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.