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\title{$\alpha $-H\"{o}lder linearization}
\author{Victoria Rayskin}
\date{}
\begin{document}
\maketitle
\begin{center}
\small University of Texas, Austin\\
Department of Mathematics\\
Austin, TX 78703\\
E-mail: {\tt rayskin@math.utexas.edu}
\end{center}
\begin{abstract}
A well known theorem of Hartman-Grobman says that a $C^2$ diffeomorphism
$f : \bf R^n \rightarrow \bf R^n $ with a hyperbolic fixed point at $0$ can be
topologically
conjugated to the linear diffeomorphism $L = df(0)$ (in a neighborhood
of $0$). On the other hand, if a non-planar map has resonance, then linearization
may not be $C^1$. A counter-example is due to P. Hartman (see \cite{H2}).
In this paper we will show that for any $\alpha \in (0,1)$ there exists an
$\alpha $-H\"{o}lder linearization in a neighborhood of $0$ for the counter-example of Hartman. No resonance
condition will be required. A linearization of a more general map will be discussed.
\end{abstract}
\section{Introduction}
Let $f$ be a $C^3$ diffeomorphism with a hyperbolic fixed point at $0$ and linear part $L$.
We will say that $f$ {\it can be linearized\/} if there exist a neighborhood $U$ of $0$ and homeomorphism $h:U \rightarrow h(U)$ with $h(0)=0$ such that $h\circ f=L\circ h$ on $U$.
A well known theorem of Hartman-Grobman says that such a diffeomorphism can be
topologically linearized (locally). See \cite{H1}. A simpler proof can be found in \cite{P}.
However, if linearization is only $C^0$, it is inconvenient for various
applications. Of course, if we will make additional assumptions on
resonance, we will obtain smooth linearization. (See \cite{S}). But
this technical assumption is unnatural for applications.
P. Hartman showed (\cite{H3}) that in the planar case, a $C^2$ diffeomorphism with a hyperbolic fixed point can be $C^1$ linearized.
G. Belitskii and S. van Strien (\cite{B}, \cite{ABZ}, \cite{v-S}) have considered this problem in higher dimensions and proved that for some $\alpha $ less then 1 there exists a local linearization in
the $\alpha $-H\"{o}lder
class. In their works this $\alpha $ depend on the resonance.
Hartman's counter-example (\cite{H2}) shows that a non-planar map may not be $C^1$ linearizable, and bounds the search for linearization.
In this paper, first, we will show that for any $\alpha \in (0,1)$ there exists an
$\alpha $-H\"{o}lder linearization of the Hartman map in a neighborhood of $0$. No resonance
condition will be required.
\\
Then, we will deduce the same conclusion for a more general map,
i.e., we are going to prove the following:
Let f be a $C^3$ diffeomorphism of $\bf R^3$ into itself with a hyperbolic fixed point at $0$,
\[
f(x)=Lx+\mu (x),\ x=(x_1, x_2, x_3),
\]
\\
such that $L$ is diagonal and $f$ leaves invariant $x$ and $y$ axis, and $(x,y)$ and $(y,z)$ planes.
Then, for any $\alpha $ arbitrarily close to but less than $1$, there exists a
neighborhood $U\subset \bf R^3 $ of the origin and a homeomorphism
$R_{\alpha}:U \rightarrow R_{\alpha}(U) $ such that $R_{\alpha } \in D^{\alpha} $ (see Definition~\ref{D^alpha} below) and
\[
R_{\alpha} \circ f = L \circ R_{\alpha}.
\]
We conclude the paper with the following conjecture:
Let f be a $C^3$ diffeomorphism of $\bf R^n$ into itself with a hyperbolic fixed point at 0,
\[
f(x)=Lx+\lambda (x),
\]
\\
such that all first-order partial derivatives of $\lambda $ at $0$ are zero.
\\
Then for any $\alpha $ arbitrarily close to but less than $1$ there exists a
neighborhood $U \subset \bf R^n $ of the origin and a homeomorphism
$R_{\alpha}:U \rightarrow R_{\alpha}(U)$ such that $R_{\alpha } \in D^{\alpha} $ and
\[
R_{\alpha} \circ f = L \circ R_{\alpha},
\]
(i.e., our result holds even without the hypotheses on $\mu $.)
This research was motivated by the study of degenerate homoclinic
crossings of stable and unstable manifolds, see
~\cite{R}.
