$\displaystyle:=\mathcal{L}^{6}\overset{\eqref{ubtr_ju_fn11}}{=}\omega\cdot\partial_{\varphi}-D_{0}+\mathcal{R}_{0},$

we prove in Subsection 10.5 that $\mathcal{R}_{0}$ indeed satisfies the
smallness condition (R2) and thus achieve the desired reduction to (1.142).

Lastly, we note that all transformations, constructed in the reduction
procedure from $\mathcal{L}_{\omega}$ to $\mathcal{L}^{6}$, are required to be
1) reversibility preserving, 2) $\frac{2\pi}{\mathtt{M}}$-translation
invariance preserving and 3) real (mapping a real-valued function to a real-
valued function), to guarantee that the quasiperiodic solution $f(t,x)$ in
Theorem 1.2 is a reversible, $\frac{2\pi}{\mathtt{M}}$-translation invariant
and real-valued solution.

####  1.2.3 Summary of the sketch and the structure of the monograph

In summary, we derive a weak Birkhoff normal form from the Hamiltonian of the
gSQG equation with two purposes: 1) extraction of an internal parameter and 2)
reduction of the size of the perturbative term, inspired by [8, 9, 52]. Once
the weak Birkhoff normal from is derived, we aim to perform a Nash-Moser
scheme, which leads us to study an approximate inverse of the linearized
operator. The search for an approximate inverse reduces to the invertibility
of the linearized operator restricted to the normal space in the spirit of the
derivation of the normal form derivation near an invariant torus in [12]. The
invertibility of the linearized operator in the normal direction will be
achieved by reduction of the operator to a constant coefficients operator (a
diagonalized operator) by means of symplectic transformations studied in [6,
16]. Once the reduction is complete, then we finally prove that the
approximate solutions constructed by the Nash-Moser iteration converge to the
desired quasiperiodic solution. Meanwhile, necessary non-resonance conditions
on the frequencies will be taken into account following the strategy in [65].

In this monograph, the above scheme has been structured as follows:

Chapter 2 is devoted to the basic definitions of the spaces and operators, and
also to the spaces we will work in. In particular, we also analyze the
different multipliers that will appear throughout the proof and the properties
of the special functions involved.

Chapter 3 explains the Hamiltonian character of the gSQG equation and recasts
it in a way which is compatible with its associated Poisson bracket. We also
perform expansions of the Hamiltonian that will later prove useful in the
upcoming chapters. Some useful invariance properties of the gSQG equation will
be considered as well.

Chapter 4 performs the calculation of the weak Birkhoff normal form, reducing
the nonlinear interactions in a way that for homogeneous terms of degree $n$,
$n\leq 5$, which ensures the sufficient smallness condition of the
perturbative term. The nonexistence of non-trivial resonances in the quartic
Hamiltonian will be investigated, which serves as a key ingredient to derive a
desired normal form.

Chapters 5 and 6 set the problem up into action-angle coordinates and split
the nonlinear operator into three different components. We summarize the
necessary conditions on the choice of the tangential sites $S$, which will be
mainly used in the measure estimate of frequency set. We also state our main
theorem there in terms of finding a zero of a nonlinear operator
$\mathcal{F}_{\omega}$.

Chapter 7 reduces the problem of understanding the linearization of
$\mathcal{F}_{\omega}$ at a given embedding into a linearization of a normal
form around the trivial embedding and proving it is invertible as an operator
between the corresponding spaces. Furthermore, the system is diagonal and the
condition can be further reduced to prove the invertibility of the normal part
of the operator.

Chapter 8 computes a more explicit representation of the aforementioned
operator, emphasizing that the calculation is not restricted to the $O(1)$
terms but also to the $O(\varepsilon)$ and $O(\varepsilon^{2})$ as well, since
these terms are not perturbative. For this purpose, we will classify the
coefficients/smoothing operators arising from the linearized operator as
special classes $\mathfrak{C}_{1},\mathfrak{C}_{2}/\mathfrak{R}$, which
reveals that the non-perturbative terms (terms of size
$O(\varepsilon),O(\varepsilon^{2})$) are independent of the approximate
solutions.

Chapter 9 collects some useful properties of symplectic transformations that
will be used in the reduction process towards a constant coefficients
operator.

In Chapter 10, we will conjugate the operator via reversible transformations
in order to reduce it to a constant-coefficient operator, modulo semilinear
and $O(\varepsilon^{3})$ parts. This is the most important piece of the
manuscript and the most demanding one. These transformations are discussed in
Chapter 9, along with their corresponding spaces.

Finally, in Chapters 11 and 12 we finalize the proof of the Theorem by using a
Nash-Moser argument and computing the estimates on the measure of set of the
non-resonant frequencies.

Appendix A contains technical lemmas related to the computations of the
different pseudo-differential operators.

Appendix B contains a rigorous construction of tangential sites satisfying all
the non-resonance conditions described in Section 5.1.

The relevant connections between sections/propositions in the proof are
summarized in Figure 1.

Homogeneous expansion of the Hamiltonian $\mathcal{H}$ (Proposition 3.2)
Derivation of a weak Birkhoff normal form (Proposition 4.4) Derivation of
parameter-dependent Hamiltonian (Proposition 5.1) Functional equation setting
and the precise statement of the main theorem (Theorem 6.5) Construction of
approximate inverse (Proposition 7.3), assuming Proposition 7.14 Nash-Moser
iteration: Convergence of approximate solutions (Proposition 11.1) Proof of
Theorem 6.5 Non-existence of nontrivial resonance in the tangential sites
(Proposition 4.4) Elimination of low-degree homogeneous terms (Proposition
4.4) Specification of quadratic terms in the normal variable (Proposition 4.4)
Linear Birkhoff normal form: Elimination of non-perturbative terms of size
$O(\varepsilon^{2})$ (Subsection 10.4) Reduction of variable coefficients of
high order operators (Subsection 10.1-10.3) Derivation of the linearized
operator in the normal direction (Proposition 8.18) Summary of the reduction
up to a sufficiently small remainder (Proposition 10.76) Full reduction to a
constant coefficients operator and the estimates of the eigenvalues
(Subsection 10.6) Proof of Prop 7.14 (Subsection 10.7) Necessary conditions on
the tangential sites (Section 5.1) and the proof of their existence (Appendix
B) Measure estimates of the set of non-resonant frequencies (Subsection 11.1)
Figure 1: Connections between relevant sections and propositions

## Acknowledgements

JGS was partially supported by NSF through Grant DMS 1763356, and by the AGAUR
project 2021-SGR-0087 (Catalunya). JGS and JP were partially supported by
MICINN (Spain) research grant number PID2021-125021NA-I00 and by the European
Research Council through ERC-StG-852741-CAPA. AI was partially supported by
NSF grant DMS-2007008. JP was partially supported by ERC Starting Grant
676675-FLIRT, and the SNF grant 212573-FLUTURA. Part of this work was carried
out while JGS and JP were visiting the Department of Mathematics at Princeton
University. We wholeheartedly thank them for their warm hospitality. This
material is based upon work supported by the National Science Foundation under
Grant No. DMS-1929284 while the authors were in residence at the Institute for
Computational and Experimental Research in Mathematics in Providence, RI,
during the program “Hamiltonian Methods in Dispersive and Wave Evolution
Equations”. This work is supported by the Spanish State Research Agency,
through the Severo Ochoa and María de Maeztu Program for Centers and Units of
Excellence in R&D (CEX2020-001084-M).

##  2 Preliminaries and notations

In this section, we recall some classical concepts and specify notations that
are used throughout the paper.

#####  2.0.0.1 Basic notations

We denote by $\mathbb{N},\mathbb{Z}$, the sets of natural numbers and
integers, respectively, and denote
$\mathbb{N}_{0}:=\mathbb{N}\cup\left\\{0\right\\}$.

For an object $O$, which possibly is a number, operator or a set, and two real
numbers $A,B$, we denote $A\leq_{O}B,$ if there exists a constant $C$ that
depends on $O$ such that $A\leq CB$. Furthermore, we denote by $C_{O},c_{O}$
or $C(O),c(o)$, positive constants that depend on $O$, which might vary from
line to line in the computations.

For a linear operator $A$ acting on a vector space $X$, we denote the action
of $A$ on $h\in X$ by $Ah$ or $A[h]$, depending on whether a a clarification
is necessary. In case where $X$ is a Hilbert space, we denote by $A^{T}$ be
the adjoint operator of $A$.

###  2.1 Function spaces and norms

#####  2.1.0.1 Usual Sobolev spaces

We denote by $L^{2}(\mathbb{T})$ the space of square-integrable
$2\pi$-periodic real-valued functions. We denote for $s\in\mathbb{N}$,

$\displaystyle L^{2}_{0}:=\left\\{u\in
L^{2}(\mathbb{T}):\int_{\mathbb{T}}u(x)dx=0\right\\},\quad
H^{s}_{0}:=\left\\{u\in
H^{s}(\mathbb{T}):\int_{\mathbb{T}}u(x)dx=0\right\\},\quad
C^{\infty}_{0}:=\cap_{s\geq 0}H^{s}_{0}.$ (2.1)

For $\varphi\in\mathbb{T}^{\nu}$, we think of a $\varphi$-dependent function
$f(\varphi)\in C^{\infty}_{0}$ as a function $f=f(\varphi,x)$ on
$\mathbb{T}^{\nu+1}$. We denote the Sobolev spaces:

$\displaystyle H^{s}_{\varphi,x}:=H^{s}(\mathbb{T}^{\nu+1}),\quad
H^{s}_{\varphi}:=H^{s}(\mathbb{T}^{\nu}),\quad H^{s}_{x}:=H^{s}(\mathbb{T}),$
(2.2) $\displaystyle L^{2}_{\varphi,x}:=L^{2}(\mathbb{T}^{\nu+1}),\quad
L^{2}_{\varphi}:=L^{2}(\mathbb{T}^{\nu}),\quad L^{2}_{x}:=L^{2}(\mathbb{T}),$
$\displaystyle C^{\infty}_{\varphi,x}:=\cap_{s\geq 0}H^{s}_{\varphi,x},\quad
C^{\infty}_{\varphi}:=\cap_{s\geq 0}H^{s}_{\varphi},\quad
C^{\infty}_{x}:=\cap_{s\geq 0}H^{s}_{x}.$

For $u\in C^{\infty}_{0}$ and $f\in C^{\infty}_{\varphi,x}$, we denote its
Fourier modes by

$\displaystyle\widehat{u}_{j}:=\frac{1}{2\pi}\int_{\mathbb{T}}u(x)e^{-\text{i}jx}dx,\quad\text{
and
}\quad\widehat{f}_{j}(l):=\frac{1}{(2\pi)^{\nu+1}}\int_{\mathbb{T}^{\nu}}\int_{\mathbb{T}}f(\varphi,x)e^{-\text{i}(l\cdot\varphi+jx)}dxd\varphi.$

so that the Fourier inversion formula reads

$\displaystyle
u(x)=\sum_{j\in\mathbb{Z}}\widehat{u}_{j}e^{\text{i}jx},\quad\text{ and }\quad
f(\varphi,x)=\sum_{l\in\mathbb{Z}^{\nu},j\in\mathbb{Z}}\widehat{f}_{j}(l)e^{\text{i}(l\cdot\varphi+jx)}.$
(2.3)

We use the norm in $H^{s}_{\varphi,x}$, defined as

$\displaystyle\rVert f\rVert_{s}:=\rVert
f\rVert_{H^{s}_{\varphi,x}}:=\sqrt{\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}|\widehat{f}_{i}(l)|^{2}},\text{ where $\langle
l,j\rangle:=\sqrt{1+|l|^{2}+|j|^{2})}$.}$ (2.4)

Throughout the paper, we consider $H^{s}_{\varphi,x}$, where the regularity
index $s$ is assumed to be $s\geq s_{0}$, unless specified, and $s_{0}$ is
fixed so that

$\displaystyle s\geq s_{0}>\frac{\nu+2}{2}.$ (2.5)

For such $s,s_{0}$, the usual Sobolev embedding theorem gives us that

$\displaystyle\rVert fg\rVert_{s}\leq_{s}\rVert f\rVert_{s}\rVert
g\rVert_{s_{0}}+\rVert f\rVert_{s_{0}}\rVert g\rVert_{s},\text{ for all
$f,g\in C^{\infty}_{\varphi,x}$}.$ (2.6)

######  Lemma 2.1.

[16, Lemma 2.2] Let $a_{0},b_{0}\geq 0$ and $p,q>0$. For all $\varepsilon>0$,
there exists a constant $C=C(\varepsilon,p,q)>0$ such that

$\rVert u\rVert_{{a_{0}+p}}\rVert v\rVert_{{b_{0}+q}}\leq\varepsilon\rVert
u\rVert_{{a_{0}+p+q}}\rVert v\rVert_{{b_{0}}}+C\rVert u\rVert_{a_{0}}\rVert
v\rVert_{{b_{0}+p+q}},\text{ for all $u,v\in C^{\infty}_{\varphi,x}$.}$

#####  2.1.0.2 $\omega$-dependent functions

For a set $\Omega\subset\mathbb{R}^{\nu}$ and a Hilbert space
$(E,\rVert\cdot\rVert_{E})$, we consider $E$-valued functions $u=u(\omega)$
that depend on $\omega\in\Omega$. For such functions, we define the norms

$\displaystyle\rVert u\rVert_{E,\Omega}^{\sup}:=\sup_{\omega\in\Omega}\rVert
u(\omega)\rVert_{E},\quad\text{ and }\quad\rVert
u\rVert_{E,\Omega}^{\text{lip}}:=\sup_{\omega_{1},\omega_{2}\in\Omega,\omega_{1}\neq\omega_{2}}\frac{\rVert
u(\omega_{1})-u(\omega_{2})\rVert_{E}}{|\omega_{1}-\omega_{2}|}.$

Given $0<\gamma<1$, we denote

$\displaystyle\rVert u\rVert_{E}^{\text{Lip}(\gamma,\Omega)}:=\rVert
u\rVert_{E,\Omega}^{\sup}+\gamma\rVert
u\rVert_{E,\Omega}^{\text{lip}},\quad\text{Lip}(\Omega,E):=\left\\{f:\Omega\mapsto
E\ |\ \rVert f\rVert_{E}^{\text{Lip}(\gamma,\Omega)}<\infty\right\\}.$ (2.7)

In case $E=\mathbb{R}^{n}$ for some $n\in\mathbb{N}$, we simply denote

$\displaystyle|u|^{\text{Lip}(\gamma,\Omega)}:=\sup_{\omega\in\Omega}|u(\omega)|+\gamma\sup_{\omega_{1},\omega_{2}\in\Omega}\frac{|u(\omega_{1})-u(\omega_{2})|}{|\omega_{1}-\omega_{2}|}.$
(2.8)

When $E=H^{s}_{\varphi,x}$, we simply denote

$\rVert u\rVert^{\text{Lip}(\gamma,\Omega)}_{s}=:\rVert
u\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s}_{\varphi,x}}.$

We have useful estimates for such $\omega$-dependent functions:

######  Lemma 2.2 (Multiplication).

[6, Lemma 2.3]

$\rVert uv\rVert_{s}^{\text{Lip}(\gamma,\Omega)}\leq_{s}\rVert
u\rVert_{s}^{\text{Lip}(\gamma,\Omega)}\rVert
v\rVert_{s_{0}}^{\text{Lip}(\gamma,\Omega)}+\rVert
u\rVert_{s_{0}}^{\text{Lip}(\gamma,\Omega)}\rVert
v\rVert_{s}^{\text{Lip}(\gamma,\Omega)},\text{ for all
$f,g\in\text{Lip}(\Omega,H^{s}_{\varphi,x})$}.$

######  Lemma 2.3 (Interpolation).

Let $p\geq q\geq 0$. Then, there exists a constant $C=C(s_{0},q,p)$ such that

$\displaystyle\rVert u\rVert_{s_{0}+q}^{\text{Lip}(\gamma,\Omega)}\leq
C\left(\rVert
u\rVert^{\text{Lip}(\gamma,\Omega)}_{s_{0}}\right)^{1-\frac{q}{p}}\left(\rVert
u\rVert_{s_{0}+p}^{\text{Lip}(\gamma,\Omega)}\right)^{\frac{q}{p}}.$

###### Proof.

Apply the Gagliardo-Nirenberg interpolation inequality to $u(\omega)$ and
$\gamma\frac{u(\omega_{1})-u(\omega_{2})}{|\omega_{1}-\omega_{2}|}$ for
$\omega,\omega_{1},\omega_{2}\in\Omega$ and take the supremum over
$\omega,\omega_{1},\omega_{2}$. ∎

######  Lemma 2.4 (Change of variables).

[8, Lemma 2.4] Let $\beta:\mathbb{T}\mapsto\mathbb{T}$ be such that
$\rVert\beta\rVert_{W^{1,\infty}(\mathbb{T})}<\frac{1}{2}$. Then,

  1. (1)

$f_{\beta}:x\mapsto x+\beta(x)$ is invertible, and there exists
$\tilde{\beta}=\tilde{\beta}(x)$ such that
$f_{\beta}^{-1}(x)=x+\tilde{\beta}(x)$.

  2. (2)

If $\beta$ depends on $\omega\in\Omega$ and $\varphi\in\mathbb{T}^{\nu}$, it
holds that
$\rVert\tilde{\beta}\rVert^{\text{Lip}(\gamma,\Omega)}_{W^{s,\infty}(\mathbb{T}^{\nu+1})}\leq_{s}\rVert\beta\rVert^{\text{Lip}(\gamma,\Omega_{1})}_{W^{s+1,\infty}(\mathbb{T}^{\nu+1})}$.

  3. (3)

Given $u\in C^{\infty}_{\varphi,x}$, it holds that denoting $(u\circ
f_{\beta})(\varphi,x):=u(\varphi,x+\beta(\varphi,x))$,

$\rVert u\circ
f_{\beta}\rVert_{s}^{\text{Lip}(\gamma,\Omega)}\leq_{s}\left(\rVert
u\rVert^{\text{Lip}(\gamma,\Omega)}_{s+1}+\rVert\beta\rVert^{\text{Lip}(\gamma,\Omega)}_{W^{s,\infty}(\mathbb{T}^{\nu+1})}\rVert
u\rVert_{s_{0}}^{\text{Lip}(\gamma,\Omega)}\right).$

###  2.2 Linear operators

Let $\mathcal{A}:C^{\infty}_{x}\mapsto C^{\infty}_{x}$ be a linear operator.
We denote $\mathcal{A}^{T}$ to be its adjoint operator, that is,

$(\mathcal{A}[u],v)_{L^{2}_{x}}:=\int_{\mathbb{T}}\mathcal{A}u(x)\overline{v}(x)dx=\int_{\mathbb{T}}u(x)\overline{\mathcal{A}^{T}v}(x)dx=(u,\mathcal{A}^{T}[v])_{L^{2}_{x}},$

where $\overline{u}(x)$ is the complex conjugation of $u(x)$. We also consider
linear operators $\mathcal{A}:C^{\infty}_{\varphi,x}\mapsto
C^{\infty}_{\varphi,x}$. Clearly, we can think of
$\mathcal{A}=\mathcal{A}(\varphi)$ as a $\varphi$-dependent linear operator in
$C^{\infty}_{x}$. Thanks to the following elementary relation:

$\displaystyle(\mathcal{A}[f],g)_{L^{2}_{\varphi,x}}:=\int_{\mathbb{T}^{\nu}}\int_{\mathbb{T}}\mathcal{A}(\varphi)\left(f(\varphi,x)\right)\overline{g}(\varphi,x)dxd\varphi=\int_{\mathbb{T}^{\nu}}\int_{\mathbb{T}}f(\varphi,x)\overline{(\mathcal{A}(\varphi))^{T}g(\varphi,x)}dxd\varphi=(f,\mathcal{A}^{T}[g])_{L^{2}_{\varphi,x}},$

we have that $\mathcal{A}^{T}(\varphi)=\mathcal{A}(\varphi)^{T}$, we simply
use the notation $\mathcal{A}^{T}$ to denote the adjoint of $\mathcal{A}$ as a
linear operator in $C^{\infty}_{\varphi,x}$ or $C^{\infty}_{x}$, without any
confusion.

For a linear operator $\mathcal{A}:C^{\infty}_{\varphi,x}\mapsto
C^{\infty}_{\varphi,x}$, we have its matrix representation as (recalling our
convention of the Fourier series expansion in (2.3))

$\displaystyle\mathcal{A}f(\varphi,x)=\sum_{j_{1},j_{2}\in\mathbb{Z},\
l,l^{\prime}\in\mathbb{Z}^{\nu}}\mathcal{A}^{j_{1}}_{j_{2}}(l-l^{\prime})\widehat{f}_{j_{1}}(l^{\prime})e^{\text{i}(l\cdot\varphi+j_{2}x)}.$
(2.9)

######  Definition 2.5.

We say that an operator (not necessarily linear) $\mathcal{A}$ is real if
$\mathcal{A}$ maps a real-valued function to a real-valued function.

####  2.2.1 Pseudo differential operators

We consider symbols $\mathfrak{a}=\mathfrak{a}(x,\xi)\in
C^{\infty}(\mathbb{T}\times\mathbb{R})$. We say $\mathfrak{a}$ is a classical
symbol of order $m$ and denote $\mathfrak{a}\in\mathcal{S}^{m}$ for some
$m\in\mathbb{R}$, if

$\displaystyle\sup_{x\in\mathbb{T}}|\partial_{x}^{s}\partial_{\xi}^{\eta}\mathfrak{a}(x,\xi)|\langle\xi\rangle^{-(m-\eta)}<\infty,\text{
for all $s,\eta\geq 0$},\text{ where
$\langle\xi\rangle:=\sqrt{1+|\xi|^{2}}$.}$ (2.10)

We denote

$\displaystyle|\mathfrak{a}|_{m,H^{s}_{x},\eta_{0}}:=\max_{0\leq\eta\leq\eta_{0}}\sup_{\xi\in\mathbb{R}}\rVert\partial_{\xi}^{\eta}\mathfrak{a}(\cdot,\xi)\rVert_{H^{s}_{x}}\langle\xi\rangle^{\eta-m}.$
(2.11)

#####  2.2.1.1 Standard quantization and Weyl quantization

For a given symbol $\mathfrak{a}$, we denote its standard quantization by

$\displaystyle
Op(\mathfrak{a})[u]=\sum_{j\in\mathbb{Z}}\mathfrak{a}(x,j)\widehat{u}_{j}e^{\text{i}jx}.$
(2.12)

We also define the Weyl quantization of $\mathfrak{a}$ by

$\displaystyle
Op^{W}(\mathfrak{a})[u]:=\sum_{{k,j}\in\mathbb{Z}}\widehat{\mathfrak{a}}\left(k-j,\frac{k+j}{2}\right)\widehat{u}_{j}e^{\text{i}kx},\text{
where
}\widehat{\mathfrak{a}}(k,\xi):=\frac{1}{2\pi}\int_{\mathbb{T}}\mathfrak{a}(x,\xi)e^{-\text{i}kx}dx.$
(2.13)

Compared to the usual quantization in (2.12), one can easily check that

$\displaystyle
Op^{W}(\mathfrak{a})=Op(\mathfrak{b})\iff\widehat{\mathfrak{a}}\left(k,\frac{k}{2}+\xi\right)=\widehat{\mathfrak{b}}\left(k,\xi\right)\text{
for all $k,\xi\in\mathbb{Z}$}.$ (2.14)

For a linear operator $A$, we say a linear operator $A\in OPS^{m}$, if there
exists a classical symbol $\mathfrak{a}\in\mathcal{S}^{m}$ such that
$A=Op(\mathfrak{a})$ or $A=Op^{W}(\mathfrak{a})$. Note that thanks to Lemma
2.6, the statement $A\in OPS^{m}$ is well-defined, whether we use the standard
quantization or the Weyl quantization.

