# Restriction of Toeplitz Operators on Their Reducing Subspaces

Authors: Anjian Xu; Yang Zou
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We study the restrictions of analytic Toeplitz operator on its minimal reducing subspaces for the unit disc and construct their models on slit domains. Furthermore, it is shown thatTzn is similar to the sum of n copies of the Bergman shift.


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## Body

## 1. Introduction

LetT be a bounded linear operator on a Hilbert space H; a subspace M of H is called aninvariant subspace of T if T(M)⊆M and areducing subspace of T if M is an invariant subspace of T and T∗. A reducing subspace M of T is calledminimal if for every reducing subspace M~ of T such that M~⊆M then either M~=M or M~=0. For a concrete bounded operator T on a separable Hilbert space H, it is important to determine invariant subspaces and reducing subspaces for T.LetD be the unit open disc in the complex plane and dA(z) be the normalized area measure on D. The Bergman space La2(D) consists of all analytic functions in the Lebesgue space L2(dA(z)). It is clear that La2(D) is a closed subspace of L2(dA(z)), and let P denote the projection from L2(dA(z)) onto La2(D). The Toeplitz operator Tφ on La2(D) with symbol φ∈L∞(dA(z)) is defined by (Tφf)(z)=P(φf)(z); it is called an analytic Toeplitz operator if φ∈H∞(D).Annth-order Blaschke product B is the analytic function on D given by(1)Bz=eiθ∏i=1nz-ai1-a¯iz,where θ is a real number and ai∈D for 1≤i≤n. A Blaschke product is very important in the theory of Hardy space.Characterization of reducing subspaces of an analytic Toeplitz operatorTφ on Bergman space has been of great interest for last two decades. Thomson [1, 2] showed that it suffices to study reducing subspace of TB for a finite Blaschke product in the case of Hardy space. It can be generalized to Bergman spaces easily.Zhu studied the reducing subspaces ofTB for a Blaschke product of order 2 firstly and showed that TB has exactly two distinct minimal reducing subspaces (cf. [3]). Motivated by this fact, Zhu conjectured that the number of minimal reducing subspaces of TB equals the order of B (cf. [3]). Guo et al. showed that in general this is not true (cf. [4]), and they found that the number of minimal reducing subspaces of TB equals the number of connected components of the Riemann surface of B(z)=B(w) when the order of B is 3,4,6. Then they conjectured that the number of minimal reducing subspaces of TB equals the number of connected components of the Riemann surface of B(z)=B(w) for any finite Blaschke product (called the refined Zhu’s conjecture, cf. [4]). Douglas et al. confirmed the conjecture in [5, 6] by using local inverses of Blaschke products [7]. Tikaradze [8] generalized a part of results in [5] to bounded smooth pseudoconvex domains in CN. Douglas and Kim [9] studied reducing subspace Tzn on Bergman space of the annulus; the case of Hardy space was summarized in [10].In [11], Douglas et al. generalized the bundle shift [12] to the case of Bergman spaces, constructed a vector bundle model for analytic Toeplitz operator Tφ on the Bergman space La2(D), and tried to build vector bundle models for restrictions of Tφ to its minimal reducing subspaces, but it is not completed. Douglas [13] studied unitary equivalence of the restrictions by computing their curvatures of corresponding geometric models. Hu et al. [14] showed that, for Tφ, there is a distinguished reducing subspace M0 such that the restriction of Tφ on M0 is the Bergman shift. In this paper, we analyze the concrete examples to see what are the possible models for these restrictions for further research.


## 2. Unitary Equivalence and Similarity of Weighted Shifts

LetS be an operator on a separable infinite-dimensional Hilbert space H. S is called a one-side weighted shift if there exist an orthonormal basis {ei},i=0,1,2,… for H and a bounded sequence {αi} of complex numbers such that Sei=αiei+1. Similarly, S is called a two-side weighted shift if there exist an orthonormal basis {ei},i=…,-2,-1,0,1,2,… for H and a bounded sequence {αi} of complex numbers such that Sei=αiei+1 for all i∈Z.Lemma 1 (see [15]).
Suppose thatS and T are two injective one-side weighted shifts with weights {αi} and {βi} respectively; then S is unitarily equivalent to T if and only if αi=βi for all i.Lemma 2 (see [15]).
Suppose thatS and T are two injective one-side weighted shifts with weights {αi} and {βi}, respectively; then S is similar to T if and only if there exist two constants M and N such that(2)0<M≤∏j=0iβjαj≤N<∞for all i.


