$\begin{split}&\Big{|}\Big{[}1-\sum_{\dagger\in\\{+,-\\}}\varphi\big{(}(y-y^{\dagger}_{\lambda})/|\lambda-b(y_{j\ast})|^{1/2}\big{)}\Big{]}h(y)\Big{|}\\\
&\lesssim\varrho_{j,k}^{\gamma_{1}-3}\varrho_{j}^{-3}(y;\Lambda)\lesssim\delta^{-1-2(2-\gamma_{2})}\varrho_{j}^{-2+2(2-\gamma_{2})}\varrho_{j,k}^{\gamma_{2}-3}(y;\Lambda).\end{split}$
(4.76)

The desired bounds then follow from the estimates (4.51).

We now turn to the proof of (4.75) for $h$ as in (4.73). The bounds (4.75) for

$h\in\Big{\\{}\partial_{y}^{2}a(y,\lambda)\partial_{y}D_{\lambda}\psi^{\ast}_{k,\nu},\partial_{y}a(y,\lambda)\partial^{2}_{y}D_{\lambda}\psi^{\ast}_{k,\nu}\Big{\\}}$
(4.77)

follows from the bounds (4.11) with $\sigma=1$ and integration by parts using
the bounds on the modified Green’s function. To establish the bound (4.74) for
$h=A_{\Theta}^{-1}\big{(}f_{0k}^{\ast\ast}(\cdot,\lambda,\nu,\alpha)\big{)}(y)$,
using the commutator relations (4.40) and noting that the terms associated
with the commutators are easier to handle, and in view of the support property
of $ab^{\prime}-1$, it suffices to prove (4.75) for

$\begin{split}&h\in
A_{\Theta}^{-1}\Big{\\{}D_{\lambda}f_{0k}^{\ast}+A_{\Theta}\Big{[}\partial_{y}^{2}a(y,\lambda)\partial_{y}\psi^{\ast}_{k,\nu}+2\partial_{y}a(y,\lambda)\partial^{2}_{y}\psi^{\ast}_{k,\nu}\Big{]}(y,\Lambda)\\\
&\quad+\Big{[}\frac{\nu}{k}\partial_{y}^{2}a(y,\lambda)(\partial_{y}^{2}-k^{2})\partial_{y}+2\frac{\nu}{k}\partial_{y}a(y,\lambda)(\partial_{y}^{2}-k^{2})\partial_{y}^{2}\Big{]}D_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)\Big{\\}}.\end{split}$
(4.78)

The bound (4.75) follows for

$h=\Big{[}\partial_{y}^{2}a(y,\lambda)\partial_{y}\psi^{\ast}_{k,\nu}+2\partial_{y}a(y,\lambda)\partial^{2}_{y}\psi^{\ast}_{k,\nu}\Big{]}(y,\Lambda)$
(4.79)

follows from the bounds on the modified Green’s function, integration by parts
and the bounds (4.4) for $\sigma=0$. The bounds (4.75) for

$h=A_{\Theta}^{-1}\Big{[}\frac{\nu}{k}\partial_{y}^{2}a(y,\lambda)(\partial_{y}^{2}-k^{2})\partial_{y}+2\frac{\nu}{k}\partial_{y}a(y,\lambda)(\partial_{y}^{2}-k^{2})\partial_{y}^{2}\Big{]}D_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)$
(4.80)

follows from equation (4.55) using argument similar to (4.30)-(4.34). It
remains to consider the term

$h=A_{\Theta}^{-1}D_{\lambda}f_{0k}^{\ast}.$ (4.81)

Recalling the identity (4.56) and skipping terms that can be bounded that are
similar as before, we focus only on proving (4.75) for

$h=A_{\Theta}^{-1}D^{2}_{\lambda}f_{0k}.$ (4.82)

Recalling the formula (4.8), using the commutator relations (4.40) and
focusing on the most singular terms, it suffices to consider the term

$h=(\partial_{y}^{2}-k^{2})D_{\lambda}^{2}\Big{[}\sum_{j=1}^{2}\varphi\big{(}(y-y_{j\ast})/\delta_{0}\big{)}\omega_{0k}(y)\Big{]}.$
(4.83)

The desired bound (4.75) for $h$ in (4.83) follow from simple calculations
using the bounds on the modified Green’s function.

The proof of Proposition 4.2 is now complete.

##  5\. Proof of the main theorem

In this section we give the proof of the main theorem 1.1, using the limiting
absorption principle, see Proposition 4.2 and Corollary 4.1, with
$\gamma_{\ast}=\gamma+\frac{2-\gamma}{2}$ for sufficiently small $\nu>0$
depending on $\gamma$, and the representation formula (1.19).

As in [41], the representation formulae (1.19) for $\omega_{0}\in
C^{\infty}(\mathbb{T}_{\mathfrak{p}})$ follow from the theory of semigroups of
sectorial operators, see section 4, Chapter II of [21]. We omit the standard
details and refer to [21]. The main task is to establish the following bounds.

######  Proposition 5.1.

Assume that $k\in\mathbb{Z}\cap[1,\infty)$ and $\omega_{0k}\in
C^{\infty}(\mathbb{T}_{\mathfrak{p}})$. Define

$\aleph:=\\{\mu\in\mathbb{C}:\real\mu\geq-\sigma_{\sharp}|\nu/k|^{1/2}\\}.$
(5.1)

For any $\mu=\alpha-i\lambda\in\aleph$, and $\nu$ small, (1.18) is solvable
for a unique solution $\omega_{k,\nu}\in H^{2}(\mathbb{T}_{\mathfrak{p}})$
which is holomorphic in $\mu$ in the interior of $\aleph$ and continuous on
$\aleph$. In addition, we have the bound

$\|\omega_{k,\nu}(\cdot,\Lambda)\|_{L^{2}(\mathbb{T}_{\mathfrak{p}})}\lesssim_{k,\nu,\omega_{0k}}\frac{1}{\langle\lambda,\alpha\rangle},$
(5.2)

and

$\|\partial_{\lambda}\omega_{k,\nu}(\cdot,\Lambda)\|_{L^{2}(\mathbb{T}_{\mathfrak{p}})}\lesssim_{k,\nu,\omega_{0k}}\frac{1}{\langle\lambda,\alpha\rangle^{2}},$
(5.3)

Proposition 5.1 follows from the limiting absorption principle, see
Proposition 4.1, which implies that there is no eigenvalue of $L_{k,\nu}$
inside $\aleph$. The absence of eigenvalues inside $\aleph$ is sufficient to
deduce (5.2)-(5.3) since we allow the bounds to depend on $\nu$ and
consequently the problem is now elliptic.

We divide the proof of Theorem 1.1 into three different regions (i)
$\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\gtrsim 1$, (ii)
$|\nu/k|^{1/4}\ll\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\ll 1$, and (iii)
$\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\ll|\nu/k|^{1/4}$. We assume that $t\geq
1$ as the case of $0<t<1$ follows from the local wellposedness of the Navier-
Stokes equations.

Denote

$M_{k}:=\big{\|}\omega_{0k}\big{\|}_{H^{3}_{k}(\mathbb{T}_{\mathfrak{p}})},$
(5.4)

and recall the definition (4.21) for $a(y,\lambda)$ with
$y\in\mathbb{T}_{\mathfrak{p}},\lambda\in\mathbb{R}$,

$a(y,\lambda):=\Big{[}1-\varphi\big{(}\frac{y-y_{1\ast}}{\delta_{2}(\lambda)}\big{)}\Big{]}\Big{[}1-\varphi\big{(}\frac{y-y_{2\ast}}{\delta_{2}(\lambda)}\big{)}\Big{]}\frac{1}{b^{\prime}(y)}.$
(5.5)

We shall use the following formulae (recall also (4.9) for the derivative
$D_{\lambda}$),

$\begin{split}\partial_{\lambda}&=D_{\lambda}-a(y,\lambda)\partial_{y},\\\
\partial_{y}\partial_{\lambda}&=\partial_{y}D_{\lambda}-\partial_{y}a(y,\lambda)\partial_{y}-a(y,\lambda)\partial_{y}^{2},\\\
\partial_{\lambda}^{2}&=D_{\lambda}^{2}-2a(y,\lambda)\partial_{y}D_{\lambda}-\partial_{\lambda}a(y,\lambda)\partial_{y}+a(y,\lambda)\partial_{y}a(y,\lambda)\partial_{y}+a^{2}(y,\lambda)\partial_{y}^{2}.\end{split}$
(5.6)

The main tools are Proposition 4.2, Corollary 4.1, and the representation
formula (1.19). More specifically we use the identities, with
$\alpha=-\sigma_{\sharp}|\nu/k|^{1/2}$, that for
$y\in\mathbb{T}_{\mathfrak{p}}$ and $t\geq 1$,

$\psi_{k}(t,y)=\frac{1}{2\pi k^{2}t^{2}}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\mathbb{R}}e^{-ik\lambda
t}\partial_{\lambda}^{2}\psi_{k,\nu}^{\ast}(y,\Lambda)\,d\lambda,$ (5.7)
$\partial_{y}\psi_{k}(t,y)=-\frac{1}{2i\pi kt}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\mathbb{R}}e^{-ik\lambda
t}\partial_{y}\partial_{\lambda}\psi_{k,\nu}^{\ast}(y,\Lambda)\,d\lambda,$
(5.8)

and

$\omega_{k}(t,y)=-\frac{1}{2\pi}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\mathbb{R}}e^{-ik\lambda
t}\omega^{\ast}_{k,\nu}(y,\Lambda)\,d\lambda.$ (5.9)

To prove Theorem 1.1, it suffices to show that for
$k\in\mathbb{Z}\cap[1,\infty)$ and $t\geq 1$,

$|\psi_{k}(t,y)|\lesssim\frac{|y-y_{1\ast}|^{\gamma-2}+|y-y_{2\ast}|^{\gamma-2}}{\langle
t\rangle^{2}}e^{-\sigma_{\sharp}\nu^{1/2}t}|k|^{-7/4}M_{k},$ (5.10)
$|\partial_{y}\psi_{k}(t,y)|\lesssim e^{-\sigma_{\sharp}\nu^{1/2}t}\langle
t\rangle^{-1}|k|^{-3/4}M_{k},$ (5.11)

and

$\big{|}\omega_{k}(t,y)\big{|}\lesssim\Big{[}\big{(}|y-y_{j\ast}|+|\nu|^{1/4}\big{)}^{\gamma}+t^{-\gamma}\Big{]}e^{-\sigma_{\sharp}\nu^{1/2}t}|k|^{-1/2-(2-\gamma)/10}M_{k}.$
(5.12)

We shall give the detailed proofs for (5.10) and (5.12) only, as the proof of
(5.11) is identical to that of (5.7). We only remark that
$\partial_{\lambda}\partial_{y}\psi_{k,\nu}^{\ast}(y,\Lambda)$ is less
singular than $\partial_{\lambda}^{2}\psi_{k,\nu}^{\ast}(y,\Lambda)$ in the
region $\lambda\in\Sigma_{j,\delta_{0}},j\in\\{1,2\\}$. Roughly speaking, when
$\lambda\in\Sigma_{j,\delta_{0}}$, taking one derivative in $y$ costs a factor
$\varrho_{j,k}^{-1}$ while taking a derivative in $\lambda$ costs a larger
factor of $\varrho_{j,k}^{-1}\varrho_{j}^{-1}$.

Since the desired bounds (5.10) and (5.12) are pointwise, in view of the
nature of the singularities of the integrand in (5.7) and (5.9), it is
convenient to divide the integrals (5.7) and (5.9) into the sum of a local
integral and a nonlocal integral, as follows. For
$y\in\mathbb{T}_{\mathfrak{p}}$, we define

$\begin{split}&\psi_{k{\rm loc}}(t,y)\\\ &:=\frac{1}{2\pi
k^{2}t^{2}}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\\{\lambda\in\mathbb{R}:\,|\lambda-b(y)|\ll\,\min_{j\in\\{1,2\\}}|b(y)-b(y_{j\ast})|+|\nu/k|^{1/2}\\}}e^{-ik\lambda
t}\partial_{\lambda}^{2}\psi_{k,\nu}^{\ast}(y,\Lambda)\,d\lambda,\end{split}$
(5.13) $\begin{split}&\psi_{k{\rm nloc}}(t,y)\\\ &:=\frac{1}{2\pi
k^{2}t^{2}}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\\{\lambda\in\mathbb{R}:\,|\lambda-b(y)|\gtrsim\,\min_{j\in\\{1,2\\}}|b(y)-b(y_{j\ast})|+|\nu/k|^{1/2}\\}}e^{-ik\lambda
t}\partial_{\lambda}^{2}\psi_{k,\nu}^{\ast}(y,\Lambda)\,d\lambda.\end{split}$
(5.14)

Similarly, we define, for $y\in\mathbb{T}_{\mathfrak{p}}$,

$\begin{split}&\omega_{k{\rm loc}}(t,y)\\\
&:=-\frac{1}{2\pi}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\\{\lambda\in\mathbb{R}:\,|\lambda-b(y)|\ll\,\min_{j\in\\{1,2\\}}|b(y)-b(y_{j\ast})|+|\nu/k|^{1/2}\\}}e^{-ik\lambda
t}\omega^{\ast}_{k,\nu}(y,\Lambda)\,d\lambda,\end{split}$ (5.15)
$\begin{split}&\omega_{k{\rm nloc}}(t,y)\\\
&:=-\frac{1}{2\pi}e^{\alpha|k|t}e^{-\nu
k^{2}t}\int_{\\{\lambda\in\mathbb{R}:\,|\lambda-b(y)|\gtrsim\,\min_{j\in\\{1,2\\}}|b(y)-b(y_{j\ast})|+|\nu/k|^{1/2}\\}}e^{-ik\lambda
t}\omega^{\ast}_{k,\nu}(y,\Lambda)\,d\lambda.\end{split}$ (5.16)

To establish (5.10) and (5.12), it suffices to prove that for
$h\in\\{\psi_{k{\rm loc}}(t,y),\psi_{k{\rm nloc}}(t,y)\\}$ and
$y\in\mathbb{T}_{\mathfrak{p}},t\geq 1$, we have the bounds

$|h(t,y)|\lesssim\frac{|y-y_{1\ast}|^{\gamma-2}+|y-y_{2\ast}|^{\gamma-2}}{\langle
t\rangle^{2}}e^{-\sigma_{\sharp}\nu^{1/2}t}|k|^{-7/4}M_{k},$ (5.17)

and that $h\in\\{\omega_{k{\rm loc}}(t,y),\omega_{k{\rm nloc}}(t,y)\\}$ and
$y\in\mathbb{T}_{\mathfrak{p}},t\geq 1$, we have the bounds for
$j\in\\{1,2\\}$,

$\big{|}h(t,y)\big{|}\lesssim\Big{[}\big{(}|y-y_{j\ast}|+|\nu|^{1/4}\big{)}^{\gamma}+t^{-\gamma/2}\Big{]}e^{-\sigma_{\sharp}\nu^{1/2}t}|k|^{-1/2-(2-\gamma)/10}M_{k}.$
(5.18)

We divide the proof of (5.17) and (5.18) into several cases, depending on the
location of $y\in\mathbb{T}_{\mathfrak{p}}$ relative to the critical points
$y_{1\ast},y_{2\ast}$.

###  5.1. The region $\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\gtrsim 1$.

We firstly summarize the following bounds for
$\psi^{\ast}_{k,\nu}(y,\Lambda)$, which follow from Proposition 4.2 and
Corollary 4.1. For $\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\gtrsim 1$ and
$\lambda\in\mathbb{R}\backslash\cup_{j=1}^{2}\Sigma_{j,\delta_{0}}$ and
$\sigma\in\\{0,1,2\\}$, we have the bounds

$\|D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(\cdot,\Lambda)\|_{H^{1}_{k}(\mathbb{T}_{\mathfrak{p}})}\lesssim|k|^{-5/2+\sigma}\langle\lambda\rangle^{-1-\sigma}M_{k}.$
(5.19)

Additionally, for $\lambda\in\Sigma_{j,\delta_{0}}$ with $j\in\\{1,2\\}$,
$\beta\in\\{0,1\\}$ and $\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\gtrsim 1$, we
have the bounds

$\begin{array}[]{lr}|k|^{-\beta}|\partial_{y}^{\beta}D_{\lambda}^{\sigma}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim|k|^{-\gamma_{\ast}-1/2+\sigma}M_{k},&\quad{\rm
if}\,\,\sigma\in\\{0,1\\},\\\
|\partial_{y}^{\beta}D_{\lambda}^{2}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim|k|^{-\gamma_{\ast}+3/2+\beta}\big{(}\delta(\lambda)\big{)}^{-1-2(2-\gamma_{\ast})}M_{k},&\quad{\rm
if}\,\,\sigma=2.\end{array}$ (5.20)

The bounds (5.19)-(5.20), together with (4.15)-(4.16) and the identities (5.6)
are sufficient for proving the desired estimates bounds (5.17) for
$h=\psi_{k{\rm nloc}}$ and (5.20) for $\omega_{k{\rm nloc}}$.

To prove the bounds (5.17) for $h=\psi_{k{\rm loc}}$, we use (4.16), the
identity (5.6), and with a slight abuse of notation denoting
$\psi_{k,\nu}^{\ast}(y,\lambda)$ as $\psi_{k,\nu}^{\ast}(y,\Lambda)$ to
emphasize the dependence on $\lambda$, to obtain that

$\psi^{\ast}_{k,\nu}(y,\lambda)-\psi^{\ast}_{k,\nu}(y,b(y))=\int_{b(y)}^{\lambda}\Big{[}D_{\tau}\psi^{\ast}_{k,\nu}(y,\tau)-a(y,\tau)\partial_{y}\psi^{\ast}_{k,\nu}(y,\tau)\Big{]}\,d\tau$
(5.21)

and for $\lambda\in\mathbb{R}$ with $|\lambda-b(y)|\ll 1$ that

$\begin{split}&\bigg{|}\omega^{\ast}_{k,\nu}(y,\Lambda)-\frac{\omega_{0k}(y)-ib^{\prime\prime}(y)\psi^{\ast}_{k,\nu}(y,b(y))}{i(\lambda-b(y))+|\nu/k|^{1/3}}\bigg{|}\\\
&\lesssim\frac{|\nu/k|^{-1/3}}{1+|\nu/k|^{-2/3}|\lambda-b(y)|^{2}}|k|^{-5/2}M_{k}+\frac{1}{|\lambda-b(y)|^{1/2}}|k|^{-3/2}M_{k}+|k|^{-1/2}M_{k}.\end{split}$
(5.22)

The desired bounds then follow from (5.22) and the identity

$(\partial_{y}^{2}-k^{2})\psi_{k,\nu}^{\ast}(y,\Lambda)=\omega^{\ast}_{k,\nu}(y,\Lambda).$
(5.23)

The bounds (5.20) for $\omega_{k{\rm loc}}$ follow from (5.22).

###  5.2. The region $|\nu/k|^{1/4}\ll\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\ll
1$.

For the sake of concreteness we assume that $|y-y_{1\ast}|\in[m/2,m]$ and
$|\nu/k|^{1/4}\ll m\ll 1$. It follows from proposition 4.2 that the following
inequalities hold. For $\sigma\in\\{0,1\\},\beta\in\\{0,1\\}$,

  * •

If $|\lambda-b(y_{1\ast})|^{1/2}\ll m$, then

$|\partial_{y}^{\beta}D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim|y-y_{1\ast}|^{-\sigma}\big{(}|y-y_{1\ast}|\wedge\frac{1}{|k|}\big{)}^{\gamma_{\ast}-\sigma-\beta}|k|^{-1/2}M_{k};$
(5.24)

  * •

If $\lambda\in\mathbb{R}\backslash\Sigma_{2,\delta_{0}}$ and
$|\lambda-b(y_{1\ast})|^{1/2}\gtrsim m$, then

$\|\partial_{y}^{\beta}D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)\|_{L^{2}(m/4,4m)}\lesssim\delta(\lambda)^{1/2-\sigma}\big{(}\delta(\lambda)\wedge\frac{1}{|k|}\big{)}^{\gamma_{\ast}-\sigma-\beta}\frac{1}{\langle\lambda\rangle^{1+\sigma}}|k|^{-1/2}M_{k};$
(5.25)

  * •

If $\lambda\in\Sigma_{2,\delta_{0}}$, then

$|\partial_{y}^{\beta}D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim|k|^{-\gamma_{\ast}-1/2+\sigma+\beta}M_{k}.$
(5.26)

For $\sigma=2,\beta\in\\{0,1\\}$, we have the following bounds.

