# Weighted Vector-Valued Holomorphic Functions on Banach Spaces

Authors: Enrique Jordá
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the weighted Banach spaces of vector-valued holomorphic functions defined on an open and connected subset of a Banach space. We use linearization results on these spaces to get conditions which ensure that a functionf defined in a subset A of an open and connected subset U of a Banach space X, with values in another Banach space E, and admitting certain weak extensions in a Banach space of holomorphic functions can be holomorphically extended in the corresponding Banach space of vector-valued functions.


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## Body

## 1. Introduction, Notation, and Preliminaries

LetE be a locally convex space. The problem of deciding when a function f:Ω⊂ℂ→E is holomorphic whenever u∘f∈H(Ω) for each u∈E′ goes back to Dunford [1], who proved that this happens when E is a Banach space. Grothendieck [2] extended the result for E being quasicomplete. Bogdanowicz [3] gives extension results through weak extension, that is, he proved between other results that if Ω1⊂Ω2⊂ℂ are two domains (open and connected subsets), E is a complex, sequentially complete, and locally convex Hausdorff space, and f:Ω1→E satisfies that u∘f admits holomorphic extension for each u∈E′, then f admits a holomorphic extension to Ω2. More recently Grosse-Erdmann, Arendt and Nikolski, Bonet, Frerick, Wengenroth, and the author have given results in this way smoothing the conditions on Ω1 and also requiring extensions of u∘f only for a proper subset H⊆E′ (cf. [4–8]). Also, Laitila and Tylli have recently discussed the difference between strong and weak definitions for important spaces of vector-valued functions [9, Section 6].Our main aim here is to analyze a weak criterion for holomorphy and to give extension results for the Banach spaces of holomorphic functions defined on a nonvoid open subsetU of a Banach space X. To obtain these extension results, we use linearization results, that is, theorems which permit to identify classes of vector valued functions defined in U and with values in E with continuous linear mappings from a certain space G and with values in E. Recent work of Beltrán [10], Carando and Zalduendo [11], and Mujica [12] is devoted to get linearization results. We use for our extension results also linearization results obtained by Bierstedt in [13, 14].Our notation for the Banach spaces, locally convex spaces, and functional analysis is standard. We refer the reader to [15–17]. For a locally convex space F which is nonnormed, we denote by F′ its topological dual. For a Banach space (E,∥·∥), the dual of E is denoted by E*. We mainly deal with Banach spaces. The absolutely convex hull of a subset C of E is denoted by Γ(C), and the closure of C is denoted by C¯. If the closure is taken with respect to other topology τ, it will be denoted by C¯τ. (E,w) and (E*,w*) are E and E* endowed with the weak (σ(E,E*)) and the weak* (σ(E*,E)) topology, respectively. The open unit ball of E will be denoted by BE. A subset M⊆E (M⊆E*) is said to be total if span(M) is (σ(E*,E)) dense. By the Hahn Banach theorem, M being total in E* is equivalent to M being separating, that is, if e∈E and u(e)=0 for all u∈M, then e=0. M⊆E* is said to be norming if M is bounded, and its associated functional qM:E→ℝ, e↦sup{|u(e)|:u∈M} defines an equivalent norm in E, that is, if the polar set M∘:={e∈E:|u(e)|≤1forallu∈M} defines an equivalent (closed) unit ball in E. It is immediate that if M is norming then it is also separating. If qM is the norm of E, then M is called 1-norming. A subset M⊂E is called norming or total when we consider E⊂E**, that is, qM defines an equivalent norm in E*. A subspace H of E* is said to determine boundedness whenever all the σ(E,H)-bounded subsets of E are (σ(E,E*)-) bounded. A subspace H⊆E* is said to be norming if BH is a norming subset of E. We give below a relation between these concepts. The result is given in [6, Proposition 7, Remark 8] in the more general context of the Fréchet spaces, though in this paper the norming subspaces are called almost norming. We give a proof for the Banach case because it is very transparent.Proposition 1 (see [6]).
LetE be a Banach space. A subspace H⊂E* is norming if and only if H¯ determines boundedness in E.Proof.
It is standard to check thatqBH=qBH¯ on E. From this, it follows that H is norming if and only if H¯ is. Assume first that H is norming. This implies that H¯ is separating on E, and then, we can consider the algebraic inclusion E↪H¯*. By the very definition, qBH¯ is the restriction of ∥·∥H¯* to E. The hypothesis H norming means that E is isomorphic to (E,qBH¯), which is a subspace of (H¯*,∥·∥H¯*). By the Uniform Boundedness Principle, the σ(E,H¯) bounded subsets of E are norm bounded in H¯*, and then, ∥·∥ bounded since the norms are supposed to be equivalent in E. Conversely, let one assume that H¯ determines boundedness in E. This implies, again by the Uniform Boundedness Principle, that the identity I:(E,∥·∥H¯*)→(E,∥·∥) is bounded. Hence, there exists C≥1 such that ∥e∥≤C∥e∥H¯*=CqBH¯(e) for each e∈E, which implies that H¯ is norming.Thus, the property of beingnorming for subspaces in E* is between weak*-dense and strongly dense.LetM=(mi)i∈I be a bounded subset of E and I an index set. Let
(1)l1(M):={∑‍x∈E:exists(ai)i∈I∈l1(I)suchthatx=∑‍aimi},
equipped with the norm which makes it isomorphic to a quotient of l1(I). We will use the following lemma, which we supposed to be well known.Lemma 2.
