Some spectral properties and convergence of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number

Pembe Ipek Al ipekpembe@gmail.com Department of Mathematics, Faculty of Sciences, Karadeniz Technical University, Trabzon, 61080 Turkey Zameddin I. Ismailov zameddin.ismailov@gmail.com Fuad Kittaneh fkitt@ju.edu.jo Department of Mathematics, The University of Jordan, Amman, Jordan Satyajit Sahoo satyajitsahoo2010@gmail.com, ssahoomath@gmail.com School of Mathematical Sciences, National Institute of Science Education and Research, Bhubaneswar, India

Abstract

In this study, some estimates are given for the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number via the $A$ -numerical radius and $A$ -Crawford number for the $A$ -bounded linear operators in any complex semi-Hilbert space, respectively. Then, some evolutions are studied for the tensor product of two operators. Lastly, some convergence properties of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number, via the $A$ -uniform convergence of operator sequences, are investigated. We also considered several examples to illustrate our results. Finally, a few applications of some operator functions classes are also given.

keywords:

Operaor semi-norm,

(A,q)

-numerical radius,

(A,q)

-Crawford number.
AMS 2010 Subject Classification: 47A05, 47A12, 47A30, 47A63.

1. Introduction

Throughout this article, $H$ denotes a complex Hilbert space endowed with the inner product $\langle\cdot,\cdot\rangle$ and associated norm $\|\cdot\|.$ Let $L(H)$ stand for the $C^{*}$ -algebra of all bounded linear operators acting on $H.$ The identity operator on $H$ will be simply denoted by $I.$ From now on, by an operator we mean a bounded linear operator acting on $H.$ Recall that an operator $T$ is called positive, written $T\geq 0,$ if $\langle Tx,x\rangle\geq 0$ for all $x\in H.$ The square root of a positive operator $T$ is denoted by $T^{1/2}.$ We denote by $|T|=(T^{*}T)^{1/2}$ the absolute value of an operator $T.$ Here, $T^{*}$ denote the adjoint of $T.$ The Crawford number of an operator $T$ is given by

c(T)=\inf\{|\langle Tx,x\rangle|:x\in H,\ \|x\|=1\}.

The usual operator norm and the numerical radius of an operator $T$ are, respectively, defined by

\|T\|=\sup\{\|Tx\|:\ x\in H,\ \|x\|=1\}

and

\omega(T)=\sup\{|\langle Tx,x\rangle|:\ x\in H,\ \|x\|=1\}.

An operator $T$ is called normal if $T^{*}T=TT^{*}.$ It is well known that the equality $\|T\|=\omega(T)$ holds for every normal operator $T.$ However, in general, the above equality fails to be true for non-normal operators. Notice that the following inequalities hold for every $T\in L(H):$

\displaystyle\frac{1}{2}\|T\|\leq\omega(T)\leq\|T\|.

(1)

The inequalities (1) play an important role in the approximation of $\omega(\cdot)$ , which have been the interest of numerous authors. The reader may consult, for example, [6, 7, 8, 13, 14, 22] and the references therein or the books [3, 11, 12].
For the rest of this paper, $A\neq 0$ will denote a bounded linear positive operator on $H.$ We are going to consider an additional semi-inner product $\langle\cdot,\cdot\rangle_{A}$ on $H$ defined by $\langle x,y\rangle_{A}=\langle Ax,y\rangle$ for all $x,y\in H,$ which induces a semi-norm $\|x\|_{A}$ on $H.$ This makes $H$ into a semi-Hilbert space. After that, we replace the operator norm with the following operator semi-norm

\|T\|_{A}=\sup\{\|Tx\|_{A}:\ x\in H,\ \|x\|_{A}=1\}

for an $A$ -bounded operator $T,$ i. e. an operator $T$ which satisfies $\|Tx\|_{A}\leq c\|x\|_{A}$ for all $x\in H$ and some constant $c>0.$ Let $L^{A}(H)$ stand for all $A$ -bounded linear operators acting on $H.$
Recently, several generalizations for the concept of the numerical radius have been introduced (see [28] and references therein). One of these generalizations is the so-called $A-$ numerical radius of an operator $T\in L^{A}(H),$ which was firstly introduced by Saddi in [26] as

\omega_{A}(T)=\sup\{|\langle Tx,x\rangle_{A}|:x\in H,\|x\|_{A}=1\}.

The related $A$ -Crawford number is defined as

c_{A}(T)=\inf\{|\langle Tx,x\rangle_{A}|:x\in H,\|x\|_{A}=1\}.

Note that, it may happen that $\omega_{A}(T)=+\infty$ for some $T\in L^{A}(H)$ (see [2]). Several results on the $A$ -numerical radius have been established by many mathematicians. See, e.g. [4, 5, 9, 10, 15, 19, 21, 24, 25, 27, 30] and references therein.
For $q\in\mathbb{C},\ |q|\leq 1,$ the $q$ -numerical range is defined as

W_{q}(T)=\{\langle Tx,y\rangle:\|x\|=\|y\|=1,\ \langle x,y\rangle=q\}

(see [16, 18, 23, 29]).
Recall that for $T\in L(H),\ q\in\mathbb{C},\ |q|\leq 1,$ the $q$ -numerical radius and $q$ -Crawford number are defined as

\displaystyle\omega_{q}(T)=\sup\{|\lambda|:\lambda\in W_{q}(T)\},\ c_{q}(T)=\inf\{|\lambda|:\lambda\in W_{q}(T)\},

respectively.
Motivated by this, we introduce a new definition.

Definition 1.

For the operator $T\in L^{A}(H),\ q\in\mathbb{C},\ 0<|q|\leq 1,$ the following set

W_{A,q}(T):=\left\{\langle Tx,y\rangle_{A}:\ \|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q\right\}

is called the $(A,q)$ -numerical range of the operator $T$ . Also, the numbers

\omega_{A,q}(T)=\sup\{|\lambda|:\lambda\in W_{A,q}(T)\}

and

c_{A,q}(T)=\inf\{|\lambda|:\lambda\in W_{A,q}(T)\},

are called the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number of the operator $T$ , respectively.

