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Find all analytic functions $f:(1,\infty)\to(1,\infty)$ satisfying $f(f(x))=x^{\ln^3x}$

@青青子衿 found two solutions $x^{\ln x},\exp\left(\frac{1}{\ln^2x}\right)$.

Is there any other solutions?

hbghlyj
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    Without more conditions there's probably infinitely many solutions. Are you looking for continuous functions? or analytic functions? – jjagmath Oct 05 '23 at 16:56
  • @jjagmath Thanks, I've edited the question. – hbghlyj Oct 05 '23 at 17:01
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    Meaning no disrespect, I wouldn't be so fast to accept. It still remains to be seen whether or not the only globally analytic functions with $g(g(x))=x^4$ are the stated ones – FShrike Oct 05 '23 at 17:26
  • @FShrike Yes, the problem is not resolved completely. – hbghlyj Oct 05 '23 at 17:39

1 Answers1

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Observe that $x^{\log^3 x} = \exp(\log^4(x))$ and the two given solutions can be written as $\exp(\log^2(x))$ and $\exp(\log^{-2}(x))$.

It's clear now that we need to conjugate by $\exp$ to work on an easier space:

Set $u = \log x$ so $x = \exp u$ and $g = \log\circ f\circ \exp$

Then $g\circ g = \log\circ f\circ f\circ \exp$ and the equation to solve transforms into $(g\circ g)(u) = u^4$ and we are looking for analytic solutions $g:(0,\infty) \to (0,\infty)$. The two given solutions are $g(u)=u^2$ and $g(u) = u^{-2}$.

At this point I'd guess that those are the only analytic solutions to $(g\circ g)(u) = u^4$ and hence the original equation has only the two given solutions.

Edit:

Conjugating even further we can transform the equation in $(h\circ h)(v)=4v$ with $h:\Bbb R \to \Bbb R$ and now it's easier to see that the only analytic solutions are $h(v)=2v$, $h(v)=-2v$.

jjagmath
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