Upper bounds for numerical radius inequalities involving off-diagonal operator matrices

TL;DR

This paper derives new upper bounds for the numerical radius of off-diagonal 2x2 operator matrices, involving functions of operators and generalized Euclidean radii, advancing the understanding of operator inequalities.

Contribution

It introduces novel upper bounds for the numerical radius of off-diagonal operator matrices using functions f and g, and explores inequalities for generalized Euclidean operator radius.

Findings

Established bounds for numerical radius involving off-diagonal blocks.

Derived inequalities using functions satisfying f(t)g(t)=t.

Extended results to generalized Euclidean operator radius.

Abstract

In this paper, we establish some upper bounds for numerical radius inequalities including of $2\times 2$ operator matrices and their off-diagonal parts. Among other inequalities, it is shown that if $T=\left[\begin{array}{cc} 0&X, Y&0 \end{array}\right]$, then \begin{align*} \omega^{r}(T)\leq 2^{r-2}\left\|f^{2r}(|X|)+g^{2r}(|Y^*|)\right\|^\frac{1}{2}\left\|f^{2r}(|Y|)+g^{2r}(|X^*|)\right\|^\frac{1}{2} \end{align*} and \begin{align*} \omega^{r}(T)\leq 2^{r-2}\left\|f^{2r}(|X|)+f^{2r}(|Y^*|)\right\|^\frac{1}{2}\left\|g^{2r}(|Y|)+g^{2r}(|X^*|)\right\|^\frac{1}{2}, \end{align*} where $X, Y$ are bounded linear operators on a Hilbert space ${\mathscr H}$, $r\geq 1$ and $f$, $g$ are nonnegative continuous functions on $[0, \infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0, \infty))$. Moreover, we present some inequalities involving the generalized Euclidean operator radius of operators $T_{1},\cdots,T_{n}$.

Equations

$${\omega }^r(T)\le 2^{r-2}{\Vert f^{2r}(|X|)+g^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert f^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}$$

$${\omega }^r(T)\le 2^{r-2}{\Vert f^{2r}(|X|)+f^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert g^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}},$$

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

$${\left(a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}^m+{\left(\frac{1}{2}\right)}^m{\left(a^{\frac{m}{2}}-b^{\frac{m}{2}}\right)}^2\le 2^{-m}(a+b{)}^m.$$

$$\omega (T)\le \frac{1}{2}\Vert |T|+|T^{\ast }|\Vert ,$$

$$\omega (T)\le \frac{1}{2}{\Vert |X|+|Y^{\ast }|\Vert }^{\frac{1}{2}}{\Vert |X^{\ast }|+|Y|\Vert }^{\frac{1}{2}}.$$

$${\omega }_p(T_1,\cdots ,T_n):=\underset{\Vert x\Vert =1}{\sup }{\left(\sum _{i=1}^n|⟨T_{i}x,x⟩{|}^p\right)}^{\frac{1}{p}}.$$

$${\omega }_p^p(T_1,\cdots ,T_n)\le \frac{1}{2}\Vert \sum _{i=1}^n\left(f^{2p}(|T_i|)+g^{2p}(|T_i^{\ast }|)\right)\Vert -\underset{\Vert x\Vert =1}{\inf }\zeta (x),$$

$$\zeta (x)=\frac{1}{2}\sum _{i=1}^n{\left({⟨f^{2p}(|T_i|)x,x⟩}^{\frac{1}{2}}-{⟨g^{2p}(|T_i^{\ast }|)x,x⟩}^{\frac{1}{2}}\right)}^2.$$

$${⟨Tx,x⟩}^r$$

$$\le ⟨T^{r}x,x⟩(\text{since}\Vert x\Vert \le 1\text{and}2r-2\ge 0).$$

$$|⟨Tx,y⟩{|}^2\le ⟨f^2(|T|)x,x⟩⟨g^2(|T^{\ast }|)y,y⟩.$$

$${\omega }^r(T)\le 2^{r-2}{\Vert f^{2r}(|X|)+g^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert f^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}$$

$${\omega }^r(T)\le 2^{r-2}{\Vert f^{2r}(|X|)+f^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert g^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}.$$

