Abstract

Let 𝑅(𝔻) be the algebra generated in Sobolev space 𝑊22(𝔻) by the rational functions with poles outside the unit disk 𝔻. In this paper, we study the similarity invariant of the multiplication operators 𝑀𝑔 in (𝑅(𝔻)), when 𝑔 is univalent analytic on 𝔻 or 𝑀𝑔 is strongly irreducible. And the commutants of multiplication operators whose symbols are composite functions, univalent analytic functions, or entire functions are studied.

1. Introduction

Let Ω be an analytic Cauchy domain in the complex plane and let 𝑊22(Ω) denote the Sobolev space, 𝑊22(Ω)=𝑓𝐿2(Ω,𝑑𝐴)thedistributionalpartialderivativesofrstandsecondorderof𝑓belongto𝐿2(Ω,𝑑𝐴),(1.1)𝑑𝐴 denotes the planar Lebesgue measure. For 𝑓,𝑔𝑊22(Ω), we define 𝑓,𝑔=|𝛼|2𝐷𝛼𝑓𝐷𝛼𝑔𝑑𝐴.(1.2) Then 𝑊22(Ω) is a Hilbert space and a Banach algebra with identity under an equivalent norm. 𝑊22(Ω) can be continuously embedded in the space 𝐶(Ω) of continuous functions on Ω by Sobolev embedding theorem.

Let 𝑅(Ω) be the subalgebra generated by the rational functions with poles outside Ω. When Ω=𝔻, the unit disc, we call 𝑅(𝔻) Sobolev disk algebra. For 𝑓𝑅(𝔻), the multiplication operator 𝑀𝑓 on 𝑅(𝔻) is defined by 𝑀𝑓(𝑔)=𝑓𝑔, 𝑔𝑅(𝔻). Then 𝑅(𝔻)=𝒜(𝑀𝑧)𝑒0, where 𝑒0 is the identity in 𝑅(𝔻) and 𝒜(𝑀𝑧)is the algebra generated by 𝑀𝑧 and identity. In fact, 𝑅(𝔻) consists of all analytic functions in 𝑊22(𝔻). We have the following properties of the space 𝑅(𝔻) and the multiplication operators on it.

Proposition 1.1 (see [1]). (i) Hilbert space 𝑅(𝔻) has an orthogonal basis {𝑒𝑛}+𝑛=0, where 𝑒𝑛=𝛽𝑛𝑧𝑛,𝛽𝑛=𝑛+13𝑛4𝑛2𝜋+2𝑛+11/2,𝑛=0,1,2,.(1.3)
(ii) As a functional Hilbert space, 𝑅(𝔻) has reproducing kernel which is 𝑘(𝑢,𝑣)=+𝑛=0𝛽2𝑛𝑢𝑛𝑣𝑛.(1.4) Then for 𝑧0𝔻, 𝑘𝑧0=𝑛=0𝛽2𝑛𝑧𝑛0𝑧𝑛.(1.5)
(iii) If 𝑓(𝑧)=+𝑛=0𝑓𝑛𝑧𝑛 is analytic on 𝔻, then 𝑓(𝑧)𝑅(𝔻) if and only if +𝑛=0|𝑓𝑛|2/𝛽2𝑛<+.
(iv) The operator 𝑀𝑓 admits the following matrix representation with respect to {𝑒𝑛}+𝑛=0: 𝑀𝑓=𝑐0𝑐1𝛽0𝛽1𝑐00𝑐2𝛽0𝛽2𝑐1𝛽1𝛽2𝑐0𝑐3𝛽0𝛽3𝑐2𝛽1𝛽3𝑐1𝛽2𝛽3𝑐0.(1.6)

Note that 𝑅(𝔻) is a subset of the disk algebra 𝐴(𝔻), hence a subset of 𝐻. Because of the special definition of the inner product and the complex behavior of the boundary value, the structure of the space 𝑅(𝔻) is much more complicated than 𝐻 or 𝐻2. For more details about the Sobolev disk algebra, the reader refers to [13].

Let be a complex separable Hilbert space and () denote the collection of bounded linear operators on . One of the basic problems in operator theory is to determine when two operators 𝐴 and 𝐵 in () are similar. A quantity (quantities) or a property (properties) 𝑃 is similarity invariant (invariants) if 𝐴 has 𝑃 and 𝐴𝐵 implies that 𝐵 has 𝑃 [2]. From this point of view, one of the basic problems in operator theory mentioned above is to determine the similarity invariants. There have already been a lot of results on the similarity invariants of operators, especially that of nonadjoint operators, which can be found in, for example, [46]. In [7], Wang et al. proved that in 𝑅(𝔻), 𝑀𝑓 is similar to 𝑀𝑧𝑛 if and only if 𝑓 is an 𝑛-Blaschke product. In this paper, we study the similarity invariant of the multiplication operators 𝑀𝑔 in (𝑅(𝔻)), when 𝑔 is univalent analytic on 𝔻 or 𝑀𝑔 is strongly irreducible.

It is well known that the commutant of a bounded linear operator or operators on a complex, separable Hilbert space plays an important role in determining the structure of this operator or these operators. The commutant of a multiplication operator on Sobolev disk algebra has been studied in the literature (see [13]). In this paper, we describe the commutant of the multiplication operator 𝑀𝑓𝑔 when 𝑔 is an 𝑛-Blaschke product. And by this result, we generalize the result which is obtained by Liu and Wang in [3]. Moreover, we study the commutants of the multiplication operators whose symbols are composite functions, univalent analytic functions, or entire functions.

2. Similarity Invariant of Some Multiplication Operators

In this section, we will characterize the similarity invariant of some multiplication operators on Sobolev disk algebra. Here, we briefly recall some background information.

For 𝑇 in (), let 𝜎(𝑇), 𝜎𝑝(𝑇),  and  𝜎𝑒(𝑇) be the spectrum, point spectrum, and essential spectrum of a bounded linearly operator 𝑇, respectively. An operator 𝐴 in () is said to be a Cowen-Douglas operator with index 𝑛 if there exists Ω, a connected open subset of complex plane , and 𝑛, a positive integer, such that(i)Ω𝜎(𝐴)={𝜆𝐴𝜆isnotinvertible}; (ii)ran(𝐴𝜆)={𝑦;(𝐴𝜆)𝑥=𝑦,𝑥}= for 𝜆 in Ω;(iii)nul(𝐴𝜆)=dimker(𝐴𝜆)=𝑛 for 𝜆 in Ω;(iv){ker(𝐴𝜆)𝜆Ω}=,where (iv) is equivalent to (iv)′ ([8]);(iv) there exists 𝜆0 in Ω, such that {ker(𝐴𝜆0)𝑘𝑘1}=.

