Reversed and Refined Berezin Number Inequalities via Buzano’s Inequality and Polar Decomposition

Abstract

The Berezin number and Berezin norm have become indispensable tools for analyzing the structure of operators on reproducing kernel Hilbert spaces, yet the gap between these quantities and the operator norm is still far from fully understood. In this paper, we narrow that gap by establishing several new reverse-type inequalities for $\mathrm{ber}\left(V\right)$ and ${\parallel V\parallel }_{\widetilde{\mathrm{ber}}}$ under natural disc-localization and accretivity conditions. Combining a refinement of the Schwarz inequality due to Dragomir with Buzano’s inequality, we further derive sharp bounds for $\mathrm{ber}\left(A^{*}TA\right)$ and for products $TA$, together with their power-type extensions. Applying the polar decomposition, we also obtain companion inequalities for ${\left|T\right|}^{\alpha }$ and the Aluthge transform. Finally, for operators with closed range, we use the Moore–Penrose inverse to derive new reverse-type Berezin inequalities that refine and complement the results obtained via the polar decomposition, and we illustrate the sharpness of these bounds with an explicit example.

1. Introduction

The numerical radius $w\left(A\right)$ of an operator A acting on a Hilbert space H is defined by

$$w\left(A\right)=\underset{∥x∥=1}{\sup }\left|⟨Ax,x⟩\right|.$$

It turns out that $w\left(·\right)$ defines a norm on the Banach algebra $B\left(H\right)$ of all bounded linear operators $A:H\to H.$

This norm is equivalent to the operator norm, as the following double inequality illustrates:

$$w\left(A\right)\le ∥A∥\le 2w\left(A\right)$$

(1)

for any $A\in B\left(H\right)$.

Reproducing kernel Hilbert spaces play a key role as a central tool across a range of disciplines, including machine learning and statistics, and they arise naturally in complex analysis, group representation theory, integral operator theory, and probability theory as well.

We now fix the notation and terminology used throughout the paper. Let $\Omega$ be a non-empty set and $\mathcal{F}(\Omega )$ denote the set of all functions from $\Omega$ to the complex field $\mathbb{C}$. A set ${\mathcal{H}}_{\Omega }\subseteq \mathcal{F}(\Omega )$ is called a reproducing kernel Hilbert space (RKHS for short) on $\Omega$ if ${\mathcal{H}}_{\Omega }$ is a Hilbert space and for every $\mu \in \Omega$, the map $E_{\mu }:{\mathcal{H}}_{\mu }\to \mathbb{C}$ is bounded, where $E_{\mu }\left(f\right)=f\left(\mu \right)$ for all $f\in {\mathcal{H}}_{\Omega }$. Let $⟨·,·⟩$ denote the inner product of ${\mathcal{H}}_{\Omega }$. When no confusion can arise, we write H for ${\mathcal{H}}_{\Omega }$ in statements about vectors, and I for the identity operator on ${\mathcal{H}}_{\Omega }$. The Riesz representation theorem guarantees that, for each $\mu \in \Omega$, there exists a unique vector $k_{\mu }\in {\mathcal{H}}_{\Omega }$ such that $h\left(\mu \right)=E_{\mu }\left(h\right)=⟨h,k_{\mu }⟩$ for every $h\in {\mathcal{H}}_{\Omega }$. The associated function $k:\Omega \times \Omega \to \mathbb{C}$ given by

$$k(z,\mu )=k_{\mu }\left(z\right)=⟨k_{\mu },k_z⟩,$$

is referred to as the kernel function of the RKHS $({\mathcal{H}}_{\Omega },⟨·,·⟩)$.

For $\mu \in \Omega$, let ${\widehat{k}}_{\mu }={\displaystyle \frac{k_{\mu }}{\parallel k_{\mu }\parallel }}$ be the normalized reproducing kernel of ${\mathcal{H}}_{\Omega }$, where $\parallel ·\parallel$ is the norm associated with the inner product defined on ${\mathcal{H}}_{\Omega }$. Notice that the following set {${\widehat{k}}_{\mu };\mu \in \Omega \}$ is a total set in ${\mathcal{H}}_{\Omega }$. For any bounded linear operator $V\in B\left({\mathcal{H}}_{\Omega }\right)$, the function $\tilde{V}$ defined on $\Omega$ by $\tilde{V}\left(\mu \right)=⟨V{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩$ is the Berezin symbol of V, which was first introduced by Berezin [1,2]. The Berezin transform of an operator provides important structural information about that operator. For more details, the reader may consult, for instance, [3] and its references or [4]. For some applications related to the Berezin symbol of operators, the interested reader is invited to consult, for example, the following references [5,6,7,8].

We recall that the Berezin number and the Berezin norm of the operator V are defined by

$$\mathrm{ber}\left(V\right):=\underset{\mu \in \Omega }{\sup }|\tilde{V}\left(\mu \right)|=\underset{\mu \in \Omega }{\sup }|⟨V{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩|$$

and

$${\parallel V\parallel }_{\mathrm{ber}}:=\sup \left\{|⟨V{\widehat{k}}_{\lambda },{\widehat{k}}_{\mu }⟩|;\lambda ,\mu \in \Omega \right\},$$

where ${\widehat{k}}_{\lambda }$, ${\widehat{k}}_{\mu }$ are two normalized reproducing kernels of the space ${\mathcal{H}}_{\Omega }$ (see [9] and references therein).

In recent years, several inequalities involving the Berezin number and Berezin norm of operators have been established (see, for instance, [10,11,12,13] and the references therein).

For $V\in B\left({\mathcal{H}}_{\Omega }\right)$, we can consider, see for example [14], the following quantity:

$${\parallel V\parallel }_{\widetilde{\mathrm{ber}}}:=\underset{\mu \in \Omega }{\sup }\parallel V{\widehat{k}}_{\mu }\parallel .$$

It is not difficult to see that

$$\mathrm{ber}\left(V\right)\le {\parallel V\parallel }_{\mathrm{ber}}\le {\parallel V\parallel }_{\widetilde{\mathrm{ber}}}\le \parallel V\parallel ,\forall V\in B\left({\mathcal{H}}_{\Omega }\right),$$

(2)

where $\parallel V\parallel$ stands for the usual operator norm of V. It is well known that the inequalities (2) are in general strict.

It is well-known that the operator norm on Hilbert spaces is submultiplicative. However, this property may not be correct in general for the Berezin norm of operators even if it is for positive operators; see [10,14].

By the definitions of $\mathrm{ber}(·)$ and ${\parallel ·\parallel }_{\mathrm{ber}}$, the following properties

$$\mathrm{ber}\left(V\right)=\mathrm{ber}\left(V^{*}\right){\mathrm{and}\parallel V\parallel }_{\mathrm{ber}}={\parallel V^{*}\parallel }_{\mathrm{ber}},$$

hold immediately for every $V\in B\left({\mathcal{H}}_{\Omega }\right)$. Here $V^{*}$ denotes the adjoint of V. It should be mentioned here that the equality ${∥V∥}_{\widetilde{\mathrm{ber}}}={∥V^{*}∥}_{\widetilde{\mathrm{ber}}}$ fails to be true for some $V\in B\left({\mathcal{H}}_{\Omega }\right)$. We also observe that $\mathrm{ber}\left(V\right)\le w\left(V\right)$.

It is also known that $w\left(V^n\right)\le w^n\left(V\right)$ for every positive integer n and the equality $w\left(V^n\right)=w^n\left(V\right)$ holds if V is a selfadjoint operator, i.e., $V^{*}=V$, (or in particular if V is positive). However, the above two properties do not hold, in general, for the Berezin number of operators.

For an account of the results involving the numerical range and radius of operators, we refer the reader to [13,15,16,17] and references therein.

It has been recently proven in [10] that if $P\ge 0$, then

$$\mathrm{ber}\left(P\right)={\parallel P\parallel }_{\mathrm{ber}},$$

(3)

however this equality does not hold, in general, for self-adjoint operators.

We also observe that if $P\ge 0,$ then

$$\mathrm{ber}\left(P\right)=\underset{\mu \in \Omega }{\sup }⟨P{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩=\underset{\mu \in \Omega }{\sup }{∥P^{1/2}{\widehat{k}}_{\mu }∥}^2={∥P^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2.$$

By taking $P={\left|A\right|}^2,$ we also have the equalities

$$\mathrm{ber}\left({\left|A\right|}^2\right)={\parallel {\left|A\right|}^2\parallel }_{\mathrm{ber}}={∥\left|A\right|∥}_{\widetilde{\mathrm{ber}}}^2$$

(4)

for all $A\in B\left({\mathcal{H}}_{\Omega }\right).$

Inequalities of this type have numerical-radius counterparts based on Buzano’s inequality and Schwarz-type refinements [18,19]. Berezin-number and Berezin-norm inequalities have been widely studied [10,20,21], including with Buzano’s inequality [22,23] and the Moore–Penrose inverse [23,24,25]. Our method is of the same kind (a vector inequality applied to the normalized kernels ${\widehat{k}}_{\mu }$, followed by a supremum over $\mu$), but the estimates differ in form: reverse-type bounds for ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)$ under disc localization (Section 2); bounds for $\mathrm{ber}\left(A^{*}TA\right)$ and $TA$ via Dragomir’s Schwarz refinement and Buzano’s inequality (Section 3); results for ${\left|T\right|}^{\alpha }$ and the Aluthge transform via polar decomposition (Section 4); and reverse-type estimates in which $M{M}^{†}$ enters through the condition $\parallel M-\lambda I\parallel \le r$ (Section 5). To the best of our knowledge, these inequalities are new and do not appear elsewhere in the literature.

The rest of the paper is organized as follows. In Section 2, we establish reverse inequalities for $\mathrm{ber}(·)$ and ${\parallel ·\parallel }_{\widetilde{\mathrm{ber}}}$ under disc-localization conditions of type (5). In Section 3, we combine Dragomir’s refinement of the Schwarz inequality with Buzano’s inequality to obtain bounds for $\mathrm{ber}\left(A^{*}TA\right)$ and for products $TA$, together with power-type extensions. In Section 4, we use the polar decomposition to derive companion inequalities for ${\left|T\right|}^{\alpha }$ and the Aluthge transform. Finally, in Section 5 we obtain reverse Berezin inequalities that involve the Moore–Penrose inverse of an operator. Section 6 contains concluding remarks.

2. Reverse Berezin Inequalities via Disc-Localization

As shown in (2), the Berezin number $\mathrm{ber}(·)$ is dominated from above by ${\parallel ·\parallel }_{\widetilde{\mathrm{ber}}}$, but the two quantities do not coincide in general. In this section we obtain reverse-type inequalities that bound the gap ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)$ from above whenever T is localized in a disc of the complex plane. The following results may be stated:

Theorem 1.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right).$ If $\lambda \in \mathbb{C}\setminus \left\{0\right\}$ and $r>0$ are such that

$${∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}\le r,$$

(5)

where I is the identity operator on ${\mathcal{H}}_{\Omega },$ then

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)\le {\displaystyle \frac{1}{2}}{\displaystyle \frac{r^2}{\left|\lambda \right|}}.$$

(6)

Proof. 

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (5) that

$$∥T{\widehat{k}}_{\mu }-\lambda {\widehat{k}}_{\mu }∥\le {∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}\le r,$$

giving that

$${∥T{\widehat{k}}_{\mu }∥}^2+{\left|\lambda \right|}^2\le 2\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]+r^2\le 2\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|+r^2.$$

(7)

Taking the supremum over $\mu \in \Omega$ in (7) we get the following inequality that is of interest in itself:

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2+{\left|\lambda \right|}^2\le 2\mathrm{ber}\left(T\right)\left|\lambda \right|+r^2.$$

(8)

Since, by the elementary inequality $2ab\le a^2+b^2$ for real $a,b$,

$$2{∥T∥}_{\widetilde{\mathrm{ber}}}\left|\lambda \right|\le {∥T∥}_{\widetilde{\mathrm{ber}}}^2+{\left|\lambda \right|}^2,$$

(9)

hence by (8) and (9) we deduce

$$2{∥T∥}_{\widetilde{\mathrm{ber}}}\left|\lambda \right|\le 2\mathrm{ber}\left(T\right)\left|\lambda \right|+r^2,$$

which is equivalent to the desired inequality (6). □

Example 1.

