# Numerical radius and Berezin number inequality

Satyabrata Majee  Indian Institute of Technology Roorkee, Department of
Mathematics, Roorkee-247 667, Uttarakhand, India<EMAIL_ADDRESS>,  Amit
Maji  Indian Institute of Technology Roorkee, Department of Mathematics,
Roorkee-247 667, Uttarakhand, India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Atanu Manna  Indian Institute of Carpet Technology,
Bhadohi-221401, Uttar Pradesh, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We study various inequalities for numerical radius and Berezin number of a
bounded linear operator on a Hilbert space. It is proved that the numerical
radius of a pure two-isometry is $1$ and the Crawford number of a pure two-
isometry is $0$. In particular, we show that for any scalar-valued non-
constant inner function $\theta$, the numerical radius and the Crawford number
of a Toeplitz operator $T_{\theta}$ on a Hardy space is $1$ and $0$,
respectively. It is also shown that numerical radius is multiplicative for a
class of isometries and sub-multiplicative for a class of commutants of a
shift. We have illustrated these results with some concrete examples. Finally,
some Hardy-type inequalities for Berezin number of certain class of operators
are established with the help of the classical Hardy’s inequality.

###### Key words and phrases:

Numerical Radius, Berezin number, Isometry, Shift, Reproducing kernel Hilbert
spaces, Hardy’s inequality

###### 2010 Mathematics Subject Classification:

47A12, 47A63

##  1\. Introduction

The concepts of numerical radius and Berezin number of an operator have been
studied extensively due to their enormous applications in engineering, quantum
computing, quantum mechanics, numerical analysis, differential equations, etc.

In the early days, one of the main goals for studying Hilbert spaces was
quadratic forms. The fundamental question about quadratic forms associated
with an operator is its numerical range. Firstly, Toeplitz [24] defined the
numerical range for matrices in 1918. Later, Wintner [25] studied the
relationship between the numerical range and the convex hull of the spectrum
of a bounded linear operator on a Hilbert space.

We denote $\mathcal{B}(\mathcal{H})$ as the $C^{*}$-algebra of all bounded
linear operators on a Hilbert space $\mathcal{H}$. For any bounded linear
operator $A$ on $\mathcal{H}$ the numerical radius, denoted by $\omega(A)$,
yields a norm on $\mathcal{B}(\mathcal{H})$. Indeed, for
$A\in\mathcal{B}(\mathcal{H})$

${\|A\|\over 2}\leq\omega(A)\leq\|A\|.$

Therefore, the operator norm and the numerical radius norm are equivalent. In
particular, if $A$ is normal (that is, $A^{*}A=AA^{*}$), then
$\omega(A)=\|A\|$. One can show that the numerical radius $\omega(\cdot)$ is
neither multiplicative nor sub-multiplicative on
$\mathcal{B}(\mathcal{H})$(see, [11]). Thus a natural question arises:

$(Q)$ Which class of operators on a Hilbert space $\omega(\cdot)$ is
multiplicative or sub-multiplicative?

Though numerical radius is not multiplicative or sub-multiplicative in
general, but it satisfies the power inequality. More precisely, for any
positive integer $n$ and $A\in\mathcal{B}(\mathcal{H})$ the following
inequality holds:

$\omega(A^{n})\leq\omega(A)^{n}$.

The above power inequality was first conjectured by Halmos and a delicate
proof was given by Berger (see [4], [11]). After that, generalizations for
polynomial or analytic functions on the unit disc have been done. In 1966,
Pearcy [22] gave elementary proof of this. However, the reverse power
inequality does not hold in general. Indeed, for a nilpotent matrix
$A=\begin{pmatrix}0&0\\\ 1&0\end{pmatrix},$ we have $\omega(A)=\frac{1}{2}$
but $\omega(A^{2})=0$. The reader is referred to [5], [10] and references
therein for various applications of numerical radius inequalities.

On the other hand, Berezin transform which connects operators with functions
plays an important role in operator theory, more specifically to study
Toeplitz operators, Hankel operators and composition operators. Another
important notion in operator theory is Berezin number. The Berezin number of
an operator $A$ is denoted by $ber(A)$ and defined on a reproducing kernel
Hilbert space $H(\Omega)$. Indeed, for any $A\in\mathcal{B}(H(\Omega))$,
$0\leq ber(A)\leq\omega(A)\leq\|A\|$ and if $ber(A)\neq 0$, then the following
inequality for $n\in\mathbb{N}$ (see Garayev et al. [8])

$ber(A^{n})\leq(\frac{w(A)}{ber(A)})^{n}ber(A)^{n}\quad\mbox{holds}.$

A lot of research work has been carried out to find the reverse power
inequality of the Berezin number of operators over the last few years and many
researchers have studied the reverse inequality by using Hardy-Hilbert
inequality. Apart from these studies, Yamancı et al.[27] considered Berezin
number operator inequalities for continuous convex functions and some certain
class of operators. For more details one can see Karaev([16], [17]), Garayev
et al. ([8], [9]), Yamancı et al.[26] and references therein. However, Coburn
[6] gave an example of concrete operator $A$ on the Bergman space such that
$ber(A^{2})>ber(A)^{2}$ holds. Motivated by the earlier studies, we here
attempt to explore the following inequality:

$ber(A)^{n}\leq\alpha ber(A^{n})$ for $A\in\mathcal{B}(H(\Omega)$ all integer
$n>1$ and some constant $\alpha>0$.

The main guiding tools used in this article are the geometry of Hilbert
spaces, Wold-von Neumann decomposition for pure two-isometry and isometry.
Using the notion of functional calculus, some reverse power inequalities for
Berezin number are also established by using the classical _discrete Hardy’s
inequality_ (See Section 2, (2.1)).

The paper is organized as follows: in Section 2 we discuss preliminaries and
some basic results. Section 3 deals with the numerical radius and Crawford
number of certain class of operators and various results on the Berezin
number. In section 4 with the help of _discrete Hardy’s inequality_ we develop
the Berezin number inequality which improves many earlier results.

##  2\. Preliminaries and Basic Results

Let $p>1$ be any real number and $\\{a_{n}\\}$ be a sequence of non-negative
real numbers. Then the discrete version of _Hardy’s inequality_ (named after
G. H. Hardy) is

(2.1)
$\displaystyle{}\displaystyle\sum_{n=1}^{\infty}\Big{(}\frac{1}{n}\sum_{k=1}^{n}a_{k}\Big{)}^{p}\leq\Big{(}\frac{p}{p-1}\Big{)}^{p}\sum_{n=1}^{\infty}{a_{n}}^{p},$

and equality holds if $a_{n}=0$ for all $n$. The constant term
$\Big{(}\frac{p}{p-1}\Big{)}^{p}$ in (2.1) is the best possible constant.
There are several generalizations and extensions which have been studied by
many researchers in different scientific works (for more details, see [12]).

For $p=2$ the above inequality (2.1) becomes

(2.2)
$\displaystyle{}\displaystyle\sum_{n=1}^{\infty}w_{n}^{H}\left(\sum_{k=1}^{n}a_{k}\right)^{2}\leq\sum_{n=1}^{\infty}{a_{n}}^{2},$

where $w_{n}^{H}=\frac{1}{4n^{2}}$. We say $w_{n}^{H}$ is the classical Hardy
weight. Recently, Keller et al. [15] obtained an improved version of discrete
Hardy inequality for $p=2$ as follows

(2.3)
$\displaystyle{}\displaystyle\sum_{n=1}^{\infty}w_{n}\Big{(}\sum_{k=1}^{n}a_{k}\Big{)}^{2}\leq\sum_{n=1}^{\infty}{a_{n}}^{2},$

where $w_{n}=2-\sqrt{1+\frac{1}{n}}-\sqrt{1-\frac{1}{n}}$ for
$n\in\mathbb{N}$. Clearly, $w_{n}>\frac{1}{4n^{2}}=w_{n}^{H}$ for each
$n\in\mathbb{N}$. Also note that
$\displaystyle\sum_{n=1}^{\infty}w_{n}^{H}=\frac{\pi^{2}}{24}$ and the series
$\displaystyle\sum_{n=1}^{\infty}w_{n}$ is convergent and converges to
$0.753045$ (approximately).

