where $\gamma$ parameterizes the boundary of a ball containing $z$ and no
other eigenvalues of $L_{\text{st}}$.

It is enough to show that $P_{z}(U_{km_{0}})=\\{0\\}$ for any
$k\in\mathbb{Z}\setminus\\{-1,1\\}$,
$z\in\textrm{spec}_{m_{0}}\,(L_{\text{st}})\cap\\{\textrm{Re}\,z>0\\}$ since
it gives

$V_{z}=P_{z}(U_{m_{0}}\cup U_{-m_{0}})\subset U_{m_{0}}\cup U_{-m_{0}}\,,$

where the second inclusion follows from the fact that $U_{m}$ is always an
invariant space of $L_{\text{s}t}$.

If $k>1$, from (A.1) we know that
$z\notin\textrm{spec}\,(L_{\text{st}},U_{km_{0}})$, hence $P_{z}(U_{km_{0}})$
is trivial. If $k<-1$, we reduce to the previous situation by observing that
$P_{z}(U_{km_{0}})=\overline{P_{\bar{z}}(U_{-km_{0}})}$.

###  A.2. Proof of Remark 4.0.3(i)

In order to show this point, given Lemma 4.9.1, we just need to prove the
following statement.

######  Lemma A.2.1.

For every fixed $\Xi\in\mathscr{C}$ there is $M_{0}>0$ such that
$\mathscr{U}_{m}$ is empty for every $m\geq M_{0}$.

Indeed, given the conclusion above we infer that $\mathscr{U}_{m}$ is empty
for every $m\geq m_{a}$ and it thus suffices to select $m_{0}$ as the largest
integer strictly smaller than $m_{a}$.

Before coming to the proof of Lemma A.2.1 we state an auxiliary fact which
will be used in the argument and which can be readily inferred from the
computations in Step 1 of the proof of Lemma 4.8.1.

######  Lemma A.2.2.

For every $0<\sigma<\tau<1$ there is a constant $C$ (depending only upon
$\sigma$ and $\tau$) such that
$B_{z}:=\mathcal{K}_{m_{0}}\circ\frac{1}{\Xi-z}$ is a bounded operator from
$C^{\sigma}$ to $C^{\tau}$ for every $z$ with $\textrm{Im}\,z>0$ and

$\|B_{z}\|_{\mathcal{L}(C^{\sigma},C^{\tau})}\leq C\,.$

###### Proof of Lemma A.2.1.

The proof will be by contradiction and thus, assuming that the statement is
false, we can select:

  * (i)

a sequence $\\{m_{j}\\}\subset[1,\infty[$ with $m_{j}\to\infty$;

  * (ii)

a sequence $\\{z_{j}\\}\subset\mathbb{C}$ with $\textrm{Im}\,z_{j}>0$;

  * (iii)

and a sequence $\\{\psi_{j}\\}\subset L^{2}(\mathbb{R})$ solving the equation

$-\frac{d^{2}\psi_{j}}{dt^{2}}+m_{j}^{2}\psi_{j}+\frac{A}{\Xi-
z_{j}}\psi_{j}=0\,.$ (A.3)

Step 1. We first prove that $\\{z_{j}\\}$ is bounded and every cluster point
must be an element of $[0,\Xi(-\infty)]$. Otherwise for a subsequence, not
relabeled, we get the estimate

$\sup\left\|\frac{A}{\Xi-z_{j}}\right\|_{L^{\infty}}=:C_{0}<\infty\,.$

By scalar multiplying (A.3) by $\psi_{j}$ and taking the real part of the
resulting equation we then conclude

$\int(|\psi_{j}^{\prime}|^{2}+m_{j}^{2}|\psi_{j}|^{2})\leq
C_{0}\int|\psi_{j}|^{2}\,,$

which clearly it is not feasible because $C_{0}<m_{j}^{2}$ for a sufficiently
large $j$ (and $\psi_{j}$ is nontrivial).

Up to subsequences we can thus assume that $z_{j}$ converges to some
$z_{0}\in[0,\Xi(-\infty)]$.

Step 2. We next analyze the cases $z_{0}=0$ and $z_{0}=\Xi(-\infty)$. The
argument is similar to that used in Section 4.6 in case (C). Let us argue
first for $z_{0}=0$. We observe that $\Xi^{-1}|A|$ belongs to
$L^{1}(]-\infty,N])$ for any fixed $N$ and that, likewise, $|\Xi-
z_{j}|^{-1}|A|$ have a uniform $L^{1}$ bound on any $]-\infty,N]$. We can then
use the Lemma 4.5.2 to normalize $\psi_{j}$ so that it is asymptotic to
$e^{m_{j}t}$ and also to write

$\psi_{j}(t)=e^{m_{j}t}(1+z_{j}(t))$

with

$|z_{j}(t)|\leq\exp\left(\frac{1}{m_{j}}\int_{-\infty}^{N}\frac{|A|}{|\Xi-
z_{j}|}\right)-1\qquad\mbox{for all $t\leq N$.}$

In particular, we have $|z_{j}(t)|\leq\frac{C(N)}{m_{j}}$ on $]-\infty,N]$. We
next scalar multiply (A.3) by $\psi_{j}$ and take the imaginary part to
conclude

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

In particular, since $\frac{A}{|\Xi-z_{j}|^{2}}$ is bounded from above by a
constant $C$ independent of $j$ on $[a,b]$ and $-\frac{A}{|\Xi-z_{j}|^{2}}$ is
bounded from below by a constant $c>0$ independent of $j$ on $[b+1,b+2]$, we
conclude

$\int_{b+1}^{b+2}|\psi_{j}|^{2}\leq\frac{C}{c}\int_{a}^{b}|\psi_{j}|^{2}\,.$

We next choose $N$ larger than $b+2$ and use the estimate
$|z_{j}(t)|\leq\frac{C(N)}{m_{j}}$ to argue that, for $j$ large enough, we
have $\frac{1}{2}e^{m_{j}t}\leq|\psi_{j}(t)|\leq 2e^{m_{j}t}$ on
$]-\infty,N]$. In particular, we infer

$\int_{b+1}^{b+2}e^{2m_{j}t}\leq C\int_{a}^{b}e^{2m_{j}t}$

provided the constant $C$ is chosen large enough (but independent of $j$) and
$j$ is large enough. The latter inequality is certainly impossible for $m_{j}$
large enough, leading to a contradiction.

The argument to exclude $z_{0}=\Xi(-\infty)$ is entirely analogous, this time
normalizing for $t\to\infty$ and reaching an inequality of type

$\int_{a-2}^{a-1}e^{-2m_{j}t}\leq C\int_{a}^{b}e^{-2m_{j}t}$

for a constant $C$ independent of $j$ and any $j$ large enough.

Step 3. We next examine the last case, that is $z_{0}=\Xi(c)$. This time we
fix a $\sigma\in\,]0,1[$ and normalize $\psi_{j}$ so that

$\|\psi_{j}\|_{C^{\sigma}}=1\,.$ (A.4)

We observe that

$\psi_{j}=-\mathcal{K}_{m_{j}}\left(\frac{A}{\Xi-z_{j}}\psi_{j}\right)\,,$

and also recall that
$\mathcal{K}_{m_{j}}(\varphi)=\frac{1}{2m_{j}}e^{-m_{j}|\cdot|}*\varphi$. We
set $m_{0}=m_{a}$ write further

$\psi_{j}=-\mathcal{K}_{m_{j}}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)\left(\mathcal{K}_{m_{0}}\left(\frac{A}{\Xi-
z_{j}}\right)\psi_{j}\right)\,.$

Recalling Lemma A.2.2, we can fix a $\tau\in]\sigma,1[$ to achieve

$\left\|\left(\mathcal{K}_{m_{0}}\left(\frac{A}{\Xi-
z_{j}}\right)\psi_{j}\right)\right\|_{C^{\tau}}\leq C$

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

We will show in the final step that

  * (Cl)

$\|\mathcal{K}_{m_{j}}\circ(-\frac{d^{2}}{dt^{2}}+m_{0}^{2})\|_{\mathcal{L}(C^{\tau},C^{\tau})}\leq
C$ for some constant $C$ independent of $k$.

In particular, we achieve

$\|\psi_{j}\|_{C^{\tau}}\leq C\,.$ (A.5)

We now wish to show that indeed $\|\psi_{j}\|_{C^{\sigma}}\leq\frac{1}{2}$ for
$j$ large enough, which obviously would be a contradiction. In order to
achieve the latter estimate we use a Littlewood-Paley decomposition. We fix a
cut-off function $\chi$ which is supported in $]\frac{1}{2},2[$, define
$\chi_{\ell}(t):=\chi(2^{-\ell}t)$ for $\ell\in\mathbb{N}\cup\\{0\\}$ and
assume that $\chi$ has been chosen so that

$\sum_{\ell\in\mathbb{N}\cup\\{0\\}}\chi_{\ell}\equiv 1\qquad\mbox{on
$[1,\infty[$}.$

We then define

$\chi_{-1}:=1-\sum_{\ell\in\mathbb{N}}\chi_{\ell}\,$

and introduce the Littlewood-Paley operator $\Delta_{\ell}$ as
$\Delta_{\ell}(\varphi)=\mathscr{F}^{-1}(\chi_{\ell}\mathscr{F}(\varphi))$,
where $\mathscr{F}$ is the Fourier transform. We finally recall that (see [18,
Section 1.4.2]), if we define

$\|\varphi\|_{X^{\sigma}}:=\sum_{\ell\geq-1}2^{\sigma\ell}\|\Delta_{\ell}\varphi\|_{L^{\infty}}\,,$

then

$C(\sigma)^{-1}\|\varphi\|_{X^{\sigma}}\leq\|\varphi\|_{C^{\sigma}}\leq
C(\sigma)\|\varphi\|_{X^{\sigma}}\,.$

We are now ready to perform our final estimate. We fix a large $N$, which will
be chosen later, and for $\ell\geq N$ we write

$\displaystyle\sum_{\ell\geq
N}2^{\sigma\ell}\|\Delta_{\ell}\psi_{j}\|_{\infty}$ $\displaystyle\leq
2^{-N(\tau-\sigma)}\sum_{\ell\geq
N}2^{\tau\ell}\|\Delta_{\ell}\psi_{j}\|_{\infty}$ $\displaystyle\leq
2^{-N(\tau-\sigma)}C(\tau)\|\psi_{j}\|_{C^{\tau}}\leq C2^{-N(\tau-\sigma)}\,,$

where the constant $C$ is independent of both $N$ and $j$. Next, for any
$\ell$ we observe that

