#  Cauchy problem for operators with triple effectively hyperbolic
characteristics
– Ivrii’s conjecture –

Tatsuo Nishitani††Department of Mathematics, Osaka University:
<EMAIL_ADDRESS>
###### Abstract

Ivrii’s conjecture asserts that the Cauchy problem is $C^{\infty}$ well-posed
for any lower order term if every singular point of the characteristic variety
is effectively hyperbolic. Effectively hyperbolic singular point is at most
triple characteristic. If every characteristic is at most double this
conjecture has been proved in 1980’. In this paper we prove the conjecture for
the remaining cases, that is for operators with triple effectively hyperbolic
characteristics.

Keywords: Ivrii’s conjecture, Bézout matrix, triple characteristic,
effectively hyperbolic, Cauchy problem, Weyl calculus, Tricomi operator.

Mathematics Subject Classification 2010: Primary 35L30, Secondary 35G10

##  1 Introduction

This paper is devoted to the Cauchy problem

(1.1)
$\left\\{\begin{array}[]{ll}Pu=D_{t}^{m}u+\sum_{j=0}^{m-1}\sum_{|\alpha|+j\leq
m}a_{j,\alpha}(t,x)D_{x}^{\alpha}D_{t}^{j}u=0,\\\\[8.0pt]
D_{t}^{j}u(0,x)=u_{j}(x),\quad j=0,\ldots,m-1\end{array}\right.$

where $t\geq 0$, $x\in{\mathbb{R}}^{d}$ and the coefficients
$a_{j,\alpha}(t,x)$ are real valued $C^{\infty}$ functions in a neighborhood
of the origin of ${\mathbb{R}}^{1+d}$ and
$D_{x}=(D_{x_{1}},\ldots,D_{x_{d}})=D$, $D_{x_{j}}=(1/i)(\partial/\partial
x_{j})$ and $D_{t}=(1/i)(\partial/\partial t)$. The Cauchy problem (1.1) is
$C^{\infty}$ well-posed at the origin for $t\geq 0$ if one can find a
$\delta>0$ and a neighborhood $U$ of the origin of ${\mathbb{R}}^{d}$ such
that (1.1) has a unique solution $u\in C^{\infty}([0,\delta)\times U)$ for any
$u_{j}(x)\in C^{\infty}({\mathbb{R}}^{d})$. We assume that the principal
symbol of $P$

$p(t,x,\tau,\xi)=\tau^{m}+\sum_{j=0}^{m-1}\sum_{|\alpha|+j=m}a_{j,\alpha}(t,x)\xi^{\alpha}\tau^{j}$

is hyperbolic for $t\geq 0$, that is there exist $\delta^{\prime}>0$ and a
neighborhood $U^{\prime}$ of the origin such that

(1.2) $p=0$ has only real roots in $\tau$ for
$(t,x)\in[0,\delta^{\prime})\times U^{\prime}$ and $\xi\in{\mathbb{R}}^{d}$

which is indeed necessary in order that the Cauchy problem (1.1) is
$C^{\infty}$ well-posed near the origin for $t\geq 0$ ([19], [22]).

In [9], Ivrii and Petkov proved that if the Cauchy problem (1.1) is
$C^{\infty}$ well posed for any lower order term near the origin for $t\geq 0$
(such operator $P$ is called strongly hyperbolic) then the Hamilton map
$F_{p}$ has a pair of non-zero real eigenvalues at every singular point of
$p=0$ located in $[0,\delta^{\prime\prime})\times
U^{\prime\prime}\times({\mathbb{R}}^{d+1}\setminus 0)$ ([9, Theorem 3]) with
some $\delta^{\prime\prime}>0$ and a neighborhood $U^{\prime\prime}$ of $x=0$.
With $X=(t,x)$, $\Xi=(\tau,\xi)$ the Hamilton map $F_{p}$ is defined by

$F_{p}(X,\Xi)=\begin{bmatrix}\displaystyle{\frac{\partial^{2}p}{\partial
X\partial\Xi}}&\displaystyle{\frac{\partial^{2}p}{\partial\Xi\partial\Xi}}\\\\[10.0pt]
\displaystyle{-\frac{\partial^{2}p}{\partial X\partial
X}}&\displaystyle{-\frac{\partial^{2}p}{\partial\Xi\partial X}}\end{bmatrix}.$

A singular point of $p=0$ where the Hamilton map $F_{p}$ has a pair of non-
zero real eigenvalues is called effectively hyperbolic ([6], [12]). He has
conjectured that the converse would be also true, that is if every singular
point of $p=0$ is effectively hyperbolic then the Cauchy problem is
$C^{\infty}$ well posed for any lower order term. If a singular point
$(t,x,\tau,\xi)$ is effectively hyperbolic then $\tau$ is a characteristic
root of multiplicity at most $2$ if $t>0$ and at most $3$ if $t=0$ ([9, Lemma
8.1]). When every multiple characteristic root is at most double, the
conjecture has been proved for some special class in [10], [21] and for the
general case in [13, 14, 15], [23, 26].

For the case when we have triple effectively hyperbolic characteristic, Ivrii
has also proved in [10] that the conjecture is true if $p$ admits a
factorization $p=q_{1}q_{2}$ near singular points with real smooth symbols
$q_{i}$, transforming the original $P$, by means of operator powers of
evolution generators, to an operator for which a parametrix can be
constructed. In this case a singular point is effectively hyperbolic if and
only if the Poisson bracket $\\{q_{1},q_{2}\\}$ does not vanish there. If
$m=3$ it is clear that, for such a factorization to exist, it is necessary
that the equation $p=0$ has a $C^{\infty}$ real root $\tau=\tau(t,x,\xi)$ near
a conic neighborhood of singular points. A typical example is

$p=q_{1}q_{2},\quad q_{1}=\tau^{2}-t|\xi|^{2},\;q_{2}=\tau$

where $q_{1}$ is the Tricomi operator (symbol). Note that $p$ has a complex
characteristic root if $t<0$. This is a common feature. In fact if
$p(0,0,\tau,\xi)=0$ has a triple characteristic root and
$F_{p}(0,0,\tau,\xi)\neq O$ then $p$ has necessarily non-real characteristic
roots in the $t<0$ side near $(0,0,\xi)$ ([9, Lemma 8.1]).

When $m=3$, without restrictions we can assume that $p$ has the form

$p=\tau^{3}-a(t,x,\xi)|\xi|^{2}\tau+b(t,x,\xi)|\xi|^{3}$

hence the condition (1.2) is reduced to $\Delta=4\,a^{3}-27\,b^{3}\geq 0$ for
$(t,x)\in[0,\delta^{\prime})\times U^{\prime}$, $|\xi|=1$. Also note that
$(0,0,0,{\bar{\xi}})$ is a triple characteristic then $(0,0,0,{\bar{\xi}})$ is
effectively hyperbolic if and only if $\partial_{t}a(0,0,{\bar{\xi}})\neq 0$
hence $a(t,x,\xi)>0$ microlocally for small $t>0$. In [2], Bernardi, Bove,
Petkov investigated the case that $p$ has a triple effectively hyperbolic
characteristic but $p$ may not be be factorized. They studied $P$ with the
principal symbol

$\displaystyle
p=\tau^{3}-(ta_{2}(t,x,\xi)+\alpha(x,\xi))|\xi|^{2}\tau+t^{2}b_{3}(t,x,\xi)|\xi|^{3}$

where $a_{2}(t,x,\xi)\geq c^{\prime}>0$, $\alpha(x,\xi)\geq 0$ and proved the
conjecture for such $P$, deriving weighted energy estimates by separating
(multiplier) operator method. Note that $\Delta\geq c\,a^{3}$ holds with some
$c>0$ for this $p$. They also proved that if $b_{3}(0,0,{\bar{\xi}})\neq 0$
then a smooth factorization $p=q_{1}q_{2}$ is possible only if
$\alpha(x,{\bar{\xi}})=0$ for all $x$ near $x=0$. This result was extended in
[28, 30] such that the conjecture is true if $\Delta\geq c\,ta^{2}$ or if
$\Delta\geq c\,t^{2}a$ with some additional conditions, where after reducing
the original equation to a first order $3\times 3$ system, a symmetrizer $S$
is constructed and used to get weighted energy estimates. These results are
concerned with the case that $p$ is strictly hyperbolic in $t>0$, while in
general case, double effectively hyperbolic characteristics in $t>0$ (where
$\Delta=0$) approaching to a triple effectively hyperbolic characteristic on
$t=0$ might exist and we must handle them. Moreover the following example
([28])

(1.3)
$p(t,x,\tau,\xi)=\tau^{3}-(t+\alpha(x))\xi^{2}\tau+(t^{m}/2-t)\sqrt{\alpha(x)}\,\xi^{3},\;\;x,\xi\in{\mathbb{R}}$

where $\alpha(x)\geq 0$ and $\sqrt{\alpha(x)}$ is smooth, suggests that it is
not enough just to study the zeros themselves of $\Delta$. Indeed since
$\Delta=(t-2\alpha)^{2}(4t+\alpha)+27t^{m+1}\alpha(1-t^{m-1}/4)$ so that
$\Delta>0$ for small $t>0$, while $\Delta=27\cdot
2^{m+1}\alpha^{m+2}(1-2^{m-3}\alpha^{m-1})$ if $t=2\alpha$ hence one has no
estimate such as $\Delta\geq c\,t^{k}(t+\alpha)^{q}$ with $c>0$ for small
$\alpha>0$ if $m>k+q-2$.

In [31] we employed a new idea which is to diagonalize the symmetrizer $S$
mentioned above so that the system is transformed to a system with a diagonal
symmetrizer. We see that three diagonal entries (the eigenvalues of $S$) are
bounded from below by $\Delta/a$, $a$, $1$ respectively and we recognize here
a close relation between the diagonal symmetrizer and two discriminants
$\Delta$ of $p=0$ and $\Delta^{\prime}(=a)$ of $\partial_{\tau}p=0$. In
example (1.3) we see $\Delta/a\geq(t-2\alpha)^{2}$ which looks like
$\tau^{2}-(\Delta/a)|\xi|^{2}$ has double effectively hyperbolic
characteristics on $t=2\alpha$ though $\Delta\neq 0$ there (see [26, 27]).
When the coefficients of $p$ depend only on $t$, the behavior of
$\Delta(t,\xi)/a(t,\xi)$ can be analyzed relatively easily. Writing
$\Delta(t,\xi)=e\prod(t-\nu_{j}(\xi))$ and dividing $[0,T]$ into subintervals
with the end points ${\mathsf{Re}}\,\nu_{j}(\xi)$ we can obtain suitable
estimates of $\Delta/a$ from below in each subinterval. In particular, in this
way, we have proved the conjecture for the $t$ dependent case [31, Theorem
4.1]. In this paper we extend this idea and apply it to the general case of
which outline is given in the next section.

For this class of operators we have always a loss of regularity then the way
to obtain microlocal energy estimates for operators of order $m$ from that of
order $2$ or $3$ and the way to prove local $C^{\infty}$ well-posed results
from such obtained microlocal energy estimates with loss of regularity is not
so straightforward.

Finally we note that if there is a triple characteristic which is not
effectively hyperbolic the Cauchy problem is not well-posed in any Gevrey
class of order $s>2$ in general, even though the subprincipal symbol vanishes
identically ([3]).

In this paper we prove

######  Theorem 1.1.

Assume (1.2). If every singular point $(0,0,\tau,\xi)$, $|(\tau,\xi)|\neq 0$
of $p=0$ is effectively hyperbolic then for any $a_{j,\alpha}(t,x)$ with
$j+|\alpha|\leq m-1$, which are $C^{\infty}$ in a neighborhood of $(0,0)$,
there exist $\delta>0$, a neighborhood $U$ of the origin and $n>0$, $\ell>0$
such that for any $s\in{\mathbb{R}}$ and any $f$ with
$t^{-n+1/2}\langle{D}\rangle^{s}f\in L^{2}((0,\delta)\times{\mathbb{R}}^{d})$
there exists $u$ with $t^{-n-1/2}\langle{D}\rangle^{-\ell+s+m-j}D_{t}^{j}u\in
L^{2}((0,\delta)\times{\mathbb{R}}^{d})$, $j=0,1,\ldots,m-1$, satisfying
$Pu=f$ in $(0,\delta)\times U$.

Here $\langle{D}\rangle$ stands for$\sqrt{1+|D|^{2}}$ and $n$, $\ell$ are
given by

$n=12\sqrt{2}\,\sup\frac{|P_{sub}(0,0,\tau,\xi)|}{e(0,0,\tau,\xi)}+{\bar{C}^{*}},\quad\ell=k\,(n+2)$

where $P_{sub}$ denotes the subprincipal symbol and $e(0,0,\tau,\xi)$ is the
positive real eigenvalue of $F_{p}(0,0,\tau,\xi)$ and the supremum is taken
over all singular points $(0,0,\tau,\xi)$ with $|(\tau,\xi)|=1$ of $p=0$ and
${\bar{C}^{*}}$ is a constant depending only on the principal symbol $p$. Here
$k$ is the maximal number of singular points $(0,0,\tau,\xi)$ of $p=0$ with
$|(\tau,\xi)|=1$ hence $k\leq[m/2]$. For more detailed estimate of
${\bar{C}^{*}}$ see (3.14) and (10.2) below. The constant $12\sqrt{2}$ may not
be the best.

######  Theorem 1.2.

Under the same assumption as in Theorem 1.1, for any $a_{j,\alpha}(t,x)$ with
$j+|\alpha|\leq m-1$, which are $C^{\infty}$ in a neighborhood of $(0,0)$,
there exist $\delta>0$ and a neighborhood $U$ of the origin such that for any
$u_{j}(x)\in C^{\infty}_{0}({\mathbb{R}}^{d})$, $j=0,1,\ldots,m-1$, there
exists $u(t,x)\in C^{\infty}([0,\delta)\times U)$ satisfying (1.1) in
$[0,\delta)\times U$. If $u(t,x)\in C^{\infty}([0,\delta)\times U)$ with
$\partial_{t}^{j}u(0,x)=0$, $j=0,1,\ldots,m-1$, satisfies $Pu=0$ in
$[0,\delta)\times U$ then $u=0$ in a neighborhood of $(0,0)$.

###### Proof.

Compute $u_{j}(x)=D_{t}^{j}u(0,x)$ for $j=m,m+1,\ldots$ from $u_{j}(x)$,
$j=0,1,\ldots,m-1$ and the equation $Pu=0$. By a Borel’s lemma there is
$w(t,x)\in C_{0}^{\infty}({\mathbb{R}}^{1+d})$ such that
$D_{t}^{j}w(0,x)=u_{j}(x)$ for all $j\in{\mathbb{N}}$. Since
$(D_{t}^{j}Pw)(0,x)=0$ for all $j\in{\mathbb{N}}$ it is clear that
$t^{-n+1/2}\langle{D}\rangle^{s}Pw\in L^{2}((0,\delta)\times{\mathbb{R}}^{d})$
for any $s$. Thanks to Theorem 1.1 there exists $v$ with
$t^{-n-1/2}\langle{D}\rangle^{-\ell+s+m-j}D_{t}^{j}v\in
L^{2}((0,\delta)\times{\mathbb{R}}^{d})$, $j=0,1,\ldots,m-1$ satisfying
$Pv=-Pw$ in $(0,\delta)\times U$. Since $D_{t}^{k}v\in
L^{2}((0,\delta)\times{\mathbb{R}}^{d})$ for any $k$ hence
$D_{t}^{j}v(0,x)=0$, $j=0,1,\ldots,m-1$ we conclude that $u=v+w$ is a desired
solution. Local uniqueness follows from Theorem 13.4 below because
$\partial_{t}^{k}u(0,x)=0$ for any $k\in{\mathbb{N}}$ by $Pu=0$. ∎

##  2 Outline of the proof of Theorem 1.1

As noted in Introduction, if a singular point $(t,x,\tau,\xi)$ of $p=0$ is
effectively hyperbolic then $\tau$ is a characteristic root of multiplicity at
most $3$. This implies that it is essential to study third order operator $P$;

(2.1) $P=D_{t}^{3}+\sum_{j=1}^{3}a_{j}(t,x,D)\langle{D}\rangle^{j}D_{t}^{3-j}$

which is a differential operator in $t$ with coefficients $a_{j}\in S^{0}$,
classical pseudodifferential operators of order $0$, where
$\langle{D}\rangle={\rm op}({(1+|\xi|^{2})^{1/2}})$. One can reduce $P$ to the
case with $a_{1}(t,x,D)=0$ and hence the principal symbol is

(2.2)
$p(t,x,\tau,\xi)=\tau^{3}-a(t,x,\xi)\langle{\xi}\rangle^{2}\tau-b(t,x,\xi)\langle{\xi}\rangle^{3}.$

All characteristic roots are real for $t\geq 0$ implies that

(2.3) $\Delta=4\,a(t,x,\xi)^{3}-27\,b(t,x,\xi)^{2}\geq
0,\quad(t,x,\xi)\in[0,T)\times U\times{\mathbb{R}}^{d}.$

Assume that $p(0,0,\tau,{\bar{\xi}})=0$ has a triple characteristic root
$\tau={\bar{\tau}}$, necessarily ${\bar{\tau}}=0$. The singular point
$(0,0,{\bar{\tau}},{\bar{\xi}})$ is effectively hyperbolic if and only if

(2.4) $\partial_{t}a(0,0,{\bar{\xi}})\neq 0$

hence one can write $a=e(t+\alpha(x,\xi))$ where $e>0$ and $\alpha\geq 0$.
From the conditions (2.3) and (2.4) the discriminant $\Delta$ is essentially a
third order polynomial in $t$. In Section 3, for regularized
$a_{\epsilon}=e(t+\alpha+\epsilon^{2})$ and the corresponding discriminant
$\Delta_{\epsilon}$ we construct a smooth $\psi_{\epsilon}$ such that

(2.5)
${\Delta_{\epsilon}}\geq{c}\min{\big{\\{}t^{2},(t-\psi_{\epsilon})^{2}\big{\\}}}\,(t+\rho_{\epsilon}),\quad
t\geq 0$

where $\rho_{\epsilon}=\alpha+\epsilon^{2}$. In Section 4, introducing a large
parameter $M$, we localize the coefficients near reference point
$(0,{\bar{\xi}})$ replacing $(x,\xi)$ by localized coordinates
$\chi(M^{2}x)x$,
$\chi(M^{2}(\xi/\langle{\xi}\rangle-{\bar{\xi}}))(\xi-{\bar{\xi}}\langle{\xi}\rangle)+{\bar{\xi}}\langle{\xi}\rangle$
and taking $\epsilon=M^{1/2}\langle{\xi}\rangle^{-1/2}$ where $\chi\in
C_{0}^{\infty}$ is $1$ near $0$. At this point related symbols are localized
near $(0,{\bar{\xi}})$ (defined in ${\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$)
and (2.5) yields

(2.6) $\Delta/a\geq
c\min{\big{\\{}t^{2},(t-\psi)^{2}+M\rho\langle{\xi}\rangle^{-1}\big{\\}}},\quad
t\geq 0.$

We also estimate such localized symbols in terms of the localized $\rho$. In
particular we show that

$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi(x,\xi)\big{|}\precsim\rho(x,\xi)^{1-|\alpha+\beta|/2}\langle{\xi}\rangle^{-|\beta|}$

where, and from now on $A\precsim B$ means that $A$ is bounded by constant,
independent of parameters, times $B$.

One of main arguments in the paper is to reduce the original equation to a
first order $3\times 3$ system with diagonal symmetrizer. With
$U={{}^{t}}(D_{t}^{2}u,\langle{D}\rangle D_{t}u,\langle{D}\rangle^{2}u)$ the
equation $Pu=f$ is reduced to $D_{t}U=A(t,x,D)\langle{D}\rangle U+B(t,x,D)U+F$
where $A,B\in S^{0}$, $F={{}^{t}}(f,0,0)$ and

$A(t,x,\xi)=\begin{bmatrix}0&a&b\\\ 1&0&0\\\ 0&1&0\end{bmatrix}.$

Let $S$ be the Bézout matrix of $p$ and $\partial p/\partial\tau$, that is

$S(t,x,\xi)=\begin{bmatrix}3&0&-a\\\ 0&2a&3b\\\ -a&3b&a^{2}\end{bmatrix}$

then $S$ is positive semidefinite and symmetrizes $A$, that is $SA$ is
symmetric. We now diagonalize $S$ by an orthogonal matrix $T$ so that
$T^{-1}ST=\varLambda$. Then with $V={\rm op}({T^{-1}})U$ the system is reduced
to, roughly

(2.7) $D_{t}V=A^{T}(t,x,D)\langle{D}\rangle V+{\tilde{B}}(t,x,D)V$

where $\varLambda$ is diagonal and symmetrizes $A^{T}(t,x,\xi)=T^{-1}AT$. This
reduction is carried out in Section 9 after examining $T(t,x,\xi)$ carefully
in Section 5.

Note that $\varLambda={\rm diag}\,(\lambda_{1},\lambda_{2},\lambda_{3})$ where
$0<\lambda_{1}<\lambda_{2}<\lambda_{3}$ are the eigenvalues of $S$. As
mentioned in Introduction a significant feature of $\lambda_{j}$ are

(2.8) $\Delta/a\precsim\lambda_{1}\precsim a^{2},\quad\lambda_{2}\simeq
a,\quad\lambda_{3}\simeq 1,\quad t\geq 0.$

Section 5 is devoted mainly to estimate derivatives of $\lambda_{j}$ and we
prove

$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\lambda_{j}\big{|}\precsim
a^{3-j-|\alpha+\beta|/2}\langle{\xi}\rangle^{-|\beta|},\quad t\geq 0,\quad
j=1,2,3$

which also gives detailed information on the derivatives of $T$. Since (2.7)
is a system with diagonal symmetrizer $\varLambda$, a natural energy would be
$\big{(}{\rm op}({\varLambda})V,V\big{)}=\sum_{j=1}^{3}\big{(}{\rm
op}({\lambda_{j}})V_{j},V_{j}\big{)}$ and (2.6) and (2.8) suggest that a
weighted energy with a scalar pseudodifferential weight ${\rm
op}({t^{-n}\phi^{-n}})$

$\phi=\omega+t-\psi,\quad\omega=\sqrt{(t-\psi)^{2}+M\rho\langle{\xi}\rangle^{-1}}$

would work, where ${\rm op}({\phi^{-n}})$ is chosen after the weight employed
for studying double effectively hyperbolic characteristics in [26] (see also
[27]), and satisfies

$\partial_{t}(t\phi)=\kappa\,(t\phi),\quad\kappa=1/t+1/\omega.$

In Section 6, to treat these weight functions, we introduce a $\sigma$
temperate (uniformly in $M$) metric

$g=M^{-1}(\langle{\xi}\rangle|dx|^{2}+\langle{\xi}\rangle^{-1}|dx|^{2})$

and prove that $\omega^{s}\in S(\omega^{s},g)$, $\phi^{s}\in S(\phi^{s},g)$
with $s\in{\mathbb{R}}$, uniformly in $t\geq 0$, estimating derivatives of
$\omega$, $\phi$. In Section 7 we prove that $\omega$, $\phi$ and
$\lambda_{j}$ are $\sigma,g$ temperate uniformly in $t\in[0,M^{-4}]$ (in this
paper such functions are called admissible weights for $g$, while $\sigma$ is
reserved for denoting a certain function). This fact enables us to apply the
Weyl calculus of pseudodifferential operators (see [8, Chapter 18]) to ${\rm
op}({\phi^{s}})$, ${\rm op}({\omega^{s}})$ and ${\rm op}({\lambda_{j}^{s}})$
with $s\in{\mathbb{R}}$, for example we have ${\rm op}({\phi^{s_{1}}}){\rm
op}({\phi^{s_{2}}})={\rm op}({\phi^{s_{1}}\\#\phi^{s_{2}}})$ where
$\phi^{s_{1}}\\#\phi^{s_{2}}\in S(\phi^{s_{1}+s_{2}},g)$. In Section 8 we
prove some basic facts on inverses and $L^{2}$ bounds of pseudodifferential
operators associated to the metric $g$ which enables us, for example, to write
${\rm op}({\phi^{s_{1}}}){\rm op}({\phi^{s_{2}}})={\rm op}({1+r}){\rm
op}({\phi^{s_{1}+s_{2}}})$ with $r\in S(M^{-1},g)$. We also give lower bounds
of ${\rm op}({\lambda_{j}})$ here.

