Families of strongly annular functions: linear structure

Revista Matem´atica Complutense manuscript No. (will be inserted by the editor) Families of strongly annular functions: linear structure Luis Bernal-Gonz´alez ·Antonio Bonilla Received: date / Accepted: date Abstract A function fholomorphic in the unit disk Dis called strongly annular if there exists a sequence of concentric circles in Dexpanding out to the unit circle such that fgoes to infinity as |z|goes to 1 through these circles. The residuality of the family of strongly annular functions in the space of holomorphic functions on Dis well known, and it is extended here to certain classes of functions. This important topological property is enriched in this paper by studying algebraic-topological properties of the mentioned family, in the modern setting of lineability. Namely, we prove that although this family is clearly nonlinear, it contains, except for the zero function, large vector subspaces as well as infinitely generated algebras. Similar results are obtained for strongly annular functions on the whole complex plane and for weighted Bergman spaces. Keywords Strongly annular functions ·entire functions ·dense-lineability · algebrability ·Bergman spaces. Mathematics Subject Classification (2010) 15A30 ·30B10 ·30B30 · 30J99 ·46E10 ·46E15. 1 Introduction Let us denote by D, as usual, the open unit disk of the complex plane C, and by H(D) the space of holomorphic functions in D, endowed with the Luis Bernal-Gonz´alez Departamento de An´alisis Matem´atico, Facultad de Matem´aticas, Universidad de Sevilla, Avda. Reina Mercedes, 41080 Sevilla, Spain Tel.: +34-954557118 Fax: +34-954557972 E-mail: lb[email protected] Antonio Bonilla Departamento de An´alisis Matem´atico, Facultad de Matem´aticas, Universidad de La Laguna, C/Astrof´ısico Francisco S´anchez, 38271 La Laguna, Tenerife, Spain 2 Bernal and Bonilla compact-open topology τc. Under τc, this space becomes an F-space, i.e., a complete metrizable topological vector space. An interesting family in H(D) is SA, formed by the so-called strongly annular functions. By definition, a function f∈H(D) belongs to SA provided that lim sup r→1 min{|f(z)|:|z|=r}= +∞. For the sake of convenience, we establish the next notation. Denote by Σthe set of all strictly increasing sequences σ={rn}n≥1⊂(0,1) with rn→1. If σ is as before, we set C(σ) := S∞ n=1 rnT, where T:= {z:|z|= 1}. Then f∈ SA if and only if there is σ={rn}n≥1∈Σsuch that limn→∞ min{|f(z)|:|z|= rn}= +∞, or equivalently, lim |z|→1 z∈C(σ) |f(z)|= +∞. There is an extensive literature on this kind of functions, see for instance [6], [12], [13], [14], [15], [16], [18], [19], [20], [25], [26], [27], [28], [29], [30] and the references contained in them. The study of SA is motivated by the search of functions in H(D) having fast radial growth. Observe that there is not any function f∈H(D) such that lim|z|→1|f(z)|= +∞. Indeed, by way of contradiction, assume that fis one of such functions. Then the set of zeros of fform a compact subset of D. By the analytic continuation principle, this set of zeros is finite. Let Pbe a polynomial whose zeros are exactly those of f, counting multiplicities. It follows that P/f ∈H(D) and, since Pis bounded on D, lim|z|→1|P(z)/f(z)|= 0. By the maximum modulus principle, P/f ≡0, which is clearly impossible. Once the existence of strongly