proof of Hartman-Grobman theorem

Lemma 1.

Let A:EE be an hyperbolic isomorphism, and let φ and ψ be ε-Lipschitz maps from E to itself such that φ(0)=ψ(0)=0. If ε is sufficiently small, then A+φ and A+ψ are topologically conjugate.

Since A is hyperbolic, we have E=EsEu and there is λ<1 (possibly changing the norm of E by an equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath box-type one), called the skewness of A, such that




Let us denote by (E~,0) the Banach spaceMathworldPlanetmath of all boundedPlanetmathPlanetmath, continuous mapsMathworldPlanetmath from E to itself, with the norm of the supremumMathworldPlanetmathPlanetmath induced by the norm of E. The operatorMathworldPlanetmath A induces a linear operator A~:E~E~ defined by (A~u)(x)=A(u(x)), which is also hyperbolic. In fact, letting E~i be the set of all maps u:E~E~ whose range is contained in Ei (for i=s,u) we have that E~=E~sE~u is a hyperbolic splitting for A~ with the same skewness as A.

From now on we denote the projection of x to Ei by xi, and the restrictionPlanetmathPlanetmath A|Ei:EiEi by Ai (i=s,u).

We will try to find a conjugation of the form I+u where uE~.

Proposition 1.

There exists ε>0 such that if φ and ψ are ε-LipschitzPlanetmathPlanetmath, then there is a unique uE~ such that


We want to find u such that


which is the same as


This can be rewriten as

uu =Au-1(uu(A+φ)+φu-ψu(I+u))
us =(Asus+ψs(I+u)-φs)(A+φ)-1,

where we use the fact that by the Lipschitz inverse mapping theorem, if Lip(φ)<1/λA-1-1 (where λ is the skewness of A) then A+φ is invertible with Lipschitz inversePlanetmathPlanetmathPlanetmath.

Now define Γ:E~E~ by

Γs(u) =(Asus+ψs(I+u)-φs)(A+φ)-1
Γu(u) =Au-1(uu(A+φ)+φu-ψu(I+u))

We assert that, if ε is small, Γ is a contraction. In fact,

Γs(u)-Γs(v)0 =(As(us-vs)+ψs(I+u)-ψs(I+v))(A+φ)-10


Γu(u)-Γu(v)0 =Au-1(uu(A+φ)-vu(A+φ)-ψu(I+u)+ψu(I+v))0

Thus, if ε<ε0min{λ,(1-λ)/λ}, Γ 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 propositionPlanetmathPlanetmath is a homeomorphism.


Using the previous proposition with φ and ψ switched, we get a unique vE~ such that


It follows that

(I+v)(I+u)(A+φ)=(I+v)(A+ψ)(I+u)=(A+φ)(I+v)(I+u). (1)

Also, the previous proposition with φ=ψ implies that that there is a unique wE~ such that


which obviously is w=0. But since (I+v)(I+u)=I+(u+v+uv) and u+v+uvE~, (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 continuousMathworldPlanetmath inverse. ∎

The two previous propositions prove the lemma.

Proposition 3.

If U is an open neighborhood of 0 and f:UE is a C1 map with f(0)=0, then for every ε>0 there is δ>0 such that φf-Df(0) is ε-Lipschitz in the ball B(0,δ).


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

Proposition 4.

There is a constant k such that if φ:B¯(0,r)E is an ε-Lipschitz map, then there is a kε-Lipschitz map φ~:EE which coincides with φ in B(0,r/2).


Let η: be a 𝒞 bump function: an infinitely differentiable map such that η(x)=1 for x<1/2 and η(x)=0 for x>1, with derivativePlanetmathPlanetmath bounded by M and |η(x)1 for all x. Now define φ~(x)=φ(x)η(x/r) (when φ(x) is not defined, we assume that it is zero). If x and y are both in B(0,r) then we have

φ~(x)-φ~(y) =φ(x)η(x/r)-φ(y)η(y/r)

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


where y* is defined as x+τ(y-x) with


This is true because φ~(y*)=0. Also, x-y*=τx-yx-y; hence


Finally, if both x and y are outside B(0,r), then φ~(x)-φ~(y)=0(M+1)x-y. Letting k=M+1 we get the desired result. ∎

Proof of the theorem. Taking the particular ψ=0 in the lemma, we observe that there is ε>0 such that for any ε-Lipschitz map φ, Df(0) is conjugate to φ+Df(0). Choose δ such that f-Df(0) is ε/k-Lipschitz in B(0,2δ). Let φ~ be the ε-Lipschitz extensionPlanetmathPlanetmath of f-Df(0) to B(0,δ) obtained from the previous proposition. We have that Df(0)+φ~ is conjugate to Df(0). But for xB(0,δ) we have Df(0)+φ~=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