A relationship: Subnormal, polynomially hyponormal and semi-weakly hyponormal weighted shifts

J. Math. Anal. Appl. 479 (2019) 703–717

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

A relationship: Subnormal, polynomially hyponormal and semi-weakly hyponormal weighted shifts ✩ Chunji Li a , Mi Ryeong Lee b,∗ , Seunghwan Baek c a

Department of Mathematics, Northeastern University, Shenyang 110819, PR China Institute of Liberal Education, Daegu Catholic University, Gyeongsan, Gyeongbuk 38430, Republic of Korea c Department of Mathematics, Kyungpook National University, Daegu 41566, Republic of Korea b

a r t i c l e

i n f o

Article history: Received 13 November 2017 Available online 17 June 2019 Submitted by R. Curto Keywords: Subnormal Polynomially hyponormal Semi-weakly hyponormal Weighted shifts

a b s t r a c t √ √ √ √ ∧ Let α(x) : x, ( a, b, c) be a one-step backward extension sequence of Stampfli’s subnormal completion, where 0 < x ≤ a < b < c, and let Wα(x) be the associated weighted shift. In this paper, we prove that Wα(x) is subnormal if and only if Wα(x) is polynomially hyponormal, which also is equivalent to that Wα(x) is completely semi-weakly hyponormal. © 2019 Elsevier Inc. All rights reserved.

1. Introduction and preliminaries Let H be a separable infinite dimensional complex Hilbert space and let B(H) be the algebra of all bounded linear operators on H. For A and B in B(H), we let [A, B] := AB − BA. A k-tuple (T1 , ..., Tk ) of bounded operators in B(H) is called hyponormal if the operator matrix ([Tj∗ , Ti ])ki,j=1 is positive on the direct sum of H ⊕ · · · ⊕ H (k-copies). An operator T ∈ B(H) is said to be (strongly) k-hyponormal if (T, ..., T k ) is hyponormal ([6], [9], [10]). Obviously, 1-hyponormal operator T ∈ B(H) is hyponormal. Recall that an operator T ∈ B(H) is subnormal if it is (unitarily equivalent to) the restriction of a normal operator to an invariant subspace. The notion of a subnormal operator was introduced by Halmos who gave its first characterization in 1950 ([24]), and it was successively simplified by Bram ([4]). It follows from this Bram-Halmos’ criterion that an operator T is subnormal if and only if T is k-hyponormal for all k ∈ N, where N is the set of natural numbers ([4]). ✩ This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2017R1D1A1A02017817). * Corresponding author. E-mail addresses: [email protected] (C. Li), [email protected] (M.R. Lee), [email protected] (S. Baek).

https://doi.org/10.1016/j.jmaa.2019.06.046 0022-247X/© 2019 Elsevier Inc. All rights reserved.

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C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

An operator T ∈ B(H) is said to be polynomially hyponormal if p(T ) is hyponormal for all complex polynomials p. For k ∈ N, an operator T is weakly k-hyponormal if for every polynomial p of degree k or less, p(T ) is hyponormal ([9], [15], [16], [22]). Sometimes weak 2- and 3-hyponormality are referred to as quadratic and cubic hyponormality, respectively ([7], [26]). It holds that every subnormal operator is polynomially hyponormal operator and every k-hyponormal operator is weakly k-hyponormal operator for each k ∈ N. In [15], Curto-Putinar proved that there exists an operator that is polynomially hyponormal but not 2-hyponormal. This solved a long standing open problem (cf. [15]): “if T ∈ B(H) is polynomially hyponormal, must T be subnormal?” Although the existence of a weighted shift which is polynomially hyponormal but not subnormal was established in [15] and [16], concrete examples of such weighted shifts have not yet been found. Recently in [21], [29] or [30], the classes of semi-weakly k-hyponormal operators have been studied in an attempt to bridge the gap between subnormality and hyponormality. An operator T is called semi-weakly k-hyponormal if T + sT k is hyponormal for all s in the set C of complex numbers ([21], [30]). An operator T is said to be completely semi-weakly hyponormal if T is semi-weakly k-hyponormal for all k ∈ N ([29]). It is obvious that every polynomially hyponormal operator is a completely semi-weakly hyponormal operator, and that weakly k-hyponormal implies semi-weakly k-hyponormal for each k ∈ N. However it is known that converse implications are not always true ([5], [21], [29], [30]). The semi-weak 3-hyponormality is referred to as semi-cubic hyponormality. Let α = {αi }∞ i=0 be a bounded weight sequence in the set R+ of the positive real numbers. The weighted shift Wα acting on 2 (N0 ), with an orthonormal basis {ei }∞ i=0 , is defined by Wα ej = αj ej+1 for all j ∈ N0 := N ∪ {0}. It is worthwhile to point out that the subnormal weighted shift Wα is closely related to the classical moment problem. By the Stieltjes theorem (cf. [3, Theorem 6.2.5]), a sequence {γn }∞ n=0 is a Stieltjes moment sequence if and only if the infinite matrices (γi+j )0≤i,j<∞ and (γi+j+1 )0≤i,j<∞ are positive 2 definite, where γ0 = 1 and γn = α02 · · · αn−1 (n ≥ 1), which is equivalent to the subnormality of Wα . Weighted shifts have played an important role in detecting properties of weak subnormality. Two particular types of weighted shift operators have often been used in this setting, namely those of Bergman type and those of recursive type found in Stampfli’s method (cf. [2], [11–14], [17–20], [23], [31], etc.). For a weighted shift Wα of Bergman type, for example in [25] α:

