Sylwester Zając. Hypercyclicity of composition operators in domains of holomorphy. Uniwersytet Jagielloński

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1 Sylwester Zając Uniwersytet Jagielloński Hypercyclicity of composition operators in domains of holomorphy Praca semestralna nr 2 (semestr zimowy 2011/12) Opiekun pracy: Paweł Domański

2 Hypercyclicity of composition operators in domains of holomorphy Sylwester Zaj c Institute of Mathematics, Faculty of Mathematics and Computer Science, Jagiellonian University, Šojasiewicza 6, Kraków, Poland sylwester.zajac@im.uj.edu.pl Abstract We characterize hypercyclic composition operators C ϕ : f f ϕ on the space of functions holomorphic on Ω, where Ω C N is a pseudoconvex domain and ϕ is a holomorphic self-mapping of Ω. In the case when all the balls with respect to the Carathéodory pseudodistance are relatively compact in Ω, we show that much simpler characterisation is possible (many natural classes of domains satisfy this conditioon, i.e. strictly pseudoconvex domains, bounded convex domains, etc.). Moreover, we show that in such a class of domains, and in simply connected or innitely connected planar domains, hypercyclicity of C ϕ implies its hereditary hypercyclicity. 1 Introduction Let Ω be a pseudoconvex domain in C N and let ϕ : Ω Ω be a holomorphic mapping. We are interested in the problem of hypercyclicity and hereditary hypercyclicity of the composition operator C ϕ : f f ϕ on the space O(Ω) of holomorphic functions f : Ω C, endowed with the usual topology of locally uniform convergence. In the case when Ω is a planar domain, a complete characterization of hypercyclicity was given by Grosse-Erdmann and Mortini in [6]. In higher dimensions the problem was considered by several authors, mostly in cases when Ω is a polydisc, an euclidean ball or the whole C N with ϕ being special (see [2] and the references in [6]). Analogous problem in spaces of real analytic functions was considered in [3] and in kernels of general (non-necessarily Cauchy-Riemann) partial dierential equations in [10]. (1) 2010 Mathematics Subject Classication. Primary: 47B33; Secondary: 32H50. Key words and phrases: Spaces of holomorphic functions; composition operator; hypercyclic operator; holomorphic convexity; Ω-convexity. Acknowledgement: This paper was prepared as a semester paper under the guidance of P. Doma«ski in the framework of joint Ph. D. program SDNM (Poland) during the author research stay at Faculty of Mathematics and Computer Science, Adam Mickiewicz University, Pozna«(Poland). 1

3 In this paper we give a characterization of hypercyclicity and hereditary hypercyclicity of C ϕ for arbitrary pseudoconvex domain Ω C N and arbitrary holomorphic mapping ϕ : Ω Ω (Theorem 3.4), using ideas developed by several authors. It is formulated in the language of holomorphic hulls of compact subsets of Ω, which in dimension one coincides with the notion of Ω-convexity used in [6]. However, our best results are contained in Sections 5, 6 and 7. A disadvantage of the conditions in Theorem 3.4 is that they require to answer the question: when a sum of two disjoint holomorphically convex sets are holomorphically convex? On the other hand, the necessary conditions (Proposition 3.1), which avoid this question, turn out to be sucient in the cases of simply connected and innitely connected planar domains (see [6, Theorem 3.21]). This fact motivated us to ask about domains in C N having the same property. We deal with this topic in Section 5, showing that it holds for Ω if the closed balls with respect do the Carathéodory pseudodistance are compact in the topology of Ω (Theorem 5.4). Such a class of domains includes many 'nice' domains, e.g. bounded convex domains, strictly pseudoconvex domains, analytic polyhedra, etc. (see Corollary 5.5). As a simple observation it follows from Theorem 3.4 that the operator C ϕ is hypercyclic if and only if it is hereditarily hypercyclic with respect to some sequence (n l ) l (Observation 3.6). As we show in Section 6, in many domains even more is true: C ϕ is hypercyclic if and only if it is hereditarily hypercyclic (Theorem 6.2). Using this result, we conclude that domains with compact Carathéodory balls have also this property. As the second conclusion, in Section 7 we show that hypercyclicity and hereditary hypercyclicity are equivalent in simply connected and innitely connected planar domains, what is an improvement of Theorem 3.21 in [6]. It remains an open problem if every hypercyclic operator C ϕ is automatically hereditarily hypercyclic. Given an increasing sequence (n l ) l N, in this paper we study the following problems: (p1) Hypercyclicity of C ϕ. (p2) Hypercyclicity of C ϕ with respect to (n l ) l. (p3) Hereditary hypercyclicity of C ϕ with respect to (n l ) l. (p4) Hereditary hypercyclicity of C ϕ. For the denition of this notions see Denition 2.1. It is clear that there hold the implications (p4) (p3) (p2) (p1). 2 Preliminaries For an open set Ω C N, by O(Ω, C M ) be denote the space of all holomorphic functions f : Ω C M, equipped with the compact-open topology, i.e. the topology of uniform convergence on compact subsets. In the case M = 1 we shortly write O(Ω) instead of O(Ω, C). For an open set Ω C M by O(Ω, Ω ) we denote the set of all holomorphic mappings from Ω to Ω ; this set is open in O(Ω, C M ). For a domain Ω C N by Ω = Ω { Ω } we denote the usual compactication of Ω by an element Ω Ω. We denote: D - the unit disc in C, D = D \ {0} and C = C \ {0}. 2

4 For a boundary point a Ω we call a holomorphic function F : Ω D a peak function for a if lim Ω z a F (z) = 1. Although usually a peak function is dened in a dierent way and requires stronger conditions, the above denition is sucient for our considerations. For a compact set K Ω by K Ω or (K) Ω we denote the holomorphic hull of the set K with respect to Ω, i.e. K Ω := {z Ω : f(z) sup f for every f O(Ω)}. K The set K is called holomorphically convex if K = K Ω ; we call such set shortly: Ω-convex. For Ω = C N we shortly write K and (K). To get more informations, see [7]. We say that a sequence (K l ) l of compact sets is an exhaustion of Ω if l K l = Ω and K l int K l+1. For pseudoconvex domains U Ω C N we say that U is a Runge domain with respect to Ω if every function from O(U) can be approximated locally uniformly on U by functions from O(Ω). For some important facts connected with this notion see [7]. We say that a sequence of holomorphic functions f n : Ω Ω is compactly divergent (in O(Ω, Ω )) if for each compact subsets K Ω, L Ω there is n 0 such that f n (K) L = for all n n 0. We say that(f n ) n is run-away (in O(Ω, Ω )) if for each compact subsets K Ω, L Ω there is n such that f n (K) L =. In the case Ω = Ω it is always enough to consider the situation when L = K. Note that the sequence (f n ) n is run-away if and only if it has a compactly divergent subsequence, and (f n ) n is compactly divergent if and only if each of its subsequences is run-away. For a domain Ω C N and points z, w Ω let c Ω(z, w) := sup{ F (z) : F O(Ω, D), F (w) = 0}, c Ω (z, w) := 1 2 log 1 + c Ω (z, w) 1 c Ω (z, w). Here c Ω is the Carathéodory pseudodistance, and c Ω is the Möbius pseudodistance in Ω. For more informations, we refer the reader to [9]. Given a set X, a mapping T : X X and an integer number n, we denote by T [n] the n-th iteration of T, i.e. the mapping T T... T (n times). Denition 2.1. Let T and T n (n N) be continuous self-mappings of a topological space X. 1. We call a point x X an universal element for (T n ) n, if the set {T n (x) : n N} is dense in X. 2. We say that the sequence (T n ) n is universal if there exists a universal element for it, and we call this sequence hereditarily universal if each of its subsequences is universal. 3. We say that the mapping T is hypercyclic (resp. hereditarily hypercyclic) with respect to an increasing sequence (n l ) l N if the sequence (T [nl] ) nl is universal (resp. hereditarily universal). 3

