Fredholm Weighted Composition Operators on Dirichlet Space

Abstract

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

1. Introduction

Let 𝐻 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 𝑀_𝜓∶𝑓→𝜓𝑓,𝑓∈𝐻. For an analytic self-mapping 𝜑 of 𝐷, the composition operator 𝐶_𝜑 defined on 𝐻 as 𝐶_𝜑𝑓=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 𝐶_𝜓,𝜑 on 𝐻 as 𝐶_𝜓,𝜑∶𝑓⟶𝜓(𝑓∘𝜑),𝑓∈𝐻.(1.1)

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 space 𝒟 that consists of analytic function 𝑓 on 𝐷 with finite Dirichlet integral: 𝐷(𝑓)=_𝐷||𝑓′||²𝑑𝐴<∞,(1.2) 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 ‖𝑓‖=||𝑓(0)||²+𝐷(𝑓)^1/2,𝑓∈𝒟.(1.3)𝒟 is a Hilbert space with inner product ⟨𝑓,𝑔⟩=𝑓(0)𝑔(0)+_𝐷𝑓^(𝑧)𝑔^(𝑧)𝑑𝐴(𝑧),𝑓,𝑔∈𝒟.(1.4) Furthermore 𝒟 is a reproducing function space with reproducing kernel 𝐾_𝜆(𝑧)=1+log11−𝜆⁻𝑧,𝜆,𝑧∈𝐷.(1.5)

Denote ℳ={𝜓∶𝜓isanalyticon𝐷,𝜓𝑓∈𝒟for𝑓∈𝒟}. ℳ is called the multiplier space of 𝒟. By the closed graph theorem, the multiplication operator 𝑀_𝜓 defined by 𝜓∈ℳ is bounded on 𝒟. For the characterization of the element in ℳ, see [11].

For analytic function 𝜓 on 𝐷 and analytic self-mapping 𝜑 of 𝐷, the weighted composition operator 𝐶_𝜓,𝜑 on 𝒟 is not necessarily bounded. Even the composition operator 𝐶_𝜑 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. Let 𝜓 and 𝜑 be analytic functions on 𝐷 with 𝜑(𝐷)⊂𝐷. Then 𝐶_𝜓,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.

If 𝜓(𝑧)=1, then the result above gives the characterization of bounded Fredholm composition operator 𝐶_𝜑 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 on 𝒟, 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 𝐶_𝜓,𝜑 is defined on 𝒟.

Proposition 2.1. Let 𝐶_𝜓,𝜑 be a bounded Fredholm operator on 𝒟. Then 𝜓 has at most finite zeroes in 𝐷 and 𝜑 is an inner function.

Proof. If 𝐶_𝜓,𝜑 is a bounded Fredholm operator, then there exist a bounded operator 𝑇 and a compact operator 𝑆 on 𝒟 such that 𝑇𝐶_𝜓,𝜑^∗=𝐼+𝑆,(2.1) where 𝐼 is the identity operator.
Since 𝐶_𝜓,𝜑^∗𝐾_𝑤(𝑧)==⟨𝐾𝐶^∗𝜓,𝜑_𝑤𝐾_𝑤,𝐾_𝑧=𝐾_𝑤,𝐶_𝜓,𝜑𝐾_𝑧=,𝜓𝐾_𝑧∘𝜑⟩=𝜓(𝑤)𝐾_𝑧(𝜑(𝑤))𝜓(𝑤)𝐾_𝜑(𝑤)(𝑧),(2.2) we have ‖𝑇‖||𝜓(𝑤)||‖‖𝐾^𝜑(𝑤)‖‖‖‖𝐾_𝑤‖‖≥‖‖𝑇𝐶^𝜓,𝜑^∗≥‖‖𝑘_𝑤𝑘_𝑤‖‖=1−‖‖‖‖𝑆−𝑘_𝑤‖‖𝑆𝑘_𝑤‖‖‖‖,(2.3) where 𝑘_𝑤=𝐾_𝑤/‖𝐾_𝑤‖ is the normalization of reproducing kernel function 𝐾_𝑤.
Since 𝑆 is compact and 𝑘_𝑤 weakly converges to 0 as |𝑤|→1, ‖𝑆𝑘_𝑤‖→0 as |𝑤|→1. It follows that there exists constant 𝑟, 0<𝑟<1, such that ‖𝑆𝑘_𝑤‖<1/2 for all 𝑤 with 𝑟<|𝑤|<1. Inequality (2.3) shows that ||𝜓(𝑤)||‖‖𝐾_𝑤‖‖≥2‖𝑇‖‖‖1𝐾_𝜑(𝑤)‖‖,𝑟<|𝑤|<1,(2.4) 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}.
Since 𝑘_𝑤 weakly converges to 0 as |𝑤|→1, ⟨𝜓,𝑘_𝑤⟩→0 as |𝑤|→1, that is, 𝜓(𝑤)‖‖𝐾_𝑤‖‖⟶0,|𝑤|⟶1.(2.5) It follows from (2.4) that ‖𝐾_𝜑(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].

