# A Similarity Invariant and the Commutant of some Multiplication Operators on the Sobolev Disk Algebra

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

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## Abstract

LetR(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 ℒ(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.


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## Body

## 1. Introduction

LetΩ be an analytic Cauchy domain in the complex plane and let W22(Ω) denote the Sobolev space,
(1.1)W22(Ω)={f∈L2(Ω,dA):thedistributionalpartialderivativesoffirstandsecondorderoffbelongtoL2(Ω,dA)},dA denotes the planar Lebesgue measure. For f,g∈W22(Ω), we define
(1.2)〈f,g〉=∑|α|≤2‍∫‍DαfDαg¯dA.
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.LetR(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 f∈R(𝔻), the multiplication operator Mf on R(𝔻) is defined by Mf(g)=fg, g∈R(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 spaceR(𝔻) has an orthogonal basis {en}n=0+∞, where
(1.3)en=βnzn,βn=[n+1(3n4-n2+2n+1)π]1/2,n=0,1,2,….
(ii) As a functional Hilbert space,R(𝔻) has reproducing kernel which is
(1.4)k(u,v)=∑n=0+∞‍βn2unv¯n.
Then for z0∈𝔻¯,
(1.5)kz0=∑n=0∞‍(βn2z¯0n)zn.
(iii) Iff(z)=∑n=0+∞fnzn is analytic on 𝔻, then f(z)∈R(𝔻) if and only if ∑n=0+∞|fn|2/βn2<+∞.
(iv) The operatorMf admits the following matrix representation with respect to {en}n=0+∞:
(1.6)Mf=(c0c1β0β1c00c2β0β2c1β1β2c0c3β0β3c2β1β3c1β2β3c0⋮⋮⋱⋱⋱).Note thatR(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 [1–3].Letℋ be a complex separable Hilbert space and ℒ(ℋ) denote the collection of bounded linear operators on ℋ. One of the basic problems in operator theory is to determine when two operators A and B in ℒ(ℋ) are similar. A quantity (quantities) or a property (properties) P is similarity invariant (invariants) if A has P and A~B 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, [4–6]. 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 ℒ(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 [1–3]). 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.ForT in ℒ(ℋ), 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 ℒ(ℋ) is said to be a Cowen-Douglas operator with index n if there exists Ω, a connected open subset of complex plane ℂ, and n, a positive integer, such that(i)
Ω⊂σ(A)={λ∈ℂ:A-λisnotinvertible};(ii)
ran(A-λ):={y;(A-λ)x=y,x∈ℋ}=ℋ for λ in Ω;(iii)
nul(A-λ):=dimker(A-λ)=n for λ in Ω;(iv)
⋁{ker(A-λ):λ∈Ω}=ℋ,where (iv) is equivalent to (iv)′ ([8]);(iv)′
there existsλ0 in Ω, such that ⋁{ker(A-λ0)k:k≥1}=ℋ.ℬ
n
(
Ω
) denotes the collection of Cowen-Douglas operators with index n.ForT∈ℒ(ℋ), the set of operators which commute with it is 𝒜′(T). That is 𝒜′(T)={A∈ℒ(ℋ):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 ℋ.Definition 2.1.
Letℋ be a Hilbert space and A,B be in ℒ(ℋ). τAB is said to be a Rosenblum operator on ℒ(ℋ) if for arbitrary C∈ℒ(ℋ),τAB(C)=AC-CB.Lemma 2.2 (see [1]).
Letf 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)
letz0∈𝔻 and f(z0)∉f(∂𝔻). Denote the component of ρs-F(Mf) containing f(z0) as Ω, then Mf*∈ℬn(Ω), where n is the number of the zeros of f(z)-f(z0) in 𝔻.Lemma 2.3 (see [10]).
Setf∈R(𝔻) and Ba(z)=(z-a)/(1-a¯z),a∈𝔻. Then Mf∘Ba~Mf.Theorem 2.4.
Letf and g be in R(𝔻) and be univalent and analytic on 𝔻. Then Mf~Mg if and only if f(𝔻)=g(𝔻).Proof.
