# Trace formulas revisited and a new representation of KdV solutions with
short-range initial data

Alexei Rybkin  Department of Mathematics and Statistics, University of Alaska
Fairbanks, PO Box 756660, Fairbanks, AK 99775<EMAIL_ADDRESS>We dedicate
this paper to Vladimir Marchenko on the occasion of his centennial birthday.
This paper is also dedicated to the memory of Vladimir Zakharov who has just
left us.

(Date: August, 2023)

###### Abstract.

We put forward a new approach to Deift-Trubowitz type trace formulas for the
1D Schrodinger operator with potentials that are summable with the first
moment (short-range potentials). We prove that these formulas are preserved
under the KdV flow whereas the class of short-range potentials is not.
Finally, we show that our formulas are well-suited to study the dispersive
smoothing effect.

###### Key words and phrases:

trace formula, KdV equation, Hankel operator.

###### 1991 Mathematics Subject Classification:

34L25, 37K15, 47B35

The author is supported in part by the NSF grant DMS-2009980.

##  1\. Introduction

We are concerned with the Cauchy problem for the Korteweg-de Vries (KdV)
equation

$\begin{cases}\partial_{t}q-6q\partial_{x}q+\partial_{x}^{3}q=0,\ \ \
x\in\mathbb{R},t\geq 0\\\ q(x,0)=q(x).\end{cases}$ (1.1)

As is well-known, (1.1) is the first nonlinear evolution PDE solved in the
seminal 1967 Gardner-Greene-Kruskal-Miura paper [7] by the method which is now
referred to as the inverse scattering transform (IST). Conceptually, the IST
is similar to the Fourier method but is based on the direct/inverse scattering
(spectral) theory for the 1D Schrödinger operator
$\mathbb{L}_{q}=-\partial_{x}^{2}+q(x)$. Explicit formulas, however, are in
short supply and trace formulas are among a few available. Historically, for
short-range potentials $q\left(x\right)$ (i.e. summable with the first moment)
such a formula (see (5.7)) was put forward by Deift-Trubowitz in [5] in the
late 70s (we call it the Deift-Trubowitz trace formula). However, no
adaptation of the trace formula (5.7) to the solution $q\left(x,t\right)$ to
(1.1) is offered in [5] and, to the best of our knowledge, it has not been
done in the literature. We emphasize that, as we show below (see Corollary
7.3), $q\left(x,t\right)$ need not be short-range for $t>0$ and therefore the
approach of [5] breaks down in a serious way.

In this contribution we put forward an elementary approach to Deift-Trubowitz
type trace formulas which is based on Hardy space arguments and Hankel
operators. This way we generate trace type formulas like (5.2), (5.4), (5.8),
(5.12) that may serve different purposes. E.g. (5.2) remains valid for certain
long-range potentials (will be done elsewhere), (5.12) is a convenient
starting point to include time evolution $q\left(x,t\right)$ under the KdV
flow (section 6), and (5.8) is well-suited for subtle analysis of the gain of
regularity (aka dispersive smoothing) phenomenon for the KdV equation (section
7). Our trace formulas are made of the essentially same ingredients as (5.7)
and therefore equivalent to the latter, which is demonstrated in Appendix.
Note that in our [12], [13], [14] we rely on the Dyson formula (aka the second
log determinant formula) for extension of the IST to initial data
$q\left(x\right)$ that is essentially arbitrary at $-\infty$ but still short-
range at $+\infty$. Comparing with the Dyson formula considerations, our trace
approach is more robust for analysis of KdV solutions (see Remark 7.2). To the
best of our knowledge Theorem 7.1 is new.

Note that for periodic potentials the trace formula was studied in great
detail the 70s by McKean-Moerbeke [21], Trubowitz [27] and many others (see
e.g. [10] for a nice historic review) before (5.7). It was generalized by
Craig in [4] in the late 80s to arbitrary bounded continuous potentials (the
so-called Craig’s trace formula). In the 90s Gesztesy et al [9] developed a
general approach to Craig type trace formulas based on the Krein trace formula
(the ”true” trace formula) under the only condition of essential boundedness
from below. The general trace formula put forward in [9] yields previously
known ones. In the 2000s we [24] introduced a new way of generating trace-type
formulas that is not based upon Krein’s trace formula but rests on the
Titchmarsh-Weyl theory for second order differential equations and asymptotics
of the Titchmarsh-Weyl m-function. The approach is quite elementary and
essentially free of any conditions. Recently, in Binder et al [1] Craig’s
trace formulas was used in the KdV context to address some open problems
related to almost periodic initial data.

The paper is organized as follows. In section 2 we introduce our notations
Section 3 is devoted to basics of Hardy spaces and Hankel operator our
approach is based upon. In section 4 we review the classical direct/inverse
scattering theory for Schrodinger operators on the line using the language of
Hankel operators. Section 5 is where our trace formulas are introduced. We do
not claim their originality but believe that the approach is new. In Section 6
we derive a representation for the solution to the KdV equation with short-
range initial data. To the best of our knowledge it is new. In the final
section 7 we demonstrate how our trace formula for the KdV is well-suited for
the analysis of dispersive smoothing. The approach builds upon our recent [14]
and suggests an effective way to understanding how the KdV flow trades the
decay of initial data for gain of regularity. In Appendix we demonstrate that
the Deift-Trubowitz trace formula is actually a ”nonlinearization” of ours.

##  2\. Notations

Our notations are quite standard:

  * •

Unless otherwise stated, all integrals are Lebesgue and, as is commonly done,
we drop limits of integration if the integral (absolutely convergent) is over
the whole line. For convergent integrals that are not absolutely convergent we
always use the Cauchy principal value

$\left(PV\right)\int=\lim_{a\rightarrow\infty}\int_{-a}^{a}.$

  * •

$\chi_{S}$ is the characteristic function of a (measurable) set $S$.

  * •

As usual, $L^{p}\left(S\right),\ 0<p\leq\infty$, is the Lebesgue space on a
(measurable) set $S$. If $S=\mathbb{R}$ then we abbreviate
$L^{p}\left(\mathbb{R}\right)=L^{p}$. We include $L^{p}$ in the family of
weighted $L^{p}$ spaces defined by

$L_{\alpha}^{p}=\left\\{f\ |\ \int\left|f\left(x\right)\right|^{p}\left\langle
x\right\rangle^{\alpha}\mathrm{d}x<\infty\right\\},\ \ \alpha>0.$

where $\left\langle x\right\rangle=\sqrt{1+x^{2}}$. The class $L_{1}^{1}$ is
basic to scattering theory for 1D Schrödinger operators (short-range
potentials).

  * •

$\left\|\cdot\right\|_{X}$ stands for a norm in a Banach space $X$. The most
common space is $X=L^{2}$. We merely write $\left\|\cdot\right\|$ in this case
and also

$\left\|f\right\|^{2}=\left\langle f,f\right\rangle\text{ where }\left\langle
f,g\right\rangle=\int f\left(x\right)\overline{g}\left(x\right)\mathrm{d}x.$

  * •

We write $x\simeq y$ if $x=Cy$ for some universal constant $C$;
$x\lesssim_{a}y$ if $x,y\geq 0$ and $x\leq C\left(a\right)y$ with a positive
$C$ dependent on $a$. We drop $a$ if $C$ is a universal constant.

  * •

We do not distinguish between classical and distributional derivatives.

  * •

A statement $A_{\pm}$ means two separate statements: $A_{-}$ and $A_{+}$.

##  3\. Hardy spaces and Hankel operators

To fix our notation we review some basics of Hardy spaces and Hankel operators
following [22].

A function $f$ analytic in
$\mathbb{C}^{\pm}=\left\\{z\in\mathbb{C}:\pm\operatorname{Im}z>0\right\\}$ is
in the Hardy space $H_{\pm}^{p}$ for some $0<p\leq\infty$ if

$\|f\|_{H_{\pm}^{p}}^{p}\overset{\operatorname*{def}}{=}\sup_{y>0}\|f(\cdot\pm
iy)\|_{p}<\infty.$

We set $H^{p}=H_{+}^{p}.$ It is a fundamental fact of the theory of Hardy
spaces that any $f\left(z\right)\in H_{\pm}^{p}$ with $0<p\leq\infty$ has non-
tangential boundary values $f\left(x\pm\mathrm{i}0\right)$ for almost every
(a.e.) $x\in\mathbb{R}$ and

$\|f\|_{H_{\pm}^{p}}=\|f\left(\cdot\pm\mathrm{i}0\right)\|_{L^{p}}=\left\|f\right\|_{L^{p}}.$
(3.1)

Classes $H_{\pm}^{\infty}$ and $H_{\pm}^{2}$ will be particularly important.
$H_{\pm}^{\infty}$ is the algebra of uniformly bounded in $\mathbb{C}^{\pm}$
functions and $H_{\pm}^{2}$ is the Hilbert space with the inner product
induced from $L^{2}$.

It is well-known that $L^{2}=H_{+}^{2}\oplus H_{-}^{2},$ the orthogonal
(Riesz) projection $\mathbb{P}_{\pm}$ onto $H_{\pm}^{2}$ being given by

$(\mathbb{P}_{\pm}f)(x)=\pm\frac{1}{2\pi\mathrm{i}}\lim_{\varepsilon\rightarrow
0+}\int\frac{f(s)\mathrm{d}s}{s-(x\pm\mathrm{i}\varepsilon)}=:\pm\frac{1}{2\pi\mathrm{i}}\int\frac{f(s)\
\mathrm{d}s}{s-(x\pm\mathrm{i}0)}.$ (3.2)

Observe that the Riesz projections can also be rewritten in the form

$\left(\mathbb{P}_{\pm}f\right)\left(x\right)=\left(\widetilde{\mathbb{P}}_{\pm}f\right)\left(x\right)\mp\frac{1}{2\pi\mathrm{i}}\int\frac{f(s)}{s+\mathrm{i}}\mathrm{d}s,$
(3.3)

where

$\left(\widetilde{\mathbb{P}}_{\pm}f\right)\left(x\right):=(x+\mathrm{i})\left(\mathbb{P}_{\pm}\frac{f}{\cdot+\mathrm{i}}\right)(x)$

is well-defined for any $f\in L^{\infty}$. This representation is very
important in what follows.

If $f\in L^{2}$ then $\mathbb{P}_{-}f$ is by definition in $H_{-}^{2}$ but of
course not in $L^{1}$. However under a stronger decay condition we have the
following statement.

