$\underbrace{-\varphi^{\prime\prime}+\mathcal{K}\varphi}_{=:\mathcal{L}\varphi}=-m^{2}\varphi\,.$
(4.25)

$\mathcal{L}$ is a self-adjoint operator on the appropriate function spaces.
Using (4.24), it is possible to design $\Xi$ such that $\mathcal{K}$ is
negative enough to ensure that $\mathcal{L}$ has a negative eigenvalue; see
Lemma 4.4.7. Let $-m_{0}^{2}$ denote the bottom of the spectrum of
$\mathcal{L}$, with the convention $m_{0}>0$. Let $\varphi_{0}$ denote an
associated eigenfunction,

$\mathcal{L}\varphi_{0}=-m_{0}^{2}\varphi_{0}\,,$ (4.26)

which we normalize to $\|\varphi_{0}\|_{L^{2}}^{2}=1$. By Sturm-Liouville
theory, the eigenvalue $-m_{0}^{2}$ is algebraically simple, and its
eigenfunction $\varphi_{0}$ does not vanish except at $\partial I$. In
summary, $(\varphi_{0},m_{0},z_{0})$ is a solution to Rayleigh’s equation. The
eigenfunction $\varphi_{0}$ is known as a _neutral mode_ , since it is neither
stable nor unstable: $\textrm{Im}z=0$. $m_{0}$ is the corresponding _neutral
wavenumber_ , and in general it may not be an integer.

We seek unstable modes $\varphi_{\varepsilon}$ nearby the neutral mode
$\varphi_{0}$ in a perturbation expansion:

$\displaystyle\varphi_{\varepsilon}$
$\displaystyle=\varphi_{0}+\varepsilon\varphi_{1}+\varepsilon^{2}\varphi_{2}+\cdots$
(4.27) $\displaystyle m_{\varepsilon}$ $\displaystyle=m_{0}+\varepsilon
m_{1}+\varepsilon^{2}m_{2}+\cdots$ $\displaystyle z_{\varepsilon}$
$\displaystyle=z_{0}+\varepsilon z_{1}+\varepsilon^{2}z_{2}+\cdots\,.$

We normalize $\|\varphi_{\varepsilon}\|_{L^{2}}^{2}=1$, which yields the
condition

$\langle\varphi_{0},\varphi_{1}\rangle=0\,.$ (4.28)

Eventually, we will see that it is also possible to fix the freedom in the
parameter $\varepsilon$ by choosing, e.g., $|m_{1}|=1$. Due to the conjugation
symmetry of Rayleigh’s equation, the perturbation expansion will
simultaneously detect stable modes.

The equation to be satisfied by $\varphi_{1}$ is

$\displaystyle-z_{1}$
$\displaystyle\Big{(}-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\Big{)}\varphi_{0}+2(\Xi-
z_{0})m_{0}m_{1}\varphi_{0}$ $\displaystyle+\underbrace{(\Xi-
z_{0})\Big{(}-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\Big{)}\varphi_{1}+A\varphi_{1}}_{=(\Xi-
z_{0})(\mathcal{L}+m_{0}^{2})\varphi_{1}}=0\,.$ (4.29)

We wish to divide by $\Xi-z_{0}$ and apply the equation (4.26) for
$\varphi_{0}$ to obtain

$z_{1}(\Xi-
z_{0})^{-1}\mathcal{K}\varphi_{0}+2m_{0}m_{1}\varphi_{0}+(\mathcal{L}+m_{0}^{2})\varphi_{1}=0\,.$
(4.30)

However, we should not manipulate terms containing $(\Xi-z_{0})^{-1}$
cavalierly; $(\Xi-z_{0})^{-1}$ is not locally integrable. Rather, we divide by
$\Xi-z_{0}\pm i\delta$ and send $\delta\to 0^{+}$. We will find that one
choice of $\pm$ will detect the stable eigenvalue, and the other choice will
detect the unstable eigenvalue. We write

$z_{1}\lim_{\delta\to 0^{+}}[(\Xi-z_{0}\pm
i\delta)^{-1}\mathcal{K}\varphi_{0}]+2m_{0}m_{1}\varphi_{0}+(\mathcal{L}+m_{0}^{2})\varphi_{1}=0\,.$
(4.31)

By the Fredholm theory, the above equation is solvable provided that

$\langle z_{1}(\Xi-z_{0}\pm
i0)^{-1}\mathcal{K}\varphi_{0}+2m_{0}m_{1}\varphi_{0},\varphi_{0}\rangle=0\,,$
(4.32)

borrowing the physics shorthand $\pm i0$ to track the regularization
procedure. By the Plemelj formula,

$\displaystyle\int(\Xi-z_{0}\pm i0)^{-1}\mathcal{K}|\varphi_{0}|^{2}\,dt=$
$\displaystyle\mp
i\pi\Xi^{\prime}(t_{0})^{-1}\mathcal{K}(t_{0})|\varphi_{0}(t_{0})|^{2}$
$\displaystyle+\textrm{pv}\,\int(\Xi-
z_{0})^{-1}\mathcal{K}|\varphi_{0}|^{2}\,dt\,.$ (4.33)

Hence,

$z_{1}\left[\mp
i\pi\Xi^{\prime}(t_{0})^{-1}\mathcal{K}(t_{0})|\varphi_{0}(t_{0})|^{2}+\textrm{pv}\,\int(\Xi-
z_{0})^{-1}\mathcal{K}|\varphi_{0}|^{2}\,dt\right]+2m_{0}m_{1}=0\,.$ (4.34)

The first term is guaranteed not to vanish. In particular, we have the
relationship

$z_{1}=\frac{2m_{0}m_{1}}{\pm
i\pi\Xi^{\prime}(t_{0})^{-1}\mathcal{K}(t_{0})|\varphi_{0}(t_{0})|^{2}-\textrm{pv}\,\int(\Xi-
z_{0})^{-1}\mathcal{K}|\varphi_{0}|^{2}\,dt}\neq 0\,,$ (4.35)

and

$\textrm{sgn}(\textrm{Im}\,{z_{1}})=\mp\textrm{sgn}\frac{\mathcal{K}(t_{0})m_{1}}{\Xi^{\prime}(t_{0})}\,.$
(4.36)

The perturbation expansion suggests the existence of a curve of unstable modes
bifurcating from the neutral mode; see Figure 4.3. For the rigorous
implementation of the ideas above to show that the existence of the unstable
branch in a neighborhood of the neutral mode we refer to Lemma 4.8.1 where
ultimately Plemelj’s formula will play a crucial role.

However, it is also important to say something about the global picture of the
curve of unstable modes and, in particular, to ensure that $m$ can be chosen
to be an integer. This is done by continuing the curve; see Proposition 4.4.5.
Heuristically, the main obstacle will be that the curve could “fall” back to
the plane $\textrm{Im}z=0$. A key part of understanding the global picture
will be to demonstrate that the only possible _neutral limiting modes_ , where
the curve touches $\textrm{Im}z=0$, are at the critical values
$z_{c}=\Xi(t_{c})$ where $A(t_{c})=0$ (otherwise, the potential $A/(\Xi-z)$
contains a singularity which will lead to a contradiction); see Proposition
4.4.1.

Figure 4.3. A schematic depiction of the emergence of stable and unstable
modes from neutral modes. The dotted curve is the corresponding curve of
stable modes, obtained by complex conjugation. Notably, the curve is not
smooth across $\textrm{Im}z=0$. The one-sided derivatives of $z$ at $m_{a}$
may be calculated as $-z_{1}$ in (4.35) with $m_{1}=-1$.

The case of rotating flows in an annulus was considered by Lin in [25], and
the reader to encouraged to compare with the approach therein.

###  4.4. Overview of the proof of Theorem 4.2.4

The rest of the chapter is devoted to proving Theorem 4.2.4. The proof will be
achieved through a careful study of Rayleigh’s stability equation (4.17) and,
in particular, the set $\mathscr{P}$ of pairs $(m,z)$ such that
$z\in\mathscr{U}_{m}$ and $m>1$, i.e.,

$\mathscr{P}:=\left\\{(m,z)\in\mathbb{R}\times\mathbb{C}:z\in\mathscr{U}_{m},m>1\right\\}\,.$
(4.37)

Given that $\Xi$ is strictly decreasing, we have

$\lim_{t\to-\infty}\Xi(t)>\Xi(a)>\Xi(b)>\lim_{t\to\infty}\Xi(t)=0$

and in order to simplify our notation we will use $\Xi(-\infty)$ for
$\lim_{t\to-\infty}\Xi(t)$ and, occasionally, $\Xi(\infty)$ for $0$.

The first step in the proof of Theorem 4.2.4 is understanding which pairs
$(m,z)$ belong to the closure of $\mathscr{P}$ and have $\textrm{Im}\,z=0$.
Solutions $(m,z,\varphi)$ to (4.17) with $(m,z)\in\overline{\mathscr{P}}$ and
$\textrm{Im}\,z=0$ are sometimes called _neutral limiting modes_ [25]. To that
end, it is convenient to introduce the following two self-adjoint operators:

$\displaystyle L_{a}$
$\displaystyle:=-\frac{d^{2}}{dt^{2}}+\frac{A(t)}{\Xi(t)-\Xi(a)}$ (4.38)
$\displaystyle L_{b}$
$\displaystyle:=-\frac{d^{2}}{dt^{2}}+\frac{A(t)}{\Xi(t)-\Xi(b)}\,.$ (4.39)

Thanks to the definition of the class $\mathscr{C}$, it is easy to see that
both functions $\frac{A(t)}{\Xi(t)-\Xi(a)}$ and $\frac{A(t)}{\Xi(t)-\Xi(b)}$
are bounded and that $\frac{A(t)}{\Xi(t)-\Xi(a)}<\frac{A(t)}{\Xi(t)-\Xi(b)}$.

Moreover, the first is negative on $]\\!-\\!\infty,b[$ and positive on
$]b,\infty[$, while the second is negative on $]\\!-\\!\infty,a[$ and positive
on $]a,\infty[$. Recall that the spectra of these operators are necessarily
real and denote by $-\lambda_{a}$ and $-\lambda_{b}$ the smallest element in
the respective spectra: observe that, by the Rayleigh quotient
characterization, $-\lambda_{a}<-\lambda_{b}$.

While _a priori_ , $\lambda_{a}=\lambda_{b}=0$ is possible, for the remainder
of this section, we work only in the setting $-\lambda_{b}<0$; this is
possible due to Proposition 4.4.6.

The following proposition characterizes the possible neutral limiting modes:

######  Proposition 4.4.1.

If $(m_{0},z)\in\overline{\mathscr{P}}$ and $\textrm{Im}\,z=0$, then either
$z=\Xi(a)$ or $z=\Xi(b)$. Moreover, in either case, if $m_{0}>1$ then
necessarily $m_{0}=\sqrt{\lambda_{a}}$ or $m_{0}=\sqrt{\lambda_{b}}$. Assume
in addition that $-\lambda_{a}<-1$. Then, for $z=\Xi(a)$, the unique $m\geq 1$
such that (4.17) has a nontrivial solution $\psi_{a}\in L^{2}$ is
$m_{a}=\sqrt{\lambda_{a}}$. Moreover, any nontrivial solution has the property
that $\psi_{a}(a)\neq 0$.

######  Remark 4.4.2.

We remark that the exact same argument applies with $b$ in place of $a$ when
$\lambda_{b}>1$, even though this fact does not play any role in the rest of
the notes.

Observe that this does not yet show that
$(m_{a},\Xi(a))\in\overline{\mathscr{P}}$ corresponds to a neutral limiting
mode. The latter property will be achieved in a second step, in which we seek
a curve of unstable modes emanating from $(m_{a},\Xi(a))$:

######  Proposition 4.4.3.

Assume $-\lambda_{a}<-1$ and let $m_{a}=\sqrt{\lambda_{a}}$. There are
positive constants $\varepsilon>0$ and $\delta>0$ with the following property:
For every $h\in]0,\delta[$, $\mathscr{U}_{m_{a}-h}\cap
B_{\varepsilon}(\Xi(a))\neq\emptyset$.

This is proved by making the formal argument in Section 4.3 rigorous.

######  Remark 4.4.4.

In fact, the argument given for the proposition proves the stronger conclusion
that $\mathscr{U}_{m_{a}-h}\cap B_{\varepsilon}(\Xi(a))$ consists of a single
point $z$, with the property that $mz$ is an eigenvalue of $\mathcal{L}_{m}$
with geometric multiplicity $1$. Moreover, the very same argument applies to
$b$ in place of $a$ and $h\in]-\delta,0[$ if $\lambda_{b}>1$.

Combined with some further analysis, in which the curve of unstable modes is
continued, the latter proposition will allow us to conclude the following:

######  Proposition 4.4.5.

Assume $-\lambda_{a}<-1$, let $m_{a}=\sqrt{\lambda_{a}}$ and
$m_{b}:=\sqrt{\max\\{1,\lambda_{b}\\}}$. Then $\mathscr{U}_{m}\neq\emptyset$
for every $m\in]m_{b},m_{a}[$.

Thus far, we have not selected our function $\Xi$: the above properties are
valid for any element in the class $\mathscr{C}$. The choice of $\Xi$ comes in
the very last step.

######  Proposition 4.4.6.

There is a choice of $\Xi\in\mathscr{C}$ with the property that
$]m_{b},m_{a}[$ contains an integer larger than $1$.

Clearly, the combination of Proposition 4.4.5 and Proposition 4.4.6 gives
Theorem 4.2.4: we first choose $\Xi$ as in Proposition 4.4.6 and hence we
select $m_{0}$ as the largest natural number which belongs to the interval
$]m_{b},m_{a}[$; the properties claimed in Theorem 4.2.4 follow then from
Proposition 4.4.5. The proof of Proposition 4.4.6 is in fact a rather
straightforward application of the following.

######  Lemma 4.4.7.

Let $m_{0}$ be any integer. Then there exists $\Xi\in\mathscr{C}$ with $a=0$
and $b=\frac{1}{2}$ such that the smallest eigenvalue of the operator $L_{a}$
is smaller than $-m_{0}^{2}$.

######  Remark 4.4.8.

A consequence of Lemma 4.4.7 is that the most unstable wavenumber $m_{0}$ can
be made arbitrarily large. Only $m_{0}\geq 2$ is necessary to prove non-
uniqueness. We warn the reader that here $m_{0}$ does not refer to the neutral
wavenumber of Section 4.3.

The rest of the chapter will be devoted to proving the Propositions 4.4.1 and
4.4.3 and Lemma 4.4.7. We finish this section by giving the simple proof of
Proposition 4.4.6

###### Proof.

For simplicity we fix $a=0$ and $b=\frac{1}{2}$ and we look at the set of
functions $\Xi$ with this particular choice of zeros for $A$. We then denote
by $L_{\Xi,a}$ the operator in (4.38). We fix an arbitrary
$\Xi_{0}\in\mathscr{C}$ and let $-\lambda(0)$ be the smallest eigenvalue of
$L_{\Xi_{0},a}$. We then consider the smallest integer $m_{0}\geq 3$ such that
$m_{0}^{2}>\lambda(0)$. By Lemma 4.4.7 there is an element
$\Xi_{1}\in\mathscr{C}$ with the property that $a=0$, $b=\frac{1}{2}$ and, if
$-\lambda(1)$ is the smallest element of the spectrum of $L_{\Xi_{1},a}$, then
$-\lambda(1)<-m_{0}^{2}$. For $\sigma\in[0,1]$ consider $L_{\Xi_{\sigma},a}$
where

$\Xi_{\sigma}=(1-\sigma)\Xi_{0}+\sigma\Xi_{1}\,$

and observe that $\Xi_{\sigma}\in\mathscr{C}$ for every $\sigma\in[0,1]$.

Since $\sigma\mapsto\Xi_{\sigma}$ is continuous in the uniform convergence, by
the Rayleigh quotient characterization we see that the smallest element
$-\lambda(\sigma)$ of the spectrum of $L_{\Xi_{\sigma},a}$ is a continuous
function of $\sigma$. There is thus one $\sigma\in[0,1[$ with
$\lambda(\sigma)=m_{0}^{2}$. Let $\sigma_{0}$ be the largest $\sigma$ with
$\lambda(\sigma)=m_{0}^{2}$. Observe now that, if we let $-\mu(\sigma_{0})$ be
the smallest eigenvalue of $L_{\Xi_{\sigma_{0}},b}$, then
$\mu(\sigma_{0})<m_{0}^{2}$. In addition, $\sigma\mapsto\mu(\sigma)$ is also
continuous and thus there is $h>0$ such that $\mu(\sigma)<m_{0}^{2}$ for all
$\sigma\in[\sigma_{0}-h,\sigma_{0}+h]$. On the other hand
$\lambda(\sigma_{0}+h)>m_{0}^{2}$. This shows that $m_{b}<m_{0}<m_{a}$ if we
choose $\Xi=\Xi_{\sigma_{0}+h}$, completing the proof of our claim. ∎

###  4.5. ODE Lemmas

An essential tool in the proofs of the Propositions 4.4.1 and 4.4.3 are the
following two ODE lemmas.

######  Lemma 4.5.1.

Let $m>0$. For every $f\in L^{2}(\mathbb{R})$ there is a unique $\psi\in
L^{2}(\mathbb{R})\cap W^{2,2}_{\text{loc}}$ s.t.

$-\frac{d^{2}\psi}{dt^{2}}+m^{2}\psi=f$ (4.40)

and it is given by

$\psi(t)=\frac{1}{2m}\int_{\mathbb{R}}e^{-m|t-\tau|}f(\tau)\,d\tau\,.$ (4.41)

###### Proof.

The lemma is a classical well-known fact. At any rate the verification that
$\psi$ as in (4.41) solves (4.40) is an elementary computation while, since
obviosuly $e^{-m|t|}\in L^{1}$, $\psi\in L^{2}$ if $f\in L^{2}$. Moreover, any
other solution $\hat{\psi}$ of (4.40) must satisfy
$\hat{\psi}(t)=\psi(t)+C_{+}e^{mt}+C_{-}e^{-mt}$ for some constants $C_{\pm}$
and the requirement $\hat{\psi}\in L^{2}$ immediately implies $C_{+}=C_{-}=0$.
∎

The second ODE Lemma is the following:

######  Lemma 4.5.2.

Let $v\in L^{1}(\mathbb{R},\mathbb{C})$. Then for every constant $c_{-}$ there
is a unique solution $y\in W^{2,1}_{\text{loc}}(\mathbb{R},\mathbb{C})$ of

$-\frac{d^{2}y}{dt^{2}}+(m^{2}+v)y=0$ (4.42)

with the property that

$\lim_{t\to-\infty}e^{-mt}y(t)=c_{-}\,.$ (4.43)

Moreover we have $y(t)=e^{mt}(c_{-}+z(t))$ for a function $z(t)$ which
satisfies the bounds

$\displaystyle|z(t)|$
$\displaystyle\leq|c_{-}|\left[\exp\left(\frac{1}{2m}\int_{-\infty}^{t}|v(s)|\,ds\right)-1\right]$
(4.44) $\displaystyle|z^{\prime}(t)|$ $\displaystyle\leq
2m|c_{-}|\left[\exp\left(\frac{1}{2m}\int_{-\infty}^{t}|v(s)|\,ds\right)-1\right]$
(4.45)

A symmetric statement, left to the reader, holds for solutions such that

$\lim_{t\to\infty}e^{mt}y(t)=c_{+}\,.$ (4.46)

Important consequences of the above Lemmas are the following:

######  Corollary 4.5.3.

If $(m,z)\in\mathscr{P}$, then the space of solutions $\varphi\in L^{2}\cap
W^{2,2}_{\text{loc}}$ of (4.17) is $1$-dimensional. Moreover for any such
$\varphi$ there is a constant $C$ with the property that

$\displaystyle|\varphi(t)|$ $\displaystyle\leq Ce^{-m|t|}\,$ (4.47)

and there are two constants $C_{+}$ and $C_{-}$ such that

$\displaystyle\lim_{t\to\infty}e^{mt}\varphi(t)$ $\displaystyle=C_{+}$ (4.48)
$\displaystyle\lim_{t\to-\infty}e^{-mt}\varphi(t)$ $\displaystyle=C_{-}\,.$
(4.49)

The constants are either both nonzero or both zero, in which case $\varphi$
vanishes identically.

The same conclusions apply if $m>1$, $z\in\\{\Xi(a),\Xi(b)\\}$ and $\varphi$
solves (4.17).

###### Proof.

Observe that $|\Xi(t)-z|\geq|\textrm{Im}\,z|$, while $A(t)=6c_{0}e^{2t}$ for
$-t$ sufficiently large and $|A(t)|\leq 2e^{-{\bar{\alpha}}t}$ for $t$
sufficiently large. In particular

$\frac{|A(t)|}{|\Xi(t)-z|}\leq Ce^{-{\bar{\alpha}}|t|}\,.$ (4.50)

First of all notice that, if $\varphi\in L^{2}\cap W^{2,2}_{\text{loc}}$
solves (4.17), by Lemma 4.5.1 (applied with $f=-\frac{A\varphi}{\Xi-z}$) we
have

$|\varphi(t)|\leq\frac{C}{2m}\int
e^{-m|t-\tau|}e^{-{\bar{\alpha}}|\tau|}|\varphi(\tau)|\,d\tau\,.$ (4.51)

Using Cauchy-Schwarz and the fact that $\varphi\in L^{2}$ we immediately
obtain that $\varphi\in L^{\infty}$, namely, that there is a constant $C$ such
that $|\varphi|\leq C$. We now prove inductively that $|\varphi(t)|\leq
C_{k}e^{-k{\bar{\alpha}}|t|/2}$ as long as $k{\bar{\alpha}}/2\leq m$. The case
$k=0$ has already been shown. Assume thus that the inequality holds for $k-1$
and that $k{\bar{\alpha}}/2\leq m$. We then observe that

$\displaystyle e^{-m|t-\tau|}e^{-{\bar{\alpha}}|\tau|}|\varphi(\tau)|$
$\displaystyle\leq
C_{k-1}e^{-m|t-\tau|-k{\bar{\alpha}}|\tau|/2}e^{-{\bar{\alpha}}|\tau|/2}$
$\displaystyle\leq
C_{k-1}e^{-k{\bar{\alpha}}(|t-\tau|+|\tau|)/2}e^{-{\bar{\alpha}}|\tau|/2}$
$\displaystyle\leq
C_{k-1}e^{-k{\bar{\alpha}}|t|/2}e^{-{\bar{\alpha}}|\tau|/2}\,.$

Inserting in (4.51) and using $e^{-{\bar{\alpha}}|\tau|/2}\in L^{1}$ we then
obtain $|\varphi(t)|\leq C_{k}e^{-k{\bar{\alpha}}|t|/2}$. Assuming now
$k{\bar{\alpha}}/2\leq m<(k+1){\bar{\alpha}}/2$ we can, likewise, bound

$e^{-m|t-\tau|}e^{-{\bar{\alpha}}|\tau|}|\varphi(\tau)|\leq
C_{k}e^{-m|t-\tau|-(k+1){\bar{\alpha}}|\tau|/2}e^{-{\bar{\alpha}}|\tau|}\leq
C_{k}e^{-m|t|}e^{-{\bar{\alpha}}|\tau|/2}$

and plugging into (4.51) one last time we conclude $|\varphi(t)|\leq
Ce^{-m|t|}$.