“⇒”: Set Mf~Mg. By Lemma 2.2, we have
(2.1)f(𝔻¯)=σ(Mf)=σ(Mg)=g(𝔻¯),f(𝕋)=σe(Mf)=σe(Mg)=g(𝕋),
where 𝕋 is the unit circle. Since f and g are univalent and analytic on 𝔻, then
(2.2)f(𝕋)=∂f(𝔻¯)=∂g(𝔻¯)=g(𝕋).
Therefore,
(2.3)f(𝔻)=f(𝔻¯)∖f(𝕋)=g(𝔻¯)∖g(𝕋)=g(𝔻).
“⇐”: Set f(𝔻)=g(𝔻)=Ω. Because g is univalent analytic from 𝔻 to Ω, g-1:Ω→𝔻 is also univalent analytic. Then g-1∘f is injective and surjective analytic function on 𝔻. If g-1∘f(z0)=0, there exists a Möbius transform φ with φz0=(z-z0)/(1-z0¯z) and a complex number c with |c|=1 such that g-1∘f(z)=cφz0 (see [11]). Therefore f(z)=g(cφz0(z)). By Lemma 2.3, Mf~Mg.Lemma 2.5 (see [3]).
Givenf∈R(𝔻), the following are equivalent:(i)
Mf*∈ℬ1(Ω);(ii)
𝒜′(Mf)={Mg:g∈R(𝔻)};(iii)
Mf∈(SI).Theorem 2.6.
Letf,g∈R(𝔻) and f is univalent analytic on 𝔻. Mf~Mg if and only if there exists a function χ=c((z-z0)/(1-z0¯z)) such that f=g∘χ, where z0∈𝔻 and c∈ℂ,|c|=1.Proof.
“⇒”: Suppose that g is not univalent on 𝔻. There exists some λ∈g(𝔻) such that the number of zeros of g(z)-λ on 𝔻 is n>1. By Lemma 2.2, Mg*∈ℬn(Ω) where Ω is a connected open subset of g(𝔻). Since Mf~Mg, we have Mf*∈ℬn(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((z-z0)/(1-z0¯z)) such that f=g∘χ, where z0∈𝔻 and c∈ℂ,|c|=1.
“⇐”: By the conditions of the theorem, g is univalent analytic on 𝔻. Since f(𝔻)=g∘χ(𝔻)=g(𝔻), we have Mf~Mg by Theorem 2.4.For any operatorT on Hilbert space ℋ and any integer n,1≤n≤∞, let T(n) denote the direct sum of n copies of T on ℋ(n)=ℋ⊕⋯⊕ℋ.Lemma 2.7 (see [2]).
LetA1,A2 be strongly irreducible Cowen-Douglas operators. Assume that A1≁A2 and T=A1(n1)⊕A2(n2), where n1,  n2 are natural numbers. Then for each maximal ideal 𝒥 of 𝒜′(T), 𝒥 must be one of the following two forms:
(2.4)(i)𝒥=(𝒥11kerτA1(n1),A2(n2)kerτA2(n2),A1(n1)𝒜′(A2(n2)));(ii)𝒥=(𝒜′(A1(n1))kerτA1(n1),A2(n2)kerτA2(n2),A1(n1)𝒥22),
where 𝒥ii is a maximal ideal of 𝒜′(Ai(ni)), i=1,2.Theorem 2.8.
Letf,g∈R(𝔻) and Mf,Mg∈(SI). The following statements are equivalent:(i)
Mf~Mg;(ii)
there existX1,X2,…,Xn∈kerτMf,Mg and Y1,Y2,…,Yn∈kerτMg,Mf such that
(2.5)X1Y1+X2Y2+⋯+XnYn=idR(𝔻),Y1X1+Y2X2+⋯+YnXn=idR(𝔻),where idR(𝔻) denotes the identity operator of R(𝔻).Proof.
(
i
)
⇒
(
ii
)
: Let Mf~Mg. Set X be invertible in ℒ(R(𝔻)) and MfX=XMg. Then X and X-1 are what we want.