In this case it is clear that

	$\displaystyle W_{A,1}(T)=W_{A}(T),\ W_{I,1}(T)=W(T),$
	$\displaystyle\omega_{A,1}(T)=\omega_{A}(T),\ c_{A,1}(T)=c_{A}(T),$
	$\displaystyle\omega_{I,1}(T)=\omega(T),\ c_{I,1}(T)=c(T),$
	$\displaystyle W_{I,q}(T)=W_{q}(T),\ \omega_{I,q}(T)=\omega_{q}(T),\ c_{I,q}(T)=c_{q}(T).$

Throughout this paper, the $(A,q)$ -numerical gap is the difference between the $A$ -operator semi-norm and $(A,q)$ -numerical radius, which is defined as

g_{\omega_{A,q}}(T):=\|T\|_{A}-\omega_{A,q}(T)\geq 0\ \text{(see, Theorem \ref{thm1} (1))}

for $T\in L^{A}(H),\ q\in\mathbb{C},\ 0<|q|\leq 1.$
Also, the $(A,q)$ -Crawford gap is the difference between the $A$ -operator semi-norm and $(A,q)$ -Crawford number, which is defined as

g_{c_{A,q}}(T):=\|T\|_{A}-c_{A,q}(T)

for $T\in L^{A}(H),\ q\in\mathbb{C},\ 0<|q|\leq 1.$
In this study, the contents are as follows. In Section 2, some estimates are given for the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number via the $A$ -numerical radius and $A$ -Crawford number, respectively, for the $A$ -bounded linear operators in any complex semi-Hilbert space. Then, some evolutions are studied for the tensor product of two operators. We also consider several examples on Hilbert space and establish $\omega_{q}(\cdot)$ only for a real number $q\in[0,1]$ to illustrate our results. In Section 3, some convergence properties of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number via $A$ -uniformly convergence of operator sequences are investigated. In Section 4, a few applications of some operator functions classes are also given.

2. Some properties of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number

Theorem 1.

For the operator $T\in L^{A}(H)$ and $q,\lambda,\mu\in\mathbb{C},\ 0<|q|\leq 1,$ the following are true
(1) $\omega_{A,q}(T)\leq\|T\|_{A}.$
(2) If $\alpha\in\mathbb{C},\ |\alpha|=1,$ then $\omega_{A,q}(\alpha T)=\omega_{A,\alpha q}(T).$
(3) If $\alpha\in\mathbb{C},\ |\alpha|=1,$ then $c_{A,q}(\alpha T)=c_{A,\alpha q}(T).$
(4) If $\gamma=\sqrt{|\lambda|^{2}+|\mu|^{2}+2Re(\lambda\overline{\mu}q)}\neq 0,$ then $\gamma\omega_{A,\frac{\lambda+\mu\overline{q}}{\gamma}}(T)\leq|\lambda|\omega_{A}(T)+|\mu|\omega_{A,q}(T^{*}).$
(5) If $\gamma=\sqrt{|\lambda|^{2}+|\mu|^{2}+2Re(\lambda\overline{\mu}q)}\neq 0,$ then $\gamma c_{A,\frac{\lambda+\mu\overline{q}}{\gamma}}(T)\leq|\lambda|\omega_{A}(T)+|\mu|c_{A,q}(T^{*}).$
(6) For $x,y\in H,\ \|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ 0<|q|\leq 1$ , it is true

\|x\pm y\|_{A}=\sqrt{2}\sqrt{1\pm Req}.

(7) $\omega_{A}(T)\leq\omega_{A,q}(T)+\sqrt{2(1-Req)}\omega_{A,\frac{1-q}{\sqrt{2(1-Req)}}}(T),\ q\neq 1.$
(8) $c_{A}(T)\leq c_{A,q}(T)+\sqrt{2(1-Req)}\omega_{A,\frac{1-q}{\sqrt{2(1-Req)}}}(T),\ q\neq 1.$

Proof.

(1) For $x,y\in H,$ $\|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,$ we have

$\displaystyle\|\langle Tx,y\rangle_{A}\|$	$\displaystyle=$	$\displaystyle\|\langle ATx,y\rangle\|$
	$\displaystyle\leq$	$\displaystyle\|\langle ATx,Tx\rangle^{1/2}\|\|\langle Ay,y\rangle^{1/2}\|$
	$\displaystyle=$	$\displaystyle\|\langle Tx,Tx\rangle_{A}^{1/2}\|\|\langle y,y\rangle_{A}^{1/2}\|$
	$\displaystyle=$	$\displaystyle\\|Tx\\|_{A}\\|y\\|_{A}$
	$\displaystyle=$	$\displaystyle\\|Tx\\|_{A}.$

Hence, it is obtained that

\omega_{A,q}(T)\leq\|T\|_{A}.

(2)-(3) These equalities are results of the following relations

\|\alpha x\|_{A}=|\alpha|\|x\|_{A}=1,

\langle\alpha x,y\rangle_{A}=\alpha\langle x,y\rangle_{A}=\alpha q,\ 0<|\alpha q|=|q|\leq 1.

(4)-(5) For the operator $T\in L^{A}(H)$ and $\lambda,\mu\in\mathbb{C},$ the following relation

	$\displaystyle\|\langle T(\lambda x+\mu y),x\rangle_{A}\|$	$\displaystyle\leq$	$\displaystyle\|\lambda\|\|\langle Tx,x\rangle_{A}\|+\|\mu\|\|\langle Ty,x\rangle_{A}\|$
		$\displaystyle=$	$\displaystyle\|\lambda\|\|\langle Tx,x\rangle_{A}\|+\|\mu\|\|\langle T^{*}x,y\rangle_{A}\|$

is true. Since for $\|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ 0<|q|\leq 1,$

$\displaystyle\\|\lambda x+\mu y\\|_{A}^{2}$	$\displaystyle=$	$\displaystyle\langle A(\lambda x+\mu y),\lambda x+\mu y\rangle$
	$\displaystyle=$	$\displaystyle\langle A(\lambda x),\lambda x\rangle+\langle A(\lambda x),\mu y\rangle+\langle A(\mu y),\lambda x\rangle+\langle A(\mu y),\mu y\rangle$
	$\displaystyle=$	$\displaystyle\|\lambda\|^{2}\langle Ax,x\rangle+\lambda\overline{\mu}\langle Ax,y\rangle+\overline{\lambda}\mu\langle Ay,x\rangle+\|\mu\|^{2}\langle Ay,y\rangle$
	$\displaystyle=$	$\displaystyle\|\lambda\|^{2}\langle x,x\rangle_{A}+\lambda\overline{\mu}\langle x,y\rangle_{A}+\overline{\lambda}\mu\langle y,x\rangle_{A}+\|\mu\|^{2}\langle y,y\rangle_{A}$
	$\displaystyle=$	$\displaystyle\|\lambda\|^{2}+\|\mu\|^{2}+\lambda\overline{\mu}q+\overline{\lambda}\mu\overline{q}$
	$\displaystyle=$	$\displaystyle\|\lambda\|^{2}+\|\mu\|^{2}+2Re\left(\lambda\overline{\mu}q\right),$

then for $\gamma=\sqrt{|\lambda|^{2}+|\mu|^{2}+2Re(\lambda\overline{\mu}q)}$ we have $\|\dfrac{\lambda x+\mu y}{\gamma}\|_{A}=1$ and $\langle\frac{\lambda x+\mu y}{\gamma},x\rangle_{A}=\frac{\lambda+\overline{\mu}q}{\gamma}.$ Hence, from the above inequality and the definitions of $A$ -numerical and $(A,q)$ -numerical radii we have

\gamma\omega_{A,\frac{\lambda+\mu\overline{q}}{\gamma}}(T)\leq|\lambda|\omega_{A}(T)+|\mu|\omega_{A,q}(T^{*}).