In Section 10 applying the Weyl calculus of pseudodifferential operators we
estimate the weighted energy

${\mathsf{Re}}\,e^{-\theta t}({\rm op}({\varLambda}){\rm
op}({t^{-n}\phi^{-n}})V,{\rm op}({t^{-n}\phi^{-n}})V)$

and derive energy estimates for any lower order term (including the term
$M\langle{D}\rangle D_{t}$ because we have added this to the original operator
at the beginning) (Proposition 10.1). In Section 11, using energy estimates
for the system coming from the adjoint operator of $P$, which is obtained
repeating exactly the same arguments, we prove an existence result of the
Cauchy problem for $P_{\bar{\xi}}$, which is the localized operator near
$(0,{\bar{\xi}})$ of the original $P$ (Theorem11.1). In Section 12, in order
to sum up such obtained solutions (which might be considered as a microlocal
solution to the Cauchy problem near $(0,{\bar{\xi}})$), we prove that the wave
front sets of such solutions propagate with finite speed (Proposition 12.4). A
more precise picture of the propagation of wave front set of solutions is also
proved applying the same arguments (Theorem 12.1). Finally, in Section 13,
using the propagation results in Section 12 we complete the proof of Theorem
1.1.

##  3 Construction of $\psi(x,\xi)$

Study third order operators $P$ of the form (2.1) with $a_{1}(t,x,D)=0$, hence
the principal symbol has the form (2.2) where $a(t,x,\xi)$ and $b(t,x,\xi)$
are homogeneous of degree $0$ in $\xi$ and assumed to satisfy (2.3) with some
$T>0$ and some neighborhood $U$ of the origin of ${\mathbb{R}}^{d}$. Assume
that $p(t,x,\tau,\xi)$ has a triple characteristic root $\tau=0$ at
$(0,0,{\bar{\xi}})$, $|{\bar{\xi}}|=1$ and $(0,0,0,{\bar{\xi}})$ is
effectively hyperbolic. It is clear that $a(0,0,{\bar{\xi}})=0$ and
$b(0,0,{\bar{\xi}})=0$. Since
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a(0,0,{\bar{\xi}})=0$ for
$|\alpha+\beta|=1$ and
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}b(0,0,{\bar{\xi}})=0$ for
$|\alpha+\beta|\leq 2$ by (2.3) (see Lemma 4.2 below) it is easy to see that

(3.1) ${\rm det}\,\big{(}\lambda-
F_{p}(0,0,0,{\bar{\xi}})\big{)}=\lambda^{2d}\big{(}\lambda^{2}-\\{\partial_{t}a(0,0,{\bar{\xi}})\\}^{2}\big{)}$

hence $(0,0,0,{\bar{\xi}})$ is effectively hyperbolic if and only if
$\partial_{t}a(0,0,{\bar{\xi}})\neq 0$. Then there is a neighborhood
${\mathcal{U}}$ of $(0,0,{\bar{\xi}})$ in which one can write

$a(t,x,\xi)=e(t,x,\xi)(t+\alpha(x,\xi))$

where $e>0$ in ${\mathcal{U}}$. Note that $\alpha(x,\xi)\geq 0$ near
${\bar{\xi}}$ because $a(t,x,\xi)\geq 0$ in $[0,T)\times
U\times{\mathbb{R}}^{d}$.

###  3.1 A perturbed discriminant

Introducing a small parameter $\epsilon$ we consider

(3.2)
$\begin{split}\tau^{3}-e(t,x,\xi)(t+\alpha(x,\xi)+\epsilon^{2})|\xi|^{2}\tau-b(t,x,\xi)\,|\xi|^{3}\\\
=\tau^{3}-a(t,x,\xi,\epsilon)|\xi|^{2}-b(t,x,\xi)|\xi|^{3}.\end{split}$

From now on we write $b(X)$ or $a(X,\epsilon)$ and so on to make clearer that
these symbols are defined in some conic (in $\xi$) neighborhood of
${\bar{X}}=(0,{\bar{\xi}})$ or $({\bar{X}},0)$. Consider the discriminant of
(3.2); $\Delta(t,X,\epsilon)=4\,a^{3}(t,X,\epsilon)-27\,b^{2}(t,X)$.

######  Lemma 3.1.

One can write

$\Delta={\tilde{e}}(t,X,\epsilon)\big{(}t^{3}+a_{1}(X,\epsilon)t^{2}+a_{2}(X,\epsilon)t+a_{3}(X,\epsilon)\big{)}$

in a neighborhood of $(0,{\bar{X}},0)$ where $a_{j}({\bar{X}},0)=0$, $j=1,2,3$
and ${\tilde{e}}>0$.

###### Proof.

It is clear that $\partial_{t}^{k}a^{3}(0,{\bar{X}},0)=0$ for $k=0,1,2$ and
$\partial_{t}^{3}a^{3}(0,{\bar{X}},0)\neq 0$. Show
$\partial_{t}b(0,{\bar{X}},0)=0$. Suppose the contrary and hence
$b(t,{\bar{X}},0)=t(b_{1}+tb_{2}(t))$ with $b_{1}\neq 0$. Since
$a(t,{\bar{X}},0)=c\,t$ with $c>0$ then
$\Delta(t,{\bar{X}},0)=4\,c^{3}\,t^{3}-27\,b(t,{\bar{X}},0)^{2}\geq 0$ leads
to a contradiction. Thus $\partial_{t}^{k}\Delta(0,{\bar{X}},0)=0$ for
$k=0,1,2$ and $\partial_{t}^{3}\Delta(0,{\bar{X}},0)\neq 0$. Then from the
Malgrange preparation theorem (e.g. [7, Theorem 7.5.5]) one can conclude the
assertion. ∎

Introducing

(3.3) $\rho(X,\epsilon)=\alpha(X)+\epsilon^{2}$

one can also write

$\displaystyle\Delta=4e^{3}(t+\rho)^{3}-27b^{2}=4e^{3}\big{\\{}(t+\rho)^{3}-27b^{2}/(4e^{3})\big{\\}}=4e^{3}\big{\\{}(t+\rho)^{3}-{\hat{b}}^{2}\big{\\}}$

with ${\hat{b}}=3\sqrt{3}\,b/2e^{3/2}$. Denoting
${\hat{b}}(t,X)=\sum_{j=0}^{2}{\hat{b}}_{j}(X)t^{j}+{\hat{b}}_{3}(t,X)t^{3}$
where ${\hat{b}}_{0}({\bar{X}})={\hat{b}}_{1}({\bar{X}})=0$ which is clear
from the proof of Lemma 3.1, one can write

(3.4)
$\begin{split}\Delta/{\tilde{e}}={\bar{\Delta}}=t^{3}+a_{1}(X,\epsilon)t^{2}+a_{2}(X,\epsilon)t+a_{3}(X,\epsilon)\\\
=E\,\big{\\{}(t+\rho)^{3}-\big{(}\sum_{j=0}^{2}{\hat{b}}_{j}(X)t^{j}+{\hat{b}}_{3}(t,X)t^{3}\big{)}^{2}\big{\\}}\end{split}$

with $E(t,X,\epsilon)=4e^{3}/{\tilde{e}}$. Here note that
$E(0,{\bar{X}},0)=1$.

######  Lemma 3.2.

There is a neighborhood $V$ of ${\bar{X}}$ such that
$\big{|}{\hat{b}}_{1}(X)\big{|}\leq 4\,\alpha^{1/2}(X)$ for $X\in V$.

###### Proof.

It is clear that $|{\hat{b}}_{0}(X)|\leq\alpha^{3/2}(X)$. If $\alpha(X)=0$
then the assertion is obvious. Assume $\alpha(X)\neq 0$. Since

(3.5)
$(t+\alpha(X))^{3}\geq\big{(}\sum_{j=0}^{2}{\hat{b}}_{j}(X)t^{j}+{\hat{b}}_{3}(t,X)t^{3}\big{)}^{2},\quad
0\leq t\leq T$

choosing $t=3\alpha(X)\leq T$ and writing $\alpha=\alpha(X)$ it follows from
(3.5) that

$\displaystyle
8\alpha^{3/2}\geq\big{|}{\hat{b}}_{0}(X)+3{\hat{b}}_{1}(X)\alpha\big{|}-C\alpha^{2}\geq
3|{\hat{b}}_{1}(X)|\alpha-C\alpha^{2}-\alpha^{3/2}$

hence the assertion is clear because $\alpha({\bar{X}})=0$. ∎

######  Lemma 3.3.

In a neighborhood of $({\bar{X}},0)$ we have
$a_{j}(X,\epsilon)=O\big{(}\rho(X,\epsilon)^{j}\big{)}$ for $j=1,2,3$. More
precisely

$\displaystyle
a_{1}(X,\epsilon)=E(0,X,\epsilon)\big{(}3\rho(X,\epsilon)-{\hat{b}}_{1}^{2}(X)\big{)}+O(\rho^{3/2}),$
$\displaystyle
a_{2}(X,\epsilon)=E(0,X,\epsilon)\big{(}3\rho^{2}(X,\epsilon)-2{\hat{b}}_{0}(X){\hat{b}}_{1}(X)\big{)}+O(\rho^{3/2}),$
$\displaystyle
a_{3}(X,\epsilon)=E(0,X,\epsilon)\big{(}\rho^{3}(X,\epsilon)-{\hat{b}}^{2}_{0}(X)\big{)}.$

###### Proof.

Since ${\bar{\Delta}}(0,X,\epsilon)\geq 0$ it follows from (3.4) that
$a_{3}(X,\epsilon)=E(0,X,\epsilon)\big{(}\rho(X,\epsilon)^{3}-{\hat{b}}_{0}(X)^{2}\big{)}\geq
0$ hence ${\hat{b}}_{0}=O(\rho^{3/2})$ and consequently
$a_{3}(X,\epsilon)=O(\rho^{3})$. From (3.4)

$\displaystyle
a_{2}(X,\epsilon)=\partial_{t}E(0,X,\epsilon)a_{3}(X,\epsilon)+E(0,X,\epsilon)\big{(}3\rho^{2}(X,\epsilon)-2{\hat{b}}_{0}(X){\hat{b}}_{1}(X)\big{)}.$

Since ${\hat{b}}_{0}(X){\hat{b}}_{1}(X)=O(\rho^{2})$ by Lemma 3.2 hence the
above equality shows the assertion for $a_{2}(X,\epsilon)$. Finally from (3.4)
again

$\displaystyle
2a_{1}(X,\epsilon)=\partial_{t}^{2}E(0,X,\epsilon)a_{3}(X,\epsilon)+2\partial_{t}E(0,X,\epsilon)\big{(}3\rho^{2}(X,\epsilon)-2{\hat{b}}_{0}(X){\hat{b}}_{1}(X)\big{)}$
$\displaystyle+2E(0,X,\epsilon)\big{(}3\rho(X,\epsilon)-{\hat{b}}_{1}(X)^{2}-2{\hat{b}}_{0}(X){\hat{b}}_{2}(X)\big{)}$

and Lemma 3.2 one concludes the assertion for $a_{1}(X,\epsilon)$. ∎

###  3.2 Lower bound of perturbed discriminant

Denote

(3.6) $\nu(X,\epsilon)=\inf\\{t\mid{\bar{\Delta}}(t,X,\epsilon)>0\\}$

and hence ${\bar{\Delta}}(\nu,X,\epsilon)=0$. First check that
$\nu(X,\epsilon)\leq 0$. Suppose the contrary $\nu(X,\epsilon)=\nu>0$. Since
${\bar{\Delta}}(t,X,\epsilon)\geq 0$ for $t\geq 0$ one can write
${\bar{\Delta}}(t)=(t-\nu)^{2}(t-{\tilde{\nu}})$ with a real ${\tilde{\nu}}$
where ${\tilde{\nu}}\neq\nu$ and ${\tilde{\nu}}\leq 0$. Therefore we have
${\bar{\Delta}}(t)>0$ in ${\tilde{\nu}}<t<\nu$ which is incompatible with the
definition of $\nu$. Write

${\bar{\Delta}}(t,X,\epsilon)=(t-\nu(X,\epsilon))(t^{2}+A_{1}(X,\epsilon)t+A_{2}(X,\epsilon))$

where $A_{1}=\nu+a_{1}$. Here we prepare following lemma.

######  Lemma 3.4.

One can find a neighborhood ${\mathcal{U}}$ of $({\bar{X}},0)$ such that for
any $(X,\epsilon)\in{\mathcal{U}}$ there is $j\in\\{1,2,3\\}$ such that
$|\nu_{j}(X,\epsilon)|\geq\rho(X,\epsilon)/9$ where
${\bar{\Delta}}(t,X,\epsilon)=\prod_{j=1}^{3}(t-\nu_{j}(X,\epsilon))$.

###### Proof.

First show that there is $1/3<\delta<1/2$ such that

(3.7)
$\max{\big{\\{}\big{|}\rho^{3}-{\hat{b}}_{0}^{2}\big{|}^{1/3},\;\;\big{|}\rho^{2}-2{\hat{b}}_{0}{\hat{b}}_{1}/3\big{|}^{1/2},\;\;\big{|}\rho-{\hat{b}}_{1}^{2}/3\big{|}\big{\\}}}\geq\delta^{2}\rho.$

In fact denoting
$f(\delta)=2(1-\delta^{6})^{1/2}(1-\delta^{2})^{1/2}/\sqrt{3}-1-\delta^{4}$ it
is easy to check that $f(1/3)>0$ and $f(1/2)<0$. Take $1/3<\delta<1/2$ such
that $f(\delta)=0$. If $|\rho^{3}-{\hat{b}}_{0}^{2}|^{1/3}<\delta^{2}\rho$ and
$|\rho-{\hat{b}}_{1}^{2}/3|<\delta^{2}\rho$ then
$|{\hat{b}}_{0}|\geq(1-\delta^{6})^{1/2}\rho^{3/2}$ and
$|{\hat{b}}_{1}|\geq\sqrt{3}(1-\delta^{2})^{1/2}\rho^{1/2}$ hence

$\displaystyle|\rho^{2}-2{\hat{b}}_{0}{\hat{b}}_{1}/3|\geq
2|{\hat{b}}_{0}{\hat{b}}_{1}|/3-\rho^{2}\geq\big{(}f(\delta)+\delta^{4}\big{)}\rho^{2}=\delta^{4}\rho^{2}.$

Thus (3.7) is proved. Thanks to Lemma 3.3, taking $E(0,{\bar{X}},0)=1$ and
$1/3<\delta$ into account, one can find a neighborhood ${\mathcal{U}}$ of
$({\bar{X}},0)$ such that

$\max{\big{\\{}3|a_{1}(X,\epsilon)|,\;(3^{3}|a_{2}(X,\epsilon)|)^{1/2},\;(3^{6}|a_{3}(X,\epsilon)|)^{1/3}\big{\\}}}\geq\rho,\quad(X,\epsilon)\in{\mathcal{U}}.$

Then the assertion follows from the relations between $\\{\nu_{i}\\}$ and
$\\{a_{i}\\}$. ∎

######  Lemma 3.5.

Denote $\nu$ defined in (3.6) by $\nu_{1}$ and by $\nu_{j}$, $j=2,3$ the other
roots of ${\bar{\Delta}}=0$. Then one can find a neighborhood ${\mathcal{U}}$
of $({\bar{X}},0)$ and $c_{i}>0$ such that

(3.8)
$\text{if}\quad\nu_{1}+a_{1}<2\,c_{1}\,\rho,\;\;(X,\epsilon)\in{\mathcal{U}}\quad\text{then}\quad|\nu_{1}-\nu_{j}|\geq
c_{2}\,\rho,\;\;j=2,3.$

In particular $\nu_{1}(X,\epsilon)$ is smooth in
${\mathcal{U}}\cap\\{\nu_{1}+a_{1}<2\,c_{1}\rho\\}$.

###### Proof.

Write

(3.9)
${\bar{\Delta}}(t)=\prod_{j=1}^{3}(t-\nu_{j})=(t-\nu_{1})((t+A_{1}/2)^{2}-D)$

so that ${\mathsf{Re}}\,\nu_{j}=-A_{1}/2$, $j=2,3$ where
$A_{1}=\nu_{1}+a_{1}$. Take $c_{1}=1/27$ and assume $A_{1}<2c_{1}\rho$. First
note that if ${\mathsf{Re}}\,\nu_{j}\geq c_{1}\rho$, $j=2,3$ it is clear that
$|\nu_{1}-\nu_{j}|\geq|\nu_{1}-{\mathsf{Re}}\,\nu_{j}|\geq{\mathsf{Re}}\,\nu_{j}\geq
c_{1}\rho$ because $\nu_{1}\leq 0$ then we may assume

(3.10) $-c_{1}\rho<{\mathsf{Re}}\,\nu_{j}=-A_{1}/2<c_{1}\rho,\quad j=2,3.$

If $D>0$ then one has $-A_{1}/2+\sqrt{D}\leq 0$. Otherwise ${\bar{\Delta}}(t)$
would be negative for some $t>0$ near $-A_{1}/2+\sqrt{D}$ which is a
contradiction. Thus $\sqrt{D}\leq A_{1}/2\leq c_{1}\,\rho$ which shows that
$|\nu_{2}|$, $|\nu_{3}|\leq|A_{1}|/2+\sqrt{D}\leq 2c_{1}\,\rho<\rho/9$ hence
$|\nu_{1}|\geq\rho/9=3c_{1}\rho$ by Lemma 3.4. Therefore
$|\nu_{1}-\nu_{j}|\geq|\nu_{1}|-|\nu_{j}|\geq c_{1}\rho$. Turn to the case
$D\leq 0$ such that $\nu_{2}$, $\nu_{3}=-A_{1}/2\pm i\sqrt{|D|}$. Thanks to
Lemma 3.4 again either $|\nu_{1}|\geq 3c_{1}\rho$ or $|\nu_{2}|=|\nu_{3}|\geq
3c_{1}\rho$. If $|\nu_{1}|\geq 3c_{1}\rho$ then it follows from (3.10) that

$|\nu_{1}-\nu_{j}|\geq|\nu_{1}+A_{1}/2|\geq|\nu_{1}|-|A_{1}|/2\geq
2c_{1}\rho.$

If $|\nu_{2}|=|\nu_{3}|\geq 3c_{1}\rho$ so that $|A_{1}|/2+\sqrt{|D|}\geq
3c_{1}\rho$ hence $\sqrt{|D|}\geq 3c_{1}\rho-|A_{1}|/2\geq 2c_{1}\rho$ which
proves $|\nu_{1}-\nu_{j}|\geq\sqrt{|D|}\geq 2c_{1}\rho$ hence the assertion. ∎

Now define $\psi(X,\epsilon)$ which plays a crucial role in our arguments
deriving weighted energy estimates. Choose $\chi(s)\in
C^{\infty}({\mathbb{R}})$ such that $0\leq\chi(s)\leq 1$ with $\chi(s)=1$ if
$s\leq 0$ and $\chi(s)=0$ for $s\geq 1$. Define

$\psi(X,\epsilon)=-\chi\Big{(}\frac{\nu_{1}+a_{1}}{2c_{1}\rho}\Big{)}\frac{\nu_{1}+a_{1}}{2},\quad\epsilon\neq
0.$

######  Proposition 3.1.

One can find a neighborhood ${\mathcal{U}}$ of $({\bar{X}},0)$ such that

(3.11)
${\bar{\Delta}}(t,X,\epsilon)\geq{\bar{c}}\min{\big{\\{}t^{2},(t-\psi(X,\epsilon))^{2}\big{\\}}}\,(t+\rho(X,\epsilon))$

holds for $(X,\epsilon)\in{\mathcal{U}}$, $\epsilon\neq 0$ and $t\in[0,T]$
where ${\bar{c}}=1/32$.

###### Proof.

Set $\delta=1/9$ in this proof. First check that one can find $c\geq{\bar{c}}$
such that

(3.12) ${\bar{\Delta}}(t,X,\epsilon)\geq c\,t^{2}(t+\rho)\quad\text{if}\quad
A_{1}=\nu_{1}+a_{1}\geq 0.$

Let $D>0$ in (3.9). It was seen in the proof of Lemma 3.5 that
$-A_{1}/2+\sqrt{D}\leq 0$ so that $\nu_{2},\nu_{3}=-A_{1}/2\pm\sqrt{D}\leq 0$.
If $|\nu_{1}|\geq\delta\rho$ then $t-\nu_{1}=t+|\nu_{1}|\geq t+\delta\rho$
hence $\delta^{-1}(t-\nu_{1})\geq t+\rho$ and $t-\nu_{i}=t+|\nu_{i}|\geq t$
then (3.12) holds with $c=\delta$. Consider the case $D\leq 0$ so that
$\nu_{2},\nu_{3}=-A_{1}/2\pm i\sqrt{|D|}$. If $|\nu_{1}|\geq\delta\rho$ then
$\delta^{-1}(t-\nu_{1})\geq t+\rho$ as above and
$|t-\nu_{i}|\geq|t+A_{1}/2|\geq t$ thus (3.12) holds with $c=\delta$. If
$|\nu_{2}|=|\nu_{3}|\geq\delta\rho$ then $A_{1}/2+\sqrt{|D|}\geq\delta\rho$.
Since

$\displaystyle(t-\nu_{2})(t-\nu_{3})\geq\big{(}t+A_{1}/2+\sqrt{|D|}\,\big{)}^{2}/2\geq(t+\delta\rho)^{2}/2\geq\delta\,t\,(t+\rho)/2$

then (3.12) holds with $c=\delta/2$.

