#  $H^{s}$ Bounds for the Derivative Nonlinear Schrödinger Equation

Hajer Bahouri  CNRS & Sorbonne Université, Laboratoire Jacques-Louis Lions
(LJLL) UMR 7598, Place Jussieu, 75005 Paris, France
<EMAIL_ADDRESS>,  Trevor M. Leslie  University of Southern
California, 3620 S. Vermont Ave., Los Angeles, CA 90089<EMAIL_ADDRESS>and
Galina Perelman  Laboratoire D’Analyse et de Mathématiques Appliquées UMR
8050, Université Paris-Est Créteil, 61, Avenue Du Général De Gaulle, 94010
Créteil Cedex, France<EMAIL_ADDRESS>
###### Abstract.

We study the derivative nonlinear Schrödinger equation on the real line and
obtain global-in-time bounds on high order Sobolev norms.

This material is based upon work supported by the National Science Foundation
under Grant No. DMS-1928930 while the authors participated in a program hosted
by the Mathematical Sciences Research Institute in Berkeley, California,
during the Spring 2021 semester.

##  1\. Introduction

We consider the Cauchy problem for the derivative nonlinear Schrödinger
equation (DNLS) on the real line $\mathbb{R}$:

(1)
$\left\\{\begin{array}[]{rcl}i\partial_{t}u+\partial_{x}^{2}u&=&-i\partial_{x}(|u|^{2}u),\\\
u\big{|}_{t=0}&=&u_{0}\in
H^{s}(\mathbb{R}),\;s\geq\frac{1}{2}.\end{array}\right.$

We remark right away that the DNLS is $L^{2}$ critical, as it is invariant
under the scaling

(2) $u(t,x)\mapsto u_{\mu}(t,x):=\sqrt{\mu}u(\mu^{2}t,\mu x),\qquad\mu>0.$

The DNLS equation was introduced by Mio-Ogino-Minami-Takeda and Mjølhus [20,
21] as a model for studying magnetohydrodynamics, and it has received a great
deal of attention from the mathematics community after being shown to be
completely integrable by Kaup-Newell [13]. The infinitely many conserved
quantities admitted by the DNLS equation play an important role in the
wellposedness theory. The first three—the mass, momentum, and energy—are as
follows.

(3) $\displaystyle M(u)$
$\displaystyle:=\int_{\mathbb{R}}|u|^{2}\,\mathrm{d}x,$ (4) $\displaystyle
P(u)$
$\displaystyle:=\operatorname{Im}\int_{\mathbb{R}}\overline{u}u_{x}\,\mathrm{d}x+\frac{1}{2}\int_{\mathbb{R}}|u|^{4}\,\mathrm{d}x,$
(5) $\displaystyle E(u)$
$\displaystyle:=\int_{\mathbb{R}}\big{(}|u_{x}|^{2}-\frac{3}{2}\operatorname{Im}(|u|^{2}u\overline{u}_{x})+\frac{1}{2}|u|^{6}\big{)}\,\mathrm{d}x.$

Before stating our main result, let us give a very brief review of what is
known about the wellposedness of the DNLS equation. More detailed overviews
can be found, for example, in the introductions of [2] and [14]. Local
wellposedness in $H^{s}(\mathbb{R})$ for $s\geq\frac{1}{2}$ was proven by
Takaoka [25], improving earlier work [22] by Ozawa. On the other hand, for
$s<\frac{1}{2}$, the uniform continuity of the data-to-solution map fails in
$H^{s}(\mathbb{R})$ [3, 26]. One can, however, close the
$\frac{1}{2}$-derivative gap between the $H^{\frac{1}{2}}$ threshold and the
critical space $L^{2}(\mathbb{R})$ by working in more general Fourier-Lebesgue
spaces, c.f. Grünrock [6] and references therein.

A line of results, due to Hayashi-Ozawa [8], Colliander-Keel-Staffilani-
Takaoka-Tao [4], Wu [28], and Guo-Wu [7], establishes global well-posedness of
the DNLS equation in $H^{s}(\mathbb{R})$ for $s\geq\frac{1}{2}$, for initial
data having mass less than $4\pi$. Another line (Pelinovsky-Saalmann-
Shimabukuro [23], Pelinovsky-Shimabukuro [24], and Jenkins-Liu-Perry-Sulem
[12, 11, 10]) uses inverse scattering techniques to establish global
wellposedness under stronger regularity and decay assumptions on the initial
data, but without a smallness requirement on the mass.

The first and third authors proved in [2] that the DNLS equation is globally
well-posed in $H^{s}(\mathbb{R})$ for $s\geq\frac{1}{2}$ and that solutions
generated from $H^{\frac{1}{2}}$ initial data remain bounded in
$H^{\frac{1}{2}}(\mathbb{R})$ for all time. There have also been some recent
works below the aforementioned $s=\frac{1}{2}$ threshold of uniform $H^{s}$
continuity with respect to initial data [3, 26]. Klaus-Schippa [17] gave
$H^{s}$ a priori estimates for $0<s<\frac{1}{2}$ in the case of small mass,
Killip-Ntekoume-Vişan [14] improved the small mass assumption to $4\pi$ and
furthermore proved a global wellposedness result in $H^{s}(\mathbb{R})$,
$\frac{1}{6}\leq s<\frac{1}{2}$, for initial data with mass less than $4\pi$.
Very recently, Harrop-Griffiths, Killip, and Vişan [9] have removed the small
mass assumption both from their $H^{s}$ a priori bounds, $0<s<\frac{1}{2}$, as
well as from their global wellposedness result in $H^{s}(\mathbb{R})$ with
$\frac{1}{6}\leq s<\frac{1}{2}$.

In this paper, we are concerned with the global-in-time boundedness of
solutions to the DNLS equation in $H^{s}$ spaces. We prove that a uniform-in-
time bound in $H^{s}(\mathbb{R})$ holds for all $s\geq\frac{1}{2}$.

######  Theorem 1.1.

Suppose $u$ is a solution to the DNLS equation with initial data $u_{0}\in
H^{s}(\mathbb{R})$, $s\geq\frac{1}{2}$. There exists a finite positive
constant $C=C(s,\|u_{0}\|_{H^{s}(\mathbb{R})})$, such that

$\sup_{t\in\mathbb{R}}\|u(t)\|_{H^{s}(\mathbb{R})}\leq
C(s,\|u_{0}\|_{H^{s}(\mathbb{R})}).$

The main idea is to build off of the $H^{s}$ bounds with $0<s<\frac{1}{2}$
from [9] and to take advantage of the complete integrability of the equation.
As in [2], [9], the present work relies heavily on the conservation of the
transmission coefficient for the spectral problem associated to the DNLS
equation. This property has already been used in many other works; of
particular relevance to us are the papers of Gérard [5], Killip-Vişan-Zhang
[16], Killip-Vişan [15], and Koch-Tataru [18], on the cubic NLS and KdV
equations.

Note that by continuity of the flow, and the preservation of the Schwartz
class under the flow, we lose nothing by restricting attention to the Schwartz
class; we will thus work exclusively with Schwartz functions for the remainder
of the manuscript. We will also suppress the time dependence when it does not
play a role.

One can easily prove Theorem 1.1 in the special case $s=1$, using the
conserved quantity $E(u)$. Indeed, simply rearranging (5) yields

$\|u\|_{\dot{H}^{1}(\mathbb{R})}^{2}=E(u)-\frac{1}{2}\|u\|_{L^{6}(\mathbb{R})}^{6}+\frac{3}{2}\int_{\mathbb{R}}\operatorname{Im}(|u|^{2}u\overline{u}_{x})\,\mathrm{d}x.$

Clearly, the last term can be bounded above in absolute value by
$\frac{1}{2}\|u\|_{\dot{H}^{1}(\mathbb{R})}^{2}+C\|u\|_{L^{6}(\mathbb{R})}^{6}$,
whence the desired bound follows by Sobolev embedding.

The higher-order Sobolev norms of integer order can be dealt with similarly,
once we have a formula for the corresponding higher-order conserved
quantities. We will show that for any nonnegative integer $\ell$, one of the
conserved quantities is equal to a constant multiple of
$\|u\|_{\dot{H}^{\ell}(\mathbb{R})}^{2}$, plus terms which are of lower order.
For noninteger $s$, we will use a sort of ‘generalized energy’, comparable to
$\|u\|_{\dot{H}^{s}(\mathbb{R})}^{2}$, that will be defined in terms of the
transmission coefficient of the DNLS spectral problem. We sketch presently the
background necessary to define these objects precisely; for more details, see,
for example, [1, 12, 11, 10, 13, 19, 24, 27].

The DNLS equation can be obtained as a compatibility condition of the
following system [13]:

(6) $\begin{split}\partial_{x}\psi&=\mathcal{U}(\lambda)\psi,\\\
\partial_{t}\psi&=\Upsilon(\lambda)\psi.\end{split}$

Here $\lambda\in\mathbb{C}$ is a spectral parameter, independent of $t$ and
$x$, and $\psi=\psi(t,x,\lambda)$ is $\mathbb{C}^{2}$-valued. The operators
$\mathcal{U}(\lambda)$ and $\Upsilon(\lambda)$ are defined by

(7) $\begin{split}\mathcal{U}(\lambda)&=-i\sigma_{3}(\lambda^{2}+i\lambda
U),\\\
\Upsilon(\lambda)&=-i(2\lambda^{4}-\lambda^{2}|u|^{2})\sigma_{3}+\begin{pmatrix}0&2\lambda^{3}u-\lambda|u|^{2}u+i\lambda
u_{x}\\\
-2\lambda^{3}\overline{u}+\lambda|u|^{2}u+i\lambda\overline{u}_{x}&0\end{pmatrix},\end{split}$

where

$\sigma_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},\qquad
U=\begin{pmatrix}0&u\\\ \overline{u}&0\end{pmatrix}.$

To be more specific about the sense in which the DNLS is a compatibility
condition, we note that $u$ satisfies the DNLS equation if and only if
$\mathcal{U}$ and $\Upsilon$ satisfy the so-called ‘zero-curvature’
representation

$\frac{\partial\mathcal{U}}{\partial t}-\frac{\partial\Upsilon}{\partial
x}+[\mathcal{U},\Upsilon]=0.$

The first equation of (6) can be written in the form

(8) $L_{u}(\lambda)\psi:=(i\sigma_{3}\partial_{x}-\lambda^{2}-i\lambda
U)\psi=0,$

which defines the scattering transform associated to the DNLS. Let us denote

$\Omega_{+}:=\\{\lambda\in\mathbb{C}:\operatorname{Im}\lambda^{2}>0\\}.$

Then given $u\in\mathcal{S}(\mathbb{R})$ and
$\lambda\in\overline{\Omega}_{+}$, there are unique solutions to (8) (the
“Jöst solutions”) exhibiting the following behavior at $\pm\infty$:

(9)
$\begin{split}\psi_{1}^{-}(x,\lambda)=e^{-i\lambda^{2}x}\left[\begin{pmatrix}1\\\
0\end{pmatrix}+o(1)\right],&\qquad\text{ as }x\to-\infty,\\\
\psi_{2}^{+}(x,\lambda)=e^{i\lambda^{2}x}\left[\begin{pmatrix}0\\\
1\end{pmatrix}+o(1)\right],&\qquad\text{ as }x\to+\infty.\end{split}$

Finally, we denote by $a_{u}(\lambda)$ the Wronskian of the Jöst solutions
defined above:111The transmission coefficient mentioned earlier is the inverse
of $a_{u}(\lambda)$.

(10) $a_{u}(\lambda)=\det(\psi_{1}^{-}(x,\lambda),\psi_{2}^{+}(x,\lambda)).$

Using the second equation in (6), it can be shown that $a_{u}(\lambda)$ is
time-independent if $u$ is a solution of (1). Furthermore, $a_{u}$ is a
holomorphic function of $\lambda$ in $\Omega_{+}$, and one may determine the
behavior of $a_{u}$ at infinity by transforming (8) into a Zakharov-Shabat
spectral problem, linear with respect to the spectral parameter, c.f. [13],
[23]. The equivalence between the two problems allows us to write

(11)
$\lim_{|\lambda|\to\infty,\,\lambda\in\overline{\Omega}_{+}}a_{u}(\lambda)=e^{-\frac{i}{2}\|u\|_{L^{2}(\mathbb{R})}^{2}}.$

For fixed $u$, we can thus define the logarithm so that

(12) $\lim_{|\lambda|\to\infty,\lambda\in\overline{\Omega}_{+}}\ln
a_{u}(\lambda)=-\frac{i}{2}\|u\|_{L^{2}(\mathbb{R})}^{2}.$

Moreover, $\ln a_{u}(\lambda)$ admits an asymptotic expansion of the following
form:

(13) $\ln
a_{u}(\lambda)=\sum_{j=0}^{\infty}\frac{E_{j}(u)}{\lambda^{2j}}\qquad\text{ as
}|\lambda|\to\infty,\;\lambda\in\overline{\Omega}_{+}.$

Since $a_{u}(\lambda)$ is time-independent, the quantities $E_{j}(u)$ are
conservation laws. They are all polynomial in $u$, $\overline{u}$, and their
derivatives. Furthermore, the $E_{j}(u)$’s inherit scaling properties from
$a_{u}(\lambda)$. That is, for $\mu>0$, the fact that
$a_{u_{\mu}}(\lambda)=a_{u}(\frac{\lambda}{\sqrt{\mu}})$ implies that
$E_{j}(u_{\mu})=\mu^{j}E_{j}(u)$, for each $j\in\mathbb{N}$. The first several
of the $E_{j}(u)$’s are (up to multiplicative constants) the conserved
quantities (3)–(5) mentioned earlier:

$E_{0}(u)=-\frac{i}{2}\|u\|_{L^{2}(\mathbb{R})}^{2}=-\frac{i}{2}M(u),\qquad
E_{1}(u)=\frac{i}{4}P(u),\qquad E_{2}(u)=-\frac{i}{8}E(u).$

For each $\ell\in\mathbb{N}^{*}$, the quantity $E_{2\ell}(u)$ can be used to
control $\|u\|_{\dot{H}^{\ell}(\mathbb{R})}^{2}$. Let us define, for $\rho$
positive sufficiently large and $L\in\mathbb{N}$,

(14) $\varphi_{L}(u,\rho)=\operatorname{Im}\left[\ln
a_{u}(\sqrt{i\rho})-\sum_{j=0}^{2L+1}\frac{E_{j}(u)}{(i\rho)^{j}}\right].$

If $u$ is a solution of the DNLS equation, then $\varphi_{L}(u,\rho)$ is time-
independent, being a sum of time-independent quantities.

In order to establish bounds on the $H^{s}$ norm of $u$, for
$s\geq\frac{1}{2}$, we will show that
$\int_{R}^{\infty}\rho^{2s-1}\varphi_{[s]}(u,\rho)\mathrm{d}\rho$ with $R>0$
large enough controls the $\dot{H}^{s}$ seminorm of $u$, in a sense to be made
precise later. Here and below, we use $[s]$ to denote the integer part of a
real number $s$.

Our proof of Theorem 1.1 relies on a good understanding of the structure of
the remainder associated to the expansion (13). Note that when
$\lambda^{2}=i\rho$, the imaginary part of this remainder (which is what we
really use) is simply $\varphi_{L}(u,\rho)$. In Section 2, we will introduce a
determinant characterization of $a_{u}(\lambda)$; we use this characterization
to formulate a technical statement (Lemma 2.1 below) on the size of the
remainder. Assuming the result of Lemma 2.1, we will prove Theorem 1.1 at the
end of Section 2. Then, in Section 3, we will prove our technical Lemma,
completing the circle of ideas. Most of the work is contained in this last
section.

