b_{1}^{2\delta-2}\mathscr{E}_{\Bbbk}\left[b_{1}^{2}b_{1}^{(2-2\delta)(1+\eta)}+b_{1}^{(2-2\delta)(1+\eta)}+b_{1}^{4}b_{1}^{-2\delta(1+\eta)}\right]\quad\lesssim
b_{1}^{2\eta(1-\delta)}\mathscr{E}_{\Bbbk}.$ (4.40)

This concludes the proof of (4.38) for $\Sigma_{0}$.

We now prove the estimate (4.38) for $\Sigma_{1}$. From the Cauchy-Schwartz
inequality, we write

$|\Sigma_{1}|\lesssim\sqrt{\mathscr{E}_{\Bbbk}}\left[\left\|\big{(}\mathbf{M}_{1}-\partial_{s}\Upsilon_{1}\big{)}_{\Bbbk}\right\|_{L^{2}}+\left\|\big{(}\mathbf{M}_{2}-\partial_{s}\Upsilon_{2}\big{)}_{\Bbbk-1}\right\|_{L^{2}}\right].$

We only deal with the second coordinate because the first one is estimated in
the same way. Let us write

$\displaystyle\mathbf{M}_{2}-\partial_{s}\Upsilon_{2}$
$\displaystyle=\sum_{k=1,\textup{odd}}^{L-1}\big{[}(b_{k})_{s}+(k-\gamma)b_{1}b_{k}-b_{k-1}\big{]}\left(T_{k}+\sum_{j=k+1,\textup{odd}}^{L+2}\frac{\partial
S_{j}}{\partial
b_{k}}\right)\chi_{B_{1}}-\left(\frac{\lambda_{s}}{\lambda}+b_{1}\right)D(\mathbf{Q}_{b})_{2}$
$\displaystyle+\left[(b_{L})_{s}+(L-\gamma)b_{1}b_{L}-\frac{d}{ds}C_{\Upsilon}\right]\left(T_{L}+\frac{\partial
S_{L+2}}{\partial
b_{L}}\right)\chi_{B_{1}}-C_{\Upsilon}\frac{d}{ds}\left[\left(T_{L}+\frac{\partial
S_{L+2}}{\partial b_{L}}\right)\chi_{B_{1}}\right].$

Since $\mathscr{L}^{\frac{L+1}{2}}T_{k}=0$ for $1\leq k\leq L$, we us the
admissibility of $\vec{T}_{k}$ to estimate

$\sum_{k=1,\textup{odd}}^{L}\left\|\big{(}\chi_{B_{1}}T_{k}\big{)}_{\Bbbk-1}\right\|_{L^{2}}^{2}\lesssim\sum_{k=1,\textup{odd}}^{L}\int_{y\sim
B_{1}}y^{2(k-\gamma-\Bbbk)}\lesssim\int_{y\sim
B_{1}}y^{2(-\gamma-\hbar-1)}\lesssim b_{1}^{2(1-\delta)(1+\eta)}.$

From the homogeneity of $\vec{S}_{k}$, we have

$\sum_{k=1,\textup{odd}}^{L}\sum_{j=k+1,\textup{odd}}^{L+2}\left\|\left(\frac{\partial
S_{j}}{\partial
b_{k}}\chi_{B_{1}}\right)_{\Bbbk-1}\right\|_{L^{2}}^{2}\lesssim\sum_{k=1,\textup{odd}}^{L}\sum_{j=k+1,\textup{odd}}^{L+2}\int_{y\leq
2B_{1}}\frac{b_{1}^{4}}{1+y^{2(-j+1+\gamma+\Bbbk)}}\lesssim
b_{1}^{2(1-\delta)(1+\eta)}.$

Similarly, since $\left|\frac{d}{ds}\chi_{b_{1}}\right|\lesssim b_{1}$ and
$\frac{\partial S_{L+2}}{\partial b_{L}}$ does not depend on $b_{L}$, we have
the estimates

$\displaystyle\left\|\big{(}D(\mathbf{Q}_{b})_{2}\big{)}_{\Bbbk-1}\right\|^{2}_{L^{2}}\lesssim
b_{1}^{2(1-\delta)(1+\eta)},\quad\left\|\left(\frac{d}{ds}\left(T_{L}+\frac{\partial
S_{L+2}}{\partial
b_{L}}\right)\chi_{B_{1}}\right)_{\Bbbk-1}\right\|_{L^{2}}^{2}\lesssim
b_{1}^{2}b_{1}^{2(1-\delta)(1+\eta)}.$

Gathering all these above estimates together with the modulation equations
(4.5), (4.9) and the estimate (4.37) yields

$\left\|\big{(}\mathbf{M}_{2}-\partial_{s}\Upsilon_{2}\big{)}_{\Bbbk-1}\right\|_{L^{2}}\lesssim
b_{1}^{1+\eta(1-\delta)}\sqrt{\mathscr{E}_{\Bbbk}}+b_{1}^{L+1+(1-\delta)(2+\eta)},$
(4.41)

which follows the estimate (4.38) for $\Sigma_{1}$.

We now turn to the proof of (4.38) for $\Sigma_{2}$. We only deal with the
second coordinate because the same proof holds for the first one. Let us write
from equation (3.9),

$\displaystyle\int\Upsilon_{2}\mathscr{L}^{\Bbbk-1}\left(\partial_{s}q_{2}+\partial_{s}\Upsilon_{2}\right)$
$\displaystyle\quad=\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}\left(\frac{\lambda_{s}}{\lambda}Dq_{2}-\mathscr{L}q_{1}-(\mathbf{\Psi}_{b})_{2}-\mathbf{M}_{2}+\partial_{s}\Upsilon_{2}+L(q_{1})-N(q_{1})\right).$

Using the admissibility of $\vec{T}_{L}$, the homogeneity of $\vec{S}_{L+2}$
and the fact that $\mathscr{L}^{\frac{L+1}{2}}T_{L}=0$, we have

$\displaystyle\left|\mathscr{L}^{\Bbbk-1}\Upsilon_{2}\right|$
$\displaystyle\lesssim|C_{\Upsilon}|\left(\frac{1}{1+y^{-L+1+\gamma+2\Bbbk-2}}\mathbf{1}_{\\{B_{1}\leq
y\leq
2B_{1}\\}}+\frac{b_{1}^{2}}{1+y^{-L+\gamma+2\Bbbk-2}}\mathbf{1}_{\\{y\leq
2B_{1}\\}}\right)$ $\displaystyle\quad\lesssim
b_{1}^{\delta-1}\sqrt{\mathscr{E}_{\Bbbk}}\frac{b_{1}^{1+\eta}}{1+y^{\Bbbk-1+\gamma+\hbar}}\mathbf{1}_{\\{y\leq
2B_{1}\\}}.$ (4.42)

From (4.42), the coercivity bound (B.3) and $-\frac{\lambda_{s}}{\lambda}\sim
b_{1}$, we estimate

$\displaystyle\left|\frac{\lambda_{s}}{\lambda}\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}Dq_{2}\right|$
$\displaystyle\quad\lesssim
b_{1}^{1+\delta+\eta}\sqrt{\mathscr{E}_{\Bbbk}}\left[\left(\int\frac{|q_{2}|^{2}}{1+y^{2(\Bbbk-1)}}\right)^{\frac{1}{2}}+\left(\int\frac{|\partial_{y}q_{2}|^{2}}{1+y^{2(\Bbbk-2)}}\right)^{\frac{1}{2}}\right]\left(\int_{y\leq
2B_{1}}\frac{1}{1+y^{2(\gamma+\hbar)}}\right)^{\frac{1}{2}}$
$\displaystyle\qquad\lesssim
b_{1}^{1+\delta+\eta}\mathscr{E}_{\Bbbk}b_{1}^{-\delta(1+\eta)}\quad\lesssim
b_{1}^{1+\eta(1-\delta)}\mathscr{E}_{\Bbbk}.$

Similarly, we have

$\displaystyle\left|\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}\mathscr{L}q_{1}\right|\lesssim
b_{1}^{\delta-1}\sqrt{\mathscr{E}_{\Bbbk}}\left(\int\frac{|(q_{1})_{2}|^{2}}{1+y^{2(\Bbbk-2)}}\right)^{\frac{1}{2}}\left(\int_{y\leq
2B_{1}}\frac{1}{1+y^{2(\gamma+\hbar+1)}}\right)^{\frac{1}{2}}\lesssim
b_{1}^{1+\eta(1-\delta)}\mathscr{E}_{\Bbbk}.$

Using (2.46) and (4.40) yields

$\displaystyle\left|\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}(\mathbf{\Psi}_{b})_{2}\right|\lesssim\left\|\big{(}(\mathbf{\Psi}_{b})_{2}\big{)}_{\Bbbk-1}\right\|_{L^{2}}\left\|\big{(}\Upsilon_{2}\big{)}_{\Bbbk-1}\right\|_{L^{2}}\lesssim
b_{1}^{1+\eta(1-\delta)}\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+(1-\delta)(1+\eta)}.$

From (4.41) and (4.40), we have

$\displaystyle\left|\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}\big{(}\mathbf{M}_{2}-\partial_{s}\Upsilon_{2}\big{)}\right|$
$\displaystyle\lesssim\left\|\big{(}\mathbf{M}_{2}-\partial_{s}\Upsilon_{2}\big{)}_{\Bbbk-1}\right\|_{L^{2}}\left\|\big{(}\Upsilon_{2}\big{)}_{\Bbbk-1}\right\|_{L^{2}}$
$\displaystyle\quad\lesssim
b_{1}^{1+2\eta(1-\delta)}\mathscr{E}_{\Bbbk}+b_{1}^{1+\eta(1-\delta)}\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+(1-\delta)(1+\eta)}.$

Note from the definition (3.11) of $L(q_{1})$ that

$|L(q_{1})|\lesssim\frac{b_{1}^{2(1-\eta)}|q_{1}|}{1+y^{2\gamma}}.$

Thus, by using (4.42) and the coercivity bound (B.3), we derive

$\left|\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}L(q_{1})\right|\lesssim
b_{1}^{1+\eta(1-\delta)}\mathscr{E}_{\Bbbk}.$

For the nonlinear term $N(q_{1})$ defined in (3.12), we note that
$|N(q_{1})|\lesssim\frac{q_{1}^{2}}{y^{2}}$, we then use (4.42), the
coercivity bound (B.1) and the bootstrap bound given in Definition 3.2 to
estimate

$\displaystyle\left|\int\mathscr{L}^{\Bbbk-1}\Upsilon_{2}N(q_{1})\right|$
$\displaystyle\lesssim
b_{1}^{\delta+\eta}\sqrt{\mathscr{E}_{\Bbbk}}\int_{y\leq
2B_{1}}\frac{q_{1}^{2}}{y^{2}(1+y^{2\Bbbk-2})}(1+y^{\Bbbk-1-\gamma-\hbar})$
$\displaystyle\quad\lesssim
b_{1}^{\delta+\eta}\sqrt{\mathscr{E}_{\Bbbk}}\mathscr{E}_{\Bbbk}b_{1}^{(-L+\gamma)(1+\eta)}\lesssim
b_{1}^{1+\eta(1-\delta)}\mathscr{E}_{\Bbbk}.$

This finishes the proof of (4.38) for $\Sigma_{2}$.

A collection of the estimates (4.20), (4.21), (4.28), (4.34), (4.35) and
(4.39) into the identity (4.16) yields the formula (4.14). This concludes the
proof of Proposition 4.5. ∎

###  4.3. Local Morawetz control.

We establish in this subsection the so-called Morawetz type identity in order
to control the local term $\mathscr{E}_{\Bbbk,\text{loc}}$ involved in the
formula (4.14). In particular, we have the following:

######  Proposition 4.6 (Local Morawetz control).

Let $0<\nu\ll 1$ and $A\gg 1$ be small and large enough constants, we define

$\phi_{A}(y)=\int_{0}^{y}\chi_{A}(\xi)\xi^{1-\nu}d\xi,$ (4.43)

and

$Mc=-\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}(q_{2})_{\Bbbk-1}.$
(4.44)

Then the following bounds hold for all $s\in[s_{0},s_{1}]$ for $s_{0}$ large
enough,

$|Mc|\leq C(A,M)\mathscr{E}_{\Bbbk},$ (4.45)

and

$\frac{d}{dt}\left[\frac{Mc}{\lambda^{2\Bbbk-d}}\right]\geq\frac{1}{\lambda^{2\Bbbk-d+1}}\left(\frac{\nu}{2N^{\nu}}\mathscr{E}_{\Bbbk,\textup{loc}}-\frac{C(M)}{A^{\nu}}\mathscr{E}_{\Bbbk}-C(A,M)\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+1+(1-\delta)(1+\eta)}\right),$
(4.46)

where the large constants $M$, $N$ and $\mathscr{E}_{\Bbbk,\textup{loc}}$ are
introduced in (3.4) and (4.15).

