# Reducing Subspaces for Toeplitz OperatorTz1k1z2k2+az¯1l1z¯2l2 on the Weighted Hardy Space over the Bidisk

Authors: Changguo Wei; Xin Ding; Yanyue Shi
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we completely characterize the reducing subspaces forTφa on weighted Hardy space ℋω2D2 under three assumptions on ω, where φa=zk+az¯l, k,l∈ℕ2,k≠l, and a∈0,1. It is shown that the coefficient a∈0,1 does not affect the reducing subspaces for Tφa. We also prove that, for every δ>0, weighted Dirichlet space Dδ2D2 is a weighted Hardy space which satisfies these assumptions. As an application, we describe the reducing subspaces for Tφa on Dδ2D2 and get the structure of commutant algebra V∗Tφa.


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## Body

## 1. Introduction

Letℳ be a closed subspace of a complex Hilbert space ℋ; ℳ is said to be an invariant subspace for A∈Bℋ if Aℳ⊂ℳ. The subspace ℳ is said to be a reducing subspace for A if both ℳ and ℳ⊥ are invariant under A. Let Pℳ be the orthogonal projection from ℋ onto ℳ; then, ℳ is a reducing subspace for A if and only if Pℳ∈V∗A, where V∗A=A,A∗′.For a bounded analytic functionφ over disk D, let Mφ denote the multiplication operator with the symbol φ on Bergman space A2D over the disk, given by Mφf=φf with f∈A2D. Sun and Wang [1] described the commutators and reducing subspaces for multiplication operator MB2 whose symbol B2 is the product of two Blaschke factors. Zhu [2] also independently characterized the reducing subspaces for multiplication operator MB2. Let B be a finite Blaschke product. The reducing subspaces for MB have been completely described with the joint efforts of many scholars (see [3, 4] and their references).On some familiar function spaces such as weighted Bergman space and Dirichlet-type space over the polydisk, the reducing subspaces for Toeplitz operatorTφ have been described when φ is a polynomial. Some details can be seen in [5–12] and their references.Gu [13] firstly described the reducing subspaces for operator-weighted shifts on weighted Hardy space ℋω2D2. In fact, many known function spaces are all weighted Hardy spaces, such as weighted Bergman space and weighted Dirichlet space. So, Gu provided a uniform way to study the reducing subspaces for multiplication operators on these function spaces. Moreover, Gu [14] characterized the reducing subspaces for Tz1k+az¯2l with a∈0,1 on ℋω2D2 under five assumptions on ω. He showed that the coefficient a does not affect the structure of reducing subspaces. Recently, Ren and Shi [15] characterized the reducing subspaces for Tz1k1z2k2+z¯1l1z¯2l2 on ℋω2D2.In this paper, we keep on considering reducing subspaces forTφ with φ=z1k1z2k2+az¯1l1z¯2l2 on ℋω2D2. As in [14], we also pay attention to the case that a∈0,1. A complete characterization is given under three assumptions on ω, which also shows that coefficient a does not affect the structure of reducing subspaces.In order to state the results more precisely, we recall some notations. Letℕ be the set of all positive integers and ℤ+ be the set of all nonnegative integers. Given a sequence of positive numbers ω=ωj:j∈ℤ+ such that liminfj⟶+∞ωj1/j≥1. Denote D=z∈ℂ:z<1. The weighted Hardy space ℋω2D2 over the bidisk with weights ω is defined by(1)ℋω2D2=fz=∑n∈ℤ+2fnzn:fn∈ℂ,f2=∑n∈ℤ+2ωnfn2<∞,where n=n1,n2∈ℤ+2andωn=ωn1ωn2. Set φ=z1k1z2k2+az¯1l1z¯2l20<a≤1; then, for every n∈ℤ+2,(2)Tφ∗Tφ−TφTφ∗zn=λnznandQnp=λn+pk+l,∀p∈ℕ.DenoteQnp≡0 if Qnp=0,∀p∈ℕ. Otherwise, we write Qnp≢0. Set(3)Δ=m1,m2∈ℤ+2:m1∈0,s1,m2∈0,k1l2−k2l1s1,k1l2≠k2l1,m1,m2∈ℤ+2:m1∈0,s1orm2∈0,s2,k1l2=k2l1,and si=gcdki,lii=1,2. For every m∈Δ, let(4)Lm=span¯zm+uk+vl:u,v∈ℤ,m+uk+vl∈ℤ+2.Whena=1, the reducing subspaces for Tz1k1z2k2+z¯1l1z¯2l2 are described in [15] under seven assumptions as follows:(P1) limp⟶+∞ωm+pk+l/ωn+pk+l=1.(P2) If there exists pj⊆ℕ such that limj⟶+∞pj=+∞ and Qnpj=0, then Qnp≡0.(P3) If Qnp≡0, then Qn+lp≢0andQn+kp≢0.(P4) If Qnp≡0, then limp⟶+∞pωn+p+1k+lωn+pk+l/ωn+pk+l+l2−1=0orlimp⟶+∞pωn+p+1k+lωn+pk+l/ωn+pk+l+k2−1=0.(P5) Let n∈Ω1andm∈Ω4. If Qnp≢0 and λn=λm, then Qmp≢0.(P6) If n≠m and Qnp≡Qmp, then the following statements hold:(i)
