On Positive Quadratic Hyponormality of a Unilateral Weighted Shift with Recursively Generated by Five Weights

Abstract

Let $1<a<b<c<d$ and ${\widehat{\alpha }}_{\left[5\right]}:={\left(1,\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d}\right)}^{\wedge }$ be a weighted sequence that is recursively generated by five weights $1,\sqrt{a},\sqrt{b},$$\sqrt{c},\sqrt{d}$. In this paper, we give sufficient conditions for the positive quadratic hyponormalities of $W_{\alpha \left(x\right)}$ and $W_{\alpha \left(y,x\right)}$, with $\alpha \left(x\right):\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}$ and $\alpha \left(y,x\right):\sqrt{y},\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}.$

1. Introduction

Let $\mathcal{H}$ be a separable, infinite dimensional, complex Hilbert space, and let $\mathcal{L}\left(\mathcal{H}\right)$ be the algebra of all bounded linear operators on $\mathcal{H}$. An operator T in $\mathcal{L}\left(\mathcal{H}\right)$ is said to be normal if $T^{*}T=T{T}^{*},$ hyponormal if $T^{*}T\ge T{T}^{*},$ and subnormal if $T={\left.N\right|}_{\mathcal{H}},$ where N is normal on some Hilbert space $K\supseteq \mathcal{H}.$ For $A,B\in \mathcal{L}\left(\mathcal{H}\right)$, let $[A,B]:=AB-BA$. We say that an n-tuple $T=(T_1,\dots ,T_n)$ of operators in $\mathcal{L}\left(\mathcal{H}\right)$ is hyponormal if the operator matrix ${\left([T_j^{*},T_i]\right)}_{i,j=1}^n$ is positive on the direct sum of n copies of $\mathcal{H}$. For arbitrary positive integer k, $T\in \mathcal{L}\left(\mathcal{H}\right)$ is (strongly) k-hyponormal if $(I,T,\dots ,T^k)$ is hyponormal. It is well known that T is subnormal if and only if T is ∞-hyponormal. An operator T in $\mathcal{L}\left(\mathcal{H}\right)$ is said to be weaklyn-hyponormal if $p\left(T\right)$ is hyponormal for any polynomial p with degree less than or equal to $n.$ An operator T is polynomially hyponormal if $p\left(T\right)$ is hyponormal for every polynomial $p$. In particular, the weak two-hyponormality (or weak three-hyponormality) is referred to as quadratical hyponormality (or cubical hyponormality, resp.) and has been considered in detail in [1,2,3,4,5,6,7,8,9].

Let ${\left\{e_n\right\}}_{n=0}^{\infty }$ be the canonical orthonormal basis for Hilbert space $l^2\left({\mathbb{Z}}_{+}\right)$, and let $\alpha :={\left\{{\alpha }_n\right\}}_{n=0}^{\infty }$ be a bounded sequence of positive numbers. Let $W_{\alpha }$ be a unilateral weighted shift defined by $W_{\alpha }{e}_n:={\alpha }_n{e}_{n+1}\left(n\ge 0\right).$ It is well known that $W_{\alpha }$ is hyponormal if and only if ${\alpha }_n\le {\alpha }_{n+1}\left(n\ge 0\right).$ The moments of $W_{\alpha }$ are usually defined by ${\gamma }_0:=1,{\gamma }_i:={\alpha }_0^2\cdots {\alpha }_{i-1}^2\left(i\ge 1\right).$ It is well known that $W_{\alpha }$ is subnormal if and only if there exists a Borel probability measure $\mu$ supported in $\left[0,{∥W_{\alpha }∥}^2\right],$ with ${∥W_{\alpha }∥}^2\in$ supp $\mu ,$ such that [10]${\gamma }_n=\int t^n$ d $\mu \left(t\right)$$\left(\forall n\ge 0\right).$ It follows from [11] (Theorem 4) that $W_{\alpha }$ is subnormal if and only if for every $k\ge 1$ and every $n\ge 0,$ the Hankel matrix:

$$A\left(n,k\right):=\left[\begin{array}{ccccc}{\gamma }_n& {\gamma }_{n+1}& {\gamma }_{n+2}& \cdots & {\gamma }_{n+k}\\ {\gamma }_{n+1}& {\gamma }_{n+2}& {\gamma }_{n+3}& \cdots & {\gamma }_{n+k+1}\\ {\gamma }_{n+2}& {\gamma }_{n+3}& {\gamma }_{n+4}& \cdots & {\gamma }_{n+k+2}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {\gamma }_{n+k}& {\gamma }_{n+k+1}& {\gamma }_{n+k+2}& \cdots & {\gamma }_{n+2k}\end{array}\right]\ge 0.$$

A weighted shift $W_{\alpha }$ is said to be recursively generated if there exists $i\ge 1$ and $\phi =\left({\phi }_0,\dots ,{\phi }_{i-1}\right)\in {\mathbb{C}}^i$ such that:

$${\gamma }_n={\phi }_{i-1}{\gamma }_{n-1}+\cdots +{\phi }_0{\gamma }_{n-i}\left(n\ge i\right),$$

where ${\gamma }_n$ is the moment of $W_{\alpha },$ i.e., ${\gamma }_0:=1,{\gamma }_i:={\alpha }_0^2\cdots {\alpha }_{i-1}^2\left(i\ge 1\right),$ equivalently,

$${\alpha }_n^2={\phi }_{i-1}+\frac{{\phi }_{i-2}}{{\alpha }_{n-1}^2}+\cdots +\frac{{\phi }_0}{{\alpha }_{n-1}^2\cdots {\alpha }_{n-i+1}^2}\left(n\ge i\right).$$