annular functions is established, the next natural step is to study the topological nature and the size of SA. This was carried out by Bonar and Carroll [13], who proved in 1975 that SA is a dense Gδ(hence residual) subset of H(D). Therefore it can be said that SA is topologically large. But, what can be asserted about its algebraic structure and size? It is plain that SA is not even a vector space. In recent years, a plethora of papers have been published stating the existence of large algebraic structures within nonlinear sets. To this respect, the following notions have been recently introduced. Assume that Xis a topological vector space and that µ is a cardinal number. Then a subset Aof Xis called •lineable [3] if A∪ {0}contains an infinite dimensional vector subspace, •µ-lineable [3] if A∪ {0}contains a µ-dimensional vector subspace, •dense-lineable or algebraically generic [7] whenever A∪{0}contains a dense vector subspace of X, •maximal dense-lineable [10] if A∪ {0}contains a dense vector subspace M of Xwith dim(M) = dim(X), and •algebrable ([4] and [5]) if Xis a function space and A∪ {0}contains some infinitely generated algebra. See also [1] and [22]. Recall that a vector space Mof functions is said to be an algebra provided that fg ∈Mif f, g ∈M. Clearly, maximal dense-lineability implies dim(X)-lineability plus dense lineability, but the converse is not true. Strongly annular functions 3 It follows from the definition that if a strongly annular function has a radial limit, the limit must be infinity. Consequently, by Fatou’s theorem, no function in the classical Hardy spaces Hp(D) := {f∈H(D) : sup0≤r<1R2π 0|f(reiθ)|pdθ <+∞} (see e.g. Rudin [31]) can belong to SA. Nevertheless, in 2007 Redett [30] was able to construct a strongly annular function in each weighted Bergman space Ap α(D) (0 <p<+∞, α > −1). Our aim in this paper is to establish that SA is not only topologically large, but also algebraically large, in the sense of the above definitions. This will be accomplished in Section 4. Section 2 will be devoted to give the necessary background. In Section 3, residuality is reinforced and examined within certain subspaces of H(D). Finally, in Sections 5 and 6 we extend our results to weighted Bergman spaces and to the space of entire functions. 2 Preliminary results A number of preliminary assertions will be used in due course. We begin with a simple observation. If Xis a separable infinite-dimensional F-space then Baire’s theorem implies that dim(X) = c, the cardinality of the continuum. Hence cis the maximal dimension allowed for any subspace of X. For instance, dim(H(D)) = c. The following statement on lineability was established in Bernal [10, Lemma 2.1], which in turn is a strengthening of Theorem 2.1 in [9]; see also Aron et al. [2, Theorem 2.2 and p. 152] for related results. Lemma 1 Assume that Xis a metrizable separable topological vector space. Suppose that Γis a family of linear subspaces of Xsuch that TS∈ΓSis dense in Xand TS∈Γ(X\S)is µ-lineable, where µis an infinite cardinal number. Then TS∈Γ(X\S)∪ {0}contains a dense µ-dimensional vector subspace. In the next elementary lemma one meets the nice notion of “stronger than”, coined by Aron, Garc´ıa, P´erez and Seoane in [2]. Lemma 2 Suppose that the following holds: (a) (X, τ0)is a topological vector space. (b) Ais a dense Gδsubset of X. (c) Yis