  * •

If $|\lambda-b(y_{1\ast})|^{1/2}\ll m$, then

$\begin{split}&|\partial_{y}^{\beta}D^{2}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\\\
&\lesssim(\delta(\lambda))^{-1-2(2-\gamma_{\ast})}|y-y_{1\ast}|^{-1+2(2-\gamma_{\ast})}\big{(}|y-y_{1\ast}|\wedge\frac{1}{|k|}\big{)}^{\gamma_{\ast}-2-\beta}|k|^{-1/2}M_{k};\end{split}$
(5.27)

  * •

If
$\lambda\in\mathbb{R}\backslash\Sigma_{2,\delta_{0}},|\lambda-b(y_{1\ast})|^{1/2}\gtrsim
m$, then

$\|\partial_{y}^{\beta}D^{2}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)\|_{L^{2}(m/4,4m)}\lesssim\delta(\lambda)^{-3/2}\big{(}\delta(\lambda)\wedge\frac{1}{|k|}\big{)}^{\gamma_{\ast}-2-\beta}\frac{1}{\langle\lambda\rangle^{3}}|k|^{-1/2}M_{k};$
(5.28)

  * •

If $\lambda\in\Sigma_{2,\delta_{0}}$, then

$|\partial_{y}^{\beta}D^{2}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim(\delta(\lambda))^{-1-2(2-\gamma_{\ast})}|k|^{-\gamma_{\ast}+3/2+\beta}M_{k}.$
(5.29)

The desired bounds (5.17) for $h=\psi_{k{\rm nloc}}(t,y)$ follow from the
bounds (5.24)-(5.29), by straightforward calculations, using the formula
(5.14).

To prove (5.17) for $h=\psi_{k{\rm loc}}(t,y)$ and (5.18), we use the identity
(5.23), the bound (4.14) which we rewrite as follows. Denoting
$\psi^{\ast}_{k,\nu}(y,\Lambda)$ as $\psi^{\ast}_{k,\nu}(y,\lambda)$ with a
slight abuse of notation to emphasize the dependence on $\lambda$, for
$\lambda\in\Sigma_{1,\delta_{0}}$ with
$|\lambda-b(y)|\ll\,|b(y)-b(y_{1\ast})|$, we have the bounds

$\begin{split}&\bigg{|}\omega^{\ast}_{k,\nu}(y,\Lambda)+\frac{ib^{\prime\prime}(y)\psi^{\ast}_{k,\nu}(y,b(y))}{i(\lambda-b(y))+|\nu/k|^{1/3}|y-y_{1\ast}|^{2/3}}\bigg{|}\\\
&\lesssim\varrho_{1,k}^{\gamma_{\ast}-1/2}(y;\Lambda)|y-y_{1\ast}|^{-1/2}\frac{|\nu/k|^{-1/3}|y-y_{1\ast}|^{1/3}}{1+(|\nu/k|^{1/3}|y-y_{1\ast}|^{2/3})^{-2}|\lambda-b(y)|^{2}}|k|^{-1/2}M_{k}\\\
&\quad+\varrho_{1,k}^{\gamma_{\ast}-3/2}(y;\Lambda)\varrho_{1}^{1/2}(y;\Lambda)|\lambda-b(y)|^{-1/2}|k|^{-1/2}M_{k}.\end{split}$
(5.30)

The bounds (5.18) for $h=\omega_{k{\rm loc}}(t,y)$ follow from (5.30).

To prove (5.18) with $h=\omega_{k{\rm nloc}}(t,y)$, we use also the following
bounds. In view of (4.53) and Proposition 4.2, we have the pointwise estimates
for $\lambda\in\Sigma_{1,\delta_{0}}$ with
$|\lambda-b(y)|\gtrsim|b(y)-b(y_{1\ast})|\\}$,

$\begin{split}&\Big{|}\omega_{k,\nu}^{\ast}(y,\Lambda)+(\partial_{y}^{2}-k^{2})\Big{\\{}\sum_{j=1}^{2}\varphi\big{(}(y-y_{j\ast})/\delta_{0}\big{)}\frac{\omega_{0k}(y)}{ib^{\prime\prime}(y)}\Big{\\}}\Big{|}\\\
&\lesssim\varrho_{j,k}^{\gamma_{\ast}}(y;\Lambda)\varrho_{j}^{-2}(y;\Lambda)|k|^{-1/2}M_{k},\end{split}$
(5.31)
$|\partial_{\lambda}\omega_{k,\nu}^{\ast}(y,\Lambda)|\lesssim\varrho_{j,k}^{\gamma_{\ast}-3/2}(y;\Lambda)\varrho_{j}^{-5/2}(y;\Lambda)|k|^{-1/2}M_{k}.$
(5.32)

Using the identity

$\partial_{\lambda}\omega_{k,\nu}^{\ast}(y,\Lambda)=\partial_{\lambda}\omega_{k,\nu}(y,\Lambda)$
(5.33)

and equation (1.18) we get that for
$\lambda\in\mathbb{R}\backslash\big{(}\cup_{j=1}^{2}\Sigma_{j,\delta_{0}}\big{)}$,

$|\partial_{\lambda}\omega_{k,\nu}^{\ast}(y,\Lambda)|\lesssim\frac{1}{\langle\lambda\rangle^{2}}|k|^{-3/2}M_{k}.$
(5.34)

For $\lambda\in\Sigma_{2,\delta_{0}}$, using (4.53) and Proposition 4.2 we get
that

$|\partial_{\lambda}\omega_{k,\nu}^{\ast}(y,\Lambda)|\lesssim|k|^{-1/2}M_{k}.$
(5.35)

Fix $\tau\in(1,\infty)$ as a parameter to be chosen below. We divide the
integral (5.16) into three regions for $\lambda\in\mathbb{R}$: (I)
$\lambda\in\Sigma_{1,\delta_{0}}$ with
$|b(y)-b(y_{1\ast})|\lesssim|\lambda-b(y)|<\tau^{2}|b(y)-b(y_{1\ast})|$; (II)
$\lambda\in\Sigma_{1,\delta_{0}}$ with
$|\lambda-b(y)|\geq\tau^{2}|b(y)-b(y_{1\ast})|$; (III)
$\lambda\in\mathbb{R}\backslash\big{[}\Sigma_{1,\delta_{0}}\cup\Sigma_{2,\delta_{0}}\big{]}$;
(IV) $\lambda\in\Sigma_{2,\delta_{0}}$.

Using the bounds (5.31)-(5.35), and integrating by parts in $\lambda$ in the
regions (II)-(IV), we obtain the bounds for $t\geq 1$,

$\begin{split}&|\omega_{k{\rm
nloc}}(t,y)|\lesssim\bigg{[}|y-y_{1\ast}|^{\gamma}+\tau^{\gamma}|y-y_{1\ast}|^{\gamma}+\frac{1}{|k|t}\tau^{\gamma-2}|y-y_{1\ast}|^{\gamma-2}\bigg{]}|k|^{-1/2-(2-\gamma)/10}M_{k}.\end{split}$
(5.36)

Optimizing in $\tau\in(1,\infty)$, the desired bounds for (5.18) with
$h=\omega_{k{\rm nloc}}(t,y)$ follows.

###  5.3. The region
$\min\\{|y-y_{1\ast}|,|y-y_{2\ast}|\\}\lesssim|\nu/k|^{1/4}$.

For the sake of concreteness, we assume that
$|y-y_{1\ast}|\lesssim|\nu/k|^{1/4}$. In this case we shall use the following
bounds, which are consequences of Proposition 4.2.

  * •

If $\lambda\in\Sigma_{1,\delta_{0}}$, $\sigma\in\\{0,1,2\\},\beta\in\\{0,1\\}$
and $\lambda\in\Sigma_{1,\delta_{0}}$,

$\begin{split}&\|\partial_{y}^{\beta}D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)\|_{L^{2}(|y-y_{1\ast}|\leq
3\delta(\Lambda)}\lesssim(\delta(\Lambda))^{1/2-\sigma}\Big{[}\delta(\Lambda)\wedge\frac{1}{|k|}\Big{]}^{\gamma_{\ast}-\sigma-\beta}|k|^{-1/2}M_{k};\end{split}$
(5.37)

  * •

If $\lambda\in\mathbb{R}\backslash\cup_{j\in\\{1,2\\}}\Sigma_{j,\delta_{0}}$,
$\sigma\in\\{0,1,2\\}$ and $\beta\in\\{0,1\\}$,

$|\partial_{y}^{\beta}D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim\langle\lambda\rangle^{-1-\sigma}|k|^{-5/2+\sigma+\beta}M_{k};$
(5.38)

  * •

If $\lambda\in\Sigma_{2,\delta_{0}}$, $\sigma\in\\{0,1\\}$ and
$\beta\in\\{0,1\\}$,

$|\partial_{y}^{\beta}D^{\sigma}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim|k|^{-\gamma_{\ast}+\sigma+\beta}|k|^{-1/2}M_{k};$
(5.39)

  * •

If $\lambda\in\Sigma_{2,\delta_{0}}$, $\sigma=2$ and $\beta\in\\{0,1\\}$,

$|\partial_{y}^{\beta}D^{2}_{\lambda}\psi^{\ast}_{k,\nu}(y,\Lambda)|\lesssim\big{(}\delta(\Lambda)\big{)}^{-1-2(2-\gamma_{\ast})}|k|^{-\gamma_{\ast}+2+\beta}.$
(5.40)

This case is somewhat simpler in fact, since, roughly speaking, we can use the
ellipticity of the Orr-Sommerfeld equation to bound derivatives without much
loss, using Lemma 3.3. Indeed, the desired bounds (5.17)-(5.18) follow from an
analogous argument as in section 5.2, using (5.37)-(5.40), but somewhat
simpler since there is no need to separately treat the region
$|\lambda-b(y)|\ll|\nu/k|^{1/2}$ which is no longer singular. We omit the
details. Theorem 1.1 is now proved.

##  6\. Proof of the limiting absorption principle (I): non-degenerate region

In this section we give the proof of the limiting absorption principle in the
non-degenerate region when
$\lambda\in\mathbb{R}\backslash\big{(}\Sigma_{1,\delta_{0}}\cup\Sigma_{2,\delta_{0}}\big{)}$
or $\alpha\geq 1$. We begin with some technical bounds on various operators
that are useful below, and then prove the limiting absorption principle using
a compactness argument.

Denote $\epsilon:=\nu/k$. Recall that $\epsilon\in(0,\frac{1}{8}]$,
$k\in\mathbb{Z}\cap[1,\infty)$, and

$(\alpha,\lambda)\in\Big{[}(-\sigma_{0}\epsilon^{1/2},\infty)\times\big{(}\mathbb{R}\setminus(\Sigma_{1,\delta_{0}}\cup\Sigma_{2,\delta_{0}})\big{)}\Big{]}\cup\Big{[}(1,\infty)\times\big{(}\Sigma_{1,\delta_{0}}\cup\Sigma_{2,\delta_{0}}\big{)}\Big{]}.$

We also recall the definition (4.1) for $T_{\Theta}$ and set for simplicity of
notations $\Delta_{k}:=\partial_{y}^{2}-k^{2}$.

###  6.1. Preliminary bounds

We define two regions in the $(\alpha,\lambda)$ plane

$\displaystyle R_{nd}$
$\displaystyle:=\big{\\{}(\alpha,\lambda)\,|\,\alpha\geq-\sigma_{0}|\epsilon|^{1/2},\lambda\in\mathbb{R}\setminus(\Sigma_{1,\delta_{0}}\cup\Sigma_{2,\delta_{0}})\big{\\}},$
(6.1) $\displaystyle R_{\alpha}$
$\displaystyle:=\big{\\{}(\alpha,\lambda)\,|\,\alpha\geq
1,\lambda\in\Sigma_{1,\delta_{0}}\cup\Sigma_{2,\delta_{0}}\big{\\}}.$ (6.2)

In $R_{nd}$, which we call the “non-degenerate region”, $\lambda$ is separated
from the critical points; in $R_{\alpha}$, which we call the “$\alpha$
dominated region”, $\alpha\geq 1$. Therefore, there is no singularity at the
critical layer in either case when we apply the operator $A_{\Theta}^{-1}$.

Throughout the section we assume that $(\alpha,\lambda)\in R_{nd}\cup
R_{\alpha}$. We begin with the following pointwise bounds.

######  Lemma 6.1.

Suppose that

$m:=\inf_{y\in\mathbb{T}_{\mathfrak{p}}}(|b(y)-\lambda|+|\alpha|+\epsilon^{\frac{1}{3}})\geq\delta_{0}/k.$

Then for any bounded function $f$ on $\mathbb{T}_{\mathfrak{p}}$,

$\displaystyle\sup_{y\in\mathbb{T}_{\mathfrak{p}}}\Big{|}\frac{1}{k^{\beta}}\partial_{y}^{\beta}T_{\Theta}f(y)\Big{|}\lesssim\frac{\norm{f}_{L^{\infty}(\mathbb{T}_{\mathfrak{p}})}}{k^{2}m},\quad{\rm
with}\,\,\beta\in\\{0,1\\}.$ (6.3)

###### Proof.

The proof is a consequence (2.3), Proposition 3.1 in the non-degenerate
regime, and the fact that $ke^{-k|y|}$ is an approximation of the identity. ∎

The bounds from Lemma 6.1 are sufficient for our purposes when
$(\alpha,\lambda)\in R_{\alpha}$. For $y\in\mathbb{T}_{\mathfrak{p}}$ such
that

$k(|b(y)-\lambda|+|\alpha|+\epsilon^{\frac{1}{3}})\ll 1,$

we need to carry out a finer analysis of the behavior near the critical layer.
To begin with, we decompose $T_{\Theta}$ as follows. Let $\varphi_{0}\in
C_{c}^{\infty}(-1,1)$ with $\varphi_{0}\equiv 1$ on $[-1/2,1/2]$. Then for any
number $L>0$, define for $y\in\mathbb{R}$,

$\varphi_{L}(y):=\varphi_{0}\left(\frac{y}{L}\right).$ (6.4)

We now decompose for any $f\in H^{1}(\mathbb{T}_{\mathfrak{p}})$,

$\displaystyle T_{\Theta}f$ $\displaystyle=\sum_{\ell=0}^{2}T_{\Theta,\ell}f$
(6.5)

where

$\displaystyle
T_{\Theta,0}f(y):=\int_{\mathbb{T}_{\mathfrak{p}}}G_{k}(y,z)A_{\Theta}^{-1}(f)(z)\Big{[}1-\sum_{\ell=1}^{2}\varphi_{\theta}(z-y_{\ell})\Big{]}\,dz$
(6.6)

and

$\displaystyle
T_{\Theta,\ell}f(y):=\int_{\mathbb{T}_{\mathfrak{p}}}G_{k}(y,z)\varphi_{\theta}(z-y_{\ell})A_{\Theta}^{-1}(f)(z)\,dz$
(6.7)

where $b(y_{\ell})=\lambda,\ell\in\\{1,2\\}$ and $\theta=\theta(\delta_{0})>0$
is a sufficiently small constant. We note that for $\lambda$ in the interior
of $b(\mathbb{T}_{\mathfrak{p}})$, the set $b^{-1}(\lambda)$ has cardinality
2.

Since the argument of $T_{\Theta,0}$ is supported away from the critical layer
we have the following pointwise bounds.

######  Lemma 6.2.

There exists $\theta_{0}\in(0,1)$, depending only on $b(y)$, such that for
$\theta\in(0,\theta_{0}]$ and any $f\in L^{\infty}(\mathbb{T}_{\mathfrak{p}})$
we have the following bounds:

$\frac{1}{k^{\beta}}\big{|}\partial_{y}^{\beta}T_{\Theta,0}f(y)\big{|}\lesssim\frac{\norm{f}_{L^{\infty}(\mathbb{T}_{\mathfrak{p}})}}{k^{2}},\quad\beta\in\\{0,1,2\\}.$
(6.8)

###### Proof.

Proposition 3.1 implies that

$\Big{|}A_{\Theta}^{-1}(f)(z)\big{(}1-\sum_{\ell=1}^{2}\varphi_{\theta}^{2}(z-y_{\ell})\big{)}\Big{|}\lesssim_{\theta_{0}}\norm{f}_{L^{\infty}(\mathbb{T}_{\mathfrak{p}})}.$
(6.9)

The proof is then an immediate consequence of Lemma 2.1 and the fact that
$ke^{-k|y|}$ is an approximation of identity. ∎

We now turn our attention to $T_{\Theta,\ell},\ell\in\\{1,2\\}$. The following
result is useful in characterizing the singular behavior near the critical
layer.

######  Lemma 6.3.

There exists $\theta_{0}\in(0,1)$, such that for
$\theta\in(0,\theta_{0}],\,\ell\in\\{1,2\\}$, the following statement holds.
Suppose that $f\in H^{1}_{k}(\mathbb{T}_{\mathfrak{p}})$ with

$M:=\norm{f}_{H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})}.$ (6.10)

We view the function $\varphi_{\theta}(y-y_{\ell})A_{\Theta}^{-1}(f)(y)$ as a
function on $\mathbb{R}$. Then

$\sup_{\xi\in\mathbb{R}}\big{|}\mathcal{F}\left(\varphi_{\theta}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f)\right)(\xi)\big{|}\lesssim M,$ (6.11)

where $\mathcal{F}(h)$ denotes the Fourier transform of $h$ and $y_{\ell}$
satisfies $b(y_{\ell})=\lambda$.

###### Proof.

With a slight abuse of notation, denote
$\varphi_{\ell}(y):=\varphi\big{(}\frac{2(y-y_{\ell})}{\theta}\big{)}$, for
$\ell\in\\{1,2\\},y\in\mathbb{R}$. Define also $w_{c,\ell}$ as the solution to

$\epsilon\partial_{y}^{2}w_{c,\ell}(y)-\alpha
w_{c,\ell}(y)+ib^{\prime}(y_{\ell})(y_{\ell}-y)w_{c,\ell}(y)=f(y_{\ell}),\quad
y\in\mathbb{R}.$ (6.12)

We can solve $w_{c,\ell}$ explicitly as

$w_{c,\ell}(y,y_{\ell})=\frac{\
f(y_{\ell})}{b^{\prime}(y_{\ell})}\left(\frac{b^{\prime}(y_{\ell})}{\epsilon}\right)^{\frac{1}{3}}W_{\ell}\left(\left(\frac{b^{\prime}(y_{\ell})}{\epsilon}\right)^{\frac{1}{3}}(y-y_{\ell})\right),$
(6.13)

where

$\widehat{W}_{\ell}(\xi)=-\sqrt{2\pi}\exp\left(\frac{\xi^{3}}{3}+\frac{\epsilon^{-\frac{1}{3}}\alpha\xi}{(b^{\prime}(y_{\ell}))^{\frac{2}{3}}}\right)\mathbbm{1}_{(-\infty,0]}(\xi),\quad\xi\in\mathbb{R}.$
(6.14)

Define $w_{R}:\mathbb{T}_{\mathfrak{p}}\to\mathbb{C}$ as

$\displaystyle
w_{R}:=A_{\Theta}^{-1}(f)-\sum_{\ell\in\\{1,2\\}}\varphi_{\ell}w_{c,\ell},$
(6.15)

where we view $\varphi_{\ell}w_{c,\ell}$ as functions on
$\mathbb{T}_{\mathfrak{p}}$. It follows from (6.12) and (6.15) that $w_{R}$
satisfies the equation for $y\in\mathbb{T}_{\mathfrak{p}}$,

$\displaystyle\epsilon\partial_{y}^{2}w_{R}(y)-\alpha
w_{R}(y)+i(\lambda-b(y))w_{R}(y)$
$\displaystyle=f(y)-\sum_{\ell\in\\{1,2\\}}\varphi_{\ell}(y)f(y_{\ell})-\epsilon\sum_{\ell\in\\{1,2\\}}\Big{[}\partial_{y}\varphi_{\ell}(y)\partial_{y}w_{c,\ell}(y)+\partial_{y}^{2}\varphi_{\ell}(y)w_{c,\ell}(y)\Big{]}$
$\displaystyle\quad+i\sum_{\ell\in\\{1,2\\}}\varphi_{\ell}(y)w_{c,\ell}(y)\Big{[}b(y_{\ell})+b^{\prime}(y_{\ell})(y-y_{\ell})-b(y)\Big{]}.$
(6.16)

In view of Proposition 3.1, we conclude that for
$y\in\mathbb{T}_{\mathfrak{p}}$,

$|w_{R}(y)|\lesssim\frac{M\epsilon^{-\frac{1}{6}}}{\langle\epsilon^{-\frac{1}{3}}|\alpha|,\epsilon^{-\frac{1}{3}}(b(y)-\lambda)\rangle^{\frac{1}{2}}}.$
(6.17)

Therefore,

$\big{|}\mathcal{F}(\varphi_{\theta}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f))(\xi)\big{|}\lesssim M,\quad\ell\in\\{1,2\\},$
(6.18)

which completes the proof. ∎

######  Remark 6.1.

Note that $\varphi_{\theta}(y-y_{\ell})A_{\Theta}^{-1}(f)(y)$ can also be
viewed as a periodic function on $\mathbb{T}_{\mathfrak{p}}$. Then the Fourier
coefficients $\big{[}\varphi_{\theta}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f)\big{]}_{l}$ are bounded up to an implicit constant
by $M$,

$\sup_{l\in\frac{2\pi\mathbb{Z}}{\mathfrak{p}}}\Big{|}\big{[}\varphi_{\theta}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f)\big{]}_{l}\Big{|}\lesssim M.$ (6.19)

Recalling the definition of $T_{\Theta,\ell}f,\ell\in\\{1,2\\}$ as
$\Delta_{k}^{-1}\varphi_{\theta}(\cdot-
y_{\ell})A_{\Theta}^{-1}(ib^{\prime\prime}f)$ and the Fourier multiplier
representation of $\Delta_{k}^{-1}$, we have as a consequence of Lemma 6.3 the
following bounds for Fourier coefficients of $T_{\Theta,\ell}f$:

######  Corollary 6.1.

Let $\norm{f}_{H_{k}^{1}}=M$. Then we have

$\big{|}[T_{\Theta,\ell}f]_{l}\big{|}\lesssim\frac{M}{l^{2}+k^{2}},\quad
k\in\mathbb{Z},\,l\in\frac{2\pi\mathbb{Z}}{\mathfrak{p}}.$ (6.20)

Finally, combining Lemma 6.1, Lemma 6.2 and Corollary 6.1, we can conclude the
following bounds for $T_{\Theta}$ on $H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})$.