## 3. Models for Restriction of Toeplitz Operators on Their Minimal Reducing Subspaces

### 3.1. The Bergman Spaces on the Slit Disc

The domainG=D∖[0,1) is called the slit disk. Let dA denote the normalized area measure on G. La2(G) is the set of analytic functions in the Lebesgue space L2(G,dA). For a nonnegative measurable function g(w) on G, we can define the weighted Bergman space with respect to g(w)dA(w) to be the set of all analytic functions in the Lebesgue space L2(G,g(w)dA). Ross studied invariant subspaces of Bergman spaces on slit domains in [16]. Aleman et al. defined and studied the Hardy space of a slit domain and in particular they studied the invariant subspace of the slit disk; one can consult [17] for details.


### 3.2. Slit Disc Models forMz2

It is easy to check that1/4|w| is a measurable function on G and that 1/4|w|dA(w) is a probability measure on G.Lemma 3.
The Bergman spaceLa2(G,1/4|w|dA(w)) contains constants and power functions {wn∣n∈N}. Furthermore, {2(2n+1)wn;n∈N} and {2n+1wn;n∈N} are orthonormal basis of La2(G,1/4dA(w)) and La2(G,1/4|w|dA(w)), respectively.Proof.
For the Bergman spaceLa2(G,1/4|w|dA(w)) and n∈N,(3)wn2=∫Gwn214wdAw=14π∫01r2ndr∫02πdθ=122n+1,so wn=1/2(2n+1).
For the Bergman spaceLa2(G,1/4dA(w)) and n∈N,(4)wn2=∫Gwn214dAw=14π∫01r2n+1dr∫02πdθ=14n+1,so wn=1/2(n+1).Lemma 4.
The multiplication operatorTw is a bounded operator on La2(G,1/4|w|dA(w)) and La2(G,1/4dA(w)).Proof.
For the Bergman spaceLa2(G,1/4|w|dA(w)), it is easy to show that(5)Tw22n+1wn=22n+1wn+1=2n+12n+322n+3wn+1.So Tw is a weighted shift with weight {2n+1/2n+3}n=0∞ and Tw≤1. Similarly, on the Bergman space La2(G,1/4dA(w)), Tw is a weighted shift with weight {n+1/n+2}n=0∞ and so bounded.We know that(6)M0=⋁zk;k≡0mod2,M1=⋁zk;k≡1mod2are the only two minimal reducing subspaces of Tz2 on La2(D).Theorem 5.
T
z
2
M
0, Tz2M1 are unitarily equivalent to the multiplication operators Tw on the following two spaces, respectively:(7)La2G,14wdAw,La2G,14dAw.Proof.
Define the following maps:(8)La2G,14dAw⟶M1V1:fw⟼zfz2La2G,14wdAw⟶M0V0:fw⟼fz2.Then V1,V0 are two isometries on M1,M0, respectively. As a fact, by changing of variable z2=w,  dz/dw=1/2w, we have(9)V1fw2=zfz22=∫Dzfz22dAz=∫Gwfw2dzdw2dAw=∫Gwfw212w2dAw=∫Gfw214dAw=fw2,so V1 is an isometry. And(10)V0fw2=fz22=∫Dfz22dAz=∫Gfw2dzdw2dAw=∫Gfw212w2dAw=∫Gfw214wdAw=fw2;it shows that V0 is an isometry. The next is to show that Tz2V1=V1Tw on La2(G,1/4dA(w)). For every function f(w)∈La2(G,1/4dA(w)), we have(11)Tz2V1fw=Tz2zfz2=z3fz2=zz2fz2=V1wfw=V1Twfw.One can show that Tz2V0=V0Tw on La2(G,1/4wdA(w)) similarly.Suppose thatφ is a Blaschke product φ=a-z/1-a¯zb-z/1-b¯z=φa(z)φb(z) with two different zeroes a and b; let kλ(z)=1-|λ|2/(1-λ¯z)2 be the normalized reproducing kernel at λ.Lemma 6 (see [3]).
Letφ=a-z/1-a¯zb-z/1-b¯z with two different zeroes a,b∈D. Let m be the geodesic midpoint between a and b. Then the operator(12)Mφ:La2D⟶La2Dhas(13)Xe=f∘φmkm;f∈La2D,f-z=fz,Xo=f∘φmkm;f∈La2D,f-z=-fz,as its only two proper reducing subspaces, where φm=m-z/1-m¯z.Theorem 7.
The restrictions ofTφ on Xe and Xo are unitarily equivalent to Tψ(w) on La2(G,1/4|w|dA(w)) and La2(G,1/4dA(w)), respectively, where ψ=φ∘φm.Proof.
Now we define the following operators:(14)La2G,14dAw⟶XoVo:fw⟼ufu2∘φmzkmz,La2G,14wdAw⟶XeVe:fw⟼fu2∘φmzkmz.Then Vo,Ve are two isometries on Xo,Xe, respectively. As a fact, by changing of variables two times, (u=φm(z), φm′(z)=1-|m|2/(1-m¯z)2=km(z) for the first time; z2=w,  dz/dw=1/2w for the second time), we have(15)Vofw2=ufu2∘φmzkmz2=∫Dufu2∘φmzkmz2dAz=∫Dufu22dAu=∫Dwfw2dzdw2dAw=∫Dwfw212w2dAw=∫Dfw214dAw=fw2,so Vo is a isometry. And(16)Vefw=fu2∘φmzkmz2=∫Dfu2∘φmzkmz2dAz=∫Dfu22dAu=∫Dfw2dzdw2dAw=∫Dfw212w2dAw=∫Dfw214wdAw=fw2;it shows that Ve is a isometry. Moreover, for any f(w)∈La2(G,1/4|w|dA(w)), we have(17)TφVefw=φzfφm2zkmz=ψφmzfφm2zkmz,where ψ(z)=φ(φm(z)) and m is the midpoint between a and b. On the other hand, (18)VeTψwfw=Veψwfw=ψφmzfφm2zkmz.Note that ψ(w) is a function in La2(G,1/4|w|dA(w)). For f(w)∈La2(G,1/4dA(w)), one can prove similarly that(19)TφVofw=VoTψwfw.