With the above definition of $Op^{W}(\mathfrak{a})$, one can easily check,
using (2.13), that

$\displaystyle Op^{W}(\mathfrak{a})^{T}=Op^{W}(\overline{\mathfrak{a}}),$
(2.15)

therefore,

$\displaystyle\text{$Op^{W}(\mathfrak{a})$ is a symmetric operator }\iff\text{
$\mathfrak{a}$ is a real-valued symbol.}$ (2.16)

The following results are classical:

######  Lemma 2.6.

We have that:

  1. (1)

(Equivalence of the quantizations) For
$\mathfrak{a},\mathfrak{b}\in\mathcal{S}^{m}$ such that
$Op^{W}(\mathfrak{a})=Op(\mathfrak{b})$, there exists $\mu=\mu(m,\eta)$ such
that for all $s\geq 0$ and $\eta\geq 0$,

$\displaystyle|\mathfrak{b}|_{m,H^{s}_{x},\eta}\leq_{\eta,s}|\mathfrak{a}|_{m,H^{s+\mu}_{x},\eta},\quad|\mathfrak{a}|_{m,H^{s}_{x},\eta}\leq_{\eta,s}|\mathfrak{b}|_{m,H^{s+\mu}_{x},\eta}.$

  2. (2)

(Action as a linear operator) Let $\mathfrak{a}\in\mathcal{S}^{m}$. There
exists $\mu=\mu(m)$ such that for all $s\geq 0$,

$\displaystyle\rVert
Op^{W}(\mathfrak{a})h\rVert_{H^{s}_{x}}\leq|\mathfrak{a}|_{m,H^{s+\mu}_{x},0}\rVert
h\rVert_{H^{m}_{x}}+|\mathfrak{a}|_{m,H^{\mu}_{x},0}\rVert
h\rVert_{H^{s+m}_{x}}.$ (2.17)

###### Proof.

Proof of (1). We prove only

$\displaystyle|\mathfrak{b}|_{m,H^{s}_{x},\eta}\leq_{\eta,s,m}|\mathfrak{a}|_{m,H^{s+\mu}_{x},\eta},$
(2.18)

since the other estimate follows in the same way. By definition of the norm in
(2.11), we have that for all $k\in\mathbb{Z},\xi\in\mathbb{R}$,

$\displaystyle|\widehat{\partial_{\xi}^{\eta}\mathfrak{a}}(k,\xi)|^{2}\langle
k\rangle^{2s}\leq\sum_{j\in\mathbb{Z}}|\widehat{\partial_{\xi}^{\eta}\mathfrak{a}}(j,\xi)|^{2}\langle
j\rangle^{2s}\leq_{s,\eta}|\mathfrak{a}|_{m,s,\eta}^{2}\langle\xi\rangle^{2(m-\eta)}.$
(2.19)

Therefore, using (2.14), we compute

$\displaystyle\rVert\partial_{\xi}^{\eta}\mathfrak{b}(\cdot,\xi)\rVert_{H^{s}_{x}}^{2}$
$\displaystyle=\sum_{k\in\mathbb{Z}}\langle
k\rangle^{2s}|\widehat{\partial_{\xi}^{\eta}\mathfrak{b}}(k,\xi)|^{2}=\sum_{k\in\mathbb{Z}}\langle
k\rangle^{2s}\left|\widehat{\partial_{\xi}^{\eta}\mathfrak{a}}(k,\xi+\frac{k}{2})\right|^{2}$
$\displaystyle\leq\sum_{k\in\mathbb{Z}}\langle
k\rangle^{2s+\mu}\left|\widehat{\partial_{\xi}^{\eta}\mathfrak{a}}(k,\xi+\frac{k}{2})\right|^{2}\langle
k\rangle^{-\mu}$
$\displaystyle\overset{\eqref{jsd2sdszzz}}{\leq_{s,\eta}}|\mathfrak{a}|_{m,s+\mu,\eta}^{2}\sum_{k\in\mathbb{Z}}\langle\xi+\frac{k}{2}\rangle^{2(m-\eta)}\langle
k\rangle^{-\mu},$ (2.20)

for any $\mu\geq 0$. By choosing $\mu:=4|m-\eta|+2$, it is clear that

$\langle\xi+\frac{k}{2}\rangle^{2(m-\eta)}\langle
k\rangle^{-\mu}\leq_{m,\eta}\begin{cases}\langle k\rangle^{2(m-\eta)}\langle
k\rangle^{-\mu}\leq_{m,\eta}\langle k\rangle^{-2|m-\eta|-2},&\text{ if
$\frac{2}{3}|k|\geq|\xi|$,}\\\ \langle\xi\rangle^{2(m-\eta)}\langle
k\rangle^{-\mu}\leq_{m,\eta}\langle\xi\rangle^{2(m-\eta)}\langle
k\rangle^{-2},&\text{ if $\frac{2}{3}|k|\leq|\xi|$}.\end{cases}$

Hence, (2.20) yields that

$\displaystyle\rVert\partial_{\xi}^{\eta}\mathfrak{b}(\cdot,\xi)\rVert^{2}_{H^{s}_{x}}$
$\displaystyle\leq_{s,m,\eta}|\mathfrak{a}|^{2}_{m,H^{s+\mu}_{x},\eta}\left(\sum_{k\in\mathbb{Z},\
\frac{2}{3}|k|\geq|\xi|}\langle
k\rangle^{-2|m-\eta|-2}+\langle\xi\rangle^{2(m-\eta)}\sum_{k\in\mathbb{Z},\
\frac{2}{3}|k|\leq 4|\xi|}\langle k\rangle^{-2}\right)$
$\displaystyle\leq_{m,\eta}|\mathfrak{a}|^{2}_{m,H^{s+\mu}_{x},\eta}\langle\xi\rangle^{2(m-\eta)},$

which proves (2.18).

Proof of (2). Let us choose $\mathfrak{b}$ so that
$Op(\mathfrak{b})=Op^{W}(\mathfrak{a})$ using (2.14). We compute

$\rVert Op^{W}(\mathfrak{a})h\rVert_{H^{s}_{x}}^{2}=\rVert
Op(\mathfrak{b})h\rVert_{H^{s}_{x}}^{2}=\sum_{k\in\mathbb{Z}}\left(\sum_{j\in\mathbb{Z}}\langle
k\rangle^{s}\widehat{\mathfrak{b}}(k-j,j)h_{j}\right)^{2}.$

Using $\langle k\rangle^{s}\leq_{s}\langle k-j\rangle^{s}+\langle
j\rangle^{s}$, we have

$\displaystyle\rVert
Op^{W}(\mathfrak{a})h\rVert_{H^{s}_{x}}^{2}\leq_{s}\sum_{k\in\mathbb{Z}}\left(\left(\sum_{j\in\mathbb{Z}}\langle
k-j\rangle^{s}|\widehat{\mathfrak{b}}(k-j,j)||h_{j}|\right)^{2}+\left(\sum_{j\in\mathbb{Z}}|\widehat{\mathfrak{b}}(k-j,j)|\langle
j\rangle^{s}|h_{j}|\right)^{2}\right)$ (2.21)

Using (2.19) with $\eta=0$, we have

$|\widehat{\mathfrak{b}}(k-j,j)|\leq|\mathfrak{b}|_{m,H^{s+2}_{x},0}\langle
k-j\rangle^{-s-2}\langle j\rangle^{m},\text{ and
}|\widehat{\mathfrak{b}}(k-j,j)|\leq|\mathfrak{b}|_{m,H^{2}_{x},0}\langle
k-j\rangle^{-2}\langle j\rangle^{m}.$

Plugging this into (2.21), we get

$\displaystyle\rVert
Op^{W}(\mathfrak{a})h\rVert_{H^{s}_{x}}^{2}\leq_{s}|\mathfrak{b}|_{m,H^{s+2}_{x},0}^{2}\sum_{k\in\mathbb{Z}}\left(\sum_{j\in\mathbb{Z}}\langle
k-j\rangle^{-2}\langle
j\rangle^{m}|h_{j}|\right)^{2}+|\mathfrak{b}|^{2}_{m,H^{2}_{x},0}\sum_{k\in\mathbb{Z}}\left(\sum_{j\in\mathbb{Z}}\langle
k-j\rangle^{-2}\langle j\rangle^{s+m}|h_{j}|\right)^{2}$ (2.22)

Let $\underline{g},\underline{h}_{m},\underline{h}_{s+m}$, be such that

$\underline{g}(x):=\sum_{k}\langle
k\rangle^{-2}e^{\text{i}kx},\quad\underline{h}_{m}(x):=\sum_{j}\langle
j\rangle^{m}|h_{j}|e^{\text{i}jx},\quad\underline{h}_{s+m}(x):=\sum_{j}\langle
j\rangle^{s+m}|h_{j}|e^{\text{i}jx}.$

Clearly,

$\rVert\underline{g}\rVert_{L^{\infty}_{x}}\leq
C,\quad\rVert\underline{h}_{m}\rVert_{L^{2}_{x}}\leq\rVert
h\rVert_{H^{m}_{x}},\quad\rVert\underline{h}_{s+m}\rVert_{L^{2}_{x}}\leq\rVert
h\rVert_{H^{s+m}_{x}},$

for some $C>0$. Hence, it follows from (2.22) that

$\displaystyle\rVert Op^{W}(\mathfrak{a})h\rVert_{H^{s}_{x}}^{2}$
$\displaystyle\leq_{s}|\mathfrak{b}|_{m,H^{s+2}_{x},0}^{2}\int_{\mathbb{T}}(\underline{g}(x)\underline{h}_{m}(x))^{2}dx+|\mathfrak{b}|_{m,H^{2}_{x},0}^{2}\int_{\mathbb{T}}(\underline{g}(x)\underline{h}_{s+m}(x))^{2}dx$
$\displaystyle\leq C\left(|\mathfrak{b}|_{m,H^{s+2}_{x},0}^{2}\rVert
h\rVert_{H^{m}_{x}}^{2}+|\mathfrak{b}|_{m,H^{2}_{x},0}^{2}\rVert
h\rVert_{H^{s+m}_{x}}^{2}\right).$

Using (1), we obtain (2.17). ∎

#####  2.2.1.2 Compositions:

For two symbols $\mathfrak{a}\in\mathcal{S}^{m}$,
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$, it follows straightforwardly from
the definitions in (2.12) and (2.13) that

$\displaystyle Op(\mathfrak{a})\circ
Op(\mathfrak{b})=Op(\mathfrak{c}),\quad\mathfrak{c}\in\mathcal{S}^{m+m^{\prime}},\quad\widehat{\mathfrak{c}}(k,\xi)=\sum_{j_{1}+j_{2}=k}\widehat{\mathfrak{a}}(j_{1},\xi+j_{2})\widehat{\mathfrak{b}}(j_{2},\xi),$
(2.23) $\displaystyle Op^{W}(\mathfrak{a})\circ
Op^{W}(\mathfrak{b})=Op^{W}(\mathfrak{c}),\quad\mathfrak{c}\in\mathcal{S}^{m+m^{\prime}},\quad\widehat{\mathfrak{c}}(k,\xi)=\sum_{j_{1}+j_{2}=k}\widehat{\mathfrak{a}}\left(j_{1},\xi+\frac{j_{2}}{2}\right)\widehat{\mathfrak{b}}\left(j_{2},\xi-\frac{j_{1}}{2}\right).$

######  Lemma 2.7.

For two symbols $\mathfrak{a}\in\mathcal{S}^{m}$,
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$, the composition of
$Op^{W}(\mathfrak{a}),Op^{W}(\mathfrak{b})$ can be expressed as

$\displaystyle Op^{W}(\mathfrak{a})\circ Op^{W}(\mathfrak{b})$
$\displaystyle=Op^{W}(\mathfrak{c}),$ (2.24)
$\displaystyle{\mathfrak{c}}(x,\xi)$
$\displaystyle=\mathfrak{a}\mathfrak{b}+\frac{1}{2\text{i}}\left\\{\mathfrak{a},\mathfrak{b}\right\\}+\sigma_{\mathfrak{a},\mathfrak{b}}^{1}+\sigma^{2}_{\mathfrak{a},\mathfrak{b}},$

where
$\left\\{\mathfrak{a},\mathfrak{b}\right\\}:=\partial_{\xi}\mathfrak{a}\partial_{x}\mathfrak{b}-\partial_{\xi}\mathfrak{b}\partial_{x}\mathfrak{a}$,
and $\sigma_{\mathfrak{a},\mathfrak{b}}^{1}\in\mathcal{S}^{m+m^{\prime}-2},\
\sigma^{2}_{\mathfrak{a},\mathfrak{b}}\in\mathcal{S}^{m+m^{\prime}-3}$ with

$\displaystyle\sigma_{\mathfrak{a},\mathfrak{b}}^{1}(x,\xi)$
$\displaystyle=\frac{1}{8}\left(-\partial_{xx}\mathfrak{a}\partial_{\xi\xi}\mathfrak{b}+2\partial_{x\xi}\mathfrak{a}\partial_{x\xi}\mathfrak{b}-\partial_{\xi\xi}\mathfrak{a}\partial_{xx}\mathfrak{b}\right),$
$\displaystyle\widehat{\sigma^{2}_{\mathfrak{a},\mathfrak{b}}}(k,\xi)$
$\displaystyle=(a_{1}b_{4}+a_{2}(b_{3}+b_{4})+a_{3}(b_{2}+b_{3}+b_{4})+a_{4}(b_{1}+b_{2}+b_{3}+b_{4})),\text{
where $a_{i},b_{i}$ are given in \eqref{fsdsdnotsds}.}$

###### Proof.

Recall the Fourier series representation of the composed symbol from (2.23):

$\displaystyle\widehat{\mathfrak{c}}(k,\xi)=\sum_{j_{1}+j_{2}=k}\widehat{\mathfrak{a}}\left(j_{1},\xi+\frac{j_{2}}{2}\right)\widehat{\mathfrak{b}}\left(j_{2},\xi-\frac{j_{1}}{2}\right).$
(2.25)

Using the Taylor expansion of smooth functions,

$\displaystyle
f(x)=\sum_{n=0}^{N}f^{(n)}(0)\frac{x^{n}}{n!}+x^{N+1}\int_{0}^{1}f^{(N+1)}(x\tau)\frac{(1-\tau)^{N}}{N!}d\tau\text{
for $x\in\mathbb{R}$},$ (2.26)

we consider the expansion of $\widehat{\mathfrak{a}},\widehat{\mathfrak{b}}$
in the variable $\xi$:

$\displaystyle\widehat{\mathfrak{a}}\left(j_{1},\xi+\frac{j_{2}}{2}\right)$
$\displaystyle=\widehat{\mathfrak{a}}(j_{1},\xi)+\widehat{\partial_{\xi}\mathfrak{a}}(j_{1},\xi)\frac{\text{i}j_{2}}{2\text{i}}+\frac{1}{2!}\widehat{\partial_{\xi\xi}\mathfrak{a}}(j_{1},\xi)\left(\frac{\text{i}j_{2}}{2\text{i}}\right)^{2}+\left(\frac{j_{2}}{2}\right)^{3}\int_{0}^{1}\widehat{\partial_{\xi\xi\xi}\mathfrak{a}}(j_{1},\xi+\tau\frac{j_{2}}{2})\frac{(1-\tau)^{2}}{2!}d\tau$
(2.27) $\displaystyle=:a_{1}+a_{2}+a_{3}+a_{4},$
$\displaystyle\widehat{\mathfrak{b}}\left(j_{2},\xi-\frac{j_{1}}{2}\right)$
$\displaystyle=\widehat{\mathfrak{b}}(j_{2},\xi)-\widehat{\partial_{\xi}\mathfrak{b}}(j_{2},\xi)\frac{\text{i}j_{1}}{2\text{i}}+\frac{1}{2!}\widehat{\partial_{\xi\xi}\mathfrak{b}}(j_{2},\xi)\left(\frac{\text{i}j_{1}}{2\text{i}}\right)^{2}-\left(\frac{j_{1}}{2}\right)^{3}\int_{0}^{1}\widehat{\partial_{\xi\xi\xi}\mathfrak{b}}(j_{2},\xi-\tau\frac{j_{1}}{2})\frac{(1-\tau)^{2}}{2!}d\tau$
$\displaystyle=:b_{1}+b_{2}+b_{3}+b_{4}.$

Then, we can collect the terms as

$\displaystyle\widehat{\mathfrak{a}}\left(j_{1},\xi+\frac{j_{2}}{2}\right)\widehat{\mathfrak{b}}\left(j_{2},\xi-\frac{j_{1}}{2}\right)$
$\displaystyle=(a_{1}+a_{2}+a_{3}+a_{4})(b_{1}+b_{2}+b_{3}+b_{4})$
$\displaystyle=a_{1}b_{1}+(a_{1}b_{2}+a_{2}b_{1})+(a_{1}b_{3}+a_{2}b_{2}+a_{3}b_{1})$
$\displaystyle\
+(a_{1}b_{4}+a_{2}(b_{3}+b_{4})+a_{3}(b_{2}+b_{3}+b_{4})+a_{4}(b_{1}+b_{2}+b_{3}+b_{4})).$

Summing up all $j_{1},j_{2}$, we have

$\displaystyle\sum_{j_{1}+j_{2}=k}a_{1}b_{1}$
$\displaystyle=\mathfrak{a}\mathfrak{b},$
$\displaystyle\sum_{j_{1}+j_{2}=k}(a_{1}b_{2}+a_{2}b_{1})$
$\displaystyle=-\frac{1}{2\text{i}}\left(\partial_{x}\mathfrak{a}\partial_{\xi}\mathfrak{b}-\partial_{\xi}\mathfrak{a}\partial_{x}\mathfrak{b}\right)$
$\displaystyle\sum_{j_{1}+j_{2}=k}(a_{1}b_{3}+a_{2}b_{2}+a_{3}b_{1})$
$\displaystyle=\frac{1}{4}\left(-\frac{1}{2}\partial_{xx}\mathfrak{a}\partial_{\xi\xi}\mathfrak{b}+\partial_{x\xi}\mathfrak{a}\partial_{x\xi}\mathfrak{b}-\frac{1}{2}\partial_{\xi\xi}\mathfrak{a}\partial_{xx}\mathfrak{b}\right)=:\sigma^{1}_{\mathfrak{a},\mathfrak{b}}.$

Denoting the rest by

$\sigma^{2}_{\mathfrak{a},\mathfrak{b}}:=\sum_{k}\sum_{j_{1}+j_{2}=k}(a_{1}b_{4}+a_{2}(b_{3}+b_{4})+a_{3}(b_{2}+b_{3}+b_{4})+a_{4}(b_{1}+b_{2}+b_{3}+b_{4})),$

we obtain the expression for $\mathfrak{c}$ in (2.24). Recalling the
definition of $\mathcal{S}^{m}$ in (2.10) and noting that
$\mathfrak{a},\mathfrak{b}$ are smooth in the variable $x$, it follows
straightforwardly that
$\sigma^{1}_{\mathfrak{a},\mathfrak{b}}\in\mathcal{S}^{m+m^{\prime}-2},\
\sigma^{2}_{\mathfrak{a},\mathfrak{b}}\in\mathcal{S}^{m+m^{\prime}-3}$ (we do
not prove quantitative estimates for
$\sigma^{1}_{\mathfrak{a},\mathfrak{b}},\sigma^{2}_{\mathfrak{a},\mathfrak{b}}$
here, but we postpone them to Lemma 2.14). ∎

######  Remark 2.8.

In case $\mathfrak{a}$ or $\mathfrak{b}$ is independent of $x$, we have
cleaner expression for the composition. Indeed, if $\mathfrak{a}(\xi)$ is
independent of $x$, then it immediately follows from (2.23) that

$\displaystyle Op^{W}(\mathfrak{a})\circ
Op^{W}(\mathfrak{b})=\mathfrak{a}\left(\xi+\frac{k}{2}\right)\widehat{\mathfrak{b}}(k,\xi),\quad
Op^{W}(\mathfrak{b})\circ
Op^{W}(\mathfrak{a})=\widehat{\mathfrak{b}}(k,\xi)\mathfrak{a}\left(\xi-\frac{k}{2}\right).$
(2.28)

This immediately implies that

$\displaystyle\partial_{x}Op^{W}(\mathfrak{a})=Op^{W}\left(\frac{1}{2}\partial_{x}\mathfrak{a}+i\xi\mathfrak{a}\right),\quad
Op^{W}(\mathfrak{a})\partial_{x}=Op^{W}\left(-\frac{1}{2}\partial_{x}\mathfrak{a}+\text{i}\xi\mathfrak{a}\right).$
(2.29)

#####  2.2.1.3 Symmetric commutator

The usual commutator of two linear operators is defined as

$\displaystyle[A,B]:=AB-BA.$ (2.30)

We also consider a slight variation:

$\displaystyle[A,B]_{x}:=A\partial_{x}B-B\partial_{x}A.$ (2.31)

Note that if $A,B$ are symmetric operators, then $[A,B]_{x}$ is also a
symmetric operator, since $\partial_{x}$ is asymmetric.

######  Lemma 2.9.

Let $A=Op^{W}(\mathfrak{a})$, $B=Op^{W}(\mathfrak{b})$ for some
$\mathfrak{a}\in\mathcal{S}^{m}$, $\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$.
Then $[A,B]_{x}\in OPS^{m+m^{\prime}}$. That is, there exists a symbol
$\mathfrak{a}\star\mathfrak{b}\in\mathcal{S}^{m+m^{\prime}}$ such that

$\displaystyle[A,B]_{x}=Op^{W}(\mathfrak{a}\star\mathfrak{b}).$ (2.32)

The symbol $\mathfrak{a}\star\mathfrak{b}$ admits an expansion

$\displaystyle(\mathfrak{a}\star\mathfrak{b})(x,\xi)$
$\displaystyle=\left\\{\mathfrak{a},\mathfrak{b}\right\\}_{x}+r_{\mathfrak{a},\mathfrak{b}},$
(2.33) $\displaystyle\left\\{\mathfrak{a},\mathfrak{b}\right\\}_{x}$
$\displaystyle:=\mathfrak{a}\partial_{x}\mathfrak{b}-\mathfrak{b}\partial_{x}\mathfrak{a}+\xi\left\\{\mathfrak{a},\mathfrak{b}\right\\}=\partial_{\xi}(\xi\mathfrak{a})\partial_{x}\mathfrak{b}-\partial_{x}\mathfrak{a}\partial_{\xi}(\xi\mathfrak{b})\in\mathcal{S}^{m+m^{\prime}},$
$\displaystyle r_{\mathfrak{a},\mathfrak{b}}$
$\displaystyle:=\left(\frac{1}{2}(\sigma^{1}_{\mathfrak{a},\partial_{x}\mathfrak{b}}+\sigma^{2}_{\mathfrak{a},\partial_{x}\mathfrak{b}})+\text{i}\sigma^{2}_{\mathfrak{a},\xi\mathfrak{b}}\right)-\left(\frac{1}{2}(\sigma^{1}_{\mathfrak{b},\partial_{x}\mathfrak{a}}+\sigma^{2}_{\mathfrak{b},\partial_{x}\mathfrak{a}})+\text{i}\sigma^{2}_{\mathfrak{b},\xi\mathfrak{a}}\right)\in\mathcal{S}^{m+m^{\prime}-2},$

where $\sigma_{\mathfrak{a},\mathfrak{b}}^{i}$ for $i=1,2$ are as in Lemma
2.7.