5 4. We say that the sequence (T n ) n is topologically transitive if for every non-empty open subsets U, V X there exists n such that T n (U) V. Below we recall two theorems which are essential for our considerations (see [5, Theorem 1 and Proposition 1]). Theorem 2.2. Let X be a separable Fréchet space. A sequence (T n ) n of continuous selfmaps of X is topologically transitive if and only if the set of its universal elements is dense in X. Moreover, if one of these conditions holds, then the set of universal elements for (T n ) n is a dense G δ -subset of X. Theorem 2.3. Let X be a separable Fréchet space. Suppose that (T n ) n is a sequence of continuous self-maps of X such that each T n has dense range and that the family (T n ) n is commuting, i.e. T n T m = T m T n, for m, n N. Then the set of universal elements for (T n ) n is empty or dense. From now we consider the case when X is the space O(Ω) and T is the composition operator C ϕ : f f ϕ for a pseudoconvex domain Ω C N and a holomorphic mapping ϕ : Ω Ω. As mentioned, we are interested in problems (p1)-(p4). It is clear that (p1) (resp. (p4)) is just (p2) (resp. (p3)) with n l = l. Also observe that if C ϕ is hypercyclic, i.e. the condition (p1) is fullled, then for any n N the map C [n] ϕ has dense range. Indeed, if f is a universal element for (C [n] ϕ ) n, then f ϕ [k] = C [n] ϕ (f ϕ [k n] ) belongs to the range [n] [n of C ϕ for each k n. It is obvious that the sequence (C l ] ϕ ) l is commuting, so as a corollary from Theorems 2.2 and 2.3, we obtain the following fact: Corollary 2.4. Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Then the following conditions are equivalent: (1) There exists an universal element for (C ϕ [n l ] ) l. (2) The set of universal elements of (C ϕ [n l ] ) l is a dense G δ -subset of O(Ω). (3) The sequence (C ϕ [n l ] ) l is topologically transitive. Since the sets W f0,k,ɛ := {f O(Ω) : f f 0 < ɛ on K}, f 0 O(Ω), ɛ > 0, K Ω compact, form a basis of the topology of O(Ω), the topological transitivity of a subsequence (C ϕ [n l ] ) l means that for every ɛ > 0, g, h holomorphic on Ω and compact K Ω there are l N and a function f holomorphic on Ω such that f g < ɛ on K and f ϕ [n l] h < ɛ on K. Assuming that the mapping ϕ is injective (as we shall see in Proposition 3.1, it is necessary even for hypercyclicity of C ϕ ), the above condition takes the form: (tt) f g < ɛ on K and f h ϕ [ n l] < ɛ on ϕ [n l] (K). 4

6 We are going to formulate our theorems in the language of holomorphic hulls of compact subsets of Ω. By this reason, let us recall some well-known properties of holomorphic hulls: Lemma 2.5. Let Ω C N be a pseudoconvex domain and let K, L Ω be compact and such that K Ω L Ω =. Then the following conditions are equivalent: (1) K L Ω = K Ω L Ω. (2) There exist open and disjoint subsets U, V Ω such that KΩ U, L Ω V and K L Ω U V. (3) There exists a holomorphic function F : Ω C such that F (K) F (L) =. For the reader's convenience, we present a sketch of the proof. Sketch of the proof. (1) (3): Consider the function f equal 0 in a neighborhood of K Ω and 1 in a neighborhood of L Ω. By (1), this function is holomorphic in a neighborhood of the Ω-convex set K L Ω, so it can be approximated on this set by functions holomorphic on Ω (see [7, Theorems and 4.3.4]). Hence there is some F O(Ω) such that F f < 1 on K L 2 Ω. This F satises (3). (3) (2): It suces to dene U := F 1 (U 0 ), V := F 1 (V 0 ), where U 0 and V 0 are some disjoint open neighborhoods of the compact sets F (K) and F (L), respectively. (2) (1): The right-to-left inclusion is obvious, so we prove the other one. Fix z 0 I := K L Ω. We can assume that z 0 U. We prove that z 0 K Ω. The characteristic function χ U of U, restricted to the set U V, is (by (2)) holomorphic in a neighborhood of the Ω-convex set I, so there exists a sequence of functions (g n ) n O(Ω) uniformly convergent to χ U on I. Hence g n 1 on I U and g n 0 on I V. For any function f O(Ω) the sequence (fg n ) n converges uniformly to f on I U and to 0 on I V, so f(z 0 ) = lim f(z 0 )g n (z 0 ) lim n n = lim sup n z K f(z)g n (z) = sup z K This implies that z 0 K Ω and nishes the proof. sup z K L f(z). f(z)g n (z) Lemma 2.6. Let Ω C N be a pseudoconvex domain and let K, L Ω be compact subsets of Ω. If K L = and the set K L is Ω-convex, then K and L are both Ω-convex. Proof. As above, we can nd a function F O(Ω) such that F < 1 on K and F 1 < on L. There is K Ω K L and ) ( ) ( 1 K Ω L F ( F 1 (K) F 1 (F (L)) F 1 2 D F ) 2 D =, so K Ω K. 5

