# Fredholm Weighted Composition Operators on Dirichlet Space

Authors: Liankuo Zhao
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

This paper characterizes bounded Fredholm, bounded invertible, and unitary weighted composition operators on Dirichlet space.


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## Body

## 1. Introduction

LetH be a Hilbert space of analytic functions on the unit disk D. For an analytic function ψ on D, we can define the multiplication operator Mψ:f→ψf,f∈H. For an analytic self-mapping φ of D, the composition operator Cφ defined on H as Cφf=f∘φ,f∈H. These operators are two classes of important operators in the study of operator theory in function spaces [1–3]. Furthermore, for ψ and φ, we define the weighted composition operator Cψ,φ on H as
(1.1)Cψ,φ:f⟶ψ(f∘φ),f∈H.Recently, the boundedness, compactness, norm, and essential norm of weighted composition operators on various spaces of analytic functions have been studied intensively, see [4–9] and so on. In this paper, we characterize bounded Fredholm weighted composition operators on Dirichlet space of the unit disk.Recall the Dirichlet space𝒟 that consists of analytic function f on D with finite Dirichlet integral:
(1.2)D(f)=∫D‍|f′|2dA<∞,
where dA is the normalized Lebesgue area measure on D. It is well known that 𝒟 is the only möbius invariant Hilbert space up to an isomorphism [10]. Endow 𝒟 with norm
(1.3)‖f‖=(|f(0)|2+D(f))1/2,f∈D.𝒟 is a Hilbert space with inner product
(1.4)〈f,g〉=f(0)g(0)¯+∫D‍f′(z)g′(z)¯dA(z),f,g∈D.
Furthermore 𝒟 is a reproducing function space with reproducing kernel
(1.5)Kλ(z)=1+log⁡11-λ-z,λ,z∈D.Denoteℳ={ψ:ψisanalyticonD,ψf∈𝒟forf∈𝒟}. ℳ is called the multiplier space of 𝒟. By the closed graph theorem, the multiplication operator Mψ defined by ψ∈ℳ is bounded on 𝒟. For the characterization of the element in ℳ, see [11].For analytic functionψ on D and analytic self-mapping φ of D, the weighted composition operator Cψ,φ on 𝒟 is not necessarily bounded. Even the composition operator Cφ is not necessarily bounded on 𝒟, which is different from the cases in Hardy space and Bergman space. See [12] for more information about the properties of composition operators acting on the Dirichlet space.The main result of the paper reads as the following.Theorem 1.1.
Letψ and φ be analytic functions on D with φ(D)⊂D. Then Cψ,φ is a bounded Fredholm operator on 𝒟 if and only if ψ∈ℳ, bounded away from zero near the unit circle, and φ is an automorphism of D.Ifψ(z)=1, then the result above gives the characterization of bounded Fredholm composition operator Cφ on 𝒟, which was obtained in [12].As corollaries, in the end of this paper one gives the characterization of bounded invertible and unitary weighted composition operator on𝒟, respectively. Some idea of this paper is derived from [4, 13], which characterize normal and bounded invertible weighted composition operator on the Hardy space, respectively.