###### Proof.

The estimate (4.45) simply follows from the coercivity bound (B.1). We aim at
proving the bound

$\frac{d}{ds}Mc\geq\frac{\nu}{2N^{\nu}}\mathscr{E}_{\Bbbk,\textup{loc}}-\frac{C(M)}{A^{\nu}}\mathscr{E}_{\Bbbk}-C(A,M)\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+1+(1-\delta)(1+\eta)},$
(4.47)

which immediately implies (4.46). Indeed, from
$-\frac{\lambda_{s}}{\lambda}\sim b_{1}$ and the bound (4.45), we have

$\displaystyle\frac{d}{dt}\left[\frac{Mc}{\lambda^{2\Bbbk-d}}\right]=\frac{1}{\lambda^{2\Bbbk-d+1}}\frac{d}{ds}Mc-\frac{(2\Bbbk-d)\lambda_{s}}{\lambda^{2\Bbbk-d+1}\lambda}Mc$
$\displaystyle\quad\geq\frac{1}{\lambda^{2\Bbbk-d+1}}\left[\frac{\nu}{2N^{\nu}}\mathscr{E}_{\Bbbk,\textup{loc}}-\frac{C(M)}{A^{\nu}}\mathscr{E}_{\Bbbk}-C(A,M)\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+1+(1-\delta)(1+\eta)}\right]-C(A,M)\frac{b_{1}}{\lambda^{2\Bbbk-d+1}}\mathscr{E}_{\Bbbk}.$

Since $b_{1}(s)\leq b_{1}(s_{0})\sim\frac{1}{s_{0}}$, we can take
$s_{0}=s_{0}(A)$ large such that $b_{1}(s_{0})\leq\frac{1}{A}$, then the
estimate (4.46) follows.

Let us give the proof of (4.47). We first claim the following:

######  Lemma 4.7 (Morawetz type identity at the linear level).

Let $A\gg 1$ and $0<\nu\ll 1$, there holds the following:

$\displaystyle\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}\mathscr{L}(q_{1})_{\Bbbk-1}$
$\displaystyle\quad-\int\big{[}\nabla\phi_{A}.\nabla(q_{2})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{2})_{\Bbbk-1}\big{]}(q_{2})_{\Bbbk-1}\quad\gtrsim\frac{\nu}{N^{\nu}}\mathscr{E}_{\Bbbk,\textup{loc}}-\frac{1}{A^{\nu}}\mathscr{E}_{\Bbbk}.$
(4.48)

###### Proof.

The proof follows exactly the same lines as Lemma 3.8 in [C.(2016b)] because
we have the same definition $\phi_{A}$ and a similar structure of the linear
operator $\mathscr{L}$. Although the potential $\frac{Z}{y^{2}}$ is different
from the one defined in [C.(2016b)], it still satisfies

$\frac{1}{2}y\partial_{y}\left(\frac{Z}{y^{2}}\right)\geq-\frac{1}{y^{2}}\left[\left(\frac{d-2}{2}\right)^{2}-\kappa(d)\right]\quad\text{for
some $\kappa(d)>0$ and $d\geq 7$},$

thanks to the asymptotic behavior (2.4) and the fact that
$\frac{(d-2)^{2}}{4}-(d-1)\geq\frac{1}{4}$ for $d\geq 7$. For that reason, we
refer the interested reader to [C.(2016b)] for details of the proof. ∎

We now use the identity (4.48) to derive the formula (4.47). We compute from
the definition of $Mc$ and the equation (3.9),

$\displaystyle\frac{d}{ds}Mc$
$\displaystyle=-\int\nabla\phi_{A}.\nabla\left(q_{2}-\frac{\lambda_{s}}{\lambda}\Lambda
q_{1}-(\mathbf{\Psi}_{b})_{1}-\mathbf{M}_{1}\right)_{\Bbbk-1}(q_{2})_{\Bbbk-1}$
$\displaystyle-\int\frac{1-\nu}{2}\Delta\phi_{A}\left(q_{2}-\frac{\lambda_{s}}{\lambda}\Lambda
q_{1}-(\mathbf{\Psi}_{b})_{1}-\mathbf{M}_{1}\right)_{\Bbbk-1}(q_{2})_{\Bbbk-1}$
$\displaystyle+\int\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}\left(\mathscr{L}q_{1}+\frac{\lambda_{s}}{\lambda}Dq_{2}+(\mathbf{\Psi}_{b})_{2}+\mathbf{M}_{2}-L(q_{1})+N(q_{1})\right)_{\Bbbk-1}$
$\displaystyle+\int\frac{1-\nu}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\left(\mathscr{L}q_{1}+\frac{\lambda_{s}}{\lambda}Dq_{2}+(\mathbf{\Psi}_{b})_{2}+\mathbf{M}_{2}-L(q_{1})+N(q_{1})\right)_{\Bbbk-1}.$
(4.49)

From the definitions of $\mathscr{A}$ and $\mathscr{A}^{*}$, the coercivity
bound (B.1), $-\frac{\lambda_{s}}{\lambda}\sim b_{1}$ and the compactness of
the support of $\nabla\phi_{A}$ and $\Delta\phi_{A}$, we have the estimate

$\displaystyle\left|\frac{\lambda_{s}}{\lambda}\int\big{[}\nabla\phi_{A}.\nabla\left(\Lambda
q_{1}\right)_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}\left(\Lambda
q_{1}\right)_{\Bbbk-1}\big{]}(q_{2})_{\Bbbk-1}\right|$
$\displaystyle+\left|\frac{\lambda_{s}}{\lambda}\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}\left(Dq_{2}\right)_{\Bbbk-1}\right|\quad\lesssim
C(A)b_{1}\mathscr{E}_{\Bbbk}.$

Again from the compactness of the support of $\nabla\phi_{A}$ and
$\Delta\phi_{A}$, we use the Cauchy-Schwartz inequality and the local bound
(2.47) for $\vec{\mathbf{\Psi}}$ to obtain the estimate

$\displaystyle\left|\int\big{[}\nabla\phi_{A}.\nabla\left((\mathbf{\Psi}_{b})_{1}\right)_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}\left((\mathbf{\Psi}_{b})_{1}\right)_{\Bbbk-1}\big{]}(q_{2})_{\Bbbk-1}\right|$
$\displaystyle+\left|\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}\left((\mathbf{\Psi}_{b})_{2}\right)_{\Bbbk-1}\right|\quad\lesssim
C(A)\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+3}.$

For the small linear term $L(q_{1})$ and the nonlinear term $N(q_{1})$, we use
the Cauchy-Schwartz inequality, the bounds (4.33) and (4.28) to estimate

$\displaystyle\left|\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}\left(L(q_{1})-N(q_{1})\right)_{\Bbbk-1}\right|\quad\lesssim
C(A)b_{1}\mathscr{E}_{\Bbbk}.$

It remains to control the modulation term. We use the fact that
$\chi_{B_{1}}T_{i}=T_{i}$ for $y\leq 2A\ll B_{1}$ and $(T_{i})_{\Bbbk-1}=0$
for $1\leq i\leq L$ to deduce that

$\displaystyle\sum_{i=1,\textup{even}}^{L}\int\big{[}\nabla\phi_{A}.\nabla\left(T_{i}\right)_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}\left(T_{i}\right)_{\Bbbk-1}\big{]}(q_{2})_{\Bbbk-1}$
$\displaystyle+\sum_{i=1,\textup{odd}}^{L}\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}\left(\chi_{B_{1}}T_{i}\right)_{\Bbbk-1}=0.$

For the term $\frac{\partial S_{j}}{b_{i}}$ for $j\geq i+1$, we recall that
$S_{j}$ is homogeneous of degree $(j,j-1,j\wedge 2,j)$. Thus,
$\left|\frac{\partial S_{j}}{b_{i}}\right|\lesssim C(A)b_{1}$ for $y\leq 2A$.
We then use Lemma 4.3 to derive the bound for $1\leq i\leq L$,

$\displaystyle\sum_{j=i+1,\textup{even}}^{L+2}\left|\big{[}(b_{i})_{s}+(i-\gamma)b_{1}b_{i}-b_{i+1}\big{]}\int\left[\nabla\phi_{A}.\nabla\left(\chi_{B_{1}}\frac{\partial
S_{j}}{\partial
b_{i}}\right)_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}\left(\chi_{B_{1}}\frac{\partial
S_{j}}{\partial b_{i}}\right)_{\Bbbk-1}\right](q_{2})_{\Bbbk-1}\right|$
$\displaystyle+\sum_{j=i+1,\textup{odd}}^{L+2}\left|\big{[}(b_{i})_{s}+(i-\gamma)b_{1}b_{i}-b_{i+1}\big{]}\int\big{[}\nabla\phi_{A}.\nabla(q_{1})_{\Bbbk-1}+\frac{(1-\nu)}{2}\Delta\phi_{A}(q_{1})_{\Bbbk-1}\big{]}\left(\chi_{B_{1}}\frac{\partial
S_{j}}{\partial b_{i}}\right)_{\Bbbk-1}\right|$ $\displaystyle\quad\lesssim
C(A,M)\left(b_{1}\mathscr{E}_{\Bbbk}+\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+2+(1-\delta)(1+\eta)}\right).$

Injecting all the above estimates and identity (4.48) to (4.49) yields the
formula (4.47). This concludes the proof of Proposition 4.6. ∎

###  4.4. Monotonicity for $\mathscr{E}_{\sigma}$.

We now in the position to derive the monotonicity formula for
$\mathscr{E}_{\sigma}$. We claim the following.

######  Proposition 4.8 (Lyapunov monotonicity for $\mathscr{E}_{\sigma}$).

Given $\hbar$, $\delta$ and $\eta$ as defined in (1.17) and (1.20). For $K\geq
1$, we assume that there is $s_{0}(K)\gg 1$ such that
$(b_{1}(s),\cdots,b_{L}(s),\vec{q}(s))\in\mathcal{S}_{K}(s)$ for
$s\in[s_{0},s_{1}]$ for some $s_{1}\geq s_{0}$. Then, the followings hold for
$s\in[s_{0},s_{1}]$:

$\displaystyle\frac{d}{dt}\left(\frac{\mathscr{E}_{\sigma}}{\lambda^{2\sigma-d}}\right)\leq\frac{b_{1}}{\lambda^{2\sigma-d+1}}\sqrt{\mathscr{E}_{\sigma}}b_{1}^{\frac{\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}\left[b_{1}^{\frac{1}{L}\left(\frac{\gamma}{2}-\frac{1}{4}\right)+\mathcal{O}\left(\frac{1}{L}\left|\sigma-\frac{d}{2}\right|\right)}+b_{1}^{\frac{\gamma-1}{2}}\right].$
(4.50)

###### Proof.

We compute from definition of $\mathscr{E}_{\sigma}$ and equation (3.14),

$\displaystyle\frac{d}{dt}\left[\frac{\mathscr{E}_{\sigma}}{\lambda^{2\sigma-d}}\right]=\frac{d}{dt}\left[\int|\nabla^{\sigma}v_{1}|^{2}+\int|\nabla^{\sigma-1}v_{2}|^{2}\right]$
$\displaystyle=2\int\nabla^{\sigma}v_{1}.\nabla^{\sigma}\left(v_{2}+\frac{1}{\lambda}(\mathcal{F}_{1})_{\lambda}\right)+2\int\nabla^{\sigma-1}v_{2}.\nabla^{\sigma-1}\left(-\mathscr{L}_{\lambda}v_{1}+\frac{1}{\lambda^{2}}(\mathcal{F}_{2})_{\lambda}\right)$
$\displaystyle\quad=-2\int\nabla^{\sigma-1}v_{2}.\nabla^{\sigma-1}\left(\frac{Z_{\lambda}}{r^{2}}v_{1}\right)+2\int\nabla^{\sigma}v_{1}.\nabla^{\sigma}\left(\frac{1}{\lambda}(\mathcal{F}_{1})_{\lambda}\right)+2\int\nabla^{\sigma-1}v_{2}.\nabla^{\sigma-1}\left(\frac{1}{\lambda^{2}}(\mathcal{F}_{2})_{\lambda}\right)$
$\displaystyle\qquad\leq
C\frac{\sqrt{\mathscr{E}_{\sigma}}}{\lambda^{2\sigma-d+1}}\left\\{\left\|\nabla^{\sigma-1}\left(\frac{Z}{y^{2}}q_{1}\right)\right\|_{L^{2}}+\left\|\nabla^{\sigma}\mathcal{F}_{1}\right\|_{L^{2}}+\left\|\nabla^{\sigma-1}\mathcal{F}_{2}\right\|_{L^{2}}\right\\},$
(4.51)

where $Z$, $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ are defined in (1.23) and
(3.9).