######  Lemma 6.4.

$T_{\Theta}:H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})\to
H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})$ satisfies the bounds

$\norm{T_{\Theta}}_{H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})\to
H^{1}_{k}(\mathbb{T}_{\mathfrak{p}})}\lesssim k^{-1/2}.$ (6.21)

Moreover, the family of operators $T_{\Theta}$ has the following “uniform
compactness” property. More precisely, assume that
$\ell,k_{\ell}\in\mathbb{Z}\cap[1,\infty)$ and that $f_{\ell}$ is a sequence
of functions satisfying
$\norm{f_{\ell}}_{H_{k_{\ell}}^{1}(\mathbb{T}_{\mathfrak{p}})}=1$. Let

$\Theta_{\ell}:=(\epsilon_{\ell},\alpha_{\ell},k_{\ell},\lambda_{\ell})\in\big{(}0,1/8\big{)}\times(-\sigma_{0}\epsilon_{\ell}^{1/2},1)\times\big{(}\mathbb{Z}\cap[1,\infty)\big{)}\times(\Sigma_{0}\setminus\cup_{j\in\\{1,2\\}}\Sigma_{j,\delta_{0}}).$

Then for any $k_{0}\in\mathbb{Z}\cap[1,\infty)$, the sequence

$T_{\Theta_{\ell}}(f_{\ell})$ (6.22)

has a convergent subsequnece in $H_{k_{0}}^{1}(\mathbb{T}_{\mathfrak{p}})$.

For applications below, we need to study the singular behavior of
$A_{\Theta}^{-1}(f)$ as $\epsilon\to 0$, which is useful to connect the
spectral assumption on the invisicd Rayleigh operator to the viscous Orr-
Sommerfed operator.

######  Lemma 6.5.

Let $f\in H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})$. Then $A_{\Theta}^{-1}(f)$ has
the following limiting behavior for $y\in\mathbb{T}_{\mathfrak{p}}$ in the
sense of distributions.

  * •

If $\alpha_{0}\neq 0$ or $\lambda_{0}\notin b(\mathbb{T}_{\mathfrak{p}})$,

$\lim_{\epsilon\to
0,\,\alpha\to\alpha_{0},\,\lambda\to\lambda_{0}}A_{\Theta}^{-1}(f)(y)=\frac{-f(y)}{\alpha_{0}+i(b(y)-\lambda_{0})};$
(6.23)

  * •

If $\lambda_{0}\in
b(\mathbb{T}_{\mathfrak{p}})\setminus\cup_{j\in\\{1,2\\}}\Sigma_{j,\delta_{0}}$,

$\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}A_{\Theta}^{-1}(f)(y)={\rm
P.V.}\frac{-f(y)}{i(b(y)-\lambda_{0})}-\pi\sum_{b(z)=\lambda_{0}}\frac{f(z)}{|b^{\prime}(z)|}\delta(y-z).$
(6.24)

###### Proof.

We only give the detailed proof of (6.24), since the other case is easier, and
indicate the necessary changes for the proof of (6.23) at the end of the
proof. Fix an even cutoff function $\Phi\in C_{0}^{\infty}(-2,2)$ with
$\Phi\equiv 1$ on $[-1,1]$. Let $\phi\in
C_{0}^{\infty}(\mathbb{T}_{\mathfrak{p}})$ be a test function. We have

$\displaystyle\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}\int_{\mathbb{T}_{\mathfrak{p}}}A_{\Theta}^{-1}(f)(y)\phi(y)\,dy$
$\displaystyle=\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}\int_{\mathbb{T}_{\mathfrak{p}}}\sum_{b(z)=\lambda}\Phi\left(\frac{y-z}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)A_{\Theta}^{-1}(f)(y)\phi(y)\,dy$
$\displaystyle\quad+\int_{\mathbb{T}_{\mathfrak{p}}}\sum_{b(z)=\lambda}\left(1-\Phi\left(\frac{y-z}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)\right)A_{\Theta}^{-1}(f)(y)\phi(y)\,dy.$
(6.25)

For $y$ in the support of $1-\Phi$, we divide (3.33) by $i(b(y)-\lambda)$ to
get that

$A_{\Theta}^{-1}(f)(y)=-\frac{f(y)}{i(b(y)-\lambda)}-\frac{\alpha
A_{\Theta}^{-1}(f)(y)}{i(b(y)-\lambda)}+\frac{\epsilon\partial_{y}^{2}A_{\Theta}^{-1}(f)(y)}{i(b(y)-\lambda)}.$
(6.26)

Proposition 3.1 then implies that

$\displaystyle\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}\int_{\mathbb{T}_{\mathfrak{p}}}\sum_{b(z)=\lambda}\left(1-\Phi\left(\frac{y-z}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)\right)A_{\Theta}^{-1}(f)(y)\phi(y)\,dy=-\text{P.V.}\int_{\mathbb{T}_{\mathfrak{p}}}\frac{f(y)\phi(y)}{i(b(y)-\lambda_{0})}\,dy.$
(6.27)

On the other hand, recalling the decomposition (6.15) in Lemma 6.3, we obtain
that

$\displaystyle\int_{\mathbb{T}_{\mathfrak{p}}}\sum_{b(z)=\lambda}\Phi\left(\frac{y-z}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)A_{\Theta}^{-1}(f)(y)\phi(y)\,dy$
$\displaystyle=\int_{\mathbb{T}_{\mathfrak{p}}}\sum_{\ell\in\\{1,2\\}}\Phi\left(\frac{y-z_{\ell}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{c,\ell}(y)\phi(y)\,dy+\int_{\mathbb{T}_{\mathfrak{p}}}\sum_{\ell\in\\{1,2\\}}\Phi\left(\frac{y-z_{\ell}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{R}(y)\phi(y)\,dy.$
(6.28)

From (6.17), it follows that

$\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}\int_{\mathbb{T}_{\mathfrak{p}}}\Phi\left(\frac{y-z_{\ell}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{R}(y)\phi(y)\,dy=0.$
(6.29)

We now focus our attention on $w_{c,1}$ with the treatment of $w_{c,2}$ being
identical. Since $w_{c,1}$ is defined on $\mathbb{R}$, we have

$\displaystyle\int_{\mathbb{R}}\Phi\left(\frac{y-z_{1}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{c,1}(y)\phi(y)\,dy$
$\displaystyle=\phi(z_{1})\int_{\mathbb{R}}\Phi\left(\frac{y-z_{1}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{c,1}(y)\,dy+\int_{\mathbb{R}}\Phi\left(\frac{y-z_{1}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{c,1}(y)(\phi(y)-\phi(z_{1}))\,dy.$
(6.30)

Since $w_{c,1}(y)(\phi(y)-\phi(z_{1}))$ is bounded, the second term converges
to 0 as $\alpha,\epsilon\to 0$. Therefore, it suffices to consider the first
term. In view of (6.13), we get that

$\int_{\mathbb{R}}\Phi\left(\frac{y-z_{1}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)w_{c,1}(y)\,dy=\int_{\mathbb{R}}\Phi\left(\frac{y-z_{1}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)\frac{\epsilon^{-\frac{1}{3}}f(z_{1})}{(b^{\prime}(z_{1}))^{\frac{2}{3}}}W(\epsilon^{-\frac{1}{3}}b^{\prime}(z_{1})(y-z_{1}))\,dy.$
(6.31)

By rescaling and Parseval’s identity, we obtain that

$\displaystyle\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}\int_{\mathbb{R}}\Phi\left(\frac{y-z_{1}}{(|\alpha|+\epsilon)^{\frac{1}{4}}}\right)\frac{\epsilon^{-\frac{1}{3}}f(z_{1})}{(b^{\prime}(z_{1}))^{\frac{2}{3}}}W(\epsilon^{-\frac{1}{3}}b^{\prime}(z_{1})(y-z_{1}))\,dy$
$\displaystyle=-\sqrt{2\pi}\varphi(z_{1})\frac{f(z_{1})}{|b^{\prime}(z_{1})|}\int_{-\infty}^{0}\widehat{\Phi}(\xi)d\xi=-\pi\varphi(z_{1})\frac{f(z_{1})}{|b^{\prime}(z_{1})|}.$
(6.32)

We now turn to the case where $\lambda_{0}\notin b(\mathbb{T}_{\mathfrak{p}})$
and $\alpha_{0}=0$. For $\lambda$ sufficiently close to $\lambda_{0}$,
$b(y)-\lambda$ does not vanish on $\mathbb{T}_{\mathfrak{p}}$. Hence, assuming
that $\lambda\notin b(\mathbb{T}_{\mathfrak{p}})$, we can write that

$A_{\Theta}^{-1}(f)(y)=-\frac{f(y)}{\alpha+i(b(y)-\lambda)}+\frac{\epsilon\partial_{y}^{2}A_{\Theta}^{-1}(f)(y)}{\alpha+i(b(y)-\lambda)}.$
(6.33)

Proposition 3.1 implies

$\displaystyle\lim_{\epsilon\to 0,\,\alpha\to
0,\,\lambda\to\lambda_{0}}\int_{\mathbb{T}_{\mathfrak{p}}}A_{\Theta}^{-1}(f)(y)\phi(y)\,dy=\int_{\mathbb{T}_{\mathfrak{p}}}\frac{f(y)\phi(y)}{i(b(y)-\lambda_{0})}\,dy.$
(6.34)

Finally, assume that $\alpha_{0}\neq 0$ and $\lambda\in
b(\mathbb{T}_{\mathfrak{p}})$. In this case we again write
$A_{\Theta}^{-1}(f)$ as in (6.33) by taking $\alpha$ sufficiently close to
$\alpha_{0}$ so that $\alpha\neq 0$. It then follows from Proposition 3.1 that

$\displaystyle\lim_{\epsilon\to
0,\,\alpha\to\alpha_{0},\,\lambda\to\lambda_{0}}\int_{\mathbb{T}_{\mathfrak{p}}}A_{\Theta}^{-1}(f)(y)\phi(y)\,dy=\int_{\mathbb{T}_{\mathfrak{p}}}\frac{f(y)\phi(y)}{\alpha_{0}+i(b(y)-\lambda_{0})}\,dy,$
(6.35)

which completes the proof of Lemma 6.5. ∎

###  6.2. The compactness argument

We are now ready to prove the limiting absorption in the non-degenerate case.

######  Proposition 6.1.

There exist $\kappa>0$ and $\epsilon_{0}\in(0,1/8)$ such that for all
$\epsilon\in(0,\epsilon_{0})$, $k\in\mathbb{Z}\cap[1,\infty)$, and
$(\alpha,\lambda)\in R_{nd}\cup R_{\alpha}$, we have the bounds

$\norm{\psi+T_{\Theta}(ib^{\prime\prime}\psi)}_{H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})}\geq\kappa\norm{\psi}_{H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})},$
(6.36)

for any $\psi\in H_{k}^{1}(\mathbb{T}_{\mathfrak{p}})$.

###### Proof.

Suppose (6.36) does not hold. Then there exists a sequence
$(\psi_{\ell},\kappa_{\ell},\Theta_{\ell})$ for
$\ell\in\mathbb{Z}\cap[1,\infty)$, with $(\kappa_{\ell},\epsilon_{\ell})\to
0+$, $\norm{\psi_{\ell}}_{H_{k_{\ell}}^{1}(\mathbb{T}_{\mathfrak{p}})}=1$,
such that

$\norm{\psi_{\ell}+T_{\Theta_{\ell}}\psi_{\ell}}_{H_{k_{\ell}}^{1}(\mathbb{T}_{\mathfrak{p}})}\leq\kappa_{\ell}.$
(6.37)

We first assume that $(\alpha_{\ell},\lambda_{\ell})\in R_{nd}$. By Lemma 6.1
and Lemma 6.4 we see that $(k_{\ell},\alpha_{\ell},\lambda_{\ell})$ are
bounded. Therefore, passing to a subsequence if necessary, we can assume that

$(\alpha_{\ell},k_{\ell},\lambda_{\ell})\to(\alpha_{0},k_{0},\lambda_{0})\in[0,\infty)\times\mathbb{Z}\cap[1,\infty)\times\overline{\mathbb{R}\setminus(\cup_{j\in\\{1,2\\}}\Sigma_{j,\delta_{0}}}).$

It also follows from Lemma 6.4 that
$T_{\Theta_{\ell}}(ib^{\prime\prime}\psi_{\ell})$ is relatively compact in
$H_{k_{0}}^{1}(\mathbb{T}_{\mathfrak{p}})$. Writing $\psi_{\ell}$ as

$\psi_{\ell}=\psi_{\ell}+T_{\Theta_{\ell}}(ib^{\prime\prime}\psi_{\ell})-T_{\Theta_{\ell}}(ib^{\prime\prime}\psi_{\ell})$
(6.38)

and passing to a subsequence if necessary, we can infer that $\psi_{\ell}$
converges _strongly_ in $H_{k_{0}}^{1}(\mathbb{T}_{\mathfrak{p}})$ to $\Psi\in
H_{k_{0}}^{1}(\mathbb{T}_{\mathfrak{p}})$ with
$\|\Psi\|_{H_{k_{0}}^{1}(\mathbb{T}_{\mathfrak{p}})}=1$.

As a consequence of (6.37), we conclude that $\Psi$ satisfies the equation
that

$\Psi+\lim_{\ell\to\infty}T_{\Theta_{\ell}}(ib^{\prime\prime}\Psi)=0.$ (6.39)

Since

$\lim_{\ell\to\infty}T_{\Theta_{\ell}}(ib^{\prime\prime}\Psi)=\lim_{\ell\to\infty}\Delta_{k_{0}}^{-1}A_{\Theta_{\ell}}(ib^{\prime\prime}\Psi),$
(6.40)

applying $-\Delta_{k_{0}}$ yields

$-\Delta_{k_{0}}\Psi-\lim_{\ell\to\infty}A_{\epsilon_{\ell}}^{-1}(ib^{\prime\prime}\Psi)=0,$
(6.41)

in the sense of distributions. Assuming that $\alpha_{0}=0$, and using Lemma
6.5, we obtain that for $y\in\mathbb{T}_{\mathfrak{p}}$,

$-\Delta_{k_{0}}\Psi(y)+\text{P.V.}\frac{b^{\prime\prime}(y)\Psi(y)}{b(y)-\lambda_{0}}+i\pi\sum_{b(z)=\lambda_{0}}\frac{b^{\prime\prime}(z)\Psi(z)}{|b^{\prime}(z)|}\delta(y-z)=0,$
(6.42)

which contradicts the main spectral assumption (1.1).

In the case when $\alpha_{0}\neq 0$, we have

$-\Delta_{k_{0}}\Psi(y)+\frac{b^{\prime\prime}(y)\Psi(y)}{b(y)-\lambda_{0}-i\alpha_{0}}=0,$
(6.43)

which would imply the existence of a discrete eigenvalue for the Rayleigh
operator which also contradicts the main spectral Assumption 1.1.

Finally, assume that $(\alpha_{\ell},\lambda_{\ell})\in R_{\alpha}$ and
therefore
$(\alpha_{0},\lambda_{0})\in[1,\infty)\times\overline{(\cup_{j\in\\{1,2\\}}\Sigma_{j,\delta_{0}})}$.
Since $\alpha_{0}\neq 0$, we can use Lemma 6.5 to once again obtain (6.43),
which leads to another contradiction. The proof of Proposition 6.1 is now
complete. ∎

##  7\. Proof of the limiting absorption principle (II): the degenerate region

In this section, we consider the case where
$\lambda\in\cup_{j\in\\{1,2\\}}\Sigma_{j,\delta_{0}}$ and $\alpha$ is small.
More precisely, we define a region $R_{dj}$ in the $(\alpha,\lambda)$ plane as

$\
R_{dj}:=\big{\\{}(\alpha,\lambda)\in\mathbb{R}^{2}\,|\,\alpha\in(-\sigma_{0}\epsilon^{1/2},1),\,\lambda\in\Sigma_{j,\delta_{0}}\big{\\}},\quad
j\in\\{1,2\\},$ (7.1)

where we recall that $\sigma_{0}$ is chosen in Proposition 3.1. In contrast to
the case when $(\alpha,\lambda)\in R_{nd}\cup R_{\alpha}$, $A_{\Theta}^{-1}$
is now “quadratically” singular near the critical layers, $b(y)=\lambda$,
using the heuristic in (1.28). Therefore $A_{\Theta}^{-1}$ is of the same
order as $\Delta_{k}$ and we can no longer treat
$T_{\Theta}=\Delta_{k}^{-1}A_{\Theta}^{-1}$ perturbatively. To address this
issue, we need to incorporate part of the nonlocal term in the main terms and
prove various weighted estimates, before we can carry out an analogous
compactness argument as in the non-degenerate case to prove the limiting
absorption argument.

###  7.1. Bounds on the operator $T_{\Theta}$.

While the definition of $T_{\Theta}$ given in (4.2) is concise, it obscures
the important cancellation between $A_{\Theta}^{-1}(ib^{\prime\prime}\psi)$
and the potential $b^{\prime\prime}(y)(b(y)-\lambda-i\alpha)^{-1}$. Therefore,
we expand $A_{\Theta}^{-1}(ib^{\prime\prime}\psi)$ further which leads to the
following decomposition for $T_{\Theta}$.

######  Lemma 7.1.

Assume that $\epsilon\in(0,1/8)$,
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{1/2},1)\times\Sigma_{j,\delta_{0}},\,j\in\\{1,2\\}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. For $f\in
H^{1}(\mathbb{T}_{\mathfrak{p}})$ we can decompose for
$y\in\mathbb{T}_{\mathfrak{p}}$,

$T_{\Theta}f(y)=-iT_{I1}f(y)-iT_{I2}f(y)-iT_{v1}f(y)-T_{v2}f(y),$ (7.2)

where

$\displaystyle
T_{I1}f(y):=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}^{j}(y,z;\Lambda)A_{\Theta}^{-1}\left[\Big{(}1-\varphi_{0}\Big{(}\frac{\cdot-
y_{j*}}{\delta_{0}}\Big{)}\Big{)}b^{\prime\prime}f\right](z)\,dz,$ (7.3a)
$\displaystyle
T_{I2}f(y):=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}^{j}(y,z;\Lambda)A_{\Theta}^{-1}\Big{(}\varphi_{0}\Big{(}\frac{\cdot-
y_{j*}}{\delta(\Lambda)}\Big{)}b^{\prime\prime}f\Big{)}(z)\,dz,$ (7.3b)
$\displaystyle T_{v1}f(y)$
$\displaystyle:=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}^{j}(y,z;\Lambda)\varphi_{0}\left(\frac{(z-y_{j*})}{\delta(\Lambda)/3}\right)A_{\Theta}^{-1}\left[\Big{(}\varphi_{0}\Big{(}\frac{\cdot-
y_{j*}}{\delta_{0}}\Big{)}-\varphi_{0}\Big{(}\frac{\cdot-
y_{j*}}{\delta(\Lambda)}\Big{)}\Big{)}b^{\prime\prime}f\right](z)\,dz,$ (7.3c)
$\displaystyle T_{v2}f(y)$
$\displaystyle:=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}^{j}(y,z;\Lambda)\left(1-\varphi_{0}\left(\frac{(z-y_{j*})}{\delta(\Lambda)/3}\right)\right)\frac{\epsilon\partial_{z}^{2}A_{\Theta}^{-1}\left[\left(\varphi_{0}\left(\frac{\cdot-
y_{j*}}{\delta_{0}}\right)-\varphi_{0}\left(\frac{\cdot-
y_{j*}}{\delta(\Lambda)}\right)\right)b^{\prime\prime}f\right](z)}{b(z)-\lambda-i\alpha}\,dz.$
(7.3d)

where $\varphi_{0}$ is the same as in the modified Green’s function, see
(2.7).

######  Remark 7.1.

The motivation for the subscripts $I$ and $v$ are “inviscid” and “viscous”
respectively. Both $T_{I1}$ and $T_{I2}$ have analogs that appear in the
inviscid case (see [38]) but $T_{v1}$ and $T_{v2}$ only appear when viscosity
is present.