In order to show that $\varphi$ is unique up to a multiplicative constants, it
suffices to show that $\lim_{t\to-\infty}e^{-mt}\varphi(t)$ exists and is
finite. Hence Lemma 4.5.2 would conclude that the solution is uniquely
determined by $C_{-}$, and that the latter must be nonzero, otherwise
$\varphi\equiv 0$. In order to show existence and finiteness of $C_{-}$
rewrite

$\varphi(t)=\frac{e^{mt}}{2m}\int_{t}^{\infty}e^{-ms}\frac{A(s)}{\Xi(s)-z}\varphi(s)\,ds+\frac{e^{-mt}}{2m}\int_{-\infty}^{t}e^{ms}\frac{A(s)}{\Xi(s)-z}\varphi(s)\,ds\,.$

Since by our estimates both $e^{-ms}\frac{A(s)}{\Xi(s)-z}\varphi(s)$ and
$e^{ms}\frac{A(s)}{\Xi(s)-z}\varphi(s)$ are integrable, we conclude that
$C_{\pm}$ exist and equal

$\displaystyle C_{\pm}=\frac{1}{2m}\int_{-\infty}^{\infty}e^{\pm
ms}\frac{A(s)}{\Xi(s)-z}\varphi(s)\,ds\,.$

As for the last sentence of the statement of the lemma, the same arguments can
be used in the case $z\in\\{\Xi(a),\Xi(b)\\}$, since the crucial point is
that, thanks to the assumption that $A(a)=A(b)=0$ and $\Xi^{\prime}(a)\neq
0\neq\Xi^{\prime}(b)$, the estimate (4.50) remains valid. ∎

###### Proof of Lemma 4.5.2.

We distinguish between the case $c_{-}\neq 0$ and $c_{-}=0$. In the case
$c_{-}\neq 0$ we can divide by $c_{-}$ and reduce the statement to $c_{-}=1$.
For the existence it suffices to look for a solution of (4.42) which satisfies
(4.43) on a half-line of type $]-\infty,T]$ for some $T$. Such solution has
then a $W^{2,1}_{\text{loc}}$ continuation on $[T,\infty[$ by standard ODE
theory. Likewise the uniqueness is settled once we can show the uniqueness
holds on $]-\infty,T]$. Observe next that, if the solution exists, we would
clearly conclude that $\frac{d^{2}y}{dt^{2}}\in L^{1}(]-\infty,T])$, hence
implying that

$\lim_{t\to-\infty}y^{\prime}(t)$

exists and is finite. On the other hand (4.43) implies that such limit must be
$0$.

Let $\tilde{y}(t)=e^{-mt}y(t)$ and observe that we are looking for a solution
of

$(e^{2mt}\tilde{y}^{\prime})^{\prime}=e^{2mt}v\tilde{y}\,.$

Integrating between $-N$ and $t$ the latter identity and letting $t\to-\infty$
we conclude

$e^{2mt}\tilde{y}^{\prime}(t)=\int_{-\infty}^{t}e^{2ms}v(s)\tilde{y}(s)\,ds\,.$
(4.52)

Divide by $e^{2mt}$ and integrate once more to reach

$\displaystyle\tilde{y}(t)-1$
$\displaystyle=-\int_{-\infty}^{t}\int_{-\infty}^{r}e^{2m(s-r)}v(s)\tilde{y}(s)\,ds\,dr$
$\displaystyle=\frac{1}{2m}\int_{-\infty}^{t}\big{(}1-e^{-2m(t-s)}\big{)}v(s)\tilde{y}(s)\,ds$

We then define the transformation

$\mathscr{F}(\tilde{y})(t)=\frac{1}{2m}\int_{-\infty}^{t}\big{(}1-e^{-2m(t-s)}\big{)}v(s)\tilde{y}(s)\,ds+1\,$
(4.53)

which we consider as a map from $L^{\infty}(]-\infty,T])$ into itself. From
our discussion we conclude that $y$ solves (4.42) and obeys (4.43) if and only
if $\tilde{y}$ is a fixed point of $\mathscr{F}$. Choosing $T$ large enough so
that $\|v\|_{L^{1}(]-\infty,T])}\leq m$ we see immediately that $\mathscr{F}$
is contraction on $L^{\infty}(]-\infty,T])$ and it thus has a unique fixed
point. We have thus showed existence and uniqueness of the solution in
question.

Observe now that $z(t)=\tilde{y}(t)-1$ and set

$Z(t):=\exp\left(\frac{1}{2m}\int_{-\infty}^{t}|v(s)|\,ds\right)-1\,.$

$Z$ solves the ODE $Z^{\prime}=\frac{|v|}{2m}Z+\frac{|v|}{2m}$ and, since
$\lim_{t\to-\infty}Z(t)=0$, the integral equation

$Z(t)=\frac{1}{2m}\int_{-\infty}^{t}|v(s)|Z(s)\,ds+\frac{1}{2m}\int_{-\infty}^{t}|v(s)|\,ds\,.$

We first want to show that $|z(t)|\leq Z(t)$ on $]-\infty,T]$. We set
$\tilde{y}_{0}:=Z+1$ and define inductively
$\tilde{y}_{i+1}=\mathscr{F}(\tilde{y}_{i})$. From the above discussion we
know that $\tilde{y}_{i}$ converges uniformly to $\tilde{y}$ and it suffices
thus to show that $|\tilde{y}_{i}-1|\leq Z$ for all $i$. By definition we have
$|\tilde{y}_{0}-1|=Z$ and thus we need to show the inductive step. We estimate

$\displaystyle|\tilde{y}_{i+1}(t)-1|$
$\displaystyle\leq\frac{1}{2m}\int_{-\infty}^{t}|v(s)||\tilde{y}_{i}(s)|\,ds$
$\displaystyle\leq\frac{1}{2m}\int_{-\infty}^{t}|v(s)|Z(s)\,ds+\frac{1}{2m}\int_{-\infty}^{t}|v(s)|\,ds=Z(t)\,,$

We have shown (4.44) on $]-\infty,T]$. In order to extend the inequality to
the whole real axis observe first that we can assume, without loss of
generality, that $\|v\|_{L^{1}(\mathbb{R})}>0$, otherwise we trivially have
$|\tilde{y}(t)-1|=Z(t)=0$ for all $t$. In particular we can select $T$ so that
all of the above holds and at the same time $\|v\|_{L^{1}(]-\infty,T])}>0$.
This implies $Z(T)>0$. Moreover, by (4.53) and
$\mathscr{F}(\tilde{y})=\tilde{y}$, either

$\displaystyle|\tilde{y}(T)-1|$
$\displaystyle<\frac{1}{2m}\int_{-\infty}^{T}|v(s)||\tilde{y}(s)|\,ds$

or $|v||\tilde{y}|$ vanishes identically on $]-\infty,T]$. In both cases we
conclude $|\tilde{y}(T)-1|<Z(T)$. Consider now $\sup\\{t\geq
T:|\tilde{y}(t)-1|<Z(t)\\}$. Such supremum cannot be a finite number $T_{0}$
because in that case we would have $|\tilde{y}(T_{0})-1|=Z(t_{0})$ while the
same argument leading to the strict inequality $|\tilde{y}(T)-1|<Z(T)$ implies
$|\tilde{y}(T_{0})-1|<Z(T_{0})$.

Having shown (4.44) we now come to (4.45). Recalling (4.52) we have

$\displaystyle z^{\prime}(t)$
$\displaystyle=\int_{-\infty}^{t}e^{-2m(t-s)}v(s)(z(s)+1)\,ds$
$\displaystyle\leq\int_{-\infty}^{t}e^{-2m(t-s)}|v(s)|Z(s)\,ds+\int_{-\infty}^{t}e^{-2m(t-s)}|v(s)|\,ds=2mZ(t)\,.$

We now come to the case $c_{-}=0$. In that case we need to show that the
unique solution is identically $0$. Arguing as for the case $c_{-}=1$ we
conclude that $\varphi$ is a fixed point of the transformation

$\mathscr{F}(\varphi)(t)=\frac{1}{2m}\int_{-\infty}^{t}\big{(}1-e^{-2m(t-s)}\big{)}v(s)\tilde{y}(s)\,ds$

Again, for a sufficiently small $T$, $\mathscr{F}$ is a contraction on
$L^{\infty}(]\\!-\infty,T])$ and hence it has a unique fixed point. Since
however $0$ is, trivially, a fixed point, we conclude that $\varphi\equiv 0$
on $]\\!-\\!\infty,T]$. Standard ODE theory implies then that $\varphi$
vanishes identically on the whole $\mathbb{R}$. ∎

###  4.6. Proof of Proposition 4.4.1

We start by showing the last statement of the proposition, namely:

  * (A)

For $z=\Xi(a)$ and under the assumption that $\lambda_{a}>1$, the unique
$m\geq 1$ such that (4.17) has a nontrivial solution $\psi_{a}\in L^{2}$ is
$m_{a}=\sqrt{\lambda_{a}}$.

Before coming to its proof we also observe that the same argument applies with
$b$ in place of $a$.

First of all observe that, for $z=\Xi(a)$, the equation (4.17), which becomes

$-\frac{d^{2}\varphi}{dt^{2}}+m^{2}\varphi+\frac{A}{\Xi-\Xi(a)}\varphi=0,$
(4.54)

has nontrivial solutions $\varphi\in W^{2,2}_{\text{loc}}\cap
L^{2}(\mathbb{R};\mathbb{C})$ if and only if it has nontrivial solution
$\varphi\in W^{2,2}_{\text{loc}}\cap L^{2}(\mathbb{R};\mathbb{R})$. That the
equation has a nontrivial solution when $m=\sqrt{\lambda_{a}}$ follows from
the classical theory of self-adjoint operators. We therefore only need to show
that the existence of a nontrivial solution is only possible for a single
$m\geq 1$. Arguing by contradiction assume there are two, $1\leq m_{1}<m_{2}$,
and denote by $\psi_{1}$ and $\psi_{2}$ the respective solutions. Then there
is a nontrivial linear combination

$\psi=C_{1}\psi_{1}+C_{2}\psi_{2}$

which vanishes on $a$. Observe that $\psi_{1}$ and $\psi_{2}$ can be
interpreted as eigenfuctions of the self-adjoint operator
$-\frac{d^{2}}{dt^{2}}+\frac{A(t)}{\Xi(t)-\Xi(a)}$ relative to distinct
eigenvalues and they are, therefore, $L^{2}$ orthogonal. Summing the
equations, multiplying by $\psi$ and integrating by parts we achieve

$\underbrace{\int\Big{(}(\psi^{\prime})^{2}+\frac{A}{\Xi-\Xi(a)}\psi^{2}\Big{)}}_{=:I}=-C_{1}^{2}m_{1}^{2}\int\psi_{1}^{2}-C_{2}^{2}m_{2}^{2}\int\psi_{2}^{2}\,.$
(4.55)

Recalling that
$A=\Xi^{\prime\prime}+2\Xi^{\prime}=(\Xi^{\prime}+2\Xi)^{\prime}$, we wish to
integrate by parts the second integrand in the left-hand side. Observe that,
because $\psi$ vanishes on $a$ and $\Xi^{\prime}(a)\neq 0$, the function
$\frac{\psi^{2}}{\Xi-\Xi(a)}$ is in fact continuously differentiable. In
particular we can write

$\int\frac{A}{\Xi-\Xi(a)}\psi^{2}=\int\left(\frac{\Xi^{\prime}+2\Xi}{(\Xi-\Xi(a))^{2}}\Xi^{\prime}\psi^{2}-2\frac{\Xi^{\prime}+2\Xi}{\Xi-\Xi(a)}\psi\psi^{\prime}\right)\,.$

Substituting it into I, we achieve

$\displaystyle I$
$\displaystyle=\int\left(\psi^{\prime}-\frac{\Xi^{\prime}}{\Xi-\Xi(a)}\psi\right)^{2}+\int\left(\frac{2\Xi\Xi^{\prime}\psi^{2}}{(\Xi-\Xi(a))^{2}}-\frac{4\Xi\psi\psi^{\prime}}{\Xi-\Xi(a)}\right)$
$\displaystyle=\int\left(\psi^{\prime}-\frac{\Xi^{\prime}}{\Xi-\Xi(a)}\psi\right)^{2}+2\int\frac{\Xi^{\prime}}{\Xi-\Xi(a)}\psi^{2}\,,$

where to reach the second line we have written the first term in the second
integral as

$-2\,\Xi\frac{d}{dt}\left(\frac{1}{\Xi-\Xi(a)}\right)\psi^{2}$

and integrated it by parts. Again thanks to the fact that $\psi$ vanishes at
$a$ we can write it as $\psi=(\Xi-\Xi(a))\eta$ and hence conclude

$\displaystyle I$ $\displaystyle=\int((\Xi-\Xi(a))\eta^{\prime})^{2}+\int
2(\Xi-\Xi(a))\Xi^{\prime}\eta^{2}$
$\displaystyle=\int((\Xi-\Xi(a))\eta^{\prime})^{2}-2\int(\Xi-\Xi(a))^{2}\eta\eta^{\prime}$
$\displaystyle=\int(\Xi-\Xi(a))^{2}(\eta^{\prime}-\eta)^{2}-\int(\Xi-\Xi(a))^{2}\eta^{2}$
$\displaystyle=\int(\Xi-\Xi(a))^{2}(\eta^{\prime}-\eta)^{2}-\int(C_{1}^{2}\psi_{2}^{2}+C_{2}^{2}\psi_{2}^{2})\,.$

Inserting the latter in (4.55) we conclude

$\int(\Xi-\Xi(a))^{2}(\eta^{\prime}-\eta)^{2}=-C_{1}^{2}(m_{1}^{2}-1)\int\psi_{1}^{2}-C_{2}^{2}(m_{2}^{2}-1)\int\psi_{2}^{2}\,.$

Observe that, since $m_{2}>1$ and $\psi_{2}$ is nontrivial, we conclude that
$C_{2}=0$. This would then imply that $\psi=C_{1}\psi_{1}$ and we can thus
assume $C_{1}=1$ in all our computations. In particular $\eta^{\prime}=\eta$,
which implies $\eta(t)=Ce^{t}$. We can now write
$\psi_{1}(t)=(\Xi(t)-\Xi(a))\eta(t)$ and given the properties of $\Xi(t)$ we
easily see that this would violate the decay at $+\infty$ that we know for
$\psi_{1}$ from Corollary 4.5.3.

######  Remark 4.6.1.

We record here a consequence of the above argument: a nontrivial solution
$\varphi$ of (4.54) necessarily satisfies $\varphi(a)\neq 0$ (and thus it must
be unique up to constant factors).

We next show that

  * (B)

If $(m_{0},z)\in\overline{\mathscr{P}}$, $m_{0}\geq 1$ and $z\in\mathbb{R}$,
then $z$ is in the closure of the range of $\Xi$.

We again argue by contradiction and assume the existence of

  * (i)

A sequence $\\{m_{j}\\}\subset\,]1,\infty[$ converging to
$m_{0}\in[1,\infty[$;

  * (ii)

A sequence $\\{z_{j}\\}\subset\mathbb{C}$ with $\textrm{Im}\,z_{j}>0$
converging to $z\in\mathbb{R}\setminus\overline{\Xi(\mathbb{R})}$;

  * (iii)

A sequence $\psi_{j}$ of nontrivial solutions of

$-\frac{d^{2}\psi_{j}}{dt^{2}}+m^{2}\psi_{j}+\frac{A}{\Xi-z_{j}}\psi_{j}=0$
(4.56)

By Corollary 4.5.3 we can normalize our functions $\psi_{j}$ so that
$\psi_{j}(t)e^{-m_{j}t}\to 1$ as $t\to-\infty$ and $\psi_{j}(t)e^{m_{j}t}\to
C_{j}\neq 0$ as $t\to\infty$. Observe also that there is a positive constant
$c_{0}$ such that $|\Xi-z_{j}|\geq c_{0}$ for all $j$ sufficiently large,
thanks to (ii). In particular, the functions $\frac{A}{\Xi-z_{j}}$ are
uniformly bounded in $L^{1}$. By Lemma 4.5.2 there is a positive $T_{0}\geq
b+1$, independent of $j$ such that

$\left|\psi_{j}(t)-C_{j}e^{-m_{j}t}\right|\leq\frac{C_{j}}{2}e^{-m_{j}t}\qquad\forall
t\geq T_{0}\,,$ (4.57)

and there is a constant $C$, independent of $j$ such that

$\|\psi_{j}\|_{L^{\infty}([a,b])}\leq C\,.$ (4.58)

Next multiply (4.56) by $\bar{\psi}_{j}$, integrate in $t$ and take the
imaginary part of the resulting equality to conclude

$\int\frac{A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}|\psi_{j}|^{2}=0\,.$
(4.59)

We might break the integral into three integrals on the regions
$]\\!-\\!\infty,a[$, $]a,b[$, and $]b,\infty[$, where the function $A$ is,
respectively, negative, positive, and negative. This gives

$-\int_{T_{0}}^{2T_{0}}\frac{A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}|\psi_{j}|^{2}\leq\int_{a}^{b}\frac{A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}|\psi_{j}|^{2}$

Now, the right-hand side of the inequality can be bounded uniformly
independently of $j$ by (4.58) and (ii). On the other hand the function
$\frac{-A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}z_{j})^{2}}$ is larger
than a positive constant $c$ independent of $j$ on $[T_{0},2T_{0}]$. Using
(4.57) we can achieve a uniform bound $|C_{j}|\leq C$ for the constants
$C_{j}$. The latter bound, combined with the estimates of Lemma 4.5.2 and the
uniform bound on $\|\frac{A}{\Xi-z_{j}}\|_{L^{1}}$ easily imply that
$\psi_{j}$ is precompact in $L^{2}$. We can thus extract a subsequence, not
relabeled, converging to a nontrivial $L^{2}$ solution $\psi$ of

$-\frac{d^{2}\psi}{dt^{2}}+m_{0}^{2}\psi+\frac{A}{\Xi-z}\psi=0\,.$ (4.60)

Without loss of generality we assume that $\psi$ is real valued, since $z$ is
real. We can thus multiply (4.60) by $\psi$ and integrate to achieve

$\int((\psi^{\prime})^{2}+m_{0}^{2}\psi^{2})+\int\frac{\Xi^{\prime\prime}+2\Xi^{\prime}}{\Xi-z}\psi^{2}=0\,.$

Integrating by parts $\int\frac{\Xi^{\prime\prime}}{\Xi-z}\psi^{2}$ we find

$\int((\psi^{\prime})^{2}+m_{0}^{2}\psi^{2})+\int\left(\frac{(\Xi^{\prime})^{2}}{(\Xi-z)^{2}}\psi^{2}-2\frac{\Xi^{\prime}}{\Xi-z}\psi^{\prime}\psi\right)+\int\frac{2\Xi^{\prime}}{\Xi-z}\psi^{2}=0\,,$

which we can rewrite as

$\int\Big{(}\Big{(}\psi^{\prime}-\frac{\Xi^{\prime}}{\Xi-z}\psi\Big{)}^{2}+m_{0}^{2}\psi^{2}\Big{)}+2\int\frac{\Xi^{\prime}}{\Xi-z}\psi^{2}=0\,.$
(4.61)

As already done in the previous paragraphs we set $\eta=\frac{\psi}{\Xi-z}$
and write the identity as

$\int\left((\Xi-z)^{2}(\eta^{\prime})^{2}+m_{0}^{2}(\Xi-z)^{2}\eta^{2}+2\Xi^{\prime}(\Xi-z)\eta^{2}\right)=0$

Integrating by parts the last term we find

$\int(\Xi-z)^{2}(\eta^{\prime}-\eta)^{2}+\int(m_{0}^{2}-1)(\Xi-z)^{2}\eta^{2}=0\,.$

We thus conclude that $m_{0}=1$ and $\eta^{\prime}=\eta$, i.e.
$\eta(t)=Ce^{t}$, but again we see that this would violate $\psi\in L^{2}$.

We next employ a suitable variation of the latter argument to show that

  * (C)

$(m_{0},0)$ and $(m_{0},\Xi(-\infty))$ do not belong to
$\overline{\mathscr{P}}$ if $m_{0}\geq 1$.

We again argue by contradiction and assume the existence of

  * (i)

A sequence $\\{m_{j}\\}\subset\,]1,\infty[$ converging to
$m_{0}\in[1,\infty[$;

  * (ii)

A sequence $\\{z_{j}\\}\subset\mathbb{C}$ with $\textrm{Im}\,z_{j}>0$
converging to $0$ or to $\Xi(-\infty)$;

  * (iii)

A sequence $\psi_{j}$ of nontrivial solutions of

$-\frac{d^{2}\psi_{j}}{dt^{2}}+m_{j}^{2}\psi_{j}+\frac{A}{\Xi-
z_{j}}\psi_{j}=0\,.$ (4.62)

We first focus on the case $z_{j}\to 0$. Normalize again the solutions so that
$\psi_{j}(t)$ is asymptotic to $e^{m_{j}t}$ for $t$ negative, and to
$C_{j}e^{-m_{j}t}$ for $t$ positive.

Observe that in this case we have $\frac{A}{\Xi}\in L^{1}(]-\infty,N])$ for
every $N$, while $\frac{A}{\Xi-z_{j}}$ enjoys a uniform $L^{1}$ bound on any
$]-\infty,N]$. We can thus apply Lemma 4.5.2 and conclude the $\psi_{j}$ can
be assumed to converge uniformly to a function $\psi$ on $]-\infty,N]$ for
every $N$ and that likewise $\psi(t)$ is asymptotic to $e^{m_{0}t}$ for $t$
negative.

