\documentclass[JEP,XML,SOM]{cedram}
\datereceived{2013-10-03}
\dateaccepted{2014-03-31}
\dateepreuves{2014-04-28}

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\begin{document}
\frontmatter
\title{$L^2$ well-posed Cauchy problems and symmetrizability of
first order systems}

\alttitle{Problèmes de Cauchy bien posés dans $L^2$ et symétrisabilité pour les systèmes du premier ordre}

\author[\initial{G.} \lastname{Métivier}]{\firstname{Guy} \lastname{Métivier}}
\address{Université de Bordeaux - CNRS,
Institut de Mathématiques de Bordeaux\\
351 Cours de la Libération, 33405 Talence Cedex, France}
\email{Guy.Metivier@math.u-bordeaux.fr}
\urladdr{http://www.math.u-bordeaux1.fr/~gmetivie/}

\begin{abstract}
The Cauchy problem for first order system $L(t, x, \partial_t, \partial_x)$ is known to be well-posed in $L^2$ when it admits a microlocal symmetrizer $S(t,x, \xi)$ which is smooth in $\xi$ and Lipschitz continuous in $(t, x)$. This paper contains three main results. First we show that a Lipschitz smoothness globally in $(t,x, \xi)$ is sufficient. Second, we show that the existence of symmetrizers with a given smoothness is equivalent to the existence of \emph{full symmetrizers} having the same smoothness. This notion was first introduced in \cite{FriLa1}. This is the key point to prove the third result saying that the existence of microlocal symmetrizer is preserved if one changes the direction of time, implying local uniqueness and finite speed of propagation.
\end{abstract}

\begin{altabstract}
Le problème de Cauchy est bien posé dans $L^2$ pour les systèmes du premier ordre $L(t, x, \partial_t, \partial_x)$ qui admettent un symétriseur microlocal $S(t,x, \xi)$ $C^\infty $ en $\xi\ne0$ et lipschitzien en $(t, x)$. Cet article contient trois principaux résultats. D'abord, il est montré qu'une régularité lipschitzienne globale en $(t,x, \xi)$ pour le symétriseur est suffisante. Ensuite, il est établi que l'existence de symétriseurs microlocaux est équivalente à l'existence de \emph{symétriseurs complets} $\Sigma(t, x, \tau, \xi)$ de même régularité, notion introduite dans \cite{FriLa1}. Cette étape est le point clé dans la démonstration du troisième résultat qui affirme que l'existence de symétriseurs microlocaux est préservée par changement de variable de temps. Un corollaire en est l'unicité locale et la vitesse finie de propagation.
\end{altabstract}

\subjclass{35L}

\keywords{Systems of partial differential equations, Cauchy problem, hyperbolicity, strong hyperbolicity, symmetrizers, energy estimate, local uniqueness, finite speed of propagation}

\altkeywords{Systèmes d'équations aux dérivées partielles, problème de Cauchy, hyperbolicité, hyperbolicité forte, symétriseurs, estimation d'énergie, unicité locale, propagation à vitesse finie}

\maketitle
\vspace*{\baselineskip}%
\tableofcontents
\mainmatter

\section{Introduction}

This paper is concerned with the well-posedness in $L^2$ of the Cauchy problem for first order $N \times N$ systems
\begin{equation}
\label{eqt}
\left\{\begin{aligned}
&Lu:= A_0(t, x) \D_t u + \sum_{j = 1}^d A_j (t, x) \D_{x_j} u + B(t,x) u = f, \qquad t > 0,
\\[-5pt]
&u_{| t = 0} = u_0.
\end{aligned}\right.
\end{equation}
The starting point is the well-known theory of hyperbolic symmetric systems in the sense of
Friedrichs (\cite{Fried1, Fried2}): if the matrices $A_j$ are Lipschitz continuous on $[0,T] \times \RR^d$, Hermitian symmetric, and if $A_0$ is positive definite with $A_0^{-1}$ bounded, then for all
$u_0 \in L^2(\RR^d)$ and $f \in L^{1} ([0, T] ; L^2(\RR^d))$,
the equation \eqref{eqt} has a unique solution
$u \in C^{0} ([0, T] ; L^2(\RR^d))$ which satisfies
\begin{equation}
\label{estim2}
\big\| u (t) \big\|_{L^2(\RR^d)} \le C \big\| u _0 \big\|_{L^2(\RR^d)}
+ C \int_0^t \big\| L u (s) \big\|_{L^2(\RR^d)} ds,
\end{equation}
for some constant $C$ independent of $u_0$. Additional properties are local uniqueness
and finite speed propagation. The question discussed in this paper is to know for which systems
these properties remain true.

For scalar equations of order $m$, the analogue would be the well-posedness in Sobolev spaces
$H^{m-1}$, for which strict hyperbolicity is necessary (\cite{IP}) and sufficient (\cite {Gard2, Leray}).
This completely settles the question for scalar equations but for systems, the situation
is much more complex.

A necessary condition has been given by V.\,Ivrii and V.\,Petkov (\cite{IP}): they have shown
that if the estimate \eqref{estim2} is valid for $u \in C^\infty_0 (]0, T[ \times \RR^d)$,
then there exists a bounded microlocal symmetrizer $S(t,x, \xi) $ for \eqref{eqt} (the precise definition is recalled below). This is equivalent to a strong form of hyperbolicity of the principal symbol, which we call \emph{strong hyperbolicity of the symbol}, namely that $L + B_1$ is hyperbolic for all matrix $B_1(t,x)$.
Of course it is stronger than hyperbolicity, which is known to be a necessary condition for the Cauchy problem to be well-posed
in $C^\infty$
(see \cite{Lax1, Miz1, Gard1} and the review paper \cite{Gard3}). In particular,
\eqref{estim2} are the best estimates in terms of regularity that one can expect for the Cauchy problem.

On the side of sufficient conditions, except in the constant coefficient case,
where the energy estimate \eqref{estim2} is easily obtained on the space-Fourier transform of the equation,
the existence of a bounded symmetrizer does not imply in general that the problem is well-posed, even in $C^\infty$,
A counterexample is given in \cite{Str} and another one is proposed in Section~\ref{sectionExamples}.
Besides the case of symmetric systems recalled above for which the symmetrizer $S(t, x)$ is independent of $\xi$,
the Cauchy problem is known to be well-posed in $L^2$
when the microlocal symmetrizer is smooth in $\xi$ and at least
Lipschitz continuous in $(t,x)$ (see \cite{Lax2, Met} and Theorem~\ref{L2WPLipold} below for a precise statement).
In this case, the energy estimates are proven using
the usual pseudo-differential calculus or the para-differential calculus when the coefficient
have limited smoothness. This covers the case of strictly hyperbolic systems and
the more general case of hyperbolic systems with constant multiplicity (e.g. \cite{Cal}, \cite{Yam}).
This also applies to the case of ``generic'' double eigenvalues, still assuming the strong hyperbolicity
of the symbol, see Theorem~\ref{theodouble} below.

The first objective of this paper is to revisit these questions under the angle of the smoothness
of the symmetrizer. We prove that
the Lipschitz continuity in $(t,x,\xi)$ for $\xi \ne 0$ of the symmetrizer $S$ is sufficient to
obtain the $L^2$ estimates and the $L^2$ well-posedness. In addition, we give examples and counterexamples showing that the Lipschitz condition is sharp.

The second main result of this paper is to prove that the existence of microlocal symmetrizer
is preserved by a change of time, as this is essential to obtain local uniqueness
and the precise description of the propagation of the support of solutions (see \cite{JMR1,Rauch}).
More surprisingly, we show that the existence of symmetrizers of a given smoothness
is equivalent to the existence of \emph{full symmetrizers} of the same smoothness, a notion introduced in \cite{FriLa1}.
This link is the key point in the proof of existence of microlocal symmetrizers in any direction of hyperbolicity.

\medskip
We now briefly present the results. Note that \eqref{estim2} applied to
$e^{\gamma t} u$, with $u \in\nobreak C^\infty_0 (]0, T[ \times \RR^d)$, implies that
\begin{equation}
\label{estim1}
\forall \gamma \ge \gamma_0, \: \quad
\gamma \big\| u \big\|_{L^2(\RR^{1+d})} \le C \big\| (L + \gamma A_0) u \big\|_{L^2(\RR^{1+d})},
\end{equation}
for some constants $C$ and $\gamma_0$ independent of $ u$.
This estimate is elliptic like: with $\chi \in C^\infty_0 (\RR^{1+d})$ and $\underline u \in \CC^N$, applying it to
\begin{equation*}
u (t, x) = e^{i \lambda (t \tau + x \xi)} \lambda^{- \alpha d/2} \chi (\lambda^{1/2} (t - t_0, x- x_0)) \underline u,
\end{equation*}
and to $\gamma = \lambda \gamma_0$, and letting $\lambda$ tend to $+ \infty$ implies
\begin{lem}
\label{lemellip}
Suppose that the coefficients of $L$ are continuous and bounded on the open set $ \Omega$ and there are constants $\gamma_0$ and $C$ such that
\begin{equation}
\label{P2b}
\gamma \big\| u \big\|_{L^2 (\Omega)} \le C
\big\| \big(L + \gamma A_0)\big) u \big\|_{L^2(\Omega)}
\end{equation}
for all $\gamma \ge \gamma_0$ and $u \in C^\infty_0(\Omega)$.
Then the principal symbol $L_1(t, x, \tau, \xi)$ of $L$ satisfies
for all $(t,x) \in \Omega$, all $\gamma \in \RR$ and $u\in \CC^N$:
\begin{equation}
\label{estresolv}
| \gamma | \big| u \big| \le
C \big| (L_1(t,x, \tau, \xi) + i \gamma A_0(t,x)) \underline u \big|.
\end{equation}
\end{lem}
There is no sign condition on $\gamma$, as seen by changing $\tilde \xi$ to $-\tilde \xi$.
When \eqref{estresolv} holds, we say that the \emph{symbol is strongly hyperbolic} in the time direction.
The condition \eqref{estresolv} has several equivalent formulations, see Section~\ref{sectionstroghyp}.
One of them is that the symbols admits a bounded symmetrizer.

\begin{defi}
\label{defmicrosym}
A microlocal symmetrizer for $L_1$ is a bounded
matrix $S(t, x, \xi)$, homogeneous of degree $0$ in $\xi\ne 0 $, such that $S(t, x, \xi) A_0(t, x)$
is symmetric and uniformly positive definite, and $S(t, x, \xi) A(t, x, \xi) $
is symmetric, where
$A(t, x, \xi) = \sum A_j(t, x) \xi_j$.
\end{defi}

Combining Lemma~\ref{lemellip} and Theorem~\ref{theostronghyp} below, we recover the
the necessary condition given in \cite{IP}:
\begin{prop}
If $L$ has continuous coefficient on the open set $ \Omega$ and there are constants $\gamma_0$ and $C$ such that \eqref{P2b} is satisfied, then
the principal symbol $L_1$ must admit a bounded symmetrizer $S(t,x, \xi)$ on
$\Omega \times \RR^d\setminus{0}$.
\end{prop}

In the constant coefficients case, the existence of a bounded symmetrizer
is also sufficient, as immediately seen
by Fourier synthesis.
For variable coefficients, this condition is
far from being sufficient for the well-posedness of the Cauchy problem~\eqref{eqt}:
in section~\ref{sectionExamples} we give an example
of a $3 \times 3$ systems in space dimension $d=2$, whose symbol
$L(x, \tau, \xi)$ is strongly hyperbolic uniformly in $x$, and such that
the Cauchy problem \eqref{eqt} is ill-posed, even locally and with $C^\infty $ data.

On the side of sufficient conditions, let us first recall
the following result:
\begin{theo}
\label{L2WPLipold}
Suppose that the coefficients $A_j \in W^{1, \infty} ([0, T] \times \RR^{d})$ and there exists a microlocal
symmetrizer $S$, homogeneous of degree $0 $ and $C^\infty$in $\xi \ne 0$ which satisfies
$\D_{t,x}^\beta \D_\xi^\alpha S \in L^\infty ([0, T] \times \RR^d \times S^{d-1})$ for all
$\alpha \in \NN^d$ and all $| \beta | \le 1$.
Then, there are constants
$C$ and $\gamma$ such that for all
$u_0 \in L^2(\RR^d)$ and $f \in L^1([0, T] ; L^2(\RR^d))$, the Cauchy problem
\eqref{eqt} has a unique solution
$u \in C^0([0, T] ; L^2(\RR^d))$ which satisfies
\begin{equation}
\label{estim2b}
\big\| u (t) \big\|_{L^2(\RR^d)} \le C e^{\gamma_0 t} \big\| u _0 \big\|_{L^2(\RR^d)}
+ C \int_0^t e^{\gamma_0 (t-s)} \big\| L u (s) \big\|_{L^2(\RR^d)} ds.
\end{equation}
\end{theo}

When the symmetrizer does not depend on $\xi$, this is Friedrichs theory, in which case
the estimate \eqref{estim2b} is easily obtained by forming the real part of the scalar product of $S Lu$
with $u$ and performing integrations by parts.
For microlocal symmetrizers, one replaces the multiplication by $S $ by the action of the
pseudodifferential operator
$S(t,x, D_x) $ (\cite{Lax2}) when the coefficients are also smooth in $x$, or by a para-differential version
when the coefficients are Lipschitz (see e.g.\ \cite{Met}). This theorem applies to hyperbolic systems with constant multiplicities which admit smooth symmetrizers.
Indeed, multiple eigenvalues of $A(t,x, \xi)$ with variable multiplicities are the
main difficulty for the construction of smooth symmetrizers. However,
we note in Proposition~\ref{theodouble} that the theorem above applies to
strongly hyperbolic systems which have only generic double eigenvalues.

The first main result of the paper extends this result to Lipschitz symmetrizers,
using a Wick quantization of the symbols.

\begin{theo}
\label{L2WPLip}
Suppose that the coefficients $A_j \in W^{2, \infty} (\RR^{d+1})$ and there exists a microlocal
symmetrizer, homogeneous of degree $0 $ in $\xi \ne 0$ and Lipschitz
continuous in $(t, x, \xi)$ on $\RR^{d+1} \times S^{d-1}$. Then, there are constants
$C$ and $\gamma$ such that for all
$u_0 \in L^2(\RR^d)$ and $f \in L^1([0, T] ; L^2(\RR^d))$, the Cauchy problem
\eqref{eqt} has a unique solution
$u \in C^0([0, T] ; L^2(\RR^d))$ which satisfies \eqref{estim2b}.
\end{theo}

This theorem is proved in Section~\ref{sectionL2WP}. In Section~\ref{sectionExamples} we discuss the existence
of Lipschitz symmetrizers. In particular, we give examples of systems which admit a Lipschitz symmetrizer
but no $C^1$ symmetrizer. We also prove that the Lipschitz condition is sharp,
in the sense that for all $\mu< 1$, there are examples of systems admitting H\"older continuous symmetrizers of order $\mu< 1$, for which the Cauchy problem with $C^\infty$ data is locally ill-posed.

\begin{rem}
In this theorem we assume that the coefficients are $W^{2, \infty}$ whereas $W^{1, \infty}$
was sufficient when the symmetrizers are smooth in $\xi$. This is due to the use of the Wick quantization.
One one hand it helps to deal with symbols which are not smooth in $\xi$. On the other hand, the symbolic
calculus is less precise, and the $W^{2, \infty} $ smoothness of the coefficient is used to prove
that in~$\mathrm{Op}^\mathrm{Wick}(S)\circ A (x, \D_x) - \mathrm{Op}^\mathrm{Wick} (i S A) $ is bounded in $L^2$.
At the present time, it is not known wether this additional smoothness which is crucial for the proof is necessary or not for the validity of the result.
\end{rem}


The second part of the paper is concerned with the
local theory of the Cauchy problem and the finite speed propagation property
for the support of the solutions.
A~classical proof of this property relies on the invariance of the assumptions
by changes of time variables, so that one can convexify the initial surface.
The existence of a local symmetrizer is clearly invariant by change of time,
as well as strict hyperbolicity or the property that the characteristic variety is smooth
with constant multiplicities. In all these cases the local theory was well established.
This invariance is not clear for the existence of smooth microlocal symmetrizers.
However, when there are smooth symmetrizers, local uniqueness and finite speed of propagation
are proved in \cite{Rauch}
using another approach based on finite difference approximation schemes and uniform estimates due to
\cite{Lax-Nir, Vaillancourt}.