$$|⟨T\mathbf{x},\mathbf{x}⟩{|}^r$$

$$=|⟨X{x}_2,x_1⟩+⟨Y{x}_1,x_2⟩{|}^r$$

$$\le {\left(|⟨X{x}_2,x_1⟩|+|⟨Y{x}_1,x_2⟩|\right)}^r(\text{by the triangular inequality})$$

$$\le \frac{2^r}{2}\left(|⟨X{x}_2,x_1⟩{|}^r+|⟨Y{x}_1,x_2⟩{|}^r\right)(\text{by the convexity}f(t)=t^r)$$

$$\displaystyle\leq\frac{2^{r}}{2}\Big{(}\left(\left\langle f^{2}(|X|)x_{2},x_{2}\right\rangle^{\frac{1}{2}}\left\langle g^{2}(|X^{*}|)x_{1},x_{1}\right\rangle^{\frac{1}{2}}\right)^{r}$$

$$\displaystyle\qquad+\left(\left\langle f^{2}(|Y|)x_{1},x_{1}\right\rangle^{\frac{1}{2}}\left\langle g^{2}(|Y^{*}|)x_{2},x_{2}\right\rangle^{\frac{1}{2}}\right)^{r}\Big{)}\qquad(\textrm{by Lemma\,\,}\ref{5})$$

$$\le \frac{2^r}{2}\left({⟨f^{2r}(|X|)x_2,x_2⟩}^{\frac{1}{2}}{⟨g^{2r}|X^{\ast }|x_1,x_1⟩}^{\frac{1}{2}}+{⟨f^{2r}(|Y|)x_1,x_1⟩}^{\frac{1}{2}}{⟨g^{2r}(|Y^{\ast }|)x_2,x_2⟩}^{\frac{1}{2}}\right)$$

$$(\text{by Lemmaref3(a)})$$

$$\le \frac{2^r}{2}{\left(⟨f^{2r}(|X|)x_2,x_2⟩+⟨g^{2r}(|Y^{\ast }|)x_2,x_2⟩\right)}^{\frac{1}{2}}$$

$$\times {\left(⟨f^{2r}(|Y|)x_1,x_1⟩+⟨g^{2r}(|X^{\ast }|)x_1,x_1⟩\right)}^{\frac{1}{2}}(\text{by the Cauchy-Schwarz inequality})$$

$$=\frac{2^r}{2}{⟨(f^{2r}(|X|)+g^{2r}(|Y^{\ast }|))x_2,x_2⟩}^{\frac{1}{2}}{⟨(f^{2r}(|Y|)+g^{2r}(|X^{\ast }|))x_1,x_1⟩}^{\frac{1}{2}}$$

$$\le \frac{2^r}{2}{\Vert f^{2r}(|X|)+g^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert f^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}\Vert x_1\Vert \Vert x_2\Vert$$

$$\le \frac{2^r}{2}{\Vert f^{2r}(|X|)+g^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert f^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}\left(\frac{\Vert x_1{\Vert }^2+\Vert x_2{\Vert }^2}{2}\right)$$

$$(\text{by the arithmetic-geometric mean inequality})$$

$$=\frac{2^r}{4}{\Vert f^{2r}(|X|)+g^{2r}(|Y^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert f^{2r}(|Y|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}.$$

$$|⟨T\mathbf{x},\mathbf{x}⟩{|}^r$$

$$=|⟨X{x}_2,x_1⟩+⟨Y{x}_1,x_2⟩{|}^r$$

$$\le {\left(|⟨X{x}_2,x_1⟩|+|⟨Y{x}_1,x_2⟩|\right)}^r(\text{by the triangular inequality})$$

$$\le \frac{2^r}{2}\left(|⟨X{x}_2,x_1⟩{|}^r+|⟨Y{x}_1,x_2⟩{|}^r\right)(\text{by the convexity}f(t)=t^r)$$

$$\le \frac{2^r}{2}({\left({⟨f^2(|X|)x_2,x_2⟩}^{\frac{1}{2}}{⟨g^2(|X^{\ast }|)x_1,x_1⟩}^{\frac{1}{2}}\right)}^r$$