### 3.3. Weighted Shift Models

Proposition 8.
The restrictions ofTz2 on its minimal reducing subspaces are one-side weighted shifts, and they are not unitarily equivalent to each other.Proof.
We know that(20)M0=∑k=0∞akz2k;∑k=0∞2k+1ak2<∞,M1=∑k=0∞bkz2k+1;∑k=0∞2k+2bk2<∞are the two minimal reducing subspaces of Tz2 on the Bergman space La2(D). It is clear that(21)Tz22k+1z2k=2k+12k+32k+3z2k+1,Tz22k+2z2k+1=k+1k+22k+4z2k+1+1,so Tz2M0 and Tz2M1 are weighted shifts with the weight sequences {2k+1/2k+3} and {k+1/k+2}, respectively. It is clear that Tz2M0 is not unitarily equivalent to Tz2M1 by Lemma 1.Proposition 9.
The restrictions ofTz2 on its minimal reducing subspaces are one-side weighted shifts, and they are similar to each other.Proof.
LetM0 and M0 be the minimal reducing subspaces of Tφ; we know that TφM1 is the weighted shift with weight {2k+1/2k+3}k=0∞ and TφM1 is the weighted shift with weight {k+1/k+2}k=0∞. By a direct computation, we have(22)∏k=0n2k+1/2k+3k+1/k+2=∏k=0n2k+1/2k+3∏k=0nk+1/k+2=n+22n+3.So for any n, we have(23)12≤∏k=0n2k+1/2k+3k+1/k+2≤23,since n+2/2n+3 is decreasing as n increases to infinity. It implies that TφM0 is similar to TφM1 by Lemma 2.Remark 10.
T
φ
M
1 is just the Bergman shift, so we know that Tφ is similar to the direct sum of two copies of the Bergman shift.