\- Estimate for the potential term: From the expansion (B.2), we have

$\int_{y\leq
1}\left|\nabla^{\sigma-1}\left(\frac{Z}{y^{2}}q_{1}\right)\right|^{2}\lesssim\mathscr{E}_{\Bbbk}.$

For $y\geq 1$, we note from the asymptotic behavior (2.4) that
$\left|\partial_{y}^{j}\left(\frac{Z}{y^{2}}\right)\right|\lesssim\frac{1}{y^{2+j}}$
for $j\in\mathbb{N}$. We then use the Leibniz rule and interpolation bound
(B.5) with $\lfloor\sigma-1\rfloor+2>\sigma$ to obtain the estimates

$\displaystyle\int_{y\geq
1}\left|\nabla^{\lfloor\sigma-1\rfloor}\left(\frac{Z}{y^{2}}q_{1}\right)\right|^{2}\lesssim\mathscr{E}_{\sigma}^{\frac{\Bbbk-\lfloor\sigma-1\rfloor-2}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{\lfloor\sigma-1\rfloor+2-\sigma}{\Bbbk-\sigma}},\quad\int_{y\geq
1}\left|\nabla^{\Bbbk-1}\left(\frac{Z}{y^{2}}q_{1}\right)\right|^{2}\lesssim\mathscr{E}_{\Bbbk}.$

By interpolation and the bootstrap bounds given in Definition (3.2), we have

$\int\left|\nabla^{\sigma-1}\left(\frac{Z}{y^{2}}q_{1}\right)\right|^{2}\leq
C\mathscr{E}_{\sigma}^{\frac{\Bbbk-\lfloor\sigma-1\rfloor-2}{\Bbbk-1-\lfloor\sigma-1\rfloor}}\mathscr{E}_{\Bbbk}^{\frac{1}{\Bbbk-1-\lfloor\sigma-1\rfloor}}\leq
CK^{2}b_{1}^{\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)+\mathcal{O}\left(\frac{1}{L}\left|\sigma-\frac{d}{2}\right|\right)}b_{1}^{\frac{2L+2(1-\delta)(1+\eta)}{L+\hbar-\lfloor\sigma-1\rfloor}}.$

Since $\left|\sigma-\frac{d}{2}\right|\ll 1$, we note that
$\lfloor\sigma-1\rfloor\geq\frac{d-3}{2}=\gamma+\hbar+\delta-\frac{3}{2}$. We
then compute the exponent

$\displaystyle\frac{2L+2(1-\delta)(1+\eta)}{L+\hbar-\lfloor\sigma-1\rfloor}$
$\displaystyle=2\left[1+\frac{(1-\delta)(1+\eta)+\lfloor\sigma-1\rfloor-\hbar}{L}+\mathcal{O}\left(\frac{1}{L^{2}}\right)\right]$
$\displaystyle\geq
2\left[1+\frac{\gamma-\frac{1}{2}+\eta(1-\delta)}{L}+\mathcal{O}\left(\frac{1}{L^{2}}\right)\right].$

Since $b_{1}(s)\sim\frac{1}{s}\leq\frac{1}{s_{0}}$, we can take
$s_{0}=s_{0}(K)$ large enough to obtain the bound

$\int\left|\nabla^{\sigma-1}\left(\frac{Z}{y^{2}}q_{1}\right)\right|^{2}\leq
b_{1}^{\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}b_{1}^{2+\frac{2\gamma-1}{2L}+\mathcal{O}\left(\frac{1}{L}\left|\sigma-\frac{d}{2}\right|\right)}.$
(4.52)

We now turn to the estimate for the last term in (4.51). We only deal with the
$\mathcal{F}_{2}$ term since the same estimate holds for $\mathcal{F}_{1}$.
Let us recall form (3.9),

$\mathcal{F}_{2}=-(\mathbf{\Psi}_{b})_{2}-\mathbf{M}_{2}+L(q_{1})-N(q_{1}).$

\- Estimate for the error term $(\mathbf{\Psi}_{b})_{2}$: we apply (2.45) with
$j=\lfloor\sigma-1\rfloor-\hbar$ and $j=\lfloor\sigma\rfloor-\hbar$ to find
that

$\displaystyle\|\nabla^{\lfloor\sigma-1\rfloor}(\mathbf{\Psi}_{b})_{2}\|_{L^{2}}^{2}$
$\displaystyle\lesssim
b_{1}^{2\lfloor\sigma-1\rfloor-2\hbar+2+2(1-\delta)-C_{L}\eta}\lesssim
b_{1}^{2\gamma+1-C_{L}\eta},$
$\displaystyle\|\nabla^{\lfloor\sigma\rfloor}(\mathbf{\Psi}_{b})_{2}\|_{L^{2}}^{2}$
$\displaystyle\lesssim
b_{1}^{2\lfloor\sigma\rfloor-2\hbar+2+2(1-\delta)-C_{L}\eta}\lesssim
b_{1}^{2\gamma+3-C_{L}\eta},$

where we used the fact that $2\lfloor\sigma-1\rfloor\geq d-3$ and
$2\lfloor\sigma\rfloor\geq d-1$ for $\left|\sigma-\frac{d}{2}\right|\ll 1$ and
$d=2\gamma+2\hbar+2\delta$ from (1.17). Note that
$\lfloor\sigma\rfloor-\lfloor\sigma-1\rfloor=1$ and
$\sigma-1-\lfloor\sigma-1\rfloor\geq\sigma-\frac{d}{2}$ for
$\left|\sigma-\frac{d}{2}\right|\ll 1$, we have by interpolation

$\displaystyle\|\nabla^{\sigma-1}(\mathbf{\Psi}_{b})_{2}\|_{L^{2}}^{2}$
$\displaystyle\lesssim
b_{1}^{(2\gamma+1-C_{L}\eta)(\lfloor\sigma\rfloor-\sigma+1)}b_{1}^{(2\gamma+3-C_{L}\eta)(\sigma-1-\lfloor\sigma-1\rfloor)}$
$\displaystyle=b_{1}^{2\gamma+1-C_{L}\eta}b_{1}^{2(\sigma-1-\lfloor\sigma-1\rfloor)}\lesssim
b_{1}^{2}b_{1}^{\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}b_{1}^{\frac{2\gamma-1}{2}},$
(4.53)

for $\eta$ and $\left|\sigma-\frac{d}{2}\right|$ small enough.

\- Estimate for the modulation term $\mathbf{M}_{2}$: From Lemma 4.3, the
admissibility of $\vec{T}_{k}$ and homogeneity of $\vec{S}_{k}$, we estimate

$\displaystyle\left\|\nabla^{\sigma-1}\mathbf{M}_{2}\right\|_{L^{2}}$
$\displaystyle\lesssim\sqrt{\mathscr{E}_{\Bbbk}}\left[\sum_{k=1,\textup{odd}}^{L}\left(\left\|\nabla^{\sigma-1}\left(\chi_{B_{1}}T_{k}\right)\right\|_{L^{2}}+\sum_{j=k+1,\textup{odd}}^{L+2}\left\|\nabla^{\sigma-1}\left(\chi_{B_{1}}\frac{\partial
S_{j}}{\partial b_{k}}\right)\right\|_{L^{2}}\right)\right]$
$\displaystyle\quad+b_{1}^{L+1+(1-\delta)(1+\eta)}\left\|\nabla^{\sigma-1}\left(D(\mathbf{Q}_{b})_{2}\right)\right\|_{L^{2}}$
$\displaystyle\qquad\lesssim\sqrt{\mathscr{E}_{\Bbbk}}B_{1}^{L-\sigma+\hbar+\delta}+b_{1}^{L+1+(1-\delta)(1+\eta)}b_{1}^{2(1-\eta)}B_{1}^{\hbar+\delta}\lesssim
b_{1}b_{1}^{\frac{\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}b_{1}^{\frac{\gamma-1}{2}},$
(4.54)

for $\eta$ and $\left|\sigma-\frac{d}{2}\right|$ small enough.

\- Estimate for the small linear term $L(q_{1})$: From the asymptotic behavior
(2.4) of $Q$ and the definition (3.11) of $L(q_{1})$, we have by Leibniz rule

$\forall j\in\mathbb{N},\quad\left|\partial_{y}^{j}(L(q_{1}))\right|\lesssim
b_{1}^{2(1-C(L)\eta)}\sum_{i=0}^{j}\frac{|\partial_{y}^{i}q_{1}|}{1+y^{2\gamma+j-i}}.$

From the bounds (B.3) and (B.5), we have the estimate

$\int|\nabla^{\lfloor\sigma-1\rfloor}L(q_{1})|^{2}\lesssim\mathscr{E}_{\sigma}^{\frac{\Bbbk-(2\gamma+\lfloor\sigma-1\rfloor)}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{2\gamma+\lfloor\sigma-1\rfloor-\sigma}{\Bbbk-\sigma}},\quad\int|\nabla^{\Bbbk-2}L(q_{1})|^{2}\lesssim\mathscr{E}_{\Bbbk}.$

By interpolation and the same computation of the exponent as for the potential
term, we obtain the estimate

$\displaystyle\left\|\nabla^{\sigma-1}L(q_{1})\right\|^{2}_{L^{2}}\lesssim
b_{1}^{2+\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}b_{1}^{\frac{\gamma-1}{2}}.$
(4.55)

\- Estimate for the nonlinear term $N(q_{1})$: We claim that

$\displaystyle\left\|\nabla^{\sigma-1}N(q_{1})\right\|^{2}_{L^{2}}\lesssim
b_{1}^{2+\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}b_{1}^{\frac{\gamma}{2L}}.$
(4.56)

For $y<1$, we use the expansion (4.24) to deduce that

$\left\|\nabla^{\sigma-1}N(q_{1})\right\|^{2}_{L^{2}(y<1)}\lesssim\mathscr{E}_{\Bbbk}.$

For $y\geq 1$, we shall control
$\left\|\nabla^{\lfloor\sigma-1\rfloor}N(q_{1})\right\|^{2}_{L^{2}(y\geq 1)}$
and $\left\|\nabla^{\lfloor\sigma\rfloor}N(q_{1})\right\|^{2}_{L^{2}(y\geq
1)}$, then obtain the result by interpolation. From (4.26), the Leibniz rule
and estimate (4.31), we write

$\displaystyle\int_{y\geq
1}|\partial_{y}^{\lfloor\sigma-1\rfloor}N(q_{1})|^{2}$
$\displaystyle\lesssim\sum_{j=0}^{\lfloor\sigma-1\rfloor}\sum_{i=0}^{j}\sum_{k=0}^{i}\int_{y\geq
1}\frac{|\partial_{y}^{k}q_{1}|^{2}|\partial_{y}^{i-k}q_{1}|^{2}|\partial_{y}^{\lfloor\sigma-1\rfloor-j}\psi|^{2}}{y^{4+2(j-i)}}$
$\displaystyle\lesssim\sum_{j=0}^{\lfloor\sigma-1\rfloor}\sum_{i=0}^{j}\sum_{k=0}^{i}\sum_{m\in\mathcal{I}}\int_{y\geq
1}\frac{|\partial_{y}^{k}q_{1}|^{2}|\partial_{y}^{i-k}q_{1}|^{2}}{y^{4+2(j-i)}}\prod_{l=1}^{\lfloor\sigma-1\rfloor-j}\left(\frac{b_{1}^{-C\eta}}{y^{2\gamma+2l}}+|\partial_{y}^{l}q_{1}|^{2}\right)^{m_{l}},$

where
$\mathcal{I}=\\{m\in\mathbb{N}^{\lfloor\sigma-1\rfloor-j}:\sum_{l=1}^{\lfloor\sigma-1\rfloor-j}lm_{l}=[\sigma-1]-j\\}$.
From the Hardy inequality (A.3) and (B.5), we have the estimate

$\displaystyle\left\|\frac{\partial_{y}^{k}q_{1}}{y^{-k+\sigma-\frac{d}{2}}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\int\frac{|\partial_{y}^{k+1}q_{1}|^{2}}{y^{2(-k-1+\sigma)}}+|\partial_{y}^{k}q_{1}(1)|^{2}\lesssim\mathscr{E}_{\sigma}\quad\text{for}\quad
k=0,\cdots,[\sigma-1].$ (4.57)

from which we derive for every $m\in\mathcal{I}$,

$\displaystyle\frac{y^{2(\lfloor\sigma-1\rfloor-j)}}{y^{\sum_{l}m_{l}(2\sigma-d)}}\prod_{l=1}^{\lfloor\sigma-1\rfloor-j}\left(\frac{b_{1}^{-C\eta}}{y^{2\gamma+2l}}+|\partial_{y}^{l}q_{1}|^{2}\right)^{m_{l}}=\prod_{l=1}^{\lfloor\sigma-1\rfloor-j}\left(\frac{b_{1}^{-C\eta}}{y^{2\gamma+2\sigma-d}}+\frac{|\partial_{y}^{l}q_{1}|^{2}}{y^{-2l+2\sigma-d}}\right)^{m_{l}}\lesssim
b_{1}^{-C\eta}.$

