Universality on higher order Hardy spaces

Universality on higher order Hardy spaces L. Bernal-Gonz´alez, A. Bonilla and M.C. Calder´on-Moreno ∗ Abstract We prove a Seidel–Walsh-type theorem about universality of a sequence of derivation-composition operators generated by automorphisms of the unit disk in the setting of the higher order Hardy spaces. Moreover, some related positive or negative assertions involving interpolating sequences and sequences between two tangent circles are established for the class of bounded functions in the unit disk. Our statements improve earlier ones due to Herzog and to the first and third authors. 2000 Mathematics Subject Classification: Primary 30E10. Secondary 30D55, 47A16, 47B38. Key words and phrases: Unit disk, higher order Hardy spaces, Seidel– Walsh theorem, bounded function, interpolating sequence, hypercyclic function, derivation-composition operator. 1 Introduction and notation In this paper, we denote by N,C,D,N0the set of positive integers, the complex plane, the open unit disk {z∈C:|z|<1}and the set N∪ {0}, respectively. The boundary of Dis the unit circle ∂D={z∈C:|z|= 1}. If G⊂Cis a domain (= nonempty connected open subset), then H(G) stands for the Fr´echet space of holomorphic functions on Gendowed with the topology of uniform convergence on compact subsets. A domain Gis said to be simply connected whenever its complement with respect to the extended plane is connected. The class A(D) is the Banach space of all functions which are continuous on the closure Dof Dand holomorphic in D, ∗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 MCYT-FEDER Project no. BFM 2002-02098. 1 endowed with the supremum norm k·k∞. If 1 ≤p < ∞, the Hardy space Hp(D) is defined as Hp(D) = {f∈H(D) : kfkp<∞}, where kfkp= sup r<11 2πZ2π 0 |f(reiθ)|pdθ1/p . Then Hp(D) becomes a Banach space if it is endowed with this norm. And H∞(D) is the space of all f∈H(D) which are bounded on D. It becomes a Banach space when endowed with k · k∞. It is well known that for every f∈Hp(D) the radial limit f(eiθ) = lim r→1f(reiθ) exists and is finite for almost all θ∈[0,2π]. In addition, kfkp=1 2πZ2π 0 |f(eiθ)|pdθ1/p . See [5] for an extensive study of Hardy spaces. For N∈N, denote the higher order Hardy space Hp N(D) = {f∈H(D) : f(N)∈Hp(D)}, which becomes a Banach space whenever it is endowed with the norm kfk=kf(N)kp+ N−1 X j=0 kf(j)k∞. For the sake of uniformity, the symbol Hp 0(D) will denote Hp(D). It is well known (see [5, Chapter 5, Exercise 9]) that if f0∈Hp(D) then f∈A(D), so the norm kfk above makes sense on Hp N(D). The polynomials are dense in Hp N(D). Higher order Hardy spaces are extensively studied in [13]. Note that Hp N(D)⊂AN−1(D) := {f∈ H(D) : f(j)∈A(D) for j= 0,1, . . . , N −1}. The group Aut(D) of automorphisms of Dis the set of M¨obius transformations {σa,k :|a|<1 = |k|}, where σa,k(z) = k·z−a 1−az . In 1941 W. Seidel and J. L. Walsh [14] 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⊂Ddepending on gsuch that f◦σan,1→g(n→ ∞) in H(G). This result