$(S(\mathcal{W},c_{n}))$ for all $n$. We start by proving that
$\\{E(u_{n})\\}$ remains bounded. Indeed, arguing by contradiction, assume
that $\\{|S_{n}^{r}|\\}$ is unbounded for every $r\in(c/\sqrt{2},1)$, where
$S_{n}^{r}$ is defined in Lemma 4.17. Let us thus fix $r\in({c}/{\sqrt{2}},1)$
to be chosen later. Observe that, by (4.63), the sequence
$\\{(\rho_{n}^{\prime})^{2}\\}$ is bounded in $L^{1}(\mathbb{R})$. Then,
applying Lemma 4.18, for every $n$ there exist $x_{n}\in S_{n}^{r}$ and
$R_{n}>0$ such that $R_{n}\to\infty$ and

$\int_{B(0,R_{n})}\rho_{n}^{\prime}(x+x_{n})^{2}dx\to 0.$ (4.69)

Now we define $\tilde{u}_{n}=u_{n}(\cdot+x_{n})$ and
$\tilde{\rho}_{n}=|\tilde{u}_{n}|$. We know from (H5) and Proposition 4.5 that
$\\{\tilde{u}_{n}\\}$ is bounded in $W^{k,\infty}(\mathbb{R};\mathbb{C})$ for
every $k\in\mathbb{N}$. As a consequence, arguing as in the proof of Lemma 2.1
and taking (4.64) into account, we deduce that $\\{\tilde{\rho}_{n}\\}$ is
bounded in $W^{k,\infty}(\mathbb{R})$ for every $k\in\mathbb{N}$. In
particular, there exists $\rho\in W^{2,\infty}(\mathbb{R})$ such that, up to a
subsequence,

$\tilde{\rho}_{n}\to\rho\quad\text{ in
}W^{2,\infty}_{\text{loc}}(\mathbb{R}).$

Moreover, thanks to (4.69), we deduce that $\rho^{\prime}\equiv 0$, so $\rho$
is a constant. Furthermore, the pointwise convergence
$\tilde{\rho}_{n}\to\rho$ leads to

$\rho_{n}(x_{n})=\tilde{\rho}_{n}(0)\to\rho.$

Therefore, since $\rho_{n}(x_{n})<r$ for all $n$, it follows that $\rho\leq
r$.

Notice that $\tilde{\rho}_{n}\in 1+H^{1}(\mathbb{R})$ satisfies the equation

$-\tilde{\rho}_{n}^{\prime\prime}+\frac{c_{n}^{2}(1-\tilde{\rho}_{n}^{4})}{4\tilde{\rho}_{n}^{3}}=\tilde{\rho}_{n}(\mathcal{W}\ast(1-\tilde{\rho}_{n}^{2}))\quad\text{
on }\mathbb{R}.$ (4.70)

We aim to pass to the limit in (4.70). In order to do so we notice that, since
$\rho$ is a constant,

$\tilde{\rho}_{n}^{\prime\prime}\to 0\quad\text{ in
}L^{\infty}_{\text{loc}}(\mathbb{R}).$

Besides, by virtue of Lemma 4.19 and (4.3), we have

$\tilde{\rho}_{n}(\mathcal{W}\ast(1-\tilde{\rho}_{n}^{2}))\to\rho(\mathcal{W}\ast(1-{\rho}^{2}))=\rho(1-\rho^{2})\widehat{\mathcal{W}}(0)\quad\text{
in }\mathcal{D}^{\prime}(\mathbb{R}).$

Thus, using that $\widehat{\mathcal{W}}(0)=1$, we can to pass to the limit in
(4.70) in $\mathcal{D}^{\prime}(\mathbb{R})$ to obtain

$c^{2}(1-\rho^{4})=4\rho^{4}(1-\rho^{2}).$

Using that $1-\rho^{4}=(1-\rho^{2})(1+\rho^{2})$, it follows that

$c^{2}(1+\rho^{2})=4\rho^{4}.$

From this equation, it is immediate to check that

$\rho^{2}=\frac{c^{2}+c\sqrt{c^{2}+16}}{8}.$

Observe that $\frac{c^{2}+c\sqrt{c^{2}+16}}{8}>\frac{c^{2}}{2}$ since
$c<\sqrt{2}$. Therefore, we can choose $\varepsilon>0$ small enough
(independent of $r$) so that

$\varepsilon+\frac{c^{2}}{2}<\frac{c^{2}+c\sqrt{c^{2}+16}}{8}=\rho^{2}\leq
r^{2}.$

Finally, if we choose $r=\sqrt{\varepsilon+\frac{c^{2}}{2}}$ and take
$\varepsilon>0$ possibly smaller so that $r\in(c/\sqrt{2},1)$, then we arrive
at a contradiction. Therefore, $\\{|S_{n}^{r}|\\}$ must be bounded for some
$r\in(c/\sqrt{2},1)$. Consequently, Lemma 4.17 implies that $\\{E(u_{n})\\}$
is bounded too.

Arguing as before, there exists $u\in\mathcal{C}^{2}(\mathbb{R};\mathbb{C})$
such that, up to a subsequence, $u_{n}\to u$ in
$W^{2,\infty}_{\text{loc}}(\mathbb{R}).$ Moreover, Lemma 4.19 implies that
$\mathcal{W}\ast(1-|u_{n}|^{2})\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}\mathcal{W}\ast(1-|u|^{2})$
in $L^{\infty}(\mathbb{R}).$ Thus, we can pass to the limit in
$(S(\mathcal{W},c_{n}))$ so that $u$ is a solution to ($S({\mathcal{W},c})$).

Let us now check that $u\in\mathcal{E}(\mathbb{R})$. Indeed, as in Lemma 4.15,
using (4.1), we have

$\displaystyle E(u_{n})$
$\displaystyle\geq\frac{1}{2}\int_{\mathbb{R}}|u_{n}^{\prime}|^{2}+\frac{1}{4}\int_{\mathbb{R}}(1-|u_{n}|^{2})^{2}-\frac{m}{16\pi}\int_{\mathbb{R}}|\xi|^{2}|\widehat{\eta}_{n}|^{2}$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}}|u_{n}^{\prime}|^{2}+\frac{1}{4}\int_{\mathbb{R}}(1-|u_{n}|^{2})^{2}-\frac{m}{2}\int_{\mathbb{R}}\rho_{n}^{2}(\rho_{n}^{\prime})^{2}.$

Hence, (H5), (4.63) and the fact that $\\{E(u_{n})\\}$ is bounded, imply that

$\frac{1}{2}\int_{\mathbb{R}}|u^{\prime}_{n}|^{2}+\frac{1}{4}\int_{\mathbb{R}}(1-|u_{n}|^{2})^{2}\leq
C,$

for all $n$ and for some $C>0$ independent of $n$. By virtue of Fatou’s lemma,

$\frac{1}{2}\int_{\mathbb{R}}|u^{\prime}|^{2}+\frac{1}{4}\int_{\mathbb{R}}(1-|u|^{2})^{2}\leq
C.$

That is, $u\in\mathcal{E}(\mathbb{R})$.

Finally, the estimate (4.64) ensures that
$u\in\mathcal{N}\mathcal{E}(\mathbb{R})$, while Proposition 4.4 implies that
$u$ is nontrivial. The proof is concluded. ∎

##  5 Nonexistence and properties of solitons

This section is devoted to the study of the Fourier transform of equation
(2.4), that is

$M_{c}(\xi)\widehat{\eta}(\xi)=\widehat{F}(\xi),\quad\text{with }\
M_{c}(\xi)=\xi^{2}+2\widehat{\mathcal{W}}(\xi)-c^{2},$ (5.1)

where

$F=2K+2\eta(\mathcal{W}\ast\eta),\quad\eta=1-|u|^{2}\quad\text{and}\quad
K=|u^{\prime}|^{2}.$

We keep this notation for $F$ and $\eta$ for the rest of the Section, and we
assume, as always, that $\mathcal{W}$ satisfies (H0). If $M_{c}>0$ a.e., we
can recast (5.1) as

$\widehat{\eta}(\xi)=L_{c}(\xi)\widehat{F}(\xi),\quad\text{with }\
L_{c}(\xi)=\widehat{\mathcal{L}}_{c}(\xi)=\frac{1}{M_{c}(\xi)}.$ (5.2)

We will see that the operator $\mathcal{L}_{c}$ plays an essential role in
order to study the regularity and asymptotic behavior at infinity of the
solitons given by Theorems 1.1 and 4.1.

We also stress that (H1) is sufficient condition for $\mathcal{L}_{c}$ to be
well defined, for $c\in[0,\sqrt{2\sigma})$. Indeed, we have
$M_{c}(\xi)\geq(1-2\kappa)\xi^{2}+2\sigma-c^{2}>0$ for a.e.
$\xi\in\mathbb{R}$, so that

$\int_{\mathbb{R}}|L_{c}(\xi)|d\xi\leq\int_{\mathbb{R}}\frac{d\xi}{(1-2\kappa)\xi^{2}+2\sigma-c^{2}}<\infty.$
(5.3)

Thus $L_{c}\in L^{1}(\mathbb{R})$ and $\mathcal{L}_{c}$ is a bounded
continuous function on $\mathbb{R}$.

We can now establish our nonexistence result for solitons with critical speed.

######  Theorem 5.1.

Assume that $\widehat{\mathcal{W}}\geq 0$ a.e. on $\mathbb{R}$ and that there
exists $\delta>0$ such that one of the following holds:

  1. (i)

$\widehat{\mathcal{W}}(\xi)=1-\xi^{2}/2$, for all $\xi\in(-\delta,\delta)$.

  2. (ii)

$\widehat{\mathcal{W}}$ is differentiable on $(-\delta,\delta)$,
$\widehat{\mathcal{W}}(0)=1$ and $\widehat{\mathcal{W}}(\xi)\not=1-\xi^{2}/2$
for a.e. $\xi\in(-\delta,\delta)$.

Then $(S(\mathcal{W},\sqrt{2}))$ admits no nontrivial solution in
$\mathcal{E}(\mathbb{R})$.

###### Proof.

Arguing by contradiction, assume that there exists a nontrivial solution
$u\in\mathcal{E}(\mathbb{R})$ to $(S(\mathcal{W},\sqrt{2}))$. Then,
Proposition 2.2 implies that (5.1) holds, i.e.

$M(\xi)\widehat{\eta}(\xi)=\widehat{F}(\xi),\quad\text{with }\
M(\xi)=\xi^{2}+2\widehat{\mathcal{W}}(\xi)-2.$ (5.4)

On the other hand, by virtue of Lemma 2.1, $\eta$ and $K$ belong to
$W^{k,p}(\mathbb{R})$ for all $k\in\mathbb{N}$ and all $p\in[2,\infty]$.

