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.

up vote 2 down vote accepted

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".

  • 1
    "increasing and tends to infinity slower than any iterated logarithm": how do you know this exists? – Jean Duchon Oct 18 '17 at 15:36
  • Everything seems great! Took me a little while to get the punchline, but it seems straight forward now. I also am curious though how we know there is a function that tends to infinity slower than any iterated logarithm. I think I saw a proof by Ramanujan a while back (but I can't be sure if it was phrased like this, or if the result was a little different). I vaguely remember it being a proof that the iterated logarithms can't partition the growth of functions (i.e: given $f$ it isn't necessarily true that there exists $n$ with $\log^{\circ n} \le f$). Do you have a reference for this fact? – user78249 Oct 18 '17 at 17:20
  • If $(a_n)_n$ is an increasing sequence growing faster than any exponential, then one can take a continuous piecewise linear function $g$ such that $g(a_n) = n$, and take $S(r) = r^{g(r)}$. For example $a_n = \ ^n n$ works (tetration operation). – js21 Oct 19 '17 at 12:20
  • any exponential --> any iterated exponential (correction) – js21 Oct 19 '17 at 12:28
  • For some reason I was thinking $S(r)$ needed to be analytic. Don't know why I thought that... I do think that's a proof of Ramanujan though, that one can construct analytic functions that do this. – user78249 Oct 28 '17 at 7:52

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