is in turn a non-Euclidean version of Birkhoff’s theorem about density of translates of certain entire functions [4]. Both results have been generalized and developed in several directions (see [9] for references), but in this paper we are mainly interested in a special one, namely, a version where composition-derivation make appearance. To this end, we need the next brief report. In 1995, A. Montes-Rodr´ıguez and the first author [3] 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∈D|an|= 1 if and only if the action of {Sn}∞ 1is properly discontinuous 2 on D, that is, given a compact subset K⊂Dthere exists m=m(K)∈Nsuch that K∩Sm(K) = ∅. In particular, if ϕ=σa,k (|a|<1, k=eiθ) and Sn=ϕ◦ · · · ◦ ϕ (ntimes), 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|. In 1995, G. Herzog [11] proves the following “Seidel–Walsh theorem for derivatives”: If Xis a Banach space of holomorphic functions on Dwith A(D)⊂X such that convergence in Ximplies compact convergence on Dand polynomials are dense in X, then for every sequence {an} ⊂ Dwith |an| → 1 (n→ ∞) the set {f∈X:{f0◦σan,1:n∈N}is dense in H(D)}is a residual subset of X. Trivially, the expression f0◦σan,1cannot be changed to f◦σan,1(just take X=A(D)). The assertion of Herzog’s result is also obviously false for X=A1(D). In 1999, the first and third authors [2] extended Herzog’s result to an operator of the form Φ(D), where Dis the differentiation operator (Df =f0) and Φ is a nonconstant polynomial, and in fact to a C-bounded sequence of polynomials. A sequence {Φn(z) = PN j=0 b(n) jzj}∞ 1 of polynomials of the same degree N∈Nis C-bounded whenever each sequence {b(n) j:n∈N}(j= 0,1,2, . . . , N) is bounded and there exists a positive constant α such that |b(n) N| ≥ αfor all n∈N. In [2] the following is shown: If Xis an F-space of holomorphic functions in Dwith A(D)⊂Xsuch that convergence in Ximplies compact convergence on Dand polynomials are dense in X, and if {Sn}∞ 1⊂Aut(D) and {Φn}∞ 1is a C-bounded sequence of polynomials, then the set {f∈X:{Φn(D)◦ Sn:n∈N}is dense in H(D)}is a residual subset of Xif and only if it is not empty if and only if the action of {Sn}∞ 1is properly discontinuous on D. All above density results can be expressed in the terminology of universality. If Xand Yare topological vector spaces, then a sequence Tn:X→Y(n∈N) of continuous linear mappings is called universal or hypercyclic whenever the set Uof elements x∈Xsuch that the orbit {Tnx}∞ 1is dense in Yis not empty. Each element of Uis said to be universal for {Tn}∞ 1. See [9] and [10] for good updated surveys about these topics. In this paper, we extend the main results of [2] and [11] to the higher order Hardy spaces Hp N(D), see Theorem 2.5 below. The conclusion is not true for H∞(D). In fact, we will see in Section 3 that interpolating sequences and sequences lying on the region between two circles which are tangent to a boundary point do not generate “good” sequences of automorphisms in order to yield universality. This improves [11, Section 4]. 