(j_{1}+k)-m_{1,\alpha}(j_{1}-k))-(m_{1,\alpha}(j_{2}+k)-m_{1,\alpha}(j_{2}-k))|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-3}|k|.$
(B.33)

Let

$\displaystyle L_{1}(j,k)$
$\displaystyle:=(j+k)(m_{1,\alpha}(j)+m_{1,\alpha}(k)+m_{1,\alpha}(j+k)-T_{\alpha})^{2}$
(B.34) $\displaystyle L_{2}(j,k)$
$\displaystyle:=m_{1,\alpha}(j)+m_{1,\alpha}(k)-T_{\alpha}.$

Since $\xi\mapsto m_{1,\alpha}(\xi)\in\mathcal{S}^{\alpha-1}$ is even, we have

$\displaystyle|L_{1}(j,k)|$
$\displaystyle\leq_{\alpha}|j|^{2\alpha-1}+|k|^{2\alpha-1},$ (B.35)
$\displaystyle|L_{2}(j,k)|$
$\displaystyle\leq_{\alpha}|j|^{\alpha-1}+|k|^{\alpha-1},$ (B.36)
$\displaystyle|L_{2}(j,k)-L_{2}(j,-k)|$ $\displaystyle=0,$ (B.37)
$\displaystyle|L_{2}(j_{1},k)-L_{2}(j_{2},k)|$
$\displaystyle\overset{\eqref{m1sess234}}{\leq_{\alpha}}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}.$
(B.38)

Since $\mathtt{B}^{N}_{21}(j,k)=L_{1}(j,k)-L_{1}(j,-k)$, we expand

$\displaystyle\mathtt{B}^{N}_{21}(j,k)$
$\displaystyle=(j+k)(L_{2}(j,k)+m_{1,\alpha}(j+k))^{2}-(j-k)(L_{2}(j,k)+m_{1,\alpha}(j-k))^{2}$
$\displaystyle=j((L_{2}(j,k)+m_{1,\alpha}(j+k))^{2}-(L_{2}(j,k)+m_{1,\alpha}(j-k))^{2})$
$\displaystyle\
+k((L_{2}(j,k)+m_{1,\alpha}(j+k))^{2}+(L_{2}(j,k)+m_{1,\alpha}(j-k))^{2})$
$\displaystyle=j\underbrace{(2L_{2}(j,k)+m_{1,\alpha}(j+k)+m_{1,\alpha}(j-k))}_{=:L_{3}(j,k)}\underbrace{(m_{1,\alpha}(j+k)-m_{1,\alpha}(j-k))}_{=:L_{4}(j,k)}$
$\displaystyle\
+2kL_{2}(j,k)^{2}+\underbrace{2kL_{2}(j,k)(m_{1,\alpha}(j+k)+m_{1,\alpha}(j-k))}_{L_{5}(j,k)}+\underbrace{k(m_{1,\alpha}(j+k)^{2}+m_{1,\alpha}(j-k)^{2})}_{=:L_{6}(j,k)}.$
(B.39)

We have that

$\displaystyle|L_{3}(j,k)|$
$\displaystyle\overset{\eqref{l1eess111s2}}{\leq_{\alpha}}|j|^{\alpha-1}+|k|^{\alpha-1},$
(B.40) $\displaystyle|L_{3}(j_{1},k)-L_{3}(j_{2},k)|$
$\displaystyle\overset{\eqref{l1eess111s4},\eqref{m1sess234}}{\leq_{\alpha}}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2},$
(B.41) $\displaystyle|L_{4}(j,k)|$
$\displaystyle\overset{\eqref{m1sess1}}{\leq_{\alpha}}|k||j|^{\alpha-2},$
(B.42) $\displaystyle|L_{4}(j_{1},k)-L_{4}(j_{2},k)|$
$\displaystyle\overset{\eqref{m1sess2}}{\leq_{\alpha}}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-3}|k|$
(B.43) $\displaystyle|L_{5}(j,k)|$
$\displaystyle\overset{\eqref{l1eess111s2}}{\leq_{\alpha}}|k|(|j|^{2\alpha-2}+|k|^{2\alpha-2}),$
(B.44) $\displaystyle|L_{5}(j_{1},k)-L_{5}(j_{2},k)|$
$\displaystyle\overset{\eqref{l1eess111s4}\eqref{m1sess234}}{\leq_{\alpha}}|k||j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}(|j_{1}|^{\alpha-1}+|k|^{\alpha-1}),$
(B.45) $\displaystyle|L_{6}(j,k)|$
$\displaystyle\leq_{\alpha}|k|(|j|^{2\alpha-2}+|k|^{2\alpha-2}),$ (B.46)
$\displaystyle|L_{6}(j_{1},k)-L_{6}(j_{2},k)|$
$\displaystyle\overset{\eqref{m1sess234}}{\leq_{\alpha}}|k|(|j_{1}|^{\alpha-1}+|j_{2}|^{\alpha-1}+|k|^{\alpha-1})|j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}.$
(B.47)

Therefore,

$\displaystyle|\mathtt{B}^{N}_{21}(j,k)|$
$\displaystyle\overset{\eqref{rksdsdssxxxb21}}{=}|j||L_{3}(j,k)||L_{4}(j,k)|+2k|L_{2}(j,k)|^{2}+|L_{5}(j,k)|+|L_{6}(j,k)|$
$\displaystyle\overset{\eqref{qkfrjfmadl1},\eqref{qkfrjfmadl4},\eqref{qkfrjfmadl5},\eqref{qkfrjfmadl7},\eqref{l1eess111s2}}{\leq_{\alpha}}|k|(|j|^{2\alpha-2}+|k|^{\alpha-1}|j|^{\alpha-1}+|k|^{2\alpha-2})$
$\displaystyle\leq_{\alpha}|k|(|j|^{2\alpha-2}+|k|^{2\alpha-2}),$

where the last inequality follows from Young’s inequality. Therefore, the
estimate (B.24) follows. Towards the estimate (B.25), we have that

$\displaystyle|\mathtt{B}^{N}_{21}(j_{1},k)-\mathtt{B}^{N}_{21}(j_{2},k)|$
$\displaystyle\overset{\eqref{rksdsdssxxxb21}}{\leq}|j_{1}-j_{2}||L_{3}(j_{1},k)||L_{4}(j_{1},k)|$
$\displaystyle\ +|j_{2}||L_{3}(j_{1},k)-L_{3}(j_{2},k)||L_{4}(j_{1},k)|$
$\displaystyle\ +|j_{2}||L_{3}(j_{2},k)||L_{4}(j_{1},k)-L_{4}(j_{2},k)|$
$\displaystyle\
+|L_{2}(j_{1},k)+L_{2}(j_{2},k)||L_{2}(j_{1},k)-L_{2}(j_{2},k)|$
$\displaystyle\
+|L_{5}(j_{1},k)-L_{5}(j_{2},k)|+|L_{6}(j_{1},k)-L_{6}(j_{2},k)|$
$\displaystyle\overset{\eqref{l1eess111s2},\eqref{qkfrjfmadl1}-\eqref{qkfrjfmadl8}}{\leq_{\alpha}}|k||j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}(|j_{1}|^{\alpha-1}+|j_{2}|^{\alpha-1}+|k|^{\alpha-1}),$

which proves (B.25).

Proof of (B.26), (B.27). Let

$\displaystyle
L_{7}(j,k):=2\lambda_{\alpha}^{\circ}(j)-\lambda_{\alpha}^{\circ}(j-k)-\lambda_{\alpha}^{\circ}(j+k).$
(B.48)

From (B.23) and (B.34), we have

$\displaystyle\mathtt{B}^{N}_{22}(j,k)=L_{1}(j,-k)L_{7}(j,k),$ (B.49)

while it holds that

$\displaystyle|L_{1}(j_{1},k)-L_{1}(j_{2},k)|$
$\displaystyle\overset{\eqref{l1ssd11whatsapp}}{\leq}|j_{1}-j_{2}||L_{2}(j_{1},k)+m_{1,\alpha}(j_{1}+k)|^{2}$
$\displaystyle\
+|j_{2}+k|\left(L_{2}(j_{1},k)+L_{2}(j_{2},k)+m_{1,\alpha}(j_{1}+k)+m_{1,\alpha}(j_{2}+k)\right)$
$\displaystyle\times\left(L_{2}(j_{1},k)-L_{2}(j_{2},k)+m_{1,\alpha}(j_{1}+k)-m_{1,\alpha}(j_{2}+k)\right)$
$\displaystyle\overset{\eqref{l1eess111s2},\eqref{l1eess111s4},\eqref{m1sess234}}{\leq_{\alpha}}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|+|k|)^{2\alpha-2}.$
(B.50)

From (B.49), we have

$\displaystyle|\mathtt{B}^{N}_{22}(j,k)|{\leq_{\alpha}}|L_{1}(j,-k)||L_{7}(j,k)|\overset{\eqref{l1eess111s1},\eqref{m1sess3}}{\leq_{\alpha}}|k|^{2}|j|^{\alpha-2}(|j|^{2\alpha-1}+|k|^{2\alpha-1}),$
(B.51)

which gives (B.26). Towards the estimate (B.27), let us assume, without loss
of generality, that $|j_{1}|\geq|j_{2}|$. Then we have that

$\displaystyle|\mathtt{B}^{N}_{22}(j_{1},k)-\mathtt{B}^{N}_{22}(j_{2},k)|$
$\displaystyle\leq|L_{1}(j_{1},-k)-L_{1}(j_{2},-k)||L_{7}(j_{1},k)|+|L_{1}(j_{2},-k)||L_{7}(j_{1},k)-L_{7}(j_{2},k)|$
$\displaystyle\overset{\eqref{rllxxxcxcxcxc1},\eqref{l1eess111s1},\eqref{m1sess3},\eqref{m2sess3}}{\leq_{\alpha}}|k|^{2}|j_{1}-j_{2}|$
$\displaystyle\times(|j_{1}|^{\alpha-2}(|j_{1}|+|j_{2}|+|k|)^{2\alpha-2}+(|j_{1}|+|j_{2}|+|k|)^{2\alpha-1}(|j_{1}|+|j_{2}|)^{\alpha-3})$
$\displaystyle\leq_{\alpha}|k|^{2}|j_{1}-j_{2}|(|j_{1}|+|j_{2}|+|k|)^{2\alpha-1}((|j_{1}|+|j_{2}|)^{\alpha-3}+|j_{1}|^{\alpha-3}).$
(B.52)

Since $j_{1},j_{2}\neq 0$, and $|j_{1}|\geq|j_{2}|$, we have
$|j_{1}|\geq_{\alpha}|j_{1}|+|j_{2}|$, therefore, (B.2) implies (B.27).

Proof of (B.28), (B.29). From (B.22), (2.76), and (2.97), we see that

$\displaystyle\mathtt{B}^{N}_{22}(j,k)=\underbrace{j(m_{1,\alpha}(j+k)-m_{1,\alpha}(j))}_{=:L_{8}(j,k)}+\underbrace{k(m_{1,\alpha}(j+k)-m_{1,\alpha}(k))}_{=:L_{9}(j,k)}.$

From (B.12), we have

$\displaystyle|L_{8}(j,k)|\leq_{\alpha}|k||j|^{\alpha-1},.$ (B.53)

We claim that

$\displaystyle|L_{8}(j_{1},k)-L_{8}(j_{2},k)|\leq_{\alpha}|j_{1}-j_{2}||k|(|j_{1}|+|j_{2}|)^{\alpha-2}.$
(B.54)

Indeed, assuming $|j_{1}|\geq|j_{2}|$, without loss of generality, we have

$\displaystyle|L_{8}(j_{1},k)-L_{8}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}||m_{1,\alpha}(j_{1}+k)-m_{1,\alpha}(j_{1})|$
$\displaystyle\
+|j_{2}||m_{1,\alpha}(j_{1}+k)-m_{1,\alpha}(j_{2}+k)+m_{1,\alpha}(j_{1})-m_{1,\alpha}(j_{2})|$
$\displaystyle\overset{\eqref{p1qosiixx1},\eqref{p1qosiixx2}}{\leq_{\alpha}}|j_{1}-j_{2}||k|(|j_{1}|^{\alpha-2}+(|j_{1}|+|j_{2}|)^{\alpha-2})$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}||k|(|j_{1}|+|j_{2}|)^{\alpha-2},$

which gives (B.54). For $L_{9}$, we compute

$\displaystyle|L_{9}(j,k)|$
$\displaystyle\leq_{\alpha}|k|(|j|^{\alpha-1}+|k|^{\alpha-1}),$ (B.55)
$\displaystyle|L_{9}(j_{1},k)-L_{9}(j_{2},k)|$
$\displaystyle\overset{\eqref{m1sess234}}{\leq}_{\alpha}|k||j_{1}-j_{2}|(|j_{1}|+|j_{2}|)^{\alpha-2}.$
(B.56)

Therefore, we have

$\displaystyle|\mathtt{B}^{N}_{22}(j,k)|$
$\displaystyle\leq_{\alpha}|k|(|j|^{\alpha-1}+|k|^{\alpha-1}),$
$\displaystyle|\mathtt{B}^{N}_{22}(j_{1},k)-\mathtt{B}^{N}_{22}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}||k|(|j_{1}|+|j_{2}|)^{\alpha-2},$

which gives (B.28), and (B.29).