In what follows, $\mathcal{H}$ stands for separable complex Hilbert space. The
set of all bounded linear operators from $\mathcal{H}$ to itself is denoted by
$\mathcal{B}(\mathcal{H})$. An operator $A\in\mathcal{B}(\mathcal{H})$ is said
to be an isometry if $A^{*}A=I_{\mathcal{H}}$ and $A$ is said to be a pure
isometry or a shift if $A^{*n}\rightarrow 0$ as $n\rightarrow\infty$ in strong
operator topology (cf. [11]). We say an operator $A$ is a co-isometry if
$A^{*}$, adjoint of $A$ is an isometry and $A$ is a co-shift if $A^{*}$ is a
shift. An operator $A\in\mathcal{B}(\mathcal{H})$ is said to be positive if
$A$ is self-adjoint, that is, $A^{*}=A$ and

$\langle Ax,x\rangle\geq 0\quad(x\in\mathcal{H}).$

The numerical range of any $A\in\mathcal{B}(\mathcal{H})$, denoted by $W(A)$,
is defined as

$W(A):=\\{\langle Ax,x\rangle:x\in\mathcal{H},\|x\|=1\\},$

the numerical radius of $A$ and the Crawford number of $A$, denoted by
$\omega(A)$ and $c(A)$, respectively as

$\omega(A)=\sup\\{|\langle Ax,x\rangle|:x\in\mathcal{H},\|x\|=1\\},$

and

$c(A)=\inf\\{|\langle Ax,x\rangle|:x\in\mathcal{H},\|x\|=1\\}.$

We say two operators $A_{1}$ on $\mathcal{H}_{1}$ and $A_{2}$ on
$\mathcal{H}_{2}$ are said to be unitarily equivalent if there exists a
unitary $U:\mathcal{H}_{1}\rightarrow\mathcal{H}_{2}$ such that
$UA_{1}=A_{2}U$. From the definition it readily follows that the numerical
radius and the Crawford number are unitarily invariant. Indeed, if $A_{1}$ on
$\mathcal{H}_{1}$ and $A_{2}$ on $\mathcal{H}_{2}$ are unitarily equivalent,
then $\omega(A_{2})=\omega(UA_{1}U^{*})=\omega(A_{1})$ and
$c(A_{2})=c(UA_{1}U^{*})=c(A_{1})$. Also $\omega(A)=\omega(A^{*})$ and
$c(A)=c(A^{*})$ for any $A\in\mathcal{B}(\mathcal{H})$.

Let $\Omega$ be a non-empty set. A reproducing kernel Hilbert space (in short
rkHs) is a Hilbert space $H(\Omega)$ of complex-valued functions on $\Omega$
such that for every point $w\in\Omega$ the point evaluation $f\mapsto f(w)$ is
a bounded linear functional on $H(\Omega)$. Since for each $w\in\Omega$ the
map $f\mapsto f(w)$ is a continuous linear functional on $H(\Omega)$, there is
a unique element $k_{w}$ of $H(\Omega)$ because of the classical Riesz’s
representation theorem such that $f(w)=\langle f,k_{w}\rangle$ for all $f\in
H(\Omega)$. The map $k:\Omega\times\Omega\rightarrow\mathbb{C}$ defined by

$k(z,w)=k_{w}(z)=\langle k_{w},k_{z}\rangle$

is called the reproducing kernel function of $H(\Omega)$. We denote
$\hat{k}_{w}=\frac{{k}_{w}}{\|{k}_{w}\|}$ for $w\in\Omega$ as the normalized
reproducing kernel of $H(\Omega)$ and the set $\\{\hat{k}_{w}:w\in\Omega\\}$
is a total set in $H(\Omega)$. If a sequence $\\{e_{n}\\}$ is an orthonormal
basis for $H(\Omega)$, then the kernel function can be written as

$k(z,w)=\displaystyle\sum_{n=0}^{\infty}e_{n}(z)\overline{e_{n}(w)}$.

For more details and references on rkHs, see [1], [21]. Let $H(\Omega)$ be a
reproducing kernel Hilbert space on $\Omega$ and let
$A\in\mathcal{B}(H(\Omega))$. Define a function $\widetilde{A}$ on $\Omega$ as

$\widetilde{A}(z)=\langle A\hat{k}_{z},\hat{k}_{z}\rangle,\quad(z\in\Omega).$

The function $\widetilde{A}(z)$ is called Berezin transform or Berezin symbol
of $A$ which was firstly introduced and studied by Berezin ([2], [3]). Using
Cauchy-Schwarz inequality, one can easily say that the Berezin transform
$\widetilde{A}$ is a bounded function on $\Omega$. Indeed,

$|\widetilde{A}(z)|\leq\|A\hat{k}_{z}\|\|\hat{k}_{z}\|\leq\|A\|\quad(z\in\Omega).$

The Berezin set and Berezin number of an operator
$A\in\mathcal{B}(H(\Omega))$, denoted by $Ber(A)$ and $ber(A)$, respectively
is defined as

$\displaystyle Ber(A)$ $\displaystyle=\\{\widetilde{A}(z):z\in\Omega\\},$
$\displaystyle ber(A)$
$\displaystyle=\displaystyle\sup_{z\in\Omega}|\widetilde{A}(z)|.$

Therefore, one can easily observe that for $A\in\mathcal{B}(H(\Omega))$

$0\leq ber(A)\leq w(A)\leq\|A\|$.

From the definition of Berezin number, it readily follows that $ber(\cdot)$ is
a semi-norm on $\mathcal{B}(H(\Omega))$.

Before proceeding further, let us recall the notion of functional calculus
[7]. Let $A$ be a normal operator on a Hilbert space $\mathcal{H}$ and
$\sigma(A)$ denote the spectrum of $A$. Let $\mathcal{C}(\sigma(A))$ and
$B_{\infty}(\sigma(A))$ be the continuous complex-valued functions and the
bounded measurable complex-valued functions on $\sigma(A)\subseteq\mathbb{C}$,
respectively. Then $B_{\infty}(\sigma(A))$ is a $C^{*}$-algebra with the
involution map defined by $f\mapsto\bar{f}$. Let
$\pi:\mathcal{C}(\sigma(A))\rightarrow\mathcal{B}(\mathcal{H})$ be a
$*$-homomorphism such that $\pi(1)=I_{\mathcal{H}}$, where $1$ is a constant
function with value one and $I_{\mathcal{H}}$ is an identity operator. Then
there exists a unique spectral measure $\mathcal{P}$ in
$(\sigma(A),\mathcal{H})$ such that

$\pi(f)=\int_{\sigma(A)}fd\mathcal{P}.$

Now the continuous functional can be extended by Borel functional calculus for
any $f\in B_{\infty}(\sigma(A))$, where

$f(A)=\displaystyle\int_{\sigma(A)}fd\mathcal{P}.$

In particular, if $A\in\mathcal{B}(\mathcal{H})$ is a positive operator, then
$\sigma(A)$ is a subset of $[0,\infty)$ and $f(A)$ is a positive operator for
any positive Borel function whose domain contains the spectrum of $A$.

The following inequalities are useful to prove our results.

######  Lemma 2.1.

([23]) Let $A$ be any positive operator on a Hilbert space $\mathcal{H}$ and
$p>0$ be any real number. Then for $x\in\mathcal{H}$,

$\langle Ax,x\rangle^{p}\leq\langle A^{p}x,x\rangle$ whenever $1\leq p<\infty$

and

$\langle Ax,x\rangle^{p}>\langle A^{p}x,x\rangle$ whenever $0<p<1$.

######  Lemma 2.2.

([13]) Let $A$ be any bounded linear operator on a Hilbert space
$\mathcal{H}$. Then for any $x,y\in\mathcal{H}$

$|\langle
Ax,y\rangle|^{2}\leq\langle|A|^{2\alpha}x,x\rangle\langle|{A^{*}}|^{2(1-\alpha)}y,y\rangle$,

where $0\leq\alpha\leq 1$.

##  3\. Numerical Radius and Berezin number

In general, it is very difficult to compute Berezin number and numerical
radius of a bounded operator (even for particular classes of operators). But
if we concentrate on the model space, then these values can be calculated for
certain classes of operators. In this section, our treatment is analytic in
nature, and we also discuss some inequalities and concrete examples to find
Berezin number, Crawford number, and numerical radius.

