Contractions without non-trivial invariant subspaces satisfying a positivity condition

Abstract

An operator $A\in B(\mathcal{H})$, the algebra of bounded linear transformations on a complex infinite dimensional Hilbert space $\mathcal{H}$, belongs to class $\mathcal{A}(n)$ (resp., $\mathcal{A}(*-n)$) if $\vert A\vert^{2}\leq\vert A^{n+1}\vert^{\frac{2}{n+1}}$ (resp., $\vert A^{*}\vert^{2}\leq \vert A^{n+1}\vert^{\frac{2}{n+1}}$) for some integer $n\geq1$, and an operator $A\in B(\mathcal{H})$ is called n-paranormal, denoted $A\in \mathcal{P}(n)$ (resp., $*-n$-paranormal, denoted $A\in \mathcal{P}(*-n)$) if $\Vert Ax\Vert ^{n+1}\leq \Vert A^{n+1}x\Vert \Vert x\Vert ^{n}$ (resp., $\Vert A^{*}x\Vert ^{n+1}\leq \Vert A^{n+1}x\Vert \Vert x\Vert ^{n}$) for some integer $n\geq 1$ and all $x \in\mathcal{H}$. In this paper, we prove that if $A\in\{\mathcal{A}(n)\cup \mathcal{P}(n)\}$ (resp., $A\in\{\mathcal{A}(*-n)\cup \mathcal{P}(*-n)\}$) is a contraction without a non-trivial invariant subspace, then A, $\vert A^{n+1}\vert^{\frac{2}{n+1}}-\vert A\vert^{2}$ and $\vert A^{n+1}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ 1$ (resp., A, $\vert A^{n+1}\vert^{\frac{2}{n+1}}-\vert A^{*}\vert^{2}$ and $\vert A^{n+2}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ 1\geq0$) are proper contractions.

1 Introduction

Let $B(\mathcal{H})$ denote the algebra of bounded linear operators on an infinite dimensional complex Hilbert space $\mathcal{H}$ with inner product $\langle\cdot, \cdot\rangle$. An operator $A\in B(\mathcal{H})$ is a class $\mathcal{A}$ (resp., ∗-class $\mathcal{A}$) operator, [1] and [2], if $\vert A\vert^{2}\leq\vert A^{2}\vert$ (resp., $\vert A^{*}\vert^{2}\leq\vert A^{2}\vert$). As a generalization of class $\mathcal{A}$, Yuan and Gao [3] have introduced the class of $\mathcal{A}(n)$ operators as follows: An operator $A\in B(\mathcal{H})$ belongs to class $\mathcal{A}(n)$ (resp., $\mathcal{A}(*-n)$) if $\vert A\vert^{2}\leq\vert A^{n+1}\vert^{\frac{2}{n+1}}$ (resp., $\vert A^{*}\vert^{2}\leq \vert A^{n+1}\vert^{\frac{2}{n+1}}$) for some integer $n\geq1$. An operator $A\in B(\mathcal{H})$ is called n-paranormal, denoted $A\in \mathcal{P}(n)$ (resp., $*-n$-paranormal, denoted $A\in \mathcal{P}(*-n)$) if $\Vert Ax\Vert ^{n+1}\leq \Vert A^{n+1}x\Vert \Vert x\Vert ^{n}$ (resp., $\Vert A^{*}x\Vert ^{n+1}\leq \Vert A^{n+1}x\Vert \Vert x\Vert ^{n}$) for some integer $n\geq 1$ and all $x \in\mathcal{H}$ is a generalization of the class of paranormal (resp., ∗-paranormal) operators (see [4]).

Recall [5] that a contraction $A\in B(\mathcal{H})$ (i.e., an operator $A\in B(\mathcal{H})$ such that $\Vert A\Vert \leq1$; equivalently, such that $\Vert Ax\Vert \leq \Vert x\Vert$ for every $x \in\mathcal{H}$) is said to be a proper contraction if $\Vert Ax\Vert < \Vert x\Vert$ for every non-zero $x \in \mathcal{H}$. A strict contraction (i.e., a contraction A such that $\Vert A\Vert < 1$) is a proper contraction, but a proper contraction is not necessarily a strict contraction. Kubrusly and Levan [5] have proved that if a hyponormal ($\Vert Ax\Vert \geq \Vert A^{*}x\Vert$) contraction A has no non-trivial invariant subspace, then

• (a) A is a proper contraction and

• (b) its self-commutator $[A^{*},A]=A^{*}A-AA^{*}$ is a strict contraction.

