# Fredholm Weighted Composition Operators on Weighted Banach Spaces of Analytic Functions of TypeH∞

Authors: M. Carmen Gómez-Collado; David Jornet
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We study Fredholm (weighted) composition operators between general weighted Banach spaces of analytic functions on the open unit disc with sup-norms.


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## Body

## 1. Introduction

LetX,Y be Banach spaces (infinite dimensional spaces). We denote by B(X,Y) the space of bounded linear operators from X to Y. A bounded linear operator T:X→Y is said to beFredholm if the spaces KerT and CoKerT=Y/ImT are finite dimensional. Every Fredholm operator has closed range. It is known that T is Fredholm if and only if T is invertible modulo the compact operators and if and only if its dual map is Fredholm. For background on Fredholm operators we refer to [1].Letφ and ψ be analytic functions on the open unit disk D of the complex plane C such that φ(D)⊆D. These maps define, on the space H(D) of analytic functions on D, the so-calledweighted composition operatorWφ,ψ by Wφ,ψ(f)=ψ(f∘φ). It combines the classical composition operator Cφ:f↦f∘φ with the pointwise multiplication operator Mψ:f↦ψ·f. These operators have been studied by different authors on various types of function spaces. For more information about composition operators, we refer the reader to the excellent monographs of Cowen and MacCluer [2] and Shapiro [3]. In his thesis Schwartz [4] showed that a composition operator Cφ on the Hardy space H2(D) is invertible if and only if φ is a conformal automorphism of the unit disk. Cima et al. [5] proved that only the invertible operators on H2(D) are Fredholm. Various authors have considered the composition operator Cφ on several Banach spaces of analytic functions and have studied when Cφ is Fredholm. See, for example, [6] for the Hardy and Bergman spaces in D, [7] for various spaces on domain in Cn, [8] for a variety of Hilbert spaces of analytic functions on domains in Cn, [9] for the space H∞(BE) of bounded analytic functions on the open unit ball of a complex Banach space E, and [10] for Banach spaces of analytic functions on D satisfying certain conditions. Recently, Zhao has given a characterization of bounded Fredholm weighted composition operator on Dirichlet spaces [11], on Hardy spaces [12], and on a class of weighted Hardy spaces [13].In this paper we consider the weighted composition operator defined on the weighted Banach spaces of holomorphic functionsHv∞ and on the smaller spaces Hv0 (see Section 2 for the definition). In this framework, Contreras and Hernández-Díaz [14] and Montes-Rodríguez [15], continuing work by [16], characterized the boundedness and compactness of Wφ,ψ between weighted Banach spaces of analytic functions Hv∞ and Hv0. We are interested in finding a characterization of Fredholm weighted composition operators in terms of properties of the symbol φ and of the multiplier function ψ.Our paper is motivated mainly by the works [10, 17]. In the first one, Bonet et al. characterized when Mψ is a Fredholm operator if it is considered between Hv∞ or Hv0. Recently, Galindo and Lindström considered in [10] composition operators on a class of Banach spaces of analytic functions defined on D and E, satisfying certain conditions, and they proved that the composition operator Cφ:E→E associated with the analytic self-map φ:D→D is Fredholm if and only if φ is a biholomorphic map and if and only if Cφ is invertible. We will present a characterization of Fredholm composition operators and Fredholm weighted composition operators when they are considered between Hv∞ or Hv0, for a general typical weight function v (see Theorem 10). Similar results to the ones presented here for more concrete weights are already established in Section 3 of [18]. Although some of the techniques are known and have already been used in the mentioned paper, we present here proofs obtained independently in a more general setting with some applications. Some particular cases of Banach spaces E considered in [10, 18] are the standard weighted Bergman spaces Aαp with α>-1 and p≥1, the little Bloch space, and the weighted Banach spaces Hvp0, 0<p<∞, where vp(z)=(1-|z|2)p. For any weight, Hv0 does not satisfy in general condition (C1) of [10] or conditions (C1) and C1′ of [18]. Hence, we complement recent work by Contreras, Galindo, Hernández-Daz, Hyvärinen, Lindström, Nieminen, and Saukko, among others.


