# Sobolev norms of $L^{2}-$ solutions to NLS

Roman V. Bessonov, Sergey A. Denisov  Roman Bessonov<EMAIL_ADDRESS>St.
Petersburg State University Universitetskaya nab. 7-9, 199034 St. Petersburg,
RUSSIA St. Petersburg Department of Steklov Mathematical Institute Russian
Academy of Sciences Fontanka 27, 191023 St.Petersburg, RUSSIA Sergey Denisov:
<EMAIL_ADDRESS>University of Wisconsin–Madison Department of Mathematics
480 Lincoln Dr., Madison, WI, 53706, USA

###### Abstract.

We apply inverse spectral theory to study Sobolev norms of solutions to the
nonlinear Schrödinger equation. For initial datum $q_{0}\in L^{2}(\mathbb{R})$
and $s\in[-1,0]$, we prove that there exists a conserved quantity which is
equivalent to $H^{s}(\mathbb{R})$-norm of the solution.

###### Key words and phrases:

Dirac operators, NLS, scattering, Sobolev norms

###### 2010 Mathematics Subject Classification:

35Q55

The work of RB in Sections 2 and 3 is supported by the Russian Science
Foundation grant 19-71-30002. The work of SD in the rest of the paper is
supported by the grant NSF DMS-2054465 and Van Vleck Professorship Research
Award. RB is a Young Russian Mathematics award winner and would like to thank
its sponsors and jury.

##  1\. Introduction

The classical defocusing nonlinear Schrödinger equation (NLS) [13, 23, 27] on
the real line has the form

$\begin{cases}i\frac{\partial q}{\partial
t}=-\frac{\partial^{2}q}{\partial\xi^{2}}+2|q|^{2}q,\\\
q\big{\rvert}_{t=0}=q_{0},\end{cases}\qquad\xi\in\mathbb{R},\quad
t\in\mathbb{R}.$ (1.1)

It is known that for sufficiently regular initial datum $q_{0}$ the unique
classical solution $q=q(\xi,t)$ exists globally in time. For example, if
$q_{0}$ lies in the Schwartz class $\mathcal{S}(\mathbb{R})$, then
$q(\cdot,t)\in\mathcal{S}(\mathbb{R})$ for all $t\in\mathbb{R}$. The long-time
asymptotics of $q$ is known [9, 26, 10]. For less regular initial datum
$q_{0}$, one can define the solution by an approximation argument (see, e.g.,
[25]):

######  Theorem 1.1.

Let $q_{0}\in L^{2}(\mathbb{R})$, and let $q_{0,n}\in\mathcal{S}(\mathbb{R})$
converge to $q_{0}$ in $L^{2}(\mathbb{R})$. Denote by $q_{n}(\xi,t)$ the
solution of (1.1) corresponding to $q_{0,n}$. We have

$\lim_{n\to+\infty}\|q_{n}(\cdot,t)-q(\cdot,t)\|_{L^{2}(\mathbb{R})}=0,\qquad
t\in\mathbb{R},$

for some function $q(\xi,t):\mathbb{R}^{2}\to\mathbb{R}$ that does not depend
on the choice of the sequence $q_{0,n}$.

The function $q$ in Theorem 1.1 is called the $L^{2}$–solution of (1.1)
corresponding to the initial datum $q_{0}\in L^{2}(\mathbb{R})$. It is clear
that such a solution is unique. The total energy of the solution is its
$L^{2}(\mathbb{R})$-norm and it is conserved in time:

$\|q(\cdot,t)\|_{L^{2}(\mathbb{R})}=\|q_{0}\|_{L^{2}(\mathbb{R})},\qquad
t\in\mathbb{R}\,.$

By Plancherel’s formula, it is equal to
$\|(\mathcal{F}q)(\cdot,t)\|_{L^{2}(\mathbb{R})}$ where $\mathcal{F}$ stands
for the Fourier transform. In this paper, we work with Sobolev spaces
$H^{s}(\mathbb{R})$, $s\in\mathbb{R}$. The $H^{s}(\mathbb{R})$-norm of a
function $f\in\mathcal{S}(\mathbb{R})$ is defined by

$\|f\|_{H^{s}(\mathbb{R})}=\left(\int_{\mathbb{R}}(1+|\eta|^{2})^{s}|(\mathcal{F}f)(\eta)|^{2}d\eta\right)^{\frac{1}{2}}\,.$
(1.2)

The space $H^{s}(\mathbb{R})$ is the completion of $\mathcal{S}(\mathbb{R})$
with respect to this norm. Equivalently, one can define it by

$H^{s}(\mathbb{R})=\\{f\in\mathcal{S}^{\prime}(\mathbb{R}):(1+|\eta|^{2})^{{\frac{s}{2}}}\mathcal{F}f\in
L^{2}(\mathbb{R})\\}\,,$

where $\mathcal{S}^{\prime}(\mathbb{R})$ is the space of tempered
distribution.

In contrast to the linear Schrödinger equation for which all Sobolev norms are
conserved, the solutions of NLS can exhibit inflation of Sobolev norm
$H^{s}(\mathbb{R})$ for $s\leqslant-\frac{1}{2}$ (see, e.g., [8, 17] for
details). Specifically, given an arbitrarily small positive $\varepsilon$ and
$s\leqslant-\frac{1}{2}$, there exists a solution $q$ to (1.1) that satisfies

$q_{0}\in\mathcal{S}(\mathbb{R}),\qquad\|q_{0}\|_{H^{s}(\mathbb{R})}\leqslant\varepsilon,\qquad\|q(\cdot,\varepsilon)\|_{H^{s}(\mathbb{R})}\geqslant\varepsilon^{-1}\,,$
(1.3)

see [8] for that construction. This result is related to the “high-to-low
frequency cascade”. It occurs when for initial datum
$q_{0}\in\mathcal{S}(\mathbb{R})$, a part of $L^{2}(\mathbb{R})$-norm of $q$,
when written on the Fourier side, moves from high to low frequencies as time
increases. The Sobolev norms with negative index $s$ can be used to capture
this phenomenon. Indeed, since $\|q(\cdot,t)\|_{L^{2}(\mathbb{R})}$ is time-
invariant and the weight $(1+\eta^{2})^{s}$ in (1.2) vanishes at infinity when
$s<0$, the transfer of $L^{2}$-norm from high to low values of frequency
$\eta$ makes the $H^{s}(\mathbb{R})$-norm grow.

For NLS, the inflation of $H^{s}(\mathbb{R})$-norm can not happen for
$s>-{\frac{1}{2}}$. In [18], Koch and Tataru discovered the set of conserved
quantities which agree with $H^{s}(\mathbb{R})$-norm up to a quadratic term
for a small value of $\|q_{0}\|_{H^{s}(\mathbb{R})}$ and $s>-\frac{1}{2}$. As
a corollary, they obtained the bounds on $\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}$
that are uniform in time:

$\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\leqslant
C(s)\left\\{\begin{matrix}\mathcal{R}+\mathcal{R}^{1+2s},&s>0,\\\
\mathcal{R}+\mathcal{R}^{\frac{1+4s}{1+2s}},&s\in(-\frac{1}{2},0)\,,\end{matrix}\right.\quad\mathcal{R}=\|q_{0}\|_{H^{s}(\mathbb{R})}\,.$
(1.4)

In [16], Killip, Vişan, and Zhang proved a similar estimate using a different
method. The estimates on the growth of $H^{s}(\mathbb{R})$-norms are related
to questions of well-posedness and ill-posedness of NLS in Sobolev classes
which have been extensively studied previously, see, e.g., [14, 18, 17, 5, 6,
7, 20].

In our paper, we use some recent results in the inverse spectral theory [2, 3,
1] to show that there are conserved quantities of NLS which agree with
$H^{s}(\mathbb{R})$-norm provided that $s\in[-1,0]$ and the value of
$\|q_{0}\|_{L^{2}(\mathbb{R})}$ is under control. We apply our analysis to
prove the following theorem.

######  Theorem 1.2.

Let $q_{0}\in L^{2}(\mathbb{R})$ and let $q=q(\xi,t)$ be the solution of (1.1)
corresponding to $q_{0}$. Then,

$C_{1}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{2s}\|q_{0}\|_{H^{s}(\mathbb{R})}\leqslant\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\leqslant
C_{2}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{-2s}\|q_{0}\|_{H^{s}(\mathbb{R})},$
(1.5)

where $\,t\in\mathbb{R}\,,\,s\in[-1,0]$, and $C_{1}$ and $C_{2}$ are two
positive absolute constants.

This result shows, in particular, that for a given function
$q_{0}:\|q_{0}\|_{L^{2}(\mathbb{R})}=1$ whose $L^{2}(\mathbb{R})$-norm is
concentrated on high frequencies, we will never see a significant part of
$L^{2}(\mathbb{R})$-norm of the solution $q$ moving to the low frequencies.
That limits the “high-to-low frequency cascade” we discussed above. The close
inspection of construction used in [8] shows that the function $q_{0}$ in
(1.3) has $H^{s}(\mathbb{R})$-norm smaller than $\varepsilon$ but its
$L^{2}(\mathbb{R})$-norm is large when $\varepsilon$ is small. Hence, the
bounds in Theorem 1.2 do not contradict the estimates in (1.3) when
$s\in[-1,-\frac{1}{2}]$. We do not know whether Theorem 1.2 holds for $s<-1$.

The main idea of the proof of Theorem 1.2 is based on the analysis of the
conserved quantity $a(z)$, $\operatorname{Im}z>0$, which is a coefficient in
the transition matrix for the Dirac equation with potential $q=q(\cdot,t)$. We
take $z=i$ and show that $\log|a(i)|$ is related to a certain quantity
$\widetilde{\mathcal{K}}_{Q}$ (see the Lemma 3.3 below) that characterizes
both size and oscillation of $q$. Using $\widetilde{\mathcal{K}}_{Q}$ in the
context of NLS is the main novelty of our work. We study
$\widetilde{\mathcal{K}}_{Q}$ and show that it is equivalent to
$H^{-1}(\mathbb{R})$ norm of $q$ with constants that depend on its
$L^{2}(\mathbb{R})$-norm. That gives the estimate (1.5) for $s=-1$ and the
intermediate range of $s\in(-1,0)$ is handled by interpolation. Our analysis
relies heavily on the recent results [2, 3, 1] that characterize Krein – de
Branges canonical systems and the Dirac operators whose spectral measures
belong to the Szegő class on the real line. We also establish the framework
that allows working with NLS in the context of well-studied Krein systems.

Notation

  * •

The symbol $I$ stands for $2\times 2$ identity matrix
$I=\left(\begin{smallmatrix}1&0\\\ 0&1\\\ \end{smallmatrix}\right)$ and symbol
$J$ stands for $J=\left(\begin{smallmatrix}0&-1\\\
1&0\end{smallmatrix}\right)$. Constant matrices $\sigma_{3}$, $\sigma_{\pm}$,
$\sigma$ are defined in (2.2).

  * •

For a measurable set $S\subset\mathbb{R}$, we say that $f\in
L^{1}_{\rm\rm{loc}}(S)$ is $f\in L^{1}(K)$ for every compact $K\subset S$.

  * •

The Fourier transform of a function $f$ is defined by

$(\mathcal{F}f)(\eta)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-i\eta
x}\,dx.$

  * •

The symbol $C$, unless we specify explicitly, denotes the absolute constant
which can change the value from formula to formula. If we write, e.g.,
$C(\alpha)$, this defines a positive function of parameter $\alpha$.

  * •

For two non-negative functions $f_{1}$ and $f_{2}$, we write $f_{1}\lesssim
f_{2}$ if there is an absolute constant $C$ such that $f_{1}\leqslant Cf_{2}$
for all values of the arguments of $f_{1}$ and $f_{2}$. We define $\gtrsim$
similarly and say that $f_{1}\sim f_{2}$ if $f_{1}\lesssim f_{2}$ and
$f_{2}\lesssim f_{1}$ simultaneously. If $|f_{3}|\lesssim f_{4}$, we will
write $f_{3}=O(f_{4})$.

  * •

Symbols $\\{e_{j}\\}$ are reserved for the standard basis in $\mathbb{C}^{2}$:
$e_{1}=\left(\begin{smallmatrix}1\\\ 0\\\ \end{smallmatrix}\right)$,
$e_{2}=\left(\begin{smallmatrix}0\\\ 1\\\ \end{smallmatrix}\right)$.

  * •

For matrix $A$, the symbol $\|A\|_{\rm HS}$ denotes its Hilbert-Schmidt norm:
$\|A\|_{\rm HS}=({\rm tr}(A^{*}A))^{\frac{1}{2}}$.

##  2\. Preliminaries

Our proof of Theorem 1.1 uses complete integrability of equation (1.1). In
that framework, (1.1) can be solved by using the method of inverse scattering
which we discuss next following [13].

###  2.1. The inverse scattering approach to NLS

Given a complex-valued function $q\in\mathcal{S}(\mathbb{R})$, define the
differential operator

$L_{q}=i\sigma_{3}\frac{d}{d\xi}+i(q\sigma_{-}-\overline{q}\sigma_{+}),$ (2.1)

where we borrow notation for constant matrices $\sigma_{3}$, $\sigma_{\pm}$
from [13]:

$\sigma_{3}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix},\qquad\sigma_{+}=\begin{pmatrix}0&1\\\
0&0\end{pmatrix},\qquad\sigma_{-}=\begin{pmatrix}0&0\\\
1&0\end{pmatrix},\qquad\sigma=\sigma_{-}+\sigma_{+}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix}.$ (2.2)

The expression $L_{q}$ is one of the forms in which the Dirac operator can be
written. In Section 3, we will introduce another form and will show how the
two are related. Let us also define

$E(\xi,\lambda)=e^{\frac{\lambda}{2i}\xi\sigma_{3}}=\begin{pmatrix}e^{\frac{\lambda}{2i}\xi}&0\\\
0&e^{-\frac{\lambda}{2i}\xi}\end{pmatrix},$

as in [13]. In the free case when $q=0$, the matrix-function $E$ solves
$L_{0}E={\frac{\lambda}{2}}E,\,E(0,\lambda)=I$. Since
$q\in\mathcal{S}(\mathbb{R})$, it decays at infinity fast and therefore one
can find two solutions $T_{\pm}=T_{\pm}(\xi,\lambda)$ such that

$L_{q}T_{\pm}=\frac{\lambda}{2}T_{\pm},\qquad
T_{\pm}=E(\xi,\lambda)+o(1),\qquad\xi\to\pm\infty,$ (2.3)

for every $\lambda\in\mathbb{R}$. These solutions are called the Jost
solutions for $L_{q}$. Since both $T_{+}$ and $T_{-}$ solve the same ODE, they
must satisfy

$T_{-}(\xi,\lambda)=T_{+}(\xi,\lambda)T(\lambda),\qquad\xi\in\mathbb{R},\qquad\lambda\in\mathbb{R},$
(2.4)

where the matrix $T=T(\lambda)$ does not depend on $\xi\in\mathbb{R}$. One can
show that it has the form

$T(\lambda)=\begin{pmatrix}a(\lambda)&\overline{b(\lambda)}\\\
b(\lambda)&\overline{a(\lambda)}\end{pmatrix},\qquad\det T=|a|^{2}-|b|^{2}=1.$
(2.5)

The matrix $T$ is called the reduced transition matrix for $L_{q}$, and the
ratio ${\bf r}_{q}=b/a$ is called the reflection coefficient for $L_{q}$. One
can obtain $T$ in a different way: let $Z_{q}=Z_{q}(\xi,\lambda)$,
$\xi\in\mathbb{R}$, $\lambda\in\mathbb{C}$ be the fundamental matrix for
$L_{q}$, that is,

$L_{q}Z_{q}=\frac{\lambda}{2}Z_{q},\qquad
Z_{q}(0,\lambda)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\
\end{smallmatrix}\right).$ (2.6)

Then, we have $Z_{q}(\xi,\lambda)=T_{\pm}(\xi,\lambda)T^{-1}_{\pm}(0,\lambda)$
and the pointwise limits

$T^{-1}_{\pm}(0,\lambda)=\lim_{\xi\to\pm\infty}E^{-1}(\xi,\lambda)Z_{q}(\xi,\lambda)$
(2.7)

exist for every $\lambda\in\mathbb{R}$. Moreover, we have
$T(\lambda)=T_{+}^{-1}(0,\lambda)T_{-}(0,\lambda)$ on $\mathbb{R}$.

The coefficients $a,b$, and ${\bf r}_{q}$ were defined for
$\lambda\in\mathbb{R}$ and they satisfy $|a|^{2}=1+|b|^{2}$, $1-|{\bf
r}_{q}|^{2}=|a|^{-2}$ for these $\lambda$. However, one can show that
$a(\lambda)$ is the boundary value of the outer function defined in
${\mathbb{C}}_{+}=\\{z\in\mathbb{C}:\;\operatorname{Im}z>0\\}$ by the formula
(see (6.22) in [13])

$a(z)=\exp\left(\frac{1}{\pi
i}\int_{\mathbb{R}}\frac{1}{\lambda-z}\log|a(\lambda)|\,d\lambda\right),\quad
z\in\mathbb{C}_{+},$

which, in view of identity $1-|{\bf r}_{q}|^{2}=|a|^{-2}$ on $\mathbb{R}$, can
be written as

$a(z)=\exp\left(-\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{1}{\lambda-z}\log(1-|{\bf
r}_{q}(\lambda)|^{2})d\lambda\right)\,.$ (2.8)

That shows, in particular, that $b$ defines both $a$ and ${\bf r}_{q}$, and
${\bf r}_{q}$ defines $a$ and $b$.

The map $q\mapsto{\bf r}_{q}$ is called the direct scattering transform and
its inverse is called the inverse scattering transform. These maps are well-
studied when $q\in\mathcal{S}(\mathbb{R})$. In particular, we have the
following result (see [13] for the proof).

######  Theorem 2.1.

The map $q\mapsto{\bf r}_{q}$ is a bijection from $\mathcal{S}(\mathbb{R})$
onto the set of complex-valued functions $\\{{\bf
r}\in\mathcal{S}(\mathbb{R}),\|{\bf r}\|_{L^{\infty}(\mathbb{R})}<1\\}$.

The scattering transform has some symmetries:

######  Lemma 2.1.

If $q\in\mathcal{S}(\mathbb{R})$ and $\lambda\in\mathbb{R}$, then

(dilation): $\displaystyle\qquad{\bf r}_{\alpha q(\alpha\xi)}(\lambda)={\bf
r}_{q(\xi)}(\alpha^{-1}\lambda),\quad\alpha>0\,,$ (conjugation):
$\displaystyle\qquad{\bf r}_{\overline{q}(\xi)}(\lambda)={\overline{{\bf
r}_{q(\xi)}(-\lambda)}}\,,$ (translation): $\displaystyle\qquad{\bf
r}_{q(\xi-\ell)}(\lambda)={\bf
r}_{q(\xi)}(\lambda)e^{-i\lambda\ell},\quad\ell\in\mathbb{R}\,,$ (modulation):
$\displaystyle\qquad{\bf r}_{e^{-i\beta\xi}q(\xi)}(\lambda)={\bf
r}_{q(\xi)}(\lambda+\beta),\quad\beta\in\mathbb{R}\,.$ (rotation):
$\displaystyle\qquad{\bf r}_{\mu q(\xi)}(\lambda)={\mu}{\bf
r}_{q(\xi)}(\lambda),\quad\mu\in\mathbb{C},\;|\mu|=1\,.$

Proof. Indeed, the direct substitution into (2.3) shows that if
$T_{\pm}(\xi,\lambda)$ are Jost solutions for $q(\xi)$, then

  * $(a)$

$T_{\pm}(\alpha\xi,\alpha^{-1}\lambda)$ are the Jost solutions for $\alpha
q(\alpha\xi)$,

  * $(b)$

$\overline{T}_{\pm}(\xi,{-\lambda})$ are the Jost solutions for
$\overline{q(\xi)}$,

  * $(c)$

$T_{\pm}(\xi-\ell,{\lambda})E(\ell,\lambda)$ are the Jost solutions for
$q(\xi-\ell)$,

  * $(d)$

$E(-\xi,\beta)T_{\pm}(\xi,\lambda+\beta)$ are the Jost solutions for
$e^{-i\beta\xi}q(\xi)$,

  * $(e)$

$\left(\begin{smallmatrix}1&0\\\
0&\mu\end{smallmatrix}\right)T_{\pm}(\xi,\lambda)\left(\begin{smallmatrix}1&0\\\
0&\overline{\mu}\end{smallmatrix}\right)$ are the Jost solutions for $\mu
q(\xi)$, $|\mu|=1$.

