# proof of Hartman-Grobman theorem

###### Lemma 1.

Let $A\mathrm{:}E\mathrm{\to}E$ be an hyperbolic isomorphism, and let $\phi $ and $\psi $ be $\epsilon $-Lipschitz maps from $E$ to itself such that $\phi \mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\psi \mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{0}$. If $\epsilon $ is sufficiently small, then $A\mathrm{+}\phi $ and $A\mathrm{+}\psi $ are topologically conjugate.

Since $A$ is hyperbolic, we have $E={E}^{s}\oplus {E}^{u}$ and there is $$
(possibly changing the norm of $E$ by an equivalent^{} box-type one), called the skewness of $A$,
such that

$$ |

and

$$\parallel x\parallel =\mathrm{max}\{\parallel {x}_{s}\parallel ,\parallel {x}_{u}\parallel \}.$$ |

Let us denote by $(\stackrel{~}{E},\parallel \cdot {\parallel}_{0})$ the Banach space^{} of all bounded^{}, continuous maps^{} from $E$
to itself, with the norm of the supremum^{} induced by the norm of $E$.
The operator^{} $A$ induces a linear operator $\stackrel{~}{A}:\stackrel{~}{E}\to \stackrel{~}{E}$ defined by
$(\stackrel{~}{A}u)(x)=A(u(x))$, which is also hyperbolic. In fact, letting
${\stackrel{~}{E}}^{i}$ be the set of all maps $u:\stackrel{~}{E}\to \stackrel{~}{E}$ whose range is contained
in ${E}^{i}$ (for $i=s,u$) we have that $\stackrel{~}{E}={\stackrel{~}{E}}^{s}\oplus {\stackrel{~}{E}}^{u}$ is a hyperbolic splitting
for $\stackrel{~}{A}$ with the same skewness as $A$.

From now on we denote the projection of $x$ to ${E}^{i}$ by ${x}_{i}$, and
the restriction^{} ${A|}_{{E}_{i}}:{E}^{i}\to {E}^{i}$ by ${A}_{i}$ ($i=s,u$).

We will try to find a conjugation of the form $I+u$ where $u\in \stackrel{~}{E}$.

###### Proposition 1.

There exists $\epsilon \mathrm{>}\mathrm{0}$ such that if $\phi $ and $\psi $ are $\epsilon $-Lipschitz^{},
then there is a unique $u\mathrm{\in}\stackrel{\mathrm{~}}{E}$ such that

$$(I+u)(A+\phi )=(A+\psi )(I+u).$$ |

###### Proof.

We want to find $u$ such that

$$A+\phi +u(A+\phi )=A+Au+\psi (I+u)$$ |

which is the same as

$$\phi +u(A+\phi )=Au+\psi (I+u).$$ |

This can be rewriten as

${u}_{u}$ | $={A}_{u}^{-1}({u}_{u}(A+\phi )+{\phi}_{u}-{\psi}_{u}(I+u))$ | ||

${u}_{s}$ | $=({A}_{s}{u}_{s}+{\psi}_{s}(I+u)-{\phi}_{s}){(A+\phi )}^{-1},$ |

where we use the fact that by the Lipschitz inverse mapping theorem, if $$
(where $\lambda $ is the skewness of $A$) then $A+\phi $ is invertible with Lipschitz inverse^{}.

Now define $\mathrm{\Gamma}:\stackrel{~}{E}\to \stackrel{~}{E}$ by

${\mathrm{\Gamma}}_{s}(u)$ | $=({A}_{s}{u}_{s}+{\psi}_{s}(I+u)-{\phi}_{s}){(A+\phi )}^{-1}$ | ||

${\mathrm{\Gamma}}_{u}(u)$ | $={A}_{u}^{-1}({u}_{u}(A+\phi )+{\phi}_{u}-{\psi}_{u}(I+u))$ |

We assert that, if $\epsilon $ is small, $\mathrm{\Gamma}$ is a contraction. In fact,

${\parallel {\mathrm{\Gamma}}_{s}(u)-{\mathrm{\Gamma}}_{s}(v)\parallel}_{0}$ | $={\parallel \left({A}_{s}({u}_{s}-{v}_{s})+{\psi}_{s}(I+u)-{\psi}_{s}(I+v)\right){(A+\phi )}^{-1}\parallel}_{0}$ | ||