###### Proof.

For $y\in\mathbb{T}_{\mathfrak{p}}$ we write

$\displaystyle
A_{\Theta}^{-1}(ib^{\prime\prime}f)=A_{\Theta}^{-1}\big{[}(1-\eta_{j,\delta})ib^{\prime\prime}g\big{]}+A_{\Theta}^{-1}\big{[}\eta_{j,\delta}ib^{\prime\prime}f\big{]},$
(7.4)

where for $y\in\mathbb{T}_{\mathfrak{p}}$,

$\eta_{j,\delta}(y):=\varphi\Big{(}\frac{y-y_{j\ast}}{\delta_{0}}\Big{)}-\varphi\Big{(}\frac{y-y_{j\ast}}{\delta(\Lambda)}\Big{)}.$
(7.5)

It follows from (3.33) that for $y$ away from the critical layer we have

$A_{\Theta}^{-1}(\eta_{j,\delta}ib^{\prime\prime}f)(y)=-\frac{\eta_{j,\delta}(y)b^{\prime\prime}(y)f(y)}{b(y)-\lambda-i\alpha}+\frac{\epsilon\partial_{y}^{2}A_{\Theta}^{-1}(\eta_{j,\delta}b^{\prime\prime}f)(y)}{b(y)-\lambda-i\alpha},$
(7.6)

therefore,

$\displaystyle\Big{(}1-\varphi\Big{(}\frac{y-y_{j*}}{\delta(\Lambda)/3}\Big{)}\Big{)}\left[A_{\Theta}^{-1}\big{(}\eta_{j,\delta}ib^{\prime\prime}f\big{)}(z)+\frac{b^{\prime\prime}(y)}{b(y)-\lambda-i\alpha}\Big{[}\varphi\Big{(}\frac{z-y_{j\ast}}{\delta_{0}}\Big{)}-\varphi\Big{(}\frac{z-y_{j\ast}}{\delta(\Lambda)}\Big{)}\Big{]}f(z)\right]$
$\displaystyle=\left(1-\varphi\Big{(}\frac{y-y_{j*}}{\delta(\Lambda)/3}\Big{)}\right)\frac{\epsilon\partial_{y}^{2}A_{\Theta}^{-1}(\eta_{j,\delta}b^{\prime\prime}f)(y)}{b(y)-\lambda-i\alpha},$
(7.7)

and

$\displaystyle\varphi\Big{(}\frac{y-y_{j*}}{\delta(\Lambda)/3}\Big{)}\left[A_{\Theta}^{-1}\big{(}\eta_{j,\delta}ib^{\prime\prime}f\big{)}(y)+\frac{b^{\prime\prime}(y)}{b(y)-\lambda-i\alpha}\Big{[}\varphi\left(\frac{y-y_{j\ast}}{\delta_{0}}\right)-\varphi\left(\frac{y-y_{j\ast}}{\delta(\Lambda)}\right)\Big{]}f(y)\right]$
$\displaystyle=\varphi\Big{(}\frac{y-y_{j*}}{\delta(\Lambda)/3}\Big{)}A_{\Theta}^{-1}\big{(}\eta_{j,\delta}ib^{\prime\prime}f\big{)}(y).$
(7.8)

Finally, we can write

$A_{\Theta}^{-1}\big{[}(1-\eta_{j,\delta})ib^{\prime\prime}f\big{]}=A_{\Theta}^{-1}\left[1-\varphi\Big{(}\frac{\cdot-
y_{j*}}{\delta_{0}}\Big{)}ib^{\prime\prime}f\right]+A_{\Theta}^{-1}\left[\varphi\Big{(}\frac{\cdot-
y_{j*}}{\delta(\Lambda)}\Big{)}ib^{\prime\prime}f\right].$ (7.9)

Combining (7.1), (7.1), and (7.9) concludes the proof. ∎

In order to prove the limiting absorption principle for $(\alpha,\lambda)\in
R_{dj}$ we will need to prove bounds for $T_{\Theta}$ on
$X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ for $(\sigma_{1},\sigma_{2})$ given
by (4.3) as well as their properties in higher regularity spaces. We summarize
all the required results in the following propositions.

######  Proposition 7.1.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{1/2},1)\times\Sigma_{j,\delta_{0}},\,j\in\\{1,2\\}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ and
$\sigma_{1}\in[0,1]$, $\sigma_{2}\in[-2,1/2]$ satisfying
$\sigma_{1}+\sigma_{2}\in(-2,1)$ we have the following bounds on $T_{\Theta}$
for $f\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$,

$\displaystyle\norm{T_{\Theta}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim_{\sigma_{1},\sigma_{2}}\frac{\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\langle
k\delta(\Lambda)\rangle^{\frac{1}{4}}},$ (7.10)

and for $k\delta\lesssim 1$ and $|y-y_{j\ast}|\gtrsim\delta_{0}$,

$\big{|}T_{\Theta}f(y)\big{|}\lesssim_{\sigma_{1},\sigma_{2}}\varrho_{j}^{-\sigma_{1}-1/4}\varrho_{j,k}^{-\sigma_{2}+1/4}(y;\Lambda)\,\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(7.11)

Furthermore,

$\displaystyle\norm{T_{v1}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}+\norm{T_{v2}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim_{\sigma_{1},\sigma_{2}}\frac{\epsilon^{1/2}}{\delta(\Lambda)^{2}}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(7.12)

In particular, in the definition (1.38), $C^{\dagger}$ is chosen so that we
have the bound

$\displaystyle\norm{T_{v1}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}+\norm{T_{v2}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\leq\frac{1}{2}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(7.13)

In order to prove the requisite compactness needed for the proof of the
limiting absorption principle, we will need a more detailed understanding of
the regularity of $T_{\Theta}$ than those implied by Proposition 7.1. The type
of regularity we can prove for $T_{\Theta}$ will depend on whether $\lambda$
is in the “intermediate regime”:
$\epsilon^{\frac{1}{2}}\ll|b(y_{j*})-\lambda|\ll 1$, or the “viscous regime”:
$|\lambda-b(y_{j*})|\lesssim\epsilon^{\frac{1}{2}}$.

We firstly consider the intermediate region.

######  Proposition 7.2.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{1/2},1)\times\Sigma_{j,\delta_{0}},\,j\in\\{1,2\\}$,
$k\in\mathbb{Z}\cap[1,\infty)$, and $k\delta\leq N$, for some $N\geq 1$. Then
for all $\epsilon\in(0,1/8)$, $\epsilon^{\frac{1}{2}}\ll|\lambda-b(y_{j*})|\ll
1$, $\sigma_{1}\in[0,1]$, $\sigma_{2}\in[-2,1/2]$ satisfying
$\sigma_{1}+\sigma_{2}\in(-2,1)$ we have the following bounds for
$y\in\mathbb{T}_{\mathfrak{p}}$.

  * •

(bounds on $T_{I1}$)

$\displaystyle\big{|}\varrho_{j,k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}f(y)\big{|}\lesssim_{N}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\varrho_{j,k}^{2}(y;\Lambda)k^{\sigma_{2}},\,\beta\in\\{0,1,2\\};$
(7.14)

  * •

(bounds on $T_{I2}$) We can write $\partial_{y}T_{I2}f$ as

$\displaystyle\partial_{y}T_{I2}f$ $\displaystyle=\Upsilon_{I2}+R_{I2},$
(7.15)

where $\Upsilon_{I2}:\mathbb{T}_{\mathfrak{p}}\to\mathbb{C}$ is compactly
supported in $\\{y:\,|y-y_{j\ast}|\leq\delta(\Lambda)\\}$,
$R_{I2}:\mathbb{T}_{\mathfrak{p}}\to\mathbb{C}$, and satisfy

$\displaystyle\sup_{\xi\in\mathbb{R}}\Big{[}(|\xi|+1)|\mathcal{F}(\Upsilon_{I2})(\xi)|\Big{]}\lesssim_{N}\frac{\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{1+\sigma_{1}+\sigma_{2}}},$
(7.16)
$\displaystyle\big{|}T_{I2}f(y)\big{|}+\varrho_{j,k}(y;\Lambda)\big{|}R_{I2}(y)\big{|}+\varrho_{j,k}^{2}(y;\Lambda)\big{|}\partial_{y}R_{I2}(y)+d_{j,k}^{-1}\Upsilon_{I2}\big{|}$
$\displaystyle\lesssim_{N}\frac{\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}}\frac{\delta(\Lambda)}{\varrho_{j}(y;\Lambda)};$
(7.17)

  * •

(bounds on $T_{v1}$)

$\displaystyle\varrho_{j,k}^{\beta}(y;\Lambda)|\partial_{y}^{\beta}T_{v1}f(y)|\lesssim_{N}\frac{\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}})\delta(\Lambda)\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho_{j}(y;\Lambda)\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}},\quad\beta\in\\{0,1,2\\},$
(7.18)

where $c_{1}>0$ is sufficiently small (and more precisely is given by Lemma
8.1 below);

  * •

(bounds on $T_{v2}$)

$\displaystyle|\varrho_{j,k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{v2}f(y)|$
$\displaystyle\lesssim_{N,\sigma_{1},\sigma_{2}}\frac{\epsilon\delta(\Lambda)\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{4}\varrho_{j}(y;\Lambda)\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}},\quad\beta\in\\{0,1\\}.$
(7.19)

We now turn to the viscous region.

######  Proposition 7.3.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{1/2},1)\times\Sigma_{j,\delta_{0}},\,j\in\\{1,2\\}$,
$k\in\mathbb{Z}\cap[1,\infty)$, and $k\delta\leq N$. Then for all
$\epsilon\in(0,1/8)$, $|\lambda-b(y_{j*})|\lesssim\epsilon^{\frac{1}{2}}$, and
$f\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ satisfying
$\sigma_{1}\in[0,1]$, $\sigma_{1}+\sigma_{2}\in(-2,1)$ the following bounds
hold for $y\in\mathbb{T}_{\mathfrak{p}}$,

  * •

(bounds on $T_{I1}$)

$\displaystyle\big{|}\varrho_{j,k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}f(y)\big{|}\lesssim_{N}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\varrho_{j,k}^{2}(y;\Lambda)k^{\sigma_{2}},\,\beta\in\\{0,1,2\\};$
(7.20)

  * •

(bounds on $T_{I2}$)

$\displaystyle\big{|}\varrho_{j,k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I2}f(y)\big{|}+\big{|}\varrho_{j,k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{v1}f(y)\big{|}$
$\displaystyle\lesssim_{N,\sigma_{1},\sigma_{2}}\frac{\norm{f}_{X^{\sigma_{1},\sigma_{2}}}}{\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}}\frac{\delta(\Lambda)}{\varrho_{j}(y;\Lambda)},\,\beta\in\\{0,1,2\\};$
(7.21)

  * •

(bounds on $T_{v1}$ and $T_{v2}$)

$\displaystyle\big{|}\varrho_{j,k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{v2}f(y)\big{|}$
$\displaystyle\lesssim_{N,\sigma_{1},\sigma_{2}}\frac{\epsilon\delta(\Lambda)\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{4}\varrho_{j}(y;\Lambda)\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}},\quad\beta\in\\{0,1\\},$
(7.22)
$\displaystyle\big{|}\varrho_{j,k}^{2}(y;\Lambda)\partial_{y}^{2}T_{v2}f(y)\big{|}$
$\displaystyle\lesssim_{N,\sigma_{1},\sigma_{2}}\frac{\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho_{j}^{\sigma_{1}}(y;\Lambda)\varrho_{j,k}^{\sigma_{2}}(y;\Lambda)}.$
(7.23)

######  Remark 7.2.

The implicit constants appearing in Propositions 7.1, 7.2 and 7.3 may grow
with $N$. However, in the application of these propositions in the proof of
the limiting absorption principle, we will only need to consider the case
where $N$ is fixed. We will only prove Propositions 7.2 and 7.3 under the
assumption that $N=1$, but an inspection of the proofs shows the general case
follows with perhaps a larger constant.

######  Remark 7.3.

The role of higher regularity is to capture the “uniform compactness” of
$T_{\Theta}$ (in the sense given in Lemma 6.4). Some care is needed here as
the function space $X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ itself has non-
trivial $\lambda$ dependence. The essential feature is that, upon rescaling by
$\delta(\Lambda)$, the higher order regularity estimates encode compactness on
the fixed space $H_{loc}^{1}(\mathbb{R})$. This is made precise in section
7.2.2.

The proofs of Propositions 7.1, 7.2, and 7.3 are very important but are
somewhat technical. The full proofs will be given in Section 8.

###  7.2. The compactness argument

Using Propositions 7.1, 7.2, and 7.3 we are now ready to prove the limiting
absorption in the degenerate case. For the ease of notations, we assume that
$y_{j\ast}=0$ and also drop the dependence of various quantities on
$j\in\\{1,2\\}$ which is fixed, when there is no possibility of confusion.

We firstly consider the intermediate region where the spectral parameter is
$|\nu/k|^{1/2}$ away from the critical value.

######  Proposition 7.4.

Fix $\gamma\in[7/4,2)$. Assume that

$(\sigma_{1},\sigma_{2})\in\\{(0,-\gamma),(1,-\gamma+1),(1-2(2-\gamma),-\gamma+2)\\},$
(7.24)

and let $\sigma_{0}$ be as in Proposition 3.1. There exists
$\epsilon_{0,I}\in(0,1/8),\kappa_{I}>0,M>0$, depending on $\gamma$, such that
for all $\epsilon\in(0,\epsilon_{0,I})$, $k\in\mathbb{Z}\cap[1,\infty)$,
$\alpha\in[-\sigma_{0}\epsilon^{1/2},1]$ , and
$\lambda\in\Sigma_{j,\delta_{0}}\setminus[-M\epsilon^{1/2},M\epsilon^{1/2}]$
with $j\in\\{1,2\\}$, we have

$\norm{f+T_{\Theta}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\geq\kappa_{I}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}$
(7.25)

for all $f\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$.

We now turn to the viscous region where $|\lambda|\lesssim|\nu/k|^{1/2}$.

######  Proposition 7.5.

Let $\gamma,(\sigma_{1},\sigma_{2}),\text{ and }M$ be as in Proposition 7.4.
There exist $\epsilon_{0,v}\in(0,1/8)$, $\kappa_{v}>0$, and
$\sigma_{0}(M)\in(0,1)$ such that for all $\epsilon\in(0,\epsilon_{0,v})$,
$k\in\mathbb{Z}\cap[1,\infty)$, $\alpha\in[-\sigma_{0}(M)\epsilon^{1/2},1]$,
and $\lambda\in[-M\epsilon^{1/2},M\epsilon^{1/2}]$ we have

$\norm{f+T_{\Theta}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\geq\kappa_{v}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}$
(7.26)

for all $f\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$.

######  Remark 7.4.

In order to prove Proposition 7.5 we will need to choose $\sigma_{0}(M)$ small
based on $M$ (see (7.65)). However, the proof of Proposition 7.4 only requires
smallness of $\sigma_{0}$ in a way such that Proposition 3.1 holds.

If we set $\kappa:=\min(\kappa_{I},\kappa_{v})$,
$\epsilon_{0}=\min(\epsilon_{0,I},\epsilon_{0,v})$, and
$\sigma_{\sharp}:=\min(\sigma_{0},\sigma_{0}(M))$ then Propositions 7.4 and
7.5 immediately imply the following result.

######  Proposition 7.6.

Fix $\gamma\in[7/4,2)$. Assume that

$(\sigma_{1},\sigma_{2})\in\\{(0,-\gamma),(1,-\gamma+1),(1-2(2-\gamma)),-\gamma+2)\\}.$
(7.27)

Then there exists $\epsilon_{0},\kappa,\sigma_{\sharp}>0$, depending on
$\gamma$, such that for all $\epsilon\in(0,\epsilon_{0,})$,
$k\in\mathbb{Z}\cap[1,\infty)$, $\alpha\in[-\sigma_{\sharp}\epsilon^{1/2},1]$,
and $\lambda\in\Sigma_{j,\delta_{0}}$ with $j\in\\{1,2\\}$, we have

$\norm{f+T_{\Theta}f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\geq\kappa\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}$
(7.28)

for all $f\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$.

In the rest of the section we give the proof of Proposition 7.4 and
Proposition 7.5, beginning with the proof of Proposition 7.4.

###### Proof of Proposition 7.4.

Suppose (7.25) does not hold. Then there exist

$(f_{\ell},\kappa_{I,\ell},\epsilon_{\ell},\lambda_{\ell},k_{\ell},\alpha_{\ell},M_{\ell}),\quad{\rm
with}\,\,(\kappa_{I,\ell},\epsilon_{\ell},M_{\ell}^{-1})\to
0+,\lambda_{\ell}\in\Sigma_{j,\delta_{0}}\setminus\big{[}-M_{\ell}\epsilon_{\ell}^{1/2},M_{\ell}\epsilon_{\ell}^{1/2}\big{]},$

and $\norm{f_{\ell}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}=1$, such that

$\norm{f_{\ell}+T_{\Theta_{\ell}}f_{\ell}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\leq\kappa_{I,\ell}.$
(7.29)

In (7.29) we have introduced $\Theta_{\ell}$ to represent dependence on
$\epsilon_{\ell},\lambda_{\ell},k_{\ell},\alpha_{\ell}$.

We claim that we may assume
$(\alpha_{\ell},M_{\ell}\epsilon_{\ell}^{1/2},\lambda_{\ell})\to 0$. To see
this, we first note that

$\liminf_{\ell\to\infty}{M_{\ell}\epsilon_{\ell}^{1/2}}<\infty,$

otherwise there is nothing to prove by our assumption on the range of the
spectral parameter $\lambda_{\ell}$. We may therefore assume, passing to
subsequences if necessary, that
$(\alpha_{\ell},M_{\ell}\epsilon_{\ell}^{1/2},\lambda_{\ell})$ converges. If
$\lim_{\ell\to\infty}\alpha_{\ell}$ or $\lim_{\ell\to\infty}\lambda_{\ell}$
are not zero, then $\lim_{\ell\to\infty}\delta(\Lambda_{\ell})>0$. In this
case we may follow similar arguments as in the proof Proposition 6.1, using
the higher order regularity proved for $T_{I1}$, $T_{I2}$ in Propositions 7.2
and 7.3, and the bounds proved for $T_{v1}$ and $T_{v2}$ on
$X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ in Proposition 7.1, which ensure the
needed convergence for $T_{\Theta_{\ell}}f_{\ell}$.

We can therefore assume that
$(\alpha_{\ell},\epsilon_{\ell},M_{\ell}\epsilon_{\ell}^{1/2},\lambda_{\ell})$
converges to 0, which implies that $\delta(\Lambda_{\ell})$ converges to 0.
Proposition 7.1 also implies that there exists a subsequence such that

$k_{\ell}\delta(\Lambda_{\ell})\lesssim 1.$

There are two main cases to consider: in the first case, a uniform-in-$\ell$
amount of $f_{\ell}$ stays $\delta_{0}$ distance away from the critical point
(the “Non-concentrated” case); in the second case, the sequence $f_{\ell}$
concentrates near the critical point (the “Concentrated” case). More
precisely, we distinguish two scenarios:

  1. (1)

The Non-Concentrated case

$\limsup_{\ell\to\infty}\big{\|}\varrho^{-\sigma_{1}}\varrho_{k_{\ell}}^{-\sigma_{2}}(\cdot;\Lambda)f_{\ell}\big{\|}_{L^{\infty}(\mathbb{T}_{\mathfrak{p}}\backslash
S_{\frac{\delta_{0}}{2}})}>0,$ (7.30)

  2. (2)

Concentrated case

$\limsup_{\ell\to\infty}\big{\|}\varrho^{-\sigma_{1}}\varrho_{k_{\ell}}^{-\sigma_{2}}(\cdot;\Lambda)f_{\ell}\big{\|}_{L^{\infty}(\mathbb{T}_{\mathfrak{p}}\backslash
S_{\frac{\delta_{0}}{2}})}=0.$ (7.31)

####  7.2.1. The non-concentrated case

It follows from (7.11), (7.29) and (7.30) that $|k_{\ell}|\lesssim 1$. We
introduce the notation

$\displaystyle T_{I1\ell}f(y)$
$\displaystyle:=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k_{\ell}}(y,z;\Lambda_{\ell})A_{\Theta_{\ell}}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}f)\,dz,$
(7.32a) $\displaystyle T_{I2\ell}f(y)$
$\displaystyle:=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k_{\ell}}(y,z;\Lambda_{\ell})A_{\Theta_{\ell}}^{-1}(\varphi_{\delta(\Lambda_{\ell})}b^{\prime\prime}f)\,dz,$
(7.32b) $\displaystyle T_{v1\ell}f(y)$
$\displaystyle:=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k_{\ell}}(y,z;\Lambda_{\ell})\varphi_{\frac{\delta(\Lambda_{\ell})}{{3}}}(z)A_{\Theta_{\ell}}^{-1}(\eta_{\delta(\Lambda_{\ell})}b^{\prime\prime}f)\,dz,$
(7.32c) $\displaystyle T_{v2\ell}f(y)$
$\displaystyle:=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k_{\ell}}(y,z;\Lambda_{\ell})(1-\varphi_{\frac{\delta(\Lambda_{\ell})}{{3}}}(z))\frac{\epsilon\partial_{z}^{2}A_{\Theta_{\ell}}^{-1}(\eta_{\delta(\Lambda_{\ell})}b^{\prime\prime}f)}{b(z)-\lambda_{\ell}-i\alpha_{\ell}}\,dz.$
(7.32d)

By Propositions 7.2 and 7.3, it follows that for all
$\lambda\in\Sigma_{j,\delta_{0}}$ with $k_{\ell}\delta(\Lambda_{\ell})\lesssim
1$ and $y\in\mathbb{T}_{\mathfrak{p}}$, we have

$\displaystyle|T_{I2}f(y)|+|T_{v1}f(y)|+|T_{v2}f(y)|$
$\displaystyle\lesssim\frac{\delta(\Lambda_{\ell})\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho(y;\Lambda_{\ell})\delta(\Lambda_{\ell})^{\sigma_{1}+\sigma_{2}}},$
(7.33)

where we used the fact that
$\epsilon_{\ell}\lesssim\delta^{4}(\Lambda_{\ell})$.