√

x,



2/3,



(n + 1)/(n + 2) (n ≥ 2),

the authors provide a formula to explain the characterizations between subnormality and k-hyponormality. Similarly, in [29] one finds an example which is not subnormal but completely semi-weakly hyponormal. Another family commonly used is the family of subnormal shifts arising from Stampfli’s subnormal completion ([32]): given positive real numbers a, b, c with a < b < c, there exists a recursively subnormal weighted shift with the initial weights a, b, c (see Section 2). The Stampfli’s subnormal completion of a, b, and c is denoted by W(√a,√b,√c)∧ , which is the main operator model of this paper. Recall that, given (m +1) positive real numbers α0 , α1 , ..., αm , the subnormal completion problem seeks necessary and sufficient conditions for the existence of a subnormal weighted shift Wα whose first (m + 1) weights are α0 , α1 , ..., αm . This subnormal completion problem is closely related to the truncated Stieltjes moment problem ([8], [9], [10], [28]). Hence the work of the present paper will be applied to the truncated Stieltjes moment problem. Since Curto introduced the k-hyponormality in 1989 ([6]), many operator theorists have studied weak subnormality: k-hyponormality and weak k-hyponormality of weighted shifts. On the other hand, Choi [5] proved that if Wα is a polynomially hyponormal weighted shift with α0 = α1 , then α0 = αk for all k ∈ N. In fact, he showed this flatness in the case of polynomially hyponormal weighted shift. This idea has contributed to the study of semi-weak k-hyponormality (cf. [1], [2], [21], [29], [30]). In fact, the gaps among the classes of subnormal, polynomially hyponormal and semi-weakly hyponormal weighted shifts have been

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

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investigated markedly via various models of weighted shifts. In√this point of view, it is worth studying the √ √ √ weighted shift Wα(x) with the weight sequence α(x) : x, ( a, b, c)∧ . Recall that Wα(x) is subnormal if and only if Wα(x) is 2-hyponormal ([18]). As we mentioned above, because a concrete example of polynomially hyponormal weighted shifts that are not also subnormal is not yet known, we were strongly motivated to study gaps between the subnormality [or 2-hyponormality] and polynomial hyponormality of weighted shifts. In this paper we discuss gaps between subnormality, 2-hyponormality, polynomial hyponormality, and completely semi-weak hyponormality of Wα(x) . This paper consists of four sections. In Section 2 we recall some standard terminology about semiweakly k-hyponormal weighted shifts and also we obtain several technical lemmas to detect the structure of semi-weakly k-hyponormal weighted shifts Wα(x) . In Section 3 we discuss the structure of semi-weakly k-hyponormal weighted shifts Wα(x) . In Section 4, we obtain an interesting result: if Wα(x) is a weighted shift with weight sequence α(x), then its subnormality [or 2-hyponormality], polynomial hyponormality, and completely semi-weak hyponormality are equivalent (see Theorem 4.3). Some of the calculations in this paper were accomplished by using the software tool Mathematica ([33]). 2. Technical lemmas We begin by recalling some standard terminology about semi-weakly k-hyponormal weighted shifts (cf. [21], [29], [30]). Throughout this paper we may consider k ≥ 3. In fact the semi-weak 2-hyponormality is quadratic hyponormality. Let Wα be a weighted shift with a weight sequence α = {αi }∞ i=0 and let Pn denote the orthogonal [k] n projection onto ∨i=0 {ei }. For n ∈ N0 , define Dn by   ∗ Dn[k] := Dn[k] (s) = Pn Wα + sWαk , Wα + sWαk Pn ⎛ [k] [k] q 0 ··· 0 z0 0 ··· ⎜ 0 . . .. ⎜ [k] [k] .. .. . 0 z1 ⎜ 0 q1 ⎜ . .. .. .. .. .. .. ⎜ . . . . . . . ⎜ . ⎜ ⎜ .. .. .. .. [k] ⎜ 0 . . . . 0 qk−2 =⎜ ⎜ [k] . . . [k] .. .. .. ⎜z 0 0 qk−1 ⎜ 0 ⎜ .. .. .. .. ⎜ 0 z [k] . . . . . . . ⎜ 1 ⎜ . .. .. .. .. .. [k] ⎜ . . . . . . qn−1 ⎝ . [k] 0 ··· 0 z n+1−k 0 ··· 0

0 .. .

⎞

⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ [k] zn+1−k ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ .. ⎟ . ⎟ ⎟ ⎟ 0 ⎠ [k] qn

(2.1)

for all s ∈ C, where [k]

[k]

[k]

[k]

qi := ui + vi |s|2 , zi :=

 [k] [k] 2 wi s¯, ui := αi2 − αi−1 ,

[k]

2 2 2 2 2 vi := αi2 αi+1 · · · αi+k−1 − αi−k αi−k+1 · · · αi−1 ,

(2.2)

[k]

2 2 2 2 wi := αi2 αi+1 · · · αi+k−2 (αi+k−1 − αi−1 )2 ,

with α−k = α−k+1 = · · · = α−1 = 0 for convenience. It is obvious that Wα is semi-weakly k-hyponormal if [k] and only if Dn (s) ≥ 0 for every s ∈ C and n ∈ N0 . Now we recall the Stampfli’s method from [10] and [32] for the subnormal completion of given finite positive real numbers. For given real numbers α0 , α1 , α2 with 0 < α0 < α1 < α2 , define

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

706

αn2 = Ψ1 +   α2 α2 α2 −α2

Ψ0 , 2 αn−1

n ≥ 2,

(2.3)

  α2 α2 −α2

2 1 2 0 where Ψ0 = − 0 α12 −α and Ψ1 = 1α2 −α . Then we obtain a recursively generated weight sequence 2 2 1 0 1 0 ∞ ∧ {αn }n=0 , and denote it by (α0 , α1 , α2 ) . It follows from [10] that

αn2

1  L := 2 2

   2 Ψ1 + Ψ1 + 4Ψ0 ,

n → ∞.

