Kyungpook Mathematical Journal 2016; 56(3): 899-910  
On Semi-cubically Hyponormal Weighted Shifts with First Two Equal Weights
Seunghwan Baek1, Il Bong Jung1, George R. Exner2, Chunji Li3
Kyungpook National University, Daegu 702-701, Korea1
Bucknell University, Lewisburg, Pennsylvania 17837, USA2
Northeastern University, Shenyang 110004, P. R. China3
Received: February 24, 2016; Accepted: April 19, 2016; Published online: September 1, 2016.
© Kyungpook Mathematical Journal. All rights reserved.

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Abstract

It is known that a semi-cubically hyponormal weighted shift need not satisfy the flatness property, in which equality of two weights forces all or almost all weights to be equal. So it is a natural question to describe all semi-cubically hyponormal weighted shifts Wα with first two weights equal. Let α:1,1,x,(u,v,w) be a backward 3-step extension of a recursively generated weight sequence with 1 < x < u < v < w and let Wα be the associated weighted shift. In this paper we characterize completely the semi-cubical hyponormal Wα satisfying the additional assumption of the positive determinant coefficient property, which result is parallel to results for quadratic hyponormality.

Keywords:

weighted shifts, hyponormality, semi-cubical hyponormality.

1. Introduction and Notation

Let ℋ be a separable infinite dimensional complex Hilbert space and let ℒ(ℋ) be the algebra of all bounded linear operators on ℋ. For A, B ∈ ℒ(ℋ), we set [A,B] := ABBA. A k-tuple T = (T1, ..., Tk) of operators on ℋ is called hyponormal if the operator matrix ([Tj*,Ti])i,j=1k is positive on the direct sum of ℋ⊕· · ·⊕ℋ (k copies). Also an operator T is said to be (strongly) k-hyponormal for each positive integer k if (I, T, ..., Tk) is hyponormal. The Bram-Halmos criterion shows that an operator T is subnormal if and only if T is k-hyponormal for all k ≥ 1 ([2], [16]). An operator T is polynomially hyponormal if for every polynomial p, p(T) is hyponormal, and T is weakly k-hyponormal if for every polynomial p of degree k or less, p(T) is hyponormal ([5],[10],[11]). In particular, weak 2-hyponormality (or weak 3-hyponormality) is referred to as quadratically hyponormal (or cubically hyponormal, respectively). For a positive integer k, an operator T ∈ ℒ(ℋ) is called semi-weakly k-hyponormal if T+sTk is hyponormal for all s ∈ ℂ ([12]). It is obvious that a weakly k-hyponormal operator is semi-weakly k-hyponormal. In particular, weak 2-hyponormality is equivalent to semi-weak 2-hyponormality.

It is well known that k-hyponormality implies weak k-hyponormality for each positive integer k. The following results provide a bridge between subnormal and hyponormal operators: subnormal ⇒ polynomially hyponormal ⇒ ··· ⇒ weakly 3-hyponormal ⇒ weakly 2-hyponormal ⇒ hyponormal. However, one does not yet have concrete examples about the converse implications for n ≥ 3; see [9], [17] and [18] for weak 2- and weak 3-hyponormalities.

J. Stampfli ([21]) proved that a subnormal weighted shift with two equal weights αn = αn+1 for some nonnegative n has the property that α1 = α2 = · · ·, which is known as the “flatness property.” Stampfli’s result has been used to attempt the construction of nonsubnormal polynomially hyponormal weighted shifts (cf. [1],[3],[4],[7],[12],[15],[17]). In [3], Choi proved that if a weighted shift Wα is polynomially hyponormal with the first two weights equal, then Wα has the flatness property. In [4], Curto obtained a quadratically hyponormal weighted shift with first two weights equal but not satisfying flatness. Also in [17], the authors showed that a weighted shift Wα with weights α:23,23,n+1n+2(n2) is not cubically hyponormal. And in [19], it was shown that if a weighted shift Wα is cubically hyponormal with first two weights equal, then Wα has flatness. However, in [12], it was proved that there exists a semi-cubically hyponormal weighted shift Wα with α0 = α1 < α2 which is not 2-hyponormal. Hence the following problem arises naturally as the analog to the question for quadratically hyponormal weighted shifts.