###### Proof.

From Lemma 2.7 and (2.29), we have that

$\displaystyle A\partial_{x}B$ $\displaystyle=Op^{W}(\mathfrak{a})\circ
Op^{W}(\frac{1}{2}\partial_{x}\mathfrak{b}+\text{i}\xi\mathfrak{b})=\frac{1}{2}Op^{W}(\mathfrak{a})\circ
Op^{W}(\partial_{x}\mathfrak{b})+\text{i}Op^{W}(\mathfrak{a})\circ
Op^{W}(\mathfrak{\xi\mathfrak{b}})$
$\displaystyle=\frac{1}{2}Op^{W}\left(\mathfrak{a}\partial_{x}\mathfrak{b}+\frac{1}{2\text{i}}\left\\{\mathfrak{a},\partial_{x}\mathfrak{b}\right\\}+(\sigma^{1}_{\mathfrak{a},\partial_{x}\mathfrak{b}}+\sigma^{2}_{\mathfrak{a},\partial_{x}\mathfrak{b}})\right)+\text{i}Op^{W}\left(\left(\mathfrak{a}\xi\mathfrak{b}+\frac{1}{2\text{i}}\left\\{\mathfrak{a},\xi\mathfrak{b}\right\\}+\sigma^{1}_{\mathfrak{a},\xi\mathfrak{b}}\right)+\sigma^{2}_{\mathfrak{a},\xi\mathfrak{b}}\right)$
$\displaystyle=Op^{W}\left(\frac{1}{2}\mathfrak{a}\partial_{x}\mathfrak{b}+\frac{1}{4\text{i}}\left\\{\mathfrak{a},\partial_{x}\mathfrak{b}\right\\}+\text{i}\mathfrak{a}\xi\mathfrak{b}+\frac{1}{2}\left\\{\mathfrak{a},\xi\mathfrak{b}\right\\}+\text{i}\sigma_{\mathfrak{a},\xi\mathfrak{b}}^{1}\right)+Op^{W}\left(\frac{1}{2}(\sigma^{1}_{\mathfrak{a},\partial_{x}\mathfrak{b}}+\sigma^{2}_{\mathfrak{a},\partial_{x}\mathfrak{b}})+\text{i}\sigma^{2}_{\mathfrak{a},\xi\mathfrak{b}}\right).$

Similarly, we have

$B\partial_{x}A=Op^{W}\left(\frac{1}{2}\mathfrak{b}\partial_{x}\mathfrak{a}+\frac{1}{4\text{i}}\left\\{\mathfrak{b},\partial_{x}\mathfrak{a}\right\\}+\text{i}\mathfrak{b}\xi\mathfrak{a}+\frac{1}{2}\left\\{\mathfrak{b},\xi\mathfrak{a}\right\\}+\text{i}\sigma_{\mathfrak{b},\xi\mathfrak{a}}^{1}\right)+Op^{W}\left(\frac{1}{2}(\sigma^{1}_{\mathfrak{b},\partial_{x}\mathfrak{a}}+\sigma^{2}_{\mathfrak{b},\partial_{x}\mathfrak{a}})+\text{i}\sigma^{2}_{\mathfrak{b},\xi\mathfrak{a}}\right)$

Using the definitions of $\left\\{\mathfrak{a},\mathfrak{b}\right\\}$ and
$\sigma_{\mathfrak{a},\mathfrak{b}}^{1}$ in Lemma 2.7, we have

$\displaystyle A\partial_{x}B-B\partial_{x}A$
$\displaystyle=Op^{W}\left(\left\\{\mathfrak{a},\mathfrak{b}\right\\}_{x}\right)+Op^{W}\left(\left(\frac{1}{2}(\sigma^{1}_{\mathfrak{a},\partial_{x}\mathfrak{b}}+\sigma^{2}_{\mathfrak{a},\partial_{x}\mathfrak{b}})+\text{i}\sigma^{2}_{\mathfrak{a},\xi\mathfrak{b}}\right)-\left(\frac{1}{2}(\sigma^{1}_{\mathfrak{b},\partial_{x}\mathfrak{a}}+\sigma^{2}_{\mathfrak{b},\partial_{x}\mathfrak{a}})+\text{i}\sigma^{2}_{\mathfrak{b},\xi\mathfrak{a}}\right)\right),$

which gives us (2.33). ∎

#####  2.2.1.4 Norms of symbols

As in Subsection 2.1, we will consider symbols that depend on
$\omega\in\Omega\subset\mathbb{R}^{\nu}$ and $\varphi\in\mathbb{T}^{\nu}$,
that is,
$\mathfrak{a}(\omega)=\mathfrak{a}(\omega,\varphi,x,\xi)\in\mathcal{S}^{m}$.
We define the norms:

$\displaystyle|\mathfrak{a}(\omega)|_{m,s,\eta_{0}}$
$\displaystyle:=\max_{0\leq\eta\leq\eta_{0}}\sup_{\xi\in\mathbb{R}}\rVert\partial_{\xi}^{\eta}\mathfrak{a}(\omega,\cdot,\cdot,\xi)\rVert_{H^{s}_{\varphi,x}}\langle\xi\rangle^{\eta-m},$
(2.34) $\displaystyle|\mathfrak{a}|_{m,s,\eta_{0}}^{\sup}$
$\displaystyle:=\sup_{\omega\in\Omega}|\mathfrak{a}(\omega)|_{m,s,\eta_{0}},$
$\displaystyle|\mathfrak{a}|_{m,s,\eta_{0}}^{\text{lip}}$
$\displaystyle:=\sup_{\omega_{1},\omega_{2}\in\Omega}\frac{|\mathfrak{a}(\omega_{1})-\mathfrak{a}(\omega_{2})|_{m,s,\eta_{0}}}{|\omega_{1}-\omega_{2}|},$
$\displaystyle|\mathfrak{a}|_{m,s,\eta_{0}}^{\text{Lip}(\gamma,\Omega)}$
$\displaystyle:=|\mathfrak{a}|_{m,s,\eta_{0}}^{\sup}+\gamma|\mathfrak{a}|_{m,s,\eta_{0}}^{\text{lip}},\text{
for some $\gamma\in(0,1)$.}$

The following lemma can be easily proved by following the proof of Lemma 2.6,
word by word. For simplicity, we omit the proof:

######  Lemma 2.10.

We have that:

  1. (1)

(Equivalence of the quantizations) For
$\mathfrak{a},\mathfrak{b}\in\mathcal{S}^{m}$ such that
$Op^{W}(\mathfrak{a})=Op(\mathfrak{b})$, there exists $\mu=\mu(m,\eta)$ such
that for all $s\geq s_{0}$ and $\eta\geq 0$,

$\displaystyle|\mathfrak{b}|_{m,s,\eta}^{\text{Lip}(\gamma,\Omega)}\leq_{\eta,s}|\mathfrak{a}|_{m,{s+\mu},\eta}^{\text{Lip}(\gamma,\Omega)},\quad|\mathfrak{a}|_{m,s,\eta}^{\text{Lip}(\gamma,\Omega)}\leq_{\eta,s}|\mathfrak{b}|_{m,{s+\mu},\eta}^{\text{Lip}(\gamma,\Omega)}.$

  2. (2)

(Action as a linear operator) Let $\mathfrak{a}\in\mathcal{S}^{m}$. There
exists $\mu=\mu(m)$ such that for all $s\geq s_{0}$,

$\displaystyle\rVert
Op^{W}(\mathfrak{a})h\rVert^{\text{Lip}(\gamma,\Omega)}_{s}\leq_{s}|\mathfrak{a}|_{m,{s+\mu},0}^{\text{Lip}(\gamma,\Omega)}\rVert
h\rVert^{\text{Lip}(\gamma,\Omega)}_{s_{0}+m}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,{s_{0}+\mu},0}\rVert
h\rVert^{\text{Lip}(\gamma,\Omega)}_{{s+m}}.$ (2.35)

In what follows, we collect useful properties of the norms for the symbols
depending on $\omega,\varphi$. We denote the Fourier modes of a symbol
$\mathfrak{a}$ in $\varphi,x$ by

$\displaystyle\widehat{\mathfrak{a}}(\omega,\varphi,j,\xi)$
$\displaystyle:=\frac{1}{2\pi}\int_{\mathbb{T}}\mathfrak{a}(\omega,\varphi,x,\xi)e^{-\text{i}jx}dx,\quad\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l,j,\xi)$
$\displaystyle:=\frac{1}{(2\pi)^{\nu+1}}\int_{\mathbb{T}^{\nu}\times\mathbb{T}}\mathfrak{a}(\omega,\varphi,x,\xi)e^{-\text{i}(l\cdot\varphi+jx)}d\varphi
dx.$ (2.36)

We first prove the estimates for a product of two symbols:

######  Lemma 2.11.

Given $\mathfrak{a}\in\mathcal{S}^{m}$,
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$ for some
$m,m^{\prime}\in\mathbb{R}$, it holds that

$\displaystyle|\mathfrak{a}\mathfrak{b}|_{m+m^{\prime},s,\eta}^{\text{Lip}(\gamma,\Omega)}$
$\displaystyle\leq_{m,m^{\prime},s,\eta}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s,\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0},\eta}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0},\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m,s,\eta}\text{
for $\gamma\in(0,1)$.}$ (2.37)

###### Proof.

We first prove

$\displaystyle|\mathfrak{a}\mathfrak{b}|_{m+m^{\prime},s,\eta_{0}}^{\sup}$
$\displaystyle\leq_{m,m^{\prime},s,\eta}|\mathfrak{a}|^{\sup}_{m,s,\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s_{0},\eta}+|\mathfrak{a}|^{\sup}_{m,s_{0},\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s,\eta}.$
(2.38)

Let $\omega$ be fixed. For simplicity, we omit the dependence on
$\varphi,x,\xi$ in the notation. For a fixed $0\leq\eta\in\mathbb{N}_{0}$, it
is clear that

$\left(\partial_{\xi}\right)^{\eta}(\mathfrak{a}(\omega)\mathfrak{b}(\omega))=\sum_{n=0}^{\eta}\binom{\eta}{n}((\partial_{\xi})^{\eta-n}\mathfrak{a}(\omega))((\partial_{\xi})^{n}\mathfrak{b}(\omega)).$

Therefore using (2.6) and splitting
$\langle\xi\rangle^{\eta-(m+m^{\prime})}\leq_{\eta,m,m^{\prime}}\langle\xi\rangle^{\eta-
n-m}\langle\xi\rangle^{n-m^{\prime}}$, we get

$\rVert\partial_{\xi}^{\eta}(\mathfrak{a}\mathfrak{b})(\omega,\cdot,\cdot,\xi)\rVert_{H^{s}_{\varphi,x}}\langle\xi\rangle^{\eta-(m+m^{\prime})}\leq_{\eta,s,m,m^{\prime}}|\mathfrak{a}(\omega)|_{m,s,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s_{0},\eta}+|\mathfrak{a}(\omega)|_{m,s_{0},\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s,\eta}.$

Therefore,

$\displaystyle|\mathfrak{a}(\omega)\mathfrak{b}(\omega)|_{m+m^{\prime},s,\eta}\leq_{\eta,s,m,m^{\prime}}|\mathfrak{a}(\omega)|_{m,s,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s_{0},\eta}+|\mathfrak{a}(\omega)|_{m,s_{0},\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s,\eta},$
(2.39)

This immediately implies (2.38) by taking the supremum in $\omega$. For the
Lipschitz dependence in (2.37), we have (denoting
$\Delta_{12}\mathfrak{a}:=\mathfrak{a}(\omega_{1})-\mathfrak{a}(\omega_{2})$),
that
$\mathfrak{a}(\omega_{1})\mathfrak{b}(\omega_{1})-\mathfrak{a}(\omega_{2})\mathfrak{b}(\omega_{2})=(\Delta_{12}\mathfrak{a})\mathfrak{b}(\omega_{1})+\mathfrak{a}(\omega_{2})\Delta_{12}\mathfrak{b}.$
Hence,

$\frac{\mathfrak{a}(\omega_{1})\mathfrak{b}(\omega_{1})-\mathfrak{a}(\omega_{2})\mathfrak{b}(\omega_{2})}{|\omega_{1}-\omega_{2}|}=\frac{\Delta_{12}\mathfrak{a}}{|\omega_{1}-\omega_{2}|}\mathfrak{b}(\omega_{1})+\mathfrak{a}_{2}(\omega_{2})\frac{\Delta_{12}\mathfrak{b}}{|\omega_{1}-\omega_{2}|}.$

Again using (2.39), we have

$\displaystyle\frac{\left|\mathfrak{a}(\omega_{1})\mathfrak{b}(\omega_{1})-\mathfrak{a}(\omega_{2})\mathfrak{b}(\omega_{2})\right|_{m+m^{\prime},s,\eta}}{|\omega_{1}-\omega_{2}|}$
$\displaystyle\quad\leq_{\eta,s,m,m^{\prime}}|\mathfrak{a}|^{\text{lip}}_{m,s,\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s_{0},\eta}+|\mathfrak{a}|^{\text{lip}}_{m,s_{0},\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s,\eta}+|\mathfrak{b}|^{\text{lip}}_{m^{\prime},s,\eta}|\mathfrak{a}|^{\sup}_{m,s_{0},\eta}+|\mathfrak{b}|^{\text{lip}}_{m^{\prime},s_{0},\eta}|\mathfrak{a}|^{\sup}_{m,s,\eta}$
$\displaystyle\quad\leq(|\mathfrak{a}|^{\text{lip}}_{m,s,\eta}+|\mathfrak{a}|^{\sup}_{m,s,\eta})(|\mathfrak{b}|^{\text{lip}}_{m^{\prime},s_{0},\eta}+|\mathfrak{b}|^{\sup}_{m^{\prime},s_{0},\eta})+(|\mathfrak{a}|^{\text{lip}}_{m,s_{0},\eta}+|\mathfrak{a}|^{\sup}_{m,s_{0},\eta})(|\mathfrak{b}|^{\text{lip}}_{m^{\prime},s,\eta}+|\mathfrak{b}|^{\sup}_{m^{\prime},s,\eta}).$

Then, the result follows by multiplying by $\gamma$ on both sides and taking
the supremum in $\omega_{1},\omega_{2}$. ∎

######  Remark 2.12.

In case $\mathfrak{a}=\mathfrak{a}(\omega,\xi)\in\mathcal{S}^{m}$ and
$\mathfrak{b}=\mathfrak{b}(\omega,\xi)\in\mathcal{S}^{m^{\prime}}$ are Fourier
multipliers, that is, they do not depend on $x,\varphi$, then (2.28) shows
that

$Op^{W}(\mathfrak{a})\circ
Op^{W}(\mathfrak{b})=Op^{W}(\mathfrak{a}\mathfrak{b}).$

Using the norms in (2.11) and (2.7), it follows straightforwardly from Lemma
2.11 that

$|\mathfrak{a}\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m,H^{0}_{x},\eta}\leq_{\eta}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,H^{0}_{x},\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m,H^{0}_{x},\eta}.$

Now, we estimate symbols obtained from a composition of two symbols. In view
of Lemma 2.7, we prove the following lemma:

######  Lemma 2.13.

Given $\omega,\varphi$ dependent symbols $\mathfrak{a}\in\mathcal{S}^{m}$,
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$ for some
$m,m^{\prime}\in\mathbb{R}$, we define $\mathfrak{c}_{\tau_{1},\tau_{2}}$ be
such that

$\widehat{\mathfrak{c}_{\tau_{1},\tau_{2}}}(\omega,\varphi,k,\xi):=\sum_{j_{1}+j_{2}=k}\widehat{\mathfrak{a}}(\omega,\varphi,j_{1},\xi+\tau_{1}j_{2})\widehat{\mathfrak{b}}(\omega,\varphi,j_{2},\xi+\tau_{2}j_{1}).$

Then, for each $\eta\in\mathbb{N}_{0}$, there exists
$\mu=\mu(m,m^{\prime},\eta)$ such that

$\displaystyle\sup_{|\tau_{1}|,|\tau_{2}|\leq
1}|\mathfrak{c}_{\tau_{1},\tau_{2}}|^{\sup}_{m+m^{\prime},s,\eta}\leq_{m,m^{\prime},s,\eta}|\mathfrak{a}|^{\sup}_{m,s+\mu,\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s_{0}+\mu,\eta}+|\mathfrak{a}|^{\sup}_{m,s_{0}+\mu,\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s+\mu,\eta}.$
(2.40)

And,

$\displaystyle\sup_{|\tau_{1}|,|\tau_{2}|\leq 1}$
$\displaystyle|\mathfrak{c}_{\tau_{1},\tau_{2}}|^{\text{Lip}(\gamma,\Omega)}_{m+m^{\prime},s,\eta}\leq_{\nu,m,m^{\prime},s,\eta}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s+\mu,\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0}+\mu,\eta}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0}+\mu,\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s+\mu,\eta}\text{
for $|\gamma|\leq 1$}.$ (2.41)

###### Proof.

Using the notation in (2.36) and recalling the norm $|\cdot|_{m,s,\eta}$ in
(2.34), it is clear that for each fixed $\omega\in\Omega$,

$\sup_{\xi\in\mathbb{R}}\sum_{j\in\mathbb{Z},\ l\in\mathbb{Z}^{\nu}}(\langle
j\rangle+\langle
l\rangle)^{2s}|\partial_{\xi}^{\eta}\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l,j,\xi)|^{2}\langle\xi\rangle^{2(\eta-m)}\leq_{m,\eta,s}|\mathfrak{a}(\omega)|^{2}_{m,s,\eta}.$

Therefore for each fixed $(l,j,\xi)$,

$\displaystyle|\partial_{\xi}^{\eta}\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l,j,\xi)|\leq_{m,s,\eta}|\mathfrak{a}(\omega)|_{m,s,\eta}\frac{\langle\xi\rangle^{m-\eta}}{(\langle
j\rangle+\langle l\rangle)^{s}},\text{ for each $s\geq s_{0}$}.$ (2.42)

Similarly, we have

$\displaystyle|\partial_{\xi}^{\eta}\widehat{\mathfrak{b}}^{\varphi,x}(\omega,l,j,\xi)|\leq_{m^{\prime},s,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s,\eta_{0}}\frac{\langle\xi\rangle^{m^{\prime}-\eta}}{(\langle
j\rangle+\langle l\rangle)^{s}},\text{ for each $s\geq s_{0}$}.$ (2.43)

Then we can write each Fourier mode of
$\partial_{\xi}^{\eta}\mathfrak{c}_{\tau_{1},\tau_{2}}$ as

$\displaystyle\partial_{\xi}^{\eta}\widehat{\mathfrak{c}_{\tau_{1},\tau_{2}}}^{\varphi,x}(\omega,l,k,\xi)$
$\displaystyle=\sum_{n=0}^{\eta}\binom{\eta}{n}\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\
l_{1}+l_{2}=l\in\mathbb{Z}^{\nu}\end{subarray}}\partial_{\xi}^{n}\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l_{1},j_{1},\xi+\tau_{1}j_{2})\partial_{\xi}^{\eta-n}\widehat{b}^{\varphi,x}(\omega,l_{2},j_{2},\xi+\tau_{2}j_{1}).$

Using that $(\langle k\rangle+\langle l\rangle)^{2s}\leq_{s}(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{2s}+(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{2s}$ for $j_{1}+j_{2}=k,\ l_{1}+l_{2}=l$, we have

$\displaystyle(\langle k\rangle$ $\displaystyle+\langle
l\rangle)^{2s}|\partial_{\xi}^{\eta}\widehat{\mathfrak{c}_{\tau_{1},\tau_{2}}}^{\varphi,x}(\omega,l,k,\xi)|^{2}$
(2.44)
$\displaystyle\leq_{s,\eta,n}\sum_{n=0}^{\eta}\left(\left(\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\
l_{1}+l_{2}=l\end{subarray}}(\langle j_{1}\rangle+\langle
l_{1}\rangle)^{s}|(\partial_{\xi})^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l_{1},j_{1},\xi+\tau_{1}j_{2})||(\partial_{\xi})^{n}\widehat{\mathfrak{b}}^{\varphi,x}(\omega,l_{2},j_{2},\xi+\tau_{2}j_{1})|\right)^{2}\right.$
$\displaystyle\left.\ +\left(\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\
l_{1}+l_{2}=l\end{subarray}}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{s}|(\partial_{\xi})^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l_{1},j_{1},\xi+\tau_{1}j_{2})||(\partial_{\xi})^{n}\widehat{\mathfrak{b}}^{\varphi,x}(\omega,l_{2},j_{2},\xi+\tau_{2}j_{1})|\right)^{2}\right).$

Using (2.42) and (2.43), we have that for some $\mu=\mu(m,m^{\prime},\eta)\geq
0$, which will be chosen later,

$\displaystyle|(\partial_{\xi})^{\eta-n}$
$\displaystyle\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l_{1},j_{1},\xi+\tau_{1}j_{2})||(\partial_{\xi})^{n}\widehat{\mathfrak{b}}^{\varphi,x}(\omega,l_{2},j_{2},\xi+\tau_{2}j_{1})|$
$\displaystyle\leq_{m,m^{\prime},\eta,s}|\mathfrak{a}(\omega)|_{m,s+\mu,\eta}\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}}{(\langle
j_{1}\rangle+\langle
l_{1}\rangle)^{s+\mu}}|\mathfrak{b}(\omega)|_{m^{\prime},s_{0}+\mu,\eta_{0}}\frac{\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{2}\rangle+\langle l_{2}\rangle)^{s_{0}+\mu}}$
$\displaystyle\leq|\mathfrak{a}(\omega)|_{m,s+\mu,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s_{0}+\mu,\eta}\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{s+\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{s_{0}+\mu}}.$ (2.45)

Similarly, the same quantity can be estimated as

$\displaystyle|(\partial_{\xi})^{\eta-n}$
$\displaystyle\widehat{\mathfrak{a}}^{\varphi,x}(\omega,l_{1},j_{1},\xi+\tau_{1}j_{2})||(\partial_{\xi})^{n}\widehat{\mathfrak{b}}^{\varphi,x}(\omega,l_{2},j_{2},\xi+\tau_{2}j_{1})|$
$\displaystyle\leq_{m,m^{\prime},\eta,s}|\mathfrak{a}(\omega)|_{m,s_{0}+\mu,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s+\mu,\eta}\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{s_{0}+\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{s+\mu}}.$ (2.46)

