1 Introduction

Grüss [17] proved the following integral inequality which gives an approximation of the integral of the product in terms of the product of the integrals as follows:

$$\begin{aligned}&\left| \frac{1}{b-a}\overset{b}{\underset{a}{\int }}f(x)g(x)\mathrm{d}(x)-\frac{1}{b-a}\overset{b}{\underset{a}{\int }}f(x)\mathrm{d}(x).\frac{1}{b-a}\overset{b}{\underset{a}{\int }}g(x)\mathrm{d}(x)\right| \\&\quad \le \frac{1}{4}(\Phi -\phi )(\Gamma -\gamma ), \end{aligned}$$

where \(f,g:[a,b]\rightarrow \mathbb {R}\) are integrable on [a, b] and satisfy the condition

$$\begin{aligned} \phi \le f(x)\le \Phi ,\gamma \le g(x)\le \Gamma \end{aligned}$$

for each \(x\in \left[ a,b\right] \) , where \(\phi ,\)\(\Phi ,\)\(\gamma ,\)\(\Gamma \) are given real constants.

Moreover, the constant \(\frac{1}{4}\) is sharp in the sense that it cannot be replaced by a smaller one.

The discrete version of the Grüss’ inequality can be found in [22] as following:

Let \(a=(a_{1},\)\(\ldots \)\(,a_{n}),\)\(b=(b_{1},\)\(\ldots \)\(,b_{n})\) be two n-tuples of real numbers such that \(r\le a_{i}\le R\) and \(s\le b_{i}\le S\) for \(i=1,\ldots ,n.\) Then, one has

$$\begin{aligned}&\left| \frac{1}{n}\underset{i=1}{\overset{n}{ {\displaystyle \sum } }}a_{i}b_{i}-\frac{1}{n}\underset{i=1}{\overset{n}{ {\displaystyle \sum } }}a_{i}.\frac{1}{n}\underset{i=1}{\overset{n}{ {\displaystyle \sum } }}b_{i}\right| \\&\quad \le \frac{1}{n}\left[ \frac{n}{2}\right] \left( 1-\frac{1}{n}\left[ \frac{n}{2}\right] \right) \left( R-r\right) \left( S-s\right) , \end{aligned}$$

where [x] denotes the integer part of \(x\in \mathbb {R}\). In fact, the presented version of the discrete Grüss’ inequality is due to Biernacki et al. [7]. For Grüss-type inequalities, we refer to [3, 8, 9, 11, 12] and references therein.

Let A be a selfadjoint linear operator on a complex Hilbert space \(\mathcal {H}.\) The Gelfand map establishes a \(*\)-isometrically isomorphism \(\Phi \) between the set C(Sp(A)) of all continuous functions defined on the spectrum of A, denoted by Sp(A), and the \(C^{*}\)-algebra \(C^{*}(A)\) generated by A and the identity operator \(1_{\mathcal {H}}\) on \(\mathcal {H}\) as follows (see for instance [14]).

For any \(f,g\in C(Sp(A))\) and any \(\alpha ,\beta \in \mathbb {C}\), we have

  1. (i)

    \(\Phi \left( \alpha f+\beta g\right) =\alpha \Phi \left( f\right) +\beta \Phi \left( g\right) ;\)

  2. (ii)

    \(\Phi \left( fg\right) =\Phi \left( f\right) \Phi \left( g\right) \) and \(\Phi \left( \overline{f}\right) =\Phi \left( f\right) ^{*};\)

  3. (iii)

    \(\left| \left| \Phi \left( f\right) \right| \right| =\left| \left| f\right| \right| :=\underset{t\in Sp(A)}{\sup }\left| f\left( t\right) \right| ;\)

  4. (iv)

    \(\Phi \left( f_{0}\right) =1_\mathcal {H}\) and \(\Phi \left( f_{1}\right) =A\), where \(f_{0}\left( t\right) =1\) and \(f_{1}\left( t\right) =t\), for \(t\in Sp(A)\).