(2.4)

Let x, a, b and c be real numbers with 0 < x ≤ a < b < c be given, where, as usual x is a real variable. In this section we characterize the semi-weak k-hyponormality of weighted shifts Wα(x) with a recursive √ √ √ √ ∧ weight sequence α(x) : x, ( a, b, c) . In particular, either a = b or b = c forces the flatness of Wα(x) ([21], [31], [29]). To avoid the trivial case, we assume 0 < x ≤ a < b < c throughout this paper. We need several technical lemmas to detect the structure of semi-weakly k-hyponormal weighted shifts. Ψ0 2 According to (2.3), we may produce a recursive weight sequence {αn }∞ (n ≥ 3), n=0 such that αn = Ψ1 + α2 n−1

where α02 = x, α12 = a, α22 = b, α32 = c, Ψ0 = − ab(c−b) and Ψ1 = b−a Lemma 2.1. Let α (x) :

√

b(c−a) b−a .

√ √ √ x, ( a, b, c)∧ be given as above. Then 2 2 αn2 αn+1 · · · αn+j = Aj αn2 + Aj−1 Ψ0 ,

n ∈ N,

(2.5)

where  Ψ1 + Aj =



 j+1  − Ψ1 − Ψ21 + 4Ψ0  , 2j+1 · Ψ21 + 4Ψ0

Ψ21 + 4Ψ0

j+1

j ∈ N0 .

(2.6)

Proof. First we consider the equation 2 2 αn2 αn+1 · · · αn+j = Aj αn2 + Bj ,

n ∈ N, j ∈ N0

(2.7)

with A0 := 1 and B0 := 0. By using (2.7) and (2.3), we get   2 2 2 αn2 αn+1 · · · αn+j+1 = Aj αn+1 + Bj αn2 = (Aj Ψ1 + Bj ) αn2 + Aj Ψ0 ,

n ∈ N, j ∈ N0 .

(2.8)

On the other hand, it follows from (2.8) that 2 2 αn2 αn+1 · · · αn+j+1 = Aj+1 αn2 + Bj+1 ,

n ∈ N, j ∈ N0 ,

which implies that Aj+1 = Ψ1 Aj + Bj and Bj+1 = Ψ0 Aj , Now we consider a real number γ = 12 (−Ψ1 + Ψ0 = 0, which produces



(2.9)

Ψ21 + 4Ψ0 ) being a solution γ for an equation y 2 + Ψ1 y −

Ψ1 = L2 − γ and Ψ0 = γL2 . Using (2.9) and (2.10), we can obtain

j ∈ N0 .

(2.10)

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

Aj+1 + γAj = Ψ1 Aj + Ψ0 Aj−1 + γAj = L2 (Aj + γAj−1 ),

707

j ∈ N.

From the first forms in (2.9) and (2.10), A0 = 1 and B0 = 0, we have A1 + γA0 = L2 , i.e., A1 = L2 − γ. Hence Aj + γAj−1 = L2 (Aj−1 + γAj−2 ) = · · · = (L2 )j−1 (A1 + γA0 ) = L2j ,

j ∈ N,

which induces Aj = L2j − γAj−1 (j ≥ 1). Therefore for all j ∈ N, Aj = L2j − γAj−1 = L2j − γL2(j−1) + γ 2 Aj−2 = · · · = L2j − γL2(j−1) + (−γ)2 L2(j−2) + · · · + (−γ)j−2 L4 + (−γ)j−1 A1 = L2j + (−γ)L2(j−1) + (−γ)2 L2(j−2) + · · · + (−γ)j−2 L4 + (−γ)j−1 L2 + (−γ)j =

L2(j+1) − (−γ)j+1 . L2 + γ

Using the values of L2 in (2.4) and γ, we can prove this lemma without difficulty. 2 Lemma 2.2. Let α (x) :

√

√ √ √ x, ( a, b, c)∧ be given as above. Then [k]

[k] u[k] n vn+k−1 = wn ,

n ≥ 2.

(2.11)

Proof. It follows from Lemma 2.1 that for all n ≥ 2, [k]

2 2 2 2 2 vn+k−1 = αn+k−1 αn+k · · · αn+2k−2 − αn−1 αn2 · · · αn+k−2 2 2 = Ak−1 (αn+k−1 − αn−1 ),

(2.12)

and 2 2 2 2 wn[k] = αn2 αn+1 · · · αn+k−2 (αn+k−1 − αn−1 )2 2 2 2 2 2 2 2 = (αn+k−1 − αn−1 )(αn2 αn+1 · · · αn+k−1 − αn−1 αn2 αn+1 · · · αn+k−2 ) 2 2 2 = (αn+k−1 − αn−1 )Ak−1 (αn2 − αn−1 ). [k]

Combining (2.12) with the definition of un , we get [k]

[k] 2 2 2 2 [k] u[k] n vn+k−1 − wn = (αn − αn−1 )Ak−1 (αn+k−1 − αn−1 ) − wn = 0.