Hence, we have

$\displaystyle\int_{y\geq
1}|\partial_{y}^{\lfloor\sigma-1\rfloor}N(q_{1})|^{2}$
$\displaystyle\lesssim\sum_{j=0}^{\lfloor\sigma-1\rfloor}\sum_{i=0}^{j}\sum_{k=0}^{i}\sum_{m\in\mathcal{I}}b_{1}^{-C\eta}\int_{y\geq
1}\frac{|\partial_{y}^{k}q_{1}|^{2}|\partial_{y}^{i-k}q_{1}|^{2}}{y^{4-2i+2\lfloor\sigma-1\rfloor-C(2\sigma-d)}}$
$\displaystyle\lesssim\sum_{j=0}^{\lfloor\sigma-1\rfloor}\sum_{i=0}^{j}\sum_{k=0}^{i}\sum_{m\in\mathcal{I}}b_{1}^{-C\eta}\left\|\frac{|\partial_{y}^{k}q_{1}|^{2}}{y^{-2k+2\sigma-d}}\right\|_{L^{\infty}(y\geq
1)}\int_{y\geq
1}\frac{|\partial_{y}^{i-k}q_{1}|^{2}}{y^{4-2i+2k+2\lfloor\sigma-1\rfloor-C(2\sigma-d)}}$
$\displaystyle\lesssim
b_{1}^{-C\eta}\mathscr{E}_{\sigma}\mathscr{E}_{\sigma}^{\frac{\Bbbk-(2+\lfloor\sigma-1\rfloor-C(2\sigma-d))}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{2+\lfloor\sigma-1\rfloor-\sigma-C(2\sigma-d)}{\Bbbk-\sigma}}\lesssim\mathscr{E}_{\sigma}^{2+\mathcal{O}\left(\frac{1}{L}\right)}b_{1}^{2\big{(}2+\lfloor\sigma-1\rfloor-\sigma\big{)}\left(1+\frac{\gamma}{2L}\right)},$

where in the last estimate, we used the bootstrap bound on
$\mathscr{E}_{\Bbbk}$ given in Definition (3.2), the smallness
$|2\sigma-d|+\eta=\mathcal{O}\left(\frac{1}{L^{2}}\right)$ from (1.19) and
(1.20) and the following algebra

$\displaystyle-C\eta+{\frac{2+\lfloor\sigma-1\rfloor-\sigma-C(2\sigma-d)}{\Bbbk-1}}\big{(}2L+2(1-\delta)(1+\eta)\big{)}$
$\displaystyle\quad=2\big{(}2+\lfloor\sigma-1\rfloor-\sigma\big{)}\left(1+\frac{\gamma}{L}\right)+\mathcal{O}\left(\frac{1}{L^{2}}\right)\geq
2\big{(}2+\lfloor\sigma-1\rfloor-\sigma\big{)}\left(1+\frac{\gamma}{2L}\right)\quad\text{for}\quad
L\gg 1.$

Similarly, we have the estimate

$\int_{y\geq
1}|\partial_{y}^{\lfloor\sigma\rfloor}N(q_{1})|^{2}\lesssim\mathscr{E}_{\sigma}^{2+\mathcal{O}\left(\frac{1}{L}\right)}b_{1}^{2\big{(}2+\lfloor\sigma\rfloor-\sigma\big{)}\left(1+\frac{\gamma}{2L}\right)}.$

By interpolation and the fact that
$\lfloor\sigma\rfloor-\lfloor\sigma-1\rfloor=1$ for $|\sigma-d/2|\ll 1$, we
have

$\displaystyle\int|\partial_{y}^{\sigma-1}N(q_{1})|^{2}$
$\displaystyle\lesssim\mathscr{E}_{\sigma}^{2+\mathcal{O}\left(\frac{1}{L}\right)}b_{1}^{2\left(1+\frac{\gamma}{2L}\right)\Big{(}\big{(}2+\lfloor\sigma-1\rfloor-\sigma\big{)}\big{(}\lfloor\sigma\rfloor-\sigma+1\big{)}+\big{(}2+\lfloor\sigma\rfloor-\sigma\big{)}\big{(}\sigma-1-\lfloor\sigma-1\rfloor\big{)}\Big{)}}$
$\displaystyle\quad=\mathscr{E}_{\sigma}^{2+\mathcal{O}\left(\frac{1}{L}\right)}b_{1}^{2\left(1+\frac{\gamma}{2L}\right)}\lesssim
b_{1}^{\frac{\ell}{\ell-\gamma}(2\sigma-d)}b_{1}^{2\left(1+\frac{\gamma}{2L}\right)},$

which concludes the proof of (4.56).

We note that the bounds (4.53) and (4.54) also hold for
$\|\nabla^{\sigma}(\mathbf{\Psi}_{b})_{1}\|^{2}_{L^{2}}$ and
$\|\nabla^{\sigma}\mathbf{M}_{1}\|^{2}_{L^{2}}$ by using the same computation.
We then inject the estimates (4.52), (4.53), (4.54), (4.55) and (4.56) into
the identity (4.51) to obtain the desired formula (4.50). This concludes the
proof of Proposition 4.8. ∎

###  4.5. Conclusion of Proposition 3.7.

We give the proof of Proposition 3.7 in this subsection in order to complete
the proof of Theorem 1.1. Note that this section corresponds to Section 6.1 of
[Merle et al.(2015)Merle, Raphaël, and Rodnianski]. Here we follow exactly the
same lines as in [Merle et al.(2015)Merle, Raphaël, and Rodnianski] and no new
ideas are needed. We divide the proof into 2 parts:

\- Part 1: Reduction to a finite dimensional problem. Assume that for a given
$K>0$ large and an initial time $s_{0}\geq 1$ large, we have
$(b_{1}(s),\cdots,b_{L}(s),\vec{q}(s))\in\mathcal{S}_{K}(s)$ for all
$s\in[s_{0},s_{1}]$ for some $s_{1}\geq s_{0}$. By using (4.5), (4.9), (4.14)
and (4.50), we derive new bounds on $\mathcal{V}_{1}(s)$, $b_{k}(s)$ for
$\ell+1\leq k\leq L$ and $\mathscr{E}_{\sigma}$ and $\mathscr{E}_{\Bbbk}$,
which are better than those defining $\mathcal{S}_{K}(s)$ (see Definition
3.2). It then remains to control
$(\mathcal{V}_{2}(s),\cdots,\mathcal{V}_{\ell}(s))$. This means that the
problem is reduced to the control of a finite dimensional function
$(\mathcal{V}_{2}(s),\cdots,\mathcal{V}_{\ell}(s))$ and then get the
conclusion $(i)$ of Proposition 3.7.

\- Part 2: Transverse crossing. We aim at proving that if
$(\mathcal{V}_{2}(s),\cdots,\mathcal{V}_{\ell}(s))$ touches

$\partial\hat{\mathcal{S}}_{K}(s):=\partial\left(-\frac{K}{s^{\frac{\eta}{2}}(1-\delta)},\frac{K}{s^{\frac{\eta}{2}}(1-\delta)}\right)^{\ell-1}$

at $s=s_{1}$, it actually leaves $\partial\hat{\mathcal{S}}_{K}(s)$ at
$s=s_{1}$ for $s_{1}\geq s_{0}$, provided that $s_{0}$ is large enough. We
then get the conclusion $(ii)$ of Proposition 3.7.

####  $\bullet\,$ Reduction to a finite dimensional problem.

We give the proof of item $(i)$ of Proposition 3.7 in this part. Given $K>0$,
$s_{0}\geq 1$ and the initial data at $s=s_{0}$ as in Definition 3.1, we
assume for all $s\in[s_{0},s_{1}]$,
$(b_{1}(s),\cdots,b_{L}(s),\vec{q}(s))\in\mathcal{S}_{K}(s)$ for some
$s_{1}\geq s_{0}$. We claim that for all $s\in[s_{0},s_{1}]$,

$\displaystyle|\mathcal{V}_{1}(s)|\leq s^{-\frac{\eta}{2}(1-\delta)},$ (4.58)
$\displaystyle|b_{k}(s)|\lesssim
s^{-(k+\eta(1-\delta))}\quad\text{for}\;\;\ell+1\leq k\leq L,$ (4.59)
$\displaystyle\mathscr{E}_{\sigma}\leq\frac{K}{2}s^{-\frac{2\ell(\sigma-\frac{d}{2})}{\ell-\gamma}}$
(4.60)
$\displaystyle\mathscr{E}_{\Bbbk}\leq\frac{K}{2}s^{-(2L+2(1-\delta)(1+\eta))},$
(4.61)

Once these estimates are proved, it immediately follows from Definition 3.2 of
$\mathcal{S}_{K}$ that if

$(b_{1}(s_{1}),\cdots,b_{L}(s),\vec{q}(s_{1}))\in\partial\mathcal{S}_{K}(s_{1}),$

then $(\mathcal{V}_{2},\cdots,\mathcal{V}_{\ell}))(s_{1})$ must be in
$\partial\hat{\mathcal{S}}_{K}(s_{1})$, which concludes the proof of item
$(i)$ of Proposition 3.7.

Before going to the proof of (4.58)-(4.61), let us compute explicitly the
scaling parameter $\lambda$. To do so, let us note from (2.52) and the a
priori bound on $\mathcal{U}_{1}$ given in Definition 3.2 that
$b_{1}(s)=\frac{c_{1}}{s}+\frac{\mathcal{U}_{1}}{s}=\frac{\ell}{(\ell-\gamma)s}+\mathcal{O}\left(\frac{1}{s^{1+c\eta}}\right).$
From the modulation equation (4.5), we have

$-\frac{\lambda_{s}}{\lambda}=\frac{\ell}{(\ell-\gamma)s}+\mathcal{O}\left(\frac{1}{s^{1+c\eta}}\right),$
(4.62)

from which we write

$\left|\frac{d}{ds}\left\\{\log\Big{(}s^{\frac{\ell}{\ell-\gamma}}\lambda(s)\Big{)}\right\\}\right|\lesssim\frac{1}{s^{1+c\eta}}.$

We now integrate by using the initial data value $\lambda(s_{0})=1$ to get

$\lambda(s)=\left(\frac{s_{0}}{s}\right)^{\frac{\ell}{\ell-\gamma}}\left[1+\mathcal{O}\left(s^{-c\eta}\right)\right]\quad\text{for}\;\;s_{0}\gg
1.$ (4.63)

This implies that

$s_{0}^{-\frac{\ell}{\ell-\gamma}}\lesssim\frac{s^{-\frac{\ell}{\ell-\gamma}}}{\lambda(s)}\lesssim
s_{0}^{-\frac{\ell}{\ell-\gamma}}.$ (4.64)

\- Improved control of $\mathscr{E}_{\Bbbk}$: We aim at using (4.14) to derive
the improved bound (4.61). From the Morawetz formula (4.47), we have

$\frac{b_{1}\mathscr{E}_{\Bbbk,\textup{loc}}}{\lambda^{2\Bbbk-d}}\leq
C(N)b_{1}\frac{d}{ds}\left[\frac{Mc}{\lambda^{2\Bbbk-d}}\right]+\frac{C(M,N)}{A^{\nu}}\frac{b_{1}\mathscr{E}_{\Bbbk}}{\lambda^{2\Bbbk-d}}+\frac{C(A,M,N)}{\lambda^{2\Bbbk-d}}\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+2+(1-\delta)(1+\eta)},$

from which and the monotonicity formula (4.14), we write

$\displaystyle\frac{d}{ds}\left\\{\frac{\mathscr{E}_{\Bbbk}}{\lambda^{2\Bbbk-d}}\left[1+\mathcal{O}(b_{1}^{\eta(1-\delta)})\right]\right\\}\leq
C(N,M)b_{1}\frac{d}{ds}\left[\frac{Mc}{\lambda^{2\Bbbk-d}}\right]$
$\displaystyle\quad+\frac{b_{1}}{\lambda^{2\Bbbk-d}}\left\\{C(A,M,N)\sqrt{\mathscr{E}_{\Bbbk}}b_{1}^{L+(1-\delta)(1+\eta)}+C(M)\left[Kb_{1}^{\frac{\ell}{\ell-\gamma}(2\sigma-d)+\eta(1-\delta)+\mathcal{O}\left(\frac{|2\sigma-d|}{L}\right)}+\frac{1}{N^{2\gamma-1}}+\frac{C(N)}{A^{\nu}}\right]\mathscr{E}_{\Bbbk}\right\\}$
$\displaystyle\quad\leq\frac{C(N,M)}{s}\frac{d}{ds}\left[\frac{Mc}{\lambda^{2\Bbbk-d}}\right]+\frac{1}{\lambda^{2\Bbbk-d}}\left[\left(C(M)(\sqrt{K}+1)s^{-(2L+1+2(1-\delta)(1+\eta))}\right)\right],$

where we used the bootstrap bounds given in Definition 3.2,
$b_{1}\sim\frac{1}{s}\leq\frac{1}{s_{0}}$ and the constants $s_{0},A,N,M,L$ is
fixed large enough. Integrating in time by using $\lambda(s_{0})=1$ and
$|Mc|\leq C(A,M)\mathscr{E}_{\Bbbk}$ yields for all $s\in[s_{0},s_{1})$,