Proof of (B.30), (B.31). From (B.22) and (B.23), we see that

$\displaystyle\mathtt{B}^{D}_{22}(j,k)=-\mathtt{B}^{D}_{21}(j,k)\mathtt{B}^{D}_{21}(j,-k).$
(B.57)

Therefore, we have

$\displaystyle|\mathtt{B}^{D}_{22}(j,k)|$
$\displaystyle\overset{\eqref{resumoxcqq5}}{\leq_{\alpha}}|k|^{2}(|j|+|k|)^{2\alpha-2},$
$\displaystyle|\mathtt{B}^{D}_{22}(j_{1},k)-\mathtt{B}^{D}_{22}(j_{2},k)|$
$\displaystyle\overset{\eqref{resumoxcqq5},\eqref{resumoxcqq6}}{\leq_{\alpha}}|j_{1}-j_{2}||k|^{2}(|j_{1}|+|j_{2}|)^{\alpha-2}(|j_{1}|+|j_{2}|+|k|)^{\alpha-1},$

which gives (B.30) and (B.31). ∎

######  Lemma B.4.

For $\alpha\in(1,2)$ and
$j,j_{1},j_{2},k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that
$|j|,|j_{1}|,|j_{2}|\neq|k|$, it holds that

$\displaystyle|\mathtt{B}_{21}(j,k)|$
$\displaystyle\leq_{\alpha}\frac{|k|}{\min\left\\{|j|,|k|\right\\}}(|j|+|k|)^{\alpha-1},$
(B.58) $\displaystyle|\mathtt{B}_{21}(j_{1},k)-\mathtt{B}_{21}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|\left(\frac{|k|}{\min\left\\{|j_{1}|,|j_{2}|,|k|\right\\}}\right)^{2}\left(\frac{(|j_{1}|+|j_{2}|)^{\alpha-2}(|j_{1}|+|j_{2}|+|k|)^{2\alpha-2}}{(|j_{1}|+|k|)^{\alpha-1}(|j_{2}|+|k|)^{\alpha-1}}\right),$
(B.59) $\displaystyle|\mathtt{B}_{22}(j,k)|$
$\displaystyle\leq_{\alpha}\left(\frac{|k|}{\min\left\\{|j|,|k|\right\\}}\right)^{2}|j|^{\alpha-2}(|j|+|k|),$
(B.60) $\displaystyle|\mathtt{B}_{22}(j_{1},k)-\mathtt{B}_{22}(j_{2},k)|$
$\displaystyle\leq_{\alpha}|j_{1}-j_{2}|\left(\frac{|k|}{\min\left\\{|j_{1}|,|j_{2}|,|k|\right\\}}\right)^{4}\left(\frac{(|j_{1}|+|j_{2}|+|k|)^{4\alpha-3}(|j_{1}|+|j_{2}|)^{\alpha-3}}{(|j_{1}|+|k|)^{2\alpha-2}(|j_{2}|+|k|)^{2\alpha-2}}\right).$
(B.61)

###### Proof.

From Lemma 2.33, we have

$|\mathtt{B}^{D}_{21}(j,k)|\geq_{\alpha}(|j|+|k|)^{\alpha-1}\min\left\\{|j|,|k|\right\\},\quad|\mathtt{B}^{D}_{22}(j,k)|\overset{\eqref{tlqkfgk1x}}{\geq_{\alpha}}(|j|+|k|)^{2\alpha-2}\min\left\\{|j|,|k|\right\\}^{2}.$

Therefore, the results (B.58)-(B.61) follow straightforwardly from Lemma B.3.
∎

######  Lemma B.5.

For $\alpha\in(1,2)$ and $j\in\mathbb{N}$, we have that

$\displaystyle|\mathtt{B}_{3}(j)|\leq_{\alpha}|j|^{\alpha-1}.$ (B.62)

###### Proof.

We first recall the definition of $\mathtt{B}_{3}$ from (B.6) and estimate the
denominator/numerator separately. For the denominator, Recalling
$\lambda_{\alpha-1}$ from (2.79), it follows from (2.69) that

$m_{1,\alpha}^{\circ}(2j)-m_{1,\alpha}^{\circ}(j)=\mathcal{C}_{\alpha}(\lambda_{\alpha-1}(2j)-\lambda_{\alpha-1}(j)),$

where $\mathcal{C}_{\alpha}$ is a positive constant given in (2.78). For
$j\in\mathbb{N}$, we see that

$\displaystyle|m_{1,\alpha}^{\circ}(2j)-m_{1,\alpha}^{\circ}(j)|=\mathcal{C}_{\alpha}\int_{j}^{2j}\partial_{\xi}\lambda_{\alpha-1}(\xi)d\xi\overset{\eqref{sdsds11taxdsd}}{\geq_{\alpha}}\int_{j}^{2j}\xi^{\alpha-2}d\xi\geq_{\alpha}j^{\alpha-1}.$
(B.63)

For the numerator, we see that $\xi\mapsto m_{1,\alpha}(\xi)$ is a Fourier
multiplier in $\mathcal{S}^{\alpha-1}$ and
$m_{1,\alpha}(j)=m_{1,\alpha}^{\circ}(j)$ (see (2.97)), therefore we have

$\displaystyle|2m_{1,\alpha}^{\circ}(j)+m_{1,\alpha}^{\circ}(2j)-T_{\alpha}|\leq_{\alpha}j^{\alpha-1}.$
(B.64)

Therefore, plugging (B.63) and (B.64) into (B.6), we obtain (B.62).

∎

Now, we start simplifying $m_{3,\alpha}^{\circ}(j,k)$ in (B.1). Since
$j\mapsto m_{3,\alpha}^{\circ}(j,k),k\mapsto m_{3,\alpha}^{\circ}(j,k)$ are
even, we will focus on $j,k>0$.

######  Lemma B.6.

For $\alpha\in(1,2)$ and $j,k\in\mathbb{Z}\backslash\left\\{0\right\\}$, we
have that

$\displaystyle
m_{3,\alpha}^{\circ}(j,k)=\frac{\mathcal{C}_{\alpha}}{\alpha(\alpha+1)}\left(|j+k|^{\alpha+1}+|j-k|^{\alpha+1}-2(|j|^{\alpha+1}+|k|^{\alpha+1})\right)+m_{3,\alpha,r}(j,k),$
(B.65)

where $\mathcal{C}_{\alpha}$ is as in (2.70), and $m_{3,\alpha,r}(j,k)$
satisfies

$\displaystyle|m_{3,\alpha,r}(j,k)|\leq_{\alpha}|k|^{2}\langle
j\rangle^{\alpha-2}+|k|^{\alpha}.$ (B.66)

Also, if $j_{1},j_{2}\in\mathbb{Z}$ and $j_{1},j_{2}\neq k$, then

$\displaystyle|m_{3,\alpha,r}(j_{1},k)-m_{3,\alpha,r}(j_{2},k)|\leq_{\alpha}(|j_{1}|+|j_{2}|)^{\alpha-2}|k|^{2}.$
(B.67)

###### Proof.

Since $j,k\mapsto m_{3,\alpha}^{\circ}(j,k)$ is even and symmetric, that is
$m_{3,\alpha}^{\circ}(j,k)=m^{\circ}_{3,\alpha}(k,j)$, we can assume, without
loss of generality, that $j,k\in\mathbb{N}$ and $j>k$.

For a natural number $k$, we decompose $(2-2\cos kx)$ as

$\displaystyle 2-2\cos kx$
$\displaystyle=(1-e^{\text{i}kx})(1-e^{-\text{i}kx})=(1-e^{\text{i}x})(1-e^{-\text{i}x})\sum_{n_{1},n_{2}=0}^{k-1}e^{\text{i}(n_{1}-n_{2})x}=(2-2\cos
x)\sum_{n_{1},n_{2}=0}^{k-1}e^{\text{i}(n_{1}-n_{2})x}$
$\displaystyle=(2-2\cos x)\sum_{n_{1},n_{2}=0}^{k-1}\cos((n_{1}-n_{2})x).$

Recalling $m^{\circ}_{3,\alpha}$ from (B.1), we have

$\displaystyle m_{3,\alpha}^{\circ}(j,k)$
$\displaystyle=\sum_{n_{1},n_{2}=0}^{k-1}\int_{\mathbb{T}}(2-2\cos
x)^{-\frac{\alpha}{2}}(2-2\cos jx)\cos((n_{1}-n_{2})x)dx$
$\displaystyle=\sum_{n_{1},n_{2}=0}^{k-1}\int_{\mathbb{T}}(2-2\cos
x)^{-\frac{\alpha}{2}}((1-\cos((j+n_{1}-n_{2})x))+(1-\cos((j-n_{1}+n_{2})x))-(2-2\cos((n_{1}-n_{2})x)))dx$
$\displaystyle=\frac{1}{2}\sum_{n_{1},n_{2}=0}^{k-1}(m_{1,\alpha}(j+(n_{1}-n_{2}))+m_{1,\alpha}(j-(n_{1}-n_{2}))-2m_{1,\alpha}(n_{1}-n_{2}))$
$\displaystyle=\sum_{n_{1},n_{2}=0}^{k-1}(m_{1,\alpha}(j+n_{1}-n_{2})-m_{1,\alpha}(n_{1}-n_{2})),$

where the second last equality follows from the definition of
$m_{1,\alpha}^{\circ}$ in (B.1) and (2.97), and the last equality follows from
evenness of $m_{1,\alpha}$ and exchange $n_{1}$ and $n_{2}$ in the summation.
Now, we recall $\lambda_{\alpha-1}$ from (2.79) so that using explicit formula
for $m_{1,\alpha}^{\circ}$ in (2.69), we have

$\displaystyle
m_{3,\alpha}^{\circ}(j,k)=\mathcal{C}_{\alpha}\sum_{n_{1},n_{2}=1}^{k-1}(\lambda_{\alpha-1}(j+(n_{1}-n_{2}))-\lambda_{\alpha-1}(n_{1}-n_{2})),$

In view of (2.82), we write

$\displaystyle
m_{3,\alpha}^{\circ}(j,k)=\mathcal{C}_{\alpha}\left(f_{1}(j,k)+f_{2}(j,k)-f_{2}(0,k)+f_{3}(j,k)-f_{3}(0,k)\right),$
(B.68)

where

$\displaystyle f_{1}(j,k)$
$\displaystyle:=\int_{0}^{k}\int_{0}^{k}(j+s-t)^{\alpha-1}-|s-t|^{\alpha-1}dsdt,$
(B.69) $\displaystyle f_{2}(j,k)$
$\displaystyle:=\sum_{n_{1},n_{2}=0}^{k-1}(\lambda_{\alpha-1}(j+n_{1}-n_{2})-|j+n_{1}-n_{2}|^{\alpha-1},$
(B.70) $\displaystyle f_{3}(j,k)$
$\displaystyle:=\sum_{n_{1},n_{2}=0}^{k-1}|j+n_{1}-n_{2}|^{\alpha-1}-\int_{0}^{k}\int_{0}^{k}|j+s-t|^{\alpha-1}dsdt.$
(B.71)

$f_{1}$ can be explicitly computed as

$\displaystyle
f_{1}(j,k)=\frac{1}{\alpha(\alpha+1)}\left((j+k)^{\alpha+1}+(j-k)^{\alpha+1}-2(j^{\alpha+1}+k^{\alpha+1})\right),\text{
for $j\geq k$}.$ (B.72)

For $f_{2}$, it follows from (2.82) that (denoting $n:=n_{1}-n_{2}$),

$|\lambda_{\alpha-1}(j+n)-|j+n|^{\alpha-1}|\leq_{\alpha}\langle
j+n\rangle^{\alpha-2},\text{ for $j\neq n\in\mathbb{N}$}.$

Therefore, we obtain

$\displaystyle|f_{2}(j,k)|\leq_{\alpha}\begin{cases}\sum_{n_{1},n_{2}=0}^{k-1}\langle
j+n_{1}-n_{2}\rangle^{\alpha-2}\leq_{\alpha}|k|^{2}\langle
j\rangle^{\alpha-2},&\text{ for $j\geq k$, $j,k\in\mathbb{N}$,}\\\
\sum_{n_{1},n_{2}=0}^{k-1}\langle
n_{1}-n_{2}\rangle^{\alpha-2}\leq_{\alpha}|k|^{\alpha},&\text{ for $j=0$,
$k\in\mathbb{N}$.}\end{cases}$ (B.73)

For $f_{3}$, we have

$\displaystyle f_{3}(j,k)$
$\displaystyle=\sum_{n_{1},n_{2}=0}^{k-1}\int_{0}^{1}\int_{0}^{1}|j+n_{1}-n_{2}|^{\alpha-1}-|j+n_{1}-n_{2}+s-t|^{\alpha-1}dsdt.$

Using

$|j+n_{1}-n_{2}|^{\alpha-1}-|j+n_{1}-n_{2}+s-t|^{\alpha-1}\leq_{\alpha}\langle
j+n_{1}-n_{2}\rangle^{\alpha-2},\text{ for $j,n_{1},n_{2}\in\mathbb{N}$,
$s,t\in[0,1]$},$

we see that

$\displaystyle|f_{3}(j,k)|\leq_{\alpha}\begin{cases}|k|^{2}\langle
j\rangle^{\alpha-2},&\text{ for $j\geq k$, $j,k\in\mathbb{N}$,}\\\
|k|^{\alpha},&\text{ for $j=0$, $k\in\mathbb{N}$.}\end{cases}$ (B.74)

Therefore, denoting
$m_{3,\alpha,r}(j,k):=\mathcal{C}_{\alpha}(f_{2}(j,k)-f_{2}(0,k)+f_{3}(j,k)-f_{3}(0,k))$,
the decomposition of $m_{3,\alpha}$ as in (B.65) follows from (B.72) and
(B.68). The estimates (B.66) and (B.67) follow immediately from (B.73) and
(B.74). ∎

######  Lemma B.7.

For $\alpha\in(1,2)$ and $j\in\mathbb{Z}$, we have that

$\displaystyle W(j)=-\frac{\mathcal{C}_{\alpha}}{2}j|j|^{\alpha-1}+W_{r}(j),$
(B.75)

where $W_{r}(j)$ satisfies

$\displaystyle|W_{r}(j)$ $\displaystyle|\leq_{\alpha}|j|,$ (B.76)
$\displaystyle|W_{r}(j)-W_{r}(k)|$ $\displaystyle\leq_{\alpha}|j-k|.$ (B.77)

###### Proof.

It follows from (2.97),(2.98),(2.76),(2.77) and (2.79) that
$\lambda_{\alpha}(j)=\mathcal{C}_{\alpha}j\lambda_{\alpha-1}(j)-\mathcal{C}_{\alpha}\lambda_{\alpha-1}(0)j$.
Therefore, we have

$\displaystyle W(j)$
$\displaystyle\overset{\eqref{defoflsksdjwwww}}{=}-\frac{1}{2}\lambda_{\alpha}(j)+\frac{T_{\alpha}}{4}j$
$\displaystyle=-\frac{\mathcal{C}_{\alpha}}{2}j\lambda_{\alpha-1}(j)+\left(\frac{\mathcal{C}_{\alpha}}{2}\lambda_{\alpha-1}(0)+\frac{T_{\alpha}}{4}\right)j$
$\displaystyle=-\frac{\mathcal{C}_{\alpha}}{2}j|j|^{\alpha-1}+W_{r}(j),$
(B.78)

where

$\displaystyle
W_{r}(j):=\frac{\mathcal{C}_{\alpha}}{2}j(|j|^{\alpha-1}-\lambda_{\alpha-1}(j))+\left(\frac{\mathcal{C}_{\alpha}}{2}\lambda_{\alpha-1}(0)+\frac{T_{\alpha}}{4}\right)j.$
(B.79)

Therefore, it suffices to show that $W_{r}$ in (B.79) satisfies the estimates
in (B.76) and (B.77). These estimates follow straightforwardly from (2.82). ∎

###  B.3 Asymptotics of $\mathbb{A}$ and $\vec{D}(\xi)$ for large $\mathtt{M}$

#####  B.3.0.1 Notation

We recall from (4.5), (4.6) and (5.30) that

$\displaystyle
S_{0}^{+}=\left\\{0<\mathtt{j}_{1}<\cdots<\mathtt{j}_{\nu}:\mathtt{j}_{1},\ldots,\mathtt{j}_{\nu}\in\mathbb{N}\right\\},$
$\displaystyle\quad
S^{+}=\left\\{0<j_{1}<\cdots<j_{\nu}:j_{i}=\mathtt{M}\mathtt{j}_{i}\right\\},$
(B.80) $\displaystyle S=\left\\{\pm j:j\in S^{+}\right\\},$
$\displaystyle\quad
S_{\mathtt{M}}^{\perp}:=\left\\{\mathtt{M}j\in\mathbb{Z}\backslash\left\\{0\right\\}:\mathtt{M}j\notin
S\right\\}.$ (B.81)

We also denote by $D_{S_{0}^{+}}$ the diagonal matrix whose $i$-th diagonal
entry is $\mathtt{j}_{i}$, that is,

$\displaystyle\left(D_{S_{0}^{+}}\right)^{i}_{i}:=\mathtt{j}_{i},\quad\left(D_{S_{0}^{+}}\right)^{i}_{j}=0,\text{
for $i,j=1,\ldots,\nu$ and $i\neq j$.}$ (B.82)

Given a matrix $B\in\mathbb{R}^{\nu\times\nu}$, we denote

$\displaystyle|B|:=\max_{i,j=1,\ldots,\nu}|B^{i}_{j}|.$ (B.83)

We also denote

$\displaystyle(\vec{u})_{k}:=1,\text{ for $k=1,\ldots,\nu$}.$ (B.84)

Given $S_{0}^{+}$, we denote by $\mathbb{A}_{S_{0}^{+}}$ be a $\nu\times\nu$
symmetric matrix defined as

$\displaystyle\left(\mathbb{A}_{S_{0}^{+}}\right)^{i}_{k}:=\begin{cases}(2^{\alpha+1}-4)\mathtt{j}_{i}^{\alpha+3},&\text{
if $i=k$,}\\\
2\mathtt{j}_{i}\mathtt{j}_{k}\left((\mathtt{j}_{i}+\mathtt{j}_{k})^{\alpha+1}+(\mathtt{j}_{i}-\mathtt{j}_{k})^{\alpha+1}-2(\mathtt{j}_{i}^{\alpha+1}+\mathtt{j}_{k}^{\alpha+1})\right),&\text{
if $i>k$,}\\\ (\mathbb{A}_{S_{0}^{+}})^{k}_{i},&\text{ if $i<k$.}\end{cases}$
(B.85)

In this subsection, we will study the asymptotics of the hypotheses
$(\mathtt{H}1)$-$(\mathtt{H}3)$ in Subsection 5.1 for large $\mathtt{M}$.

