# A question on $J(f)$ and $J(f')$

I was confused by the following question for a long time:

Does there exists a transcendental entire function $f$ such that $J(f)\cap J(f')=\emptyset$ ?

where $J(f)$, ($J(f')$) is the Julia set of $f$ $(f').$

Edit: Following the same style, one may also ask Does there exists a transcendental entire function $f$ such that $I(f)\cap I(f')=\emptyset$ ?

where $I(f)$, ($I(f')$) is the escaping set of $f$ $(f').$ $I(f)\neq \emptyset$ was first proved by Eremenko using Wiman-Valiron method. See $http://www.math.purdue.edu/~eremenko/dvi/banach.pdf for details. Any comments will be appreciated. • Probably it does not. But why would one care about such a question? – Alexandre Eremenko Jul 30 '15 at 14:12 • @ Alexandre Eremenko I remembered someday I read a proof of some theorem related comparison functional property between$f$and$f'$, in a Nevanlinna theory book. I just ask myself what it will be when compare its dynamical property. – yaoxiao Jul 30 '15 at 14:19 • I do not think anyone asked this question before, and I conjecture that$J(f)$and$J(f')$always have common points. – Alexandre Eremenko Jul 30 '15 at 14:26 • @AlexandreEremenko Dear professor, what's your opinion about$I(f)\cap I(f')$, thanks? – yaoxiao Jul 30 '15 at 14:30 • same opinion. In fact all 4 sets must have non-empty intersection. – Alexandre Eremenko Jul 30 '15 at 14:32 ## 1 Answer In my opinion, the question is completely arbitrary. There is no reason to expect relationships between the dynamics of$f$and that of its derivative. Their relationship will even change under affine conjugacy. This being said, the answer to the question is positive in that such a function$f$does indeed exist; i.e. the conjecture discussed in the comments below the questionis false. Indeed, for a counterexample consider$\newcommand{\eps}{\varepsilon}f(z)=\eps e^{-z}$. Then$f'(z) = -\eps e^{-z}=-f(z)$. For sufficiently small$\eps$, the right half plane$H_R$belongs to the basin of attraction of an attracting fixed point, for both maps. Hence the Julia sets are both contained in the left half plane$H_L$. However, clearly$f^{-1}(H_L)\cap(f')^{-1}(H_L)=f^{-1}(H_L)\cap f^{-1}(H_R)=\emptyset$. Hence$J(f)\cap J(f')=\emptyset$. In particular also$I(f)\cap I(f')=\emptyset$. EDIT: If you are willing to let$f$be a polynomial, then here is an even simpler example. Take$f(z)=z^3/3 + c$. Then$f'(z)=z^2$, so$J(f')$is the unit circle. For large$c$, clearly the closed unit disc belongs to the basin of$\infty$of$f$, and hence does not intersect$J(f)$. (A simple calculation shows that$c=4$will suffice.) Of course, the same argument will work for any family of the form$f(z)=p(z)+c$, with$p$a fixed polynomial of degree at least$3$. (The latter assumption is there to ensure that dynamics of$f'\$ is nontrivial.)

• Thank you, professor. I know the polynomial example earlier. While, I was surprised by your such a elegant transcendental example. – yaoxiao Aug 3 '15 at 6:27