$\displaystyle\mathscr{E}_{\Bbbk}(s)$ $\displaystyle\leq
C(M)\lambda(s)^{2\Bbbk-d}\left[\mathscr{E}_{\Bbbk}(s_{0})+\left(\sqrt{K}+1\right)\int_{s_{0}}^{s}\frac{\tau^{-(2L+1+2(1-\delta)(1+\eta))}}{\lambda(\tau)^{2\Bbbk-d}}d\tau\right]$
$\displaystyle\quad+C(N,M)\left[\frac{Mc(s)}{s}-\frac{Mc(s_{0})}{s_{0}}\lambda(s)^{2\Bbbk-d}\right]+C(M,N)\lambda(s)^{2\Bbbk-d}\int_{s_{0}}^{s}\frac{|Mc(\tau)|}{\tau^{2}\lambda(\tau)^{2\Bbbk-d}}$
$\displaystyle\quad\leq
C(M)\lambda(s)^{2\Bbbk-d}\left[\mathscr{E}_{\Bbbk}(s_{0})+\left(\sqrt{K}+1\right)\int_{s_{0}}^{s}\frac{\tau^{-(2L+1+2(1-\delta)(1+\eta))}}{\lambda(\tau)^{2\Bbbk-d}}d\tau\right],$

for $s_{0}$ large enough. Using (4.64), we estimate

$\displaystyle\lambda(s)^{2\Bbbk-d}\int_{s_{0}}^{s}\frac{\tau^{-(2L+1+2(1-\delta)(1+\eta))}}{\lambda(\tau)^{2\Bbbk-d}}d\tau$
$\displaystyle\quad\lesssim
s^{-\frac{\ell(2\Bbbk-d)}{\ell-\gamma}}\int_{s_{0}}^{s}\tau^{\frac{\ell(2\Bbbk-d)}{\ell-\gamma}-(2L+1+2(1-\delta)(1+\eta))}d\tau\lesssim
s^{-(2L+2(1-\delta)(1+\eta))}.$

Here we used the fact that the integral is divergent because

$\frac{\ell(2\Bbbk-d)}{\ell-\gamma}-[2L+1+2(1-\delta)(1+\eta)]=\frac{2\gamma
L}{\ell-\gamma}+\mathcal{O}_{L\to+\infty}(1)\gg-1.$

Using again (4.64) and the initial bound (3.19), we estimate

$\lambda(s)^{2\Bbbk-d}\mathscr{E}_{\Bbbk}(s_{0})\leq\left(\frac{s_{0}}{s}\right)^{\frac{\ell(2\Bbbk-d)}{\ell-\gamma}}s_{0}^{-\frac{10L\ell}{\ell-\gamma}}\lesssim
s^{-(2L+2(1-\delta)(1+\eta))},$

for $L$ large enough. Therefore, we obtain

$\mathscr{E}_{\Bbbk}(s)\leq
C\left(\sqrt{K}+1\right)s^{-(2L+2(1-\delta)(1+\eta))}\leq\frac{K}{2}s^{-(2L+2(1-\delta)(1+\eta))},$

for $K=K(M)$ large enough. This concludes the proof of (4.61).

\- Improved control of $\mathscr{E}_{\sigma}$. We can improve the control of
$\mathscr{E}_{\sigma}$ by using the monotonicity formula (4.50). Indeed, we
see from (4.50) that there exists a small constant $0<\epsilon\ll 1$ such that

$\frac{d}{ds}\left[\frac{\mathscr{E}_{\sigma}}{\lambda^{2\sigma-d}}\right]\leq\frac{b_{1}\sqrt{\mathscr{E}_{\sigma}}}{\lambda^{2\sigma-d}}b_{1}^{\frac{\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)+\epsilon},$

from which we obtain

$\mathscr{E}_{\sigma}(s)\leq\lambda^{2\sigma-d}\left[\mathscr{E}_{\sigma}(s_{0})+\sqrt{K}\int_{s_{0}}^{s}\frac{b_{1}}{\lambda^{2\sigma-d}}b_{1}^{\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)+\epsilon}\right].$

We estimate from the initial bound (B.5) on $\mathscr{E}_{\sigma}(0)$ and
(4.62),

$\mathscr{E}_{\sigma}(s_{0})\lambda^{2\sigma-d}(s)\leq
Cs_{0}^{-\frac{10L\ell}{\ell-\gamma}}\left(\frac{s_{0}}{s}\right)^{\frac{\ell}{\ell-\gamma}\left(2\sigma-d\right)}\leq
s^{-\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}.$

Note that
$\frac{b_{1}}{\lambda^{2\sigma-d}}b_{1}^{\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)+\epsilon}\leq\frac{1}{s^{1+\epsilon}}$,
the integral is convergent. Thus, we have the estimate

$\sqrt{K}\int_{s_{0}}^{s}\frac{b_{1}}{\lambda^{2\sigma-d}}b_{1}^{\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)+\epsilon}\leq
C\sqrt{K}s^{-\frac{2\ell}{\ell-\gamma}\left(\sigma-\frac{d}{2}\right)}.$

Hence, by choosing $K$ large enough such that $1+C\sqrt{K}\leq\frac{K}{2}$, we
deduce the improve bound (4.60).

\- Control of the stable modes $b_{k}$’s. We now close the control of the
stable modes $(b_{\ell+1},\cdots,b_{L})$, in particular, we prove (4.59). We
first treat the case when $k=L$. Let

$\tilde{b}_{L}=b_{L}+\frac{\left<\mathscr{H}^{L}\vec{q},\chi_{B_{0}}\Lambda\vec{Q}\right>}{\left<\chi_{B_{0}}\Lambda
Q,\Lambda
Q+(-1)^{\frac{L-1}{2}}\mathscr{L}^{\frac{L-1}{2}}\left(\frac{\partial
S_{L+2}}{\partial b_{L}}\right)\right>},$

then from (4.12), (4.13) and (4.61),

$|\tilde{b}_{L}-b_{L}|\lesssim
b_{1}^{-(1-\delta)}\sqrt{\mathscr{E}_{\Bbbk}}\lesssim
b_{1}^{L+\eta(1-\delta)}.$

Hence, we have from the improved modulation equation (4.9),

$\displaystyle|(\tilde{b}_{L})_{s}+(L-\gamma)b_{1}\tilde{b}_{L}|$
$\displaystyle\lesssim
b_{1}|\tilde{b}_{L}-b_{L}|+b_{1}^{\delta}\left[C(M)\sqrt{\mathscr{E}_{\Bbbk}}+b_{1}^{L+(1-\delta)}\right]\lesssim
b_{1}^{L+1+\eta(1-\delta)}.$

This follows

$\left|\frac{d}{ds}\left\\{\frac{\tilde{b}_{L}}{\lambda^{L-\gamma}}\right\\}\right|\lesssim\frac{b_{1}^{L+1+\eta(1-\delta)}}{\lambda^{L-\gamma}}.$

Integrating this identity in time from $s_{0}$ and recalling that
$\lambda(s_{0})=1$ yields

$\displaystyle\tilde{b}_{L}(s)\lesssim
C\lambda(s)^{L-\gamma}\left(\tilde{b}_{L}(s_{0})+\int_{s_{0}}^{s}\frac{b_{1}(\tau)^{L+1+\eta(1-\delta)}}{\lambda(\tau)^{L-\gamma}}d\tau\right).$

Using the fact that $b_{1}(s)\sim\frac{1}{s}$, the initial bounds (3.18) and
(3.19) together with (4.64), we estimate

$\lambda(s)^{L-\gamma}\tilde{b}_{L}(s_{0})\lesssim\left(\frac{s_{0}}{s}\right)^{\frac{\ell(L-\gamma)}{\ell-\gamma}}\left(s_{0}^{-\frac{5\ell(L-\gamma)}{\ell-\gamma}}+s_{0}^{\eta(1-\delta)}s_{0}^{-\frac{5L\ell}{\ell-\gamma}}\right)\lesssim
s^{-L-\eta(1-\delta)},$

and

$\displaystyle\lambda(s)^{L-\gamma}\int_{s_{0}}^{s}\frac{b_{1}(\tau)^{L+1+\eta(1-\delta)}}{\lambda(\tau)^{L-\gamma}}d\tau$
$\displaystyle\lesssim
s^{-\frac{\ell(L-\gamma)}{\ell-\gamma}}\int_{s_{0}}^{s}\tau^{\frac{\ell(L-\gamma)}{\ell-\gamma}-L-1-\eta(1-\delta)}d\tau$
$\displaystyle\quad\lesssim s^{-L-\eta(1-\delta)}.$

Therefore,

$b_{L}(s)\lesssim|\tilde{b}_{L}(s)|+|\tilde{b}_{L}(s)-b_{L}(s)|\lesssim
s^{-L-\eta(1-\delta)},$

which concludes the proof of (4.59) for $k=L$. Now we will propagate this
improvement that we found for the bound of $b_{L}$ to all $b_{k}$ for all
$\ell+1\leq k\leq L-1$. To do so we do a descending induction where the
initialization is for $k=L$. Let assume the bound

$|b_{k}|\lesssim b_{1}^{k+\eta(1-\delta)},$

for $k+1$ and let’s prove it for $k$. Indeed, from (4.5) and the induction
bound, we have

$\left|(b_{k})_{s}-(k-\gamma)\frac{\lambda_{s}}{\lambda}b_{k}\right|\lesssim
b_{1}^{L+1}+|b_{k+1}|\lesssim b_{1}^{k+1+\eta(1-\delta)},$

which follows

$\left|\frac{d}{ds}\left\\{\frac{b_{k}}{\lambda^{k-\gamma}}\right\\}\right|\lesssim\frac{b_{1}^{k+1+\eta(1-\delta)}}{\lambda^{k-\gamma}}.$

Integrating this identity in time as for the case $k=L$, we end-up with

$\displaystyle b_{k}(s)$ $\displaystyle\lesssim
C\lambda(s)^{k-\gamma}\left(b_{k}(s_{0})+\int_{s_{0}}^{s}\frac{b_{1}(\tau)^{k+1+\eta(1-\delta)}}{\lambda(\tau)^{k-\gamma}}d\tau\right)$
$\displaystyle\quad\lesssim s^{-k-\eta(1-\delta)},$

where we used the initial bound (3.18), (4.64) and $k\geq\ell+1$. This
concludes the proof of (4.59).

\- Control of the stable mode $\mathcal{V}_{1}$. We recall from (2.52) and
(3.16) that

$b_{k}=b_{k}^{e}+\frac{\mathcal{U}_{k}}{s^{k}},\quad 1\leq
k\leq\ell,\quad\mathcal{V}=P_{\ell}\mathcal{U},$

where $P_{\ell}$ diagonalize the matrix $A_{\ell}$ with spectrum (2.55). From
(2.53), and (4.5), we estimate for $1\leq k\leq\ell-1$,

$|s(\mathcal{U}_{k})_{s}-(A_{\ell}\mathcal{U})_{k}|\lesssim
s^{k+1}|(b_{k})_{s}+(k-\gamma)b_{1}b_{k}-b_{k+1}|+|\mathcal{U}|^{2}\lesssim
s^{-L+k}+|\mathcal{U}|^{2},$

From (2.54), (4.5) and the improved bound (4.59), we have

$|s(\mathcal{U}_{\ell})_{s}-(A_{\ell}\mathcal{U})_{\ell}|\lesssim
s^{\ell+1}\left(|(b_{k})_{s}+(k-\gamma)b_{1}b_{\ell}-b_{\ell+1}|+|b_{\ell+1}|\right)+|\mathcal{U}|^{2}\lesssim
s^{-\eta(1-\delta)}+|\mathcal{U}|^{2}.$

Using the diagonalization (2.55), we obtain

$s\mathcal{V}_{s}=D_{\ell}\mathcal{V}+\mathcal{O}(s^{-\eta(1-\delta)}).$
(4.65)

Using (2.55) again yields the control of the stable mode $\mathcal{V}_{1}$:

$|(s\mathcal{V}_{1})_{s}|\lesssim s^{-\eta(1-\delta)}.$

Thus from the initial bound (3.18),

$|s^{\eta(1-\delta)}\mathcal{V}_{1}(s)|\leq\left(\frac{s_{0}}{s}\right)^{1-\eta(1-\delta)}s_{0}^{\eta(1-\eta)}\mathcal{V}_{1}(s_{0})+1\lesssim
s_{0}^{\eta(1-\delta)},$

which yields (4.58) for $s_{0}\geq s_{0}(\eta)$ large enough.