#####  B.3.0.2 Asymptotic analysis for the hypothesis $(\mathtt{H}1)$

######  Lemma B.8.

$\mathbb{A}$ in (5.10) admits a decomposition

$\displaystyle\mathbb{A}=\frac{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}{192(\alpha+1)}\left(\mathbb{A}_{S^{+}_{0}}+\mathbb{B}_{1}\right),$
(B.86)

where $\mathbb{B}_{1}\in\mathbb{R}^{\nu\times\nu}$ satisfies

$\displaystyle|\mathbb{B}_{1}|\leq_{\alpha,S_{0}^{+}}\mathtt{M}^{-1}.$ (B.87)

###### Proof.

We recall the decomposition of $\mathbb{A}$ from (B.8):

$\displaystyle\mathbb{A}=\frac{\alpha\pi}{192}\mathbb{A}_{1}+\frac{\alpha\pi}{192}\mathbb{A}_{2}.$
(B.88)

We claim that

$\displaystyle|\mathbb{A}_{2}|\leq_{\alpha,S^{+}_{0}}\mathtt{M}^{\alpha+1}.$
(B.89)

In view of the definition of $\mathbb{A}_{2}$ in (B.8), we estimate
$A_{2}(j_{i},j_{k})$: For $j_{1},\ldots,j_{\nu}\in S^{+}$, it holds that for
$i,k=1,\ldots,\nu$,

$\displaystyle|\mathtt{B}_{1}(j_{i},j_{k})|$
$\displaystyle\overset{\eqref{b1estimate11sx}}{\leq_{\alpha,S_{0}^{+}}}\mathtt{M}^{\alpha-1},\quad|\mathtt{B}_{2}(j_{i},j_{k})|\overset{\eqref{b1deco1},\eqref{rlaqkqskfkcs1},\eqref{rlaqkqskfkcs3}}{\leq_{\alpha,S_{0}^{+}}}\mathtt{M}^{\alpha-1},\quad|\mathtt{B}_{3}(j_{i})|\overset{\eqref{lemxcxcb3}}{\leq_{\alpha,S_{0}^{+}}}\mathtt{M}^{\alpha-1}.$
(B.90)

Plugging these estimates into $A_{2}$ in (B.7), we have
$|A_{2}(j_{i},j_{k})|\leq_{\alpha,S_{0}^{+}}\mathtt{M}^{\alpha+1}$ for all
$j_{i},j_{k}\in S^{+}$. Therefore, plugging this into the definition of
$\mathbb{A}_{2}$ in (B.8), we obtain (B.89).

Now, we estimate $\mathbb{A}_{1}$. In view of the definition of
$\mathbb{A}_{1}$ in (B.8), (B.7) and the decomposition of
$m_{3,\alpha}^{\circ}$ in (B.65), we have

$\displaystyle\mathbb{A}_{1}$
$\displaystyle=\frac{\mathcal{C}_{\alpha}}{\alpha(\alpha+1)}\mathbb{A}_{11}+\mathbb{A}_{12},\text{
where
$\mathcal{C}_{\alpha}=-\frac{2\pi\Gamma(1-\alpha)}{\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(1-\frac{\alpha}{2}\right)}$,}$
(B.91)

where $\mathbb{A}_{11},\mathbb{A}_{12}$ are symmetric matrices defined by

$\displaystyle\left(\mathbb{A}_{11}\right)^{i}_{k}$
$\displaystyle:=\begin{cases}(2^{\alpha+1}-4)j_{i}^{\alpha+3},&\text{ if
$i=k$,}\\\
2j_{i}j_{k}\left((j_{i}+k_{i})^{\alpha+1}+(j_{i}-k_{i})^{\alpha+1}-2(j_{i}^{\alpha+1}+k_{i}^{\alpha+1})\right),&\text{
if $i>k$,}\\\ \end{cases}$ (B.92)
$\displaystyle\left(\mathbb{A}_{12}\right)^{i}_{k}$
$\displaystyle:=\begin{cases}j_{i}^{2}m_{3,\alpha,r}(j_{i},j_{i}),&\text{ if
$i=k$},\\\ 2j_{i}j_{k}m_{3,\alpha,r}(j_{i},j_{k})&\text{ if
$i>k$,}\end{cases}$ (B.93)

for $j_{1},\ldots,j_{\nu}\in S^{+}$. From (B.66) and (B.80), we have

$\displaystyle|\mathbb{A}_{12}|\leq_{\alpha,S^{+}_{0}}\mathtt{M}^{\alpha+2}.$
(B.94)

For $\mathbb{A}_{11}$, it follows from (B.85) and (B.80) that

$\displaystyle\mathbb{A}_{11}=\mathtt{M}^{\alpha+3}\mathbb{A}_{S^{+}_{0}}.$
(B.95)

Thus, we have

$\displaystyle\mathbb{A}$
$\displaystyle\overset{\eqref{amsjxcxcxc2}}{=}\frac{\alpha\pi}{192}(\mathbb{A}_{1}+\mathbb{A}_{2})\overset{\eqref{a1furtherdocms}}{=}\frac{\mathcal{C}_{\alpha}\pi}{192(\alpha+1)}\left(\mathbb{A}_{11}+\frac{\alpha(\alpha+1)}{\mathcal{C}_{\alpha}}(\mathbb{A}_{12}+\mathbb{A}_{2})\right)\overset{\eqref{Alsjjxca1sd}}{=:}\frac{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}{192(\alpha+1)}\left(\mathbb{A}_{S^{+}_{0}}+\mathbb{B}_{1}\right),$
(B.96)

where

$\displaystyle\mathbb{B}_{1}:=\frac{\alpha(\alpha+1)}{\mathcal{C}_{\alpha}\mathtt{M}^{3+\alpha}}(\mathbb{A}_{12}+\mathbb{A}_{2}).$
(B.97)

Then, the estimate (B.87) follows from (B.94) and (B.89). ∎

######  Lemma B.9.

If $\mathbb{A}_{S_{0}^{+}}$ is invertible, then there exists
$\mathtt{M}_{0}=\mathtt{M}(\alpha,\nu,S_{0}^{+})$ such that if
$\mathtt{M}\geq\mathtt{M}_{0}$, then $\mathbb{A}$ is invertible and satisfies

$\displaystyle\mathbb{A}^{-1}=\frac{192(\alpha+1)}{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}\left(\mathbb{A}_{S^{+}_{0}}^{-1}+\mathbb{B}_{2}\right)$
(B.98)

where $\mathbb{B}_{2}\in\mathbb{R}^{\nu\times\nu}$ satisfies

$\displaystyle|\mathbb{B}_{2}|\leq_{\alpha,\nu,S_{0}^{+}}\mathtt{M}^{-1}.$
(B.99)

for some constants
$c_{1}=c_{1}(\alpha,\nu),c_{2}=c_{2}(\nu,\alpha,S^{+}_{0})>0$.

###### Proof.

In view of (B.86), we need to show that
$\mathbb{A}_{S^{+}_{0}}+\mathbb{B}_{1}$ is invertible. Using the invertibility
of $\mathbb{A}_{S_{0}^{+}}$, we can write

$\mathbb{A}_{S^{+}_{0}}+\mathbb{B}_{1}=\mathbb{A}_{S_{0}^{+}}\left(I+\left(\mathbb{A}_{S_{0}^{+}}\right)^{-1}\mathbb{B}_{1}\right),$

and (B.87) gives us that

$\displaystyle\left|\left(\mathbb{A}_{S_{0}^{+}}\right)^{-1}\mathbb{B}_{1}\right|\leq_{\alpha,\nu,S_{0}^{+}}\mathtt{M}^{-1}.$
(B.100)

Therefore, for sufficiently large $\mathtt{M}$, Gershgorin circle theorem
tells us that the matrix
$I+\left(\mathbb{A}_{S_{0}^{+}}\right)^{-1}\mathbb{B}_{1}$ is invertible and
we can find a matrix $\mathbb{B}_{2}$ such that

$(\mathbb{A}_{S^{+}_{0}}+\mathbb{B}_{1})^{-1}=\mathbb{A}_{S^{+}_{0}}^{-1}+\mathbb{B}_{2},$

that is, (B.98) is satisfied. The estimate in (B.99) follows from the estimate
(B.100) with the Neumann series of
$\left(I+\left(\mathbb{A}_{S_{0}^{+}}\right)^{-1}\mathbb{B}_{1}\right)^{-1}$.
∎

#####  B.3.0.3 Asymptotic analysis for the hypothesis $(\mathtt{H}2)$

We will first decompose $\vec{D}(\xi)$ depending on the dependence on
$\mathtt{M}$.

######  Lemma B.10.

For each $\xi\in S_{\mathtt{M}}^{\perp}$, $\vec{D}(\xi)$ admits a
decomposition,

$\displaystyle\vec{D}(\xi)=\frac{\alpha}{16}\vec{m}_{3,1}(\xi)+\vec{m}_{4}(\xi),$
(B.101)

where

$\displaystyle(\vec{m}_{3,1}(\xi))_{k}$
$\displaystyle:=m_{3,\alpha}^{\circ}(\xi,j_{k})j_{k},\text{ for $j_{k}\in
S^{+}$ and for $k=1,\ldots,\nu$,}$ (B.102)

and $\vec{m}_{4}(\xi)$ satisfies

$\displaystyle|\vec{m}_{4}(\xi)|$
$\displaystyle\leq_{\alpha,S^{+}_{0},\nu}\mathtt{M}|\xi|^{\alpha-1},$ (B.103)
$\displaystyle|\vec{m}_{4}(\xi_{1})-\vec{m}_{4}(\xi_{2})|$
$\displaystyle\leq_{\alpha,S^{+}_{0},\nu}\mathtt{M}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2},\text{
for $\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}$.}$ (B.104)

###### Proof.

Given $\xi\in S_{\mathtt{M}}^{\perp}$, let us denote by $i_{\xi}$ the integer
such that

$\displaystyle\mathtt{M}i_{\xi}=\xi.$ (B.105)

In view of (B.10), we define

$\displaystyle(\vec{m}_{4}(\xi))_{k}$
$\displaystyle:=-\frac{\alpha}{16}j_{k}\mathtt{B}_{1}(\xi,j_{k})+\frac{\alpha^{2}}{32}j_{k}\mathtt{B}_{2}(\xi,j_{k}),$
(B.106)

for $j_{1},\ldots,j_{k}\in S^{+}$, so that we have the decomposition as in
(B.101). Now, it suffices to prove the estimates (B.103) and (B.104).

We first prove (B.103). Indeed, we have

$\displaystyle|\mathtt{B}_{1}(\xi,j_{k})|$
$\displaystyle\overset{\eqref{b1estimate11sx}}{\leq_{\alpha}}(|\xi|^{\alpha-1}+|j_{k}|^{\alpha-1})\overset{\eqref{xidsd2sdsdw1sd},\eqref{stanxcxcx11}}{=}\mathtt{M}^{\alpha-1}(|i_{\xi}|^{\alpha-1}+|\mathtt{j}_{k}|^{\alpha-1})\leq_{\alpha,S_{0}^{+}}\mathtt{M}^{\alpha-1}|i_{\xi}|^{\alpha-1}\overset{\eqref{xidsd2sdsdw1sd}}{=}|\xi|^{\alpha-1}.$
(B.107)

Similarly, we have

$\displaystyle|\mathtt{B}_{21}(\xi,j_{k})|$
$\displaystyle\overset{\eqref{rlaqkqskfkcs1}}{\leq_{\alpha}}|\xi|^{\alpha-1}+|j_{k}|^{\alpha-1}\leq_{\alpha,S^{+}_{0}}|\xi|^{\alpha-1},$
$\displaystyle|\mathtt{B}_{22}(\xi,j_{k})|$
$\displaystyle\overset{\eqref{rlaqkqskfkcs3}}{\leq_{\alpha}}|\xi|^{\alpha-2}(|\xi|+|j_{k}|)\leq_{\alpha,S^{+}_{0}}|\xi|^{\alpha-1},$

which implies

$\displaystyle|\mathtt{B}_{2}(\xi,j_{k})|\overset{\eqref{b1deco1}}{\leq_{\alpha,S^{+}_{0}}}|\xi|^{\alpha-1}.$
(B.108)

Plugging (B.107) and (B.108) into (B.106), we obtain (B.103).

