# Super attracting fixed points have no fractional iteration

My question is really easy to state, but I'm having trouble hitting the final nail in the coffin in a proof of the result. The question concerns fractional iterations of holomorphic functions, for clarity I'll define what I mean.

Suppose we have a holomorphic function $f(\xi)$ which takes the unit disk $D$ to itself. Further let us assume it fixes zero--where in addition $f^{\circ n}(\xi) \to 0$ for all $\xi$ in the unit disk as $n\to \infty$.

A fractional iteration (defined in the sense I am concerned with) is a holomorphic function $f(z,\xi)$ (holomorphic in $z$ and $\xi$) such that $f(z,\xi) : S \times D \to D$ where $S$ is open and connected, and furthermore $S$ is closed under addition of its elements. Additionally we are given the semigroup property

$$f(z_0,f(z_1,\xi)) = f(z_0 + z_1,\xi)$$

and the initial conditions $f(1,\xi) = f(\xi)$ and $f(z,0) = 0$.

In the case where $f'(\xi_0) = \lambda$ and $0 < | \lambda| < 1$, it is possible to produce countably infinite versions of these complex iterations where $S$ is some half plane containing $\mathbb{R}^+$ and $S$ depends on our choice of which fractional iteration we are speaking of. But this is not the case I am concerned with. Instead I am concerned with the case where $\lambda = 0$, i.e: the case where $0$ is a super attracting fixed point. My intuition is telling me that no such fractional iteration exists in such a case.

To explain my reasoning, simply consider $f(\xi) = \xi^2$ wherein a natural choice for an iteration is $f(z,\xi) = \xi^{2^z}$, however this function is no longer holomorphic in a neighbourhood of $0$ in $\xi$. We get branch cuts speaking simply.

Extending this idea, let us assume without loss of generality the Taylor expansion of $f$ starts with

$$f(\xi) = \xi^n + ...$$

so that $$f(f(...(k\,times)...f(\xi) = \xi^{n^k} + ...$$

giving the notion that a suitable $f(z,\xi)$ will probably start out

$$f(z,\xi) = \xi^{n^z} +...$$

implying no fractional iteration exists. However this is far from a proof. I've boiled the question down into one single idea. Assume that such a fractional iteration exists and consider the Böttcher function. For those who may not remember this, or know of it, it is defined by the following limit

$$F(\xi) = \lim_{k\to\infty} \sqrt[n^k]{f^{\circ k}(\xi)}$$

wherein $F:D \to D$ and

$$F(f(\xi)) = F(\xi)^n$$

Therein the final result will follow if we can show that our candidate fractional iteration $f(z,\xi)$ satisfies

$$F(f(z,\xi)) = F(\xi)^{n^z}$$

for some branch of this multivalued function, therein showing that $f(z,\xi)$ cannot be holomorphic in a neighbourhood of zero. This reduces the question to something smaller, that there exists no fractional iteration of $\xi^n$--but even showing this is causing me grief. I feel like I'm missing something small, but I'm not sure what I am missing.

All in all, I'm asking either for a proof or a reference or suggestions on where to look to solve this odd looking problem. It would also be really neat if someone could find a counter example to this statement--an example of a holomorphic function with a super attracting fixed point with a fractional iteration attached to it.

• It seems to me that you only need to consider the Taylor expansion that your "fractional iterate" would have near the origin, to get your contradiction. Dec 6 '16 at 22:45
• @LasseRempe-Gillen I was considering that, but was having difficulty actually constructing the contradiction, so I thought maybe it would be more straightforward with the Bottcher coordinate, as it behaves like the first term of the Taylor series.
– user78249
Dec 6 '16 at 23:50

I suppose this result (non-existence of fractional iterates) is known and very old. From continuity $z\mapsto f(z,\zeta)$ follows that the fixed point at $0$ must be independent of $z$.

Suppose, for example that $f(\zeta)=a\zeta^2+\ldots$ Taking $z=1/2$ and denoting $g(\zeta)=f(1/2,\zeta)$ we obtain $f=g\circ g$. This $f$ is locally $2$-to-$1$ near zero. So what can be the local valency of $g$?? Contradiction.

If $f=a\zeta^m$, use $z=1/k$ such that there is no integer $n$ with $n^k=m$.

If I do not mistake, this fact is due to Lucian Emile Böttcher (very beginning of 20th century). But I am lazy to look at his papers, most of them not easily available.

• Oh, that is a very good and simple proof. I knew I was missing something. I was constantly assuming a Taylor series expansion, and trying to create a contradiction through some algebraic manipulation, forcing the solution to be $\xi^{2^z}$. I never thought of this topological (would you call this topological?) proof. That's much cleaner. And, I assumed this is an old result, but it's very hard to find literature on fractional iterations. I'm always caught in a bog of unrelated subjects.
– user78249
Dec 7 '16 at 0:41
• @Joseph O'Rourke: Thanks. I am using Windows. Of course I can go to the "German keyboard online" and paste and copy, but as I said I am too lazy:-) Dec 7 '16 at 1:17
• Lol, I just googled Bottcher and knew there'd be a wikipedia article where they typefaced it properly and copied and pasted the typeface.
– user78249
Dec 7 '16 at 2:09
• @james.nixon This is completely equivalent to a Taylor series consideration. Dec 11 '16 at 23:00
• I realize that now, except I never considered making a contradiction by fixing $z$ and looking at a single case, I tried to create a contradiction by forcing the first coefficient to be $\xi^{n^z}$ for all $z$. I knew it was something obvious I was missing. In a sense, I was looking at it globally in $z$.
– user78249
Dec 11 '16 at 23:31