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\noindent\parbox{2.9cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}}
\noindent\parbox{4.9in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 1 (2007), no. 2, 154--175\\
$\frac{\rule{4.5in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
\textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\
$\frac{{}}{\rule{4.5in}{0.05in}}$}\\[.5in]}

\title[Inequalities for the norm and numerical radius]{A survey of some
recent inequalities for the norm and numerical radius of operators
in Hilbert spaces}

\author[S.S. Dragomir]{Sever S. Dragomir$^1$}

\address{$^1$ School of Computer Science and Mathematics,
Victoria University of Technology, PO Box 14428, Melbourne City,
Victoria 8001, Australia.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{sever.dragomir@vu.edu.au}}

\dedicatory{This paper is dedicated to Professor Themistocles. M.
Rassias.\\
\vspace{.5cm}{\rm  Submitted by F. Kittaneh}}

\subjclass[2000]{47A12}

\keywords{Numerical range, numerical radius, bounded linear
operator, Hilbert space.}

\date{Received: 8 March 2007; Accepted: 1 November 2007.}

\begin{abstract}
Some recent inequalities for the norm and the numerical radius of
linear operators in Hilbert spaces are surveyed.
\end{abstract}

\maketitle


\section{Introduction}

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

\begin{enumerate}
\item[(i)] $W\left( \alpha I+\beta T\right) =\alpha +\beta W\left( T\right) $
for $\alpha ,\beta \in \mathbb{C};$

\item[(ii)] $W\left( T^{\ast }\right) =\left\{ \bar{\lambda},\lambda \in
W\left( T\right) \right\} ,$ where $T^{\ast }$ is the \emph{adjoint
operator }of $T;$

\item[(iii)] $W\left( U^{\ast }TU\right) =W\left( T\right) $ for any \emph{%
unitary} operator $U.$
\end{enumerate}

The following classical fact about the geometry of the numerical range \cite[%
p. 4]{GR} may be stated:

\begin{theorem}[Toeplitz-Hausdorff]
The numerical range of an operator is convex.
\end{theorem}

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

\begin{theorem}[Spectral inclusion]
The spectrum of an operator is contained in the closure of its numerical
range.
\end{theorem}

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

\begin{theorem}
The following statements hold true:

\begin{enumerate}
\item[(i)] $T$ is self-adjoint iff $W\left( T\right) $ is real;

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

\item[(iii)] If $W\left( T\right) =\left[ m,M\right] ,$ then $m,M\in \sigma
\left( T\right) .$
\end{enumerate}
\end{theorem}

The \emph{numerical radius} $w\left( T\right) $ of an operator $T$
on $H$
is given by \cite[p. 8]{GR}:%
\begin{equation}
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}
\end{equation}%
Obviously, by (\ref{1.1}), for any $x\in H$ one has%
\begin{eqnarray*}
\left\vert \left\langle Tx,x\right\rangle \right\vert \leq w\left(
T\right) \left\Vert x\right\Vert ^{2}.
\end{eqnarray*}

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.,

\begin{enumerate}
\item[(i)] $w\left( T\right) \geq 0$ for any $T\in B\left( H\right) $ and $%
w\left( T\right) =0$ if and only if $T=0;$

\item[(ii)] $w\left( \lambda T\right) =\left\vert \lambda \right\vert
w\left( T\right) $ for any $\lambda \in \mathbb{C}$ and $T\in B\left(
H\right) ;$

\item[(iii)] $w\left( T+V\right) \leq w\left( T\right) +w\left( V\right) $
for any $T,V\in B\left( H\right) .$
\end{enumerate}

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

\begin{theorem}[Equivalent norm]
For any $T\in B\left( H\right) $ one has%
\begin{equation}
w\left( T\right) \leq \left\Vert T\right\Vert \leq 2w\left( T\right) .
\label{1.3}
\end{equation}
\end{theorem}

Let us now look at two extreme cases of the inequality (\ref{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 \emph{%
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 \emph{point spectrum }of $T.$

The following results hold \cite[p. 10]{GR}:

\begin{theorem}
\label{t1}We have

\begin{enumerate}
\item[(i)] If $w\left( T\right) =\left\Vert T\right\Vert ,$ then $r\left(
T\right) =\left\Vert T\right\Vert .$

\item[(ii)] If $\lambda \in W\left( T\right) $ and $\left\vert \lambda
\right\vert =\left\Vert T\right\Vert ,$ then $\lambda \in \sigma _{p}\left(
T\right) .$
\end{enumerate}
\end{theorem}

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 \cite[p. 11]{GR})%
\begin{equation*}
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\} .
\end{equation*}

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

It is well-known that the two-dimensional shift%
\begin{equation*}
S_{2}=\left[
\begin{array}{ll}
0 & 0 \\
1 & 0%
\end{array}%
\right] ,
\end{equation*}%
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 \cite[p. 11]%
{GR}:

\begin{theorem}
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 $S_{2}.$
\end{theorem}

For other results on numerical radius, see \cite{H}, Chapter 11.

We recall some classical results involving the numerical radius of two
linear operators $A,B.$

The following general result for the product of two operators holds \cite[p.
37]{GR}:

\begin{theorem}
\label{t1.1}If $A,B$ are two bounded linear operators on the Hilbert space $%
\left( H,\left\langle \cdot ,\cdot \right\rangle \right) ,$ then%
\begin{eqnarray*}
w\left( AB\right) \leq 4w\left( A\right) w\left( B\right) .
\end{eqnarray*}
In the case that $AB=BA,$ then%
\begin{eqnarray*}
w\left( AB\right) \leq 2w\left( A\right) w\left( B\right) .
\label{1.5}
\end{eqnarray*}
\end{theorem}

The following results are also well known \cite[p. 38]{GR}.

\begin{theorem}
\label{t1.2}If $A$ is a unitary operator that commutes with another operator
$B,$ then%
\begin{eqnarray}
w\left( AB\right) \leq w\left( B\right). \label{1.6}
\end{eqnarray}
If $A$ is an isometry and $AB=BA,$ then {\rm(\ref{1.6})} also holds
true.
\end{theorem}

We say that $A$ and $B$ \emph{double commute} if $AB=BA$ and
$AB^{\ast }=B^{\ast }A.$

The following result holds \cite[p. 38]{GR}.

\begin{theorem}[Double commute]
\label{t1.3}If the operators $A$ and $B$ double commute, then%
\begin{eqnarray*}
w\left( AB\right) \leq w\left( B\right) \left\Vert A\right\Vert .
\end{eqnarray*}
\end{theorem}

As a consequence of the above, we have \cite[p. 39]{GR}:

\begin{corollary}
\label{c1.1}Let $A$ be a normal operator commuting with $B.$ Then%
\begin{eqnarray*}
w\left( AB\right) \leq w\left( A\right) w\left( B\right) .
\end{eqnarray*}
\end{corollary}

For other results and historical comments on the above see \cite[p. 39--41]%
{GR}. For more results on the numerical radius, see \cite{H}.

The main aim of this paper is to survey some inequalities for the norm and
the numerical radius of bounded linear operators in complex Hilbert spaces
obtained by the author in a sequence of recent works. For the sake of
completeness and since not all involved results have yet been published,
detailed proofs are given as well.

\section{Reverse inequalities for one operator}

The following results may be stated \cite{SSD3}:

\begin{theorem}
\label{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%
\begin{equation}
\left\Vert T-\lambda I\right\Vert \leq r,  \label{2.1}
\end{equation}%
where $I:H\rightarrow H$ is the identity operator on $H,$ then%
\begin{equation}
\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}
\end{equation}
\end{theorem}

\begin{proof}
For $x\in H$ with $\left\Vert x\right\Vert =1,$ we have from (\ref{2.1}) that%
\begin{equation*}
\left\Vert Tx-\lambda x\right\Vert \leq \left\Vert T-\lambda I\right\Vert
\leq r,
\end{equation*}%
giving%
\begin{equation}
\left\Vert Tx\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}\leq 2\textrm{%
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}
\end{equation}%
Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (\ref{2.3}%
) we get the following inequality that is of interest in itself:%
\begin{equation}
\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}
\end{equation}%
Since, obviously,%
\begin{equation}
\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}
\end{equation}%
hence by (\ref{2.4}) and (\ref{2.5}) we deduce the desired inequality (\ref%
{2.2}).
\end{proof}

\begin{remark}
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 (\ref{2.2}). Indeed,
by Theorem \ref{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 \ref%
{t.1}, $\lambda =1,$ $r=1,$ then we get in both sides of (\ref{2.2}) the
same quantity $\frac{1}{2}.$
\end{remark}

\begin{problem}
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 $%
\left\Vert T-\lambda I\right\Vert \leq \left\vert \lambda \right\vert ^{%
\frac{1}{2}}.$
\end{problem}

The following corollary may be stated \cite{SSD3}:

\begin{corollary}
\label{c.1}Let $A:H\rightarrow H$ be a bounded linear operator and $\varphi
,\phi \in \mathbb{C}$ with $\phi \neq -\varphi ,\varphi .$ If%
\begin{equation}
\textrm{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}
\end{equation}%
then%
\begin{equation}
\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}
\end{equation}
\end{corollary}

\begin{proof}
Utilising the fact that in any Hilbert space the following two statements
are equivalent:

\begin{enumerate}
\item[(i)] $\textrm{Re}\left\langle Z-x,x-z\right\rangle \geq 0,$ $x,z,Z\in H;$