Before moving on, let us establish a few notational conventions that we wish
to add to the ones introduced above. First of all, we use the following
normalization for the Fourier transform:

$\widehat{f}(\zeta)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{ix\zeta}f(x)\,\mathrm{d}x.$

The symbol $\mathbb{N}$ will denote the nonnegative integers, and
$\mathbb{N}^{*}=\mathbb{N}\backslash\\{0\\}$. We will use $\|\cdot\|_{2}$ to
denote the Hilbert-Schmidt norm, and $\|\cdot\|$ will denote the operator norm
on $L^{2}(\mathbb{R})$. And we will use the following shorthand for
derivatives:

$D=-i\partial_{x},\qquad\mathcal{L}_{0}=i\sigma_{3}\partial_{x}.$

Whenever $2\leq p<\infty$, we will use $s^{*}(p)$ to denote the Sobolev
exponent $s^{*}(p)=\frac{1}{2}-\frac{1}{p}$ such that the embedding
$H^{s^{*}(p)}(\mathbb{R})\hookrightarrow L^{p}(\mathbb{R})$ holds.

Finally, we set notation for the following subset of $\Omega_{+}$:

$\Gamma_{\delta}=\\{\lambda\in\Omega_{+}:\delta<\arg(\lambda^{2})<\pi-\delta\\}.$

This notation will be useful in some of the intermediate steps we use to prove
Theorem 1.1, as our estimates will frequently depend on
$\frac{|\lambda|^{2}}{\operatorname{Im}\lambda^{2}}$ (which is $\leq
C(\delta)$ on $\Gamma_{\delta}$). However, the value of $\delta>0$ will be
inconsequential for our final steps, where we will take $\lambda^{2}$ to be
pure imaginary. Therefore, for simplicity of presentation, we will fix
$\delta>0$ once and for all and suppress dependence on $\delta$ in all bounds
below.

##  2\. Proof of the Main Result

###  2.1. The Determinant Characterization of $a_{u}(\lambda)$

An important property of $a_{u}(\lambda)$ is the fact that it can be realized
as a perturbation determinant:

(15) $a_{u}(\lambda)^{2}=\det(I-T_{u}(\lambda)^{2}),$

where

$T_{u}(\lambda)=i\lambda(\mathcal{L}_{0}-\lambda^{2})^{-1}U,\qquad\lambda\in\Omega_{+}.$

The operator $T_{u}(\lambda)$ is Hilbert-Schmidt, with

(16)
$\|T_{u}(\lambda)\|_{2}^{2}=\frac{|\lambda|^{2}}{\operatorname{Im}(\lambda^{2})}\|u\|_{L^{2}(\mathbb{R})}^{2}.$

As a consequence of (15), we may write222This series expansion of $\ln
a_{u}(\lambda)$ is consistent with the definition (12).

(17) $\ln
a_{u}(\lambda)=-\sum_{k=1}^{\infty}\frac{\operatorname{Tr}(T_{u}(\lambda)^{2k})}{2k},\qquad\text{
if }\|T_{u}(\lambda)\|<1.$

This series will converge whenever $\lambda\in\Gamma_{\delta}$ has large
enough modulus; indeed, using the explicit kernel of
$(\mathcal{L}_{0}-\lambda^{2})^{-1}$, it can easily be shown that for any
$p>2$, we have

(18)
$\|T_{u}(\lambda)\|\lesssim\frac{|\lambda|\|u\|_{L^{p}(\mathbb{R})}}{\operatorname{Im}(\lambda^{2})^{1-\frac{1}{p}}},\qquad\lambda\in\Omega_{+},\;u\in
L^{p}(\mathbb{R}).$

In particular, we can find $R_{0}=R_{0}(\|u\|_{H^{\frac{1}{3}}(\mathbb{R})})$
such that $\|T_{u}(\lambda)\|\leq\frac{1}{2}$ for all
$\lambda\in\Gamma_{\delta}$ satisfying $|\lambda|^{2}\geq R_{0}$. We will fix
the notation $R_{0}$ for use below.

As we shall see later, each term of the series (17) can be expanded in powers
of $\lambda^{-2}$:

(19)
$-\frac{\operatorname{Tr}(T_{u}(\lambda)^{2k})}{2k}=\sum_{j=k-1}^{\infty}\frac{\mu_{j,k}(u)}{\lambda^{2j}}.$

According to (13) and (17), the $E_{j}(u)$’s should then satisfy

(20) $E_{j}(u)=\sum_{k=1}^{j+1}\mu_{j,k}(u).$

We will use the following notation for the remainders after truncation of the
expansions (17) and (19):

(21) $\ln
a_{u}(\lambda)=-\sum_{k=1}^{2L+2}\frac{\operatorname{Tr}(T_{u}(\lambda)^{2k})}{2k}+\tau_{L}^{*}(u,\lambda),\qquad
L\in\mathbb{N};$ (22)
$-\frac{\operatorname{Tr}(T_{u}(\lambda)^{2k})}{2k}=\sum_{j=k-1}^{2L+1}\frac{\mu_{j,k}(u)}{\lambda^{2j}}+\tau^{k}_{L}(u,\lambda),\qquad
k\in\\{1,\ldots,2L+2\\},\;L\in\mathbb{N}.$

The primary difficulty of the proof of Theorem 1.1—and indeed, the subject of
Lemma 2.1—is the understanding of the size and structure of the remainder
terms $\tau^{k}_{L}(u,\lambda)$, and to a lesser extent, the $\mu_{j,k}(u)$’s.
On the other hand, for $\lambda\in\Gamma_{\delta}$ with large enough modulus,
it is easy to bound the $\tau_{L}^{*}(u,\lambda)$’s. For example, if
$\|T_{u}(\lambda)\|\leq\frac{1}{2}$, then

(23) $\begin{split}|\tau_{L}^{*}(u,\lambda)|&=\bigg{|}\ln
a_{u}(\lambda)+\sum_{k=1}^{2L+2}\frac{\operatorname{Tr}T_{u}^{2k}(\lambda)}{2k}\bigg{|}\leq\sum_{k=2L+3}^{\infty}\|T_{u}(\lambda)\|^{2k-2}\|T_{u}(\lambda)\|_{2}^{2}\\\
&\lesssim\|T_{u}(\lambda)\|^{4L+4}\|T_{u}(\lambda)\|_{2}^{2}\lesssim\frac{\|u\|_{H^{s^{*}(p)}(\mathbb{R})}^{4L+4}\|u\|_{L^{2}(\mathbb{R})}^{2}}{|\lambda|^{(4L+4)(1-\frac{2}{p})}},\quad
2<p<\infty,\;s^{*}(p)=\frac{1}{2}-\frac{1}{p}.\end{split}$

The following table summarizes the various relationships among the quantities
introduced above and will be helpful to keep track of the numerology. More
precise information about the $\mu_{j,k}(u)$’s and $\tau_{L}^{k}(u,\lambda)$’s
will be provided below.

$\begin{array}[]{rcccccccccccccc}&&-\dfrac{\operatorname{Tr}T_{u}^{2}(\lambda)}{2}&&-\dfrac{\operatorname{Tr}T_{u}^{4}(\lambda)}{4}&&-\dfrac{\operatorname{Tr}T_{u}^{6}(\lambda)}{6}&&\cdots&&-\dfrac{\operatorname{Tr}T_{u}^{4L+2}(\lambda)}{4L+2}&&-\dfrac{\operatorname{Tr}T_{u}^{4L+4}(\lambda)}{4L+4}&&\\\
\cline{3-14}\cr\ln
a_{u}(\lambda)=&&\mu_{0,1}(u)&&&&&&&&&&&\vline&E_{0}(u)\\\\[2.84544pt]
&+&\dfrac{\mu_{1,1}(u)}{\lambda^{2}}&+&\dfrac{\mu_{1,2}(u)}{\lambda^{2}}&&&&&&&&&\vline&\dfrac{E_{1}(u)}{\lambda^{2}}\\\
&+&\dfrac{\mu_{2,1}(u)}{\lambda^{4}}&+&\dfrac{\mu_{2,2}(u)}{\lambda^{4}}&+&\dfrac{\mu_{2,3}(u)}{\lambda^{4}}&&&&&&&\vline&\dfrac{E_{2}(u)}{\lambda^{4}}\\\
&&\vdots&&\vdots&&\vdots&&\ddots&&&&&\vline&\vdots\\\
&+&\dfrac{\mu_{2L,1}(u)}{\lambda^{4L}}&+&\dfrac{\mu_{2L,2}(u)}{\lambda^{4L}}&+&\dfrac{\mu_{2L,3}(u)}{\lambda^{4L}}&+&\cdots&+&\dfrac{\mu_{2L,2L+1}(u)}{\lambda^{4L}}&&&\vline&\dfrac{E_{2L}(u)}{\lambda^{4L}}\\\
&+&\dfrac{\mu_{2L+1,1}(u)}{\lambda^{4L+2}}&+&\dfrac{\mu_{2L+1,2}(u)}{\lambda^{4L+2}}&+&\dfrac{\mu_{2L+1,3}(u)}{\lambda^{4L+2}}&+&\cdots&+&\dfrac{\mu_{2L+1,2L+1}(u)}{\lambda^{4L+2}}&+&\dfrac{\mu_{2L+1,2L+2}(u)}{\lambda^{4L+2}}&\vline&\dfrac{E_{2L+1}(u)}{\lambda^{4L+2}}\\\\[8.5359pt]
\cline{14-15}\cr&+&\tau_{L}^{1}(u,\lambda)&+&\tau_{L}^{2}(u,\lambda)&+&\tau_{L}^{3}(u,\lambda)&+&\cdots&+&\tau_{L}^{2L+1}(u,\lambda)&+&\tau_{L}^{2L+2}(u,\lambda)&+&\tau_{L}^{*}(u,\lambda)\end{array}$

###  2.2. Structure of the Traces

In this section, we record all the information about the traces that we need
in order to prove our main result. We deal first with the easy case of
$\operatorname{Tr}T_{u}(\lambda)^{2}$, about which we need more explicit
information. A straightforward computation gives us

(24)
$\operatorname{Tr}T_{u}^{2}(\lambda)=2i\lambda^{2}\int_{\mathbb{R}}\frac{|\widehat{u}(\zeta)|^{2}}{\zeta+2\lambda^{2}}\mathrm{d}\zeta.$

We determine the expansion of $\operatorname{Tr}T_{u}^{2}(\lambda)$ by simply
substituting into (24) the identity

$\frac{2\lambda^{2}}{\zeta+2\lambda^{2}}=\sum_{j=0}^{2L+1}\left(-\frac{\zeta}{2\lambda^{2}}\right)^{j}+\frac{\zeta}{\zeta+2\lambda^{2}}\left(\frac{\zeta}{2\lambda^{2}}\right)^{2L+1},\qquad
L\in\mathbb{N},$

to obtain

(25)
$-\frac{\operatorname{Tr}T_{u}^{2}(\lambda)}{2}=\sum_{j=0}^{2L+1}\frac{1}{\lambda^{2j}}\cdot\underbrace{\frac{i}{(-2)^{j+1}}\int_{\mathbb{R}}\zeta^{j}|\widehat{u}(\zeta)|^{2}\mathrm{d}\zeta}_{=:\mu_{j,1}(u)}-\underbrace{\frac{i}{4^{L+1}\lambda^{4L+2}}\int_{\mathbb{R}}\frac{\zeta^{2L+2}|\widehat{u}(\zeta)|^{2}}{\zeta+2\lambda^{2}}\mathrm{d}\zeta}_{=:\tau_{L}^{1}(u,\lambda)},\qquad
L\in\mathbb{N}.$

Now we state our main Lemma, which describes the structure of the other
$\mu_{j,k}(u)$’s and $\tau_{L}^{k}(u,\lambda)$’s.

######  Lemma 2.1.

For any $k\in\mathbb{N}^{*}$, $L\in\mathbb{N}$, the traces
$\operatorname{Tr}T_{u}^{2k}(\lambda)$ admit the decomposition (22). The
$\mu_{j,k}(u)$’s and $\tau_{L}^{k}(u,\lambda)$’s satisfy the properties below,
where for any $n\in\mathbb{N}$ we denote $\sigma(n)=\max\\{n,\frac{1}{3}\\}$.

  * •

Each $\mu_{j,k}(u)$ is a homogeneous polynomial of degree $2k$ in $u$,
$\overline{u}$, and their derivatives; it is homogeneous with respect to the
natural scaling. We have

(26)
$\begin{split}|\mu_{2\ell,2}(u)|&\lesssim\|u\|_{H^{\sigma(\ell-1)}(\mathbb{R})}^{3}\|u\|_{H^{\ell}(\mathbb{R})},\qquad\ell\in\mathbb{N}^{*},\\\
|\mu_{2\ell,k}(u)|&\lesssim\|u\|_{H^{\sigma(\ell-1)}(\mathbb{R})}^{2k},\qquad\qquad\quad\;\;\ell\in\mathbb{N}^{*},\;k\in\\{3,\ldots,2\ell+1\\},\\\
|\mu_{2\ell+1,k}(u)|&\lesssim\|u\|_{H^{\ell}(\mathbb{R})}^{2k},\qquad\qquad\qquad\quad\;\ell\in\mathbb{N}^{*},\;k\in\\{2,\ldots,2\ell+2\\}.\end{split}$

  * •

For $|\lambda|^{2}>R_{0}$, $\lambda\in\Gamma_{\delta}$, we have the following
bounds:

(27) $\displaystyle|\tau_{L}^{2}(u,\lambda)|$
$\displaystyle\lesssim_{\alpha}\frac{\|u\|_{H^{\sigma(L)}(\mathbb{R})}^{3}\|u\|_{H^{L+\alpha}(\mathbb{R})}}{|\lambda|^{4L+2+2\alpha}},$
$\displaystyle L\in\mathbb{N},\;0\leq\alpha<1;$ (28)
$\displaystyle|\tau_{L}^{k}(u,\lambda)|$
$\displaystyle\lesssim\frac{\|u\|_{H^{L}(\mathbb{R})}^{2k}}{|\lambda|^{4L+4}},$
$\displaystyle L\in\mathbb{N}^{*},\;k\in\\{3,\ldots,2L+2\\}.$

We postpone the proof of the Lemma until Section 3.

###  2.3. Proof of Theorem 1.1

In this section, we will prove Theorem 1.1, assuming the result of Lemma 2.1.
For $s\in\mathbb{N}^{*}$, the conclusion follows easily from Lemma 2.1,
together with (20), (25), and an induction argument; we provide the details
presently. Actually, the case $s=1$ was already proved in the Introduction.
Therefore, let us turn to our inductive hypothesis. For $k=1,\ldots,\ell-1$,
we assume that the following bound holds.

(29) $\sup_{t\in\mathbb{R}}\|u(t)\|_{H^{k}(\mathbb{R})}\leq
C(k,\|u_{0}\|_{H^{k}(\mathbb{R})}).$

We will prove that the same bound holds with $k=\ell\geq 2$.

First of all, for any integer $\ell\geq 2$, and any time $t$, we have

$\displaystyle\|u(t)\|_{\dot{H}^{\ell}(\mathbb{R})}^{2}$
$\displaystyle=C(\ell)\mu_{2\ell,1}(u(t))$ $\displaystyle\text{ by
}\eqref{e:trtu2expanded}$
$\displaystyle=C(\ell)\left[E_{2\ell}(u(t))-\sum_{k=2}^{2\ell+1}\mu_{2\ell,k}(u(t))\right]$
$\displaystyle\text{ by }\eqref{e:Ejmu}$ $\displaystyle\leq
C(\ell)E_{2\ell}(u_{0})+\frac{1}{2}\|u(t)\|_{\dot{H}^{\ell}(\mathbb{R})}^{2}+C(\ell,\|u_{0}\|_{H^{\ell-1}(\mathbb{R})}).$

To pass to the last line, we used time-independence of $E_{2\ell}(u(t))$, the
bounds (26), and our inductive hypothesis (29) (with $k=\ell-1$). Finally,
using that

$E_{2\ell}(u_{0})=\sum_{k=1}^{2\ell+1}\mu_{2\ell,k}(u_{0})\leq
C(\ell,\|u_{0}\|_{H^{\ell}(\mathbb{R})}),$

we get

$\sup_{t\in\mathbb{R}}\|u(t)\|_{\dot{H}^{\ell}(\mathbb{R})}\leq
C(\ell,\|u_{0}\|_{H^{\ell}(\mathbb{R})}),$

which finishes the induction argument, and thus the proof of Theorem 1.1 for
$s\in\mathbb{N}^{*}$.