Let us show that $\widehat{F}$ is continuous and $\widehat{F}(0)>0$. Indeed,
from (2.6) and from the fact that $\eta(\pm\infty)=0$, we deduce the existence
of constants $R,C>0$ such that

$|u^{\prime}(x)|^{2}\leq C(\eta(x)^{2}+\eta^{\prime}(x)^{2})\quad\text{for all
}x\in\mathbb{R}\setminus[-R,R].$

Hence $K\in L^{1}(\mathbb{R})$ and, in turn, $F\in L^{1}(\mathbb{R})$ and
$\widehat{F}$ is continuous. Also, it follows from (5.4) that we can assume
that $M\widehat{\eta}$ is also continuous. Moreover, since $u$ is not trivial
and $\widehat{\mathcal{W}}\geq 0$ a.e., Plancherel’s identity yields

$\widehat{F}(0)=\int_{\mathbb{R}}F(x)dx=2\int_{\mathbb{R}}|u^{\prime}|^{2}+\frac{1}{\pi}\int_{\mathbb{R}}\widehat{\mathcal{W}}(\xi)|\widehat{\eta}(\xi)|^{2}d\xi>0.$
(5.5)

If assumption (i) holds, we deduce that $M\widehat{\eta}=0$ on
$(-\delta,\delta)$, so that, by (5.4), $\widehat{F}(0)=0$, which contradicts
(5.5).

If assumption (ii) holds, then $M$ is differentiable on $(-\delta,\delta)$,
$M(0)=0$ and $M(\xi)\not=0$ for a.e. $\xi\in(-\delta,\delta)$. Therefore,

$\widehat{\eta}(\xi)=L(\xi)\widehat{F}(\xi)\quad\text{a.e.
}\xi\in(-\delta,\delta),$ (5.6)

where $L=1/M$. Let us show now that $L\not\in
L^{1}((-\tilde{\delta},\tilde{\delta}))$ for every
$\tilde{\delta}\in(0,\delta)$ small enough. Expanding $M$ around zero leads to

$M(\xi)=M^{\prime}(0)\xi+o(\xi^{2}),\quad\text{ for all
}\xi\in(-\tilde{\delta},\tilde{\delta}),$

for some $\tilde{\delta}\in(0,\delta)$. Thus, taking $\tilde{\delta}$ even
smaller if necessary,

$|M(\xi)|\leq(|M^{\prime}(0)|+1)|\xi|,\quad\text{ for all
}\xi\in(-\tilde{\delta},\tilde{\delta}).$

In consequence,

$\int_{-\tilde{\delta}}^{\tilde{\delta}}L(\xi)d\xi\geq\frac{1}{|M^{\prime}(0)|+1}\int_{-\tilde{\delta}}^{\tilde{\delta}}\frac{d\xi}{|\xi|}=\infty.$

In particular, $L\not\in L^{2}((-\tilde{\delta},\tilde{\delta}))$. Hence,
taking (5.6) into account, in order $\widehat{\eta}$ to belong to
$L^{2}(\mathbb{R})$, it is necessary that $\widehat{F}(0)=0$. This is again a
contradiction with (5.5). ∎

We can prove now the nonexistence result stated in the Introduction.

###### Proof of Theorem 1.5.

Let $\delta>0$ be such that
$\widehat{\mathcal{W}}\in\mathcal{C}^{2}((-\delta,\delta))$. Recall that
$\widehat{\mathcal{W}}(0)=1$ and $(\widehat{\mathcal{W}})^{\prime}(0)=0$. If
$(\widehat{\mathcal{W}})^{\prime\prime}(\xi)=-1$ for all $(-\delta,\delta)$,
then $\widehat{\mathcal{W}}(\xi)=1-\xi^{2}/2$, for $\xi\in(-\delta,\delta)$,
so that we are in the case (ii) of Theorem 5.1.

Assume now that $(\widehat{\mathcal{W}})^{\prime\prime}(0)\neq-1$. Then, by
decreasing $\delta$ if necessary, we deduce by continuity that
$(\widehat{\mathcal{W}})^{\prime\prime}>-1$ on $(-\delta,\delta)$, or
$(\widehat{\mathcal{W}})^{\prime\prime}<-1$ on $(-\delta,\delta)$. On the
other hand, by Taylor’s theorem, and using that $\widehat{\mathcal{W}}$ is
even, we deduce that for any $\xi\in(-\delta,\delta)$, there exists
$\tilde{\xi}\in(-\delta,\delta)$ such that

$\widehat{\mathcal{W}}(\xi)=1+(\widehat{\mathcal{W}})^{\prime\prime}(\tilde{\xi})\frac{\xi^{2}}{2}.$

Thus we are in the case (ii) of Theorem 5.1, and the conclusion follows. ∎

###  5.1 Decay at infinity

We assume now that $M_{c}>0$ a.e. on $\mathbb{R}$ so that (5.2) holds a.e.
Notice also that if $L_{c}\in\mathcal{S}^{\prime}(\mathbb{R})$, then

$D^{k}\eta=\mathcal{L}_{c}\ast D^{k}F,\quad\text{ for all }k\in\mathbb{N}.$
(5.7)

This equation will be the key for analyzing the decay of the solutions
$u\in\mathcal{E}(\mathbb{R})$ to ($S({\mathcal{W},c})$) as $x\to\pm\infty$.
More precisely, it will be provided the decay of $\eta=1-|u|^{2}$ at infinity.
We start by showing that we can also recover limits of $u$ at $\pm\infty$.
First, we need to establish the integrability of $\eta$, which means that $u$
has finite mass.

######  Lemma 5.2.

Let $c\geq 0$ and let $u\in\mathcal{E}(\mathbb{R})$ be a solution to
($S({\mathcal{W},c})$). If $\mathcal{L}_{c}\in L^{1}(\mathbb{R})$, then
$\eta\in W^{k,1}(\mathbb{R})$ for every $k\in\mathbb{N}$.

###### Proof.

In the case $k=0$, we argue as in the proof of Theorem 5.1 to prove that $F\in
L^{1}(\mathbb{R})$. Then, if $\mathcal{L}_{c}\in L^{1}(\mathbb{R})$, Young’s
inequality applied to (5.7) with $k=0$ implies that $\eta\in
L^{1}(\mathbb{R})$. The case $k\geq 1$ may be tackled similarly by taking into
account that
$F^{\prime}=4\eta^{\prime}\mathcal{W}\ast\eta+2\eta\mathcal{W}\ast\eta^{\prime}$,
so the successive derivatives of $F$ have the form
$D^{k}F=\sum_{j=1}^{2k}a_{j}b_{j},$ where $a_{j},b_{j}\in L^{2}(\mathbb{R})$.
∎

For any $u\in\mathcal{E}(\mathbb{R})$, the limits $\lim_{x\to\pm}u(x)$ do not
exist in general (see [26]). The following result shows that, if $u$ solves
($S({\mathcal{W},c})$), then they do exist whenever $u$ presents no vortices
and $\eta=1-|u|^{2}$ has finite mass.

######  Proposition 5.3.

Let $c>0$ and let $u=\rho e^{i\theta}\in\mathcal{N}\mathcal{E}(\mathbb{R})$ be
a solution to ($S({\mathcal{W},c})$). Assume that $\eta\in L^{1}(\mathbb{R})$.
Then the following limits exist and are finite:

$\theta(\pm\infty)=\theta(0)+\frac{c}{2}\int_{0}^{\pm\infty}\frac{\eta}{1-\eta}.$
(5.8)

In particular,

$u(+\infty)=e^{i\theta(+\infty)},\quad u(-\infty)=e^{i\theta(-\infty)}.$ (5.9)

######  Remark 5.4.

On the one hand, we recall that Lemma 5.2 provides sufficient conditions that
assure that $\eta\in L^{1}(\mathbb{R})$. On the other hand, we stress the fact
that the limits $u(\pm\infty)$, if they exist, may be different from each
other. In fact, from Proposition 5.3 it is easy to see that
$u(+\infty)=u(-\infty)$ if and only if
$\int_{\mathbb{R}}\frac{\eta}{1-\eta}=0$.

###### Proof of Proposition 5.3.

By Proposition 2.4, $\theta$ satisfies (2.12). By integrating, we have

$\theta(x)-\theta(0)=\frac{c}{2}\int_{0}^{x}\left(\frac{1}{\rho(y)^{2}}-1\right)dy=\frac{c}{2}\int_{0}^{x}\frac{\eta(y)}{1-\eta(y)}dy,\quad\text{
for all }x\in\mathbb{R}.$

Therefore,

$\theta(+\infty)\coloneqq\lim_{x\to\infty}\theta(x)=\theta(0)+\int_{0}^{\infty}\frac{\eta}{1-\eta},\quad\theta(-\infty)\coloneqq\lim_{x\to-\infty}\theta(x)=\theta(0)-\int_{-\infty}^{0}\frac{\eta}{1-\eta}.$

Since $\inf_{\mathbb{R}}(1-\eta)=\inf_{\mathbb{R}}\rho^{2}>0$ and $\eta\in
L^{1}(\mathbb{R})$, it follows that both limits $\theta(+\infty)$ and
$\theta(-\infty)$ are finite. Hence, we deduce directly (5.9) from the fact
that $\rho(\pm\infty)=1$. ∎

In the rest of the subsection we will adapt the Bona–Li theory in [13, 14] to
our equation. First, we recall the following technical result proved in [50].

######  Lemma 5.5 ([50]).

For any $0<\ell<m$ and $\varepsilon>0$, the following inequality holds,

$\int_{\mathbb{R}}\frac{e^{\ell|x|}}{(1+\varepsilon
e^{|x|})^{m}e^{m|x-y|}}dx\leq B\frac{e^{\ell|y|}}{(1+\varepsilon
e^{|y|})^{m}},\quad\text{ for all }y\in\mathbb{R},$ (5.10)

where $B=(\min\\{\ell,m-\ell\\})^{-1}$.

###### Proof of Theorem 1.4.

First, we point out that the result holds true for $\ell=0$. Indeed, (1.14)
and Hölder’s inequality yield $\mathcal{L}_{c}\in L^{1}(\mathbb{R})$. Thus, by
virtue of Lemmas 2.1 and 5.2, $D^{k}\eta\in L^{1}(\mathbb{R})\cap
L^{\infty}(\mathbb{R})$ for every $k\in\mathbb{N}$. We will then focus only on
the case $\ell\in(0,m)$.