Let ${\mathcal{H}}_{\Omega }=H^2\left(\mathbb{D}\right)$ be the Hardy space of the unit disc $\Omega =\mathbb{D}$, with the orthonormal basis ${\left\{z^n\right\}}_{n\ge 0}$, reproducing kernel

$$k_{\mu }\left(z\right)={\displaystyle \frac{1}{1-\overline{\mu }z}},{\parallel k_{\mu }\parallel }^2={\displaystyle \frac{1}{1-{\left|\mu \right|}^2}},$$

and normalized kernel

$${\widehat{k}}_{\mu }\left(z\right)=\sqrt{1-{\left|\mu \right|}^2}{\displaystyle \sum _{n=0}^{\infty }}{\overline{\mu }}^n{z}^n.$$

Let D be the rank-one operator

$$Df=2⟨f,z⟩z,i.e.,D=2z\otimes z.$$

Since

$$|⟨{\widehat{k}}_{\mu },z⟩{|}^2={(1-|\mu |}^2{\left)\right|\mu |}^2,$$

putting $s={\left|\mu \right|}^2\in [0,1)$ gives

$$⟨D{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩=2s(1-s),{\parallel D{\widehat{k}}_{\mu }\parallel }^2=4s(1-s).$$

Moreover,

$$\begin{array}{cc}\hfill \parallel (D-I){\widehat{k}}_{\mu }{\parallel }^2& =\parallel D{\widehat{k}}_{\mu }{\parallel }^2-2\mathrm{Re}⟨D{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩+{\parallel {\widehat{k}}_{\mu }\parallel }^2\hfill \\ \hfill & =4s(1-s)-4s(1-s)+1=1.\hfill \end{array}$$

Hence

$$\mathrm{ber}\left(D\right)=\underset{0\le s<1}{\max }2s(1-s)={\displaystyle \frac{1}{2}},$$

while

$${\parallel D\parallel }_{\widetilde{\mathrm{ber}}}=\underset{0\le s<1}{\max }2\sqrt{s(1-s)}=1\mathrm{and}{\parallel D-I\parallel }_{\widetilde{\mathrm{ber}}}=1.$$

Therefore, for $\lambda =1$ and $r=1$, inequality (5) is satisfied, and

$${\parallel D\parallel }_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(D\right)={\displaystyle \frac{1}{2}}={\displaystyle \frac{r^2}{2\left|\lambda \right|}},$$

so the equality holds in (6).

The following corollary may be stated:

Corollary 1.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right)$ and $\phi ,\psi \in \mathbb{C}$ with $\psi \notin \left\{-\phi ,\phi \right\}.$ If

$$\mathrm{Re}⟨\psi x-Tx,Tx-\phi x⟩\ge 0\mathrm{for}\mathrm{any}x\in H,∥x∥=1,$$

(10)

then

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)\le {\displaystyle \frac{1}{4}}{\displaystyle \frac{{\left|\psi -\phi \right|}^2}{\left|\psi +\phi \right|}}.$$

(11)

Proof. 

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

(i)

$\mathrm{Re}⟨u-x,x-z⟩\ge 0,$ $x,z,u\in H;$

(ii)

$∥x-{\displaystyle \frac{z+u}{2}}∥\le {\displaystyle \frac{1}{2}}∥u-z∥,$

We deduce that (10) is equivalent to

$$∥Tx-{\displaystyle \frac{\psi +\phi }{2}}·Ix∥\le {\displaystyle \frac{1}{2}}\left|\psi -\phi \right|$$

(12)

for any $x\in H$, $∥x∥=1,$ which in its turn is equivalent to the operator norm inequality:

$$∥T-{\displaystyle \frac{\psi +\phi }{2}}·I∥\le {\displaystyle \frac{1}{2}}\left|\psi -\phi \right|.$$

(13)

Since ${\parallel V\parallel }_{\widetilde{\mathrm{ber}}}\le \parallel V\parallel ,$ for any $V\in B\left({\mathcal{H}}_{\Omega }\right),$ then

$${∥T-{\displaystyle \frac{\psi +\phi }{2}}·I∥}_{\widetilde{\mathrm{ber}}}\le {\displaystyle \frac{1}{2}}\left|\psi -\phi \right|.$$

Now, applying Theorem 1 for $\lambda ={\displaystyle \frac{\phi +\psi }{2}}$ and $r={\displaystyle \frac{1}{2}}\left|\phi -\psi \right|,$ we deduce the desired result (11). □

Remark 1.

We say that an operator $B:H\to H$ is accretive, if $\mathrm{Re}⟨Bx,x⟩\ge 0$ for every $x\in H.$ One may observe that the Assumption (10) above is then equivalent to the fact that the operator $\left(T^{*}-\overline{\phi }I\right)\left(\psi I-T\right)$ is accretive.

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

Corollary 2.

Let $\phi ,\psi \in \mathbb{C}$ with $\psi \notin \left\{-\phi ,\phi \right\}$ and $T\in B\left({\mathcal{H}}_{\Omega }\right).$ If $\left(T^{*}-\overline{\phi }I\right)\left(\psi I-T\right)$ is self-adjoint and

$$\left(T^{*}-\overline{\phi }I\right)\left(\psi I-T\right)\ge 0$$

(14)

in the operator order, then

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)\le {\displaystyle \frac{1}{4}}{\displaystyle \frac{{\left|\psi -\phi \right|}^2}{\left|\psi +\phi \right|}}.$$

(15)

Corollary 3.

Assume that $T,\lambda ,r$ are as in Theorem 1. If, in addition,

$$\left|\left|\lambda \right|-\mathrm{ber}\left(T\right)\right|\ge \rho ,$$

(16)

for some $\rho \ge 0,$ then

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le r^2-{\rho }^2.$$

(17)

Proof. 

From (8) of Theorem 1, we have

$$\begin{array}{cc}\hfill {∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2& \le r^2-{\left[\mathrm{ber}\left(T\right)\right]}^2+2\mathrm{ber}\left(T\right)\left|\lambda \right|-{\left|\lambda \right|}^2\hfill \\ \hfill & =r^2-{\left(\left|\lambda \right|-\mathrm{ber}\left(T\right)\right)}^2.\hfill \end{array}$$

(18)

The desired inequality follows from (16). □

Remark 2.

In particular, if ${∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}\le r$, and if we take $\left|\lambda \right|=\mathrm{ber}\left(T\right)$, $\lambda \in \mathbb{C}$, then we get

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le r^2.$$

(19)

The following result may be stated as well.

Theorem 2.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right)$ and $\lambda \in \mathbb{C}\setminus \left\{0\right\},$ $r>0$ with $\left|\lambda \right|>r.$ If

$${∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}\le r,$$

(20)

then

$$\sqrt{1-{\displaystyle \frac{r^2}{{\left|\lambda \right|}^2}}}\le {\displaystyle \frac{\mathrm{ber}\left(T\right)}{{∥T∥}_{\widetilde{\mathrm{ber}}}}}\left(\le 1\right).$$

(21)

Proof. 

From (8) of Theorem 1, we have

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2+{\left|\lambda \right|}^2-r^2\le 2\left|\lambda \right|\mathrm{ber}\left(T\right),$$

which implies, on dividing with $\sqrt{{\left|\lambda \right|}^2-r^2}>0$, that

$${\displaystyle \frac{{∥T∥}_{\widetilde{\mathrm{ber}}}^2}{\sqrt{{\left|\lambda \right|}^2-r^2}}}+\sqrt{{\left|\lambda \right|}^2-r^2}\le {\displaystyle \frac{2\left|\lambda \right|\mathrm{ber}\left(T\right)}{\sqrt{{\left|\lambda \right|}^2-r^2}}}.$$

(22)

By the elementary inequality

$$2{∥T∥}_{\widetilde{\mathrm{ber}}}\le {\displaystyle \frac{{∥T∥}_{\widetilde{\mathrm{ber}}}^2}{\sqrt{{\left|\lambda \right|}^2-r^2}}}+\sqrt{{\left|\lambda \right|}^2-r^2}$$

(23)

and by (22) we deduce that

$${∥T∥}_{\widetilde{\mathrm{ber}}}\le {\displaystyle \frac{\mathrm{ber}\left(T\right)\left|\lambda \right|}{\sqrt{{\left|\lambda \right|}^2-r^2}}},$$

which is equivalent to (21). □

Remark 3.

Squaring (21), we get the inequality

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le {\displaystyle \frac{r^2}{{\left|\lambda \right|}^2}}{∥T∥}_{\widetilde{\mathrm{ber}}}^2.$$

(24)

The following corollary holds.

Corollary 4.

Let $\phi ,\psi \in \mathbb{C}$ with $\mathrm{Re}\left(\psi \overline{\phi }\right)>0.$ If $T\in B\left({\mathcal{H}}_{\Omega }\right)$ is such that either (10) or (14) holds true, then:

$${\displaystyle \frac{2\sqrt{\mathrm{Re}\left(\psi \overline{\phi }\right)}}{\left|\psi +\phi \right|}}\le {\displaystyle \frac{\mathrm{ber}\left(T\right)}{{∥T∥}_{\widetilde{\mathrm{ber}}}}}\left(\le 1\right)$$

(25)

and

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le {\left|{\displaystyle \frac{\psi -\phi }{\psi +\phi }}\right|}^2{∥T∥}_{\widetilde{\mathrm{ber}}}^2.$$

(26)

Proof. 

If we consider $\lambda ={\displaystyle \frac{\psi +\phi }{2}}$ and $r={\displaystyle \frac{1}{2}}\left|\psi -\phi \right|,$ then

$${\left|\lambda \right|}^2-r^2={\left|{\displaystyle \frac{\psi +\phi }{2}}\right|}^2-{\left|{\displaystyle \frac{\psi -\phi }{2}}\right|}^2=\mathrm{Re}\left(\psi \overline{\phi }\right)>0.$$

Now, on applying Theorem 2, we deduce the desired result. □

The following result may be of interest as well.

Theorem 3.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right)$ and $\lambda \in \mathbb{C}\setminus \left\{0\right\}$, $r>0$ with $\left|\lambda \right|>r.$ If

$${∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}\le r,$$

(27)

then

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le {\displaystyle \frac{2r^2}{\left|\lambda \right|+\sqrt{{\left|\lambda \right|}^2-r^2}}}\mathrm{ber}\left(T\right).$$

(28)

Proof. 

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ From the proof of Theorem 1, we have that

$${∥T{\widehat{k}}_{\mu }∥}^2+{\left|\lambda \right|}^2\le 2\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]+r^2$$

(29)

for $\mu \in \Omega$.

If we divide (29) by $\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|$ (which is positive, since $\left|\lambda \right|>r$, (29) gives $2\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]\ge {\left|\lambda \right|}^2-r^2>0$, hence $⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\ne 0$) then we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{∥T{\widehat{k}}_{\mu }∥}^2}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}\hfill \\ \hfill & \le {\displaystyle \frac{2\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}+{\displaystyle \frac{r^2}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\displaystyle \frac{\left|\lambda \right|}{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}\hfill \end{array}$$

(30)

for $\mu \in \Omega$.

If we subtract in (30) the same quantity ${\displaystyle \frac{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}{\left|\lambda \right|}}$ from both sides, then we get

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{∥T{\widehat{k}}_{\mu }∥}^2}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\displaystyle \frac{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}{\left|\lambda \right|}}\hfill \\ \hfill & \le {\displaystyle \frac{2\mathrm{Re}\left[\overline{\lambda }⟨{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}+{\displaystyle \frac{r^2}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\displaystyle \frac{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}{\left|\lambda \right|}}-{\displaystyle \frac{\left|\lambda \right|}{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}\hfill \\ \hfill & ={\displaystyle \frac{2\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\displaystyle \frac{{\left|\lambda \right|}^2-r^2}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\displaystyle \frac{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}{\left|\lambda \right|}}\hfill \\ \hfill & ={\displaystyle \frac{2\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\left({\displaystyle \frac{\sqrt{{\left|\lambda \right|}^2-r^2}}{\sqrt{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}}-{\displaystyle \frac{\sqrt{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}{\sqrt{\left|\lambda \right|}}}\right)}^2\hfill \\ \hfill & -2{\displaystyle \frac{\sqrt{{\left|\lambda \right|}^2-r^2}}{\left|\lambda \right|}}\hfill \end{array}$$

(31)

for $\mu \in \Omega$.

