# Isomorphisms on Weighed Banach Spaces of Harmonic and Holomorphic Functions

Authors: Enrique Jordá; Ana María Zarco
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

For an arbitrary open subsetU⊂ℝd or U⊆ℂd and a continuous function v:U→]0,∞[ we show that the space hv0(U) of weighed harmonic functions is almost isometric to a (closed) subspace of c0, thus extending a theorem due to Bonet and Wolf for spaces of holomorphic functions Hv0(U) on open sets U⊂ℂd. Inspired by recent work of Boyd and Rueda, we characterize in terms of the extremal points of the dual of hv0(U) when hv0(U) is isometric to a subspace of c0. Some geometric conditions on an open set U⊆ℂd and convexity conditions on a weight v on U are given to ensure that neither Hv0(U) nor hv0(U) are rotund.


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## Body

## 1. Introduction, Notation, and Preliminaries

LetU⊆ℝd or U⊆ℂd be a nonempty open set. A weight on U is a function v:U→ℝ which is strictly positive, continuous, and bounded. Let h(U) denote the space of complex valued harmonic functions on U. For a weightv the weighed Banach spaces of harmonic functions with weightv are defined by
(1)hv(U):={f∈h(U):∥f∥v:=supz∈Gv(z)|f(z)|<∞},hv0(U):={f∈h(U):vfvanishesatinfinityonU}.A functiong:U→ℂ is said to vanish at infinity on U if for each ε>0 there exists a compact subset K⊂U such that |g(x)|≤ε for each x∈U∖K.IfU⊆ℂd then the weighed Banach spaces of holomorphic functions Hv(U) and Hv0(U) are defined in the same way. The unit disc and the unit circle in ℂ are denoted by 𝔻 and Γ, respectively. U is said to be balanced if λz∈U for each z∈U and each λ∈𝔻. A weight v on a balanced open subset U⊆ℂn is said to be radial whenever v(λz)=v(z) for each λ∈Γ.Weighed Banach spaces of holomorphic functions with weighed supremum norms and composition operators between them have been studied by Bierstedt et al., Bonet et al., Contreras and Hernández-Díaz, García et al., Montes Rodríguez, and others; see [1–8] and the references therein.Spaces of harmonic functions have been investigated by Shields and Williams [9], in connection with the growth of the harmonic conjugate of a function. In [10], they proved results of duality for weighed spaces of harmonic functions on the open unit disk. Lusky also considers weighed spaces of harmonic functions in [11, 12], where the isomorphism classes in the case of radial weights on the disk are determined. In [13] the authors have studied associated weights in weighed Banach spaces of harmonic functions and composition operators with holomorphic symbol between weighed spaces of pluriharmonic functions.In [14–16] Boyd and Rueda investigate the geometry of hv0(U) and Hv0(U) where U is an open set of the complex plane. Moreover they examined the extreme points of hv0(U)* and of Hv0(U)* for U⊆ℂd.In this work we prove in the second section that, for any weightv on an open subset U of ℝd, the weighed Banach space hv0(U) of harmonic functions with vf vanishing at infinity on U is always isomorphic to a subspace of c0 thus extending a result due to Bonet and Wolf [17] for holomorphic functions. Further, given 0<ε<1 one can get a linear mapping T:hv0(U)→c0 such that (1-ε)∥f∥v≤∥T(f)∥≤∥f∥v for each f∈hv0(U); that is, hv0(U) is almost isometric to a subspace of c0. We show that for radial weights in balanced open subsets the isomorphism cannot be isometric, generalizing [14, Theorem 16, Corollary 17] to spaces of harmonic functions defined on not necessarily bounded sets. We also show in this section that if the unit ball of hv0(U)* has a big quantity of extremal points, then hv0(U) is not isometric to any subspace of c0.In the third section we mainly focus ourselves on weighed spaces of holomorphic and harmonic functions defined on balanced subsetsU⊆ℂd and with radial weights. We study the extremal points of the unit sphere of Hv0(U)* in the