I have a continuous function $q:\mathbb{R}^+ \to \mathbb{R}^+$. An interesting property of this function is that

$$F(s) = \frac{e^{-q(s)}q(s+1)}{1-e^{s-q(s)}}$$

which also takes $\mathbb{R}^+ \to \mathbb{R}^+$, can be analytically continued to an entire function on $\mathbb{C}$.

Because of this, I wonder:

Must $q$ be analytic on $\mathbb{R}^+$?

If it helps, $F(s)$ satisfies the flippant equation:

$$F(s) = e^{F(s-1)e^{s-1}}$$

  • 1
    $\begingroup$ Are you sure there is no typo? Why not write it easier as $F(s) = \dfrac{ q(s+1)}{e^{ q(s)}-e^{s }}$ then ? But no way to derive your flippant equation. $\endgroup$
    – Wolfgang
    Feb 16, 2019 at 9:34
  • $\begingroup$ It's not hard to see that any continuous $\bar{q}: [0, 1] \rightarrow \mathbb{R}^+$ such that $q(1) = F(0) (e^{q(0)} - 1)$ can be extended to some continuous $q: \mathbb{R}^{\geq 0} \rightarrow \mathbb{R}$ that satisfies the whole equation, by induction on the integral part of $s$. $\endgroup$
    – user44191
    Feb 16, 2019 at 11:36
  • $\begingroup$ Is your flippant equation actually a functional equation? $\endgroup$
    – LSpice
    Mar 9, 2020 at 3:39

1 Answer 1


No. $F$ does not put enough constraints on $q$.

Suppose that $F$ is a given analytic function, and let $q_0:[0, 1]\to \mathbb R_+$ be continuous. Extend $q_0$ to a function on $\mathbb R_+$ using the recurrence $$q(s+1) = (e^{q(s)}-e^s)F(s).$$ In general, $q$ will not be continuous or positive. One an write down conditions for that, but they are not needed to answer this question.

Let $q$ be one function that satisfies all assumptions. Let $q'_0=q|_{(0,1]} + g$, where $g$ is a non-negative continuous function on $[0,1]$ with $g'(0)=g'(1)=0$, CBS extend $q'_0$ to a function on $\mathbb R_+$ as sketched above. If $q$ is analytic but $g$ not, then $q'$ will not be analytic either.

  • $\begingroup$ For example, take $F=0$. $\endgroup$ Feb 16, 2019 at 13:39
  • $\begingroup$ @AlexandreEremenko you are right, that is the easiest example, unless the OP wants $F$ to be strictly positive. $\endgroup$
    – jarauh
    Feb 16, 2019 at 17:28

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