KYUNGPOOK Math. J. 2019; 59(2): 315-324
The Flatness Property of Local-cubically Hyponormal Weighted Shifts
Seunghwan Baek and Hyunjin Do, Mi Ryeong Lee,Chunji Li
Department of Mathematics, Kyungpook National University, Daegu 41566, Korea
e-mail : seunghwan@knu.ac.kr and jinjinyy@daum.net

Institute of Liberal Education, Daegu Catholic University, Gyeongsan, Gyeongbuk 38430, Korea
e-mail : leemr@cu.ac.kr

Department of Mathematics, Northeastern University, Shenyang 110819, P. R. China
e-mail : lichunji@mail.neu.edu.cn
* Corresponding Author.
Received: October 16, 2018; Accepted: February 20, 2019; Published online: June 23, 2019.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In this note we introduce a local-cubically hyponormal weighted shift of order θ with $0≤θ≤π2$, which is a new notion between cubic hyponormality and quadratic hyponormality of operators. We discuss the property of flatness for local-cubically hyponormal weighted shifts.

Keywords: quadratically hyponormal, cubically hyponormal, weighted shifts.
1. Introduction and Preliminaries

Let ℋ be a separable, infinite dimensional, complex Hilbert space and let B(ℋ) denote the algebra of all bounded linear operators on ℋ. An operator TB(ℋ) is said to be normal if T*T = TT*, hyponormal if T*TTT*, and subnormal if there exists a normal operator N on some Hilbert space such that T = N|. To discuss gaps between hyponormality and subnormality, several classes of operators have been introduced, for example, k-hyponormal and weakly k-hyponormal operators (cf. [3]), whose definitions will be given below. An n-tuple (T1, ···, Tn) of operators in B(ℋ) is hyponormal if the operator matrix $([Tj*,Ti])i,j=1n$ is positive on the direct sum of n copies of ℋ, where [X, Y] = XYYX for X, YB(ℋ). An n-tuple (T1, ···, Tn) is weakly hyponormal if λ1T1 + ··· + λnTn is hyponormal for every λi ∈ ℂ, i = 1, ···, n, where ℂ is the set of complex numbers. For k ≥ 1 and TB(ℋ), T is k-hyponormal if (I, T, ···, Tk) is hyponormal. An operator TB(ℋ) is weakly k-hyponormal if (T, T2, ···, Tk) is weakly hyponormal. It is well known that subnormal ⇒ k-hyponormal ⇒ weakly k-hyponormal, for every k ≥ 1. In particular, weak 2- and weak 3-hyponormality are often referred to as quadratic- and cubic-hyponormality [5, 6, 8, 9, 10].

The weighted shifts have played a fundamental role in studying properties of weak subnormality. Indeed the flatness of weighted shift operators makes an important role to detect the structure of k-hyponormality and weak k-hyponormality of weighted shifts [2, 4, 5, 6]. In 1966, Stampfli [13] proved that if a weighted shift Wα with a weight sequence $α={αn}n=0∞$ is subnormal with αk = αk+1 for some k ≥ 0, then α1 = α2 = ···. In 1990, Curto [4] showed that Stampfli’s result on flatness holds in the case of 2-hyponormality. It is well known that the associated weighted shift Wα with a weight sequence $α:2/3,2/3,(k+1)/(k+2) (k≥2)$ is quadratically hyponormal but not 2-hyponormal [3, 4]. Recently Cho-Lee-Li [11] showed such the property of flatness also preserves when Wα is a cubically hyponormal weighted shift. Also, in [7] they proved that a semi-cubically hyponormal weighted shift Wα with αk = αk+1 for some k ≥ 1 has α1 = α2 = ···, i.e., it has flatness property. This paper is a continuation of such study.

This paper consists of four sections. In Section 2 we recall the definition of cubically hyponormal weighted shifts and obtain a recursive formula of determinants for a pentadiagonal matrix corresponding to its operator. In Section 3 we introduce a new notion between cubic hyponormality and quadratic hyponormality. For $0<θ≤π2$, we define a local-cubically hyponormal of order θ, which will serve as a bridge of gaps between quadratic hyponormality and semi-cubic hyponormality. In Section 4, we discuss the flatness of local-cubic hyponormal weighted shifts of some order θ. Finally, we prove that a local-cubically hyponormal weighted shift Wα with first three equal weights of order θ satisfies the flatness property.