Problem 1.1

Describe all semi-cubically hyponormal weighted shifts Wα with first two weights equal.

In [12], Do-Exner-Jung-Li characterized the semi-cubical hyponormality of the weighted shift Wα(x) with positive determinant coefficients (p.d.c. – definition reviewed below), where α(x):x,x,k+1k+2(k2) is a weight sequence with Bergman tail. In this paper we describe the semi-cubical hyponormality of the weighted shifts having the p.d.c. property but with recursive tails. More precisely, for a three step backward extended weight sequence α:1,1x,(u,v,w) with 1 < x < u < v < w, where (u,v,w) is the Stampfli (recursively generated) subnormal completion of u, v,w (cf. [21]), we characterize completely the semi-cubical hyponormality of Wα with p.d.c. Note that, by the nature of a recursive tail, the one and two step backward extensions are special cases of the three step backward extension (see Remark 3.5).

For the reader’s convenience, we recall the Stampfli subnormal completion (cf. [6],[21]). For given numbers α0, α1, α2 with 0 < α0 < α1 < α2, define

αn2=Ψ1+Ψ0αn-12for all n3,

where Ψ0=-α02α12(α22-α12)α12-α02 and Ψ1=α12(α22-α02)α12-α02. Then we may obtain a weight sequence {αn}n=0 generated recursively by (1.1), which is usually denoted by (α0, α1, α2) (for example, see [21]); the associated shift is subnormal. It follows from [6] that

αnL:12(Ψ1+Ψ12+4Ψ0)1/2as n.

The organization of this paper is as follows. In Section 2 we recall some terminology concerning semi-cubically hyponormal weighted shifts. In Section 3 we characterize the semi-cubic hyponormality of weighted shifts Wα with p.d.c., where α:1,1,x,(u,v,w) with 1 < x < u < v < w, and then consider a related example.

Throughout this paper, ℝ+, ℕ, and ℕ0 are the sets of nonnegative real numbers, positive integers, and nonnegative integers, respectively.

2. Preliminaries

We recall some standard terminology for semi-cubically hyponormal weighted shifts (cf. [12]). Let ℓ2(ℕ0) be the space of square summable sequences in ℂ and let {ei}i=0 be an orthonormal basis of ℓ2(ℕ0). For a weight sequence α={αi}i=0 in ℝ+, the associated weighted shift Wα acting on ℓ2(ℕ0) is semi-cubically hyponormal if

D(s):=[(Wα+sWα3)*,Wα+sWα3]0,         s.

In fact, the condition in (2.1) is equivalent to a simpler one, as in the following proposition whose proof comes from [8, Prop. 1].

Proposition 2.1

Let Wαbe a weighted shift with a weight sequence α={αi}i=0 in+. Then Wαis semi-weakly n-hyponormal if and only if Wα+tWαn is hyponormal for all t ≥ 0.

Proof

It is sufficient to show the necessity. For any s ∈ ℂ, we may take nonnegative real numbers t and θ such that s = tei(n–1)θ. Recall that there exists a unitary operator U such that UWαU* = eWα. Then

U(Wα+sWαn)U*=UWαU*+s(UWαU*)n=e-iθ(Wα+tWαn),

so the inequality for all t ≥ 0 suffices to yield (2.1).

