How prove this they have a common fixed point

Question

If two continuous mappings $f$ and $g$ of an interval into ifself conumute,that is $$f(g(x))=g(f(x))$$,then they have a common fixed point?

This problem is from Mathemmatical Analysis (Zorich) PP169,Thank you everyone.

Answer 1

In the textbook by Zorich published in 2002 this exercise states the following:

If two continuous maps $f$ and $g$ of a line segment into itself commutate, i.e. $f\circ g = g \circ f$, then they have a common fixed point.

In the latest version of the textbook by Zorich (2015) this exercise is modified to almost the opposite statement:

If two continuous maps $f$ and $g$ of a line segment into itself commutate, i.e. $f\circ g = g \circ f$, then they do not always have a common fixed point, though for linear maps and polynomials in general we always have such a point.

Here is the proof for linear functions.

Let $f(x), g(x)$ are commuting linear functions. Suppose that none of them is constant or identical (otherwise our task is trivial). Then they intersect the line $e(x) = x$ in a single point. This means that both of them has a single fixed point.

Let $f(x_0) = x_0$. Suppose, that $g(x_0) = x_1 \neq x_0$. Then $f(x_1) = f(g(x_0)) = g(f(x_0)) = g(x_0) = x_1$ and $f$ has two fixed points, which is impossible for non-identical linear functions.

For polynomials I can only give some hints, because I do not have a complete proof.

Let $L = [a, b] \subset \mathbb{R},\ f = f|_L:L \to L, g = g|_L:L \to L$ are polynomial functions, $F$ and $G$ are sets of fixed points of $f$ and $g$ respectively.

1) Suppose, we have $f(x_i) = x_i \in F$ and $g(x_i) = x_j$. Then we apply the same logic, as above: $f(x_j) = f(g(x_i)) = g(f(x_i)) = g(x_i) = x_j$. This means, that $g(x_i) \in F$ for any $x_i \in F$, i.e. $g(F) \subset F$. Similarly, $\ f(G) \subset G$.

2) So if $|F| = 1$ or $|G| = 1$, then we are done. Let $|F| > 1$ and $|G| > 1$.

3) $f$ is a polynomial $\Rightarrow$ it has a finite degree $\Rightarrow$ it has a finite amount of extremums $\Rightarrow$ if $f \not \equiv x$, then $|F|$ is finite. So as $|G|$. We will not consider the case when $f \equiv x$ or $g \equiv x$ because it is trivial.

4) As $g(F) \subset F$ and $|F|$ is finite, then $g$ has a periodic point in it. This means, that $g^n(x_i) = x_i$. So $f$ and $g^n$ have a common fixed point. So as $g$ and $f^m$.

Answer 2

I can't comment, so I will post it as answer. Additional condition is that interval must have endpoints, i.e. it must be closed interval. For now I can't prove this exercise.

I suppose that map $x - \frac{1}{x}$ doesn't have fixed point in $(0,1)$ interval, and it commute with identity map, $x$, so it really doesn't work for open intervals.

Answer 3

This is only a hint, because the question is seriously lacking any sort of clarity!!!

I would have liked to have used the "usual trick" from measure theory. In three steps: first I would have proven it for constant functions, then I would have shown it for simple functions, finally for general functions.

Step 1

Suppose $g(x) = c$ where $c\in \mathbb{R}$ is a constant. And $f$ is a continuous function. From the assumption, that they commute, it follows that:

$$f(c)=f(g(x))=g(f(x))=c$$ $$g(c)=c$$

So they share a fixed point.

Step 2

Clean up your question!

Your statement isn't true in general!