From (5.7), Hölder’s inequality and (1.14), we deduce the following estimate,

$|\eta(x)|\leq\int_{\mathbb{R}}|\mathcal{L}_{c}(x-y)|e^{m|x-y|}\frac{|F(y)|}{e^{m|x-y|}}dy\leq
C_{1}^{\frac{1}{q}}\left(\int_{\mathbb{R}}\frac{|F(y)|^{q}}{e^{qm|x-y|}}dy\right)^{\frac{1}{q}},$
(5.11)

where $C_{1}=\|e^{m|\cdot|}\mathcal{L}_{c}\|_{L^{p}(\mathbb{R})}^{q}$. We will
prove next that $e^{\ell|\cdot|}\eta\in L^{q}(\mathbb{R})$ and
$e^{\ell|\cdot|}\eta^{\prime}\in L^{q}(\mathbb{R})$ for all $\ell\in(0,m)$. In
order to do so, let $\ell\in(0,m)$ and, for all $\varepsilon\in(0,1]$, let us
consider the functions

$h_{\varepsilon}(x)=\frac{e^{\ell|x|}}{(1+\varepsilon
e^{|x|})^{m}}|\eta(x)|,\quad\tilde{h}_{\varepsilon}(x)=\frac{e^{\ell|x|}}{(1+\varepsilon
e^{|x|})^{m}}|\eta^{\prime}(x)|.$

Since $\eta,\eta^{\prime}\in L^{\infty}(\mathbb{R})$ and $\ell<m$, it is clear
that $h_{\varepsilon},\tilde{h}_{\varepsilon}\in L^{q}(\mathbb{R})$. Let us
take now $r\in(0,q)$ and $R>1$. Using (5.11) and Hölder’s inequality with
exponents $\frac{q}{q-r}$ and $\frac{q}{r}$, we deduce that

$\displaystyle\int_{\\{|x|>R\\}}$
$\displaystyle|h_{\varepsilon}(x)|^{q}dx=\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q-r}\frac{e^{\ell
r|x|}}{(1+\varepsilon e^{|x|})^{rm}}|\eta(x)|^{r}dx$ $\displaystyle\leq
C_{1}^{\frac{r}{q}}\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q-r}\frac{e^{\ell
r|x|}}{(1+\varepsilon
e^{|x|})^{rm}}\left(\int_{\mathbb{R}}\frac{|F(y)|^{q}}{e^{qm|x-y|}}dy\right)^{\frac{r}{q}}dx$
$\displaystyle\leq
C_{1}^{\frac{r}{q}}\left(\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx\right)^{\frac{q-r}{q}}\left(\int_{\\{|x|>R\\}}\frac{e^{\ell
q|x|}}{(1+\varepsilon
e^{|x|})^{qm}}\left(\int_{\mathbb{R}}\frac{|F(y)|^{q}}{e^{qm|x-y|}}dy\right)dx\right)^{\frac{r}{q}}.$

From the previous inequality, one gets directly

$\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx\leq
C_{1}\int_{\\{|x|>R\\}}\frac{e^{\ell q|x|}}{(1+\varepsilon
e^{|x|})^{qm}}\left(\int_{\mathbb{R}}\frac{|F(y)|^{q}}{e^{qm|x-y|}}dy\right)dx.$

Now, by Fubini’s theorem and Lemma 5.5, we derive

$\displaystyle\int_{\\{|x|>R\\}}$ $\displaystyle|h_{\varepsilon}(x)|^{q}dx\leq
C_{1}\int_{\mathbb{R}}|F(y)|^{q}\left(\int_{\\{|x|>R\\}}\frac{e^{\ell
q|x|}}{(1+\varepsilon e^{|x|})^{qm}e^{qm|x-y|}}dx\right)dy$ $\displaystyle\leq
C_{1}$ $\displaystyle\int_{\\{|y|>R\\}}|F(y)|^{q}\frac{Be^{\ell
q|y|}}{(1+\varepsilon e^{|y|})^{qm}}dy$ $\displaystyle+C_{1}$
$\displaystyle\int_{\\{|y|\leq R\\}}|F(y)|^{q}\int_{\\{|x|>R\\}}\frac{e^{\ell
q|x|}}{(1+\varepsilon e^{|x|})^{qm}e^{qm|x-y|}}dxdy.$

We will now estimate the last two integrals. On the one hand, using the
inequality $||x|-|y||\leq|x-y|$, we obtain

$\displaystyle\int_{\\{|y|\leq R\\}}$
$\displaystyle|F(y)|^{q}\int_{\\{|x|>R\\}}\frac{e^{\ell q|x|}}{(1+\varepsilon
e^{|x|})^{qm}e^{qm|x-y|}}dxdy\leq\|F\|_{L^{\infty}(\mathbb{R})}^{q}\int_{\\{|y|\leq
R\\}}\int_{\\{|x|>R\\}}\frac{e^{\ell q|x|}}{e^{qm|x-y|}}dxdy$
$\displaystyle\leq\|F\|_{L^{\infty}(\mathbb{R})}^{q}\left(\int_{\\{|y|\leq
R\\}}e^{qm|y|}dy\right)\left(\int_{\\{|x|>R\\}}e^{-(m-\ell)q|x|}dx\right)\coloneqq
C_{2}/C_{1}.$

On the other hand, recall that, by Lemma 2.1,
$\eta(\pm\infty)=\eta^{\prime}(\pm\infty)=(\mathcal{W}\ast\eta)(\pm\infty)=0.$
Hence, equation (2.6) in Proposition 2.2 implies that, for any fixed
$\delta>0$ we may choose $R>1$ large enough so that

$|F(y)|^{q}\leq\delta|\eta(y)|^{q}+\delta|\eta^{\prime}(y)|^{q},\quad\text{
for all }|y|>R.$ (5.12)

Therefore,

$\int_{\\{|y|>R\\}}|F(y)|^{q}\frac{Be^{\ell q|y|}}{(1+\varepsilon
e^{|y|})^{qm}}dy\leq\delta
B\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\delta
B\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx.$

In sum,

$\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx\leq\delta
BC_{1}\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\delta
BC_{1}\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx+C_{2}.$ (5.13)

We will now derive a similar estimate for $\tilde{h}_{\varepsilon}$. Indeed,
from (5.7) with $k=1$ and using (2.5), it follows that

$\eta^{\prime}=\mathcal{L}_{c}\ast\left(4\eta^{\prime}\mathcal{W}\ast\eta+2\eta\mathcal{W}\ast\eta^{\prime}\right).$

Notice that $(\mathcal{W}\ast\eta^{\prime})(\pm\infty)=0$ too. Hence, taking
$R>1$ larger if necessary, we deduce that

$\left|4\eta^{\prime}(y)\mathcal{W}\ast\eta(y)+2\eta(y)\mathcal{W}\ast\eta^{\prime}(y)\right|^{q}\leq\delta|\eta(y)|^{q}+\delta|\eta^{\prime}(y)|^{q},\quad\text{
for all }|y|>R.$

This estimate allows us to follow the same arguments as we did for
$h_{\varepsilon}$ in order to deduce

$\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx\leq\delta
BC_{1}\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx+\delta
BC_{1}\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+C_{3},$ (5.14)

where
$C_{3}=C_{2}\|F^{\prime}\|^{q}_{L^{\infty}(\mathbb{R})}/\|F\|^{q}_{L^{\infty}(\mathbb{R})}$.
Taking now $\delta\in(0,1/2BC_{1})$, it follows directly from (5.13) and
(5.14) that

$\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx\leq
C_{4},$

where $C_{4}=(C_{2}+C_{3})/(1-2\delta BC_{1})$. By virtue of Fatou’s lemma, we
take limits as $\varepsilon$ tends to zero and obtain

$\int_{\\{|x|>R\\}}e^{\ell q|x|}|\eta(x)|^{q}dx+\int_{\\{|x|>R\\}}e^{\ell
q|x|}|\eta^{\prime}(x)|^{q}dx\leq C_{4}.$

In conclusion, $e^{\ell|\cdot|}\eta\in L^{q}(\mathbb{R})$,
$e^{\ell|\cdot|}\eta^{\prime}\in L^{q}(\mathbb{R})$.

Notice that both $e^{\ell x}\eta(x)$ and $e^{-\ell x}\eta(x)$ belong to
$W^{1,q}(\mathbb{R})$. Indeed, from what we have already proved, it is clear
that

$e^{\pm\ell x}\eta(x)\in L^{q}(\mathbb{R}),\quad\left(e^{\pm\ell
x}\eta(x)\right)^{\prime}=\left(\pm\ell\eta(x)+\eta^{\prime}(x)\right)e^{\pm\ell
x}\in L^{q}(\mathbb{R}).$

Hence, the Sobolev’s embedding theorem implies that $e^{\ell|\cdot|}\eta\in
L^{\infty}(\mathbb{R})$ and $\lim_{x\to\pm\infty}e^{\ell|x|}\eta(x)=0$.

We have just proved the result for $k=0$. Taking $k=2$ in (5.7), we deduce
analogously as before that $\eta^{\prime\prime}=\mathcal{L}_{c}\ast
F^{\prime\prime}$, where $F^{\prime\prime}$ satisfies

$|F^{\prime\prime}(y)|^{q}\leq\delta|\eta(y)|^{q}+\delta|\eta^{\prime}(y)|^{q}+\delta|\eta^{\prime\prime}(y)|^{q},\quad\text{
for all }|y|>R.$

Following the same process as above, and using the estimates we already have
for $e^{\ell|\cdot|}\eta$ and $e^{\ell|\cdot|}\eta^{\prime}$, we prove the
result for $k=1$. The complete proof follows easily by induction. ∎

Conditions (1.14) in Theorem 1.4 is not easy to check, since the operator
$\mathcal{L}_{c}$ is not simple to compute in general. For this reason, we
recall the following Paley–Wiener theorem that provides sufficient conditions
on $L_{c}$ that we will use when applying Theorem 1.4 to our examples in
Section 6. We refer to Theorem 5.4.2 in [40] or Theorem IX.13 in [52] for
details.

######  Theorem 5.6.

Let $T\in L^{2}(\mathbb{R})$. Then $e^{b|x|}T\in L^{2}(\mathbb{R})$ for all
$b<a$, if and only if $\widehat{T}$ has an analytical continuation to the
strip $\\{z\in\mathbb{C}:|z|<a\\}$ with the property that for each
$\zeta\in\mathbb{R}$ with $|\zeta|<a$, $\widehat{T}(\cdot+i\zeta)\in
L^{2}(\mathbb{R})$ and for any $b<a$,

$\sup_{|\zeta|\leq
b}\lVert\widehat{T}(\cdot+i\zeta)\rVert_{L^{2}(\mathbb{R})}<\infty.$ (5.15)

We now tackle the algebraic decay, whose proof will follow similar lines to
that of Theorem 1.4. We will employ the following lemma proved in [14].

######  Lemma 5.7.

For every $m,\ell\in\mathbb{R}$ such that $m>1$ and $0<\ell<m-1$, there exists
$B>0$ such that the following inequality holds:

$\int_{\mathbb{R}}\frac{|x|^{\ell}}{(1+\varepsilon|x|)^{m}(1+|x-y|)^{m}}dx\leq\frac{B|y|^{\ell}}{(1+\varepsilon|y|)^{m}},\quad\text{
for all }y\in\mathbb{R}\text{ and for all }\varepsilon\in(0,1].$ (5.16)

######  Theorem 5.8.

Let $c\geq 0$ and let $u\in\mathcal{E}(\mathbb{R})$ be a solution to
($S({\mathcal{W},c})$). Assume that $L_{c}\in\mathcal{S}^{\prime}(\mathbb{R})$
and

$(1+|\cdot|)^{s}\mathcal{L}_{c}\in L^{p}(\mathbb{R})\text{ for some
}p\in(1,\infty],\,\,s>1-\frac{1}{p}.$ (5.17)

Setting $q=p^{\prime}$ and $\eta=1-|u|^{2}$, we have

$|\cdot|^{\ell}D^{k}\eta\in L^{q}(\mathbb{R})\cap
L^{\infty}(\mathbb{R}),\quad\lim_{x\to\pm\infty}|x|^{\ell}D^{k}\eta(x)=0,\quad\text{
for all }\ell\in\Big{(}0,s-1+\frac{1}{p}\Big{)},\,\,k\in\mathbb{N}.$