(
ii
)
⇒
(
i
)
: Since Mf and Mg are in (SI), we have Mf* and Mg* that are strongly irreducible and in ℬ1(Ω) by Lemma 2.5. Computations show
(2.6)𝒜′(Mf*⊕Mg*)=(𝒜′(Mf*)kerτMf*,Mg*kerτMg*,Mf*𝒜′(Mg*)).
Suppose that Mf*≁Mg*. By Lemma 2.7, each maximal ideal 𝒥 of 𝒜′(Mf*⊕Mg*) must be one of the following two forms
(2.7)𝒥=(𝒥11kerτMf*,Mg*kerτMg*,Mf*𝒜′(Mg*))or𝒥=(𝒜′(Mf*)kerτMf*,Mg*kerτMg*,Mf*𝒥22),
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
(2.8)(0Yi*Xi*0)∈𝒥,i=1,2,…,n(0Yi*Xi*0)(0Yi*Xi*0)=(Yi*Xi*00Xi*Yi*)∈𝒥.
It follows that
(2.9)∑i=1n‍(Yi*Xi*00Xi*Yi*)=(idR(𝔻)00idR(𝔻))∈𝒥.
This contradicts that 𝒥 is a maximal ideal. So Mf*~Mg* and Mf~Mg.


## 3. The Commutant Algebra of Multiplication Operator

In [3], Liu and Wang give the following proposition.Proposition 3.1.
Letf(z)=znh(z)∈R(𝔻), h(z)≠0, z∈𝔻¯. Then 𝒜′(Mf)=𝒜′(Mzn)∩𝒜′(Mh).LetBn(z)=∏k=1n(z-ak)/(1-a¯kz), (|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)h(z). The following theorem is obtained, and by this result, the above proposition is generalized.Theorem 3.2.
Letf=Bnh∈R(𝔻) with h∈R(𝔻) where h does not vanish on 𝔻¯. If there exists 0≠λ∈ℂ such that h-λ can be divided by each (z-ak)/(1-a¯kz), then 𝒜′(Mf)=𝒜′(MBn)∩𝒜′(Mh).To prove the above theorem, we need the following lemmas.Lemma 3.3 (see [7]).
Giveng∈R(𝔻),Mg~Mzn(~(Mz)n) if and only if g is an n-Blaschke product.Lemma 3.4 (see [12]).
LetN be a nilpotent operator on H and let X0=λ+N,0≠λ∈ℂ. If B,A0,A1,…∈B(H) satisfy ∥Ak∥≤M(k=0,1,2,…) and AkX0=X0Ak-1+B,(k=1,2,3…), then A0=A1=A2=⋯.Now we will prove Theorem3.2.Proof.
Fromh∈R(𝔻), we have Mf*=MBn*Mh*. Then, from Lemma 3.3, there exists an invertible operator X∈ℒ(R(𝔻),⊕1nR(𝔻)) such that XMBn*X-1=⊕1nMz*. It follows that
(3.1)A=XMBn*X-1XMh*X-1=(⊕1nMz*)T,
where T=XMh*X-1∈𝒜′(⊕1nMz*). So we only need to prove
(3.2)𝒜′(A)=𝒜′(⊕1nMz*)∩𝒜′(T).
Since(3.3)⊕1nR(𝔻)=ker(⊕1nMz*)⊕[ker(⊕1nMz*)2-ker(⊕1nMz*)]⊕[ker(⊕1nMz*)3-ker(⊕1nMz*)2]⊕⋯⊕[ker(⊕1nMz*)k-ker(⊕1nMz*)k-1]⊕⋯
we have
(3.4)⊕1nMz*=(0β0β1In00⋯00β1β2In0⋯000β2β3In⋯⋮⋮⋮⋮⋱),
where In is an n×n identity matrix. Computations show that G∈𝒜′(⊕1nMz*) if and only if
(3.5)G=(G1β0β1G12β0β2G13β0β3G14⋯β0βk-1G1,k⋯0G1β1β2G12β1β3G13⋯β1βk-1G1,k-1⋯00G1β2β3G12⋯β2βk-1G1,k-2⋯⋮⋮⋮⋮⋱⋮⋯0000⋯βk-2βk-1G1,2⋯⋮⋮⋮⋮⋯⋮⋱).
