Is there an interesting (graphical) intuition about what the condition $f\circ g=g\circ f$ implies?

Question

I have a few questions about continuous functions $f$ and $g$ that satisfy the condition

$$f\circ g=g\circ f\tag{1}$$

(1) Is there an interesting (graphical) intuition about what this condition implies? I have to think a lot to come up with an example that isn't inverse functions. Is there some insight that would permit coming up with examples easier?

(2) Given a function such as $f(x)=1+2x$ how do we find a $g$ that together with $f$ satisfies $(1)$?

(3) If we impose the additional condition that $0\leq f,g\leq 1$ for all $x\in [0,1]$, what are some examples in which

(3a) $f(1)\neq g(1)$?

(3b) in particular, $f(1)$ and $g(1)$ are in $(0,1)$ and $f(1)\neq g(1)$?

Here are examples I came up with that aren't inverses.

$f(x)=\sqrt{x}$, $g(x)=x^4$, $f(g(x))=g(f(x))=x^2$

Here we have a case of (3a)

$f(x)=x$, $g(x)=1-x$, $f(g(x))=g(f(x))=1-x$

I haven't been able to come up with a case (3b).

Some Context

This question came up while solving a problem in Chapter 22, "Infinite Sequences", from Spivak's Calculus.

The task there was to prove that if

• $f$ and $g$ are continuous functions on $[0,1]$
• $f\circ g=g\circ f$
• $0\leq f(x),g(x)\leq 1$ for all $x\in [0,1]$
• $f$ increasing

then $f$ and $g$ have a common fixed point.

I was able to prove this, but even so I don't feel like I have a good feel for what the assumptions mean.

Hence, my questions.

Answer 1

Fairly obvious examples include

• $f(x)=x^a$, $g(x)=x^b$, $f\circ g(x)=g\circ f(x)=x^{ab}$;
• $f(x)=x+a$, $g(x)=x+b$, $f\circ g(x)=g\circ f(x)=x+a+b$;
• $f(x)=ax$, $g(x)=bx$, $f\circ g(x)=g\circ f(x)=abx$, etc.

Is there some insight that would permit coming up with examples easier?

I think anything which is homogeneous fits the bill. As soon as you go into polynomials it starts to fail, so $f(x)=ax+b$ and $g(x)=cx+d$ fail since $f(g(x))=a(cx+d)+b\neq g(f(x))=c(ax+b)+d$ unless $ad+b=cb+d$.

Answer 2

I first propose answers to the 3 questions, then give a general way to construct a $g$ that commutes with a given $f$ (with some conditions), while still defining freely $g$ on $[0,f(0))$.
Note that for any $f$ there are solutions such as $g=f \circ f$, $g=f \circ f \circ f$, etc.

Question 1
Here is a graphical interpretation. Sorry for the manual drawing.

Draw the graphs of $f, g, f^{-1}, g^{-1}$. For simplicity of drawing I assume $f$ and $g$ to be $[0,1] \to [0,1]$. $f^{-1}$ and $g^{-1}$ are symmetrics of $f$ and $g$ by the first diagonal.

Then $f(g(x))$ and $g(f(x))$ are constructed as follows:

$f$ and $g$ commute if $f(g(x))$ and $g(f(x))$ reach the same abscissa; which is almost but not quite the case on the drawing.

In addition, because this is true $\forall x$, there is a relation between the slopes of 4 points in the diagram: the slope of $f$ on abscissa $x$ divided by the slope of $g^{-1}$ on abscissa $g(f(x))$ equals the slope of $g$ on abscissa $x$ divided by the slope of $f^{-1}$ on abscissa $f(g(x))$. Which is in fact the differentiation of $f(g(x))=g(f(x))$:
$g'(x)f'(g(x))=f'(x)g'(f(x))$
$g'(x)/(f^{-1})'(f(g(x))=f'(x)/(g^{-1})'(g(f(x))$

Question 2
For $f(x)=1+2x$, we can look for $g(x)=ax+b$ that commutes with $f$:
$1+2(ax+b)=a(1+2x)+b$
$b=a-1$
Hence $\forall a, g(x)=ax+a-1$ commutes with $f$.

Question 3
An example fulfilling (3b) is easy with affine functions:
for $f(x)=\frac {x+1} 3$, any $g(x)=ax+\frac {1-a} 2$ commutes with $f$.
So we can choose $g(x)=\frac x 4 + \frac 3 8$, so that $\forall x \in [0,1], 0 < g(x) < 1$, and $g(1) \ne f(1)$.

In general
If $f$ is a strictly increasing continuous function with $\forall x, f(x) > x$, it is possible to build continuous functions $g$ that commute with $f$ while choosing $g$ freely on $[0, f(0))$. Here is how.

We'll call $f_n=f(f_{n-1})$ with $f_0=0$. $(f_n)_{n\in\mathbb{N}}$ is a strictly increasing sequence. This makes a partition of $\mathbb{R}^+$ into intervals $[f_n, f_{n+1})$.

First, define $g$ on $[f_0,f_1)$ as any continuous strictly increasing function, such that $g(f_1)=f(g(f_0))$. We'll call $g_n=g(f_n)$.
Then $g_n=g(f(f_{n-1}))$ which must be equal to $f(g(f_{n-1}))=f(g_{n-1})$, so $(g_n)_{n\in\mathbb{N}}$ is defined by $g_0=g(f_0)$ (freely chosen) and $g_n=f(g_{n-1})$.

Then we define $g$ by recurrence on intervals $[f_n, f_{n+1})$:
$g(x)=f(g(f^{-1}(x)))$
As $x \in [f_n, f_{n+1})=[f(f_{n-1}),f(f_n))$,
$f^{-1}(x) \in [f_{n-1},f_n)$,
$g(f^{-1}(x))$ has been defined at the previous recurrence step,
so $g(x)=f(g(f^{-1}(x)))$ is correctly defined.

$g$ is continuous in $(f_n, f_{n+1})$ by recurrence: it is constructed as continuous on $(f_0, f_1)$, and continuity on $(f_{n-1}, f_n)$ implies continuity on $(f_n, f_{n+1})$ by composition of continuous functions: $g(x)=f(g(f^{-1}(x)))$.
For the same reason it is continuous at right of $f_n$.
$g$ is continuous at left of $f_n$ by recurrence: we have chosen $g(f_1)=f(g(f_0))=f(g(f^{-1}(f1)))$, and continuity at left of $f_{n-1}$ implies continuity at left of $f_n$ by the definition $g(x)=f(g(f^{-1}(x)))$.

Then $\forall x, g(f(x))=f(g(f^{-1}(f(x))))=f(g(x))$: $f$ and $g$ commute.

This has defined $g$ on $\mathbb{R}^+$. $g$ can be defined on $\mathbb{R}^-$ by the same procedure, going downwards.