## 2. Notation and Preliminaries

LetD be the open unit disk in the complex plane C and let us denote by H(D) the set of analytic functions from D into C. A weight function v:D→R+ is a radial bounded continuous function that is nonincreasing with respect to |z| such that lim|z|→1-v(z)=0. In the literature these types of weight functions are calledtypical weight functions. The weighted Banach spaces of analytic functions are defined as follows: (1)Hv∞=Hv∞D≔f∈HD:fv=supz∈D⁡vzfz<∞,Hv0=Hv0D≔f∈Hv∞D:limz→1⁡vzfz=0,endowed with the norm ·v. The function v≡1 is not a weight function according to our assumptions. In this case Hv∞=H∞ and Hv0={0}.Many results on weighted spaces of analytic functions are formulated in terms of the so-calledassociated weight function which is defined by the formula(2)v~z≔1sup⁡fz:f∈Hv∞,fv≤1=1δzHv∞∗,z∈D,where δz denotes the point evaluation of z. By [19], we know that v~ is also a weight function; we have that v~≥v>0, and for each z∈D we can find fz∈Hv∞, fzv≤1, such that |fz(z)|=1/v~(z). It is well known that Hv~∞ is isometrically isomorphic to Hv∞ and (Hv0)∗∗ is isometric to Hv∞, where the inclusion map is the canonical injection from a Banach space into its bidual. We refer to [19–21] for more information about these spaces.Forφ,ψ∈H(D) with φ(D)⊆D and E=Hv∞ or Hv0, we consider the composition operator (3)Cφ:E⟶Ef⟼Cφf=f∘φ,and the weighted composition operator is defined as follows: (4)Wφ,ψ:E⟶Ef⟼ψf∘φ,provided it is well defined. We always assume that given a weighted composition operator Wφ,ψ there exists a point z∈D such that ψ(z)≠0.In case the composition operatorCφ is bounded and ψ is a bounded analytic function on the unit disk, ψ∈H∞, the multiplication operator Mψ:Hv∞→Hv∞, defined as Mψf=ψ·f, is bounded ([17, Proposition  2.1]), and the weighted composition operator Wφ,ψ:Hv∞→Hv∞ is also bounded and can be decomposed as Wφ,ψ=MψCφ, where Mψ:Hv∞→Hv∞ and Cφ:Hv∞→Hv∞.By Schwarz Lemma, ifφ(0)=0, then |φ(z)|≤|z| for every z∈D and therefore Cφ is bounded on every space Hv∞. If v satisfies the Lusky condition L1infn⁡v~1-2-n-1v~1-2-n>0,then Theorem  2.3 in [16] ensures that Cφ is bounded on Hv∞ and Hv0. For instance, the standard weights vp(z)=(1-|z|)p,p>0, are weight functions which have L1.The point evaluations play an important role in our results. Our first two lemmas are known. We give a short proof for the reader’s convenience.Lemma 1.
For everyz∈D, the point evaluation δz:Hv∞→C, f↦δz(f)=f(z), is continuous and δz(Hv0)∗=δz(Hv∞)∗.Proof.
Iff∈Hv∞, fix z∈D: (5)fz=1vzvzfz≤1vzsupξ∈D⁡vξfξ≤1vzfv.Then δz∈(Hv∞)∗.
For allu∈(Hv∞)∗, it is clear that u(Hv0)∗≤u(Hv∞)∗. On the other hand, (6)δzHv∞∗=sup⁡fz:fv≤1,f∈Hv∞.Under the hypothesis that v is a typical weight we know that the closed unit ball of Hv0 is dense in the closed unit ball of Hv∞ for the compact open topology (see [20]). Then if f∈Hv∞, fv≤1, there exists (gi)i∈N,gi∈Hv0,giv≤1 such that gi→f in the compact open topology c0. Then gi(z)→f(z). Hence δz(gi)→δz(f). Since |δz(gi)|≤δz(Hv0)∗, we obtain (7)δzHv∞∗≤δzHv0∗.Lemma 2.
Ifv is a weight function, then (i)
lim|z|→1δz(Hv∞)∗=lim|z|→1δz(Hv0)∗=∞,(ii)
limz→1(δzf/δz(Hv∞)∗)=0 for all f∈Hv0.Proof.
(i) By [16, Proposition 1.1], we have limr→1v~(r)=0. From formula (2) and Lemma 1 we obtain the conclusion.
(ii) LetP be a polynomial; then (8)limz→1⁡δzPδzHv∞∗≤sup⁡Pz:z∈D¯limz→1⁡1δzHv∞∗=0.Since {δz/δz(Hv∞)∗:z∈D} is equicontinuous in (Hv0)∗ and the polynomials are dense in Hv0, we have limz→1(δzf/δz(Hv∞)∗)=0 for all f∈Hv0.ForX,Y Banach spaces (infinite dimensional spaces) a bounded linear operator T:X→Y is said to beFredholm if the spaces KerT and CoKerT=Y/ImT are finite dimensional. Note that a consequence of the Fredholmness is that the condition codim(ImT)=dimY/ImT<∞ implies that the range of T is closed. It is known that T:X→Y is Fredholm if and only if T∗:Y∗→X∗ is Fredholm and if and only if there are S∈B(Y,X),K1∈K(X), and K2∈K(Y) such that ST=IX+K1,  TS=IY+K2, where K(X) denote the set of compact operators on X. As a consequence, invertible operators are Fredholm and compact operators cannot be Fredholm. We refer to [1, Chapter III] for the proofs of these results.