\section{Linearization Theorem}
In 1960, P. Hartman presented an example (\cite{H2}) of a map (flow) with resonance (see the map $f$ in Theorem~\ref{Hart-thm})
that
does not have a $C^1$ linearization. First, we are going to show that this map can
be linearized in the class $D^{\alpha}$ (see the definition below) with $\alpha$
less than but arbitrarily close to 1.\\
Using this fact, we will show how to obtain the same result for a more general map with a hyperbolic fixed point (with no resonance assumption). See Theorem~\ref{main-lin-thm}.
\begin{definition}\label{D^alpha}
Let $\Phi $ be a homeomorphism from $\bf R^n$ to $\bf R^n$, $\Phi (0) =0$.
We say that $\Phi \in D^{\alpha}$, ($0< \alpha < 1$), if there exists a positive
constant $A $ such that
\[
\frac {\vert \Phi ( x) - \Phi ( y) \vert}
{{\vert x - y \vert}^{\alpha}} \leq A,
\]
and
\[
\frac {\vert {\Phi}^{-1} ( x) - {\Phi}^{-1} ( y) \vert}
{{\vert x - y \vert}^{\alpha}} \leq A,
\]
for all sufficiently small $\vert x \vert \not= \vert y \vert$, $x$, $y$ different from $0$.
\end{definition}
The theorem presented below for Hartman's map will be generalized in the next section.
\begin{theorem}\label{Hart-thm}
Let $f$ be the mapping of $R^3$ into itself
\[
f:\left( \begin{array}{c}
x\\
y\\
z
\end{array}
\right)
\mapsto
\left( \begin{array}{c}
ax\\
ac(y+xz)\\
cz
\end{array}
\right)
\]
where $c<10.$\\
Let
\[\alpha = \max \left( c \left( \frac{1}{ac} \right)^{1-\delta},
\left( {\frac{1}{ ac }}\right)^{\delta}\right).
\]
Then
\[
\Vert Pg-Ph {\Vert}_{\delta} \leq \alpha \Vert g-h {\Vert}_{\delta},
\]
with $ \alpha <1.$\\
Therefore $P$ contracts.
\\
The proof for the operator $P'$ is similar to the one above.
QED
\end{proof}
\begin{lemma}\label{complete}
$E$ is complete with respect to $\Vert . {\Vert}_{\delta}.$
\end{lemma}
\begin{proof}
Take a sequence $\{ g_{n} \} \subset E$, $n =1, 2,... $, which is Cauchy
with respect to $\Vert . {\Vert }_{\delta } .$\\
Let $C_0$ be the set of uniformly bounded continuous functions.\\
Then
\[
\Vert g_n -g_m {{\Vert}}_{{{C}_{0}}}=
\]
\[
\displaystyle\sup_{q_1 \in {\bf R^3} }
\vert g_{n}(q_{1})-g_{m}(q_{1})\vert \leq
\]
\[
\displaystyle\sup_{q_1 \in {\bf R^3} }
\frac
{{\vert g_{n}(q_{1})-g_{m}(q_{1})\vert}}
{{\vert x_{1} z_{1} {\vert}^{1-\delta }}}=
\]
\[
\displaystyle\sup_{q_1 \in {\bf R^3} }
\frac
{{\vert g_{n}(q_{1})-g_{m}(q_{1})-g_{n}(0)+g_{m}(0)\vert }}
{{\vert x_{1} z_{1}-0\cdot 0 {\vert}^{1-\delta }}}\leq
\]
\[
\displaystyle\sup_{q_{1},q_{2} \in {\bf R^3} }
\frac
{{\vert g_{n}(q_{1})-g_{m}(q_{1})-g_{n}(q_{2})+g_{m}(q_{2})\vert }}
{{\vert x_{1} z_{1}-x_{2} z_{2} {\vert}^{1-\delta }}} < \epsilon ,
\]
for all $x_{1}z_{1} \ne x_{2}z_{2}.$\\
Therefore $\{ g_{n} \} $ is Cauchy in $C_0$ with respect to the supremum norm and converges in $C_0$ to a limit $g$.
Fix an arbitrarily small $\epsilon >0.$
Then there exists $N$ such that for all $n, m >N$
\[
\displaystyle\sup_{{x_{1}z_{1}\ne x_{2}z_{2}}\atop {q_{1},q_{2} \in {\bf R^3} }}
\frac
{{\vert g_{n}(q_{1})-g_{m}(q_{1})-g_{n}(q_{2})+g_{m}(q_{2})\vert }}
{{\vert x_{1} z_{1}-x_{2} z_{2} {\vert}^{1-\delta }}} < \epsilon .
