Under what conditions, if any, can I state the following?
If $f(f(x))$ is a linear fractional transformation, then so is $f(x)$.
Compositional roots of linear fractional transformations
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real-analysis
complex-analysis
analysis
mobius-transformation
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0What kind of conditions are you looking for? – 2017-01-06
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0Any, really. Context:I asked students if it's ever true that $f^{-1}(x) = \frac{1}{f(x)}$, as students often accidentally do. Composing with $f(x)$ inside, I arrived at the condition $f(f(x)) = 1/x$. Trying to work on $f(x)$ at all from there. – 2017-01-06
1 Answers
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For your original question (in the comments):
Suppose $f(f(x)) = 1/x.$ Notice that $x\rightarrow 1/x$ is an involution, so as an element of $\mathbb{S}_\mathbb{R},$ it can be written as an (uncountable) product of disjoint transpositions, plus a fixed point at $1.$. What could the cycle type of $f$ be ($f$ has to be $1-1,$ since $f\circ f$ is)? The answer is: product of 4-cycles, and a fixed point. So (let's assume for simplicity that the domain of your function is $(0, \infty)$), then group numbers in $(0, 1)$ into pairs $(a, b)$ in your favorite way, and then let $f(a) = b, f(b) = 1/a, f(1/a) = 1/b, f(1/b)=a.$ Can you make it continuous? Probably.
For an arbitrary linear fractional, a similar approach works. Whether you can find $f$ to be continuous is, of course, a whole'nother matter. (and, I just realized that I had reading comprehension issues - you want $f$ to be linear fractional - that depends on the dynamics of $f\circ f,$ just as above.
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0Thanks, that helps some. Can you delineate/link to what $\mathbb{S}_\mathbb{R}$ is? – 2017-01-06
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0$\mathbb{S}_{\mathbb{R}} is just the group of bijections from $\mathbb{R}$ to itself (of course, it does not have to be all of $\mathbb{R}.$ – 2017-01-06