# Reducing Subspaces of the Dual Truncated Toeplitz Operator

Authors: Yinyin Hu; Jia Deng; Tao Yu; Liu Liu; Yufeng Lu
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We define the dual truncated Toeplitz operators and give some basic properties of them. In particular, spectrum and reducing subspaces of some special dual truncated Toeplitz operator are characterized.


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## Body

## 1. Introduction

LetD denote the open unit disk in the complex plane C and T denote the unit circle. As usual, L2 denotes the Hilbert space of Lebesgue square integral functions on T with the inner product: (1)u,v=∫Tuv¯dm,where u,v∈L2, m is the normalized Lebesgue measure. As usual, H2 will be identified with the subspace of L2 consisting of the functions whose Fourier coefficients with negative indices vanish. Let P denote the projection from L2 to H2, which is given explicitly by the Cauchy integral: (2)Pfλ=∫Tfξ1-λ¯ξdmξ,λ∈D.For φ∈L∞, the Toeplitz operator Tφ on H2, with symbol φ, is defined by (3)Tφf=Pφf,where f∈H2. Let S denote the unilateral shift operator on H2. Its adjoint, the backward shift, is given by (4)S∗fz=fz-f0z.For the remainder of the paper,u will denote a nonconstant inner function. The subspace Ku2=H2⊖uH2 is a proper nontrivial invariant subspace of S∗, the most general one by the well-known theorem of A. Beurling [1]. Truncated Toeplitz operators (TTO) are compressions of the standard Toeplitz operators on the Hardy space H2 to its coinvariant, the so-called model space Ku2. Let Pu denote the orthogonal projection from L2 onto the subspace Ku2. For φ∈L2, the truncated Toeplitz operator Aφ with symbol φ is defined by (5)Aφf=Puφf,on the dense subset Ku2∩L∞ of the space Ku2. The symbol is not unique [2].Although the truncated Toeplitz operators share many fundamental properties of classical Toeplitz operators on the Hardy space, they differ in many crucial ways. For example, compact Toeplitz operators on Hardy space are zero, but there are many nonzero compact truncated Toeplitz operators. In part motivated by several of the problems posed in the aforementioned article [2], the area has undergone vigorous development during the past several years [3–22]. While several of the initial questions raised by Sarason have now been resolved, the study of truncated Toeplitz operators has nevertheless proven to be fertile ground, spawning both new questions and unexpected results.Recently, Gu [23] defined truncated Hankel operator (THO) as the compression of Hankel operator to invariant subspaces for the backward shift and proved a number of algebraic properties of them. Some of the properties in his paper reveal the relation between the THOs and TTOs. Later, Kang and Kim [24] characterized the pairs of truncated Hankel operator on the model spaces Ku2 whose products result in truncated Toeplitz operators when the inner function u has a certain symmetric property.In this paper, we will define dual truncated Toeplitz operators and introduce some algebraic properties of them. As is well known, the orthogonal complement ofKu2, Ku2⊥=uH2⊕H¯02, which H02 is the subspace of H2 with f(0)=0. Let φ∈L∞, and the dual truncated Toeplitz operator Bφ on Ku2⊥ with symbol φ is defined by (6)Bφf=I-Pufφ,where f∈Ku2⊥. For λ∈D, the kernel function in Ku2⊥ for the functional of evaluation at λ will be denoted by Kλu, which is (7)Kλuz=λz¯1-λz¯+uλ¯uz1-λ¯z.The paper is organized as follows. In Section2, we give some basic properties of dual truncated Toeplitz operator. In Section 3, we characterize the spectrum and reducing subspaces of some special dual truncated Toeplitz operator.


## 2. Basic Properties

An operatorA on Ku2⊥ is called C-symmetric if CAC=A∗, where C is an antiunitary involution on L2 defined by (Cf)(ξ)=u(ξ)ξfξ¯(ξ=1). In fact, it is easily seen that C maps uH2 to H¯02 and H¯02 to uH2.Lemma 1.
Dual truncated Toeplitz operators are C-symmetric.Proof.
Letf be in L∞ with Bf bounded. For g and g in Ku2⊥, we have (8)CBfCg,h=Ch,BfCg=∫uξξ¯hξ¯fξ¯uξ¯ξgξdmξ=∫hξ¯gξfξ¯dmξ=Bf¯g,h=Bf∗g,h.This completes the proof.In the following, we will discuss the boundedness and compactness of dual Truncated Toeplitz operators.Theorem 2.