######  Lemma 3.1.

If $\left\langle x\right\rangle f\left(x\right)\in L^{2}$ then

$\left(PV\right)\int\mathbb{P}_{-}f=\frac{1}{2}\int f.$ (3.4)

###### Proof.

Note first that if $\left\langle x\right\rangle f\left(x\right)\in L^{2}$ then
$f$ is of course integrable as one sees from

$\int\left|f\right|=\int\left|\left\langle x\right\rangle
f\left(x\right)\right|\frac{\mathrm{d}x}{\left\langle
x\right\rangle}\leq\left\|\left\langle\cdot\right\rangle
f\right\|\left\|\left\langle\cdot\right\rangle^{-1}\right\|<\infty.$

It follows then that for a finite $a>0$ we have

$\displaystyle\int_{-a}^{a}\mathbb{P}_{-}f$
$\displaystyle=\left\langle\mathbb{P}_{-}f,\chi_{\left|\cdot\right|\leq
a}\right\rangle=\left\langle f,\mathbb{P}_{-}\chi_{\left|\cdot\right|\leq
a}\right\rangle\text{ \ (by (\ref{regul proj}))}$ $\displaystyle=\left\langle
f,(\cdot+\mathrm{i})\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|\leq
a}}{\cdot+\mathrm{i}}\right\rangle+\frac{1}{2\pi\mathrm{i}}\int\frac{\chi_{\left|\cdot\right|\leq
a}}{s-\mathrm{i}}\mathrm{d}s\ \int f$ (3.5)
$\displaystyle=\left\langle(\cdot-\mathrm{i})f,\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|\leq
a}}{\cdot+\mathrm{i}}\right\rangle+\left(\frac{1}{2}-\frac{1}{\pi}\arctan\frac{1}{a}\right)\int
f.$

Here we have used

$\displaystyle\frac{1}{2\pi\mathrm{i}}\int\frac{\chi_{\left|\cdot\right|\leq
a}}{s-\mathrm{i}}\mathrm{d}s$
$\displaystyle=\frac{1}{2\pi\mathrm{i}}\int_{-a}^{a}\frac{\mathrm{d}s}{s-\mathrm{i}}=\frac{1}{2}-\frac{1}{\pi}\arctan\frac{1}{a}$
(3.6) $\displaystyle\rightarrow\frac{1}{2},\ \ \ a\rightarrow+\infty.$

Since $\chi_{\left|\cdot\right|\leq a}=1-\chi_{\left|\cdot\right|>a}$ and
$1/\left(x+\mathrm{i}\right)\in H^{2}$ we have

$\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|\leq
a}}{\cdot+\mathrm{i}}=\mathbb{P}_{-}\frac{1}{\cdot+\mathrm{i}}-\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|>a}}{\cdot+\mathrm{i}}=-\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|>a}}{\cdot+\mathrm{i}}$

and therefore

$\left\langle(\cdot-\mathrm{i})f,\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|\leq
a}}{\cdot+\mathrm{i}}\right\rangle=-\left\langle(\cdot-\mathrm{i})f,\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|>a}}{\cdot+\mathrm{i}}\right\rangle.$

It follows that

$\displaystyle\left|\left\langle(\cdot-\mathrm{i})f,\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|\leq
a}}{\cdot+\mathrm{i}}\right\rangle\right|$
$\displaystyle\leq\left\|(\cdot-\mathrm{i})f\right\|\left\|\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|>a}}{\cdot+\mathrm{i}}\right\|$
$\displaystyle\leq\left\|\left\langle\cdot\right\rangle
f\right\|\left\|\frac{\chi_{\left|\cdot\right|>a}}{\cdot+\mathrm{i}}\right\|\rightarrow
0,\ a\rightarrow\infty.$

This means that

$\lim_{a\rightarrow\infty}\left\langle(\cdot-\mathrm{i})f,\mathbb{P}_{-}\frac{\chi_{\left|\cdot\right|\leq
a}}{\cdot+\mathrm{i}}\right\rangle=0$

and we can pass in (3.5) as $a\rightarrow\infty$:

$\lim_{a\rightarrow\infty}\int_{-a}^{a}\mathbb{P}_{-}f=\frac{1}{2}\int f.$

∎

We now define the Hankel operator on $H^{2}$. Let $(\mathbb{J}f)(x)=f(-x)$ be
the operator of reflection. Given $\varphi\in L^{\infty}$ the operator
$\mathbb{H}(\varphi):H^{2}\rightarrow H^{2}$ given by the formula

$\mathbb{H}(\varphi)f=\mathbb{JP}_{-}\varphi f,\ \ \ f\in H_{+}^{2},$ (3.7)

is called the Hankel __ operator with symbol $\varphi$. Clearly
$\left\|\mathbb{H}(\varphi)\right\|\leq\left\|\varphi\right\|_{L^{\infty}}$,
$\mathbb{H}(\varphi)$ is self-adjoint if
$(\mathbb{J}\varphi)(x)=\overline{\varphi\left(x\right)}$ (this is always our
case), $\mathbb{H}(\varphi)=0$ if $\varphi$ is a constant, and

$\displaystyle\mathbb{H}(\varphi)$
$\displaystyle=\mathbb{H}(\widetilde{\mathbb{P}}_{-}\varphi)$ (3.8)
$\displaystyle=\mathbb{H}(\mathbb{P}_{-}\varphi)\text{ (if }\varphi\in
L^{2}\cap L^{\infty}\text{).}$

The relevance of the Hankel operator in our setting is on the surface as the
Marchenko operator, the cornerstone of the IST, is a Hankel operator. However,
while in the literature on integrable systems it is rarely used in the form
(3.7), we find it particularly convenient due, among others, to the property
(3.8), which is less transparent in the integral representation.

Finally we note that reliance on the theory of Hankel operator in the study of
completely integrable systems has recently picked up momentum (see e.g. [2],
[6], [8], [11], [19] and the references cited therein).

##  4\. Overview of short-range scattering

Unless otherwise stated all facts are taken from [20]. Through this section we
assume that $q$ is short-range, i.e. $q\in L_{1}^{1}$. Associate with $q$ the
full line Schrödinger operator $\mathbb{L}_{q}=-\partial_{x}^{2}+q(x)$. As is
well-known, $\mathbb{L}_{q}$ is self-adjoint on $L^{2}$ and its spectrum
consists of $J$ simple negative eigenvalues $\\{-\kappa_{j}^{2}:1\leq j\leq
J\\},$ called bound states ($J=0$ if there are no bound states), and two fold
absolutely continuous component filling $\left(0,\infty\right)$. There is no
singular continuous spectrum. Two linearly independent (generalized)
eigenfunctions of the a.c. spectrum $\psi_{\pm}(x,k),\;k\in\mathbb{R}$, can be
chosen to satisfy

$\psi_{\pm}(x,k)=e^{\pm\mathrm{i}kx}+o(1),\;\partial_{x}\psi_{\pm}(x,k)\mp\mathrm{i}k\psi_{\pm}(x,k)=o(1),\
\ x\rightarrow\pm\infty.$ (4.1)

The function $\psi_{\pm}$, referred to as right/left Jost solution of the
Schrödinger equation

$\mathbb{L}_{q}\psi=k^{2}\psi,$ (4.2)

is analytic for $\operatorname{Im}k>0$. It is convenient to introduce

$y_{\pm}\left(k,x\right):=e^{\mp\mathrm{i}kx}\psi_{\pm}\left(x,k\right)-1,$

($1+y_{\pm}\left(k,x\right)$ is sometimes referred to as the Faddeev
function), which is $H^{2}$ for each $x$. Since $q$ is real,
$\overline{\psi}_{\pm}$ also solves (4.2) and one can easily see that the
pairs $\\{\psi_{+},\overline{\psi}_{+}\\}$ and
$\\{\psi_{-},\overline{\psi}_{-}\\}$ form fundamental sets for (4.2). Hence
$\psi_{\mp}$ is a linear combination of
$\\{\psi_{\pm},\overline{\psi}_{\pm}\\}$. We write this fact as follows

$T(k)\psi_{\mp}(x,k)=\overline{\psi_{\pm}(x,k)}+R_{\pm}(k)\psi_{\pm}(x,k),\ \
\ k\in\mathbb{R}\text{,}$ (4.3)

where $T$ and $R_{\pm},$ are called transmission, right/left reflection
coefficients respectively. The function $T\left(k\right)$ is meromorphic for
$\operatorname{Im}k>0$ with simple poles at
$\left(\mathrm{i}\kappa_{j}\right)$ and continuous for $\operatorname{Im}k=0$.
Generically, $T\left(0\right)=0$. The reflection coefficient
$R_{\pm}\left(k\right)\in L^{2}$ but need not admit be analytic.

In the context of the IST Zakharov-Faddeev trace formulas [29] (conservation
laws) play very important role. For Schwarz potentials $q$ they are infinitely
many. Explicitly,

$\frac{8}{\pi}\int\log\left(1-\left|R\left(k\right)\right|^{2}\right)^{-1}\mathrm{d}k=\int
q+{\displaystyle\sum}\kappa_{n}\text{ \ \ (first trace formula)}$ (4.4)
$\frac{8}{\pi}\int
k^{2}\log\left(1-\left|R_{\pm}\left(k\right)\right|^{2}\right)^{-1}\mathrm{d}k=\int
q^{2}-\frac{16}{3}{\displaystyle\sum}\kappa_{n}^{3}\text{ \ \ (second trace
formula)}$ (4.5)

It is shown in the recent [15] that (4.4) holds for any $q\in L^{1}$, each
term being finite. Since $\left|R_{\pm}\left(k\right)\right|\leq 1$ and

$\log\left(1-\left|R_{\pm}\left(k\right)\right|^{2}\right)^{-1}\geq\left|R_{\pm}\left(k\right)\right|^{2},$

one concludes that $R_{\pm}\left(k\right)\in L^{2}$ for $q\in L^{1}$. The
second one (4.5) holds for $q\in L^{1}\cap L^{2}$ [17] and readily implies
that $kR_{\pm}\left(k\right)\in L^{2}$. Note, that Zakharov-Faddeev trace
formulas are not directly related to the trace formulas we discuss in
Introduction but they are also related to the trace of some operators.