The second main theorem of this paper asserts that the existence
of a Lipschitz \lcr\resp $C^\infty$\rcr symmetrizer is preserved by change of timelike directions.
This is a key step for establishing a local theory, starting with local uniqueness,
finite speed of propagation and ending with the sharp description of the
propagation of support as stated in \cite{JMR1, Rauch}.

Let $\tilde x $ denote the space-time variables $(t, x)$ and set accordingly
$\tilde \xi = (\tau, \xi)$ Assuming that $L_1(\tilde x, \tilde \xi) $ is hyperbolic in the time direction
$(1,0) \in \RR^{1+d}$, denote by $\Gamma_{\tilde x}$ the cone of hyperbolic directions
that is the component of $(1,0)$ in $\{\tilde \xi: \det L_1(\tilde x, \tilde \xi) \ne 0\} $.

\begin{theo}
\label{invt}
Suppose that the coefficients $A_j $ are Lipschitz continuous \lcr\resp $C^\infty$\rcr on $\RR^{d+1}$ and that there exists a microlocal
symmetrizer $S(t, x, \xi)$, homogeneous of degree $0 $ in $\xi \ne 0$ and Lipschitz
continuous \lcr\resp in $C^\infty$\rcr $(t, x, \xi)$ on $\RR^{d+1} \times S^{d-1}$.
Then, for any time-like direction $\tilde \nu \in \Gamma_{t,x}$, the symbol
$L(t, x, \tilde \nu)^{-1} A(t,x, \xi)$ admits a Lipschitz \lcr\resp $C^\infty$\rcr symmetrizer.
\end{theo}

\begin{cor}
Under the assumptions of Theorem~$\ref{L2WPLip}$, the Cauchy problem for~$L$ with
initial data on any space like hyperplane is well-posed in $L^2$.
\end{cor}

In Theorem~\ref{theomajext}, we prove that one can choose symmetrizers which also depend
smoothly on $\tilde \nu$. As said above, with Theorems~\ref{L2WPLipold} and \ref{L2WPLip}, this implies
local uniqueness, and finite speed of propagation. Together with the Lipschitz dependence
of the cone of propagation implied by Proposition~\ref{proplipvp}, this allows to
the results on the precise propagation of support stated in \cite{JMR1, Rauch}.
We refer the reader to these papers for precise statements.

The proof of this theorem is based on an intrinsic characterization of the existence of
Lipschitz symmetrizers which uses the notion of \emph{full symmetrizers} introduced by
K.O.\,Friedrichs and P.\,Lax \cite{FriLa1}:

\begin{defi}
A full symmetrizer for \eqref{eqt} is a bounded
matrix $\widetilde S(t, x, \tau, \xi)$, homogeneous of degree $0$ in $(\tau, \xi)\ne 0 $, such that
$\widetilde S(t, x, \tau, \xi) L(t, x,\tau, \xi) $
is symmetric.

The matrix $\widetilde S$ is said to be positive in the direction $\tilde \nu$, if
$ \re \widetilde S(t, x, \tau, \xi) L(t, x, \tilde \nu)$ is positive definite
on $\ker L(t, x, \tau, \xi)$ for all $(t,x, \tau, \xi)$.
\end{defi}
Of course the condition is nontrivial only near characteristic points, but says nothing about
hyperbolicity. Our third main theorem is the following:

\begin{theo}
Suppose that $L$ is hyperbolic in the time direction. Then, $L$ admits a continuous \lcr\resp Lipschitz\rcr
microlocal symmetrizer $S(t,x, \xi)$ if and only if it admits a continuous \lcr\resp Lipschitz\rcr
full symmetrizer $\widetilde S (t, x, \tau, \xi)$ which is positive in the time direction.

In this case, $\widetilde S$ is positive in any direction of hyperbolicity $\tilde \nu$.
\end{theo}

\section{Lipschitz symmetrizability is sufficient for the $L^2$ well-posedness}
\label{sectionL2WP}

The goal of this section is to prove Theorems~\ref{L2WPLip}.
We consider a system
\begin{equation}
\label{geq}
L u = \sum_{j = 0}^d A_j (\tilde x) \D_{\tilde x_j} u
\end{equation}
with coefficients $A_j$ which are at least $W^{1, \infty} ([0; T] \times \RR^d)$.

\subsection{Wave packets and localization}

For $u \!\in\! L^2 (\RR^s)$, $\lambda> 0$ and $B \!\in\! L^\infty (\RR^d \times\nobreak \RR^d)$, let
\begin{equation}
\label{defWB}
W_{\lambda, B} u (x, \xi) = \frac{1}{(2 \pi) ^{\sfrac{d}{2}}}
\Big(\frac{\lambda}{\pi}\Big)^{\sfrac{d}{4}} \int e^{i (x - y) \xi - \mez \lambda | x- y|^2}
B(x, y) u (y) \,dy.
\end{equation}

\begin{lem}
\label{action1}
The operator $W_{\lambda, B}$ is bounded from $L^2(\RR^d)$ to
$L^2(\RR^{2d})$ and
\begin{equation}
\big\| W_{\lambda, B} u \big\|_{L^2(\RR^d \times \RR^d)}
\le \big\| B \big\|_{L^\infty} \big\| u \big\|_{L^2 (\RR^d)}.
\end{equation}
Moreover, if $ B (x, z) \equiv \Id$, $W_\lambda:= W_{\lambda, \Id} $ is isometric from
$L^2(\RR^d)$ into $L^2(\RR^d \times \RR^d)$.
\end{lem}
\begin{proof}
Let $\sF$ denote the Fourier transform and
$
v_x (y) = B(x, y) e^{- \mez \lambda | x - y |^2} u (y)
$. Then
$$
W_{\lambda, B} u (x, \xi) = \frac{1}{(2 \pi) ^{\sfrac{d}{2}}}
\Big(\frac{\lambda}{\pi}\Big)^{\sfrac{d}{4}}
e^{i \lambda x \xi} \sF \big(v_x \big) (\xi).
$$
Therefore
\begin{multline*}
\int \big| W_B u (x, \xi) \big|^2 \,dx \,d\xi =
\Big(\frac{\lambda}{\pi}\Big)^{\sfrac{d}{2}} \int \big| v_x(y) \big|^2 \,dx \,dy
\\
=\Big(\frac{\lambda}{\pi}\Big)^{\sfrac{d}{2}}
\int e^{- \lambda | x - y |^2} \big| B(x, y) u (y) \big|^2 \,dx \,dy
\le \big\| B \big\|^2_{L^\infty}
\ \big\| u \big\|^2_{L^2 (\RR^d)}.
\end{multline*}
When $B = 1$, the inequality is an equality.
\end{proof}

We will adapt the scale $\lambda$ to the size of the
frequency $| \xi |$. One has
$$
\begin{aligned}
W_\lambda u (x, \xi) =
(2 \pi) ^{- d}
\Big(\frac{1} {\pi \lambda}\Big)^{\sfrac{d}{4}} \int e^{i x \eta - \frac{1} {2\lambda} | \xi - \eta |^2} \hat u (\eta) \,d\eta.
\end{aligned}
$$
This shows that for a fixed $\xi$, $W_\lambda u (\, \cdot, \xi)$ is the inverse Fourier transform of
$
w_\xi (\eta) = \Big(\frac{1} {\pi \lambda} \Big)^{\sfrac{d}{4}} \ e^{- \frac{1} {2\lambda} | \xi - \eta |^2} \hat u (\eta).
$
Therefore
\begin{equation}
\label{rel2}
\begin{aligned}
\int \big| W_\lambda u (x, \xi) \big|^2 \,dx & = (2\pi)^{- d} \int \big| w_\xi (\eta) \big|^2 \,d\eta
\\
& = (2\pi)^{- d} \Big(\frac{1} {\pi \lambda} \Big)^{\sfrac{d}{2}}
\int e^{- \frac{1} {\lambda} | \xi - \eta |^2} | \hat u (\eta) |^2 \,d\eta.
\end{aligned}
\end{equation}
Integrating in $\xi$, we recover the isometry of $W_\lambda$,
but the important point is that we use \eqref{rel2} to localize in
$| \xi |$.

Consider a dyadic partition of unity
\begin{equation}
1 = \vp_0 (\xi) + \sum_{j= 1}^\infty \theta_j (\xi)
\end{equation}
with $\vp_0 \in C^\infty_0(\RR^d)$, supported in $\{| \xi | \le 2 \}$ and equal to one
on $\{| \xi | \le 1\}$,
$\theta_j (\xi) = \vp_j (\xi) - \vp_{j-1} (\xi)$ and
$\vp_j(\xi) = \vp_0 (2^j \xi)$ for $j \ge 1$.
To unify notations, we set $\theta_0 = \vp_0$ and for $j \ge 0$ we denote by
define $\Theta_j $ the operator
$$
\Theta_j u = \sF^{-1} \big(\theta_j \hat u\big),
$$
so that
\begin{equation}
u = \sum_{j= 0}^\infty \Theta_j u.
\end{equation}

\begin{prop}
\label{proplocal1}
For all $n$, $m$ and $\alpha$, there is a constant $C$ such that for all $j \ge 0$
\begin{equation}
\big\| | \xi|^m (1 - \vp_{j+2}) W_{2^j}
\D_y^\alpha \Theta_j u \big\|_{L^2(\RR^{2d})} \le
C 2^{-j n} \big\| \Theta_j u \big\|_{L^2(\RR^d)}.
\end{equation}
\end{prop}

\begin{proof}
By \eqref{rel2}
\begin{multline*}
\big\| (1 - \vp_{j+2}) W_{2^j} \D_y^\alpha \Theta_j u \big\|^2_{L^2(\RR^{2d})}\\
= (2\pi)^{- d} \Big(\frac{1} {\pi 2^j} \Big)^{\sfrac{d}{2}}
\int e^{- 2^{-j} | \xi - \eta |^2} (1 - \vp_{j+2}(\xi))^2 | \eta^\alpha |^2
(\theta_j (\eta))^2 | \hat u (\eta) |^2 \,d\eta \,d\xi.
\end{multline*}
On the support of $ (1 - \vp_{j+2}(\xi))
\theta_j (\eta)$, one has $ |\xi | \ge 2^{j+2} $, $| \eta | \le 2^{j+1}$ so that
\hbox{$| \xi - \eta | \ge \mez | \xi| $} and therefore
$
2^{-j} | \xi - \eta |^2 \ge \mez 2^{-j} | \xi - \eta |^2 + \mez | \xi |.
$
Hence,
\begin{align*}
\big\| 2^{jn}& |\xi|^m (1 - \vp_{j+2}) W_{2^j} \D_y^\alpha \Theta_j u \big\|^2_{L^2(\RR^{2d})}
\\
&\le (2\pi)^{- d} \Big(\frac{1} {\pi 2^j} \Big)^{\sfrac{d}{2}}
2^{j n} 2^{(j+1) | \alpha |} e^{- 2^{j}} \int | \xi|^{2m} e^{- \frac{1}{2} | \xi |}
e^{- \mez 2^{-j} | \xi - \eta |^2} 
| \theta_j (\eta) \hat u (\eta) |^2 \,d\eta \,d\xi
\\
&\le C \big\| \Theta_j u \big\|_{L^2(\RR^d)}^2.\qedhere
\end{align*}
\end{proof}

\begin{cor}
\label{cor103}
There is a constant $C$ such that for all $j \ge 0$,
\begin{equation}
\big\| (1 - \vp_{j+2}) W_{2^j}\Theta_j \D_{x_k} u \big\|_{L^2(\RR^{2d})} \le
C \big\| \Theta_j u \big\|_{L^2(\RR^d)}.
\end{equation}
\end{cor}

\subsection{The main estimate}

Let
\begin{equation}
A(x, \D_x) = \sum_{j=1}^d A_j(x) \D_{x_j}, \qquad A(x, \xi) = \sum_{j=1}^d \xi_j A_j(x).
\end{equation}
We assume that we are given a matrix $S(s, \xi)$, homogeneous of degree $0$ in $\xi$ such that
$S(x, \xi) A(x, \xi)$ is Hermitian symmetric.