With this notation, we define

$$\begin{aligned} f\left( A\right) :=\Phi \left( f\right) \text { for all }f\in C(Sp(A)) \end{aligned}$$

and it is called the continuous functional calculus for the selfadjoint operator A.

If A is a selfadjoint operator and f is a real-valued continuous function on Sp(A),  then \(f(t)\ge 0\) for any \(t\in Sp(A)\) implies that \(f(A)\ge 0\) on \(\mathcal {H}\). Therefore, if f and g are real-valued functions on Sp(A), then the following basic property holds:

$$\begin{aligned} f(t)\ge g(t)\text { for any }t\in Sp(A)\text { implies that }f(A)\ge g(A) \end{aligned}$$
(1)

in the operator order of \(B(\mathcal {H}).\)

Let \(\Omega \) be an arbitrary set. Denote by \(\mathcal {F}\left( \Omega \right) \) the set of all complex-valued functions on \(\Omega \). A reproducing kernel Hilbert space (RKHS for short) on the set \(\Omega \) is a Hilbert space \(\mathcal {H=H}\left( \Omega \right) \subset \mathcal {F}\left( \Omega \right) \) with a function \(k_{\lambda }:\Omega \mathcal {\times }\Omega \rightarrow \mathcal {H}\), which is called the reproducing kernel enjoying the reproducing property \(k_{\lambda }:=k\left( .,\lambda \right) \in \mathcal {H}\) for all \(\lambda \in \Omega \) and \(f(\lambda )=\left\langle f,k_{\lambda }\right\rangle _{\mathcal {H}}\) holds for all \(\lambda \in \Omega \) and all \(f\in \mathcal {H}\) (see [24]). As it is known (see [2, 24]),

$$\begin{aligned} k_{\lambda }\left( z\right) =\sum \limits _{n=0}^{\infty }\overline{e_{n}\left( \lambda \right) }e_{n}\left( z\right) \end{aligned}$$

for any orthonormal basis \(\left\{ e_{n}\left( z\right) \right\} _{n\ge 0}\) of the space \(\mathcal {H}\left( \Omega \right) .\)

Let \(\widehat{k}_{\lambda }=\frac{k_{\lambda }}{\left\| k_{\lambda }\right\| }\) be the normalized reproducing kernel of the space \(\mathcal {H}\). For any bounded linear operator A on \(\mathcal {H}\), the Berezin transform of A is the function \(\widetilde{A}\) defined by (see [23])

$$\begin{aligned} \widetilde{A}(\lambda ):=\left\langle A\widehat{k}_{\lambda },\widehat{k}_{\lambda }\right\rangle _{\mathcal {H}}\ (\lambda \in \Omega ). \end{aligned}$$

The Berezin set and the Berezin number for operator A are defined by (see [19, 20])

$$\begin{aligned} {\mathrm{Ber}}\left( A\right) :=\left\{ \widetilde{A}\left( \lambda \right) :\lambda \in \Omega \right\} \text { and }{\mathrm{ber}}\left( A\right) :=\sup \left\{ \left| \widetilde{A}\left( \lambda \right) \right| :\lambda \in \Omega \right\} , \end{aligned}$$

respectively. Recently, some Berezin number inequalities have been obtained by authors [5, 15, 16, 25,26,27].

