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Let $f,g: I \to I := [0,1]$ be continous functions satisfying $f \circ g = g \circ f$. Does there exist $x_0 \in I$ such that $f(x_0) = g(x_0)$ ?

Background: In a homework the problem was posed with $g=\operatorname{id}$ (where it can easily be solved with the help of the intermediate value theorem). The lecturer said the stronger statement above is true, but he didn't know a proof. I googled a little around, but could only find something about the "commuting function problem" (existence of a common fixed point of $f$ and $g$) that is known to be false.

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    $\begingroup$ Yes. If not then $f(x) < g(x)$ (or the reverse inequality) by the intermediate value theorem. Then $f$ maps the minimum fixed point of $g$ to a smaller fixed point. Contradiction. $\endgroup$
    – George Lowther
    Commented Sep 4, 2011 at 17:40
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    $\begingroup$ Btw, there was a question some time ago about the multidimensional generalisation of this. That is an open problem. $\endgroup$
    – George Lowther
    Commented Sep 4, 2011 at 17:42
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    $\begingroup$ Here is the link mathoverflow.net/questions/3332/… $\endgroup$
    – Gjergji Zaimi
    Commented Sep 4, 2011 at 17:44
  • $\begingroup$ Thank you all very much for the proof and the link and the quick reply. $\endgroup$
    – tomasz
    Commented Sep 4, 2011 at 17:58
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    $\begingroup$ Well known; e.g. it's problem 518 (the last of the book) of Bernard Gelbaum's Problems in Analysis. $\endgroup$
    – Pietro Majer
    Commented Sep 4, 2011 at 22:15

1 Answer 1

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Yes. The set of fixed points of $g$ is closed, nonempty, and is mapped into itself by $f$. Letting $a\le b$ be, respectively, the minimum and maximum fixed points of $g$, we have $f(a)\ge a=g(a)$ and $f(b)\le b=g(b)$. So, by the intermediate value theorem, there is an $x\in[a,b]$ with $f(x)=g(x)$.

Also, to reiterate the points made in the comments, this is a difficult problem for more general domains. The case of commuting maps on the closed disc has been asked before, and is still open. In fact, even the case of commuting maps on the simple triod (i.e., a capital 'T') appears to be an open problem, according to the contributed problem from Jeff Norden here (Commuting, coincidence-point-free maps on a triod).

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  • $\begingroup$ I see, I mixed two distinct problems: (a) common value is true for $n=1,$ unknown for $n \geq 2,$ (b) common fixed point is false for $n=1$ and we have nothing in print about $n \geq 2$, these all being on the closed unit ball in $\mathbb R^n$ so existence of fixed points for each map is automatic $\endgroup$
    – Will Jagy
    Commented Sep 4, 2011 at 19:50
  • $\begingroup$ @Will: (b) is false for $n\ge2$. Just take a counterexample for the $n=1$ case and extend to $I^n$ by making the maps fix the coordinates $x_i$ ($i=2,\ldots,n$). $\endgroup$
    – George Lowther
    Commented Sep 4, 2011 at 19:56

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