Class $\mathcal{A}$ operators A satisfy the property that if A is a contraction with no non-trivial invariant subspace, then the non-negative operator $D_{0}=\vert A^{2}\vert-\vert A\vert^{2}$ is a proper contraction, and hence of the class $C_{00}$ of contractions [6]. Since $\langle \vert A^{2}\vert x,x\rangle\leq\langle\vert A^{2}\vert^{2}x,x\rangle ^{\frac{1}{2}} \Vert x\Vert$ (by the Hölder-McCarthy inequality: if $T\in B(\mathcal{H})$ is a non-negative (i.e., ≥0) operator, then $\langle T^{\lambda}x,x\rangle\leq\langle Tx,x \rangle^{\lambda} \Vert x\Vert ^{2(1-\lambda)}$ for all $0<\lambda\leq1$), if $A\in \mathcal{A}$, then $\Vert Ax\Vert ^{2}\leq \Vert A^{2}x\Vert \Vert x\Vert$ for all $x\in\mathcal{H}$. Thus class $\mathcal{A}$ operators are paranormal. Paranormal operators $A\in B(\mathcal{H})$ are characterized by the positivity condition $\vert A^{2}\vert^{2}-2{\lambda}\vert A\vert^{2}+{\lambda}^{2}\geq0$ for all real $\lambda >0$. Choosing $\lambda=1$, it follows that class $\mathcal{A}$ (also, paranormal) operators A satisfy the positivity property $D_{1}= \vert A^{2}\vert^{2}- 2\vert A\vert^{2}+ 1\geq0$. If we now choose A to be a contraction without a non-trivial invariant subspace, then $D_{1}$ (along with A) is a proper contraction [5].

Positivity properties of the type satisfied by class $\mathcal{A}$ operators are satisfied by other classes of Hilbert space operators, some of them generalizations of the class $\mathcal{A}$ and others distinct from class $\mathcal{A}$.

It is easily seen (we prove so in Section 2) that class $\mathcal{A}(n)$ and class $\mathcal{P}(n)$ satisfy the positivity property that

$$\bigl\vert A^{n+1}\bigr\vert ^{2}- {\frac{n+1}{n}} \vert A\vert^{2}+ 1\geq0 $$

and class $\mathcal{A}(*-n)$ and class $\mathcal{P}(*-n)$ satisfy the positivity properties

$$\bigl\vert A^{n+1}\bigr\vert ^{2}- {\frac{n+1}{n}} \bigl\vert A^{*}\bigr\vert ^{2}+ 1\geq0 $$

and

$$\bigl\vert A^{n+2}\bigr\vert ^{2}- {\frac{n+1}{n}} \vert A\vert^{2}+ 1\geq0. $$

We prove in the following that there is a method as regards the ‘proper contraction property satisfied by the operators $D_{0}$ and $D_{1}$ above’. We prove that if an $A\in\{\mathcal{A}(n)\cup \mathcal{P}(n)\}$ (resp., $A\in\{\mathcal{A}(*-n)\cup \mathcal{P}(*-n)\}$) is a contraction without a non-trivial invariant subspace, then A, $\vert A^{n+1}\vert^{\frac{2}{n+1}}-\vert A\vert^{2}$ and $\vert A^{n+1}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ 1$ (resp., A, $\vert A^{n+1}\vert^{\frac{2}{n+1}}-\vert A^{*}\vert^{2}$ and $\vert A^{n+1}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ 1\geq0$) are proper contractions.

2 Results

We begin with the following lemma. Let $A\in B(\mathcal{H})$.

Lemma 2.1

• (i) $A\in \mathcal{P}(n)\cup \mathcal{P}(*-n)$ if and only if

  $$\bigl\vert A^{n+1}\bigr\vert ^{2}-(n+1) \lambda^{n} \vert B\vert^{2}+ n \lambda^{n+1} \geq0,\quad\textit{all } \lambda> 0, $$

  where $B=A$ if $A\in \mathcal{P}(n)$ and $B=A^{*}$ if $A\in \mathcal{P}(*-n)$.

• (ii) If $A\in \mathcal{P}(*-n)$, then $A\in \mathcal{P}(n+1)$.

Proof

(i) If we let $\alpha=\Vert A^{n+1}x\Vert ^{2}$ and $\beta=\beta_{1}= \cdots=\beta_{n}= \lambda^{n+1}\Vert x\Vert ^{2}$ for real $\lambda > 0$, then the generalized arithmetic-geometric inequality $\alpha\beta_{1}\beta_{2} \cdots\beta_{n}\leq (\frac{\alpha+\beta_{1}+\beta_{2}+ \cdots+\beta_{n}}{n+1})^{n+1}$ [7], p.17, says that

$${\lambda}^{n(n+1)}\bigl\Vert A^{n+1}x\bigr\Vert ^{2} \Vert x\Vert ^{2n}\leq\biggl(\frac{\Vert A^{n+1}x\Vert ^{2}+ n{\lambda}^{n+1}\Vert x\Vert ^{2}}{n+1} \biggr)^{n+1}. $$