Now, it is left to use the formula (2.4) which defines $T$. A computation
using (2.5) shows how $a$ and $b$ change under symmetries $(a)$–$(e)$. For
example, the translation does not change $a$ and it multiplies $b$ by
$e^{-i\lambda l}$. The modulation $e^{-i\beta\xi}q(\xi)$, however, gives
$a_{e^{-i\beta\xi}q(\xi)}(\lambda)=a_{q(\xi)}(\lambda+\beta)$. Then, the claim
follows from the definition of the reflection coefficient ${\bf r}_{q}=b/a$. ∎

The next result (see formula $(7.5)$ in [13]), along with the previous
theorem, shows how the inverse scattering transform can be used to solve
(1.1).

######  Theorem 2.2.

Let $q_{0}\in\mathcal{S}(\mathbb{R})$ and let ${\bf r}_{q_{0}}={\bf
r}_{q_{0}}(\lambda)$ be the reflection coefficient of $L_{q_{0}}$. Define the
family

${\bf r}(\lambda,t)=e^{-i\lambda^{2}t}{\bf
r}_{q_{0}}(\lambda),\qquad\lambda\in\mathbb{R},\quad t\in\mathbb{R}.$ (2.9)

For each $t\in\mathbb{R}$, let $q=q(\xi,t)$ be the potential in the previous
theorem generated by ${\bf r}(\lambda,t)$. Then, $q=q(\xi,t)$ is the unique
classical solution of (1.1) with the initial datum $q_{0}$. Moreover, for
every $t\in\mathbb{R}$, the function $\xi\mapsto q(\xi,t)$ lies in
$\mathcal{S}(\mathbb{R})$.

The solutions to NLS equation

$i\frac{\partial q}{\partial
t}=-\frac{\partial^{2}q}{\partial\xi^{2}}+2|q|^{2}q$ (2.10)

behave in an explicit way under some transformations. Specifically, we have

  * $(a)$

Dilation: if $q(\xi,t)$ solves (2.10), then $\alpha q(\alpha\xi,\alpha^{2}t)$
solves (2.10) for every $\alpha\neq 0$.

  * $(b)$

Time reversal: if $q(\xi,t)$ solves (2.10), then $\overline{q}(\xi,-t)$ solves
(2.10). In particular, if $q_{0}$ is real-valued, then $q(\xi,t)=q(\xi,-t)$.

  * $(c)$

Translation: if $q(\xi,t)$ solves (2.10), then $q(\xi-\ell,t)$ solves (2.10)
for every $\ell\in\mathbb{R}$.

  * $(d)$

Modulation or Galilean symmetry: if $q(\xi,t)$ solves (2.10), then $e^{iv\xi-
iv^{2}t}q(\xi-2vt,t)$ solves (2.10) for every $v\in\mathbb{R}$.

  * $(e)$

Rotation: if $q(\xi,t)$ solves (2.10), then $\mu q(\xi,t)$ solves (2.10) for
every $\mu\in\mathbb{C},|\mu|=1$.

These properties can be checked by direct calculation (see, e.g., formula
(1.19) in [14] for $(d)$) and a simple inspection shows that the bound (1.5)
is consistent with all these transformations. The statements of Theorem 2.2
and Lemma 2.1 are consistent with these symmetries as well.

Now, we can explain the idea behind the proof of the Theorem 1.2.

The idea of the proof for Theorem 1.2. One can proceed as follows. First, we
assume that $q_{0}\in\mathcal{S}(\mathbb{R})$ and notice that conservation of
$|{\bf r}(\lambda,t)|$, $\lambda\in\mathbb{R}$, guaranteed by (2.9), yields
that $\log|a(i,t)|$ is conserved, where $a(z,t)$ is defined for
$z\in\mathbb{C}_{+}$ by (2.8). Separately, for every Dirac operator $L_{q}$
with $q\in L^{2}(\mathbb{R})$, we show that $\log|a(i)|$ is equivalent to some
explicit quantity $\widetilde{\mathcal{K}}_{Q}$ that involves $q$. That
quantity was introduced and studied in [2, 3, 1]: it resembles the matrix
Muckenhoupt $A_{2}(\mathbb{R})$ condition and it is equivalent to
$H^{-1}(\mathbb{R})$ norm of $q$ provided that $\|q\|_{L^{2}(\mathbb{R})}$ is
under control, e.g., $\|q\|_{L^{2}(\mathbb{R})}<C$ with some fixed $C$.
Putting things together, we see that Sobolev $H^{-1}(\mathbb{R})$ norm of
$q(\cdot,t)$ does not change much in time provided that the bound
$\|q(\cdot,t)\|_{L^{2}(\mathbb{R})}<C$ holds. Since
$\|q(\cdot,t)\|_{L^{2}(\mathbb{R})}=\|q_{0}\|_{L^{2}(\mathbb{R})}$ is time-
invariant, we arrive to the statement of Theorem 1.2 for
$q_{0}\in\mathcal{S}(\mathbb{R})$ and $s=-1$. For $s=0$, the claim of Theorem
1.2 is trivial. The intermediate range of $s\in(-1,0)$ is handled by
interpolation using Galilean invariance of NLS. The general case when
$q_{0}\in L^{2}(\mathbb{R})$ follows by a density argument if one uses the
stability of $L^{2}$-solutions guaranteed by Theorem 1.1.

There are other methods that use conserved quantities that agree with negative
Sobolev norms. The paper [16] uses a representation of $\log|a(i)|$ through a
perturbation determinant. Then, the analysis of the perturbation series allows
the authors of [16] to obtain estimates similar to (1.4). It is conceivable
that this approach can provide results along the same lines as Theorem 1.2.

To focus on the Dirac operator with $q\in L^{2}(\mathbb{R})$, we first
consider this operator on half-line $\mathbb{R}_{+}$ in connection to Krein
systems that were introduced in [19].

###  2.2. Operator $L_{q}$ and Krein system

Let $A:\mathbb{R}_{+}\to\mathbb{C}$ be a function on the positive half-line
$\mathbb{R}_{+}=[0,+\infty)$ such that

$\int_{0}^{r}|A(\xi)|\,d\xi<\infty,$

for every $r\geqslant 0$. Recall that we denote the set of such functions by
$L^{1}_{\rm{loc}}(\mathbb{R}_{+})$. The Krein system (see the formula $(4.52)$
in [12]) with the coefficient $A$ has the form

$\begin{cases}P^{\prime}(\xi,\lambda)=i\lambda
P(\xi,\lambda)-\overline{A(\xi)}P_{*}(\xi,\lambda),&P(0,\lambda)=1,\\\
P^{\prime}_{*}(\xi,\lambda)=-A(\xi)P(\xi,\lambda),&P_{*}(0,\lambda)=1,\end{cases}$
(2.11)

where the derivative is taken with respect to $\xi\in\mathbb{R}_{+}$ and
$\lambda\in\mathbb{C}$. Let also

$\begin{cases}\widehat{P}^{\prime}(\xi,\lambda)=i\lambda\widehat{P}(\xi,\lambda)+\overline{A(\xi)}\widehat{P}_{*}(\xi,\lambda),&\widehat{P}(0,\lambda)=1,\\\
\widehat{P}^{\prime}_{*}(\xi,\lambda)=A(\xi)\widehat{P}(\xi,\lambda),&\widehat{P}_{*}(0,\lambda)=1,\end{cases}$
(2.12)

denote the so-called dual Krein system (see Corollary 5.7 in [12]). Set

$Y(\xi,\lambda)=e^{-i\lambda\xi}\begin{pmatrix}P(2\xi,\lambda)&i\widehat{P}(2\xi,\lambda)\\\
P_{*}(2\xi,\lambda)&-i\widehat{P}_{*}(2\xi,\lambda)\end{pmatrix}.$ (2.13)

The matrix-function $Z_{q}$, which was defined in (2.6) for
$q\in\mathcal{S}(\mathbb{R})$, makes sense if we assume that $q\in L^{1}_{\rm
loc}(\mathbb{R})$. In the next lemma, we relate $Y$ to $Z_{q}$.

######  Lemma 2.2.

Let $q\in L^{1}_{\rm{loc}}(\mathbb{R})$, $A(2\xi)=-\overline{q(\xi)}/2$ on
$\mathbb{R}_{+}$, and $Y$ be the corresponding matrix-valued function defined
by (2.13). Then, $Z_{q}(\xi,2\lambda)=\sigma
Y(\xi,\lambda)Y^{-1}(0,\lambda)\sigma$ for $\xi\geqslant 0$ and
$\lambda\in\mathbb{C}$.

Proof. The proof is a computation. We have

$\displaystyle L_{\bar{q}}Y(\xi,\lambda)=$
$\displaystyle\lambda\sigma_{3}Y(\xi,\lambda)+i\sigma_{3}e^{-i\lambda\xi}\frac{d}{d\xi}\begin{pmatrix}P(2\xi,\lambda)&i\widehat{P}(2\xi,\lambda)\\\
P_{*}(2\xi,\lambda)&-i\widehat{P}_{*}(2\xi,\lambda)\end{pmatrix}+i(\bar{q}\sigma_{-}-q\sigma_{+})Y(\xi,\lambda),$
$\displaystyle=$ $\displaystyle
2i\sigma_{3}e^{-i\lambda\xi}\begin{pmatrix}i\lambda
P(2\xi,\lambda)-\overline{A(2\xi)}P_{*}(2\xi,\lambda)&-\lambda\widehat{P}(2\xi,\lambda)+i\overline{A(2\xi)}\widehat{P}_{*}(2\xi,\lambda)\\\
-A(2\xi)P(2\xi,\lambda)&-iA(2\xi)\widehat{P}(2\xi,\lambda)\end{pmatrix}$
$\displaystyle+i(\bar{q}\sigma_{-}-q\sigma_{+}-i\lambda\sigma_{3})Y(\xi,\lambda).$

Notice, that

$\displaystyle
i(\bar{q}\sigma_{-}-q\sigma_{+}-i\lambda\sigma_{3})Y(\xi,\lambda)$
$\displaystyle=ie^{-i\lambda\xi}\begin{pmatrix}-i\lambda&-q\\\
\bar{q}&i\lambda\end{pmatrix}\begin{pmatrix}P(2\xi,\lambda)&i\widehat{P}(2\xi,\lambda)\\\
P_{*}(2\xi,\lambda)&-i\widehat{P}_{*}(2\xi,\lambda)\end{pmatrix}$
$\displaystyle=ie^{-i\lambda\xi}\begin{pmatrix}-i\lambda
P(2\xi,\lambda)-qP_{*}(2\xi,\lambda)&\lambda\widehat{P}(2\xi,\lambda)+iq\widehat{P}_{*}(2\xi,\lambda)\\\
\bar{q}P(2\xi,\lambda)+i\lambda
P_{*}(2\xi,\lambda)&i\bar{q}\widehat{P}(2\xi,\lambda)+\lambda\widehat{P}_{*}(2\xi,\lambda)\end{pmatrix}\,.$

Using relation $2A(2\xi)+\bar{q}(\xi)=0$, we obtain

$L_{\bar{q}}Y(\xi,\lambda)=ie^{-i\lambda\xi}\begin{pmatrix}i\lambda
P(2\xi,\lambda)&-\lambda\widehat{P}(2\xi,\lambda)\\\ i\lambda
P_{*}(2\xi,\lambda)&\lambda\widehat{P}_{*}(2\xi,\lambda)\end{pmatrix}=-\lambda
Y(\xi,\lambda).$

Since $\sigma\sigma_{3}\sigma=-\sigma_{3}$ and
$\sigma\sigma_{\pm}\sigma=\sigma_{\mp}$, one has $\sigma
L_{\bar{q}}\sigma=-L_{q}$. Therefore,

$L_{q}(\sigma Y(\xi,\lambda)\sigma)=\lambda(\sigma Y(\xi,\lambda)\sigma).$

It follows that matrix-valued functions $Z_{q}(\xi,2\lambda)$ and $\sigma
Y(\xi,\lambda)Y^{-1}(0,\lambda)\sigma$ solve the same Cauchy problem and thus
$Z_{q}(\xi,2\lambda)=\sigma Y(\xi,\lambda)Y^{-1}(0,\lambda)\sigma$, as
required. ∎

######  Lemma 2.3.

Let $q\in L^{1}_{\rm{loc}}(\mathbb{R})$, let $A(2\xi)=q(-\xi)/2$ on
$\mathbb{R}_{+}$, and let $Y$ be the corresponding matrix-valued function
defined by (2.13). Then,
$Z_{q}(-\xi,2\lambda)=Y(\xi,\lambda)Y^{-1}(0,\lambda)$ for $\xi\geqslant 0$
and $\lambda\in\mathbb{C}$.

Proof. Recall that matrices $\sigma_{3}$, $\sigma_{\pm}$, $\sigma$ are defined
in (2.2). Using relations $\sigma\sigma_{3}\sigma=-\sigma_{3}$ and
$\sigma\sigma_{\pm}\sigma=\sigma_{\mp}$, we see that
$L_{\tilde{q}}\widetilde{Z}_{q}=\frac{\lambda}{2}\widetilde{Z}_{q}$, where
$\tilde{q}(\xi)=-\overline{q(-\xi)}$ and
$\widetilde{Z}_{q}(\xi,\lambda)=\sigma Z_{q}(-\xi,\lambda)\sigma$. Then,
previous lemma applies to $\tilde{q}$,
$Z_{\tilde{q}}(\xi,2\lambda)=\widetilde{Z}_{q}(\xi,2\lambda)$ and
$A(2\xi)=-\overline{\tilde{q}(\xi)}/2=q(-\xi)/2$. It gives
$\widetilde{Z}_{q}(\xi,2\lambda)=\sigma
Y(\xi,\lambda)Y^{-1}(0,\lambda)\sigma$. Returning to $Z_{q}$, we get
$Z_{q}(-\xi,2\lambda)=Y(\xi,\lambda)Y^{-1}(0,\lambda)$. ∎

Given $q\in L^{2}(\mathbb{R})$, we define the continuous analogs of Wall
polynomials (see [15] and Section 7 in [12]) by

$\mathfrak{A}^{\pm}=\frac{P^{\pm}_{*}+\widehat{P}^{\pm}_{*}}{2},\qquad\mathfrak{A}^{\pm}_{*}=\frac{P^{\pm}+\widehat{P}^{\pm}}{2},\qquad\mathfrak{B}^{\pm}=\frac{P^{\pm}_{*}-\widehat{P}^{\pm}_{*}}{2},\qquad\mathfrak{B}^{\pm}_{*}=\frac{P^{\pm}-\widehat{P}^{\pm}}{2},$
(2.14)

where $P^{\pm}$, $P^{\pm}_{*}$, $\widehat{P}^{\pm}$, $\widehat{P}^{\pm}_{*}$
are the solutions of systems (2.11), (2.12) for the coefficient
$A^{+}(\xi)=-\overline{q(\xi/2)}/2$ from Lemma 2.2 and the coefficient
$A^{-}(\xi)=q(-\xi/2)/2$ from Lemma 2.3, correspondingly. Functions $P^{\pm}$,
$P^{\pm}_{*}$, $\widehat{P}^{\pm}$, $\widehat{P}^{\pm}_{*}$ are continuous
analogs of polynomials orthogonal on the unit circle, they depend on two
parameters: $\xi\in\mathbb{R}_{+}$ and $\lambda\in\mathbb{C}$ and they satisfy
identities (see formula (4.32) in [12]):

$P_{*}^{\pm}(\xi,\lambda)=e^{i\xi\lambda}\overline{P^{\pm}(\xi,\lambda)},\qquad\widehat{P}_{*}^{\pm}(\xi,\lambda)=e^{i\xi\lambda}\overline{\widehat{P}^{\pm}(\xi,\lambda)}$
(2.15)

for real $\lambda$.

We will use the following result (see Lemma 2 in [11] which contains a
stronger statement).

######  Theorem 2.3.

Let $A\in L^{2}(\mathbb{R}_{+})$, and let $P$, $P_{*}$ be the solutions of
system (2.11) for the coefficient $A$. Then, the limit

$\Pi(\lambda)=\lim_{\xi\to+\infty}P_{*}(\xi,\lambda)$ (2.16)

exists for every $\lambda\in\mathbb{C}_{+}$. That function $\Pi$ is outer in
$\mathbb{C}_{+}$. If $\lambda\in\mathbb{R}$, the convergence in (2.16) holds
in the Lebesgue measure on $\mathbb{R}$ where $\Pi(\lambda)$ now denotes the
non-tangential boundary value of $\Pi$.

That theorem allows us to define

$\mathfrak{a}^{\pm}(\lambda)=\lim_{\xi\to+\infty}\mathfrak{A}^{\pm}(\xi,\lambda),\qquad\mathfrak{b}^{\pm}(\lambda)=\lim_{\xi\to+\infty}\mathfrak{B}^{\pm}(\xi,\lambda)$
(2.17)

for every $\lambda\in\mathbb{C}_{+}$ and for almost every
$\lambda\in\mathbb{R}$. Moreover, Corollary 12.2 in [12] gives

$|\mathfrak{a}^{\pm}(\lambda)|^{2}=1+|\mathfrak{b}^{\pm}(\lambda)|^{2}$ (2.18)

for a.e. $\lambda\in\mathbb{R}$. For every $\lambda\in\mathbb{C}_{+}$, we
define

$a(\lambda)=\mathfrak{a}^{+}(\lambda)\mathfrak{a}^{-}(\lambda)-\mathfrak{b}^{+}(\lambda)\mathfrak{b}^{-}(\lambda)\,.$

######  Proposition 2.1.

The function $a$ is outer in $\mathbb{C}_{+}$.

Proof. We can write

$a=\mathfrak{a}^{+}\mathfrak{a}^{-}(1-s^{+}s^{-}),\quad
s^{\pm}=\frac{\mathfrak{b}^{\pm}}{\mathfrak{a}^{\pm}}\,.$

It is known that $\mathfrak{a}^{\pm}$ are outer (see the formulas (12.9) and
(12.29) in [12]) and that $s^{\pm}$ satisfy $|s^{\pm}|<1$ in $\mathbb{C}_{+}$.
The function $1-s^{+}s^{-}$ has a positive real part in $\mathbb{C}_{+}$ and
so is an outer function. That shows that $a$ is a product of three outer
functions and hence it is outer itself. ∎

######  Proposition 2.2.

Let $q\in L^{2}(\mathbb{R})$ and let $Z_{q}$ be defined by (2.6). Then, the
limits in (2.7) exist in the Lebesgue measure on $\mathbb{R}$. The matrix
$T(\lambda)=T_{+}^{-1}(0,\lambda)T_{-}(0,\lambda)$ has the form (2.5) where

$a=\mathfrak{a}^{+}\mathfrak{a}^{-}-\mathfrak{b}^{+}\mathfrak{b}^{-},\qquad
b=\mathfrak{a}^{-}\overline{\mathfrak{b}^{+}}-\mathfrak{b}^{-}\overline{\mathfrak{a}^{+}}\,,$
(2.19)

and $\mathfrak{a}^{\pm}$, $\mathfrak{b}^{\pm}$ are defined Lebesgue almost
everywhere on $\mathbb{R}$ by the convergence in (2.17) in measure.

