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.