By definition $A\in\{\mathcal{P}(n)\cup \mathcal{P}(*-n)\}$ if and only if

$$\Vert Bx\Vert ^{n+1}\leq\bigl\Vert A^{n+1}x\bigr\Vert \Vert x\Vert ^{n} $$

(where $B=A$ if $A\in \mathcal{P}(n)$ and $B=A^{*}$ if $A\in \mathcal{P}(*-n)$). Thus

$$(n+1){\lambda}^{n}\Vert Bx\Vert ^{2}\leq(n+1){ \lambda}^{n}\bigl\Vert A^{n+1}x\bigr\Vert ^{\frac{2}{n+1}} \Vert x\Vert ^{\frac {2n}{n+1}}\leq\bigl\Vert A^{n+1}x\bigr\Vert ^{2}+ n{\lambda}^{n+1}\Vert x\Vert ^{2} $$

for all $\lambda> 0$ and all $x\in\mathcal{H}$. Equivalently, if $A\in \mathcal{P}(n)\cup \mathcal{P}(*-n)$, then $\vert A^{n+1}\vert^{2}-(n+1)\lambda^{n} \vert B\vert^{2}+ n \lambda ^{n+1}\geq0$ for all $\lambda> 0$.

To see the sufficiency, let $\lambda\rightarrow0$ in

$$\Vert Bx\Vert ^{2}\leq{\frac{1}{(n+1){\lambda}^{n}}} \bigl\Vert A^{n+1}x\bigr\Vert ^{2}+ {\frac{n}{n+1}} {\lambda} \Vert x\Vert ^{2},\quad x\in\mathcal{H}, $$

if $\Vert A^{n+1}x\Vert =0$ (when it is seen that $\Vert Bx\Vert =0$) and let $\lambda= (\frac{\Vert A^{n+1}x\Vert }{\Vert x\Vert } )^{\frac{2}{n+1}}$ otherwise (when it follows that $\Vert Bx\Vert \leq (\Vert A^{n+1}x\Vert \Vert x\Vert ^{n})^{\frac{1}{n+1}}$, $x\in\mathcal{H}$).

(ii) If $A\in \mathcal{P}(*-n)$, then, for all $x\in\mathcal{H}$,

$$\begin{aligned} &\Vert Ax\Vert ^{2(n+1)}= \bigl\langle A^{*}Ax,x \bigr\rangle ^{n+1}\leq\bigl\Vert A^{*}Ax\bigr\Vert ^{n+1}\Vert x \Vert ^{n+1}\leq\bigl\Vert A^{n+2}x\bigr\Vert \Vert Ax \Vert ^{n}\Vert x\Vert ^{n+1} \\ &\quad \Longrightarrow\quad \Vert Ax\Vert ^{n+2}\leq\bigl\Vert A^{n+2}x\bigr\Vert \Vert x\Vert ^{n+1}, \end{aligned}$$

i.e., $A\in \mathcal{P}(n+1)$. □

It is immediate from Lemma 2.1 that the operators $A\in \mathcal{P}(n)$ (resp., $A\in \mathcal{P}(*-n)$) satisfy the positivity property, henceforth denoted property $Q_{\lambda}(n)$ (resp., property $Q_{\lambda}(*-n)$), that

$$\bigl\vert A^{n+1}\bigr\vert ^{2}- (n+1){ \lambda}^{n} \vert B\vert^{2}+ n{\lambda}^{n+1} \geq0 $$

for all $\lambda> 0$. (Here, as above, $B=A$ if $A\in \mathcal{P}(n)$ and $B=A^{*}$ if $A\in \mathcal{P}(*-n)$.) We prove that the operators $A\in \mathcal{A}(n)$ (resp., $A\in \mathcal{A}(*-n)$) also satisfy property $Q_{\lambda}(n)$ (resp., $Q_{\lambda}(*-n)$). The following lemma, the Hölder-McCarthy inequality, is well known.

Lemma 2.2

If $A\in B(\mathcal{H})$, then the following properties hold:

• (1) $\langle A^{\lambda}x,x\rangle\geq\langle Ax,x\rangle^{\lambda} \Vert x\Vert ^{2(1-\lambda)}$ for any $\lambda>1$ and any vector x.

• (2) $\langle A^{\lambda}x,x\rangle\leq\langle Ax,x\rangle^{\lambda} \Vert x\Vert ^{2(1-\lambda)}$ for any $\lambda\in (0,1]$ and any vector x.

Lemma 2.3

The operators $A\in \mathcal{A}(n)$ (resp., $A\in \mathcal{A}(*-n)$) satisfy property $Q_{\lambda}(n)$ (resp., property $Q_{\lambda}(*-n)$).

