Linear structure of the weighted holomorphic non-extendibility

Linear structure of the weighted holomorphic non-extendibility L. Bernal-Gonz´alez Abstract In this paper, it is proved that, for any domain Gof the complex plane, there exist an infinite-dimensional closed linear submanifold M1and a dense linear submanifold M2with maximal algebraic dimension in the space H(G) of holomorphic functions on Gsuch that Gis the domain of holomorphy of every nonzero member of fof M1or M2 and, in addition, the growth of fnear each boundary point is as fast as prescribed. 2000 Math. Subject Classification: Primary 30D40. Secondary 30B40, 30E10, 30H05, 46B15. Key words and phrases: holomorphic non-extendibility, infinite-di- mensional closed linear manifold, maximal algebraic dimension, spaceable set. Research partially supported by Plan Andaluz de Investigaci´on de la Junta de Andaluc´ıa FQM-127 and by DGES Grants BFM2003-03893- C02-01 and MTM2004-21420-E. 1 Introduction and notation Throughout this paper, the following standard terminology and notation will be used. The symbols N,C,D,Tdenote, respectively, the set of positive integers, the complex plane, the open unit disk {z∈C:|z|<1}, and the unit circle {z∈C:|z|= 1}. If a∈Cand r > 0 then B(a, r) (B(a, r), resp.) denotes the open (closed, resp.) euclidean ball with center aand radius r; in particular, B(0,1) = D. For points a, b of C, the line segment joining a with bis [a, b]. If A⊂Cthen A(A0, ∂A, resp.) denotes its closure (interior, 1 boundary, resp.) in C. Moreover, if z0∈Cthen d(z0, A) := inf{|z0−z|: z∈A}. A domain is a nonempty open subset of C. If Gis a domain, then H(G) denotes the Fr´echet space (= completely metrizable locally convex space) of holomorphic functions on G, endowed with the topology of uniform convergence on compacta. In particular, H(G) is a Baire space. Finally, if a∈Gand f∈H(G) then ρ(f, a) represents the radius of convergence of the Taylor series of fwith center at a. It is well known that ρ(f, a)≥d(a, ∂G). In 1884 Mittag-Leffler (see [9, Chapter 10]) discovered that for any domain Gthere exists a function f∈H(G) having Gas its domain of holomorphy. Recall that Gis said to be a domain of holomorphy for fif fis holomorphic exactly at G, that is, f∈H(G) and fis analytically non-extendible across ∂G or, more precisely, ρ(f, a) = d(a, ∂G) for all a∈G. Note that this implies that fhas no holomorphic extension on any domain containing Gstrictly. Both properties are equivalent if, for instance, Gis a Jordan domain, but the equivalence is not general (for instance, consider G:= C\(−∞,0] and f:= the principal branch of the logarithm on G). By He(G) we denote the subclass of functions which are holomorphic exactly at G. Hence, the Mittag-Leffler result mentioned above says that He(G)6=∅for any domain G. In 1933 Kierst and Szpilrajn [12] showed that at least for G=Dthe property discovered by Mittag-Leffler is generic, in the sense that He(D) is not only nonempty but even residual –hence dense– in H(D), that is, its complement in H(D) is of first category. Recently, Kahane ([11, Theorem 3.1 and following remarks]; see also [10, Proposition 1.7.6] and [4, Theorem 3.1]) has observed that Kierst-Szpilrajn’s theorem can be extended to every domain Gand to rather general topological vector spaces X⊂H(G) (including the full space X=H(G)); indeed, under