$\Delta_{\ell}\psi_{j}=\mathcal{K}_{m_{j}}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)\underbrace{\left(\Delta_{\ell}\left(\mathcal{K}_{m_{0}}\left(\frac{A}{\Xi-z}\psi_{j}\right)\right)\right)}_{=:\Gamma_{\ell,j}}\,.$

Now

$\|\Gamma_{\ell,j}\|_{L^{\infty}}\leq
C2^{-\ell\sigma}\left\|\mathcal{K}_{m_{0}}\left(\frac{A}{\Xi-z}\psi_{j}\right)\right\|_{C^{\sigma}}\leq
C2^{-\ell\sigma}\,.$

On the other hand, because of the frequency localization, we have

$\Delta_{\ell}\psi_{j}=\mathcal{K}_{m_{j}}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)\circ(\Delta_{\ell-1}+\Delta_{\ell}+\Delta_{\ell+1})(\Gamma_{\ell,j})$

and the estimate

$\left\|\mathcal{K}_{m_{j}}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)\circ\Delta_{\ell}\right\|_{\mathcal{L}(L^{\infty},L^{\infty})}\leq\frac{C}{m_{j}^{2}}\left(2^{2\ell}+m_{0}^{2}\right)\,.$

We can therefore write the estimate

$\displaystyle\|\psi_{j}\|_{C^{\sigma}}$
$\displaystyle\leq\frac{C}{m_{j}^{2}}\sum_{\ell=-1}^{N}(2^{2\ell}+m_{0}^{2})+C2^{-N(\tau-\sigma)}$
$\displaystyle\leq\frac{CN}{m_{j}^{2}}\left(2^{2N}+m_{0}^{2}\right)+C2^{-N(\tau-\sigma)}\,,$

where the constants $C$ are independent of $N$ and $j$. In particular, we fix
first $N$ large enough to get $C2^{-N(\tau-\sigma)}\leq\frac{1}{4}$ and we
then choose $m_{j}$ large enough so that

$\frac{CN}{m_{j}^{2}}\left(2^{2N}+m_{0}^{2}\right)\leq\frac{1}{4}\,.$

These two estimates imply $\|\psi_{j}\|_{C^{\sigma}}\leq\frac{1}{2}$,
contradicting the normalization (A.4).

Step 4. To complete the proof of the Lemma we need to show (Cl). We first
write

$T_{m,\ell}:=\Delta_{\ell}\circ\mathcal{K}_{m}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)\,.$

The operator $T_{m,\ell}$ is the convolution with a kernel $K_{m,\ell}$ whose
Fourier symbol is given by
$\chi_{\ell}(\xi)\frac{|\xi|^{2}+m_{0}^{2}}{|\xi|^{2}+m^{2}}$. Hence, for
$\ell\geq 0$ we have

$K_{m,\ell}(t)=\frac{1}{2\pi}\int\chi\left(\frac{\xi}{2^{\ell}}\right)\frac{|\xi|^{2}+m_{0}^{2}}{|\xi|^{2}+m^{2}}e^{i\xi
t}\,d\xi\,.$

and

$(-it)^{k}K_{m,\ell}(t)=\frac{1}{2\pi}\int\frac{d^{k}}{d\xi^{k}}\left(\chi\left(\frac{\xi}{2^{\ell}}\right)\frac{|\xi|^{2}+m_{0}^{2}}{|\xi|^{2}+m^{2}}\right)e^{it\xi}\,d\xi\,.$

In particular, we easily conclude

$\||t|^{k}K_{m,\ell}\|_{L^{\infty}}\leq C(k)2^{\ell(1-k)}\,,$

for a constant $C(k)$ independent of both $m\geq 1$ and $\ell$, but which
depends on $k$. From the latter we can estimate

$\displaystyle\|K_{m,\ell}\|_{L^{1}}$ $\displaystyle\leq\int_{|t|\leq
2^{-\ell}}|K_{m,\ell}(s)|\,ds+\int_{|t|\geq
2^{-\ell}}\frac{|s^{2}K_{m,\ell}(s)|}{|s|^{2}}\,ds$ $\displaystyle\leq
C+C2^{-\ell}\int_{2^{-\ell}}^{\infty}\frac{1}{s^{2}}\,ds\leq C\,.$

For $\ell=-1$ we just conclude, likewise

$\||t|^{k}K_{m,-1}\|_{L^{\infty}}\leq C(k)$

for a constant $C(k)$ independent of $m$, but depending of $k$. Once again
using the cases $k=0$ and $2$ of the latter inequality we achieve

$\|K_{m,-1}\|_{L^{1}}\leq C\,.$

We have thus bounded all $\|K_{m,\ell}\|_{L^{1}}$ with a universal constant
$C$ independent of both $m\geq 1$ and $\ell\in\mathbb{N}\cup\\{-1\\}$. In
particular, since
$\|T_{m,\ell}\|_{\mathcal{L}(L^{\infty},L^{\infty})}=\|K_{m,\ell}\|_{L^{1}}$
and

$\mathcal{K}_{m}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)=\sum_{\ell\geq-1}T_{m,\ell}=\sum_{\ell\geq-1}T_{m,\ell}\circ(\Delta_{\ell-1}+\Delta_{\ell}+\Delta_{\ell+1})\,,$

we can estimate

$\displaystyle\left\|\mathcal{K}_{m}\circ\left(-\frac{d^{2}}{dt^{2}}+m_{0}^{2}\right)(\varphi)\right\|_{C^{\sigma}}$
$\displaystyle\leq
C(\sigma)\sum_{\ell\geq-1}2^{\sigma\ell}\|T_{m,\ell}(\varphi)\|_{L^{\infty}}$
$\displaystyle=C(\sigma)\sum_{\ell\geq-1}2^{\sigma\ell}\|T_{m,\ell}(\Delta_{\ell}\varphi)\|_{L^{\infty}}$
$\displaystyle\leq
C(\sigma)\sum_{\ell\geq-1}2^{\sigma\ell}\|\Delta_{\ell}\varphi\|_{L^{\infty}}$
$\displaystyle\leq C(\sigma)\|\varphi\|_{C^{\sigma}}\,.$

This completes the proof of (Cl) and hence of the entire Lemma. ∎

###  A.3. Proof of Theorem 4.0.4

In [40], Vishik claims the following improved version of Theorem 4.2.4, which
would immediately imply Theorem 4.0.4.

######  Theorem A.3.1.

There are a function $\Xi\in\mathscr{C}$ and an integer $m_{0}\geq 2$ such
that $\mathscr{U}_{m}=\emptyset$ for any integer $m>m_{0}$ and
$\mathscr{U}_{m_{0}}$ consists of a single $z$. Moreover, the algebraic
multiplicity of $m_{0}z$ as an eigenvalue of $\mathcal{L}_{m_{0}}$ is $1$.

Vishik’s suggested proof of Theorem A.3.1 builds upon Proposition 4.4.1 and
the following improved versions of Proposition 4.4.3 and Proposition 4.4.5.

######  Proposition A.3.2.

Assume $-\lambda_{a}<-1$ and let $m_{a}=\sqrt{\lambda_{a}}$. Then there exist
$\varepsilon>0$ and $\delta>0$ with the following property. For every
$h\in]0,\delta[$, $\mathscr{U}_{m_{a}-h}\cap
B_{\varepsilon}(\Xi(a))=\\{z_{m_{a}-h}\\}$, where $(m_{a}-h)z_{m_{a}-h}$ is an
eigenvalue of $\mathcal{L}_{m_{a}-h}$ with algebraic multiplicity $1$.

In [40] Vishik only gives the argument that $\mathscr{U}_{m_{a}-h}\cap
B_{\varepsilon}(\Xi(a))$ contains a single element $z_{m_{a}-h}$ and the
corresponding eigenspace of $(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}$ has
dimension $1$ (i.e. its geometric multiplicity is $1$, cf. Remark 4.4.4).
However it is essential to have the algebraic multiplicity equal to $1$ in
order to complete his suggested argument. After we pointed out to him the gap
in his paper, he suggested in [41] the proof of Proposition A.3.2 reported
below. Before coming to it, we point out that a spectral perturbation argument
as in the proof of Lemma 4.9.1 (which we outline anyway below) easily implies
the following.

######  Proposition A.3.3.

Assume $-\lambda_{a}<-1$ and let

$\displaystyle m_{a}$ $\displaystyle:=\sqrt{\lambda_{a}}\,,$ $\displaystyle
m_{b}$ $\displaystyle:=\max\\{1,\sqrt{\lambda_{b}}\\}\,.$

Then $\mathscr{U}_{m}$ consists of a single element $z_{m}$ for every
$m\in]m_{b},m_{a}[$ and moreover the algebraic multiplicity of $z_{m}$ as an
eigenvalue of $m^{-1}\mathcal{L}_{m}$ is $1$.

Taking the previous proposition for granted, we just need a choice of $\Xi$
for which $\lambda_{a}>1$ and $]m_{b},m_{a}[$ contains an integer, which is
guaranteed by Lemma 4.4.7, and we conclude Theorem A.3.1.

We now explain how to prove Proposition A.3.3. From Proposition 4.4.5 and
Lemma A.2.1 we know that $\mathscr{U}_{m}\neq\emptyset$, for every
$m\in]m_{b},m_{a}[$, and $\mathscr{U}_{m}=\emptyset$ for $m\geq m_{a}$.
Moreover, Remark 4.9.2 implies that the sum of the algebraic multiplicities of
$z\in\mathscr{U}_{m}$, as eigenvalues of $m^{-1}\mathcal{L}_{m}$, is constant
for $m\in]m_{b},m_{a}[$. Hence, to conclude we just need to prove that the
latter is $1$ for some $m\in]m_{b},m_{a}[$. To that aim we show that for any
$\varepsilon>0$ there exists $\delta>0$ such that
$\mathscr{U}_{m_{a}-h}=\mathscr{U}_{m_{a}-h}\cap B_{\varepsilon}(\Xi(a))$ for
any $h\in]0,\delta[$. This is enough for our purposes since, together with
Proposition A.3.2, it gives $\mathscr{U}_{m_{a}-h}=\mathscr{U}_{m_{a}-h}\cap
B_{\varepsilon}(\Xi(a))=\\{z_{m_{a}-h}\\}$ where $z_{m_{a}-h}$ is an
eigenvalue of $(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}$ with algebraic
multiplicity $1$.