As done previously we multiply the equation (4.62) by $\bar{\psi}_{j}$,
integrate, and take the imaginary part. In particular we gain the inequality

$\int_{b}^{\infty}\frac{A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}|\psi_{j}|^{2}\leq-\int_{a}^{b}\frac{A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}|\psi_{j}|^{2}\,.$

Since $z_{j}\to 0$ and the range of $\Xi$ on $[a,b]$ is bounded away from $0$,
we conclude that the right-hand side is uniformly bounded. In particular,
passing to the limit we conclude that

$\Xi^{-2}A|\psi|^{2}\in L^{1}([b,\infty[)\,.$ (4.63)

Observe however that

$\lim_{t\to\infty}\frac{A(t)}{\Xi(t)}=\lim_{t\to\infty}\frac{-{\bar{\alpha}}e^{-{\bar{\alpha}}t}}{c_{1}e^{-2t}+\frac{1}{2-{\bar{\alpha}}}e^{-{\bar{\alpha}}t}}=-{\bar{\alpha}}(2-{\bar{\alpha}})\,.$

In particular we conclude that $\psi\in L^{2}$. Moreover, we can write

$\frac{A}{\Xi}=-{\bar{\alpha}}(2-{\bar{\alpha}})+B$

for a function $B$ which belongs to $L^{1}([T,\infty[)$ for every $T$. We thus
have that

$-\frac{d^{2}\psi}{dt^{2}}+(m_{0}^{2}-{\bar{\alpha}}(2-{\bar{\alpha}}))\psi+B\psi=0\,.$

Recalling that $0<{\bar{\alpha}}<1$ and $m_{0}\geq 1$, we have
$m_{0}^{2}-{\bar{\alpha}}(2-{\bar{\alpha}})>0$ and we can therefore apply
Lemma 4.5.2 to conclude that, for
$\bar{m}:=\sqrt{m_{0}^{2}-{\bar{\alpha}}(2-{\bar{\alpha}})}$

$\lim_{t\to\infty}e^{\bar{m}t}\psi(t)$

exists, it is finite, and nonzero. Observe however that (4.63) forces
$e^{{\bar{\alpha}}t}|\psi|^{2}\in L^{1}$, which in particular implies that
$\bar{m}>\frac{{\bar{\alpha}}}{2}$ We next argue as in the derivation of
(4.61) to get

$\int\Big{(}\Big{(}\psi^{\prime}-\frac{\Xi^{\prime}}{\Xi}\psi\Big{)}^{2}+m_{0}^{2}\psi^{2}\Big{)}+2\int\frac{\Xi^{\prime}}{\Xi}\psi^{2}=0\,.$

We again set $\psi=\Xi\eta$ and observe that, by our considerations, $\eta$
decays exponentially at $-\infty$, while it is asymptotic to
$e^{({\bar{\alpha}}-\bar{m})t}$ at $+\infty$. We rewrite the latter identity
as

$\int(\Xi^{2}(\eta^{\prime})^{2}+m_{0}^{2}\Xi^{2}\eta^{2}+2\Xi\Xi^{\prime}\eta^{2})=0\,.$

We wish to integrate by parts the latter term to find

$\int(\Xi^{2}(\eta^{\prime}-\eta)^{2}+(m_{0}^{2}-1)\Xi^{2}\eta^{2})=0\,.$
(4.64)

Since we have exponential decay of $\eta$ at $-\infty$, while at $+\infty$
$\eta$ might grow, the latter integration by parts need some careful
justification. First of all we notice that $\Xi\Xi^{\prime}\eta^{2}$ decays
exponentially at $+\infty$ and thus, since the other two integrands are
positive, we can write

$\int(\Xi^{2}(\eta^{\prime})^{2}+m_{0}^{2}\Xi^{2}\eta^{2}+2\Xi\Xi^{\prime}\eta^{2})=\lim_{N\to\infty}\int_{-\infty}^{N}(\Xi^{2}(\eta^{\prime})^{2}+m_{0}^{2}\Xi^{2}\eta^{2}+2\Xi\Xi^{\prime}\eta^{2})\,.$

Next, we can integrate by parts the third integrand (before passing to the
limit) to write

$\displaystyle\int_{-\infty}^{N}(\Xi^{2}(\eta^{\prime})^{2}+m_{0}^{2}\Xi^{2}\eta^{2}+2\Xi\Xi^{\prime}\eta^{2})$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{N}(\Xi^{2}(\eta^{\prime}-\eta)^{2}+(m_{0}^{2}-1)\Xi^{2}\eta^{2})+\Xi^{2}(N)\eta^{2}(N)\,.$

Since $\Xi(N)\eta(N)$ converges to $0$ exponentially, passing into the limit
we conclude (4.64). As before this would imply $m_{0}=1$ and $\eta(t)=Ce^{t}$
at $+\infty$, while we have have already argued that $\eta$ is asymptotic to
$e^{(\bar{\alpha}-\bar{m})t}$, which is a contradiction because
$\bar{\alpha}-\bar{m}\leq\bar{\alpha}<1$.

We next tackle the case $z_{j}\to\Xi(-\infty)$. This time we observe that
$\frac{A}{\Xi-z_{j}}$ enjoys a uniform $L^{1}$ bound on $[T,\infty[$ for every
$T$ and we thus normalize the functions $\psi_{j}$ so that $\psi_{j}(t)$ is
asymptotic to $e^{-m_{j}t}$ for $t\to\infty$. Arguing as above, we assume that
$\psi_{j}$ converges uniformly on all $[T,\infty[$ to a $\psi$ which is
asymptotic to $e^{-m_{0}t}$ and solves

$-\frac{d^{2}\psi}{dt^{2}}+m_{0}^{2}\psi+\frac{A}{\Xi-\Xi(-\infty)}\psi=0\,.$
(4.65)

As above we can assume that $\psi$ is real valued. Moreover, this time we
infer (with the same method used to prove (4.63))

$(\Xi-\Xi(-\infty))^{-2}A\psi^{2}\in L^{1}(\mathbb{R})$ (4.66)

This time observe that, for $t$ sufficiently negative,
$\frac{A(t)}{\Xi(t)-\Xi(-\infty)}=8$. In particular we can explicitly solve
the equation as

$\psi(t)=C_{1}e^{-t\sqrt{m_{0}^{2}+8}}+C_{2}e^{t\sqrt{m_{0}^{2}+8}}$

when $t$ is sufficiently negative. However, if $C_{1}$ were positive, (4.66)
would not hold. In particular we infer exponential decay at $-\infty$. We can
now argue as for the case $z_{j}\to 0$: we multiply (4.65) by $\psi$,
integrate in time and perform an integration by part to infer

$\int\Big{(}\Big{(}\psi^{\prime}-\frac{\Xi^{\prime}}{\Xi-\Xi(-\infty)}\psi\Big{)}^{2}+m_{0}\psi^{2}\Big{)}+2\int\frac{\Xi^{\prime}}{\Xi-\Xi(-\infty)}\psi^{2}=0\,.$

We then introduce $\eta$ so that $\psi=(\Xi-\Xi(-\infty))\eta$. This time we
infer exponential decay for $\eta$ at both $\infty$ and $-\infty$. Arguing as
above we rewrite the last identity as

$\int((\Xi-\Xi(-\infty))^{2}(\eta^{\prime}-\eta)^{2}+(m_{0}^{2}-1)(\Xi-\Xi(-\infty))^{2}\eta^{2})=0\,,$

reaching again a contradiction.

In order to complete the proof of the proposition we need to show

  * (D)

If $(m_{0},\Xi(c))\in\overline{\mathscr{P}}$ and $m_{0}>1$, then either $c=a$
or $c=b$ and moreover we have, respectively, $m_{0}=\sqrt{\lambda_{a}}$ or
$m_{0}=\sqrt{\lambda_{b}}$.

As before we argue by contradiction and assume the existence of

  * (i)

A sequence $\\{m_{j}\\}\subset\,]1,\infty[$ converging to
$m_{0}\in\,]1,\infty[$;

  * (ii)

A sequence $\\{z_{j}\\}\subset\mathbb{C}$ with $\textrm{Im}\,z_{j}>0$
converging to $\Xi(c)$ for some $c\not\in\\{a,b\\}$;

  * (iii)

A sequence $\psi_{j}$ of nontrivial solutions of

$-\frac{d^{2}\psi_{j}}{dt^{2}}+m_{j}^{2}\psi_{j}+\frac{A}{\Xi-
z_{j}}\psi_{j}=0\,.$ (4.67)

This time we normalize the $\psi_{j}$’s so that

$\int(|\psi_{j}^{\prime}|^{2}+m_{j}^{2}|\psi_{j}|^{2})=1\,.$ (4.68)

By Lemma 4.5.2 we know that $\psi_{j}(t)$ is asymptotic to
$\rho_{j}^{\pm}e^{\mp m_{j}t}$ for $t\to\pm\infty$, where
$\rho_{j}^{\pm}\in\mathbb{C}\setminus\\{0\\}$. Since $\Xi(c)$ has a positive
distance from both $0$ and $\Xi(-\infty)$, we can apply Lemma 4.5.2 to achieve
uniform times $T_{\pm}$ with the properties that

$\displaystyle\left|\psi_{j}(t)-\rho_{j}^{+}e^{-m_{j}t}\right|$
$\displaystyle\leq\frac{|\rho_{j}^{+}|}{2}e^{-m_{j}t}\qquad\qquad\forall t\geq
T_{+}\,,$ (4.69)
$\displaystyle\left|\psi_{j}(t)-\rho_{j}^{-}e^{m_{j}t}\right|$
$\displaystyle\leq\frac{|\rho_{j}^{-}|}{2}e^{m_{j}t}\qquad\qquad\forall t\leq
T_{-}\,.$ (4.70)

Combining the latter inequalities with (4.68) we conclude that
$\sup_{j}|\rho_{j}^{\pm}|<\infty$, and in particular $\\{\psi_{j}\\}_{j}$ is
tight in $L^{2}$, i.e. for every $\varepsilon>0$ there is $N=N(\varepsilon)$
such that

$\sup_{j}\int_{|t|\geq N}|\psi_{j}|^{2}<\varepsilon\,.$

The latter bound combined with (4.68) implies, up to extraction of a
subsequence which we do not relabel, the strong $L^{2}$ convergence of
$\psi_{j}$ to a function $\psi$. Thanks to Sobolev embedding, the convergence
is uniform on any compact set and, moreover, $\psi\in C^{1/2}$.

Arguing as for (4.59) we infer

$\int\frac{A}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}|\psi_{j}|^{2}=0$
(4.71)

The latter bound implies $\psi(c)=0$. In fact first we observe that
$\frac{A}{|\Xi-z_{j}|^{2}}|\psi_{j}|^{2}$ converges in $L^{1}$ on
$\mathbb{R}\setminus]c-\delta,c+\delta[$ for every $\delta$. Choosing
$\delta>0$ so that $|A(t)-A(c)|\leq\frac{|A(c)|}{2}$ for
$t\in[c-\delta,c+\delta]$ and recalling that $|A(c)|>0$, we easily infer that

$\sup_{j}\int_{c-h}^{c+h}\frac{|\psi_{j}|^{2}}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}<\infty\qquad\forall
h<\delta\,.$

If $\psi(c)$ were different from $0$, we can select a positive $h<\delta$ and
a positive $c_{0}$ with the property that $|\psi(t)|^{2}\geq 2c_{0}$ for all
$t\in[c-h,c+h]$. In particular, for a large enough $j$ we infer
$|\psi_{j}(t)|^{2}\geq c$ for all $t\in[c-\delta,c+\delta]$. But then we would
conclude

$\sup_{j}\int_{c-h}^{c+h}\frac{1}{(\Xi-\textrm{Re}\,z_{j})^{2}+(\textrm{Im}\,z_{j})^{2}}<\infty\,.$

Since the denominator converges to $(\Xi-\Xi(c))^{2}$, this is clearly not
possible.

We now wish to pass in the limit in (4.67) to derive that

$-\psi^{\prime\prime}+m_{0}^{2}\psi+\frac{A}{\Xi-\Xi(c)}\psi=0\,,$ (4.72)

where we notice that, thanks to $\psi(c)=0$ and the Hölder regularity of
$\psi$, the function $\frac{A}{\Xi-\Xi(c)}\psi$ is indeed in $L^{p}$ for every
$p<2$. We thus understand the equation distributionally. The equation clearly
passes to the limit outside the singularity $c$ of the denominator and thus we
just need to pass it to the limit distributionally in some interval
$]c-h,c+h[$. We write the third term as

$\displaystyle\frac{A}{\Xi-z_{j}}\psi_{j}$
$\displaystyle=\left(\frac{d}{dt}\ln(\Xi-
z_{j})\right)\frac{A}{\Xi^{\prime}}\psi_{j}$
$\displaystyle=\frac{d}{dt}\left(\ln(\Xi-
z_{j})\frac{A}{\Xi^{\prime}}\psi_{j}\right)-\ln(\Xi-
z_{j})\frac{A}{\Xi^{\prime}}\psi^{\prime}_{j}-\ln(\Xi-
z_{j})\frac{d}{dt}\left(\frac{A}{\Xi^{\prime}}\right)\psi_{j}\,.$

Observe that we can define the logarithm unequivocally because $\Xi$ is real
valued and $\textrm{Im}\,z_{j}>0$.

Next, we remark that:

  * (i)

$\frac{A}{\Xi^{\prime}}$ is smooth in $]c-h,c+h[$;

  * (ii)

$\ln(\Xi-z_{j})$ converges strongly 111Since $\ln(\Xi-z_{j})$ converges
uniformly to $\ln(\Xi-\Xi(c))$ on any compact set which does not contain $c$,
in order to reach the conclusion it suffices to prove a uniform $L^{q}$ bound
on the functions, for every $q<\infty$. This can be easily concluded as
follows. Choose an interval $[c-h,c+h]$ and recall that $\Xi$ does not change
sign on it. For each $j$ large enough we then find a unique
$c_{j}\in[c-h,c+h]$ such that $\Xi(c_{j})=\textrm{Re}\,z_{j}$. Using the mean
value theorem we easily conclude that
$|\Xi(t)-z_{j}|\geq|\Xi(t)-\Xi(c_{j})|\geq C^{-1}|t-c_{j}|$ for every
$t\in[c-h,c+h]$, where $C^{-1}=\min\\{|\Xi^{\prime}(t)|:c-h\leq t\leq c+h\\}$.
to $\ln(\Xi-\Xi(c))$ in $L^{q}(]c-h,c+h[)$ for every $q<\infty$;

  * (iii)

$\psi_{j}^{\prime}\to\psi^{\prime}$ weakly in $L^{2}$, while $\psi_{j}\to\psi$
uniformly.

We thus conclude that $\frac{A}{\Xi-z_{j}}\psi_{j}$ converges distributionally
to

$\frac{d}{dt}\left(\ln(\Xi-\Xi(c))\frac{A}{\Xi^{\prime}}\psi\right)-\ln(\Xi-\Xi(c))\frac{A}{\Xi^{\prime}}\psi^{\prime}-\ln(\Xi-\Xi(c))\frac{d}{dt}\left(\frac{A}{\Xi^{\prime}}\right)\psi\,.$

Using now that $\psi\in W^{1,2}$ and $\psi(c)=0$ we can rewrite the latter
distribution as

$\frac{A}{\Xi-\Xi(c)}\psi$

and hence conclude the validity of (4.72).

Observe next that from (4.72) we infer $\psi^{\prime\prime}\in L^{p}$ for
every $p<2$, which in turn implies that $\psi$ is indeed
$C^{1,\kappa}_{\text{loc}}$ for every $\kappa<\frac{1}{2}$. In turn this
implies that $\frac{A}{\Xi-\Xi(c)}\psi$ is continuous at $c$, so that in
particular $\psi$ is twice differentiable. We thus can argue as for the
derivation of (4.61) and get

$\int\Big{(}\Big{(}\psi^{\prime}-\frac{\Xi^{\prime}}{\Xi-\Xi(c)}\psi\Big{)}^{2}+m_{0}^{2}\psi^{2}\Big{)}+2\int\frac{\Xi^{\prime}}{\Xi-\Xi(c)}\psi^{2}=0\,.$
(4.73)

Once again we can set $\psi=(\Xi-\Xi(c))\eta$ and observe that $\eta\in
W^{1,2}$, to rewrite the latter identity as

$\int((\Xi-\Xi(c))^{2}(\eta^{\prime}-\eta)^{2}+(m_{0}^{2}-1)(\Xi-\Xi(c))^{2}\eta^{2})=0\,,$

inferring that $\eta=0$.

We thus have concluded that $\psi$ vanishes identically, but this is not yet a
contradiction since the normalization (4.68) and the strong $L^{2}$
convergence does not ensure that $\psi$ is nontrivial. In order to complete
our argument, note first that, by the monotonicity of $\Xi$, for each $j$
large enough there is a unique $c_{j}$ such that
$\Xi(c_{j})=\textrm{Re}\,z_{j}$. We then multiply the equation (4.67) by
$\bar{\psi}_{j}-\overline{\psi_{j}(c_{j})}$ to obtain

$\int\Big{(}|\psi_{j}^{\prime}|^{2}+m_{j}^{2}\psi_{j}(\bar{\psi}_{j}-\overline{\psi_{j}(c_{j})})+\frac{A}{\Xi-
z_{j}}\psi_{j}(\bar{\psi}_{j}-\overline{\psi_{j}(c_{j})})\Big{)}=0\,.$

Note that $c_{j}$ must converge to $c$ and that the integrals

$\int\psi_{j}(\bar{\psi}_{j}-\overline{\psi_{j}(c_{j})})$

converges to $0$ because $\psi_{j}-\psi_{j}(c_{j})$ converges to $0$ uniformly
and, thanks to the uniform exponential decay of $\psi_{j}$, the latter are
uniformly bounded in $L^{1}$. For the same reason the first integral in the
sum

$\int_{|t-c|\geq h}\frac{A}{\Xi-
z_{j}}\psi_{j}(\bar{\psi}_{j}-\overline{\psi_{j}(c_{j})})+\int_{|t-c|\leq
h}\frac{A}{\Xi-z_{j}}\psi_{j}(\bar{\psi}_{j}-\overline{\psi_{j}(c_{j})})$
(4.74)

converges to $0$ for every fixed $h$. On the other hand,
$|\frac{A(t)}{\Xi(t)-z_{j}}||\psi_{j}(t)-\psi_{j}(c_{j})|\leq
C|t-c_{j}|^{-1/2}$ and thus the second integrand in (4.74) converges to $0$ as
well. We thus conclude that the $L^{2}$ norm of $\psi^{\prime}_{j}$ converges
to $0$ as well. This however contradicts the normalization (4.68).

###  4.7. Proof of Proposition 4.4.3: Part I

We set $m_{0}=m_{a}$, $z_{0}=\Xi(a)$, and we fix a $\psi_{0}$ solution of

$-\frac{d^{2}\psi_{0}}{dt^{2}}+m_{0}^{2}\psi_{0}+\frac{A}{\Xi-
z_{0}}\psi_{0}=0$

with $L^{2}$ norm equal $1$. Since the operator is self-adjoint we will indeed
assume that $\psi_{0}$ is real. We then define the projector
$P_{0}:L^{2}(\mathbb{R};\mathbb{C})\to\\{\kappa\psi_{0}:\kappa\in\mathbb{C}\\}$
as

$P_{0}(\psi)=\langle\psi,\psi_{0}\rangle\psi_{0}\,.$

Observe that $P_{0}$ is self-adjoint. Next, in a neighborhood of
$(m_{0},z_{0})$ we will look for solutions of (4.17) by solving

$\left\\{\begin{array}[]{l}-\psi^{\prime\prime}+m^{2}\psi+\frac{A}{\Xi-z}\psi+P_{0}(\psi)=\psi_{0}\\\
\\\ \langle\psi,\psi_{0}\rangle=1\end{array}\right.$ (4.75)

which we can rewrite as

$\left\\{\begin{array}[]{l}-\psi^{\prime\prime}+m_{0}^{2}\psi+\frac{A}{\Xi-
z_{0}}\psi+P_{0}(\psi)=A\left(((\Xi-z_{0})^{-1}-(\Xi-z)^{-1})\psi\right)\\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+(m_{0}^{2}-m^{2})\psi+\psi_{0}\\\
\\\ \langle\psi,\psi_{0}\rangle=1\end{array}\right.$ (4.76)

Next we observe that the operator $-\frac{d^{2}}{dt^{2}}+m_{0}^{2}$,
considered as a closed unbounded self-adjoint operator in $L^{2}$ (with domain
$W^{2,2}$) has an inverse $\mathcal{K}_{m_{0}}\colon L^{2}\to L^{2}$ which is
a bounded operator. We thus rewrite (4.76) as

$\left\\{\begin{array}[]{ll}\underbrace{\psi+\mathcal{K}_{m_{0}}\left(\frac{A}{\Xi-
z_{0}}\psi+P_{0}(\psi)\right)}_{=:T(\psi)}\\\
\qquad\qquad=\underbrace{\mathcal{K}_{m_{0}}\left(\left(A\left((\Xi-
z_{0})^{-1}-(\Xi-z)^{-1}\right)+(m_{0}^{2}-m^{2})\right)\psi\right)}_{=:-\mathcal{R}_{m,z}(\psi)}+\mathcal{K}_{m_{0}}(\psi_{0})\\\
\\\ \langle\psi,\psi_{0}\rangle=1\,.\end{array}\right.$ (4.77)

The proof of Proposition 4.4.3 will then be broken into two pieces. In this
section we will show the first part, which we can summarize in the following

######  Lemma 4.7.1.

For every $\mu>0$, if $(m,z)$ sufficiently close to $(m_{0},z_{0})$ and
$\textrm{Im}\,z\geq\mu|\textrm{Re}\,(z-z_{0})|$ then there is a unique
$\psi=\psi(m,z)\in L^{2}(\mathbb{R})$ solving

$T(\psi)+\mathcal{R}_{m,z}(\psi)=\mathcal{K}_{m_{0}}(\psi_{0})\,.$ (4.78)

We recognize this as, essentially, the first step in the Lyapunov-Schmidt
reduction but written for the solution $\psi$ rather than the perturbation
$\tilde{\psi}=\psi-\psi_{0}$. (The second step will be to the solve one-
dimensional equation (4.86) enforcing $\langle\psi,\psi_{0}\rangle=1$.)

Before coming to the proof of Lemma 4.7.1, we single out two important
ingredients.

######  Lemma 4.7.2.

$T$ is a bounded operator with bounded inverse on the spaces $L^{2}$ and
$C^{\sigma}$, for any $\sigma\in]0,1[$.

###### Proof.

Recall that the operator $\mathcal{K}_{m}$ is given by the convolution with
$\frac{1}{2m}e^{-m|\cdot|}$. Observe that $\mathcal{K}_{m}$ is well-defined on
$C^{\sigma}$ and so is the multiplication by $\frac{A}{\Xi-z_{0}}$, since the
latter is a smooth functions with bounded derivatives, and the operator
$P_{0}(\psi)=\langle\psi,\psi_{0}\rangle\psi_{0}$: for the latter we just need
to check that $\psi\overline{\psi_{0}}$ is integrable, which follows from the
exponential decay of $\psi_{0}$, cf. Corollary 4.5.3.

In this first step we prove that $T$ is a bounded operator with bounded
inverse in the spaces $L^{2}(\mathbb{R})$ and $C^{\sigma}(\mathbb{R})$.

Recall that $\frac{A}{\Xi-z_{0}}=\frac{A}{\Xi-\Xi(a)}$ is indeed a bounded
smooth function (thanks to the structural assumptions on $\Xi$: in particular
recall that $\Xi^{\prime}(a)\neq 0$ and $A(a)=0$, which implies that
$\frac{A}{\Xi-\Xi(a)}$ is in fact smooth at $a$). Moreover the function and
its derivatives decay exponentially at $\pm\infty$. It follows therefore that
$\psi\mapsto\mathcal{K}_{m_{0}}(\frac{A}{\Xi-z_{0}}\psi+P_{0}(\psi))$ is a
compact operator, both on $L^{2}$ and on $C^{\sigma}$. Thus $T$ is a Fredholm
operator with index $0$. We thus just need to check that the kernel is $0$ in
order to conclude that it is invertible with bounded inverse. In both cases we
need to show that the equation

$-\frac{d^{2}\psi}{dt^{2}}+m_{0}^{2}\psi+\frac{A}{\Xi-\Xi(a)}\psi+P_{0}(\psi)=0$
(4.79)

has only the trivial solution. Observe that the kernel $V$ of the operator
$\psi\mapsto-\frac{d^{2}\psi}{dt^{2}}+m_{0}^{2}\psi+\frac{A}{\Xi-\Xi(a)}\psi$
is $1$-dimensional by Lemma 4.5.2 and Corollary 4.5.3. In particular $V$ is
generated by $\psi_{0}$. Since the operator $P_{0}$ is the orthogonal
projection onto $V$ and $-\frac{d^{2}}{dt^{2}}+m_{0}^{2}+\frac{A}{\Xi-\Xi(a)}$
is self-adjoint, the kernel of
$-\frac{d^{2}}{dt^{2}}+m_{0}^{2}+\frac{A}{\Xi-\Xi(a)}+P_{0}$ in $L^{2}$ must
be trivial.

