On the Sequential (p, δ, τ)-Numerical Radius Function of Operator Sequence

Abstract

In this article, firstly, some basic properties of the sequential $(p,\delta ,\tau )$-numerical radius function are investigated. The relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its coordinate operators are analyzed. Then, the relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its real and imaginary parts are presented. Finally, this analysis is extended to the case in which the coordinate operators are sectorial, providing additional insight into the structural behavior of the sequential $(p,\delta ,\tau )$-numerical radius function. The obtained results are generalized to some well-known famous results about the numerical radius function from the recent literature. Also, an important contribution is made to the existing literature via different and useful results.

1. Introduction

The concepts of numerical range and numerical radius of an operator play a very significant role in pure and applied mathematics and have been studied extensively due to their applications in engineering, quantum computing, quantum mechanics, numerical analysis, differential equations, fluid dynamics, the geometry of Banach space, etc. (see, [1,2,3]).

Throughout this article, $\mathcal{H}$ denotes a complex Hilbert space endowed with the inner product $\langle ·,·\rangle$ and associated norm $\parallel ·\parallel .$ Let $\mathfrak{B}\left(\mathcal{H}\right)$ and ${\mathfrak{B}}_{\infty }\left(\mathcal{H}\right)$ stand for the $C^{\ast }$-algebra of all bounded linear operators and compact operators acting on $\mathcal{H},$ respectively. The classical numerical radius and the usual operator norm of a linear bounded operator T are, respectively, given by

$$\omega \left(T\right)=sup\left\{\right|\langle Tx,x\rangle |:x\in \mathcal{H},\parallel x\parallel =1\}$$

(see, [4]) and

$$\parallel T\parallel =sup\{\parallel Tx\parallel :x\in \mathcal{H},\parallel x\parallel =1\}.$$

In the literature there is one important formula from Yamazaki [5] for the calculation of numerical radius $\omega \left(T\right)$ of $T\in \mathfrak{B}\left(\mathcal{H}\right)$:

$$\omega \left(T\right)=\underset{t\in \mathbb{R}}{sup}\parallel Re\left(e^{it}T\right)\parallel .$$

There is another formula for the computation of generalized numerical radius:

$${\omega }_N\left(T\right)=\underset{t\in \mathbb{R}}{sup}N\left(Re\left(e^{it}T\right)\right),$$

where N is the given norm of $T\in \mathfrak{B}\left(H\right)$ [6].

Finding the exact value of $\omega \left(T\right)$ for an arbitrary $T\in \mathfrak{B}\left(\mathcal{H}\right)$ is not an easy task, except in some special cases. Therefore, researchers in this field have devoted a considerable amount of time and effort to find lower and upper bounds of $\omega \left(T\right).$

Recall that for any $T\in \mathfrak{B}\left(\mathcal{H}\right)$ the numerical radius $\omega \left(T\right)$ is a norm on $\mathfrak{B}\left(\mathcal{H}\right)$ and among the most basic bounds of $\omega \left(T\right)$ is the following two-sided inequality

$${\displaystyle \frac{1}{2}}\parallel T\parallel \le \omega \left(T\right)\le \parallel T\parallel$$

(1)

(see, e.g., [7]). The right hand side of (1) becomes an equality when T is normal, while the left side becomes an equality when $T^2=0.$

Sharpening (1) has received considerable attention in the literature. We refer the reader to [8,9,10,11,12,13,14,15,16] as a sample of recent attempts to obtain sharper and better bounds of $\omega \left(T\right)$.

In [9], the numerical radius of accretive operators (matrices) was investigated, where several bounds were established, including both extensions of known results and newly derived inequalities. This work appears to have stimulated further interest in the study of the numerical radius of accretive and related classes of operators.

On the other hand, Bhunia et al. [10] focused on refining the inequalities in (1), with the Cartesian decomposition playing a central role in these refinements. In particular, the authors employed this decomposition to improve the first inequality in the following celebrated bound

$${\displaystyle \frac{1}{4}}\parallel T^{\ast }T+T{T}^{\ast }\parallel \le {\omega }^2\left(T\right)\le {\displaystyle \frac{1}{2}}\parallel T^{\ast }T+T{T}^{\ast }\parallel ,$$

(2)

which was originally proved in [13] as a refinement of both inequalities in (1).

Earlier, it was shown in [12] that

$$\omega \left(A\right)\le {\displaystyle \frac{1}{2}}∥\left|A\right|+|A^{\ast }|∥,$$

which is among the sharpest refinements of the second inequality in (1). In fact, this inequality is sharper than the upper bound in (2); however, the symmetric structure of (2) makes it particularly appealing and widely cited.

Further developments were presented in [14], where various relations between an operator and its Cartesian decomposition were utilized to derive several improvements of both (1) and (2). In that work, the study of accretive–dissipative operators led to replacing the factor ${\displaystyle \frac{1}{2}}$ in (1) with ${\displaystyle \frac{1}{\sqrt{2}}}$, yielding a significant improvement for this class of operators. More recently, ref. [11] showed that this improved factor is in fact valid for all accretive operators, thereby strengthening numerous known results, particularly those in [8]. It is worth noting that the primary aim in [8] was to establish interpolated inequalities that provide a geometric perspective on several classical bounds.

Now, in alignement with our work, we proceed to the definitions of $\delta$-numerical range and $\delta$-numerical radius.

Definition 1

([17]). For the operator $T\in \mathfrak{B}\left(\mathcal{H}\right)$ and $0\le \delta \le \parallel T\parallel ,$ the δ-numerical range is defined as

$$W_{\delta }\left(T\right)=cl\left(\left\{\langle Tx,x\rangle :x\in \mathcal{H},\parallel x\parallel =1,\parallel Tx\parallel \ge \delta \right\}\right),$$

where $cl(·)$ denotes the closure of a set in the complex plane.

It is readily seen that $W_{\delta }\left(T\right)$ is a closed subset of the closure of the classical numerical range, and that

$$W_0\left(T\right)=\bigcap _{\delta <\parallel T\parallel }{W}_{\delta }\left(T\right).$$

Furthermore, by a slight modification of a theorem from Dekker [18], it follows that $W_{\delta }\left(T\right)$ is connected. An interesting question is whether $W_{\delta }\left(T\right)$ is convex. This is indeed the case when T is normal or when the underlying Hilbert space is two-dimensional.

Definition 2.

For the operator $T\in \mathfrak{B}\left(\mathcal{H}\right)$ and $0\le \delta \le \parallel T\parallel ,$ the δ-numerical radius function is defined as

$${\omega }_{\delta }\left(T\right)=sup\left\{\left|\lambda \right|:\lambda \in W_{\delta }\left(T\right)\right\}.$$

From the definitions of the $\delta$-numerical radius function, it follows immediately that ${\omega }_0\left(T\right)=\omega \left(T\right).$

Now, let us give the following two new definitions.

Definition 3.

Let $1\le p<\infty$ and $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ satisfying

$${\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n\parallel }^p<\infty .$$

The following definitions are introduced:

1. 

${\parallel \mathbb{A}x\parallel }_p={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}},x\in \mathcal{H};$

2. 

${\parallel \mathbb{A}\parallel }_p=sup\left\{{\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}:x\in \mathcal{H},\parallel x\parallel =1\right\};$

3. 

$\langle \mathbb{A}x,x\rangle =\left(\langle A_{n}x,x\rangle \right)\in {\mathbb{C}}^{\infty }\left(\mathbb{C}\right),x\in \mathcal{H};$

4. 

${\left|\langle \mathbb{A}x,x\rangle \right|}_p={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}},x\in \mathcal{H}.$

Definition 4.

Let $1\le p<\infty$ and $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ satisfying

$${\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n\parallel }^p<\infty .$$

The sequential $(p,\delta ,\tau )$-numerical radius of the operator sequence $\mathbb{A}=\left(A_n\right)$ is defined by

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)=sup\left\{{\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}:\parallel x\parallel =1,\delta \le {\parallel \mathbb{A}x\parallel }_p\le \tau \right\}$$

for $0\le \delta \le \tau \le {\parallel \mathbb{A}\parallel }_p.$

We note that the proposed definition recovers several well-known numerical radii as special cases:

(1)

If $\mathbb{A}=\left(A_n\right),A_m=0,m\ge 2$ and $\delta =0,\tau ={\parallel \mathbb{A}\parallel }_p=\parallel \mathbb{A}\parallel ,$ then this definition coincides with the definition of classical numerical radius (see [7]);

(2)

If $A\in \mathfrak{B}\left(\mathcal{H}\right),\mathbb{A}=(A,A^{\ast }A,0,0\dots ),p=2,\delta =0,\tau ={\parallel A\parallel }_p,$ then this definition coincides with the definition of Davis–Wielandt radius of the operator A (see [19,20]);

(3)

If $\mathbb{A}=\left(A_n\right),A_m=0,m\ge 2$ and $0\le \delta \le {\parallel \mathbb{A}\parallel }_p=\parallel \mathbb{A}\parallel$ and $\tau ={\parallel \mathbb{A}\parallel }_p=\parallel \mathbb{A}\parallel ,$ then this definition coincides with the definition of $\delta$- numerical radius (see [21]);

(4)

If $\mathbb{A}=\left(A_n\right),A_m=0,m\ge n\in \mathbb{N}$ and $p=2,\delta =0,\tau ={\parallel \mathbb{A}\parallel }_p,$ then this definition coincides with the definition of p- numerical radius (see [22]);

(5)

If $\mathbb{A}=\left(A_n\right),A_m=0,m\ge n\in \mathbb{N}$ and $p=2,0\le \delta \le {\parallel \mathbb{A}\parallel }_p,\tau ={\parallel \mathbb{A}\parallel }_p,$ then this definition coincides with the definition of joint $\delta$-numerical radius (see [23]).

Example 1.