Some of the calculations in this paper were obtained through computer experiments using the software tool Mathematica [15].

2. Determinants for a Pentadiagonal Matrix

Let Wα be a weighted shift with a weight sequence $α={αi}i=0∞$. We recall that a weighted shift Wα is said to be cubically hyponormal if

$[(Wα+sWα2+tWα3)*,Wα+sWα2+tWα3]≥0 for all s,t∈ℂ.$

In particular, if we take s = 0 in (2.1), the weighted shift Wα is semi-cubically hyponormal [1, 7]. Also, if t = 0 in (2.1), then Wα is quadratically hyponormal. Let Pn denote the orthogonal projection onto $∨i=0n{ei}$. For n ≥ 0 and s, t ∈ ℂ, define

$Dn:=Dn(s,t)≡Pn[(Wα+sWα2+tWα3)*,Wα+sWα2+tWα3]Pn=(q0r0z0000 r¯0q1r1z100 z¯0r¯1q2r2z20 0z¯1r¯2q3r3z3⋱ ⋱⋱⋱⋱⋱⋱0 ⋱⋱⋱⋱⋱zn-2 ⋱⋱⋱⋱rn-1 0z¯n-2r¯n-1qn)$

where

$qn:=αn2-αn-12+(αn2αn-12-αn-22αn-12)∣s∣2+(αn2αn+12αn+22-αn-32αn-22αn-12)∣t∣2,rn:=αn(αn+12-αn-12)s¯+αn(αn+12αn+22-αn-12αn-22)st¯,zn:=αnαn+1(αn+22-αn-12)t¯,$

with α−3 = α−2 = α−1 = 0. It is obvious that if Wα is cubically hyponormal if and only if Dn(s, t) ≥ 0 for every s, t ∈ ℂ and n ≥ 0. To detect the positivity of Dn(s, t) for every s, t ∈ ℂ and n ≥ 0, we usually utilize the Sylvester’s criterion, which is called the nested determinants test (see [5, p.213]). We consider the determinant for the matrix Dn, dndn(s, t): = det Dn. By simple computations, we get

$d0=q0, d1=q0q1-∣r0∣2,d2=q2d1-∣r1∣2q0-q1∣z0∣2+r0r1z¯0+z0r¯0r¯1,d3=q3d2-∣r2∣2d1+(∣z0∣2-q0q2)∣z1∣2+ 2q0Re(r1r2z¯1)-2Re(r0r¯2z¯0z1),d4=q4d3-∣r3∣2d2+(2Re(r2r3z¯2)-q3∣z2∣2) d1-d0(2Re(r1r¯3z¯1z2)-∣z1∣2∣z2∣2)+2Re(r0r3z¯0z1z¯2).$

In order to find the recursive formula of the determinant dn for any n ≥ 4, we consider the following pentadiagonal submatrices in (2.2) as follows

$Φn-2=(q0r0z00 r¯0q1r1z1⋱ z¯0r¯1q2r2⋱0 ⋱⋱⋱⋱zn-50 0z¯n-6r¯n-5qn-4rn-4zn-4 0z¯n-5r¯n-4qn-3rn-3 00z¯n-3r¯n-2),$$Ψn-2=(q0r0z00⋯ r¯0q1r1z1⋱ z¯0r¯1q2⋱⋱0 ⋱⋱⋱⋱zn-50 0z¯n-6r¯n-5qn-4rn-40 0z¯n-5r¯n-4qn-3rn-3 0z¯n-4r¯n-3rn-2),$

where qi, ri and zi (i ≥ 0) are given in (2.3). In particular, we give initial matrices of Φi and Ψi for i = 0, 1 as follows: Φ0: = (0), Ψ0: = (r0) (which are considered as 1 × 1 matrices for our convenience),

$Φ1:=(q0r0z¯0r¯1) and Ψ1:=(q0z0r¯0r1).$

Now we denote φj and ψj for the determinants of the matrices Φj and Ψj (j ≥ 0) in (2.5) and (2.6), respectively. Then it is obvious that ψj = φ̄j for all j ≥ 0. Moreover we can have the following Lemma 2.1.