Plugging these two estimates into (2.44) (Plugging (2.45) into the first
summation and plugging (2.2.1.4) into the second summation into (2.44)), we
obtain

$\displaystyle(\langle k\rangle$ $\displaystyle+\langle
l\rangle)^{2s}|(\partial_{\xi})^{\eta}\widehat{\mathfrak{c}_{\tau_{1},\tau_{2}}}^{\varphi,x}(\omega,l,k,\xi)|^{2}$
$\displaystyle\
\leq_{m,m^{\prime},s,\eta}\left(\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\
l_{1}+l_{2}=l\end{subarray}}|\mathfrak{a}(\omega)|_{m,s+\mu,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s_{0}+\mu,\eta}\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{s_{0}+\mu}}\right)^{2}$ $\displaystyle\
+\left(\sum_{\begin{subarray}{c}j_{1}+j_{2}=k,\\\
l_{1}+l_{2}=l\end{subarray}}|\mathfrak{a}(\omega)|_{m,s_{0}+\mu,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s+\mu,\eta}\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{s_{0}+\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{\mu}}\right)^{2}$

Using the Cauchy-Schwartz inequality, we get (using that
$\sum_{j,l}\left(\frac{1}{\langle j\rangle+\langle
l\rangle}\right)^{s_{0}}<_{s_{0}}1$, thanks to (2.5)),

$\displaystyle\sum_{k\in\mathbb{Z},\ l\in\mathbb{Z}^{\nu}}(\langle k\rangle$
$\displaystyle+\langle
l\rangle)^{2s}|(\partial_{\xi})^{\eta}\widehat{\mathfrak{c}_{\tau_{1},\tau_{2}}}^{\varphi,x}(\omega,l,k,\xi)|^{2}$
(2.47)
$\displaystyle\leq_{m,m^{\prime},\eta,s}(|\mathfrak{a}(\omega)|_{m,s+\mu,\eta_{0}}|\mathfrak{b}(\omega)|_{m^{\prime},s_{0}+\mu,\eta})^{2}\sum_{j_{1},j_{2}\in\mathbb{Z},\
l_{1},l_{2}\in\mathbb{Z}^{\nu}}\left(\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{\mu}}\right)^{2}$ $\displaystyle\
+\left(|\mathfrak{a}(\omega)|_{m,s_{0}+\mu,\eta}|\mathfrak{b}(\omega)|_{m^{\prime},s+\mu,\eta}\right)^{2}\sum_{j_{1},j_{2}\in\mathbb{Z},\
l_{1},l_{2}\in\mathbb{Z}^{\nu}}\left(\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{\mu}}\right)^{2}.$

Now, we estimate the summations in the above estimate. Since
$|\tau_{1}|,|\tau_{2}|\leq 1$, we have that

$\displaystyle\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}\leq_{m,m^{\prime},\eta_{0}}\langle\xi\rangle^{m+m^{\prime}-\eta},\text{
for $|j_{1}|,|j_{2}|\leq\frac{|\xi|}{2}$}.$ (2.48)

If $|j_{1}|>\frac{|\xi|}{2}$ or $|j_{2}|>\frac{|\xi|}{2}$, then one can choose
large enough $\mu$ depending only on $m,m^{\prime},\eta$ such that

$\displaystyle\sum_{\begin{subarray}{c}j_{1},j_{2}\in\mathbb{Z},\\\
\text{$|j_{1}|\geq\frac{|\xi|}{2}$ or $|j_{2}|\geq\frac{|\xi|}{2}$},\\\
l_{1},l_{2}\in\mathbb{Z}^{\nu}\end{subarray}}\left(\frac{\langle\xi+\tau_{1}j_{2}\rangle^{m-(\eta-n)}\langle\xi+\tau_{2}j_{1}\rangle^{m^{\prime}-n}}{(\langle
j_{1}\rangle+\langle l_{1}\rangle)^{\mu}(\langle j_{2}\rangle+\langle
l_{2}\rangle)^{\mu}}\right)^{2}\leq_{m,m^{\prime},\eta}\langle\xi\rangle^{2(m+m^{\prime}-\eta)}.$

Combining this with (2.48) and taking the supremum in $\omega$, the estimate
in (2.47) yields (2.40), Now we turn to (2.41). it is clear that (denoting
$\Delta_{12}\mathfrak{a}:=\mathfrak{a}(\omega_{1})-\mathfrak{a}(\omega_{2})$),

$\frac{\Delta_{12}\mathfrak{c}_{\tau_{1},\tau_{2}}}{|\omega_{1}-\omega_{2}|}=\frac{\Delta_{12}\mathfrak{a}}{|\omega_{1}-\omega_{2}|}\mathfrak{b}(\omega_{1})+\mathfrak{a}(\omega_{2})\frac{\Delta_{12}\mathfrak{b}}{|\omega_{1}-\omega_{2}|}.$

Applying (2.40) to
$\frac{\Delta_{12}\mathfrak{a}}{|\omega_{1}-\omega_{2}|}\mathfrak{b}(\omega_{1})$
and
$\mathfrak{a}(\omega_{2})\frac{\Delta_{12}\mathfrak{b}}{|\omega_{1}-\omega_{2}|}$,
we obtain that for each $\omega_{1},\omega_{2}$,

$\displaystyle|$
$\displaystyle\frac{\Delta_{12}\mathfrak{c}_{\tau_{1},\tau_{2}}}{|\omega_{1}-\omega_{2}|}|_{m+m^{\prime},s,\eta_{0}}$
(2.49)
$\displaystyle\quad\leq_{\nu,m,m^{\prime},s,\eta}|\mathfrak{a}|^{\text{lip}}_{m,s+\mu,\eta}|\mathfrak{b}|^{\sup}_{m^{\prime},s_{0}+\mu,\eta}+|\mathfrak{a}|^{\text{lip}}_{m,s_{0}+\mu,\eta}|\mathfrak{b}|^{\sup}_{m,s+\mu,\eta}$
$\displaystyle\quad\
+|\mathfrak{a}|^{\sup}_{m,s+\mu,\eta}|\mathfrak{b}|^{\text{lip}}_{m,s_{0}+\mu,\eta}+|\mathfrak{a}|^{\sup}_{m,s_{0}+\mu,\eta}|\mathfrak{b}|^{\text{lip}}_{m,s+\mu,\eta}$
$\displaystyle\quad\leq(|\mathfrak{a}|^{\text{lip}}_{m,s+\mu,\eta}+|\mathfrak{a}|^{\sup}_{m,s+\mu,\eta})(|\mathfrak{b}|^{\text{lip}}_{m^{\prime},s_{0}+\mu,\eta}+|\mathfrak{b}|^{\sup}_{m^{\prime},s_{0}+\mu,\eta})$
$\displaystyle\
+(|\mathfrak{a}|^{\text{lip}}_{m,s_{0}+\mu,\eta}+|\mathfrak{a}|^{\sup}_{m,s_{0}+\mu,\eta})(|\mathfrak{b}|^{\text{lip}}_{m^{\prime},s+\mu,\eta}+|\mathfrak{b}|^{\sup}_{m^{\prime},s+\mu,\eta}).$

Noting that $|\gamma|<1$ and taking the supremum in $\omega_{1},\omega_{2}$,
we get

$\gamma|\mathfrak{c}_{\tau_{1},\tau_{2}}|^{\text{lip}}_{m+m^{\prime},s,\eta}\leq_{m,m^{\prime},\eta}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s+\mu,\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0}+\mu,\eta}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0}+\mu,\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s+\mu,\eta}.$

Combining this with (2.40) and recalling the definition of the norm in (2.7),
the desired estimate (2.41) follows. ∎

######  Lemma 2.14.

Given $\mathfrak{a}\in\mathcal{S}^{m}$,
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$ for some
$m,m^{\prime}\in\mathbb{R}$, let
$\sigma^{1}_{\mathfrak{a},\mathfrak{b}},\sigma^{2}_{\mathfrak{a},\mathfrak{b}}$
be the symbols given in (2.24). Then, we have
$\sigma_{\mathfrak{a},\mathfrak{b}}^{1}\in\mathcal{S}^{m+m^{\prime}-2},\sigma^{2}_{\mathfrak{a},\mathfrak{b}}\in\mathcal{S}^{m+m^{\prime}-3}$.
Furthermore, for each $\eta\in\mathbb{N}_{0}$, there exist
$\mu=\mu({m,m^{\prime},\eta})$ and an absolute constant $N>0$, such that

$\displaystyle|\sigma^{1}_{\mathfrak{a},\mathfrak{b}}|_{m+m^{\prime}-2,s,\eta}^{\text{Lip}(\gamma,\Omega)}$
$\displaystyle\leq_{m,m^{\prime},s,\eta}|\mathfrak{a}|_{m,s_{0}+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}|\mathfrak{b}|_{m^{\prime},s+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}+|\mathfrak{a}|_{m,s+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}|\mathfrak{b}|_{m^{\prime},s_{0}+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)},$
$\displaystyle|\sigma^{2}_{\mathfrak{a},\mathfrak{b}}|_{m+m^{\prime}-3,s,\eta}^{\text{Lip}(\gamma,\Omega)}$
$\displaystyle\leq_{m,m^{\prime},s,\eta}|\mathfrak{a}|_{m,s_{0}+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}|\mathfrak{b}|_{m^{\prime},s+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}+|\mathfrak{a}|_{m,s+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}|\mathfrak{b}|_{m^{\prime},s_{0}+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}.$

Consequently, $r_{\mathfrak{a},\mathfrak{b}}\in\mathcal{S}^{m+m^{\prime}-2}$
in (2.33) satisfies

$|r_{\mathfrak{a},\mathfrak{b}}|_{m+m^{\prime}-2,s,\eta}^{\text{Lip}(\gamma,\Omega)}\leq_{m,m^{\prime},s,\eta}|\mathfrak{a}|_{m,s_{0}+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}|\mathfrak{b}|_{m^{\prime},s+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}+|\mathfrak{a}|_{m,s+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}|\mathfrak{b}|_{m^{\prime},s_{0}+\mu,\eta+N}^{\text{Lip}(\gamma,\Omega)}$

###### Proof.

It follows straightforwardly from expressions for
$\sigma^{1}_{\mathfrak{a},\mathfrak{b}},\sigma^{2}_{\mathfrak{a},\mathfrak{b}}$
in Lemma 2.7 and the estimates in Lemma 2.11, and Lemma 2.13. ∎

######  Lemma 2.15.

Let $\mathfrak{a}\in\mathcal{S}^{m}$ and
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$ for some
$m,m^{\prime}\in\mathbb{R}$. For each $\eta\in\mathbb{N}_{0}$, there exist
$\mu=\mu(m,m^{\prime},\eta)$ and an absolute constant $N>0$ such that the
followings hold true:

  1. (1)

For $\mathfrak{c}$ such that $Op^{W}(\mathfrak{a})\circ
Op^{W}(\mathfrak{b})=Op^{W}(\mathfrak{c})$, we have $\mathfrak{c}\in
S^{m+m^{\prime}}$ and for all $s\geq s_{0}$,

$\displaystyle|\mathfrak{c}|^{\text{Lip}(\gamma,\Omega)}_{m+m^{\prime},s,\eta}\leq_{m,m^{\prime},\eta,s}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s+\mu,\eta+N}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0}+\mu,\eta+N}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0}+\mu,\eta+N}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s+\mu,\eta+N}.$

  2. (2)

For $\mathfrak{c}$ such that
$[Op^{W}(\mathfrak{a}),Op^{W}(\mathfrak{b})]=Op^{W}(\mathfrak{c})$, we have
$\mathfrak{c}\in\mathcal{S}^{m+m^{\prime}-1}$ and for all $s\geq s_{0}$,

$\displaystyle|\mathfrak{c}|^{\text{Lip}(\gamma,\Omega)}_{m+m^{\prime}-1,s,\eta}\leq_{m,m^{\prime},\eta,s}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s+\mu,\eta+N}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0}+\mu,\eta+N}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0}+\mu,\eta+N}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s+\mu,\eta+N}.$

  3. (3)

For $\mathfrak{c}$ such that
$[Op^{W}(\mathfrak{a}),Op^{W}(\mathfrak{b})]_{x}=Op^{W}(\mathfrak{c})$, we
have $\mathfrak{c}\in\mathcal{S}^{m+m^{\prime}}$ and for all $s\geq s_{0}$,

$\displaystyle|\mathfrak{c}|^{\text{Lip}(\gamma,\Omega)}_{m+m^{\prime},s,\eta}\leq_{m,m^{\prime},\eta,s}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s+\mu,\eta+N}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0}+\mu,\eta+N}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0}+\mu,\eta+N}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s+\mu,\eta+N}.$

###### Proof.

The result follows from Lemma 2.13, Lemma 2.7, Lemma 2.11, Lemma 2.9 and Lemma
2.14. ∎

For (1) of Lemma 2.15, we will obtain a finer estimate in case that
$\mathfrak{a}$ or $\mathfrak{b}$ is supported on a finite number of Fourier
modes:

######  Lemma 2.16.

Let $\mathfrak{a}\in\mathcal{S}^{m}$ and
$\mathfrak{b}\in\mathcal{S}^{m^{\prime}}$. Assume that
$\mathfrak{b}=\mathfrak{b}(\omega,\varphi,x,\xi)$ is supported on a finite
number of Fourier modes, then we can estimate the symbol of the composition of
$Op(\mathfrak{a})$ and $Op(\mathfrak{b})$ without any loss of derivatives.
More precisely, if

$\displaystyle\widehat{\mathfrak{b}}(\omega,l,k,\xi)=0,\text{ if $|l|+|k|\geq
N$ for some $N\geq 0$.}$ (2.50)

then, for a symbol $\mathfrak{c}$ such that $Op(\mathfrak{a})\circ
Op(\mathfrak{b})=Op(\mathfrak{c})$, we have that

$\displaystyle|\mathfrak{c}|^{\text{Lip}(\gamma,\Omega)}_{m+m^{\prime},s,\eta}\leq_{m,m^{\prime},s,\eta,N}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s,\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s_{0},\eta}+|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{m,s_{0},\eta}|\mathfrak{b}|^{\text{Lip}(\gamma,\Omega)}_{m^{\prime},s,\eta},\text{
for all $s\geq s_{0}$.}$ (2.51)

###### Proof.

The proof is similar to the one of Lemma 2.13. Using (2.23), we have (omitting
$\omega$ dependence for simplicity),

$\partial_{\xi}^{\eta}\widehat{\mathfrak{c}}^{\varphi,x}(l,k,\xi)=\sum_{n=0}^{\eta}C_{n}\sum_{j_{1}+j_{2}=k,\
l_{1}+l_{2}=l}\partial_{\xi}^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(l_{1},j_{1},\xi+{j_{2}})\partial_{\xi}^{n}\widehat{\mathfrak{b}}^{\varphi,x}(l_{2},j_{2},\xi).$

Hence,

$\displaystyle|\partial_{\xi}^{\eta}\widehat{\mathfrak{c}}^{\varphi,x}(l,k,\xi)|^{2}(\langle
l\rangle+\langle k\rangle)^{2s}$
$\displaystyle\leq\sum_{n=0}^{\eta}C_{n}\left|\sum_{j_{1}+j_{2}=k,\
l_{1}+l_{2}=l}\partial_{\xi}^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(l_{1},j_{1},\xi+{j_{2}})\partial_{\xi}^{n}\widehat{\mathfrak{b}}^{\varphi,x}(l_{2},j_{2},\xi)\right|^{2}(\langle
l\rangle+\langle k\rangle)^{2s}$
$\displaystyle\leq_{s,\eta}\sum_{n=0}^{\eta}C_{n}\left|\sum_{j_{1}+j_{2}=k,\
l_{1}+l_{2}=l}(\langle l_{1}\rangle+\langle
j_{1}\rangle)^{s}|\partial_{\xi}^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(l_{1},j_{1},\xi+{j_{2}})||\partial_{\xi}^{n}\widehat{\mathfrak{b}}^{\varphi,x}(l_{2},j_{2},\xi)|\right|^{2}$
$\displaystyle\ +\sum_{n=0}^{\eta}C_{n}\left|\sum_{j_{1}+j_{2}=k,\
l_{1}+l_{2}=l}|\partial_{\xi}^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(l_{1},j_{1},\xi+{j_{2}})|(\langle
l_{2}\rangle+\langle
j_{2}\rangle)^{s}|\partial_{\xi}^{n}\widehat{\mathfrak{b}}^{\varphi,x}(l_{2},j_{2},\xi)|\right|^{2}$
$\displaystyle=A_{1}+A_{2}.$ (2.52)

For $A_{1}$, we have that

$\displaystyle\sum_{l\in\mathbb{Z}^{\nu},k\in\mathbb{Z}}A_{1}$
$\displaystyle\leq\sum_{n=0}^{\eta}C_{n}\sum_{k,l}\left|\sum_{j_{1}+j_{2}=k,\
l_{1}+l_{2}=l}(\langle l_{1}\rangle+\langle
j_{1}\rangle)^{s}|\partial_{\xi}^{\eta-n}\widehat{\mathfrak{a}}^{\varphi,x}(l_{1},j_{1},\xi+{j_{2}})|(\langle
l_{2}\rangle+\langle
j_{2}\rangle)^{s_{0}}|\partial_{\xi}^{n}\widehat{\mathfrak{b}}^{\varphi,x}(l_{2},j_{2},\xi)|(\langle
l_{2}\rangle+\langle j_{2}\rangle)^{-s_{0}}\right|^{2}$
$\displaystyle\leq_{s_{0}}\sum_{n=0}^{\eta}C_{n}\sum_{j_{1},j_{2}\in\mathbb{Z},\
l_{1},l_{2}\in\mathbb{Z}^{\nu}}\left((\langle l_{1}\rangle+\langle
j_{1}\rangle)^{s}|\widehat{\partial_{\xi}^{\eta-n}\mathfrak{a}}(l_{1},j_{1},\xi+{j_{2}})|(\langle
l_{2}\rangle+\langle
j_{2}\rangle)^{s_{0}}|\widehat{\partial_{\xi}^{n}\mathfrak{b}}(l_{2},j_{2},\xi)|\right)^{2}$
$\displaystyle\leq\sum_{n=0}^{\eta}C_{n}\sum_{l_{2}\in\mathbb{Z}^{\nu},j_{2}\in\mathbb{Z}}\rVert\partial_{\xi}^{\eta-n}\mathfrak{a}(\cdot,\cdot,\xi+j_{2})\rVert_{s}^{2}\left((\langle
l_{2}\rangle+\langle
j_{2}\rangle)^{s_{0}}|\widehat{\partial_{\xi}^{n}\mathfrak{b}}(l_{2},j_{2},\xi)|\right)^{2}$
$\displaystyle\overset{\eqref{xxxxx12323sd}}{\leq}\sum_{n=0}^{\eta}C_{n}\sum_{|l_{2}|+|j_{2}|\leq
N}|\mathfrak{a}|_{m,s,\eta}^{2}\langle\xi+j_{2}\rangle^{2(m-(\eta-n))}\left((\langle
l_{2}\rangle+\langle
j_{2}\rangle)^{s_{0}}|\widehat{\partial_{\xi}^{n}\mathfrak{b}}(l_{2},j_{2},\xi)|\right)^{2}$
$\displaystyle\leq_{N,m,\eta}\sum_{n=0}^{\eta}C_{n}|\mathfrak{a}|_{m,s,\eta}^{2}\langle\xi\rangle^{2(m-(\eta-n))}|\mathfrak{b}|_{m^{\prime},s_{0},\eta}^{2}\langle\xi\rangle^{2(m^{\prime}-n)}$
$\displaystyle\leq_{\eta}|\mathfrak{a}|^{2}_{m,s,\eta_{0}}|\mathfrak{b}|^{2}_{m^{\prime},s_{0},\eta}\langle\xi\rangle^{2(m+m^{\prime}-\eta)}.$
(2.53)

where the second inequality follows from the Cauchy-Schwarz inequality, and
the fourth and fifth inequalities follow from the definition of the norm
$|\cdot|_{m,s,\eta}$. The same computations give us that

$\sum_{l\in\mathbb{Z}^{\nu},k\in\mathbb{Z}}A_{2}\leq_{s,N,\eta}|\mathfrak{a}|^{2}_{m,s_{0},\eta}|\mathfrak{b}|^{2}_{m^{\prime},s,\eta}\langle\xi\rangle^{2(m+m^{\prime}-\eta)}.$

Plugging this and (2.53) into (2.52),

$\rVert\partial_{\xi}^{\eta}\mathfrak{c}(\cdot,\cdot,\xi)\rVert_{s}^{2}\leq_{s,N,m,m^{\prime},\eta}\left(|\mathfrak{a}|_{m,s,\eta}|\mathfrak{b}|_{m^{\prime},s_{0},\eta}+|\mathfrak{a}|_{m,s_{0},\eta}|\mathfrak{b}|_{m^{\prime},s,\eta}\right)^{2}\langle\xi\rangle^{2(m+m^{\prime}-\eta)},$

which implies

$|\mathfrak{c}|_{m+m^{\prime},s,\eta}\leq_{s,N,m,m^{\prime},\eta}|\mathfrak{a}|_{m,s,\eta}|\mathfrak{b}|_{m^{\prime},s_{0},\eta}+|\mathfrak{a}|_{m,s_{0},\eta}|\mathfrak{b}|_{m^{\prime},s,\eta}.$

The Lipschitz dependence on $\omega$ can be proved following the same argument
as in the proof of Lemma 2.13 (especially following the computations in
(2.49)). Then, (2.51) follows immediately. ∎

####  2.2.2 Lipschitz-tame operators

We recall from [16, Section 2], [52, Section 2] the notion of the Lipschitz-
tame estimates of linear operators.

######  Definition 2.17 ($\mu$-tame operators).

For $\mu\geq 0$ and $\mathtt{S}\gg s_{0}$, a linear operator
$\mathcal{A}:C^{\infty}_{\varphi,x}\mapsto C^{\infty}_{\varphi,x}$ is said to
be $\mu$-tame up to $\mathtt{S}$, if there exists a non-decreasing function
$[s_{0},\mathtt{S}]\ni s\mapsto\mathfrak{M}_{\mathcal{A}}(\mu,s)\geq 0$, such
that

$\displaystyle\rVert\mathcal{A}h\rVert_{s}\leq\mathfrak{M}_{\mathcal{A}}(\mu,s)\rVert
h\rVert_{s_{0}+\mu}+\mathfrak{M}_{\mathcal{A}}(\mu,s_{0})\rVert
h\rVert_{s+\mu},\text{ for all $h\in C^{\infty}_{\varphi,x}$.}$

$\mathtt{S}$ can be possibly chosen to be $\infty$, in which case, we simply
say that the operator $\mathcal{A}$ is $\mu$-tame.