Proof. If $q\in L^{2}(\mathbb{R})$, the fundamental matrix $Z_{q}$ and the
continuous Wall polynomials (2.14) are related by the formula

$\displaystyle
Z_{q}(\xi,2\lambda)=\begin{cases}e^{-i\lambda\xi}\begin{pmatrix}\mathfrak{A}^{+}(2\xi,\lambda)&\mathfrak{B}^{+}(2\xi,\lambda)\\\
\mathfrak{B}^{+}_{*}(2\xi,\lambda)&\mathfrak{A}^{+}_{*}(2\xi,\lambda)\end{pmatrix},&\xi\geqslant
0,\\\
e^{i\lambda\xi}\begin{pmatrix}\mathfrak{A}^{-}_{*}(-2\xi,\lambda)&\mathfrak{B}^{-}_{*}(-2\xi,\lambda)\\\
\mathfrak{B}^{-}(-2\xi,\lambda)&\mathfrak{A}^{-}(-2\xi,\lambda)\end{pmatrix},&\xi<0.\end{cases}$
(2.20)

Indeed, it is enough to use Lemma 2.2, Lemma 2.3, and the fact that
$Y^{-1}(0,\lambda)=\frac{1}{2}\left(\begin{smallmatrix}1&1\\\
-i&i\end{smallmatrix}\right)$. Our next step is to prove that the limit

$T_{+}^{-1}(0,2\lambda)=\lim_{\xi\to+\infty}E^{-1}(\xi,2\lambda)Z_{q}(\xi,2\lambda)$
(2.21)

exists in Lebesgue measure when $\lambda\in\mathbb{R}$. From (2.15), we obtain

$\displaystyle E^{-1}(\xi,2\lambda)Z_{q}(\xi,2\lambda)=\begin{pmatrix}1&0\\\
0&e^{-2i\lambda\xi}\end{pmatrix}\begin{pmatrix}\mathfrak{A}^{+}(2\xi,\lambda)&\mathfrak{B}^{+}(2\xi,\lambda)\\\
\mathfrak{B}^{+}_{*}(2\xi,\lambda)&\mathfrak{A}^{+}_{*}(2\xi,\lambda)\end{pmatrix}=\begin{pmatrix}\mathfrak{A}^{+}(2\xi,\lambda)&\mathfrak{B}^{+}(2\xi,\lambda)\\\
\overline{\mathfrak{B}^{+}}(2\xi,\lambda)&\overline{\mathfrak{A}^{+}}(2\xi,\lambda)\end{pmatrix},$

for every $\xi\geqslant 0$ and $\lambda\in\mathbb{R}$. Similarly,

$\displaystyle
E^{-1}(-\xi,2\lambda)Z_{q}(-\xi,2\lambda)=\begin{pmatrix}e^{-2i\lambda\xi}&0\\\
0&1\end{pmatrix}\begin{pmatrix}\mathfrak{A}^{-}_{*}(2\xi,\lambda)&\mathfrak{B}^{-}_{*}(2\xi,\lambda)\\\
\mathfrak{B}^{-}(2\xi,\lambda)&\mathfrak{A}^{-}(2\xi,\lambda)\end{pmatrix}=\begin{pmatrix}\overline{\mathfrak{A}^{-}}(2\xi,\lambda)&\overline{\mathfrak{B}^{-}}(2\xi,\lambda)\\\
\mathfrak{B}^{-}(2\xi,\lambda)&\mathfrak{A}^{-}(2\xi,\lambda)\end{pmatrix}.$

Hence, the limits

$T^{-1}_{\pm}(0,2\lambda)=\lim_{\xi\to\pm\infty}E^{-1}(\xi,2\lambda)Z_{q}(\xi,2\lambda)$
(2.22)

exist in Lebesgue measure on $\mathbb{R}$ by Theorem 2.3. Moreover,

$\displaystyle T(2\lambda)$
$\displaystyle=T_{+}^{-1}(0,2\lambda)T_{-}(0,2\lambda)=\begin{pmatrix}\mathfrak{a}^{+}(\lambda)&\mathfrak{b}^{+}(\lambda)\\\
\overline{\mathfrak{b}^{+}(\lambda)}&\overline{\mathfrak{a}^{+}(\lambda)}\end{pmatrix}\begin{pmatrix}\overline{\mathfrak{a}^{-}(\lambda)}&\overline{\mathfrak{b}^{-}(\lambda)}\\\
\mathfrak{b}^{-}(\lambda)&\mathfrak{a}^{-}(\lambda)\end{pmatrix}^{-1}$
$\displaystyle\stackrel{{\scriptstyle\eqref{sa5}}}{{=}}\begin{pmatrix}\mathfrak{a}^{+}(\lambda)&\mathfrak{b}^{+}(\lambda)\\\
\overline{\mathfrak{b}^{+}(\lambda)}&\overline{\mathfrak{a}^{+}(\lambda)}\end{pmatrix}\begin{pmatrix}\mathfrak{a}^{-}(\lambda)&-\overline{\mathfrak{b}^{-}(\lambda)}\\\
-\mathfrak{b}^{-}(\lambda)&\overline{\mathfrak{a}^{-}(\lambda)}\end{pmatrix}=\begin{pmatrix}a(\lambda)&\overline{b(\lambda)}\\\
b(\lambda)&\overline{a(\lambda)}\end{pmatrix}$

and the proposition follows. ∎

We end this section with a few remarks on reflection coefficients of
potentials in $L^{2}(\mathbb{R})$. We have $|a|^{2}-|b|^{2}=1$ almost
everywhere on $\mathbb{R}$ due to the fact that $\det T_{\pm}(0,\lambda)=1$
almost everywhere on $\mathbb{R}$. That can also be established directly using
(2.18). Proposition 2.2 then allows to define the reflection coefficient ${\bf
r}_{q}=b/a$ for every $q\in L^{2}(\mathbb{R})$. The Lemma 2.1 holds for ${\bf
r}_{q}$ in that case as well. However, not all results about scattering
transform can be generalized from the case $q\in\mathcal{S}(\mathbb{R})$ to
$q\in L^{2}(\mathbb{R})$. For example, scattering transform is injective on
$\mathcal{S}(\mathbb{R})$ by Theorem 2.1, but it is not longer so when
extended to $L^{2}(\mathbb{R})$ (see Example 6.1 in Appendix).

##  3\. Another form of Dirac operator, $q\in L^{2}(\mathbb{R})$, and the
entropy function.

Suppose $q\in L^{2}(\mathbb{R})$. The alternative to $L_{q}$ form of writing
Dirac operator on the line is given by an expression

$\mathcal{D}_{Q}:X\mapsto JX^{\prime}+QX,\qquad
Q=\begin{pmatrix}-\operatorname{Im}q&-\operatorname{Re}q\\\
-\operatorname{Re}q&\operatorname{Im}q\end{pmatrix}\,.$ (3.1)

$\mathcal{D}_{Q}$ is densely defined self-adjoint operator on the Hilbert
space $L^{2}(\mathbb{R},\mathbb{C}^{2})$ of functions
$X:\mathbb{R}\to\mathbb{C}^{2}$ such that
$\|X\|_{L^{2}(\mathbb{R},\mathbb{C}^{2})}^{2}=\int_{\mathbb{R}}\|X(\xi)\|^{2}_{\mathbb{C}^{2}}\,d\xi$
is finite. $\mathcal{D}_{Q}$ and $L_{q}$ defined in (2.1) are related by a
simple formula:

$\mathcal{D}_{Q}=\Sigma
L_{q}\Sigma^{-1},\quad\Sigma=\frac{1}{\sqrt{2}}\left(\begin{matrix}1&1\\\
-i&i\end{matrix}\right),\quad\Sigma^{-1}=\frac{1}{\sqrt{2}}\left(\begin{matrix}1&i\\\
1&-i\end{matrix}\right)\,.$

One way to study $\mathcal{D}_{Q}$ is to focus on Dirac operators on half-line
$\mathbb{R}_{+}$ first. Given $q\in L^{2}(\mathbb{R}_{+})$, we define
$\mathcal{D}^{+}_{Q}$ on $L^{2}(\mathbb{R}_{+},\mathbb{C}^{2})$ by

$\mathcal{D}^{+}_{Q}:X\mapsto JX^{\prime}+QX,\qquad
Q=\begin{pmatrix}-\operatorname{Im}q&-\operatorname{Re}q\\\
-\operatorname{Re}q&\operatorname{Im}q\end{pmatrix}$ (3.2)

on the dense subset of absolutely continuous functions $X\in
L^{2}(\mathbb{R}_{+},\mathbb{C}^{2})$ such that $\mathcal{D}^{+}_{Q}X\in
L^{2}(\mathbb{R}_{+},\mathbb{C}^{2})$, $X(0)=\left(\begin{smallmatrix}*\\\
0\end{smallmatrix}\right)$. We will call $\mathcal{D}^{+}_{Q}$ the Dirac
operator defined on the positive half-line with boundary conditions
$X(0)=\left(\begin{smallmatrix}*\\\ 0\end{smallmatrix}\right)$ or simply the
half-line Dirac operator. Set $A(\xi)=-\overline{q(\xi/2)}/2$ for
$\xi\in\mathbb{R}_{+}$, and let $P(\xi,\lambda)$, $P_{*}(\xi,\lambda)$ be the
solutions of Krein system (2.11) generated by $A$. The Krein system with
coefficient $A$ and Dirac equation (3.2) are related as follows (see the proof
of Lemma 6.1 in Appendix): if $N_{Q}$ solves the Cauchy problem
$JN^{\prime}_{Q}(\xi,\lambda)+Q(\xi)N_{Q}(\xi,\lambda)=\lambda
N_{Q}(\xi,\lambda)$, $N_{Q}(0,\lambda)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\
\end{smallmatrix}\right)$, then

$N_{Q}(\xi,\lambda)=\frac{e^{-i\lambda\xi}}{2}\left(\begin{matrix}1&1\\\
i&-i\end{matrix}\right)\left(\begin{matrix}\mathfrak{A}_{*}^{+}(2\xi,\lambda)&\mathfrak{B}_{*}^{+}(2\xi,\lambda)\\\
\mathfrak{B}^{+}(2\xi,\lambda)&\mathfrak{A}^{+}(2\xi,\lambda)\end{matrix}\right)\left(\begin{matrix}1&-i\\\
1&i\end{matrix}\right)\,,$

where the continuous Wall polynomials
$\mathfrak{A}^{+},\mathfrak{B}^{+},\mathfrak{A}_{*}^{+},\mathfrak{B}^{+}_{*}$
were defined in (2.14). The Weyl function of the operator
$\mathcal{D}^{+}_{Q}$ coincides (see Lemma 6.1 in Appendix) with

$m_{Q}(z)=\lim_{\xi\to+\infty}i\frac{\widehat{P}_{*}(\xi,z)}{P_{*}(\xi,z)},\qquad
z\in\mathbb{C}_{+}.$ (3.3)

It is known (see Theorem 7.3 in [12]) that the limit above exists for every
$z\in\mathbb{C}_{+}$ and defines an analytic function of Herglotz-Nevanlinna
class in $\mathbb{C}_{+}$. The latter means that
$m_{Q}(\mathbb{C}_{+})\subset\mathbb{C}_{+}$. In the next theorem,
$\operatorname{Im}m_{Q}(\lambda)$ denotes the nontangential boundary value on
$\mathbb{R}$ which exists Lebesgue almost everywhere. It is understood as a
nonnegative function $g=\operatorname{Im}m$ on $\mathbb{R}$ and it satisfies
$g/(1+\lambda^{2})\in L^{1}(\mathbb{R})$.

######  Theorem 3.1.

Let $q\in L^{2}(\mathbb{R}_{+})$ and let $Q$, $\mathcal{D}^{+}_{Q}$, $m_{Q}$
be defined by (3.2), (3.3). Denote by $N_{Q}$ the solution of the Cauchy
problem $JN^{\prime}_{Q}(\xi)+Q(\xi)N_{Q}(\xi)=0$,
$N_{Q}(0)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\ \end{smallmatrix}\right)$,
and set $\mathcal{H}_{Q}=N_{Q}^{*}N_{Q}$. Define also

$\displaystyle\mathcal{K}^{+}_{Q}$
$\displaystyle=\log\operatorname{Im}m_{Q}(i)-\frac{1}{\pi}\int_{\mathbb{R}}\log\operatorname{Im}m_{Q}(\lambda)\frac{d\lambda}{\lambda^{2}+1},$
(3.4) $\displaystyle\widetilde{\mathcal{K}}^{+}_{Q}$
$\displaystyle=\sum_{k=0}^{+\infty}\left(\det\int_{k}^{k+2}\mathcal{H}_{Q}(\xi)\,d\xi-4\right).$
(3.5)

Then, we have

$c_{1}\mathcal{K}_{Q}^{+}\leqslant\widetilde{\mathcal{K}}_{Q}^{+}\leqslant
c_{2}e^{c_{2}\mathcal{K}_{Q}^{+}}$ (3.6)

for some positive absolute constants $c_{1}$, $c_{2}$.

Proof. Lemma 6.1 in Appendix shows that $m_{Q}$ coincides with the Weyl
function for the canonical system with Hamiltonian $\mathcal{H}_{Q}$. Then,
the bounds in (3.6) follow from the Theorem 1.2 in [3] (see also Corollary 1.4
in [3]).∎

The quantity $\mathcal{K}^{+}_{Q}$ will be called the entropy of the Dirac
operator on $\mathbb{R}_{+}$. We now turn to (3.1) to define the entropy for
the Dirac operator on the whole line. Take $q\in L^{2}(\mathbb{R})$ and let
$A^{+}(\xi)=-\overline{q(\xi/2)}/2$ and $A^{-}(\xi)=q(-\xi/2)/2$,
$\xi\in\mathbb{R}_{+}$ be the coefficients of Krein systems associated to
restrictions of $q$ to the half-lines $\mathbb{R}_{+}$ and $\mathbb{R}_{-}$.
As in (3.3), the half-line Weyl functions $m_{\pm}$ satisfy

$m_{\pm}(z)=\lim_{\xi\to+\infty}i\frac{\widehat{P}_{*}^{\pm}(\xi,z)}{P_{*}^{\pm}(\xi,z)},\qquad
z\in\mathbb{C}_{+}.$ (3.7)

These Weyl functions $m_{\pm}$ can be used to construct the spectral
representation for the Dirac operator. Let

$m(z)=-\frac{1}{m_{+}(z)+m_{-}(z)}\begin{pmatrix}-2m_{+}(z)m_{-}(z)&m_{+}(z)-m_{-}(z)\\\
m_{+}(z)-m_{-}(z)&2\end{pmatrix},\qquad z\in\mathbb{C}_{+}.$ (3.8)

Using $\operatorname{Im}m_{\pm}(z)>0$, one can show that
$\operatorname{Im}m(z)$ is a positive definite matrix for
$z\in\mathbb{C}_{+}$. In other words, $m$ is the matrix-valued Herglotz
function. Therefore, there exists a unique matrix-valued measure $\rho$ taking
Borel subsets of $\mathbb{R}$ into $2\times 2$ nonnegative matrices such that

$m(z)=\alpha+\beta
z+\frac{1}{\pi}\int_{\mathbb{R}}\left(\frac{1}{\lambda-z}-\frac{\lambda}{\lambda^{2}+1}\right)\,d\rho(\lambda),\qquad
z\in\mathbb{C}_{+},$

where $\alpha$, $\beta$ are constant $2\times 2$ real matrices,
$\beta\geqslant 0$. The importance of $\rho$ becomes clear when we recall the
spectral decomposition for $\mathcal{D}_{Q}$. Specifically, let $N_{Q}(\xi,z)$
be the solution of the Cauchy problem

$J\frac{\partial}{\partial\xi}N_{Q}(\xi,z)+Q(\xi)N_{Q}(\xi,z)=z{N}_{Q}(\xi,z),\quad\quad
N_{Q}(0,z)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\
\end{smallmatrix}\right),\quad z\in\mathbb{C},\quad\xi\in\mathbb{R}\,.$ (3.9)

Then, the mapping

$\mathcal{F}_{\mathcal{D}_{Q}}:X\mapsto\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}N_{Q}^{*}(\xi,\lambda)X(\xi)\,d\xi,\qquad\lambda\in\mathbb{R},$
(3.10)

initially defined on the set of compactly supported smooth functions
$X:\mathbb{R}\to\mathbb{C}^{2}$, extends (see Appendix) to the unitary
operator between the Hilbert spaces $L^{2}(\mathbb{R},\mathbb{C}^{2})$ and
$L^{2}(\rho)$,

$\displaystyle L^{2}(\rho)$
$\displaystyle=\Bigl{\\{}Y:\mathbb{R}\to\mathbb{C}^{2}:\;\|Y\|^{2}_{L^{2}(\rho)}=\int_{\mathbb{R}}Y^{*}(\lambda)\,d\rho(\lambda)\,Y(\lambda)<\infty\Bigr{\\}}.$

Moreover, $\mathcal{D}_{Q}$ is unitary equivalent to the operator of
multiplication by the independent variable in $L^{2}(\rho)$ and the unitary
equivalence is given by the operator $\mathcal{F}_{Q}$. In fact, these
properties of $\rho$ will not be used in the paper, we mention them only to
motivate the following definition. Let us define the entropy function
$\mathcal{K}_{Q}(z)$ by

$\mathcal{K}_{Q}(z)=-\frac{1}{\pi}\int_{\mathbb{R}}\log(\det\rho_{ac}(\lambda))\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda,\qquad
z\in\mathbb{C}_{+},$ (3.11)

where $\rho_{ac}$ denotes the absolutely continuous part of the spectral
measure $\rho$ and it satisfies $\rho_{ac}(\lambda)=\lim_{\varepsilon\to
0,\varepsilon>0}\operatorname{Im}m(\lambda+i\varepsilon)$ for a.e.
$\lambda\in\mathbb{R}$. The quantity $\mathcal{K}_{Q}$ will play a crucial
role in our considerations. We first relate it to the coefficient $a$ of the
reduced transition matrix $T$ which was introduced in Proposition 2.2.

######  Lemma 3.1.

We have $\det\rho_{ac}(\lambda)=|a(\lambda)|^{-2}$ for almost all
$\lambda\in\mathbb{R}$. In particular, $\mathcal{K}_{Q}(z)=2\log|a(z)|$ for
all $z\in\mathbb{C}_{+}$.

Proof. From the definition (or see page 59 in [21]), one has

$\det\operatorname{Im}m(z)=4\frac{\operatorname{Im}m_{+}(z)\operatorname{Im}m_{-}(z)}{|m_{+}(z)+m_{-}(z)|^{2}},\qquad
z\in\mathbb{C}_{+}.$ (3.12)

Substituting expressions for

$m_{\pm}(z)=\lim_{\xi\to+\infty}i\frac{\widehat{P}_{*}^{\pm}(\xi,z)}{P^{\pm}(\xi,z)}=i\frac{\mathfrak{a}^{\pm}(z)-\mathfrak{b}^{\pm}(z)}{\mathfrak{a}^{\pm}(z)+\mathfrak{b}^{\pm}(z)},\qquad
z\in\mathbb{C}_{+},$

into (3.12), we obtain

$\displaystyle\det\rho_{ac}(\lambda)$
$\displaystyle=\lim_{\varepsilon\to+0}\det\operatorname{Im}m(\lambda+i\varepsilon)$
$\displaystyle=\lim_{\varepsilon\to+0}\frac{(|\mathfrak{a}^{+}(\lambda+i\varepsilon)|^{2}-|\mathfrak{b}^{+}(\lambda+i\varepsilon)|^{2})(|\mathfrak{a}^{-}(\lambda+i\varepsilon)|^{2}-|\mathfrak{b}^{-}(\lambda+i\varepsilon)|^{2})}{|\mathfrak{a}^{+}(\lambda+i\varepsilon)\mathfrak{a}^{-}(\lambda+i\varepsilon)-\mathfrak{b}^{+}(\lambda+i\varepsilon)\mathfrak{b}^{-}(\lambda+i\varepsilon)|^{2}}$
$\displaystyle=\frac{1}{|a(\lambda)|^{2}},$

for almost every $\lambda\in\mathbb{R}$ and the first claim of the lemma
follows. Then, the second claim is immediate because $a$ is an outer function
as we showed in Proposition 2.1.∎

Consider again the half-line entropy functions

$\mathcal{K}_{Q}^{\pm}(z)=\log\operatorname{Im}m_{\pm}(z)-\frac{1}{\pi}\int_{\mathbb{R}}\log\operatorname{Im}m_{\pm}(\lambda)\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda,\qquad
z\in\mathbb{C}_{+}.$

We see that $\mathcal{K}_{Q}^{+}(i)$ coincides with the entropy (3.4) for the
restriction of $Q$ to $\mathbb{R}_{+}$ (that explains why we use the same
notation for the two objects), and
$\mathcal{K}^{-}_{Q}(z)=\mathcal{K}^{+}_{Q_{-}}(z)$ for the potential

$Q_{-}(\xi)=\begin{pmatrix}-\operatorname{Im}q(-\xi)&\operatorname{Re}q(-\xi)\\\
\operatorname{Re}q(-\xi)&\operatorname{Im}q(-\xi)\end{pmatrix},\qquad\xi\in\mathbb{R}_{+}.$

Our plan now is to relate $\mathcal{K}^{\pm}_{Q}(i)$ with $\mathcal{K}_{Q}(i)$
and then use the fact that the full line entropy $\mathcal{K}_{Q}(i)$ is
conserved, see Lemma 3.1. That will eventually lead to the proof of Theorem
1.2.