### 3.4. For Reducing Subspace ofMzn

Theorem 11.
The slit disc models for restrictions ofTzn are the Tw on the following n spaces:(24)La2G,1n2w2n-k-1/ndAw,k=0,1,…,n-1.Proof.
Let(25)Mkn=⋁zm;m≡kmodn,k=0,1,…,n-1.We define the following n operators, for k=0,1,…,n-1:(26)La2G,1n2w2n-k-1/ndAw⟶MknVkn:fw⟼zkfzn,and then we can show that Vkn are isometries; as a fact, by changing of variable w=zn, dz/dw=1/nwn-1n, so(27)Vknfw2=zkfzn2=∫Dzkfzn2dAz=∫Dwk/nfw2dzdw2dAw=∫Dw2k/nfw21nwn-1n2dAw=∫Dfw21n2w2n-k-1/ndAw=fw2.Moreover, for any f(w)∈La2(G,1/n2|w|2(n-k-1)/ndA(w)), we have(28)TznVknfw=Tznzkfzn=zn+kfzn=zkznfzn=VknTwfw.Remark 12.
r
α
d
A
(
z
) is a finite measure on G for α>-2, so 1/n2|w|2(n-k-1)/ndA(w) is a finite measure on G.Proposition 13.
The restrictions ofTzn on its minimal reducing subspaces are one-side weighted shifts, and they are not unitarily equivalent to each other.Proof.
For the Bergman spaceLa2(G,1/n2w2(n-k-1)/ndA(w)) and any j≥0, we have(29)wj2=∫Gwj21n2w2n-k-1/ndAw=1n2π∫02πdθ∫01r2j+1-2n-k-1/ndr=1nnj+k+1,that is, w=1/n(nj+k+1). It implies that(30)Twnnj+k+1wj=nj+k+1nj+1+k+1nnj+1+k+1wj+1,that is, Tw, and so Tφ|Mkn is the weighted shift with weight {nj+k+1/n(j+1)+k+1}, and so Tw is bounded and TφMkn is not unitarily equivalent to TφMjn when k≠j.Proposition 14.
The restrictions ofTzn on its minimal reducing subspaces are one-side weighted shifts, and they are similar to each other.Proof.
T
φ
M
k
n and TφMln are the one-side weighted shift with weights {nj+k+1/n(j+1)+k+1} and {nj+l+1/n(j+1)+l+1}, where 0≤k<l≤n-1. So, for any m,(31)∏j=0mnj+k+1/nj+1+k+1nj+l+1/nj+1+l+1=∏j=0mnj+k+1/nj+1+k+1∏j=0mnj+l+1nj+1+l+1=k+1l+1nm+1+l+1nm+1+k+1.Then,(32)0<k+1l+1≤∏j=0mnj+k+1/nj+1+k+1nj+l+1/nj+1+l+1≤k+1l+1n+l+1n+k+1,since n(m+1)+l+1/n(m+1)+k+1 is decreasing as m increases to infinity. It implies that Tzn|Mkn is similar to Tzn|Mln for any k≠l.Remark 15.
For any0≤l≤n-1, TznMln is similar to TznM0n which is the Bergman shift, and so Tzn is similar to the direct sum of n copies of the Bergman shift.


## 3.1. The Bergman Spaces on the Slit Disc

The domainG=D∖[0,1) is called the slit disk. Let dA denote the normalized area measure on G. La2(G) is the set of analytic functions in the Lebesgue space L2(G,dA). For a nonnegative measurable function g(w) on G, we can define the weighted Bergman space with respect to g(w)dA(w) to be the set of all analytic functions in the Lebesgue space L2(G,g(w)dA). Ross studied invariant subspaces of Bergman spaces on slit domains in [16]. Aleman et al. defined and studied the Hardy space of a slit domain and in particular they studied the invariant subspace of the slit disk; one can consult [17] for details.