Therefore,
(3.6)T=(T1β0β1T12β0β2T13β0β3T14⋯β0βk-1T1,k⋯0T1β1β2T12β1β3T13⋯β1βk-1T1,k-1⋯00T1β2β3T12⋯β2βk-1T1,k-2⋯⋮⋮⋮⋮⋱⋮⋯0000⋯βk-2βk-1T1,2⋯⋮⋮⋮⋮⋯⋮⋱).
Setgk=(z-ak)/(1-ak¯z),1≤k≤n, and Bn=g1g2⋯gn. Then for each gk, there exists φk∈R(𝔻) such that h-λ=gkφk,1≤k≤n. So
(3.7)T=XMh*X-1=XMgk*X-1XMφk*X-1+λ¯id⊕1nR(𝔻)=VkSk+λ¯id⊕1nR(𝔻),
where Vk=XMgk*X-1,Sk=XMφk*X-1. 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(⊕1nMz*) and
(3.8)ker(⊕1nMz*)=kerXMBn*X-1=kerV1V2⋯Vn
for all x∈ker(⊕1nMz*),
(3.9)(T-λ¯id⊕1nR(𝔻))nx=V1S1V2S2⋯VnSnx=S1S2⋯SnV1V2⋯Vnx=0.
Therefore, T1-λ¯In is nilpotent operator. Now we set N=T1-λ¯In, that is T1=λ¯In+N. So
(3.10)A=⊕1nMz*T=(0β0β1(λ¯In+N)β0β2T12β0β3T13⋯β0βkT1,k⋯00β1β2(λ¯In+N)β1β3T12⋯β1βkT1,k-1000β2β3(λ¯In+N)⋯β2βkT1,k-2⋯⋮⋮⋮⋮⋱⋮⋯0000⋯βk-1βk(λ¯In+N)⋯⋮⋮⋮⋮⋯⋮⋱).
So we only need to prove that𝒜′(A)⊂𝒜′(⊕1nMz*). In fact, if Q∈𝒜′(A)⊂𝒜′(⊕1nMz*),
(3.11)QA=AQ⇒QT(⊕1nMz*)=T(⊕1nMz*)Q⇒QT(⊕1nMz*)=TQ(⊕1nMz*)⇒(QT-TQ)(⊕1nMz*)=0⇒(⊕1nMz)(T*Q*-Q*T*)=0.
It follows from ker(⊕1nMz)={0} that T*Q*=Q*T*. Namely, TQ=QT.
We are now in need to prove that𝒜′(A)⊂𝒜′(⊕1nMz*). Suppose that Q∈𝒜′(A). Since h does not vanish on 𝔻¯, ranMh¯=R(𝔻) and so kerT={0}. Because A=(⊕1nMz*)T=T(⊕1nMz*),
(3.12)kerA=ker(⊕1nMz*),kerAk=ker(⊕1nMz*)k.
It follows from QA=AQ that kerA and kerAk are both the invariant subspaces of Q. Since
(3.13)⊕1nR(𝔻)=kerA⊕(kerA2⊖kerA)⊕(kerA3⊖kerA2)⊕⋯⊕(kerAk⊖kerAk-1)⊕⋯,A admits the matrix representation (3.5) with the above decomposition. So
(3.14)Q=(Q1Q12Q13Q14⋯0Q2Q23Q24⋯00Q3Q34⋯⋮⋮⋮⋮⋮).
From QA=AQ, we have
(3.15)(0β0β1Q1(λ¯In+N)β0β2Q1T12+β1β2Q12(λ¯In+N)⋯00β1β2Q2(λ¯In+N)⋯⋮⋮⋮⋮)=(0β0β1(λ¯In+N)Q2β0β1(λ¯In+N)Q23+β0β2T12Q3⋯00β1β2(λ¯In+N)Q3⋯⋮⋮⋮⋮).