###### Proof.

First, from (5.7) with $k=0$, Hölder’s inequality and (5.17), we deduce that

$|\eta(x)|\leq
C_{1}^{\frac{1}{q}}\left(\int_{\mathbb{R}}\frac{|F(y)|^{q}}{(1+|x-y|)^{sq}}dy\right)^{\frac{1}{q}},$

where $C_{1}=\|(1+|\cdot|)^{s}\mathcal{L}_{c}\|_{L^{p}(\mathbb{R})}^{q}$. Now,
for $\ell\in\left(0,s-1+1/p\right)$ and $\varepsilon\in(0,1]$, we consider the
functions

$h_{\varepsilon}(x)=\frac{|x|^{\ell}}{(1+\varepsilon|x|)^{s}}|\eta(x)|,\quad\tilde{h}_{\varepsilon}(x)=\frac{|x|^{\ell}}{(1+\varepsilon|x|)^{s}}|\eta^{\prime}(x)|.$

Let us take $R>1$. Arguing as in the proof of Theorem 1.4 and using Lemma 5.7,
we obtain the estimate

$\displaystyle\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx$ $\displaystyle\leq
C_{1}\int_{\\{|y|>R\\}}|F(y)|^{q}\frac{B|y|^{\ell
q}}{(1+\varepsilon|y|)^{sq}}dy$ (5.18) $\displaystyle+C_{1}\int_{\\{|y|\leq
R\\}}|F(y)|^{q}\int_{\\{|x|>R\\}}\frac{|x|^{\ell
q}}{(1+\varepsilon|x|)^{sq}(1+|x-y|)^{sq}}dxdy.$

On the one hand, since $\ell q<sq-1$, then the function
$x\mapsto\frac{|x|^{\ell q}}{(|x|+1-R)^{sq}}$ belongs to $L^{1}((R,\infty))$.
Hence, the inequality $||x|-|y||\leq|x-y|$ leads to

$\displaystyle\int_{\\{|y|\leq R\\}}$
$\displaystyle|F(y)|^{q}\int_{\\{|x|>R\\}}\frac{|x|^{\ell
q}}{(1+\varepsilon|x|)^{sq}(1+|x-y|)^{sq}}dxdy$ $\displaystyle\leq
2R\|F\|_{L^{\infty}(\mathbb{R})}^{q}\int_{\\{|x|>R\\}}\frac{|x|^{\ell
q}}{(|x|+1-R)^{sq}}dx\coloneqq C_{2}/C_{1}.$

On the other hand, a shown in the proof of Theorem 1.4, for any fixed
$\delta>0$, we may choose $R>1$ large enough so that (5.12) holds. Therefore,

$\int_{\\{|y|>R\\}}|F(y)|^{q}\frac{B|y|^{\ell
q}}{(1+\varepsilon|y|)^{sq}}dy\leq\delta
B\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\delta
B\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx.$

In sum, by (5.18),

$\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx\leq\delta
BC_{1}\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\delta
BC_{1}\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx+C_{2}.$

Reasoning as in the proof of Theorem 1.4, we derive the analogous estimate for
$\tilde{h}_{\varepsilon}$:

$\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx\leq\delta
BC_{1}\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\delta
BC_{1}\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx+C_{3},$

where
$C_{3}=C_{2}\|F^{\prime}\|_{L^{\infty}(\mathbb{R})}^{q}/\|F\|_{L^{\infty}(\mathbb{R})}^{q}$.
Combining the last two inequalities and taking $\delta\in(0,1/2BC_{1})$ yields

$\int_{\\{|x|>R\\}}|h_{\varepsilon}(x)|^{q}dx+\int_{\\{|x|>R\\}}|\tilde{h}_{\varepsilon}(x)|^{q}dx\leq
C_{4},$

where $C_{4}=(C_{2}+C_{3})/(1-2\delta BC_{1})$. By virtue of Fatou’s lemma, we
take limits as $\varepsilon$ tends to zero, and obtain

$\int_{\\{|x|>R\\}}|x|^{\ell q}|\eta(x)|^{q}dx+\int_{\\{|x|>R\\}}|x|^{\ell
q}|\eta^{\prime}(x)|^{q}dx\leq C_{4}.$

Equivalently, $|\cdot|^{\ell}\eta\in L^{q}(\mathbb{R})$ and
$|\cdot|^{\ell}\eta^{\prime}\in L^{q}(\mathbb{R})$.

Let us now consider a function $\varphi:\mathbb{R}\to(0,\infty)$ that is of
class $\mathcal{C}^{1}$ and satisfies that $\varphi(x)=|x|^{\ell}$ for every
$|x|>1$. At this point, it is clear that $\varphi\eta\in W^{1,q}(\mathbb{R})$.
Hence, the Sobolev’s embedding implies that $\varphi\eta\in
L^{\infty}(\mathbb{R})$ and $\lim_{x\to\pm\infty}\varphi(x)\eta(x)=0$. This
proves the result for $k=0$. As in the proof of Theorem 1.4, the rest of the
proof follows by induction. ∎

Conditions (5.17) in Theorem 5.8 can be difficult to verify. The next
corollary provides sufficient (and easy to check) conditions on $\mathcal{W}$
that guarantee (5.17) and, in turn, algebraic decay of finite energy traveling
waves.

######  Corollary 5.9.

Assume that $\mathcal{W}$ satisfies (H1) and that weakly differentiable up to
order $s\in\mathbb{N}\setminus\\{0\\}$, with

$D^{s}\widehat{\mathcal{W}}\in L^{\infty}(\mathbb{R}).$ (5.19)

Let $c\in[0,\sqrt{2\sigma})$ and let $u\in\mathcal{E}(\mathbb{R})$ be a
solution to ($S({\mathcal{W},c})$). Then,

$|\cdot|^{\ell}D^{k}\eta\in L^{2}(\mathbb{R})\cap
L^{\infty}(\mathbb{R}),\quad\lim_{x\to\pm\infty}|x|^{\ell}D^{k}\eta(x)=0,\quad\text{
for all }\ell\in(0,s-{1}/{2}),\,\,k\in\mathbb{N}.$

###### Proof.

By (5.3), since (H1) holds, we see $L_{c}$ is bounded with $L_{c}\in
L^{1}(\mathbb{R})$, so that $L_{c}\in L^{2}(\mathbb{R})$. Using also (5.19),
one may verify that $D^{s}L_{c}\in L^{1}(\mathbb{R})\cap
L^{\infty}(\mathbb{R})$ too. In particular, $D^{s}L_{c}\in L^{2}(\mathbb{R})$.
Then, applying Fourier transform, $|\cdot|^{s}\mathcal{L}_{c}\in
L^{2}(\mathbb{R})$. Hence, since $\mathcal{L}_{c}\in L^{\infty}(\mathbb{R})$,
it follows that (5.17) holds for $p=2$ and we can apply Theorem 5.8. ∎

###  5.2 Analyticity

Let us recall that for $H\in\mathcal{S}^{\prime}(\mathbb{R})$, the associated
multiplier operator $\mathcal{H}$ is defined by

$\widehat{\mathcal{H}(\varphi)}(\xi)=H(\xi)\widehat{\varphi}(\xi),\quad\text{
for all }\varphi\in\mathcal{S}(\mathbb{R}).$

We say that $H$ is an $L^{p}$-_multiplier_ , with $p\in[1,\infty]$, if there
exists $\alpha>0$ such that

$\|\mathcal{H}(\varphi)\|_{L^{p}(\mathbb{R})}\leq\alpha\|\varphi\|_{L^{p}(\mathbb{R})},\quad\text{
for all }\varphi\in L^{p}(\mathbb{R}).$

The smallest $\alpha>0$ for which the previous inequality holds is the norm of
the multiplier, and it is denoted by $\|\mathcal{H}\|_{p}$. For instance, by
assumption (H0), $\mathcal{W}$ is an $L^{2}$-multiplier and, by (1.23),
$\|\mathcal{W}\|_{2}=\|\widehat{\mathcal{W}}\|_{L^{\infty}(\mathbb{R})}$.

We recall the so-called Hörmander–Mikhlin multiplier theorem in [32, 48] (see
also [44]) adapted to our one-dimensional setting, as follows.

######  Theorem 5.10 ([32, 48]).

Let $H:\mathbb{R}\to\mathbb{R}$ be a weakly differentiable function and
suppose that there exists $M>0$ such that

$\sup\\{|\xi^{k}D^{k}H(\xi)|\ :\ \xi\in\mathbb{R}\setminus\\{0\\},\
k\in\\{0,1\\}\\}\leq M.$ (5.20)

Then $H$ is an $L^{p}$-multiplier for every $p\in(1,\infty)$. Moreover, there
exists a constant $C_{p}>0$, depending only on $p$, such that

$\lVert\mathcal{H}\rVert_{p}\leq C_{p}M.$

Assume that $\mathcal{W}$ satisfies (H1), and including also the limit case
$\kappa=1/2,$ and that $\widehat{\mathcal{W}}$ is (weakly) differentiable. We
will apply this theorem to the function

$H_{c}(\xi)=\frac{-\xi^{2}}{\xi^{2}+2\widehat{\mathcal{W}}(\xi)-c^{2}},\quad\text{
for }c\in[0,\sqrt{2\sigma}).$

Observe that then $H_{c}\in L^{\infty}(\mathbb{R})$ and that

$\xi
H^{\prime}_{c}(\xi)=\frac{2\xi^{3}\big{(}\widehat{\mathcal{W}}\big{)}^{\prime}(\xi)-4\xi^{2}\widehat{\mathcal{W}}(\xi)+2c^{2}\xi^{2}}{(\xi^{2}+2\widehat{\mathcal{W}}(\xi)-c^{2})^{2}}.$

Therefore, $\xi\mapsto\xi H^{\prime}_{c}(\xi)$ is a bounded function if

$\big{|}\big{(}\widehat{\mathcal{W}}\big{)}^{\prime}(\xi)\big{|}\leq
C(|\xi|+1)\quad\text{a.e. }\xi\in\mathbb{R},$ (5.21)

for some $C>0$. In this case, using Theorem 5.10 we conclude that $H_{c}$ is
an $L^{p}$-multiplier for every $p\in(1,\infty)$. More precisely, for every
$p\in(1,\infty)$, there exists a constant $\alpha_{p}>0$ such that

$\|\mathcal{H}_{c}(\varphi)\|_{L^{p}(\mathbb{R})}\leq\alpha_{p}\|\varphi\|_{L^{p}(\mathbb{R})},\quad\text{
for all }\varphi\in L^{p}(\mathbb{R}),$ (5.22)

where $\widehat{\mathcal{H}}_{c}=H_{c}$.

Let $u\in\mathcal{E}(\mathbb{R})$ be a solution to ($S({\mathcal{W},c})$). We
will exploit (5.7) in order to prove that $\eta$ is a real analytic function.
First, we prove a technical lemma.