Assume for contradiction the existence of a sequence
$(m_{j})_{j\in\mathbb{N}}$ in $]m_{b},m_{a}[$ converging to $m_{a}$ such that
there are $z_{j}\in\mathscr{U}_{m_{j}}$ with $|z_{j}-\Xi(a)|>\varepsilon$ for
some $\varepsilon>0$. Up to extracting a subsequence, we may assume $z_{j}\to
z$ for some $z\in\mathbb{C}$ with $|z-\Xi(a)|\geq\varepsilon$. Proposition
4.4.1 implies that the imaginary part of $z$ is positive. Arguing as in the
first step of proof of Proposition 4.4.1 we can prove that
$z\in\mathscr{U}_{m_{a}}$ and reach a contradiction.

###  A.4. Proof of Proposition A.3.2

The proof of Proposition A.3.2 can be reduced to the following weaker version
using Remark 4.9.2 and the argument outlined in the previous paragraph.

######  Proposition A.4.1.

Assume $-\lambda_{a}<-1$ and let $m_{a}=\sqrt{\lambda_{a}}$. Let $h$ and
$\varepsilon$ be sufficiently small so that Proposition 4.4.1 and Remark 4.4.4
apply, namely $\mathscr{U}_{m_{a}-h}\cap
B_{\varepsilon}(\Xi(a))=\\{z_{m_{a}-h}\\}$, where $z_{m_{a}-h}$ is an
eigenvalue of $(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}$ with geometric
multiplicity $1$. Then, if $h$ is chosen possibly smaller, the algebraic
multiplicity of $z_{m_{a}-h}$ is also $1$.

We now come to the proof of the latter, which is the heart of the matter.
First of all, we introduce a suitable transformation of the space
$\mathcal{H}$ (which we recall is the domain of the operator
$\mathcal{L}_{m}$, defined in (4.1)). We introduce the Hilbert space

$\mathcal{H}^{e}:=\left\\{f:\mathbb{R}\to\mathbb{C}\,:\,\int|f(t)|^{2}e^{-2t}\,dt<\infty\right\\}$

and the isometry $T:\mathcal{H}\to\mathcal{H}^{e}$ given by

$\gamma(r)\mapsto e^{2t}\gamma(e^{t})\,.$

Rather than considering the operator $\mathcal{L}_{m}$ on $\mathcal{H}$, it
turns out to be more convenient to consider the operator
$T\circ\mathcal{L}_{m}\circ T^{-1}$ on $\mathcal{H}^{e}$. Since the spectra of
the two operators coincide, with a slight abuse of notation we will keep
writing $\mathcal{L}_{m}$ in place of $T\circ\mathcal{L}_{m}\circ T^{-1}$, and
we will keep $\mathscr{U}_{m}$ to denote the point spectrum of
${m^{-1}}T\circ\mathcal{L}_{m}\circ T^{-1}$ in the upper half plane.

Simple computations show that the operator $\mathcal{L}_{m}$ is given, on
$\mathcal{H}^{e}$ by

$\mathcal{L}_{m}(\alpha)=m\Xi\alpha-mA\varphi$

where $\varphi$ is the unique $\mathcal{H}^{e}$ solution of

$\varphi^{\prime\prime}-m^{2}\varphi=\alpha\,.$

We can now come to the main idea behind the simplicity of $z_{m_{a}-h}$, which
is borrowed from [41]. A prominent role is played by the adjoint of
$\mathcal{L}_{m}$ (considered as a bounded linear operator from
$\mathcal{H}^{e}$ into itself): for the latter we will use the notation
$\mathcal{L}_{m}^{\star}$.

######  Lemma A.4.2.

Suppose that $h$ and $\varepsilon$ are sufficiently small such that

$\\{z_{m_{a}-h}\\}=\mathscr{U}_{m_{a}-h}\cap B_{\varepsilon}(\Xi(a))$

and $z_{m_{a}-h}$ has geometric multiplicity $1$ in
$\textrm{spec}\,((m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h},\mathcal{H}^{e})$. Let
$\alpha_{h}\in\mathcal{H}^{e}\setminus\\{0\\}$ be such that
$(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}(\alpha_{h})-z_{m_{a}-h}\alpha_{h}=0$. If
$h$ is small enough, then there is $\beta_{h}\in\mathcal{H}^{e}$ such that
$(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}^{\star}(\beta_{h})-\bar{z}_{m_{a}-h}\beta_{h}=0$
and

$\langle\alpha_{h},\beta_{h}\rangle_{\mathcal{H}^{e}}=\int\alpha_{h}(t)\bar{\beta}_{h}(t)\,e^{-2t}dt\neq
0\,.$ (A.6)

Let us show how the latter implies Proposition A.4.1. Assume $z_{m_{a}-h}$
were an element of
$\textrm{spec}\,((m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h},\mathcal{H}^{e})\cap
B_{\varepsilon}(\Xi(a))$ with geometric multiplicity $1$ and algebraic
multiplicity larger than $1$: our goal is to show that $h$ cannot be too
small. The properties just listed mean that the following bounded operator on
$\mathcal{H}^{e}$,

$L_{h}:=(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}-z_{m_{a}-h}\,,$

has a 1-dimensional kernel, $0$ is in its point spectrum, and $0$ has
algebraic multiplicity strictly larger than $1$. These properties imply that
any element $\alpha_{h}$ in the kernel of $L_{h}$ (i.e. any eigenfunction of
$(m_{a}-h)^{-1}\mathcal{L}_{m_{a}-h}$ with eigenvalue $z_{m_{a}-h}$) is in the
image of $L_{h}$. Fix one such element $\alpha_{h}$ and let $\eta_{h}$ be such
that $L_{h}(\eta_{h})=\alpha_{h}$. If $h$ is small enough, we can fix
$\beta_{h}$ as in Lemma A.4.2, and observe that it is in the kernel of the
adjoint operator $L^{\star}_{h}$. We then must have

$0\neq\int\alpha_{h}\bar{\beta}_{h}\,e^{-2t}dt=\int
L_{h}(\eta_{h})\bar{\beta}_{h}\,e^{-2t}dt=\int\eta_{h}\overline{L_{h}^{\star}(\beta_{h})}\,e^{-2t}dt=0\,,$

which is not possible.

###### Proof of Lemma A.4.2.

We begin by proving the following claim:

  * (Cl)

For any $z\in\mathcal{U}_{m}$, with $m>1$, such that
$m^{-1}\mathcal{L}_{m}(\alpha_{z})-z\alpha_{z}=0$, there exists
$\beta_{z}\in\mathcal{H}^{e}$ such that

$m^{-1}\mathcal{L}_{m}^{\star}(\beta_{z})-\bar{z}\beta_{z}=0\,,$ (A.7)

and

$\left\langle\alpha_{z},\beta_{z}\right\rangle_{\mathcal{H}^{e}}=\int_{\mathbb{R}}\frac{A(t)}{(\Xi(t)-z)^{2}}\varphi_{z}(t)^{2}\,dt\,,$
(A.8)

where $\varphi_{z}$ is the unique solution in $\mathcal{H}^{e}$ of
$\varphi_{z}^{\prime\prime}-m^{2}\varphi_{z}=\alpha_{z}$.

To that aim we first observe that the adjoint of $\mathcal{L}_{m}$ in
$\mathcal{H}^{e}$ is given by

$\mathcal{L}_{m}^{\star}(\alpha)=m(\Xi\alpha+e^{2t}\mathcal{K}_{m}(A\alpha
e^{-2t}))\,,$ (A.9)

where $\mathcal{K}_{m}$ is the inverse of $-\frac{d^{2}}{dt^{2}}+m^{2}$ as a
closed unbounded self-adjoint operator in $L^{2}(\mathbb{R})$. Notice that
$\mathcal{L}^{\star}$ is well defined because $e^{-t}\alpha\in
L^{2}(\mathbb{R})$ and $A\sim e^{2t}$ as $t\to-\infty$.

We now observe that, if $z\in\mathcal{U}_{m}$,
$m^{-1}\mathcal{L}_{m}(\alpha_{z})=z\alpha_{z}$, and $\beta_{z}$ is defined by

$\beta_{z}:=e^{2t}\frac{\bar{\varphi}_{z}}{\Xi-\bar{z}}=-e^{2t}\frac{\mathcal{K}_{m}(\bar{\alpha}_{z})}{\Xi-\bar{z}}\,,$

then

$m^{-1}\mathcal{L}_{m}^{\star}(\beta_{z})=\bar{z}\beta_{z}\,.$ (A.10)

Notice that $\beta_{z}\in W^{2,2}_{\textrm{loc}}\cap\mathcal{H}^{e}$ decays
exponentially fast at $\infty$ thanks to the bound $|\varphi_{z}(t)|\leq
Ce^{-m|t|}$, for every $t\in\mathbb{R}$, proven in Lemma 4.5.3. Let us now
verify (A.10): We first observe that

$\alpha_{z}=\frac{A\varphi_{z}}{\Xi-z}\,,$ (A.11)

hence

$\displaystyle m^{-1}\mathcal{L}_{m}^{\star}(\beta_{z})$
$\displaystyle=\Xi\beta_{z}+e^{2t}\mathcal{K}_{m}\left(\frac{A\bar{\varphi}_{z}}{\Xi-\bar{z}}\right)=\Xi\beta_{z}+e^{2t}\mathcal{K}_{m}(\bar{\alpha}_{z})$
$\displaystyle=\Xi\beta_{z}-(\Xi-\bar{z})\beta_{z}=\bar{z}\beta_{z}\,.$

It is now immediate to conclude (A.8).