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
(3.16)βn+1βn(λ¯In+N)Qn+1,n+2=βnβn-1Qn,n+1(λ¯In+N)+(Q1T12-T12Q1).
It follows from Lemma 3.4 that
(3.17)βn+1βnQn+1,n+2=βnβn-1Qn,n+1.
Setting Q12′=(β1/β0)Q12, we have
(3.18)Qn,n+1=β1βn-1β0βnQ12=βn-1βnQ12′.
Inductively, if Qn,n+k-1=(βn-1/βn+k-2)Q1,k′, where Q1,k′=(βk-1/β0)Q1k, we need to prove Qn,n+k=(βn-1/βn+k-1)Q1,k+1′. Comparing the (n,n+k+1) entries of QA and AQ, we have
(3.19)βn+kβn(λ¯In+N)Qn+1,n+k+1=βn+k-1βn-1Qn,n+k(λ¯In+N)+(T12Q1,k′-Q1,k′T12)+(T13Q1,k-1′-Q1,k-1′T13)+⋯+(T1,k+1Q1-Q1T1,k+1).
Therefore, by Lemma 3.4,
(3.20)βn+kβnQn+1,n+k+1=βn+k-1βn-1Qn,n+k.
Computations show Qn,n+k=(βn-1/βn+k-1)Q1,k+1′. Since Q is the form of (3.5), Q∈𝒜′(⊕1nMz*). So 𝒜′(Mf)=𝒜′(MBn)∩𝒜′(Mh).Corollary 3.5.
Letf=Bnh∈R(𝔻) with h∈R(𝔻) where h does not vanish on 𝔻¯. If there exists 0≠λ∈ℂ that h-λ can be divided by Bn, then 𝒜′(Mf)=𝒜′(MBn)∩𝒜′(Mh).By the following lemma, we discuss the commutant of the multiplication operators whose symbols are composite functions inR(𝔻).Lemma 3.6.
ForT in ℒ(R(𝔻)) and f in R(𝔻) the following are equivalent:(i)
T∈𝒜′(Mf);(ii)
for allα∈𝔻, T*kα⊥(f-f(α))R(𝔻);(iii)
there is a setJ⊂𝔻 such that ∑α∈J(1-|α|)=∞ and for all α∈J, T*kα⊥(f-f(α))R(𝔻).Proof.
(
i
)
⇒
(
ii
) Let T∈𝒜′(Mf). For all g∈R(𝔻) and α∈𝔻,we have
(3.21)〈(f-f(α))g,T*kα〉=〈TMfg,kα>-f(α)<Tg,kα〉=〈MfTg,kα〉-f(α)〈Tg,kα〉=f(α)(Tg)(α)-f(α)(Tg)(α)=0.
(
ii
)
⇒
(
iii
) Let J=𝔻.
(
iii
)
⇒
(
i
) Let T∈ℒ(R(𝔻)) with T*kα⊥(f-f(α))R(𝔻) for all α∈J. For g∈R(𝔻) and α∈J, we have
(3.22)0=〈(f-f(α))g,T*kα〉=(TMfg)(α)-f(α)(Tg)(α)=(TMfg)(α)-(MfTg)(α).
SinceJ is not a Blaschke sequence, this means TMfg=MfTg. Therefore TMf=MfT.Lemma 3.7 (see [13]).
Supposef:G→Ω is a surjective analytic function and for each ξ∈Ω, n(ξ) is the number of points in f-1(ξ). Then
(3.23)∫G|f′|2dA=∫Ωn(ξ)dA(ξ).Proposition 3.8.
Letf be in R(𝔻) and f is analytic on 𝔻¯. Suppose for each ξ∈f(𝔻), there are n points in f-1(ξ). Then for g∈R(f(𝔻)), we have 𝒜′(Mf)⊂𝒜′(Mg∘f).Proof.
