Does there exist a function $f:\mathbb{C} \rightarrow \mathbb{C}$ such that $f\circ f = \exp$? (We have no continuity hypothesis on $f$.)
This problem was given as an oral exam at the ENS (a school in France). I don't have the clues that the examiner might have given to the candidate.
I know a proof of this in the case where $f$ is continuous (see https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-compositional-square-root). I quote the proof here :
$f(z)$ cannot take all values, for then so does $f(f(z))$, while $e^z$ omits zero. If $f(z)$ omits a value, that has to be zero, for $e^z$ takes all other values. Thus there is an entire function $h(z)$ such that $f(z) = e^{h(z)}$ (the complex plane is simply connected). Now $ e^{h(e^{h(z)})} = e^z $ and so $h(e^{h(z)}) = z + 2{\pi}ik$ for some fixed integer $k$. Since the right hand side takes all values, so does the left hand side. So $h(z)$ takes the two values $0$ and $2{\pi}i$, say $h(a) = 0$ and $h(b) = 2{\pi}i$. Now $a + 2{\pi}ik = h(e^{h(a)}) = h(e^0) = h(e^{2{\pi}i}) = h(e^{h(b)}) = b + 2{\pi}ik$ and so $a = b$. Contradiction!
The problem is if we don't suppose $f$ is continuous, then we can't directly say $k$ is fixed. However the rest of the proof is still correct and seems a good way to start.
Can someone help?