Let $\psi \in C^\infty_0 (\RR^d)$, and for
$\lambda \ge 1$ introduce $\psi_ \lambda (\xi) = \psi (\lambda^{-1} \xi)$.

\begin{prop}
\label{mainestloc}
Suppose that the coefficients $A_j $ belong to $W^{2, \infty} (\RR^d)$ and that
$S \in W^{1, \infty} (\RR^d \times S^{d-1})$. Then, there is a constant $C$ such that for all
$\lambda \ge 1 $ and $u$
\begin{equation}
\label{energy5}
\Big| \re\Big(\psi_{\lambda} S W_{\lambda} u, W_{\lambda} A(x, \D_x) u \big)_{L^2(\RR^{2d})}\Big|
\le C \big\| u \|^2_{L^2}.
\end{equation}

\end{prop}

\begin{proof}
Let $S_\lambda = \vp_\lambda S$ and
consider the self-adjoint operator
$
\Sigma_\lambda = W_\lambda^* S _\lambda W_ \lambda$:
$$
\Sigma_\lambda u (x) = \kappa \lambda^{\sfrac{d}{2}}
\int e^{\Phi_\lambda (x, y, z, \xi)} S_\lambda (z, \xi) u (y) \,dz \,d\xi \,dy,
$$
where $\kappa$ is a normalization factor and
$$
\Phi_\lambda (x, y, z, \xi) = i (x - y) \xi - \mez \lambda (| x - z|^2 + | y-z|^2).
$$

The estimate to prove is
\begin{equation}
\label{commut}
\Big| \re \big(A(x, \D_x)^* \Sigma_\lambda u, u \big)_{L^2}\Big| \le C \big\| u \big\|_{L^2}^2.
\end{equation}
One can replace $A(x, \D_x)^*$ by $ \tilde A^*:= \sum - A_j^* \D_{x_j}$ since the difference
is bounded in $L^2$ with norm $ O(\sup_j \| A_j \|_{W^{1, \infty}}) $. One has
$$
A_j^* \D_j \Sigma_\lambda u (x) = \kappa
\int e^{\Phi_\lambda} A_j^* (x) \big(i \xi_j - \lambda (x_j - z_j) \big) S_\lambda (z, \xi) u (y) \,dz \,d\xi \,dy.
$$
Therefore
\begin{equation}
\label{comm36}
\tilde A^* (x, \D_x) \Sigma_\lambda = W_\lambda ^* (-i A^* S_\lambda) W_\lambda + \sum_j (R_j^1 + R_j^2),
\end{equation}
where
\begin{align*}
R_j^1 u(x) &= i \kappa \int e^{\Phi_\lambda} \xi_j \big(A_j^* (z) - A_j^*(x)) \big) S_\lambda (z, \xi) u (y) \,dz \,d\xi \,dy,\\
R_j^2 u(x) &= \kappa \int e^{\Phi_\lambda} \lambda (x_j - z_j) A_j^*(x) \big) S_\lambda (z, \xi) u (y) \,dz \,d\xi \,dy.
\end{align*}
Because $S$ is a symmetrizer for $\sum A_j \xi_j$, the matrix
$i A^*(x, \xi) S_\lambda$ is skew-symmetric and thus the real part
of the first term in \eqref{comm36} vanishes and it is sufficient to
show that the remainders $R_j^{1, 2}$ are bounded in $L^2$. Write
\begin{equation}
A_j^* (x) - A_j^*(z) = \sum_k (x_k - z_k) \tilde A_{j,k}(x, z),
\end{equation}
with $\tilde A_{j,k} \in W^{1, \infty}$, and use that
$$
\begin{aligned}
2 \lambda (x_k - z_k) = \D_{z_k} \Phi_\lambda - i \lambda \D_{\xi_k} \Phi_\lambda:= Z_k \Phi_\lambda.
\end{aligned}
$$
Integrating by parts in $(z, \xi)$ yields that $R_j^1 = \sum_k R^1_{j, k} $ with
$$
R_{j,k}^1 u(x) = - \frac{i \kappa}{\lambda} \int e^{\Phi_\lambda} Z_k^*
\big(\xi_j A_{j,k}^* S_\lambda \big) u (y) \,dz \,d\xi \,dy.
$$
Note that
$
\frac{1}{\lambda} Z_k^*
\big(\xi_j A_{j,k}^* S_\lambda \big) $ is a sum a terms of the form $ B_{l} (x, z) S_l (z, \xi, \lambda)
$
where the~$B_l$ and $S_l$ are uniformly bounded. Therefore
$R_{j,k}^1$ is of the sum of the operators
$W^*_{\lambda, B_{l}} S_l W _\lambda $ where the definition of
$W_{\lambda, B_l}$ is given in \ref{defWB}.
Lemma~\ref{action1} implies that the $R_j^1$ are uniformly in $\lambda$.

The analysis of $ R_j^2 $ is similar and the proof of the proposition is complete.
\end{proof}

\subsection{Proof of Theorem~\ref{L2WPLip}}

Suppose that the operator $L$ in \eqref{geq} has $W^{2, \infty} $ coefficients.
Without loss of generality, multiplying $L$ on the left by $A_0^{-1}$,
we assume that the coefficient of $D_t $ is $A_0 = \Id$ so that
$L = \D_t + A(t, x, \D_x)$.
We are given a Lipschitz symmetrizer
$S (t, x, \xi)$ which is uniformly positive definite and such that
\begin{equation}
S, \ \D_{t, x} S, \ | \xi | \D_\xi S \in L^\infty.
\end{equation}
Consider the energy
\begin{equation}
\label{energy}
\cE_t (u) = \sum_{j=0}^\infty \big(S(t) W_{2^j} \Theta_j u, W_{2^j} \Theta_j u \big)_{L^2(\RR^{2d})}.
\end{equation}

\begin{lem}
There are constants $C \ge c > 0$ such that
\begin{equation*}
c \big\| u \big\|_{L^2(\RR^d)}^2 \le \cE (u) \le C\big\| u \big\|_{L^2(\RR^d)}^2.
\end{equation*}
\end{lem}
\begin{proof}
Because $S(t, x, \xi) $ is positive definite bounded from above and from below,
\begin{equation*}
\cE_t (u) \approx \sum_{j=0}^\infty \big\| W_{2^j} \Theta_j u \big\|^2_{L^2(\RR^{2d})}
= \sum_{j=0}^\infty \big\| \Theta_j u \big\|^2_{L^2(\RR^{2d})} \approx
\big\| u \big\|^2_{L^2(\RR^{2d})}.\qedhere
\end{equation*}
\end{proof}

The theorem follows from the energy estimate.

\begin{prop} If $u $ satisfies $L u = f$,
then
\begin{equation}
\frac{d}{dt} \cE_t (u(t)) \le C\Big(
\big\| f (t) \big\|_{L^2(\RR^{2d})} \big\| u (t) \big\|_{L^2(\RR^{2d})} + \big\| u(t) \big\|^2_{L^2(\RR^{2d})}
\Big).
\end{equation}
\end{prop}

\begin{proof}
One has
\begin{equation*}
\begin{aligned}
\frac{d}{dt} \cE_t (u(t)) & = (\D_t \cE) (u (t)) +
2 \re \widetilde \cE_t (u(t), \D_t u (t)))
\\
& = (\D_t \cE) (u (t)) +
2 \re \widetilde \cE_t (u(t), f (t))) - 2 \re \widetilde \cE_t (u(t), A u (t))),
\end{aligned}
\end{equation*}
where $\D_t \cE_t$ is the expression \eqref{energy} with $S$ replaced by $\D_t S$ and
$\widetilde \cE$ is the bilinear version of $\cE$. Because
$\D_t S \in L^\infty$, the first term
is $O (\| u(t) \|_{L^2}^2)$.
Similarly, the second term
is $O (\| u(t) \|_{L^2} \| f(t) \|_{L^2})$ and it remains to prove that
\begin{equation}
\label{mainestim}
\big| \re \widetilde \cE_t \big(u(t), A u (t)\big) \big| \le
C \big\| u(t) \big\|_{L^2}^2.
\end{equation}
For simplicity we drop the time from the notations, $t$ being a parameter
and all the estimates below being uniform in $t$.

The expression to consider is
\begin{equation}
\label{energy2}
\widetilde \cE (u, A u) = \sum_{j=0}^\infty \big(S W_{2^j} \Theta_j u, W_{2^j} \Theta_j A u \big)_{L^2(\RR^{2d})}.
\end{equation}
Corollary~\ref{cor103} implies that
\begin{equation*}
\Big| \big((1- \vp_{j+2}) S W_{2^j} \Theta_j u, W_{2^j} \Theta_j \widetilde A u \big)_{L^2(\RR^{2d})}
\Big| \lesssim
\sum_k \big\| \Theta_j u \big\|_{L^2}
\big\| \Theta_j (A_k u) \big\|_{L^2},
\end{equation*}
and the sum over $j$ of these terms is $ O (\| u \|^2_{L^2})$.
Next, we replace
$\Theta_j (A u) $ by $ A \Theta_j u $ using the following lemma.

\begin{lem}
\label{lemcomloc}
The $ g_j = [ A, \Theta_j ] u$ satisfy
\begin{equation}
\label{114}
\sum \big\| g_j (t) \big\|^2_{L^2} \lesssim \big\| u (t) \big\|^2_{L^2}.
\end{equation}
\end{lem}

Since $S(x, \xi)$ is bounded and since the $W_\lambda$ are isometries one has
\begin{equation*}
\label{energy2b}
\Big| \sum_{j=0}^\infty \big(\vp_{j+2} S W_{2^j} \Theta_j u, W_{2^j} g_j \big)_{L^2(\RR^{2d})}
\Big| \lesssim \sum_{j=0}^\infty \big\| \Theta_j u \|_{L^2} \big\| g_j \big\|_{L^2}
\lesssim \big\| u \|^2_{L^2}.
\end{equation*}
Summing up, we have proved that
\begin{equation*}
\widetilde \cE (u, A u) =
\sum_{j=0}^\infty \big(\vp_{j+2} S W_{2^j} \Theta_j u, W_{2^j} A \Theta_j u \big)_{L^2(\RR^{2d})}
+ O(\big\| u \big\|^2_{L^2}).
\end{equation*}
By Proposition~\ref{mainestloc}, the real part of the sum is
$O (\sum \| \Theta_j u\|_{L^2}^2) = O(\| u \|_{L^2}^2)$ finishing the proof of
\ref{mainestim} and of the proposition.
\end{proof}

\begin{proof}[Proof of Lemma~$\ref{lemcomloc}$]
Using the para-differential calculus (see e.g. \cite{Met}), one has
\begin{equation*}
A (x, \D_x) = T_{i A} + R,
\end{equation*}
where $T_{i A}$ is a para-differential operator of symbol
$i A(x, \xi)$ and $R$ is bounded from~$L^2$ to~$L^2$. Thus
\begin{equation*}
g_j = [ T_A, \Theta_j ] u + R \Theta_j u - \Theta_j Ru.
\end{equation*}
The last two terms satisfy \eqref{114}. The symbolic calculus implies that the
$[ T_{i A}, \Theta_j ] $ are uniformly bounded in $L^2$ since the coefficients of
$A$ belong to $W^{1, \infty}$. Moreover, the spectral properties
of the para-differential calculus shows that
$ [ T_{i A}, \Theta_j ] = [ T_{i A}, \Theta_j ] \widetilde \Theta_j $ where the cut-off function
$ \tilde \theta_j $ is supported in the annulus
$\{2^{j - n} \le | \xi | \le 2^{j+n} \} $ for some fixed $n$ if $j \ge 1$
and in the ball $\{| \xi | \le 2^{n} \} $ if $j = 0$. Therefore
$ g'_j = [ T_{i A}, \Theta_j ] u $ satisfies
$$
\big\| g'_j \big\|_{L^2} \lesssim \big\| \widetilde \Theta_j u \big\|_{L^2}
$$
and thus \eqref{114}.
\end{proof}

\section{Examples and counterexamples}
\label{sectionExamples}

In this section, we discuss the existence
of symmetrizers of limited smoothness. The case of generic double eigenvalues is
specific, as shown in the next subsection. But for eigenvalues of higher order, it is easy to construct
examples of systems with symmetrizers which necessarily have no, or a limited, smoothness.

Second, we show on an example that Lipschitz smoothness is sharp, even for well-posedness
in $C^\infty$.

\subsection{Example of non smooth symmetrizers}

In space dimension two consider, near the origin, a system of the form
\begin{equation}
\label{explegen0}
L_0 (x, \D_t, \D_x, \D_y) = \cL_0 (\D_t, \D_x, x \D_y) = \D_t + A \D_x + x B \D_y,
\end{equation}
with $\cL_0(\tau, \xi, \eta)$ strictly hyperbolic. Consider next a perturbation
\begin{equation}
\label{explegena}
L_a (x, \D_t, \D_x, \D_y) = L_0 (x, \D_t, \D_x, \D_y) + x a(x) C \D_y
= \cL (a(x), \D_t, \D_x, x \D_y).
\end{equation}
We will give explicit examples below.
For $a$ small, $\cL(a, \tau,\xi, \eta)$ is still strictly hyperbolic and therefore it
has smooth symmetrizers $\cS(a, \xi, \eta)$ for $(\xi, \eta) \ne (0, 0)$, providing
bounded symmetrizers for $L_a (x, \tau, \xi, \eta) $
\begin{equation}
\label{symgena}
S(a, x, \xi, \eta) = \cS(a, \xi, x \eta)
\end{equation}
for $(x, \xi) \ne (0,0) $. On the unit sphere $\xi^2 + \eta^2 = 1$, they are smooth when
$(x, \xi) \ne (0,0) $. The definition of $S$ can be extended at $(x, \xi) = (0,0) $,
but in general they have a singularity there.

\begin{lem}
Suppose in addition that $\cL_0$ is symmetric. Then, for $a$ small, there is a symmetrizer
$\cS$ of the form
\begin{equation}
\label{syma}
\cS (a, \xi, \eta) = \Id + a \cS_1(a, \xi, \eta),
\end{equation}
with $\cS_1$ homogeneous of degree $0$ in $(\xi, \eta)$ and smooth
in $ (a, \xi, \eta)$ for $(\xi, \eta) $ in the unit sphere $\xi^2 + \eta^2 = 1$.
\end{lem}

\begin{proof}
The spectral projectors $\Pi_j(a, \xi, \eta)$ are smooth in
$ (a, \xi, \eta)$ for $(\xi, \eta) $ in the unit sphere $\xi^2 + \eta^2 = 1$ and
$\cS = \sum \Pi_j^* \Pi_j$ is a symmetrizer. Since $\cL_0$ is symmetric, the $\Pi_j$ are symmetric when $a= 0$ and therefore
$\cS (0, \xi, \eta) =\Id$, implying \eqref{syma}.
\end{proof}

Substituting in \eqref{symgena} implies the following corollary.

\begin{cor}
\label{corholdsym}
If $\cL_0$ is symmetric and strictly hyperbolic and $a(x) = | x |^\alpha$ with $0 < \alpha < 1$,
$L_a$ admits H\"older continuous
symmetrizers $S (a, x, \xi, \eta)$ of class $C^\alpha$.

If $a(x) = x$, it admits a Lipschitz symmetrizer.
\end{cor}

\begin{exam}
Consider
\begin{equation}
\label{exple1}
L_a = \D_t +
\begin{pmatrix}
0 & \D_x + x a \D_y & \ x \D_y
\\ \D_x - x a \D_y & 0 & 0
\\
x (1 + a^2)) \D_y & 0 & 0
\end{pmatrix}.
\end{equation}
In this case, $\det \cL(a, \tau, \xi, \eta) = \tau (\tau^2 - \xi^2 - \eta^2)$ is always strictly hyperbolic.
\end{exam}

\begin{lem}
If $a \ne 0 $ is a constant, there are bounded symmetrizers $S (x, \xi, \eta)$ for~$L_a$, but no continuous
symmetrizers at $(x, \xi) = (0, 0) $ when $\eta = 1$.
\end{lem}

\begin{proof}
Fix $\eta = 1$. If $ S (x, \xi) $ is a symmetrizer, then its complex conjugate is also a symmetrizer,
so that $ S + \overline {S}$ is a symmetrizer. Thus, it is sufficient to consider
the case where $ S$ has real coefficients $s_{j, k}$. The symmetry condition reads
\begin{equation*}
\begin{aligned}
(\xi + a x) s_{11}& = (\xi - a x) s_{22} + (1+a^2) x s_{23}
\\
\eta s_{11} & = (\xi - a x) s_{23} + (1+a^2) x s_{33}
\\
\eta s_{12} & = (\xi + a x) s_{13}.
\end{aligned}
\end{equation*}
The third condition is independent of the first two, it only involves $s_{12} $ and $s_{13}$, and
is trivially satisfied by $s_{12}= s_{13} = 0$.

