The bounded analytic semiHyper-operators

[JmsNxn]

Has it fixed points?

The only fixed point of \( \uparrow \) I can think of are constant functions. Since

\( \uparrow \phi(z) = \phi^{\circ z}(1) \)

What about the successor function?

Oh yes it is a fixed point in the general sense, but not in the space I'm concerning myself. So yes, the successor function is a fixed point. But the successor function isn't bounded in the right half plane. I kind of implicitly meant this when I wrote that. You are right though.

The space in question is \(|\phi(z)| \le M\) for \(\Re(z) \ge 0\). I apologize for the confusion.