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.

enter image description here

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

Here we have a case of (3a)

enter image description here

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

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.

  • 5
    $\begingroup$ +1 If $f \circ g = g \circ f$, then we say $f$ and $g$ "commute". This is a property that is frequently useful in abstract algebra, and indeed other fields where functions are iterated (such as fixed point theory). The concept is very well-studied, but I don't know of any nice way to spot such functions visually from their graphs (though I can't claim to be an expert). Also, the function $f(x) = x$, the identity function, commutes with everything, so I wouldn't read too much into the second example! $\endgroup$ Oct 22, 2022 at 8:28
  • 2
    $\begingroup$ There seem to be this post linked to many others which could serve as an entry point math.stackexchange.com/q/11431/399263 $\endgroup$
    – zwim
    Oct 22, 2022 at 11:15
  • 1
    $\begingroup$ See also M. Zdun, A note on commutable functions. $\endgroup$
    – Kurt G.
    Oct 22, 2022 at 11:36
  • 2
    $\begingroup$ Under the name of "permutable" functions such $f\circ g=g\circ f$ have intrigued many mathematicians since the days of Ritt 1923 if not earlier see for example this paper. A fairly general class of examples is obtained from $f_a(x)=F(F^{-1}(x)+a)$ where $F$ is an invertible function and $a$ a real parameter. This gives an entire semigroup of permutable functions $f_a\circ f_b=f_b\circ f_a=f_{a+b}$. $\endgroup$
    – Kurt G.
    Oct 22, 2022 at 14:14

2 Answers 2


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 $ad+b=cb+d$.

  • $\begingroup$ What do you mean by "homogeneous" here? $\endgroup$
    – MJD
    Oct 22, 2022 at 14:44
  • $\begingroup$ I mean homogeneous in the usual sense rather than a specific mathematical definition. OP was asking how to guess examples of this condition, so I gave some examples which I'd thought of. I think those are all "homogeneous" in the sense of the dictionary. wordnik.com/words/homogeneous $\endgroup$ Oct 22, 2022 at 23:00
  • $\begingroup$ Just a small remark. Your 2nd example are inhomogeneous polynomials :) . $\endgroup$
    – Kurt G.
    Oct 23, 2022 at 4:00

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:

example drawing

$f$ 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 $f(g(x))=g(f(x))$:

Question 2
For $f(x)=1+2x$, we can look for $g(x)=ax+b$ that commutes with $f$:
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 $[f_n, f_{n+1})$:
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.

  • $\begingroup$ This seems like manually going through the process of, given the functions, determining if they commute, correct? $\endgroup$
    – evianpring
    Oct 23, 2022 at 0:36
  • $\begingroup$ @evianpring Yes. This is similar to what you could have done on your drawings: checking that $f$ and $g$ commute for each $x$. The only little improvement is that, by using also $f^{-1}$ and $g^{-1}$ graphs, getting $f(g(x))$ and $g(f(x))$ is slightly simpler. $\endgroup$ Oct 23, 2022 at 7:46
  • $\begingroup$ The issue I have is that this doesn't seem to answer the main questions posed, which have to do with general insights about functions that satisfy $f\circ g=g\circ f$. Unfortunately, your answer also doesn't provide the examples that were requested. $\endgroup$
    – evianpring
    Oct 23, 2022 at 11:25
  • $\begingroup$ @evianpring I have now completed the answer. Note however that this problem has a large literature, as was pointed out in the comments, and no "silver-bullet" answer, $\endgroup$ Oct 23, 2022 at 19:47

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