Similarly, from the above inequality and the definitions of $A$ -numerical radius and $(A,q)$ -Crawford number it is implies

\gamma c_{A,\frac{\lambda+\mu\overline{q}}{\gamma}}(T)\leq|\lambda|\omega_{A}(T)+|\mu|c_{A,q}(T^{*}).

(6) For $x,y\in H,\ \|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ 0<|q|\leq 1$ , it is true

$\displaystyle\\|x\pm y\\|^{2}_{A}$	$\displaystyle=$	$\displaystyle\langle A(x\pm y),x\pm y\rangle$	(2)
	$\displaystyle=$	$\displaystyle\langle Ax,x\rangle\pm\langle Ax,y\rangle\pm\langle Ay,x\rangle+\langle Ay,y\rangle$
	$\displaystyle=$	$\displaystyle\\|x\\|_{A}^{2}\pm 2Re\langle x,y\rangle_{A}+\\|y\\|_{A}^{2}$
	$\displaystyle=$	$\displaystyle 2\pm 2Req.$

Hence, we have

\|x\pm y\|_{A}=\sqrt{2}\sqrt{1\pm Req}.

(7) For any $x,y\in H,\ \|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ q\neq 1,$ we have

	$\displaystyle\|\langle Tx,x\rangle_{A}\|$	$\displaystyle\leq$	$\displaystyle\|\langle Tx,y\rangle_{A}\|+\|\langle Tx,x-y\rangle_{A}\|$
		$\displaystyle=$	$\displaystyle\|\langle Tx,y\rangle_{A}\|+\sqrt{2(1-Req)}\|\langle Tx,\frac{x-y}{\sqrt{2(1-Req)}}\rangle_{A}\|.$

Then, from (6), we have $\|\frac{x-y}{\sqrt{2(1-Req)}}\|_{A}=1$ and $\langle x,\frac{x-y}{\sqrt{2(1-Req)}}\rangle=\frac{1-q}{\sqrt{2(1-Req)}}.$ Hence, from the previous relation and the definitions of $A$ -numerical and $(A,q)$ -numerical radii we have

\omega_{A}(T)\leq\omega_{A,q}(T)+\sqrt{2(1-Req)}\omega_{A,\frac{1-q}{\sqrt{2(1-Req)}}}(T).

(8) This claim can be proved similarly.

∎

Note: From the claims (7) and (8) of Theorem 1, if $q=1,$ then

\omega_{A,q}(T)=\omega_{A}(T),\ c_{A,q}(T)=c_{A}(T)

for any operator $T\in L^{A}(H).$

Note: From Theorem 1, the following is true

(1-\sqrt{2(1-Req)})\omega_{A}(T)\leq\omega_{A,q}(T)

for any $T\in L^{A}(H)$ and $q\in\mathbb{C},\ 0<|q|\leq 1.$

Theorem 2.

For the operator $T\in L^{A}(H)$ and $q\in\mathbb{C},\ 0<|q|\leq 1,$ the following is true

2|Req|\omega_{A}(T)\leq\omega_{A,q}(T)+\omega_{A,\overline{q}}(T)\leq 2\omega_{A}(T)+2\sqrt{2}\sqrt{1-Req}\|T\|_{A}.

Proof.

For $x,y\in H,\ \|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ 0<|q|\leq 1,$ it is clear that

$\displaystyle\langle T(x\pm y),x\pm y\rangle_{A}$	$\displaystyle=$	$\displaystyle\langle AT(x\pm y),x\pm y\rangle$
	$\displaystyle=$	$\displaystyle\langle ATx,x\rangle\pm\langle ATx,y\rangle\pm\langle ATy,x\rangle+\langle ATy,y\rangle$
	$\displaystyle=$	$\displaystyle\langle Tx,x\rangle_{A}\pm\langle Tx,y\rangle_{A}\pm\langle Ty,x\rangle_{A}+\langle Ty,y\rangle_{A}.$

Since,

\displaystyle|\langle T(x\pm y),x\pm y\rangle_{A}|\leq|\langle Tx,x\rangle_{A}|+|\langle Tx,y\rangle_{A}|+|\langle Ty,x\rangle_{A}|+|\langle Ty,y\rangle_{A}|,

then for $-1\leq Req<0$ , we have

2(1-Req)|\langle T\left(\frac{x-y}{\sqrt{2(1-Req)}}\right),\frac{x-y}{\sqrt{2(1-Req)}}\rangle_{A}|\leq 2\omega_{A}(T)+\omega_{A,q}(T)+\omega_{A,\overline{q}}(T).

Consequently,

2(1-Req)\omega_{A}(T)\leq 2\omega_{A}(T)+\omega_{A,q}(T)+\omega_{A,\overline{q}}(T).

That is, for $q\in\mathbb{C},\ 0<|q|\leq 1,\ -1\leq Req<0,$

-2Req\omega_{A}(T)\leq\omega_{A,q}(T)+\omega_{A,\overline{q}}(T).

Similarly, for the $0\leq Req\leq 1$ , we have

2(1+Req)|\langle T\left(\frac{x+y}{\sqrt{2(1+Req)}}\right),\frac{x+y}{\sqrt{2(1+Req)}}\rangle|\leq 2\omega_{A}(T)+\omega_{A,q}(T)+\omega_{A,\overline{q}}(T).

Then

2(1+Req)\omega_{A}(T)\leq 2\omega_{A}(T)+\omega_{A,q}(T)+\omega_{A,\overline{q}}(T).

Hence, we have

2Req\omega_{A}(T)\leq\omega_{A,q}(T)+\omega_{A,\overline{q}}(T).

Consequently, from these facts, it is established that

2|Req|\omega_{A}(T)\leq\omega_{A,q}(T)+\omega_{A,\overline{q}}(T)

for $q\in\mathbb{C},\ 0<|q|\leq 1.$
On the other hand, for $x,y\in H,\ \|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ 0<|q|\leq 1,$ it is clear that

|\langle Tx,y\rangle_{A}|\leq|\langle Tx,x\rangle_{A}|+|\langle Tx,y-x\rangle_{A}|.

Then, we have

	$\displaystyle\omega_{A,q}(T)$	$\displaystyle\leq$	$\displaystyle\omega_{A}(T)+\\|T\\|_{A}\\|x-y\\|_{A}$
		$\displaystyle\leq$	$\displaystyle\omega_{A}(T)+\sqrt{2(1-Req)}\\|T\\|_{A}.$

Similarly, we get

\omega_{A,\overline{q}}(T)\leq\omega_{A}(T)+\sqrt{2(1-Re\overline{q})}\|T\|_{A}.

Then, from the first part of this theorem we have

\omega_{A,q}(T)+\omega_{A,\overline{q}}(T)\leq 2\omega_{A}(T)+2\sqrt{2}\sqrt{1-Req}\|T\|_{A}.

∎

Lemma 1.