By the Embedding Theory of Sobolev space,g∘f∈C(𝔻¯). Therefore
(3.24)∫𝔻‍|g(f(z))|2dA(z)≤π∥g∥∞.
By Lemma 3.7,
(3.25)∫𝔻‍|[g(f(z))]′|2dA(z)=∫𝔻‍|g′(f(z))|2|f′(z)|2dA(z)=n∫f(𝔻)‍|g′(ω)|2dA(ω)≤n∥g∥R(f(𝔻)).
Since f is analytic on 𝔻, f′(z), f′′(z) is bounded on 𝔻. Hence
(3.26)∫𝔻‍|g′′(f(z))|2|f′(z)|4dA(z)≤∥f′(z)∥∞∫𝔻‍|g′′(f(z))|2|f′(z)|2dA(z)≤n∥f′(z)∥∞∥g∥R(f(𝔻)),∫𝔻‍|g′(f(z))|2|f′′(z)|2dA(z)≤∥f′′(z)∥∞∥g∥R(f(𝔻)).
Therefore,
(3.27)∫𝔻‍|[g(f(z))]′′|2dA(z)<∞.
By (3.24), (3.25), and (3.27), we have g∘f∈R(𝔻).
For allβ∈f(𝔻), because ((g(z)-g(β))/(z-β))∈R(f(𝔻)), ((g∘f(z)-β)/(f(z)-β))∈R(𝔻). For all α∈𝔻(3.28)g(f(z))-g(f(α))=g(f(z))-g(f(α))f(z)-f(α)(f(z)-f(α)),
so that
(3.29)(g(f(z))-g(f(α)))R(𝔻)⊂(f(z)-f(α))R(𝔻).
SetT∈𝒜′(Mf). By Lemma 3.6, for all α∈𝔻, T*kα⊥(f-f(α))R(𝔻). Hence, T*kα⊥(g(f(z))-g(f(α)))R(𝔻) and we have T∈𝒜′(Mg∘f).Corollary 3.9.
IfBn(z)=∏i=1n(z-ai)/(1-ai¯z), (ai≠aj,i≠j,|ai|<1) and f∈R(𝔻), 𝒜′(MBn)⊂𝒜′(Mf∘Bn).Proposition 3.10.
Letf be in R(𝔻) and Bn is an n-Blaschke product. If 𝒜′(Mf)=𝒜′(Mz)={Mg;g∈R(𝔻)}, then 𝒜′(Mf∘Bn)=𝒜′(MBn).Proof.
From [3], we know that 𝒜′(Mz)={Mg;g∈R(𝔻)}. By Lemma 3.3, MBn~⊕1nMz. Then there exists an invertible operator X in ℒ(R(𝔻),⊕1nR(𝔻)) such that XMBnX-1=⊕1nMz. Since Mf∘Bn=f(MBn), we have XMf∘BnX-1=⊕1nMf. Therefore, we will only prove that 𝒜′(⊕1nMf)=𝒜′(⊕1nMz). Set T∈𝒜′(⊕1nMf) and
(3.30)T=(T11T12⋯T1nT12T13⋯T2n⋮⋮⋯⋮)R(𝔻)R(𝔻)⋮.
Since
(3.31)T(⊕1nMf)=(T11T12⋯T1nT12T13⋯T2n⋮⋮⋯⋮)(Mf0⋯0Mf⋯0⋱)=(⊕1nMf)T=(Mf0⋯0Mf⋯0⋱)(T11T12⋯T1nT12T13⋯T2n⋮⋮⋯⋮),
we have TijMf=MfTij for i,j=1,2,…. So Tij∈𝒜′(Mf)=𝒜′(Mz) and 𝒜′(⊕1nMf)⊂𝒜′(⊕1nMz). Similarly, 𝒜′(⊕1nMz)=𝒜′(⊕1nMf). Hence 𝒜′(Mf∘Bn)=𝒜′(MBn).Letf be an injective function in R(𝔻) and Ω=f(𝔻). Then for each z0∈𝔻, it is obvious that f(z0) is not in f(𝕋). Denote the component of ρs-F(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;f∈R(𝔻)}=𝒜′(Mz). So the following corollary is obtained.Corollary 3.11.
