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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)) which is a contradiction. Similarly $g(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?

2 Answers 2

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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.

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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.

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    Bounded and monotonic, as you noted, since $f$ is monotonic and all live in $[a,b]$. Well, continuity is in the hypothesis, so it might be handled as related to convergence and similar stuffs.2012-11-17