I want to prove the following property of Fatou and Julia set.
Let $R$ be non-constant rational map, $g$ be Moebius map, and let $S = gRg^{-1}$. Then $F(S) = g(F(R))$ and $J(S) = g(J(R))$.
I tried to prove it by the inclusion of the sets. I proved the inclusion from one side. But I am not able to show other inclusion. My proof:
The set $g(F(R))$ contains the images of points $F(R)$ under the map $g$. That is if $z_0 \in F(R)$ then $g(z_0) \in g(F(R))$. First we prove $g(F(R)) \subset F(S)$. Observe if $z_0 \in F(R)$ we have for every $\epsilon > 0$ then there exist $\delta > 0$ such that $d_c(R^{n}(z), R^{n}(z_0)) < \epsilon$ whenever $d_c(z , z_0) < \delta$ for all $n \in \mathbb{N}$. Let $g(z_0) \in g(F(R))$ and we have \begin{align*} d_c(S^{n}g(z), S^{n}g(z_0)) &= d_c(gR^{n}(z), gR^{n}(z_0)) \quad \text{Since} \; S^{n}g = gR^{n}\\ & \leq M d_c(R^{n}(z), R^{n}(z_0)) \quad \text{Since} \; g \; \text{satisfies Lipschitz condition with constant} \; M.\\ & < M \frac{\epsilon}{M} < \epsilon \quad \text{Since} \; z_0 \in F(R). \end{align*} Hence $g(z_0) \in F(S)$ implies $g(F(R)) \subset F(S)$.