𝑛(Ω) denotes the collection of Cowen-Douglas operators with index 𝑛.

For 𝑇(), the set of operators which commute with it is 𝒜(𝑇). That is 𝒜(𝑇)={𝐴()𝐴𝑇=𝑇𝐴}. Operator 𝑇 is strongly irreducible if there is no nontrivial idempotent in 𝒜(𝑇) [8, 9]. Denote (𝑆𝐼) the set of all strongly irreducible operators on .

Definition 2.1. Let be a Hilbert space and 𝐴,𝐵 be in (). 𝜏𝐴𝐵 is said to be a Rosenblum operator on () if for arbitrary 𝐶(),𝜏𝐴𝐵(𝐶)=𝐴𝐶𝐶𝐵.

Lemma 2.2 (see [1]). Let 𝑓 be in 𝑅(𝔻), then(i)𝜎(𝑀𝑓)=𝑓(𝔻);(ii)𝜎𝑒(𝑀𝑓)=𝑓(𝜕𝔻), where 𝜕𝔻 denotes the boundary of the unit disc 𝔻;(iii)let 𝑧0𝔻 and 𝑓(𝑧0)𝑓(𝜕𝔻). Denote the component of 𝜌𝑠𝐹(𝑀𝑓) containing 𝑓(𝑧0) as Ω, then 𝑀𝑓𝑛(Ω), where 𝑛 is the number of the zeros of 𝑓(𝑧)𝑓(𝑧0) in 𝔻.

Lemma 2.3 (see [10]). Set 𝑓𝑅(𝔻) and 𝐵𝑎(𝑧)=(𝑧𝑎)/(1𝑎𝑧),𝑎𝔻. Then 𝑀𝑓𝐵𝑎𝑀𝑓.

Theorem 2.4. Let 𝑓 and 𝑔 be in 𝑅(𝔻) and be univalent and analytic on 𝔻. Then 𝑀𝑓𝑀𝑔 if and only if 𝑓(𝔻)=𝑔(𝔻).

Proof. ”: Set 𝑀𝑓𝑀𝑔. By Lemma 2.2, we have 𝑓𝔻𝑀=𝜎𝑓𝑀=𝜎𝑔=𝑔𝔻,𝑓(𝕋)=𝜎𝑒𝑀𝑓=𝜎𝑒𝑀𝑔=𝑔(𝕋),(2.1) where 𝕋 is the unit circle. Since 𝑓 and 𝑔 are univalent and analytic on 𝔻, then 𝑓(𝕋)=𝜕𝑓𝔻=𝜕𝑔𝔻=𝑔(𝕋).(2.2) Therefore, 𝑓(𝔻)=𝑓𝔻𝑓(𝕋)=𝑔𝔻𝑔(𝕋)=𝑔(𝔻).(2.3)
”: Set 𝑓(𝔻)=𝑔(𝔻)=Ω. Because 𝑔 is univalent analytic from 𝔻 to Ω, 𝑔1Ω𝔻 is also univalent analytic. Then 𝑔1𝑓 is injective and surjective analytic function on 𝔻. If 𝑔1𝑓(𝑧0)=0, there exists a Möbius transform 𝜑 with 𝜑𝑧0=(𝑧𝑧0)/(1𝑧0𝑧) and a complex number 𝑐 with |𝑐|=1 such that 𝑔1𝑓(𝑧)=𝑐𝜑𝑧0 (see [11]). Therefore 𝑓(𝑧)=𝑔(𝑐𝜑𝑧0(𝑧)). By Lemma 2.3, 𝑀𝑓𝑀𝑔.

Lemma 2.5 (see [3]). Given 𝑓𝑅(𝔻), the following are equivalent:(i)𝑀𝑓1(Ω);(ii)𝒜(𝑀𝑓)={𝑀𝑔𝑔𝑅(𝔻)};(iii)𝑀𝑓(𝑆𝐼).

Theorem 2.6. Let 𝑓,𝑔𝑅(𝔻) and 𝑓 is univalent analytic on 𝔻. 𝑀𝑓𝑀𝑔 if and only if there exists a function 𝜒=𝑐((𝑧𝑧0)/(1𝑧0𝑧)) such that 𝑓=𝑔𝜒, where 𝑧0𝔻 and 𝑐,|𝑐|=1.

Proof. ”: Suppose that 𝑔 is not univalent on 𝔻. There exists some 𝜆𝑔(𝔻) such that the number of zeros of 𝑔(𝑧)𝜆 on 𝔻 is 𝑛>1. By Lemma 2.2, 𝑀𝑔𝑛(Ω) where Ω is a connected open subset of 𝑔(𝔻). Since 𝑀𝑓𝑀𝑔, we have 𝑀𝑓𝑛(Ω). This contradicts to 𝑀𝑓 that is a strongly irreducible operator (see Lemma 2.5). So 𝑔 is univalent analytic on 𝔻. By the proof of Theorem 2.4, there exists a function 𝜒=𝑐((𝑧𝑧0)/(1𝑧0𝑧)) such that 𝑓=𝑔𝜒, where 𝑧0𝔻 and 𝑐,|𝑐|=1.
”: By the conditions of the theorem, 𝑔 is univalent analytic on 𝔻. Since 𝑓(𝔻)=𝑔𝜒(𝔻)=𝑔(𝔻), we have 𝑀𝑓𝑀𝑔 by Theorem 2.4.

For any operator 𝑇 on Hilbert space and any integer 𝑛,1𝑛, let 𝑇(𝑛) denote the direct sum of 𝑛 copies of 𝑇 on (𝑛)=.