[20] Suppose $0\leq q\leq 1$ and $T\in M_{2}(\mathbb{C})$ . Then $T$ is unitarily similar to $e^{it}\begin{pmatrix}\gamma&a\\ b&\gamma\end{pmatrix}$ for some $0\leq t\leq 2\pi$ and $0\leq b\leq a$ . Also,

\displaystyle W_{q}(T)=e^{it}\left\{\gamma q+r\left((c+pd)\cos(s)+i(d+pc)\sin(s)\right):0\leq r\leq 1,0\leq s\leq 2\pi\right\},

with $c=\frac{a+b}{2},d=\frac{a-b}{2}~\mbox{and}~p=\sqrt{1-q^{2}}.$

Remark 1.

Setting $A=I$ in Theorem 2, and consider $q\in[0,1]$ , we have

\displaystyle 2q\omega(T)\leq 2\omega_{q}(T)\leq 2\omega(T)+2\sqrt{2}\sqrt{1-q}\|T\|.

(3)

Example 1.

Let $T=\begin{pmatrix}0&\frac{1}{70}\\ 0&0\end{pmatrix}$ and $q\in[0,1]$ . Using Lemma 1, we get

W_{q}(T)=\left\{\frac{re^{is}}{140}(1+\sqrt{1-q^{2}}):0\leq r\leq 1,0\leq s\leq 2\pi\right\}.

So,

\omega_{q}(T)=\frac{1}{140}(1+\sqrt{1-q^{2}}).

Since $T^{2}=0$ , we have $\omega(T)=\frac{\|T\|}{2}=\frac{1}{140}$ . From the Remark 1, we get $2q\omega(T)=\frac{q}{70}$ , $2\omega_{q}(T)=\frac{1}{70}(1+\sqrt{1-q^{2}})$ , and using the fact $\|T\|=\frac{1}{70}$ , the value of $2\omega(T)+2\sqrt{2}\sqrt{1-q}\|T\|=\frac{1}{70}(1+2\sqrt{2}\sqrt{1-q^{2}}).$ Figure 1 compares the upper and lower bounds of $2\omega_{q}(T)$ .

Refer to caption — Figure 1: Comparision of $\omega_{q}(T)$ with the upper and lower bounds (3) for Example 1.

Theorem 3.

If $A_{1}\in L(H_{1}),\ A_{2}\in L(H_{2}),\ A_{1}\geq 0,\ A_{2}\geq 0$ and $T_{1}\in L^{A_{1}}(H_{1}),\ T_{2}\in L^{A_{2}}(H_{2}).$ Then, for $A=A_{1}\otimes A_{2},\ T=T_{1}\otimes T_{2},$ we have

c_{A,q}(T)\leq c_{A_{1},q_{1}}(T_{1})c_{A_{2},q_{2}}(T_{2})\leq\omega_{A_{1},q_{1}}(T_{1})\omega_{A_{2},q_{2}}(T_{2})\leq\omega_{A,q}(T),

where, $q_{1},q_{2}\in\mathbb{C},\ 0<|q_{1}|\leq 1,\ 0<|q_{2}|\leq 1,\ q=q_{1}q_{2}.$

Proof.

For any $x=x_{1}\otimes x_{2},\ y=y_{1}\otimes y_{2}\in H_{1}\otimes H_{2},$ we have $\langle x,y\rangle_{A}=\langle x_{1},y_{1}\rangle_{A_{1}}\langle x_{2},y_{2}\rangle_{A_{2}}=q_{1}q_{2},$ $\|x\|_{A}=\|x_{1}\|_{A_{1}}\|x_{2}\|_{A_{2}},\ \|y\|_{A}=\|y_{1}\|_{A_{1}}\|y_{2}\|_{A_{2}}$ and

$\displaystyle\langle Tx,y\rangle_{A}$	$\displaystyle=$	$\displaystyle\langle(A_{1}\otimes A_{2})(T_{1}x_{1}\otimes T_{2}x_{2}),y_{1}\otimes y_{2}\rangle$
	$\displaystyle=$	$\displaystyle\langle A_{1}T_{1}x_{1}\otimes A_{2}T_{2}x_{2},y_{1}\otimes y_{2}\rangle$
	$\displaystyle=$	$\displaystyle\langle T_{1}x_{1},y_{1}\rangle_{A_{1}}\langle T_{2}x_{2},y_{2}\rangle_{A_{2}}.$

If $\|x_{1}\|_{A_{1}}=\|x_{2}\|_{A_{2}}=\|y_{1}\|_{A_{1}}=\|y_{2}\|_{A_{2}}=1,$ then for $\|x\|_{A}=\|y\|_{A}=1$ and $\langle x,y\rangle_{A}=q,\ q=q_{1}q_{2}$ , we have

W_{A_{1},q_{1}}(T_{1})W_{A_{2},q_{2}}(T_{2})\subset W_{A,q}(T).

On the other hand, since

			$\displaystyle\sup\left\{\|\langle T_{1}x_{1},y_{1}\rangle_{A_{1}}\|\|\langle T_{2}x_{2},y_{2}\rangle_{A_{2}}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle=$	$\displaystyle\sup\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle\leq$	$\displaystyle\sup\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x\\|_{A}=\\|y\\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ q=q_{1}q_{2}\right\}$

and

			$\displaystyle\inf\left\{\|\langle T_{1}x_{1},y_{1}\rangle_{A_{1}}\|\|\langle T_{2}x_{2},y_{2}\rangle_{A_{2}}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle=$	$\displaystyle\inf\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle\geq$	$\displaystyle\inf\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x\\|_{A}=\\|y\\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ q=q_{1}q_{2}\right\},$

then we get

\omega_{A_{1},q_{1}}(T_{1})\omega_{A_{2},q_{2}}(T_{2})\leq\omega_{A,q}(T),

c_{A_{1},q_{1}}(T_{1})c_{A_{2},q_{2}}(T_{2})\geq c_{A,q}(T).

∎

Using an argument similar to that used in the proof of the previous theorem, it is easy to prove the following result.

Corollary 1.

(1) If $c_{A_{1},q_{1}}(T_{1})>0,$ then $W_{A_{2},q_{2}}(T_{2})\subset W_{A,q}(T)W_{A_{1},q_{1}}^{-1}(T_{1}).$
(2) If $c_{A_{2},q_{2}}(T_{2})>0$ , then $W_{A_{1},q_{1}}(T_{1})\subset W_{A,q}(T)W_{A_{2},q_{2}}^{-1}(T_{2})$ .
Consequently,
if $c_{A_{1},q_{1}}(T_{1})>0,$ then $\omega_{A_{2},q_{2}}(T_{2})\leq\frac{\omega_{A,q}(T)}{c_{A_{1},q_{1}}(T_{1})},$
if $c_{A_{2},q_{2}}(T_{2})>0,$ then $\omega_{A_{1},q_{1}}(T_{1})\leq\frac{\omega_{A,q}(T)}{c_{A_{2},q_{2}}(T_{2})},$
if $\omega_{A_{1},q_{1}}(T_{1})>0,$ then $c_{A_{2},q_{2}}(T_{2})\leq\frac{c_{A,q}(T)}{\omega_{A_{1},q_{1}}(T_{1})},$
if $\omega_{A_{2},q_{2}}(T_{2})>0,$ then $c_{A_{1},q_{1}}(T_{1})\leq\frac{c_{A,q}(T)}{\omega_{A_{2},q_{2}}(T_{2})}.$
Here, for $S=\{z\in\mathbb{C}:z\neq 0\},\ S^{-1}=\left\{\dfrac{1}{z}:z\in S\right\}.$

3. Some convergence properties of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number via operator sequences

Definition 2.