The identities (4.3) are totally elementary but serve as a basis for inverse
scattering theory and for this reason they are commonly referred to as basic
scattering relations. As is well-known (see, e.g. [20]), the triple
$\\{R_{\pm},(\kappa_{j},c_{\pm,j})\\}$, where
$c_{\pm,j}=\left\|\psi_{\pm}(\cdot,\mathrm{i}\kappa_{j})\right\|^{-1}$,
determines $q$ uniquely and is called the scattering data for
$\mathbb{L}_{q}$. We emphasize that in order to come from a $L_{1}^{1}$
potential the scattering data $\\{R_{\pm},(\kappa_{n},c_{\pm,n})\\}$ must
satisfy some conditions known as Marchenko’s characterization [20]. The actual
process of solving the inverse scattering problem necessary for the IST is
historically based on the Marchenko theory (also knows as Faddeev-Marchenko or
Gelfand-Levitan-Marchenko). In fact, this procedure is quite transparent from
the Hankel operator point of view. Indeed, replacing $\psi_{\pm}$ in (4.3)
with $y_{\pm}$ and applying the operator $\mathbb{JP}_{-}$, a straightforward
computation [12] leads to

$y_{\pm}+\mathbb{H}(\varphi_{\pm})y_{\pm}=-\mathbb{H}(\varphi_{\pm})1,\text{
(Marchenko's equation)}$ (4.6)

where $\mathbb{H}(\varphi_{\pm})$ is the Hankel operator (3.7) with symbol

$\varphi_{\pm}\left(k,x\right)={\displaystyle\sum_{n=1}^{N}}\frac{-\mathrm{i}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}}{k-\mathrm{i}\kappa_{n}}+R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx},$ (4.7)

and $\mathbb{H}(\varphi_{\pm})1$ is understood as

$\mathbb{H}(\varphi_{\pm})1=\mathbb{JP}_{-}\varphi_{\pm}=\mathbb{P}_{+}\mathbb{J}\varphi_{\pm}=\mathbb{P}_{+}\overline{\varphi}_{\pm}.$

We call (4.6) the Marchenko equation as its Fourier image is the Marchenko
integral equation. It is proven in [12, Theorem 8.2] that
$I+\mathbb{H}(\varphi_{\pm})$ is positive definite and therefore

$y_{\pm}=-\left[I+\mathbb{H}(\varphi_{\pm})\right]^{-1}\mathbb{H}(\varphi_{\pm})1\in
H^{2}.$ (4.8)

Thus, given data $\\{R_{\pm},(\kappa_{j},c_{\pm,j})\\}$ we compute
$\varphi_{\pm}$ by (4.7) and form the Hankel operator
$\mathbb{H}(\varphi_{\pm})$. The function $y_{\pm}\left(k,x\right)$ is found
by (4.8). The potential $q\left(x\right)$ can then be recovered in a few ways.
Our method is, of course, to apply a suitable trace formula, which we derive
in the next section.

Since many of our proofs below are based on limiting arguments we need to
understand in what sense scattering data converges as we approximate $q$ in
the $L_{1}^{1}$. In particular the following statement plays an important
role.

######  Proposition 4.1.

If $q_{n}\left(x\right)$ converges in $L_{1}^{1}$ to $q\left(x\right)$ then
the sequence of reflection coefficients $R_{\pm,n}\left(k\right)$
corresponding to $q_{n}\left(x\right)$ converges in $L^{2}$ to
$R_{\pm}\left(k\right).$

###### Proof.

We consider the $+$ case only and we suppress $+$ sign. We use the following a
priori estimates (see e.g. [5])

$\left|y_{-}\left(x,k\right)\right|\lesssim_{q}\left\langle
x\right\rangle/\left\langle k\right\rangle$ (4.9)
$\left|y_{-}\left(x,k\right)-y_{-,n}\left(x,k\right)\right|\lesssim_{q}\left\langle
x\right\rangle\left\|q-q_{n}\right\|_{L_{1}^{1}}$ (4.10)
$\left|T\left(k\right)-T_{n}\left(k\right)\right|\lesssim_{q}\left|k\right|^{-1}\left\|q-q_{n}\right\|_{L_{1}^{1}}.$
(4.11)

Consider $\left\|R-R_{n}\right\|^{2}$ and rewrite it as ($\varepsilon$ is any)

$\displaystyle\left\|R-R_{n}\right\|^{2}$
$\displaystyle=\left\|\left(R-R_{n}\right)\chi_{\left|\cdot\right|\leq\varepsilon}\right\|^{2}+\left\|\left(R-R_{n}\right)\chi_{\left|\cdot\right|>\varepsilon}\right\|^{2}$
(4.12) $\displaystyle\leq
8\varepsilon+\left\|\left(R-R_{n}\right)\chi_{\left|\cdot\right|>\varepsilon}\right\|^{2}.$

It follows from the general formula [5]

$R\left(k\right)=\frac{T\left(k\right)}{2\mathrm{i}k}\int
e^{-2\mathrm{i}kx}q\left(x\right)\left(1+y_{-}\left(x,k\right)\right)\mathrm{d}x$
(4.13)

that

$\displaystyle R\left(k\right)-R_{n}\left(k\right)$
$\displaystyle=\frac{T\left(k\right)-T_{n}\left(k\right)}{2\mathrm{i}k}\int
e^{-2\mathrm{i}kx}q\left(x\right)\left(1+y_{-}\left(x,k\right)\right)\mathrm{d}x$
$\displaystyle+\frac{T_{n}\left(k\right)}{2\mathrm{i}k}\int
e^{-2\mathrm{i}kx}\left(q\left(x\right)-q_{n}\left(x\right)\right)\mathrm{d}x$
$\displaystyle+\frac{T_{n}\left(k\right)}{2\mathrm{i}k}\int
e^{-2\mathrm{i}kx}q\left(x\right)\left(y_{-}\left(x,k\right)-y_{-,n}\left(x,k\right)\right)\mathrm{d}x$
$\displaystyle=I_{1}\left(k\right)+I_{2}\left(k\right)+I_{3}\left(k\right)$

and hence

$\left\|\left(R-R_{n}\right)\chi_{\left|\cdot\right|>\varepsilon}\right\|\leq\left\|I_{1}\chi_{\left|\cdot\right|>\varepsilon}\right\|+\left\|I_{2}\chi_{\left|\cdot\right|>\varepsilon}\right\|+\left\|I_{3}\chi_{\left|\cdot\right|>\varepsilon}\right\|.$

Estimate each term separately. For
$\left\|I_{1}\chi_{\left|\cdot\right|>\varepsilon}\right\|$ we have

$\displaystyle\left\|I_{1}\chi_{\left|\cdot\right|>\varepsilon}\right\|^{2}$
$\displaystyle\lesssim_{q}\int\left|q\right|\left(1+\left|y_{-}\left(x,k\right)\right|\right)\mathrm{d}x\cdot\int_{\left|k\right|>\varepsilon}\left|\frac{T\left(k\right)-T_{n}\left(k\right)}{k}\right|^{2}\mathrm{d}k$
$\displaystyle\lesssim_{q}\left\|q-q_{n}\right\|_{L_{1}^{1}}^{2}\int_{\left|k\right|>\varepsilon}k^{-4}\mathrm{d}k\text{\
\ (by (\ref{est for y}),(\ref{est for T-Tn}))}$
$\displaystyle\lesssim\varepsilon^{-3}\left\|q-q_{n}\right\|_{L_{1}^{1}}^{2}.$

Thus

$\left\|I_{1}\chi_{\left|\cdot\right|>\varepsilon}\right\|\lesssim_{q}\varepsilon^{-3/2}\left\|q-q_{n}\right\|_{L_{1}^{1}}.$

For $\left\|I_{2}\chi_{\left|\cdot\right|>\varepsilon}\right\|$ we have

$\displaystyle\left\|I_{2}\chi_{\left|\cdot\right|>\varepsilon}\right\|^{2}$
$\displaystyle\leq\left\|q-q_{n}\right\|_{L^{1}}^{2}\int_{\left|k\right|>\varepsilon}\left|\frac{T_{n}\left(k\right)}{k}\right|^{2}\mathrm{d}k$
$\displaystyle\lesssim\frac{1}{\varepsilon}\left\|q-q_{n}\right\|_{L_{1}^{1}}^{2}$

and hence

$\left\|I_{2}\chi_{\left|\cdot\right|>\varepsilon}\right\|\lesssim_{q}\varepsilon^{-1/2}\left\|q-q_{n}\right\|_{L^{1}}.$

Finally for $\left\|I_{2}\chi_{\left|\cdot\right|>\varepsilon}\right\|$ one
has in a similar manner

$\displaystyle\left\|I_{3}\chi_{\left|\cdot\right|>\varepsilon}\right\|$
$\displaystyle\leq_{q}\varepsilon^{-1/2}\sup_{k}\left|\int
e^{-2\mathrm{i}kx}q\left(x\right)\left(y_{-}\left(x,k\right)-y_{-,n}\left(x,k\right)\right)\mathrm{d}x\right|$
$\displaystyle\leq_{q}\varepsilon^{-1/2}\left\|q-q_{n}\right\|_{L_{1}^{1}}\
\text{\ (by (\ref{est for y-yn}))}$

and hence

$\left\|I_{3}\chi_{\left|\cdot\right|>\varepsilon}\right\|\lesssim_{q}\varepsilon^{-1/2}\left\|q-q_{n}\right\|_{L_{1}^{1}}.$

One can now sees that each
$\left\|I_{j}\chi_{\left|\cdot\right|>\varepsilon}\right\|$, $j=1,2,3,$
vanishes as $\left\|q-q_{n}\right\|_{L_{1}^{1}}$ does and hence, since
$\varepsilon$ is arbitrary, it follows from (4.12) that

$\left\|R-R_{n}\right\|\rightarrow 0,n\rightarrow\infty.$

∎

Note that the question in what sense the reflection coefficient converges when
we approximate the potential in a certain way is a subtle one [23].

Finally we observe that $\psi_{\pm},y_{\pm},T,R_{\pm}$ as functions of $k$
(momentum) satisfy

$\left(\mathbb{J}f\right)\left(k\right)=f\left(-k\right)=\overline{f}\left(k\right)\text{
\ \ \ (symmetry property).}$ (4.14)

##  5\. Trace formulas

In this section we put forward a new approach to generate Deift-Trubowitz type
trace formulas. It is based on Hardy spaces and Hankel operators.