Forϕ∈L∞, we have Bϕ=ϕ∞=r(Bϕ).Proof.
Letσap(T) denote the approximate point spectrum of the bounded linear operator T. It is well known that multiplication operators Mϕ on the Hardy space H2, Mϕ=ϕ∞=r(Mϕ), and σ(Mϕ)=σap(Mϕ).
Assuming thatλ∈σap(Mϕ), then there exists a sequence {fn} of functions in L2 with fn=1 such that (9)Mϕ-λfn⟶0asn⟶∞.By removing a “tail” of small norm and renormalizing, for each n there is gn of norm 1 that has only a finite number of nonzero Fourier coefficients corresponding to positive indices and satisfy fn-gn≤1/n. Then (10)Mϕ-λgn≤Mϕ-λfn+Mϕ-λfn-gn⟶0asn⟶∞.Let W∗=Me-iθ, and, for each n, there exists a positive integer Mn such that W∗Mngn is in H¯02. Since W∗ is unitary and commute with Mϕ, it follows that (11)Mϕ-λW∗Mngn=W∗MnMϕ-λgn=Mϕ-λgnand (12)W∗Mngn=gn=1.For each n, define hn=W∗Mngn. Then each hn is in H¯02, (13)hn=1andMϕ-λhn⟶0.Since (14)Bϕ-λhn=I-PuMϕ-λhn≤Mϕ-λhn,we have (Bϕ-λ)hn→0, which implies that λ∈σap(Bϕ). Thus we get that r(Mϕ)≤r(Bϕ). Hence (15)Mϕ=rMϕ≤rBϕ≤Bϕ.On the other hand, (16)Bϕ≤I-PuMϕ≤Mϕ,which completes the proof.In this paper, letek denote the orthonormal basis. For k≥0, we have ek=u(z)zk. For k<0, we get ek=z¯-k.Theorem 3.
The only compact dual truncated Toeplitz operator is zero.Proof.
LetBf denote a compact dual Truncated Toeplitz operator. For nonnegative integers s and t, since {es+n} converges weakly to zero as n→+∞, and it is obtained that (17)Bfes+n,et+n⟶0asn⟶+∞.Let f=∑k=-∞+∞akeikθ, where ak is the k-th Fourier coefficient of f, and it follows that (18)limn→+∞Bfes+n,et+n=limn→+∞fes+n,et+n=limn→+∞∑k=-∞+∞akeikθuzzs+n,uzzt+n=at-s.Hence at-s=0 for all nonnegative integers s and t. So ak=0 for all integer k which implies that f=0. Thus Bf=0.With the above theorem, we get the following corollary.Corollary 4.
A dual truncated Toeplitz operator is self-adjoint if and only if the symbol is real-valued almost everywhere.


## 3. Spectrum and Reducing Subspace

In this section, we discuss the spectrum and reducing subspace of some special dual truncated Toeplitz operator. Letσ(T) denote the spectrum of the linear operator T and σe(T) denote the essential spectrum and σp(T) denote the point spectrum.Theorem 5.
For dual Truncated Toeplitz operatorBz, if u(0)≠0, then (19)σBz=σeBz=T,σpBz=∅.Proof.
It is clear that the spectrum ofBz is contained in D¯. First, we will show that how the dual Truncated Toeplitz operator Bz act on the orthonormal basis {en}. A direct computation gives(20)Bzz¯n=z¯n-1,n≥2,Bzz¯=u0¯u,Bzuzzn=uzzn+1,n≥0.From the above, if u(0)=1, Bz is the bilateral shift. By corollary 24.4 in [25], we have the spectrum of Bz that is the unit circle. In the following, assume that u(0)≠1. Then Bz is a special weight shift with the sequence of scalar {αn,n∈Z}, which αn=1 for n≠-1 and αn=u(0)¯ for n=-1.