Given an initial segment of weights $\alpha :{\alpha }_0,\dots ,{\alpha }_{2k}\left(k\ge 0\right),$ there is a canonical procedure to generate a sequence (denoted $\widehat{\alpha }$) in such a way that $W_{\widehat{\alpha }}$ is a recursively-generated shift having $\alpha$ as an initial segment of weight. In particular, given an initial segment of weights $\alpha :\sqrt{a},\sqrt{b},\sqrt{c}$ with $0<a<b<c,$ we obtain ${\phi }_0=-\frac{ab\left(c-b\right)}{b-a}$ and ${\phi }_1=\frac{b\left(c-a\right)}{b-a}.$

In [12,13], Curto-Putinar proved that there exists an operator that is polynomially hyponormal, but not two-hyponormal. Although the existence of a weighted shift, which is polynomially hyponormal, but not subnormal, was established in [12,13], a concrete example of such weighted shifts has not been found yet. Recently, the authors in [14] proved that the subnormality is equivalent to the polynomial hyponormality for recursively-weighted shift $W_{\alpha }$ with $\alpha :\sqrt{x},{\left(\sqrt{a},\sqrt{b},\sqrt{c}\right)}^{\wedge }.$ Based on this, in this paper, we have to consider the weighted shift operator with five generated elements.

The organization of this paper is as follows. In Section 2, we recall some terminology and notations concerning the quadratic hyponormality and positive quadratic hyponormality of unilateral weighted shifts $W_{\alpha }$. In Section 3, we give some results on the unilateral weighted shifts with recursively generated by five weights ${\widehat{\alpha }}_{\left[5\right]}:={\left(1,\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d}\right)}^{\wedge }\left(1<a<b<c<d\right)$. In Section 4, we consider positive quadratic hyponormalities of $W_{\alpha }$ with weights $\alpha :\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}$ and $\alpha :\sqrt{y},\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}.$ In Section 5, we give more results on the positive quadratic hyponormality for any unilateral weighted shift $W_{\alpha }$. In Section 6, we present the conclusions.

2. Preliminaries and Notations

Recall that a weighted shift $W_{\alpha }$ is quadratically hyponormal if $W_{\alpha }+s{W}_{\alpha }^2$ is hyponormal for any $s\in \mathbb{C}$ [2], i.e., $D\left(s\right):=[{(W_{\alpha }+s{W}_{\alpha }^2)}^{*},W_{\alpha }+s{W}_{\alpha }^2]\ge 0,$ for any $s\in \mathbb{C}.$ Let ${\left\{e_i\right\}}_{i=0}^{\infty }$ be an orthonormal basis for $\mathcal{H}$, and let $P_n$ be the orthogonal projection on ${\vee }_{i=0}^n\left\{e_i\right\}.$ For $s\in \mathbb{C}$, we let:

$$\begin{array}{ccc}\hfill D_n\left(s\right)& =& P_n[{(W_{\alpha }+s{W}_{\alpha }^2)}^{*},W_{\alpha }+s{W}_{\alpha }^2]P_n\hfill \\ & =& \left[\begin{array}{cccccc}{q}_0& r_0& 0& \cdots & 0& 0\\ \overline{r_0}& q_1& r_1& \cdots & 0& 0\\ 0& \overline{r_1}& q_2& \ddots & 0& 0\\ ⋮& ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& \ddots & q_{n-1}& r_{n-1}\\ 0& 0& 0& \cdots & \overline{r_{n-1}}& q_n\end{array}\right],\hfill \end{array}$$

where:

$$\begin{array}{ccc}\hfill q_k& :& =u_k+{\left|s\right|}^2{v}_k,r_k:=w_k\overline{s},\hfill \\ \hfill u_k& :& ={\alpha }_k^2-{\alpha }_{k-1}^2,v_k:={\alpha }_k^2{\alpha }_{k+1}^2-{\alpha }_{k-2}^2{\alpha }_{k-1}^2,\hfill \\ \hfill w_k& :& ={\alpha }_k^2{({\alpha }_{k+1}^2-{\alpha }_{k-1}^2)}^2,\mathrm{for}k\ge 0,\hfill \end{array}$$

and ${\alpha }_{-1}={\alpha }_{-2}:=0.$ Hence, $W_{\alpha }$ is quadratically hyponormal if and only if $D_n\left(s\right)\ge 0$ for every $s\in \mathbb{C}$ and every $n\ge 0.$ Hence, we consider $d_n\left(·\right):=det{D}_n\left(·\right)$, which is a polynomial in $t:={\left|s\right|}^2$ of degree $n+1,$ with Maclaurin expansion $d_n\left(t\right):={\sum }_{i=0}^{n+1}c\left(n,i\right)t^i.$ It is easy to find the following recursive relations [2]:

$$\left\{\begin{array}{c}{d}_0\left(t\right)=q_0,\hfill \\ d_1\left(t\right)=q_0{q}_1-{\left|r_0\right|}^2,\hfill \\ d_{n+2}\left(t\right)=q_{n+2}{d}_{n+1}\left(t\right)-{\left|r_{n+1}\right|}^2{d}_n\left(t\right)\left(n\ge 0\right).\hfill \end{array}\right.$$

Furthermore, we can obtain the following:

$$\begin{array}{ccc}\hfill c\left(0,0\right)& =& u_0,c\left(0,1\right)=v_0,\hfill \\ \hfill c\left(1,0\right)& =& u_1{u}_0,c\left(1,1\right)=u_1{v}_0+u_0{v}_1-w_0,c\left(1,2\right)=v_1{v}_0,\hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill c\left(n+2,i\right)& =& u_{n+2}c\left(n+1,i\right)+v_{n+2}c\left(n+1,i-1\right)-w_{n+1}c\left(n,i-1\right)\hfill \\ & & \left(n\ge 0,\mathrm{and}0\le i\le n+1\right).\hfill \end{array}$$

In particular, for any $n\ge 0,$ we have:

$$c\left(n,0\right)=u_0{u}_1\cdots u_n,c\left(n,n+1\right)=v_0{v}_1\cdots v_n.$$

Furthermore, we can obtain the following results.

