A New Seminorm for d-Tuples of A-Bounded Operators and Their Applications

Abstract

The aim of this paper was to introduce and investigate a new seminorm of operator tuples on a complex Hilbert space $\mathcal{H}$ when an additional semi-inner product structure defined by a positive (semi-definite) operator A on $\mathcal{H}$ is considered. We prove the equality between this new seminorm and the well-known A-joint seminorm in the case of A-doubly-commuting tuples of A-hyponormal operators. This study is an extension of a well-known result in [Results Math 75, 93(2020)] and allows us to show that the following equalities $r_A\left(\mathbf{T}\right)={\omega }_A\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_A$ hold for every A-doubly-commuting d-tuple of A-hyponormal operators $\mathbf{T}=(T_1,\dots ,T_d)$. Here, $r_A\left(\mathbf{T}\right),{\parallel \mathbf{T}\parallel }_A$, and ${\omega }_A\left(\mathbf{T}\right)$ denote the A-joint spectral radius, the A-joint operator seminorm, and the A-joint numerical radius of $\mathbf{T}$, respectively.

1. Introduction

In functional analyses, many authors have studied the tuples of operators. For example, we refer to [1,2,3,4,5] and the references therein.

Consider a complex Hilbert space $(\mathcal{H},\langle ·,·\rangle )$, where the norm induced by $\langle ·,·\rangle$ is denoted by $\parallel ·\parallel$. The set $\mathbb{B}\left(\mathcal{H}\right)$ denotes the $C^{*}$-algebra of all bounded linear operators acting on $\mathcal{H}$ with identity $I_{\mathcal{H}}$ (or shortly I). If $\mathcal{H}$ is n-dimensional, we identify $\mathbb{B}\left(\mathcal{H}\right)$ with the space ${\mathcal{M}}_n$ of all $n\times n$ matrices with entries in the complex field and denote its identity by $I_n$. In what follows, by an operator, we mean a bounded linear operator. We will mention some specific notions of an operator, i.e., the null space of every operator T is denoted by $\mathcal{N}\left(T\right)$, its range by $\mathcal{R}\left(T\right)$, and $T^{*}$ is the adjoint of T. An operator $T\in \mathbb{B}\left(\mathcal{H}\right)$ is said to be positive if $\langle Tx,x\rangle \ge 0$ for all $x\in \mathcal{H}$. We write $T\ge 0$ if T is positive. If $T\ge 0$, then $T^{1/2}$ means the square root of T. The commutator of two operators $T,S\in \mathbb{B}\left(\mathcal{H}\right)$ is defined as $[T,S]:=TS-ST$. It is easy to see that $[T-\lambda I,S-\mu I]=[T,S]$, for every $\lambda ,\mu \in \mathbb{C}$ and $T,S\in \mathbb{B}\left(\mathcal{H}\right)$. Recall that $T\in \mathbb{B}\left(\mathcal{H}\right)$ is called normal (respectively hyponormal) if $[T^{*},T]=0$ (respectively, $[T^{*},T]\ge 0$).

Next, we present some inequalities related to operators that we need in the future. First, we give the classical Schwarz inequality for a positive operator $T\in \mathbb{B}\left(\mathcal{H}\right)$:

$${\left|\langle Tx,y\rangle \right|}^2\le \langle Tx,x\rangle \langle Ty,y\rangle ,$$

(1)

where $x,y\in \mathcal{H}$.

In [6], Halmos obtains a result similar to the inequality above

$$\begin{array}{c}\hfill \left|〈Tx,x〉\right|\le {〈\left|T\right|x,x〉}^{1/2}{〈\left|T^{*}\right|x,x〉}^{1/2}\end{array}$$

for every $T\in \mathbb{B}\left(\mathcal{H}\right)$ and for any $x,y\in \mathcal{H}$. In [7], Kato proves a Schwarz-type inequality (1), which generalizes the inequality of Halmos:

$${\left|\langle Tx,y\rangle \right|}^2\le {\langle \left|T\right|}^{2\theta }x,x\rangle \langle |T^{*}{|}^{2(1-\theta )}y,y\rangle$$

(2)

for all operators $T\in \mathbb{B}\left(\mathcal{H}\right)$, for every vector $x,y\in \mathcal{H}$, and $\theta \in [0,1]$. McCarthy [8] gives an important inequality in the theory of operators as follows:

Lemma 1 

(Theorem 1.4 in [8]). Let $T\in \mathbb{B}\left(\mathcal{H}\right)$ be a positive operator and $x\in \mathcal{H}$ satisfy $\parallel x\parallel =1$. Then, for $r\ge 1$,

$$\begin{array}{c}\hfill {〈Tx,x〉}^r\le 〈T^{r}x,x〉.\end{array}$$

For $0\le r\le 1$, the above inequality is reversed.

In what follows, we assume that A is a positive nonzero operator that defines the following positive semi-definite sesquilinear form:

$$\begin{array}{cc}\hfill {\langle ·,·\rangle }_A:\mathcal{H}\times \mathcal{H}& \to \mathbb{C}\hfill \\ \hfill (x,y)& \mapsto {\langle x,y\rangle }_A:=\langle Ax,y\rangle =\langle A^{1/2}x,A^{1/2}y\rangle .\hfill \end{array}$$

The seminorm induced by ${\langle ·,·\rangle }_A$ is given by ${\parallel x\parallel }_A=\sqrt{{\langle x,x\rangle }_A},\forall x\in \mathcal{H}$. It can be seen that ${\parallel ·\parallel }_A$ is a norm on $\mathcal{H}$ if and only if A is injective, and the semi-Hilbert space $(\mathcal{H},\parallel ·{\parallel }_A)$ is complete if and only if $\mathcal{R}\left(A\right)$ is closed in $\mathcal{H}$.

Definition 1 

([9]). An operator $S\in \mathbb{B}\left(\mathcal{H}\right)$ is called an A-adjoint of $T\in \mathbb{B}\left(\mathcal{H}\right)$, if we have ${\langle Tx,y\rangle }_A={\langle x,Sy\rangle }_A$ ($AS=T^{*}A$) for every $x,y\in \mathcal{H}$.

The existence of an A-adjoint operator is not guaranteed. Thus, we denote by ${\mathbb{B}}_A\left(\mathcal{H}\right)$ the set of all operators that admit A-adjoints. Using Douglas’ theorem [10], we obtain the following:

$${\mathbb{B}}_A\left(\mathcal{H}\right)=\left\{T\in \mathbb{B}\left(\mathcal{H}\right);\mathcal{R}\left(T^{*}A\right)\subseteq \mathcal{R}\left(A\right)\right\}$$

and

$${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)=\left\{T\in {\mathbb{B}\left(\mathcal{H}\right);\exists c>0;\parallel Tx\parallel }_A\le c{\parallel x\parallel }_A,\forall x\in \mathcal{H}\right\}.$$

When $T\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$, we say that T is A-bounded. The sets ${\mathbb{B}}_A\left(\mathcal{H}\right)$ and ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ are subalgebras of $\mathbb{B}\left(\mathcal{H}\right)$, which are neither closed nor dense in $\mathbb{B}\left(\mathcal{H}\right)$. Moreover, the inclusions

$${\mathbb{B}}_A\left(\mathcal{H}\right)\subseteq {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)\subseteq \mathbb{B}\left(\mathcal{H}\right)$$

hold. We have equality if A is injective and $\overline{\mathcal{R}\left(A\right)}=\mathcal{R}\left(A\right)$, where $\overline{\mathcal{R}\left(A\right)}$ means the closure of $\mathcal{R}\left(A\right)$ in the norm topology of $\mathcal{H}$ (see [11]). Further, ${\langle ·,·\rangle }_A$ gives the following seminorm on ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$

$${\displaystyle {\parallel T\parallel }_A:=\underset{\begin{array}{c}x\in \overline{\mathcal{R}\left(A\right)},\\ x\ne 0\end{array}}{sup}\frac{{\parallel Tx\parallel }_A}{{\parallel x\parallel }_A}=\underset{\begin{array}{c}x\in \mathcal{H},\\ {\parallel x\parallel }_A=1\end{array}}{sup}{\parallel Tx\parallel }_A=\underset{\begin{array}{c}x,y\in \mathcal{H},\\ {\parallel x\parallel }_A={\parallel y\parallel }_A=1\end{array}}{sup}\left|{\langle Tx,y\rangle }_A\right|<\infty }$$

(3)

(see [12] and the references therein). It is useful to note that if $T\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$, then $T\left(\mathcal{N}\right(A\left)\right)\subseteq \mathcal{N}\left(A\right)$. Further, ${\parallel TS\parallel }_A\le {\parallel T\parallel }_A{\parallel S\parallel }_A$ for any $T,S\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$. Let $X^{†}$ denote the Moore–Penrose pseudo-inverse of an operator X (for more details concerning this operator, see [11]). Following [11], we have: $T\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ implies that $A^{1/2}T{(A^{1/2})}^{†}\in \mathbb{B}\left(\mathcal{H}\right)$ and

$${\parallel T\parallel }_A=\parallel A^{1/2}T{(A^{1/2})}^{†}\parallel .$$

(4)

In 2012, Saddi [13] introduced the A-numerical radius of an operator $T\in \mathbb{B}\left(\mathcal{H}\right)$ by

$$\begin{array}{cc}\hfill {\omega }_A\left(T\right)& :=sup\left\{{|\langle Tx,x\rangle }_A{|;x\in \mathcal{H},\parallel x\parallel }_A=1\right\}.\hfill \end{array}$$

In 2020, the concept of the A-spectral radius of A-bounded operators was defined in [14] as follows:

$${\displaystyle r_A\left(T\right):=\underset{n\ge 1}{inf}\parallel T^n{\parallel }_A^{\frac{1}{n}}=\underset{n\to \infty }{lim}{\parallel T^n\parallel }_A^{\frac{1}{n}}.}$$

(5)

Note that ${\parallel T\parallel }_A$ and ${\omega }_A\left(T\right)$ may equal $+\infty$ for some $T\in \mathbb{B}\left(\mathcal{H}\right)$ (see [14]). However, the following relation shows that ${\parallel ·\parallel }_A$ and ${\omega }_A(·)$ are equivalent seminorms on ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$:

$$r_A\left(T\right)\le max\left\{\frac{1}{2}{\parallel T\parallel }_A,r_A\left(T\right)\right\}\le {\omega }_A\left(T\right)\le {\parallel T\parallel }_A.$$

(6)

For the proof of (6), we refer to the following references [12,14]. If $A=I$, then the classical definitions of the operator norm, numerical radius, and spectral radius for Hilbert space operators are obtained and are simply denoted by $\parallel ·\parallel$, $\omega (·)$ and $r(·)$.