\]
Taking the limit as $m$ approaches $\infty$, we will obtain
\[\frac
{{\vert g_{n}(q_{1})-g(q_{1})-g_{n}(q_{2})+g(q_{2})\vert }}
{{\vert x_{1} z_{1}-x_{2} z_{2} {\vert}^{1-\delta }}} \leq \epsilon
\]
for all $q_{1}, q_{2} \in \bf R^3$, $x_{1}z_{1}\ne x_{2}z_{2} $.\\
This implies that $\Vert g_{n}-g {\Vert}_{\delta} \leq \epsilon .$\\
One can prove directly (without using property of the norm ), that
\[
\Vert g {\Vert}_{\delta} \leq \Vert g_{n}-g {\Vert}_{\delta} + \Vert g_{n}{\Vert}_{\delta} .
\]
Then $\Vert g {\Vert}_{\delta} \leq \epsilon +1 $ for arbitrarily small $\epsilon .$
\\
Thus, $\Vert g {\Vert}_{\delta} \leq 1.$
Clearly, g is continuous, bounded and $g(0)=0$.
Therefore $E$ is complete with respect to $\Vert . {\Vert }_{\delta } .$
\\
QED
\end{proof}
Thus, we have contracting operators $P$ and $P'$ that act on the complete space $E_{\delta}$ and map this space into itself. (See Lemma~\ref{onto}, Lemma~\ref{contract}, Lemma~\ref{complete}). By the Fixed Point Theorem there exist fixed points $g \in E_{\delta}$ and $g'\in E_{\delta}$ such that $Pg=g$ and $Pg'=g'$.
We now propose that $id+g$ and $id+g'$ are the locally conjugating homeomorphisms we sought.
\begin{lemma}\label{inverse}
Assume
\[
(id+g) \circ f_{\epsilon} = L \cdot (id+g) , \ \ g \in E_{\delta}
\]
and
\[
(id+g') \circ L = f_{\epsilon} \circ (id+g') , \ \ g' \in E_{\delta}
\]
\\
Then the $R_{\alpha}=id+g $ ($1-\delta = \alpha $) is a homeomorphism, $R_{\alpha}\in D^{\alpha}$, and $R_{\alpha}^{-1}=id+g'$.
\end{lemma}
\begin{proof}
For simplicity we will denote $R_{\alpha}$ by $R.$
\\
Then $R= id + g$, $g$ is in $E_{\delta}.$
\\
Let $h' =id + g'$, $g'$ is in $E_{\delta}.$
\\
First we need to show that $R^{-1} =id + g'=h'.$
\\
To show this, first we will prove that $h\circ h'= id$.
\\
We will show that $L\circ (h\circ h')= (h\circ h')\circ L$.
\[
L\circ (h\circ h')= h\circ f\circ f^{-1}\circ h'\circ L = (h\circ h')\circ L.
\]
By uniqueness of the globally conjugating map, $h\circ h'= id$.
\\
Similarly, $h'\circ h= id$.
\\
This implies that $R^{-1} =id + g'=h'.$
To demonstrate, that $R_{\alpha} \in D^{\alpha}$ we will prove that
$\frac{\vert g(q_1)-g(q_2)\vert }{\vert q_1-q_2{\vert }^{1-\delta}}$ and
$\frac{\vert g'(q_1)-g'(q_2)\vert }{\vert q_1-q_2{\vert }^{1-\delta}}$
are bounded above for sufficiently small $\vert q_1 \vert \not= \vert q_2 \vert$.
\\
Suppose that $\vert q_1 \vert \leq 1$, $\vert q_1 \vert \leq 1$.
\[
\vert x_{1}z_{1}-x_{2}z_{2}\vert \leq
\vert x_{1}\vert \cdot \vert z_{1}-z_{2}\vert +
\vert z_{2}\vert \cdot \vert x_{1}-x_{2}\vert \leq
\]
\[
2 \vert q_{1}-q_{2}\vert
\]
Then
\[
\frac{\vert g(q_1)-g(q_2)\vert }{\vert q_1-q_2{\vert }^{1-\delta}} \leq
\]
\[
\left( 2 \right)^{1-\delta }\cdot
\frac{\vert g(q_1)-g(q_2)\vert }{\vert x_{1}z_{1}-x_{2}z_{2}{\vert }^{1-\delta}}
\leq
\left( 2 \right)^{1-\delta }.
\]
Similarly one can show the same property for $g'.$
\\
Therefore $R_{\alpha} \in D^{\alpha}.$
QED
\end{proof}
\section{Discussion of Further Results}
Let $f$ be the mapping of $R^3$ into itself
\[
f:\left( \begin{array}{c}
x\\
y\\
z
\end{array}
\right)
\mapsto
\left( \begin{array}{c}
(a+o(1))x\\
(ac+o(1))(y+xz),\ \ \mbox{as} \vert x\vert,\ \vert y\vert,\ \vert z\vert \longrightarrow 0\\
(c+o(1))z
\end{array}
\right)
\]
where $c<1