7 3 General results We start this section with formulating some necessary conditions. In fact, they appear in several papers. Proposition 3.1. Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Suppose that C ϕ is hypercyclic w.r.t. (n l ) l. Then: (c1) The mapping ϕ is injective. (c2) The image ϕ(ω) is a Runge domain w.r.t. Ω. (c3) The sequence (ϕ [n l] ) l is run-away. Remark 3.2. Note that the rst condition implies that ϕ is a biholomorphism on its image, which is a classical result ([8, Theorem 2.2.1]). Also note that the second condition means that the set ϕ(k) is Ω-convex for each compact Ω-convex subset K Ω ([7, Theorem 4.3.3]). This immediately implies that for any integer number n the set ϕ [n] (K) also is Ω-convex. [n Proof of Proposition 3.1. Let f be a universal function for (C l ] ϕ ) l. The rst part is obvious. For the second we need to prove that the restrictions g ϕ(ω), g O(Ω), are dense in O(ϕ(Ω)). Let h O(ϕ(Ω)). Then h ϕ is holomorphic on Ω, so there is a sequence (l k ) k such that f ϕ [n l ] k h ϕ on Ω. Hence f ϕ [n l 1] k h on ϕ(ω), as the mapping ϕ is a biholomorphism on its image. We prove the third part. Let K Ω be compact. For each j N there exists l j such that f ϕ [n l ] j j 1 on K. This implies j inf f(z) = inf f z ϕ [n l ] j z K ϕ[n l ] j (z) j 1 (K) j > sup f(z) for big j, z K so ϕ [n l j ] (K) K =. Necessary conditions for hereditary universality are given by the following twin proposition: Proposition 3.3. Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Suppose that C ϕ is hereditarily hypercyclic w.r.t. (n l ) l. Then: (h1) The mapping ϕ is injective. (h2) The image ϕ(ω) is a Runge domain w.r.t. Ω. (h3) The sequence (ϕ [n l] ) l is compactly divergent. Proof. The rst two parts follow from the previous proposition. For the last one it is enough to recall that a sequence of holomorphic mappings is compactly divergent if and only if each of its subsequences is run-away. 6

8 It is a natural to ask whether the necessary conditions given by Propositions 3.1 and 3.3 are sucient. In Section 5 we give some classes of domains where this is indeed true. But in general this is not so, as we can see using a simple example Ω = D and ϕ(z) = 1 2 z (then by Theorems 3.4 or 7.1 the operator C ϕ is not hypercyclic, although it satises the conditions (c1), (c2), (c3) and (h1), (h2), (h3)). Theorem 3.4. Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Then: (1) The operator C ϕ is hypercyclic w.r.t. (n l ) l if and only if ϕ is injective and for every Ω-convex compact subset K Ω there exists l such that K ϕ [n l] (K) = and the set K ϕ [n l] (K) is Ω-convex. (2) The operator C ϕ is hereditarily hypercyclic w.r.t. (n l ) l if and only if ϕ is injective and for every Ω-convex compact subset K Ω there exists l 0 such that K ϕ [nl] (K) = and the set K ϕ [nl] (K) is Ω-convex for any l l 0. Remark 3.5. Note that it suces to prove the above conditions only for some exhaustion (K l ) l of Ω with K l being Ω-convex. This fact is extensively used throughout this paper. Indeed, if K Ω is an arbitrary compact and Ω-convex set, then taking l 0 such that K K l0 we get that for l l 0 the following implication holds: if n l is such that K l ϕ [nl] (K l ) = and the set K l ϕ [nl] (K l ) is Ω-convex, then the same condition holds with K l replaced by K. This holds by Lemma 2.5, (3): there exists a function that separates K l and ϕ [nl] (K l ), and so it separates K and ϕ [nl] (K). Proof of Theorem 3.4. First we prove (1). Necessity. Suppose that C ϕ is hypercyclic w.r.t. (n l ) l. Fix K. By Remark 3.2 we get that the set ϕ [nl] (K) is Ω-convex. Using condition (tt) for g 0, h 1, ɛ = 1 we get that there are F O(Ω, C) and l N 2 such that F (K) 1D and F 2 (ϕ[nl] (K)) 1 + 1D. This implies that K and 2 ϕ[nl] (K) are disjoint, and by Lemma 2.5 the sum K ϕ [nl] (K) is Ω-convex. Suciency. We prove that the condition (tt) is satised. Fix a compact set K Ω, a number ɛ > 0 and holomorphic functions g, h : Ω C. Take l such that K ϕ [nl] (K) = and the set I := K ϕ [nl] (K) is Ω-convex. The function f dened as g in a neighborhood of K and h ϕ [ n l] in a neighborhood of ϕ [nl] (K) is well-dened and holomorphic in a neighborhood of an Ω-convex set I, so it can be approximated by functions holomorphic on Ω (see [7, Theorems and 4.3.4]). This implies that there exists a function f O(Ω) such that f g < ɛ on K and f h ϕ [ nl] < ɛ on ϕ [nl] (K). Observe, that (2) follows immediately from (1). Indeed, the condition in (2) does not hold if and only if there is an Ω-convex subset K Ω and an increasing sequence (l k ) k such that for each k the sets K and ϕ [n l ] k (K) are not disjoint or their sum is not Ω-convex. On the other hand, C ϕ is not hereditarily hypercyclic w.r.t. (n l ) l if and only if for some increasing sequence (l k ) k it is not hypercyclic w.r.t. (n lk ) k. In view of (1), these conditions are equivalent. From the above theorem it follows an interesting observation: Observation 3.6. Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Then the operator C ϕ is hypercyclic w.r.t. (n l ) l if and only if (n l ) l has a subsequence for which C ϕ is hereditarily hypercyclic. 7

9 Proof. Indeed, if C ϕ is hypercyclic w.r.t. (n l ) l and (K µ ) µ is a sequence of Ω-convex compact sets which exhausts Ω, then for each µ there is l µ such that K µ ϕ [n lµ ] (K µ ) = and the set K µ ϕ [n lµ ] (K µ ) is Ω-convex. We may assume that the sequence (l µ ) µ is increasing. Then, by Theorem 3.4, the operator C ϕ is hereditarily hypercyclic w.r.t. the sequence (n lµ ) µ (argue as in Remark 3.5). Remark 3.7. In many 'nice' domains Ω there holds a stronger implication than the above: for any ϕ O(Ω, Ω) hypercyclicity of C ϕ implies its hereditary hypercyclicity. We deal with this topic in Section 6. Corollary 3.8. Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Then: (1) The operator C ϕ is hypercyclic w.r.t. (n l ) l if and only if ϕ is injective, ϕ(ω) is a Runge domain w.r.t. Ω and for every Ω-convex compact subset K Ω there are l N and F O(Ω) such that the sets (F (K)) and ( F (ϕ [n l] (K)) ) are disjoint. (2) The operator C ϕ is hereditarily hypercyclic w.r.t. (n l ) l if and only if ϕ is injective, ϕ(ω) is a Runge domain w.r.t. Ω and for every Ω-convex compact subset K Ω there exists l 0 such that for any l l 0 there is F O(Ω, D) for which the sets (F (K)) and ( F (ϕ [n l] (K)) ) are disjoint. Proof. Suciency in both parts follows immediately from Theorem 3.4. Indeed, if the sets (F (K)) and ( F (ϕ [n l] (K)) ) Ω are disjoint, then K and ϕ [n l] (K) are also disjoint and, by Lemma 2.5, their sum is Ω-convex (since ϕ(ω) is a Runge domain in Ω, ϕ [n l] (K) is Ω-convex if K is so). Necessity. In both parts, using Theorem 3.4 for given K we get suitable l or l 0. Now using Lemma 2.5 for the sets K and ϕ [n l] (K) we get a required function F. Remark 3.9. Observe that from the above corollary it follows that (for an injective ϕ with ϕ(ω) being a Runge domain w.r.t. Ω) to get hypercyclicity of C ϕ w.r.t. (n l ) l it suces to prove condition (tt) for g 0, h 1 and ɛ = 1, i.e. to prove that for each 2 compact Ω-convex subset K Ω there is a number l and a function f O(Ω) such that f < 1 2 on K and f 1 < 1 2 on ϕ[n l] (K). Theorem Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Suppose that ϕ is injective and that ϕ(ω) is a Runge domain w.r.t. Ω. If there exists a point z 0 Ω such that lim c Ω(z 0, ϕ [nl] (z 0 )) =, l then the operator C ϕ is hereditarily hypercyclic w.r.t. (n l ) l. Note that the limit condition in the assumptions means that there exists a sequence (F l ) l O(Ω, D) such that F l (z 0 ) = 0 and F l (ϕ [n l] (z 0 )) 1 as l. 8

