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Are there real-valued functions $f$ and $g$ which are neither each other's inverses, the identity, nor linear, yet exhibit the behaviour $$f\circ g = g \circ f?$$

Examples such as $f(x) = 2x$ and $g(x)=3x$ are "trivial" in this sense.

Moreover, given a function $f$, can one go about obtaining an example of a function $g$ which commutes with $f$?

I suppose this would be similar to fixed point iteration? E.g. if $f\colon\mathbb R\smallsetminus\{1\}\to\mathbb R$ is defined by $f(x) = 2x/(1-x)$, I would need a function $g$ such that $$g(x) = f^{-1}\circ g\circ f =\frac{g(\frac{2x}{1-x})}{2+g(\frac{2x}{1-x})},$$ so maybe choosing an appropriate "starting function" $g_0$ and finding a fixed point of $g_{n+1} \mapsto f^{-1}\circ g_n\circ f$ would be a possible strategy, but I can't seem to find a suitable $g_0$.

Luke Collins
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    $f(x) = x^n, g(x) = x^m$ works for reals $n, m$. – balddraz May 04 '19 at 13:14
  • https://math.stackexchange.com/questions/1194115/when-does-function-composition-commute and https://math.stackexchange.com/questions/2126116/an-example-of-two-functions-commuting-with-each-other and https://math.stackexchange.com/questions/1664492/when-actually-fgx-gfx-holds and https://math.stackexchange.com/questions/1391401/is-is-true-that-f-circ-g-g-circ-f-implies-f-is-linear-or-g-is-linear and probably a few more. – Gerry Myerson May 04 '19 at 13:29
  • How about $f(x)=2x$ and $g$ a discontinuous solution of the additivity equation $g(x+y)=g(x)+g(y)$? Is that too trivial for you? Howe trivial can it be if you need the axiom of choice to prove it exists? – bof May 04 '19 at 14:37

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