Since

$$\mathrm{Re}\left[\overline{\lambda }⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right]\le \left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|$$

and

$${\left({\displaystyle \frac{\sqrt{{\left|\lambda \right|}^2-r^2}}{\sqrt{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}}-{\displaystyle \frac{\sqrt{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}{\sqrt{\left|\lambda \right|}}}\right)}^2\ge 0,$$

by (31) we get

$${\displaystyle \frac{{∥T{\widehat{k}}_{\mu }∥}^2}{\left|\lambda \right|\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}}-{\displaystyle \frac{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}{\left|\lambda \right|}}\le {\displaystyle \frac{2\left(\left|\lambda \right|-\sqrt{{\left|\lambda \right|}^2-r^2}\right)}{\left|\lambda \right|}},$$

which gives the inequality

$${∥T{\widehat{k}}_{\mu }∥}^2\le {\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2+2\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|\left(\left|\lambda \right|-\sqrt{{\left|\lambda \right|}^2-r^2}\right)$$

(32)

for $\mu \in \Omega .$

Taking the supremum over $\mu \in \Omega ,$ we get

$$\begin{array}{cc}\hfill {∥T∥}_{\widetilde{\mathrm{ber}}}^2& \le \sup \left\{{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2+2\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|\left(\left|\lambda \right|-\sqrt{{\left|\lambda \right|}^2-r^2}\right)\right\}\hfill \\ \hfill & \le \sup \left\{{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2\right\}+2\left(\left|\lambda \right|-\sqrt{{\left|\lambda \right|}^2-r^2}\right)\sup \left\{\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|\right\}\hfill \\ \hfill & ={\left[\mathrm{ber}\left(T\right)\right]}^2+2\left(\left|\lambda \right|-\sqrt{{\left|\lambda \right|}^2-r^2}\right)\mathrm{ber}\left(T\right),\hfill \end{array}$$

which is equivalent to (28), because $2\left(\left|\lambda \right|-\sqrt{{\left|\lambda \right|}^2-r^2}\right)=2r^2/\left(\left|\lambda \right|+\sqrt{{\left|\lambda \right|}^2-r^2}\right)$. □

Corollary 5.

Let $\phi ,\psi \in \mathbb{C}$ with $\mathrm{Re}\left(\psi \overline{\phi }\right)>0.$ If $T\in B\left({\mathcal{H}}_{\Omega }\right)$ is such that either (10) or (14) hold true, then:

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le \left[\left|\psi +\phi \right|-2\sqrt{\mathrm{Re}\left(\psi \overline{\phi }\right)}\right]\mathrm{ber}\left(T\right).$$

(33)

Remark 4.

If $M\ge m>0$ are such that either $\left(T^{*}-mI\right)\left(MI-T\right)$ is accretive, or, sufficiently, $\left(T^{*}-mI\right)\left(MI-T\right)$ is self-adjoint and

$$\left(T^{*}-mI\right)\left(MI-T\right)\ge 0\mathit{in}\mathit{the}\mathit{operator}\mathrm{order},$$

(34)

then, by (25) we have:

$$\left(1\le \right){\displaystyle \frac{{∥T∥}_{\widetilde{\mathrm{ber}}}}{\mathrm{ber}\left(T\right)}}\le {\displaystyle \frac{M+m}{2\sqrt{mM}}},$$

(35)

which is equivalent to

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)\le {\displaystyle \frac{{\left(\sqrt{M}-\sqrt{m}\right)}^2}{2\sqrt{mM}}}\mathrm{ber}\left(T\right),$$

(36)

while from (28) we have

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left(\mathrm{ber}\left(T\right)\right)}^2\le {\left(\sqrt{M}-\sqrt{m}\right)}^2\mathrm{ber}\left(T\right).$$

(37)

Also, the inequality (11) becomes

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)\le {\displaystyle \frac{1}{4}}{\displaystyle \frac{{\left(M-m\right)}^2}{M+m}}.$$

(38)

3. Berezin Inequalities via Buzano’s Inequality

We now turn from the disc-localization approach of Section 2 to a different technique, based on Buzano’s inequality, which allows us to control $\mathrm{ber}\left(A^{*}TA\right)$ and the Berezin norm of products $TA$ and $T^{*}A$ without any localization assumption on T. By making use of the celebrated Buzano’s inequality, we can state the following result as well:

Theorem 4.

Let $A,T\in B\left({\mathcal{H}}_{\Omega }\right)$. Then

$${\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2\le {\displaystyle \frac{1}{2}}\left({∥TA∥}_{\widetilde{\mathrm{ber}}}{∥T^{*}A∥}_{\widetilde{\mathrm{ber}}}+\mathrm{ber}\left(A^{*}{T}^{2}A\right)\right){∥A∥}_{\widetilde{\mathrm{ber}}}^2.$$

(39)

Proof. 

We need the following refinement of Schwarz’s inequality obtained by S.S. Dragomir in 1985 [26] [Theorem 2] (see also [27,28]):

$$∥a∥∥b∥\ge \left|⟨a,b⟩-⟨a,e⟩⟨e,b⟩\right|+\left|⟨a,e⟩⟨e,b⟩\right|\ge \left|⟨a,b⟩\right|,$$

(40)

provided $a,b,e$ are vectors in H and $∥e∥=1.$

Observing that

$$\left|⟨a,b⟩-⟨a,e⟩⟨e,b⟩\right|\ge \left|⟨a,e⟩⟨e,b⟩\right|-\left|⟨a,b⟩\right|,$$

hence by the first inequality in (40) we deduce

$${\displaystyle \frac{1}{2}}\left(∥a∥∥b∥+\left|⟨a,b⟩\right|\right)\ge \left|⟨a,e⟩⟨e,b⟩\right|.$$

(41)

This inequality was obtained in a different way earlier by M. L. Buzano in [29].

If we replace e with $y/∥y∥$ for $y\ne 0$ we get

$${\displaystyle \frac{1}{2}}\left(∥a∥∥b∥+\left|⟨a,b⟩\right|\right){∥y∥}^2\ge \left|⟨a,y⟩⟨y,b⟩\right|.$$

Now, in (41), choose $y=x$, $a=Tx$ and $b=T^{*}x$ to get

$${\displaystyle \frac{1}{2}}\left(∥Tx∥∥T^{*}x∥+\left|⟨T^{2}x,x⟩\right|\right){∥x∥}^2\ge {\left|⟨Tx,x⟩\right|}^2$$

(42)

for any $x\in H.$

Moreover, if we replace x with $Az$, $z\in H,$ then we get

$${\left|⟨A^{*}TAz,z⟩\right|}^2\le {\displaystyle \frac{1}{2}}\left(∥TAz∥∥T^{*}Az∥+\left|⟨A^{*}{T}^{2}Az,z⟩\right|\right){∥Az∥}^2$$

(43)

for all $z\in H.$

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (43) that

$${\left|⟨A^{*}TA{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2\le {\displaystyle \frac{1}{2}}\left(∥TA{\widehat{k}}_{\mu }∥∥T^{*}A{\widehat{k}}_{\mu }∥+\left|⟨A^{*}{T}^{2}A{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|\right){∥A{\widehat{k}}_{\mu }∥}^2$$

for $\mu \in \Omega .$

By taking the supremum over $\mu \in \Omega ,$ and using,

$$\underset{\mu }{\sup }\left(fg\right)\le \underset{\mu }{\sup }f\underset{\mu }{\sup }g$$

and

$$\underset{\mu }{\sup }(f+g)\le \underset{\mu }{\sup }f+\underset{\mu }{\sup }g$$

for nonnegative functions $f,g$, we get

$$\begin{array}{cc}\hfill & {\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2\hfill \\ \hfill & =\underset{\mu \in \Omega }{\sup }{\left|⟨A^{*}TA{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\underset{\mu \in \Omega }{\sup }\left[\left(∥TA{\widehat{k}}_{\mu }∥∥T^{*}A{\widehat{k}}_{\mu }∥+\left|⟨A^{*}{T}^{2}A{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|\right){∥A{\widehat{k}}_{\mu }∥}^2\right]\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left(\underset{\mu \in \Omega }{\sup }∥TA{\widehat{k}}_{\mu }∥\underset{\mu \in \Omega }{\sup }∥T^{*}A{\widehat{k}}_{\mu }∥+\underset{\mu \in \Omega }{\sup }\left|⟨A^{*}{T}^{2}A{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|\right)\hfill \\ \hfill & \times \underset{\mu \in \Omega }{\sup }{∥A{\widehat{k}}_{\mu }∥}^2\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left({∥TA∥}_{\widetilde{\mathrm{ber}}}{∥T^{*}A∥}_{\widetilde{\mathrm{ber}}}+\mathrm{ber}\left(A^{*}{T}^{2}A\right)\right){∥A∥}_{\widetilde{\mathrm{ber}}}^2,\hfill \end{array}$$

which gives the desired result (39). □

Corollary 6.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right)$. Then

$${\left[\mathrm{ber}\left(T\right)\right]}^2\le {\displaystyle \frac{1}{2}}({∥T∥}_{\widetilde{\mathrm{ber}}}{∥T^{*}∥}_{\widetilde{\mathrm{ber}}}+\mathrm{ber}(T^2\left)\right).$$

(44)

The proof follows by (39) on taking $A=I$.

The following three examples show that the bound (44) in Corollary 6 can be attained with equality, and that it can also be strict.

Example 2.

In Example 1, D is self-adjoint, $D^2=4z\otimes z$ and $\mathrm{ber}\left(D^2\right)={\sup }_{s}4s(1-s)=1$; therefore

$${[ber\left(D\right)]}^2=\frac{1}{4}<1=\frac{1}{2}\left({\parallel D\parallel }_{\widetilde{\mathrm{ber}}}{\parallel D^{*}\parallel }_{\widetilde{\mathrm{ber}}}+\mathrm{ber}\left(D^2\right)\right),$$

i.e., the inequality of Corollary 6 is strict for this operator.

Example 3.

Let ${\mathbb{H}}_{\Omega }={\mathbb{C}}^2$ with the standard inner product and $\Omega =\{1,2\}$, so the normalized reproducing kernels are $e_1,e_2$. For $a,b\in \mathbb{R}$, let

$$T=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right).$$

Since T is self-adjoint, $T^{*}=T$, and $T^2=\mathrm{diag}(a^2,b^2)$. Because T acts diagonally on the reproducing-kernel basis we have

$$\mathrm{ber}\left(T\right)=\max \left\{\right|a|,|b\left|\right\},$$

$${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}=\parallel T^{*}{\parallel }_{\widetilde{\mathrm{ber}}}=\max \left\{\right|a|,|b\left|\right\},$$

and

$$\mathrm{ber}\left(T^2\right)=\max \{a^2,b^2\}={\left(\max \left\{\right|a|,|b\left|\right\}\right)}^2.$$

Hence the right-hand side of (44) equals

$$\frac{1}{2}\left(\max {\left\{\right|a|,|b\left|\right\}}^2+\max {\left\{\right|a|,|b\left|\right\}}^2\right)=\max {\left\{\right|a|,|b\left|\right\}}^2={\left[\mathrm{ber}\left(T\right)\right]}^2,$$

so equality holds for every diagonal self-adjoint T. Taking $a=2,b=1$ gives the concrete instance ${\left[\mathrm{ber}\left(T\right)\right]}^2=4=\frac{1}{2}(2·2+4)$.

Example 4.

On the same space, let

$$T=\left(\begin{array}{cc}1& 2\\ 0& 3\end{array}\right).$$

Here $\mathrm{ber}\left(T\right)=\max \{1,3\}=3$, so ${\left[\mathrm{ber}\left(T\right)\right]}^2=9$. A direct computation gives $T^2=\left(\begin{array}{cc}1& 8\\ 0& 9\end{array}\right)$, so $\mathrm{ber}\left(T^2\right)=9$. Since $T^{*}=\left(\begin{array}{cc}1& 0\\ 2& 3\end{array}\right)$, the columns of T have norms 1 and $\sqrt{13}$, and those of $T^{*}$ have norms $\sqrt{5}$ and 3; thus, ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}=\sqrt{13}$ and $\parallel T^{*}{\parallel }_{\widetilde{\mathrm{ber}}}=3$. The right-hand side of (44) is $\frac{1}{2}(3\sqrt{13}+9)\approx 9.908$, while the left-hand side is 9. This non-self-adjoint example shows the bound is strict in general.