By Proposition 2.1, Wα is semi-cubically hyponormal if and only if D(s) ≥ 0 for all s ∈ ℝ+. Observe that

D(s)=(q00z000q10z1z00q20z1),         s+,

where for all k ∈ ℕ0,

qk:=uk+vks2,zk:=wks,uk:=αk2-αk-12,vk:=αk2αk+12αk+22-αk-32αk-22αk-12,wk:=αk2αk+12(αk+22-αk-12)2,

with α−3 = α−2 = α−1 = 0. Consider two submatrices

D(1)(s)=(q0z00z0q2z200z2q4z40z4)and D2(s)=(q1z10z1q3z300z3q5z50z5)

and observe that D(s) = D(1)(s) ⊕ D(2)(s), s ∈ ℝ+. Define

Dn(1)(t)=(q0z00z0q2z200z2q4z40z4q2n),Dn(2)(t)=(q1z10z1q3z300z3q5z50z5q2n+1),

where t = s2. Then Wα is semi-cubically hyponormal if and only if Dn(j)(t)0 for all n ≥ 0, j = 1, 2.

To detect the positivity of Dn(j)(t) in (2.2), we consider a matrix with the form below:

Mn(t)=(qˇ0rˇ00rˇ0qˇ1rˇ100rˇ1qˇ2rˇ20rˇ20qˇn-1rˇn-10rˇn-1qˇn),

where k : = k + kt, rˇk:=wˇkt(k0), and k ≥ 0, k ≥ 0, k ≥ 0, t ≥ 0. (We take the approach in [6] for what follows.) Then

dn(t):=det Mn(t)=i=0n+1c(n,i)ti,

and it follows from [6] that

c(0,0)=uˇ0,c(0,1)=vˇ0,c(1,0)=uˇ0uˇ1,c(1,1)=uˇ1vˇ0+uˇ0vˇ1-wˇ0,c(1,2)=vˇ1vˇ0,c(n,i)=uˇnc(n-1,i)+vˇnc(n-1,i-1)-wˇn-1c(n-2,i-1),c(n,n+1)=vˇ0vˇ1vˇn,n2,0in,

with c(−n,i) := 0 for all n, i ∈ ℕ. Suppose that nn+1 = n (n ≥ 2); this yields that

c(n,i)={vˇnvˇ2c(1,2),if i=n+1,uˇnc(n-1,n)+vˇnvˇ3ρ,if i=n,uˇnc(n-1,n-1)+vˇnvˇ3τ,if i=n-1,uˇnc(n-1,i)if 0in-2,

for all n ≥ 3, where

ρ:=vˇ2c(1,1)-wˇ1c(0,1)and τ:=vˇ2c(1,0)-wˇ1c(0,0).

To detect the positivity of Dn(j)(t), j = 1,2, we consider

dn(j)(t):=det Dn(j)(t)=i=0n+1cj(n,i)ti,

for j = 1, 2. Note that if cj(n, n+1) > 0 and cj(n, i) ≥ 0 for all n ≥ 0 with 0 ≤ in, then every matrix Dn(j)(t) is obviously positive for all n ≥ 0 and t > 0. Recall that Wα has positive determinant coefficients (p.d.c.) (and is therefore semi-cubically hyponormal) if all coefficients in dn(j)(t) are nonnegative and the cj(n, n + 1) are strictly positive for all n ∈ ℤ+ and j = 1, 2 (cf. [12, Def. 2.2]).

The following is the crucial lemma which can be obtained from the proof of Theorem 4.3 in [6], and which we will apply in succession to Dn(1)(t) and Dn(2)(t).

Lemma 2.2

Under the notation above, we suppose that ǔnn+1 = n (n ≥ 2). Then the determinant of Mn(t) in (2.3) has non-negative coefficients c(n, i) for all n and i if and only if the following conditions hold:

  • c(1, 1), c(2, 1), c(2, 2), c(3, 2) are all positive,

  • Γn:=vˇ2vˇ1vˇ0+vˇnuˇnρ0, for all n ≥ 3,

  • Ωn:=vˇ2vˇ1vˇ0+vˇn-1uˇn-1ρ+vˇn-1uˇn-1vˇnuˇnτ0, for all n ≥ 4.