A sequence $(T_{n})\subset L^{A}(H)$ is said to converge uniformly to $T\in L^{A}(H),$ if for any $\epsilon>0,$ there exists a positive integer $N$ such that

\|T_{n}-T\|_{A}<\epsilon\ \text{ for all}\ n>N.

Theorem 4.

For any $T\in L^{A}(H)$ and $q\in\mathbb{C},\ 0<|q|\leq 1$ , the following are true:
(1) $|\omega_{A,q}(T)-\omega_{A}(T)|\leq\sqrt{2(1-Req)}\|T\|_{A}.$
(2) If $q_{n}\in\mathbb{C},\ 0<|q_{n}|\leq 1$ and $Req_{n}\rightarrow 1,\ n\rightarrow\infty,$ then

\lim\limits_{n\rightarrow\infty}\omega_{A,q_{n}}(T)=\omega_{A}(T).

(3) If the operator sequences $(T_{n})$ converges uniformly to an operator $T$ in $L^{A}(H),$ then

\lim\limits_{n\rightarrow\infty}\omega_{A,q}(T_{n})=\omega_{A,q}(T).

Proof.

(1) For any $x,y\in H$ with the properties $\|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ 0<|q|\leq 1,$ it is valid that

$\displaystyle\|\langle Tx,y\rangle_{A}\|$	$\displaystyle=$	$\displaystyle\|\langle ATx,y\rangle\|$
	$\displaystyle\leq$	$\displaystyle\|\langle ATx,x\rangle\|+\|\langle ATx,y-x\rangle\|$
	$\displaystyle\leq$	$\displaystyle\|\langle Tx,x\rangle_{A}\|+\langle ATx,Tx\rangle^{1/2}\langle A(y-x),y-x\rangle^{1/2}$
	$\displaystyle=$	$\displaystyle\|\langle Tx,x\rangle_{A}\|+\\|Tx\\|_{A}\\|y-x\\|_{A}.$

Now, using inequality (2) and the previous estimate, we get

|\langle Tx,y\rangle_{A}|\leq|\langle Tx,x\rangle_{A}|+\sqrt{2}\sqrt{1-Req}\|T\|_{A}.

This means that

\omega_{A,q}(T)\leq\omega_{A}(T)+\sqrt{2}\sqrt{1-Req}\|T\|_{A}.

On the other hand, similarly from the following inequality,

|\langle Tx,x\rangle_{A}|\leq|\langle Tx,y\rangle_{A}|+|\langle Tx,x-y\rangle_{A}|,

it can be easily to shown that

\omega_{A}(T)\leq\omega_{A,q}(T)+\sqrt{2}\sqrt{1-Req}\|T\|_{A}.

Consequently, it is established that

|\omega_{A,q}(T)-\omega_{A}(T)|\leq\sqrt{2(1-Req)}\|T\|_{A}.

(2) This is a result of the property of the first proposition of this theorem.
(3) From the definition of the $(A,q)$ -numerical radius, we have

\omega_{A,q}(T+S)\leq\omega_{A,q}(T)+\omega_{A,q}(S)

for $T,S\in L^{A}(H)$ and $q\in\mathbb{C},\ 0<|q|\leq 1.$ So, the following claim is true

|\omega_{A,q}(T)-\omega_{A,q}(S)|\leq\omega_{A,q}(T\pm S).

Hence, for any $n\geq 1$ , from the claim (1) of Theorem 1, it is obtained that

|\omega_{A,q}(T_{n})-\omega_{A,q}(T)|\leq\omega_{A,q}(T_{n}-T)\leq\|T_{n}-T\|_{A}.

From the last relation and the uniform convergence of the sequence $(T_{n})$ to $T$ in $L^{A}(H),$ it follows that

\lim\limits_{n\rightarrow\infty}\omega_{A,q}(T_{n})=\omega_{A,q}(T).

∎

Remark 2.

Setting $A=I$ in Theorem 4 (1), and considering $q\in[0,1]$ , we have

\displaystyle|\omega_{q}(T)-\omega(T)|\leq\sqrt{2(1-q)}\|T\|,~\mbox{for any}~T\in L(H).

(4)

Here we consider some examples only for a real number $q\in[0,1]$ .

Example 2.

Let $T=\begin{pmatrix}0&\frac{1}{24}\\ 0&0\end{pmatrix}$ and $q\in[0,1]$ . Using Lemma 1, we get

W_{q}(T)=\left\{\frac{re^{is}}{48}(1+\sqrt{1-q^{2}}):0\leq r\leq 1,0\leq s\leq 2\pi\right\}.

So,

\omega_{q}(T)=\frac{1}{48}(1+\sqrt{1-q^{2}}).

By Remark 2 and the fact that $\|T\|=\frac{1}{24}$ , we see that the right side of Remark 2 is $\sqrt{2(1-q)}\|T\|=\frac{\sqrt{2(1-q)}}{24}$ . Since $T^{2}=0$ , we have $\omega(T)=\frac{\|T\|}{2}=\frac{1}{48}$ . Therefore, the left side of Remark 2 is $|\omega_{q}(T)-\omega(T)|=\frac{\sqrt{1-q^{2}}}{48}.$ Figure 2 compares $|\omega_{q}(T)-\omega(T)|=\frac{\sqrt{1-q^{2}}}{48}$ and the upper bound $\frac{\sqrt{2(1-q)}}{24}$ .

Example 3.

Let us consider $T=\frac{1}{20}I$ and $q\in[0,1]$ . In this case, $\omega_{q}(T)=\frac{1}{20}q.$ Using $\omega(T)=\|T\|=\frac{1}{20},$ the left side of Remark 2 is $|\omega_{q}(T)-\omega(T)|=\frac{1}{20}(q-1)$ , while the right side of Remark 2 is $\sqrt{2(1-q)}\|T\|=\frac{\sqrt{2(1-q)}}{20}$ . Figure 3 compares $|\omega_{q}(T)-\omega(T)|=\frac{1}{20}(q-1)$ and the upper bound $\frac{\sqrt{2(1-q)}}{20}$ .

Example 4.

Let $T=\begin{pmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}$ and $q\in[\frac{1}{2},1]$ . Using [17, Proposition 4.3], we get

\omega_{q}(T)=\frac{1}{8}\sqrt{27+18q-13q^{2}+(9+7q)\sqrt{(1-q)(9+7q)}}.