IfQn+lp≡Qm+lp, then Qn+lp≢0 and Qnp≢0.(ii)
IfQn+kp≡Qm+kp, then Qn+kp≢0 and Qnp≢0.(P7) Let m,n∈Δ and n≠m. If ωm+k=ωn+k and ωm+hk+l=ωn+hk+l for h∈ℤ+, then zn∉Lm.GivenA∈V∗Tφ. For every β∈Δ, set(5)Azα=∑p∈ℤcpzβ+pk+l+qz,where q⊥zβ+pk+l. The equalities ATφ∗zα=Tφ∗Azα and ATφzα=TφAzα imply that(6)aAzα+l=czβ−k+∑p∈ℤacp+cp+1ωβ+p+1k+lωβ+p+1k+l−kzβ+pk+l+l+Tφ∗qz,Azα+k=c′zβ−l+∑p∈ℤcp+acp+1ωβ+p+1k+lωβ+p+1k+l−lzβ+pk+l+k+Tφqz,where α∉p,q:p≥k1,q≥k2∪p,q:p≥l1,q≥l2 and c,c′∈ℂ. It seems that the results in [15] can be generalized to the case that φ=z1k1z2k2+az¯1l1z¯2l20<a<1 as soon as replacing (P4) by the following.(P4′) If Qnp≡0, then limp⟶+∞pωn+p+1k+lωn+pk+l/a2ωn+pk+l+l2−1=0orlimp⟶+∞pωn+p+1k+lωn+pk+l/a2ωn+pk+l+k2−1=0.However, (P1) and (P4′) imply that a2=limp⟶+∞ωn+p+1k+lωn+pk+l/ωn+pk+l+l2=1, which contradicts with 0<a<1. We find that p∈ℕ:Qnp=0 is a finite set at most on some classical spaces, such as weighted Dirichlet space. So, this paper is based on the following three assumptions:(A1) limp⟶+∞ωm+pk+l/ωn+pk+l=1.(A2) If p∈ℕ:Qnp=0 is nonempty, then it is a finite set.(A3) Let m,n∈Δ and n≠m. If ωm+k=ωn+k and ωm+hk+l=ωn+hk+l for h∈ℤ+, then zn∈Lm.Our main results are as follows: ifω satisfies (A1) and (A2), then the reducing subspace for Tzk+az¯l with a∈0,1 generated by zm is Lm (Theorem 1); if ω satisfies (A1), (A2), and (A3), then Lm is minimal (Theorem 2); for every δ>0, Dδ2D2 is a weighted Hardy space satisfying (A1), (A2), and (A3). In addition, we get the structure of reducing subspaces for Tzk+az¯l on Dδ2D2 (Theorem 3) and the algebraic structure of V∗Tzk+az¯l (Theorem 5).This paper is arranged as follows: in Section2, we pay attention to the properties of Qn and give the expression of the element in commutant algebra; in Section 3, we characterize the structure of reducing subspaces for Tzk+az¯l; in Section 4, we check (A1), (A2), and (A3) on weighted Dirichlet space Dδ2D2δ>0. This part contains a large number of skilled calculations.


## 2. The Expression of the Element in Commutant Algebra

Throughout this paper, letk=k1,k2,l=l1,l2∈ℕ2,k≠l, and α=α1,α2,β=β1,β2∈ℤ+2. Let φz=zk+az¯l=z1k1z2k2+az¯1l1z¯2l2 where 0<a<1. On weighted Hardy space ℋω2D2, Toeplitz operator Tφ with nonanalytic symbol φ is defined to be(7)Tφ=Tzk+az¯l=Mzk+aMzl∗,where Mzk and Mzl are multiplication operators and Mzl∗ is the adjoint of Mzl. Let V∗Tφ be the commutant algebra of the von Neumann algebra generated by Tφ. In this section, we mainly consider the properties of A∈V∗Tφ under assumptions (A1) and (A2). Clearly, we can write A as follows:(8)Azα=∑p∈Δα,βcpzβ+pk+l+qz,where cp∈ℂ∖0,(9)Δα,β=p∈ℤ:Azα,zβ+pk+l≠0,(10)q∈Hβ0=span¯zm:m≠β+pk+l,p∈ℤ,m∈ℤ+2.We will prove that the number of elements inΔα,β is no more than 1 under proper assumptions. To show the details, we recall several notes.Denote byα⪰β if α1≥β1 and α2≥β2. Otherwise, denote by α⋡β. Clearly, ℤ+2=∪i=14Ωi where(11)Ω1=n∈ℤ+2:n⋡k,n⋡l,Ω2=n∈ℤ+2:n⪰k,n⋡l,Ω3=n∈ℤ+2:n⋡k,n⪰l,Ω4=n∈ℤ+2:n⪰k,n⪰l.A direct computation gives(12)Tφzn=zn+k,n⋡l,zn+k+aωnωn−lzn−l,n⪰l,Tφ∗zn=azn+l,n⋡k,azn+l+ωnωn−kzn−k,n⪰k.Let(13)T=Tφ∗Tφ−TφTφ∗.It is not difficult to check that(14)Tzn=λnzn,where(15)λn=ωn+kωn−a2ωn+lωn,n∈Ω1,ωn+kωn−ωnωn−k−a2ωn+lωn,n∈Ω2,ωn+kωn−a2ωn+lωn+a2ωnωn−l,n∈Ω3,ωn+kωn−ωnωn−k−a2ωn+lωn+a2ωnωn−l,n∈Ω4.Denote by(16)Qnp=λn+pk+l,∀p∈ℕ.Lemma 1.