a vector subspace of Xand τ1is a topology on Ysuch that (Y, τ1)is a topological vector space and τ1is finer than τ0|Y. (d) There is a τ1-dense subset Dof Ysuch that Ais stronger than D, that is, A+D⊂A. (e) A∩Y6=∅. Then A∩Yis a dense Gδsubset of (Y, τ1). Proof According to (b), there are τ0-open sets Gn(n≥1) with A=T∞ n=1 Gn. Then A∩Y=T∞ n=1(Gn∩Y), which is a τ1-Gδsubset of Ybecause of (c). From (e), there is x0∈A∩Yand, by (d), x0+D⊂A∩Y. But x0+Dis τ1-dense in Y. Consequently, the same holds for A∩Y. 4 Bernal and Bonilla If ϕ:D→(0,+∞) is continuous and σ∈Σ, we define SA(ϕ) := f∈H(D) : lim sup r→1 min{|f(z)| ϕ(z):|z|=r}= +∞ and SA(ϕ, σ) := (f∈H(D) : lim |z|→1 z∈C(σ) |f(z)| ϕ(z)= +∞). Then it is plain that SA(ϕ) = Sσ∈ΣSA(ϕ, σ) and that SA(1) = SA. The following assertion will be employed to study dense-lineability. Recall that if A⊂Cthen f∈H(A) means that there is an open set G=G(f)⊃A such that f∈H(G) := {holomorphic functions on G}. Lemma 3 Assume that ϕ:D→(0,+∞)is a continuous function satisfying lim |z|→1 log ϕ(z) log 1 1−|z| = +∞.(1) If f∈ SA(ϕ)and g∈H(D)\ {0}then fg ∈ SA. Proof Fix f, g as in the statement. Then we can choose a connected open set with G⊃D,g∈H(G) and g6≡ 0 in G. From the analytic continuation principle one derives that there are only finitely many zeros of gon D. Hence we can assume that gpossesses zeros z1, . . . , zpin Dand zeros w1, . . . , wqon T, with respective multiplicities m1, . . . , mp,n1, . . . , nq(other cases are easier to handle). Then g=PQh, where h∈H(G), hlacks zeros in Dand P(z) := Qp k=1(z−zk)mk,Q(z) := Qq k=1(z−wk)nk. By hypothesis, f∈ SA(ϕ, σ) for some sequence σ= (rn)∈Σ. Let n0be such that rn>max{|z1|,...,|zp|} for all n≥n0, and choose α, β > 0 with |h(z)|> α (z∈D) and |z−zk|> β (|z|=rn, n ≥n0;k= 1, . . . , p). If |z|=rnwith n≥n0we have |f(z)g(z)|=|h(z)||P(z)||Q(z)|ϕ(z)·|f(z)| ϕ(z) > α p Y k=1 |z−zk|mk q Y k=1 (1 − |z|)nkϕ(z)·|f(z)| ϕ(z) > αβdegree (P)(1 − |z|)degree (Q)ϕ(z)·|f(z)| ϕ(z). By (1), lim|z|→1(1 − |z|)Nϕ(z)=+∞for all N∈N:= {1,2,3, . . . }. But recall that lim |z|→1 z∈C(σ) |f(z)|/ϕ(z) = +∞. Therefore lim |z|→1 z∈C(σ) |f(z)g(z)|= +∞, that is, fg ∈ SA. Finally, Lemma 4 will be needed to examine dense-lineability in the context of entire functions. Lemma 4 If fis an entire function that is not a polynomial then the family {fα:α > 0}is linearly independent, where we have set fα(z) := f(αz). Strongly annular functions 5 Proof Let f(z) = P∞ n=0 anznand suppose, by way of contradiction, that there is a finite linear combination PN k=1 ckfαk= 0, where αk>0, ck∈C(k= 1, . . . , N) and not all the ckare zero. We can assume that N≥2, α1< α2< · · · < αNand cN6= 0. Then an(c1αn 1+· · · +cNαn N) = 0 (n∈N). Since fis not a polynomial, one can find a sequence {n1< n2<· · · < nj<· · · } ⊂ N such that c1αnj 1+· · · +cNαnj N= 0 (j∈N). Therefore 1 = − N−1 X k=1 ckc−1 N(αkα−1 N)nj−→ 0 (j→ ∞). This is the desired contradiction.  