In order to simplify our notation we use $\mathcal{L}_{h}$, $z_{h}$ and
$m_{h}$ in place of $\mathcal{L}_{m_{a}-h}$, $z_{a-h}$ and $m_{a}-h$. Given
$\alpha_{h}\in\mathcal{H}^{e}$ as in the statement of the Lemma we denote by
$\varphi_{h}$ the unique $L^{2}$ solution of
$\varphi_{h}^{\prime\prime}-m_{h}^{2}\varphi_{h}=\alpha_{h}$. We now can apply
(Cl) above to find $\beta_{h}\in\mathcal{H}^{e}$ which solves
$m_{h}^{-1}\mathcal{L}_{h}^{\star}(\beta_{h})=\bar{z}_{h}\beta_{h}$ and such
that

$\left\langle\alpha_{h},\beta_{h}\right\rangle_{\mathcal{H}^{e}}=\int_{\mathbb{R}}\frac{A(t)}{(\Xi(t)-z_{h})^{2}}\varphi_{h}(t)^{2}\,dt\,.$
(A.12)

To conclude the proof it suffices to show that, after appropriately
normalizing the functions $\alpha_{h}$ (i.e. after multiplying them by an
appropriate constant factor, which might depend on $h$) we have

$\lim_{h\to
0}\left\langle\alpha_{h},\beta_{h}\right\rangle_{\mathcal{H}^{e}}=c\neq 0\,.$
(A.13)

Note that for the latter conclusion, which we will prove in the next two
steps, we will use the assumption that $\alpha_{h}\neq 0$.

Step 1: We show that, up to multiplication of $\alpha_{h}$ by a suitable
constant factor (which might vary with $h$), $\varphi_{h}\to\varphi$ in
$W^{1,2}$ and in $C^{1,\alpha}$, as $h\to 0$, where $\varphi\in W^{2,\infty}$
is a nontrivial solution to

$-\frac{d^{2}\varphi}{dt^{2}}+m_{a}^{2}\varphi+\frac{A}{\Xi-\Xi(a)}\varphi=0\,.$
(A.14)

By Remark 4.6.1, the nontriviality of $\varphi$ implies $\varphi(a)\neq 0$,
and hence, up to multiplication by another constant factor, we will assume,
without loss of generality, that $\varphi(a)=1$.

Recall that $\varphi_{h}$ solves the equation

$-\frac{d^{2}\varphi_{h}}{dt^{2}}+m_{h}^{2}\varphi_{h}+\frac{A}{\Xi-
z_{h}}\varphi_{h}=0\,.$ (A.15)

For the moment, let us normalize the functions so that

$\int(|\varphi_{h}^{\prime}|^{2}+m_{h}^{2}|\varphi_{h}|^{2})=1\,,$ (A.16)

as in (4.68). We then can argue as for the bounds (4.69) and (4.70) to derive
the existence of constants $C$ and $\beta>2$ (independent of $h$) such that

$\left|\varphi_{h}(t)\right|\leq Ce^{-\beta|t|}\quad\forall t\,.$ (A.17)

Recalling Section 4.8, we know that $z_{h}=\Xi(a)+c(a)h+o(h)$, where $c(a)$ is
a complex number with positive imaginary part, which we denote by $d(a)$.
Using the monotonicity of $\Xi$ we can write $z_{h}=\Xi(t(h))+i(d(a)h+o(h))$
for some $t(h)$ which satisfies the bound $|t(h)-a|\leq Ch$ for some positive
constant $C$. In particular, using the mean value theorem and the fact that
the derivative of $\Xi$ does not vanish on $[a-1,a+1]$, we get

$|\Xi(t)-z_{h}|^{2}\geq C^{-1}(|t-t(h)|^{2}+|h|^{2})\qquad\forall
t\in[a-1,a+1]\,,$

where $C$ is some positive constant. Next, using that $|t(h)-a|\leq Ch$, we
conclude the estimate

$|\Xi(t)-z_{h}|\geq C^{-1}|t-a|\qquad\forall t\in[a-1,a+1]\,,$

with a constant $C$ independent of $h$. Since $a$ is a zero of $A$, we finally
conclude that the functions

$\frac{A(t)}{\Xi(t)-z_{h}}$

are in fact uniformly bounded, independently of $h$. Using the latter estimate
and (A.17) we thus infer that

$|\varphi_{h}^{\prime\prime}(t)|\leq Ce^{-\beta|t|}\,.$ (A.18)

In particular, upon extraction of a subsequence, we can assume that the
$\varphi_{h}$ converge to a function $\varphi$ strongly in $W^{1,2}$, weakly
in $W^{2,\infty}$, and hence strongly in $C^{1,\alpha}$ for every $\alpha<1$.
In particular, because of the normalization (A.16), $\varphi$ is a nontrivial
$W^{2,\infty}$ function, satisfying the same exponential decay as in (A.17)
and (A.18). Moreover, given the bound on the functions
$\frac{A(t)}{\Xi(t)-z_{h}}$, $\varphi$ is in fact a solution of

$-\frac{d^{2}\varphi}{dt^{2}}+m_{a}^{2}\varphi+\frac{A}{\Xi-\Xi(a)}\varphi=0\,.$
(A.19)

Recalling Remark 4.6.1, $\varphi(a)\neq 0$ and $\varphi$ is unique up to a
constant factor. In particular, we must have

$\liminf_{h\downarrow 0}|\varphi_{h}(a)|>0,$ (A.20)

otherwise for a suitable subsequence we would have convergence to a nontrivial
solution $\varphi$ for which $\varphi(a)=0$. Because of (A.20) we can use the
different normalization $\varphi_{h}(a)=1$, which in turn implies that
$\varphi_{h}$ converges (without extracting subsequences) to the unique
$W^{2,2}$ solution $\varphi$ of (A.19) which satisfies $\varphi(a)=1$.

Step 2: We prove that

$\lim_{h\to
0}\,\textrm{Im}\int_{\mathbb{R}}\frac{A(t)}{(\Xi(t)-z_{h})^{2}}\varphi_{h}(t)^{2}\,dt=\frac{2A^{\prime}(a)\varphi(a)^{2}}{d(a)\Xi^{\prime}(a)^{2}}\int_{\mathbb{R}}\frac{s^{2}}{(1+s^{2})^{2}}ds\,.$
(A.21)

Recalling that $z_{h}=\Xi(t(h))+i(d(a)h+o(h))$, we write

$\displaystyle\quad\;\textrm{Im}\,\left[\frac{A}{(\Xi-
z_{h})^{2}}\varphi_{h}^{2}\right]$
$\displaystyle=\textrm{Im}\,\left[\frac{A((\Xi-\Xi(t(h)))+i(d(a)h+o(h)))^{2}}{((\Xi-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2})^{2}}(\textrm{Re}\,\varphi_{h}+i\textrm{Im}\,\varphi_{h})^{2}\right]$
$\displaystyle=\frac{2(d(a)h+o(h))A(\Xi-\Xi(t(h)))}{((\Xi-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2})^{2}}(\mathrm{Re}\,\varphi_{h}^{2}-\textrm{Im}\,\varphi_{h}^{2})$
$\displaystyle\qquad+\frac{2A}{(\Xi-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2}}\textrm{Re}\,\varphi_{h}\textrm{Im}\,\varphi_{h}$
$\displaystyle\qquad-\frac{4(d(a)h+o(h))^{2}A}{((\Xi-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2})^{2}}\textrm{Re}\,\varphi_{h}\textrm{Im}\,\varphi_{h}$
$\displaystyle=:I_{h}+II_{h}+III_{h}\,.$

To ease notation we set

$f_{h}:=\textrm{Re}\,\varphi_{h}^{2}-\textrm{Im}\,\varphi_{h}^{2}\,,\quad
g_{h}:=\textrm{Re}\,\varphi_{h}\,\textrm{Im}\varphi_{h}\,,$ (A.22)

and observe that $f_{h}\to\varphi^{2}$, $g_{h}\to 0$ as $h\to 0$, where the
convergence is, in both cases, in the strong topologies of $L^{2}$ and of
$C^{\alpha}$, for every $\alpha<1$. We will show below that:

$\displaystyle\lim_{h\to 0}\int I_{h}$
$\displaystyle=\frac{2A^{\prime}(a)\varphi(a)^{2}}{d(a)\Xi^{\prime}(a)^{2}}\int_{\mathbb{R}}\frac{s^{2}}{(1+s^{2})^{2}}ds=:L(a)\,,$
(A.23) $\displaystyle\lim_{h\to 0}\int II_{h}$ $\displaystyle=0\,,$ (A.24)
$\displaystyle\lim_{h\to 0}\int III_{h}$ $\displaystyle=0\,.$ (A.25)

Considering that none of the numbers $\Xi^{\prime}(a)$, $A^{\prime}(a)$,
$\varphi(a)$, and $d(a)$ vanish, $L(a)\neq 0$. This implies (A.13) and
concludes the proof. We next study separately the three limits above.

Proof of (A.23). There exists $\delta>0$ and $r>0$ such that for any $h$
sufficiently small one has $|\Xi(t)-\Xi(t(h))|>\delta$ for all
$t\in\mathbb{R}\setminus(a-r/2,a+r/2)$. This implies that

$\lim_{h\to 0}\int_{\mathbb{R}\setminus(t(h)-r,t(h)+r)}I_{h}=0\,,$ (A.26)

hence, we are left with

$\lim_{h\to
0}2hd(a)\int_{t(h)-r}^{t(h)+r}\frac{A(t)(\Xi(t)-\Xi(t(h))}{((\Xi(t)-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2})^{2}}f_{h}(t)\,dt\,.$
(A.27)

We change variables according to $t=t(h)+sh$:

$\displaystyle C(h)$
$\displaystyle:=2\int_{-\frac{r}{h}}^{\frac{r}{h}}s\left(\frac{A(t(h)+sh)}{h}\frac{\Xi(t(h)+sh)-\Xi(t(h))}{sh}\right)\times$
$\displaystyle\left(s^{2}\left(\frac{\Xi(t(h)+sh)-\Xi(t(h))}{sh}\right)^{2}+(d(a)+o(h)/h)^{2}\right)^{-2}f_{h}(t(h)+sh)\,ds\,.$

Notice that, for any $s\in\mathbb{R}$, we have

$\lim_{h\to 0}\frac{\Xi(t(h)+sh)-\Xi(t(h))}{sh}=\Xi^{\prime}(a)\,.$ (A.28)

Moreover the monotonicity of $\Xi$ implies that

$1/C\leq\left|\frac{\Xi(t(h)+sh)-\Xi(t(h))}{sh}\right|\leq C\quad\text{for any
$s\in(-r/h,r/h)$}\,.$ (A.29)

Notice that

$\frac{A(t(h)+sh)}{h}=\frac{A(t(h)+sh)-A(t(h))}{h}+\frac{A(t(h))-A(a)}{h}\,,$
(A.30)

hence, up to extracting a subsequence $h_{i}\to 0$, we have

$\frac{A(t(h_{i})+sh_{i})}{h_{i}}\to A^{\prime}(a)s+x$ (A.31)

for some $x\in\mathbb{R}$ (recall that $|t(h)-a|\leq Ch$ and note that $x$
might depend on the subsequence).

