$f\circ g=g\circ f$ + 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 ?
 A: Let $z$ be a fixed point of $f$. Then $$f(g(z))=g(f(z))=g(z),$$ so also $g(z)$ is a fixed point of $f$. Now if they were unique, then $g(z)=z$. In general, $g$ induces a permutation map of the fixed points of $f$, now argue with monotonicity... 

The analogy should be the case of eigenvectors of commuting matrices, if $v$ is an eigenvector of $A$ and $AB=BA$, then $B^kv$ are all eigenvectors of $A$ with the same eigenvalue, so there is a minimal polynomial with $p(B)v=0$, with degree smaller than the dimension of the eigenspace of $A$ containing $v$, and then it gets more complicated.

Consider the sequence $g^k(z)$ and its limit points. Resp. if avoiding limits, start with the smallest fixed point of $f$ and consider the least upper bound of the sequence.

Perhaps an even better idea is to consider the fixed points of $h=f∘g=g∘f$. The set 
$$
B=\{x:h(x)\le x\}
$$ 
is non-empty, bounded and contains all potential fixed points. As before, 
$$
z=\inf B
$$ 
actually is one of the fixed points. Now both $f(z)$ and $g(z)$ are also fixed points of $h$, since $h(f(z))=f(g(f(z)))=f(h(z))=f(z)$ etc. using commutativity of the functions $f$ and $g$, so must be contained in $B$. Which means by the construction of $z$ as greatest lower bound of $B$
$$
f(z)\ge z\text{ and }g(z)\ge z.
$$
By monotonicity, $g(z)\le g(f(z))=h(z)=z$ and similarly $f(z)\le z$ follow, so $z$ is also a common fixed point of $f$ and $g$.
A: 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.