We next establish a power-type version of the preceding inequalities:

Theorem 5.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right)$ and $\lambda \in \mathbb{C}\setminus \left\{0\right\}.$ If $∥T∥\le \left|\lambda \right|,$ then

$${∥T∥}_{\widetilde{\mathrm{ber}}}^{2s}+{\left|\lambda \right|}^{2s}\le 2{\left|\lambda \right|}^s{∥T∥}_{\widetilde{\mathrm{ber}}}^{s-1}\mathrm{ber}\left(T\right)+s^2{\left|\lambda \right|}^{2s-2}{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2,$$

(45)

where $s\ge 1.$

Also,

$${∥T∥}_{\widetilde{\mathrm{ber}}}^s\le {∥T∥}_{\widetilde{\mathrm{ber}}}^{s-1}\mathrm{ber}\left(T\right)+{\displaystyle \frac{1}{2}}{s}^2{\left|\lambda \right|}^{s-2}{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2.$$

(46)

Proof. 

We use the following inequality for vectors in inner product spaces due to Goldstein, Ryff and Clarke [30]:

$$\begin{array}{cc}\hfill & {∥a∥}^{2s}+{∥b∥}^{2s}-2{∥a∥}^s{∥b∥}^s{\displaystyle \frac{\mathrm{Re}⟨a,b⟩}{∥a∥∥b∥}}\hfill \\ \hfill & \le \left\{\begin{array}{cc}{s}^2{∥a∥}^{2s-2}{∥a-b∥}^2\hfill & \mathrm{if}s\ge 1,\hfill \\ \hfill & \\ {∥b∥}^{2s-2}{∥a-b∥}^2\hfill & \hfill \mathrm{if}s<1,\end{array}\right.\hfill \end{array}$$

(47)

provided $s\in \mathbb{R}$ and $a,b\in H$ with $∥a∥\ge ∥b∥.$

Now, let $x\in H$ with $∥x∥=1.$ From the hypothesis of the theorem, we have that $∥Tx∥\le \left|\lambda \right|∥x∥$, and applying (47) for the choices $a=\lambda x$, $∥x∥=1$, $b=Tx$, we get

$${∥Tx∥}^{2s}+{\left|\lambda \right|}^{2s}-2{∥Tx∥}^{s-1}{\left|\lambda \right|}^s\left|⟨Tx,x⟩\right|\le s^2{\left|\lambda \right|}^{2s-2}{∥Tx-\lambda x∥}^2$$

(48)

for any $x\in H$, $∥x∥=1$ and $s\ge 1.$

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (48) that

$$\begin{array}{cc}\hfill & {∥T{\widehat{k}}_{\mu }∥}^{2s}+{\left|\lambda \right|}^{2s}\hfill \\ \hfill & \le 2{∥T{\widehat{k}}_{\mu }∥}^{s-1}{\left|\lambda \right|}^s\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|+s^2{\left|\lambda \right|}^{2s-2}{∥T{\widehat{k}}_{\mu }-\lambda {\widehat{k}}_{\mu }∥}^2.\hfill \end{array}$$

Taking the supremum over $\mu \in \Omega$, we get

$$\begin{array}{cc}\hfill & {∥T∥}_{\widetilde{\mathrm{ber}}}^{2s}+{\left|\lambda \right|}^{2s}\hfill \\ \hfill & =\underset{\mu \in \Omega }{\sup }\left({∥T{\widehat{k}}_{\mu }∥}^{2s}+{\left|\lambda \right|}^{2s}\right)\hfill \\ \hfill & \le \underset{\mu \in \Omega }{\sup }\left[2{∥T{\widehat{k}}_{\mu }∥}^{s-1}{\left|\lambda \right|}^s\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|+s^2{\left|\lambda \right|}^{2s-2}{∥T{\widehat{k}}_{\mu }-\lambda {\widehat{k}}_{\mu }∥}^2\right]\hfill \\ \hfill & \le 2{\left|\lambda \right|}^s\underset{\mu \in \Omega }{\sup }{∥T{\widehat{k}}_{\mu }∥}^{s-1}\underset{\mu \in \Omega }{\sup }\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|+s^2{\left|\lambda \right|}^{2s-2}\underset{\mu \in \Omega }{\sup }{∥T{\widehat{k}}_{\mu }-\lambda {\widehat{k}}_{\mu }∥}^2\hfill \\ \hfill & =2{\left|\lambda \right|}^s{∥T∥}_{\widetilde{\mathrm{ber}}}^{s-1}\mathrm{ber}\left(T\right)+s^2{\left|\lambda \right|}^{2s-2}{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2,\hfill \end{array}$$

which proves (45).

By the elementary inequality $2cd\le c^2+d^2$, $c,d\ge 0,$ we also have that

$$2{\left|\lambda \right|}^s{∥T∥}_{\widetilde{\mathrm{ber}}}^s\le {∥T∥}_{\widetilde{\mathrm{ber}}}^{2s}+{\left|\lambda \right|}^{2s}.$$

By (45) we then get

$$2{\left|\lambda \right|}^s{∥T∥}_{\widetilde{\mathrm{ber}}}^s\le 2{\left|\lambda \right|}^s{∥T∥}_{\widetilde{\mathrm{ber}}}^{s-1}\mathrm{ber}\left(T\right)+s^2{\left|\lambda \right|}^{2s-2}{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2,$$

which, by division with $2{\left|\lambda \right|}^s$, gives the desired result (46). □

Remark 5.

If we take $s=1$ in (46), then we get

$${∥T∥}_{\widetilde{\mathrm{ber}}}\le \mathrm{ber}\left(T\right)+{\displaystyle \frac{1}{2\left|\lambda \right|}}{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2.$$

For $s=2$ we also get

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2\le {∥T∥}_{\widetilde{\mathrm{ber}}}\mathrm{ber}\left(T\right)+2{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2.$$

We observe that, if we take $\lambda =\beta ∥T∥$ with $\beta \in \mathbb{C}\setminus \left\{0\right\}$ and $\left|\beta \right|=1,$ then by (45) we get

$${∥T∥}_{\widetilde{\mathrm{ber}}}^{2s}+{∥T∥}^{2s}\le 2{∥T∥}^s{∥T∥}_{\widetilde{\mathrm{ber}}}^{s-1}\mathrm{ber}\left(T\right)+s^2{∥T∥}^{2s-2}{∥T-\beta ∥T∥I∥}_{\widetilde{\mathrm{ber}}}^2,$$

where $s\ge 1$, while from (46) we get

$${∥T∥}_{\widetilde{\mathrm{ber}}}^s\le {∥T∥}_{\widetilde{\mathrm{ber}}}^{s-1}\mathrm{ber}\left(T\right)+{\displaystyle \frac{1}{2}}{s}^2{∥T∥}^{s-2}{∥T-\beta ∥T∥I∥}_{\widetilde{\mathrm{ber}}}^2.$$

For $s=1$ in this last inequality, we have that

$${∥T∥}_{\widetilde{\mathrm{ber}}}\le \mathrm{ber}\left(T\right)+{\displaystyle \frac{1}{2}}∥T∥{∥{\displaystyle \frac{T}{∥T∥}}-\beta I∥}_{\widetilde{\mathrm{ber}}}^2$$

while for $s=2$, we get

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2\le {∥T∥}_{\widetilde{\mathrm{ber}}}\mathrm{ber}\left(T\right)+2{∥T-\beta ∥T∥I∥}_{\widetilde{\mathrm{ber}}}^2.$$

Moreover, we can state the following bounds for the Berezin norm ${∥TA∥}_{\widetilde{\mathrm{ber}}}:$

Theorem 6.

Let $A,T\in B\left({\mathcal{H}}_{\Omega }\right)$ with ${\left|A\right|}^2\ge mI$ in the operator order and $m>0.$ Then for any $\alpha \in \left[0,1\right]$ and $t\in \mathbb{R}$ one has the inequality:

$$\begin{array}{cc}\hfill {∥TA∥}_{\widetilde{\mathrm{ber}}}^2& \le \left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\displaystyle \frac{{\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2}{m}}\hfill \\ \hfill & +\alpha {∥\left(tI-T\right)A∥}_{\widetilde{\mathrm{ber}}}^2+\left(1-\alpha \right){∥\left(itI-T\right)A∥}_{\widetilde{\mathrm{ber}}}^2.\hfill \end{array}$$

(49)

Unlike its original (unconditional) form, this estimate is no longer valid for an arbitrary $A\in B\left({\mathcal{H}}_{\Omega }\right)$: the lower bound ${\left|A\right|}^2\ge mI$, $m>0$ is essential, since it is precisely what allows ${∥Az∥}^2\ge m{∥z∥}^2$ to control the normalization $x=y/∥y∥$ in the proof below. The unrestricted case $A=I$ remains available as a special case, since one may then take $m=1$.

Proof. 

We use the following inequality obtained by S. S. Dragomir in [28]:

$$\begin{array}{cc}\hfill & \left[\alpha {∥tb-a∥}^2+\left(1-\alpha \right){∥itb-a∥}^2\right]{∥b∥}^2\hfill \\ \hfill & \ge {∥a∥}^2{∥b∥}^2-{\left[\left(1-\alpha \right)\mathrm{Im}⟨a,b⟩+\alpha \mathrm{Re}⟨a,b⟩\right]}^2\left(\ge 0\right)\hfill \end{array}$$

to get:

$$\begin{array}{cc}\hfill {∥a∥}^2{∥b∥}^2& \le {\left[\left(1-\alpha \right)\mathrm{Im}⟨a,b⟩+\alpha \mathrm{Re}⟨a,b⟩\right]}^2\hfill \\ \hfill & +\left[\alpha {∥tb-a∥}^2+\left(1-\alpha \right){∥itb-a∥}^2\right]{∥b∥}^2\hfill \\ \hfill & \le \left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\left|⟨a,b⟩\right|}^2\hfill \\ \hfill & +\left[\alpha {∥tb-a∥}^2+\left(1-\alpha \right){∥itb-a∥}^2\right]{∥b∥}^2\hfill \end{array}$$

(50)

for any $a,b\in H$, $\alpha \in \left[0,1\right]$ and $t\in \mathbb{R}$.