suitable conditions on X, he obtains that He(G)∩Xis residual in X. In other words, He(G)∩Xis topologically large in X. Recently, we have proved [4] for the case G=Dthat under adequate hypotheses a topological vector space X⊂H(D) satisfies that He(D)∩Xis also algebraically large, in the sense that the last subset contains –except for zero– some “large” (= dense, or closed infinite-dimensional) linear manifold. Again, the case X=H(D) is covered. Note that the fact that He(G) is not a linear manifold increases the interest in this matter. As for a general domain G, Aron, Garc´ıa and Maestre [1, Theorem 8] had already proved in 2001 that H(G) contains a dense linear manifold M1as well as a closed infinite- 2 dimensional linear manifold M2such that Mi\ {0} ⊂ He(G) (i= 1,2). In fact, their result extends to any domain of holomorphy in CN(see also [4, Theorem 5.1] for an independent, different proof in the ‘dense’ case with N= 1), and the manifolds Mi(i= 1,2) are even ideals. In the terminology of [8], a subset Sof a linear topological space Eis spaceable whenever S∪{0}contains some closed infinite-dimensional subspace in E(see [8] and [2] for nice, recent examples of spaceable sets). Therefore, under this convention, it has been demonstrated in [1, Theorem 8] that He(G) is spaceable in H(G). Nevertheless, the approach in [1, Theorem 8] does not give any information about how fast the functions in M1or M2can grow near the boundary. In [4, note after Theorem 5.1] it is suggested how this can be proved for the manifold M1(‘dense’ case) in H(G), with G⊂C. Hence, it is natural to ask the following: Given any prescribed (‘weight’) function ϕ:G→(0,+∞), is the set Sϕ:= {f∈He(G) : lim sup z→t |f(z)|/ϕ(z)=+∞for all t∈∂G} spaceable in H(G)? The main aim in this paper is to furnish an affirmative answer to this question. This will be obtained in Section 2. Finally, in Section 3 we will complete this study by showing the existence of a dense linear submanifold Mwith maximal algebraic dimension –that is, dim (M) = χ:= the cardinality of the continuum– such that M\ {0} ⊂ Sϕ, where ϕis a given weight function as above. 2 Spaceability of the weighted non-extendibility Before establishing our main result, an auxiliary statement about basic sequences is needed. Let us consider the Hilbert space L2(T) of all (Lebesgue classes of) measurable functions f:T→Cwith finite quadratic norm kfk2:= (R2π 0|f(eiθ)|2dθ 2π)1/2. Since (zn)∞ n=−∞ is an orthonormal basis of L2(T), we have that (zn)n≥1is a basic sequence in L2(T). Recall that two basic sequences (xn)n≥1,(yn)n≥1in a Banach space (E, k · k) are said to be equivalent if, for every sequence (an)n≥1of scalars, the series P∞ n=1 anxnconverges if and only if the series P∞ n=1 anynconverges. This happens (see [3, 3 page 108]) if and only if there exist two constants m, M ∈(0,+∞) such that, for every finite sequence (aj)j=1,...,J of scalars, we have m J X j=1 ajxj≤ J X j=1 ajyj≤M J X j=1 ajxj.(1) Lemma 2.1. Assume that Gis a domain with D⊂Gand that (fj)j≥1⊂ H(G)is a sequence such that it is a basic sequence in L2(T)that is equivalent to (zj)j≥1. If {hl:= PJ(l) j=1 cj,lfj}l≥1is a sequence in span (fj)j≥1converging in H(G), then sup l∈N J(l) X j=1 |cj,l|2<+∞.