The numerical range and numerical radius of A in \(\mathcal {B}\left( \mathcal {H}\right) \) are, respectively, defined by

$$\begin{aligned} W\left( A\right) :=\left\{ \left\langle Af,f\right\rangle :f\in \mathcal {H}\text {, }\left\| f\right\| =1\right\} \text { and }w\left( A\right) :=\sup \left\{ \left| z\right| :z\in W\left( A\right) \right\} . \end{aligned}$$

The Berezin set and the Berezin number have a relationship with the numerical range and the numerical radius as follows:

$$\begin{aligned} {\mathrm{Ber}}\left( A\right) \subset W\left( A\right) \text { and }{\mathrm{ber}}\left( A\right) \le w\left( A\right) \le \left\| A\right\| . \end{aligned}$$

For the numerical radius and its applications, we refer to [1, 4, 6, 10, 13, 21], and references therein. The numerical radius inequality for the product of two operators is following:

$$\begin{aligned} w\left( AB\right) \le 4w\left( A\right) w\left( B\right) \end{aligned}$$

for the bounded linear operators A, B on the Hilbert space \(\mathcal {H}\). In that case that \(AB=BA\), then

$$\begin{aligned} w\left( AB\right) \le 2w\left( A\right) w\left( B\right) \end{aligned}$$

(see [18] for detailed information). So, the following questions are natural:

Is it true that the above inequality is also provided for Berezin number of operators? For which operator classes, there exists a number\(C>0\)such that

$$\begin{aligned} {\mathrm{ber}}\left( AB\right) \le C{\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) \,? \end{aligned}$$
(2)

In this paper, we study inequality (2) by using the transform \(C_{\alpha ,\beta }\left( A\right) \) on reproducing kernel Hilbert spaces (RKHS). Moreover, we give Grüss-type inequalities for selfadjoint operators in RKHS.

2 Berezin Number Inequalities for Two Operators

Let \(\alpha ,\beta \in \mathbb {C}\) and let \(A\in \mathcal {B}\left( \mathcal {H} \right) \) be a bounded linear operator. We define the following transform [11]

$$\begin{aligned} C_{\alpha ,\beta }\left( A\right) :=\left( A^{*}-\overline{\alpha }I\right) \left( \beta I-A\right) , \end{aligned}$$

where \(A^{*}\) denotes the adjoint of A. The transform \(C_{\alpha ,\beta }\left( .\right) \) has some interesting properties for \(A,B\in \mathcal {B}\left( \mathcal {H}\right) \) and \(\alpha ,\beta \in \mathbb {C}\) as following:

  1. (i)

    \(C_{\alpha ,\beta }\left( I\right) :=\left( 1-\overline{\alpha }\right) \left( \beta -1\right) I\) and \(C_{\alpha ,\alpha }\left( A\right) :=-\left( \alpha I-T\right) ^{*}\left( \alpha I-A\right) \).

  2. (ii)

    \(\left[ C_{\alpha ,\beta }\left( A\right) \right] ^{*}=C_{\beta ,\alpha }\left( A\right) \) and \(C_{\overline{\beta } ,\overline{\alpha }}\left( A^{*}\right) -C_{\alpha ,\beta }\left( A\right) =A^{*}A-AA^{*}\).

A bounded linear operator A on the RKHS \(\mathcal {H}\) is said to be accretive if \({\text {Re}}\widetilde{A}\left( \lambda \right) \ge 0\) for any \(\lambda \in \Omega \). Using this property, we have

$$\begin{aligned} {\text {Re}}\widetilde{C_{\alpha ,\beta }\left( A\right) }\left( \lambda \right) ={\text {Re}}\widetilde{C_{\beta ,\alpha }\left( A\right) }\left( \lambda \right) =\frac{1}{4}\left| \beta -\alpha \right| ^{2}-\left\| \left( A-\frac{\beta +\alpha }{2}I\right) \widehat{k} _{\lambda }\right\| ^{2} \end{aligned}$$

for any scalars \(\alpha ,\beta \in \mathbb {C}\) and \(\lambda \in \Omega \). So we can give a simple result.

