Brin-Stuck proof of Hadamard-Perron theorem I'm currently studying Brin and Stuck book "Introduction to dynamical systems", especially the part of Hadamard-Perron theorem. I'm actually stuck on some arguments presented by the authors, that usually don't go explaining with length what they do (which makes the book both beautiful and somewhat hard to follow). Once they've introduced the graph transformation, they go on to prove that this transformation is a contraction. Here's the proof : 
I'm really looking for some help to write formally and prove some of the claims:


*

*That if $\epsilon$ is small enough, then the tangent vectors to $f_n(c_u)$ lies in the cone $K^u_L(n+1)$ and the tangent vectors to the graph of $\phi_{n+1}$ lies in $K^s_L(n+1)$ ?

*The calculation made at the end. I don't fathom how they are done, honestly. I did some calculation on my side, which yields similar results but I believe them to be false.


My infinite gratitude to any of you for helping me out.

I believe the following two hypothesis should be used:


*

*The angle between $E^s(n)$ and $E^u(n)$ are uniformly bounded away from $0$,

*The sequence $(df_n(\cdot))_n$ is equicontinous. 

 A: Let me be specific and illustrate the type of inequalities that you need to use.

For simplicity of the writing I will assume that $f_n=f$ for all $n$, but it is really the same because of the uniform conditions. Consider the cone
  $$
C =\{(x,y) \in E^s \times E^u : \lVert x \rVert \le L \lVert y \rVert \}.
$$
  Let me show that for any sufficiently small $\varepsilon>0$, we have $df C \subset C$ on the ball $B_\varepsilon(0)$.

Given $(x,y) \in E^s \times E^u$, write
$$
df(x,y) =(A x +dg(x,y), B y +dh(x,y) ).
$$
We have
$$
\lVert A x +dg(x,y) \rVert \le \lambda \lVert x \rVert + c \lVert (x,y) \rVert
$$
and
$$
\lVert B y +dh(x,y) \rVert \ge \lambda^{-1} \lVert y \rVert -c \lVert (x,y) \rVert,
$$
where $c$ is some constant that tends to $0$ when $\varepsilon\to0$.
For $(x,y) \in C$, we have $\lVert x\rVert \le L \lVert y \rVert$ and so
$$
\lVert Ax +dg(x,y) \rVert \le \lambda L \lVert y \rVert + c (1+L) \lVert y\rVert
$$
and
$$
L \lVert B y +dh(x,y) \rVert \ge \lambda^{-1} L \lVert y \rVert -c L (1+L) \lVert y \rVert.
$$
For $\varepsilon$ sufficiently small so that $c$ is sufficiently we get
$$\lVert A x +dg(x,y) \rVert \le L\lVert B y +dh(x,y)\rVert,$$ and we are done.