$$+{\left({⟨g^2(|Y|)x_1,x_1⟩}^{\frac{1}{2}}{⟨f^2(|Y^{\ast }|)x_2,x_2⟩}^{\frac{1}{2}}\right)}^r)(\text{by Lemma}\text{ref}5)$$

$${\omega }^r(T)\le 2^{r-2}{\Vert |X{|}^{2rp}+|Y^{\ast }{|}^{2r(1-p)}\Vert }^{\frac{1}{2}}{\Vert |Y{|}^{2rp}+|X^{\ast }{|}^{2r(1-p)}\Vert }^{\frac{1}{2}}$$

$${\omega }^r(T)\le 2^{r-2}{\Vert |X{|}^{2rp}+|Y^{\ast }{|}^{2rp}\Vert }^{\frac{1}{2}}{\Vert |Y{|}^{2r(1-p)}+|X^{\ast }{|}^{2r(1-p)}\Vert }^{\frac{1}{2}}.$$

$$\omega (T)\le \frac{1}{2}{\Vert |X|+|Y^{\ast }|\Vert }^{\frac{1}{2}}{\Vert |Y|+|X^{\ast }|\Vert }^{\frac{1}{2}},$$

$${\omega }^r(X)\le 2^{r-2}\Vert f^{2r}(|X|)+g^{2r}(|X^{\ast }|)\Vert$$

$${\omega }^r(X)\le 2^{r-2}{\Vert f^{2r}(|X|)+f^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}{\Vert g^{2r}(|X|)+g^{2r}(|X^{\ast }|)\Vert }^{\frac{1}{2}}.$$

Full text

[TABLE]

and

[TABLE]

where $X,Y$ are bounded linear operators on a Hilbert space ${\mathscr{H}}$, $r\geq 1$ and $f$, $g$ are nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$. Moreover, we present some inequalities involving the generalized Euclidean operator radius of operators $T_{1},\cdots,T_{n}$.

1. Introduction

Let ${\mathbb{B}}(\mathscr{H})$ denote the $C^{*}$-algebra of all bounded linear operators on a Hilbert space ${\mathscr{H}}$. In the case when ${\rm dim}{\mathscr{H}}=n$, we identify ${\mathbb{B}}({\mathscr{H}})$ with the matrix algebra $\mathbb{M}_{n}$ of all $n\times n$ matrices with entries in the complex field. An operator $A\in{\mathbb{B}}(\mathscr{H})$ is said to be contraction, if $A^{*}A\leq I$. The numerical radius of $T\in{\mathbb{B}}({\mathscr{H}})$ is defined by

[TABLE]

It is well known that $\omega(\,\cdot\,)$ defines a norm on ${\mathbb{B}}({\mathscr{H}})$, which is equivalent to the usual operator norm. In fact, $\frac{1}{2}\|\,\cdot\,\|\leq\omega(\,\cdot\,)\leq\|\,\cdot\,\|$; see [9]. An important inequality for $\omega(A)$ is the power inequality stating that $\omega(A^{n})\leq\omega(A)^{n}\,\,(n=1,2,\cdots)$. For further information about the properties of numerical radius inequalities we refer the reader to [1, 5, 13] and references therein. Let ${\mathscr{H}_{1}},{\mathscr{H}_{2}}$ be Hilbert spaces, and consider the direct sum ${\mathscr{H}}={\mathscr{H}_{1}}\oplus{\mathscr{H}_{2}}$. With respect to this decomposition, every operator $T\in{\mathbb{B}}({\mathscr{H}})$ has a $2\times 2$ operator matrix representation $T=[T_{ij}]$ with entries $T_{ij}\in{\mathbb{B}}({\mathscr{H}_{j}},{\mathscr{H}_{i}})$, the space of all bounded linear operators from ${\mathscr{H}_{j}}$ to ${\mathscr{H}_{i}}\,\,(1\leq i,j\leq 2)$. Operator matrices provide a usual tool for studying Hilbert space operators, which have been extensively studied in the literatures. Let $A\in{\mathbb{B}}({\mathscr{H}_{1}},{\mathscr{H}_{1}})$, $B\in{\mathbb{B}}({\mathscr{H}_{2}},{\mathscr{H}_{1}})$, $C\in{\mathbb{B}}({\mathscr{H}_{1}},{\mathscr{H}_{2}})$ and $D\in{\mathbb{B}}({\mathscr{H}_{2}},{\mathscr{H}_{2}})$. The operator \left[\begin{array}[]{cc}A&0\\ 0&D\end{array}\right] is called the diagonal part of \left[\begin{array}[]{cc}A&B\\ C&D\end{array}\right] and \left[\begin{array}[]{cc}0&B\\ C&0\end{array}\right] is the off-diagonal part.

