A New Family of Semi-Norms Between the Berezin Radius and the Berezin Norm

Abstract

A functional Hilbert space is the Hilbert space ℋ of complex-valued functions on some set $\Theta \subseteq \mathbb{C}$ such that the evaluation functionals $\varphi _{\tau }\left ( f\right ) =f\left ( \tau \right )$, $\tau \in \Theta$, are continuous on ℋ. The Berezin number of an operator $X$ is defined by $\mathbf{ber}(X)=\underset{\tau \in {\Theta } }{\sup }\big \vert \widetilde{X}(\tau )\big \vert = \underset{\tau \in {\Theta } }{\sup }\big \vert \langle X\hat{k}_{\tau },\hat{k}_{\tau }\rangle \big \vert$, where the operator $X$ acts on the reproducing kernel Hilbert space ${\mathscr{H}}={\mathscr{H}(}\Theta )$ over some (non-empty) set $\Theta$. In this paper, we introduce a new family involving means $\Vert \cdot \Vert _{\sigma _{t}}$ between the Berezin radius and the Berezin norm. Among other results, it is shown that if $X\in {\mathscr{L}}({\mathscr{H}})$ and $f$, $g$ are two non-negative continuous functions defined on $[0,\infty )$ such that $f(t)g(t) = t,\,(t\geqslant 0)$, then

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma }\leqslant \textbf{ber}\left (\frac{1}{4}(f^{4}( \vert X\vert )+g^{4}(\vert X^{*}\vert ))+\frac{1}{2}\vert X\vert ^{2} \right ) \end{aligned}$$

and

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma }\leqslant \frac{1}{2}\sqrt{\textbf{ber} \left (f^{4}(\vert X\vert )+g^{2}(\vert X\vert ^{2})\right ) \textbf{ber}\left (f^{2}(\vert X\vert ^{2})+g^{4}(\vert X^{*}\vert ) \right )}, \end{aligned}$$

where $\sigma$ is a mean dominated by the arithmetic mean $\nabla$.

1 Introduction

Let $\mathscr{L}\left ( \mathscr{H}\right )$ be the $C^{*}$-algebra of all bounded linear operators defined on a complex Hilbert space $\left ( \mathscr{H},\left \langle \cdot ,\cdot \right \rangle \right )$ with the identity operator $I_{\mathscr{H}}$ in $\mathscr{L}\left ( \mathscr{H}\right )$. When $\mathscr{H} = \mathbb{C}^{n}$, we identify $\mathscr{L}\left ( \mathscr{H}\right )$ with the algebra $\mathcal{M}_{n}(\mathbb{C})$ of $n$-by-$n$ complex matrices.

A functional Hilbert space is the Hilbert space of complex-valued functions on some set $\Theta \subseteq \mathbb{C}$ such that the evaluation functionals $\varphi _{\tau}\left ( f\right ) =f\left ( \tau \right )$, $\tau \in \Theta$, are continuous on ℋ. Then, by the Riesz representation theorem there is a unique element $k_{\tau}\in \mathscr{H}$ such that $f\left ( \tau \right ) =\left \langle f,k_{\tau}\right \rangle$ for all $f\in \mathscr{H}$ and every $\tau \in \Theta$. The function $k$ on $\Theta \times \Theta$ defined by $k\left ( z,\tau \right ) =k_{\tau}\left ( z\right )$ is called the reproducing kernel of ℋ, see [2, 4, 5, 17] and references therein. It was shown that $k_{\tau}\left ( z\right )$ can be represented by

$$\begin{aligned} k_{\tau}\left ( z\right ) =\sum \limits _{n=1}^{\infty}{ \overline {e_{n}\left ( \tau \right ) }e_{n}\left ( z\right ) } \end{aligned}$$