Let $A\in\mathcal{B}(\mathcal{H})$. The positive square root of $A$, denoted
by $|A|$, is defined as $|A|=\sqrt{A^{*}A}$ is a self-adjoint operator on
$\mathcal{H}$. Now for any $x\in\mathcal{H}$,

$\||A|\|^{2}=\sup_{\|x\|=1}\||A|x\|^{2}=\sup_{\|x\|=1}|\langle
A^{*}Ax,x\rangle|=\|A^{*}A\|=\|A\|^{2}.$

Therefore

$\||A|\|=\|A\|=\|A^{*}\|=\||A^{*}|\|.$

Since $|A|$ and $|A^{*}|$ are self-adjoint operators,

$\omega(|A|)=\||A|\|=\|A\|=\||A^{*}|\|=\omega(|A^{*}|).$

Hence for any operator $A\in\mathcal{B}(\mathcal{H})$,

$\omega(A)\leq\omega(|A|).$

Now for any real number $p>0$ and $x\in\mathcal{H}$ with $\|x\|=1$,

$\displaystyle|\langle Ax,x\rangle|^{p}\leq\|Ax\|^{p}=\langle
Ax,Ax\rangle^{p\over 2}=\langle A^{*}Ax,x\rangle^{p\over 2}.$

Therefore

$\sup_{\|x\|=1}|\langle Ax,x\rangle|^{p}\leq\sup_{\|x\|=1}\langle
A^{*}Ax,x\rangle^{p\over 2}=\|A^{*}A\|^{p\over
2}=\|A\|^{p}=\||A|\|^{p}=\||A|^{p}\|,$

where the last equality follows from functional calculus for the positive
operator $|A|$. Hence

$\omega(A)^{p}\leq\omega(|A|^{p}).$

From the above deliberation we have the following:

######  Lemma 3.1.

Let $A\in\mathcal{B}(\mathcal{H})$ and $p>0$ be any real number. Then

$\omega(A)^{p}\leq\omega(|A|^{p}).$

######  Theorem 3.2.

Let $A_{i},X\in\mathcal{B}(\mathcal{H})$ for $i=1,\ldots,n$ and $X$ be any
positive operator. Then

$\omega({\sqrt{\sum_{i=1}^{n}A_{i}^{*}XA_{i}}})={\sqrt{\omega(\sum_{i=1}^{n}A_{i}^{*}XA_{i})}}.$

###### Proof.

Suppose that $A_{i},X\in\mathcal{B}(\mathcal{H})$ for $i=1,\ldots,n$ and $X$
is any positive operator. Then $\sum_{i=1}^{n}A_{i}^{*}XA_{i}$ is a positive
operator as for any $h\in\mathcal{H}$

$\displaystyle\langle(\sum_{i=1}^{n}A_{i}^{*}XA_{i})h,h\rangle=\sum_{i=1}^{n}\langle
A_{i}^{*}XA_{i}h,h\rangle=\sum_{i=1}^{n}\langle XA_{i}h,A_{i}h\rangle\geq 0.$

Therefore

$\displaystyle\omega({\sqrt{\sum_{i=1}^{n}A_{i}^{*}XA_{i}}})=\sup_{\|h\|=1}|\langle{\sqrt{\sum_{i=1}^{n}A_{i}^{*}XA_{i}}}h,h\rangle|$
$\displaystyle=\|{\sqrt{\sum_{i=1}^{n}A_{i}^{*}XA_{i}}}\|$
$\displaystyle=\|{\sum_{i=1}^{n}A_{i}^{*}XA_{i}}\|^{1\over 2}$
$\displaystyle=(\sup_{\|h\|=1}|\langle{\sum_{i=1}^{n}A_{i}^{*}XA_{i}}h,h\rangle|)^{1\over
2}$ $\displaystyle=\sqrt{\omega(\sum_{i=1}^{n}A_{i}^{*}XA_{i})}.$

Here the above third equality follows from functional calculus for a positive
operator. This proves the desire inequality.

The numerical radius $\omega(\cdot)$ is a norm on $\mathcal{B}(\mathcal{H})$
but $\omega(\cdot)$ is not multiplicative, that is,
$\omega(RT)\neq\omega(R)\omega(T)$ (even for normal operators)
$R,T\in\mathcal{B}(\mathcal{H})$. For example we consider
$T,R\in\mathcal{B}(\mathbb{C}^{2})$, defined as $T(z,w)=(z,0)$ and
$R(z,w)=(0,w)$. Then $\omega(T)=\|T\|=1=\|R\|=\omega(R)$ but $\omega(RT)=0$.
Since ${\|T\|\over 2}\leq\omega(T)\leq\|T\|$ for
$T\in\mathcal{B}(\mathcal{H})$ (see [11]),

$\omega(RT)\leq 4\omega(R)\omega(T)$

for any $R,T\in\mathcal{B}(\mathcal{H})$. For a normal operator $T$,
$\omega(T)=\|T\|$ and hence if $R,T$ are normal, then

$\omega(RT)\leq\omega(R)\omega(T).$

Therefore $\omega(\cdot)$ is sub-multiplicative in case of normal operators,
but not in general. Here we attempt to give an answer (partially) to the
natural question raised in Section 1.

We will firstly study model spaces and discuss the classical Wold-von Neumann
decomposition for an isometry (see [18], [19]).

######  Theorem 3.3.

Let $V$ be an isometry on a Hilbert space $\mathcal{H}$. Then $\mathcal{H}$
decomposes as a direct sum of $V$-reducing subspaces
$\mathcal{H}_{s}=\displaystyle{\mathop{\oplus}_{m=0}^{\infty}}V^{m}\mathcal{W}$
and $\mathcal{H}_{u}=\mathcal{H}\ominus\mathcal{H}_{s}$ and

$V=\begin{bmatrix}V_{s}&0\\\
0&V_{u}\end{bmatrix}\in\mathcal{B}(\mathcal{H}_{s}\oplus\mathcal{H}_{u}),$

where $\mathcal{W}=\mathcal{H}\ominus V\mathcal{H}$,
$V_{s}=V|_{\mathcal{H}_{s}}$ is a shift operator and
$V_{u}=V|_{\mathcal{H}_{u}}$ is a unitary operator.

The Hardy space, denoted by $H^{2}(\mathbb{D})$, is the Hilbert space of all
square summable holomorphic functions on the unit disc $\mathbb{D}$ (cf.
[18]). The Hardy space is also a reproducing kernel Hilbert space
corresponding to the Szegö kernel

$k(z,w)=\frac{1}{1-z\bar{w}}\quad\quad(z,w\in\mathbb{D}).$

For any Hilbert space $\mathcal{E}$, $H^{2}_{\mathcal{E}}(\mathbb{D})$ denotes
the $\mathcal{E}$-valued Hardy space with reproducing kernel
$k(z,w)I_{\mathcal{E}}$, and
$H^{\infty}_{\mathcal{B}(\mathcal{E})}(\mathbb{D})$ denotes as the space of
$\mathcal{B}(\mathcal{E})$-valued bounded holomorphic functions on
$\mathbb{D}$. The multiplication operator $M_{z}$ by the coordinate function
$z$ on $H^{2}_{\mathcal{E}}(\mathbb{D})$ is defined by

$(M_{z}f)(w)=wf(w)\quad\quad\quad(f\in
H^{2}_{\mathcal{E}}(\mathbb{D}),w\in\mathbb{D}).$

The Wold-von Neumann construction yields an analytic description of an
isometry as follows: Let $V$ be an isometry on $\mathcal{H}$, and let
$\mathcal{H}=\mathcal{H}_{s}\oplus\mathcal{H}_{u}$ be the Wold-von Neumann
orthogonal decomposition of $V$. Define

$\Pi:\mathcal{H}_{s}\oplus\mathcal{H}_{u}\rightarrow
H^{2}_{\mathcal{W}}(\mathbb{D})\oplus\mathcal{H}_{u}$

by

$\Pi(V^{m}\eta\oplus f)=z^{m}\eta\oplus f\quad\quad(m\geq
0,f\in\mathcal{H}_{u},\eta\in\mathcal{W}).$

Then $\Pi$ is a unitary and

$\Pi(V_{s}\oplus V_{u})=(M_{z}\oplus V_{u})\Pi,$

that is, $V$ on $\mathcal{H}$ and $M_{z}\oplus V_{u}$ on
$H^{2}_{\mathcal{W}}(\mathbb{D})\oplus\mathcal{H}_{u}$ are unitarily
equivalent. In particular, if $V$ is a shift, then $\mathcal{H}_{u}=\\{0\\}$
and hence

$\Pi V=M_{z}\Pi.$

Therefore, an isometry $V$ on $\mathcal{H}$ is a shift operator if and only if
$V$ is unitarily equivalent to $M_{z}$ on $H^{2}_{\mathcal{E}}(\mathbb{D})$,
where $\dim\mathcal{E}=\dim(\mathcal{H}\ominus V\mathcal{H})$. We call
$(\Pi,M_{z})$ as the Wold-von Neumann decomposition of the isometry $V$. Let
$C$ be a bounded linear operator on $H^{2}_{\mathcal{E}}(\mathbb{D})$. Then
$C\in\\{M_{z}\\}^{{}^{\prime}}$, commutant of $M_{z}$ (that is,
$CM_{z}=M_{z}C$), if and only if (cf. [18])

$C=M_{\Theta},$

for some $\Theta\in H^{\infty}_{\mathcal{B}(\mathcal{E})}(\mathbb{D})$, where
$(M_{\Theta}f)(w)=\Theta(w)f(w)$ for all $f\in
H^{2}_{\mathcal{E}}(\mathbb{D})$ and $w\in\mathbb{D}$. Let $V$ be a pure
isometry on a Hilbert space $\mathcal{H}$, and let
$C\in\\{V\\}^{{}^{\prime}}$. Suppose that $(\Pi,M_{z})$ is the Wold-von
Neumann decomposition of the pure isometry $V$ on $\mathcal{H}$, and
$\mathcal{W}=\mathcal{H}\ominus V\mathcal{H}$. Then $\Pi
C\Pi^{*}\in\mathcal{B}(H^{2}_{\mathcal{W}}(\mathbb{D}))$ and $(\Pi
C\Pi^{*})M_{z}=M_{z}(\Pi C\Pi^{*})$ as $CV=VC$. Therefore

$\Pi C\Pi^{*}=M_{\Theta}$

for some $\Theta\in H^{\infty}_{\mathcal{B}(\mathcal{W})}(\mathbb{D})$.

