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This question is the last exercise of chapter 2 in Lan Wen`s Differential Dynamical system. (Exercise 2.12)

let $E$ a finite-dimensional normed vector space and $p \in E$ be a hyperbolic fixed point of $f$. Given any positive integer $m$, prove there is a neighborhood $V$ of $p$ such that any period point of $f$ in $V-{p}$ has a period greater than $m$.

here $U$ is open subset of $E$ and $ f: U\longrightarrow E $ is $C^k$ and local Diffeomorphism.

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You can do it by contradiction.

Suppose this is not true. You will find a $M$ and a sequence of point $p_k \to p$ of periodic point converging to $p$ with period less than $M$.

Taking a sub-sequence, they can have the same period $m \leq M$ so they are fixed for $f^m$. Notice that $p$ is still a hyperbolic fixed point of $f^m$.

Now taking again a sub-sequence, they can approch $p$ in the same direction, that is $\frac{p-p_k}{\| p -p_k \|} \to v \in E$.

Computing the differential in the $v$ direction for $f^m$ with this sequence will give you a $0$ and this cannnot be since it is suppose hyperbolic.

EtienneBfx
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  • Dear @EtienneBfx thanks for your answer. I can not understand your fourth paragraph. Do you take two sub-sequence that approch $p$ in the same direction? and How? – Hossein Khayam Jan 29 '21 at 16:50
  • Sorry I am not fluent in english and that might be confusing. You should take only one sequence that approch $p$ in a given direction. If you note $x_k=\frac{p-p_k}{| p- p_k | }$ then all $x_k$ are in the sphere of vector of norm $1$ which is compact. You take a sub-sequence converging to an accumulation point. – EtienneBfx Feb 01 '21 at 10:43
  • @EtienneBfx Would you mind explaining more about your answer? Why could we take a subsequence of $p_k$ with the same period $m$? – Reza Yaghmaeian Jan 04 '22 at 16:02
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    All $p_k$ are periodic point with period less that $M$, so there is only a finite number of possibilities for their period $0,1,2,...,M$. So there is at least one subsequence of $p_k$ with the same period. – EtienneBfx Jan 05 '22 at 16:55