IfΔα,β is nonempty, β⋡k+l, then Δα,β is a finite set.Proof.
Sinceωj>0 for j∈ℤ+, we have λα,λβ∈ℝ. For every A∈V∗Tφ, it is easy to check that(17)λαAzα,zβ=λαAzα,zβ=ATzα,zβ=TAzα,zβ=Azα,Tzβ=λβAzα,zβ.
So,Δα,β∖0⊆p∈ℕ:Qβp=λα. If Δα,β is infinite, there exists pjj=1+∞ such that pj∈Δα,β and limj⟶+∞pj=+∞. Then, Qβpj=λα,∀j∈ℕ. (A1) shows that Qβpj=λα=limj⟶∞Qβpj=0, which contradicts with (A2).
Fors∈ℤ+, if Δα+sk+l,β is nonempty, β⋡k+l, by the above lemma, we can set(18)Mα,βs=maxΔα+sk+l,β,mα,βs=minΔα+sk+l,β.Lemma 2.
Letα,β⋡k+l and s∈ℤ+. Then, Mα,βs+1=Mα,βs+1.Proof.
Let(19)Azα+sk+l=ctzβ+tk+l+fz,where f⊥zβ+tk+l, t=Mα,βs, and ct≠0. ATφ∗Tφzα+sk+l=Tφ∗TφAzα+sk+l implies that(20)Aazα+s+1k+l+ρzα+sk+l+ηzα+s−1k+l=actzβ+t+1k+l+Fz+Gz,where F∈span¯zβ+pk+l:p≤t,β+pk+l⪰0, G∈Hβ0, ρ>0, η≥0, and η=0 if and only if α+s−1k+l⪰0. We prove this lemma by induction.
Ifs=0, then η=0 in equation (20). Since F+G⊥zβ+t+1k+l and Azα⊥zβ+t+1k+l, we have(21)Azα+k+l,zβ+t+1k+l=ctzβ+t+1k+l2≠0.
By equation (20), we get Mα,β1=t+1=Mα,β0+1.
Suppose the statement holds whens=N−1N≥1. If s=N, η>0 in equation (20). By hypothesis, Mα,βN−1=Mα,βN−1=t−1, which implies that Azα+N−1k+l⊥zβ+t+1k+l. Together with F+G⊥zβ+t+1k+l and Azα+Nk+l⊥zβ+t+1k+l, equation (20) shows that(22)Azα+N+1k+l,zβ+t+1k+l=ctzβ+t+1k+l2≠0.
So,Mα,βN+1=t+1=Mα,βN+1.Lemma 3.
Supposeα⋡k+l and A∈V∗Tφ. If Δα,β is nonempty, then CardΔα,β=1. Moreover, if Δα,β=p0, then Δα+hk+l,β=p0+h for any h∈ℤ+.Proof.
Without loss of generality, we may assumeβ⋡k+l. By Lemma 1, let p0=Mα,β0. For every p∈ℤ+, as in Lemma 1, we can prove that h∈ℤ+:Azα+hk+l,zβ+pk+l≠0 is a finite set at most. Set(23)Ep0=∪0≤p≤p0h∈ℤ+:Azα+hk+l,zβ+pk+l≠0.
Then,0∈Ep0, and Ep0 is finite. Let h0=maxEp0; we prove the following statements hold by induction:(i)
For everyh∈ℤ+, mα,βh0+h+q>p0+h, ∀q∈ℕ.(ii)
For everyh∈ℤ+, mα,βh0+h+1=p0+h+1.Step 1.