Choosing in (50) $a=Tx$, $b=x$, $x\in H$, $∥x∥=1,$ we get

$$\begin{array}{cc}\hfill {∥Tx∥}^2& \le \left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\left|⟨Tx,x⟩\right|}^2\hfill \\ \hfill & +\alpha {∥tx-Tx∥}^2+\left(1-\alpha \right){∥itx-Tx∥}^2.\hfill \end{array}$$

(51)

Now, if we take $x=y/∥y∥,$ with $y\in H,$ then we also get

$$\begin{array}{cc}\hfill {∥Ty∥}^2& \le [{\left(1-\alpha \right)}^2+{\alpha }^2]{\displaystyle \frac{{\left|⟨Ty,y⟩\right|}^2}{{∥y∥}^2}}\hfill \\ \hfill & +\alpha {∥ty-Ty∥}^2+\left(1-\alpha \right){∥ity-Ty∥}^2\hfill \end{array}$$

for all $y\in H,$ which gives for $y=Az$ that

$$\begin{array}{cc}\hfill {∥TAz∥}^2& \le [{\left(1-\alpha \right)}^2+{\alpha }^2]{\displaystyle \frac{{\left|⟨A^{*}TAz,z⟩\right|}^2}{{∥Az∥}^2}}\hfill \\ \hfill & +\alpha {∥\left(tI-T\right)Az∥}^2+\left(1-\alpha \right){∥\left(itI-T\right)Az∥}^2\hfill \\ \hfill & \le [{\left(1-\alpha \right)}^2+{\alpha }^2]{\displaystyle \frac{{\left|⟨A^{*}TAz,z⟩\right|}^2}{m{∥z∥}^2}}\hfill \\ \hfill & +\alpha {∥\left(tI-T\right)Az∥}^2+\left(1-\alpha \right){∥\left(itI-T\right)Az∥}^2\hfill \end{array}$$

(52)

for $z\in H$, $z\ne 0,$ since ${∥Az∥}^2\ge m{∥z∥}^2$ for $z\in H$, $z\ne 0.$

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (52) that

$$\begin{array}{cc}\hfill {∥TA{\widehat{k}}_{\mu }∥}^2& \le \left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\displaystyle \frac{{\left|⟨A^{*}TA{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2}{m}}\hfill \\ \hfill & +\alpha {∥\left(tI-T\right)A{\widehat{k}}_{\mu }∥}^2+\left(1-\alpha \right){∥\left(itI-T\right)A{\widehat{k}}_{\mu }∥}^2.\hfill \end{array}$$

By taking the supremum over $\mu \in \Omega ,$ we obtain

$$\begin{array}{cc}\hfill {∥TA∥}_{\widetilde{\mathrm{ber}}}^2& =\underset{\mu \in \Omega }{\sup }{∥TA{\widehat{k}}_{\mu }∥}^2\le \left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\displaystyle \frac{{\sup }_{\mu \in \Omega }{\left|⟨A^{*}TA{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2}{m}}\hfill \\ \hfill & +\alpha \underset{\mu \in \Omega }{\sup }{∥\left(tI-T\right)A{\widehat{k}}_{\mu }∥}^2+\left(1-\alpha \right)\underset{\mu \in \Omega }{\sup }{∥\left(itI-T\right)A{\widehat{k}}_{\mu }∥}^2\hfill \\ \hfill & =\left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\displaystyle \frac{{\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2}{m}}\hfill \\ \hfill & +\alpha {∥\left(tI-T\right)A∥}_{\widetilde{\mathrm{ber}}}^2+\left(1-\alpha \right){∥\left(itI-T\right)A∥}_{\widetilde{\mathrm{ber}}}^2,\hfill \end{array}$$

which proves (49). □

Remark 6.

Let $A,T$ be as in Theorem 6. Then from (49) for $\alpha ={\displaystyle \frac{1}{2}},0,1$ respectively, and for any $t\in \mathbb{R}$ one has the inequalities:

$${∥TA∥}_{\widetilde{\mathrm{ber}}}^2\le {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{{\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2}{m}}+{∥\left(T-tI\right)A∥}_{\widetilde{\mathrm{ber}}}^2+{∥\left(T-itI\right)A∥}_{\widetilde{\mathrm{ber}}}^2\right],$$

$${∥TA∥}_{\widetilde{\mathrm{ber}}}^2\le {\displaystyle \frac{{\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2}{m}}+{∥\left(T-itI\right)A∥}_{\widetilde{\mathrm{ber}}}^2$$

and

$${∥TA∥}_{\widetilde{\mathrm{ber}}}^2\le {\displaystyle \frac{{\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2}{m}}+{∥\left(T-tI\right)A∥}_{\widetilde{\mathrm{ber}}}^2.$$

These imply that

$$\begin{array}{cc}\hfill {∥TA∥}_{\widetilde{\mathrm{ber}}}^2& \le {\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left[\mathrm{ber}\left(A^{*}TA\right)\right]}^2}{m}}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\underset{t\in \mathbb{R}}{\inf }\left[{∥\left(T-tI\right)A∥}_{\widetilde{\mathrm{ber}}}^2+{∥\left(T-itI\right)A∥}_{\widetilde{\mathrm{ber}}}^2\right].\hfill \end{array}$$

We observe from (49) for $A=I$ and $m=1,$ we get for $\alpha \in \left[0,1\right]$ that

$$\begin{array}{cc}\hfill {∥T∥}_{\widetilde{\mathrm{ber}}}^2& \le \left[{\left(1-\alpha \right)}^2+{\alpha }^2\right]{\left[\mathrm{ber}\left(T\right)\right]}^2\hfill \\ \hfill & +\alpha {∥tI-T∥}_{\widetilde{\mathrm{ber}}}^2+\left(1-\alpha \right){∥itI-T∥}_{\widetilde{\mathrm{ber}}}^2.\hfill \end{array}$$

For $\alpha =1/2$ we get

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2\le {\displaystyle \frac{1}{2}}\left({\left[\mathrm{ber}\left(T\right)\right]}^2+{∥tI-T∥}_{\widetilde{\mathrm{ber}}}^2+{∥itI-T∥}_{\widetilde{\mathrm{ber}}}^2\right),$$

for $\alpha =0$ we obtain

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2\le {\left[\mathrm{ber}\left(T\right)\right]}^2+{∥itI-T∥}_{\widetilde{\mathrm{ber}}}^2,$$

while for $\alpha =1$ we derive that

$${∥T∥}_{\widetilde{\mathrm{ber}}}^2\le {\left[\mathrm{ber}\left(T\right)\right]}^2+{∥tI-T∥}_{\widetilde{\mathrm{ber}}}^2$$

for all $t\in \mathbb{R}$.

Theorem 7.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right).$ Then

$$\left(0\le \right){∥T∥}_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le \underset{\lambda \in \mathbb{C}}{\inf }{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2$$

(53)

and, if $T\ne 0$,

$$\left(0\le \right)1-{\displaystyle \frac{{\left[\mathrm{ber}\left(T\right)\right]}^2}{{∥T∥}_{\widetilde{\mathrm{ber}}}^2}}\le \underset{\lambda \in \mathbb{C}}{\inf }{∥\lambda T-I∥}_{\widetilde{\mathrm{ber}}}^2.$$

(54)

Proof. 

We know that for any $a,b\in H$, $b\ne 0,$ (see, for instance, [31]) the bound

$$\underset{\lambda \in \mathbb{C}}{\inf }{∥a-\lambda b∥}^2={\displaystyle \frac{{∥a∥}^2{∥b∥}^2-{\left|⟨a,b⟩\right|}^2}{{∥b∥}^2}}$$

actually implies that

$${∥a∥}^2{∥b∥}^2-{\left|⟨a,b⟩\right|}^2\le {∥b∥}^2{∥a-\lambda b∥}^2$$

(55)

for any $a,b\in H$ and $\lambda \in \mathbb{C}$.

Now if in (55) we choose $a=Tx$, $b=x$, $x\in H$, $∥x∥=1,$ then we obtain

$${∥Tx∥}^2-{\left|⟨Tx,x⟩\right|}^2\le {∥Tx-\lambda x∥}^2$$

(56)

for any $\lambda \in \mathbb{C}$.

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (56) that

$${∥T{\widehat{k}}_{\mu }∥}^2\le {\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2+{∥T{\widehat{k}}_{\mu }-\lambda {\widehat{k}}_{\mu }∥}^2$$

and if we take the supremum over $\mu \in \Omega$, then we get the desired result (53).

If we take $a=x$, $b=Tx$ in (55), then we obtain

$${∥Tx∥}^2\le {\left|⟨Tx,x⟩\right|}^2+{∥Tx∥}^2{∥x-\lambda Tx∥}^2$$

(57)

for any $x\in H$, $∥x∥=1$ and $\lambda \in \mathbb{C}$.

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (57) that

$${∥T{\widehat{k}}_{\mu }∥}^2\le {\left|⟨T{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|}^2+{∥T{\widehat{k}}_{\mu }∥}^2{∥{\widehat{k}}_{\mu }-\lambda T{\widehat{k}}_{\mu }∥}^2$$

and if we take the supremum over $\mu \in \Omega$, we get ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}^2\le {\left[\mathrm{ber}\left(T\right)\right]}^2+{\parallel T\parallel }_{\widetilde{\mathrm{ber}}}^2{\parallel \lambda T-I\parallel }_{\widetilde{\mathrm{ber}}}^2$ for every $\lambda \in \mathbb{C}$. Since $T\ne 0$ and $\left\{{\widehat{k}}_{\mu }\right\}$ is total, ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}>0$; dividing by ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}^2$ and taking the infimum over $\lambda$, we obtain the desired result (54). □

Remark 7.

We have for $\lambda \ne 0$ that

$$\underset{\zeta \in \mathbb{C}}{\inf }{∥\zeta T-I∥}_{\widetilde{\mathrm{ber}}}^2\le {∥{\displaystyle \frac{1}{\lambda }}T-I∥}_{\widetilde{\mathrm{ber}}}^2={\displaystyle \frac{1}{{\left|\lambda \right|}^2}}{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2.$$

(58)

So, if

$${∥T-{\displaystyle \frac{\psi +\phi }{2}}·I∥}_{\widetilde{\mathrm{ber}}}\le {\displaystyle \frac{1}{2}}\left|\psi -\phi \right|\mathrm{for}\psi \ne \pm \phi$$

(59)

then by (54) and (58) for $\lambda ={\displaystyle \frac{\psi +\phi }{2}}$ we get

$$\left(0\le \right)1-{\displaystyle \frac{{\left[\mathrm{ber}\left(T\right)\right]}^2}{{∥T∥}_{\widetilde{\mathrm{ber}}}^2}}\le {\displaystyle \frac{1}{{\left|\frac{\psi +\phi }{2}\right|}^2}}{∥T-{\displaystyle \frac{\psi +\phi }{2}}I∥}_{\widetilde{\mathrm{ber}}}^2\le {\displaystyle \frac{\frac{1}{4}{\left|\psi -\phi \right|}^2}{{\left|\frac{\psi +\phi }{2}\right|}^2}},$$

which gives that

$$\left(0\le \right)1-{\displaystyle \frac{{\left[\mathrm{ber}\left(T\right)\right]}^2}{{∥T∥}_{\widetilde{\mathrm{ber}}}^2}}\le {\left|{\displaystyle \frac{\psi -\phi }{\psi +\phi }}\right|}^2,$$

provided that (59) is valid. This shows that the inequality (26) holds in fact without the restriction $\mathrm{Re}\left(\psi \overline{\phi }\right)>0.$ Here $T\ne 0$ is assumed in the two displayed ratio inequalities, in accordance with Theorem 7; for $T=0$, the inequality (26) is trivial.

Theorem 8.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right).$ Then for $p\ge 2$ and $\lambda \in \mathbb{C}\setminus \left\{0\right\}$ we have that

$${∥T∥}_{\widetilde{\mathrm{ber}}}^{p/2}\le {\displaystyle \frac{1}{4{\left|\lambda \right|}^{p/2}}}\left({∥T+\lambda I∥}_{\widetilde{\mathrm{ber}}}^p+{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^p\right).$$

(60)

In particular,

$${∥T∥}_{\widetilde{\mathrm{ber}}}\le {\displaystyle \frac{1}{4\left|\lambda \right|}}\left({∥T+\lambda I∥}_{\widetilde{\mathrm{ber}}}^2+{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^2\right).$$

Proof. 

