# proof of Hartman-Grobman theorem

###### Lemma 1.

Let $A\colon E\to E$ be an hyperbolic isomorphism, and let $\varphi$ and $\psi$ be $\varepsilon$-Lipschitz maps from $E$ to itself such that $\varphi(0)=\psi(0)=0$. If $\varepsilon$ is sufficiently small, then $A+\varphi$ and $A+\psi$ are topologically conjugate.

Since $A$ is hyperbolic, we have $E=E^{s}\oplus E^{u}$ and there is $\lambda<1$ (possibly changing the norm of $E$ by an equivalent     box-type one), called the skewness of $A$, such that

 $\|A|_{E^{s}}\|<\lambda,\quad\|A^{-1}|_{E^{u}}\|<\lambda$

and

 $\|x\|=\max\{\|x_{s}\|,\|x_{u}\|\}.$

Let us denote by $(\tilde{E},\|\cdot\|_{0})$ 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 $\tilde{A}:\tilde{E}\to\tilde{E}$ defined by $(\tilde{A}u)(x)=A(u(x))$, which is also hyperbolic. In fact, letting $\tilde{E}^{i}$ be the set of all maps $u\colon\tilde{E}\to\tilde{E}$ whose range is contained in $E^{i}$ (for $i=s,\,u$) we have that $\tilde{E}=\tilde{E}^{s}\oplus\tilde{E}^{u}$ is a hyperbolic splitting for $\tilde{A}$ with the same skewness as $A$.

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

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

###### Proposition 1.

There exists $\varepsilon>0$ such that if $\varphi$ and $\psi$ are $\varepsilon$-Lipschitz  , then there is a unique $u\in\tilde{E}$ such that

 $(I+u)(A+\varphi)=(A+\psi)(I+u).$
###### Proof.

We want to find $u$ such that

 $A+\varphi+u(A+\varphi)=A+Au+\psi(I+u)$

which is the same as

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

This can be rewriten as

 $\displaystyle u_{u}$ $\displaystyle=A_{u}^{-1}(u_{u}(A+\varphi)+\varphi_{u}-\psi_{u}(I+u))$ $\displaystyle u_{s}$ $\displaystyle=(A_{s}u_{s}+\psi_{s}(I+u)-\varphi_{s})(A+\varphi)^{-1},$

where we use the fact that by the Lipschitz inverse mapping theorem, if $\operatorname{Lip}(\varphi)<1/\lambda\leq\|A^{-1}\|^{-1}$ (where $\lambda$ is the skewness of $A$) then $A+\varphi$ is invertible with Lipschitz inverse   .

Now define $\Gamma:\tilde{E}\to\tilde{E}$ by

 $\displaystyle\Gamma_{s}(u)$ $\displaystyle=(A_{s}u_{s}+\psi_{s}(I+u)-\varphi_{s})(A+\varphi)^{-1}$ $\displaystyle\Gamma_{u}(u)$ $\displaystyle=A_{u}^{-1}(u_{u}(A+\varphi)+\varphi_{u}-\psi_{u}(I+u))$

We assert that, if $\varepsilon$ is small, $\Gamma$ is a contraction. In fact,

 $\displaystyle\left\|\Gamma_{s}(u)-\Gamma_{s}(v)\right\|_{0}$ $\displaystyle=\left\|\left(A_{s}(u_{s}-v_{s})+\psi_{s}(I+u)-\psi_{s}(I+v)% \right)(A+\varphi)^{-1}\right\|_{0}$ $\displaystyle\leq\|\tilde{A}_{s}\|\cdot\|(u_{s}-v_{s})(A+\varphi)^{-1}\|_{0}+% \|(\psi_{s}(I+u)-\psi_{s}(I+v))(A+\varphi)^{-1}\|_{0}$ $\displaystyle\leq\lambda\|u_{s}-v_{s}\|_{0}+\varepsilon\|u-v\|_{0}$ $\displaystyle\leq(\lambda+\varepsilon)\|u-v\|_{0}$

and

 $\displaystyle\|\Gamma_{u}(u)-\Gamma_{u}(v)\|_{0}$ $\displaystyle=\left\|A_{u}^{-1}(u_{u}(A+\varphi)-v_{u}(A+\varphi)-\psi_{u}(I+u% )+\psi_{u}(I+v))\right\|_{0}$ $\displaystyle\leq\|\tilde{A}_{u}^{-1}\|\cdot\left(\|u_{u}(A+\varphi)-v_{u}(A+% \varphi)\|_{0}+\|\psi_{u}(I+u)-\psi_{u}(I+v)\|_{0}\right)$ $\displaystyle\leq\lambda\left(\|u_{u}-v_{u}\|_{0}+\varepsilon\|u-v\|_{0}\right)$ $\displaystyle\leq\lambda(1+\varepsilon)\|u-v\|_{0}.$