By Remark 2 and the fact that $\|T\|=1$ , we see that the right side of inequality (4) is $\sqrt{2(1-q)}\|T\|=\sqrt{2(1-q)}$ . Also, we know that $\omega(T)=\frac{1}{\sqrt{2}}$ . Therefore, the left side of inequality (4) is $|\omega_{q}(T)-\omega(T)|=\frac{1}{8}\sqrt{27+18q-13q^{2}+(9+7q)\sqrt{(1-q)(9+7q)}}-\frac{1}{\sqrt{2}}.$ Figure 4 compares $|\omega_{q}(T)-\omega(T)|$ and the upper bound $\sqrt{2(1-q)}$ .

Theorem 5.

For any $T\in L^{A}(H)$ and $q\in\mathbb{C},\ 0<|q|\leq 1,$ the following are true:
(1) $|c_{A,q}(T)-c_{A}(T)|\leq\sqrt{2(1-Req)}\|T\|_{A}.$
(2) If $q_{n}\in\mathbb{C},\ 0<|q_{n}|\leq 1$ and $Req_{n}\rightarrow 1,\ n\rightarrow\infty,$ then

\lim\limits_{n\rightarrow\infty}c_{A,q_{n}}(T)=c_{A}(T).

(3) For any $T,S\in L^{A}(H)$ and $q\in\mathbb{C},\ 0<|q|\leq 1,$ we have

|c_{A,q}(T)-c_{A,q}(S)|\leq\omega_{q}(T\pm S).

(4) If the operator sequence $(T_{n})$ converges uniformly to an operator $T$ in $L^{A}(H),$ then

\lim\limits_{n\rightarrow\infty}c_{A,q}(T_{n})=c_{A,q}(T).

Proof.

(1) For $x,y\in H,$ $\|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,$ we have

$\displaystyle\|\langle Tx,y\rangle_{A}\|$	$\displaystyle\geq$	$\displaystyle\|\langle Tx,x\rangle_{A}+\langle Tx,y-x\rangle_{A}\|$
	$\displaystyle=$	$\displaystyle\|\langle Tx,x\rangle_{A}\|-\|\langle Tx,y-x\rangle_{A}\|$
	$\displaystyle=$	$\displaystyle\|\langle Tx,x\rangle_{A}\|-\|\langle ATx,y-x\rangle\|$
	$\displaystyle\geq$	$\displaystyle\|\langle Tx,x\rangle_{A}\|-\|\langle ATx,Tx\rangle\|^{1/2}\langle A(y-x),y-x\rangle^{1/2}$
	$\displaystyle=$	$\displaystyle\|\langle Tx,x\rangle_{A}\|-\\|Tx\\|_{A}\langle y-x,y-x\rangle^{1/2}_{A}$
	$\displaystyle\geq$	$\displaystyle\|\langle Tx,x\rangle_{A}\|-\\|T\\|_{A}\sqrt{2}\sqrt{1-Req}.$

From this and from the definition of the Crawford number, we have

c_{A,q}(T)\geq c_{A}(T)-\|T\|_{A}\sqrt{2}\sqrt{1-Req},

that is,

c_{A,q}(T)-c_{A}(T)\geq-\|T\|_{A}\sqrt{2}\sqrt{1-Req}.

Similarly, by the previous method, it can be shown that

c_{A,q}(T)-c_{A}(T)\leq\|T\|_{A}\sqrt{2}\sqrt{1-Req}.

Consequently, it is true

|c_{A,q}(T)-c_{A}(T)|\leq\sqrt{2(1-Req)}\|T\|_{A}.

(2) The validity of this result is clear from the first claim.
(3) For any $x,y\in H,$ $\|x\|_{A}=\|y\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ q\in\mathbb{C},\ 0<|q|\leq 1$ , we have

	$\displaystyle\|\langle Tx,y\rangle_{A}\|$	$\displaystyle=$	$\displaystyle\|\langle(T+S)x,y\rangle_{A}-\langle Sx,y\rangle_{A}\|$
		$\displaystyle\geq$	$\displaystyle\|\langle(T+S)x,y\rangle_{A}\|-\|\langle Sx,y\rangle_{A}\|.$

From the last relation, we have

c_{A,q}(T)\geq c_{A,q}(T+S)-\omega_{A,q}(S),

that is,

c_{A,q}(T+S)-c_{A,q}(T)\leq\omega_{A,q}(S).

In a similar way, it can be established that

-\omega_{A,q}(S)\leq c_{A,q}(T+S)-c_{A,q}(T).

Consequently, from the previous results, we have

|c_{A,q}(T+S)-c_{A,q}(T)|\leq\omega_{A,q}(S).

Hence, it is clear that

|c_{A,q}(T)-c_{A,q}(S)|\leq\omega_{A,q}(T-S).

(4) The validity of this result follows from the last claim and the inequality

\omega_{A,q}(T_{n}-T)\leq\|T_{n}-T\|_{A},\ n\geq 1.

(See the first claim of Theorem 1.) ∎

Corollary 2.

For the operator sequence $(T_{n})$ in $L^{A}(H),$ which converges uniformly to an operator $T\in L^{A}(H),$ the following are true:

\lim\limits_{n\rightarrow\infty}g_{\omega_{A,q}}(T_{n})=g_{\omega_{A,q}}(T)

and

\lim\limits_{n\rightarrow\infty}g_{c_{A,q}}(T_{n})=g_{c_{A,q}}(T).

Proof.

Indeed, for the gaps, we have

$\displaystyle\|g_{\omega_{A,q}}(T_{n})-g_{\omega_{A,q}}(T)\|$	$\displaystyle=$	$\displaystyle\|\left(\\|T_{n}\\|_{A}-\omega_{A,q}(T_{n})\right)-\left(\\|T\\|_{A}-\omega_{A,q}(T)\right)\|$
	$\displaystyle\leq$	$\displaystyle\\|T_{n}-T\\|_{A}+\|\omega_{A,q}(T_{n})-\omega_{A,q}(T)\|$
	$\displaystyle\leq$	$\displaystyle 2\\|T_{n}-T\\|_{A},\ n\geq 1$

and

	$\displaystyle\|g_{c_{{A,q}}}(T_{n})-g_{c_{{A,q}}}(T)\|$	$\displaystyle\leq$	$\displaystyle\\|T_{n}-T\\|_{A}+\|c_{A,q}(T_{n})-c_{A,q}(T)\|$
		$\displaystyle\leq$	$\displaystyle 2\\|T_{n}-T\\|_{A},\ n\geq 1.$

Since $\|T_{n}-T\|_{A}\rightarrow 0,\ n\rightarrow\infty,$ then the validity of the corollary is clear. ∎

4. Some applications

In this section, some applications of our results will be given.

Application 1 Let $H_{1}$ and $H_{2}$ be two complex Hilbert spaces, $A_{1}\in L(H_{1}),\ A_{2}\in L(H_{2}),\ A_{1}\geq 0,\ A_{2}\geq 0$ and $A=A_{1}\oplus A_{2}.$ Consider the following operator sequences

T_{n}=S_{n}\oplus M_{n},\ T_{n}\in L^{A}\left(H_{1}\oplus H_{2}\right),\ n\geq 1.