3. A Special Case of Semi-cubical Hyponormality with p.d.c

Let α:1,1,x,(u,v,w) with 1 < x < u < v < w. To determine when Wα is a semi-cubically hyponormal weighted shift with p.d.c., we will use Lemma 2.2 for j = 1, 2 considering conditions (i), (ii), and (iii) in Lemma 2.2. We will denote the coefficients of the determinants c1(n, i) and c2(n, i) with the obvious meaning, and distinguish other quantities between j = 1 and j = 2 with superscripts (for example, ρ(1) and ρ(2)). Note first that it follows from Lemma 3.1 of [20] that we have both un(1)vn+1(1)=wn(1) and un(2)vn+1(2)=wn(2) for n ≥ 2.

3.1. The p.d.c. condition for Dn(1)

The following (n + 1) × (n + 1) matrix is the expression of Dn(1):

Dn(1)=(u0+v0tw0t000w0tu2+v2tw2t000w2tu4+v4tw4t000w4tu6+v6tw2n-2t000w2n-2tu2n+v2nt).

Since c1(1, 1) = (uv − 1)x > 0 and positivity conditions for the coefficients c1(2, 1), c1(2, 2), c1(3, 2) can be obtained by some direct computations, we will concentrate our consideration on (ii) and (iii) in Lemma 2.2.

Set ηn=vnun. Then we may prove that ηnU; see [20, Lemma 3.3], where

U=(Ψ12+Ψ0)22Ψ02(2Ψ0+Ψ12+Ψ14Ψ0+Ψ12),

where the values Ψ0 and Ψ1 are those associated with the Stampfli completion.

Lemma 3.1

With the notation above, we have that Γn(1):=v4v2v0+η2nρ(1)0 for all n ≥ 3, if and only if one of the following conditions holds:

  • ρ(1) ≥ 0,

  • ρ(1) < 0 and v4v2v0 + (1) ≥ 0,

where ρ(1) = v4c1(1, 1) − w2c1(0, 1) is as in (2.5).

Proof

We consider first Γn(1)=v4v2v0+η2nρ(1), where ρ(1) is as in (2.5). In this case, to check the positivity of Γn(1), we define

ΔΓn(1)=Γn+1(1)-Γn(1).

Then ΔΓn(1)=(ηn+2-ηn)ρ(1). If ρ(1) ≥ 0, since {ηn} is increasing, ΔΓn(1) is positive, i.e., Γn(1) is increasing in n, so to detect the positivity of Γn(1) for n ≥ 3, it is enough to consider only the positivity of Γ3(1). On the other hand, if ρ(1) < 0, then ΔΓn(1) is negative. So Γn(1) is decreasing in n, and to detect the positivity of Γn(1) for n ≥ 3, it is enough to examine the positivity of the limit of Γn(1). Since ηnU, we can obtain

limnΓn(1)=v4v2v0+Uρ(1).

Hence the proof is complete.

Lemma 3.2

With above notation, we have that Ωn(1):=v4v2v0+η2n-2ρ(1)+η2n-2η2nτ(1)0 for n ≥ 4 if and only if one of the following conditions holds:

  • s ≥ 0 and v4v2v0 + η6ρ(1) + η6η8τ(1) ≥ 0,

  • s < 0 and v4v2v0 + (1) + U2τ(1) ≥ 0,

where τ(1) = v4c1(1, 0) − w2c1(0, 0) as in (2.5) and

s=ρ(1)+η8η10-η6η8η8-η6τ(t).

Proof. To consider the positivity of Ωn(1), we first define ΔΩn(1)=Ωn+1(1)-Ωn(1) for any n ∈ ℕ. Then

ΔΩn(1)=(η2n-η2n-2)ρ(1)+(η2nη2n+2-η2n-2η2n)τ(1)=(η2n-η2n-2)(ρ(1)+η2nη2n+2-η2n-2η2nη2n-η2n-2τ(1)).