######  Lemma 3.2.

Let $q\in L^{2}(\mathbb{R})$ and let $q_{\ell}(\xi)=q(\xi-\ell)$, where
$\ell\in\mathbb{R}$ and $\xi\in\mathbb{R}$. Denote by $Q_{\ell}$ the matrix-
function in (3.1) corresponding to $q_{\ell}$. Then,
$\mathcal{K}^{+}_{Q_{\ell}}(z)\to\mathcal{K}_{Q}(z)$,
$\mathcal{K}^{-}_{Q_{\ell}}(z)\to 0$ as $\ell\to+\infty$ for every
$z\in\mathbb{C}_{+}$.

Proof. Take $z\in\mathbb{C}_{+}$. We have

$\mathcal{K}^{+}_{Q_{\ell}}(z)+\mathcal{K}^{-}_{Q_{\ell}}(z)=\log\Bigl{(}\operatorname{Im}m_{\ell,+}(z)\operatorname{Im}m_{\ell,-}(z)\Bigr{)}-\frac{1}{\pi}\int_{\mathbb{R}}\log\Bigl{(}\operatorname{Im}m_{\ell,+}(\lambda)\operatorname{Im}m_{\ell,-}(\lambda)\Bigr{)}\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda,$

for the corresponding Weyl functions $m_{\ell,\pm}$. We also have

$\log|m_{\ell,+}(z)+m_{\ell,-}(z)|^{2}=\frac{1}{\pi}\int_{\mathbb{R}}\log|m_{\ell,+}(\lambda)+m_{\ell,-}(\lambda)|^{2}\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda$

by the mean value theorem for harmonic functions. From (3.12), it follows that

$\mathcal{K}^{+}_{Q_{\ell}}(z)+\mathcal{K}^{-}_{Q_{\ell}}(z)=\log\left(4\frac{\operatorname{Im}m_{\ell,+}(z)\operatorname{Im}m_{\ell,-}(z)}{|m_{\ell,+}(z)+m_{\ell,-}(z)|^{2}}\right)+\mathcal{K}_{Q_{\ell}}(z).$

Notice that $\mathcal{K}_{Q_{\ell}}(z)$ does not depend on $\ell\in\mathbb{R}$
because the coefficient $a$ in Lemma 3.1 for the potential $Q_{\ell}$ does not
depend on $\ell$. So, we only need to show that

$\mathcal{K}^{-}_{Q_{\ell}}(z)\to 0\quad\,\,{\rm
and}\,\,\quad\log\left(4\frac{\operatorname{Im}m_{\ell,+}(z)\operatorname{Im}m_{\ell,-}(z)}{|m_{\ell,+}(z)+m_{\ell,-}(z)|^{2}}\right)\to
0,$

when $\ell\to+\infty$ and $z\in\mathbb{C}_{+}$. The second relation follows
from $m_{\ell,+}(z)\to i$, $m_{\ell,-}(z)\to i$, which hold because $q_{\ell}$
tends to zero weakly in $L^{2}(\mathbb{R})$ as $\ell\to+\infty$ and
$\|q_{\ell}\|_{L^{2}(\mathbb{R})}=\|q\|_{L^{2}(\mathbb{R})}$ (see Lemma 6.2 in
Appendix). Moreover, relation $m_{\ell,-}(z)\to i$ implies that
$\mathcal{K}^{-}_{Q_{\ell}}(z)\to 0$ if and only if

$\frac{1}{\pi}\int_{\mathbb{R}}\log\operatorname{Im}m_{\ell,-}(\lambda)\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda\to
0\,.$ (3.13)

In the rest of the proof, we will show (3.13). Let $\mathfrak{a}_{\ell}^{-}$,
$\mathfrak{b}_{\ell}^{-}$ be the limits of continuous Wall polynomials
corresponding to $Q_{\ell}^{-}$. Consider
$s_{\ell}^{-}=\mathfrak{b}_{\ell}^{-}/\mathfrak{a}_{\ell}^{-}$. The formula
(12.57) in [12] gives

$s_{\ell}^{-}(z)=\frac{1+im_{\ell,-}(z)}{1-im_{\ell,-}(z)}\,,\qquad{m_{\ell,-}(z)=i\frac{\mathfrak{a}_{\ell}^{-}(z)-\mathfrak{b}^{-}_{\ell}(z)}{\mathfrak{a}^{-}_{\ell}(z)+\mathfrak{b}_{\ell}^{-}(z)}}.$

It implies that
$\operatorname{Im}m_{\ell,-}(\lambda)=|\mathfrak{a}^{-}_{\ell}(\lambda)+\mathfrak{b}^{-}_{\ell}(\lambda)|^{-2}$
when $\lambda\in\mathbb{R}$ and that $s^{-}_{\ell}(z)\to 0$ when
$\ell\to+\infty$ and $z\in\mathbb{C}_{+}$. Now, we can write

$\displaystyle\frac{1}{\pi}\int_{\mathbb{R}}\log\operatorname{Im}m_{\ell,-}(\lambda)\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda$
$\displaystyle=\frac{1}{\pi}\int_{\mathbb{R}}\log\left(\frac{1}{|\mathfrak{a}^{-}_{\ell}(\lambda)+\mathfrak{b}^{-}_{\ell}(\lambda)|^{2}}\right)\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda$
$\displaystyle=\log\frac{1}{|\mathfrak{a}^{-}_{\ell}(z)+\mathfrak{b}^{-}_{\ell}(z)|^{2}}=\log\frac{1}{|\mathfrak{a}_{\ell}^{-}(z)|^{2}}+\log\frac{1}{|1+s_{\ell}^{-}(z)|^{2}}\,.$

So, it remains to show that $|\mathfrak{a}_{\ell}^{-}(z)|^{2}\to 1$ as
$\ell\to+\infty$. That holds because
$\|q_{\ell,-}\|_{L^{2}(\mathbb{R}_{+})}\to 0$ as $\ell\to+\infty$ and

$\|q_{\ell,-}\|_{L^{2}(\mathbb{R}_{+})}^{2}=\frac{1}{\pi}\int_{\mathbb{R}}\log|\mathfrak{a}_{\ell}^{-}(\lambda)|^{2}\,d\lambda\geqslant\frac{\operatorname{Im}z}{\pi}\int_{\mathbb{R}}\log|\mathfrak{a}_{\ell}^{-}(\lambda)|^{2}\frac{\operatorname{Im}z}{|\lambda-z|^{2}}\,d\lambda=\operatorname{Im}z\cdot\log|\mathfrak{a}_{\ell}^{-}(z)|^{2}\geqslant
0,$

where the first equality follows from
$\|q_{\ell,-}\|_{L^{2}(\mathbb{R}_{+})}^{2}=2\|A_{\ell,-}\|_{L^{2}(\mathbb{R}_{+})}^{2}$
and the formula $(12.2)$ in [12]. Thus, (3.13) holds and we are done. ∎

As an immediate corollary of Theorem 3.1 and Lemma 3.2, we get the following
estimate.

######  Lemma 3.3.

Let $q\in L^{2}(\mathbb{R})$. Denote by $N_{Q}$ the solution of the Cauchy
problem $JN_{Q}^{\prime}(\xi)+Q(\xi)N_{Q}(\xi)=0$,
$N_{Q}(0)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\ \end{smallmatrix}\right)$,
and set $\mathcal{H}_{Q}=N_{Q}^{*}N_{Q}$. Consider

$\mathcal{K}_{Q}=\mathcal{K}_{Q}(i),\qquad\widetilde{\mathcal{K}}_{Q}=\sum_{k\in\mathbb{Z}}\left(\det\int_{k}^{k+2}\mathcal{H}_{Q}(\xi)\,d\xi-4\right).$
(3.14)

Then, we have

$c_{1}\mathcal{K}_{Q}\leqslant\widetilde{\mathcal{K}}_{Q}\leqslant
c_{2}\mathcal{K}_{Q}e^{c_{2}\mathcal{K}_{Q}}$ (3.15)

for some positive absolute constants $c_{1}$, $c_{2}$.

Proof. By Lemma 3.2, we have
$\mathcal{K}_{Q}=\lim_{\ell\to+\infty}\mathcal{K}^{+}_{Q_{\ell}}$. It remains
to substitute $Q_{\ell}$ into the estimate (3.6) and take the limit as
$\ell\to+\infty$ for $\ell\in\mathbb{Z}$. ∎

##  4\. Proof of Theorem 1.2

The following result will play a crucial role in what follows. We postpone its
proof to the next section.

######  Theorem 4.1.

Suppose $q\in L^{2}(\mathbb{R})$ and let $N_{Q}$ satisfy
$JN_{Q}^{\prime}+QN_{Q}=0,N_{Q}(0)=I$, where
$Q=\left(\begin{smallmatrix}-\operatorname{Im}q&-\operatorname{Re}q\\\
-\operatorname{Re}q&\operatorname{Im}q\end{smallmatrix}\right).$ Then,

$\displaystyle
e^{-C_{1}R}\|q\|_{H^{-1}(\mathbb{R})}^{2}\lesssim\widetilde{\mathcal{K}}_{Q}\lesssim
e^{C_{2}R}\|q\|^{2}_{H^{-1}(\mathbb{R})}\,,$ (4.1)

where $R=\|q\|_{L^{2}(\mathbb{R})}$ and $C_{1}$, $C_{2}$ are two positive
absolute constants.

Proof of Theorem 1.2 in the case $s=-1$. First, assume that
$q_{0}\in\mathcal{S}(\mathbb{R})$ and let $q(\xi,t)$ be the solution of (1.1)
with the initial datum $q_{0}$. We want to prove that

$C_{1}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{-2}\|q_{0}\|_{H^{-1}(\mathbb{R})}\leqslant\|q(\cdot,t)\|_{H^{-1}(\mathbb{R})}\leqslant
C_{2}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{2}\|q_{0}\|_{H^{-1}(\mathbb{R})}\,.$
(4.2)

We have $\|q(\cdot,t)\|_{L^{2}(\mathbb{R})}=\|q_{0}\|_{L^{2}(\mathbb{R})}$ for
all $t$, see formula $(4.33)$ in [13]. Let $a(z,t)$ denote the coefficient in
the matrix (2.5) given by $q(\xi,t)$. For each $t\in\mathbb{R}$, define $Q$ by
(3.2). Let $\widetilde{\mathcal{K}}_{Q}(t)$ be as in Lemma 3.3 and
$\mathcal{K}_{Q}(z,t)$ be defined by (3.11). Formulas (2.8) and (2.9) show
that $a({z},t)$ is constant in $t$ and Lemma 3.1 says that
$\mathcal{K}_{Q}(z,t)$ is constant in $t$ as well. The bound (3.15) yields

$c_{1}\mathcal{K}_{Q}(i,0)\leqslant\widetilde{\mathcal{K}}_{Q}(t)\leqslant
c_{2}\mathcal{K}_{Q}(i,0)e^{c_{2}\mathcal{K}_{Q}(i,0)}\,.$ (4.3)

Assume first that $R=\|q_{0}\|_{L^{2}(\mathbb{R})}\leqslant 1$. Taking $t=0$
in (4.3) and applying (4.1) to $q_{0}$, we get $\mathcal{K}_{Q}(i,0)\lesssim
1$ since $\|q_{0}\|_{H^{-1}(\mathbb{R})}\leqslant R\leqslant 1$. Hence, in
that case (4.3) can be written as
$\widetilde{\mathcal{K}}_{Q}(t)\sim\mathcal{K}_{Q}(i,0)\,.$ By (4.1),
$\|q(\cdot,t)\|_{H^{-1}(\mathbb{R})}^{2}\sim\widetilde{\mathcal{K}}_{Q}(t)$,
and so
$\|q(\cdot,t)\|_{H^{-1}(\mathbb{R})}^{2}\sim\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})}^{2}$.

If $R=\|q_{0}\|_{L^{2}(\mathbb{R})}>1$, we use dilation. Consider
$q_{\alpha}(\xi,t)=\alpha q(\alpha\xi,\alpha^{2}t)$ which solves the same
equation and notice that
$\|q_{\alpha}\|_{L^{2}(\mathbb{R})}=\alpha^{{\frac{1}{2}}}R\,.$

Let $\alpha=\alpha_{c}=R^{-2}<1$ making
$\|q_{\alpha_{c}}\|_{L^{2}(\mathbb{R})}=1$. Then, for the Sobolev norm, we get

$\|q_{\alpha}(\cdot,t)\|_{H^{-1}(\mathbb{R})}=\alpha^{{\frac{1}{2}}}\left(\int_{\mathbb{R}}\frac{1}{1+\alpha^{2}\eta^{2}}|(\mathcal{F}q)(\eta,{\alpha^{2}}t)|^{2}d\eta\right)^{{\frac{1}{2}}}\,.$
(4.4)

Since

$\frac{1}{1+\eta^{2}}\leqslant\frac{1}{1+\alpha_{c}^{2}\eta^{2}}\leqslant\frac{1}{\alpha_{c}^{2}(1+\eta^{2})},$
(4.5)

one has

$\alpha_{c}^{{\frac{1}{2}}}\|{q(\cdot,\alpha_{c}^{2}t)}\|_{H^{-1}(\mathbb{R})}{\leqslant}\|q_{\alpha_{c}}(\cdot,{t})\|_{H^{-1}(\mathbb{R})}{\leqslant}\alpha_{c}^{-{\frac{1}{2}}}\|{q(\cdot,\alpha_{c}^{2}t)}\|_{H^{-1}(\mathbb{R})}\,.$

In particular, at $t=0$ we get

$\alpha_{c}^{{\frac{1}{2}}}\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})}\leqslant\|q_{\alpha_{c}}(\cdot,0)\|_{H^{-1}(\mathbb{R})}\leqslant\alpha_{c}^{-{\frac{1}{2}}}\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})}\,.$

Since $\|q_{\alpha_{c}}(\cdot,0)\|_{L^{2}(\mathbb{R})}=1$, one can apply the
previous bounds to obtain

$\|q_{\alpha_{c}}(\cdot,t)\|_{H^{-1}(\mathbb{R})}\sim\|q_{\alpha_{c}}(\cdot,0)\|_{H^{-1}(\mathbb{R})}\,.$

Then,

$\displaystyle\alpha_{c}^{{\frac{1}{2}}}\|q(\cdot,\alpha_{c}^{2}t)\|_{H^{-1}(\mathbb{R})}$
$\displaystyle\lesssim\|q_{\alpha_{c}}(\cdot,0)\|_{H^{-1}(\mathbb{R})}\lesssim\alpha_{c}^{-{\frac{1}{2}}}\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})},$
$\displaystyle\alpha_{c}^{-{\frac{1}{2}}}\|q(\cdot,\alpha_{c}^{2}t)\|_{H^{-1}(\mathbb{R})}$
$\displaystyle\gtrsim\|q_{\alpha_{c}}(\cdot,0)\|_{H^{-1}(\mathbb{R})}\gtrsim\alpha_{c}^{{\frac{1}{2}}}\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})}.$

Recalling that $\alpha_{c}=R^{-2}$, we obtain

$R^{-2}\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})}\lesssim\|q(\cdot,t)\|_{H^{-1}(\mathbb{R})}\lesssim
R^{2}\|q(\cdot,0)\|_{H^{-1}(\mathbb{R})}\,$

for all $t\in\mathbb{R}$. Finally, having proved (4.2) for
$q_{0}\in\mathcal{S}(\mathbb{R})$, it is enough to use Theorem 1.1 to extend
(4.2) to $q_{0}\in L^{2}(\mathbb{R})$.∎

Our next goal is to prove the estimate

$C_{1}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{2s}\|q_{0}\|_{H^{s}(\mathbb{R})}\leqslant\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\leqslant
C_{2}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{-2s}\|q_{0}\|_{H^{s}(\mathbb{R})},$
(4.6)

where $\,t\in\mathbb{R}\,,\,s\in(-1,0]$, and $C_{1}$ and $C_{2}$ are positive
absolute constants. For $s=0$, this bound is trivial. To cover $s\in(-1,0)$,
we will need some auxiliary results first. One of the basic properties of NLS
which we discussed in the Introduction has to do with modulation: if
$q(\xi,t)$ solves (2.10), then $\widetilde{q}_{v}(\xi,t)=e^{iv\xi-
iv^{2}t}q(\xi-2vt,t)$ solves (2.10) for every $v\in\mathbb{R}$.

######  Lemma 4.1.

Let $q_{0}\in L^{2}(\mathbb{R})$, $t\in\mathbb{R}$. Then,

$\|\tilde{q}_{v}(\cdot,t)\|_{H^{-1}(\mathbb{R})}^{2}=\int_{\mathbb{R}}\frac{|(\mathcal{F}q)(\eta,t)|^{2}}{1+(\eta+v)^{2}}\,d\eta.$

Proof. It is clear that
$\|e^{-iv^{2}t}f\|_{H^{-1}(\mathbb{R})}=\|f\|_{H^{-1}(\mathbb{R})}$ for every
$f\in H^{-1}(\mathbb{R})$ and $t\in\mathbb{R}$, because $e^{-iv^{2}t}$ is a
unimodular constant. We have
$\mathcal{F}(e^{iv\xi}q(\xi-2vt,t))(\eta)=(\mathcal{F}q(\xi,t))(\eta-v)e^{-2ivt(\eta-v)}$,
$\eta\in\mathbb{R}$. Since $|e^{-2ivt(\eta-v)}|=1$, it only remains to change
the variable of integration in

$\|\widetilde{q}_{v}\|_{H^{-1}(\mathbb{R})}=\int_{\mathbb{R}}\frac{|(\mathcal{F}q(\xi,t))(\eta-v)|^{2}}{1+\eta^{2}}\,d\eta$

to get the statement of the lemma. ∎

The next result is a standard property of convolutions.