We also consider Lipschitz dependence of the operators on the frequency
$\omega\in\Omega\subset\mathbb{R}^{\nu}$. To estimate such operators, we
recall the following definition:

######  Definition 2.18 (Lip-$\mu$-tame operators).

Let $\mathcal{A}=\mathcal{A}(\omega):C^{\infty}_{\varphi,x}\mapsto
C^{\infty}_{\varphi,x}$ be a $\omega$-dependent linear operator, for
$\omega\in\Omega\subset\mathbb{R}^{\nu}$. We denote

$\Delta_{12}\mathcal{A}:=\frac{\mathcal{A}(\omega_{1})-\mathcal{A}(\omega_{2})}{|\omega_{1}-\omega_{2}|}.$

For $\mu\geq 0$, $\mathtt{S}\gg s_{0}$ and $\gamma\in(0,1)$, the operator
$\mathcal{A}$ is said to be Lip-$\mu$-tame up to $\mathtt{S}$, if there exists
a non-decreasing function $[s_{0},\mathtt{S}]\ni
s\mapsto\mathfrak{M}^{\gamma}_{\mathcal{A}}(\mu,s)\geq 0$, such that

$\displaystyle\sup_{\omega\in\Omega}\rVert\mathcal{A}(\omega)h\rVert_{s}+\gamma\sup_{\omega_{1}\neq\omega_{2}\in\Omega}\rVert\Delta_{12}\mathcal{A}h\rVert_{s}\leq\mathfrak{M}^{\gamma}_{\mathcal{A}}(\mu,s)\rVert
h\rVert_{s_{0}+\mu}+\mathfrak{M}^{\gamma}_{\mathcal{A}}(\mu,s_{0})\rVert
h\rVert_{s+\mu},\text{ for all $h\in C^{\infty}_{\varphi,x}$.}$

$\mathtt{S}$ can be possibly chosen to be $\infty$, in which case, we simply
say that the operator $\mathcal{A}$ is Lip-$\mu$-tame.

#####  2.2.2.1 Majorant operators and modulo-tame operators

We recall the notion of majorant functions and majorant operators:

######  Definition 2.19 (Majorant functions/operators).

Given a function $u\in C^{\infty}_{\varphi,x}$, we define its majorant as

$\displaystyle\underline{u}(\varphi,x):=\sum_{l\in\mathbb{Z}^{\nu},\
j\in\mathbb{Z}}|u_{j}(l)|e^{\text{i}(l\cdot\varphi+jx)}.$ (2.54)

Given a linear operator $\mathcal{A}$, we define its majorant as

$\displaystyle\underline{\mathcal{A}}h:=\sum_{j_{1},j_{2}\in\mathbb{Z},\
l,l^{\prime}\in\mathbb{Z}^{\nu}}|\mathcal{A}_{j_{2}}^{j_{1}}(l-l^{\prime})|h_{j_{1}}(l^{\prime})e^{\text{i}(l\cdot\varphi+j_{2}x)}.$
(2.55)

Using the notion of majorant operators, we define modulo-tame operators:

######  Definition 2.20 ($\mu$-modulo tame operators).

For $\mu\geq 0$ and $\mathtt{S}\gg s_{0}$, a linear operator
$\mathcal{A}:C^{\infty}_{\varphi,x}\mapsto C^{\infty}_{\varphi,x}$ is said to
be $\mu$-modulo tame up to $\mathtt{S}$, if there exists a non-decreasing
function $[s_{0},\mathtt{S}]\ni
s\mapsto\mathfrak{M}^{\sharp}_{\mathcal{A}}(\mu,s)\geq 0$, such that

$\displaystyle\rVert\underline{\mathcal{A}}h\rVert_{s}\leq\mathfrak{M}^{\sharp}_{\mathcal{A}}(\mu,s)\rVert
h\rVert_{s_{0}+\mu}+\mathfrak{M}^{\sharp}_{\mathcal{A}}(\mu,s_{0})\rVert
h\rVert_{s+\mu},\text{ for all $h\in C^{\infty}_{\varphi,x}$.}$

$\mathtt{S}$ can be possibly chosen to be $\infty$, in which case, we simply
say that the operator $\mathcal{A}$ is $\mu$-tame.

######  Remark 2.21.

Clearly, we have that

$\displaystyle\rVert\underline{u}\rVert_{s}=\rVert u\rVert_{s},$ (2.56)

while we cannot have
$\mathfrak{M}^{\sharp}_{{\mathcal{A}}}(\mu,s)\leq\mathfrak{M}_{\mathcal{A}}(\mu,s)$
in general. In Section 10, we will estimate tame constants of
$[\partial_{\varphi}^{\vec{b}}(\mathcal{A}),\partial_{x}]$ for sufficiently
large vector $\vec{b}$, as well as $\mathcal{A}$, to bound the modulo tame
constant by tame constant (see Lemma 2.27).

######  Definition 2.22 (Lip-$\mu$-modulo tame operators).

Let $\mathcal{A}=\mathcal{A}(\omega):C^{\infty}_{\varphi,x}\mapsto
C^{\infty}_{\varphi,x}$ be a $\omega$-dependent linear operator, for
$\omega\in\Omega\subset\mathbb{R}^{\nu}$. For $\mu\geq 0$, $\mathtt{S}\gg
s_{0}$ and $\gamma\in(0,1)$, the operator $\mathcal{A}$ is said to be
Lip-$\mu$-modulo tame up to $\mathtt{S}$, if there exists a non-decreasing
function $[s_{0},\mathtt{S}]\ni
s\mapsto\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(\mu,s)\geq 0$, such that

$\displaystyle\sup_{\omega\in\Omega}\rVert\underline{\mathcal{A}(\omega)}h\rVert_{s}+\gamma\sup_{\omega_{1}\neq\omega_{2}\in\Omega}\rVert\underline{\Delta_{12}\mathcal{A}}h\rVert_{s}\leq\mathfrak{M}^{{}^{\sharp},\gamma}_{\mathcal{A}}(\mu,s)\rVert
h\rVert_{s_{0}+\mu}+\mathfrak{M}^{{}^{\sharp},\gamma}_{\mathcal{A}}(\mu,s_{0})\rVert
h\rVert_{s+\mu},\text{ for all $h\in C^{\infty}_{\varphi,x}$.}$

$\mathtt{S}$ can be possibly chosen to be $\infty$, in which case, we simply
say that the operator $\mathcal{A}$ is Lip-$\mu$-modulo tame.

#####  2.2.2.2 Properties of the tame operators

In what follows, we list useful properties of the tame operators. The proofs
can be found in [16, Section 2] and [52, Section 2].

We use the following notations: For $\varphi\in\mathbb{T}^{\nu}$, and
$\vec{b}\in\mathbb{N}_{0}^{\nu}$ and $l\in\mathbb{Z}^{\nu}$, we denote

$\partial_{\varphi}^{\vec{b}}:=\partial_{\varphi_{1}}^{\vec{b}_{1}}\ldots\partial_{\varphi_{\nu}}^{\vec{b}_{\nu}},\quad
l^{\vec{b}}:=\Pi_{i=1}^{\nu}(l_{i})^{\vec{b}_{i}},\text{ where
$l=(l_{1},\ldots,l_{\nu}),\ \vec{b}=(\vec{b}_{1},\ldots,\vec{b}_{\nu})$.}$

Given $N\in\mathbb{R}^{+}$, and a linear operator $\mathcal{A}$, let us denote
by $\Pi_{\leq N}\mathcal{A}$, the operator defined as

$\displaystyle(\Pi_{N}\mathcal{A})^{j_{1}}_{j_{2}}(l):=\begin{cases}\mathcal{A}^{j_{1}}_{j_{2}}(l),&\text{
if $|l|\leq N$},\\\ 0,&\text{ otherwise},\end{cases}$ (2.57)

where $\mathcal{A}^{j_{1}}_{j_{2}}(l)$ is as in (2.9), and we denote
$\Pi_{>N}:=I-\Pi_{\leq N}$.

######  Lemma 2.23.

[54, Lemma A.1][16, Lemma 2.25] Let $\mathcal{A},\mathcal{B}$ be
Lip-$\mu_{\mathcal{A}}$-tame and Lip-$\mu_{\mathcal{B}}$-tame operators up to
$\mathtt{S}$. Then, $\mathcal{A}\circ\mathcal{B}$ is
Lip-$\left(\mu_{A}+\mu_{B}\right)$-tame operator up to $\mathtt{S}$ with tame
constant

$\mathfrak{M}_{\mathcal{A}\circ\mathcal{B}}^{\gamma}(\mu_{A}+\mu_{B},s)\leq\mathfrak{M}_{\mathcal{A}}^{\gamma}(\mu_{A},s)\mathfrak{M}_{\mathcal{B}}^{\gamma}(\mu_{\mathcal{B}},s_{0}+\mu_{A})+\mathfrak{M}_{\mathcal{A}}^{\gamma}(\mu_{A},s_{0})\mathfrak{M}_{\mathcal{B}}^{\gamma}(\mu_{B},s_{0}+\mu_{A})$

If $\mu_{\mathcal{A}}=\mu_{\mathcal{B}}=0$, then we have that for
$\vec{b}\in\mathbb{N}_{0}^{\nu}$,

$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A}\mathcal{B})}(0,s)$
$\displaystyle\leq
C(\vec{b})\left(\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A})}(0,s)\mathfrak{M}^{\sharp,\gamma}_{\mathcal{B}}(0,s_{0})+\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A})}(0,s_{0})\mathfrak{M}^{\sharp,\gamma}_{\mathcal{B}}(0,s)\right.$
(2.58) $\displaystyle\left.\
+\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s)\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{B})}(0,s_{0})+\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s_{0})\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{B})}(0,s)\right)$

######  Lemma 2.24.

[54, Lemma A.2] Let $\mathcal{A}$ be a Lip-$\mu$-tame operator up to
$\mathtt{S}$ for some $\mu\geq 0$ and let $h=h(\omega)$ be an
$\omega$-dependent function in $C^{\infty}_{\varphi,x}$. Then,

$\displaystyle\rVert\mathcal{A}h\rVert_{s}^{\text{Lip}(\gamma,\Omega)}\leq_{s}\mathfrak{M}_{\mathcal{A}}^{\gamma}(\mu,s)\rVert
h\rVert_{s_{0}+\mu}^{\text{Lip}(\gamma,\Omega)}+\mathfrak{M}_{\mathcal{A}}^{\gamma}(\mu,s_{0})\rVert
h\rVert_{s+\mu}^{\text{Lip}(\gamma,\Omega)}.$

######  Lemma 2.25 (Pseudo differential operator).

[16, Lemma 2.21] Let
$\mathfrak{a}=\mathfrak{a}(\omega,\varphi,x,\xi)\in\mathcal{S}^{0}$ be a
symbol that depends on $\omega,\varphi$. For $s\geq s_{0}$,
$\mathcal{A}:=Op^{W}(\mathfrak{a})$ is a Lip-$0$-tame operator with tame
constant

$\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s)\leq_{s}|\mathfrak{a}|^{\text{Lip}(\gamma,\Omega)}_{0,s,0}.$

######  Lemma 2.26.

[16, Lemma 2.27, 2.28] For $\vec{\mathtt{b}}\in\mathbb{N}_{0}^{\nu}$, it holds
that

$\displaystyle\mathfrak{M}^{\sharp}_{\Pi_{>N}\mathcal{A}}(0,s)\leq
N^{-|\mathtt{b}|}\mathfrak{M}^{\sharp}_{\partial_{\varphi}^{\vec{b}}\mathcal{A}}(0,s),\quad\mathfrak{M}^{\sharp}_{\Pi_{>N}\mathcal{A}}(0,s)\leq\mathfrak{M}^{\sharp}_{\mathcal{A}}(0,s),$
(2.59)
$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\Pi_{>N}\mathcal{A}}(0,s)\leq
N^{-|\mathtt{b}|}\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}\mathcal{A}}(0,s),\quad\mathfrak{M}^{\sharp,\gamma}_{\Pi_{>N}\mathcal{A}}(0,s)\leq\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s).$
(2.60)

######  Lemma 2.27.

(Lipschitz-tame to Lipschitz-modulo-tame) Let $\mathtt{b}_{0}\in\mathbb{N}$ be
fixed. For each $\vec{\mathtt{b}}\in\mathbb{N}^{\nu}$ such that
$|\vec{\mathtt{b}}|\leq s_{0}+\mathtt{b}_{0}$, assume that
$\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})$ and
$[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]$ are
Lip-$0$-tame operators up to $\mathtt{S}$ with tame constants
$\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s),\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s)$
for $s\in[s_{0},\mathtt{S}]$. Then, for all $\vec{b}\in\mathbb{N}^{\nu}$ such
that $|\vec{b}|\leq\mathtt{b}_{0}$,
$\partial_{\varphi}^{\vec{b}}(\mathcal{A})$ is Lipschitz-0-modulo tame up to
$\mathtt{S}$ and

$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A})}(0,s)\leq_{s_{0},\mathtt{b}_{0}}\sum_{\begin{subarray}{c}\vec{\mathtt{b}}\in\mathbb{N}^{\nu},\\\
|\vec{\mathtt{b}}|\leq
s_{0}+\mathtt{b}_{0}\end{subarray}}\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)+\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s)\right).$
(2.61)

###### Proof.

To simplify the notation, let us denote for fixed
$\omega,\omega_{1},\omega_{2}\in\Omega$,

$\displaystyle\Delta_{12}^{k}\mathcal{A}:=\begin{cases}\mathcal{A}(\omega),&\text{
if $k=0$,}\\\
\gamma\frac{\mathcal{A}(\omega_{1})-\mathcal{A}(\omega_{2})}{|\omega_{1}-\omega_{2}|},&\text{
if $k=1$.}\end{cases}$ (2.62)

First, we test the operator $\mathcal{A}$ with a monomial
$h(\varphi,x):=e^{\text{i}(l^{\prime}\cdot\varphi+j^{\prime}x)}$, then it
follows from the definition of
$\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)$
that

$\displaystyle\rVert\Delta_{12}^{k}\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})h\rVert_{s}^{2}$
$\displaystyle\leq
C\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)\right)^{2}\rVert
h\rVert_{s_{0}}^{2}+\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s_{0})\right)^{2}\rVert
h\rVert_{s}^{2}\right)$
$\displaystyle=C\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)\right)^{2}\langle
l^{\prime},j^{\prime}\rangle^{2s_{0}}+\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s_{0})\right)^{2}\langle
l^{\prime},j^{\prime}\rangle^{2s}\right),$

Using the matrix representation of a linear operator in (2.9), we have

$\left(\Delta_{12}^{k}\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})\right)^{j^{\prime}}_{j}(l-l^{\prime})=\text{i}^{|\vec{\mathtt{b}}|}(l-l^{\prime})^{\vec{\mathtt{b}}}(\Delta_{12}^{k}\mathcal{A})^{j^{\prime}}_{j}(l-l^{\prime}).$

Therefore, the above inequality for
$\rVert\Delta_{12}^{k}\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})h\rVert_{s}^{2}$
gives us

$\displaystyle\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}|(l-l^{\prime})^{\vec{\mathtt{b}}}|^{2}\left|\left(\Delta_{12}^{k}\mathcal{A}\right)^{j^{\prime}}_{j}(l-l^{\prime})\right|^{2}\leq
C\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)\right)^{2}\langle
l^{\prime},j^{\prime}\rangle^{2s_{0}}+\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s_{0})\right)^{2}\langle
l^{\prime},j^{\prime}\rangle^{2s}\right).$ (2.63)

Similarly, using the matrix representation,

$([\Delta_{12}^{k}\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}])^{j^{\prime}}_{j}(l-l^{\prime})=\text{i}^{|\vec{\mathtt{b}}|+1}(l-l^{\prime})^{\vec{\mathtt{b}}}(j-j^{\prime})(\Delta_{12}^{k}\mathcal{A})^{j^{\prime}}_{j}(l-l^{\prime}),$

and using that
$[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]$ is also
Lip-$0$-tame, we have that

$\displaystyle\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}$ $\displaystyle\langle
l,j\rangle^{2s}|(l-l^{\prime})^{\vec{\mathtt{b}}}|^{2}|j-j^{\prime}|^{2}\left|\left(\Delta_{12}^{k}\mathcal{A}\right)^{j^{\prime}}_{j}(l-l^{\prime})\right|^{2}$
$\displaystyle\leq
C\left(\left(\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s)\right)^{2}\langle
l^{\prime},j^{\prime}\rangle^{2s_{0}}+\left(\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s_{0})\right)^{2}\langle
l^{\prime},j^{\prime}\rangle^{2s}\right).$ (2.64)

Now, we recall the majorant operator from (2.55) and see that for all $h\in
C^{\infty}_{\varphi,x}$,

$\displaystyle\rVert\underline{\Delta_{12}^{k}\partial_{\varphi}^{\vec{b}}(\mathcal{A}})h\rVert_{s}^{2}$
$\displaystyle=\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}\left(\sum_{(l^{\prime},j^{\prime})\in\mathbb{T}^{\nu+1}}|(\Delta_{12}^{k}\partial_{\varphi}^{\vec{b}}(\mathcal{A}))^{j^{\prime}}_{j}(l-l^{\prime})|h_{j^{\prime}}(l^{\prime})\right)^{2}$
$\displaystyle\leq\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}\left(\sum_{(l^{\prime},j^{\prime})\in\mathbb{T}^{\nu+1}}|(\Delta_{12}^{k}\mathcal{A})^{j^{\prime}}_{j}(l-l^{\prime})||(l-l^{\prime})^{\vec{b}}|\langle
l-l^{\prime}\rangle^{s_{0}}\langle
j-j^{\prime}\rangle|h_{j^{\prime}}(l^{\prime})|\frac{1}{\langle
l-l^{\prime}\rangle^{s_{0}}\langle j-j^{\prime}\rangle}\right)^{2}$
$\displaystyle\leq_{s_{0}}\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s}\sum_{(l^{\prime},j^{\prime})\in\mathbb{T}^{\nu+1}}|(\Delta_{12}^{k}\mathcal{A})^{j^{\prime}}_{j}(l-l^{\prime})|^{2}\langle
l-l^{\prime}\rangle^{2(s_{0}+|\vec{b}|)}\langle
j-j^{\prime}\rangle^{2}|h_{j^{\prime}}(l^{\prime})|^{2}$
$\displaystyle=\sum_{(l^{\prime},j^{\prime})\in\mathbb{Z}^{\nu+1}}|h_{j^{\prime}}(l^{\prime})|^{2}\sum_{(l,j)\in\mathbb{T}^{\nu+1}}|(\Delta_{12}^{k}\mathcal{A})^{j^{\prime}}_{j}(l-l^{\prime})|^{2}\langle
l-l^{\prime}\rangle^{2(s_{0}+|\vec{b}|)}\langle j-j^{\prime}\rangle^{2}\langle
l,j\rangle^{2s},$ (2.65)

where the first inequality follows from the Cauchy-Schwarz inequality with
(2.5). Since $|\vec{b}|\leq\mathtt{b}_{0}$, we have

$\displaystyle\langle l-l^{\prime}\rangle^{2(s_{0}+|\vec{b}|)}\langle
j-j^{\prime}\rangle^{2}$
$\displaystyle\leq_{s_{0}}\sum_{\begin{subarray}{c}\vec{\mathtt{b}}\in\mathbb{N}^{\nu},\\\
|\vec{\mathtt{b}}|\leq
s_{0}+\mathtt{b}_{0}\end{subarray}}\left(|(l-l^{\prime})^{\vec{\mathtt{b}}}|^{2}+|(l-l^{\prime})^{\vec{\mathtt{b}}}|^{2}|j-j^{\prime}|^{2}\right),$

therefore, (2.63), (2.64) and (2.65) give us that

$\displaystyle\rVert\underline{\Delta_{12}^{k}\partial_{\varphi}^{\vec{b}}(\mathcal{A}})h\rVert_{s}^{2}$
$\displaystyle\leq_{s_{0},\mathtt{b}_{0}}\sum_{\begin{subarray}{c}\vec{\mathtt{b}}\in\mathbb{N}^{\nu},\\\
|\vec{\mathtt{b}}|\leq
s_{0}+\mathtt{b}_{0}\end{subarray}}\left(\sum_{(l^{\prime},j^{\prime})\in\mathbb{T}^{\nu+1}}|h_{j^{\prime}}(l^{\prime})|^{2}\left(\langle
l^{\prime},j^{\prime}\rangle^{2s_{0}}\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)\right)^{2}+\left(\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s)\right)^{2}\right)\right.\right.$
$\displaystyle\left.\left.+\langle
l^{\prime},j^{\prime}\rangle^{2s}\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s_{0})\right)^{2}+\left(\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s_{0})\right)^{2}\right)\right)\right)$
$\displaystyle\leq\sum_{\begin{subarray}{c}\vec{\mathtt{b}}\in\mathbb{N}^{\nu},\\\
|\vec{\mathtt{b}}|\leq
s_{0}+\mathtt{b}_{0}\end{subarray}}\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s)\right)^{2}+\left(\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s)\right)^{2}\right)\rVert
h\rVert_{s_{0}}$ $\displaystyle\
+\sum_{\begin{subarray}{c}\vec{\mathtt{b}}\in\mathbb{N}^{\nu},\\\
|\vec{\mathtt{b}}|\leq
s_{0}+\mathtt{b}_{0}\end{subarray}}\left(\left(\mathfrak{M}^{\gamma}_{\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A})}(0,s_{0})\right)^{2}+\left(\mathfrak{M}^{\gamma}_{[\partial_{\varphi}^{\vec{\mathtt{b}}}(\mathcal{A}),\partial_{x}]}(0,s_{0})\right)^{2}\right)\rVert
h\rVert_{s}.$

Recalling the definition of the Lip-$0$-modulo tame operator in Definition
2.22, we obtain (2.61). ∎

######  Lemma 2.28.

[16, Lemma 2.26] Let $\vec{b}\in\mathbb{N}^{\nu}_{0}$ and
$\Phi:=I+\mathcal{A}$ where
$\mathcal{A},\partial_{\varphi}^{\vec{b}}(\mathcal{A})$ are Lip-$0$-modulo
tame operators up to $\mathtt{S}$ with modulo-tame constants
$\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s),\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A})}(0,s)$.
Then, there exist constants $C=C(|\vec{b}|)$ and $\delta=\delta(|\vec{b}|)>0$
such that if $\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s_{0})\leq\delta$,
then $\Phi$ is invertible and $\tilde{\mathcal{A}}:=\Phi^{-1}-I$ satisfies

$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\tilde{\mathcal{A}}}(0,s)$
$\displaystyle\leq C\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s),$ (2.66)
$\displaystyle\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\tilde{\mathcal{A}})}(0,s)$
$\displaystyle\leq
C\left(\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A})}(0,s)+\mathfrak{M}^{\sharp,\gamma}_{\partial_{\varphi}^{\vec{b}}(\mathcal{A})}(0,s_{0})\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s)\right),\text{
for $s\in[s_{0},\mathtt{S}]$.}$

Furthermore, the same statement holds true, replacing the Lip-$0$-modulo-tame
constant by the $0$-modulo tame constant.

