This question, although appearing deceptively easy, has resisted many attacks against it. The question, being simple to state, is something rather non-trivial that is rather crucial towards more general work. Rather than working around the question (which is kind of possible in the scenario I'm in), I thought I'd post it here to see if someone else has a crack at it and it breaks.


Consider a function $\phi$ that is holomorphic on an open set $G$ sending to $G$. Let us assume that $\phi(0) = 0$ and $0<|\phi'(0)| < 1$ and $\phi^{\circ n}(\xi) \to 0$ as $n\to\infty$ for all $\xi \in G$. Then necessarily the sum


$$\sum\_{n=0}^\infty \phi^{\circ n}(\xi)$$ 


converges normally on compact subsets of $G$. This is simple to show, as for all compacts of $G$, for all $\epsilon > 0$ there exists $N$ such when $n > N$ 


$$|\frac{\phi^{\circ n}(\xi)}{\phi'(0)^n} - \Psi(\xi)| < \epsilon$$ 


so that


$$|\phi^{\circ n}(\xi)| < |\phi'(0)|^n (||\Psi(\xi)||\_\infty + \epsilon)<Cr^n\,\,\,\,\,0 < r < 1$$


Now the question I pose is a slight alteration of this problem. Let us assume $\phi$ is holomorphic on $G$ sending $G$ to itself, $\phi(0) = 0$ and $\phi'(0) = 1$. Let us take $\mathcal{S}$ to be the set such that $\phi^{\circ n}(\xi) \to 0$ as $n \to \infty$ for all $\xi \in \mathcal{S}$.


Does


$$\sum\_{n=0}^\infty \phi^{\circ n}(\xi)$$


converge for $\xi \in \mathcal{S}$? Convergence can be pointwise, normal, or uniform. Pointwise would be sufficient but normal or uniform would be even better.


I do believe this follows (it seems to work in most cases that I've tried). Letting $\eta = e^{1/e}$ for example, if $f(x) = \eta^{x+e} - e$ then $f(0) = 0$ and $f'(0) = 1$. The line $(-\infty, 0]$ satisfies $f^{\circ n}(x) \to 0$ for all $x \in \mathbb{R}$.


The function $$\sum\_{n=0}^\infty \eta^{\eta^{...(n\,times)...^{\eta^{x}}}}-e$$ converges for $x \in (-\infty,e]$.


I thought using the contraction property might work, that $|\phi(\xi)| \le |\xi|$ but I haven't gotten much luck. Namely for all $1 >|\xi| > \delta$ we have a $0 < q\_\delta < 1$ such that $|\phi(\xi)| \le q\_\delta|\xi|$ but as $\delta \to 0$ it is necessary $q\_\delta \to 1$ so that $|\phi^{\circ n}(\xi)| < (\prod\_{i=1}^n q\_{\delta\_i}) |\xi|$ with $|\phi(\xi)| > \delta\_{i+1}$ when $1>|\xi| > \delta\_i$. therefore we would need to show $\sum\_{n=0}^\infty \prod\_{i=1}^n q\_{\delta\_i} < \infty$, and I can't think of any plausible manner of proving this.


Any help, comments, suggestions or further questions on what else I can say are welcome and greatly appreciated. Thanks a bunch, I hope someone has an answer or suggestion for this.



Summability of iterates of analytic function
No. For example take $f(z)=z-z^2/2$ and $\xi = 1$. If the successive images $z\_0=1$, $z\_{n+1}=z\_n-z\_n^2/2$ were (absolutely, since $0<z\_n\le 1$) summable, then $z\_{n+1}=(1-a\_n)z\_n$ with $a\_n\in\ell^1$, $0<a\_n<1$, but in this scenario $z\_{n+1}=\prod\_{k=1}^n(1-a\_k)$ doesn't even go to zero.