way that Boyd and Rueda initiated in [14, 15]. More precisely, we give conditions on U and v under which one can find in the unit sphere of Hv0(U)* a big quantity of extremal points. We show that these conditions imply that Hv0(U) is not rotund.Our notation for functional analysis and Banach space theory is standard. We refer to [18]. For a Banach space X we denote by BX and SX its unit ball and its unit sphere, respectively. A (real or complex) Banach space X is said to be rotund whenever ∥(x+y)/2∥<1 for ∥x∥=∥y∥=1 and x≠y. A vector x in the unit sphere SX of a Banach space X is called extremal if there are not y,z∈SX such that y≠z and x=(1/2)(y+z), and it is called exposed if there is x*∈X* such that ∥x*∥=1, x*(x)=1, and x*(y)≠1 for each y∈SX∖{x}. If X is complex this last condition is equivalent to x*(x)=1 and Rex*(y)<1 for y∈BX∖{x}, where Rez denotes the real part of a complex number z. If x is exposed, then x is also extremal. If X is a dual Banach space and the functional exposing x can be taken in the predual, then x is called weak*exposed. For x,y∈ℝd (or ℂd) we denote by 〈x,y〉 its canonical scalar product. The Euclidean norm is denoted by ∥·∥. When we consider other norms on a finite dimensional space they are denoted by |·|, making an abuse of notation because the modulus of a complex number is denoted also by |·|.For a Hausdorff locally compact topological spaceQ we denote by Q^={Q∪∞} its Alexandroff compactification. The boundary of a subset U of a topological space is denoted by ∂U. If U⊆ℝd is an open subset then the space hv0(U) is formed by the harmonic functions on U such that vf can be extended continuously to U^ by (vf)(∞)=0. For an open set U⊆ℝd, the isometry between hv0(U) and a subspace of C(U^) by means of f↦vf permits to use the same argument as in [14, Proposition 1] to show that the extreme points of hv0(U)* are contained in the set
(2){λv(z)δz:λ∈Γ,z∈U},
where δz denotes the evaluation at z. The following definitions are due to Boyd and Rueda [14–16]. The harmonic v-boundary is defined as follows:
(3)bv(U)≔{z∈U:v(z)δzisextremalintheunitsphereofhv0(U)*00000000isextremalintheunitsphereofhv0(U)*}.
The set of harmonic v-peak points is defined as
(4)pv(U):={z∈U:v(z)δzisweak*exposedintheunitsphereofhv0(U)*00000000isweak*exposedintheunitsphereofhv0(U)*}.Remark 1.
The same argument as in [15, Theorem 6] shows that to check that a functional of the form v(x0)δx0 is weak* exposed it is enough to show that there exists f in the unit ball of hv0(U) such that v(x0)f(x0)=1, v(x)Ref(x)<1 for each x∈U∖{x0}.IfU⊆ℂd then the holomorphic v-boundary and the set of holomorphic v-peak points are defined analogously, and they are denoted by Bv(U) and Pv(U), respectively. Since Hv0(U)⊂hv0(U), in this case, we have
(5)Pv(U)⊆Bv(U)⊆U,
and also
(6)Pv(U)⊆pv(U)⊆bv(U)⊆U.We finish this section with this lemma also due to Boyd and Rueda.Lemma 2 (see [14, 15]).
LetU⊆ℝd(U⊆ℂd) be a nonempty open set, and let v be a weight which vanishes at infinity on U. (a)
For anyλ∈Γ and z∈U the functional λv(z)δz is extremal in the unit sphere of hv0(U)*  (is extremal in the unit sphere of Hv0(U)*) if and only if z∈bv(U)(z∈Bv(U)).(b)
IfU⊆ℂd is balanced, v is radial and λ∈Γ then λz∈bv(U) for each z∈bv(U)(λz∈Bv(U) for each z∈Bv(U)).


## 2. Isomorphisms onhv0(U)

We extend below the main theorem of [17] to weighed spaces of harmonic functions. The proof is analogous to the proof given by Bonet and Wolf. One only has to take in account that there are Cauchy type inequalities which are valid for the derivatives of harmonic functions [19, 2.4]. We include it for the sake of completeness. The argument was inspired by one used by Kalton and Werner [20, Corollary 4.9] to show that the little Bloch space B0 embeds almost isometrically in c0. Lusky showed in [21] that whenever v is a radial weight on 𝔻, then hv0(𝔻) is isomorphic to a subspace of c0.Theorem 3.