Another important vector-valued reproducing kernel Hilbert space is the
Dirichlet space $D_{\mathcal{E}}(\mu)$ which is defined as the space of all
$\mathcal{E}$-valued holomorphic functions $f$ in the unit disc $\mathbb{D}$
with the finite norm

$\|f\|_{\mu}^{2}=\|f\|^{2}_{H^{2}(\mathbb{D})}+\int_{\mathbb{D}}\langle
P[\mu](z)f^{{}^{\prime}}(z),f^{{}^{\prime}}(z)\rangle dA(z),$

where $dA$ stands for the normalized Lebesgue area measure on $\mathbb{D}$ and
$P[\mu]$ is the Poisson integral of the positive
$\mathcal{B}(\mathcal{E})$-valued operator measure on the unit circle. The
multiplication operator $M_{z}^{D_{\mathcal{E}}(\mu)}$ stands for the
Dirichlet shift on the space $D_{\mathcal{E}}(\mu)$ (see [20] for more
details). An operator $V\in\mathcal{B}(\mathcal{H})$ is said to be a two-
isometry if

$\|V^{2}h\|^{2}+\|h\|^{2}=2\|Vh\|^{2}\quad(\forall~{}h\in\mathcal{H}).$

Thus every isometry is a two-isometry. Let $V$ be a pure two-isometry on
$\mathcal{H}$, that is
$\mathcal{H}_{u}=\cap_{n=0}^{\infty}V^{n}\mathcal{H}=\\{0\\}$. Then the Wold-
von Neumann decomposition for the two isometry gives an analytic
representation. More precisely, the unitary
$\tilde{\Pi}:\mathcal{H}\rightarrow D_{\mathcal{E}}(\mu)$ defined by

$(\tilde{\Pi}h)(z)=\sum_{n=0}^{\infty}(P_{\mathcal{E}}L^{n}h)z^{n}\quad(h\in\mathcal{H},z\in\mathbb{D}),$

where $P_{\mathcal{E}}$ is the projection of $\mathcal{H}$ onto
$\mathcal{E}=\mathcal{H}\ominus V\mathcal{H}$, $L=(V^{*}V)^{-1}V^{*}$ is the
left inverse of $V$, and
$\tilde{\Pi}V=M_{z}^{D_{\mathcal{E}}(\mu)}\tilde{\Pi}$.

We are now in a position to state our results:

######  Theorem 3.4.

Let $V$ be a shift or a co-shift on a Hilbert space $\mathcal{H}$. Then the
Crawford number of $V$ is $0$.

###### Proof.

Suppose that $V$ is a shift on a Hilbert space $\mathcal{H}$. We here prove
only for shift case. If $V$ is a co-shift, then we take adjoint of $V$,
$V^{*}$ which becomes a shift.

Let $(\Pi,M_{z})$ be the Wold-von Neumann decomposition of the shift $V$. That
means the map $\Pi:\mathcal{H}\rightarrow H^{2}_{\mathcal{E}}(\mathbb{D})$,
where $\mathcal{E}=\mathcal{H}\ominus V\mathcal{H}$ is a unitary such that

$\Pi V\Pi^{*}=M_{z}.$

Therefore

$\displaystyle c(M_{z})=c(\Pi V\Pi^{*})=c(V)\leq\|V\|=1.$

Now consider the normalized kernel function
$\hat{k}(\cdot,w)=\frac{{k}(\cdot,w)}{\|{k}(\cdot,w)\|}$ of
$H^{2}(\mathbb{D})$. Also the set
$\\{\hat{k}(\cdot,w)\eta:w\in\mathbb{D},\eta\in\mathcal{E}\\}$ is a total set
in $H^{2}_{\mathcal{E}}(\mathbb{D})$. Therefore from the definition of
Crawford number of an operator, we obtain

$\displaystyle 0\leq c(M_{z})$
$\displaystyle\leq\inf_{w\in\mathbb{D},\|\eta\|=1}|\langle
M_{z}\hat{k}(\cdot,w)\eta,\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\inf_{w\in\mathbb{D},\|\eta\|=1}|\langle\hat{k}(\cdot,w)\eta,M_{z}^{*}\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\inf_{w\in\mathbb{D},\|\eta\|=1}|\bar{w}||\langle\hat{k}(\cdot,w)\eta,\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\inf_{w\in\mathbb{D}}|{w}|$ $\displaystyle=0.$

Thus

$c(V^{*})=c(V)=c(M_{z})=0.$

This completes the proof.

######  Remark 3.5.

Let $\theta\in H^{\infty}(\mathbb{D})$ be a non-constant inner function. Then
the analytic Toeplitz operator $T_{\theta}\in\mathcal{B}(H^{2}(\mathbb{D}))$
is a shift (cf. [14]). Hence

$\omega(T_{\theta}^{*n})=\omega(T_{\theta}^{n})=1$

and

$c(T_{\theta}^{*n})=c(T_{\theta}^{n})=0\quad(n\in\mathbb{N}).$

######  Remark 3.6.

Suppose $\theta\in H^{\infty}(\mathbb{D})$. If $\theta(z_{0})=0$ for some
point $z_{0}\in\mathbb{D}$, then

$c(T_{\theta}^{n})=0\quad(n\in\mathbb{N}).$

For an example, consider $\theta(z)=1$ for $z\in\mathbb{D}$. Then clearly
$\theta\in H^{\infty}(\mathbb{D})$ and $\theta(z)\neq 0$ for any $z$ in
$\mathbb{D}$. Also $c(T_{\theta}^{n})=1$ for any $n\in\mathbb{N}$.

######  Theorem 3.7.

Let $V$ be a pure two-isometry on a Hilbert space $\mathcal{H}$. Then the
numerical radius of $V$ is $1$ and the Crawford number of $V$ is $0$.

###### Proof.

Suppose that $V$ is a pure two-isometry on $\mathcal{H}$. Let
$(\tilde{\Pi},M_{z}^{D_{\mathcal{E}}(\mu)})$ be the Wold-von Neumann
decomposition of $V$. Then
$\tilde{\Pi}V\tilde{\Pi}^{*}=M_{z}^{D_{\mathcal{E}}(\mu)}.$ Thus

$\displaystyle\omega(M_{z}^{D_{\mathcal{E}}(\mu)})=\omega(\tilde{\Pi}V\tilde{\Pi}^{*}).$

Again consider the normalized kernel function
$\hat{k}(\cdot,w)=\frac{{k}(\cdot,w)}{\|{k}(\cdot,w)\|}$ of $D(\mu)$. Also the
set $\\{\hat{k}(\cdot,w)\eta:w\in\mathbb{D},\eta\in\mathcal{E}\\}$ is a total
set in $D_{\mathcal{E}}(\mu)$. Therefore

$\displaystyle\omega(M_{z}^{D_{\mathcal{E}}(\mu)})$
$\displaystyle=\sup_{w\in\mathbb{D},\|\eta\|=1}|\langle
M_{z}^{D_{\mathcal{E}}(\mu)}\hat{k}(\cdot,w)\eta,\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\sup_{w\in\mathbb{D},\|\eta\|=1}|\bar{w}||\langle\hat{k}(\cdot,w)\eta,\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\sup_{w\in\mathbb{D}}|{w}|$ $\displaystyle=1.$

Hence

$\omega(V)=\omega(\tilde{\Pi}V\tilde{\Pi}^{*})=\omega(M_{z}^{D_{\mathcal{E}}(\mu)})=1.$

For the case of Crawford number, the same lines of the above proof yields

$c(V)=c(\tilde{\Pi}V\tilde{\Pi}^{*})=c(M_{z}^{D_{\mathcal{E}}(\mu)})=\inf_{w\in\mathbb{D}}|w|=0.$

This completes the proof.