There is no restriction in assuming that $s_{22} = 1$. Setting
$s'_{11} = s_{11} -1 $, $s'_{33} = (1+a^2) s_{33} -s_{11} $, one must have
\begin{equation}
\label{pds}
\begin{aligned}
(\xi + a x) s'_{11}& = -2 a x + (1+a^2) x s_{23},
\\
x s'_{33} & = - (\xi - a x) s_{23}.
\end{aligned}
\end{equation}
Suppose that the coefficients are continuous at $(x,\xi)= (0,0)$. Then taking $x = 0$
and $\xi \ne 0$ in the equations above, dividing by $\xi$ and letting $\xi$ tend to $0$ implies that
$s_{2,3} (0, 0) = s'_{11} (0, 0) = 0$. Taking $\xi = 0$ dividing by $x$ and letting $x$ tend
to $0$ implies that $ s'_{11} (0, 0) = - 2 +(1 +a^2) s'_{23}(0, 0)$
and $s'_{33}(0,0) = a s_{2,3} (0, 0) $.
These conditions can be met only if $a = 0$.
\end{proof}

When $a(x) = x$, by Corollary~\ref{corholdsym}, there is a Lipschitz symmetrizer, but it turns out
that in this specific case, one can construct a $C^\infty$ symmetrizer.
The next example shows that this is not always the case.

\begin{exam}
Consider the $4 \times 4 $ system with symbol
\begin{equation}\arraycolsep3.5pt
L_a = \D_t + \begin{pmatrix} \Omega & a J \\ 0 & 2 \Omega \end{pmatrix},
\qquad \Omega = \begin{pmatrix} \xi & x\eta \\ x\eta & - \xi \end{pmatrix},
\qquad J = \begin{pmatrix} x \eta & 0\\ 0 & 0 \end{pmatrix}.
\end{equation}
By Corollary~\ref{corholdsym}, when $a(x) = x$, $L_a$ has a Lipschitz symmetrizer but
\end{exam}

\begin{lem}
When $a(x) = x$, there are no $C^1$ symmetrizers for $L_a$.
\end{lem}

\begin{proof}
Fix $\eta = 1$. Suppose that $S (\xi, x) $ is a $C^1$ symmetrizer near $(x, \xi) = (0,0)$.
We can assume that $S$ has real coefficients. Using the block notation
\begin{equation}
S = \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22}\end{pmatrix},
\end{equation}
the symmetry conditions imply
\begin{equation}\arraycolsep3.5pt
\label{exple2b}
\Omega S_{12} - 2 S_{12} \Omega = x^2 J_0 S_{11},
\qquad J_0 = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}.
\end{equation}
This is a linear system in $S_{12}$ and since
$\Omega $ and $2 \Omega$ have no common eigenvalue it has a unique solution.

If $S_{12}$ is $C^1$ near the origin, plugging its Taylor expansion
$\Sigma_0 + x \Sigma_1 + \xi \Sigma_2$ in \eqref{exple2b}
and using the notation $\Omega = x \Omega_1 + \xi \Omega_2$, yields at first order
\begin{equation*}
\Omega_1 \Sigma_0 - 2 \Sigma_0 \Omega_1 = \Omega_2 \Sigma_0 - 2 \Sigma_0 \Omega_2 = 0,
\end{equation*}
which implies that $\Sigma_0 = 0$. The term in $\xi^2$ is
\begin{equation*}
\Omega_2 \Sigma_2 - 2 \Sigma_2 \Omega_2 = 0,
\end{equation*}
showing that $\Sigma_2 = 0$. The term in $x \xi$ is then
\begin{equation*}
\Omega_2 \Sigma_1 - 2 \Sigma_1 \Omega_2 = 0,
\end{equation*}
implying that $\Sigma_1 \!=\! 0$, which is incompatible with the equation given by the term in~$x^2$:
\begin{equation*}
\Omega_1 \Sigma_1 - 2 \Sigma_1 \Omega_1 = - 2 J_0 S_{11} (0, 0) \ne 0
\end{equation*}
since $S_{11}(0,0)$ must be positive definite.
\end{proof}

\Subsection{Existence of smooth symmetrizers for generic double eigenvalues}
\label{subsec32}

Consider a symbol $ \tau \Id + A(a, \xi) $ which is strongly hyperbolic in the time direction, thus admitting a bounded
symmetrizer $S(a, \xi)$.
At $(\underline a, \underline \xi)$, $\underline \xi \ne 0$, the characteristic polynomial $p(a, \tau, \xi) = \det (\tau \Id + A(a, \xi)) $
has roots $\underline \tau_j$ of multiplicity $m_j$. Near this point, it
can be smoothly factored as
\begin{equation}
\label{factor}
p(a, \tau, \xi) = \prod_j p_j (a, \tau, \xi),
\end{equation}
with $p_j $ of order $m_j $.

\begin{ass} In a neighborhood of
$(\underline a, \underline \xi)$, $ \underline \xi \ne 0$, the roots of $p$ are either of constant multiplicity
or of multiplicity at most two.

In the second case, we assume that the multiplicity is two on a smooth manifold~$\cM$. Denoting by
$p_j$ the corresponding factor in \eqref{factor}, we further assume that either

(i) $\cM$ has codimension one and the discriminant of $p_j$ vanishes on $\cM$ at finite order,

\noindent or

(ii) $\cM$ has codimension two and the discriminant of $p_j$ vanishes on $\cM$ exactly at order two.
\end{ass}

\begin{theo}
\label{theodouble}
Under these assumptions, there is a smooth symmetrizer $S(a, \xi)$ on
a neighborhood of
$(\underline a, \underline \xi)$.
\end{theo}

\begin{proof}
The construction is local in $ \rho = (a, \xi)$ and one can perform a block reduction
of $A$ near $\underline \rho$ and it is sufficient to construct a symmetrizer
for each block. They are either diagonal and thus symmetric, or of dimension two.
Eliminating the trace, it is therefore sufficient to consider matrices
\begin{equation}\arraycolsep3.5pt
A (\rho) = \begin{pmatrix} - a & b \\ c & a \end{pmatrix}.
\end{equation}
The hyperbolicity condition is that the discriminant
$\Delta = a^2 + b c $ is real and non-negative.
Strong hyperbolicity holds if and only if there exists $\eps > 0$ such that
\begin{equation}
\label{stronghyp22}
\Delta = a^2 + b c \ge \eps (|a| ^2 + | b |^2 + |c|^2).
\end{equation}
Our assumption is that $\Delta$ vanishes on a manifold $\cM$,
at finite order if $\codim \cM = 1$ and at order two if $\codim \cM = 2$.
\begin{enumeratea}
\item
If $\Delta $ vanishes at finite order on a manifold of codimension $1$ of equation $\{\vp = 0\}$, then
\eqref{stronghyp22} implies that for some integer $k$,
\begin{equation}
A = \vp^k A_r
\end{equation}
with $\det A_r \ne 0 $ and still strongly hyperbolic. Thus $A_r$ has distinct real eigenvalues and is
therefore smoothly diagonalizable.

\item
Suppose that $\Delta$ vanishes exactly at second order on $\cM$
given by the equations $\{\vp = \psi = 0\}$. This means that
$\Delta \ge \eps_1 (\vp^2 + \psi^2)$.
Together with \eqref{stronghyp22}, this implies that $A$ vanishes
on $\cM$ and that
\begin{equation}
A = \vp A_1 + \psi A_2,
\end{equation}
and $A_1$ and $A_2$ have distinct real eigenvalues at $\underline \rho$. We can smoothly conjugate $A_1$
to a real diagonal and traceless form and changing $\vp$ we are reduced to the case where
$$\arraycolsep3.5pt
A_1 = \begin{pmatrix}
-1 &0 \\
0 & 1
\end{pmatrix}, \qquad A_2 = \begin{pmatrix}
- a_2 & b_2 \\
c_2 & a_2
\end{pmatrix}.
$$
Moreover, changing $\vp$ to $\vp - \re a_2 \psi$, we can assume that $\re a_2 = 0$.
Since $\Delta $ is real
\begin{equation}
\label{realdelta}
2 \vp \im a_2 + \psi \im (b_2 c_2) = 0.
\end{equation}
Moreover, \eqref{stronghyp22} implies for $\vp = 0$
\begin{equation}
\label{positiv22}
\re (b_2 c_2) > (\im a_2)^2,
\end{equation}
and this remains true in a neighborhood of $\underline \rho$. In particular
$b_2 (\underline \rho) \ne 0$ and conjugating by a diagonal matrix with diagonal entries
$b_2 / | b_2 |$ and $1$ changes $b_2$ into $|b_2 | $, meaning that we can assume that
$b_2$ is real. Having performed these reductions, one easily checks using \eqref{realdelta} that
\begin{equation}
\begin{pmatrix}
\re c_2 & i \im a_2 \\
- i \im a_2 & b_2
\end{pmatrix}
\end{equation}
is a smooth symmetrizer for $\vp A_1 + \psi A_2$, which is positive definite by \eqref{positiv22}.\qedhere
\end{enumeratea}
\end{proof}

\subsection{Ill-posedness for non-Lipschitz symmetrizers}

Consider the system \eqref{exple1} with $a= a(x) = | x|^\alpha$. For $\eta$ large and $\beta $ to be determined, we look for solutions of $L_a U = 0$
of the form
\begin{equation}
\label{illp1}
U(t, x, y) = e^{i \beta \sqrt \eta\, t + i y \eta} \begin{pmatrix}
u(\sqrt \eta\, x) \\ v(\sqrt \eta\, x) \\ w(\sqrt \eta\, x)
\end{pmatrix}.
\end{equation}
With $\eps = \eta^{-\alpha/2}$,
the equation $LU = 0$ is equivalent to
\begin{equation}
\label{illp2}
v (x) = \frac{i}{\beta} (\D_x - i \eps x a) u (x), \qquad w
= - \frac{1}{\beta} (1 + \eps^2 a^2) u (x),
\end{equation}
and the scalar equation for the first component is
\begin{equation*}
\big(\beta^2 + (\D_x + i \eps x a) (\D_x - i \eps x a (x)) - x^2 (1+ \eps^2 a^2) \big) u = 0,
\end{equation*}
that is, since $\D_x(xa) = (\alpha+1) a$,
\begin{equation}
\label{ill3}
\big(\beta^2 + \D_x ^2 - x^2 - i \eps (\alpha+1) a \big) u = 0.
\end{equation}
The example has been cooked up precisely to get an eigenvalue problem for
a perturbation of the harmonic oscillator.

When $\alpha = 0$, $\eps = 1$ and $u(x) = e^{- \mez x^2} $ is a a solution when
\begin{equation}
\label{ill4}
\beta^2 = i -1.
\end{equation}
Choosing the root with negative imaginary part, this yields exact solutions of $L_a U = 0$ of the form
\begin{equation}
\label{illp5}
U_\lambda (t, x, y) = \int e^{i \beta \sqrt \eta\, t + i y \eta} e^{- \mez \eta x^2}
(U_0 + \sqrt \eta\, x U_1) \vp (\eta / \lambda) \,d\eta
\end{equation}
with constant vectors $U_0 $ and $U_1$ not equal to $0$ and
$\vp \in C^\infty_0 (\RR)$ with support in the interval $[1, 2]$.
The exponential growth of
$e^{i t \sqrt \eta\, \beta}$ implies that there is no control
of any $H^{-s}$ norm at positive time by an $H^{s'}$ norm of the initial data.
This can be localized in $(x, y)$ and

\begin{prop}
When $a = 1$, $L_a$ has bounded symmetrizers but the Cauchy problem for \eqref{exple1} is ill-posed in $L^2$ as well as in
$C^\infty$.
\end{prop}

Consider now the case $\alpha \in ]0, 1[$. By standard perturbation theory the eigenvalue problem
\eqref{ill3} has a solution
\begin{equation}
\label{ill5}
u = e^{- \mez x^2} + \eps u_1, \qquad \beta^2 = 1 + i \eps \lambda_1 + O(\eps^2)
\end{equation}
with
\begin{equation}
\label{ill6}
\lambda_1 \int e^{- x^2} \,dx = (\alpha+1) \int a(x) e^{- x^2} \,dx > 0,
\end{equation}
so that $\lambda_1 > 0$. Therefore one can choose
$
\beta = - 1 - \frac{i}{2} \eps \lambda_1 + O(\eps^2)
$, so that
\begin{equation}
\im (\beta \sqrt \eta) \sim \mez \lambda_1 \eta^{\mez (1- \alpha)} < 0,
\end{equation}
which is arbitrarily large if $\alpha < 1$. This provides solutions of $L_a U = 0$,
with exponentially amplified $L^2$ norms implying the following proposition.

\begin{prop}
When $a = | x|^\alpha$ with $0< \alpha < 1$, $L_a$ has $C^\alpha$ symmetrizers
but the Cauchy problem for \eqref{exple1} is ill-posed in $L^2$.
\end{prop}

\section{Strong hyperbolicity of first order symbols}
\label{sectionstroghyp}

In this section we introduce the notion of strong hyperbolicity and show that it is equivalent
to the existence of symmetrizers. Next we discuss the existence of smooth
symmetrizers. We show that these notions are preserved by a change of the time direction.
For the convenience of the reader, we postpone to the appendix
the proof of several independent results on matrices.

\subsection{Basic properties}
We denote by $\tilde x \in \RR^{1+d}$ the time-space variables and by
$\tilde \xi $ the dual variables.
We consider $N \times N$ first order system systems $ \sum_{j=0}^d A_j \D_{\tilde x_j} + B $.
Their characteristic determinant is $p(\tilde \xi) = \det\big(\sum_{j=0}^d i {\tilde \xi_j}A_j + B\big) $,
the principal part of which is $p_N (\tilde \xi) = \det\big(\sum_{j=0}^d i {\tilde \xi_j}A_j \big) $.

\skpt
\begin{defi}
\begin{enumeratei}
\item
$ \sum_{j=0}^d A_j \D_{\tilde x_j} + B $ is said to be hyperbolic
in the direction $\nu \in \RR^{1+d} $ if
$p_N (\nu) \ne 0$ and there is $\gamma_0$ such that
$p(i \tau \nu + \tilde \xi) \ne 0 $ for all $\xi \in \RR^{1+d}$ and all real
$\tau$ such that $| \tau | > \gamma_0$.

\item
$L = \sum_{j=0}^d A_j \D_{\tilde x_j} $ is strongly hyperbolic in the direction $\nu$ if and only if for all matrix $B$, $L+B$ is hyperbolic in the direction $\nu$.
\end{enumeratei}
\end{defi}

The classical definition of hyperbolicity is that the roots of $p(i \tau \nu + \xi) \ne 0 $ are located in $\tau < \gamma_0$. But, since hyperbolicity in the direction $\nu$ implies hyperbolicity in the direction $- \nu$, the definition above is equivalent to the usual one.

\begin{prop}
\label{invhyp1}
$L = \sum_{j=0}^d A_j \D_{\tilde x_j} $ is strongly hyperbolic in the direction $\nu$ if and only if
there is a constant $C $ such that
\begin{enumeratei}
\item
for all $\tilde \xi\in \RR^{1+d} $ and all
matrix $B$, the roots of $\det \big(L(\tilde \xi + \lambda \nu) + B\big) = 0 $ are located in the strip
$| \im \lambda | \le C | B|$.

