Let $E$ be a Banach space and $f:E\to E$ be a continuous map. By $f^n$ we denote the $n$-th iterate of $f$, i.e. $f^n:=\underbrace{f\circ f\circ\cdots \circ f}\_{\text{n times}}$. Let $x\_0$ denote a fixed point of $f^n$. 


1) Is it true that the question as to when $x\_0$ is a fixed point of $f$ has been resolved in the case of $E$ being reflexive? (I would be also thankful for some reference.)


2) What are considered to be the main obstacles in the case of $E$ being not reflexive when dealing with the above question (besides the obvious)? (As far as I understand, the problem is still open for general Banach spaces. Again, I would be thankful for a reference.)


A side remark: I realize that my questions may appear a bit too general. I am also aware of the fact that the field (fixed-point theory in banach spaces) is overflowed with publications of questionable value/quality/contribution like no other, which makes things even harder particularly for someone who is coming from a different field (e.g. number theory). Thus MO seems to be the only reasonable place to get a clear and compact answer from someone who has the overlook over the field.


Thanks!



When is a fixed point of f^n a fixed point of f?
Actually I can't see the role of reflexivity for an answer to the question as it is, unless further properties on $f$ are assumed. 


In general, to start with the obvious case: if $f$ is a contraction, it has a fixed point, which is also the unique fixed point of the contraction $f^n$, therefore it is $x\_0$, so $x\_0$ is a fixed point of $f$. Of course, if $f$ is not a contraction, like e.g. the map $x\mapsto -x$, a fixed point of $f^n$ need not be a fixed point of $f$. Also, in a certain sense, the case of contractions essentially covers all cases where the answer is affirmative: by a result of Bessaga of around '60, if a map $f$ on a set $S$ is such that for all $n\in \mathbb{N}$ the iterated $f^n$ has a unique fixed point, then $f$ is a contraction with respect to a suitable complete metric on $S$. 