Thus, if $\varepsilon<\varepsilon_{0}\doteq\min\{\lambda,(1-\lambda)/\lambda\}$, $\Gamma$ has Lipschitz constant smaller than $1$, so it is a contraction. Hence $u$ exists and is unique. ∎

###### Proof.

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

 $(I+v)(A+\psi)=(A+\varphi)(I+v).$

It follows that

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

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

 $(I+w)(A+\varphi)=(A+\varphi)(I+w),$

which obviously is $w=0$. But since $(I+v)(I+u)=I+(u+v+uv)$ and $u+v+uv\in\tilde{E}$, (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 $0$ and $f\colon U\to E$ is a $\mathcal{C^{1}}$ map with $f(0)=0$, then for every $\varepsilon>0$ there is $\delta>0$ such that $\varphi\doteq f-Df(0)$ is $\varepsilon$-Lipschitz in the ball $B(0,\delta)$.

###### Proof.

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

###### Proposition 4.

There is a constant $k$ such that if $\varphi\colon\overline{B}(0,r)\to E$ is an $\varepsilon$-Lipschitz map, then there is a $k\varepsilon$-Lipschitz map $\tilde{\varphi}\colon E\to E$ which coincides with $\varphi$ in $B(0,r/2)$.

###### Proof.

Let $\eta\colon\mathbb{R}\to\mathbb{R}$ be a $\mathcal{C}^{\infty}$ bump function: an infinitely differentiable map such that $\eta(x)=1$ for $x<1/2$ and $\eta(x)=0$ for $x>1$, with derivative  bounded by $M$ and $|\eta(x)\|\leq 1$ for all $x\in\mathbb{R}$. Now define $\tilde{\varphi}(x)=\varphi(x)\eta(\|x\|/r)$ (when $\varphi(x)$ is not defined, we assume that it is zero). If $x$ and $y$ are both in $B(0,r)$ then we have

 $\displaystyle\|\tilde{\varphi}(x)-\tilde{\varphi}(y)\|$ $\displaystyle=\big{\|}\varphi(x)\eta(\|x\|/r)-\varphi(y)\eta(\|y\|/r)\big{\|}$ $\displaystyle\leq\big{\|}(\varphi(x)-\varphi(y))\eta(\|x\|/r)\big{\|}+\big{\|}% \varphi(y)(\eta(\|x\|/r)-\eta(\|y\|/r))\big{\|}$ $\displaystyle\leq\varepsilon\|x-y\|+\|\varphi(y)-\varphi(0)\|\cdot\big{\|}\eta% (\|x\|/r)-\eta(\|y\|/r)\big{\|}$ $\displaystyle\leq\varepsilon\|x-y\|+\varepsilon\|y\|(M\|x-y\|/r)$ $\displaystyle\leq(M+1)\varepsilon\|x-y\|;$

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

 $\|\tilde{\varphi}(x)-\tilde{\varphi}(y)\|=\|\tilde{\varphi}(x)-\tilde{\varphi}% (y^{*})\|,$

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 $\tilde{\varphi}(y^{*})=0$. Also, $\|x-y^{*}\|=\tau\|x-y\|\leq\|x-y\|$; hence

 $\|\tilde{\varphi}(x)-\tilde{\varphi}(y)\|=\|\tilde{\varphi}(x)-\tilde{\varphi}% (y^{*})\|\leq(M+1)\varepsilon\|x-y^{*}\|\leq(M+1)\varepsilon\|x-y\|.$

Finally, if both $x$ and $y$ are outside $B(0,r)$, then $\|\tilde{\varphi}(x)-\tilde{\varphi}(y)\|=0\leq(M+1)\|x-y\|$. 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 $\varepsilon>0$ such that for any $\varepsilon$-Lipschitz map $\varphi$, $Df(0)$ is conjugate to $\varphi+Df(0)$. Choose $\delta$ such that $f-Df(0)$ is $\varepsilon/k$-Lipschitz in $B(0,2\delta)$. Let $\tilde{\varphi}$ be the $\varepsilon$-Lipschitz extension  of $f-Df(0)$ to $B(0,\delta)$ obtained from the previous proposition. We have that $Df(0)+\tilde{\varphi}$ is conjugate to $Df(0)$. But for $x\in B(0,\delta)$ we have $Df(0)+\tilde{\varphi}=f$, so that $f$ is locally conjugate to $Df(0)$.

Title proof of Hartman-Grobman theorem ProofOfHartmanGrobmanTheorem 2013-03-22 14:25:29 2013-03-22 14:25:29 Koro (127) Koro (127) 7 Koro (127) Proof msc 37C25