2 Universal functions in higher order Hardy spaces Before establishing the main result of this section, we need the four following statements, which can be found respectively in [8, Satz 1.2.2 and Satz 1.4.2] (see also [9, 3 Proposition 6]) and [2, Lemmas 1, 2, 3]. Theorem 2.1. Let X,Ybe metrizable topological vector spaces with Xcomplete and Yseparable, and let Λ = {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 Λis a residual subset of X. (b) The set of universal elements for Λis a dense subset of X. (c) The set {(x, Ln(x)) : x∈X, n ∈N}is dense in X×Y. If, in addition, there is a dense subset Cof Xsuch that limn→∞ Ln(x)exists for all x∈C, then (a), (b) and (c) are equivalent to (d) The set of universal elements for Λis not empty. Lemma 2.2. Let {Φn(z) = N X j=0 a(n) jzj}∞ n=1 be a sequence of polynomials with the same degree N∈N0such that every sequence {a(n) j:n∈N}(j= 0,1, . . . , N)is bounded. Then there is a subsequence {Φnk:k∈N}and a polynomial Psatisfying that Φnk(D)ϕ→P(D)ϕ(k→ ∞)in H(C)for every entire function ϕ. Lemma 2.3. Assume that Gand Ωare two domains of Cand that H: Ω →C, Hk: Ω →C(k∈N),Ψ : G→C,Ψk:G→C(k∈N)are functions satisfying the following properties: (i) Hktends to H(k→ ∞)uniformly on compacts sets in Ω. (ii) Ψktends to Ψ (k→ ∞)uniformly on compacts sets in G. (iii) Ψ(G)⊂Ω. (iv) Ψ is continuous on Gand His continuous on Ω. Then Hk◦Ψk→H◦Ψ (k→ ∞)uniformly on compact subsets in G. Lemma 2.4. Let G⊂Cbe a simply connected domain, a∈G,F∈H(G)and, for each k∈N, (IkF)(z) = Zz a (z−ξ)k−1 (k−1)! F(ξ)dξ (z∈G), where the integration is taken along any rectifiable curve in Gjoining ato z. If we set I0F=F, then IkFis well-defined for every k∈ {0,1,2, . . .},IkF∈H(G)and (IkF)(j)=Ik−jFfor j∈ {0,1, . . . , k}. 4 For the sake of completeness, we extend (with the same definition) the notion of C-boundedness to the case N= 0, that is, when all Φnare constant. Theorem 2.5. Assume that N∈N0. Let Xbe an F-space of holomorphic functions on Dhaving the following properties: (a) Convergence in Ximplies convergence on D. (b) Hp N(D)⊂Xfor some p∈[1,∞). (c) The polynomials are dense in X. Assume that {Sn}∞ 1is a sequence of automorphisms of Dand that {Φn}∞ 1is a Cbounded sequence of polynomials of degree N. Denote Tn= Φn(D) (n∈N)and consider the set U:= {f∈X:{(Tnf)◦Sn:n∈N}is dense in H(D)}. Then Uis a residual set of Xif and only if Uis not empty if and only if the action of {Sn}∞ 1is properly discontinuous on D. Proof. Assume that N∈N. We have that Φn(z) = N X j=0 b(n) jzj(n∈N), |b(n) j| ≤ Bj< +∞(j= 0,1, . . . , N;n∈N) and there exists α > 0 such that |b(n) N| ≥ α(n∈N). Each function Snhas the form Sn=σan,kn, where |an|<1 = |kn|for all n∈N. Suppose that the action of {Sn}∞ 1is not properly discontinuous on D, so (see Section 1) lim supn→∞ |an|<1. If r∈(0,1) then (see the proof of Theorem 4 in [2]) we can associate it a number c∈(0,1) such that |Sn(z)| ≤ cfor all n∈N. Given f∈H(D), the set [ n∈N [(Φn(D)f)◦Sn]({|z| ≤ r}) is bounded, because it is contained in the disk {|z| ≤ s}where s:= N X j=0 