Why $f^n(0)$ (where $n$ is the number of iterations of the function $f(z)=z^2+c$) will always tend to the attractive fixpoint $z^*$ as $n\rightarrow \infty$ if $abs(f'(z^*))<1?$ I know how to demonstrate that a fixpoint is attractive if $f'(z^*)<1,$ but not that $f^n(0)$ always tends to it.
Why all the sequences associated to the values of the main cardioid of the Mandelbrot set tend to the attractive fixpoint?
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complex-analysis
fixed-point-iteration
1 Answers
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The main cardioid $C=\{\,\frac{2u-u^2}4\mid u\in\Bbb D\,\}$ is precisely the set of all points $c\in \Bbb C$ for which $f\colon z\mapsto z^2+c$ has a fixed point $z^*$ with $|f'(z^*)|<1$. Note that this is a connected set.
Let $$A=\{\,c\in C\mid \lim_{n\to\infty}f_c^{\circ n}(0)\text{ exists and is an attractive fixed point of }f_c\,\}. $$
If $c_0\in A$ and $z_0^*=\lim f_{c_0}^{\circ n}(0)$, then there exists $q<1$ and $r>0$ such that $|f_{c_0}'(z)| Let $c_\infty\in C$ and $c_n\in A$ a sequence with $c_n\to c_\infty$.
Let $z_\infty^*$ and $z_n^*$ be the corresponding attractive fixed points. Then it follows that also $z_n^*\to z_\infty^*$. One can show that then one actually has $c_\infty\in C$ (though it seems I'm too tired right now to succeed in that). It follows that $A$ is also a (relatively) closed subset of $C$.
As $C$ is connected and clearly $0\in A$, it follows that $A=C$, as desired.