By proposition 27.7(c) in [22], we have σe(Bz)=T. Then we will see the point spectrum of Bz. Suppose that f∈uH2⊕H¯02, f=∑n=0+∞u(z)anzn+∑m=1+∞bmz¯m. Let λ∈D¯, and it is obtained that (21)Bzf=∑n=0+∞uzanzn+1+∑m=2+∞bmz¯m-1+b1u0¯uzand (22)λf=λ∑n=0+∞uzanzn+λ∑m=1+∞bmz¯m.From above, we get that Bzf=0 implies f=0 almost everywhere. Suppose that Bzf=λf, we have (23)bm=λbm-1,m≥2,b1u0¯=a0λ,m=1,an=λan+1,n≥0.Since λ≤1 and f∈H2, we have that f=0 almost everywhere. Hence we have that the point spectrum of Bz is ∅.
It suffices for us to show that any point inD is not the spectrum of the dual truncated Toeplitz operator Bz. Since it is known that σe(Bz)=T, for any λ∈D, it follows that λI-Bz is Fredholm operator. Let U~ be an operator defined on uH2⊕H¯02 as follows (24)U~f=Mzf,f∈uH2⊕z¯H¯02,u,f=z¯,where f is in uH2⊕H¯02. It is clear that U~~ is an unitary operator and σ(U~)=T. Hence λI-U~ is invertible. It is obtained that ind(λI-U~)=0. A direct computation gives that (25)λI-Bz=λI-U~+1-u0¯u⊗z¯,hence we have ind(λI-Bz)=0. As σp(Bz)=∅, it follows that (26)dimkerλI-Bz=dimkerλI-Bz∗=0.So we get that λI-Bz is invertible and λ∉σ(Bz). This completes the proof.In the rest of this section, we characterize the structure of lattice of reducing subspaces forBz2 and Bz. To do so, we need to give some notations.LetS be a bounded linear operator on a Hilbert space H. A closed subspace M is said to be a reducing subspace for S, if SM⊆M and SM⊥⊆M⊥. Or equivalently, M is a reducing subspace for S if and only if SPM=PMS and S∗PM=PMS∗, where PM is the orthogonal projection from H onto M.In addition,M is called minimal if there is no nonzero reducing subspace N contained in M properly. The operator S is said to be completely reducible if its lattice of reducing subspaces has no nonzero minimal elements [26].By some easy computations, we have the following equations:(27)Bz2z¯j=I-Puz¯j-2=z¯j-2,j>2,a1¯u+a0¯uz,j=1,a0¯u,j=2.(28)Bz2uzj=uzj+2,j≥0.(29)Bz2∗z¯j=z¯j+2,j≥1.(30)Bz2∗uzj=uzj-2,j≥2,a1z¯+a0z¯2,j=0,a0z¯,j=1.Moreover, we have(31)Bz2∗Bz2z¯j=a12+a02z¯+a0a1¯z¯2,j=1,a0¯a1z¯+a02z¯2,j=2,z¯j,j≥3.(32)Bz2∗Bz2uzi=uzi,i≥0.(33)Bz2Bz2∗z¯j=z¯j,j≥1.(34)Bz2Bz2∗uzi=a12+a02u+a0¯a1uz,i=0,a0a1¯u+a02uz,i=1,uzi,i≥2.Letu(z)=∑n=0∞anzn. Let [f] denote the reducing subspace generated by f, that is, the minimal reducing subspace containing f. Denote by Z+, N all the nonnegative integers and positive integers, respectively. Write T∈S,S∗′ if TS=ST and TS∗=S∗T.Theorem 6.
Ifa0a1≠0 and a12≠(1-a02)2, then Bz2 has no nontrivial reducing subspaces on (Ku2)⊥.Proof.
Suppose there exists a reducing subspaceM⊆(Ku2)⊥ for Bz2 and PM denote the orthogonal projection from (Ku2)⊥ onto M. Then PM∈Bz2,Bz2∗′. Moreover, from (32) and (34), we have(35)a0a1¯PMu=PMBz2Bz2∗-a02Bz2∗Bz2uz=Bz2Bz2∗-a02Bz2∗Bz2PMuz,(36)a0¯a1PMuz=PMBz2Bz2∗-a02+a12Bz2∗Bz2u=Bz2Bz2∗-a02+a12Bz2∗Bz2PMu.In the following, we shall prove that M=(Ku2)⊥ or M={0}.