The classical Young inequality says that if $p,q>1$ such that $\frac{1}{p}+\frac{1}{q}=1$, then $ab\leq\frac{a^{p}}{p}+\frac{b^{q}}{q}$ for positive real numbers $a,b$. In [3], the authors showed that a refinement of the scalar Young inequality as follows $\left(a^{\frac{1}{p}}b^{\frac{1}{q}}\right)^{m}+r_{0}^{m}\left(a^{\frac{m}{2}}-b^{\frac{m}{2}}\right)^{2}\leq\left(\frac{a}{p}+\frac{b}{q}\right)^{m},$ where $r_{0}=\min\{\frac{1}{p},\frac{1}{q}\}$ and $m=1,2,\cdots$. In particular, if $p=q=2$, then

[TABLE]

It has been shown in [8], that if $T\in{\mathbb{B}}({\mathscr{H}})$, then

[TABLE]

where $|T|=(T^{*}T)^{\frac{1}{2}}$ is the absolute value of $T$. Recently [2], the authors extended this inequality for off-diagonal operator matrices of the form T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}) as follows

[TABLE]

Let $T_{1},T_{2},\cdots,T_{n}\in{\mathbb{B}}({\mathscr{H}})$. The functional $\omega_{p}$ of operators $T_{1},\cdots,T_{n}$ for $p\geq 1$ is defined in [11] as follows

[TABLE]

If $p=2$, then we have the Euclidean operator radius of $T_{1},\cdots,T_{n}$ which was defined in [10]. In [13], the authors showed that an upper bound for the functional $\omega_{p}$

[TABLE]

where $T_{i}\in{\mathbb{B}}({\mathscr{H}})\,\,(i=1,2,\cdots,n)$, $f$, $g$ are nonnegative continuous functions on $[0,\infty)$ such that $f(t)g(t)=t\,(t\in[0,\infty))$, $p\geq 1$ and

[TABLE]

In this paper, we show some inequalities involving powers of the numerical radius for off-diagonal parts of $2\times 2$ operator matrices. In particular, we extend inequalities (1.2) and (1.3) for nonnegative continuous functions $f$, $g$ on $[0,\infty)$ such that $f(t)g(t)=t\,(t\in[0,\infty))$. Moreover, we present some inequalities including the generalized Euclidean operator radius $\omega_{p}$.

2. main results

To prove our first result, we need the following lemmas.

Lemma 2.1**.**

[6, 14*]** Let $X\in{\mathbb{B}}({\mathscr{H}})$. Then

$(a)\,\,\omega(X)=\underset{\theta\in\mathbb{R}}{\max}\left\|\textrm{Re}\left(e^{i\theta}X\right)\right\|=\underset{\theta\in\mathbb{R}}{\max}\left\|\textrm{Im}\left(e^{i\theta}X\right)\right\|.$

(b)\,\,\omega\left(\left[\begin{array}[]{cc}0&X\\ X&0\end{array}\right]\right)=\omega(X).**

The next lemma follows from the spectral theorem for positive operators and Jensen inequality; see [7].

Lemma 2.2**.**

*Let $T\in{\mathbb{B}}({\mathscr{H}})$, $T\geq 0$ and $x\in{\mathscr{H}}$ such that $\|x\|\leq 1$. Then

$(a)\,\,\left\langle Tx,x\right\rangle^{r}\leq\left\langle T^{r}x,x\right\rangle$ for $r\geq 1.$

$(b)\,\,\left\langle T^{r}x,x\right\rangle\leq\left\langle Tx,x\right\rangle^{r}$ for $0<r\leq 1$.

