I think I can clarify some confusion. Daniel is looking for the functor:

$$

\iota f = F\\

$$

Where \(F(s+1) = f(F(s))\).

Him asking if this is analytic, is a tad incorrect. But a very reasonable question. As it asks if there is some analytic operation which sends \(f\) to \(F\). This would mean we have something like:

$$

\begin{align}

F &= \int_\gamma h(z,f)\,dz\\

F &= \sum_{n=0}^\infty q_n f^n\\

\end{align}

$$

Where by, \(F\) is a function of \(f\). We can write this \(F[f]\). And we ask that:

$$

\lim_{\epsilon \to 0}\frac{F[f + \epsilon h] - F[f]}{\epsilon} = G[f]\\

$$

Is a convergent limit, for an arbitrary test function \(h\). This is common in functional analysis, and I believe that is what Daniel is asking. And by saying analytic, he's basically just abusing the language, but it means the same thing. It is analytic of the function argument, not of the argument of the function.

As an answer, YES DANIEL. If \(1 < b < \eta\), then the successorship operator is "analytic"! This actually isn't that hard to prove! (Schrodinger mechanics save us along every step).

EDIT::::

Proof :

Take:

$$

F[f](s) = f^{\circ s}(1) = \frac{d^{s}}{dw^{s}}\Big{|}_{w=0} \sum_{n=0}^\infty f^{\circ n}(1)\frac{w^n}{n!}\\

$$

The operator \(\frac{d^{s}}{dw^{s}}\Big{|}_{w=0}\) is a bounded, and hence continuous, operator on Ramanujan's space--it's just a fancy Mellin transform. If we sub for \(f_\epsilon = f + \epsilon h\), for some entire function \(h\), and \(\epsilon\) is arbitrarily small, then:

$$

F[f_\epsilon](s) = (f+\epsilon h)^{\circ s}(1) \\

$$

Then this converges just as well (though the domain problem can be tricky). Additionally, it's really just busy work to show that:

$$

\frac{F[f_\epsilon](s) - F[f](s)}{\epsilon} = G[f] + O (\epsilon)\\

$$

By which, YESS, the functor "taking us to the super function" is "analytic". Though your language is wrong, Daniel

To those of you who may still be confused. This is essentially the statement, that if \(h\) is an entire function (for our purposes with \(1 < b < \eta\), it is bounded on \(\Re(z) > 0\)), \(\epsilon\) is arbitrarily small; that:

$$

(f+\epsilon h)^{\circ s}(1) = f^{\circ s}(1) + O(\epsilon)\\

$$

So that epsilon shifts in the base function we are taking the super operator of, creates epsilon shifts in the super function. We can write this differentially, and since this is in many ways a complex "functional/functor" derivative, we can call it "analytic".

This shit is much more common in physics, and Laplacian stuff, or old Eulerian mechanics, but still even then they would say "F[f] is analytic in f"--though it means something a bit different to them. But essentially the statement is that \(F[f]\) is "complex differentiable" in "f", therefore it is "analytic". This is absolutely true. Daniel's just chosen weird language.

A good example can be written using \(1 < b < \eta\). Let's write:

$$

f(z) = b \uparrow^n z\\

$$

And:

$$

f_\epsilon(z) = (b+\epsilon) \uparrow^n z\\

$$

One can rigorously show that:

$$

\frac{f_\epsilon(z) - f(z)}{\epsilon} \to G(z)\\

$$

As \(\epsilon \to 0\). Which is essentially the statement that:

$$

f_\epsilon(z) = f(z) + \epsilon G(z) + O(\epsilon^2)\\

$$

Which is largely considered the statement of differentiability. But we are complex oriented, and differentiability for complex variables is "analytic".

And now when \(n \mapsto n+1\), the same result appears. Whereby:

$$

F[f_\epsilon] = (b+\epsilon)\uparrow^{n+1} z\\

$$

And not only does this display analytic behaviour in \(z\), it also displays analytic behaviour like \(f_\epsilon \to f\)--when taking limits in this manner. And we can say:

$$

\begin{align}

F[f_\epsilon] &= F[f] + O(\epsilon)\\

(b+\epsilon) \uparrow^{n+1} z &= b \uparrow^{n+1} z + O(\epsilon)\\

\end{align}

$$

And if we take:

$$

G[f](z) = \lim_{\epsilon \to 0}\frac{F[f_\epsilon](z) - F[f](z)}{\epsilon}\\

$$

Then:

$$

F[f_\epsilon](z) = F[f](z) + \epsilon G[f](z) + O(\epsilon^2)\\

$$

I still agree entirely with everything Mphlee, and Bo said. They're totally correct in their analysis, Daniel was not clear. But I'm pretty sure this is what he was asking, he just lacked the vocabulary to ask it.

TL;DR

The super function operator is analytic (if you choose an older meaning of analytic)--but I only show it for the bounded case.