Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Periodic orbits

Suppose that $f$ is a continuous map from $\mathbb R$ to $\mathbb R$, which satisfies $f(f(x)) = x$ for each $x \in \mathbb{R}$.

Does $f$ necessarily have a fixed point?

share|improve this question
1  
Yes. Check this math.stackexchange.com/questions/233246/periodic-orbits/… –  Amr Dec 22 '12 at 15:49
    
Thanks for your help... –  salar_ve Dec 22 '12 at 16:06
add comment

marked as duplicate by Amr, Alexander Gruber, Hagen von Eitzen, draks ..., tomasz Dec 22 '12 at 20:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers

Here's a somewhat simpler (in my opinion) argument. It's essentially the answer in Amr's link given in the first comment to the question, but simplified a bit to treat just the present question, not a generalization. Start with any $a\in\mathbb R$. If we're very lucky, $f(a)=a$ and we're done. If we're not that lucky, let $b=f(a)$; by assumption $a=f(f(a))=f(b)$. Since we weren't lucky, $a\neq b$. Suppose for a moment that $a<b$. Then the function $g$ defined by $g(x)=f(x)-x$ is positive at $a$ and negative at $b$, so, by the intermediate value theorem, it's zero at some $c$ (between $a$ and $b$). That means $f(c)=c$, and we have the desired fixed point, under the assumption $a<b$. The other possibility, $b<a$, is handled by the same argument with the roles of $a$ and $b$ interchanged.

As user1551 noted, we need $f(f(x))=x$ for only a single $x$, since we can then take that $x$ as the $a$ in the argument above.

share|improve this answer
add comment

Yes. here is an outline of how to show this:

$\ \ \ \ $1) Show that the range of $f$ is $\Bbb R$.

$\ \ \ \ $2) Show that $f$ is one-to-one; hence, monotone by continuity.

Assume $f$ is not the identity function (otherwise the result is trivial). Then:

$\ \ \ \ $3) Show that $f$ is strictly decreasing.

Then, using 1) and 3):

$\ \ \ \ $4) Show that the function $g$ defined by $g(x)=f(x)-x$ has a (unique) zero.

Conclude that $f$ has a (unique) fixed point.

share|improve this answer
add comment

Yes. This is a direct consequence of Sharkovsky's theorem. And the requirement $f(f(x))=x$ for each $x\in\mathbb{R}$ is too strong. Actually we only need $f(f(x))=x$ for some $x\in\mathbb{R}$ to guarantee that $f$ has a fixed point.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.