######  Lemma 5.11.

Assume that there exist $\sigma\in(0,1]$ and $\kappa\in[0,1/2]$ such that
$\widehat{\mathcal{W}}(\xi)\geq\sigma-\kappa\xi^{2}$ a.e. on $\mathbb{R}$.
Assume in addition that $\widehat{\mathcal{W}}$ is weakly differentiable and
that there exists $C>0$ such that (5.21) holds. Let $c\in[0,\sqrt{2\sigma})$
and let $u\in\mathcal{E}(\mathbb{R})$ be a solution to ($S({\mathcal{W},c})$).
Let us denote

$\mu_{k}\coloneqq\max\\{\|D^{j}F^{\prime}\|_{L^{q}(\mathbb{R})}:j=0,\dots,k\\},\quad\text{
for all }k\in\mathbb{N}.$

Then there exist $\beta,\gamma>0$, depending on $\eta$ only through
$\|D^{j}\eta\|_{L^{2}(\mathbb{R})}$ and
$\|D^{j}\eta\|_{L^{\infty}(\mathbb{R})}$ for $j=0,1,2$, such that

$\displaystyle\|D^{j}\eta\|_{L^{\infty}(\mathbb{R})}\leq\beta\mu_{k}\quad$
$\displaystyle\text{ for all }k\in\mathbb{N},\,\,\text{ for all
}j=0,\dots,k+2,$ (5.23)
$\displaystyle\|D^{j}\eta\|_{L^{2}(\mathbb{R})}\leq\beta\mu_{k}\quad$
$\displaystyle\text{ for all }k\in\mathbb{N},\,\,\text{ for all
}j=0,\dots,k+2,$ (5.24) $\displaystyle\mu_{k}\leq\gamma^{k}k^{k-1}\quad$
$\displaystyle\text{ for all
}k\in\mathbb{N}\setminus\\{0\\},\qquad\mu_{0}\leq\frac{\gamma}{24\omega\beta^{2}}-1,$
(5.25)

where $\omega=\|\widehat{\mathcal{W}}\|_{L^{\infty}(\mathbb{R})}$.

###### Proof.

We start by proving (5.23) and (5.24). These estimates hold true for $k=0$ by
simply choosing

$\beta\geq\frac{\max\\{\|D^{j}\eta\|_{L^{\infty}(\mathbb{R})},\|D^{j}\eta\|_{L^{2}(\mathbb{R})}:j=0,1,2\\}}{\|F^{\prime}\|_{L^{2}(\mathbb{R})}}.$

Let us take $k\geq 1$ and $j\in\\{3,\dots,k+2\\}$. By (5.7), we have

$\eta^{\prime\prime\prime}=\mathcal{H}_{c}(F^{\prime})\quad\text{ on
}\mathbb{R},\text{ with
}F^{\prime}=4\eta^{\prime}(\mathcal{W}\ast\eta)+2\eta(\mathcal{W}\ast\eta^{\prime}).$
(5.26)

By using also (5.22) with $p=2$, it follows that

$\|D^{j}\eta\|_{L^{2}(\mathbb{R})}\leq\alpha_{2}\|D^{j-3}F^{\prime}\|_{L^{2}(\mathbb{R})}\leq\alpha_{2}\mu_{k}.$

Moreover, by invoking the Sobolev’s embedding (see Remark 3.4), we obtain

$\displaystyle\|D^{j}\eta\|_{L^{\infty}(\mathbb{R})}$
$\displaystyle\leq\frac{1}{2}\left(\|D^{j}\eta\|_{L^{2}(\mathbb{R})}+\|D^{j+1}\eta\|_{L^{2}(\mathbb{R})}\right)$
$\displaystyle\leq\frac{\alpha_{2}}{2}\left(\|D^{j-3}F^{\prime}\|_{L^{2}(\mathbb{R})}+\|D^{j-2}F^{\prime}\|_{L^{2}(\mathbb{R})}\right)\leq\alpha_{2}\mu_{k}.$

Therefore, we take $\beta\geq\alpha_{2}$ so that (5.23) and (5.24) follow.

As far as (5.25) is concerned, we will prove it by induction. Indeed, it holds
true for $k=1$ if one chooses $\gamma\geq\mu_{1}$. Let us assume as induction
hypothesis that there exists $\tilde{k}\in\mathbb{N}\setminus\\{0,1\\}$ such
that (5.25) holds for every $k\leq\tilde{k}$. Next we compute for
$k=\tilde{k}$, taking (5.7) into account,

$\displaystyle\|D^{k+1}$ $\displaystyle
F^{\prime}\|_{L^{2}(\mathbb{R})}=\|D^{k+1}\left(2\eta(\mathcal{W}\ast\eta^{\prime})+4\eta^{\prime}(\mathcal{W}\ast\eta)\right)\|_{L^{2}(\mathbb{R})}$
$\displaystyle=\|2D^{k}\left(\eta^{\prime}(\mathcal{W}\ast\eta^{\prime})+\eta(\mathcal{W}\ast\eta^{\prime\prime})\right)+4D^{k}\left(\eta^{\prime}(\mathcal{W}\ast\eta^{\prime})+\eta^{\prime}(\mathcal{W}\ast\eta^{\prime\prime})\right)\|_{L^{2}(\mathbb{R})}$
$\displaystyle\leq 2\sum_{j=0}^{k}{k\choose j}(\|D^{j+1}\eta(\mathcal{W}\ast
D^{k-j+1}\eta)\|_{L^{2}(\mathbb{R})}+\|D^{j}\eta(\mathcal{W}\ast
D^{k-j+2}\eta)\|_{L^{2}(\mathbb{R})})$ $\displaystyle+4\sum_{j=0}^{k}{k\choose
j}(\|(\mathcal{W}\ast
D^{j+1}\eta)D^{k-j+1}\eta\|_{L^{2}(\mathbb{R})}+\|(\mathcal{W}\ast
D^{j}\eta)D^{k-j+2}\eta\|_{L^{2}(\mathbb{R})}).$

Using (1.23), (5.23) and (5.24), we deduce that

$\displaystyle\|D^{k+1}$ $\displaystyle F^{\prime}\|_{L^{2}(\mathbb{R})}\leq
2\omega\sum_{j=0}^{k}{k\choose
j}(\|D^{j+1}\eta\|_{L^{\infty}(\mathbb{R})}\|D^{k-j+1}\eta\|_{L^{2}(\mathbb{R})}+\|D^{j}\eta\|_{L^{\infty}(\mathbb{R})}\|D^{k-j+2}\eta\|_{L^{2}(\mathbb{R})})$
$\displaystyle+4\omega\sum_{j=0}^{k}{k\choose
j}(\|D^{j+1}\eta\|_{L^{2}(\mathbb{R})}\|D^{k-j+1}\eta\|_{L^{\infty}(\mathbb{R})}+\|D^{j}\eta\|_{L^{2}(\mathbb{R})}\|D^{k-j+2}\eta\|_{L^{\infty}(\mathbb{R})})$
$\displaystyle\leq 12\omega\beta^{2}\sum_{j=0}^{k}{k\choose
j}\mu_{j}\mu_{k-j}.$

The induction hypothesis leads to

$\displaystyle\|D^{k+1}$ $\displaystyle F^{\prime}\|_{L^{2}(\mathbb{R})}\leq
12\omega\beta^{2}\Bigg{(}2\mu_{0}\mu_{k}+\gamma^{k}\sum_{j=1}^{k-1}{k\choose
j}j^{j-1}(k-j)^{k-j-1}\Bigg{)}$
$\displaystyle=12\omega\beta^{2}\Bigg{(}2(\mu_{0}\mu_{k}-\gamma^{k}k^{k-1})+\gamma^{k}\sum_{j=0}^{k}\frac{k!}{j!(k-j)!}j^{(j-1)^{+}}(k-j)^{(k-j-1)^{+}}\Bigg{)},$

where we adopt the convention $0^{0}=1$. Now, a combinatorial lemma due to
Kahane [37] implies that

$\|D^{k+1}F^{\prime}\|_{L^{2}(\mathbb{R})}\leq
12\omega\beta^{2}\left(2(\mu_{0}\mu_{k}-\gamma^{k}k^{k-1})+4\gamma^{k}k^{k-1}\right).$

Using the induction hypothesis again and choosing $\gamma>0$ large enough so
that $\mu_{0}\leq\frac{\gamma}{24\omega\beta^{2}}-1$, we deduce that

$\|D^{k+1}F^{\prime}\|_{L^{2}(\mathbb{R})}\leq
24\omega\beta^{2}(\mu_{0}+1)\gamma^{k}k^{k-1}\leq\gamma^{k+1}k^{k-1}.$

In conclusion,

$\mu_{k+1}=\max\\{\mu_{k},\|D^{k+1}F^{\prime}\|_{L^{2}(\mathbb{R})}\\}\leq\max\\{\gamma^{k}k^{k-1},\gamma^{k+1}k^{k-1}\\}=\gamma^{k+1}k^{k-1}\leq\gamma^{k+1}(k+1)^{k},$

which completes the proof. ∎

We are thus led to the following analyticity of $\eta$ as follows.

######  Theorem 5.12.

Under the hypotheses of Lemma 5.11, for every solution
$u\in\mathcal{E}(\mathbb{R})$ to ($S({\mathcal{W},c})$) with
$c\in[0,\sqrt{2\sigma})$, there exists $r>0$ such that $\eta=1-|u|^{2}$ and
$u$ have analytic extensions to the strip
$\mathcal{S}_{r}=\\{z\in\mathbb{C}:|\operatorname{Im}z|<r\\}$. If
$c\in(0,\sqrt{2\sigma})$, then $u$ is real analytic on $\mathbb{R}$, in the
sense that $\operatorname{Re}(u)$ and $\operatorname{Im}(u)$ are real analytic
on $\mathbb{R}$.

###### Proof.

We need to prove that the Taylor series expansion about any point
$x_{0}\in\mathbb{R}$ converges with radius of convergence $r>0$ independent of
$x_{0}$. Indeed, let $I_{r}=[x_{0}-r,x_{0}+r]$, then, by Taylor’s theorem,

$\eta(x)-\sum_{k=0}^{n}\frac{D^{k}\eta(x_{0})}{k!}(x-x_{0})^{k}=\frac{D^{n+1}\eta(\zeta)}{(n+1)!}(x-x_{0})^{n+1}$
(5.27)

for every $x\in I_{r}$ and for some $\zeta\in I_{r}$. Now we deduce from Lemma
5.11 that

$|D^{k}\eta(\zeta)|\leq\beta\mu_{k}\leq\beta\gamma^{k}k^{k-1}$

for every $k\in\mathbb{N}\setminus\\{0\\}$. Since

$\big{(}\frac{\gamma^{k}k^{k-1}}{k!}\big{)}^{1/k}\to\gamma e,\quad\text{as
}k\to\infty,$

we conclude that the left-hand side of (5.27) goes to zero as $n\to\infty$ and
that the radius of convergence satisfies $r\geq(\gamma e)^{-1}$. In addition,
$\eta$ has an analytic extensions to the strip $\mathcal{S}_{r}$.

In the case that $c\in(0,\sqrt{2\sigma})$, by Proposition 2.2, we have
$\sup_{\mathbb{R}}\eta<1$, so that $\rho=|u|=\sqrt{1-u^{2}}$ is a real
analytic function, as a composition of real analytic functions. Also, $\theta$
given by (2.14) is real analytic as the integral of a real analytic function.
Consequently, $\operatorname{Re}(u)=\rho\cos(\theta)$ and
$\operatorname{Im}(u)=\rho\sin(\theta)$ are real analytic functions on
$\mathbb{R}$. ∎

######  Remark 5.13.

As pointed out by Corollary 4.1.5 in [14], the fact that $\eta$ has an
analytic extension to the strip $\mathcal{S}_{r}$ implies the following
exponential decay of its Fourier transform:

$\int_{\mathbb{R}}|\hat{\eta}(\xi)|^{2}e^{2\mu|\xi|}d\xi<\infty,\quad\text{
for all }\mu\in(0,r).$

We end this section by proving Corollary 1.6 as a consequence of Theorem 5.12.