It remains to consider the situation where $s\notin\mathbb{N}^{*}$. We start
by recording the characterization of $\varphi_{L}(u,\rho)$ in terms of the
remainders $\tau_{L}^{k}(u,\sqrt{i\rho})$, and we also set notation for the
quadratic part of $\varphi_{L}(u,\rho)$. We also note that the case $L=0$ is
included in the definition.

(30) $\displaystyle\varphi_{L}(u,\rho)$
$\displaystyle=\operatorname{Im}\left[\ln
a_{u}(\sqrt{i\rho})-\sum_{j=0}^{2L+1}\frac{E_{j}(u)}{(i\rho)^{j}}\right]=\operatorname{Im}\left[\sum_{k=1}^{2L+2}\tau_{L}^{k}(u,\sqrt{i\rho})+\tau_{L}^{*}(u,\sqrt{i\rho})\right],$
$\displaystyle L\in\mathbb{N},$ (31) $\displaystyle\varphi_{L,0}(u,\rho)$
$\displaystyle=\operatorname{Im}\tau^{1}_{L}(u,\sqrt{i\rho})=\frac{(-1)^{L}}{2^{2L+1}\rho^{2L}}\int_{\mathbb{R}}\frac{\zeta^{2L+2}|\widehat{u}(\zeta)|^{2}}{\zeta^{2}+4\rho^{2}}\mathrm{d}\zeta,$
$\displaystyle L\in\mathbb{N}.$

The conclusion of Theorem 1.1 for noninteger $s\geq\frac{1}{2}$ will be
deduced from the following two Lemmas.

######  Lemma 2.2.

Suppose $u\in\mathcal{S}(\mathbb{R})$, $s>0$, $s\notin\mathbb{N}^{*}$, and
$R>0$. Then the following comparison holds.

(32)
$\int_{\mathbb{R}_{+}}\rho^{2s-1}|\varphi_{[s],0}(u,\rho)|\mathrm{d}\rho\lesssim_{s}\|u\|_{\dot{H}^{s}(\mathbb{R})}^{2}\lesssim_{s}\int_{R}^{\infty}\rho^{2s-1}|\varphi_{[s],0}(u,\rho)|\mathrm{d}\rho+R^{2(s-[s])}\|u\|_{\dot{H}^{[s]}(\mathbb{R})}^{2}.$

###### Proof.

Let us define the function $f_{\nu}:\mathbb{R}\to\mathbb{R}$, for $0<\nu<1$,
by $f_{\nu}(z)=\frac{|z|^{2\nu-1}}{1+z^{2}}$. Note that $f_{\nu}\in
L^{1}(\mathbb{R})$ for this range of $\nu$.

We make a direct substitution of the formula (31) for
$\varphi_{[s],0}(u,\rho)$ into the left side of (32), then we switch the order
of integration. Continuing the computation yields

$\displaystyle\int_{R}^{\infty}\rho^{2s-1}|\varphi_{[s],0}(u,\rho)|\mathrm{d}\rho$
$\displaystyle=\frac{1}{2^{2[s]+1}}\int_{\mathbb{R}}\zeta^{2[s]+2}|\widehat{u}(\zeta)|^{2}\int_{R}^{\infty}\frac{\rho^{2(s-[s])-1}}{\zeta^{2}+4\rho^{2}}\mathrm{d}\rho\,\mathrm{d}\zeta$
$\displaystyle=\frac{1}{2^{2s+1}}\int_{\mathbb{R}}|\zeta|^{2s}|\widehat{u}(\zeta)|^{2}\int_{\frac{2R}{|\zeta|}}^{\infty}f_{s-[s]}(z)\,\mathrm{d}z\,\mathrm{d}\zeta$
$\displaystyle=\frac{1}{4^{s+1}}\|f_{s-[s]}\|_{L^{1}(\mathbb{R})}\|u\|_{\dot{H}^{s}(\mathbb{R})}^{2}-\frac{1}{2}\int_{\mathbb{R}}\left|\frac{\zeta}{2}\right|^{2s}|\widehat{u}(\zeta)|^{2}\int_{0}^{\frac{2R}{|\zeta|}}f_{s-[s]}(z)\,\mathrm{d}z\,\mathrm{d}\zeta.$

We estimate the second term on the right by means of the trivial replacement
$\frac{1}{1+z^{2}}\leq 1$:

$\displaystyle\frac{1}{2}\int_{\mathbb{R}}\left|\frac{\zeta}{2}\right|^{2s}|\widehat{u}(\zeta)|^{2}\int_{0}^{\frac{2R}{|\zeta|}}f_{s-[s]}(z)\,\mathrm{d}z\,\mathrm{d}\zeta$
$\displaystyle\leq\frac{1}{2}\int_{\mathbb{R}}\left|\frac{\zeta}{2}\right|^{2s}|\widehat{u}(\zeta)|^{2}\int_{0}^{\frac{2R}{|\zeta|}}z^{2(s-[s])-1}\,\mathrm{d}z\,\mathrm{d}\zeta=\frac{R^{2(s-[s])}}{s-[s]}\cdot\frac{\|u\|_{\dot{H}^{[s]}(\mathbb{R})}^{2}}{4^{[s]+1}}.$

The comparison (32) follows. ∎

######  Lemma 2.3.

Suppose $u\in\mathcal{S}(\mathbb{R})$, $s>0$, $s\notin\mathbb{N}^{*}$.
Denoting $\beta=\max\\{[s],\,\frac{s+[s]+1}{4([s]+1)},\,\frac{1}{3}\\}$, we
have

(33)
$|\varphi_{[s]}(u,\rho)-\varphi_{[s],0}(u,\rho)|\leq\frac{C(s,\|u\|_{H^{\beta}(\mathbb{R})})}{\rho^{s+[s]+1}}(\|u\|_{H^{s}(\mathbb{R})}+1),\quad\forall\rho\geq
R_{0}.$

###### Proof.

Choose $p>2$ to solve $2([s]+1)(1-\frac{2}{p})=s+[s]+1$. (Note that
$s^{*}(p)=\frac{s+[s]+1}{4([s]+1)}$ for this choice of $p$.) Then for
$\rho>R_{0}$, we have

$\displaystyle|\varphi_{[s]}(u,\rho)-\varphi_{[s],0}(u,\rho)|\leq\sum_{k=2}^{2[s]+2}|\tau_{[s]}^{k}(u,\sqrt{i\rho})|+|\tau^{*}_{[s]}(u,\sqrt{i\rho})|$
$\displaystyle\text{ by }\eqref{e:fLdef},\eqref{e:fL0def}$ $\displaystyle\leq
C(s)\bigg{[}\frac{\|u\|_{H^{\beta}(\mathbb{R})}^{3}\|u\|_{H^{s}(\mathbb{R})}}{\rho^{s+[s]+1}}+\sum_{k=3}^{2[s]+2}\frac{\|u\|_{H^{[s]}(\mathbb{R})}^{2k}}{\rho^{2[s]+2}}+\frac{\|u\|_{H^{s^{*}(p)}(\mathbb{R})}^{4[s]+4}\|u\|_{L^{2}(\mathbb{R})}^{2}}{\rho^{(2[s]+2)(1-\frac{2}{p})}}\bigg{]}$
$\displaystyle\text{ by
}\eqref{e:tau2bound},\eqref{e:taubound},\eqref{e:tau*}$
$\displaystyle\leq\frac{C(s,\|u\|_{H^{\beta}(\mathbb{R})})}{\rho^{s+[s]+1}}(\|u\|_{H^{s}(\mathbb{R})}+1).$

In the second line, we understand the sum over $k$ to be empty if $[s]=0$. ∎

The conclusion of Theorem 1.1 for noninteger $s\geq\frac{1}{2}$ follows from
Lemmas 2.2 and 2.3, the time-independence of the quantity
$\varphi_{[s]}(u,\rho)$ for solutions of the DNLS equation, and the bound

(34) $\sup_{t\in\mathbb{R}}\|u(t)\|_{H^{\beta}(\mathbb{R})}\leq
C(\beta,\|u_{0}\|_{H^{\beta}(\mathbb{R})}),$

where $\beta=\max\\{[s],\frac{s+[s]+1}{4([s]+1)},\frac{1}{3}\\}$ is as in the
statement of Lemma 2.3. The bound (34) follows from our induction argument if
$s>1$ and from the result of Harrop-Griffiths, Killip, and Vişan [9] if
$\frac{1}{2}\leq s<1$.

Let us give the remaining details of the proof of Theorem 1.1 presently. For
any $t\in\mathbb{R}$, we have

$\displaystyle\|u(t)\|_{\dot{H}^{s}(\mathbb{R})}^{2}$
$\displaystyle\lesssim_{s}\int_{R_{0}}^{\infty}\rho^{2s-1}|\varphi_{[s],0}(u(t),\rho)|\mathrm{d}\rho+R_{0}^{2(s-[s])}\|u(t)\|_{H^{[s]}(\mathbb{R})}^{2}$
$\displaystyle\leq
C(s)\int_{R_{0}}^{\infty}\rho^{2s-1}|\varphi_{[s]}(u(t),\rho)|\mathrm{d}\rho+C(s,R_{0},\|u(t)\|_{H^{\beta}(\mathbb{R})})(\|u(t)\|_{H^{s}(\mathbb{R})}+1)$
$\displaystyle\leq
C(s)\int_{R_{0}}^{\infty}\rho^{2s-1}|\varphi_{[s]}(u_{0},\rho)|\mathrm{d}\rho+C(s,\|u_{0}\|_{H^{s}(\mathbb{R})})(\|u(t)\|_{H^{s}(\mathbb{R})}+1)$
$\displaystyle\leq\frac{1}{2}\|u(t)\|^{2}_{H^{s}(\mathbb{R})}+C(s,\|u_{0}\|_{H^{s}(\mathbb{R})}),$

which establishes the desired conclusion. Note that the first line in the
calculation above is simply the upper bound in Lemma 2.2. To pass from the
first line to the second, we use Lemma 2.3, followed by the lower bound of
Lemma 2.2. We use (34) and the time independence of $\varphi_{[s]}(u(t),\rho)$
to pass to the third line. Finally, we justify the last line by noting that

$\int_{R_{0}}^{\infty}\rho^{2s-1}|\varphi_{[s]}(u_{0},\rho)|\mathrm{d}\rho\lesssim_{s}C(s,\|u_{0}\|_{H^{s}(\mathbb{R})}),$

which follows from an application of Lemma 2.3, followed by the lower bound in
Lemma 2.2.

##  3\. Proof of Lemma 2.1

###  3.1. Outline of the Proof

In this section, we expand each $\operatorname{Tr}(T_{u}^{2k}(\lambda))$ in
powers of $\lambda^{-2}$, up to a specified order, and we establish bounds on
the remainders, in order to prove our key Lemma 2.1. In Section 3.2, we
consider the case $L=0$, which is easy to treat explicitly but does not fit
naturally into our argument for the other cases. When $L\geq 1$, we follow the
strategy of [5], deducing the expansions of the traces from the expansion of
the resolvent $L_{u}(\lambda)^{-1}$. The relationship between $T_{u}(\lambda)$
and $L_{u}(\lambda)$ is the following:

(35) $L_{u}(\lambda)=(\mathcal{L}_{0}-\lambda^{2})(I-T_{u}(\lambda)).$

Therefore,

(36)
$L_{u}(\lambda)^{-1}=(I-T_{u}(\lambda))^{-1}(\mathcal{L}_{0}-\lambda^{2})^{-1}=\sum_{n=0}^{\infty}\underbrace{T_{u}(\lambda)^{n}(\mathcal{L}_{0}-\lambda^{2})^{-1}}_{=:\mathcal{R}_{n}},\qquad\|T_{u}(\lambda)\|<1.$

The point is that

(37) $T_{u}^{2k}(\lambda)=i\lambda\mathcal{R}_{2k-1}U.$

Thus, the part of $L_{u}^{-1}(\lambda)$ that is of relevance to us is
$\mathcal{R}_{2k-1}$, i.e., the term in the expansion (36) that is homogeneous
of degree $2k-1$ in $u,\overline{u}$. In particular, we seek an expansion of
$\lambda\mathcal{R}_{2k-1}$ in powers of $\lambda^{-2}$, up to order
$\lambda^{4L+2}$ for a given $L\in\mathbb{N}^{*}$, and a good understanding of
the remainder term.

Our strategy will be to examine the symbol $R(x,\zeta)$ of the
pseudodifferential operator $L_{u}(\lambda)^{-1}$. In Section 3.3, we will
expand the diagonal and antidiagonal parts $R^{d}(x,\zeta)$ and
$R^{a}(x,\zeta)$ of $R(x,\zeta)$ in powers of $\lambda^{-2}$, determining
recursively the form of each term of the expansion. Homogeneity considerations
will then give us the desired expansion of $\lambda\mathcal{R}_{2k-1}$ (and
thus of $\operatorname{Tr}T_{u}^{2k}(\lambda)$) in powers of $\lambda^{-2}$.
In Section 3.4, we identify the $\mu_{j,k}(u)$’s from (22) and separate them
from the remainder term. In Section 3.5 we estimate the remainder term,
finishing the proof of the Lemma. The final Section 3.6 consists of the proof
by induction of a technical result stated in Section 3.3.1, on the form of the
terms of the expansions for $R^{d}$ and $R^{a}$.

###  3.2. Case $L=0$

Let us note first of all that the desired decomposition in the case $L=0$
reads

$\ln
a_{u}(\lambda)=[\underbrace{\mu_{0,1}(u)+\lambda^{-2}\mu_{1,1}(u)+\tau_{0}^{1}(u,\lambda)}_{=-\frac{1}{2}\operatorname{Tr}T_{u}^{2}(\lambda)}]+[\underbrace{\lambda^{-2}\mu_{1,2}(u)+\tau_{0}^{2}(u,\lambda)}_{=-\frac{1}{4}\operatorname{Tr}T_{u}^{4}(\lambda)}]+\tau_{0}^{*}(u,\lambda).$

(See the table in Section 2.1.) The only term which we have not already
understood is $\tau_{0}^{2}(u,\lambda)$; in order to treat it, we decompose
$T_{u}^{4}(\lambda)$ explicitly as follows. A computation (the details of
which are contained, for instance, in [2]) tells us that

$\operatorname{Tr}T_{u}^{4}(\lambda)=i(2\lambda^{2})^{2}\int_{\mathbb{R}}\overline{u}(x)\big{(}(D+2\lambda^{2})^{-1}u(x)\big{)}^{2}(D-2\lambda^{2})^{-1}\overline{u}(x)\,\mathrm{d}x.$

Then, making a few simple manipulations, we can bring the right side of the
equation above into the following form.

$\displaystyle\operatorname{Tr}T_{u}^{4}(\lambda)$
$\displaystyle=\frac{i}{-2\lambda^{2}}\int_{\mathbb{R}}\overline{u}(x)\bigg{[}u(x)-(D+2\lambda^{2})^{-1}Du(x)\bigg{]}^{2}\big{[}\overline{u}(x)-(D-2\lambda^{2})^{-1}D\overline{u}(x)\big{]}\,\mathrm{d}x$
$\displaystyle=-\frac{i}{2\lambda^{2}}\bigg{[}\int_{\mathbb{R}}|u(x)|^{4}\,\mathrm{d}x-\int_{\mathbb{R}}|u|^{2}u(x)(D-2\lambda^{2})^{-1}D\overline{u}(x)\,\mathrm{d}x-2\int_{\mathbb{R}}|u|^{2}\overline{u}(x)(D+2\lambda^{2})^{-1}Du(x)\,\mathrm{d}x$
$\displaystyle\qquad\qquad+2\int_{\mathbb{R}}|u(x)|^{2}(D+2\lambda^{2})^{-1}Du(x)(D-2\lambda^{2})^{-1}D\overline{u}(x)\,\mathrm{d}x+\int_{\mathbb{R}}((D+2\lambda^{2})^{-1}Du(x))^{2}\overline{u}(x)^{2}\,\mathrm{d}x$
$\displaystyle\qquad\qquad-\int_{\mathbb{R}}\overline{u}(x)((D+2\lambda^{2})^{-1}Du(x))^{2}((D-2\lambda^{2})^{-1}D\overline{u}(x))\,\mathrm{d}x\bigg{]}$
$\displaystyle=-\frac{4}{\lambda^{2}}\underbrace{\left[\frac{i}{8}\|u\|_{L^{4}(\mathbb{R})}^{4}\right]}_{=\mu_{1,2}(u)}-4\tau_{0}^{2}(u,\lambda).$

To estimate $\tau_{0}^{2}(u,\lambda)$, we use the following simple Lemma, the
proof of which we omit.