If $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, then by Douglas’s theorem [10] there exists a unique solution, given by $T^{{♯}_A}$, of the following problem

$$\begin{array}{c}\hfill AX=T^{*}A,\mathcal{R}\left(X\right)\subseteq \overline{\mathcal{R}\left(A\right)}.\end{array}$$

Note that $T^{{♯}_A}=A^{†}{T}^{*}A$, where $A^{†}$ is the Moore–Penrose pseudo-inverse of A (see [11]). If $T,S\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, then $TS,\alpha T+\beta S\in {\mathbb{B}}_A\left(\mathcal{H}\right)$ for every $\alpha ,\beta \in \mathbb{R}$ and we have ${\left(TS\right)}^{{♯}_A}=S^{{♯}_A}{T}^{{♯}_A}$ and ${(\alpha T+\beta S)}^{{♯}_A}=\overline{\alpha }{T}^{{♯}_A}+\overline{\beta }{S}^{{♯}_A}$. Moreover, if $P_{\overline{\mathcal{R}\left(A\right)}}$ denotes the orthogonal projection onto $\overline{\mathcal{R}\left(A\right)}$, then for a given $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, we have $T^{{♯}_A}\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, ${(T^{{♯}_A})}^{{♯}_A}=P_{\overline{\mathcal{R}\left(A\right)}}T{P}_{\overline{\mathcal{R}\left(A\right)}}$ and ${({(T^{{♯}_A})}^{♯A})}^{♯A}=T^{{♯}_A}$. For more details about the operator $T^{{♯}_A}$, one can see [9,11,15]. Furthermore, we recall that an operator T is said to be A-positive if $AT$ is a positive operator and we write $T{\ge }_{A}0$. It can be observed that A-positive operators are in ${\mathbb{B}}_A\left(\mathcal{H}\right)$. For $T,S\in \mathbb{B}\left(\mathcal{H}\right)$, the notation $T{\ge }_{A}S$ means $T-S{\ge }_{A}0$. When $A=I$, then $T{\ge }_{I}S$ will simply be denoted by $T\ge S$.

The structure of this paper is organized as follows: in Section 2, we give some notions that characterize a d-tuple of operators $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$. In Section 3, we introduce a new joint norm of tuples of operators that generalizes the joint norm given in (12) and define the class of doubly-commuting tuples of hyponormal operators acting on an A-weighted Hilbert space, where A is a positive operator that is not assumed to be invertible. We proved a generalization of the well-known result due to G. Popescu [16]. We also present an inequality that characterizes the Euclidean norm of an operator tuple $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$. In Section 4, we give several characterizations related to the operators from ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ and the operators from $\mathbb{B}\left(\mathbf{R}\left(A^{1/2}\right)\right)$. For an A-doubly-commuting d-tuple of hyponormal operators, we prove the equalities ${\parallel \mathbf{T}\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_A$ and $r_A\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_A={\omega }_A\left(\mathbf{T}\right).$ The motivation for our investigation comes from a recent paper [17].

2. Preliminaries

To prepare the framework in which we will work, we present in this section some notions and notations that will be useful in this paper.

Let $\mathbb{N}$ and ${\mathbb{N}}^{*}$ denote the set of nonnegative and positive integers, respectively. Let $d\in {\mathbb{N}}^{*}$ and $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$ be a d-tuple of operators. If $[T_i,T_j]=0$ for all $i,j\in \{1,\dots ,d\}$, then $\mathbf{T}$ is said to be a commuting tuple. Moreover, if $\mathbf{T}$ is a commuting d-tuple of operators and $[T_i^{*},T_j]=0$ for every $1\le i\ne j\le d$, then it is called a doubly-commuting operator tuple.

In the next definition, we recall two important classes of operators in semi-Hilbert spaces.

Definition 2 

([14]). An operator $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$ is called

(i) 

A-normal if $[T^{{♯}_A},T]=0$;

(ii) 

A-hyponormal if $[T^{{♯}_A},T]{\ge }_{A}0$.

For some results concerning the above two classes of operators, see [14] and the references therein. For $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, the equalities

$$r_A\left(T\right)={\omega }_A\left(T\right)={\parallel T\parallel }_A$$

(7)

hold for the class of A-normal, A-hyponormal, and A-positive operators (see [14]). Since $T^{{♯}_A}T{\ge }_{A}0$ and $T{T}^{{♯}_A}{\ge }_{A}0$, then an application of the second equality in (7) together with the last equality in (3) shows that

$$\parallel T^{{♯}_A}{T\parallel }_A=\parallel T{T}^{{♯}_A}{\parallel }_A={\parallel T\parallel }_A^2={\parallel T^{{♯}_A}\parallel }_A^2.$$

(8)

Now, associated with a d-tuple of operators, $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$ (not necessarily commuting), the following quantities

$${\displaystyle {\parallel \mathbf{T}\parallel }_A:=sup\left\{\sqrt{{\displaystyle \sum _{j=1}^d{\parallel T_{j}x\parallel }_A^2}};x\in \mathcal{H},{\parallel x\parallel }_A=1\right\},}$$

and

$${\displaystyle {\omega }_A\left(\mathbf{T}\right):=sup\left\{\sqrt{{\displaystyle \sum _{j=1}^d{\left|{\langle T_{j}x,x\rangle }_A\right|}^2}};x\in \mathcal{H},{\parallel x\parallel }_A=1\right\}}$$

are defined in [12]. If $T_j\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ for all $j\in \{1,\dots ,d\}$, then one can verify that ${\parallel ·\parallel }_A$ and ${\omega }_A(·)$ two seminorms on ${\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$. Notice that ${\omega }_A\left(\mathbf{T}\right)$ and ${\parallel \mathbf{T}\parallel }_A$ are called the A-joint numerical radius and the A-joint operator seminorm of $\mathbf{T}$, respectively.

In [18], H. Baklouti et al. introduced the concept of the A-joint spectral radius associated with a d-tuple of commuting operators $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ as follows

$${\displaystyle r_A\left(\mathbf{T}\right):=\underset{n\in {\mathbb{N}}^{*}}{inf}{∥{\displaystyle \sum _{\begin{array}{c}\left|\alpha \right|=n,\\ \alpha \in {\mathbb{N}}^d\end{array}}\frac{n!}{\alpha !}{\left({\mathbf{T}}^{{♯}_A}\right)}^{\alpha }{\mathbf{T}}^{\alpha }}∥}_A^{\frac{1}{2n}}=\underset{n\to \infty }{lim}{∥{\displaystyle \sum _{\begin{array}{c}\left|\alpha \right|=n,\\ \alpha \in {\mathbb{N}}^d\end{array}}\frac{n!}{\alpha !}{\left({\mathbf{T}}^{{♯}_A}\right)}^{\alpha }{\mathbf{T}}^{\alpha }}∥}_A^{\frac{1}{2n}},}$$

(9)

where ${\mathbf{T}}^{{♯}_A}=(T_1^{{♯}_A},\dots ,T_d^{{♯}_A})$. Moreover, for the multi-index $\alpha =({\alpha }_1,\dots ,{\alpha }_d)\in {\mathbb{N}}^d$, we will use the following notations:

$${\mathbf{T}}^{\alpha }:=\prod _{k=1}^d{T}_k^{{\alpha }_k},\left|\alpha \right|:=\sum _{j=1}^d\left|{\alpha }_j\right|\mathrm{and}\alpha !:=\prod _{k=1}^d{\alpha }_k!.$$

We mention here that the second equality in (9) has also been proved by Baklouti et al. in [18]. Notice that for every commuting operator tuple $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$, we have

$$r_A\left(\mathbf{T}\right)\le max\left\{\frac{1}{2\sqrt{d}}{\parallel \mathbf{T}\parallel }_A,r_A\left(\mathbf{T}\right)\right\}\le {\omega }_A\left(\mathbf{T}\right)\le {\parallel \mathbf{T}\parallel }_A$$

(10)

(see Theorem 2.4 in [12] and Theorem 2.2 in [19]). In [19], it is stated that if $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ is any d-tuple of commuting A-normal operators, then

$$r_A\left(\mathbf{T}\right)={\omega }_A\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_A.$$

(11)

One of the main targets of this work is to establish the equalities in (11) for a new class of multivariable operators.

Next, for $A=I$, we define $r_I\left(\mathbf{T}\right)$, ${\omega }_I\left(\mathbf{T}\right)$, and ${\parallel \mathbf{T}\parallel }_I$ which will simply be denoted by $r\left(\mathbf{T}\right)$, $\omega \left(\mathbf{T}\right)$ and $\parallel \mathbf{T}\parallel$, respectively. Thus, we obtain

$${\displaystyle \parallel \mathbf{T}\parallel :=sup\left\{\sqrt{{\displaystyle \sum _{j=1}^d{\parallel T_{j}x\parallel }^2}};x\in \mathcal{H},\parallel x\parallel =1\right\},}$$

and

$${\displaystyle \omega \left(\mathbf{T}\right):=sup\left\{\sqrt{{\displaystyle \sum _{j=1}^d{\left|\langle T_{j}x,x\rangle \right|}^2}};x\in \mathcal{H},\parallel x\parallel =1\right\}.}$$

The last equality is given in [20] by M. Chō and M. Takaguchi and is the Euclidean operator radius of an operator tuple $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$, see also [16].

For $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$, G. Popescu defined in [16] the following quantity

$${\displaystyle {\parallel \mathbf{T}\parallel }_e:=\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\parallel {\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d\parallel ,}$$

(12)

where ${\mathbb{B}}_d$ denotes the open unit ball of ${\mathbb{C}}^d$ with respect to the Euclidean norm, i.e.,

$${\mathbb{B}}_d:=\left\{\lambda =({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{C}}^d{;\parallel \lambda \parallel }_2^2:=\sum _{j=1}^d{\left|{\lambda }_j\right|}^2<1\right\}.$$

It is clear that we can change ${\mathbb{B}}_d$ with its closure in (12) without changing the value of ${\parallel \mathbf{T}\parallel }_e$. Note that ${\parallel ·\parallel }_e$ defines a norm on $\mathbb{B}{\left(\mathcal{H}\right)}^d$. Moreover, in [17], the following equality is established:

$$\parallel \mathbf{T}\parallel ={\parallel \mathbf{T}\parallel }_e$$

(13)

for every doubly-commuting d-tuple of hyponormal operators $\mathbf{T}$. It is important to mention that G. Popescu proved in [16] that the following inequalities hold

$$\frac{1}{\sqrt{d}}\sqrt{∥{\displaystyle \sum _{j=1}^d{T}_j{T}_j^{*}}∥}\le {\parallel \mathbf{T}\parallel }_e\le \sqrt{∥{\displaystyle \sum _{j=1}^d{T}_j{T}_j^{*}}∥}$$

(14)

for any d-tuple $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$. Furthermore, it has been shown in [16] that the constants $\frac{1}{\sqrt{d}}$ and 1 are the best choices possible.