## 2. Proof of the Main Result

In the following,ψ and φ denote analytic functions on D with φ(D)⊂D. It is easy to verify that ψ∈𝒟 if Cψ,φ is defined on 𝒟.Proposition 2.1.
LetCψ,φ be a bounded Fredholm operator on 𝒟. Then ψ has at most finite zeroes in D and φ is an inner function.Proof.
IfCψ,φ is a bounded Fredholm operator, then there exist a bounded operator T and a compact operator S on 𝒟 such that
(2.1)T(Cψ,φ)*=I+S,
where I is the identity operator.
Since(2.2)(Cψ,φ)*Kw(z)=〈Cψ,φ*Kw,Kz〉=〈Kw,Cψ,φKz〉=〈Kw,ψKz∘φ〉=ψ(w)Kz(φ(w))¯=ψ(w)¯Kφ(w)(z),
we have
(2.3)‖T‖|ψ(w)|‖Kφ(w)‖‖Kw‖≥‖T(Cψ,φ)*kw‖≥‖kw‖-‖Skw‖=1-‖Skw‖,
where kw=Kw/∥Kw∥ is the normalization of reproducing kernel function Kw.
SinceS is compact and kw weakly converges to 0 as |w|→1, ∥Skw∥→0 as |w|→1. It follows that there exists constant r, 0<r<1, such that ∥Skw∥<1/2 for all w with r<|w|<1. Inequality (2.3) shows that
(2.4)|ψ(w)|‖Kw‖≥12‖T‖‖Kφ(w)‖,r<|w|<1,
which implies that ψ has no zeroes in {w∈D,r<|w|<1}, and, hence, ψ has at most finite zeroes in {w∈D,|w|≤r}.
Sincekw weakly converges to 0 as |w|→1, 〈ψ,kw〉→0 as |w|→1, that is,
(2.5)ψ(w)‖Kw‖⟶0,|w|⟶1.
It follows from (2.4) that ∥Kφ(w)∥=(1+log(1/(1-|φ(w)|2)))1/2→∞ and hence |φ(w)|→1 as |w|→1, that is, φ is an inner function.For the proof of the following lemma, we cite Carleson's formula for the Dirichlet integral [14].Letf∈𝒟, f=BSF be the canonical factorization of f as a function in the Hardy space, where B=∏j=1∞‍(a-j/|aj|)((aj-z)/(1-a-jz)), is a Blaschke product, S is the singular part of f and F is the outer part of f. Then
(2.6)D(f)=∫T‍∑n=1∞‍Pαn(ξ)|f(ξ)|2|dξ|2π+∬T2|ζ-ξ|2|f(ξ)|2dμ(ζ)|dξ|2π+∬T(e2u(ζ)-e2u(ξ))(u(ζ)-u(ξ))|ζ-ξ|2|dζ|2π|dξ|2π,
where T is the unit circle, u(ξ)=log|f(ξ)|, Pα(ξ) is the Poisson kernel, and μ is the singular measure corresponding to S.Lemma 2.2.
LetCψ,φ be a bounded operator on 𝒟, ψ=BF with B a finite Blaschke product. Then CF,φ is bounded.Proof.
LetMB be the multiplication operator on 𝒟. Then Cψ,φ=MBCF,φ. Since B is a finite Blaschke product, by the Carleson's formula, we have
(2.7)D(ψ(f∘φ))=D(BF(f∘φ))≥D(F(f∘φ)),f∈D.
Since∥f∥2=|f(0)|2+D(f),f∈𝒟, by the inequality above it is easy to verify that CF,φ is bounded on 𝒟 if Cψ,φ is bounded.Lemma 2.3.
LetF be an analytic function on D with zero-free. If CF,φ is a bounded Fredholm operator on 𝒟, then φ is univalent.Proof.
Ifφ(a)=φ(b) for a,b∈D with a≠b, by a similar reasoning as [1, Lemma 3.26], there exist infinite sets {an} and {bn} in D which is disjoint such that φ(an)=φ(bn). Hence,
(2.8)(CF,φ)*(KanF(an)¯-KbnF(bn)¯)=0,
which contradicts to that kernel of (CF,φ)* is finite dimensional.Corollary 2.4.
IfCψ,φ is a bounded Fredholm operator on 𝒟, then φ is an automorphism of D and ψ∈ℳ.Proof.
By Proposition2.1, ψ has the factorization of BF with B a finite Blaschke product and F zero free in D. By Lemma 2.2, CF,φ is a bounded operator on 𝒟. Since Cψ,φ=MBCF,φ and MB is a Fredholm operator, CF,φ is a Fredholm operator also. By Proposition 2.1 and Lemma 2.3, φ is an univalent inner function, it follows from [1, Corollary 3.28] that φ is an automorphism of D.
SinceCψ,φCφ-1=Mψ, Mψ is a bounded multiplication operator on 𝒟, which implies that ψ∈ℳ.The following lemmas is well-known. It is easy to verify by the factMψ*Kw=ψ(w)¯Kw also.Lemma 2.5.