\item[(ii)] $\left\Vert x-\frac{z+Z}{2}\right\Vert \leq \frac{1}{2}%
\left\Vert Z-z\right\Vert ,$
\end{enumerate}

we deduce that (\ref{2.6}) is equivalent to%
\begin{eqnarray*}
\left\Vert Ax-\frac{\phi +\varphi }{2}\cdot Ix\right\Vert \leq \frac{1}{2}%
\left\vert \phi -\varphi \right\vert
\end{eqnarray*}
for any $x\in H,$ $\left\Vert x\right\Vert =1,$ which in its turn is
equivalent with the operator norm inequality:%
\begin{eqnarray*}
\left\Vert A-\frac{\phi +\varphi }{2}\cdot I\right\Vert \leq \frac{1}{2}%
\left\vert \phi -\varphi \right\vert .
\end{eqnarray*}
Now, applying Theorem \ref{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 (\ref{2.7}).
\end{proof}

\begin{remark}
Following \cite[p. 25]{GR}, we say that an operator $B:H\rightarrow
H$ is accretive, if $\textrm{Re}\left\langle Bx,x\right\rangle \geq
0$ for any $x\in H.$ One may observe that the assumption (\ref{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
accretive.
\end{remark}

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

\begin{corollary}
\label{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%
\begin{equation}
\left( A^{\ast }-\bar{\varphi}I\right) \left( \phi I-A\right) \geq 0
\label{2.10}
\end{equation}%
in the operator order, then%
\begin{eqnarray*}
\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 }.
\end{eqnarray*}
\end{corollary}

The following result may be stated as well:

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

\begin{proof}
From (\ref{2.4}) of Theorem \ref{t.1}, we have%
\begin{align}
\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} \notag\\
& =r^{2}-\left( \left\vert \lambda \right\vert -w\left( T\right) \right)
^{2}.  \notag
\end{align}%
On utilising (\ref{2.4}) and (\ref{2.12a}) we deduce the desired inequality (%
\ref{2.13}).
\end{proof}

\begin{remark}
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%
\begin{eqnarray*}
\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left(
T\right) \leq r^{2}.
\end{eqnarray*}
\end{remark}

The following result may be stated as well.

\begin{theorem}
\label{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%
\begin{eqnarray*}
\left\Vert T-\lambda I\right\Vert \leq r,
\end{eqnarray*}%
then%
\begin{equation}
\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}
\end{equation}
\end{theorem}

\begin{proof}
From (\ref{2.4}) of Theorem \ref{t.1}, we have%
\begin{equation*}
\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) ,
\end{equation*}
which implies, on dividing with $\sqrt{\left\vert \lambda
\right\vert ^{2}-r^{2}}>0$ that
\begin{equation}
\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}
\end{equation}
By the elementary inequality%
\begin{eqnarray*}
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}}
\end{eqnarray*}
and by (\ref{3.3}) we deduce%
\begin{equation*}
\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}}},
\end{equation*}
which is equivalent to (\ref{3.2}).
\end{proof}

\begin{remark}
Squaring (\ref{3.2}), we get the inequality%
\begin{eqnarray*}
\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}.
\end{eqnarray*}
\end{remark}

\begin{remark}
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 (\ref{3.2}%
) would produce a refinement of this classic fact only in the case when%
\begin{equation*}
\frac{1}{2}\leq \left( 1-\frac{r^{2}}{\left\vert \lambda \right\vert ^{2}}%
\right) ^{\frac{1}{2}},
\end{equation*}%
which is equivalent to $r/\left\vert \lambda \right\vert \leq \sqrt{3}/2.$
\end{remark}

The following corollary holds \cite{SSD3}.

\begin{corollary}
\label{c.4}Let $\varphi ,\phi \in \mathbb{C}$ with
$\textrm{Re}\left( \phi \bar{\varphi}\right) >0.$ If $T:H\rightarrow
H$ is a bounded linear operator
such that either {\rm(\ref{2.6})} or {\rm(\ref{2.10})} holds true, then:%
\begin{equation}
\frac{2\sqrt{\textrm{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}
\end{equation}%
and%
\begin{eqnarray*}
\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}.
\end{eqnarray*}
\end{corollary}

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

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

The following result may be of interest as well \cite{SSD3}.

\begin{theorem}
\label{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%
\begin{eqnarray*}
\left\Vert T-\lambda I\right\Vert \leq r,
\end{eqnarray*}%
then%
\begin{equation}
\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}
\end{equation}
\end{theorem}

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

If we divide (\ref{4.2a}) by $\left\vert \lambda \right\vert \left\vert
\left\langle Tx,x\right\rangle \right\vert ,$ (which, by (\ref{4.2a}), is
positive) then we obtain%
\begin{equation}
\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\textrm{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}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

If we subtract in (\ref{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{align}
& \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\textrm{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\textrm{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
\end{align}
\begin{align*}
& =\frac{2\textrm{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}\frac{\left\vert \left\langle
Tx,x\right\rangle \right\vert }{\left\vert \lambda \right\vert }  \notag \\
& \qquad -2\frac{\sqrt{\left\vert \lambda \right\vert
^{2}-r^{2}}}{\left\vert \lambda \right\vert }.  \notag
\end{align*}%
Since%
\begin{equation*}
\textrm{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
\end{equation*}%
and%
\begin{equation*}
\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
\end{equation*}%
hence by (\ref{4.4}) we get%
\begin{equation*}
\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 }
\end{equation*}%
which gives the inequality%
\begin{eqnarray*}
\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)
\end{eqnarray*}%
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{align*}
\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{align*}%
which is clearly equivalent to (\ref{4.2}).
\end{proof}

\begin{corollary}
\label{c.5}Let $\varphi ,\phi \in \mathbb{C}$ with
$\textrm{Re}\left( \phi \bar{\varphi}\right) >0.$ If $A:H\rightarrow
H$ is a bounded linear operator
such that either {\rm(\ref{2.6})} or {\rm(\ref{2.10})} hold true, then:%
\begin{eqnarray*}
\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{\textrm{Re}\left( \phi \bar{\varphi}\right) }\right] w\left(
A\right) .
\end{eqnarray*}
\end{corollary}

\begin{remark}
If $M\geq m>0$ are such that either $\left( A^{\ast }-mI\right) \left(
MI-A\right) $ is accretive, or, sufficiently, $\left( A^{\ast }-mI\right)
\left( MI-A\right) $ is self-adjoint and%
\begin{eqnarray*}
\left( A^{\ast }-mI\right) \left( MI-A\right) \geq 0\quad \text{in
the operator order,}
\end{eqnarray*}%
then, by (\ref{3.6}) we have:%
\begin{eqnarray*}
\left( 1\leq \right) \frac{\left\Vert A\right\Vert }{w\left(
A\right) }\leq \frac{M+m}{2\sqrt{mM}},
\end{eqnarray*}%
which is equivalent to%
\begin{eqnarray*}
\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) ,
\end{eqnarray*}%
while from (\ref{4.2}) we have%
\begin{eqnarray*}
\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) .
\end{eqnarray*}%
Also, the inequality (\ref{2.7}) becomes%
\begin{eqnarray*}
\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}.
\end{eqnarray*}
\end{remark}

\bigskip

\section{Other inequalities for one operator}

The following result may be stated as well \cite{SSD5}:

\begin{theorem}
\label{t.1b}Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $
be a Hilbert space and $T:H\rightarrow H$ a bounded linear operator on $H.$
Then%
\begin{equation}
w^{2}\left( T\right) \leq \frac{1}{2}\left[ w\left( T^{2}\right) +\left\Vert
T\right\Vert ^{2}\right] .  \label{2.1b}
\end{equation}%
The constant $\frac{1}{2}$ is best possible in {\rm(\ref{2.1b})}.
\end{theorem}

\begin{proof}
We need the following refinement of Schwarz's inequality obtained by the
author in 1985 \cite[Theorem 2]{SSD0} (see also \cite{DS} and \cite{SSD2}):%
\begin{equation}
\left\Vert a\right\Vert \left\Vert b\right\Vert \geq \left\vert \left\langle
a,b\right\rangle -\left\langle a,e\right\rangle \left\langle
e,b\right\rangle \right\vert +\left\vert \left\langle a,e\right\rangle
\left\langle e,b\right\rangle \right\vert \geq \left\vert \left\langle
a,b\right\rangle \right\vert ,  \label{2.2b}
\end{equation}%
provided $a,b,e$ are vectors in $H$ and $\left\Vert e\right\Vert =1.$

Observing that%
\begin{equation*}
\left\vert \left\langle a,b\right\rangle -\left\langle a,e\right\rangle
\left\langle e,b\right\rangle \right\vert \geq \left\vert \left\langle
a,e\right\rangle \left\langle e,b\right\rangle \right\vert -\left\vert
\left\langle a,b\right\rangle \right\vert ,
\end{equation*}%
hence by the first inequality in (\ref{2.2b}) we deduce%
\begin{equation}
\frac{1}{2}\left( \left\Vert a\right\Vert \left\Vert b\right\Vert
+\left\vert \left\langle a,b\right\rangle \right\vert \right) \geq
\left\vert \left\langle a,e\right\rangle \left\langle e,b\right\rangle
\right\vert .  \label{2.3b}
\end{equation}%
This inequality was obtained in a different way earlier by M.L. Buzano in
\cite{B}.