Turn to the case $A_{1}<0$. Since $\psi=-(\nu_{1}+a_{1})/2>0$, one can write
${\bar{\Delta}}(t)=(t-\nu_{1})((t-\psi)^{2}-D)$. Note that $D\leq 0$ otherwise
$\psi+\sqrt{D}>0$ would be a positive simple root of ${\bar{\Delta}}(t)$ and a
contradiction. Then $(t-\psi)^{2}-D=(t-\psi)^{2}+|D|\geq(t-\psi)^{2}$.
Consider the case $|\nu_{1}|\geq\delta\rho$. Recalling
$t-\nu_{1}=t+|\nu_{1}|\geq\delta(t+\rho)$ we get

(3.13) ${\bar{\Delta}}(t,X,\epsilon)\geq c\,(t-\psi)^{2}(t+\rho)$

with $c=\delta$. When $|\nu_{2}|=|\nu_{3}|=\big{|}\psi\pm
i\sqrt{|D|}\,\big{|}=\sqrt{\psi^{2}+|D|}\geq\delta\rho$ one has

$(t-\nu_{2})(t-\nu_{3})=(t-\psi)^{2}+|D|\geq\big{(}|t-\psi|+\sqrt{|D|}\,\big{)}^{2}/2.$

Assume $\psi\geq\sqrt{|D|}$ so that $\sqrt{2}\,\psi\geq\delta\rho$. For $0\leq
t\leq\psi/2$ hence $t\leq|t-\psi|$ and $\psi/2\leq|t-\psi|$ one has

$\displaystyle(1-\gamma)|t-\psi|+\gamma|t-\psi|\geq(1-\gamma)t+\gamma\psi/2\geq\delta(2\sqrt{2}+\delta)^{-1}(t+\rho)$

with $\gamma=2\sqrt{2}/(2\sqrt{2}+\delta)$. Since
$|t-\psi|+\sqrt{|D|}\geq|t-\psi|\geq t$ and $|t-\nu_{1}|=t+|\nu_{1}|\geq t$ it
is clear that (3.12) holds with $c=\delta/(2\sqrt{2}+\delta)$. For $\psi/2\leq
t$ such that $|t-\psi|\leq t$ one sees

$t-\nu_{1}\geq t=(1-\gamma)t+\gamma
t\geq(1-\gamma)t+\gamma\psi/2\geq\delta(2\sqrt{2}+\delta)^{-1}(t+\rho)$

and hence $(t-\nu_{1})((t-\psi)^{2}+|D|)\geq c\,(t+\rho)(t-\psi)^{2}$ which is
(3.13) with $c=\delta/(2\sqrt{2}+\delta)$. Next assume $\sqrt{|D|}\geq\psi$ so
that $\sqrt{2}\sqrt{|D|}\geq\delta\rho$. For $0\leq t\leq\psi/2$ one has
$|t-\psi|\geq t$ and hence $|t-\psi|+\sqrt{|D|}\geq
t+\delta\rho/\sqrt{2}\geq(\delta/\sqrt{2})(t+\rho)$. Noting
$|t-\nu_{1}|=t+|\nu_{1}|\geq t$ it is clear that (3.12) holds with
$c=\delta/2\sqrt{2}$. For $\psi/2\leq t$ we see that

$|t-\psi|+\sqrt{|D|}\geq t-|\psi|+\sqrt{|D|}\geq
t,\quad|t-\psi|+\sqrt{|D|}\geq\sqrt{|D|}\geq\delta\rho/\sqrt{2}$

which shows that
$|t-\psi|+\sqrt{|D|}\geq\delta(\sqrt{2}+\delta)^{-1}(t+\rho)$. Recalling
$|t-\nu_{1}|=t+|\nu_{1}|\geq t$ again one has (3.12) with
$c=\delta/2(\sqrt{2}+\delta)$. Thus choosing
${\bar{c}}=1/32<1/(18\sqrt{2}+2)=\delta/2(\sqrt{2}+\delta)$ the proof is
completed. ∎

######  Lemma 3.6.

One can find a neighborhood ${\mathcal{U}}$ of $({\bar{X}},0)$ and $C^{*}>0$
such that

(3.14) $\frac{|\partial_{t}{\Delta}(t,X,\epsilon)|}{\Delta(t,X,\epsilon)}\leq
C^{*}\Big{(}\frac{1}{t}+\frac{1}{|t-\psi|+\sqrt{a}\,\epsilon\,}\Big{)},\quad(X,\epsilon)\in{\mathcal{U}},\;\;\epsilon>0$

holds for $t\in(0,T]$.

###### Proof.

It will suffice to show (3.14) for ${\Delta}(t,X,\sqrt{2}\epsilon)$ which we
denote by ${\tilde{\Delta}}(t,X,\epsilon)$. It is clear that

${\tilde{\Delta}}=\Delta+4e^{3}\big{(}3(t+\rho)^{2}\epsilon^{2}+3(t+\rho)\epsilon^{4}+\epsilon^{6}\big{)}=\Delta+\Delta_{r}.$

Writing
${\tilde{\Delta}}={\tilde{e}}\big{(}{\bar{\Delta}}+{\bar{\Delta}}_{r}\big{)}$
with $\Delta_{r}={\tilde{e}}{\bar{\Delta}}_{r}$ it suffices to show the
assertion for ${\bar{\Delta}}+{\bar{\Delta}}_{r}$ instead of
${\tilde{\Delta}}$. Note that

(3.15) $|\partial_{t}{\bar{\Delta}}_{r}|/\bar{\Delta}_{r}\leq
C\big{(}1+1/(t+\rho)\big{)}\leq C^{\prime}/t$

always holds. Write ${\bar{\Delta}}=\prod_{j=1}^{3}(t-\nu_{j})$ and note that
$\partial_{t}{\bar{\Delta}}/{\bar{\Delta}}=\sum_{j=1}^{3}(t-\nu_{j})^{-1}$.
When $A_{1}\geq 0$ we see from the proof of Proposition 3.1 that
$|t-\nu_{j}|\geq t$ hence
$\big{|}\partial_{t}{\bar{\Delta}}/{\bar{\Delta}}\big{|}\leq 3/t$. Therefore
one has

$\displaystyle|\partial_{t}{\tilde{\Delta}}|/{\tilde{\Delta}}\leq|\partial_{t}{\bar{\Delta}}|/\big{(}{\bar{\Delta}}+{\bar{\Delta}}_{r}\big{)}+|\partial_{t}{\bar{\Delta}}_{r}|/\big{(}{\bar{\Delta}}+{\bar{\Delta}}_{r}\big{)}$
$\displaystyle\leq|\partial_{t}{\bar{\Delta}}|/{\bar{\Delta}}+|\partial_{t}{\bar{\Delta}}_{r}|/{\bar{\Delta}}_{r}$

which proves the assertion. Let $A_{1}<0$ then
${\bar{\Delta}}=(t-\nu_{1})((t-\psi)^{2}-D)$ where $\psi>0$ and $D\leq 0$ as
seen in the proof of Proposition 3.1. If $|D|\geq a\epsilon^{2}$

$|t-\psi|(|t-\psi|+\sqrt{a}\,\epsilon)\leq\sqrt{2}((t-\psi)^{2}+|D|)$

which shows the assertion since $|t-\nu_{1}|=t+|\nu_{1}|\geq t$. Similarly if
$|t-\psi|\geq\sqrt{a}\,\epsilon$ one has
$|t-\psi|(|t-\psi|+\sqrt{a}\,\epsilon)\leq 2(t-\psi)^{2}\leq
2((t-\psi)^{2}+|D|)$ hence the assertion. If $|D|<a\epsilon^{2}$ and
$|t-\psi|<\sqrt{a}\,\epsilon$ it follows that

$|\partial_{t}{\bar{\Delta}}|\leq(t-\psi)^{2}+|D|+2|t-\nu_{1}||t-\psi|\leq
2a\epsilon^{2}+Ca^{3/2}\epsilon$

because $|t-\nu_{1}|\leq Ca$. In view of $C{\bar{\Delta}}_{r}\geq
a^{2}\epsilon^{2}$ one concludes that

$\displaystyle|\partial_{t}{\bar{\Delta}}|/\big{(}{\bar{\Delta}}+{\bar{\Delta}}_{r}\big{)}\leq|\partial_{t}{\bar{\Delta}}|/{\bar{\Delta}}_{r}\leq
C\,\big{(}2a\epsilon^{2}+Ca^{3/2}\epsilon\big{)}/(a^{2}\epsilon^{2})$
$\displaystyle\leq C\Big{(}\frac{1}{a}+\frac{1}{\sqrt{a}\,\epsilon}\Big{)}\leq
C^{\prime}\Big{(}\frac{1}{t}+\frac{1}{|t-\psi|+\sqrt{a}\,\epsilon}\Big{)}$

which together with (3.15) proves the assertion. ∎

##  4 Localized symbols

In the preceding Sections 3.1 and 3.2 all symbols we have studied are defined
in some conic (in $\xi$) neighborhood of $(X,\epsilon)=({\bar{X}},0)$ or
$X={\bar{X}}$. In this section we define symbols on
${\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$ which localizes such symbols around
$(X,\epsilon)=({\bar{X}},0)$ or $X={\bar{X}}$ with a parameter $M$.

###  4.1 Localization via localized coordinates functions

Let ${\bar{X}}=(0,{\bar{\xi}})$ with $|{\bar{\xi}}|=1$. Let $\chi(s)\in
C^{\infty}({\mathbb{R}})$ be equal to $1$ in $|s|\leq 1$, vanishes in $|s|\geq
2$ such that $0\leq\chi(s)\leq 1$. Define $y(x)$ and $\eta(\xi)$ by

$\displaystyle
y_{j}(x)=\chi(M^{2}x_{j})x_{j},\quad\eta_{j}(\xi)=\chi(M^{2}(\xi_{j}\langle{\xi}\rangle_{\gamma}^{-1}-{\bar{\xi}}_{j}))(\xi_{j}-{\bar{\xi}}_{j}\langle{\xi}\rangle_{\gamma})+{\bar{\xi}}_{j}\langle{\xi}\rangle_{\gamma}$

for $j=1,2,\ldots,d$ with
$\langle{\xi}\rangle_{\gamma}=(\gamma^{2}+|\xi|^{2})^{1/2}$ where $M$ and
$\gamma$ are large positive parameters constrained

(4.1) $\gamma\geq M^{5}.$

It is easy to see that
$(1-CM^{-2})\langle{\xi}\rangle_{\gamma}\leq|\eta|\leq(1+CM^{-2})\langle{\xi}\rangle_{\gamma}$
and

(4.2) $|y|\leq CM^{-2},\quad|\eta/|\eta|-{\bar{\xi}}|\leq CM^{-2}$

with some $C>0$ so that $(y,\eta)$ is contained in a conic neighborhood of
$(0,{\bar{\xi}})$, shrinking with $M$. Note that $(y,\eta)=(x,\xi)$ on the
conic neighborhood of $(0,{\bar{\xi}})$

(4.3) $W_{M}=\big{\\{}(x,\xi)\mid|x|\leq
M^{-2},\;|\xi_{j}/|\xi|-{\bar{\xi}}_{j}|\leq M^{-2}/2,\;|\xi|\geq\gamma
M\big{\\}}$

since
$\big{|}\xi_{j}/\langle{\xi}\rangle_{\gamma}-{\bar{\xi}}_{j}\big{|}\leq\big{|}\xi_{j}/\langle{\xi}\rangle_{\gamma}-\xi_{j}/|\xi|\big{|}+\big{|}\xi_{j}/|\xi|-{\bar{\xi}}_{j}\big{|}\leq
M^{-2}$ if $(x,\xi)\in W_{M}$ where $\delta_{ij}$ is the Kronecker’s delta.
Let $f(X,\epsilon)$, $h(X)$ be smooth functions in a conic neighborhood of
$({\bar{X}},0)$, ${\bar{X}}$ respectively which are homogeneous of degree $0$
in $\xi$ then we define localized symbols of $f(x,\xi)$, $h(x,\xi)$ of
$f(X,\epsilon)$, $h(X)$ by

$\displaystyle f(x,\xi)=f(y(x),\eta(\xi),\epsilon(\xi)),\quad
h(x,\xi)=h(y(x),\eta(\xi))$

with $\epsilon(\xi)=M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$ or
$\epsilon(\xi)=\sqrt{2}M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$. In view of
(4.2) such extended symbols are defined on
${\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$, taking $M$ large if necessary. Let

$G=M^{4}(|dx|^{2}+\langle{\xi}\rangle_{\gamma}^{-2}|d\xi|^{2}).$

Then it is easy to see

(4.4) $y_{j}\in
S(M^{-2},G),\;\;\eta_{j}-{\bar{\xi}}_{j}\langle{\xi}\rangle_{\gamma}\in
S(M^{-2}\langle{\xi}\rangle_{\gamma},G),\;\;\epsilon(\xi)\in S(M^{-2},G)$

for $j=1,\ldots,d$. To avoid confusions we denote $|\eta(\xi)|$ by $[\xi]$
hence

(4.5) $[\xi]\in
S(\langle{\xi}\rangle_{\gamma},G),\quad[\xi]\langle{\xi}\rangle_{\gamma}^{-1}-1\in
S(M^{-2},G).$

######  Lemma 4.1.

Let $f(X,\epsilon)$ be a smooth function in a conic neighborhood of
$({\bar{X}},0)$ which is homogeneous of degree $0$ in $\xi$. If
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\partial_{\epsilon}^{k}f({\bar{X}},0)=0$
for $0\leq|\alpha+\beta|+k<r$ then
$f(x,\xi)=f(y(x),\eta(\xi),\epsilon(\xi))\in S(M^{-2r},G)$. Let $h(X)$ be a
smooth function in a conic neighborhood of ${\bar{X}}$ which is homogeneous of
degree $0$ in $\xi$. Then $h(x,\xi)-h(0,{\bar{\xi}})\in S(M^{-2},G)$.

###### Proof.

We prove the first assertion. By the Taylor formula one can write

$\displaystyle
f(y,\eta,\epsilon)=\sum_{|\alpha+\beta|+k=r}\frac{1}{\alpha!\beta!k!}y^{\alpha}(\eta-{\bar{\xi}}\langle{\xi}\rangle_{\gamma})^{\beta}\epsilon^{k}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\partial_{\epsilon}^{k}f(0,{\bar{\xi}}\langle{\xi}\rangle_{\gamma},0)$
$\displaystyle+(r+1)\sum_{|\alpha+\beta|+k=r+1}\Big{[}\frac{1}{\alpha!\beta!k!}y^{\alpha}(\eta-{\bar{\xi}}\langle{\xi}\rangle_{\gamma})^{\beta}\epsilon^{k}$
$\displaystyle\times\int_{0}^{1}(1-\theta)^{r}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\partial_{\epsilon}^{k}f(\theta
y,\theta(\eta-{\bar{\xi}}\langle{\xi}\rangle_{\gamma})+{\bar{\xi}}\langle{\xi}\rangle_{\gamma},\theta\epsilon)d\theta\Big{]}.$

It is clear that

$y^{\alpha}(\eta-{\bar{\xi}}\langle{\xi}\rangle_{\gamma})^{\beta}\epsilon^{k}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\partial_{\epsilon}^{k}f(0,{\bar{\xi}},0)\langle{\xi}\rangle_{\gamma}^{-|\beta|}\in
S(M^{-2r},G)$

for $|\alpha+\beta|+k=r$ in view of (4.4). Since
$\langle{\xi}\rangle_{\gamma}/C\leq|\theta(\eta-{\bar{\xi}}\langle{\xi}\rangle_{\gamma})+{\bar{\xi}}\langle{\xi}\rangle_{\gamma}|\leq
C\langle{\xi}\rangle_{\gamma}$ the integral belongs to
$S(\langle{\xi}\rangle_{\gamma}^{-|\beta|},G)$ hence the second term on the
right-hand side is in $S(M^{-2r-2},G)$ thus the assertion. ∎

###  4.2 Estimate of localized symbols

From now on it is assumed that all constants are independent of $M$ and
$\gamma$. As explained before we write $A\precsim B$ if $A$ is bounded by
constant, independent of parameters, $M$ and $\gamma$, times $B$. Let
$\rho(x,\xi)$ be the localized symbol of ${\rho}(X,\epsilon)$ with
$\epsilon=M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$ so that

$\rho(x,\xi)=\alpha(x,\xi)+M\langle{\xi}\rangle_{\gamma}^{-1}.$

From Lemma 4.1 we see $\rho\in S(M^{-4},G)$ hence
$|\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\rho|\precsim\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
for $|\alpha+\beta|=2$. Since $\rho\geq 0$ it follows from the Glaeser’s
inequality that

(4.6)
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\rho\big{|}\precsim\sqrt{\rho}\,\langle{\xi}\rangle_{\gamma}^{-|\beta|},\qquad|\alpha+\beta|=1.$

######  Lemma 4.2.

Assume that $f(X,\epsilon)$ is smooth and homogeneous of degree $0$ in $\xi$
in a conic neighborhood of $({\bar{X}},0)$ and satisfies $|f(X,\epsilon)|\leq
C{\rho}(X,\epsilon)^{n}$ with some $n>0$ there. For the localized symbol
$f(x,\xi)$ there is $C_{\alpha\beta}>0$ such that

(4.7) $\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}f(x,\xi)\big{|}\leq
C_{\alpha\beta}\rho(x,\xi)^{n-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}.$

###### Proof.

From the assumption it follows that
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\partial_{\epsilon}^{k}f(0,{\bar{\xi}},0)=0$
for $|\alpha+\beta|+k<2n$ hence Lemma 4.1 shows that $f(x,\xi)\in
S(M^{-4n},G)$. Therefore for $|\alpha+\beta|\geq 2n$ one sees
$\big{|}\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}f(x,\xi)\big{|}\leq
CM^{2|\alpha+\beta|-4n}\leq
C\big{(}C_{0}\rho^{-1}\big{)}^{|\alpha+\beta|/2-n}$ because $M^{4}\leq
C_{0}\rho^{-1}$. Hence (4.7) holds for $|\alpha+\beta|\geq 2n$. Assume
$|\alpha+\beta|\leq 2n-1$. Writing $X=(x,\xi)$,
$Y=(y,\eta\langle{\xi}\rangle_{\gamma})$ and applying the Taylor formula to
obtain

(4.8)
$\begin{split}\big{|}f(X+sY)\big{|}=\Big{|}\sum_{j=0}^{2n-1}\frac{s^{j}}{j!}d^{j}f(X;Y)+\frac{s^{2n}}{(2n)!}d^{2n}f(X+s\theta
Y;Y)\Big{|}\\\ \leq
C\Big{(}\sum_{j=0}^{2n-1}\frac{s^{j}}{j!}d^{j}\rho(X;Y)+\frac{s^{2n}}{(2n)!}d^{2n}\rho(X+s\theta^{\prime}Y;Y)\Big{)}^{n}\end{split}$

where
$d^{j}f(X;Y)=\sum_{|\alpha+\beta|=j}(j!/\alpha!\beta!)\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)y^{\alpha}\eta^{\beta}\langle{\xi}\rangle_{\gamma}^{|\beta|}$,
$0<\theta,\theta^{\prime}<1$. If $\rho(x,\xi)=0$ then
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\rho(x,\xi)=0$ for
$|\alpha+\beta|=1$ because $\rho\geq 0$ and then it follows from (4.8) that
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}f(x,\xi)=0$ for
$|\alpha+\beta|\leq 2n-1$ hence (4.7). We fix a small $s_{0}>0$. If
$\rho(x,\xi)\geq s_{0}$ then one has

$\displaystyle\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}f(x,\xi)\langle{\xi}\rangle_{\gamma}^{|\beta|}\big{|}\leq
C_{\alpha\beta}\leq
C_{\alpha\beta}s_{0}^{-n+|\alpha+\beta|/2}\rho^{n-|\alpha+\beta|/2}$

for $|\alpha+\beta|\leq 2n-1$ which proves (4.7). Assume
$0<\rho(x,\xi)<s_{0}$. Note that

$\big{|}d^{2n}f(X+s\theta Y;Y)\big{|}\leq C,\quad
d^{2n}\rho(X+s\theta^{\prime}Y;Y)\leq C\rho(X)^{1-n}$

for any $|(y,\eta)|\leq 1/2$. Indeed the first one is clear from $f(x,\xi)\in
S(M^{-4n},G)$. To check the second one it is enough to note that
$\rho(x,\xi)\in S(M^{-4},G)$ and

(4.9) $M^{-4+2j}\leq(C_{0}\rho^{-1})^{j/2-1},\;\;j\geq
2,\quad\sqrt{2}\langle{\xi+\theta\langle{\xi}\rangle_{\gamma}\eta}\rangle_{\gamma}\geq\langle{\xi}\rangle_{\gamma}/2$

for $|\eta|\leq 1/2$ and $|\theta|<1$. Take $s=\rho(X)^{1/2}$ in (4.8) to get

$\displaystyle\Big{|}\sum_{j=0}^{2n-1}\frac{1}{j!}d^{j}f(X;Y)\rho(X)^{j/2}\Big{|}\leq
C\Big{(}\sum_{j=0}^{2n-1}\frac{1}{j!}d^{j}\rho(X;Y)\rho(X)^{j/2}\Big{)}^{n}+C\rho(X)^{n}$

where the right-hand side is bounded by $C\rho(X)^{n}$ for $|d\rho(X;Y)|\leq
C^{\prime}\rho(X)^{1/2}$ in view of (4.6) and (4.9) for $j\geq 3$. Replacing
$(y,\eta)$ by $s(y,\eta)$, $|(y,\eta)|=1/2$, $0<|s|<1$ one obtains

$\Big{|}\sum_{j=1}^{2n-1}\frac{s^{j}}{j!}d^{j}f(X;Y)\frac{\rho(X)^{j/2}}{\rho(X)^{n}}\Big{|}\leq
C_{1}.$

Since two norms $\sup_{|s|\leq 1}|p(s)|$ and $\max{\\{|c_{j}|\\}}$ on the
vector space consisting of all polynomials $p(s)=\sum_{j=0}^{2n-1}c_{j}s^{j}$
are equivalent one obtains $|d^{j}f(X;Y)|\leq B^{\prime}\rho(X)^{n-j/2}$.
Since $|(y,\eta)|=1/2$ is arbitrary one concludes (4.7). ∎

######  Lemma 4.3.

Let $s\in{\mathbb{R}}$. Then
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\rho^{s}\big{|}\precsim\rho^{s-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.

###### Proof.

When $s=1$ the assertion follows from Lemma 4.2. Since

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\rho^{s}=\sum
C_{\alpha^{(j)}\beta^{(j)}}\rho^{s}\,\big{(}\partial_{x}^{\alpha^{(1)}}\partial_{\xi}^{\beta^{(1)}}\rho/\rho\big{)}\cdots\big{(}\partial_{x}^{\alpha^{(k)}}\partial_{\xi}^{\beta^{(k)}}\rho/\rho\big{)}$

the proof for the case $s\in{\mathbb{R}}$ is clear. ∎

######  Lemma 4.4.

Let $a_{j}(x,\xi)$ be the localized symbol of $a_{j}(X,\epsilon)$ then

$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a_{j}(x,\xi)\big{|}\precsim\rho(x,\xi)^{j-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},\quad
j=1,2,3.$

###### Proof.

The assertion follows from Lemmas 3.3 and 4.2. ∎

For the localized symbol $\psi(x,\xi)$ of $\psi(X,\epsilon)$ with
$\epsilon=M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$ we have

######  Lemma 4.5.

One has
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi(x,\xi)\big{|}\precsim\rho(x,\xi)^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.

