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

$$\parallel x\parallel =\max \{\parallel x_s\parallel ,\parallel x_u\parallel \}.$$

Let us denote by $(\tilde{E},\parallel \cdot {\parallel }_0)$ 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 $\tilde{A}:\tilde{E}\to \tilde{E}$ 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 $u:\tilde{E}\to \tilde{E}$ whose range is containedin $E^i$ (for $i=s,u$) we have that $\tilde{E}={\tilde{E}}^s\oplus {\tilde{E}}^u$ is a hyperbolic splittingfor $\tilde{A}$ with the same skewness as $A$.

From now on we denote the projection of $x$ to $E^i$ by $x_i$, andthe 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 $u\in \tilde{E}$.

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 $\mathrm{\Gamma }:\tilde{E}\to \tilde{E}$ 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$

$\le \parallel {\tilde{A}}_s\parallel \cdot {\parallel (u_s-v_s){(A+\phi )}^{-1}\parallel }_0+{\parallel ({\psi }_s(I+u)-{\psi }_s(I+v)){(A+\phi )}^{-1}\parallel }_0$

$\le \lambda {\parallel u_s-v_s\parallel }_0+\epsilon {\parallel u-v\parallel }_0$

$\le (\lambda +\epsilon ){\parallel u-v\parallel }_0$

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$

$\le \parallel {\tilde{A}}_u^{-1}\parallel \cdot \left({\parallel u_u(A+\phi )-v_u(A+\phi )\parallel }_0+{\parallel {\psi }_u(I+u)-{\psi }_u(I+v)\parallel }_0\right)$

$\le \lambda \left({\parallel u_u-v_u\parallel }_0+\epsilon {\parallel u-v\parallel }_0\right)$

$\le \lambda (1+\epsilon ){\parallel u-v\parallel }_0.$

Thus, if , $\mathrm{\Gamma }$ has Lipschitz constantsmaller 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$v\in \tilde{E}$ 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 thatthat 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.∎

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 thefact 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, thenthere 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 $\eta :ℝ\to ℝ$ be a ${𝒞}^{\mathrm{\infty }}$ bump function: an infinitelydifferentiable 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 $x\in ℝ$.Now define $\tilde{\phi }(x)=\phi (x)\eta (\parallel x\parallel /r)$ (when $\phi (x)$ is notdefined, we assume that it is zero).If $x$ and $y$ are both in $B(0,r)$ then we have

$\parallel \tilde{\phi }(x)-\tilde{\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

$$\parallel \tilde{\phi }(x)-\tilde{\phi }(y)\parallel =\parallel \tilde{\phi }(x)-\tilde{\phi }(y^{*})\parallel ,$$

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 $\tilde{\phi }(y^{*})=0$. Also, $\parallel x-y^{*}\parallel =\tau \parallel x-y\parallel \le \parallel x-y\parallel$; hence

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

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

Proof of the theorem.Taking the particular $\psi =0$ in the lemma, we observe thatthere 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 $Df(0)$.