Proof

The proof is a simple consequence of an application of Lemma 2.2: If $A\in \mathcal{A}(n)\cup \mathcal{A}(*-n)$ and the operator B is defined as above, then, for all $x\in\mathcal{H}$,

$$\begin{aligned} & \bigl\langle \vert B\vert^{2}x,x\bigr\rangle \leq\bigl\langle \bigl\vert A^{n+1}\bigr\vert ^{\frac{2}{n+1}}x,x\bigr\rangle \leq \bigl\langle \bigl\vert A^{n+1}\bigr\vert ^{2}x,x\bigr\rangle ^{\frac{1}{n+1}} \Vert x\Vert ^{\frac{2n}{n+1}}, \end{aligned}$$

i.e., $A\in \mathcal{A}(n)$ implies $A\in \mathcal{P}(n)$ and $A\in \mathcal{A}(*-n)$ implies $A\in \mathcal{P}(*-n)$. Consequently the operators $A\in \mathcal{A}(n)$ satisfy property $Q_{\lambda}(n)$ and the operators $A\in \mathcal{A}(*-n)$ satisfy property $Q_{\lambda}(*-n)$. □

It is clear from Lemma 2.2 that the operators $A\in \mathcal{P}(*-n)$ satisfy property $Q_{\lambda}(n+1)$ (i.e., if $A\in \mathcal{P}(*-n)$, then $\vert A^{n+2}\vert^{2}-(n+2)\lambda^{n+1} \vert B\vert^{2}+ (n+1) \lambda^{n+2}\geq0$ for all $\lambda>0$).

Given an operator $A\in B(\mathcal{H})$, let B denote either A or $A^{*}$ (exclusive ‘or’), and let

$$M= \bigl\{ x\in\mathcal{H}: \Vert Bx\Vert =\Vert B\Vert \Vert x\Vert = \Vert A\Vert \Vert x\Vert \bigr\} . $$

Lemma 2.4

If $A\in\{\mathcal{P}(n)\cup \mathcal{P}(*-n)\}$, then M is a closed subspace of $\mathcal{H}$ such that $A(M)\subseteq M$.

Proof

M being the null space of the operator $\vert B\vert^{2}-\Vert A\Vert ^{2}$ is a closed subspace of $\mathcal{H}$. Define the operator B as before by letting $B=A$ whenever $A\in \mathcal{P}(n)$ and $B=A^{*}$ whenever $A\in \mathcal{P}(*-n)$. Let $x\in M$, and let $A\in\{\mathcal{P}(n)\cup \mathcal{P}(*-n)\}$. Then

$$\begin{aligned} \Vert Bx\Vert ^{2} \leq& \bigl\langle \bigl\vert A^{n+1}\bigr\vert ^{2}x,x \bigr\rangle ^{\frac{1}{n+1}} \Vert x\Vert ^{\frac{2n}{n+1}} = \bigl\Vert A^{n+1}x\bigr\Vert ^{\frac{2}{n+1}}\Vert x\Vert ^{\frac {2n}{n+1}} \\ \leq& \Vert A\Vert ^{2}\Vert x\Vert ^{2}= \Vert Bx\Vert ^{2}, \end{aligned}$$

and hence

$$\begin{aligned} &\Vert Bx\Vert ^{2}= \bigl\Vert A^{n+1}x\bigr\Vert ^{\frac{2}{n+1}}\Vert x\Vert ^{\frac{2n}{n+1}}= \Vert A\Vert ^{2} \Vert x\Vert ^{2} \\ &\quad \Longleftrightarrow\quad \Vert Bx\Vert ^{n+1}= \bigl\Vert A^{n+1}x\bigr\Vert \Vert x\Vert ^{n}= \Vert A\Vert ^{n+1}\Vert x\Vert ^{n+1}. \end{aligned}$$

But then (for all $x\in M$ and $A\in \mathcal{P}(n)\cup \mathcal{P}(*-n)$)

$$\begin{aligned} \Vert Bx\Vert ^{n+1} = & \bigl\Vert A^{n+1}x\bigr\Vert \Vert x\Vert ^{n} \\ \leq& \Vert A\Vert \bigl\Vert A^{n}x\bigr\Vert \Vert x \Vert ^{n}= \bigl\Vert A^{n}x\bigr\Vert \Vert x\Vert ^{n-1}\Vert Bx\Vert \\ \cdots& \\ \leq& \Vert Ax\Vert \Vert Bx\Vert ^{n}\leq \Vert Bx\Vert ^{n+1}, \end{aligned}$$

which implies

$$\Vert Bx\Vert ^{m}= \bigl\Vert A^{m}x\bigr\Vert \Vert x\Vert ^{m-1}, \quad \mbox{all integers } 1\leq m\leq n+1. $$

In particular,

$$\Vert Bx\Vert = \Vert Ax\Vert \quad \mbox{and}\quad \bigl\Vert A^{2}x \bigr\Vert \Vert x\Vert =\Vert Bx\Vert ^{2}. $$