###### Proof.

The estimates (2.66) can be proved in the same way as in [16, Lemma 2.26],
just replacing the $\mathcal{D}^{k_{0}}$-$0$-modulo-tame constant by the
Lip-$0$-tame constant. To obtain the result for the $0$-modulo-tame estimate,
we can simply apply the estimates (2.66) for an operator that does not depend
on $\omega$, noting that by definition,
$\mathfrak{M}^{\sharp}_{\mathcal{A}}(0,s)=\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s)$,
if $\mathcal{A}$ is independent of $\omega$. ∎

######  Lemma 2.29.

Let $\mathcal{A}$ be a Lip-$0$-modulo-tame operator up to $\mathtt{S}$, with
tame constant $\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s)$. We define

$r(\omega,j):=\begin{cases}\mathcal{A}^{j}_{j}(0),&\text{ for
$j\in\mathbb{Z}\backslash\left\\{0\right\\}$,}\\\ 0,&\text{
otherwise}.\end{cases}$

Then, it holds that

$|r|^{\text{Lip}(\gamma,\Omega)}_{0,0,0}:=\sup_{\omega,\omega_{1},\omega_{2}\in\Omega,\
j\in\mathbb{Z}\backslash\left\\{0\right\\}}|r(\omega,j)|+\gamma\frac{|r(\omega_{1},j)-r(\omega_{2},j)|}{|\omega_{1}-\omega_{2}|}\leq_{s_{0}}\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s_{0}).$

###### Proof.

Using the notation (2.62), let us denote

$\Delta^{k}_{12}r(j):=\begin{cases}r(\omega,j),&\text{ if $k=0$,}\\\
\gamma\frac{r(\omega_{1},j)-r(\omega_{2},j)}{|\omega_{1}-\omega_{2}|},&\text{
if $k=1$},\end{cases}$

so that $\Delta^{k}_{12}r(j)=\Delta_{12}^{k}\mathcal{A}^{j}_{j}$. Plugging
$l^{\prime}=0,\vec{\mathtt{b}}=0,s=s_{0}$ in (2.63), we have

$\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s_{0}}\left|\left(\Delta_{12}^{k}\mathcal{A}\right)^{j^{\prime}}_{j}(l)\right|^{2}\leq_{s_{0}}\left(\mathfrak{M}^{\gamma}_{\mathcal{A}}(0,s_{0})\right)^{2}\langle
j^{\prime}\rangle^{2s_{0}},$

while we obviously have

$\langle
j^{\prime}\rangle^{2s_{0}}\left|\left(\Delta^{k}_{12}\mathcal{A}\right)^{j^{\prime}}_{j^{\prime}}(0)\right|^{2}\leq\sum_{(l,j)\in\mathbb{Z}^{\nu+1}}\langle
l,j\rangle^{2s_{0}}\left|\left(\Delta_{12}^{k}\mathcal{A}\right)^{j^{\prime}}_{j}(l)\right|^{2}.$

Therefore, combining the above two inequalities, we obtain
$|\Delta^{k}_{12}r(j)|\leq_{s_{0}}\mathfrak{M}^{\gamma}_{\mathcal{A}}(0,s_{0})\leq\mathfrak{M}^{\sharp,\gamma}_{\mathcal{A}}(0,s_{0})$,
which proves the desired estimate, taking the supremum in
$\omega,\omega_{1},\omega_{2}$. ∎

###  2.3 Modified fractional Laplacians: $\Lambda^{\alpha-1}$ and
$\Upsilon^{\alpha-3}$.

For $f\in C^{\infty}_{0}$, we define

$\displaystyle\Lambda^{\alpha-1}f(x):=\int_{\mathbb{T}}(2-2\cos(x-y))^{-\frac{\alpha}{2}}(f(x)-f(y))dy,\quad\Upsilon^{\alpha-3}f(x):=\int_{\mathbb{T}}(2-2\cos(x-y))^{1-\frac{\alpha}{2}}f(y)dy.$
(2.67)

We denote the multipliers of $\Lambda^{\alpha-1}$ and $\Upsilon^{\alpha-3}$ by

$\displaystyle m^{\circ}_{1,\alpha}(j):=\int_{\mathbb{T}}(2-2\cos
y)^{-\frac{\alpha}{2}}(1-e^{-\text{i}jy})dy,\quad
m^{\circ}_{2,\alpha}(j):=\int_{\mathbb{T}}(2-2\cos
y)^{1-\frac{\alpha}{2}}e^{-\text{i}jy}dy.$ (2.68)

More explicitly, We have that [26, Lemma 2.6 and Eq. (2.1)]

$\displaystyle m^{\circ}_{1,\alpha}(j):$
$\displaystyle=\mathcal{C}_{\alpha}\left(\frac{\Gamma\left(|j|+\frac{\alpha}{2}\right)}{\Gamma\left(1+|j|-\frac{\alpha}{2}\right)}-\frac{\Gamma\left(\frac{\alpha}{2}\right)}{\Gamma\left(1-\frac{\alpha}{2}\right)}\right),\quad
m_{2,\alpha}^{\circ}(j):=2\pi\frac{(-1)^{j}\Gamma(3-\alpha)}{\Gamma(2+|j|-\frac{\alpha}{2})\Gamma(2-|j|-\frac{\alpha}{2})},$
(2.69)

where

$\displaystyle\mathcal{C}_{\alpha}:=-\frac{2\pi\Gamma(1-\alpha)}{\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(1-\frac{\alpha}{2}\right)}>0.$
(2.70)

With $m^{\circ}_{1,\alpha},m^{\circ}_{2,\alpha}$ above, we have

$\displaystyle\Lambda^{\alpha-1}f(x)=\sum_{j\in\mathbb{Z}}{m^{\circ}_{1,\alpha}}(j)\widehat{f}_{j}e^{\text{i}jx},\quad\Upsilon^{\alpha-3}f(x)=\sum_{j\in\mathbb{Z}}m^{\circ}_{2,\alpha}(j)\widehat{f}_{j}e^{\text{i}jx}.$
(2.71)

One can show that using the asymptotic expansion of Gamma functions and
Euler’s reflection formula [100, Theorem 2.1] that for
$\alpha\in(0,2)\backslash\left\\{1\right\\}$,

$\displaystyle\lim_{|j|\to\infty}m^{\circ}_{1,\alpha}(j)|j|^{-(\alpha-1)}=C_{1,\alpha},\quad\lim_{|j|\to\infty}m^{\circ}_{2,\alpha}(j)|j|^{-(\alpha-3)}=C_{2,\alpha},$
(2.72)

for some constants $C_{1,\alpha},C_{2,\alpha}$.

We will also consider the weighted operators:

$\displaystyle\Lambda^{\alpha-1}_{a}h:=\int_{\mathbb{T}}(2-2\cos(x-y))^{-\frac{\alpha}{2}}a(x,y)(h(x)-h(y))dy,$
(2.73)
$\displaystyle\Upsilon^{\alpha-3}_{a}h:=\int_{\mathbb{T}}(2-2\cos(x-y))^{1-\frac{\alpha}{2}}a(x,y)h(y)dy.$

Therefore defining for each $j\in\mathbb{Z}$,

$\displaystyle\mathfrak{a}_{\Lambda}(x,j)$
$\displaystyle:=\int_{\mathbb{T}}(2-2\cos
y)^{-\frac{\alpha}{2}}a(x,x-y)(1-e^{-\text{i}jy})dy,$ (2.74)
$\displaystyle\mathfrak{a}_{\Upsilon}(x,j)$
$\displaystyle:=\int_{\mathbb{T}}(2-2\cos
y)^{1-\frac{\alpha}{2}}a(x,x-y)e^{-\text{i}jy}dy,$

we have that

$\displaystyle\Lambda_{a}^{\alpha-1}h(x)=\sum_{j\in\mathbb{Z}}\mathfrak{a}_{\Lambda}(x,j)\widehat{h}_{j}e^{\text{i}jx},\quad\Upsilon_{a}^{\alpha-3}h(x)=\sum_{j\in\mathbb{Z}}\mathfrak{a}_{\Upsilon}(x,j)\widehat{h}_{j}e^{\text{i}jx}.$
(2.75)

In the rest of this subsection, we aim to find smooth symbols to represent the
operators $\Lambda^{\alpha-1},\Upsilon^{\alpha-3}$, In order to use the lemmas
introduced in Subsection 2.2. More precisely, we will extend the above symbols
to be defined on $\mathbb{T}\times\mathbb{R}$.

#####  2.3.0.1 Analysis of $\Lambda^{\alpha-1}$:

We study the operator $\Lambda^{\alpha-1}$. We mainly focus on the properties
of the multiplier:

$\displaystyle\lambda^{\circ}_{\alpha}(j):=jm^{\circ}_{1,\alpha}(j),\text{ for
$j\in\mathbb{Z}$.}$ (2.76)

Indeed, (2.71) tells us that the operator $\partial_{x}\Lambda^{\alpha-1}$ can
be characterized by the symbol $i\lambda^{\circ}_{\alpha}(j)$.

Let us denote

$\displaystyle\lambda^{\circ}_{\alpha-1}(j):=\frac{\Gamma(|j|+\frac{\alpha}{2})}{\Gamma(1+|j|-\frac{\alpha}{2})},$
(2.77)

so that we can write $\lambda^{\circ}_{\alpha}$ as, recalling
$m^{\circ}_{1,\alpha}$ from (2.69),

$\displaystyle\lambda^{\circ}_{\alpha}(j)=\mathcal{C}_{\alpha}j(\lambda^{\circ}_{\alpha-1}(j)-\lambda^{\circ}_{\alpha-1}(0))=\mathcal{C}_{\alpha}j\left(\frac{\Gamma(|j|+\frac{\alpha}{2})}{\Gamma(1+|j|-\frac{\alpha}{2})}-\frac{\Gamma(\frac{\alpha}{2})}{\Gamma(1-\frac{\alpha}{2})}\right).$
(2.78)

Note that $\lambda^{\circ}_{\alpha-1}$ is defined only on the integers. Recall
that the function $z\mapsto\Gamma(z)$ is holomorphic on
$\left\\{z\in\mathbb{C}:\text{Re}(z)>0\right\\}$. Since we always assume that
$\alpha\in(1,2)$, $\lambda^{\circ}_{\alpha-1}$ can be naturally extended to a
smooth function on $\mathbb{R}\backslash\left\\{0\right\\}$ as:

$\displaystyle\lambda_{\alpha-1}(\xi):=\frac{\Gamma(|\xi|+\frac{\alpha}{2})}{\Gamma(1+|\xi|-\frac{\alpha}{2})}\text{
for $\xi\in\mathbb{R}$.}$ (2.79)

######  Lemma 2.30.

For each $\eta\in\mathbb{N}_{0}$, it holds that

$\displaystyle\sup_{\xi\in\mathbb{R},|\xi|\geq\frac{1}{2}}\partial_{\xi}^{\eta}\lambda_{\alpha-1}(\xi)\langle\xi\rangle^{-(\alpha-1)+\eta}\leq_{\alpha,\eta}1.$
(2.80)

###### Proof.

We prove it by usual induction. For $\eta=0$, we recall Gautschi’s inequality
[101, Remark 2.1.1]:

$\displaystyle x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+s)^{1-s},\text{ for
all $x>0$ and $s\in(0,1)$.}$ (2.81)

Plugging $x=\frac{\alpha}{2}-1+|\xi|>0$ for $|\xi|\geq\frac{1}{2}$ and
$s=(2-\alpha)\in(0,1)$, we have that

$\displaystyle\left(|\xi|-\frac{2-\alpha}{2}\right)^{\alpha-1}<\lambda_{\alpha-1}(\xi)<\left(|\xi|+\frac{2-\alpha}{2}\right)^{\alpha-1},\text{
for all $\xi\in\mathbb{R}$},$ (2.82)

This proves (2.80) when $\eta=0$.

Now we assume that (2.80) holds true for some $\eta\geq 0$ and aim to prove it
for $\partial_{\xi}^{\eta+1}\lambda_{\alpha-1}$. Since
$\xi\mapsto\lambda_{\alpha-1}$ is even, let us assume, without loss of
generality, that $\xi>0$ to obtain (2.80). We denote by $\psi^{(\eta)}(\xi)$
be the $\eta$-th polygamma function:

$\displaystyle\psi^{(\eta)}(\xi):=-\int_{0}^{1}\frac{t^{\xi-1}}{1-t}(\log{t})^{\eta}dt,\quad\psi^{(0)}(\xi):=\log\xi-\frac{1}{2\xi}-2\int_{0}^{\infty}\frac{t}{(t^{2}+\xi^{2})(e^{2\pi
t}-1)}dt,\text{ for $\eta\in\mathbb{N}$, $0<\xi\in\mathbb{R}$}.$ (2.83)

We collect necessary properties of the polygamma functions [73, Lemma 1]:

$\displaystyle\psi^{(0)}(\xi)$
$\displaystyle=\frac{\Gamma^{\prime}(\xi)}{\Gamma(\xi)}\text{ for all
$\xi>0$.}$ (2.84) $\displaystyle\psi^{(\eta)}(\xi)$
$\displaystyle=\left(\frac{d}{d\xi}\right)^{\eta}\psi^{(0)}(\xi),\text{ for
$\eta,\xi\geq 0$.}$ (2.85)
$\displaystyle\frac{(\eta-1)!}{\xi^{\eta}}+\frac{\eta!}{2\xi^{\eta+1}}$
$\displaystyle\leq(-1)^{\eta+1}\psi^{(\eta)}(\xi)\leq\frac{(\eta-1)!}{\xi^{\eta}}+\frac{\eta!}{\xi^{\eta+1}},\text{
for $\eta\geq 1$, $\xi>0$.}$ (2.86)

From (2.86) and the definition of $\psi^{(0)}(\xi)$ in (2.83), it is clear
that

$\displaystyle|\psi^{(\eta)}(\xi+\alpha/2)-\psi^{(\eta)}(\xi+1-\alpha/2)|\leq_{\eta}\langle\xi\rangle^{-\eta-1},\text{
for $\xi\geq\frac{1}{2}$ and $\eta\in\mathbb{N}_{0}$.}$ (2.87)

Also, using (2.84) and $\lambda_{\alpha-1}$ in (2.79), we have that

$\displaystyle\frac{d}{d\xi}\lambda_{\alpha-1}(\xi)=\lambda_{\alpha-1}(\xi)(\psi^{(0)}(\xi+\frac{\alpha}{2})-\psi^{(0)}(\xi+1-\frac{\alpha}{2})),\text{
for $\xi>0$}.$ (2.88)

Therefore, it is straightforward that

$\displaystyle\left|\left(\frac{d}{d\xi}\right)^{\eta+1}\lambda_{\alpha-1}(\xi)\right|$
$\displaystyle=\left|\left(\frac{d}{d\xi}\right)^{\eta}\left(\lambda_{\alpha-1}(\xi)(\psi^{(0)}(\xi+\frac{\alpha}{2})-\psi^{(0)}(\xi+1-\frac{\alpha}{2}))\right)\right|$
$\displaystyle\overset{\eqref{poly3}}{=}\left|\sum_{k=0}^{\eta}\binom{\eta}{k}\left(\left(\frac{d}{d\xi}\right)^{k}(\lambda_{\alpha-1})(\xi)(\psi^{(\eta-k)}(\xi+\alpha/2)-\psi^{(\eta-k)}(\xi+1-\alpha/2))\right)\right|$
$\displaystyle\overset{\eqref{cartmanandbunny}}{\leq}_{\eta}\sum_{k=0}^{\eta}\binom{\eta}{k}\left|\left(\frac{d}{d\xi}\right)^{k}(\lambda_{\alpha-1})(\xi)\right|\langle\xi\rangle^{-(\eta-k+1)}$
$\displaystyle\overset{\eqref{kyleandcartman12}}{\leq}_{\eta}\sum_{k=0}^{\eta}\binom{\eta}{k}\langle\xi\rangle^{(\alpha-1)-k}\langle\xi\rangle^{-(\eta-k+1)}$
$\displaystyle\leq_{\eta}\langle\xi\rangle^{(\alpha-1)-(\eta+1)},$

which gives us (2.80) for $\partial_{\xi}^{(\eta+1)}\lambda_{\alpha-1}$. This
finishes the proof. ∎

######  Lemma 2.31.

For $\xi\geq\frac{1}{2}$, $\xi\mapsto\lambda_{\alpha-1}(\xi)$ is strictly
monotone-increasing and

$\displaystyle\partial_{\xi}\lambda_{\alpha-1}(\xi)\geq_{\alpha}\xi^{\alpha-2},\quad\text{
for $\xi\geq\frac{1}{2}$}.$ (2.89)

###### Proof.

From (2.82), we have that $\lambda_{\alpha-1}(\xi)\geq_{\alpha}\xi^{\alpha-1}$
for $\xi>\frac{1}{2}$. Also, from (2.86), we have
$\psi^{(1)}(\xi)\geq\frac{1}{\xi}+\frac{1}{2\xi^{2}}$ for $\xi>\frac{1}{2}$,
especially (since $\alpha>1$) (2.85) and the mean-value theorem tells us that

$\displaystyle\psi^{(0)}(\xi+\frac{\alpha}{2})-\psi^{(0)}(\xi+1-\frac{\alpha}{2})\geq_{\alpha}\frac{1}{\xi}.$

Therefore, the result follows from (2.88). ∎

Now, we study the multipliers $\lambda^{\circ}_{\alpha}$ and
$\lambda_{\alpha-1}$ at integer points.

######  Lemma 2.32.

For all $j,k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that $j\neq k$, it
holds that

$\displaystyle|\lambda^{\circ}_{\alpha}(j)-\lambda^{\circ}_{\alpha}(k)|\geq_{\alpha}|j-k|\left(|j|^{\alpha-1}+|k|^{\alpha-1}\right).$
(2.90)

###### Proof.

Note that $j\mapsto\lambda_{\alpha}^{\circ}(j)$ is odd. Therefore, without
loss of generality, we only need to prove (2.90) for the cases: $j>0>k$ and
$j>k>0$.

Proof for $j>0>k$. We first observe from $\psi^{(0)}$ in (2.83) that
$\xi\mapsto\psi^{(0)}(\xi)$ is strictly increasing for $\xi>0$. Since
$\Gamma(\xi)$ is strictly positive for $\xi>0$, we have
$\lambda_{\alpha-1}(\xi)>0$. Especially, (2.88) tells us that
$\lambda_{\alpha-1}$ is strictly increasing function for $\xi>0$. Therefore,
we have

$\displaystyle\lambda_{\alpha-1}(\xi)>\lambda_{\alpha-1}(0)>0,\text{ for all
$\xi\geq 0$.}$ (2.91)

Thus, for $j>0>k$, it follows from (2.78) that

$\lambda_{\alpha}^{\circ}(j)-\lambda_{\alpha}^{\circ}(k)\geq\mathcal{C}_{\alpha}\left(j\left(\lambda_{\alpha-1}(j)-\lambda_{\alpha-1}(0)\right)+|k|\left(\lambda_{\alpha-1}(k)-\lambda_{\alpha-1}(0)\right)\right)\geq_{\alpha}j^{\alpha}+|k|^{\alpha},$

where the last inequality follows from Lemma (2.31). The above inequality
certainly implies (2.90).

Proof for $j>k>0$. It follows from (2.91) and

$\lambda_{\alpha}^{\circ}(j)-\lambda_{\alpha}^{\circ}(k)=\mathcal{C}_{\alpha}\int_{k}^{j}\left(\lambda_{\alpha-1}(\xi)-\lambda_{\alpha-1}(0)\right)+\xi\partial_{\xi}\lambda_{\alpha-1}(\xi)d\xi\geq\mathcal{C}_{\alpha}\int_{k}^{j}\xi\lambda_{\xi}\lambda_{\alpha-1}(\xi)d\xi.$

Using Lemma 2.31, we have

$\lambda_{\alpha}^{\circ}(j)-\lambda_{\alpha}^{\circ}(k)\geq_{\alpha}j^{\alpha}-k^{\alpha},$

which implies (2.90). ∎

######  Lemma 2.33.

For all $j,k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that $j+k\neq 0$,
it holds that

$|\lambda^{\circ}_{\alpha}(j+k)-\lambda^{\circ}_{\alpha}(j)-\lambda^{\circ}_{\alpha}(k)|\geq_{\alpha}(\max\left\\{|j|,|k|\right\\})^{\alpha-1}\min\left\\{|j|,|k|\right\\}.$

###### Proof.

In view of (2.78), it suffices to show that

$\displaystyle|(j+k)\lambda_{\alpha-1}(j+k)-j\lambda_{\alpha-1}(j)-k\lambda_{\alpha-1}(k)|\geq_{\alpha}(\max\left\\{|j|,|k|\right\\})^{\alpha-1}\min\left\\{|j|,|k|\right\\}.$
(2.92)

Furthermore, we will assume, without loss of generality, that $j>0$. Under
this assumption, we consider three cases: $j=k>0,\ j>k>0,\ k>j>0$, and
$j>0>k$.

Case $j=k>0$: In this case, we have

$\displaystyle|(j+k)\lambda_{\alpha-1}(j+k)-j\lambda_{\alpha-1}(j)-k\lambda_{\alpha-1}(k)|=2k(\lambda_{\alpha-1}(2k)-\lambda_{\alpha-1}(k)).$
(2.93)

If $k=1$, then it follows from Lemma 2.31 that

$\displaystyle
2k(\lambda_{\alpha-1}(2k)-\lambda_{\alpha-1}(k))\geq_{\alpha}1=k^{\alpha}.$
(2.94)

If $k\geq 2$, then, using (2.82), we have

$\displaystyle\lambda_{\alpha-1}(2k)$
$\displaystyle>\left(2k-\frac{2-\alpha}{2}\right)^{\alpha-1}>k^{\alpha-1}\left(2-\frac{2-\alpha}{2k}\right)^{\alpha-1}\overset{k\geq
2}{>}k^{\alpha-1}\left(\frac{7}{4}\right)^{\alpha-1},$
$\displaystyle\quad\lambda_{\alpha-1}(k)$
$\displaystyle<\left(k+\frac{2-\alpha}{2}\right)^{\alpha-1}\leq
k^{\alpha-1}\left(1+\frac{(2-\alpha)}{2k}\right)^{\alpha-1}\overset{k\geq
2}{\leq}k^{\alpha-1}\left(\frac{5}{4}\right)^{\alpha-1}.$

Hence, it follows that

$\lambda_{\alpha-1}(2k)-\lambda_{\alpha-1}(k)>_{\alpha}k^{\alpha-1}.$

Plugging this and (2.94) into (2.93), we get

$\displaystyle|(j+k)\lambda_{\alpha-1}(j+k)-j\lambda_{\alpha-1}(j)-k\lambda_{\alpha-1}(k)|>_{\alpha}k^{\alpha},\text{
if $k=j>0$},$ (2.95)

which gives (2.92).

Case $j>k>0$: Using Lemma 2.31, we compute

$\displaystyle|(j+k)\lambda_{\alpha-1}(j+k)-j\lambda_{\alpha-1}(j)-k\lambda_{\alpha-1}(k)|$
$\displaystyle=j\left(\lambda_{\alpha-1}(j+k)-\lambda_{\alpha}(j)\right)+k\left(\lambda_{\alpha-1}(j+k)-\lambda_{\alpha-1}(k)\right)$
$\displaystyle>_{\alpha}j\int_{j}^{j+k}\partial_{\xi}\lambda_{\alpha-1}(\xi)d\xi$
$\displaystyle>_{\alpha}j|j+k|^{\alpha-2}k.$ (2.96)

Then, we can further compute

$\displaystyle
j|j+k|^{\alpha-2}k\geq_{\alpha}\begin{cases}jk^{\alpha-1}\geq_{\alpha}j^{\alpha-1}k,&\text{
if $k<j\leq 2k$,}\\\ jj^{\alpha-2}k\geq_{\alpha}j^{\alpha-1}k,&\text{ if
$j>2k$.}\end{cases}$

Plugging this into (2.3.0.1), we obtain (2.92).