3. New Joint Seminorm for Operator Tuples

In this section, we aim to introduce and investigate a new joint seminorm for d-tuples of A-bounded operators. An alternative and easy proof of a well-known result due to G. Popescu [16] is established.

First, we introduce the following definition, which is a natural generalization of (12).

Definition 3. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$. The A-Euclidean seminorm of the d-tuple of A-bounded operators $\mathbf{T}$ is given by

$${\displaystyle {\parallel \mathbf{T}\parallel }_{e,A}:=\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\parallel {\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d\parallel }_A.}$$

In the next proposition, we state some connections between the seminorms ${\parallel ·\parallel }_{e,A}$ and ${\parallel ·\parallel }_A$.

Proposition 1. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$ be the d-tuple of the operators. Then, the following assertions hold:

(1) 

${\parallel \mathbf{T}\parallel }_{e,A}\le {\parallel \mathbf{T}\parallel }_A$;

(2) 

If $T_k\in {\mathbb{B}}_A\left(\mathcal{H}\right)$ for all $k\in \{1,\dots ,d\}$, then $\parallel {\mathbf{T}}^{{♯}_A}{\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_{e,A}$ and

$$\frac{1}{\sqrt{d}}max\left\{{\parallel \mathbf{T}\parallel }_A,{\parallel {\mathbf{T}}^{{♯}_A}\parallel }_A\right\}\le {\parallel \mathbf{T}\parallel }_{e,A},$$

(15)

where ${\mathbf{T}}^{{♯}_A}=(T_1^{{♯}_A},\dots ,T_d^{{♯}_A})$.

Proof. 

(1) Let $x\in \mathcal{H}$ and $\lambda =({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d$. Then, by applying the Cauchy–Schwarz inequality (in short (C–S)) and making several calculations, we deduce that

$$\begin{array}{cc}\hfill {∥\sum _{j=1}^d{\lambda }_j{T}_{j}x∥}_A^2& =\langle \left(\sum _{j=1}^d{\lambda }_j{T}_j\right)x,\left(\sum _{k=1}^d{\lambda }_k{T}_k\right)x{\rangle }_A\hfill \\ \hfill & =\sum _{j=1}^d\sum _{k=1}^d{\lambda }_j\overline{{\lambda }_k}{\langle T_{j}x,T_{k}x\rangle }_A\hfill \\ \hfill & \le \sum _{j=1}^d\sum _{k=1}^d|{\lambda }_j|\times |{\lambda }_k|\times \parallel T_j{x\parallel }_A{\parallel T_{k}x\parallel }_A\hfill \\ \hfill & ={\left(\sum _{j=1}^d|{\lambda }_j|\times \parallel T_j{x\parallel }_A\right)}^2.\hfill \end{array}$$

By applying the inequality (C–S) again, we obtain the following inequality

$${∥\sum _{k=1}^d{\lambda }_k{T}_{k}x∥}_A^2\le {\parallel \lambda \parallel }_2^2\left(\sum _{j=1}^d{\parallel T_{j}x\parallel }_A^2\right).$$

Then, by taking the supremum over all $x\in \mathcal{H}$ with ${\parallel x\parallel }_A=1$, we find

$${∥\sum _{k=1}^d{\lambda }_k{T}_k∥}_A\le {\parallel \lambda \parallel }_2{∥\mathbf{T}∥}_A.$$

So, the desired inequality is proved by taking the supremum over all $\lambda \in {\mathbb{B}}_d$.

(2) The fact that $\parallel {\mathbf{T}}^{{♯}_A}{\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_{e,A}$ follows trivially since $\parallel X^{{♯}_A}{\parallel }_A={\parallel X\parallel }_A$ for all $X\in {\mathbb{B}}_A\left(\mathcal{H}\right)$. Now, in order to prove (15), we need to recall from [16] the following facts: if we denote by ${\mathbb{S}}_d$ the unit sphere of ${\mathbb{C}}^d$ and $\sigma$ the rotation-invariant positive Borel measure on ${\mathbb{S}}_d$ for which $\sigma \left({\mathbb{S}}_d\right)=1$, then for all $\mu =({\mu }_1,\dots ,{\mu }_d)\in {\mathbb{C}}^d$, we have

$${\int }_{{\mathbb{S}}_d}{\left|{\mu }_k\right|}^{2}d\sigma \left(\mu \right)=\frac{1}{d},\forall k\in \{1,\dots ,d\}\mathrm{and}{\int }_{{\mathbb{S}}_d}{\mu }_i\overline{{\mu }_j}d\sigma \left(\mu \right)=0,\forall 1\le i\ne j\le d.$$

(16)

Now, let ${\overline{\mathbb{B}}}_d$ denote the closed unit ball of ${\mathbb{C}}^d$. It is clear that

$$\begin{array}{cc}\hfill {\parallel \mathbf{T}\parallel }_{e,A}^2& =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{∥\sum _{j=1}^d{\lambda }_j{T}_j∥}_A^2\hfill \\ \hfill & =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\overline{\mathbb{B}}}_d}{sup}{∥\sum _{j=1}^d{\lambda }_j{T}_j∥}_A^2.\hfill \end{array}$$

Further, by using (8), we see that

$$\begin{array}{cc}\hfill {\parallel \mathbf{T}\parallel }_{e,A}^2& =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\overline{\mathbb{B}}}_d}{sup}{∥{\left(\sum _{j=1}^d{\lambda }_j{T}_j\right)}^{{♯}_A}\left(\sum _{j=1}^d{\lambda }_j{T}_j\right)∥}_A\hfill \\ \hfill & =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\overline{\mathbb{B}}}_d}{sup}\underset{{\parallel x\parallel }_A=1}{sup}\langle \left(\sum _{j=1}^d\overline{{\lambda }_j}{T}_j^{{♯}_A}\right)\left(\sum _{j=1}^d{\lambda }_j{T}_j\right)x,x{\rangle }_A\hfill \\ \hfill & =\underset{{\parallel x\parallel }_A=1}{sup}\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\overline{\mathbb{B}}}_d}{sup}\sum _{i,j=1}^d\overline{{\lambda }_i}{\lambda }_j\langle T_i^{{♯}_A}{T}_{j}x,x{\rangle }_A.\hfill \end{array}$$

On the other hand, since $\sigma \left({\mathbb{S}}_d\right)=1$, then it follows that

$$\begin{array}{cc}\hfill \underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\overline{\mathbb{B}}}_d}{sup}\sum _{i,j=1}^d\overline{{\lambda }_i}{\lambda }_j{\langle T_i^{{♯}_A}{T}_{j}x,x\rangle }_A& ={\int }_{{\mathbb{S}}_d}\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\overline{\mathbb{B}}}_d}{sup}\sum _{i,j=1}^d\overline{{\lambda }_i}{\lambda }_j{\langle T_i^{{♯}_A}{T}_{j}x,x\rangle }_{A}d\sigma \left(\mu \right)\hfill \\ \hfill & \ge {\int }_{{\mathbb{S}}_d}\sum _{i,j=1}^d\overline{{\mu }_i}{\mu }_j{\langle T_i^{{♯}_A}{T}_{j}x,x\rangle }_{A}d\sigma \left(\mu \right),\hfill \end{array}$$

for all $x\in \mathcal{H}$. This implies that, through (16),

$$\begin{array}{cc}\hfill {\parallel \mathbf{T}\parallel }_{e,A}^2& \ge \underset{{\parallel x\parallel }_A=1}{sup}{\int }_{{\mathbb{S}}_d}\sum _{i,j=1}^d{\mu }_i\overline{{\mu }_j}{\langle T_i^{{♯}_A}{T}_{j}x,x\rangle }_{A}d\sigma \left(\mu \right)\hfill \\ \hfill & =\frac{1}{d}\underset{{\parallel x\parallel }_A=1}{sup}\sum _{i=1}^d{\langle T_i^{{♯}_A}{T}_{i}x,x\rangle }_A=\frac{1}{d}{\parallel \mathbf{T}\parallel }_A^2.\hfill \end{array}$$

This proves that

$$\frac{1}{d}{\parallel \mathbf{T}\parallel }_A\le {\parallel \mathbf{T}\parallel }_{e,A}.$$

(17)

By replacing $T_k$ by $T_k^{{♯}_A}$ in (17) and then using the fact that $\parallel {\mathbf{T}}^{{♯}_A}{\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_{e,A}$, we have

$$\frac{1}{d}\parallel {\mathbf{T}}^{{♯}_A}{\parallel }_A\le {\parallel \mathbf{T}\parallel }_{e,A}.$$

(18)

Combining (17) together with (18) yields (15) as desired. □

Remark 1. 

(1) If $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$, then clearly ${\displaystyle \sum _{k=1}^d{T}_k^{{♯}_A}{T}_k{\ge }_{A}0}$. Hence, a direct application of (7) shows that

$${\parallel \mathbf{T}\parallel }_A=\sqrt{{∥{\displaystyle \sum _{j=1}^d{T}_j^{{♯}_A}{T}_j}∥}_A}.$$

(19)

It should be mentioned here that the equality $\parallel {\mathbf{T}}^{{♯}_A}{\parallel }_A={\parallel \mathbf{T}\parallel }_A$ may not be correct even if $\mathbf{T}$ is a commuting operator tuple. Indeed, let us consider the following matrices in ${\mathcal{M}}_3$: $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 1\end{array}\right)$, $T_1=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$ and $T_2=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$. We remark that $[T_1,T_2]=0$. Furthermore, by using the fact that $T_k^{{♯}_A}=A^{†}{T}_k^{*}A$ with $k\in \{1,2\}$, it can be seen that $T_1^{{♯}_A}=\left(\begin{array}{ccc}0& 0& 0\\ \frac{1}{2}& 0& 0\\ 0& 0& 0\end{array}\right)$ and $T_2^{{♯}_A}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 1& 0& 0\end{array}\right)$. Further, by applying (19) and (4), we can show that

$$\parallel (T_1,T_2){\parallel }_A=\sqrt{{∥T_1^{{♯}_A}{T}_1+T_2^{{♯}_A}{T}_2∥}_A}=1$$

and

$$\parallel (T_1^{{♯}_A},T_2^{{♯}_A}){\parallel }_A=\sqrt{{∥{(T_1^{{♯}_A})}^{{♯}_A}{T}_1^{{♯}_A}+{(T_2^{{♯}_A})}^{{♯}_A}{T}_2^{{♯}_A}∥}_A}=\frac{\sqrt{5}}{2}.$$

(2) In virtue of Proposition 1, we infer that ${\parallel ·\parallel }_A$ and ${\parallel ·\parallel }_{e,A}$ are equivalent seminorms on ${\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$.

