$f\circ g=g\circ f$ + increasing $\Rightarrow$ common fixed point.

Question

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 ?

Idea (not sure that works here) : Arguing by contradiction, assume that $h(x)=f(x)-g(x)$ is strictly positive on the compact $[a,b]$. Then there exist $r>0$ such that for all $x\in[a,b]$ we have $f(x)-g(x)\geq r>0$ and one can prove that $f(x)^n-g(x)^n\geq nr$. Unfortunately we don't have $f^{n}(x)\in [0,1]$ but perhaps someone can fixed it. — , May 12 '14 at 15:29
I'm not sure I agree with your initial assumption : $f-g$ may tend to $0$ even on a compact set... — Gabriel Romon, May 12 '14 at 16:01
I am just using the fact : If a function $f$ does not vanish on $I$ then $f$ is of constant sign. Oh my bad I haven't seen the assumption that $f$ and $g$ are increasing, which is not necessary.. — , May 12 '14 at 16:08
You are right, it's a very interesting question. I will search. — , May 12 '14 at 16:15

Lutz Lehmann · Accepted Answer · 2019-02-19T15:12:03.013

4

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 $g$.

edited Feb 19 '19 at 15:12

answered May 12 '14 at 12:22

Lutz Lehmann

126,666

I see why $g$ acts injectively over fixed points of $f$, but why is it surjective ? – Gabriel Romon May 12 '14 at 12:35
That is actually a good question. Perhaps "permutation" was premature. – Lutz Lehmann May 12 '14 at 12:39
See my answer below. I really don't understand your third paragraph. – Gabriel Romon May 12 '14 at 16:46
It is the same idea, just using a different function of the family of available functions. A strange anomaly seems to be that you avoid using "supremum" and "infimum" for the least upper bound and greatest lower bound? – Lutz Lehmann May 12 '14 at 18:40
LUB and GLB just convey way more sense than infimum and supremum. As I consider them to be very elaborate concepts, I prefer not to reduce them to infima or suprema. Just a fantasy of mine I guess. – Gabriel Romon May 12 '14 at 18:46
@LutzL If $f$ and $g$ aren't monotonically increasing, $ f$ and $g$ have a common fixed point? – Li Taiji Feb 19 '19 at 14:53

score 0 · Answer 2 · answered May 12 '14 at 16:45

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.

$f\circ g=g\circ f$ + increasing $\Rightarrow$ common fixed point.

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