The pointwise inequality (7.33) directly implies that

$\lim_{\ell\to\infty}\left(\norm{T_{I2\ell}f_{\ell}}_{L^{p}(\mathbb{T}_{\mathfrak{p}})}+\norm{T_{v1\ell}f_{\ell}}_{L^{p}(\mathbb{T}_{\mathfrak{p}})}+\norm{T_{v2\ell}f_{\ell}}_{L^{p}(\mathbb{T}_{\mathfrak{p}})}\right)=0,$
(7.34)

for $p>1$ with $(\gamma-1)p<1$.

We now show that we can extract a convergent sequence whose limit satisfies
Rayleigh’s equation. By weak compactness of $L^{p}$, we can find $f_{*}\in
L^{p}(\mathbb{T}_{\mathfrak{p}})$ such that $f_{\ell}\rightharpoonup f_{*}$.
This weak convergence can be upgraded to strong convergence using (7.29),
(7.34), and the higher regularity of $T_{I1\ell}$. Passing $\ell\to\infty$, we
obtain that

$f_{*}+\lim_{\ell\to\infty}T_{I1\ell}f_{*}=0\quad\text{in
}L^{p}(\mathbb{T}_{\mathfrak{p}}).$ (7.35)

Using Lemma 4.1, we obtain that $f_{*}$ is bounded and moreover, for
$y\in\mathbb{T}_{\mathfrak{p}}$

$|f_{*}(y)|=|\lim_{\ell\to\infty}T_{I1\ell}f_{*}(y)|\lesssim y^{2}.$ (7.36)

To compute $\lim_{\ell\to\infty}T_{I1\ell}f_{*}$, we observe that
$\mathcal{G}_{\ell}$ and
$A_{\ell}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}f_{*})$ have convergent
subsequences in $L^{\infty}(\mathbb{T}_{\mathfrak{p}})$ from Arzela-Ascoli,
Lemma 2.1, and Lemma 6.5. Therefore

$\lim_{\ell\to\infty}T_{I1\ell}f_{*}=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{*}(y,z)A_{*}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}f_{*})\,dz$
(7.37)

where we have defined
$\mathcal{G}_{*}:=\lim_{\ell\to\infty}\mathcal{G}_{\ell}$ and
$A_{*}^{-1}:=\lim_{\ell\to\infty}A_{\ell}^{-1}$. By the proof of Lemma 6.5 we
have

$A_{*}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}f_{*})=\frac{(1-\varphi_{\delta_{0}}(y))b^{\prime\prime}(y)f_{*}(y)}{b(y)}$
(7.38)

Additionally, for $z\neq 0$,

$\mathcal{L}_{*}\mathcal{G}_{*}(y,z):=\left(-\partial_{y}^{2}+k_{*}^{2}+\frac{b^{\prime\prime}(y)}{b(y)}\varphi_{\delta_{0}}(y)\right)\mathcal{G}_{*}(y,z)=\delta(y-z)$
(7.39)

in the sense of distributions. Applying $\mathcal{L}_{*}$ to (7.35) gives

$\left(-\partial_{y}^{2}+k_{*}^{2}+\frac{b^{\prime\prime}(y)}{b(y)}\right)f_{*}(y)=0,$
(7.40)

for $y\in\mathbb{T}_{\mathfrak{p}}$, in the sense of distributions. This
contradicts our main spectral assumption 1.1.

####  7.2.2. The concentrated case

For the remainder of the proof we adopt the shorthand
$\delta_{\ell}=\delta(\Lambda_{\ell})$ with a slight abuse of notation when
$\ell\in\\{0,1,2\\}$. Here our main concern is when $\ell\to\infty$ so there
should be no possibility of confusion.

By (7.13), (7.29), (7.31) and Lemma 4.1, we can find an $N$ such that for
$\ell\geq N$ we have

$\norm{f_{\ell}+T_{\Theta_{\ell}}f_{\ell}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}+\norm{(T_{v1\ell}+T_{v2\ell})f_{\ell}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}+\norm{T_{I1\ell}f_{\ell}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\leq\frac{2}{3},$
(7.41)

which implies that

$\norm{T_{I2\ell}f_{\ell}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\geq\frac{1}{3}.$
(7.42)

Since $T_{I2\ell}f_{\ell}=T_{I2\ell}(\varphi_{2\delta_{\ell}}f_{\ell})$ and
$T_{I2\ell}$ are uniformly bounded on
$X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ by Proposition 7.1, it follows from
(7.42) that there exists $\theta>0$ which is independent of $\ell$, such that

$\displaystyle\delta_{\ell}^{\sigma_{1}+\sigma_{2}}\Big{[}\delta_{\ell}^{-\frac{1}{2}}\Big{(}\norm{f_{\ell}}_{L^{2}(S_{\delta_{\ell}})}+\norm{\delta_{\ell}\partial_{y}f_{\ell}}_{L^{2}(S_{\delta_{\ell}})}\Big{)}+\norm{f_{\ell}}_{L^{\infty}(S_{4\delta_{\ell}}\setminus
S_{\delta_{\ell}})}+\norm{\delta_{\ell}\partial_{y}f_{\ell}}_{L^{\infty}(S_{4\delta_{\ell}}\setminus
S_{\delta_{\ell}})}\Big{]}\geq\theta.$ (7.43)

In view of the fact that a non-vanishing amount of $f_{\ell}$ concentrates
near the critical point, we use a blow-up argument. To this end, define the
re-scaled quantities

$\displaystyle(Y,Z)$
$\displaystyle:=\left(\frac{y}{\delta_{\ell}},\frac{z}{\delta_{\ell}}\right)\quad(y,z)\in\mathbb{T}_{\mathfrak{p}}^{2},$
(7.44a) $\displaystyle B_{\ell}(Y)$ $\displaystyle:=\delta_{\ell}^{-2}b(y),$
(7.44b) $\displaystyle F_{\ell}(Y)$
$\displaystyle:=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}f_{\ell}(y),$ (7.44c)
$\displaystyle G_{\ell}(Y,Z)$
$\displaystyle:=\delta_{\ell}^{-1}\mathcal{G}_{k_{\ell}}(y,z;\Lambda_{\ell}),$
(7.44d) $\displaystyle\widetilde{A}_{\ell}^{-1}(Q_{\ell})(Y)$
$\displaystyle:=\delta_{\ell}^{2+\sigma_{1}+\sigma_{2}}A_{\Theta_{\ell}}^{-1}(q)(y),\quad\text{
where }Q_{\ell}(Y):=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}q(y).$ (7.44e)

It will be convenient to extend the rescaled functions so that they are
defined on $\mathbb{R}$. For the $Y$ values not in the range of
$\frac{y}{\delta_{\ell}},y\in[-\mathfrak{p}/2,\mathfrak{p}/2]$, we simply
extend all of the functions as zero. In this way, we can use all of the
previously obtained bounds to control the sequence.

It follows from rescaling that $G_{\ell}$ satisfies for
$|Y|<\mathfrak{p}/(2\delta_{\ell})$ and $|Z|<\mathfrak{p}/(2\delta_{\ell})$,

$-\partial_{Y}^{2}G_{\ell}(Y,Z)+(k_{\ell}\delta_{\ell})^{2}G_{\ell}(Y,Z)+\frac{B_{\ell}^{\prime\prime}(Y)(\varphi(\delta_{0}^{-1}\delta_{\ell}Y)-\varphi(Y))}{B_{\ell}(Y)-(\lambda_{\ell}+i\alpha_{\ell})\delta_{\ell}^{-2}}G_{\ell}(Y,Z)=\delta(Y-Z),$
(7.45)

and that $\widetilde{A}_{\ell}^{-1}(Q_{\ell})(Y)$ satisfies for
$|Y|<\mathfrak{p}/(2\delta_{\ell})$ and $|Z|<\mathfrak{p}/(2\delta_{\ell})$,

$\frac{\epsilon_{\ell}}{\delta_{\ell}^{4}}\partial_{Y}^{2}\widetilde{A}_{\ell}^{-1}(Q_{\ell})(Y)-i\left(B_{\ell}(Y)-\frac{\lambda_{\ell}-i\alpha_{\ell}}{\delta_{\ell}^{2}}\right)\widetilde{A}_{\ell}^{-1}(Q_{\ell})(Y)=Q_{\ell}(Y).$
(7.46)

Crucially, (7.43) and (7.44) give uniform-in-$\ell$ lower bounds on $F_{\ell}$
in $H^{1}[-1,1]$ intersection with

$W^{1,\infty}\big{(}[-4,4]\setminus[-1,1]\big{)}.$

Finally, we define, with (7.44a),

$\displaystyle K_{\ell}(Y)$
$\displaystyle:=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}T_{\Theta_{\ell}}f_{\ell}(y),$
(7.47a) $\displaystyle K_{1\ell}(Y)$
$\displaystyle:=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}T_{I1\ell}f_{\ell}(y),$
(7.47b) $\displaystyle K_{2\ell}(Y)$
$\displaystyle:=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}T_{I2\ell}f_{\ell}(y),$
(7.47c) $\displaystyle K_{3\ell}(Y)$
$\displaystyle:=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}T_{v1\ell}f_{\ell}(y),$
(7.47d) $\displaystyle K_{4\ell}(Y)$
$\displaystyle:=\delta_{\ell}^{\sigma_{1}+\sigma_{2}}T_{v2\ell}f_{\ell}(y).$
(7.47e)

Using (7.44) and (7.32), we obtain that

$\displaystyle K_{1\ell}(Y)$
$\displaystyle=\int_{\mathbb{R}}G_{\ell}(Y,Z)\widetilde{A}_{\ell}^{-1}\left(\left(1-\varphi_{\frac{\delta_{0}}{\delta_{\ell}}}\right)B_{\ell}^{\prime\prime}F_{\ell}\right)(Z)\,dZ,$
(7.48a) $\displaystyle K_{2\ell}(Y)$
$\displaystyle=\int_{\mathbb{R}}G_{\ell}(Y,Z)\widetilde{A}_{\ell}^{-1}\left(\varphi
B_{\ell}^{\prime\prime}F_{\ell}\right)(Z)\,dZ,$ (7.48b) $\displaystyle
K_{3\ell}(Y)$
$\displaystyle=\int_{\mathbb{R}}G_{\ell}(Y,Z)\varphi\left(\frac{Z}{3}\right)\widetilde{A}_{\ell}^{-1}\left(\Phi(\delta_{\ell}\cdot)B_{\ell}^{\prime\prime}F_{\ell}\right)(Z)\,dZ,$
(7.48c) $\displaystyle K_{4\ell}(Y)$
$\displaystyle=\int_{\mathbb{R}}G_{\ell}(Y,Z)\left(1-\varphi\left(\frac{Z}{3}\right)\right)\frac{\epsilon_{\ell}}{\delta_{\ell}^{4}}\frac{\partial_{Z}^{2}\widetilde{A}_{\ell}^{-1}(\eta_{\delta}(\delta_{\ell}\cdot)B_{\ell}^{\prime\prime}F_{\ell})(Z)}{B_{\ell}(Z)-\lambda_{\ell}\delta_{\ell}^{-2}-i\alpha_{\ell}\delta_{\ell}^{-2}}\,dZ.$
(7.48d)

Propositions 7.2 and 7.3 imply for $\beta\in\\{0,1\\}$ and
$1\leq|Y|\ll\delta_{\ell}^{-1}$ that

$\displaystyle\min\\{1+|Y|,(k_{\ell}\delta_{\ell})^{-1}\\}^{\beta}|\partial_{Y}^{\beta}K_{1\ell}(Y)|$
$\displaystyle\lesssim\delta_{\ell}^{2-|\sigma_{2}|}(1+|Y|^{2}),$ (7.49a)
$\displaystyle\min\\{1+|Y|,(k_{\ell}\delta_{\ell})^{-1}\\}^{\beta}|\partial_{Y}^{\beta}K_{2\ell}(Y)|$
$\displaystyle\lesssim\frac{1}{1+|Y|},$ (7.49b)
$\displaystyle\min\\{1+|Y|,(k_{\ell}\delta_{\ell})^{-1}\\}^{\beta}|\partial_{Y}^{\beta}K_{3\ell}(Y)|$
$\displaystyle\lesssim\frac{\exp\left(-c_{0}\delta_{\ell}^{2}\epsilon_{\ell}^{-\frac{1}{2}}\right)}{1+|Y|},$
(7.49c)
$\displaystyle\min\\{1+|Y|,(k_{\ell}\delta_{\ell})^{-1}\\}^{\beta}|\partial_{Y}^{\beta}K_{4\ell}(Y)|$
$\displaystyle\lesssim\frac{\epsilon_{\ell}}{\delta_{\ell}^{4}(1+|Y|)}.$
(7.49d)

From the higher order regularity estimates proved in Propositions 7.2 and 7.3
it follows that $K_{\ell}$ is relatively compact in $H^{1}([-1,1])$ and
$W_{{loc}}^{1,\infty}(\mathbb{R}\setminus[-1,1])$. We conclude that $K_{\ell}$
has a convergent subsequence in $H_{\rm loc}^{1}(\mathbb{R})$, the limit of
which we call $K_{*}$. Combined with (7.29) and (7.44), this implies that
there exists a subsequence such that $F_{\ell}$ converges to $F_{*}$ in
$H_{\rm loc}^{1}(\mathbb{R})$. From the convergence of $F_{\ell},K_{\ell}$ and
(7.29) we have that

$F_{*}+K_{*}=0.$ (7.50)

From the spatial decay of $K_{\ell}$ in (7.49) we can deduce that $F_{*}$ is
in $H^{1}(\mathbb{R})$. To further understand the limiting solution $F_{*}$,
we observe that

$\lim_{\ell\to\infty}\frac{\epsilon_{\ell}}{\delta_{\ell}^{4}}\lesssim\lim_{\ell\to\infty}\frac{1}{M_{\ell}^{2}}=0.$
(7.51)

Therefore, from (7.49), for $Y,Z\in\mathbb{R}$,

$K_{*}=\lim_{\ell\to\infty}K_{2\ell}=\lim_{\ell\to\infty}\int_{\mathbb{R}}G_{*}(Y,Z)\widetilde{A}_{\ell}^{-1}\left(\varphi
b^{\prime\prime}(0)F_{*}\right)\,dZ,$ (7.52)

where $G_{*}$ satisfies for $Y,Z\in\mathbb{R}$,

$\displaystyle\widetilde{\mathcal{L}_{*}}G_{*}:=-\partial_{Y}^{2}G_{*}(Y,Z)+k_{*}^{2}G_{*}(Y,Z)+\frac{b^{\prime\prime}(0)(1-\varphi(Y))}{\frac{b^{\prime\prime}(0)}{2}Y^{2}-\lambda_{*}-i\alpha_{*}}G_{*}(Y,Z)=\delta(Y-Z),$
(7.53)

with $\lambda_{*}:=\lim_{\ell\to\infty}\lambda_{\ell}\delta_{\ell}^{-2}$,
$\alpha_{*}:=\lim_{\ell\to\infty}\alpha_{\ell}\delta_{\ell}^{-2}$ and
$k_{*}:=\lim_{\ell\to\infty}k_{\ell}\delta_{\ell}$ (passing to subsequences if
necessary to ensure that all sequences converge). Given the definition of
$\delta(\Lambda)$, one of the following must be true:

  1. (1)

$|\alpha_{*}|>0$,

  2. (2)

$|\lambda_{*}|>0$.

By using the re-scaled bounds for the relevant functions, we can follow the
proof of Lemma 6.5, and obtain that $\widetilde{A}_{\ell}^{-1}(Q)(Y)$ has the
following limiting behavior:

  * •

If $\alpha_{*}\neq 0$ or if $\frac{\lambda_{*}}{b^{\prime\prime}(0)}<0$, then
in the sense of distributions for $Y\in\mathbb{R}$,

$\lim_{\delta_{\ell}^{-4}\epsilon_{\ell}\to
0,\,\delta_{\ell}^{-2}\alpha_{\ell}\to\alpha_{*},\,\delta_{\ell}^{-2}\lambda_{\ell}\to\lambda_{*}}\widetilde{A}_{\ell}(Q)(Y)=\frac{-Q(Y)}{\alpha_{*}+i(\frac{b^{\prime\prime}(0)Y^{2}}{2}-\lambda_{*})},$
(7.54)

  * •

If $\alpha_{*}=0$ and $\frac{\lambda_{*}}{b^{\prime\prime}(0)}>0$, then in the
sense of distributions for $Y\in\mathbb{R}$,

$\displaystyle\lim_{\delta_{\ell}^{-4}\epsilon_{\ell}\to
0,\,\delta_{\ell}^{-2}\alpha_{\ell}\to
0,\,\delta_{\ell}^{-2}\lambda_{\ell}\to\lambda_{*}}\widetilde{A}_{\ell}(Q)(Y)=\text{P.V.}\frac{-Q(Y)}{i(\frac{b^{\prime\prime}(0)Y^{2}}{2}-\lambda_{*})}+\pi\sum_{\frac{b^{\prime\prime}(0)c^{2}}{2}=\lambda_{*}}\frac{Q(c)}{|b^{\prime\prime}(0)c|}\delta(Y-c).$
(7.55)

If $\alpha_{*}\neq 0$ or if $\frac{\lambda_{*}}{b^{\prime\prime}(0)}<0$, then
(7.54) implies that for $Y\in\mathbb{R}$,

$\lim_{\ell\to\infty}\int_{\mathbb{R}}G_{*}(Y,Z)\widetilde{A}_{\ell}^{-1}\left(\varphi
b^{\prime\prime}(0)F_{*}\right)\,dZ=\int_{\mathbb{R}}G_{*}(Y,Z)\frac{\varphi(Z)b^{\prime\prime}(0)F_{*}(Z)}{\frac{b^{\prime\prime}(0)Z^{2}}{2}-\lambda_{*}-i\alpha_{*}}\,dZ.$
(7.56)

Applying $\widetilde{\mathcal{L}}_{*}$ to (7.50) we get that

$-\partial_{Y}^{2}F_{*}+k_{*}^{2}F_{*}+\frac{b^{\prime\prime}(0)}{\frac{b^{\prime\prime}(0)}{2}Y^{2}-\lambda_{*}-i\alpha_{*}}F_{*}=0,$
(7.57)

for $Y\in\mathbb{R}$ in the sense of distributions. This contradicts the well-
known fact that the linearized operator associated with the Poiseuille flow
has no discrete eigenvalues.

Now assume that $\alpha_{*}=0$ and
$\frac{\lambda_{*}}{b^{\prime\prime}(0)}>0$. Using (7.55) we obtain that for
$Y\in\mathbb{R}$,

$F_{*}(Y)+\int_{\mathbb{R}}\frac{G_{*}(Y,Z)(1-\varphi(Z))b^{\prime\prime}(0)F(Z)}{\frac{b^{\prime\prime}(0)}{2}Z^{2}-\lambda_{*}}dZ+i\pi\sum_{\frac{b^{\prime\prime}(0)c^{2}}{2}=\lambda_{*}}\frac{G_{*}(Y,c)F_{*}(c)}{|b^{\prime\prime}(0)c|}=0,$
(7.58)

which implies that for $Y\in\mathbb{R}$,

$-\partial_{Y}^{2}F_{*}(Y)+k_{*}^{2}F_{*}(Y)+\text{P.V.}\frac{b^{\prime\prime}(0)}{\frac{b^{\prime\prime}(0)}{2}Y^{2}-\lambda_{*}}F_{*}(Y)+i\pi\sum_{\frac{b^{\prime\prime}(0)c^{2}}{2}=\lambda_{*}}\frac{F_{*}(c)}{|b^{\prime\prime}(0)c|}\delta(Y-c)=0,$
(7.59)

in the sense of distributions. Multiplying $\overline{F_{\ast}}$ with (7.59),
integrating over $\mathbb{R}$ and taking the imaginary part, we see that
$F_{\ast}(c)=0$. This again contradicts with the fact that the linearized
operator associated with the Poiseuille flow has no discrete eigenvalues. The
proof is now complete. ∎

Now that we have quantified the boundary between the intermediate viscous
regime through $M$, we can give the proof of Proposition 7.5.

