# Can an entire function have every root function?

My question is an amalgamation of two previous questions. The first question I'd like to draw attention to is here. It asks whether there can exist a non trivial semigroup defined on $\mathbb{C}$

$$\phi(s,z):\mathbb{C}_{\Re(s) > 0}\times\mathbb{C} \to \mathbb{C}$$ $$\phi(s_0, \phi(s_1,z)) = \phi(s_0+s_1,z)$$

The answer is no, because such a semigroup would necessarily have a repelling fixed point for some $s_0$, and each function commutes with each other, and an entire function with a repelling fixed point can only have a countable number of functions that commute with it; which forms a contradiction. A big thanks goes to Alexandre Eremenko for referencing I.N. Baker and solving this.

The second question is found here and asks whether an entire function $g$ of finite order can have a composite square-root. The answer is yes. Again, another big thanks goes to Alexandre Eremenko for referencing I.N. Baker again, and solving this.

A great question which hybridizes the above two questions, but laying insoluble in both instances:

For an entire function $f:\mathbb{C} \to \mathbb{C}$, can there exist a sequence of entire functions $\{f_n\}_{n=1}^\infty$ such that: $$f_n(f_n(...(n\,times)...f_n(z))) = f_n^{\circ n}(z) = (f_n \circ f_n \circ ...(n\,times)...\circ f_n)(z) = f(z)$$

This first answer fails to prove this can't happen, this is only a countable list of functions that commute with $f$, so the proof provided by Baker does not work. The second answer suggests this might be possible, but only refers to a square root of $f$, not an arbitrary $n$'th root for all $n$.

Essentially I'm asking this because I've learnt that if $f:\mathbb{C} \to \mathbb{C}$ there is no valid way of constructing

$$(f \circ f \circ...(s\,times)...\circ f)(z) : \mathbb{C}_{\Re(s) > 0}\times\mathbb{C} \to \mathbb{C}$$

so I'm wondering if we can weaken this to, sure there is no semigroup, but there can be

$$(f\circ f\circ... (q\,times)...\circ f)(z):\mathbb{Q} \times \mathbb{C} \to \mathbb{C}$$

My money is on the fact this is impossible and I'm crossing my fingers that Alexandre Eremenko knows why because he's read so much Baker.

• This is a conjecture (perhaps of Baker) which is still unsolved. – Alexandre Eremenko Mar 27 '17 at 20:57
• That makes this sound even more intriguing. Is it his conjecture that no such functions $f$ exist other than linear functions, or that there could be such a function? Just to be sure I'm on the winning team. – user78249 Mar 27 '17 at 21:11
• The conjecture is that such a function does not exist, except an affine one, so the chance that you loose your money is small. – Alexandre Eremenko Mar 27 '17 at 21:20
• @AlexandreEremenko Do you know what some attempts at proving this are? The only way I can think is showing that if $q_k$ is a cauchy sequence of rationals, then $f^{\circ q_k}(z)$ uniformly converges on compact subsets of $\mathbb{C}$, giving a semigroup on an uncountable domain. I think this is rather too simple though, professional attempts are probably more clever than that. – user78249 Mar 28 '17 at 0:07
• read Baker's early papers. On compositions of entire functions, his works represent the current state of things, more or less. – Alexandre Eremenko Mar 28 '17 at 0:08