Lemma 1.

Let $\rho :=v_2\left(u_0{v}_1-w_0\right)+v_0\left(u_1{v}_2-w_1\right).$ Then, for any $n\ge 4,$ we have:

$$\begin{array}{ccc}\hfill c\left(n,n\right)& =& u_{n}c\left(n-1,n\right)+\left(u_{n-1}{v}_n-w_{n-1}\right)c\left(n-2,n-1\right)\hfill \\ & & +\sum _{i=1}^{n-3}{v}_n{v}_{n-1}\cdots v_{i+3}\left(u_{i+1}{v}_{i+2}-w_{i+1}\right)c\left(i,i+1\right)+v_n{v}_{n-1}\cdots v_3\rho .\hfill \end{array}$$

Lemma 2.

Let $\tau :=u_0\left(u_1{v}_2-w_1\right).$ Then, for any $n\ge 4,$ we have:

$$\begin{array}{ccc}\hfill c\left(n,n-1\right)& =& u_{n}c\left(n-1,n-1\right)+\left(u_{n-1}{v}_n-w_{n-1}\right)c\left(n-2,n-2\right)\hfill \\ & & +\sum _{i=1}^{n-3}{v}_n{v}_{n-1}\cdots v_{i+3}\left(u_{i+1}{v}_{i+2}-w_{i+1}\right)c\left(i,i\right)+v_n{v}_{n-1}\cdots v_3\tau .\hfill \end{array}$$

Lemma 3.

For any $n\ge 5$ and $0\le i\le n-2,$ we have:

$$\begin{array}{ccc}\hfill c\left(n,i\right)& =& u_{n}c\left(n-1,i\right)+\left(u_{n-1}{v}_n-w_{n-1}\right)c\left(n-2,i-1\right)\hfill \\ & & +\sum _{j=1}^{n-3}{v}_n{v}_{n-1}\cdots v_{j+3}\left(u_{j+1}{v}_{j+2}-w_{j+1}\right)c\left(j,j+i-n+1\right)\hfill \\ & & +v_n{v}_{n-1}\cdots v_{5}c\left(i-n+5,0\right)\left(u_{i-n+6}{v}_{i-n+7}-w_{i-n+6}\right).\hfill \end{array}$$

To detect the positivity of $d_n\left(t\right),$ we need the following concept.

Definition 1.

Let $\alpha :{\alpha }_0,{\alpha }_1,\dots$ be a positive weight sequence. We say that $W_{\alpha }$ is positively quadratically hyponormal if $c\left(n,i\right)\ge 0$ for all $n,i\ge 0$ with $0\le i\le n+1,$ and $c\left(n,n+1\right)>0$ for all $n\ge 0$ [2].

Positive quadratic hyponormality implies quadratic hyponormality, but the converse is false [15]. In addition, the authors in [15] showed that the positive quadratic hyponormality is equivalent to the quadratic hyponormality for recursively-generated weighted shift $W_{\alpha }$ with $\alpha :\sqrt{x},{\left(\sqrt{a},\sqrt{b},\sqrt{c}\right)}^{\wedge }$ (here, $0<x\le a<b<c$).

3. Recursive Relation of $W_{{\widehat{\alpha }}_{\left[5\right]}}$

Given the initial segment of weights $\alpha :1,\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d}$ with $1<a<b<c<d,$ we obtain the moments:

$${\gamma }_0={\gamma }_1=1,{\gamma }_2=a,{\gamma }_3=ab,{\gamma }_4=abc,{\gamma }_5=abcd.$$

Let:

$$V_0=\left(\begin{array}{c}{\gamma }_0\hfill \\ {\gamma }_1\hfill \\ {\gamma }_2\hfill \end{array}\right),V_1=\left(\begin{array}{c}{\gamma }_1\hfill \\ {\gamma }_2\hfill \\ {\gamma }_3\hfill \end{array}\right),V_2=\left(\begin{array}{c}{\gamma }_2\hfill \\ {\gamma }_3\hfill \\ {\gamma }_4\hfill \end{array}\right),$$

and we assume that $V_0,V_1,V_2$ are linearly independent, i.e.,

$$det\left(V_0,V_1,V_2\right)=detA(0,2)\ne 0.$$

Then, there exist three nonzero numbers ${\phi }_0,{\phi }_1,{\phi }_2$, such that:

$$\left(\begin{array}{c}{\gamma }_3\hfill \\ {\gamma }_4\hfill \\ {\gamma }_5\hfill \end{array}\right)={\phi }_0\left(\begin{array}{c}{\gamma }_0\hfill \\ {\gamma }_1\hfill \\ {\gamma }_2\hfill \end{array}\right)+{\phi }_1\left(\begin{array}{c}{\gamma }_1\hfill \\ {\gamma }_2\hfill \\ {\gamma }_3\hfill \end{array}\right)+{\phi }_2\left(\begin{array}{c}{\gamma }_2\hfill \\ {\gamma }_3\hfill \\ {\gamma }_4\hfill \end{array}\right).$$