Assume that

$$A_1=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),A_1:{\mathbb{R}}^2\to {\mathbb{R}}^2,A_2=\left(\begin{array}{cc}0& 0\\ 0& 2\end{array}\right),A_2:{\mathbb{R}}^2\to {\mathbb{R}}^2,$$

and $\mathbb{A}=(A_1,A_2,\dots ),p=2.$ Then

$${\parallel \mathbb{A}\parallel }_2=\underset{x\in S_1\left({\mathbb{R}}^2\right)}{sup}{\left({\displaystyle \parallel }{A}_1{x\parallel }^2+{\parallel A_{2}x\parallel }^2\right)}^{\frac{1}{2}}=\underset{x\in S_1\left({\mathbb{R}}^2\right)}{sup}{\left(x_1^2+4x_2^2\right)}^{\frac{1}{2}}=\underset{x\in S_1\left({\mathbb{R}}^2\right)}{sup}{\left(1+3x_2^2\right)}^{\frac{1}{2}}=2.$$

On the other hand, for $\delta \le \parallel \mathbb{A}x\parallel \le \tau \le 2$ and $x\in {\mathbb{R}}^2$ from the last relation

$${\delta }^2\le \parallel A_1{x\parallel }^2+{\parallel A_{2}x\parallel }^2\le {\tau }^2,$$

i.e.,

$${\delta }^2\le x_1^2+4x_2^2\le {\tau }^2,$$

it is obtained that

$${\displaystyle \frac{{\delta }^2-1}{3}}\le x_2^2\le {\displaystyle \frac{{\tau }^2-1}{3}}.$$

Therefore,

$$\begin{array}{cc}\hfill {\omega }_{(2,\delta ,\tau )}^2\left(\mathbb{A}\right)& =sup\left\{{\langle A_{1}x,x\rangle }^2+{\langle A_{2}x,x\rangle }^2:x\in S_1\left({\mathbb{R}}^2\right),\delta \le \parallel \mathbb{A}x\parallel \le \tau \right\}\hfill \\ \hfill & =sup\left\{x_1^4+4x_2^4:x\in S_1\left({\mathbb{R}}^2\right),\delta \le \parallel \mathbb{A}x\parallel \le \tau \right\}\hfill \\ \hfill & =sup\left\{1+7x_2^2\left(x_2^2-{\displaystyle \frac{2}{7}}\right):x\in S_1\left({\mathbb{R}}^2\right),\delta \le \parallel \mathbb{A}x\parallel \le \tau \right\}\hfill \\ \hfill & =1+7\left({\displaystyle \frac{{\tau }^2-1}{3}}\right)\left({\displaystyle \frac{{\tau }^2-1}{3}}-{\displaystyle \frac{2}{7}}\right)\hfill \\ \hfill & ={\displaystyle \frac{7{\tau }^4-20{\tau }^2+16}{3}}\hfill \\ \hfill & ={\displaystyle \frac{7}{3}}{\left({\tau }^2-{\displaystyle \frac{10}{7}}\right)}^2+{\displaystyle \frac{4}{7}}.\hfill \end{array}$$

Throughout this paper, we consider operator sequences $\mathbb{A}=\left(A_n\right)\subset \mathfrak{B}\left(\mathcal{H}\right)$ and adopt the following conventions for the parameters $\delta$ and $\tau$:

(1)

If $\delta <0$ and $0\le \tau \le {\parallel \mathbb{A}\parallel }_p,$ then ${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\omega }_{(p,0,\tau )}\left(\mathbb{A}\right).$

(2)

If $0\le \delta \le {\parallel \mathbb{A}\parallel }_p$ and ${\tau >\parallel \mathbb{A}\parallel }_p,$ then ${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\omega }_{(p,\delta ,\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{A}\right).$

(3)

If $\delta <0$ and ${\tau >\parallel \mathbb{A}\parallel }_p,$ then ${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\omega }_{(p,0,\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{A}\right).$

Also for any operator sequences $\mathbb{A}=\left(A_n\right)\subset \mathfrak{B}\left(\mathcal{H}\right),1\le p<\infty$ and $t,s\in \mathbb{R}$, $-\infty <t\le {\parallel \mathbb{A}\parallel }_p$, $0\le s<\infty$ it will be defined

$${\mathrm{\Delta }}_{(p,t,s)}\left(\mathbb{A}\right)=\left\{x\in S_1\left(\mathcal{H}\right):t\le \parallel \mathbb{A}x\parallel \le s\right\},$$

where $S_1\left(\mathcal{H}\right)$ denotes the unit sphere of the Hilbert space $\mathcal{H}.$

Definition 5.

For any fixed $-{\displaystyle \frac{\pi }{2}}<\gamma \le \alpha <{\displaystyle \frac{\pi }{2}},$

(i) 

$S(\gamma ,\alpha )=\left\{r{e}^{i\phi }:r\ge 0,\gamma \le \phi \le \alpha \right\}\subset \mathbb{C}$ is a sector with a vertex at the origin and angles γ and $\alpha .$

(ii) 

The operator $T:\mathcal{H}\to \mathcal{H},T\in \mathfrak{B}\left(\mathcal{H}\right)$ in any Hilbert space $\mathcal{H}$ is said to be $(\gamma ,\alpha )$-sectorial if $W\left(T\right)\subseteq S(\gamma ,\alpha ).$

(iii) 

The class of all $(\gamma ,\alpha )$-sectorial operators on Hilbert space $\mathcal{H}$ is denoted by $Se{c}_{(\gamma ,\alpha )}\left(\mathcal{H}\right).$

It is clear that, if $A\in Se{c}_{\alpha }\left(\mathcal{H}\right),0\le \alpha \le {\displaystyle \frac{\pi }{2}},$ then A is accretive. Remember that an operator $A\in \mathfrak{B}\left(\mathcal{H}\right)$ is called accretive if $Re\langle Ax,x\rangle \ge 0$ for all $x\in \mathcal{H}.$ It is called dissipative if $Im\langle Ax,x\rangle \ge 0$ for all $x\in \mathcal{H}.$

This work is organized as follows: In Section 2, fundamental properties of the sequential $(p,\delta ,\tau )$-numerical radius function are investigated. The relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its coordinate operators are then analyzed. In addition, the relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its real and imaginary parts are presented. Section 3 extends the study to the case in which the coordinate operators are sectorial.

The main goal of this work is to develop and generalize several well-known classical inequalities related to the numerical radius, as studied in the recent literature (see, e.g., [7,8,13,19,20,21,22,23,24,25,26]), with particular emphasis on obtaining refined lower and upper bounds for the numerical radius.

In this direction, we introduce a sequential $(p,\delta ,\tau )$-numerical radius framework for operator sequences, which provides a unified setting for studying numerical radius-type quantities and allows the derivation of both new inequalities and extensions of several known results.

Moreover, this approach provides a unified perspective on operator sequences and generates improved spectral estimates for the numerical radius.

To be more precise, we prove several fundamental properties of the sequential $(p,\delta ,\tau )$-numerical radius function for operator sequences. One of our main results is to establish precise relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and those of its coordinate operators. This generalizes the well-known inequality due to Drnovšek and Müller [24,25], which states that for $\mathbb{A}=(A_1,\dots ,A_n,0,0,\dots )\in \mathfrak{B}\left(\mathcal{H}\right)$,

$${\displaystyle \frac{1}{2\sqrt{n}}}\parallel \mathbb{A}\parallel \le \omega \left(\mathbb{A}\right).$$

Another main contribution of this work is to establish relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its real and imaginary parts. This generalizes the well-known inequality due to Kittaneh [13], which states that for $A\in \mathfrak{B}\left(\mathcal{H}\right)$,

$${\displaystyle \frac{1}{4}}\parallel A^{\ast }A+A{A}^{\ast }\parallel \le {\omega }^2\left(A\right)\le {\displaystyle \frac{1}{2}}\parallel A^{\ast }A+A{A}^{\ast }\parallel .$$

Furthermore, we extend our analysis to the case of sectorial coordinate operators, obtaining new structural insights and sharper estimates. The results obtained in this section generalize the inequalities due to Bedrani et al. [9], which state that for $A\in Se{c}_{\alpha }\left(\mathcal{H}\right)$, $0<\alpha <{\displaystyle \frac{\pi }{2}}$,

$$\omega \left(A\right)\le sec\left(\alpha \right)\omega \left(ReA\right)\mathrm{and}\omega \left(A\right)\le csc\left(\alpha \right)\omega \left(ImA\right).$$

Moreover, our results improve the inequalities due to Sammour et al. [8], which state that for $A\in Se{c}_{\alpha }\left(\mathcal{H}\right)$, $0\le \alpha <{\displaystyle \frac{\pi }{2}}$,

$$\parallel A\parallel \le \sqrt{1+{sin}^2\left(\alpha \right)}\omega \left(A\right),$$

and which state that for $A,B\in \mathfrak{B}\left(\mathcal{H}\right)$ satisfying

$$W\left(A\right),W\left(B\right)\subset \left\{r{e}^{i\phi }:r\ge 0,0<\gamma <\alpha <{\displaystyle \frac{\pi }{2}}\right\},$$

$$\omega \left(AB\right)\le ({cos}^2\gamma +{sin}^2\alpha )\omega \left(A\right)\omega \left(B\right).$$

The main advantage of the sequential $(p,\delta ,\tau )$-numerical radius framework is that it gives a unified way to study different numerical radius-type quantities for operator sequences. It also allows us to recover many known results as special cases and to obtain new inequalities and improved estimates.

As a consequence, the results presented in this paper both generalize and significantly improve upon previously known results, while also contributing new and useful techniques to the study of numerical radius-type functions.

Note that each operator $T\in \mathfrak{B}\left(\mathcal{H}\right)$ can be expressed in the Cartesian decomposition form as $T=ReT+iImT,$ where $ReT={\displaystyle \frac{T+T^{\ast }}{2}}$ and $ImT={\displaystyle \frac{T-T^{\ast }}{2i}}.$ Here, $T^{\ast }$ denotes the adjoint of $T.$ Throughout this study we denote by $\left|T\right|={\left(T^{\ast }T\right)}^{1/2}$ the absolute value of an operator $T\in \mathfrak{B}\left(\mathcal{H}\right).$

2. Sequential $\mathbf{(}\mathit{p}\mathbf{,}\mathsf{\delta }\mathbf{,}\mathsf{\tau }\mathbf{)}$-Numerical Radius

In this section, some important properties of the sequential $(p,\delta ,\tau )$-numerical radius function will be investigated.