Collecting all the estimates above, and using the dominated convergence
theorem we deduce that, along the subsequence $h_{i}$,

$\lim_{i\to\infty}C(h_{i})=2\int_{\mathbb{R}}\frac{s(A^{\prime}(a)s+x)\Xi^{\prime}(a)}{(s^{2}\Xi^{\prime}(a)^{2}+d(a)^{2})^{2}}\varphi(a)^{2}\,ds=\frac{2A^{\prime}(a)\varphi(a)^{2}}{d(a)\Xi^{\prime}(a)^{2}}\int_{\mathbb{R}}\frac{s^{2}}{(1+s^{2})^{2}}ds\,.$
(A.32)

Observe that the limit does not depend on $x$ and hence does not depend on the
chosen subsequence.

Proof of (A.25). Arguing as we did for $I_{h}$ and using that $g_{h}\to 0$ in
$C^{1/2}$, as $h\to 0$, we easily deduce that

$\lim_{h\to 0}\int III_{h}=0\,.$ (A.33)

Proof of (A.24). We need to show that

$\lim_{h\to
0}\int_{\mathbb{R}}\frac{A(t)}{(\Xi(t)-\Xi(t(h))^{2}+(d(a)h+o(h))^{2}}g_{h}(t)\,dt=0\,.$
(A.34)

Observe that
$G_{h}:=g_{h}/|\varphi_{h}|^{2}=|\varphi_{h}|^{-2}\textrm{Re}\,\varphi_{h}\textrm{Im}\,\varphi_{h}$,
and in particular, $|G_{h}|\leq\frac{1}{2}$. Moreover there exists $r>0$ such
that $G_{h}\to 0$ in $(a-r,a+r)$ in the $C^{\alpha}$ topology for every
$\alpha<1$ (here, we are using that
$|\varphi_{h}(a)|^{2}\to|\varphi(a)|^{2}\neq 0$ as $h\to 0$). We write

$\displaystyle\int_{\mathbb{R}}$
$\displaystyle\frac{A(t)|\varphi_{h}(t)|^{2}}{(\Xi(t)-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2}}G_{h}(t)\,dt$
(A.35)
$\displaystyle=\int_{\mathbb{R}}\frac{A(t)|\varphi_{h}(t)|^{2}}{(\Xi(t)-\Xi(t(h))^{2}+(d(a)h+o(h))^{2}}(G_{h}(t)-G_{h}(t(h)))dt\,,$
(A.36)

where we took advantage of the identity

$\int_{\mathbb{R}}\frac{A(t)|\varphi_{h}(t)|^{2}}{(\Xi(t)-\Xi(t(h))^{2}+(d(a)h+o(h))^{2}}\,dt=0\,,$
(A.37)

proven in (4.59).

Arguing as we did for $I_{h}$ we can reduce the problem to show

$\displaystyle\lim_{h\to 0}$
$\displaystyle\int_{t(h)-r}^{t(h)+r}\frac{A(t)|\varphi_{h}(t)|^{2}}{(\Xi(t)-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2}}(G_{h}(t)-G_{h}(t(h)))\,dt=0\,.$
(A.38)

We split the integral to the sum of

$\displaystyle J_{1}(h)$
$\displaystyle:=\int_{t(h)-r}^{t(h)+r}\frac{(A(t)-A(t(h))|\varphi_{h}(t)|^{2}}{(\Xi(t)-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2}}(G_{h}(t)-G_{h}(t(h)))\,dt$
$\displaystyle J_{2}(h)$
$\displaystyle:=A(t(h))\int_{t(h)-r}^{t(h)+r}\frac{|\varphi_{h}(t)|^{2}}{(\Xi(t)-\Xi(t(h)))^{2}+(d(a)h+o(h))^{2}}(G_{h}(t)-G_{h}(t(h)))\,dt\,.$

Next observe that, in the interval that interests us, the following
inequalities hold provided $r$ and $h$ are sufficiently small:

$\displaystyle|A(t)-A(t(h))|$ $\displaystyle\leq C|t-t(h)|$
$\displaystyle|A(t(h)|$ $\displaystyle=|A(t(h))-A(a)|\leq C|t(h)-a|\leq Ch$
$\displaystyle|G_{h}(t)-G_{h}(t(h))|$
$\displaystyle\leq\|G_{h}\|_{C^{1/2}(a-r,a+r)}|t-t(h)|^{1/2}$
$\displaystyle|\Xi(t)-\Xi(t(h))|$ $\displaystyle\geq C^{-1}|t-t(h)|$
$\displaystyle(d(a)h+o(h))^{2}$ $\displaystyle\geq C^{-1}h^{2}\,.$

Since $\|\varphi_{h}\|_{L^{\infty}}\leq C$, we can change variable in the
integrals to $\sigma=t-t(h)$ and estimate them as follows:

$\displaystyle|J_{1}(h)|$ $\displaystyle\leq
C\|G_{h}\|_{C^{1/2}(a-r,a+r)}\int_{-r/2}^{r/2}\sigma^{-1/2}\,d\sigma\leq
C\|G_{h}\|_{C^{1/2}}r^{1/2}\,,$ $\displaystyle|J_{2}(h)|$ $\displaystyle\leq
Ch\int_{-\infty}^{\infty}\frac{\sigma^{1/2}}{\sigma^{2}+C^{-1}h^{2}}\,d\sigma=Ch^{1/2}\int\frac{\tau^{1/2}}{\tau^{2}+C^{-1}}d\tau\leq
Ch^{1/2}\,.$

Clearly $J_{2}(h)\to 0$, while $J_{1}(h)\to 0$ because
$\|G_{h}\|_{C^{1/2}(a-r,a+r)}\to 0$. ∎

##  Appendix B Proofs of technical statements

###  B.1. Proof of Remark 1.0.2

More generally, we will show here that, for any $q_{0}\in[1,2[$ and
$q_{1}\in]2,\infty]$, it is true that

$\lVert K_{2}*\omega\rVert_{L^{\infty}(\mathbb{R}^{2};\mathbb{R}^{2})}\leq
C(q_{0},q_{1})(\lVert\omega\rVert_{L^{q_{0}}(\mathbb{R}^{2})}+\lVert\omega\rVert_{L^{q_{1}}(\mathbb{R}^{2})})$

for all $\omega\in L^{q_{0}}\cap L^{q_{1}}$. Indeed, passing to polar
coordinates, one sees that $K_{2}|_{B_{1}}\in
L^{q_{1}^{*}}(B_{1};\mathbb{R}^{2})$ and $K_{2}|_{\mathbb{R}^{2}\setminus
B_{1}}\in L^{q_{0}^{*}}(\mathbb{R}^{2}\setminus B_{1};\mathbb{R}^{2})$, where
$q_{0}^{*},q_{1}^{*}$ are given by $\frac{1}{q_{i}}+\frac{1}{q_{i}^{*}}=1$ for
$i\in\\{0,1\\}$. Hölder’s inequality implies that for any
$x\in\mathbb{R}^{2}$,

$\begin{split}\lvert(K_{2}*\omega)(x)\rvert&=\lvert((K_{2}\mathbf{1}_{B_{1}})*\omega)(x)+((K_{2}(1-\mathbf{1}_{B_{1}}))*\omega)(x)\rvert\\\
&\leq\lVert
K_{2}\rVert_{L^{q_{1}^{*}}(B_{1})}\lVert\omega\rVert_{L^{q_{1}}(\mathbb{R}^{2})}+\lVert
K_{2}\rVert_{L^{q_{0}^{*}}(\mathbb{R}^{2}\setminus
B_{1})}\lVert\omega\rVert_{L^{q_{0}}(\mathbb{R}^{2})}\\\ &\leq
C(q_{0},q_{1})(\lVert\omega\rVert_{L^{q_{0}}(\mathbb{R}^{2})}+\lVert\omega\rVert_{L^{q_{1}}(\mathbb{R}^{2})}).\end{split}$

Since $x$ is arbitrary, this achieves a proof of the claim above.

###  B.2. Proof of Theorem 1.0.3

Existence. The existence argument is a classical density argument. Take any
sequence $(\omega_{0}^{(n)})_{n\in\mathbb{N}}$ of functions in $L^{1}\cap
C^{\infty}_{c}$ that converges strongly in $L^{1}$ to $\omega_{0}$.
Analogously, pick a sequence of smooth functions $(f_{n})_{n\in\mathbb{N}}$ in
$C^{\infty}_{c}(\mathbb{R}^{2}\times[0,T])$ converging in
$L^{1}(\mathbb{R}^{2}\times[0,T])$ to $f$ and satisfying the bound
$\|f_{n}(\cdot,t)\|_{L^{\infty}}\leq\|f(\cdot,t)\|_{L^{\infty}}$ for a.e. $t$.
Then, let $\omega^{(n)}$ denote the solution of the corresponding Cauchy
problem of the Euler equations in vorticity form. The existence of such
solutions is a classical well-known fact, see for instance [30, Theorem A].
Following Remark 1.0.4, these solutions satisfy all the à priori estimates
needed in Proposition 2.2.3. Therefore, following the proof of Proposition
2.2.3, one obtains, in the limit $n\to\infty$, a solution $\omega\in
L^{\infty}([0,T];L^{1}\cap L^{\infty})$ of the given Cauchy problem.
Furthermore, since the à priori estimates of Remark 1.0.4 are uniform in $n$,
one gets $K_{2}*\omega\in L^{\infty}([0,T];L^{2})$.

######  Remark B.2.1.

The proof of Proposition 2.2.3 has a fixed force $f$ but a straightforward
adaptation of the arguments handles the case above, namely with a sequence of
forces $(f_{n})_{n\in\mathbb{N}}$ that converges in
$L^{1}(\mathbb{R}^{2}\times[0,T])$ to a given $f$. More precisely the only
difference occurs in the term $I_{4}(k)$ of (B.6), which anyway enjoys
convergence to the same limit.

Uniqueness. The uniqueness proof needs two important facts. The first is a
well-known ODE inequality, whose short proof is given, for the reader’s
convenience, at the end of the section.

