Some examples of Baker domains (Q2882655)

Let \(f:\mathbb{C}\to\mathbb{C}\) be a transcendental entire function. A Baker domain of \(f\) is a maximal open connected set \(U\subset\mathbb{C}\) such that \(f^k(U)\subset U\) for some \(k\geq 1\) and such that \(f^n(z)\to\infty\) for every \(z\in U\), as \(n\to\infty\).NEWLINENEWLINEBy a theorem of \textit{I. N. Baker} [``The domains of normality of an entire function'', Ann. Acad. Sci. Fenn., Ser. A I 1, 277--283 (1975; Zbl 0329.30019)], \(U\) must be simply-connected. Following \textit{N. Fagella} and \textit{C. Henriksen} [``Deformation of entire functions with Baker domains'', Discrete Contin. Dyn. Syst. 15, No. 2, 379--394 (2006; Zbl 1112.37034)], a Baker domain is called hyperbolic, simply parabolic or doubly parabolic depending on whether the Riemann surface \(U/f\) is conformally isomorphic to an annulus, a punctured disk, or a punctured plane. (It follows from a theorem of \textit{C. C. Cowen} [``Iteration and the solution of functional equations for functions analytic in the unit disk'', Trans. Am. Math. Soc. 265, 69--95 (1981; Zbl 0476.30017)] that one of these alternatives must always hold.)NEWLINENEWLINEFagella and Henriksen [loc. cit.] asked whether there can be a simply parabolic Baker domain in which the function \(f\) is not univalent. The authors answer this question in the affirmative. The idea of the proof is to start with a well-known example of a simply parabolic domain, due to Herman, and to carefully modify the function using quasiconformal surgery so as to make the function non-injective.NEWLINENEWLINEFurthermore, the construction ensures that \(\partial U\) contains a Jordan arc that is accessible from \(U\), so that the Riemann map \(\phi:\mathbb{D}\to U\), where \(\mathbb{D}\) denotes the unit disk, extends continuously to an arc of the unit circle. In particular, this gives a negative answer to a question of \textit{I. N. Baker} and \textit{P. Domínguez} [``Boundaries of unbounded Fatou components of entire functions'', Ann. Acad. Sci. Fenn., Math. 24, No. 2, 437--464 (1999; Zbl 0935.30020)], who asked whether, for a Baker domain on which \(f\) is not univalent, there is always a dense set of angles where \(\phi\) has radial limit \(\infty\). \textit{D. Bargmann} [``Iteration of inner functions and boundaries of components of the Fatou set'', Lond. Math. Soc. Lect. Note Ser. 348, 1--36 (2008; Zbl 1168.30007)] showed that the answer is positive in the case of doubly parabolic domains (where the map is never univalent).NEWLINENEWLINEThe authors also show how to modify the construction to obtain a hyperbolic Baker domain with the same properties. (There are previously known examples of hyperbolic Baker domains on which the function is not univalent, but these are not counterexamples to the question of Baker and Domínguez.)