## 3. Fredholm Weighted Composition Operators

Recall that the weighted composition operatorWφ,ψ:Hv∞→Hv∞ is bounded if and only supz∈D(ψzvz/v~φz) is finite ([14, Proposition  3.1]) and the operator Wφ,ψ:Hv0→Hv0 is bounded if and only if ψ∈Hv0 and supz∈D(ψzvz/v~φz) is finite ([14, Proposition  3.2]).Lemma 3.
Letψ∈Hv0 and let φ∈H(D) satisfy φ(D)⊆D. The operator Wφ,ψ:Hv∞→Hv∞ is Fredholm if and only if Wφ,ψ:Hv0→Hv0 is Fredholm.Proof.
It is a consequence of the fact that(Hv0)∗∗=Hv∞ and that Wφ,ψ:Hv∞→Hv∞ coincides with the biadjoint map of Wφ,ψ:Hv0→Hv0 whenever both operators are well defined.Lemma 4.
Letφ∈H(D), φ(D)⊆D, and ψ∈Hv∞. If Wφ,ψ:Hv∞→Hv∞ is Fredholm, then there can be at most finitely many points at which ψ vanishes.Proof.
We first prove thatδa∈(Im(Wφ,ψ))⊥ for every zero a of ψ. To do this suppose that ψ(a)=0 and g∈Im(Wφ,ψ). Then there exists f∈Hv∞ such that 〈δa,g〉=ψ(a)(f∘φ)(a)=0. Hence δa∈(Im(Wφ,ψ))⊥.
Now, suppose that there exists an infinite sequence{zi}i=1∞ such that ψ(zi)=0. The sequence {δzi}i=1∞ is linearly independent in (Hv∞)∗ (see [17] pp. 178) and {δzi}i=1∞⊆(Im(Wφ,ψ))⊥. Then dim(Im(Wφ,ψ))⊥=∞.
On the other hand, sinceWφ,ψ is Fredholm, Wφ,ψ∗ is also Fredholm; hence dim(Im(Wφ,ψ)⊥)=dim(Ker(Wφ,ψ∗))<∞, which is a contradiction.Proposition 5.
Letφ:D→D be analytic function and ψ∈Hv∞. If Wφ,ψ:Hv∞→Hv∞ is Fredholm, then φ is injective.Proof.
Our proof starts with the observation that for every point inD there is a neighborhood where φ is injective. Otherwise, there exists x∈D and there are disjoint infinite sequences {um}m and {vm}m⊆D such that limmum=limmvm=x and φ(um)=φ(vm). By Lemma 4, ψ has only a finite number of zeros in D; then we can assume ψ(um)ψ(vm)≠0. For every m∈N, we may define ϑm≔δum/ψ(um)-δvm/ψ(vm)∈(Hv∞)∗. Then {ϑm}m⊆KerWφ,ψ∗. Indeed, (9)Wφ,ψ∗ϑmf=δumψum-δvmψvm,Wφ,ψf=1ψumψumf∘φum-1ψvmψvmf∘φvm=fφum-fφvm=0.Moreover, {ϑm}m is an infinite linearly independent sequence in (Hv∞)∗. Indeed, if we suppose (10)αi1ϑi1+⋯+αikϑik=0inHv∞∗,then for all polynomial P we obtain (11)αi1Pui1ψui1-Pvi1ψvi1+⋯+αikPuikψuik-Pvikψvik=0.For every j∈{1,2,…,k} it is sufficient to take a polynomial Pj such that (12)Pjuim=Pjvim=0∀m∈1,2,…,k∖jPjuij=1,Pjvij=0to conclude αij=0.
By assuming thatWφ,ψ is Fredholm, then Wφ,ψ∗ is Fredholm and dim(KerWφ,ψ∗) is finite. This fact contradicts the fact that {ϑm}m is an infinite linearly independent sequence in KerWφ,ψ∗.