Bjsup |z|≤c |f(j)(z)|<+∞. Thus, the set {(Φn(D)f)◦Sn:n∈N}cannot be dense in H(D). Hence Uis empty. Now, the only property to be proved is that Uis residual whenever supn∈N|an|= 1. Let us prove it first in the case X=Hp N(D). Define the mappings Ln:Hp N(D)→H(D) (n∈N) 5 by Ln(f)=(Tnf)◦Sn. Each Lnis linear and continuous, because the fact gj→0 (j→ ∞) in Hp N(D) implies g(ν) j→0 (j→ ∞) compactly in D(ν= 0,1, . . . , N). If we prove that the set G={(f, Ln(f)) : f∈Hp N(D), n ∈N} is dense in Hp N(D)×H(D) then an application of Theorem 2.1 would yield the conclusion in this case. Since the polynomials are dense in Hp N(D) and in H(D), it is sufficient to prove that given a compact subset K⊂D, two polynomials p,qand ε∈(0,1), there exist g∈Hp N(D) and n0∈Nsuch that kp−gk< ε and |q(z)−Ln0g(z)|< ε (z∈K). From the facts supn∈N|an|= 1 and |kn|= 1 for every n∈N, we can suppose with no loss of generality, by taking a subsequence if necessary, that there is a point γ∈∂D such that Sn→γ(n→ ∞) uniformly on compact subsets of D. Consider the function a(z) = 1 + γz 2(z∈C). This is a “peak-function” at γfor Din the sense that a(γ) = 1 and |a(z)|<1 for all z∈D\ {γ}(see [6, page 189]). Let β= 1 + kqk∞+ N X j=0 Bjkp(j)k∞and choose m∈Nsuch that m > 2β(1 + PN j=0 Bj)(N+ 1) αε (1) and kamkp<αε β(N+ 1).(2) The latter inequality is possible by the Lebesgue Bounded Convergence Theorem. Moreover, since Sn→γin H(D) and a(γ) = 1, there exists n0∈Nsuch that sup z∈K |1−[a(Sn0(z))]m|<ε 2β.(3) Consider the function F(z) = a(z)m b(n0) N [q(S−1 n0(z)) −(Φn0(D)p)(z)]. 6 With the notation of Lemma 2.4, take a= 0 and define the function h=INF on the domain G=|an0|−1D. Then h∈H(|an0|−1D) and so h∈Hp N(D). If z∈D and j∈ {0,1, . . . , N −1}then we have |h(j)(z)|= 1 b(n0) NZz 0 (z−ξ)N−1−j (N−1−j)! 1 + γξ 2m [q(S−1 n0(ξ)) −(Φn0(D)p)(ξ)] dξ = 1 b(n0) NZ1 0 zN−1−j(1 −t)N−1−jz (N−1−j)! 1 + γzt 2m [q(S−1 n0(zt)) −(Φn0(D)p)(zt)] dt <1 αZ1 0 1·(1 + t)m 2m·β dt =β α2m(m+ 1)(2m+1 −1) <2β αm <ε (N+ 1)(1 + PN j=0 Bj), because of (1). Hence kh(j)k∞<ε (N+ 1)(1 + PN j=0 Bj)for j∈ {0,1, . . . , N −1}.(4) As for h(N), we obtain that kh(N)kp=kFkp = 1 2π|b(n0) N|pZ2π 0 |a(eiθ)m|p· |q(S−1 n0(eiθ)) −(Φn0(D)p)(eiθ)|pdθ!1/p ≤β αkamkp<ε N+ 1 by (2). Thus khk=kh(N)kp+ N−1 X j=0 kh(j)k∞< ε. Define g=p+h. Then g∈Hp N(D) and kg−pk=khk< ε. Moreover, q(z)−(Ln0g)(z) = q(z)−(Ln0p)(z)−(Ln0h)(z) =q(z)−(Φn0(D)p)(Sn0(z)) −b(n0) Nh(N)(Sn0(z)) − N−1 X j=0 b(n0) jh(j)(Sn0(z)) =q(z)−(Φn0(D)p)(Sn0(z)) −a(Sn0(z))m[q(z)−(Φn0(D)p)(Sn0(z))] 7 − N−1 X j=0 b(n0) jh(j)(Sn0(z)) = (1 −a(Sn0(z))m)(q(z)−(Φn0(D)p)(Sn0(z))) − N−1 X j=0 b(n0) jh(j)(Sn0(z)) for all z∈ |an0|−1D. Now, if z∈Kthen we get |q(z)−(Ln0g)(z)| ≤ |1−a(Sn0(z))m|·|q(z)−(Φn0(D)p)(Sn0(z))|+ N−1 X j=0 Bjkh(j)k∞ <ε 2β·β+ε 2(1 + PN j=0 Bj)· N−1 X j=0 Bj< ε because of (3) and (4). Thus the closure of Gcontains the set {(p, q) : p, q polynomials}, which is dense in Hp N(D)×H(D), so Gis also dense, as required. Now, if Xis an F-space as in the hypothesis, then the mappings Ln:X→H(D) (n∈N) are continuous by (a). As