### Lemma 2.1

For any j ≥ 0, it holds that:

$φj+1=r¯j+1dj-z¯jψj and ψj+1=rj+1dj-zjφj,$

where dn denotes for the determinant of the matrix Dn in (2.2).

We now give the recursive formula of determinant of pentadiagonal matrix in (2.2) which improves the main theorem in [14]; see Corollary 2.3 below.

### Theorem 2.2

Let the matrix Dn be given by (2.2) and let dn = det Dn (n ≥ 1). Then the following recursive formula holds:

$dn=qndn-1-∣rn-1∣2dn-2+(2Re(z¯n-2rn-1rn-2)-qn-1∣zn-2∣2) dn-3-(2Re(r¯n-3z¯n-3rn-2rn-1zn-3)-∣zn-3∣2∣zn-2∣2) dn-4+2Re(r¯n-1z¯n-3zn-2zn-4φn-4)$

with di ≡ 1 and φi ≡ 1 for i = −3, −2, −1.

Proof

It is obvious from (2.4) that the recursive formula of dn holds for n ≤ 3. To prove this theorem, we use the pentadiagonal submatrix in (2.5) for all n ≥ 4. It follows from computations that

$dn=qndn-1-rn-1φn-1+zn-2(r¯n-1φn-2-qn-1z¯n-2dn-3+∣zn-3∣2z¯n-2dn-4).$

Applying Lemma 2.1 to (2.7) with φn−1, we have

$dn=qndn-1-rn-1(r¯n-1dn-2-z¯n-2ψn-2)+zn-2r¯n-1φn-2-∣zn-2∣2qn-1dn-3+∣zn-3∣2∣zn-2∣2dn-4=qndn-1-∣rn-1∣2dn-2-qn-1∣zn-2∣2dn-3+∣zn-3∣2∣zn-2∣2dn-4+zn-2r¯n-1φn-2+rn-1z¯n-2ψn-2,$

for all n ≥ 4. Now we apply Lemma 2.1 to (2.7) with φn−2 and ψn−2, respectively, we can obtain that

$dn=qndn-1-∣rn-1∣2dn-2+(2Re(z¯n-2rn-1rn-2)-qn-1∣zn-2∣2) dn-3+∣zn-3∣2∣zn-2∣2dn-4-zn-3z¯n-2rn-1φn-3-z¯n-3zn-2r¯n-1ψn-3.$

Repeating the above methods and from ψj = φ̄j, we have the recurrence formula for dn for n ≥ 4.

In particular, by considering Theorem 2.2 with real numbers rn and zn, we can obtain the recursive formula for dn in [14] easily as following.

### Corollary 2.3

Let the matrix Dn be given by (2.2) and let n ≥ 1. Then

$dn=(qn-rn-1zn-2rn-2) dn-1-(rn-12-qn-1rn-1zn-2rn-2) dn-2-zn-2(qn-1zn-2-rn-2rn-1)dn-3+zn-32zn-2(zn-2-qn-2rn-1rn-2) dn-4+zn-42zn-32zn-2rn-1rn-2dn-5,$

where di ≡ 1 for i = −4, −3, −2, −1.

3. Basic Construction

Recall that cubic hyponormality implies semi-cubic hyponormality, and semi-cubic hyponormality and quadratic hyponormality are independent one from another [12]. We now define a new notion, namely local-cubic hyponormality between cubic hyponormality and quadratic hyponormality.

### Definition 3.1

Let $α={αi}i=0∞$ be a sequence of positive real numbers and let Wα be the associated weighted shift with a sequence α. For $θ∈[0,π2]$, a weighted shift Wα is called a local-cubically hyponormal of order θ if $Wα+s(cosθ)Wα2+s(sinθ)Wα3$ is hyponormal for all s ∈ ℂ, i.e.,

$[(Wα+s cos θWα2+s sin θWα3)*,Wα+s cos θWα2+s sin θWα3]≥0, s∈ℂ.$

### Remark 3.2

Let Wα be the weighted shift with a weight sequence $α={αi}i=0∞$. Then the following assertions hold.