Now, choose in (\ref{2.3b}), $e=x,$ $\left\Vert x\right\Vert =1,$ $a=Tx$ and
$b=T^{\ast }x$ to get%
\begin{equation}
\frac{1}{2}\left( \left\Vert Tx\right\Vert \left\Vert T^{\ast }x\right\Vert
+\left\vert \left\langle T^{2}x,x\right\rangle \right\vert \right) \geq
\left\vert \left\langle Tx,x\right\rangle \right\vert ^{2}  \label{2.4b}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Taking the supremum in (\ref{2.4b}) over $x\in H,$ $\left\Vert x\right\Vert
=1,$ we deduce the desired inequality (\ref{2.1}).

Now, if we assume that (\ref{2.1b}) holds with a constant $C>0,i.e.,$%
\begin{equation}
w^{2}\left( T\right) \leq C\left[ w\left( T^{2}\right) +\left\Vert
T\right\Vert ^{2}\right]  \label{2.4.1b}
\end{equation}%
for any $T\in B\left( H\right) ,$ then if we choose $T$ a normal operator
and use the fact that for normal operators we have $w\left( T\right)
=\left\Vert T\right\Vert $ and $w\left( T^{2}\right) =\left\Vert
T^{2}\right\Vert =\left\Vert T\right\Vert ^{2},$ then by (\ref{2.4.1b}) we
deduce that $2C\geq 1$ which proves the sharpness of the constant.
\end{proof}

\begin{remark}
From the above result (\ref{2.1b}) we obviously have%
\begin{align}
w\left( T\right) &\leq \left\{ \frac{1}{2}\left[ w\left(
T^{2}\right)
+\left\Vert T\right\Vert ^{2}\right] \right\} ^{1/2}\notag \\
&\leq \left\{ \frac{1}{2}%
\left( \left\Vert T^{2}\right\Vert +\left\Vert T\right\Vert
^{2}\right) \right\} ^{1/2}\leq \left\Vert T\right\Vert \notag
\end{align}%
and%
\begin{align}
w\left( T\right)& \leq \left\{ \frac{1}{2}\left[ w\left(
T^{2}\right)
+\left\Vert T\right\Vert ^{2}\right] \right\} ^{1/2}\notag\\
&\leq \left\{ \frac{1}{2}%
\left( w^{2}\left( T\right) +\left\Vert T\right\Vert ^{2}\right)
\right\} ^{1/2}\leq \left\Vert T\right\Vert ,\notag
\end{align}
that provide refinements for the first inequality in (\ref{1.3}).
\end{remark}

The following result may be stated \cite{SSD5}.

\begin{theorem}
\label{t.2b}Let $T:H\rightarrow H$ be a bounded linear operator on the
Hilbert space $H$ and $\lambda \in \mathbb{C}\backslash \left\{ 0\right\} .$
If $\left\Vert T\right\Vert \leq \left\vert \lambda \right\vert ,$ then%
\begin{equation}
\left\Vert T\right\Vert ^{2r}+\left\vert \lambda \right\vert ^{2r}\leq
2\left\Vert T\right\Vert ^{r-1}\left\vert \lambda \right\vert ^{r}w\left(
T\right) +r^{2}\left\vert \lambda \right\vert ^{2r-2}\left\Vert T-\lambda
I\right\Vert ^{2},  \label{2.5b}
\end{equation}%
where $r\geq 1.$
\end{theorem}

\begin{proof}
We use the following inequality for vectors in inner product spaces due to
Goldstein, Ryff and Clarke \cite{GRC} (see also \cite{SSD1}):%
\begin{multline}
\left\Vert a\right\Vert ^{2r}+\left\Vert b\right\Vert
^{2r}-2\left\Vert a\right\Vert ^{r}\left\Vert b\right\Vert
^{r}\frac{\textrm{Re}\left\langle a,b\right\rangle }{\left\Vert
a\right\Vert \left\Vert b\right\Vert }
\label{2.6b} \\
\leq \left\{
\begin{array}{ll}
r^{2}\left\Vert a\right\Vert ^{2r-2}\left\Vert a-b\right\Vert ^{2} & \text{%
if \ }r\geq 1, \\
&  \\
\left\Vert b\right\Vert ^{2r-2}\left\Vert a-b\right\Vert ^{2} & \text{if \ }%
r<1,%
\end{array}%
\right.
\end{multline}%
provided $r\in \mathbb{R}$ and $a,b\in H$ with $\left\Vert a\right\Vert \geq
\left\Vert b\right\Vert .$

Now, let $x\in H$ with $\left\Vert x\right\Vert =1.$ From the hypothesis of
the theorem, we have that $\left\Vert Tx\right\Vert \leq \left\vert \lambda
\right\vert \left\Vert x\right\Vert $ and applying (\ref{2.6b}) for the
choices $a=\lambda x,$ $\left\Vert x\right\Vert =1,$ $b=Tx,$ we get%
\begin{equation}
\left\Vert Tx\right\Vert ^{2r}+\left\vert \lambda \right\vert
^{2r}-2\left\Vert Tx\right\Vert ^{r-1}\left\vert \lambda \right\vert
^{r}\left\vert \left\langle Tx,x\right\rangle \right\vert \leq
r^{2}\left\vert \lambda \right\vert ^{2r-2}\left\Vert Tx-\lambda
x\right\Vert ^{2}  \label{2.7b}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1$ and $r\geq 1.$

Taking the supremum in (\ref{2.7b}) over $x\in H,$ $\left\Vert x\right\Vert
=1,$ we deduce the desired inequality (\ref{2.5b}).
\end{proof}

The following result may be stated as well \cite{SSD5}:

\begin{theorem}
\label{t.3b}Let $T:H\rightarrow H$ be a bounded linear operator on the
Hilbert space $\left( H,\left\langle \cdot ,\cdot \right\rangle \right) .$
Then for any $\alpha \in \left[ 0,1\right] $ and $t\in \mathbb{R}$ one has
the inequality:%
\begin{multline*}
\left\Vert T\right\Vert ^{2}\leq \left[ \left( 1-\alpha \right)
^{2}+\alpha ^{2}\right] w^{2}\left( T\right) +\alpha \left\Vert
T-tI\right\Vert ^{2}+\left( 1-\alpha \right) \left\Vert
T-itI\right\Vert ^{2}.
\end{multline*}
\end{theorem}

\begin{proof}
We use the following inequality obtained by the author in \cite{SSD2}:%
\begin{multline*}
\left[ \alpha \left\Vert tb-a\right\Vert ^{2}+\left( 1-\alpha \right)
\left\Vert itb-a\right\Vert ^{2}\right] \left\Vert b\right\Vert ^{2} \\
\geq \left\Vert a\right\Vert ^{2}\left\Vert b\right\Vert ^{2}-\left[ \left(
1-\alpha \right) \textrm{Im}\left\langle a,b\right\rangle +\alpha \textrm{Re}%
\left\langle a,b\right\rangle \right] ^{2}\left( \geq 0\right)
\end{multline*}%
to get:%
\begin{align}
\left\Vert a\right\Vert ^{2}\left\Vert b\right\Vert ^{2}& \leq \left[ \left(
1-\alpha \right) \textrm{Im}\left\langle a,b\right\rangle +\alpha \textrm{Re}%
\left\langle a,b\right\rangle \right] ^{2}  \label{2.10b} \\
& \qquad +\left[ \alpha \left\Vert tb-a\right\Vert ^{2}+\left( 1-\alpha
\right) \left\Vert itb-a\right\Vert ^{2}\right] \left\Vert b\right\Vert ^{2}
\notag \\
& \leq \left[ \left( 1-\alpha \right) ^{2}+\alpha ^{2}\right] \left\vert
\left\langle a,b\right\rangle \right\vert ^{2}  \notag \\
& \qquad +\left[ \alpha \left\Vert tb-a\right\Vert ^{2}+\left( 1-\alpha
\right) \left\Vert itb-a\right\Vert ^{2}\right] \left\Vert b\right\Vert ^{2}
\notag
\end{align}%
for any $a,b\in H,$ $\alpha \in \left[ 0,1\right] $ and $t\in \mathbb{R}$.