###### Proof of Proposition 7.5.

The proof proceeds identically to that of Proposition 7.4 except in the
concentrated case in section 7.2.2. For
$\lambda\in[-M\epsilon^{1/2},M\epsilon^{1/2}]$ we no longer have (7.51). Since
$\frac{\epsilon_{\ell}}{\delta_{\ell}^{4}}$ is bounded, it has a convergent
subsequence, the limit of which we denote as $\epsilon_{*}$. If
$\epsilon_{*}=0$, then we argue as in Proposition 7.4. Therefore, we assume
that $\epsilon_{*}>0$. From (7.50), $K_{*}$ is now of the form

$\displaystyle K_{*}$
$\displaystyle=\lim_{\ell\to\infty}(K_{2\ell}+K_{3\ell}+K_{4\ell})$
$\displaystyle=\int_{\mathbb{R}}G_{*}(Y,Z)\widetilde{A}_{*}^{-1}\left(\varphi
b^{\prime\prime}(0)F_{*}\right)dZ+\int_{\mathbb{R}}G_{*}(Y,Z)\varphi\left(\frac{Z}{3}\right)\widetilde{A}_{*}^{-1}\left(b^{\prime\prime}(0)F_{*}\right)dZ$
$\displaystyle+\int_{\mathbb{R}}G_{*}(Y,Z)\left(1-\varphi\left(\frac{Z}{3}\right)\right)\epsilon_{*}\frac{\partial_{Z}^{2}\widetilde{A}_{*}^{-1}(b^{\prime\prime}(0)F_{*})}{\frac{b^{\prime\prime}(0)}{2}Z^{2}-\lambda_{*}-i\alpha_{*}}\,dZ.$
(7.60)

Using the regularity of $K_{*}$, we can bootstrap the regularity of $F_{*}$ to
prove that $F_{*}\in H^{2}(\mathbb{R})$ and satisfies for $Y\in\mathbb{R}$,

$\displaystyle\epsilon_{*}\partial_{Y}^{2}W_{*}-\alpha_{*}W_{*}-i\left(\frac{b^{\prime\prime}(0)Y^{2}}{2}-\lambda_{*}\right)W_{*}+ib^{\prime\prime}(0)F_{*}=0,$
$\displaystyle W_{*}=(\partial_{Y}^{2}-|k_{*}|^{2})F_{*}.$ (7.61)

Our goal is to derive a contradiction by showing the solution of (7.2.2) must
be zero. To this end, we use a direct energy method. Multiplying by
$\overline{W_{*}}$ and integrating over $\mathbb{R}$, we obtain that

$\displaystyle\epsilon_{*}\int_{\mathbb{R}}|\partial_{Y}W_{*}(Y)|^{2}dY+\alpha_{*}\int_{\mathbb{R}}|W_{*}(Y)|^{2}dY=0,$
$\displaystyle\int_{\mathbb{R}}\left(\frac{1}{2}Y^{2}-\frac{\lambda_{*}}{b^{\prime\prime}(0)}\right)|W_{*}(Y)|^{2}dY+\int_{\mathbb{R}}|\partial_{Y}F_{*}(Y)|^{2}+|k_{*}F_{*}(Y)|^{2}dY=0.$
(7.62)

If $\frac{\lambda_{*}}{b^{\prime\prime}(0)}\leq 0$ we have a contradiction so
we assume that it is positive. Similarly, if $\alpha_{*}\geq 0$ we immediately
have a contradiction so we also assume that
$-\sigma_{0}(M)\epsilon_{*}^{1/2}\leq\alpha_{*}<0$. Combining (7.2.2) with the
inequality

$\norm{f}_{L^{2}(\mathbb{R})}^{2}\leq
2\norm{Yf}_{L^{2}(\mathbb{R})}\norm{\partial_{Y}f}_{L^{2}(\mathbb{R})},$
(7.63)

we obtain that

$\norm{W_{*}}_{L^{2}(\mathbb{R})}^{2}\leq
2\left(\frac{\lambda_{*}|\alpha_{*}|}{\epsilon_{*}b^{\prime\prime}(0)}\right)^{\frac{1}{2}}\norm{W_{*}}_{L^{2}(\mathbb{R})}^{2}.$
(7.64)

This is a contradiction because

$\left|\frac{\lambda_{*}|\alpha_{*}|}{\epsilon_{*}b^{\prime\prime}(0)}\right|\leq\frac{M\sigma_{0}(M)}{|b^{\prime\prime}(0)|},$
(7.65)

and we can choose $\sigma_{0}(M)>0$ sufficiently small so that
$M\sigma_{0}(M)<\frac{|b^{\prime\prime}(0)|}{4}$.

∎

##  8\. Proof of Proposition 7.1 \- Proposition 7.3

In the section we provide detailed proofs for Proposition 7.1 \- Proposition
7.3. Recall that $\epsilon\in(0,1/8),j\in\\{1,2\\}$, $(\alpha,\lambda)\in
R_{dj}$ (see (7.1) for the definition of $R_{dj}$), and
$k\in\mathbb{Z}\cap[1,\infty)$. Since $j$ and $y_{j\ast}$ are fixed, we often
drop the dependence of various quantities on $j$ and assume that
$y_{j\ast}=0$, when there is no possibility of confusion. We also allow the
implied constants to depend on the indices $\sigma_{1},\sigma_{2}$.

Before proving the bounds on $T_{\Theta}$, we first prove the following
technical lemma which will be useful for quantifying the heuristic that
$A_{\Theta}^{-1}(f)$ is exponentially localized to the support of $f$. Recall
the definition (3.34) for $d_{j,k}$, which we denote as $d_{k}$ below.

######  Lemma 8.1.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{1/2},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Furthermore, suppose that $c\in[2,4]$, and
that $D\subset\mathbb{T}_{\mathfrak{p}}$ is an interval. Then for all
$\epsilon\in(0,1/8)$ and $f\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ with
$\text{supp}(f)\subset D$, $\sigma_{1}\geq 0,\,\sigma_{2}\in\mathbb{R},$ the
following bounds hold.

  * •

For $y\in
S_{\frac{\delta(\Lambda)}{c}},D\subset\mathbb{T}_{\mathfrak{p}}\setminus
S_{\delta(\Lambda)}$, we have

$\big{|}A_{\Theta}^{-1}(f)(y)\big{|}\lesssim_{\sigma_{1},\sigma_{2}}\norm{f}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\frac{\exp\big(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}}\big{missing})}{d_{k}^{\sigma_{2}}\delta(\Lambda)^{2+\sigma_{1}}};$
(8.1)

  * •

For $y\in\mathbb{T}_{\mathfrak{p}}\setminus S_{c\delta(\Lambda)},D\subset
S_{\delta(\Lambda)}$, we have

$\begin{split}|A_{\Theta}^{-1}(f)(y)|&\lesssim\norm{f}_{L^{\infty}(S_{\delta(\Lambda)})}\epsilon^{-1/2}\exp(-c_{1}|y|\delta(\Lambda)\epsilon^{-\frac{1}{2}}),\end{split}$
(8.2)

where $c_{1}>0$ is a small constant independent of
$(\epsilon,\alpha,k,\lambda)$.

###### Proof.

We first prove (8.1). For $z\in\text{supp}(f)$ and $y\in
S_{\frac{\delta(\Lambda)}{c}}$ we have
$|y-z|=|y-\text{sgn}(y)\delta(\Lambda)|+|\text{sgn}(y)\delta(\Lambda)-z|$.
Then, we can write

$\exp(-L(y,z)|y-z|)=\exp(-L(y,z)\left|y-\text{sgn}(y)\delta(\Lambda)\right|)\times\exp(-L(y,z)\left|z-\text{sgn}(y)\delta(\Lambda)\right|)$
(8.3)

where $L(y,z)$ represents the coefficient of $|y-z|$ in the exponential in
(3.4) and (3.3). For $z\in\mathbb{T}_{\mathfrak{p}}\setminus
S_{\delta(\Lambda)}$ and for $y_{0}$ satisfying $b(y_{0})=\lambda$,
$|y_{0}|\leq\frac{\delta(\Lambda)}{c}$ we have $|b(z)-\lambda|\gtrsim|b(z)|$
and $L(y,z)\gtrsim\frac{\delta(\Lambda)}{\epsilon^{\frac{1}{2}}}$, with an
implicit constants only depending on $b$. This implies there exists $c_{1}$,
depending only on $b$ such that

$\displaystyle\exp(-L(y,z)\left|y-\text{sgn}(y)\delta(\Lambda)\right|)$
$\displaystyle\lesssim\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}}),\quad
y\in S_{\frac{\delta(\Lambda)}{c}},$
$\displaystyle\exp(-L(y,z)\left|z-\text{sgn}(y)\delta(\Lambda)\right|)$
$\displaystyle\lesssim\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\left|z-\text{sgn}(y)\delta(\Lambda)\right|),\quad
z\in\mathbb{T}_{\mathfrak{p}}\setminus S_{\delta}(\Lambda),$ (8.4)

and consequently,

$\displaystyle|A_{\Theta}^{-1}(f)(y)|\lesssim\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}})\int|f(z)|\epsilon^{-\frac{1}{2}}\delta(\Lambda)^{-1}\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\left|z-\text{sgn}(y)\delta(\Lambda)\right|)\,dz,$
$\displaystyle\quad y\in\mathbb{T}_{\mathfrak{p}}.$ (8.5)

The bound (8.1) will follow from (8) if

$\int_{\mathbb{T}_{\mathfrak{p}}}\frac{\epsilon^{-\frac{1}{2}}\delta(\Lambda)}{\varrho^{\sigma_{1}}(z;\Lambda)\varrho_{k}^{\sigma_{2}}(z;\Lambda)}\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\left|z-\text{sgn}(y)\delta(\Lambda)\right|)\,dz\lesssim\frac{1}{\delta(\Lambda)^{\sigma_{1}}d_{k}^{\sigma_{2}}}.$
(8.6)

To prove (8.6) we consider several cases. First, assume that $k\delta\geq 1$.
Then $\varrho_{k}^{\sigma_{2}}(z;\Lambda)=\frac{1}{k^{\sigma_{2}}}$.
Therefore, it is enough to only consider $\varrho^{\sigma_{1}}(z,\Lambda)$.
Using that $y\in S_{\frac{\delta(\Lambda)}{c}}$ and
$\varrho(z,\Lambda)\geq\delta(\Lambda),$ (8.6) follows. Similar considerations
hold for $k\delta(\Lambda)\leq 1$ and $\sigma_{2}\geq 0$. Now, consider the
case where $k\delta(\Lambda)\leq 1$ and $\sigma_{2}<0$. In this situation

$\frac{1}{\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y,\Lambda)}\lesssim\frac{\varrho^{|\sigma_{2}|}(y;\Lambda)}{\varrho^{\sigma_{1}}(y;\Lambda)}\lesssim\frac{\varrho^{|\sigma_{2}|}(y;\Lambda)}{\delta(\Lambda)^{\sigma_{1}}}.$
(8.7)

Therefore, it’s sufficient to prove that

$\int_{\mathbb{T}_{\mathfrak{p}}}\epsilon^{-\frac{1}{2}}\delta(\Lambda)\varrho_{k}^{|\sigma_{2}|}(z;\Lambda)\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\left|z-\text{sgn}(y)\delta(\Lambda)\right|)\,dz\lesssim\delta(\Lambda)^{|\sigma_{2}|}.$
(8.8)

Since
$c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}})$
is an approximation of the identity and
$\frac{\epsilon^{\frac{1}{2}}}{\delta(\Lambda)}\leq\delta(\Lambda)$, (8.8)
follows, completing the proof of (8.1).

For the proof of (8.2) we proceed as before except with the roles of $y$ and
$z$ reversed from (8). This yields

$\displaystyle\exp(-L(y,z)\left|y-\text{sgn}(y)\delta(\Lambda)\right|)$
$\displaystyle\lesssim\exp(-c_{1}\delta(\Lambda)|y|\epsilon^{-\frac{1}{2}}),\quad
y\in\mathbb{T}_{\mathfrak{p}}\setminus S_{c\delta(\Lambda)},$
$\displaystyle\exp(-L(y,z)\left|z-\text{sgn}(y)\delta(\Lambda)\right|)$
$\displaystyle\lesssim\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\left|z-\text{sgn}(y)\delta(\Lambda)\right|),\quad
z\in S_{\delta(\Lambda)}.$ (8.9)

Then, for $y\in\mathbb{T}_{\mathfrak{p}}\setminus S_{c\delta(\Lambda)}$

$\displaystyle|A_{\Theta}^{-1}(f)(y)|\lesssim$
$\displaystyle\exp(-c_{1}\delta(\Lambda)|y|\epsilon^{-\frac{1}{2}})\norm{f}_{L^{\infty}(S_{\delta(\Lambda)})}\epsilon^{-\frac{2}{3}}\min(\delta(\Lambda),\epsilon^{\frac{1}{4}})^{-\frac{1}{3}}$
(8.10)
$\displaystyle\times\int_{\mathbb{T}_{\mathfrak{p}}}\exp(-c_{1}\delta(\Lambda)\epsilon^{-\frac{1}{2}}\left|z-\text{sgn}(y)\delta(\Lambda)\right|)\,dz,$
(8.11)

which implies the desired bounds. ∎

We will prove of Propositions 7.1 \- 7.3 by proving the needed bounds for each
$T_{I1},T_{I2},T_{v1}$, and $T_{v2}$ individually. Each of the following
subsections is dedicated to one of the operators and proves the bounds needed
for the proof of the propositions.

###  8.1. Bounds on $T_{I1}$

Recall that the definition of $T_{I1}$ with $y_{1*}=0$ is

$T_{I1}g(y)=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz,\quad
y\in\mathbb{T}_{\mathfrak{p}}.$ (8.12)

Since the $(1-\varphi_{\delta_{0}})b^{\prime\prime}g$ is supported
approximately $\delta_{0}$ away from the critical layer we can essentially
view $A_{\Theta}^{-1}$ as a bounded multiplication operator. However, in order
to avoid the logarithmic loss in $\delta(\Lambda)$ for
$\mathcal{G}_{k}(y,z;\Lambda)$ applied to bounded functions, we use Lemma 8.1
to view $A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)$ as if it
was supported $\delta_{0}$ distance away from the origin. The following Lemma
implies Propositions 7.2 and 7.3 for $T_{I1}$.

######  Lemma 8.2.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ and
$g\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ with
$\sigma_{1},\sigma_{2}\in\mathbb{R}$ we have the following bounds for all
$y\in\mathbb{T}_{\mathfrak{p}}$ and $\beta\in\\{0,1,2\\}$,

$\displaystyle\big{|}\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}g(y)\big{|}\lesssim_{\sigma_{1},\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\varrho_{k}^{2}(y;\Lambda)k^{\sigma_{2}}\times\begin{cases}k^{-1},&{\rm
if}\,\,\varrho_{k}(y;\Lambda)=\varrho(y,\Lambda),\\\ 1,&{\rm
if}\,\,\varrho_{k}(y;\Lambda)=\frac{1}{k}.\end{cases}$ (8.13)

###### Proof.

First, suppose that $\varrho_{k}(y;\Lambda)=\varrho(y;\Lambda)$. Decompose
$T_{I1}g$ as

$\displaystyle\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz$
$\displaystyle=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{\frac{\delta_{0}}{3}}(z)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz$
(8.14)
$\displaystyle+\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{\frac{\delta_{0}}{3}}(z))A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz.$

Proposition 3.1 and (8.1) with $\delta=\delta_{0}$ imply

$\displaystyle|(1-\varphi_{\frac{\delta_{0}}{3}}(z))A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)|$
$\displaystyle\lesssim_{\delta_{0},\sigma_{1},\sigma_{2}}k^{\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})},$
$\displaystyle|\varphi_{\frac{\delta_{0}}{3}}(z)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)|$
$\displaystyle\lesssim_{\delta_{0},\sigma_{1},\sigma_{2}}\exp(-c_{1}\delta_{0}^{2}\epsilon^{-\frac{1}{2}})k^{\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(8.15)

Lemma 2.1 combined with (8.1) for $\beta\in\\{0,1\\}$ yields

$\displaystyle\left|\int_{\mathbb{T}_{\mathfrak{p}}}\varrho^{\beta}(y;\Lambda)\partial_{y}^{\beta}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{\frac{\delta_{0}}{3}}(z)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz\right|$
$\displaystyle\lesssim_{\delta_{0},\sigma_{1},\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\varrho^{2}(y;\Lambda)\exp(-c_{1}\delta_{0}^{2}\epsilon^{-\frac{1}{2}})\int_{\mathbb{T}_{\mathfrak{p}}}\frac{k^{\sigma_{2}}\varrho_{k}(z,\Lambda)}{\varrho^{2}(z,\Lambda)}\,dz$
$\displaystyle\lesssim_{\delta_{0},\sigma_{1},\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\varrho^{2}(y;\Lambda)k^{\sigma_{2}-1}\delta(\Lambda)^{-1}\exp(-c_{1}\delta_{0}^{2}\epsilon^{-\frac{1}{2}}).$
(8.16)

Similarly, for $\beta\in\\{0,1\\}$ we obtain the bound

$\displaystyle\left|\int_{\mathbb{T}_{\mathfrak{p}}}\varrho^{\beta}(y;\Lambda)\partial_{y}^{\beta}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{\frac{\delta_{0}}{3}}(z))A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz\right|\lesssim_{\delta_{0},\sigma_{1},\sigma_{2}}\frac{\varrho^{2}(y;\Lambda)}{k^{1-\sigma_{2}}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(8.17)

Since $\delta(\Lambda)\gtrsim\epsilon^{\frac{1}{4}}$ it follows

$\delta(\Lambda)^{-1}\exp(-c_{1}\delta_{0}^{2}\epsilon^{-\frac{1}{2}})\lesssim
1,$ (8.18)

giving (8.13) for $\varrho_{k}(y;\Lambda)=\varrho(y;\Lambda)$ and
$\beta\in\\{0,1\\}$. For $\varrho_{k}(y;\Lambda)=\frac{1}{k}$, we bound
$A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)$ as

$|A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)|\lesssim_{\delta_{0},\sigma_{1},\sigma_{2}}k^{\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(8.19)

Combining (8.19) with Lemma 2.1 then yields (8.13) for $\beta\in\\{0,1\\}$.
For $\beta=2$, we have

$\displaystyle\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}^{2}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz$
$\displaystyle=-A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(y)+(V(y)+k^{2})\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}((1-\varphi_{\delta_{0}})b^{\prime\prime}g)(z)\,dz.$
(8.20)

By (8.19), the rest of the proof proceeds as for $\beta=0$. ∎

The following corollary implies Proposition 7.1 for $T_{I1}$.

######  Corollary 8.1.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ and
$\sigma_{1}\in[0,1]$, $\sigma_{2}\in[-2,1/2]$, we have the following bounds
for $g\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$,

$\norm{T_{I1}g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k^{\frac{1}{2}}}.$
(8.21)

###### Proof.

Recalling the definition of $X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$, it is
sufficient to prove that for all $y\in\mathbb{T}_{\mathfrak{p}}$ and
$\beta\in\\{0,1\\}$ we have

$\displaystyle\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)|\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}g(y)|\lesssim\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k^{\frac{1}{2}}}.$
(8.22)

Since $\mathbb{T}_{\mathfrak{p}}$ is bounded and $\sigma_{1}\geq 0$ it is
enough to prove

$\displaystyle\varrho_{k}^{\sigma_{2}}(y;\Lambda)|\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}g(y)|\lesssim\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k^{\frac{1}{2}}}.$
(8.23)

We first consider the case $\varrho_{k}(y;\Lambda)=\varrho(y;\Lambda)$. From
(8.13) it follows that

$\displaystyle\varrho^{\sigma_{1}}(y;\Lambda)|\varrho^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}g(y)|\lesssim\varrho^{\sigma_{1}+2}(y;\Lambda)k^{\sigma_{2}-1}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k^{\frac{1}{2}}},$
(8.24)

where we used that $\sigma_{2}\in[-2,1/2]$. Now assume that
$\varrho_{k}(y;\Lambda)=\frac{1}{k}$; (8.13) yields

$\displaystyle
k^{-\sigma_{2}}|\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I1}g(y)|\lesssim\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k^{2}},$
(8.25)

which completes the proof. ∎

###  8.2. Bounds on $T_{I2}$

Recall that the definition of $T_{I2}$ with $y_{1*}=0$ is given for
$y\in\mathbb{T}_{\mathfrak{p}}$ by

$T_{I2}g(y)=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.$
(8.26)

Unlike the other terms in the decomposition of $T_{\Theta}$, the kernel of
$T_{I2}$ exhibits singular behavior due to the critical layer. This requires a
more careful analysis near the critical layer than the pointwise estimates
given by the fundamental solution bounds (see Lemma 8.5). However, the Airy
fundamental solution bounds are still useful, even for
$|y|\lesssim\delta(\Lambda)$, provided that we have the bound

$|A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(y)|\lesssim\frac{\norm{g}_{L^{\infty}(S_{\delta})}}{\varrho^{2}(y;\Lambda)},$
(8.27)

with the implicit constant independent of $\alpha,\epsilon,k,\lambda$.
Comparing this with the bounds in Proposition 3.1, we determine that (8.27)
will happen in the following three situations:

$\displaystyle\delta(\Lambda)$ $\displaystyle\lesssim|\alpha|^{\frac{1}{2}},$
(8.28a) $\displaystyle\delta(\Lambda)$
$\displaystyle\lesssim\epsilon^{\frac{1}{4}},$ (8.28b)
$\displaystyle\delta(\Lambda)$
$\displaystyle\lesssim|b(y)-\lambda|^{1/2}\,\,{\rm
for}\,\,y\in\mathbb{T}_{\mathfrak{p}},\quad{\rm and}\,\,\lambda\notin
b(\mathbb{T}_{\mathfrak{p}}),$ (8.28c)

with implicit constants independent of $\epsilon,\alpha,k,\lambda$. We refer
to the situations in (8.28) as _regularly degenerate_. In these situations, we
have the following lemma.