In order to argue that the kernel is $0$ on $C^{\sigma}$ we apply a variation
of the same idea: first we observe that if $\psi$ is a $C^{\sigma}$ solution
of (4.79), then $\frac{A}{\Xi-\Xi(a)}\psi+P_{0}(\psi)$ is also in $C^{\sigma}$
and hence $\psi^{\prime\prime}\in C^{\sigma}$. Observe also that the operator
is self-adjoint and thus we can assume that $\psi$ is real-valued. We then
multiply both sides of (4.79) by $\bar{\psi}_{0}$, integrate by parts and use
the fact that $\psi_{0}$ is in the kernel of the self-adjoint operator
$-\frac{d^{2}}{dt^{2}}+m_{0}^{2}+\frac{A}{\Xi-\Xi(a)}$ to conclude that
$(\langle\psi,\psi_{0}\rangle)^{2}=0$. But then $\psi$ is a bounded solution
of $-\frac{d\psi}{dt^{2}}+m_{0}^{2}\psi+\frac{A}{\Xi-\Xi(a)}\psi=0$. Given
that $\frac{A}{\Xi-\Xi(a)}\psi$ is a product of an exponentially decaying
function and a bounded function, we conclude that
$-\frac{d^{2}\psi}{dt^{2}}+m_{0}^{2}\psi$ is an exponentially decaying
function $f$. We thus have
$\psi=\mathcal{K}_{m}(f)+C_{1}e^{-m_{0}t}+C_{2}e^{m_{0}t}$ for two constants
$C_{1}$ and $C_{2}$. However $\mathcal{K}_{m}(f)$ decays exponentially at both
$\pm\infty$ and thus, given that $\psi$ is bounded, we must have
$C_{1}=C_{2}=0$. In particular $\psi$ decays exponentially at both $\pm\infty$
and so it is an $L^{2}$ function. But we already saw that every $L^{2}$
solution is trivial. ∎

######  Lemma 4.7.3.

For every constant $\mu>0$ we define the cone
$C_{\mu}:=\\{z:\textrm{Im}z\geq\mu|\textrm{Re}\,(z-z_{0})|\\}$. Then

$\lim_{z\in C_{\mu},(m,z)\to(m_{0},z_{0})}\|\mathcal{R}_{m,z}\|_{O}=0\,,$
(4.80)

where $\|L\|_{O}$ is the operator norm of $L$ when considered as a bounded
operator from $L^{2}$ to $L^{2}$.

###### Proof.

Clearly, it suffices to show that

$\lim_{z\in C_{\mu},z\to z_{0}}\|\mathcal{K}_{m_{0}}\circ(A/(\Xi-z)-A/(\Xi-
z_{0}))\|_{O}=0\,.$ (4.81)

We can rewrite the operator as

$\psi\mapsto\mathcal{K}_{m_{0}}\Big{(}\frac{A(z-z_{0})}{(\Xi-z)(\Xi-
z_{0})}\psi\Big{)}\,.$

First of all observe that the operators

$\psi\mapsto L_{z}(\psi)=\frac{A(z-z_{0})}{(\Xi-z)(\Xi-z_{0})}\psi$

are bounded in the operator norm uniformly in $z\in C_{\mu}$ by a constant
$M$. Moreover, we can see that the adjoint operator is given by
$L_{z}^{*}(\psi)=\frac{A(\bar{z}-z_{0})}{(\Xi-\bar{z})(\Xi-z_{0})}\psi$
converges strongly in $L^{2}$ to $0$: indeed the functions
$\frac{A(\bar{z}-z_{0})}{(\Xi-\bar{z})(\Xi-z_{0})}$ are uniformly bounded and
they converge to $0$ on $\mathbb{R}\setminus\\{a\\}$. We now use an argument
entirely similar to that used in the proof of Lemma 3.3.2: given any
$\varepsilon>0$ we fix the orthogonal projection $P_{N}$ onto a finite-
dimensional subspace $V$ of $L^{2}$ with the property that
$\|\mathcal{K}_{m_{0}}\circ P_{N}-\mathcal{K}_{m_{0}}\|_{O}$ is smaller than
$\frac{\varepsilon}{2M}$. We then argue that for $|z-z_{0}|$ sufficiently
small $P_{N}\circ L_{z}$ has operator norm smaller than
$\frac{\varepsilon}{2}$. Having chosen an orthonormal base
$\psi_{1},\ldots,\psi_{N}$ for $V$, we recall that

$P_{N}(\psi)=\sum_{i}\langle\psi_{i},\psi\rangle\psi_{i}\,$

(where $\langle\cdot,\cdot\rangle$ denotes the scalar product in $L^{2}$).
Therefore our claim amounts to show that

$|\langle\psi_{i},L_{z}(\psi)\rangle|\leq\frac{\varepsilon}{2N}$

for $z$ sufficiently close to $z_{0}$ and every $\psi$ with
$\|\psi\|_{L^{2}}\leq 1$. For the latter we use

$|\langle\psi_{i},L_{z}(\psi)\rangle|=|\langle
L_{z}^{*}(\psi_{i}),\psi\rangle|\leq\|L_{z}^{*}(\psi_{i})\|_{L^{2}}\,.$

∎

###### Proof of Lemma 4.7.1.

We rewrite the equation that we want to solve as

$\psi+T^{-1}\circ\mathcal{R}_{m,z}(\psi)=T^{-1}\circ\mathcal{K}_{m_{0}}(\psi_{0})\,.$

Note that $P_{0}(\psi_{0})=\psi_{0}$. Furthermore, since $\mathcal{K}_{m_{0}}$
is, by definition, the inverse operator of
$-\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}+m_{0}^{2}\operatorname{Id}$,

$\mathcal{K}_{m_{0}}^{-1}\Big{(}\psi_{0}+\mathcal{K}_{m_{0}}\Big{(}\frac{A}{\Xi-
z_{0}}\psi_{0}\Big{)}\Big{)}=-\psi_{0}^{\prime\prime}+m_{0}^{2}\psi_{0}+\frac{A}{\Xi-
z_{0}}\psi_{0}=0.$

Therefore,

$\psi_{0}+\mathcal{K}_{m_{0}}\Big{(}\frac{A}{\Xi-z_{0}}\psi_{0}\Big{)}=0.$

In combination with the definition of $T$ in (4.77), we get

$T(\psi_{0})=\psi_{0}+\mathcal{K}_{m_{0}}\Big{(}\frac{A}{\Xi-
z_{0}}\psi_{0}+\psi_{0}\Big{)}=\mathcal{K}_{m_{0}}(\psi_{0}),$

in other words,

$T^{-1}\circ\mathcal{K}_{m_{0}}(\psi_{0})=\psi_{0}\,.$ (4.82)

Therefore, (4.78) becomes

$(\operatorname{Id}+T^{-1}\circ\mathcal{R}_{m,z})(\psi)=\psi_{0}\,,$ (4.83)

so the existence of a unique solution is guaranteed as soon as
$\|T^{-1}\circ\mathcal{R}_{m,z}\|_{O}<1$. ∎

######  Remark 4.7.4.

In the remaining part of the proof of Proposition 4.4.3 we will take advantage
of the representation of $\psi$ as a function of $\psi_{0}$ through the
Neumann series coming from (4.83). More precisely, our proof of Lemma 4.7.1
leads to the following representation:

$\psi=\psi_{0}-(T^{-1}\circ\mathcal{R}_{m,z})(\psi_{0})+\sum_{k=2}^{\infty}(-1)^{k}(T^{-1}\circ\mathcal{R}_{m,z})^{k}(\psi_{0})\,.$
(4.84)

###  4.8. Proof of Proposition 4.4.3: Part II

We now complete the proof of Proposition 4.4.3. The positive parameter $\mu>0$
in Lemma 4.7.1 will have to be chosen sufficiently small: its choice will be
specified in a few paragraphs, while for the moment we assume it to be fixed.
We set $m_{0}=m_{a}$ and $z_{0}=\Xi(a)$. Thus, for each $(m_{0}+h,z)$ in a set

$U_{\delta,\mu}:=\\{|h|<\delta,|z-z_{0}|<\delta,\textrm{Im}z>\mu|\textrm{Re}\,(z-z_{0})|\\}$

we know that that there is a solution $\psi=\psi(m_{0}+h,z)$ of (4.78) which
moreover satisfies the expansion (4.84). We then define the function

$H(h,z):=\langle\psi(m_{0}+h,z),\psi_{0}\rangle\,,$ (4.85)

and obviously we are looking for those $z$ which solve

$H(h,z)=1\,.$ (4.86)

The main point of our analysis is the following

######  Lemma 4.8.1.

The function $H$ is holomorphic in $z$ and moreover

$H(h,z)=1-2m_{a}h+c(a)(z-z_{0})+o(|z-z_{0}|+|h|)$ (4.87)

where $c(a)$ is a complex number with $\textrm{Im}\,c(a)>0$.

The coefficient $c(a)$ is computed in (4.8) and is, in fact, $1/z_{1}$ from
(4.35) with $\textrm{Im}z_{1}<0$ and $m_{1}=-(2m_{0})^{-1}$. The formal
computation involving the Plemelj formula to obtain (4.35) will appear in the
calculation of $c(a)$ below.

Given Lemma 4.8.1, consider now $\xi(h)$ which we obtain by solving $c(a)(\xi-
z_{0})=2m_{a}h$, namely,

$\xi(h)=\frac{2m_{a}h}{c(a)}+z_{0}=\frac{2m_{a}h}{|c(a)|^{2}}\overline{c(a)}+z_{0}\,.$

The idea behind the latter definition is that, if the term $o(|z-z_{0}|+|h|)$
vanished identically, $z=\xi(h)$ would be the solution of $H(h,z)=1$. Even
though $o(|z-z_{0}|+|h|)$ does not vanish, we nonetheless expect that the
solution $z$ of $H(h,z)=1$ is relatively close to $\xi(h)$.

Since $\textrm{Im}\,c(a)>0$, $\xi(h)$ has positive imaginary part if $h<0$. In
particular we have

$\textrm{Im}\,\xi(h)\geq\gamma|h|\qquad\forall h<0\,.$

where $\gamma$ is a positive constant. We then rewrite

$H(h,z)=1+c(a)(z-\xi(h))+\underbrace{o(|\xi(h)-z_{0}|+h)}_{=:r(h)}+o(|z-\xi(h)|)\,.$

Consider the disk $D_{h}:=\\{|z-\xi(h)|\leq 2\beta|h|\\}$, for a suitably
chosen constant $\beta>0$. We will show below that adjusting the constants
$\mu$ and $\beta$ suitably, the disk will be in the domain of the holomorphic
function $H(\cdot,z)$. Leaving this aside for the moment, by Rouché Theorem,
if we choose $h$ sufficiently small the set $H(h,D_{h})$ contains a disk of
radius $|c(a)|\beta h$ centered at $1+r(h)$. But then for $h$ sufficiently
small we also have $|r(h)|\leq\frac{|c(a)|\beta h}{2}$ and so we conclude that
$1\in H(h,D_{h})$, namely that there is a point $z(h)$ in the disk $D_{h}$
which is mapped in $1$ by $H(h,\cdot)$. This would then complete the proof of
Proposition 4.4.3 if we were able to prove that $\textrm{Im}\,z(h)>0$. We
therefore need to show that $D_{h}$ is in the domain of $H(h,\cdot)$, namely,

$\textrm{Im}\,z\geq\mu|\textrm{Re}\,(z-z_{0})|\,\qquad\forall z\in D_{h}\,.$

We first estimate

$\textrm{Im}\,z\geq\,\textrm{Im}\,\xi(h)-2\beta|h|\geq(\gamma-2\beta)|h|\,.$

Then

$|\textrm{Re}\,z-z_{0}|\leq|\xi(h)-z_{0}|+|z-\xi(h)|\leq\left(\frac{2m_{a}}{|c(a)|}+2\beta\right)|h|\,.$
(4.88)

We thus conclude that

$\textrm{Im}\,z(h)\geq\frac{\gamma-2\beta}{2m_{a}|c(a)|^{-1}+2\beta}|\textrm{Re}\,(z(h)-z_{0})|\,.$
(4.89)

Thus it suffices to choose $\beta=\frac{\gamma}{3}$ and
$\mu=\frac{\gamma}{6m_{a}|c(a)|^{-1}+2\gamma}$. This guarantees at the same
time the existence of a solution and the fact that $z(h)$ has positive
imaginary part when $h<0$ (which results from combining (4.88) and (4.89).

In order to complete the proof of Proposition 4.4.3 we therefore just need to
show Lemma 4.8.1.

###### Proof of Lemma 4.8.1.

In order to show holomorphicity we just need to show that, for each fixed $z$,

$z\mapsto\sum_{k=0}^{\infty}(-T^{-1}\circ\mathcal{R}_{m,z})^{k}$

is holomorphic. Since the series converges in the operator norm, it suffices
to show that each map $z\mapsto(-T^{-1}\circ\mathcal{R}_{m,z})^{k}$ is
holomorphic for every $k$, for which indeed it suffices to show that
$z\mapsto\mathcal{R}_{m,z}$ is holomorphic. This is however obvious from the
explicit formula. We therefore now come to the Taylor expansion (4.87).

Step 1 We will show here that

$\|\mathcal{R}_{m_{0}+h,z}\|_{\mathcal{L}(C^{\sigma})}\leq
C(\sigma)(|h|+|z-z_{0}|)\,$ (4.90)

for every $\sigma\in]0,1[$, where $\|L\|_{\mathcal{L}(C^{\sigma})}$ is the
operator norm of a bounded linear operator $L$ on $C^{\sigma}$. The estimate
will have the following consequence. First of all using (4.84) and
$\|\psi_{0}\|_{L^{2}}^{2}=1$ we expand

$H(h,z)=1-\langle
T^{-1}\circ\mathcal{R}_{m_{0}+h,z}(\psi_{0}),\psi_{0}\rangle+\underbrace{\sum_{k=2}^{\infty}\langle(-T^{-1}\circ\mathcal{R}_{m_{0}+h,z})^{k}(\psi_{0}),\psi_{0}\rangle}_{=:R_{1}(z,h)}\,.$
(4.91)

Hence using (4.90) we estimate

$\displaystyle|R_{1}(z,h)|$
$\displaystyle\leq\sum_{k=2}^{\infty}\|(-T^{-1}\circ\mathcal{R}_{m_{0}+h,z})^{k}(\psi_{0})\|_{\infty}\|\psi_{0}\|_{L^{1}}$
$\displaystyle\leq
C\sum_{k=2}^{\infty}(\|T^{-1}\|_{C^{\sigma}}\|\mathcal{R}_{m_{0}+h,z}\|_{\mathcal{L}(C^{\sigma})})^{k}\|\psi_{0}\|_{C^{\sigma}}\|\psi_{0}\|_{L^{1}}$
$\displaystyle=o(|h|+|z-z_{0}|)\,,$ (4.92)

for some fixed $\sigma$. In order to show (4.90) we write

$\mathcal{R}_{m_{0}+h,z}(\psi)=(z-z_{0})\mathcal{K}_{m_{0}}\Big{(}\frac{1}{\Xi-z}\Big{(}\frac{A}{\Xi-
z_{0}}\psi\Big{)}\Big{)}+(2m_{0}h+h^{2})\mathcal{K}_{m_{0}}(\psi)\,.$

Since $\frac{A}{\Xi-z_{0}}$ is smooth, it suffices to show that the operators
$B_{z}:=\mathcal{K}_{m_{0}}\circ\frac{1}{\Xi-z}$ are uniformly bounded in
$\mathcal{L}(C^{\sigma})$. We first fix a smooth cut-off function $\varphi\in
C^{\infty}_{c}(]a-2,a+2[)$ which equals $1$ on $[a-1,a+1]$ and write

$B_{z}=B_{z}^{1}+B_{z}^{2}:=\mathcal{K}_{m_{0}}\circ\Big{(}\frac{1-\varphi}{\Xi-z}\Big{)}+\mathcal{K}_{m_{0}}\circ\Big{(}\frac{\varphi}{\Xi-z}\Big{)}\,.$

But since $(1-\varphi)/(\Xi-z)$ enjoys a uniform bound in $C^{k}$, it is easy
to conclude that $\|B^{1}_{z}\|_{\mathcal{L}(C^{\sigma})}$ is bounded
uniformly in $z$. We thus need to bound

$\displaystyle B^{2}_{z}(\psi)(t)$ $\displaystyle=\frac{1}{2m_{0}}\int
e^{-m_{0}|t-s|}\frac{\varphi(s)}{\Xi(s)-z}\psi(s)\,ds\,.$

We first bound $\|B^{2}_{z}\|_{L^{\infty}}$. We write $z=x+iy$ and, since $x$
is close to $a$, we select the only $a^{\prime}$ such that $\Xi(a^{\prime})=x$
and write

$\displaystyle B^{2}_{z}(\psi)(t)$
$\displaystyle=\frac{1}{2m_{0}}\underbrace{\int
e^{-m_{0}|t-s|}\frac{\varphi(s)(\psi(s)-\psi(a^{\prime}))}{(\Xi(s)-\Xi(a^{\prime}))-iy}\,ds}_{=:I_{1}(t)}$
$\displaystyle\qquad+\frac{\psi(a^{\prime})}{2m_{0}}\underbrace{\int
e^{-m_{0}|t-s|}\frac{\varphi(s)}{\Xi(s)-z}\,ds}_{=:I_{2}(t)}$

Writing $\frac{1}{\Xi-z}=\frac{1}{\Xi^{\prime}}\frac{d}{dt}\ln(\Xi-z)$ we can
integrate by parts to get

$\displaystyle I_{2}(t)$ $\displaystyle=-\underbrace{\int
m_{0}\frac{t-s}{|t-s|}e^{-m_{0}|t-s|}(\Xi^{\prime}(s))^{-1}\ln(\Xi(s)-z)\varphi(s)\,ds}_{=:I_{2,1}(t)}$
$\displaystyle\qquad-\underbrace{\int
e^{-m_{0}|t-s|}\ln(\Xi(s)-z)\frac{d}{ds}((\Xi^{\prime})^{-1}\varphi)(s)\,ds}_{=:I_{2,2}(t)}$

and use the uniform bound for $\ln(\Xi(s)-z)$ in $L^{1}([a-2,a+2])$ to
conclude that $|I_{2,1}|$ and $|I_{2,2}|$ are both bounded uniformly. As for
$I_{1}$, note that, on any compact interval $K$ around $a^{\prime}$, we have,
since $\Xi^{\prime}$ is continuous and $\Xi^{\prime}<0$,

$C(K):=\inf_{x\in K}|\Xi^{\prime}(x)|=-\max_{x\in K}\Xi^{\prime}(x)>0.$

Therefore by the mean value theorem, for all $s\in K$, there exists a
$\iota=\iota(s)\in K$ such that

$\displaystyle\lvert\Xi(s)-\Xi(a^{\prime})-iy\rvert$
$\displaystyle=\sqrt{y^{2}+(\Xi(s)-\Xi(a^{\prime}))^{2}}$
$\displaystyle>\lvert\Xi(s)-\Xi(a^{\prime})\rvert=\lvert
s-a^{\prime}\rvert\lvert\Xi^{\prime}(\iota)\rvert\geq\lvert s-a^{\prime}\rvert
C(K).$

By the definition of the Hölder semi-norm, we thus have, for all $s\in K$,

$\left|\frac{\psi(s)-\psi(a^{\prime})}{\Xi(s)-\Xi(a^{\prime})-iy}\right|\leq\frac{\|\psi\|_{C^{\sigma}}}{C(K)|s-a^{\prime}|^{1-\sigma}},$

which is integrable. Furthermore, outside of $K$ the integrand of $I_{1}$ is
bounded and decays exponentially, therefore one can uniformly bound $I_{1}$.

We next wish to bound the seminorm

$[B^{2}_{z}(\psi)]_{\sigma}:=\sup_{t\neq
t^{\prime}}\frac{|B^{2}_{z}(\psi)(t)-B^{2}_{z}(\psi)(t^{\prime})|}{|t-t^{\prime}|^{\sigma}}\,.$

We write

$B^{2}_{z}(\psi)(t)-B^{2}_{z}(\psi)(t^{\prime})=(I_{1}(t)-I_{1}(t^{\prime}))+\psi^{\prime}(a)(I_{2}(t)-I_{2}(t^{\prime}))\,.$

Using that $|e^{-m_{0}|t-s|}-e^{-m_{0}|t^{\prime}-s|}|\leq C|t-t^{\prime}|$ we
can bound

$|I_{1}(t)-I_{1}(t^{\prime})|\leq
C|t-t^{\prime}|\int|\varphi(s)|\frac{|\psi(s)-\psi(a^{\prime})|}{|\Xi(s)-\Xi(a^{\prime})|}\,ds\leq
C\|\psi\|_{C^{\sigma}}|t-t^{\prime}|\,.$

Similarly we can write

$|I_{2,2}(t)-I_{2,2}(t^{\prime})|\leq
C|t-t^{\prime}|\int\left|\ln(\Xi(s)-z)\frac{d}{ds}((\Xi^{\prime})^{-1}\varphi)(s)\right|\,ds\leq
C|t-t^{\prime}|\,.$

Next denoting the function $(\Xi^{\prime}(s))^{-1}\varphi(s)\ln(\Xi(s)-z)$ by
$B(s)$ we assume $t>t^{\prime}$ and write further

$\displaystyle I_{2,1}(t)-I_{2,1}(t^{\prime})$
$\displaystyle=m_{0}\Big{(}\underbrace{\int_{t}^{\infty}e^{-m_{0}(s-t)}B(s)\,ds-\int_{t^{\prime}}^{\infty}e^{-m_{0}(s-t^{\prime})}B(s)\,ds}_{=:J_{+}(t,t^{\prime})}\Big{)}$
$\displaystyle\quad-
m_{0}\Big{(}\underbrace{\int_{-\infty}^{t}e^{-m_{0}(t-s)}B(s)\,ds-\int_{-\infty}^{t^{\prime}}e^{-m_{0}(t^{\prime}-s)}B(s)\,ds}_{=:J_{-}(t,t^{\prime})}\Big{)}\,.$

Then we choose $p=\frac{1}{\sigma}$, let $p^{\prime}$ be the dual exponent and
estimate

$\displaystyle|J_{+}(t,t^{\prime})|$ $\displaystyle\leq
C|t-t^{\prime}|\int_{t}^{\infty}|B(s)|\,ds+\int_{t^{\prime}}^{t}|B(s)|\,ds$
$\displaystyle\leq
C|t-t^{\prime}|\|B\|_{L^{1}}+|t-t^{\prime}|^{\sigma}\|B\|_{L^{p^{\prime}}}\,.$

A similar estimate for $J_{-}(t,t^{\prime})$ finally shows the existence of a
constant $C$ such that

$|B^{2}_{z}(\psi)(t)-B^{2}_{z}(\psi)(t^{\prime})|\leq
C\|\psi\|_{C^{\sigma}}\left(|t-t^{\prime}|+|t-t^{\prime}|^{\sigma}\right)\,.$

Clearly this implies

$|B^{2}_{z}(\psi)(t)-B^{2}_{z}(\psi)(t^{\prime})|\leq
C\|\psi\|_{C^{\sigma}}|t-t^{\prime}|^{\sigma}\qquad\mbox{if
$|t-t^{\prime}|\leq 1$.}$

On the other hand we can trivially bound

$|B^{2}_{z}(\psi)(t)-B^{2}_{z}(\psi)(t^{\prime})|\leq
2\|B^{2}_{z}(\psi)\|_{\infty}\leq
C\|\psi\|_{C^{\sigma}}|t-t^{\prime}|^{\sigma}\quad\mbox{if $|t-t^{\prime}|\geq
1$.}$

Step 2. In this second step we compute

$\displaystyle\langle T^{-1}\mathcal{R}_{m,z}(\psi_{0}),\psi_{0}\rangle$
$\displaystyle=\langle
T^{-1}\circ\mathcal{K}_{m_{0}}\left(A((\Xi-z)^{-1}-(\Xi-
z_{0})^{-1})\psi_{0}\right),\psi_{0}\rangle$
$\displaystyle\qquad+(2m_{0}h+h^{2})\langle
T^{-1}\circ\mathcal{K}_{m_{0}}(\psi_{0}),\psi_{0}\rangle\,.$

Recalling (4.82) (and using that $\mathcal{K}_{m_{0}}$ is self-adjoint) we
rewrite the expression as

$\displaystyle\langle T^{-1}\mathcal{R}_{m,z}(\psi_{0}),\psi_{0}\rangle$
$\displaystyle=(z-z_{0})\langle
T^{-1}\circ\mathcal{K}_{m_{0}}\big{(}A(\Xi-z)^{-1}(\Xi-
z_{0})^{-1}\psi_{0}\big{)},\psi_{0}\rangle$
$\displaystyle\qquad+2m_{a}h+h^{2}$ $\displaystyle=(z-z_{0})\langle
A(\Xi-z)^{-1}(\Xi-
z_{0})^{-1}\psi_{0},\mathcal{K}_{m_{0}}\circ(T^{*})^{-1}(\psi_{0})\rangle$
$\displaystyle\qquad+2ma_{h}+h^{2}$
$\displaystyle=(z-z_{0})\underbrace{\langle A(\Xi-z)^{-1}(\Xi-
z_{0})^{-1}\psi_{0},\psi_{0}\rangle}_{=:G(z)}+2m_{a}h+h^{2}\,.$ (4.93)

We thus want to show that the following limit exists and to compute its
imaginary part:

$\displaystyle-c(a)$ $\displaystyle:=\lim_{\textrm{Im}\,z>0,z\to\Xi(a)}G(z)$
$\displaystyle=\lim_{\textrm{Im}\,z>0,z\to\Xi(a)}\int\frac{1}{\Xi(s)-z}|\psi_{0}(s)|^{2}\frac{A(s)}{\Xi(s)-\Xi(a)}\,ds\,.$
(4.94)

Observe indeed that inserting $G(z)=-c(a)+o(1)$ in (4.93) and taking into
account (4.91) and (4.92) we conclude that (4.87) holds.