The following corollary provides a generalization and improvement of the well-known result due to G. Popescu [16].

Corollary 1. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ be a d-tuple of operators. Then, the inequality

$$\frac{1}{\sqrt{d}}max\left\{\alpha ,\beta \right\}\le {\parallel \mathbf{T}\parallel }_{e,A}\le min\left\{\alpha ,\beta \right\}$$

(20)

holds, where $\alpha =\sqrt{{∥{\displaystyle \sum _{k=1}^d{T}_k{T}_k^{{♯}_A}}∥}_A}$ and $\beta =\sqrt{{∥{\displaystyle \sum _{k=1}^d{T}_k^{{♯}_A}{T}_k}∥}_A}.$

Proof. 

By applying Proposition 1 together with (19), we deduce that

$$\frac{1}{\sqrt{d}}\sqrt{{∥{\displaystyle \sum _{k=1}^d{T}_k^{{♯}_A}{T}_k}∥}_A}\le {\parallel \mathbf{T}\parallel }_{e,A}\le \sqrt{{∥{\displaystyle \sum _{k=1}^d{T}_k^{{♯}_A}{T}_k}∥}_A}.$$

(21)

By replacing $T_k$ by $T_k^{{♯}_A}$ in (21), we can see that

$$\frac{1}{\sqrt{d}}\sqrt{{∥{\left({\displaystyle \sum _{k=1}^d{T}_k{T}_k^{{♯}_A}}\right)}^{{♯}_A}∥}_A}\le {\parallel {\mathbf{T}}^{{♯}_A}\parallel }_{e,A}\le \sqrt{{∥{\left({\displaystyle \sum _{k=1}^d{T}_k{T}_k^{{♯}_A}}\right)}^{{♯}_A}∥}_A},$$

from which we have

$$\frac{1}{\sqrt{d}}\sqrt{{∥{\displaystyle \sum _{k=1}^d{T}_k{T}_k^{{♯}_A}}∥}_A}\le {\parallel \mathbf{T}\parallel }_{e,A}\le \sqrt{{∥{\displaystyle \sum _{k=1}^d{T}_k{T}_k^{{♯}_A}}∥}_A}.$$

(22)

A combination of (21) together with (22) yields (20) as desired. □

Remark 2. 

Note that the following equality

$${∥{\displaystyle \sum _{j=1}^d{T}_j^{{♯}_A}{T}_j}∥}_A={∥{\displaystyle \sum _{j=1}^d{T}_j{T}_j^{{♯}_A}}∥}_A$$

may not be correct for some d-tuple of operators $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ even if $A=I$. Indeed, we consider the following matrices in ${\mathcal{M}}_2$: $A=I_2$, $T_1=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ and $T_2=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. It is not difficult to check that

$$∥{\displaystyle \sum _{j=1}^2{T}_j{T}_j^{*}}∥=1\ne 2=∥{\displaystyle \sum _{j=1}^2{T}_j^{*}{T}_j}∥.$$

In the next theorem, we give a new formula of ${\parallel \mathbf{T}\parallel }_{A,e}$ for $\mathbf{T}\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$, which allows us to prove that ${\parallel ·\parallel }_{A,e}$ and ${\parallel ·\parallel }_A$ are two equivalent seminorms on ${\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$. Notice that our new techniques provide an alternative and easy proof of the inequalities (14), which were first proved in [16].

Theorem 1. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$. Then, the equality

$${\displaystyle {\parallel \mathbf{T}\parallel }_{A,e}=sup\left\{\sqrt{\sum _{j=1}^d{\left|{\langle T_{j}x,y\rangle }_A\right|}^2};x,y\in {\mathcal{H},\parallel x\parallel }_A={\parallel y\parallel }_A=1\right\}}$$

(23)

holds.

Proof. 

By using (3), we see that

$$\begin{array}{cc}\hfill {\parallel \mathbf{T}\parallel }_{e,A}& {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\parallel {\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d\parallel }_A}\hfill \\ \hfill & {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}sup\left\{{∥{\displaystyle \sum _{k=1}^d{\lambda }_k{T}_{k}x}∥}_A;x\in \mathcal{H},{\parallel x\parallel }_A=1\right\}}\hfill \\ \hfill & {\displaystyle =sup\left\{{\displaystyle \underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{∥{\displaystyle \sum _{k=1}^d{\lambda }_k{T}_{k}x}∥}_A;x\in \mathcal{H},{\parallel x\parallel }_A=1}\right\}.}\hfill \end{array}$$

(24)

Moreover, recall from [19] that for complex numbers $z_1,\dots ,z_d$, we have

$$\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left|\sum _{k=1}^d{\lambda }_k{z}_k\right|=\sqrt{\sum _{k=1}^d{\left|z_k\right|}^2}.$$

(25)

Now, let $x,y\in \mathcal{H}$. By using (25), we have

$$\begin{array}{cc}\hfill \sqrt{{\displaystyle \sum _{j=1}^d{\left|{\langle T_{j}x,y\rangle }_A\right|}^2}}& =\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left|\sum _{j=1}^d{\lambda }_j{\langle T_{j}x,y\rangle }_A\right|\hfill \\ \hfill & =\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left|\langle \left(\sum _{j=1}^d{\lambda }_j{T}_{j}x\right),y{\rangle }_A\right|\hfill \end{array}$$

Hence, by taking the supremum over all $y\in \mathcal{H}$ with ${\parallel y\parallel }_A=1$ in the last equality we have

$$\begin{array}{cc}\hfill \underset{\begin{array}{c}y\in \mathcal{H},\\ {\parallel y\parallel }_A=1\end{array}}{sup}\sqrt{{\displaystyle \sum _{k=1}^d{\left|{\langle T_{k}x,y\rangle }_A\right|}^2}}& =\underset{\begin{array}{c}y\in \mathcal{H},\\ {\parallel y\parallel }_A=1\end{array}}{sup}\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left|\langle \left(\sum _{k=1}^d{\lambda }_k{T}_{k}x\right),y{\rangle }_A\right|.\hfill \end{array}$$

This yields that

$$\begin{array}{cc}\hfill \underset{\begin{array}{c}y\in \mathcal{H},\\ {\parallel y\parallel }_A=1\end{array}}{sup}\sqrt{{\displaystyle \sum _{k=1}^d{\left|{\langle T_{k}x,y\rangle }_A\right|}^2}}& =\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left[\underset{\begin{array}{c}y\in \mathcal{H},\\ {\parallel y\parallel }_A=1\end{array}}{sup}\left|\langle \left(\sum _{k=1}^d{\lambda }_k{T}_{k}x\right),y{\rangle }_A\right|\right].\hfill \end{array}$$

(26)

On the other hand, it is not difficult to check that

$${\displaystyle sup\left\{{|\langle u,v\rangle }_A{|;v\in \mathcal{H},\parallel v\parallel }_A=1\right\}={\parallel u\parallel }_A,\forall u\in \mathcal{H}.}$$

(27)

Thus, by using (26) and (27), we obtain

$$\begin{array}{c}\hfill \underset{\begin{array}{c}y\in \mathcal{H},\\ {\parallel y\parallel }_A=1\end{array}}{sup}\sqrt{{\displaystyle \sum _{k=1}^d{\left|{\langle T_{k}x,y\rangle }_A\right|}^2}}=\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{∥\sum _{k=1}^d{\lambda }_k{T}_{k}x∥}_A.\end{array}$$

(28)

Combining (28) together with (24) yields (23) as required, and, hence, the proof is complete. □

Remark 3. 

By letting $A=I$ in (23), we obtain a well-known result established by Dragomir in Theorem 9 in [21], and when the 2-tuple is $(T,T^{{♯}_A})$, where $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, we obtain a recent result in [22].

The following corollary is an application of Theorem 1 and provides an improvement of the results given in Proposition 1 since ${\mathbb{B}}_A\left(\mathcal{H}\right)\subseteq {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$. Moreover, the new Formula (23) enables us to derive an alternative and easy proof of the inequalities (14).

Corollary 2. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$ be a d-tuple of operators. Then,

$$\frac{1}{\sqrt{d}}{\parallel \mathbf{T}\parallel }_A\le {\parallel \mathbf{T}\parallel }_{e,A}\le {\parallel \mathbf{T}\parallel }_A.$$

(29)

Proof. 

By using (23) and then applying the inequality (C–S), we easily prove the second inequality in (29). Now, let $x\in \mathcal{H}$ be such that ${\parallel x\parallel }_A=1$. Assume that $T_{k}x\notin \mathcal{N}\left(A\right)$ for all $k\in \{1,\dots ,d\}$ and let

$${\displaystyle y_k=\frac{T_{k}x}{\parallel T_k{x\parallel }_A},\forall k\in \{1,\dots ,d\}.}$$

(If $T_{k_0}x\in \mathcal{N}\left(A\right)$ for some $k_0\in \{1,\dots ,d\}$, we choose $y_{k_0}=x$). We clearly have

$$\parallel y_k{\parallel }_A=1\mathrm{and}|{\langle T_{k}x,y_k\rangle }_A{|}^2={\parallel T_{k}x\parallel }_A^2,\forall k\in \{1,\dots ,d\}.$$

Thus, by applying (23), we have

$${\parallel \mathbf{T}\parallel }_{A,e}^2\ge \sum _{k=1}^d|{\langle T_{k}x,y_k\rangle }_A{|}^2\ge |{\langle T_{1}x,y_1\rangle }_A{|}^2={\parallel T_{1}x\parallel }_A^2.$$

Similarly, we prove that ${\parallel \mathbf{T}\parallel }_{A,e}^2\ge {\parallel T_{i}x\parallel }_A^2$ for all $i\in \{1,\dots ,d\}$. This yields

$${d\parallel \mathbf{T}\parallel }_{A,e}^2\ge \sum _{k=1}^d{\parallel T_{k}x\parallel }_A^2.$$

Therefore, by taking the supremum over all $x\in \mathcal{H}$ with ${\parallel x\parallel }_A=1$ in the last inequality, we have

$${\parallel \mathbf{T}\parallel }_{A,e}\ge \frac{1}{\sqrt{d}}{\parallel \mathbf{T}\parallel }_A.$$

Hence, the proof is complete. □

Remark 4. 

By letting $A=I$ in (29) and then replacing $T_k$ with $T_k^{*}$ for all $k\in \{1,\dots ,d\}$ we easily obtain the inequalities (14) that have been already established by G. Popescu in [16] by using a different argument.

To establish our next result, we require the following lemma.

Lemma 2. 

