formal solution of $f(f(z)) = g(z)$ is analytic?

182 Views Asked by At

In this question I said with an analytic function $g(z) = z+\sum_{n=2}^\infty b_n z^n$ the equation $$f(f(z)) = g(z)$$ has a formal solution $$f(z) = \sum_{n=1}^\infty c_n z^n, \qquad c_1 = 1, \qquad c_m = \frac{1}{2}(b_m - \sum_{n=2}^{m-1} c_n\sum_{\sum_{l=1}^n k_l = m}\prod_{l=1}^n c_{k_l})$$

But proving $f(z)$ is analytic, that is $c_n = \mathcal{O}(R^n)$ is not so easy.

Can you help proving it, or find some conditions for $f(z)$ to be analytic ? (in particular when $g(z) = \sin(z)$)

2

There are 2 best solutions below

5
On

I see Gottfried has answered in various places. Good.

There is a real-valued $C^\infty$ solution, and this is $C^\omega$ except at the origin. It is in Gevrey class at the origin. To be specific, the solution can be extended to a holomorphic solution in a funny diamond shaped open region with two vertices at real $0$ and $ \pi. $ This should be thought of a the intersection of a sector between rays beginning at the origin, with a sector facing backwards ending at $\pi.$ The solution simply cannot be extended around the origin; this is quite visible in Ecalle's method, where a logarithm is an essential part of things.

The best books with relevant material are Milnor Dynamics in One Complex Variable and Kuczma, Choczewski, and Ger Iterative Functional Equations.

0
On

I've just looked at the 1967 article "Non-Embeddable Functions with a Fixpoint of Multiplier 1" of I.N.Baker (Math. Zeitschr. 99, 377-- 384 (1967)) and it seems, that it answers -at least to a big part- your question. The keyword "Embeddable" means here that such function can (or cannot, when "not embeddable") have an analytic/continuous iteration, which is analoguous to your question, whether/when fractional iterates have power series with nonzero convergence radius. I've just taken a screen shot of the second page of the article, perhaps this is useful:
image