Let g(z) be an n-degree polynomial (n≥2). Inspired by Sarason’s result, we introduce the operator T1 defined by the multiplication operator Mg plus the weighted Volterra operator Vg on the Bergman space. We show that the operator T1 is similar to Mg on some Hilbert space Sg2(D). Then for g(z)=zn, by using matrix manipulations, the reducing subspaces of the corresponding operator T2 on the Bergman space are characterized.

NNSF of China113711191. Introduction

The invariant subspace and reducing subspace problems are interesting and important themes in operator theory. The conjecture is that every bounded linear operator T on a separable Hilbert space H has a nontrivial closed invariant subspace. A closed linear nontrivial subspace X of H is called an invariant subspace for T if X is different from {0} and H such that TX⊂X. If X and X⊥ are both invariant subspaces for T, then X is said to be a reducing subspace for T. The invariant subspace and reducing subspace problems on the Hardy space and the Bergman space have been studied extensively in the literature. We mention here the papers [1–13] and the books [14–17] which include a lot of the information on the corresponding operator theory.

Let D be the unit disk in the complex plane C, and let La2(D) denote the Bergman space of analytic functions which belongs to L2(D), where L2(D) is the space of square integrable functions on D. It is well known that La2(D) is a Hilbert space. If f∈La2(D), then (1)f22=∫Dfz2dAz,where dA is the normalized area measure on D, and (2)dAz=1πdxdy=rπdrdθ.For f,g∈La2(D), let (3)fz=∑k=0∞akzk,gz=∑k=0∞bkzk,and then the inner product of f and g is defined by (4)f,g=∫Dfzgz¯dAz=∑k=0∞akb¯kk+1.In this inner product, La2(D) has an orthonormal basis {ek}k=0∞, where (5)ekz=k+1zk,k=0,1,….Let H∞(D) denote the algebra of bounded analytic functions on D. For φ∈H∞(D), Mφ is an analytic multiplication operator on the Bergman space defined by (6)Mφf=φf,forf∈La2D.Mφ is a bounded linear operator on La2(D) with (7)Mφ=φ∞=supφz:z∈D.

Over the years it has been shown that many familiar classes of operators do have invariant subspaces. The lattice of shift operator acting on the Hardy space is completely described by Beurling’s Theorem [4]. Sarason (see [11]) characterized all closed invariant subspaces of the Volterra operator (8)Vfx=∫0xfydy,forf∈L20,1,0<y≤x<1.In [1], Aleman characterized boundedness and compactness of the integral operator (9)Tgfz=∫0zg′wfwdwbetween Hardy space Hp and Hq for p,q>0. In [2], using Beurling’s Theorem, Aleman and Korenblum studied the complex Volterra operator in the Hardy space H2(D) defined by (10)Vfz=∫0zfwdw.Then they characterized the lattice of closed invariant subspaces of V. Sarason (see [11]) studied the lattice of closed invariant subspaces of Mx+V acting on L2(0,1). Montes-Rodriguez et al. (see [9]) and Cowen et al. (see [5]) used the idea of Sarason to study the invariant subspaces of certain classes of composition operators on the Hardy space. Following Sarason’s work, Čučković and Paudyal (see [6]) characterized the lattice of closed invariant subspaces of the shift plus complex Volterra operator on the Hardy space. In their paper, the operator T is defined by (11)Tfz=zfz+∫0zfwdw,forf∈H2D,z∈D.Ball (see [3]) and Nordgren (see [10]) studied the problem of determining the reducing subspaces for an analytic Toeplitz operator on the Hardy space. In [12], Stessin and Zhu gave a complete description of the weighted unilateral shift operator of finite multiplicity on some Hilbert spaces type I and type II. In [13], Zhu described the properties of the commutant of analytic Toeplitz operators with inner function symbols on the Hardy space and the Bergman space and characterized the reducing subspaces of a class of multiplication operators. In 2011, Douglas and Kim in [7] studied the reducing subspaces for an analytic multiplication operator Mzn on the Bergman space Aα2(Ar) of the annulus Ar.

Based on the above works, for an n-degree polynomial g(z)(n≥2), we introduce the operator T1 defined by the multiplication operator Mg plus the weighted Volterra operator Vg on the Bergman space. We show that the operator T1 is similar to Mg on some Hilbert space Sg2(D). Then for g(z)=zn, by using matrix manipulations, the reducing subspaces of corresponding operator T2 on the Bergman space are characterized.