Letf be a univalent analytic function in R(𝔻) and Bn be an n-Blaschke  product. Then 𝒜′(MBn)=𝒜′(Mf∘Bn).Lemma3.6 shows if f is in R(𝔻) and z0 is in 𝔻, then
(3.32)⋁{T*kz0:T∈𝒜′(Mf)}⊂[(f-f(z0))R(𝔻)]⊥=kerMf-f(z0)*.
Easy examples show that kerMf-f(z0)*⊂⋁{T*kz0:T∈𝒜′(Mf)} is not true. The following proposition shows that if f(z)=zn, ⋁{T*kz0:T∈𝒜′(Mf)}=kerMf-f(z0)*.Lemma 3.12 (see [1]).
Letf∈R(𝔻),Mf*∈ℬn(Ω),z0∈D1:=f-1(Ω), and
(3.33)f(z)-f(z0)=(z-z0)h1(z-z1)h2⋯(z-zl)hl+1gz0(z),
where {zi}i=1l⊂D1 are pairwise distinct, ∑i=1l+1hi=n,gz0(z)≠0,z∈𝔻¯. Choose kzi1,…,kzihi+1-1∈R(𝔻) such that
(3.34)Mz-zi*kzi1=kzi,…,Mz-zi*kzihi+1-1=kzihi+1-2,(0≤i≤l).
Then there exists a linearly independent set
(3.35)Kz0:={kz0,kz01,…,kz0h1-1,kz1,…,kz1h2-1,…,kzlhl+1-1},
such that kerMf-f(z0)*=⋁Kz0.Letn≥2 and ω be the nth root of 1, that is, ω∈ℂ and ωn=1. Let Δn denote the Vandermonde determinant of order n:
(3.36)Δn=|111⋯11ωω2⋯ωn-1⋮⋮⋮⋮1ωn-1ω2(n-1)⋯ω(n-1)2|.
For 1≤i, j≤n, the (i,j)-cofactor will be denoted by Δij.Lemma 3.13 (see [1]).
A
∈
𝒜
′
(
M
z
n
) if and only if for all g∈R(𝔻) and z≠0,
(3.37)(Ag)(z)=∑i=1n‍αi(z)g(ωi-1z),
where αi(z)=(∑j=1nΔij(hj/zj-1))/Δn, for some {hj}j=1n in R(𝔻).Proposition 3.14.
For allz0∈𝔻, ⋁{T*kz0:T∈𝒜′(Mzn)}=[(zn-z0n)R(𝔻)]⊥.Proof.
By Lemma3.6, ⋁{T*kz0:T∈𝒜′(Mzn)}⊂[(zn-z0n)R(𝔻)]⊥. Now we prove that [(zn-z0n)R(𝔻)]⊥⊂⋁{T*kz0:T∈𝒜′(Mzn)}.
Since(3.38)zn-z0n=(z-z0)(z-ωz0)⋯(z-ωn-1z0),
by Lemma 3.12,
(3.39)[(zn-z0n)R(𝔻)]⊥=kerMzn-z0n*=⋁{kz0,kωz0,…,kωn-1z0}.
Set f=a1kz0+a2kωz0+⋯+ankωn-1z0∈kerMzn-z0n* with a1,a2,…,an∈ℂ. For all g∈R(𝔻), we define an operator T:R(𝔻)↦R(𝔻) as follows:
(3.40)Tg(z)=a1¯g(z)+a2¯g(ωz)+⋯+an¯g(ωn-1z).
By Lemma 3.13, T∈𝒜′(Mzn). For all λ∈𝔻,
(3.41)T*kz0(λ)=〈T*kz0,kλ〉=〈kz0,Tkλ〉=〈kz0,a1¯kλ(z)+a2¯kλ(ωz)+⋯+an¯kλ(ωn-1z)〉=a1〈kλ(z),kz0〉¯+a2〈kλ(ωz),kz0〉¯+⋯+an〈kλ(ωn-1z),kz0〉¯=a1kλ(z0)¯+a2kλ(ωz0)¯+⋯+ankλ(ωn-1z0)¯.