LetM⊆BE be a norming subset. Then, the injection of l1(M) in E is an onto isomorphism.Proof.
The hypothesis onM yields that there exist c,C>0 such that
(2)cB¯E*⊆M∘⊆CB¯E*.
Hence, we take polars and apply the bipolar theorem to get
(3)(1C)B¯E⊆Γ(M)¯⊆(1c)B¯E.
Let CE be the equivalent open unit ball in E such that C¯E=Γ(M)¯. We define T:l1(I)→E, (ai)i↦∑i∈I‍aimi. T is clearly bounded. Moreover, Γ(M)⊆T(Bl1(I)), and then,
(4)CE⊆Γ(M)¯⊆T(Bl1(I))¯.
We get from the Schauder lemma [16, Lemma 3.9] that T is open and then surjective. We conclude from the very definition of l1(M).Remark 3.
If we assume in Lemma2 that M is 1-norming, then the isomorphism is an isometry.We see below that if the bounded subsetM is not norming, then the assertion is not true in general.Remark 4 (Bonet).
The assertion in Lemma2 implies that if M⊆E is bounded, then x∈Γ(M)¯ if and only if for each ε>0 there exist sequences (αn)n∈(1+ε)Bl1 and (mn)n⊂M such that x=∑n‍αnmn. This is not in general true if we assume that M is only to be bounded. Valdivia showed (see [17, Example 3.2.21]) that in every infinite dimensional Banach space there is an absolutely convex bounded subset B⊂E which is not closed such that EB:=(span(B),pB) is a Banach space with closed unit ball B, where
(5)pB(x):=inf{λ>0,x∈λB},x∈span(B).
From EB being Banach, we conclude l1(B)⊆EB, but from the proof given in [17, Example 3.2.21], it follows that B¯ is not included in span(B).LetU be a connected open subset of a Banach space X. The space of all the holomorphic functions on U is denoted by H(U). The compact open topology on H(U) denoted by τc. (H(U),τc) is a semi-Montel space; that is, each closed and bounded subset is compact. A subset A⊂U is called U-bounded if it is bounded and the distance of A to the complementary of A is positive. For U=X, U-bounded means simply bounded. If U is the unit ball of a Banach space, A is U-bounded if and only if it is contained in a ball of radius r<1. If E is a Banach space, the space of E-valued holomorphic functions on U is denoted by H(U,E). We refer to [18] for the precise definitions. A weightv:U:→]0,∞[ is a continuous function which is strictly positive. According to [19], we say that a weight v on U satisfies the property (I) whenever it is bounded below in each U-bounded subset A of U. The weighted Banach spaces of holomorphic functions are defined as
(6)Hv(U):={f∈H(U):supx∈Uv(x)|f(x)|<∞}Hv0(U):={f∈H(U):vfvanishesatinfinityonU-boundedsets}.
Recall that a function g:U→ℝ is said to vanish at infinity on U-bounded sets when for each ε>0 there exists a U-bounded subset A such that |g(x)|<ε for x∈U∖A. Hv(U) is always continuously embedded in (H(U),τc). If v satisfies (I), then Hv(U) is continuously embedded in the space of holomorphic functions of bounded typeHb(U), which is H(U) endowed with the Fréchet topology of uniform convergence on U-bounded sets.Analogously, for a Banach spaceE, we define the weighted spaces of vector-valued functions as
(7)Hv(U,E):={f∈H(U,E):supx∈Uv(x)∥f(x)∥<∞}Hv0(U,E):={f∈H(U,E):v∥f∥vanishesatinfinityonU-boundedsets}.
During all the work, our model spaces will be Hv(U) and Hv0(U). But we will deal with general closed subspaces Av(U) of Hv(U) and their corresponding vector-valued analogues Av(U,E) (which will be defined in the following section) in order to consider important subspaces as they are the spaces 𝒫(mX) of homogeneous polynomials of degree m and, in case of U being bounded, the algebras A(U) and Au(U) of holomorphic and bounded functions which are continuous and uniformly continuous on U¯, respectively.LetAv(U) be a subspace of Hv(U). A subset A⊆U is said to be a set of uniqueness for Av(U) if each f∈Av(U) which vanishes at A is identically null. A set A⊆X is said to be sampling for Av(U) if there exists some constant C≥1 such that, for every f∈Av(U),
(8)supx∈Xv(x)|f(x)|≤Csupa∈Av(a)|f(a)|.In caseAv(U) is an algebra the constant, C can be always taken 1 and, according to Globevnik, the sampling sets are called boundaries [20–22]. If A⊂U and MA={v(x)δx:x∈A}⊆Av(U)*, it follows from the definitions that MA⊆BAv*, A is sampling if and only if MA is norming, and A is a set of uniqueness if and only if MA is total.The sampling sets (as well as interpolation sets) of the weighted spaceA-p(𝔻) (i.e., Hv(𝔻) for v(z)=(1-z)p, p>0) were characterized by Seip in [23] in terms of certain densities.