Lemma 1

For \(A\in \mathcal {B}\left( \mathcal {H}\left( \Omega \right) \right) \) and complex numbers \(\alpha ,\beta \), the following statements are equivalent:

  1. (i)

    The transforms \(C_{\alpha ,\beta }\left( A\right) \) and \(C_{\overline{\alpha },\overline{\beta }}\left( A^{*}\right) \) are accretive;

  2. (ii)

    \(\left\| A\widehat{k}_{\lambda }-\dfrac{\beta +\alpha }{2}\widehat{k}_{\lambda }\right\| \le \dfrac{1}{2}\left| \beta -\alpha \right| \) and \(\left\| A^{*}\widehat{k}_{\lambda }-\dfrac{\overline{\beta }+\overline{\alpha }}{2} \widehat{k}_{\lambda }\right\| \le \dfrac{1}{2}\left| \beta -\alpha \right| \)for any \(\lambda \in \Omega \).

Theorem 1

Let \(C_{\alpha ,\beta }\left( A\right) \) and \(C_{\gamma ,\delta }\left( B\right) \) be accretive transform for \(A,B\in \mathcal {B}\left( \mathcal {H}\right) \) and \(\alpha ,\beta ,\gamma ,\delta \in \mathbb {C}\). Then,

$$\begin{aligned} {\mathrm{ber}}\left( BA\right) \le 3{\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) +\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| . \end{aligned}$$

Proof

By hypothesis, \(C_{\alpha ,\beta }\left( A\right) \) and \(C_{\gamma ,\delta }\left( B\right) \) are accretive, and then, from Lemma 1 we get \(\left\| A\widehat{k}_{\lambda }-\dfrac{\beta +\alpha }{2}\widehat{k}_{\lambda }\right\| \le \dfrac{1}{2}\left| \beta -\alpha \right| \) and \(\left\| B^{*}\widehat{k}_{\lambda }-\dfrac{\overline{\gamma }+\overline{\delta }}{2} \widehat{k}_{\lambda }\right\| \le \dfrac{1}{2}\left| \overline{\gamma }-\overline{\delta }\right| \) for any \(\lambda \in \Omega \).

Using the Schwarz inequality, we get that

$$\begin{aligned}&\left| \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta } -\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\rangle \right| \nonumber \\&\quad \le \left\| A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda }\right\| \left\| B^{*}\widehat{k}_{\eta }-\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\| \end{aligned}$$
(3)

for all \(\lambda ,\eta \in \Omega \).

Since \(\left\| f-\left\langle f,\widehat{k}_{\lambda }\right\rangle \widehat{k}_{\lambda }\right\| =\inf \limits _{\phi \in \mathbb {C}}\left\| f-\phi \widehat{k}_{\lambda }\right\| \) for any \(f\in \mathcal {H}\) and \(\lambda \in \Omega \), we have

$$\begin{aligned} \left\| A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda }\right\| \le \left\| A\widehat{k}_{\lambda } -\dfrac{\beta +\alpha }{2}\widehat{k}_{\lambda }\right\| \le \dfrac{1}{2}\left| \beta -\alpha \right| \end{aligned}$$

and

$$\begin{aligned} \left\| B^{*}\widehat{k}_{\eta }-\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\| \le \left\| B^{*}\widehat{k}_{\eta } -\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\| \le \dfrac{1}{2}\left| \gamma -\delta \right| \end{aligned}$$

for all \(\lambda ,\eta \in \Omega \). Hence, we have

$$\begin{aligned} \left\| A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda }\right\| \left\| B^{*}\widehat{k}_{\eta }-\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\| \le \dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| \end{aligned}$$
(4)

for all \(\lambda ,\eta \in \Omega \). On the other hand,

$$\begin{aligned}&\left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta }-\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\rangle \\&\quad =\left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle +\widetilde{A}\left( \lambda \right) \widetilde{B}\left( \eta \right) \left\langle \widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \\&\quad \quad \; -\widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle -\left\langle A\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \widetilde{B}\left( \eta \right) \end{aligned}$$

for all \(\lambda ,\eta \in \Omega \). Taking the modulus in the above equality, we have