For 0≤i≤n-1,
(3.42)kλ(ωiz0)¯=∑m=0∞‍βm2(λ¯)m(ωiz0)m¯=∑m=0∞‍βm2(ωiz0¯)mλm=kωiz0(λ).
Therefore,
(3.43)T*kz0(λ)=a1kz0(λ)+a2kωz0(λ)+⋯+ankωn-1z0(λ)=f(λ),
that is T*kz0=f. Then f∈⋁{T*kz0:T∈𝒜′(Mzn)} and we have [(zn-z0n)R(𝔻)]⊥⊂⋁{T*kz0:T∈𝒜′(Mzn)}.Easy examples show that, in general, the converse of Proposition3.8 is false. But the following case is an exception. To prove it, we need the following lemma.Lemma 3.15.
Letf=h(zn) be in R(𝔻) and h, analytic on 𝔻. Then h is in R(𝔻).Proof.
Sinceh is analytic on 𝔻, we have h(z)=∑m=0∞hmzm, hence,
(3.44)f(z)=h(zn)=∑m=0∞‍hmznm.
From f being in R(𝔻), we have
(3.45)∑m=0∞‍|hm|2βnm2<+∞.
Because {βn} is monotonically decreasing, βm≥βnm for all m≥0. So
(3.46)∑m=0∞‍|hm|2βm2<∑m=0∞‍|hm|2βnm2<+∞,
and this shows that h is in R(𝔻).Proposition 3.16.
Iff∈R(𝔻) and 𝒜′(Mzn)⊂𝒜′(Mf), then there exists h being in R(𝔻) such that f=h(zn).Proof.
By Proposition3.14, for all z0∈𝔻, we have
(3.47)[(zn-z0n)R(𝔻)]⊥=⋁{T*kz0:T∈𝒜′(Mzn)}⊂⋁{T*kz0:T∈𝒜′(Mf)}⊂[(f-f(z0))R(𝔻)]⊥.
For each λ∈𝔻, we can find z0∈𝔻 such that z0n=λ. We define h on 𝔻 by h(λ)=f(z0) and h is well defined. Indeed, set z0n=z1n=λ. Then
(3.48)kz1∈[(zn-z0n)R(𝔻)]⊥⊂[(f-f(z0))R(𝔻)]⊥.
Hence f(z1)=f(z0).
For0≠λ0∈𝔻, we have z0≠0. Therefore,
(3.49)limλ→λ0h(λ)-h(λ0)λ-λ0=limz→z0f(z)-f(z0)zn-z0n=limz→z0f(z)-f(z0)z-z0z-z0zn-z0n=f′(z0)nz0n-1.
If λ=0, z0=0. Since
(3.50)znR(𝔻)⊃(f-f(0))R(𝔻),
there exists g∈R(𝔻) such that f-f(0)=zng. Hence,
(3.51)limλ→0h(λ)-h(0)λ-0=limz→0f(z)-f(0)zn=limz→0zng(z)zn=g(0).
So h is analytic on 𝔻. By Lemma 3.15, we have h∈R(𝔻) and f(z)=h(zn).For eachf∈R(𝔻) and a∉f(∂𝔻), η(f(∂𝔻),a) denote the winding number of f(∂𝔻) at a. Define
(3.52)s=k(f)=inf{η(f(∂𝔻),a):η(f(∂𝔻),a)≠0}.Proposition 3.17.
Iff∈R(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 h such that f(z)=h(zs) and k(h)=1. Since h is an entire function, h, h′, and h′′ are all bounded and analytic on 𝔻¯. So h∈R(𝔻). By k(h)=1, there is only one zero of h-λ in 𝔻 for some λ. By Lemma 2.2, Mh*∈ℬ1(Ω). By Proposition 3.10, 𝒜′(Mf)=𝒜′(Mh(zs))=𝒜′(Mzs).



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*Source: 378217-2012-09-02.xml*