---
abstract: 'Some elementary inequalities providing upper bounds for the difference of the norm and the numerical radius of a bounded linear operator on Hilbert spaces under appropriate conditions are given.'
address: |
    School of Computer Science and Mathematics\
    Victoria University of Technology\
    PO Box 14428, Melbourne City\
    Victoria 8001, Australia.
author:
- 'S.S. Dragomir'
date: '29 August, 2005'
title: Reverse Inequalities for the Numerical Radius of Linear Operators in Hilbert Spaces 
---

Introduction
============

Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ be a complex Hilbert space. The *numerical range* of an operator $T$ is the subset of the complex numbers $\mathbb{C}$ given by [@GR p. 1]:$$W\left( T\right) =\left\{ \left\langle Tx,x\right\rangle ,\ x\in H,\
\left\Vert x\right\Vert =1\right\} .$$The following properties of $W\left( T\right) $ are immediate:

1.  $W\left( \alpha I+\beta T\right) =\alpha +\beta W\left( T\right) $ for $\alpha ,\beta \in \mathbb{C};$

2.  $W\left( T^{\ast }\right) =\left\{ \bar{\lambda},\lambda \in
    W\left( T\right) \right\} ,$ where $T^{\ast }$ is the *adjoint operator* of $T;$

3.  $W\left( U^{\ast }TU\right) =W\left( T\right) $ for any *unitary* operator $U.$

The following classical fact about the geometry of the numerical range [@GR p. 4] may be stated:

The numerical range of an operator is convex.

An important use of $W\left( T\right) $ is to bound the *spectrum* $%
\sigma \left( T\right) $ of the operator $T$ [@GR p. 6]:

The spectrum of an operator is contained in the closure of its numerical range.

The self-adjoint operators have their spectra bounded sharply by the numerical range [@GR p. 7]:

The following statements hold true:

1.  $T$ is self-adjoint iff $W\left( T\right) $ is real;

2.  If $T$ is self-adjoint and $W\left( T\right) =\left[ m,M\right] $ (the closed interval of real numbers $m,M$), then $\left\Vert T\right\Vert
    =\max \left\{ \left\vert m\right\vert ,\left\vert M\right\vert \right\} .$

3.  If $W\left( T\right) =\left[ m,M\right] ,$ then $m,M\in \sigma
    \left( T\right) .$

The *numerical radius* $w\left( T\right) $ of an operator $T$ on $H$ is given by [@GR p. 8]:$$w\left( T\right) =\sup \left\{ \left\vert \lambda \right\vert ,\lambda \in
W\left( T\right) \right\} =\sup \left\{ \left\vert \left\langle
Tx,x\right\rangle \right\vert ,\left\Vert x\right\Vert =1\right\} .
\label{1.1}$$Obviously, by (\[1.1\]), for any $x\in H$ one has$$\left\vert \left\langle Tx,x\right\rangle \right\vert \leq w\left( T\right)
\left\Vert x\right\Vert ^{2}.  \label{1.2}$$

It is well known that $w\left( \cdot \right) $ is a norm on the Banach algebra $B\left( H\right) $ of all bounded linear operators $T:H\rightarrow
H,$ i.e.,

1.  $w\left( T\right) \geq 0$ for any $T\in B\left( H\right) $ and $%
    w\left( T\right) =0$ if and only if $T=0;$

2.  $w\left( \lambda T\right) =\left\vert \lambda \right\vert
    w\left( T\right) $ for any $\lambda \in \mathbb{C}$ and $T\in B\left(
    H\right) ;$

3.  $w\left( T+V\right) \leq w\left( T\right) +w\left( V\right) $ for any $T,V\in B\left( H\right) .$

This norm is equivalent with the operator norm. In fact, the following more precise result holds [@GR p. 9]:

For any $T\in B\left( H\right) $ one has$$w\left( T\right) \leq \left\Vert T\right\Vert \leq 2w\left( T\right) .
\label{1.3}$$