We use the following inequality obtained by Dragomir and Sándor in [27]:

$${∥a+b∥}^p+{∥a-b∥}^p\ge 2\left({∥a∥}^p+{∥b∥}^p\right)$$

(61)

for any $a,b\in H$ and $p\ge 2.$

Now, if we choose $a=Tx$, $b=\lambda x$, $\lambda \in \mathbb{C}\setminus \left\{0\right\}$ then we get

$${∥Tx+\lambda x∥}^p+{∥Tx-\lambda x∥}^p\ge 2\left({∥Tx∥}^p+{\left|\lambda \right|}^p\right)$$

(62)

for any $x\in H$, $∥x∥=1.$

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (62) that

$${∥T{\widehat{k}}_{\mu }∥}^p+{\left|\lambda \right|}^p\le {\displaystyle \frac{1}{2}}\left({∥T{\widehat{k}}_{\mu }+\lambda {\widehat{k}}_{\mu }∥}^p+{∥T{\widehat{k}}_{\mu }-\lambda {\widehat{k}}_{\mu }∥}^p\right).$$

If we take the supremum over $\mu \in \Omega ,$ we get that

$${∥T∥}_{\widetilde{\mathrm{ber}}}^p+{\left|\lambda \right|}^p\le {\displaystyle \frac{1}{2}}\left({∥T+\lambda I∥}_{\widetilde{\mathrm{ber}}}^p+{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^p\right).$$

(63)

Since

$$2{\left|\lambda \right|}^{p/2}{∥T∥}_{\widetilde{\mathrm{ber}}}^{p/2}\le {∥T∥}_{\widetilde{\mathrm{ber}}}^p+{\left|\lambda \right|}^p,$$

then by (63) we get

$$2{\left|\lambda \right|}^{p/2}{∥T∥}_{\widetilde{\mathrm{ber}}}^{p/2}\le {\displaystyle \frac{1}{2}}\left({∥T+\lambda I∥}_{\widetilde{\mathrm{ber}}}^p+{∥T-\lambda I∥}_{\widetilde{\mathrm{ber}}}^p\right),$$

which gives the desired result (60). □

4. Berezin Inequalities via Polar Decomposition

We now combine the Buzano-type inequalities of Section 3 with the polar decomposition of an operator in order to obtain reverse Berezin inequalities for the operator modulus ${\left|T\right|}^{\alpha }$ and for the Aluthge transform of T. Every bounded linear operator $T:H\to H$ on a complex Hilbert space H can be uniquely factored as:

$$T=U\left|T\right|$$

The operator $\left|T\right|$ is defined via the continuous functional calculus as the unique positive square root of $T^{*}T$. The operator U is a partial isometry. It acts as an isometry on the orthogonal complement of the kernel (null space) of T, mapping it isomorphically onto the closure of the range of T.

We need the following fact:

Lemma 1.

If $\phi :[0,\parallel T\parallel ]\to \mathbb{C}$ is a continuous function satisfying $\phi \left(0\right)=0$, then:

$$U\phi \left(\right|T\left|\right)U^{*}=\phi \left(\right|T^{*}\left|\right),$$

(64)

where $|T^{*}|={\left(T{T}^{*}\right)}^{1/2}$.

Proof. 

For the sake of completeness we give a simple proof as follows.

We first prove that

$$U\left|T\right|U^{*}=\left|T^{*}\right|.$$

Observe that

$$\left(U\right|T|U^{*}{)}^2=\left(U\right|T|U^{*}\left)\right(U\left|T\right|U^{*})=U|T|\left(U^{*}U\right)\left|T\right|U^{*}.$$

Recall that $U^{*}U=P_{\overline{\mathrm{ran}}\left(\right|T\left|\right)}$ represents the orthogonal projection onto $\overline{\mathrm{ran}}\left(\right|T\left|\right)=\ker {\left(\right|T\left|\right)}^{\perp }$. Since the operator $\left|T\right|$ vanishes identically on its own kernel, then

$$\left(U^{*}U\right)\left|T\right|=\left|T\right|.$$

Substituting back, gives

$$\left(U\right|T|U^{*}{)}^2=U\left|T\right|\left|T\right|U^{*}=U{\left|T\right|}^2{U}^{*}.$$

Further, we have that

$${U\left|T\right|}^2{U}^{*}=U\left(T^{*}T\right)U^{*}.$$

We also have that $T=U\left|T\right|$ and $T^{*}=\left|T\right|U^{*}$ and then

$$U\left(T^{*}T\right)U^{*}=\left(U\right|T\left|\right)\left(\right|T|U^{*})=T{T}^{*}.$$

Hence, we have verified that:

$$\left(U\right|T|U^{*}{)}^2=T{T}^{*}$$

Because $\left|T\right|\ge 0$, the operator $U\left|T\right|U^{*}$ is positive. By the unique existence of the positive square root of the positive operator $T{T}^{*}$, we conclude:

$$U\left|T\right|U^{*}=\sqrt{T{T}^{*}}=\left|T^{*}\right|.$$

We extend the property to higher powers using induction for all integers $n\ge 1$. Assume that the identity holds for $n=k$,

$${U\left|T\right|}^k{U}^{*}={\left|T^{*}\right|}^k.$$

We want to prove it for $n=k+1$:

$${U\left|T\right|}^{k+1}{U}^{*}={U\left|T\right|}^k\left|T\right|U^{*}.$$

We insert the projection $U^{*}U$ directly before the final $\left|T\right|$ term, namely,

$${U\left|T\right|}^{k+1}{U}^{*}={U\left|T\right|}^k\left(U^{*}U\right)\left|T\right|U^{*}={\left(U\right|T|}^k{U}^{*}\left)\right(U\left|T\right|U^{*}).$$

Applying the induction hypothesis yields:

$${\left(U\right|T|}^k{U}^{*}\left)\right(U\left|T\right|U^{*})=|T^{*}{|}^k|T^{*}|=|T^{*}{|}^{k+1}.$$

Therefore, by mathematical induction, ${U\left|T\right|}^n{U}^{*}={\left|T^{*}\right|}^n$ holds for all integers $n\ge 1$.

Let $p\left(t\right)={\sum }_{k=1}^n{a}_k{t}^k$ be an arbitrary polynomial satisfying $p\left(0\right)=0$. We obtain:

$$Up\left(\right|T\left|\right)U^{*}=U\left({\displaystyle \sum _{k=1}^n}{a}_k{\left|T\right|}^k\right)U^{*}={\displaystyle \sum _{k=1}^n}{a}_k\left({U\left|T\right|}^k{U}^{*}\right)={\displaystyle \sum _{k=1}^n}{a}_k|T^{*}{|}^k=p\left(\right|T^{*}\left|\right).$$

Let $\phi :[0,\parallel T\parallel ]\to \mathbb{C}$ be a continuous function satisfying $\phi \left(0\right)=0$. By the Stone–Weierstrass Approximation Theorem, there exists a sequence of polynomials ${\left\{p_m\right\}}_{m=1}^{\infty }$ with $p_m\left(0\right)=0$ that converges uniformly to $\phi$ on the compact domain $[0,\parallel T\parallel ]$. The continuous functional calculus assures that uniform convergence of a sequence of continuous functions implies the norm convergence of their corresponding operators:

$$\parallel p_m\left(\right|T\left|\right)-\phi \left(\right|T\left|\right)\parallel \to 0\mathrm{and}\parallel p_m\left(\right|T^{*}\left|\right)-\phi \left(\right|T^{*}\left|\right)\parallel \to 0\mathrm{as}m\to \infty$$

Since U and $U^{*}$ are bounded linear operators, they preserve limits under the operator norm topology:

$$U\phi \left(\right|T\left|\right)U^{*}=U\left(\underset{m\to \infty }{\lim }{p}_m\left(\right|T\left|\right)\right)U^{*}=\underset{m\to \infty }{\lim }U{p}_m\left(\right|T\left|\right)U^{*}$$

Substituting our polynomial identity derived above yields:

$$\underset{m\to \infty }{\lim }U{p}_m\left(\right|T\left|\right)U^{*}=\underset{m\to \infty }{\lim }{p}_m\left(\right|T^{*}\left|\right)=\phi \left(\right|T^{*}\left|\right).$$

This completes the proof. □

For the power function $\phi \left(t\right)=t^{\alpha }$, $\alpha >0$, we then have that

$$U{\left|T\right|}^{\alpha }{U}^{*}={\left|T^{*}\right|}^{\alpha }\mathrm{for}\alpha >0.$$

(65)

Theorem 9.

Let $T\in B\left({\mathcal{H}}_{\Omega }\right).$ For $\alpha >0$ we have that

$${∥{\left|T\right|}^{\alpha }∥}_{\mathrm{ber}}^2\le {\displaystyle \frac{1}{2}}\left({∥T∥}^{2\alpha }+{∥{\left|T\right|}^{2\alpha }∥}_{\mathrm{ber}}\right).$$

(66)

In particular,

$${∥\left|T\right|∥}_{\mathrm{ber}}^2\le {\displaystyle \frac{1}{2}}\left({∥T∥}^2+{∥{\left|T\right|}^2∥}_{\mathrm{ber}}\right)$$

and

$${∥{\left|T\right|}^2∥}_{\mathrm{ber}}^2\le {\displaystyle \frac{1}{2}}\left({∥T∥}^4+{∥{\left|T\right|}^4∥}_{\mathrm{ber}}\right).$$

Proof. 

Let $V=U\left|V\right|$ be the polar decomposition of V. From the inequality (39) for $A=U^{*}$ and $T={\left|V\right|}^{\alpha }$, we get

$$\begin{array}{cc}\hfill & {\left[\mathrm{ber}\left(U{\left|V\right|}^{\alpha }{U}^{*}\right)\right]}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({∥{\left|V\right|}^{\alpha }{U}^{*}∥}_{\widetilde{\mathrm{ber}}}{∥{\left|V\right|}^{\alpha }{U}^{*}∥}_{\widetilde{\mathrm{ber}}}+\mathrm{ber}\left(U{\left|V\right|}^{2\alpha }{U}^{*}\right)\right){∥U^{*}∥}_{\widetilde{\mathrm{ber}}}^2.\hfill \end{array}$$

(67)

Since, by (65), $U{\left|V\right|}^{\alpha }{U}^{*}={\left|V^{*}\right|}^{\alpha }$ and $U{\left|V\right|}^{2\alpha }{U}^{*}={\left|V^{*}\right|}^{2\alpha },$ then we get from (67) that

$${\left[\mathrm{ber}\left({\left|V^{*}\right|}^{\alpha }\right)\right]}^2\le {\displaystyle \frac{1}{2}}\left({∥{\left|V\right|}^{\alpha }{U}^{*}∥}_{\widetilde{\mathrm{ber}}}^2+\mathrm{ber}\left({\left|V^{*}\right|}^{2\alpha }\right)\right){∥U^{*}∥}_{\widetilde{\mathrm{ber}}}^2.$$

(68)

This inequality is of interest in itself as well.

Since ${\left|V^{*}\right|}^{\alpha },{\left|V^{*}\right|}^{2\alpha }\ge 0,$ then by (3) we have that

$$\mathrm{ber}\left({\left|V^{*}\right|}^{\alpha }\right)={∥{\left|V^{*}\right|}^{\alpha }∥}_{\mathrm{ber}}\mathrm{and}\mathrm{ber}\left({\left|V^{*}\right|}^{2\alpha }\right)={∥{\left|V^{*}\right|}^{2\alpha }∥}_{\mathrm{ber}}.$$

Since

$${∥U^{*}∥}_{\widetilde{\mathrm{ber}}}^2\le {∥U^{*}∥}^2\le 1$$

and

$${∥{\left|V\right|}^{\alpha }{U}^{*}∥}_{\widetilde{\mathrm{ber}}}^2\le {∥{\left|V\right|}^{\alpha }∥}^2{∥U^{*}∥}_{\widetilde{\mathrm{ber}}}^2={∥V∥}^{2\alpha }{∥U^{*}∥}_{\widetilde{\mathrm{ber}}}^2\le {∥V∥}^{2\alpha },$$

then by (68) we get

$${∥{\left|V^{*}\right|}^{\alpha }∥}_{\mathrm{ber}}^2\le {\displaystyle \frac{1}{2}}\left({∥V∥}^{2\alpha }+{∥{\left|V^{*}\right|}^{2\alpha }∥}_{\mathrm{ber}}\right).$$

If we take $V=T^{*}$ in this last inequality, we derive

$${∥{\left|T\right|}^{\alpha }∥}_{\mathrm{ber}}^2\le {\displaystyle \frac{1}{2}}\left({∥T^{*}∥}^{2\alpha }+{∥{\left|T\right|}^{2\alpha }∥}_{\mathrm{ber}}\right),$$

which is the desired result. □

In the following examples, we discuss the sharpness of (66). Let ${\mathbb{H}}_{\Omega }={\mathbb{C}}^2$, $\Omega =\{1,2\}$.

Example 5.

Let

$$T=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right).$$

with $a,b\ge 0$. Then $T=\left|T\right|$, so ${\parallel \left|T\right|\parallel }_{\mathrm{ber}}=\max \{a,b\}=\parallel T\parallel$, and

$${\left|T\right|}^2=\left(\begin{array}{cc}{a}^2& 0\\ 0& b^2\end{array}\right).$$

gives ${\parallel \left|T\right|}^2{\parallel }_{\mathrm{ber}}=\max {\{a,b\}}^2$. Both sides of (66) (with $\alpha =1$) equal $\max {\{a,b\}}^2$, so the equality holds for every positive diagonal T.