######  Lemma 4.2.

Let $\gamma\in(-\frac{1}{2},1]$ and set $a_{k}=\frac{1}{(1+k^{2})^{\gamma}}$
for $k\in\mathbb{Z}$. We have

$\sum_{k\in\mathbb{Z}}\frac{a_{k}}{1+(\eta-k)^{2}}\sim
C_{\gamma}\frac{1}{(1+\eta^{2})^{\gamma}},\qquad\eta\in\mathbb{R}.$

Proof. After comparing the sum to an integral, it is enough to show that

$\int_{\mathbb{R}}\frac{du}{(1+u^{2})^{\gamma}(1+(\eta-u)^{2})}\sim
C_{\gamma}\frac{1}{(1+\eta^{2})^{\gamma}}\,.$

The function on the left-hand side is even and continuous in $\eta$ and
$\gamma$, so we can assume that $\eta>1$. Then,

$\int_{|\eta-u|<0.5\eta}\frac{du}{(1+u^{2})^{\gamma}(1+(\eta-u)^{2})}\sim\frac{1}{(1+\eta^{2})^{\gamma}},\quad\int_{|\eta-u|>0.5\eta}\frac{du}{(1+u^{2})^{\gamma}(1+(\eta-u)^{2})}\lesssim\mathcal{I}_{1}+\mathcal{I}_{2}\,,$

where

$\mathcal{I}_{1}=\int_{u<-\eta/2,\;{u>3\eta/2}}\frac{du}{(1+u^{2})^{\gamma}(1+(\eta-u)^{2})}\lesssim\int_{-\infty}^{-\eta/2}\frac{du}{u^{2+2\gamma}}+\int_{3\eta/2}\frac{du}{u^{2+2\gamma}}\leqslant
C_{\gamma}\eta^{-1-2\gamma}\,,$
$\mathcal{I}_{2}=\int_{|u|<\eta/2}\frac{du}{(1+u^{2})^{\gamma}(1+(\eta-u)^{2})}\lesssim\eta^{-2}\int_{|u|<\eta/2}\frac{du}{(1+u^{2})^{\gamma}}\lesssim\eta^{-2\gamma}\,.$

Combining these bounds proves the lemma. ∎

Proof of Theorem 1.2, the case $s\in(-1,0)$. We can again assume that
$q_{0}\in\mathcal{S}(\mathbb{R})$. Recall the estimate (1.5) for $s=-1$:

$C_{1}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{-2}\|q_{0}\|_{H^{-1}(\mathbb{R})}\leqslant\|q(\cdot,t)\|_{H^{-1}(\mathbb{R})}\leqslant
C_{2}(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{2}\|q_{0}\|_{H^{-1}(\mathbb{R})}\,.$
(4.7)

According to Lemma 4.1, we have

$\|\tilde{q}_{v}(\cdot,t)\|_{H^{-1}(\mathbb{R})}^{2}=\int_{\mathbb{R}}\frac{|(\mathcal{F}q(\cdot,t))(\eta)|^{2}}{1+(v+\eta)^{2}}\,d\eta$
(4.8)

for $\tilde{q}_{v}(\xi,t)=e^{iv\xi-iv^{2}t}q(\xi-2vt,t)$. Let $a_{k}$,
$k\in\mathbb{Z}$, be the coefficients from Lemma 4.2 with $\gamma=-s$. Then,
(4.8) and Lemma 4.2 imply

$\sum_{k\in\mathbb{Z}}a_{k}\|\tilde{q}_{k}(\cdot,t)\|_{H^{-1}(\mathbb{R})}^{2}\sim
C_{s}\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}^{2}\,.$ (4.9)

In particular, taking $t=0$ gives

$\sum_{k\in\mathbb{Z}}a_{k}\|\tilde{q}_{k}(\cdot,0)\|_{H^{-1}(\mathbb{R})}^{2}\sim
C_{s}\|q_{0}\|_{H^{s}(\mathbb{R})}^{2}\,.$ (4.10)

We now apply (4.7) to $\tilde{q}_{k}$ and use (4.9) and (4.10) to get

$C_{1}(s)(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{-2}\|q_{0}\|_{H^{s}(\mathbb{R})}\leqslant\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\leqslant
C_{2}(s)(1+\|q_{0}\|_{L^{2}(\mathbb{R})})^{2}\|q_{0}\|_{H^{s}(\mathbb{R})}\,.$
(4.11)

If $R=\|q_{0}\|_{L^{2}(\mathbb{R})}\leqslant 1$, we have the statement of our
theorem. If $R=\|q_{0}\|_{L^{2}(\mathbb{R})}>1$, we use dilation
transformation like in the previous proof for $s=-1$. Consider
$q_{\alpha}(\xi,t)=\alpha q(\alpha\xi,\alpha^{2}t)$ which solves the same
equation and notice that
$\|q_{\alpha}\|_{L^{2}(\mathbb{R})}=\alpha^{\frac{1}{2}}R\,.$ Let
$\alpha=\alpha_{c}=R^{-2}<1$ making
$\|q_{\alpha_{c}}\|_{L^{2}(\mathbb{R})}=1$. Then, for the Sobolev norm, we
have

$\|q_{\alpha}(\cdot,t)\|_{H^{s}(\mathbb{R})}=\alpha^{\frac{1}{2}}\left(\int_{\mathbb{R}}\frac{1}{(1+\alpha^{2}\eta^{2})^{|s|}}|(\mathcal{F}q)(\eta,{\alpha^{2}}t)|^{2}d\eta\right)^{\frac{1}{2}}\,.$

From (4.5),

$\frac{1}{(1+\eta^{2})^{|s|}}\leqslant\frac{1}{(1+\alpha_{c}^{2}\eta^{2})^{|s|}}\leqslant\frac{1}{\alpha_{c}^{2|s|}(1+\eta^{2})^{|s|}}\,.$

Then, one has

$\alpha_{c}^{\frac{1}{2}}\|{q(\cdot,\alpha_{c}^{2}t)}\|_{H^{s}(\mathbb{R})}{\leqslant}\|q_{\alpha_{c}}(\cdot,{t})\|_{H^{s}(\mathbb{R})}{\leqslant}\alpha_{c}^{\frac{1}{2}-|s|}\|{q(\cdot,\alpha_{c}^{2}t)}\|_{H^{s}(\mathbb{R})}\,.$

In particular, taking $t=0$ gives us

$\alpha_{c}^{\frac{1}{2}}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}\leqslant\|q_{\alpha_{c}}(\cdot,0)\|_{H^{s}(\mathbb{R})}\leqslant\alpha_{c}^{\frac{1}{2}-|s|}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}\,.$

Now $\|q_{\alpha_{c}}(\cdot,0)\|_{L^{2}(\mathbb{R})}=1$ and we can apply the
previous bounds to get

$\|q_{\alpha_{c}}(\cdot,t)\|_{H^{s}(\mathbb{R})}\sim\|q_{\alpha_{c}}(\cdot,0)\|_{H^{s}(\mathbb{R})}\,.$

Then,

$\displaystyle\alpha_{c}^{\frac{1}{2}}\|q(\cdot,\alpha_{c}^{2}t)\|_{H^{s}(\mathbb{R})}$
$\displaystyle\lesssim\|q_{\alpha_{c}}(\cdot,0)\|_{H^{s}(\mathbb{R})}\lesssim\alpha_{c}^{\frac{1}{2}-|s|}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})},$
$\displaystyle\alpha_{c}^{\frac{1}{2}-|s|}\|q(\cdot,\alpha_{c}^{2}t)\|_{H^{s}(\mathbb{R})}$
$\displaystyle\gtrsim\|q_{\alpha_{c}}(\cdot,0)\|_{H^{s}(\mathbb{R})}\gtrsim\alpha_{c}^{\frac{1}{2}}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}.$

Recalling that $\alpha_{c}=R^{-2}=\|q_{0}\|_{L^{2}(\mathbb{R})}^{-2}$, we
obtain

$\|q_{0}\|_{L^{2}(\mathbb{R})}^{-2|s|}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}\lesssim\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\lesssim\|q_{0}\|_{L^{2}(\mathbb{R})}^{2|s|}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}\,$

for all $t\in\mathbb{R}$. ∎

Our approach also provides the bounds for some positive Sobolev norms. The
following proposition slightly improves (1.4) when $s\in[0,\frac{1}{2})$,
$\|q_{0}\|_{H^{s}(\mathbb{R})}$ is large, and $\|q_{0}\|_{L^{2}(\mathbb{R})}$
is much smaller than $\|q_{0}\|_{H^{s}(\mathbb{R})}$.

######  Proposition 4.1.

Let $q_{0}\in\mathcal{S}(\mathbb{R})$ and let $q=q(\xi,t)$ be the solution of
(1.1) corresponding to $q_{0}$. Then, for each $s\in[0,\frac{1}{2})$, we get

$\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\sim C_{s}\|q_{0}\|_{H^{s}(\mathbb{R})}$
(4.12)

if $\|q_{0}\|_{L^{2}(\mathbb{R})}\leqslant 1$ and

$\|q(\cdot,t)\|_{H^{s}(\mathbb{R})}\lesssim
C_{s}(\|q_{0}\|_{L^{2}(\mathbb{R})}^{1+2s}+\|q_{0}\|_{H^{s}(\mathbb{R})})$
(4.13)

if $\|q_{0}\|_{L^{2}(\mathbb{R})}>1$.

Proof. In the case when $\|q\|_{L^{2}(\mathbb{R})}\leqslant 1$, the proof of
proposition repeats the arguments given above to get (4.11) except that the
constants in the inequalities depend on $s$ and can blow up when
$s\to\frac{1}{2}$. Suppose $\|q\|_{L^{2}(\mathbb{R})}\geqslant 1$. Then, for
the Sobolev norm, we have

$\|q_{\alpha}(\cdot,t)\|_{H^{s}(\mathbb{R})}=\alpha^{\frac{1}{2}}\left(\int_{\mathbb{R}}{(1+\alpha^{2}\eta^{2})^{s}}|(\mathcal{F}q)(\eta,{\alpha^{2}}t)|^{2}d\eta\right)^{\frac{1}{2}}\,.$

Take $\alpha=\alpha_{c}$ and write the following estimate for the integral
above:

$\displaystyle\int_{\mathbb{R}}{\left(1+\frac{\eta^{2}}{R^{4}}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta\sim\int_{-R^{2}}^{R^{2}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta+R^{-4s}\int_{|\eta|>R^{2}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta$
$\displaystyle\lesssim
R^{2}+R^{-4s}\int_{\mathbb{R}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta\,.$

We use $\|q_{\alpha_{c}}(\cdot,t)\|_{L^{2}(\mathbb{R})}=1$ and (4.11) to get
$\|q_{\alpha_{c}}(\cdot,t)\|_{H^{s}(\mathbb{R})}\sim
C_{s}\|q_{\alpha_{c}}(\cdot,0)\|_{H^{s}(\mathbb{R})}$. The previous estimate
for $t=0$ yields $\|q_{\alpha_{c}}(\cdot,0)\|_{H^{s}(\mathbb{R})}\lesssim
1+R^{-1-2s}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}.$ Hence,
$\|q_{\alpha_{c}}(\cdot,t)\|_{H^{s}(\mathbb{R})}\leqslant
C_{s}(1+R^{-1-2s}\|q(\cdot,0)\|_{H^{s}(\mathbb{R})}).$ We can write a lower
bound

$\|q_{\alpha_{c}}(\cdot,t)\|_{H^{s}(\mathbb{R})}^{2}=R^{-2}\int_{\mathbb{R}}{\left(1+\frac{\eta^{2}}{R^{4}}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta\gtrsim
R^{-2-4s}\int_{|\eta|>R^{2}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta$

so

$\int_{|\eta|>R^{2}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta\leqslant
C_{s}(R^{2+4s}+\|q(\cdot,0)\|^{2}_{H^{s}(\mathbb{R})}).$

Writing the integral as a sum of two:

$\int_{\mathbb{R}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta=\int_{|\eta|>R^{2}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta+\int_{|\eta|<R^{2}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta\,$

and estimating each of them, we get a bound which holds for all $t$:

$\int_{\mathbb{R}}{\left(1+\eta^{2}\right)^{s}}|(\mathcal{F}q)(\eta,{\alpha_{c}^{2}}t)|^{2}d\eta\leqslant
C_{s}(R^{2+4s}+\|q(\cdot,0)\|^{2}_{H^{s}(\mathbb{R})})\,.$

That is the required upper bound (4.13). ∎

##  5\. Oscillation and Sobolev space $H^{-1}(\mathbb{R})$.

In this part of the paper, our goal is to prove the Theorem 4.1. Let us recall
its statement.

######  Theorem 5.1.

Suppose that $q\in L^{2}(\mathbb{R})$ and let $N_{Q}$ satisfy
$JN^{\prime}_{Q}+QN_{Q}=0,N_{Q}(0)=I$, where
$Q=\left(\begin{smallmatrix}-\operatorname{Im}q&-\operatorname{Re}q\\\
-\operatorname{Re}q&\operatorname{Im}q\end{smallmatrix}\right).$ Then,

$\displaystyle
e^{-C_{1}R}\|q\|_{H^{-1}(\mathbb{R})}^{2}\lesssim\widetilde{\mathcal{K}}_{Q}\lesssim
e^{C_{2}R}\|q\|^{2}_{H^{-1}(\mathbb{R})}\,,$ (5.1)

where $R=\|q\|_{L^{2}(\mathbb{R})}$ and $C_{1}$, $C_{2}$ are two positive
absolute constants.

Theorem 5.1 is of independent interest in the spectral theory of Dirac
operators. For example, Lemma 3.3 shows that $\|q\|_{L^{2}(\mathbb{R})}$ and
$\|q\|_{H^{-1}(\mathbb{R})}$ control the size of $\mathcal{K}_{Q}$.

The strategy of the proof is the following. In the next subsection, we show
that $H^{-1}(\mathbb{R})$-norm of any function can be characterized through
BMO-like condition for its “antiderivative”. In Subsection 5.2, we consider
solution to Cauchy problem $JN^{\prime}+QN=0,N(0)=I$ on the interval $[0,1]$
where zero-trace symmetric $Q$ and study the quantity
$\det\int_{0}^{1}N^{*}Ndx$, which represents a single term in the sum for
$\widetilde{\mathcal{K}}_{Q}$. The results in Subsection 5.3 show that small
value of $\widetilde{\mathcal{K}}_{Q}$ guarantees that the “local” $H^{-1}$
norm of $Q$ is also small. This rough estimate is used in the proof of Theorem
4.1 which is contained in Subsection 5.4.

###  5.1. One property of Sobolev space $H^{-1}(\mathbb{R})$

Observe that a function $f\in L^{2}(\mathbb{R})$ belongs to the Sobolev space
$H^{-1}(\mathbb{R})$ if and only if

$\int_{\mathbb{R}}\left|\int_{\mathbb{R}}f(y)\chi_{\mathbb{R}_{+}}(x-y)e^{-(x-y)}\,dy\right|^{2}\,dx<\infty.$
(5.2)

Moreover, the last integral is equal to $\|f\|_{H^{-1}(\mathbb{R})}^{2}.$
Indeed, recall that $\mathcal{F}f$ stands for the Fourier transform of $f$:

$(\mathcal{F}f)(\eta)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-i\eta
x}\,dx\,.$

Then, from Plancherel’s identity and formula

$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}_{+}}e^{-x}e^{-ix\eta}dx=\frac{1}{\sqrt{2\pi}}\frac{1}{1+i\eta},$

we obtain

$\|f\|_{H^{-1}(\mathbb{R})}^{2}=2\pi\|(\mathcal{F}f)\mathcal{F}(\chi_{\mathbb{R}_{+}}e^{-x})\|^{2}_{L^{2}(\mathbb{R})}=\int_{\mathbb{R}}\frac{|(\mathcal{F}f)(\eta)|^{2}}{1+\eta^{2}}\,d\eta$

by properties of convolutions. We will need the following proposition.

######  Proposition 5.1.

Suppose that $f\in L^{1}_{\rm loc}(\mathbb{R})\cap H^{-1}(\mathbb{R})$. Let
$g$ be an absolutely continuous function on $\mathbb{R}$ such that
$g^{\prime}=f$ almost everywhere on $\mathbb{R}$. Then,

$c_{1}\|f\|_{H^{-1}(\mathbb{R})}^{2}\leqslant\sum_{k\in\mathbb{Z}}\int_{I_{k}}|g-\langle
g\rangle_{I_{k}}|^{2}\,dx\leqslant c_{2}\|f\|_{H^{-1}(\mathbb{R})}^{2},$ (5.3)

where $I_{k}=[k,k+2]$, $\langle g\rangle_{I}=\frac{1}{|I|}\int_{I}g(x)\,dx$,
and the positive constants $c_{1}$ and $c_{2}$ are universal.

Proof. Take a function $f\in L^{1}_{\rm loc}(\mathbb{R})\cap
H^{-1}(\mathbb{R})$, and let $g$ be an absolutely continuous function on
$\mathbb{R}$ such that $g^{\prime}=f$ almost everywhere on $\mathbb{R}$.
Assume first that $f$ has a compact support. The integral under the sum does
not change if we add a constant to $g$, so we can suppose without loss of
generality that

$g(x)=\int_{-\infty}^{x}f(s)\,ds,\qquad x\in\mathbb{R}.$

Upper bound. Given $f$, define $o_{f}$ by

$o_{f}(x)=e^{-x}\int_{-\infty}^{x}f(y)e^{y}dy$

and recall (see (5.2)) that

$\|f\|_{H^{-1}(\mathbb{R})}=\|o_{f}\|_{L^{2}(\mathbb{R})}.$ (5.4)

Moreover,

$o^{\prime}_{f}+o_{f}=f\,.$ (5.5)

For each interval $I_{k}$, we use (5.5) for the corresponding term in the sum
(5.3):

$\displaystyle\int_{I_{k}}\left|\int_{k}^{x}fdx_{1}-\frac{1}{2}\int_{k}^{k+2}\left(\int_{k}^{x_{1}}f(x_{2})dx_{2}\right)dx_{1}\right|^{2}dx$
$\displaystyle=\int_{I_{k}}\left|\int_{k}^{x}o(x_{1})dx_{1}+o(x)-\frac{1}{2}\int_{k}^{k+2}\left(o(x_{1})+\int_{k}^{x_{1}}o(x_{2})dx_{2}\right)dx_{1}\right|^{2}dx$
$\displaystyle\lesssim\int_{I_{k}}|o|^{2}dx$

after the Cauchy-Schwarz inequality is applied. Summing these estimates in
$k\in\mathbb{Z}$ and using (5.4), we get the upper bound in (5.3) for
compactly supported $f$.

Lower bound. Integration by parts gives

$\displaystyle\int_{-\infty}^{x}f(y)e^{-(x-y)}\,dy$
$\displaystyle=\int_{-\infty}^{x}f(s)\,ds-\int_{-\infty}^{x}\left(\int_{-\infty}^{y}f(s)\,ds\right)\,e^{-(x-y)}\,dy$
$\displaystyle=g(x)-\int_{-\infty}^{x}g(y)\,e^{-(x-y)}\,dy$
$\displaystyle=\int_{-\infty}^{x}(g(x)-g(y))\,e^{-(x-y)}\,dy.$

Therefore,

$\displaystyle\int_{\mathbb{R}}\left|\int_{-\infty}^{x}f(y)e^{-(x-y)}\,dy\right|^{2}\,dx$
$\displaystyle\leqslant\sum_{k\in\mathbb{Z}}\int_{k}^{k+2}\left(\int_{-\infty}^{x}|g(x)-g(y)|^{2}e^{-(x-y)}\,dy\right)\,dx$
$\displaystyle\lesssim\sum_{k\in\mathbb{Z}}\sum_{j\leqslant
k}e^{-(k-j)}\int_{k}^{k+2}\int_{j}^{j+2}|g(x)-g(y)|^{2}\,dx\,dy\,.$

Using the inequality $(x+y+z)^{2}\leqslant 3(x^{2}+y^{2}+z^{2})$, we continue
the estimate:

$...\lesssim\sum_{k\in\mathbb{Z}}\sum_{j\leqslant
k}e^{-(k-j)}\left(\int_{I_{k}}|g-\langle g\rangle_{I_{k}}|^{2}dx+|\langle
g\rangle_{I_{j}}-\langle g\rangle_{I_{k}}|^{2}+\int_{I_{j}}|g-\langle
g\rangle_{I_{j}}|^{2}dx\right).$

Since

$\sum_{k\in\mathbb{Z}}\sum_{j\leqslant
k}e^{-(k-j)}\left(\int_{I_{k}}|g-\langle
g\rangle_{I_{k}}|^{2}dx+\int_{I_{j}}|g-\langle
g\rangle_{I_{j}}|^{2}dx\right)\lesssim\sum_{k\in\mathbb{Z}}\int_{I_{k}}|g-\langle
g\rangle_{I_{k}}|^{2}dx\,,$

we are left with estimating

$\sum_{k\in\mathbb{Z}}\sum_{j\leqslant k}e^{-(k-j)}|\langle
g\rangle_{I_{j}}-\langle g\rangle_{I_{k}}|^{2}.$

Applying the Cauchy-Schwarz inequality for the telescoping sum

$\langle g\rangle_{I_{k}}-\langle
g\rangle_{I_{j}}=\sum_{s=j+1}^{k}\Bigl{(}\langle g\rangle_{I_{s}}-\langle
g\rangle_{I_{s-1}}\Bigr{)}\,,$

we get

$|\langle g\rangle_{I_{j}}-\langle
g\rangle_{I_{k}}|^{2}\leqslant(k-j)\sum_{j\leqslant s\leqslant k-1}|\langle
g\rangle_{I_{s}}-\langle g\rangle_{I_{s+1}}|^{2}.$

Then,

$\sum_{k\in\mathbb{Z}}\sum_{j\leqslant k}e^{-(k-j)}(k-j)\sum_{j\leqslant
s\leqslant k-1}|\langle g\rangle_{I_{s}}-\langle
g\rangle_{I_{s+1}}|^{2}=\sum_{s\in\mathbb{Z}}|\langle g\rangle_{I_{s}}-\langle
g\rangle_{I_{s+1}}|^{2}\sum_{k,j:\;j\leqslant s\leqslant k-1}(k-j)e^{-(k-j)}.$

We have

$\displaystyle\sum_{k,j:\;j\leqslant s\leqslant k-1}(k-j)e^{-(k-j)}$
$\displaystyle=\sum_{j\leqslant s}\sum_{m\geqslant 1}(s+m-j)e^{-(s+m-j)}$
$\displaystyle=\sum_{j\leqslant 0}\sum_{m\geqslant
1}(m-j)e^{-(m-j)}=\sum_{j\geqslant 0}\sum_{m\geqslant 1}(m+j)e^{-(m+j)}.$

The last sum is finite and does not depend on index $s$. Now, the estimate

$|\langle g\rangle_{I_{s}}-\langle g\rangle_{I_{s+1}}|^{2}=\int_{I_{s}\cap
I_{s+1}}|\langle g\rangle_{I_{s}}-g+g-\langle
g\rangle_{I_{s+1}}|^{2}dx\leqslant 2\int_{I_{s}}|g-\langle
g\rangle_{I_{s}}|^{2}dx+2\int_{I_{s+1}}|g-\langle g\rangle_{I_{s+1}}|^{2}dx$

proves that

$\sum_{k\in\mathbb{Z}}\sum_{j\leqslant k}e^{-(k-j)}|\langle
g\rangle_{I_{j}}-\langle
g\rangle_{I_{k}}|^{2}\lesssim\sum_{s\in\mathbb{Z}}\int_{I_{s}}|g-\langle
g\rangle_{I_{s}}|^{2}dx\,.$

Hence, the lower bound in (5.3) holds for compactly supported $f$.