######  Remark 3.8.

Let $V$ be an isometry on a Hilbert space $\mathcal{H}$. Then it is a two-
isometry with $\|V\|=1$. Now the spectral radius formula gives that the
numerical radius of $V$ is $1$. However, here we provide a new proof by the
help of the analytic structure of an isometry. Clearly,

$\displaystyle\omega(V)=\sup_{\|x\|=1}|\langle Vx,x\rangle|\leq\|V\|=1.$

Let $(\Pi,M_{z})$ be the Wold-von Neumann decomposition of the isometry $V$.
Then

$\Pi V\Pi^{*}=M_{z}\oplus U,$

where $U$ is the unitary part of $V$. Thus

$\displaystyle\omega(M_{z}\oplus U)=\omega(\Pi V\Pi^{*})=\omega(V)\leq 1.$

Again consider the normalized kernel function
$\hat{k}(\cdot,w)=\frac{{k}(\cdot,w)}{\|{k}(\cdot,w)\|}$ of
$H^{2}(\mathbb{D})$. Also the set
$\\{\hat{k}(\cdot,w)\eta:w\in\mathbb{D},\eta\in\mathcal{E}\\}$ is a total set
in $H^{2}_{\mathcal{E}}(\mathbb{D})$. Therefore

$\displaystyle\omega(M_{z}\oplus U)$
$\displaystyle\geq\sup_{w\in\mathbb{D},\|\eta\|=1}|\langle(M_{z}\oplus
U)(\hat{k}(\cdot,w)\eta\oplus 0),\hat{k}(\cdot,w)\eta\oplus 0\rangle|$
$\displaystyle=\sup_{w\in\mathbb{D},\|\eta\|=1}|\langle\hat{k}(\cdot,w)\eta,M_{z}^{*}\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\sup_{w\in\mathbb{D},\|\eta\|=1}|\bar{w}||\langle\hat{k}(\cdot,w)\eta,\hat{k}(\cdot,w)\eta\rangle|$
$\displaystyle=\sup_{w\in\mathbb{D}}|{w}|$ $\displaystyle=1.$

Hence

$\omega(V)=\omega(\Pi V\Pi^{*})=\omega(M_{z}\oplus U)=1.$

If $V$ is a co-isometry, then $V^{*}$ is an isometry. Thus from the above we
have

$\omega(V)=\omega(V^{*})=1.$

The above analytic proof yields that the numerical radius of a two-isometry
with unit norm is $1$.

######  Corollary 3.9.

The numerical radius $\omega(\cdot)$ is multiplicative on the class of
isometries on a Hilbert space.

###### Proof.

Let $R,T$ be two isometries on a Hilbert space $\mathcal{H}$. Then $RT$ is an
isometry on $\mathcal{H}$ as

$(RT)^{*}(RT)=T^{*}(R^{*}R)T=(T^{*}I_{\mathcal{H}})T=T^{*}T=I_{\mathcal{H}}.$

Therefore from the above Remark 3.8, we have

$\omega(RT)=1~{}~{}~{}\mbox{and}~{}~{}~{}\omega(R)=1=\omega(T).$

Hence

$\omega(RT)=\omega(R)\omega(T).$

This finishes the proof.

######  Remark 3.10.

Let $T\in\mathcal{B}(\mathcal{H})$ be any isometry. Then the above Corollary
gives $\omega(T^{n})=\omega(T)^{n}$ for any $n\in\mathbb{N}$. It is a
noteworthy to mention that for a normal operator
$T\in\mathcal{B}(\mathcal{H})$, $\omega(T^{n})=\omega(T)^{n}$ for any
$n\in\mathbb{N}$ as

$\omega(T^{n})=\|T^{n}\|=\|T\|^{n}=\omega(T)^{n}.$

######  Lemma 3.11.

Let $T$ be a shift on a Hilbert space $\mathcal{H}$ and let $R$ be a commutant
of $T$. Then the numerical radius of $R$ is $\|R\|$.

###### Proof.

Let $T$ be a shift on a Hilbert space $\mathcal{H}$. Then $T$ is unitarily
equivalent to $M_{z}$ on $H^{2}_{\mathcal{E}}(\mathbb{D})$, where
$\mathcal{E}=Ker(T^{*})$. That means there exists a unitary
$\Pi:\mathcal{H}\rightarrow H^{2}_{\mathcal{E}}(\mathbb{D})$ such that

$\Pi T\Pi^{*}=M_{z}.$

Now if $R$ is any commutant of $T$, that is $RT=TR$, then

$(\Pi R\Pi^{*})M_{z}=M_{z}(\Pi R\Pi^{*}).$

Thus $\Pi R\Pi^{*}$ is an analytic Toeplitz operator on
$H^{2}_{\mathcal{E}}(\mathbb{D})$ and hence $\Pi R\Pi^{*}=M_{\Phi}$ for some
$\Phi\in H^{\infty}_{\mathcal{B}(\mathcal{E})}(\mathbb{D})$. Therefore

$\omega(R)=\omega(\Pi
R\Pi^{*})=\omega(M_{\Phi})=\sup_{\lambda\in\mathbb{D}}\|\Phi^{*}(\lambda)\|=\|M_{\Phi}\|=\|R\|.$

The next result gives a partial answer of the question raised in the previous
section.

######  Theorem 3.12.

Let $T_{1},T_{2}\in\mathcal{B}(\mathcal{H})$ such that $T_{1}T_{2}=T_{2}T_{1}$
and $T_{1}T_{2}$ be a shift on $\mathcal{H}$. Then
$\omega(T_{1}T_{2})\leq\omega(T_{1})\omega(T_{2})$.

###### Proof.

Let $T=T_{1}T_{2}$ be a shift on $\mathcal{H}$ with $T_{1}T_{2}=T_{2}T_{1}$.
Since $T$ is a shift, Wold decomposition yields a unitary
$\Pi:\mathcal{H}\rightarrow H^{2}_{\mathcal{E}}(\mathbb{D})$, where
$\mathcal{E}=Ker(T^{*})$ such that

$\Pi T\Pi^{*}=M_{z}.$

Now

$TT_{1}=(T_{1}T_{2})T_{1}=T_{1}(T_{1}T_{2})=T_{1}T.$

Similarly,

$TT_{2}=T_{2}T.$

Therefore from the above Lemma 3.11, we have

$\omega(T_{i})=\|T_{i}\|\quad(i=1,2).$

Thus

$\omega(T)=\omega(T_{1}T_{2})\leq\|T_{1}\|\|T_{2}\|=\omega(T_{1})\omega(T_{2}).$

Now we discuss some results on Berezin number of an operator on a reproducing
kernel Hilbert space.

Let $H(K,\mathbb{D})$ be a reproducing kernel Hilbert space with kernel
function $K$ defined as $K(\cdot,w)(z)=K(z,w)$ for all $z\in\mathbb{D}$. We
say that $H(K,\mathbb{D})$ is an analytic Hilbert space if the multiplication
operator $M_{z}$ by the coordinate function $z$ on $H(K,\mathbb{D})$, defined
by $M_{z}f=zf$ for all $f\in H(K,\mathbb{D})$ is a contraction. Also

$M_{z}^{*}K(\cdot,w)=\bar{w}K(\cdot,w)\quad(w\in\mathbb{D}).$

Let $\hat{K}(\cdot,w)$ for $w\in\mathbb{D}$ be the normalized kernel function.
Then

$\displaystyle ber(M_{z})$ $\displaystyle=\sup\\{|\langle
M_{z}\hat{K}(\cdot,w),\hat{K}(\cdot,w)\rangle|:w\in\mathbb{D}\\}$
$\displaystyle=\sup\\{|\langle\hat{K}(\cdot,w),M_{z}^{*}\hat{K}(\cdot,w)\rangle|:w\in\mathbb{D}\\}$
$\displaystyle=\sup\\{|\langle\hat{K}(\cdot,w),\bar{{w}}\hat{K}(\cdot,w)\rangle|:w\in\mathbb{D}\\}$
$\displaystyle=\sup\\{|{{w}}|:w\in\mathbb{D}\\}$ $\displaystyle=1.$

Again

$\displaystyle 1\geq\omega(M_{z})$ $\displaystyle=\sup\\{|\langle
M_{z}f,f\rangle|:\|f\|=1\\}$ $\displaystyle\geq\sup\\{|\langle
M_{z}\hat{K}(\cdot,w),\hat{K}(\cdot,w)\rangle|:w\in\mathbb{D}\\}$
$\displaystyle=1.$

To summarize the above, we have the following:

######  Theorem 3.13.

Let $H(K,\mathbb{D})$ be an analytic reproducing kernel Hilbert space over the
unit disc $\mathbb{D}$ with the kernel function $K$ and $M_{z}$ be the
multiplication operator defined by the coordinate function $z$ on
$H(K,\mathbb{D})$. Then $ber(M_{z})=1$ and $\omega(M_{z})=1$.

Recall that $H(\Omega)$ be a rkHs on $\Omega$ and $\hat{k}_{\lambda}$ for
$\lambda\in\Omega$ normalized kernel function of $H(\Omega)$. For
$T\in\mathcal{B}(H(\Omega))$,

$\|T\|_{ber}=\sup_{\lambda\in\Omega}\|T\hat{k}_{\lambda}\|.$

Here we have the following result:

######  Theorem 3.14.

Let $H(\Omega_{i})$ be a reproducing kernel Hilbert space on $\Omega_{i}$ and
$T_{i}\in\mathcal{B}(H(\Omega_{i}))$ for $i=1,2$.Then $\|T_{1}\otimes
T_{2}\|_{ber}=\|T_{1}\|_{ber}\|T_{2}\|_{ber}$.