With the same constant $C$, this condition is equivalent to
\item
for all $(\gamma, \tilde \xi, u) \in \RR \times \RR^{1+d} \times \CC^N$:
\begin{equation}
\label{resolvp1}
\big| \gamma u \big| \le C \big| L(\tilde \xi + i \gamma \nu) u \big|.
\end{equation}
\end{enumeratei}
\end{prop}

Other equivalent formulations can be deduced from Proposition~\ref{propstronghyp} below.

\skpt
\begin{proof}
\begin{enumeratea}
\item
By homogeneity, (ii) is equivalent to the condition
\begin{equation}
\label{eqC2prime}
\qquad \big| \im \lambda | \ge C \quad \Longrightarrow \quad
\big| L(\tilde \xi + \lambda \nu)^{-1} \big| \le 1.
\end{equation}
By Lemma~\ref{leminv} below, this is equivalent to the condition
that for all matrices $B$ such that $| B| < 1$, $L(\tilde \xi + \lambda \nu) + B$ is invertible
when $| \im \lambda | \ge C $. This is equivalent to saying that the roots of
$\det\big(L(\tilde \xi + \lambda \nu) + B \big) = 0$ are contained in $\{| \im \lambda | < C\} $.
By homogeneity, this is equivalent to (i).

\item
Note that \eqref{resolvp1} applied to $\tilde \xi=0$
implies that $L(\nu) $ is invertible. It is then clear that (i) implies
strong hyperbolicity. Conversely, assume that $L$ is strongly hyperbolic.
Consider the matrix $ B_{j, k}$ with all entries equal to zero, except the entry
of indices $(j, k)$ equal to one. Then
\begin{equation*}
\det\Big(L(\tilde \xi) + B_{j,k}\Big) =
\det L(\tilde \xi) + m_{j k}(\xi),
\end{equation*}
where $m_{j,k}$ is the cofactor of indices $(j, k)$ in the matrix $L(\tilde \xi)$.
Following Theorem 12.4.6 in \cite{Ho1}, the hyperbolicity condition
implies that there is a constant $C$ such that
\begin{equation*}
\label{estimqp}
\big| m_{j, k} (\tilde \xi + i \nu) \big| \le C \big| \det L (\tilde \xi + i \nu) \big|.
\end{equation*}
Since
$L(\tilde \xi + i \nu)^{-1} = (\det L(\tilde \xi + i \nu))^{-1} \widetilde M(\tilde \xi + i \nu)$ where
$\widetilde M$ is the matrix with entries $(-1)^{j+k} m_{k,j}$,
this implies that there is another constant $C$ such that \eqref{resolvp1} is satisfied for $\gamma =1$. By homogeneity, it is also satisfied for all
$ \gamma$.\qedhere
\end{enumeratea}
\end{proof}

When $p$ is hyperbolic in the direction $\nu$, the component of
$\nu $ in the set $\{p _N \ne 0 \}$ is a convex open cone, which we denote by $\Gamma (\nu) $,
and $p$ is hyperbolic in any direction $\nu' \in \Gamma (\nu)$.
This property is also true for strong hyperbolicity. We give a quantitative
version of this result, as we will need it later on.

\begin{lem}
\label{lemopencone}
Suppose that $L$ is hyperbolic in the direction $\nu$. For all $\nu'\in \Gamma(\nu)$,
the ball centered at $\nu'$ of radius $\eps: = |p_N(\nu') | / K |\nu'|^{N-1}$ is contained in
$\Gamma(\nu)$, where
$ K = \max_{| \tilde \xi | \le 2} | \na_{\tilde \xi} \det L(\tilde \xi) | $.
\end{lem}

\begin{proof}
By homogeneity, one can assume that
$|\nu' | = 1$. In this case,
$|p_N(\nu '')| > | p_N(\nu')| - K | \nu'- \nu"| $ if $\nu" - \nu'| \le 1$.
Noticing that $| p_N (\nu') | = | \nu' \na_{\tilde \xi} p_N(\nu') | \le K$, this implies that
$|p_N(\nu '')| > 0 $ if $|\nu" - \nu'| \le \eps $.
\end{proof}

\begin{prop}
\label{propinvnu}
Suppose that $L(\tilde \xi) $ satisfies \eqref{resolvp1} and let $\nu' \in \Gamma(\nu) $ such that
$\big| \det L(\nu') \big| \ge c > 0$. Then
\begin{equation}
\label{resolvp2}
\big| \gamma u \big| \le C_1 \big| L(\tilde \xi + i \gamma \nu') u \big|,
\end{equation}
with $C_1 = K C | \nu| / c | \nu'| $ and
$ K = \max_{| \tilde \xi | \le 2} | \na_{\tilde \xi} \det L(\tilde \xi) | $.
\end{prop}

\begin{proof}
Let $p (\tilde \xi) = \det (L(\tilde \xi) + B)$ and
$p_N = \det L(\xi)$ its principal part.
Suppose that $|\nu | = | \nu' | = 1$. The general case follows immediately.
By Proposition~\eqref{invhyp1}, one has
\begin{equation}
\label{eq501}
p(\tilde \xi + i \gamma \nu) \ne 0 \quad \mathrm{for} \ | \gamma | > C | B|.
\end{equation}

We choose $\nu'' = \nu' - \eps \nu $ with $\eps = c / K$.
By Lemma~\ref{lemopencone} $\nu'' \in \Gamma(\nu)$ and following
\cite{Gard1, Ho1}, (see e.g. \cite{Ho1} vol 2, chap 12), one has
\begin{equation}
\label{eq51}
p(\tilde \xi + i \gamma \nu + i \sigma \nu'') \ne 0 \quad \mathrm{for} \ \gamma > C, \ \sigma \ge 0,
\end{equation}
and also for $ \gamma < - C |B| $ and $\sigma \le 0$.
Indeed, all the roots of $ p_N(t \nu + \nu'')$ are real and negative:
\begin{equation}
\label{eq50}
p_N (t\nu + \nu'') = 0
\quad \Longrightarrow\quad t < 0.
\end{equation}
By \eqref{eq501},
$p(\tilde \xi + i \gamma \nu + z\nu'') = 0$ has no root on the real axis, so that
the number of roots in $\{\im z \ge 0 \} $ is independent of
$\tilde \xi$ and $\gamma > C | B| $. Taking $\tilde \xi = 0$ and letting $\gamma $ tend to
$+ \infty$, \eqref{eq50} implies that this number is equal to zero, hence \eqref{eq51}.
The proof for $\gamma \le - C | B|$ and $\sigma \le 0$ is similar.

Substituting $\nu'' = \nu' - \eps \nu$ in \eqref{eq51} and choosing $\gamma = \eps' \sigma$
we conclude that
$$
p(\tilde \xi + i \sigma \nu') \ne 0
$$
if $ \eps | \sigma | > C | B| $. Applying again Proposition~\ref{invhyp1},
\eqref{resolvp2} follows.
\end{proof}

In most applications, the coefficients of the system $L$ and even the direction $\nu$ may depend
on parameters, such as the space time variables, the unknown itself etc. The direction $\nu$ itself
can be seen as a parameter.
This leads to consider families of systems, $L(a, \D_{\tilde x})$ and directions $\nu_a$, depending on
parameters $a \in \cA$. Their symbol is
\begin{equation}
\label{gesymb}
L (a, {\tilde \xi}) = \sum_{j = 0}^d \tilde \xi_j A_j (a).
\end{equation}
When considering such families, we always assume that
the matrices $A_j(a)$ and the directions $\nu_a$ are uniformly bounded.

\begin{defi}
We say that the family $ L(a, \, \cdot\,)$ is uniformly strongly hyperbolic in the direction $\nu_a$ for
$a \in \cA$ if,
\begin{enumeratei}
\item
$c_\cA:= \inf_{a \in \cA}| \det L(a, \nu_a) | > 0$,

\item
the equivalent conditions (i) and (ii) of Proposition~\ref{invhyp1} are satisfied with a constant
$C$ independent of $a \in \cA$.
\end{enumeratei}
\end{defi}

The next result is an immediate consequence of Proposition~\ref{propinvnu}.
It shows that one can enlarge the set of strongly hyperbolic directions, preserving uniformity:
let $\Gamma_a$ denote the component of $\nu_a$ in $\{\det L(a, \tilde \xi) \ne 0\}$; for $c \in{} ] 0, c_{\cA}] $ and $C >0$,
introduce the set
\begin{equation}
\label{tildecA}
\widetilde \cA = \big\{(a, \nu) \mid a \in \cA,\, \nu \in \Gamma_a,\, | \nu | \le C,\, | \det L(a, \nu) | \ge c
\big\}.
\end{equation}

\begin{prop}
\label{prophyppol}
Suppose that $ L(a, \, \cdot\,)$ is uniformly strongly hyperbolic in the direction $\nu_a$ for
$a \in \cA$.
Then $ L(a, \, \cdot\,)$ is uniformly strongly hyperbolic in the direction $\nu$, for
$(a, \nu) \in \widetilde \cA$.
\end{prop}

\subsection{Symmetrizers}

We start with the notion of \emph{full symmetrizer} introduced in \cite{FriLa1}.

\begin{defi}
A full symmetrizer for $L(\tilde \xi) = \sum \tilde \xi_j A_j$ is a bounded
matrix ${\bf S} (\tilde \xi) $, homogeneous of degree $0$
on $\RR^{1+d} \moins\{0\}$, such that
$\bS(\tilde \xi) L(\tilde \xi)$ is self-adjoint.
It is positive in the direction $\nu \ne 0$ if there is a constant $c> 0$ such that
for all $\tilde \xi \ne 0$:
\begin{equation}
\label{posfullsym}
u \in \ker L(\tilde \xi) \quad \Longrightarrow \quad
\re \big(\bS(\tilde \xi) L(\nu) u, u \big) \ge c | u|^2.
\end{equation}
Given a family of systems $L(a, \, \cdot)$ and directions $\nu_a$, a bounded family
of full symmetrizers $\bS(a, \, \cdot \,) $ for $a\in \cA$ is said to be uniformly positive
in the direction $\nu_a$ if the constant $c$ above can be chosen independent of $a$.
\end{defi}

In \eqref{posfullsym}, $ (f, u) $ denotes the Hermitian scalar product in $\CC^N$.
More intrinsically, it should be thought as the antiduality between covectors
$f \in \VV^* $ and vectors $u \in \VV $, where $\VV$ is a vector space of dimension $N$,
so that the adjoint $P^*$ of the operator~$P $ from $\VV$ to $\VV$ satisfies
$(f, Pu) = (P^*f, u)$. In this spirit, the symbol $L(\tilde \xi)$ must be thought as linear mapping from
a vector space $\VV$ to another vector space $\WW$ and the symmetrizer $S (\tilde \xi)$ maps
$\WW $ to $\VV^*$ so that the antiduality $ (S L u, u)$ makes sense.

Note that outside a conical neighborhood of the
the characteristic variety, the symmetrizer can be chosen arbitrarily and thus
contains no information.

\smallskip
A different and more familiar notion of symmetrizer depends on the choice
of a time direction $\nu$. Choosing a space $\EE $ such that $ \RR^{1+ d} = \EE \oplus \RR \nu$
the symmetrizer is seen as a function of frequencies $\xi \in \EE$.
Since the open cone $\Gamma(\nu)$ is strictly convex, one can also require that
$\Gamma(\nu) \cap \EE = \emptyset$.
In a more intrinsic definition, it can be seen as a symmetrizer
invariant by translation in the direction $\nu$, or defined on $\RR^{1+d} / \RR \nu$.
To avoid technicalities, we choose the first option choosing a space $\EE$.
and when considering families $(L(a), \nu_a)$, we assume that we can choose $\EE$
in such a way that there is a compact set $K$ such that
\begin{equation}
\label{unifdecomp}
\forall a \in \cA,\ \nu_a \in K \quad \mathrm{and}\quad K \cap \EE = \emptyset.
\end{equation}
This condition can always be met locally. In particular, uniformly in $a \in \cA$:
\begin{equation}
\label{unifdecomp2}
c (| \xi | + | \tau|) \le | \xi + \tau \nu_a | \le C (| \xi | + | \tau|), \qquad \xi \in \EE, \ \tau \in \RR.
\end{equation}

\begin{defi}
A symmetrizer for $L(\tilde \xi) = \sum \tilde \xi_j A_j$ in the direction $\nu$ is a bounded
matrix $S(\xi) $, homogeneous of degree $0$ in $\xi \in \EE$ such that
$S(\xi) L(\xi)$ and $S(\xi) L(\nu)$ are
self-adjoint for all $ \xi $ and there is $c > 0$ such that:
\begin{equation}
\label{possym}
\forall \xi \in \EE\moins\{0\}, \ \forall u \in \CC^N, \quad \big(S(\xi) L(\nu) u, u \big) \ge c | u|^2.
\end{equation}
Given a family of systems $L(a, \, \cdot)$ and directions $\nu_a$ satisfying \eqref{unifdecomp}, a uniform family
of symmetrizers $S(a, \, \cdot \,) $ for $\{L(a, \cdot), \nu_a\} $, $ a\in \cA$, is a bounded family $S(a, \cdot)$ of symmetrizers
for $L(a, \cdot)$ in the direction $ \nu_a$, such that the constant $c$ can be chosen independent of $a$.
\end{defi}

\begin{rem}
\label{remdefsym}
$\bS (\xi + \tau \nu) = S(\xi)$, is almost a full symmetrizer, except that it not necessarily
defined on the line $\RR \nu$. But \eqref{possym} implies that
$L(\nu) $ is invertible and one can always choose
$\bS(\nu)$ so that
$\bS(\nu) L(\nu)$ is positive definite.
This modification can be extended to a conical neighborhood of $\nu$
where $L (\tilde \xi))$ remains invertible. This construction
obviously preserves positivity.
This remains true for families and uniformity can be preserved.
\end{rem}

The existence of symmetrizers is equivalent to strong hyperbolicity, in the
following sense.

\begin{theo}
\label{theostronghyp}
Consider a family $\{L(a, \, \cdot\,), \nu_a, a \in \cA\} $.
\begin{enumeratei}
\item
Assuming \eqref{unifdecomp}, $L(a,\cdot)$ is strongly hyperbolic in the direction $\nu_a$
if and only if there exists a uniform family of symmetrizers $S(a, \cdot) $.
\item
$L(a,\cdot)$ is strongly hyperbolic in the direction $\nu_a$ if and only if
\begin{enumerate}
\item
it is hyperbolic in the direction $\nu_a$ and $\inf_{a \in \cA} | \det L(a, \nu_a) | > 0$,
\item
there is a bounded family of full symmetrizer $\bS(a, \cdot) $ which is
uniformly positive in the direction $\nu_a$.
\end{enumerate}
\end{enumeratei}
\end{theo}

\skpt
\begin{proof}
\begin{enumeratei}
\item
If $L(a, \cdot) $ is uniformly strongly hyperbolic in the direction $\nu_a$,
then $L(a, \nu_a)$ and $L(a, \nu_a)^{-1}$ are uniformly bounded,
Similarly, \eqref{possym} implies that $L(a, \nu_a)^{-1}$ is bounded.