It is clear that (i) holds forh=0 from the definition of h0. And we can set(24)Azα+h0+1k+l=d1zβ+p0+1k+l+f1z+g1z,where d1∈ℂ,f1∈span¯zβ+hk+l:h≥p0+2, and g1∈Hβ0. By ATφ∗Tφzα+h0+1k+l=Tφ∗TφAzα+h0+1k+l, we have(25)Aazα+h0+2k+l+ρzα+h0+1k+l+ηzα+h0k+l=ad1ωβ+p0+1k+lωβ+p0k+lzβ+p0k+l+F1z+G1z,where ρ,η>0,F1∈span¯zβ+hk+l:h≥p0+1,andG1∈Hβ0. Since h0+1,h0+2∉Ep0, we have mα,βh0+i≥p0+1,i=1,2. So, equation (25) deduces that(26)Aηzα+h0k+l=ad1ωβ+p0+1k+lωβ+p0k+lzβ+p0k+l+F1˜z+G1˜z,where F1˜∈span¯zβ+hk+l:h≥p0+1andG1˜∈Hβ0. It means that mα,βh0≥p0. On the contrary, the definition of h0 shows that mα,βh0≤p0. So,(27)mα,βh0=p0,and d1≠0 from equation (26). Hence, equation (24) implies that mα,βh0+1=p0+1. So, (ii) holds for h=0.Step 2.
Suppose (i) and (ii) hold forh≤N; then, mα,βh0+N+q>p0+N, ∀q∈ℕ. So, mα,βh0+N+1+q>p0+N. We can set(28)Azα+h0+N+1+qk+l=d2zβ+p0+N+1k+l+f2z+g2z,where d2∈ℂ,f2∈span¯zβ+hk+l:h≥p0+N+2,andg2∈Hβ0. By ATφ∗Tφzα+h0+N+1+qk+l=Tφ∗TφAzα+h0+N+1+qk+l, we get(29)Aazα+h0+N+2+qk+l+ρ′zα+h0+N+1+qk+l+η′zα+h0+N+qk+l=ad2ωβ+p0+N+1k+lωβ+p0+Nk+lzβ+p0+Nk+l+F2z+G2z,where ρ′,η′>0,F2∈span¯zβ+hk+l:h≥p0+N+1,andG2∈Hβ0. Since (i) holds for h=N and F2+G2 is orthogonal to zβ+p0+Nk+l, we obtain ad2ωβ+p0+N+1k+l/ωβ+p0+Nk+lzβ+p0+Nk+l,zβ+p0+Nk+l=0. This means that d2=0. So, (i) holds for h=N+1 from equation (28). In particular, mα,βh0+N+2≥p0+N+2. Replacing h0 and p0 by h0+N+1 and p0+N+1 in equation (24), respectively, we can prove that mα,βh0+N+2=p0+N+2. So, (ii) holds for h=N+1.
Equation (27) and (ii) show that mα,βh0+h=p0+h, ∀h∈ℤ+. Together with Lemma 2, we get(30)mα,βh0+h=p0+h=Mα,β0+h=Mα,βh,∀h∈ℤ+.
So,(31)λα+h0+hk+l=λβ+p0+hk+l=λα+hk+l,∀h∈ℤ+.
Ifh0≠0, then λα+h0k+l=λα+nh0k+l=limn⟶+∞Qαnh0=0 from (A1), which contradicts with (A2). Therefore, equation (30) implies that mα,βh=p0+h=Mα,βh, ∀h∈ℤ+. So, we get the desired results.Lemma 4.
Letα∈Ω1 and A∈V∗Tφ. Then, Azα=∑β∈Ω1λβ=λαcβzβ.Proof.
Suppose there exists someβ∈Ω2∪Ω3∪Ω4 such that Azα,zβ≠0. Lemma 3 shows that Azα=cβzβ+hz, where cβ≠0 and h∈Hβ0. Now, we discuss in two cases.
Case one:β∈Ω2∪Ω4. By ATφ∗zα=Tφ∗Azα, we have Aazα+l=acβzβ+l+cβωβ/ωβ−kzβ−k+H1z, where H1∈Hβ0. Hence, Δα+l,β−k=p∈ℤ:Azα+l,zβ−k+pk+l≠0=0,1. This is a contradiction to Lemma 3.
Case two:β∈Ω3. By ATφzα=TφAzα, we have Aazα+k=cβzβ+k+acβωβ/ωβ−lzβ−l+H2z, where H2∈Hβ0. Hence, Δα+k,β−l=p∈ℤ:Azα+k,zβ−l+pk+l≠0=0,1. This is also a contradiction.
So,Azα⊥span¯zβ:β∈Ω2∪Ω3∪Ω4; then, Azα=∑β∈Ω1λβ=λαcβzβ.


