# Smoothness in Ecalle's method for fractional iterates

Some four years ago I answered my own question on fractional iteration, concluding that there is a half iterate of sine, that is $f(f(x)) = \sin x,$ which is real analytic for $0 < x < \pi$ but cannot be extended to the complexes in a neighborhood of the origin, there is a stubborn logarithm term. Does the formal power series solution to $f(f(x))= \sin( x)$ converge?

Yesterday I did the same thing for $x + x^2,$ where the conclusion is analyticity for $x > 0$ and continuity at $0.$ Having had practice, i made a better tutorial of this answer. How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$?

Let's see, I put a bunch of related material at WEB PAGE. In particular, Jean Ecalle's method is on pages 346-347 and 351-352, references page 517; he did publish as part of Theorie iterative: Introduction a la theorie des invariants holomorphs, J. Math. Pures Appl. vol. 54 (1975), pages 183-258, I have not seen that one...

So, here is the question, very natural I think. I have some holomorphic target function $g(z)$ which is real valued along the real line, then has fixpoint $g(0) = 0$ with $g'(0) = 1,$ which is what makes the problem difficult. Then I have a real $C^1$ function $f(x)$ with $f(0) = 0, f'(0) = 1$ and $f(f(x)) = g(x)$ along the real line, constructed with Ecalle's method, so $f$ is holomorphic on segments containing $0 < x < A$ and $-B < x < 0.$ IS IT TRUE that the real function $f(x)$ is $C^\infty?$ I would like that.

Put another way, one may easily find the formal power series for a half iterate around $0,$ that is what I did four years ago. It turns out that this series has radius of convergence $0,$ this is Theorem 8.5.3 on page 347 of the KCG book. It would be nice if the formal power series displayed derivatives of all orders for the half iterate, and if Ecalle's solution matched all of that at $0.$

I guess I should not paste these here, but I just put all the directly relevant bits from the book by Kuczma, Choczewski, and Ger onto two jpegs, I can email those. Came out a little smaller than I wanted, don't know why the scanner did that; clear, though.

• It’s not in any sense a deep fact or result, but the corresponding $p$-adic question for the $p$-adic analytic function $x+x^2$ is easy to solve. The $1/n$-th iterate exists and is analytic in the open unit disk of $\mathbb C_p$ as long as $n$ is prime to $p$. – Lubin Aug 30 '14 at 21:20
• @Lubin, thanks. Reminds me of the results of Baker and his student Liverpool, (and evidently Ecalle, independent) that either a function has no fractional iterates, or a $1/n$th fractional iterate and nothing smaller, or a $w$-th iterate for any nonzero complex number $w,$ in which case the thing was conjugate to a Moebius transformation in the first place. – Will Jagy Aug 30 '14 at 21:27
• Please let me see your jpegs. I think my email occurs in my profile-page. – Gottfried Helms Aug 31 '14 at 6:53
• Everything is fine. Thank you very much! – Gottfried Helms Aug 31 '14 at 19:34
• @WillJagy, the $p$-adic situation is most nearly parallel to the last clause in your response to my comment. (But not precisely!) Anyhow, this is a case where $p$-adic analysis is much less deep than complex. – Lubin Sep 1 '14 at 2:04

I heard back from Prof. Ecalle. If we have real analytic $f(x)$ with $A \neq 0$ and $$f(x) = x + A x^{p+1} + o(x^{p+1}),$$ then there are natural fractional iterates $g(x),$ any rational or real order, and these are $C^\infty$ at the origin and of Gevrey class $1/p,$ which result from $$g\left( x^{1/p} \right)$$ being the Laplace transform of an analytic function with no more than exponential growth at $\infty.$
The Gevrey bound is: $$\left| \frac{g^{(n)}(0)}{n!} \right| < C_0 \; \; C_1^n \; \; (n/p)!$$ where the last item might need the Gamma function if $n/p$ is not an integer.
For references, he gave item 7 at http://www.math.u-psud.fr/~ecalle/publi.html and then Example 2 (with $\nu = 1$ in (2.5.18)) on pages 106-107 of item 19, editor Dana Schlomiuk,1993. Google books show page 106 but hides page 107. Not sure the name Gevrey comes up in either publication.