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?

  • $\begingroup$ (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
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    $\begingroup$ What can you say about $f$ monotonicity? $\endgroup$ – mathcounterexamples.net Oct 7 '15 at 20:19
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    $\begingroup$ @WillJagy $x \mapsto 1/x$ is not a continuous map $\mathbb{R}\rightarrow\mathbb{R}$ $\endgroup$ – Thomas Oct 7 '15 at 20:34
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    $\begingroup$ 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

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.

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

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 $g(x):=f(x)-x$.
  • $\begingroup$ 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
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    $\begingroup$ I think the OP ran away. Or fell asleep $\endgroup$ – Will Jagy Oct 7 '15 at 21:01

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