Theorem 1.

Let $\mathbb{A}=\left(A_n\right),\mathbb{B}=\left(B_n\right)$ be two operator sequences in $\mathfrak{B}\left(\mathcal{H}\right).$

(1) 

If $\delta \le {\parallel \lambda \mathbb{A}\parallel }_p\le \tau$, $\lambda \in \mathbb{C},\lambda \ne 0$, then

$${\omega }_{(p,\delta ,\tau )}\left(\lambda \mathbb{A}\right)=\left|\lambda \right|{\omega }_{\left(p,\frac{\delta }{\left|\lambda \right|},\frac{\tau }{\left|\lambda \right|}\right)}\left(\mathbb{A}\right).$$

(2) 

If $\delta \le {\parallel \mathbb{A}+\mathbb{B}\parallel }_p\le \tau ,$ then

$${\omega }_{(p,\delta ,\tau )}(\mathbb{A}+\mathbb{B})\le {\omega }_{\left(p,\delta -{\parallel \mathbb{B}\parallel }_p,\tau +{\parallel \mathbb{B}\parallel }_p\right)}\left(\mathbb{A}\right)+{\omega }_{\left(p,\delta -{\parallel \mathbb{A}\parallel }_p,\tau +{\parallel \mathbb{A}\parallel }_p\right)}\left(\mathbb{B}\right).$$

(3) 

If $\mathbb{A}=\left(A_n\right)$ is a normal operator sequence in $\mathfrak{B}\left(\mathcal{H}\right),$ then for any $\delta \le {\parallel \mathbb{A}\parallel }_p\le \tau ,$

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\omega }_{(p,\delta ,\tau )}\left({\mathbb{A}}^{\ast }\right).$$

(4) 

If $\mathbb{A}=\left(A_n\right)$ is a hyponormal operator sequence in $\mathfrak{B}\left(\mathcal{H}\right),$ then for any $0\le \delta \le \parallel {\mathbb{A}}^{\ast }{\parallel }_f$

$${\omega }_{(p,\delta ,\tau )}\left({\mathbb{A}}^{\ast }\right)\le {\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right).$$

Proof. 

(1)

For any $x\in \mathcal{H},$ it is clear that

$${\left|\langle \lambda \mathbb{A}x,x\rangle \right|}_p={\left|\lambda \right|\left|\langle \mathbb{A}x,x\rangle \right|}_p$$

(3)

On the other hand, for $x\in {\mathrm{\Delta }}_{(p,\delta ,\lambda )}\left(\lambda \mathbb{A}\right),\lambda \in \mathbb{C}\setminus \left\{0\right\},$ i.e.,

$$\delta \le {\parallel \lambda \mathbb{A}x\parallel }_p\le \tau$$

it is obtained that

$${\displaystyle \frac{\delta }{\left|\lambda \right|}}\le {\parallel \mathbb{A}x\parallel }_p\le {\displaystyle \frac{\tau }{\left|\lambda \right|}}.$$

This means that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\lambda \mathbb{A}\right)={\mathrm{\Delta }}_{\left(p,\frac{\delta }{\left|\lambda \right|},\frac{\tau }{\left|\lambda \right|}\right)}\left(\mathbb{A}\right).$$

(4)

Therefore, from (3) and (4) it is found that

$${\omega }_{(p,\delta ,\tau )}\left(\lambda \mathbb{A}\right)=\left|\lambda \right|{\omega }_{\left(p,\frac{\delta }{\left|\lambda \right|},\frac{\tau }{\left|\lambda \right|}\right)}\left(\mathbb{A}\right),$$

for $\delta \le {\parallel \lambda \mathbb{A}\parallel }_p\le \tau .$

(2)

For any $x\in \mathcal{H},$ it is easy to see that

$${\left|\langle (\mathbb{A}+\mathbb{B})x,x\rangle \right|}_p\le {\left|\langle \mathbb{A}x,x\rangle \right|}_p+{\left|\langle \mathbb{B}x,x\rangle \right|}_p.$$

(5)

On the other hand, if $x\in S_1\left(\mathcal{H}\right)$ and $\delta \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_p\le \tau ,$ then

$$\delta \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_p\le {\parallel \mathbb{A}x\parallel }_p+{\parallel \mathbb{B}\parallel }_p$$

and

$$\delta \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_p\le {\parallel \mathbb{A}\parallel }_p+{\parallel \mathbb{B}x\parallel }_p.$$

Hence,

$$\delta -{\parallel \mathbb{B}\parallel }_p\le {\parallel \mathbb{A}x\parallel }_p\mathrm{and}\delta -{\parallel \mathbb{A}\parallel }_p\le {\parallel \mathbb{B}x\parallel }_p.$$

(6)

Also, if $x\in S_1\left(\mathcal{H}\right)$ and ${\parallel (\mathbb{A}+\mathbb{B})x\parallel }_p\le \tau ,$ then

$$\begin{array}{cc}\hfill {\parallel \mathbb{A}x\parallel }_p& \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_p+{\parallel \mathbb{B}\parallel }_p\le \tau +{\parallel \mathbb{B}\parallel }_p,\hfill \\ \hfill {\parallel \mathbb{B}x\parallel }_p& \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_p+{\parallel \mathbb{A}\parallel }_p\le \tau +{\parallel \mathbb{A}\parallel }_p.\hfill \end{array}$$

(7)

Therefore, from (6) and (7) it is obtained that for $x\in S_1\left(\mathcal{H}\right)$

$$\delta -{\parallel \mathbb{B}\parallel }_p\le {\parallel \mathbb{A}x\parallel }_p\le \tau +{\parallel \mathbb{B}\parallel }_p\mathrm{and}\delta -{\parallel \mathbb{A}\parallel }_p\le {\parallel \mathbb{B}x\parallel }_p\le \tau +{\parallel \mathbb{A}\parallel }_p.$$

Consequently,

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}(\mathbb{A}+\mathbb{B})\subset {\mathrm{\Delta }}_{(p,\delta -\parallel \mathbb{B}{\parallel }_p,\tau +\parallel \mathbb{B}{\parallel }_p)}\left(\mathbb{A}\right)\cap {\mathrm{\Delta }}_{(p,\delta -\parallel \mathbb{A}{\parallel }_p,\tau +\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{B}\right).$$

(8)

Finally, from (5) and (8) it is obtained that

$${\omega }_{(p,\delta ,\tau )}(\mathbb{A}+\mathbb{B})\le {\omega }_{\left(p,\delta -{\parallel \mathbb{B}\parallel }_p,\tau +{\parallel \mathbb{B}\parallel }_p\right)}\left(\mathbb{A}\right)+{\omega }_{\left(p,\delta -{\parallel \mathbb{A}\parallel }_p,\tau +{\parallel \mathbb{A}\parallel }_p\right)}\left(\mathbb{B}\right).$$

(3)

For any $x\in \mathcal{H}$

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p={\left|\langle {\mathbb{A}}^{\ast }x,x\rangle \right|}_p.$$

(9)

As well as for any $x\in S_1\left(\mathcal{H}\right)$

$$\delta \le {\parallel \mathbb{A}x\parallel }_p={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n^{\ast }x\parallel }^p\right)}^{\frac{1}{p}}={\parallel {\mathbb{A}}^{\ast }x\parallel }_p\le \tau .$$

From these relations it is established that

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\omega }_{(p,\delta ,\tau )}\left({\mathbb{A}}^{\ast }\right).$$

(4)

From the equality (9) and

$$\delta \le \parallel {\mathbb{A}}^{\ast }{x\parallel }_p={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n^{\ast }x\parallel }^p\right)}^{\frac{1}{p}}\le {\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}={\parallel \mathbb{A}x\parallel }_p,x\in S_1\left(\mathcal{H}\right),$$

it is obtained that

$${\omega }_{(p,\delta ,\tau )}\left({\mathbb{A}}^{\ast }\right)\le {\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right).$$

□

Remark 1.

Let $\mathbb{A}=\left(A_n\right),A_m=0,m>n,p=2.$

(1) 

If $\delta =0$ and $\tau \ge {\parallel \mathbb{A}\parallel }_2,$ then Theorem 1 coincides with well-known result found in [26].

(2) 

If $0\le \delta \le {\parallel \mathbb{A}\parallel }_2$ and $\tau \ge {\parallel \mathbb{A}\parallel }_2,$ then Theorem 1 is a generalization which has a corresponding result in [23].

Now, we provide a useful auxiliary lemma.

Lemma 1

([27]). If $T\in \mathfrak{B}\left(\mathcal{H}\right),T\ge 0,$ then for any $x\in T$,

$${\parallel Tx\parallel }^2\le \parallel T\parallel \langle Tx,x\rangle .$$

Theorem 2.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $0\le \delta \le {\parallel \mathbb{A}x\parallel }_p\le \tau .$ Then

$${\omega }_{(p,\delta ,\tau )}^2\left(\mathbb{A}\right)\le a{\omega }_{(\frac{p}{2},\delta ,\tau )}\left(\right|\mathbb{A}\left|\right),$$

where $a=\underset{n\ge 1}{sup}\parallel |A_n|\parallel .$

Proof. 