In order to compute $c(a)$ we observe first an

$\phi(s):=|\psi_{0}(s)|^{2}\frac{A(s)}{\Xi(s)-\Xi(a)}$

is smooth and decays exponentially. We thus rewrite

$G(z)=\int\frac{1}{\Xi(s)-z}\phi(s)\,ds\,.$

Next we decompose $z$ into its real and imaginary part as $z=x+iy$ and observe
that

$\displaystyle\lim_{\textrm{Im}\,z>0,z\to\Xi(a)}\textrm{Re}\,G(z)$
$\displaystyle=\lim_{x\to\Xi(a),y\downarrow
0}\int\frac{\Xi(s)-x}{(\Xi(s)-x)^{2}+y^{2}}\phi(s)\,ds$

Here we are only interested in showing that the limit exists and we thus fix a
cut-off function $\varphi\in C^{\infty}_{c}(]a-2,a+2[)$, identically $1$ on
$[a-1,a+1]$ and split the integral into

$\displaystyle\textrm{Re}\,G(z)$
$\displaystyle=\int\frac{\Xi(s)-x}{(\Xi(s)-x)^{2}+y^{2}}\phi(s)\varphi(s)\,ds$
$\displaystyle\qquad+\int\frac{\Xi(s)-x}{(\Xi(s)-x)^{2}+y^{2}}\phi(s)(1-\varphi(s))\,ds\,.$

The second integral has a limit, while in order to show that the first has a
limit we write

$\frac{\Xi(s)-x}{(\Xi(s)-x)^{2}+y^{2}}=\frac{1}{2\Xi^{\prime}(s)}\frac{d}{ds}\ln((\Xi(s)-x)^{2}+y^{2})\,.$

We then integrate by parts and use the fact that $\ln((\Xi(s)-x)^{2}+y^{2})$
converges to $2\ln|(\Xi(s)-\Xi(a)|$ strongly in $L^{q}([a-2,a+2])$ for every
$q$ to infer the existence of the limit of the first integral.

As for the imaginary part we write instead

$\displaystyle\lim_{\textrm{Im}\,z>0,z\to\Xi(a)}\textrm{Im}\,G(z)$
$\displaystyle=\lim_{x\to\Xi(a),y\downarrow
0}\int\frac{y}{(\Xi(s)-x)^{2}+y^{2}}\phi(s)\,ds\,.$ (4.95)

We wish to show that the latter integral converges to

$I=\phi(a)\int\frac{ds}{(\Xi^{\prime}(a))^{2}s^{2}+1}=\frac{\pi\phi(a)}{|\Xi^{\prime}(a)|}\,.$
(4.96)

On the other hand
$\phi(a)=|\psi_{0}(a)|^{2}A^{\prime}(a)(\Xi^{\prime}(a))^{-1}$. Since
$A^{\prime}(a)>0$ and $\Xi^{\prime}(a)<0$, we conclude that $c(a)$ exists and
it is a complex number with positive imaginary part, which completes the proof
of the lemma.

It remains to show the convergence of (4.95) to (4.96). First observe that for
each $x$ sufficiently close to $\Xi(a)$ there is a unique
$a^{\prime}=\Xi^{-1}(x)$ such that $\Xi(a^{\prime})=x$. Changing variables
($s$ becomes $a^{\prime}+s$), the integral in (4.95) becomes

$\int\frac{y}{(\Xi(a^{\prime}+s)-x)^{2}+y^{2}}\phi(a^{\prime}+s)\,ds\,$ (4.97)

and we wish to show that its limit is $I$ as $(a^{\prime},y)\to(a,0)$. Next,
fix any $\delta>0$ and observe that

$\lim_{y\to
0}\int_{|s|\geq\delta}\frac{y}{(\Xi(a^{\prime}+s)-x)^{2}+y^{2}}\phi(a^{\prime}+s)\,ds=0$

uniformly in $a^{\prime}\in[a-1,a+1]$. We therefore define

$I(\delta,a^{\prime},y):=\int_{-\delta}^{\delta}\frac{y}{(\Xi(a^{\prime}+s)-x)^{2}+y^{2}}\phi(a^{\prime}+s)\,ds$

and we wish to show that, for every $\varepsilon>0$ there is a $\delta>0$ such
that

$\limsup_{(a^{\prime},y)\downarrow(a,0)}\left|I(\delta,a^{\prime},y)-I\right|\leq
C\varepsilon\,,$ (4.98)

where $C$ is a geometric constant. We rewrite

$I(\delta,a^{\prime},y)=\int_{-\delta y^{-1}}^{\delta
y^{-1}}\frac{\phi(a^{\prime}+ys)}{y^{-2}(\Xi(a^{\prime}+ys)-\Xi(a^{\prime}))^{2}+1}\,ds\,.$

Fix now $\varepsilon$ and observe that, since $\Xi^{\prime}$ and $\phi$ are
continuous, if $\delta$ is chosen sufficiently small, then

$\displaystyle((\Xi^{\prime}(a))^{2}-\varepsilon^{2})s^{2}\leq
y^{-2}(\Xi(a^{\prime}+ys)-\Xi(a^{\prime}))^{2}\leq((\Xi^{\prime}(a))^{2}+\varepsilon^{2})s^{2}$
(4.99) $\displaystyle|\phi(a^{\prime}+ys)-\phi(a)|\leq\varepsilon\,.$ (4.100)

for all $|a^{\prime}-a|<\delta$ and $y|s|\leq\delta$. Choosing $\varepsilon>0$
so that $\varepsilon\leq\frac{|\Xi^{\prime}(a)|}{2}$ we easily see that, when
$|a^{\prime}-a|<\delta$, we have

$\left|I(\delta,a^{\prime},y)-\phi(a)\int_{-\delta y^{-1}}^{\delta
y^{-1}}\frac{ds}{(\Xi^{\prime}(a))^{2}s^{2}+1}\right|\leq C\varepsilon\,.$

In particular, as $y\downarrow 0$, we conclude (4.98). ∎

###  4.9. Proof of Proposition 4.4.5

We reduce the proof of Proposition 4.4.5 to the following lemma.

######  Lemma 4.9.1.

Consider $G:=\\{m>1,m\neq m_{a},m_{b}:\mathscr{U}_{m}\neq\emptyset\\}$. Then
$G$ is relatively open and relatively closed in
$]1,\infty[\setminus\\{m_{a},m_{b}\\}$.

Proposition 4.4.5 is an obvious consequence of the latter lemma and of
Proposition 4.4.3: Lemma 4.9.1 implies that $G$ is the union of connected
components of $[1,\infty[\setminus\\{m_{a},m_{b}\\}$. On the other hand the
connected component $]m_{b},m_{a}[$ intersects $G$ because of Proposition
4.4.3 and thus it is contained in $G$. We thus complete the proof of
Proposition 4.4.5 showing Lemma 4.9.1

###### Proof of Lemma 4.9.1.

We start with some preliminary considerations. Fix an interval
$[c,d]\subset]1,\infty[\setminus\\{m_{a},m_{b}\\}$. Recalling Proposition
4.1.4 we know that, since the operator norm of $\mathcal{L}_{m}$ is bounded
uniformly in $m\in[c,d]$,

  * (a)

There is $R>0$ such that $\mathscr{U}_{m}\subset B_{R}(0)$ for all
$m\in[c,d]$.

However it also follows from Proposition 4.4.1 that

  * (b)

There is a $\delta>0$ such that
$\mathscr{U}_{m}\subset\\{\textrm{Im}\,z>\delta\\}$.

Step 1. We first prove that $G$ is relatively closed. To that end we fix a
sequence $m_{j}\to m\in]1,\infty[\setminus\\{m_{a},m_{b}\\}$ such that $m_{j}$
belongs to $G$. Without loss of generality we can assume
$\\{m_{j}\\}\subset[c,d]\subset]1,\infty[\setminus\\{m_{a},m_{b}\\}$. For each
$m_{j}$ we can then consider $z_{j}\in\mathscr{U}_{m_{j}}$, which by (a) and
(b) we can assume to converge to some $z\in\mathbb{C}$ with positive imaginary
part. We then let $\psi_{j}$ be a sequence of nontrivial elements in $L^{2}$
such that

$-\psi_{j}^{\prime\prime}+m_{j}^{2}\psi_{j}+\frac{A}{\Xi-z_{j}}\psi_{j}=0\,,$
(4.101)

and normalize them to $\|\psi_{j}\|_{L^{2}}=1$ Since
$\textrm{Im}\,z_{j}\geq\delta>0$, the sequence of functions $\frac{A}{\Xi-
z_{j}}$ enjoy uniform bounds in the spaces $L^{1}$ and $C^{k}$. We can then
argue as in Section 4.6 to find that

  * (i)

$\|\psi_{j}^{\prime}\|_{L^{2}}$ enjoy a uniform bound;

  * (ii)

There are uniformly bounded nonzero constants $\\{C^{\pm}_{j}\\}$ with the
property that $\psi_{j}$ is asymptotic to $C^{\pm}_{j}e^{\mp m_{j}t}$ and
$\pm\infty$;

  * (iii)

There is a $T_{0}>0$ independent of $j$ with the property that

$|\psi_{j}(t)-C^{\pm}_{j}e^{\mp m_{j}t}|\leq\frac{|C^{\pm}_{j}|}{2}e^{\mp
m_{j}t}\qquad\forall\pm t>T_{0}\,.$

These three properties together imply that a subsequence, not relabeled,
converges strongly in $L^{2}$ to some $\psi$. Passing into the limit in
(4.101) we conclude that

$-\psi^{\prime\prime}+m^{2}\psi+\frac{A}{\Xi-z}\psi=0\,.$

This shows that $z\in\mathscr{U}_{m}$, i.e. that $m\in G$.

Step 2. Here we show that $G$ is relatively open. To that end we consider some
sequence $m_{j}\to m\in]1,\infty[\setminus\\{m_{a},m_{b}\\}$ with the property
that $m_{j}\not\in G$ and we show that $m\not\in G$. By (a) and (b) above, it
suffices to show that the domain

$\Delta:=\\{|z|<R:\textrm{Im}\,z>\delta\\}$

does not contain any element of $\textrm{spec}\,(\mathcal{L}_{m})$. Observe
first that, since we know that it does not intersect $\gamma=\partial\Delta$,
the distance between $\gamma$ and any element in
$\textrm{spec}\,(\mathcal{L}_{m})$ is larger than a positive constant
$\varepsilon$. Recalling that the spectrum on the upper half complex space is
discrete, we have that

$P_{m}:=\int_{\gamma}(\mathcal{L}_{m}-z)^{-1}\,dz$

is a projection on a finite-dimensional space which contains all eigenspaces
of the elements
$z\in\textrm{spec}\,(\mathcal{L}_{m})\cap\Delta=m\mathscr{U}_{m}=\\{mz:z\in\mathscr{U}_{m}\\}$.
And since all such elements belong to the discrete spectrum,
$\mathscr{U}_{m}=\emptyset$ if and only if $P_{m}=0$. On the other hand

$P_{m_{j}}:=\int_{\gamma}(\mathcal{L}_{m_{j}}-z)^{-1}\,dz$

equals $0$ precisely because $m_{j}\not\in G$. We thus just need to show that
$P_{m_{j}}$ converges to $P_{m}$ to infer that $m\in G$. The latter follows
from the following observations:

  * (i)

Since $\gamma$ is a compact set and does not intersect the spectrum of
$\mathcal{L}_{m}$, there is a constant $M$ such that
$\|(\mathcal{L}_{m}-z)^{-1}\|_{O}\leq M$ for all $z\in\gamma$;

  * (ii)

$\mathcal{L}_{m_{j}}$ converges to $\mathcal{L}_{m}$ in the operator norm;

  * (iii)

Writing

$\displaystyle\qquad(\mathcal{L}_{m_{j}}-z)^{-1}$
$\displaystyle=(\textrm{Id}+(\mathcal{L}_{m}-z)^{-1}(\mathcal{L}_{m_{j}}-\mathcal{L}_{m}))^{-1}(\mathcal{L}_{m}-z)^{-1}\,,$

when $\|\mathcal{L}_{m_{j}}-\mathcal{L}_{m}\|_{\mathcal{L}}\leq\frac{1}{2M}$
we can use the Neumann series for the inverse to infer

$\sup_{z\in\gamma}\|(\mathcal{L}_{m_{j}}-z)^{-1}-(\mathcal{L}_{m}-z)^{-1}\|_{O}\leq
C\|\mathcal{L}_{m}-\mathcal{L}_{m_{j}}\|_{O}\,,$

for some constant $C$ independent of $j$.

We then conclude that $P_{m_{j}}$ converges to $P_{m}$ in the operator norm. ∎

######  Remark 4.9.2.

An immediate outcome of the argument above is that the sum of the algebraic
multiplicities of $z\in\mathscr{U}_{m}$, as eigenvalues of
$m^{-1}\mathcal{L}_{m}$, is constant on any connected component of
$]\\!-\\!\infty,\infty[\setminus\\{m_{a},m_{b}\\}$. Indeed, it coincides with
the rank of the operator $P_{m}$ defined in Step 2.

###  4.10. Proof of Lemma 4.4.7

We start by observing that

$\Xi^{\prime}(t):=\int_{-\infty}^{t}e^{-2(t-\tau)}A(\tau)\,d\tau\,.$

In fact, since $A=\Xi^{\prime\prime}+2\Xi$, the identity follows from the
classical solution formula for first order ODEs with constant coefficients
once we show that the right hand side and the left hand side coincide for all
sufficiently negative $t$. The latter can be verified with a direct
computation using that $\Xi^{\prime}(t)=-2c_{0}2^{2t}$ for all $t$
sufficiently negative.

We next read the conditions for $\Xi\in\mathscr{C}$ in terms of $A$ to find
that it suffices to impose

  * (i)

$A(t)=-8c_{0}e^{2t}$ for all $t$ sufficiently negative;

  * (ii)

$A(t)=-{\bar{\alpha}}e^{-{\bar{\alpha}}t}$ for all $t\geq\ln 2$;

  * (iii)

There are exactly two zeros $a<b$ of $A$ and $A^{\prime}(a)>0$,
$A^{\prime}(b)<0$;

  * (iv)

$\int_{-\infty}^{t}e^{-2(t-\tau)}A(\tau)d\tau<0$ for every $t$.

Note indeed that, by imposing $\Xi(\infty)=0$, with a simple computation we
can determine $\Xi(-\infty)$ as

$\Xi(-\infty)=-\int_{-\infty}^{\infty}\Xi^{\prime}(\tau)\,d\tau=-\frac{1}{2}\int_{-\infty}^{\infty}A(\tau)\,d\tau\,$

and moreover (using $\Xi^{\prime}<0$, which follows from (iv)) we conclude
that the conditions (i)-(iv) ensure

$\int_{-\infty}^{\infty}A(\tau)\,d\tau<0\,.$

We next fix $a=0$ and $b=\frac{1}{2}$ and rather than imposing (iv) we impose
the two conditions

  * (v)

$\int_{-\infty}^{0}e^{2\tau}A(\tau)d\tau=-1$;

  * (vi)

$\max A\leq\frac{1}{e}$.

Observe, indeed, that (iv) is equivalent to

$\int_{-\infty}^{t}e^{2\tau}A(\tau)\,d\tau<0$

and that, since $A$ is negative on $]-\infty,0[$ and $]\frac{1}{2},\infty[$,
the integral on the left-hand side is maximal for $t=\frac{1}{2}$. We then can
use (v) and (vi) to estimate

$\int_{-\infty}^{\frac{1}{2}}e^{2\tau}A(\tau)\,d\tau\leq-1+\frac{e}{2}\max
A\leq-\frac{1}{2}\,.$

We next recall that, by the Rayleigh criterion,

$-\lambda_{a}=\min_{\|\psi\|_{L^{2}}=1}\langle\psi,L_{a}\psi\rangle=\min_{\|\psi\|_{L^{2}}=1}\int\Big{(}|\psi^{\prime}|^{2}+\frac{A}{\Xi-\Xi(0)}|\psi|^{2}\Big{)}\,.$

We test the right-hand side with

$\psi(t):=\left\\{\begin{array}[]{ll}0&\mbox{for $|t|\geq\frac{1}{2}$}\\\ \\\
\sqrt{2}\cos(\pi t)&\mbox{for $|t|\leq\frac{1}{2}$.}\end{array}\right.$

We therefore get

$-\lambda_{a}\leq
2\pi^{2}+2\int_{-1/2}^{1/2}\frac{A(t)}{\Xi(t)-\Xi(0)}\cos^{2}\pi t\,dt\,.$
(4.102)

Next, for any fixed positive constant $B>0$, we impose that $A(t)=Bt$ on the
interval $]-\sqrt{B}^{-1},0]$ and we then continue it smoothly on $[0,\infty[$
so to satisfy (ii), (iii), and (v) on $[0,\infty[$ (the verification that this
is possible is rather simple). We also can continue it smoothly on
$]\\!-\\!\infty,-\sqrt{B}^{-1}]$ so to ensure (i). In order to ensure (v) as
well we just need to show that

$\int_{-\sqrt{B}^{-1}}^{0}e^{2\tau}A(\tau)\,d\tau\geq-\frac{1}{2}\,.$

The latter is certainly ensured by

$\int_{-\sqrt{B}^{-1}}^{0}e^{2\tau}A(\tau)\,d\tau\geq\int_{-\sqrt{B}^{-1}}^{0}A(\tau)\,d\tau=-\frac{1}{2}\,.$

Now, observe that
$\Xi^{\prime}(0)=\int_{-\infty}^{0}e^{2\tau}A(\tau)\,d\tau=-1$. For
$t\in\,]-\sqrt{B}^{-1},0[$ we wish to estimate $\Xi^{\prime}(t)$ and to do it
we compute

$\displaystyle|\Xi^{\prime}(t)-\Xi^{\prime}(0)|$
$\displaystyle=\left|e^{-2t}\int_{-\infty}^{t}e^{2\tau}A(\tau)\,d\tau-\int_{-\infty}^{0}e^{2\tau}A(\tau)\,d\tau\right|$
$\displaystyle\leq
e^{-2t}\left|\int_{t}^{0}e^{2\tau}A(\tau)\,d\tau\right|+(e^{-2t}-1)\left|\int_{-\infty}^{0}e^{2\tau}A(\tau)\,d\tau\right|$
$\displaystyle\leq\frac{e^{2\sqrt{B}^{-1}}}{2}+(e^{2\sqrt{B}^{-1}}-1)\leq\frac{3}{4}\,,$

which can be ensured by taking $B$ sufficiently large. In particular
$-\frac{1}{4}\geq\Xi^{\prime}(t)\geq-2$ for $t\in\,]-\sqrt{B}^{-1},0[$. We
thus conclude that

$-2t\leq\Xi(t)-\Xi(0)\leq-\frac{t}{4}\qquad\forall
t\in\,]-\sqrt{B}^{-1},0[\,.$

In turn the latter can be used to show

$\frac{A(t)}{\Xi(t)-\Xi(0)}\leq-\frac{B}{2}\qquad\forall
t\in\,]-\sqrt{B}^{-1},0[\,.$

Since $\frac{A}{\Xi-\Xi(0)}$ is otherwise negative on
$]-\frac{1}{2},\frac{1}{2}[$, we conclude

$-\lambda_{a}\leq 2\pi^{2}-2\int_{-\sqrt{B}^{-1}}^{0}B\cos^{2}\pi t\,dt\,.$
(4.103)

By taking $B$ large enough we can ensure that $\cos^{2}\pi t\geq\frac{1}{2}$
on the interval $]-\sqrt{B}^{-1},0[$. In particular we achieve

$-\lambda_{a}\leq 2\pi^{2}-\sqrt{B}\,.$

Since we can choose $\sqrt{B}$ as large as we wish, the latter inequality
completes the proof of the lemma.

##  Chapter 5 Nonlinear theory

This final chapter will prove Theorem 2.5.1 and hence complete the argument
leading to Theorem 1.0.1. To that end we fix a choice of $\bar{\Omega}$,
$\bar{V}$, $m$ and $\eta$ as given by Theorem 2.4.2, where $\bar{a}>0$ is a
large parameter whose choice will be specified only later. We moreover fix a
suitable subspace $X$ of $L^{2}_{m}$ and we will prove an estimate
corresponding to (2.46) in this smaller space.

######  Definition 5.0.1.

We denote by $X$ the subspace of elements $\Omega\in L^{2}_{m}$ for which the
following norm is finite:

$\|\Omega\|_{X}:=\|\Omega\|_{L^{2}}+\||x|\nabla\Omega\|_{L^{2}}+\|\nabla\Omega\|_{L^{4}}\,.$
(5.1)

The above norm has two features which will play a crucial role in our
estimates. The first feature is that it ensures an appropriate decay of the
$L^{2}$ norm of $D\Omega$ on the complements of large disks
$\mathbb{R}^{2}\setminus B_{R}$. The second feature is that it allows to bound
the $L^{\infty}$ norm of $\Omega$ and $\nabla(K_{2}*\Omega)$ and to give a
bound on the growth of $K_{2}*\Omega$ at infinity. More precisely, we have the
following:

######  Proposition 5.0.2.

For all $\kappa\in]0,1[$, there is a constant $C(\kappa)>0$ such that the
following estimates hold for every $m$-fold symmetric $\Omega\in X$:

$\displaystyle|\nabla(K_{2}*\Omega)(x)|+|\Omega(x)|$
$\displaystyle\leq\frac{C(\kappa)}{1+|x|^{1-\kappa}}\|\Omega\|_{X}\qquad\forall
x\in\mathbb{R}^{2}$ (5.2) $\displaystyle|K_{2}*\Omega(x)|$ $\displaystyle\leq
C\|\Omega\|_{X}\min\\{|x|,1\\}\qquad\forall x\in\mathbb{R}^{2}\,.$ (5.3)

The aim of this chapter is therefore to give the bound

$\|\Omega_{\text{per},k}(\cdot,\tau)\|_{X}\leq
e^{\tau(a_{0}+\delta_{0})}\qquad\qquad\qquad\forall\tau\leq\tau_{0}\,$ (5.4)

for some appropriately chosen constants $\delta_{0}>0$ and $\tau_{0}<0$,
independent of $k$. Of course the main difficulty will be to give the explicit
estimate (5.4). However a first point will be to show that the norm is indeed
finite for every $\tau\geq-k$. This will be a consequence of the following:

######  Lemma 5.0.3.

Provided $a_{0}$ is large enough, the eigenfunction $\eta$ of Theorem 2.4.2
belongs to $C^{2}(\mathbb{R}^{2}\setminus\\{0\\})$ and satisfies the pointwise
estimates

$|D^{j}\eta(x)|\leq C(1+|x|)^{-m-2-j-{\bar{\alpha}}}\qquad\forall x,\forall
j\in\\{0,1,2\\}\,.$ (5.5)

In particular $\eta\in C^{1}(\mathbb{R}^{2})$ and its first derivatives are
Lipschitz (namely, $\eta\in W^{2,\infty}(\mathbb{R}^{2})$). Moreover
$K_{2}*\eta\in L^{2}$.