Proof.

Let $r\geq 1$ and $x\in{\mathscr{H}}$ such that $\|x\|\leq 1$. Fix $u=\frac{x}{\|x\|}$. Using the McCarty inequality we have $\left\langle Tu,u\right\rangle^{r}\leq\left\langle T^{r}u,u\right\rangle$, whence

[TABLE]

Hence, we get the first inequality. The proof of the second inequality is similar. ∎

Lemma 2.3**.**

[7, Theorem 1]** Let $T\in{\mathbb{B}}({\mathscr{H}})$ and $x,y\in{\mathscr{H}}$ be any vectors. If $f$, $g$ are nonnegative continuous functions on $[0,\infty)$ which are satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$, then

[TABLE]

Now, we are in position to demonstrate the main results of this section by using some ideas from [2, 13].

Theorem 2.4**.**

Let T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}), $r\geq 1$ and $f$, $g$ be nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$. Then

[TABLE]

and

[TABLE]

Proof.

Let \mathbf{x}=\left[\begin{array}[]{cc}x_{1}\\ x_{2}\end{array}\right]\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}} be a unit vector (i.e., $\|x_{1}\|^{2}+\|x_{2}\|^{2}=1$). Then

[TABLE]

Hence, we get the first inequality. Now, applying this fact

[TABLE]

and a similar argument to the proof of the first inequality we have the second inequality and this completes the proof of the theorem. ∎

Theorem 2.4 includes a special case as follows.

Corollary 2.5**.**

Let T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}), $0\leq p\leq 1$ and $r\geq 1$. Then

[TABLE]

and

[TABLE]

Proof.

The result follows immediately from Theorem 2.4 for $f(t)=t^{p}$ and $g(t)=t^{1-p}\,\,(0\leq p\leq 1)$. ∎

Remark 2.6*.*

Taking $f(t)=g(t)=t^{\frac{1}{2}}\,(t\in[0,\infty))$ and $r=1$ in Theorem 2.4, we get (see [2, Theorem 4])

[TABLE]

where T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}).

If we put $Y=X$ in Theorem 2.4, then by using Lemma 2.1(b) we get an extension of Inequality (1.2).

Corollary 2.7**.**

Let $X\in{\mathbb{B}}({\mathscr{H}})$, $r\geq 1$ and $f$, $g$ be nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$. Then

[TABLE]

and

[TABLE]

Corollary 2.8**.**

Let $X,Y\in{\mathbb{B}}({\mathscr{H}})$ and $0\leq p\leq 1$. Then

[TABLE]

and

[TABLE]

for $r\geq 1$.

Proof.

It follows from the power inequality $\omega^{\frac{1}{2}}\left(T^{2}\right)\leq\omega\left(T\right)$ that

[TABLE]

The required result follows from Corollary 2.5. ∎

Corollary 2.9**.**

Let $X,Y\in{\mathbb{B}}({\mathscr{H}})$ and $r\geq 1$. Then

[TABLE]

In particular, if $X$ and $Y$ are normal operators, then

[TABLE]

Proof.

Applying Lemma 2.1(a) and Corollary 2.5 (for $p=\frac{1}{2}$), we have

[TABLE]

where T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]. Similarly,

[TABLE]

Hence we get the desired result. For the particular case, observe that $\left|Y^{\ast}\right|=$ $\left|Y\right|$ and $|X^{\ast}|=|X|.$ ∎

Remark 2.10*.*

It should be mentioned here that inequality $\left(\text{\ref{8}}\right)$, which has been given earlier, is a generalized form of the well-known inequality (see [4]): if $A$ and $B$ are normal operators, then

[TABLE]

The normality of  $X$ and $Y$ are necessary that means Inequality (2.3) is not true for arbitrary operators $X$ and $Y$; see [12]

In the next theorem, we show another upper bound for numerical radius involving off-diagonal operator matrices.

Theorem 2.11**.**

Let T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}), $r\geq 1$ and $f$, $g$ be nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$. Then

[TABLE]

and

[TABLE]

where $\frac{1}{p}+\frac{1}{q}=1$ and $p\geq 1$.