Lemma 2.7 (see [2]). Let 𝐴1,𝐴2 be strongly irreducible Cowen-Douglas operators. Assume that 𝐴1̸𝐴2 and 𝑇=𝐴(𝑛1)1𝐴(𝑛2)2, where 𝑛1,  𝑛2 are natural numbers. Then for each maximal ideal 𝒥 of 𝒜(𝑇), 𝒥 must be one of the following two forms: 𝒥(i)𝒥=11ker𝜏𝐴1)1(𝑛,𝐴2)2(𝑛ker𝜏𝐴2)2(𝑛,𝐴1)1(𝑛𝒜𝐴(𝑛2)2;𝒜(ii)𝒥=𝐴(𝑛1)1ker𝜏𝐴1)1(𝑛,𝐴2)2(𝑛ker𝜏𝐴2)2(𝑛,𝐴1)1(𝑛𝒥22,(2.4) where 𝒥𝑖𝑖 is a maximal ideal of 𝒜(𝐴(𝑛𝑖)𝑖), 𝑖=1,2.

Theorem 2.8. Let 𝑓,𝑔𝑅(𝔻) and 𝑀𝑓,𝑀𝑔(𝑆𝐼). The following statements are equivalent:(i)𝑀𝑓𝑀𝑔;(ii)there exist 𝑋1,𝑋2,,𝑋𝑛ker𝜏𝑀𝑓,𝑀𝑔 and 𝑌1,𝑌2,,𝑌𝑛ker𝜏𝑀𝑔,𝑀𝑓 such that 𝑋1𝑌1+𝑋2𝑌2++𝑋𝑛𝑌𝑛=𝑖𝑑𝑅(𝔻),𝑌1𝑋1+𝑌2𝑋2++𝑌𝑛𝑋𝑛=𝑖𝑑𝑅(𝔻),(2.5)where 𝑖𝑑𝑅(𝔻) denotes the identity operator of 𝑅(𝔻).

Proof. (i)(ii) Let 𝑀𝑓𝑀𝑔. Set 𝑋 be invertible in (𝑅(𝔻)) and 𝑀𝑓𝑋=𝑋𝑀𝑔. Then 𝑋 and 𝑋1 are what we want.
(ii)(i) Since 𝑀𝑓 and 𝑀𝑔 are in (𝑆𝐼), we have 𝑀𝑓 and 𝑀𝑔 that are strongly irreducible and in 1(Ω) by Lemma 2.5. Computations show 𝒜𝑀𝑓𝑀𝑔=𝒜𝑀𝑓ker𝜏𝑀𝑓,𝑀𝑔ker𝜏𝑀𝑔,𝑀𝑓𝒜𝑀𝑔.(2.6) Suppose that 𝑀𝑓̸𝑀𝑔. By Lemma 2.7, each maximal ideal 𝒥 of 𝒜(𝑀𝑓𝑀𝑔) must be one of the following two forms 𝒥𝒥=11ker𝜏𝑀𝑓,𝑀𝑔ker𝜏𝑀𝑔,𝑀𝑓𝒜𝑀𝑔𝒜or𝒥=𝑀𝑓ker𝜏𝑀𝑓,𝑀𝑔ker𝜏𝑀𝑔,𝑀𝑓𝒥22,(2.7) where 𝒥11 and 𝒥22 are the maximal ideals of 𝒜(𝑀𝑓) and 𝒜(𝑀𝑔), respectively. We can assume that 𝒥 admits the first form. Then 0𝑌𝑖𝑋𝑖0𝒥,𝑖=1,2,,𝑛0𝑌𝑖𝑋𝑖00𝑌𝑖𝑋𝑖0=𝑌𝑖𝑋𝑖00𝑋𝑖𝑌𝑖𝒥.(2.8) It follows that 𝑛𝑖=1𝑌𝑖𝑋𝑖00𝑋𝑖𝑌𝑖=𝑖𝑑𝑅(𝔻)00𝑖𝑑𝑅(𝔻)𝒥.(2.9) This contradicts that 𝒥 is a maximal ideal. So 𝑀𝑓𝑀𝑔 and 𝑀𝑓𝑀𝑔.

3. The Commutant Algebra of Multiplication Operator

In [3], Liu and Wang give the following proposition.

Proposition 3.1. Let 𝑓(𝑧)=𝑧𝑛(𝑧)𝑅(𝔻), (𝑧)0, 𝑧𝔻. Then 𝒜(𝑀𝑓)=𝒜(𝑀𝑧𝑛)𝒜(𝑀).

Let 𝐵𝑛(𝑧)=𝑛𝑘=1(𝑧𝑎𝑘)/(1𝑎𝑘𝑧), (|𝑎𝑘|<1,𝑘=1,2,,𝑛) be 𝑛-Blaschke product. Considering 𝑧𝑛 is a special 𝑛-Blaschke product, we study the commutant of 𝑀𝑓 where 𝑓(𝑧)=𝐵𝑛(𝑧)(𝑧). The following theorem is obtained, and by this result, the above proposition is generalized.

Theorem 3.2. Let 𝑓=𝐵𝑛𝑅(𝔻) with 𝑅(𝔻) where does not vanish on 𝔻. If there exists 0𝜆 such that 𝜆 can be divided by each (𝑧𝑎𝑘)/(1𝑎𝑘𝑧), then 𝒜(𝑀𝑓)=𝒜(𝑀𝐵𝑛)𝒜(𝑀).

To prove the above theorem, we need the following lemmas.

Lemma 3.3 (see [7]). Given 𝑔𝑅(𝔻),𝑀𝑔𝑀𝑧𝑛((𝑀𝑧)𝑛) if and only if 𝑔 is an 𝑛-Blaschke product.

Lemma 3.4 (see [12]). Let 𝑁 be a nilpotent operator on 𝐻 and let 𝑋0=𝜆+𝑁,0𝜆. If 𝐵,𝐴0,𝐴1,𝐵(𝐻) satisfy 𝐴𝑘𝑀(𝑘=0,1,2,) and 𝐴𝑘𝑋0=𝑋0𝐴𝑘1+𝐵,(𝑘=1,2,3), then 𝐴0=𝐴1=𝐴2=.

Now we will prove Theorem 3.2.

