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Contractions without non-trivial invariant subspaces satisfying a positivity condition

Journal of Inequalities and Applications20162016:116

https://doi.org/10.1186/s13660-016-1058-4

  • Received: 17 November 2015
  • Accepted: 30 March 2016
  • Published:

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.

Keywords

  • class \(\mathcal{A}(n)\) operator
  • class \(\mathcal{A}(*-n)\) operator
  • \(\mathcal{P}(n)\) operator
  • class \(\mathcal{P}(*-n)\) operator
  • contraction
  • proper contraction
  • strongly stable

MSC

  • 47B20
  • 47A10

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.

Declarations

Acknowledgements

The authors are grateful to the referees for their valuable comments and suggestions for improvement of the paper. The third author was supported by the Incheon National University Research Grant in 2014.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
8 Redwood Grove, Northfield Avenue, Ealing, London, W5 4SZ, United Kingdom
(2)
Department of Mathematics Education, Seoul National University of Education, Seoul, 137-742, Korea
(3)
Department of Mathematics, Incheon National University, Incheon, 406-772, Korea

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© Duggal et al. 2016

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