LetU⊂ℝd be a nonempty open set, let v be a weight on U, and let ε>0. There exists an isomorphism T:hv0(U)→c0 such that (1-ε)∥f∥v≤∥T(f)∥≤∥f∥v for each f∈hv0(U).Proof.
Let(Kj)j be a fundamental sequence of compact subsets of U, and let ε>0. We claim that there exists a sequence of pairwise disjoint finite subsets (Fj)j such that Fj⊂Aj:=Kj∖Kj-1∘ and satisfying that for each f∈hv0 with ∥f∥v=1(7)supx∈Ajv(x)|f(x)|≤ε+supy∈Fjv(y)|f(y)|.
This implies that if we denoteF:=∪jFj, then we have
(8)supx∈Fv(x)|f(x)|≤∥f∥v≤(1+ε)supx∈Fv(x)|f(x)|,
and consequently, if we write F as a sequence (wj)j, then the linear mapping T:hv0(U)→c0, f↦(v(wj)f(wj))j is injective; it has closed range and norm no greater than one, and its inverse T-1 defined on a closed subspace of c0 has norm less than or equal to 1+ε. Since this is true for each ε>0, this is equivalent to our statement.
We proceed to show the claim. For eachj∈ℕ we define
(9)Mj:=supx∈Kjv(x),aj:=min(12d(Kj,ℝN∖Kj+1∘),1),
and we select x(j)∈Kj such that v(x(j))=minx∈Kjv(x). For each f∈hv0(U) with ∥f∥v=1 and each j∈ℕ, we have
(10)supx∈Kj|f(x)|≤1v(x(j)).
If x∈Kj and ζ∈D(x,aj) (here D(x,aj) denotes the closed ball with center x and radius aj), then from the definition of aj we deduce that ζ∈Kj+1, and hence, together with the previous inequality,
(11)|f(ζ)|≤1v(x(j+1)).
By Cauchy’s inequalities for harmonic functions given in [19, 2.4], there exists a positive number C (independent of x) such that, for each ζ∈D(x,aj), the ith derivative of f satisfies
(12)|Dif(ζ)|≤Cajv(x(j+1)),i=1,2,…,d.
By compactness we can get a finite subset Fj of Aj such that
(13)Aj⊂⋃x∈Fj‍{x′∈U:|x′-x|<δj,|v(x′)-v(x)|<δj},
where δj has been chosen satisfying δj<aj and
(14)(1v(x(j))+Mj+1dCajv(x(j+1)))δj<ε.
Consequently, for each x∈Aj, there exists w∈Fj with |w-x|<δj and |v(w)-v(x)|<δj. On the other hand, D(x,δj)⊂D(x,aj)⊂Kj+1 for each x=(x1,x2,…,xd)∈Aj. We write w=(w1,w2,…,wd) to get
(15)|f(x)|=|f(x1,…,xd)|≤|f(x1,…,xd)-f(w1,x2,…,xd)|+|f(w1,x2,…,xd)-f(w1,w2,…,xd)|+⋯+|f(w1,w2,…,wd)|.
We apply the mean value theorem to each difference and (12) to get
(16)|f(x)|≤Cajv(x(j+1))d|x-w|+|f(w)|≤dδjCajv(x(j+1))+|f(w)|.
Now, sincew∈Kj+1,
(17)v(x)|f(x)|≤|v(x)-v(w)||f(x)|+v(w)|f(x)|≤δj1v(x(j))+v(w)(dδjCajv(x(j+1))+|f(w)|)≤δjv(x(j))+Mj+1dδjCajv(x(j+1))+v(w)|f(w)|≤ε+v(w)|f(w)|≤ε+maxw∈Fjv(w)|f(w)|.
Therefore,(18)supx∈Ajv(x)|f(x)|≤ε+maxw∈Fjv(w)|f(w)|.Again as in [17], since each infinite dimensional subspace of c0 contains a complemented subspace isomorphic to c0, the following result follows.Corollary 4.