For any vectors $x_1,x_2,\dots ,x_d$ in $\mathcal{H}$ and for arbitrary complex numbers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_d$, with ${\lambda }_i\ne 0$, $i=\overline{1,d}$, we have

$$\sum _{i=1}^d{\parallel x_i\parallel }_A^2\ge \frac{\parallel {\displaystyle \sum _{i=1}^d}{\lambda }_i{x}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^d}{\left|{\lambda }_i\right|}^2}+\underset{i,j\in \{1,\dots ,d\}}{max}\frac{\parallel \overline{{\lambda }_i}{x}_j-\overline{{\lambda }_j}{x}_i{\parallel }_A^2}{|{\lambda }_i{|}^2+{\left|{\lambda }_j\right|}^2}$$

for any $d\ge 2$.

Proof. 

We use, as in [23] or [24], the technique of the monotony of a sequence. Consider the sequence

$$S_d=\sum _{i=1}^d{\parallel x_i\parallel }_A^2-\frac{\parallel {\displaystyle \sum _{i=1}^d}{\lambda }_i{x}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^d}{\left|{\lambda }_i\right|}^2},d\ge 1.$$

By studying the monotony of sequence $S_k,k\le d$, we have

$$S_{k+1}-S_k={\parallel x_{k+1}\parallel }_A^2+\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^k}{\left|{\lambda }_i\right|}^2}-\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i+{\lambda }_{k+1}{x}_{k+1}{\parallel }_A^2}{{\displaystyle \sum _{i=1}^k}|{\lambda }_i{|}^2+{\left|{\lambda }_{k+1}\right|}^2}.$$

For two vectors $x,y\in \mathcal{H}$ and for complex numbers $\lambda ,\mu \ne 0$, the following equality holds:

$$\frac{{\parallel x\parallel }_A^2}{{\left|\lambda \right|}^2}+\frac{{\parallel y\parallel }_A^2}{{\left|\mu \right|}^2}-\frac{{\parallel x+y\parallel }_A^2}{{\left|\lambda \right|}^2+{\left|\mu \right|}^2}=\frac{{\parallel \left|\mu \right|}^{2}x-{\left|\lambda \right|}^2{y\parallel }_A^2}{{\left|\lambda \right|}^2{\left|\mu \right|}^2{\left(\right|\lambda |}^2+{\left|\mu \right|}^2)}.$$

(30)

Since the term on the right side of equality (30) is positive, then we have

$$\frac{{\parallel x\parallel }_A^2}{{\left|\lambda \right|}^2}+\frac{{\parallel y\parallel }_A^2}{{\left|\mu \right|}^2}\ge \frac{{\parallel x+y\parallel }_A^2}{{\left|\lambda \right|}^2+{\left|\mu \right|}^2}.$$

(31)

Now, using the inequality from (31), we have:

$$\parallel x_{k+1}{\parallel }_A^2+\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^k}{\left|{\lambda }_i\right|}^2}=\frac{\parallel {\lambda }_{k+1}{x}_{k+1}{\parallel }_A^2}{|{\lambda }_{k+1}{|}^2}+\frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^k}{\left|{\lambda }_i\right|}^2}\ge \frac{\parallel {\displaystyle \sum _{i=1}^k}{\lambda }_i{x}_i+{\lambda }_{k+1}{x}_{k+1}{\parallel }_A^2}{{\displaystyle \sum _{i=1}^k}|{\lambda }_i{|}^2+{\left|{\lambda }_{k+1}\right|}^2}.$$

It is easy to see that $S_{k+1}-S_k\ge 0$, that is, the sequence $S_k$ is increasing. Therefore, we deduce that

$$S_d\ge S_{d-1}\ge \dots \ge S_2\ge S_1=0.$$

However, by applying relation (30) for $x={\lambda }_1{x}_1$, $y={\lambda }_2{x}_2$, $\lambda ={\lambda }_1$ and $\mu ={\lambda }_2$, we obtain

$$\begin{array}{cc}\hfill S_2& =\parallel x_1{\parallel }_A^2+{\parallel x_2\parallel }_A^2-\frac{\parallel {\lambda }_1{x}_1+{\lambda }_2{x}_2{\parallel }_A^2}{|{\lambda }_1{|}^2+{\left|{\lambda }_2\right|}^2}\hfill \\ \hfill & =\frac{\parallel \overline{{\lambda }_1}{x}_2-\overline{{\lambda }_2}{x}_1{\parallel }_A^2}{|{\lambda }_1{|}^2+{\left|{\lambda }_2\right|}^2}.\hfill \end{array}$$

Taking into account that we can rearrange the terms of the two sequences, we obtain the inequality:

$$S_d=\sum _{i=1}^d{\parallel x_i\parallel }_A^2-\frac{\parallel {\displaystyle \sum _{i=1}^d}{\lambda }_i{x}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^d}{\left|{\lambda }_i\right|}^2}\ge S_2=\frac{\parallel \overline{{\lambda }_1}{x}_2-\overline{{\lambda }_2}{x}_1{\parallel }_A^2}{|{\lambda }_1{|}^2+{\left|{\lambda }_2\right|}^2}.$$

Consequently, we deduce the inequality of the statement. □

We are now able to establish the following result.

Theorem 2. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$ be a d-tuple of operators. Then, the inequality

$${\parallel \mathbf{T}\parallel }_A\ge {\parallel \mathbf{T}\parallel }_{e,A}+\underset{i,j\in \{1,\dots ,d\}}{max}\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\underset{{\parallel x\parallel }_A=1}{inf}{\parallel (\overline{{\lambda }_i}{T}_j-\overline{{\lambda }_j}{T}_i)x\parallel }_A^2$$

(32)

holds for any $d\ge 2$.

Proof. 

In Lemma 2, set $x_i=T_{i}x$ for all $i\in \{1,\dots ,d\}$, then

$$\sum _{i=1}^d{\parallel T_{i}x\parallel }_A^2\ge \frac{\parallel {\displaystyle \sum _{i=1}^d}{\lambda }_i{T}_{i}x{\parallel }_A^2}{{\displaystyle \sum _{i=1}^d}{\left|{\lambda }_i\right|}^2}+\underset{i,j\in \{1,\dots ,d\}}{max}\frac{\parallel \overline{{\lambda }_i}{T}_{j}x-\overline{{\lambda }_j}{T}_i{x\parallel }_A^2}{|{\lambda }_i{|}^2+{\left|{\lambda }_j\right|}^2}.$$

(33)

First, we take the supremum over all $x\in \mathcal{H}$ with ${\parallel x\parallel }_A=1$ in relation (33), we deduce

$${\parallel \mathbf{T}\parallel }_A\ge \frac{\parallel {\displaystyle \sum _{i=1}^d}{\lambda }_i{T}_i{\parallel }_A^2}{{\displaystyle \sum _{i=1}^d}{\left|{\lambda }_i\right|}^2}+\underset{i,j\in \{1,\dots ,d\}}{max}\underset{{\parallel x\parallel }_A=1}{inf}\frac{\parallel (\overline{{\lambda }_i}{T}_j-\overline{{\lambda }_j}{T}_i){x\parallel }_A^2}{|{\lambda }_i{|}^2+{\left|{\lambda }_j\right|}^2}.$$

(34)

Therefore, it we take the supremum over all $({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d$ in relation (34), then we find the inequality of the statement. □

Remark 5. 

By letting $A=I$ in (32), we obtain

$$\parallel \mathbf{T}\parallel \ge {\parallel \mathbf{T}\parallel }_e+\underset{i,j\in \{1,\dots ,d\}}{max}\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\underset{\parallel x\parallel =1}{inf}{\parallel (\overline{{\lambda }_i}{T}_j-\overline{{\lambda }_j}{T}_i)x\parallel }^2$$

for any d-tuple of operators $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$ and $d\ge 2$.

Next, we will present a result that characterizes the Euclidean norm of an operator tuple $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$.

Proposition 2. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$ be a d-tuple of operators. The following inequality holds:

$${\parallel \mathbf{T}\parallel }_e\le {\parallel \mathbf{T}\parallel }^{\theta }{\parallel {\mathbf{T}}^{*}\parallel }^{1-\theta },$$

(35)

where $\theta \in [0,1].$ Here ${\mathbf{T}}^{*}=(T_1^{*},\dots ,T_d^{*})$.

Proof. 

First, we will prove a radon-type inequality,

$$\sum _{k=1}^d{a}_k^{\theta }{b}_k^{1-\theta }\le {\left(\sum _{k=1}^d{a}_k\right)}^{\theta }{\left(\sum _{k=1}^d{b}_k\right)}^{1-\theta },$$

(36)

for every $a_k\ge 0$ and $b_k>0$ with $k\in \{1,\dots ,d\}$. If we apply the Jensen inequality for the function $f\left(x\right)=x^{\theta }$, which is concave for $\theta \in [0,1]$, we deduce

$$\frac{{\sum }_{k=1}^d{b}_k{\left({\displaystyle \frac{a_k}{b_k}}\right)}^{\theta }}{{\sum }_{k=1}^d{b}_k}\le {\left(\frac{{\sum }_{k=1}^d{b}_k\left({\displaystyle \frac{a_k}{b_k}}\right)}{{\sum }_{k=1}^d{b}_k}\right)}^{\theta },$$

which is equivalent to inequality (36). In [25], Dragomir applied Hölder’s inequality for this. For $\theta \ge 1$, the function $f\left(x\right)=x^{\theta }$ is convex and the inequality sign in (36) is flipped, obtaining the classical Radon inequality

$$\sum _{k=1}^d\frac{a_k^{\theta }}{b_k^{\theta -1}}\ge \frac{{\left({\sum }_{k=1}^d{a}_k\right)}^{\theta }}{{\left({\sum }_{k=1}^d{b}_k\right)}^{\theta -1}}.$$

Thus, we have

$$\begin{array}{cc}\hfill \sum _{k=1}^d{\left|\langle T_{k}x,y\rangle \right|}^2& \stackrel{Kato}{\le }\sum _{k=1}^d\langle |T_k{|}^{2\theta }x,x\rangle \langle |T_k^{*}{|}^{2(1-\theta )}y,y\rangle \hfill \\ \hfill & \stackrel{McCarthy}{\le }\sum _{k=1}^d\langle |T_k{|}^2{x,x\rangle }^{\theta }\langle |T_k^{*}{{|}^{2}y,y\rangle }^{1-\theta }\hfill \\ \hfill & \stackrel{\left(36\right)}{\le }{\left(\sum _{k=1}^d\langle |T_k{|}^{2}x,x\rangle \right)}^{\theta }{\left(\sum _{k=1}^d\langle |T_k^{*}{|}^{2}y,y\rangle \right)}^{1-\theta }\hfill \\ \hfill & ={\left[{\left(\sum _{k=1}^d{\parallel T_{k}x\parallel }^2\right)}^{1/2}\right]}^{2\theta }{\left[{\left(\sum _{k=1}^d{\parallel T_k^{*}y\parallel }^2\right)}^{1/2}\right]}^{2(1-\theta )}.\hfill \end{array}$$

Therefore, we deduce

$${\left(\sum _{k=1}^d{\left|\langle T_{k}x,y\rangle \right|}^2\right)}^{1/2}\le {\left[{\left(\sum _{k=1}^d{\parallel T_{k}x\parallel }^2\right)}^{1/2}\right]}^{\theta }{\left[{\left(\sum _{k=1}^d{\parallel T_k^{*}y\parallel }^2\right)}^{1/2}\right]}^{1-\theta }.$$

(37)

Consequently, by taking the supremum over $x,y\in \mathcal{H}$ with $\parallel x\parallel =\parallel y\parallel =1$ in inequality (37) and taking into account the equality from Theorem 1 for $A=I$, we obtain the desired result. □

Remark 6. 