###### Proof.

Suppose $H(\Omega_{i})$ is a reproducing kernel Hilbert space on $\Omega_{i}$
and $T_{i}\in\mathcal{B}(H(\Omega_{i}))$ for $i=1,2$. Then $T_{1}\otimes
T_{2}\in\mathcal{B}(H(\Omega_{1})\otimes H(\Omega_{2}))$. Let
$\hat{k}_{\lambda}$ for $\lambda\in\Omega_{1}$ and $\hat{k}_{\mu}$ for
$\mu\in\Omega_{2}$ be the normalized kernel functions for $H(\Omega_{1})$ and
$H(\Omega_{2})$, respectively. Then $\hat{k}_{\lambda}\otimes\hat{k}_{\mu}$
for $\lambda\in\Omega_{1}$, $\mu\in\Omega_{2}$ is the normalized kernel
function for $H(\Omega_{1})\otimes H(\Omega_{2})$. Therefore

$\displaystyle\|T_{1}\otimes T_{2}\|_{ber}$
$\displaystyle=\sup_{\lambda\in\Omega_{1},\mu\in\Omega_{2}}\|(T_{1}\otimes
T_{2})(\hat{k}_{\lambda}\otimes\hat{k}_{\mu})\|$
$\displaystyle=\sup_{\lambda\in\Omega_{1},\mu\in\Omega_{2}}\|T_{1}\hat{k}_{\lambda}\otimes
T_{2}\hat{k}_{\mu}\|$
$\displaystyle=(\sup_{\lambda\in\Omega_{1}}\|T_{1}\hat{k}_{\lambda}\|)(\sup_{\mu\in\Omega_{2}}\|T_{2}\hat{k}_{\mu}\|)$
$\displaystyle=\|T_{1}\|_{ber}\|T_{2}\|_{ber}.$

######  Lemma 3.15.

Let $H(\Omega)$ be any reproducing kernel Hilbert space on $\Omega$ and let
$A,X\in\mathcal{B}(\mathcal{H}(\Omega))$ and $X$ be any positive operator.
Then $ber({\sqrt{A^{*}XA}})<\sqrt{ber(A^{*}XA})$.

###### Proof.

Suppose $A,X\in\mathcal{B}(H(\Omega))$ and $X$ is positive operator. Then
$A^{*}XA$ is a positive operator on $H(\Omega)$. Let $\hat{k}_{\lambda}$ for
$\lambda\in\Omega$ be the normalized kernel function for $H(\Omega)$. Then

$\displaystyle ber(\sqrt{A^{*}XA}))$
$\displaystyle=\sup_{\lambda\in\Omega}|\langle\sqrt{A^{*}XA}\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle|$
$\displaystyle<\sup_{\lambda\in\Omega}|\langle{A^{*}XA}\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle|^{1\over
2}$ $\displaystyle=\sqrt{ber(A^{*}XA}).$

Here the second inequality comes from Lemma 2.1. This proves the desired
inequality.

We will give some concrete operators for which Berezin number, numerical
radius, and Crawford number are calculated.

######  Example 3.16.

Let $\\{e_{1},e_{2}\\}$ be the standard orthonormal basis for
$\mathbb{C}^{2}$. Consider $\mathbb{C}^{2}$ as a rkHs on the set $\\{1,2\\}$.
Then $e_{1},e_{2}$ (identified a complex pair as a function) are the kernel
functions defined by

$e_{i}(j)=\begin{cases}1&~{}\mbox{if}~{}i=j\\\ 0&~{}\mbox{if}~{}i\neq
j\end{cases}$

for $i,j\in\\{1,2\\}$. Let $U:\mathbb{C}^{2}\rightarrow\mathbb{C}^{2}$ be a
linear map defined by

$\displaystyle U(e_{1})$ $\displaystyle=e_{2}$ $\displaystyle U(e_{2})$
$\displaystyle=e_{1}.$

Then clearly $U$ is a self-adjoint and unitary operator on the rkHs
$\mathbb{C}^{2}$. From the definition we have

$ber(U)=\sup\\{|\langle Ue_{1},e_{1}\rangle|,|\langle
Ue_{2},e_{2}\rangle|\\}=0,$ $\|U\|_{ber}=\sup\\{\|Ue_{i}\|:i=1,2\\}=1,$

and

$\omega(U)=\|U\|=1,~{}~{}c(U)=0.$

######  Example 3.17.

Let $\\{e_{n}:n\in\mathbb{Z}_{+}\\}$ be a standard orthonormal basis for
$\ell^{2}(\mathbb{Z}_{+})$ and $\bm{w}=\\{w_{n}\\}_{n\in\mathbb{Z}_{+}}$ be a
bounded sequence of complex numbers. Consider $\ell^{2}(\mathbb{Z}_{+})$ as a
rkHs on the set $\mathbb{Z}_{+}$. With the identification a sequence as a
function, the kernel function $e_{n}$ for $n\in\mathbb{Z}_{+}$ defined as

$e_{n}(m)=\begin{cases}1&~{}\mbox{if}~{}n=m\\\ 0&~{}\mbox{if}~{}n\neq
m\end{cases}$

for $m,n\in\mathbb{Z}_{+}$. Let
$S_{\bm{w}}:\ell^{2}(\mathbb{Z}_{+})\rightarrow\ell^{2}(\mathbb{Z}_{+})$ be a
weighted right shift operator defined by

$\displaystyle S_{\bm{w}}(x_{0},x_{1},x_{2},\ldots)$
$\displaystyle=(0,w_{0}x_{0},w_{1}x_{1},w_{2}x_{2},\ldots).$

From the definition one can easily compute

$ber(S_{\bm{w}})=\sup\\{|\langle
S_{\bm{w}}e_{i},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}=\sup\\{|\langle
w_{i}e_{i+1},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}=0,$
$\|S_{\bm{w}}\|_{ber}=\sup\\{\|S_{\bm{w}}e_{i}\|:i\in\mathbb{Z}_{+}\\}=\sup_{n}|w_{n}|=\|S_{\bm{w}}\|.$

In particular, if we take the sequence
$\bm{w}=\\{w_{n}\\}_{n\in\mathbb{Z}_{+}}$ as $w_{n}=1$ for all $n$. Then the
operator $S_{\bm{w}}$ denoted as $S$, is called a unilateral shift on the rkHs
$\ell^{2}(\mathbb{Z}_{+})$ over the set $\mathbb{Z}_{+}$. Then we have

$ber(S)=\sup\\{|\langle
Se_{i},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}=\sup\\{|\langle
e_{i+1},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}=0,$
$\|S\|_{ber}=\sup\\{\|Se_{i}\|:i\in\mathbb{Z}_{+}\\}=1=\|S\|$

and by Theorem 3.4 & Remark 3.8, we have

$\omega(S)=1,~{}~{}c(S)=0.$

######  Example 3.18.

Consider $\ell^{2}(\mathbb{Z}_{+})$ as rkHs with kernel function $e_{n}$ for
$n\in\mathbb{Z}_{+}$ as before. Let $\\{\lambda_{n}\\}$ be a bounded sequence
of scalars such that $\lambda_{n}\rightarrow 0$ and
$D:\ell^{2}(\mathbb{Z}_{+})\rightarrow\ell^{2}(\mathbb{Z}_{+})$ be a diagonal
operator defined by $De_{n}=\lambda_{n}e_{n}$ for $n\in\mathbb{Z}_{+}$. It is
well known that $D$ is a compact normal operator. Now

$ber(D)=\sup\\{|\langle
De_{i},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}=\sup\\{|\langle\lambda_{i}e_{i},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}=\sup_{i}{|\lambda_{i}|},$

$\|D\|_{ber}=\sup\\{\|De_{i}\|:i\in\mathbb{Z}_{+}\\}=\sup_{i}{|\lambda_{i}|}=\|D\|.$

Also

$\omega(D)=\|D\|.$

and

$\displaystyle 0\leq c(D)$ $\displaystyle=\inf\\{|\langle
Dx,x\rangle|:x\in\ell^{2}(\mathbb{Z}_{+}),\|x\|=1\\}$
$\displaystyle\leq\inf\\{|\langle De_{i},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}$
$\displaystyle=\inf\\{|\langle\lambda_{i}e_{i},e_{i}\rangle|:i\in\mathbb{Z}_{+}\\}$
$\displaystyle=\inf_{i}{|\lambda_{i}|}=0.$

Therefore $c(D)=0$.