Now if $A\in \mathcal{P}(n)$ (so that $B=A$), then

$$\begin{aligned} &\bigl\Vert A^{2}x\bigr\Vert \Vert x\Vert = \Vert Ax\Vert ^{2}= \Vert A\Vert ^{2}\Vert x\Vert ^{2}= \Vert A\Vert \Vert Ax\Vert \Vert x\Vert \\ &\quad \Longrightarrow\quad\bigl\Vert A^{2}x\bigr\Vert = \Vert A \Vert \Vert Ax\Vert \quad\Longrightarrow\quad A(M)\subseteq M, \end{aligned}$$

and if $A\in \mathcal{P}(*-n)$ (so that $B=A^{*}$), then (using Lemma 2.1(ii))

$$\begin{aligned} &\Vert Ax\Vert =\Vert Bx\Vert =\bigl\Vert A^{*}x\bigr\Vert =\Vert A \Vert \Vert x\Vert \\ &\quad\Longrightarrow\quad\bigl\{ x\in \mathcal{H}: \Vert Ax\Vert =\Vert A\Vert \Vert x\Vert \bigr\} = \bigl\{ x\in\mathcal{H}: \bigl\Vert A^{*}x\bigr\Vert = \Vert A\Vert \Vert x\Vert \bigr\} , \end{aligned} $$

and hence $A(M)\subseteq M$. □

Corollary 2.5

If a contraction $A\in\{\mathcal{P}(n)\cup \mathcal{P}(*-n)\}$ has no non-trivial invariant subspace, then A is a proper contraction.

Proof

If $A\in\{\mathcal{P}(n)\cup \mathcal{P}(*-n)\}$, then $AM\subseteq M$. Now if A is not a proper contraction, then (it is not a strict contraction, hence) $\Vert A\Vert =\Vert A^{*}\Vert =1$. Hence, since A has no non-trivial invariant subspace, $M= \{0\}$ (for the reason that if $M=\mathcal{H}$, then either A or $A^{*}$ is an isometry and isometries have non-trivial invariant subspaces). Consequently, $\Vert Ax\Vert \leq \Vert A\Vert \Vert x\Vert < \Vert x\Vert$ for all $x\in\mathcal{H}$, i.e., A is a proper contraction. □

We say in the following that an operator $A\in B(\mathcal{H})$ satisfies the positivity condition:

($D_{1}$):

if the operator $D_{1}= \vert A^{n+1}\vert^{\frac{2}{n+1}}- \vert A\vert^{2}\geq0$,

($D_{2}$):

if the operator $D_{2}= \vert A^{n+1}\vert^{\frac{2}{n+1}}- \vert A^{*}\vert^{2}\geq 0$, and

($D_{3}$):

if the operator $D_{3}= \vert A^{n+1}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ 1\geq0$.

It is evident from the definition of operators $A\in \mathcal{A}(n)$ (resp., $A\in \mathcal{A}(*-n)$) that $\mathcal{A}(n)$ operators satisfy condition ($D_{1}$) (resp., $\mathcal{A}(*-n)$ operators satisfy condition ($D_{2}$)). If we choose $0< \lambda= {\frac{1}{\sqrt[n]{n}}}$ in $Q_{\lambda}(n)$, then (since $\vert A^{n+1}\vert^{2}-{\frac{n+1}{n}}\vert A\vert^{2}+ {\frac{1}{\sqrt[n]{n}}}\leq\vert A^{n+1}\vert^{2}- {\frac {n+1}{n}}\vert A\vert^{2}+ 1$ for all integers $n\geq1$) operators $A\in \mathcal{P}(n)$ are seen to satisfy positivity condition ($D_{3}$). Again, if we choose $0< \lambda=\sqrt[n+1]{\frac{n+1}{n(n+2)}}$, then the fact that $\mathcal{P}(*-n)$ operators satisfy property $Q_{\lambda}(n+1)$ implies that $\vert A^{n+2}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ 1\geq \vert A^{n+2}\vert^{2}- {\frac{n+1}{n}}\vert A\vert^{2}+ (n+1)({\frac{n+1}{n(n+2)}})^{\frac{n+2}{n+1}}\geq0$; in particular, $\mathcal{P}(*-n)$ contractions A satisfy positivity condition ($D_{3}$).

Remark 2.6

An interesting class of operators, which contains many a familiar class of operators (such as p-hyponormal operators, $0< p\leq1$, w-hyponormal operators and class $\mathcal{A}$ operators) considered by a large number of authors in the recent past, is that of the class $\mathcal{A}(s,t)$ operators $A\in B(\mathcal {H})$ defined by the positivity condition $\vert A^{*}\vert^{2t}\leq (\vert A^{*}\vert^{t}\vert A\vert^{2s}\vert A^{*}\vert^{t})^{\frac {t}{s+t}}$, $0< s,t$ [8]. Class $\mathcal{A}(s,t)$ operators satisfy the property that $A\in \mathcal{A}(s,t)$ implies $A\in \mathcal{A}(\alpha,\beta)$ for every $\alpha\geq s$ and $\beta\geq t$ ([8], Theorem 4). Hence, if $0< s,t\leq1$, then every $A\in \mathcal{A}(s,t)$ is an $\mathcal{A}(1,1)= \mathcal{A}$ operator ([8], Theorem 3). Consequently the operators $A\in \mathcal{A}(s,t)$, $0< s,t\leq1$, satisfy positivity conditions ($D_{1}$) and ($D_{3}$) (with $n=1$).

Lemma 2.7

If an operator $A\in B(\mathcal{H})$ is a contraction such that $D_{i}\geq0$, $1\leq i\leq3$, for an $i= i_{0}$, then $D_{i_{0}}$ is a contraction.