######  Lemma B.2.2.

Let $T>0$ and let $E:[0,T]\to[0,\infty[$ be a differentiable function
satisfying

$\dot{E}(t)\leq pME(t)^{1-1/p}\quad\text{ and }\quad E(0)=0$ (B.1)

for some fixed $M>0$. Then $E(t)\leq(Mt)^{p}$ for all $t\in[0,T]$.

The second is the classical Calderón-Zygmund $L^{p}$ estimate, where we need
the sharp $p$-dependence of the corresponding constant. This fact is also well
known, cf. for instance [29, Formula (8.45), page 322]).

######  Lemma B.2.3.

For every $p_{0}>1$ there is a constant $c(p_{0})$ with the following
property. If $v=K_{2}*\omega$ for some $\omega\in L^{1}\cap
L^{p}(\mathbb{R}^{2})$ with $p\in[p_{0},\infty[$, then $\lVert
Dv\rVert_{L^{p}}\leq pc\lVert\omega\rVert_{L^{p}}$.

Now, let $v_{1}=K_{2}*\omega_{1},v_{2}=K_{2}*\omega_{2}$ be two solutions of
(1.1) satisfying the assumptions of Theorem 1.0.3 and note that
$w:=v_{1}-v_{2}$ solves

$\partial_{t}w+(v_{1}\cdot\nabla)w+(w\cdot\nabla)v_{2}=-\nabla(p_{1}-p_{2})$
(B.2)

(where $p_{1},p_{2}$ are the pressures corresponding to $v_{1}$ and $v_{2}$).
Clearly

$E(t):=\int_{\mathbb{R}^{2}}|w(x,t)|^{2}\,\mathrm{d}x\leq
2\int_{\mathbb{R}^{2}}|v_{1}(x,t)|^{2}\,\mathrm{d}x+2\int_{\mathbb{R}^{2}}|v_{2}(x,t)|^{2}\,\mathrm{d}x<\infty$

is a bounded function on $[0,T]$.

We scalar multiply (B.2) with $w$, integrate by parts and use the divergence
free conditions of $v_{1},v_{2}$, and $w$ to conclude and

$\displaystyle\dot{E}(t)=$
$\displaystyle-2\int_{\mathbb{R}^{2}}((w\cdot\nabla)v_{2})w\,\mathrm{d}x\leq
2\int_{\mathbb{R}^{2}}|w(x,t)|^{2}\lvert Dv_{2}(x,t)\rvert\,\mathrm{d}x$
$\displaystyle\leq$ $\displaystyle 2\lVert\nabla
v_{2}(\cdot,t)\rVert_{L^{p}}\lVert w(\cdot,t)\rVert_{L^{\infty}}^{2/p}\lVert
w(\cdot,t)\rVert_{L^{2}}^{2-2/p}\,.$

Using Remark 1.0.2, we also have

$\begin{split}&\sup_{t\in[0,T]}\lVert
w(\cdot,t)\rVert_{L^{\infty}}\leq\sup_{t\in[0,T]}(\lVert
v_{1}(\cdot,t)\rVert_{L^{\infty}}+\lVert v_{2}(\cdot,t)\rVert_{L^{\infty}})\\\
&\hskip 19.91684pt\leq
C\sup_{t\in[0,T]}(\lVert\omega_{1}(\cdot,t)\rVert_{L^{1}}+\lVert\omega_{1}(\cdot,t)\rVert_{L^{\infty}}+\lVert\omega_{2}(\cdot,t)\rVert_{L^{1}}+\lVert\omega_{2}(\cdot,t)\rVert_{L^{\infty}})<\infty\,.\end{split}$

Next fix any $p\geq 2$. From Lemma B.2.3 and the classical $L^{p}$
interpolation we conclude

$\lVert Dv_{2}(\cdot,t)\rVert_{L^{p}}\leq
pc\lVert\omega_{2}(\cdot,t)\rVert_{L^{p}}\leq
pc\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{1})}^{1/p}\lVert\omega_{2}(\cdot,t)\rVert_{L^{\infty}([0,T];L^{\infty})}^{1-1/p}.$

Therefore, $\dot{E}(t)\leq pM_{p}E(t)^{1-1/p}$ with

$\displaystyle M_{p}$ $\displaystyle=2\lVert
w\rVert_{L^{\infty}([0,T];L^{\infty})}^{2/p}c\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{1})}^{1/p}\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{\infty})}^{1-1/p}$
$\displaystyle\leq 2c\left({\textstyle{\frac{1}{p}}}\lVert
w\rVert_{L^{\infty}([0,T];L^{\infty})}^{2}\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{1})}+\left(1-{\textstyle{\frac{1}{p}}}\right)\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{\infty})}\right)$
$\displaystyle\leq 2c(\lVert
w\rVert_{L^{\infty}([0,T];L^{\infty})}^{2}\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{1})}+\lVert\omega_{2}\rVert_{L^{\infty}([0,T];L^{\infty})})=:M<\infty\,.$

We can thus apply Lemma B.2.2 to obtain that
$E(t)\leq(M_{p}t)^{p}\leq(Mt)^{p}$. In particular, for any $t\leq\frac{1}{2M}$
we have $E(t)\leq\frac{1}{2^{p}}$ and we can let $p$ tend to $\infty$ to infer
$E(t)=0$. Since the same estimates apply to any translation
$\tilde{E}(t):=E(t+t_{0})$ of the function $E$, we immediately conclude that
$E$ vanishes identically, namely that $v_{1}=v_{2}$ on
$\mathbb{R}^{2}\times[0,T]$.

###### Proof of Lemma B.2.2.

Fix an arbitrary $t_{0}\leq T$ and note that if $E(t_{0})=0$ there is nothing
to show. Hence assume $E(t_{0})>0$ and set $a:=\sup\\{t:E(t)=0\text{ and
}t\leq t_{0}\\}$ (note that the set is nonempty because $E(0)=0$). $E(a)=0$ by
continuity of $E$ and clearly $E(t)>0$ for all $t\in]a,t_{0}]$. Therefore, we
can divide (B.1) by $E(t)^{1-1/p}$ to obtain that $\dot{E}(t)E^{1/p-1}(t)\leq
pM$ for all $t\in]a,t_{0}]$. Integrating both sides gives

$\int_{a}^{t_{0}}\dot{E}(t)E^{1/p-1}(t)\leq pM(t_{0}-a)\,.$ (B.3)

But the left hand side equals $pE^{1/p}(t_{0})-pE^{1/p}(a)=pE^{1/p}(t_{0})$,
from which we infer $E^{1/p}(t_{0})\leq M(t_{0}-a)\leq Mt_{0}$. ∎

###  B.3. Proof of Proposition 2.2.3

Recall first the following classical metrizability result of weak∗ topologies
of separable Banach spaces.

######  Lemma B.3.1 (Metrizability Lemma).

Let $X$ be a separable Banach space and let $K\subset X^{*}$ be
weakly∗-compact. Then $K$ is metrizable in the weak∗ topology inherited from
$X^{*}$ and a metric that induces this topology is given by

$d(l,\tilde{l})=\sum_{n=1}^{\infty}2^{-n}\min\\{1,|l(x_{n})-\tilde{l}(x_{n})|\\},$
(B.4)

where $(x_{n})_{n\in\mathbb{N}}$ is any sequence in $X$ such that
$\\{x_{n}:n\in\mathbb{N}\\}$ is dense in $X$.

Now on to the proof of Proposition 2.2.3. We will prove convergence of the
$\omega_{\varepsilon,k}$ to a $\omega_{\varepsilon}$ for fixed $\varepsilon$
and $k\to\infty$ in the space $C([0,T];K)$, where $K:=\\{u\in
L^{q}(\mathbb{R}^{2}):\lVert u\rVert_{L^{q}}\leq R\\}$ is equipped with the
weak∗ topology inherited from $L^{q}_{\text{w}}$. (We will talk about the
choice of $q$ later.) Here, $R$ is the uniform bound obtained in (2.24) of
Corollary 2.2.2. Note that since every $L^{q}$ space is reflexive, one can
work just as well with the weak topology on $K$. Let
$(\phi_{n})_{n\in\mathbb{N}}$ be a sequence of smooth functions such that
$\\{\phi_{n}:n\in\mathbb{N}\\}$ is dense in every $L^{q}$. The metric given by
(B.4) now induces the topology of $K$, and it does not depend on $q$.
Therefore, using the uniform bound (2.24), we conclude that _the choice of $q$
does not matter_. It is sufficient to prove the statement of Proposition 2.2.3
for one fixed $q\in]1,p]$ in order to prove it for all $q\in]1,p]$.

######  Claim B.3.2.

$\omega_{\varepsilon,k}$, seen as functions from $[0,T]$ to $K$, are
equicontinuous (for simplicity we define each
$\omega_{\varepsilon,k}(\cdot,t)$ on the interval $[0,t_{k}]$ as constantly
equal to $\omega_{\varepsilon,k}(\cdot,t_{k})$).