for any orthonormal basis $\{e_{n}\}_{n\geq 1}$ of ℋ, see [30]. For example, for the Hardy-Hilbert space $\mathscr{H}^{2}=\mathscr{H}^{2}\left ( \mathbb{D}\right )$ over the unit disc $\mathbb{D}=\left \{ z\in \mathbb{C}:\vert z\vert <1\right \}$, $\{z^{n}\}_{n\geq 1}$ is an orthonormal basis, therefore the reproducing kernel of $\mathscr{H}^{2}$ is the function $k_{\tau}\left ( z\right ) =\sum \limits _{n=1}^{\infty}{ \overline{\tau _{n}}z^{n}}=\left ( 1-\overline{\tau}z\right ) ^{-1}$, $\tau \in \mathbb{D}$. Let $\widehat{k}_{\tau}= \frac{{k_{\tau}}}{{\left \Vert {k_{\tau}}\right \Vert }}$ be the normalized reproducing kernel of the space ℋ. For a given a bounded linear operator $X$ on ℋ, the Berezin symbol (or Berezin transform) of $X$ is the bounded function $\widetilde{X}$ on $\Theta$ defined by

$$\begin{aligned} \widetilde{X}\left ( \tau \right ) =\left \langle {X\widehat{k}_{ \tau }\left ( z\right ) ,\widehat{k}_{\tau}\left ( z\right ) }\right \rangle ,\text{ }\tau \in \Theta . \end{aligned}$$

An important property of the Berezin symbol is that for all $X,Y\in \mathscr{L}\left ( \mathscr{H}\right )$, if $\widetilde{X}\left ( \tau \right ) =\widetilde{Y}\left ( \tau \right )$ for all $\tau \in \Theta$, then $X=Y$ (at least when ℋ consists of analytic functions, see Zhu [31]). For more details, see [3, 6, 8–10, 12–14, 16, 18–29]. So, the map $X\rightarrow \widetilde{X}$ is injective [15]. The Berezin set and the Berezin number(radius) of an operator $X$ are defined, respectively, by

$$\begin{aligned} \textbf{Ber}\left ( X\right ) =\left \{ {\widetilde{X}\left ( \tau \right ) :\tau \in \Theta}\right \} =\mathrm{Range}\left ( { \widetilde{X}}\right ) \end{aligned}$$

and

$$\begin{aligned} \textbf{ber}\left ( X\right ) =\sup \left \{ {\left \vert \gamma \right \vert :\gamma \in \mathrm{Ber}\left ( X\right ) }\right \} = \sup _{\tau \in \Theta }\left \vert \widetilde{X}\left ( \tau \right ) \right \vert . \end{aligned}$$

The Berezin norm of an operator $X\in \mathscr{L}\left ( \mathscr{H}\right )$ is defined by

$$\begin{aligned} \left \Vert X\right \Vert _{\mathrm{ber}}:=\sup _{\tau \in \Theta} \big\Vert {X\widehat{k}_{\tau}}\big\Vert . \end{aligned}$$

For $X,Y \in {\mathscr{L}}({\mathscr{H}})$, it is clear from the above definitions of the Berezin radius (or the Berezin number) and the Berezin norm that the following properties hold:

• (1) $\textbf{ber}( t X)=\vert t\vert \textbf{ber}(X)$ for all $t\in \mathbb{C}$;

• (2) $\textbf{ber}(X+Y)\leqslant \textbf{ber}(X)+\textbf{ber}(Y)$;

• (3) $\textbf{ber}(X)\leqslant \left \Vert X\right \Vert _{\mathrm{ber}}$ and $\textbf{ber}(X)=\textbf{ber}(X^{*})$;

• (4) $\left \Vert t X\right \Vert _{\mathrm{ber}}=\vert t\vert \left \Vert X\right \Vert _{\mathrm{ber}}$ for all $t\in \mathbb{C}$;

• (5) $\left \Vert X+Y\right \Vert _{\mathrm{ber}}\leq \left \Vert X\right \Vert _{\mathrm{ber}}+\left \Vert Y\right \Vert _{\mathrm{ber}}$.

