Compacity of the Julia sets

Let $P$ be a polynomial of degree at least $2$ and let $z\in\mathbb{C}$ . We define the sequence $(u_n(z))_n$ by $u_0(z):=z$ and $u_{n+1}(z)=P(u_n(z))$ .

We denote by $K$ the set of all $z$ such that the sequence $(u_n(z))_n$ is bounded.

Show that $K$ is compact.

In the image below, such a set $K$ has been drawn for $P(X)=X^2+c$ . This is what is called a Julia set. It can be shown that a sequence $(u_n(z))_n$ for this polynomial is necessarily bounded by $2$ if it is bounded. To construct a Julia set and obtain this beautiful gradient, we look at the first index from which the sequence exceeds the critical value $2$ . The larger this index is, the closer the color associated with the parameter $z$ is to white, whereas if this index is small, the color tends toward black. The choice of the intermediate colors is purely aesthetic.

The well-known Mandelbrot sets, on the other hand, correspond to finding the values of $c$ for which our sequence is bounded, while fixing the initial condition $u_0$ .

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Solution by AndurilFR

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Notons que, par définition, $u_n(z)=P^{\circ n}(z)$ où $f^{\circ n}:=f\circ \dots\circ f$ (composée $n$ fois).

Quand $\vert x\vert$ tend vers l’infini, $\vert x\vert = o(P(x))$ car $P$ est de degré au moins $2$ .
Ainsi il existe $M>0$ tel que si $\vert x\vert>M$ , alors $\vert P(x)\vert>2\vert x\vert$ .
Donc si $|z|>M$ , en itérant cette inégalité, $(P^{\circ n}(z))_{n \in \mathbb{N}}$ n’est pas bornée (minorée en norme par $2^{n}M$ ).
D’où $K\subset \overline{B(0,M)}$ donc $K$ est bornée.

De plus, avec ce qui précède, $K$ est l’ensemble des $z$ tels que pour tout $n\in\N$ , $|P^{n}(z)|$ est inférieur à $M$ .
Ainsi $K=\bigcap_{n\in\N} u_n^{-1}(\overline{B(0,M)})$ donc c’est une intersection de fermés (car les $u_n$ sont continues en tant que polynômes et la boule fermée de rayon $M$ est fermée). Donc $K$ est un fermé.

Ainsi, $K$ est fermé borné et on est en dimension finie donc c’est un compact.

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