Proof. From 𝑅(𝔻), we have 𝑀𝑓=𝑀𝐵𝑛𝑀. Then, from Lemma 3.3, there exists an invertible operator 𝑋(𝑅(𝔻),𝑛1𝑅(𝔻)) such that 𝑋𝑀𝐵𝑛𝑋1=𝑛1𝑀𝑧. It follows that 𝐴=𝑋𝑀𝐵𝑛𝑋1𝑋𝑀𝑋1=𝑛1𝑀𝑧𝑇,(3.1) where 𝑇=𝑋𝑀𝑋1𝒜(𝑛1𝑀𝑧). So we only need to prove 𝒜(𝐴)=𝒜𝑛1𝑀𝑧𝒜(𝑇).(3.2)
Since 𝑛1𝑅(𝔻)=ker𝑛1𝑀𝑧ker𝑛1𝑀𝑧2ker𝑛1𝑀𝑧ker𝑛1𝑀𝑧3ker𝑛1𝑀𝑧2ker𝑛1𝑀𝑧𝑘ker𝑛1𝑀𝑧𝑘1(3.3) we have 𝑛1𝑀𝑧=0𝛽0𝛽1𝐼𝑛𝛽00001𝛽2𝐼𝑛𝛽00002𝛽3𝐼𝑛,(3.4) where 𝐼𝑛 is an 𝑛×𝑛 identity matrix. Computations show that 𝐺𝒜(𝑛1𝑀𝑧) if and only if 𝐺𝐺=1𝛽0𝛽1𝐺12𝛽0𝛽2𝐺13𝛽0𝛽3𝐺14𝛽0𝛽𝑘1𝐺1,𝑘0𝐺1𝛽1𝛽2𝐺12𝛽1𝛽3𝐺13𝛽1𝛽𝑘1𝐺1,𝑘100𝐺1𝛽2𝛽3𝐺12𝛽2𝛽𝑘1𝐺1,𝑘2𝛽0000𝑘2𝛽𝑘1𝐺1,2.(3.5) Therefore, 𝑇𝑇=1𝛽0𝛽1𝑇12𝛽0𝛽2𝑇13𝛽0𝛽3𝑇14𝛽0𝛽𝑘1𝑇1,𝑘0𝑇1𝛽1𝛽2𝑇12𝛽1𝛽3𝑇13𝛽1𝛽𝑘1𝑇1,𝑘100𝑇1𝛽2𝛽3𝑇12𝛽2𝛽𝑘1𝑇1,𝑘2𝛽0000𝑘2𝛽𝑘1𝑇1,2.(3.6)
Set 𝑔𝑘=(𝑧𝑎𝑘)/(1𝑎𝑘𝑧),1𝑘𝑛, and 𝐵𝑛=𝑔1𝑔2𝑔𝑛. Then for each 𝑔𝑘, there exists 𝜑𝑘𝑅(𝔻) such that 𝜆=𝑔𝑘𝜑𝑘,1𝑘𝑛. So 𝑇=𝑋𝑀𝑋1=𝑋𝑀𝑔𝑘𝑋1𝑋𝑀𝜑𝑘𝑋1+𝜆𝑖𝑑𝑛1𝑅(𝔻)=𝑉𝑘𝑆𝑘+𝜆𝑖𝑑𝑛1𝑅(𝔻),(3.7) where 𝑉𝑘=𝑋𝑀𝑔𝑘𝑋1,𝑆𝑘=𝑋𝑀𝜑𝑘𝑋1. For 𝑘=1,2,3,,𝑛,𝑗=1,2,3,,𝑛,𝑀𝜑𝑘, and 𝑀𝑔𝑗 pairwise commute. Hence, 𝑉𝑘,𝑆𝑗 pairwise commute too. Since 𝑇1=𝑇|ker(𝑛1𝑀𝑧) and ker𝑛1𝑀𝑧=ker𝑋𝑀𝐵𝑛𝑋1=ker𝑉1𝑉2𝑉𝑛(3.8) for all 𝑥ker(𝑛1𝑀𝑧), 𝑇𝜆𝑖𝑑𝑛1𝑅(𝔻)𝑛𝑥=𝑉1𝑆1𝑉2𝑆2𝑉𝑛𝑆𝑛𝑥=𝑆1𝑆2𝑆𝑛𝑉1𝑉2𝑉𝑛𝑥=0.(3.9) Therefore, 𝑇1𝜆𝐼𝑛 is nilpotent operator. Now we set 𝑁=𝑇1𝜆𝐼𝑛, that is 𝑇1=𝜆𝐼𝑛+𝑁. So 𝐴=𝑛1𝑀𝑧𝑇=0𝛽0𝛽1𝜆𝐼𝑛𝛽+𝑁0𝛽2𝑇12𝛽0𝛽3𝑇13𝛽0𝛽𝑘𝑇1,𝑘𝛽001𝛽2𝜆𝐼𝑛𝛽+𝑁1𝛽3𝑇12𝛽1𝛽𝑘𝑇1,𝑘1𝛽0002𝛽3𝜆𝐼𝑛𝛽+𝑁2𝛽𝑘𝑇1,𝑘2𝛽0000𝑘1𝛽𝑘𝜆𝐼𝑛.+𝑁(3.10)
So we only need to prove that 𝒜(𝐴)𝒜(𝑛1𝑀𝑧). In fact, if 𝑄𝒜(𝐴)𝒜(𝑛1𝑀𝑧), 𝑄𝐴=𝐴𝑄𝑄𝑇𝑛1𝑀𝑧=𝑇𝑛1𝑀𝑧𝑄𝑄𝑇𝑛1𝑀𝑧=𝑇𝑄𝑛1𝑀𝑧(𝑄𝑇𝑇𝑄)𝑛1𝑀𝑧=0𝑛1𝑀𝑧𝑇𝑄𝑄𝑇=0.(3.11) It follows from ker(𝑛1𝑀𝑧)={0} that 𝑇𝑄=𝑄𝑇. Namely, 𝑇𝑄=𝑄𝑇.
We are now in need to prove that 𝒜(𝐴)𝒜(𝑛1𝑀𝑧). Suppose that 𝑄𝒜(𝐴). Since does not vanish on 𝔻, ran𝑀=𝑅(𝔻) and so ker𝑇={0}. Because 𝐴=(𝑛1𝑀𝑧)𝑇=𝑇(𝑛1𝑀𝑧), ker𝐴=ker𝑛1𝑀𝑧,ker𝐴𝑘=ker𝑛1𝑀𝑧𝑘.(3.12) It follows from 𝑄𝐴=𝐴𝑄 that ker𝐴 and ker𝐴𝑘 are both the invariant subspaces of 𝑄. Since 𝑛1𝑅(𝔻)=ker𝐴ker𝐴2ker𝐴ker𝐴3ker𝐴2ker𝐴𝑘ker𝐴𝑘1,(3.13)𝐴 admits the matrix representation (3.5) with the above decomposition. So 𝑄𝑄=1𝑄12𝑄13𝑄140𝑄2𝑄23𝑄2400𝑄3𝑄34.(3.14) From 𝑄𝐴=𝐴𝑄, we have 0𝛽0𝛽1𝑄1𝜆𝐼𝑛𝛽+𝑁0𝛽2𝑄1𝑇12+𝛽1𝛽2𝑄12𝜆𝐼𝑛𝛽+𝑁001𝛽2𝑄2𝜆𝐼𝑛=0𝛽+𝑁0𝛽1𝜆𝐼𝑛𝑄+𝑁2𝛽0𝛽1𝜆𝐼𝑛𝑄+𝑁23+𝛽0𝛽2𝑇12𝑄3𝛽001𝛽2𝜆𝐼𝑛𝑄+𝑁3.(3.15) Comparing the (𝑛,𝑛+1) entries of both sides, we have 𝑄1=𝑄2=𝑄3= by Lemma 3.4. Comparing the (𝑛,𝑛+2) entries of 𝑄𝐴 and 𝐴𝑄, we have 𝛽𝑛+1𝛽𝑛𝜆𝐼𝑛𝑄+𝑁𝑛+1,𝑛+2=𝛽𝑛𝛽𝑛1𝑄𝑛,𝑛+1𝜆𝐼𝑛+𝑄+𝑁1𝑇12𝑇12𝑄1.(3.16) It follows from Lemma 3.4 that 𝛽𝑛+1𝛽𝑛𝑄𝑛+1,𝑛+2=𝛽𝑛𝛽𝑛1𝑄𝑛,𝑛+1.(3.17) Setting 𝑄12=(𝛽1/𝛽0)𝑄12, we have 𝑄𝑛,𝑛+1=𝛽1𝛽𝑛1𝛽0𝛽𝑛𝑄12=𝛽𝑛1𝛽𝑛𝑄12.(3.18) Inductively, if 𝑄𝑛,𝑛+𝑘1=(𝛽𝑛1/𝛽𝑛+𝑘2)𝑄1,𝑘, where 𝑄1,𝑘=(𝛽𝑘1/𝛽0)𝑄1𝑘, we need to prove 𝑄𝑛,𝑛+𝑘=(𝛽𝑛1/𝛽𝑛+𝑘1)𝑄1,𝑘+1. Comparing the (𝑛,𝑛+𝑘+1) entries of 𝑄𝐴 and 𝐴𝑄, we have 𝛽𝑛+𝑘𝛽𝑛𝜆𝐼𝑛𝑄+𝑁𝑛+1,𝑛+𝑘+1=𝛽𝑛+𝑘1𝛽𝑛1𝑄𝑛,𝑛+𝑘𝜆𝐼𝑛+𝑇+𝑁12𝑄1,𝑘𝑄1,𝑘𝑇12+𝑇13𝑄1,𝑘1𝑄1,𝑘1𝑇13𝑇++1,𝑘+1𝑄1𝑄1𝑇1,𝑘+1.(3.19) Therefore, by Lemma 3.4, 𝛽𝑛+𝑘𝛽𝑛𝑄𝑛+1,𝑛+𝑘+1=𝛽𝑛+𝑘1𝛽𝑛1𝑄𝑛,𝑛+𝑘.(3.20) Computations show 𝑄𝑛,𝑛+𝑘=(𝛽𝑛1/𝛽𝑛+𝑘1)𝑄1,𝑘+1. Since 𝑄 is the form of (3.5), 𝑄𝒜(𝑛1𝑀𝑧). So 𝒜(𝑀𝑓)=𝒜(𝑀𝐵𝑛)𝒜(𝑀).

