Recently when I read a paper about Fatou component, I met the following theorem which cited in Professor Eremenko's paper "on the iteration of entire functions"

If Fatou set has a Multiply connected Fatou component implies every component of F(f) is bounded.

it is really known theorem proved by I.N Baker in his paper in "The domains of normality of an entire function". However I really did know how to find this paper. any comments and materials about this theorem will be very appreciated. thanks.

  • $\begingroup$ You do not specify which class of functions you are talking about: rational, polynomials, entire, meromorphic. $\endgroup$ – Alexandre Eremenko Sep 9 '12 at 1:41

The correct statement is this. If the Fatou set $F(f)$ of an ENTIRE function has a multiply connected component, then all components of $F(f)$ are bounded. The idea of the proof is the following. Let $D$ be a multiply connected component. Then Maximum Principle implies that that the iterates tend to infinity in $D$, and if $\gamma$ is a non-trivial closed curve in $D$ then the index of the image of $\gamma$ under the iterates of $f$ tends to infinity. This implies that all other components (except $D$ itself) must be bounded. It remains to prove that $D$ is bounded. Here one uses Baker's theorem that an invariant component of $F(f)$ must be simply connected. This implies that $D$ must be wandering, that is some images of $\gamma$ "surround" $D$, so $D$ is bounded.

Baker's theorem that I refer to is published in the Finnish Annals in 1975. But his proof is reproduced in the survey of Eremenko and Lyubich, as theorem 4.3. The survey is available on Internet: http://www.math.purdue.edu/~eremenko/dvi/AAsurveyp3.pdf (This is the third part. For parts 1-2 replace the 3 in the address by 1 or 2.)

P.S. It is hard to believe that the Finnish Annals are not available in some university library in China.

  • $\begingroup$ Thank you professor Eremenko. Your survey article is really great help, thanks. $\endgroup$ – yaoxiao Sep 14 '12 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.