In both case, $ A(a, \xi) \!=\! L(a, \nu_a)^{-1} L(a, \xi)$ for $| \xi | \!=\! 1$ is bounded, and
strong hyperbo\-licity is equivalent to the existence of a constant $C$ such that
for all $\lambda,a, \xi \!\in\! \EE $ and~$u$:
\begin{equation}
\label{resolvestp11}
| \im \lambda | \big| u \big| \le C \big| A(a, \xi) u - \lambda u \big|.
\end{equation}
By Proposition~\ref{propstronghyp} this is equivalent to the existence of
a symmetric matrix
$S_A(a, \xi)$, bounded and uniformly positive definite, such that $S_A A$ is symmetric.
This is equivalent to the condition that $S (a, \xi) = S_A(a, \xi) L(a, \nu_a)^{-1}$ is a symmetrizer
for $L(a, \cdot) $ bounded and uniformly positive in the direction $\nu_a$.

\item
Strong hyperbolicity implies the existence of a symmetrizer, thus of a full positive symmetrizer by Remark~\ref{remdefsym}. Hence it only remains to prove the converse part of (ii).

Let $\bS(a, \cdot) $ be a full symmetrizer for $L(a, \cdot) $, positive in the direction
$\nu_a$. Suppose in addition that $L(a, \cdot)$ is hyperbolic in this direction,
so that $L(a, \nu)$ is invertible. Then, Proposition~\ref{propPSP} implies that
when $\ker L(a, \tilde \xi) \ne \{0\}$, $0$ is a
semi-simple eigenvalue of $ A(a, \tilde \xi) = L(a, \nu_a)^{-1} L a, \tilde \xi)$. Moreover, the spectral projectors,
that is the projectors on $\ker A = \ker L $ parallel to the range of $A$, are uniformly bounded.
Applied to $\tilde \xi + \tau \nu$, this implies that all the real eigenvalues of
$A(a, \tilde \xi)$ are semi-simple and all the corresponding spectral projectors are uniformly
bounded. Since $L$ is hyperbolic, all the eigenvalues are real and with Proposition~\ref{propstronghyp}
this implies that \eqref{resolvestp11}
is satisfied.\qedhere
\end{enumeratei}
\end{proof}

\subsection{Smooth symmetrizers}

We now consider a family $\{L(a, \, \cdot\,), \nu_a, a \!\in\!\cA\} $ where~$\cA$ is an open set of
some space $\RR^m$. We assume that the condition \eqref{unifdecomp} is satisfied.

\begin{theo}
\label{theomaj}
Suppose that the coefficients of $L(a, \cdot)$ are continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\cA$ and that the mapping $a \mapsto \nu_a$ is
continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr. Then, there exists a full symmetrizer $\bS(a, \tilde \xi)$
which is continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\cA \times S^d$ and uniformly positive in the direction $\nu_a$,
if and only if there is a symmetrizer $S(a, \xi)$
which is continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\cA \times S^{d-1}$ and uniformly positive in the direction $\nu_a$.
\end{theo}

\begin{proof}
By Remark~\ref{remdefsym} passing from a symmetrizer to a full symmetrizer
is immediate. The converse statement follows from a more general
result given in Theorem~\ref{theomajext} where the construction
is extended to other directions $\nu \in \Gamma(\nu_a)$.
\end{proof}

\subsection{Invariance by change of time}

Proposition~\ref{prophyppol} shows that strong hyperboli\-city, thus the existence of bounded symmetrizers or
of full symmetrizers, extends from~$\nu_a$ to all directions in the cone of hyperbolicity $\Gamma_a$,
preserving uniformity in sets such as \eqref{tildecA}.
We now prove that this is also true for smooth symmetrizers.
The key point, is to prove that for a continuous full symmetrizer,
positivity extends from $\nu_a$ to $\Gamma (\nu_a)$.

\begin{prop}
\label{propextpos}
Consider a family $\{L(a, \, \cdot\,), \nu_a, a \in \cA\} $ and assume that
$L(a, \cdot)$ is uniformly strongly hyperbolic in the direction $\nu_a$.
For $c>0$ and $C$ given, define $\widetilde \cA$ as in \eqref{tildecA}.

Suppose that $\bS(a, \cdot)$ is a full symmetrizer of $L(a, \cdot)$ which depends continuously
on $ \tilde \xi \in \RR^{1+d} \moins\{0\}$, such that $\bS(a, \cdot) $ is uniformly positive in the direction
$\nu_a$. Then, $ \bS(a, \cdot) $ is uniformly positive in the direction $\nu$ for
$(a, \nu) \in \widetilde \cA$.
\end{prop}

\begin{proof}
Since $\bS(a, \tilde \xi + s \eta) L(a, \tilde \xi + s\tilde \eta)$ is symmetric,
for $u$ and $v$ in $\ker L(a, \tilde \xi)$ one has
\begin{equation*}
\big(\bS(a, \tilde \xi + s \eta) L(a,\tilde \eta) u, v \big)
= \big(u, \bS(a, \tilde \xi + s \eta) L(a,\tilde \eta) v \big).
\end{equation*}
Letting $s$ tend to $0$, shows that for all $\eta$,
the matrices $\bS(a, \tilde \xi) L(a,\tilde \eta) $ are symmetric on $\ker L(a, \tilde \xi)$.

By Lemma~\ref{lemopencone}, there is $\eps$ such that for all $(a, \nu) \in \cP$,
the ball centered at $\nu$ and radius $\eps$ is contained in $\Gamma_a(\nu_a)$.
Therefore, there is $t_0 \in ]0, 1[$ such that for all $(a, \nu)\in \cP$, there is
$\nu' \in \Gamma_a(\nu_a)$ on the line joining $\nu_a$ and $\nu$ such that
$\nu = t \nu_a + (1-t) \nu'$ with $t \in [t_0, 1[$.
By Proposition~\ref{propinvnu}, $L(a, \dot) $ is strongly hyperbolic
in the direction $\nu'$, implying that $L(a, \nu')^{-1} L(a, \tilde \xi)$ has
only real an semi-simple eigenvalues.
Therefore, the result follows from Proposition~\ref{positivcone} applied to
$J_t = (1-t) L(a, \nu_a) + t L(a, \nu) $.
\end{proof}

We are now ready to prove that the existence of a regular full symmetrizer
implies the existence of a symmetrizer, having the same smoothness,
in all directions $\nu\!\in\!\Gamma (\nu_a)$. In particular, this finishes the proof of
Theorem~\ref{theomaj}.
Consider a strongly hyperbolic family $\{L(a, \, \cdot\,), \nu_a, a \in \cA\} $. Assume that the condition \eqref{unifdecomp} holds.
For $c > 0$ and $C >0$ and $\cO$ an open neighborhood of $K$ such that $\overline \cO \cap \EE = \emptyset$
let
\begin{equation}
\label{tildecA0}
\widetilde \cA_0 = \big\{(a, \nu) \mid a \in \cA,\, \nu \in \Gamma_a \cap \cO,\, | \nu | \le C,\, | {\det}\, L(a, \nu) | \ge c
\big\}.
\end{equation}

\begin{theo}
\label{theomajext}

Suppose that the coefficients of $L(a, \cdot)$ are continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\cA$ and that the mapping $a \mapsto \nu_a$ is
continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr. Suppose that $S(a, \xi)$ is a uniform bounded family of symmetrizers for $\{L(a, \, \cdot\,)$
in the directions $\nu_a$ for $a\in \cA$, which is continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\cA \times \EE$. Then, there exist a continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr uniform family of symmetrizers $S (a, \nu, \xi) $ for $L(a, \cdot)$ in the direction
$\nu$, which is continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\widetilde \cA_0 \times \EE$.
\end{theo}

\begin{proof}
For $\tilde a = (a, \nu) \in \widetilde \cA_0$, consider
\begin{equation*}
\tilde L(\tilde a, \tau, \xi) = \tau L(a, \nu) + L(a, \xi) = L (a, \xi + \tau \nu).
\end{equation*}
Then $\tilde \bS (\tilde a, \tau, \xi) = \bS(a, \xi + \tau \nu)$ symmetrizes $\tilde L(\tilde a, \tau, \xi)$.
By \eqref{unifdecomp2}, $\tilde L$ and $\tilde \bS$ are
continuous, \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr functions on $\widetilde \cA_0 \times \RR \times S^{d-1} $.
The positivity condition
\begin{equation}
\re \big(\tilde \bS (\tilde a, \tau, \xi) L(a, \nu) u, u \big) \ge c | u|^2
\end{equation}
on $\ker \tilde L(\tilde a, \tau, \xi) = \ker L (a, \xi + \tau \nu)$ follows from Proposition~\ref{propextpos}
and the construction of a symmetrizer
$S(a, \nu, \xi)$, with the same smoothness as $\tilde \bS$, is given by Theorem~\ref{theomajmod}.
\end{proof}

\section{Appendix A: Strongly hyperbolic matrices}
\label{SecStrHypMat}

We collect here the various technical results on matrices which have been used
in the previous section. Changing slightly the notations, for instance including $\xi$
or $\nu$ among the parameters, we consider a family of $N \times N$ matrices, $A(a)$ depending on parameters $ a \in \Omega$
where $\Omega$ is an open subset of $\RR^n$. We denote by $\Sigma(a)$ the spectrum of
$A(a)$.

\Subsection{Definition and properties}

\begin{prop}
\label{propstronghyp}
The following properties are equivalent
\begin{enumeratei}
\item
There is a real $C_1$ such that
\begin{equation}
\label{eqC1}
\forall t \in \RR, \forall a \in \Omega,\quad
\big| e^{i t A(a)} \big| \le C_1.
\end{equation}
\item
All the the eigenvalues $\lambda$ of $A(a)$ are real and semi-simple and
there is a real~$C_2$ such that all the
eigen-projectors $\Pi_\lambda (a)$ satisfy
\begin{equation}
\label{eqC3}
\forall a \in \Omega, \quad
\big| \Pi_\lambda (a) \big| \le C_2.
\end{equation}
\item
$A(a)- \lambda \Id $ is invertible when $\im \lambda \ne 0$ and
there is a real $C_3$ such that
\begin{equation}
\label{eqC2}
\forall \lambda \notin \RR \ \forall a \in \Omega, \quad
\big| \big(A(a)- \lambda \Id \big)^{-1} \big| \le C_3 \big| \im \lambda |^{-1}.
\end{equation}
\item
There are positive definite
matrices $S(a)$ and there are constants
$C_4$ and $c_4 > 0$ such that for all $a \in \Omega$, $S(a) A(a) $ is symmetric,
and
\begin{equation}
\big| S(a) \big| \le C_4, \quad S(a) \ge c_4 \Id.
\end{equation}
\item
There is a real $C_5$ such that for all matrix $B$, all $a\in \Omega$ and all
$\rho \in \RR$, the eigenvalues of $\rho A(a) + B$ are located in
$\{| \im \lambda | < C_5 | B | \}$.
\end{enumeratei}
\end{prop}

\skpt
\begin{proof}
\begin{enumeratea}
\item
(ii) implies that $A(a)$ has the spectral decomposition
$A = \sum \lambda_j \Pi_j $ with real~$\lambda_j $'s.
Thus \eqref{eqC3} implies that
$\big| e^{i t A)} \big| = \big| \sum e^{it \lambda_j} \Pi_j\big| \le N C_2$.

Conversely, (i) implies that the eigenvalues $\lambda_j$ of $A(a)$ are real, and semi-simple
and thus that $A(a) = \sum \lambda_j \Pi_j $.
Moreover,
$$
\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T e^{i t (A(a) - \lambda_j \Id)} dt =
\sum_k \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T e^{i t (\lambda_k - \lambda_j \Id)} \Pi_k dt = \Pi_j.
$$
Thus, $| \Pi_j | \le C_1 $ if \eqref{eqC1} is true.

\item
Suppose that (ii) is satisfied so that $A = \sum \lambda_j \Pi_j$ and
$\Id = \sum \Pi_j$. Then
\begin{equation}
S (a) = \sum \Pi_j^* \Pi_j
\end{equation}
is positive definite, satisfies $S \ge N^{-1} \Id$, $| S | \le N C^2_2$, and
$S A = \sum \lambda_j \Pi_j^* \Pi_j $ is self-adjoint.

\smallskip
If (iv) holds then, with $\epsilon = \mathrm{sign} (\gamma)$,
$$
\begin{aligned}
c_4 | \gamma | \big| u\big|^2 \le \re \epsilon \big(S (- i A + \gamma \Id) u, u \big)
\le C_4 \big| (A + i \gamma) u \big| \ \big| u \big|,
\end{aligned}
$$
implying (iii) with $C_3 = C_4/ c_4$.

\smallskip
If (iii) is satisfied, then the eigenvalues of $A(a)$ are real, and semi-simple, for if there were a nondiagonal block in the Jordan's decomposition of $A - \lambda_j \Id $, the norm of
$ (A - (\lambda_j - i \gamma) \Id)^{-1} $ would be at least of order $\gamma^{-2}$
when $\gamma \to 0$. Thus $A = \sum \lambda_j \Pi_j$ and
$$
\lim_{\gamma \to 0} i \gamma \big(A - (\lambda_j - i \gamma) \Id)^{-1} =
\sum_k \lim_{\gamma \to 0} \frac{i \gamma}{(\lambda_k - \lambda_j + i \gamma} \Pi_k = \Pi_j,
$$
hence $| \Pi_j | \le \ C_3 $.

\item
By homogeneity, (iii) is equivalent to the condition
\begin{equation*}
\forall a \in \Omega, \forall \rho \in \RR, \qquad \big| \im \lambda | \ge C_3 \quad \Longrightarrow \quad
\big| \big(\rho A(a)- \lambda \Id \big)^{-1} \big| \le 1.
\end{equation*}
By Lemma~\ref{leminv} below, this is equivalent to the condition
that for all matrix $B$ such that $| B| < 1$, $\rho A - \lambda \Id + B$ is invertible
when $| \im \lambda | \ge C_3$, meaning that the spectrum of
$\rho A + B$ is contained in $\{| \im \lambda | < C_3$.
By homogeneity, this is equivalent to (v) with $C_5 = C_3$.

The proof of the proposition is now complete.\qedhere
\end{enumeratea}
\end{proof}

\begin{lem}
\label{leminv}
The matrix $A$ is invertible with $| A^{-1}| \le \kappa $ if and only if
$A+B$ is invertible for all $B$ such that $|B | < \kappa ^{-1} $.
\end{lem}

\begin{proof}
If $| A^{-1}| \le \kappa $, then $ A + B = A^{-1} (\Id + A^{-1} B)$ is invertible
for all $B$ such that $| A^{-1} B| \le \kappa | B| < 1$.

