Is it possible to define a function $f$, from positive real to positive real such that $f(f(x)) = {1 \over x}$?
The motivation comes from $1990$ IMO problem $4$, which one step involves defining such a function over the positive rationals. In the countable rational space one can use the trick that divides numbers into two countable lists and map one to another using different functions.
A comment in the video of that problem suggests a new problem: whether the function can be extended to positive real numbers? Basically we need to find a way to partition the reals into two sets and have two functions mapping one set to another set in a bijective way and the composition of the two functions in either order will give ${1 \over x}$. My thought is this looks impossible but I cannot prove it.
Best Answer
Given a function $f:\mathbb R_{>0}\to\mathbb R_{>0}$, define a function $g:\mathbb R\to \mathbb R$ by $g(x)=\log f(e^x)$. Then $f$ satisfies the functional equation $f(f(x))=\frac1x$ if and only if $g$ satisfies the functional equation $g(g(x))=-x$. The latter equation is treated in great detail here:
Find a real function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = -x$?