If we take $\theta =1$ in relation (35), then ${\parallel \mathbf{T}\parallel }_e\le \parallel \mathbf{T}\parallel$ and for $\theta =0$ in the same relation, we obtain ${\parallel \mathbf{T}\parallel }_e\le \parallel {\mathbf{T}}^{*}\parallel$. Using the Kittaneh–Manasrah inequality [26] and inequality (35), we found the following inequality:

$${\parallel \mathbf{T}\parallel }_e\le \theta \parallel \mathbf{T}\parallel +(1-\theta )\parallel {\mathbf{T}}^{*}\parallel -min\{\theta ,1-\theta \}{\left(\sqrt{\parallel \mathbf{T}\parallel }-\sqrt{\parallel {\mathbf{T}}^{*}\parallel }\right)}^2$$

for all $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$ a d-tuple of positive operators.

4. $\mathit{A}$-Doubly-Commuting Tuples of $\mathit{A}$-Hyponormal Operators

In this section, we give several characterizations related to the operators from ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ and the operators from $\mathbb{B}\left(\mathbf{R}\left(A^{1/2}\right)\right)$. For an A-doubly-commuting d-tuple of hyponormal operators, we proved the equalities ${\parallel \mathbf{T}\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_A$ and $r_A\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_A={\omega }_A\left(\mathbf{T}\right).$

Let us introduce the following definition.

Definition 4. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$. The d-tuple $\mathbf{T}$ is said to be A-doubly commuting if:

(i) 

it is commuting, i.e., $[T_i,T_k]=0$ for all $i,k\in \{1,\dots ,d\}$,

(ii) 

$T_i^{{♯}_A}{T}_k=T_k{T}_i^{{♯}_A}$ for all $1\le i\ne k\le d$.

In this section, we will study the connection between ${\parallel \mathbf{T}\parallel }_A$ and ${\parallel \mathbf{T}\parallel }_{e,A}$, when $\mathbf{T}$ is a d-tuple of A-doubly-commuting tuples of A-hyponormal operators. For this purpose, we need to recall some aspects: the semi-inner product ${\langle ·,·\rangle }_A$ induces an inner product on the quotient space $\mathcal{H}/\mathcal{N}\left(A\right)$ is given by

$$(\overline{x},\overline{y})=\langle Ax,y\rangle ,$$

for any $\overline{x}=x+\mathcal{N}\left(A\right),\overline{y}=y+\mathcal{N}\left(A\right)\in \mathcal{H}/\mathcal{N}\left(A\right)$. We remark that $\left(\mathcal{H}/\mathcal{N}\left(A\right),(·,·)\right)$ is not complete unless $\mathcal{R}\left(A\right)$ is closed in $\mathcal{H}$. However, L. de Branges and J. Rovnyak [27] proved that the completion of $\mathcal{H}/\mathcal{N}\left(A\right)$ is isometrically isomorphic to the Hilbert space $\mathbf{R}\left(A^{1/2}\right):=\left(\mathcal{R}\left(A^{1/2}\right),{\langle ·,·\rangle }_{\mathbf{R}\left(A^{1/2}\right)}\right)$, where ${\langle ·,·\rangle }_{\mathbf{R}(A^{1/2})}$ is given by

$${\langle A^{1/2}\delta ,A^{1/2}\xi \rangle }_{\mathbf{R}(A^{1/2})}:=\langle P_{\overline{\mathcal{R}\left(A\right)}}\delta ,P_{\overline{\mathcal{R}\left(A\right)}}\xi \rangle ,\forall \delta ,\xi \in \mathcal{H}.$$

It is obvious that ${\parallel ·\parallel }_{\mathbf{R}(A^{1/2})}$ stands for the norm induced by the inner product ${\langle ·,·\rangle }_{\mathbf{R}(A^{1/2})}$. We mention here that, in view of Proposition 2.1. in [15], we have $\mathcal{R}\left(A\right)$ is dense in $\mathbf{R}(A^{1/2})$. Further, since $\mathcal{R}\left(A\right)\subseteq \mathcal{R}(A^{1/2})$, then we observe that

$$\begin{array}{c}\hfill {\langle A\delta ,A\xi \rangle }_{\mathbf{R}(A^{1/2})}={\langle \delta ,\xi \rangle }_A,\forall \delta ,\xi \in \mathcal{H},\end{array}$$

(38)

which implies that

$${\parallel A\xi \parallel }_{\mathbf{R}(A^{1/2})}={\parallel \xi \parallel }_A,\forall x\in \mathcal{H}.$$

(39)

In the next proposition, we give an interesting connection between operators in ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ and operators in $\mathbb{B}(\mathbf{R}(A^{1/2}))$. Furthermore, we summarize in the same proposition some useful properties.

Proposition 3 

([14,15,19,28]). Let $T\in \mathbb{B}\left(\mathcal{H}\right)$. Then there exists $\tilde{T}\in \mathbb{B}\left(\mathbf{R}\left(A^{1/2}\right)\right)$ such that $\tilde{T}{Z}_A=Z_{A}T$ if and only if $T\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$. In such cases $\tilde{T}$ is unique. Here $Z_A:\mathcal{H}\to \mathbf{R}\left(A^{1/2}\right)$ is defined by $Z_{A}x=Ax$. Furthermore, for every $T,S\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ and $\alpha ,\beta \in \mathbb{C}$, we have the following properties:

(1) 

${\parallel T\parallel }_A={\parallel \tilde{T}\parallel }_{\mathbb{B}(\mathbf{R}(A^{1/2}))}$ and $r_A\left(T\right)=r(\tilde{T})$.

(2) 

If $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$, then $\widetilde{T^{♯A}}={(\tilde{T})}^{*}$ and $\tilde{{\left(T^{{♯}_A}\right)}^{{♯}_A}}=\tilde{T}.$

(3) 

If $T_k\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ and ${\eta }_k\in \mathbb{C}$ for all $k\in \{1,\dots ,d\}$, then we have

(i) 

${\displaystyle \sum _{k=1}^d{\eta }_k{T}_k\in \mathbb{B}(\mathbf{R}(A^{1/2}))}$ and ${\displaystyle \widetilde{{\displaystyle \sum _{k=1}^d{\eta }_k{T}_k}}=\sum _{k=1}^d{\eta }_k\widetilde{T_k}}$;

(ii) 

${\displaystyle \prod _{k=1}^d{\eta }_k{T}_k\in \mathbb{B}(\mathbf{R}(A^{1/2}))}$ and ${\displaystyle \widetilde{{\displaystyle \prod _{k=1}^d{\eta }_k{T}_k}}=\prod _{k=1}^d{\eta }_k\widetilde{T_k}}$.

(4) 

If $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ such that $T_i{T}_j=T_j{T}_i$ for all $i,j\in \{1,\dots ,d\}$, then $r_A\left(\mathbf{T}\right)=r(\tilde{\mathbf{T}})$, where $\tilde{\mathbf{T}}=(\widetilde{T_1},\dots ,\widetilde{T_d})$.

The following lemma is also useful in proving our results in this section.

Lemma 3. 

Let $T,S\in {\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$. Then, the following assertions hold:

(i) 

If $T{\ge }_{A}S$, then $\tilde{T}\ge \tilde{S}$;

(ii) 

If $T\in {\mathbb{B}}_A\left(\mathcal{H}\right)$ is A-hyponormal, then $\tilde{T}$ is hyponormal on $\mathbf{R}(A^{1/2})$.

Proof. 

(1) Let $x\in \mathcal{H}$. By applying (38) together with Proposition 3 (3), we have

$$\begin{array}{cc}\hfill {〈(T-S)x,x〉}_A& ={〈A(T-S)x,Ax〉}_{\mathbf{R}(A^{1/2})}\hfill \\ \hfill & ={\langle \widetilde{(T-S)}Ax,Ax\rangle }_{\mathbf{R}(A^{1/2})}\hfill \\ \hfill & ={\langle (\tilde{T}-\tilde{S})Ax,Ax\rangle }_{\mathbf{R}(A^{1/2})}\ge 0,\hfill \end{array}$$

where the last inequality follows by using the fact that $T{\ge }_{A}S$. Now, by using the fact that $\mathcal{R}\left(A\right)$ is dense in $\mathbf{R}(A^{1/2})$, we can check that

$${\langle (\tilde{T}-\tilde{S})A^{1/2}x,A^{1/2}x\rangle }_{\mathbf{R}(A^{1/2})}\ge 0,\forall x\in \mathcal{H}.$$

Hence, $\tilde{T}-\tilde{S}$ is a positive operator on the Hilbert space $\mathbf{R}(A^{1/2})$. Therefore, $\tilde{T}\ge \tilde{S}$ as required.

(2) Since T is A-hyponormal, then $T^{{♯}_A}T{\ge }_{A}T{T}^{{♯}_A}$. Hence, by the first assertion of this lemma, we deduce that $\widetilde{T^{{♯}_A}T}\ge \widetilde{T{T}^{{♯}_A}}$. By Proposition 3, this implies that ${(\tilde{T})}^{*}\tilde{T}\ge \tilde{T}{(\tilde{T})}^{*}$. Therefore, $\tilde{T}$ is an hyponormal operator on $\mathbf{R}(A^{1/2})$. □

Now, we are able to prove the following result.

Theorem 3. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ be an A-doubly-commuting d-tuple of A-hyponormal operators. Then, the following equality

$${\parallel \mathbf{T}\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_A$$

(40)

holds.

Proof. 