Towards (B.104), we denote by $i_{\xi,1},i_{\xi_{2}}$ the integers such that

$\displaystyle\mathtt{M}i_{\xi,1}=\xi_{1},\quad\mathtt{M}i_{\xi,2}=\xi_{2}.$
(B.109)

We have that

$\displaystyle|\mathtt{B}_{1}(\xi_{1},j_{k})-\mathtt{B}_{1}(\xi_{2},j_{k})|$
$\displaystyle\overset{\eqref{bestimate11sx1}}{\leq_{\alpha}}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$
(B.110)

For $\mathtt{B}_{21}$, we consider to cases: 1) $|i_{\xi,1}|\geq 2|i_{\xi,2}|$
or $|i_{\xi,2}|\geq 2|i_{\xi,1}|$ and
$2)\frac{1}{2}|i_{\xi,2}|\leq|i_{\xi,1}|\leq 2|i_{\xi,2}|$. In the first case,
assuming $|i_{\xi,1}|\geq 2|i_{\xi,2}|$, without loss of generality, we have

$\displaystyle|\mathtt{B}_{21}(\xi_{1},j_{k})-\mathtt{B}_{21}(\xi_{2},j_{k})|$
$\displaystyle\leq|\mathtt{B}_{21}(\xi_{1},j_{k})|+|\mathtt{B}_{21}(\xi_{2},j_{k})|$
$\displaystyle\overset{\eqref{rlaqkqskfkcs1}}{\leq_{\alpha,S_{0}^{+}}}|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1}\leq_{\alpha,S^{+}_{0}}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2},$

where the last inequality follows from that $|i_{\xi,1}|\geq 2|i_{\xi,2}|$
implies $|\xi_{1}|\geq 2|\xi_{2}|$. For the second case, assuming
$\frac{1}{2}|i_{\xi,2}|\leq|i_{\xi,1}|\leq 2|i_{\xi,2}|$, we have

$\displaystyle|\mathtt{B}_{21}(\xi_{1},j_{k})-\mathtt{B}_{21}(\xi_{2},j_{k})|$
$\displaystyle\overset{\eqref{rlaqkqskfkcs2},\eqref{xjasd1xs1},\eqref{stanxcxcx11}}{\leq_{\alpha}}\mathtt{M}^{\alpha-2}|\xi_{1}-\xi_{2}|\left(\frac{(|i_{\xi,1}|+|i_{\xi,2}|)^{\alpha-2}(|i_{\xi,1}|+|i_{\xi,2}|+|\mathtt{j}_{k}|)^{2\alpha-2}}{(|i_{\xi,1}|+|\mathtt{j}_{k}|)^{\alpha-1}(|i_{\xi,2}|+|\mathtt{j}_{k}|)^{\alpha-1}}\right)$
$\displaystyle\leq_{\alpha,S^{+}_{0}}\mathtt{M}^{\alpha-2}|\xi_{1}-\xi_{2}|(|i_{\xi,1}|+|i_{\xi,2}|)^{\alpha-2}$
$\displaystyle\overset{\eqref{xjasd1xs1}}{=}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$

Therefore, in either case, we have

$\displaystyle|\mathtt{B}_{21}(\xi_{1},j_{k})-\mathtt{B}_{21}(\xi_{2},j_{k})|\leq_{\alpha,S^{+}_{0}}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$
(B.111)

Similarly, if $|i_{\xi,1}\geq 2|i_{\xi,2}|$ or $|i_{\xi_{2}}|\geq
2|i_{\xi,1}|$, we have

$\displaystyle|\mathtt{B}_{22}(\xi_{1},j_{k})-\mathtt{B}_{22}(\xi_{2},j_{k})|\leq|\mathtt{B}_{22}(\xi_{1},j_{k})|+|\mathtt{B}_{22}(\xi_{2},j_{k})|\overset{\eqref{rlaqkqskfkcs3}}{\leq}_{\alpha}|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1}\leq_{\alpha}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2},$

and if $\frac{1}{2}|i_{\xi,2}|\leq|i_{\xi,1}|\leq 2|i_{\xi,2}|$, we have

$\displaystyle|\mathtt{B}_{22}(\xi_{1},j_{k})-\mathtt{B}_{22}(\xi_{2},j_{k})|\overset{\eqref{rlaqkqskfkcs4}}{\leq_{\alpha,S^{+}_{0}}}\mathtt{M}^{\alpha-2}|\xi_{1}-\xi_{2}|(|i_{\xi,1}|+|i_{\xi,2}|)^{\alpha-2}\leq_{\alpha}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$

Therefore, in either case, we have
$|\mathtt{B}_{22}(\xi_{1},j_{k})-\mathtt{B}_{22}(\xi_{2},j_{k})|\leq_{\alpha,S^{+}_{0}}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}$.
Plugging this and (B.111) into (B.21), we have

$\displaystyle|\mathtt{B}_{2}(\xi_{1},j_{k})-\mathtt{B}_{2}(\xi_{2},j_{k})|\leq_{\alpha,S^{+}_{0}}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$
(B.112)

Then, the estimate (B.104) follows straightforwardly from (B.106), (B.110) and
(B.112). ∎

In view of (B.65), we can further decompose $\vec{m}_{3,1}$ for sufficiently
large $\xi$:

######  Lemma B.11.

For $\xi\in S_{\mathtt{M}}^{\perp}$ such that
$|\xi|\geq\sup\left\\{j_{k}:j_{k}\in S^{+}\right\\}$, $\vec{m}_{3,1}(\xi)$
admits a decomposition,

$\displaystyle\vec{m}_{3,1}(\xi)=\frac{\mathcal{C}_{\alpha}}{\alpha(\alpha+1)}\vec{m}_{3,2}(\xi)+\vec{m}_{3,3}(\xi),$
(B.113)

where

$\displaystyle(\vec{m}_{3,2}(\xi))_{k}=(|\xi+j_{k}|^{\alpha+1}+|\xi-
j_{k}|^{\alpha+1}-2(|\xi|^{\alpha+1}+|j_{k}|^{\alpha+1}))j_{k},$ (B.114)

and $\vec{m}_{3,3}(\xi)$ satisfies

$\displaystyle|(\vec{m}_{3,3}(\xi))_{k}|$
$\displaystyle\leq_{\alpha,S^{+}_{0}}\mathtt{M}^{2}|\xi|^{\alpha-1},$ (B.115)
$\displaystyle|(\vec{m}_{3,3}(\xi_{1}))_{k}-(\vec{m}_{3,3}(\xi_{2}))_{k}|$
$\displaystyle\leq_{\alpha,S^{+}_{0}}\mathtt{M}^{2}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$
(B.116)

###### Proof.

For $\xi\in S^{\perp}$ such that $|\xi|\geq\sup\left\\{j_{k}:j_{k}\in
S^{+}\right\\}$, we define (using that
$m_{3,\alpha}^{\circ}(\xi,k)=m_{3,\alpha}^{\circ}(-\xi,k)$, which follows from
its definition in (B.1)),

$\displaystyle(\vec{m}_{3,3}(\xi))_{k}:=m_{3,\alpha,r}(\xi,j_{k})j_{k},$
(B.117)

so that (B.102), (B.114) and (B.65) yield the decomposition (B.113).
Therefore, it suffices to show that $\vec{m}_{3,3}$ in (B.117) satisfies the
estimates in (B.115) and (B.116). Since $\xi\mapsto
m_{3,\alpha}^{\circ}(\xi,k),\vec{m}_{3,2}(\xi)$ are even, we only need to show
the estimate for $\xi>0$. For positive $\xi,\xi_{1},\xi_{2}\in S^{\perp}$ such
that $\xi,\xi_{1},\xi_{2}\geq\sup\left\\{|j|:j\in S\right\\}$, we have

$\displaystyle|m_{3,\alpha,r}(\xi,j_{k})|$
$\displaystyle\overset{\eqref{estimate1m3alax3}}{\leq_{\alpha}}|j_{k}|^{2}|\xi|^{\alpha-2}+|j_{k}|^{\alpha}\overset{\eqref{xidsd2sdsdw1sd},\eqref{stanxcxcx11}}{\leq_{\alpha,S^{+}_{0}}}\mathtt{M}^{\alpha}|i_{\xi}|^{\alpha-2}=\mathtt{M}^{2}|\xi|^{\alpha-2}\leq\mathtt{M}|\xi|^{\alpha-1},$
$\displaystyle|m_{3,\alpha,r}(\xi_{1},j_{k})-m_{3,\alpha,r}(\xi_{2},j_{k})|$
$\displaystyle\overset{\eqref{estimate1m3alax4}}{\leq_{\alpha,S^{+}_{0}}}\mathtt{M}^{2}|\xi_{1}+\xi_{2}|^{\alpha-2}=\mathtt{M}\frac{\mathtt{M}}{|\xi_{1}-\xi_{2}|}|\xi_{1}-\xi_{2}|(\xi_{1}+\xi_{2})^{\alpha-2}\leq\mathtt{M}|\xi_{1}-\xi_{2}|(\xi_{1}+\xi_{2})^{\alpha-2},$

where the two last equalities in the estimates follow from that
$\xi,\xi_{1},\xi_{2}$ are $\mathtt{M}$-multiples of natural numbers.
Therefore, plugging these estimates into (B.117), we obtain the estimates
(B.115) and (B.116). ∎

######  Lemma B.12.

For $\xi\in S_{\mathtt{M}}^{\perp}$ such that $|\xi|\geq
2\sup\left\\{j_{k}:j_{k}\in S^{+}\right\\}$, $\vec{m}_{3,2}(\xi)$ admits a
decomposition,

$\displaystyle(\vec{m}_{3,2}(\xi))_{k}=\alpha(\alpha+1)j_{k}^{3}|\xi|^{\alpha-1}+(\vec{m}_{3,r}(\xi))_{k},$
(B.118)

where $(\vec{m}_{3,r}(\xi))_{k}$ satisfies

$\displaystyle|(\vec{m}_{3,r}(\xi))_{k}|$
$\displaystyle\leq_{\alpha}j_{k}^{4}|\xi|^{\alpha-2},$ (B.119)
$\displaystyle|(\vec{m}_{3,r}(\xi_{1}))_{k}-(\vec{m}_{3,r}(\xi_{2}))_{k}|$
$\displaystyle\leq_{\alpha}j_{k}^{4}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-3},\text{
for $|\xi_{1}|,|\xi_{2}|\geq 2\sup\left\\{|j_{k}|:j_{k}\in S\right\\}$ }$
(B.120)

###### Proof.

In view of (B.114), we rewrite $j_{k}(|\xi+j_{k}|^{\alpha+1}+|\xi-
j_{k}|^{\alpha+1}-2(|\xi|^{\alpha+1}+|j_{k}|^{\alpha+1}))$ as

$\displaystyle j_{k}(|\xi$ $\displaystyle+j_{k}|^{\alpha+1}+|\xi-
j_{k}|^{\alpha+1}-2(|\xi|^{\alpha+1}+|j_{k}|^{\alpha+1}))$
$\displaystyle=\alpha(\alpha+1)j_{k}^{3}|\xi|^{\alpha-1}+\alpha(\alpha+1)(\alpha-1)j_{k}|\xi|^{\alpha+1}\int_{0}^{\frac{j_{k}}{|\xi|}}\int_{-t}^{t}\int_{0}^{s}(1+u)^{\alpha-2}dudsdt$
$\displaystyle=:\alpha(\alpha+1)j_{k}^{3}|\xi|^{\alpha-1}+(\vec{m}_{3,r}(\xi))_{k},$
(B.121)

which gives us the decomposition in (B.118). It suffices to check the
estimates (B.119) and (B.120). For (B.119), we have that (since $|\xi|\geq
2\sup\left\\{|j_{k}|:j_{k}\in S\right\\}$, the integrand is strictly bounded),

$\displaystyle\left|\int_{0}^{\frac{j_{k}}{|\xi|}}\int_{-t}^{t}\int_{0}^{s}(1+u)^{\alpha-2}dudsdt\right|\leq_{\alpha}\int_{0}^{\frac{j_{k}}{|\xi|}}\int_{-t}^{t}\int_{0}^{s}1dudsdt=\frac{1}{3}j_{k}^{3}|\xi|^{-3}.$
(B.122)

Plugging this into the definition of $(\vec{m}_{3,r}(\xi))_{k}$ in (B.3.0.3),
we obtain (B.119). To show (B.120), we first assume that $\xi_{1}$ and
$\xi_{2}$ have opposite signs. Then, we have

$\displaystyle|(\vec{m}_{3,r}(\xi_{1}))_{k}-(\vec{m}_{3,r}(\xi_{2}))_{k}|\overset{\eqref{rlqsjsdsxc1}}{\leq_{\alpha}}j_{k}^{4}(|\xi_{1}|^{\alpha-2}+|\xi_{2}|^{\alpha-2}),$
(B.123)

while it holds that

$\displaystyle|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-3}\geq_{\alpha}(|\xi_{1}|+|\xi_{2}|)(|\xi_{1}|+|\xi_{2}|)^{\alpha-3}\geq_{\alpha}(|\xi_{1}|+|\xi_{2}|)^{\alpha-2},$
(B.124)

therefore we obtain (B.120). Now, we assume that $\xi_{1},\xi_{2}>0$. Without
loss of generality, we can assume that $\xi_{1}>\xi_{2}>0$. If $\xi_{1}\geq
2\xi_{2}$, we still have (B.123), and the inequalities in (B.124) hold true as
well, therefore we obtain (B.120). Lastly, we assume

$\displaystyle 0<\xi_{2}<\xi_{1}<2\xi_{2}.$ (B.125)

Then we have

$\displaystyle|(\vec{m}_{3,r}(\xi_{1}))_{k}-(\vec{m}_{3,r}(\xi_{2}))_{k}|$
$\displaystyle\overset{\eqref{def_fsjsd1mmees}}{\leq_{\alpha}}j_{k}(\xi_{1}^{\alpha+1}-\xi_{2}^{\alpha+1})\int_{0}^{\frac{j_{k}}{|\xi_{2}|}}\int_{-t}^{t}\int_{0}^{s}(1+u)^{\alpha-2}dudsdt$
$\displaystyle\
+j_{k}\xi_{2}^{\alpha+1}\int_{\frac{j_{k}}{|\xi_{1}|}}^{\frac{j_{k}}{|\xi_{2}|}}\int_{-t}^{t}\int_{0}^{s}(1+u)^{\alpha-2}dudsdt$
$\displaystyle\overset{\eqref{skksdxcxc11}}{\leq_{\alpha}}j_{k}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha}+|\xi_{2}|^{\alpha})j_{k}^{3}|\xi_{2}|^{-3}+j_{k}\xi_{2}^{\alpha+1}\int_{\frac{j_{k}}{|\xi_{1}|}}^{\frac{j_{k}}{|\xi_{2}|}}t^{2}dt$
$\displaystyle\overset{\eqref{auusdsxcxc1}}{\leq_{\alpha}}j_{k}^{4}|\xi_{1}-\xi_{2}||\xi_{1}+\xi_{2}|^{\alpha-3}+j_{k}^{4}\xi_{2}^{\alpha+1}\left(\frac{1}{\xi_{2}^{3}}-\frac{1}{\xi_{1}^{3}}\right)$
$\displaystyle\leq_{\alpha}j_{k}^{4}|\xi_{1}-\xi_{2}||\xi_{1}+\xi_{2}|^{\alpha-3},$

which gives (B.120). ∎

In view of (5.32), let us denote

$\displaystyle\mathtt{A}(\xi):=W(\xi)-\frac{\pi}{6}\xi\vec{D}(\xi)\cdot\mathbb{A}^{-1}\overline{\omega},\text{
for $\xi\in S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}$}.$ (B.126)

We simplify the expression of $\mathtt{A}$:

######  Lemma B.13.

Suppose $\mathbb{A}_{S_{0}^{+}}$ in (B.85) is invertible. For $\xi\in
S_{\mathtt{M}}^{\perp}$ such that $|\xi|\geq 2\max\left\\{|j_{k}|:j_{k}\in
S\right\\}$, and for sufficiently large $\mathtt{M}$, depending on
$\alpha,\nu,S_{0}^{+}$, we have