• Wα is local-cubically hyponormal of order 0 if and only if it is quadratically hyponormal.

• Wα is local-cubically hyponormal of order $π2$ if and only if it is semi-cubically hyponormal.

### Proposition 3.3

Let$α:23,23,{n+1n+2}n=2∞$and let Wα be the associated weighted shift. Then there exists a subinterval J of$(0,π2)$such that, for any θJ, Wα can not be local-cubically hyponormal of order θ.

Proof

It is well-known that Wα is a quadratically hyponormal ([3]). Thus Wα is a local-cubically hyponormal of order 0 obviously. By [7, Coro.3.7], Wα is local-cubically hyponormal of order $π2$ obviously. Now we may consider t = s tan θ in (2.1) for s ∈ ℂ{0} and $θ∈(0,π2)$. From direct computations for the determinant of the matrix D5 in (2.2) with $s=1421000$, we can obtain that

$d5(θ):=det D5=50419966796875×1029∑k=012ηk tank θ,$

where

• η0 = 40670274983630949477389024, η1 = −574431783234080178778288000,

• η2 = 2047797391244271274743389024, η3 = −205644859254080233768212000,

• η4 = 452898903830369167032899466, η5 = −81082893891528184974414000,

• η6 = 74975691886145969632638027, η7 = −1732543363600864086621000,

• η8 = 4077348011641935048888990, η9 = −39613104620739962784000,

• η10 = 292868801146745266253193, η11 = 0, η12 = 7910241828954011318430.

We can check that d5 < 0 at $θ=9π200$, which proves this proposition.

In Proposition 3.3, the weighted shift Wα with a weight $α:23,23,{n+1n+2}n=2∞$ is not local-cubically hyponormal of some order in $(0,π2)$. We do not know whether there exists $θ∈(0,π2)$ such that Wα is local-cubically hyponormal of order θ. We suggest a problem as following.

### Problem 3.4

Let $α:23,23,{n+1n+2}n=2∞$ and let Wα be the associated weighted shift. For arbitrary $θ∈(0,π2)$, is it true that Wα is not a local-cubically hyponormal weighted shift of order θ?

4. Flatness

Recall that if $α:2/3,2/3,(k+1)/(k+2) (k≥2)$, then it is well-known that the associated weighted shift Wα is quadratically hyponormal, i.e., the local-cubical hyponormality with order 0 does not preserve the flatness. It follows from [7, Lemma 3.10] that if Wα is a local-cubically hyponormal weighted shift of order $π2$ with α1 = α2, then α1 = α2 = ···. Hence the following natural question arises.

Question: For$θ∈(0,π2)$, suppose that Wα is local-cubically hyponormal weighted shift of order θ such that α0 = α1 = α2. Is it true α0 = α1 = α2 = ···?

The answer of this question is affirmative (see Theorem 4.2). Before proving Theorem 4.2, we give an elementary lemma of the outer propagation for reader’s convenience.

### Lemma 4.1

Let$α={αi}i=0∞$be a sequence of positive real numbers and let Wα be the associated weighted shift with a sequence α. If Wα is local-cubically hyponormal of order θ, then the weighted shift Wαwith a weight sequence$α′={αn}n=k∞$is also local-cubically hyponormal of order θ, k ≥ 1.

Note that Lemma 4.1 can be extended to the case of any weakly n-hyponormal weighted shifts.

### Theorem 4.2

For$θ∈(0,π2)$, suppose that a weighted shift Wα is local-cubically hyponormal of order θ such that α0 = α1 = α2. Then α0 = α1 = α2 = α3 = ···, i.e. Wα satisfies the flatness property.