$\le \parallel {\stackrel{~}{A}}_{s}\parallel \cdot {\parallel ({u}_{s}-{v}_{s}){(A+\phi )}^{-1}\parallel}_{0}+{\parallel ({\psi}_{s}(I+u)-{\psi}_{s}(I+v)){(A+\phi )}^{-1}\parallel}_{0}$ | |||

$\le \lambda {\parallel {u}_{s}-{v}_{s}\parallel}_{0}+\epsilon {\parallel u-v\parallel}_{0}$ | |||

$\le (\lambda +\epsilon ){\parallel u-v\parallel}_{0}$ |

and

${\parallel {\mathrm{\Gamma}}_{u}(u)-{\mathrm{\Gamma}}_{u}(v)\parallel}_{0}$ | $={\parallel {A}_{u}^{-1}({u}_{u}(A+\phi )-{v}_{u}(A+\phi )-{\psi}_{u}(I+u)+{\psi}_{u}(I+v))\parallel}_{0}$ | ||

$\le \parallel {\stackrel{~}{A}}_{u}^{-1}\parallel \cdot \left({\parallel {u}_{u}(A+\phi )-{v}_{u}(A+\phi )\parallel}_{0}+{\parallel {\psi}_{u}(I+u)-{\psi}_{u}(I+v)\parallel}_{0}\right)$ | |||

$\le \lambda \left({\parallel {u}_{u}-{v}_{u}\parallel}_{0}+\epsilon {\parallel u-v\parallel}_{0}\right)$ | |||

$\le \lambda (1+\epsilon ){\parallel u-v\parallel}_{0}.$ |

Thus, if $$, $\mathrm{\Gamma}$ has Lipschitz constant smaller than $1$, so it is a contraction. Hence $u$ exists and is unique. ∎

###### Proposition 2.

The map $u$ from the previous proposition^{} is a homeomorphism.

###### Proof.

Using the previous proposition with $\phi $ and $\psi $ switched, we get a unique $v\in \stackrel{~}{E}$ such that

$$(I+v)(A+\psi )=(A+\phi )(I+v).$$ |

It follows that

$$(I+v)(I+u)(A+\phi )=(I+v)(A+\psi )(I+u)=(A+\phi )(I+v)(I+u).$$ | (1) |

Also, the previous proposition with $\phi =\psi $ implies that that there is a unique $w\in \stackrel{~}{E}$ such that

$$(I+w)(A+\phi )=(A+\phi )(I+w),$$ |

which obviously is $w=0$. But since $(I+v)(I+u)=I+(u+v+uv)$ and $u+v+uv\in \stackrel{~}{E}$,
(1) implies that $w=u+v+uv$ is a solution of the above equation, so that $u+v+uv=0$ and $(I+v)(I+u)=I$. In a similar way, we see that $(I+u)(I+v)=I$. Hence $I+u$ is invertible, with continuous^{} inverse.
∎

The two previous propositions prove the lemma.

###### Proposition 3.

If $U$ is an open neighborhood of $\mathrm{0}$ and $f\mathrm{:}U\mathrm{\to}E$ is a ${\mathrm{C}}^{\mathrm{1}}$ map with $f\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{0}$, then for every $\epsilon \mathrm{>}\mathrm{0}$ there is $\delta \mathrm{>}\mathrm{0}$ such that $\phi \mathrm{\doteq}f\mathrm{-}D\mathit{}f\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}$ is $\epsilon $-Lipschitz in the ball $B\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}\delta \mathrm{)}$.

###### Proof.

This is a direct consequence of the mean value inequality and the fact that $D\phi $ is continuous and $D\phi (0)=0$. ∎

###### Proposition 4.

There is a constant $k$ such that if $\phi \mathrm{:}\overline{B}\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}r\mathrm{)}\mathrm{\to}E$ is an $\epsilon $-Lipschitz map, then there is a $k\mathit{}\epsilon $-Lipschitz map $\stackrel{\mathrm{~}}{\phi}\mathrm{:}E\mathrm{\to}E$ which coincides with $\phi $ in $B\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}r\mathrm{/}\mathrm{2}\mathrm{)}$.