######  Remark 3.19.

Let $T\in\mathcal{B}(\mathcal{H})$ be a compact normal operator on an infinite
dimensional Hilbert space $\mathcal{H}$. Then from the spectral theorem
(cf.[7]), we have a countable non zero distinct eigenvalues
$\\{\lambda_{n}\\}$ such that $\lambda_{n}\rightarrow 0$ and a sequence of
orthonormal eigenvectors $\\{f_{n}\\}$ such that

$Tx=\sum_{n=1}^{\infty}{\lambda_{n}}\langle x,f_{n}\rangle
f_{n}\quad(x\in\mathcal{H}).$

We have

$\displaystyle c(T)$ $\displaystyle=\inf\\{|\langle
Tx,x\rangle|:x\in\mathcal{H},\|x\|=1\\}$ $\displaystyle\leq\inf\\{|\langle
Tf_{n},f_{n}\rangle|:n\in\mathbb{N}\\}$
$\displaystyle=\inf\\{|\langle\lambda_{n}f_{n},f_{n}\rangle|:n\in\mathbb{N}\\}$
$\displaystyle=\inf_{n}{|\lambda_{n}|}=0.$

Therefore the Crawford number of a compact normal operator $T$ is zero.

######  Remark 3.20.

It is a noteworthy to mention that the unilateral shift $S$ on
$\ell^{2}(\mathbb{Z}_{+})$ and the multiplication operator $M_{z}$ by the the
coordinate function $z$ on $H^{2}(\mathbb{D})$ are unitarily equivalent.
Therefore $\omega(S)=1=\omega(M_{z})$. On the other hand let
$\hat{k}(\cdot,w)$ for $w\in\mathbb{D}$ be the normalized Szegö kernel
function on $H^{2}(\mathbb{D})$. Then

$\displaystyle ber(M_{z})$ $\displaystyle=\sup\\{|\langle
M_{z}\hat{k}(\cdot,w),\hat{k}(\cdot,w)\rangle|:w\in\mathbb{D}\\}$
$\displaystyle=\sup\\{|{{w}}|:w\in\mathbb{D}\\}=1.$

But from the above Example 3.17, we have

$ber(S)=0.$

Therefore Berezin number does not satisfy the unitarily invariant property.

##  4\. Hardy-Type Inequality for Berezin number

In this section, we establish some Berezin number inequalities of a certain
class of operators with the help of classical Hardy inequality and functional
calculus.

######  Theorem 4.1.

Let $p>1$ be any real number and $N$ be any fixed natural number and let $f$
be a positive continuous function defined on $\Delta\subset(0,\infty)$. Let
$A$ be a positive operator on a Hilbert space $\mathcal{H}$ such that the
spectrum of $A$ lies in $\Delta$. Then for all $x\in\mathcal{H}$

$\Big{(}1+N^{p-1}\displaystyle\sum_{n=N+1}^{\infty}\frac{1}{n^{p}}\Big{)}\langle
f(A)x,x\rangle^{p}<\Big{(}\frac{p}{p-1}\Big{)}^{p}\langle f(A)^{p}x,x\rangle.$

###### Proof.

Let $N$ be a fixed natural number and let $\\{a_{n}\\}$ be a sequence of non-
negative real numbers such that $a_{n}=0$ for $n>N$. Then the inequality (2.1)
yields

(4.1) $\displaystyle
a_{1}^{p}+(\frac{1}{2}\displaystyle\sum_{k=1}^{2}a_{k})^{p}+\cdots+(\frac{1}{N}\displaystyle\sum_{k=1}^{N}a_{k})^{p}+(\displaystyle\sum_{k=1}^{N}a_{k})^{p}\\{\displaystyle\sum_{n={N+1}}^{\infty}\frac{1}{n^{p}}\\}<(\frac{p}{p-1})^{p}(\sum_{k=1}^{N}a_{k}^{p}).$

Since $A\in\mathcal{B}(\mathcal{H})$ is a positive operator and $f$ is a
continuous function on $\Delta\subset(0,\infty)$ with the spectrum of $A$ lies
in $\Delta$, we can use the continuous functional calculus. Therefore
$f(A)\in\mathcal{B}(\mathcal{H})$ makes sense and for any $x\in\mathcal{H}$,
$\langle f(A)x,x\rangle\geq 0$. Now replacing $a_{k}$ by $\langle
f(A)x,x\rangle$ for $k=1,2,\ldots,N$ in the above inequality (4.1), we obtain

(4.2) $\displaystyle{}N\langle f(A)x,x\rangle^{p}+{N}^{p}\langle
f(A)x,x\rangle^{p}\Big{\\{}\displaystyle\sum_{n=N+1}^{\infty}\frac{1}{n^{p}}\Big{\\}}$
$\displaystyle<N\Big{(}\frac{p}{p-1}\Big{)}^{p}\langle f(A)x,x\rangle^{p}.$

Applying Lemma 2.1 in the right hand side of the above inequality (4.2), we
have

$\displaystyle\langle
f(A)x,x\rangle^{p}\Big{(}1+N^{p-1}\displaystyle\sum_{n=N+1}^{\infty}\frac{1}{n^{p}}\Big{)}$
$\displaystyle<\Big{(}\frac{p}{p-1}\Big{)}^{p}\langle f(A)^{p}x,x\rangle.$

This completes the proof.

In particular, when $p=2$ and $N=2$, that is a sequence $\\{a_{n}\\}$ of non-
negative real numbers such that $a_{n}=0$ for $n>2$ then from the above
theorem we have the following result.

######  Corollary 4.2.

Let $f$ be a positive continuous function defined on $\Delta\subset(0,\infty)$
and $A$ be a positive operator on $\mathcal{H}$ such that the spectrum of $A$
lies in $\Delta$. Then for all $x\in\mathcal{H}$

$\langle f(A)x,x\rangle^{2}<\frac{24}{2\pi^{2}-9}\langle f(A)^{2}x,x\rangle.$

######  Remark 4.3.

Now if we choose $N=3$, that is first three terms of the sequence
$\\{a_{n}\\}$ are non-zero, and the rests are zero, then applying the similar
techniques used in Theorem 4.1, the following power inequality is obtained:

$\Big{(}2+3^{p}(\displaystyle\sum_{n=3}^{\infty}1/n^{2})\Big{)}\langle
f(A)x,x\rangle^{p}<3(\frac{p}{p-1})^{p}~{}\langle f(A)^{p}x,x\rangle$.

In particular, for $p=2$, we deduce the following inequality

$\langle f(A)x,x\rangle^{2}<\frac{48}{6\pi^{2}-37}~{}\langle
f(A)^{2}x,x\rangle\approx 2.1604~{}\langle f(A)^{2}x,x\rangle$.

######  Remark 4.4.

For the sake of simplicity, we shall discuss operator inequalities for Berezin
numbers by taking Hardy’s inequality for a scalar sequence $\\{a_{n}\\}$ with
two non-zero positive numbers and rests are zero, i.e., $a_{n}=0$ for $n>2$.

We now focus operators on reproducing kernel Hilbert spaces $H(\Omega)$ and we
obtain the following inequalities for Berezin number with the help of Berezin
transform.

######  Theorem 4.5.

Let $p>1$ be any real number and let $f$ be a positive continuous function
defined on $\Delta\subset(0,\infty)$. Let $A$ be a positive operator on a
reproducing kernel Hilbert space $H(\Omega)$ such that the spectrum of $A$
lies in $\Delta$. Then the below inequality

$\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}ber^{p}(f(A))<\Big{(}\frac{p}{p-1}\Big{)}^{p}~{}ber(f(A)^{p})~{}holds~{}true.$

In particular, we have the following inequality for $p=2$

$ber^{2}(f(A))<\frac{24}{2\pi^{2}-9}~{}ber(f(A)^{2}).$

###### Proof.

Suppose $p>1$ is any real number and $\\{a_{n}\\}$ is a sequence of non-
negative real numbers such that $a_{n}=0$ for $n>2$. Then from the above
Theorem 4.1 for any $x\in H(\Omega)$, we have

(4.3)
$\displaystyle{}\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}\langle
f(A)x,x\rangle^{p}<\Big{(}\frac{p}{p-1}\Big{)}^{p}\langle f(A)^{p}x,x\rangle.$

Take $x=\hat{k}_{\lambda}$ for $\lambda\in\Omega$, in the above inequality, we
get

$\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}\langle
f(A)\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle^{p}<\Big{(}\frac{p}{p-1}\Big{)}^{p}\langle
f(A)^{p}\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle.$

Hence, we have

$\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}\langle
f(A)\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle^{p}<\Big{(}\frac{p}{p-1}\Big{)}^{p}\sup_{\lambda\in\Omega}|\langle
f(A)^{p}\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle|.$

Therefore by using the definition of Berezin number, we obtain the desired
inequality.

Now if we put $p=2$, then the following inequality for Berezin number is
obtained from the above

$ber^{2}(f(A))<\frac{24}{2\pi^{2}-9}~{}ber(f(A)^{2})\approx
2.2349~{}ber(f(A)^{2})$.

######  Remark 4.6.

Using the Hilbert-Hardy inequality Yamancı and Gürdal [26] obtained a strict
inequality for Berezin number as follows:

$ber^{2}(f(A))<\frac{36\pi^{2}}{53}~{}ber(f(A)^{2})\approx
6.7039~{}ber(f(A)^{2})$.