###### Proof of Corollary 1.6.

Let $\mathfrak{q}\in(0,\mathfrak{q}_{*})$ and let $u=\rho
e^{i\theta}\in\mathcal{N}\mathcal{E}(\mathbb{R})$ be the nontrivial solution
to ($S({\mathcal{W},c})$), given by Theorem 1 in [22], satisfying

$E(u)=E_{\min}(\mathfrak{q}).$ (5.28)

Arguing as in the proof of Proposition 3.12 in [22], we see that there exists
$a_{0}\in\mathbb{R}$ such that

$\frac{1}{2}\int_{a_{0}}^{\infty}(1-\rho^{2})\theta^{\prime}=\frac{\mathfrak{q}}{2},$

which allows us to define the following function

$\tilde{u}(x):=\tilde{\rho}(x)e^{i\tilde{\theta}(x)}=\rho(x-a_{0})e^{i(\theta(x-a_{0})-\theta(-a_{0}))}.$

Notice that $\tilde{u}$ is nothing but $u$ multiplied by the constant of
modulus one $e^{i\theta(-a_{0})}$ and translated in the space variable, so
$\tilde{u}$ is still satisfies (5.28), i.e. it is a solution to the
minimization problem. Moreover, $\tilde{u}$ satisfies that
$\tilde{\theta}(0)=0$ and

$\frac{1}{2}\int_{0}^{\infty}(1-\tilde{\rho}^{2})\tilde{\theta}^{\prime}=\frac{\mathfrak{q}}{2}.$
(5.29)

Furthermore,

$\frac{1}{2}\int_{-\infty}^{0}(1-\tilde{\rho}^{2})\tilde{\theta}^{\prime}=p(v)-\frac{1}{2}\int_{0}^{\infty}(1-\tilde{\rho}^{2})\tilde{\theta}^{\prime}=\frac{\mathfrak{q}}{2}.$
(5.30)

For notational simplicity, we continue to write $u$, $\rho$ and $\theta$ for
$\tilde{u}$, $\tilde{\rho}$ and $\tilde{\theta}$. By using the reflection
operators $T^{\pm}$ and $S^{\pm}$ introduced in the proof of Proposition 3.12
in [22], and the fact that $\rho$ and $\theta$ are continuous, it follows that
the functions

$u^{\pm}=(T^{\pm}\rho)e^{iS^{\pm}\theta}$

belong to $\mathcal{N}\mathcal{E}(\mathbb{R})$. Bearing in mind (5.29) and
(5.30), we obtain that $p(u^{\pm})=\mathfrak{q},$ which implies that
$E_{\textup{min}}(\mathfrak{q})\leq E(u^{\pm}).$

On the other hand, as in Proposition 3.12 in [22], we get

$E(u^{+})+E(u^{-})=2E_{\textup{k}}(u)+E_{\textup{p}}(u^{+})+E_{\textup{p}}(u^{-})\quad\text{
and }\quad E_{\textup{p}}(u^{+})+E_{\textup{p}}(u^{-})\leq
2E_{\textup{p}}(u).$

Since $u$ satisfies (5.28), we deduce that

$E_{\textup{min}}(\mathfrak{q})\leq\frac{E(u^{+})+E(u^{-})}{2}\leq
E(u)=E_{\textup{min}}(\mathfrak{q}).$

Hence,

$\frac{E(u^{+})+E(u^{-})}{2}=E(u)=E_{\textup{min}}(\mathfrak{q}).$

Observe that

$E_{\textup{min}}(\mathfrak{q})\leq E(u^{+})=2E(u)-E(u^{-})\leq
E_{\textup{min}}(\mathfrak{q}).$

In consequence, $E(u^{+})=E(u^{-})=E(u)=E_{\textup{min}}(\mathfrak{q})$. This
shows that $u^{\pm}$ and $u$ are solutions to the minimization problem (1.5)
and therefore, $u^{\pm}$ and $u$ satisfy ($S({\mathcal{W},c})$), for some $c$
depending on $\mathfrak{q}$. By virtue of Theorem 5.12, we have that
$|u^{\pm}|^{2}$ and $|u|^{2}$ are real analytic functions. Thus, since
$|u^{+}|=|u|$ in $\mathbb{R}_{+}$, then $|u^{+}|=|u|$ in $\mathbb{R}$. This
proves that $\rho=|u|$ is even. On the other hand, from (2.12), from the
symmetry of $\rho$ and from the fact that $\theta(0)=0$, we derive

$\theta(x)=\frac{c}{2}\int_{0}^{x}\left(\frac{1}{\rho(y)^{2}}-1\right)dy=-\frac{c}{2}\int_{0}^{-x}\left(\frac{1}{\rho(y)^{2}}-1\right)dy=-\theta(-x).$

This concludes the proof. ∎

##  6 Proofs of the examples

###### Proof of Theorem 1.7.

The existence of a solution $u$ for every $c\in(0,\sqrt{2})$ is an immediate
consequence of Theorem 4.1. Also, since $\widehat{\mathcal{W}}_{\alpha,\beta}$
fulfills (5.17), $\eta=1-|u|^{2}$ is real analytic by Theorem 5.12. The
nonexistence of finite energy solutions follows from Theorem 1.8 and the fact
that
$(\widehat{\mathcal{W}}_{\alpha,\beta})^{\prime\prime}(0)=4\alpha\beta^{-2}(\beta-2\alpha)^{-1}\neq-1$.

It remains to prove the exponential decay. By explicit computations, we can
find for some $\beta_{1},\beta_{2}>0$, depending only on $\alpha$, $\beta$ and
$c$ such that

$\mathcal{L}_{c}(x)=\alpha_{1}e^{-\beta_{1}|x|}+\alpha_{2}e^{-\beta_{2}|x|},\quad\text{
for all }x\in\mathbb{R},\quad\text{with
}\alpha_{1}=\frac{\beta^{2}-\beta_{1}^{2}}{2\beta_{1}(\beta_{2}^{2}-\beta_{1}^{2})},\
\alpha_{2}=\frac{\beta_{2}^{2}-\beta^{2}}{2\beta_{2}(\beta_{2}^{2}-\beta_{1}^{2})}.$

Thus, $\mathcal{L}_{c}$ satisfies the condition (1.14) in Theorem 1.4 with
$m=\min\\{\beta_{1},\beta_{2}\\}$ and $p=\infty$. ∎

###### Proof of Theorem 1.8.

It is clear that (H0) holds for the three potentials. Notice that
$2-\cos(\lambda\xi)\geq 1$, for $\xi\in\mathbb{R}$, and using the elementary
inequalities $e^{x}\geq 1+x$ and $\sin(x)/x\geq 1-x^{2}/6$, for
$x\in\mathbb{R}$,

$\displaystyle e^{-\lambda\xi^{2}}\geq 1-{\lambda}\xi^{2}\quad\text{ and
}\quad\frac{\sin(\lambda\xi)}{\lambda\xi}\geq
1-\frac{\lambda^{2}\xi^{2}}{6},\quad\text{ for all }\xi\in\mathbb{R}.$ (6.1)

Hence, (H1) is satisfied, with $(\sigma,\kappa)=(1,0)$,
$(\sigma,\kappa)=(1,\lambda)$ and $(\sigma,\kappa)=(1,\lambda^{2}/3)$, in case
(i), (ii) and (iii), respectively. In particular, in the three cases we have

$M_{c}(\xi)=\xi^{2}+2\widehat{\mathcal{W}}_{\lambda}(\xi)-c^{2}\geq
2-c^{2}+\xi^{2}(1-2\kappa)>0,\ \text{ for all }\xi\in\mathbb{R}\text{ and
}c\in(0,\sqrt{2}),$ (6.2)

and the existence of solutions is given by Theorem 1.1. The analyticity of
$\eta=1-|u|^{2}$ follows from Theorem 5.12.

The nonexistence of finite energy solutions follows from Theorem 1.5 and the
fact that in the case (i) we have
$(\widehat{\mathcal{W}}_{\lambda})^{\prime\prime}(0)=\lambda^{2}$, while in
the case (ii),
$(\widehat{\mathcal{W}}_{\lambda})^{\prime\prime}(0)=-2\lambda$.

To prove the exponential decay, in view of (6.2), we deduce that in all the
cases $\widehat{\mathcal{W}}_{\lambda}$ can be extended as an analytic
function on $\mathbb{C}$. Hence, we only need to verify that for fixed
$c\in(0,\sqrt{2})$ and $\lambda$, we can find a constant
$\delta=\delta(c,\lambda)>0$ such that
$M_{c}(z)=z^{2}+2\widehat{\mathcal{W}}_{\lambda}(z)-c^{2}$ does not vanish on
the strip
$\mathcal{S}_{\delta}:=\\{z\in\mathbb{C}:|\operatorname{Im}z|<\delta\\}$ and
that $L_{c}(z)=(M_{c}(z))^{-1}$ satisfies the integrability condition in
(5.15). This will imply that $e^{\delta|\cdot|}\mathcal{L}_{c}\in
L^{2}(\mathbb{R})$, so that the decay follows by invoking Theorem 1.4 with
$p=2$.

Let us show that there is $\delta=\delta(c,\lambda)\in(0,1)$ such that

$|M_{c}(\xi+iw)|\geq\delta,\quad\text{ for all }|w|\leq\delta,\text{ for all
}\xi\in\mathbb{R}.$ (6.3)

Arguing by contradiction, we get the existence of sequences
$\delta_{n}\in(0,1)$, $|w_{n}|\leq 1$, $\xi_{n}\in\mathbb{R}$, with
$\delta_{n}\to 0$, $w_{n}\to 0$, and such that

$\displaystyle T(\xi_{n},w_{n})$ $\displaystyle=o_{n}(1),\quad\text{ with
}T(\xi,w):=\xi^{2}-w^{2}+2\operatorname{Re}(\widehat{\mathcal{W}}_{\lambda}(\xi+iw))-c^{2},$
(6.4) $\displaystyle G(\xi_{n},w_{n})$ $\displaystyle=o_{n}(1),\quad\text{
with }G(\xi,w):=2\xi
w+2\operatorname{Im}(\widehat{\mathcal{W}}_{\lambda}(\xi+iw)).$ (6.5)

By using the explicit expressions for $\widehat{\mathcal{W}}_{\lambda}$, it is
easy to check that in the case (i), we have

$\displaystyle|\widehat{\mathcal{W}}_{\lambda}(\xi+iw)|=|2-\cos(\lambda\xi)\cosh(\lambda
w)+i\sin(\lambda\xi)\sinh(\lambda w)|\leq 2+\cosh(\lambda
w)+\sinh(\lambda|w|).$ (6.6)

In the case (ii), we get
$|\widehat{\mathcal{W}}_{\lambda}(\xi+iw)|=|e^{-\lambda(\xi^{2}-w^{2}+2i\xi
w)}|\leq e^{\lambda w^{2}}$, while in the case (iii),

$|\widehat{\mathcal{W}}_{\lambda}(\xi+iw)|=\frac{|\sin(\lambda\xi)\cosh(\lambda
w)+i\cos(\lambda\xi)\sinh(\lambda
w)|}{\lambda\sqrt{\xi^{2}+w^{2}}}\leq\cosh(\lambda
w)+\frac{\sinh(\lambda|w|)}{\lambda|w|}.$

Hence, $\widehat{\mathcal{W}}_{\lambda}$ is bounded on the strip
$\mathcal{S}_{1}$, that is, there is $K>0$ such that
$|\widehat{\mathcal{W}}_{\lambda}(\xi+iw)|\leq K$, for all
$\xi+iw\in\mathcal{S}_{1}$. Therefore, we infer from (6.4) that
$\\{\xi_{n}\\}$ is bounded, so that there are $\xi_{*}\in\mathbb{R}$ and
subsequence, that we do not relabel, such that $\xi_{n}\to\xi_{*}$. In this
manner, passing to the limit in (6.4), we conclude that $T(\xi_{*},0)=0$, i.e.
$M_{c}(\xi_{*})=0$, which contradicts (6.2). The proof of (6.3) is completed.

