Dynamics of slowly growing entire functions (Q2730738)

For an entire function \(f\), not a linear polynomial, let \(f^n\) denote the \(n\)-th iterate. The Fatou set \(\mathcal{F}(f)\) is the maximal open set in which \((f^n)\) is normal, while the complement of \(\mathcal{F}(f)\) in \(\mathbb C\) is the Julia set \(\mathcal{J}(f)\). It is well known that \(\mathcal{J}(f)\) is non-empty, perfect and completely invariant under \(f\).NEWLINENEWLINEThe growth of \(f\) is measured by \(M(r,f) := \max{\{\, | f(z)| : | z| =r \,\}}\). If \(\log{M(r,f)}=O(\log{r})\) as \(r\to\infty\), then \(f\) is a polynomial. Since \(\log{M(r,f)}\) is a convex increasing function of \(\log{r}\) the statement that there are transcendental entire functions of arbitrarily slow growth with a property \(P\) means that for a given function \(A(r)\) which is positive and increases to \(\infty\) as \(r\to\infty\) on \([0,\infty)\), there exists a transcendental entire function \(f\) such that \(f\) has the property \(P\) and satisfies \(\log{M(r,f)}<A(r)\log{r}\) for all sufficiently large \(r\).NEWLINENEWLINESufficiently slowly growing transcendental entire functions have many properties with polynomials in common. But the dynamical properties of iterates of such functions may be very different. The author first gives a survey of known results. For every transcendental entire function the Julia set is unbounded. There exist transcendental entire functions of arbitrarily slow growth whose Fatou set contains a multiply connected and bounded wandering component. Also, there are such functions \(f\) with \(\mathcal{J}(f)=\mathbb C\).NEWLINENEWLINEThe main result of this article states that there exists a transcendental entire function \(f\) of arbitrarily slow growth such that \(0\) is an attracting fixed point of \(f\) (which gives \(\mathcal{J}(f) \neq \mathbb C\)), every component \(G\) of \(\mathcal{F}(f)\) is simply connected and bounded, \(f^n \to 0\) as \(n\to\infty\) in \(G\), and \(\mathcal{J}(f)\) is a connected subset of \(\mathbb C\). Furthermore, \(\mathcal{J}(f)\) contains a dense subset of buried points which do not belong to the boundary of any component of \(\mathcal{F}(f)\).