Proof

Without loss of generality, we may assume that α0 = α1 = α2 = 1, and put α3 := 1 + h, α4 := 1 + h + k, α5 := 1 + h + k + ℓ and α6 := 1 + h + k + ℓ + m with h, k, ℓ, m ≥ 0. As in the proof of Proposition 3.3, we substitute t = s tan θ for $θ∈(0,π2)$. Estimating the determinant of the matrix D4 in (2.2), we have

$d4(s,h,k,ℓ,m)=s4∑n=06cnsn$

with

$c0=-h3(2+h)3(1+h+k)2 tan4 θ,c1=4h3(2+h)3(1+h+k)2 tan4 θ,$

and cn are some polynomials in h, k, ℓ and m for n = 2, …, 6; in fact these are fairly lengthy polynomials (see Appendix for their expressions). The readers can find them by direct computation with a computer program. By the local-cubic hyponormality of Wα, obviously it holds that

$d4(s,h,k,ℓ,m)≥0, s≥0.$

From (4.2) we have c0 ≥ 0. Since h ≥ 0, by (4.1) h = 0. Hence α3 = 1. By Lemma 4.1, we get αi = 1 for all i ≥ 4. Hence the proof is complete.

Acknowledgements

The authors would like to express their gratitude to the referee for careful reading of the paper and helpful comments.

Appendix

Expressions of polynomials cn in the proof of Theorem 4.2.

• For n = 2, $c2=-h(2+h) tan2 θ∑i=02c2i tan2i θ,$

where $c20=6h2(2+h)2(1+h+k)2, c21=∑j=02φ21jℓj, c22=(1+h+k)2∑j=02φ22jmj,$

$φ210:=Σj=06φ210jkj$, ϕ211 := −2(1 + h + k)3(2h + h2 + 2k + 2hk)(4 + 6h + 3h2 + 2k + 2hk),

ϕ212 := −(1 + h + k)2(2h + h2 + 2k + 2hk)(4 + 6h + 3h2 + 2k + 2hk),

$φ220:=Σi=04φ220iℓi$, ϕ221 := −2(2h + h2 + k + hk)(2 + 2h + h2 + k + hk)(1 + h + k + ℓ)3,

ϕ222 := −(2h + h2 + k + hk)(2 + 2h + h2 + k + hk)(1 + h + k + ℓ)2,

ϕ2100 := −h2(1 + h)2(2 + h)2(7 + 6h + 3h2), ϕ2101 := −4h(1 + h)3(2 + h)(8 + 10h + 5h2),

ϕ2102 := −2(1 + h)2(4 + 6h + 3h2)(4 + 18h + 9h2), ϕ2103 := −4(1 + h)(16 + 72h + 112h2 + 76h3 + 19h4),

ϕ2104 := −56 − 232h − 352h2 − 236h3 − 59h4, ϕ2105 := −24(1 + h)3, ϕ2106 := −4(1 + h)2,

$φ2200:=Σj=06φ2200jkj$,

ϕ2201 := 2(1 + h + k)3(−6h−3h2 + 8h3 + 12h4 + 6h5 + h6 −4k −12hk −12h2k −4h3k −2k2 −4hk2 −2h2k2),

ϕ2202 := (1 + h + k)2(−22h−27h2−8h3 + 8h4 + 6h5 + h6−12k−36hk−36h2k−12h3k−6k2−12hk2−6h2k2),

ϕ2203 := −4(1 + h + k)(2h + h2 + k + hk)(2 + 2h + h2 + k + hk), ϕ2204 := −(2h + h2 + k + hk)(2 + 2h + h2 + k + hk),

ϕ22000 := h3(1 + h)2(2 + h)3(2 + 2h + h2), ϕ22001 := 2(1 + h)3(−1 − 8h + 20h3 + 25h4 + 12h5 + 2h6),

ϕ22002 := 3(1 + h)2(3 + 4h + 2h2)(−1 − 4h + 2h2 + 4h3 + h4),

ϕ22003 := 4(1 + h)(−4 − 18h − 21h2 − 4h3 + 9h4 + 6h5 + h6),

ϕ22004 := −14 − 58h − 81h2 − 44h3h4 + 6h5 + h6, ϕ22005 := −6(1 + h)3, ϕ22006 := −(1 + h)2.