(2) Proof. Observe first that, since Dis a compact subset of G, convergence in H(G) is stronger than convergence in L2(T)-norm. Therefore (hl)l≥1con- verges in L2(T), so the sequence (khlk2)l≥1is bounded, say khlk2≤α(l∈N). Let xj, yj,k · k be respectively the function z7→ zj, the function fjand the norm k·k2. Then, by (1), we get for every l∈Nthat m2 J(l) X j=1 |cj,l|2=m2 J(l) X j=1 cj,lzj 2 2 ≤ J(l) X j=1 cj,lfj 2 2 =khlk2 2≤α2. Hence (2) is satisfied because the supremum is not greater that α2/m2. Now, our main assertion about non-extendibility can be established. Theorem 2.2. Let G⊂Cbe a domain and ϕ:G→(0,+∞)be a function. Then Sϕis spaceable in H(G). Proof. We must prove the existence of an infinite-dimensional closed linear manifold Min H(G) such that M\{0}⊂Sϕ. The case G=Cbeing trivial, we may assume G6=C. We denote by G∗the one-point compactification of G. Recall that in G∗the whole boundary ∂G collapses to a unique point, say ω. Without loss of generality, it can be supposed that D⊂G. We are going to choose countably many pairwise disjoint sequences {a(k, n) : n∈N}(k∈N) of distinct points of G\Dsuch that each of them has no accumulation point in Gand every prime end (see [5, Chapter 9]) of ∂G 4 is an accumulation point of each such sequence. The last property means, more precisely, the following: For every k∈N, every a∈Gand every r > d(a, ∂G), the intersection of {a(k, n) : n∈N}with the connected component of B(a, r)∩Gcontaining ais infinite. In particular, every point t∈∂G would be an accumulation point of each sequence {a(k, n) : n∈N}. Let us show how such a family of sequences can be constructed. We begin with k= 1. Let {cj:j∈N}be a dense countable subset of G. For each j∈Nchoose bj∈∂G such that |bj−cj|=d(cj, ∂G). For every j∈Nlet {d1,j,l :l∈N}be a sequence of points in [cj, bj]\Dsuch that |d1,j,l −bj|<1/(1 + j+l) (j, l ∈N). Then we choose as {a(1, n) : n∈N} a one-fold sequence (without repetitions) consisting of all distinct points of the set {d1,j,l :j, l ∈N}. It is easy to check that {a(1, n) : n∈N}satisfies the required property. In a second step –that is, for k= 2– we can select for every j∈Na sequence {d2,j,l :l∈N}of points of [cj, bj]\(D∪ {a(1, n) : n∈N}) such that, in addition, |d2,j,l −bj|<1/(2 + j+l) (j, l ∈N); this is possible due to the denumerability of {a(1, n) : n∈N}. Again, we define {a(2, n) : n∈N}as a sequence consisting of all distinct points of the set {d2,j,l :j, l ∈N}; it satisfies evidently the required prime end property. It is now clear that this process can be repeated inductively, so yielding the desired disjoint family {{a(k, n) : n∈N}:k∈N}. Secondly, let us consider the subset A:= D∪B⊂G, where B:= {a(k, n) : k, n ∈N}. Recall that for each k∈Nthe sequence {a(k, n) : n∈ N}is an enumeration of the distinct points of a certain subset {dk,j,l :j, l ∈ N} ⊂ Gsatisfying |dk,j,l −bj|<1 k+j+l(j, l ∈N).(3) We have that Ais relatively closed in G. Indeed, the set of accumulation points of Ain Gis just D(which is included in A) because the set of accumulation points of Bin Gis empty. Let us explain why this is so. Assume, by way of contradiction, that z0∈Gis an accumulation point of B. Then there is a sequence of distinct points (dk(n),j(n),l(n))n≥1in Btending to z0. Then the set {(k(n), j(n), l(n)) : n∈N}is infinite, so at least one of the sets of positive integers {k(n) : n∈N},{j(n) : n∈N},{l(n) : n∈N}is infinite, hence unbounded. Therefore the sequence (k(n) + j(n) + l(n))n≥1is