$$\begin{aligned}&\left| \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta } -\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\rangle \right| \\&\quad \ge \left| \left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| -\left| \widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle +\left\langle A\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \widetilde{B}\left( \eta \right) -\widetilde{A}\left( \lambda \right) \widetilde{B}\left( \eta \right) \left\langle \widehat{k}_{\lambda } ,\widehat{k}_{\eta }\right\rangle \right| \\&\quad \ge \left| \left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| -\left| \widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| -\left| \left\langle A\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \widetilde{B}\left( \eta \right) \right| -\left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \eta \right) \left\langle \widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| , \end{aligned}$$

which is equivalent to

$$\begin{aligned}&\left| \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta } -\widetilde{B^{*}}\left( \eta \right) \widehat{k}_{\eta }\right\rangle \right| \nonumber \\&\quad \quad +\left| \widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| +\left| \left\langle A\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \widetilde{B}\left( \eta \right) \right| +\left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \eta \right) \left\langle \widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| \nonumber \\&\quad \ge \left| \left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| \end{aligned}$$
(5)

for all \(\lambda ,\eta \in \Omega \). So we have for \(\lambda =\eta \) from \(\left( 3\right) \)-\(\left( 5\right) \)

$$\begin{aligned}&\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| +\left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \lambda \right) \right| +\left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \lambda \right) \right| \nonumber \\&\quad +\left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \lambda \right) \right| \ge \left| \widetilde{BA}\left( \lambda \right) \right| . \end{aligned}$$
(6)

Taking the supremum in \(\left( 6\right) \) over \(\lambda \in \Omega \), we get that

$$\begin{aligned} {\mathrm{ber}}\left( BA\right) \le 3{\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) +\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| . \end{aligned}$$

This gives the desired result. \(\square \)

Now, we consider a different approach in the following result.

Theorem 2

Let \(C_{\alpha ,\beta }\left( A\right) \) and \(C_{\gamma ,\delta }\left( B\right) \) be accretive transform for \(A,B\in \mathcal {B}\left( \mathcal {H}\right) \) and \(\alpha ,\beta ,\gamma ,\delta \in \mathbb {C}\). Then,

$$\begin{aligned} {\mathrm{ber}}\left( BA\right) \le {\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) +\dfrac{1}{4}\left| \beta -\alpha \right| \left( \left| \gamma -\delta \right| +\left| \gamma +\delta \right| \right) . \end{aligned}$$

Proof

We can state the following inequality from the Schwarz inequality and the assumptions

$$\begin{aligned} \left| \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta } -\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\rangle \right|&\le \left\| A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda }\right\| \left\| B^{*} \widehat{k}_{\eta }-\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\| \nonumber \\&\le \left\| A\widehat{k}_{\lambda }-\dfrac{\beta +\alpha }{2}\widehat{k}_{\lambda }\right\| \left\| B^{*}\widehat{k}_{\eta }-\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\| \nonumber \\&\le \dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| \end{aligned}$$
(7)

for all \(\lambda ,\eta \in \Omega \).

Since

$$\begin{aligned}&\left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta }-\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\rangle \\&\quad =\left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle -\widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle -\dfrac{\gamma +\delta }{2}\left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \end{aligned}$$

on taking the modulus in this inequality, we obtain

$$\begin{aligned}&\left| \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta } -\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\rangle \right| \nonumber \\&\quad \ge \left| \left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| -\left| \widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| -\left| \dfrac{\gamma +\delta }{2}\right| \left| \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \right| \end{aligned}$$
(8)

for all \(\lambda ,\eta \in \Omega \). Then, we have for \(\lambda =\eta \in \Omega \) from (7) and (8)

$$\begin{aligned} \left| \widetilde{BA}\left( \lambda \right) \right|&\le \left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \lambda \right) \right| +\left| \dfrac{\gamma +\delta }{2}\right| \left| \widetilde{A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) }\right| +\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| \\&\le \left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \lambda \right) \right| +\left| \dfrac{\gamma +\delta }{2}\right| \left\| A\widehat{k}_{\lambda }-\dfrac{\beta +\alpha }{2}\widehat{k}_{\lambda }\right\| +\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| \\&\le \left| \widetilde{A}\left( \lambda \right) \widetilde{B}\left( \lambda \right) \right| +\dfrac{1}{4}\left| \beta -\alpha \right| \left( \left| \gamma -\delta \right| +\left| \gamma +\delta \right| \right) . \end{aligned}$$