## 2. Banach Subspaces ofHv(U) Which Are Dual Spaces

Let one considerAv(U)⊆Hv(U) to be a subspace with compact closed unit ball for τc. Notice that this condition implies that Av(U) is norm closed. We define the Banach space of vector-valued functions in a weak sense:
(9)Av(U,E):={f:U→E:u∘f∈Av(U)∀u∈E*}.
Since weakly holomorphic functions are holomorphic and weakly bounded sets are bounded, it follows that for Av(U)=Hv(U) this definition agrees with the strong definition given previously. Following the same steps as in [12, Theorem 2.1] (also in [24, Lemma 10]), we get that Av(U,E) can be identified with L(GAv,E), GAv being the predual of Av(U) that exists by the Dixmier-Ng Theorem [25].Remark 5.
In [25], it is shown that GAv consists of all the functionals y∈Av(U)* such that y restricted to BAv is τc continuous. Let A⊆U. If we denote MA:={δx:x∈A}⊂BGAv, we have that MA is separating in Av(U) if and only if span(MA) is (weakly) dense if and only if A is a set of uniqueness for Av(U). Analogously, MA is norming in Av(U) if and only if A is sampling for Av(U).Proposition 6.
LetAv(U) be a subspace of Hv(U) with τc compact closed unit ball. Then, f∈Av(U,E) if and only if there exists T∈L(GAv,E) such that f(x)=T(δx). Moreover, the correspondence is an isometry.Proof.
IfT∈L(GAv,E), we define f(x):=T(δx). Since T is continuous and Gv*=Av(U), it follows that u∘f∈Av(U) for each u∈E*, and then, f∈Av(U,E) by the very definition.
Conversely, letf∈Av(U,E). We set MU:={v(x)δx:x∈U}. Since MU is 1-norming in Av(U), we apply Lemma 2 and Remark 3, to get that GAv=l1(M)={∑x∈U‍αxv(x)δx:∑x∈U‍|αx|<∞}. We see that T(δx):=f(x) defines a linear mapping on Gv. If ∑x∈U‍αxδx=0, then, for each u∈E*,
(10)〈u,T(∑x∈U‍αxδx)〉=〈u,∑x∈U‍αxf(x)〉=〈u∘f,∑x∈U‍αxδx〉=0,
and then, T is well defined. Moreover, since BGv={∑x∈U‍αxv(x)δx:∑x∈U‍|αx|<1}, it is easy to compute ∥T∥=sup{v(x)∥f(x)∥:x∈U}=∥f∥v.Now, we are going to show that there are more natural spaces with compact unit ball for the compact open topology. To do this, we present a general result of complemented subspaces in the Fréchet spaces of analytic functions which could be of independent interest. We state it for Fréchet instead of Banach to include the important spaceHb(U). For Hv(U) and Hb(U) with v radial and U balanced, it is done by García et al. in [19, Proposition 3, Example 14].Theorem 7.
LetF(U) be a Fréchet space of holomorphic functions on U such that F(U)↪(H(U),τc) continuously. If, for m∈ℕ, 𝒫(mX)⊂F(U), then 𝒫(mX) endowed with its norm topology is a complemented subspace of F(U). If F(U)=Hv(U), then Bv¯∩𝒫(mX) is compact in (H(U),τc).Proof.
Let one assume without loss of generality that0∈U. For f∈H(U), we denote Pmf∈𝒫(mX) the m-homogeneous polynomial such that
(11)f(x)=f(0)+∑k∈ℕ‍Pkf(x).
Let τp be denoted by the topology in 𝒫(mX) of pointwise convergence on B¯X. The projection
(12)Pm:(H(U),τc)→(P(Xm),τp),f↦Pmf,
is continuous. We checked it. Let f∈H(U), and let (fi)i∈I be a net convergent to f in (H(U),τc). Let r>0 such that the closed ball D(0,r)⊂E, and let u∈E with ∥u∥=1. For i∈I, we define gi(z):=fi(zu)∈H∞(Br), Br being the ball with radius r in ℂ. Let g(z):=f(zu). We have that (gi)i∈I converges to g in H∞(Br). We conclude from the continuity of the evaluations of the derivatives in this last space and
(13)Pmfi(u)=gi(m)(0)m!m∈ℕ0,i∈I.
