# Fredholm Weighted Composition Operator on Weighted Hardy Space

Authors: Liankuo Zhao
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

This paper gives a unified characterization of Fredholm weighted composition operator on a class of weighted Hardy spaces.


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## Body

## 1. Introduction

Letℋ be a Hilbert space of analytic functions on the open unit disk 𝔻. For ψ∈ℋ and analytic self-map φ of 𝔻, the weighted composition operator Cψ,φ on ℋ is defined as
(1)Cψ,φf=ψ(φ∘f),f∈ℋ.
It is natural to ask for which ψ and φ the weighted composition operator Cψ,φ is bounded and compact on ℋ. In the past ten years, this problem on various Hilbert spaces of analytic functions has been studied extensively. It seems there is no unified ways to solve the problem since different spaces have very distinguished properties; see [1–10] for the characterization of bounded and compact weighted composition operators on different function spaces. In [11], the invertibility of weighted composition operator on the classical Hardy space is characterized, as an extension, in [12, 13]; the Fredholmness of weighted composition operator on Dirichlet space and weighted Dirichlet space is characterized, respectively. The results show that the Fredholmness and invertibility of weighted composition operator are closely related to properties of the reproducing kernel functions. Inspired by the characterization of Fredholm composition operator in [6], in this paper, we give a unified characterization of Fredholm weighted composition operator on a class of weighted Hardy space.Now we follow some notations as in [14].ℋ is called a weighted Hardy space if the monomials 1,z,z2,… constitute a complete orthogonal set of nonzero vectors in ℋ. Denote β(n)=∥zn∥; then for f∈ℋ with f(z)=∑n=0∞‍f^(n)zn,
(2)∥f∥2=∑n=0∞‍|f^(n)|2β(n)2.
So we use H2(β) to denote the weighed Hardy space with weighted sequence {β(n)}.H
2
(
β
) is called automorphism invariant if, for any automorphism map φ of 𝔻,
(3)f∘φ∈H2(β),f∈H2(β).
That is, the composition operator Cφ is bounded on H2(β) if φ is an automorphism map of 𝔻. It is well known that many familiar function spaces such as Hardy space, Bergman space, and Dirichlet space are automorphism-invariant weighted Hardy space. In [14, p. 127], an example of nonautomorphism-invariant weighted Hardy space is presented.Our main result reads as follows.Theorem 1.
LetH2(β) be an automorphism-invariant weighted Hardy space such that
(4)∑n=0∞‍n2jβ(n)2=∞
for some nonnegative integer j. If Cψ,φ is a bounded Fredholm operator on H2(β), then there exist constants c, r with c>0, 0<r<1 such that |ψ(z)|≥c for r<|z|<1, and φ is an automorphism of  𝔻.As an application, in the end of this paper, we give a complete characterization of Fredholm and invertible weighed composition operator on𝒟α (-∞<α<∞). Recall that 𝒟α is a class of important weighed Hardy space with β(n)=(n+1)α.