3 Residuality We start with a refinement of the residuality of the family SA. In fact, we can fix the sequence of radii supporting big values of |f|as well as the rate of growth so that residuality is kept. As usual, we denote B(a, r) = {z∈C:|z−a|< r} and B(a, r) = {z∈C:|z−a| ≤ r}(a∈C, r > 0). Theorem 1 Let be prescribed a continuous function ϕ:D→(0,+∞)and a sequence σ∈Σ. Then the set SA(ϕ, σ)is residual in H(D). Consequently, SA(ϕ)is also residual in H(D). Proof Let σ= (rn), so that 0 < r1< r2<· · · → 1. For every pair m, n ∈N we denote Sm,n := {f∈H(D) : |f(z)|> nϕ(z) for all z∈rmT}. If we set Sn=Sm≥nSm,n (n∈N) then one can express SA(ϕ, σ) = ∞ \ n=1 Sn. For each compact set K⊂Dand each continuous function fon Dwe set kfkK:= sup{|f(z)|:z∈K}and m(f, K) := min{|f(z)|:z∈K}. A basic open neighborhood of a function g∈H(D) has the form V(g, K, ε) = {h∈ H(D) : kh−gkK< ε}, where ε > 0 and Kis a compact subset of D. Fix m, n ∈N. If g∈ Sm,n then δ:= m(|g| − nϕ, rmT)>0. If h∈ V(g, rmT, δ) then we have for all z∈rmTthat −|h(z)|+|g(z)|≤|h(z)−g(z)|< m(|g| − nϕ, rmT), so |h(z)|>|g(z)| − m(|g| − nϕ, rmT)≥ |g(z)|−|g(z)|+nϕ(z) = nϕ(z). Hence V(g, rmT, δ)⊂ Sm,n, which proves that Sm,n is open. Therefore every Snis open. By Baire’s theorem it is enough to show that each Snis dense. To this end, fix a basic open set V(g, K, ε). Choose m≥max{n, 3}such that K⊂B(0, rm−2). Since rm−2< rm−1< rm, we can select p∈Nsatisfying rm−2 rm−1p<ε kϕkrmT and rm rm−1p> n +kgkrmT kϕkrmT . 6 Bernal and Bonilla Define f(z) := g(z)+(z/rm−1)pkϕkrmT. Then kf−gkK≤(rm/rm−1)pkϕkrmT< ε, so f∈V(g, K, ε). Furthermore, for all z∈rmT, |f(z)|>(rm/rm−1)pkϕkrmT− |g(z)| > nkϕkrmT+kgkrmT− |g(z)| ≥ nϕ(z). Thus, f∈V(g, K, ε)∩ Sn, which proves the density of Sn. The last result will be of help in Section 4 to find algebraic genericity inside SA. Remark 1 With minor modifications in the proof, one can obtain the following enhancement of Theorem 1. Let ϕ:D→(0,+∞) is continuous and 0 < s1< r1< s2< r2<· · · < sn< rn<· · · −→ 1. If we set A:= S∞ n=1{z:sn<|z|< rn}and SA(ϕ, A) := f∈H(D) : lim |z|→1 z∈A |f(z)| ϕ(z)= +∞, then SA(ϕ, A) is residual in H(D). Observe that for each function f∈ SA(ϕ) there is a set A=A(f) as before such that f∈ SA(ϕ, A); indeed, it suffices to apply the continuity of f. Note also that sequences (rn),(sn) can be selected so as to their corresponding set Ais rather large, in the sense that its radial boundary density ω-dens(A) is maximal (that is, equal to 1). Here ωdens(A) := limr→1 λ(A∩ {z:r < |z|<1}) π(1 −r2), whenever this limit exists, where λdenotes bidimensional Lebesgue measure. For related results (with different classes of functions), see Belna and Redett [8]. We finish this section by extending residuality to other spaces of holomorphic functions in D. Among these well-behaved spaces, we find the weighted Bergman spaces Ap α(D). For every p∈(0,+∞) and every α∈(−1,+∞) the space Ap α(D) is defined (see e.g. [23]) as the class of functions f∈H(D) for which kfkp,α := Z ZD |f(z)|p(1 − |z|)αdxdymin{1,1/p} <+∞. It becomes a separable F-space under the F-norm k·kp,α. If p≥1 (p= 2, resp.), k·kp,α even makes Ap α(D) a Banach (Hilbert, resp.) space. For α= 0 one obtains the classical Bergman spaces Ap(D) = {f∈H(D) : RRD|f(z)|pdxdy < +∞}. Theorem 2 Assume that Yis a Baire topological vector space with Y⊂H(D) such that Yis endowed with a topology τwhich is finer that τc|Y. Let ϕ:D→ (0,+∞)be continuous, and σ∈Σ. We have: (a) If SA ∩Y6=∅and there is a dense subset Dof Ysuch that each function f∈ D is bounded on D, then SA ∩ Yis residual in Y. (b) If SA(ϕ)∩Y6=∅and there is a dense subset Dof Ysuch that f/ϕ is bounded on Dfor each f∈ D, then SA(ϕ)∩Yis residual in Y. Strongly annular functions 7 (c) If SA(ϕ, σ)∩Y6=∅and there is a dense subset Dof Ysuch that f/ϕ is bounded on Dfor each f∈ D, then SA(ϕ, σ)∩Yis residual in Y. Proof