Proof

Let $D_{i}=R_{i}^{2}$, let $x\in\mathcal{H}$ and let $R_{i}^{m}x=y_{i}$. Then

$$\begin{aligned}& \begin{aligned} \bigl\langle D_{1}^{m+1}x,x \bigr\rangle &= \bigl\langle \bigl\vert A^{n+1}\bigr\vert ^{\frac{2}{n+1}}y_{1},y_{1} \bigr\rangle - \Vert Ay_{1}\Vert ^{2} \\ & \leq\bigl\Vert A^{n+1}y_{1}\bigr\Vert ^{\frac{2}{n+1}} \Vert y_{1}\Vert ^{\frac {2n}{n+1}}- \Vert Ay_{1}\Vert ^{2} \\ & \leq \Vert y_{1}\Vert ^{2}= \bigl\langle D_{1}^{m}x,x \bigr\rangle \quad(\mbox{case }i_{0}=1), \end{aligned} \\& \begin{aligned} \bigl\langle D_{2}^{m+1}x,x \bigr\rangle &= \bigl\langle \bigl\vert A^{n+1}\bigr\vert ^{\frac {2}{n+1}}y_{2},y_{2} \bigr\rangle - \bigl\Vert A^{*}y_{2}\bigr\Vert ^{2} \\ & \leq\bigl\Vert A^{n+1}y_{2}\bigr\Vert ^{\frac{2}{n+1}} \Vert y_{2}\Vert ^{\frac{2n}{n+1}}- \bigl\Vert A^{*}y_{2} \bigr\Vert ^{2} \\ &\leq \Vert y_{2}\Vert ^{2}= \bigl\langle D_{2}^{m}x,x \bigr\rangle \quad(\mbox{case }i_{0}=2), \end{aligned} \\& \begin{aligned} \bigl\langle D_{3}^{m+1}x,x \bigr\rangle &= \bigl\langle \bigl\vert A^{n+1}\bigr\vert ^{2}y_{3},y_{3} \bigr\rangle - {\frac{n+1}{n}} \Vert Ay_{3}\Vert ^{2}+ \Vert y_{3}\Vert ^{2} \\ & = \bigl\Vert A^{n+1}y_{3}\bigr\Vert ^{2}- { \frac{n+1}{n}}\Vert Ay_{3}\Vert ^{2}+ \Vert y_{3}\Vert ^{2} \\ & \leq \Vert Ay_{3}\Vert ^{2}- {\frac{n+1}{n}} \Vert Ay_{3}\Vert ^{2}+\Vert y_{3}\Vert ^{2} \\ &= \Vert y_{3}\Vert ^{2}- {\frac{1}{n}}\Vert Ay_{3}\Vert ^{2} \\ & \leq \Vert y_{3}\Vert ^{2}= \bigl\langle D_{3}^{m}x,x \bigr\rangle \quad(\mbox{case }i_{0}=3). \end{aligned} \end{aligned}$$

Hence, in either of the cases $i_{0}=1,2$ and 3, $D_{i_{0}}$ is a contraction. □

We remark here that it is in general false that if $A\in B(\mathcal{H})$ is a $\mathcal{P}(n)$ (or $\mathcal{P}(*-n)$) contraction, then the positive operator $D=\vert A^{n+1}\vert^{2}-(n+1)\lambda^{n} \vert A\vert ^{2}+ n \lambda ^{n+1}\geq0$ (resp., $\vert A^{n+1}\vert^{2}-(n+1)\lambda^{n} \vert A^{*}\vert^{2}+ n \lambda^{n+1}\geq0$), all $\lambda> 0$, characterizing $\mathcal{P}(n)$ (resp., $\mathcal{P}(*-n)$) operators is a contraction. Consider for example the forward unilateral shift $U\in B(\mathcal{H})$. Trivially, $\alpha U \in \mathcal{P}(1)$ is a (proper) contraction for all positive $\alpha< 1$. The operator $D= \vert{\alpha}^{2}U^{2}\vert ^{2}-2\lambda \vert{\alpha}U\vert^{2}+ {\lambda}^{2}= {\alpha}^{4}- 2{\alpha }^{2}{\lambda}+ {\lambda}^{2}= (\alpha^{2}-\lambda)^{2}> 1$ for all $\lambda> 1+{\alpha}^{2}$. It is possible that, for contractions $A\in \mathcal{P}(*-n)$, the positive operator $D= \vert A^{n+1}\vert^{2}- {\frac {n+1}{n}}\vert A^{*}\vert^{2}+ 1$ is a contraction. We have, however, not been able to prove this.

The conclusion that $D_{i_{0}}$ is a contraction in Lemma 2.7 implies that the sequence $\{D_{i_{0}}^{p}\}_{1}^{\infty}$ being a monotonic decreasing bounded sequence of non-negative operators converges to a projection $P_{i_{0}}$.