## 3. Reducing Subspaces forTφ on the Weighted Hardy Space

In this section, we mainly consider the structure ofzm where m∈Δ and zm is the reducing subspace for Tφ on ℋω2D2 generated by zm. Clearly, Lm=span¯zm+uk+vl:u,v∈ℤ,m+uk+vl∈ℤ+2 is a reducing subspace for Tφ with m∈ℤ+2. Let m=m+uk+vl∈ℤ+2:u,v∈ℤ and(32)Δ=m1,m2∈ℤ+2:m1∈0,s1,m2∈0,k1l2−k2l1s1,k1l2≠k2l1,m1,m2∈ℤ+2:m1∈0,s1orm2∈0,s2,k1l2=k2l1,where si=gcdki,lii=1,2. Then, ℤ+2=∪m∈Δm and ℋω2D2=⊕m∈ΔLm. Let(33)Ω=u,v∈ℤ2:m+uk+vl∈ℤ+2,Ω˜=u,v∈Ω:zm+uk+vl∈zm;cn=minc∈ℤ+:m+ck⪰nl,dn=mind∈ℤ+:m+dl⪰nk,∀n∈ℕ.The following theorem is similar as Theorem 3.2 in [15]. However, the addition of the coefficient a changes the expressions of Tφ and Tφ∗. For the reader’s convenience, we show the details of the proof.Theorem 1.
Letℋω2D2 be a weighted Hardy space satisfying assumptions (A1) and (A2). Then, Lm=zm for every m∈Δ.Proof.
To showLm⊆zm, we only need to prove Ω˜=Ω. Obviously, Ω˜⊆Ω. Since cn and dn are strictly monotonically increasing series (see Theorem 3.1 in [15]),(34)Ω=∪n=1+∞−n+1,cn×−n+1,dn∩Ω.
To showΩ⊂Ω˜, we will prove the following statements hold for every n∈ℕ by induction.(T1) −n+1,cn×−n+1,dn∩Ω⊆Ω˜(T2) cn,−n∈Ω˜(T3) −n,dn∈Ω˜
We mainly use the following two important equalities:(35)Tφzm+uk+vl=zm+u+1k+vl+aωm+uk+vlωm+uk+v−1lzm+uk+v−1lwithu,v−1∈Ω,(36)Tφ∗zm+uk+vl=azm+uk+v+1l+ωm+uk+vlωm+u−1k+vlzm+u−1k+vlwithu−1,v∈Ω.Step 1.
We prove (T1), (T2), and (T3) hold when n=1. By calculating, we know(37)Tφjzm=zm+jk∈zm,0≤j≤c1,Tφ∗jzm=ajzm+jl∈zm,0≤j≤d1.
Thus,0,c1×0⊆Ω˜ and 0×0,d1⊆Ω˜.
Ifd1=0, then (T1) holds. If d1≥1, equation (35) implies that 1×1,d1∈Ω˜ since u,v∈0×1,d1⊆Ω˜ and u,v−1∈0×0,d1−1⊆Ω˜. Similarly, it is easy to check that c1×1,d1⊆Ω˜. Thus, 0,c1×1,d1⊆Ω˜. Together with 0,c1×0⊆Ω˜, we obtain 0,c1×0,d1⊆Ω˜. (T1) holds for n=1.
LetP be the orthogonal projection from ℋω2D2 into zm. Since c1,0∈Ω˜, we have(38)Tφzm+c1k=zm+c1+1k+aωm+c1kωm+c1k−lzm+c1k−l=Pzm+c1+1k+aωm+c1kωm+c1k−lPzm+c1k−l.
So,(39)Pzm+c1+1k−zm+c1+1k=aωm+c1kωm+c1k−lzm+c1k−l−Pzm+c1k−l.
By the definition ofcn, we know m+c1k−l∈Ω1 and m+c1+1k∈Ω4. Lemma 4 implies that Pzm+c1k−l,zm+c1+1k=0. Therefore, Pzm+c1k−l=zm+c1k−l∈zm from equation (39). So, (T2) holds for n=1. The proof of (T3) is similar to that of (T2). We can prove Pzm−k+d1l=zm−k+d1l∈zm, which implies (T3) holds for n=1.Step 2.
Suppose (T1), (T2), and (T3) hold for 1≤n≤pp∈ℕ, and then we prove that (T1), (T2), and (T3) also hold for n=p+1. Since cp,−p,−p,dp∈Ω˜,(40)Tφjzm+cpk−pl=zm+cpk−pl+jk∈zm,0≤j≤cp+1−cp,Tφ∗izm−pk+dpl=aizm−pk+dpl+il∈zm,0≤i≤dp+1−dp,so cp,cp+1×−p⊆Ω˜ and −p×dp,dp+1⊆Ω˜.
By equation (36), u,v∈cp+1,cp+1×−p⊆Ω˜ and u−1,v∈cp,cp+1×−p⊆Ω˜ imply that(41)cp+1,cp+1×−p+1⊆Ω˜.
Also,cp+1,cp+1×−p+j⊆Ω˜ for 1≤j≤dp+p. Therefore, cp+1,cp+1×−p,dp⊆Ω˜.
By equation (35), u,v∈−p×dp+1,dp+1⊆Ω˜ and u,v−1∈−p×dp,dp+1⊆Ω˜ imply that(42)−p+1×dp+1,dp+1⊆Ω˜.
Using the same technique, we finally get−p+1,cp+1×dp+1,dp+1⊆Ω˜. By hypothesis to (T1), we obtain −p,cp+1×−p,dp+1∩Ω⊆Ω˜. So, (T1) holds for n=p+1.