Apply Lemma 2 with X:= H(D), τ0:= τc,τ1:= τand D:= D. In the situation of (c), the proof of Theorem 1 reveals that A:= SA(ϕ, σ) is a Gδ subset of X. Clearly A+D⊂A. By Lemma 2, SA(ϕ, σ)∩Yis a dense Gδsubset of Y. Since Yis Baire, SA(ϕ, σ)∩Yis residual in Y. Hence (c) is proved. Under the hypotheses of (b), there must exist s∈Σsuch that SA(ϕ, s)∩Y6=∅. From (c), SA(ϕ, s)∩Yis residual in Y. Since SA(ϕ, s)⊂ SA(ϕ), this larger set is also residual, which proves (b). Part (a) is the special case of (b) when one takes ϕ≡1.  Corollary 1 If p∈(0,+∞)and α∈(−1,+∞)then SA ∩ Ap α(D)is residual in Ap α(D). Proof By Redett’s result [30], SA∩Ap α(D)6=∅. Just apply Theorem 2(a) with Y:= Ap α(D) and D:= {polynomials}, and take into account that convergence in Ap α(D) implies convergence in each compact subset of D[23, Prop. 1.1] and that the polynomials form a dense subset of Ap α(D) [23, Prop. 1.3].  4 Lineability of SA We proceed to study the lineability of SA and of subfamilies of it. By span(Y) we denote the linear span of a family Yof functions, while hfiwill stand for the span of {f}, that is, the set {λf :λ∈C}. Theorem 3 SA is maximal dense-lineable in H(D). Proof Consider the function ϕ(z) := exp 1 1−|z|(z∈D). According to Theorem 1, we can select a function f0∈ SA(ϕ). Consider the functions eα(z) := exp(αz) (α > 0) and the set M:= span{eαf0:α > 0}. It is clear that Mis a vector subspace of H(D). Moreover, dim(M) = c. Indeed, since the cardinality of (0,+∞) is c, it is enough to prove the linear independence of the functions eαf0(α > 0). For this, consider a nontrivial linear combination N X j=1 ajeαjf0= 0 where, without loss of generality, we can assume that 0 < α1< α2<· · · < αN and aN6= 0. Since f06≡ 0, the analytic continuation principle guarantees the existence of an open interval I⊂(−1,1) such that f0(x)6= 0 for all x∈I. Then, after dividing by f0and transposing terms, we get aN= −PN−1 j=1 aje(αj−αN)x(x∈I). Now, the analytic continuation principle comes 8 Bernal and Bonilla again in our help, yielding that the last equality holds for all x∈R. Letting x→+∞, we have e(αj−αN)x→0 (j= 1, . . . , N −1), hence aN= 0, a contradiction. Therefore the functions eαf0(α > 0) are independent. Fix f∈M\ {0}. Then f=f0g, where gis a nonzero linear combination PN j=1 ajeαjas before. Since, obviously, ϕsatisfies (1) and g∈H(D)\ {0}, it follows from Lemma 3 that M\ {0} ⊂ SA, whence SA is c-lineable. To conclude, take X:= H(D) and Γ:= {hfi+P:f∈H(D)\SA}, where P:= {polynomials}. Since 0 /∈ SA and the sum of a polynomial and of a function in H(D)\SA stays in H(D)\SA (i.e. H(D)\SA is stronger than P), we have on one hand that TS∈ΓS=P, which is dense in X, and on the other hand that TS∈Γ(X\S) = SA, which is c-lineable. According to Lemma 1, SA ∪ {0}contains a dense c-dimensional vector subspace or, that is the same, SA is maximal dense-lineable.  In the next assertion, we settle algebrability. Theorem 4 SA is algebrable. Proof For f∈H(D) the standard notation M(f, r) := max{|f(z)|:|z|=r} (0 < r < 1) will be used. We start with a function f1∈ SA. Then there is a sequence of radii σ= (rn)∈Σsuch that limn→∞ min{|f1(z)|:z∈rnT}= +∞. Hence f1∈ SA(ϕ0, σ), where ϕ0≡1. Let ϕ1(z) := exp M(f1,|z|). According to Theorem 1, we can select a function f2∈ SA(ϕ1, σ). By induction, assume that for some N≥2 the functions f1, . . . , fN−1, ϕ0, . . . , ϕN−2have been