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proof of Hartman-Grobman theorem

Lemma 1 Let $A\colon E\to E$ be an hyperbolic isomorphism, and let $\phi$ and $\psi$ be $\epsilon$ -Lipschitz maps from $E$ to itself such that $\phi(0)=\psi(0)=0$ . If $\epsilon$ is sufficiently small, then $A+\phi$ 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 $(\tilde Au)(x) = A(u(x))$ , 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 $\epsilon>0$ such that if $\phi$ and $\psi$ are $\epsilon$ -Lipschitz, then there is a unique $u\in \tilde E$ 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

$\displaystyle u_u$	$\displaystyle = A_u^{-1}(u_u(A+\varphi ) + \varphi _u - \psi_u(I+u))$
$\displaystyle u_s$	$\displaystyle = (A_su_s + \psi_s(I+u) - \varphi _s)(A+\varphi )^{-1},$

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

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

$\displaystyle \Gamma_s(u)$	$\displaystyle = ( A_su_s + \psi_s(I+u) - \varphi _s)(A+\varphi )^{-1}$
$\displaystyle \Gamma_u(u)$	$\displaystyle = A_u^{-1}(u_u(A+\varphi ) + \varphi _u - \psi_u(I+u))$

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

$\displaystyle \left\Vert\Gamma_s(u) - \Gamma_s(v)\right\Vert _0$	$\displaystyle = \left\Vert\left(A_s(u_s-v_s) + \psi_s(I+u) - \psi_s(I+v)\right)(A+\varphi )^{-1}\right\Vert _0$
	$\displaystyle \leq \Vert\tilde A_s\Vert\cdot\Vert(u_s-v_s)(A+\varphi )^{-1}\Vert _0 + \Vert(\psi_s(I+u) - \psi_s(I+v))(A+\varphi )^{-1}\Vert _0$
	$\displaystyle \leq \lambda\Vert u_s-v_s\Vert _0 + \varepsilon \Vert u-v\Vert _0$
	$\displaystyle \leq (\lambda+\varepsilon ) \Vert u-v\Vert _0$

and

$\displaystyle \Vert\Gamma_u(u)-\Gamma_u(v)\Vert _0$	$\displaystyle = \left\Vert A_u^{-1}(u_u(A+\varphi ) - v_u(A+\varphi ) - \psi_u(I+u) + \psi_u(I+v))\right\Vert _0$
	$\displaystyle \leq \Vert\tilde A_u^{-1}\Vert\cdot\left( \Vert u_u(A+\varphi )-v_u(A+\varphi )\Vert _0 + \Vert\psi_u(I+u) - \psi_u(I+v)\Vert _0\right)$
	$\displaystyle \leq \lambda \left( \Vert u_u - v_u\Vert _0 + \varepsilon \Vert u - v\Vert _0 \right)$
	$\displaystyle \leq \lambda(1 + \varepsilon )\Vert u-v\Vert _0.$

Thus, if $\epsilon < \epsilon_0 \doteq \min\{\lambda, (1-\lambda)/\lambda\}$ , $\Gamma$ has Lipschitz constant smaller than $1$ , so it is a contraction. Hence $u$ exists and is unique. $\qedsymbol$

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 $v\in \tilde E$ such that $$(I+v)(A+\psi) = (A+\phi)(I+v).$$ It follows that

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

(1)

Also, the previous proposition with $\phi = \psi$ implies that that there is a unique $w\in\tilde E$ 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 $u+v+uv\in \tilde E$ , (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. $\qedsymbol$

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 $\epsilon>0$ there is $\delta>0$ such that $\phi\doteq f-Df(0)$ is $\epsilon$ -Lipschitz in the ball $B(0,\delta)$ .

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

Proposition 4 There is a constant $k$ such that if $\phi\colon \overline B(0,r)\to E$ is an $\epsilon$ -Lipschitz map, then there is a $k\epsilon$ -Lipschitz map $\tilde \phi\colon E\to E$ which coincides with $\phi$ in $B(0,r/2)$ .

Proof. Let $\eta\colon \R \to \R$ 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 $x\in \R$ . Now define $\tilde \phi(x) = \phi(x)\eta(\|x\|/r)$ (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

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

if $x$ is in $B(0,r)$ and $y$ is not, then $$\|\tilde\phi(x)-\tilde\phi(y)\| = \|\tilde\phi(x) - \tilde\phi(y^*)\|,$$ 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 $\tilde\phi(y^*)=0$ . Also, $\|x-y^*\| = \tau\|x - y\| \leq \|x-y\|$ ; hence $$\|\tilde\phi(x)-\tilde\phi(y)\| = \|\tilde\phi(x) - \tilde\phi(y^*)\|\leq (M+1)\epsilon\|x-y^*\| \leq (M+1)\epsilon\|x-y\|.$$ Finally, if both $x$ and $y$ are outside $B(0,r)$ , then $\|\tilde\phi(x)-\tilde\phi(y)\| = 0 \leq (M+1)\|x-y\|$ . Letting $k=M+1$ we get the desired result. $\qedsymbol$

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 $\tilde\phi$ be the $\epsilon$ -Lipschitz extension of $f-Df(0)$ to $B(0,\delta)$ obtained from the previous proposition. We have that $Df(0)+\tilde\phi$ is conjugate to $Df(0)$ . But for $x\in B(0,\delta)$ we have $Df(0)+\tilde\phi = f$ , so that $f$ is locally conjugate to $Df(0)$ .

proof of Hartman-Grobman theorem is owned by Koro.

How to Cite This Entry

Koro. "proof of Hartman-Grobman theorem" (version 4). PlanetMath.org. Freely available at http://planetmath.org/ProofOfHartmanGrobmanTheorem.html.

Classification

AMS MSC:

37C25 (Dynamical systems and ergodic theory :: Smooth dynamical systems: general theory :: Fixed points, periodic points, fixed-point index theory)

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