Corollary 3.5. Let 𝑓=𝐵𝑛𝑅(𝔻) with 𝑅(𝔻) where does not vanish on 𝔻. If there exists 0𝜆 that 𝜆 can be divided by 𝐵𝑛, then 𝒜(𝑀𝑓)=𝒜(𝑀𝐵𝑛)𝒜(𝑀).

By the following lemma, we discuss the commutant of the multiplication operators whose symbols are composite functions in 𝑅(𝔻).

Lemma 3.6. For 𝑇 in (𝑅(𝔻)) and 𝑓 in 𝑅(𝔻) the following are equivalent:(i)𝑇𝒜(𝑀𝑓);(ii)for all 𝛼𝔻, 𝑇𝑘𝛼(𝑓𝑓(𝛼))𝑅(𝔻);(iii)there is a set 𝐽𝔻 such that 𝛼𝐽(1|𝛼|)= and for all 𝛼𝐽, 𝑇𝑘𝛼(𝑓𝑓(𝛼))𝑅(𝔻).

Proof. (i)(ii) Let 𝑇𝒜(𝑀𝑓). For all 𝑔𝑅(𝔻) and 𝛼𝔻,we have (𝑓𝑓(𝛼))𝑔,𝑇𝑘𝛼=𝑇𝑀𝑓𝑔,𝑘𝛼>𝑓(𝛼)<𝑇𝑔,𝑘𝛼=𝑀𝑓𝑇𝑔,𝑘𝛼𝑓(𝛼)𝑇𝑔,𝑘𝛼=𝑓(𝛼)(𝑇𝑔)(𝛼)𝑓(𝛼)(𝑇𝑔)(𝛼)=0.(3.21)
(ii)(iii) Let 𝐽=𝔻.
(iii)(i) Let 𝑇(𝑅(𝔻)) with 𝑇𝑘𝛼(𝑓𝑓(𝛼))𝑅(𝔻) for all 𝛼𝐽. For 𝑔𝑅(𝔻) and 𝛼𝐽, we have 0=(𝑓𝑓(𝛼))𝑔,𝑇𝑘𝛼=𝑇𝑀𝑓𝑔=(𝛼)𝑓(𝛼)(𝑇𝑔)(𝛼)𝑇𝑀𝑓g(𝑀𝛼)𝑓(𝑇𝑔𝛼).(3.22)
Since 𝐽 is not a Blaschke sequence, this means 𝑇𝑀𝑓𝑔=𝑀𝑓𝑇𝑔. Therefore 𝑇𝑀𝑓=𝑀𝑓𝑇.