Now, take any $f\in L^{1}_{\rm loc}(\mathbb{R})\cap H^{-1}(\mathbb{R})$. The
definition (1.2) of $H^{-1}(\mathbb{R})$ implies that $\mathcal{F}f$ can be
written as $(1+i\eta)(\mathcal{F}o)$ for some function $o\in
L^{2}(\mathbb{R})$. Moreover, this map $f\mapsto o$ is a bijection between
$H^{-1}(\mathbb{R})$ and $L^{2}(\mathbb{R})$ and
$\|f\|_{H^{-1}(\mathbb{R})}=\|o\|_{L^{2}(\mathbb{R})}$. Taking the inverse
Fourier transform of identity $\mathcal{F}f=(1+i\eta)(\mathcal{F}o)$, one gets
a formula $f=o+o^{\prime}$ where $o^{\prime}$ is understood as a derivative in
$\mathcal{S}^{\prime}(\mathbb{R})$. Since $f\in L^{1}_{\rm loc}(\mathbb{R})$
and $o\in L^{2}(\mathbb{R})$, we have $o^{\prime}\in L^{1}_{\rm
loc}(\mathbb{R})$ and, therefore, $o$ is absolutely continuous on $\mathbb{R}$
with the derivative equal to $f-o$. Now, take $o_{n}(x)=o(x)\mu_{n}(x)$ and
define the corresponding $f_{n}=o_{n}+o_{n}^{\prime}$. Here, $\mu_{n}(x)$ is
even and

$\mu_{n}(x)=\left\\{\begin{array}[]{ll}1,&0\leqslant x<n,\\\
n+1-x,&x\in[n,n+1),\\\ 0,&x\geqslant n+1\,.\end{array}\right.$

Then, $\\{o_{n}\\}\to o$ in $L^{2}(\mathbb{R})$ and so $\\{f_{n}\\}\to f$ in
$H^{-1}(\mathbb{R})$ because the mapping $f\mapsto o$ is unitary from
$H^{-1}(\mathbb{R})$ onto $L^{2}(\mathbb{R})$. Also, each $f_{n}$ is compactly
supported and $\\{f_{n}\\}$ converges to $f$ uniformly on every finite
interval. Define $g_{n}=\int_{0}^{x}f_{n}ds,g=\int_{0}^{x}fds$, and write
(5.3) for $f_{n}$. The estimate on the right gives

$\sum_{|k|\leqslant N}\int_{I_{k}}|g_{n}-\langle
g_{n}\rangle_{I_{k}}|^{2}\,dx\leqslant
c_{2}\|f_{n}\|_{H^{-1}(\mathbb{R})}^{2}$

for each $N\in\mathbb{N}$. Sending $n\to\infty$, the bound

$\sum_{|k|\leqslant N}\int_{I_{k}}|g-\langle
g\rangle_{I_{k}}|^{2}\,dx\leqslant c_{2}\|f\|_{H^{-1}(\mathbb{R})}^{2}$

appears. Taking $N\to\infty$, one has the right estimate in (5.3). In
particular, it shows that the sum in (5.3) converges. By construction,

$\sum_{k\in\mathbb{Z}}\int_{I_{k}}|g_{n}-\langle
g_{n}\rangle_{I_{k}}|^{2}\,dx=\sum_{-n\leqslant k\leqslant
n-2}\int_{I_{k}}|g-\langle g\rangle_{I_{k}}|^{2}\,dx+\epsilon_{n}\,,$

where $\epsilon_{n}$ is a sum of integrals over
$I_{-n-2},I_{-n-1},I_{n-1},I_{n}$. Since $o\in L^{2}(\mathbb{R})$,

$\lim_{n\to\infty}\int_{I_{k}}|g_{n}-\langle
g_{n}\rangle_{I_{k}}|^{2}\,dx=0,\quad k\in\\{-n-2,-n-1,n-1,n\\}\,.$

Hence, $\lim_{n\to\infty}\epsilon_{n}=0$ and, taking $n\to\infty$ in
inequality

$c_{1}\|f_{n}\|_{H^{-1}(\mathbb{R})}^{2}\leqslant\sum_{k\in\mathbb{Z}}\int_{I_{k}}|g_{n}-\langle
g_{n}\rangle_{I_{k}}|^{2}\,dx\,,$

one gets the left bound in (5.3). Since all antiderivatives are different by a
constant and the integral in (5.3) does not change if we add a constant to
$g$, the proof is finished. ∎

###  5.2. Auxiliary perturbative results for a single interval

Notice that for any real symmetric $2\times 2$ matrix $Q$ with zero trace, we
have that $V=JQ$ is also real, symmetric and has zero trace. The converse
statement is true as well. Hence, the equation $JN_{Q}^{\prime}+QN_{Q}=0$ in
Theorem 5.1, which is equivalent to $N_{Q}^{\prime}=JQN_{Q}$, can be written
as $N_{Q}^{\prime}=VN_{Q}$ with $V$ having the same properties as $Q$. Let
$U_{+}(x,y)$ denote the solution to

$\frac{d}{dx}U_{+}(x,y)=V(x)U_{+}(x,y),\quad U_{+}(y,y)=I$

and $U_{-}(x,y)$ denote the solution to

$\frac{d}{dx}U_{-}(x,y)=-V(x)U_{-}(x,y),\quad U_{-}(y,y)=I\,.$

######  Lemma 5.1.

Suppose $N^{\prime}=VN,N(0)=I$, where $V$ is real-valued, $V\in L^{1}[0,1]$,
$V=V^{*}$, and ${\rm tr}\,V=0$. Then, for $\mathcal{H}=N^{*}N$, we have

$\displaystyle\det\int_{0}^{1}\mathcal{H}(\xi)\,d\xi=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}{\rm
tr}\,\Bigl{(}U^{*}_{+}(x,y)U_{+}(x,y)\Bigr{)}\,dxdy=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}\|U_{+}(x,y)\|^{2}_{\rm
HS}\,dxdy,$ (5.6)
$\displaystyle\det\int_{0}^{1}\mathcal{H}(\xi)\,d\xi-1=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}\left\|\Bigl{(}U_{+}(x,y)-U_{-}(x,y)\Bigr{)}e_{1}\right\|^{2}dxdy\,.$
(5.7)

Proof. Notice that $N,U_{+},U_{-}\in{\rm SL}(2,\mathbb{R})$ and that every
matrix $A\in{\rm SL}(2,\mathbb{R})$ satisfies

$JA^{*}=A^{-1}J,\qquad AJ=J(A^{*})^{-1}\,.$ (5.8)

Also, for any real $2\times 2$ matrix $B$, we have

$\det B=\langle JBe_{1},Be_{2}\rangle=-\langle JBe_{2},Be_{1}\rangle\,.$

Hence,

$\displaystyle\mathcal{I}:=\det\int_{0}^{1}\mathcal{H}(\xi)d\xi=\int_{0}^{1}\int_{0}^{1}\langle
JN^{*}(x)N(x)e_{1},N^{*}(y)N(y)e_{2}\rangle dxdy$
$\displaystyle=-\int_{0}^{1}\int_{0}^{1}\langle
JN^{*}(x)N(x)e_{2},N^{*}(y)N(y)e_{1}\rangle dxdy\,.$

For the second integrand, we have

$\langle JN^{*}(x)N(x)e_{1},N^{*}(y)N(y)e_{2}\rangle=\langle
N^{*}(y)N(y)JN^{*}(x)N(x)e_{1},e_{2}\rangle\,.$

Then, identities (5.8) imply

$N^{*}(y)N(y)JN^{*}(x)N(x)=N^{*}(y)J(N^{*}(y))^{-1}N^{*}(x)N(x)=J(N(y))^{-1}(N^{*}(y))^{-1}N^{*}(x)N(x)$

and, since $Je_{1}=e_{2}$ and $J^{*}=-J$,

$\langle
JN^{*}(x)N(x)e_{1},N^{*}(y)N(y)e_{2}\rangle=\langle(N(y))^{-1}(N^{*}(y))^{-1}N^{*}(x)N(x)e_{1},e_{1}\rangle\,.$

Similarly, $\langle
JN^{*}(x)N(x)e_{2},N^{*}(y)N(y)e_{1}\rangle=-\langle(N(y))^{-1}(N^{*}(y))^{-1}N^{*}(x)N(x)e_{2},e_{2}\rangle\,.$
Hence,

$\displaystyle\mathcal{I}=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}\sum_{j=1}^{2}\langle(N(y))^{-1}(N^{*}(y))^{-1}N^{*}(x)N(x)e_{j},e_{j}\rangle
dxdy=\hskip 85.35826pt$ $\displaystyle\frac{1}{2}\int_{0}^{1}\int_{0}^{1}{\rm
tr}\Bigl{(}(N(y))^{-1}(N^{*}(y))^{-1}N^{*}(x)N(x)\Bigr{)}dxdy=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}{\rm
tr}\Bigl{(}(N^{*}(y))^{-1}N^{*}(x)N(x)(N(y))^{-1}\Bigr{)}dxdy\,.$

Now, we use the formula $N(x)(N(y))^{-1}=U_{+}(x,y)$ to rewrite the last
expression as

$\mathcal{I}=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}{\rm
tr}\Bigl{(}U^{*}_{+}(x,y)U_{+}(x,y)\Bigr{)}dxdy\,.$

Finally, (5.7) follows from $U_{+}(x,y)\in{\rm SL}(2,\mathbb{R})$ by direct
inspection after one uses the identities $JU_{+}(x,y)J=-U_{-}(x,y)$ and ${\rm
tr}(A^{*}A)-2=\|(A+JAJ)e_{1}\|^{2}$, which holds for every $A\in{\rm
SL}(2,\mathbb{R})$. ∎

Remark. The integrand in (5.6) is symmetric: ${\rm
tr}\Bigl{(}U^{*}_{+}(x,y)U_{+}(x,y)\Bigr{)}={\rm
tr}\Bigl{(}U^{*}_{+}(y,x)U_{+}(y,x)\Bigr{)}$ because
$U_{+}(x,y)=U^{-1}_{+}(y,x)$ and $U_{+}(x,y)\in{\rm SL}(2,\mathbb{R})$. Notice
also, that

${\rm
tr}\Bigl{(}U^{*}_{+}(x,y)U_{+}(x,y)\Bigr{)}=\lambda_{x,y}^{2}+\lambda_{x,y}^{-2}\geqslant
2\,,$

where $\lambda_{x,y}$ is an eigenvalue of $U^{*}_{+}(x,y)U_{+}(x,y)$ which
explains why the left-hand side in (5.7) is nonnegative.

######  Lemma 5.2.

Suppose real-valued matrix-function $V=\left(\begin{smallmatrix}v_{1}&v_{2}\\\
v_{2}&-v_{1}\end{smallmatrix}\right)$ is defined on $[0,1]$ and satisfies
$\|V\|_{L^{1}[0,1]}<\infty$. Consider $\mathcal{H}=N^{*}N$, where $N$:
$N^{\prime}=VN,N(0)=I$. Then,

$\det\int_{0}^{1}\mathcal{H}\,dx-1\lesssim\|V\|^{2}_{L^{1}[0,1]}\exp(C\|V\|_{L^{1}[0,1]})\,.$
(5.9)

Proof. The integral equation for $N$ is

$N=I+\int_{0}^{x}VNds\,.$ (5.10)

By Gronwall’s inequality,

$\|N(x)\|\leqslant\exp\left(\int_{0}^{x}\|V(s)\|ds\right)\leqslant\exp(\|V\|_{L^{1}[0,1]}).$
(5.11)

Iteration of (5.10) gives

$N=I+\int_{0}^{x}Vdx_{1}+\int_{0}^{x}V(x_{1})\left(\int_{0}^{x_{1}}V(x_{2})N(x_{2})dx_{2}\right)dx_{1}\,.$

Then,

$\int_{0}^{1}N^{*}Ndx=I+2\int_{0}^{1}\left(\int_{0}^{x}V(x_{1})dx_{1}\right)dx+R,\;\;\|R\|\lesssim\|V\|^{2}_{L^{1}[0,1]}\exp(C\|V\|_{L^{1}[0,1]})\,.$

Since $\text{tr}\,V=0$, the identity $\det(I+A)=1+\text{tr}A+\det A\,,$ which
holds for all $2\times 2$ matrices $A$, gives

$\det\int_{0}^{1}\mathcal{H}dx-1\lesssim\|V\|^{2}_{L^{1}[0,1]}\exp(C\|V\|_{L^{1}[0,1]})\,.$

∎

######  Lemma 5.3.

Suppose real-valued symmetric matrix-functions $V$ and $O$ are defined on
$[0,1]$ and satisfy

$\displaystyle V=\left(\begin{matrix}v_{1}&v_{2}\\\
v_{2}&-v_{1}\end{matrix}\right)=O+O^{\prime},O=O^{*}=\left(\begin{matrix}o_{1}&o_{2}\\\
o_{2}&-o_{1}\end{matrix}\right)\,,$ (5.12)
$\displaystyle\delta:=\|O\|_{L^{2}[0,1]}<\infty\,,$ (5.13) $\displaystyle
d:=\|O^{\prime}\|_{L^{2}[0,1]}<\infty\,.$ (5.14)

Consider $\mathcal{H}=N^{*}N$ where $N^{\prime}=VN,N(0)=I$. Then, we have

$\det\int_{0}^{1}\mathcal{H}\,dx-1=4\sum_{j=1}^{2}\int_{0}^{1}|g_{j}-\langle
g_{j}\rangle|^{2}\,dx+r,\,|r|\lesssim\delta^{2.5}\exp(C(d+\delta))\,,$ (5.15)

where

$g_{j}:=\int_{0}^{x}v_{j}\,dx$ (5.16)

and $C$ is an absolute positive constant. An analogous result holds if $O$ and
$V$ are related by $V=O-O^{\prime}$.

Proof. We will use the formula (5.7) for our analysis. Fix $y\in[0,1]$ and
take $U_{+}(x,y)$ and $U_{-}(x,y)$ which solve
$\frac{d}{dx}U_{+}(x,y)=V(x)U_{+}(x,y),U_{+}(y,y)=I$ and
$\frac{d}{dx}U_{-}(x,y)=-V(x)U_{-}(x,y),U_{-}(y,y)=I$. Iterating the
corresponding integral equations, one gets

$\displaystyle
U_{+}(x,y)=I+\int_{y}^{x}Vdx_{1}+\int_{y}^{x}V\int_{y}^{x_{1}}Vdx_{2}dx_{1}+\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}Vdx_{3}dx_{2}dx_{1}+$
$\displaystyle\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}V\int_{y}^{x_{3}}Vdx_{4}dx_{3}dx_{2}dx_{1}+\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}V\int_{y}^{x_{3}}Vf_{+}dx_{4}dx_{3}dx_{2}dx_{1},$
$\displaystyle f_{+}(x_{4})=\int_{y}^{x_{4}}V(s)U_{+}(s,y)ds\,.$
$\displaystyle
U_{-}(x,y)=I-\int_{y}^{x}Vdx_{1}+\int_{y}^{x}V\int_{y}^{x_{1}}Vdx_{2}dx_{1}-\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}Vdx_{3}dx_{2}dx_{1}+$
$\displaystyle\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}V\int_{y}^{x_{3}}Vdx_{4}dx_{3}dx_{2}dx_{1}-\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}V\int_{y}^{x_{3}}Vf_{-}dx_{4}dx_{3}dx_{2}dx_{1},$
$\displaystyle f_{-}(x_{4})=\int_{y}^{x_{4}}V(s)U_{-}(s,y)ds\,.$

Taking $U_{+}(x,y)-U_{-}(x,y)$ as in (5.7) leaves us with

$\displaystyle\frac{U_{+}(x,y)-U_{-}(x,y)}{2}=\int_{y}^{x}Vdx_{1}+\mathcal{I}_{1}+\mathcal{I}_{2}\,,$
(5.17)
$\displaystyle\mathcal{I}_{1}=\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}Vdx_{3}dx_{2}dx_{1},$
(5.18)
$\displaystyle\mathcal{I}_{2}=\int_{y}^{x}V\int_{y}^{x_{1}}V\int_{y}^{x_{2}}V\int_{y}^{x_{3}}V(f_{+}+f_{-})dx_{4}dx_{3}dx_{2}dx_{1}\,.$
(5.19)

Recall that $V=O+O^{\prime}$ where $O$ satisfies (5.13) and (5.14). These
assumptions are to be used in the following proposition. On
$\mathbb{R}_{+}^{2}$, we define the partial order

$\left[\begin{array}[]{c}x_{1}\\\
x_{2}\end{array}\right]\leqslant\left[\begin{array}[]{c}y_{1}\\\
y_{2}\end{array}\right]$

by requiring that $x_{1}\leqslant y_{1}$ and $x_{2}\leqslant y_{2}$.

######  Proposition 5.2.

Suppose a matrix-function $O$ is defined on $[0,1]$ and denote

$\displaystyle\delta=\|O\|_{L^{2}[0,1]},\,d=\|O^{\prime}\|_{L^{2}[0,1]}.$
(5.20)

Let an operator $G_{(y)}$ be given by:
$F\mapsto(G_{(y)}F)(x)=\int_{y}^{x}(O+O^{\prime})Fds$ where $y\in[0,1]$ and a
matrix-function $F$, defined on $[0,1]$, satisfies
$\|F\|_{L^{\infty}[0,1]}<\infty$ and $\|F^{\prime}\|_{L^{2}[0,1]}<\infty$.
Then,

$\left[\begin{array}[]{l}\|G_{(y)}F\|_{L^{\infty}[0,1]}\\\
\|(G_{(y)}F)^{\prime}\|_{L^{2}[0,1]}\end{array}\right]\leqslant
C\mathcal{M}\left[\begin{array}[]{c}\|F\|_{L^{\infty}[0,1]}\\\
\|F^{\prime}\|_{L^{2}[0,1]}\end{array}\right]\,,\quad\mathcal{M}=\left(\begin{array}[]{ll}\delta+\sqrt{\delta
d}&\delta\\\ \delta+d&0\end{array}\right)\,,$ (5.21)

where $C$ is an absolute positive constant, the norms and derivatives are
computed with respect to $x$.