2. The Similarity of the Operator <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M80"><mml:mrow><mml:msub><mml:mrow><mml:mi>T</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

For an n-degree polynomial g(z)(n≥2), the operator T1 is defined by (12)T1fz=Mgzfz+Vgfz=Mgzfz+∫0zg′wfwdw,forf∈La2D.

To prove our result, we introduce the space Sg2(D) defined by (13)Sg2D=h∈HD:h0=0,Dhg′∈La2D,where H(D) is the space of holomorphic functions on the unit disk, and D=d/dz is the differentiation operator. From the definition of Sg2(D), for h∈Sg2(D), we have (14)∫DDhz2dAz=∫Dg′zDhzg′z2dAz≤g′z∞2Dhzg′z22.So Dh(z)∈La2(D). It can be shown that, for any holomorphic function h with h(0)=0 and Dh(z)∈La2(D), then h22≤Dh22. So the norm of Sg2(D) is defined by (15)hSg2D2=h22+Dh22.Corresponding inner product is given by (16)h1,h2Sg2D=h1,h2+Dh1,Dh2.

In the following, we suppose that g′(z) has no zero points on D¯. This condition guarantees that Sg2(D) is closed under the given norm.

Theorem 1.

Let Vg be the weighted Volterra operator on La2(D). Then the following statements hold:

Range of Vg is Sg2(D).

Vg is a bounded isomorphism from La2(D) to Sg2(D), and its inverse is (1/g′z)D.

The operator T1 acting on La2(D) is similar under Vg to the multiplication operator Mg on Sg2(D).

Proof.

(i) Let h be in the range of Vg, and then there exists f∈La2(D) such that (17)hz=Vgfz=∫0zg′wfwdw.So Dh(z)=g′(z)f(z), h(0)=0, and Dh(z)/g′(z)∈La2(D). Therefore, h(z)∈Sg2(D). Conversely, suppose that h∈Sg2(D); then (18)VgDhg′zz=∫0zg′wDhg′wwdw=hz-h0=hz.Hence h belongs to the range of Vg.

(ii) First we want to show Vg is a bounded operator on La2(D). For f∈La2(D), we have (19)VgfSg2D2=Vgf22+DVgf22≤2DVgf22≤2g′z∞2f22.Clearly Vg is linear. To show Vg is one to one, suppose that f1,f2∈La2(D), satisfying that (20)∫0zg′wf1wdw=∫0zg′wf2wdw,∀z∈D.Differentiating both sides, we obtain that f1(z)=f2(z) and hence Vg is one to one. From the definition of Vg we have that Vg((1/g′z)Dh)(z)=h(z), for h∈Sg2(D), and (1/g′z)D(Vgf)(z)=f(z), for f∈La2(D).

Therefore, Vg is a bounded bijective linear operator from La2(D) onto Sg2(D), and Vg-1=(1/g′)D.

(iii) Suppose f belongs to La2(D), and (Vgf)(z)=h(z). Note that (21)T1fz=Mgfz+Vgfz=gzh′zg′z+hz.Now applying Vg on both sides of the above equality, we obtain that (22)VgT1fz=∫0zg′wh′wg′wgw+hwdw=∫0zgwh′w+g′whwdw=∫0zDgwhwdw=gzhz=MgVgfz.So VgT1=MgVg and VgT1Vg-1=Mg. That is to say, Vg transforms the operator T1 into the multiplication operator Mg on Sg2(D).

3. The Reducing Subspaces of the Operator <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M159"><mml:mrow><mml:msub><mml:mrow><mml:mi>T</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>

In this section, for fixed n≥2, we consider the case of g(z)=zn. That is, the operator T2 is defined by (23)T2fz=Mznfz+Vznfz=Mznfz+∫0znwn-1fwdw,forf∈La2D.Since the n-shift operator is a contraction operator, and the operator Vzn is a bounded operators, T2 is a bounded operator on La2(D).

Let A′(T) denote the commutant of T, that is, A′(T)={S∈L(H)∣TS=ST}, where L(H) represents the collection of all bounded linear operators on a Hilbert space H. Then we have the following lemma.

Lemma 2.

Let La2(D) be the Bergman space. If P is a bounded operator on La2(D), then P∈A′(Mz) if and only if P admits the following matrix representation: (24)P=p1100⋯0⋯p21p110⋯0⋯p31γ1γ0γ1γ2p21p11⋯0⋯⋮⋮⋮⋱⋮⋯pk1γ1γ0γk-2γk-1pk-1,1γ2γ0γk-3γk-1pk-2,1⋯p11⋯⋮⋮⋮⋮⋮⋱with respect to the orthonormal basis {ek(z)=γkzk}k=0∞ of La2(D), where pjk∈C(j,k≥1) and γk=k+1. Moreover, if P is a projection, then P∈A′(Mz) if and only if P=I or 0.