######  Theorem 5.1.

Suppose that $q\in L^{1}$ and

$Q_{+}\left(x\right):=\int_{x}^{\infty}q\left(s\right)\mathrm{d}s,Q_{-}\left(x\right):=\int_{-\infty}^{x}q\left(s\right)\mathrm{d}s.$

Let $\psi_{\pm}(x,k)$ be right/left Jost solution and

$y_{\pm}\left(k,x\right)=e^{\mp\mathrm{i}kx}\psi_{\pm}(x,k)-1.$

If for all real $x$

$2\mathrm{i}ky_{\pm}\left(k,x\right)+Q_{\pm}\left(x\right)\in H^{2},$ (5.1)

then for any $\alpha>0$ and a.e. $x$

$q\left(x\right)=\mp\frac{2}{\pi}\partial_{x}\int\operatorname{Re}\frac{y_{\pm}\left(k,x\right)}{k+\mathrm{i}\alpha}k\mathrm{d}k\text{
(trace formula).}$ (5.2)

If for every real $x$

$y_{\pm}\left(\cdot,x\right)\in H^{2}$ (5.3)

then (5.2) simplifies to read

$q\left(x\right)=\mp\frac{2}{\pi}\partial_{x}\int\operatorname{Re}y_{\pm}\left(k,x\right)\mathrm{d}k.$
(5.4)

###### Proof.

Note first that both Jost solutions exist for $q\in L^{1}$ (not only for
$L_{1}^{1}$). Multiplying (5.1) by
$\mathrm{i}/\left(k+\mathrm{i}\alpha\right)\in H^{2}$ ($\alpha>0$) and
recalling that a product of two $H^{2}$ functions is in $H^{1}$, we have

$\frac{2k}{k+\mathrm{i}\alpha}y_{\pm}\left(k,x\right)-\frac{\mathrm{i}}{k+\mathrm{i}\alpha}Q_{\pm}\left(x\right)\in
H^{1}.$ (5.5)

But it is well-known that

$f\in H^{1}\Longrightarrow\int f\left(k+\mathrm{i}0\right)\mathrm{d}k=0$ (5.6)

and therefore

$\int\left[\frac{2k}{k+\mathrm{i}\alpha}y_{\pm}\left(k,x\right)-\frac{\mathrm{i}}{k+\mathrm{i}\alpha}Q_{\pm}\left(x\right)\right]\mathrm{d}k=0.$

For its real part we have

$\int\left\\{\operatorname{Re}\left[\frac{2k}{k+\mathrm{i}\alpha}y_{\pm}\left(k,x\right)\right]-\frac{\alpha}{k^{2}+\alpha^{2}}Q_{\pm}\left(x\right)\right\\}\mathrm{d}k=0,$

which can be rearranged to read

$\pi
Q_{\pm}\left(x\right)=2\int\operatorname{Re}\left[\frac{y_{\pm}\left(k,x\right)}{k+\mathrm{i}\alpha}\right]k\mathrm{d}k$

and (5.2) follows upon differentiating in $x$.

We show now (5.4). To this end, we just split (5.2) as

$\displaystyle\int\operatorname{Re}\frac{ky_{\pm}\left(k,x\right)}{k+\mathrm{i}\alpha}\mathrm{d}k$
$\displaystyle=\int\operatorname{Re}y_{\pm}\left(k,x\right)\mathrm{d}k$
$\displaystyle+\alpha\operatorname{Im}\int\frac{y_{\pm}\left(k,x\right)}{k+\mathrm{i}\alpha}\mathrm{d}k$

and observe that by (5.6) the second integral on the right hand side is zero
(both $y_{\pm}$ and $1/\left(k+\mathrm{i}\alpha\right)$ are in $H^{2}$). ∎

######  Remark 5.2.

Under the condition $q\in L_{1}^{1}$ the following formula is proven in [5]
(only $+$ sign is considered):

$\displaystyle q\left(x\right)$
$\displaystyle=-4{\displaystyle\sum_{n=1}^{N}}\kappa_{n}c_{+,n}^{2}\psi_{+}\left(x,\mathrm{i}\kappa_{n}\right)^{2}$
(5.7) $\displaystyle+\frac{2\mathrm{i}}{\pi}\left(PV\right)\int
R_{+}\left(k\right)\psi_{+}\left(x,k\right)^{2}k\mathrm{d}k.\text{ (Deift-
Trubowitz trace formula)}$

Visually it is very different from (5.4) ($\psi_{+}\left(x,k\right)$ appears
in (5.7) squared whereas in (5.4) it does not). One can however show that
(5.4) implies (5.7). We demonstrate this fact in Appendix. Theorem 5.1 is an
extension of (5.7) as it accepts certain singularities of
$\psi_{\pm}\left(k,x\right)$ at $k=0$. The latter may occur if $q\notin
L_{1}^{1}$. Thus following the terminology of [5] we may refer to our (5.2)
and (5.4) as trace formulas.

The next statement offers a version of (5.7) that is linear with respect to
the Jost solution $\psi_{\pm}$.

######  Corollary 5.3.

Suppose $q\in L_{1}^{1}\cap L^{2}$ and let
$\left\\{R_{\pm},\kappa_{n},c_{\pm,n}\right\\}$ be its scattering data. Then

$q\left(x\right)=\pm\partial_{x}\left\\{2{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp\kappa_{n}x}\psi_{\pm}(x,\mathrm{i}\kappa_{n})+\frac{1}{\pi}\int
e^{\pm\mathrm{i}kx}R_{\pm}\left(k\right)\psi_{\pm}(x,k)\mathrm{d}k\right\\}.$
(5.8)

###### Proof.

As is well-known (see e.g. [20]), for $q\in L^{1}$

$y_{\pm}\left(k,x\right)=\frac{\mathrm{i}}{2k}Q_{\pm}\left(x\right)+O\left(k^{-2}\right),\
\ \ k\rightarrow\pm\infty,$ (5.9)

and furthermore $y_{\pm}\left(k,x\right)$ is bounded at $k=0$ for $q\in
L_{1}^{1}$. It immediately follows that the condition (5.1) is satisfied.
Also, the condition (5.3) holds due to (4.8). Therefore (5.4) holds for short-
range $q$. To show (5.8) we turn to the Marchenko (4.6). Applying the operator
of reflection $\mathbb{J}$ to this equation and recalling the symmetry
property (4.14) we have

$\overline{y}_{\pm}+\mathbb{P}_{-}(\varphi_{\pm}y_{\pm})=-\mathbb{P}_{-}\varphi_{\pm},$

which together with (4.6) yield

$2\operatorname{Re}y_{\pm}=-\mathbb{JP}_{-}(\varphi_{\pm}y_{\pm})-\mathbb{P}_{-}(\varphi_{\pm}y_{\pm})-\mathbb{JP}_{-}\varphi_{\pm}-\mathbb{P}_{-}\varphi_{\pm}.$

Since obviously

$\int_{-a}^{a}\mathbb{J}f=\int_{-a}^{a}f$

we have

$\int_{-a}^{a}\operatorname{Re}y_{\pm}=-\int_{-a}^{a}\mathbb{P}_{-}\varphi_{\pm}-\int_{-a}^{a}\mathbb{P}_{-}\left(\varphi_{\pm}y_{\pm}\right).$
(5.10)

Consider each term on the right hand side of (5.10). Observing that
$\left(k-\mathrm{i}\kappa_{n}\right)^{-1}\in H_{-}^{2}$ and hence
$\mathbb{P}_{-}\left(k-\mathrm{i}\kappa_{n}\right)^{-1}=\left(k-\mathrm{i}\kappa_{n}\right)^{-1}$,
we have

$\displaystyle\mathbb{P}_{-}\varphi_{\pm}$
$\displaystyle=\mathbb{P}_{-}\left[{\displaystyle\sum_{n=1}^{N}}\frac{-\mathrm{i}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}}{k-\mathrm{i}\kappa_{n}}\right]+\mathbb{P}_{-}\left[R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\right]$
$\displaystyle={\displaystyle\sum_{n=1}^{N}}\frac{-\mathrm{i}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}}{k-\mathrm{i}\kappa_{n}}+\mathbb{P}_{-}\left[R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\right]$

and thus

$\displaystyle\int_{-a}^{a}\mathbb{P}_{-}\varphi_{\pm}$
$\displaystyle=\int_{-a}^{a}{\displaystyle\sum_{n=1}^{N}}\frac{-\mathrm{i}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}}{k-\mathrm{i}\kappa_{n}}\mathrm{d}k+\int_{-a}^{a}\mathbb{P}_{-}\left[R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\right]\mathrm{d}k$
$\displaystyle=-\mathrm{i}{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}\int_{-a}^{a}\frac{\mathrm{d}k}{k-\mathrm{i}\kappa_{n}}+\int_{-a}^{a}\mathbb{P}_{-}\left[R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\right]\mathrm{d}k.$ (5.11)

Pass in (5.11) now to the limit as $a\rightarrow\infty$. By (3.6)

$\lim_{a\rightarrow\infty}\int_{-a}^{a}\frac{\mathrm{d}k}{k-\mathrm{i}\kappa_{n}}=\lim_{a\rightarrow\infty}\int_{-a}^{a}\frac{\mathrm{d}k}{k-\mathrm{i}}=\pi\mathrm{i}$

and hence substituting this into (5.11) one has

$\displaystyle\left(PV\right)\int_{-a}^{a}\mathbb{P}_{-}\varphi_{\pm}$
$\displaystyle=\pi{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}$
$\displaystyle+\left(PV\right)\int_{-a}^{a}\mathbb{P}_{-}\left[R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\right]\mathrm{d}k.$

It remains to evaluate the integral on the right hand side. As we have shown
in Section 4, $\left\langle k\right\rangle R_{\pm}\left(k\right)\in L^{2}$. By
Lemma 3.1 then

$\left(PV\right)\int_{-a}^{a}\mathbb{P}_{-}\left[R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\right]\mathrm{d}k=\frac{1}{2}\int R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\mathrm{d}k$

and finally

$\displaystyle\left(PV\right)\int\mathbb{P}_{-}\varphi_{\pm}$
$\displaystyle=\pi{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}$ $\displaystyle+\frac{1}{2}\int R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\mathrm{d}k.$

Similarly,

$\left(PV\right)\int\mathbb{P}_{-}\left(\varphi_{\pm}y_{\pm}\right)=\frac{1}{2}\int\varphi_{\pm}y_{\pm}$

and from (5.10) we obtain

$\displaystyle\int\operatorname{Re}y_{\pm}$
$\displaystyle=-\int\mathbb{P}_{-}\varphi_{\pm}-\int\mathbb{P}_{-}\left(\varphi_{\pm}y_{\pm}\right)$
$\displaystyle=-\pi{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}-\frac{1}{2}\int R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\mathrm{d}k$
$\displaystyle-\frac{1}{2}\int\varphi_{\pm}y_{\pm}.$