Hence the proof is complete. 2 v [k]

∞ To simplify notation, we denote ηn := n[k] (n ≥ 0), where the sequences {un }∞ n=0 and {vn }n=0 are un given as in (2.2). Lemma 2.2 yields the following results. [k]

[k]

[k]

√ √ √ √ Lemma 2.3. Let α (x) : x, ( a, b, c)∧ be given as above. Define a function φ(; m) for integers  ≥ 2 and m ≥ 0 by φ(; 0) = q and

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

708

φ(; m) = q −

z2 q+k−1 −

.

2 z+k−1

..

..

.

q+(m−1)(k−1) −

.

z2 +(m−1)(k−1) q+m(k−1)

Suppose k ≥ 3. Then for integers  ≥ 2 and m ≥ 0, φ(; m) = v t + [k]

u , 1 + η+k−1 t + η+k−1 η+2(k−1) t2 + · · · + η+k−1 · · · η+m(k−1) tm

[k]

[k]

[k]

[k]

where qi := qi , zi := zi , ui := ui , vi := vi , ηi := ηi

and t = |s|2 .

Proof. We will prove this lemma by the mathematical induction. It follows from Lemma 2.2 with qn = [k] [k] [k] qn , zn = zn and ηn = ηn as in (2.2) that qn −

zn2 qn+(k−1)

un vn+(k−1) t un+(k−1) + vn+(k−1) t un , n ≥ 2. = vn t + 1 + ηn+(k−1) t = un + vn t −

(2.13)

For a sufficiently large number n ∈ N, it follows from (2.13) that qn−(k−1) −

2 zn−(k−1)

qn −

2 zn

= vn−(k−1) t +

qn+(k−1)

un−(k−1) . 1 + ηn t + ηn ηn+(k−1) t2

Similarly, we have qn−2(k−1) −

2 zn−2(k−1)

qn−(k−1) −

2 zn−(k−1) 2 zn n+(k−1)

qn − q

= un−2(k−1) + vn−2(k−1) t − = vn−2(k−1) t +

un−2(k−1) vn−(k−1) t vn−(k−1) t +

un−(k−1) 1+ηn t+ηn ηn+(k−1) t2

un−2(k−1) . 1 + ηn−(k−1) t + ηn−(k−1) ηn t2 + ηn−(k−1) ηn ηn+(k−1) t3

Continuing this process with  = n − m(k − 1) (m ≥ 1), we can obtain the lemma. 2 [k]

We now discuss properties of ηn , which will play a crucial role to characterize the semi-weak k-hyponormality of Wα(x) . Proposition 2.4. Let α (x) : [k]

√

√ √ √ x, ( a, b, c)∧ be given as above. Then

[k]

(i) ηn ≥ ηn−1 for all n ≥ k + 1, (ii)

[k] limn→∞ ηn

[k]

=Q

:= Ak−1 Δk , where Δk :=

 k L4 1− − Ψ 0 4

L 1+ Ψ

.

0

[k]

Proof. It follows from (2.12) and the definition of ui that     [k] [k] 2 vn[k] = Ak−1 αn2 − αn−k = Ak−1 u[k] n + un−1 + · · · + un−k+1 ,

n ≥ 1.

(2.14)

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

709

From (2.3), we have

[k]

un−1 [k]

=

un

2 2 2 2 2 2 αn−1 αn−1 αn−2 − αn−2 − αn−2 αn−1 = = , 2 Ψ0 0 αn2 − αn−1 −Ψ0 − αΨ 2 α2 n−1

n ≥ 2.

(2.15)

n−2

∞

Since {αj }j=0 is non-decreasing, by (2.15), we have that, for any i ≥ 1 and n ≥ k + 1,

[k]

un−i [k]

[k]

=

un

[k]

un ≥

[k]

[k]

un−1 un−2 [k]

un−1

···

un−i

=

[k]

2 4 4 4 2 αn−i αn−i+1 · · · αn−2 αn−1 αn−i−1 i

(−Ψ0 )

un−i+1

2 4 4 4 2 αn−i−1 αn−i · · · αn−3 αn−2 αn−i−2

=

i

(−Ψ0 )

[k]

[k]

[k]

[k]

un−2 un−3 un−1 un−2

[k]

···

un−i−1 [k]

[k]

=

un−i−1

un−i

[k]

.

un−1

According to (2.14), we obtain that 

[k]

ηn[k]

=

vn

= Ak−1

[k]

un

 ≥ Ak−1

[k]

1+

un−1

+

un [k]

1+

[k]

un−2 [k]

un−1

[k]

[k]

un−2 [k]

un

+ ··· + [k]

[k]

+

un−3 [k]

un−1

+ ··· +

un−k

un−k+1



[k]

un



[k]

un−1

[k]

=

vn−1 [k]

un−1

[k]

= ηn−1

and 

[k]

lim ηn[k] = lim

n→∞

n→∞

vn

[k]

un 

= Ak−1

= lim Ak−1 n→∞

[k]

1+

un−1 [k]

un

[k]

[k]

+

un−2 [k]

un

L4 L8 L4(k−1) 1− + 2 + ··· + Ψ0 Ψ0 (−Ψ0 )k−1

+ ··· +

un−k+1



[k]

un

 = Ak−1 Δk .

Therefore the proof is complete. 2 3. Structures of semi-weak k-hyponormality Let x, a, b and c with 0 < x ≤ a < b < c be given as usual and let Wα(x) be a weighted shift with a recursive √ √ √ √ weight sequence α(x) : x, ( a, b, c)∧ . It is worth recalling that Wα is semi-weakly k-hyponormal if and [k] [k] only if Dn (t) ≥ 0 for every t ∈ R+ and n ∈ N0 ([1, Proposition 2.1]). Then Dn (s) of (2.1) can be represented in a slightly simplified form with t := |s|2 ≥ 0 as follows

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

710

⎛

[k]

q0

⎜ ⎜ ⎜ 0 ⎜ . ⎜ . ⎜ . ⎜ ⎜ ⎜ 0 [k] Φn (t) := ⎜ ⎜ [k] ⎜ −z ⎜ 0 ⎜ ⎜ 0 ⎜ ⎜ . ⎜ . ⎝ . 0 [k] qi

[k] where = ui x = (x0 , . . . , xn )T

[k] + vi t and in Rn+1 as +

[k]

qk−2

[k]

0

0

−z1 .. .

