# Diffeomorphism with only hyperbolic periodic points has finitely many periodic points (Morse-Smale)

Came across the following question during a course Chaotic Dynamical Systems:

If a diffeomorphism $$f:I\to I$$ is Morse-Smale (i.e. has only hyperbolic periodic points), then it has finitely many periodic points. ($$I:=[0,1]$$)

I assume the prove is done by contradiction that a Morse-Smale diffeo can't have infinitely many periodic points for then it should have a non-hyperbolic fixed point; i.e. $$\exists p\in I$$ s.t. $$|(f^n)'(p)|=1$$ by the Mean Value Theorem (somehow). The chapter where this question is asked talks about structural stability in the sense of $$C^r-\epsilon$$-distance of two conjugate functions. So I expect the prove to contain some structural stability properties. I have no clue how to use this $$C^r-$$distance in proving this. Can anyone help me out?

Special note: $$d_{C^r}(f,g)=\sup_{x\in\mathbb{R}}\{|f-g|,|f^{(1)}-g^{(1)}|,...,|f^{(r)}-g^{(r)}|\}<\epsilon$$ for some $$\epsilon>0$$ must imply $$f$$~$$g$$. Also, I just discovered that a map near a hyperbolic fixed point is always locally topologically conjugate to it's derivative. Maybe this also holds for maps near hyperbolic periodic points?

• Yeah, Grobman-Hartman is a way to go. When you are dealing with compact phase space, a sequence of periodic points must converge somewhere. This should be a periodic point again and it is non-isolated, but still hyperbolic by our assumptions. However, if diffeomorphism is Morse-Smale, by Grobman-Hartman any fixed (or periodic) point must have a neigbourhood that doesn't contain any other periodic point, hence a contradiction. – Evgeny Oct 7 '18 at 9:58
• Thanks for your answer! Why is that periodic point non-isolated? Is it because this infinite sequence is converging in an arbitrary small nbh to that fixed point? If yes, how do you know that this infinite sequence of arbitrary close periodic points exists? – Algebear Oct 7 '18 at 17:29
• @Evgeny Why does it follow that the sequence must converge to a periodic point? – John B Oct 7 '18 at 20:43
• @JohnB I think I've rushed with this statement. It works if there is a sequence of periodic points with the same period, but even in that case it just proves that there is a finite number of periodic points of each period. And there are counterexamples to this arrangement, so I don't think this will work as it was intended. – Evgeny Oct 8 '18 at 5:14
• @GuusPalmer The initial idea was the following. Suppose that there is a sequence of fixed points of the same period. In that case you can extract a subsequence that converges to some point. Just by continuity it would be a periodic point of the same period. By assumption (all periodic points are hyperbolic) this point is hyperbolic, but since it is a limit point, it has a periodic point in an arbitrarily small neighbourhood, which contradicts Grobman-Hartman. The caveat is that I can't guarantee that such sequence exists, you are right. – Evgeny Oct 8 '18 at 5:21

## 1 Answer

I might as well answer my own question myself by now. To finish the prove, one must note that a diffeo is monotone increasing or if it's decreasing, then its second iterate will be increasing. So w.l.o.g. let $$f$$ be monotonely increasing. Since $$f$$ has infinitely many periodic points, it must have infinitely many fixed points for it should cross the line $$y=x$$ infinitely many times. Now, as I stated in the comments with some help of others, there must exist an infinite sequence of periodic points converging to some fixed point. But hyperbolic fixed points must have a neighbourhood such that this nbh does not contain any other periodic points. Hence, a contradiction since $$f$$ should only have hyperbolic periodic points by assumption. We conclude that Morse-Smale diffeos have only finitely many periodic points. $$\Box$$.