Taking the supremum over \(\lambda \in \Omega \) in the above inequality, we have

$$\begin{aligned} {\mathrm{ber}}\left( BA\right) \le {\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) +\dfrac{1}{4}\left| \beta -\alpha \right| \left( \left| \gamma -\delta \right| +\left| \gamma +\delta \right| \right) . \end{aligned}$$

This proves the theorem. \(\square \)

By using arguments in above theorem, we can get the following result.

Corollary 1

Let \(C_{\alpha ,\beta }\left( A\right) \) and \(C_{\gamma ,\delta }\left( B\right) \) be accretive transform for \(A,B\in \mathcal {B}\left( \mathcal {H}\right) \) and \(\alpha ,\beta ,\gamma ,\delta \in \mathbb {C}\). Then,

$$\begin{aligned} {\mathrm{ber}}\left( BA\right) \le {\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) +\left| \gamma +\delta \right| {\mathrm{ber}}\left( A\right) +\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| . \end{aligned}$$

Proof

Indeed, we have that

$$\begin{aligned} \left\langle A\widehat{k}_{\lambda }-\widetilde{A}\left( \lambda \right) \widehat{k}_{\lambda },B^{*}\widehat{k}_{\eta }-\dfrac{\overline{\gamma }+\overline{\delta }}{2}\widehat{k}_{\eta }\right\rangle&=\left\langle BA\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle -\widetilde{A}\left( \lambda \right) \left\langle B\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \\&\quad \; -\dfrac{\gamma +\delta }{2}\left\langle A\widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle +\dfrac{\gamma +\delta }{2}\widetilde{A}\left( \lambda \right) \left\langle \widehat{k}_{\lambda },\widehat{k}_{\eta }\right\rangle \end{aligned}$$

for all \(\lambda ,\eta \in \Omega \). Taking the supremum on \(\lambda =\eta \in \Omega \) and using the same arguments in the proof of the above theorem, we get

$$\begin{aligned} {\mathrm{ber}}\left( BA\right) \le {\mathrm{ber}}\left( A\right) {\mathrm{ber}}\left( B\right) +\left| \gamma +\delta \right| {\mathrm{ber}}\left( A\right) +\dfrac{1}{4}\left| \beta -\alpha \right| \left| \gamma -\delta \right| \end{aligned}$$

for the operators \(A,B\in \mathcal {B}\left( \mathcal {H}\right) \). \(\square \)

3 Grüss-Type Inequality

Now, we give a Grüss-type inequality for selfadjoint operators on a RKHS \(\mathcal {H=H(}\Omega \mathcal {)}\).

Theorem 3

Let \(A\in \mathcal {B}\left( \mathcal {H}\right) \) be a selfadjoint operator and assume that \(Sp\left( A\right) \subseteq \left[ m,M\right] \) for some scalars \(m<M\). If f and g are continuous on [m, M], then

$$\begin{aligned}&\widetilde{f\left( A\right) g\left( A\right) }\left( \mu \right) -\widetilde{f\left( A\right) }\left( \mu \right) \widetilde{g\left( A\right) }\left( \lambda \right) -\frac{\gamma +\Gamma }{2}\left[ \widetilde{g\left( A\right) }\left( \mu \right) -\widetilde{g\left( A\right) }\left( \lambda \right) \right] \nonumber \\&\quad \le \frac{1}{2}.\left( \Gamma -\gamma \right) \left[ \left| \left| g\left( A\right) \right| \right| ^{2}+\left( \widetilde{g\left( A\right) }\left( \lambda \right) \right) ^{2}-2\widetilde{g\left( A\right) }\left( \lambda \right) \widetilde{g\left( A\right) }\left( \mu \right) \right] ^{1/2} \end{aligned}$$
(9)