Letψ∈ℳ. Then Mψ is an invertible operator on 𝒟 if and only if ψ is invertible in ℳ.Lemma 2.6.
Letψ∈ℳ. Then Mψ is a Fredholm operator on 𝒟 if and only if ψ is bounded away from the unit circle.Now we give the proof of Theorem1.1.Proof of Theorem1.1.
IfCψ,φ is a bounded Fredholm operator on 𝒟, by Corollary 2.4, ψ∈ℳ and φ is an automorphism of D. Since Cφ is invertible, Mψ is a Fredholm operator. So ψ is bounded away form the unit circle follows from Lemma 2.6.
On the other hand, ifψ∈ℳ and bounded away from the unit circle, then Mψ is a bounded Fredholm operator on 𝒟. If φ is an automorphism of D, then Cφ is invertible. Hence Cψ,φ=MψCφ is a bounded Fredholm operator on 𝒟.As corollaries, in the following, we characterize bounded invertible and unitary weighted composition operators on𝒟.Corollary 2.7.
Letψ and φ be analytic functions on D with φ(D)⊂D. Then Cψ,φ is a bounded invertible operator on 𝒟 if and only if ψ∈ℳ, invertible in ℳ, and φ is an automorphism of D.Proof.
Since a bounded invertible operator is a bounded Fredholm operator, the proof is similar to the proof of Theorem1.1.Corollary 2.8.
Letψ and φ be analytic functions on D with φ(D)⊂D. Cψ,φ is a bounded operator on 𝒟. Then Cψ,φ is a unitary operator if and only if ψ is a constant with |ψ|=1 and φ is a rotation of D.Proof.
IfCψ,φ is a unitary operator, then it must be an invertible operator. By Corollary 2.7, φ is an automorphism of D and ψ is invertible in ℳ.
Letn be nonnegative integer, en(z)=zn,z∈D. A unitary is also an isometry, so we have
(2.9)‖ψ‖=‖Cψ,φe0‖=‖e0‖=1,(2.10)‖ψφn‖=‖Cψ,φen‖=‖en‖=n,n≥1.
Letα∈D such that φ(α)=0. Since φ is an automorphism of D, φn is a finite Blaschke product with zero α of order n. By Carleson’s formula for Dirichlet integral, we have
(2.11)D(ψφn)=n∫T‍Pα(ξ)|ψ(ξ)|2|dξ|2π+D(ψ).
Hence,
(2.12)n=‖ψφn‖2=|ψ(0)φ(0)n|2+D(ψφn)=|ψ(0)φ(0)n|2+n∫T‍Pα(ξ)|ψ(ξ)|2|dξ|2π+D(ψ),n≥1.
That is,
(2.13)1=|ψ(0)φ(0)n|2n+∫T‍Pα(ξ)|ψ(ξ)|2|dξ|2π+D(ψ)n,n≥1.
Let n→∞, then 1=∫T‍Pα(ξ)|ψ(ξ)|2(|dξ|/2π).
By (2.12), we have D(ψ)=0 and |ψ(0)φ(0)|=0. By (2.9), we obtain ψ is a constant with |ψ|=1, which implies that φ(0)=0, that is, φ is a rotation of D.
The sufficiency is easy to verify.Remark 2.9.
The key step in the proof of the main result is to analyze zeros of the symbolψ and univalency of φ. The following result pointed out by the referee gives a simple characterization of the symbols ψ and φ for the bounded Fredholm operator Cψ,φ on 𝒟.Proposition 2.10.
Letψ and φ be analytic functions on D with φ(D)⊂D. Cψ,φ is a bounded Fredholm operator on 𝒟. Then ψ has only finitely many zeros in D and φ is univalent.Proof.
Ifψ(a)=0 for a∈D, then Cψ,φ*Ka=ψ(a)¯Kφ(a)=0, which implies that Ka is in the kernel of Cψ,φ*. Thus if ψ had infinitely many zeros, the kernel of Cψ,φ* would be infinite dimensional and hence this operator would not be Fredholm.
Ifφ(a)=φ(b) for a,b∈D with a≠b, by a similar reasoning as [1, Lemma 3.26], there exist infinite sets {an} and {bn} in D which is disjoint such that φ(an)=φ(bn). Since ψ has only finitely many zeros in D, we can choose infinitely many an and bn such that ψ(an)≠0, ψ(bn)≠0. Hence,
(2.14)(Cψ,φ)*(Kanψ(an)¯-Kbnψ(bn)¯)=0.
Since Cψ,φ is a Fredholm operator, φ must be univalent.



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*Source: 970729-2012-08-26.xml*