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On Connected Preimages of Simply-Connected Domains Under Entire Functions
Geometric and Functional Analysis ( IF 2.4 ) Pub Date : 2019-08-02 , DOI: 10.1007/s00039-019-00488-2
Lasse Rempe-Gillen , Dave Sixsmith

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 \({U\subset \mathbb{C}}\) 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.

中文翻译:

整体函数下简单连通域的连通原像

f是一个超越性的完整函数,而令\({U,V \ subset \ mathbb {C}} \)是不相交的简单连接域。必须一个\({F ^ { - 1}(U)} \)\({F ^ { - 1}(V)} \)被断开?1970年,贝克(Baker)隐式地对这个问题给出了肯定的答案,以证明先验的完整函数不能具有两个不相交的完全不变的域。(如果\({f ^ {-1}(U)= U} \,则\({U \ subset \ mathbb {C}} \\)f下完全不变。)。)朱利安·杜瓦尔(Julien Duval)最近观察到,贝克的论点包含一个缺陷,该论点也用于后来贝克因结果的推广和扩展中。我们证明对上述问题的回答是否定的。因此此缺陷无法修复。确实,对于函数\({f(z)= e ^ z + z} \),存在无限多个成对不相交的简单连接域的集合,每个域都具有连接的原像。通过给出具有无限多个极点且具有相同性质的先验亚纯函数的示例,我们还回答了Eremenko的一个长期问题。此外,我们表明存在一个具有上述属性的函数f,从而使奇异值Sf)有界;换句话说,f属于Eremenko–Lyubich类。另一方面,如果Sf)是有限的(或强加了某些附加假设),则许多原始结果确实成立。为了方便研究社区,我们还在贝克的证明中包括对错误的描述,以及其他受影响论文的摘要。
更新日期:2019-08-02
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