Inserting this into (5.4) one has

$\displaystyle q\left(x\right)$
$\displaystyle=\mp\frac{2}{\pi}\partial_{x}\int\operatorname{Re}y_{\pm}\left(k,x\right)\mathrm{d}k$
$\displaystyle=\pm\partial_{x}\left\\{2{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}+\int R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\frac{\mathrm{d}k}{\pi}+\int\varphi_{\pm}\left(k,x\right)y_{\pm}\left(k,x\right)\frac{\mathrm{d}k}{\pi}\right\\}.$

It remains to evaluate the last integral on the right hand side. By the Cauchy
formula

$\displaystyle\int\varphi_{\pm}y_{\pm}$
$\displaystyle=\int{\displaystyle\sum_{n=1}^{N}}\frac{-\mathrm{i}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}}{k-\mathrm{i}\kappa_{n}}y_{\pm}\left(k,x\right)$
$\displaystyle+\int R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}y_{\pm}\left(k,x\right)\mathrm{d}k$
$\displaystyle=2\pi{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}y_{\pm}\left(\mathrm{i}\kappa_{n},x\right)+\int
R_{\pm}\left(k\right)e^{\pm 2\mathrm{i}kx}y_{\pm}\left(k,x\right)\mathrm{d}k$

and hence

$q\left(x\right)=\pm\partial_{x}\left\\{2{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp
2\kappa_{n}x}\left[1+y_{\pm}\left(\mathrm{i}\kappa_{n},x\right)\right]+\int
R_{\pm}\left(k\right)e^{\pm
2\mathrm{i}kx}\left[1+y_{\pm}\left(k,x\right)\right]\frac{\mathrm{d}k}{\pi}\right\\}.$

Recalling that $y_{\pm}\left(k,x\right)=e^{\mp\mathrm{i}kx}\psi_{\pm}(x,k)-1$
we finally obtain

$q\left(x\right)=\pm\partial_{x}\left\\{2{\displaystyle\sum_{n=1}^{N}}c_{\pm,n}^{2}e^{\mp\kappa_{n}x}\psi_{\pm}(x,\mathrm{i}\kappa_{n})+\int
e^{\pm\mathrm{i}kx}R_{\pm}\left(k\right)\psi_{\pm}(x,k)\frac{\mathrm{d}k}{\pi}\right\\},$

which is (5.8). ∎

######  Remark 5.4.

It follows from (5.9) that $y_{\pm}\left(k,x\right)\notin L^{1}$ but
$\operatorname{Re}y_{\pm}\left(k,x\right)\in L^{1}$.

An important corollary of Theorem 5.1 is the following

######  Theorem 5.5.

Suppose that $q\in L_{1}^{1}$. Let $\left\\{R,\kappa_{n},c_{n}\right\\}$ be
its right scattering data and $\psi_{0}(x,k)$ be the right Jost solution
corresponding to the data $\left\\{R,\varnothing\right\\}$. Denote

$\boldsymbol{\Psi}_{0}\left(x\right)=\left(\psi_{0}\left(x,\mathrm{i}\kappa_{n}\right)\right),\
\ \ \boldsymbol{C}=\operatorname*{diag}\left(c_{n}^{2}\right).$

Then

$\displaystyle q\left(x\right)$ $\displaystyle=q_{0}\left(x\right)$ (5.12)
$\displaystyle+2\partial_{x}\boldsymbol{\Psi}_{0}\left(x\right)\left(\boldsymbol{C}^{-1}+\int_{x}^{\infty}\boldsymbol{\Psi}_{0}\left(s\right)^{T}\boldsymbol{\Psi}_{0}\left(s\right)\mathrm{d}s\right)^{-1}\boldsymbol{\Psi}_{0}\left(x\right)^{T},$

where $q_{0}\left(x\right)$ admits the following representations

$\displaystyle q_{0}\left(x\right)$
$\displaystyle=2\partial_{x}\left\\{\int\operatorname{Re}\left[1-e^{-\mathrm{i}kx}\psi_{0}(x,k)\right]\frac{\mathrm{d}k}{\pi}\right\\}$
$\displaystyle=\partial_{x}\left(PV\right)\int
e^{\mathrm{i}kx}R\left(k\right)\psi_{0}(x,k)\frac{\mathrm{d}k}{\pi}$
$\displaystyle=\partial_{x}^{2}\int\frac{e^{2\mathrm{i}kx}-1}{2\mathrm{i}k}R\left(k\right)\frac{\mathrm{d}k}{\pi}+\partial_{x}\int
e^{2\mathrm{i}kx}R\left(k\right)y_{0}\left(k,x\right)\frac{\mathrm{d}k}{\pi}.$

###### Proof.

We merely combine the formula (5.4) from Theorem 5.1 and the version of the
binary Darboux transformation from our [26]

$q\left(x\right)=q_{0}\left(x\right)-2\partial_{x}^{2}\log\det\left(\boldsymbol{C}^{-1}+\int_{x}^{\infty}\boldsymbol{\Psi}_{0}^{T}\left(s\right)\boldsymbol{\Psi}_{0}\left(s\right)\right),$

where $q_{0}\left(x\right)$ is the potential corresponding to
$\left\\{R,\varnothing\right\\}$. Indeed, by the Jacobi formula on
differentiation of determinants one has

$\displaystyle\partial_{x}\log\det\left(\boldsymbol{C}^{-1}+\int_{x}^{\infty}\boldsymbol{\Psi}_{0}^{T}\left(s\right)\boldsymbol{\Psi}_{0}\left(s\right)\right)$
$\displaystyle=-\boldsymbol{\Psi}_{0}\left(x\right)\left(\boldsymbol{C}^{-1}+\int_{x}^{\infty}\boldsymbol{\Psi}_{0}\left(s\right)^{T}\boldsymbol{\Psi}_{0}\left(s\right)\mathrm{d}s\right)^{-1}\boldsymbol{\Psi}_{0}\left(x\right)^{T}.$

∎

We will demonstrate below that the trace formula (5.12) is convenient for
limiting arguments. Of course, a similar formula holds for the left scattering
data.

##  6\. Trace formula and KdV solutions

In this section we show that our trace formulas yield new representations for
solutions to the KdV equation with short-range initial data. Note that the
condition $q\left(x\right)\in L_{1}^{1}$ alone does not guaranty that
$q\left(x,t\right)\in L_{1}^{1}$ for $t>0$ (see Corollary 7.3) and therefore
(5.4) does not apply. We cannot even be sure that (5.1) holds for
$q\left(x,t\right).$ To overcome the problems we employ some limiting
arguments. Through the rest of the paper we use the following convenient
notation

$\xi_{x,t}\left(k\right):=\exp\mathrm{i}\left(8k^{3}t+2kx\right).$

While highly oscillatory on the real line, this function has a rapid decay
along $\mathbb{R}+\mathrm{i}a$ for any $a>0$.

######  Theorem 6.1.

If $q\left(x\right)\in L_{1}^{1}$ and $\left\\{R,\kappa_{j},c_{j}\right\\}$
are the associated right scattering data then the Cauchy problem for the KdV
equation (1.1) with initial data $q\left(x\right)$ can be represented by

$\displaystyle q\left(x,t\right)$ $\displaystyle=q_{0}\left(x,t\right)$ (6.1)
$\displaystyle+2\partial_{x}\boldsymbol{\Psi}_{0}\left(x,t\right)\left(\boldsymbol{C}\left(t\right)^{-1}+\int_{x}^{\infty}\boldsymbol{\Psi}_{0}\left(s,t\right)^{T}\boldsymbol{\Psi}_{0}\left(s,t\right)\mathrm{d}s\right)^{-1}\boldsymbol{\Psi}_{0}\left(x,t\right)^{T},$

where

$\boldsymbol{\Psi}_{0}\left(x,t\right)=\left(\psi_{0}\left(x,t,\mathrm{i}\kappa_{j}\right)\right),\
\ \ \boldsymbol{C}\left(t\right)=\left(c_{j}\exp 8\kappa_{j}^{3}t\right),$
$\psi_{0}\left(x,t,k\right)=e^{\mathrm{i}kx}\left[1+y_{0}\left(k,x,t\right)\right],$

$y_{0}\left(\cdot,x,t\right)$ is the $H^{2}$ solution of the singular integral
equation

$y+\mathbb{H}(R\xi_{x,t})y=-\mathbb{H}(R\xi_{x,t})1,$
$q_{0}\left(x,t\right)=\partial_{x}\left\\{\left(PV\right)\int
R\left(k\right)\xi_{x,t}\left(k\right)\frac{\mathrm{d}k}{\pi}+\int
R\left(k\right)\xi_{x,t}\left(k\right)y_{0}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}\right\\}.$
(6.2)

###### Proof.

For Schwarz $q\left(x\right)$ there is nothing to prove as $q\left(x,t\right)$
is also a Schwarz function. Since KdV is well-posed in any Sobolev space
$H^{-\varepsilon}$ with $0<\varepsilon\leq 1$ (see e.g. [18]) and
$L_{1}^{1}\subset H^{-\varepsilon}$, for any sequence of (real) Schwarz
functions $q_{n}\left(x\right)$ approximating $q\left(x\right)$ in $L_{1}^{1}$
the sequence of $q_{n}\left(x,t\right)$ converges in $H^{-\varepsilon}$ to
$q\left(x,t\right)$, the solution to (1.1) with the initial profile
$q\left(x\right)$. Thus, we only need to compute
$\lim_{n\rightarrow\infty}q_{n}\left(x,t\right)$. Note that convergence of
norming constants is somewhat inconvenient to deal with but results of our
recent [26] offers a simple detour of this circumstance. Take the scattering
data $\left\\{R,\varnothing\right\\}$ (i.e. no bound states) and construct by
(5.4) the corresponding potential

$q_{0}\left(x\right)=-2\partial_{x}\int\operatorname{Re}y_{0}\left(k,x\right)\mathrm{d}k.$

Since by construction $\mathbb{L}_{q_{0}}$ is positive, $q_{0}$ is the Miura
transformation

$q_{0}\left(x\right)=\partial_{x}r\left(x\right)+r\left(x\right)^{2}$

of some real $r\in L_{\operatorname*{loc}}^{2}$ [16]. Choose a sequence
$\left(r_{n}\right)$ of Schwarz function such that the sequence
$q_{0,n}=r_{n}^{2}+\partial_{x}r_{n}$ approximates $q_{0}$ in $L_{1}^{1}$. As
is well-known, each $R_{n}\left(k\right)$ is also Schwarz and so is
$q_{0,n}\left(x,t\right)$ for $t\geq 0$. Therefore, by (5.8) and recalling
that $\psi_{n}=e^{\mathrm{i}kx}\left(1+y_{n}\right)$ we have