###### Proof.

For $\tilde{\omega},\hat{\omega}\in L^{q}(\mathbb{R}^{2})$, let

$d_{i}(\tilde{\omega},\hat{\omega})\overset{\text{Def.}}{=}\left\lvert\int_{\mathbb{R}^{2}}(\tilde{\omega}-\hat{\omega})\phi_{i}\mathrm{d}x\right\rvert.$

Since each $\omega_{\varepsilon,k}$ solves the Euler equations in vorticity
form, we can estimate

$\displaystyle\;\;d_{i}(\omega_{\varepsilon,k}(t,\cdot),\omega_{\varepsilon,k}(s,\cdot))$
$\displaystyle=\left\lvert\int_{\mathbb{R}^{2}}\int_{s}^{t}\partial_{\tau}\omega_{\varepsilon,k}(\sigma,x)\phi_{i}(x)\,d\sigma\,dx\right\rvert$
$\displaystyle=\left\lvert\int_{\mathbb{R}^{2}}\int_{s}^{t}((K_{2}*\omega_{\varepsilon,k})\cdot\nabla)\omega_{\varepsilon,k}(x,\sigma)+f(x,\sigma)\phi_{i}(x)\,d\sigma\,dx\right\rvert$
$\displaystyle\leq\lVert\nabla\phi_{i}\rVert_{L^{\infty}(\mathbb{R}^{2})}\int_{\mathbb{R}^{2}}\int_{s}^{t}\lvert
K_{2}*\omega_{\varepsilon,k}\rvert\lvert\omega_{\varepsilon,k}\rvert\,d\sigma\,dx$
$\displaystyle\qquad+\lVert\phi_{i}\rVert_{L^{\infty}(\mathbb{R}^{2})}\int_{\mathbb{R}^{2}}\int_{s}^{t}|f(x,\sigma)|\,d\sigma\,dx$
$\displaystyle\leq
C(\lVert\nabla\phi_{i}\rVert_{\infty}+\|\phi_{i}\|_{\infty})\lvert
s-t\rvert^{\gamma}$ (B.5)

whenever $t\geq s\geq t_{k}$, where $\gamma=\gamma(\alpha,{\bar{\alpha}})$.
Indeed, the last inequality follows from $f\in L^{r}_{t}L^{1}_{x}$ for some
$r=r(\alpha,{\bar{\alpha}})>1$ (see the proof of Lemma 2.1.2). Let
$\tilde{\varepsilon}>0$. We can find a $N\in\mathbb{N}$ (depending on
$\tilde{\varepsilon}$) such that
$\sum_{n=N^{+}1}^{\infty}2^{-n}\leq\frac{\tilde{\varepsilon}}{2}$. If

$\lvert
t-s\rvert\leq\left(\frac{\tilde{\varepsilon}}{2NC\max_{i\in\\{1,\dots,N\\}}(\lVert\nabla\phi_{i}\rVert_{\infty}+\|\phi_{i}\|_{\infty})}\right)^{\frac{1}{\gamma}},$

where $C$ is the constant from (B.5), then, by the bound in (B.5), we get

$d(\omega_{\varepsilon,k}(t,\cdot),\omega_{\varepsilon,k}(s,\cdot))\leq\frac{\tilde{\varepsilon}}{2}+\sum_{i=1}^{N}\frac{\tilde{\varepsilon}}{2N}=\tilde{\varepsilon}.\qed$

By the Banach-Alaoglu theorem, bounded subsets are relatively compact in the
weak∗ topology. Therefore, using reflexivity, for every $t\in[0,T]$, the
bounded set $\\{\omega_{\varepsilon,k}(\cdot,t):k\in\mathbb{N}\\}$ is also
relatively compact in $L^{q}_{\text{w}}$.

Therefore, using Arzelà-Ascoli, we can conclude that there exists a
subsequence of $(\omega_{\varepsilon,k})_{k\in\mathbb{N}}$, not relabeled,
that converges in $C([0,T];L_{\text{w}}^{q})$, for every $q$, to the same
$\omega_{\varepsilon}\in C([0,T];L_{\text{w}}^{q})$.

######  Claim B.3.3.

The $\omega_{\varepsilon}$ is a solution of the Euler equations in vorticity
formulation.

###### Proof.

We have, for every $k\in\mathbb{N}$ and $\phi\in
C_{\text{c}}^{\infty}(\mathbb{R}^{2}\times[0,T])$ with $\phi(\cdot,T)=0$, (cf.
(1.3))

$\displaystyle 0=$
$\displaystyle\underbrace{\int_{\mathbb{R}^{2}}\omega_{\varepsilon,k}(x,t_{k})\phi(x,t_{k})\,\mathrm{d}x}_{=:I_{1}(k)}+\underbrace{\int_{t_{k}}^{T}\int_{\mathbb{R}^{2}}\
\omega_{\varepsilon,k}(x,t)\partial_{t}\phi(x,t)\,\mathrm{d}x\,\mathrm{d}t}_{=:I_{2}(k)}$
$\displaystyle+\underbrace{\int_{t_{k}}^{T}\int_{\mathbb{R}^{2}}\omega_{\varepsilon,k}(x,t)((K_{2}*_{x}\omega_{\varepsilon,k})(x,t)\cdot\nabla)\phi(x,t)\,\mathrm{d}x\,\mathrm{d}t}_{=:I_{3}(k)}$
$\displaystyle+\underbrace{\int_{t_{k}}^{T}\int_{\mathbb{R}^{2}}f(x,t)\phi(x,t)\,\mathrm{d}x\,\mathrm{d}t}_{=:I_{4}(k)}=0\,.$
(B.6)

The term $I_{4}(k)$ converges to

$\int_{\mathbb{R}^{2}\times[0,T]}f(x,t)\phi(x,t)\,\mathrm{d}x\,\mathrm{d}t\,.$

By the convergence of the $\omega_{\varepsilon,k}$,

$\lim_{k\to\infty}I_{2}(k)=\int_{\mathbb{R}^{2}\times[0,T]}\omega_{\varepsilon}(x,t)\partial_{t}\phi(x,t)\,\mathrm{d}x\,\mathrm{d}t.$

By the definition of the initial condition of $\omega_{\varepsilon,k}$ (cf.
(2.20)), $\omega_{\varepsilon,k}(\cdot,t_{k})$ converges strongly in
$L^{1}(\mathbb{R}^{2})$ to
$\tilde{\omega}(\cdot,0)=\omega_{0}=\omega_{\varepsilon}(\cdot,0)$. Therefore,

$\lim_{k\to\infty}I_{1}(k)=\int_{\mathbb{R}^{2}}\omega_{\varepsilon}(x,0)\phi(x,0)\,\mathrm{d}x.$

It therefore only remains to prove the convergence of $I_{3}$, for which we
will require yet another claim.

######  Claim B.3.4.

For every $r\in[2,\infty[$ and every $t\in[0,T]$, the set
$\\{v_{\varepsilon,k}(\cdot,t):k\in\mathbb{N}\\}$ is compact in $L^{r}(B_{R})$
for every $R>0$.

###### Proof.

From (2.24), we know that
$\|v_{\varepsilon,k}(\cdot,t)\|_{L^{2}(\mathbb{R}^{2})}\leq C$ for some
constant $C$ that is independent of $t$. Recall that
$v_{\varepsilon,k}=\nabla^{\bot}\psi_{\varepsilon,k}$, where
$\psi_{\varepsilon,k}$ solves
$\Delta\psi_{\varepsilon,k}=\omega_{\varepsilon,k}$. Therefore, using the
Calderón-Zygmund inequality, one gets

$\lVert\nabla v_{\varepsilon,k}(\cdot,t)\rVert_{L^{2}}\leq
C\lVert\omega_{\varepsilon,k}(\cdot,t)\rVert_{L^{2}}.$

Since the $L^{2}$ norms of the $\omega_{\varepsilon,k}(\cdot,t)$ are uniformly
bounded, we can conclude that

$\sup_{k\in\infty}\lVert\nabla
v_{\varepsilon,k}(\cdot,t)\rVert_{L^{2}}<\infty.$

Hence we conclude the compactness in $L^{r}(B_{R})$ from Rellich’s Theorem. ∎

Therefore, the $v_{\varepsilon,k}(\cdot,t)$ converge to
$v_{\varepsilon}(\cdot,t)$ strongly in every $L^{r}(B_{R})$ with
$r\in[2,\infty[$. Moreover, thanks to (2.24), we can apply the dominated
convergence theorem by Lebesgue to conclude that $v_{\varepsilon,k}\to
v_{\varepsilon}$ as $k\to\infty$ in the space $L^{1}([0,T];L^{r}(B_{R}))$ for
every $r\in[2,\infty[$.

By definition,

$\omega_{\varepsilon,k}(v_{\varepsilon,k}\cdot\nabla)\phi-\omega_{\varepsilon}(v_{\varepsilon}\cdot\nabla)\phi=\omega_{\varepsilon,k}(v_{\varepsilon,k}-v_{\varepsilon})\cdot\nabla\phi+(\omega_{\varepsilon,k}-\omega_{\varepsilon})v_{\varepsilon}\cdot\nabla\phi\,.$

We thus rewrite

$\displaystyle I_{3}(k)$
$\displaystyle=\int_{0}^{T}\int_{B_{R}}\omega_{\varepsilon,k}(v_{\varepsilon,k}-v_{\varepsilon})\cdot\nabla\phi\,dx\,dt+\int_{0}^{T}\underbrace{\int_{B_{R}}(\omega_{\varepsilon,k}-\omega_{\varepsilon})v_{\varepsilon}\cdot\nabla\phi\,dx}_{=:J_{k}(t)}\,dt\,.$
(B.7)

Observe first that, for each fixed $t$,

$\lim_{k\to\infty}J_{k}(t)=0\,,$

since $\omega_{\varepsilon,k}(\cdot,t)-\omega_{\varepsilon}(\cdot,t)$
converges weakly to $0$ in $L^{2}$, while
$v_{\varepsilon}(\cdot,t)\cdot\nabla\phi(\cdot,t)$ is a fixed $L^{2}$
function. On the other hand

$|J_{k}(t)|\leq\|\nabla\phi(\cdot,t)\|_{L^{\infty}}(\|\omega_{\varepsilon,k}(\cdot,t)\|_{L^{2}}+\|\omega_{\varepsilon}(\cdot,t)\|_{L^{2}})\|v_{\varepsilon}(\cdot,t)\|_{L^{2}}\,.$

Therefore the second integral in (B.7) converges to $0$. The first integral
can be bounded by

$\|\nabla\phi\|_{L^{\infty}}\|v_{\varepsilon,k}-v_{\varepsilon}\|_{L^{1}([0,T],L^{2}(B_{R}))}\|\omega_{\varepsilon,k}\|_{L^{\infty}([0,T],L^{2}(B_{R}))}$

and converges to $0$ as well. ∎

###  B.4. Proof of Lemma 2.4.1

Consider $\vartheta\in L^{2}_{m}\cap\mathscr{S}$ for $m\geq 2$ and let
$v:=K_{2}*\vartheta$. We first claim that

$\int_{B_{R}}v=0\qquad\qquad\mbox{for every $R>0$.}$ (B.8)

With (B.8) at our disposal, since
$\|Dv\|_{L^{2}(\mathbb{R}^{2})}=\|\vartheta\|_{L^{2}(\mathbb{R}^{2})}$, we use
the Poincaré inequality to conclude

$R^{-1}\|v\|_{L^{2}(B_{R})}+\|Dv\|_{L^{2}(B_{R})}\leq
C\|\vartheta\|_{L^{2}(\mathbb{R}^{2})}$ (B.9)

for a geometric constant $C$. This is then enough to infer the remaining
conclusions of the lemma.