Proof. Let $b=\|F\|_{L^{\infty}[0,1]},c=\|F^{\prime}\|_{L^{2}[0,1]}$. Write

$O^{*}(x)O(x)-O^{*}(y)O(y)=\int_{y}^{x}((O^{*})^{\prime}O+O^{*}O^{\prime})ds\,.$
(5.22)

Then,

$\|O(x)\|^{2}=\max_{\|\xi\|_{\mathbb{C}^{2}}\leqslant
1}\|O(x)\xi\|^{2}_{\mathbb{C}^{2}}=\max_{\|\xi\|_{\mathbb{C}^{2}}\leqslant
1}\langle
O^{*}(x)O(x)\xi,\xi\rangle\stackrel{{\scriptstyle\eqref{lak}}}{{\leqslant}}\|O(y)\|^{2}+2\int_{0}^{1}\|O^{\prime}(s)\|\cdot\|O(s)\|ds\,.$

Applying Cauchy-Schwarz inequality to the integral, integrating in $y$ from
$0$ to $1$ and maximizing in $x$ gives

$\|O\|_{L^{\infty}[0,1]}\lesssim\delta+(d\delta)^{\frac{1}{2}}\,.$ (5.23)

Then,

$(G_{(y)}F)(x)=\int_{y}^{x}OFds+O(x)F(x)-O(y)F(y)-\int_{y}^{x}OF^{\prime}ds$

and the estimate for the first coordinate in (5.21) follows from Cauchy-
Schwarz inequality and (5.23). Since $(G_{(y)}F)^{\prime}=(O+O^{\prime})F\,,$
we get
$\|(G_{(y)}F)^{\prime}\|_{L^{2}[0,1]}\leqslant(\|O\|_{L^{2}[0,1]}+\|O^{\prime}\|_{L^{2}[0,1]})\|F\|_{L^{\infty}[0,1]}$
and the bound for the second coordinate in (5.21) is obtained. ∎

Continuation of the proof of Lemma 5.3. We apply the proposition to
$\mathcal{I}_{1}$ three times with the initial choice of $F$: $F=I$. That
gives rise to taking the third power of matrix $\mathcal{M}$:
$\mathcal{M}^{3}$, applying it to $(1,0)^{t}$, and looking at the first
coordinate. As the result, one has
$\|\mathcal{I}_{1}\|_{L^{\infty}[0,1]}\lesssim\delta^{\frac{3}{2}}(\delta+d)^{\frac{3}{2}}$.
Therefore,

$\|\mathcal{I}_{1}e_{1}\|_{L^{\infty}([0,1]^{2})}\lesssim\delta^{\frac{3}{2}}\exp(\delta+d).$
(5.24)

Similarly, we consider $\mathcal{I}_{2}$ and use the previous proposition four
times making the first choice of $F$ as $F=f_{+}+f_{-}$. Applying the bound
(5.11) to $U_{+}$ and $U_{-}$, we get
$\|f_{+}+f_{-}\|_{L^{\infty}[0,1]}\lesssim(\delta+d)\exp(\delta+d),\|f^{\prime}_{+}+f_{-}^{\prime}\|_{L^{2}[0,1]}\lesssim(\delta+d)\exp(\delta+d)$.
This time, we compute the fourth power of matrix
$\mathcal{M}:\mathcal{M}^{4}$, apply it to vector
$(\delta+d)\exp(\delta+d)(1,1)^{t}$, and look at the first coordinate. In the
end, one has

$\|\mathcal{I}_{2}e_{1}\|_{L^{\infty}([0,1]^{2})}\lesssim\delta^{2}\exp(C(d+\delta))\,.$
(5.25)

The first term in (5.17) can be written as

$\int_{y}^{x}Vds=\int_{y}^{x}Ods+O(x)-O(y)$

and

$\left\|\int_{y}^{x}Ods+O(x)-O(y)\right\|_{L^{2}([0,1]^{2})}\lesssim\delta\,.$
(5.26)

For any three vectors $v_{1},v_{2}$ and $v_{3}$ in $\mathbb{R}^{2}$, we have
an estimate

$|\|v_{1}+v_{2}+v_{3}\|-\|v_{1}\||\leqslant\|v_{2}+v_{3}\|\leqslant\|v_{2}\|+\|v_{3}\|\,,$

which follows from the triangle inequality. Multiplying with

$\|v_{1}+v_{2}+v_{3}\|+\|v_{1}\|\leqslant 2\|v_{1}\|+\|v_{2}\|+\|v_{3}\|\,,$

we get

$|\|v_{1}+v_{2}+v_{3}\|^{2}-\|v_{1}\|^{2}|\leqslant
2\|v_{1}\|(\|v_{2}\|+\|v_{3}\|)+(\|v_{2}\|+\|v_{3}\|)^{2}\,.$

Applying it to (5.17) gives

$\displaystyle\left|\frac{1}{4}\|(U_{+}(x,y)-U_{-}(x,y))e_{1}\|^{2}-\left\|\left(\int_{y}^{x}Vds\right)e_{1}\right\|^{2}\right|\hskip
113.81102pt$
$\displaystyle\lesssim\left\|\left(\int_{y}^{x}Vds\right)e_{1}\right\|\cdot(\|\mathcal{I}_{1}e_{1}\|+\|\mathcal{I}_{2}e_{1}\|)+\|\mathcal{I}_{1}e_{1}\|^{2}+\|\mathcal{I}_{2}e_{1}\|^{2}\,.$

Taking $L^{1}([0,1]^{2})$ norm in variables $x$ and $y$ of both sides and
using (5.24), (5.25), (5.26) and the Cauchy-Schwartz inequality gives

$\frac{1}{4}\int_{0}^{1}\int_{0}^{1}\|(U_{+}(x,y)-U_{-}(x,y))e_{1}\|^{2}dxdy=\int_{0}^{1}\int_{0}^{1}\left\|\left(\int_{y}^{x}Vds\right)e_{1}\right\|^{2}dxdy+r,\,\,|r|\lesssim\delta^{2.5}\exp(C(d+\delta))\,.$

Recalling the definition (5.16), we get

$\left\|\left(\int_{y}^{x}Vds\right)e_{1}\right\|^{2}=\sum_{j=1}^{2}(g_{j}(x)-g_{j}(y))^{2}$

so

$\frac{1}{2}\int_{0}^{1}\int_{0}^{1}\|(U_{+}(x,y)-U_{-}(x,y))e_{1}\|^{2}dxdy=4\sum_{j=1}^{2}\int_{0}^{1}|g_{j}-\langle
g_{j}\rangle|^{2}dx+r,\,\,|r|\lesssim\delta^{2.5}\exp(C(d+\delta))\,.$

Lemma 5.3 is proved. ∎

Remark. All statements in this subsection can be easily adjusted to any
interval but the constants in the inequalities will depend on the size of that
interval.

###  5.3. Rough bound when $\widetilde{\mathcal{K}}_{Q}$ is small

######  Lemma 5.4.

Suppose an absolutely continuous function $f$ is defined on $[0,1]$ and
satisfies

$\displaystyle f\in L^{2}[0,1],\quad f^{\prime}=l_{1}+l_{2},\quad l_{1}\in
L^{1}[0,1],\quad l_{2}\in L^{2}[0,1]\,.$ (5.27)

Then,
$\|f\|_{L^{\infty}[0,1]}\leqslant\sqrt{\delta^{2}+2(\delta\tau+\epsilon(\tau+\epsilon+\delta))},$
where
$\delta=\|f\|_{L^{2}[0,1]},\,\epsilon=\|l_{1}\|_{L^{1}[0,1]},\,\tau=\|l_{2}\|_{L^{2}[0,1]}\,.$

Proof. There is $\xi\in[0,1]$ such that $|f(\xi)|\leqslant\delta$ and

$|f(x)-f(\xi)|\leqslant\left|\int_{\xi}^{x}f^{\prime}ds\right|\leqslant\tau+\epsilon\,.$

Thus, $\|f\|_{L^{\infty}[0,1]}\leqslant\tau+\epsilon+\delta$. Then, writing

$f^{2}(x)-f^{2}(y)=2\int_{y}^{x}ff^{\prime}ds\,,$

integrating in $y$ and maximizing in $x$, we get

$\|f\|^{2}_{L^{\infty}[0,1]}\leqslant\delta^{2}+2(\delta\tau+\epsilon(\tau+\epsilon+\delta))\,.$

∎

Suppose $Q$ is real-valued, symmetric matrix-function on $\mathbb{R}$ with
zero trace and $\|Q\|_{L^{2}(\mathbb{R})}<\infty$. Define
$\mathcal{H}_{Q}=N^{*}N$, where $N:N^{\prime}=JQN,N(0)=I$. Notice that
$\det\int_{n}^{n+2}S^{*}\mathcal{H}_{Q}Sdx=\det\int_{n}^{n+2}\mathcal{H}_{Q}dx$
for every constant matrix $S\in{\rm SL}(2,\mathbb{R})$. Therefore, we can
apply Lemma 5.2 to each interval $[n,n+2]$ by choosing $S=N^{-1}(n)$ and get
an estimate which explains how $\|Q\|_{L^{2}(\mathbb{R})}$ controls
$\widetilde{\mathcal{K}}_{Q}$:

$\widetilde{\mathcal{K}}_{Q}=\sum_{n\in\mathbb{Z}}\left(\det\int_{n}^{n+2}\mathcal{H}_{Q}dx-4\right)\lesssim\sum_{n\in\mathbb{Z}}\|Q\|^{2}_{L^{2}[n,n+2]}\exp(C\|Q\|_{2})\lesssim\|Q\|^{2}_{L^{2}(\mathbb{R})}\exp(C\|Q\|_{2})\,.$

The next lemma shows that $\widetilde{\mathcal{K}}_{Q}$ controls the
convolution of $Q$ with the exponential.

######  Lemma 5.5.

Suppose $Q$ is real-valued, symmetric $2\times 2$ matrix-function on
$\mathbb{R}$ with zero trace and entries in $L^{2}(\mathbb{R})$. Define
$\mathcal{H}_{Q}=N^{*}N$ where $N:N^{\prime}=JQN,N(0)=I$ and assume that
$\widetilde{\mathcal{K}}_{Q}<\infty$. If $O:=e^{x}\int_{x}^{\infty}e^{-s}Qds$,
then
$\|O\|_{L^{\infty}(\mathbb{R})}\lesssim\exp(C(\|Q\|_{L^{2}(\mathbb{R})}+\widetilde{\mathcal{K}}_{Q}))\widetilde{\mathcal{K}}_{Q}^{{\frac{1}{4}}}$
where $C$ is a positive absolute constant.

Proof. Let $R=\|Q\|_{L^{2}(\mathbb{R})}$ and $E=\widetilde{\mathcal{K}}_{Q}$.
We split the proof into several steps.

1\. Bound for a single interval $[0,1]$. The definitions (3.5) and (3.14)
imply that $\widetilde{\mathcal{K}}_{Q}^{+}\leqslant E$. From Theorem 1.2 and
Theorem 3.2 in [3], we know that $\mathcal{H}_{Q}$ admits the following
factorization on $\mathbb{R}_{+}$: $\mathcal{H}_{Q}=G^{*}WG$ where $G$ and $W$
satisfy conditions:

$\displaystyle
G^{\prime}=J(v_{1}+v_{2})G,\quad\|v_{1}\|_{L^{1}(\mathbb{R}_{+})}\lesssim
E,\quad\|v_{2}\|_{L^{2}(\mathbb{R}_{+})}\lesssim E^{{\frac{1}{2}}}\,,$ (5.28)
$\displaystyle\det G=1,\quad v_{1}+v_{2}=(v_{1}+v_{2})^{*}\,,$ (5.29)

and

$\displaystyle W\geqslant 0,\quad\det
W=1,\quad\|\text{tr}\,W-2\|_{L^{1}(\mathbb{R}_{+})}\lesssim E\,.$

Since $\|\text{tr}\,W-2\|_{L^{1}[0,1]}\lesssim E$, we have
$\|\lambda+\lambda^{-1}-2\|_{L^{1}[0,1]}\lesssim E$, where $\lambda$ is the
largest eigenvalue of $W$. If one denotes
$p=\text{tr}\,W-2=\lambda+\lambda^{-1}-2$, then

$\lambda=\frac{2+p+\sqrt{4p+p^{2}}}{2}\,,\quad\lambda^{-1}=\frac{2+p-\sqrt{4p+p^{2}}}{2}\,.$
(5.30)

In particular, that yields

$\int_{0}^{1}\|W\|\,dx\lesssim 1+E\,.$ (5.31)

The given conditions on $Q$ and (5.11) yield

$\|N(x)\|,\|N^{-1}(x)\|\lesssim\exp(CR),\quad x\in[0,1]\,,$

where the second estimate follows from the first since $\det N=1$. The
Hamiltonian $\mathcal{H}_{Q}=N^{*}N$ is absolutely continuous on
$\mathbb{R}_{+}$ and

$0<\exp(-CR){I}\lesssim\mathcal{H}_{Q}(x)\lesssim\exp(CR){I}$ (5.32)

on $[0,1]$. We claim that $\|G(0)\|\lesssim\exp(C(R+E))$ and that
$\|G^{-1}(0)\|\lesssim\exp(C(R+E))$. Indeed, if $X$ satisfies
$X^{\prime}=J(v_{1}+v_{2})X$ and $X(0)=I$, then $G=XG(0)$. Moreover, given
conditions on $v_{1}$ and $v_{2}$ and $\det X=1$, we have

$\|X(x)\|\lesssim\exp(CE),\qquad\|X^{-1}(x)\|\lesssim\exp(CE)$ (5.33)

uniformly on $[0,1]$. Identity $\mathcal{H}_{Q}=G^{*}(0)X^{*}WXG(0)$ yields

$(G^{*}(0))^{-1}\mathcal{H}_{Q}(G(0))^{-1}=X^{*}WX\,.$

Taking an arbitrary $\xi\in\mathbb{C}^{2}$ with $\|\xi\|_{\mathbb{C}^{2}}=1$,
we get

$\displaystyle\|G^{-1}(0)\xi\|^{2}\stackrel{{\scriptstyle\eqref{tot}}}{{\lesssim}}\exp(CR)\int_{0}^{1}\langle\mathcal{H}_{Q}G^{-1}(0)\xi,G^{-1}(0)\xi\rangle
dx\hskip 85.35826pt$ $\displaystyle=\exp(CR)\int_{0}^{1}\langle
WX\xi,X\xi\rangle
dx\stackrel{{\scriptstyle\eqref{P2p55}+\eqref{tot1}}}{{\lesssim}}\exp(C(R+E))\,,$

which implies $\|G^{-1}(0)\|\lesssim\exp(C(R+E))$. We also have
$\|G(0)\|\lesssim\exp(C(R+E))$ since $\det G=1$ and the claim is proved.
Finally, we have

$\|G(x)\|\lesssim\exp(C(R+E)),\quad\|G^{-1}(x)\|\lesssim\exp(C(R+E))$

for $x\in[0,1]$ since $G=XG(0)$.

Next, let us study $W$ and $W^{\prime}$. Since $W=(G^{*})^{-1}N^{*}NG^{-1}$,
one has $\|W\|\lesssim\exp(C(R+E))$ on $x\in[0,1]$. Recall that $W\geqslant 0$
and $\det W=1$, so

$\exp(-C(R+E)){I}\lesssim W\lesssim\exp(C(R+E)){I},\quad x\in[0,1]\,.$

Since $\lambda$ is the largest eigenvalue of $W$ and
$\lambda\lesssim\exp(C(R+E))$, then (5.30) yields
$\|\lambda-1\|_{L^{2}[0,1]}\lesssim E^{{\frac{1}{2}}}\exp(C(R+E))$ and
$\|\lambda^{-1}-1\|_{L^{2}[0,1]}\lesssim E^{{\frac{1}{2}}}\exp(C(R+E))$.
Introduce $\Upsilon=W-I\,.$ The matrix $\Upsilon$ is unitarily equivalent to
$\left(\begin{smallmatrix}\lambda-1&0\\\
0&1/\lambda-1\end{smallmatrix}\right)$ and that gives

$\|\Upsilon\|_{L^{2}[0,1]}\lesssim E^{{\frac{1}{2}}}\exp(C(R+E))\,.$ (5.34)

We need to study $\Upsilon^{\prime}$, which is equal to $W^{\prime}$. To do
so, notice that

$\displaystyle
2N^{*}JQN=\mathcal{H}_{Q}^{\prime}=G^{*}J(v_{1}+v_{2})WG+G^{*}WJ(v_{1}+v_{2})G+G^{*}W^{\prime}G\,.$
(5.35)

Hence,

$\Upsilon^{\prime}=W^{\prime}=F_{1}+F_{2}\,,$

where

$F_{1}=-J(v_{1}+v_{2})W-WJ(v_{1}+v_{2}),\quad
F_{2}=2(G^{*})^{-1}N^{*}JQNG^{-1}.$

The previously obtained estimates give us

$\|F_{1}\|_{L^{1}[0,1]}\lesssim
E^{{\frac{1}{2}}}\exp(C(R+E)),\quad\|F_{2}\|_{L^{2}[0,1]}\lesssim\exp(C(R+E))\,.$
(5.36)

Now, we use (5.34), (5.36) to apply the previous lemma to each component of
$\Upsilon$ to obtain

$\|\Upsilon\|_{L^{\infty}[0,1]}\lesssim E^{{\frac{1}{4}}}\exp(C(R+E))\,.$
(5.37)

The formula (5.35) also gives an expression for $Q$:

$Q=-J(H_{1}+H_{2})\,,$

where

$H_{1}=0.5(N^{*})^{-1}(G^{*}J(v_{1}+v_{2})WG+G^{*}WJ(v_{1}+v_{2})G)N^{-1}$

and

$H_{2}=0.5(N^{*})^{-1}(G^{*}\Upsilon^{\prime}G)N^{-1}\,.$

Since $\|H_{1}\|_{L^{1}[0,1]}\lesssim E^{{\frac{1}{2}}}\exp(C(R+E))$, we have

$\left\|e^{x}\int_{x}^{1}e^{-s}H_{1}ds\right\|_{L^{\infty}[0,1]}\lesssim
E^{{\frac{1}{2}}}\exp(C(R+E))\,.$

For smooth matrix-functions $u_{1}$, $u_{2}$, $u_{3}$, we have

$\int^{1}_{x}u_{1}u^{\prime}_{2}u_{3}\,ds=u_{1}u_{2}u_{3}\Bigr{\rvert}^{1}_{x}-\int^{1}_{x}u^{\prime}_{1}u_{2}u_{3}\,ds-\int^{1}_{x}u_{1}u_{2}u_{3}^{\prime}\,ds.$

Then,

$\displaystyle 2e^{x}\int_{x}^{1}e^{-s}H_{2}ds=\hskip 142.26378pt$
$\displaystyle
e^{x}\left(e^{-1}(N^{*}(1))^{-1}G^{*}(1)\Upsilon(1)G(1)(N(1))^{-1}-e^{-x}(N^{*}(x))^{-1}G^{*}(x)\Upsilon(x)G(x)(N(x))^{-1}\right)$
$\displaystyle-e^{x}\int_{x}^{1}(e^{-s}(N^{*}(s))^{-1}G^{*})^{\prime}\Upsilon
GN^{-1}ds-e^{x}\int_{x}^{1}e^{-s}(N^{*}(s))^{-1}G^{*}\Upsilon(GN^{-1})^{\prime}ds\,.$

Since $\|(N^{-1})^{\prime}\|_{L^{2}[0,1]}\lesssim\exp(C(R+E))$ and
$\|G^{\prime}\|_{L^{1}[0,1]}\lesssim\exp(C(R+E))$, we have

$\left\|e^{x}\int_{x}^{1}e^{-s}H_{2}ds\right\|_{L^{\infty}[0,1]}\lesssim\|\Upsilon\|_{L^{\infty}[0,1]}\exp(C(R+E))\stackrel{{\scriptstyle\eqref{tot9}}}{{\lesssim}}E^{{\frac{1}{4}}}\exp(C(R+E))\,.$

Summing up, we get

$\left\|e^{x}\int_{x}^{1}e^{-s}Qds\right\|_{L^{\infty}[0,1]}\lesssim
E^{{\frac{1}{4}}}\exp(C(R+E))\,.$ (5.38)

2\. Handling all intervals $[n,n+1],n\in\mathbb{Z}$. Take any
$n\in\mathbb{Z}$. Our immediate goal is to show the bound

$\left\|e^{x}\int_{x}^{n+1}e^{-s}Qds\right\|_{L^{\infty}[n,n+1]}\lesssim
E^{{\frac{1}{4}}}\exp(C(R+E))$ (5.39)

analogous to (5.38) but written for interval $[n,n+1]$. To this end, take the
Hamiltonian $\mathcal{H}^{(n)}(x):=\mathcal{H}_{Q}(x+n)$ defined on
$\mathbb{R}_{+}$. For the corresponding $\widetilde{\mathcal{K}}_{(n)}^{+}$,
we get $\widetilde{\mathcal{K}}_{(n)}^{+}\leqslant E$ as follows from its
definition. Since the $\widetilde{\mathcal{K}}$-characteristics of the
Hamiltonians $\mathcal{H}$ and $S^{*}\mathcal{H}S$ are equal for every
constant matrix $S\in{\rm SL}(2,\mathbb{R})$, we can instead consider
$\widehat{\mathcal{H}}^{(n)}=\widehat{N}^{*}\widehat{N}$ where
$\widehat{N}^{\prime}=JQ(x+n)\widehat{N},\widehat{N}(0)=I$. Using the
arguments in step 1 for $\widehat{\mathcal{H}}^{(n)}$, we get (5.39).

3\. Summing up. Denote
$O_{n}(x)=e^{x}\int_{x}^{\infty}e^{-s}Q\cdot\chi_{n<s<n+1}ds$ and notice that
$O=\sum_{n\in\mathbb{Z}}O_{n}$. Then, since $O_{n}(x)=0$ for $x>n+1$ and
$\|O_{n}(x)\|\lesssim e^{x-n}\|O_{n}\|_{L^{\infty}[n,n+1]}$ for $x<n$, we get

$\|O(x)\|\leqslant\sum_{n\in\mathbb{Z}}\|O_{n}(x)\|\lesssim
E^{{\frac{1}{4}}}\exp(C(R+E))\sum_{n\geqslant 0}e^{-n}\sim
E^{{\frac{1}{4}}}\exp(C(R+E))$

as follows from (5.39). That finishes the proof of Lemma 5.5. ∎

###  5.4.  Proof of Theorem 4.1

Denote $E=\widetilde{\mathcal{K}}_{Q}$, $O=e^{x}\int_{x}^{\infty}e^{-s}Qds$,
and recall that
$\|O\|_{L^{2}(\mathbb{R})}\sim\|Q\|_{H^{-1}(\mathbb{R})}\leqslant\|Q\|_{L^{2}(\mathbb{R})}$.