10 Proof. Since every subsequence of (n l ) l satisfy the same assumptions, it suces to prove hypercyclicity. Take (F l ) l as above. Using the Montel theorem and passing to a subsequence we may assume that F l ϕ [n l] G and F l H for some functions G, H O(Ω) with G(Ω), H(Ω) D. Since G(z 0 ) = 1 and H(z 0 ) = 0, if follows from the maximum principle that G 1 and H(Ω) D. Fix a compact Ω-convex subset K Ω. There is an α (0, 1) so that H(K) αd, so for big l there holds F l (K) αd. On the other hand, F l (ϕ [n l] (K)) 1 + (1 α)d, because F l ϕ [n l] 1. This implies (F l (K)) ( F l (ϕ [n l] (K)) ) αd (1+(1 α)d) =. Now apply Corollary 3.8. Remark The assumption lim l c Ω (z 0, ϕ [n l] (z 0 )) = in Theorem 3.10 is in particular fullled in each of the following situations: (1) If there exists a function F O(Ω, D) such that F ϕ [n l] (z 0 ) 1 as l. (2) If for some point c Ω the sequence (ϕ [n l] ) l converges to a constant c, and there exists a peak function for c and Ω. Given a sequence (ϕ l ) l of holomorphic self-maps of a domain Ω C N, we say that a sequence (C ϕl ) l is B-universal if the sequence (C ϕl B(Ω) ) l is universal as a sequence of mappings from B(Ω) to B(Ω), where B(Ω) := O(Ω, D) (endowed with the topology induced from O(Ω)). In [4, Theorem 3.5 and Corollary 3.7] and [6, Theorem 2.4] the authors proved that, under certain assumptions, B-universality of sequences of composition operators (C ϕl ) l implies its universality, where ϕ l are holomorphic self-maps of Ω. Restricting to the case of iterations, i.e. ϕ l = ϕ [n l], and to Ω being pseudoconvex, we give some improvement of their results: Corollary Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. Suppose that ϕ is injective and that ϕ(ω) is a Runge domain w.r.t. Ω. If the sequence (C ϕ [n l ] ) l is B-universal, then it is universal, i.e. the operator C ϕ is hypercyclic w.r.t. (n l ) l. Proof. Let f B(Ω) be a universal function for (C ϕl B(Ω) ) l, and let (K µ ) µ be an exhaustion of Ω. For every µ there is some number l µ such that f ϕ [n lµ ] (1 1 µ ) < 1 µ on K µ. This gives f ϕ [n lµ ] 1 on Ω as µ. By Remark 3.11, C ϕ is hereditarily hypercyclic w.r.t. (n lµ ) µ and hence hypercyclic w.r.t. (n l ) l. We nish this section with some simple consequence of Theorem 3.4: Corollary Let Ω C N be a pseudoconvex domain, ϕ O(Ω, Ω) and let (n l ) l N be an increasing sequence. For M N introduce the operator C ϕ,m : O(Ω, C M ) f f ϕ O(Ω, C M ). is hyper- If C ϕ is hypercyclic (resp. hereditarily hypercyclic) w.r.t. (n l ) l, then C ϕ,m cyclic (resp. hereditarily hypercyclic) w.r.t. (n l ) l for every M. 9

11 Proof. It is enough to consider the case of hypercyclicity. Obviously ϕ is injective. We [n prove that the sequence (C l ] ϕ,m ) l ) is topologically transitive (observe that Corollary 2.4 holds in fact also for the operator C ϕ,m ). Fix a number ɛ > 0, a compact Ω-convex subset K Ω and functions g 1,..., g M, h 1,..., h M O(Ω). By Theorem 3.4, there is l N such that the sets K and ϕ [nl] (K) are disjoint and their sum is Ω-convex. We need to show that there exist functions f 1,..., f M O(Ω) such that for j = 1,..., M there is f j g j < ɛ on K and f j h j ϕ [ n l] < ɛ on ϕ [n l] (K). But, as in the proof of Theorem 3.4, this follows from the fact that the functions f j dened as g j in a neighborhood of K and as h j ϕ [ n l] in a neighborhood of ϕ [n l] (K) can be approximated on the Ω-convex set K ϕ [n l] (K) by functions holomorphic on Ω. 4 Convergent subsequences When we resolve Ω-convexity of the set K ϕ [n l] (K), as it is required in Theorem 3.4, we imagine the set ϕ [n l] (K) as a compact set lying very close to the boundary of Ω (as it is disjoint with the set K, which is usually big). Thus, it becomes natural to consider the situation when the sets ϕ [n l] (K) tend to some limit compact set L(K) Ω. There arises a question: what conditions should satisfy L(K) to give hypercyclicity of C ϕ w.r.t. (n l ) l? In this section we give a partial answer. Observe that the assumption that ϕ [n l] (K) tends to L(K) as l implies that the sequence (ϕ [n l] ) l is compactly divergent and bounded on compact subsets of Ω, so by the Montel theorem it has a convergent subsequence (which is also compactly divergent in O(Ω, Ω)). Passing to that subsequence we get ϕ [n l] ϕ on Ω for some holomorphic mapping ϕ : Ω C N with ϕ(ω) Ω, so in fact we can study the situation when (ϕ [n l] ) l has a convergent subsequence which is compactly divergent in O(Ω, Ω). Note also that the situation considered in this section occurs always when Ω is bounded, since in virtue of the Montel theorem we can always choose a convergent subsequence. Theorem 4.1. Let Ω, Ω 0 C N be pseudoconvex domains, Ω Ω 0, and let ϕ : Ω Ω be an injective holomorphic mapping such that the image ϕ(ω) is a Runge domain w.r.t. Ω. Suppose that for an increasing sequence (n l ) l N the subsequence (ϕ [n l] ) l converges to some holomorphic mapping ϕ : Ω C N such that ϕ(ω) Ω 0 Ω. If for each compact Ω-convex subset K Ω the set K ( ϕ(k)) Ω0 is Ω 0 -convex and ( ϕ(k)) Ω0 Ω, then C ϕ is hereditarily hypercyclic w.r.t. (n l ) l. Remark 4.2. Note that the assumptions of the above theorem imply that the domain Ω is a Runge domain w.r.t. Ω 0. It follows directly from Lemma 2.6. Proof of Theorem 4.1. We use Theorem 3.4. Fix a compact Ω-convex subset K Ω and set L := ( ϕ(k)) Ω0. Let U Ω, V Ω 0 be open, disjoint and such that K U, L V. Since U V is an open neighborhood of Ω 0 -convex set K L, we can nd an analytic polyhedron (w.r.t. Ω 0 ) contained in U V and containing K L ([7, Lemma 5.3.7]). Therefore we may assume that U V = (f 1,..., f ν ) 1 (D ν ) for some functions f 1,..., f ν 10