For any $x\in {\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)$ it is clear that

$$\begin{array}{cc}\hfill {\delta }^p& \le {\parallel \mathbb{A}x\parallel }_p^p\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{\langle A_n^{\ast }{A}_{n}x,x\rangle }^{\frac{p}{2}}\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{|{\left(A_n^{\ast }{A}_n\right)}^{\frac{1}{2}}x|}^p\hfill \\ \hfill & ={\parallel \left|\mathbb{A}\right|x\parallel }^p\hfill \\ \hfill & \le {\tau }^p.\hfill \end{array}$$

Hence,

$$\delta \le \parallel \left|\mathbb{A}\right|x\parallel \le \tau ,$$

this shows that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\right|\mathbb{A}\left|\right).$$

(10)

On the other hand, from Lemma 1, for $x\in {\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)$ it is clear that

$$\begin{array}{cc}\hfill {\left|\langle \mathbb{A}x,x\rangle \right|}_p^p& ={\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{\langle A_n^{\ast }{A}_{n}x,x\rangle }^{\frac{p}{2}}\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{\left(\parallel |A_n{|x\parallel }^2\right)}^{\frac{p}{2}}\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}{\left(\parallel |A_n|\parallel \langle |A_n|x,x\rangle \right)}^{\frac{p}{2}}\hfill \\ \hfill & \le \underset{n\ge 1}{sup}\parallel |A_n{|\parallel }^{\frac{p}{2}}{\displaystyle \sum _{n=1}^{\infty }}\langle |A_n{|x,x\rangle }^{\frac{p}{2}}\hfill \\ \hfill & =a^{\frac{p}{2}}{|\langle |\mathbb{A}|x,x\rangle |}_{\frac{p}{2}}^{\frac{p}{2}},\hfill \end{array}$$

(11)

where $a=\underset{n\ge 1}{sup}\parallel |A_n|\parallel .$ Thus, from (10) and (11) it is obtained that

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le a^{\frac{p}{2}}{\omega }_{(\frac{p}{2},\delta ,\tau )}^{\frac{p}{2}}\left(\right|\mathbb{A}\left|\right),$$

where $a=\underset{n\ge 1}{sup}\parallel |A_n|\parallel .$ □

Example 2.

Assume that

$$A_n=\left(\begin{array}{cc}0& a_n\\ a_n& 0\end{array}\right),a_n\ge 0,n\ge 1,$$

and let $\mathbb{A}=\left(A_n\right),p=2.$ Assume also that $0\le \delta \le {\parallel \mathbb{A}x\parallel }_2\le \tau ,\underset{n\ge 1}{sup}\left|a_n\right|<\infty ,{\displaystyle \sum _{n=1}^{\infty }}{a}_n^2<\infty .$

In this case,

$$\left|A_n\right|=\left(\begin{array}{cc}{a}_n& 0\\ 0& a_n\end{array}\right),n\ge 1.$$

Then, by Theorem 2 it is obtained that

$${\omega }_{(2,\delta ,\tau )}^2\left(\mathbb{A}\right)\le \underset{n\ge 1}{sup}\left|a_n\right|{\omega }_{(1,\delta ,\tau )}\left(\right|\mathbb{A}\left|\right).$$

In a special case, if $\delta =0,\tau ={\displaystyle \sum _{n=1}^{\infty }}{a}_n^2,$ then from the last inequality we have

$${\omega }_{(2,0,\tau )}^2\left(\mathbb{A}\right)\le \underset{n\ge 1}{sup}\left|a_n\right|{\displaystyle \sum _{n=1}^{\infty }}{a}_n^2.$$

The following two theorems describe the relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its coordinate operators.

Theorem 3.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $0\le {\delta }_n\le {\tau }_n,n\ge 1.$ Then

$${\omega }_{({\delta }_n,{\tau }_n)}\left(A_n\right)\le {\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right),$$

where $\delta ={\left({\displaystyle \sum _{n=1}^{\infty }}{\delta }_n^p\right)}^{\frac{1}{p}}$ and $\tau ={\left({\displaystyle \sum _{n=1}^{\infty }}{\tau }_n^p\right)}^{\frac{1}{p}}.$

Proof. 

It is clear that for $n\ge 1$ and $x\in \mathcal{H},$

$${|\langle A_{n}x,x\rangle |}^p\le {\displaystyle \sum _{n=1}^{\infty }}{|\langle A_{n}x,x\rangle |}^p={\left|\langle \mathbb{A}x,x\rangle \right|}_p^p.$$

(12)

On the other hand, since for any $x\in {\mathrm{\Delta }}_{({\delta }_n,{\tau }_n)}\left(A_n\right),n\ge 1$

$${\displaystyle \sum _{n=1}^{\infty }}{\delta }_n^p\le {\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\le {\displaystyle \sum _{n=1}^{\infty }}{\tau }_n^p,$$

then

$${\mathrm{\Delta }}_{({\delta }_n,{\tau }_n)}\left(A_n\right)\subset {\mathrm{\Delta }}_{(\delta ,\tau )}\left(\mathbb{A}\right),$$

(13)

where $\delta ={\left({\displaystyle \sum _{n=1}^{\infty }}{\delta }_n^p\right)}^{\frac{1}{p}}$ and $\tau ={\left({\displaystyle \sum _{n=1}^{\infty }}{\tau }_n^p\right)}^{\frac{1}{p}}.$

Thus, from (12) and (13) it is obtained that for $n\ge 1$

$${\omega }_{({\delta }_n,{\tau }_n)}\left(A_n\right)\le {\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right),$$

where $\delta ={\left({\displaystyle \sum _{n=1}^{\infty }}{\delta }_n^p\right)}^{\frac{1}{p}}$ and $\tau ={\left({\displaystyle \sum _{n=1}^{\infty }}{\tau }_n^p\right)}^{\frac{1}{p}}.$ □

From Theorem 3, we obtain the next results.

Corollary 1.

Let $\mathbb{A}=\left(A_n\right),A_m=0,m>n,p=2,$ and $\delta ={\left({\displaystyle \sum _{m=1}^n}{\delta }_m^p\right)}^{\frac{1}{p}}=0$ and $\tau ={\left({\displaystyle \sum _{m=1}^n}{\tau }_m^p\right)}^{\frac{1}{p}}\ge {\parallel \mathbb{A}\parallel }_p$. Then, for any $m\ge 1$

$${\displaystyle \frac{1}{m}}{\displaystyle \sum _{n=1}^m}{\omega }_{({\delta }_n,{\tau }_n)}\left(A_n\right)\le {\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right).$$

Corollary 2.

Let $\mathbb{A}=(A_1,\dots ,A_n,0,0,\dots ).$ Then

$${\displaystyle \frac{1}{\sqrt{n}}}{\left({\displaystyle \sum _{m=1}^n}{\omega }^2\left(A_n\right)\right)}^{\frac{1}{p}}\le \omega \left(\mathbb{A}\right).$$

On the other hand, by using the following well-known inequality

$$\frac{1}{2}\parallel A_m\parallel \le \omega \left(A_m\right),1\le m\le n$$

it is obtained that

$${\displaystyle \frac{1}{2\sqrt{n}}}{\left({\displaystyle \sum _{m=1}^n}{\parallel A_m\parallel }^2\right)}^{\frac{1}{2}}\le \omega \left(\mathbb{A}\right).$$

Also, it is clear that

$${\parallel \mathbb{A}\parallel }_2\le {\left({\displaystyle \sum _{m=1}^n}{\parallel A_m\parallel }^2\right)}^{\frac{1}{2}}.$$

From this inequality and the previous relation it is established that

$${\displaystyle \frac{1}{2\sqrt{n}}}\parallel \mathbb{A}\parallel \le \omega \left(\mathbb{A}\right)$$

which was obtained in [24,25].

Theorem 4.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $\delta \le \tau .$ Then

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le {\displaystyle \sum _{n=1}^{\infty }}{\omega }_{({\delta }_n,\tau )}^p\left(A_n\right),$$

where ${\delta }_n=\delta -{\alpha }_n,{\alpha }_n={\left({\displaystyle \sum _{\begin{array}{c}m=1\\ m\ne n\end{array}}^{\infty }}{\parallel A_m\parallel }^p\right)}^{\frac{1}{p}},n\ge 1.$

Proof. 

For any $x\in {\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)$

$$\delta \le {\parallel \mathbb{A}x\parallel }_p={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}\le {\left(\parallel A_n{x\parallel }^p+{\alpha }_n^p\right)}^{\frac{1}{p}}\le \parallel A_{n}x\parallel +{\alpha }_n,$$

where ${\alpha }_n={\left({\displaystyle \sum _{\begin{array}{c}m=1\\ m\ne n\end{array}}^{\infty }}{\parallel A_m\parallel }^p\right)}^{\frac{1}{p}},n\ge 1.$ Hence,

$${\delta }_n=\delta -{\alpha }_n\le \parallel A_{n}x\parallel ,x\in \mathcal{H},n\ge 1.$$

(14)

Also, for any $n\ge 1$ from the last estimate

$$\parallel A_{n}x\parallel \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}\le \tau ,x\in \mathcal{H}$$

and previous relation (14) it is established that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\subset {\displaystyle \bigcap _{n=1}^{\infty }}{\mathrm{\Delta }}_{({\delta }_n,\tau )}\left(A_n\right).$$

(15)

On the other hand, for $x\in \mathcal{H}$ it is clear that

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p={\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p.$$

(16)

Thus, from (15) and (16) it is obtained that

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le {\displaystyle \sum _{n=1}^{\infty }}{\omega }_{({\delta }_n,\tau )}^p\left(A_n\right),$$

where ${\delta }_n=\delta -{\alpha }_n,{\alpha }_n={\left({\displaystyle \sum _{\begin{array}{c}m=1\\ m\ne n\end{array}}^{\infty }}{\parallel A_m\parallel }^p\right)}^{\frac{1}{p}},n\ge 1.$ □

The following corollary follows from Theorem 4.

Corollary 3.

Let $\mathbb{A}=(A_1,\dots ,A_n,0,0,\dots )$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $\delta \le \tau .$ Then

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\le {\left({\displaystyle \sum _{m=1}^n}{\parallel A_m\parallel }^p\right)}^{\frac{1}{p}}.$$

Also, from this, if $n=1$ (i.e., $\mathbb{A}=A\in \mathfrak{B}\left(\mathcal{H}\right)$), $\delta =0,\tau >\parallel A\parallel ,$ then we obtain the well-known inequality

$$\omega \left(A\right)\le \parallel A\parallel .$$

Theorem 5.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $0\le \delta \le \tau \le {\parallel \mathbb{A}\parallel }_p.$ Then

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le {\left({\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|\parallel }^p\right)}^{\frac{1}{2}}{\omega }_{(p,\delta ,\tau )}^{\frac{p}{2}}\left(\right|\mathbb{A}\left|\right).$$

Proof. 