######  Remark 4.7.

Some reverse inequalities for Berezin numbers are obtained by Garayev et
al.[8]. Indeed, they obtain the following inequality (see [8], Proposition 1):

$\displaystyle ber(f(A))^{2}\leq\frac{3(8\pi-3)}{8}ber(f(A)^{2}).$

Further in another result (in fact Proposition 2 of [8]), the authors
presented the better approximate of the above result as follows:

$\displaystyle ber(A)^{2}\leq\frac{24\pi}{17}ber(A^{2}),$

where $A\in\mathcal{B}(H(\Omega))$ is a self adjoint operator with spectrum
contained in $\Delta\subset(0,\infty)$.

######  Remark 4.8.

In another work of Garayev et al. [9], the following power inequality of
Berezin number for a positive operator $A$ and a continuous function $f$ is
obtained by using Hilbert-Hardy inequality which is as follows:

$\displaystyle ber(f(A))^{p}\leq\alpha ber(f(A)^{p}),$

where
$\alpha\equiv\alpha(p,q)=2\Big{[}\big{(}\frac{3}{2}\big{)}^{p}+1\Big{]}^{-1}(pq)^{p}>1$
and $\frac{1}{p}+\frac{1}{q}=1$ with $p,q>1$. Hence for $p=2$, we get

$\displaystyle ber(f(A))^{2}\leq\frac{128}{13}ber(f(A)^{2}).$

######  Remark 4.9.

Let $A\in\mathcal{B}(\mathcal{H}(\Omega))$ be a self-adjoint operator and the
spectrum of $A$ contained in $\Delta\subset(0,\infty)$. Then for any positive
continuous function $f$ defined on $\Delta$, Garayev et al. [9] obtained an
inequality given as follows:

$\displaystyle ber(f(A))^{2}\leq 4(pq-1)ber(f(A)^{2}),$

where $\frac{1}{p}+\frac{1}{q}=1$ with $p,q>1$.

Before we proceed further we establish an inequality for Berezin number for
operators by using the recently improved version of the _discrete Hardy’s
inequality_ (2.2). In fact the statement of our result reads as follows:

######  Corollary 4.10.

Let $f$ be a positive continuous function defined on
$\Delta\subset(0,\infty)$. Let $A$ be a positive operator on a reproducing
kernel Hilbert space $H(\Omega)$ such that the spectrum of $A$ lies in
$\Delta$. Then the inequality

$ber^{2}(f(A))<1.593852~{}ber(f(A)^{2})~{}\mbox{holds}.$

###### Proof.

Using the improved version of the _discrete Hardy inequality_ (2.2) in the
proof of Theorem 4.5, we obtain the following inequality

$ber^{2}(f(A))<\displaystyle\frac{2}{w_{1}+4\sum_{n=1}^{\infty}w_{n+1}}ber(f(A)^{2}),$

where for $n\in\mathbb{N}$,
$w_{n}=2-\sqrt{1+\frac{1}{n}}-\sqrt{1-\frac{1}{n}}$ and
$\displaystyle\sum_{n=1}^{\infty}w_{n}=0.753045~{}(approx.)$. This proves the
desired inequality.

Let $H(\Omega)$ be a reproducing kernel Hilbert space and
$A\in\mathcal{B}(H(\Omega))$. Let $x\in H(\Omega)$ with $\|x\|=1$. Using the
Hardy’s inequality (2.1) and replacing first two nonzero terms by $|\langle
Ax,x\rangle|$ and rest of the terms are zero, we obtain the following
inequality:

(4.4) $\displaystyle{}|\langle
Ax,x\rangle|^{p}{(1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}})}<(\frac{p}{p-1})^{p}|\langle
Ax,x\rangle|^{p}.$

Applying the mixed Cauchy-Schwartz inequality Lemma 2.2 on the right hand side
term, we get

$\displaystyle|\langle Ax,x\rangle|^{p}$
$\displaystyle\leq\Big{(}\langle|A|^{2\alpha}x,x\rangle^{\frac{1}{2}}\langle|A^{*}|^{2(1-\alpha)}x,x\rangle^{\frac{1}{2}}\Big{)}^{p}$
$\displaystyle\leq\Big{(}\frac{\langle|A|^{2\alpha}x,x\rangle+\langle|A^{*}|^{2(1-\alpha)}x,x\rangle}{2}\Big{)}^{p}$
$\displaystyle\leq\frac{1}{2}\Big{(}\langle|A|^{2\alpha}x,x\rangle^{p}+\langle|A^{*}|^{2(1-\alpha)}x,x\rangle^{p}\Big{)}$
$\displaystyle\leq\frac{1}{2}(\langle|A|^{2p\alpha}x,x\rangle+\langle|A^{*}|^{2p(1-\alpha)}x,x\rangle)$
$\displaystyle=\frac{1}{2}\langle(|A|^{2p\alpha}+|A^{*}|^{2p(1-\alpha)})x,x\rangle,$

where $\alpha\in[0,1]$. Now take $x=\hat{k}_{\lambda}$, a normalized kernel
function for $\lambda\in\Omega$, we have from the inequality (4.4)

$|\langle
A\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle|^{p}{(1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}})}<\frac{1}{2}(\frac{p}{p-1})^{p}\langle(|A|^{2p\alpha}+|A^{*}|^{2p(1-\alpha)})\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle$

From the definition of Berezin number inequality, we obtain

$ber^{p}(A){(2+2^{p}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}})}<(\frac{p}{p-1})^{p}\left(ber(|A|^{2p\alpha})+ber(|A^{*}|^{2p(1-\alpha)})\right).$

To summarize the above, we have the following Hardy-type Berezin number
inequality:

######  Theorem 4.11.

Let $p>1$ be any real number and $A$ be any bounded operator on a reproducing
kernel Hilbert space $H(\Omega)$. Then the following inequality

$ber^{p}(A){(2+2^{p}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}})}<(\frac{p}{p-1})^{p}\left(ber(|A|^{2p\alpha})+ber(|A^{*}|^{2p(1-\alpha)})\right)~{}\mbox{holds}$

for $\alpha\in[0,1]$.

######  Proposition 4.12.

Let $p>1$ be any real number. Suppose that $f:\mathbb{R}\rightarrow(0,\infty)$
is a continuous and convex function and $A$ is a self adjoint operator on a
reproducing kernel Hilbert space $H(\Omega)$. Then

(4.5)
$\displaystyle{}f^{p}(berA)\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}$
$\displaystyle<\Big{(}\frac{p}{p-1}\Big{)}^{p}ber(f^{p}(A)).$

In particular,

(4.6) $\displaystyle{}f^{2}(berA)<\frac{24}{2\pi^{2}-9}ber(f^{2}(A)).$

###### Proof.

Suppose that $p>1$ is any real number and $\\{a_{n}\\}$ is a sequence of non-
negative real numbers such that $a_{n}=0$ for $n>2$. Then from the Hardy’s
inequality and replacing $a_{1},a_{2}$ by $f(t)$ for $t\in\mathbb{R}$, we get

$\displaystyle
f^{p}(t)\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}$
$\displaystyle<\Big{(}\frac{p}{p-1}\Big{)}^{p}f^{p}(t).$

Since $A$ is a self adjoint operator on $H(\Omega)$, $\langle
A\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle\in\mathbb{R}$, where
$\hat{k}_{\lambda}$ for $\lambda\in\Omega$ is the normalized kernel vector.
Now choose $t=\langle
A\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle=\widetilde{A}(\lambda)$ in the
above inequality and using the convexity of $f$, we have

$f^{p}(\widetilde{A}(\lambda))\leq\widetilde{f^{p}(A)}(\lambda)$

which yields

$\displaystyle
f^{p}(\widetilde{A}(\lambda))\Big{(}1+2^{p-1}\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^{p}}\Big{)}$
$\displaystyle<\Big{(}\frac{p}{p-1}\Big{)}^{p}\widetilde{f^{p}(A)}(\lambda).$

This proves the desired inequality.

In particular for $p=2$, we obtain

$\displaystyle f^{2}(berA)<\frac{24}{2\pi^{2}-9}ber(f^{2}(A)).$

######  Remark 4.13.

In [27] (Corollary 2.2) and [28] (Theorem 2.1 (ii)), the authors obtained
following inequalities for convex function:

$\displaystyle
f^{2}(berA)<\Big{(}\frac{12}{7}\pi-\frac{3}{14}\Big{)}ber(f^{2}(A)).$

and

$\displaystyle f^{2}(berA)<\frac{15}{4}ber(f^{2}(A)),$

respectively.

######  Remark 4.14.

Using the improved discrete Hardy inequality (2.3) and Proposition 4.10, we
can obtain improved Hardy-type Berezin number inequalities (4.5) and (4.6).

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