###### Proof.

Since Lemma 4.2 is not available for ${\psi}(X,\epsilon)$ because it is not
defined for $\epsilon=0$ then we show the assertion directly. Let
$\nu_{1}(x,\xi)$, $a_{1}(x,\xi)$ and ${\bar{\Delta}}(t,x,\xi)$ be localized
symbols of $\nu_{1}(X,\epsilon)$, $a_{1}(X,\epsilon)$ and
${\bar{\Delta}}(t,X,\epsilon)$ with
$\epsilon=M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$ and hence
${\bar{\Delta}}(\nu_{1}(x,\xi),x,\xi)=0$. Note that
$|\partial_{t}{\bar{\Delta}}(\nu_{1},x,\xi)|\geq
4\,c_{2}^{2}\,\rho^{2}(x,\xi)$ if
$\nu_{1}(x,\xi)+a_{1}(x,\xi)<2c_{1}\rho(x,\xi)$ thanks to Lemma 3.5. Starting
with

$\partial_{t}{\bar{\Delta}}(\nu_{1},x,\xi)\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\nu_{1}+\partial_{x}^{\alpha}\partial_{\xi}^{\beta}{\bar{\Delta}}(\nu_{1},x,\xi)=0,\quad|\alpha+\beta|=1$

a repetition of the same argument in Lemma 5.3 below together with Lemma 4.4
shows

(4.10)
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\nu_{1}\big{|}\precsim\rho^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},\quad\nu_{1}+a_{1}<2c_{1}\rho.$

Here we have used $|\nu_{1}|\precsim\rho$ which also follows from Lemma 4.4.
Using (4.10) and Lemmas 4.3 and 4.4 the assertion follows easily. ∎

###  4.3 Estimate of discriminant

Let $\alpha(x,\xi)$, $a(t,x,\xi)$, $b(t,x,\xi)$, $e(t,x,\xi)$ be localized
symbols of $\alpha(X)$, $a(t,X)$, $b(t,X)$, $e(t,X)$ so that
$\tau^{3}-e(t,x,\xi)(t+\alpha(x,\xi))[\xi]^{2}\tau+b(t,x,\xi)[\xi]^{3}$ is now
defined on ${\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$ and coincides with the
original $p$ in a conic neighborhood $W_{M}$ of $(0,{\bar{\xi}})$. We add a
term $2Me(t,x,\xi)\langle{\xi}\rangle_{\gamma}^{-1}[\xi]^{2}$ to this;

(4.11)
${\hat{p}}=\tau^{3}-e(t+\alpha+2M\langle{\xi}\rangle_{\gamma}^{-1})[\xi]^{2}\tau-b\,[\xi]^{3}$

where we denote

(4.12)
$a_{M}(t,x,\xi)=e(t,x,\xi)(t+\alpha(x,\xi)+2M\langle{\xi}\rangle_{\gamma}^{-1})$

which is the localized symbol of $a(t,X,\epsilon)=a(t,X)+\epsilon^{2}$ with
$\epsilon=\sqrt{2}M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$. Consider the
discriminant

(4.13)
$\begin{split}\Delta_{M}(t,x,\xi)=4\,e^{3}\big{(}t+\alpha+2M\langle{\xi}\rangle_{\gamma}^{-1}\big{)}^{3}-27\,b^{2}\\\
=4\,e^{3}\big{(}t+\alpha+M\langle{\xi}\rangle_{\gamma}^{-1})^{3}-27\,b^{2}+\Delta_{r}(t,x,\xi)\end{split}$

where, recalling
$\alpha(x,\xi)+M\langle{\xi}\rangle_{\gamma}^{-1}=\rho(x,\xi)$, we have

$\displaystyle\Delta_{r}=4e^{3}\big{(}3(t+\rho)^{2}M\langle{\xi}\rangle_{\gamma}^{-1}+3(t+\rho)M^{2}\langle{\xi}\rangle_{\gamma}^{-2}+M^{3}\langle{\xi}\rangle_{\gamma}^{-3}\big{)}$
$\displaystyle=12e^{3}\big{(}c_{1}(x,\xi)t^{2}+c_{2}(x,\xi)t+c_{3}(x,\xi)\big{)}\geq
12e^{3}M(t+\rho)^{2}\langle{\xi}\rangle_{\gamma}^{-1}.$

It is clear that $c_{j}(x,\xi)$ verifies
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}c_{j}\big{|}\precsim\rho^{j-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.
Let $\Delta(t,x,\xi)$, ${\bar{\Delta}}(t,x,\xi)$ be localized symbols of
$\Delta(t,X,\epsilon)$, ${\bar{\Delta}}(t,X,\epsilon)$ with
$\epsilon=M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$. Thanks to Proposition
3.1 one has
${\bar{\Delta}}(t,x,\xi)\geq{\bar{c}}\,\min{\\{t^{2},(t-\psi)^{2}\\}}(t+\rho)$.
Noting that $\Delta(t,x,\xi)={\tilde{e}}\,{\bar{\Delta}}$ we see

(4.14)
$\begin{split}\Delta(t,x,\xi)={\tilde{e}}\,{\bar{\Delta}}\geq{\tilde{e}}\,{\bar{c}}\,\min{\\{t^{2},(t-\psi)^{2}\\}}(t+\rho)\\\
\geq\big{(}{\tilde{e}}/e\big{)}{\bar{c}}\,\min{\\{t^{2},(t-\psi)^{2}\\}}e(t+\rho).\end{split}$

Therefore choosing a constant ${\bar{\nu}}>0$ such that
$12\,e^{2}\geq\big{(}{\tilde{e}}/e\big{)}{\bar{c}}\,{\bar{\nu}}$ one obtains
from (4.13), (4.14) that

(4.15)
$\begin{split}\Delta_{M}\geq\big{(}{\tilde{e}}/e\big{)}{\bar{c}}\,\min{\big{\\{}t^{2},(t-\psi)^{2}\big{\\}}}e(t+\rho)+12\,e^{3}(t+\rho)^{2}M\langle{\xi}\rangle_{\gamma}^{-1}\\\
\geq\big{(}{\tilde{e}}/e\big{)}{\bar{c}}\,\big{(}\min{\big{\\{}t^{2},(t-\psi)^{2}\big{\\}}}+{\bar{\nu}}(t+\rho)M\langle{\xi}\rangle_{\gamma}^{-1}\big{)}e(t+\rho)\\\
\geq\big{(}{\tilde{e}}/e\big{)}{\bar{c}}\,\min{\big{\\{}t^{2},(t-\psi)^{2}+{\bar{\nu}}M\rho\langle{\xi}\rangle_{\gamma}^{-1}\big{\\}}}e(t+\rho),\qquad
t\geq 0.\end{split}$

######  Proposition 4.1.

One can write

$\Delta_{M}=e\big{(}t^{3}+a_{1}(x,\xi)t^{2}+a_{2}(x,\xi)t+a_{3}(x,\xi)\big{)}$

where $0<e\in S(1,G)$ uniformly in $t$ and
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a_{j}\big{|}\precsim\rho^{j-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.
Moreover there exist ${\bar{\nu}}>0$ and $c>0$ such that

(4.16)
$\frac{\Delta_{M}}{a_{M}}\geq\frac{\tilde{e}}{2e}\,{\bar{c}}\,\min{\big{\\{}t^{2},(t-\psi)^{2}+{\bar{\nu}}M\rho\langle{\xi}\rangle_{\gamma}^{-1}\big{\\}}},\;\;\frac{\Delta_{M}}{a_{M}}\geq
c\,M\langle{\xi}\rangle_{\gamma}^{-1}\,a_{M}$

for $0\leq t\leq T$ where $\psi$ and $\rho$ satisfy
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi\big{|}$,
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\rho\big{|}\precsim\rho^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.

###### Proof.

Choosing $\epsilon=\sqrt{2}M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$ and
applying Lemma 3.1 one can write $\Delta_{M}$ as a third order polynomial in
$t$, up to non-zero factor and can estimate the coefficients thanks to Lemmas
3.3 and 4.2 in terms of $\alpha+2M\langle{\xi}\rangle_{\gamma}^{-1}$. Noting
$\rho(x,\xi)\leq\alpha(x,\xi)+2M\langle{\xi}\rangle_{\gamma}^{-1}\leq
2\rho(x,\xi)$ we have the desired estimates for $a_{j}$. The assertion (4.16)
follows from (4.15) for
$a_{M}=e(t+\rho+M\langle{\xi}\rangle_{\gamma}^{-1/2})\leq 2\,e(t+\rho)$. The
estimates for $\psi$ and $\rho$ are nothing but Lemmas 4.3 and 4.5 with
$\epsilon=M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$. ∎

######  Remark 4.1.

Denoting ${\bar{e}}=e(0,0,{\bar{\xi}})=\partial_{t}a(0,0,{\bar{\xi}})$ it is
clear from (3.1) that

${\bar{e}}\;\;\text{ is the nonzero positive real eigenvalue
of}\;\;F_{p}(0,0,0,{\bar{\xi}})$

and the coefficient of the right-hand side of (4.16) is
${\tilde{e}}\,{\bar{c}}/(2e)=2\,{\bar{e}}^{2}\,{\bar{c}}(1+O(M^{-2}))$. On the
other hand, denoting the subprincipal symbol of $P$ by $P_{sub}$ and
$b_{3}(0,0,{\bar{\xi}})$ by ${\bar{b}}_{3}$, it is easy to see that

(4.17) $P_{sub}(0,0,0,{\bar{\xi}})={\bar{e}}/(2i)+{\bar{b}}_{3}.$

######  Lemma 4.6.

With ${\bar{e}}=e(0,0,{\bar{\xi}})$ we have

$\big{|}\partial_{t}b\big{|}\leq(1+CM^{-2})\big{(}2\sqrt{2/3}\big{)}{\bar{e}}\sqrt{a_{M}},\quad
0\leq t\leq M^{-2}.$

###### Proof.

Write $b=\beta_{0}(x,\xi)+t\beta_{1}(x,\xi)+t^{2}\beta_{3}(t,x,\xi)$. Setting
$t=0$ in $27b^{2}\leq 4a^{3}$ it is clear that
$|\beta_{0}|\leq(2/3\sqrt{3}){\bar{e}}^{3/2}(1+CM^{-2})\alpha^{3/2}$. We first
check that

(4.18) $|\beta_{1}|\leq(1+CM^{-2})(2/\sqrt{3}){\bar{e}}^{3/2}\sqrt{\alpha}.$

If $\alpha(x,\xi)=0$ then $\beta_{1}(x,\xi)=0$ hence (4.18) is clear. If
$\alpha(x,\xi)>0$ taking $t=3\alpha$ and noting
$e(3\alpha,x,\xi)\leq(1+CM^{-2}){\bar{e}}+C\alpha$ it follows that

$\displaystyle 3\alpha|\beta_{1}|\leq
2(4^{3/2}(1+CM^{-2}){\bar{e}}^{3/2}/3\sqrt{3})\alpha^{3/2}+|\beta_{0}|+C\alpha^{2}$
$\displaystyle\leq(6/\sqrt{3})(1+CM^{-2}){\bar{e}}^{3/2}\alpha^{3/2}+C\alpha^{2}\leq(6/\sqrt{3})(1+CM^{-2}){\bar{e}}^{3/2}\alpha^{3/2}$

because $\alpha\leq CM^{-4}$ which proves (4.18). Since
$|\partial_{t}b|\leq|\beta_{1}|+Ct$ we see that
$|\partial_{t}b|\leq(1+CM^{-2})(2/\sqrt{3}){\bar{e}}^{3/2}\sqrt{\alpha}+CM^{-2}\sqrt{t}$
thus the proof is immediate. ∎

##  5 Bézout matrix and diagonal symmetrizer

Add $-2M{\rm op}({e(t,x,\xi)\langle{\xi}\rangle_{\gamma}^{-1}})[D]^{2}D_{t}$
to the principal part and subtract the same one from the lower order part so
that the operator is left to be invariant;

$\displaystyle{\hat{P}}=D_{t}^{3}-a_{M}(t,x,D)[D]^{2}D_{t}-b(t,x,D)\,[D]^{3}+b_{1}(t,x,D)D_{t}^{2}$
$\displaystyle+\big{(}b_{2}(t,x,D)+d_{M}(t,x,D))[D]D_{t}+b_{3}(t,x,D)[D]^{2}$

where $b_{j}(t,x,\xi)\in S(1,G)$ and
$d_{M}(t,x,\xi)=2M(e\langle{\xi}\rangle_{\gamma}^{-1})\\#[\xi]\in S(M,G)$.
Here from Lemma 4.1 and (4.5) it follows that

(5.1) $d_{M}(t,x,\xi)-2M{\bar{e}}\in S(M^{-1},g).$

With $U={{}^{t}}(D_{t}^{2}u,[D]D_{t}u,[D]^{2}u)$ the equation ${\hat{P}}u=f$
is transformed to

(5.2) $D_{t}U=A(t,x,D)[D]U+B(t,x,D)U+F$

where $F={{}^{t}}(f,0,0)$ and

$A(t,x,\xi)=\begin{bmatrix}0&a_{M}&b\\\ 1&0&0\\\ 0&1&0\end{bmatrix},\quad
B(t,x,\xi)=\begin{bmatrix}b_{1}&b_{2}+d_{M}&b_{3}\\\ 0&0&0\\\
0&0&0\end{bmatrix}.$

Let $S$ be the Bézout matrix of ${\hat{p}}$ and
$\partial{\hat{p}}/\partial\tau$, that is

$S(t,x,\xi)=\begin{bmatrix}3&0&-a_{M}\\\ 0&2a_{M}&3b\\\
-a_{M}&3b&a_{M}^{2}\end{bmatrix}$

then $S$ is positive semidefinite and symmetrizes $S$, that is $SA$ is
symmetric which is easily examined directly, though this is a special case of
a general fact (see [16], [29]).

###  5.1 Eigenvalues of Bézout matrix

To simplify notation denote

$\sigma(t,x,\xi)=t+\alpha(x,\xi)+2M\langle{\xi}\rangle_{\gamma}^{-1}=t+\rho(x,\xi)+M\langle{\xi}\rangle_{\gamma}^{-1}$

hence $a_{M}(t,x,\xi)=e(t,x,\xi)\sigma(t,x,\xi)$ and
$(1-CM^{-2}){\bar{e}}\,\sigma\leq a_{M}\leq(1+CM^{-2}){\bar{e}}\,\sigma$. In
what follows we assume that $t$ varies in the interval

$0\leq t\leq M^{-4}.$

Since $\rho\in S(M^{-4},G)$ it is clear that $\sigma(t,x,\xi)\in S(M^{-4},G)$.

######  Lemma 5.1.

We have
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\sigma\big{|}\precsim\sigma^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.
In particular $\sigma\in S(\sigma,g)$.

###### Proof.

It is clear from (4.6) that
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\sigma\big{|}\precsim\sqrt{\sigma}\,\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
for $|\alpha+\beta|=1$. For $|\alpha+\beta|\geq 2$ it results
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\sigma\big{|}\precsim
M^{2|\alpha+\beta|-4}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\precsim\sigma^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
from $\rho\in S(M^{-4},G)$ since $C\sigma^{-1}\geq M^{4}$. The second
assertion is clear from $\sigma^{-1}\leq M^{-1}\langle{\xi}\rangle_{\gamma}$.
∎

######  Corollary 5.1.

Let $s\in{\mathbb{R}}$. Then
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\sigma^{s}\big{|}\precsim\sigma^{s-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.
In particular $\sigma^{s}\in S(\sigma^{s},g)$.

######  Definition 5.1.

To simplify notations we denote by ${\mathcal{C}}(\sigma^{s})$ the set of
symbols $r(t,x,\xi)$ satisfying
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}r\big{|}\precsim\sigma^{s-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$.

It is clear that ${\mathcal{C}}(\sigma^{s})\subset S(\sigma^{s},g)$ because
$\sigma^{-|\alpha+\beta|/2}\leq
M^{-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{|\alpha+\beta|/2}$. It is
also clear that if $p\in{\mathcal{C}}(\sigma^{s})$ with $s>0$ then
$(1+p)^{-1}-1\in{\mathcal{C}}(\sigma^{s})$.

######  Lemma 5.2.

One has $a_{M}^{s}\in{\mathcal{C}}(\sigma^{s})$ for $s\in{\mathbb{R}}$,
$b\in{\mathcal{C}}(\sigma^{3/2})$, $\partial_{t}a_{M}\in{\mathcal{C}}(1)$ and
$\partial_{t}b\in{\mathcal{C}}(\sqrt{\sigma})$.

###### Proof.

The first assertion is clear from Corollary 5.1 because $a_{M}=e\sigma$ and
$e\in S(1,G)$, $1/C\leq e\leq C$. To show the second assertion, recalling that
$b(t,x,\xi)$ is the localized symbol of $b(t,X)$, write

(5.3)
$\begin{split}b(t,x,\xi)={b}(0,y(x),\eta(\xi))+\partial_{t}{b}(0,y(x),\eta(\xi))t\\\
+\int_{0}^{1}(1-\theta)\partial_{t}^{2}{b}(\theta
t,y(x),\eta(\xi))d\theta\cdot t^{2}.\end{split}$

Since $\partial_{x}^{\alpha}\partial_{\xi}^{\beta}{b}(0,0,{\bar{\xi}})=0$ for
$|\alpha+\beta|\leq 2$ and $\partial_{t}{b}(0,0,{\bar{\xi}})=0$ then thanks to
Lemma 4.1 one has ${b}(0,y(x),\eta(\xi))\in S(M^{-6},G)$ and
$\partial_{t}{b}(0,y(x),\eta(\xi))\in S(M^{-2},G)$. Since $0\leq t\leq M^{-4}$
we conclude that $b(t,x,\xi)\in S(M^{-6},G)$. Since $|b|\leq C\sigma^{3/2}$
and $\sigma\in S(M^{-4},G)$ a repetition of the same arguments proving Lemma
4.2 shows the second assertion. The third assertion is clear because
$\partial_{t}a_{M}=e+(\partial_{t}e)\sigma$. As for the last assertion, recall
Lemma 4.6 that $|\partial_{t}b|\leq Ca_{M}^{1/2}\leq C^{\prime}\sigma^{1/2}$.
Noting $\partial_{t}b\in S(M^{-2},G)$ which results from (5.3) one sees
$|\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\partial_{t}b|\precsim
M^{2|\alpha+\beta|-2}\precsim\sigma^{1/2-|\alpha+\beta|/2}$ for
$|\alpha+\beta|\geq 1$ hence the assertion. ∎

Recall [31, Proposition 2.1]

######  Proposition 5.1.

Let
$0\leq\lambda_{1}(t,x,\xi)\leq\lambda_{2}(t,x,\xi)\leq\lambda_{3}(t,x,\xi)$ be
the eigenvalues of $S(t,x,\xi)$. There exist $M_{0}$ and $K>0$ such that one
has for $M\geq M_{0}$

$\displaystyle\Delta_{M}/(6a_{M}+2a_{M}^{2}+2a_{M}^{3})\leq\lambda_{1}\leq\big{(}2/3+Ka_{M}\big{)}\,a_{M}^{2},$
$\displaystyle(2-Ka_{M})\,a_{M}\leq\lambda_{2}\leq(2+Ka_{M})\,a_{M},$
$\displaystyle 3\leq\lambda_{3}\leq 3+Ka_{M}^{2}.$

###### Proof.

Since $a_{M}=e\,\sigma$ and $\sigma\in S(M^{-4},G)$ then for any
${\bar{\epsilon}}>0$ there is $M_{0}$ such that
$e\,M_{0}^{-4}\leq{\bar{\varepsilon}}$. Then the assertion follows from [31,
Proposition 2.1]. ∎

######  Corollary 5.2.

The eigenvalues $\lambda_{i}(t,x,\xi)$ are smooth in
$(0,M^{-4}]\times{\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$.

###  5.2 Estimates of eigenvalues

First we prove

######  Lemma 5.3.

One has $\lambda_{j}\in{\mathcal{C}}(\sigma^{3-j})$ for $j=1,2,3$.

Denote $q(\lambda)={\rm det}\,(\lambda I-S)$ so that

(5.4)
$q(\lambda)=\lambda^{3}-(3+2a_{M}+a_{M}^{2})\lambda^{2}+(6a_{M}+2a_{M}^{2}+2a_{M}^{3}-9b^{2})\lambda-\Delta_{M}.$

Note that
$\partial_{\lambda}q(\lambda_{i})\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\lambda_{i}+\partial_{x}^{\alpha}\partial_{\xi}^{\beta}q(\lambda_{i})=0$
for $|\alpha+\beta|=1$. Let us write
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}=\partial_{x,\xi}^{\alpha,\beta}$
for simplicity. We show by induction on $|\alpha+\beta|$ that

(5.5)
$\begin{split}\partial_{\lambda}q(\lambda_{i})\partial^{\alpha,\beta}_{x,\xi}\lambda_{i}=\sum_{2|\mu+\nu|+s\geq
2}C_{\mu,\nu,\gamma^{(j)},\delta^{(j)},s}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}q(\lambda_{i})\\\
\times\big{(}\partial_{x,\xi}^{\gamma^{(1)},\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)},\delta^{(s)}}\lambda_{i}\big{)}\end{split}$

where $\mu+\sum\gamma^{(i)}=\alpha$, $\nu+\sum\delta^{(j)}=\beta$ and
$|\gamma^{(i)}+\delta^{(j)}|\geq 1$. The assertion $|\alpha+\beta|=1$ is
clear. Suppose that (5.5) holds for $|\alpha+\beta|=m$. With $|e+f|=1$ after
operating $\partial_{x,\xi}^{e,f}$ to (5.5) the resulting left-hand side is

$\displaystyle\partial_{\lambda}q(\lambda_{i})\partial_{x,\xi}^{\alpha+e,\beta+f}\lambda_{i}$
$\displaystyle-\sum_{2|\mu+\nu|+s\geq
2}C_{\mu,\nu,\gamma^{(j)},\delta^{(j)},s}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}q(\lambda_{i})\big{(}\partial_{x,\xi}^{\gamma^{(1)},\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)},\delta^{(s)}}\lambda_{i}\big{)}$

while the resulting right-hand side is

$\displaystyle\sum
C_{\ldots}\partial_{x,\xi}^{\mu+e,\nu+f}\partial_{\lambda}^{s}q(\lambda_{i})\big{(}\partial_{x,\xi}^{\gamma^{(1)},\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)},\delta^{(s)}}\lambda_{i}\big{)}$
$\displaystyle+\sum
C_{\ldots}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s+1}q(\lambda_{i})\big{(}\partial_{x,\xi}^{e,f}\lambda_{i}\big{)}\big{(}\partial_{x,\xi}^{\gamma^{(1)},\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)},\delta^{(s)}}\lambda_{i}\big{)}$
$\displaystyle+\sum_{j=1}^{s}\sum
C_{\ldots}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}q(\lambda_{i})\big{(}\partial_{x,\xi}^{\gamma^{(1)},\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(j)}+e,\delta^{(j)}+f}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)},\delta^{(s)}}\lambda_{i}\big{)}$

which can be written as

$\sum_{2|\mu+\nu|+s\geq
2}C_{\mu,\nu,\gamma^{(j)},\delta^{(j)},s}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}q(\lambda_{i})\big{(}\partial_{x,\xi}^{\gamma^{(1)},\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)},\delta^{(s)}}\lambda_{i}\big{)}$

where $\mu+\sum\gamma^{(i)}=\alpha+e$, $\nu+\sum\delta^{(i)}=\beta+f$ and
$|\gamma^{(j)}+\delta^{(j)}|\geq 1$. Therefore we conclude (5.5). In order to
estimate $\partial_{x,\xi}^{\alpha,\beta}\lambda_{i}$ one needs to estimate
$\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}q(\lambda_{i})$.