######  Lemma 8.3.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Furthermore, assume that at least one of
the cases in $\eqref{reg:degen}$ holds. Then for all $\epsilon\in(0,1/8)$,
$g\in L^{\infty}(S_{\delta})$ with ${\rm supp}\,g\subseteq S_{\delta}$, and
all $y\in\mathbb{T}_{\mathfrak{p}}$ the following bounds hold,

  * •

If $k\delta(\Lambda)\leq 1$, then we have

$\varrho_{k}^{\beta}(y;\Lambda)|\partial_{y}^{\beta}T_{I2}g(y)|\lesssim\frac{\delta(\Lambda)\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}}{\varrho(y;\Lambda)}\left[\frac{\varrho_{k}(y;\Lambda)}{\varrho(y;\Lambda)}\right]^{1/4},\quad\beta\in\\{0,1,2\\};$
(8.29)

  * •

If $k\delta(\Lambda)\geq 1$, then

$\displaystyle|T_{I2}g(y)|+\Big{|}\frac{1}{k}\partial_{y}T_{I2}g(y)\Big{|}\lesssim\frac{\norm{g}_{L^{\infty}(S_{\delta})}}{k^{2}\varrho^{2}(y;\Lambda)}.$
(8.30)

###### Proof.

We provide the details only for (8.29) since the proof of (8.30) is simpler.
We decompose $T_{I2}$ as

$\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{3\delta(\Lambda)}(z)A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz+\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{3\delta(\Lambda)}(z))A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.$
(8.31)

By Lemma 2.1 it follows that for $\beta\in\\{0,1\\}$

$\displaystyle\Big{|}\varrho_{k}^{\beta}(y;\Lambda)\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}^{\beta}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{3\delta(\Lambda)}(z))A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz\Big{|}$
$\displaystyle\lesssim\big{(}\delta\wedge\frac{1}{k}\big{)}\min\Big{\\{}\frac{\delta}{\varrho(y;\Lambda)},\,e^{-k\varrho(y;\Lambda)}\Big{\\}}\int_{\mathbb{T}_{\mathfrak{p}}}(1-\varphi_{3\delta(\Lambda)}(z))|A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)|\,dz.$
(8.32)

On the support of $(1-\varphi_{3\delta(\Lambda)}(z))$, (8.2) implies

$\begin{split}&|A_{\Theta}^{-1}(\varphi_{\delta}(\Lambda)b^{\prime\prime}g)(z)|\\\
&\lesssim\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}\epsilon^{-1/2}\exp(-\frac{1}{2}c_{1}|z|\delta(\Lambda)\epsilon^{-\frac{1}{2}})\exp(-\frac{c_{1}}{2}\frac{\delta(\Lambda)^{2}}{\epsilon^{\frac{1}{2}}}),\end{split}$
(8.33)

where $c_{1}$ is as in Lemma 8.1. Since

$\displaystyle\int_{\mathbb{T}_{\mathfrak{p}}}(1-\varphi_{3\delta(\Lambda)}(z))\delta(\Lambda)\,\epsilon^{-1/2}\exp(-\frac{c_{1}}{2}|z|\delta(\Lambda)\epsilon^{-\frac{1}{2}})\,dz\lesssim
1,$ (8.34)

The desired bound (8.29) for $\beta\in\\{0,1\\}$ follows from (8.33)-(8.34)
and (8.2).

For the second derivatives, (2.7) yields

$\displaystyle\partial_{y}^{2}\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}(\varphi_{\delta(\Lambda))}b^{\prime\prime}g)(z)\,dz$
$\displaystyle=-A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(y)+(V(y)+k^{2})\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.$
(8.35)

The desired bound (8.29) for $\beta=2$ then follows from (8.2) and
(8.33)-(8.34). For the proof of (8.30), we do not decompose $T_{I2}$: We
directly use (8.27), Lemma 2.1, and that $ke^{-k|z|}$ is an approximation of
the identity. ∎

Due to Lemma 8.3, we can assume the following conditions hold for the
remaining bounds on $T_{I2}$ :

$\displaystyle|\lambda|\gg\epsilon^{\frac{1}{2}}\text{ with an implicit
constant independent of }\epsilon,\alpha,k,\lambda,$ (8.36a)
$\displaystyle\lambda\in b(\mathbb{T}_{\mathfrak{p}}),$ (8.36b)
$\displaystyle\delta(\Lambda)\approx\delta_{1}(\lambda),$ (8.36c)

where we recall that
$\delta_{1}(\lambda)=8\sqrt{|\lambda|/|b^{\prime\prime}(0)|}$. We seek to
isolate the part of
$A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)$ that is
supported at a distance closer to the critical layer than the length scale
$d_{k}$ of the modified Green’s function. On the complement of this region,
the smallness from the Green’s function can balance the size of
$A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)$ exactly as in
the “regularly degenerate” situation covered in Lemma 8.3.

Following (6.5), we decompose $T_{I2}$ as

$\displaystyle T_{I2}g$ $\displaystyle=\sum_{\ell=0}^{2}T_{I2\ell}g.$ (8.37)

We define $T_{I20}$ and $T_{I2\ell},\ell\in\\{1,2\\}$ as

$\displaystyle
T_{I20}g(y):=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\Big{(}1-\sum_{\ell=1}^{2}\varphi_{\theta
d_{k}}^{2}(z-y_{\ell})\Big{)}\,dz,$ (8.38)

and

$\displaystyle
T_{I2\ell}g(y):=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{\theta
d_{k}}^{2}(z-y_{\ell})A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz,$
(8.39)

where $\ell\in\\{1,2\\}$, $b(y_{\ell})=\lambda$ and $\theta\in(0,1)$ is a
constant only depending on $b(y)$ (see Lemma 8.5). Since
$\delta(\Lambda)=\delta_{1}(\lambda)$, for $\theta$ sufficiently small,
$\varphi_{\theta d_{k}}(y-y_{1})$ and $\varphi_{\theta d_{k}}(y-y_{2})$ have
disjoint supports. Using a similar proof as for Lemma 8.3, we obtain the
following bounds for $T_{I20}$.

######  Lemma 8.4.

Assume that,
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
$k\in\mathbb{Z}\cap[1,\infty)$, and let $\theta$ to be fixed in the range
given by Lemma 8.5. Then for all $\epsilon\in(0,1/8)$ and $g\in
L^{\infty}(S_{\delta(\Lambda)})$ we have the following bounds for all
$y\in\mathbb{T}_{\mathfrak{p}}$.

  * •

If $k\delta(\Lambda)\leq 1$, then

$\displaystyle\varrho_{k}^{\beta}(y;\Lambda)|\partial_{y}^{\beta}T_{I20}g(y)|\lesssim\frac{\delta(\Lambda)\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}}{\varrho(y;\Lambda)}\left[\frac{\varrho_{k}(y;\Lambda)}{\varrho(y;\Lambda)}\right]^{1/4},\quad{\rm
for}\,\,\beta\in\\{0,1,2\\},$ (8.40)

  * •

If $k\delta(\Lambda)\geq 1$, then

$\displaystyle\Big{|}\frac{1}{k^{\beta}}\partial_{y}^{\beta}T_{I20}g(y)\Big{|}\lesssim\frac{\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}}{k\varrho(y;\Lambda)},\quad{\rm
for}\,\,\beta\in\\{0,1\\}.$ (8.41)

###### Proof.

We only consider the case where $k\delta(\Lambda)\geq 1$ since the case for
$k\delta(\Lambda)\leq 1$ is identical to Lemma 8.3. We have the following:

$\displaystyle|A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)(1-\sum_{j=1}^{2}\varphi_{\theta
d_{k}}^{2}(z-y_{\ell}))|\lesssim\begin{cases}\frac{k\norm{g}_{L^{\infty}(S_{\delta}(\Lambda))}}{\delta(\Lambda)}&|z|\leq\delta(\Lambda)\\\
\frac{\norm{g}_{L^{\infty}(S_{\delta}(\Lambda))}}{\varrho^{2}(z,\Lambda)}&|z|\geq\delta(\Lambda)\end{cases}$
(8.42)

Combining these estimates yields

$\Big{|}A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)(1-\sum_{j=1}^{2}\varphi_{\theta
d_{k}}^{2}(z-y_{\ell}))\Big{|}\lesssim\frac{k\norm{g}_{L^{\infty}(S_{\delta}(\Lambda))}}{\varrho(z,\Lambda)}.$
(8.43)

Therefore, by Lemma 2.1 we have

$\Big{|}\frac{1}{k^{\beta}}\partial_{y}^{\beta}T_{I20}g(y)\Big{|}\lesssim\frac{\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}}{k\varrho(y;\Lambda)},\quad{\rm
for}\,\,\beta\in\\{0,1\\},$ (8.44)

which proves the result. ∎

For the region that lies in the support of $\varphi_{\theta
d_{k}}(y-y_{\ell})$ with $b(y_{\ell})=\lambda$, the size of
$A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)$ is too large to
be controlled by the pointwise bounds for the modified Green’s function
directly. As in the invisicd case we expect that integrating by parts to
“remove” a derivative from $A_{\Theta}^{-1}$ and placing it on
$\mathcal{G}_{k}(y,z;\Lambda)$ would be beneficial to gain from the larger
scale of variation of $\mathcal{G}_{k}(y,z;\Lambda)$ (see [38], [39]). The
following lemma makes this precise using a frequency based characterization of
the inverse of the generalized Airy operator.

######  Lemma 8.5.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Furthermore, assume that
$\delta(\Lambda)\approx\delta_{1}(\lambda)$, $\lambda\in
b(\mathbb{T}_{\mathfrak{p}})$, and $\epsilon^{\frac{1}{2}}\ll|\lambda|\ll 1$
with implicit constants independent of $(\epsilon,\alpha,k,\lambda)$. Then
there exists $\theta_{0}\in(0,1/2)$, depending only on $b(y)$, such that for
all $\theta\in(0,\theta_{0}]$ and $\epsilon\in(0,1/8)$ the following statement
holds. Suppose that $f:\mathbb{T}_{\mathfrak{p}}\to\mathbb{C}$ satisfies

$\norm{f}_{L^{2}(S_{\delta(\Lambda)})}+d_{k}\norm{\partial_{y}f}_{L^{2}(S_{\delta(\Lambda)})}\leq
M,$ (8.45)

and is 0 outside of $S_{\delta(\Lambda)}$. We view the function

$\varphi_{\theta d_{k}}(y-y_{\ell})A_{\Theta}^{-1}(f)(y),$ (8.46)

as a function on $\mathbb{R}$. Then

$\big{|}\sup_{\xi\in\mathbb{R}}\mathcal{F}\left(\varphi_{\theta d_{k}}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f)\right)(\xi)\big{|}\lesssim\frac{M}{d_{k}^{\frac{1}{2}}\delta(\Lambda)},$
(8.47)

and

$\norm{(1+d_{k}\partial_{y})^{-1}\left(\varphi_{\theta d_{k}}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f)\right)}_{L^{2}(\mathbb{R})}\lesssim\frac{M}{d_{k}\delta(\Lambda)},$
(8.48)

where $y_{\ell}$ satisfies $b(y_{\ell})=\lambda$.

The following inequality will be useful in the proof of Lemma 8.5:

$\norm{f}_{L^{\infty}(I)}\lesssim|I|^{-\frac{1}{2}}\norm{f}_{L^{2}(I)}+|I|^{\frac{1}{2}}\norm{\partial_{y}f}_{L^{2}(I)}$
(8.49)

where $I$ is an interval and $|\cdot|$ denotes the length of the interval.

###### Proof.

Assuming (8.47), by Plancherel’s theorem we have

$\displaystyle\norm{(1+d_{k}\partial_{y})^{-1}\left(\varphi_{\theta
d_{k}}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f)\right)}_{L^{2}(\mathbb{R})}=\norm{(1-id_{k}\xi)^{-1}\mathcal{F}(\varphi_{\theta
d_{k}}(\cdot-y_{\ell})A_{\Theta}^{-1}(f))(\xi)}_{L_{\xi}^{2}(\mathbb{R})}$
$\displaystyle\lesssim\frac{M}{d_{k}^{\frac{1}{2}}\delta(\Lambda)}\norm{(1-id_{k}\xi)^{-1}}_{L_{\xi}^{2}(\mathbb{R})}\lesssim\frac{M}{d_{k}\delta(\Lambda)^{\frac{1}{2}}}.$

Therefore, it suffices to prove $\eqref{eq:Airy:freq:pw}$. We proceed as in
Lemma 6.3. For non-critical values of $\lambda$, $b^{-1}(\lambda)$ has
cardinality 2: $b^{-1}(\lambda)=\\{y_{1},y_{2}\\}$. Define
$\varphi_{\ell}(y):=\varphi\left(\frac{2(y-y_{\ell})}{\theta d_{k}}\right)$
for $\ell\in\\{1,2\\}$. Choose $\theta_{0}>0$ so that the supports of
$\varphi_{1}$ and $\varphi_{2}$ are separated by distance $\theta_{0}d_{k}$.
Furthermore, define $w_{c,\ell}$ as the solution to the equation for
$y\in\mathbb{R}$,

$\epsilon\partial_{y}^{2}w_{c,\ell}(y)-\alpha
w_{c,\ell}(y)+ib^{\prime}(y_{\ell})(y_{\ell}-y)w_{c,\ell}(y)=f(y_{\ell}).$
(8.50)

We can solve for $w_{c,\ell}$ explicitly as

$w_{c,\ell}(y,y_{\ell})=\frac{\
f(y_{\ell})}{b^{\prime}(y_{\ell})}\Big{(}\frac{b^{\prime}(y_{\ell})}{\epsilon}\Big{)}^{\frac{1}{3}}W\Big{(}\Big{(}\frac{b^{\prime}(y_{\ell})}{\epsilon}\Big{)}^{\frac{1}{3}}(y-y_{\ell})\Big{)},\quad
y\in\mathbb{R},$ (8.51)

where $W$ is given in the Fourier variable $\xi\in\mathbb{R}$ as

$\widehat{W}(\xi)=-\sqrt{2\pi}\exp\left(\frac{\xi^{3}}{3}+\frac{\epsilon^{-\frac{1}{3}}\alpha\xi}{(b^{\prime}(y_{\ell}))^{\frac{2}{3}}}\right)\mathbbm{1}_{(-\infty,0]}(\xi).$
(8.52)

We define $w_{R}:\mathbb{T}_{\mathfrak{p}}\to\mathbb{C}$ as

$\displaystyle
w_{R}:=A_{\Theta}^{-1}(f)-\sum_{\ell\in\\{1,2\\}}\varphi_{\ell}w_{c,\ell},$
(8.53)

where we have view $\varphi_{\ell}w_{c,\ell}$ as a function on
$\mathbb{T}_{\mathfrak{p}}$. $w_{R}$ can be bounded similarly as in Lemma 3.4
(see (3.56)), and is more regular than $W$. Therefore, we obtain from (8.52)
that

$\big{|}\mathcal{F}(\varphi_{\theta d_{k}}(\cdot-
y_{\ell})A_{\Theta}^{-1}(f))(\xi)\big{|}\lesssim\frac{M}{d_{k}^{\frac{1}{2}}\delta(\Lambda)}\quad
j\in\\{1,2\\}$ (8.54)

which completes the proof. ∎

For the remainder of the section on $T_{I2}$ we will take $\theta$ to be
fixed. Before proceeding with the proof of the bounds for $T_{I2j}$, it will
be convenient to introduce the following notation. For $\ell\in\\{1,2\\}$
define for $y\in\mathbb{R}$,

$\Xi_{\ell}(y):=(1+d_{k}\partial_{y})^{-1}\left[\varphi_{\theta d_{k}}(\cdot-
y_{\ell})A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)\right].$
(8.55)

where $b(y_{\ell})=\lambda$ and $\varphi_{\theta d_{k}}(y-y_{\ell})$ is as in
Lemma 8.5. We now proceed with obtaining the remaining bounds for
$T_{I2\ell}$.

######  Lemma 8.6.

Assume that $\ell\in\\{1,2\\}$,
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Furthermore, assume that
$\delta(\Lambda)=\delta_{1}(\lambda)$, $\lambda\in
b(\mathbb{T}_{\mathfrak{p}})$, and $\epsilon^{\frac{1}{2}}\ll|\lambda|\ll 1$
with implicit constants independent of $\epsilon,\alpha,k,\lambda$. Take
$\theta$ to be fixed in the range given by Lemma 8.5 Then for all
$\epsilon\in(0,1/8)$ and $g\in L^{\infty}(S_{\delta(\Lambda)})$the following
bounds hold for all $y\in\mathbb{T}_{\mathfrak{p}}$,

  * •

If $k\delta(\Lambda)\leq 1$, then

$\begin{split}&|T_{I2\ell}g(y)|+\varrho_{k}(y;\Lambda)\big{|}\partial_{y}T_{I2\ell}g(y)+d_{k}\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)\big{|}\\\
&\lesssim\left[\frac{\varrho_{k}(y;\Lambda)}{\varrho(y;\Lambda)}\right]^{1/4}\frac{\delta(\Lambda)}{\varrho(y;\Lambda)}\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})},\\\
&\varrho_{k}^{2}(y,\Lambda)\big{|}\partial_{y}\big{(}\partial_{y}T_{I2j}g(y)+d_{k}\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)\big{)}+\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)\big{|}\\\
&\lesssim\left[\frac{\varrho_{k}(y;\Lambda)}{\varrho(y;\Lambda)}\right]^{1/4}\frac{\delta(\Lambda)}{\varrho(y;\Lambda)}\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}.\end{split}$
(8.56)

  * •

If $k\delta(\Lambda)\geq 1$, then

$\displaystyle|T_{I2\ell}g(y)|+\frac{1}{k}\big{|}\partial_{y}T_{I2\ell}g(y)+d_{k}\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)\big{|}$
$\displaystyle\lesssim\frac{\exp\big(-k\max(|y|-\delta_{1}(\lambda),0)\big{missing})}{k\delta(\Lambda)}\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})}.$
(8.57)

###### Proof.

First, observe that $T_{I2\ell}$ can be written as

$\displaystyle T_{I2\ell}g(y)$
$\displaystyle=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{\theta
d_{k}}^{2}(z-y_{\ell})A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz$
$\displaystyle=\int_{\mathbb{T}_{\mathfrak{p}}}(1-d_{k}\partial_{z})\left(\varphi_{\theta
d_{k}}(z-y_{\ell})\mathcal{G}_{k}(y,z;\Lambda)\right)\Xi_{\ell}(z)\,dz.$
(8.58)

Similarly, $\partial_{y}T_{I2\ell}$ can be written as

$\displaystyle\partial_{y}T_{I2\ell}g(y)=\partial_{y}\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{\theta
d_{k}}^{2}(z-y_{\ell})A_{\Theta}^{-1}(\varphi_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz$
$\displaystyle=\partial_{y}\int_{\mathbb{T}_{\mathfrak{p}}}(1-d_{k}\partial_{z})\left(\varphi_{\theta
d_{k}}(z-y_{\ell})\mathcal{G}_{k}(y,z;\Lambda)\right)\Xi_{\ell}(z)\,dz,$
$\displaystyle=d_{k}\int_{\mathbb{T}_{\mathfrak{p}}}\varphi_{\theta
d_{k}}(z-y_{\ell})\partial_{y}^{2}\mathcal{G}_{k}(y,z;\Lambda)\Xi_{\ell}(z)\,dz$
$\displaystyle-
d_{k}\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}(\partial_{y}+\partial_{z})\left(\varphi_{\theta
d_{k}}(z-y_{\ell})\mathcal{G}_{k}(y,z;\Lambda)\right)\Xi_{\ell}(z)\,dz$
$\displaystyle+\int_{\mathbb{T}_{\mathfrak{p}}}\left(\varphi_{\theta
d_{k}}(z-y_{\ell})\partial_{y}\mathcal{G}_{k}(y,z;\Lambda)\right)\Xi_{\ell}(z)\,dz.$
(8.59)

The equation for $\mathcal{G}_{k}(y,z;\Lambda)$ implies

$\displaystyle\int_{\mathbb{T}_{\mathfrak{p}}}\varphi_{\theta
d_{k}}(z-y_{\ell})\partial_{y}^{2}\mathcal{G}_{k}(y,z;\Lambda)\Xi_{\ell}(z)\,dz$
$\displaystyle=-\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)+(V(y)+k^{2})\int_{\mathbb{T}_{\mathfrak{p}}}\left(\varphi_{\theta
d_{k}}(z-y_{\ell})\mathcal{G}_{k}(y,z;\Lambda)\right)\Xi_{\ell}(z)\,dz.$
(8.60)

To summarize, we write

$\displaystyle\partial_{y}T_{I2\ell}g(y)=-d_{k}\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)+\int_{\mathbb{T}_{\mathfrak{p}}}R_{\ell}(y,z)\Xi_{\ell}(z)\,dz,$
(8.61)

where $\text{supp}(R_{\ell}(y,\cdot))\subset\left[y_{\ell}-\theta
d_{k},y_{\ell}-\theta d_{k}\right]$, and

$\displaystyle|R_{\ell}(y,z)|\lesssim\begin{cases}e^{-k|y-z|},\quad&d_{k}=\frac{1}{k},\\\
\min\Big{\\{}\frac{\delta(\Lambda)}{\varrho(y;\Lambda)},e^{-k|y-z|}\Big{\\}},&d_{k}=\delta(\Lambda).\end{cases}$
(8.62)

Therefore,

$\displaystyle|T_{I2\ell}g(y)|+\varrho_{k}(y)\left|\partial_{y}T_{I2\ell}g(y)+d_{k}\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)\right|$
$\displaystyle\lesssim\begin{cases}\frac{\exp{-k\max\big{(}|y|-\delta_{1}(\lambda),0\big{)}}}{k\delta(\Lambda)}\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})},&{\rm
if}\,\,d_{k}=\frac{1}{k},\\\
\left[\frac{\varrho_{k}(y;\Lambda)}{\varrho(y;\Lambda)}\right]^{1/4}\frac{\delta(\Lambda)}{\varrho(y;\Lambda)}\norm{g}_{L^{\infty}(S_{\delta(\Lambda)})},&{\rm
if}\,\,d_{k}=\delta(\Lambda),\end{cases}$ (8.63)

as desired. To derive (8.56), observe that (8.2) and (8.2) imply

$\displaystyle\partial_{y}\big{(}\partial_{y}T_{I2\ell}g+d_{k}\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)\big{)}$ $\displaystyle=-\varphi_{\theta
d_{k}}(y-y_{\ell})\Xi_{\ell}(y)+(V(y)+k^{2})\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\Xi_{\ell}(z)\,dz$
$\displaystyle\quad+d_{k}\partial_{y}\Big{(}\big{(}V(y)+k^{2}\big{)}\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\Xi_{\ell}(z)\,dz\Big{)}$
(8.64) $\displaystyle\quad-
d_{k}\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}^{2}(\partial_{y}+\partial_{z})\left(\varphi_{\theta
d_{k}}(z-y_{\ell})\mathcal{G}_{k}(y,z;\Lambda)\right)\Xi_{\ell}(z)\,dz,$
(8.65)

from which (8.56) follows. ∎

Combining Lemmas 8.3, 8.4, and 8.6 we get the following bounds for $T_{I2}$ on
$X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$.