$\displaystyle\mathtt{A}(\xi)=-\frac{\mathcal{C}_{\alpha}}{2}\left(1-2\alpha(\alpha+1)I_{S_{0}^{+}}\right)\xi|\xi|^{\alpha-1}+m_{\mathtt{A},r}(\xi),$
(B.127)

where

$\displaystyle I_{S_{0}^{+}}$
$\displaystyle:=\vec{u}\cdot\left(\left(D_{S_{0}^{+}}\right)^{-\alpha}\cdot\mathbb{A}_{S^{+}_{0}}\left(D_{S^{+}_{0}}\right)^{-3}\right)^{-1}\vec{u},\quad(\vec{u})_{k}=1,\text{
for $k=1,\ldots,\nu$},$ (B.128)

and $m_{\mathtt{A},r}(\xi)$ satisfies

$\displaystyle|m_{\mathtt{A},r}(\xi)|$
$\displaystyle{\leq_{\alpha,\nu,S_{0}^{+}}}\mathtt{M}^{1-\alpha}|\xi|^{\alpha}+\mathtt{M}|\xi|^{\alpha-1},$
(B.129) $\displaystyle|m_{\mathtt{A},r}(\xi_{1})-m_{\mathtt{A},r}(\xi_{2})|$
$\displaystyle{\leq_{\alpha,\nu,S_{0}^{+}}}|\xi_{1}-\xi_{2}|(\mathtt{M}^{1-\alpha}(|\xi_{1}|+|\xi_{2}|)^{\alpha-1}+\mathtt{M}(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}).$
(B.130)

###### Proof.

Recalling $W$ from (5.31), we have

$\displaystyle-\frac{1}{2}\lambda_{\alpha}(\xi)+\frac{T_{\alpha}}{4}\xi=W(\xi)\overset{\eqref{wksjdsdsxcsd}}{=}-\frac{\mathcal{C}_{\alpha}}{2}\xi|\xi|^{\alpha-1}+W_{r}(\xi).$
(B.131)

We choose a sufficiently large $\mathtt{M}$ so that Lemma B.9 gives us that
$\mathbb{A}$ is invertible. In this case, it follows from (B.98) that

$\displaystyle\mathbb{A}^{-1}=\frac{192(\alpha+1)}{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}\left(\mathbb{A}_{S^{+}_{0}}^{-1}+\mathbb{B}_{2}\right),$
(B.132)

where

$\displaystyle|\mathbb{B}_{2}|\overset{\eqref{B2dsdx11sx}}{\leq}_{\alpha,\nu,S_{0}^{+}}\mathtt{M}^{-1}.$
(B.133)

Also, we recall $\overline{\omega}$ from (5.9) so that we can write
$(\overline{\omega})_{k}\overset{\eqref{defoflsksdjwwww}}{=}W(j_{k}),$ for
$j_{1},\ldots,j_{\nu}\in S^{+}$. In view of (B.75), we write
$\overline{\omega}$ as (using the notations in (B.80) and (B.82))

$\displaystyle\overline{\omega}$
$\displaystyle=-\frac{\mathcal{C}_{\alpha}}{2}\mathtt{M}^{\alpha}\left(\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u}+\vec{w}\right),$
(B.134)

where

$\displaystyle(\vec{u})_{k}=1,\text{ for
$k=1,\ldots,\nu$},\quad(\vec{w})_{k}=-\frac{2\mathtt{M}^{-\alpha}}{C_{\alpha}}W_{r}(j_{k}),\text{
for $j_{k}\in S^{+}$.}$ (B.135)

For the size of $\vec{w}$, we have

$\displaystyle|\vec{w}|\overset{\eqref{wxxrxxcxcxw2}}{\leq}_{\alpha}\mathtt{M}^{-\alpha}\sum_{k=1}^{\nu}|j_{k}|\overset{\eqref{stanxcxcx11}}{\leq}_{\alpha,S_{0}^{+}}\mathtt{M}^{1-\alpha}.$
(B.136)

Then, we compute

$\displaystyle\mathbb{A}^{-1}\overline{\omega}$
$\displaystyle\overset{\eqref{Absdsdxc},\eqref{delfjscxcxcs}}{=}-\frac{96(\alpha+1)}{\pi\mathtt{M}^{3}}\left(\mathbb{A}_{S^{+}_{0}}^{-1}+\mathbb{B}_{2}\right)\left(\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u}+\vec{w}\right)$
$\displaystyle=-\frac{96(\alpha+1)}{\pi\mathtt{M}^{3}}\left(\mathbb{A}_{S^{+}_{0}}^{-1}\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u}+\vec{v}\right),$
(B.137)

where

$\displaystyle\vec{v}:=\left(\mathbb{A}_{S_{0}^{+}}\right)^{-1}\vec{w}+\mathbb{B}_{2}\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u}+\mathbb{B}_{2}\vec{w}.$
(B.138)

For $\vec{v}$, we estimate its size as,

$\displaystyle|\vec{v}|$
$\displaystyle\overset{\eqref{bk2jsdsdad},\eqref{shhxcxca11}}{\leq}_{\alpha,\nu,S_{0}^{+}}\mathtt{M}^{1-\alpha}.$
(B.139)

For $\vec{D}(\xi)$, we simplify it only for $|\xi|\geq
2\max\left\\{j_{k}:j_{k}\in S^{+}\right\\}$, since in the other case, its
expression is not necessary. Assuming $|\xi|\geq 2\max\left\\{j_{k}:j_{k}\in
S^{+}\right\\}$, we have

$\displaystyle\vec{D}(\xi)$
$\displaystyle\overset{\eqref{decompoasd}}{=}\frac{\alpha}{16}(\vec{m}_{3,1}(\xi))_{k}+(\vec{m}_{4}(\xi))_{k}$
$\displaystyle\overset{\eqref{m3sdjsdjxcxcsds}}{=}\frac{\mathcal{C}_{\alpha}}{16(\alpha+1)}(\vec{m}_{3,2}(\xi))_{k}+\frac{\alpha}{16}(\vec{m}_{3,3}(\xi))_{k}+(\vec{m}_{4}(\xi))_{k}$
$\displaystyle\overset{\eqref{tlqkfrowhwrkxcxza}}{=}\frac{\alpha\mathcal{C}_{\alpha}}{16}j_{k}^{3}|\xi|^{\alpha-1}+\frac{\mathcal{C}_{\alpha}}{16(\alpha+1)}(\vec{m}_{3,r}(\xi))_{k}+\frac{\alpha}{16}(\vec{m}_{3,3}(\xi))_{k}+(\vec{m}_{4}(\xi))_{k},$
$\displaystyle=:\frac{\alpha\mathcal{C}_{\alpha}}{16}\left(j_{k}^{3}|\xi|^{\alpha-1}+\left(\frac{\alpha\mathcal{C}_{\alpha}}{16}\right)^{-1}\left(\frac{\mathcal{C}_{\alpha}}{16(\alpha+1)}(\vec{m}_{3,r}(\xi))_{k}+\frac{\alpha}{16}(\vec{m}_{3,3}(\xi))_{k}+(\vec{m}_{4}(\xi))_{k}\right)\right).$
(B.140)

Therefore, using the notation for $D_{S_{0}^{+}}$ and $\vec{u}$ in (B.82) and
(B.84), we can write

$\displaystyle\vec{D}(\xi)=\frac{\alpha\mathcal{C}_{\alpha}\mathtt{M}^{3}}{16}\left(|\xi|^{\alpha-1}(D_{S_{0}^{+}})^{3}\vec{u}+\vec{m}_{5}(\xi)\right),\text{
for $|\xi|\geq 2\max\left\\{j_{k}:j_{k}\in S^{+}\right\\}$},$ (B.141)

where

$(\vec{m}_{5}(\xi))_{k}:=\frac{1}{\mathtt{M}^{3}}\left(\frac{\alpha\mathcal{C}_{\alpha}}{16}\right)^{-1}\left(\frac{\mathcal{C}_{\alpha}}{16(\alpha+1)}(\vec{m}_{3,r}(\xi))_{k}+\frac{\alpha}{16}(\vec{m}_{3,3}(\xi))_{k}+(\vec{m}_{4}(\xi))_{k}\right)$

For the estimates of $\vec{m}_{5}$, we have that

$\displaystyle|\vec{m}_{5}(\xi)|$
$\displaystyle\leq_{\alpha}\mathtt{M}^{-3}(|\vec{m}_{3,r}(\xi)|+|\vec{m}_{3,3}(\xi)|+|\vec{m}_{4}(\xi)|)$
$\displaystyle\overset{\eqref{rlqsjsdsxc1}}{\leq_{\alpha,\nu,S_{0}^{+}}}\mathtt{M}|\xi|^{\alpha-2}+\mathtt{M}^{-3}|\vec{m}_{3,3}(\xi)|+\mathtt{M}^{-3}|\vec{m}_{4}(\xi)|$
$\displaystyle\overset{\eqref{sdjjwkkksdxcxsd}}{\leq_{\alpha,S_{0}^{+},\nu}}\mathtt{M}|\xi|^{\alpha-2}+\mathtt{M}^{-1}|\xi|^{\alpha-1}+\mathtt{M}^{-3}|\vec{m}_{4}(\xi)|$
$\displaystyle\overset{\eqref{mrmsidx1}}{\leq_{\alpha,S_{0}^{+},\nu}}\mathtt{M}|\xi|^{\alpha-2}+\mathtt{M}^{-1}|\xi|^{\alpha-1}.$
(B.142)

For $\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}$ such that
$|\xi_{1},\xi_{2}|\geq 2\max\left\\{j_{k}:j_{k}\in S^{+}\right\\}$, we also
have

$\displaystyle|\vec{m}_{5}(\xi_{1})-\vec{m}_{5}(\xi_{2})|$
$\displaystyle\leq_{\alpha}\mathtt{M}^{-3}\left(|\vec{m}_{3,r}(\xi_{1})-\vec{m}_{3,r}(\xi_{2})|+|\vec{m}_{3,3}(\xi_{1})-\vec{m}_{3,3}(\xi_{2})|+|\vec{m}_{4}(\xi_{1})-\vec{m}_{4}(\xi_{2})|\right)$
$\displaystyle\overset{\eqref{rlqsjsdsxc2},\eqref{sdjjwkkksdxcxsd1},\eqref{mrmsidx2}}{\leq_{\alpha,\nu,S_{0}^{+}}}\mathtt{M}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-3}+\mathtt{M}^{-1}|\xi_{1}-\xi_{2}|(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}.$
(B.143)

We combine (B.3.0.3) and (B.141) and obtain (for $|\xi|\geq
2\max\left\\{j_{k}:j_{k}\in S^{+}\right\\}$),

$\displaystyle\frac{\pi}{6}\xi\vec{D}(\xi)\cdot\mathbb{A}^{-1}\overline{\omega}$
$\displaystyle=-{\alpha(\alpha+1)\mathcal{C}_{\alpha}}\xi\left(|\xi|^{\alpha-1}(D_{S_{0}^{+}})^{3}\vec{u}+\vec{m}_{5}(\xi)\right)\cdot\left(\mathbb{A}_{S^{+}_{0}}^{-1}\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u}+\vec{v}\right)$
$\displaystyle=:-\alpha(\alpha+1)\mathcal{C}_{\alpha}I_{S_{0}^{+}}\xi|\xi|^{\alpha-1}+m_{6}(\xi),$
(B.144)

where

$\displaystyle I_{S_{0}^{+}}$
$\displaystyle:=(D_{S_{0}^{+}})^{3}\vec{u}\cdot\mathbb{A}_{S^{+}_{0}}^{-1}\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u},$
(B.145) $\displaystyle m_{6}(\xi)$
$\displaystyle:=-{\alpha(\alpha+1)\mathcal{C}_{\alpha}}\left(\xi|\xi|^{\alpha-1}(D_{S_{0}^{+}})^{3}\vec{u}\cdot\vec{v}+\xi\vec{m}_{5}(\xi)\cdot\left(\mathbb{A}_{S^{+}_{0}}^{-1}\left(D_{S^{+}_{0}}\right)^{\alpha}\vec{u}+\vec{v}\right)\right).$
(B.146)

For the size of $m_{6}$, we have

$\displaystyle|m_{6}(\xi)|$
$\displaystyle\overset{\eqref{rjjjsdmu1},\eqref{vsiz2s}}{\leq_{\alpha,\nu,S_{0}^{+}}}\mathtt{M}^{1-\alpha}|\xi|^{\alpha},$
(B.147) $\displaystyle|m_{6}(\xi_{1})-m_{6}(\xi_{2})|$
$\displaystyle\overset{\eqref{vsiz2s},\eqref{rjjjsdmu2}}{\leq_{\alpha,\nu,S_{0}^{+}}}|\xi_{1}-\xi_{2}|(\mathtt{M}^{1-\alpha}(|\xi_{1}|+|\xi_{2}|)^{\alpha-1}+\mathtt{M}(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}).$
(B.148)

Combining (B.3.0.3) with (B.131) and (B.126), we obtain (B.127), with

$m_{\mathtt{A},r}(\xi):=W_{r}(\xi)-m_{6}(\xi).$

Then the estimates (B.129) and (B.130) follow from (B.147), (B.148), (B.76)
and (B.77). ∎

#####  B.3.0.4 Asymptotic analysis for the hypothesis $(\mathtt{H}3)$

We derive more explicit expression for the matrix
$\mathbb{B}_{\xi_{1},\xi_{2}}$ and $\mathbb{C}_{\xi_{1},\xi_{2}}$ in (5.33).
We denote

For $\xi,\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}$, we denote by
$i_{\xi},i_{\xi,1},i_{\xi,2}$, the integers such that
$\mathtt{M}i_{\xi}=\xi,\mathtt{M}i_{\xi,1}=\xi_{1},\mathtt{M}i_{\xi,2}=\xi_{2}$,
(B.149) $\displaystyle
f(x,y):=|x+y|^{1+\alpha}+|x-y|^{1+\alpha}-2(|x|^{1+\alpha}+|y|^{1+\alpha}).$
(B.150)

######  Lemma B.14.

For $\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}$ such
that $\xi_{1}\neq\xi_{2}$, we have that

$\displaystyle\mathbb{C}_{\xi_{1},\xi_{2}}$
$\displaystyle=\frac{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}{192(\alpha+1)}(\mathbb{A}_{S_{0}^{+}}-2(D_{S_{0}^{+}})^{3}\vec{F}(\xi_{1},\xi_{2})\vec{u}^{T}(D_{S_{0}^{+}})^{\alpha})+W_{r,*}(\xi_{1},\xi_{2}),$
(B.151)

where $\vec{F}$ is defined as

$\displaystyle(\vec{F}(\xi_{1},\xi_{2}))_{k}:=\begin{cases}\frac{f(i_{\xi,1},\mathtt{j}_{k})}{|i_{\xi,1}|^{\alpha-1}\mathtt{j}_{k}^{2}},&\text{
if $\xi_{1}\neq 0,\xi_{2}=0$},\\\ (\vec{F}(\xi_{2},\xi_{1}))_{k},&\text{ if
$\xi_{1}=0,\xi_{2}\neq 0$},\\\
\frac{\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{k})-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{k})\right)}{(i_{\xi,1}|i_{\xi,1}|^{\alpha-1}-i_{\xi,2}|i_{\xi,2}|^{\alpha-1})\mathtt{j}_{k}^{2}},&\text{
if $\xi_{1},\xi_{2}\neq 0$}.\end{cases}$ (B.152)

Furthermore, $W_{r,*}(\xi_{1},\xi_{2})$ satisfies

$\displaystyle|W_{r,*}(\xi_{1},\xi_{2})|\leq_{\alpha,S_{0}^{+},\nu}\mathtt{M}^{4}.$
(B.153)

###### Proof.

For $\xi\in S_{\mathtt{M}}^{\perp}$, we derive expressions for $\vec{D}(\xi)$,
$W(\xi)$ and $\overline{\omega}$ in (5.33) without assuming $|\xi|$ is
dominating the size of the tangential site $S$.

