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Compositional universality in the N-dimensional ball

Bernal González, Luis; Bonilla Ramírez, Antonio Lorenzo; Calderón Moreno, María del Carmen

Abstract

It is proved in this note that a sequence of automorphisms on the N-dimensional unit ball acts properly discontinuously if and only if its corresponding sequence of composition operators is universal on the Hardy space of such ball, and if and only if there exists a dense linear manifold of universal functions. Our result completes earlier ones by several authors.

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Analysis 26, 365–372 (2006) / DOI 10.1524/anly.2006.26.99.365 c Oldenbourg Wissenschaftsverlag, M¨ unchen 2006 Compositional universality in the N-dimensional ball L. Bernal-Gonz´ alez, A. Bonilla, M. C. Calder´ on-Moreno Received: November 29, 2005 Summary:It isproved inthisnote that asequence ofautomorphisms on the N-dimensional unitball acts properly discontinuously if and only if its corresponding sequence of composition operators is universal on the Hardy space of such ball, and if and only if there exists a dense linear manifold of universal functions. Our result completes earlier ones by several authors. 1 Introduction and notation Throughoutthis paper,we will denoteby N,C,Dthe set of positive integers,the complex plane and the open unit disk {z∈C:|z|<1}, respectively. The boundary of Dis the unit circle ∂D={z∈C:|z|=1}.IfN∈Nand G⊂CNis an open subset, then H(G)stands for the Fr´ echet space of holomorphic functions on Gendowed with the topology of uniform convergenceon compact subsets. A domain G⊂CNis a nonempty connected open subset of CN. A domain G⊂Cis said to be simply connected whenever its complement with respect to the extended complex plane is connected. If 1 ≤p<∞, the Hardy space Hp(D)is defined as Hp(D)={f∈H(D):fp<∞}, where fp=sup 0<r<11 2π2π 0|f(riθ)|pdθ1/p .ThenHp(D)becomes a Banach space if it is endowed with this norm. See [10] for an extensive study of Hardy spaces. The group Aut(D) of automorphisms of Dis the set of M¨ obius transformations {σa,k:|a|<1=|k|},whereσa,k(z)=k·z−a 1−az.In1941,W.SeidelandJ.L.Walsh[16] established the existence of a function f∈H(D)such that, given a simply connected domain G⊂Dand a function g∈H(G), there is a sequence {an}∞ 1⊂Ddependingon g such that f◦σan,1→g(n→∞)inH(G).This resultis in turn a non-Euclideanversion of Birkhoff’s theorem about density of translates of certain entire functions [3]. Recall AMS 2000 subject classifications: Primary: 47B33; Secondary: 32A35, 47A16 Key words and phrases: Hardy space, Seidel–Walsh theorem, composition operator, N-dimensional ball, algebraically generic universality Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM 366 Bernal-Gonz´ alez -- Bonilla -- Calder´ on-Moreno that if Gis a domain in CNthen it is said that the action of a sequence {Sn}∞ 1⊂Aut(G) is properly discontinuous if and only if, given a compact subset K⊂G, there exists m=m(K)∈Nsuch that K∩Sm(K)=∅. In 1995, A. Montes-Rodr´ıguez and the first author [2] extended Seidel–Walsh’s theorem by showing that if {Sn=σan,kn:n∈N}⊂ Aut(D), then the set {f∈H(D):{f◦Sn}is dense in H(D)}is not empty if and only if it is residual if and only if supn∈N|an|=1 if and only if the action of {Sn}∞ 1is properly discontinuous on D. In particular, if ϕ=σa,k(|a|<1, k=iθ)andSn=ϕ◦ ··· ◦ϕ(n times), then the set {f∈H(D):{f◦Sn}is dense in H(D)}is not empty if and only if it is residual if and only if ϕhas no fixed point in Dif and only if |sin(θ/2)|≤|a|. If ϕis a holomorphic selfmapping on D, then the composition mapping Cϕ:f∈ Hp(D)→ f◦ϕ∈Hp(D)is a well-defined linear operator on Hp(D)(see [19, pages 220–221]). In [5, Theorem 2.3 and Proposition 0.1] (see also [17, Chapter 7]) Bourdon and Shapiro showed that