######  Lemma 8.7.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{0}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$,
$\sigma_{1}\in[0,1]$, and $\sigma_{2}\in[-2,1/2]$ satisfying
$\sigma_{1}+\sigma_{2}\leq 1$, we have the following bounds for $g\in
X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$,

$\norm{T_{I2}g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\langle
k\delta(\Lambda)\rangle^{\frac{1}{2}}},$ (8.66)

and for $y\in\mathbb{T}_{\mathfrak{p}}$ with $|y|\gtrsim\delta_{0}$,

$\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)\big{|}\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I2}g(y)\big{|}\lesssim_{\sigma_{1},\sigma_{2}}\varrho^{\sigma_{1}-1/4}(y;\Lambda)\varrho_{k}^{\sigma_{2}+1/4}(y;\Lambda)\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}.$
(8.67)

###### Proof.

We focus on (8.66), since (8.67) is easier as $|y|\gtrsim\delta_{0}$. We first
assume that we are in the regularly degenerate case so that one of the cases
in (8.28) holds. From Lemma 8.3 we have for $\beta\in\\{0,1\\}$,

$\displaystyle\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)|\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{I2}g(y)|\lesssim\begin{cases}\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)\frac{\delta(\Lambda)\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}\varrho(y;\Lambda)},&{\rm
if}\,\,k\delta(\Lambda)\leq 1,\\\
\varrho^{\sigma_{1}}(y;\Lambda)\frac{(k\delta(\Lambda))^{\frac{1}{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k^{2}\varrho^{2}(y;\Lambda)\delta(\Lambda)^{\sigma_{1}}},&{\rm
if}\,\,k\delta(\Lambda)\geq 1.\end{cases}$ (8.68)

Using the fact that $\sigma_{1}+\sigma_{2}\leq 1$ when $k\delta(\Lambda)\leq
1$ and that $\sigma_{1}\leq 1$ when $k\delta(\Lambda)\geq 1$ implies the
desired bound. Now consider the singularly degenerate case (8.36). For $g\in
X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ we have from Lemma that 8.5

$\delta(\Lambda)^{-\frac{1}{2}}d_{k}^{\sigma_{2}+1}\delta(\Lambda)^{\sigma_{1}}\norm{d_{k}\varphi_{\theta
d_{k}}(\cdot-
y_{\ell})\Xi_{\ell}}_{L^{2}(\mathbb{T}_{\mathfrak{p}})}\lesssim\begin{cases}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})},&{\rm
if}\,\,k\delta(\Lambda)\leq 1,\\\
\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{k\delta(\Lambda)},&{\rm
if}\,\,k\delta(\Lambda)\geq 1,\end{cases}$ (8.69)

where $d_{k}\varphi_{\theta d_{k}}(\cdot-y_{\ell})\Xi_{\ell}$ is as in Lemma
8.6. Finally, (• ‣ 8.6) implies that for $y\in\mathbb{T}_{\mathfrak{p}}$

$\displaystyle\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)\big{|}T_{I2j}g(y)\big{|}+\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{1+\sigma_{2}}(y;\Lambda)\big{|}\partial_{y}T_{I2j}g(y)+d_{k}\varphi_{\theta
d_{k}}(\cdot-y_{\ell})\Xi_{\ell}(y)\big{|}$
$\displaystyle\lesssim\begin{cases}\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)\frac{\delta(\Lambda)\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}\varrho(y;\Lambda)},&\quad{\rm
if}\,\,k\delta(\Lambda)\leq 1,\\\
\varrho^{\sigma_{1}}(y;\Lambda)\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{(k\delta(\Lambda))^{\frac{1}{2}}\delta(\Lambda)^{\sigma_{1}}}\times\exp{-k\max(|y|-\delta_{1}(\lambda),0)},&\quad{\rm
if}\,\,k\delta(\Lambda)\geq 1.\end{cases}$ (8.70)

The fact that $\sigma_{1}+\sigma_{2}\leq 1$ when $k\delta(\Lambda)\leq 1$
implies the desired bound. ∎

###  8.3. Bounds on $T_{v1}$

Recall the definition of $T_{v1}$ (assuming that $y_{1*}=0$) for
$y\in\mathbb{T}_{\mathfrak{p}}$,

$T_{v1}g(y)=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\varphi_{\frac{\delta(\Lambda)}{3}}(z)A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.$
(8.71)

Heuristically, we expect that $T_{v1}$ to be very small since for functions
$g$ that are supported at distances greater than $\delta(\Lambda)$ away from
the critical layer, $A_{\Theta}^{-1}(g)(y)$ is exponentially small for
$|y|<\delta(\Lambda)$ from Lemma 8.1. Crucially, Lemma 8.1 also has the
correct scaling between $\delta(\Lambda)$ and $\epsilon$: when
$\delta(\Lambda)\approx\epsilon^{1/4}$, the exponential term is no longer
small enough to view $T_{v1}$ as negligible and we need higher regularity
estimates for $T_{v1}$ in order to prove the required compactness for the
limiting absorption principle.

######  Lemma 8.8.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ and
$g\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ with
$\sigma_{1}\in[0,1],\sigma_{2}\in[-2,1/2]$ satisfying
$\,\sigma_{1}+\sigma_{2}\in(-2,1)$, the following bounds hold for all
$y\in\mathbb{T}_{\mathfrak{p}}$.

  * •

For $\beta\in\\{0,1,2\\}\text{ and }k\delta(\Lambda)\leq 1$,

$\displaystyle\varrho_{k}^{\beta}(y;\Lambda)|\partial_{y}^{\beta}T_{v1}g(y)|\lesssim\frac{\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}})\delta(\Lambda)\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho(y;\Lambda)\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}};$
(8.72)

  * •

For $\beta\in\\{0,1\\}$, we also have

$\displaystyle\varrho_{k}^{\beta}(y;\Lambda)|\partial_{y}^{\beta}T_{v1}g(y)|\lesssim\begin{cases}\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}})\varrho^{-\sigma_{1}-1/4}\varrho_{k}^{-\sigma_{2}+1/4}(y;\Lambda)\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})},&k\delta(\Lambda)\lesssim
1,\\\
k^{\sigma_{2}}\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}})\varrho^{-\sigma_{1}}(y;\Lambda)\big{(}k\delta(\Lambda)\big{)}^{-2}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})},&k\delta(\Lambda)\gtrsim
1,\end{cases}$ (8.73)

where $c_{1}$ is as in (8.1).

######  Corollary 8.2.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ with
$\sigma_{1}\in[0,1]$, $\sigma_{2}\in[-2,1/2]$ satisfying
$\sigma_{1}+\sigma_{2}\in(-2,1)$ the following bounds hold for $g\in
X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$,

$\displaystyle\norm{T_{v1}g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim_{\sigma_{1},\sigma_{2}}\begin{cases}\frac{\epsilon\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{4}}&k\delta(\Lambda)\leq
1\\\ \frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\langle
k\delta(\Lambda)\rangle^{2}}&k\delta(\Lambda)\geq 1,\end{cases}$ (8.74)

where the implicit constants are is uniform in $(\alpha,\epsilon,k,\lambda)$.

###### Proof.

From (8.1) we have for $z\in\mathbb{T}_{\mathfrak{p}}$,

$\Big{|}\varphi_{\frac{\delta(\Lambda)}{3}}(z)A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\Big{|}\lesssim_{\sigma_{1},\sigma_{2}}\exp(-c_{1}\delta(\Lambda)^{2}\epsilon^{-\frac{1}{2}})\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{d_{k}^{\sigma_{1}}\delta(\Lambda)^{2+\sigma_{2}}}.$
(8.75)

The proof of (8.72) then follows from Lemma 2.1. For (8.73), observe that
since $\sigma_{1}+\sigma_{2}<1$ and $k\delta(\Lambda)\leq 1$ it follows for
$y\in\mathbb{T}_{\mathfrak{p}}$ that

$\frac{\delta(\Lambda)\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)}{\varrho(y;\Lambda)\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}}\lesssim
1.$ (8.76)

For the case of $k\delta(\Lambda)\geq 1$, we use that $ke^{-k|z|}$ is an
approximation of the identity. The proof of corollary 8.2 is immediate. ∎

###  8.4. Bounds on $T_{v2}$

Recall from (7.3) the definition of $T_{v2}$ assuming that $y_{1*}=0$,

$T_{v2}g(y)=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z))\frac{\epsilon\partial_{z}^{2}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)}{b(z)-\lambda-i\alpha}\,dz.$
(8.77)

For $\epsilon^{\frac{1}{2}}\ll|\lambda|$, we expect that $T_{v2}$ to be small
since $A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)$ acts like
multiplication by $\frac{1}{|b(y)-\lambda|+\delta(\Lambda)^{2}}$. However,
some care is needed in the case when $|\lambda|\lesssim\epsilon^{\frac{1}{2}}$
in order to ensure the correct behavior for $|y|\gg\delta(\Lambda)$ (see
(8.79)). Unlike $T_{v1}$, the kernel of $T_{v2}$ is not localized to
$S_{\delta(\Lambda)}$ which complicates matters since it is no longer
sufficient to simply consider the size of the kernel on $S_{\delta(\Lambda)}$.
Capturing the optimal decay in the kernel away from $S_{\delta(\Lambda)}$ is
the reason why we integrate by parts in (8.4) and (8.82).

######  Lemma 8.9.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ and
$g\in X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ with
$\sigma_{1}\in[0,1],\,\sigma_{1}+\sigma_{2}\in(-2,1)$ the following bounds
hold for all $y\in\mathbb{T}_{\mathfrak{p}}$.

  * •

For all values of $k\delta(\Lambda)$ and all $\lambda\in\Sigma_{j,\delta_{0}}$
we have

$\displaystyle|\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{v2}g(y)|\lesssim_{\sigma_{1},\sigma_{2}}\frac{\epsilon^{1/2}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{2}\varrho^{\sigma_{1}+1/2}(y;\Lambda)\varrho_{k}^{\sigma_{2}-1/2}(y;\Lambda)},\quad\beta\in\\{0,1\\};$
(8.78)

  * •

For $k\delta(\Lambda)\leq 1$,

$\displaystyle\big{|}\varrho_{k}^{\beta}(y;\Lambda)\partial_{y}^{\beta}T_{v2}g(y)\big{|}$
$\displaystyle\lesssim_{\sigma_{1},\sigma_{2}}\frac{\epsilon^{1/2}\delta(\Lambda)\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)^{2}\varrho(y;\Lambda)\delta(\Lambda)^{\sigma_{1}+\sigma_{2}}},\quad\beta\in\\{0,1\\};$
(8.79)

  * •

For $|\lambda|\lesssim\epsilon^{\frac{1}{2}}$,

$\displaystyle\big{|}k^{-\beta}\partial_{y}^{\beta}T_{v2}g(y)\big{|}$
$\displaystyle\lesssim_{\sigma_{1},\sigma_{2}}\frac{1}{\big{(}k\delta(\Lambda)\big{)}^{1/2}}\frac{\epsilon^{\frac{1}{4}}k^{\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)\varrho^{\sigma_{1}}(y;\Lambda)},\quad
k\delta(\Lambda)\geq 1,\,\,\beta\in\\{0,1\\},$ (8.80a)
$\displaystyle\big{|}\varrho_{k}^{2}(y;\Lambda)\partial_{y}^{2}T_{v2}g(y)\big{|}$
$\displaystyle\lesssim_{\sigma_{1},\sigma_{2}}\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho^{\sigma_{1}+1/2}(y;\Lambda)\varrho_{k}^{\sigma_{2}-1/2}(y,\Lambda)},\quad
k\delta(\Lambda)\leq 1.$ (8.80b)

######  Remark 8.1.

For $|\epsilon|^{\frac{1}{2}}\ll|\lambda|\ll 1$, (8.78) is sufficient to view
$T_{v2}$ as negligible as part of the proof of the limiting absorption
principle. However, when $|\lambda|\lesssim\epsilon^{\frac{1}{2}}$, we need to
treat $T_{v2}$ like $T_{I2}$ and prove higher regularity estimates and
favorable dependence on $k\delta(\Lambda)$.

###### Proof.

Integration by parts in (8.77) yields that $T_{v2}$ can be written as

$\displaystyle T_{v2}g(y)$
$\displaystyle=\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z))\frac{\epsilon\partial_{z}^{2}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)}{b(z)-\lambda-i\alpha}\,dz$
$\displaystyle=-\epsilon\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{z}\mathcal{G}_{k}(y,z;\Lambda)\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz$
$\displaystyle-\epsilon\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\partial_{z}\left(\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\right)\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.$
(8.81)

Similarly, $\partial_{y}T_{{}_{v}2}$ can be written for
$y\in\mathbb{T}_{\mathfrak{p}}$ as

$\begin{split}\partial_{y}T_{v2}g(y)&=\partial_{y}\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)(1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z))\frac{\epsilon\partial_{z}^{2}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)}{b(z)-\lambda-i\alpha}\,dz\\\
&=-\epsilon\partial_{y}\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{z}\mathcal{G}_{k}(y,z;\Lambda)\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz\\\
&-\epsilon\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}\mathcal{G}_{k}(y,z;\Lambda)\partial_{z}\left(\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\right)\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz,\end{split}$
(8.82)

and using (2.7), as

$\begin{split}\partial_{y}T_{v2}g(y)=&-\epsilon\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(y)}{b(y)-\lambda-i\alpha}\partial_{y}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(y)\\\
&+\epsilon\big{(}V(y)+k^{2}\big{)}\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz\\\
&-\epsilon\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}(\partial_{z}+\partial_{y})\mathcal{G}_{k}(y,z;\Lambda)\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz\\\
&-\epsilon\int_{\mathbb{T}_{\mathfrak{p}}}\partial_{y}\mathcal{G}_{k}(y,z;\Lambda)\partial_{z}\left(\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\right)\partial_{z}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.\end{split}$

Since $\eta_{\delta(\Lambda)}$ is supported a distance of approximately
$\delta(\Lambda)$ away from 0, Lemma 3.3 implies that for
$y\in\mathbb{T}_{\mathfrak{p}}$,

$\displaystyle\left|\epsilon\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(y)}{b(y)-\lambda-i\alpha}\partial_{y}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(y)\right|\lesssim\frac{\epsilon^{1/2}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho^{3+\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)}.$
(8.83)

The bounds (8.79)-(8.79) then follow from (8.83) and Lemma 4.1.

We now consider the case $|\lambda|\lesssim\epsilon^{\frac{1}{2}}$. We no
longer integrate by parts to remove a derivative from
$\partial_{y}^{2}A_{\Theta}^{-1}$ since the scale of variation of
$\mathcal{G}_{k}(y,z;\Lambda)$ is comparable or smaller than that of
$A_{\Theta}^{-1}$. Lemma 3.3 implies that for $y\in\mathbb{T}_{\mathfrak{p}}$,

$\left|\epsilon\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(y)}{b(y)-\lambda-i\alpha}\partial_{y}^{2}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(y)\right|\lesssim\frac{\epsilon^{\frac{1}{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho^{3+\sigma_{1}}(y;\Lambda)\varrho_{k}^{1+\sigma_{2}}(y;\Lambda)},$
(8.84)

which, in combination with Lemma 4.1 and (8.84), implies that for
$k\delta(\Lambda)\geq 1$,

$\displaystyle\left|k^{-\beta}\partial_{y}^{\beta}T_{v2}g(y)\right|\lesssim\frac{1}{\big{(}k\delta(\Lambda)\big{)}^{1/2}}\frac{\epsilon^{\frac{1}{4}}k^{\sigma_{2}}\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\delta(\Lambda)\varrho^{\sigma_{1}}(y;\Lambda)},\quad\beta\in\\{0,1\\}.$
(8.85)

For the second derivative estimates, we simply use the identity that

$\displaystyle\partial_{y}^{2}T_{v2}g(y)=$
$\displaystyle-\epsilon\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(y)}{b(y)-\lambda-i\alpha}\partial_{y}^{2}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(y)$
$\displaystyle+(V(y)+k^{2})\int_{\mathbb{T}_{\mathfrak{p}}}\mathcal{G}_{k}(y,z;\Lambda)\frac{1-\varphi_{\frac{\delta(\Lambda)}{{3}}}(z)}{b(z)-\lambda-i\alpha}\epsilon\partial_{z}^{2}A_{\Theta}^{-1}(\eta_{\delta(\Lambda)}b^{\prime\prime}g)(z)\,dz.$
(8.86)

Therefore for $k\delta(\Lambda)\leq 1$

$\varrho_{k}^{2}(y;\Lambda)|\partial_{y}^{2}T_{v2}g(y)|\lesssim_{\sigma_{1},\sigma_{2}}\frac{\norm{g}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}}{\varrho^{\sigma_{1}}(y;\Lambda)\varrho_{k}^{\sigma_{2}}(y;\Lambda)},$
(8.87)

which concludes the proof. ∎

The following is an immediate corollary of Lemma 8.9.

######  Corollary 8.3.

Assume that
$(\alpha,\lambda)\in(-\sigma_{0}\epsilon^{\frac{1}{2}},1)\times\Sigma_{j,\delta_{0}}$,
and $k\in\mathbb{Z}\cap[1,\infty)$. Then for all $\epsilon\in(0,1/8)$ the
following bounds hold for $T_{v2}:X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})\to
X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})$ with $\sigma_{1}\in[0,1]$,
$\sigma_{1}+\sigma_{2}\in(-2,1)$.

$\displaystyle\norm{T_{v2}}_{X^{\sigma_{1},\sigma_{2}}(\mathfrak{M})}\lesssim_{\sigma_{1},\sigma_{2}}\begin{cases}\frac{\epsilon^{1/2}}{\delta(\Lambda)^{2}},&{\rm
if}\,\,\lambda\in\Sigma_{j,\delta_{0}},\\\
\frac{\epsilon^{\frac{1}{4}}}{\delta(\Lambda)\langle
k\delta(\Lambda)\rangle^{1/2}},&{\rm
if}\,\,|\lambda|\lesssim\epsilon^{\frac{1}{2}}\text{ and }k\delta(\Lambda)\geq
1.\end{cases}$ (8.88)

Acknowledgement

We thank Peter Constantin for bringing the paper [55] to our attention. RB
would also like to thank Dallas Albritton, Tarek Elgindi, and Vlad Vicol for
helpful discussions and encouragement.

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