Let us now look at two extreme cases of the inequality (\[1.3\]). In the following $r\left( t\right) :=\sup \left\{ \left\vert \lambda \right\vert
,\lambda \in \sigma \left( T\right) \right\} $ will denote the *spectral radius* of $T$ and $\sigma _{p}\left( T\right) =\left\{ \lambda \in
\sigma \left( T\right) ,\ Tf=\lambda f\text{ \ for some \ }f\in H\right\} $ the *point spectrum* of $T.$

The following results hold [@GR p.10]:

\[t1\]We have

1.  If $w\left( T\right) =\left\Vert T\right\Vert ,$ then $r\left(
    T\right) =\left\Vert T\right\Vert .$

2.  If $\lambda \in W\left( T\right) $ and $\left\vert \lambda
    \right\vert =\left\Vert T\right\Vert ,$ then $\lambda \in \sigma _{p}\left(
    T\right) .$

To address the other extreme case $w\left( T\right) =\frac{1}{2}\left\Vert
T\right\Vert ,$ we can state the following sufficient condition in terms of (see [@GR p. 11])$$R\left( T\right) :=\left\{ Tf,\ f\in H\right\} \quad \text{and}\quad R\left(
T^{\ast }\right) :=\left\{ T^{\ast }f,\ f\in H\right\} .$$

\[t.e\]If $R\left( T\right) \perp R\left( T^{\ast }\right) ,$ then $%
w\left( T\right) =\frac{1}{2}\left\Vert T\right\Vert .$

It is well-known that the two-dimensional shift$$S_{2}=\left[ 
\begin{array}{ll}
0 & 0 \\ 
1 & 0%
\end{array}%
\right] ,$$has the property that $w\left( T\right) =\frac{1}{2}\left\Vert T\right\Vert
. $

The following theorem shows that some operators $T$ with $w\left( T\right) =%
\frac{1}{2}\left\Vert T\right\Vert $ have $S_{2}$ as a component [GR]{}:

If $w\left( T\right) =\frac{1}{2}\left\Vert T\right\Vert $ and $T$ attains its norm, then $T$ has a two-dimensional reducing subspace on which it is the shift $S_{2}.$

For other results on numerical radius, see [@H], Chapter 11.

The main aim of the present paper is to point out some upper bounds for the nonnegative difference$$\left\Vert T\right\Vert -w\left( T\right) \qquad \left( \left\Vert
T\right\Vert ^{2}-\left( W\left( T\right) \right) ^{2}\right)$$under appropriate assumptions for the bounded linear operator $%
T:H\rightarrow H.$

The Results
===========

The following results may be stated:

\[t.1\]Let $T:H\rightarrow H$ be a bounded linear operator on the complex Hilbert space $H.$ If $\lambda \in \mathbb{C}\backslash \left\{
0\right\} $ and $r>0$ are such that$$\left\Vert T-\lambda I\right\Vert \leq r,  \label{2.1}$$where $I:H\rightarrow H$ is the identity operator on $H,$ then$$\left( 0\leq \right) \left\Vert T\right\Vert -w\left( T\right) \leq \frac{1}{%
2}\cdot \frac{r^{2}}{\left\vert \lambda \right\vert }.  \label{2.2}$$

For $x\in H$ with $\left\Vert x\right\Vert =1,$ we have from (\[2.1\]) that$$\left\Vert Tx-\lambda x\right\Vert \leq \left\Vert T-\lambda I\right\Vert
\leq r,$$giving$$\left\Vert Tx\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}\leq 2\func{%
Re}\left[ \overline{\lambda }\left\langle Tx,x\right\rangle \right]
+r^{2}\leq 2\left\vert \lambda \right\vert \left\vert \left\langle
Tx,x\right\rangle \right\vert +r^{2}.  \label{2.3}$$Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (\[2.3\]) we get the following inequality that is of interest in itself:$$\left\Vert T\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}\leq
2w\left( T\right) \left\vert \lambda \right\vert +r^{2}.  \label{2.4}$$Since, obviously,$$\left\Vert T\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}\geq
2\left\Vert T\right\Vert \left\vert \lambda \right\vert ,  \label{2.5}$$hence by (\[2.4\]) and (\[2.5\]) we deduce the desired inequality ([2.2]{}).

