$f,g:[a,b]\to [a,b]$ are continuous, $f\circ g=g\circ f$ and $f$ is 1-1. Show that $\exists t\in [a,b]$ so that $f(t)=g(t)=t$

Question

Using the Intermediate Value Theorem on $h(x)=f(x)-x$ and $r(x)=g(x)-x$ I can easily show that $\exists t_f,t_g\in[a,b]$ so that $f(t_f)=t_f$ and $g(t_g)=t_g$. It remains to show that $t_f=t_g$. Since $f$ is 1-1 and continuous, $f$ is monotone.

Case 1: $f$ is strictly decreasing. If $g(t_f)>t_f$ then $f(g(t_f))<f(t_f)=t_f\Rightarrow g(f(t_f))<t_f\Rightarrow g(t_f)<t_f$ which is a contradiction. Similarly $g(t_f)<t_f$ leads to a contradiction and so $g(t_f)=t_f$.

Case 2:$f$ is strictly increasing. Here is where I am stuck :( Any hints?

Answer 1

Consider only the $t_g$'s. It is possible that there are more of them, and not necessarily all would also be fixed point of $f$.

Let $t_0$ be a fixed point of $g$, and consider $$t_{n+1}:= f(t_n)$$ Then, by induction $$g(t_{n+1}) = f\circ g\circ f^{-1}(f(t_n))= f(g(t_n))=f(t_n)=t_{n+1}$$ it remains a fixed point of $g$. I think, you can figure out the rest.

Answer 2

Set $p(x):=f(x)-g(x)$. If $p(x)=0$ haven't root, then we can assume that $p(x)>0$ for any $x$, since $f$ and $g$ is continous on compact set, so there exiats an element $x_{0}$ such that $\min p(x)=p(x_{0})>0$, so we have $$f(x)>g(x)+p(x_{0})$$ for all $x$. Now set $f(x)$ instead of $x$, since $f(g(x))=g(f(x))$, from last inequality we obtain $$f^{2}(x)>g(f(x))+p(x_{0})=f(g(x))+p(x_{0})>g(x)+2p(x_{0})$$ and by induction we can obtain $$f^{n}(x)>g(x)+np(x_{0})$$ since $p(x_{0})>0$, its implies that $f$ is unbounded function, where its a contradiction.