Proof.

Denote the orthonormal basis of La2(D) by {ek(z)=γkzk}k=0∞. Note that (25)Mzek=zγkzk=γkγk+1ek+1. So the operator Mz admits the following matrix representation with respect to the above basis: (26)Mz=00⋯00⋯γ0γ10⋯00⋯0γ1γ2⋯00⋯⋮⋮⋮⋮⋮⋯00⋯γk-1γk0⋯⋮⋮⋮⋮⋱⋱.Suppose that P has the following matrix representation with respect to the orthonormal basis of La2(D): (27)P=p11p12p13⋯p1k⋯p21p22p23⋯p2k⋯p31p32p33⋯p3k⋯⋮⋮⋮⋮⋮pk1pk2pk3⋯pkk⋯⋮⋮⋮⋮⋮⋱.From MzP=PMz, we have (28)00⋯0⋯γ0γ1p11γ0γ1p12⋯γ0γ1p1k⋯γ1γ2p21γ1γ2p22⋯γ1γ2p2k⋯⋮⋮⋮⋮γk-2γk-1pk-1,1γk-2γk-1pk-1,2⋯γk-2γk-1pk-1,k⋯⋮⋮⋮⋮⋱=γ0γ1p12γ1γ2p13⋯γk-2γk-1p1k⋯γ0γ1p22γ1γ2p23⋯γk-2γk-1p2k⋯⋮⋮⋮⋮⋯γ0γ1pk2γ1γ2pk3⋯γk-2γk-1pkk⋯⋮⋮⋮⋮⋱.Thus (29)pij=0,i<j,pii=p11,i=2,3,…,γj+k-1γj-1pj+k,j=γj+k-2γj-2pj+k-1,j-1,j≥2,k≥1.

So we obtain (30)P=p1100⋯0⋯p21p110⋯0⋯p31γ1γ0γ1γ2p21p11⋯0⋯⋮⋮⋮⋱⋮⋯pk1γ1γ0γk-2γk-1pk-1,1γ2γ0γk-3γk-1pk-2,1⋯p11⋯⋮⋮⋮⋮⋮⋱.

Conversely, if P admits the matrix representation (30) with respect to the above basis, simple computation shows that MzP=PMz. So P∈A′(Mz). Moreover, if P is a projection, we deduce that P∈A′(Mz) if and only if P has the following form: (31)P=p11p11⋱p11⋱,where p11=1 or 0.

Remark 3.

In fact, since Mz is irreducible on La2(D), any projection in A′(Mz) is I or 0.

The following lemma will be used in the proof of main theorem.

Lemma 4.

Let Hj=span¯{enk+j:k≥0}(j=0,1,…,n-1). Then

{enk+j}k=0∞ forms an orthonormal basis of Hj.

La2(D)=H0⊕H1⊕⋯⊕Hn-1.

Hj is a reducing subspace of T2.

Proof.

(i) and (ii) are obvious. We only need to show (iii). Note that (32)T2enk+jz=Mznenk+jz+Vznenk+jz=znγnk+jznk+j+∫0znwn-1γnk+jwnk+jdw=1+nnk+1+jγnk+jγnk+1+jenk+1+j.So we have T2Hj⊂Hj and (33)T2H0⊕H1⊕⋯⊕Hj-1⊕Hj+1⊕⋯⊕Hn-1⊂H0⊕H1⊕⋯⊕Hj-1⊕Hj+1⊕⋯⊕Hn-1,as desired.

Set T2j=T2Hj(j=0,1,…,n-1). Then we have the following theorem.

Theorem 5.

If Q:La2(D)→La2(D) is a projection, then (34)QT20T21⋱T2,n-1=T20T21⋱T2,n-1Qif and only if(35)Q=Q0Q1⋱Qn-1,where Qi(i=0,1,…,n-1) is IHi or 0.

Proof.