######  Lemma 5.4.

For any $s\in{\mathbb{N}}$ and $\alpha,\beta$ it holds that

$\displaystyle|\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}^{s}q(\lambda_{j})|\precsim\sigma^{4-j-(3-j)s-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},\;\;j=1,2,$
$\displaystyle|\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}^{s}q(\lambda_{3})|\precsim\sigma^{-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}.$

###### Proof.

From Proposition 5.1 and (5.4) one sees that

$\displaystyle|q(\lambda_{i})|\precsim|\lambda_{i}|^{2}+|a_{M}||\lambda_{i}|+|a_{M}|^{3},$
$\displaystyle|\partial_{x,\xi}^{\alpha,\beta}q(\lambda_{i})|\precsim\big{(}|\partial_{x,\xi}^{\alpha,\beta}a_{M}|+|\partial_{x,\xi}^{\alpha,\beta}b^{2}|\big{)}|\lambda_{i}|+|\partial_{x,\xi}^{\alpha,\beta}a_{M}^{3}|+|\partial_{x,\xi}^{\alpha,\beta}b^{2}|,\quad|\alpha+\beta|\geq
1$

because $|\Delta_{M}|\precsim a_{M}^{3}$ and $|b|\precsim a_{M}^{3/2}$.
Therefore thanks to Proposition 5.1 and Lemma 5.2 one obtains the assertions
for the case $s=0$. Since

$\displaystyle|\partial_{\lambda}q(\lambda_{i})|\precsim|\lambda_{i}|+|a_{M}|,\qquad|\partial_{\lambda}^{s}q(\lambda_{i})|\precsim
1,\;\;s\geq 2,$
$\displaystyle|\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}q(\lambda_{i})|\precsim|\partial_{x,\xi}^{\alpha,\beta}a_{M}||\lambda_{i}|+|\partial_{x,\xi}^{\alpha,\beta}a_{M}|+|\partial_{x,\xi}^{\alpha,\beta}b^{2}|,\quad|\alpha+\beta|\geq
1,$
$\displaystyle|\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}^{2}q(\lambda_{i})|\precsim|\partial_{x,\xi}^{\alpha,\beta}a_{M}|,\quad\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}^{s}q(\lambda_{i})=0,\;\;s\geq
3,\quad|\alpha+\beta|\geq 1$

the assertions for the case $s\geq 1$ are clear by Proposition 5.1 and Lemma
5.2. ∎

Proof of Lemma 5.3: Since $\partial_{\lambda}q(\lambda_{i})=\prod_{k\neq
i}(\lambda_{i}-\lambda_{k})$ it follows from Proposition 5.1 that

(5.6) $6a_{M}(1-Ca_{M})\leq\big{|}\partial_{\lambda}q(\lambda_{i})\big{|}\leq
6a_{M}(1+Ca_{M}),\;i=1,2,\;\;\partial_{\lambda}q(\lambda_{3})\simeq 1.$

Then for $|\alpha+\beta|=1$ one has

$\big{|}\partial_{x,\xi}^{\alpha,\beta}\lambda_{j}\big{|}\precsim\big{|}\partial_{x,\xi}^{\alpha,\beta}q(\lambda_{j})/\partial_{\lambda}q(\lambda_{j})\big{|}\precsim\sigma^{3-j-1/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},\;\;j=1,2,3$

by Lemma 5.4 with $s=0$. Assume that
$\big{|}\partial_{x,\xi}^{\alpha,\beta}\lambda_{j}\big{|}\precsim\sigma^{3-j-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$,
$j=1,2,3$ holds for $|\alpha+\beta|\leq m$. Lemma 5.4 and (5.5) show that

$\displaystyle\big{|}\partial_{\lambda}q(\lambda_{1})\partial_{x,\xi}^{\alpha,\beta}\lambda_{1}\big{|}\precsim\sum\sigma^{3-2s-|\mu+\nu|/2}\sigma^{2-|\gamma^{(1)}+\delta^{(1)}|/2}\cdots\sigma^{2-|\gamma^{(s)}+\delta^{(s)}|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
$\displaystyle\precsim\sum\sigma^{3-|\mu+\nu|/2}\sigma^{-|\gamma^{(1)}+\delta^{(1)}|/2}\cdots\sigma^{-|\gamma^{(s)}+\delta^{(s)}|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\precsim\sigma^{3-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}.$

This together with (5.6) proves the estimate for $\lambda_{1}$. The same
arguments show the assertion for $\lambda_{2}$. The estimate for $\lambda_{3}$
is clear from (5.5) because of (5.6). Thus we have the assertion for
$|\alpha+\beta|=m+1$ and the proof is completed by induction on
$|\alpha+\beta|$. ∎

######  Lemma 5.5.

One has $\partial_{t}\lambda_{1}\in{\mathcal{C}}(\sigma)$,
$\partial_{t}\lambda_{2}\in{\mathcal{C}}(1)$ and
$\partial_{t}\lambda_{3}\in{\mathcal{C}}(1)$.

###### Proof.

First examine that
$\partial_{\lambda}q(\lambda_{i})\partial_{x,\xi}^{\alpha,\beta}\partial_{t}\lambda_{i}$
can be written as

(5.7)
$\begin{split}\sum_{|\alpha^{\prime}+\beta^{\prime}|<|\alpha+\beta|}C_{\ldots}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s+1}q(\lambda_{i})\big{(}\partial_{x,\xi}^{\alpha^{\prime},\beta^{\prime}}\partial_{t}\lambda_{i}\big{)}\big{(}\partial_{x,\xi}^{\gamma^{(1)}+\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)}+\delta^{(s)}}\lambda_{i}\big{)}\\\
+\sum
C_{\ldots}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}\partial_{t}q(\lambda_{i})\big{(}\partial_{x,\xi}^{\gamma^{(1)}+\delta^{(1)}}\lambda_{i}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)}+\delta^{(s)}}\lambda_{i}\big{)}\end{split}$

where $\alpha^{\prime}+\mu+\sum\gamma^{(i)}=\alpha$,
$\beta^{\prime}+\nu+\sum\delta^{(i)}=\beta$ and
$|\gamma^{(i)}+\delta^{(i)}|\geq 1$. Indeed (5.7) is clear when
$|\alpha+\beta|=0$ from
$\partial_{\lambda}q(\lambda_{i})\partial_{t}\lambda_{i}+\partial_{t}q(\lambda_{i})=0$.
Differentiating this by $\partial_{x,\xi}^{e,f}$ and repeating the same
arguments proving (5.5) one obtains (5.7) by induction. To prove Lemma 5.5
first check that

(5.8)
$|\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}^{s}\partial_{t}q(\lambda_{j})|\precsim\sigma^{3-j-(3-j)s-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},\;\;j=1,2,3.$

In fact from

(5.9)
$\partial_{t}q(\lambda)=-\partial_{t}(2a_{M}+a_{M}^{2})\lambda^{2}+\partial_{t}(6a_{M}+2a_{M}^{2}+2a_{M}^{3}-9b^{2})\lambda-\partial_{t}\Delta_{M}$

it follows that $|\partial_{t}q(\lambda_{i})|\precsim\lambda_{i}+\sigma^{2}$
and
$|\partial_{x,\xi}^{\alpha,\beta}\partial_{t}q(\lambda_{i})|\precsim(\lambda_{i}+\sigma^{2})\sigma^{-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
for $|\alpha+\beta|\geq 1$ in view of Lemma 5.2 and hence the assertion for
$s=0$. Since
$|\partial_{x,\xi}^{\alpha,\beta}\partial_{\lambda}^{s}\partial_{t}q(\lambda_{i})|\precsim\sigma^{-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
for $s\geq 1$ the assertion can be proved. We now show Lemma 5.5 for
$\lambda_{1}$ by induction on $|\alpha+\beta|$. Assume

(5.10)
$\big{|}\partial_{x,\xi}^{\alpha,\beta}\partial_{t}\lambda_{1}\big{|}\precsim\sigma^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}.$

It is clear from (5.6) and (5.8) that (5.10) holds for $|\alpha+\beta|=0$.
Assume that (5.10) holds for $|\alpha+\beta|\leq m$. For $|\alpha+\beta|=m+1$,
thanks to the inductive assumption, Lemma 5.4 and Lemma 5.3 it follows that

$\displaystyle\sum_{|\alpha^{\prime}+\beta^{\prime}|<|\alpha+\beta|}\big{|}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s+1}q(\lambda_{1})\big{(}\partial_{x,\xi}^{\alpha^{\prime},\beta^{\prime}}\partial_{t}\lambda_{1}\big{)}\big{(}\partial_{x,\xi}^{\gamma^{(1)}+\delta^{(1)}}\lambda_{1}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)}+\delta^{(s)}}\lambda_{1}\big{)}\big{|}$
$\displaystyle\precsim\sum\sigma^{3-2(s+1)-|\mu+\nu|/2}\sigma^{1-|\alpha^{\prime}+\beta^{\prime}|/2}\sigma^{2-|\gamma^{(1)}+\delta^{(1)}|/2}\cdots\sigma^{2-|\gamma^{(s)}+\delta^{(s)}|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$

which is bounded by
$\sigma^{2-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$. On the
other hand one sees

$\displaystyle\sum\big{|}\partial_{x,\xi}^{\mu,\nu}\partial_{\lambda}^{s}\partial_{t}q(\lambda_{1})\big{(}\partial_{x,\xi}^{\gamma^{(1)}+\delta^{(1)}}\lambda_{1}\big{)}\cdots\big{(}\partial_{x,\xi}^{\gamma^{(s)}+\delta^{(s)}}\lambda_{1}\big{)}\big{|}$
$\displaystyle\preceq\sum\sigma^{2-2s-|\mu+\nu|/2}\sigma^{2-|\gamma^{(1)}+\delta^{(1)}|/2}\cdots\sigma^{2-|\gamma^{(s)}+\delta^{(s)}|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\precsim\sigma^{2-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$

in view of (5.8) and Lemma 5.3. This proves that (5.10) holds for
$|\alpha+\beta|=m+1$ and hence for all $\alpha$, $\beta$. As for
$\lambda_{2}$, $\lambda_{3}$ the proof is similar. ∎

###  5.3 Eigenvectors of Bézout matrix

We sometimes denote by ${\mathcal{C}}(\sigma^{s})$ a function belonging to
${\mathcal{C}}(\sigma^{s})$. If we write $n_{ij}$ for the $(i,j)$-cofactor of
$\lambda_{k}I-S$ then ${}^{t}(n_{j1},n_{j2},n_{j3})$ is, if non-trivial, an
eigenvector of $S$ corresponding to $\lambda_{k}$. We take $k=1$, $j=3$ and
hence

$\begin{bmatrix}a_{M}(2\,a_{M}-\lambda_{1})\\\ 3\,b(\lambda_{1}-3)\\\
(\lambda_{1}-3)(\lambda_{1}-2\,a_{M})\end{bmatrix}=\begin{bmatrix}\ell_{11}\\\
\ell_{21}\\\ \ell_{31}\end{bmatrix}$

is an eigenvector of $S$ corresponding to $\lambda_{1}$ and therefore

${\bf t}_{1}=\begin{bmatrix}t_{11}\\\ t_{21}\\\
t_{31}\end{bmatrix}=\frac{1}{d_{1}}\begin{bmatrix}\ell_{11}\\\ \ell_{21}\\\
\ell_{31}\end{bmatrix},\quad
d_{1}=\sqrt{\ell_{11}^{2}+\ell_{21}^{2}+\ell_{31}^{2}}$

is a unit eigenvector of $S$ corresponding to $\lambda_{1}$. Thanks to
Proposition 5.1 and recalling $b\in{\mathcal{C}}(\sigma^{3/2})$ it is clear
that
$d_{1}=\sqrt{36\,a_{M}^{2}+{\mathcal{C}}(\sigma^{3})}=6a_{M}(1+{\mathcal{C}}(\sigma))$.
Therefore since $\ell_{11}={\mathcal{C}}(\sigma^{2})$,
$\ell_{21}={\mathcal{C}}(\sigma^{3/2})$ and
$\ell_{31}=6\,a+{\mathcal{C}}(\sigma^{2})$ we have

${\bf t}_{1}=\begin{bmatrix}t_{11}\\\ t_{21}\\\
t_{31}\end{bmatrix}=\begin{bmatrix}a_{M}/3+{\mathcal{C}}(\sigma^{2})\\\
-3b/(2a_{M})+{\mathcal{C}}(\sigma)\\\ 1+{\mathcal{C}}(\sigma)\end{bmatrix}.$

Similarly choosing $k=2,j=2$ and $k=3,j=1$

$\begin{bmatrix}-3a_{M}b\\\
(\lambda_{2}-3)(\lambda_{2}-a_{M}^{2})-a_{M}^{2}\\\
3b(\lambda_{2}-3)\end{bmatrix}=\begin{bmatrix}\ell_{12}\\\ \ell_{22}\\\
\ell_{32}\end{bmatrix},\;\;\begin{bmatrix}(\lambda_{3}-2a_{M})(\lambda_{3}-a_{M}^{2})-9b^{2}\\\
-3a_{M}b\\\
-a_{M}(\lambda_{3}-2a_{M})\end{bmatrix}=\begin{bmatrix}\ell_{13}\\\
\ell_{23}\\\ \ell_{33}\end{bmatrix}$

are eigenvectors of $S$ corresponding to $\lambda_{2}$ and $\lambda_{3}$
respectively and

${\bf t}_{j}=\begin{bmatrix}t_{1j}\\\ t_{2j}\\\
t_{3j}\end{bmatrix}=\frac{1}{d_{j}}\begin{bmatrix}\ell_{1j}\\\ \ell_{2j}\\\
\ell_{3j}\end{bmatrix},\quad
d_{j}=\sqrt{\ell_{1j}^{2}+\ell_{2j}^{2}+\ell_{3j}^{2}}$

are unit eigenvectors of $S$ corresponding to $\lambda_{j}$, $j=2,3$. Thanks
to Proposition 5.1 it is easy to see
$d_{2}=3\lambda_{2}(1+{\mathcal{C}}(\sigma))$ and
$d_{3}=\lambda_{3}^{2}(1+{\mathcal{C}}(\sigma))$. Then repeating the same
arguments one concludes

$\begin{bmatrix}t_{12}\\\ t_{22}\\\
t_{32}\end{bmatrix}=\begin{bmatrix}{\mathcal{C}}(\sigma^{3/2})\\\
-1+{\mathcal{C}}(\sigma)\\\
-3b/\lambda_{2}+{\mathcal{C}}(\sigma)\end{bmatrix},\qquad\begin{bmatrix}t_{13}\\\
t_{23}\\\ t_{33}\end{bmatrix}=\begin{bmatrix}1+{\mathcal{C}}(\sigma)\\\
{\mathcal{C}}(\sigma^{5/2})\\\
-a_{M}/\lambda_{3}+{\mathcal{C}}(\sigma^{2})\end{bmatrix}.$

Now $T=({\bf t}_{1},{\bf t}_{2},{\bf t}_{3})=(t_{ij})$ is an orthogonal matrix
which diagonalizes $S$;

$\varLambda=T^{-1}ST={{}^{t}}TST=\begin{bmatrix}\lambda_{1}&0&0\\\
0&\lambda_{2}&0\\\ 0&0&\lambda_{3}\end{bmatrix}.$

Note that $\varLambda A^{T}$ is symmetric. Summarize what we have proved in

######  Lemma 5.6.

Let $T$ be defined as above. Then there is $M_{0}$ such that $T$ has the form

$\displaystyle
T=\begin{bmatrix}a_{M}/3+{\mathcal{C}}(\sigma^{2})&{\mathcal{C}}(\sigma^{3/2})&1+{\mathcal{C}}(\sigma)\\\
-3b/(2a_{M})+{\mathcal{C}}(\sigma)&-1+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sigma^{5/2})\\\
1+{\mathcal{C}}(\sigma)&-3b/\lambda_{2}+{\mathcal{C}}(\sigma)&-a_{M}/\lambda_{3}+{\mathcal{C}}(\sigma^{2})\\\
\end{bmatrix}$
$\displaystyle=\begin{bmatrix}{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sigma^{3/2})&1+{\mathcal{C}}(\sigma)\\\
{\mathcal{C}}(\sigma^{1/2})&-1+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sigma^{5/2})\\\
1+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sigma^{1/2})&{\mathcal{C}}(\sigma)\\\
\end{bmatrix},\quad M\geq M_{0}.$

In particular $T,\;T^{-1}\in S(1,g)$.

######  Lemma 5.7.

We have

$\displaystyle\partial_{t}T=\begin{bmatrix}\partial_{t}(a_{M}/3)+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sigma^{1/2})&{\mathcal{C}}(1)\\\
-\partial_{t}(3b/2a_{M})+{\mathcal{C}}(1)&{\mathcal{C}}(1)&{\mathcal{C}}(\sigma^{3/2})\\\
{\mathcal{C}}(1)&-\partial_{t}(3b/\lambda_{2})+{\mathcal{C}}(1)&-\partial_{t}(a_{M}/\lambda_{3})+{\mathcal{C}}(\sigma)\\\
\end{bmatrix}$
$\displaystyle=\begin{bmatrix}{\mathcal{C}}(1)&{\mathcal{C}}(\sigma^{1/2})&{\mathcal{C}}(1)\\\
{\mathcal{C}}(\sigma^{-1/2})&{\mathcal{C}}(1)&{\mathcal{C}}(\sigma^{3/2})\\\
{\mathcal{C}}(1)&{\mathcal{C}}(\sigma^{-1/2})&{\mathcal{C}}(1)\\\
\end{bmatrix},\quad M\geq M_{0}.$

###### Proof.

Note that every entry of $T$ is a function in $a_{M}$, $b$ and $\lambda_{j}$.
Then the assertion is clear from Lemmas 5.2 and 5.5. ∎

From Lemma 5.6 it follows that

(5.11)
$\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}T=\begin{bmatrix}{\mathcal{C}}(\sqrt{\sigma})&{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sqrt{\sigma})\\\
{\mathcal{C}}(1)&{\mathcal{C}}(\sqrt{\sigma})&{\mathcal{C}}(\sigma^{2})\\\
{\mathcal{C}}(\sqrt{\sigma})&{\mathcal{C}}(1)&{\mathcal{C}}(\sqrt{\sigma})\\\
\end{bmatrix},\quad|\alpha+\beta|=1.$

######  Lemma 5.8.

There is $M_{0}$ such that $A^{T}=T^{-1}AT$ has the form

$A^{T}=\begin{bmatrix}{\mathcal{C}}(\sqrt{\sigma})&-1+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sqrt{\sigma})\\\
\lambda_{1}\,{\mathcal{C}}(\sigma^{-1})&{\mathcal{C}}(\sqrt{\sigma})&-1+{\mathcal{C}}(\sigma)\\\
\lambda_{1}\,{\mathcal{C}}(\sqrt{\sigma})&\lambda_{2}\,{\mathcal{C}}(1)&{\mathcal{C}}(\sigma^{5/2})\\\
\end{bmatrix},\quad M\geq M_{0}.$

###### Proof.

Writing $A^{T}=({\tilde{a}}_{ij})$ it is clear that

${\tilde{a}}_{ij}=t_{1i}\,a_{M}\,t_{2j}+t_{1i}\,b\,t_{3j}+t_{2i}t_{1j}+t_{3i}t_{2j}$

from which the assertions for ${\tilde{a}}_{ij}$, $j\geq i$ follows easily.
Therefore one sees

$\varLambda
A^{T}=\begin{bmatrix}\lambda_{1}{\mathcal{C}}(\sqrt{\sigma})&\lambda_{1}(-1+{\mathcal{C}}(\sigma))&\lambda_{1}{\mathcal{C}}(\sqrt{\sigma})\\\
\lambda_{2}{\tilde{a}}_{21}&\lambda_{2}{\tilde{a}}_{22}&\lambda_{2}(-1+{\mathcal{C}}(\sigma))\\\
\lambda_{3}{\tilde{a}}_{31}&\lambda_{3}{\tilde{a}}_{32}&\lambda_{3}{\tilde{a}}_{33}\\\
\end{bmatrix}.$

Since $\varLambda A^{T}$ is symmetric it follows immediately

${\tilde{a}}_{31}=\lambda_{1}{\mathcal{C}}(\sqrt{\sigma})/\lambda_{3},\;{\tilde{a}}_{32}=\lambda_{2}(-1+{\mathcal{C}}(\sigma))/\lambda_{3},\;{\tilde{a}}_{21}=\lambda_{1}(-1+{\mathcal{C}}(\sqrt{\sigma}))/\lambda_{2}$

which proves the assertion because $1/\lambda_{3}\in{\mathcal{C}}(1)$ and
$1/\lambda_{2}\in{\mathcal{C}}(\sigma^{-1})$. ∎

######  Corollary 5.3.

There is $M_{0}$ such that $A^{T}=T^{-1}AT$ has the form

$A^{T}=\begin{bmatrix}{\mathcal{C}}(\sqrt{\sigma})&-1+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sqrt{\sigma})\\\
{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sqrt{\sigma})&-1+{\mathcal{C}}(\sigma)\\\
{\mathcal{C}}(\sigma^{5/2})&{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sigma^{5/2})\\\
\end{bmatrix},\quad M\geq M_{0}.$

######  Corollary 5.4.

We have

$\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}A^{T}=\begin{bmatrix}{\mathcal{C}}(1)&{\mathcal{C}}(\sqrt{\sigma})&{\mathcal{C}}(1)\\\
{\mathcal{C}}(\sqrt{\sigma})&{\mathcal{C}}(1)&{\mathcal{C}}(\sqrt{\sigma})\\\
{\mathcal{C}}(\sigma^{2})&{\mathcal{C}}(\sqrt{\sigma})&{\mathcal{C}}(\sigma^{2})\\\
\end{bmatrix},\quad|\alpha+\beta|=1.$

###### Proof.

The proof is clear since
$\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(-1+{\mathcal{C}}(\sigma))={\mathcal{C}}(\sqrt{\sigma})$.
∎

Finally consider $T^{-1}(\partial_{t}T)$. Note that $\langle{\partial_{t}{\bf
t}_{i},{\bf t}_{j}}\rangle+\langle{{\bf t}_{i},\partial_{t}{\bf
t}_{j}}\rangle=0$ so that $(\partial_{t}T^{-1})T$ is antisymmetric. From
Lemmas 5.6 and 5.7 one has

(5.12)
$T^{-1}(\partial_{t}T)=\begin{bmatrix}0&-\partial_{t}(3b/2a_{M})+{\mathcal{C}}(1)&{\mathcal{C}}(1)\\\
\partial_{t}(3b/2a_{M})+{\mathcal{C}}(1)&0&{\mathcal{C}}(\sqrt{\sigma})\\\
\partial_{t}(a_{M}/3)+{\mathcal{C}}(\sigma)&{\mathcal{C}}(\sqrt{\sigma})&0\\\
\end{bmatrix}.$

For later use we estimate $(2,1)$-th and $(3,1)$-th entries of
$T^{-1}(\partial_{t}T)$. Recalling
$a_{M}=e(t+\alpha+2M\langle{\xi}\rangle_{\gamma}^{-1})$ and $0\leq t\leq
M^{-4}$ it is clear $\partial_{t}a_{M}-{\bar{e}}\in S(M^{-2},g)$. Taking
$|b^{2}/a_{M}^{3}|\leq 4/27$ into account, thanks to Lemma 4.6 it follows that

(5.13) $\begin{split}\big{|}\sqrt{a_{M}}\partial_{t}(3b/2a_{M})\big{|}\leq
3\big{(}\big{|}\partial_{t}b/\sqrt{a_{M}}\big{|}+\big{|}b/a_{M}^{3/2}\big{|}|\partial_{t}a_{M}|\big{)}/2\\\
\leq(1+CM^{-2})\big{(}(1+3\sqrt{2})/\sqrt{3}\big{)}\,{\bar{e}}.\end{split}$

##  6 Metric $g$ and estimates of $\omega$ and $\phi$

Introduce the metric

$g=g_{(x,\xi)}=M^{-1}\big{(}\langle{\xi}\rangle_{\gamma}|dx|^{2}+\langle{\xi}\rangle_{\gamma}^{-1}|d\xi|^{2}\big{)}$

which is a basic metric with which we work in this paper. Note that
$S(M^{s},G)\subset S(M^{s},g)$ because
$M^{s+2|\alpha+\beta|}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\leq
M^{s}M^{-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}$
in view of $\langle{\xi}\rangle_{\gamma}\geq\gamma\geq M^{5}$. The metric $g$
is temperate (see [8, Chapter 18]) uniformly in $\gamma\geq M^{5}\geq 1$ which
will be checked in Section 7.