There are not infinite dimensional subspaces ofhv0(U) which are reflexive.In order to get examples where the isomorphism above cannot be an isometry, we will consider a condition which is inspired by the results of Boyd and Rueda [14, Theorem 16, Corollary 17] for weighed spaces defined on bounded open sets of ℂd. First we observe that for any weight v for which hv0(U) contains the polynomials it happens that the functionals {δz:z∈U}⊆hv0(U)* are linearly independent. If U is bounded this is the case for any weight v which tends to zero at the boundary. If U=ℝd, then the condition is satisfied when v=g(|x|), with |·| being a norm in ℝd and g:[0,∞[→]0,∞[ being a rapidly decreasing continuous function, that is, with limt→∞tng(t)=0 for any n∈ℕ. First we extend easily [14, Lemma 10] to unbounded domains.Lemma 5.
LetU⊂ℝd be a nonempty open set, and let v be a continuous strictly positive weight which vanishes at infinity on U. If hv0(U) contains the polynomials, then the map μ:U→(hv0(U)*,w*), and z↦v(z)δz is a homeomorphism onto its image.Proof.
Since the polynomials belong tohv0(U), we get that {δz:z∈U}⊂hv0(U)* are linearly independent functionals. We consider the map μ^:U^→(hv0(U)*,w*), z→v(z)δz if z∈U and μ^(∞)=0. The very definition of hv0(U) implies that μ^ is continuous, and it is injective by the linear independence of the evaluations. By the compactness of U^ it follows that μ^ is a homeomorphism, and hence also its restriction μ to U.Proposition 6.
LetU⊂ℝd(U⊂ℂd) be a nonempty open set, and let v be a weight for which hv0(U)(Hv0(U)) contains the polynomials. Assume that bv(U)(Bv(U)) is not discrete. Then hv0(U)(Hv0(U)) cannot be isometric to any subspace of c0.Proof.
We give only the proof for the harmonic case. Denote byℕ^=ℕ∪{∞} the Alexandroff compactification of ℕ. Let T:hv0(U)→c0 be an isometry onto its image. Let F:=T(hv0(U))⊆c0. Since c0 is the subspace of C(ℕ^) formed by the functions which vanish at ∞, the extremal points of the unit sphere of F* are contained in {λδn:λ∈Γ,n∈ℕ} (cf. [22, Lemma V.8.6]). Let Q:F→hv0(U) be the inverse isometry of  T. The transpose linear mapping Qt:hv0(U)*→F* is also an isometry, and it maps extremal points to extremal points. Thus for each z∈bv(U) there exists λz∈Γ and n(z)∈ℕ such that Qt(v(z)δz)=λzδn(z). From the linear independence of {δz:z∈U} in hv0*, we conclude that n(z)≠n(w) for z,w∈U with z≠w, and then bv(U) is countable and {λzδn(z):z∈bv(U)} is discrete for the weak* topology, since it is a sequence in the sphere of F* which is σ(F*,F) convergent to 0. But Qt is a weak*-weak* homeomorphism. Hence we conclude from Lemma 5 that bv(U) is discrete.Now we apply the set of extremal points of the unit ball of a dual Banach space which is never empty and Lemma2 to get the following result.Corollary 7.
IfU⊆ℂd is a balanced open set and v is a radial weight which vanishes at infinity on U, then neither Hv0(U) nor hv0(U) is isometric to a subspace of c0.Although we do not have an example of weightv on U such that bv(U) is a discrete subset of U, we see below that we can reformulate the problem of finding an example of space hv0(U) which can be isometrically embedded in c0 in terms of the possible existence of an open set U⊆ℝd and a weight v such that bv(U) is not discrete. Of course the analogue result is also true for the holomorphic case. In view of Lemma 2 this is not possible for radial weights in balanced domains.Proposition 8.
IfU⊆ℝd is an open set and v is a weight on U such that hv0(U) contains the polynomials and bv(U) is discrete, then hv0(U) embeds isometrically in c0.Proof.
Since the closed unit ballD of hv0(U)* is the weak*closure of the absolutely convex hull of the extremal points of the sphere, Lemma 2(a) implies that D is in fact the closure of the absolutely convex hull of {v(z)δz:z∈U}. Thus this implies that bv(U) cannot be finite, and since it is discrete we have that bv(U)=(xn)n with xn tending to infinity on U when n goes to infinity. We get now that D is the weak* closure of the absolutely convex hull of {v(xn)δxn:n∈ℕ}, and hence, for each f∈hv0(U), it holds:
(19)∥f∥v:=supn∈ℕv(xn)|f(xn)|.