Hence, by the closed graph theorem, we get that the mapPm:F(U)→𝒫(mX), f↦Pmf is continuous. Since the map is by hypothesis surjective and restricted to 𝒫(mX) that is the identity, it also follows that 𝒫(mX) is closed in F(U). Thus, the inclusion i:=Pm|𝒫(mX) is an isomorphism. Hence, the inverse of the inclusion j:=i-1:𝒫(mX)→F(U) satisfies that Pm∘j is the identity in 𝒫(mX). We apply [26, Chapter 2, Section 7, Proposition 3] to conclude that 𝒫(mX) is complemented in F(U).
We check now thatBv¯∩𝒫(mX) is compact for the topology of pointwise convergence on U. Let (fi)i∈I be a net in 𝒫(mX)∩Bv such that it is convergent to f∈Bv pointwise in U. Assume without loss of generality that V:=U∩BX is nonempty. The net (fi)i∈I is a bounded net in 𝒫(mX) which is Cauchy for the topology of pointwise convergence in V. This topology is Hausdorff and weaker than the topology of pointwise convergence in BX. Since (𝒫(mX),∥·∥m) is a dual space [18, Proposition 1.17], the topology of pointwise convergence on BX is relatively compact restricted to the bounded sets in (𝒫(mX),∥·∥m) and then agrees in the bounded sets with the topology of pointwise convergence on V. Moreover, Bv¯∩𝒫(mX) is bounded in (𝒫(mX),∥·∥m), and hence, we get that (fi)i is convergent to g∈𝒫(mX) pointwise in X. Since U⊆X, we get f=g. We have proved that Bv¯∩𝒫(mX) is closed in Bv¯ for the topology of pointwise convergence in U, and then it is compact.For spacesHv(U) containing 𝒫(mX), we have that (𝒫(mX),∥·∥v) is a subspace which is complemented and it is isomorphic to 𝒫(mX) endowed with its natural norm ∥·∥m. Moreover, (𝒫(mX),∥·∥v) has a compact unit ball for the topology of pointwise convergence in U, and hence, it is a dual Banach space because of Dixmier-Ng theorem [25]. We denote by Gvm the predual of (𝒫(mX),∥·∥v) and by Gm the predual of (𝒫(Xm),∥·∥m) obtained in [12, Theorem 2.4]. In Gm, the subset M:={δx:x∈BX} is norming and then spans a (σ(Gm,𝒫(mX))-) dense subspace. The same applies for Mv:={v(x)δx:x∈U} in (𝒫(mX),∥·∥v). Both M and Mv are formed by functionals which are linearly independent by [11, Proposition 1]. We check below that there is a natural isomorphism between Gm and Gvm.Proposition 8.
Letv be a weight on U such that 𝒫(mX)∈Hv(U). The predual Gvm of (𝒫(mX),∥·∥v) is isomorphic to the predual Gm of (𝒫(mX),∥·∥m) canonically, that is, there exists T:Gvm→Gm such that T(δx)=δx for each x∈U.Proof.
LetMv be as defined previously. If we defineT1:span(Mv)→Gm, by means of δx↦δx, we have that T is well defined since δx=∥x∥mδx/∥x∥∈span(M), it is (weakly) continuous, and then, it can be extended to T^1:Gvm→Gm. If we consider now span(Mv) as a subspace of Gm it is (weakly) dense since open sets are sets of uniqueness in 𝒫(mX). The linear map T2:span(Mv)→Gvm, δx↦δx is again (weakly) continuous, and hence that we get, an extension T^2:Gm→Gvm. T^2∘T^1:Gvm→Gvm is a continuous linear mapping, and then, it is the identity since both coincide in span(Mv). Moreover, T1(Gvm) has dense range in Gm. Hence, T^1 is an onto isomorphism by [26, Chapter 2, Section 7, Proposition 3].From the linearization of these dual Banach subspaces ofHv(U), one can get easily an extension of the Blaschke-type result for vector-valued functions [4, Theorem 2.5] generalized in [7, Corollary 4.2]. The proof that we give is strongly based on the Banach-Steinhaus principle.Proposition 9.
LetAv(U) be a subspace of Hv(U) which has a τc-compact closed unit ball, let A⊆U be a set of uniqueness for Av(U), and let E be a Banach space. If (fi)i∈I is a bounded net in Av(U,E) such that (fi(x))i∈I is convergent for each x∈A, then (fi)i∈I is convergent to a function f∈Av(U,E) uniformly on the compact subsets of U.Proof.
Let(Ti)i∈I be the sequence of operators in L(GAv,E) such that fi(x)=Ti(δx) for each x∈U. Let Mv={v(x)δx:x∈U}⊆Gv(U). Mv is a 1-norming subset of Gv, that is,
(14)Γ(Mv)¯=Mv∘∘=BAv∘=BGAv.
By hypothesis, there exists C>0 such that
(15)Ti(Mv)={v(x)fi(x):x∈U}⊂CBEforeachi∈I.