In the recent paper [7], the authors defined the $t$-Berezin norm on ${\mathscr{L}}({\mathscr{H}})$ as follows:

$$\begin{aligned} \Vert X\Vert _{ t-\textbf{ber}}=\underset{\tau \in \Theta }{\sup}\sqrt{ t\big\vert \widetilde{X}\left ( \tau \right ) \big\vert ^{2}+(1- t)\big\Vert X\hat{k}_{\tau}\big\Vert ^{2}}. \end{aligned}$$

The $t$-Berezin norm is also a norm on ${\mathscr{L}}({\mathscr{H}})$ for $t\in [0, 1)$, and for $t=1$ it is a norm if the functional Hilbert space has the Ber property, i.e., for any two operators $X, Y\in {\mathscr{L}}({\mathscr{H}})$ such that $\widetilde{X}(\tau )=\widetilde{Y}(\tau )$ for all $\tau \in \Theta$, we have $X=Y$. Hence, the $t$-Berezin norm is a norm in the familiar functional for Hilbert spaces, for instance Hardy and Bergman spaces. The $t$-Berezin norm satisfies the following inequalities:

$$\begin{aligned} \mathbf{ber}(X)\leqslant \Vert X\Vert _{ t-\textbf{ber}}\leqslant \Vert X\Vert _{\mathbf{ber}} \quad \textrm{for}\,\, t\in [0, 1]. \end{aligned}$$

A binary function $\sigma$ on $[0,+\infty )$ is called a mean, if the following conditions are satisfied:

• (i) If $a\leqslant b$, then $a\leqslant a\,\sigma \,b \leqslant b$;

• (ii) $a\leqslant c$ and $b\leqslant d$ imply $a\,\sigma \,b\leq c\,\sigma \,d$;

• (iii) $\sigma$ is continuous in both variables;

• (iv) $t(a\,\sigma b)\leq (ta)\sigma (tb)\,\,(t>0)$.

For instance, if $\mu \in (0,1)$, the weighted geometric mean is $a\sharp _{\mu }b=a^{1-\mu}b^{\mu}$. The case $\mu =1/2$ gives rise to the geometric mean $a\sharp b$. A mean $\sigma$ is symmetric if $a\,\sigma \,b=b\,\sigma \,a$ for all positive numbers $a$, $b$. For a symmetric mean $\sigma$, a parametrized mean $\sigma _{t},\,\, 0 \leq t\leq 1$ is called an interpolational path for $\sigma$ if it satisfies

• (1) $a\,\sigma _{0} \,b=a$, $a\,\sigma _{1/2}\,b=a\,\sigma \,b$, and $a\,\sigma _{1}\,b=b$;

• (2) $(a\,\sigma _{p} \,b)\sigma (a\sigma _{q}\,b)=a\sigma _{\frac{p+q}{2}} \,b$ for all $p,q\in [0,1]$;

• (3) The map $t\in [0,1]\mapsto a\,\sigma _{t}\,b$ is continuous for each $a$ and $b$;

• (4) $\sigma _{t}$ is increasing in each of its components for $t\in [0, 1]$.

It is easy to see that the set of all $r\in [0, 1]$ satisfying

$$\begin{aligned} (a\sigma _{p} b)\sigma _{r}(a\sigma _{q}b)=a\sigma _{rp+(1-r)q}b \end{aligned}$$

(1.1)

for all $p$, $q$ is a convex subset of $[0, 1]$ including 0 and 1. For instance, the power means

$$\begin{aligned} a \,m_{r}b=\left (\frac{a^{r}+b^{r}}{2}\right )^{\frac{1}{r}}\qquad (r \in [-1,1]) \end{aligned}$$

are some typical interpolational means. Their interpolational paths are

$$\begin{aligned} a\,m_{r,t}b=\left ({(1-t)a^{r}+t b^{r}}\right )^{\frac{1}{r}}\qquad (t \in [0,1]). \end{aligned}$$

In particular, $a\,m_{1,t}b=a\nabla _{t} b=(1-t)a+tb$ is the weighted arithmetic mean, $a\,m_{0,t}b=a\sharp _{t} b=a^{1-t}b^{t}$ is the weighted geometric mean and $a\,m_{-1,t}b=a!_{t} b=\left ((1-t)a^{-1}+tb^{-1}\right )^{-1}$ is the weighted harmonic mean. It is well-known that $a!_{t}b\leqslant a\sharp _{t} b\leqslant a\nabla _{t} b$ for positive numbers $a$ and $b$ and $t\in [0, 1]$. For more information about means, see [25] and references therein.

