Does there exist mappings $f,g:[1,2]\to[1,2]$ such that $f\circ g=g\circ f$ but for no $x\in[1,2]$ we have $f(x)=g(x)=x?$
I am trying to figure out such mapping for quite a long time but failed. For all such pair I found a common fixed point.
Please help.
Seeing how arbitrary maps are allowed, there is plenty to choose from. One example would be (unless I have misunderstood the conditions) $$f(x) = g(x) = \begin{cases} 3 - x & x \not\in \{ 1, 1.5 \} \\ 1 & x = 1.5 \\ 1.5 & x = 1 \end{cases}:$$ First of all, since $f = g$, we obviously have $f \circ g = g \circ f$. Furthermore, suppose $f(x) = x$. If $x = 1$, $f(x) = 1.5 \neq 1 = x$; if $x = 1.5$, $f(x) = 1 \neq 1.5 = x$. So, $x \not\in \{ 1, 1.5 \}$, thus $f(x) = 1 - x$. As we have assumed $f(x) = x$, $x = 3-x$, i.e., $x=1.5$ - but we have just excluded that, contradiction! Thus, there is no $x \in [1,2]$ such that $f(x) = x$.