Choosing in (\ref{2.10b}) $a=Tx,$ $b=x,$ $x\in H,$ $\left\Vert x\right\Vert
=1,$ we get%
\begin{multline}
\left\Vert Tx\right\Vert ^{2}\leq \left[ \left( 1-\alpha \right) ^{2}+\alpha
^{2}\right] \left\vert \left\langle Tx,x\right\rangle \right\vert ^{2}
\label{2.11b} \\
+\alpha \left\Vert tx-Tx\right\Vert ^{2}+\left( 1-\alpha \right) \left\Vert
itx-Tx\right\Vert ^{2}.
\end{multline}%
Finally, taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (%
\ref{2.11b}), we deduce the desired result.
\end{proof}

The following particular cases may be of interest \cite{SSD5}.

\begin{corollary}
\label{c.4b}For any $T$ a bounded linear operator on $H,$ one has:%
\begin{equation}
\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left( T\right) \leq
\left\{
\begin{array}{l}
\inf\limits_{t\in \mathbb{R}}\left\Vert T-tI\right\Vert ^{2} \\
\\
\inf\limits_{t\in \mathbb{R}}\left\Vert T-itI\right\Vert ^{2}%
\end{array}%
\right.  \label{2.12b}
\end{equation}%
and%
\begin{eqnarray*}
\left\Vert T\right\Vert ^{2}\leq \frac{1}{2}w^{2}\left( T\right) +\frac{1}{2}%
\inf\limits_{t\in \mathbb{R}}\left[ \left\Vert T-tI\right\Vert
^{2}+\left\Vert T-itI\right\Vert ^{2}\right] .
\end{eqnarray*}
\end{corollary}

\begin{remark}
The inequality (\ref{2.12b}) can in fact be improved taking into account
that for any $a,b\in H,$ $b\neq 0,$ (see for instance \cite{SSD4}) the bound%
\begin{equation*}
\inf_{\lambda \in \mathbb{C}}\left\Vert a-\lambda b\right\Vert ^{2}=\frac{%
\left\Vert a\right\Vert ^{2}\left\Vert b\right\Vert ^{2}-\left\vert
\left\langle a,b\right\rangle \right\vert ^{2}}{\left\Vert b\right\Vert ^{2}}
\end{equation*}%
actually implies that%
\begin{equation}
\left\Vert a\right\Vert ^{2}\left\Vert b\right\Vert ^{2}-\left\vert
\left\langle a,b\right\rangle \right\vert ^{2}\leq \left\Vert b\right\Vert
^{2}\left\Vert a-\lambda b\right\Vert ^{2}  \label{2.14b}
\end{equation}%
for any $a,b\in H$ and $\lambda \in \mathbb{C}$.

Now if in (\ref{2.14b}) we choose $a=Tx,$ $b=x,$ $x\in H,$ $\left\Vert
x\right\Vert =1,$ then we obtain%
\begin{eqnarray*}
\left\Vert Tx\right\Vert ^{2}-\left\vert \left\langle
Tx,x\right\rangle \right\vert ^{2}\leq \left\Vert Tx-\lambda
x\right\Vert ^{2}
\end{eqnarray*}%
for any $\lambda \in \mathbb{C}$, which, by taking the supremum over
$x\in
H, $ $\left\Vert x\right\Vert =1$, implies that%
\begin{eqnarray*}
\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left(
T\right) \leq \inf_{\lambda \in \mathbb{C}}\left\Vert T-\lambda
I\right\Vert ^{2}.
\end{eqnarray*}
\end{remark}

\begin{remark}
If we take $a=x,$ $b=Tx$ in (\ref{2.14b}), then we obtain%
\begin{equation}
\left\Vert Tx\right\Vert ^{2}\leq \left\vert \left\langle Tx,x\right\rangle
\right\vert ^{2}+\left\Vert Tx\right\Vert ^{2}\left\Vert x-\mu Tx\right\Vert
^{2}  \label{2.17b}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1$ and $\mu \in \mathbb{C}$.
Now, if we take the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (%
\ref{2.17b}), then we get%
\begin{eqnarray*}
\left( 0\leq \right) \left\Vert T\right\Vert ^{2}-w^{2}\left(
T\right) \leq \left\Vert T\right\Vert ^{2}\inf_{\mu \in
\mathbb{C}}\left\Vert I-\mu T\right\Vert ^{2}.
\end{eqnarray*}
\end{remark}

Finally and from a different view point we may state \cite{SSD5}:

\begin{theorem}
\label{t.4b}Let $T:H\rightarrow H$ be a bounded linear operator on $H.$ If $%
p\geq 2,$ then:%
\begin{equation}
\left\Vert T\right\Vert ^{p}+\left\vert \lambda \right\vert ^{p}\leq \frac{1%
}{2}\left( \left\Vert T+\lambda I\right\Vert ^{p}+\left\Vert T-\lambda
I\right\Vert ^{p}\right) ,  \label{2.19b}
\end{equation}%
for any $\lambda \in \mathbb{C}$.
\end{theorem}

\begin{proof}
We use the following inequality obtained by Dragomir and S\'{a}ndor in \cite%
{DS}:%
$$\left\Vert a+b\right\Vert ^{p}+\left\Vert a-b\right\Vert ^{p}\geq
2\left( \left\Vert a\right\Vert ^{p}+\left\Vert b\right\Vert
^{p}\right)$$
for any $a,b\in H$ and $p\geq 2$.

Now, if we choose $a=Tx,$ $b=\lambda x,$ then we get%
\begin{equation}
\left\Vert Tx+\lambda x\right\Vert ^{p}+\left\Vert Tx-\lambda x\right\Vert
^{p}\geq 2\left( \left\Vert Tx\right\Vert ^{p}+\left\vert \lambda
\right\vert ^{p}\right)  \label{2.21b}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Taking the supremum in (\ref{2.21b}) over $x\in H,$ $\left\Vert x\right\Vert
=1,$ we get the desired result (\ref{2.19b}).
\end{proof}

\begin{remark}
For $p=2,$ we have the simpler result:%
\begin{equation*}
\left\Vert T\right\Vert ^{2}+\left\vert \lambda \right\vert ^{2}\leq \frac{1%
}{2}\left( \left\Vert T+\lambda I\right\Vert ^{2}+\left\Vert
T-\lambda I\right\Vert ^{2}\right)
\end{equation*}%
for any $\lambda \in \mathbb{C}$. This can easily be obtained from the
parallelogram identity as well.
\end{remark}

\section{Reverse inequalities for two operators}

The following result may be stated \cite{SSD6}:

\begin{theorem}
\label{t2.1c}Let $A,B:H\rightarrow H$ be two bounded linear operators on the
Hilbert space $\left( H,\left\langle \cdot ,\cdot \right\rangle \right) .$
If $r>0$ and%
\begin{equation}
\left\Vert A-B\right\Vert \leq r,  \label{2.1c}
\end{equation}%
then%
\begin{equation}
\left\Vert \frac{A^{\ast }A+B^{\ast }B}{2}\right\Vert \leq w\left( B^{\ast
}A\right) +\frac{1}{2}r^{2}.  \label{2.2c}
\end{equation}
\end{theorem}

\begin{proof}
For any $x\in H,$ $\left\Vert x\right\Vert =1,$ we have from (\ref{2.1c})
that%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2}\leq 2\textrm{Re}%
\left\langle Ax,Bx\right\rangle +r^{2}.  \label{2.3c}
\end{equation}%
However%
\begin{align*}
\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2}& =\left\langle
\left( A^{\ast }A\right) x,x\right\rangle +\left\langle \left( B^{\ast
}B\right) x,x\right\rangle \\
& =\left\langle \left( A^{\ast }A+B^{\ast }B\right) x,x\right\rangle
\end{align*}%
and by (\ref{2.3c}) we obtain%
\begin{equation}
\left\langle \left( A^{\ast }A+B^{\ast }B\right) x,x\right\rangle \leq
2\left\vert \left\langle \left( B^{\ast }A\right) x,x\right\rangle
\right\vert +r^{2}  \label{2.4c}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (\ref%
{2.4c}) we get%
\begin{equation}
w\left( A^{\ast }A+B^{\ast }B\right) \leq 2w\left( B^{\ast }A\right) +r^{2}
\label{2.5c}
\end{equation}%
and since the operator $A^{\ast }A+B^{\ast }B$ is self-adjoint, hence
\begin{equation*}
w\left( A^{\ast }A+B^{\ast }B\right) =\left\Vert A^{\ast }A+B^{\ast
}B\right\Vert
\end{equation*}%
and by (\ref{2.5c}) we deduce the desired inequality (\ref{2.2c}).
\end{proof}

\begin{remark}
We observe that, from the proof of the above theorem, we have the
inequalities%
\begin{equation}
0\leq \left\Vert \frac{A^{\ast }A+B^{\ast }B}{2}\right\Vert -w\left( B^{\ast
}A\right) \leq \frac{1}{2}\left\Vert A-B\right\Vert ^{2},  \label{2.6c}
\end{equation}%
provided that $A,B$ are bounded linear operators in $H.$

The second inequality in (\ref{2.6c}) is obvious while the first inequality
follows by the fact that%
\begin{align*}
\left\langle \left( A^{\ast }A+B^{\ast }B\right) x,x\right\rangle
&=\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2} \\
&\geq 2\left\Vert Ax\right\Vert \left\Vert Bx\right\Vert \geq
2\left\vert \left\langle \left( B^{\ast }A\right) x,x\right\rangle
\right\vert
\end{align*}%
for any $x\in H.$
\end{remark}

The inequality (\ref{2.2c}) is obviously a reach source of particular
inequalities of interest.

