Uniform convergence of a sequence of polynomial logarithm Let $P\in \Bbb{C}[X]$ of degree $d\ge 2$. For $n\in \Bbb{N}$ (include $O$). Denote by $P^n$ the $n$-th composition and $g_n: z\mapsto \frac{1}{d^n}\log(\max \{1,\vert P^n\vert\})$. Show that $(g_n)_{n\in \Bbb{N}}$ converges uniformly on $ \Bbb{C}$.
I tried to use the classic theorem of convergence but it seems that we need a trick to solve it. Please suggest any hint how can I solve this.
 A: Let $G$ be a function define as follow,
$$ \begin{array}{ccccc}
f & : & \Bbb{C} & \to & \Bbb{R} \\
 & & z & \mapsto & \lvert \log(\max(1,\lvert P(z)\rvert))-d\log(\max(1,\lvert z)\rvert ))\rvert \\
\end{array}$$
As $z\mapsto \max(1,\lvert P(z)\rvert)$ and $z\mapsto \max(1,\lvert z\rvert)$ are continuous and $[1,+\infty)$-valued. So $G$ is defined and continuous.
In fact $G$ is bounded below. 
Proof. Arguing by contradiction, assume there exist a sequence $(z_q)\in \Bbb{C}^{\Bbb{N}}$ such that $G(z_q)\rightarrow +\infty$. If $z_q$ had a bounded subsequence, that would be valued in a compact. Then by continuity of $G$, its image by $G$ would be bounded below, which is nonsense. Therefore $\lvert z_q \rvert=+\infty$. Denote $l$ the leading coefficient of $P$. Clearly, $\lvert P(z_q) \rvert\sim \lvert l\rvert\lvert z_q \rvert^d$ so that from a certain rank $k_0$,
$$
G(z_q)=\lvert\log \lvert P(z_q)\rvert-d\log \lvert z_q\rvert \rvert=\log\frac{\lvert P(z_q)\rvert}{\lvert z_q\rvert^d}\rightarrow \log \lvert c\rvert. \square
$$
Let $n\in \Bbb{N^*}$,
$$
\lvert g_{n+1}(z)-g_n(z)\rvert=\frac{1}{d^{n+1}}G(P_n(z))\le \frac{\lVert G\rVert_\infty}{d^{n+1}}
$$
Finally, the series $\sum (g_{n+1}(z)-g_n(z))$ converges normally and and the conclusion follows.
