# function $\varphi$ such that $\varphi (\varphi (u))=\exp(u)$

Here it's cited:

the existence of the holomorphic function $\varphi$ such that $\varphi (\varphi (u))=\exp(u)$ had been demonstrated in 1950 by Hellmuth Kneser.

However I can't find the definition of the function $\varphi(u)$ anywhere.

Does this function truly exist? And if so, What is the function?

• Yes, it exists, since its existence has been proven by Kneser. Are you asking instead if we know it explicitly (constructive proof), or if the proof of its existence is non-constructive? – Clement C. Nov 10 '17 at 18:09
• @ClementC. Sorry should have been more clear, I'm wondering what the function actually is defined as. Or if it only exists abstractly. – Graviton Nov 10 '17 at 18:10
• I guess the main question then is: do you speak German? (to read the original article), or do you speak French? (to read the review on MathSciNet). – Clement C. Nov 10 '17 at 18:11
• In the latter, it is written " The solutions of the functional equation $\varphi\circ \varphi =f$, where $f$ is given and $\varphi$ is the unknown function, is of the form $$\varphi(x) = \psi^{-1}\!\left(\psi(x)+\frac{1}{2}\beta\right)$$ where $\psi$ is a solution of the Abel equation $\psi(f(x)) = \psi(x)+\beta$ where $f$ and $\beta$ are given; the inverse function $\psi^{-1}$ needing to be suitably defined. [...]" – Clement C. Nov 10 '17 at 18:15
• There is a thorough discussion on MathOverflow: f(f(x))=exp(x)-1 and other functions “just in the middle” between linear and exponential – user357151 Nov 12 '17 at 4:14