## 2. Fredholm Weighted Composition Operator onH2(β)

Recall thatH2(β) has reproducing kernel function
(5)Kλ(z)=∑n=0∞‍λ¯zβ(n)2,λ,z∈𝔻.
That is, for f∈H2(β),
(6)f(λ)=〈f,Kλ〉.Let(7)Kλ[j]=∑n=j∞‍n!(n-j)!λ¯n-jznβ(n)2,
then Kλ[j] is the jth derivative evaluation kernel function at λ in H2(β) [14], that is, for f∈H2(β),
(8)f(j)(λ)=〈f,Kλ[j]〉.Firstly we give some discussion about weighted composition operatorCψ,φ acting on Kλ[j]. The following lemmas is well known and easy to verify.Lemma 2.
ConsiderCψ,φ*Kλ=ψ(λ)¯Kφ(λ).Lemma 3.
LetKz(i)(λ)=Kλ[i](z)¯.By Fań di Bruno’s formula,(Kz∘φ)(i)(λ) can be expressed as linear combination of
(9)Kz(m)(φ(λ))∏k=1i‍(φ(k)(λ))mk,
where m=∑k=1i‍mk and ∑k=1i‍kmk=i with nonnegative integer mk.Since(10)Cψ,φ*Kλ[j](z)=〈Cψ,φ*Kλ[j],Kz〉=〈Kλ[j],Cψ,φKz〉=(ψKz∘φ)(j)¯(λ),(11)(ψKz∘φ)(j)¯(λ)=∑i=0j‍j!i!(j-i)!ψ(j-i)¯(λ)(Kz∘φ)(i)¯(λ),
it follows from Lemma 3 and (9) that Cψ,φ*Kλ[j] can be expressed as
(12)Cψ,φ*Kλ[j]=∑i=0j‍Fi(ψ,φ)¯(λ)Kφ(λ)[i],
where Fi(ψ,φ) is algebraic combination of ψ(m) and φ(k) with 0≤m,  k≤j.Since(13)Kφ(λ)[m](z)=∑n=i∞‍n!(n-m)!φ(λ)¯(n-m)znβ(n)2,
the degree of nonzero coefficient Kφ(λ)[m](z) is at least m and the coefficient of zk for k≥m is
(14)k!(k-m)!φk-m(λ)¯β(k)2,
so by (12) the coefficient of zn in Cψ,φ*Kλ[j] for 0≤n≤j is
(15)∑i=0n‍Fi(ψ,φ)¯(λ)n!(n-i)!φn-i(λ)¯β(n)2.On the other hand,(16)(ψKz∘φ)(j)¯(λ)=ψ(λ)Kz(φ(λ))¯(j)=∑n=0∞‍ψ(λ)φn(λ)¯(j)znβ(n)2.
So by (10) we have
(17)Cψ,φ*Kλ[j](z)=∑n=0∞‍ψ(λ)φn(λ)¯(j)znβ(n)2.Comparing (12) and (17) with the aid of (15), we obtain
(18)F0(ψ,φ)(λ)=ψ(j)(λ),F1(ψ,φ)(λ)+F0(ψ,φ)(λ)φ(λ)=(ψ(λ)φ(λ))(j),⋮Fi(ψ,φ)(λ)+Fi-1(ψ,φ)(λ)φ(λ)+⋯+F1(ψ,φ)(λ)1(i-1)!φi-1(λ)+F0(ψ,φ)(λ)1i!φi(λ)=1i!(ψ(λ)φi(λ))(j),⋮Fj(ψ,φ)(λ)+Fj-1(ψ,φ)(λ)φ(λ)+⋯+F1(ψ,φ)(λ)1(j-1)!φj-1(λ)+F0(ψ,φ)(λ)1j!φj(λ)=1j!(ψ(λ)φj(λ))(j).By the above reasoning, we obtain the following lemma.Lemma 4.
Consider(19)Cψ,φ*Kλ[j]=∑i=0j‍Fi(ψ,φ)¯(λ)Kφ(λ)[i],
where Fi(ψ,φ) is linear combination of (ψφl)(j)φm with 0≤l, m≤j.Note that in [15, Lemma 2], a different form of Cψ,φ*Kλ[j] is given.Now we give the characterization of Fredholm weighted composition operator onH2(β).Proposition 5.
IfCψ,φ is a bounded Fredholm operator on H2(β), then ψ has only finite zeroes in 𝔻, and φ is univalent.Proof.
Letλ∈𝔻 such that ψ(λ)=0. By Lemma 2,
(20)Cψ,φ*Kλ=0,
which implies that if ψ(λ)=0; then Kλ∈kerCψ,φ*. Since Cψ,φ is a Fredholm operator, dim(kerCψ,φ*) is finite; it follows that ψ has only finite zeroes in 𝔻.
Ifφ(a)=φ(b) for some a,b∈𝔻 with a≠b, then there exists infinitely disjoint subset {an}, {bn} of  𝔻 such that φ(an)=φ(bn). Since ψ has only finite zeroes in 𝔻, we can assume ψ(an)ψ(bn)≠0. By Lemma 2, we have
(21)Cψ,φ*(Kanψ(an)¯-Kbnψ(bn)¯)=0,
which implies that Kan/ψ(an)¯-Kbn/ψ(bn)¯∈ker(Cψ,φ*), a contradiction to the Fredholmness of Cψ,φ*. So φ must be univalent.Theorem 6.
LetH2(β) be a weighted Hardy space of 𝔻 such that
(22)∑n=0∞‍n2jβ(n)2=∞
for some nonnegative integer j. If Cψ,φ is a bounded Fredholm operator on H2(β), then φ is an automorphism of 𝔻.Proof.
Letj be the minimal nonnegative integer such that
(23)∑n=0∞‍n2jβ(n)2=∞.
Then ∥Kλ[j]∥→∞ as |λ|→1-. As in the proof of [14, Theorem 2.17],
(24)Kλ[j]∥Kλ[j]∥weaklyconvergesto0asλ⟶1-.
And for 0≤i<j,  ∥Kλ[i]∥ is uniformly bounded for λ∈𝔻.
Sinceψφi=Cψ,φ(zi), it follows that ψφi∈H(β). So
(25)(ψφi)(j)(λ)=〈ψφi,Kλ[j]〉,lim|λ|→1-(ψφi)(j)(λ)∥Kλ[j]∥=0.
By induction and the fact that|φ(λ)|<1, it follows from Lemma 4 that
(26)lim|λ|→1-Fi(ψ,φ)(λ)∥Kλ[j]∥=0
for 0≤i≤j.
By (19), we have
(27)lim|λ|→1-∥Cψ,φ*Kλ[j]∥∥Kλ[j]∥≤∑i=0j‍lim|λ|→1-|Fi(ψ,φ)(λ)¯|∥Kλ[j]∥∥Kφ(λ)[i]∥=lim|λ|→1-|Fj(ψ,φ)(λ)¯|∥Kλ[j]∥∥Kφ(λ)[j]∥.
As in [6, Theorem 2], we can obtain that φ is onto, otherwise
(28)lim|λ|→1-∥Cψ,φ*Kλ[j]∥∥Kλ[j]∥=0.
contradicts the Fredholmness of Cψ,φ. Combined with Proposition 5, it follows that φ is an automorphism of 𝔻.Theorem6 gives the necessary condition of φ for the weighted composition operator Cψ,φ to be Fredholm on H2(β) with condition (22). Furthermore, if H2(β) is automorphism invariant, then the multiplication operator Mψ on H2(β) must be bounded Fredholm since in this case Cφ is invertible and
(29)Mψ=Cψ,φCφ-1.
So in the following, we consider the condition for a multiplication operator Mψ on H2(β) to be Fredholm.Denoteℳ={ψisanalyticin𝔻,ψf∈H2(β),f∈H2(β)}. For ψ∈ℳ, the multiplication operator Mψ on H2(β) is bounded. ℳ is called the multiplier space of H2(β).Lemma 7.
Letψ∈ℳ. Then Mψ*Kλ[j]=∑i=0j‍Cjiψ(j-i)(λ)¯Kλ[i].Proposition 8.
LetH2(β) be a weighted Hardy space such that
(30)∑n=0∞‍n2jβ(n)2=∞
for some nonnegative integer j. If Mψ is a bounded Fredholm operator on H2(β), then there exist c>0 and 0<r<1 such that |ψ(z)|≥c for r<|z|<1.Proof.
IfMψ is a Fredholm operator, then there exist bounded operator T and compact operator S such that
(31)TMψ*=I+K,
where I is identity. Acting on Kλ[j]/∥Kλ[j]∥, we have
(32)∑i=0j‍Cji|ψ(j-i)(λ)|∥Kλ[i]∥∥Kλ[j]∥∥T∥≥∥TMψ*Kλ[j]∥=∥Kλ[j]∥Kλ[j]∥+SKλ[j]∥Kλ[j]∥∥≥1-∥SKλ[j]∥Kλ[j]∥∥.
Let |λ|→1-; then |ψ(λ)|≥1/∥T∥, since ∥S(Kλ[j]/∥Kλ[j]∥)∥→0.The following example shows that generallyH2(β) is not automorphism invariant even if H2(β) satisfies condition (22).Example 9.
Fixa∈𝔻(a≠0). Let β(2k)=1 and β(2k-1)=|a|-(2k-1) for k=1,2,…. It is easy to verify that
(33)∑n=0∞‍1β(n)2=∞,∑n=0∞‍|β(n)|2|an|2=∞,
which means that ∥φa∥=∞ and hence φa∉H2(β), where φa(z)=(a-z)/(1-a¯z) is an automorphism of 𝔻.Now, we give the proof of the main result of Theorem1.Proof.
By Theorem6, φ is an automorphism of 𝔻. Since H2(β) is automorphism invariant, Cφ is bounded and invertible with Cφ-1=Cφ-1. It follows that Mψ=Cψ,φCφ-1 is a bounded Fredholm operator on H2(β). By Proposition 8, there exist c>0 and 0<r<1 such that |ψ(z)|≥c for r<|z|<1.