Let 𝑓∈𝒟, 𝑓=𝐵𝑆𝐹 be the canonical factorization of 𝑓 as a function in the Hardy space, where 𝐵=∏^∞𝑗=1(𝑎⁻_𝑗/|𝑎_𝑗|)((𝑎_𝑗−𝑧)/(1−𝑎⁻_𝑗𝑧)), is a Blaschke product, 𝑆 is the singular part of 𝑓 and 𝐹 is the outer part of 𝑓. Then 𝐷(𝑓)=_T^∞_𝑛=1𝑃_𝛼𝑛(𝜉)||𝑓(𝜉)||²||𝑑𝜉||2𝜋+_T2||𝜁−𝜉||²||𝑓(𝜉)||²𝑑𝜇(𝜁)||𝑑𝜉||+2𝜋_T𝑒^2𝑢(𝜁)−𝑒^2𝑢(𝜉)(𝑢(𝜁)−𝑢(𝜉))||𝜁−𝜉||²||𝑑𝜁||2𝜋||𝑑2𝜉𝜋||,(2.6) where T is the unit circle, 𝑢(𝜉)=log|𝑓(𝜉)|, 𝑃_𝛼(𝜉) is the Poisson kernel, and 𝜇 is the singular measure corresponding to 𝑆.

Lemma 2.2. Let 𝐶_𝜓,𝜑 be a bounded operator on 𝒟, 𝜓=𝐵𝐹 with 𝐵 a finite Blaschke product. Then 𝐶_𝐹,𝜑 is bounded.

Proof. Let 𝑀_𝐵 be the multiplication operator on 𝒟. Then 𝐶_𝜓,𝜑=𝑀_𝐵𝐶_𝐹,𝜑. Since 𝐵 is a finite Blaschke product, by the Carleson's formula, we have 𝐷(𝜓(𝑓∘𝜑))=𝐷(𝐵𝐹(𝑓∘𝜑))≥𝐷(𝐹(𝑓∘𝜑)),𝑓∈𝒟.(2.7)
Since ‖𝑓‖²=|𝑓(0)|²+𝐷(𝑓),𝑓∈𝒟, by the inequality above it is easy to verify that 𝐶_𝐹,𝜑 is bounded on 𝒟 if 𝐶_𝜓,𝜑 is bounded.

Lemma 2.3. Let 𝐹 be an analytic function on 𝐷 with zero-free. If 𝐶_𝐹,𝜑 is a bounded Fredholm operator on 𝒟, then 𝜑 is univalent.

Proof. If 𝜑(𝑎)=𝜑(𝑏) for 𝑎,𝑏∈𝐷 with 𝑎≠𝑏, by a similar reasoning as [1, Lemma 3.26], there exist infinite sets {𝑎_𝑛} and {𝑏_𝑛} in 𝐷 which is disjoint such that 𝜑(𝑎_𝑛)=𝜑(𝑏_𝑛). Hence, 𝐶_𝐹,𝜑^∗⎛⎜⎜⎝𝐾_𝑎𝑛𝐹𝑎_𝑛−𝐹𝐾𝑏^𝑏𝑛_𝑛⎞⎟⎟⎠=0,(2.8) which contradicts to that kernel of (𝐶_𝐹,𝜑)^∗ is finite dimensional.

Corollary 2.4. If 𝐶_𝜓,𝜑 is a bounded Fredholm operator on 𝒟, then 𝜑 is an automorphism of 𝐷 and 𝜓∈ℳ.

Proof. By Proposition 2.1, 𝜓 has the factorization of 𝐵𝐹 with 𝐵 a finite Blaschke product and 𝐹 zero free in 𝐷. By Lemma 2.2, 𝐶_𝐹,𝜑 is a bounded operator on 𝒟. Since 𝐶_𝜓,𝜑=𝑀_𝐵𝐶_𝐹,𝜑 and 𝑀_𝐵 is a Fredholm operator, 𝐶_𝐹,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.
Since 𝐶_𝜓,𝜑𝐶_𝜑−1=𝑀_𝜓, 𝑀_𝜓 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 fact 𝑀^∗𝜓𝐾_𝑤=𝜓(𝑤)𝐾_𝑤 also.

Lemma 2.5. Let 𝜓∈ℳ. Then 𝑀_𝜓 is an invertible operator on 𝒟 if and only if 𝜓 is invertible in ℳ.

Lemma 2.6. Let 𝜓∈ℳ. Then 𝑀_𝜓 is a Fredholm operator on 𝒟 if and only if 𝜓 is bounded away from the unit circle.

Now we give the proof of Theorem 1.1.

Proof of Theorem 1.1. If 𝐶_𝜓,𝜑 is a bounded Fredholm operator on 𝒟, by Corollary 2.4, 𝜓∈ℳ and 𝜑 is an automorphism of 𝐷. Since 𝐶_𝜑 is invertible, 𝑀_𝜓 is a Fredholm operator. So 𝜓 is bounded away form the unit circle follows from Lemma 2.6.
On the other hand, if 𝜓∈ℳ and bounded away from the unit circle, then 𝑀_𝜓 is a bounded Fredholm operator on 𝒟. If 𝜑 is an automorphism of 𝐷, then 𝐶_𝜑 is invertible. Hence 𝐶_𝜓,𝜑=𝑀_𝜓𝐶_𝜑 is a bounded Fredholm operator on 𝒟.