1\. Lower bound. Define $\delta_{n}=\|O\|_{L^{2}[n,n+1]}$. By Lemma 5.5, we
know that $\sup_{n}\delta_{n}\lesssim E^{{\frac{1}{4}}}\exp(C(R+E))$. Next, we
apply Lemma 5.3 to each interval $[n,n+2]$. The remainder $r_{n}$ in that
lemma allows the estimate

$r_{n}\lesssim(\delta_{n}+\delta_{n+1})^{2.5}\exp(C(\delta_{n}+\delta_{n+1}+R)),\quad
n\in\mathbb{Z}\,.$

For each $R>0$ and $\eta>0$, we can find a positive $E_{0}(R,\eta)$ such that
$E\in(0,E_{0}(R,\eta))$ implies that the remainder $r_{n}$ is smaller than
${\eta}(\delta_{n}^{2}+\delta_{n+1}^{2})$ uniformly in all $n$. For example,
one can take

$E_{0}(R,\eta)\sim e^{-C_{\eta}R},$ (5.40)

where $C_{\eta}$ is a sufficiently large positive number that depends on
$\eta$. Therefore, for such $E$ and some positive constant $c$ independent of
$\eta$, we have

$\displaystyle\sum_{n\in\mathbb{Z}}(c-\eta)\delta_{n}^{2}\lesssim\sum_{n\in\mathbb{Z}}\left(\det\int_{n}^{n+2}\mathcal{H}_{Q}dx-4\right)\lesssim\sum_{n\in\mathbb{Z}}(c+\eta)\delta_{n}^{2},$

where the Proposition 5.1 has been applied to the terms
$\int_{n}^{n+2}|g_{j}-\langle g_{j}\rangle|^{2}\,dx$ in the right-hand side of
(5.15), adjusted to the interval $[n,n+2]$, to show that they are comparable
to $\delta_{n}^{2}+\delta_{n+1}^{2}$. Taking $\eta=c/2$, we see that

$E=\sum_{n\in\mathbb{Z}}\left(\det\int_{n}^{n+2}\mathcal{H}_{Q}dx-4\right)\sim\sum_{n\in\mathbb{Z}}\delta_{n}^{2}\sim\|O\|_{L^{2}(\mathbb{R})}^{2},$

in the case $E\leqslant E_{0}(R,\frac{c}{2})$. If $E>E_{0}(R,\frac{c}{2})$,
one uses inequality $\|O\|_{L^{2}(\mathbb{R})}\lesssim R$ to get

$e^{-CR}\|O\|^{2}_{L^{2}(\mathbb{R})}\lesssim\frac{E_{0}(R,\frac{c}{2})}{1+R^{2}}\|O\|^{2}_{L^{2}(\mathbb{R})}\lesssim
E\,,$ (5.41)

which holds for some positive absolute constant $C$ due to (5.40). That
provides the required lower bound.

2\. Upper bound. Let $\delta_{n}=\|O\|_{L^{2}[n,n+1]}\,.$ For given value of
$R$, apply Lemma 5.3 and Proposition 5.1 to each interval $[n,n+2]$. That
gives

$E\lesssim\sum_{n\in\mathbb{Z}}\delta^{2}_{n}e^{C(R+\delta_{n})}$

with an absolute constant $C$. Since
$\sum_{n\in\mathbb{Z}}\delta_{n}^{2}\sim\|q\|_{H^{-1}(\mathbb{R})}^{2}$ and
$\|q\|_{H^{-1}(\mathbb{R})}\lesssim R$, one has

$E\lesssim\|q\|_{H^{-1}(\mathbb{R})}^{2}e^{C(R+\|q\|_{H^{-1}(\mathbb{R})})}\lesssim\|q\|_{H^{-1}(\mathbb{R})}^{2}e^{C_{2}R}\,.$

∎

##  6\. Appendix

Here we collect some auxiliary results used in the main text.

1. We start with an example that shows that the scattering transform is not injective when defined on $q\in L^{2}(\mathbb{R})$. This is an analog of Lemma 17 in [24].

######  Example 6.1.

There exist potentials $q_{1},q_{2}\in L^{2}(\mathbb{R})$ such that $q_{1}\neq
q_{2}$ in $L^{2}(\mathbb{R})$ but we have ${\bf r}_{q_{1}}={\bf r}_{q_{2}}$
a.e. on $\mathbb{R}$ for their reflection coefficients. In other words, the
scattering transform $q\mapsto{\bf r}_{q}$ is not injective on
$L^{2}(\mathbb{R})$.

Proof. Let us consider

$\mathfrak{a}_{1}^{+}=1,\quad\mathfrak{b}_{1}^{+}=0,\quad\mathfrak{a}_{1}^{-}=\mathfrak{a},\quad\mathfrak{b}_{1}^{-}=\mathfrak{b},$

and

$\mathfrak{a}_{2}^{+}=\mathfrak{a},\quad\mathfrak{b}_{2}^{+}=\mathfrak{b},\quad\mathfrak{a}_{2}^{-}=1,\quad\mathfrak{b}_{2}^{-}=0,$

where $\mathfrak{a}=1+i/x$, $\mathfrak{b}=i/x$. Note that

$\int_{\mathbb{R}}\log(1-|s_{k}^{\pm}(x)|^{2})\,dx>-\infty,\qquad
s_{k}^{\pm}=\frac{\mathfrak{b}^{\pm}}{\mathfrak{a}_{k}^{\pm}},\qquad k=1,2.$

Theorem 12.11 in [12] says that for every contractive analytic function $s$ on
$\mathbb{C}_{+}$ whose boundary values on $\mathbb{R}$ satisfy
$\log(1-|s|^{2})\in L^{1}(\mathbb{R})$ there exists a unique coefficient $A\in
L^{2}(\mathbb{R}_{+})$ such that
$s=\lim_{\xi\to+\infty}\frac{\mathfrak{B}(\xi,\lambda)}{\mathfrak{A}(\xi,\lambda)}$,
$\lambda\in\mathbb{C}_{+}$ for the continuous Wall polynomials generated by
$A$. Moreover, we have

$2\pi\|A\|_{L^{2}(\mathbb{R}_{+})}^{2}=\|\log(1-|s|^{2})\|_{L^{1}(\mathbb{R}_{+})}.$
(6.1)

Applying this result, we see that there exist functions
$A_{1}^{\pm},A_{2}^{\pm}\in L^{2}(\mathbb{R}_{+})$ such that
$\mathfrak{a}^{\pm}_{1,2}$, $\mathfrak{b}_{1,2}^{\pm}$ are the limits of their
continuous Wall polynomials. Now define potentials $q_{1,2}\in
L^{2}(\mathbb{R})$ by relations

$A_{1,2}^{+}(\xi)=-\overline{q_{1,2}(\xi/2)}/2,\qquad
A_{1,2}^{-}(\xi)=q_{1,2}(-\xi/2)/2,\qquad\xi\in\mathbb{R}_{+}.$

From Proposition 2.2, we conclude that the coefficients $a_{1,2}$, $b_{1,2}$
for these potentials satisfy

$a_{1}=\mathfrak{a}=a_{2},\qquad
b_{1}=-\mathfrak{b}=\overline{\mathfrak{b}}=b_{2},$

on $\mathbb{R}\setminus\\{0\\}$. Hence, ${\bf r}_{q_{1}}={\bf r}_{q_{2}}$ on
$\mathbb{R}\setminus\\{0\\}$. On the other hand, we have $A_{1}^{+}=0$ and
$A_{2}^{-}=0$ by construction. It follows that
$\mathop{\mathrm{supp}}\nolimits q_{1}\subset(-\infty,0]$ and
$\mathop{\mathrm{supp}}\nolimits q_{2}\subset[0,+\infty)$. Since $q_{1}$ and
$q_{2}$ are nonzero (they have a nonzero $L^{2}(\mathbb{R})$-norm as follows
from (6.1)), that yields $q_{1}\neq q_{2}$ in $L^{2}(\mathbb{R})$. ∎

2. Next, we outline how to prove that the spectral representation for the Dirac operator $\mathcal{D}_{Q}$, defined by relation (3.1), is given by the Weyl-Titchmarsh transform (3.10). To this end, we will use the corresponding result for canonical Hamiltonian systems proved in [21].

At first, we note that if $\mathcal{H}_{Q}=N_{Q}^{*}N_{Q}$ is the Hamiltonian
from Theorem 3.1, then $\det\mathcal{H}_{Q}=1$ on $\mathbb{R}$ and the
operator $V:X\mapsto N_{Q}^{-1}X$ is unitary from
$L^{2}(\mathbb{R},\mathbb{C}^{2})$ onto the Hilbert space

$L^{2}(\mathcal{H}_{Q})=\Bigl{\\{}Y:\mathbb{R}\to\mathbb{C}^{2}:\;\|Y\|_{L^{2}(\mathcal{H}_{Q},\mathbb{R})}^{2}=\int_{\mathbb{R}}\langle\mathcal{H}_{Q}(\xi)Y(\xi),Y(\xi)\rangle_{\mathbb{C}^{2}}\,d\xi<\infty\Bigr{\\}}.$

Moreover, $V\mathcal{D}_{Q}V^{-1}$ coincides with the operator
$\mathcal{D}_{\mathcal{H}_{Q}}:Y\mapsto\mathcal{H}^{-1}JY^{\prime}$ of the
canonical Hamiltonian system generated by the Hamiltonian $\mathcal{H}_{Q}$.
Thus, the operator $\mathcal{D}_{Q}$ on $L^{2}(\mathbb{R},\mathbb{C}^{2})$ is
unitary equivalent to the operator $\mathcal{D}_{\mathcal{H}_{Q}}$ on
$L^{2}(\mathcal{H}_{Q})$. Let $\widetilde{M}$ be the solution of Cauchy
problem

$J\widetilde{M}^{\prime}(\xi,z)=z\mathcal{H}_{Q}(\xi)\widetilde{M}(\xi,z),\qquad\widetilde{M}(0,z)=\left(\begin{smallmatrix}1&0\\\
0&1\\\ \end{smallmatrix}\right),$ (6.2)

where $z\in\mathbb{C}$, $\xi\in\mathbb{R}$, and the differentiation is taken
with respect to $\xi\in\mathbb{R}$. The Weyl-Titchmarsh transform for
$\mathcal{D}_{\mathcal{H}_{Q}}$ is defined by

$\mathcal{F}_{\mathcal{D}_{\mathcal{H}_{Q}}}:Y\mapsto\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\widetilde{M}(\xi,\lambda)^{*}\mathcal{H}_{Q}(\xi)Y(\xi)\,d\xi$

on a dense subset of $L^{2}(\mathcal{H}_{Q})$ of smooth compactly supported
functions. This operator is unitary from $L^{2}(\mathcal{H}_{Q})$ onto the
space $L^{2}(\rho)$ defined in the same way as at the beginning of Section 3.
Specifically, we let $m_{\pm}$ be the half-line Weyl functions of
$\mathcal{H}_{Q}$ and define $\rho$ as the representing measure for the
matrix-valued Herglotz function $m$ in (3.8). It was proved in Theorem 3.21 in
[21] that
$\mathcal{F}_{\mathcal{D}_{\mathcal{H}_{Q}}}\mathcal{D}_{\mathcal{H}_{Q}}\mathcal{F}_{\mathcal{D}_{\mathcal{H}_{Q}}}^{-1}$
coincides with the operator of multiplication by the independent variable in
$L^{2}(\rho)$. We also have

$\mathcal{F}_{\mathcal{D}_{\mathcal{H}_{Q}}}(VX)=\mathcal{F}_{\mathcal{D}_{{Q}}}X,\qquad
X\in L^{2}(\mathbb{R},\mathbb{C}^{2}).$

Thus, we only need to check that the Weyl functions $m_{\pm}$ used in Section
3 coincide with the half-line Weyl functions of the Hamiltonian
$\mathcal{H}_{Q}$. For the $\mathbb{R}_{+}$-Weyl functions this follows from
Lemma 6.1 below. Comparing the formulas for $A^{+}$, $A^{-}$ in the beginning
of Section 3, we see that the Weyl function $m_{-}$ for $\mathcal{D}_{Q}$
corresponds to the Weyl function $m_{+}$ for $\mathcal{D}_{\widetilde{Q}}$
where $\widetilde{Q}(\xi)=\sigma_{3}Q(-\xi)\sigma_{3}$. Similarly, in the
setting of canonical Hamiltonian systems, the Weyl function $m_{-}$ for
$\mathcal{D}_{\mathcal{H}_{Q}}$ coincides with the Weyl function $m_{+}$ of
$\mathcal{D}_{\widetilde{\mathcal{H}}_{Q}}$,
$\widetilde{\mathcal{H}}_{Q}(\xi)=\sigma_{3}\mathcal{H}(-\xi)\sigma_{3}$.
Therefore, the statement for $A^{-}$ follows from Lemma 6.1 below and from the
relation
$\widetilde{\mathcal{H}}_{Q}=\sigma_{3}\mathcal{H}_{Q}\sigma_{3}=(\sigma_{3}N_{Q}^{*}\sigma_{3})(\sigma_{3}N_{Q}\sigma_{3})=\mathcal{H}_{\widetilde{Q}}$.

######  Lemma 6.1.

Let $q\in L^{2}(\mathbb{R}_{+})$. Define

$Q(\xi)=\begin{pmatrix}-\operatorname{Im}q(\xi)&-\operatorname{Re}q(\xi)\\\
-\operatorname{Re}q(\xi)&\operatorname{Im}q(\xi)\end{pmatrix},\qquad
A(\xi)=-\overline{q(\xi/2)}/2,\qquad\xi\in\mathbb{R}_{+}.$

Let $N_{Q}$ be defined by
$JN^{\prime}_{Q}(\xi,\lambda)+Q(\xi)N_{Q}(\xi,\lambda)=\lambda
N_{Q}(\xi,\lambda)$, $N_{Q}(0,\lambda)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\
\end{smallmatrix}\right)$. Consider the Hamiltonian
$\mathcal{H}_{Q}=N_{Q}^{*}(\xi,0)N_{Q}(\xi,0)$ on $\mathbb{R}_{+}$ and let
$\widetilde{M}=\left(\begin{smallmatrix}\widetilde{M}_{11}&\widetilde{M}_{12}\\\
\widetilde{M}_{21}&\widetilde{M}_{22}\end{smallmatrix}\right)$ be defined by
$J\widetilde{M}^{\prime}(\xi,z)=z\mathcal{H}_{Q}(\xi)\widetilde{M}(\xi,z)$,
$\widetilde{M}(0,z)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\
\end{smallmatrix}\right)$. Let, finally, $P$, $P_{*}$, $\widehat{P}$,
$\widehat{P}_{*}$ be the solutions to Krein systems (2.11), (2.12) for the
coefficient $A$ on $\mathbb{R}_{+}$. Then,

$\lim_{\xi\to+\infty}\frac{\widetilde{M}_{22}(\xi,z)}{\widetilde{M}_{21}(\xi,z)}=\lim_{\xi\to+\infty}\frac{(N_{Q})_{22}(\xi,z)}{(N_{Q})_{21}(\xi,z)}=\lim_{\xi\to+\infty}i\frac{\widehat{P}_{*}(\xi,z)}{P_{*}(\xi,z)},\qquad
z\in\mathbb{C}_{+}.$ (6.3)

In other words, the function $m_{+}$ in (3.7) is the half-line Weyl function
for the operators $\mathcal{D}_{\mathcal{H}_{Q}}$, $\mathcal{D}_{Q}$.

Proof. The formula

$\lim_{\xi\to+\infty}\frac{\widetilde{M}_{22}(\xi,z)}{\widetilde{M}_{21}(\xi,z)}=\lim_{\xi\to+\infty}\frac{(N_{Q})_{22}(\xi,z)}{(N_{Q})_{21}(\xi,z)}$

for $\mathcal{D}_{Q}$ and $\mathcal{D}_{\mathcal{H}_{Q}}$ is well-known and
can be derived from the analysis of Weyl circles by using identity
$N_{Q}(\xi,\lambda)=N_{Q}(\xi,0)\widetilde{M}(\xi,\lambda)$ and the invariance
of Weyl circles under transforms generated by $J$-unitary matrices (in our
setting, the $J$-unitary matrix is $N_{Q}(\xi,0)$: we have
$N_{Q}^{*}(\xi,0)JN_{Q}(\xi,0)=J$ on $\mathbb{R}$). See, e.g., [4] or Section
8 in [22] for more details on Weyl circles for canonical Hamiltonian systems.
Thus, we focus on the second identity in (6.3) and define

$X(\xi,z)=e^{-i\xi
z}\begin{pmatrix}\displaystyle\frac{P(2\xi,z)+P_{*}(2\xi,z)}{2}&\displaystyle\frac{\widehat{P}(2\xi,z)-\widehat{P}_{*}(2\xi,z)}{2i}\\\
\displaystyle\frac{P_{*}(2\xi,z)-P(2\xi,z)}{2i}&\displaystyle\frac{\widehat{P}(2\xi,z)+\widehat{P}_{*}(2\xi,z)}{2}\end{pmatrix},\qquad\xi\in\mathbb{R},\quad
z\in\mathbb{C}\,.$

Differentiating, one obtains $JX^{\prime}+QX=zX$,
$X(0,z)=\left(\begin{smallmatrix}1&0\\\ 0&1\\\ \end{smallmatrix}\right)$. It
follows that $X(\xi,z)=N_{Q}(\xi,z)$. In particular, we have

$(N_{Q})_{22}=e^{-i\xi
z}\frac{\widehat{P}(2\xi,z)+\widehat{P}_{*}(2\xi,z)}{2},\qquad(N_{Q})_{21}=e^{-i\xi
z}\frac{P_{*}(2\xi,z)-P(2\xi,z)}{2i}.$

Since $P(\xi,z)\to 0$, $P_{*}(\xi,z)\to\Pi(z)\neq 0$ as $\xi\to+\infty$ (see
Theorem 12.1 in [12]), and analogous relations hold for $\widehat{P}$ and
$\widehat{P}_{*}$, we have

$\lim_{\xi\to+\infty}\frac{(N_{Q})_{22}(\xi,z)}{(N_{Q})_{21}(\xi,z)}=\lim_{\xi\to+\infty}i\frac{\widehat{P}_{*}(\xi,z)}{P_{*}(\xi,z)},\qquad
z\in\mathbb{C}_{+}.$

The lemma is proved. ∎

3. Lemma 6.1 and some known results for canonical systems can be used to show that weak convergence of potentials of the Dirac operator implies convergence of the corresponding Weyl functions.

######  Lemma 6.2.

Suppose $\\{q_{\ell}\\}_{\ell>0}$ is a bounded sequence in
$L^{2}(\mathbb{R}_{+})$ which converges to zero weakly. Let $Q_{\ell}$ be the
associated matrix-functions defined as in Lemma 6.1. Then, the sequence of
corresponding Weyl functions $\\{m_{\ell,+}\\}$ converges to $i$ locally
uniformly in $\mathbb{C}_{+}$ when $\ell\to+\infty$.

Proof. For $\ell>0$, denote by $\mathcal{H}_{Q_{\ell}}$ the Hamiltonian
generated by $Q_{\ell}$ as in Lemma 6.1. Then, $m_{\ell,+}$ is the Weyl
function for the half-line operators $\mathcal{D}_{\mathcal{H}_{Q_{\ell}}}$
and $\mathcal{D}_{Q_{\ell}}$. Since
$\sup_{\ell>0}\|q_{\ell}\|_{L^{2}(\mathbb{R}_{+})}<\infty$ and $q_{\ell}$
converge to zero weakly in $L^{2}(\mathbb{R}_{+})$ as $\ell\to+\infty$, the
Hamiltonians $\mathcal{H}_{Q_{\ell}}$ tend to the identity matrix
$\mathcal{H}_{0}=\left(\begin{smallmatrix}1&0\\\ 0&1\\\
\end{smallmatrix}\right)$ uniformly on compact subsets on $\mathbb{R}_{+}$.
Then, their Weyl functions $m_{+,\ell}$ tend to the Weyl function $m_{+}=i$ of
the Hamiltonian $\mathcal{H}_{0}$ locally uniformly in $\mathbb{C}_{+}$ by
Theorem $5.7$ $(b)$ in [21]. ∎

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