If the operator $T:H\rightarrow H$ is such that $R\left( T\right) \perp
R\left( T^{\ast }\right) ,$ $\left\Vert T\right\Vert =1$ and $\left\Vert
T-I\right\Vert \leq 1$, then the equality case holds in (\[2.2\]). Indeed, by Theorem \[t.e\], we have in this case $w\left( T\right) =\frac{1}{2}%
\left\Vert T\right\Vert =\frac{1}{2}$ and since we can choose in Theorem [t.1]{}, $\lambda =1,$ $r=1,$ then we get in both sides of (\[2.2\]) the same quantity $\frac{1}{2}.$

Find the bounded linear operators $T:H\rightarrow H$ with $\left\Vert
T\right\Vert =1,$ $R\left( T\right) \perp R\left( T^{\ast }\right) $ and $%
\left\Vert T-\lambda I\right\Vert \leq \left\vert \lambda \right\vert ^{%
\frac{1}{2}}.$

The following corollary may be stated:

\[c.1\]Let $A:H\rightarrow H$ be a bounded linear operator and $\varphi
,\phi \in \mathbb{C}$ with $\phi \neq -\varphi ,\varphi .$ If$$\func{Re}\left\langle \phi x-Ax,Ax-\varphi x\right\rangle \geq 0\quad \text{%
for any}\quad x\in H,\ \left\Vert x\right\Vert =1  \label{2.6}$$then$$\left( 0\leq \right) \left\Vert A\right\Vert -w\left( A\right) \leq \frac{1}{%
4}\cdot \frac{\left\vert \phi -\varphi \right\vert ^{2}}{\left\vert \phi
+\varphi \right\vert }.  \label{2.7}$$

Utilising the fact that in any Hilbert space the following two statements are equivalent:

1.  $\func{Re}\left\langle Z-x,x-z\right\rangle \geq 0,$ $x,z,Z\in H;$

2.  $\left\Vert x-\frac{z+Z}{2}\right\Vert \leq \frac{1}{2}%
    \left\Vert Z-z\right\Vert ,$

we deduce that (\[2.6\]) is equivalent to$$\left\Vert Ax-\frac{\phi +\varphi }{2}\cdot Ix\right\Vert \leq \frac{1}{2}%
\left\vert \phi -\varphi \right\vert  \label{2.8}$$for any $x\in H,$ $\left\Vert x\right\Vert =1,$ which in its turn is equivalent with the operator norm inequality:$$\left\Vert A-\frac{\phi +\varphi }{2}\cdot I\right\Vert \leq \frac{1}{2}%
\left\vert \phi -\varphi \right\vert .  \label{2.9}$$Now, applying Theorem \[t.1\] for $T=A,$ $\lambda =\frac{\varphi +\phi }{2}
$ and $r=\frac{1}{2}\left\vert \Gamma -\gamma \right\vert ,$ we deduce the desired result (\[2.7\]).

Following [@GR p. 25], we say that an operator $B:H\rightarrow H$ is accreative, if $\func{Re}\left\langle Bx,x\right\rangle \geq 0$ for any $%
x\in H.$ One may observe that the assumption (\[2.6\]) above is then equivalent with the fact that the operator $\left( A^{\ast }-\bar{\varphi}%
I\right) \left( \phi I-A\right) $ is accreative.