Note thatφ cannot be constant. We are now in position to show that φ is injective. Assume φ is not injective. Then there are x,y∈D, x≠y, such that φ(x)=φ(y)=:ω. By our observation above there are Bx=B(x,r1) and By=B(y,r2), Bx∩By=∅, such that φ|Bx and φ|By are injective. Then φ(Bx) and φ(By) are open in D. Hence φ(Bx)∩φ(By) is open in D and it is an open neighborhood of ω. Therefore (13)∃um⊆Bx,∃vm⊆Bysuchthatφum=φvm.As in the first part of the proof, we can assume that ψ(um)ψ(vm)≠0. We define {ϑm} as above. It is an infinite linearly independent sequence in KerWφ,ψ∗. A contradiction since Wφ,ψ∗ is Fredholm.Remark 6.
Forφ∈H(D) with φ(D)⊂r0D (0<r0<1) the operator Wφ,ψ:Hv∞→Hv∞ is compact for every ψ∈Hv∞. Then Wφ,ψ is not Fredholm. In fact, the associated weight v~ is also radial and decreasing. Hence v~(φ(z))≥v~(r0) for all z∈D and (14)supz∈D⁡ψzvzv~φz≤1v~r0supz∈D⁡ψzvz<∞.From [14, Proposition  3.1], Wφ,ψ is bounded. Since φ(D)⊂r0D, with 0<r0<1, by [16, Corollary  3.2] we obtain that the composition operator Cφ is compact and Cφ(Hv∞)⊆H∞. Then the operator Wφ,ψ can be written as Wφ,ψ=Mψ~Cφ, where Mψ~:H∞→Hv∞, Mψ~(f)=ψf. Since Cφ∈K(Hv∞,H∞) and Mψ~∈B(H∞,Hv∞), we deduce that Wφ,ψ is compact.Proposition 7.
Letφ:D→D be an analytic function and ψ∈Hv0. If Wφ,ψ:Hv∞→Hv∞ or Wφ,ψ:Hv0→Hv0 is Fredholm, then φ is an automorphism.Proof.
By Lemma3, it is enough to consider only the case Wφ,ψ:Hv0→Hv0. If Wφ,ψ is Fredholm then Wφ,ψ∗ is also a Fredholm operator. From [1, Chapter III, Theorem  13] there are a bounded operator S∈B((Hv0)∗,(Hv0)∗) and a compact operator K∈K((Hv0)∗), such that SWφ,ψ∗=I+K. For z∈D, we consider νz=δz/δz(Hv∞)∗. By Lemma 2, νz weakly converges to 0 in (Hv0)∗ as |z| goes to 1. Since K is a compact operator, Kνz→0 as |z|→1. Hence there is r, 0<r<1, such that Kνz<1/2 for all z∈D with r<|z|<1. Since(15)Wφ,ψ∗δz=ψzδφz,we obtain that (16)1-Kνz=νz-Kνz≤νz+Kνz=SWφ,ψ∗νz≤Sψzδφzδz.For z∈D with r<|z|<1, we have(17)Sψzδφzδz≥1-Kνz≥1-12=12.Then for z∈D with r<|z|<1, (18)12δφz≤Sψzδz.From Lemma 2, for f=ψ, we know that limz→1(ψz/δz)=0. Hence, by (2), (19)limz→1⁡v~φz=limz→1⁡1δφz=0.It follows that lim|z|→1|φ(z)|=1 and then φ is an inner function.
By [2, Corollary  3.28] it suffices to prove that φ is univalent to show that φ is an automorphism. Since Wφ,ψ:Hv0→Hv0 is Fredholm, from Lemma 3 we obtain that Wφ,ψ:Hv∞→Hv∞ is Fredholm. By Proposition 5, we get that φ is injective and the proof is complete.Lemma 8.
Ifφ is an automorphism on D and Cφ:Hv0→Hv0 is bounded, then Cφ-1:Hv0→Hv0 is bounded and Cφ-1=Cφ-1.Proof.