######  Lemma 6.1.

One has

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi\in
S(M^{-(|\alpha+\beta|-1)/2}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2},g),\quad|\alpha+\beta|\geq
1.$

###### Proof.

It is enough to remark

$\displaystyle\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi\big{|}\precsim\rho^{1/2}\rho^{-(|\alpha+\beta|-1)/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\precsim\rho^{1/2}(M^{-1}\langle{\xi}\rangle_{\gamma})^{(|\alpha+\beta|-1)/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$

for $|\alpha+\beta|\geq 1$. ∎

###  6.1 Estimate $\omega$ by metric $g$

Taking Proposition 4.1 into account we introduce a preliminary weight

$\omega(t,x,\xi)=\sqrt{(t-\psi(x,\xi))^{2}+{\bar{\nu}}M\rho\langle{\xi}\rangle_{\gamma}^{-1}}.$

Since the exact value of ${\bar{\nu}}>0$ is irrelevant in the following
arguments so we assume ${\bar{\nu}}=1$ from now on. In what follows we work
with symbols depending on $t$ where $t$ varies in some fixed interval $[0,T]$
and it is assumed that all constants are independent of $t\in[0,T]$ and
$\gamma$, $M$ if otherwise stated. Now $A\precsim B$ implies that $A$ is
bounded by constant, independent of $t$, $M$ and $\gamma$, times $B$.

######  Lemma 6.2.

One has

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\omega^{s}\in
S\big{(}M^{-(|\alpha+\beta|-1)/2}\omega^{s}\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2+(|\alpha|-|\beta|)/2},g\big{)},\quad|\alpha+\beta|\geq
1.$

###### Proof.

Recall $\omega^{2}=(t-\psi)^{2}+M\rho\langle{\xi}\rangle_{\gamma}^{-1}$. Note
that for $|\alpha+\beta|\geq 2$

$\displaystyle\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(t-\psi)^{2}\big{|}\precsim\omega|\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi|+\sum|\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}\psi||\partial^{\alpha^{\prime\prime}}\partial_{\xi}^{\beta^{\prime\prime}}\psi|$
$\displaystyle\precsim\omega^{2}\big{\\{}\omega^{-1}\rho^{1/2}\rho^{-(|\alpha+\beta|-1)/2}+\omega^{-2}\rho\rho^{-(|\alpha+\beta|-2)/2}\big{\\}}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
$\displaystyle\precsim\omega^{2}(\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2})M^{-(|\alpha+\beta|-1)/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}$

since $\rho\geq M\langle{\xi}\rangle_{\gamma}^{-1}$ and
$\omega\geq\sqrt{M}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$. When
$|\alpha+\beta|=1$ it is clear

$\displaystyle\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(t-\psi)^{2}\big{|}\precsim\omega\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}=\omega^{2}(\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2})\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}.$

Next it is easy to see that for $|\alpha+\beta|\geq 1$

$\displaystyle\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(M\rho\langle{\xi}\rangle_{\gamma}^{-1})\big{|}\precsim
M\rho\langle{\xi}\rangle_{\gamma}^{-1}\rho^{-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
$\displaystyle\precsim\omega^{2}\big{(}M\omega^{-2}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1}\big{)}\big{(}M^{-1}\langle{\xi}\rangle_{\gamma}\big{)}^{(|\alpha+\beta|-1)/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
$\displaystyle\precsim\omega^{2}(\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2})M^{-(|\alpha+\beta|-1)/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}$

because $\omega\geq\sqrt{M}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\geq
M\langle{\xi}\rangle_{\gamma}^{-1}$. Therefore one concludes that

$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\omega^{2}\big{|}\precsim\omega^{2}(\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2})M^{-(|\alpha+\beta|-1)/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}$

which proves the assertion for $s=2$. For general $s$ noting

$\displaystyle\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(\omega^{2})^{s/2}\big{|}\precsim\sum_{|\alpha^{i}+\beta^{i}|\geq
1}\big{|}(\omega^{2})^{s/2}\big{(}\partial_{x}^{\alpha^{1}}\partial_{\xi}^{\beta^{1}}\omega^{2}/\omega^{2}\big{)}\cdots\big{(}\partial_{x}^{\alpha^{l}}\partial_{\xi}^{\beta^{l}}\omega^{2}/\omega^{2}\big{)}\big{|}$

the proof is immediate since
$\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\leq M^{-1/2}\leq 1$.
∎

######  Corollary 6.1.

We have $\omega^{s}\in S(\omega^{s},g)$ for $s\in{\mathbb{R}}$.

###  6.2 Estimate $\phi$ by metric $g$

Introduce a wight function which plays a crucial role in deriving energy
estimates

$\phi(t,x,\xi)=\omega(t,x,\xi)+t-\psi(x,\xi).$

If $t-\psi(x,\xi)\geq 0$ then $\phi\geq\omega=\omega^{2}/\omega\geq
M\rho\langle{\xi}\rangle_{\gamma}^{-1}/\omega$ and if $t-\psi(x,\xi)\geq 0$ we
see $\phi=M\rho\langle{\xi}\rangle_{\gamma}^{-1}/(\omega+|t-\psi|)\geq\rho
M\langle{\xi}\rangle_{\gamma}^{-1}/(2\omega)$ hence

(6.1) $\phi(t,x,\xi)\geq M\rho\langle{\xi}\rangle_{\gamma}^{-1}/(2\,\omega).$

######  Lemma 6.3.

There is $C>0$ such that $\phi(t,x,\xi)\geq
M\langle{\xi}\rangle_{\gamma}^{-1}/C$.

###### Proof.

When $t-\psi(x,\xi)\geq 0$ then $\phi\geq\omega\geq
M^{1/2}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\geq
M\langle{\xi}\rangle_{\gamma}^{-1}$ is obvious for $\rho\geq
M\langle{\xi}\rangle_{\gamma}^{-1}$. Assume $t-\psi(x,\xi)<0$ then $0\leq
t<\psi(x,\xi)\leq\delta\rho(x,\xi)$ with some $\delta>0$ by Lemma 4.5.
Noticing that $|t-\psi(x,\xi)|=\psi(x,\xi)-t\leq\delta\rho(x,\xi)$ we have
$\omega^{2}(t,x,\xi)\leq\delta^{2}\rho^{2}+M\rho\langle{\xi}\rangle_{\gamma}^{-1}\leq\delta^{2}\rho^{2}+\rho^{2}=(\delta^{2}+1)\rho^{2}$.
Now the proof is immediate from (6.1). ∎

######  Lemma 6.4.

We have $\phi\in S(\phi,g)$.

###### Proof.

Let $|\alpha+\beta|=1$ and write

(6.2)
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\phi=\frac{-\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi}{\omega}\phi+\frac{\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(M\rho\langle{\xi}\rangle_{\gamma}^{-1})}{2\omega}=\phi_{\alpha\beta}\phi+\psi_{\alpha\beta}.$

From Corollary 6.1 and Lemma 4.3 it follows that

$\displaystyle\big{|}\partial_{x}^{\mu}\partial_{\xi}^{\nu}\big{(}\psi_{\alpha\beta}\big{)}\big{|}\precsim\omega^{-1}M\rho\langle{\xi}\rangle_{\gamma}^{-1}M^{-|\alpha+\beta+\mu+\nu|/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha+\mu|-|\beta+\nu|)/2}$
$\displaystyle\precsim\phi
M^{-|\alpha+\beta+\mu+\nu|/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha+\mu|-|\beta+\nu|)/2}$

in view of (6.1). On the other hand thanks to Lemma 6.1 and Corollary 6.1 it
follows that
$|\partial_{x}^{\mu}\partial_{\xi}^{\nu}\phi_{\alpha\beta}|\precsim
M^{-|\alpha+\beta+\mu+\nu|/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha+\mu|-|\beta+\nu|)/2}$.
Hence using (6.2) the assertion is proved by induction on $|\alpha+\beta|$. ∎

We refine this lemma.

######  Lemma 6.5.

One has

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\phi\in
S(\phi\,M^{-(|\alpha+\beta|-1)/2}\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2},g),\quad|\alpha+\beta|\geq
1.$

###### Proof.

From Lemma 6.1 one has $\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\psi\in
S(\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2},g)$
for $|\alpha+\beta|=1$ hence $\phi_{\alpha\beta}\in
S(\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2},g)$
for $|\alpha+\beta|=1$ by Lemma 6.2. From Lemma 6.2 it follows that

$\big{|}\partial_{x}^{\mu}\partial_{\xi}^{\nu}\big{(}\psi_{\alpha\beta}\big{)}\big{|}\precsim\omega^{-1}\rho^{1/2}M\langle{\xi}\rangle_{\gamma}^{-1-|\beta|}M^{-|\mu+\nu|/2}\langle{\xi}\rangle_{\gamma}^{(|\mu|-|\nu|)/2}$

for $|\alpha+\beta|=1$ because
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(M\rho\langle{\xi}\rangle_{\gamma}^{-1})\in
S(M\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1-|\beta|},g)$. Thanks to Lemma
6.3 one sees $M\langle{\xi}\rangle_{\gamma}^{-1}\leq C\phi(t,x,\xi)$ and hence

$\psi_{\alpha\beta}\in
S(\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}\phi,g),\quad|\alpha+\beta|=1.$

Since $\phi\in S(\phi,g)$ by Lemma 6.4 we conclude the assertion from (6.2). ∎

##  7 $\phi$ and $\lambda_{j}$ are admissible weights for $g$

Write $z=(x,\xi)$ and $w=(y,\eta)$. It is clear that

$g^{\sigma}_{z}=M\big{(}\langle{\xi}\rangle_{\gamma}|dx|^{2}+\langle{\xi}\rangle_{\gamma}^{-1}|d\xi|^{2}\big{)}=M^{2}g_{z}$

where
$g_{z}^{\sigma}(t_{1},t_{2})=\sup|\langle{t_{2},s_{1}}\rangle-\langle{t_{1},s_{2}}\rangle|^{2}/g_{z}(s_{1},s_{2})$
(see [8, Chapter 18]). Note that $|\xi-\eta|\leq
c\,\langle{\xi}\rangle_{\gamma}$ with $0<c<1$ implies

$(1-c)\langle{\xi}\rangle_{\gamma}/\sqrt{2}\leq\langle{\eta}\rangle_{\gamma}\leq\sqrt{2}\,(1+c)\langle{\xi}\rangle_{\gamma}.$

If $g_{z}(w)<c$ then
$|\xi-\eta|^{2}<c\,M\langle{\xi}\rangle_{\gamma}=c\,M\langle{\xi}\rangle_{\gamma}^{-1}\langle{\xi}\rangle_{\gamma}^{2}\leq
c\,\langle{\xi}\rangle_{\gamma}^{2}$ then

$g_{z}(X)/C\leq g_{w}(X)\leq Cg_{z}(X),\quad
X\in{\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$

with $C$ independent of $\gamma\geq M^{5}\geq 1$ that is $g_{z}$ is slowly
varying uniformly in $\gamma\geq M^{5}\geq 1$. Similarly noting that
$|\xi-\eta|\geq(\gamma+|\xi|)/2\geq\langle{\xi}\rangle_{\gamma}/2$ if
$\langle{\eta}\rangle_{\gamma}\leq\langle{\xi}\rangle_{\gamma}/2\sqrt{2}$ and
$|\xi-\eta|\geq(\gamma+|\eta|)/2\geq\langle{\eta}\rangle_{\gamma}/2$ if
$\langle{\eta}\rangle_{\gamma}\geq 2\sqrt{2}\langle{\xi}\rangle_{\gamma}$ it
is clear that

(7.1)
$\frac{\langle{\xi}\rangle_{\gamma}}{\langle{\eta}\rangle_{\gamma}}+\frac{\langle{\eta}\rangle_{\gamma}}{\langle{\xi}\rangle_{\gamma}}\leq
C\big{(}1+\langle{\eta}\rangle_{\gamma}^{-1}|\xi-\eta|^{2}\big{)}\leq
C(1+g_{w}^{\sigma}(z-w)\big{)}$

hence $g_{w}(X)\leq Cg_{z}(X)\big{(}1+g_{w}^{\sigma}(z-w)\big{)}$, that is $g$
is a temperate metric uniformly in $\gamma\geq 0$ and $M\geq 1$ (see [8,
Chapter 18]). It is clear from (7.1) that

(7.2) $g_{z}^{\sigma}(z-w)\leq C\big{(}1+g_{w}^{\sigma}(z-w)\big{)}^{2}.$

###  7.1 $\rho$ and $\sigma$ are admissible weights for $g$

We adapt the same convention as in Sections 6, 5 even to weights for $g$ so
that we omit to say uniformly in $t\in[0,M^{-4}]$.

######  Lemma 7.1.

$\rho$ is an admissible weight for $g$.

###### Proof.

First study $\rho^{1/2}$. Assume
$g_{z}(w)=M^{-1}\langle{\xi}\rangle_{\gamma}\big{(}|y|^{2}+\langle{\xi}\rangle_{\gamma}^{-2}|\eta|^{2}\big{)}<c\;(<1/2)$
so that $M^{-1}\langle{\xi}\rangle_{\gamma}^{-1}|\eta|^{2}<c$ hence
$|\eta|<c\langle{\xi}\rangle_{\gamma}$ for
$M\langle{\xi}\rangle_{\gamma}^{-1}\leq 1$ hence

(7.3)
$\langle{\xi+s\eta}\rangle_{\gamma}/C\leq\langle{\xi}\rangle_{\gamma}\leq
C\langle{\xi+s\eta}\rangle_{\gamma}$

where $C$ is independent of $|s|\leq 1$. Lemma 4.3 shows

$|\rho^{1/2}(z+w)-\rho^{1/2}(z)|\leq
C\big{(}|y|+\langle{\xi+s\eta}\rangle_{\gamma}^{-1}|\eta|\big{)}\leq
CM^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}g_{z}^{1/2}(w).$

Since $\rho(z)\geq M\langle{\xi}\rangle_{\gamma}^{-1}$ this yields

(7.4) $|\rho^{1/2}(z+w)-\rho^{1/2}(z)|\leq C\rho^{1/2}(z)g_{z}^{1/2}(w).$

Choosing $c$ such that $C\,c<1/2$ one has
$\big{|}\rho(z+w)/\rho(z)-1\big{|}<1/2$ which implies
$\rho^{1/2}(z+w)/2\leq\rho^{1/2}(z)\leq 3\,\rho^{1/2}(z+w)/2$ that is
$\rho^{1/2}$ is $g$ continuous hence so is $\rho$. Note that
$M\langle{\xi}\rangle_{\gamma}^{-1}\leq\rho(z)\leq CM^{-4}\leq C$. If
$|\eta|\geq c\,\langle{\xi}\rangle_{\gamma}/2$ then $g_{z}^{\sigma}(w)\geq
Mc^{2}\langle{\xi}\rangle_{\gamma}/4$ and $g_{z}^{\sigma}(w)\geq Mc|\eta|/2$
thus

$\rho(z+w)\leq C\leq C\langle{\xi}\rangle_{\gamma}\rho(z)\leq
C^{\prime}\rho(z)(1+g_{z}^{\sigma}(w)).$

If $|\eta|\leq c\langle{\xi}\rangle_{\gamma}$ then (7.4) gives

(7.5) $\rho^{1/2}(z+w)\leq C\rho^{1/2}(z)(1+g_{z}(w))^{1/2}\leq
C\rho^{1/2}(z)(1+g_{z}^{\sigma}(w))^{1/2}$

hence, in view of (7.2), $\rho$ is an admissible weight. ∎

######  Lemma 7.2.

$\sigma$ is an admissible weight for $g$ and $\sigma\in S(\sigma,g)$.

###### Proof.

Since $\rho(z)+M\langle{\xi}\rangle_{\gamma}^{-1}$ is admissible for $g$ by
Lemma 7.1 then it is clear that so is
$\sigma=t+\rho(z)+M\langle{\xi}\rangle_{\gamma}^{-1}$ for $t\geq 0$. The
second assertion is clear from
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\sigma\big{|}\precsim\sigma^{1-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\precsim\sigma(M^{-1}\langle{\xi}\rangle_{\gamma})^{|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
for $\sigma\geq M\langle{\xi}\rangle_{\gamma}^{-1}$. ∎

###  7.2 $\omega$ and $\phi$ are admissible weights for $g$

We start with showing

######  Lemma 7.3.

$\omega$ and $\phi$ are $g$ continuous.

###### Proof.

Denote $f=t-\psi$ and $h=M^{1/2}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$
so that $\omega^{2}=f^{2}+h^{2}$. Note that

(7.6)
$\begin{split}|\omega(z+w)-\omega(z)|=|\omega^{2}(z+w)-\omega^{2}(z)|/|\omega(z+w)+\omega(z)|\\\
\leq 2|f(z+w)-f(z)|+2|h(z+w)-h(z)|\end{split}$

because $|f(z+w)+f(z)|/|\omega(z+w)+\omega(z)|\leq 2$,
$|h(z+w)+h(z)|/|\omega(z+w)+\omega(z)|\leq 2$. Assume $g_{z}(w)<c\;(\leq 1/2)$
hence (7.3). It is assumed that constants $C$ may change from line to line but
independent of $\gamma\geq M^{5}\geq 1$. Noting
$|f(z+w)-f(z)|=|\psi(z+w)-\psi(z)|$ it follows from Lemma 6.1 that

(7.7) $\begin{split}|f(z+w)-f(z)|\leq
C\rho^{1/2}(z+sw)\big{(}|y|+\langle{\xi+s\eta}\rangle_{\gamma}^{-1}|\eta|\big{)}\\\
\leq
C\rho^{1/2}(z+sw)\big{(}|y|+\langle{\xi}\rangle_{\gamma}^{-1}|\eta|\big{)}\leq
CM^{1/2}\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}g_{z}^{1/2}(w)\end{split}$

since $\rho$ is $g$ continuous. Noting that $\omega(z)\geq
M^{1/2}\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}$ it results

(7.8) $|f(z+w)-f(z)|\leq C\omega(z)g_{z}^{1/2}(w).$

Similar argument shows that $|h(z+w)-h(z)|\leq
CM^{1/2}\langle{\xi}\rangle_{\gamma}^{-1}g_{z}^{1/2}(w)$. Since $\omega(z)\geq
M\langle{\xi}\rangle_{\gamma}^{-1}$ we have $|h(z+w)-h(z)|\leq
CM^{-1/2}\omega(z)g_{z}^{1/2}(w)$. Therefore from (7.6) one has
$|\omega(z+w)-\omega(z)|\leq C\omega(z)g_{z}^{1/2}(w)$. Choosing $c$ such that
$C\,c<1/2$ we conclude that $\omega$ is $g$ continuous. Next consider
$\phi=\omega+f$. Write

(7.9) $\begin{split}&\phi(z+w)-\phi(z)\\\
=&\frac{(f(z+w)-f(z))(\phi(z+w)+\phi(z))+h^{2}(z+w)-h^{2}(z)}{\omega(z+w)+\omega(z)}.\end{split}$

Since $\omega(z+w)/C\leq\omega(z)\leq C\,\omega(z+w)$, decreasing $c>0$ if
necessary, which together with (7.8) gives
$|f(z+w)-f(z)|/(\omega(z+w)+\omega(z))\leq Cg_{z}^{1/2}(w)$. Recalling
$h^{2}(z)=M\rho(z)\langle{\xi}\rangle_{\gamma}^{-1}$ and repeating similar
arguments one sees

(7.10) $\begin{split}|h^{2}(z+w)-h^{2}(z)|\leq
CM^{1/2}\rho(z)\langle{\xi}\rangle_{\gamma}^{-1}g_{z}^{1/2}(w)\end{split}$

for $\rho^{1/2}(z)\geq M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}$. Taking
(6.1) into account it follows from (7.10) that

$|h^{2}(z+w)-h^{2}(z)|/(\omega(z+w)+\omega(z))\leq C\phi(z)g_{z}^{1/2}(w).$

Combining these estimates we obtain from (7.9) that

$\big{|}\phi(z+w)/\phi(z)-1\big{|}\leq
C\big{|}\phi(z+w)/\phi(z)+1\big{|}g_{z}^{1/2}(w)+Cg_{z}^{1/2}(w)$

which gives $\phi(z)/C\leq\phi(z+w)\leq C\,\phi(z)$ choosing $c>0$ small,
showing that $\phi$ is $g$ continuous. ∎

######  Lemma 7.4.

$\omega$ and $\phi$ are admissible weights for $g$ and $\omega\in
S(\omega,g)$, $\phi\in S(\phi,g)$.

###### Proof.

Note that $\langle{\xi}\rangle_{\gamma}^{-1}\leq
M\langle{\xi}\rangle_{\gamma}^{-1}\leq\sqrt{M}\sqrt{\rho}\langle{\xi}\rangle_{\gamma}^{-1/2}\leq\omega\leq
CM^{-4}\leq C$. Assume $|\eta|\geq c\,\langle{\xi}\rangle_{\gamma}$ hence
$g_{z}^{\sigma}(w)\geq Mc^{2}\langle{\xi}\rangle_{\gamma}\geq
c^{2}\langle{\xi}\rangle_{\gamma}$. Therefore

(7.11) $\omega(z+w)\leq C\leq C\langle{\xi}\rangle_{\gamma}\omega(z)\leq
C^{\prime}\omega(z)(1+g^{\sigma}_{z}(w)).$

Assume $|\eta|\leq c\,\langle{\xi}\rangle_{\gamma}$ and note that (7.5). Then
checking the proof of Lemma 7.3 we see that $|f(z+w)-f(z)|\leq
C\omega(z)(1+g_{z}^{\sigma}(w))$ and $|h(z+w)-h(z)|\leq
C\omega(z)(1+g_{z}^{\sigma}(w))^{1/2}$. Then (7.11) follows from (7.6) which
proves that $\omega$ is admissible for $g$. Turn to $\phi$. From Lemma 6.3 it
follows that

$\langle{\xi}\rangle_{\gamma}^{-1}/C\leq
M\langle{\xi}\rangle_{\gamma}^{-1}/C\leq\phi(z)=\omega(z)+f(z)\leq CM^{-4}\leq
C.$

If $|\eta|\geq\langle{\xi}\rangle_{\gamma}/2$ then $g_{z}^{\sigma}(w)\geq
M\langle{\xi}\rangle_{\gamma}/4\geq\langle{\xi}\rangle_{\gamma}/4$ hence

$\displaystyle\phi(z+w)\leq C\leq C^{2}\langle{\xi}\rangle_{\gamma}\phi(z)\leq
C\phi(z)(1+g_{z}^{\sigma}(w)).$

Assume $|\eta|\leq\langle{\xi}\rangle_{\gamma}/2$ so that (7.3) holds. From
(7.5) and (7.7) it results that

$|f(z+w)-f(z)|\leq
C\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}(1+g_{z}^{\sigma}(w)).$

Recalling (7.5) and $M^{2}g_{z}(w)=g_{z}^{\sigma}(w)$ the same arguments
obtaining (7.10) shows that $|h^{2}(z+w)-h^{2}(z)|\leq
C\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-3/2}(1+g_{z}^{\sigma}(w))$.
Taking these into account (7.9) yeilds

(7.12) $\begin{split}|\phi(z+w)-\phi(z)|\leq
C\Big{(}\frac{\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}}{\omega(z+w)+\omega(z)}\big{(}\phi(z+w)+\phi(z)\big{)}\\\
+\frac{\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-3/2}}{\omega(z+w)+\omega(z)}\Big{)}(1+g_{z}^{\sigma}(w)).\end{split}$

Applying Lemma 6.3 to (7.12) to obtain

$\displaystyle|\phi(z+w)-\phi(z)|\leq
C\big{(}\phi(z+w)+2\phi(z)\big{)}\frac{\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}}{\omega(z+w)+\omega(z)}(1+g_{z}^{\sigma}(w)).$

If
$\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}(1+g_{z}^{\sigma}(w))\big{/}(\omega(z+w)+\omega(z))<1/4$
then it follows

$\big{|}\phi(z+w)/\phi(z)-1\big{|}\leq\big{(}\phi(z+w)/\phi(z)+2\big{)}/4$

from which we have $\phi(z+w)\leq 2\phi(z)\leq 5\phi(z+w)$. If

$\rho^{1/2}(z)\langle{\xi}\rangle_{\gamma}^{-1/2}(1+g_{z}^{\sigma}(w))\big{/}(\omega(z+w)+\omega(z))\geq
1/4$

we have $32(1+g_{z}^{\sigma}(w))^{2}\geq
4\langle{\xi}\rangle_{\gamma}\omega(z+w)\omega(z)\big{/}\rho(z)\geq\phi(z+w)\big{/}\phi(z)$
by (6.1) and an obvious inequality $\phi(z+w)\leq 2\,\omega(z+w)$. Thus we
conclude that $\phi$ is admissible for $g$. ∎

###  7.3 $\lambda_{j}$ are admissible weights for $g$

######  Lemma 7.5.

Assume that $\lambda\in{\mathcal{C}}(\sigma^{2})$ and $\lambda\geq
cM\sigma\langle{\xi}\rangle_{\gamma}^{-1}$ with some $c>0$. Then $\lambda$ is
an admissible weight for $g$.