In order to achieve (B.8) observe first that $v=\nabla^{\perp}h$, where $h$ is
the unique potential-theoretic solution of $\Delta h=\vartheta$, given by
$h=K*\vartheta$ with $K(x)=\frac{1}{2\pi}\log|x|$. Since $K(R_{\theta}x)=K(x)$
and $\vartheta(x)=\vartheta(R_{2\pi/m}x)$, it follows that
$h(R_{2\pi/m}x)=h(x)$, i.e. $h$ is $m$-fold symmetric. Therefore
$R_{-2\pi/m}\nabla h(R_{2\pi/m}x)=\nabla h(x)$. In particular, integrating in
$x$ and using that the rotation is a measure-preserving transformation of the
disk, we conclude

$\int_{B_{R}}\nabla h=R_{2\pi/m}\int_{B_{R}}\nabla h\,,$

and thus,

$\int_{B_{R}}\nabla h=\frac{1}{m}\sum_{k=0}^{m-1}R_{2k\pi/m}\int_{B_{R}}\nabla
h\,.$

However, since $m\geq 2$, $\sum_{k=0}^{m-1}R_{2k\pi/m}=0$, showing that
$\int_{B_{R}}\nabla h=0$.

######  Remark B.4.1.

We next show that it is not possible to find a continuous extension of the
operator $L^{2}\cap\mathscr{S}\ni\vartheta\mapsto
K_{2}*\vartheta\in\mathscr{S}^{\prime}$ to the whole $L^{2}$. First of all we
observe that, if such an extension exists, it then needs to coincide with
$K_{2}*\vartheta$ when $\vartheta\in L^{1}\cap L^{2}$. We next exhibit a
sequence of divergence free vector fields $\\{v_{k}\\}\subset W^{1,1}\cap
W^{1,2}$ with the property that $\omega_{k}=\operatorname{curl}v_{k}$ converge
to $0$ strongly in $L^{2}$ but $v_{k}$ converge locally to a constant vector
field $v_{0}\neq 0$. In order to do this, we first define the following
functions $\phi_{k}$ on the positive real axis:

$\phi_{k}(r):=\left\\{\begin{array}[]{ll}1+\frac{3}{4\ln k}&\mbox{for
$r\leq\frac{k}{2}$}\\\ \\\ 1+\frac{3}{4\ln k}-\frac{1}{k^{2}\ln
k}\left(r-\frac{k}{2}\right)^{2}&\mbox{for $\frac{k}{2}\leq r\leq k$}\\\ \\\
1+\frac{1}{2\ln k}-\frac{1}{\ln k}\ln\frac{r}{k}&\mbox{for $k\leq r\leq
k^{2}$}\\\ \\\ \frac{1}{2k^{4}\ln k}(r-2k^{2})^{2}&\mbox{for $k^{2}\leq r\leq
2k^{2}$}\\\ \\\ 0&\mbox{for $r\geq 2k^{2}$}\,.\end{array}\right.$

Observe that $\phi_{k}$ is $C^{1}$ and its derivative is Lipschitz. Next we
define the stream functions

$\psi_{k}(x)=-\phi_{k}(|x|)v_{0}^{\perp}\cdot x$

and the vector field $v_{k}(x)=\nabla^{\perp}\psi_{k}(x)$. By construction
$v_{k}$ is divergence free, compactly supported, and Lipschitz. In particular,
it belongs to $W^{1,p}$ for every $p$. Moreover, $v_{k}$ equals
$(1+\frac{3}{4\ln k})v_{0}$ on $B_{k/2}$ and it thus follows that, as
$k\to\infty$, $v_{k}$ converges locally to the constant vector field $v_{0}$.
It remains to check that $\operatorname{curl}v_{k}=\Delta\psi_{k}$ converges
to $0$ strongly in $L^{2}$. We compute

$\Delta\psi_{k}=-\underbrace{v_{0}^{\perp}\cdot
x\,\Delta(\phi_{k}(|x|))}_{=:f_{k}}-\underbrace{\nabla(\phi_{k}(|x|))\cdot
v_{0}^{\perp}}_{=:g_{k}}\,$

and we seek to bound $f_{k}$ and $g_{k}$ pointwise. For what concerns $f_{k}$
observe that $\Delta\phi_{k}$ vanishes on $|x|\leq\frac{k}{2}$, $k\leq|x|\leq
k^{2}$, and $2k^{2}\leq|x|$. On the remaining regions, using the formula for
the Laplacian in polar coordinates, we can estimate

$|f_{k}(x)|\leq|v_{0}||x|(|\phi^{\prime\prime}(|x|)|+|x|^{-1}|\phi^{\prime}(|x|)|)\,.$

In particular, we conclude

$|f_{k}(|x|)|\leq\frac{C}{|x|\ln k}\,,$

for a constant $C$ independent of $k$. As for $g_{k}$, it vanishes for
$|x|\leq\frac{k}{2}$ and $|x|\geq k^{2}$, and where it does not vanish we have
the estimate

$|g_{k}(x)|\leq|v_{0}||\psi^{\prime}(|x|)|\leq\frac{C}{|x|\ln k}\,,$

again for a constant $C$ independent of $k$. Passing to polar coordinates, we
can thus estimate

$\displaystyle\|\Delta\psi_{k}\|^{2}_{L^{2}(\mathbb{R}^{2})}\leq$
$\displaystyle\frac{C}{(\ln
k)^{2}}\int_{k/2}^{2k^{2}}\frac{1}{r}\,dr=\frac{C}{(\ln
k)^{2}}\left(\ln(2k^{2})-\ln{\textstyle{\frac{k}{2}}}\right)=\frac{C\ln(4k)}{(\ln
k)^{2}}\,.$

## Index

  * $\alpha$ §2.1
  * $\beta$ §2.1
  * $\varepsilon$ Theorem 2.2.1
  * $\zeta$ §2.1
  * $\eta$ item (a)
  * $\lambda_{a}$ §4.4
  * $\lambda_{b}$ §4.4
  * $\Xi$ §4.10, §4.2
  * $\xi$ §2.3
  * $\tau$ §2.3
  * $\chi$ §2.1
  * $\Omega$ §2.3
  * $\omega$ Chapter 1
  * $\omega_{0}$ §2.1
  * $\bar{\Omega}$ §2.1, §4.1, §4.2
  * $\bar{\omega}$ item 1
  * $\omega_{\varepsilon}$ Theorem 2.2.1
  * $\Omega_{\varepsilon,k}$ §2.3
  * $\omega_{\varepsilon,k}$ §2.2
  * $\Omega_{\text{per},k}$ §2.3
  * $\Omega_{r}$ §2.3
  * $\tilde{\Omega}$ §2.3
  * $\tilde{\omega}$ §2.1
  * $A$ §4.10, §4.2
  * $a_{0}$ item (b)
  * $b_{0}$ item (b)
  * $\mathscr{C}$ §4.2
  * $e^{\tau L_{\text{ss}}}$ Theorem 3.0.4
  * $F$ §2.3
  * $f$ Chapter 1, §2.1
  * $g$ §2.1, §4.2
  * $\mathcal{H}$ §4.1
  * $K_{2}$ Chapter 1
  * $\mathcal{K}_{m}$ §4.7
  * $\mathscr{K}$ §3.1, §4.1
  * $\mathscr{K}_{m}$ §4.1, §4.1
  * $L^{2}_{m}$ §2.4
  * $L^{2}_{m}(\mathbb{R})$ §3.1
  * $L_{a}$ §4.4
  * $L_{b}$ §4.4
  * $\mathcal{L}_{m}$ §4.1, §4.1
  * $L_{\text{ss}}$ §2.3, §2.4
  * $L_{\text{st}}$ Chapter 3
  * $m_{a}$ §4.4
  * $m_{b}$ §4.4
  * $P_{0}$ §4.7
  * $\mathscr{P}$ §4.4
  * $P_{z}$ item (c)
  * $\mathcal{R}_{m,z}$ §4.7
  * $S_{1}$ §3.1, §4.1
  * $S_{2}$ §3.1
  * $\mathcal{S}_{m}$ §4.1, §4.1
  * $\textrm{spec}_{m}$ §2.4, Chapter 3
  * $\textrm{spec}_{m}(\cdot,U_{j})$ Chapter 4
  * $\mathscr{S}$ Remark 1.0.5
  * $\mathscr{S}^{\prime}$ Remark 1.0.5
  * $T$ §4.7
  * $t_{k}$ §2.2
  * $U_{k}$ Chapter 3
  * $U_{k}(\mathbb{R})$ §3.1
  * $\mathscr{U}_{m}$ §4.2
  * $V$ §2.3
  * $v$ Chapter 1
  * $v_{0}$ §2.1
  * $\bar{V}$ §2.1
  * $V_{r}$ §2.3
  * $\tilde{V}$ §2.3
  * $\tilde{v}$ §2.1
  * $z_{0}$ §2.3
  * algebraic multiplicity Theorem A.3.1, item (c’), Remark 3.0.2, item (ii), §4.3, Remark 4.9.2
  * Biot-Savart kernel Chapter 1
  * Cauchy problem Chapter 1
  * critical layer §4.3
  * critical value §4.3
  * Euler equations Chapter 1
  * external force Chapter 1
  * force, external Chapter 1
  * geometric multiplicity §A.3, Chapter 3, Remark 4.4.4
  * Lyapunov Schmidt reduction §4.7
  * neutral mode §4.3, §4.4
  * neutral wavenumber §4.3
  * Plemelj’s formula §4.3, §4.8
  * Rayleigh’s inflection point theorem §4.3
  * Rayleigh’s stability equation §4.2, §4.3, §4.4
  * resolvent §2.4
  * Riesz projector item (c)
  * Rotationally symmetric function space §2.4
  * Rotationally symmetric function space, inner product §3.1
  * Schwartz space $\mathscr{S}$ Remark 1.0.5
  * Self-similar operator §2.3, §2.4
  * Semigroup Chapter 3
  * Semigroup, growth bound Theorem 3.0.4
  * shear flow §4.3
  * Similarity variables §2.3
  * Solution, weak Chapter 1
  * space $\mathscr{S}^{\prime}$ of tempered distributions Remark 1.0.5
  * spectrum §2.4, Chapter 3, Chapter 4
  * tempered distribution Remark 1.0.5
  * velocity Chapter 1
  * vorticity Chapter 1
  * Yudovich uniqueness theorem Chapter 1

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