Case $k>j>0$: Changing the role of $j,k$ in the case where $j>k>0$ above, we
get (2.92).

Case $j>0>k$: In this case, we replace $k$ by $-k$, then (2.92) follows from
the above cases. ∎

Finally we summarize the results for the multipliers $m_{1,\alpha}$ and
$\lambda_{\alpha}$, which are proper extensions of $m_{1,\alpha}^{\circ}$ and
$\lambda_{\alpha}^{\circ}$ to $\mathbb{R}$.

######  Proposition 2.34.

There exist real-valued classical symbols
$\lambda_{\alpha}\in\mathcal{S}^{\alpha},m_{1,\alpha}\in\mathcal{S}^{\alpha-1}$
such that

  1. (1)

It holds that and $\Lambda^{\alpha-1}=Op^{W}(m_{1,\alpha})$ and
$\partial_{x}\Lambda^{\alpha-1}=Op^{W}(\text{i}\lambda_{\alpha})$ as linear
operators on $C^{\infty}_{0}$ (set of smooth periodic functions with zero-
average. See (2.1)). That is, it holds that

$\displaystyle
m_{1,\alpha}(j)=m_{1,\alpha}^{\circ}(j)=\mathcal{C}_{\alpha}\left(\frac{\Gamma\left(|j|+\frac{\alpha}{2}\right)}{\Gamma\left(1+|j|-\frac{\alpha}{2}\right)}-\frac{\Gamma\left(\frac{\alpha}{2}\right)}{\Gamma\left(1-\frac{\alpha}{2}\right)}\right),\text{
for all $j\in\mathbb{Z}\backslash\left\\{0\right\\}$,}$ (2.97)

and

$\displaystyle\lambda_{\alpha}(\xi)=\xi m_{1,\alpha}(\xi),\text{ for all
$\xi\in\mathbb{R}$},$ (2.98)

  2. (2)

$\xi\mapsto m_{1,\alpha}(\xi)$ is strictly positive, even and monotone
increasing.

  3. (3)

For all $\xi\in\mathbb{R}$,
$\langle\xi\rangle^{\alpha-1}\leq_{\alpha}|\partial_{\xi}\lambda_{\alpha}(\xi)|\leq_{\alpha}\langle\xi\rangle^{\alpha-1}$,
therefore $\frac{1}{\partial_{\xi}\lambda_{\alpha}}\in\mathcal{S}^{1-\alpha}$.

  4. (4)

For all $j,k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that $j+k\neq 0$,
it holds that

$|\lambda_{\alpha}(j+k)-\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|\geq_{\alpha,k}|j|^{\alpha-1}.$

  5. (5)

For all $j,k\in\mathbb{Z}\backslash\left\\{0\right\\}$ such that $j\neq k$, it
holds that

$\displaystyle|j-k|\left(|j|^{\alpha-1}+|k|^{\alpha-1}\right)\leq_{\alpha}|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|\leq_{\alpha}|j-k|\left(|j|^{\alpha-1}+|k|^{\alpha-1}\right).$
(2.99)

###### Proof.

We recall from (2.79) that $\lambda_{\alpha-1}$ is well-defined for all real
numbers and smooth everywhere except for $0$. In view of (2.69), we define
$m_{1,\alpha}$ on $\mathbb{R}$ so that $m_{1,\alpha}(\xi)$ is strictly
positive, even, smooth on $\mathbb{R}$, and monotone increasing for
$0<\xi<\frac{1}{2}$ and

$m_{1,\alpha}(\xi):=\mathcal{C}_{\alpha}\left(\lambda_{\alpha-1}(\xi)-\lambda_{\alpha-1}(0)\right),\text{
if $|\xi|\geq\frac{1}{2}$,}$

and even and smooth in $\mathbb{R}$. Such an extension is possible since
$\lambda_{\alpha-1}$ is even, smooth, strictly increasing and
$\lambda_{\alpha-1}(\xi)-\lambda_{\alpha-1}(0)>0$ for $\xi\geq\frac{1}{2}$,
and which follows from (2.91), Lemma 2.31, Lemma 2.30, and
$\mathcal{C}_{\alpha}$ defined in (2.70) is a strictly positive constant for
each $\alpha\in(1,2)$. Once $m_{1,\alpha}(\xi)$ is defined for all
$\xi\in\mathbb{R}$, we define $\lambda_{\alpha}:=\xi m_{1,\alpha}$ for all
$\xi\in\mathbb{R}$.

By its definition, we have $m_{1,\alpha}=m_{1,\alpha}^{\circ}$ and
$\lambda_{\alpha}=\lambda_{\alpha}^{\circ}$ on
$\mathbb{Z}\backslash\left\\{0\right\\}$, therefore we have
$Op^{W}(m_{1,\alpha})=\Lambda^{\alpha-1}$ and
$Op^{W}(\text{i}\xi\lambda_{\alpha})=\partial_{x}\Lambda^{\alpha-1}$ on
$C^{\infty}_{0}$. Thanks to Lemma 2.30, we have that
$m_{1,\alpha}\in\mathcal{S}^{\alpha-1}$ and
$\lambda_{\alpha}\in\mathcal{S}^{\alpha}$ are classical symbols. Threfore (1)
and (2) follow. Since $m_{1,\alpha}$ is strictly positive and monotone
increasing,
$\partial_{\xi}\lambda_{\alpha}(\xi)=m_{1,\alpha}(\xi)+\xi\partial_{\xi}m_{1,\alpha}(\xi)>0$
for all $\xi\in\mathbb{R}$. Therefore, (3) follows from Lemma 2.31. Again,
since $\lambda_{\alpha}=\lambda^{\circ}_{\alpha}$ on
$\mathbb{Z}\backslash\left\\{0\right\\}$, (4) and the lower bound in (5)
follow from Lemma 2.32 and Lemma 2.33, respectively.

Now, it suffices to prove the upper bound in (2.99). Using
$\lambda_{\alpha}\in\mathcal{S}^{\alpha}$ and assuming $j>k$, without loss of
generality, we have

$|\lambda_{\alpha}(j)-\lambda_{\alpha}(k)|=\left|\int_{k}^{j}\partial_{\xi}(\lambda_{\alpha}(\xi))d\xi\right|\leq_{\alpha}\left|\int_{k}^{j}|\xi|^{\alpha-1}d\xi\right|\leq_{\alpha}|j-k|(|j|^{\alpha-1}+|k|^{\alpha-1}),$

which gives the upper bound in (2.99). ∎

######  Lemma 2.35.

For each $k\in\mathbb{Z}\backslash\left\\{0\right\\}$, there exists a
classical symbol $\kappa_{k,1-\alpha}(\xi)\in\mathcal{S}^{1-\alpha}$ such that

$\displaystyle\kappa_{k,1-\alpha}(\xi)=\frac{1}{\lambda_{\alpha}(\xi)-\lambda_{\alpha}(\xi-k)-\lambda_{\alpha}(k)},\text{
for all $\xi\in\mathbb{Z}\backslash\left\\{0,k\right\\}$ }.$ (2.100)

###### Proof.

Thanks to (4) of Proposition 2.34 (replacing $j$ by $-j$), we see that
$\kappa_{k,1-\alpha}(\xi)$ is well defined for all $\xi\in\mathbb{R}$ except
for $\xi=0$ and $\xi=k$ for which the denominator vanishes. Therefore, we can
construct $\kappa_{k,1-\alpha}(\xi)$ for $\xi\in\mathbb{R}$ in such a way that

$\displaystyle\kappa_{k,1-\alpha}(\xi)=\begin{cases}\frac{1}{\lambda_{\alpha}(\xi)-\lambda_{\alpha}(\xi-k)-\lambda_{\alpha}(k)},&\text{
if $|\xi-0|\geq\frac{1}{2}$ or $|\xi-k|\geq\frac{1}{2}$ },\\\ 0,&\text{ if
$\xi=0$ or $\xi=k$},\end{cases}$

and it is smooth in $\mathbb{R}$. In order to see
$\kappa_{k,1-\alpha}(\xi)\in\mathcal{S}^{1-\alpha}$, we only need to focus on
large $\xi$, that is, it is sufficient to show that

$\displaystyle\lim_{|\xi|\to\infty}\partial_{\xi}^{\eta}\kappa_{k,1-\alpha}(\xi)\langle\xi\rangle^{-(1-\alpha)+\eta}\leq_{k,\eta,\alpha}1.$
(2.101)

This follows straightforwardly from the usual induction argument, using that
$\lambda_{\alpha}\in\mathcal{S}^{\alpha}$. Without loss of generality, let us
assume that $k>0$. We will show

$\displaystyle\sup_{\xi\to\infty}\partial_{\xi}^{\eta}\kappa_{k,1-\alpha}(\xi)\langle\xi\rangle^{-(1-\alpha)+\eta}\leq_{k,\eta,\alpha}1.$
(2.102)

Note that the other necessary estimate
$\sup_{\xi\to-\infty}\partial_{\xi}^{\eta}\kappa_{k,1-\alpha}(\xi)\langle\xi\rangle^{-(1-\alpha)+\eta}\leq_{k,\eta}1$
follows in the same way, hence we will omit the proof.

Towards (2.102), we see that for $\xi\gg 0$ and $\eta=0$,

$\displaystyle|\kappa_{k,1-\alpha}(\xi)\langle\xi\rangle^{-(1-\alpha)}|=\left|\frac{\langle\xi\rangle^{\alpha-1}}{\lambda_{\alpha}(\xi)-\lambda_{\alpha}({\xi-k})-\lambda_{\alpha}(k)}\right|\leq\frac{1}{\left|\frac{\int_{0}^{1}\partial_{\xi}\lambda_{\alpha}(\xi+(t-1)k)dt}{\langle\xi\rangle^{\alpha-1}}\right|-\frac{\lambda_{\alpha}(k)}{\langle\xi\rangle^{\alpha-1}}}.$
(2.103)

Since $\xi\mapsto m_{1,\alpha}$ is monotone increasing for $\xi\geq 0$ ((2) of
Proposition 2.34), we see from (2.98) that

$\lim_{\xi\to\infty}\left|\frac{\int_{0}^{1}\partial_{\xi}\lambda_{\alpha}(\xi+(t-1)k)dt}{\langle\xi\rangle^{\alpha-1}}\right|=\lim_{\xi\to\infty}\frac{\int_{0}^{1}|\partial_{\xi}\lambda_{\alpha}(\xi+(t-1)k)|dt}{\langle\xi\rangle^{\alpha-1}}\geq_{\alpha}1,$

where the last inequality follows from (3) of Proposition 2.34. Therefore,
taking $\xi$ to $\infty$ in (2.103), we obtain (2.102), when $\eta=0$. For
$\eta+1\geq 1$, we have that for $\xi\geq k+1$,

$\displaystyle|\partial_{\xi}^{\eta+1}\kappa_{k,1-\alpha}(\xi)|$
$\displaystyle=|\partial_{\xi}^{\eta}(|\kappa_{k,1-\alpha}(\xi)|^{2}(\partial_{\xi}\lambda_{\alpha}(\xi)-\partial_{\xi}\lambda_{\alpha}(\xi-k)))|$
$\displaystyle=\left|\partial_{\xi}^{\eta}\left(|\kappa_{k,1-\alpha}(\xi)|^{2}\int_{0}^{1}\partial_{\xi}^{2}\lambda_{\alpha}(\xi+(t-1)k)dt\right)\right|$
$\displaystyle=\left|\sum_{\eta_{1}+\eta_{2}=\eta}C_{\eta_{1},\eta_{2}}\partial_{\xi}^{\eta_{1}}(\kappa_{k,1-\alpha}(\xi)^{2})\int_{0}^{1}\partial_{\xi}^{\eta_{2}+2}\lambda_{\alpha}(\xi+(t-1)k)dt\right|.$

Using the induction hypothesis (2.102), we have
$|\partial_{\xi}^{\eta_{1}}(\kappa_{k,1-\alpha}(\xi)^{2})|\leq_{\eta,\alpha,k}\langle\xi\rangle^{2(1-\alpha)-\eta_{1}}$,
while (3) of Proposition 2.34 gives us that
$|\partial_{\xi}^{\eta_{2}+2}\lambda_{\alpha}(\xi+(t-1)k)|\leq_{\eta,k,\alpha}\langle\xi\rangle^{\alpha-\eta_{2}-2}$.
Therefore, we have that

$|\partial_{\xi}^{\eta+1}\kappa_{k,1-\alpha}(\xi)|\leq_{\eta,\alpha,k}\langle\xi\rangle^{-\alpha-\eta}=\langle\xi\rangle^{(1-\alpha)-(\eta+1)}.$

This proves (2.102) for $\eta+1$. ∎

#####  2.3.0.2 Analysis of $\Upsilon_{a}^{\alpha-3}$

We study the operator $\Upsilon_{a}^{\alpha-3}$ defined in (2.73), for some
$a=a(\omega,\varphi,x,y)$ and $\omega\in\Omega$ and
$(\varphi,x)\in\mathbb{T}^{\nu}\times\mathbb{T}$. Especially, we aim to
express the operator $\Upsilon^{\alpha-3}_{a}$ as a pseudo differential
operator with a classical symbol.

Using (2.12) and (2.73), it is easy to see that

$\displaystyle\Upsilon^{\alpha-3}_{a}=Op(\mathfrak{a}),\text{ where
}\mathfrak{a}(\omega,\varphi,x,j):=\int_{\mathbb{T}}(2-2\cos
y)^{1-\frac{\alpha}{2}}a(\omega,\varphi,x,x-y)e^{-\text{i}jy}dy\text{ for
$j\in\mathbb{Z}$}.$ (2.104)

Here, the “symbol”, $\mathfrak{a}(\omega,\varphi,x,j),$ is defined only for
$j\in\mathbb{Z}$. In order to make a use of the lemmas studied in Subsection
2.2, we aim to find an extension $\mathfrak{a}(\omega,\varphi,x,\xi)$, which
is well-defined for all $\xi\in\mathbb{R}$, so that the symbol can be measured
by the norms in (2.34). Of course, a particular choice of an extension does
not change the operator, since our pseudo differential operators that we
consider in this paper always act on periodic functions.

To this end, we pick $\psi_{1},\psi_{2}$ to be smooth non-negative bump
functions on $\mathbb{R}$ such that $\text{supp}(\psi_{1})\in(-1/4,1/4)$ with
$\int_{\mathbb{R}}\psi_{1}(\xi)d\xi=1$, and
$\text{supp}(\psi_{2})\in(-\pi/2,\pi/2)$ with $\psi_{2}(y)=1$ for
$y\in(-\pi/3,\pi/3)$. For such $\psi_{1},\psi_{2}$, we define
$S_{\Upsilon}(a)(\omega,\varphi,x,\xi)$ as

$\displaystyle S_{1,\Upsilon}(a)(\omega,\varphi,x,\xi)$
$\displaystyle:=\int_{-\pi}^{\pi}(2-2\cos
y)^{1-\frac{\alpha}{2}}a(\omega,\varphi,x,x-y)\psi_{2}(y)e^{-\text{i}\xi
y}dy,$ (2.105) $\displaystyle
S_{\text{step},\Upsilon}(a)(\omega,\varphi,x,\xi)$
$\displaystyle:=\int_{-\pi}^{\pi}(2-2\cos
y)^{1-\frac{\alpha}{2}}a(\omega,\varphi,x,x-y)(1-\psi_{2}(y))e^{-\text{i}jy}dy,$
(2.106) $\displaystyle\text{ for a unique $j\in\mathbb{Z}$ such that
$\xi\in(j-1/2,j+1/2]$},$ $\displaystyle
S_{2,\Upsilon}(a)(\omega,\varphi,x,\xi)$
$\displaystyle:=\psi_{1}*S_{\text{step},\Upsilon}(a):=\int_{\mathbb{R}}\psi_{1}(\xi-\xi_{0})S_{\text{step},\Upsilon}(a)(\omega,\varphi,x,\xi_{0})d\xi_{0},$
(2.107) $\displaystyle S_{\Upsilon}(a)(\omega,\varphi,x,\xi)$
$\displaystyle:=S_{1,\Upsilon}(a)(\omega,\varphi,x,\xi)+S_{2,\Upsilon}(a)(\omega,\varphi,x,\xi).$
(2.108)

Note that $S_{\Upsilon}(a)$ is smooth in $\varphi,x,\xi$, if so is $a$, and

$S_{\Upsilon}(a)(\omega,\varphi,x,j)=\int_{\mathbb{T}}(2-2\cos
y)^{1-\frac{\alpha}{2}}a(\omega,\varphi,x,x-y)e^{-\text{i}jy}dy,$

when $\xi=j$ for some $j\in\mathbb{Z}$. Therefore, (2.104) tells us that

$\displaystyle Op(S_{\Upsilon}(a))=\Upsilon^{\alpha-3}_{a}.$ (2.109)

Now, we aim to prove that $S_{\Upsilon}(a)$ is a classical symbol in
$\mathcal{S}^{\alpha-3}$, that is,
$\partial_{\xi}^{\eta}(S_{\Upsilon}(a))\langle\xi\rangle^{-(\alpha-3)+\eta}$
is bounded for each $\eta\in\mathbb{N}_{0}$. We will consider $S_{1,\Upsilon}$
and $S_{2,\Upsilon}$ in (2.108) separately.

We first analyze the symbol $S_{1,\Upsilon}(a)$.

######  Lemma 2.36.

For each $\eta\in\mathbb{N}_{0}$, there exists $\mu=\mu(\eta,\nu)\geq 0$ such
that

$\displaystyle|S_{1,\Upsilon}(a)|^{\text{Lip}(\gamma,\Omega)}_{\alpha-3,s,\eta}\leq_{s,\eta,\alpha,\nu}\rVert
a\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s+\mu}(\mathbb{T}^{\nu+2})},\text{ for
all $s\geq s_{0}$.}$ (2.110)

###### Proof.

Let $\eta\in\mathbb{N}_{0}$ and $s\geq s_{0}$ be fixed, and let us choose
$s_{1},s_{2}\in\mathbb{N}_{0}$ such that $s_{1}+s_{2}=s$. For
$\omega,\omega_{1},\omega_{2}\in\Omega$, we denote

$\displaystyle\Delta_{12}^{k}a(\varphi,x,y)$
$\displaystyle:=\begin{cases}a(\omega,\varphi,x,y),&\text{ if $k=0$,}\\\
\gamma\frac{a(\omega_{1},\varphi,x,y)-a(\omega_{2},\varphi,x,y)}{|\omega_{1}-\omega_{2}|},&\text{
if $k=1$},\end{cases}$ (2.111)
$\displaystyle\Delta_{12}^{k}S_{1,\Upsilon}(a)(\varphi,x,\xi)$
$\displaystyle:=\begin{cases}S_{1,\Upsilon}(a)(\omega,\varphi,x,y),&\text{ if
$k=0$,}\\\
\gamma\frac{S_{1,\Upsilon}(a)(\omega_{1},\varphi,x,\xi)-S_{1,\Upsilon}(a)(\omega_{2},\varphi,x,\xi)}{|\omega_{1}-\omega_{2}|},&\text{
if $k=1$}.\end{cases}$ (2.112)

Since $a\mapsto S_{1,\Upsilon}(a)$ is linear, we have that

$\displaystyle\Delta_{12}^{k}(\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}S_{1,\Upsilon}(a))(\varphi,x,\xi)=S_{1,\Upsilon}(\Delta_{12}^{k}(\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}a))(\varphi,x,\xi),$
(2.113)

while (2.105) tells us that

$\displaystyle
S_{1,\Upsilon}(\Delta_{12}^{k}(\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}a))(\varphi,x,\xi)=\int_{-\pi}^{\pi}|y|^{2-\alpha}\underbrace{\left(\frac{2-2\cos
y}{|y|^{2}}\right)^{1-\frac{\alpha}{2}}\psi_{2}(y)\Delta^{k}_{12}\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}(a(\varphi,x,x-y))}_{=:f(\varphi,x,y)}e^{-\text{i}\xi
y}dy.$ (2.114)

Since $\text{supp}(\psi_{2})\subset(-\pi/2,\pi/2)$, it holds that
$\text{supp}(f(\varphi,x,\cdot))\subset(-\pi/2,\pi/2)$. Therefore, applying
Lemma 2.37, which will be proved after the proof of this Lemma, we get

$\displaystyle\left|\partial_{\xi}^{\eta}(S_{1,\Upsilon}(\Delta_{12}^{k}(\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}a)))(\varphi,x,\xi)\right|\leq_{\alpha,\eta}\rVert
f(\varphi,x,\cdot)\rVert_{H^{\eta+3}(\mathbb{T})}\langle\xi\rangle^{\alpha-3-\eta}.$
(2.115)

From the definition of $f$ in (2.114), it is clear that (using the usual
Sobolev embedding theorem) we can choose $\mu(\nu)>0$ such that

$\displaystyle\rVert
f(\varphi,x,\cdot)\rVert_{H^{\eta+3}(\mathbb{T})}\leq_{s,\nu}\rVert
a\rVert_{H^{s+\mu(\nu)+\eta+3}(\mathbb{T}^{\nu+2})}^{\text{Lip}(\gamma,\Omega)}.$
(2.116)

Therefore,

$\displaystyle\rVert\partial_{\xi}^{\eta}\Delta^{k}_{12}(S_{1,\Upsilon}(a))(\cdot,\cdot,\xi)\rVert_{s}$
$\displaystyle\leq_{s}\sum_{\begin{subarray}{c}s_{1}+s_{2}=s,\\\
s_{1},s_{2}\in\mathbb{N}_{0}\end{subarray}}\rVert\partial_{\xi}^{\eta}\Delta_{12}^{k}\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}S_{1,\Upsilon}(a)\rVert_{L^{2}_{\varphi,x}}$
$\displaystyle\overset{\eqref{rjjsdsd1sdsd}}{=}\sum_{\begin{subarray}{c}s_{1}+s_{2}=s,\\\
s_{1},s_{2}\in\mathbb{N}_{0}\end{subarray}}\rVert\partial_{\xi}^{\eta}S_{1,\Upsilon}(\Delta_{12}^{k}\partial_{\varphi}^{s_{1}}\partial_{x}^{s_{2}}a)\rVert_{L^{2}_{\varphi,x}}$
$\displaystyle\overset{\eqref{whattobys},\eqref{fsjsdpwosmds}}{\leq_{s,\alpha,\eta,\nu}}\rVert
a\rVert_{H^{s+\mu(\nu)+\eta+3}(\mathbb{T}^{\nu+2})}^{\text{Lip}(\gamma,\Omega)}\langle\xi\rangle^{\alpha-3-\eta}.$

Therefore, replacing $\mu(\nu)+\eta+3$ by $\mu(\nu,\eta)$, we obtain (2.110).
∎

######  Lemma 2.37.

Let $f\in C^{\infty}_{c}(\mathbb{-\pi,\pi})$ be such that
$\text{supp}(f)\subset(-\frac{\pi}{2},\frac{\pi}{2})$. We denote

$\mathfrak{b}(\xi):=\int_{-\pi}^{\pi}|y|^{2-\alpha}f(y)e^{-\text{i}\xi
y}dy,\text{ for $(x,\xi)\in\mathbb{T}\times\mathbb{R}$.}$

Then, for each $\eta\in\mathbb{N}_{0}$, $\mathfrak{b}$ satisfies

$\displaystyle\sup_{\xi\in\mathbb{R}}|\partial_{\xi}^{\eta}{\mathfrak{b}}(\xi)|\langle\xi\rangle^{-(\alpha-3)+\eta}\leq_{\alpha,\eta}\rVert
f\rVert_{H^{\eta+3}(\mathbb{T})},\text{ for all $\eta\in\mathbb{N}_{0}$.}$
(2.117)

###### Proof.

Since (2.117) concerns only large $|\xi|$, we will assume that $|\xi|\geq 3$.
We also assume that $\xi>0$, since the case where $\xi<0$ can be done in the
same manner. We argue by induction.