··· .. . .. . .. . .. . .. . .. .

···

0

−zn+1−k

0 [k]

q1 ..

.

..

.

0 [k]

0 .. ..

−z0

0

.

0

.

..

qk−1 .. . .. .

−z1 .. . .. . .. . .. . .. .

0

···

[k]

.

[k]

..

.

..

.

[k]

··· .. . .. . .. . .. . .. . [k]

qn−1 0

⎞

0 .. .

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ [k] −zn+1−k ⎟ ⎟ , n ∈ N, ⎟ ⎟ 0 ⎟ ⎟ .. ⎟ . ⎟ ⎟ ⎟ 0 ⎠ [k] qn

 [k] [k] = wi t. We may consider the quadratic form for Φn (t) at a vector follows: [k] zi

Fn[k] (x0 , x1 , ..., xn , t) := Φ[k] n (t)x, x , where ·, · denotes the inner product in Rn+1 . By Lemma 2.2, it turns out that Fn[k] (x0 , ..., xn , t) = f [k] (x0 , ..., xk , t) +

n 

[k] ui x2i

+

n−k+1  

[k] ui xi

2  [k] − tvi+k−1 xi+k−1 , n ≥ k + 1,

(3.1)

i=2

i=n−k+2

where f [k] (x0 , ..., xk , t) :=

1 

[k]

qi x2i − 2

i=0

1  

[k]

twi xi xi+k−1 + t

i=0

k 

[k]

vi x2i .

i=2

We now introduce the following result which plays an important role in proving the main theorem. Note that it is also a generalization of results in [27] and [30]. For a special case, see [27, Lemma 4.2, 4.3] for the case k = 2 [i.e., the case of quadratic hyponormality], and [30, Lemma 3.4] for the case k = 3 [i.e., [k] [k] [k] semi-cubic hyponormality]. To prove the following result, we will use two formulae, qi = ui + vi t for [k] [k] [k] t ∈ R+ in (2.2) and ui+2 vi+k+1 = wi (i ≥ 2) in (2.11). Lemma 3.1. Let α(x) :

√

√ √ √ x, ( a, b, c)∧ and n ≥ k + 1. Then the following conditions are equivalent:

[k]

(i) Fn (x0 , ..., xn , t) ≥ 0 for any xi and t ∈ R+ (i = 0, 1, . . . , n); (ii) for any xi and t ∈ R+ (i = 0, 1, ..., k), f [k] (x0 , ..., xk , t) +

r+2 

P (; m + 1)x2 +

=2

k 

P (; m)x2 ≥ 0,

=r+3

1 where m =  n−2 k−1  − 1 (· means the floor function ), r = 0, 1, ..., k − 2 and [k]

P (i; j) =

1

x = max{n ∈ Z : n ≤ x}.

ui [k]

[k]

[k]

[k]

[k]

1 + ηi+k−1 t + ηi+k−1 ηi+2(k−1) t2 + · · · + ηi+k−1 · · · ηi+j(k−1) tj

.

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

[k]

711

[k]

Proof. For brevity, we write Fn := Fn (x0 , ..., xn , t) and f [k] := f [k] (x0 , ..., xk , t), and also we let by [k] [k] [k] [k] [k] qi := qi , zi := zi , ui := ui , vi := vi , and ηi := ηi . For n ≥ k + 1, we give a new expression as n = (m + 1)(k − 1) + r + 2 (m ∈ N0 , 0 ≤ r ≤ k − 2). Now for fixed number k, we set sk (j; r) := j(k − 1) + r + 2, j ∈ N0 , 0 ≤ r ≤ k − 2, with sk (−1; r) = 2 for convenience. For the case of m = 0, i.e. n = k + r + 1 (0 ≤ r ≤ k − 2), it follows from [k] (2.2) and (2.11) that the quadratic form Fn in (3.1) induces the following expression

[k]

Fn[k] = Fk+r+1 = f [k] +

k+1+r 

ui x2i +

i=r+3

= f [k] +

r+2   √

r+2 

√ (ui x2i − 2 twi xi xi+k−1 + tvi+k−1 x2i+k−1 )

i=2

qi+k−1 xi+k−1 − √

i=2

2

zi qi+k−1

For convenience, taking xi ∈ R+ such that xi+k−1 = following implication:

xi

ui −

i=2

zi qi+k−1 xi

k 

[k]

Fk+r+1 (x0 , ..., xn , t) ≥ 0 =⇒ f [k] +

r+2  

+



zi2 qi+k−1

k 

x2i +

ui x2i .

i=r+3

(2 ≤ i ≤ r + 2, 0 ≤ r ≤ k − 2), we obtain the

ui x2i +

r+2  

i=r+3

ui −

i=2



zi2

x2i ≥ 0.

qi+k−1

For the general case, we suppose that m ≥ 1, i.e., n = (m + 1)(k − 1) + r + 2 (0 ≤ r ≤ k − 2). Set, for 0 ≤ j ≤ m, 



 φ(i + k − 1; j)xi+k−1 − 

sk (m−j;r)

Γj :=

i=sk (m−j−1;r)+1

2

zi φ(i + k − 1; j)

xi

.