Lemma 3.7 (see [13]). Suppose 𝑓𝐺Ω is a surjective analytic function and for each 𝜉Ω, 𝑛(𝜉) is the number of points in 𝑓1(𝜉). Then 𝐺||𝑓||2𝑑𝐴=Ω𝑛(𝜉)𝑑𝐴(𝜉).(3.23)

Proposition 3.8. Let 𝑓 be in 𝑅(𝔻) and 𝑓 is analytic on 𝔻. Suppose for each 𝜉𝑓(𝔻), there are 𝑛 points in 𝑓1(𝜉). Then for 𝑔𝑅(𝑓(𝔻)), we have 𝒜(𝑀𝑓)𝒜(𝑀𝑔𝑓).

Proof. By the Embedding Theory of Sobolev space, 𝑔𝑓𝐶(𝔻). Therefore 𝔻||||𝑔(𝑓(𝑧))2𝑑𝐴(𝑧)𝜋𝑔.(3.24) By Lemma 3.7, 𝔻||[]𝑔(𝑓(𝑧))||2𝑑𝐴(𝑧)=𝔻||𝑔(||𝑓(𝑧))2||𝑓(||𝑧)2𝑑𝐴(𝑧)=𝑛𝑓(𝔻)||𝑔||(𝜔)2𝑑𝐴(𝜔)𝑛𝑔𝑅(𝑓(𝔻)).(3.25) Since 𝑓 is analytic on 𝔻, 𝑓(𝑧), 𝑓(𝑧) is bounded on 𝔻. Hence 𝔻||𝑔(||𝑓(𝑧))2||𝑓(||𝑧)4𝑓𝑑𝐴(𝑧)(𝑧)𝔻||𝑔(||𝑓(𝑧))2||𝑓(||𝑧)2𝑓𝑑𝐴(𝑧)𝑛(𝑧)𝑔𝑅(𝑓(𝔻)),𝔻||𝑔||(𝑓(𝑧))2||𝑓||(𝑧)2𝑓𝑑𝐴(𝑧)(𝑧)𝑔𝑅(𝑓(𝔻)).(3.26) Therefore, 𝔻||[]𝑔(𝑓(𝑧))||2𝑑𝐴(𝑧)<.(3.27) By (3.24), (3.25), and (3.27), we have 𝑔𝑓𝑅(𝔻).
For all 𝛽𝑓(𝔻), because ((𝑔(𝑧)𝑔(𝛽))/(𝑧𝛽))𝑅(𝑓(𝔻)), ((𝑔𝑓(𝑧)𝛽)/(𝑓(𝑧)𝛽))𝑅(𝔻). For all 𝛼𝔻𝑔(𝑓(𝑧))𝑔(𝑓(𝛼))=𝑔(𝑓(𝑧))𝑔(𝑓(𝛼))(𝑓(𝑧)𝑓(𝛼)𝑓(𝑧)𝑓(𝛼)),(3.28) so that (𝑔(𝑓(z))𝑔(𝑓(𝛼)))𝑅(𝔻)(𝑓(𝑧)𝑓(𝛼))𝑅(𝔻).(3.29)
Set 𝑇𝒜(𝑀𝑓). By Lemma 3.6, for all 𝛼𝔻, 𝑇𝑘𝛼(𝑓𝑓(𝛼))𝑅(𝔻). Hence, 𝑇𝑘𝛼(𝑔(𝑓(𝑧))𝑔(𝑓(𝛼)))𝑅(𝔻) and we have 𝑇𝒜(𝑀𝑔𝑓).

Corollary 3.9. If 𝐵𝑛(z)=𝑛𝑖=1(𝑧𝑎𝑖)/(1𝑎𝑖𝑧), (𝑎𝑖𝑎𝑗,𝑖𝑗,|𝑎𝑖|<1) and 𝑓𝑅(𝔻), 𝒜(𝑀𝐵𝑛)𝒜(𝑀𝑓𝐵𝑛).

Proposition 3.10. Let 𝑓 be in 𝑅(𝔻) and 𝐵𝑛 is an 𝑛-Blaschke product. If 𝒜(𝑀𝑓)=𝒜(𝑀𝑧)={𝑀𝑔;𝑔𝑅(𝔻)}, then 𝒜(𝑀𝑓𝐵𝑛)=𝒜(𝑀𝐵𝑛).

Proof. From [3], we know that 𝒜(𝑀𝑧)={𝑀𝑔;𝑔𝑅(𝔻)}. By Lemma 3.3, 𝑀𝐵𝑛𝑛1𝑀𝑧. Then there exists an invertible operator 𝑋 in (𝑅(𝔻),𝑛1𝑅(𝔻)) such that 𝑋𝑀𝐵𝑛𝑋1=𝑛1𝑀𝑧. Since 𝑀𝑓𝐵𝑛=𝑓(𝑀𝐵𝑛), we have 𝑋𝑀𝑓𝐵𝑛𝑋1=𝑛1𝑀𝑓. Therefore, we will only prove that 𝒜(𝑛1𝑀𝑓)=𝒜(𝑛1𝑀𝑧). Set 𝑇𝒜(𝑛1𝑀𝑓) and 𝑇𝑇=11𝑇12𝑇1𝑛𝑇12𝑇13𝑇2𝑛𝑅(𝔻)𝑅(𝔻).(3.30) Since 𝑇𝑛1𝑀𝑓=𝑇11𝑇12𝑇1𝑛𝑇12𝑇13𝑇2𝑛𝑀𝑓00𝑀𝑓=0𝑛1𝑀𝑓𝑀𝑇=𝑓00𝑀𝑓𝑇011𝑇12𝑇1𝑛𝑇12𝑇13𝑇2𝑛,(3.31) we have 𝑇𝑖𝑗𝑀𝑓=𝑀𝑓𝑇𝑖𝑗 for 𝑖,𝑗=1,2,. So 𝑇𝑖𝑗𝒜(𝑀𝑓)=𝒜(𝑀𝑧) and 𝒜(𝑛1𝑀𝑓)𝒜(𝑛1𝑀𝑧). Similarly, 𝒜(𝑛1𝑀𝑧)=𝒜(𝑛1𝑀𝑓). Hence 𝒜(𝑀𝑓𝐵𝑛)=𝒜(𝑀𝐵𝑛).