Lemma 2.8

If $D_{i}$ ($i=1,2,3$) is the non-negative contraction of Lemma  2.7 with $\lim_{p\rightarrow\infty}{D_{i_{0}}^{p}}= P_{i_{0}}$ for an $i= i_{0}$, then $AP_{i_{0}}= 0$ if $i_{0}=1, 3$ and $A^{*}P_{i_{0}}=0$ if $i_{0}=2$.

Proof

Letting $D_{i_{0}}=R_{i_{0}}^{2}$ and $R_{i_{0}}^{m}x=y_{i_{0}}$ for $x\in\mathcal{H}$ and $1\leq i_{0}\leq3$, we have

$$\begin{aligned}& \begin{aligned} \Vert y_{1}\Vert ^{2}- \Vert R_{1}y_{1}\Vert ^{2}&= \Vert y_{1}\Vert ^{2}- \bigl\langle \bigl\vert A^{n+1} \bigr\vert ^{\frac{2}{n+1}}y_{1},y_{1}\bigr\rangle + \Vert Ay_{1}\Vert ^{2} \\ & \geq \Vert y_{1}\Vert ^{2}-\bigl\Vert A^{n+1}y_{1}\bigr\Vert ^{\frac {2}{n+1}}\Vert y_{1}\Vert ^{\frac{2n}{n+1}}+ \Vert Ay_{1}\Vert ^{2} \\ &\geq \Vert Ay_{1}\Vert ^{2}\quad(\mbox{case } i_{0}=1), \end{aligned} \\& \begin{aligned} \Vert y_{2}\Vert ^{2}- \Vert R_{2}y_{2}\Vert ^{2}&= \Vert y_{2}\Vert ^{2}- \bigl\langle \bigl\vert A^{n+1} \bigr\vert ^{\frac{2}{n+1}}y_{2},y_{2}\bigr\rangle + \bigl\Vert A^{*}y_{2}\bigr\Vert ^{2} \\ & \geq \Vert y_{2}\Vert ^{2}-\bigl\Vert A^{n+1}y_{2}\bigr\Vert ^{\frac {2}{n+1}}\Vert y_{2} \Vert ^{\frac{2n}{n+1}}+ \bigl\Vert A^{*}y_{2}\bigr\Vert ^{2} \\ &\geq\bigl\Vert A^{*}y_{1}\bigr\Vert ^{2} \quad( \mbox{case } i_{0}=2),\quad\mbox{and} \end{aligned} \\& \begin{aligned} \Vert y_{3}\Vert ^{2}- \Vert R_{3}y_{3}\Vert ^{2}&= \Vert y_{3}\Vert ^{2}- \bigl\langle \bigl\vert A^{n+1} \bigr\vert ^{2}y_{3},y_{3}\bigr\rangle + { \frac{n}{n+1}}\Vert Ay_{3}\Vert ^{2}- \Vert y_{3}\Vert ^{2} \\ &= -\bigl\Vert A^{n+1}y_{3}\bigr\Vert ^{2} \Vert y_{3}\Vert + {\frac {n+1}{n}}\Vert Ay_{3} \Vert ^{2} \\ & \geq{\frac{1}{n}}\Vert Ay_{3}\Vert ^{2}\quad( \mbox{case }i_{0}=3). \end{aligned} \end{aligned}$$

Let the operator B stand for A if $i_{0}=1$ or 3, and let $B=A^{*}$ if $i_{0}=2$. Letting $a=1$ if $i_{0}=1$ or 2, and $a={1/n}$ if $i_{0}=3$, we then have

$$a \sum_{m=0}^{p}{\bigl\Vert BR_{i_{0}}^{m}x\bigr\Vert ^{2}}\leq\sum _{m=0}^{p}{\bigl\Vert R_{i_{0}}^{m}x \bigr\Vert ^{2}}- \sum_{m=0}^{p}{ \bigl\Vert R_{i_{0}}^{m+1}x\bigr\Vert ^{2}}= \Vert x\Vert ^{2}-\bigl\Vert R_{i_{0}}^{p+1}x\bigr\Vert ^{2}\leq \Vert x\Vert ^{2} $$

for every $x\in\mathcal{H}$ and integer $p\geq0$. The positive integer n being fixed, it follows that $\Vert BR_{i_{0}}^{p}x\Vert \rightarrow0$ as $p\rightarrow\infty$; hence

$$0= \lim_{p\rightarrow\infty}{BR_{i_{0}}^{2p}x}= B\lim _{p\rightarrow\infty}{D_{i_{0}}^{p}x}=B P_{i_{0}} $$

for every $x\in\mathcal{H}$. Consequently, $AP_{i_{0}}=0$ if $i_{0}=1, 3$ and $A^{*}P_{i_{0}}=0$ if $i_{0}=2$. □

Recall that $T\in B(\mathcal{H})$ is a $C_{0\cdot}$-contraction (resp., $C_{1\cdot}$-contraction) if $\Vert T^{n} x\Vert$ converges to 0 for all $x\in\mathcal{H}$ (resp., does not converge to 0 for all non-trivial $x\in\mathcal{H}$); T is of class $C_{\cdot0}$, or $C_{\cdot1}$, if $T^{*}$ is of class $C_{0\cdot}$, respectively $C_{1\cdot}$. All combinations are allowed, leading to the classes $C_{00}$, $C_{01}$, $C_{10}$, and $C_{11}$ of contractions ([9], p.72). We say that a contraction $T\in B(\mathcal{H})$ is strongly stable if $T^{n}$ converges strongly to the 0 operator as $n\rightarrow\infty$.