By using Lemma 1, for any $x\in S_1\left(\mathcal{H}\right)$

$$\begin{array}{cc}\hfill {\left|\langle \mathbb{A}x,x\rangle \right|}_p^p& ={\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{\langle A_n^{\ast }{A}_{n}x,x\rangle }^{\frac{p}{2}}\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|x\parallel }^p\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|\parallel }^{\frac{p}{2}}\langle |A_n{|x,x\rangle }^{\frac{p}{2}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|\parallel }^p\right)}^{\frac{1}{2}}{\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{2}}\hfill \\ \hfill & ={\left({\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|\parallel }^p\right)}^{\frac{1}{2}}{|\langle |\mathbb{A}|x,x\rangle |}_p^{\frac{p}{2}}.\hfill \end{array}$$

(17)

On the other hand, for $x\in S_1\left(\mathcal{H}\right),$ since

$${\delta }^p\le {\parallel \mathbb{A}x\parallel }_p^p={\displaystyle \sum _{n=1}^{\infty }}\parallel A_n{x\parallel }^p={\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|x\parallel }^p={\parallel \left|\mathbb{A}\right|x\parallel }_p^p\le {\tau }^p,$$

then

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)={\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\right|\mathbb{A}\left|\right).$$

(18)

Thus, from (17) and (18) it is obtained that

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le {\left({\displaystyle \sum _{n=1}^{\infty }}\parallel |A_n{|\parallel }^p\right)}^{\frac{1}{2}}{\omega }_{(p,\delta ,\tau )}^{\frac{p}{2}}\left(\right|\mathbb{A}\left|\right).$$

□

Now, let us present a result that establishes the relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the sequential $(p,\delta ,\tau )$-numerical radii of its real and imaginary parts.

Theorem 6.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$, $2\le p<\infty ,$ and $0\le \delta \le \tau \le {\parallel \mathbb{A}\parallel }_p.$ Then

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le {\omega }_{\left(\frac{p}{2},{\delta }^2-\alpha ,{\tau }^2+\alpha \right)}^{\frac{p}{2}}\left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right),$$

where $\alpha ={\left({\displaystyle \sum _{n=1}^{\infty }}{\alpha }_n^{\frac{p}{2}}\right)}^{\frac{2}{p}}$, ${\alpha }_n=\parallel Re{A}_{n}Im{A}_n-Im{A}_{n}Re{A}_n\parallel ,n\ge 1.$

Proof. 

For any $x\in S_1\left(\mathcal{H}\right)$ and $n\ge 1$

$$\begin{array}{cc}\hfill |\langle A_{n}x,x\rangle {|}^p& ={\left(|\langle Re{A}_{n}x,x\rangle {|}^p+{\left|\langle Im{A}_{n}x,x\rangle \right|}^p\right)}^{\frac{p}{2}}\hfill \\ \hfill & \le {\left(\parallel Re{A}_n{x\parallel }^p+{\parallel Im{A}_{n}x\parallel }^p\right)}^{\frac{p}{2}}\hfill \\ \hfill & =|\langle \left(R{e}^2{A}_n+I{m}^2{A}_n\right)x,x\rangle {|}^{\frac{p}{2}}.\hfill \end{array}$$

Hence,

$${\displaystyle \sum _{n=1}^{\infty }}{|\langle A_{n}x,x\rangle |}^p\le {\displaystyle \sum _{n=1}^{\infty }}{\left|\langle \left(R{e}^2{A}_n+I{m}^2{A}_n\right)x,x\rangle \right|}^{\frac{p}{2}}.$$

Therefore,

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p\le {\left|\langle \left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right)x,x\rangle \right|}_{\frac{p}{2}}^{\frac{p}{2}}.$$

(19)

On the other hand, for $x\in S_1\left(\mathcal{H}\right)$ and $n\ge 1,$ it follows that

$$\begin{array}{cc}\hfill {\displaystyle \parallel }{A}_n{x\parallel }^p& =|\langle A_n^{\ast }{A}_{n}x,x\rangle {|}^{\frac{p}{2}}\hfill \\ \hfill & =|\langle {(Re{A}_n+iIm{A}_n)}^{\ast }(Re{A}_n+iIm{A}_n)x,x\rangle {|}^{\frac{p}{2}}\hfill \\ \hfill & =|\langle (Re{A}_n-iIm{A}_n)(Re{A}_n+iIm{A}_n)x,x\rangle {|}^{\frac{p}{2}}\hfill \\ \hfill & =|\langle \left((R{e}^2{A}_n+I{m}^2{A}_n)+i(Re{A}_{n}Im{A}_n-Im{A}_{n}Re{A}_n)\right)x,x\rangle {|}^{\frac{p}{2}}\hfill \\ \hfill & =|\langle (R{e}^2{A}_n+I{m}^2{A}_n)x,x\rangle +i\langle (Re{A}_{n}Im{A}_n-Im{A}_{n}Re{A}_n)x,x\rangle {|}^{\frac{p}{2}}\hfill \\ \hfill & \le {\left(\langle (R{e}^2{A}_n+I{m}^2{A}_n)x,x\rangle +\left|\langle (Re{A}_{n}Im{A}_n-Im{A}_{n}Re{A}_n)x,x\rangle \right|\right)}^{\frac{p}{2}}\hfill \\ \hfill & \le {\left(\langle (R{e}^2{A}_n+I{m}^2{A}_n)x,x\rangle +\parallel Re{A}_{n}Im{A}_n-Im{A}_{n}Re{A}_n\parallel \right)}^{\frac{p}{2}}\hfill \\ \hfill & ={\left(\langle (R{e}^2{A}_n+I{m}^2{A}_n)x,x\rangle +{\alpha }_n\right)}^{\frac{p}{2}}\hfill \end{array}$$

where ${\alpha }_n=\parallel Re{A}_{n}Im{A}_n-Im{A}_{n}Re{A}_n\parallel ,n\ge 1.$

Hence, for $x\in S_1\left(\mathcal{H}\right)$ and $0\le \delta \le {\parallel \mathbb{A}x\parallel }_p$ it follows that

$$\begin{array}{cc}\hfill {\delta }^2& \le {\parallel \mathbb{A}x\parallel }_p^2\hfill \\ \hfill & ={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{|\langle \left(R{e}^2{A}_n+I{m}^2{A}_n\right)x,x\rangle +{\alpha }_n|}^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle \left(R{e}^2{A}_n+I{m}^2{A}_n\right)x,x\rangle \right|}^{\frac{p}{2}}\right)}^{\frac{2}{p}}+{\left({\displaystyle \sum _{n=1}^{\infty }}{\alpha }_n^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel \left(R{e}^2{A}_n+I{m}^2{A}_n\right)x\parallel }^{\frac{p}{2}}\right)}^{\frac{2}{p}}+\alpha \hfill \\ \hfill & =\parallel \left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right){x\parallel }_{\frac{p}{2}}+\alpha ,\hfill \end{array}$$

where $\alpha ={\left({\displaystyle \sum _{n=1}^{\infty }}{\alpha }_n^{\frac{p}{2}}\right)}^{\frac{2}{p}}.$

Consequently, from the last relation it is obtained that for $x\in S_1\left(\mathcal{H}\right)$

$${\delta }^2-\alpha \le {\parallel \left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right)x\parallel }_{\frac{p}{2}}.$$

As well as this, if $x\in S_1\left(\mathcal{H}\right)$ and $n\ge 1,$ then

$$\parallel \left(R{e}^2{A}_n+I{m}^2{A}_n\right){x\parallel }^{\frac{p}{2}}\le {\left(\parallel A_n{x\parallel }^2+{\alpha }_n\right)}^{\frac{p}{2}}.$$

Hence, for $x\in S_1\left(\mathcal{H}\right)$

$$\begin{array}{cc}\hfill {\displaystyle {\displaystyle \parallel }}\left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right){x\parallel }_{\frac{p}{2}}& ={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel \left(R{e}^2{A}_n+I{m}^2{A}_n\right)x\parallel }^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\left(\parallel A_n{x\parallel }^2+{\alpha }_n\right)}^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{2}{p}}+{\left({\displaystyle \sum _{n=1}^{\infty }}{\alpha }_n^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & ={\parallel \mathbb{A}x\parallel }_p^2+\alpha .\hfill \end{array}$$

Consequently, if $x\in S_1\left(\mathcal{H}\right)$ and ${\parallel \mathbb{A}x\parallel }_p\le \tau ,$

$$\parallel \left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right){x\parallel }_{\frac{p}{2}}\le {\tau }^2+\alpha .$$

Therefore, from these calculations it is found that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\subset {\mathrm{\Delta }}_{\left(\frac{p}{2},{\delta }^2-\alpha ,{\tau }^2+\alpha \right)}\left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right).$$

(20)

Thus, from (19) and (20), it is obtained that

$${\omega }_{(p,\delta ,\tau )}^p\left(\mathbb{A}\right)\le {\omega }_{\left(\frac{p}{2},{\delta }^2-\alpha ,{\tau }^2+\alpha \right)}^{\frac{p}{2}}\left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right).$$

□

Example 3.

