Must $q$ be analytic?

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:

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

• 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. – Wolfgang Feb 16 at 9:34
• 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 $s$. – user44191 Feb 16 at 11:36

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