Sinceφ is an automorphism, it is known that there are θ∈R and a∈D such that φ(z)=eiθφa(z), where φa(z)=(a-z)/(1-a¯z). Then φ-1(z)=φa(e-iθz). The Möbius transformation φa satisfies that φa-1=φa. Now applying that v~ is radial, we obtain (20)supω∈D⁡v~ωv~φ-1ω=supz∈D⁡v~φzv~z=supz∈D⁡v~φazv~z=supz∈D⁡v~zv~φaz=supz∈D⁡v~zv~φz.As Cφ:Hv0→Hv0 is bounded, by [16, Proposition  2.1] we obtain that Cφ-1:Hv0→Hv0 is bounded. An easy computation shows that Cφ-1=Cφ-1.We observe that Lemma8 is also true if we replace Hv0 by Hv∞.Proposition 9.
Letφ:D→D be an analytic function and ψ∈Hv0. If Cφ and Wφ,ψ are bounded on Hv0 or Hv∞ and Wφ,ψ is also Fredholm, then Cφ and Mψ are Fredholm.Proof.
By Lemma3 we only need to consider the case E=Hv0. Proposition 7 implies that φ is an automorphism. Hence by the lemma above Cφ-1=Cφ-1:Hv0→Hv0 is bounded and is Fredholm. Thus the operator Mψ=Cφ-1Wφ,ψ:Hv0→Hv0 is well defined, bounded, and Fredholm.We obtain a characterization of Fredholm weighted composition operators by applying the results above.Theorem 10.
Letφ:D→D be an analytic function and ψ∈Hv0. If Cφ is bounded on Hv0 or Hv∞, then the following assertions are equivalent:(1)
Wφ,ψ:Hv0→Hv0 is Fredholm.(2)
Wφ,ψ:Hv∞→Hv∞ is Fredholm.(3)
Mψ:Hv0→Hv0 and Cφ:Hv0→Hv0 are Fredholm.(4)
Mψ:Hv∞→Hv∞ and Cφ:Hv∞→Hv∞ are Fredholm.(5)
ψ∈H∞, ∃ɛ>0:|ψ(z)|≥ɛ∀1>|z|≥1-ɛ, and φ is an automorphism.Proof.
We start with the observation thatCφ is bounded on Hv0 if and only if it is bounded on Hv∞ ([16, Proposition  2.1]). (1)⇔(2) It is Lemma 3. (1)⇒(3) and (2)⇒(4) are exactly Proposition 9. (3)⇒(1) and (4)⇒(2) follow by applying [1, Chapter III, Theorem  5]. (3)⇒(5) The condition on ψ follows by [17, Proposition  2.4]. That φ is an automorphism is deduced by Proposition 7 taking ψ:D→C, ψ(z)=1. (5)⇒(3) follows from [17, Proposition  2.4], Lemma 8, and [1, Chapter III, Theorem  13].Our next result covers the case of composition operators onHvp0, with vp(z)=(1-|z|2)p and 0<p<∞ considered in [10].Corollary 11.
Letφ:D→D be an analytic function. If Cφ is bounded on Hv0 or Hv∞, then the following assertions are equivalent:(1)
Cφ:Hv0→Hv0 is Fredholm.(2)
Cφ:Hv∞→Hv∞ is Fredholm.(3)
φ is an automorphism.(4)
Cφ:Hv0→Hv0 is invertible.(5)
Cφ:Hv∞→Hv∞ is invertible.Proof.
(
1
)
⇔
(
2
)
⇔
(
3
) is a consequence of Theorem 10 taking ψ:D→C, ψ(z)=1. (3)⇒(4) and (3)⇒(5) follow by Lemma 8. (4)⇒(1) and (5)⇒(2) are a consequence of every invertible operator being Fredholm.


