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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.

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    $\begingroup$ Probably it does not. But why would one care about such a question? $\endgroup$ – Alexandre Eremenko Jul 30 '15 at 14:12
  • $\begingroup$ @ 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. $\endgroup$ – yaoxiao Jul 30 '15 at 14:19
  • $\begingroup$ I do not think anyone asked this question before, and I conjecture that $J(f)$ and $J(f')$ always have common points. $\endgroup$ – Alexandre Eremenko Jul 30 '15 at 14:26
  • $\begingroup$ @AlexandreEremenko Dear professor, what's your opinion about $I(f)\cap I(f')$, thanks? $\endgroup$ – yaoxiao Jul 30 '15 at 14:30
  • $\begingroup$ same opinion. In fact all 4 sets must have non-empty intersection. $\endgroup$ – Alexandre Eremenko Jul 30 '15 at 14:32
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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.)

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

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