Thus,
(16)Ti(BGAv)=Ti(Γ(Mv)¯)⊂Γ(Ti(Mv))¯⊆CB¯E,
for each i∈I. By Remark 5, the subset MA:={δx:x∈A} is total in GAv. Since (Ti)i∈I is equicontinuous, the topology of pointwise convergence on GAv coincides with the topology of pointwise convergence in MA by [27, 39.4(1)]. Thus, (Ti)i∈I is pointwise convergent to T∈L(GAv,E). The convergence is uniform on the compact subsets of GAv by [27, 39.4(2)]. If K⊂U is compact, then {δx:x∈K} is compact in GAv. This follows from the observation that U→GAv, x↦δx is (weakly) holomorphic and then continuous.Proposition9 and Theorem 7 yield that the Banach-Steinhaus theorem stated as in [27, 39.4(1)] can be extended to the space of vector-valued polynomials P(mX,E). Bochnak and Siciak showed [28, Theorem 2] that the uniform boundedness principle also is valid for polynomials.The following results are extensions of those obtained in [7] by Frerick et al. for spaces of bounded holomorphic and harmonic functions on open subsets of finite-dimensional subspaces with values in locally convex spaces. Our results are valid for spaces of functions defined on an open and connected subset U of a Banach space X, but we restrict to the case of Banach-valued functions. The proofs that we give here are simpler. The next theorem extends [7, Theorem 2.2].Theorem 10.
Letv be a weight on U, let Av(U) be a subspace of Hv(U) with τc-compact closed unit ball, let A be a set of uniqueness for Av(U), let E be a Banach space, and let H⊆E* be a subspace which determines boundedness in E. If f:A→E is a function such that u∘f admits an extension fu∈Av(U) for each u∈H, then f admits a unique extension F∈Av(U,E).Proof.
LetFA be the span of {δx:x∈A}. The hypothesis implies that FA is σ(GAv,Av(U)) dense, and then, it is dense in norm. The map T:FA→E, δx↦f(x) is well defined since H is separating. Let x=∑i=1k‍αiδxi be an element in the unit ball BFA of FA, and let u∈H. We compute:
(17)|〈T(x),u〉|=|〈∑i=1k‍αif(xi),u〉|=|〈∑i=1k‍αiδxi,fu〉|≤∥fu∥v.
Since this is true for eachx∈BFA, we conclude that T(BFA) is σ(E,H) bounded and then norm bounded by hypothesis. Thus, T:FA→E is a bounded linear mapping. Since FA is dense in GAv, we can extend T to T^:GAv→E. We conclude by Proposition 6.The following result is a generalization of [6, Theorem 1(ii)].Theorem 11.
Letv be a weight on U, let Av(U) be a subspace of Hv(U) with τc-compact closed unit ball, let A be a set of uniqueness for Av(U), let E be a Banach space, and let H⊆E* be a norming subspace. If f:A→E is a function such that u∘f admits an extension fu∈Av(U) for each u∈H such that (fu)u∈BH is bounded in Av(U), then f admits a unique extension F∈Av(U,E).Proof.
Ifu∈BH¯ and (un)n⊂BH tend to u, then (fun)n is a bounded sequence such that (fun(x))n converges to u(f(x)) for each x∈A. Proposition 9 yields that there exists fu∈Av(U) such that (fun(x))n tends to fu(x) for each x∈U. The conclusion is a consequence of Proposition 1 and Theorem 10.We now study the problem of extending functions which admit extensions for functionals in a subspaceH of E* which we assume only to be σ(E*,E) dense. In this case, we require that A is quite large. This is symmetric with the problem studied by Gramsch [29], Grosse-Erdmann [8], and Bonet et al. [5]. The next theorem is an extension to our context of [7, Theorem 3.2].Theorem 12.
Letv be a weight on U, let Av(U) be a subspace of Hv(U) with τc-compact unit ball, and let A be a sampling set for Av(U). Let E be a Banach space, and let H be a σ(E*,E)-dense subspace of E*. If f:A→E is a function such supa∈Av(a)∥f(a)∥<∞ and such that u∘f admits an extension fu∈Av(U) for each u∈H, then there exists a unique extension F∈Av(U,E) of f.Proof.
The setMA:={v(x)δx:x∈A}⊆GAv is norming for Av(U); hence, we apply Lemma 2 to get that GAv is isomorphic to l1(MA). This means that for each x∈GAv, there exists α:=(αn)n∈l1 and (xn)n⊂A such that
(18)x=∑n‍αnv(xn)δxn.
The open unit ball B1 in GAv for the norm which makes this space isometric to l1(MA) is formed by the vectors x∈GAv such that the sequence (αn)n in the previous representation can be taken in the open unit ball of l1. We define that T:GAv→E, x↦∑n‍αnv(xn)f(xn). Since {v(a)f(a):a∈A} is bounded by hypothesis, the series is convergent. Moreover, if ∑n‍αnv(xn)δxn=0, then for each u∈H(19)u(∑n‍αnv(xn)f(xn))=〈∑n‍αnv(xn)δxn,fu〉=0.
