Fredholm Weighted Composition Operators on Dirichlet Space · space of D. By the closed graph...

Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2012, Article ID 970729, 7 pagesdoi:10.1155/2012/970729

Research ArticleFredholm Weighted Composition Operators onDirichlet Space

Liankuo Zhao

School of Mathematics and Computer Science, Shanxi Normal University, Linfen 041004, China

Correspondence should be addressed to Liankuo Zhao, [email protected]

Received 3 June 2012; Accepted 3 August 2012

Academic Editor: Henryk Hudzik

Copyright q 2012 Liankuo Zhao. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

This paper characterizes bounded Fredholm, bounded invertible, and unitary weighted composi-tion operators on Dirichlet space.

1. Introduction

Let H be a Hilbert space of analytic functions on the unit disk D. For an analytic function ψon 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. Theseoperators are two classes of important operators in the study of operator theory in functionspaces [1–3]. Furthermore, for ψ and ϕ, we define the weighted composition operator Cψ,ϕ onH as

Cψ,ϕ : f −→ ψ(f ◦ ϕ), f ∈ H. (1.1)

Recently, the boundedness, compactness, norm, and essential norm of weightedcomposition 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 compositionoperators on Dirichlet space of the unit disk.

2 International Journal of Mathematics and Mathematical Sciences

Recall the Dirichlet space D that consists of analytic function f on D with finiteDirichlet 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 onlymobius 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) +

∫

D

f ′(z)g ′(z)dA(z), f, g ∈ D. (1.4)

Furthermore D is a reproducing function space with reproducing kernel

Kλ(z) = 1 + log1

1 − λz, λ, z ∈ D. (1.5)

Denote M = {ψ : ψ is analytic on D, ψf ∈ D for f ∈ D}. M is called the multiplierspace of D. By the closed graph theorem, the multiplication operator Mψ defined by ψ ∈ Mis 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 weightedcomposition operator Cψ,ϕ on D is not necessarily bounded. Even the composition operatorCϕ is not necessarily bounded on D, which is different from the cases in Hardy space andBergman space. See [12] for more information about the properties of composition operatorsacting 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 Cψ,ϕ is a boundedFredholm operator on D if and only if ψ ∈ M, bounded away from zero near the unit circle, and ϕ isan automorphism of D.

If ψ(z) = 1, then the result above gives the characterization of bounded Fredholmcomposition operator Cϕ on D, which was obtained in [12].

As corollaries, in the end of this paper one gives the characterization of boundedinvertible and unitary weighted composition operator on D, respectively. Some idea of thispaper is derived from [4, 13], which characterize normal and bounded invertible weightedcomposition 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 verifythat ψ ∈ D if Cψ,ϕ is defined on D.

International Journal of Mathematics and Mathematical Sciences 3

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

Proof. If Cψ,ϕ is a bounded Fredholm operator, then there exist a bounded operator T and acompact operator S on D such that

T(Cψ,ϕ

)∗ = I + S, (2.1)

where I is the identity operator.Since

(Cψ,ϕ

)∗Kw(z) =

⟨C∗ψ,ϕKw,Kz

⟩=⟨Kw,Cψ,ϕKz

⟩

=⟨Kw, ψKz ◦ ϕ

⟩= ψ(w)Kz

(ϕ(w)

)

= ψ(w)Kϕ(w)(z),

(2.2)

we have

‖T‖∣∣ψ(w)∣∣∥∥Kϕ(w)

∥∥

‖Kw‖ ≥ ∥∥T(Cψ,ϕ

)∗kw

∥∥

≥ ‖kw‖ − ‖Skw‖= 1 − ‖Skw‖,

(2.3)

where kw = Kw/‖Kw‖ is the normalization of reproducing kernel function Kw.Since S 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 allwwith r < |w| < 1.Inequality (2.3) shows that

∣∣ψ(w)∣∣

‖Kw‖ ≥ 12‖T‖∥∥Kϕ(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 finitezeroes in {w ∈ D, |w| ≤ r}.

Since kw weakly converges to 0 as |w| → 1, 〈ψ, kw〉 → 0 as |w| → 1, that is,

ψ(w)‖Kw‖ −→ 0, |w| −→ 1. (2.5)

It follows from (2.4) that ‖Kϕ(w)‖ = (1 + log(1/(1 − |ϕ(w)|2)))1/2 → ∞ and hence |ϕ(w)| → 1as |w| → 1, that is, ϕ is an inner function.

For the proof of the following lemma, we cite Carleson’s formula for the Dirichletintegral [14].


Let f ∈ D, f = BSF be the canonical factorization of f as a function in the Hardy space,where B =

∏∞j=1(aj/|aj |)((aj − z)/(1 − ajz)), is a Blaschke product, S is the singular part of f

and F is the outer part of f . Then

D(f)=∫

T

∞∑

n=1

Pαn(ξ)∣∣f(ξ)

∣∣2 |dξ|

2π+∫∫

T

2

|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

,

(2.6)

where T is the unit circle, u(ξ) = log |f(ξ)|, Pα(ξ) is the Poisson kernel, and μ is the singularmeasure corresponding to S.