Since $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d\subseteq {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$, then by applying Proposition 3 we deduce that for each $k\in \{1,\dots ,d\}$ there exists $\widetilde{T_k}\in \mathbb{B}(\mathbf{R}(A^{1/2}))$ such that $Z_A{T}_k=\widetilde{T_k}{Z}_A$. Another application of Proposition 3 shows that

$$\begin{array}{cc}\hfill {\parallel \mathbf{T}\parallel }_{e,A}& {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\parallel {\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d\parallel }_A}\hfill \\ \hfill & {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\parallel \widetilde{{\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d}\parallel }_{\mathbb{B}(\mathbf{R}(A^{1/2}))}}\hfill \\ \hfill & {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\parallel {\lambda }_1\widetilde{T_1}+\dots +{\lambda }_d\widetilde{T_d}\parallel }_{\mathbb{B}(\mathbf{R}(A^{1/2}))},}\hfill \end{array}$$

and hence,

$${\parallel \mathbf{T}\parallel }_{e,A}={\parallel \tilde{\mathbf{T}}\parallel }_e,$$

(41)

where $\tilde{\mathbf{T}}=(\widetilde{T_1},\dots ,\widetilde{T_d})$. On the other hand, we observe that the A-joint seminorm of $\mathbf{T}$ can be written as: ${\parallel \mathbf{T}\parallel }_A={sup\{\parallel \mathit{\lambda }\parallel }_2;\mathit{\lambda }=({\lambda }_1,\dots ,{\lambda }_d)\in {\mathsf{\Omega }}_A\left(\mathbf{T}\right)\}$ where

$${\mathsf{\Omega }}_A\left(\mathbf{T}\right):=\left\{\left(\parallel T_1{x\parallel }_A,\dots ,{\parallel T_{d}x\parallel }_A\right);x\in \mathcal{H},{\parallel x\parallel }_A=1\right\}.$$

If $A=I$, we simply denote ${\mathsf{\Omega }}_I(·)$ by $\mathsf{\Omega }(·)$. In particular, we observe that

$$\mathsf{\Omega }(\tilde{\mathbf{T}})=\left\{(\parallel \widetilde{T_1}{y\parallel }_{\mathbf{R}(A^{1/2})},\dots ,{\parallel \widetilde{T_d}y\parallel }_{\mathbf{R}(A^{1/2})});y\in \mathbf{R}(A^{1/2}),{\parallel y\parallel }_{\mathbf{R}(A^{1/2})}=1\right\}.$$

Now, by using the decomposition $\mathcal{H}=\mathcal{N}\left(A\right)\oplus \overline{\mathcal{R}\left(A^{1/2}\right)}$ together with (39), we obtain

$$\begin{array}{cc}\hfill {\mathsf{\Omega }}_A\left(\mathbf{T}\right)& =\left\{\left(\parallel T_1{x\parallel }_A,\dots ,{\parallel T_{d}x\parallel }_A\right);x\in \overline{\mathcal{R}\left(A^{1/2}\right)};{\parallel x\parallel }_A=1\right\}\hfill \\ \hfill & =\left\{(\parallel A{T}_1{x\parallel }_{\mathbf{R}(A^{1/2})},\dots ,{\parallel A{T}_{d}x\parallel }_{\mathbf{R}(A^{1/2})});x\in \overline{\mathcal{R}(A^{1/2})},{\parallel Ax\parallel }_{\mathbf{R}(A^{1/2})}=1\right\},\hfill \end{array}$$

from which

$${\mathsf{\Omega }}_A\left(\mathbf{T}\right)=\left\{(\parallel \widetilde{T_1}{Ax\parallel }_{\mathbf{R}(A^{1/2})},\dots ,{\parallel \widetilde{T_d}Ax\parallel }_{\mathbf{R}(A^{1/2})});x\in \overline{\mathcal{R}(A^{1/2})},{\parallel Ax\parallel }_{\mathbf{R}(A^{1/2})}=1\right\}.$$

(42)

This immediately implies that

$${\mathsf{\Omega }}_A\left(\mathbf{T}\right)\subseteq \mathsf{\Omega }(\tilde{\mathbf{T}})$$

(43)

Furthermore, it can be seen that

$$\begin{array}{cc}\hfill & \mathsf{\Omega }(\tilde{\mathbf{T}})\hfill \\ \hfill & =\left\{\left(\parallel \widetilde{T_1}{A}^{1/2}{x\parallel }_{\mathbf{R}(A^{1/2})},\dots ,{\parallel \widetilde{T_d}{A}^{1/2}x\parallel }_{\mathbf{R}(A^{1/2})}\right);x\in \mathcal{H},{\parallel A^{1/2}x\parallel }_{\mathbf{R}(A^{1/2})}=1\right\}\hfill \\ \hfill & =\left\{\left(\parallel \widetilde{T_1}{A}^{1/2}{x\parallel }_{\mathbf{R}(A^{1/2})},\dots ,{\parallel \widetilde{T_d}{A}^{1/2}x\parallel }_{\mathbf{R}(A^{1/2})}\right);x\in \overline{\mathcal{R}(A^{1/2})},{\parallel A^{1/2}x\parallel }_{\mathbf{R}(A^{1/2})}=1\right\}.\hfill \end{array}$$

Now, let $\mathbf{\lambda }=({\lambda }_1,\dots ,{\lambda }_d)\in \mathsf{\Omega }(\tilde{\mathbf{T}})$. Then, there exists $x\in \overline{\mathcal{R}\left(A^{1/2}\right)}$ satisfying

$$\parallel A^{1/2}{x\parallel }_{\mathbf{R}(A^{1/2})}=1\mathrm{and}{\lambda }_i={\parallel \widetilde{T_i}{A}^{1/2}x\parallel }_{\mathbf{R}(A^{1/2})},\forall i\in \{1,\dots ,d\}.$$

(44)

Since the subspace $\mathcal{R}\left(A\right)$ is dense in $\mathbf{R}(A^{1/2})$, then there exists a sequence $\left\{{\xi }_n\right\}$, which may be assumed to be in $\overline{\mathcal{R}(A^{1/2})}$ (because of the fact that $\mathcal{H}=\mathcal{N}\left(A\right)\oplus \overline{\mathcal{R}(A^{1/2})}$) such that ${\displaystyle A^{1/2}x=\underset{n\to +\infty }{lim}A{\xi }_n}$. Thus, by taking (44) into consideration, it holds that

$$\underset{n\to +\infty }{lim}\parallel A{\xi }_n{\parallel }_{\mathbf{R}(A^{1/2})}=1\mathrm{and}{\lambda }_i=\underset{n\to +\infty }{lim}{\parallel \widetilde{T_i}A{\xi }_n\parallel }_{\mathbf{R}(A^{1/2})},$$

(45)

for all $i\in \{1,\dots ,d\}$. Now, set ${\theta }_n:=\frac{{\xi }_n}{\parallel A{\xi }_n{\parallel }_{\mathbf{R}(A^{1/2})}}$. Clearly, we have $\parallel A{\theta }_n{\parallel }_{\mathbf{R}(A^{1/2})}=1$. Further, by using (45), it can be checked that

$$\underset{n\to +\infty }{lim}{\parallel \widetilde{T_i}A{\theta }_n\parallel }_{\mathbf{R}(A^{1/2})}={\lambda }_i,\forall i\in \{1,\dots ,d\}.$$

This implies that, through (42), $\mathbf{\lambda }\in \overline{{\mathsf{\Omega }}_A\left(\mathbf{T}\right)}$ and, therefore,

$$\mathsf{\Omega }(\tilde{\mathbf{T}})\subseteq \overline{{\mathsf{\Omega }}_A\left(\mathbf{T}\right)}.$$

(46)

From (43) and (46), we deduce that $\overline{\mathsf{\Omega }(\tilde{\mathbf{T}})}=\overline{{\mathsf{\Omega }}_A\left(\mathbf{T}\right)}$. Therefore, we infer that

$${\parallel \mathbf{T}\parallel }_A=\parallel \tilde{\mathbf{T}}\parallel .$$

(47)

On the other hand, since $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ is an A-doubly-commuting d-tuple of A-hyponormal operators, then

$$T_i{T}_j=T_j{T}_i,\forall i,j\in \{1,\dots ,d\},T_i^{{♯}_A}{T}_j=T_j{T}_i^{{♯}_A},\forall 1\le i\ne j\le d.$$

$$\mathrm{and}{T}_k^{{♯}_A}{T}_k{\ge }_A{T}_k{T}_k^{{♯}_A},\forall k\in \{1,\dots ,d\}.$$

Therefore, by applying Proposition 3 together with Lemma 3 (ii), we have

$$\widetilde{T_i}\widetilde{T_j}=\widetilde{T_j}\widetilde{T_i},\forall i,j\in \{1,\dots ,d\}\mathrm{and}{(\widetilde{T_i})}^{*}\widetilde{T_j}=\widetilde{T_j}{(\widetilde{T_i})}^{*},\forall 1\le i\ne j\le d.$$

$$\mathrm{and}{(\widetilde{T_k})}^{*}\widetilde{T_k}\ge \widetilde{T_k}{(\widetilde{T_k})}^{*},\forall k\in \{1,\dots ,d\}.$$

Hence, $\tilde{\mathbf{T}}=(\widetilde{T_1},\dots ,\widetilde{T_d})$ is a d-tuple of doubly-commuting hyponormal operators on the Hilbert space $\mathbf{R}(A^{1/2})$. Therefore, by (13), we have

$$\parallel \tilde{\mathbf{T}}\parallel =\parallel \tilde{\mathbf{T}}{\parallel }_e.$$

This completes the proof by taking (41) and (47) into consideration. □

Remark 7. 

Note that the converse of Theorem 3 need not be correct as shown in the next example.

Example 1. 

Let us consider the same matrices in ${\mathcal{M}}_3$ given in Remark 1, i.e., $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 1\end{array}\right)$, $T_1=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$ and $T_2=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$. By Remark 1, we have $\parallel (T_1,T_2){\parallel }_A=1$. Further, we see that

$$\parallel (T_1,T_2){\parallel }_{e,A}=\underset{({\lambda }_1,{\lambda }_2)\in {\mathbb{B}}_2}{sup}{∥{\lambda }_1{T}_1+{\lambda }_2{T}_2∥}_A=\underset{({\lambda }_1,{\lambda }_2)\in {\overline{\mathbb{B}}}_2}{sup}{∥{\lambda }_1{T}_1+{\lambda }_2{T}_2∥}_A,$$

(48)

where ${\overline{\mathbb{B}}}_2$ means the closed unit ball of ${\mathbb{C}}^2$. So, by using (4) and making direct calculations, we show that

$$\parallel (T_1,T_2){\parallel }_{e,A}=\underset{|{\lambda }_1{|}^2+{\left|{\lambda }_1\right|}^2\le 1}{sup}\left(\underset{{\left|x\right|}^2+{\left|y\right|}^2+{\left|z\right|}^2=1}{sup}\left|\frac{{\lambda }_1}{\sqrt{2}}y+{\lambda }_{2}z\right|\right).$$

By making use of the Cauchy–Schwarz inequality, it can be easily checked that $\parallel (T_1,T_2){\parallel }_{e,A}\le 1$. On the other hand, by using (48) and then (4), we see that

$$\parallel (T_1,T_2){\parallel }_{e,A}\ge {\parallel T_2\parallel }_A=1,$$

from which $\parallel (T_1,T_2){\parallel }_{e,A}=1$. So,

$$\parallel (T_1,T_2){\parallel }_A={\parallel (T_1,T_2)\parallel }_{e,A}=1.$$

However, it can be verified that $T_1^{{♯}_A}{T}_2\ne T_2{T}_1^{{♯}_A}$ and this $(T_1,T_2)$ is not an A-doubly-commuting 2-tuple of A-hyponormal operators.