$\displaystyle q_{0,n}\left(x,t\right)$ $\displaystyle=\partial_{x}\int
R_{n}\left(k\right)\xi_{x,t}\left(k\right)\left[1+y_{n}\left(k,x,t\right)\right]\frac{\mathrm{d}k}{\pi}$
$\displaystyle=\partial_{x}^{2}\int\frac{\xi_{x,t}\left(k\right)-1}{2\mathrm{i}k}R_{n}\left(k\right)\frac{\mathrm{d}k}{\pi}+\partial_{x}\int
R_{n}\left(k\right)\xi_{x,t}\left(k\right)y_{n}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}$
$\displaystyle=:q_{0,n}^{\left(1\right)}\left(x,t\right)+q_{0,n}^{\left(2\right)}\left(x,t\right)$

Here we have used a well-known regularization of the Fourier integral. This
representation is convenient for passing to the limit as $n\rightarrow\infty$.
By Proposition 4.1 the sequence of reflection coefficients $R_{n}$ converges
in $L^{2}$ to $R$. Paring this sequence, if needed, we may assume that
$R_{n}\rightarrow R$ a.e. Clearly $R_{n}\xi_{x,t}\rightarrow R\xi_{x,t}$ a.e.
too. But then, as is well-known (can also be easily shown), the corresponding
sequence of Hankel operators $\mathbb{H}(R_{n}\xi_{x,t})$ converges to
$\mathbb{H}(R\xi_{x,t})$ in the strong operator topology. Since (all)
$I+\mathbb{H}(R_{n}\xi_{x,t})$ and $I+\mathbb{H}(R\xi_{x,t})$ are positive
definite [12, Theorem 8.2] for all $x,t$ we conclude that in $H^{2}$

$\displaystyle y_{n}$
$\displaystyle=-\left(I+\mathbb{H}(R_{n}\xi_{x,t})\right)^{-1}\mathbb{H}(R_{n}\xi_{x,t})1$
$\displaystyle\rightarrow-\left(I+\mathbb{H}(R\xi_{x,t})\right)^{-1}\mathbb{H}(R\xi_{x,t})1=:y_{0}\
\ \ n\rightarrow\infty,$

where $y_{0}\left(\cdot,x,t\right)\in H^{2}$. Therefore, for all $x,t$

$\int\frac{\xi_{x,t}\left(k\right)-1}{2\mathrm{i}k}R_{n}\left(k,t\right)\frac{\mathrm{d}k}{\pi}\rightarrow\int\frac{\xi_{x,t}\left(k\right)-1}{2\mathrm{i}k}R\left(k,t\right)\frac{\mathrm{d}k}{\pi}$

and

$\int
R\left(k\right)\xi_{x,t}\left(k\right)y_{n}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}\rightarrow\int
R\left(k\right)\xi_{x,t}\left(k\right)y_{0}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}.$

Thus we conclude that for each $t\geq 0$

$w^{\ast}-\lim_{n\rightarrow\infty}q_{0,n}^{\left(1\right)}\left(x,t\right)=\partial_{x}^{2}\int\frac{\xi_{x,t}\left(k\right)-1}{2\mathrm{i}k}R\left(k\right)\frac{\mathrm{d}k}{\pi}$
$w^{\ast}-\lim_{n\rightarrow\infty}q_{0,n}^{\left(2\right)}\left(x,t\right)=\partial_{x}\int
R_{n}\left(k\right)\xi_{x,t}\left(k\right)y_{n}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}$

and (6.2) follows. Performing the binary Darboux transformation [26] we arrive
at (6.1). ∎

######  Remark 6.2.

Performing in (6.1) the inverse binary Darboux transformation [26], we can
conclude that we also have

$q\left(x,t\right)=-\frac{2}{\pi}\partial_{x}\int\operatorname{Re}y\left(k,x,t\right)\mathrm{d}k$
(6.3)

but cannot claim that this integral is absolutely convergent as it was in
(5.4). This of course would be true if the asymptotic (5.9) held for
$q\left(x,t\right)$. The problem with (5.9) is that the error in (5.9) depends
on $\left\|q\left(\cdot,t\right)\right\|_{L^{1}}$, which need not be finite.
We however conjecture that the integral in (6.3) is indeed absolutely
convergent but we can no longer use tools and estimates from the short-range
scattering theory.

######  Corollary 6.3.

The trace Deift-Trubowitz trace formula (5.7) holds for $t>0$ (not only for
$t=0$).

###### Proof.

Indeed, the approximating sequence $q_{n}\left(x,t\right)$ that corresponds to
the sequence $\left\\{R_{n},\kappa_{j},c_{j}\right\\}$ where $R_{n}$ is the
same as constructed in the proof of Theorem 6.1 will do the job. The only
question is why the first term in (5.7) holds for $t>0$. This easily follows
from our arguments. Indeed, since $y_{n}\left(\cdot,x,t\right)\rightarrow
y\left(\cdot,x,t\right)$ in $H^{2}$ we also have uniform convergence for
$\operatorname{Im}k>0$ on compacts. Therefore,
$\psi_{n}\left(\mathrm{i}\kappa_{j},x,t\right)\rightarrow\psi\left(\mathrm{i}\kappa_{j},x,t\right)$
for all $x,t$. ∎

######  Remark 6.4.

Extension of the Deift-Trubowitz trace formula (5.7) to KdV solution would be
a hard problem back in the 70s as the breakthrough in the understanding of
wellposedness in the $L^{2}$ based Sobolev spaces with negative indexes only
occurred after the seminal 1993 Bourgain paper [3], where wellposedness was
proven in $L^{2}$. With no well-posedness at hand we cannot use limiting
arguments even if $q\left(x\right)\in L_{1}^{1}\cap L^{2}$.

##  7\. How KdV trades decay for smoothness

The goal of this section is to show how the results of the previous section
could be useful in understanding the phenomenon of dispersive smoothing (aka
gain of regularity).

######  Theorem 7.1.

If $q\left(x\right)\in L_{1}^{1}\cap L^{2}$ then $q\left(x,t\right)\in
L_{\operatorname*{loc}}^{\infty}\cap L^{2}$ for $t>0$.

###### Proof.

We first note that we may assume that the negative spectrum is absent. I.e.
$\mathbb{L}_{q}$ is positive. Split our initial profile as

$q=q_{-}+q_{+},\ \ \ q_{\pm}:=\left.q\right|_{\mathbb{R}_{\pm}}.$

We may assume that $\mathbb{L}_{q_{+}}$ is positive as possible appearance of
a negative eigenvalue could only lead to minor technical complications. We use
the following representation from [14]:

$R\left(k\right)=\phi_{1}\left(k\right)+\phi_{2}\left(k\right)+A\left(k\right),$
(7.1)

where

$\phi_{1}\left(k\right)=\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\widehat{q}\left(k\right),$
$\phi_{2}\left(k\right)=\frac{T_{0}\left(k\right)}{\left(2\mathrm{i}k\right)^{2}}\widehat{p}\left(k\right),$
$\widehat{f}\left(k\right):=\int_{0}^{\infty}e^{-2\mathrm{i}ks}p\left(s\right)\mathrm{d}s;$

$T_{0}\in H^{\infty}$ is the transmission coefficient for $q_{+}$; $p$ is the
derivative of an absolutely continuous function and

$\left|p\left(x\right)\right|\lesssim_{\left\|q_{+}\right\|_{L^{1}}}\left|q\left(x\right)\right|+C\int_{x}^{\infty}\left|q\right|,\
\ x\geq 0;$ (7.2)

and $A\in H^{\infty}$ (which form is not important). Note that
$\dfrac{T_{0}\left(k\right)}{2\mathrm{i}k}$ remains bounded at $k=0$ as well
as
$\dfrac{T_{0}\left(k\right)}{\left(2\mathrm{i}k\right)^{2}}\widehat{p}\left(k\right)$.

It follows from (6.2) that

$\displaystyle q_{0}\left(x,t\right)$
$\displaystyle=\partial_{x}\left\\{\left(PV\right)\int
R\left(k\right)\xi_{x,t}\left(k\right)\frac{\mathrm{d}k}{\pi}+\int
R\left(k\right)\xi_{x,t}\left(k\right)y_{0}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}\right\\}$
$\displaystyle=\partial_{x}\left(PV\right)\int
R\left(k\right)\xi_{x,t}\left(k\right)\frac{\mathrm{d}k}{\pi}$
$\displaystyle+\int
R\left(k\right)\left[\partial_{x}\xi_{x,t}\left(k\right)\right]y_{0}\left(k,x,t\right)\frac{\mathrm{d}k}{\pi}$
$\displaystyle+\int
R\left(k\right)\xi_{x,t}\left(k\right)\left[\partial_{x}y_{0}\left(k,x,t\right)\right]\frac{\mathrm{d}k}{\pi}$
$\displaystyle=:q_{1}\left(x,t\right)+q_{2}\left(x,t\right)+q_{3}\left(x,t\right).$
(7.3)

Consider each term separately. By (7.1) for $q_{1}\left(x,t\right)$ we have

$q_{1}\left(x,t\right)=q_{11}\left(x,t\right)+q_{12}\left(x,t\right)+q_{13}\left(x,t\right)$

where

$q_{1n}\left(x,t\right):=\partial_{x}\left(PV\right)\int\xi_{x,t}\left(k\right)\phi_{n}\left(k\right)\frac{\mathrm{d}k}{\pi},\
\ \ n=1,2$

and

$q_{13}\left(x,t\right)=\partial_{x}\left(PV\right)\int\xi_{x,t}\left(k\right)A\left(k\right)\frac{\mathrm{d}k}{\pi}.$

The simplest term is $q_{13}$. Since $\xi_{x,t}\left(k+\mathrm{i}a\right)$
(and all its $x$-derivatives) rapidly decays along $\mathbb{R+}\mathrm{i}a$
for any $a>0$ we deform the contour of integration to $\mathbb{R+}\mathrm{i}a$
that provides a rapid convergence of the integral (the original integral need
not be absolutely convergent).