####  $\bullet\,$ Transverse crossing.

We give the proof of item $(ii)$ of Proposition 3.7 in this part. We compute
from (4.65) and (2.55) at the exit time $s=s_{1}$:

$\displaystyle\frac{1}{2}\frac{d}{ds}\left(\sum_{k=2}^{\ell}|s^{\frac{\eta}{2}(1-\delta)}\mathcal{V}_{k}(s)|^{2}\right)_{\big{|}_{s=s_{1}}}$
$\displaystyle\quad=\left(s^{\eta(1-\delta)-1}\sum_{k=2}^{\ell}\left[\frac{\eta}{2}(1-\delta)\mathcal{V}_{k}^{2}(s)+s\mathcal{V}_{k}(\mathcal{V}_{k})_{s}\right]\right)_{\big{|}_{s=s_{1}}}$
$\displaystyle\qquad=\left(s^{\eta(1-\delta)-1}\Bigg{[}\sum_{k=2}^{\ell}\left[\frac{k\gamma}{k-\gamma}+\frac{\eta}{2}(1-\delta)\right]\mathcal{V}_{k}^{2}(s)+\mathcal{O}\left(\frac{1}{s^{\frac{3}{2}\eta(1-\delta)}}\right)\Bigg{]}\right)_{\big{|}_{s=s_{1}}}$
$\displaystyle\quad\qquad\geq\frac{1}{s_{1}}\left[c(d,\ell)\sum_{k=2}^{\ell}|s_{1}^{\frac{\eta}{2}(1-\delta)}\mathcal{V}_{k}(s_{1})|^{2}+\mathcal{O}\left(\frac{1}{s_{1}^{\frac{\eta}{2}(1-\delta)}}\right)\right]$
$\displaystyle\qquad\qquad\geq\frac{1}{s_{1}}\left[c(d,\ell)+\mathcal{O}\left(\frac{1}{s_{1}^{\frac{\eta}{2}(1-\delta)}}\right)\right]>0,$

where we used item $(i)$ of Proposition 3.7 in the last step. This completes
the proof of Proposition 3.7.

##  Appendix A Coercivity of the adapted norms.

In this section we aim at proving the coercivity of the adapted norms
$\mathscr{E}_{\Bbbk}$ defined by (see (1.26))

$\mathscr{E}_{\Bbbk}:=\|\vec{q}\|^{2}_{\Bbbk}=\int_{\mathbb{R}^{d}}q_{1}\mathscr{L}^{\Bbbk}q_{1}+\int_{\mathbb{R}^{d}}q_{2}\mathscr{L}^{\Bbbk-1}q_{2}=\int_{\mathbb{R}^{d}}|(q_{1})_{\Bbbk}|^{2}+\int_{\mathbb{R}^{d}}|(q_{2})_{\Bbbk-1}|^{2},$

where we exploit the notation

$f_{0}=f,\quad f_{2k+1}=\mathscr{A}f_{2k}=\mathscr{A}\mathscr{L}^{k}f,\quad
f_{2k+2}=\mathscr{A}^{*}f_{2k+1}=\mathscr{L}^{k+1}f\quad\text{for}\quad
k\in\mathbb{N}.$

To do so, we first recall some results in [Ghoul et al.(2018a)Ghoul, Ibrahim,
and Nguyen] concerning the coercivity estimates for the operator
$\mathscr{A}$, $\mathscr{A}^{*}$ under some suitable orthogonality condition.
As a consequence, we then obtain the coercivity of $\mathscr{E}_{\Bbbk}$.

We recall from Lemma 4.1 that the direction $\vec{\Phi}_{M}$ defined in (3.4)
is of the form

$\vec{\Phi}_{M}=\binom{\Phi_{M}}{0}\quad\text{with}\quad\Phi_{M}=\sum_{k=0}^{\frac{L-1}{2}}(-1)^{k}c_{2k,M}\mathscr{L}^{k}(\chi_{M}\Lambda
Q),$ (A.1)

where $L\gg 1$ is an odd integer. We denote by $\mathcal{D}_{rad}$ as the set
of all radially symmetric functions. For simplicity, we write

$\int f:=\int_{0}^{+\infty}f(y)y^{d-1}dy.$

We have the following:

######  Lemma A.1 (Hardy inequalities).

Let $d\geq 7$ and $f\in\mathcal{D}_{rad}$, then
$(i)$ (Control near the origin)

$\int_{y\leq
1}\frac{|\partial_{y}f|^{2}}{y^{2i}}\geq\frac{(d-2-2i)^{2}}{4}\int_{y\leq
1}\frac{f^{2}}{y^{2+2i}}-C(d)f^{2}(1),\quad i=0,1,2.$

$(ii)$ (Control away from the origin for the non-critical exponent) Let
$\alpha>0,\alpha\neq\frac{d-2}{2}$, then

$\displaystyle\int_{y\geq
1}\frac{|\partial_{y}f|^{2}}{y^{2\alpha}}\geq\left(\frac{d-(2\alpha+2)}{2}\right)^{2}\int_{y\geq
1}\frac{f^{2}}{y^{2+2\alpha}}-C(\alpha,d)f^{2}(1),$ (A.2)
$\displaystyle\int_{y\geq
1}\frac{|\partial_{y}f|^{2}}{y^{2\alpha}}\geq\left(\frac{d-(2\alpha+2)}{2}\right)^{2}\left\|\frac{f}{y^{\alpha+1-\frac{d}{2}}}\right\|^{2}_{L^{\infty}(y\geq
1)}-C(\alpha,d)f^{2}(1),$ (A.3)

$(iii)$ (Control away from the origin for the critical exponent) Let
$\alpha=\frac{d-2}{2}$, then

$\int_{y\geq
1}\frac{|\partial_{y}f|^{2}}{y^{2\alpha}}\geq\frac{1}{4}\int_{y\geq
1}\frac{f^{2}}{y^{2+2\alpha}(1+\log y)^{2}}-C(d)f^{2}(1).$

$(iv)$ (Weighted Hardy inequality) For any $\mu>0$, $k\geq 2$ be an integer
and $1\leq j\leq k-1$,

$\int\frac{|\partial_{y}^{j}f|^{2}}{1+y^{\mu+2(k-j)}}\lesssim_{j,\mu}\int\frac{|\partial_{y}^{k}f|^{2}}{1+y^{\mu}}+\int\frac{f^{2}}{1+y^{\mu+2k}}.$
(A.4)

###### Proof.

See Lemma B.1 in [Merle et al.(2015)Merle, Raphaël, and Rodnianski]. ∎

We have the following coercivity of $\mathscr{A}^{*}$ and $\mathscr{A}$:

######  Lemma A.2 (Weight coercivity of $\mathscr{A}^{*}$).

Let $\alpha\geq 0$, and $f\in\mathcal{D}_{rad}$ satisfying

$i=0,1,2,\quad\int\frac{|\partial_{y}f|^{2}}{y^{2i}(1+y^{2\alpha})}+\int\frac{f^{2}}{y^{2i+2}(1+y^{2\alpha})}<+\infty,$

then

$i=0,1,2,\quad\int\frac{|\mathscr{A}^{*}f|^{2}}{y^{2i}(1+y^{2\alpha})}\geq
c_{\alpha}\left(\int\frac{|\partial_{y}f|^{2}}{y^{2i}(1+y^{2\alpha})}+\int\frac{f^{2}}{y^{2i+2}(1+y^{2\alpha})}\right),$
(A.5)

for some $c_{\alpha}>0$.

###### Proof.

See Lemma A.2 in [Ghoul et al.(2018a)Ghoul, Ibrahim, and Nguyen]. ∎

We also have the following coercivity of $\mathscr{A}$.

######  Lemma A.3 (Weight coercivity of $\mathscr{A}$).

Let $p\geq 0$ and $i=0,1,2$ such that $|2p+2i-(d-2-2\gamma)|\neq 0$, where
$\gamma\in(1,2]$ is defined by (1.10). For all $f\in\mathcal{D}_{rad}$ with

$\int\frac{|\partial_{y}f|^{2}}{y^{2i}(1+y^{2p})}+\int\frac{f^{2}}{y^{2i+2}(1+y^{2p})}<+\infty,$

and

$\left<f,\Phi_{M}\right>=0\quad\text{if}\quad 2i+2p>d-2\gamma-2,$ (A.6)

where $\Phi_{M}$ is defined in (3.4), we have

$\int\frac{|\mathscr{A}f|^{2}}{y^{2i}(1+y^{2p})}\gtrsim\int\frac{|\partial_{y}f|^{2}}{y^{2i}(1+y^{2p})}+\int\frac{f^{2}}{y^{2i+2}(1+y^{2p})}.$
(A.7)

###### Proof.

See Lemma A.3 in [Ghoul et al.(2018a)Ghoul, Ibrahim, and Nguyen]. ∎

From the coercivity estimates of $\mathscr{A}$ and $\mathscr{A}^{*}$ given in
Lemmas A.3 and A.2, we can turn to the core of this part: the coercivity of
the adapted norm $\mathscr{E}_{\Bbbk}$. In particular, we have the following.

######  Lemma A.4 (Coercivity of $\mathscr{E}_{\Bbbk}$).

Let $L\gg 1$ be an odd integer and $\Bbbk$ be defined as in (1.18), there
exists a constant $c=c(L,M)>0$ such that for all radially symmetric vector
function $\vec{q}$ satisfying

$\sum_{k=0}^{\Bbbk-1}\int\frac{|(q_{1})_{k}|^{2}}{y^{2}(1+y^{2\Bbbk-2-2k})}+\sum_{k=0}^{\Bbbk-2}\int\frac{|(q_{2})_{k}|^{2}}{y^{2}(1+y^{2\Bbbk-4-2k)}}<+\infty,$
(A.8)

and

$\big{\langle}\vec{q},\mathscr{H}^{*i}\vec{\Phi}_{M}\big{\rangle}=0\quad\text{for}\quad
0\leq i\leq L,$ (A.9)

there holds:

$\sum_{k=0}^{\Bbbk-1}\int\frac{|(q_{1})_{k}|^{2}}{y^{2}(1+y^{2\Bbbk-2-2k})}+\sum_{k=0}^{\Bbbk-2}\int\frac{|(q_{2})_{k}|^{2}}{y^{2}(1+y^{2\Bbbk-4-2k)}}\leq
c\mathscr{E}_{\Bbbk}.$ (A.10)

###### Proof.

By (2.18), we see that the condition (A.9) is equivalent to

$\big{\langle}\mathscr{L}^{i}q_{1},\Phi_{M}\big{\rangle}=0\quad\text{and}\quad\big{\langle}\mathscr{L}^{i}q_{2},\Phi_{M}\big{\rangle}=0\quad\text{for}\;\;0\leq
i\leq\frac{L-1}{2}.$ (A.11)

Recall that

$\mathscr{E}_{\Bbbk}=\int|(q_{1})_{\Bbbk}|^{2}+\int|(q_{2})_{\Bbbk-1}|^{2}.$
(A.12)

We will write indifferently $q$ to denote $q_{1}$ and $q_{2}$, and try to
control the term of the form $\int|q_{k}|^{2}$ with $k=\Bbbk$ or $k=\Bbbk-1$.
Let us rewrite

$q_{k}=\mathscr{A}q_{k-1}\;\text{or}\;q_{k}=\mathscr{A}^{*}q_{k-1},$

and apply Lemma A.3 or Lemma A.2 with $i=p=0$ to find that

$\int|q_{k}|^{2}\gtrsim\int\frac{|q_{k-1}|^{2}}{y^{2}}.$

If $k-1=0$, we are done, if not, we repeat this step again by writing

$q_{k-1}=\mathscr{A}^{*}q_{k-2}\;\text{or}\;q_{k-1}=\mathscr{A}q_{k-2},$

and so forth. Note that $\frac{1}{y^{2}}\gtrsim\frac{1}{1+y^{2}}$ and that the
orthogonal condition in Lemma A.3 is fulfilled thanks to (A.11). Then,
applying Lemma A.3 or Lemma A.2 with the appropriate values of $i$ and $p$
would give