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

Proof. Let MB be the multiplication operator on D. Then Cψ,ϕ = MBCF,ϕ. Since B is a finiteBlaschke 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 CF,ϕ

is bounded on D if Cψ,ϕ is bounded.

Lemma 2.3. Let F be an analytic function onD with zero-free. IfCF,ϕ is a bounded Fredholm operatoron 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], thereexist infinite sets {an} and {bn} in D which is disjoint such that ϕ(an) = ϕ(bn). Hence,

(CF,ϕ

)∗(

Kan

F(an)− Kbn

F(bn)

)

= 0, (2.8)

which contradicts to that kernel of (CF,ϕ)∗ is finite dimensional.

Corollary 2.4. If Cψ,ϕ is a bounded Fredholm operator on D, then ϕ is an automorphism of D andψ ∈ M.

Proof. By Proposition 2.1, ψ has the factorization of BF with B a finite Blaschke product and Fzero free in D. By Lemma 2.2, CF,ϕ is a bounded operator on D. Since Cψ,ϕ =MBCF,ϕ andMB

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 automorphismof D.

Since Cψ,ϕCϕ−1 = Mψ , Mψ is a bounded multiplication operator on D, which impliesthat ψ ∈ M.

The following lemmas is well-known. It is easy to verify by the factM∗ψKw = ψ(w)Kw

also.


Lemma 2.5. Let ψ ∈ M. ThenMψ is an invertible operator on D if and only if ψ is invertible inM.

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

Now we give the proof of Theorem 1.1.

Proof of Theorem 1.1. If Cψ,ϕ is a bounded Fredholm operator on D, by Corollary 2.4, ψ ∈ Mand ϕ is an automorphism of D. Since Cϕ is invertible, Mψ is a Fredholm operator. So ψ isbounded 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 Mψ is abounded Fredholm operator onD. If ϕ is an automorphism ofD, then Cϕ is invertible. HenceCψ,ϕ =MψCϕ is a bounded Fredholm operator on D.

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

Corollary 2.7. Let ψ and ϕ be analytic functions on D with ϕ(D) ⊂ D. Then Cψ,ϕ is a boundedinvertible operator on D if and only if ψ ∈ M, invertible inM, and ϕ is an automorphism of D.

Proof. Since a bounded invertible operator is a bounded Fredholm operator, the proof issimilar to the proof of Theorem 1.1.

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

Proof. If Cψ,ϕ is a unitary operator, then it must be an invertible operator. By Corollary 2.7, ϕis an automorphism of D and ψ is invertible inM.

Let n be nonnegative integer, en(z) = zn, z ∈ D. A unitary is also an isometry, so wehave

∥∥ψ∥∥ =

∥∥Cψ,ϕe0∥∥ = ‖e0‖ = 1, (2.9)

∥∥ψϕn∥∥ =

∥∥Cψ,ϕen∥∥ = ‖en‖ =

√n, n ≥ 1. (2.10)

Let α ∈ D such that ϕ(α) = 0. Since ϕ is an automorphism of D, ϕn is a finite Blaschkeproduct with zero α of order n. By Carleson’s formula for Dirichlet integral, we have

D(ψϕn

)= n

∫

TPα(ξ)

∣∣ψ(ξ)∣∣2 |dξ|

2π+D

(ψ). (2.11)

Hence,

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.

(2.12)


That is,

1 =

∣∣ψ(0)ϕ(0)n

∣∣2

n+∫

TPα(ξ)

∣∣ψ(ξ)

∣∣2 |dξ|

2π+D(ψ)

n, n ≥ 1. (2.13)

Let n → ∞, then 1 =∫T Pα(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 constantwith |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 simplecharacterization of the symbols ψ and ϕ for the bounded Fredholm operator Cψ,ϕ on D.

Proposition 2.10. Let ψ and ϕ be analytic functions on D with ϕ(D) ⊂ D. Cψ,ϕ is a boundedFredholm 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,ϕKa = ψ(a)Kϕ(a) = 0, which implies that Ka is in

the kernel of C∗ψ,ϕ. Thus if ψ had infinitely many zeros, the kernel of C∗

ψ,ϕ would be infinitedimensional 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], thereexist infinite sets {an} and {bn} in D which is disjoint such that ϕ(an) = ϕ(bn). Since ψ hasonly finitely many zeros in D, we can choose infinitely many an and bn such that ψ(an)/= 0,ψ(bn)/= 0. Hence,

(Cψ,ϕ

)∗(

Kan

ψ(an)− Kbn

ψ(bn)

)

= 0. (2.14)

Since Cψ,ϕ is a Fredholm operator, ϕmust be univalent.

Acknowledgments

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

References

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