Since H is separating, T is well defined. Moreover, the hypothesis of boundedness of {v(a)f(a):a∈A} implies that T(B1) is bounded. Hence, we conclude by Proposition 6.Remark 13.
If we considerf:B(0,1/2)→l1, z→(zn)n, we have that f∈H(𝔻,l1); hence, f(B(0,1/2)) is relatively compact in l1. Moreover, it is immediate that u∘f admits an extension to H∞(𝔻) for each u∈φ (the space of sequences which are zero but finitely many components), φ is σ(l∞,l1) dense (even norming since it is dense in c0), and B(0,1/2) is a set of uniqueness for H∞(𝔻). However, f∉H∞(𝔻,l1) since ∥f(z)∥1=|z|/(1-|z|) for each z∈𝔻. This shows that the hypothesis in Theorems 10 and 12 is optimal, that is, for the conditions on the set A where the functions are defined and in the subspace H for which functionals, we have weak extensions that cannot be simultaneously relaxed, and also the condition of boundedness in the extensions in Theorem 11 can not be dropped.


## 3. General Banach Subspaces ofHv(U)

For arbitrary Banach spacesAv(U)⊂Hv(U) with no assumption on the unit ball, the equivalence between the weak and the strong definitions does not hold in general. We discuss it below. We consider the space Hv0(U), and we define
(20)Hv0(U,E)w:={f:U→E:u∘f∈Hv0(U)∀u∈E*}.
A Banach space E is said to satisfy the Schur property if every sequence (xn)n in E which is weakly convergent is also norm convergent. The well-known theorem of Schur asserts that l1 satisfies this property.Proposition 14.
IfE is a Banach space with the Schur property, then Hv0(U,E)=Hv0(U,E)w.Proof.
Suppose that there existsf∈Hv0(U,E)w∖Hv0(U,E). Then, there exist c>0 and (xn)n going to infinity on U-bounded sets such that v(xn)∥f(xn)∥>c and u∘f∈Hv0(U) for all u∈E*. This last condition implies that
(21){v(xn)f(xn):n∈ℕ}
is (weakly) convergent to zero, a contradiction.We see below that the situation differs for function with values in the general Banach spaces.Example 15.
Assume thatX is finite dimensional and Hv0(U) is infinite dimensional. Then, Hv0(U,c0)⊊Hv0(U,c0)w.Proof.
First, we proceed similarly as in [30, Lemma 21] to get a sequence (fn)n in Bv0 such that (fn)n converges to 0 in τc, and there exists δ>0 such that ∥fn∥v≥δ for all n∈ℕ. Since Hv0(U) is infinite dimensional, there is δ>0 and (gn)n∈Bv0 such that ∥gn-gk∥v>δ for n≠k. We apply that (Hv(U),τc) is metrizable and Bv¯  is τc compact to get that (Bv0,τc) is relatively sequentially compact. Hence, we can extract a subsequence of (gn)n which is Cauchy for τc, and we denote again by (gn)n. Defining fn:=(gn-gn+1)/2, we get the desired sequence.
We considerf:U→c0, z→(fn(z))n. Let u=(un)n∈l1 be arbitrary. Since (fn)n⊂Bv0, the series ∑n‍unfn is convergent in Hv0(U). Hence, f∈Hv0(U,c0)w. The convergence of (fn)n for the compact open topology implies that for each K⊂U there exists n0=n0(K) such that
(22)sup{|fn0(z)|:z∈K}<δ2max{v(x):x∈K}.
Since ∥fn0∥v≥δ, we obtain that there exists z0∈U∖K such that
(23)v(z0)∥f(z0)∥≥v(z0)|fn0(z0)|≥δ2.
Thus, f∉Hv0(U,c0).Example 16.
Assume thatU is the unit ball of a Banach space X, g:[0,1]→]0,∞[ continuous with g(t)>0 for 0≤t<1 and g(1)=0 and v(x)=g(∥x∥) for x∈U. Then, Hv0(U,c0)w⊊Hv0(U,c0).Proof.
The hypothesis onv implies that for each f∈Hv the Taylor polynomials Pnf of the development at zero converge to f in (H(U),τb). If we consider the Cesàro means
(24)Cn(f)=1n+1∑i=0n‍(∑k=0i‍Pn(f)),
then Cn(f)∈Bv0 for each f∈Bv ([31, Proposition 1.2], [19, Proposition 4]) and Cn(f)→f in (H(U),τb). If f∈Bv∖Bv0, then (Cn(f))n is not Cauchy in Hv(U), since Hv0(U) is closed. Hence, there are ε>0 and a subsequence (gk)k:=(Cnk(f))k such that
(25)∥gk-gk+1∥v>ε.