The term $q_{12}$ is also easy. Indeed, since $\phi_{2}$ is clearly in $L^{1}$
we have

$\displaystyle q_{12}\left(x,t\right)$
$\displaystyle=\partial_{x}\int\phi_{2}\left(k\right)\xi_{x,t}\left(k\right)\frac{\mathrm{d}k}{\pi}$
$\displaystyle=\int\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\widehat{p}\left(k\right)\xi_{x,t}\left(k\right)\frac{\mathrm{d}k}{\pi}.$

It remains to show that this integral is absolutely convergent. It follows
from (7.2) that

$\displaystyle\left\|p\right\|$
$\displaystyle\lesssim_{\left\|q_{+}\right\|_{L^{1}}}\left\|q\right\|+C\left(\int_{0}^{\infty}\left(\int_{x}^{\infty}\left|q\right|\right)^{2}\mathrm{d}x\right)^{1/2}$
$\displaystyle\leq\left\|q\right\|+C\left(\int_{0}^{\infty}x\left|q\left(x\right)\right|\left(\int_{x}^{\infty}\left|q\right|\right)\mathrm{d}x\right)^{1/2}\
\ \ \ \ \left(q\in L_{1}^{1}\right)$
$\displaystyle\leq\left\|q\right\|+C\left\|q\right\|_{L^{1}}^{1/2}\left\|q\right\|_{L_{1}^{1}}^{1/2}<\infty$

and hence $\widehat{p}\in L^{2}$. Therefore, $q_{12}\left(x,t\right)$ is
locally bounded for $t\geq 0$. (In fact, continuous).

Consider the remaining term $q_{11}\left(x,t\right)$. In order to proceed we
need first to regularize the improper integral. It cannot be done by merely
deforming $\mathbb{R}$ to $\mathbb{R+}\mathrm{i}a$ as is done for $q_{3}$
since $\widehat{q}\left(k\right)$ need not admit analytic continuation into
the upper half plane. To detour this circumstance we define
$\widehat{q}\left(k\right)$ by

$\widehat{q}\left(\overline{k}\right)=\int_{0}^{\infty}e^{-2\mathrm{i}\overline{k}s}q\left(s\right)\mathrm{d}s,\
\ \ \operatorname{Im}k\geq 0,$

and apply the Cauchy-Green formula for the strip $0\leq\operatorname{Im}k\leq
1$. We have

$\displaystyle q_{11}\left(x,t\right)$
$\displaystyle=\partial_{x}\left(PV\right)\int\xi_{x,t}\left(k\right)\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\widehat{q}\left(\overline{k}\right)\frac{\mathrm{d}k}{\pi}$
$\displaystyle=\partial_{x}\int_{0\leq\operatorname{Im}k\leq
1}\xi_{x,t}\left(k\right)\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\partial_{\overline{k}}\widehat{q}\left(\overline{k}\right)\frac{\mathrm{d}u\mathrm{d}v}{\pi^{2}}\text{
\ \ }(k=u+\mathrm{i}v)$
$\displaystyle+\partial_{x}\int_{\mathbb{R+}\mathrm{i}}\xi_{x,t}\left(k\right)\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\widehat{q}\left(\overline{k}\right)\frac{\mathrm{d}k}{\pi}$
$\displaystyle=:q_{111}\left(x,t\right)+q_{112}\left(x,t\right).$

The second term $q_{112}\left(x,t\right)$ is treated the same way as
$q_{13}\left(x,t\right)$ and one immediately concludes that
$q_{112}\left(x,t\right)$ are bounded (in fact, smooth) for $t>0$. Turn to
$q_{111}$. We rearrange it by observing that the double integral is absolutely
convergent and the order of integration may be interchanged:

$\displaystyle q_{111}\left(x,t\right)$
$\displaystyle=\partial_{x}\int_{0\leq\operatorname{Im}k\leq
1}\xi_{x,t}\left(k\right)\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\left[\int_{0}^{\infty}\left(-2\mathrm{i}s\right)e^{-2\mathrm{i}\overline{k}s}q\left(s\right)\mathrm{d}s\right]\frac{\mathrm{d}u\mathrm{d}v}{\pi^{2}}$
$\displaystyle=-\partial_{x}\int_{0}^{\infty}sq\left(s\right)\left[\int_{0\leq\operatorname{Im}k\leq
1}\xi_{x,t}\left(k\right)\frac{T_{0}\left(k\right)}{k}e^{-2\mathrm{i}\overline{k}s}\frac{\mathrm{d}u\mathrm{d}v}{\pi^{2}}\right]\mathrm{d}s$
$\displaystyle=-\mathrm{i}\int_{0}^{\infty}\left[\int_{0\leq\operatorname{Im}k\leq
1}e^{-2vs}\xi_{x-s,t}\left(k\right)T_{0}\left(k\right)\frac{\mathrm{d}u\mathrm{d}v}{\pi^{2}}\right]sq\left(s\right)\mathrm{d}s,\
\ \ (\overline{k}=k-2\mathrm{i}v)$
$\displaystyle\simeq\int_{0}^{\infty}\left\\{\int_{0}^{1}e^{-2vs}I\left(s-x,t\right)\mathrm{d}v\right\\}sq\left(s\right)\mathrm{d}s,$

where

$\displaystyle I\left(s-x,t\right)$
$\displaystyle:=\int_{\mathbb{R}+\mathrm{i}v}\xi_{x-s,t}\left(k\right)T_{0}\left(k\right)\mathrm{d}k$
$\displaystyle=\int_{\mathbb{R}+\mathrm{i}}\xi_{x-s,t}\left(k\right)T_{0}\left(k\right)\mathrm{d}k$

is independent of $v\geq 0$. Thus

$\displaystyle q_{111}\left(x,t\right)$
$\displaystyle\simeq\partial_{x}\int_{0}^{\infty}\left\\{\int_{0}^{1}e^{-2vs}\mathrm{d}v\right\\}I\left(s-x,t\right)sq\left(s\right)\mathrm{d}s$
$\displaystyle=\int_{0}^{\infty}\frac{1-e^{-2s}}{2}I\left(s-x,t\right)q\left(s\right)\mathrm{d}s$
$\displaystyle=\int_{0}^{\infty}I\left(s-x,t\right)\frac{1-e^{-2s}}{2}q\left(s\right)\mathrm{d}s.$

It remains to study the behavior of $I\left(s-x,t\right)$ as
$s\rightarrow+\infty$. Since $T_{0}\left(k\right)=1+O\left(k^{-1}\right)$ as
$k\rightarrow\infty$ and $x$ is fixed we only need to worry about

$I_{0}\left(s,t\right)=\int_{\mathbb{R}+\mathrm{i}}\xi_{-s,t}\left(k\right)\mathrm{d}k,$

which is closely related to the Airy function. For the reader convenience we
offer a direct treatment. Rewrite

$\xi_{-s,t}\left(k\right)=\exp\mathrm{i}\left[\Omega
S\left(\lambda\right)\right],$

where $S\left(\lambda\right)=\lambda^{3}/3-\lambda$ and

$\omega=2\left(s/3t\right)^{1/2},\Omega=3t\left(s/3t\right)^{3/2},\lambda=k/\omega$

Noticing that we need not adjust the contour of integration, we then have

$I_{0}\left(s,t\right):=\omega\int_{\mathbb{R}+\mathrm{i}}e^{\mathrm{i}\Omega
S\left(\lambda\right)}\mathrm{d}\lambda.$ (7.4)

Apparently, the phase $S\left(\lambda\right)=$ $\lambda^{3}/3-\lambda$ has
stationary points at $\lambda=\pm 1$ and we need to deform the contour in
(7.4) to pass through points $\lambda=\pm 1$. We denote such a contour
$\Gamma$. To apply the steepest descent we need to make sure that
$\exp\mathrm{i}\left[\omega S\left(\lambda\right)\right]$ decay on $\Gamma$
away from $\pm 1$. To this end $\Gamma$ must be in the lower half plane
between points $-1$ and $1$. Noticing that
$\Omega=\left(3t/8\right)\omega^{3}$, $\omega=O\left(s^{1/2}\right)$ by the
steepest descent method (see e.g. [28]) one has

$\displaystyle I_{0}\left(s,t\right)$
$\displaystyle=\omega\int_{\Gamma}e^{\mathrm{i}\Omega
S\left(\lambda\right)}\mathrm{d}\lambda=\omega O\left(\Omega^{-1/2}\right),\ \
\ \Omega\rightarrow+\infty,$ $\displaystyle=O\left(s^{-1/4}\right),\ \ \
s\rightarrow+\infty.$

Thus $q_{111}\left(x,t\right)$ is bounded for $t>0$ (even if $q\left(x\right)$
decays slower than $L^{1}$). All four pieces $q_{1}\left(x,t\right)$ is made
of are bounded and so is $q_{1}\left(x,t\right)$.

There is now only one term $q_{3}$ left in (7.3) to analyze. We are done if we
show that $\partial_{x}y_{0}\in H^{2}$. Differentiating

$y_{0}+\mathbb{H}y_{0}=-\mathbb{H}1,\ \ \
\mathbb{H}:=\mathbb{H}\left(R\xi_{x,t}\right),$

in $x$ one has

$\partial_{x}y_{0}+\mathbb{H}\left(\partial_{x}y_{0}\right)=-\partial_{x}\mathbb{H}1-\left(\partial_{x}\mathbb{H}\right)y_{0}.$

Thus

$\partial_{x}y_{0}=-\left(I+\mathbb{H}\right)^{-1}\left[\left(\partial_{x}\mathbb{H}\right)1+\left(\partial_{x}\mathbb{H}\right)y_{0}\right].$

It follows that we only need to show that
$\left(\partial_{x}\mathbb{H}\right)1\in H^{2}$ and $\partial_{x}\mathbb{H}$
is a bounded operator. Note first that

$\left(\partial_{x}\mathbb{H}\right)=\mathbb{H}\left(2\mathrm{i}kR\xi_{x,t}\right).$

Since $kR\left(k\right)\in L^{2}$ (from the second Zakharov-Faddeev trace
formula),

$\left(\partial_{x}\mathbb{H}\right)1=\mathbb{JP}_{-}\left(2\mathrm{i}kR\left(k\right)\xi_{x,t}\left(k\right)\right)\in
H^{2}$

as desired. The proof of boundedness of $\left(\partial_{x}\mathbb{H}\right)$
is a bit more complicated. By (7.1) we have