Proof.

If \mathbf{x}=\left[\begin{array}[]{cc}x_{1}\\ x_{2}\end{array}\right]\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}} is a unit vector, then by a similar argument to the proof of Theorem 2.4 we have

[TABLE]

Let $\alpha=\frac{\left\|\left(f^{2r}(|X|)+g^{2r}(|Y^{*}|)\right)^{p}\right\|^{\frac{1}{2}}}{p}$ and $\beta=\frac{\left\|\left(f^{2r}(|Y|)+g^{2r}(|X^{*}|)\right)^{q}\right\|^{\frac{1}{2}}}{q}$. It follows from

[TABLE]

and Inequality (2) that we deduce

[TABLE]

Taking the supremum over all unit vectors $\mathbf{x}\in\mathscr{H}_{1}\oplus\mathscr{H}_{2}$ we get the first inequality. Now, according to inequality (2) and the same argument in the proof of the first inequality, we obtain the second inequality. ∎

Remark 2.12*.*

If T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}) and $\frac{1}{p}+\frac{1}{q}=1$, then by using Theorem 2.4 and the Young inequality we obtain the inequalities

[TABLE]

and

[TABLE]

where $r\geq 1$ and $f$, $g$ are nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$. Now, Theorem 2.11 shows some other upper bounds for $\omega(T)$.

In the special case of Theorem 2.11 for $Y=X$ and $p=q=2$, we have the next result.

Corollary 2.13**.**

Let $X\in{\mathbb{B}}({\mathscr{H}})$, $r\geq 1$ and $f$, $g$ be nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t\,(t\in[0,\infty))$. Then

[TABLE]

and

[TABLE]

Applying Inequality (1.1) we obtain the following theorem.

Theorem 2.14**.**

Let T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}) and $f$, $g$ be nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t$ $(t\in[0,\infty))$. Then for $r\geq 1$

[TABLE]

where

[TABLE]

Proof.

Let \mathbf{x}=\left[\begin{array}[]{cc}x_{1}\\ x_{2}\end{array}\right]\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}} be a unit vector. Then

[TABLE]

Taking the supremum over all unit vectors \mathbf{x}=\left[\begin{array}[]{cc}x_{1}\\ x_{2}\end{array}\right]\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}} we get the desired inequality. ∎

If we put $Y=X$ in Theorem 2.14, then we get next result.

Corollary 2.15**.**

Let $X\in{\mathbb{B}}({\mathscr{H}})$ and $f$, $g$ be nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t$ $(t\in[0,\infty))$. Then for $r\geq 1$

[TABLE]

where

[TABLE]

Remark 2.16*.*

If \mathbf{x}=\left[\begin{array}[]{cc}x_{1}\\ x_{2}\end{array}\right]\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}} is a unit vector, then by using the inequality

[TABLE]

and the same argument in the proof if Theorem 2.14 we get the following inequality

[TABLE]

where T=\left[\begin{array}[]{cc}0&X\\ Y&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}}), $f$, $g$ are nonnegative continuous functions on $[0,\infty)$ satisfying the relation $f(t)g(t)=t$ $(t\in[0,\infty))$, $r\geq 1$ and

[TABLE]

3. Some upper bounds for $\omega_{p}$

In this section, we obtain some upper bounds for $\omega_{P}$. We first show the following theorem.

Theorem 3.1**.**

Let \widetilde{S}_{i}=\left[\begin{array}[]{cc}A_{i}&0\\ 0&B_{i}\end{array}\right],\widetilde{T}_{i}=\left[\begin{array}[]{cc}0&X_{i}\\ Y_{i}&0\end{array}\right] and \widetilde{U}_{i}=\left[\begin{array}[]{cc}C_{i}&0\\ 0&D_{i}\end{array}\right] be operators matrices in {\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}})$$\,\,(1\leq i\leq n) such that $A_{i},B_{i},C_{i}$ and $D_{i}$ are contractions. Then

[TABLE]

and

[TABLE]

where $p\geq 1$.

Proof.