for any \(\lambda ,\mu \in \Omega \), where \(\gamma =\underset{t\in \left[ m,M\right] }{\min }f\left( t\right) \), \(\Gamma =\underset{t\in \left[ m,M\right] }{\max }f\left( t\right) \)

Proof

Indeed, we have the identity

$$\begin{aligned}&\langle \left( f\left( A\right) -\xi .1_{\mathcal {H}}\right) \left( g\left( A\right) -\langle g\left( A\right) \widehat{k}_{\lambda } ,\widehat{k}_{\lambda }\rangle .1_{\mathcal {H}}\right) \widehat{k}_{\mu },\widehat{k}_{\mu }\rangle \nonumber \\&\quad =\langle f\left( A\right) g\left( A\right) \widehat{k}_{\mu } ,\widehat{k}_{\mu }\rangle -\xi .\left[ \langle g\left( A\right) \widehat{k}_{\mu },\widehat{k}_{\mu }\rangle -\langle g\left( A\right) \widehat{k}_{\lambda },\widehat{k}_{\lambda }\rangle \right] \nonumber \\&\quad \quad \;-\langle g\left( A\right) \widehat{k}_{\lambda },\widehat{k}_{\lambda }\rangle \langle f\left( A\right) \widehat{k}_{\mu },\widehat{k}_{\mu } \rangle \end{aligned}$$
(10)

for each \(\xi \in \mathbb {R}\) and \(\lambda ,\mu \in \Omega \).

Taking the modulus in (10), we obtain

$$\begin{aligned}&\left| \widetilde{f\left( A\right) g\left( A\right) }\left( \mu \right) -\xi .\left[ \widetilde{g\left( A\right) }\left( \mu \right) -\widetilde{g\left( A\right) }\left( \lambda \right) \right] -\widetilde{g\left( A\right) }\left( \lambda \right) \widetilde{f\left( A\right) }\left( \mu \right) \right| \nonumber \\&\quad =\left| \left\langle \left( g\left( A\right) -\widetilde{g\left( A\right) }\left( \lambda \right) .1_{\mathcal {H}}\right) \widehat{k}_{\mu },(f\left( A\right) -\xi .1_{{H}})\widehat{k}_{\mu }\right\rangle \right| \nonumber \\&\quad \le \left\| g\left( A\right) \widehat{k}_{\mu }-\widetilde{g\left( A\right) }\left( \lambda \right) \widehat{k}_{\mu }\rangle \right\| \left\| f\left( A\right) \widehat{k}_{\mu }-\xi \widehat{k}_{\mu }\right\| \nonumber \\&\quad =\left[ \left\| g\left( A\right) \widehat{k}_{\mu }\right\| ^{2}+\left( \widetilde{g\left( A\right) }\left( \lambda \right) \right) ^{2}-2\widetilde{g\left( A\right) }\left( \lambda \right) \widetilde{g\left( A\right) }\left( \mu \right) \right] ^{\frac{1}{2}}\nonumber \\&\quad \quad \times \left\| \left( f\left( A\right) \widehat{k}_{\mu } -\lambda \widehat{k}_{\mu }\right) \right\| \nonumber \\&\quad \le \left[ \left\| g\left( A\right) \right\| ^{2}+\left( \widetilde{g\left( A\right) }\left( \lambda \right) \right) ^{2} -2\widetilde{g\left( A\right) }\left( \lambda \right) \widetilde{g\left( A\right) }\left( \mu \right) \right] ^{\frac{1}{2}}\nonumber \\&\quad \quad \times \left\| f\left( A\right) -\lambda .1_{\mathcal {H}}\right\| \end{aligned}$$
(11)

for any \(\lambda ,\mu \in \Omega \).