######  Lemma 3.1.

The following estimates hold, for $\lambda\in\Gamma_{\delta}$.

  * •

If $0\leq\alpha_{1}\leq\alpha_{2}\leq 1$, then

(38) $\left\|(D\pm
2\lambda^{2})^{-1}Du\right\|_{\dot{H}^{\alpha_{1}}(\mathbb{R})}\lesssim_{\alpha_{2}-\alpha_{1}}\frac{\|u\|_{\dot{H}^{\alpha_{2}}(\mathbb{R})}}{(2\operatorname{Im}(\lambda^{2}))^{\alpha_{2}-\alpha_{1}}},\quad\forall\,u\in
H^{\alpha_{2}}(\mathbb{R}).$

  * •

If $2\leq p<\infty$, then

(39) $\left\|(D\pm
2\lambda^{2})^{-1}Du\right\|_{L^{p}(\mathbb{R})}\lesssim_{p}\|u\|_{H^{s^{*}(p)}(\mathbb{R})},\quad\forall\,u\in
H^{s^{*}(p)}(\mathbb{R}).$

We estimate one of the terms defining $\tau_{0}^{2}(u,\lambda)$ explicitly;
the others can be dealt with in an entirely similar way.

$\displaystyle\left|\frac{1}{\lambda^{2}}\int_{\mathbb{R}}\overline{u}(x)((D+2\lambda^{2})^{-1}Du(x))^{2}((D-2\lambda^{2})^{-1}D\overline{u}(x))\,\mathrm{d}x\right|$
$\displaystyle\quad\leq\frac{1}{|\lambda|^{2}}\|u\|_{L^{6}(\mathbb{R})}\|(D+2\lambda^{2})^{-1}Du\|_{L^{6}(\mathbb{R})}^{2}\|(D-2\lambda^{2})^{-1}D\overline{u}\|_{L^{2}(\mathbb{R})}\lesssim_{\alpha}\frac{\|u\|_{H^{\frac{1}{3}}(\mathbb{R})}^{3}\|u\|_{H^{\alpha}(\mathbb{R})}}{|\lambda|^{2+2\alpha}}.$

We conclude that $\tau_{0}^{2}(u,\lambda)$ satisfies the required bound,
finishing the case $L=0$.

###  3.3. Expanding the Resolvent

####  3.3.1. Formal Expansion of $R^{a}$ and $R^{d}$

As stated above, for $L\geq 1$ we seek an expansion of the symbol of
$L_{u}^{-1}(\lambda)$, in powers of $\lambda^{-2}$. That is, we seek to
understand $R(x,\zeta)$ in the expression

(40) $L_{u}^{-1}(\lambda)f=\frac{1}{\sqrt{2\pi}}\int\mathrm{d}\zeta
e^{ix\zeta}R(x,\zeta)\widehat{f}(\zeta).$

The identity $L_{u}(\lambda)R(x,D)=I$ implies

(41)
$i\sigma_{3}\partial_{x}R(x,\zeta)-(\zeta\sigma_{3}+\lambda^{2})R(x,\zeta)-i\lambda
U(x)R(x,\zeta)=I.$

Introducing the new variable $p=\frac{\zeta}{\lambda^{2}}$, this reads

(42)
$i\sigma_{3}\partial_{x}R(x,\zeta)-\lambda^{2}(p\sigma_{3}+1)R(x,\zeta)-i\lambda
U(x)R(x,\zeta)=I.$

We split $R$ into its diagonal and antidiagonal parts $R^{d}$ and $R^{a}$,
respectively,

$R(x,\zeta)=R^{d}(x,\zeta)+R^{a}(x,\zeta),$

and we also split equation (42) accordingly:

(43)
$i\sigma_{3}\partial_{x}R^{d}(x,\zeta)-\lambda^{2}(p\sigma_{3}+1)R^{d}(x,\zeta)-i\lambda
U(x)R^{a}(x,\zeta)=I;$ (44)
$i\sigma_{3}\partial_{x}R^{a}(x,\zeta)-\lambda^{2}(p\sigma_{3}+1)R^{a}(x,\zeta)-i\lambda
U(x)R^{d}(x,\zeta)=0.$

Setting the notation

$R^{d}(x,\zeta)=\sum_{k\geq 0}\frac{1}{\lambda^{2+2k}}R^{d}_{k}(x,p),\qquad
R^{a}(x,\zeta)=\sum_{k\geq 0}\frac{1}{\lambda^{3+2k}}R^{a}_{k}(x,p),$

we rewrite (43) and (44) in expanded form:

(45) $\displaystyle I$
$\displaystyle=-(p\sigma_{3}+1)R_{0}^{d}+\sum_{k=1}^{\infty}\frac{i\sigma_{3}\partial_{x}R_{k-1}^{d}-(p\sigma_{3}+1)R_{k}^{d}-iUR_{k-1}^{a}}{\lambda^{2k}};$
(46) $\displaystyle 0$
$\displaystyle=-(p\sigma_{3}+1)R_{0}^{a}-iUR_{0}^{d}+\sum_{k=1}^{\infty}\frac{i\sigma_{3}\partial_{x}R_{k-1}^{a}-(p\sigma_{3}+1)R_{k}^{a}-iUR_{k}^{d}}{\lambda^{2k}}.$

We thus obtain the recursive system (47)–(49) below.

(47) $R_{0}^{d}(x,p)=-\frac{p\sigma_{3}-1}{p^{2}-1},\qquad
R_{0}^{a}(x,p)=-\frac{iU}{p^{2}-1},$ (48)
$R_{k}^{d}(x,p)=\frac{1}{p^{2}-1}\big{[}-iUR_{k-1}^{a}(x,p)+i\partial_{x}R_{k-1}^{d}(x,p)\sigma_{3}\big{]}(p\sigma_{3}-1),\qquad\qquad\qquad\qquad
k\geq 1,$ (49)
$\begin{split}R_{k}^{a}(x,p)&=\frac{1}{p^{2}-1}\big{[}iUR_{k}^{d}(x,p)+i\partial_{x}R_{k-1}^{a}(x,p)\sigma_{3}\big{]}(p\sigma_{3}+1)\\\
&=\frac{1}{p^{2}-1}\big{[}U^{2}R_{k-1}^{a}(x,p)-U\partial_{x}R_{k-1}^{d}(x,p)\sigma_{3}+i\partial_{x}R_{k-1}^{a}(x,p)\sigma_{3}(p\sigma_{3}+1)\big{]},\end{split}\quad
k\geq 1.$

Note that we used the formula for $R_{k}^{d}(x,p)$ to pass to the second line
in the formula for $R_{k}^{a}(x,p)$. We also used several times the fact that
$\sigma_{3}A=-A\sigma_{3}$ for any antidiagonal matrix.

We use the computations above to clarify the form of the $R_{k}^{d}$’s and
$R_{k}^{a}$’s; the precise statement is contained in the following Lemma.

######  Lemma 3.2.

The $R_{k}^{d}$’s and $R_{k}^{a}$’s take the following form:

(50) $R_{k}^{d}(x,p)=\sum_{r=1}^{k}R_{k,r}^{d}(x,p),\qquad k\geq 1,$ (51)
$R_{k}^{a}(x,p)=\sum_{r=0}^{k}R_{k,r}^{a}(x,p),\qquad k\geq 0,$

where the entries of the $R_{k,r}^{d}$’s and $R_{k,r}^{a}$’s are homogeneous
polynomials of degrees $2r$ and $2r+1$, respectively, in $u,\overline{u}$, and
their derivatives. More specifically, setting
$Q_{\gamma}=\partial_{x}^{\gamma_{1}}U\cdots\partial_{x}^{\gamma_{n}}U$, for
$\gamma\in\mathbb{N}^{n}$, we have

(52) $\displaystyle R^{d}_{k,r}(x,p)$
$\displaystyle=\frac{1}{(p^{2}-1)^{k+1}}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r}\\\
|\gamma|=k-r\end{subarray}}Q_{\gamma}(x)P_{|\gamma|}(p)(p\sigma_{3}-1),$ (53)
$\displaystyle R^{a}_{k,r}(x,p)$
$\displaystyle=\frac{1}{(p^{2}-1)^{k+1}}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r+1}\\\
|\gamma|=k-r\end{subarray}}Q_{\gamma}(x)P_{|\gamma|}(p).$

Here and below we use the notation $P_{n}$ to denote any diagonal matrix whose
diagonal entries are polynomials in $p$ having degree at most $n$.

We postpone the proof of this Lemma until Section 3.6, so as not to interrupt
the flow of ideas.

####  3.3.2. The Truncated Expansion, and a Formula for $\mathcal{R}_{2m-1}$

For a fixed $N\in\mathbb{N}^{*}$, we set the following notation. (Later we
will set $N=2L$.)

(54)
$\begin{split}R^{(N)}(x,p)&=\underbrace{\sum_{k=0}^{N}\frac{R_{k}^{d}(x,p)}{\lambda^{2+2k}}}_{=:R_{d}^{(N)}(x,p)}+\underbrace{\sum_{k=0}^{N-1}\frac{R_{k}^{a}(x,p)}{\lambda^{3+2k}}}_{=:R_{a}^{(N)}(x,p)}\\\
&=\underbrace{\frac{R_{0}^{d}(x,p)}{\lambda^{2}}}_{=:R^{(N)}_{d,0}(x,p)}+\sum_{r=1}^{N}\underbrace{\sum_{k=r}^{N}\frac{R_{k,r}^{d}(x,p)}{\lambda^{2+2k}}}_{=:R^{(N)}_{d,r}(x,p)}+\sum_{r=0}^{N-1}\underbrace{\sum_{k=r}^{N-1}\frac{R_{k,r}^{a}(x,p)}{\lambda^{3+2k}}}_{=:R^{(N)}_{a,r}(x,p)}.\end{split}$

The symbol $R^{(N)}(x,p)$ is a truncated expansion of $R(x,p)$ in inverse
powers of $\lambda$, having diagonal and antidiagonal parts $R^{(N)}_{d}$,
$R^{(N)}_{a}$, respectively. The point of this definition is that, using Lemma
3.2, we know that $R^{(N)}_{d,r}$ is homogeneous of degree $2r$ in $u$,
$\overline{u}$, and their derivatives, while $R^{(N)}_{a,r}$ is homogeneous of
degree $2r+1$ in these quantities. Expanding $R^{(N)}$ according to (54) and
applying the recursive identities (45)–(46), we see that $R^{(N)}(x,p)$
satisfies

$[i\sigma_{3}\partial_{x}-\lambda^{2}(p\sigma_{3}+1)-i\lambda
U(x)]R^{(N)}(x,p)=I+Y^{(N)}(x,p),$

where $Y^{(N)}(x,p)=Y^{(N)}_{d}(x,p)+Y^{(N)}_{a}(x,p)$,

$Y^{(N)}_{d}(x,p)=\frac{1}{\lambda^{2+2N}}i\sigma_{3}(\partial_{x}R_{N}^{d})(x,p),\quad
Y^{(N)}_{a}(x,p)=-\frac{1}{\lambda^{1+2N}}R_{N}^{a}(x,p)(p\sigma_{3}-1).$

This implies

(55)
$L_{u}^{-1}(\lambda)=R^{(N)}(x,\lambda^{-2}D)-L_{u}^{-1}(\lambda)Y^{(N)}(x,\lambda^{-2}D).$

Recall that $\mathcal{R}_{2m-1}$ is the term in the expansion (36) which is
homogeneous of order $2m-1$ in $u$, $\overline{u}$, and their derivatives. On
the other hand, the portion of $R^{(N)}$ which is of this homogeneity is
precisely $R^{(N)}_{a,m-1}$. Combining these considerations with (55), we see
that $\mathcal{R}_{2m-1}$ is the difference between $R^{(N)}_{a,m-1}$ and the
part of $L_{u}^{-1}(\lambda)Y^{(N)}(x,\lambda^{-2}D)$ that is homogeneous of
degree $2m-1$ in $u$, $\overline{u}$, and their derivatives. Using the
expansion (36) to isolate this part, we obtain:

(56)
$\begin{split}\mathcal{R}_{2m-1}&=R_{a,m-1}^{(N)}(x,\lambda^{-2}D)-\frac{1}{\lambda^{2+2N}}\sum_{\begin{subarray}{c}k+r^{\prime}=m-1\\\
k\geq 0,\;1\leq r^{\prime}\leq
N\end{subarray}}T_{u}(\lambda)^{2k+1}(\mathcal{L}_{0}-\lambda^{2})^{-1}(\mathcal{L}_{0}R_{N,r^{\prime}}^{d})(x,\lambda^{-2}D)\\\
&\quad-\frac{1}{\lambda^{1+2N}}\sum_{\begin{subarray}{c}k+r^{\prime}=m-1\\\
k\geq 0,\;0\leq r^{\prime}\leq
N\end{subarray}}T_{u}(\lambda)^{2k}(\mathcal{L}_{0}-\lambda^{2})^{-1}R_{N,r^{\prime}}^{a}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1).\end{split}$

###  3.4. Extracting the $\mu_{j,m}(u)$’s

Combining (56) with (37), (54), and pulling out inverse powers of $\lambda$,
we easily find the following formula for
$\operatorname{Tr}T_{u}^{2m}(\lambda)$ with $m\geq 2$, truncated at $N=2L$.

(57)
$\begin{split}\operatorname{Tr}(T_{u}^{2m}(\lambda))=&\;\sum_{j=m-1}^{2L-1}\frac{1}{\lambda^{2j+2}}\operatorname{Tr}[iUR_{j,m-1}^{a}(x,\lambda^{-2}D)]\\\
&+\sum_{\begin{subarray}{c}k+r=m-1\\\ k\geq 0,\;1\leq r\leq
2L\end{subarray}}\frac{(-1)^{k}}{\lambda^{4L+4+2k}}\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+2}(\mathcal{L}_{0}R_{2L,r}^{d})(x,\lambda^{-2}D)\big{]}\\\
&+\sum_{\begin{subarray}{c}k+r=m-1\\\ k\geq 0,\;0\leq r\leq
2L\end{subarray}}\frac{i(-1)^{k+1}}{\lambda^{4L+2+2k}}\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+1}R_{2L,r}^{a}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)\big{]}.\end{split}$

We will refer to the three sums above as $I$, $II$, and $III$, respectively.

We now identify the coefficients $\mu_{j,m}(u)$’s and verify that they satisfy
the properties claimed in Lemma 2.1. The claimed homogeneity properties will
be clear from the formulas that we derive below; we will just need to verify
the bounds (26). The latter are also straightforward to verify but will
require us to use the structure of the $R^{d}_{k,r}$’s and $R^{a}_{k,r}$’s
from (52)–(53).

The first sum has the form

$-2m\sum_{j=m-1}^{2L-1}\frac{\mu_{j,m}(u)}{\lambda^{2j}}$

with

(58)
$\begin{split}\mu_{j,m}(u)&=-\frac{i}{2m\lambda^{2}}\operatorname{Tr}\big{[}UR_{j,m-1}^{a}(x,\lambda^{-2}D)\big{]}\\\
&=-\frac{i}{4m\pi}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2m-1}\\\
|\gamma|=j-(m-1)\end{subarray}}\operatorname{Tr}\left[\int
U(x)Q_{\gamma}(x)\,\mathrm{d}x\int\frac{P_{|\gamma|}(\tfrac{\zeta}{\lambda^{2}})}{((\tfrac{\zeta}{\lambda^{2}})^{2}-1)^{j+1}}\frac{\mathrm{d}\zeta}{\lambda^{2}}\right].\end{split}$

Note that ‘$\operatorname{Tr}$’ denotes an operator trace in the first line,
whereas it refers to the $2\times 2$ matrix trace in the second and third
lines. We will use the notation ‘$\operatorname{Tr}$’ similarly in what
follows without further comment.

