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What's the official name of "commuting" functions $f:\mathbb{R}\rightarrow \mathbb{R}$, $g:\mathbb{R}^k\rightarrow \mathbb{R}$ with

$$f(g(x_1,\dots x_k)) = g(f(x_1),\dots, f(x_k))$$

For $k=1$ an invertible function $f$ and its inverse $f^{-1}$ with $f(f^{-1}(x)) = f^{-1}(f(x)) = x$ do "commute", but other examples are around:

  • $f(x) = ax$ and $g(x,y) = x + y$

  • $f(x) = x^a$ and $g(x,y) = xy$

My main question is: Is it the case, that for a given $g(x,y)$ there is essentially one $f(x)$ that commutes with $g$? If yes: How is it proved? Which requirements have to be posed on $g$ to be able to prove it?

A generalization would be functions $f:\mathbb{R}^k\rightarrow \mathbb{R}$, $g:\mathbb{R}^k\rightarrow \mathbb{R}$ with

$$f(g(x^1_1,\dots x^1_k),\dots , g(x^k_1,\dots x^k_k)) = g(f(x^1_1,\dots x^k_1),\dots , f(x^1_k,\dots x^k_k))$$

for which

  • $f(x,y) = a(x-y)$ and $g(x,y) = x + y$

  • $f(x,y) = \frac{x}{y}$ and $g(x,y) = xy$

are an example.

Hans-Peter Stricker
  • 18,159
  • The general composition of functions is $h(\underline x)=g(f_1(\underline x),\ldots,f_k(\underline x))$. This is what I use in my class on ''Computability Theory''. – Wuestenfux Sep 24 '18 at 08:33
  • This is not exactly what I asked for. – Hans-Peter Stricker Sep 24 '18 at 08:50
  • There are interesting solutions for $k=1$ already, with the assumption that the functions $f$ and $g$ are polynomials. See https://math.stackexchange.com/questions/1899400. But I don't know how to get all solutions even for $k=1$. – Batominovski Sep 24 '18 at 08:58

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