Indeed, if we assume, for $\lambda \in \mathbb{C}$ and a bounded linear
operator $T,$ that we have
\begin{eqnarray*}
\left\Vert T-\lambda T^{\ast }\right\Vert \leq r,
\end{eqnarray*}%
for a given positive number $r,$ then by (\ref{2.6c}) we deduce the
inequality%
\begin{eqnarray*}
0\leq \left\Vert \frac{T^{\ast }T+\left\vert \lambda \right\vert
^{2}TT^{\ast }}{2}\right\Vert -\left\vert \lambda \right\vert
w\left( T^{2}\right) \leq \frac{1}{2}r^{2}.
\end{eqnarray*}

Now, if we assume that for $\lambda \in \mathbb{C}$ and a bounded linear
operator $V$ we have that%
\begin{eqnarray*}
\left\Vert V-\lambda I\right\Vert \leq r,
\end{eqnarray*}%
where $I$ is the identity operator on $H,$ then by (\ref{2.2c}) we
deduce
the inequality%
\begin{equation*}
0\leq \left\Vert \frac{V^{\ast }V+\left\vert \lambda \right\vert ^{2}I}{2}%
\right\Vert -\left\vert \lambda \right\vert w\left( V\right) \leq \frac{1}{2}%
r^{2}.
\end{equation*}

As a dual approach, the following result may be noted as well \cite{SSD6}:

\begin{theorem}
\label{t2.2c}Let $A,B:H\rightarrow H$ be two bounded linear operators on the
Hilbert space $H.$ Then%
\begin{equation}
\left\Vert \frac{A+B}{2}\right\Vert ^{2}\leq \frac{1}{2}\left[ \left\Vert
\frac{A^{\ast }A+B^{\ast }B}{2}\right\Vert +w\left( B^{\ast }A\right) \right]
.  \label{2.7c}
\end{equation}
\end{theorem}

\begin{proof}
We obviously have%
\begin{align*}
\left\Vert Ax+Bx\right\Vert ^{2}& =\left\Vert Ax\right\Vert ^{2}+2\textrm{Re}%
\left\langle Ax,Bx\right\rangle +\left\Vert Bx\right\Vert ^{2} \\
& \leq \left\langle \left( A^{\ast }A+B^{\ast }B\right) x,x\right\rangle
+2\left\vert \left\langle \left( B^{\ast }A\right) x,x\right\rangle
\right\vert
\end{align*}%
for any $x\in H.$

Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1,$ we get%
\begin{align*}
\left\Vert A+B\right\Vert ^{2}& \leq w\left( A^{\ast }A+B^{\ast }B\right)
+2w\left( B^{\ast }A\right) \\
& =\left\Vert A^{\ast }A+B^{\ast }B\right\Vert +2w\left( B^{\ast }A\right) ,
\end{align*}%
from where we get the desired inequality (\ref{2.7c}).
\end{proof}

\begin{remark}
The inequality (\ref{2.7c}) can generate some interesting particular results
such as the following inequality%
\begin{eqnarray*}
\left\Vert \frac{T+T^{\ast }}{2}\right\Vert ^{2}\leq
\frac{1}{2}\left[
\left\Vert \frac{T^{\ast }T+TT^{\ast }}{2}\right\Vert +w\left( T^{2}\right) %
\right] ,
\end{eqnarray*}%
holding for each bounded linear operator $T:H\rightarrow H.$
\end{remark}

The following result may be stated as well \cite{SSD6}.

\begin{theorem}
\label{t2.3c}Let $A,B:H\rightarrow H$ be two bounded linear operators on the
Hilbert space $H$ and $p\geq 2.$ Then%
\begin{equation}
\left\Vert \frac{A^{\ast }A+B^{\ast }B}{2}\right\Vert ^{\frac{p}{2}}\leq
\frac{1}{4}\left[ \left\Vert A-B\right\Vert ^{p}+\left\Vert A+B\right\Vert
^{p}\right] .  \label{2.8c}
\end{equation}
\end{theorem}

\begin{proof}
We use the following inequality for vectors in inner product spaces obtained
by Dragomir and S\'{a}ndor in \cite{DS}:%
\begin{equation}
2\left( \left\Vert a\right\Vert ^{p}+\left\Vert b\right\Vert ^{p}\right)
\leq \left\Vert a+b\right\Vert ^{p}+\left\Vert a-b\right\Vert ^{p}
\label{2.9c}
\end{equation}%
for any $a,b\in H$ and $p\geq 2.$

Utilising (\ref{2.9c}) we may write%
\begin{equation}
2\left( \left\Vert Ax\right\Vert ^{p}+\left\Vert Bx\right\Vert ^{p}\right)
\leq \left\Vert Ax+Bx\right\Vert ^{p}+\left\Vert Ax-Bx\right\Vert ^{p}
\label{2.10c}
\end{equation}%
for any $x\in H.$

Now, observe that%
\begin{equation*}
\left\Vert Ax\right\Vert ^{p}+\left\Vert Bx\right\Vert ^{p}=\left(
\left\Vert Ax\right\Vert ^{2}\right) ^{\frac{p}{2}}+\left( \left\Vert
Bx\right\Vert ^{2}\right) ^{\frac{p}{2}}
\end{equation*}%
and by the elementary inequality:%
\begin{equation*}
\frac{\alpha ^{q}+\beta ^{q}}{2}\geq \left( \frac{\alpha +\beta }{2}\right)
^{q},\quad \alpha ,\beta \geq 0\text{ \ and \ }q\geq 1
\end{equation*}%
we have%
\begin{align}
\left( \left\Vert Ax\right\Vert ^{2}\right) ^{\frac{p}{2}}+\left( \left\Vert
Bx\right\Vert ^{2}\right) ^{\frac{p}{2}}& \geq 2^{1-\frac{p}{2}}\left(
\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2}\right) ^{\frac{p%
}{2}}  \label{2.11c} \\
& =2^{1-\frac{p}{2}}\left[ \left\langle \left( A^{\ast }A+B^{\ast }B\right)
x,x\right\rangle \right] ^{\frac{p}{2}}.  \notag
\end{align}%
Combining (\ref{2.10c}) with (\ref{2.11c}) we get%
\begin{eqnarray*}
\frac{1}{4}\left[ \left\Vert Ax-Bx\right\Vert ^{p}+\left\Vert
Ax+Bx\right\Vert ^{p}\right] \geq \left\vert \left\langle \left( \frac{%
A^{\ast }A+B^{\ast }B}{2}\right) x,x\right\rangle \right\vert ^{\frac{p}{2}}
\end{eqnarray*}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$ Taking the supremum over $%
x\in H,$ $\left\Vert x\right\Vert =1,$ and taking into account that
\begin{equation*}
w\left( \frac{A^{\ast }A+B^{\ast }B}{2}\right) =\left\Vert \frac{A^{\ast
}A+B^{\ast }B}{2}\right\Vert ,
\end{equation*}%
we deduce the desired result (\ref{2.8c}).
\end{proof}

\begin{remark}
If $p=2,$ then we have the inequality:%
\begin{eqnarray*}
\left\Vert \frac{A^{\ast }A+B^{\ast }B}{2}\right\Vert \leq \left\Vert \frac{%
A-B}{2}\right\Vert ^{2}+\left\Vert \frac{A+B}{2}\right\Vert ^{2},
\end{eqnarray*}%
for any $A,B$ bounded linear operators. This result can also be
obtained directly on utilising the parallelogram identity.

We also should observe that for $A=T$ and $B=T^{\ast },$ $T$ a normal
operator, the inequality (\ref{2.8c}) becomes%
\begin{equation*}
\left\Vert T\right\Vert ^{p}\leq \frac{1}{4}\left[ \left\Vert T-T^{\ast
}\right\Vert ^{p}+\left\Vert T+T^{\ast }\right\Vert ^{p}\right] ,
\end{equation*}%
where $p\geq 2.$
\end{remark}

The following result may be stated as well \cite{SSD6}.

\begin{theorem}
\label{t2.4c}Let $A,B:H\rightarrow H$ be two bounded linear operators on the
Hilbert space $H$ and $r\geq 1.$ If $A^{\ast }A\geq B^{\ast }B$ in the
operator order or, equivalently, $\left\Vert Ax\right\Vert \geq \left\Vert
Bx\right\Vert $ for any $x\in H,$ then:%
\begin{multline}
\left\Vert \frac{A^{\ast }A+B^{\ast }B}{2}\right\Vert ^{r}\\\leq
\left\Vert
A\right\Vert ^{r-1}\left\Vert B\right\Vert ^{r-1}w\left( B^{\ast }A\right) +%
\frac{1}{2}r^{2}\left\Vert A\right\Vert ^{2r-2}\left\Vert A-B\right\Vert
^{2}.  \label{2.14c}
\end{multline}
\end{theorem}

\begin{proof}
We use the following inequality for vectors in inner product spaces due to
Goldstein, Ryff and Clarke \cite{GRC}:%
\begin{equation}
\left\Vert a\right\Vert ^{2r}+\left\Vert b\right\Vert ^{2r}\leq
2\left\Vert a\right\Vert ^{r-1}\left\Vert b\right\Vert
^{r-1}\textrm{Re}\left\langle a,b\right\rangle +r^{2}\left\Vert
a\right\Vert ^{2r-2}\left\Vert a-b\right\Vert ^{2},  \label{2.15c}
\end{equation}%
where $r\geq 1,$ $a,b\in H$ and $\left\Vert a\right\Vert \geq \left\Vert
b\right\Vert .$