Defining hk:=gk-gk+1, we have that (hk)k tends to 0 in Hb(U). Proceeding as in Example 15, we obtain that h:U→c0,x↦(hk(x))k satisfies h∈Hv0(U,c0)w∖Hv0(U,c0).
The proof is complete sinceHv(U)∖Hv0(U) is never empty. We checked it. If U=𝔻, then Hv(𝔻) is the bidual of Hv0(𝔻) and this last space is not reflexive [32, 33]. Hence, there exists f0∈Hv(𝔻)∖Hv0(𝔻) with ∥f0∥v≤1. For U arbitrary, we consider x0∈X such that ∥x0∥=1 and u∈X* such that ∥u∥=1 and u(x0)=1. Define f:U→ℂ by f(x)=f0(u(x)). Since g is nonincreasing, we have
(26)v(x)|f(x)|=g(∥x∥)|f0(u(x))|≤g(|u(x)|)|f0(u(x))|≤1.
Since f0∉Hv0(𝔻), there exist c>0 and a sequence (zn)n of complex numbers smaller than 1 such that limn|zn|=1 and
(27)v(znx0)|f(znx0)|=g(|zn|)|f0(zn)|>c,
hence, f∉Hv0(U).Thus, on the contrary that with the concrete examples of dual spacesAv(U) considered in the previous section (Hv(U) and 𝒫(mX)), in Hv0(U) the definition of the corresponding spaces of vector-valued functions in the weak sense are not consistent with the natural definition. For linearization for these spaces with the weak definition, we refer to the work of Carando and Zalduendo [11].In view of Proposition14, one could expect that the analogous extensions of Theorems 10 and 12 are possible for Hv0(U,E) when E is required to have the Schur property. This is not the case as the following example shows.Example 17.
LetU be the unit ball of a Banach space X, and let v(x)=1-∥x∥ for x∈U. Fix x0∈X with ∥x0∥=1 and h∈BX* with h(x0)=1. Consider that f:U:→l1, x↦(h(x)n)n, then the following applies. (a)
f∈Hv(U,l1) and for each 0<r<1(28)∥f∥v=supx∈U∖rUv(x)∥f(x)∥1=1.
Hence, f∉Hv0(U,l1).(b)
u∘f∈Hv0(U) for each u=(un)n∈c0.Proof.
To prove (a), we observe thatp(t)=t/(1-t) is increasing for t∈[0,1[; hence,
(29)supx∈Uv(x)∥f∥1=supx∈U(1-∥x∥)|h(x)|1-|h(x)|=supx=tx0,0<t<1(1-∥x∥)|h(x)|1-|h(x)|=supx=tx0,0<t<1|h(x)|=supx=tx0,r<t<1(1-∥x∥)|h(x)|1-|h(x)|=1.
Let u=(un)n∈c0. Let ε>0 and n0∈ℕ such that n>n0 implies |un|<ε/2. For each x∈U, since |h(x)|<1, we have
(30)v(x)|∑n>n0‍unh(x)n|≤(1-∥x∥)ε2∑n∈ℕ‍|h(x)|n≤ε2.
Let 0<r<1 such that for each r<t<1(31)(1-t)∑n≤n0‍|un|<ε2.
From (30), (31), and |h(x)|<1, we obtain that
(32)supx∈U∖rUv(x)|u∘f(x)|=supx∈U∖rU(1-∥x∥)∑n∈ℕ‍|unh(x)n|≤ε.Remark 18.
The same computation as in Example17(b) shows that for 1<p<∞ and v(x)=(1-∥x∥)(1/p) the function f:U→lp, x↦(h(x)n)n satisfies that f∈Hv0(U,lp)w∖Hv0(U,lp).


## 4. Spaces of Weighted Compact Range Vector-Valued Holomorphic Functions

In this section, we consider the natural extension to the weighted case of the vector-valued compact holomorphic functions introduced by Aron and Schottenloher in [34] by means of the weak definition, that is, for an open and connected subset U of a Banach space X, a closed subspace Av(U) of Hv(U), and a Banach space E, we define that
(33)Avc(U,E)={f∈Hv(U,E):(vf)(U)isrelativelycompactandu∘f∈Av(U)∀u∈E*}.In caseX is finite dimensional, the space Hv0(U,E) is the space of holomorphic functions such that f is continuous in the Alexandroff compactification U∪{∞} of U and f(∞)=0. Hence, Hv0(U,E)=Hv0c(U,E) in this case. If X is infinite dimensional, the inclusion Hv0c(U,E)⊂Hv0(U,E) is strict in general. Observe that if U is the unit ball and v vanishes at ∞ on U, then I|U∈Hv0(U,X).We check that this (weak) definition agrees with the natural definition whenU is the unit ball of X, v=1, and Av(U)=Au(U) the space of the holomorphic and uniformly continuous functions on U, that is, we want to show that
(34)Auc(U,E):={f∈H(U,E):fisuniformlycontinuousandf(U)isrelativelycompact}={f:U→E:u∘f∈Au(U)foreachu∈E*andf(U)isrelativelycompact}.Assume thatf∈H(U,E) satisfies that u∘f∈Au(U) for each u∈E*. Given ε>0, since f(U) is relatively compact, there exists a weak neighbourhood V of 0 such that
(35)(f(U)-f(U))∩V⊆(f(U)-f(U))∩B(0,ε).
