4
$\begingroup$

Does a continuous mapping $f\colon \mathbb R \to \mathbb R$ which satisfies $f(f(x))=x$ for each $x \in \mathbb R$ necessarily have a fixed point?

$\endgroup$
  • $\begingroup$ (Yes) Have you tried something to prove it? $\endgroup$ – Silvia Ghinassi Oct 7 '15 at 20:07
  • $\begingroup$ Ciao (your name looks like it's coming from my country). Anyway, in class (it's first year PhD math for economists) we covered cases where a function is non decreasing and then Tarski theorem makes sense. Browel tells me that if a function maps from a compact and convex space to another compact and convex space it has a fixed point. Should I go in that direction? I'm kinda lost. And I have 4 more exercises like that one. Please...a direction....Thanks a lot $\endgroup$ – Gian Luca Oct 7 '15 at 20:13
  • 1
    $\begingroup$ What can you say about $f$ monotonicity? $\endgroup$ – mathcounterexamples.net Oct 7 '15 at 20:19
  • 2
    $\begingroup$ @WillJagy $x \mapsto 1/x$ is not a continuous map $\mathbb{R}\rightarrow\mathbb{R}$ $\endgroup$ – Thomas Oct 7 '15 at 20:34
  • 1
    $\begingroup$ Alright, second try: Q1: if we always have $f(x) > x,$ what does that say about $f(f(x))?$ Q2:if we always have $f(x) < x,$ what does that say about $f(f(x))?$ $\endgroup$ – Will Jagy Oct 7 '15 at 20:39
3
$\begingroup$

Suppose $f(a) = b$ for some $a \neq b$, then $f(b) = f(f(a)) = a$. Define $g(x) = f(x) -x$, and we have $$g(a) = f(a) - a = b-a$$ and $$g(b) = f(b) - b = a-b$$ from $g$ is continuous, by intermediate value theorem, there exists $c$ between $a,b$ such that $$g(c) = f(c) - c = 0$$ because $b-a$ and $a-b$ have opposite signs.

$\endgroup$
  • $\begingroup$ Thanks. This looks elegant. Yes, brilliant $\endgroup$ – Gian Luca Oct 8 '15 at 14:38
2
$\begingroup$

Extended hints:

  1. We can assume that $f(0)\neq0$. Without loss of generality we can then also assume that $f(0)>0$. This is because if $f(f(x))=x$ for all reals $x$, then also $F(F(x))=x$ for all $x$, where I define $F(x)=-f(-x)$. Furthermore, $f(-x)=-x$ iff $F(x)=x$, so if one has a fixed point so does the other. All this amounts to is that we can study $F$ instead of $f$ to get $f(0)>0$.
  2. Look at the restriction of $f$ to the interval $[0,f(0)]$. Notice that $f$ maps the endpoints of this interval to each other. Plot the graphs of both $f$ and the identity function $id(x)=x$. Why must they intersect in this interval? Bolzano's theorem (or intermediate value theorem) on $g(x):=f(x)-x$.
$\endgroup$
  • $\begingroup$ The same idea as in Will Jagy's comment. Specified a bit. $\endgroup$ – Jyrki Lahtonen Oct 7 '15 at 20:39
  • $\begingroup$ If step 1. feels strange, you can skip it and split the treatment in step 2 into two cases. You study the restriction of $f$ to either $[0,f(0)]$ or $[f(0),0]$ depending on the sign of $f(0)$. $\endgroup$ – Jyrki Lahtonen Oct 7 '15 at 20:54
  • 1
    $\begingroup$ I think the OP ran away. Or fell asleep $\endgroup$ – Will Jagy Oct 7 '15 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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