Utilising (\ref{2.15c}) we can state that:%
\begin{multline}
\left\Vert Ax\right\Vert ^{2r}+\left\Vert Bx\right\Vert ^{2r}
\label{2.16c} \\
\leq 2\left\Vert Ax\right\Vert ^{r-1}\left\Vert Bx\right\Vert
^{r-1}\left\vert \left\langle Ax,Bx\right\rangle \right\vert
+r^{2}\left\Vert Ax\right\Vert ^{2r-2}\left\Vert Ax-Bx\right\Vert
^{2},
\end{multline}%
for any $x\in H.$

As in the proof of Theorem \ref{t2.3c}, we also have%
\begin{equation}
2^{1-r}\left[ \left\langle \left( A^{\ast }A+B^{\ast }B\right)
x,x\right\rangle \right] ^{r}\leq \left\Vert Ax\right\Vert ^{2r}+\left\Vert
Bx\right\Vert ^{2r},  \label{2.17c}
\end{equation}%
for any $x\in H.$

Therefore, by (\ref{2.16c}) and (\ref{2.17c}) we deduce%
\begin{multline}
\left[ \left\langle \left( \frac{A^{\ast }A+B^{\ast }B}{2}\right)
x,x\right\rangle \right] ^{r}  \label{2.18c} \\
\leq \left\Vert Ax\right\Vert ^{r-1}\left\Vert Bx\right\Vert
^{r-1}\left\vert \left\langle Ax,Bx\right\rangle \right\vert +\frac{1}{2}%
r^{2}\left\Vert A\right\Vert ^{2r-2}\left\Vert Ax-Bx\right\Vert ^{2}
\end{multline}%
for any $x\in H.$

Taking the supremum in (\ref{2.18c}) we obtain the desired result (\ref%
{2.14c}).
\end{proof}

\begin{remark}
Following \cite[p. 156]{GR}, we recall that the bounded linear operator $V$
is hyponormal, if
\begin{equation*}
\left\Vert V^{\ast }x\right\Vert \leq \left\Vert Vx\right\Vert \text{ for
all }x\in H.
\end{equation*}%
Now, if we choose in (\ref{2.14c}) $A=V$ and $B=V^{\ast },$ then, on taking
into account that for hyponormal operators $w\left( V^{2}\right) =\left\Vert
V\right\Vert ^{2},$ we get the inequality%
\begin{eqnarray*}
\left\Vert \frac{V^{\ast }V+VV^{\ast }}{2}\right\Vert ^{r}\leq
\left\Vert
V\right\Vert ^{2r-2}\left[ \left\Vert V\right\Vert ^{2}+\frac{1}{2}%
r^{2}\left\Vert V-V^{\ast }\right\Vert ^{2}\right] ,
\end{eqnarray*}%
holding for any hyponormal operator $V$ and any $r\geq 1.$
\end{remark}

\section{Further inequalities for an invertible operator}

In this section we assume that $B:H\rightarrow H$ is an invertible bounded
linear operator and let $B^{-1}:H\rightarrow H$ be its inverse. Then,
obviously,%
\begin{equation}
\left\Vert Bx\right\Vert \geq \frac{1}{\left\Vert B^{-1}\right\Vert }%
\left\Vert x\right\Vert \quad \text{for any \ }x\in H,  \label{3.1c}
\end{equation}%
where $\left\Vert B^{-1}\right\Vert $ denotes the norm of the inverse $%
B^{-1}.$

The following result holds true \cite{SSD6}:

\begin{theorem}
\label{t3.1c}Let $A,B:H\rightarrow H$ be two bounded linear operators on $H$
and $B$ is invertible such that, for a given $r>0,$%
\begin{equation}
\left\Vert A-B\right\Vert \leq r.  \label{3.2c}
\end{equation}%
Then:%
\begin{equation}
\left\Vert A\right\Vert \leq \left\Vert B^{-1}\right\Vert \left[ w\left(
B^{\ast }A\right) +\frac{1}{2}r^{2}\right] .  \label{3.3c}
\end{equation}
\end{theorem}

\begin{proof}
The condition (\ref{3.2c}) is obviously equivalent to:%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2}\leq 2\textrm{Re}%
\left\langle \left( B^{\ast }A\right) x,x\right\rangle +r^{2}  \label{3.4c}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Since, by (\ref{3.1c}),%
\begin{equation*}
\left\Vert Bx\right\Vert ^{2}\geq \frac{1}{\left\Vert B^{-1}\right\Vert ^{2}}%
\left\Vert x\right\Vert ^{2},\quad x\in H
\end{equation*}%
and $\textrm{Re}\left\langle \left( B^{\ast }A\right)
x,x\right\rangle \leq \left\vert \left\langle \left( B^{\ast
}A\right) x,x\right\rangle
\right\vert ,$ hence by (\ref{3.4c}) we get%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}+\frac{\left\Vert x\right\Vert ^{2}}{\left\Vert
B^{-1}\right\Vert ^{2}}\leq 2\left\vert \left\langle \left( B^{\ast
}A\right) x,x\right\rangle \right\vert +r^{2}  \label{3.5c}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (\ref%
{3.5c}), we have%
\begin{equation}
\left\Vert A\right\Vert ^{2}+\frac{1}{\left\Vert B^{-1}\right\Vert ^{2}}\leq
2w\left( B^{\ast }A\right) +r^{2}.  \label{3.6c}
\end{equation}%
By the elementary inequality%
\begin{equation}
\frac{2\left\Vert A\right\Vert }{\left\Vert B^{-1}\right\Vert }\leq
\left\Vert A\right\Vert ^{2}+\frac{1}{\left\Vert B^{-1}\right\Vert ^{2}}
\label{3.7c}
\end{equation}%
and by (\ref{3.6c}) we then deduce the desired result (\ref{3.3c}).
\end{proof}

\begin{remark}
If we choose above $B=\lambda I,$ $\lambda \neq 0,$ then we get the
inequality%
\begin{eqnarray*}
\left( 0\leq \right) \left\Vert A\right\Vert -w\left( A\right) \leq \frac{1}{%
2\left\vert \lambda \right\vert }r^{2},
\end{eqnarray*}%
provided $\left\Vert A-\lambda I\right\Vert \leq r.$ This result has
been obtained in \cite{SSD3}.

Also, if we assume that $B=\lambda A^{\ast },$ $A$ is invertible, then we
obtain%
\begin{eqnarray*}
\left\Vert A\right\Vert \leq \left\Vert A^{-1}\right\Vert \left[
w\left( A^{2}\right) +\frac{1}{2\left\vert \lambda \right\vert
}r^{2}\right] ,
\end{eqnarray*}%
provided $\left\Vert A-\lambda A^{\ast }\right\Vert \leq r,$
$\lambda \neq 0. $
\end{remark}

The following result may be stated as well \cite{SSD6}:

\begin{theorem}
\label{ta.2c}Let $A,B:H\rightarrow H$ be two bounded linear operators on $H.$
If $B$ is invertible and for $r>0,$%
\begin{equation}
\left\Vert A-B\right\Vert \leq r,  \label{a.1c}
\end{equation}%
then%
\begin{equation}
\left( 0\leq \right) \left\Vert A\right\Vert \left\Vert B\right\Vert
-w\left( B^{\ast }A\right) \leq \frac{1}{2}r^{2}+\frac{\left\Vert
B\right\Vert ^{2}\left\Vert B^{-1}\right\Vert ^{2}-1}{\left\Vert
B^{-1}\right\Vert ^{2}}.  \label{a.2c}
\end{equation}
\end{theorem}

\begin{proof}
The condition (\ref{a.1c}) is obviously equivalent to%
\begin{equation*}
\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2}\leq 2\textrm{Re}%
\left\langle Ax,Bx\right\rangle +r^{2}
\end{equation*}%
for any $x\in H,$ which is clearly equivalent to%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}\leq 2\textrm{Re}%
\left\langle B^{\ast }Ax,x\right\rangle +r^{2}+\left\Vert B\right\Vert
^{2}-\left\Vert Bx\right\Vert ^{2}.  \label{a.3c}
\end{equation}%
Since%
\begin{equation*}
\textrm{Re}\left\langle B^{\ast }Ax,x\right\rangle \leq \left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert ,\quad
\left\Vert Bx\right\Vert ^{2}\geq \frac{1}{\left\Vert
B^{-1}\right\Vert ^{2}}\left\Vert x\right\Vert ^{2}
\end{equation*}%
and%
\begin{equation*}
\left\Vert Ax\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}\geq 2\left\Vert
B\right\Vert \left\Vert Ax\right\Vert
\end{equation*}%
for any $x\in H,$ hence by (\ref{a.3c}) we get%
\begin{eqnarray*}
2\left\Vert B\right\Vert \left\Vert Ax\right\Vert \leq 2\left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert
+r^{2}+\frac{\left\Vert B\right\Vert ^{2}\left\Vert
B^{-1}\right\Vert ^{2}-1}{\left\Vert B^{-1}\right\Vert ^{2}}
\end{eqnarray*}%
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 deduce
the desired result (\ref{a.2c}).
\end{proof}

\begin{remark}
If we choose in Theorem \ref{ta.2c}, $B=\lambda A^{\ast },$ $\lambda \neq 0,$
$A$ is invertible, then we get the inequality:%
\begin{eqnarray*}
\left( 0\leq \right) \left\Vert A\right\Vert ^{2}-w\left(
A^{2}\right) \leq \frac{1}{2\left\vert \lambda \right\vert
}r^{2}+\left\vert \lambda \right\vert \cdot \frac{\left\Vert
A\right\Vert ^{2}\left\Vert A^{-1}\right\Vert ^{2}-1}{\left\Vert
A^{-1}\right\Vert ^{2}}
\end{eqnarray*}%
provided $\left\Vert A-\lambda A^{\ast }\right\Vert \leq r.$
\end{remark}