Since \(\gamma =\underset{t\in \left[ m,M\right] }{\min }f\left( t\right) \ \)and \(\Gamma =\underset{t\in \left[ m,M\right] }{\max }f\left( t\right) \), by the property (1) we have that \(\gamma \le \widetilde{f\left( A\right) }\left( \mu \right) \le \Gamma \) for each \(\mu \in \Omega \) which is obviously equivalent to

$$\begin{aligned} \left| \widetilde{f\left( A\right) }\left( \mu \right) -\frac{\gamma +\Gamma }{2}\left\| \widehat{k}_{\mu }\right\| ^{2}\right| \le \frac{1}{2}\left( \Gamma -\gamma \right) \end{aligned}$$

or with

$$\begin{aligned} \left| \widetilde{\left( f\left( A\right) -\frac{\gamma +\Gamma }{2}1_{\mathcal {H}}\right) }\left( \mu \right) \right| \le \frac{1}{2}\left( \Gamma -\gamma \right) \end{aligned}$$

for each \(\mu \in \Omega \).

Taking the supremum in this inequality, we get

$$\begin{aligned} \left\| f\left( A\right) -\frac{\gamma +\Gamma }{2}.1_{\mathcal {H} }\right\| \le \frac{1}{2}\left( \Gamma -\gamma \right) , \end{aligned}$$

which together with the inequality (11) applied for \(\xi =\frac{\gamma +\Gamma }{2}\) produces the desired results. \(\square \)

As a special case of the above theorem, we can give the following result.

Corollary 2

Let \(A\in \mathcal {B}\left( \mathcal {H}\right) \) be a selfadjoint operator and assume that \(Sp\left( A\right) \subseteq \left[ m,M\right] \) for some scalars \(m<M\). Then,

$$\begin{aligned} {\mathrm{ber}}\left( f^{2}\left( A\right) -\left( f\left( A\right) \right) ^{2}\right) \le \frac{1}{2}\left( \Gamma -\gamma \right) \left[ \left\| f\left( A\right) \right\| ^{2}-{\mathrm{ber}}\left( \left( f\left( A\right) \right) ^{2}\right) \right] ^{\frac{1}{2}} \end{aligned}$$

for each \(\lambda \in \Omega \), where \(\gamma =\underset{t\in \left[ m,M\right] }{\min }f\left( t\right) \), \(\Gamma =\underset{t\in \left[ m,M\right] }{\max }f\left( t\right) \).

Proof

Taking \(f=g\) and \(\lambda =\mu \) in (9), then

$$\begin{aligned}&\left| \widetilde{f^{2}\left( A\right) }\left( \lambda \right) -\left( \widetilde{f\left( A\right) }\left( \lambda \right) \right) ^{2}\right| \\&\quad \le \frac{1}{2}\left( \Gamma -\gamma \right) \left[ \left\| f\left( A\right) \right\| ^{2}-\left( \widetilde{f\left( A\right) }\left( \lambda \right) \right) ^{2}\right] ^{\frac{1}{2}} \end{aligned}$$

for all \(\lambda \in \Omega \). Taking the supremum on \(\lambda \in \Omega \) in above inequality, we have

$$\begin{aligned} {\mathrm{ber}}\left( f^{2}\left( A\right) -\left( f\left( A\right) \right) ^{2}\right) \le \frac{1}{2}\left( \Gamma -\gamma \right) \left[ \left\| f\left( A\right) \right\| ^{2}-{\mathrm{ber}}\left( \left( f\left( A\right) \right) ^{2}\right) \right] ^{\frac{1}{2}} \end{aligned}$$

for any selfadjoint operator \(A\in \mathcal {B}\left( \mathcal {H}\right) \). This proves the theorem. \(\square \)