The equality in Example 5 is not surprising, since T is already positive; the next example shows that (66) can still be an equality even when T is far from self-adjoint.

Example 6.

Let

$$T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right).$$

Then

$$T^{*}T=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)=\left|T\right|$$

and ${\parallel \left|T\right|\parallel }_{\mathrm{ber}}=1$. Also $\parallel T\parallel =1$ (its only nonzero singular value) and ${\parallel \left|T\right|}^2{\parallel }_{\mathrm{ber}}=1$. Thus $1=\frac{1}{2}(1+1)$: the equality holds even for this non-self-adjoint nilpotent shift.

The two preceding examples might suggest that (66) is always an equality; the following example shows this is not the case in general.

Example 7.

Finally, let

$$T=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right).$$

Here $T^{*}T=\left(\begin{array}{cc}4& 2\\ 2& 2\end{array}\right)$, with eigenvalues $3\pm \sqrt{5}$, so ${\parallel T\parallel }^2=3+\sqrt{5}\approx 5.236$ and ${\parallel \left|T\right|}^2{\parallel }_{\mathrm{ber}}=\mathrm{ber}\left(T^{*}T\right)=4$. Diagonalizing $T^{*}T$ to compute its square root gives $\left|T\right|\approx \left(\begin{array}{cc}1.897& 0.633\\ 0.633& 1.265\end{array}\right)$, so ${\parallel \left|T\right|\parallel }_{\mathrm{ber}}\approx 1.897$, and the left-hand side is $\approx 3.60$. The right-hand side is $\frac{1}{2}(5.236+4)\approx 4.618$. This example shows the inequality is strict in general.

We close this section with the following bound for the Aluthge transform:

Theorem 10.

Let $V\in B\left({\mathcal{H}}_{\Omega }\right)$ and $V=U\left|V\right|$ be the polar decomposition of V. If $\tilde{V}$ is the Aluthge transform of $V,$ namely, $\tilde{V}:={\left|V\right|}^{1/2}U{\left|V\right|}^{1/2},$ then

$$\begin{array}{cc}\hfill & {\left[\mathrm{ber}\left(\tilde{V}\right)\right]}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({∥U{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}{∥U^{*}{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}+\mathrm{ber}\left({\left|V\right|}^{1/2}{U}^2{\left|V\right|}^{1/2}\right)\right){∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2+\mathrm{ber}\left({\left|V\right|}^{1/2}{U}^2{\left|V\right|}^{1/2}\right)\right){∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2+{∥{\left|V\right|}^{1/2}{U}^2{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}\right){∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}+{∥V∥}^{1/2}\right){∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^3.\hfill \end{array}$$

(69)

Proof. 

Let $V=U\left|V\right|$ be the polar decomposition of $V.$ From the inequality (39) for $A={\left|V\right|}^{1/2}$ and $T=$U

$$\begin{array}{cc}\hfill & {\left[\mathrm{ber}\left({\left|V\right|}^{1/2}U{\left|V\right|}^{1/2}\right)\right]}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({∥U{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}{∥U^{*}{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}+\mathrm{ber}\left({\left|V\right|}^{1/2}{U}^2{\left|V\right|}^{1/2}\right)\right){∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}^2.\hfill \end{array}$$

(70)

We observe that ${\left|V\right|}^{1/2}U{\left|V\right|}^{1/2}=\tilde{V}$, and by (70) we get (69).

Also, we have that

$${∥U{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}\le ∥U∥{∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}={∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}},$$

which proves the second inequality in (69).

The third inequality follows from $\mathrm{ber}\left(X\right)\le {\parallel X\parallel }_{\mathrm{ber}}\le {\parallel X\parallel }_{\widetilde{\mathrm{ber}}}$, see (2), and the last inequality follows by the fact that

$$\begin{array}{cc}\hfill {∥{\left|V\right|}^{1/2}{U}^2{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}& \le ∥{\left|V\right|}^{1/2}{U}^2∥{∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}\le ∥{\left|V\right|}^{1/2}∥∥U^2∥{∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}\hfill \\ \hfill & \le {∥V∥}^{1/2}{∥{\left|V\right|}^{1/2}∥}_{\widetilde{\mathrm{ber}}}.\hfill \end{array}$$

□

5. Reverse Berezin Inequalities via the Moore–Penrose Inverse

In this final section, we complement the disc-localization and polar-decomposition techniques of Section 2, Section 3 and Section 4 by involving, for operators with closed range, the Moore–Penrose inverse in reverse Berezin-type inequalities. This provides a further tool, alternative to the ones used above, for estimating the gap between $\mathrm{ber}(·)$, ${\parallel ·\parallel }_{\mathrm{ber}}$, and ${\parallel ·\parallel }_{\widetilde{\mathrm{ber}}}$. A fundamental concept in operator theory is the Moore–Penrose inverse, which generalizes the notion of operator inverses to more general settings. In a related direction, Ren, Ighachane and Benchiheb [32] recently obtained forward-type Berezin number bounds for operators admitting a Moore–Penrose inverse, using interpolational means and McCarthy-type inequalities; the results below are of a complementary, reverse-type nature, derived instead from the disc-localization condition $\parallel M-\lambda I\parallel \le r$.

For an operator M∈$B\left(H\right)$, the Moore–Penrose inverse of M, denoted by $M^{†}$, is the unique operator satisfying the following four Penrose equations [33],

$$M{M}^{†}M=M,M^{†}M{M}^{†}=M^{†},{\left(M{M}^{†}\right)}^{*}=M{M}^{†},{\left(M^{†}M\right)}^{*}=M^{†}M.$$

These conditions ensure the existence and uniqueness of $M^{†}$ whenever M has a closed range, i.e., range $\left(M\right)$ is closed in $H.$

We define the set of all operators in $\mathcal{B}\left(H\right)$ that admit a Moore–Penrose inverse as follows:

$$\mathcal{CR}\left(H\right)=\{M\in \mathcal{B}(H\left)\right|\mathrm{range}\left(M\right)\mathrm{is}\mathrm{closed}\}.$$

For any $M\in \mathcal{CR}\left(H\right)$, its Moore–Penrose inverse $M^{†}$ is also a bounded operator in $\mathcal{B}\left(H\right)$.

For an operator $M\in \mathcal{CR}\left(H\right)$, the Moore–Penrose inverse satisfies the following important properties:

• Self-adjoint property: If M is self-adjoint, i.e., $M=M^{*},$ then $M^{†}$ is also selfadjoint and ${\left(M^{†}\right)}^{*}=M^{†};$
• Involutivity: The Moore–Penrose inverse satisfies the equality ${\left(M^{†}\right)}^{†}=M;$
• Order reversal: If $M_1,M_2\in \mathcal{CR}\left(H\right)$ with $M_1\le M_2$ in the operator order, then $M_2^{†}\le M_1^{†},$ provided $M_1,M_2$ are positive and boundedly invertible, in which case $M_i^{†}=M_i^{-1}$, and this is the classical operator-antitonicity of the inverse (see [34]). This property is recalled for completeness only and is not used below.

Note that $M{M}^{†}$ is the orthogonal projection onto $\mathrm{ran}\left(M\right)$ and $M{M}^{†}M=M$; in particular $M{M}^{†}\ge 0$, $\parallel M{M}^{†}\parallel \le 1$, and $M{M}^{†}=I$ when M is invertible. These are the only properties of $M^{†}$ used in the proof of Theorem 11.

We have the following inequalities:

Theorem 11.

Let $M\in \mathcal{CR}\left({\mathcal{H}}_{\Omega }\right).$ If $\lambda \in \mathbb{C}\setminus \left\{0\right\}$ and $r>0$ are such that

$$∥M-\lambda I∥\le r,$$

then

$$\begin{array}{cc}\hfill {∥{\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}∥}_{\mathrm{ber}}& \le 2\mathrm{ber}\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)+r^2\hfill \\ \hfill & \le 2\left|\lambda \right|\mathrm{ber}\left(M\right)+r^2.\hfill \end{array}$$

(71)

Also, we have that

$${∥\left|M\right|∥}_{\widetilde{\mathrm{ber}}}^2\le 2\mathrm{ber}\left[\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M-{\displaystyle \frac{1}{2}}{\left|\lambda \right|}^{2}M{M}^{†}\right]+r^2.$$

(72)

Proof. 

We observe that for $M\in \mathcal{CR}\left(H\right)$ we have that

$$\begin{array}{cc}\hfill ∥M-\lambda M{M}^{†}∥& =∥M{M}^{†}M-\lambda M{M}^{†}∥=∥M{M}^{†}\left(M-\lambda I\right)∥\hfill \\ \hfill & \le ∥M{M}^{†}∥∥M-\lambda I∥\le r.\hfill \end{array}$$

For $x\in H$ with $∥x∥=1$ we have that

$${∥Mx-\lambda M{M}^{†}x∥}^2\le {∥M-\lambda M{M}^{†}∥}^2\le r^2.$$

(73)

Since

$$\begin{array}{cc}\hfill {∥Mx-\lambda M{M}^{†}x∥}^2& ={∥Mx∥}^2-2\mathrm{Re}\left[\overline{\lambda }⟨Mx,M{M}^{†}x⟩\right]+{\left|\lambda \right|}^2{∥M{M}^{†}x∥}^2\hfill \\ \hfill & =⟨{\left|M\right|}^{2}x,x⟩-2\mathrm{Re}\left[\overline{\lambda }⟨{\left(M{M}^{†}\right)}^{*}Mx,x⟩\right]+{\left|\lambda \right|}^2⟨{\left|M{M}^{†}\right|}^{2}x,x⟩\hfill \\ \hfill & =⟨{\left|M\right|}^{2}x,x⟩-2\mathrm{Re}\left[\overline{\lambda }⟨M{M}^{†}Mx,x⟩\right]+{\left|\lambda \right|}^2⟨{\left(M{M}^{†}\right)}^{2}x,x⟩\hfill \\ \hfill & =⟨{\left|M\right|}^{2}x,x⟩-2\mathrm{Re}\left[\overline{\lambda }⟨Mx,x⟩\right]+{\left|\lambda \right|}^2⟨M{M}^{†}x,x⟩.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathrm{Re}\left[\overline{\lambda }⟨Mx,x⟩\right]& =\mathrm{Re}\left[\left(\mathrm{Re}\lambda -i\mathrm{Im}\lambda \right)\left(\mathrm{Re}⟨Mx,x⟩+i\mathrm{Im}⟨Mx,x⟩\right)\right]\hfill \\ \hfill & =\mathrm{Re}\lambda \mathrm{Re}⟨Mx,x⟩+\mathrm{Im}\lambda \mathrm{Im}⟨Mx,x⟩\hfill \\ \hfill & =\mathrm{Re}\lambda ⟨\mathrm{Re}Mx,x⟩+\mathrm{Im}\lambda ⟨\mathrm{Im}Mx,x⟩\hfill \\ \hfill & =⟨\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)x,x⟩,\hfill \end{array}$$

then by (73) we get that

$$⟨{\left|M\right|}^{2}x,x⟩+{\left|\lambda \right|}^2⟨M{M}^{†}x,x⟩\le 2⟨\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)x,x⟩+r^2,$$

(74)

namely,

$$\begin{array}{cc}\hfill ⟨\left({\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}\right)x,x⟩& \le 2⟨\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)x,x⟩+r^2\hfill \\ \hfill & \le 2\left|\lambda \right|\left|⟨Mx,x⟩\right|+r^2\hfill \end{array}$$

(75)

for $x\in H$ with $∥x∥=1,$ since $\mathrm{Re}z\le \left|z\right|$ for every $z\in \mathbb{C}$; hence, $\mathrm{Re}\left[\overline{\lambda }⟨Mx,x⟩\right]\le \left|\lambda \right|\left|⟨Mx,x⟩\right|.$

Let ${\widehat{k}}_{\mu }$ be a normalized reproducing kernel of ${\mathcal{H}}_{\Omega }.$ For $\mu \in \Omega ,$ we have from (75) that

$$\begin{array}{ccc}\hfill ⟨\left({\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}\right){\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩& \le & 2⟨\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right){\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩+r^2\hfill \\ & \le & 2\left|\lambda \right|\left|⟨M{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|+r^2.\hfill \end{array}$$

If we take the supremum over $\mu \in \Omega$, we get that

$$\begin{array}{cc}\hfill \mathrm{ber}\left({\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}\right)& \le 2\mathrm{ber}\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)+r^2\hfill \\ \hfill & \le 2\left|\lambda \right|\mathrm{ber}\left(M\right)+r^2\hfill \end{array}$$

(76)

Since ${\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}\ge 0$, we have by (3) that

$$\mathrm{ber}\left({\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}\right)={∥{\left|M\right|}^2+{\left|\lambda \right|}^{2}M{M}^{†}∥}_{\mathrm{ber}}$$

and by (76) we get (71).