For brevity, we set

sn(x,u,v,w):=ρ(1)+η2nη2n+2-η2n-2η2nη2n-η2n-2τ(1).

We will claim that sn(x, u, v,w) is constant in n ≥ 4, i.e., sn is independent of n for n ≥ 4. The original idea for the proof comes from [14, Lemma 4.3]; see more recently [13]. Define

Qn:=ηnηn+2-ηn-2ηnηn-ηn-2.

Observe that with α:1,1,x,(u,v,w),

ηn(α)=ηn-3((u,v,w)),         n6.

Then it follows from (3.1) and (3.2) that

Qn(α)=Qn-3((u,v,w)),n8.

A direct computation shows that

Q5((u,v,w))=Q6((u,v,w));

in fact, we can confirm that its value is

v(2u2w-u2v+2uv2-4uvw+vw2)(uv2+u2w-3uvw+vw2)2u2(v-u)4(w-v)2.

Put p=α62, where α6 is the 6th term of α. Then it is well known that

(u,v,w)=u,(v,w,p),

which implies that

Qn((u,v,w))=Qn-1((v,w,p)),n5.

If we mimic the proof of [14, Lemma 4.3], we get

Q7((u,v,w))=Q6((u,v,w)).

Repeating this argument, we obtain

Qn((u,v,w))=Q5((u,v,w)),for all n5.

Hence Qn is constant in n, n ≥ 8, and so sn(x, u, v,w) is constant in n for n ≥ 4.

Now, we set s = s(x, u, v,w) := s4(x, u, v,w). By (3.1), obviously

ΔΩn(1)=(η2n-η2n-2)s,         n4.

Repeating the method in the proof of Lemma 3.1, we obtain that Ωn(1)0 for n ≥ 4 if and only if either (i) s ≥ 0 and Ω4(1)=v4v2v0+η6ρ(1)+η6η8τ(1)0 or (ii) s < 0 and limnΩn(1)=v4v2v0+Uρ(1)+U2τ(1)0.

Hence the proof is complete.

3.2. The p.d.c. condition for Dn(2)

The following (n + 1) × (n + 1) matrix is Dn(2):

Dn(2)=(u1+v1tw1t000w1tu3+v3tw3t000w3tu5+v5tw5t000w5tu7+v7tw2n-1t000w2n-1tu2n+1+v2n+1t).

As in the case of Dn(1), conditions for the positivity of the coefficients c2(1, 1), c2(2, 1), c2(2, 2), c2(3, 2) can be obtained by some direct computations. One may also compute that

c2(2,1)=(w-v)c2(1,1)   and   c2(3,2)=u2(w-u)3w(v-u)(uv-2uw+w2)c2(2,2).

It is straightforward to verify that u2(w-u)3w(v-u)(uv-2uw+w2) is positive, and it results from the above that c2(2, 1) and c2(1, 1) have the same sign, as do c2(3, 2) and c2(2, 2).

Lemma 3.3

With the above notation, the following conditions are equivalent:

  • Γn(2):=v5v3v1+η2n+1ρ(2)0for all n ≥ 3,

  • Ωn(2):=v5v3v1+η2n-1ρ(2)0for all n ≥ 4,

  • one of the following conditions holds:

    • (iii-a) ρ(2) ≥ 0,

    • (iii-b) ρ(2) < 0 and v5v3v1 + (2) ≥ 0,

    where ρ(2) = v5c2(1, 1) − w3c2(0, 1) is as in (2.5).

Proof
(i) ⇔ (iii)

Since the proof is exactly that of Lemma 3.1, we omit it here.

(i) ⇔ (ii)

Recall that Ωn(2):=v5v3v1+η2n-1ρ(2)+η2n-1η2n+1τ(2). Since

τ(2)=v5c2(1,0)-w3c2(0,0)=u1(v5u3-w3)=0,

Ωn(2)=Γn-1(2) for all n ≥ 4. Hence the proof is complete.