As corollaries, in the following, we characterize bounded invertible and unitary weighted composition operators on 𝒟.

Corollary 2.7. Let 𝜓 and 𝜑 be analytic functions on 𝐷 with 𝜑(𝐷)⊂𝐷. Then 𝐶_𝜓,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 Theorem 1.1.

Corollary 2.8. Let 𝜓 and 𝜑 be analytic functions on 𝐷 with 𝜑(𝐷)⊂𝐷. 𝐶_𝜓,𝜑 is a bounded operator on 𝒟. Then 𝐶_𝜓,𝜑 is a unitary operator if and only if 𝜓 is a constant with |𝜓|=1 and 𝜑 is a rotation of 𝐷.

Proof. If 𝐶_𝜓,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.
Let 𝑛 be nonnegative integer, 𝑒_𝑛(𝑧)=𝑧^𝑛,𝑧∈𝐷. A unitary is also an isometry, so we have ‖𝜓𝜑_𝑛‖𝜓‖=‖‖𝐶_𝜓,𝜑𝑒₀‖‖=‖‖𝑒₀‖‖=1,(2.9)‖=‖‖𝐶_𝜓,𝜑𝑒_𝑛‖‖=‖‖𝑒_𝑛‖‖=√𝑛,𝑛≥1.(2.10)
Let 𝛼∈𝐷 such that 𝜑(𝛼)=0. Since 𝜑 is an automorphism of 𝐷, 𝜑^𝑛 is a finite Blaschke product with zero 𝛼 of order 𝑛. By Carleson’s formula for Dirichlet integral, we have 𝐷(𝜓𝜑^𝑛)=𝑛_T𝑃_𝛼(𝜉)||𝜓(𝜉)||²||𝑑𝜉||2𝜋+𝐷(𝜓).(2.11) Hence, 𝑛==||‖𝜓𝜓(𝜑0^𝑛)‖𝜑²(0=)_𝑛||𝜓(0)𝜑(0)^𝑛||²+𝐷(𝜓𝜑^𝑛)||²+𝑛_T𝑃_𝛼(𝜉)||𝜓(𝜉)||²||𝑑𝜉||2𝜋+𝐷(𝜓),𝑛≥1.(2.12) That is, 1=||𝜓(0)𝜑(0)^𝑛||²𝑛+_T𝑃_𝛼(𝜉)||𝜓(𝜉)||²||𝑑𝜉||2𝜋+𝐷𝑛(𝜓),𝑛≥1.(2.13) Let 𝑛→∞, then 1=∫_T𝑃_𝛼(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 symbol 𝜓 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 𝐶_𝜓,𝜑 on 𝒟.

Proposition 2.10. Let 𝜓 and 𝜑 be analytic functions on 𝐷 with 𝜑(𝐷)⊂𝐷. 𝐶_𝜓,𝜑 is a bounded Fredholm operator on 𝒟. Then 𝜓 has only finitely many zeros in 𝐷 and 𝜑 is univalent.

Proof. If 𝜓(𝑎)=0 for 𝑎∈𝐷, then 𝐶^∗𝜓,𝜑𝐾_𝑎=𝜓(𝑎)𝐾_𝜑(𝑎)=0, which implies that 𝐾_𝑎 is in the kernel of 𝐶^∗𝜓,𝜑. Thus if 𝜓 had infinitely many zeros, the kernel of 𝐶^∗𝜓,𝜑 would be infinite dimensional and hence this operator would not be Fredholm.
If 𝜑(𝑎)=𝜑(𝑏) for 𝑎,𝑏∈𝐷 with 𝑎≠𝑏, by a similar reasoning as [1, Lemma 3.26], there exist infinite sets {𝑎_𝑛} and {𝑏_𝑛} in 𝐷 which is disjoint such that 𝜑(𝑎_𝑛)=𝜑(𝑏_𝑛). Since 𝜓 has only finitely many zeros in 𝐷, we can choose infinitely many 𝑎_𝑛 and 𝑏_𝑛 such that 𝜓(𝑎_𝑛)≠0, 𝜓(𝑏_𝑛)≠0. Hence, 𝐶_𝜓,𝜑^∗⎛⎜⎜⎝𝐾_𝑎𝑛𝜓𝑎_𝑛−𝜓𝐾𝑏^𝑏𝑛_𝑛⎞⎟⎟⎠=0.(2.14) Since 𝐶_𝜓,𝜑 is a Fredholm operator, 𝜑 must be univalent.

Acknowledgments

Thanks are for referee for many helpful suggestions which promote the author to think the related issues deeply. This work is supported by YSF of Shanxi (2010021002-2) and NSFC (11201274).