We will derive expressions for (see (5.33) and (B.9)),

$\displaystyle\mathbb{C}_{\xi_{1},\xi_{2}}=\mathbb{A}-\frac{\pi}{6}(\mathbb{B})_{\xi_{1},\xi_{2}}.$
(B.154)

For $\xi\vec{D}(\xi)$ and $k=1,\ldots,\nu$ and $\xi\in
S_{\mathtt{M}}^{\perp}$, we have

$\displaystyle\xi(\vec{D}(\xi))_{k}$
$\displaystyle\overset{\eqref{decompoasd}}{=}\frac{\alpha}{16}(\xi\vec{m}_{3,1}(\xi))_{k}+(\xi\vec{m}_{4}(\xi))_{k}$
$\displaystyle\overset{\eqref{m3dscsxcsw}}{=}\frac{\alpha}{16}\xi
m^{\circ}_{3,\alpha}(\xi,j_{k})j_{k}+(\xi\vec{m}_{4}(\xi))_{k}$
$\displaystyle\overset{\eqref{m3deccsdx},\eqref{fesjsdxc}}{=}\frac{\mathcal{C}_{\alpha}}{16(\alpha+1)}\xi
f(\xi,j_{k})j_{k}+\xi m_{3,\alpha,r}(\xi,j_{k})+(\xi\vec{m}_{4}(\xi))_{k}$
$\displaystyle\overset{\eqref{ixusdkksdw},\eqref{stanxcxcx11}}{=}\frac{\mathcal{C}_{\alpha}\mathtt{M}^{\alpha+3}}{16(\alpha+1)}i_{\xi}f(i_{\xi},\mathtt{j}_{k})\mathtt{j}_{k}+\underbrace{\xi
m_{3,\alpha,r}(\xi,j_{k})+(\xi\vec{m}_{4}(\xi))_{k}}_{=:(\vec{m}_{7}(\xi))_{k}}.$
(B.155)

Therefore, for $\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}$, we have

$\displaystyle\xi_{1}(\vec{D}(\xi_{1}))_{k}-\xi_{2}(\vec{D}(\xi_{2}))_{k}=\frac{\mathcal{C}_{\alpha}\mathtt{M}^{\alpha+3}}{16(\alpha+1)}\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{k})\mathtt{j}_{k}-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{k})\mathtt{j}_{k}\right)+(\vec{m}_{7}(\xi_{1})-\vec{m}_{7}(\xi_{2}))_{k}.$
(B.156)

For $\vec{m}_{7}$ in (B.3.0.4), we estimate it as

$\displaystyle|\vec{m}_{7}(\xi)|$
$\displaystyle\overset{\eqref{estimate1m3alax3},\eqref{mrmsidx1}}{\leq_{\alpha,S_{0}^{+},\nu}}\mathtt{M}^{1+\alpha}|i_{\xi}|^{\alpha-1},$
(B.157) $\displaystyle|\vec{m}_{7}(\xi_{1})-\vec{m}_{7}(\xi_{2})|$
$\displaystyle\overset{\eqref{estimate1m3alax4},\eqref{mrmsidx2}}{\leq_{\alpha,S_{0}^{+},\nu}}\mathtt{M}^{1+\alpha}|i_{\xi,1}-i_{\xi,2}|(|i_{\xi,1}|+|i_{\xi,2}|)^{\alpha-1}.$
(B.158)

Now, we consider $W(\xi)$. From (B.75) and (B.76), we have

$\displaystyle\frac{1}{W(\xi)}=\frac{1}{-\frac{\mathcal{C}_{\alpha}}{2}\xi|\xi|^{\alpha-1}+W_{r}(\xi)}=:-\frac{2}{\mathcal{C}_{\alpha}}\frac{1}{\xi|\xi|^{\alpha-1}}+W_{r,1}(\xi)=-\frac{2}{\mathcal{C}_{\alpha}}\mathtt{M}^{-\alpha}\frac{1}{i_{\xi}|i_{\xi}|^{\alpha-1}}+W_{r,1}(\xi),$
(B.159)

where $W_{r,1}(\xi)$ satisfies

$\displaystyle|W_{r,1}(\xi)|\leq_{\alpha,S_{0}^{+},\nu}|\xi|^{1-2\alpha}=\mathtt{M}^{1-2\alpha}|i_{\xi}|^{1-2\alpha}.$
(B.160)

Similarly, we have

$\displaystyle\frac{1}{W(\xi_{1})-W(\xi_{2})}$
$\displaystyle=-\frac{2}{\mathcal{C}_{\alpha}}\frac{1}{\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1}}+W_{r,2}(\xi_{1},\xi_{2})$
$\displaystyle=-\frac{2}{\mathcal{C}_{\alpha}}\mathtt{M}^{-\alpha}\frac{1}{i_{\xi,1}|i_{\xi,1}|^{\alpha-1}-i_{\xi,2}|i_{\xi,2}|^{\alpha-1}}+W_{r,2}(\xi_{1},\xi_{2}),$
(B.161)

where (using (B.77))

$\displaystyle|W_{r,2}(\xi_{1},\xi_{2})|\leq_{\alpha,S_{0}^{+},\nu}\mathtt{M}^{1-2\alpha}\frac{1}{|i_{\xi,1}-i_{\xi,2}|(i_{\xi,1}|+|i_{\xi,2}|)^{2\alpha-2}}.$
(B.162)

Therefore, we have

$\displaystyle\frac{\xi(\vec{D}(\xi))_{k}}{W(\xi)}$
$\displaystyle\overset{\eqref{djhhsdiixcxcsw1},\eqref{sdsdsddpqodlqkfrj1}}{=}-\frac{\mathtt{M}^{3}}{8(\alpha+1)}\frac{i_{\xi}f(i_{\xi},\mathtt{j}_{k})\mathtt{j}_{k}}{i_{\xi}|i_{\xi}|^{\alpha-1}}+(\vec{W}_{r,3}(\xi))_{k},$
(B.163)
$\displaystyle\frac{\xi_{1}(\vec{D}(\xi_{1}))_{k}-\xi_{2}(\vec{D}(\xi_{2}))_{k}}{W(\xi_{1})-W(\xi_{2})}$
$\displaystyle\overset{\eqref{djhhsdiixcxcsw2},\eqref{qkfrjfskdcxc1}}{=}-\frac{\mathtt{M}^{3}}{8(\alpha+1)}\frac{\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{k})-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{k})\right)\mathtt{j}_{k}}{\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1}}+(\vec{W}_{r,4}(\xi_{1},\xi_{2}))_{k},$
(B.164)

where $(\vec{W}_{r,3}(\xi))_{k}$ and $(\vec{W}_{r,4}(\xi_{1},\xi_{2}))_{k}$
satisfy

$\displaystyle|\vec{W}_{r,3}(\xi)|$
$\displaystyle\leq_{\alpha,\nu,S_{0}^{+}}\mathtt{M}^{4-\alpha}|i_{\xi}|^{1-\alpha},$
(B.165) $\displaystyle|\vec{W}_{r,4}(\xi_{1},\xi_{2})|$
$\displaystyle\leq_{\alpha,\nu,S_{0}^{+}}\mathtt{M}^{4-\alpha},$ (B.166)

which come from the contributions of (B.158) and (B.162). ∎

Therefore, combining these with (B.134), we have

$\displaystyle\frac{\xi(\vec{D}(\xi))_{i}}{W(\xi)}(\overline{\omega})_{k}$
$\displaystyle=\frac{\mathcal{C}_{\alpha}\mathtt{M}^{\alpha+3}}{16(\alpha+1)}\frac{i_{\xi}f(i_{\xi},\mathtt{j}_{i})\mathtt{j}_{i}}{i_{\xi}|i_{\xi}|^{\alpha-1}}\mathtt{j}_{k}^{\alpha}+(W_{r,5}(\xi))^{i}_{k},$
(B.167)
$\displaystyle\frac{\xi_{1}(\vec{D}(\xi_{1}))_{i}-\xi_{2}(\vec{D}(\xi_{2}))_{i}}{W(\xi_{1})-W(\xi_{2})}(\overline{\omega})_{k}$
$\displaystyle=\frac{\mathcal{C}_{\alpha}\mathtt{M}^{\alpha+3}}{16(\alpha+1)}\frac{\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{i})-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{i})\right)\mathtt{j}_{i}}{\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1}}\mathtt{j}_{k}^{\alpha}+(W_{r,6}(\xi))^{i}_{k},$
(B.168)

where $(W_{r,5}(\xi))^{i}_{k},(W_{r,6}(\xi))^{i}_{k}$ satisfy (from
(B.165),(B.166) and (B.136))

$\displaystyle|W_{r,5}(\xi))^{i}_{k}|$
$\displaystyle\leq_{\alpha,S_{0}^{+},\nu}\mathtt{M}^{4}|i_{\xi}|^{1-\alpha}\leq\mathtt{M}^{4},$
(B.169) $\displaystyle|W_{r,6}(\xi))^{i}_{k}|$
$\displaystyle\leq_{\alpha,S_{0}^{+},\nu}\mathtt{M}^{4}.$ (B.170)

Therefore, using (B.86), (B.87), (B.154) and (5.33) we obtain

$\displaystyle\left(\mathbb{C}_{\xi,0}\right)^{i}_{k}$
$\displaystyle=\frac{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}{192(\alpha+1)}\left(\left(\mathbb{A}_{S_{0}^{+}}\right)^{i}_{k}-2\frac{i_{\xi}f(i_{\xi},\mathtt{j}_{i})\mathtt{j}_{i}}{i_{\xi}|i_{\xi}|^{\alpha-1}}\mathtt{j}_{k}^{\alpha}\right)+\left(W_{r,7}(\xi)\right)^{i}_{k},$
(B.171) $\displaystyle\left(\mathbb{C}_{\xi_{1},\xi_{2}}\right)^{i}_{k}$
$\displaystyle=\frac{\mathcal{C}_{\alpha}\pi\mathtt{M}^{\alpha+3}}{192(\alpha+1)}\left(\left(\mathbb{A}_{S_{0}^{+}}\right)^{i}_{k}-2\frac{\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{i})-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{i})\right)\mathtt{j}_{i}}{i_{\xi,1}|i_{\xi,1}|^{\alpha-1}-i_{\xi,2}|i_{\xi,2}|^{\alpha-1}}\mathtt{j}_{k}^{\alpha}\right)+\left(W_{r,8}\right)^{i}_{k},$
(B.172)

where

$\displaystyle|\left(W_{r,7}(\xi)\right)^{i}_{k}|,|\left(W_{r,8}(\xi)\right)^{i}_{k}|\leq_{\alpha,S_{0}^{+},\nu}\mathtt{M}^{4}.$
(B.173)

Therefore, using the notation $\vec{F}$ in (B.152), we obain (B.152). The
estimate (B.153) follows from (B.171) and (B.172).

###  B.4 Choice of tangential sites

The goal of this subsection is to prove Proposition B.16, which shows that
there are infinitely many choices of $S_{0}^{+}$ and $\mathtt{M}$ so that the
hypotheses described in Subsection 5.1 are satisfied.

######  Lemma B.15.

For each $2\leq\nu\in\mathbb{N}$ and $\alpha\in(1,2)$, there exist infinite
number of choices of
$S^{+}_{0}=\left\\{0<\mathtt{j}_{1}<\cdots<\mathtt{j}_{\nu}\right\\}$ such
that

  1. (1)

The tangential sites $S$ constructed by (4.6) satisfy S.1 and S.2.

  2. (2)

$\mathbb{A}_{S^{+}_{0}}$ is invertible.

  3. (3)

$1-2\alpha(\alpha+1)I_{S_{0}^{+}}\neq 0$.

  4. (4)

For all $\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}$ such
that $\xi_{1}\neq\xi_{2}$, it holds that

$\displaystyle\min_{\vec{U}\in\mathbb{R}^{\nu},\
|\vec{U}|=1}\left|\vec{U}\cdot\left(\mathbb{A}_{S_{0}^{+}}-2(D_{S_{0}^{+}})^{3}\vec{F}(\xi_{1},\xi_{2})\vec{u}^{T}(D_{S_{0}^{+}})^{\alpha}\right)\vec{U}\right|\geq_{\alpha,\nu,S_{0}^{+}}1.$
(B.174)

###### Proof.

We look for $S_{0}^{+}$ such that $\mathtt{j}_{1}=1$, that is,

$\displaystyle
S_{0}^{+}=\left\\{1<\mathtt{j}_{2}<\cdots<\mathtt{j}_{\nu}\right\\}.$ (B.175)

We denote

$\displaystyle
R_{S_{0}^{+}}:=\max_{i<k}\left\\{\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right\\}<1.$
(B.176)

We will show that if $S_{0}^{+}$ of the form (B.175) is chosen so that
$R_{S_{0}^{+}}$ is sufficiently small, then (2)-(4) are satisfied. From the
conditions S.1 and S.2, and the asymptotic properties of $\kappa_{j}$ that
follow from 5.31 and Lemma B.7, it is straightforward to see that such a
choice of $S_{0}^{+}$ with small $R_{S_{0}^{+}}$ can be made so that (1) holds
true as well. Therefore, we will focus on the proof of (2)-(4).

Proof of (2). From (B.85) and (B.150), we rewrite $\mathbb{A}_{S_{0}^{+}}$ as

$\displaystyle\mathbb{A}_{S_{0}^{+}}=(2^{\alpha+1}-4)(D_{S_{0}^{+}})^{\frac{\alpha+3}{2}}\left(I+\frac{2}{2^{\alpha+1}-4}L\right)(D_{S_{0}^{+}})^{\frac{\alpha+3}{2}},$
(B.177)

where

$\displaystyle(L)^{i}_{k}=\begin{cases}0,&\text{ if $i=k$,}\\\
f\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}},\frac{\mathtt{j}_{k}}{\mathtt{j}_{i}}\right),&\text{
if $i\neq k$.}\end{cases}$ (B.178)

Note that for $0\neq x\in\mathbb{R}$, recalling $f$ from (B.150), we see that

$f(x,x^{-1})=|x+x^{-1}|^{1+\alpha}+|x-x^{-1}|^{1+\alpha}-2(x^{1+\alpha}+x^{-(1+\alpha)})$

satisfies

$\displaystyle|f(x,x^{-1})|\leq_{\alpha}\begin{cases}|x|^{\alpha-3},&\text{ if
$|x|>1$},\\\ |x|^{3-\alpha},&\text{ if $|x|<1$.}\end{cases}$ (B.179)

Therefore, (B.178) tells us that

$\displaystyle|L^{i}_{k}|\leq_{\alpha}\begin{cases}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{3-\alpha}\leq
R_{S_{0}^{+}}^{3-\alpha},&\text{ if $i>k$},\\\
\left(\frac{\mathtt{j}_{k}}{\mathtt{j}_{i}}\right)^{3-\alpha}\leq
R_{S_{0}^{+}}^{3-\alpha},&\text{ if $i<k$}.\end{cases}$ (B.180)

Since $2^{\alpha+1}-4\neq 0$ for $\alpha>1$, we have that
$I+\frac{2}{2^{\alpha+1}-4}L$ is invertible, if $R_{S_{0}^{+}}$ is
sufficiently small depending on $\nu$. Therefore from (B.177), we see that
$\mathbb{A}_{S_{0}^{+}}$ is invertible.