Using (2.2) and (2.11), we have 

sk (m+1;r)

Fn[k] = f [k] +

ui x2i +

sk (m;r) 



ui xi −



2 tvi+k−1 xi+k−1

i=2

i=sk (m;r)+1



sk (m;r)

= f [k] + Γ0 −

√

i=sk (m−1;r)+1



sk (m−1;r)

zi2 qi+k−1

x2i +



sk (m;r)

tvi+k−1 x2i+k−1 +

i=2

i=2

Repeating this process, we obtain 

sk (m−1;r)

Fn[k] = f [k] + Γ0 + Γ1

−

i=sk (m−2;r)+1



sk (m−2;r)

+

zi2 x2 φ(i + k − 1; 1) i 

sk (m−2;r)

qi+k−1 x2i+k−1 −

i=2

Continuing the above process, we can obtain

i=2

2zi xi xi+k−1 +

k  i=2

ui x2i .

ui x2i .

(3.2)

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

712

Fn[k] = f [k] + Γ0 + Γ1 + Γ2 + · · · + Γm−1 −

k+r+1  i=r+3

r+2 r+2 k    zi2 x2i + qi+k−1 x2i+k−1 − 2zi xi xi+k−1 + ui x2i φ(i + k − 1; m − 1) i=2 i=2 i=2

+ Γ0 + Γ1 + Γ2 + · · · + Γm−1 + Γm    r+2  k   zi2 zi2 x2i + x2i . ui − ui − + φ(i + k − 1; m) φ(i + k − 1; m − 1) i=2 i=r+3

=f

[k]

On the other hand, the function φ(; j) as in Lemma 2.3 can be represented by φ(; j) = q −

z2 , φ( + k − 1; j − 1)

 ≥ 2, j ≥ 0.

Using qi = ui + tvi , we have the following equations ui −

zi2 = φ(i; m + 1) − tvi , φ(i + k − 1; m)

2 ≤ i ≤ r + 2,

ui −

zi2 = φ(i; m) − tvi , φ(i + k − 1; m − 1)

r + 3 ≤ i ≤ k,

and

which implies that, for  ≥ 2, P (; m) := φ(; m) − tv =

u 1 + η+k−1 t + η+k−1 η+2(k−1) t2 + · · · + η+k−1 · · · η+m(k−1) tm

and P (; m + 1)(:= φ(; m + 1) − tv ). Since each xi is arbitrary, we may take a number xi+k−1 with xi+k−1 =

zi xi , φ(i + k − 1; j)

2 ≤ i,

0 ≤ j ≤ m.

So it forces Γj = 0 in (3.2) for all 0 ≤ j ≤ m. Hence we obtain the implication: Fn[k] ≥ 0 =⇒ f [k] +

r+2 

P (; m + 1)x2 +

=2

k 

P (; m)x2 ≥ 0.

=r+3

[k]

For the converse implication, note that Fn (x0 , ..., xn , t) for n ≥ k + 1 can be represented by the sum f [k] (x0 , ..., xk , t) +

r+2 

P (; m + 1)x2 +

=2

k 

P (; m)x2

=r+3 [k]

and the quadratic forms in (3.1). Hence it is obvious that Fn (x0 , ..., xn , t) ≥ 0 for all xi , t ∈ R+ , 0 ≤ i ≤ n, and n ≥ k + 1. 2 [k]

Applying the argument in [27] to the quadratic form of Fn in Lemma 3.1, we obtain the following result via Proposition 2.4.

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

Proposition 3.2. Let α(x) :

√

713

√ √ √ x, ( a, b, c)∧ and n ≥ k + 1. Then the following conditions are equivalent:

(i) Wα(x) is semi-weakly k-hyponormal; [k] (ii) Fn (x0 , x1 , ..., xn , t) ≥ 0 for any xi ∈ R+ and t >

1 ; Q[k] > Q1[k] .

(iii) f [k] (x0 , ..., xk , t) ≥ 0 for any x0 , ..., xk ∈ R+ and t

To obtain the dominating number of x, we consider, for k ≥ 3, √    a, Θ[k] , hk := min

(3.3)

where Θ[k] :=

2 2 2 αk2 αk+1 · · · α2k−1 (α12 α22 · · · αk2 + Q[k] a) − Q[k] α12 α22 · · · αk−1 αk4 . 2 2 2 α12 α22 · · · αk−1 (α12 α22 · · · αk2 + aQ[k] − 2αk2 Q[k] ) + Q[k] αk2 αk+1 · · · α2k−1

(3.4)

Then we obtain a characterization for semi-weakly k-hyponormality of Wα(x) . √ √ √ √ Proposition 3.3. Let α(x) : x, ( a, b, c)∧ and n ≥ k + 1. Then Wα(x) is semi-weakly k-hyponormal if and only if 0 < x ≤ ( hk )2 , i.e.,  hk = (sup{x : Wα(x) is semi-weakly k-hyponormal})1/2 . Proof. Observe that the corresponding symmetric matrix Ω[k] (t) with respect to f [k] (x0 , x1 , ..., xk , t) satisfies   f [k] (x0 , x1 , ..., xk , t) = Ω[k] (t)(x0 , x1 , ..., xk )T , (x0 , x1 , ..., xk )T . By changing the basis of Rk+1 , we see that the matrix Ω[k] (t) is unitary equivalent to the following ⎛