Since $\lambda$ is presumed to lie in $\Gamma_{\delta}$, a comparison of the
degrees in the numerator and denominator ensures that the integrals over
$\zeta$ are finite and their values are independent of $\lambda$. The total
number of derivatives in the $x$-integrals is $j-(m-1)$, and we distribute
these so that the highest order of the derivatives that fall on a single $U$
is as small as possible. We list the bounds on $\mu_{j,m}(u)$ according to $m$
and the parity of $j$. In each case below, $\ell$ is a strictly positive
integer.

  * •

If $j=2\ell$ is even and $m=2$, then there are $2\ell-1$ derivatives; thus

$|\mu_{2\ell,2}(u)|\lesssim\|u\|_{H^{\ell}(\mathbb{R})}\|u\|_{H^{\sigma(\ell-1)}(\mathbb{R})}^{3},$

where we recall the notation $\sigma(n)=\max\\{n,\frac{1}{3}\\}$.

  * •

If $j=2\ell$ is even and $m\geq 3$, then there are at most $2(\ell-1)$ total
derivatives. This establishes the following bounds:

$\begin{split}|\mu_{2,3}(u)|&\lesssim\|u\|_{H^{\frac{1}{3}}(\mathbb{R})}^{6},\\\
|\mu_{2\ell,m}(u)|&\lesssim\|u\|_{H^{\ell-1}(\mathbb{R})}^{2m},\qquad\ell\geq
2,\;m\in\\{3,\ldots,2\ell+1\\}.\end{split}$

  * •

If $j=2\ell+1$ is odd and $m\geq 2$, then there are at most $2\ell$
derivatives, so
$|\mu_{2\ell+1,m}(u)|\lesssim\|u\|_{H^{\ell}(\mathbb{R})}^{2m}$.

Let us remark that the formula (58) determines $\mu_{j,m}(u)$ for all $m\geq
2$ and all $j\geq m-1$, not just for those $\mu_{j,m}(u)$’s that appear in the
sum $I$. Therefore, the above bounds on the $\mu_{j,m}(u)$’s complete our
proof of the estimates (26). However, in order to determine the remainders
$\tau_{L}^{m}(u,\lambda)$, we still need to extract $\mu_{2L,m}(u)$ and
$\mu_{2L+1,m}(u)$ from the sums $II$ and $III$. To this end, we remove the
parts of $II$ and $III$ which are of order $\lambda^{-4L}$ and
$\lambda^{-4L-2}$; these will correspond to $\mu_{2L,m}(u)$ and
$\mu_{2L+1,m}(u)$, respectively. We deal first with $\mu_{2L,m}(u)$; the only
term expected to be relevant is the $k=0$ term in $III$, namely

(59)
$-\frac{i}{\lambda^{4L+2}}\operatorname{Tr}\big{[}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,m-1}^{a}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)\big{]}.$

To extract the part of this expression that is really of order
$\lambda^{-4L}$, we commute the operator
$(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}$ with $R_{2L,m-1}^{a}(x,\lambda^{-2}D)$.
We will have to do something similar several times below, so let us pause to
write down a more general formula. Let $A$ denote an antidiagonal operator
with symbol $A(x,\zeta)$ and similarly let $B$ denote a diagonal operator with
symbol $B(x,\zeta)$. Then a simple application of the product rule gives the
following operator identities.

(60) $\displaystyle A(\mathcal{L}_{0}+\lambda^{2})^{-1}$
$\displaystyle=-(\mathcal{L}_{0}-\lambda^{2})^{-1}A+(\mathcal{L}_{0}-\lambda^{2})^{-1}(\mathcal{L}_{0}A)(\mathcal{L}_{0}+\lambda^{2})^{-1};$
(61) $\displaystyle B(\mathcal{L}_{0}-\lambda^{2})^{-1}$
$\displaystyle=\;\;\,(\mathcal{L}_{0}-\lambda^{2})^{-1}B+(\mathcal{L}_{0}-\lambda^{2})^{-1}(\mathcal{L}_{0}B)(\mathcal{L}_{0}-\lambda^{2})^{-1}.$

Using (60) with $A=R_{2L,m-1}^{a}$, the expression (59) becomes

(62)
$\begin{split}\underbrace{\frac{i}{\lambda^{4L+2}}\operatorname{Tr}\big{[}UR_{2L,m-1}^{a}(x,\lambda^{-2}D)\big{]}}_{-2m\mu_{2L,m}\lambda^{-4L}}-\frac{i}{\lambda^{4L+4}}\operatorname{Tr}\big{[}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}R_{2L,m-1}^{a})(x,\lambda^{-2}D)\big{]}.\end{split}$

To extract $\mu_{2L+1,m}(u)$, we need to determine the part of $II$ and $III$
that is of order $2L+1$ in $\lambda^{-2}$. There are three quantities we need
to consider:

  * •

The second term in (62):

(63)
$-\frac{i}{\lambda^{4L+4}}\operatorname{Tr}\big{[}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}R_{2L,m-1}^{a})(x,\lambda^{-2}D)\big{]}$

  * •

The $k=0$ term in $II$:

(64)
$\frac{1}{\lambda^{4L+4}}\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2}(\mathcal{L}_{0}R_{2L,m-1}^{d})(x,\lambda^{-2}D)\big{]}$

  * •

The $k=1$ term in $III$:

(65)
$\frac{i}{\lambda^{4L+4}}\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{3}R_{2L,m-2}^{a}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)\big{]}.$

We deal with each of these in turn, denoting their contributions to
$\mu_{2L+1,m}$ by $\mu_{2L+1,m}^{(1)}$, $\mu_{2L+1,m}^{(2)}$, and
$\mu_{2L+1,m}^{(3)}$, respectively. To put (63) in the desired form, we simply
apply (60) again, this time with $A=\mathcal{L}_{0}R_{2L,m-1}^{a}$. The result
is

(66)
$\begin{split}&\frac{i}{\lambda^{4L+4}}\operatorname{Tr}\big{[}U(\mathcal{L}_{0}R_{2L,m-1}^{a})(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}\big{]}\\\
&-\frac{i}{\lambda^{4L+6}}\operatorname{Tr}\big{[}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(D^{2}R_{2L,m-1}^{a})(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}\big{]}.\end{split}$

Thus

$\displaystyle\mu_{2L+1,m}^{(1)}=-\frac{i}{2m\lambda^{2}}\operatorname{Tr}\big{[}U(\mathcal{L}_{0}R_{2L,m-1}^{a})(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}\big{]}.$

Next, we look at (64). We perform two commutations, using $A=U$ in (60), then
$B=U^{2}$ in (61) to obtain

(67)
$\begin{split}[U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}]^{2}&=-(\lambda^{-4}D^{2}-1)^{-1}U^{2}\\\
&\qquad+\lambda^{-2}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}\\\
&\qquad-\lambda^{-2}(\lambda^{-4}D^{2}-1)^{-1}(\mathcal{L}_{0}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}.\end{split}$

Substituting this into (64) yields

(68)
$\begin{split}&-\frac{1}{\lambda^{4L+4}}\operatorname{Tr}\big{[}U^{2}(\mathcal{L}_{0}R_{2L,m-1}^{d})(x,\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-1}\big{]}\\\
&+\frac{1}{\lambda^{4L+6}}\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}R_{2L,m-1}^{d})(x,\lambda^{-2}D)\big{]}\\\
&-\frac{1}{\lambda^{4L+6}}\operatorname{Tr}\big{[}(\lambda^{-4}D^{2}-1)^{-1}(\mathcal{L}_{0}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}R_{2L,m-1}^{d})(x,\lambda^{-2}D)\big{]}.\end{split}$

We take

$\displaystyle\mu_{2L+1,m}^{(2)}=\frac{1}{2m\lambda^{2}}\operatorname{Tr}\big{[}U^{2}(\mathcal{L}_{0}R_{2L,m-1}^{d})(x,\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-1}\big{]}.$

Finally, we look at (65). Proceeding as in (67) but commuting one more time,
we get

$\displaystyle(\lambda^{-2}\mathcal{L}_{0}+1)[U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}]^{3}$
$\displaystyle=(\lambda^{-4}D^{2}-1)^{-1}U^{3}$
$\displaystyle\qquad+\lambda^{-2}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}$
$\displaystyle\qquad-\lambda^{-2}(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}$
$\displaystyle\qquad-\lambda^{-2}(\lambda^{-4}D^{2}-1)^{-1}(\mathcal{L}_{0}U^{3})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}.$

Substituting the above into (65) yields

(69)
$\begin{split}&\frac{i}{\lambda^{4L+4}}\operatorname{Tr}\big{[}U^{3}R_{2L,m-2}^{a}(x,\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-1}\big{]}\\\
&+\frac{i}{\lambda^{4L+6}}\operatorname{Tr}\big{[}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,m-2}^{a}(x,\lambda^{-2}D)\big{]}\\\
&-\frac{i}{\lambda^{4L+6}}\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,m-2}^{a}(x,\lambda^{-2}D)\big{]}\\\
&-\frac{i}{\lambda^{4L+6}}\operatorname{Tr}\big{[}(\lambda^{-4}D^{2}-1)^{-1}(\mathcal{L}_{0}U^{3})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,m-2}^{a}(x,\lambda^{-2}D)\big{]}.\end{split}$

Thus

$\displaystyle\mu_{2L+1,m}^{(3)}=-\frac{i}{2m\lambda^{2}}\operatorname{Tr}\big{[}U^{3}R_{2L,m-2}^{a}(x,\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-1}\big{]}.$

A short calculation involving (49) confirms that the three quantities
identified above sum to $\mu_{2L+1,m}$ as defined in (58):

$\mu_{2L+1,m}^{(1)}+\mu_{2L+1,m}^{(2)}+\mu_{2L+1,m}^{(3)}=\mu_{2L+1,m}.$

###  3.5. Estimating the Remainder

The final step of the proof is to estimate the remainder terms, which we group
together into $\tau^{m}_{L}(u,\lambda)$. This expression is a sum of the
following terms:

  * •

The $k\geq 1$ terms of $II$ and the $k\geq 2$ terms of $III$, where $II$ and
$III$ denote (as above) the second and third sums in the decomposition (57).
We refer to these as the ‘Type 1’ remainder terms.

  * •

The terms in (66), (68), and (69) where $\lambda^{-4L-6}$ appears (six terms
total). We refer to these as the ‘Type 2’ remainder terms.

####  3.5.1. Type 1 Remainder Terms

We begin with the two sums. We want to show that the following expression is
bounded by $\|u\|_{H^{L}(\mathbb{R})}^{2m}$:

(70) $\begin{split}&\sum_{\begin{subarray}{c}k+r=m-1\\\ k\geq 1,\;1\leq r\leq
2L\end{subarray}}\frac{(-1)^{k}}{\lambda^{2k}}\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+2}(\mathcal{L}_{0}R_{2L,r}^{d})(x,\lambda^{-2}D)\big{]}\\\
&+\sum_{\begin{subarray}{c}k+r=m-1\\\ k\geq 2,\;0\leq r\leq
2L\end{subarray}}\frac{i(-1)^{k+1}}{\lambda^{2(k-1)}}\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+1}R_{2L,r}^{a}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)\big{]}.\end{split}$

By virtue of (52), we can write

$\displaystyle(\mathcal{L}_{0}R^{d}_{2L,r})(x,p)=\frac{1}{(p^{2}-1)^{2L+1}}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r}\\\
|\gamma|=2L-r+1\end{subarray}}Q_{\gamma}(x)P_{|\gamma|}(p).$

Thus, the traces in the first sum in (70) may be written as a sum of terms of
the form

(71)
$\displaystyle\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+2}Q_{\gamma}(x)P_{2L-r+1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}\big{]}$

with $\gamma\in\mathbb{N}^{2r},\,\,|\gamma|=2L-r+1\leq 2L$. Integrating by
parts repeatedly in the above expression until no derivative of order larger
than $L$ falls on any single $U$, we rewrite the expression (71) as a sum of
terms of the form

$\operatorname{Tr}\big{[}(\partial_{x}^{\eta_{1}}U)(x)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}\dots(\partial_{x}^{\eta_{2k+2}}U)(x)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}Q_{\gamma}(x)P_{2L-r+1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}\big{]},$

with $\eta=(\eta_{1},\dots\eta_{2k+2})\in\mathbb{N}^{2k+2}$,
$\gamma=(\gamma_{1},\dots,\gamma_{2r})\in\mathbb{N}^{2r}$ satisfying
$|\eta|+|\gamma|=2L-r+1$ and $\max\limits_{p,q}(\eta_{p},\gamma_{q})\leq L$.
Thus, the expression (71) can be bounded by
$|\lambda|^{2}\|u\|_{H^{L}(\mathbb{R})}^{2m}$, and therefore the first sum in
(70) by $\|u\|_{H^{L}(\mathbb{R})}^{2m}$.

We deal with the second sum in (70) in essentially the same way. Invoking
(53), we may write

$R^{a}_{2L,r}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)=\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r+1}\\\
|\gamma|=2L-r\end{subarray}}Q_{\gamma}(x)P_{|\gamma|+1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}.$

Thus

$\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+1}R_{2L,r}^{a}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)\big{]}$

is a sum of terms of the form

(72)
$\operatorname{Tr}\big{[}(U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1})^{2k+1}Q_{\gamma}(x)P_{2L-r+1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}\big{]},$

where $\gamma\in\mathbb{N}^{2r+1}$ with $|\gamma|=2L-r\leq 2L$. As before, we
integrate by parts repeatedly to rewrite the expression (72) as a sum of terms
of the form

$\operatorname{Tr}\big{[}(\partial_{x}^{\eta_{1}}U)(x)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}\dots(\partial_{x}^{\eta_{2k+1}}U)(x)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}Q_{\gamma}(x)P_{2L-r+1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}\big{]},$

with $\eta=(\eta_{1},\dots\eta_{2k+1})\in\mathbb{N}^{2k+1}$,
$\gamma=(\gamma_{1},\dots,\gamma_{2r+1})\in\mathbb{N}^{2r+1}$ satisfying
$|\eta|+|\gamma|=2L-r$ and $\max\limits_{p,q}(\eta_{p},\gamma_{q})\leq L$.
This allows us to bound (72) by $|\lambda|^{2}\|u\|_{H^{L}(\mathbb{R})}^{2m}$,
thus completing the desired estimates on the Type 1 remainder terms.

####  3.5.2. Type 2 Remainder Terms

We now deal with the Type 2 remainder terms (the terms in (66), (68), and (69)
where $\lambda^{-4L-6}$ appears). When $m\geq 3$, the total number of
derivatives falling on the $U$’s is $2L$; therefore we can bound all these
terms by $|\lambda|^{-4L-4}\|u\|_{H^{L}(\mathbb{R})}^{2m}$ by arguing exactly
as we did for the Type 1 terms. To complete the proof of Lemma 2.1, it thus
remains to consider the Type 2 remainder terms with $m=2$. In this case, some
of the $U$’s appear to be overloaded with derivatives, and we need an
additional estimate. We state the following Lemma in terms of the ‘overloaded’
part of the Type 2 remainder term from (66), but the same manipulations will
yield the bound we need for the other Type 2 remainders.

######  Lemma 3.3.

The following estimate holds, for any $j\in\mathbb{N}$ and all
$\alpha\in[0,1]$.