$\int|q_{k}|^{2}\gtrsim\int\frac{|q_{k-1}|^{2}}{y^{2}}\gtrsim\cdots\gtrsim\int\frac{|q_{1}|^{2}}{y^{2}(1+y^{2k-2})}\gtrsim\int\frac{|q|^{2}}{y^{2}(1+y^{2k-1})}.$

This concludes the proof of Lemma A.4. ∎

##  Appendix B Interpolation bounds.

We derive in this section interpolation bounds on $\vec{q}$ which are the
consequence of the coercivity property given in Lemma A.4. We have the
following:

######  Lemma B.1 (Interpolation bounds).

Suppose that $\mathscr{E}_{\Bbbk}$ and $\mathscr{E}_{\sigma}$ satisfy the
bootstrap bounds in Definition 3.2 and that $\vec{q}$ satisfies the orthogonal
condition (3.3), there holds:
$(i)$ Weighted bounds for $\vec{q}$:

$\int|(q_{1})_{\Bbbk}|^{2}+\int|(q_{2})_{\Bbbk-1}|^{2}+\sum_{i=0}^{\Bbbk-1}\int\frac{|(q_{1})_{i}|^{2}}{y^{2}(1+y^{2\Bbbk-2i-2})}+\sum_{i=0}^{\Bbbk-1}\int\frac{|(q_{2})_{i}|^{2}}{y^{2}(1+y^{2\Bbbk-2i-4})}\leq
c(M)\mathscr{E}_{\Bbbk}.$ (B.1)

$(ii)$ Expansion near the origin: for $y<1$,

$\vec{q}=\sum_{i=1}^{\Bbbk}c_{i}\vec{T}_{\Bbbk-i}+\vec{r},\quad|c_{i}|\lesssim\sqrt{\mathscr{E}_{\Bbbk}},$
(B.2)

where $\vec{T}_{k}$ is defined as in (2.19) for all $k\in\mathbb{N}$, and
$\vec{r}$ satisfies the bounds

$\sum_{i=0}^{\Bbbk-1}|y^{i}\partial_{y}^{i}r_{1}|\lesssim
y^{\Bbbk-\frac{d}{2}}\sqrt{\mathscr{E}_{\Bbbk}},\quad\sum_{i=0}^{\Bbbk-2}|y^{i}\partial_{y}^{i}r_{2}|\lesssim
y^{\Bbbk-1-\frac{d}{2}}\sqrt{\mathscr{E}_{\Bbbk}}.$

$(iii)$ Weighted bounds for $\partial_{y}^{i}\vec{q}$:

$\sum_{i=0}^{\Bbbk}\int\frac{|\partial_{y}^{i}q_{1}|^{2}}{1+y^{2\Bbbk-2i}}+\sum_{i=0}^{\Bbbk-1}\int\frac{|\partial_{y}^{i}q_{2}|^{2}}{1+y^{2\Bbbk-2i-2}}\leq
c(M)\mathscr{E}_{\Bbbk},$ (B.3)

hence,

$\|\vec{q}\,\|_{\dot{H}^{\Bbbk}\times\dot{H}^{\Bbbk-1}}^{2}\leq
c(M)\mathscr{E}_{\Bbbk}\quad\text{and}\quad\|\vec{q}\,\|_{\dot{H}^{\beta}\times\dot{H}^{\beta-1}}^{2}\leq
c(M)\mathscr{E}_{\sigma}^{\frac{\Bbbk-\beta}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{\beta-\sigma}{\Bbbk-\sigma}}\quad\text{for}\quad\sigma\leq\beta\leq\Bbbk.$
(B.4)

Moreover, for $j\in\mathbb{N}$ and $p>0$ satisfying $\sigma\leq j+p\leq\Bbbk$,
we have

$\int_{y\geq 1}\frac{|\partial_{y}^{j}q_{1}|^{2}}{y^{2p}}\leq
c(M)\mathscr{E}_{\sigma}^{\frac{\Bbbk-(j+p)}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{(j+p)-\sigma}{\Bbbk-\sigma}}.$
(B.5)

$(iv)$ Weighted $L^{\infty}$ control: Let $a>0$ satisfying
$\sigma-\frac{d}{2}\leq p\leq\Bbbk-\frac{d}{2}$, then

$\left\|\frac{q_{1}}{y^{p}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\mathscr{E}_{\sigma}^{\frac{\Bbbk-\frac{d}{2}-p}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{p-\sigma+\frac{d}{2}}{\Bbbk-\sigma}}.$
(B.6)

Let $j\in\mathbb{N}^{*}$ and $p\geq 0$ such that $1\leq
j\leq\Bbbk-\frac{d}{2}$ and $0\leq p\leq\Bbbk-j-\frac{d}{2}$, then

$\left\|\frac{\partial_{y}^{j}q_{1}}{y^{p}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\mathscr{E}_{\sigma}^{\frac{\Bbbk-
j-p-\frac{d}{2}}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{j+\frac{d}{2}-\sigma}{\Bbbk-\sigma}\left(1-\frac{p}{\Bbbk-j-\frac{d}{2}}\right)+\frac{p}{\Bbbk-j-\frac{d}{2}}}.$
(B.7)

Moreover, if $1\leq j+p\ll L$, then

$\left\|\frac{\partial_{y}^{j}q_{1}}{y^{p}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim
b_{1}^{2(j+p)+\frac{2\gamma(j+p)}{L}+\mathcal{O}\left(\frac{1}{L^{2}}\right)}\mathscr{E}_{\sigma}^{1+\mathcal{O}\left(\frac{1}{L}\right)}.$
(B.8)

###### Proof.

$(i)$ The estimate (B.1) directly follows from Lemma A.4.

$(ii)$ Without loss of generality, we assume that $\hbar$ is an even integer
so that $\Bbbk=L+\hbar+1$ is also an even integer. By (2.20), the expansion
(B.2) is equivalent to

$q_{1}=\sum_{i=1}^{\frac{\Bbbk}{2}}c_{2i}T_{\Bbbk-2i}+r_{1}=\sum_{i=1}^{\frac{\Bbbk}{2}}c_{2i}\phi_{\frac{\Bbbk}{2}-i}+r_{1},\quad
q_{2}=\sum_{i=1}^{\frac{\Bbbk}{2}-1}c_{2i+1}T_{\Bbbk-2i-1}+r_{2}=\sum_{i=1}^{\frac{\Bbbk}{2}-1}c_{2i+1}\phi_{\frac{\Bbbk}{2}-i-1}+r_{2},$
(B.9)

where we recall that $T_{2k}=T_{2k+1}=\phi_{k}$ with $\phi_{k}$ being defined
as in (1.32). We only deal with the expansion of $q_{1}$ because the same
proof holds for $q_{2}$. We claim that for $1\leq m\leq\frac{\Bbbk}{2}$,
$(q_{1})_{\Bbbk-2m}$ admits the Taylor expansion at the origin

$(q_{1})_{\Bbbk-2m}=\sum_{i=1}^{m}c_{i,m}\phi_{m-i}+(r_{1})_{m},$ (B.10)

with the bounds

$|c_{i,m}|\lesssim\sqrt{\mathscr{E}_{\Bbbk}},$
$\sum_{j=0}^{2m-1}|\partial_{y}^{j}(r_{1})_{m}|\lesssim
y^{2m-\frac{d}{2}-j}\sqrt{\mathscr{E}_{\Bbbk}},\quad\text{for}\quad y<1.$

The expansion (B.2) for $q_{1}$ then follows from (B.10) with
$m=\frac{\Bbbk}{2}$.

We proceed by induction in $m$ for the proof of (B.10). For $m=1$, we write
from the definition (2.7) of $\mathscr{A}^{*}$,

$(e_{1})_{1}(y)=(q_{1})_{\Bbbk-1}(y)=\frac{1}{y^{d-1}\Lambda
Q}\int_{0}^{y}(q_{1})_{\Bbbk}\Lambda Qx^{d-1}dx+\frac{d_{1}}{y^{d-1}\Lambda
Q}.$

Note from (B.1) that
$\int\frac{|(q_{1})_{\Bbbk-1}|^{2}}{y^{2}}\lesssim\mathscr{E}_{\Bbbk}$ and
from (2.5) that $\Lambda Q\sim y$ as $y\to 0$, we deduce that $d_{1}=0$. Using
the Cauchy-Schwartz inequality, we derive the pointwise estimate

$|(e_{1})_{1}(y)|\leq\frac{1}{y^{d}}\left(\int_{0}^{y}|(q_{1})_{\Bbbk}|^{2}x^{d-1}dx\right)^{\frac{1}{2}}\left(\int_{0}^{y}x^{2}x^{d-1}dx\right)^{\frac{1}{2}}\lesssim
y^{-\frac{d}{2}+1}\sqrt{\mathscr{E}_{\Bbbk}}\quad\text{for}\;y<1.$

We remark from (B.1) that there exists $a\in(1/2,1)$ such that

$|(e_{1})_{1}(a)|=|(q_{1})_{\Bbbk-1}(a)|^{2}\lesssim\int_{y<1}|(q_{1})_{\Bbbk-1}|^{2}\mathscr{E}_{\Bbbk}.$

We then define

$(r_{1})_{1}(y)=-\Lambda Q\int_{a}^{y}\frac{(e_{1})_{1}}{\Lambda Q}dx,$

and obtain from the pointwise estimate of $(e_{1})_{1}$,

$|(r_{1})_{1}(y)|\lesssim
y^{-\frac{d}{2}+2}\sqrt{\mathscr{E}_{\Bbbk}}\quad\text{for}\;\;y<1.$

By construction and the definition (2.6) of $\mathscr{A}$, we have

$\mathscr{L}(r_{1})_{1}=\mathscr{A}^{*}(q_{1})_{\Bbbk-1}=(q_{1})_{\Bbbk}=\mathscr{L}(q_{1})_{\Bbbk-2}.$

Recall that $\text{span}(\mathscr{L})=\\{\Lambda Q,\Gamma\\}$ where $\Gamma$
admits the singular behavior (2.10). From (B.1), we have
$\int\frac{|(q_{1})_{\Bbbk-2}|^{2}}{y^{2}}\lesssim\mathscr{E}_{\Bbbk}<+\infty$.
This implies that there exists $c_{1,1}\in\mathbb{R}$ such that

$(q_{1})_{\Bbbk-2}=c_{1,1}\Lambda Q+(r_{1})_{1}=c_{1,1}\phi_{0}+(r_{1})_{1}.$

Moreover, from (B.1) there exists $a\in(1/2,1)$ such that

$|(q_{1})_{\Bbbk-2}(a)|^{2}\lesssim\int_{y<1}|(q_{1})_{\Bbbk-2}|^{2}\lesssim\mathscr{E}_{\Bbbk},$

which follows

$|c_{1,1}|\lesssim\sqrt{\mathscr{E}_{2\Bbbk}},\quad|(q_{1})_{\Bbbk-2}|\lesssim
y^{-\frac{d}{2}+2}\sqrt{\mathscr{E}_{2\Bbbk}}\quad\text{for}\;y<1.$

Since $\mathscr{A}(r_{1})_{1}=(e_{1})_{1}$, we then write from the definition
(2.6) of $\mathscr{A}$,

$|\partial_{y}(r_{1})_{1}|\lesssim|(e_{1})_{1}|+\left|\frac{(r_{1})_{1}}{y}\right|\lesssim
y^{-\frac{d}{2}+1}\sqrt{\mathscr{E}_{\Bbbk}}\quad\text{for}\;y<1.$

This concludes the proof of (B.10) for $m=1$.