One relevant outcome of Lemma 5.0.3 is that the bound (5.4) holds at least for
$\tau$ sufficiently close to $-k$, given that
$\Omega_{\text{per},k}(\cdot,-k)\equiv 0$. The main point of (5.4) is then
that we will be able to deduce the following estimates.

######  Lemma 5.0.4.

Under the assumptions of Theorem 2.5.1 there is a constant $C_{0}$
(independent of $k$) such that the following holds. Assume that
$\bar{\tau}\leq 0$ is such that for all $\tau\in[-k,\bar{\tau}]$ we have the
estimate

$\|\Omega_{\text{per},k}(\cdot,\tau)\|_{X}\leq e^{(a_{0}+\delta_{0})\tau}.$
(5.6)

Then

$\displaystyle\|\Omega_{\text{per},k}(\cdot,\bar{\tau})\|_{L^{2}}$
$\displaystyle\leq C_{0}e^{(a_{0}+\delta_{0}+1/2)\bar{\tau}}\,,$ (5.7)
$\displaystyle\||x|D\Omega_{\text{per},k}(\cdot,\bar{\tau})\|_{L^{2}}$
$\displaystyle\leq C_{0}e^{(a_{0}+2\delta_{0})\bar{\tau}}\,,$ (5.8)
$\displaystyle\|D\Omega_{\text{per},k}(\cdot,\bar{\tau})\|_{L^{4}}$
$\displaystyle\leq C_{0}e^{(a_{0}+2\delta_{0})\bar{\tau}}\,.$ (5.9)

With the above lemma we easily conclude (5.4) (and hence Theorem 2.5.1).
Indeed denote by $\tau_{k}$ the largest non-positive time such that

$\|\Omega_{\text{per},k}(\cdot,\tau)\|_{X}\leq
e^{(a_{0}+\delta_{0})\tau}\qquad\forall\tau\in[-k,\tau_{k}]\,.$ (5.10)

Then we must have

$\|\Omega_{\text{per},k}(\cdot,\tau_{k})\|_{X}=e^{(a_{0}+\delta_{0})\tau_{k}}\,.$
(5.11)

On the other hand, summing the three estimates (5.7), (5.8), and (5.9) we
conclude

$\|\Omega_{\text{per},k}(\cdot,\tau_{k})\|_{X}\leq\bar{C}e^{(a_{0}+2\delta_{0})\tau_{k}}$
(5.12)

for some constant $\bar{C}$ independent of $k$. However (5.11) and (5.12) give
$e^{\delta_{0}\tau_{k}}\geq\bar{C}^{-1}$, i.e.
$\tau_{k}\geq-\frac{1}{\delta_{0}}\ln\bar{C}$, implying that (5.4) holds with
$\tau_{0}:=-\frac{1}{\delta_{0}}\ln\bar{C}$.

After proving Proposition 5.0.2 and Lemma 5.0.3, we will dedicate two separate
sections to the three estimates (5.7), (5.8), and (5.9). The first estimate,
which we will call baseline estimate, will differ substantially from the other
two, and to it we will dedicate a section. In order to accomplish the gain in
the exponent in (5.7) we will use crucially the information on the semigroup
which comes from Theorem 2.4.2 and Theorem 3.0.4, namely, that the growth
bound $\omega(L_{ss})$ is precisely $a_{0}$ (i.e., the growth achieved by
$\Omega_{\text{lin}}$). Since, however, the terms in the Duhamel’s formula
depend on derivatives, we need to invoke an a priori control of them, which is
present in the norm $\|\cdot\|_{X}$. Indeed, one such term experiencing the
derivative loss arise from the nonlinearity is the following:

$\int_{-k}^{\tau}e^{(\tau-s)L_{\textrm{ss}}}[(K_{2}*\Omega_{\text{per},k})\cdot\nabla\Omega_{\text{per},k}](\cdot,s)\,ds\,.$
(5.13)

Note that $\|\cdot\|_{X}$ also includes the weighted $L^{2}$ norm
$\||x|D\Omega\|_{L^{2}}$ because we encounter a term where
$D\Omega_{\text{per},k}$ is multiplied by the function $V^{r}$, which grows at
$\infty$ like $|x|^{1-{\bar{\alpha}}}$ when ${\bar{\alpha}}\in]0,2[$. In order
to close the argument we then need to control the $L^{4}$ norm and the
weighted $L^{2}$ norm of $D\Omega_{\text{per},k}$. The latter estimates will
not be accomplished through a growth bound on the semigroup $e^{\tau L_{ss}}$
(which would invoke controls on yet higher derivatives), but rather through
some careful energy estimates. The structure of the problem will then enter
crucially, since the term
$(K_{2}*\Omega_{\text{per},k})\cdot\nabla\bar{\Omega}$ which we need to bound
in the energy estimates will take advantage of the improved baseline estimate
on the $L^{2}$ norm. The above term, which is responsible for the creation of
unstable eigenvalues, actually _gains a derivative_. Finally, there is one
remaining difficulty when estimating $D\Omega_{\text{per},k}$ due to the
transport term. Namely, differentiating the equation in cartesian coordinates
contributes a term $(\partial_{i}\bar{V})\cdot\nabla\Omega_{\text{per},k}$,
which could destabilize the estimates. We exploit the structure of the problem
again in a crucial way by estimating angular and radial derivatives, rather
than derivatives in cartesian coordinates, and by estimating the angular
derivatives first and the radial derivatives after.

###  5.1. Proof of Proposition 5.0.2

We start by bounding $|\Omega(x)|$. Since $W^{1,4}(B_{2})$ embeds in
$C^{1/2}$, the bound $|\Omega(x)|\leq C\|\Omega\|_{X}$ is true for every $x\in
B_{2}$. Consider further $R:=\frac{|x|}{2}\geq 1$ and define
$u_{R}(y):=\Omega(Ry)$. In particular let $B:=B_{1}(\frac{x}{R})$ and notice
that

$\displaystyle\|u_{R}\|_{L^{2}(B)}$
$\displaystyle=R^{-1}\|\Omega\|_{L^{2}(B_{R}(x))}\leq R^{-1}\|\Omega\|_{X}\,,$
(5.14) $\displaystyle\|Du_{R}\|_{L^{2}(B)}$
$\displaystyle=\|D\Omega\|_{L^{2}(B_{R}(x))}\leq
R^{-1}\||\cdot|D\Omega\|_{L^{2}(B_{R}(x))}\leq R^{-1}\|\Omega\|_{X}\,,$ (5.15)
$\displaystyle\|Du_{R}\|_{L^{4}(B)}$
$\displaystyle=R^{1/2}\|D\Omega\|_{L^{4}(B_{R}(x))}\leq
R^{1/2}\|\Omega\|_{X}\,.$ (5.16)

By interpolation, for $\frac{1}{p}=\frac{\lambda}{2}+\frac{1-\lambda}{4}$ we
have

$\|Du_{R}\|_{L^{p}(B)}\leq C\|\Omega\|_{X}R^{-\lambda+(1-\lambda)/2}\,.$

Choosing $p$ very close to $2$, but larger, we achieve
$\|Du_{R}\|_{L^{p}(B)}\leq CR^{-1+\kappa}\|\Omega\|_{X}$. Since the average of
$u_{R}$ over $B$ is smaller than $\|u_{R}\|_{L^{2}(B)}\leq
CR^{-1}\|\Omega\|_{X}$, from Poincaré we conclude $\|u_{R}\|_{W^{1,p}(B)}\leq
CR^{-1+\kappa}\|\Omega\|_{X}$. In particular using the embedding of $W^{1,p}$
in $L^{\infty}$ we conclude

$\|u_{R}\|_{L^{\infty}(B)}\leq C(p)R^{-1+\kappa}\|\Omega\|_{X}\,.$ (5.17)

Since however $|\Omega(x)|\leq\|u_{R}\|_{L^{\infty}(B)}$, we reach the
estimate

$|\Omega(x)|\leq\frac{C}{|x|^{1-\kappa}}\|\Omega\|_{X}\,.$ (5.18)

Note that the constant $C$ depends on $\kappa$: $\kappa$ is positive, but a
very small choice of it forces to choose a $p$ very close to $2$, which in
turn gives a dependence of the constant $C(p)$ in (5.17) on $\kappa$.

We next come to the estimates for $\nabla K_{2}*\Omega$. First of all, observe
that

$\displaystyle\|\nabla K_{2}*\Omega\|_{L^{2}}\leq$ $\displaystyle
C\|\Omega\|_{L^{2}}\leq C\|\Omega\|_{X}$
$\displaystyle\|D^{2}K_{2}*\Omega\|_{L^{2}}+\|D^{2}K_{2}*\Omega\|_{L^{4}}\leq$
$\displaystyle C\|D\Omega\|_{L^{2}}+C\|D\Omega\|_{L^{4}}\leq C\|\Omega\|_{X}$

and thus $|\nabla K_{2}*\Omega(x)|\leq C\|\Omega\|_{X}$ follows for every
$x\in B_{4}$. Consider now $|x|\geq 4$, set $R:=\frac{|x|}{4}$ and let
$\varphi\in C^{\infty}_{c}(B_{2R}(x))$ be a cut-off function identically equal
to $1$ on $B_{R}(x)$ and $\psi\in C^{\infty}_{c}(B_{2R})$ equal to $1$ on
$B_{R}$. We choose them so that
$\|D^{k}\psi\|_{C^{0}}+\|D^{k}\varphi\|_{C^{0}}\leq C(k)R^{-k}$.

We split

$\nabla K_{2}*\Omega=\nabla K_{2}*(\varphi\Omega)+\nabla
K_{2}*(\psi\Omega)+\nabla
K_{2}*((1-\varphi-\psi)\Omega)\,=:F_{1}+F_{2}+F_{3}\,.$

We have

$\displaystyle\|F_{1}\|_{L^{2}}$ $\displaystyle\leq
C\|\varphi\Omega\|_{L^{2}}\leq C\|\Omega\|_{X}$
$\displaystyle\|DF_{1}\|_{L^{2}}$ $\displaystyle\leq
C\|D(\varphi\Omega)\|_{L^{2}}\leq CR^{-1}\|\Omega\|_{L^{2}}+C\|\varphi
D\Omega\|_{L^{2}}\leq CR^{-1}\|\Omega\|_{X}$ $\displaystyle\|DF_{1}\|_{L^{4}}$
$\displaystyle\leq C\|\Omega\|_{X}\,.$

The argument used above implies then $|F_{1}(x)|\leq
C(\kappa)|x|^{\kappa-1}\|\Omega\|_{X}$. As for estimating $F_{2}$ we observe
that $F_{2}$ is harmonic outside $B_{2R}$. On the other hand
$\|F_{2}\|_{L^{2}}\leq C\|\Omega\|_{X}$. Using the mean-value inequality for
harmonic functions we then get

$|F_{2}(x)|\leq\frac{1}{\pi(2R)^{2}}\int_{B_{2R}(x)}|F_{2}|\leq\frac{C}{R}\|\Omega\|_{X}\,.$

As for $F_{3}$ we write, using the bound on $|\Omega|$ and $|\nabla
K(x-y)|\leq C|x-y|^{-2}$,

$\displaystyle|F_{3}(x)|$
$\displaystyle\leq\int_{\mathbb{R}^{2}\setminus(B_{2R}(x)\cup
B_{2R})}\frac{C(\kappa)\|\Omega\|_{X}}{|x-y|^{2}|y|^{1-\kappa}}$
$\displaystyle\leq\int_{(\mathbb{R}^{2}\setminus
B_{2R})\cap\\{|x-y|\geq|y|\\}}\frac{C(\kappa)\|\Omega\|_{X}}{|y|^{3-\kappa}}$
$\displaystyle\qquad+\int_{(\mathbb{R}^{2}\setminus
B_{2R}(x))\cap\\{|y|\geq|x-y|\\}}\frac{C(\kappa)\|\Omega\|_{X}}{|x-y|^{3-\kappa}}$
$\displaystyle\leq\frac{C(\kappa)\|\Omega\|_{X}}{R^{1-\kappa}}\,.$

Recalling that $K_{2}*\Omega(0)=0$, integrating (5.2) on the segment with
endpoints $0$ and $x$ we conclude (5.3) for $|x|\leq 2$. In order to prove the
bound when $|x|\geq 1$, fix a point $y$ with $3R:=|y|\geq 1$. Let $\varphi$ be
a radial cut-off function which is identically equal to $1$ on $B_{R}(0)$, is
supported in $B_{2R}(0)$ and whose gradient is bounded by $CR^{-1}$. We then
write

$K_{2}*\Omega=K_{2}*(\varphi\Omega)+K_{2}*((1-\varphi)\Omega)\,.$ (5.19)

Since the distance between $y$ and the support of $\varphi\Omega$ is larger
than $R$, we can estimate

$|K_{2}*(\varphi\Omega)(y)|\leq\frac{C}{R}\int|\varphi\Omega|\leq
C\|\Omega\|_{L^{2}}\leq C\|\Omega\|_{X}\,.$ (5.20)

Next observe that, by Calderon-Zygmund,

$\displaystyle\|D^{2}K_{2}*(\Omega(1-\varphi))\|_{L^{2}}$
$\displaystyle=\|D((1-\varphi)\Omega)\|_{L^{2}}$
$\displaystyle\leq\frac{C}{R}\|\Omega\|_{L^{2}}+C\|D\Omega\|_{L^{2}(\mathbb{R}^{2}\setminus
B_{R})}$ $\displaystyle\leq\frac{C}{R}\|\Omega\|_{X}\,.$

Since $(1-\varphi)\Omega$ belongs to $L^{2}_{m}$, the average over $B$ of
$K_{2}*((1-\varphi)\Omega)$ equals $0$. Hence we conclude from Poincaré
inequality and Calderon-Zygmund that

$\displaystyle\|K_{2}*((1-\varphi)\Omega)\|_{L^{2}(B)}$ $\displaystyle\leq
CR\|DK_{2}*((1-\varphi)\Omega)\|_{L^{2}(B)}$ $\displaystyle\leq
CR\|((1-\varphi)\Omega)\|_{L^{2}(B)}\,.$ $\displaystyle\leq
CR\|\Omega\|_{X}\,.$

From Gagliardo-Nirenberg interpolation inequality applied on $B$ we have

$\displaystyle\|K_{2}*((1-\varphi)\Omega)\|_{L^{\infty}(B)}$
$\displaystyle\leq
C\|D^{2}K_{2}*(\Omega(1-\varphi))\|_{L^{2}}^{1/2}\|K_{2}*((1-\varphi)\Omega)\|^{1/2}_{L^{2}(B)}$
$\displaystyle\qquad+\frac{C}{R}\|K_{2}*((1-\varphi)\Omega)\|_{L^{2}(B)}\leq
C\|\Omega\|_{X}.$

###  5.2. Proof of Lemma 5.0.3

The proof of the estimates for $\eta$ and its derivatives can be found below.

As for the second conclusion of the Lemma, observe that, by going back to the
solutions $\omega_{\varepsilon,k}$, it follows from the regularity and decay
of the initial data in (2.20) (just proved in the above lemma) and the
regularity and decay of the forcing term $f$ for positive times, that
$\omega_{\varepsilon,k}\in C([t_{k},T],X)$ for every $T>t_{k}$. Given the
explicit transformation from $\omega_{\varepsilon,k}$ to
$\Omega_{\varepsilon,k}$ we conclude that $\Omega_{\varepsilon,k}\in
C([-k,T],X)$ for every $T>-k$. Since the same regularity is enjoyed by
$\tilde{\Omega}$ on $[-k,T]$ (the latter is in fact smooth and compactly
supported on $\mathbb{R}^{2}\times[-k,T]$) and by $\Omega_{\text{lin}}$, we
infer that
$\Omega_{\text{per},k}=\Omega_{\varepsilon,k}-\tilde{\Omega}-\Omega_{\text{lin}}$
belongs to $C([-k,T],X)$.

###### Proof of the first part of Lemma 5.0.3.

Consider $\eta\in L^{2}_{m}$, for $m\geq 2$ as in Theorem 2.4.2 and write it
as $\eta(\theta,r)=\vartheta(r)e^{ikm\theta}$ when $b_{0}\neq 0$ or
$\eta(\theta,r)=\vartheta(r)e^{ikm\theta}+\bar{\vartheta}(r)e^{-ikm\theta}$ if
$b_{0}=0$ (where $\bar{\vartheta}$ denotes the complex conjugate of
$\vartheta$). In both cases $\vartheta(r)e^{ikm\theta}$ is an eigenfunction
and through this property we will show that it satisfies the estimates of the
Lemma. We can therefore assume without loss of generality that
$\eta(x)=\vartheta(r)e^{ikm\theta}$. We will also see that the argument leads
in fact to estimates (5.5) with $km$ replacing $m$ and hence without loss of
generality we assume $k=1$. Furthermore an outcome of the argument below is
that $\eta$ is smooth except possibly at the origin. Note moreover that after
having shown the pointwise bounds (5.5) for $\eta$ outside of the origin, the
$W^{2,\infty}$ regularity of $\eta$ follows immediately: indeed $\eta$ and
$D\eta$ can be continuously extended to the origin by setting them equal to
$0$, hence showing that $\eta\in C^{1}(B_{1})$, while the uniform bound of
$|D^{2}\eta|$ in $B_{1}\setminus\\{0\\}$ easily implies that $D\eta$ is
Lipschitz.

Step 1. Exponential coordinates. We recall that the distribution $K_{2}*\eta$
is well defined, according to Lemma 2.4.1, and its action on a test function
in $\mathscr{S}$ is given by integrating the scalar product of the test with a
function $v\in W^{1,2}_{\text{loc}}(\mathbb{R}^{2},\mathbb{C}^{2})$, cf. the
proof of Lemma 2.4.1 in the appendix. It follows from the argument given there
that $R_{2\pi/m}v(R_{2\pi/m}x)=v(x)$. Given that $\textrm{div}\,V=0$,
$-V^{\perp}$ is the gradient of a continuous function, which is determined by
its value $\psi$ at the origin. $\psi$ inherits the symmetry and can thus be
written as $\psi(\theta,r)=f(r)e^{im\theta}$. We thus conclude that

$\displaystyle\vartheta=$ $\displaystyle
f^{\prime\prime}+\frac{1}{r}f^{\prime}-\frac{m^{2}}{r^{2}}f$ (5.21)
$\displaystyle v=$
$\displaystyle\frac{f^{\prime}}{r}\frac{\partial}{\partial\theta}-\frac{imf}{r}\frac{\partial}{\partial
r}\,.$ (5.22)

Observe moreover that $v\in W^{1,2}_{\text{loc}}$ implies
$f^{\prime\prime},\frac{f^{\prime}}{r}$, $\frac{f}{r^{2}}\in
L^{2}_{\text{loc}}$. Therefore $f$ is determined by (5.21) and the boundary
conditions

  * (a)

$f(0)=0$;

  * (b)

$\int_{0}^{1}\frac{|f^{\prime}(r)|^{2}}{r}\,dr<\infty$.