3.3. The main theorem

We now give the main theorem of this paper. Combining results in Subsections 3.1 and 3.2, we obtain the following theorem with the above notation.

Theorem 3.4

Let α:1,1,x,(u,v,w)with 1 < x < u < v < w. Then Wαis a semi-cubically hyponormal weighted shift with p.d.c. if and only if the following conditions hold:

  • c1(2, 1), c1(2, 2), c1(3, 2), c2(1, 1), c2(2, 2) are nonnegative,

  • one of the following conditions holds:

    • (ii-a) ρ(1) ≥ 0,

    • (ii-b) ρ(1) < 0 and v4v2v0 + (1) ≥ 0,

  • one of the following conditions holds:

    • (iii-a) s ≥ 0 and v4v2v0 + η6ρ(1) + η6η8τ(1) ≥ 0,

    • (iii-b) s < 0 and v4v2v0 + (1) + U2τ(1) ≥ 0,

  • one of the following conditions holds:

    • (iv-a) ρ(2) ≥ 0,

    • (iv-b) ρ(2) < 0 and v5v3v1 + (2) ≥ 0.

Remark 3.5

According to the construction of Stampfli’s completion from three values u, v and w, Theorem 3.4 can yield a characterization of semi-cubical hyponormality of a backward 2-step weighted shift Wα with Stampfli’s completion tail, where α:1,1,(x,u,v) with 1 < x < u < v, because if we choose w=u(v2+ux-2vx)v(u-x) for our backward 3-step extension, then α:1,1,x,(u,v,w) produces the same weighted shift Wα. Of course, the case of 1-step extension can be considered similarly; this case was studied in [20, Th. 3.7].

Remark 3.6

Using some technical computations, we believe it is possible to characterize the semi-cubical hyponormality of backward n-step extended weighted shifts Wα with Stampfli’s completion tail. We presume their proofs will be intricate.

In what follows, we consider an example related to Theorem 3.4.

Example 3.7

Let α:1,1,x,(111100,112100,113100), where x is a real variable with 1<x<111100. We obtain a range of x for semi-cubical hyponormality with p.d.c. of Wα.

  • c1(2, 1) ≥ 0 ⇔ 69375x2 − 158688x + 90160 ≤ 0, c1(2,2)0x195776152625,

    c1(3, 2) ≥ 0 ⇔ 1149751875x2 − 2603155192x + 1462768720 ≤ 0,

    c2(1, 1) ≥ 0 ⇔ 122 − 111x ≥ 0,

    c2(2, 2) ≥ 0 ⇔ 4648721 − 4082025x ≥ 0.

    Thus we get (i) holds if and only if 4(19836-2538771)69375<x<122111.

  • Since 1x<111100, we can check without difficulty that ρ(1) > 0; i.e., (ii) always holds for 1x<111100.

  • One computes that s0ϕ(x)=1800279654375x2-4093200707344x+23141001150400;

    a computation shows that this yields also v4v2v0 + η6ρ(1) + η6η8τ(1) ≥ 0. Hence (iii-a) holds δx<111100, where δ ≈ 1.053 is the smallest root of ϕ(x) = 0. And v4v2v0+Uρ(1)+U2τ(1)0φ(x)=a3x3+a2x2-a1x+a00,

    where aj are positive real numbers. We can check easily that φ(x) has roots δ3 < 0 < δ2 < δ1, with δ2 ≈ 1.036. Assembling these computations gives s<0and v4v2v0+Uρ(1)+U2τ(1)0δ2x<δ.

    Therefore we get (iii) holds if and only if δ2x<111100.

  • holds for 1<x<111100, because further computation shows v5v3v1 + η7ρ(2) ≥ 0 and v5v3v1 + (2) ≥ 0 for 1<x<111100.

Combining the analyses of (i)–(iv) above, we get that Wα is semi-cubically hyponormal with p.d.c. if and only if 4(19836-2538771)69375<x<122111.

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