Let(37)PMu=∑k=1∞ekz¯k+∑n=0∞cnuzn,PMuz=∑k=1∞bkz¯k+∑n=0∞dnuzn.Taking the two equalities above into (35) and (36), along with (31), (32), (33) and (34), we deduce(38)a0a1¯∑k=1∞ekz¯k+∑n=0∞cnuzn=1-a02∑k=3∞bkz¯k+1-a02∑n=2∞dnuzn+a12d0+a0a1¯d1u+a0¯a1d0uz+b1-a02a02+a12b1-a02a0¯a1b2z¯+1-a04b2-a02a0a1¯b1z¯2,and(39)a0¯a1∑k=1∞bkz¯k+∑n=0∞dnuzn=1-a02+a12∑k=3∞ekz¯k+1-a02+a12∑n=2∞cnuzn+a0a1¯c1u+a0¯a1c0-a12c1uz+e1-a02+a122e1-a0¯a1a02+a12e2z¯+e2-a02a02+a12e2-a0a1¯a02+a12e1z¯2.Comparing the coefficients of z¯k(k≥3) in (38) and (39), we get(40)a0a1¯ek=1-a02bk,a0¯a1bk=1-a02-a12ek.Likewise, comparing the coefficients of uzn(n≥2), we obtain(41)a0a1¯cn=1-a02dn,a0¯a1dn=1-a02-a12cn.If (1-a02)(1-a02-a12)=0, then immediately we have ek=bk=0 for k≥3 and cn=dn=0 for n≥2; or else, if (1-a02)(1-a02-a12)≠0, it follows from (40) and (41) that ek=bk=0 for k≥3 and cn=dn=0 for n≥2 since a12≠(1-a02)2.
Considering the coefficients ofu,uz,z¯,z¯2, we have(42)c1=a0¯a1a0a1¯d0,c0=d0,d1=1-a12a0a1¯d0,and(43)a0a1¯e1=1-a02a02+a12b1-a02a0¯a1b2,a0¯a1b1=1-a02+a122e1-a02+a12a0¯a1e2,a0a1¯e2=1-a04b2-a0a1¯a02b1,a0¯a1b2=1-a02a02+a12e2-a0a1¯a02+a12e1,where (42) holds since a0a1¯≠0. By some computations, from (43), we obtain (44)a0a1¯2a02+a12b1=1-a02a02+a12b2,and (45)1-a02+a122-a02a12b1=a0¯a12a02+a12b2.
If[1-a02(a02+a12)][1-(a02+a12)2-a02a12]=0, obviously we have b1=b2=e1=e2=0; if not, since a02a12(2a02+a12)2=[1-a02a02+a12][1-(a02+a12)2-a02a12] if and only if a12=(1-a02)2, along with the assumption a12≠(1-a02)2, we also have b1=b2=e1=e2=0.
Associated with (42), we deduce (46)PMu=d0u+a0¯a1a0a1¯d0uz,PMuz=d0u+1-a12a0a1¯d0uz.If d0=0, then u,uz∈M⊥; otherwise, we have (47)PMu-PMuz=a0¯a1a0a1¯-1+a12a0a1¯d0uz∈M,which proves that u,uz∈M.
Supposeu,uz∈M. Then the following statements hold:(i)
uzj∈M,j∈Z+ since Bz2uzj=uzj+2,j∈Z+.(ii)
Bz2∗(uz)=a0z¯ and Bz2∗u=a1z¯+a0z¯2 show that z¯,z¯2∈M.(iii)
On the basis of (ii), we havez¯j∈M,j∈N since Bz2∗z¯j=z¯j+2,j∈N.
These give that(Ku2)⊥⊆M, which shows that (Ku2)⊥=M. Similarly, we can demonstrate (Ku2)⊥=M⊥ if we assume u,uz∈M⊥, and thus M={0}.
The proof is complete.Theorem 7.
Ifa0a1≠0 and a12=(1-a02)2, then (i)
ifM is a reducing subspace for Bz2 and 〈PMu,u〉=0, then M={0};(ii)
if1-a02=a0a1¯, then Bz2 has no nontrivial reducing subspaces;(iii)
if1-a02≠a0a1¯, then (48)N=span¯uz2n-1-a02a0a1¯uz2n+1,-1-a02a02a1¯z¯2n+1+a0z¯2n+2:n∈Z+is a minimal reducing subspace forBz2.Proof.