Perhaps a more convenient sufficient condition in terms of positive operators is the following one:

\[c.2\]Let $\varphi ,\phi \in \mathbb{C}$ with $\phi \neq -\varphi
,\varphi $ and $A:H\rightarrow H$ a bounded linear operator in $H.$ If $%
\left( A^{\ast }-\bar{\varphi}I\right) \left( \phi I-A\right) $ is self-adjoint and$$\left( A^{\ast }-\bar{\varphi}I\right) \left( \phi I-A\right) \geq 0
\label{2.10}$$in the operator order, then$$\left( 0\leq \right) \left\Vert A\right\Vert -w\left( A\right) \leq \frac{1}{%
4}\cdot \frac{\left\vert \phi -\varphi \right\vert ^{2}}{\left\vert \phi
+\varphi \right\vert }.  \label{2.11}$$

The following result may be stated as well:

\[c.3\]Assume that $T,\lambda ,r$ are as in Theorem \[t.1\]. If, in addition, there exists $\rho \geq 0$ such that$$\left\vert \left\vert \lambda \right\vert -w\left( T\right) \right\vert \geq
\rho ,  \label{2.12a}$$then$$\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) \leq
r^{2}-\rho ^{2}.  \label{2.13}$$

From (\[2.4\]) of Theorem \[t.1\], we have$$\begin{aligned}
\left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) & \leq r^{2}-w^{2}\left(
T\right) +2w\left( T\right) \left\vert \lambda \right\vert -\left\vert
\lambda \right\vert ^{2}  \label{2.14} \\
& =r^{2}-\left( \left\vert \lambda \right\vert -w\left( T\right) \right)
^{2}.  \notag\end{aligned}$$On utilising (\[2.4\]) and (\[2.12a\]) we deduce the desired inequality (\[2.13\]).

In particular, if $\left\Vert T-\lambda I\right\Vert \leq r$ and $\left\vert
\lambda \right\vert =w\left( T\right) ,$ $\lambda \in \mathbb{C}$, then$$\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) \leq
r^{2}.  \label{2.15}$$

The following result may be stated as well.

\[t.2\]Let $T:H\rightarrow H$ be a nonzero bounded linear operator on $H$ and $\lambda \in \mathbb{C}\setminus \left\{ 0\right\} ,$ $r>0$ with $%
\left\vert \lambda \right\vert >r.$ If$$\left\Vert T-\lambda I\right\Vert \leq r,  \label{3.1}$$then$$\sqrt{1-\frac{r^{2}}{\left\vert \lambda \right\vert ^{2}}}\leq \frac{w\left(
T\right) }{\left\Vert T\right\Vert }\quad \left( \leq 1\right) .  \label{3.2}$$

From (\[2.4\]) of Theorem \[t.1\], we have$$\left\Vert T\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}-r^{2}\leq
2\left\vert \lambda \right\vert w\left( T\right) ,$$which implies, on dividing with $\sqrt{\left\vert \lambda \right\vert
^{2}-r^{2}}>0$ that$$\frac{\left\Vert T\right\Vert ^{2}}{\sqrt{\left\vert \lambda \right\vert
^{2}-r^{2}}}+\sqrt{\left\vert \lambda \right\vert ^{2}-r^{2}}\leq \frac{%
2\left\vert \lambda \right\vert w\left( T\right) }{\sqrt{\left\vert \lambda
\right\vert ^{2}-r^{2}}}.  \label{3.3}$$By the elementary inequality$$2\left\Vert T\right\Vert \leq \frac{\left\Vert T\right\Vert ^{2}}{\sqrt{%
\left\vert \lambda \right\vert ^{2}-r^{2}}}+\sqrt{\left\vert \lambda
\right\vert ^{2}-r^{2}}  \label{3.4}$$and by (\[3.3\]) we deduce$$\left\Vert T\right\Vert \leq \frac{w\left( T\right) \left\vert \lambda
\right\vert }{\sqrt{\left\vert \lambda \right\vert ^{2}-r^{2}}},$$which is equivalent to (\[3.2\]).