Since xn tends to infinity on U, we have that the linear map
(20)T:hv0(U)⟶c0,f⟼(v(xn)f(xn))n
is an isometry.


## 3. Geometry of Weighed Banach Spaces of Holomorphic and Harmonic Functions and Their Duals

In [15, Theorem 29] Boyd and Rueda showed that if U⊆ℂd is a balanced bounded open set and v is a radial weight vanishing at infinity on U such that Pv(U)=Bv(U) then Hv0(U) is not rotund and then neither hv0(U) is. This condition is trivially satisfied when Pv(U)=U. To the best of our knowledge there are not so far concrete examples of open subsets U⊂ℂd and weights v on U such that Bv(U)∖Pv(U)≠∅. In the concrete examples of spaces Hv0(U) given in [15] where Bv(U) is calculated the equality Bv(U)=Pv(U) is always satisfied. With a similar proof to the one of [15, Theorem 29], we present below a new condition for spaces Hv0(U) with U balanced and v radial depending only on the size of Pv(U) which ensures that Hv0(U) is not rotund.Proposition 9.
LetU⊆ℂd be the open unit ball for a norm |·| in ℂd, and let v=g(|x|) for g:[0,1]→[0,∞[ be a nonincreasing continuous function with g(1)=0. If there exists z0∈∂U such that the set {tz0:0<t<1}⊆Pv(U), then Hv0(U) is not rotund and then neither hv0(U) is.Proof.
Letφ be a linear map on ℂd such that ∥φ2∥v=1. Since vφ2 is continuous on U^ and vanishes at infinity, there exists z1∈U, z1≠0 such that
(21)1=∥φ2∥v=v(z1)φ2(z1)=v(-z1)φ2(-z1).
Since |z0|=1, we apply the theorem of Hahn-Banach to get y∈ℂd such that 〈z0,y〉=1 and |〈z,y〉|≤|z| for each z∈ℂd. Let r0:=|z1|. We define
(22)h:U⟶ℂ,z⟶φ2(〈z,y〉r0z1).
We have
(23)v(r0z0)h(r0z0)=g(r0)φ2(z1)=g(|z1|)φ2(z1)=v(z1)φ2(z1)=1.
Sinceφ is linear we also get
(24)v(-r0z0)h(-r0z0)=g(r0)φ2(-z1)=v(-z1)φ2(-z1)=1.
For z∈U arbitrary, we use that |z1/r0|=1, |〈z,y〉|≤|z|, and g is not increasing to compute
(25)v(z)|h(z)|=g(|z|)|h(z)|≤g(|〈z,y〉|)|h(z)|=v(〈z,y〉r0z1)|φ2(〈z,y〉r0z1)|≤∥φ2∥v=1.
Thus h is in the unit ball of Hv0(U) and
(26)v(r0z0)δr0z0(h)=v(-r0z0)δ(-r0z0)(h)=1.
Since r0z0 is assumed to be a peak point we can find f in the unit ball of Hv0(U) different from h such that v(r0z0)f(r0z0)=1 and v(z)Re(f(z))<1 for each z∈U∖{r0z0}. Hence we conclude from
(27)∥f+h2∥v=v(r0z0)(f(r0z0)+h(r0z0))2=1
that Hv0(U) is not rotund.Letg:[0,1]→[0,∞[ be a function such that g(1)=0, g is decreasing, g is twice differentiable in ]0,1[, and log(1/g(t)) is convex. In [15, Corollary 12] the authors showed that for a weight v defined in the Euclidean unit ball U of ℂd by v(x)=g(∥x∥) the equality Pv(U)=U holds, and then Hv0(U) cannot be rotund by [15, Theorem 29]. Also if U1⊆ℂd and U2⊆ℂk are the Euclidean unit balls and v1 and v2 satisfy this last condition for certain g1 and g2, then v:U1×U2→]0,∞[, v(x,y)=v1(x)v2(y) also satisfies Pv(U)=U (cf. [15, Proposition 26]). We present below new examples of spaces Hv0(U) which cannot be rotund. In the required assumptions, besides considering a wider class of open sets, we remove from the weight the condition of differentiability given by Boyd and Rueda as we did in [13, Theorem 1] to give conditions under which v=v~, v~ being the associated weight introduced by Bierstedt et al. in [23]. The same technique permits us to get concrete radial weights v(x)=g(∥x∥) defined in the Euclidean ball U of ℝd with d odd such that the spaces hv0(U) are not isometric to any subspace of c0. If d=2k, then the assertion is a consequence of Corollary 7 since U is the Euclidean unit ball of ℂk and Hv0(U)⊂hv0(U).Proposition 10.
