Deducing asymptotically stability if $|f(x)| <|x|$ let f be a continuous map such that $f(0)=0$ and $|f(x)|<|x|$ whenever $0<|x|<\epsilon$ for some $\epsilon$>0
a) show that the sequence $f^{n}(x)$ has a limit point $x_{*}$
b) show that $|f(x_{*})|=|x_{*}|$
c) Deduce that $x{*}=0$, and hence that 0 is an asymptotically stable fixed point. 
Does anyone know how to do this or have any idea on how to start?
 A: Hints (which only show that $0$ is locally asymptotically stable with region of attraction $(-\varepsilon,\varepsilon)$):
Suppose that $\bar{x}\in(-\varepsilon,\varepsilon)$ and consider the sequences $a_1:=\bar{x}$, $a_{n+1}:=f(a_n)$ and $b_n:=|a_n|$. A monotonically decreasing sequence converges if and only if it is bounded. Show that $(b_n)$ is monotonically decreasing and bounded and conclude that $b_n\rightarrow b_*$ as $n\rightarrow\infty$.
Show that $b_*\in[0,\varepsilon)$ (use monotonicity of $(b_n)$ and the fact that a convergent sequence in a closed set converges inside the set).
Use the sequence definition of a continuous function and $(b_n)$ to show that $f(b_*)=b_*$.
Using the above and the fact that $|f(x)|<|x|$ if $x\in(-\varepsilon,\varepsilon)$, argue by contradiction that $b_*=0$. What does this imply about $a_n$ as $n\rightarrow\infty$?
I think that pretty much does it (unless I messed up somewhere). Let me know if you want me to flesh the whole solution out in detail. Also note that the above proves $b)\Rightarrow c)\Rightarrow a)$ - I couldn't think of way to start by proving $a)$ first...