before, we can assume that {Sn}∞ 1tends to a point γ∈∂Din H(D) and, by Lemma 2.2, we may suppose with no loss of generality that there is a polynomial Psuch that Tnϕ→P(D)ϕ(n→ ∞) in H(C) for every entire function ϕ. Fix a polynomial ϕand apply Lemma 2.3 on G=D, Ω = C, Ψn=Sn, Ψ = the constant γ,Hn=Tnϕand H=P(D)ϕ. We obtain Lnϕ= (Tnϕ)◦Sn→ (P(D)ϕ)(γ) (n→ ∞) in H(D), so (Lnϕ) converges in H(D) for every polynomial ϕ. Since (b) is satisfied, the set Uis not empty. Finally, since Xsatisfied (c), an application of Theorem 2.1 with Y=H(D) and C={polynomials}yields that Uis a residual subset of X. The remaining case N= 0 is much easier and left to the reader. The proof is complete. Note that the condition (b) in Theorem 2.5 is in some sense optimal, because the result is not true for X⊂ {f∈H(D) : f(N)∈H∞(D)}(and, nevertheless, Hp N(D)⊂ {f∈H(D) : f(N−1) ∈A(D)}⊂{f∈H(D) : f(N−1) ∈H∞(D)}). Indeed, if some member fof Xwere universal for {Ln}∞ 1, where Lnf=f(N)◦Sn(that is, we are taking Φn(z) = zNfor all n), we would get that {f(N)◦Sn:n∈N}is dense in H(D), which is not possible since that set is bounded. 8 3 Bounded universal functions In [11, Section 4] it is shown that there is a sequence {an} ⊂ Dwith limn→∞ |an|= 1 such that the set {f∈H∞(D) : {f0◦Sn:n∈N}is dense in H(D)}is not empty, but not dense in H∞(D), where Sn=σan,1(n∈N). We furnish in this section an extension of this result into two directions. Recall that a sequence {bn}∞ 1⊂Dis said to be an interpolating sequence for H∞(D) if and only if for every bounded sequence {cn}∞ 1there exists f∈H∞(D) such that f(bn) = cn(n∈N) (see [5, Chapter 9], [6, Chapters 7 and 10], [12, Chapter 10] for a rather complete study of interpolating sequences). Note that |bn| → 1 (n→ ∞) is a necessary condition for {bn}∞ 1to be interpolating (in fact, P(1 − |bn|) is convergent; see [5, page 150]). Theorem 3.1. Assume that {Sn=σan,kn}∞ 1(|an|<1 = |kn|for all n∈N)is a sequence of automorphisms of Dsuch that {Sn(0)}∞ 1is an interpolating sequence. Then the set U={f∈H∞(D) : {f0◦Sn}∞ 1is dense in H(D)} is nonempty, but not dense in H∞(D). Proof. Since {knan}∞ 1is interpolating then, as noted above, limn→∞ |an|=1= limn→∞ |knan|, hence {Sn}∞ 1acts properly discontinuously on D. Consequently, Theorem 3 in [11] (or our Theorem 2.5 with N= 1 and Φn(z) = zfor all n∈N) guarantees that {f∈A(D) : {f0◦Sn}∞ 1is dense in H(D)}is nonempty, whence Uis also nonempty because A(D)⊂H∞(D). Let us show that Uis not dense in H∞(D). Denote bn=−knan(n∈N). It is trivial that {bn}∞ 1is also an interpolating sequence. Consider the Blaschke product B(z) = ∞ Y 1 |bn| bn ·bn−z 1−bnz, where the nth factor should be changed to zif bn= 0. The condition P(1 − |bn|)< +∞guarantees the normal convergence of the infinite product on each compact subset of D, so B∈H∞(D). A straightforward calculation shows that |B0(bn)|=1 1− |bn|2Y j6=n bj−bn 1−bjbn . By Carleson’s theorem [6, pages 284–294], the sequence {bn}∞ 1is uniformly separated, that is, there is a constant ε > 0 such that Y j6=n bj−bn 1−bjbn ≥ε(n∈N). 9