• For n = 3, $c3=-2h(2+h) tan θ∑i=02c3i tan2i θ,$

where $c30=-2h2(2+h)2(1+h+k)2, c31=∑j=02φ31jℓj, c32=(1+h+k)2∑i=02φ32imi,$

$φ310:=Σj=06φ310jkj$,

ϕ311 := 2(1 + h + k)3(14h + 27h2 + 20h3 + 5h4 + 12k + 36hk + 36h2k + 12h3k + 6k2 + 12hk2 + 6h2k2),

ϕ312 := (1 + h + k)2(14h + 27h2 + 20h3 + 5h4 + 12k + 36hk + 36h2k + 12h3k + 6k2 + 12hk2 + 6h2k2),

$φ320:=Σj=04φ320jℓj$,

ϕ321 := 4(2h + h2 + k + hk)(2 + 2h + h2 + k + hk)(1 + h + k + ℓ)3,

ϕ322 := 2(2h + h2 + k + hk)(2 + 2h + h2 + k + hk)(1 + h + k + ℓ)2,

ϕ3100 := h2(1 + h)2(2 + h)2(11 + 10h + 5h2), ϕ3101 := 2h(1 + h)3(2 + h)(25 + 32h + 16h2),

ϕ3102 := (1 + h)2(48 + 298h + 485h2 + 336h3 + 84h4), ϕ3103 := 4(1 + h)(24 + 110h + 171h2 + 116h3 + 29h4),

ϕ3104 := 84 + 350h + 531h2 + 356h3 + 89h4, ϕ3105 := 36(1 + h)3, ϕ3106 := 6(1 + h)2,

$φ3200:=Σi=06φ3200iki, φ3201:=2(1+h+k)Σi=04φ3201iki,φ3202:=Σi=04φ3202iki$,

ϕ32000 := h2(1 + h)2(2 + h)2(3 + 2h + h2), ϕ32001 := 2h(1 + h)3(2 + h)(7 + 8h + 4h2),

ϕ32002 := (1 + h)2(8 + 86h + 139h2 + 96h3 + 24h4), ϕ32003 := 4(1 + h)(6 + 34h + 53h2 + 36h3 + 9h4),

ϕ32004 := 26 + 114h + 173h2 + 116h3 + 29h4, ϕ32005 := 12(1 + h)3, ϕ32006 := 2(1 + h)2,

ϕ32010 := 3h(1 + h)2(2 + h)(2 + 2h + h2), ϕ32011 := 2(1 + h)(2 + 26h + 41h2 + 28h3 + 7h4),

ϕ32012 := 18 + 90h + 137h2 + 92h3 + 23h4, ϕ32013 := 16(1 + h)3, ϕ32014 := 4(1 + h)2,

ϕ32020 := 11h(1 + h)2(2 + h)(2 + 2h + h2), ϕ32021 := 2(1 + h)(10 + 90h + 137h2 + 92h3 + 23h4),

ϕ32022 := 58 + 282h + 425h2 + 284h3 + 71h4, ϕ32023 := 48(1 + h)3, ϕ32024 := 12(1 + h)2.

• For n = 4, $c4=∑i=04c4i tan2i θ,$

where

• c40 = h(2 + h)(2h + h2 + k + hk)(2 + 2h + h2 + k + hk),

• c41 = h(1 + h)2(2 + h)(1 + h + k)2(h + k + ℓ)(2 + h + k + ℓ),

• c42 = (1 + h)2(h + k)(2 + h + k)τ1τ2,

• c43 = (1 + h)4(1 + h + k)2(2h + h2 + k + hk)(2 + 2h + h2 + k + hk)(1 + h + k + ℓ)2,

• c44 = (1 + h)4(1 + h + k)2τ1τ2,

τ1 := 3h + 3h2 + h3 + 2k + 4hk + 2h2k + k2 + hk2 + ℓ + 2hℓ + h2ℓ + kℓ + hkℓ,

τ2 := 2 + 3h + 3h2 + h3 + 2k + 4hk + 2h2k + k2 + hk2 + ℓ + 2hℓ + h2ℓ + kℓ + hkℓ.

For cases n = 5 and n = 6, we can have some polynomials c5 and c6 in h, k, ℓ, m. But we leave these expressions to the interested readers.

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