By (T1), we have cp+1,−p∈Ω˜. Notice that m+cp+1k−p+1l∈Ω1 and m+cp+1+1k−pl∈Ω4. Put u=cp+1 and v=−p in equation (35). As in Step 1, it is easy to prove that(43)Pzm+cp+1k−p+1l=zm+cp+1k−p+1l∈zm.
That is, (T2) holds for n=p+1. Similarly, by −p,dp+1∈Ω˜ and equation (36), we can prove that (T3) holds for n=p+1. So, we finish the proof.Lemma 5.
Letm,n∈Ω1 and n≠m. If Azm,zn≠0, then ωm+k=ωn+k and ωm+hk+l=ωn+hk+l for every h∈ℤ+.Proof.
SinceAzm,zn≠0, Lemma 3 shows that Δm+pk+l,n=p,∀p∈ℤ+. Set(44)Azm+pk+l=apzn+pk+l+fpz,where complex number ap≠0 and fp∈Hβ0. We claim that(45)ap=aq,∀p,q∈ℤ+.
For everyN∈ℕ, ATφ∗Tφzm+Nk+l=Tφ∗TφAzm+Nk+l deduces that(46)Aazm+N+1k+l+ρzm+Nk+l+aωm+Nk+lωm+N−1k+lzm+N−1k+l=aNazn+N+1k+l+ηzn+Nk+l+aωn+Nk+lωn+N−1k+lzn+N−1k+l+Tφ∗TφfNz,for some ρ,η>0. Since Azm+Nk+l,Azm+N−1k+l, and Tφ∗TφfN are all orthogonal to zn+N+1k+l, we have(47)Azm+N+1k+l=aNzn+N+1k+l+gz,for some g∈Hβ0. Comparing with equation (44) for n=N+1, there is aN=aN+1. So, it is easy to get the claim by mathematical induction.
Furthermore, equation (46) implies that(48)ωm+pk+lωm+p−1k+l=ωn+pk+lωn+p−1k+l,∀p∈ℕ.
Letp⟶+∞; then,(49)ωmωn=ωm+pk+lωn+pk+l=limp⟶+∞ωm+pk+lωn+pk+l=1.
That is,ωm=ωnandωm+pk+l=ωn+pk+l for every p∈ℕ.
Whenp=0, Azm=a0zn+f0z. By ATφ∗Tφzm=Tφ∗TφAzm, we have(50)Aazm+k+l+ωm+kωmzm=a0azn+k+l+ωn+kωnzn+Tφ∗Tφf0z.
By the expression ofAzm+k+l and Azm, we get ωm+k/ωm=ωn+k/ωn. Therefore, ωm+k=ωn+k.Theorem 2.
Letℋω2D2 be a weighted Hardy space satisfying assumptions (A1), (A2), and (A3). Then, for every m∈Δ, Lm=zm is a minimal reducing subspace for Tφ.Proof.
Assume a nonzero subspaceℳ⊆Lm is a reducing subspace for Tφ. Let Pℳ be the orthogonal projection from ℋω2D2 into ℳ; then, Pℳ∈V∗Tφ. By Lemma 4, set(51)Pℳzm=∑n∈Ω1λn=λmanzn∈ℳ⊆Lm.
Lemma5 and (A3) show that if n≠m, then zn⊥Lm. So, Pℳzm=amzm for some am∈ℂ. If am=0, then Lm⊥ℳ, which contradicts with ℳ≠0. Hence, am≠0 and Lm=zm⊆ℳ.


## 4. Results on the Weighted Dirichlet Space

The weighted Dirichlet spaceDδ2D2δ>0 is defined as follows:(52)Dδ2D2=ℋω2D2withω=ωn=n1+1δn2+1δ,n∈ℤ+2.So, we can get a more specific formula aboutQnp:(53)Qnp=∏i=12ni+ki+pki+li+1δni+pki+li+1δ−∏i=12ni+pki+li+1δni−ki+pki+li+1δ−a2∏i=12ni+li+pki+li+1δni+pki+li+1δ+a2∏i=12ni+pki+li+1δni−li+pki+li+1δ,p∈ℕ.It is clear that the weightω of Dδ2D2 satisfies (A1). The following lemmas show that ω satisfies (A2) and (A3). Then, the results in Section 3 hold.Lemma 6.
Forn∈ℤ+2, p∈ℕ:Qnp=0 is a finite set at most.Proof.
Supposep∈ℕ:Qnp=0 is an infinite set; then, there exists pjj=1+∞ such that Qnpj=0 and limj⟶+∞pj=+∞. By a direct computation, we get Qnp=0 if and only if(54)∏i=12ni+1+pki+li−liδ∏i=12ni+1+pki+li−kiδ⋅ni+1+pki+li+kiδ−∏i=12ni+1+pki+li2δ=a2∏i=12ni+1+pki+li−kiδ∏i=12ni+1+pki+li−liδ⋅ni+1+pki+li+liδ−∏i=12ni+1+pki+li2δ.