already determined. Then we define ϕN−1(z) := exp M(fN−1,|z|) and, again by Theorem 1, one can choose a function fN∈ SA(ϕN−1, σ). Therefore we obtain a sequence of functions (fn)⊂H(D) such that fn∈ SA(ϕn−1, σ) (n≥1), where ϕ0≡1 and ϕj(z)≡ exp M(fj,|z|) (j≥1). Define Mas the algebra generated by the functions fn (n∈N). Our task is to show that (fn) is a minimal system of generators of Mand that each nonzero member of Mbelongs to SA. In order to achieve the first part of the task, it is enough to prove that for each N≥2 the function fNis not algebraically generated by f1, . . . , fN−1. To do this, assume by way of contradiction that for some N≥2 there exists a polynomial P(z1, . . . , zN−1) in N−1 variables without constant term such that fN=P(f1, . . . , fN−1). Denote by mthe number of monomials forming P, by αthe maximum of the moduli of the coefficients of P, and by pthe degree of P. Then |fN(z)|=|P(f1(z), . . . , fN−1(z))| ≤ mαN−1 Y j=1 (1 + M(fj,|z|))p(z∈D). From the construction of f1, . . . , fNwe can choose n0∈Nsatisfying M(fN−1,|z|)≥M(fj,|z|) (j= 1, . . . , N −1) and Strongly annular functions 9 |fN(z)| ≥ exp M(fN−1,|z|)≥2mα(1 + M(fN−1,|z|))pN for all z∈A:= Sn≥n0rnT. It follows that, if z∈A, 1≤mα(1 + M(fN−1,|z|))pN exp M(fN−1,|z|)≤1 2, so providing the desired contradiction. Finally, fix f∈M\ {0}. Then there is N∈Nand a nonzero polynomial P(z1, . . . , zN) (without constant term, but this is inmaterial) such that f= P(f1, . . . , fN). We proceed by induction on N. If N= 1 then f=P(f1) = Pm k=0 akfk 1, say, where m∈Nand am∈C\ {0}. If m= 1 then it is trivial that f∈ SA. If m≥2 then, for z∈C(σ), we have |f(z)|≥|am||f1(z)|m1− m−1 X k=0 |ak||f1(z)|k−m−→ +∞(|z| → 1) because |f1(z)| → +∞as z∈C(σ),|z| → 1. Therefore f∈ SA(1, σ). Assume now that, for some N≥2, any nonconstant polynomial Qof N−1 variables satisfies Q(f1, . . . , fN−1)∈ SA(1, σ). Let f=P(f1, . . . , fN), where Pis as in the beginning of this paragraph. Then there are m∈Nand polynomials Q1, . . . , Qmof N−1 variables with Qm6≡ 0 such that f= Pm k=0 Qk(f1, . . . , fN−1)fk N. By the induction hypothesis, either Qmis constant or Qm(f1, . . . , fN−1)∈ SA(1, σ). Choose n0∈Nso large that fN(z)6= 0 if |z|=rnand n≥n0. Then |f(z)|=|Qm(f1(z), . . . , fN−1(z))fN(z)m| · 1− m−1 X k=0 Qk(f1, . . . , fN−1) fN(z)m−k for such points z. But, in view of the exponential growth of fNwith respect to f1, . . . , fN−1on Sn≥n0rnT, one gets that the last sum tends to zero as |z| → 1 (z∈C(σ)). Moreover, it is plain that |Qm(f1, . . . , fN−1)fm N|tends to +∞along C(σ). Hence f∈ SA(1, σ). This completes induction and shows that M\ {0} ⊂ SA(1, σ)⊂ SA, which had to be proved.  Remark 2 With slight modifications of the proofs of Theorems 3–4, it is not difficult to demonstrate the following improvement: Assume that ϕ:D→ (0,+∞) is continuous and that σ∈Σ. Then SA(ϕ, σ) (and so SA(ϕ)) is maximal dense-lineable and algebrable. Remark 3 This paper deals with special unbounded analytic (hence continuous) functions on Dor Cunder the focus of lineability. To this respect, families of unbounded continuous functions on more general topological spaces have been already studied from this point of view. Namely, Garc´ıa, Mart´ın and Seoane [21, Theorem 4.1] have recently proved that, in every non-compact metric space Ω, the set of all continuous unbounded real functions defined on it is algebrable.