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

  1. (a)

    A is a proper contraction and

  2. (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

  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 \(A\in \mathcal{P}(*-n)\).

  2. (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. (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. (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.