Squaring (\[3.2\]), we get the inequality$$\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) \leq 
\frac{r^{2}}{\left\vert \lambda \right\vert ^{2}}\left\Vert T\right\Vert
^{2}.  \label{3.5}$$

Since for any bounded linear operator $T:H\rightarrow H$ we have that $%
w\left( T\right) \geq \frac{1}{2}\left\Vert T\right\Vert ,$ hence (\[3.2\]) would produce a refinement of this classic fact only in the case when$$\frac{1}{2}\leq \left( 1-\frac{r^{2}}{\left\vert \lambda \right\vert ^{2}}%
\right) ^{\frac{1}{2}},$$which is equivalent to $r/\left\vert \lambda \right\vert \leq \sqrt{3}/2.$

The following corollary holds.

\[c.4\]Let $\varphi ,\phi \in \mathbb{C}$ with $\func{Re}\left( \phi 
\bar{\varphi}\right) >0.$ If $T:H\rightarrow H$ is a bounded linear operator such that either (\[2.6\]) or (\[2.10\]) holds true, then:$$\frac{2\sqrt{\func{Re}\left( \phi \bar{\varphi}\right) }}{\left\vert \phi
+\varphi \right\vert }\leq \frac{w\left( T\right) }{\left\Vert T\right\Vert }%
\left( \leq 1\right)  \label{3.6}$$and$$\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) \leq
\left\vert \frac{\phi -\varphi }{\phi +\varphi }\right\vert ^{2}\left\Vert
T\right\Vert ^{2}.  \label{3.7}$$

If we consider $\lambda =\frac{\phi +\varphi }{2}$ and $r=\frac{1}{2}%
\left\vert \phi -\varphi \right\vert ,$ then $\left\vert \lambda \right\vert
^{2}-r^{2}=\left\vert \frac{\phi +\varphi }{2}\right\vert ^{2}-\left\vert 
\frac{\phi -\varphi }{2}\right\vert ^{2}=\func{Re}\left( \phi \bar{\varphi}%
\right) >0.$ Now, on applying Theorem \[t.2\], we deduce the desired result.

If $\left\vert \phi -\varphi \right\vert \leq \frac{\sqrt{3}}{2}\left\vert
\phi +\varphi \right\vert ,$ $\func{Re}\left( \phi \bar{\varphi}\right) >0,$ then (\[3.6\]) is a refinement of the inequality $w\left( T\right) \geq 
\frac{1}{2}\left\Vert T\right\Vert .$

The following result may be of interest as well.

\[t.3\]Let $T:H\rightarrow H$ be a nonzero bounded linear operator on $H$ and $\lambda \in \mathbb{C}\backslash \left\{ 0\right\} ,$ $r>0$ with $%
\left\vert \lambda \right\vert >r.$ If$$\left\Vert T-\lambda I\right\Vert \leq r,  \label{4.1}$$then$$\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) \leq 
\frac{2r^{2}}{\left\vert \lambda \right\vert +\sqrt{\left\vert \lambda
\right\vert ^{2}-r^{2}}}w\left( T\right) .  \label{4.2}$$

From the proof of Theorem \[t.1\], we have$$\left\Vert Tx\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}\leq 2\func{%
Re}\left[ \overline{\lambda }\left\langle Tx,x\right\rangle \right] +r^{2}
\label{4.2a}$$for any $x\in H,$ $\left\Vert x\right\Vert =1.$

If we divide (\[4.2a\]) by $\left\vert \lambda \right\vert \left\vert
\left\langle Tx,x\right\rangle \right\vert ,$ (which, by (\[4.2a\]), is positive) then we obtain$$\frac{\left\Vert Tx\right\Vert ^{2}}{\left\vert \lambda \right\vert
\left\vert \left\langle Tx,x\right\rangle \right\vert }\leq \frac{2\func{Re}%
\left[ \overline{\lambda }\left\langle Tx,x\right\rangle \right] }{%
\left\vert \lambda \right\vert \left\vert \left\langle Tx,x\right\rangle
\right\vert }+\frac{r^{2}}{\left\vert \lambda \right\vert \left\vert
\left\langle Tx,x\right\rangle \right\vert }-\frac{\left\vert \lambda
\right\vert }{\left\vert \left\langle Tx,x\right\rangle \right\vert }
\label{4.3}$$for any $x\in H,$ $\left\Vert x\right\Vert =1.$

