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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.

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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.

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  • $\begingroup$ Took me a while to make sense of this, but now I get it. I didn't realize you were using absolute convergence of the infinite product, $\endgroup$ – user78249 Jul 30 '16 at 1:56
  • $\begingroup$ Yes, an easy way of seeing it is to take the logarithm of the product, which will converge (and in particular not go to $-\infty$). $\endgroup$ – Christian Remling Jul 30 '16 at 2:00
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    $\begingroup$ Even easier: let $a_n = 1/z_n$; then $a_{n+1} = 2a_n^2 / (2a_n-1)$, with together with $a_0=1$ implies $1/2 < a_{n+1}-a_n \leq 1$ $-$ whence $a_n \leq n+1$ and $z_n \geq 1/(n+1)$ whose sum diverges. (We could get $z_n \geq 1/(n+1)$ directly, but using $a_n$ feels like a better explanation of the harmonic decay.) $\endgroup$ – Noam D. Elkies Jul 30 '16 at 4:10
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To put the existing answer in context, you will never get summability if your function is holomorphic at $0$ and $f'(0)=1$. In particular, the example given in your question is actually also not summable.

Indeed, in this case $0$ is what is called a parabolic fixed point. Write $f(z) = z + az^{m+1} + \dots.$ Then the local behaviour near $0$ is well-understood: there are $m$ "attracting directions", along which orbits are attracted to the fixed point, and between them $m$ "repelling directions", where orbits are locally pushed away.

By conjugation, we may assume that $a=-1$, and consider the attracting direction along the positive real axis. Near this direction, the change of coordinate $w=1/z^m$ yields a function asymptotic to $w\mapsto w+1$. [Indeed, by Fatou's theorem, it is conjugate to this function within a certain neighbourhood of the attracting direction, but this is not really required here.]

So you see that orbits approaching zero will do so like $z_k=\operatorname{const}\cdot k^{-1/m}$ as $k\to\infty$. As $m\geq 1$, this sequence is not summable.

(This is essentially Noam's point in the comment to Christian's answer, but in the general setting.)

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  • $\begingroup$ Ahh yes, this makes sense. To cauterize the fault in my assumption I'll just switch everything to $\ell^2$ as opposed to $\ell^1$. $\endgroup$ – user78249 Aug 30 '16 at 16:08

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