12 holomorphic on Ω 0. For l big enough we have ϕ [n l] (K) V Ω. Set I := K ϕ [n l] (K). Remark 4.2 implies that Î Ω = ÎΩ 0 U V (the last inclusion follows from the fact that U V is an analytic polyhedrons in Ω 0 ). Therefore ÎΩ = ÎΩ Ω U (V Ω), so by Lemma 2.5 the set I is Ω-convex. Theorem 4.3. Let Ω 0 C N be a pseudoconvex domain, c R, u P SH(Ω 0 ) and let Ω be a connected component of the set {u < c}. Suppose that near a point z 0 Ω 0 Ω the function u is strictly plurisubharmonic. Let ϕ : Ω Ω be an injective holomorphic mapping with ϕ(ω) being a Runge domain w.r.t. Ω. If for an increasing sequence (n l ) l N the subsequence (ϕ [nl] ) l is convergent to a holomorphic mapping ϕ : Ω C N such that z 0 ϕ(ω), then C ϕ is hereditarily hypercyclic w.r.t. (n l ) l. This theorem works in the case when the image of the limit function ϕ contains a point of strict convexity of Ω, i.e. a point z 0 Ω for which there exists a strictly plurisubharmonic local dening function for Ω. Proof. Let z 1 Ω be such that ϕ(z 1 ) = z 0. The function u ϕ is plurisubharmonic and bounded from above by c. In the point z 1 it achieves its local maximum, so it is constant and equal to c near z 1. Therefore for each z near z 1 and X C N we have 0 = 2 (u ϕ) λ λ (z + λx) = λ=0 j,k 2 u z j z k ( ϕ(z)) Y j Y k, where Y = ϕ (z)(x). But the sum on the right side is positive if Y 0, because u is strictly plurisubharmonic near the boundary point ϕ(z). This implies ϕ (z)(x) = 0 for every X, so ϕ is constant and equal to z 0. In particular, ϕ(ω) Ω 0 Ω. Observe that Ω is a Runge domain w.r.t. Ω 0. Indeed, if K Ω is compact, connected and Ω-convex, then K Ω0 = K P SH(Ω0 ) {u < c}, so K Ω0 Ω. Here K P SH(Ω0 ) denotes the plurisubharmonic hull of K w.r.t. Ω 0 ([7, Denition 2.6.6]); it is equal to K Ω0 when Ω 0 is pseudoconvex ([7, Theorem 4.3.4]). In virtue of Theorem 4.1 it suces to prove that the set K {z 0 } is Ω 0 -convex if K Ω is compact and Ω-convex. Such a set K is Ω 0 -convex, so there is f O(Ω 0 ) with f(z 0 ) > sup z K f(z). Now, Lemma 2.5 for L = {z 0 } does the job, because f satises the condition (3). 5 Simply characterization of hypercyclicity The general characterization of hypercyclicity given by Theorem 3.4 requires resolving Ω- convexity of the sets K ϕ [n] (K), while characterization in C, given by [6, Theorem 3.21] (see also Theorem 7.2 in this paper), requires only satisfying the necessary conditions (c1), (c2), (c3), which is much simpler. Hence there arises a natural question about domains in C N in which that necessary conditions are also sucient for hypercyclicity. A simple example Ω := D \ {0} with the mapping ϕ(z) := 1 z shows that there are domains 2 where this is not true. In this section we present some natural classes of domains which supports this 'simply characterization'. 11

13 Denition 5.1. Let Ω C N be a pseudoconvex domain. By saying that '(c1), (c2), (c3) are sucient for hypercyclicity in Ω' (resp. '(h1), (h2), (h3) are sucient for hereditary hypercyclicity in Ω') we mean that the operator C ϕ is hypercyclic w.r.t. (n l ) l (resp. hereditarily hypercyclic w.r.t. (n l ) l ) for every mapping ϕ O(Ω, Ω) and every increasing sequence (n l ) l satisfying the conditions (c1), (c2), (c3) (resp. (h1), (h2), (h3)). Let us begin with a general fact: Proposition 5.2. Let Ω C N be a pseudoconvex domain. Then (c1), (c2), (c3) are sucient for hypercyclicity in Ω if and only if (h1), (h2), (h3) are sucient for hereditary hypercyclicity in Ω. Since the notions dened above are equivalent, we formulate the following denition: Denition 5.3. A pseudoconvex domain Ω C N is called hypercyclic if the conditions (c1), (c2), (c3) are sucient for hypercyclicity in Ω. Proof of Proposition 5.2. Suppose (c1), (c2), (c3) are sucient for hypercyclicity in Ω and choose ϕ and (n l ) l satisfying (h1), (h2), (h3). Then every subsequence of (n l ) l satises (c1), (c2), (c3), so every subsequence of (C ϕ [n l ] ) l is universal. Conversely, suppose that (h1), (h2), (h3) are sucient for hereditary hypercyclicity in Ω and choose ϕ and (n l ) l satisfying (c1), (c2), (c3). Since the sequence (ϕ [n l] ) l is runaway, it has a compactly divergent subsequence. This subsequence satises (h1), (h2), (h3), so (C ϕ [n l ] ) l is universal, because it has a hereditarily universal subsequence. Theorem 5.4. If Ω C N is a pseudoconvex domain for which there exists a point z 0 Ω so that lim Ω z Ω c Ω (z, z 0 ) =, then Ω is hypercyclic. Note that the assumption lim Ω z Ω c Ω (z, z 0 ) = means that all the balls w.r.t. the Carathéodory pseudodistance are relatively compact in Ω. If it holds for some z 0, then it holds for every z 0 Ω (this is an easy consequence of the triangle inequality). Proof. Choose ϕ and (n l ) l satisfying (c1), (c2), (c3). Since (ϕ [n l] ) l is run-away, by passing to a subsequence we may assume that it is compactly divergent in O(Ω, Ω). We have ϕ [n l] (z 0 ) Ω, so c Ω (z 0, ϕ [n l] (z 0 )). Now Theorem 3.10 does the job. Although most of the classes of domains which we consider below are bounded, note that this assumption is not required in the theorem above. Corollary 5.5. The following domains satisfy the assumptions of Theorem 5.4 and hence are hypercyclic: (1) A bounded pseudoconvex domain Ω C N such that for each point a Ω there is a function f a O(Ω, D) such that f a (z) 1 as z a, z Ω. (2) A bounded strictly pseudoconvex domain in C N. (3) A bounded convex domain in C N. 12