Let 𝑓 be an injective function in 𝑅(𝔻) and Ω=𝑓(𝔻). Then for each 𝑧0𝔻, it is obvious that 𝑓(𝑧0) is not in 𝑓(𝕋). Denote the component of 𝜌𝑠𝐹(𝑀𝑓) containing 𝑓(𝑧0) as Ω, then 𝑧0 is the only zero point of 𝑓(𝑧)𝑓(𝑧0) in 𝔻. By Lemma 2.2, 𝑀𝑓 is a Cowen-Douglas operator with index 1. By Lemma 2.5, we have 𝒜(𝑀𝑓)={𝑀𝑓;𝑓𝑅(𝔻)}=𝒜(𝑀𝑧). So the following corollary is obtained.

Corollary 3.11. Let 𝑓 be a univalent analytic function in 𝑅(𝔻) and 𝐵𝑛 be an 𝑛-Blaschke product. Then 𝒜(𝑀𝐵𝑛)=𝒜(𝑀𝑓𝐵𝑛).

Lemma 3.6 shows if 𝑓 is in 𝑅(𝔻) and 𝑧0 is in 𝔻, then 𝑇𝑘𝑧0𝑇𝒜𝑀𝑓𝑧𝑓𝑓0𝑅(𝔻)=ker𝑀𝑧𝑓𝑓0.(3.32) Easy examples show that ker𝑀𝑓𝑓(𝑧0){𝑇𝑘𝑧0𝑇𝒜(𝑀𝑓)} is not true. The following proposition shows that if 𝑓(𝑧)=𝑧𝑛, {𝑇𝑘𝑧0𝑇𝒜(𝑀𝑓)}=ker𝑀𝑓𝑓(𝑧0).

Lemma 3.12 (see [1]). Let 𝑓𝑅(𝔻),𝑀𝑓𝑛(Ω),𝑧0𝐷1=𝑓1(Ω), and 𝑧𝑓(𝑧)𝑓0=𝑧𝑧01𝑧𝑧12𝑧𝑧𝑙𝑙+1𝑔𝑧0(𝑧),(3.33) where {𝑧𝑖}𝑙𝑖=1𝐷1 are pairwise distinct, 𝑙+1𝑖=1𝑖=𝑛,𝑔𝑧0(𝑧)0,𝑧𝔻. Choose 𝑘1𝑧𝑖,,𝑘i+1𝑧1𝑖𝑅(𝔻) such that 𝑀𝑧𝑧𝑖𝑘1𝑧𝑖=𝑘𝑧𝑖,,𝑀𝑧𝑧𝑖𝑘𝑖+1𝑧1𝑖=𝑘𝑖+1𝑧2𝑖,(0𝑖𝑙).(3.34) Then there exists a linearly independent set 𝐾𝑧0𝑘=𝑧0,𝑘1𝑧0,,𝑘1𝑧10,𝑘𝑧1,,𝑘2𝑧11,,𝑘𝑙+1𝑧1𝑙,(3.35) such that ker𝑀𝑓𝑓(𝑧0)=𝐾𝑧0.

Let 𝑛2 and 𝜔 be the 𝑛th root of 1, that is, 𝜔 and 𝜔𝑛=1. Let Δ𝑛 denote the Vandermonde determinant of order 𝑛: Δ𝑛=||||||||||||11111𝜔𝜔2𝜔𝑛11𝜔𝑛1𝜔2(𝑛1)𝜔(𝑛1)2||||||||||||.(3.36) For 1𝑖, 𝑗𝑛, the (𝑖,𝑗)-cofactor will be denoted by Δ𝑖𝑗.

Lemma 3.13 (see [1]). 𝐴𝒜(𝑀𝑧𝑛) if and only if for all 𝑔𝑅(𝔻) and 𝑧0, (𝐴𝑔)(𝑧)=𝑛𝑖=1𝛼𝑖𝜔(𝑧)𝑔𝑖1𝑧,(3.37) where 𝛼𝑖(𝑧)=(𝑛𝑗=1Δ𝑖𝑗(𝑗/𝑧𝑗1))/Δ𝑛, for some {𝑗}𝑛𝑗=1 in 𝑅(𝔻).

Proposition 3.14. For all 𝑧0𝔻, {𝑇𝑘𝑧0𝑇𝒜(𝑀𝑧𝑛)}=[(𝑧𝑛𝑧𝑛0)𝑅(𝔻)].

Proof. By Lemma 3.6, {𝑇𝑘𝑧0𝑇𝒜(𝑀𝑧𝑛)}[(𝑧𝑛𝑧𝑛0)𝑅(𝔻)]. Now we prove that [(𝑧𝑛𝑧𝑛0)𝑅(𝔻)]{𝑇𝑘𝑧0𝑇𝒜(𝑀𝑧𝑛)}.
Since 𝑧𝑛𝑧𝑛0=𝑧𝑧0𝑧𝜔𝑧0𝑧𝜔𝑛1𝑧0,(3.38) by Lemma 3.12, 𝑧𝑛𝑧𝑛0𝑅(𝔻)=ker𝑀𝑧𝑛𝑧𝑛0=𝑘𝑧0,𝑘𝜔𝑧0,,𝑘𝜔𝑛1𝑧0.(3.39) Set 𝑓=𝑎1𝑘𝑧0+𝑎2𝑘𝜔𝑧0++𝑎𝑛𝑘𝜔𝑛1𝑧0ker𝑀𝑧𝑛𝑧𝑛0 with 𝑎1,𝑎2,,𝑎𝑛. For all 𝑔𝑅(𝔻), we define an operator 𝑇𝑅(𝔻)𝑅(𝔻) as follows: 𝑇𝑔(𝑧)=𝑎1𝑔(𝑧)+𝑎2𝑔(𝜔𝑧)++𝑎𝑛𝑔𝜔𝑛1𝑧.(3.40) By Lemma 3.13, 𝑇𝒜(𝑀𝑧𝑛). For all 𝜆𝔻, 𝑇𝑘𝑧0𝑇(𝜆)=𝑘𝑧0,𝑘𝜆=𝑘𝑧0,𝑇𝑘𝜆=𝑘𝑧0,𝑎1𝑘𝜆(𝑧)+𝑎2𝑘𝜆(𝜔𝑧)++𝑎𝑛𝑘𝜆𝜔𝑛1𝑧=𝑎1𝑘𝜆(𝑧),𝑘𝑧0+𝑎2𝑘𝜆(𝜔𝑧),𝑘𝑧0++𝑎𝑛𝑘𝜆𝜔𝑛1𝑧,𝑘𝑧0=𝑎1𝑘𝜆𝑧0+𝑎2𝑘𝜆𝜔𝑧0++𝑎𝑛𝑘𝜆𝜔𝑛1𝑧0.(3.41) For 0𝑖𝑛1, 𝑘𝜆𝜔𝑖𝑧0=𝑚=0𝛽2𝑚𝜆𝑚𝜔𝑖𝑧0𝑚=𝑚=0𝛽2𝑚𝜔𝑖𝑧0𝑚𝜆𝑚=𝑘𝜔𝑖𝑧0(𝜆).(3.42) Therefore, 𝑇𝑘𝑧0(𝜆)=𝑎1𝑘𝑧0(𝜆)+𝑎2𝑘𝜔𝑧0(𝜆)++𝑎𝑛𝑘𝜔𝑛1𝑧0(𝜆)=𝑓(𝜆),(3.43) that is 𝑇𝑘𝑧0=𝑓. Then 𝑓{𝑇𝑘𝑧0𝑇𝒜(𝑀𝑧𝑛)} and we have [(𝑧𝑛𝑧𝑛0)𝑅(𝔻)]{𝑇𝑘𝑧0𝑇𝒜(𝑀𝑧𝑛)}.