⎞ ⎛ ⎞   [k] [k] [k]   q − w0 t − w1 t ⎠ ⊕ ⎝ 1 ⎠ ⊕ Diag tv2[k] , ..., tv [k] . k−2 [k] [k] [k] [k] w0 t tvk−1 tvk − w1 t

[k] q 0

⎝ −

[k]

By the definitions of vi ≥ 0 for all k and i, we show that [k]

[k]

det Ω[k] (t) ≥ 0 ⇐⇒ d1 (t) · d2 (t) ≥ 0, [k]

[k] [k]

[k]

[k]

[k] [k]

[k]

where d1 (t) = tq0 vk−1 − tw0 and d2 (t) = tq1 vk − tw1 . Observe that  2 2  [k] [k] [k] 2 2 2 d1 (t) = v0 vk−1 t2 + tα02 αk−1 αk αk+1 · · · α2k−2 − α12 α22 · · · αk−1 > 0,

t > 0,

  [k] [k] [k] and d2 (t) = t g1 (x)t + g2 (x) , where   [k] 2 4 , g1 (x) = αk2 α12 α22 · · · α2k−1 − xα14 α24 · · · αk−1   [k] 2 2 − α12 α22 · · · αk−1 (αk4 + xα12 − 2xαk2 ). g2 (x) = α12 − x αk2 · · · α2k−1 Since 0 < x ≤ a and the weight sequence {αi }∞ i=0 is increasing, we have g1 (x) ≥ 0, and so d2 (t) ≥ 0 for [k]

all t >

1 , Q[k]

which is equivalent to −

[k] g2 (x) [k] g1 (x)

≤

1 . Q[k]

[k]

Hence d2 (t) ≥ 0 for t >

[k]

1 Q[k]

if and only if x ≤ Θ[k] .

Therefore x ≤ ( hk )2 , which proves this proposition. 2 Note that the extremal values  h2 and  h3 in (3.3) appeared in [10, Theorem 4.3] and [30, p. 1050], respectively.

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

714

4. Main theorems In this section we formulate some interesting properties about the limiting values of the sequences {Ak } and {Θ[k] } in Section 3. √ √ √ Let a, b and c with 0 < a < b < c be given and let ( a, b, c)∧ be the Stampfli’s recursively generated weight sequence as in Section 2. Then the value L as in (2.4) can be obtained by taking a limit of Ai from (2.6) as follows. Theorem 4.1. Let the sequence {Aj }j≥0 be as in Lemma 2.1. Then L2 = lim

j→∞

Proof. For simplicity, we denote β± := Ψ1 ±

Aj+1 . Aj

(4.1)



Ψ21 + 4Ψ0 (double signs in the same order, respectively).  j Then β+ > β− > 0 and β+ /2 = L2 by (2.4). Using (2.6) and ββ− → 0 (j → ∞), we have +  j+2  j+2 β+ − β2− 2 Aj+1 lim = lim  j+1  j+1 j→∞ Aj j→∞ β+ − β2− 2 = lim

j→∞

2 β+

 j+2 1 − ββ− β+ + 2   j+1  = 2 = L . 1 − ββ− +

Hence the proof is complete. 2  Recall that H2 :=

ab(c−b) a2 −2ab+bc [k]

 1/2 = sup{x : Wα(x) is 2-hyponormal} ([10]). This extremal value also

can be represented with Θ , which will be used to prove our main theorem (Theorem 4.3). Theorem 4.2. For k ≥ 3, let {Θ[k] } be given as above in (3.4). Then H22 = lim Θ[k] . k→∞

Proof. To obtain the limit of sequence {Θ[k] } in (3.4), we denote  2  [k] 2 C1 := α12 αk2 αk+1 · · · α2k−1 − α22 · · · αk2 , [k]

2 2 2 + αk2 αk+1 · · · α2k−1 − 2α12 α22 · · · αk2 . C2 := α14 α22 · · · αk−1 [k]

Using the definition of v1 in (2.2), Θ[k] can be simplified by [k]

Θ[k] =

[k]

2 2 v1 αk2 αk+1 · · · α2k−1 + Q[k] C1 [k]

[k]

2 v1 α12 α22 · · · αk−1 + Q[k] C2

.

For k ≥ 3, it follows from (2.5) that [k]

v1 = Ak−1 α12 + Ak−2 Ψ0 , [k] C1

2 = α12 αk2 Ak−2 (αk+1 − α22 ),

(4.2) (4.3)

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

[k]

2 C2 = Ak−2 {αk2 αk+1 − α12 (αk2 − α22 − α12 )}.

715

(4.4)

Using Q[k] = Ak−1 Δk of Proposition 2.4(ii) and (4.2)-(4.4), we can modify the formula Θ[k] in (3.4) as follows [k]

Θ[k] =

θ1

[k]

,

(4.5)

θ2 where [k]

2 θ1 = (Ak−1 α12 + Ak−2 Ψ0 )(Ak−1 αk2 + Ak−2 Ψ0 ) + Ak−1 Δk α12 αk2 Ak−2 (αk+1 − α22 ) [k]

2 θ2 = (Ak−1 α12 + Ak−2 Ψ0 )(Ak−2 α12 + Ak−3 Ψ0 ) + Ak−1 Δk Ak−2 {αk2 αk+1 − α12 (αk2 − α22 − α12 )}.