###### Proof.

Consider $\sqrt{\lambda}$. Assume $g_{z}(w)<c$ and hence
$\langle{\xi+s\eta}\rangle_{\gamma}\approx\langle{\xi}\rangle_{\gamma}$. Since
$\sqrt{\lambda}\in{\mathcal{C}}(\sigma)$ it follows that

(7.13) $\begin{split}\big{|}\sqrt{\lambda(z+w)}-\sqrt{\lambda(z)}\big{|}\leq
C\sqrt{\sigma(z+sw)}M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}g_{z}^{1/2}(w)\end{split}$

with $|s|<1$ which is bounded by
$C^{\prime}\sqrt{\sigma(z)}M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}g_{z}^{1/2}(w)$
since $\sigma$ is $g$ continuous. By assumption $\lambda(z)\geq
cM\sigma(z)\langle{\xi}\rangle_{\gamma}^{-1}$ one has

$\big{|}\sqrt{\lambda(z+w)}-\sqrt{\lambda(z)}\big{|}\leq
C^{\prime\prime}\sqrt{\lambda(z)}\,g_{z}^{1/2}(w).$

Choosing $c>0$ such that $C^{\prime\prime}\sqrt{c}<1$ shows that
$\sqrt{\lambda(z)}$ is $g$ continuous and so is $\lambda(z)$. From
$cM^{2}\langle{\xi}\rangle_{\gamma}^{-2}\leq
cM\sigma\langle{\xi}\rangle_{\gamma}^{-1}\leq\lambda\leq
C^{\prime}\sigma^{2}\leq C^{\prime}M^{-4}$ one sees

$c_{1}M\langle{\xi}\rangle_{\gamma}^{-1}\leq
c_{1}M^{1/2}\sigma^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\leq\sqrt{\lambda(z)}\leq
C.$

If $|\eta|\geq\langle{\xi}\rangle_{\gamma}/2$ hence $g_{z}^{\sigma}(w)\geq
M\langle{\xi}\rangle_{\gamma}/4$ then

$\displaystyle\sqrt{\lambda(z+w)}\leq C\leq
C(c_{1}M)^{-1}\langle{\xi}\rangle_{\gamma}\sqrt{\lambda(z)}\leq
C^{\prime}\sqrt{\lambda(z)}\,g_{z}^{\sigma}(w).$

If $|\eta|\leq\langle{\xi}\rangle_{\gamma}/2$, noting that $\sigma(z+w)\leq
C\,\sigma(z)(1+g_{z}^{\sigma}(w))$, it follows from (7.13)

$\displaystyle\big{|}\sqrt{\lambda(z+w)}-\sqrt{\lambda(z)}\big{|}\leq
C\sqrt{\lambda(z)}(1+g_{z}^{\sigma}(w))$

which proves that $\sqrt{\lambda}$ is admissible weight for $g$ and hence so
is $\lambda$. ∎

######  Lemma 7.6.

Assume that $\lambda\in{\mathcal{C}}(\sigma)$ and $\lambda\geq
cM\langle{\xi}\rangle_{\gamma}^{-1}$ with some $c>0$. Then $\lambda$ is an
admissible weight for $g$. If $\lambda\in{\mathcal{C}}(1)$ and $\lambda\geq c$
with some $c>0$ then $\lambda$ is an admissible weight for $g$.

###### Proof.

It is enough to repeat the proof of Lemma 7.5. ∎

######  Lemma 7.7.

Assume $\lambda\in{\mathcal{C}}(\sigma^{2})$ and $\lambda\geq
cM\sigma\langle{\xi}\rangle_{\gamma}^{-1}$ with some $c>0$. Then

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\lambda\in
S(\sqrt{\sigma}\sqrt{\lambda}\langle{\xi}\rangle_{\gamma}^{-|\beta|},g),\quad|\alpha+\beta|=1.$

In particular $\lambda\in S(\lambda,g)$.

###### Proof.

From $\lambda\in{\mathcal{C}}(\sigma^{2})$ we have
$|\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\lambda|\leq
C\,\sigma$ for $|\alpha+\beta|=2$. Since $\lambda\geq 0$, thanks to the
Glaeser’s inequality one has
$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\lambda\big{|}\leq
C^{\prime}\,\sqrt{\sigma}\,\sqrt{\lambda}\,\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
for $|\alpha+\beta|=1$. For $|\alpha^{\prime}+\beta^{\prime}|\geq 1$ note that

$\displaystyle\big{|}\partial_{x}^{\alpha^{\prime}}\partial_{\xi^{\prime}}^{\beta^{\prime}}\big{(}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\lambda\big{)}\big{|}\precsim\sigma^{1-(|\alpha^{\prime}+\beta^{\prime}|-1)/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\langle{\xi}\rangle_{\gamma}^{-|\beta^{\prime}|}$
$\displaystyle\precsim\sigma
M^{-|\alpha^{\prime}+\beta^{\prime}|/2}M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha^{\prime}|-|\beta^{\prime}|)/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$
$\displaystyle\precsim\sqrt{\sigma}M^{-|\alpha^{\prime}+\beta^{\prime}|/2}\sqrt{\lambda}\,\langle{\xi}\rangle_{\gamma}^{(|\alpha^{\prime}|-|\beta^{\prime}|)/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}$

because $\lambda\geq cM\sigma\langle{\xi}\rangle_{\gamma}^{-1}$ and
$\sigma\geq M\langle{\xi}\rangle_{\gamma}^{-1}$ which proves the first
assertion. Noting $\sqrt{\sigma}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\leq
CM^{-1/2}\sqrt{\lambda}\,\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}$
it is clear that $\lambda\in S(\lambda,g)$. ∎

######  Lemma 7.8.

Assume that $\lambda\in{\mathcal{C}}(\sigma)$ and $\lambda\geq
cM\langle{\xi}\rangle_{\gamma}^{-1}$ with some $c>0$. Then $\lambda\in
S(\lambda,g)$. If $\lambda\in{\mathcal{C}}(1)$ and $\lambda\geq c$ with some
$c>0$. Then $\lambda\in S(\lambda,g)$.

###### Proof.

It suffices to repeat the proof of Lemma 7.7. ∎

######  Corollary 7.1.

For $s\in{\mathbb{R}}$ we have $\lambda_{j}^{s}\in S(\lambda_{j}^{s},g)$,
$j=1,2,3$.

Define

$\kappa=\frac{1}{t}+\frac{1}{\omega}=\frac{t+\omega}{t\,\omega},\quad t>0.$

######  Lemma 7.9.

$\kappa$ is an admissible weight for $g$ and $\kappa^{s}\in S(\kappa^{s},g)$
for $s\in{\mathbb{R}}$.

###### Proof.

Since $\omega^{-1}$ is admissible for $g$ it is clear that so is
$\kappa=t^{-1}+\omega^{-1}$. Noting that $\omega^{-1}\in S(\omega^{-1},g)$ and
$\omega^{-1}\leq\kappa$ it is also clear that

$\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\kappa\big{|}=\big{|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\omega^{-1}\big{|}\precsim
M^{-|\alpha+\beta|/2}\kappa\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2}$

for $|\alpha+\beta|\geq 1$ which proves $\kappa\in S(\kappa,g)$. ∎

######  Lemma 7.10.

One has

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\kappa^{s}\in
S\big{(}M^{-(|\alpha+\beta|-1)/2}\kappa^{s}\omega^{-1}\rho^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2+(|\alpha|-\beta|)/2},g\big{)},\;\;|\alpha+\beta|\geq
1.$

###### Proof.

Since
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\kappa^{s}=\kappa^{s-1}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\kappa$
it is enough to show the case $s=1$. The proof for the case $s=1$ follows
easily from Lemma 6.2. ∎

######  Lemma 7.11.

Denote
${\bar{\varepsilon}}=\sqrt{3}/{\bar{e}}\,{\bar{c}}^{1/2}=4\sqrt{6}/{\bar{e}}$.
There is $C>0$ such that

$\frac{1}{\kappa\lambda_{1}}\leq{\bar{\varepsilon}}^{2}(1+CM^{-4})\kappa,\qquad\frac{1}{\sigma^{2}\kappa}\leq\kappa.$

###### Proof.

In view of Propositions 4.1 and 5.1 one sees

$\lambda_{1}\geq(1/{\bar{\varepsilon}}^{2})(1-CM^{-4})\min{\big{\\{}t^{2},\omega^{2}\big{\\}}}.$

Denote $c={\bar{\varepsilon}}^{2}(1-CM^{-4})^{-1}$. If $\omega^{2}\geq t^{2}$
and hence $\lambda_{1}\geq t^{2}/c$ then $1/\lambda_{1}\leq c/t^{2}$ which
shows that

$\frac{1}{\kappa\lambda_{1}}\leq\frac{c}{\kappa
t^{2}}=\frac{c\,t\,\omega}{(t+\omega)t^{2}}=\frac{c\,\omega}{(t+\omega)t}\leq\frac{c(t+\omega)}{t\omega}=c\kappa.$

If $t^{2}\geq\omega^{2}$ and hence $\lambda_{1}\geq\omega^{2}/c$ then
$1/\lambda_{1}\leq c/\omega^{2}$ and hence

$\frac{1}{\kappa\lambda_{1}}\leq\frac{c}{\kappa\omega^{2}}=\frac{c\,t\,\omega}{(t+\omega)\omega^{2}}=\frac{c\,t}{(t+\omega)\omega}\leq\frac{c(t+\omega)}{t\,\omega}=c\kappa$

thus the first assertion. To show the second assertion it suffices to note
$\sigma\geq t$ and then $\sigma^{2}(t+\omega)^{2}\geq t^{2}(t+\omega)^{2}\geq
t^{2}\omega^{2}$. ∎

##  8 Lower bounds of ${\rm op}({\lambda_{j}})$

###  8.1 Some preliminary lemmas

Introduce a metric
${\bar{g}}=\langle{\xi}\rangle_{\gamma}|dx|^{2}+\langle{\xi}\rangle_{\gamma}^{-1}|d\xi|^{2}$
independent of $M$ so that $g=M^{-1}\,{\bar{g}}$. We start with

######  Lemma 8.1.

Let $m$ be an admissible weight for $g$ and $p\in S(m,g)$ satisfy $p\geq c\,m$
with some constant $c>0$. Then $p^{-1}\in S(m^{-1},g)$ and there exist
$k,{\tilde{k}}\in S(M^{-1},g)$ such that

$\displaystyle
p\\#p^{-1}\\#(1+k)=1,\;\;(1+k)\\#p\\#p^{-1}=1,\;\;p^{-1}\\#(1+k)\\#p=1,$
$\displaystyle
p^{-1}\\#p\\#(1+{\tilde{k}})=1,\;\;(1+{\tilde{k}})\\#p^{-1}\\#p=1,\;\;p\\#(1+{\tilde{k}})\\#p^{-1}=1.$

###### Proof.

In this proof every constant is independent of $\gamma\geq 1$ and $M$. It is
clear that $p^{-1}\in S(m^{-1},g)$. Write $p\\#p^{-1}=1-r$ where $r\in
S(M^{-1},g)$. Since

$|r|^{(l)}_{S(1,{\bar{g}})}=\sup_{|\alpha+\beta|\leq
l,(x,\xi)\in{\mathbb{R}}^{2d}}\big{|}\langle{\xi}\rangle_{\gamma}^{(|\beta|-|\alpha|)/2}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}r\big{|}\leq
C_{l}M^{-1}$

from the $L^{2}$-boundedness theorem (see [8, Theorem 18.6.3]) we have $\|{\rm
op}(r)\|\leq CM^{-1}$. Therefore for large $M$ there exists the inverse
$(1-{\rm op}({r}))^{-1}$ which is given by
$1+\sum_{\ell=1}^{\infty}r^{\\#\ell}\in S(1,{\bar{g}})$. (see [1], [20]).
Denote $k=\sum_{\ell=1}^{\infty}r^{\\#\ell}\in S(1,{\bar{g}})$ and prove $k\in
S(M^{-1},g)$. It can be seen from the proof (e.g. [20]) that for any
$l\in{\mathbb{N}}$ one can find $C_{l}>0$, independent of $\gamma$, such that

$|k|^{(l)}_{S(1,{\bar{g}})}\leq C_{l}$

because $|k|^{(l)}_{S(1,{\bar{g}})}$ depends only on $l$,
$|r|^{(l^{\prime})}_{S(1,{\bar{g}})}$ with some $l^{\prime}=l^{\prime}(l)$ and
structure constants of ${\bar{g}}$ which is independent of $\gamma$. Note that
$k$ satisfies $(1-r)\\#(1+k)=1$, that is

(8.1) $k=r+r\\#k.$

Since $r\in S(M^{-1},g)$ it follows from (8.1) that
$\big{|}k\big{|}^{(l)}_{S(1,{\bar{g}})}\leq C_{l}M^{-1}$. Assume that

(8.2)
$\sup\big{|}\langle{\xi}\rangle_{\gamma}^{(|\beta|-|\alpha|)/2}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}k\big{|}\leq
C_{\alpha,\beta,\nu}M^{-1-l/2},\quad|\alpha+\beta|\geq l$

for $0\leq l\leq\nu$. Let $|\alpha+\beta|\geq\nu+1$ and note that

$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}k=\partial_{x}^{\alpha}\partial_{\xi}^{\beta}r+\sum
C_{\cdots}\big{(}\partial_{x}^{\alpha^{\prime\prime}}\partial_{\xi}^{\beta^{\prime\prime}}r\big{)}\\#\big{(}\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}k\big{)}$

where $\alpha^{\prime}+\alpha^{\prime\prime}=\alpha$ and
$\beta^{\prime}+\beta^{\prime\prime}=\beta$. From the assumption (8.2) we have
$\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}k\in
S(M^{-1-|\alpha^{\prime}+\beta^{\prime}|/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha^{\prime}|-|\beta^{\prime}|)/2},{\bar{g}})$
if $|\alpha^{\prime}+\beta^{\prime}|\leq\nu$ and
$\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}k\in
S(M^{-1-\nu/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha^{\prime}|-|\beta^{\prime}|)/2},{\bar{g}})$
if $|\alpha^{\prime}+\beta^{\prime}|\geq\nu+1$. Since $r\in S(M^{-1},g)$ one
has

$\big{(}\partial_{x}^{\alpha^{\prime\prime}}\partial_{\xi}^{\beta^{\prime\prime}}r\big{)}\\#\big{(}\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}k\big{)}\in
S(M^{-1-(\nu+2)/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2},{\bar{g}})$

which implies that (8.2) holds for $0\leq l\leq\nu+1$ and hence for all $\nu$
by induction on $\nu$. This proves that $k\in S(M^{-1},g)$. The proof of the
assertions for ${\tilde{k}}$ is similar. ∎

Here recall [27, Lemmas 3.1.6, 3.1.7].

######  Lemma 8.2.

Let $q\in S(1,g)$ satisfy $q\geq c$ with a constant $c$ independent of $M$.
Then there is $C>0$ such that

$\big{(}{\rm op}({q})u,u)\geq(c-CM^{-1/2})\|u\|^{2}.$

###### Proof.

One can assume $c=0$. We see $q(x,\xi)+M^{-1/2}$ is an admissible weight for
${\bar{g}}$ and $(q+M^{-1/2})^{1/2}\in S((q+M^{-1/2})^{1/2},{\bar{g}})$.
Moreover $\partial_{x}^{\alpha}\partial_{\xi}^{\beta}(q+M^{-1/2})^{1/2}\in
S(M^{-1/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha|-|\beta|)/2},{\bar{g}})$ for
$|\alpha+\beta|=1$. Therefore

$q+M^{-1/2}=(q+M^{-1/2})^{1/2}\\#(q+M^{-1/2})^{1/2}+r,\quad r\in
S(M^{-1},{\bar{g}})$

which proves the assertion. ∎

######  Lemma 8.3.

Let $q\in S(1,g)$ then there is $C>0$ such that

$\|{\rm op}({q})u\|\leq\big{(}\sup{|q|}+CM^{-1/2}\big{)}\|u\|.$

######  Lemma 8.4.

Let $m>0$ be an admissible weight for $g$ and $m\in S(m,g)$. Then there is
$C>0$ such that

$({\rm op}({m})u,u)\geq(1-CM^{-2})\|{\rm op}({\sqrt{m}})u\|^{2}.$

If $q\in S(m,g)$ then there is $C>0$ such that

$\big{|}({\rm
op}({q})u,u)\big{|}\leq\big{(}\sup{\big{(}|q|/m\big{)}}+CM^{-1/2}\big{)}\|{\rm
op}({\sqrt{m}\,})u\|^{2}.$

###### Proof.

First note that $m^{\pm 1/2}$ are admissible weights and $m^{\pm 1/2}\in
S(m^{\pm 1/2},g)$. Write

${\tilde{q}}=(1+k)\\#m^{-1/2}\\#q\\#m^{-1/2}\\#(1+{\tilde{k}})\in S(1,g)$

where $m^{1/2}\\#(1+k)\\#m^{-1/2}=1$ and
$m^{-1/2}\\#(1+{\tilde{k}})\\#m^{1/2}=1$ such that
$m^{1/2}\\#{\tilde{q}}\\#m^{1/2}=q$. Since $k$, ${\tilde{k}}\in S(M^{-1},g)$
one sees that ${\tilde{q}}=qm^{-1}+r$ with $r\in S(M^{-1},g)$. Thanks to Lemma
8.3 we have $\|{\rm
op}({qm^{-1}})v\|\leq(\sup{\big{(}|q|/m\big{)}}+CM^{-1/2})\|v\|$ hence
$\big{|}({\rm op}({q})u,u)\big{|}$ is bounded by

$\displaystyle\big{|}({\rm op}({qm^{-1}}){\rm op}({m^{1/2}})u,{\rm
op}({m^{1/2}})u)|+CM^{-1}\|{\rm op}({m^{1/2}})u\|^{2}$

which proves the second assertion. The first assertion follows from the second
since $m=m^{1/2}\\#m^{1/2}+r$ with $r\in S(M^{-2}m,g)$. ∎

######  Lemma 8.5.

Let $m_{i}>0$ be two admissible weights for $g$ and assume that $m_{i}\in
S(m_{i},g)$ and $m_{2}\leq C\,m_{1}$ with $C>0$. Then there is $C^{\prime}>0$
such that

$\big{\|}{\rm op}({m_{2}})u\big{\|}\leq C^{\prime}\big{\|}{\rm
op}({m_{1}})u\big{\|}.$

###### Proof.

Write ${\tilde{m}}_{2}=m_{2}\\#m_{1}^{-1}\\#(1+k)\in S(1,g)$ such that
$m_{2}={\tilde{m}}_{2}\\#m_{1}$ with $k\in S(M^{-1},g)$. Then $\|{\rm
op}({m_{2}})u\|=\|{\rm op}({{\tilde{m}}_{2}}){\rm op}({m_{1}})u\|\leq
C^{\prime}\|{\rm op}({m_{1}})u\|$ proves the assertion. ∎

###  8.2 Lower bounds of ${\rm op}({\lambda_{j}})$

######  Lemma 8.6.

There exist $C>0$ and $M_{0}$ such that

$\displaystyle{\mathsf{Re}}\,\big{(}{\rm
op}({\lambda_{j}\\#\kappa})u,u\big{)}\geq(1-CM^{-2})\big{\|}{\rm
op}({\kappa^{1/2}\lambda_{j}^{1/2}})u\big{\|}^{2},\quad M\geq M_{0}.$

###### Proof.

Since $\lambda_{j}\\#\kappa=\kappa\lambda_{j}+r_{j1}+r_{j2}$ where $r_{j1}$ is
pure imaginary and $r_{j2}\in S(M^{-2}\kappa\lambda_{j},g)$ the assertion
follows from Lemma 8.4. ∎

######  Lemma 8.7.

There exist $c>$ and $M_{0}$ such that

$\displaystyle\big{(}{\rm op}({\lambda_{1}})u,u\big{)}\geq c\,\big{\|}{\rm
op}({\lambda_{1}^{1/2}})u\big{\|}^{2}+cM^{2}\big{\|}\langle{D}\rangle_{\gamma}^{-1}u\big{\|}^{2},\quad
M\geq M_{0}.$

###### Proof.

From Propositions 4.1 and 5.1 it follows that $\lambda_{1}\geq
c^{\prime}\,M\sigma\langle{\xi}\rangle_{\gamma}^{-1}$ with some
$c^{\prime}>0$. Write

${\lambda}_{1}-c\,M\sigma\langle{\xi}\rangle_{\gamma}^{-1}=\lambda_{1}/2+(\lambda_{1}/2-c\,M\sigma\langle{\xi}\rangle_{\gamma}^{-1})$

where $c>0$ is chosen so that
${\tilde{\lambda}}_{1}=\lambda_{1}/2-c\,M\sigma\langle{\xi}\rangle_{\gamma}^{-1}\geq
c_{1}M\sigma\langle{\xi}\rangle_{\gamma}^{-1}$ with $c_{1}>0$. Note that
${\tilde{\lambda}}_{1}\in{\mathcal{C}}(\sigma^{2})$ since
$M\sigma\langle{\xi}\rangle_{\gamma}^{-1}\in{\mathcal{C}}(\sigma^{2})$. Thanks
to Lemmas 7.5 and 7.7 it follows that ${\tilde{\lambda}}_{1}\in
S({\tilde{\lambda}}_{1},g)$ and ${\tilde{\lambda}}_{1}$ is an admissible
weight for $g$. Thus we have $({\rm
op}({{\tilde{\lambda}}_{1}})u,u)\geq(1-CM^{-2})\|{\rm
op}({{\tilde{\lambda}}_{1}^{1/2}})u\|^{2}\geq 0$ if $M\geq\sqrt{C}$ by Lemma
8.4. Since $M^{2}\langle{\xi}\rangle_{\gamma}^{-2}\leq
M\sigma\langle{\xi}\rangle_{\gamma}^{-1}$ it follows from Lemma 8.5 that

$M\|\langle{D}\rangle_{\gamma}^{-1}u\|^{2}\leq C\|{\rm
op}({\sigma^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2}})u\|^{2}.$

Therefore the proof follows from Lemma 8.4. ∎

Similar arguments prove the following lemma.