A straightforward calculation shows that:

$$\begin{array}{ccc}\hfill {\phi }_0& =& \frac{ab(a{b}^2-2abc+b{c}^2+acd-bcd)}{a^2-2ab+a{b}^2+bc-abc},\hfill \\ \hfill {\phi }_1& =& \frac{-ab(ab-ac-bc+b{c}^2+cd-bcd)}{a^2-2ab+a{b}^2+bc-abc},\hfill \\ \hfill {\phi }_2& =& \frac{b(a^2-ab-ac+abc+cd-acd)}{a^2-2ab+a{b}^2+bc-abc}.\hfill \end{array}$$

Thus:

$${\gamma }_{n+1}={\phi }_0{\gamma }_{n-2}+{\phi }_1{\gamma }_{n-1}+{\phi }_2{\gamma }_n\left(n\ge 5\right),$$

i.e.,

$${\alpha }_n^2={\phi }_2+\frac{{\phi }_1}{{\alpha }_{n-1}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2}\left(n\ge 5\right).$$

(1)

By (1), we can obtain a recursively-generated weighted shift, and we set it as ${\widehat{\alpha }}_{\left[5\right]}$. In this case, we call the weighted shift operator $W_{{\widehat{\alpha }}_{\left[5\right]}}$ with rank three.

Proposition 1.

$W_{{\widehat{\alpha }}_{\left[5\right]}}$ with rank three is subnormal if and only if:

(1) $1<a<b,$

(2) $c>\frac{a}{b}\left(\frac{{\left(b-1\right)}^2}{a-1}+1\right),$

(3) $d>\frac{b}{c}\left(\frac{{\left(c-a\right)}^2}{b-a}+a\right).$

Proof. 

See [16], Example 3.6. □

Proposition 2.

If $W_{{\widehat{\alpha }}_{\left[5\right]}}$ with rank three is subnormal, then ${\phi }_0>0,{\phi }_1<0$ and ${\phi }_2>0.$

Proof. 

By Proposition 1, we know that:

$$\begin{array}{ccc}\hfill {ϝ}_1& :& =a{b}^2-2ab+bc+a^2-abc<0,\hfill \\ \hfill {ϝ}_2& :& =a{b}^2+b{c}^2-2abc+acd-bcd<0.\hfill \end{array}$$

Thus, ${\phi }_0>0.$ Since:

$$ab-ac-bc+b{c}^2+c\left(1-b\right)d<\left(c-b\right)\frac{{ϝ}_1}{b-a}<0,$$

and:

$$a^2-ab-ac+abc+c\left(1-a\right)d<\left(c-a\right)\frac{{ϝ}_1}{b-a}<0,$$

we have ${\phi }_1<0$ and ${\phi }_2>0.$ The proof is complete. □

Proposition 3.

Let $\frac{u_{n-1}}{u_n}={\beta }_n$$\left(n\ge 5\right).$ Then:

$${\beta }_n=-\frac{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}{{\phi }_1{\alpha }_{n-3}^2+{\phi }_0+{\phi }_0{\beta }_{n-1}}.$$

(2)

Proof. 

Since:

$$\begin{array}{ccc}\hfill u_n& =& {\alpha }_n^2-{\alpha }_{n-1}^2\hfill \\ & =& -\frac{{\phi }_1}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2}{u}_{n-1}-\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}\left(u_{n-1}+u_{n-2}\right)\hfill \\ & =& -\left(\frac{{\phi }_1}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}\right)u_{n-1}-\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}{u}_{n-2},\hfill \end{array}$$

so:

$$\begin{array}{ccc}\hfill 1& =& -\left(\frac{{\phi }_1}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}\right)\frac{u_{n-1}}{u_n}-\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}\frac{u_{n-2}}{u_n}\hfill \\ & =& -\left(\frac{{\phi }_1}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}\frac{u_{n-2}}{u_{n-1}}\right)\frac{u_{n-1}}{u_n}.\hfill \end{array}$$

Thus, we have:

$$\begin{array}{ccc}\hfill {\beta }_n& =& -\frac{1}{\left(\frac{{\phi }_1}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}+\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}{\beta }_{n-1}\right)}\hfill \\ & =& -\frac{{\alpha }_{n-1}^2{\alpha }_{n-2}^2{\alpha }_{n-3}^2}{{\phi }_1{\alpha }_{n-3}^2+{\phi }_0+{\phi }_0{\beta }_{n-1}}.\hfill \end{array}$$

Thus, we have our conclusion. □

Since ${lim}_{n\to \infty }{\alpha }_n^2=L^2$, we let ${lim}_{n\to \infty }{\beta }_n:={\beta }^{*},$ and by (2), we have ${\beta }^{*}=-\frac{L^6}{{\phi }_1{L}^2+{\phi }_0+{\phi }_0{\beta }^{*}};$ hence:

$${\beta }^{*}=-\frac{1}{2}-\frac{{\phi }_1}{2{\phi }_0}{L}^2+\frac{1}{2{\phi }_0}\sqrt{\left({\phi }_1^2-4{\phi }_0{\phi }_2\right)L^4-2{\phi }_0{\phi }_1{L}^2-3{\phi }_0^2}.$$

(3)

4. Main Results

First, we give the following result (cf. [11], Corollary 5).

Proposition 4.