By (6.3), the function

$L_{c}(\xi+iw)=\frac{1}{M_{c}(\xi+iw)}=\frac{1}{T(\xi,w)+iG(\xi,w)}$

defines an analytic function on the strip $\mathcal{S}_{\delta}$. Also, for
all $|w|\leq\delta\leq 1$, we infer the estimate

$|L_{c}(\xi+iw)|\leq\begin{cases}\dfrac{1}{\xi^{2}-3-2K},&\text{ if
}\xi^{2}\geq 4+2K,\\\ \delta^{-1},&\text{ otherwise}.\end{cases}$

Consequently, $\sup_{|w|\leq\delta}\lVert
L_{c}(\cdot+iw)\rVert_{L^{2}(\mathbb{R})}<\infty,$ which completes the proof
of the exponential decay.

It is left to prove the existence of $u_{c}$ for every $c\in(0,\sqrt{2})$ in
the case (i) for $\lambda\leq\sqrt{2/3}.$ To do so, it is enough to verify
that the hypotheses of Theorem 4.1 hold. It is clear that (H2) and (H4) are
satisfied. In order to check (H5), let us denote
$\mu_{\lambda}=-\frac{1}{4}\big{(}\delta_{-\lambda}+\delta_{\lambda}\big{)}$,
so that $\mathcal{W}_{\lambda}=2(\delta_{0}+\mu_{\lambda}$). Thus
$\mu_{\lambda}^{+}=0$ and $\|\mu_{\lambda}^{-}\|=1/2<1$, so that Proposition
4.5 implies that (H5) holds with $V_{0}(c)=\sqrt{1+c^{2}/4}$.

Finally, we show that (H3) is fulfilled, at least for
$\lambda\in(0,\sqrt{2/3}]$. Indeed, in this case, let us set
$\mathfrak{s}=\min_{x\in\mathbb{R}}(\sin(x)/x)\in(-1,0)$ and
$m_{\lambda}=-\mathfrak{s}\lambda^{2}\in(0,2/3)$. Thus

$\big{(}\widehat{\mathcal{W}}\big{)}^{\prime}(\xi)=\lambda\sin(\lambda\xi)\geq-
m_{\lambda}\xi,\quad\text{ for all }\xi\geq 0.$

Furthermore, with this choice of $m_{\lambda}$, we get, for all
$c\in(0,\sqrt{2})$,

$m_{\lambda}V_{0}(c)^{2}\leq(3/2)m_{\lambda}<1.$

Hence, we can apply Theorem 4.1. ∎

######  Remark 6.1.

A careful study of the functions $T(\xi,w)$ and $G(\xi,w)$ in the proof of
Theorem 1.8 should lead to the sharp exponential decay of the solitons.

######  Remark 6.2.

Notice that if $\mathcal{W}_{\lambda}$ is given by (1.19), then
$(\widehat{\mathcal{W}}_{\lambda})^{\prime\prime}(0)=-\lambda^{2}/3$, so that
$(\widehat{\mathcal{W}}_{\lambda})^{\prime\prime}(0)\neq-1$, for all
$\lambda\in(0,\sqrt{3})$. However, we cannot apply Theorem 1.8 due to the
change of sign of $\mathcal{W}_{\lambda}$.

###### Proof of Theorem 1.9.

For $\kappa\in(0,1/2)$, it is obvious that $\mathcal{W}_{\kappa}$ satisfies
(H0) and (H1), so we can use Theorem 1.1. For $\kappa=1/2$, we can apply
Corollary 1.2. Therefore, for any $\kappa\in(0,1/2)$, we conclude the
existence of nontrivial solutions to $(S(\mathcal{W}_{\kappa},c))$ in
$\mathcal{N}\mathcal{E}(\mathbb{R})$ for almost every $c\in(0,\sqrt{2})$.
Moreover, since $(\widehat{\mathcal{W}}_{\kappa})^{\prime}(\xi)=-2\kappa|\xi|$
for $|\xi|<1/\sqrt{\kappa}$, and
$(\widehat{\mathcal{W}}_{\kappa})^{\prime}(\xi)=0$, for
$|\xi|>1/\sqrt{\kappa}$, we can apply Theorem 5.12 to obtain that
$\eta=1-|u|^{2}$ is real analytic. In addition, since $\mathcal{W}_{\kappa}$
fulfills condition (i) if $\kappa=1/2$ in Theorem 5.1, and condition (ii)
otherwise, we get the nonexistence for $c=\sqrt{2}$.

It is left to prove the algebraic decay of $\eta$. Remark that we can apply
Corollary 5.9 with $s=1$, but we can get a better decay by computing
explicitly $\mathcal{L}_{c}$. In fact, since $L_{c}\in L^{1}(\mathbb{R})$,
then

$\mathcal{L}_{c}(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{ix\xi}L_{c}(\xi)d\xi=\frac{1}{2\pi}\int_{-\alpha}^{\alpha}\frac{\cos(x\xi)}{(1-2\kappa)\xi^{2}+2-c^{2}}d\xi+\frac{1}{2\pi}\int_{\\{|\xi|>\alpha\\}}\frac{\cos(x\xi)}{\xi^{2}-c^{2}}d\xi,$

where we have used that $L_{c}$ is even. Now, after applying integration by
parts twice, we get

$\mathcal{L}_{c}(x)=\frac{1}{x^{2}}(A\cos(\alpha x)+g(x))\quad\text{ for all
}x\not=0,\ \text{where }A=\frac{4\alpha\kappa}{\pi(\alpha^{2}-c^{2})^{2}}$

and

$g(x)=-\frac{2}{\pi}\int_{\alpha}^{\infty}\frac{(3\xi^{2}+c^{2})\cos(x\xi)}{(\xi^{2}-c^{2})^{3}}d\xi-\frac{2(1-2\kappa)}{\pi}\int_{0}^{\alpha}\frac{(3(1-2\kappa)\xi^{2}+2-c^{2})\cos(x\xi)}{((1-2\kappa)\xi^{2}+2-c^{2})^{3}}d\xi.$

Observe that $g\in L^{\infty}(\mathbb{R})$. We are not interested in the value
$\mathcal{L}_{c}(0)$. We simply remark that, since $L_{c}\in
L^{1}(\mathbb{R})$, it follows that $\mathcal{L}_{c}\in
L^{\infty}(\mathbb{R})$ and therefore, $(1+|\cdot|^{2})\mathcal{L}_{c}\in
L^{\infty}(\mathbb{R}).$ Consequently, the decay in (1.22) follows by applying
Theorem 5.8 with $s=2$ and $p=\infty$. ∎

######  Acknowledgments.

The authors acknowledge support from the Labex CEMPI (ANR-11-LABX-0007-01). A.
de Laire was also supported by the ANR project ODA (ANR-18-CE40-0020-01). S.
López-Martínez was also supported by PGC2018-096422-B-I00 (MCIU/AEI/FEDER, UE)
and Junta de Andalucía FQM-116. S. López-Martínez would like to thank the
members of the Laboratoire Paul Painlevé (Université de Lille) and of the team
PARADYSE (Inria Lille - Nord Europe) for their support and hospitality during
his postdoc stay, where this work was carried out.

## Appendix

We include here the proof of the deformation lemma.

###### Proof of Lemma 3.3.

For $j=1,2,3$, let us denote

$A_{j}=J_{c}^{-1}([\gamma-2\varepsilon,\gamma+2\varepsilon])\cap
Z_{\delta_{j}}.$

Since these are closed sets in $H^{1}(\mathbb{R})$, we may define a functional
$\psi:H^{1}(\mathbb{R})\to\mathbb{R}$ of class $\mathcal{C}^{1}$ such that
$0\leq\psi(v)\leq 1$ for every $v\in H^{1}(\mathbb{R})$ and

$\psi(v)=\left\\{\begin{array}[]{l l}1&\text{ for all }v\in A_{2},\\\ 0&\text{
for all }v\in H^{1}(\mathbb{R})\setminus A_{1}.\end{array}\right.$

Let us now consider the vector field $\varphi:H^{1}(\mathbb{R})\to
H^{1}(\mathbb{R})$ given by

$\varphi(v)=\left\\{\begin{array}[]{l
l}\displaystyle-\psi(v)\frac{J^{\prime}_{c}(v)}{\|J^{\prime}_{c}(v)\|_{H^{-1}(\mathbb{R})}}&\text{
for all }v\in A_{1},\\\ 0&\text{ for all }v\in H^{1}(\mathbb{R})\setminus
A_{1}.\end{array}\right.$

Clearly, $\varphi\in\mathcal{C}^{1}(H^{1}(\mathbb{R}))$ (see Lemma 3.1) and
$\|\varphi(v)\|_{H^{1}(\mathbb{R})}\leq 1$ for every $v\in H^{1}(\mathbb{R})$.

For any $v\in H^{1}(\mathbb{R})$, we consider the Cauchy problem

$\begin{cases}w^{\prime}(t)=\varphi(w(t)),\quad\text{ for all }t\geq 0,\\\
w(0)=v.\end{cases}$

The classical ODE theory, the Cauchy problem has a unique solution
$w(\cdot,v)\in H^{1}(\mathbb{R})$ defined in $[0,+\infty)$. Let us show that
$w(t,\mathcal{NV}(\mathbb{R}))\subset\mathcal{NV}(\mathbb{R})$ for every
$t\geq 0$. Indeed, let $v\in\mathcal{NV}(\mathbb{R})$. Clearly,
$w(0,v)=v\in\mathcal{NV}(\mathbb{R})$. Moreover, since $w(\cdot,v)$ is
continuous and $v\in\mathcal{NV}(\mathbb{R})$, there exists $\tilde{t}>0$ such
that $w(t,v)\in\mathcal{NV}(\mathbb{R})$ for every $t\in[0,\tilde{t})$. Let us
assume by contradiction that
$s\coloneqq\sup\\{\tilde{t}>0:w(t,v)\in\mathcal{NV}(\mathbb{R})\,\,\forall
t\in[0,\tilde{t})\\}<+\infty$. Then,
$w(s,v)\in\partial\mathcal{NV}(\mathbb{R})$. In particular, $w(s,v)\in
H^{1}(\mathbb{R})\setminus A_{1}$, so $\varphi(w(s,v))=0$. Actually, since
$H^{1}(\mathbb{R})\setminus A_{1}$ is open, then there exists
$\tilde{s}\in(0,s)$ such that $w(t,v)\in H^{1}(\mathbb{R})\setminus A_{1}$ for
every $t\in[s-\tilde{s},s]$. Therefore, $\varphi(w(t,v))=0$ for every
$t\in[s-\tilde{s},s]$. That is, $w^{\prime}(t,v)=0$ for every
$t\in[s-\tilde{s},s]$, so $w$ must be constant in $[s-\tilde{s},s]$ and, in
consequence, $w(s-\tilde{s},v)=w(s,v)$. But this is a contradiction since, by
definition of $s$, it is necessary that
$w(s-\tilde{s},v)\in\mathcal{NV}(\mathbb{R})$.