14 (4) A relatively compact analytic polyhedron w.r.t. a pseudoconvex domain Ω 0 C N, i.e. a domain Ω which is relatively compact in Ω 0 and is a connected component of the set (f 1,..., f m ) 1 (D m ) for some functions f 1,..., f m O(Ω 0 ). For hypercyclic domains in C see Theorem 7.2. Proof. (1): Suppose that for some sequence (z n ) n Ω, z n Ω and some constant there is c Ω (z 0, z n ) < M. Since Ω is bounded, we have z n Ω and (passing to a subsequence) we may assume that z n a for some boundary point a. Therefore f a (z n ) 1, so c Ω (z 0, z n ) ; a contradiction. (2): It follows from [9, Theorem ]. (3): This is a consequence of (1). Indeed, if a Ω, then there is a real linear functional L : C N R such that L(a) = 0 and L < 0 on Ω. We have L = Re L for some complex linear functional L. Dening f(z) := e L(z) iim L(a) we obtain a peak function for a. (4): This is also an immediate consequence of (1). 6 Hereditary hypercyclicity of C ϕ In this section we deal a bit more with hereditary hypercyclicity. As it was said in Observation 3.6, hypercyclicity of C ϕ is equivalent to its hereditary hypercyclicity w.r.t. some increasing sequence (n l ) l N. But, as we shall see in this section, there are domains in which for every mapping ϕ O(Ω, Ω) hypercyclicity of C ϕ gives its hereditary hypercyclicity, which is the strongest of conditions (p1) - (p4). First, let us recall the following fact (see [1, Theorem 2.4.3]): Theorem 6.1. Let Ω C N be a taut domain and let ϕ O(Ω, Ω). If the sequence (ϕ [n] ) n is not compactly divergent in O(Ω, Ω), then it is relatively compact in O(Ω, Ω). We say that a domain Ω C N is taut if every sequence (f n ) n O(D, Ω) is compactly divergent in O(D, Ω) or has a subsequence convergent to an element of O(D, Ω); it is well known that every taut domain in C N is pseudoconvex. The reader can nd more informations in [9]. We prove the following: Theorem 6.2. Let Ω C N be a hypercyclic taut domain and let ϕ : Ω Ω be a holomorphic mapping. If the operator C ϕ is hypercyclic, then it is hereditarily hypercyclic. Note that tautness does not imply hypercyclicity of a domain, and vice versa. The examples are simply: D is taut, but not hypercyclic, while C is hypercyclic, but not taut (see Theorem 7.2). Proof. Hypercyclicity of C ϕ implies that ϕ satises the conditions (c1), (c2), (c3) with n l = l. The last condition implies that (ϕ [n] ) n cannot be relatively compact in O(Ω, Ω), because it has a compactly divergent subsequence. Therefore by Theorem 6.1 the sequence (ϕ [n] ) n is compactly divergent, so ϕ satises the conditions (h1), (h2), (h3) with n l = l. Then, as Ω is hypercyclic, the operator C ϕ must be hereditarily hypercyclic. Combining Theorems 5.4 and 6.2 we get: 13

15 Theorem 6.3. Let Ω C N be a pseudoconvex domain for which there exists a point z 0 Ω so that lim c Ω (z, z 0 ) =. Ω z Ω Then Ω is taut and hypercyclic. In particular, if ϕ O(Ω, Ω) is such that C ϕ if hypercyclic, then C ϕ is hereditarily hypercyclic. Remark 6.4. By Corollary 5.5, for all the domains listed there hypercyclicity of C ϕ implies its hereditary hypercyclicity. For domains in C having this property see Theorem 7.2. Example 8.5 shows that there are taut domains which does not satisfy the assumptions of Theorem 6.3, but satisfy its conclusion. Proof of Theorem 6.3. In virtue of Theorems 5.4 and 6.2 it suces to prove that Ω is taut. Actually, this is a well-known fact, but we could not nd it in the literature in that form, so for the reader's convenience we present a sketch of proof. Fix a sequence (f n ) n O(D, Ω) and suppose that it is not compactly divergent. Passing to a subsequence (during the proof we do it few times) we may assume that there are compact sets K D and M Ω such that f n (K) M for every n. We show that for each compact subset L D the set I L := n=1 f n(l) is relatively compact in Ω. Suppose that for some L it is not. Hence there is a sequence (w k ) k I L such that w k Ω as k. There is w k = f nk (z k ) for some sequences (z k ) k L and (n k ) k N. Since for each m the sum m n=1 f n(l) is relatively compact in Ω, the set {n k : k N} is innite. Therefore we may assume that the sequence (n k ) k is increasing and, in addition, that z k tends to some z 0 L. For every k there is a function F k O(Ω, D) such that F k (z 0 ) = 0 and F k f nk (z k ) = F k (w k ) = c Ω(z 0, w k ). By the Montel theorem, we may assume that F k G on Ω and F k f nk H on D for some functions G O(Ω), H O(D), with G(Ω), H(D) D. We have G(z 0 ) = 0 and H(z 0 ) = lim k F k f nk (z k ) = lim k c Ω(z 0, w k ) = 1. In view of the maximum principle, H 1 and G(Ω) D. Since f nk (K) M for each k, there is min F k f nk = min F k min F k max F k max F k. K f nk (K) f nk (K) M f nk (K) M M The left side tends to 1, while the right side tends to max M G, which is less than 1; a contradiction. The set I L is relatively compact in Ω for every compact set L, so the sequence (f n ) n is locally uniformly bounded on D. By the Montel theorem is has a subsequence convergent to a map f O(D, C N ). The inclusion f(d) Ω follows again from the fact that I L Ω. Proposition 6.5. Let Ω C N be a pseudoconvex domain and ϕ O(Ω, Ω). 14