From (74) we also have that

$$\begin{array}{cc}\hfill ⟨{\left|M\right|}^{2}x,x⟩& \le 2⟨\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)x,x⟩-{\left|\lambda \right|}^2⟨M{M}^{†}x,x⟩+r^2\hfill \\ \hfill & =2⟨\left[\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)-{\displaystyle \frac{1}{2}}{\left|\lambda \right|}^{2}M{M}^{†}\right]x,x⟩+r^2\hfill \\ \hfill & \le 2\left|⟨\left[\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)-{\displaystyle \frac{1}{2}}{\left|\lambda \right|}^{2}M{M}^{†}\right]x,x⟩\right|+r^2\hfill \end{array}$$

for $x\in H$ with $∥x∥=1.$

If in this inequality we take $x={\widehat{k}}_{\mu },\mu \in \Omega$, where ${\widehat{k}}_{\mu }$ is a normalized reproducing kernel of ${\mathcal{H}}_{\Omega },$ then we get

$$⟨{\left|M\right|}^2{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\le 2\left|⟨\left[\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)-{\displaystyle \frac{1}{2}}{\left|\lambda \right|}^{2}M{M}^{†}\right]{\widehat{k}}_{\mu },{\widehat{k}}_{\mu }⟩\right|+r^2$$

and by taking the supremum over $\mu \in \Omega$,

$$\mathrm{ber}\left({\left|M\right|}^2\right)\le 2\mathrm{ber}\left[\left(\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M\right)-{\displaystyle \frac{1}{2}}{\left|\lambda \right|}^{2}M{M}^{†}\right]+r^2.$$

(77)

Since $\mathrm{ber}\left({\left|M\right|}^2\right)={∥\left|M\right|∥}_{\widetilde{\mathrm{ber}}}^2,$ then by (77) we obtain (72). □

Remark 8.

Put $Q=\mathrm{Re}\lambda \mathrm{Re}M+\mathrm{Im}\lambda \mathrm{Im}M=\mathrm{Re}\left(\overline{\lambda }M\right)$. The hypothesis $\parallel M-\lambda I\parallel \le r$ alone does not imply $Q\ge 0$, and then ${\parallel Q\parallel }_{\mathrm{ber}}\le \left|\lambda \right|\mathrm{ber}\left(M\right)$ may fail (for instance, $M=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ on ${\mathbb{C}}^2$ with $\Omega =\{1,2\}$, $\lambda =1$, $r=2$: here $\mathrm{ber}\left(M\right)=0$ but ${\parallel Q\parallel }_{\mathrm{ber}}=1$). This is why (71) is stated with $\mathrm{ber}\left(Q\right)$. If, however, $\left|\lambda \right|>r$, then for every unit vector x,

$$⟨Qx,x⟩=\mathrm{Re}\left[\overline{\lambda }⟨Mx,x⟩\right]={\left|\lambda \right|}^2+\mathrm{Re}\left[\overline{\lambda }⟨(M-\lambda I)x,x⟩\right]\ge \left|\lambda \right|\left(\right|\lambda |-r)>0,$$

so $Q\ge 0$, hence $\mathrm{ber}\left(Q\right)={\parallel Q\parallel }_{\mathrm{ber}}$ by (3), and (71) holds with ${\parallel Q\parallel }_{\mathrm{ber}}$ in place of $\mathrm{ber}\left(Q\right)$.

By taking $\lambda ={\displaystyle \frac{\phi +\psi }{2}}$ and $r=\left|{\displaystyle \frac{\phi -\psi }{2}}\right|$ in Theorem 11 we can also state that:

Corollary 7.

Let $M\in \mathcal{CR}\left({\mathcal{H}}_{\Omega }\right)$ and $\phi ,\psi \in \mathbb{C}$ with $\psi \notin \left\{-\phi ,\phi \right\}.$ If

$$\mathrm{Re}⟨\psi x-Mx,Mx-\phi x⟩\ge 0\mathrm{for}\mathrm{any}x\in H,∥x∥=1,$$

then

$$\begin{array}{cc}\hfill & {∥{\left|M\right|}^2+{\left|{\displaystyle \frac{\phi +\psi }{2}}\right|}^{2}M{M}^{†}∥}_{\mathrm{ber}}\hfill \\ \hfill & \le \mathrm{ber}\left[\left(\mathrm{Re}\phi +\mathrm{Re}\psi \right)\mathrm{Re}M+\left(\mathrm{Im}\phi +\mathrm{Im}\psi \right)\mathrm{Im}M\right]+{\left|{\displaystyle \frac{\phi -\psi }{2}}\right|}^2\hfill \\ \hfill & \le \left|\phi +\psi \right|\mathrm{ber}\left(M\right)+{\left|{\displaystyle \frac{\phi -\psi }{2}}\right|}^2\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {∥\left|M\right|∥}_{\widetilde{\mathrm{ber}}}^2& \le \mathrm{ber}\left[\left(\mathrm{Re}\phi +\mathrm{Re}\psi \right)\mathrm{Re}M+\left(\mathrm{Im}\phi +\mathrm{Im}\psi \right)\mathrm{Im}M-{\left|{\displaystyle \frac{\phi +\psi }{2}}\right|}^{2}M{M}^{†}\right]\hfill \\ \hfill & +{\left|{\displaystyle \frac{\phi -\psi }{2}}\right|}^2.\hfill \end{array}$$

Remark 9.

Let $M\in \mathcal{CR}\left({\mathcal{H}}_{\Omega }\right).$ If $S\ge s>0$ are such that either $\left(M^{*}-sI\right)\left(SI-M\right)$ is accretive, or, sufficiently, $\left(M^{*}-sI\right)\left(SI-M\right)$ is self-adjoint and

$$\left(M^{*}-sI\right)\left(SI-M\right)\ge 0\mathit{in}\mathit{the}\mathit{operator}\mathrm{order},$$

then, by Corollary 7 we have that

$$\begin{array}{cc}\hfill {∥{\left|M\right|}^2+{\left({\displaystyle \frac{s+S}{2}}\right)}^{2}M{M}^{†}∥}_{\mathrm{ber}}& \le \left(s+S\right)\mathrm{ber}\left(\mathrm{Re}M\right)+{\left({\displaystyle \frac{S-s}{2}}\right)}^2\hfill \\ \hfill & \le \left(s+S\right)\mathrm{ber}\left(M\right)+{\left({\displaystyle \frac{S-s}{2}}\right)}^2.\hfill \end{array}$$

and

$${∥\left|M\right|∥}_{\widetilde{\mathrm{ber}}}^2\le {\displaystyle \frac{s+S}{2}}\mathrm{ber}\left[2\mathrm{Re}M-{\displaystyle \frac{s+S}{2}}M{M}^{†}\right]+{\left({\displaystyle \frac{S-s}{2}}\right)}^2.$$

(78)

Here $\lambda ={\displaystyle \frac{s+S}{2}}>r={\displaystyle \frac{S-s}{2}}$ because $s>0$, so $\mathrm{Re}M\ge sI\ge 0$ and $\mathrm{ber}\left(\mathrm{Re}M\right)={\parallel \mathrm{Re}M\parallel }_{\mathrm{ber}}$; thus, the first bound may equivalently be written with ${\parallel \mathrm{Re}M\parallel }_{\mathrm{ber}}$.

Assume that $I\le M\le 3I$; then, by (78) we get for $s=1$ and $S=3$ that

$${∥M∥}_{\widetilde{\mathrm{ber}}}^2=\mathrm{ber}\left(M^2\right)\le 4\mathrm{ber}\left(M-M{M}^{†}\right)+1.$$

(79)

We will show that this inequality is sharp. Here we take ${\mathcal{H}}_{\Omega }=H$ with $\Omega$ equal to the unit sphere of H, so that ${\widehat{k}}_{\mu }$ ranges over all unit vectors of H and $\mathrm{ber}(·)$ reduces to the numerical radius $w(·)$; in particular, $\mathrm{ber}\left(P\right)=\parallel P\parallel$ for every positive P, which we use below.

Consider the matrix $M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right),$ which has the characteristic polynomial $\det (M-\lambda I)={(2-\lambda )}^2-1=0$, giving the spectrum $\sigma \left(M\right)=\{1,3\}$. Because both eigenvalues fit within the closed spectrum $[1,3]$, the operator ordering $I\le M\le 3I$ is verified.

Compute the matrix square $M^2$:

$$M^2=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}5& 4\\ 4& 5\end{array}\right).$$

The eigenvalues of this positive-definite matrix are given by $\det (M^2-\mu I)={(5-\mu )}^2-16=0,$ namely, $\mu =5\pm 4$. The maximum eigenvalue is ${\mu }_{\max }=9$. Thus:

$$\mathrm{LHS}=\mathrm{ber}\left(M^2\right)=9.$$

Also,

$$M-I=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)-\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right).$$

The eigenvalues are $\gamma =1\pm 1$, meaning $\sigma (M-I)=\{0,2\}$. The Berezin number matches the dominant boundary spectral point, ${\gamma }_{\max }=2$. Calculating the full expression gives:

$$\mathrm{RHS}=4\mathrm{ber}(M-I)+1=4\left(2\right)+1=9.$$

These prove that the inequality (79) is sharp. This shows that the equality holds in (79) for this example, so neither the coefficient 4 nor the additive constant 1 can be decreased while the other is held fixed; in this sense, the constants in (79) cannot be improved.

6. Conclusions

In this paper we established several families of reverse-type and refined inequalities for the Berezin number $\mathrm{ber}(·)$, the Berezin norm ${\parallel ·\parallel }_{\mathrm{ber}}$, and the associated quantity ${\parallel ·\parallel }_{\widetilde{\mathrm{ber}}}$ on reproducing kernel Hilbert spaces. The main contributions can be summarized as follows.

• Under disc-localization conditions, we showed that the gap ${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}-\mathrm{ber}\left(T\right)$ is bounded above by quantities depending only on the center and radius of the disc (Theorems 1–3 and their corollaries), and that equality in Theorem 1 is attained on the Hardy space (Example 1).
• Combining Buzano’s inequality with Dragomir’s refinement of the Schwarz inequality, we derived an upper bound for $\mathrm{ber}\left(A^{*}TA\right)$ (Theorem 4), together with power-type estimates (Theorem 5) and the bound

  $${\parallel T\parallel }_{\widetilde{\mathrm{ber}}}^2-{\left[\mathrm{ber}\left(T\right)\right]}^2\le \underset{\lambda }{\inf }{\parallel T-\lambda I\parallel }_{\widetilde{\mathrm{ber}}}^2$$
  (Theorem 7).
• Using the polar decomposition, we transferred these estimates to ${\left|T\right|}^{\alpha }$ and to the Aluthge transform of T (Theorems 9 and 10).
• For operators with closed range, we used the projection $M{M}^{†}$ to obtain reverse-type estimates under the disc condition $\parallel M-\lambda I\parallel \le r$ (Theorem 11); a concluding example shows that equality can occur in (79), so that neither constant in (79) can be decreased while the other is held fixed.

Taken together, these results refine and unify several known Berezin-number and Berezin-norm inequalities. Equality is attained in several of the displayed estimates, as shown by the corresponding examples; in particular, the constants identified explicitly in those results cannot be improved under the stated hypotheses. It would be of interest to investigate whether analogous disc-localization and reverse-type estimates hold for other operator radii and to explore applications of these bounds to Toeplitz and Hankel operators on reproducing kernel Hilbert spaces.