Proof for $\eta=0$. For $\eta=0$, the integration by parts gives us that

$\displaystyle\mathfrak{b}(\xi)$
$\displaystyle=\frac{1}{\text{i}\xi}\int_{-\pi}^{\pi}((2-\alpha)y|y|^{-\alpha}f(y)+|y|^{2-\alpha}f^{\prime}(y)e^{-\text{i}\xi
y})dy$
$\displaystyle=-\frac{\text{i}}{\xi}\int_{-\pi}^{\pi}(2-\alpha)y|y|^{-\alpha}f(y)e^{-\text{i}\xi
y}dy-\frac{1}{\xi^{2}}\int_{-\pi}^{\pi}\frac{d}{dy}(|y|^{2-\alpha}f^{\prime}(y))e^{-\text{i}\xi
y}dy$
$\displaystyle=:-\frac{\text{i}}{\xi}\int_{-\pi}^{\pi}(2-\alpha)y|y|^{-\alpha}f(y)e^{-\text{i}\xi
y}dy+\mathfrak{b}_{1}(\xi).$

Since $\alpha\in(1,2)$, the second integral already satisfies (2.117), that is

$\displaystyle\sup_{\xi\in\mathbb{R}}|\mathfrak{b}_{1}(\xi)|\langle\xi\rangle^{-(\alpha-3)}\leq\sup_{\xi\in\mathbb{R}}\langle\xi\rangle^{1-\alpha}\rVert
f\rVert_{W^{2,\infty}(\mathbb{T})}\leq\rVert f\rVert_{H^{3}(\mathbb{T})}.$

For the first integral, using the change of variables ($\xi y\to y$), we see
that

$\displaystyle\frac{\text{i}}{\xi}\int_{-\pi}^{\pi}y|y|^{-\alpha}f(y)e^{-\text{i}\xi
y}dy=\text{i}|\xi|^{\alpha}\xi^{-3}\int_{-\xi\pi}^{\xi\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy.$
(2.118)

We choose $j\in\mathbb{\mathbb{Z}}$ such that $\xi\in(j-1/2,j+1/2]$. Then, we
can write the integral above as

$\displaystyle\int_{-\xi\pi}^{\xi\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy=\int_{\pi}^{j\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy+\int_{-j\pi}^{-\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy+\mathfrak{b}_{2}(\xi),$

where $\sup_{\xi\in\mathbb{R}}|\mathfrak{b}_{2}(\xi)|\leq C\rVert
f\rVert_{L^{\infty}(\mathbb{T})}$ for some $C>0$. In view of (2.118), it
suffices to show that

$\displaystyle\left|\int_{\pi}^{j\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy+\int_{-j\pi}^{-\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy\right|\leq_{\alpha}\rVert
f\rVert_{H^{3}(\mathbb{T})}.$ (2.119)

To this end, we write the first integral as

$\displaystyle\int_{\pi}^{j\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy$
$\displaystyle=\int_{\pi}^{j\pi}y|y|^{-\alpha}f(y/\xi)\frac{1}{-\text{i}}\frac{d}{dy}(e^{-\text{i}y})dy$
$\displaystyle=\frac{1}{\text{i}}\int_{\pi}^{j\pi}\left((1-\alpha)y^{-\alpha}f(y/\xi)+\frac{1}{\xi}y^{1-\alpha}f^{\prime}(y/\xi)\right)e^{-\text{i}y}dy+(y^{1-\alpha}f(y/\xi)\frac{1}{-\text{i}}\bigg{|}_{y=\pi}^{y=j\pi},$

therefore,

$\left|\int_{\pi}^{j\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy\right|\leq_{\alpha}\left(\rVert
f\rVert_{W^{1,\infty}(\mathbb{T})}\frac{j}{|\xi|}\right)\leq C\rVert
f\rVert_{H^{3}(\mathbb{T})},$

where the last inequality follows from $\xi\in(j-1/2,j+1/2]$ and the Sobolev
embedding theorem. Similarly, we obtain
$\left|\int_{-j\pi}^{-\pi}y|y|^{-\alpha}f(y/\xi)e^{-\text{i}y}dy\right|\leq_{\alpha}\rVert
f\rVert_{H^{3}(\mathbb{T})}$ which gives (2.119).

Proof for $\eta\geq 1$. For $\eta\geq 1$, integration by parts gives us that

$\displaystyle\partial_{\xi}^{\eta}\mathfrak{b}(\xi)=-\frac{\text{i}}{\xi}\int_{-\pi}^{\pi}(2-\alpha+\eta)|y|^{2-\alpha}y^{\eta-1}f(y)+|y|^{2-\alpha}y^{\eta-1}(yf^{\prime}(y))e^{-\text{i}\xi
y}dy.$ (2.120)

Using the induction hypothesis, we have that

$\sup_{\xi\in\mathbb{R}}\left|\int_{-\pi}^{\pi}(2-\alpha+\eta)|y|^{2-\alpha}y^{\eta-1}f(y)+|y|^{2-\alpha}y^{\eta-1}(yf^{\prime}(y))e^{-\text{i}\xi
y}dy\right|\langle\xi\rangle^{-(\alpha-3)+\eta-1}\leq_{\alpha,\eta}\rVert
f\rVert_{H^{\eta+3}(\mathbb{T})}.$

Taking account of the factor $\xi^{-1}$ in (2.120), this proves the desired
estimate (2.117). ∎

Now, we turn to $S_{2,\Upsilon}(a)$ in (2.107).

######  Lemma 2.38.

For each $\eta\in\mathbb{N}_{0}$, there exists $\mu=\mu(\eta)\geq 0$ such that

$\displaystyle|S_{2,\Upsilon}|^{\text{Lip}(\gamma,\Omega)}_{\alpha-3,s,\eta}\leq_{s,\eta,\alpha,\nu}\rVert
a\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s+\mu}(\mathbb{T}^{\nu+2})},\text{ for
all $s\geq s_{0}$}.$

###### Proof.

Let us fix $\xi$ and let $j$ be the unique integer such that
$\xi\in(j-1/2,j+1/2]$. From (2.106), (2.107) and noting that
$\xi\mapsto\psi_{1}(\xi)$ is supported on $\xi\in(-1/4,1/4)$, it suffices to
show that

$\displaystyle\rVert
S_{\text{step},\Upsilon}(a)(\cdot,\cdot,\cdot,j)\rVert_{s}^{\text{Lip}(\gamma,\Omega)}\leq_{\alpha,\nu,s}\rVert
a\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s+\mu}(\mathbb{T}^{\nu+2})}\langle
j\rangle^{-\eta},\text{ for all $s\geq s_{0}$ and $\eta\in\mathbb{N}_{0}$},$
(2.121)

for some $\mu=\mu(\eta)>0$. To this end, we use the notation in (2.111). For
$s_{1},s_{2}\in\mathbb{N}_{0}$ such that $s_{1}+s_{2}=s$, we have

$\displaystyle\Delta^{k}_{12}\partial_{x}^{s_{1}}\partial_{\varphi}^{s_{2}}S_{\text{step},\Upsilon}(a)(\varphi,x,j)$
$\displaystyle=S_{\text{step},\Upsilon}(\Delta^{k}_{12}\partial_{x}^{s_{1}}\partial_{\varphi}^{s_{2}}a)=\int_{-\pi}^{\pi}\Delta^{k}_{12}f(\varphi,x,y)e^{-\text{i}jy}dy,$
$\displaystyle\Delta^{k}_{12}f(\varphi,x,y)$ $\displaystyle:=(2-2\cos
y)^{1-\frac{\alpha}{2}}(1-\psi_{1}(y))\partial_{x}^{s_{1}}\partial_{\varphi}^{s_{2}}(\Delta^{k}_{12}a(\varphi,x,x-y)).$
(2.122)

From this, we note that
$\Delta^{k}_{12}\partial_{x}^{s_{1}}\partial_{\varphi}^{s_{2}}S_{\text{step},\Upsilon}(a)$
is the $j$-th Fourier coefficient of the smooth map
$y\mapsto\Delta^{k}_{12}f(\varphi,x,y)$ for each fixed $\varphi,x$, since
$\psi_{2}(y)=1$ for $y\in(-\pi/3,\pi/3)$ and
$\text{supp}(\psi_{2})\in(-\pi/2,\pi/2)$ . Therefore,

$\displaystyle|\Delta^{k}_{12}\partial_{x}^{s_{1}}\partial_{\varphi}^{s_{2}}S_{\text{step},\Upsilon}(a)(\varphi,x,j)|$
$\displaystyle\leq\rVert\Delta^{k}_{12}f(\varphi,x,\cdot)\rVert_{H^{\eta}(\mathbb{T})}\langle
j\rangle^{-\eta}\leq\rVert\Delta^{k}_{12}f\rVert_{H^{s+\mu(\nu)+\eta}(\mathbb{T}^{\nu+2})}\langle
j\rangle^{-\eta},$ (2.123)

for some $\mu(\nu)>0$, where the last inequality follows from the usual
Sobolev embedding theorem. From (2.122), it is clear that

$\rVert\Delta^{k}_{12}f\rVert_{H^{s+\mu(\nu)+\eta}(\mathbb{T}^{\nu+2})}\leq_{\nu}\rVert\Delta^{k}_{12}a\rVert_{H^{s+\mu(\nu)+\eta}(\mathbb{T}^{\nu+2})}\leq\rVert
a\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s+\mu(\nu)+\eta}(\mathbb{T}^{\nu+2})}.$

Therefore, combining this with (2.123), we get

$\displaystyle\rVert
S_{\text{step},\Upsilon}(a)(\cdot,\cdot,\cdot,j)\rVert^{\text{Lip}(\gamma,\Omega)}_{s}$
$\displaystyle\leq\sup_{\omega,\omega_{1},\omega_{2}\in\Omega}\sum_{s_{1}+s_{2}=s}|\Delta^{k}_{12}\partial_{x}^{s_{1}}\partial_{\varphi}^{s_{2}}S_{\text{step},\Upsilon}(a)(\cdot,\cdot,j)|_{L^{\infty}(\mathbb{T}^{\nu+1})}$
$\displaystyle\leq_{s}\rVert
a\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s+\mu(\nu)+\eta}(\mathbb{T}^{\nu+2})}\langle
j\rangle^{-\eta},$

which proves (2.121). ∎

The following proposition follows straightforwardly from Lemma 2.36, Lemma
2.38, (2.108) and (2.109):

######  Lemma 2.39.

Given a function $a=a(\omega,\varphi,x,y)$, $S_{\Upsilon}(a)$ defined in
(2.108) is a classical symbol in $S^{\alpha-3}$ and it satisfies

$Op(S_{\Upsilon}(a))=\Upsilon^{\alpha-3}_{a},$

with the estimates

$\displaystyle|S_{\Upsilon}(a)|^{\text{Lip}(\gamma,\Omega)}_{\alpha-3,s,\eta}\leq_{s,\eta,\alpha,\nu}\rVert
a\rVert^{\text{Lip}(\gamma,\Omega)}_{H^{s+\mu}(\mathbb{T}^{\nu+2})},\text{ for
all $s\geq s_{0}$.}$ (2.124)

###  2.4 Hamiltonian structure in $L^{2}_{0}$

Given a Hamiltonian $H:L^{2}_{0}\mapsto\mathbb{R}\cup\left\\{\infty\right\\}$,
we define the $L_{0}^{2}$-gradient of $H$ at $f\in L^{2}_{0}$, $\nabla H(f)$,
as the unique vector field such that

$\displaystyle\int_{\mathbb{T}}\nabla_{L^{2}_{0}}H(f)(x)g(x)dx=\frac{d}{dt}H(f+tg)\bigg{|}_{t=0}.$
(2.125)

Similarly, we define the $L^{2}_{0}$-Hessian of $H$ at $f$, $\nabla^{2}H(f)$,
as the unique linear operator such that

$\displaystyle\int_{\mathbb{T}}\nabla_{L^{2}_{0}}^{2}H(f)[h](x)g(x)dx=\frac{d^{2}}{dsdt}H(f+th+sg)\bigg{|}_{s=t=0}.$
(2.126)

Clearly, the gradient and the Hessian of $H$ above are well-defined thanks to
the classical Riesz representation theorem.

We consider the symplectic form $\sigma$ on $L^{2}_{0}$, given by

$\displaystyle\sigma(f,g):=\int_{\mathbb{T}}\left(\partial_{x}^{-1}f\right)(x)g(x)dx.$
(2.127)

We define the Hamiltonian vector field $X_{H}:L^{2}_{0}\mapsto L^{2}_{0}$ as
the unique vector field such that

$\displaystyle(\nabla_{L^{2}_{0}}H(f),h)_{L^{2}}=\sigma(X_{H}(f),h),$

Therefore, from (2.127), we see that the Hamiltonian vector field can be
written as

$X_{H}(f)=\partial_{x}\nabla_{L^{2}_{0}}H(f).$

Let $\Phi:L^{2}_{0}\mapsto L^{2}_{0}$ be a symplectic transformation, that is,

$\sigma(\Phi(f),\Phi(g))=\sigma(f,g),\quad\text{ for all $f,g\in L^{2}_{0}$.}$

Using the definition of $\sigma$ in (2.127), it is clear that a linear
operator $\Phi$ is a symplectic transformation if and only if

$\displaystyle\Phi^{T}\partial_{x}^{-1}\Phi=\partial_{x}^{-1}.$ (2.128)

#####  2.4.0.1 Hamiltonian operators

We define a class of linear operators on $C^{\infty}_{\varphi,x}$, generated
by a Hamiltonian.

######  Definition 2.40.

Let $M=M(\varphi):L^{2}_{0}\mapsto L_{0}^{2}$ be a $\varphi$-dependent linear
operator. We say that a linear operator
$\mathcal{L}:C^{\infty}_{\varphi,x}\mapsto C^{\infty}_{\varphi,x}$ of the form

$\displaystyle\mathcal{L}[h]=\omega\cdot\partial_{\varphi}h-\partial_{x}M[h],$
(2.129)

is Hamiltonian, if $M(\varphi):H_{S^{\perp}}\mapsto H_{S^{\perp}}$ is
symmetric, that is $M(\varphi)=M(\varphi)^{T}$ for each
$\varphi\in\mathbb{T}^{\nu}$.

Let $\Phi=\Phi(\varphi)$ be a symplectic transformation for each $\varphi$.
The conjugation of the linear operator of the form in (2.129) with $\Phi$ is
also Hamiltonian, that is,

$\displaystyle\Phi^{-1}\mathcal{L}\Phi=\omega\partial_{\varphi}-\partial_{x}{N},\quad
N=\omega\cdot\partial_{\varphi}(\Phi^{T})\partial_{x}^{-1}\Phi+\Phi^{T}M\Phi.$

Indeed, $N$ is symmetric, since differentiating (2.128), we have
$\omega\cdot\partial_{\varphi}(\Phi^{T})\partial_{x}^{-1}\Phi+\Phi^{T}\partial_{x}^{-1}\omega\cdot\partial_{\varphi}(\Phi)=0$.
Therefore we have the following:

######  Lemma 2.41.

If $\mathcal{L}$ is a Hamiltonian operator and $\Phi$ is a symplectic
transformation, then $\Phi^{-1}\mathcal{L}\Phi$ is also a Hamiltonian
operator.

#####  2.4.0.2 Homogeneous Hamiltonians

Typically, a Hamiltonian $H$ in $L^{2}_{0}$ is not well-defined everywhere in
$L^{2}_{0}$ but its $L^{2}_{0}$-gradient, $\nabla_{L^{2}_{0}}H:f\mapsto\nabla
H(f)$, can be defined as a nonlinear operator between two Sobolev spaces. We
specify an elementary notion arising from the $L^{2}_{0}$-gradient of a
Hamiltonian.

######  Definition 2.42.

We say a $k$-linear map $A:(C^{\infty}_{x})^{k}\mapsto C^{\infty}_{x}$ is
$\mu$-regular, if

$\displaystyle\sup_{f=(f_{1},\ldots,f_{k})\in(C^{\infty}_{x})^{k}}\frac{\rVert
A(f)\rVert_{H^{s}_{x}}}{\sum_{i=1}^{k}\left(\rVert
f_{i}\rVert_{H^{s+\mu}_{x}}\Pi_{j=1,\ j\neq i}^{k}\rVert
f_{j}\rVert_{H^{\mu}_{x}}\right)}<\infty,\text{ for all $s\geq 0$.}$

For example, a trilinear map $A(f):=\partial_{x}^{2}(f)f^{2}$, is
$\mu$-regular, for some $\mu\geq 0$, which can be seen from (2.6). We also
specify the notion of a homogeneous Hamiltonian in $L^{2}_{0}$:

######  Definition 2.43.

We say a Hamiltonian is $\mu$-regular if $H$ admits a homogeneous expansion,

$H:=\sum_{k=2}^{\infty}H_{k},$

where $H_{k}$ is homogeneous of degree $k$ such that each
$\nabla_{L^{2}_{0}}H_{k}$, which is a $(k-1)$-linear map, is $\mu$-regular.

###  2.5 Time-reversible Hamiltonians

In this subsection, we briefly recall some basic properties of a reversible
Hamiltonian system.

######  Definition 2.44.

Let $X$ be a Hilbert space and $\rho:X\mapsto X$ be an involution, that is,
$\rho\circ\rho(f)=f,$ for all $f\in X$. We say that a Hamiltonian $H$ on $X$
is reversible (or time-reversible) with respect to $\rho$, if $H\circ\rho=H$.
We also say that $H$ is reversibility preserving if $H\circ\rho=-H$.
Furthermore, we say that a transformation $\Phi:X\mapsto X$ is reversibility
preserving if $\rho\circ\Phi\circ\rho=\Phi$.

#####  2.5.0.1 Reversible Hamiltonian on $L^{2}_{0}$.

We denote by $\rho:L^{2}_{0}\mapsto L^{2}_{0}$, the involution given as

$\displaystyle\rho(f)(x):=f(-x).$ (2.130)

######  Lemma 2.45.

Let $G$ be a reversibility preserving Hamiltonian in $L^{2}_{0}$ with respect
to an involution $\rho$, and let $\Phi_{G}^{t}$ be the unique flow map
determined by the Hamiltonian PDE at time $t$:

$\displaystyle f_{t}=X_{G}(f).$ (2.131)

Then, $\Phi^{t}_{G}$ is a reversibility preserving map.

###### Proof.

Since $G\circ\rho=-G$, differentiating $G(\rho(f))=-G(f)$ in $f$, we have
$\rho\left(\nabla_{L^{2}_{0}}G(\rho(f))\right)=-\nabla_{L^{2}_{0}}G(f).$
Hence, taking $\partial_{x}$ on both sides, we see that

$\displaystyle\rho X_{G}(\rho f)=X_{G}(f),\text{ for all $f\in L^{2}_{0}$}.$
(2.132)

In order to show that $\Phi_{G}^{t}$ is reversibility preserving, we need to
show that

$\displaystyle\rho\Phi^{t}_{G}(\rho(f))=\Phi^{t}_{G}(f).$ (2.133)

By its definition, $\Phi_{G}^{t}$ solves
$\partial_{t}\Phi^{t}_{G}(f)=X_{G}(\Phi^{t}_{G}(f))$ with $\Phi_{G}^{0}(f)=f$.
Furthermore, $\rho\Phi_{G}^{t}(\rho f)$ also solves

$\displaystyle\partial_{t}\left(\rho\Phi_{G}^{t}(\rho f)\right)=\rho
X_{G}(\Phi_{G}^{t}(\rho
f))\overset{\eqref{Hamitlon_rev}}{=}X_{G}(\rho\Phi_{G}^{t}(\rho f)),\text{
with }\rho\Phi^{t}_{G}(\rho f)|_{t=0}=\rho\circ\rho(f)=f.$

Since the flow map $\Phi^{t}_{G}$ is unique, we have $\rho\Phi_{G}^{t}(\rho
f)=f$ and this implies (2.133).∎

######  Definition 2.46.

[6, Definition 2.2] Let

$\displaystyle X:=\left\\{f\in
C^{\infty}_{\varphi,x}:f(-\varphi,-x)=f(\varphi,x)\right\\},\quad
Y:=\left\\{f\in C^{\infty}_{\varphi,x}:f(-\varphi,-x)=-f(\varphi,x)\right\\}$
(2.134)

We say that a linear operator $\mathcal{A}:C^{\infty}_{\varphi,x}\mapsto
C^{\infty}_{\varphi,x}$ is

  1. 1.

reversible if $X\mapsto Y$ and $Y\mapsto X$.

  2. 2.

reversibility preserving $X\mapsto X,\ Y\mapsto Y$.

######  Definition 2.47.

We say a symbol $\mathfrak{a}$ is a reversible symbol, if
$\partial_{x}Op^{W}(\mathfrak{a})$ is a reversible operator. We say
$\mathfrak{a}$ is a reversibility preserving symbol, if
$\partial_{x}Op^{W}(\mathfrak{a})$ is a reversibility preserving operator.

In view of the real operators (Definition 2.5), symmetric operators, and
reversible/reversibility preserving operators (Definition 2.47), one can
easily show from the definition of Weyl quantization in (2.13) that

$\displaystyle\text{$\partial_{x}Op^{W}(\mathfrak{a})$ is a real operator if
$\mathfrak{a}(\varphi,x,\xi)=\overline{a}(\varphi,x,-\xi)$},$ (2.135)
$\displaystyle\text{$Op^{W}(\mathfrak{a})$ is a symmetric operator if
$\mathfrak{a}(\varphi,x,\xi)=\overline{\mathfrak{a}}(\varphi,x,\xi)$},$
$\displaystyle\text{$\partial_{x}Op^{W}(\mathfrak{a})$ is a reversible
operator if $\mathfrak{a}(\varphi,x,\xi)=\mathfrak{a}(-\varphi,-x,-\xi)$}.$

For a general linear operator $\mathcal{A}$, we have, using the notation in
(2.9),