The following result may be stated as well \cite{SSD6}.

\begin{theorem}
\label{tb.1c}Let $A,B:H\rightarrow H$ be two bounded linear operators on $H.$
If $B$ is invertible and for $r>0$ we have%
\begin{equation}
\left\Vert A-B\right\Vert \leq r<\left\Vert B\right\Vert ,  \label{b.1c}
\end{equation}%
then%
\begin{equation}
\left\Vert A\right\Vert \leq \frac{1}{\sqrt{\left\Vert B\right\Vert
^{2}-r^{2}}}\left( w\left( B^{\ast }A\right) +\frac{\left\Vert B\right\Vert
^{2}\left\Vert B^{-1}\right\Vert ^{2}-1}{2\left\Vert B^{-1}\right\Vert ^{2}}%
\right) .  \label{b.2c}
\end{equation}
\end{theorem}

\begin{proof}
The first part of condition (\ref{b.1c}) is obviously equivalent to%
\begin{equation*}
\left\Vert Ax\right\Vert ^{2}+\left\Vert Bx\right\Vert ^{2}\leq 2\textrm{Re}%
\left\langle Ax,Bx\right\rangle +r^{2}
\end{equation*}%
for any $x\in H,$ which is clearly equivalent to%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}-r^{2}\leq 2\textrm{%
Re}\left\langle B^{\ast }Ax,x\right\rangle +\left\Vert B\right\Vert
^{2}-\left\Vert Bx\right\Vert ^{2}.  \label{b.3c}
\end{equation}%
Since%
\begin{gather*}
\textrm{Re}\left\langle B^{\ast }Ax,x\right\rangle \leq \left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert , \\
\left\Vert Bx\right\Vert ^{2}\geq \frac{1}{\left\Vert B^{-1}\right\Vert ^{2}}%
\left\Vert x\right\Vert ^{2}
\end{gather*}%
and, by the second part of (\ref{b.1c}),%
\begin{equation*}
\left\Vert Ax\right\Vert ^{2}+\left\Vert B\right\Vert ^{2}-r^{2}\geq 2\sqrt{%
\left\Vert B\right\Vert ^{2}-r^{2}}\left\Vert Ax\right\Vert ,
\end{equation*}%
for any $x\in H,$ hence by (\ref{b.3c}) we get%
\begin{equation}
2\left\Vert Ax\right\Vert \sqrt{\left\Vert B\right\Vert ^{2}-r^{2}}\leq
2\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert +\frac{%
\left\Vert B\right\Vert ^{2}\left\Vert B^{-1}\right\Vert ^{2}-1}{\left\Vert
B^{-1}\right\Vert ^{2}}  \label{b.4c}
\end{equation}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Taking the supremum over $x\in H,$ $\left\Vert x\right\Vert =1$ in (\ref%
{b.4c}), we deduce the desired inequality (\ref{b.2c}).
\end{proof}

\begin{remark}
The above Theorem \ref{tb.1c} has some particular cases of interest. For
instance, if we choose $B=\lambda I,$ with $\left\vert \lambda \right\vert
>r,$ then (\ref{b.1c}) is obviously fulfilled and by (\ref{b.2c}) we get%
\begin{eqnarray*}
\left\Vert A\right\Vert \leq \frac{w\left( A\right) }{\sqrt{1-\left( \frac{r%
}{\left\vert \lambda \right\vert }\right) ^{2}}},
\end{eqnarray*}%
provided $\left\Vert A-\lambda I\right\Vert \leq r.$ This result has
been obtained in \cite{SSD3}.

On the other hand, if in the above we choose $B=\lambda A^{\ast }$ with $%
\left\Vert A\right\Vert \geq \frac{r}{\left\vert \lambda \right\vert }$ \ $%
\left( \lambda \neq 0\right) ,$ then by (\ref{b.2c}) we get%
\begin{eqnarray*}
\left\Vert A\right\Vert \leq \frac{1}{\sqrt{\left\Vert A\right\Vert
^{2}-\left( \frac{r}{\left\vert \lambda \right\vert }\right)
^{2}}}\left[ w\left( A^{2}\right) +\left\vert \lambda \right\vert
\cdot \frac{\left\Vert A\right\Vert ^{2}\left\Vert A^{-1}\right\Vert
^{2}-1}{2\left\Vert A^{-1}\right\Vert ^{2}}\right] ,
\end{eqnarray*}%
provided $\left\Vert A-\lambda A^{\ast }\right\Vert \leq r.$
\end{remark}

The following result may be stated as well \cite{SSD6}.

\begin{theorem}
\label{t3.2c}Let $A,B$ and $r$ be as in Theorem {\rm\ref{t3.1c}}. Moreover, if%
\begin{equation}
\left\Vert B^{-1}\right\Vert <\frac{1}{r},  \label{3.7.1c}
\end{equation}%
then%
\begin{equation}
\left\Vert A\right\Vert \leq \frac{\left\Vert B^{-1}\right\Vert }{\sqrt{%
1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}}w\left( B^{\ast }A\right) .
\label{3.8c}
\end{equation}
\end{theorem}

\begin{proof}
Observe that, by (\ref{3.6c}) we have%
\begin{equation}
\left\Vert A\right\Vert ^{2}+\frac{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}{%
\left\Vert B^{-1}\right\Vert ^{2}}\leq 2w\left( B^{\ast }A\right) .
\label{3.9c}
\end{equation}%
Utilising the elementary inequality%
\begin{equation}
2\frac{\left\Vert A\right\Vert }{\left\Vert B^{-1}\right\Vert }\sqrt{%
1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}\leq \left\Vert A\right\Vert ^{2}+%
\frac{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}{\left\Vert B^{-1}\right\Vert
^{2}},  \label{3.10c}
\end{equation}%
which can be stated since (\ref{3.7.1c}) is assumed to be true, hence by (%
\ref{3.9c}) and (\ref{3.10c}) we deduce the desired result (\ref{3.8c}).
\end{proof}

\begin{remark}
If we assume that $B=\lambda A^{\ast }$ with $\lambda \neq 0$ and $A$ an
invertible operator, then, by applying Theorem \ref{t3.2c}, we get the
inequality:%
\begin{eqnarray*}
\left\Vert A\right\Vert \leq \frac{\left\Vert A^{-1}\right\Vert
w\left( A^{2}\right) }{\sqrt{\left\vert \lambda \right\vert
^{2}-r^{2}\left\Vert A^{-1}\right\Vert ^{2}}},
\end{eqnarray*}%
provided $\left\Vert A-\lambda A^{\ast }\right\Vert \leq r$ and
$\left\Vert A^{-1}\right\Vert \leq \frac{\left\vert \lambda
\right\vert }{r}.$
\end{remark}

The following result may be stated as well.

\begin{theorem}
\label{t3.3c}Let $A,B:H\rightarrow H$ be two bounded linear operators. If $%
r>0$ and $B$ is invertible with the property that $\left\Vert A-B\right\Vert
\leq r$ and%
\begin{equation}
\frac{1}{\sqrt{r^{2}+1}}\leq \left\Vert B^{-1}\right\Vert <\frac{1}{r},
\label{3.11c}
\end{equation}%
then%
\begin{equation}
\left\Vert A\right\Vert ^{2}\leq w^{2}\left( B^{\ast }A\right) +2w\left(
B^{\ast }A\right) \cdot \frac{\left\Vert B^{-1}\right\Vert -\sqrt{%
1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}}{\left\Vert B^{-1}\right\Vert }.
\label{3.12c}
\end{equation}
\end{theorem}

\begin{proof}
Let $x\in H,$ $\left\Vert x\right\Vert =1.$ Then by (\ref{3.5c}) we have%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}+\frac{1}{\left\Vert B^{-1}\right\Vert ^{2}}%
\leq 2\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert +r^{2},
\label{3.13c}
\end{equation}%
and since%
\begin{equation*}
\frac{1}{\left\Vert B^{-1}\right\Vert ^{2}}-r^{2}>0,
\end{equation*}%
we can conclude that $\left\vert \left\langle B^{\ast }Ax,x\right\rangle
\right\vert >0$ for any $x\in H,$ $\left\Vert x\right\Vert =1.$