If we subtract in (\[4.3\]) the same quantity $\frac{\left\vert
\left\langle Tx,x\right\rangle \right\vert }{\left\vert \lambda \right\vert }
$ from both sides, then we get$$\begin{aligned}
& \frac{\left\Vert Tx\right\Vert ^{2}}{\left\vert \lambda \right\vert
\left\vert \left\langle Tx,x\right\rangle \right\vert }-\frac{\left\vert
\left\langle Tx,x\right\rangle \right\vert }{\left\vert \lambda \right\vert }
\label{4.4} \\
& \leq \frac{2\func{Re}\left[ \overline{\lambda }\left\langle
Tx,x\right\rangle \right] }{\left\vert \lambda \right\vert \left\vert
\left\langle Tx,x\right\rangle \right\vert }+\frac{r^{2}}{\left\vert \lambda
\right\vert \left\vert \left\langle Tx,x\right\rangle \right\vert }-\frac{%
\left\vert \left\langle Tx,x\right\rangle \right\vert }{\left\vert \lambda
\right\vert }-\frac{\left\vert \lambda \right\vert }{\left\vert \left\langle
Tx,x\right\rangle \right\vert }  \notag \\
& =\frac{2\func{Re}\left[ \overline{\lambda }\left\langle Tx,x\right\rangle %
\right] }{\left\vert \lambda \right\vert \left\vert \left\langle
Tx,x\right\rangle \right\vert }-\frac{\left\vert \lambda \right\vert
^{2}-r^{2}}{\left\vert \lambda \right\vert \left\vert \left\langle
Tx,x\right\rangle \right\vert }-\frac{\left\vert \left\langle
Tx,x\right\rangle \right\vert }{\left\vert \lambda \right\vert }  \notag \\
& =\frac{2\func{Re}\left[ \overline{\lambda }\left\langle Tx,x\right\rangle %
\right] }{\left\vert \lambda \right\vert \left\vert \left\langle
Tx,x\right\rangle \right\vert }-\left( \frac{\sqrt{\left\vert \lambda
\right\vert ^{2}-r^{2}}}{\sqrt{\left\vert \lambda \right\vert \left\vert
\left\langle Tx,x\right\rangle \right\vert }}-\frac{\sqrt{\left\vert
\left\langle Tx,x\right\rangle \right\vert }}{\sqrt{\left\vert \lambda
\right\vert }}\right) ^{2}-2\frac{\sqrt{\left\vert \lambda \right\vert
^{2}-r^{2}}}{\left\vert \lambda \right\vert }.  \notag\end{aligned}$$Since$$\func{Re}\left[ \overline{\lambda }\left\langle Tx,x\right\rangle \right]
\leq \left\vert \lambda \right\vert \left\vert \left\langle
Tx,x\right\rangle \right\vert$$and$$\left( \frac{\sqrt{\left\vert \lambda \right\vert ^{2}-r^{2}}}{\sqrt{%
\left\vert \lambda \right\vert \left\vert \left\langle Tx,x\right\rangle
\right\vert }}-\frac{\sqrt{\left\vert \left\langle Tx,x\right\rangle
\right\vert }}{\sqrt{\left\vert \lambda \right\vert }}\right) ^{2}\geq 0$$hence by (\[4.4\]) we get$$\frac{\left\Vert Tx\right\Vert ^{2}}{\left\vert \lambda \right\vert
\left\vert \left\langle Tx,x\right\rangle \right\vert }-\frac{\left\vert
\left\langle Tx,x\right\rangle \right\vert }{\left\vert \lambda \right\vert }%
\leq \frac{2\left( \left\vert \lambda \right\vert -\sqrt{\left\vert \lambda
\right\vert ^{2}-r^{2}}\right) }{\left\vert \lambda \right\vert }$$which gives the inequality$$\left\Vert Tx\right\Vert ^{2}\leq \left\vert \left\langle Tx,x\right\rangle
\right\vert ^{2}+2\left\vert \left\langle Tx,x\right\rangle \right\vert
\left( \left\vert \lambda \right\vert -\sqrt{\left\vert \lambda \right\vert
^{2}-r^{2}}\right)   \label{4.5}$$for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1,$ we get$$\begin{aligned}
\left\Vert T\right\Vert ^{2}& \leq \sup \left\{ \left\vert \left\langle
Tx,x\right\rangle \right\vert ^{2}+2\left\vert \left\langle
Tx,x\right\rangle \right\vert \left( \left\vert \lambda \right\vert -\sqrt{%
\left\vert \lambda \right\vert ^{2}-r^{2}}\right) \right\}  \\
& \leq \sup \left\{ \left\vert \left\langle Tx,x\right\rangle \right\vert
^{2}\right\} +2\left( \left\vert \lambda \right\vert -\sqrt{\left\vert
\lambda \right\vert ^{2}-r^{2}}\right) \sup \left\{ \left\vert \left\langle
Tx,x\right\rangle \right\vert \right\}  \\
& =w^{2}\left( T\right) +2\left( \left\vert \lambda \right\vert -\sqrt{%
\left\vert \lambda \right\vert ^{2}-r^{2}}\right) w\left( T\right) ,\end{aligned}$$which is clearly equivalent to (\[4.2\]).