Suppose there exists a reducing subspaceM⊆(Ku2)⊥ for Bz2 and PM denote the orthogonal projection from (Ku2)⊥ onto M. Then PM∈Bz2,Bz2∗′.
The assumption in (i) indicates thatu∈M⊥. Bz2Bz2∗u=(a12+a02)u+a0¯a1uz implies uz∈M⊥. By the analysis in the proof of Theorem 6, we have M={0}.
SupposeM is a reducing subspace for Bz2 such that 〈PMu,u〉≠0. Let (49)PMu=∑k=1∞ekz¯k+∑n=0∞cnuzn,PMuz=∑k=1∞bkz¯k+∑n=0∞dnuzn,where c0≠0. By (40), (41), (43) and the assumption a12=(1-a02)2, we obtain (50)ek=1-a02a0a1¯bkfork≥1,cn=1-a02a0a1¯dnforn≥2.Therefore, (51)PMu=1-a02a0a1¯∑k=1∞bkz¯k+c0u+c1uz+1-a02a0a1¯∑n=2∞dnuzn,PMuz=∑k=1∞bkz¯k+∑n=0∞dnuzn.
If1-a02=a0a1¯, then combining (42) with the assumption a12=(1-a02)2, we obtain (52)PMu-PMuz=c1-d1uz=a12a0a1¯d0uz=1-a02d0uz∈M.Since a0a1≠0 and a12=(1-a02)2, we have 1-a02≠0. Because c0=0 if and only if d0=0, surely we have uz∈M. Bz2∗Bz2(uz)=a0a1¯u+a02uz shows that u∈M. By the analysis in the proof of Theorem 7, we get M=(Ku2)⊥. Associated with (i), (ii) is established.
If1-a02≠a0a1¯, from (31), one can easily get that (53)Bz2Bz2∗u-1-a02a0a1¯uz=a04u-1-a02a0a1¯uz.Suppose M0 is a reducing subspace for Bz2 contained in M. Set (54)PM0u-1-a02a0a1¯uz=∑n=0∞αn-1-a02a02a1¯z¯2n+1+a0z¯2n+2+∑m=0∞βmuz2m-1-a02a0a1¯uz2m+1.PM0Bz2Bz2∗=Bz2Bz2∗PM0 indicates that (55)a04PM0u-1-a02a0a1¯uz=∑n=0∞αn-1-a02a02a1¯z¯2n+1+a0z¯2n+2+∑m=1∞βmuz2m-1-a02a0a1¯uz2m+1+a04u-1-a02a0a1¯uz.a0≠1 shows αn=0 for n∈Z+ and βm=0 for m∈N. Thus, PM0(u-((1-a02)/a0a1¯)uz)=β0(u-((1-a02)/a0a1¯)uz). Therefore, if β0≠0, then [u-((1-a02)/a0a1¯)uz]=M⊆M0, forcing M0=M; otherwise, if β0=0, similarly we have M0={0}. Thus, (iii) holds.Theorem 8.
Ifu is an inner function, then the following statements hold: (i)
Ifa0≠0, and a1=0, then M is a minimal reducing subspace for Bz2 if and only if M has the following forms:
(1)
M=span¯{z¯2n,uz2(n-1):n∈N};(2)
M=span¯{z¯2n+1,uz2n+1:n∈Z+};(3)
M=span¯{c1z¯2k-1+c2z¯2k,c1uz2k-2+c2uz2k-1:k∈N}, where c1,c2∈C satisfying c1-c12=c22, c1c2≠0.(ii)
Ifa0=0, a1≠0 and a1≠1, then Bz2 has 3 minimal reducing subspaces.(iii)
Ifa0=a1=0 and if M is a reducing subspace for Bz2, then M is minimal if and only if (56)M=span¯c1z¯2k-1+c2z¯2k:k∈N,or(57)M=span¯c3uz2k+c4uz2k-1:k∈N,wherec1,c2,c3,c4∈C satisfying c1-c12=c22 and c3-c32=c42.Proof.
SupposeM⊆(Ku2)⊥ is a reducing subspace for Bz2 and let PM denote the orthogonal projection from (Ku2)⊥ onto M. Then PM∈Bz2,Bz2∗′.