Assume that

$$A_{n}x=a_{n}x,a_n\in \mathbb{C},A_n:\mathbb{C}\to \mathbb{C},n\ge 1$$

where $a_n=b_n+i{c}_n,b_n,c_n\in \mathbb{R},n\ge 1.$ Let $p=2,\mathbb{A}=\left(A_n\right),$ and assume that ${\displaystyle \sum _{n=1}^{\infty }}{\left|a_n\right|}^2<\infty$, $\delta =0$, $\tau ={\parallel \mathbb{A}\parallel }_2={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|a_n\right|}^2\right)}^{\frac{1}{2}}.$ Then,

$$R{e}^2\mathbb{A}+I{m}^2\mathbb{A}=(b_n^2+c_n^2).$$

Hence, from (6) it is obtained that

$${\omega }_{(2,0,\parallel \mathbb{A}{\parallel }_2)}\left(\mathbb{A}\right)\le {\omega }_{(1,0,\parallel \mathbb{A}{\parallel }_2^2)}(b_n^2+c_n^2)={\left({\displaystyle \sum _{n=1}^{\infty }}(b_n^2+c_n^2)\right)}^{\frac{1}{2}}.$$

From Theorem 6, we obtain the next results.

Corollary 4.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right),$ the operator $A_n\in \mathfrak{B}\left({\mathcal{H}}_n\right)$ is normal for any $n\ge 1$, $2\le p<\infty ,$ and $0\le \delta \le \tau \le {\parallel \mathbb{A}\parallel }_p.$ Then

$${\omega }_{(p,\delta ,\tau )}^2\left(\mathbb{A}\right)\le {\omega }_{\left(\frac{p}{2},{\delta }^2,{\tau }^2\right)}\left(R{e}^2\mathbb{A}+I{m}^2\mathbb{A}\right).$$

Remark 2.

For any $n=1,\dots$ since

$$R{e}^2{A}_n+I{m}^2{A}_n=\frac{1}{2}\left(A_n^{\ast }{A}_n+A_n{A}_n^{\ast }\right),$$

then from (1) of Theorem 1 and Theorem 6 it can be rewritten in the equivalent form

$${\omega }_{(p,\delta ,\tau )}^2\left(\mathbb{A}\right)\le {\displaystyle \frac{1}{2}}{\omega }_{\left(\frac{p}{2},2({\delta }^2-\alpha ),2({\tau }^2+\alpha )\right)}({\mathbb{A}}^{\ast }\mathbb{A}+\mathbb{A}{\mathbb{A}}^{\ast })$$

Then, if we take $n=1,\delta =0,\tau >\parallel A\parallel ,A\in \mathfrak{B}\left(\mathcal{H}\right),$ then it is true that

$${\omega }_{(p,0,\tau )}^2\left(A\right)={\omega }^2\left(A\right)\le {\displaystyle \frac{1}{2}}\parallel A^{\ast }A+A{A}^{\ast }\parallel ,$$

and this result has been found in [13].

Remark 3.

If $\mathbb{A}=\left(A_n\right),\underset{n\ge 1}{inf}\parallel A_n\parallel >0,$ for any $n\ge 1$ $A_n\ge 0$ and $2\le p<\infty$ in Theorem 6, then from Lemma 1 it is clear that

$$\begin{array}{cc}\hfill {\delta }^2& \le {\parallel \mathbb{A}x\parallel }_p^2\hfill \\ \hfill & ={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}\parallel A_n{\parallel }^{\frac{p}{2}}{\left|\langle A_{n}x,x\rangle \right|}^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le \underset{n\ge 1}{sup}\parallel A_n\parallel {\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le \lambda {\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^{\frac{p}{2}}\right)}^{\frac{2}{p}}\hfill \\ \hfill & \le \lambda \left({\displaystyle \sum _{n=1}^{\infty }}\langle A_{n}x,x\rangle \right)\hfill \\ \hfill & =\lambda \parallel \langle {\displaystyle \sum _{n=1}^{\infty }}{A}_{n}x,x\rangle \parallel \hfill \\ \hfill & =\lambda \parallel {\displaystyle \sum _{n=1}^{\infty }}{A}_{n}x\parallel ,x\in S_1\left(\mathcal{H}\right)\hfill \end{array}$$

where $\lambda =\underset{n\ge 1}{sup}\parallel A_n\parallel .$ Hence, if for any $x\in S_1\left(\mathcal{H}\right)$, then the following holds

$${\delta }^2\le {\parallel \mathbb{A}x\parallel }_p^2\le \lambda \parallel {\displaystyle \sum _{n=1}^{\infty }}{A}_{n}x\parallel .$$

(21)

On the other hand, for any $x\in S_1\left(\mathcal{H}\right)$ and $\parallel \mathbb{A}x\parallel \le \tau$

$$\underset{n\ge 1}{inf}\parallel A_n\parallel {\displaystyle \sum _{n=1}^{\infty }}\parallel A_{n}x\parallel \le {\displaystyle \sum _{n=1}^{\infty }}\parallel A_n\parallel \parallel A_{n}x\parallel \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n\parallel }^q\right)}^{\frac{1}{q}}{\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}$$

i.e.,

$$\parallel {\displaystyle \sum _{n=1}^{\infty }}{A}_{n}x\parallel \le {\left(\underset{n\ge 1}{inf}\parallel A_n\parallel \right)}^{-1}{\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n\parallel }^q\right)}^{\frac{1}{q}}{\parallel \mathbb{A}x\parallel }_p\le \sigma \tau ,$$

(22)

where $\sigma ={\left(\underset{n\ge 1}{inf}\parallel A_n\parallel \right)}^{-1}{\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_n\parallel }^q\right)}^{\frac{1}{q}}$ and $q={\displaystyle \frac{p}{p-1}}.$

Thus, from (21) and (22) it is obtained that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\subset {\mathrm{\Delta }}_{\left(1,\frac{{\delta }^2}{\lambda },\sigma \tau \right)}\left({\displaystyle \sum _{n=1}^{\infty }}{A}_n\right).$$

(23)

In addition, since $A_n\ge 0,n\ge 1,$ then for any $2\le p<\infty$ and $x\in \mathcal{H}$

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p={\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\le {\left({\displaystyle \sum _{n=1}^{\infty }}\langle A_{n}x,x\rangle \right)}^p={\langle {\displaystyle \sum _{n=1}^{\infty }}{A}_{n}x,x\rangle }^p.$$

(24)

Then, from (23) and (24) we obtain

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\le {\omega }_{\left(1,\frac{{\delta }^2}{\left|\lambda \right|},\sigma \tau \right)}\left({\displaystyle \sum _{n=1}^{\infty }}{A}_n\right).$$

Now, we will present an important result.

Theorem 7.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right).$ Then,

$${\omega }_{(p,\delta ,\tau )}\left(Re\mathbb{A}\right)\le {\omega }_{(p,0,\parallel \mathbb{A}\parallel )}\left(\mathbb{A}\right),for\delta \le \tau \le \parallel Re\mathbb{A}\parallel$$

and

$${\omega }_{(p,\delta ,\tau )}\left(Im\mathbb{A}\right)\le {\omega }_{(p,0,\parallel \mathbb{A}\parallel )}\left(\mathbb{A}\right),for\delta \le \tau \le \parallel Im\mathbb{A}\parallel .$$

Proof. 

For any $x\in \mathcal{H},$ it is clear that

$${\left|\langle Re\mathbb{A}x,x\rangle \right|}_p={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle Re{A}_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}\le {\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}={\left|\langle \mathbb{A}x,x\rangle \right|}_p,$$

and similarly

$${\left|\langle Im\mathbb{A}x,x\rangle \right|}_p\le {\left|\langle \mathbb{A}x,x\rangle \right|}_p.$$

On the other hand, it is easy to see that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(Re\mathbb{A}\right)\subset S_1\left(\mathcal{H}\right)\mathrm{for}\delta \le \tau \le {\parallel Re\mathbb{A}\parallel }_p,$$

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(Im\mathbb{A}\right)\subset S_1\left(\mathcal{H}\right)\mathrm{for}\delta \le \tau \le {\parallel Im\mathbb{A}\parallel }_p,$$

From these relations, we obtain

$${\omega }_{(p,\delta ,\tau )}\left(Re\mathbb{A}\right)\le {\omega }_{(p,0,\parallel \mathbb{A}\parallel )}\left(\mathbb{A}\right),for\delta \le \tau \le \parallel Re\mathbb{A}\parallel$$

and

$${\omega }_{(p,\delta ,\tau )}\left(Im\mathbb{A}\right)\le {\omega }_{(p,0,\parallel \mathbb{A}\parallel )}\left(\mathbb{A}\right),for\delta \le \tau \le \parallel Im\mathbb{A}\parallel .$$

□

From Theorem 7, we obtain the next results.

Corollary 5.

Let $\mathbb{A}=(A,0,0,\dots ),A\in \mathfrak{B}\left(\mathcal{H}\right)$, $\delta =0$ and $\tau \ge max\left\{{\parallel ReA\parallel }_p,{\parallel ImA\parallel }_p\right\}.$ Then,

$$\omega (R{e}^{2}A+I{m}^{2}A)\le \omega \left(R{e}^{2}A\right)+\omega \left(I{m}^{2}A\right)\le {\omega }^2\left(A\right)+{\omega }^2\left(A\right)\le 2{\omega }^2\left(A\right).$$

Therefore, from the last relation it is found that for $A\in \mathfrak{B}\left(\mathcal{H}\right)$

$${\displaystyle \frac{1}{4}}\parallel A^{\ast }A+A{A}^{\ast }\parallel \le {\omega }^2\left(A\right).$$

This result has been found by Kittaneh in [13].

3. Sectorial Case of Coordinate Operators

In this section, the sequential $(p,\delta ,\tau )$-numerical radius function will be investigated in the case when the coordinate operators are sectorial.