If Q∈A′(diag(T20,T21,…,T2,n-1)), note that La2(D)=H0⊕H1⊕⋯⊕Hn-1, and then the operator Q can be decomposed in the following form: (36)Q=Q00Q01⋯Q0,n-1Q10Q11⋯Q1,n-1⋮⋮⋮⋮Qn-1,0Qn-1,1⋯Qn-1,n-1,where Qij:Hj→Hi(i,j=0,1,…,n-1). The condition Q∈A′(diag(T20,T21,…,T2,n-1)) yields that (37)Q00T20Q01T21⋯Q0,n-1T2,n-1Q10T20Q11T21⋯Q1,n-1T2,n-1⋮⋮⋮⋮Qn-1,0T20Qn-1,1T21⋯Qn-1,n-1T2,n-1=T20Q00T20Q01⋯T20Q0,n-1T21Q10T21Q11⋯T21Q1,n-1⋮⋮⋮⋮T2,n-1Qn-1,0T2,n-1Qn-1,1⋯T2,n-1Qn-1,n-1.Suppose that Qij has the following matrix representation with respect to the orthonormal basis of Hj;(38)Qij=q11ijq12ijq13ij⋯q1kij⋯q21ijq22ijq23ij⋯q2kij⋯q31ijq32ijq33ij⋯q3kij⋯⋮⋮⋮⋮⋮qk1ijqk2ijqk3ij⋯qkkij⋯⋮⋮⋮⋮⋮⋱,where qt1+1,t2+1ij=Qijet2n+j,et1n+i(t1,t2=0,1,…).

From (32), we know the operator T2j admits the following matrix representation with respect to the basis of Hj;(39)T2j=00⋯00⋯1+nn+jγjγn+j0⋯00⋯01+n2n+jγn+jγ2n+j⋯00⋯⋮⋮⋮⋮⋮⋯00⋯1+nnk+1+jγnk+jγnk+1+j0⋯⋮⋮⋮⋮⋱⋱.

Case 1 (i=j). From (37), we get QjjT2j=T2jQjj(j=0,1,…,n-1). Applying Lemma 2, we know Qjj has the form similar to (30). Note that Q is a projection, so Qjj=Qjj∗. Hence Qjj=q11jjIHj(q11jj∈R).

Case 2 (i≠j). From (37), we have that T2iQij=QijT2j. This equality is equivalent to (40)00·0·1+nn+iγiγn+iq11ij1+nn+iγiγn+iq12ij·1+nn+iγiγn+iq1kij·1+n2n+iγn+iγ2n+iq21ij1+n2n+iγn+iγ2n+iq22ij·1+n2n+iγn+iγ2n+iq2kij·····1+nk-1n+iγk-2n+iγk-1n+iqk-1,1ij1+nk-1n+iγk-2n+iγk-1n+iqk-1,2ij·1+nk-1n+iγk-2n+iγk-1n+iqk-1,kij······=1+nn+jγjγn+jq12ij1+n2n+jγn+jγ2n+jq13ij·1+nk-1n+jγk-2n+jγk-1n+jq1kij·1+nn+jγjγn+jq22ij1+n2n+jγn+jγ2n+jq23ij·1+nk-1n+jγk-2n+jγk-1n+jq2kij·1+nn+jγjγn+jq32ij1+n2n+jγn+jγ2n+jq33ij·1+nk-1n+jγk-2n+jγk-1n+jq3kij······1+nn+jγjγn+jqk2ij1+n2n+jγn+jγ2n+jqk3ij·1+nk-1n+jγk-2n+jγk-1n+jqkkij······.So we obtain (41)Qij=q11ij000⋯q21ij1+n/n+i1+n/n+jγiγjγn+jγn+iq11ij00⋯q31ij1+n/2n+i1+n/n+jγn+jγjγn+iγ2n+iq21ij1+n/n+i1+n/n+j1+n/2n+i1+n/2n+jγiγjγ2n+jγ2n+iq11ij0⋯⋮⋮⋮⋮⋮.On the other hand, from (37) we also have T2jQji=QjiT2i. In the same way as for T2iQij=QijT2j, we get (42)Qji=q11ji000⋯q21ji1+n/n+j1+n/n+iγjγiγn+iγn+jq11ji00⋯q31ji1+n/2n+j1+n/n+iγn+iγiγn+jγ2n+jq21ji1+n/n+j1+n/n+i1+n/2n+j1+n/2n+iγjγiγ2n+iγ2n+jq11ji0⋯⋮⋮⋮⋮⋮.Q is a projection that yields Qij∗=Qji. Thus we have ql1ij=0(l=2,3,…). Solving the system of equations (43)q¯11ij=q11ji,1+n/n+i1+n/n+jγiγjγn+jγn+iq¯11ij=1+n/n+j1+n/n+iγjγiγn+iγn+jq11ji,we get q11ij=0. Therefore, Qij=0. Q2=Q implies that (44)Q=Q0Q1⋱Qn-1, where Qi(i=0,1,…,n-1) is IHi or 0.