To show (i), firstly we show that(58)M1=span¯z¯2n,uz2n-1:n∈N,M2=span¯z¯2n+1,uz2n+1:n∈Z+are two minimal reducing subspaces for Bz2. Obviously, (59)Ku2⊥=M1⨁M2,and M1,M2 are reducing subspaces generated by u,uz, respectively. Moreover, they are both minimal for Bz2. In fact, suppose there exists a reducing subspace M0⊆M1 for Bz2. Then PM0M1:M1→M0 and PM0∈Bz2,Bz2∗′. Write (60)PM0u=∑n=1∞bnz¯2n+∑k=1∞ckuz2k-1.By (33), (34), along with Bz2Bz2∗PM0u=PM0Bz2Bz2∗u, we get (61)∑n=1∞bnz¯2n+c1a02u+∑k=2∞ckuz2k-1=a02PM0u.It follows that (62)bn=a02bn,n∈N,andck=a02ck,k∈Nandk≠1,which forces that bn=0 for n∈N, and ck=0 for k∈N and k≠1 since a0≠1. Therefore, we have PM0u=c1u. If c1=0, then u∈M0⊥, showing M0⊥=[u]=M1, that is, M0={0}; if not, c1≠0 shows that u∈M0, implying M0=[u]=M1. In conclusion, M1 is minimal for Bz2. Using a similar method, we can also prove M2 is minimal for Bz2.
Next, suppose(63)PMz¯=∑k=1∞ekz¯k+∑k=0∞fkuzk,(64)PMz¯2=∑k=1∞ek′z¯k+∑k=0∞fk′uzk.By the assumption, we have (65)a02PMz¯=PMBz2∗Bz2z¯=Bz2∗Bz2PMz¯.Along with (63) and the fact that a0=u0<1, we have ek=0 for k≥3 and fk=0 for k≥0. Thus, we obtain that(66)PMz¯=e1z¯+e2z¯2.Similarly, we get(67)PMz¯2=e1′z¯+e2′z¯2.Case 1 (dete1e2e1′e2′≠0). Then, from (66) and (67), we have z¯,z¯2∈M. By the analysis above, we obtain that M1 and M2 are the only two minimal reducing subspaces for Bz2.
Case 2 (dete1e2e1′e2′=0). Again from (66) and (67), we have (68)PMz¯2,z¯=e1′=z¯2,PMz¯=e2¯,that is, e1′=e2¯. Moreover, 〈PMz¯,PMz¯2〉=〈PMz¯,z¯2〉 shows that e2′=1-e¯1. By the assumption, we have e1 is real and e1-e12=e22.
Ife1′=e2=0, then e1-e12=0. That is, e1=0or1. Suppose e1=0, then e2′=1. By (66) and (67), we have z¯∈M⊥ and z¯2∈M. Otherwise, e1=1, e2′=0 imply that z¯∈M and z¯2∈M⊥. In this way, we still obtain that M1 and M2 are the only two minimal reducing subspaces for Bz2.
On the other hand, ife1′e2≠0, it is clear that e1e2′≠0. Immediately, we have PMz¯2=e1z¯+e2z¯2=(e2¯/e1)PMz¯, where e1-e12=e22, e1e2≠0. Then, M contains a minimal reducing subspace like (69)M=span¯e1z¯2k-1+e2z¯2k,e1uz2k-2+e2uz2k-1:k∈N.Furthermore, M is a minimal reducing subspace for Bz2 if and only if (70)M=span¯e1z¯2k-1+e2z¯2k,e1uz2k-2+e2uz2k-1:k∈N,where e1-e12=e22, e1e2≠0.
The assumption in Condition (ii) and a similar method indicate that(71)PNz¯=e1z¯+e2z¯2,PNz¯2=e2′z¯2,where N is supposed to be a reducing subspace for Bz2.