Remark 8. 

According to our proof in Theorem 3, we remark that the equality

$$\parallel (T_1,\dots ,T_d){\parallel }_A=\parallel (\widetilde{T_1},\dots ,\widetilde{T_d})\parallel ,$$

(49)

holds for every d-tuple of operators $(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$. Here, $\widetilde{T_k}\in \mathbb{B}(\mathbf{R}(A^{1/2}))$ and verify $Z_A{T}_k=\widetilde{T_k}{Z}_A$ for all $k\in \{1,\dots ,d\}$. Note that (49) provides an improvement of a result of the second author in [19] since ${\mathbb{B}}_A\left(\mathcal{H}\right)$ is in general a proper subspace of ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$.

In order to derive an important consequence from Theorem 3, we first introduce the following definition that is inspired by the work of G. Popescu [16].

Definition 5. 

For $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$, we define a new A-joint numerical radius and a new A-joint spectral radius of $\mathbf{T}$ by setting

$${\displaystyle r_{e,A}\left(\mathbf{T}\right)=\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{r}_A({\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d),}$$

(50)

and

$${\displaystyle {\omega }_{e,A}\left(\mathbf{T}\right)=\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\omega }_A({\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d).}$$

Remark 9. 

If $A=I$, then $r_{e,I}(·)$ will simply be denoted by $r_e(·)$. Further, it is worth mentioning that the equality

$$r\left(\mathbf{T}\right)=r_e\left(\mathbf{T}\right)$$

(51)

holds for every commuting operator tuple $\mathbf{T}=(T_1,\dots ,T_d)\in \mathbb{B}{\left(\mathcal{H}\right)}^d$ (see Theorem 2.1 in [29] or [2]).

Now, as an application of Theorem 3, we state the following result.

Theorem 4. 

Let $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ be an A-doubly-commuting d-tuple of A-hyponormal operators. Then

$$r_A\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_A={\omega }_A\left(\mathbf{T}\right).$$

Proof. 

Since $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d$ is a d-tuple of A-doubly-commuting A-hyponormal operators, then in particular we have

$$T_k^{{♯}_A}{T}_l=T_l{T}_k^{{♯}_A},\forall 1\le k\ne l\le d\mathrm{and}{T}_m^{{♯}_A}{T}_m{\ge }_A{T}_m{T}_m^{{♯}_A},$$

(52)

for every $m\in \{1,\dots ,d\}.$ Now, for $\mathbf{\lambda }=({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{C}}^d$, we let $S_{\mathbf{\lambda }}:={\sum }_{m=1}^d{\lambda }_m{T}_m$. Clearly, $S_{\mathbf{\lambda }}\in {\mathbb{B}}_A\left(\mathcal{H}\right)$. By making simple calculations and using (52), we see that

$$\begin{array}{cc}\hfill S_{\mathbf{\lambda }}^{{♯}_A}{S}_{\mathbf{\lambda }}& =\sum _{i=1}^d\sum _{j=1}^d{\lambda }_j\overline{{\lambda }_i}{T}_i^{{♯}_A}{T}_j\hfill \\ \hfill & =\sum _{i=1}^d{\left|{\lambda }_i\right|}^2{T}_i^{{♯}_A}{T}_i+\sum _{i=1}^d\sum _{\genfrac{}{}{0pt}{}{j=1,}{j\ne i}}^d{\lambda }_j\overline{{\lambda }_i}{T}_i^{{♯}_A}{T}_j\hfill \\ \hfill & {\ge }_A\sum _{i=1}^d{\left|{\lambda }_i\right|}^2{T}_i{T}_i^{{♯}_A}+\sum _{i=1}^d\sum _{\genfrac{}{}{0pt}{}{j=1,}{j\ne i}}^d{\lambda }_j\overline{{\lambda }_i}{T}_j{T}_i^{{♯}_A}=S_{\lambda }{S}_{\lambda }^{{♯}_A}.\hfill \end{array}$$

Hence, $S_{\mathbf{\lambda }}$ is an A-hyponormal operator. This implies that, by (7),

$$r_A\left(S_{\mathbf{\lambda }}\right)={\omega }_A\left(S_{\mathbf{\lambda }}\right)={\parallel S_{\mathbf{\lambda }}\parallel }_A,$$

for all $\mathbf{\lambda }=({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{C}}^d$. If we take the supremum over all $\lambda \in {\mathbb{B}}_d$ in the last equalities, then we obtain

$$r_{e,A}\left(\mathbf{T}\right)={\omega }_{e,A}\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_{e,A}.$$

(53)

Taking into account relation (25), we have

$$\begin{array}{cc}\hfill {\omega }_A\left(\mathbf{T}\right)& {\displaystyle =\underset{\begin{array}{c}x\in \mathcal{H},\\ {\parallel x\parallel }_A=1\end{array}}{sup}\sqrt{\sum _{j=1}^d{\left|{\langle T_{j}x,x\rangle }_A\right|}^2}=\underset{\begin{array}{c}x\in \mathcal{H},\\ {\parallel x\parallel }_A=1\end{array}}{sup}\left(\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left|\sum _{j=1}^d{\lambda }_j{\langle T_{j}x,x\rangle }_A\right|\right).}\hfill \end{array}$$

This implies that

$$\begin{array}{cc}\hfill {\omega }_A\left(\mathbf{T}\right)& {\displaystyle =\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}\left[\underset{\begin{array}{c}x\in \mathcal{H},\\ {\parallel x\parallel }_A=1\end{array}}{sup}\left|{\langle \left(\sum _{j=1}^d{\lambda }_j{T}_j\right)x,x\rangle }_A\right|\right]=\underset{({\lambda }_1,\cdots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\omega }_A\left(\sum _{j=1}^d{\lambda }_j{T}_j\right).}\hfill \end{array}$$

This proves

$${\omega }_A\left(\mathbf{T}\right)={\omega }_{e,A}\left(\mathbf{T}\right).$$

(54)

In view of Theorem 3, we have ${\parallel \mathbf{T}\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_A$. So, all that remains to be proven is that $r_A\left(\mathbf{T}\right)=r_{e,A}\left(\mathbf{T}\right)$. Since we have $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_A{\left(\mathcal{H}\right)}^d\subseteq {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$, then by applying Proposition 3 we deduce that for each $j\in \{1,\dots ,d\}$ there exists $\widetilde{T_j}\in \mathbb{B}(\mathbf{R}(A^{1/2}))$ such that $Z_A{T}_j=\widetilde{T_j}{Z}_A$. Hence, another application of Proposition 3 shows that

$$\begin{array}{cc}\hfill r_{e,A}\left(\mathbf{T}\right)& {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{r}_A\left({\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d\right)}\hfill \\ \hfill & {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}r\left(\widetilde{{\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d}\right)}\hfill \\ \hfill & {\displaystyle =\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}r\left({\lambda }_1\widetilde{T_1}+\dots +{\lambda }_d\widetilde{T_d}\right),}\hfill \end{array}$$

and so,

$$r_{e,A}\left(\mathbf{T}\right)=r_e(\tilde{\mathbf{T}}),\mathrm{where}\tilde{\mathbf{T}}=(\widetilde{T_1},\dots ,\widetilde{T_d}).$$

(55)

Since $\mathbf{T}$ is an A-doubly-commuting operator tuple, then it is commuting. So, similar to the proof of Theorem 3, we find that $\tilde{\mathbf{T}}$ is a commuting d-tuple of operators in the Hilbert space $\mathbf{R}(A^{1/2})$. Therefore, by (51), we conclude that $r_e(\tilde{\mathbf{T}})=r(\tilde{\mathbf{T}})$. Further, by Proposition 3 (4), we have $r_A\left(\mathbf{T}\right)=r(\tilde{\mathbf{T}})$. Hence, by taking (55) into account, we deduce that

$$r_A\left(\mathbf{T}\right)=r_{e,A}\left(\mathbf{T}\right),$$

(56)

as desired. Thus, combining (53) with (54), (56) and (40) yields the desired result and the proof is complete. □

5. Conclusions

In this paper, we introduced a definition that is a generalization of (12). For $\mathbf{T}=(T_1,\dots ,T_d)\in {\mathbb{B}}_{A^{1/2}}{\left(\mathcal{H}\right)}^d$, the A-Euclidean seminorm of $\mathbf{T}$ is given by

$${\displaystyle {\parallel \mathbf{T}\parallel }_{e,A}:=\underset{({\lambda }_1,\dots ,{\lambda }_d)\in {\mathbb{B}}_d}{sup}{\parallel {\lambda }_1{T}_1+\dots +{\lambda }_d{T}_d\parallel }_A.}$$

Consequently, our objective was to study a new joint norm of tuples of operators which generalizes the joint norm given in (12) and define the class of doubly-commuting tuples of hyponormal operators acting on an A-weighted Hilbert space, where A is a positive operator that is not assumed to be invertible. The motivation for our investigation comes from the recent paper [17].

This article was structured as follows: In Section 3, we investigated a new joint seminorm for d-tuples of A-bounded operators. An alternative and easy proof of a well-known result due to G. Popescu [16] was established. In Section 4, we give several characterizations related to the operators from ${\mathbb{B}}_{A^{1/2}}\left(\mathcal{H}\right)$ and the operators from $\mathbb{B}\left(\mathbf{R}\left(A^{1/2}\right)\right)$. For the A-doubly-commuting d-tuple of hyponormal operators, we proved the equalities ${\parallel \mathbf{T}\parallel }_{e,A}={\parallel \mathbf{T}\parallel }_A$ and $r_A\left(\mathbf{T}\right)={\parallel \mathbf{T}\parallel }_A={\omega }_A\left(\mathbf{T}\right).$

In this paper, the ideas and methodologies used may serve as a starting point for future studies in this field. We will look for other connections of these seminorms for d-tuples of A-bounded operators by studying other possible characterizations. In future work, we will generalize the results given a countable collection of operators.