(73)
$\|(\mathcal{L}_{0}+\lambda^{2})^{-1}D^{j+1}U(\mathcal{L}_{0}-\lambda^{2})^{-1}\|_{2}\leq
C(\alpha)|\lambda|^{-1-2\alpha}\|u\|_{H^{j+\alpha}(\mathbb{R})}.$

###### Proof.

Denoting
$T=(\mathcal{L}_{0}+\lambda^{2})^{-1}D^{j+1}U(\mathcal{L}_{0}-\lambda^{2})^{-1}$,
we readily compute as follows:

$\displaystyle\|T\|_{2}^{2}$
$\displaystyle=\frac{1}{\pi}\iint\frac{|\widehat{D^{j+1}u}(\zeta_{1}-\zeta_{2})|^{2}}{|\zeta_{1}-\lambda^{2}|^{2}|\zeta_{2}-\lambda^{2}|^{2}}\mathrm{d}\zeta_{1}\mathrm{d}\zeta_{2}$
$\displaystyle=\frac{1}{\pi}\int|\zeta_{1}|^{2}|\widehat{D^{j}u}(\zeta_{1})|^{2}\left(\int\frac{\mathrm{d}\zeta_{2}}{|\zeta_{1}+\zeta_{2}-\lambda^{2}|^{2}|\zeta_{2}-\lambda^{2}|^{2}}\right)\mathrm{d}\zeta_{1}$
$\displaystyle=\frac{2}{\operatorname{Im}\lambda^{2}}\int\frac{|\zeta_{1}|^{2}|\widehat{D^{j}u}(\zeta_{1})|^{2}}{|\zeta_{1}+2i\operatorname{Im}\lambda^{2}|^{2}}\mathrm{d}\zeta_{1}\leq
C(\alpha)|\lambda|^{-2-4\alpha}\|u\|_{H^{j+\alpha}(\mathbb{R})}^{2}.$

This completes the proof. ∎

With the above Lemma at our disposal, we now return to the estimation of the
Type $2$ remainder terms for $m=2$; we deal first with the one that appears in
(66). Omitting the prefactor $\frac{-i}{\lambda^{4L+6}}$, the quantity under
consideration is

$\operatorname{Tr}\big{[}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(D^{2}R_{2L,1}^{a})(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}\big{]},$

which we write as

(74)
$\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(D^{2}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,1}^{a}(x,\lambda^{-2}D)\big{]}.$

Proceeding as we did for the Type 1 terms, we rewrite this expression as a sum
of terms of the form

(75)
$\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(D^{\eta+2}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}Q_{\gamma}(x)P_{2L-1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}],$

with $\eta\in\mathbb{N}$,
$\gamma=(\gamma_{1},\gamma_{2},\gamma_{3})\in\mathbb{N}^{3}$,
$\eta+|\gamma|=2L-1$, $\eta\leq L-1$, and
$\max(\gamma_{1},\gamma_{2},\gamma_{3})\leq L$. The expression (75) can be
bounded by

$\displaystyle\|(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(D^{\eta+2}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}\|_{2}\|Q_{\gamma}(x)P_{2L-1}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}\|_{2}.$

By virtue of Lemma 3.3, the above can bounded in turn by

$\displaystyle
C(\alpha)|\lambda|^{4-2\alpha}\|u\|_{H^{L+\alpha}(\mathbb{R})}\|u\|_{H^{L}(\mathbb{R})}^{3},$

for any $\alpha\in[0,1]$.

The next quantity we treat is the third term in (68); we want to estimate

$\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}R_{2L,1}^{d})(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}\big{]}.$

After an integration by parts we are left with the expression

$-\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(D^{2}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,1}^{d}(x,\lambda^{-2}D)(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}\big{]}.$

that can be treated in exactly the same way as (74). We note only the
modification to Lemma 3.3 that we use, namely

$\|(\mathcal{L}_{0}-\lambda^{2})^{-1}(D^{L+1}U^{2})(\mathcal{L}_{0}-\lambda^{2})^{-1}\|_{2}\lesssim_{\alpha}|\lambda|^{-1-2\alpha}\|U^{2}\|_{H^{L+\alpha}(\mathbb{R})}\lesssim_{\alpha}|\lambda|^{-1-2\alpha}\|u\|_{H^{L+\alpha}(\mathbb{R})}\|u\|_{H^{L}(\mathbb{R})}.$

We next consider the second term in (68):

$\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}R_{2L,1}^{d})(x,\lambda^{-2}D)\big{]},$

where as usual we have suppressed the prefactor $\lambda^{-4L-6}$. We start by
rewriting it as the sum

(76)
$\begin{split}-&\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(D^{2}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,1}^{d}(x,\lambda^{-2}D)\big{]}\\\
&+\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,1}^{d}(x,\lambda^{-2}D)\big{]}.\end{split}$

For the first term here we proceed exactly as before: substituting (52) and
integrating by parts we rewrite it as a sum of expressions of the form

$\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(D^{\eta_{1}+2}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(D^{\eta_{2}}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}Q_{\gamma}(x)P_{2L}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}],$

with $\eta=(\eta_{1},\eta_{2})\in\mathbb{N}^{2}$,
$\gamma=(\gamma_{1},\gamma_{2})\in\mathbb{N}^{2}$, $|\eta|+|\gamma|=2L-1$,
$\max(\eta_{1},\eta_{2})\leq L-1$, and $\max(\gamma_{1},\gamma_{2})\leq~{}L$.
We estimate the above by

$\displaystyle\|(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(D^{\eta_{1}+2}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}\|_{2}\|(D^{\eta_{2}}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}\|\|Q_{\gamma}(x)P_{2L}(\lambda^{-2}D)(\lambda^{-4}D^{2}-1)^{-2L-1}\|_{2},$

which can in turn be bounded by

$\displaystyle
C(\alpha)|\lambda|^{4-2\alpha}\|u\|_{H^{L+\alpha}(\mathbb{R})}\|u\|_{H^{L}(\mathbb{R})}^{3}.$

To treat the second term in (76) we distinguish the cases $L=1$ and $L\geq 2$.
In the case of $L=1$ we estimate this expression by

$\displaystyle\|(\lambda^{-2}\mathcal{L}_{0}+1)^{-1}(\mathcal{L}_{0}U)\|_{2}\|(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}U)\|\|(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,1}^{d}(x,\lambda^{-2}D)\|_{2}$
$\displaystyle\lesssim|\lambda|^{2+\frac{2}{p}}\|u\|_{H^{1}(\mathbb{R})}^{3}\|Du\|_{L^{p}(\mathbb{R})},\quad
2\leq p\leq\infty.$

Putting $\sigma=\frac{\alpha}{2}\in[0,\frac{1}{2}[$ and choosing $p$ such that
$\sigma=\frac{1}{2}-\frac{1}{p}$, we get the bound

$|\lambda|^{3-2\sigma}\|u\|_{H^{1}(\mathbb{R})}^{3}\|u\|_{H^{1+\sigma}(\mathbb{R})}\leq|\lambda|^{4-2\alpha}\|u\|_{H^{1}(\mathbb{R})}^{3}\|u\|_{H^{1+\alpha}(\mathbb{R})}.$

If $L\geq 2$, we can proceed as for the Type 1 remainder terms and bound the
second term in (76) by $|\lambda|^{2}\|u\|_{H^{L}(\mathbb{R})}^{4}$. This
finishes our considerations of Type 2 remainder terms coming from (68).

The last group of Type 2 remainder terms comes from (69). Omitting the common
prefactor, the quantity of interest is

(77)
$\begin{split}&\operatorname{Tr}\big{[}(\mathcal{L}_{0}U)(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,0}^{a}(x,\lambda^{-2}D)\big{]}\\\
&-\operatorname{Tr}\big{[}(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}(\mathcal{L}_{0}U^{2})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}U(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,0}^{a}(x,\lambda^{-2}D)\big{]}\\\
&-\operatorname{Tr}\big{[}(\lambda^{-4}D^{2}-1)^{-1}(\mathcal{L}_{0}U^{3})(\lambda^{-2}\mathcal{L}_{0}-1)^{-1}R_{2L,0}^{a}(x,\lambda^{-2}D)\big{]},\end{split}$

where $R_{2L,0}^{a}(x,p)=\partial_{x}^{2L}U(x)P_{2L}(p)(p^{2}-1)^{-2L-1}$. No
new ideas are involved in the estimation of these terms; we simply integrate
by parts $L-1$ times to keep $L+1$ derivatives on $U$ coming from
$R_{2L,0}^{a}$ and then apply Lemma 3.3. We omit the remaining details for
these terms. Having now established the required bounds on the remainder
$|\tau_{L}^{m}(u,\lambda)|$, we have completed the proof of Lemma 2.1, modulo
the proof of Lemma 3.2 below.

###  3.6. Proof of Lemma 3.2

We argue by induction. The base cases are easy to verify explicitly:

$R_{0}^{a}(x,p)=\frac{1}{p^{2}-1}\cdot U(-i)=R_{0,0}^{a}(x,p)$
$R_{1}^{d}(x,p)=\frac{1}{(p^{2}-1)^{2}}\cdot
U^{2}\cdot(-1)\cdot(p\sigma_{3}-1)=R_{1,1}^{d}(x,p).$

Using (48)–(49) along with our inductive hypothesis, we may write, for $k\geq
1$,

$\displaystyle R_{k}^{d}(x,p)$
$\displaystyle=\frac{1}{(p^{2}-1)^{k+1}}\bigg{[}\sum_{r=0}^{k-1}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r+1}\\\
|\gamma|=(k-1)-r\end{subarray}}-iUQ_{\gamma}(x)P_{|\gamma|}(p)$
$\displaystyle\hskip
71.13188pt+\sum_{r=1}^{k-1}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r}\\\
|\gamma|=(k-1)-r\end{subarray}}i\partial_{x}Q_{\gamma}(x)P_{|\gamma|}(p)(p\sigma_{3}-1)\sigma_{3}\bigg{]}(p\sigma_{3}-1).$

We check that the inner sums (together with the common factors of
$(p^{2}-1)^{-k-1}$ and $(p\sigma_{3}-1)$) can be absorbed into
$R_{k,r+1}^{d}(x,p)$ and $R_{k,r}^{d}(x,p)$, respectively.

  * •

When $\gamma\in\mathbb{N}^{2r+1}$ and $|\gamma|=(k-1)-r$, the term
$iUQ_{\gamma}(x)P_{|\gamma|}(p)$ can be absorbed into $R_{k,r+1}^{d}(x,p)$:

    * –

First, $UQ_{\gamma}=Q_{(0,\gamma)}$, with $(0,\gamma)\in\mathbb{N}^{2(r+1)}$
and $|(0,\gamma)|=|\gamma|=k-(r+1)$.

    * –

Second, $\deg P_{|\gamma|}(p)\leq(k-1)-r=k-(r+1)$.

  * •

When $\gamma\in\mathbb{N}^{2r}$ and $|\gamma|=(k-1)-r$, the term
$\partial_{x}Q_{\gamma}(x)P_{|\gamma|}(p)$ can be absorbed into
$R_{k,r}^{d}(x,p)$:

    * –

First, $\partial_{x}Q_{\gamma}$ is a sum of $Q_{\gamma^{\prime}}$’s, with
$|\gamma^{\prime}|=|\gamma|+1=k-r$;

    * –

Second, $\deg P_{|\gamma|}(p)(p\sigma_{3}-1)\leq[(k-1)-r]+1=k-r$.

The formula for $R_{k}^{a}(x,p)$ may be verified in exactly the same way. We
write

$\displaystyle R_{k}^{a}(x,p)$
$\displaystyle=\frac{1}{(p^{2}-1)^{k+1}}\bigg{[}\sum_{r=0}^{k-1}\sum_{\begin{subarray}{c}\gamma\in\mathbb{N}^{2r+1}\\\
|\gamma|=(k-1)-r\end{subarray}}U^{2}Q_{\gamma}(x)P_{|\gamma|}(p)+i\partial_{x}Q_{\gamma}(x)P_{|\gamma|}(p)\sigma_{3}(p\sigma_{3}+1)$
$\displaystyle\hskip
85.35826pt-\sum_{r=1}^{k-1}\sum_{\begin{subarray}{c}\eta\in\mathbb{N}^{2r}\\\
|\eta|=(k-1)-r\end{subarray}}U\partial_{x}Q_{\eta}(x)P_{|\eta|}(p)(p\sigma_{3}-1)\sigma_{3}\bigg{]}.$

As above, we perform the routine verifications of the numerology as follows.

  * •

When $\gamma\in\mathbb{N}^{2r+1}$ and $|\gamma|=(k-1)-r$, the term
$U^{2}Q_{\gamma}(x)P_{|\gamma|}(x)$ can be absorbed into $R^{a}_{k,r+1}(x,p)$:

    * –

First, $U^{2}Q_{\gamma}=Q_{(0,0,\gamma)}$, with
$(0,0,\gamma)\in\mathbb{N}^{2(r+1)+1}$, $|(0,0,\gamma)|=|\gamma|=k-(r+1)$;

    * –

Second, $\deg P_{|\gamma|}(p)\leq(k-1)-r=k-(r+1)$.

  * •

When $\gamma\in\mathbb{N}^{2r+1}$ and $|\gamma|=(k-1)-r$, the term
$i\partial_{x}Q_{\gamma}(x)P_{|\gamma|}(p)\sigma_{3}(p\sigma_{3}+1)$ can be
absorbed into $R^{a}_{k,r}(x,p)$. When $\eta\in\mathbb{N}^{2r}$ and
$|\eta|=(k-1)-r$, the same is true of
$U\partial_{x}Q_{\eta}(x)P_{|\eta|}(p)(p\sigma_{3}-~{}1)\sigma_{3}$.

    * –

First, both $\partial_{x}Q_{\gamma}$ and $U\partial_{x}Q_{\eta}$ are sums of
$Q_{\gamma^{\prime}}$’s, with $\gamma^{\prime}\in\mathbb{N}^{2r+1}$ and
$|\gamma^{\prime}|=k-r$.

    * –

Second, $P_{|\gamma|}(p)\sigma_{3}(p\sigma_{3}+1)$ and
$P_{|\eta|}(p)(p\sigma_{3}-1)\sigma_{3}$ each have degree at most $k-r$.

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The Subelliptic Heat Kernel of the Octonionic Anti-De Sitter Fibration

The Subelliptic Heat Kernel
of the Octonionic Anti-De Sitter Fibration

Fabrice BAUDOIN and Gunhee CHO

F. Baudoin and G. Cho

Department of Mathematics, University of Connecticut,
196 Auditorium Road, Storrs, CT 06269-3009, USA<EMAIL_ADDRESS><EMAIL_ADDRESS>
Received July 31, 2020, in final form January 29, 2021; Published online
February 10, 2021

In this note, we study the sub-Laplacian of the $15$-dimensional octonionic
anti-de Sitter space which is obtained by lifting with respect to the anti-de
Sitter fibration the Laplacian of the octonionic hyperbolic space
$\mathbb{O}H^{1}$. We also obtain two integral representations for the
corresponding subelliptic heat kernel.

sub-Laplacian; 15-dimensional octonionic anti-de Sitter space; the anti-de
Sitter fibration

58J35; 53C17

##  1 Introduction and results

In this note we study the sub-Laplacian and the corresponding sub-Riemannian
heat kernel of the octonionic anti-de Sitter fibration

$\mathbb{S}^{7}\hookrightarrow{\rm
AdS}^{15}(\mathbb{O})\rightarrow\mathbb{O}H^{1}.$

This paper follows the previous works [2, 3, 10] which respectively concerned:

  1. 1.

The complex anti-de Sitter fibrations:

$\mathbb{S}^{1}\hookrightarrow{\rm
AdS}^{2n+1}(\mathbb{C})\rightarrow{\mathbb{C}H}^{n}.$

  2. 2.

The quaternionic anti-de Sitter fibrations:

$\mathbb{S}^{3}\hookrightarrow{\rm
AdS}^{4n+3}(\mathbb{H})\rightarrow{\mathbb{H}H}^{n}.$

The 15-dimensional anti-de Sitter fibration is the last model space that
remained to be studied of a sub-Riemannian manifold arising from a $H$-type
semi-Riemannian submersion over a rank-one symmetric space, see the Table 3 in
[4].