$\mathbb{H=H}_{1}+\mathbb{H}_{2}+\mathbb{H}_{3}$

where

$\mathbb{H}_{n}:=\mathbb{H}\left(\phi_{n}\xi_{x,t}\right),n=1,2,\ \ \
\mathbb{H}_{3}:=\mathbb{H}\left(A\xi_{x,t}\right).$

For $n=1,2$ both $\mathbb{H}_{n}$ admit a direct differentiation in $x$.
Indeed, one can easily see that

$\partial_{x}\mathbb{H}_{n}=\mathbb{H}\left(\phi_{n}\partial_{x}\xi_{x,t}\right)=\mathbb{H}\left(2\mathrm{i}k\phi_{n}\xi_{x,t}\right),n=1,2.$

Since $q,p\in L^{1}$

$2\mathrm{i}k\phi_{1}\left(k\right)=T_{0}\left(k\right)\widehat{q}\left(k\right)\in
L^{\infty}$ (7.5)

and

$2\mathrm{i}k\phi_{2}\left(k\right)=\frac{T_{0}\left(k\right)}{2\mathrm{i}k}\widehat{p}\left(k\right)\in
L^{\infty}$

and hence the operators $\partial_{x}\mathbb{H}\left(\phi_{n}\right),n=1,2,$
are bounded. To differentiate $\mathbb{H}_{3}$ we need first to use (3.8). One
has

$\mathbb{H=H}\left(\mathbb{\bar{P}}_{-}\left(R\xi_{x,t}\right)\right).$

But

$\displaystyle\mathbb{P}_{-}\left[A\left(k\right)\xi_{x,t}\left(k\right)\right]$
$\displaystyle=-\frac{1}{2\mathrm{i}\pi}\int\frac{A\left(\lambda\right)\xi_{x,t}\left(\lambda\right)}{\lambda-\left(k-\mathrm{i}0\right)}\mathrm{d}\lambda$
$\displaystyle=-\frac{1}{2\mathrm{i}\pi}\int_{\mathbb{R}+\mathrm{i}}\frac{A\left(\lambda\right)\xi_{x,t}\left(\lambda\right)}{\lambda-k}\mathrm{d}\lambda,$

where the integral is absolutely convergent, and therefore we may
differentiate under the integral sign

$\partial_{x}\mathbb{P}_{-}\left[A\left(k\right)\xi_{x,t}\left(k\right)\right]=-\frac{1}{2\mathrm{i}\pi}\int_{\mathbb{R}+\mathrm{i}}\frac{2\mathrm{i}\lambda
A\left(\lambda\right)\xi_{x,t}\left(\lambda\right)}{\lambda-k}\mathrm{d}\lambda,$

which is well-defined and bounded. Consequently, $\partial_{x}\mathbb{H}_{3}$
is a bounded operator and so is $\partial_{x}\mathbb{H}$. Thus, indeed
$\partial_{x}y_{0}\in H^{2}$. ∎

######  Remark 7.2.

Theorem 7.1 of [13], which proof is based on the Dyson formula, relates
smoothness of $q\left(x,t\right)$ with the decay of $q\left(x\right)$. In
particular, it follows from that result that if $q\left(x\right)\in
L_{3/2}^{1}\cap L^{2}$ then $q\left(x,t\right)\in
L_{\operatorname*{loc}}^{\infty}\cap L^{2}$ for $t>0.$ Stronger decay is due
to the fact the Dyson formula involves $\det\left(I+\mathbb{H}\right)$, which
use requires to analyze differentiability of $\mathbb{H}$ in trace norm. (The
latter is also technically much more involved. It was our attempt to dispose
of trace norm considerations that led us to our trace formulas, which require
uniform norms only.

The following important consequence directly follows from Theorem 7.1 and
invariance of the KdV with respect to
$\left(x,t\right)\rightarrow\left(-x,-t\right)$.

######  Corollary 7.3.

The class $L_{1}^{1}$ is not preserved under the KdV flow.

###### Proof.

Suppose to the contrary that $L_{1}^{1}$ is preserved under the KdV flow. I.e.
if $q\left(x\right)\in L_{1}^{1}$ then $q\left(x,t\right)\in L_{1}^{1}$ for
any $t$. Take $q\left(x\right)\in L_{1}^{1}\cap L^{2}$ but
$q\left(x\right)\notin L^{\infty}$ and fix $t_{0}>0$ By Theorem 7.1,
$q_{0}\left(x\right):=q\left(x,t_{0}\right)\in
L_{\operatorname*{loc}}^{\infty}\cap L^{2}$. Take $q_{0}\left(x\right)$ as new
initial data. By our assumption it is also in $L_{1}^{1}$. Thus
$q_{0}\left(x\right)\in L_{\operatorname*{loc}}^{\infty}\cap L_{1}^{1}\cap
L^{2}$. But this leads us to a contradiction as
$q_{0}\left(x,t_{0}\right)=q\left(-x\right)$ was not assumed locally bounded.
∎

In the conclusion we mention that much more general and precise statements can
be made regarding how the KdV solutions gain regularity (smoothness) in
exchange for loss of decay. We plan on showing elsewhere how the results of
[13], [14], and [25] may be improved to optimal statements.

##  8\. Appendix

We demonstrate that the Deift-Trubowitz trace formula is actually a
”nonlinearization” of our trace formulas. Assume for simplicity that there are
no bound states (non-empty negative spectrum merely complicates the
computations) and do our computation for the $+$ sign only. The reader who has
been able to get to this point should be able to follow the calculations
below. Denoting $\mathbb{H=H}\left(R\xi_{x,t}\right)$, $h:=\mathbb{H}1$,
$1_{a}:=\chi_{\left|\cdot\right|\leq a}$, we have

$\displaystyle\pi q$
$\displaystyle=-2\partial_{x}\int\operatorname{Re}y=\frac{1}{\pi}\partial_{x}\int\operatorname{Re}\left(I+\mathbb{H}\right)^{-1}h$
$\displaystyle=2\partial_{x}\int\operatorname{Re}\left(I+\mathbb{H}\right)^{-1}h$
$\displaystyle=-2\int\left(I+\mathbb{H}\right)^{-1}\left(\partial_{x}\mathbb{H}\right)\mathbb{\left(I+\mathbb{H}\right)}^{-1}h+2\int\left(I+\mathbb{H}\right)^{-1}\partial_{x}h$
$\displaystyle=:q_{1}+q_{2}.$

For $q_{1}$ we have

$q_{1}=-2\lim_{a\rightarrow\infty}\left\langle\left(I+\mathbb{H}\right)^{-1}\left(\partial_{x}\mathbb{H}\right)\mathbb{\left(I+\mathbb{H}\right)}^{-1}h,1_{a}\right\rangle.$

For the inner product one has

$\displaystyle\left\langle\left(I+\mathbb{H}\right)^{-1}\left(\partial_{x}\mathbb{H}\right)\mathbb{\left(I+\mathbb{H}\right)}^{-1}h,\mathbb{P}_{+}1_{a}\right\rangle$
$\displaystyle=\left\langle\left(\partial_{x}\mathbb{H}\right)\mathbb{\left(I+\mathbb{H}\right)}^{-1}h,\left(I+\mathbb{H}\right)^{-1}\mathbb{P}_{+}1_{a}\right\rangle$
$\displaystyle=-\left\langle\left(\partial_{x}\mathbb{H}\right)y,\left(I+\mathbb{H}\right)^{-1}\mathbb{P}_{+}1_{a}\right\rangle$
$\displaystyle=-\left\langle\left(\partial_{x}\mathbb{H}\right)y,\mathbb{P}_{+}\left[1_{a}-\left(I+\mathbb{H}\right)^{-1}\mathbb{HP}_{+}1_{a}\right]\right\rangle$
$\displaystyle=-\left\langle\left(\partial_{x}\mathbb{H}\right)y,\mathbb{P}_{+}1_{a}\right\rangle+\left\langle\left(\partial_{x}\mathbb{H}\right),\left(I+\mathbb{H}\right)^{-1}\mathbb{HP}_{+}1_{a}\right\rangle.$

Passing to the limit yields

$q_{1}=2\int\left(\partial_{x}\mathbb{H}\right)y+\left\langle\left(\partial_{x}\mathbb{H}\right)y,y\right\rangle.$

One may now see how ”nonlinear” dependence on $y$ in (5.7) comes about.
Indeed, the second term
$\left\langle\left(\partial_{x}\mathbb{H}\right)y,y\right\rangle$ is a
quadratic form. For $q_{2}$ we similarly have

$\displaystyle q_{2}$
$\displaystyle=2\int\left(I+\mathbb{H}\right)^{-1}\partial_{x}h$
$\displaystyle=2\lim_{a\rightarrow\infty}\left\langle\left(I+\mathbb{H}\right)^{-1}\left(\partial_{x}h\right),\mathbb{P}_{+}1_{a}\right\rangle$
$\displaystyle=2\lim_{a\rightarrow\infty}\left\langle\partial_{x}h,\mathbb{P}_{+}1_{a}-\left(I+\mathbb{H}\right)^{-1}\mathbb{HP}_{+}1_{a}\right\rangle$
$\displaystyle=\int\partial_{x}h+\left\langle\partial_{x}h,y\right\rangle.$

Since

$\partial_{x}\mathbb{H}f\left(k\right)\mathbb{=}2\mathrm{i}\mathbb{JP}_{-}\left[kR\left(k\right)e^{2\mathrm{i}kx}f\left(k\right)\right]$

we have

$\displaystyle\pi q$ $\displaystyle=q_{1}+q_{2}$
$\displaystyle=\left\langle\left(\partial_{x}\mathbb{H}\right)y,y\right\rangle+2\int\left(\partial_{x}\mathbb{H}\right)y+\left\langle\partial_{x}h,y\right\rangle+2\int\partial_{x}h$
$\displaystyle=2\mathrm{i}\int
kR\left(k\right)e^{2\mathrm{i}kx}y\left(k,x\right)^{2}\mathrm{d}k$
$\displaystyle+4\mathrm{i}\int
kR\left(k\right)e^{2\mathrm{i}kx}y\left(k,x\right)\mathrm{d}k+2\mathrm{i}\int
kR\left(k\right)e^{2\mathrm{i}kx}\mathrm{d}k$ $\displaystyle=2\mathrm{i}\int
kR\left(k\right)e^{2\mathrm{i}kx}\left[1+y\left(k,x\right)\right]^{2}\mathrm{d}k$
$\displaystyle=2\mathrm{i}\int
kR\left(k\right)\psi\left(x,k\right)^{2}\mathrm{d}k$

and (5.7) with $c_{n}=0$ follows.

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