if ϕ∈Aut(D)and Sn=ϕ◦ ··· ◦ϕ(ntimes), then the set U={f∈H2(D):{f◦Sn}is dense in H2(D)}is residual in H2(D)if and only if it is not empty if and only if ϕhas no fixed point in D. A key idea of the proof is the fact that the sequence of iterates converges in H(D)to a point γ∈∂D, called the “Denjoy–Wolff point”: see [6] or [17, page 78], for instance. For N∈N, denoteby BNtheunitballinCN,thatis,BN={z=(z1,...,zN)∈CN: N j=1|zj|2<1}.Ifz=(z1,...,zn)and w=(w1,...,wN)then |z|=( N  j=1 |zj|2)1/2 and <z,w>will stand for the expression <z,w>= N  j=1 zjwj.Ifa∈BN, denote ϕa(z)=a−Pa(z)−(1−|a|2)1/2Qa(z) 1−<z,a>, where Pais the projection onto the space [a]spanned by a, i.e., Pa(z)=<z,a> |a|2a,and Qa=I−Pa, projection onto the orthogonal complement of [a]. Note that ϕa∈H(BN) (BN:= the closed unit ball) and that ϕinterchanges the values a,0. It is known (see, for instance, [15, Chapter 2] or [8, Theorem 2.72]) that if ϕ∈Aut(BN)then there exist a unitary transformation Mof CN(i.e., a linear operator that preserves inner products) and a point a∈BNsuch that ϕ=M◦ϕa(in fact, a=ϕ−1(0)). In [14], F. Le´ onSaavedra proves that if {Sn}∞ 1⊂Aut(BN), then there exists f∈H(BN)such that {f◦Sn:n∈N}is dense in H(BN)if and only if the action of {Sn}∞ 1is properly discontinuouson BN[14, Theorem5] if andonlyif limsup n→∞ |ϕ−1 n(0)|=1 [14, Proposition 2], see also [7, Theorem 3]. He also provides an analogous result for the unit polydisk DN={z=(z1,...,zN)∈CN:|zj|<1(j=1,...,N)}. Note that density in H(BN)does not imply density in Hp(BN)(see definition and some properties in Section 2 below), because convergence here is stronger than convergence in H(BN). In 2001, Xiaoman, Guangfu and Kunyu [18, Theorems 2–3] have shown—for the case p=2— that if ϕ∈Aut(BN)and ϕhas no fixed point in BNthen there exists f∈H2(BN) such that the set {f◦ϕn:n∈N}is dense in H2(BN),whereϕn=ϕ◦ ··· ◦ϕ(nfold). On the other hand, the second and third authors proved in [4, Theorem 3.1] that if Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM Compositional universality in the N -dimensional ball 367 {Sn}∞ 1⊂Aut(DN), then there exists f∈Hp(DN)such that {f◦Sn:n∈N}is dense in Hp(DN)if and only if the action of {Sn}∞ 1is properly discontinuous on BN. Note that this does not imply a correspondingresult on Hp(BN)since the ball and the polydisk are not biholomorphically isomorphic if N>1. All above results about density can be expressed in the language of universality. If X and Yare topologicalspaces, a sequence Tn:X→Y(n∈N) of continuousmappingsis called universalwheneverthe set Uofelements x∈Xsuch that the orbit {Tnx}∞ 1is dense in Yis not empty. Each element of Uis said to be universal for {Tn}∞ 1.IfXand Yare topological vector spaces and each Tnis linear then the words hypercyclic and universal are synonymous. See [12] and [13] for good updated surveys about these topics. In the Theorem 2.3 of this note we prove that properly discontinuous action is necessary and sufficient for a sequence of automorphisms on BNto generate a universal sequence of composition operators on Hp(BN). In addition, the universality is even obtained in an algebraically generic way. Therefore, our statement extends or completes the above ones due to the aforementioned authors. 2 Universality on Hp(BN) Before establishing the main result of this section, we need the following assertion, which can be found in [11, Satz 1.2.2 and Satz 1.4.2], see also [12, Proposition 6]. Theorem 2.1 Let X, Y be metrizable topological vector spaces with X complete and Y separable, and let ={Ln}∞ 1be a sequence of continuous linear operators from X to Y. Then the following statements are equivalent: (a) The set of universal elements for is a residual subset of X. (b) The set of universal elements for is a dense subset of X. (c) The set {(x,Ln(x)) :x∈X,n∈N}is dense in X ×Y. If, in addition,there isa dense subset C of X such thatlimn→∞ Ln(x)exists forall x ∈C, then (a), (b) and (c) are equivalent to (d) The set of universal elements for is not empty. In addition, we will make use of the following result due to the first author (see [1, Theorem2]),which tellsusthat,underadequateconditions,universalityis analgebraically—notonlytopologically—genericproperty.RecallthatasequenceTn:X→Y(n∈N) ofcontinuouslinear mappingsbetweentwo topologicalvectorspaces issaid tobe densely hereditarily hypercyclic (DHHC) if, for each subsequence {n(1)<n(2)<···}⊂N,the sequence {Tn(k)}∞ 1has a dense set of hypercyclic vectors. Theorem 2.2 Assume that X, Y are two separable metrizable topological vector spaces. If Tn:X→Y(n∈N)is a DHHC sequence, then there is a dense linear submanifold M in X all of whose nonzero vectors are universal for {Tn}∞ 1. Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM 368 Bernal-Gonz´ alez -- Bonilla -- Calder´ on-Moreno We also need some background about Hardy spaces in the N-dimensional unit ball (see specially [8] and [15]). For 0 <p<+∞ the Hardy space onBNis defined [8, pages 20–26] as Hp(BN):= f∈H(BN):fp:= sup 0<r<1SN |fr(ξ)|pdσN(ξ)1/p <+∞, where fr(ξ) =f(rξ) (0 <r<1) and σNis the positive Borel measure on SN:= [the unit sphere in CN]={z=(z1,...,zN):N 1|zj|2=1}that is rotation invariant (i.e. invariant under the unitary group) and normalized so that σN(SN)=1. If p≥1, Hp(BN)is a Banach space for the norm · p.If f∈Hp(BN), then the radial limit f(ξ) =limr→1f(rξ) exists σN-almost everywhere and fp=SN |f(ξ)|pdσN(ξ)1/p . As in the one-dimensional case, convergence in this norm implies uniform convergence on compacta, and the set of polynomialsin the variables z1,...,zNis dense in Hp(BN). In general, for N>1, a composition operator is not bounded as an operator from Hp(BN)into itself. But in [9] it is proved that at least operators on Hp(BN)induced by linear fractional maps—in particular, by automorphisms of BN—are bounded. We are now ready to state the promised version of the Seidel–Walsh theorem for Hardy spaces on the N-dimensional ball. Theorem 2.3 Let {Sn}∞ 1⊂Aut(BN)and p ∈[1,+∞). Consider the set U={f∈Hp(BN):{f◦Sn:n∈N}is dense in Hp(BN)}. Then the following properties are equivalent: (a) Uis residual in Hp(BN). (b) Uis not empty. (c) U∪{0}contains a dense linear submanifold of Hp(BN). (d) The action of {Sn}∞ 1is properly discontinuous on BN. Proof: That (a) implies (b) is trivial. If (b) holds then there exists f∈H(BN)such that the orbit {f◦Sn:n∈N}is dense in H(BN), because convergence in Hp(BN)implies compact convergence.Then (d) is satisfied by [14, Proposition 1]. For the part “(d) implies (a)”, we will use Theorem 2.1. Recall that the set of polynomials is dense in Hp(BN). Fix two polynomials p(z),q(z)and a number ε>0. We should find a function g∈Hp(BN)and a positive integer n0such that p−gp<ε and q−(g◦Sn0)p<ε. (2.1) Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM Compositional universality in the N -dimensional ball 369 We first prove that there exists a point γ∈SNsuch that, for some sequence n(1)< n(2)<··· <n(j)<··· of positive integers, Sn(j)(z)→γ(j→∞) for all z∈BN exceptforatmostonepointonSN.Forthis,notethatforeachn∈N,Sn=Un◦ϕan,where Unis a unitary map and an∈BN. Since the action of {Sn}∞ 1is properly discontinuous, we have by [14, Proposition 2] that supn∈N|an|=1, so there exists a point γ1∈SNand a sequence n(1)<n(2)<··· <n(j)<··· of positive integers such that an(j)→γ1 (j→∞). Since very Unis unitary, |Un(z)|=|z|for all z∈CNand all n∈N; hence {Un:n∈N}is uniformly bounded on compact sets in CNand, by Montel’s theorem and by taking a new subsequence if necessary, {Un(j)}∞ 1converges to an entire function U, uniformly on BN. Trivially, |U(z)|=|z|for all z∈CN.Defineγ:= U(γ1).Then |γ|=|γ1|=1, that