Similarly to the complex and quaternionic case, the sub-Laplacian is defined
as the lift on ${\rm AdS}^{15}(\mathbb{O})$ of the Laplace–Beltrami operator
of the octonionic hyperbolic space $\mathbb{O}H^{1}$. However, in the complex
and quaternionic case the Lie group structure of the fiber played an important
role that we can not use here, since the fiber $\mathbb{S}^{7}$ is not a
group. Instead, we make use of some algebraic properties of $\mathbb{S}^{7}$
that were already pointed out and used by the authors in [1] for the study of
the octonionic Hopf fibration:

$\mathbb{S}^{7}\hookrightarrow\mathbb{S}^{15}\rightarrow\mathbb{O}P^{1}.$

Let us briefly describe our main results. Due to the cylindrical symmetries of
the fibration, the heat kernel of the sub-Laplacian only depends on two
variables: the variable $r$ which is the Riemannian distance on
$\mathbb{O}H^{1}$ (the starting point is specified with inhomogeneous
coordinate in Section 3) and the variable $\eta$ which is the Riemannian
distance starting at a pole on the fiber $\mathbb{S}^{7}$. We prove in
Proposition 3.1 that in these coordinates, the radial part of the sub-
Laplacian $\tilde{L}$ writes

$\tilde{L}=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}+{\tanh^{2}r}\left(\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}\right).$

As a consequence of this expression for the sub-Laplacian, we are able to
derive two equivalent formulas for the heat kernel. The first formula, see
Proposition 4.1, reads as follows: for $r\geq 0$, $\eta\in[0,\pi)$, $t>0$

$\displaystyle p_{t}(r,\eta)=\int_{0}^{\infty}s_{t}(\eta,{\rm
i}u)q_{t,15}(\cosh r\cosh u)\sinh^{6}u\,{\rm d}u,$

where $s_{t}$ is the heat kernel of the Jacobi operator

$\tilde{\triangle}_{\mathbb{S}^{7}}=\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}$

with respect to the measure $\sin^{6}\eta\,{\rm d}\eta$, and where $q_{t,15}$
is the Riemannian heat kernel on the $15$-dimensional real hyperbolic space
$\mathbb{H}^{15}$ given in (4.1). The second formula, see Proposition 4.2,
writes as follows:

$\displaystyle
p_{t}(r,\eta)=\int_{0}^{\pi}\int_{0}^{\infty}G_{t}(\eta,\varphi,u)q_{t,9}(\cosh
r\cosh u)\sin^{5}\varphi\,{\rm d}u\,{\rm d}\varphi,$

where $q_{t,9}$ is Riemannian heat kernel on the $9$-dimensional hyperbolic
space $\mathbb{H}^{9}$ and $G_{t}(\eta,\varphi,u)$ is given in (4.3).

Similarly to [2, 3, 10], it might be expected that explicit integral
representations of the heat kernel might be used to study small-time
asymptotics, inside and outside of the cut-locus. Integral representations of
heat kernels can also be used to obtain sharp heat kernel estimates, see [7].
Those applications of the heat kernel representations we obtain will possibly
be addressed in a future research project.

##  2 The octonionic anti-de Sitter fibration

Let

$\mathbb{O}=\left\\{x=\sum_{j=0}^{7}x_{j}e_{j},\,x_{j}\in\mathbb{R}\right\\},$

be the division algebra of octonions (see [9] for explicit representations of
this algebra). We recall that the multiplication rules are given by

$\displaystyle e_{i}e_{j}=e_{j}\qquad\text{if}\quad i=0,$ $\displaystyle
e_{i}e_{j}=e_{i}\qquad\text{if}\quad j=0,$ $\displaystyle
e_{i}e_{j}=-\delta_{ij}e_{0}+\epsilon_{ijk}e_{k}\qquad\text{otherwise},$

where $\delta_{ij}$ is the Kronecker delta and $\epsilon_{ijk}$ is the
completely antisymmetric tensor with value $1$ when
$ijk=123,145,176,246,257,347,365$ (also see [1]). The octonionic norm is
defined for $x\in\mathbb{O}$ by

$||x||^{2}=\sum_{j=0}^{7}x^{2}_{j}.$

The octonionic anti-de Sitter space ${\rm AdS}^{15}(\mathbb{O})$ is the
quadric defined as the pseudo-hyperbolic space by:

${\rm
AdS}^{15}(\mathbb{O})=\big{\\{}(x,y)\in\mathbb{O}^{2},\,||(x,y)||^{2}_{\mathbb{O}}=-1\big{\\}},$

where

$||(x,y)||^{2}_{\mathbb{O}}:=||x||^{2}-||y||^{2}.$

In real coordinates we have $x=\sum_{j=0}^{7}x_{j}e_{j}$,
$y=\sum_{j=0}^{7}y_{j}e_{j}$, and the pseudo-norm can be written as

$x_{0}^{2}+\cdots+x_{7}^{2}-y_{0}^{2}-\cdots-y_{7}^{2}.$

As such, ${\rm AdS}^{15}(\mathbb{O})$ is embedded in the flat $16$-dimensional
space $\mathbb{R}^{8,8}$ endowed with the Lorentzian real signature $(8,8)$
metric

${\rm d}s^{2}={\rm d}x_{0}^{2}+\dots+{\rm d}x_{7}^{2}-{\rm
d}y_{0}^{2}-\cdots-{\rm d}y_{7}^{2}.$

Consequently, ${\rm AdS}^{15}(\mathbb{O})$ is naturally endowed with a pseudo-
Riemannian structure of signature $(8,7)$.

Let $\mathbb{O}H^{1}$ denote the octonionic hyperbolic space. The map
$\pi\colon{\rm AdS}^{15}(\mathbb{O})\rightarrow\mathbb{O}H^{1}$, given by
$(x,y)\mapsto[x:y]=y^{-1}x$ is a pseudo-Riemannian submersion with totally
geodesic fibers isometric to the seven-dimensional sphere $\mathbb{S}^{7}$.
Notice that, as a topological manifold, $\mathbb{O}H^{1}$ can therefore be
identified with the unit open ball in $\mathbb{O}$. The pseudo-Riemannian
submersion $\pi$ yields the octonionic anti-de Sitter fibration

$\mathbb{S}^{7}\hookrightarrow{\rm
AdS}^{15}(\mathbb{O})\rightarrow\mathbb{O}H^{1}.$

For further information on semi-Riemannian submersions over rank-one symmetric
spaces, we refer to [6].

##  3 Cylindrical coordinates and radial part of the sub-Laplacian

The sub-Laplacian $L$ on ${\rm AdS}^{15}(\mathbb{O})$ we are interested in is
the horizontal Laplacian of the Riemannian submersion $\pi\colon{\rm
AdS}^{15}(\mathbb{O})\to\mathbb{O}H^{1}$, i.e., the horizontal lift of the
Laplace–Beltrami operator of $\mathbb{O}H^{1}$. It can be written as

$L=\square_{{\rm AdS}^{15}(\mathbb{O})}+\triangle_{\mathcal{V}},$ (3.1)

where $\square_{{\rm AdS}^{15}(\mathbb{O})}$ is the d’Alembertian, i.e., the
Laplace–Beltrami operator of the pseudo-Riemannian metric and
$\triangle_{\mathcal{V}}$ is the vertical Laplacian. Since the fibers of $\pi$
are totally geodesic and isometric to $\mathbb{S}^{7}\subset{\rm
AdS}^{15}(\mathbb{O})$, we note that $\square_{{\rm AdS}^{15}(\mathbb{O})}$
and $\triangle_{\mathcal{V}}$ are commuting operators, and we can identify

$\triangle_{\mathcal{V}}={\triangle}_{\mathbb{S}^{7}}.$ (3.2)

The sub-Laplacian $L$ is associated with a canonical sub-Riemannian structure
on ${\rm AdS}^{15}(\mathbb{O})$ which is of $H$-type, see [4].

To study $L$, we introduce a set of coordinates that reflect the cylindrical
symmetries of the octonionic unit sphere which provides an explicit local
trivialization of the octonionic anti-de Sitter fibration. Consider the
coordinates $w\in\mathbb{O}H^{1}$, where $w$ is the inhomogeneous coordinate
on $\mathbb{O}H^{1}$ given by $w=y^{-1}x$, with $x,y\in{\rm
AdS}^{15}(\mathbb{O})$. Consider the north pole $p\in\mathbb{S}^{7}$ and take
$Y_{1},\dots,Y_{7}$ to be an orthonormal frame of $T_{p}\mathbb{S}^{7}$. Let
us denote $\exp_{p}$ the Riemannian exponential map at $p$ on
$\mathbb{S}^{7}$. Then the cylindrical coordinates we work with are given by

$(w,\theta_{1},\dots,\theta_{7})\mapsto\left(\frac{\exp_{p}\big{(}\sum_{i=1}^{7}\theta_{i}Y_{i}\big{)}w}{\sqrt{1-{\rho}^{2}}},\frac{\exp_{p}\big{(}\sum_{i=1}^{7}\theta_{i}Y_{i}\big{)}}{\sqrt{1-{\rho}^{2}}}\right)\in{\rm
AdS}^{15}(\mathbb{O}),$

where $\rho={\|w\|}$ and
$\|\theta\|=\sqrt{\theta_{1}^{2}+\cdots+\theta_{7}^{2}}<\pi$.

A function $f$ on ${\rm AdS}^{15}(\mathbb{O})$ is called radial cylindrical if
it only depends on the two coordinates $(\rho,\eta)\in[0,1)\times[0,\pi]$
where $\eta=\sqrt{\sum_{i=1}^{7}\theta^{2}_{i}}$. More precisely $f$ is radial
cylindrical if there exists a function $g$ so that

$f\left(\frac{\exp_{p}\big{(}\sum_{i=1}^{7}\theta_{i}Y_{i}\big{)}w}{\sqrt{1-{\rho}^{2}}},\frac{\exp_{p}\big{(}\sum_{i=1}^{7}\theta_{i}Y_{i}\big{)}}{\sqrt{1-{\rho}^{2}}}\right)=g(\rho,\eta).$

We denote by $\mathcal{D}$ the space of smooth and compactly supported
functions on $[0,1)\times[0,\pi)$. Then the radial part of $L$ is defined as
the operator $\widetilde{L}$ such that for any $f\in\mathcal{D}$, we have

$L(f\circ\psi)=\big{(}\widetilde{L}f\big{)}\circ\psi.$ (3.3)

We now compute $\widetilde{L}$ in cylindrical coordinates.

######  Proposition 3.1.

The radial part of the sub-Laplacian on ${\rm AdS}^{15}(\mathbb{O})$ is given
in the coordinates $(r,\eta)$ by the operator

$\widetilde{L}=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}+{\tanh^{2}r}\left(\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}\right),$

where $r=\tanh^{-1}\rho$ is the Riemannian distance on $\mathbb{O}H^{1}$ from
the origin.

###### Proof.

Note that the radial part of the Laplace–Beltrami operator on the octonionic
hyperbolic space $\mathbb{O}H^{1}$ is

$\widetilde{\triangle}_{\mathbb{O}H^{1}}=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth
r+7\tanh r)\frac{{\partial}}{\partial{r}},$

and the radial part of the Laplace–Beltrami operator on $\mathbb{S}^{7}$ is

$\widetilde{\triangle}_{\mathbb{S}^{7}}=\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}.$
(3.4)

Since the octonionic anti-de Sitter fibration defines a totally geodesic
submersion with base space $\mathbb{O}H^{1}$ and fiber $\mathbb{S}^{7}$, the
semi-Riemannian metric on ${\rm AdS}^{15}(\mathbb{O})$ is locally given by a
warped product between the Riemannian metric of $\mathbb{O}H^{1}$ and the
Riemannian metric on $\mathbb{S}^{7}$. Hence the radial part of the
d’Alembertian becomes

$\widetilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}+g(r)\left(\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}\right),$
(3.5)

for some smooth function $g$ to be computed.

On the other hand, from the isometric embedding ${\rm
AdS}^{15}(\mathbb{O})\subset\mathbb{O}\times\mathbb{O}$, the d’Alembertian on
${\rm AdS}^{15}(\mathbb{O})$ is a restriction of the d’Alembertian on
$\mathbb{O}\times\mathbb{O}\simeq\mathbb{R}^{8,8}$ in the sense that for a
smooth $f\colon{\rm AdS}^{15}(\mathbb{O})\to\mathbb{R}$

$\square_{{\rm
AdS}^{15}(\mathbb{O})}f=\square_{\mathbb{O}\times\mathbb{O}}f^{*}_{/{\rm
AdS}^{15}(\mathbb{O})},$

where
$\square_{\mathbb{O}\times\mathbb{O}}=\sum_{i=0}^{7}\big{(}\frac{\partial^{2}}{\partial
x_{i}^{2}}-\frac{\partial^{2}}{\partial y_{i}^{2}}\big{)}$ and for
$x,y\in\mathbb{O}$ such that $\|y\|^{2}-\|x\|^{2}>0$,
$f^{*}(x,y)=f\big{(}\frac{x}{\sqrt{\|y\|^{2}-\|x\|^{2}}},\frac{y}{\sqrt{\|y\|^{2}-\|x\|^{2}}}\big{)}$.
For the specific choice of the function $f(x,y)=y_{1}$, one easily computes
that $\square_{\mathbb{O}\times\mathbb{O}}f^{*}_{/{\rm
AdS}^{15}(\mathbb{O})}(x,y)=15y_{1}$, thus

$\square_{{\rm AdS}^{15}(\mathbb{O})}f(x,y)=15y_{1}.$

For the point with coordinates

$\left(\frac{\exp_{p}\big{(}\sum_{i=1}^{7}\theta_{i}Y_{i}\big{)}w}{\sqrt{1-{\rho}^{2}}},\frac{\exp_{p}\big{(}\sum_{i=1}^{7}\theta_{i}Y_{i}\big{)}}{\sqrt{1-{\rho}^{2}}}\right)\in{\rm
AdS}^{15}(\mathbb{O})$

one has

$y_{1}=\frac{\cos\eta}{\sqrt{1-{\rho}^{2}}}=\cosh r\cos\eta.$

We therefore deduce that

$\widetilde{\square}_{{\rm AdS}^{15}(\mathbb{O})}(\cosh r\cos\eta)=15\cosh
r\cos\eta.$

Using the formula (3.5), after a straightforward computation, this yields
$g(r)=-\frac{1}{\cosh^{2}r}$ and therefore

$\displaystyle\widetilde{\square}_{{\rm AdS}^{15}(\mathbb{O})}$
$\displaystyle=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}-\frac{1}{\cosh^{2}r}\left(\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}\right)$
$\displaystyle=\widetilde{\triangle}_{\mathbb{O}H^{1}}-\frac{1}{\cosh^{2}r}\widetilde{\triangle}_{\mathbb{S}^{7}}.$

Finally, to conclude, one notes that the sub-Laplacian $L$ is given by the
difference between the Laplace–Beltrami operator of ${\rm
AdS}^{15}(\mathbb{O})$ and the vertical Laplacian. Therefore by (3.1) and
(3.2),

$\displaystyle\widetilde{L}=\tilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}+\tilde{\triangle}_{\mathbb{S}^{7}}=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth
r+7\tanh
r)\frac{{\partial}}{\partial{r}}+{\tanh^{2}r}\left(\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}\right).$
∎

######  Remark 3.2.

As a consequence of the previous result, we can check that the Riemannian
measure of ${\rm AdS}^{15}(\mathbb{O})$ in the coordinates $(r,\eta)$, which
is the symmetric and invariant measure for $\tilde{L}$ is given by

${\rm
d}\overline{\mu}=\frac{{\pi}^{7}}{90}\sinh^{7}r\cosh^{7}r\sin^{6}\eta\,{\rm
d}r\,{\rm d}\eta.$ (3.6)

(See also Remark 2 in [1], which corresponds to the case of the octonionic
Hopf fibration.)