Conversely, if (45)Q=Q0Q1⋱Qn-1,where Qi(i=0,1,…,n-1) is IHi or 0, it is obvious that Q∈A′(diag(T20,T21,…,T2,n-1)).

From [14], we know that determining the reducing subspaces of T2 is equivalent to finding the projection in the commutant of T2. Thus we have the following conclusion.

Theorem 6 (main theorem).

Let La2(D) be the Bergman space. For n≥2, the operator T2 has 2n reducing subspaces with minimal reducing subspaces H0,H1,…,Hn-1.

Proof.

Suppose that QT2=T2Q for a projection Q. Note that T2La2(D)=T20⊕T21⊕⋯⊕T2,n-1. Hence(46)QT20T21⋱T2,n-1=T20T21⋱T2,n-1Q.By Theorem 5, the projection operator(47)Q=Q0Q1⋱Qn-1,where Qi(i=0,1,…,n-1) is IHi or 0. Applying Lemma 4, we have that the reducing subspaces of T2 are (48)c0H0⊕c1H1⊕⋯⊕cn-1Hn-1,whereci=0or1i=0,1,…,n-1,and the minimal reducing subspaces are H0,H1,…,Hn-1.

4. Some Consequences

In this section, we use the characterization of the reducing subspaces of T2 to obtain a description of reducing subspaces of Mzn on Sn2(D), which is similar to T2.

The space Sn2(D) is defined by (49)Sn2D=f∈HD:Df∈La2D,fl0=0,l=0,1,…,n-1.If f(z)∈Sn2(D), it follows that f(z)=∑k=0∞akzn+k, and then Df(z)=∑k=0∞(n+k)akzn+k-1. Since Df∈La2(D), we have Df(z)22=∑k=0∞(n+k)ak2, and (50)∫Dfz2dAz=∑k=0∞ak2n+k+1≤∑k=0∞n+kak2=Dfz22.So f(z)∈La2(D). The norm of Sn2(D) is defined by (51)fSn2D2=f22+Df22.Corresponding inner product is given by (52)f,gSn2D=f,g+Df,Dg.

The proof of the following theorem is similar to the proof of Theorem 1, so we omit it.

Theorem 7.

Let Vzn be the weighted Volterra operator on La2(D). Then the following statements hold:

Range of Vzn is Sn2(D).

Vzn is a bounded isomorphism from La2(D) to Sn2(D), and its inverse is (1/nzn-1)D.

(Remark: Note (1/nzn-1)D acting on Sn2(D), and from the definition of Sn2(D), we know that z=0 is a removable singular point of (1/nzn-1)Dg(z), for g(z)∈Sn2(D). We can define 1/nzn-1Dgzz=0 by limz→0(1/nzn-1)Dg(z). Then z=0 can be viewed as an analytic point of (1/nzn-1)Dg(z)).

The operator T2 acting on La2(D) is similar under Vzn to the multiplication operator Mzn on Sn2(D).

Note that Sn2(D)=span¯{zk+n:k≥0}; from (52), we have (53)zk+n,zm+nSn2D=zk+n,zm+n+k+nzk+n-1,m+nzm+n-1=k+nk+n+1+1k+n+1,k=m,0,k≠m.Thus, the orthonormal basis of Sn2(D) is given by (54)fk,nz=k+n+1k+nk+n+1+1zk+nk=0∞.

In order to characterize the reducing subspaces of Mzn on Sn2(D), we need the following lemma which is similar to Lemma 4.

Lemma 8.

Let Lj=span¯{fnk+j,n:k≥0}(j=0,1,…,n-1). Then

{fnk+j,n}k=0∞ forms an orthonormal basis of Lj.

Sn2(D)=L0⊕L1⊕⋯⊕Ln-1.

Lj is a reducing subspace of Mzn.

Corollary 9.

Let Sn2(D) be described as the above. For n≥2, the multiplication operator Mzn has 2n reducing subspaces with minimal reducing subspaces L0,L1,…,Ln-1.

Proof.

The proof of Corollary 9 is similar to Theorems 5 and 6. Here we omit it.

Remark 10.

In fact, Sn2(D) is a Hilbert space of type I which is considered by Stessin and Zhu in [12]. So Lemma 8 and Corollary 9 also follow from Stessin and Zhu’s paper [12].

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This research is supported by NNSF of China (11371119).

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