It follows from a simple computation that

Θ[k] =



   Ak−2 2 2  2 k−2 k−2 2 α12 + A αk2 + A Ak−1 Ψ0 Ak−1 Ψ0 + Ak−1 α1 αk αk+1 − α2     . A Ak−2 Ak−2 2 + Ak−3 Ψ 2 α2 2 (α2 − α2 − α2 )} Ψ {α − α α12 + Ak−2 α + 0 0 1 1 2 1 k k+1 k A A A k−1 k−1 k−2 k−1 1 Δk

1 Δk

By (4.1), we obtain we can obtain

Aj Aj+1

→

1 L2 .

Using Δk  ∞ (∵ L2 > −Ψ0 ), αn2  L2 and definitions of αi (i = 1, 2),

lim Θ[k] =

k→∞

aL4 − abL2 . L4 − aL2 − ab + a2

Since the value L2 is a positive root of a quadratic equation, y 2 − Ψ1 y − Ψ0 = 0, we obtain L4 = Ψ1 L2 + Ψ0 . From the definitions of Ψ0 and Ψ1 , we conclude easily that lim Θ[k] =

k→∞

a(Ψ1 L2 + Ψ0 ) − abL2 ab (c − b) . = 2 Ψ1 L2 + Ψ0 − aL2 − ab + a2 a − 2ab + bc

Hence the proof is complete. 2 Combining Proposition 3.3, Theorem 4.2 and the result of [18, Theorem 1.3], we obtain the following main theorem of the paper. √ √ √ √ Theorem 4.3. Let α(x) : x, ( a, b, c)∧ be as usual and let Wα(x) be the weighted shift with a sequence α(x). Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)

Wα(x) is subnormal; Wα(x) is polynomially hyponormal; Wα(x) is completely semi-weakly hyponormal; √ x≤ hk for all positive integer k; √ x ≤ H2 , i.e., Wα(x) is 2-hyponormal.

Proof. The implications (i) =⇒ (ii) =⇒ (iii) are straightforward. For (iii) =⇒ (iv) use Proposition 3.3. For (iv) =⇒ (v) use Theorem 4.2. The implication (v) =⇒ (i) is well known (see [18, Theorem 1.3]). 2

C. Li et al. / J. Math. Anal. Appl. 479 (2019) 703–717

716

We verify by example that Theorem 4.3 holds by revisiting an example from [10, Example 4.5]. √ √ √ Example 4.4. Consider a weighted shift Wα(x) with α(x) : x, (1, 2, 3)∧ for 0 < x ≤ 1 (cf. [10, Exam√ ple 4.5]). By direct computations, we get a recursively generated formula  hk = min{1, Θ[k] } as in (3.3) with the equation Θ[k] =

2 (Ak−1 − 2Ak−2 )(Ak−1 αk2 − 2Ak−2 ) + Q[k] αk2 Ak−2 (αk+1 − 2) 2 2 [k] (Ak−1 − 2Ak−2 )(Ak−2 − 2Ak−3 ) + Q Ak−2 {αk (αk+1 − 1) − 1}

(4.6)

√  √ √ √  as in (4.5), where Q[k] = Ak−1 Δk with Ak = 42 (2 + 2)k+1 − (2 − 2)k+1 and Δk = 12 ( 2 − √ 1/2 √   1   1 1 1) (3 + 2 2)k − 1 . We recall that  (358 − 64 2) ≈ 0.85628 and  h2 = 337 h3 = 17 411 (108047 − √ 1/2 ≈ 0.82520 are known in [10] and [30, Example 3.8], respectively. In view of Theorems 4.1 19208 2) and 4.2, we can confirm that

√ 2 L4 − 2L2 Ak+1 = . = 2 + 2 = L2 and lim Θ[k] = 4 2 k→∞ Ak k→∞ L −L −2+1 3 lim

For k ≥ 4, we can estimate  hk by the recursively generated formula of (4.6), for example, √ 1/2  1 10042022 − 1756160 2  ≈ 0.817891, h4 = 17 39097 √ 1/2  476166568777 − 83009555136 2  ≈ 0.816733, h5 = 537848475889 √ 1/2  211988219954074247 − 36931731007479000 2  ≈ 0.816537, ..., etc. h6 = 239634222108214827 We close this paper with a remark on the backward extensions of Stampfli’s subnormal completion. Remark 4.5. Using the technical computations in this paper, it is possible to detect implications among three properties of subnormality, polynomial hyponormality and completely semi-weak hyponormality of √ √ √ √ √ Wα(x1 ,··· ,xn ) with weight sequence α(x1 , · · · , xn ) : x1 , · · · , xn , ( a, b, c)∧ . We leave this to the interested reader. Acknowledgment The authors would like to express their gratitude to the referee for careful reading of the paper and helpful comments. References [1] S. Baek, G. Exner, I.B. Jung, C. Li, On semi-cubically hyponormal weighted shifts with first two equal weights, Kyungpook Math. J. 56 (2016) 899–910. [2] S. Baek, G. Exner, I.B. Jung, C. Li, Semi-cubic hyponormality of weighted shifts with Stampfli recursive tail, Integral Equations Operator Theory 88 (2017) 229–248. [3] C. Berg, J. Christensen, P. Ressel, Harmonic Analysis on Semigroups, Springer, Berlin, 1984. [4] J. Bram, Subnormal operators, Duke Math. J. 22 (1955) 15–94. [5] Y.B. Choi, A propagation of quadratically hyponormal weighted shifts, Bull. Korean Math. Soc. 37 (2000) 347–352. [6] R. Curto, Joint hyponormality: a bridge between hyponormality and subnormality, Proc. Sympos. Math. 51 (1990) 69–91. [7] R. Curto, Quadratically hyponormal weighted shifts, Integral Equations Operator Theory 13 (1990) 49–66. [8] R. Curto, L. Fialkow, Recursiveness, positivity, and truncated moment problems, Houston J. Math. 17 (1991) 603–635.

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