Conversely, if $A$ is not invertible or if $| A^{-1}| > \kappa $, there is $\underline u$ such that
$| \underline u| = 1$ and $| A \underline u | < \kappa^{-1} $. Pick a linear form
$\ell$ such that $\ell (\underline u) = 1$ and $| \ell| = 1$.
Then the matrix $B$ defined by $B u = \ell(u) A \underline u$
satisfies $| B | = | A \underline u| < \kappa^{-1} $ but $A-B$ is not invertible since
$\underline u$ is in its kernel.
\end{proof}

\Subsection{Lipschitz dependence of the eigenvalues}

\begin{ass}
\label{assSHM}
The family $\{A(a), a \in \Omega\}$ of $N \times N$ matrices,
is uniformly strongly hyperbolic in the sense that the equivalent properties of Proposition
$\ref{propstronghyp}$ are satisfied.
\end{ass}

\begin{prop}
\label{proplipvp} Suppose that $A(\cdot) \in W^{1, \infty} (\Omega)$ satisfies Assumption~$\ref{assSHM}$.
Denote by $\lambda_j(a)$, $1 \le j \le N$, the eigenvalues of $A(a)$, labelled in the increasing order
and repeated accordingly to their multiplicity.
Then, the functions $\lambda_j$ belong to $ W^{1, \infty} (\Omega)$.
\end {prop}

\begin{proof}
The continuity of the roots of a polynomial with respect to the coefficients is well-known.
The Lipschitz smoothness with respect to parameters of the roots of hyperbolic polynomials is true in general, provided that the coefficients are smooth enough (see \cite{Bron}). The proposition says that when the polynomial is the characteristic determinant of a
strongly hyperbolic system, the Lipschitz smoothness of the coefficients is sufficient.

Fix $\underline a \in \Omega$ and an eigenvalue
$\underline \lambda= \lambda_p (\underline a) = \lambda_{p+ m} (\underline a) $ of $A(\underline a)$
of multiplicity $m+1$. Let $\delta > 0$ denote the distance of $\underline \lambda$ to the remainder part
of the spectrum of $A(\underline a)$. By~Assumption~\ref{assSHM}, there
is $C$ which depends only on an upper bound of the norms of the spectral projectors, thus independent of
$\underline a$, such that
\begin{equation*}
\forall z \in \CC, \
| z - \underline \lambda | \le \delta/2, \qquad \big| \big(A(\underline a) - z \Id \big)^{-1} \big| \le C | z - \underline \lambda | ^{-1}.
\end{equation*}
Therefore, $A - z \Id$ is invertible when $ | A - A(\underline a) | < | z - \underline \lambda | \le \delta/2C$.

Let $\Omega_1 \subset \Omega $ denote a convex open neighborhood of $\underline a$. Because $A \in W^{1, \infty} (\Omega)$
for~$a$ and $a' \in \Omega_1$ there holds
\begin{equation}
\label{LipestA}
\big| A(a) - A(a') \big| \le K | a - a '|,
\end{equation}
with $K = \big\| \nabla_a A \big\|_{L^\infty(\Omega)}$.
Therefore,
$ A(a) - z \Id$ is invertible if $a \in \Omega_1$ and
$ C K | a - \underline a | \le | z - \underline \lambda | \le \delta/2$.
By Rouché's theorem, this implies that
$A(a)$ has $m+1$ eigenvalues (counted with their multiplicity) in the disk
$\{| \lambda - \underline \lambda | \le K C | a - \underline a|\}$. They must be real by assumption, and by continuity
they are $\{\lambda_p(a), \dots, \lambda_{p+m} (a) \}$. Hence,
$ | \lambda_j(a) - \underline \lambda | \le K C | a - \underline a|$ for $p\le j\le p+m$,
provided that $a \in \Omega_1$ and
$ K C | a - \underline a | < \delta /2$.

Gluing these estimates together, we have proved that there are constants $C$ and $K$ such that:
for all $\underline a \in \Omega$, there is a convex neighborhood $\omega$ of $\underline a$, such that
for $a \in \omega $ and all $j$,
$$
| \lambda_j (a) - \lambda_j (\underline a) | \le C K | a - \underline a|.
$$
The following independent lemma implies that
$ \big| \nabla_a \lambda_j \big|_{L^\infty(\Omega)} \le CK$ and the proposition follows.
\end{proof}

\begin{lem}
\label{lemLip}
Suppose that $K $ is a positive real number and $f$ is a function defined on the open set $\Omega \subset \RR^n$ such that
for all $ \underline a \in \Omega$, there is a neighborhood $\omega$ of $ \underline a$, such that
\begin{equation}
\label{youpi}
\forall a \in \omega, \quad | f(a) - f(\underline a) | \le K | a - \underline a|.
\end{equation}
Then $ \big\| \nabla_a f \big\|_{L^\infty(\Omega)} \le K$.
\end{lem}
\begin{proof}
Note that \eqref{youpi} implies that $f$ is continuous at $\underline a$, thus $f$ is continuous.

We first shown that for all convex open set
$\Omega_1 \subset \Omega$
the inequality
\begin{equation}
\label{youpipi}
| f(b) - f(a) | \le K | b - a|
\end{equation}
is satisfied for all $a$ and $b$ in $\Omega_1$.
Indeed, let $\sT$ denote the set of real numbers
$t \in [0, 1]$ such that
\begin{equation}
\label{youpi1}
\forall s \in [0, t], \quad | f(a + s (b- a)) - f(a) | \le K s | b - a|.
\end{equation}
By assumption, the property \eqref{youpipi} is satisfied on a
neighborhood of $a$, implying that~$\sT$ is not empty.
By definition $\sT$ is an interval, and by continuity of
$f$ it is closed. Using the assumption \eqref{youpi} near $a +t (b-a)$, implies that
$\sT$ is open so that $T = 1$ and \eqref{youpipi} is proved.

This implies that $f$ is Lipschitz continuous on $\Omega_1$ and that
$ \big\| \nabla_a f \big\|_{L^\infty(\Omega_1)} \le K$. Since this is true for all balls $\Omega_1 \subset \Omega$,
the lemma follows.
\end{proof}

\Subsection{Lipschitz dependence of the eigen-projectors}

\begin{prop}
\label{proplipproj}
Suppose that $A(\cdot) \in W^{1, \infty} (\Omega)$ satisfies Assumption~$\ref{assSHM}$.
Let $\underline a \in \Omega$ and consider
$ \underline \Lambda:= \{\lambda_j (\underline a), j \in J \}$ a subset of the spectrum
of $A(\underline a)$.
Let $ \underline \Lambda ' = \Sigma(\underline a) \moins \underline \Lambda
= \{\lambda_j (\underline a), j \in J' \}$ and define
\begin{equation}
\delta = \mathrm{dist} (\underline \Lambda, \underline \Lambda') =
\min_{(j,j') \in J,\times J'} \big| \lambda_j (\underline a) - \lambda_{j'} (\underline a) \big| > 0.
\end{equation}
Let $\omega$ be a neighborhood of $\underline a $
such that $|\lambda_j (a) - \lambda_j(\underline a) | \le \delta/4$ for all $j\in \{1, \dots, N\}$
and $a \in \omega$. With
$ \Lambda (a) = \{\lambda_j (a), j \in J \}$,
consider the
spectral projector
\begin{equation}
\Pi_\Lambda (a) = \sum_ {\lambda \in \Lambda (a)} \Pi_\lambda (a).
\end{equation}
Then,
$\Pi_\Lambda (a) $ is continuous on $\omega$ and
\begin{equation}
\label{estlipproj}
\big| \Pi_\Lambda (a) - \Pi_\Lambda (a') \big| \le C K \delta^{-1} \big| a - a'\big|,
\end{equation}
where $C$ depends only on an upper bound of the norms of the spectral projectors of $A(\cdot)$
and $K = \big\| \nabla_a A \big\|_{L^\infty(\Omega)}$.
\end{prop}

\begin{proof} There are finitely many Jordan curves $\Gamma_k$ in the complex domain, of total length less than $C \delta$, such that $|z - \lambda_j(\underline a) | \ge \delta /2$ for all
$z \in \cup \Gamma_k $ and all $j \in J $, surrounding
$\Lambda$ so that
$$
\Pi_\Lambda (\underline a) = \sum_k \frac{1}{2i \pi} \int_{\Gamma_k} (z \Id - A(\underline a))^{-1} \,dz.
$$
This formula extends to $a \in \omega$.
Moreover, using the estimate
\begin{multline*}
\big| (z \Id - A(a))^{-1} - (z \Id - A(a'))^{-1} \big|
\\
\le \big| (z \Id - A(a))^{-1} \big| \, \big| A(a) - A(a') \big| \, \big| (z \Id - A(a'))^{-1} \big|
\\
 \le C K \delta^{-2} \big| a - a' \big|
\end{multline*}
for $z \in \bigcup \Gamma_k$, implies \eqref{estlipproj}.
\end{proof}

\subsection{A piece of functional calculus}

We study the smoothness of $f(A(a))$, given the smoothness of $f$ and $A$.
We extend the analysis to vector or matrix valued functions $S(\lambda, a)$ using the following definition
\begin{equation}
\label{functcalc}
S_A(a) = \sum_{\lambda \in \Sigma(a)} S(\lambda, a) \Pi _\lambda (a).
\end{equation}

\begin{theo}
\label{theoFC}
Suppose that $A(\cdot)$ is continuous \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on
$\Omega$ and satisfies Assumption~$\ref{assSHM}$. Suppose that $S(\lambda, a)$ is continuous \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr
on $\RR \times \Omega$. Then
$S_A(a)$
is continuous \lcr\resp $W^{1, \infty}$\rcr \lcr\resp $C^\infty$\rcr on $\Omega$.
\end{theo}

\begin{proof}[\textup{(1)} The $C^\infty$ case]
If $S$ were holomorphic in $\lambda$ one would have
$$
S_A(a) = \frac{1}{2i \pi} \int_{\partial D} S (z, a) (z \Id - A(a))^{-1} \,dz,
$$
where $D $ is a rectangle $[-R, R] + i [-\delta, \delta ]$ containing $\Sigma (a) $ in its interior,
implying the result since $(z \Id - A(a))^{-1}$ is smooth in $a$ for $z \in \D D$.
In the $C^\infty$ case, we modify this proof considering
an almost holomorphic extension of $S$ in the variable $\lambda$. It is a $C^\infty$ function in
$(z, a) \in \CC\times \Omega $ such that, for $z$ in bounded sets,
\begin{equation}
\D_{\overline z} S (z, a) = O(| \im z |^\infty).
\end{equation}
Since the result is local, we can assume that $\Omega $ is bounded and fix $R$ such that for
$ a \in \Omega$, the spectrum of $ A(a)$ is contained in
$\{| z | \le R \}$. Let $D$ denote the disc of radius $R+1$ in $\CC$. Then,
\[
S_A (a) = \frac{1}{2i \pi} \int_{\partial D} S (z, a) (z \Id - A(a))^{-1} \,dz
+ \frac{1}{2i \pi} \int_D \D_{\overline z} S (z, a) (z \Id - A(a))^{-1} \,dz \,d \overline z.
\]
The first integral is $C^\infty$ in $a$ as explained above.
In the second, we note that
for all $m$, $ \D_{\overline z} S (z, a) = (\im z)^m R_m (z, a)$ with $R_m$ smooth and
$(\im z)^m (z \Id - A(a))^{-1}$ has $m-1$ uniformly bounded derivatives in $a$, for $z \in D \moins \RR$,
implying that the second integral is $C^{m-1}$ with respect to $a \in \Omega$.
\end{proof}

\begin{proof}[\textup{(2)} Continuity] Fix $\underline a \in \Omega$ and denote by $ \mu_j, 1 \le j\le m, $ the \emph{distinct} eigenvalues
of $A(\underline a)$ and introduce $\delta = \min_{j \ne k} | \mu_j - \mu_k| $.
For $a$ in a neighborhood $\omega $ of $\underline a$, the spectrum of $A(a)$ is
contained in nonoverlapping intervals $I_j ={} ] \mu_j - \delta /4, \mu_j + \delta /4[$.
Write
\begin{equation}
\label{decomp}
S_A(a) = \sum_j S(\mu_j, \underline a) \Pi_{I_j}(a)
+ \sum_j \sum_{\lambda \in I_j \cap \Sigma(a)}\Big(S(\lambda, a) - S(\mu_j, \underline a)\Big)
\Pi _ \lambda(a),
\end{equation}
with
\begin{equation}
\Pi_j(a) = \sum_{\lambda \in I_j \cap \Sigma(a)} \Pi _ \lambda(a).
\end{equation}
Using a uniform bound for the $\Pi_\lambda(a)$ and the continuity of the eigenvalues at $\underline a$ one concludes that the second sum in \eqref{decomp} tends to $0$
as $a $ tends to $\underline a$. By Proposition~\ref{proplipproj}, $\Pi_j$ is continuous and hence
$S_A$ is continuous at $\underline a$.
\end{proof}

\begin{proof}[\textup{(3)} Lipschitz continuity]
By Lemma~\ref{lemLip} it is sufficient to prove that there is a positive constant $K $ such that
for all $ \underline a \in \Omega$, there is a neighborhood $\omega$ of $ \underline a$, such that
\begin{equation}
\label{youpi2}
\forall a \in \omega, \qquad | S_A(a) - S_A(\underline a) | \le K | a - \underline a|.
\end{equation}
The proof starts as in (2) with the decomposition \eqref{decomp}. Shrinking $\omega $ if necessary,
the Lipschitz continuity of
$S$ and of the eigenvalues implies that for $a \in \omega$ and
$ \lambda \in I_j \cap \Sigma(a)$
$$
\Big| S(\lambda, a) - S(\mu_j, \underline a)\Big| \le C | a - \underline a|,
$$
where $C$ depends only $ \big\| \nabla_a A \big\|_{L^\infty(\Omega)}$ and
$ \big\| \nabla_{\lambda,a} S \big\|_{L^\infty(\RR \times \Omega)}$.
Since the $\Pi_\lambda$ are uniformly bounded, this implies that
the second sum in \eqref{decomp} is uniformly $O(|a- \underline a|)$ so that it remains
to prove that, with $S_j = S(\mu_j, \underline a)$ and
$ p_j(a) = \Pi_{j}(a) -
\Pi_{j}(\underline a)\big) $, one has
\begin{equation}
\label{est513}
\Big| \sum_j S_j p_j(a)\Big| \le K | a - \underline a|,
\end{equation}
with $K$ independent of $\underline a$ and $\omega$. Since $S \in W^{1, \infty}$, the $S_j $ satisfy
\begin{equation}
\label{hypvarS}
\big| S_j \big| \le K_1, \qquad \big| S_j - S_k \big| \le K_1 | \mu_j - \mu_k |.
\end{equation}
Moreover, Proposition~\ref{proplipproj} implies that
for all $J \subset \{1, \dots, m \}$ with $J \neq \emptyset$ and $J \neq \{1, \dots, m \}$,
\begin{equation}
P_J = \sum_{j \in J} p_j (a)
\end{equation}
satisfies, with $\eps = | a - \underline a|$:
\begin{equation}
\label{deltaJ}
\big| P_J \big| \le K_2 \eps\Big(\min_{j \in J, k \notin J} | \mu_j - \mu_k |\Big)^{-1}.
\end{equation}
Moreover, when $J = \{1, \dots, m \}$,
\begin{equation}
\label{deltaJmax}
P_{\{1, \dots, m \}} = 0.
\end{equation}
The next lemma implies \eqref{est513}, finishing the proof of the proposition.
\end{proof}

\begin{lem}
\label{mainlem} There is a constant $C_m $ which depends only on $m$, such that
for all $S_j$ and $p_j$, $1\le j\le m$, satisfying \eqref{hypvarS} \eqref{deltaJ}and \eqref{deltaJmax}, the
sum $S = \sum S_j p_j$ satisfies
\begin{equation}
\label{estS}
\big| S \big| \le C_m K_1 K_2 \eps.
\end{equation}
\end{lem}
\begin{proof}
By homogeneity, we can assume that $K_1 = K_2 = 1$.
The proof is by induction on $m$. When $m =1$, the condition \eqref{hypvarS} reduces to $|S_{1} | \le 1$,
the condition \eqref{deltaJ} is void and $ p_{1} = 0 $.