Theorem 8.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $0\le \delta \le \tau \le {\parallel \mathbb{A}\parallel }_p.$

(1) 

If $A_n\in Se{c}_{{\alpha }_n}\left(\mathcal{H}\right),0\le {\alpha }_n<{\displaystyle \frac{\pi }{2}},n\ge 1$ and ${\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n<{\displaystyle \frac{\pi }{2}},$ then

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\le sec\left({\alpha }_{sup}\right){\omega }_{(p,{\delta }_r,{\tau }_r)}\left(Re\mathbb{A}\right),$$

where ${\delta }_r=\delta -{\parallel Im\mathbb{A}\parallel }_p$ and ${\tau }_r=\tau +{\parallel Im\mathbb{A}\parallel }_p.$

(2) 

If $A_n\in Se{c}_{{\alpha }_n}\left(\mathcal{H}\right),0<{\alpha }_n<{\displaystyle \frac{\pi }{2}},n\ge 1$ and ${\alpha }_{inf}=\underset{n\ge 1}{inf}{\alpha }_n>0,$ then

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\le csc\left({\alpha }_{inf}\right){\omega }_{(p,{\delta }_i,{\tau }_i)}\left(Im\mathbb{A}\right),$$

where ${\delta }_i=\delta -{\parallel Re\mathbb{A}\parallel }_p$ and ${\tau }_i=\tau +{\parallel Re\mathbb{A}\parallel }_p.$

Proof. 

(1)

For each $x\in S_1\left(\mathcal{H}\right)$ it follows that

$$\begin{array}{cc}\hfill \delta \le {\parallel \mathbb{A}x\parallel }_p& ={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{m}x\parallel }^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}|\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x{\parallel |}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}\parallel Re{A}_{n}x{\parallel |}^p\right)}^{\frac{1}{p}}+{\parallel Im\mathbb{A}\parallel }_p.\hfill \end{array}$$

Then, it is clear that for each $x\in S_1\left(\mathcal{H}\right)$

$$\delta -{\parallel Im\mathbb{A}\parallel }_p\le {\parallel Re\mathbb{A}x\parallel }_p,$$

and setting ${\delta }_r=\delta -{\parallel Im\mathbb{A}\parallel }_p$ yields

$${\delta }_r\le {\parallel Re\mathbb{A}x\parallel }_p.$$

On the other hand, for $x\in S_1\left(\mathcal{H}\right)$

$${\parallel Re\mathbb{A}x\parallel }_p\le {\parallel \mathbb{A}x\parallel }_p+{\parallel Im\mathbb{A}\parallel }_p\le \tau +{\parallel Im\mathbb{A}\parallel }_p={\tau }_r.$$

Hence,

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\subset {\mathrm{\Delta }}_{(p,{\delta }_r,{\tau }_r)}\left(Re\mathbb{A}\right).$$

(25)

As well as this, for $x\in \mathcal{H}$

$$\begin{array}{cc}\hfill {\left|\langle \mathbb{A}x,x\rangle \right|}_p& ={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & ={\left({\displaystyle \sum _{n=1}^{\infty }}{\left(|\langle Re{A}_{n}x,x\rangle {|}^2+{\left|\langle Im{A}_{n}x,x\rangle \right|}^2\right)}^{\frac{p}{2}}\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\left(\left(1+{tan}^2\left({\alpha }_n\right)\right){\left|\langle Re{A}_{n}x,x\rangle \right|}^2\right)}^{\frac{p}{2}}\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left(1+{tan}^2\left({\alpha }_{sup}\right)\right)}^{\frac{1}{2}}{\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle Re{A}_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & =sec\left({\alpha }_{sup}\right){\left|\langle Re\mathbb{A}x,x\rangle \right|}_p,\hfill \end{array}$$

(26)

where ${\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n.$

Thus, from (25) and (26) it is obtained that

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\le sec\left({\alpha }_{sup}\right){\omega }_{(p,{\delta }_r,{\tau }_r)}\left(Re\mathbb{A}\right),$$

where ${\delta }_r=\delta -{\parallel Im\mathbb{A}\parallel }_p$ and ${\tau }_r=\tau +{\parallel Im\mathbb{A}\parallel }_p.$

(2)

Assume that $x\in {\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right).$ For any $x\in S_1\left(\mathcal{H}\right),$ it follows that

$$\begin{array}{cc}\hfill \delta \le {\parallel \mathbb{A}x\parallel }_p& ={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\left(\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel \right)}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel Re{A}_{n}x\parallel }^p\right)}^{\frac{1}{p}}+{\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel Im{A}_{n}x\parallel }^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\parallel Re\mathbb{A}\parallel }_p+{\parallel Im\mathbb{A}x\parallel }_p.\hfill \end{array}$$

Hence, from the last calculations it is obtained that for $x\in S_1\left(\mathcal{H}\right)$

$$\delta -{\parallel Re\mathbb{A}\parallel }_p\le {\parallel Im\mathbb{A}x\parallel }_p,$$

and setting ${\delta }_i=\delta -{\parallel Re\mathbb{A}\parallel }_p$ yields

$${\delta }_i\le {\parallel Im\mathbb{A}x\parallel }_p.$$

(27)

On the other hand, for any $x\in S_1\left(\mathcal{H}\right)$

$${\parallel Im\mathbb{A}x\parallel }_p\le {\parallel \mathbb{A}x\parallel }_p+{\parallel Re\mathbb{A}x\parallel }_p\le \tau +{\parallel Re\mathbb{A}x\parallel }_p$$

and setting ${\tau }_i=\tau +{\parallel Re\mathbb{A}\parallel }_p$ yields

$${\tau }_i\ge {\parallel Im\mathbb{A}x\parallel }_p.$$

(28)

Consequently, from (27) and (28) it is established that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\subset {\mathrm{\Delta }}_{(p,{\delta }_i,{\tau }_i)}\left(Im\mathbb{A}\right)$$

(29)

Also, since $A_n\in Se{c}_{{\alpha }_n}\left(\mathcal{H}\right),0<{\alpha }_n<{\displaystyle \frac{\pi }{2}},n\ge 1$ it is clear that for $x\in \mathcal{H}$

$$|\langle Re{A}_{n}x,x\rangle |\le cot\left({\alpha }_n\right)\left|\langle Im{A}_{n}x,x\rangle \right|.$$

Using this fact for $x\in \mathcal{H}$ it is obtained that

$$\begin{array}{cc}\hfill {\left|\langle \mathbb{A}x,x\rangle \right|}_p& ={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & ={\left({\displaystyle \sum _{n=1}^{\infty }}{\left(|\langle Re{A}_{n}x,x\rangle {|}^2+{\left|\langle Im{A}_{n}x,x\rangle \right|}^2\right)}^{\frac{p}{2}}\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}{\left(\left(1+{cot}^2\left({\alpha }_n\right)\right){\left|\langle Im{A}_{n}x,x\rangle \right|}^2\right)}^{\frac{p}{2}}\right)}^{\frac{1}{p}}\hfill \\ \hfill & \le {\left(1+{cot}^2\left({\alpha }_{inf}\right)\right)}^{\frac{1}{2}}{\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\langle Im{A}_{n}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & =csc\left({\alpha }_{inf}\right){\left|\langle Im\mathbb{A}x,x\rangle \right|}_p,\hfill \end{array}$$

(30)

where ${\alpha }_{inf}=\underset{n\ge 1}{inf}{\alpha }_n.$

Thus, from (29) and (30) it is found that

$${\omega }_{(p,\delta ,\tau )}\left(\mathbb{A}\right)\le csc\left({\alpha }_{inf}\right){\omega }_{(p,{\delta }_i,{\tau }_i)}\left(Im\mathbb{A}\right),$$

where ${\delta }_i=\delta -{\parallel Re\mathbb{A}\parallel }_p$ and ${\tau }_i=\tau +{\parallel Re\mathbb{A}\parallel }_p.$

□

Example 4.

Assume that

$$A_n=\left(\begin{array}{cc}{a}_n+i{b}_n& 0\\ 0& a_n+i{b}_n\end{array}\right):{\mathbb{C}}^2\left(\mathbb{C}\right),n\ge 1,$$

where $a_n,b_n\in \mathbb{R},a_n\ge 1,$ and $|b_n|\le \sqrt{3}.$ Let $\mathbb{A}=\left(A_n\right),p=2.$ Then,

$${\parallel \mathbb{A}\parallel }_2={\left({\displaystyle \sum _{n=1}^{\infty }}{a}_n^2+b_n^2\right)}^{\frac{1}{2}}$$

and

$${\displaystyle \frac{{\left|\langle Im\mathbb{A}x,x\rangle \right|}_2}{{\parallel \langle Re\mathbb{A}x,x\rangle \parallel }_2}}\le \sqrt{3}=tan\left({\displaystyle \frac{\pi }{3}}\right),$$

which implies that $\mathbb{A}\in Se{c}_{\left(\frac{\pi }{3}\right)}\left({\mathbb{C}}^2\right).$ Define

$${\delta }_r=\delta -{\parallel Im\mathbb{A}\parallel }_p\mathrm{and}{\tau }_r=\tau +{\parallel Im\mathbb{A}\parallel }_p$$

where ${\parallel Im\mathbb{A}\parallel }_p={\left({\displaystyle \sum _{n=1}^{\infty }}{b}_n^2\right)}^{\frac{1}{2}}.$ Therefore, by (1) of Theorem 8 it is obtained that

$${\omega }_{(2,\delta ,\tau )}\left(\mathbb{A}\right)\le 2{\omega }_{(2,{\delta }_r,{\tau }_r)}\left(Re\mathbb{A}\right)=2{\left({\displaystyle \sum _{n=1}^{\infty }}{a}_n^2\right)}^{\frac{1}{2}}.$$

If we take $p=2,\delta =0,$ and $\tau \ge {\parallel \mathbb{A}\parallel }_2$ in Theorem 8, we obtain the following corollary.

Corollary 6.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right).$

(1) 

If $A_m=0,m>n,A_k\in Se{c}_{{\alpha }_k}\left(\mathcal{H}\right),0\le {\alpha }_k<{\displaystyle \frac{\pi }{2}},k=1,\dots ,n,$ then

$$\omega \left(\mathbb{A}\right)\le sec\left({\alpha }_{sup}\right)\omega \left(Re\mathbb{A}\right),$$

where ${\alpha }_{sup}=\underset{1\le k\le n}{sup}{\alpha }_k.$

(2) 

If $A_m=0,m>n,A_k\in Se{c}_{{\alpha }_k}\left(\mathcal{H}\right),0<{\alpha }_k<{\displaystyle \frac{\pi }{2}},k=1,\dots ,n,$ then

$$\omega \left(\mathbb{A}\right)\le csc\left({\alpha }_{inf}\right)\omega \left(Im\mathbb{A}\right),$$

where ${\alpha }_{inf}=\underset{1\le k\le n}{inf}{\alpha }_k.$

Remark 4.