Let(55)vmt=∏i=12mi+1ki+lit+1δ,where m=m1,m2 and mi≥−maxki,li with i=1,2. Then, there exists r>0 such that vm∈C∞−r,r. Denote by(56)gmt=vn−mtvn+mt−vn2t,m=l,k,fk,lt=vn−ltgkt,fl,kt=vn−ktglt.
Thus, fortj=1/pj,(57)Qnpj=0⇔vn−ltjgktj=a2vn−ktjgltj⇔fk,ltj=a2fl,ktj.
Set(58)xi=kiki+li,yi=liki+li,i=1,2,(59)ei=ni+1ki+li,ai=ei−xi,bi=ei−yi,ci=ei+xi,di=ei+yi,i=1,2.
A careful computation gives(60)limt⟶0vn−kt=1,limt⟶0vn−k′t=δa1+a2,limt⟶0vn−k″t=δδ−1a1+a22+2δa1a2,limt⟶0v‴n−kt=δδ−1δ−2a1+a23+6δδ−1a1+a2a1a2,limt⟶0gkt=limt⟶0gk′t=limt⟶0fk,lt=limt⟶0fk,l′t=0,(61)limt⟶0gk″t=limt⟶0fk,l″t=−2δx12+x22,limt⟶0f‴k,lt=limt⟶03vn−l′tgk″t+g‴kt.
It follows thatlimt⟶0fk,lt−a2fl,kt=limt⟶0fk,l−a2fl,k′t=0. By L’Hospital’s rule, we have(62)limt⟶0fk,lt−a2fl,ktt2=limt⟶0fk,l−a2fl,k′​′t2=0.
That is,(63)x12+x22=a2y12+y22,(64)limt⟶0gk−a2gl″t=0.
Moreover, equation (60) shows that limt⟶0gkt−a2glt=limt⟶0gk−a2gl′t=0. By L’Hospital’s rule again, we get(65)limt⟶0gkt−a2gltt3=limt⟶0gk−a2gl‴t6=0.
That is,(66)limt⟶0g‴kt=limt⟶0a2g‴lt.
Similarly,(67)limt⟶0fk,lt−a2fl,ktt3=limt⟶0fk,l−a2fl,k‴t6=0.
Together with equations (61), (64), and (66), we get limt⟶0vn−l′t=limt⟶0vn−k′t. That is, y1+y2=x1+x2. Notice that x1+y1=x2+y2; we obtain x1=y2 and x2=y1. Combining with equation (63), we have a=1, which contradicts with a∈0,1.Lemma 7.
Letm,n∈Δ and n≠m. If ωm+k=ωn+k and ωm+hk+l=ωn+hk+l for every h∈ℤ+, then(68)n1=l1l2m2+1−1,n2=l2l1m1+1−1.Proof.
By the definition ofωn and the equalities ωm=ωn and ωm+k=ωn+k, we have(69)m1+1m2+1=n1+1n2+1,m1+k1+1m2+k2+1=n1+k1+1n2+k2+1.
Therefore,k1m2−n2+k2m1−n1=0. Thus, m1=n1 if and only if m2=n2. Similarly, by ωm+k=ωn+k and ωm+k+l=ωn+k+l, we can prove that l1m2−n2+l2m1−n1=0. Since n≠m, we get n1≠m1, n2≠m2, and k1l2=k2l1. Furthermore, by k1m2+1−n2+1=k2n1+1−m1+1 and equation (69), we get the desired results.
The above lemma shows thatω satisfies (A3). In fact, if zn∈Lm, then there exist u,v∈ℤ such that n1=l1/l2m2+1−1=m1+uk1+vl1 and n2=l2/l1m1+1−1=m2+uk2+vl2. It is easy to see that uk1+vl1=uk2+vl2=0. Then, n1=m1andn2=m2, which contradict with n≠m.
Sinceω satisfies (A1), (A2), and (A3), Theorem 2 implies the following result.Corollary 1.
Letk,l∈ℕ2,k≠l. Then, for every m∈Δ, Lm=zm is a minimal reducing subspace for Tφ on Dδ2D2δ>0.Combining Lemmas4, 5, and 7, we characterize Azm for A∈V∗Tφ and m∈Δ.Corollary 2.
LetA∈V∗Tφ and m∈Δ. Then,(i)
Ifk1l2≠k2l1, then Azm=czm, where c∈ℂ.(ii)
Ifk1l2=k2l1, then Azm=azm+bzm′, where m′=l1/l2m2+1−1,l2/l1m1+1−1, a,b∈ℂ. In particular, if m′∉ℤ+2, then b=0.Recall that two reducing subspacesℳ1 and ℳ2 for Tφ are called unitarily equivalent if there exists a linear operator U on DδD2 such that U∣ℳ1 is unitary from ℳ1 onto ℳ2, U|ℳ1⊥=0, and U commutes with both Tφ and Tφ∗. Together with Corollary 2, we get the following theorem, which gives a complete description of the reducing subspaces for Tφ.Theorem 3.
