$f\circ g=g\circ f$ + increasing $\Rightarrow$ common fixed point. 
Let $f,g:\mathbb [a,b] \to \mathbb [a,b]$ be monotonically increasing functions
   such that $f\circ g=g\circ f$
Prove that $f$ and $g$ have a common fixed point.

I found this problem in a problem set, it's quite similar to this Every increasing function from a certain set to itself has at least one fixed point  but I can't solve it.
I think it's one of those tricky problems where you need to consider a given set and use the LUB... I think $\{x \in [a,b]/ x < f(x) \text{and} x< g(x) \}$ is a good one.
Any hint ?

Let $A=\{x \in [a,b]/ x \leq f(x) \; \text{and} \; x \leq g(x) \}$


*

*$a\in A$

*let $u=\sup A$

*Let us prove that $f(u)$ and $g(u)$ are upper bounds for $A$
Indeed let $x\in A$.
Then $x\leq u$. Hence $f(x) \leq f(u)$, thus $x\leq f(x) \leq f(u)$ and finally $x\leq f(u)$ 
In the same way, $x\leq g(u)$ 


*

*Therefore, by LUB definition, $\color{red}{ u\leq f(u)}$ and $\color{red}{ u\leq g(u)}$

*Then, $f(u) \leq f(g(u))=g(f(u))$. Thus $u\leq f(u)\leq g(f(u))$ and then $u\leq g(f(u))$

*But in the same way, $u\leq f(f(u))$
Therefore, $f(u) \in A$ (see the last two last inequalities in the last two bulleted points)
Similarly, $g(u) \in A$.


*

*By LUB definition, $\color{red}{ f(u) \leq u}$ and $\color{red}{ g(u) \leq u}$


$u$ is a common fixed point.