Dividing in (\ref{3.13c}) with $\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert >0,$ we obtain%
\begin{equation}
\frac{\left\Vert Ax\right\Vert ^{2}}{\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert }\leq 2+\frac{r^{2}}{\left\vert \left\langle
B^{\ast }Ax,x\right\rangle \right\vert }-\frac{1}{\left\Vert
B^{-1}\right\Vert ^{2}\left\vert \left\langle B^{\ast }Ax,x\right\rangle
\right\vert }.  \label{3.14c}
\end{equation}%
Subtracting $\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert $
from both sides of (\ref{3.14c}), we get%
\begin{align}
& \frac{\left\Vert Ax\right\Vert ^{2}}{\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert }-\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert \notag \\
& \leq 2-\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert -%
\frac{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}{\left\vert \left\langle
B^{\ast }Ax,x\right\rangle \right\vert \left\Vert B^{-1}\right\Vert ^{2}}
\notag \\
& =2-\frac{2\sqrt{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}}{\left\Vert
B^{-1}\right\Vert }\notag \\
& \qquad -\left( \sqrt{\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert }-\frac{\sqrt{1-r^{2}\left\Vert
B^{-1}\right\Vert ^{2}}}{\left\Vert B^{-1}\right\Vert
\sqrt{\left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert }}\right) ^{2}  \notag \\
& \leq 2\left( \frac{\left\Vert B^{-1}\right\Vert -\sqrt{1-r^{2}\left\Vert
B^{-1}\right\Vert ^{2}}}{\left\Vert B^{-1}\right\Vert }\right) ,  \notag
\end{align}%
which gives:%
\begin{equation}
\left\Vert Ax\right\Vert ^{2}\leq \left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert ^{2}+2\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert \frac{\left\Vert B^{-1}\right\Vert -\sqrt{%
1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}}{\left\Vert B^{-1}\right\Vert }.
\label{3.16c}
\end{equation}%
We also remark that, by (\ref{3.11c}) the quantity%
\begin{equation*}
\left\Vert B^{-1}\right\Vert -\sqrt{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}%
\geq 0,
\end{equation*}%
hence, on taking the supremum in (\ref{3.16c}) over $x\in H,$ $\left\Vert
x\right\Vert =1,$ we deduce the desired inequality.
\end{proof}

\begin{remark}
It is interesting to remark that if we assume $\lambda \in \mathbb{C}$ with $%
0<r\leq \left\vert \lambda \right\vert \leq \sqrt{r^{2}+1}$ and $\left\Vert
A-\lambda I\right\Vert \leq r,$ then by (\ref{3.2c}) we can state the
following inequality:%
\begin{eqnarray*}
\left\Vert A\right\Vert ^{2}\leq \left\vert \lambda \right\vert
^{2}w\left( A^{2}\right) +2\left\vert \lambda \right\vert \left(
1-\sqrt{\left\vert \lambda \right\vert ^{2}-r^{2}}\right) w\left(
A\right) .
\end{eqnarray*}%
Also, if $\left\Vert A-A^{\ast }\right\Vert \leq r,$ $A$ is invertible and $%
\frac{1}{\sqrt{r^{2}+1}}\leq \left\Vert A^{-1}\right\Vert \leq \frac{1}{r},$
then, by (\ref{3.12c}) we also have%
\begin{eqnarray*}
\left\Vert A\right\Vert ^{2}\leq w^{2}\left( A^{2}\right) +2w\left(
A^{2}\right) \cdot \frac{\left\Vert A^{-1}\right\Vert -\sqrt{%
1-r^{2}\left\Vert A^{-1}\right\Vert ^{2}}}{\left\Vert A^{-1}\right\Vert }.
\end{eqnarray*}
\end{remark}

One can also prove the following result \cite{SSD6}.

\begin{theorem}\label{t3.4c}
Let $A,B :H\rightarrow H$ be two bounded linear operators. If $ r>0$
and $B$ is invertible with the property that $\left\Vert
A-B\right\Vert \leq r$ and $\left\Vert B^{-1}\right\Vert \leq
\frac{1}{r},$ then
\begin{align}
(0& \leq )\left\Vert A\right\Vert ^{2}\left\Vert B\right\Vert
^{2}-w^{2}\left( B^{\ast }A\right)  \label{3.17c} \\
& \leq 2w\left( B^{\ast }A\right) \cdot \frac{\left\Vert
B\right\Vert }{\left\Vert B^{-1}\right\Vert }\left( \left\Vert
B\right\Vert \left\Vert B^{-1}\right\Vert -\sqrt{1-r^{2}\left\Vert
B^{-1}\right\Vert ^{2}}\right) . \notag
\end{align}
\end{theorem}
\begin{proof}
We subtract the quantity $\frac{\left\vert \left\langle B^{\ast
}Ax,x\right\rangle \right\vert }{\left\Vert B\right\Vert ^{2}}$ from
both sides of (\ref{3.14c}) to obtain
\begin{align}
0& \leq \frac{\left\Vert Ax\right\Vert ^{2}}{\left\vert \left\langle
B^{\ast }Ax,x\right\rangle \right\vert }-\frac{\left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert }{\left\Vert
B\right\Vert ^{2}}  \notag
\\
& \leq 2-\frac{\left\vert \left\langle B^{\ast }Ax,x\right\rangle
\right\vert }{\left\Vert B\right\Vert ^{2}}-\frac{1-r^{2}\left\Vert
B^{-1}\right\Vert ^{2}}{\left\vert \left\langle B^{\ast }Ax,x\right\rangle
\right\vert \left\Vert B^{-1}\right\Vert ^{2}}  \notag  \\
& =2-2\cdot \frac{\sqrt{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}}{
\left\Vert B\right\Vert \left\Vert B^{-1}\right\Vert }\notag \\
&\qquad -\left( \frac{\sqrt{%
\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert
}}{\left\Vert
B\right\Vert }-\frac{\sqrt{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}}{\sqrt{%
\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert }\left\Vert
B^{-1}\right\Vert }\right) ^{2}  \notag\\
& \leq 2\cdot \frac{\left( \left\Vert B\right\Vert \left\Vert
B^{-1}\right\Vert -\sqrt{1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}\right) }{%
\left\Vert B\right\Vert \left\Vert B^{-1}\right\Vert },  \notag
\end{align}%
which is equivalent with%
\begin{align}
(0& \leq )\left\Vert Ax\right\Vert ^{2}\left\Vert B\right\Vert
^{2}-\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert ^{2}
\label{3.19c} \\
& \leq 2\frac{\left\Vert B\right\Vert }{\left\Vert B^{-1}\right\Vert }%
\left\vert \left\langle B^{\ast }Ax,x\right\rangle \right\vert \left(
\left\Vert B\right\Vert \left\Vert B^{-1}\right\Vert -\sqrt{%
1-r^{2}\left\Vert B^{-1}\right\Vert ^{2}}\right)  \notag
\end{align}%
for any $x\in H,$ $\left\Vert x\right\Vert =1.$

The inequality (\ref{3.19c}) also shows that $\left\Vert B\right\Vert
\left\Vert B^{-1}\right\Vert \geq \sqrt{1-r^{2}\left\Vert B^{-1}\right\Vert
^{2}}$ and then, by (\ref{3.19c}), we get%
\begin{multline}
\left\Vert Ax\right\Vert ^{2}\left\Vert B\right\Vert ^{2}\leq \left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert ^{2}  \label{3.20c} \\
+2\frac{\left\Vert B\right\Vert }{\left\Vert B^{-1}\right\Vert }\left\vert
\left\langle B^{\ast }Ax,x\right\rangle \right\vert \left( \left\Vert
B\right\Vert \left\Vert B^{-1}\right\Vert -\sqrt{1-r^{2}\left\Vert
B^{-1}\right\Vert ^{2}}\right)
\end{multline}%
for any $x\in X,$ $\left\Vert x\right\Vert =1.$

Taking the supremum in (\ref{3.20c}) we deduce the desired inequality (\ref%
{3.17c}).
\end{proof}

\begin{remark}
The above Theorem \ref{t3.4c} has some particular instances of interest as
follows. If, for instance, we choose $B=\lambda I$ with $\left\vert \lambda
\right\vert \geq r>0$ and $\left\Vert A-\lambda I\right\Vert \leq r,$ then
by (\ref{3.17c}) we obtain the inequality%
\begin{align}
(0& \leq )\left\Vert A\right\Vert ^{2}-w^{2}\left( A\right)\leq
2\left\vert \lambda \right\vert w\left( A\right) \left( 1-\sqrt{1-
\frac{r^{2}}{\left\vert \lambda \right\vert ^{2}}}\right) .  \notag
\end{align}%
Also, if $A$ is invertible, $\left\Vert A-\lambda A^{\ast }\right\Vert \leq
r $ and $\left\Vert A^{-1}\right\Vert \leq \frac{\left\vert \lambda
\right\vert }{r},$ then by (\ref{3.17c}) we can state:%
\begin{align}
(0& \leq )\left\Vert A\right\Vert ^{4}-w^{2}\left( A^{2}\right)\notag\\
& \leq 2\left\vert \lambda \right\vert w\left( A^{2}\right) \cdot \frac{%
\left\Vert A\right\Vert }{\left\Vert A^{-1}\right\Vert }\left( \left\Vert
A\right\Vert \left\Vert A^{-1}\right\Vert -\sqrt{1-\frac{r^{2}}{\left\vert
\lambda \right\vert ^{2}}\left\Vert A^{-1}\right\Vert ^{2}}\right) .  \notag
\end{align}
\end{remark}

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\end{document}

.