Assume that $S_{n}\rightarrow S$ and $\ M_{n}\rightarrow M$ uniformly for some $S$ and $M$ in $L^{A_{1}}(H_{1})$ and $L^{A_{2}}(H_{2}),$ respectively. Also, let

T=S\oplus M.

It is clear that $\|T\|_{A}=\max\left\{\|S\|_{A_{1}},\|M\|_{A_{2}}\right\}.$
For any $q\in\mathbb{C},\ |q|\leq 1,$ it is clear that

\displaystyle W_{A_{1},q}(S)\subset W_{A,q}(T),\ W_{A_{2},q}(M)\subset W_{A,q}(T).

(5)

Therefore, $\max\left\{\omega_{A_{1},q}(S),\omega_{A_{2},q}(M)\right\}\leq\omega_{A,q}(T).$ Hence, it is easy to see that

	$\displaystyle g_{\omega_{A,q}}(T)=\\|T\\|_{A}-\omega_{A,q}(T)$
	$\displaystyle\leq\max\left\{\\|S\\|_{A_{1}},\\|M\\|_{A_{2}}\right\}-\max\left\{\omega_{A_{1},q}(S),\omega_{A_{2},q}(M)\right\}$
	$\displaystyle=\left(\frac{\\|S\\|_{A_{1}}+\\|M\\|_{A_{2}}}{2}+\|\frac{\\|S\\|_{A_{1}}-\\|M\\|_{A_{2}}}{2}\|\right)-\left(\frac{\omega_{A_{1},q}(S)+\omega_{A_{2},q}(M)}{2}+\|\frac{\omega_{A_{1},q}(S)-\omega_{A_{2},q}(M)}{2}\|\right)$
	$\displaystyle\leq\frac{\\|S\\|_{A_{1}}-\omega_{A_{1},q}(S)}{2}+\frac{\\|M\\|_{A_{2}}-\omega_{A_{2},q}(M)}{2}+\|\frac{\\|S\\|_{A_{1}}-\\|M\\|_{A_{2}}}{2}-\frac{\omega_{A_{1},q}(S)-\omega_{A_{2},q}(M)}{2}\|$
	$\displaystyle=\frac{g_{\omega_{A_{1},q}}(S)+g_{\omega_{A_{2},q}}(M)}{2}+\|\frac{g_{\omega_{A_{1},q}}(S)-g_{\omega_{A_{2},q}}(M)}{2}\|$
	$\displaystyle=\max\left\{g_{\omega_{A_{1},q}}(S),g_{\omega_{A_{2},q}}(M)\right\}$

for any $q\in\mathbb{C},\ |q|\leq 1.$ That is, for any $q\in\mathbb{C},\ |q|\leq 1$ , it is established that

g_{\omega_{A,q}}(T)\leq\max\left\{g_{\omega_{A_{1},q}}(S),g_{\omega_{A_{2},q}}(M)\right\}.

Then, from the Corollary 2, we have

\lim\limits_{n\rightarrow\infty}g_{\omega_{A,q}}(T_{n})\leq\max\left\{g_{\omega_{A_{1},q}}(S),g_{\omega_{A_{2},q}}(M)\right\}.

On the other hand, from (5), it is true that

c_{{A,q}}(T)\leq inf\left\{c_{{A_{1},q}}(S),c_{{A_{2},q}}(M)\right\}

for any $q\in\mathbb{C},\ |q|\leq 1.$ Then,

	$\displaystyle g_{c_{A,q}}(T)=\\|T\\|_{A}-c_{A,q}(T)$
	$\displaystyle\geq\max\left\{\\|S\\|_{A_{1}},\\|M\\|_{A_{2}}\right\}-\inf\left\{c_{A_{1},q}(S),c_{A_{2},q}(M)\right\}$
	$\displaystyle=\left(\frac{\\|S\\|_{A_{1}}+\\|M\\|_{A_{2}}}{2}+\|\frac{\\|S\\|_{A_{1}}-\\|M\\|_{A_{2}}}{2}\|\right)-\left(\frac{c_{A_{1},q}(S)+c_{A_{2},q}(M)}{2}-\|\frac{c_{A_{1},q}(S)-c_{A_{2},q}(M)}{2}\|\right)$
	$\displaystyle\geq\frac{\\|S\\|_{A_{1}}-c_{A_{1},q}(S)}{2}+\frac{\\|M\\|_{A_{2}}-c_{A_{2},q}(M)}{2}+\|\frac{\\|S\\|_{A_{1}}-\\|M\\|_{A_{2}}}{2}-\frac{c_{A_{1},q}(S)-c_{A_{2},q}(M)}{2}\|$
	$\displaystyle=\frac{g_{c_{A_{1},q}}(S)+g_{c_{A_{2},q}}(M)}{2}+\|\frac{g_{c_{A_{1},q}}(S)-g_{c_{A_{2},q}}(M)}{2}\|$
	$\displaystyle=\max\left\{g_{c_{A_{1},q}}(S),g_{c_{A_{2},q}}(M)\right\}.$

Then, by Corollary 2, we have

\lim\limits_{n\rightarrow\infty}g_{c_{A,q}}(T_{n})\geq\max\left\{g_{c_{A_{1},q}}(S),g_{c_{A_{2},q}}(M)\right\}

for any $q\in\mathbb{C},\ |q|\leq 1.$
Application 2 Let $\psi\in C[0,1],\ \psi\geq 0,\ \varphi_{n}\in C[0,1],\ n\geq 1,\ \varphi_{n}\rightarrow 1$ uniformly on $[0,1]$ and

Af(x)=\psi(x)f(x),\ f\in L_{2}(0,1),

A:L_{2}(0,1)\rightarrow L_{2}(0,1),

T_{n}f(x)=\varphi_{n}(x)f(x),\ f\in L_{2}^{\psi}(0,1),

where,

L_{2}^{\psi}(0,1)=\left\{f:(0,1)\rightarrow\mathbb{C}:f\ \text{Lebesque measurable function and}\ \int\limits_{0}^{1}\psi(x)|f(x)|^{2}dx<\infty\right\}.

Then, the operator sequences $T_{n},\ n\geq 1,$ converges uniformly to the identity operator $I$ in $L_{2}^{\psi}(0,1).$ Hence, for any $q\in\mathbb{C},\ |q|\leq 1,$ by Corollary 2, we have

\lim\limits_{n\rightarrow\infty}g_{\omega_{A,q}}(T_{n})=g_{\omega_{A,q}}(T)=1-|q|,

\lim\limits_{n\rightarrow\infty}g_{c_{A,q}}(T_{n})=g_{c_{A,q}}(T)=1-|q|.

Application 3 A prominent function class $(\Lambda_{\theta})_{+}$ will be given (see [1]). Let $\theta$ be a modulus of continuity, i. e., $\theta$ is a non-decreasing continuous function on $[0,\infty)$ such that $\theta(0)=0,\ \theta(x)>0$ for $x>0,$ and

\theta(x+y)\leq\theta(x)+\theta(y),\ x,y\in[0,\infty).