Easy examples show that, in general, the converse of Proposition 3.8 is false. But the following case is an exception. To prove it, we need the following lemma.

Lemma 3.15. Let 𝑓=(𝑧𝑛) be in 𝑅(𝔻) and , analytic on 𝔻. Then is in 𝑅(𝔻).

Proof. Since is analytic on 𝔻, we have (𝑧)=𝑚=0𝑚𝑧𝑚, hence, 𝑓(𝑧)=(𝑧𝑛)=𝑚=0𝑚𝑧𝑛𝑚.(3.44) From 𝑓 being in 𝑅(𝔻), we have 𝑚=0||𝑚||2𝛽2𝑛𝑚<+.(3.45) Because {𝛽𝑛} is monotonically decreasing, 𝛽𝑚𝛽𝑛𝑚 for all 𝑚0. So 𝑚=0||𝑚||2𝛽2𝑚<𝑚=0||𝑚||2𝛽2𝑛𝑚<+,(3.46) and this shows that is in 𝑅(𝔻).

Proposition 3.16. If 𝑓𝑅(𝔻) and 𝒜(𝑀𝑧𝑛)𝒜(𝑀𝑓), then there exists being in 𝑅(𝔻) such that 𝑓=(𝑧𝑛).

Proof. By Proposition 3.14, for all 𝑧0𝔻, we have 𝑧𝑛𝑧𝑛0𝑅(𝔻)=𝑇𝑘𝑧0𝑇𝒜𝑀𝑧𝑛𝑇𝑘𝑧0𝑇𝒜𝑀𝑓𝑧𝑓𝑓0𝑅(𝔻).(3.47) For each 𝜆𝔻, we can find 𝑧0𝔻 such that 𝑧𝑛0=𝜆. We define on 𝔻 by (𝜆)=𝑓(𝑧0) and is well defined. Indeed, set 𝑧𝑛0=𝑧𝑛1=𝜆. Then 𝑘𝑧1𝑧𝑛𝑧𝑛0𝑅(𝔻)𝑧𝑓𝑓0𝑅(𝔻).(3.48) Hence 𝑓(𝑧1)=𝑓(𝑧0).
For 0𝜆0𝔻, we have 𝑧00. Therefore, lim𝜆𝜆0𝜆(𝜆)0𝜆𝜆0=lim𝑧𝑧0𝑓𝑧(𝑧)𝑓0𝑧𝑛𝑧𝑛0=lim𝑧𝑧0𝑓𝑧(𝑧)𝑓0𝑧𝑧0𝑧𝑧0𝑧𝑛𝑧𝑛0=𝑓𝑧0𝑛𝑧0𝑛1.(3.49) If 𝜆=0, 𝑧0=0. Since 𝑧𝑛𝑅(𝔻)(𝑓𝑓(0))𝑅(𝔻),(3.50) there exists 𝑔𝑅(𝔻) such that 𝑓𝑓(0)=𝑧𝑛𝑔. Hence, lim𝜆0(𝜆)(0)𝜆0=lim𝑧0𝑓(𝑧)𝑓(0)𝑧𝑛=lim𝑧0𝑧𝑛𝑔(𝑧)𝑧𝑛=𝑔(0).(3.51) So is analytic on 𝔻. By Lemma 3.15, we have 𝑅(𝔻) and 𝑓(𝑧)=(𝑧𝑛).

For each 𝑓𝑅(𝔻) and 𝑎𝑓(𝜕𝔻), 𝜂(𝑓(𝜕𝔻),𝑎) denote the winding number of 𝑓(𝜕𝔻) at 𝑎. Define 𝑠=𝑘(𝑓)=inf{𝜂(𝑓(𝜕𝔻),𝑎)𝜂(𝑓(𝜕𝔻),𝑎)0}.(3.52)

Proposition 3.17. If 𝑓𝑅(𝔻) is a nonconstant entire function and 𝑠=𝑘(𝑓), then 𝒜(𝑀𝑓)=𝒜(𝑀𝑧𝑠).

Proof. By Theorem  1 in [14], there exists an entire function such that 𝑓(𝑧)=(𝑧𝑠) and 𝑘()=1. Since is an entire function, , , and are all bounded and analytic on 𝔻. So 𝑅(𝔻). By 𝑘()=1, there is only one zero of 𝜆 in 𝔻 for some 𝜆. By Lemma 2.2, 𝑀1(Ω). By Proposition 3.10, 𝒜(𝑀𝑓)=𝒜(𝑀(𝑧𝑠))=𝒜(𝑀𝑧𝑠).

Acknowledgment

The research is supported by NSFC 11001078.