##  4 Integral representations of the subelliptic heat kernel

In this section, we give two integral representations of the subelliptic heat
kernel associated with $\tilde{L}$. We denote by $p_{t}(r,\eta)$ the heat
kernel of $\tilde{L}$ issued from the point $r=\eta=0$ with respect to the
measure (3.6). We remark that studying the subelliptic heat kernel associated
with $\tilde{L}$ is enough to study the heat kernel of $L$, because due to
(3.3) the heat kernel $h_{t}(w,\theta)$ of $L$ issued from the point with
cylindric coordinates $w=0$, $\theta=0$ is then given by

$h_{t}(w,\theta)=p_{t}\big{(}\tanh^{-1}\|w\|,\|\theta\|\big{)}.$

###  4.1 First integral representation

We denote by $s_{t}$ the heat kernel of the operator

$\tilde{\triangle}_{\mathbb{S}^{7}}=\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}$

with respect to the reference measure $\sin^{6}\eta\,{\rm d}\eta$. The
operator $\tilde{\triangle}_{\mathbb{S}^{7}}$ belongs to the family of Jacobi
diffusion operators which have been extensively studied in the literature, see
for instance the appendix in [5] and the references therein. In particular,
the spectrum of $\tilde{\triangle}_{\mathbb{S}^{7}}$ is given by

$\mathbf{Sp}\big{(}{-}\tilde{\triangle}_{\mathbb{S}^{7}}\big{)}=\\{m(m+6),\,m\in\mathbb{N}\\},$

and the eigenfunction corresponding to the eigenvalue $m(m+6)$ is
$P^{5/2,5/2}_{m}(\cos\eta)$ where $P^{5/2,5/2}_{m}$ is the Jacobi polynomial

$P_{m}^{5/2,5/2}(x)=\frac{(-1)^{m}}{2^{m}m!\big{(}1-x^{2}\big{)}^{5/2}}\frac{{\rm
d}^{m}}{{\rm d}x^{m}}\big{(}1-x^{2}\big{)}^{5/2+m}.$

As a consequence, one has the following spectral decomposition for the heat
kernel:

$\displaystyle
s_{t}(\eta,u)=\frac{1}{{\pi}}\sum_{m=0}^{+\infty}\frac{2^{4m+7}m!(m+5)![(m+3)!]^{2}}{(2m+6)!(2m+5)!}{\rm
e}^{-m(m+6)t}P_{m}^{5/2,5/2}(\cos\eta)P_{m}^{5/2,5/2}(\cos u).$

######  Proposition 4.1.

For $r\geq 0$, $\eta\in[0,\pi]$, and $t>0$ we have

$p_{t}(r,\eta)=\int_{0}^{\infty}s_{t}(\eta,{\rm i}u)q_{t,15}(\cosh r\cosh
u)\sinh^{6}u\,{\rm d}u,$

where

$q_{t,15}(\cosh s):=\frac{{\rm e}^{-49t}}{(2\pi)^{7}\sqrt{4\pi
t}}\left(-\frac{1}{\sinh s}\frac{{\rm d}}{{\rm d}s}\right)^{7}{\rm
e}^{-s^{2}/4t}$ (4.1)

is the Riemannian heat kernel on the $15$-dimensional real hyperbolic space
${\mathbb{H}}^{15}$.

###### Proof.

Since $\pi\colon{\rm AdS}^{15}(\mathbb{O})\to\mathbb{O}H^{1}$ is a (semi-
Riemannian) totally geodesic submersion, the operators
$\widetilde{{\square}}_{{\rm AdS}^{15}(\mathbb{O})}$ and
$\widetilde{\triangle}_{\mathbb{S}^{7}}$ commute. Thus

${\rm e}^{t\widetilde{L}}={\rm e}^{t(\widetilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}+\widetilde{\triangle}_{\mathbb{S}^{7}})}={\rm
e}^{t\widetilde{\triangle}_{\mathbb{S}^{7}}}{\rm
e}^{t\widetilde{\square}_{{\rm AdS}^{15}(\mathbb{O})}}.$

We deduce that the heat kernel of $\tilde{L}$ can be written as

$p_{t}(r,\eta)=\int_{0}^{\pi}s_{t}(\eta,u)p^{\widetilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}}_{t}(r,u)\sin^{6}u\,{\rm d}u,$ (4.2)

where $s_{t}$ is the heat kernel of (3.4) with respect to the measure
$\sin^{6}\eta\,{\rm d}\eta$, $\eta\in[0,\pi)$, and
$p^{\widetilde{\square}_{{\rm AdS}^{15}(\mathbb{O})}}_{t}(r,u)$ the heat
kernel at $(0,0)$ of $\widetilde{\square}_{{\rm AdS}^{15}(\mathbb{O})}$ with
respect to the measure in (3.6), i.e.,

${\rm d}\mu(r,u)=\frac{{\pi}^{7}}{90}\sinh^{7}r\cosh^{7}r\sin^{6}u\,{\rm
d}r\,{\rm d}u,\qquad r\in[0,\infty),\qquad u\in[0,\pi].$

In order to write (4.2) more precisely, let us consider the analytic change of
variables $\tau\colon(r,\eta)\rightarrow(r,{\rm i}\eta)$ that will be applied
on functions of the type $f(r,\eta)=h(r){\rm e}^{-{\rm i}\lambda\eta}$, with
$h$ smooth and compactly supported on $[0,\infty)$ and $\lambda>0$. Then as we
saw in the proof of Proposition 3.1 one can see that

$\widetilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}(f\circ\tau)=\big{(}\widetilde{\triangle}_{\mathbb{H}^{15}}f\big{)}\circ\tau,$

where

$\widetilde{\triangle}_{\mathbb{H}^{15}}=\widetilde{\triangle}_{\mathbb{O}H^{1}}+\frac{1}{\cosh^{2}r}\widetilde{\triangle}_{P},\qquad\widetilde{\triangle}_{P}=\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\coth\eta\frac{{\partial}}{\partial{\eta}}.$

Then, one deduces

${\rm e}^{t\widetilde{L}}(f\circ\tau)={\rm
e}^{t\widetilde{\triangle}_{\mathbb{S}^{7}}}{\rm
e}^{t\widetilde{\square}_{{\rm AdS}^{15}(\mathbb{O})}}(f\circ\tau)={\rm
e}^{t\widetilde{\triangle}_{\mathbb{S}^{7}}}\big{(}\big{(}{\rm
e}^{t{\widetilde{\triangle}_{\mathbb{H}^{15}}}}f\big{)}\circ\tau\big{)}=\big{(}{\rm
e}^{-t\widetilde{\triangle}_{P}}{\rm
e}^{t{\widetilde{\triangle}_{\mathbb{H}^{15}}}}f\big{)}\circ\tau.$

Now, since for every $f(r,\eta)=h(r){\rm e}^{-{\rm i}\lambda\eta}$,

$\big{(}{\rm e}^{t\widetilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}}f\big{)}(0,0)=\big{(}{\rm
e}^{t{\widetilde{\triangle}_{\mathbb{H}^{15}}}}\big{)}\big{(}f\circ\tau^{-1}\big{)}(0,0),$

one deduces that for a function $h$ depending only on $u$,

$\int_{0}^{\pi}h(u)p^{\widetilde{\square}_{{\rm
AdS}^{15}(\mathbb{O})}}_{t}(r,u)\sin^{6}u\,{\rm d}u=\int_{0}^{\infty}h(-{\rm
i}u)q_{t,15}(\cosh r\cosh u)\sinh^{6}u\,{\rm d}u.$

Therefore, coming back to (4.2), one infers that using the analytic extension
of $s_{t}$ one must have

$\displaystyle\int_{0}^{\pi}s_{t}(\eta,u)p^{\square_{{\rm
AdS}^{15}(\mathbb{O})}}_{t}(r,u)\sin^{6}u\,{\rm
d}u=\int_{0}^{\infty}s_{t}(\eta,-{\rm i}u)q_{t,15}(\cosh r\cosh
u)\sinh^{6}u\,{\rm d}u,$

where $q_{t,15}$ is the Riemannian heat kernel on the real hyperbolic space
$\mathbb{H}^{15}$ given in (4.1). ∎

###  4.2 Second integral representation

######  Proposition 4.2.

For $r\geq 0$, $\eta\in[0,\pi]$, and $t>0$ we have

$\displaystyle
p_{t}(r,\eta)=\int_{0}^{\pi}\int_{0}^{\infty}G_{t}(\eta,\varphi,u)q_{t,9}(\cosh
r\cosh u)\sin^{5}\varphi\,{\rm d}u\,{\rm d}\varphi.$

where $q_{t,9}$ is the $9$-dimensional Riemannian heat kernel on the
hyperbolic space $\mathbb{H}^{9}$:

$\displaystyle q_{t,9}(\cosh s):=\frac{{\rm e}^{-16t}}{(2\pi)^{4}\sqrt{4\pi
t}}\left(\frac{1}{\sinh s}\frac{{\rm d}}{{\rm d}s}\right)^{4}{\rm
e}^{-s^{2}/4t},$

and

$\displaystyle G_{t}(\eta,\varphi,u)=\frac{15}{8}\sum_{m\geq 0}{\rm
e}^{-(m(m+6)+33)t}(\cos\eta+{\rm i}\sin\eta\cos\varphi)^{m}\cosh((m+3)u).$
(4.3)

###### Proof.

The strategy of the following method appeals to some results proved in [8].
Firstly, we decompose the subelliptic heat kernel in the $\eta$ variable with
respect to the basis of normalized eigenfunctions of
$\tilde{\triangle}_{S^{7}}=\frac{{\partial}^{2}}{\partial{\eta}^{2}}+6\cot\eta\frac{{\partial}}{\partial{\eta}}$.
Accordingly,

$p_{t}(r,\eta)=\sum_{m\geq 0}f_{m}(t,r)h_{m}(\eta),$

where for each $m$, $h_{m}$ is given by

$h_{m}(\eta)=\frac{15}{16}\int_{0}^{\pi}{(\cos\eta+{\rm
i}\sin\eta\cos\varphi)}^{m}{\sin^{5}\varphi}\,{\rm d}\varphi$

and $f_{m}(t,\cdot)$ solves the following heat equation

$\displaystyle\frac{\partial}{\partial t}f_{m}(t,r)$
$\displaystyle=\left(\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}-m(m+6){\tanh^{2}r}\right)f_{m}(t,r)$
$\displaystyle=\left(\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}+\frac{m(m+6)}{\cosh^{2}r}-m(m+6)\right)f_{m}(t,r).$

We consider then the operator

$L_{m}:=\frac{{\partial}^{2}}{\partial{r}^{2}}+(7\coth r+7\tanh
r)\frac{{\partial}}{\partial{r}}+\frac{m(m+6)}{\cosh^{2}r}+49,$

which was studied in [8, p. 229]. From [8, Theorem 2], with
$\alpha=3+\frac{m}{2},\beta=-\frac{m}{2}$, we deduce that the solution to the
wave Cauchy problem associated with the subelliptic Laplacian is given $f\in
C^{\infty}_{0}\big{(}\mathbb{O}H^{1}\big{)}$ by

$\cos\big{(}s\sqrt{-L_{m}}\big{)}(f)(w)=\frac{-\sinh
s}{(2\pi)^{4}}\left(\frac{1}{\sinh s}\frac{{\rm d}}{{\rm
d}s}\right)^{4}\int_{\mathbb{O}H^{1}}K_{m}(s,w,y)f(y)\frac{{\rm
d}y}{\big{(}1-||y||^{2}\big{)}^{8}},$

where

$\displaystyle K_{m}(s,w,y)=\frac{(1-\overline{\langle
w,y\rangle})^{3+m/2}}{(1-\langle
w,y\rangle)^{m/2}}\frac{1}{\cosh^{3}(d(w,y))\sqrt{\cosh^{2}(s)-\cosh^{2}(d(w,y))}}$
$\displaystyle\hphantom{K_{m}(s,w,y)=}{}\times{{}_{2}}F_{1}\left(m+3,-m-3,\frac{1}{2};\frac{\cosh(d(w,y))-\cosh(s)}{2\cosh(d(w,y))}\right),$

where ${}_{2}F_{1}$ is the Gauss hypergeometric function and ${\rm d}y$ stands
for the Lebesgue measure in $\mathbb{R}^{8}$. Using the spectral formula

${\rm e}^{tL}=\frac{1}{\sqrt{4\pi t}}\int_{\mathbb{R}}{\rm
e}^{-s^{2}/(4t)}\cos\big{(}s\sqrt{-L}\big{)}\,{\rm d}s,$

which holds for any non positive self-adjoint operator, we deduce that the
solution to the heat Cauchy problem associated with $L_{m}$:

$\displaystyle{\rm e}^{tL_{m}}(f)(w)=\frac{{\rm
e}^{-m(m+6)t-7^{2}t}}{\sqrt{4\pi t}(2\pi)^{4}}\int_{\mathbb{R}}{\rm d}s(-\sinh
s){\rm e}^{-s^{2}/(4t)}$ $\displaystyle\hphantom{{\rm
e}^{tL_{m}}(f)(w)=}{}\times\left(\frac{1}{\sinh s}\frac{{\rm d}}{{\rm
d}s}\right)^{4}\int_{\mathbb{O}H^{1}}K_{m}(s,w,y)f(y)\frac{{\rm
d}y}{\big{(}1-||y||^{2}\big{)}^{8}}.$

Performing integration by parts $4$-times,

$\displaystyle\int_{\mathbb{R}}{\rm d}s(-\sinh s)\left(\frac{1}{\sinh
s}\frac{{\rm d}}{{\rm d}s}\right)^{4}{\rm
e}^{-s^{2}/(4t)}\int_{\mathbb{O}H^{1}}K_{m}(s,w,y)f(y)\frac{{\rm
d}y}{\big{(}1-||y||^{2}\big{)}^{8}}$
$\displaystyle\qquad{}=\int_{\mathbb{O}H^{1}}f(y)\frac{{\rm
d}y}{(1-||y||^{2})^{8}}\int_{\mathbb{R}}{\rm d}s(-\sinh
s)K_{m}(s,w,y)\left(\frac{1}{\sinh s}\frac{{\rm d}}{{\rm d}s}\right)^{4}{\rm
e}^{-s^{2}/4t}$ $\displaystyle\qquad{}=2\int_{\mathbb{O}H^{1}}f(y)\frac{{\rm
d}y}{\big{(}1-||y||^{2}\big{)}^{8}}\int_{d(w,y)}^{\infty}{\rm
d}(\cosh(s))K_{m}(s,w,y)\left(\frac{1}{\sinh s}\frac{{\rm d}}{{\rm
d}s}\right)^{4}{\rm e}^{-s^{2}/4t}.$

Thus we get

${\rm e}^{tL_{m}}(f)(0)=2{\rm
e}^{-(m(m+6)+33)t}\int_{\mathbb{O}H^{1}}f(y)\frac{{\rm
d}y}{\big{(}1-||y||^{2}\big{)}^{8}}\int_{d(0,y)}^{\infty}{\rm d}(\cosh
s)K_{m}(s,0,y)q_{t,9}(\cosh s).$

As a result, the subelliptic heat kernel of $L_{m}$ reads

$\displaystyle\frac{{\rm
d}y}{\big{(}1-||y||^{2}\big{)}^{8}}\int_{d(0,y)}^{\infty}{\rm d}(\cosh
s)K_{m}(s,0,y)q_{t,9}(\cosh s)$ $\displaystyle\qquad{}={\rm
d}r\sinh^{7}r\cosh^{7}r\int_{r}^{\infty}{\rm d}(\cosh
s)K_{m}(s,0,y)q_{t,9}(\cosh s).$

By changing the variable $\cosh s=\cosh r\cosh u$ for $u\geq 0$, the last
expression becomes

${\rm
d}r\sinh^{7}r\cosh^{7}r\int_{0}^{\infty}{{}_{2}}F_{1}\left(m+3,-m-3,\frac{1}{2};\frac{1-\cosh
u}{2}\right)q_{t,9}(\cosh r\cosh u)\,{\rm d}u.$

Therefore $p_{t}(r,\eta)$ has the integral representation

$2\sum_{m\geq 0}{\rm
e}^{-(m(m+6)+33)t}h_{m}(\eta)\int_{0}^{\infty}{{}_{2}}F_{1}\left(m+3,-m-3,\frac{1}{2};\frac{1-\cosh
u}{2}\right)q_{t,9}(\cosh r\cosh u)\,{\rm d}u.$

Now, notice that ${}_{2}F_{1}\big{(}m+3,-m-3,\frac{1}{2};\frac{1-\cosh
u}{2}\big{)}$ is simply the Cheybyshev polynomial of the first kind

$T_{m+3}(x)={{}_{2}}F_{1}\left(m+3,-m-3,\frac{1}{2};\frac{1-x}{2}\right),$

for all $x\in\mathbb{C}$. Therefore, one has

${}_{2}F_{1}\left(m+3,-m-3,\frac{1}{2};\frac{1-\cosh
u}{2}\right)=T_{m+3}(\cosh u)=\cosh((m+3)u),$

and the proof is over. ∎

### Acknowledgements

F.B. is partially funded by the NSF grant DMS-1901315.

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