16 (1) If C ϕ is hypercyclic, then it is hypercyclic with respect to each increasing arithmetic sequence of natural numbers. (2) If C ϕ is hereditarily hypercyclic with respect to some increasing arithmetic sequence of natural numbers, then it is hereditarily hypercyclic. Proof. (1): Choose a sequence of the form (lµ + ν) l, where µ N, µ > 0 and ν {0,..., µ 1}. The fact that C ϕ if hypercyclic w.r.t (lµ) l follows from [5, Theorem 8] (this theorem states even more: the sequences (C ϕ [l] ) l and (C ϕ [lµ] ) l have the same set of universal elements). Let f be a universal function for (C ϕ [lµ] ) l. Take h O(Ω). The function h ϕ [ ν] is holomorphic on ϕ [ν] (Ω) which is a Runge domain w.r.t. Ω. Therefore it can be approximated by elements of O(Ω), so for some increasing sequence (l k ) k N there is f ϕ [l kµ] h ϕ [ ν] on ϕ [ν] (Ω), as k. Hence f ϕ [l kµ+ν] h on Ω. (2): Assume that C ϕ is hereditarily hypercyclic w.r.t. a sequence (lµ + ν) l, where µ, ν are as above. It suces to show that for each r {0,..., µ 1} \ {ν} the operator C ϕ is hereditarily hypercyclic w.r.t. (lµ + r) l. Fix r and an increasing sequence (l k ) k N. We have µ + r ν > 0. As previously, given h O(Ω) we approximate the function h ϕ [ν r µ] on ϕ [µ+r ν] (Ω) by functions from O(Ω). The sequence (C ϕ [(l k 1)µ+ν] ) k is universal, so there is an increasing sequence (k m ) m N such that f ϕ [(l km 1)µ+ν] h ϕ [ν r µ] on ϕ [µ+r ν] (Ω) as m. This implies f ϕ [l km µ+r] h on Ω, because the mapping ϕ [µ+r ν] : Ω ϕ [µ+r ν] (Ω) is a biholomorphism. Problem 6.6. Do there exist a pseudoconvex domain Ω C N and a holomorphic map ϕ : Ω Ω such that C ϕ is hypercyclic but non-hereditarily hypercyclic? 7 One-dimensional case In [6, Theorems 3.19 and 3.21] Grosse-Erdmann and Mortini gave a complete characterization of hypercyclicity in one-dimensional case, which after a slight reformulation takes the following form: Theorem 7.1. Let (ϕ l ) l be a sequence of injective holomorphic self-maps of a domain Ω C. Then: (1) If Ω is simply connected, then (C ϕl ) l is universal if and only if (ϕ l ) l is run-away. (2) If Ω is nitely connected but not simply connected, then (C ϕl ) l is never universal. (3) If Ω is innitely connected, then (C ϕl ) l is universal if and only if for every compact Ω-convex subset K Ω and for every l 0 there is l l 0 such that ϕ l (K) is Ω-convex and ϕ l (K) K =. 15

17 Although Grosse-Erdmann and Mortini dened Ω-convexity in dierent way (they said that a compact subset K Ω of a domain Ω C is Ω-convex if every hole of K contains a point of C\Ω), our denition agrees with their in dimension one. Here by hole of K we mean a bounded connected component of C \ K. Equivalence of both denitions follows from [7, Theorems and 1.3.3]. As a corollary from our considerations and from the above theorem applied to mappings ϕ l := ϕ [n l] we obtain: Theorem 7.2. Let Ω C be a simply connected or an innitely connected domain. Then: (1) Ω is hypercyclic. (2) If for a mapping ϕ O(Ω, Ω) the operator C ϕ is hypercyclic, then it is hereditarily hypercyclic. Proof. Let Ω be simply connected or innitely connected. Part (1) follows directly from Theorem 7.1. We prove (2). It is known that a domain Ω C is taut if and only if the set C \ Ω has at least two points (see [9, Remark (d)]), so if Ω C, then Ω is taut and Theorem 6.2 does the job. It remains to consider the case Ω = C. Fix ϕ for which C ϕ is hypercyclic. By the Picard theorem the set C \ ϕ(c) contains at most one point. But since ϕ is a homeomorphism on its image, there must be ϕ(c) = C and so ϕ is an automorphism of C. Therefore ϕ is an ane endomorphism. Now it suces to use [2, Theorem 3.1]: it says (in particular) that for ϕ being an ane endomorphism of C N, the composition operator C ϕ : O(C N ) O(C N ) is hypercyclic if and only if it is hereditarily hypercyclic. As an interesting observation (but not connected with the topic of this paper), it follows from Corollary 5.5 and Theorem 7.2 a well-known fact that: Remark 7.3. No holomorphic self-mapping of an annulus in C satisfy all the conditions (c1), (c2) and (c3). 8 Examples Let us begin with the following observation: Proposition 8.1. Let a C N 2 and let D C N 1, Ω C N 1 C N 2 be pseudoconvex domains such that D {a} Ω D C N 2. Let ϕ : Ω Ω be a holomorphic mapping of the form ϕ(z, w) = (σ(z), ψ(z, w)), (z, w) Ω, where σ : D D and ψ : Ω C N 2. If ϕ is injective, the image ϕ(ω) is a Runge domain w.r.t. Ω and the operator C σ is hypercyclic (resp. hereditarily hypercyclic) w.r.t. an increasing sequence (n l ) l N, then C ϕ is also hypercyclic (resp. hereditarily hypercyclic) w.r.t. (n l ) l. 16