Proof of (3). We first derive more careful estimates for the inverse of
$I+\frac{2}{2^{\alpha+1}-4}L$. Using a Neumann series, we write

$\displaystyle\left(I+\frac{2}{2^{\alpha+1}-4}L\right)^{-1}=I+\sum_{n=1}^{\infty}\left(-\frac{2}{2^{\alpha+1}-4}\right)^{n}L^{n}=:I-\tilde{L}$
(B.181)

We claim that for all $n\geq 1$,

$\displaystyle\begin{cases}|({L}^{n})^{i}_{k}|\leq\left(3\nu
R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\right)^{n-1}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}},&\text{
if $k\geq i$},\\\ |({L}^{n})^{i}_{k}|\leq\left(3\nu
R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\right)^{n-1}\left(\frac{\mathtt{j}_{k}}{\mathtt{j}_{i}}\right)^{\frac{3-\alpha}{2}},&\text{
if $k\leq i$}.\end{cases}$ (B.182)

Since $L$ is a symmetric matrix, we can assume that $k\geq i$ to prove the
claim. If $n=1$, then the estimate follows immediately from (B.180) and
(B.178). Assuming (B.181) holds for some $n\geq 1$, we estimate
$(L^{n+1})^{i}_{k}$ for $k\geq i$ as

$\displaystyle|(L^{n+1})^{i}_{k}|$
$\displaystyle\leq\sum_{j=1}^{\nu}|(L^{n})^{j}_{k}||L^{i}_{j}|$
$\displaystyle=\sum_{1\leq j\leq i}|(L^{n})^{j}_{k}||L^{i}_{j}|+\sum_{i<j\leq
k}|(L^{n})^{j}_{k}||L^{i}_{j}|+\sum_{k<j\leq\nu}^{\nu}|(L^{n})^{j}_{k}||L^{i}_{j}|$
$\displaystyle\overset{\eqref{inmpiisdsdsd1},\eqref{liksohsds1}}{\leq}\left(3\nu
R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\right)^{n-1}\left(\sum_{1\leq j\leq
i}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{i}}\right)^{3-\alpha}\right.$
$\displaystyle\left.+\sum_{i<j\leq
k}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{j}}\right)^{3-\alpha}+\sum_{k<j\leq\nu}\left(\frac{\mathtt{j}_{k}}{\mathtt{j}_{j}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{j}}\right)^{3-\alpha}\right)$
(B.183)

Using that $0<\mathtt{j}_{1}<\cdots<\mathtt{j}_{\nu}$, we have

$\displaystyle\sum_{1\leq j\leq
i}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{i}}\right)^{3-\alpha}$
$\displaystyle\leq\nu
R_{S_{0}^{+}}^{3-\alpha}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}},$
$\displaystyle\sum_{i<j\leq
k}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{j}}\right)^{3-\alpha}$
$\displaystyle=\sum_{i<j\leq
k}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{j}}{\mathtt{j}_{i}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{j}}\right)^{3-\alpha}\leq\nu
R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}},$
$\displaystyle\sum_{k<j\leq\nu}\left(\frac{\mathtt{j}_{k}}{\mathtt{j}_{j}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{j}}\right)^{3-\alpha}$
$\displaystyle=\sum_{k<j\leq\nu}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}\mathtt{j}_{k}^{2}}{\mathtt{j}_{j}^{3}}\right)^{\frac{3-\alpha}{2}}\leq\nu
R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}}.$

Plugging these estimates into (B.4), we obtain (B.182) for $n+1$, therefore,
the claim is proved.

From the definition of $\tilde{L}$ in (B.181), we see that (B.182) implies

$\displaystyle|\tilde{L}^{i}_{k}|\leq_{\alpha,\nu}R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\min\left\\{\left(\frac{\mathtt{j}_{i}}{\mathtt{j}_{k}}\right)^{\frac{3-\alpha}{2}},\left(\frac{\mathtt{j}_{k}}{\mathtt{j}_{i}}\right)^{\frac{3-\alpha}{2}}\right\\}.$
(B.184)

Recalling from (B.128), we have that

$\displaystyle I_{S_{0}^{+}}$
$\displaystyle=\vec{u}\cdot\left((D_{S_{0}^{+}})^{-\alpha}\mathbb{A}_{S_{0}^{+}}(D_{S_{0}^{+}})^{-3}\right)^{-1}\vec{u}$
$\displaystyle=\vec{u}\cdot\left((D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}(D_{S_{0}^{+}})^{-\frac{\alpha+3}{2}}\mathbb{A}_{S_{0}^{+}}(D_{S_{0}^{+}})^{-\frac{\alpha+3}{2}}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\right)^{-1}\vec{u}$
$\displaystyle=\vec{u}\cdot(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\left((D_{S_{0}^{+}})^{-\frac{\alpha+3}{2}}\mathbb{A}_{S_{0}^{+}}(D_{S_{0}^{+}})^{-\frac{\alpha+3}{2}}\right)^{-1}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}$
$\displaystyle\overset{\eqref{rkksdsdsdsaldksgotj}}{=}(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{u}\cdot((2^{\alpha+1}-4)I+2L)^{-1}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}$
$\displaystyle\overset{\eqref{ltidlejssd2}}{=}(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{u}\cdot\left((2^{\alpha+1}-4)^{-1}(I-\tilde{L})\right)(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}$
$\displaystyle=(2^{\alpha+1}-4)^{-1}\left(|\vec{u}|^{2}+(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{u}\cdot\tilde{L}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}\right)$
$\displaystyle\overset{\eqref{Idjsdxc1}}{=}(2^{\alpha+1}-4)^{-1}\left(\nu+\sum_{i,k}\mathtt{j}_{k}^{\frac{3-\alpha}{2}}\tilde{L}^{i}_{k}\mathtt{j}_{i}^{\frac{\alpha-3}{2}}\right)$
$\displaystyle\overset{\eqref{shhsdasdjwdsd}}{\geq}(2^{\alpha+1}-4)^{-1}\left(\nu-
c_{\alpha,\nu,S_{0}^{+}}R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}\right),$

for some constant $c_{\alpha,\nu,S_{0}^{+}}>0$. Therefore, choosing
$S_{0}^{+}$ so that $R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}$ is sufficiently
small, we have

$I_{S_{0}^{+}}>(2^{\alpha+1}-4)^{-1}\frac{\nu}{2}\overset{\nu\geq
2}{\geq}(2^{\alpha+1}-4)^{-1}.$

Since $\alpha\in(1,2)$, we see that

$2\alpha(\alpha+1)I_{S_{0}^{+}}>2\alpha(\alpha+1)(2^{\alpha+1}-4)^{-1}\geq 3,$

which implies that $1-2\alpha(\alpha+1)I_{S_{0}^{+}}\neq 0$.

Proof of (4). From the definition of $f$ in (B.150), the following properties
can be verified straightforwardly: For each $j\in\mathbb{N}$,

$\displaystyle f(\xi,j)>0,\text{ for all $\xi,j\neq 0$},$ (B.185)
$\displaystyle\xi\mapsto\frac{f(\xi,j)}{|\xi|^{\alpha-1}}\text{ is even and
monotone increasing in $\xi>0$,}$ (B.186)
$\displaystyle\frac{f(\xi,j)}{|\xi|^{\alpha-1}j^{2}}\leq_{\alpha}1\text{ and
}\frac{\xi_{1}f(\xi_{1},j)-\xi_{2}f(\xi_{2},j)}{(\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1})j^{2}}\leq_{\alpha}1,\text{
for all $\xi,\xi_{1},\xi_{2}\neq 0$ and }$ (B.187)

With the above properties of $f$, we will estimate
$(\vec{F}(\xi_{1},\xi_{2}))_{k}$ in (B.152). For the case where one of
$\xi_{1},\xi_{2}$ equals to $0$, we can assume without loss of generality that
$\xi_{1}\neq 0$ and $\xi_{2}=0$, since
$\vec{F}(\xi_{1},\xi_{2})=\vec{F}(\xi_{2},\xi_{1})$. In such case, we have

$\displaystyle(\vec{F}(\xi_{1},\xi_{2}))_{k}\overset{\eqref{vecosdF}}{=}\frac{f(i_{\xi,1},\mathtt{j}_{k})}{{|i_{\xi,1}|^{\alpha-1}\mathtt{j}_{k}^{2}}}\overset{\eqref{fprossd2}}{\geq}\frac{f(2,\mathtt{j}_{k})}{\mathtt{j}_{k}^{2}},$
(B.188)

where the last equality follows from $|{i}_{\xi,1}|>\mathtt{j}_{1}=1$ for
$\xi_{1}\in S^{\perp}$. For an upper boud, we also have

$\displaystyle|(\vec{F}(\xi_{1},\xi_{2}))_{k}|=\left|\frac{f(i_{\xi,1},\mathtt{j}_{k})}{{|i_{\xi,1}|^{\alpha-1}\mathtt{j}_{k}^{2}}}\right|\overset{\eqref{fprossd3}}{\leq_{\alpha}}1.$
(B.189)

Now, let us assume $\xi_{1},\xi_{2}\neq 0$. In this case, it follows from the
explicit formula for $\vec{F}$ in (B.152) and (B.186) that we can without loss
of generality that $\xi_{1}>|\xi_{2}|>0$. For such $\xi_{1},\xi_{2}$, it
follows from (B.185) and (B.186) that

$\displaystyle(\vec{F}(\xi_{1},\xi_{2}))_{k}\overset{\eqref{vecosdF}}{=}\frac{\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{k})-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{k})\right)}{(i_{\xi,1}|i_{\xi,1}|^{\alpha-1}-i_{\xi,2}|i_{\xi,2}|^{\alpha-1})\mathtt{j}_{k}^{2}}\geq\frac{i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{k})}{i_{\xi,1}|i_{\xi,1}|^{\alpha-1}\mathtt{j}_{k}^{2}}\overset{\eqref{fprossd2}}{\geq}\frac{f(2,\mathtt{j}_{k})}{\mathtt{j}_{k}^{2}},$
(B.190)

where the first inequality follows from the elementary fact that
$\frac{b-d}{a-c}\geq\frac{b}{a}$ for $a,b,c,d>0$ such that $a>c$ and
$\frac{b}{a}>\frac{d}{c}$. For an upper bound, we also have

$\displaystyle|(\vec{F}(\xi_{1},\xi_{2}))_{k}|\overset{\eqref{vecosdF}}{=}\left|\frac{\left(i_{\xi,1}f(i_{\xi,1},\mathtt{j}_{k})-i_{\xi,2}f(i_{\xi,2},\mathtt{j}_{k})\right)}{(i_{\xi,1}|i_{\xi,1}|^{\alpha-1}-i_{\xi,2}|i_{\xi,2}|^{\alpha-1})\mathtt{j}_{k}^{2}}\right|\overset{\eqref{fprossd3}}{\leq_{\alpha}}1.$
(B.191)

Therefore, combining the above bounds for all $\xi_{1},\xi_{2}\in
S^{\perp}\left\\{0\right\\}$ such that $\xi_{1}\neq\xi_{2}$, we have

$\displaystyle(\vec{F}(\xi_{1},\xi_{2}))_{k}$
$\displaystyle\overset{\eqref{xssd22111},\eqref{jjrjsdssdsdsd1}}{\geq}\frac{f(2,\mathtt{j}_{k})}{\mathtt{j}_{k}^{2}}>0,$
(B.192) $\displaystyle|(\vec{F}(\xi_{1},\xi_{2}))_{k}|$
$\displaystyle\overset{\eqref{xssd22112},\eqref{jjrjsdssdsdsd2}}{\leq_{\alpha}}1.$
(B.193)

Towards (B.174), we write

$\displaystyle\mathbb{A}_{S_{0}^{+}}-2(D_{S_{0}^{+}})^{3}\vec{F}(\xi_{1},\xi_{2})\vec{u}^{T}(D_{S_{0}^{+}})^{\alpha}$
$\displaystyle=(D_{S_{0}^{+}})^{\frac{\alpha+3}{2}}\left((D_{S_{0}^{+}})^{-\frac{\alpha+3}{2}}\mathbb{A}_{S_{0}^{+}}(D_{S_{0}^{+}})^{-\frac{\alpha+3}{2}}-2(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{F}(\xi_{1},\xi_{2})\vec{u}^{T}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\right)(D_{S_{0}^{+}})^{\frac{\alpha+3}{2}}$
$\displaystyle\overset{\eqref{rkksdsdsdsaldksgotj}}{=}(2^{\alpha+1}-4)(D_{S_{0}^{+}})^{\frac{\alpha+3}{2}}\underbrace{\left(I+\frac{2}{2^{\alpha+1}-4}L-\frac{2}{2^{\alpha+1}-4}(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{F}(\xi_{1},\xi_{2})\vec{u}^{T}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\right)}_{=:\mathbb{B}_{S_{0}^{+}}(\xi_{1},\xi_{2})}(D_{S_{0}^{+}})^{\frac{\alpha+3}{2}}$
(B.194)

Using the norm in (B.83), it follows from (B.180) and (B.193) that

$\displaystyle|\mathbb{B}_{S_{0}^{+}}(\xi_{1},\xi_{2})|\leq_{\alpha,\nu,S_{0}^{+}}1$
(B.195)

where the upper bound is independent of $\xi_{1},\xi_{2}$. This implies that
the largest eigenvalue of $\mathbb{B}_{S_{0}^{+}}$ is bounded. Furthermore, we
recall the matrix determinant Lemma:

$\displaystyle\text{det}(A+\vec{U}\vec{V}^{T})=(1+\vec{V}\cdot
A^{-1}\vec{U})\text{det}(A),\text{ for $\vec{U},\vec{V}\in\mathbb{R}^{\nu},\
A\in\mathbb{R}^{\nu\times\nu}$}.$ (B.196)