###### Proof.

Let $\eta :\mathbb{R}\to \mathbb{R}$ be a ${\mathcal{C}}^{\mathrm{\infty}}$ bump function: an infinitely
differentiable map such that
$\eta (x)=1$ for $$ and $\eta (x)=0$ for $x>1$, with derivative^{} bounded by $M$ and $|\eta (x)\parallel \le 1$ for all $x\in \mathbb{R}$.
Now define $\stackrel{~}{\phi}(x)=\phi (x)\eta (\parallel x\parallel /r)$ (when $\phi (x)$ is not
defined, we assume that it is zero).
If $x$ and $y$ are both in $B(0,r)$ then we have

$\parallel \stackrel{~}{\phi}(x)-\stackrel{~}{\phi}(y)\parallel $ | $=\parallel \phi (x)\eta (\parallel x\parallel /r)-\phi (y)\eta (\parallel y\parallel /r)\parallel $ | ||

$\le \parallel (\phi (x)-\phi (y))\eta (\parallel x\parallel /r)\parallel +\parallel \phi (y)(\eta (\parallel x\parallel /r)-\eta (\parallel y\parallel /r))\parallel $ | |||

$\le \epsilon \parallel x-y\parallel +\parallel \phi (y)-\phi (0)\parallel \cdot \parallel \eta (\parallel x\parallel /r)-\eta (\parallel y\parallel /r)\parallel $ | |||

$\le \epsilon \parallel x-y\parallel +\epsilon \parallel y\parallel (M\parallel x-y\parallel /r)$ | |||

$\le (M+1)\epsilon \parallel x-y\parallel ;$ |

if $x$ is in $B(0,r)$ and $y$ is not, then

$$\parallel \stackrel{~}{\phi}(x)-\stackrel{~}{\phi}(y)\parallel =\parallel \stackrel{~}{\phi}(x)-\stackrel{~}{\phi}({y}^{*})\parallel ,$$ |

where ${y}^{*}$ is defined as $x+\tau (y-x)$ with

$$\tau =sup\{t:x+t(y-x)\in E\setminus B(0,r)\}$$ |

This is true because $\stackrel{~}{\phi}({y}^{*})=0$. Also, $\parallel x-{y}^{*}\parallel =\tau \parallel x-y\parallel \le \parallel x-y\parallel $; hence

$$\parallel \stackrel{~}{\phi}(x)-\stackrel{~}{\phi}(y)\parallel =\parallel \stackrel{~}{\phi}(x)-\stackrel{~}{\phi}({y}^{*})\parallel \le (M+1)\epsilon \parallel x-{y}^{*}\parallel \le (M+1)\epsilon \parallel x-y\parallel .$$ |

Finally, if both $x$ and $y$ are outside $B(0,r)$, then $\parallel \stackrel{~}{\phi}(x)-\stackrel{~}{\phi}(y)\parallel =0\le (M+1)\parallel x-y\parallel $. Letting $k=M+1$ we get the desired result. ∎

Proof of the theorem.
Taking the particular $\psi =0$ in the lemma, we observe that
there is $\epsilon >0$ such that for any $\epsilon $-Lipschitz map $\phi $,
$Df(0)$ is conjugate to $\phi +Df(0)$.
Choose $\delta $ such that $f-Df(0)$ is $\epsilon /k$-Lipschitz in $B(0,2\delta )$.
Let $\stackrel{~}{\phi}$ be the $\epsilon $-Lipschitz extension^{} of $f-Df(0)$ to
$B(0,\delta )$ obtained from the previous proposition. We have that
$Df(0)+\stackrel{~}{\phi}$ is conjugate to $Df(0)$. But for $x\in B(0,\delta )$ we have
$Df(0)+\stackrel{~}{\phi}=f$, so that $f$ is locally conjugate to $Df(0)$.

Title | proof of Hartman-Grobman theorem |
---|---|

Canonical name | ProofOfHartmanGrobmanTheorem |

Date of creation | 2013-03-22 14:25:29 |

Last modified on | 2013-03-22 14:25:29 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 7 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 37C25 |