\[c.5\]Let $\varphi ,\phi \in \mathbb{C}$ with $\func{Re}\left( \phi 
\bar{\varphi}\right) >0.$ If $A:H\rightarrow H$ is a bounded linear operator such that either (\[2.6\]) or (\[2.10\]) hold true, then:$$\left( 0\leq \right) \left\Vert A\right\Vert ^{2}-w^{2}\left( A\right) \leq 
\left[ \left\vert \phi +\varphi \right\vert -2\sqrt{\func{Re}\left( \phi 
\bar{\varphi}\right) }\right] w\left( A\right) .  \label{4.6}$$

If $M\geq m>0$ are such that either $\left( A^{\ast }-mI\right) \left(
MI-A\right) $ is accreative, or, sufficiently, $\left( A^{\ast }-mI\right)
\left( MI-A\right) $ is self-adjoint and$$\left( A^{\ast }-mI\right) \left( MI-A\right) \geq 0\quad \text{in the
operator order,}  \label{4.8}$$then, by (\[3.6\]) we have:$$\left( 1\leq \right) \frac{\left\Vert A\right\Vert }{w\left( A\right) }\leq 
\frac{M+m}{2\sqrt{mM}},  \label{4.9}$$which is equivalent to$$\left( 0\leq \right) \left\Vert A\right\Vert -w\left( A\right) \leq \frac{%
\left( \sqrt{M}-\sqrt{m}\right) ^{2}}{2\sqrt{mM}}w\left( A\right) ,
\label{4.10}$$while from (\[4.2\]) we have$$\left( 0\leq \right) \left\Vert A\right\Vert ^{2}-w^{2}\left( A\right) \leq
\left( \sqrt{M}-\sqrt{m}\right) ^{2}w\left( A\right) .  \label{4.11}$$Also, the inequality (\[2.7\]) becomes$$\left( 0\leq \right) \left\Vert A\right\Vert -w\left( A\right) \leq \frac{1}{%
4}\cdot \frac{\left( M-m\right) ^{2}}{M+m}.  \label{4.12}$$

[9]{} K.E. GUSTAFSON and D.K.M. RAO, *Numerical Range,* Springer-Verlag, New York, Inc., 1997.

P.R. HALMOS, *A Hilbert Space Problem Book,* Springer-Verlag, New York, Heidelberg, Berlin, Second edition, 1982.