Let $W_{\alpha }$ be any unilateral weighted shift. Then, $W_{\alpha }$ is two-hyponormal if and only if ${\theta }_k:=u_k{v}_{k+1}-w_k\ge 0,$ $\forall k\in \mathbb{N}.$

It is well-known that if $W_{\alpha }$ is two-hyponormal or positively quadratically hyponormal, then $W_{\alpha }$ is quadratically hyponormal. By Proposition 4 and Lemma 1∼3, we have the following result.

Theorem 1.

Let $W_{\alpha }$ be any unilateral weighted shift. If $W_{\alpha }$ is 2-hyponormal, then $W_{\alpha }$ is positively quadratically hyponormal.

4.1. The Positive Quadratic Hyponormality of $W_{\alpha \left(x\right)}$

Let $\alpha \left(x\right):\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}$ with $0<x\le 1,$ and we consider the following $\left(n+1\right)\times \left(n+1\right)$ matrix:

$$D_n=\left[\begin{array}{ccccc}{u}_0+v_{0}t& \sqrt{w_{0}t}& 0& \cdots & 0\\ \sqrt{w_{0}t}& u_1+v_{1}t& \sqrt{w_{1}t}& \cdots & 0\\ 0& \sqrt{w_{1}t}& u_2+v_{2}t& \ddots & 0\\ ⋮& ⋮& \ddots & \ddots & \sqrt{w_{n-1}t}\\ 0& 0& \cdots & \sqrt{w_{n-1}t}& u_n+v_{n}t\end{array}\right].$$

(4)

Let $d_n=det{D}_n={\sum }_{i=0}^{n+1}c\left(n,i\right)t^i.$ Then:

Lemma 4.

$c(n,i)\ge 0,n=0,1,2,$ and $0\le i\le n+1$.

Proof. 

In fact, $c(1,1)=x\left(a-x\right)>0,$ $c(2,1)=ax\left(1-x\right)\left(b-1\right)>0,$ and:

$$\begin{array}{ccc}\hfill c(2,2)& =& ax\left(\left(ab-1\right)-\left(b-1\right)x\right)\hfill \\ & >& ax\left(\left(ab-a\right)-\left(b-1\right)x\right)\hfill \\ & =& ax\left(a-x\right)\left(b-1\right)>0.\hfill \end{array}$$

Thus, we have our conclusion. □

Lemma 5.

Assume that ${\theta }_k:=u_k{v}_{k+1}-w_k\ge 0$ for $k\ge 2.$ Then, for $n\ge 3,0\le i\le n+1,$ we have:

$$c(n,i)\ge u_{n}c(n-1,i)+v_n\cdots v_3[v_{2}c(1,i-n+1)-w_{1}c(0,i-n+1)].$$

Proof. 

For $n=3,0\le i\le 4,$

$$\begin{array}{ccc}\hfill c(3,i)& =& u_{3}c(2,i)+v_{3}c(2,i-1)-w_{2}c(1,i-1)\hfill \\ & =& u_{3}c(2,i)+({\mathbf{u}}_{\mathbf{2}}{\mathbf{v}}_3-{\mathbf{w}}_2)c(1,i-1)+v_3[v_{2}c(1,i-2)-w_{1}c(0,i-2)]\hfill \\ & \ge & u_{3}c(2,i)+v_3[v_{2}c(1,i-2)-w_{1}c(0,i-2)].\hfill \end{array}$$

By the inductive hypothesis, we have our result. □

Thus, if ${\theta }_k:=u_k{v}_{k+1}-w_k\ge 0$ for $k\ge 2,$ then by Lemma 1∼3 and Lemma 5, for $n\ge 3,$ we have:

$$c(n,i)\ge \left\{\begin{array}{ccc}{v}_n\cdots v_{2}c(1,2),\hfill & \mathrm{for}\hfill & i=n+1,\hfill \\ u_{n}c(n-1,n)+v_n\cdots v_3\rho ,\hfill & \mathrm{for}\hfill & i=n,\hfill \\ u_{n}c(n-1,n-1)+v_n\cdots v_3\tau ,\hfill & \mathrm{for}\hfill & i=n-1,\hfill \\ u_{n}c(n-1,i),\hfill & \mathrm{for}\hfill & 0\le i\le n-2,\hfill \end{array}\right.$$

where $\rho =ax\left(b-1\right)\left(a-x\right)>0,$ $\tau =x\left(a\left(b-a\right)-\left(ab-2a+1\right)x\right).$ Therefore, when $n\ge 3,$ we have:

$$\left\{\begin{array}{c}c(n,n+1)>0,\hfill \\ c(n,n)>u_{n}c(n-1,n)+v_n\cdots v_3\rho \ge 0,\hfill \\ c(n,i)\ge u_n\cdots u_{i+2}c(i+1,i)(n\ge 3,0\le i\le n-2).\hfill \end{array}\right.$$

To complete our analysis of the coefficients $c(n,i),$ it suffices to determine the values of x for which $c(n,n-1)\ge 0$$(n\ge 3).$

Lemma 6.

$c(n,n-1)\ge 0$ (for any $n\ge 3),$ if $c(3,2)\ge 0,c(4,3)\ge 0$ and $A_n:=v_0{v}_1{v}_2{u}_n{u}_{n-1}+u_n{v}_{n-1}\rho +v_n{v}_{n-1}\tau \ge 0$$(n\ge 5).$

Proof. 