The following theorem is our main result.

Theorem 2.9

If a contraction $0\neq A\in B(\mathcal{H})$ has no non-trivial invariant subspace, and if A satisfies the positivity condition ($D_{i}$) ($i=1,2,3$) for an $i= i_{0}$, then $D_{i_{0}}$ is a strongly stable (hence $C_{00}$) proper contraction.

Proof

Start by recalling that A has a non-trivial invariant subspace if and only if $A^{*}$ does. The hypotheses imply that the sequence $\{D_{i_{0}}^{p}\}$ of non-negative contractions converges to a projection $P_{i_{0}}$ such that $AP_{i_{0}}= 0$ whenever $i_{0}= 1$ or 3 and $A^{*}P_{i_{0}}=0$ whenever $i_{0}= 2$. Equivalently, $P_{i_{0}}A^{*}= 0$ whenever $i_{0}= 1$ or 3 and $P_{i_{0}}A=0$ whenever $i_{0}= 2$. Thus $P^{-1}_{i_{0}}(0)$ is a non-zero invariant subspace for $A^{*}$ in the case in which $i_{0}= 1$ or 3, and $P_{2}^{-1}(0)$ is a non-zero invariant subspace for A in the case in which $i_{0}= 2$. Hence we must have $P_{i}^{-1}(0)=\mathcal{H}$ for every choice of $i_{0}$ ($= 1,2,3$), and then the sequence $\{D_{i_{0}}^{p}\}$ converges strongly to the 0 operator. Since strong stability coincides with proper contractiveness for non-negative operators [10], $D_{i_{0}}$ is a proper contraction (of the class $C_{00}$ of contractions). □

Remark 2.10

A generalization of the class $\mathcal{A}(n)$ of operators in $B(\mathcal{H})$ is obtained by considering operators $A\in B(\mathcal{H})$ for which

$$D_{m(n)}={A^{*}}^{m}\bigl(\bigl\vert A^{n+1}\bigr\vert ^{\frac{2}{n+1}}-\vert A\vert^{2}\bigr)A^{m}\geq0 $$

for integers $m\geq1$. (Similar generalizations of the classes $\mathcal{A}(*-n)$, $\mathcal{P}(n)$, and $\mathcal{P}(*-n)$ are obtained by considering ${A^{*}}^{m}(\vert A^{n+1}\vert^{\frac{2}{n+1}}-\vert A^{*}\vert^{2})A^{m}\geq0$, ${A^{*}}^{m}(\vert A^{n+1}\vert^{2}- 2\lambda\vert A\vert^{2}+ {\lambda}^{2})A^{m}\geq0$, and ${A^{*}}^{m}(\vert A^{n+1}\vert^{2}- 2\lambda\vert A^{*}\vert^{2}+ {\lambda}^{2})A^{m}\geq0$, respectively.) Calling this class of operators A the class of m-quasi($\mathcal{A}(n)$) operators, denoted $A\in m\mbox{-}\mathrm {Q}(\mathcal{A}(n))$, it is seen that the operators $A\in m\mbox{-}\mathrm {Q}(\mathcal{A}(n))$ have an upper triangular matrix representation

$$A=\left ( \textstyle\begin{array}{@{}cc@{}} A_{1} & C\\ 0 & A_{2} \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} {\overline{\operatorname {ran}A^{m}}}\\ {A^{*m}}^{-1}(0) \end{array}\displaystyle \right ), $$

where $A_{1}\in \mathcal{A}(n)$ and $A_{2}^{*}$ is $(m+1)$-nilpotent. An argument similar to that used above (cf. [3] to see the minor changes in detail that are required) shows that, if A is a contraction, then the sequence $\{D_{m(n)}^{p}\}$ of positive operators converges strongly to a projection P such that $A^{m+1}P=0$. The operators $A\in m\mbox{-}\mathrm {Q}(\mathcal{A}(n))$ are not normaloid. If a contraction $A\in m\mbox{-}\mathrm {Q}(\mathcal{A}(n))$ has no non-trivial invariant subspaces, then A is a quasi-affinity (i.e., A is injective and has a dense range), and hence $A\in \mathcal{A}(n)$. Thus Theorem 2.9 has the following analog for contractions $A\in m\mbox{-}\mathrm {Q}(\mathcal{A}(n))$: If a contraction $A\in m\mbox{-}\mathrm {Q}(\mathcal{A}(n))$ has no non-trivial invariant subspace, then A is a proper contraction such that $D_{m(n)}\in C_{00}$ is a strongly stable contraction.