Letk,l∈ℕ2,k≠l. If ℳ is a reducing subspace for Tφ on DδD2δ>0, then ℳ is the orthogonal sum of some minimal reducing subspaces. Moreover, ℳ is a minimal reducing subspace for Tφ if and only if ℳ has the form as follows:(i)
Ifk1l2≠l1k2, then ℳ=Lm for some m∈Δ.(ii)
Ifk1l2=l1k2, then there exist m∈Δ and s,t∈ℂ such that ℳ=ℳst where ℳst are defined by(70)ℳst=span¯szm+ik+jl+tzm′+ik+jl:i,j∈ℤ,m+ik+jl∈ℤ+2,withm′=l1/l2m2+1−1,l2/l1m1+1−1. In particular, if m′∉ℤ+2, then t=0.Proof.
Forn∈Δ, if zn⊥ℳ, Corollary 1 shows that zn=Ln⊥ℳ. There is m∈Δ such that Pℳzm≠0. If k1l2≠l1k2, Corollary 2 (i) implies that Pℳzm=czm with c≠0. And then, Lm⊆ℳ. If k1l2=l1k2, Corollary 2 (ii) implies that there are s,t∈ℂ such that Pℳzm=szm+tzm′, where m′=l1/l2m2+1−1,l2/l1m1+1−1. If m′∉ℤ+2, then t=0. If m′∈ℤ+2, we claim that ℳst=szm+tzm′. Therefore, ℳst⊆ℳ. In fact, we only need to prove there is a linear operator U on DδD2 such that U|Lm is unitary from Lm onto ℳs,t, UTφ=TφU, and UTφ∗=Tφ∗U. That is, Lm and ℳst are unitarily equivalent.
LetU|Lm⊥=0 and(71)Uzm+ik+jl=1s2+t2szm+ik+jl+tzm′+ik+jl,for any i,j∈ℤ such that m+ik+jl∈ℤ+2. By k1l2=k2l1, a computation gives ωm′+ik+jl=ωm+ik+jl. Thus, Uzm+ik+jl=zm+ik+jl. It remains to prove UTφ=TφU. Referring to Theorem 3.7 of [6], we can get(72)m+ik+jl∈ℤ+2⟺m′+ik+jl∈ℤ+2,where m′ is defined as in (ii). Then, m⪰l implies that m′⪰l. In this case,(73)UTφzm=Uzm+k=1s2+t2szm+k+tzm′+k,TφUzm=Tφ1s2+t2szm+tzm′=1s2+t2szm+k+tzm′+k.
Similarly,m⪰l implies that m′⪰l; then,(74)UTφzm=Uzm+k+aωmωm−lzm−l=1s2+t2szm+k+tzm′+k+saωmωm−lzm−l+taωmωm−lzm′−l,TφUzm=Tφ1s2+t2szm+tzm′=1s2+t2szm+k+saωmωm−lzm−l+tzm′+k+taωm′ωm′−lzm′−l.
Sinceωm=ωm′ and ωm−l=ωm′−l, we obtain UTφ=TφU. The rest of the proof is obvious.
Letm,m′ be as in the above theorem, and let U|Lm⊥=0 and U:Lm⟶Lm′ be a linear operator defined by(75)Uzm+ik+jl=zm′+ik+jl.
It is not difficult to get the following theorem.Theorem 4.
Letk,l∈ℕ2,k≠l. Suppose m,m′∈Δ; then, the following statements hold:(i)
Ifk1l2≠k2l1, then Lm and Lm′ are unitarily equivalent if and only if m′=m.(ii)
Ifk1l2=k2l1, then Lm and Lm′ are unitarily equivalent if and only if m′=m or m′=l1/l2m2+1−1,l2/l1m1+1−1. In particular, if m′∉ℤ+2, then Lm and Lm′ are unitarily equivalent if and only if m′=m.By the above theorem and Corollary 8.2.6 in [4], we can characterize V∗Tφ. Since the proof of the following theorem is the same as that in [6, 15], we omit the details.Theorem 5.
Letk,l∈ℕ2,k≠l. Then, V∗Tφ is a type I von Neumann algebra. Furthermore, the following statements hold:(i)
Ifk1l2≠k2l1, then V∗Tφ is abelian and is ∗−isomorphic to ⊕i=1jℂ, where j=k1l2−k2l1.(ii)
Ifk1l2=k2l1 and s=s1,s2 with si=gcdki,lii=1,2, then V∗Tφ=V∗Tzs and V∗Tφ is never abelian. Moreover, if s1=s2=r, then V∗Tφ is ∗−isomorphic to(76)⊕j=1∞M2ℂ⊕⊕i=1rℂ.
Ifs1≠s2, then V∗Tφ is ∗−isomorphic to the direct sum of countably many M2ℂ⊕ℂ.



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