For $n\ge 4,$ by Lemma 2, we have:

$$\begin{array}{ccc}\hfill c(n,n-1)& \ge & u_{n}c(n-1,n-1)+v_n\cdots v_3\tau \hfill \\ & \ge & u_n[u_{n-1}c(n-2,n-1)+v_{n-1}\cdots v_3\rho ]+v_n\cdots v_3\tau ,\hfill \end{array}$$

and since $c(n-2,n-1)=v_{n-2}\cdots v_0,$ we get:

$$c(n,n-1)\ge u_n(u_{n-1}{v}_{n-2}\cdots v_0+v_{n-1}\cdots v_3\rho )+v_n\cdots v_3\tau .$$

If $n\ge 5,$ we can factor $v_{n-2}\cdots v_3$ to get:

$$\begin{array}{ccc}\hfill c(n,n-1)& \ge & v_{n-2}\cdots v_3(v_0{v}_1{v}_2{u}_n{u}_{n-1}+u_n{v}_{n-1}\rho +v_n{v}_{n-1}\tau )\hfill \\ & =& v_{n-2}\cdots v_3{A}_n.\hfill \end{array}$$

Hence, we have our result. □

Let:

$$x_n:=sup\{x:c(n,n-1)\ge 0\mathrm{in}{W}_{\alpha }\}(n\ge 3).$$

By direct computations, we have:

$$\begin{array}{ccc}\hfill x_3& =& \frac{a{\theta }_2+a\left(b-a\right)v_3+a\left(ab-1\right)u_3}{{\theta }_2+a\left(b-1\right)u_3+\left(-2a+ab+1\right)v_3},\hfill \\ \hfill x_4& =& \frac{a\left(ab-1\right){\theta }_3+a\left(u_4+v_4\right){\theta }_2+\left(a^2\left(b-1\right)u_4+v_{4}a\left(b-a\right)\right)v_3+a^{2}b{u}_3{u}_4}{a\left(b-1\right){\theta }_3+v_4{\theta }_2+\left(\left(ab-2a+1\right)v_4+u_{4}a\left(b-1\right)\right)v_3+a{u}_3{u}_4}.\hfill \end{array}$$

For $n\ge 5,$ a calculation using the specific form of $v_0,v_1,v_2,\rho$ and $\tau$ shows that:

$$\begin{array}{ccc}\hfill A_n& =& [a^{2}b{u}_n{u}_{n-1}+a^2(b-1)u_n{v}_{n-1}+a(b-a)v_n{v}_{n-1}\hfill \\ & & -\left(a{u}_n{u}_{n-1}+a(b-1)u_n{v}_{n-1}+(ab+1-2a)v_n{v}_{n-1}\right)x]x,\hfill \end{array}$$

it follows that:

$$x_n=\frac{a^{2}b{u}_n{u}_{n-1}+a^2(b-1)u_n{v}_{n-1}+a(b-a)v_n{v}_{n-1}}{a{u}_n{u}_{n-1}+a(b-1)u_n{v}_{n-1}+(ab+1-2a)v_n{v}_{n-1}}.$$

Let $z_n:=\frac{v_n}{u_n}$$(u_n\ne 0)$. Then, for $n\ge 5,$

$$x_n=\frac{a^{2}b+a^2(b-1)z_{n-1}+a(b-a)z_n{z}_{n-1}}{a+a(b-1)z_{n-1}+(ab+1-2a)z_n{z}_{n-1}}.$$

(5)

Lemma 7.

${lim}_{n\to \infty }{z}_n=K:=(1+{\beta }^{*})\left({\phi }_2-\frac{{\phi }_0}{L^4}{\beta }^{*}\right),$ where ${\beta }^{*}$ as in (3).

Proof. 

Since:

$${\alpha }_{n+1}^2={\phi }_2+\frac{{\phi }_1}{{\alpha }_n^2}+\frac{{\phi }_0}{{\alpha }_n^2{\alpha }_{n-1}^2}(n\ge 5),$$

from which it follows that:

$${\alpha }_n^2{\alpha }_{n+1}^2={\phi }_2{\alpha }_n^2+{\phi }_1+\frac{{\phi }_0}{{\alpha }_{n-1}^2}.$$

Thus:

$$v_n={\phi }_2(u_n+u_{n-1})-\frac{{\phi }_0(u_{n-1}+u_{n-2})}{{\alpha }_{n-1}^2{\alpha }_{n-3}^2}.\left(n\ge 5\right)$$

(6)

Thus, we have (for $n\ge 5$):

$$z_n={\phi }_2(1+{\beta }_n)-\frac{{\phi }_0}{{\alpha }_{n-1}^2{\alpha }_{n-3}^2}{\beta }_n(1+{\beta }_{n-1}).$$

Since ${\alpha }_n^2\to L^2,$ we have:

$$\underset{n\to \infty }{lim}{z}_n=(1+{\beta }^{*})\left({\phi }_2-\frac{{\phi }_0}{L^4}{\beta }^{*}\right).$$

Thus, we have our conclusion. □

Let:

$$f(z,w):=\frac{a^{2}b+a^2(b-1)z+a(b-a)zw}{a+a(b-1)z+(ab+1-2a)zw}.$$

By Lemma 7 and the fact in [2] (p. 399), if $z_n$ is increasing, then we know that ${\left\{x_n\right\}}_{n\ge 5}$ in (5) is decreasing and ${inf}_{n\ge 5}{x}_n=f(K,K).$ Thus, we have the following result.

Theorem 2.