## 4. Application to Composition Operators on Weighted Bloch Spaces

In this section, weighted composition operators onHv∞ are used to characterize Fredholm composition operators on weighted Bloch type spaces. These spaces are defined as follows: (21)Bv∞=f∈HD:supz∈D⁡vzf′z<∞,Bv0=f∈Bv∞:limz→1⁡vzf′z=0.Bv∞ and Bv0 become Banach spaces under the norm (22)fv=f0+supz∈D⁡vzf′z.For instance, if v(z)=(1-|z|2)α, where α>0, then we have theBloch type spaces. In particular, for α=1 the spaces Bv∞ and Bv0 are the usualBloch space and thelittle Bloch space.We also consider the Banach spaces defined by(23)B~v∞=f∈HD:f0=0,fv=supz∈D⁡vzf′z<∞,B~v0=f∈B~v∞:limz→1⁡vzf′z=0.These spaces are also calledweighted Bloch spaces.The mapS:B~v∞→Hv∞ given by Sf=f′ and S-1:Hv∞→B~v∞ defined by S-1f=F, where F′=f, are isometric isomorphisms. The same is true if we replace the symbol ∞ by 0.Our next result forv(z)=1-|z|2 can be seen in [10].Corollary 12.
Letφ:D→D be an analytic function such that Cφ:Bv0→Bv0 is bounded. The following assertions are equivalent:(1)
Cφ:Bv0→Bv0 is Fredholm.(2)
φ is automorphism.(3)
Cφ:Bv0→Bv0 is invertible.Proof.
(
1
)
⇔
(
2
) We take α=φ(0) and consider ϕ=φα∘φ, where φα is the Möbius transformation φα(z)=(α-z)/(1-α¯z). Then Cφα:Bv0→Bv0 is an isomorphism ([14, Lemma  6.2]). Hence Cφ:Bv0→Bv0 is Fredholm if and only if Cϕ=Cφ∘Cφα:Bv0→Bv0 is Fredholm. Since ϕ(0)=0, by Schwarz Lemma we have that Cϕ:Hv0→Hv0 is bounded, and Cϕ:Bv0→Bv0 is bounded if and only if Cϕ:B~v0→B~v0 is bounded. Moreover, since Bv0=B~v0⊕C, Cϕ:Bv0→Bv0 is Fredholm if and only if Cϕ:B~v0→B~v0 is Fredholm. This is equivalent to Wϕ,ϕ′:Hv0→Hv0 being Fredholm, because Cϕ=S-1∘Wϕ,ϕ′∘S, where S is the isomorphism given by S:B~v0→Hv0, Sf=f′. Finally, every automorphism on D is of the form μ(z)=eiθ(a-z/1-a¯z) with θ∈R and a∈D, and applying Theorem 10, we obtain that Wϕ,ϕ′ is Fredholm if and only if ϕ is an automorphism, which is equivalent to φ being an automorphism. (2)⇒(3) follows by a similar argument to the proof of Lemma 8, applying in this case [22, Proposition  3]. (3)⇒(1) is a consequence of [1, Chapter III, Theorem  13].



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