## 3.2. Slit Disc Models forMz2

It is easy to check that1/4|w| is a measurable function on G and that 1/4|w|dA(w) is a probability measure on G.Lemma 3.
The Bergman spaceLa2(G,1/4|w|dA(w)) contains constants and power functions {wn∣n∈N}. Furthermore, {2(2n+1)wn;n∈N} and {2n+1wn;n∈N} are orthonormal basis of La2(G,1/4dA(w)) and La2(G,1/4|w|dA(w)), respectively.Proof.
For the Bergman spaceLa2(G,1/4|w|dA(w)) and n∈N,(3)wn2=∫Gwn214wdAw=14π∫01r2ndr∫02πdθ=122n+1,so wn=1/2(2n+1).
For the Bergman spaceLa2(G,1/4dA(w)) and n∈N,(4)wn2=∫Gwn214dAw=14π∫01r2n+1dr∫02πdθ=14n+1,so wn=1/2(n+1).Lemma 4.
The multiplication operatorTw is a bounded operator on La2(G,1/4|w|dA(w)) and La2(G,1/4dA(w)).Proof.
For the Bergman spaceLa2(G,1/4|w|dA(w)), it is easy to show that(5)Tw22n+1wn=22n+1wn+1=2n+12n+322n+3wn+1.So Tw is a weighted shift with weight {2n+1/2n+3}n=0∞ and Tw≤1. Similarly, on the Bergman space La2(G,1/4dA(w)), Tw is a weighted shift with weight {n+1/n+2}n=0∞ and so bounded.We know that(6)M0=⋁zk;k≡0mod2,M1=⋁zk;k≡1mod2are the only two minimal reducing subspaces of Tz2 on La2(D).Theorem 5.
T
z
2
M
0, Tz2M1 are unitarily equivalent to the multiplication operators Tw on the following two spaces, respectively:(7)La2G,14wdAw,La2G,14dAw.Proof.
Define the following maps:(8)La2G,14dAw⟶M1V1:fw⟼zfz2La2G,14wdAw⟶M0V0:fw⟼fz2.Then V1,V0 are two isometries on M1,M0, respectively. As a fact, by changing of variable z2=w,  dz/dw=1/2w, we have(9)V1fw2=zfz22=∫Dzfz22dAz=∫Gwfw2dzdw2dAw=∫Gwfw212w2dAw=∫Gfw214dAw=fw2,so V1 is an isometry. And(10)V0fw2=fz22=∫Dfz22dAz=∫Gfw2dzdw2dAw=∫Gfw212w2dAw=∫Gfw214wdAw=fw2;it shows that V0 is an isometry. The next is to show that Tz2V1=V1Tw on La2(G,1/4dA(w)). For every function f(w)∈La2(G,1/4dA(w)), we have(11)Tz2V1fw=Tz2zfz2=z3fz2=zz2fz2=V1wfw=V1Twfw.One can show that Tz2V0=V0Tw on La2(G,1/4wdA(w)) similarly.Suppose thatφ is a Blaschke product φ=a-z/1-a¯zb-z/1-b¯z=φa(z)φb(z) with two different zeroes a and b; let kλ(z)=1-|λ|2/(1-λ¯z)2 be the normalized reproducing kernel at λ.Lemma 6 (see [3]).
Letφ=a-z/1-a¯zb-z/1-b¯z with two different zeroes a,b∈D. Let m be the geodesic midpoint between a and b. Then the operator(12)Mφ:La2D⟶La2Dhas(13)Xe=f∘φmkm;f∈La2D,f-z=fz,Xo=f∘φmkm;f∈La2D,f-z=-fz,as its only two proper reducing subspaces, where φm=m-z/1-m¯z.Theorem 7.
The restrictions ofTφ on Xe and Xo are unitarily equivalent to Tψ(w) on La2(G,1/4|w|dA(w)) and La2(G,1/4dA(w)), respectively, where ψ=φ∘φm.Proof.
Now we define the following operators:(14)La2G,14dAw⟶XoVo:fw⟼ufu2∘φmzkmz,La2G,14wdAw⟶XeVe:fw⟼fu2∘φmzkmz.Then Vo,Ve are two isometries on Xo,Xe, respectively. As a fact, by changing of variables two times, (u=φm(z), φm′(z)=1-|m|2/(1-m¯z)2=km(z) for the first time; z2=w,  dz/dw=1/2w for the second time), we have(15)Vofw2=ufu2∘φmzkmz2=∫Dufu2∘φmzkmz2dAz=∫Dufu22dAu=∫Dwfw2dzdw2dAw=∫Dwfw212w2dAw=∫Dfw214dAw=fw2,so Vo is a isometry. And(16)Vefw=fu2∘φmzkmz2=∫Dfu2∘φmzkmz2dAz=∫Dfu22dAu=∫Dfw2dzdw2dAw=∫Dfw212w2dAw=∫Dfw214wdAw=fw2;it shows that Ve is a isometry. Moreover, for any f(w)∈La2(G,1/4|w|dA(w)), we have(17)TφVefw=φzfφm2zkmz=ψφmzfφm2zkmz,where ψ(z)=φ(φm(z)) and m is the midpoint between a and b. On the other hand, (18)VeTψwfw=Veψwfw=ψφmzfφm2zkmz.Note that ψ(w) is a function in La2(G,1/4|w|dA(w)). For f(w)∈La2(G,1/4dA(w)), one can prove similarly that(19)TφVofw=VoTψwfw.