Let one assume that V:={e∈E:|ui(e)|<γ,1≤i≤k}. Since ui∘f is uniformly continuous on U for 1≤i≤k, there exists δ>0 such that ∥x-y∥<δ implies f(x)-f(y)∈V, and therefore, ∥f(x)-f(y)∥<ε.Given two locally convex spacesF and E, we denote by FεE its ε-product of Schwartz, that is, the space of all linear and continuous mappings ℒe(Fco′,E), endowed with the topology of uniform convergence on the equicontinuous subsets of F′.  Fco′ is F′ endowed with the topology τco of uniform convergence on the convex compact subsets of F. The ε-product is symmetric by means of the transpose mapping [27, 43.3(3)]. In case E and F being Banach spaces, T:F*→E belongs to FεE if and only if T is a compact operator which is weak*-weak continuous by [27, 43.3(2)]. The next theorem is the analogous of Theorem 12 in the case of general Banach spaces of functions, not necessarily dual Banach spaces. However, the techniques used here are different. The proof is analogous to the one given by Bierstedt and Holtmanns in [35] when the linearization result is obtained in a much more general context, but we only require the function to be defined in a sampling set.Theorem 19.
LetAv(U) be a closed subspace of Hv(U), and let A⊂U be a sampling set for Hv(U). Let E be a Banach space, and let H be a weak*-dense subspace of E*. The following are equivalent. (i)
f:A→E satisfies that (vf)(A) is relatively compact in E, and u∘f admits an extension fu∈Av(U) for each u∈H.(ii)
The linear mappingT:H→Av(U),  u→fu admits an extension T^∈Av(U)εE.(iii)
f can be extended to F∈Avc(U,E).Proof.
Iff satisfies (i), then the linear mapping T:H→Av(U),  u↦fu is τco - ∥·∥ continuous, since the absolute convex hull of f(A) is relatively compact in E, and the uniform convergence on A defines an equivalent topology in Av(U). By the Hahn-Banach theorem, H is dense in E* endowed with any topology t such that E=(E*,σ(E*,E))′=(E*,t)′, that is, which respects the duality of E* and E, in particular, for Eco*:=(E*,τco). Since Av(U) is a Banach space, we can extend it to T^∈ℒ(Eco*,Av(U))=EεAv(U).
If (ii) is satisfied, then the transposeT^t:Av(U)*→E is weak*-weak continuous, and T^t maps the unit ball of Av(U)* to a relatively compact subset of E ([27, 43.3(2),(3)]). We define f(x)=T^t(δx), since u∘f(x)=〈T(u),δx)〉 for each x∈U, and for each u∈E*, we have that u∘f∈Av(U). We conclude since {v(x)δx:x∈U} is in the unit ball of Av(U)*.
Finally, that (iii) implies (i) is trivial.Observe that, settingA=U in Theorem 19, we obtain a linearization of the space Avc(U,E), which also can be obtained as a consequence of the much more general linearization result given by Bierstedt in [14, Bemerkung 3.1] and [15, Corollary 3.94]. Example 15 shows that Theorems 10 and 12 cannot be stated avoiding the condition of relative compactness on the range for general Banach spaces of holomorphic functions.We finish showing that the weak definition given in this section forHv0c(U,E) is consistent with the natural one, that is,
(36)Hv0c(U,E)={f∈Hv0(U,E):(vf)(U)isrelativelycompactHv0}={Hv0f∈Hv(U,E):(vf)(U)isrelativelycompactandu∘f∈Hv0(U)∀u∈E*}.We use a similar argument to the one used by Bierstedt in [13, page 200] in a more general setting, including our case when X is finite dimensional (i.e., putting compact instead of U-bounded in the definition of Hv0(U)). If f:U→E satisfies that u∘f∈Hv0(U) for each u∈E*, then by the previous theorem, there exists T:Hv0(U)*→E defined by T(δx)=f(x) which is weak*-weak continuous and such that T(Bv0) is relatively compact. This implies that the restriction of T to Bv0 is weak*-norm continuous. Let ε>0. There exists a weak* 0-neighbourhood V in Hv0(U)* such that ∥T(y)∥<ε for every y∈V. Let {f1,…,fk}⊆Hv0(U) be such that V={u∈Hv0(U)*:|u(fi)|<1,1≤i≤k}. There exists a U-bounded subset K such that v(x)|fi(x)|≤1 for each x∈U∖K. This yields that {v(x)δx:x∈U∖K}∈V, and consequently, v(x)∥f(x)∥=∥T(v(x)δx)∥≤ε for every x∈U∖K.



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