We assume that the lemma is proved up to order $m- 1 \ge1$ and we prove it at the order $m$.
Let
\begin{equation}
\delta:= \min_{j \ne k} | \mu_j - \mu_k | > 0.
\end{equation}
Permuting the indices we can assume that the infimum is attained
for $(j, k) = (m-\nobreak1, m)$, which means that
\begin{equation}
\label{relabel}
\forall j \ne k, \quad | \mu_{m-1} - \mu_m | \le | \mu_j - \mu_k |.
\end{equation}
We split $S$ in two terms:
\begin{equation}
\label{splitS}
S = \widetilde S \ + \ (S_m - S_{m-1}) p_m,
\end{equation}
with
\begin{equation}
\label{varS1}
\widetilde S = \sum_{j= 1}^{m-1} S_j \tilde p_j,
\end{equation}
where
$ \tilde p_j = p_j $ when $ j \le m - 2$ and $ \tilde p_{m-1} = p_{m-1} + p_m $.

The condition \eqref{deltaJ} applied to $ J = \{m\}$ and \eqref{relabel} imply that
$| p_m | \le \eps \delta^{-1} $ while \eqref{hypvarS} and \eqref{relabel} imply that
$| S_{m} - S_{m-1} | \le \delta$. This shows that the second term in \eqref{splitS} satisfies
$| (S_m - S_{m-1}) p_m | \le \eps $.

We now check that
the induction hypothesis can be applied to $ \widetilde S $.
The condition \eqref{hypvarS} is clear, so we only have to show
that the conditions \eqref{deltaJ}
and \eqref{deltaJmax} are satisfied for the $\tilde p_j $.

Consider a non empty subset $\widetilde J \subset \{1, \dots, m-1\}$. Then
$\widetilde P_{\widetilde J} = P_{J} $ with
\begin{enumerate}
\item
$ J = \widetilde J \cup\{m\}$ if $m-1 \in \widetilde J$,

\item
$ J = \widetilde J $ if $m-1 \notin \widetilde J$.
\end{enumerate}
In particular, $\widetilde P_{\{1, \dots, m-1\}} = P_{\{1, \dots, m\}} = 0$
so that \eqref{deltaJmax} for $\widetilde S$ is satisfied.

Suppose that $\widetilde J \ne \{1, \dots, m-1\}$ and let $J$ be as above. Introduce also
$\widetilde K = \{1, \dots, m-1\} \moins \widetilde J$ and
$K = \{1, \dots, m \} \moins J$. One has $K = \widetilde K$ in case (1) and
$K = \widetilde K \cup\{m \}$ in case (2). By assumption, we know that
\begin{equation*}
\big| \widetilde P_{\tilde J} \big|
= \big| P_{J} \big| \le \eps\Big(\min_{j \in J, k \in K}
| \mu_j - \mu_k |\Big)^{-1}.
\end{equation*}
We claim that
\begin{equation}
\label{verif1}
\min_{j \in \tilde J,\, k \in \tilde K} | \mu _j - \mu_k |
\le 2
\min_{j \in J, k \in K} | \mu_j - \mu_k |.
\end{equation}
Indeed, if it is true, it implies that
\begin{equation}
\label{deltaJt}
\big| \widetilde P_{\tilde J} \big| \le 2 \eps\Big(\min_{j \in \tilde J
, k \in \tilde K} | \mu_j - \mu_k |\Big)^{-1},
\end{equation}
so that the induction hypothesis is satisfied for $\widetilde S$ with $\eps$ replaced by $2 \eps$. Thus
$
\big| \widetilde S \big| \le\nobreak 2 C_{m-1} \eps
$ and \eqref{estS} follows with $C_m = 1 + 2 C_{m-1}$.

Therefore, to complete the proof, it remains to prove the claim \eqref{verif1}.
In this estimate, $\tilde J$ and $\tilde K$ play symmetric roles, and therefore
we can assume that $m \in J$, that is, $m-1 \in \tilde J$.
Comparing the sets $\tilde J \times \tilde K$ and $J \times K$, we see that the only nontrivial case concerns
$ | \mu_m- \mu_k | $ when $k \in K = \tilde K$. In this case, since
$m-1 \in \tilde J$, the claim follows from the inequality
\begin{equation*}
| \mu_{m-1}- \mu_k | \le
| \mu_{m}- \mu_k | + | \mu_{m-1}- \mu_m | \le 2 | \mu_m- \mu_k |,
\end{equation*}
where we have used \eqref{relabel}.
The proof of the lemma is now complete.
\end{proof}

\section{Appendix B: Symmetrizable matrices}

\Subsection{Positivity of symmetrizers and bounds}

\begin{prop}
\label{propPSP}
Suppose that $L$ and $\bS$ are matrices such that $\bS L$ is Hermitian symmetric.
Suppose that $J $ is an invertible matrix such that
\begin{equation}
\label{positiv1}
\forall u \in \ker L, \quad \re \big(\bS J u, u \big) \ge c | u |^2.
\end{equation}
Then, $0$ is a semi-simple eigenvalue of $J^{-1}L $ and the associated eigen-projector
$\Pi $ satisfies
\begin{equation}
\label{eqPiSPi}
\Pi^* \bS J \Pi = \Pi^* \bS J
\end{equation}
and
\begin{equation}
\label{normePi}
\big| \Pi \big| \le | \Sigma J |/ c\,.
\end{equation}
Moreover,
\begin{equation}
\label{ortho}
\bS f \in (\ker L) ^\perp \quad \Longleftrightarrow \quad f \in \mathrm{range} (L).
\end{equation}
\end{prop}

\begin{proof}
Let $\bK $ and $\bR$ denote respectively the kernel and the range of $L$.
The identity $(\bS Lu, v) = (u, \bS L v) $ implies
that $\bS \bR \subset \bK^\perp$ and hence $\bR \subset \bS^{-1}(\bK^\perp) $.
Next~we note that \eqref{positiv1} implies that if
$u\in \bK$ and $\bS J u \in \bK^\perp$, then $u= 0$, so that
\hbox{$ J \bK \cap \bS^{-1} (\bK^\perp) = \{0\}$}:
\begin{equation}
\label{tagada}
\bR \subset \bS^{-1}(\bK^\perp), \qquad J \bK \cap \bS^{-1} (\bK^\perp) = \{0\}.
\end{equation}
In particular $J \bK \cap \bR = \{0\}$ and $\bK \cap J^{-1} \bR = \{0\}$.
This means that the kernel $\bK$ of $A_J - \tau \Id$ has a trivial intersection
with the range $J^{-1} \bR $ of $A_J - \tau\Id$, that is that
$\tau $ is a semi-simple eigenvalue of $A_J$.
Moreover, since $\dim \bR + \dim J \bK = N $, \eqref{tagada} implies that
$\bR = \bS^{-1}(\bK^\perp) $, that is \eqref{ortho}.

In the splitting $u = \Pi u + (\Id- \Pi) u$, $\Pi u \in \bK $ and there is $v$
such that $(\Id- \Pi) u= J^{-1} Lv$. Therefore $(\bS L v, \Pi u) = 0$ and
$\big(\bS J \Pi u, \Pi u) = \big(\bS J u, \Pi u)$. Hence,
$$
c \big| \Pi u \big|^2 \le \re \big(\bS J \Pi u, \Pi u) = \re \big(\bS J u, \Pi u)
\le \big| \bS J u \big| \ \big| \Pi u \big|,
$$
and \eqref{normePi} follows.

For $f \in \bK^\perp$ and $u\in \CC^N$ one has $(\Pi^* f, u) = (f, \Pi u) = 0$.
Thus $ \bK^\perp \subset \ker \Pi^*$ and indeed $ \bK^\perp = \ker \Pi^*$
since the two spaces have the same dimension.
For all $u$, $(\Id - \Pi) u \in J^{-1} \bR$,
hence $ \bS J (\Id - \Pi) u \in \bK^\perp$ and therefore
$\Pi^* \Sigma J (\Id - \Pi) u = 0$ that is \eqref{eqPiSPi}.
\end{proof}

\begin{prop}
\label{positivcone}
Suppose that $L$ and $\bS$ ares matrices such that $\bS L$ is Hermitian symmetric.
We now assume that we are given two matrices
$J_0$ and $J_1$ such that, for all $t \in [0,1]$,
$J_t = (1-t) J_0 + t J_1$ is invertible, and for all $u$ and $v$ in $\ker $,
\begin{equation}
\label{symlin1}
\big(\bS J_t u, v \big) = \big(u, \bS J_t v \big).
\end{equation}
Suppose that $J_0$ satisfies
\begin{equation}
\label{positiv10}
\forall u \in \bK_a, \quad \big(\bS J_0 u, u \big) \ge c | u |^2,
\end{equation}
and suppose that for $t \in [0, 1[$, $0$ is a semi-simple eigenvalue of $J_t^{-1} L $.
Then for all $t \in [0,1]$,
\begin{equation}
\label{positiv11}
\forall u \in \ker L, \quad \big(\bS J_t u, u \big) \ge (1-t) c | u |^2.
\end{equation}
\end{prop}

\begin{proof}
The assumption \eqref{symlin1} means that the restriction of $\bS J_t $ to $\ker L$ is
symmetric. By \eqref{positiv10}, it is positive definite for $t = 0$. By continuity, it remains positive
as long as it remains definite. It is indefinite when there is a $u\in \ker L$, $u \ne 0$ such that
$\bS J_t u (\in \ker L)^\perp$. By \eqref{ortho}, this would imply that
$ u\ne 0$ would belong both to the kernel and to the range
of $J_t^{-1} L$, contradicting the assumption that $0$ is a semi-simple eigenvalue.
By continuity, $\bS J_1$ is nonnegative and \eqref{positiv11} follows.
\end{proof}

\subsection{From full symmetrizers to symmetrizers}

We consider here $N \times N$ matrices
\begin{equation}
L(\tau, a) = \tau J(a) - A (a),
\end{equation}
which depend on parameters $a$ in an open set $\Omega $ and $\tau \in \RR$.
We always assume that $J(a) $ is invertible. We link the spectral properties
of $A_J(a):= J(a)^{-1} A(a)$ to the existence of symmetrizers and full symmetrizers
of $L(\tau, a)$.

\begin{ass}
\label{assfullsym}
We assume that the matrices $J(a)$, $J(a)^{-1} $ are uniformly bounded and that
there are uniformly bounded matrices
$\bS (\tau, a)$ is such that for all $\tau$ and $a$,
$\bS(\tau, a) L(\tau, a)$ is Hermitian symmetric and
\begin{equation}
\label{positivbis}
\forall u \in \ker L(\tau, a), \quad \re \big(\bS (\tau, a) J(a) u, u \big) \ge c | u |^2.
\end{equation}
where $c$ is independent of $a$ and $\tau$.

We further assume that
all the complex roots in $\tau$ of $\det L(\tau, a) = 0$ are real.
\end{ass}

Proposition~\ref{propPSP} implies that the eigenvalues $\tau$ of $J(a)^{-1} A(a)$ are real and semi-simple and that the corresponding
eigen-projectors are uniformly bounded:

\begin{cor}
\label{cor62}
Under Assumption~\ref{assfullsym}, the family $A_J(a)$ is uniformly strongly hyperbolic in the sense of Assumption~$\ref{assSHM}$.
\end{cor}

\begin{theo}
\label{theomajmod}
In addition to Assumption~$\ref{assfullsym}$, suppose that
$J$, $A$ and $\bS$ are continuous \lcr\resp Lipschitz continuous\rcr \lcr\resp $C^\infty$\rcr in $a \in \Omega$ and $\tau \in \RR$. Then there is a bounded and
continuous \lcr\resp Lipschitz continuous\rcr \lcr\resp $C^\infty$\rcr matrix $S(\cdot)$ on $\Omega$ such that
\begin{equation}
S(a) J(a) = \big(S(a) J(a) \big)^* \ge c_1 \Id, \qquad S(a) A (a) = \big(S(a) A(a) \big)^*,
\end{equation}
with $c_1 > 0$ independent of $a$.
\end{theo}

\begin{proof} Multiplying $L$ by $J^{-1}$ reduces to the case $J = \Id$.
A symmetrizer is
\begin{equation}
\label{defiS}
S(a) = \sum_{\tau \in \Sigma(a)} \Pi(\tau, a)^* \bS (\tau, a) \Pi(\tau, a),
\end{equation}
where $\Sigma(a) \subset \RR $ denotes the spectrum of
$A(a)$.

For $u$ and $v$ in $\ker L(\tau, a)$, one has $L(\tau + \sigma s, a) u = \sigma J(a) u $ and a similar expression for $v$.
The symmetry implies
\begin{equation*}
\big(\bS(\tau + \sigma \nu) J (a) u, v \big) = \big(u, \bS(\tau + \sigma u)J (a) v \big),
\end{equation*}
and letting $\sigma$ tend to zero implies
\begin{equation}
\label{symlin2}
\big(\bS(\tau, a) J(a) u, v \big) = \big(u, \bS (\tau, a) J(a) v \big).
\end{equation}

This shows that each term of the sum \eqref{defiS} is symmetric and
$S$ is symmetric. Moreover, by \eqref{positivbis} and Corollary~\ref{cor62}, it is uniformly bounded and uniformly positive.
By construction, $S(a) A(a)$ is symmetric and
by \eqref{eqPiSPi} and symmetry, one has
\begin{equation}
S(a) = \sum_{\tau \in \Sigma(a)} \Pi(\tau, a)^* \bS (\tau, a) =
\sum_{\tau \in \Sigma(a)} \bS^* (\tau, a) \Pi(\tau, a).
\end{equation}
Theorem~\ref{theoFC} implies that $S $ is continuous \lcr\resp Lipschitz continuous\rcr
\lcr\resp $C^\infty$\rcr in $a$, finishing the proof of the theorem.
\end{proof}

\backmatter
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