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.


  • $\begingroup$ I think retagging this to "non-linear-functional-analysis" and/or "fixed-point-theory" would be more suitable, but unfortunately I cannot create new tags. $\endgroup$ – user3014 Jul 18 '10 at 22:15

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 $S$.

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  • $\begingroup$ It is a trivial remark (not due to me) that if a mapping $ f $ on a set $ X $ (no metric required) has a unique fixed point, then this is also a fixed point for any mapping which commutes with $f$. This implies immediately that a mapping on a complete metric space has a fixed point whenever some iterate is a contraction. No continuity assumption is required. $\endgroup$ – jbc Sep 18 '12 at 19:48
  • $\begingroup$ (or, also: if $f^n$ is a contraction on a complete metric space, it has a unique fixed point $x$; but $f(x)$ is a fixed point of $f^n$ too, so by the uniqueness $x=f(x)$.) $\endgroup$ – Pietro Majer Jan 10 '13 at 14:31

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