TITLE: Have commuting functions a common value ? 
QUESTION [7 upvotes]: 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.

REPLY [13 votes]: 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).