# Showing that some maps are topologically conjugated

Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be an orientation-preserving contracting $C^1$ diffeomorphism. Prove that it is topologically conjugate to the linear map $x \mapsto \frac{x}{2}$.

My work: Let $A : \mathbb{R}^n \to \mathbb{R}^n, A x = \frac{x}{2}$. Then $det(A - \lambda I) = \Big(\frac{1}{2} - \lambda\Big)^n.$ Hence $\; det(A - \lambda I) = 0 \Longrightarrow \lambda = \frac{1}{2}$, i.e $A$ is hyperbolic (because A has no eigenvalues of absolute value = 1).

To show that $f$ is topologically conjugate to $A$, I think that we have to apply the Hartman-Grobman Theorem, but I don't know how. To apply the local form of this theorem, the fixed point $p$ of $f$ should be hyperbolic and $A$ should be $Df_p$, but I don't know how to find out from the above assumptions if $p$ is hyperbolic and $A = Df_p$ ($f$ is contracting, hence $f$ has a fixed point). To apply the global form of the theorem, $A$ should be the linear part of $f$ and, again, I don't know how to check this.

Can someone help me? Thank you!

• Two comments that should help: note that it suffices to solve the problem in some domain $U$ such that $A(U)\subset U$ (and so only the local version is needed); note that a linear map is topologically conjugate to $x/2$ if and only if all its eigenvalues are inside the unit circle. Also, my first comment shows that we don't need something as strong as the Grobman-Hartman theorem (do call it this, the two proved the theorem independently, so the order should be alphabetical), although you will miss some usual properties then. – John B Sep 1 '17 at 23:53