(observe that condition (b) implies $f^{\prime}(0)=0$ when $f\in C^{1}$ and we
can therefore interpret it as a surrogate for a boundary condition). We recall
also that we have the estimate $\|D^{2}\psi\|_{L^{2}}\leq
C\|\vartheta\|_{L^{2}}<\infty$ and owing to $\int_{B_{R}}D\psi=0$, by Poincaré
we achieve

$\|D\psi\|_{L^{p}(B_{R})}\leq C(p)R^{2/p}\,.$

Using Morrey’s embedding and the fact that $\psi(0)=0$ we conclude, in turn

$|\psi(x)|\leq C\|D\psi\|_{L^{p}(B_{|x|})}|x|^{1-2/p}\leq C|x|\,.$

In particular we conclude

$|f(r)|\leq Cr\,.$ (5.23)

The equation satisfied by $\eta$ can thus be written in terms of the function
$f$ as

$\left(1+\frac{r}{\alpha}\frac{d}{dr}-z_{0}-im\beta\zeta\right)\left(f^{\prime\prime}+\frac{f^{\prime}}{r}-\frac{m^{2}}{r^{2}}f\right)-\frac{imf\beta
g^{\prime}}{r}=0\,,$ (5.24)

where $g$ is the smooth function such that $\bar{\Omega}(x)=g(|x|)$ (in
particular $g$ is constant in a neighborhood of the origin, and equals
$r^{-\alpha}$ for $r\geq 2$) and $\zeta$ is given by the formula (2.8). We
next set $s=\ln r$ and in particular

$\displaystyle\tilde{g}(s)$ $\displaystyle=g(e^{s})$ (5.25) $\displaystyle
h(s)$ $\displaystyle=f(e^{s})$ (5.26) $\displaystyle\tilde{\zeta}(s)$
$\displaystyle=\zeta(e^{s})\,.$ (5.27)

Note that

$\vartheta(e^{s})=e^{-2s}(h^{\prime\prime}(s)-m^{2}h(s))=:e^{-2s}\Gamma(s)\,.$
(5.28)

In these new coordinates we observe that the claim of the lemma corresponds
then to showing that $\vartheta\in C^{2}_{\text{loc}}(\mathbb{R})$ and

$\displaystyle|\Gamma(s)|+|\Gamma^{\prime}(s)|+|\Gamma^{\prime\prime}(s)|$
$\displaystyle\leq Ce^{4s}\qquad$ $\displaystyle\forall s\leq 0\,,$ (5.29)
$\displaystyle|\Gamma(s)|+|\Gamma^{\prime}(s)|+|\Gamma^{\prime\prime}(s)|$
$\displaystyle\leq Ce^{-(m+{\bar{\alpha}})s}\qquad$ $\displaystyle\forall
s\geq 0\,.$ (5.30)

In order to achieve the latter estimates we will need the following bounds on
$\tilde{g}^{\prime}$, $\tilde{g}^{\prime\prime}$, $\tilde{\zeta}$, and
$\tilde{\zeta}^{\prime}$, which we record here and can be easily checked:

$\displaystyle|\tilde{g}(s)|+|\tilde{g}^{\prime}(s)|+|\tilde{\zeta}(s)|+|\tilde{\zeta}^{\prime}(s)|$
$\displaystyle\leq Ce^{-{\bar{\alpha}}s}\qquad$ $\displaystyle\forall s\geq
0\,,$ (5.31)
$\displaystyle|\tilde{g}(s)-g(0)|+|\tilde{g}^{\prime}(s)|+|\tilde{\zeta}(s)-\zeta(0)|+|\tilde{\zeta}^{\prime}(s)|$
$\displaystyle\leq Ce^{2s}\qquad$ $\displaystyle\forall s\leq 0\,.$ (5.32)

We observe next that, by (5.23)

$|h(s)|\leq Ce^{s}\,.$ (5.33)

Step 2. The equation for $h$. In these new coordinates the equation (5.24)
becomes then

$\left(1+\frac{1}{\alpha}\frac{d}{ds}-z_{0}-im\beta\tilde{\zeta}\right)\left(e^{-2s}(h^{\prime\prime}-m^{2}h)\right)-imhe^{-2s}\tilde{g}^{\prime}=0\,,$
(5.34)

which we simplify as

$\left[\frac{d}{ds}-\alpha\left(im\beta\tilde{\zeta}+z_{0}-1+\frac{2}{\alpha}\right)\right](h^{\prime\prime}-m^{2}h)-i\alpha
mh\tilde{g}^{\prime}=0\,.$ (5.35)

We then define the integrating factor

$I(s)=\exp\left[-\alpha\int_{0}^{s}\left(im\beta\tilde{\zeta}+z_{0}-1+\frac{2}{\alpha}\right)\,d\sigma\right]$

We can thus write

$\frac{d}{ds}\left[I(h^{\prime\prime}-m^{2}h)\right]=i\alpha
mIh\tilde{g}^{\prime}\,.$

Given that $z_{0}=a_{0}+ib_{0}$,

$|I(s)|\leq Ce^{-(2+\alpha(a_{0}-1))s}$ (5.36)

and in particular, by (5.31)

$|Ih\tilde{g}^{\prime}|(s)\leq Ce^{-(1+\alpha
a_{0}+{({\bar{\alpha}}-\alpha)})s}\,.$ (5.37)

This implies that the latter is an integrable function on every halfline
$[s,\infty[$ so that we can write

$\Gamma(s)=h^{\prime\prime}(s)-m^{2}h(s)=-\alpha
I(s)^{-1}\int_{s}^{\infty}imIh\tilde{g}^{\prime}\,.$ (5.38)

Since $\Gamma(s)=e^{2s}\vartheta(e^{s})$ and $\vartheta\in L^{2}(rdr)$,
$e^{-s}\Gamma\in L^{2}(ds)$. We claim in particular that $e^{-m|s|}\Gamma(s)$
is integrable. Indeed:

$\int_{\mathbb{R}}|\Gamma(s)|e^{-m|s|}\,ds\leq\|e^{-s}\Gamma\|_{L^{2}(\mathbb{R})}\left(\int_{\mathbb{R}}e^{-2(m|s|-s)}\,ds\right)^{1/2}<\infty\,.$

We claim then that for the function $h$ we have the formula

$h(s)=-\frac{1}{2m}e^{-ms}\int_{-\infty}^{s}e^{ms^{\prime}}\Gamma(s^{\prime})ds^{\prime}-\frac{1}{2m}e^{ms}\int_{s}^{\infty}e^{-ms^{\prime}}\Gamma(s^{\prime})\,ds^{\prime}\,.$
(5.39)

In order to prove the identity denote the right hand side by $H$ and observe
that it is a solution of the same ODE satisfied by $H$, namely,
$H^{\prime\prime}-m^{2}H=\Gamma$. Hence $(\frac{d^{2}}{ds^{2}}-m^{2})(H-h)=0$,
which implies that $H(s)-h(s)=C_{1}e^{ms}+C_{2}e^{-ms}$. On the other hand,
using the information that $e^{-s}\Gamma\in L^{2}$, it can be readily checked
that $H(s)=o(e^{m|s|})$ at both $\pm\infty$. Since this property is shared by
$h$, thanks to the bound (5.33), we conclude that $C_{1}e^{ms}+C_{2}e^{-ms}$
must be $o(e^{m|s|})$ at $\pm\infty$, implying $C_{1}=C_{2}=0$.

Step 3. Estimates at $+\infty$. In this step we give bounds for the asymptotic
behavior of $h$ at $+\infty$. We recall (5.31) and hence observe that

$|\Gamma(s)|\leq Ce^{(2+\alpha
a_{0}-\alpha)s}\int_{s}^{\infty}|h(\sigma)|e^{-(2+\alpha
a_{0}+{({\bar{\alpha}}-\alpha)})\sigma}\,d\sigma\,,$ (5.40)

for $s$ positive. On the other hand, for $s\geq 0$ we can also write from
(5.39)

$|h(s)|\leq
Ce^{-ms}+Ce^{-ms}\int_{0}^{s}e^{m\sigma}|\Gamma(\sigma)|\,d\sigma+Ce^{ms}\int_{s}^{\infty}e^{-m\sigma}|\Gamma(\sigma)|\,d\sigma\,,$
(5.41)

Starting with the information $|h(s)|\leq Ce^{s}$ for $s>0$, we then infer
from (5.40) that $|\Gamma(s)|\leq Ce^{(1-{\bar{\alpha}})s}$ for $s>0$. In turn
plugging the latter into (5.41) we infer $|h(s)|\leq Ce^{(1-{\bar{\alpha}})s}$
for $s>0$. The latter, plugged into (5.40) turns into $|\Gamma(s)|\leq
Ce^{(1-2{\bar{\alpha}})s}$ for $s>0$. We then can keep iterating this
procedure. The bootstrap argument can be repeated until we reach the largest
integer $k$ such that $(1-k{\bar{\alpha}})>-m$: one last iteration of the
argument gives then

$|h(s)|\leq Ce^{-ms}$ (5.42)

and hence, inserting one last time in (5.40)

$|\Gamma(s)|\leq Ce^{-(m+{\bar{\alpha}})s}\,.$ (5.43)

In order to estimate the first and second derivatives of $\Gamma$ we observe
that

$\frac{I^{\prime}}{I}=-\alpha(im\beta\tilde{\zeta}+z_{0}-1+2\alpha)$

and we compute explicitly

$\displaystyle\Gamma^{\prime}$
$\displaystyle=\alpha(im\beta\tilde{\zeta}+z_{0}-1+2\alpha)\Gamma+imh\tilde{g}^{\prime}$
(5.44) $\displaystyle\Gamma^{\prime\prime}$ $\displaystyle=\alpha
im\beta\tilde{\zeta}^{\prime}\Gamma+\alpha(im\beta\tilde{\zeta}+z_{0}-1+2\alpha)\Gamma^{\prime}+imh\tilde{g}^{\prime\prime}+imh^{\prime}\tilde{g}^{\prime}\,.$
(5.45)

From (5.44) and the bounds (5.43),(5.42), and (5.31), we immediately conclude

$|\Gamma^{\prime}(s)|\leq Ce^{-(m+{\bar{\alpha}})s}\,.$ (5.46)

As for the second derivative, using (5.46), (5.43),(5.42), and (5.31), we
conclude

$|\alpha
im\beta\tilde{\zeta}^{\prime}\Gamma+\alpha(im\beta\tilde{\zeta}+z_{0}-1+2\alpha)\Gamma^{\prime}+imh\tilde{g}^{\prime\prime}|\leq
Ce^{-(m+{\bar{\alpha}})s}\,.$

In order to estimate the term $imh^{\prime}\tilde{g}^{\prime}$ we
differentiate (5.39) to infer

$h^{\prime}(s)=\frac{1}{2}e^{-ms}\int_{-\infty}^{s}e^{ms^{\prime}}\Gamma(s^{\prime})ds^{\prime}-\frac{1}{2}e^{ms}\int_{s}^{\infty}e^{-ms^{\prime}}\Gamma(s^{\prime})\,ds^{\prime}\,.$
(5.47)

We then derive $|h^{\prime}(s)|\leq Ce^{-ms}$ using the same argument for
bounding $h$. In turn, combined again with (5.31) we conclude
$|imh^{\prime}\tilde{g}^{\prime}(s)|\leq Ce^{-(m+{\bar{\alpha}})s}$, hence
completing the proof of (5.30).

Step 4. Estimates at $-\infty$. For the bound at $-\infty$ we use instead
(5.32) (which we observe holds for positive $s$ as well). This leads to the
inequality

$|\Gamma(s)|\leq
Ce^{(2+\alpha(a_{0}-1))s}\int_{s}^{\infty}e^{-\alpha(a_{0}-1)\sigma}|h(\sigma)|\,d\sigma$
(5.48)

In this argument we assume that $a_{0}$ is selected very large, depending on
$m$. In turn we estimate $h$ for negative $s$ by

$|h(s)|\leq
Ce^{ms}+Ce^{ms}\int_{s}^{0}e^{-m\sigma}|\Gamma(\sigma)|\,d\sigma+Ce^{-ms}\int_{-\infty}^{s}e^{m\sigma}|\Gamma(\sigma)|\,d\sigma\,.$
(5.49)

Observe now that we have $|h(s)|\leq Ce^{s}$ for every $s$. Inserting this
bound in (5.48) and assuming that $a_{0}$ is large enough we conclude
$|\Gamma(s)|\leq Ce^{3s}$. In turn we can insert the latter bound in (5.49) to
conclude $|h(s)|\leq C(e^{ms}+e^{3s})$. Since $m\geq 2$ we can then conclude
$|h(s)|\leq Ce^{2s}$ and inserting it in (5.48) we conclude $|\Gamma(s)|\leq
Ce^{4s}$.

For the first and second derivatives we use the formulae (5.44), (5.44), and
(5.47) and argue as above to conclude
$|\Gamma^{\prime}(s)|+|\Gamma^{\prime\prime}(s)|\leq Ce^{4s}$.

Step 5. Estimate on $v=K_{2}*\eta$. Recall that we can explicitly compute
$K_{2}*\eta=\nabla^{\perp}\psi$ and that we already argued that $v\in
W^{1,2}_{loc}$. Hence to complete the proof we need to show

$\int_{\\{|x|\geq
1\\}}|\nabla\psi|^{2}\,dx=\int_{1}^{\infty}\left((f^{\prime}(r))^{2}+\frac{(f(r))^{2}}{r^{2}}\right)\,r\,dr<\infty\,.$
(5.50)

Recall next that $f(r)=h(\ln r)$ and $f^{\prime}(r)=\frac{1}{r}h^{\prime}(\ln
r)$. Thus the exponential decays for $h$ proved in the previous steps imply
$f(r)\leq C(1+r)^{-m}$ and $f^{\prime}(r)\leq C(1+r)^{-m-1}$. Since $m\geq 2$,
this clearly implies (5.50) and completes the proof. ∎

###  5.3. Proof of the baseline $L^{2}$ estimate

In this section we prove (5.7). In order to simplify the notation, from now on
we will use $\Omega$ in place $\Omega_{\text{per},k}$. We recall next equation
(2.47)

$\displaystyle(\partial_{\tau}-L_{\text{ss}})\Omega=$
$\displaystyle-\underbrace{(V_{\text{lin}}\cdot\nabla)\Omega}_{=:\mathscr{F}_{1}}-\underbrace{(V_{r}\cdot\nabla)\Omega}_{=:\mathscr{F}_{2}}-\underbrace{(V\cdot\nabla)\Omega_{\text{lin}}}_{=:\mathscr{F}_{3}}+\underbrace{(V\cdot\nabla)\Omega_{r}}_{=:\mathscr{F}_{4}}+\underbrace{(V\cdot\nabla)\Omega}_{=:\mathscr{F}_{5}}$
$\displaystyle-\underbrace{(V_{\text{lin}}\cdot\nabla)\Omega_{\text{lin}}}_{=:\mathscr{F}_{6}}-\underbrace{(V_{r}\cdot\nabla)\Omega_{\text{lin}}}_{=:\mathscr{F}_{7}}-\underbrace{(V_{\text{lin}}\cdot\nabla)\Omega_{r}}_{=:\mathscr{F}_{8}}\,.$
(5.51)

We then define $\mathscr{F}:=-\sum_{i=1}^{8}\mathscr{F}_{i}$. Recalling
Theorem 3.0.4 and the fact that $\Omega(\cdot,-k)=0$, we estimate via
Duhamel’s formula

$\|\Omega(\cdot,\bar{\tau})\|_{L^{2}}\leq
C(\varepsilon)\int_{-k}^{\bar{\tau}}e^{(a_{0}+\varepsilon)(\bar{\tau}-s)}\|\mathscr{F}(\cdot,s)\|_{L^{2}}\,ds\,.$
(5.52)

We next estimate the $L^{2}$ norms of the various $\mathscr{F}_{i}$. In order
to keep our notation simpler we use $\|\cdot\|_{2}$ for $\|\cdot\|_{L^{2}}$
and $\|\cdot\|_{\infty}$ for $\|\cdot\|_{L^{\infty}}$. $\mathscr{F}_{1}$ is
simple:

$\|\mathscr{F}_{1}(\cdot,s)\|_{2}\leq\|V_{\text{lin}}(\cdot,s)\|_{\infty}\|D\Omega(\cdot,s)\|_{L^{2}}\leq
Ce^{a_{0}s}e^{(a_{0}+\delta_{0})s}\leq Ce^{(2a_{0}+\delta_{0})s}\,.$ (5.53)

As for $\mathscr{F}_{2}$ we use the fact that

$\displaystyle\int|\mathscr{F}_{2}(\xi,\tau)|^{2}\,d\xi$ $\displaystyle\leq
C\int_{|\xi|\geq
e^{-\tau/\alpha}}|\xi|^{2-{2{\bar{\alpha}}}}|D\Omega(\xi,\tau)|^{2}\,d\xi$
$\displaystyle\leq
Ce^{{\frac{2{\bar{\alpha}}}{\alpha}}\tau}\int|\xi|^{2}|D\Omega(\xi,\tau)|^{2}\,d\xi\leq
Ce^{{\frac{2{\bar{\alpha}}}{\alpha}}\tau}\|\Omega(\cdot,\tau)\|^{2}_{X}\,.$

We hence conclude

$\|\mathscr{F}_{2}(\cdot,\tau)\|_{L^{2}}\leq
Ce^{(a_{0}+{\frac{{\bar{\alpha}}}{\alpha}}+\delta_{0})\tau}\leq
Ce^{(a_{0}+1+\delta_{0})\tau}\,.$ (5.54)

As for $\mathscr{F}_{3}$, for every fixed $\tau$ with $\kappa=\frac{1}{2}$ we
can use Proposition 5.0.2 to conclude

$\displaystyle\|\mathscr{F}_{3}(\cdot,\tau)\|_{L^{2}}$
$\displaystyle\leq\|V(\cdot,\tau)\|_{L^{\infty}}\|\nabla\Omega_{\text{lin}}(\cdot,\tau)\|_{L^{2}}$
(5.55) $\displaystyle\leq
C\|\Omega(\cdot,\tau)\|_{X}\|\nabla\Omega_{\text{lin}}(\cdot,\tau)\|_{L^{2}}$
(5.56) $\displaystyle\leq Ce^{(2a_{0}+{\delta_{0}})\tau}\,.$ (5.57)

To estimate $\mathscr{F}_{4}$ we recall that

$\Omega_{r}(\xi,\tau)=\beta(1-\chi(e^{\tau/\alpha}(\xi))\bar{\Omega}(\xi)+e^{\tau/\alpha}(\beta\zeta(|\xi|)|\xi|)\chi^{\prime}(e^{\tau/\alpha}\xi)\,.$
(5.58)

Differentiating the latter identity we get:

$\displaystyle|\nabla\Omega_{r}(\xi,\tau)|\leq$ $\displaystyle
C\mathbf{1}_{|\xi|\geq e^{-\tau/\alpha}}|D\bar{\Omega}|(\xi)$
$\displaystyle+Ce^{\tau/\alpha}(|\bar{\Omega}|(\xi)+|D(\zeta(|\xi|)|\xi|))\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}$
$\displaystyle+Ce^{2\tau/\alpha}(|\zeta(\xi)||\xi|)\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}$ $\displaystyle\leq$ $\displaystyle C\mathbf{1}_{|\xi|\geq
e^{-\tau/\alpha}}|\xi|^{-1-{\bar{\alpha}}}$
$\displaystyle+C(e^{\tau/\alpha}|\xi|^{-{\bar{\alpha}}}+e^{2\tau/\alpha}|\xi|^{1-{\bar{\alpha}}})\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}$ (5.59)

where we are assuming that $\textrm{spt}\,(\chi)\subset B_{R}$. We next use
Proposition 5.0.2 with $\kappa=\alpha/2$ to get
$\|V(\cdot,\tau)\|_{L^{\infty}}\leq C\|\Omega(\cdot,\tau)\|_{X}\leq
Ce^{(a_{0}+\delta_{0})\tau}$. In particular we can estimate

$\displaystyle\int|\mathscr{F}_{4}(\xi,\tau)|^{2}d\xi\leq$ $\displaystyle
Ce^{2(a_{0}+\delta_{0})\tau}\int_{e^{-\tau/\alpha}}^{\infty}r^{-1-{2{\bar{\alpha}}}}dr$
$\displaystyle+Ce^{2(a_{0}+\delta_{0}+1/\alpha)\tau}\int_{e^{-\tau/\alpha}}^{e^{-\tau/\alpha}R}r^{1-{2{\bar{\alpha}}}}\,dr$
$\displaystyle+Ce^{2(a_{0}+\delta_{0}+2/\alpha)\tau}\int_{e^{-\tau/\alpha}}^{e^{-\tau/\alpha}R}r^{3-{2{\bar{\alpha}}}}\,dr\leq
Ce^{(2a_{0}+2\delta_{0}+{2})\tau}\,.$

We thus conclude

$\|\mathscr{F}_{4}(\cdot,\tau)\|_{2}\leq Ce^{(a_{0}+\delta_{0}+1)\tau}\,.$
(5.60)

For $\mathscr{F}_{5}$ we use again $\|V(\cdot,\tau)\|_{L^{\infty}}\leq
Ce^{(a_{0}+\delta_{0})\tau}$ to get

$\|\mathscr{F}_{5}(\cdot,\tau)\|_{2}\leq
Ce^{(a_{0}+\delta_{0})\tau}\|D\Omega(\cdot,\tau)\|_{2}\leq
Ce^{2(a_{0}+\delta_{0})\tau}\,.$ (5.61)

$\mathscr{F}_{6}$ follows easily from Lemma 5.0.3

$\|\mathscr{F}_{6}(\cdot,\tau)\|_{2}\leq\|\Omega_{\text{lin}}(\cdot,\tau)\|_{\infty}\|\Omega_{\text{lin}}(\cdot,\tau)\|_{2}\leq
Ce^{2(a_{0}+\delta_{0})\tau}\,.$ (5.62)

$\mathscr{F}_{7}$ and $\mathscr{F}_{8}$ can be easily estimated using the
explicit formula for $\Omega_{r}$ and the decay estimates given by Lemma 5.0.3
for $\Omega_{\text{lin}}$, in particular they enjoy an estimate which is
better than (5.60), i.e.

$\|\mathscr{F}_{7}(\cdot,\tau)\|_{2}+\|\mathscr{F}_{8}(\cdot,\tau)\|_{2}\leq
Ce^{(a_{0}+\delta_{0}+1/2)\tau}\,.$ (5.63)

Assuming that $a_{0}$ is sufficiently large we then achieve the estimate

$\|\mathscr{F}(\cdot,\tau)\|_{2}\leq Ce^{(a_{0}+\delta_{0}+1/2)\tau}\,.$
(5.64)

Inserting in (5.52) we choose $\varepsilon<1/2+\delta_{0}$ and we then achieve

$\|\Omega(\cdot,\bar{\tau})\|_{2}\leq
Ce^{(a_{0}+\varepsilon)\bar{\tau}}\int_{-k}^{\bar{\tau}}e^{(\delta_{0}+1/2-\varepsilon)s}\,ds\leq
Ce^{(a_{0}+\delta_{0}+1/2)\bar{\tau}}\,.$ (5.65)

In fact, observe that the argument just given implies the stronger conclusion

$\|\Omega(\cdot,\tau)\|_{2}\leq
Ce^{(a_{0}+\delta_{0}+1/2)\tau}\,,\qquad\forall\tau\in[-k,\bar{\tau}]\,.$
(5.66)

###  5.4. Estimates on the first derivative

In this section we prove (5.8) and (5.9). The proof will be achieved via
$L^{2}$ and $L^{4}$ energy estimates, where we will differentiate (2.47) first
with respect to the angular variable and then with respect to the radial
variable. We start rewriting (2.47) as

$\displaystyle\partial_{\tau}\Omega-\Omega+\left(\left(-\frac{\xi}{\alpha}+\beta\bar{V}+V_{r}+V_{\text{lin}}+V\right)\cdot\nabla\right)\Omega$
$\displaystyle=-\beta(V\cdot\nabla)\bar{\Omega}-(V\cdot\nabla)\Omega_{\text{lin}}-(V\cdot\nabla)\Omega_{r}-(V_{\text{lin}}\cdot\nabla)\Omega_{\text{lin}}$
$\displaystyle\qquad-(V_{r}\cdot\nabla)\Omega_{\text{lin}}-(V_{\text{lin}}\cdot\nabla)\Omega_{r}$
$\displaystyle=:\mathscr{G}\,.$ (5.67)

We next differentiate in $\theta$. In order to simplify our notation we will
write $\theta$ in the subscript (or eventually $,\theta$ if there is already
another subscript). We also recall that $\Omega_{r}$, $\bar{\Omega}$ are
radial functions, while $(V_{r}\cdot\nabla)$ and $(\bar{V}\cdot\nabla)$ are
angular derivatives times a function which is radial, and $\xi\cdot\nabla$ is
a radial derivative times a radial function. So we can write

$\displaystyle\partial_{\tau}\Omega_{\theta}-\Omega_{\theta}+\left(\left(-\frac{\xi}{\alpha}+\beta\bar{V}+V_{r}+V_{\text{lin}}+V\right)\cdot\nabla\right)\Omega_{\theta}$
$\displaystyle=$
$\displaystyle\mathscr{G}_{\theta}-(V_{\text{lin},\theta}\cdot\nabla)\Omega-(V_{\theta}\cdot\nabla)\Omega=:\mathscr{H}_{1}\,$
(5.68)
$\displaystyle\partial_{\tau}\frac{\Omega_{\theta}}{r}-\left(\frac{1}{\alpha}+1\right)\frac{\Omega_{\theta}}{r}+\left(\left(-\frac{\xi}{\alpha}+\beta\bar{V}+V_{r}+V_{\text{lin}}+V\right)\cdot\nabla\right)\frac{\Omega_{\theta}}{r}$
$\displaystyle=$
$\displaystyle\frac{1}{r}\mathscr{G}_{\theta}-\frac{1}{r}(V_{\text{lin},\theta}\cdot\nabla)\Omega-\frac{1}{r}(V_{\theta}\cdot\nabla)\Omega+\Omega_{\theta}((V_{\text{lin}}+V)\cdot\nabla)\frac{1}{r}=:\mathscr{H}_{2}\,.$
(5.69)

We then multiply by $\Omega_{\theta}$ the first equation and integrate by
parts the terms on the left-hand side to conclude

$\displaystyle\frac{d}{d\tau}\frac{1}{2}\|\Omega_{\theta}(\cdot,\tau)\|_{2}^{2}$
$\displaystyle=\left(1-\frac{1}{\alpha}\right)\|\Omega_{\theta}(\cdot,\tau)\|_{2}^{2}+\int\mathscr{H}_{1}(\xi,\tau)\Omega_{\theta}(\xi,\tau)$
$\displaystyle\leq\|\mathscr{H}_{1}(\cdot,\tau)\|_{2}\|\Omega_{\theta}(\cdot,\tau)\|_{2}$
(5.70)

Likewise we multiply the second identity by $(\frac{1}{r}\Omega_{\theta})^{3}$
and integrate by parts to achieve

$\displaystyle\frac{d}{d\tau}\frac{1}{4}\|r^{-1}\Omega_{\theta}(\cdot,\tau)\|_{4}^{4}$
$\displaystyle=\left(1+\frac{1}{2\alpha}\right)\|r^{-1}\Omega_{\theta}(\cdot,\tau)\|_{4}^{4}+\int\mathscr{H}_{2}(\xi,\tau)(r^{-1}\Omega_{\theta}(\xi,\tau))^{3}$

which implies

$\|r^{-1}\Omega_{\theta}(\cdot,\tau)\|_{4}\leq\int_{-k}^{\tau}e^{\left(1+\frac{1}{2\alpha}\right)(\tau-\hat{\tau})}\|\mathscr{H}_{2}(\cdot,\hat{\tau})\|_{4}\,d\hat{\tau}\,\quad\forall\tau\in[-k,\bar{\tau}]\,.$
(5.71)

We next wish to estimate the two integrals in the right-hand sides of both
equations.

