Assume $f$ is an entire non-polynomial function of arbitrarily small exponential order ('zero'th order' if you're into calling it that). Is it possible that for all $n$ we have


$$|f^{\circ n}(z)| < M\_ne^{|z|}$$


Where $f^{\circ n}$ is the $n$-fold iterate of $f$. This seems a little wonky to actually happen but I'm struggling to find a counter example. I am asking because I suspect it requires a deep result, just like proving there exists a zero'th order function which when composed with itself can be a function of exponential order. Last time, I got an answer for that question here, hopefully someone can aid me in this case as well.


I'm wondering if this could be another uniqueness criterion for polynomials in complex analysis.


Any help would be greatly appreciated.



Can iterates of a non-polynomial function be bounded by an exponential indefinitely?
It is sufficient to find a non-polynomial entire function $f$ such that $T(r) = \sup\_{|z| \leq r} |f(z)|$ satisfies $T^{\circ n}(r) = O\_n(e^r)$. Let $r \in \mathbb{R}\_+ \mapsto S(r) \in \mathbb{R}^\*\_+$ be a continuous function such that $\log S(r) / \log r$ is increasing and tends to infinity slower than any iterated logarithm. Then $S^{\circ n}(r) = O\_n(e^r)$ and we just have to construct a non-polynomial $f$ such that $T(r) \leq S(r)$.


For each $n \geq 0$ there is some $a\_n > 0$ such that $a\_n r^n \leq S(r) / (en!)$ for each $r \geq 0$. Then 
$$
f(z) = \sum\_{n \geq 0} a\_n z^n
$$
satisfies $|f(z)| \leq S(|z|)$ and is not a polynomial. Thus the answer to your question is "yes".