## 3. Application

In the section, we give a complete characterization of bounded Fredholm and bounded invertible weighted composition operator on𝒟α.First, we cite some well-known results of𝒟α. Denotes ℳα by the multiplier space of 𝒟α.(1)
Forα>1, 𝒟α is an algebra. α=1 is the classical Dirichlet space. For α<1,  𝒟α is equivalent to the Hilbert space with reproducing kernel function Kλ(z)=1/(1-λ¯z)1-α. α=0 is the classical Hardy space. For α<0, 𝒟α=La2(𝔻,(1-|z|2)-1-αdA) with equivalent norms [16, 17].(2)
Forα>1,  ℳα=Dα. For α≤0, ℳα=H∞(𝔻). (For more information about ℳα, see [18].)(3)
Letψ∈ℳα. Mψ is Fredholm if and only if there exist c>0, 0<r<1, |ψ(z)|≥c,  and r<|z|<1. Mψ is invertible if and only if there exists c>0 such that |ψ(z)|≥c, |z|<1.(4)
𝒟α is automorphism invariant.By Theorem1, (29), and the results (3), (4) above, we have the following corollaries.Corollary 10.
C
ψ
,
φ is a bounded Fredholm operator on 𝒟α if and only if ψ∈ℳα, and there exist c>0 and 0<r<1 such that |ψ(z)|≥c for r<|z|<1;  φ is automorphism of 𝔻.Corollary 11.
C
ψ
,
φ is a bounded invertible operator on 𝒟α if and only if ψ∈ℳα, and there exists c>0 such that |ψ(z)|≥c for |z|<1; φ is automorphism of 𝔻.



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