We now assume that (B.10) holds for $j=1,\cdots,m$ for some $m\geq 1$, and
prove that (B.10) holds for $j=1,\cdots,m+1$. The term $(r_{1})_{m+1}$ is
built as follows:

$(r_{1})_{m+1}=-\Lambda Q\int_{a}^{y}\frac{(e_{1})_{m+1}}{\Lambda Q}dx,$

where $a\in(1/2,1)$ and

$(e_{1})_{m+1}=\frac{1}{y^{d-1}\Lambda Q}\int_{0}^{y}(r_{1})_{m}\Lambda
Qx^{d-1}dx.$

We now use the induction hypothesis to estimate

$\displaystyle|(e_{1})_{m+1}|$ $\displaystyle=\left|\frac{1}{y^{d-1}\Lambda
Q}\int_{0}^{y}(r_{1})_{m}\Lambda Qx^{d-1}dx\right|$
$\displaystyle\quad\lesssim\frac{1}{y^{d}}\sqrt{\mathscr{E}_{\Bbbk}}\int_{0}^{y}x^{2m+\frac{d}{2}}dx\lesssim
y^{2m-\frac{d}{2}+1}\sqrt{\mathscr{E}_{\Bbbk}}.$

Then, we have

$\displaystyle|(r_{1})_{m+1}|=\left|\Lambda
Q\int_{a}^{y}\frac{(e_{1})_{m+1}}{\Lambda Q}dx\right|$ $\displaystyle\lesssim
y^{2m-\frac{d}{2}+2}\sqrt{\mathscr{E}_{\Bbbk}}.$

By construction, we have

$\mathscr{L}(r_{1})_{m+1}=r_{m}.$

From the induction hypothesis and the definition (2.19) of $T_{k}$, we write

$\mathscr{L}(q_{1})_{\Bbbk-2(m+1)}=(q_{1})_{\Bbbk-2m}=\sum_{i=1}^{m}c_{i,m}\phi_{m-i}+(r_{1})_{m}=\sum_{i=1}^{m}c_{i,m}\mathscr{L}\phi_{m+1-i}+\mathscr{L}r_{m+1}.$

The singularity (2.10) of $\Gamma$ at the origin and the bound
$\int_{y<1}\frac{|(q_{1})_{\Bbbk-2(m+1)}|^{2}}{y^{2}}\lesssim\mathscr{E}_{\Bbbk}$
implies that

$(q_{1})_{\Bbbk-2(m+1)}=\sum_{i=1}^{m}c_{i,m}\phi_{m+1-i}+c_{m+1,m}\Lambda
Q+(r_{1})_{m+1}.$

From (B.1), we see that there exists $a\in(1/2,1)$ such that

$|(q_{1})_{\Bbbk-2(m+1)}(a)|^{2}\lesssim\int_{y<1}|(q_{1})_{\Bbbk-2(m+1)}|^{2}\lesssim\mathscr{E}_{\Bbbk},$

from which we derive the bound
$|c_{m+1,m}|\lesssim\sqrt{\mathscr{E}_{\Bbbk}}$. For the estimate on
$\partial_{y}^{j}(r_{1})_{m+1}$, we note that by construction

$(r_{1})_{m+2}:=\mathscr{A}(r_{1})_{m+1}=(e_{1})_{m+1},\quad(r_{1})_{m+3}:=\mathscr{L}(r_{1})_{m+1}=(r_{1})_{m},$
$(r_{1})_{m+4}:=\mathscr{A}\mathscr{L}(r_{1})_{m+1}=\mathscr{A}(r_{1})_{m}=(e_{1})_{m},\quad(r_{1})_{m+5}:=\mathscr{L}^{2}(r_{1})_{m+1}=(r_{1})_{m-1},\cdots$
$(r_{1})_{3m+1}:=\mathscr{L}^{m}(r_{1})_{m+1}=(r_{1})_{1},\quad(r_{1})_{3m+2}:=\mathscr{A}\mathscr{L}^{m}(r_{1})_{m+1}=(e_{1})_{1}.$

A brute force computation using the definitions of $\mathscr{A}$ and
$\mathscr{A}^{*}$ and the asymptotic behavior (2.8) ensure that for any
function $f$, we have

$\partial_{y}^{j}f=\sum_{i=0}^{j}P_{i,j}f_{i}\quad\text{with}\quad|P_{i,j}|\lesssim\frac{1}{y^{j-i}}.$
(B.11)

Hence, we have for $0\leq j\leq 2m+1$ and $y<1$,

$\displaystyle|\partial_{y}^{j}r_{m+1}|$
$\displaystyle\lesssim\sum_{i=0}^{j}\frac{|r_{m+1+i}|}{y^{j-i}}\lesssim\sqrt{\mathscr{E}_{\Bbbk}}\sum_{i=0}^{j}\frac{y^{2m+2-i-\frac{d}{2}}}{y^{j-i}}\lesssim
y^{2m+2-\frac{d}{2}-j}\sqrt{\mathscr{E}_{\Bbbk}}.$

This concludes the proof of (B.10) as well as (B.2).

$(iii)$ We use (B.11), (B.1) and the expansion (B.2) to estimate

$\displaystyle\sum_{i=0}^{\Bbbk}\int\frac{|\partial_{y}^{i}q_{1}|^{2}}{1+y^{2\Bbbk-2i}}+\sum_{i=0}^{\Bbbk-1}\int\frac{|\partial_{y}^{i}q_{2}|^{2}}{1+y^{2\Bbbk-2i-2}}$
$\displaystyle\lesssim\mathscr{E}_{\Bbbk}+\sum_{i=0}^{\Bbbk-1}\int_{y<1}|\partial_{y}^{i}q_{1}|^{2}+\sum_{i=0}^{\Bbbk-2}\int_{y<1}|\partial_{y}^{i}q_{2}|^{2}+\sum_{i=0}^{\Bbbk-1}\int_{y>1}\frac{|\partial_{y}^{i}(q_{1})|^{2}}{y^{2\Bbbk-2i}}+\sum_{i=0}^{\Bbbk-2}\int_{y>1}\frac{|\partial_{y}^{i}(q_{2})|^{2}}{y^{2\Bbbk-2i-2}}$
$\displaystyle\lesssim\mathscr{E}_{\Bbbk}+\sum_{i=0}^{\Bbbk-1}\sum_{j=0}^{i}\int_{y>1}\frac{|(q_{1})_{j}|^{2}}{y^{2\Bbbk-2j}}+\sum_{i=0}^{\Bbbk-2}\sum_{j=0}^{i}\int_{y>1}\frac{|(q_{2})_{j}|^{2}}{y^{2\Bbbk-2j-2}}\lesssim\mathscr{E}_{\Bbbk},$

which concludes the proof of (B.3). The estimate (B.4) simply follows from an
interpolation. The estimate (B.5) follows from (A.2) and the interpolation
(B.4).

$(iv)$ We apply the Hardy inequality (A.3) to $q_{1}$ with $\alpha=\sigma-1$,
the bound (B.5) with $j=1$ and $p=\sigma-1$, and the expansion (B.2) to find
that

$\left\|\frac{q_{1}}{y^{\sigma-\frac{d}{2}}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\int_{y\geq
1}\frac{|\partial_{y}q_{1}|^{2}}{y^{2(\sigma-1)}}+|q_{1}(1)|^{2}\lesssim\mathscr{E}_{\sigma}.$

Similarly, we have

$\left\|\frac{q_{1}}{y^{\Bbbk-\frac{d}{2}}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\int_{y\geq
1}\frac{|\partial_{y}q_{1}|^{2}}{y^{2(\Bbbk-1)}}+|q_{1}(1)|^{2}\lesssim\mathscr{E}_{\Bbbk}.$

An interpolation of the two estimates yields the bound (B.6).

For $1\leq j\leq\Bbbk-\frac{d}{2}$, we have from Sobolev and the bound (B.4),

$\|\nabla^{j}q_{1}\|^{2}_{L^{\infty}}+\|\nabla^{\frac{d}{2}+j}q_{1}\|^{2}_{L^{2}}\lesssim\mathscr{E}_{\sigma}^{\frac{\Bbbk-j-\frac{d}{2}}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{j+\frac{d}{2}-\sigma}{\Bbbk-\sigma}}.$
(B.12)

We apply (A.3) to $\partial_{y}^{j}q_{1}$ with $\alpha=\Bbbk-j-1$, then use
(B.5) and (B.2) to estimate

$\left\|\frac{\partial_{y}^{j}q_{1}}{y^{\Bbbk-j-\frac{d}{2}}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\int_{y\geq
1}\frac{|\partial_{y}^{j+1}q_{1}|^{2}}{y^{2\Bbbk-2j-2}}+|\partial_{y}^{j}q_{1}(1)|^{2}\lesssim\mathscr{E}_{\Bbbk}.$

We interpolate for $0\leq p\leq\Bbbk-j-\frac{d}{2}$,

$\displaystyle\left\|\frac{\partial_{y}^{j}q_{1}}{y^{p}}\right\|_{L^{\infty}(y\geq
1)}$ $\displaystyle\lesssim\mathscr{E}_{\sigma}^{\frac{\Bbbk-
j-p-\frac{d}{2}}{\Bbbk-\sigma}}\mathscr{E}_{\Bbbk}^{\frac{j+\frac{d}{2}-\sigma}{\Bbbk-\sigma}\left(1-\frac{p}{\Bbbk-j-\frac{d}{2}}\right)+\frac{p}{\Bbbk-j-\frac{d}{2}}}.$

If $1\leq j+p\ll L$, then we have

$\frac{\Bbbk-
j-p-\frac{d}{2}}{\Bbbk-\sigma}=1+\mathcal{O}\left(\frac{1}{L}\right).$

Recall from (1.19) and (1.20) that
$|\sigma-d/2|+\eta=\mathcal{O}\left(\frac{1}{L^{2}}\right)$, we compute the
exponent

$\displaystyle\left[\frac{j+\frac{d}{2}-\sigma}{\Bbbk-\sigma}\left(1-\frac{p}{\Bbbk-j-\frac{d}{2}}\right)+\frac{p}{\Bbbk-j-\frac{d}{2}}\right](2L+2(1-\delta)(1+\eta))$
$\displaystyle\quad=2(j+p)+\frac{2\gamma(j+p)}{L}+\mathcal{O}\left(\frac{1}{L^{2}}\right).$

This yields

$\left\|\frac{\partial_{y}^{j}q_{1}}{y^{\Bbbk-j-\frac{d}{2}}}\right\|^{2}_{L^{\infty}(y\geq
1)}\lesssim\mathscr{E}_{\sigma}^{1+\mathcal{O}\left(\frac{1}{L}\right)}b_{1}^{2(j+p)+\frac{2\gamma(j+p)}{L}+\mathcal{O}\left(\frac{1}{L^{2}}\right)}.$

This concludes the proof of (B.7) and (B.8) as well as the proof of Lemma B.1.
∎

For the estimates of the linear and commutator terms in derivation of the
monotonicity formula (4.14), we need the following Leibniz rule for
$\mathscr{L}^{k}$ whose proof can be found in [Ghoul et al.(2018a)Ghoul,
Ibrahim, and Nguyen], Lemma C.1:

######  Lemma B.2 (Leibniz rule for $\mathscr{L}^{k}$).

Let $\phi$ be a smooth function and $k\in\mathbb{N}$, we have

$\mathscr{L}^{k+1}(\phi
f)=\sum_{m=0}^{k+1}f_{2m}\phi_{2k+2,2m}+\sum_{m=0}^{k}f_{2m+1}\phi_{2k+2,2m+1},$
(B.13)

and

$\mathscr{A}\mathscr{L}^{k}(\phi
f)=\sum_{m=0}^{k}f_{2m+1}\phi_{2k+1,2m+1}+\sum_{m=0}^{k}f_{2m}\phi_{2k+1,2m},$
(B.14)

where
\- for $k=0$,

$\displaystyle\phi_{1,0}=-\partial_{y}\phi,\quad\phi_{1,1}=\phi,$
$\displaystyle\phi_{2,0}=-\partial_{y}^{2}\phi-\frac{d-1+2V}{y}\partial_{y}\phi,\quad\phi_{2,1}=2\partial_{y}\phi,\quad\phi_{2,2}=\phi,$

\- for $k\geq 1$

$\displaystyle\phi_{2k+1,0}=-\partial_{y}\phi_{2k,0},$
$\displaystyle\phi_{2k+1,2i}=-\partial_{y}\phi_{2k,2i}-\phi_{2k,2i-1},\quad
1\leq i\leq k,$
$\displaystyle\phi_{2k+1,2i+1}=\phi_{2k,2i}+\frac{d-1+2V}{y}\phi_{2k,2i+1}-\partial_{y}\phi_{2k,2i+1},\quad
0\leq i\leq k-1,$ $\displaystyle\phi_{2k+1,2k+1}=\phi_{2k,2k}=\phi,$
$\displaystyle\phi_{2k+2,0}=\partial_{y}\phi_{2k+1,0}+\frac{d-1+2V}{y}\phi_{2k+1,0},$
$\displaystyle\phi_{2k+2,2i}=\phi_{2k+1,2i-1}+\partial_{y}\phi_{2k+1,2i}+\frac{d-1+2V}{y}\phi_{2k+1,2i},\quad
1\leq i\leq k,$
$\displaystyle\phi_{2k+2,2i+1}=-\phi_{2k+1,2i}+\partial_{y}\phi_{2k+1,2i+1},\quad
0\leq i\leq k,$ $\displaystyle\phi_{2k+2,2k+2}=\phi_{2k+1,2k+1}=\phi.$

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