Solve $g(g(x))=f(x)$ If $f(x)$ is a continuous and monotonically increasing function on an interval $(0,∞)$ and $f(x)>0$ for every $x>0$, then does there always exist a continuous and monotonically increasing function $g(x)$ on $(0,∞)$ so that for every $x\in (0,∞),g(x)\in (0,∞)$ and $g(g(x))=f(x)$?
If $f(x)=c~x^k~(c>0,k>0),$and $g(x)=r~x^{\sqrt{k}},$where $c=r^{\sqrt{k}+1},$then $g(g(x))=f(x)~(x>0).$
But how to solve it in general conditions?Thanks in advance!
 A: EDDDITTT: the answer is YES. This is Theorem 11.2.2 on pages 424-423 of the KCG book mentioned below. I think I will put the jpeg at the end so that my text is readable.
A preliminary answer. The book you want is Iterative Functional Equations by Kuczma, Choczewski, and Ger. I got a used copy at a very reasonable price. 
I put several items about the same problem for real analytic functions at BAKER. The person who initiated this was Helmuth Kneser, father of Martin Kneser. He showed that there was a real analytic solution solving $g(g(x)) = e^x$ on the entire real line, meaning that it extends to a holomorphic function on an open set surrounding the real axis. Note that the question of commutation is exactly right: Baker re-worded the problem into one of commuting functions and formal power series. For the most difficult case an elaborate procedure was given by Ecalle in about 1973, see QUESTION and my own ANSWER.  
The main obstruction in the continuous case is that you cannot have fixed points of the target function with negative slope. However, you have said monotonic increasing. In case you have something like $x+ \sin x$ I think there is material in the KCG book for solving on each interval between fixpoints, but i will need to check that. If your function has no fixpoints or, say, only one, I think you can do it. But there may be no simple recipe or formula, just an existence result.  
=-=-=-=-=-=-=-=

=-=-=-=-=-=-=-=