Assume that $W_{{\widehat{\alpha }}_{\left[5\right]}}$ with rank three is subnormal. Let $\alpha \left(x\right):\sqrt{x},{\widehat{\alpha }}_{\left[5\right]},$ and let:

$$h_2^{+}:=sup\{x:W_{\alpha \left(x\right)}\mathit{be}\mathit{positively}\mathit{quadratically}\mathit{hyponormal}\}.$$

If $z_n:=\frac{v_n}{u_n}\left(n\ge 5\right)$ is increasing, then:

$$h_2^{+}\ge min\left\{1,x_3,x_4,\frac{a^{2}b+a^2(b-1)K+a(b-a)K^2}{a+a(b-1)K+(ab+1-2a)K^2}\right\},$$

where:

$$\begin{array}{ccc}\hfill x_3& =& \frac{a{\theta }_2+a\left(b-a\right)v_3+a\left(ab-1\right)u_3}{{\theta }_2+a\left(b-1\right)u_3+\left(-2a+ab+1\right)v_3},\hfill \\ \hfill x_4& =& \frac{a\left(ab-1\right){\theta }_3+a\left(u_4+v_4\right){\theta }_2+\left(a^2\left(b-1\right)u_4+a\left(b-a\right)v_4\right)v_3+a^{2}b{u}_3{u}_4}{a\left(b-1\right){\theta }_3+v_4{\theta }_2+\left(a\left(b-1\right)u_4+\left(ab-2a+1\right)v_4\right)v_3+a{u}_3{u}_4},\hfill \\ \hfill K& =& (1+{\beta }^{*})\left({\phi }_2-\frac{{\phi }_0}{L^4}{\beta }^{*}\right).\hfill \end{array}$$

Remark 1.

By (6), we know that if ${\beta }_n$ is increasing, then so is $z_n.$ Hence, our problems are as follows.

Problem 1.

Let $\alpha :={\left\{{\alpha }_n\right\}}_{n=0}^{\infty }$ be any unilateral weighted sequence. If $W_{\alpha }$ is subnormal, is ${\beta }_n$ increasing or not? In particular, what is the answer for subnormal $W_{{\widehat{\alpha }}_{\left[5\right]}}$ with rank three?

Example 1.

Let $\alpha \left(x\right):\sqrt{x},{\left(1,\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\right)}^{\wedge }.$ Then, ${\phi }_0=6,{\phi }_1=-18,{\phi }_2=9,L^2\approx 6.2899,$ and $K\approx 32.118,$ $x_3=\frac{17}{18}\approx 0.94444,x_4=\frac{226}{249}\approx 0.90763,f(K,K)\approx 0.77512.$ We obtain $h_2^{+}⪆0.77512.$ That is, if $0<x⪅0.77512,$ then $W_{\alpha \left(x\right)}$ is positively quadratically hyponormal. Numerically, we know that ${\beta }_n$ and $z_n$ are all increasing. See the following

Table 1. Numerical data for ${\beta }_n$ and $z_n$ in Example 1.

n	7	8	9	10	11	12
${\beta }_n$	$1.9851$	$2.3929$	$2.6008$	$2.6882$	$2.7218$	$2.7343$
$z_n$	$25.597$	$29.120$	$30.909$	$31.660$	$31.949$	$32.056$
n	13	14	15	16	17	18
${\beta }_n$	$2.7389$	$2.7406$	$2.7412$	$2.7414$	$2.7415$	$2.7415$
$z_n$	$32.096$	$32.11$	$32.116$	$32.117$	$32.118$	$32.118$

.

4.2. The Positive Quadratic Hyponormality of $W_{\alpha \left(y,x\right)}$

Let $\alpha \left(y,x\right):\sqrt{y},\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}.$ We also consider the matrix as in (4), and let $d_n=det{D}_n={\sum }_{i=0}^{n+1}c\left(n,i\right)t^i.$ Then:

$$\begin{array}{ccc}\hfill c\left(1,1\right)& =& xy\left(1-y\right)>0,\hfill \\ \hfill c\left(2,1\right)& =& y\left(x-y\right)\left(a-x\right)>0,\hfill \\ \hfill c\left(2,2\right)& =& xy\left(a-ay+xy-x^2\right)\ge xy\left(1-y\right)\left(a-x\right)>0,\hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill c\left(3,1\right)& \ge & u_{3}c(2,1)>0,\hfill \\ \hfill c\left(3,2\right)& =& y\left(\left(x-a\right)\left(x^2-2x+ab\right)y+x\left(\left(1-a\right)x^2+a\left(1-b\right)x+a\left(ab-1\right)\right)\right),\hfill \\ \hfill c\left(3,3\right)& \ge & u_{3}c(2,3)+v_3\rho >0.\hfill \end{array}$$

Since:

$$\begin{array}{ccc}\hfill \rho & =& xy\left(1-y\right)\left(a-x\right)>0,\hfill \\ \hfill \tau & =& y\left(\left(2x-x^2-a\right)y+x\left(a-1\right)\right),\hfill \end{array}$$

we can similarly show that $c(n,n-1)\ge 0$ for all $n\ge 3$, if $c(3,2)\ge 0,c(4,3)\ge 0$ and $y\le \frac{x\left(\left(a-1\right)K^2+\left(a-x\right)K+ax\right)}{\left(a-2x+x^2\right)K^2+\left(a-x\right)xK+x^3},$ where $K=(1+{\beta }^{*})\left({\phi }_2-\frac{{\phi }_0}{L^4}{\beta }^{*}\right)$. Thus, we have the following result.

Theorem 3.