Letg:[0,1]→[0,∞[ be a continuous decreasing function such that g(1)=0 and log(1/g) is strictly convex in [0,1[.(a)
LetU⊆ℂdbe the unit ball for a norm|·|andv:U→]0,∞[,x↦g(|x|). ThenHv0(U)is not rotund (and then neitherhv0(U)is).(b)
LetU⊆ℝdbe the Euclidean unit ball andv(x)=g(∥x∥)forx∈U. Thenhv0(U)is not isometric to any subspace ofc0.Proof.
Leth(r):=log(1/g(r)), 0≤r<1. Since h is increasing and strictly convex, for each 0≤r0<1, the map
(28)[0,1[∖{r0}⟶]0,∞[,r⟼(h(r)-h(r0))(r-r0)
is increasing. Hence for each 0≤r0<1 there exists α0≥0 which depends on r0 such that
(29)h(r)-h(r0)>α0(r-r0)
for all r∈[0,1[∖{r0}. Now we compute
(30)sup0≤r<1,r≠r0g(r)exp(α0r)=exp(sup0<r<1,r≠r0(-h(r)+α0r))<exp(sup0≤r<1,r≠r0(-α0(r-r0)000000-h(r0)+α0r)sup0≤r<1,r≠r0(-α0)=g(r0)exp(α0r0).
We prove (a). Straszewicz’s theorem [24, Theorem 7.89] together with the Krein-Milman theorem imply that there exists z0∈∂U which is exposed. Let y∈ℂd such that 〈z0,y〉=1 and Re(〈z,y〉)<|z| if z∈U∖{rz0:0≤r<1}. Let 0≤r0<1 be arbitrary, and let α0 be as above. The function f:U→ℂ defined by f0(z):=(exp(α0〈z,y〉))/(g(r0)exp(α0r0)) is holomorphic and v(r0z0)f0(r0z0)=1. It follows immediately from (30) that if 0≤r<1, r≠r0(31)v(rz0)|f0(rz0)|=g(r)exp(α0r)g(r0)exp(α0r0)<1.
Ifz∈U∖{rz0:0≤r<1}, then we use (30) and Re(〈z,y〉)<|z| to get
(32)v(z)|f0(z)|=g(|z|)exp(α0Re(〈z,y〉))g(r0)exp(α0r0)<g(|z|)exp(α0|z|)g(r0)exp(α0r0)≤1.
Hence f0 peaks v(r0z0)δr0z0. We conclude that {rz0:0≤r<1}⊆Pv(U) since r0 is arbitrary. We apply Proposition 9 to get that Hv0(U) is not rotund.
To prove (b) we considerx0∈ℝd with 0<∥x0∥=r0<1 and get α0 again as above. Let A be an orthogonal transformation such that A(x0)=(r0,0,…,0). For x=(x1,…,xd) we take the harmonic function h(x):=exp(α0x1)cos(α0x2) and we define f0(x):=h∘A(x) for x∈U, which is harmonic; see [19, Chapter 1]. A similar argument as in (a) permits us to conclude that v(x0)f0(x0)=1 and v(x)|f(x)|<1 for each x∈U∖{x0}. Hence pv(U)=U, and therefore hv0(U) is not isometrically embedded in c0 by Proposition 6.Remark 11.
(a) An inspection of the proofs of Propositions9 and 10 shows that the statements remain true when U=ℂd and g:[0,∞[→]0,∞[ is a decreasing continuous function which satisfies that log(1/g) is strictly convex and g(t)=o(e-at) when t goes to infinity for each a>0.
(b) For any increasing functionh:[0,1[→[0,∞[ which is strictly convex and satisfies limt→1h(t)=∞ (or h:[0,∞[→]0,∞[ with eat=o(eh(t)) when t goes to infinity for each a>0), the function g(t)=e-h(t) satisfies the hypothesis of Propositions 9 and 10. The examples given in [15, Example 13] of weights in the Euclidean unit ball can be obtained using this general method.



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