Note that there is always a reducing subspaceN satisfying 〈PNz¯2,z¯2〉≠0. Thus, PNz¯2=e2′z¯2 shows that z¯2∈N. Therefore, N1=[z¯2]=span¯{z¯2n:n∈N} is definitely a minimal reducing subspace for Bz2. Hence, z¯∈N1⊥ shows that if N is a reducing subspace such that 〈PNz¯,z¯〉≠0, then PNz¯=e1z¯, e1≠0. Similarly, N2=[z¯]=span¯{z¯2n+1,uz2n:n∈Z+} is a minimal reducing subspace for Bz2. Denote by N3=(Ku2)⊥⊖(N1⊕N2)=span¯{uz2n+1:n∈Z+}. We shall prove N3 is also minimal. Assume N0⊆N3 is a reducing subspace for Bz2 and (72)PN0uz=∑k=0∞fkuz2k+1.From (34), we see that Bz2Bz2∗(uz)=0. Therefore, (73)0=PN0Bz2Bz2∗uz=Bz2Bz2∗PN0uz=f0a12u+∑k=2∞fkuzk,which forces that f0=f2=f3=⋯=0 and thus PN0(uz)=f1uz. Hence, we have N0={0} or N0=[uz]=N3.
By the similar way in (ii), the statement in (iii) shows that(74)K1=span¯z¯2n-1:n∈N,K2=span¯z¯2n:n∈N,K3=span¯uz2n:n∈Z+,K4=span¯uz2n+1:n∈Z+are four minimal reducing subspaces for Bz2. Using a similar method as in (i), along with (27)-(34), we obtain that (75)PMz¯=e1z¯+e2z¯2,PMz¯2=e1′z¯+e2′z¯2,PMu=f1u+f2uz,PMuz=f1′u+f2′uz,where e1,e2,e1′,e2′,f1,f2,f1′,f2′∈C. A similar discussion as in (i) leads to the desired conclusion.Theorem 9.
Ifa0≠0, a1=0 and a0=1, then Bz2 is completely reducible.Proof.
Under the assumptions, along with (31)-(34), Bz2 is a normal operator on (Ku2)⊥. By the Spectral Theorem of the normal operators, the range of E({λ}) is the eigenspace ker(Bz2-λ), where E is the spectral measure for Bz2. Thus, for every λ∈C, ker(Bz2-λ)={0} if and only if Bz2 has no minimal reducing subspaces, that is, Bz2 is completely reducible.
Givenλ∈C, suppose there exists f∈(Ku2)⊥ such that(76)Bz2f=λf.Assume(77)f=∑n=1∞unz¯n+∑k=0∞vnuzn.Taking this into (76), we get (78)∑n=3∞unz¯n-2+a0¯u2u+u1uz+∑k=0∞vnuzn+2=λ∑n=1∞unz¯n+∑k=0∞vnuzn,from which we can obtain (79)u2n-1=λnu1,u2n=λnu2forn∈N,and (80)v2n+1=1λn+1u1a0¯,v2n=1λn+1u2a0¯forn∈Z+.Then, (77) turns to be (81)f=u1∑n=1∞λnz¯2n-1+u2∑n=1∞λnz¯2n+u1a0¯∑k=0∞1λn+1uz2n+1+u2a0¯∑k=0∞1λn+1uz2n.Thus, we must have u1=u2=0, or f(Ku2)⊥ is infinite. Therefore, we conclude f≡0. Then, the desired result follows.Remark 10.
Notice that the assumption in Theorem9 implies that u is a constant function with module 1, and hence the conclusion in this theorem is trivial.Remark 11.
If the condition in Theorem8 (ii) is changed to a0=0 and a1=1, then the result is different. In fact, N2=span¯{uz2n+1:n∈Z+}, N3=span¯{z¯2n:n∈N} are still both minimal for Bz2. As given in the proof of Theorem 8, N1=span¯{z¯2n+1,uz2n:n∈Z+}. Since a0=0, a1=1, Bz2N1 is normal on N1. By the same method in Theorem 9, we obtain that Bz2 is completely reducible when restricted to N1.The proofs in Theorems6–9 apply to characterize the structure of the reducing subspaces for Bz. By some computations, we have (82)Bzz¯j=z¯j-1,j>1a0¯u,j=1,Bzuzj=uzj+1,j≥0,Bz∗z¯j=z¯j+1,j≥1,Bz∗u=a0z¯,j=0uzj-1,j≥1.Moreover,(83)BzBz∗z¯j=z¯jforj≥1,BzBz∗uzj=a02u,j=0uzj,j≥1.It is easy to verify that if a0≠1, then Bz has no nontrivial reducing subspaces.Ifa0=1, by (83), Bz is a normal operator on (Ku2)⊥. Theorem 5 gives σpBz=∅. Associated with the proof in Theorem 9, Bz is completely reducible on (Ku2)⊥



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