Let $\alpha \left(y,x\right):\sqrt{y},\sqrt{x},{\widehat{\alpha }}_{\left[5\right]}.$ If $z_n:=\frac{v_n}{u_n}\left(n\ge 5\right)$ is increasing, and:

(1) $0<x\le min\left\{a-\sqrt{a\left(a-1\right)},\frac{ab\left(c-1\right)-\sqrt{ab\left(a-1\right)\left(a^{2}b+b{c}^2-ab-ac+2a^2-a^3-abc\right)}}{a\left(a-1\right)+b\left(c-a\right)}\right\},$

(2) $0<y\le min\left\{x,f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right\},$ where:

$$\begin{array}{ccc}\hfill f_1\left(x\right)& =& \frac{x\left(a\left(ab-1\right)+a\left(1-b\right)x-\left(a-1\right)x^2\right)}{\left(a-x\right)\left(ab-2x+x^2\right)},\hfill \\ \hfill f_2\left(x\right)& =& -\frac{x{p}_1\left(x\right)}{p_2\left(x\right)},\mathit{with}:\hfill \\ & & \begin{array}{cc}\hfill p_1\left(x\right)& =\left(a^{2}b+bc-a^2-abc\right)x^2+\left(a^{2}b-2ab+a^2-a{b}^{2}c+abc\right)x\hfill \\ & +\left(c{a}^2{b}^2-2a^{3}b+2a^{2}b-cab\right),\hfill \end{array}\hfill \\ & & \begin{array}{cc}\hfill p_2\left(x\right)& =\left(bc-ab-a+a^2\right)x^3+\left(a-a^{2}b+3ab-2bc-abc\right)x^2\\ & +\left(a^{3}b-3a^{2}b-a^2+ca{b}^2+2cab\right)x+a^{2}b\left(a-bc\right),\hfill \end{array}\hfill \\ \hfill f_3\left(x\right)& =& \frac{x\left(\left(a-1\right)K^2+\left(a-x\right)K+ax\right)}{\left(a-2x+x^2\right)K^2+\left(a-x\right)xK+x^3},K=(1+{\beta }^{*})\left({\phi }_2-\frac{{\phi }_0}{L^4}{\beta }^{*}\right),\hfill \end{array}$$

then $W_{\alpha \left(y,x\right)}$ is positively quadratically hyponormal.

Example 2.

Let $a=2,b=3,c=4,d=5.$ If $0<x\le 2-\sqrt{2}\approx 0.58579,$ $y\le \frac{x\left(K^2+\left(2-x\right)K+2x\right)}{\left(x^2-2x+2\right)K^2+\left(2x-x^2\right)K+x^3}$ with $K\approx 32.118,$ then $W_{\alpha \left(y,x\right)}$ is positively quadratically hyponormal. See the following Figure 1.

5. More Results

From the above discussions, we obtain the following criteria for any unilateral weighted shifts.

Proposition 5.

Let $\alpha \left(x\right):\sqrt{x},1,\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d},\sqrt{{\alpha }_5},\sqrt{{\alpha }_6},\dots ,$ and $\alpha :1,\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d},\sqrt{{\alpha }_5},\sqrt{{\alpha }_6},\dots$ be subnormal weighted shifts. Let:

$$h_2^{+}:=sup\{x:W_{\alpha \left(x\right)}\mathit{be}\mathit{positively}\mathit{quadratically}\mathit{hyponormal}\}.$$

If $z_n=\frac{v_n}{u_n}\left(n\ge 5\right)$ is increasing and $z_n\to K$ (as $n\to \infty$), then:

$$h_2^{+}\ge min\left\{1,x_3,x_4,\frac{a^{2}b+a^2(b-1)K+a(b-a)K^2}{a+a(b-1)K+(ab+1-2a)K^2}\right\},$$

where:

$$\begin{array}{ccc}\hfill x_3& =& \frac{a{\theta }_2+a\left(b-a\right)v_3+a\left(ab-1\right)u_3}{{\theta }_2+a\left(b-1\right)u_3+\left(-2a+ab+1\right)v_3},\hfill \\ \hfill x_4& =& \frac{a\left(ab-1\right){\theta }_3+a\left(u_4+v_4\right){\theta }_2+\left(a^2\left(b-1\right)u_4+a\left(b-a\right)v_4\right)v_3+a^{2}b{u}_3{u}_4}{a\left(b-1\right){\theta }_3+v_4{\theta }_2+\left(a\left(b-1\right)u_4+\left(ab-2a+1\right)v_4\right)v_3+a{u}_3{u}_4}.\hfill \end{array}$$

By Proposition 5, we can have the following results, but we omit the concrete computations.

Example 3.

(1) Let $\alpha \left(x\right):\sqrt{x},\sqrt{\frac{2}{3}},\sqrt{\frac{3}{4}},\sqrt{\frac{4}{5}},\dots .$ Then, $h_2^{+}=\frac{2}{3}$ (cf. [11], Proposition 7).

(2) Let $\alpha \left(x\right):\sqrt{x},\sqrt{\frac{5}{8}},\sqrt{\frac{3}{4}},\sqrt{\frac{4}{5}},\dots .$ Then, $h_2^{+}=\frac{1945}{3136}$ (cf. [15], Theorem 3.7).

(3) Let $\alpha \left(x\right):\sqrt{x},1,{\left(\sqrt{2},\sqrt{2.1},\sqrt{12.1}\right)}^{\wedge }.$ Then, $h_2^{+}⪆0.16682.$

(4) Let $\alpha \left(x\right):\sqrt{x},\sqrt{\frac{n}{n+1}·\frac{1}{2}·\frac{2^{n+1}-1}{2^n-1}}.$ Then, $h_2^{+}=\frac{3}{4}$ (cf. [16], Example 3.4).

6. Conclusions

In this work, we study a weighed shift operator for which the weights are recursively generated by five weights. We give sufficient conditions of the positive quadratic hyponormalities. Next, it is worth studying the cubic hyponormality, semi-weak k-hyponormalities, and so on.