is, γ∈SN. Let us denote, for the sake of simplicity, Rj=Sn(j)and bj=an(j)(j∈N). If z=(z1,...,zn)∈CNthen z:= (z1,...,zN). We now show that Rj(z)→γ(j→∞) for all z∈BN\{γ1}. We have that, for very z∈BN, |Rj(z)−γ|=|(Un(j)◦ϕbj)(z)−U(γ1)| ≤|Un(j)(γ1)−U(γ1)|+|Un(j)(ϕbj(z)) −Un(j)(γ1)| =|Un(j)(γ1)−U(γ1)|+|ϕbj(z)−γ1|, because each Un(j)is unitary. Now, the first term of the last hand side tends to zero as j→∞. On the other hand, |ϕbj(z)−γ1| = bj−Pbj(z)−1−|bj|21/2Qbj(z) 1−<z,bj>−γ1 ≤ bj−<z,bj> |bj|2bj 1−<z,bj>−γ1 +1−|bj|21/2|Qbj(z)| |1−<z,bj>| ≤ bj−<z,bj>bj 1−<z,bj>−γ1 + <z,bj> |bj|2bj−<z,bj>bj 1−<z,bj> +1−|bj|21/2 |1−<z,bj>| ≤|bj−γ1|+1 |bj|2+11−|bj|21/2 |1−<z,bj>|→0(j→∞) forallz∈BN\{γ1},becausebj→γ1,|bj|2→1and1−<z,bj>→1−<z,γ 1>= 0 (j→∞). Therefore Sn(j)→γ(j→∞) pointwise on BN\{γ1}. Again with no loss of generality we can assumethat Sn→γ(n→∞) on thatset and, in particular, Sn(z)→γ (n→∞)forσN-almost very point of SN. Consider the peak function for γdefined as a(z)=1+<z,γ> 2, Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM 370 Bernal-Gonz´ alez -- Bonilla -- Calder´ on-Moreno that is, a(z)is continuous on BN, holomorphic in BN,a(γ) =1and|a(z)|<1 for all z∈BN\{γ}. Choose a positive integer msuch that amp<ε p∞+q∞ ,(2.2) whichispossiblebecauseoftheLebesgueBoundedConvergenceTheorem.Hereh∞:= max{|h(z)|:z∈BN}for any continuous function hon BN.Sincea(z)is continuous and a(γ) =1, we have that 1−[a(Sn(z))]m→0(n→∞) forσN–almost very pointof SN. Again by the LebesgueBounded ConvergenceTheorem, we derive that 1−[a(Sn(·))]mp→0(n→∞), so there is n0∈Nwith 1−[a(Sn0(·))]mp<ε p∞+q∞ .(2.3) Let us define g(z)=p(z)+a(z)mqS−1 n0(z)−p(z). Then gis continuous on BN,sog∈Hp(BN). We have, from (2.2) and (2.3), that p−gp=  a(z)mqS−1 n0(z)−p(z)  p<ε and q(z)−(g◦Sn0)(z)p = q(z)−p(Sn0(z)) −a(Sn0(z))m[q(z)−p(Sn0(z))] p = 1−[a(Sn0(z))]m(q(z)−p(Sn0(z)) p<ε. Consequently, (2.1) is satisfied. An application of Theorem 2.1 with X=Y= Hp(BN)and Ln=CSn(n∈N) yields the desired result. Hence (a), (b) and (d) are equivalent. Finally, it is evident that (c) implies (b). Thus, it is enough to show that (d) implies (c). Again by [14], we have that if (d) holds then supn∈N|an|=1. Therefore there exists a sequence {m(1)<m(2)<···} ⊂ Nwith |am(j)|→1as j→∞. But this yields |an(j)|→1 (so supn∈N|an(j)|=1) for every subsequence {n(1)<n(2)<···} ⊂ {m(1)<m(2)<···}. Once more, from [14] it is derived that the action of {CSn(j)}∞ 1 is properly discontinuous, so each subsequence {CSn(j)}∞ 1of {CSm(j)}∞ 1has a dense set of hypercyclic vectors. Consequently, {CSm(j)}∞ 1is DHHC. By Theorem 2.2 (applied on X=Y=Hp(BN)and Tj=CSm(j)), there exists a dense linear submanifold Mof Hp(BN)such that M\{0}is included in the set U1of universal vectors for {CSm(j)}∞ 1. But, clearly, U⊃U1. Hence U⊃M\{0}, and we are done.  In particular, Theorem 2.3 can be applied when Snis the nth-iterate of a single ϕ∈Aut(BN)such that ϕhas no fixed point in BNbecause, in this case, the iterates of ϕ Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM Compositional universality in the N -dimensional ball 371 tend uniformly on compacta to a point of the unit sphere (the “Denjoy–Wolff point” of ϕ): see [8, page108]. As mentionedin Section 1, this last result—exceptfor the algebraic genericity—hasbeen obtained for the case p=2 by Xiaoman et al. [18] by using a rather different approach. Acknowledgements. The first and third authors have been partially supported by the Plan Andaluz de Investigaci´ on de la Junta de Andaluc´ıa FQM-127 and by DGES Grant BFM2003-03893-C02-01. The second author has been partially supported by MEC and FEDER MTM2005-07347. 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Reina Mercedes 41080 Sevilla Spain [email protected] Brought to you by | Biblioteca de la Universidad de Sevilla Authenticated Download Date | 6/1/16 2:04 PM