Abstract
Let f be a transcendental entire function, and let {U,V subset mathbb{C}} be disjoint simply-connected domains. Must one of {f^{-1}(U)} and {f^{-1}(V)} be disconnected? In 1970, Baker implicitly gave a positive answer to this question, in order to prove that a transcendental entire function cannot have two disjoint completely invariant domains. (A domain {Usubset mathbb{C}} is completely invariant under f if {f^{-1}(U)=U}.) It was recently observed by Julien Duval that Baker's argument, which has also been used in later generalisations and extensions of Baker's result, contains a flaw. We show that the answer to the above question is negative; so this flaw cannot be repaired. Indeed, for the function {f(z)= e^z+z}, there is a collection of infinitely many pairwise disjoint simply-connected domains, each with connected preimage. We also answer a long-standing question of Eremenko by giving an example of a transcendental meromorphic function, with infinitely many poles, which has the same property. Furthermore, we show that there exists a function f with the above properties such that additionally the set of singular values S(f) is bounded; in other words, f belongs to the Eremenko–Lyubich class. On the other hand, if S(f) is finite (or if certain additional hypotheses are imposed), many of the original results do hold. For the convenience of the research community, we also include a description of the error in Baker's proof, and a summary of other papers that are affected.
Highlights
Must one of f −1(U ) and f −1(V ) be disconnected? In 1970, Baker implicitly gave a positive answer to this question, in order to prove that a transcendental entire function cannot have two disjoint completely invariant domains. (A domain U ⊂ C is completely invariant under f if f −1(U ) = U .) It was recently observed by Julien Duval that Baker’s argument, which has been used in later generalisations and extensions of Baker’s result, contains a flaw
Almost half a century ago, Baker [Bak70] proved that a transcendental entire function cannot have two disjoint completely invariant domains; in particular, the Fatou set of such a function has either one or infinitely many connected components. (Since we do not focus on dynamics in this paper, we refer to [Ber93] for background and definitions.) in 2016 Julien Duval observed that there is a flaw in Baker’s proof
It follows that the question of whether a transcendental entire function can have two disjoint completely invariant domains remains open
Summary
Almost half a century ago, Baker [Bak70] proved that a transcendental entire function cannot have two disjoint completely invariant domains; in particular, the Fatou set of such a function has either one or infinitely many connected components. (Since we do not focus on dynamics in this paper, we refer to [Ber93] for background and definitions.) in 2016 Julien Duval observed that there is a flaw in Baker’s proof. (Since we do not focus on dynamics in this paper, we refer to [Ber93] for background and definitions.) in 2016 Julien Duval observed that there is a flaw in Baker’s proof It follows that the question of whether a transcendental entire function can have two disjoint completely invariant domains remains open. There is a transcendental entire function f ∈ B such that there is an infinite sequence (Uj)∞ j=1 of pairwise disjoint simplyconnected domains such that f −1(Uj) is connected for all j. There exists a transcendental entire function f and pairwise disjoint -connected domains U and V , each with connected preimage and each containing exactly one asymptotic value of f. The proof uses Baker’s flawed argument, and would again imply that any -connected domain with connected preimage contains all such values.
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