Applying this to $\mathbb{B}_{S_{0}^{+}}$, we get

$\displaystyle\text{det}(\mathbb{B}_{S_{0}^{+}}(\xi_{1},\xi_{2}))$
$\displaystyle=\left(1-\frac{2}{2^{\alpha+1}-4}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}\cdot\left(I+\frac{2}{2^{\alpha+1}-4}L\right)^{-1}(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{F}(\xi_{1},\xi_{2})\right)\text{det}\left(I+\frac{2}{2^{\alpha+1}-4}L\right)$
$\displaystyle\overset{\eqref{ltidlejssd2}}{=}\left(1-\frac{2}{2^{\alpha+1}-4}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}\cdot(I-\tilde{L})(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{F}(\xi_{1},\xi_{2})\right)\text{det}\left(I+\frac{2}{2^{\alpha+1}-4}L\right).$
(B.197)

We can further compute

$\displaystyle\frac{2}{2^{\alpha+1}-4}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}\cdot(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{F}(\xi_{1},\xi_{2})$
$\displaystyle\overset{\eqref{vecudefsd}}{=}\frac{2}{2^{\alpha+1}-4}\sum_{k=1}^{\nu}(\vec{F}(\xi_{1},\xi_{2}))_{k}$
$\displaystyle\overset{\eqref{sjsjdjsdad1sxfF}}{\geq}\frac{2}{2^{\alpha+1}-4}\sum_{k=1}^{\nu}\frac{f(2,\mathtt{j}_{k})}{\mathtt{j}_{k}^{2}}$
$\displaystyle\overset{\eqref{fprossd1},\eqref{forms1}}{\geq}\frac{2}{2^{\alpha+1}-4}{f(2,1)}$
$\displaystyle\overset{\eqref{fesjsdxc}}{=}\frac{2(3^{1+\alpha}-2^{2+\alpha}-1)}{2^{\alpha+1}-4}$
$\displaystyle\geq 3,$ (B.198)

where the last inequality can be straightforwardly proved using that
$\alpha\in(1,2)$. Also, we have

$\displaystyle\left|\frac{2}{2^{\alpha+1}-4}(D_{S_{0}^{+}})^{\frac{\alpha-3}{2}}\vec{u}\cdot\tilde{L}(D_{S_{0}^{+}})^{\frac{3-\alpha}{2}}\vec{F}(\xi_{1},\xi_{2})\right|$
$\displaystyle\leq_{\alpha}\sum_{k,i=1}^{\nu}\mathtt{j}_{i}^{\frac{\alpha-3}{2}}|\tilde{L}^{i}_{k}|\mathtt{j}_{k}^{\frac{3-\alpha}{2}}|(\vec{F}(\xi_{1},\xi_{2}))_{k}|$
$\displaystyle\overset{\eqref{shhsdasdjwdsd},\eqref{sjsjdjsdad1sxfF2}}{\leq_{\alpha,\nu}}R_{S_{0}^{+}}^{\frac{3-\alpha}{2}}$
$\displaystyle\leq 1,$ (B.199)

for sufficiently small $R_{S_{0}^{+}}$. Plugging this and (B.198) into (B.197)
and using that $I+\frac{2}{2^{\alpha+1}-4}L$ is invertible, we see that

$\displaystyle|\text{det}(\mathbb{B}_{S_{0}^{+}}(\xi_{1},\xi_{2}))|\geq_{\alpha,\nu}1,$
(B.200)

independently of $\xi_{1},\xi_{2}$. Combining this with (B.195), we see that

$\displaystyle\min_{\vec{U}\in\mathbb{R}^{\nu},\
|\vec{U}|=1}\left|\vec{U}\cdot\mathbb{B}_{S_{0}^{+}}(\xi_{1},\xi_{2})\vec{U}\right|\geq_{\alpha,\nu,S_{0}^{+}}1.$

In view of (B.194), this implies (B.174). ∎

######  Proposition B.16.

For each $2\leq\nu\in\mathbb{N}$ and $\alpha\in(1,2)$, there exist infinite
number of choices of $S_{0}^{+}$ such that for sufficiently large $\mathtt{M}$
depending on $\alpha,\nu,S_{0}^{+}$, the following hold:

  1. (1)

The tangential sites $S$ constructed by (4.6) satisfies S.1 and S.2.

  2. (2)

$(\mathtt{H}1)$ holds true. That is, $\mathbb{A}$ in (5.10) is invertible.

  3. (3)

$(\mathtt{H}2)$ holds true. That is, there exists a constant
$C_{\mathtt{H}2}({S}_{0}^{+},\nu,\alpha,\mathtt{M})>0$ such that if
$\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}$ and
$\max\left\\{|\xi_{1}|,|\xi_{2}|\right\\}\geq C_{\mathtt{H}2}$, then

$\displaystyle|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1})\leq_{\alpha,\nu,S}\left|\left(W(\xi_{1})-\frac{\pi}{6}\xi_{1}\vec{D}(\xi_{1})\cdot\mathbb{A}^{-1}\overline{\omega}\right)-\left(W(\xi_{2})-\frac{\pi}{6}\xi_{2}\vec{D}(\xi_{2})\cdot\mathbb{A}^{-1}\overline{\omega}\right)\right|.$
(B.201)

  4. (4)

$(\mathtt{H}3)$ holds true. That is,
$\text{det}(\mathbb{C}_{\xi_{1},\xi_{2}})\neq 0$ for all $\xi_{1},\xi_{2}\in
S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}$.

###### Proof.

Let us choose $S_{0}^{+}$ such that (1)-(4) in Lemma B.15 are satisfied. For
such a choice of $S_{0}^{+}$, item (1) follows trivially from (1) of Lemma
B.15.

Proof of (2). (2) follows from (2) of Lemma B.15 and (B.9).

Proof of (3). It follows from (B.126) that (B.201) is equivalent to

$\displaystyle|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-2})\leq_{\alpha,\nu,S}\left|\mathtt{A}(\xi_{1})-\mathtt{A}(\xi_{2})\right|.$
(B.202)

From (3) in Lemma B.15 and Lemma B.13, there exist two constants
$C_{1}(\alpha,\nu,S_{0}^{+})\neq 0\ ,C_{2}(\alpha,\nu,S_{0}^{+})>0$ such that
for $|\xi|\geq 2\text{max}_{S^{+}}:=2\max\left\\{j_{k}:j_{k}\in
S^{+}\right\\}$,

$\displaystyle\mathtt{A}(\xi)$
$\displaystyle=C_{1}\xi|\xi|^{\alpha-1}+m_{\mathtt{A},r}(\xi),$ (B.203)
$\displaystyle|m_{\mathtt{A},r}(\xi)|$
$\displaystyle{\leq}C_{2}\left(\mathtt{M}^{1-\alpha}|\xi|^{\alpha}+\mathtt{M}|\xi|^{\alpha-1}\right),$
(B.204) $\displaystyle|m_{\mathtt{A},r}(\xi_{1})-m_{\mathtt{A},r}(\xi_{2})|$
$\displaystyle{\leq}C_{2}|\xi_{1}-\xi_{2}|(\mathtt{M}^{1-\alpha}(|\xi_{1}|+|\xi_{2}|)^{\alpha-1}+\mathtt{M}(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}).$
(B.205)

We also see from (B.126) and (11.101) that for all $\xi,\xi_{1},\xi_{2}\in
S_{\mathtt{M}}^{\perp}\cup\left\\{0\right\\}$,

$\displaystyle|\mathtt{A}(\xi)|$ $\displaystyle\leq C_{3}|\xi|^{\alpha},$
(B.206)

for some $C_{3}=C_{3}(\alpha,\nu,S_{0}^{+},\mathtt{M})>0$. In addition, it is
straightforward to see that

$\displaystyle|\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1}|\geq
c_{\alpha}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1}),$
(B.207)

for some $c_{\alpha}>0$, that depends on only $\alpha\in(1,2)$. Without loss
of generality, we can assume that

$\displaystyle c_{\alpha}<1.$ (B.208)

To prove (B.202), let us choose $\mathtt{M}$ and $C_{\mathtt{H}2}$ so that

$\displaystyle
c_{\alpha}|C_{1}|-C_{2}\mathtt{M}^{1-\alpha}\geq\frac{c_{\alpha}|C_{1}|}{2},\quad
C_{\mathtt{H}2}\geq\max\left\\{\frac{4C_{2}\mathtt{M}}{|C_{1}|c_{\alpha}},2\text{max}_{S^{+}},\frac{16C_{3}\text{max}_{S+}}{|C_{1}|c_{\alpha}}\right\\},$
(B.209)

which is possible since $C_{1}$ and $C_{2}$ depend on $\alpha,\nu,S_{0}^{+}$
only. Note that without loss of generality, we can assume that
$|\xi_{1}|\geq|\xi_{2}|$, so that

$\displaystyle|\xi_{1}|\geq C_{\mathtt{H}2}.$ (B.210)

Then we consider two cases: 1) $|\xi_{2}|\leq 2\text{max}_{S^{+}}$ and 2)
$|\xi_{2}|\geq 2\text{max}_{S^{+}}$.

In the first case, we have that

$\displaystyle|\mathtt{A}(\xi_{2})|\overset{\eqref{rlkksdowow1}}{\leq}C_{3}|\xi_{2}|^{\alpha}\leq
2\text{max}_{S^{+}}C_{3}.$ (B.211)

while $\mathtt{A}(\xi_{1})$ can be estimated as

$\displaystyle|\mathtt{A}(\xi_{1})|$
$\displaystyle\overset{\eqref{rjjsdsdsdsd1},\eqref{xisd1alsdwd}}{\geq}|C_{1}||\xi_{1}|^{\alpha}-|m_{\mathtt{A},r}(\xi_{1})|$
$\displaystyle\overset{\eqref{skkksdhsd2123},\eqref{csdjsjd1}}{\geq}\left(c_{\alpha}|C_{1}|-C_{2}\mathtt{M}^{1-\alpha}\right)|\xi_{1}|^{\alpha}-C_{2}\mathtt{M}|\xi_{1}|^{\alpha-1}$
$\displaystyle\overset{\eqref{cjjjsdiidwda},\eqref{xisd1alsdwd}}{\geq}\frac{c_{\alpha}|C_{1}|}{2}|\xi_{1}|^{\alpha}-\frac{c_{\alpha}|C_{1}|}{4}|\xi_{1}|^{\alpha}$
$\displaystyle\geq\frac{c_{\alpha}|C_{1}|}{4}|\xi_{1}|^{\alpha}.$ (B.212)

Therefore, we have

$\displaystyle|\mathtt{A}(\xi_{1})-\mathtt{A}(\xi_{2})|\geq|\mathtt{A}(\xi_{1})|-|\mathtt{A}(\xi_{2})|\overset{\eqref{skjsdsdxi2222s},\eqref{largemsdjxck1}}{\geq}\frac{c_{\alpha}|C_{1}|}{4}|\xi_{1}|-2\text{max}_{S^{+}}C_{3}\overset{\eqref{cjjjsdiidwda},\eqref{xisd1alsdwd}}{\geq}\frac{c_{\alpha}|C_{1}|}{8}|\xi_{1}|,$
(B.213)

while, it holds that

$\displaystyle|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1})\overset{{|\xi_{1}|\geq|\xi_{2}|}}{\leq}4|\xi_{1}|^{\alpha}.$
(B.214)

Combining this with (B.213), we obtain

$\displaystyle|\mathtt{A}(\xi_{1})-\mathtt{A}(\xi_{2})|\geq\frac{c_{\alpha}|C_{1}|}{16}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1}),\text{
if $|\xi_{2}|\leq 2\text{max}_{S^{+}}$.}$ (B.215)

In the second case, assuming $|\xi_{2}|\geq 2\text{max}_{S+}$, we have

$\displaystyle|\mathtt{A}(\xi_{1})-\mathtt{A}(\xi_{2})|$
$\displaystyle\overset{\eqref{rjjsdsdsdsd1}}{\geq}|C_{1}||\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1}|-|m_{\mathtt{A},r}(\xi_{1})-m_{\mathtt{A},r}(\xi_{2})|$
$\displaystyle\overset{\eqref{skkksdhsd2223}}{\geq}|C_{1}||\xi_{1}|\xi_{1}|^{\alpha-1}-\xi_{2}|\xi_{2}|^{\alpha-1}|-C_{2}|\xi_{1}-\xi_{2}|(\mathtt{M}^{1-\alpha}(|\xi_{1}|+|\xi_{2}|)^{\alpha-1}+\mathtt{M}(|\xi_{1}|+|\xi_{2}|)^{\alpha-2})$
$\displaystyle\overset{\eqref{alpsdoosd1},\eqref{cjjjsdiidwda}}{\geq}\frac{|C_{1}|c_{\alpha}}{2}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1})-C_{2}\mathtt{M}(|\xi_{1}|+|\xi_{2}|)^{\alpha-2}$
$\displaystyle\geq\frac{|C_{1}|c_{\alpha}}{2}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1})-\frac{C_{2}\mathtt{M}}{|\xi_{1}|}(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1})$
$\displaystyle\overset{\eqref{cjjjsdiidwda},\eqref{xisd1alsdwd}}{\geq}\frac{|C_{1}|c_{\alpha}}{4}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1}).$
(B.216)

Combining this with (B.215), we obtain

$\displaystyle|\mathtt{A}(\xi_{1})-\mathtt{A}(\xi_{2})|\geq\frac{|C_{1}|c_{\alpha}}{16}|\xi_{1}-\xi_{2}|(|\xi_{1}|^{\alpha-1}+|\xi_{2}|^{\alpha-1}),\text{
if $\max\left\\{|\xi_{1}|,|\xi_{2}|\right\\}\geq C_{\mathtt{H}2}$ and
$\xi_{1},\xi_{2}\in S_{\mathtt{M}}^{\perp}\left\\{0\right\\}$}$ (B.217)

This proves (B.202).

Proof of (4). From (B.174),and (B.151), we see that for some
$C_{1}(\alpha,\nu,S_{0}^{+}),C_{2}(\alpha,\nu,S_{0}^{+})>0$,

$\displaystyle\min_{\vec{U}\in\mathbb{R}^{\nu},\
|\vec{U}|=1}\left|\vec{U}\cdot\mathbb{C}_{\xi_{1},\xi_{2}}\vec{U}\right|\geq
C_{1}\mathtt{M}^{\alpha+3}-|\vec{U}\cdot W_{r,*}\vec{U}|$ $\displaystyle\geq
C_{1}\mathtt{M}^{\alpha+3}-C_{2}\mathtt{M}^{4}.$

Since $\alpha>1$, we can choose $\mathtt{M}$ large enough, depending on
$\alpha,\nu,S_{0}^{+}$ so that

$\min_{\vec{U}\in\mathbb{R}^{\nu},\
|\vec{U}|=1}\left|\vec{U}\cdot\mathbb{C}_{\xi_{1},\xi_{2}}\vec{U}\right|\geq\frac{C_{1}}{2}{\mathtt{M}}^{\alpha+3}.$

Therefore, $\text{det}(\mathbb{C}_{\xi_{1},\xi_{2}})\neq 0$. ∎

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Javier Gómez-Serrano
---
Department of Mathematics
Brown University
314 Kassar House, 151 Thayer St.
Providence, RI 02912, USA
Email<EMAIL_ADDRESS>
Alexandru D. Ionescu
Department of Mathematics
Princeton University
705 Fine Hall, Washington Rd,
Princeton, NJ 08544, USA
e-mail<EMAIL_ADDRESS>
Jaemin Park
Department of Mathematics and Computer Science
University of Basel