## 3.3. Weighted Shift Models

Proposition 8.
The restrictions ofTz2 on its minimal reducing subspaces are one-side weighted shifts, and they are not unitarily equivalent to each other.Proof.
We know that(20)M0=∑k=0∞akz2k;∑k=0∞2k+1ak2<∞,M1=∑k=0∞bkz2k+1;∑k=0∞2k+2bk2<∞are the two minimal reducing subspaces of Tz2 on the Bergman space La2(D). It is clear that(21)Tz22k+1z2k=2k+12k+32k+3z2k+1,Tz22k+2z2k+1=k+1k+22k+4z2k+1+1,so Tz2M0 and Tz2M1 are weighted shifts with the weight sequences {2k+1/2k+3} and {k+1/k+2}, respectively. It is clear that Tz2M0 is not unitarily equivalent to Tz2M1 by Lemma 1.Proposition 9.
The restrictions ofTz2 on its minimal reducing subspaces are one-side weighted shifts, and they are similar to each other.Proof.
LetM0 and M0 be the minimal reducing subspaces of Tφ; we know that TφM1 is the weighted shift with weight {2k+1/2k+3}k=0∞ and TφM1 is the weighted shift with weight {k+1/k+2}k=0∞. By a direct computation, we have(22)∏k=0n2k+1/2k+3k+1/k+2=∏k=0n2k+1/2k+3∏k=0nk+1/k+2=n+22n+3.So for any n, we have(23)12≤∏k=0n2k+1/2k+3k+1/k+2≤23,since n+2/2n+3 is decreasing as n increases to infinity. It implies that TφM0 is similar to TφM1 by Lemma 2.Remark 10.
T
φ
M
1 is just the Bergman shift, so we know that Tφ is similar to the direct sum of two copies of the Bergman shift.


## 3.4. For Reducing Subspace ofMzn

Theorem 11.
The slit disc models for restrictions ofTzn are the Tw on the following n spaces:(24)La2G,1n2w2n-k-1/ndAw,k=0,1,…,n-1.Proof.
Let(25)Mkn=⋁zm;m≡kmodn,k=0,1,…,n-1.We define the following n operators, for k=0,1,…,n-1:(26)La2G,1n2w2n-k-1/ndAw⟶MknVkn:fw⟼zkfzn,and then we can show that Vkn are isometries; as a fact, by changing of variable w=zn, dz/dw=1/nwn-1n, so(27)Vknfw2=zkfzn2=∫Dzkfzn2dAz=∫Dwk/nfw2dzdw2dAw=∫Dw2k/nfw21nwn-1n2dAw=∫Dfw21n2w2n-k-1/ndAw=fw2.Moreover, for any f(w)∈La2(G,1/n2|w|2(n-k-1)/ndA(w)), we have(28)TznVknfw=Tznzkfzn=zn+kfzn=zkznfzn=VknTwfw.Remark 12.
r
α
d
A
(
z
) is a finite measure on G for α>-2, so 1/n2|w|2(n-k-1)/ndA(w) is a finite measure on G.Proposition 13.
The restrictions ofTzn on its minimal reducing subspaces are one-side weighted shifts, and they are not unitarily equivalent to each other.Proof.
For the Bergman spaceLa2(G,1/n2w2(n-k-1)/ndA(w)) and any j≥0, we have(29)wj2=∫Gwj21n2w2n-k-1/ndAw=1n2π∫02πdθ∫01r2j+1-2n-k-1/ndr=1nnj+k+1,that is, w=1/n(nj+k+1). It implies that(30)Twnnj+k+1wj=nj+k+1nj+1+k+1nnj+1+k+1wj+1,that is, Tw, and so Tφ|Mkn is the weighted shift with weight {nj+k+1/n(j+1)+k+1}, and so Tw is bounded and TφMkn is not unitarily equivalent to TφMjn when k≠j.Proposition 14.
The restrictions ofTzn on its minimal reducing subspaces are one-side weighted shifts, and they are similar to each other.Proof.
T
φ
M
k
n and TφMln are the one-side weighted shift with weights {nj+k+1/n(j+1)+k+1} and {nj+l+1/n(j+1)+l+1}, where 0≤k<l≤n-1. So, for any m,(31)∏j=0mnj+k+1/nj+1+k+1nj+l+1/nj+1+l+1=∏j=0mnj+k+1/nj+1+k+1∏j=0mnj+l+1nj+1+l+1=k+1l+1nm+1+l+1nm+1+k+1.Then,(32)0<k+1l+1≤∏j=0mnj+k+1/nj+1+k+1nj+l+1/nj+1+l+1≤k+1l+1n+l+1n+k+1,since n(m+1)+l+1/n(m+1)+k+1 is decreasing as m increases to infinity. It implies that Tzn|Mkn is similar to Tzn|Mln for any k≠l.Remark 15.
For any0≤l≤n-1, TznMln is similar to TznM0n which is the Bergman shift, and so Tzn is similar to the direct sum of n copies of the Bergman shift.



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*Source: 2190418-2017-09-07.xml*