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I need to estimate $|x - f^{-1}(g^{-1}(f(g(x))))|$ for various values of $x$ for two smooth invertible functions $f$ and $g$ on $\mathbb{R}$ (actually some other spaces, but $\mathbb{R}$ will do.) Are there any general results on bounds for this sort of measure of non-commutativity of functions that I can look up to see what to expect and how the proofs go?

Under what conditions would repeated application of $f^{-1} \circ g^{-1} \circ f \circ g$ to $x$ revert to $x$ in finitely many steps or converge to it in the limit?

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  • $\begingroup$ Not much... but I would suggest that some insight might be gained by playing with graphs: plot $y=x$ and examples of $g, f$ in the same $xy$-plane; notation: $fx=f(x), gx=g(x)$; then pick a point $(x,0)$ on the $x$-axis; go up/down to $(x,gx)$, turn left/right to $(gx,gx)$, up/down to $(gx,fgx)$, left/right to $(g^{-1}fgx, fgx)$, up/down to $(g^{-1}fgx,g^{-1}fgx)$, left/right to $(f^{-1}g^{-1}fgx,g^{-1}fgx)$, and finally up/down to $(f^{-1}g^{-1}fgx,0)$. Using Geogebra to create some piece-wise quadratic (or linear) $f,g$ one can then modify them and see how the final point moves around. $\endgroup$ Dec 19, 2021 at 23:08

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This is really only an observation. If $f$ and $g$ are linear, the condition that a fixed point exists is $f(0)-g(0) + f'(0)g(0)-f(0)g'(0) = 0$. If this is satisfied, every point $x$ satisfies $(f^{-1})(g^{-1})fg(x) = x$. I would therefore suggest restricting to the case where $f(0) = g(0) = 0$. In simple non-linear examples I have observed only the two possibilities (a) 0 is a unique fixed point and (b) $x$ is a fixed point for every $x$. In case (a) you could look at when the sequence $x_{n+1}= (f^{-1})(g^{-1})fg(x_n)$ converges to 0.

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  • $\begingroup$ Since this is not an answer, it seems more appropriate as a comment. $\endgroup$
    – LSpice
    Feb 5, 2022 at 17:11
  • $\begingroup$ Sorry. I read "Comments are used to ask for clarification or to point out problems in the post. " and this did not fit exactly to what I wanted to add...Should I delete what I wrote? $\endgroup$
    – Derek
    Mar 8, 2022 at 14:35
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    $\begingroup$ I would say no; I think it fits better as a comment, but it's better to have it as an answer than to delete it. $\endgroup$
    – LSpice
    Mar 8, 2022 at 15:55
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I assume that by "finitely many steps" you refer to the function composition operation. If so, the general direction at which you can look is probably the theory of attractive fixed points of discrete non-autonomous dynamical systems. For example look at Navascués - New equilibria of non-autonomous discrete dynamical systems or Cánovas - On $\omega$-limit sets of non-autonomous discrete systems, or Cánovas - Recent results on non-autonomous discrete systems (just to indicate a few among many works).

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