We summarize the relevant estimates in the Lemma 5.4.1 below. Note that they
imply

$\displaystyle\|\Omega_{\theta}(\cdot,\hat{\tau})\|_{2}$ $\displaystyle\leq
Ce^{(a_{0}+2\delta_{0})\hat{\tau}}\,\qquad$
$\displaystyle\forall\hat{\tau}\in[-k,\bar{\tau}]\,,$ (5.72)
$\displaystyle\|r^{-1}\Omega_{\theta}(\cdot,\hat{\tau})\|_{4}$
$\displaystyle\leq Ce^{(a_{0}+2\delta_{0})\hat{\tau}}\qquad$
$\displaystyle\forall\hat{\tau}\in[-k,\bar{\tau}]\,,$ (5.73)

provided $a_{0}$ is chosen big enough.

######  Lemma 5.4.1.

Under the assumptions of Lemma 5.0.4 we have

$\displaystyle\|D\mathscr{G}(\cdot,\tau)\|_{4}$ $\displaystyle\leq
Ce^{(a_{0}+2\delta_{0})\tau}$ (5.74)
$\displaystyle\|rD\mathscr{G}(\cdot,\tau)\|_{2}$ $\displaystyle\leq
Ce^{(a_{0}+2\delta_{0})\tau}$ (5.75)
$\displaystyle\||DV_{\text{lin}}||\nabla\Omega|(\cdot,\tau)\|_{4}+\||DV||\nabla\Omega|(\cdot,\tau)\|_{4}$
$\displaystyle\leq Ce^{(a_{0}+2\delta_{0})\tau}$ (5.76)
$\displaystyle\|r|DV_{\text{lin}}||\nabla\Omega|(\cdot,\tau)\|_{2}+\|r|DV||\nabla\Omega|(\cdot,\tau)\|_{2}$
$\displaystyle\leq Ce^{(a_{0}+2\delta_{0})\tau}$ (5.77)
$\displaystyle\|r^{-1}V_{\text{lin}}D\Omega(\cdot,\tau)\|_{4}+\|r^{-1}VD\Omega(\cdot,\tau)\|_{4}$
$\displaystyle\leq Ce^{(a_{0}+2\delta_{0})\tau}\,.$ (5.78)

###### Proof.

Proof of (5.74) and of (5.75). We break the terms as

$\displaystyle\|D\mathscr{G}\|_{4}\leq$ $\displaystyle
C\|DVD\bar{\Omega}\|_{4}+C\|VD^{2}\bar{\Omega}\|_{4}+\|DVD\Omega_{\text{lin}}\|_{4}$
$\displaystyle+\|VD^{2}\Omega_{\text{lin}}\|_{4}+C\|DVD\Omega_{r}\|_{4}+C\|VD^{2}\Omega_{r}\|_{4}$
$\displaystyle+C\|DV_{\text{lin}}D\Omega_{\text{lin}}\|_{4}+C\|V_{\text{lin}}D^{2}\Omega_{\text{lin}}\|_{4}+\|DV_{r}D\Omega_{\text{lin}}\|_{4}$
$\displaystyle+\|V_{r}D^{2}\Omega_{\text{lin}}\|_{4}+\|DV_{\text{lin}}D\Omega_{r}\|_{4}+\|V_{\text{lin}}D^{2}\Omega_{r}\|_{4}\,,$
(5.79)

and

$\displaystyle\|rD\mathscr{G}\|_{2}\leq$ $\displaystyle
C\|rDVD\bar{\Omega}\|_{2}+C\|rVD^{2}\bar{\Omega}\|_{2}+\|rDVD\Omega_{\text{lin}}\|_{2}$
$\displaystyle+\|rVD^{2}\Omega_{\text{lin}}\|_{2}+C\|rDVD\Omega_{r}\|_{2}+C\|VrD^{2}\Omega_{r}\|_{2}$
$\displaystyle+C\|rDV_{\text{lin}}D\Omega_{\text{lin}}\|_{2}+C\|rV_{\text{lin}}D^{2}\Omega_{\text{lin}}\|_{2}+\|rDV_{r}D\Omega_{\text{lin}}\|_{2}$
$\displaystyle+\|rV_{r}D^{2}\Omega_{\text{lin}}\|_{2}+\|rDV_{\text{lin}}D\Omega_{r}\|_{2}+\|rV_{\text{lin}}D^{2}\Omega_{r}\|_{2}\,.$
(5.80)

The terms involving $\Omega$ and $\bar{\Omega}$ is where we use the baseline
$L^{2}$ estimate. Observe that

$\|\Omega(\cdot,\tau)\|_{4}\leq\|\Omega(\cdot,\tau)\|_{\infty}^{1/2}\|D\Omega(\cdot,\tau)\|^{1/2}_{2}\leq
Ce^{(a_{0}+\delta_{0}+1/4)\tau}\,$ (5.81)

and, by Calderón-Zygmund,

$\|DK_{2}*\Omega(\cdot,\tau)\|_{4}\leq C\|\Omega(\cdot,\tau)\|_{4}\leq
Ce^{(a_{0}+\delta_{0}+1/4)\tau}\,.$ (5.82)

Next we estimate

$\displaystyle\|DVD\bar{\Omega}(\cdot,\tau)\|_{4}$
$\displaystyle\leq\|D\bar{\Omega}(\cdot,\tau)\|_{\infty}\|DV(\cdot,\tau)\|_{4}$
$\displaystyle\leq C\|\Omega(\cdot,\tau)\|_{4}$ $\displaystyle\leq
Ce^{(a_{0}+\delta_{0}+1/4)\tau}\,,$ (5.83)
$\displaystyle\|rDVD\bar{\Omega}(\cdot,\tau)\|_{L^{2}}$
$\displaystyle\leq\|rD\bar{\Omega}(\cdot,\tau)\|_{\infty}\|DV(\cdot,\tau)\|_{2}$
$\displaystyle\leq C\|\Omega(\cdot,\tau)\|_{L^{2}}$ $\displaystyle\leq
Ce^{(a_{0}+\delta_{0}+1/2)\,\tau}\,.$ (5.84)

Next, recalling Lemma 2.4.1 we get

$\|V(\cdot,\tau)\|_{L^{2}(B_{R})}\leq CR\|\Omega(\cdot,\tau)\|_{2}\leq
CRe^{(a_{0}+\delta_{0}+1/2)\tau}\,.$ (5.85)

However, using $\int_{B_{R}}V(\cdot,\tau)=0$, we can in fact estimate also

$\|V(\cdot,\tau)\|_{L^{4}(B_{R})}\leq CR^{1/2}\|\Omega(\cdot,\tau)\|_{2}\leq
CR^{1/2}e^{(a_{0}+\delta_{0}+1/2)\tau}\,.$ (5.86)

In particular we can infer

$\|(1+|\cdot|)^{-(1+\varepsilon)}V(\cdot,\tau)\|_{2}\leq
C(\varepsilon)e^{(a_{0}+\delta_{0}+1/2)\tau}$ (5.87)

and

$\|(1+|\cdot|)^{-(1/2+\varepsilon)}V(\cdot,\tau)\|_{4}\leq
C(\varepsilon)e^{(a_{0}+\delta_{0}+1/2)\tau}$ (5.88)

for every positive $\varepsilon$. On the other hand, given that
$|D^{2}\bar{\Omega}(\xi)|\leq C(1+|\xi|)^{-2-{\bar{\alpha}}}$, we easily infer

$\displaystyle\|VD^{2}\bar{\Omega}(\cdot,\tau)\|_{2}$ $\displaystyle\leq
C\|(1+|\cdot|)^{-{1}}V(\cdot,\tau)\|_{4}\leq Ce^{(a_{0}+\delta_{0}+1/2)\tau}$
(5.89) $\displaystyle\|rVD^{2}\bar{\Omega}(\cdot,\tau)\|_{2}$
$\displaystyle\leq C\|(1+|\cdot|)^{-1-{\bar{\alpha}}}V(\cdot,\tau)\|_{2}\leq
Ce^{(a_{0}+\delta_{0}+1/2)\tau}\,.$ (5.90)

From now on we will not handle the terms with the weight $r$ as the proof is
entirely analogous: we will just focus on the $L^{4}$ estimates and leave to
the reader the computations with the weight.

For the two quadratic terms in $V_{\text{lin}}$ we can use Lemma 5.0.3 to
achieve

$\|DV_{\text{lin}}D\Omega_{\text{lin}}(\cdot,\tau)\|_{4}+\|V_{\text{lin}}D^{2}\Omega_{\text{lin}}(\cdot,\tau)\|_{4}\leq
Ce^{2a_{0}\tau}\,.$ (5.91)

Likewise we can estimate

$\|DVD\Omega_{\text{lin}}(\cdot,\tau)\|_{4}+\|VD^{2}\Omega_{\text{lin}}\|_{4}\leq
Ce^{a_{0}\tau}\|\Omega(\cdot,\tau)\|_{X}\leq Ce^{(2a_{0}+\delta_{0})\tau}\,,$
(5.92)

(where for the second term we use the decay at infinity of
$D^{2}\Omega_{\text{lin}}$ to compensate the moderate growth of $V$, the
argument is the same as for (5.89) and we do not repeat it here). Observe next
that, by (5.59), $\|D\Omega_{r}(\cdot,\tau)\|_{\infty}\leq C$ for $\tau\leq
0$. Hence the term $DVD\Omega_{r}$ can be estimated as in (5.83):

$\|DVD\Omega_{r}(\cdot,\tau)\|_{4}\leq C\|DV(\cdot,\tau)\|_{4}\leq
Ce^{(a_{0}+\delta_{0}+1/4)\tau}\,.$ (5.93)

As for the other term, differentiating once more and arguing as for (5.59) we
get:

$\displaystyle|D^{2}\Omega_{r}(\xi,\tau)|\leq$ $\displaystyle
C|\xi|^{-2-{\bar{\alpha}}}\mathbf{1}_{|\xi|\geq e^{-\tau/\alpha}}$
$\displaystyle+(e^{\tau/\alpha}|\xi|^{-1-{\bar{\alpha}}}+e^{2\tau/\alpha}|\xi|^{-{\bar{\alpha}}}+e^{3\tau/\alpha}|\xi|^{1-{\bar{\alpha}}})\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}$ $\displaystyle\leq$ $\displaystyle
C|\xi|^{-2-{\bar{\alpha}}}\mathbf{1}_{|\xi|\geq
e^{-\tau/\alpha}}+Ce^{3\tau/\alpha}|\xi|^{1-{\bar{\alpha}}}\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}\,.$ (5.94)

We can thus argue similarly as for (5.89) to conclude

$\displaystyle\|VD^{2}\bar{\Omega}_{r}(\cdot,\tau)\|_{4}$ $\displaystyle\leq
C\|(1+|\cdot|)^{-3/2}V(\cdot,\tau)\|_{4}+Ce^{3\tau/\alpha}\|V\|_{L^{4}(B_{Re^{-\xi/\tau}})}$
$\displaystyle\leq Ce^{(a_{0}+\delta_{0}+1/4)\tau}\,.$ (5.95)

In order to handle the remaining three terms, we recall that, by Lemma 5.0.3,

$\displaystyle\|V_{\text{lin}}(\cdot,\tau)\|_{\infty}$ $\displaystyle\leq
Ce^{a_{0}\tau}$ (5.96) $\displaystyle|DV_{\text{lin}}(\xi,\tau)|$
$\displaystyle\leq Ce^{a_{0}\tau}|\xi|^{-2-{\bar{\alpha}}}$ (5.97)
$\displaystyle|D^{k}\Omega_{\text{lin}}(\xi,\tau)|$ $\displaystyle\leq
Ce^{a_{0}\tau}|\xi|^{-2-k-{\bar{\alpha}}}\,.$ (5.98)

On the other hand, owing to the computations in this and the previous section
we can also write

$\displaystyle|V_{r}(\xi,\tau)|$ $\displaystyle\leq
C|\xi|^{1-{\bar{\alpha}}}\mathbf{1}_{|\xi|\geq e^{-\tau/\alpha}}$
$\displaystyle|DV_{r}(\xi,\tau)|+|\Omega_{r}(\xi,\tau)|$ $\displaystyle\leq
C|\xi|^{-{\bar{\alpha}}}\mathbf{1}_{|\xi|\geq
e^{-\tau/\alpha}}+Ce^{\tau/\alpha}|\xi|^{1-{\bar{\alpha}}}\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}$ $\displaystyle|D^{k}\Omega_{r}(\xi,\tau)|$
$\displaystyle\leq C|\xi|^{-k-{\bar{\alpha}}}\mathbf{1}_{|\xi|\geq
e^{-\tau/\alpha}}$
$\displaystyle\qquad+Ce^{k\tau/\alpha}|\xi|^{1-{\bar{\alpha}}}\mathbf{1}_{e^{-\tau/\alpha}R\geq|\xi|\geq
e^{-\tau/\alpha}}\,.$

Integrating the estimates in the respective domain we easily get

$\|DV_{r}D\Omega_{\text{lin}}\|_{4}+\|V_{r}D^{2}\Omega_{\text{lin}}\|_{4}+\|DV_{\text{lin}}D\Omega_{r}\|_{4}+\|V_{\text{lin}}D^{2}\Omega_{r}\|_{4}\leq
Ce^{(a_{0}+1)\tau}\,.$ (5.99)

Remaining estimates. The two terms (5.76) and (5.77) have already been covered
in the argument above. It remains to handle (5.78). Notice that, by Lemma
5.0.3 and Proposition 5.0.2 we have

$\displaystyle\|r^{-1}V(\cdot,\tau)\|_{\infty}$ $\displaystyle\leq
C\|\Omega(\cdot,\tau)\|_{X}\leq Ce^{(a_{0}+\delta_{0})\tau}$ (5.100)
$\displaystyle\|r^{-1}V_{\text{lin}}(\cdot,\tau)\|_{\infty}$
$\displaystyle\leq Ce^{a_{0}\tau}\,.$ (5.101)

We thus conclude easily

$\displaystyle\|r^{-1}VD\Omega(\cdot,\tau)\|_{4}$ $\displaystyle\leq
Ce^{(a_{0}+\delta_{0})\tau}\|D\Omega(\cdot,\tau)\|_{4}\leq
Ce^{2(a_{0}+\delta_{0})\tau}\,$ (5.102)
$\displaystyle\|r^{-1}V_{\text{lin}}D\Omega(\cdot,\tau)\|_{4}$
$\displaystyle\leq Ce^{a_{0}\tau}\|D\Omega(\cdot,\tau)\|_{4}\leq
Ce^{(2a_{0}+\delta_{0})\tau}\,.$ (5.103)

∎

We next differentiate in $r$ (5.67) in order to achieve similar identities to
(5.68) and (5.69). This time, given the ambiguity with $V_{r}$ and
$\Omega_{r}$, we write $,r$ in the subscript to denote the radial derivative
of any function.

$\displaystyle\partial_{\tau}\Omega_{,r}+\left(1-\frac{1}{\alpha}\right)\Omega_{,r}+\left(\left(-\frac{\xi}{\alpha}+\beta\bar{V}+V_{r}+V_{\text{lin}}+V\right)\cdot\nabla\right)\Omega_{,r}$
$\displaystyle=$
$\displaystyle\mathscr{G}_{,r}-(V_{\text{lin},r}\cdot\nabla)\Omega-(V_{,r}\cdot\nabla)\Omega-\beta(\bar{V}_{,r}\cdot\nabla)\Omega-(V_{r,r}\cdot\nabla)\Omega$
(5.104)
$\displaystyle\partial_{\tau}r\Omega_{,r}+r\Omega_{,r}+\left(\left(-\frac{\xi}{\alpha}+\beta\bar{V}+V_{r}+V_{\text{lin}}+V\right)\cdot\nabla\right)(r\Omega_{,r})$
$\displaystyle=$ $\displaystyle
r\mathscr{G}_{,r}-r(V_{\text{lin},r}\cdot\nabla)\Omega-r(V_{,r}\cdot\nabla)\Omega-r\beta(\bar{V}_{,r}\cdot\nabla)\Omega-r(V_{r,r}\cdot\nabla)\Omega$
$\displaystyle+\Omega_{,r}(V_{\text{lin}}+V)\cdot\nabla r\,.$ (5.105)

Multiplying by $(\Omega_{,r})^{3}$ and $r\Omega_{,r}$ respectively, and using
the estimates (5.74) and (5.75), we achieve, in the respective cases:

$\displaystyle\frac{d}{d\tau}\|\Omega_{,r}(\cdot,\tau)\|_{4}$
$\displaystyle\leq
Ce^{(a_{0}+2\delta_{0})\tau}+C\|DV_{\text{lin}}D\Omega(\cdot,\tau)\|_{4}+C\|DVD\Omega(\cdot,\tau)\|_{4}$
$\displaystyle\qquad\qquad+C\|\bar{V}_{,r}\cdot\nabla\Omega(\cdot,\tau)\|_{4}+\|DV_{r}D\Omega(\cdot,\tau)\|_{4}\,,$
(5.106) $\displaystyle\frac{d}{d\tau}\|r\Omega_{,r}(\cdot,\tau)\|_{2}$
$\displaystyle\leq
Ce^{(a_{0}+2\delta_{0})\tau}+C\|rV_{\text{lin},r}D\Omega(\cdot,\tau)\|_{2}$
$\displaystyle+C\|r(V_{,r}\cdot\nabla)\Omega(\cdot,\tau)\|_{2}+C\|r\bar{V}_{,r}\cdot\nabla\Omega(\cdot,\tau)\|_{2}$
$\displaystyle+\|r(V_{r,r}\cdot\nabla)\Omega(\cdot,\tau)\|_{2}+\|V_{\text{lin}}D\Omega\|_{2}+\|VD\Omega\|_{2}\,.$
(5.107)

Note next that

$\displaystyle\|DV_{\text{lin}}D\Omega(\cdot,\tau)\|_{4}+\|DVD\Omega(\cdot,\tau)\|_{4}$
(5.108)
$\displaystyle\leq(\|DV_{\text{lin}}(\cdot,\tau)\|_{\infty}+\|DV(\cdot,\tau)\|_{\infty})\|D\Omega(\cdot,\tau)\|_{4}\leq
Ce^{(2a_{0}+\delta_{0})\tau}\,,$

and likewise

$\displaystyle\|rDV_{\text{lin}}D\Omega(\cdot,\tau)\|_{2}$
$\displaystyle+\|rDVD\Omega(\cdot,\tau)\|_{2}$
$\displaystyle\leq(\|DV_{\text{lin}}(\cdot,\tau)\|_{\infty}+\|DV(\cdot,\tau)\|_{\infty})\|rD\Omega(\cdot,\tau)\|_{2}$
$\displaystyle\leq Ce^{(2a_{0}+\delta_{0})\tau}\,,$ (5.109)

The terms $(\bar{V}_{,r}\cdot\nabla)\Omega$ and
$r(\bar{V}_{,r}\cdot\nabla)\Omega$ can be bounded observing that they involve
only the angular derivative and that $\bar{V}_{,r}$ is bounded. Since the
angular derivative has already been estimated (this is the reason for
estimating it before estimating the radial derivative), we get

$\displaystyle\|\bar{V}_{,r}\cdot\nabla\Omega(\cdot,\tau)\|_{4}$
$\displaystyle\leq C\|r^{-1}\Omega_{\theta}(\cdot,\tau)\|_{4}\leq
Ce^{(a_{0}+2\delta_{0})\tau}\,.$ (5.110)
$\displaystyle\|r\bar{V}_{,r}\cdot\nabla\Omega(\cdot,\tau)\|_{2}$
$\displaystyle\leq C\|\Omega_{\theta}(\cdot,\tau)\|_{2}\leq
Ce^{(a_{0}+2\delta_{0})\tau}\,.$ (5.111)

As for $DV_{r}D\Omega$ and $rDV_{r}D\Omega$ we observe that
$\|DV_{r}\|_{\infty}\leq e^{\tau}$ and thus we easily get

$\displaystyle\|DV_{r}D\Omega(\cdot,\tau)\|_{4}$ $\displaystyle\leq
Ce^{\tau}\|D\Omega(\cdot,\tau)\|_{4}\leq Ce^{(a_{0}+\delta_{0}+1)\tau}$
(5.112) $\displaystyle\|rDV_{r}D\Omega(\cdot,\tau)\|_{2}$ $\displaystyle\leq
Ce^{\tau}\|rD\Omega(\cdot,\tau)\|_{2}\leq Ce^{(a_{0}+\delta_{0}+1)\tau}\,.$
(5.113)

We finally need to estimate $\|V_{\text{lin}}D\Omega\|_{2}$ and
$\|VD\Omega\|_{2}$, but we observe that this has already been done in the
previous section, since they correspond to the terms $\mathscr{F}_{1}$ and
$\mathscr{F}_{4}$ in (5.51), cf. (5.53) and (5.60). Summarizing, we conclude

$\displaystyle\frac{d}{d\tau}\|\Omega_{,r}(\cdot,\tau)\|_{2}$
$\displaystyle\leq Ce^{(a_{0}+\delta_{0}+1/2)\tau}$ (5.114)
$\displaystyle\frac{d}{d\tau}\|r\Omega_{,r}(\cdot,\tau)\|_{2}$
$\displaystyle\leq Ce^{(a_{0}+\delta_{0}+1/2)\tau}\,,$ (5.115)

which we then integrate between $0$ and $\bar{\tau}$ to achieve the desired
conclusion.

##  Appendix A A more detailed spectral analysis

###  A.1. From Remark 4.0.3(i) to Remark 3.0.2(c)

Let us assume the validity of 4.0.3(i) and prove Remark 3.0.2(c). Let
$m_{0}\geq 2$ be the integer such that

$\displaystyle\textrm{spec}\,(L_{\text{st}},U_{m_{0}})\cap\\{\textrm{Re}\,z>0\\}\neq\emptyset\,,$
(A.1)
$\displaystyle\textrm{spec}\,(L_{\text{st}},U_{m})\cap\\{\textrm{Re}\,z>0\\}=\emptyset\,\quad\text{for
any $m>m_{0}$}\,.$ (A.2)

We show that Remark 3.0.2(c) holds with $m=m_{0}$.

For any $z\in\textrm{spec}_{m_{0}}\,(L_{\text{st}})\cap\\{\textrm{Re}\,z>0\\}$
we denote by $V_{z}:=P_{z}(L^{2}_{m_{0}})$ the image of the Riesz projector

$P_{z}=\frac{1}{2\pi i}\int_{\gamma}(w-L_{\text{st}})^{-1}dw\,,$