######  Lemma 8.8.

There exist $c>0$ and $M_{0}$ such that

$\displaystyle\big{(}{\rm op}({\lambda_{2}})u,u\big{)}\geq c\,\big{\|}{\rm
op}({\lambda_{2}^{1/2}})u\big{\|}^{2}+c\,M\big{\|}\langle{D}\rangle_{\gamma}^{-1/2}u\big{\|}^{2},\quad
M\geq M_{0},$ $\displaystyle\big{(}{\rm op}({\lambda_{3}})u,u\big{)}\geq
c\,\big{\|}u\big{\|}^{2},\quad M\geq M_{0}.$

Summarize what we have proved in

######  Proposition 8.1.

There exist $c>0$, $C>0$ and $M_{0}$ such that

$\displaystyle{\mathsf{Re}}\,({\rm
op}({\varLambda\\#\kappa})W,W)\geq(1-CM^{-2})\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}})W\|^{2},$
$\displaystyle{\mathsf{Re}}\,({\rm op}({\varLambda})W,W)\geq c\,(\|{\rm
op}({\varLambda^{1/2}})W\|^{2}+\|{\rm op}({{\mathcal{D}}})W\|^{2})$

for $M\geq M_{0}$ where ${\mathcal{D}}={\rm
diag}\big{(}M\langle{\xi}\rangle_{\gamma}^{-1},M^{1/2}\langle{\xi}\rangle_{\gamma}^{-1/2},1\big{)}$.

##  9 System with diagonal symmetrizer

Diagonalizing the Bézout matrix introduced in Section 5 we reduce the system
(5.2) to a system with a diagonal symmetrizer.

######  Lemma 9.1.

Let $p\in{\mathcal{C}}(\sigma^{k})$ then
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}p\in
S(\sigma^{k-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},g)$.

###### Proof.

The proof is clear from

$\displaystyle\big{|}\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}(\partial_{x}^{\alpha}\partial_{\xi}^{\beta}p)\big{|}\precsim\sigma^{k-|\alpha^{\prime}+\beta^{\prime}+\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta^{\prime}+\beta|}$
$\displaystyle\precsim\sigma^{k-|\alpha+\beta|/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|}\sigma^{-|\alpha^{\prime}+\beta^{\prime}|/2}\langle{\xi}\rangle_{\gamma}^{-|\alpha^{\prime}+\beta^{\prime}|/2}\langle{\xi}\rangle_{\gamma}^{(|\alpha^{\prime}|-|\beta^{\prime}|)/2}$

for $\sigma\geq\rho\geq M\langle{\xi}\rangle_{\gamma}^{-1}$. ∎

######  Lemma 9.2.

Let $p\in{\mathcal{C}}(\sigma^{k})$ and $q\in{\mathcal{C}}(\sigma^{\ell})$.
Then

$p\\#p-p^{2}\in S(\sigma^{2k-2}\langle{\xi}\rangle_{\gamma}^{-2},g),\quad
p\\#q-pq\in S(\sigma^{k+\ell-1}\langle{\xi}\rangle_{\gamma}^{-1},g).$

###### Proof.

The assertions follows from Lemma 9.1 and the Weyl calculus of
pseudodifferential operators. ∎

In what follows to simplify notations sometimes we abbreviate $S(m,g)$ to
$S(m)$ where $m$ is admissible for $g$. Since $a\in{\mathcal{C}}(\sigma)$,
$b\in{\mathcal{C}}(\sigma^{3/2})$ one sees $A\\#[\xi]=A(t,x,\xi)[\xi]+R$ with
$R$ whose first row is $(0,S(\sigma^{1/2}),S(\sigma))$ for
$\partial_{\xi}^{\beta}[\xi]\in S(1,g)$ by (4.5). Moving $R$ to $B$ we denote
$L=D_{t}-{\rm op}({{\tilde{A}}})-{\rm op}({B})$ where

(9.1) ${\tilde{A}}=\begin{bmatrix}0&a&b\\\ 1&0&0\\\
0&1&0\end{bmatrix}[\xi],\quad
B=\begin{bmatrix}b_{1}&b_{2}+d_{M}+S(\sigma^{1/2})&b_{3}+S(\sigma)\\\ 0&0&0\\\
0&0&0\end{bmatrix}$

and transform $L$ to another system using $T$ introduced in Section 5.3. Note
that $T^{-1}\\#T=I-R$ with $R\in S(M^{-1},g)$. Thanks to Lemma 8.1 there is
$K\in S(M^{-1},g)$ such that $(I-R)\\#(I+K)=I=(I+K)\\#(I-R)$ and hence

$T^{-1}\\#T\\#(I+K)=I,\quad(I+K)\\#T^{-1}\\#T=I,\quad T\\#(I+K)\\#T^{-1}=I.$

Therefore one can write

(9.2) $L\,{\rm op}({T})={\rm op}({T})\,{\tilde{L}}$

where ${\tilde{L}}=D_{t}-{\rm op}({(I+K)\\#T^{-1}\\#({\tilde{A}}+B)\\#T})+{\rm
op}({(I+K)\\#T^{-1}\\#(D_{t}T)})$.

######  Lemma 9.3.

Notations being as above. Then $K\in
S(M^{-1}\langle{\xi}\rangle_{\gamma}^{-1},g)$.

###### Proof.

Write $T=(t_{ij})$ then $T^{-1}\\#T=(\sum_{k=1}^{3}t_{ki}\\#t_{kj})$. Denote

$\sum_{k=1}^{3}t_{ki}\\#t_{kj}=\delta_{ij}+r_{ij}.$

Taking Lemma 5.6 into account, we see $r_{ii}\in
S(\sigma^{-1}\langle{\xi}\rangle_{\gamma}^{-2},g)\subset
S(M^{-1}\langle{\xi}\rangle_{\gamma}^{-1},g)$ and $r_{ij}\in
S(\sigma^{1/2}\langle{\xi}\rangle_{\gamma}^{-1},g)\subset
S(M^{-2}\langle{\xi}\rangle_{\gamma}^{-1},g)$ for $i\neq j$ thanks to Lemma
9.2 hence $R\in S(M^{-1}\langle{\xi}\rangle_{\gamma}^{-1},g)$. Since $K\in
S(M^{-1},g)$ satisfies $K=R+R\\#K$ we conclude the assertion. ∎

Therefore $K\\#T^{-1}\\#({\tilde{A}}+B)\\#T\in S(M^{-1},g)$ is clear. Hence

${\tilde{L}}=D_{t}-{\rm
op}({T^{-1}\\#({\tilde{A}}+B)\\#T-T^{-1}\\#(D_{t}T)})+{\rm
op}({S(M^{-1},g)}).$

In view of Lemmas 5.6 and 5.7 it follows from Lemma 9.2 that

(9.3) $\begin{split}&T^{-1}\\#(\partial_{t}T)=T^{-1}\partial_{t}T\\\
&+\begin{bmatrix}S(\sigma^{-1}\langle{\xi}\rangle_{\gamma}^{-1})&S(\sigma^{-1/2}\langle{\xi}\rangle_{\gamma}^{-1})&S(\langle{\xi}\rangle_{\gamma}^{-1})\\\
S(\sigma^{-1/2}\langle{\xi}\rangle_{\gamma}^{-1})&S(\sigma^{-1}\langle{\xi}\rangle_{\gamma}^{-1})&S(\sigma^{-1/2}\langle{\xi}\rangle_{\gamma}^{-1})\\\
S(\langle{\xi}\rangle_{\gamma}^{-1})&S(\sigma^{-1/2}\langle{\xi}\rangle_{\gamma}^{-1})&S(\langle{\xi}\rangle_{\gamma}^{-1})\\\
\end{bmatrix}\end{split}$

hence $T^{-1}\\#(\partial_{t}T)=T^{-1}\partial_{t}T+S(M^{-1},g)$ because
$\sigma\geq M\langle{\xi}\rangle_{\gamma}^{-1}$.

Turn to study $T^{-1}\\#{\tilde{A}}\\#T$. Noting that
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a\in
S(\sigma^{1/2}\langle{\xi}\rangle_{\gamma}^{-|\beta|},g)$ and
$\partial_{x}^{\alpha}\partial_{\xi}^{\beta}b\in
S(\sigma\langle{\xi}\rangle_{\gamma}^{-|\beta|},g)$ for $|\alpha+\beta|=1$ and
$\partial_{\xi}^{\beta}[\xi]\in S(1,g)$, $|\beta|=1$ we have

$T^{-1}\\#{\tilde{A}}=T^{-1}{\tilde{A}}+R,\quad
R=\begin{bmatrix}S(1)&S(M^{-2})&S(M^{-6})\\\ S(M^{-2})&S(1)&S(M^{-8})\\\
S(M^{-8})&S(M^{-2})&S(M^{-6})\end{bmatrix}.$

Therefore $T^{-1}\\#{\tilde{A}}\\#T=(T^{-1}{\tilde{A}})\\#T+R_{1}$ with

$R_{1}=R\\#T=\begin{bmatrix}S(M^{-4})&S(M^{-2})&S(1)\\\
S(M^{-2})&S(1)&S(M^{-2})\\\ S(M^{-4})&S(M^{-2})&S(M^{-8})\end{bmatrix}.$

Note that

$T^{-1}{\tilde{A}}=\begin{bmatrix}C(\sigma^{1/2})&1+C(\sigma)&C(\sigma^{5/2})\\\
-1+C(\sigma)&C(\sigma^{1/2})&C(\sigma^{3})\\\
C(\sigma^{5/2})&C(\sigma)&C(\sigma^{3/2})\end{bmatrix}[\xi]$

and hence

$\langle{\xi}\rangle_{\gamma}^{|\beta|}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\big{(}T^{-1}{\tilde{A}}\big{)}=\begin{bmatrix}S(1)&S(1)&S(M^{-8})\\\
S(1)&S(1)&S(M^{-10})\\\
S(M^{-8})&S(M^{-2})&S(M^{-4})\end{bmatrix},\quad|\alpha+\beta|=1.$

Then thanks to (5.11) one sees

$(T^{-1}{\tilde{A}})\\#T=T^{-1}{\tilde{A}}T+R_{2},\quad
R_{2}=\begin{bmatrix}S(1)&S(M^{-2})&S(M^{-2})\\\ S(1)&S(M^{-2})&S(M^{-2})\\\
S(M^{-2})&S(M^{-4})&S(M^{-6})\end{bmatrix}.$

Thus we obtain $T^{-1}\\#{\tilde{A}}\\#T=T^{-1}{\tilde{A}}T+R_{1}+R_{2}$ where

$R_{1}+R_{2}=\begin{bmatrix}S(1)&S(M^{-2})&S(M^{-2})\\\
S(1)&S(M^{-2})&S(M^{-2})\\\ S(M^{-2})&S(M^{-4})&S(M^{-6})\end{bmatrix}.$

Recall $B$ given by (9.1). Since $d_{M}\in S(M,g)$ one sees by Lemma 5.6 that

$T^{-1}\\#B=\begin{bmatrix}S(\sigma)&S(M\sigma)&S(\sigma)\\\
S(\sigma^{3/2})&S(M\sigma^{3/2})&S(\sigma^{3/2})\\\
b_{1}+S(\sigma)&b_{2}+d_{M}+S(\sigma^{1/2})&b_{3}+S(\sigma)\end{bmatrix}$

because $\sigma\leq CM^{-4}$. Thus we conclude that $T^{-1}\\#B\\#T$ is
written

(9.4) $\begin{bmatrix}S(\sigma)&S(M\sigma)&S(\sigma)\\\
S(\sigma^{3/2})&S(M\sigma^{3/2})&S(\sigma^{3/2})\\\
b_{3}+S(M\sigma^{1/2})&-b_{2}-d_{M}+S(\sigma^{1/2})&b_{1}+S(\sigma)\end{bmatrix}.$

Noting $b_{3}(t,x,\xi)-{\bar{b}}_{3}\in S(M^{-2},g)$ we can summarize what we
have proved in

######  Proposition 9.1.

One can write $L\cdot{\rm op}({T})={\rm op}({T})\cdot{\tilde{L}}$ where

$\displaystyle{\tilde{L}}=D_{t}-{\rm
op}({{\mathcal{A}}+{\mathcal{B}}}),\quad{\mathcal{A}}=(T^{-1}AT)[\xi]\,,\quad{\mathcal{B}}={\mathcal{B}}_{1}-T^{-1}D_{t}T$
$\displaystyle{\mathcal{B}}_{1}=T^{-1}\\#B\\#T=\begin{bmatrix}S(1)&S(1)&S(1)\\\
S(1)&S(1)&S(1)\\\
{\bar{b}}_{3}+S(M^{-1})&-2M{\bar{e}}+S(M^{-1})&S(1)\end{bmatrix}.$

##  10 Weighted energy estimates

###  10.1 Energy form

Let $w=t\phi(t,x,\xi)$ and consider the energy with scalar weight ${\rm
op}({w^{-n}})$;

${\mathcal{E}}(V)=e^{-\theta t}\big{(}{\rm op}({\varLambda}){\rm
op}({w^{-n}})V,{\rm op}({w^{-n}})V\big{)}$

where $\theta>0$ is a large positive parameter and $n$ is fixed such that

(10.1)
$n>(4\sqrt{2}\,)\,|3\,{\bar{b}}_{3}+i{\bar{e}}|/{\bar{e}}+C^{*}+2+8(1+3\sqrt{2})$

where $C^{*}$ is given by (3.14). It is clear from (4.17) that (10.1) follows
from

(10.2)
$n>12\sqrt{2}\,\,\frac{|P_{sub}(0,0,0,{\bar{\xi}})|}{{\bar{e}}}+{\bar{C}^{*}},\quad{\bar{C}^{*}}=C^{*}+10+32\sqrt{2}.$

Note that $\partial_{t}\phi=\omega^{-1}\phi$ and hence

$\partial_{t}w^{-n}=-n\Big{(}\frac{1}{t}+\frac{1}{\omega}\Big{)}w^{-n}=-n\kappa\,w^{-n}.$

Recall that $V$ satisfies

(10.3) $\partial_{t}V={\rm
op}({i{\mathcal{A}}+i{\mathcal{B}}})V+F,\quad{\mathcal{B}}={\mathcal{B}}_{1}-T^{-1}D_{t}T.$

Noting that $\Lambda$ is real and diagonal hence ${\rm
op}({\varLambda})^{*}={\rm op}({\varLambda})$ one has

(10.4) $\begin{split}\frac{d}{dt}{\mathcal{E}}=-\theta e^{-\theta
t}\big{(}{\rm op}({\varLambda}){\rm op}({w^{-n}})V,{\rm
op}({w^{-n}})V\big{)}\\\ -2n{\mathsf{Re}}\,e^{-\theta t}\big{(}{\rm
op}({\varLambda}){\rm op}({\kappa w^{-n}})V,{\rm op}({w^{-n}})V)\\\
+e^{-\theta t}\big{(}{\rm op}({\partial_{t}\Lambda}){\rm op}({w^{-n}})V,{\rm
op}({w^{-n}})V\big{)}\\\ +2{\mathsf{Re}}\,e^{-\theta t}\big{(}{\rm
op}({\varLambda}){\rm op}({w^{-n}})({\rm
op}({i{\mathcal{A}}+i{\mathcal{B}}})V+F),{\rm op}({w^{-n}})V).\end{split}$

Consider ${\rm op}({\phi^{-n}}){\rm op}({\varLambda}){\rm
op}({\kappa\phi^{-n}})={\rm
op}({\phi^{-n}\\#\varLambda\\#(\kappa\phi^{-n})})$. Since $\kappa$ and
$\phi^{-n}$ are admissible weights for $g$ one has
$\kappa\\#\phi^{-n}=\kappa\phi^{-n}-r$ with $r\in S(M^{-1}\kappa\phi^{-n},g)$.
Let ${\tilde{r}}=r\\#\phi^{n}\\#(1+k)\in S(M^{-1}\kappa,g)$ such that
$r={\tilde{r}}\\#\phi^{-n}$ and hence
$\kappa\phi^{-n}=(\kappa+{\tilde{r}})\\#\phi^{-n}$. Thus we have

$\displaystyle{\mathsf{Re}}\,\big{(}{\rm op}({\varLambda}){\rm op}({\kappa
w^{-n}})V,{\rm op}({w^{-n}})V\big{)}\geq{\mathsf{Re}}\,\big{(}{\rm
op}({\varLambda\\#\kappa}){\rm op}({w^{-n}})V,{\rm op}({w^{-n}})V\big{)}$
$\displaystyle-\big{|}\big{(}{\rm op}({\varLambda\\#{\tilde{r}}}){\rm
op}({w^{-n}})V,{\rm op}({w^{-n}})V)\big{|}.$

Since $\lambda_{j}\\#{\tilde{r}}\in S(M^{-1}\kappa\lambda_{j},g)$ thanks to
Lemma 8.4 the second term on the right-hand side is bounded by $CM^{-1}\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}}){\rm op}({w^{-n}})V\|$. Applying
Proposition 8.1, denoting $W_{j}={\rm op}({w^{-n}})V_{j}$, one can conclude
that

$\displaystyle{\mathsf{Re}}\,\big{(}{\rm op}({\varLambda}){\rm op}({\kappa
w^{-n}})V,{\rm op}({w^{-n}})V\big{)}\geq(1-CM^{-1})\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}})W\|^{2},$
$\displaystyle{\mathsf{Re}}\,\big{(}{\rm op}({\varLambda}){\rm
op}({w^{-n}})V,{\rm op}({w^{-n}})V\big{)}\geq c\big{(}\|{\rm
op}({\varLambda^{1/2}})W\|^{2}+\|{\rm op}({{\mathcal{D}}})W\|^{2}\big{)}$

for $M\geq M_{0}$.

######  Definition 10.1.

To simplify notations we denote

$\displaystyle{\mathcal{E}}_{1}(V)=\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}}){\rm op}({w^{-n}})V\|^{2}=t^{-2n}\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}}){\rm op}({\phi^{-n}})V\|^{2},$
$\displaystyle{\mathcal{E}}_{2}(V)=\|{\rm op}({\varLambda^{1/2}}){\rm
op}({w^{-n}})V\|^{2}=t^{-2n}\|{\rm op}({\varLambda^{1/2}}){\rm
op}({\phi^{-n}})V\|^{2}.$

Now we summarize

######  Lemma 10.1.

One can find $C>0$, $c>0$ and $M_{0}$ such that

$\displaystyle n\,{\mathsf{Re}}\,\big{(}{\rm op}({\varLambda}){\rm op}({\kappa
w^{-n}})V,{\rm op}({w^{-n}})V\big{)}+\theta\,{\mathsf{Re}}\,\big{(}{\rm
op}({\varLambda}){\rm op}({w^{-n}})V,{\rm op}({w^{-n}})V\big{)}$
$\displaystyle\geq
n(1-CM^{-1}){\mathcal{E}}_{1}(V)+c\,\theta{\mathcal{E}}_{2}(V),\quad M\geq
M_{0}.$

###  10.2 Term $({\rm op}({\varLambda}){\rm op}({w^{-n}}){\rm
op}({{\mathcal{B}}})V,{\rm op}({w^{-n}})V)$

Recall that $\lambda_{i}\in S(\lambda_{i},g)$ and $\lambda_{1}\leq
C\sigma\lambda_{2}\leq C\sigma^{2}\lambda_{3}$ with some $C>0$. We show

######  Lemma 10.2.

Let $W={\rm op}({\phi^{-n}})V$. Then we have

$\displaystyle\big{|}({\rm op}({\lambda_{i}}){\rm
op}({b})W_{j},W_{i})\big{|}\leq
CM^{-2}{\mathcal{E}}_{1}(V)+CM^{2}{\mathcal{E}}_{2}(V),\;\;b\in
S(\sigma^{-1/2},g),\;j\geq i,$ $\displaystyle|({\rm op}({\lambda_{3}}){\rm
op}({b})W_{2},W_{3})|\leq
CM^{-2}{\mathcal{E}}_{1}(V)+CM^{2+2l}{\mathcal{E}}_{2}(V),\;\;b\in
S(M^{l},g),$ $\displaystyle|({\rm op}({\lambda_{3}}){\rm
op}({b})W_{1},W_{3})|\leq\big{(}\sqrt{3}\,{\bar{\varepsilon}}\|{\rm
op}({b})\|+CM^{-1/2}\big{)}{\mathcal{E}}_{1}(V),\;\;b\in S(1,g),$
$\displaystyle|({\rm op}({\lambda_{2}}){\rm
op}({b})W_{1},W_{2})|\leq({\bar{\varepsilon}}\|{\rm
op}({\lambda_{2}^{1/2}b})\|+CM^{-1/2}\big{)}{\mathcal{E}}_{1}(V),\;\;b\in
S(\sigma^{-1/2},g).$

###### Proof.

Let $b\in S(\sigma^{-1/2},g)$. Noting $\lambda_{i}^{1/2}\lambda_{j}^{-1/2}\in
S(\sigma^{(j-i)/2},g)$ one can write

$\displaystyle
r=(1+k)\\#(\kappa^{-1/2}\lambda_{i}^{-1/2})\\#(\lambda_{i}\\#b)\\#\lambda_{j}^{-1/2}\\#(1+{\tilde{k}})\in
S(\sigma^{(j-i)/2},g),\quad j\geq i$

for $\sigma\kappa\geq 1$, such that
$(\kappa^{1/2}\lambda_{i}^{1/2})\\#r\\#\lambda_{j}^{1/2}=\lambda_{i}\\#b$.
Then we have

$|({\rm op}({\lambda_{i}}){\rm op}({b})W_{j},W_{i})|\leq M^{-2}\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}})W\|^{2}+CM^{2}\|{\rm
op}({\varLambda^{1/2}})W\|^{2}$

for $j\geq i$. Let $b\in S(M^{l},g)$ and denote

$r=(1+k)\\#(\kappa^{-1/2}\lambda_{2}^{-1/2})\\#(\lambda_{3}\\#b)\\#\lambda_{3}^{-1/2}\\#(1+{\tilde{k}})$

such that
$(\kappa^{1/2}\lambda_{2}^{1/2})\\#r\\#\lambda_{3}^{1/2}=\lambda_{3}\\#b$.
Since $r\in S(\kappa^{-1/2}\lambda_{3}^{1/2}\lambda_{2}^{-1/2},g)\subset
S(1,g)$ in view of Lemma 7.11 then $\big{|}({\rm op}({\lambda_{3}}){\rm
op}({b})W_{2},W_{3})\big{|}$ is bounded by

$\displaystyle CM^{-2}\|{\rm
op}({\kappa^{1/2}\varLambda^{1/2}})W\|^{2}+CM^{2+2l}\|{\rm
op}({\varLambda^{1/2}})W\|^{2}.$

We check $({\rm op}({\lambda_{3}}){\rm op}({b})W_{1},W_{3})$ for $b\in
S(1,g)$. Noting $\kappa^{-1}\lambda_{1}^{-1/2}\in S(1,g)$ by Lemma 7.11 write

$r=(1+k)\\#(\kappa^{-1/2}\lambda_{1}^{-1/2})\\#(\lambda_{3}\\#b)\\#(\kappa^{-1/2}\lambda_{3}^{-1/2})\\#(1+{\tilde{k}})\in
S(1,g)$

such that
$(\kappa^{1/2}\lambda_{1}^{1/2})\\#r\\#(\kappa^{1/2}\lambda_{3}^{1/2})=\lambda_{3}\\#b$.
Since $k,\;{\tilde{k}}\in S(M^{-1},g)$ it is easy to see that
$r=(\lambda_{3}^{1/2}\lambda_{1}^{-1/2}\kappa^{-1})\\#b+{\tilde{r}}$ with
${\tilde{r}}\in S(M^{-1/2},g)$. By Proposition 5.1 and Lemma 7.11 one sees
that
$\big{|}\lambda_{3}^{1/2}\lambda_{1}^{-1/2}\kappa^{-1}\big{|}\leq\sqrt{3}\,{\bar{\varepsilon}}+CM^{-4}$
hence

$\displaystyle|({\rm op}({\lambda_{3}}){\rm op}({b})W_{1},W_{3})|=|({\rm