On the other hand, for any $v\in H^{1}(\mathbb{R})$ and $t\geq 0$, we have

$\|w(t,v)-v\|_{L^{\infty}(\mathbb{R})}\leq\|w(t,v)-v\|_{H^{1}(\mathbb{R})}\leq\left\|\int_{0}^{t}\varphi(w(s,v))ds\right\|_{H^{1}(\mathbb{R})}\leq
t.$

Observe that, in the previous inequality, we have used that the norm of the
continuous embedding $H^{1}(\mathbb{R})\subset L^{\infty}(\mathbb{R})$ is
equal to one (see Remark 3.4). Hence, we deduce that
$w(t,Z_{\delta_{3}})\subset Z_{\delta_{2}}$ for every
$t\leq\delta_{3}-\delta_{2}$.

Let us define $h:[0,1]\times\mathcal{NV}(\mathbb{R})\to H^{1}(\mathbb{R})$ by

$h(t,v)=w\left((\delta_{3}-\delta_{2})t,v\right),\quad\text{ for all
}(t,v)\in[0,1]\times\mathcal{NV}(\mathbb{R}).$

We have already verified that
$h([0,1]\times\mathcal{NV}(\mathbb{R}))\subset\mathcal{NV}(\mathbb{R})$.
Furthermore, with this definition, items (i) and (iii) are obviously
satisfied. On the other hand, if $v\in\mathcal{NV}(\mathbb{R})\setminus
A_{1}$, then $\varphi(v)=0$, so $w(t)=v$ is the unique solution to the Cauchy
problem and item (ii) is satisfied too.

As far as item (iv) is concerned, let $v\in\mathcal{NV}(\mathbb{R})$. Since
$w(t,v)\in\mathcal{NV}(\mathbb{R})$ for every $t\geq 0$, then the function
$J_{c}(w(\cdot,v))$ is differentiable and

$\frac{d}{dt}J_{c}(w(t,v))=\langle
J^{\prime}_{c}(w(t,v)),w^{\prime}(t,v)\rangle=\langle
J^{\prime}_{c}(w(t,v)),\varphi(w(t,v))\rangle\leq 0.$

Thus, $J_{c}(w(\cdot,v))$ is nonincreasing and item (iv) holds true.

Lastly, we check item (v). Indeed, let $v\in J_{c}^{\gamma+\varepsilon}\cap
Z_{\delta_{3}}$. If there exists $t\in\left[0,\delta_{3}-\delta_{2}\right)$
such that $J_{c}(w(t,v))<\gamma-\varepsilon$, then item (iv) implies that
$J_{c}\left(w\left(\delta_{3}-\delta_{2},v\right)\right)<\gamma-\varepsilon$,
so $w\left(\delta_{3}-\delta_{2},v\right)\in J_{c}^{\gamma-\varepsilon}\cap
Z_{\delta_{2}}$. Otherwise, for every
$t\in\left[0,\delta_{3}-\delta_{2}\right)$, one has

$\gamma-\varepsilon\leq J_{c}(w(t,v))\leq
J_{c}(w(0,v))=J_{c}(v)\leq\gamma+\varepsilon.$

In particular, $w(t,v)\in A_{2}$ for every
$t\in\left[0,\delta_{3}-\delta_{2}\right)$. In addition, by the definition of
$\varphi$, we derive

$\displaystyle J_{c}\left(w\left(\delta_{3}-\delta_{2},v\right)\right)$
$\displaystyle=J_{c}(v)+\int_{0}^{\delta_{3}-\delta_{2}}\frac{d}{dt}J_{c}(w(t,v))dt$
$\displaystyle=J_{c}(v)+\int_{0}^{\delta_{3}-\delta_{2}}\langle
J_{c}^{\prime}(w(t,v)),\varphi(w(t,v))\rangle dt$
$\displaystyle=J_{c}(v)-\int_{0}^{\delta_{3}-\delta_{2}}\|J_{c}^{\prime}(t,v))\|_{H^{-1}(\mathbb{R})}dt$
$\displaystyle\leq\gamma+\varepsilon-(\delta_{3}-\delta_{2})\frac{2\varepsilon}{\delta_{3}-\delta_{2}}=\gamma-\varepsilon.$

This concludes the proof. ∎

We include also a technical lemma needed in the proof of Proposition 3.9. Even
though the proof of the lemma is elementary, it is not straightforward and a
sort of uniform continuity of the functional $J_{c}$ is required.

######  Lemma 6.3.

Let $c>0$ and fix $\delta_{c}\in(0,1)$, $R_{c}>0$ and $\gamma_{c}>0$. For
every $\alpha>0$ and $\delta\in(0,1)$, we consider the set

$X_{\alpha,\delta}=J_{c}^{-1}((\gamma_{c}-\alpha,\gamma_{c}+\alpha))\cap
Z_{\delta},$

where $Z_{\delta}$ is defined by (3.5), i.e.

$Z_{\delta}=\\{v\in\mathcal{NV}(\mathbb{R}):\,\,\|v\|_{H^{1}(\mathbb{R})}\leq
R_{c}+1-\delta,\,\,v\leq 1-\delta\text{ on }\mathbb{R}\\}.$

Let us also denote

$I_{\alpha,\delta}=\inf\\{\|J_{c}^{\prime}(v)\|_{H^{-1}(\mathbb{R})}:\,\,v\in
X_{\alpha,\delta}\\}.$

Assume that there exists $\alpha>0$ such that $I_{\alpha,\delta_{c}}>0$. Then
there exists $\bar{\delta}\in(0,\delta_{c})$ such that
$I_{\alpha,\bar{\delta}}>0$.

###### Proof.

Let us take $\delta\in(0,\delta_{c})$. Observe that $Z_{\delta_{c}}\subset
Z_{\delta}$. Recall that $J_{c}\in\mathcal{C}^{2}(\mathcal{NV}(\mathbb{R}))$,
see Lemma 3.1. It is simple to check from (3.2) that
$\|J^{\prime\prime}_{c}(v)\|\leq C$ for every $v\in Z_{\delta}$, where $C>0$
depends only on $R_{c}$ and $\delta$. Hence, since $Z_{\delta}$ is connected
and convex, the Mean Value theorem implies that $J_{c}^{\prime}$ is Lipschitz
in $Z_{\delta}$, with Lipschitz constant denoted by $l_{\delta}>0$. In
consequence, for every $\varepsilon>0$, if we take
$\beta=\varepsilon/l_{\delta}$, then the following holds for any $v,w\in
Z_{\delta}$ satisfying $\|v-w\|_{H^{1}(\mathbb{R})}<\beta$,

$\big{|}\|J^{\prime}_{c}(v)\|_{H^{-1}(\mathbb{R})}-\|J^{\prime}_{c}(w)\|_{H^{-1}(\mathbb{R})}\big{|}\leq\|J^{\prime}_{c}(v)-J^{\prime}_{c}(w)\|_{H^{-1}(\mathbb{R})}\leq
l_{\delta}\|v-w\|_{H^{1}(\mathbb{R})}<\varepsilon.$

In the previous inequality, we take $\varepsilon=I_{\alpha,\delta_{c}}/2$.
Therefore, if $v\in Z_{\delta}$, $w\in X_{\alpha,\delta_{c}}$ and
$\|v-w\|_{H^{1}(\mathbb{R})}\leq\beta$, then

$\|J_{c}^{\prime}(v)\|_{H^{-1}(\mathbb{R})}>\|J_{c}^{\prime}(w)\|_{H^{-1}(\mathbb{R})}-\frac{I_{\alpha,\delta_{c}}}{2}\geq\frac{I_{\alpha,\delta_{c}}}{2}>0.$

We have proved that

$\|J^{\prime}_{c}(v)\|_{H^{-1}(R)}\geq\frac{I_{\alpha,\delta_{c}}}{2}>0\quad\text{for
all }v\in\mathcal{X},$ (6.7)

where

$\mathcal{X}=\\{v\in
X_{\alpha,\delta}:\,\,\text{dist}(v,X_{\alpha,\delta_{c}})<\beta\\}.$

Notice that

$X_{\alpha,\delta_{c}}\subset\mathcal{X}\subset X_{\alpha,\delta}.$

Using (6.7), the proof of the lemma will be finished as soon as we show that
there exists $\bar{\delta}\in(\delta,\delta_{c})$ such that

$X_{\alpha,\delta_{c}}\subset X_{\alpha,\bar{\delta}}\subset\mathcal{X}\subset
X_{\alpha,\delta}.$

In order to do so, let $\bar{\delta}\in(\delta,\delta_{c})$ to be chosen
later, and let $v\in X_{\alpha,\bar{\delta}}$. For some $\lambda>0$ to be
chosen later too, we aim to prove that

$\lambda v\in X_{\alpha,\delta_{c}}\quad\text{ and }\quad\|v-\lambda
v\|_{H^{1}(\mathbb{R})}<\beta,$ (6.8)

which implies that $v\in\mathcal{X}$.

On the one hand, simple computations show that a sufficient condition for
$\lambda v\in Z_{\delta_{c}}$ and $\|v-\lambda v\|_{H^{1}(\mathbb{R})}<\beta$
is

$1-\frac{\beta}{R_{c}+1-\bar{\delta}}<\lambda<\frac{1-\delta_{c}}{1-\bar{\delta}}.$
(6.9)

Observe that $\lambda>0$ can be chosen so that (6.9) holds whenever
$\bar{\delta}$ is close enough to $\delta_{c}$.

On the other hand, it is left to prove that $J_{c}(\lambda
v)\in(\gamma_{c}-\alpha,\gamma_{c}+\alpha)$. Indeed, since $J_{c}$ is
uniformly continuous in $Z_{\delta}$ and
$J_{c}(v)\in(\gamma_{c}-\alpha,\gamma_{c}+\alpha)$, it follows that there
exists $\lambda_{0}\in(0,1)$, independent of $v$, such that $J_{c}(\lambda
v)\in(\gamma_{c}-\alpha,\gamma_{c}+\alpha)$ for every
$\lambda\in(\lambda_{0},1]$. Now we take $\bar{\delta}$ even closer to
$\delta_{c}$ so that $\lambda_{0}<\frac{1-\delta_{c}}{1-\bar{\delta}}$. Thus,
for every $\lambda\in(\lambda_{0},1)$ satisfying (6.9), we have that (6.8)
holds. The proof is finished. ∎

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