If we take $n=1$ in Corollary 6, then for $A\in Se{c}_{\alpha }\left(\mathcal{H}\right),0<\alpha <{\displaystyle \frac{\pi }{2}}$

$$\omega \left(A\right)\le sec\left(\alpha \right)\omega \left(ReA\right)\mathrm{and}\omega \left(A\right)\le csc\left(\alpha \right)\omega \left(ImA\right),$$

which was proved in [9].

Theorem 9.

Let $\mathbb{A}=\left(A_n\right)$ be an operator sequence in $\mathfrak{B}\left(\mathcal{H}\right)$ and $A_n\in \mathfrak{B}\left(\mathcal{H}\right),$$n\ge 1$ be an accretive–dissipative operator such that

$$W\left(A_n\right)\subset \left\{r{e}^{i\phi }:r\ge 0,0<{\gamma }_m\le \phi \le {\alpha }_m<{\displaystyle \frac{\pi }{2}}\right\},\mathrm{for}\mathrm{each}n\ge 1.$$

Then,

(1) 

for any $\delta \le \tau \le {\parallel Re\mathbb{A}\parallel }_p,$

$${\omega }_{(p,\delta ,\tau )}\left(Re\mathbb{A}\right)\le cos\left({\gamma }_{inf}\right){\omega }_{(p,0,\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{A}\right),$$

where ${\gamma }_{inf}=\underset{n\ge 1}{inf}{\gamma }_n,$

(2) 

for any $\delta \le \tau \le {\parallel Im\mathbb{A}\parallel }_p,$

$${\omega }_{(p,\delta ,\tau )}\left(Im\mathbb{A}\right)\le sin\left({\alpha }_{sup}\right){\omega }_{(p,0,\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{A}\right),$$

where ${\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n.$

Proof. 

(1)

For any $x\in \mathcal{H}$ it is clear that

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p={\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p\le {\displaystyle \sum _{n=1}^{\infty }}{\left(|\langle Re{A}_{n}x,x\rangle {|}^2+{\left|\langle Im{A}_{n}x,x\rangle \right|}^2\right)}^{\frac{p}{2}}.$$

On the other hand, since

$$|\langle Im{A}_{n}x,x\rangle |\ge tan\left({\gamma }_n\right)\left|\langle Re{A}_{n}x,x\rangle \right|,n\ge 1$$

and the tangent function is monotonically increasing on $\left[0,{\displaystyle \frac{\pi }{2}}\right),$ then

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p\ge {\displaystyle \sum _{n=1}^{\infty }}{(1+{tan}^2\left({\gamma }_n\right))}^{\frac{p}{2}}|\langle Re{A}_{n}x,x\rangle {|}^p\ge {(1+{tan}^2\left({\gamma }_{inf}\right))}^{\frac{p}{2}}{\left|\langle Re\mathbb{A}x,x\rangle \right|}_p,$$

i.e., for $x\in \mathcal{H}$

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p\ge sec\left({\gamma }_{inf}\right)\left)\right|\langle Re\mathbb{A}x,x\rangle {|}_p,$$

(31)

where ${\gamma }_{inf}=\underset{n\ge 1}{inf}{\delta }_n.$

As well as this, for any $\delta \le \tau \le {\parallel Re\mathbb{A}x\parallel }_p$ it is clear that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(Re\mathbb{A}\right)\subset S_1\left(\mathcal{H}\right).$$

(32)

Hence, from (31) and (32) it is found that

$${\omega }_{(p,\delta ,\tau )}\left(Re\mathbb{A}\right)\le cos\left({\gamma }_{inf}\right){\omega }_{(p,0,\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{A}\right),$$

(2)

For any $x\in \mathcal{H}$ it is clear that

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p={\displaystyle \sum _{n=1}^{\infty }}{\left|\langle A_{n}x,x\rangle \right|}^p={\displaystyle \sum _{n=1}^{\infty }}{\left(|\langle Re{A}_{n}x,x\rangle {|}^2+{\left|\langle Im{A}_{n}x,x\rangle \right|}^2\right)}^{\frac{p}{2}}.$$

On the other hand, since

$$|\langle Im{A}_{n}x,x\rangle |\le tan\left({\alpha }_n\right)\left|\langle Re{A}_{n}x,x\rangle \right|,n\ge 1$$

and the cotangent function is monotonically decreasing on $\left(0,{\displaystyle \frac{\pi }{2}}\right),$ then

$${\left|\langle \mathbb{A}x,x\rangle \right|}_p^p\ge {\displaystyle \sum _{n=1}^{\infty }}{(1+{cot}^2\left({\alpha }_m\right))}^{\frac{p}{2}}|\langle Im{A}_{n}x,x\rangle {|}^p\ge {csc}^p\left({\alpha }_{sup}\right){\left|\langle Im\mathbb{A}x,x\rangle \right|}_p^p,$$

i.e., for $x\in \mathcal{H}$

$${\left|\langle Im\mathbb{A}x,x\rangle \right|}_p\le sin\left({\alpha }_{sup}\right){\left|\langle \mathbb{A}x,x\rangle \right|}_p,$$

(33)

where ${\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n.$

As well as this, for any $\delta \le \tau \le {\parallel Im\mathbb{A}x\parallel }_p$ it is clear that

$${\mathrm{\Delta }}_{(p,\delta ,\tau )}\left(Im\mathbb{A}\right)\subset S_1\left(\mathcal{H}\right).$$

(34)

Hence, from (33) and (34) it is found that

$${\omega }_{(p,\delta ,\tau )}\left(Im\mathbb{A}\right)\le sin\left({\alpha }_{sup}\right){\omega }_{(p,0,\parallel \mathbb{A}{\parallel }_p)}\left(\mathbb{A}\right).$$

□

Corollary 7.

Under the condition of Theorem 9, if $\delta \le \tau \le min\left\{{\parallel Re\mathbb{A}\parallel }_p,{\parallel Im\mathbb{A}\parallel }_p\right\},$ then

$${\omega }_{(p,\delta ,\tau )}^2\left(Re\mathbb{A}\right)+{\omega }_{(p,\delta ,\tau )}^2\left(Im\mathbb{A}\right)\le \left({cos}^2\left({\gamma }_{inf}\right)+{sin}^2\left({\alpha }_{sup}\right)\right){\omega }_{(p,0,\parallel \mathbb{A}{\parallel }_p)}^2\left(\mathbb{A}\right).$$

Corollary 8.

Under the condition of Theorem 9, if $A_m=0,m>n,\delta =0,$ and $\tau \ge min\left\{{\parallel Re\mathbb{A}\parallel }_p,{\parallel Im\mathbb{A}\parallel }_p\right\},$ then from Corollary 7

$${\displaystyle \frac{{\parallel Re\mathbb{A}\parallel }_p^2+{\parallel Im\mathbb{A}\parallel }_p^2}{{cos}^2\left({\gamma }_{inf}\right)+{sin}^2\left({\alpha }_{sup}\right)}}\le {\omega }_p^2\left(\mathbb{A}\right).$$

In addition, if $\mathbb{A}=(A,0,0,\dots ),A\in \mathfrak{B}\left(\mathcal{H}\right),$ then from the last relation it is obtained that for $0<\gamma <\alpha <{\displaystyle \frac{\pi }{2}}$

$${\displaystyle \frac{{\parallel ReA\parallel }^2+{\parallel ImA\parallel }^2}{{cos}^2\left(\gamma \right)+{sin}^2\left(\alpha \right)}}\le {\omega }^2\left(A\right).$$

On the other hand, it is well known that from [8]

$${\parallel A\parallel }^2\le {\parallel ReA\parallel }^2+{\parallel ImA\parallel }^2.$$

Therefore, from the previous inequality it is obtained that

$$\parallel A\parallel \le \sqrt{{cos}^2\left(\gamma \right)+{sin}^2\left(\alpha \right)}\omega \left(A\right).$$

(35)

It must be noted that in the matrix case of $A\in Se{c}_{\alpha }\left(\mathcal{H}\right),$$0\le \alpha <{\displaystyle \frac{\pi }{2}}$ the following main result was obtained in [8],

$$\parallel A\parallel \le \sqrt{1+{sin}^2\left(\alpha \right)}\omega \left(A\right).$$

Therefore, the previous estimate shows that in the accretive–dissipative case the last inequality can be improved.

It must be noted that for $A,B\in \mathfrak{B}\left(\mathcal{H}\right)$ with condition

$$W\left(A\right),W\left(B\right)\subset \left\{r{e}^{i\phi }:r\ge 0,0<\gamma <\alpha <{\displaystyle \frac{\pi }{2}}\right\}$$

from (35), the following interpolated inequality is implied

$$\omega \left(AB\right)\le ({cos}^2\gamma +{sin}^2\alpha )\omega \left(A\right)\omega \left(B\right).$$

This shows that the analogous result in [8] can be improved.

4. Conclusions

In this study, we investigated several fundamental properties of the sequential $(p,\delta ,\tau )$-numerical radius function. First, we examined the relationships between the sequential $(p,\delta ,\tau )$-numerical radius of an operator sequence and the corresponding numerical radii of its coordinate operators. We further analyzed the connections between the sequential $(p,\delta ,\tau )$-numerical radius and the numerical radii of the real and imaginary parts of the associated operator sequences.

Finally, these results were extended to the case where the coordinate operators are sectorial, which provided additional structural insights into the behavior of the sequential $(p,\delta ,\tau )$-numerical radius function. The obtained results generalize several well-known classical results on the numerical radius function in the recent literature and contribute new and useful perspectives to the existing theory.

Future research may focus on extending these results to more general classes of operators, as well as exploring possible applications of the sequential $(p,\delta ,\tau )$-numerical radius in other areas of functional analysis and operator theory.