For any unit vector \mathbf{x}=\left[\begin{array}[]{cc}x_{1}\\ x_{2}\end{array}\right]\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}} we have

[TABLE]

[TABLE]

Taking the supremum over all unit vectors $\mathbf{x}\in{\mathscr{H}_{1}\oplus\mathscr{H}_{2}}$ we obtain the first inequality. Using the inequality

[TABLE]

and a similar fashion in the proof of the first inequality we reach the second inequality. ∎

In the special case of Theorem 3.1 for $A_{i}=B_{i}=C_{i}=D_{i}=I\,\,(1\leq i\leq n)$ we have the next result.

Corollary 3.2**.**

Let T_{i}=\left[\begin{array}[]{cc}0&X_{i}\\ Y_{i}&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}})\,\,(1\leq j\leq n). Then

[TABLE]

and

[TABLE]

for $p\geq 1$.

If we put $f(t)=g(t)=t^{\frac{1}{2}}\,(t\in[0,\infty))$, then we get the next result.

Corollary 3.3**.**

Let T_{i}=\left[\begin{array}[]{cc}0&X_{i}\\ Y_{i}&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}})\,\,(1\leq j\leq n). Then

[TABLE]

for $p\geq 1$.

Theorem 3.4**.**

Let T_{i}=\left[\begin{array}[]{cc}A_{i}&B_{i}\\ C_{i}&D_{i}\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}}\oplus{\mathscr{H}_{2}})\,\,(1\leq i\leq n) and $p\geq 1$. Then

[TABLE]

In particular,

[TABLE]

Proof.

Let \mathbf{x}=\left[\begin{array}[]{c}x_{1}\\ x_{2}\end{array}\right] be a unit vector in ${\mathscr{H}_{1}\oplus\mathscr{H}_{2}}$. Then

[TABLE]

Thus,

[TABLE]

This completes the proof. ∎

For $A_{i}=D_{i}$ and $B_{i}=C_{i}\,\,(1\leq i\leq n)$ we get the following result.

Corollary 3.5**.**

Let T_{i}=\left[\begin{array}[]{cc}\pm A_{i}&\pm B_{i}\\ \pm B_{i}&\pm A_{i}\end{array}\right]\be an operator matrix with $A_{i},B_{i}\in\mathbb{B}(\mathscr{H})$ $\,(1\leq i\leq n)$. Then for all $p\geq 1$,

[TABLE]

In particular, if $A,B\in\mathbb{B}(\mathscr{H})$, then

[TABLE]

If we take $B_{i}=C_{i}=0\,\,(1\leq i\leq n)$ in Theorem 3.4, then we get the following inequality.

Corollary 3.6**.**

Let T_{i}=\left[\begin{array}[]{cc}A_{i}&0\\ 0&D_{i}\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}})\,\,(1\leq i\leq n). Then for all $p\geq 1$,

[TABLE]

For $C_{i}=D_{i}=0\,\,(1\leq i\leq n)$ we obtain a result that generalize and refine the inequality \omega\left(\left[\begin{array}[]{cc}A&B\\ 0&0\end{array}\right]\right)\leq\omega(A)+\frac{\left\|B\right\|}{2}.

Corollary 3.7**.**

Let T_{i}=\left[\begin{array}[]{cc}A_{i}&B_{i}\\ 0&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}_{1}\oplus\mathscr{H}_{2}})\,\,(1\leq i\leq n) and $p\geq 1$. Then

[TABLE]

In particular,

[TABLE]

If we put $A_{i}=D_{i}=0\,\,(1\leq i\leq n)$, then we deduce

Corollary 3.8**.**

Let T_{i}=\left[\begin{array}[]{cc}0&B_{i}\\ C_{i}&0\end{array}\right]\in{\mathbb{B}}({\mathscr{H}}_{1}\oplus{\mathscr{H}}_{2})\,(1\leq i\leq n) and $p\geq 1$. The

[TABLE]

In particular, if $B\in\mathbb{B}(\mathscr{H}_{2},\mathscr{H}_{1})$ and $C\in\mathbb{B}(\mathscr{H}_{1},\mathscr{H}_{2})$, then

[TABLE]

Acknowledgement. The first author would like to thank the Tusi Mathematical Research Group (TMRG).