estimate of iterates of a polynomial Let $P\in\mathbb R[x]$ of degree $d\ge2$. I want to prove that for all $x\in\mathbb R$, but maybe for a set with zero Lebesgue measure, one has $\lim_{n\to+\infty}\limits\frac{\ln(P^{[n]}(x))}{d^n\ln(|x|)}=1$, where $P^{[n]}(x)$ is defined inductively by $P^{[n+1]}(x)=P(P^{[n]}(x))$ and $P^{[0]}(x)=x$. For $P(x)=x^d$, it is true obviously. But in the general case, I did not manage to prove it.
Thanks in advance for a solution or any hint.
 A: Fix an integer $d \geq 2$. Given a polynomial $P \in \mathbb{R}[x]$ of degree $d$ and $x \in \mathbb{R}$, it is in fact unlikely that $$\lim_{n \rightarrow +\infty} \frac{\log\left( \left\lvert P^{[n]}(x) \right\rvert \right)}{d^{n}} = \log\left( \lvert x \rvert \right) \, \text{.}$$

*

*For example, consider the polynomial $$P(x) = (x -1)^{d} +1 \in \mathbb{R}[x] \, \text{.}$$ For every $n \geq 0$, we have $P^{[n]}(x) = (x -1)^{d^{n}} +1$. Therefore, for every $x \in \mathbb{R}$, we have $$\lim_{n \rightarrow +\infty} \frac{\log\left( \left\lvert P^{[n]}(x) \right\rvert \right)}{d^{n}} = \begin{cases} 0 & \text{if } x \in (0, 2] \text{ or } (x = 0 \text{ and } d \text{ is even})\\ -\infty & \text{if } x = 0 \text{ and } d \text{ is odd}\\ \log\left( \lvert x -1 \rvert \right) & \text{if } x \in (-\infty, 0) \cup (2, +\infty) \end{cases} \, \text{.}$$


*In fact, assume that $P \in \mathbb{R}[x]$ has degree $d$ and there exists $x_{0} \in (-1, 0) \cup (0, 1)$ such that $$\lim_{n \rightarrow +\infty} \frac{\log\left( \left\lvert P^{[n]}\left( x_{0} \right) \right\rvert \right)}{d^{n}} = \log\left( \left\lvert x_{0} \right\rvert \right) \, \text{,}$$ and let us prove that $P(x) = \pm x^{d}$. Write $$P(x) = \sum_{j = 0}^{d} a_{j} x^{j} \, \text{,} \quad \text{with} \quad a_{0}, \dotsc, a_{d} \in \mathbb{R} \, \text{,} \quad \text{and} \quad m = \min\left\lbrace j \in \lbrace 0, \dotsc, d \rbrace : a_{j} \neq 0 \right\rbrace \, \text{,}$$ and let us prove that $m = d$ and $a_{d} = \pm 1$. We have $$\log\left( \left\lvert P(x) \right\rvert \right) = m \log\left( \lvert x \rvert \right) +\log\left( \left\lvert \sum_{j = m}^{d} a_{j} x^{j -m} \right\rvert \right) = m \log\left( \lvert x \rvert \right) +O_{0}(1) \, \text{,}$$ and hence $$\lim_{\substack{x \rightarrow 0\\ x \neq 0}} \frac{\log\left( \left\lvert P(x) \right\rvert \right)}{d \log\left( \lvert x \rvert \right)} = \frac{m}{d} \, \text{.}$$ For $n \geq 0$, set $$x_{n} = P^{[n]}\left( x_{0} \right) \in \mathbb{R} \quad \text{and} \quad u_{n} = \frac{\log\left( \left\lvert x_{n} \right\rvert \right)}{d^{n}} \in [-\infty, +\infty) \, \text{.}$$ Since $\lim\limits_{n \rightarrow +\infty} u_{n} = \log\left( \left\vert x_{0} \right\rvert \right)$ and $\log\left( \left\lvert x_{0} \right\rvert \right) \in (-\infty, 0)$ by hypothesis, we have $$x_{n} \neq 0 \text{ for large } n \, \text{,} \quad \lim_{n \rightarrow +\infty} x_{n} = 0 \quad \text{and} \quad \lim_{n \rightarrow +\infty} \frac{\log\left( \left\lvert P\left( x_{n} \right) \right\rvert \right)}{d \log\left( \left\lvert x_{n} \right\rvert \right)} = \lim_{n \rightarrow +\infty} \frac{u_{n +1}}{u_{n}} = 1 \, \text{.}$$ Therefore, we have $\frac{m}{d} = 1$, and hence $P(x) = a_{d} x^{d}$. It follows by induction that $$P^{[n]}(x) = a_{d}^{\frac{d^{n} -1}{d -1}} x^{d^{n}}$$ for all $n \geq 0$, and in particular $$\lim_{n \rightarrow +\infty} \frac{\log\left( \left\lvert P^{[n]}\left( x_{0} \right) \right\rvert \right)}{d^{n}} = \lim_{n \rightarrow +\infty} \left( \log\left( \left\lvert x_{0} \right\rvert \right) +\frac{1 -\frac{1}{d^{n}}}{d -1} \log\left( \left\lvert a_{d} \right\rvert \right) \right) = \log\left( \left\lvert x_{0} \right\rvert \right) +\frac{\log\left( \left\lvert a_{d} \right\rvert \right)}{d -1} \, \text{.}$$  Therefore, we have $a_{d} = \pm 1$, and hence $P(x) = \pm x^{d}$.


*Finally, let me mention that this is very related to the notion of Green function of a polynomial in complex dynamics. Given a polynomial $P \in \mathbb{C}[z]$ of degree $d$ and $z \in \mathbb{C}$, set $$g_{P}(z) = \lim_{n \rightarrow +\infty} \frac{\log^{+}\left( \left\lvert P^{[n]}(z) \right\rvert \right)}{d^{n}} \, \text{,} \quad \text{where} \quad \log^{+} = \max\lbrace \log, 0 \rbrace \, \text{.}$$ This gives a well-defined map $g_{P} \colon \mathbb{C} \rightarrow \mathbb{R}_{\geq 0}$. Now, define the filled-in Julia set of $P$ to be $$K_{P} = \left\lbrace z \in \mathbb{C} : \left( P^{[n]}(z) \right)_{n \geq 0} \text{ is bounded} \right\rbrace \, \text{,}$$ which is a compact subset of $\mathbb{C}$. The Green function $g_{P}$ of $P$ has the following properties:

*

*the map $g_{P}$ is continuous on $\mathbb{C}$;

*for every $z \in \mathbb{C}$, we have $g_{P}(z) = 0$ if and only if $z \in K_{P}$;

*the map $g_{P}$ is harmonic on $\mathbb{C} \setminus K_{P}$;

*we have $$g_{P}(z) = \log\left( \lvert z \rvert \right) +\frac{1}{d -1} \log\left( \left\lvert a_{d} \right\rvert \right) +o_{\infty}(1) \, \text{,} \quad \text{where} \quad P(z) = \sum_{j = 0}^{d} a_{j} z^{j} \, \text{.}$$
Furthermore, these properties characterize the map $g_{P}$.
