I'm following the book Fractal Geometry, by K. Falconer, and in Chapter 14, Theorem 14.10 he proves that $J(f)=J_0(f)$, for $f $ polynomial and $J(f) = $ {closure of the set of repelling periodic points of $f$} and $J_0(f) = $ {points on which the family of iterates of $f$ isn't normal}.
In the second part of the proof, he considers: $$K=\{w\in J_0(f);\,\exists z\neq w;\,f(z)=w\wedge f'(z)\neq0\}$$ He then tries to show $K\subset J(f)$ as follows: Taking $w\in K$, $\exists V$ neighbourhood of $w$, and $f^{-1}:V\rightarrow \mathbb{C}$ local inverse. He then defines: $$h_k(z) = \frac{f^k(z)-z}{f^{-1}(z)-z}$$ In $V$ and uses the fact that it isn't normal to conclude it must assume the value $0$ or $1$, so that there exists a periodic point $\overline{z}\in V$. It then affirms that implies $w \in J(f)$.
My question is: the last implication requires $\overline{z}$ to be a $\textbf{repelling} $ periodic point, but I can't prove that.
My attempt was as follows:
If $|f'(\overline z)|^k<1$, then there would exist a neighbourhood of $\overline{z},\, U\subset V$, where $f^k(z)\rightarrow \overline{z},\,z\in U$. But then by analicity, $f^k(z)\rightarrow \overline{z}$ in all compact subsets of $V$ containing $U$. But by hypothesis $f^k$ isn't normal in $V$ (since it is a negihbourhood of $w$), so that $f^k$ may not converge in every compact subset of $V$,...., but this argument doesn't include the compact subsets of $V$ that don't contain $U$... Nor does it work if $|f'(\overline z)|^k=1$...
Any ideias?

  • $\begingroup$ what's $\exists \not = w$? $\endgroup$ – mathworker21 Sep 21 '18 at 17:58
  • $\begingroup$ @mathworker21 My bad, fixed it $\endgroup$ – MathNewbie Sep 21 '18 at 18:22

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