Iteration of mapping with nested iterates in logarithm

Question

Consider the mapping $$ g: [0, 1) \longrightarrow \mathbb{R}, x \longmapsto g(x) := \frac{x}{1-\ln|x|} $$ It is easily seen that $g$ is a bijective continuous mapping from $[0, 1)$ onto itself with $g(0) = 0$ well-defined. I'm concerned with the iteration of $g$ with itself, i.e. for $n \in \mathbb{N}_0$, consider $$ g^{[0]}(x) := x, \\ g^{[1]}(x) := g(x), \\ g^{[2]}(x) := g(g(x)),\\ \vdots \\ g^{[n]}(x) := g(g(g(\dots g(x) \dots))) \; \; (n \text{ times}) $$ where $g^{[n]}$ denotes the n-fold iterate of $g$. Via induction, one can see that

$$ g^{[n]}(x) = \frac{x}{\prod \limits_{k=0}^{n-1} \left( 1 -\ln\left| g^{[k]}(x) \right| \right)} \tag{1} $$

holds for all $x \in [0,1)$ and all $n \in \mathbb{N}_0$. In other words, the $n$-fold iterate $g^{[n]}$ can be expressed using the former $k$-fold iterates $g^{[k]}$ for $k = 0, \dots, n-1$. This "nesting" of iterates (especially in combination with the logarithm) makes it quite hard for me to analyze the expression in (1), to be honest. Basically, I'm having three questions on this:

1. Is there some theory or concepts on things like nested mappings similar to the situation given above? Note that I'm aware of iteration theory in complex analysis (also called complex dynamics), but I'm not able to see how this fits in here...note that $g$ may also be considered as a mapping of the complex unit disk $\mathbb{D} := \{ |z| < 1 \}$ (and its closure) into $\mathbb{C}$, of course.
2. Can anything be said if $n$ tends to infinity, i.e. what "is" $g^{[\infty]}$ if it exists? In other words, does the product in (1) converge in a certain sense (pointwise, [locally] uniformly)?
3. Is there a way of estimating the expression in (1) (with respect to the absolute value)?

Thanks in advance for any help!

Answer 1

2. Can anything be said if $n$ tends to infinity?

$0 \le g(x) < x$ for all $x \in [0, 1)$, so for fixed $x$, the sequence $g^{[n]}(x)$ is decreasing and bounded below, and therefore convergent.

$$ g(g^{[n]}(x)) = g^{[n+1]}(x) $$ and the continuity of $g$ implies that the limit is a fixed-point of $g$, so we have $$ \lim_{n \to \infty } g^{[n]}(x) = 0 $$ pointwise in $[0, 1)$.

And since $g$ is increasing, the convergence is uniform on each compact interval $[0, r] \subset [0, 1)$: $$ 0 \le g^{[n]}(x) \le g^{[n]}(r) \to 0 \text{ for } n \to \infty. $$

The convergence is not uniform on $[0, 1)$ because for every $n$ $$ \sup \{ g^{[n]}(x) \mid 0 \le x < 1 \} = \lim_{x \to 1 } g^{[n]}(x) = 1 \, . $$

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3. Is there a way of estimating the expression in (1)?

A simple estimate would be $$ g^{[n]}(x) = \frac{x}{\prod \limits_{k=0}^{n-1} \left( 1 -\ln g^{[k]}(x) \right)} \le \frac{x}{\prod \limits_{k=0}^{n-1} \left( 2 - g^{[k]}(x) \right)} \le \frac{x}{(2-x)^n} \, , $$ using the “well-known” inequality $\ln x \le x-1$.