also unbounded, whence k(n) + j(n) + l(n)>2/d(z0, ∂G) for infinitely many 5 n∈N. Consequently, |dk(n),j(n),l(n)−z0|≥|z0−bj(n)|−|dk(n),j(n),l(n)−bj(n)| ≥d(z0, ∂G)−1 k(n) + j(n) + l(n)>d(z0, ∂G) 2 for infinitely many n∈N, which is absurd. Thus, Ais closed in G. But note that G∗\Ais connected as well as locally connected at ω, because Dis compact (so it is “far” from ω, and we can suppose that the basic connected neighborhoods of ωdo not intersect D), G\Dis connected and Bis countable (so deleting Bfrom G\Dmakes no influence in connectedness or local connectedness). Let us consider, for every N∈N, the function gN:A→Cdefined as gN(z) =    zNif z∈D, n(1 + ϕ(a(N, n))) if z=a(N, n) and n∈N, 0 if z=a(k, n) and k, n ∈Nwith k6=N. Observe that gNis continuous on Aand holomorphic on A0(= D). Then the Arakelian approximation theorem (see [7, pages 136–144]) guarantees the existence of a function fN∈H(G) such that |fN(z)−gN(z)|<1 3Nfor all z∈A. Consequently, one obtains |fN(z)−zN|<1 3Nfor all z∈D,(4) |fN(a(N, n)) −n(1 + ϕ(a(N, n)))|<1 for all n∈N,and (5) |fN(a(k, n))|<1 3Nfor all n∈Nand all k∈N\ {N}.(6) Finally, we define the sought-after linear manifold Mby M:= closureH(G)(span {fN:N∈N}). It is clear that Mis a closed linear manifold in H(G). On the other hand, we have from (4) that kfN−ϕNk2<3−Nfor all N∈N(where ϕN(z) := zN). By using this last inequality as well as the fact P∞ N=1 3−N<1 together with 6 the basis perturbation theorem [6, page 46, Theorem 9], we can derive that (fN)N≥1is a basic sequence in L2(T). Indeed, let (e∗ n)n≥1be the sequence of coefficient functionals corresponding to the basic sequence (zn)n≥1. Since ke∗ nk2= 1 (n∈N), one obtains ∞ X N=1 ke∗ nk2kfN−ϕNk<1. Therefore the perturbation theorem applies because (ϕN)N≥1is a basic sequence. Since (fN)N≥1is a basic sequence, we get that, in particular, the functions fN(N∈N) are linearly independent. Hence Mhas infinite dimension. It remains to show that M\ {0}⊂Sϕ. Fix f∈M\ {0}. Since the convergence in H(G) is stronger that the convergence in L2(T), we have that (the restriction to Tof) fis in f M:= closureL2(T)(span{fN:N∈N}). Therefore fhas a (unique) representation f=P∞ j=1 cjfjin L2(T), because (fN)N≥1is a basic sequence in this space. But f6= 0, so there is N∈Nwith cN6= 0. On the other hand, there is a sequence {hl:= PJ(l) j=1 cj,lfj}l≥1in span{fj:j∈N}(without loss of generality, we can assume that J(l)≥N for all l) that converges to fcompactly in G. By Lemma 2.1, C:= sup l∈N J(l) X j=1 |cj,l|2<+∞. But (hl)l≥1also converges to fin L2(T), so the continuity of each projection P∞ j=1 djfj∈f M7→ dm∈C(m∈N) yields that liml→∞ cN,l =cN. In particular, there exists l0∈Nsuch |cN,l| ≥ |cN| 2for all l≥l0.(7) Let us fix n∈N. Since the singleton {a(N, n)}is a compact subset of G, we get the existence of a positive integer l=l(n)≥l0such that |hl(a(N, n)) −f(a(N, n))|<1.