18 Proof. It is enough to consider the case of hypercyclicity. Fix K Ω compact and Ω- convex, and set L := (π C N 2 (K)) D. By Corollary 3.8, there are l N and f O(D) such that ( f(σ [n l ] (L)) ) (f(l)) =. Dene g(z, w) := f(z) for (z, w) Ω. We have ϕ [n l] (K) σ [n l] (L) C N 2, so ( g(ϕ [n l ] (K)) ) (g(k)) ( f(σ [n l] (L)) ) (f(l)) =, so C ϕ is hypercyclic w.r.t. (n l ) l (use Corollary 3.8). We get an immediate corollary: Corollary 8.2. For any j = 1,..., m, let Ω j C N j be a pseudoconvex domain and let ϕ j O(Ω j, Ω j ). Suppose that for every j the mapping ϕ j is injective and its image is a Runge domain w.r.t. Ω j. Dene Ω := Ω 1... Ω m and ϕ := ϕ 1... ϕ m. If there exists j 0 such that the operator C ϕj0 is hypercyclic w.r.t. an increasing sequence (n l ) l N, then the operator C ϕ is hypercyclic w.r.t. (n l ) l. Example 8.3. Put Ω = D D. Then, although D does not have any self-mappings with hypercyclic composition operator, the domain Ω has many of them, e.g. every mapping of the form ϕ(z, w) = (ψ(z), w) for ψ O(D, D) such that C ψ is hypercyclic. Example 8.4. Consider the Hartogs domain given by Ω := {(z, ζ) D C : ζ e u(z) < 1}, where D C N is pseudoconvex and u : D [, ) is plurisubharmonic, u. It is well-known that such a domain Ω is pseudoconvex. Consider the following situation, which is in a certain sense opposite to that in Proposition 8.1: let ϕ be of the form where ψ O(Ω, C). Suppose that C ϕ is hypercyclic. Then: ϕ(z, ζ) = (z, ψ(z, ζ)), (1) The function u is equal to log G for some G O(D, C ), and hence the domain Ω is biholomorphic to D D. (2) The operator C ϕ is hereditarily hypercyclic. Put ψ z := ψ(z, ), Ω z = {ζ C : ζ < e u(z) } for z D. Then ψ z O(Ω z, Ω z ) and Ω z is a disc when u(z), and the whole C in the opposite case. We have ϕ [n] (z, ζ) = (z, ψ z [n] (ζ)) for each n N. Since ϕ is injective and the sequence (ϕ [n] ) n is run-away, for each z D the mapping ψ z is injective and the sequence (ψ [n] z ) n is run-away. Therefore, as Ω z is simply connected, in view of Theorems 7.1 and 7.2, the operator C ψz is hereditarily hypercyclic for every z. 17

19 We prove (1). As the sequence (ϕ [n] ) n is run-away, it has a compactly divergent subsequence, say (ϕ [µ l] ) l. Set τ l (z) := ψ z [µ l ] (0) for z D. The sequence (id D, τ l ) l is compactly divergent in O(D, Ω), because ϕ [µ l] (z, 0) = (z, τ l (z)). Let W D be a domain. As u is upper semi continuous, it is bounded from above on W, so there is an ɛ > 0 such that W ɛd Ω. As (id D, τ l ) l is compactly divergent, we get that for l greater than some l 0 the functions τ l are zero-free on W and the sequence ( 1 τ l ) l l0 is uniformly bounded on W. Using the above conclusion and the Montel theorem, we shall prove that the sequence ( 1 τ l ) l has a subsequence convergent locally uniformly on D to a function H O(D). Indeed, let (W p ) p be a sequence of domains such that W p W p+1 and p W p = D. We apply the Montel theorem. There is an increasing sequence (lp) 1 1 p N such that τ l 1 p converges to a function H 1 O(W 1 ) locally uniformly on W 1. Next, there is (lp) 2 p (lp) 1 p such that 1 τ l 2 p such that 1 τ q lp converges to H 2 O(W 2 ) on W 2. Doing so, for every q we get a sequence (l q p) p converges to H q O(W q ) on W q, and (lp) q p (lp q 1 ) p. We have H q1 = H q2 on W q1 W q2, and hence we may dene a function H O(D) as H(z) = H q (z) for z W q. For each q there is l q such that H 1 τ lq < 1 q on W q. We may assume that the sequence (l q ) q is increasing. Then the sequence 1 τ lq converges to H locally uniformly on D. As it was said, the sequence (id D, τ l ) l is compactly divergent in O(D, Ω). Therefore τ l (z) e u(z) as l for each z D, and hence H(z) = e u(z) for z D. Observe that H is zero-free on D. Indeed, as u, there is H 0. If H(z) = 0 for some z, then by the Hurwitz theorem 1 τ lp has a zero near z for p big enough, what is a contradiction. The mapping Ω (z, ζ) (z, ζh(z)) D D is a required biholomorphism between Ω and D D. We prove (2). Via the above biholomorphic equivalence we may assume that Ω = D D. Fix an increasing sequence (n l ) l N. The sequence (ϕ [nl] ) nl is locally uniformly bounded on Ω, so passing to a subsequence we get ϕ [nl] ϕ. There is ϕ(z, ζ) = (z, ψ(z, ζ)) for some ψ O(Ω, C) such that ψ(ω) D. Therefore ψ is equal to some constant c D. Putting f(z, ζ) := 1 ζ allows us to apply Remark The part (2) is c proved. Note that in general there exist mappings ϕ of the form considered above for which the operator C ϕ is hypercyclic. Indeed, put D := D. Via conformal equivalence we may replace the set D D with Ω := H + H +, where H + is the right half-plane in C, i.e. H + := {Re > 0}. Now set Φ(z, w) := (z, w + z) and ϕ := Φ Ω. It is clear that ϕ(ω) Ω, ϕ is an injection and the sequence (ϕ [n] ) n is compactly divergent in O(Ω, Ω). The mapping Φ is an automorphism of C 2 and Ω is a Runge domain in C 2, so the image Φ(Ω) is also a Runge domain in C 2 and hence in Ω. Therefore the conditions (h1), (h2), (h3) are satised for ϕ. By Theorem 6.3, the domain Ω is hypercyclic and the operator C ϕ is hereditarily hypercyclic. 18

20 Example 8.5. Let Ω 0 C be an innitely connected domain and let a Ω 0. Dene Ω := Ω 0 \{a}. Then by Theorem 7.2 the domain Ω satises the conclusion of Theorem 6.3, but it does not satisfy the assumption that lim Ω z Ω c Ω (z, z 0 ) =. This follows from the classical Riemann extension theorem: there is c Ω = c Ω0 on Ω, Ω and c Ω (z 0, z) c Ω0 (z 0, a) < as z a, for any z 0. References [1] M. Abate, Iteration theory of holomorphic maps on taut manifolds, Mediterranean Press, [2] L. Bernal-Gonzalez, Universal entire functions for ane endomorphisms in C N, J. Math. Anal. Appl. 305 (2005) [3] J. Bonet, P. Doma«ski, Hypercyclic composition operators on spaces of real analytic functions, Math. Proc. Cambridge Phil. Soc. (to appear). [4] P. Gorkin, F. León-Saavedra, R. Mortini, Bounded universal functions in one and several complex variables, Math. Z. 258 (2008), [5] K. Grosse-Erdmann, Universal families and hypercyclic operators, Bull. Amer. Math. Soc. (N. S.) 36 (1999), [6] K. Grosse-Erdmann, R. Mortini, Universal functions for composition operators with non-automorphic symbol, J. Anal. Math. 107 (2009), [7] L. Hörmander, An introduction to complex analysis in several variables (3rd edition), North-Holland, Amsterdam (1990). [8] P. Jakóbczak, M. Jarnicki, Lectures on holomorphic functions of several complex variables, [9] M. Jarnicki, P. Pug, Invariant distances and metrics in complex analysis, Walter de Gruyter & Co., [10] T. Kalmes, M. Niess, Universal zero solutions of linear partial dierential operators, Studia Math. 198 (1) (2010),