In this case, $(\Lambda_{\theta})_{+}$ is defined as

(\Lambda_{\theta})_{+}:=\left\{f\in A(\mathbb{D}):|f|_{\Lambda_{\theta}}=\sup_{\begin{subarray}{c}a,b\in\mathbb{D}\\ a\neq b\end{subarray}}\frac{|f(a)-f(b)|}{\theta(|a-b|)}<\infty\right\},

where $\mathbb{D}:=\{z\in\mathbb{C}:|z|<1\}$ is the unit disc and $A(\mathbb{D})$ is the class of analytic functions on $\mathbb{D}.$
Also, when $\theta$ is given, $\theta_{*}$ will be defined as

\theta_{*}(x):=x\int\limits_{x}^{\infty}\frac{\theta(t)}{t^{2}}dt,\ x>0.

Remark that $\lim\limits_{x\rightarrow 0^{+}}\theta_{*}(x)=0.$

Theorem 6.

[1] For every modulus of continuity $\theta,$ for arbitrary contractions $S$ and $T,$ and $f\in(\Lambda_{\theta})_{+},$ there exists a constant $c>0$ such that

\|f(S)-f(T)\|\leq c|f|_{\Lambda_{\theta}}\theta_{*}(\|S-T\|).

Now, we will examine how the results in this section vary for operator functions.

Theorem 7.

Let $(T_{n}),\ n\geq 1$ in $L^{A}(H)$ be a sequence of contraction operators, which converges uniformly to an operator $T\in L^{A}(H)$ and $f\in(\Lambda_{\theta})_{+}.$ In this case,

\omega_{A,q}(f(T))=\lim\limits_{n\rightarrow\infty}\omega_{A,q}(f(T_{n})),\ q\in\mathbb{C},\ 0<|q|\leq 1,

c_{A,q}(f(T))=\lim\limits_{n\rightarrow\infty}c_{A,q}(f(T_{n})),\ q\in\mathbb{C},\ 0<|q|\leq 1

are true.

Proof.

Let $(T_{n}),\ n\geq 1$ in $L^{A}(H)$ be a sequence of contraction operators, which converges uniformly to an operator $T\in L^{A}(H)$ and $f\in(\Lambda_{\theta})_{+}.$ In this situation, $T$ is a contraction operator. Also, since $f\in(\Lambda_{\theta})_{+},$ then there exists $c>0$ such that $\|f(T)-f(T_{n})\|\leq c|f|_{\Lambda_{\theta}}\theta_{*}(\|T-T_{n}\|),\ n\geq 1,$ by Theorem 3. Thus, since $\lim\limits_{x\rightarrow 0^{+}}\theta_{*}(\|T-T_{n}\|)=0,$ then the operator sequence $(f(T_{n}))$ converges uniformly to $f(T).$ Therefore, the validity of the assertions of this theorem under the corresponding conditions is obvious from Theorems 4 and 5. ∎

Conflict of interest: The authors state that there is no conflict of interest.

Data availability: Data sharing not applicable to the present paper as no data sets were generated or analyzed during the current study.

Funding: Not applicable.

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$\displaystyle\|\langle Tx,y\rangle_{A}\|$	$\displaystyle=$	$\displaystyle\|\langle ATx,y\rangle\|$
	$\displaystyle\leq$	$\displaystyle\|\langle ATx,Tx\rangle^{1/2}\|\|\langle Ay,y\rangle^{1/2}\|$
	$\displaystyle=$	$\displaystyle\|\langle Tx,Tx\rangle_{A}^{1/2}\|\|\langle y,y\rangle_{A}^{1/2}\|$
	$\displaystyle=$	$\displaystyle\\|Tx\\|_{A}\\|y\\|_{A}$
	$\displaystyle=$	$\displaystyle\\|Tx\\|_{A}.$

	$\displaystyle\|\langle T(\lambda x+\mu y),x\rangle_{A}\|$	$\displaystyle\leq$	$\displaystyle\|\lambda\|\|\langle Tx,x\rangle_{A}\|+\|\mu\|\|\langle Ty,x\rangle_{A}\|$
		$\displaystyle=$	$\displaystyle\|\lambda\|\|\langle Tx,x\rangle_{A}\|+\|\mu\|\|\langle T^{*}x,y\rangle_{A}\|$

	$\displaystyle\|\langle Tx,x\rangle_{A}\|$	$\displaystyle\leq$	$\displaystyle\|\langle Tx,y\rangle_{A}\|+\|\langle Tx,x-y\rangle_{A}\|$
		$\displaystyle=$	$\displaystyle\|\langle Tx,y\rangle_{A}\|+\sqrt{2(1-Req)}\|\langle Tx,\frac{x-y}{\sqrt{2(1-Req)}}\rangle_{A}\|.$

			$\displaystyle\sup\left\{\|\langle T_{1}x_{1},y_{1}\rangle_{A_{1}}\|\|\langle T_{2}x_{2},y_{2}\rangle_{A_{2}}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle=$	$\displaystyle\sup\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle\leq$	$\displaystyle\sup\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x\\|_{A}=\\|y\\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ q=q_{1}q_{2}\right\}$

			$\displaystyle\inf\left\{\|\langle T_{1}x_{1},y_{1}\rangle_{A_{1}}\|\|\langle T_{2}x_{2},y_{2}\rangle_{A_{2}}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle=$	$\displaystyle\inf\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x_{1}\\|_{A_{1}}=\\|y_{1}\\|_{A_{1}}=1,\ \langle x_{1},y_{1}\rangle_{A_{1}}=q_{1},\right.$
			$\displaystyle\ \ \ \ \ \ \left.\\|x_{2}\\|_{A_{2}}=\\|y_{2}\\|_{A_{2}}=1,\ \langle x_{2},y_{2}\rangle_{A_{2}}=q_{2}\right\}$
		$\displaystyle\geq$	$\displaystyle\inf\left\{\|\langle Tx,y\rangle_{A}\|:\ \\|x\\|_{A}=\\|y\\|_{A}=1,\ \langle x,y\rangle_{A}=q,\ q=q_{1}q_{2}\right\},$

Some spectral properties and convergence of the (A,q)(A,q)-numerical radius and (A,q)(A,q)-Crawford number

Abstract

keywords:

1. Introduction

Definition 1.

2. Some properties of the (A,q)(A,q)-numerical radius and (A,q)(A,q)-Crawford number

Theorem 1.

Proof.

Theorem 2.

Proof.

Lemma 1.

Remark 1.

Example 1.

Theorem 3.

Proof.

Corollary 1.

3. Some convergence properties of the (A,q)(A,q)-numerical radius and (A,q)(A,q)-Crawford number via operator sequences

Definition 2.

Theorem 4.

Proof.

Remark 2.

Example 2.

Example 3.

Example 4.

Theorem 5.

Proof.

Corollary 2.

Proof.

4. Some applications

Theorem 6.

Theorem 7.

Proof.

References

Some spectral properties and convergence of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number

2. Some properties of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number

3. Some convergence properties of the $(A,q)$ -numerical radius and $(A,q)$ -Crawford number via operator sequences