(8) By using (5), (6), (7), (8), the triangle inequality and the Cauchy-Schwarz inequality, we obtain |f(a(N, n))|≥|hl(a(N, n))| − 1 7 ≥ |cN,lfN(a(N, n))| − J(l) X j=1 j6=N |cj,lfj(a(N, n))| − 1 ≥|cN| 2(n(1 + ϕ(a(N, n))) −1) − J(l) X j=1 j6=N |cj,l|1 3j−1 ≥|cN| 2(n(1+ϕ(a(N, n)))−1)− ∞ X j=1 (1 3j)2!1/2   J(l) X j=1 j6=N |cj,l|2   1/2 −1 ≥|cN| 2(n(1 + ϕ(a(N, n))) −1) −C1/2−1. Consequently, limn→∞ f(a(N, n)) = ∞= limn→∞ f(a(N, n))/ϕ(a(N, n)). The second equality shows that lim supz→t|f(z)|/ϕ(z) = +∞for all t∈∂G, because each boundary point is a limit point of (zn:= a(N, n))n≥1. Now, it is time to use the prime end approximation property of the sequence (zn). Suppose, by way of contradiction, that f6∈ Sϕ. Then f6∈ He(G), so there must be a point c∈Gsuch that ρ(f, c)> d(c, ∂G). Choose rwith d(c, ∂G)< r < ρ(f, c). By the construction of the sequences (a(k, n))n≥1(k∈N), there exists a sequence {n1< n2<· · · } ⊂ Nfor which znj∈G∩B(c, r) (j∈N). Finally, the sum S(z) of the Taylor series of fwith center cis bounded on B(c, r). But S=fon G∩B(c, r), so S(znj) = f(znj)→ ∞ (j→ ∞), which is absurd. This contradiction finishes the proof. 3 Manifolds with maximal algebraic dimension We conclude this note with a theorem that completes our Theorem 2.2 as well as Theorem 5.1 in [4] and (in the one-dimensional case) Theorem 8 in [1]. Specifically, we are able to construct –for a prescribed function ϕ:G→(0,+∞)– a linear submanifold M⊂H(G) with M\{0} ⊂ Sϕthat is not only dense, but even it satisfies dim (M) = χ(notice that the dense linear manifold Mwhose construction is suggested in [4, note following Theorem 5.1] was only of countably infinite dimension; in the opposite direction, the 8 dense manifold Xprovided in [1, Theorem 8] does satisfy dim(X) = χ, but the fact X\ {0}⊂Sϕdoes not hold). Observe that, as an easy consequence of Baire’s category theorem and of the fact that H(G) is infinite-dimensional, metrizable, separable and complete, we have dim (H(G)) = χ. Hence χis the maximal algebraic dimension which is permitted for the submanifolds of H(G). For instance, the linear manifold Mconstructed in the proof of Theorem 2.2 satisfies dim (M) = χ(because it is a closed subspace of H(G), so Mis also infinite-dimensional, metrizable, separable and complete) but it is not dense. Theorem 3.1. Let G⊂Cbe a domain and ϕ:G→(0,+∞)be a function. Then there is a dense linear manifold Min H(G)such that dim (M) = χ and M\ {0}⊂Sϕ. Proof. Again, the case G=Cis trivial, so we suppose G6=C. First, we consider pairwise disjoint sequences {a(k, n) : n∈N}(k∈N), and then we select a sequence {fN:N∈N} ⊂ H(G). This is made exactly as in the proof of Theorem 2.2, with the sole exception that instead of (5) we have |fN(a(N, n)) −n1/2(1 + ϕ(a(N, n)))|<1 for all n∈N.(9) In other words, with the notation of the proof of Theorem 2.2 we would define gN(a(N, n)) := n1/2(1 + ϕ(a(N, n))) (N, n ∈N) before the application of Arakelian’s theorem. The key point will be that n1/2tends to infinity as n→ ∞, but less rapidly than any power nN(N∈N). Let us define M1:= closureH(G)(span {fN:N∈N}). Therefore we obtain as in the proof of Theorem 2.2 that M1\ {0} ⊂ Sϕ. As observed at the beginning of this section, we have dim (M1) = χ. Second, fix an increasing sequence {Kn:n∈N}of compact subsets of G such that each compact subset of Gis contained in some Knand each component of the complement of every Kncontains some connected component of the complement of G(see [13, Chapter 13]). Choose a dense countable subset {ψn:n∈N}of H(G). Now consider for each N∈Nthe set AN:= KN∪ {a(k, n) : k, n ∈N}. In a similar way to the proof of Theorem 5.2 in [4], we have that ANis closed in Gand that G∗\ANis connected and locally connected at ω. The function hN:AN→Cdefined as hN(z) = ψN(z) if z∈KN, nN(1 + ϕ(a(k, n))) if z=a(k, n) (k, n ∈N) and z6∈ KN 9