In this paper, we define a new quantity and establish some related results. The main ideas of this paper are stimulated by [7] and [11].

2 Main Results

We begin this section with the following definition.

Definition 2.1

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $\sigma _{t}$ be an interpolational path of a symmetric mean $\sigma$. We define

$$\begin{aligned} \Vert X\Vert _{\sigma _{t}}=\underset{\tau \in \Theta }{\sup}\left \{\sqrt{\big\vert \widetilde{X}(\tau )\big\vert ^{2}\,\, \sigma _{t}\,\,\big\Vert X\widehat{k}_{\tau}\big\Vert ^{2} }\right \}\,\, \textrm{for}\,\,\,0\leqslant t\leqslant 1. \end{aligned}$$

Example 2.2

We consider the Hardy-Hilbert space, $\mathscr{H}^{2}\left ( \mathbb{D}\right )$, defined over the unit disc $\mathbb{D}=\left \{ z\in \mathbb{C}:\vert z\vert <1\right \}$ as follows:

$$ \mathscr{H}^{2}\left ( \mathbb{D}\right )=\{f: \mathbb{D}\to \mathbb{C}:f(z)=\sum _{n=0}^{\infty}a_{n}z^{n} \:\textrm{and}\: \sum _{n=0}^{\infty }|a_{n}|^{2}< \infty \}. $$

The inner product on $\mathscr{H}^{2}\left ( \mathbb{D}\right )$ is defined by $\langle f, g\rangle =\sum _{n=0}^{\infty} a_{n}\overline{b_{n}}$, for $f(z)=\sum _{n=0}^{\infty}a_{n}z^{n}$ and $g(z)=\sum _{n=0}^{\infty}b_{n}z^{n}$ and $\{z^{n}\}_{n\geq 1}$ forms an orthonormal basis. We can identify $\mathscr{H}^{2}\left ( \mathbb{D}\right )$ with $l^{2}(\mathbb{N})$, since

$$ f(z)=\sum _{n=0}^{\infty}a_{n}z^{n} \leftrightarrow (a_{0}, a_{1}, a_{2}, \ldots ). $$

Therefore, the reproducing kernel of $\mathscr{H}^{2}\left ( \mathbb{D}\right )$ is given by the function $k_{\tau}\left ( z\right ) =\sum \limits _{n=1}^{\infty}{ \overline{\tau}^{n}z^{n}}=\left ( 1-\overline{\tau}z\right ) ^{-1}$, $\tau \in \mathbb{D}$ and

$$ \|k_{\tau}\|^{2}=\langle k_{\tau}, k_{\tau}\rangle =k_{\tau}(\tau )= \left ( 1-\overline{\tau}\tau \right ) ^{-1}=\left ( 1-|\tau |^{2} \right ) ^{-1}, $$

for any $\tau \in \mathbb{D}$.

On $l^{2}(\mathbb{N})$, we consider the unilateral shift operator $U$ defined by

$$ U(a_{0},a_{1},a_{2},a_{3},\ldots )=(0,a_{0},a_{1},a_{2},\ldots ) $$

for $(a_{0},a_{1},a_{2}, a_{3},\ldots )\in l^{2}(\mathbb{N})$. Thus, for any $\tau \in \mathbb{D}$, we have

$$\begin{aligned} |\widetilde{U}\left ( \tau \right )|^{2} =&\left |\left \langle {U \widehat{k}_{\tau }\left ( z\right ) ,\widehat{k}_{\tau}\left ( z \right ) }\right \rangle \right |^{2} =\frac{1}{\|k_{\tau}\|^{4}} \left |\left \langle {U{k}_{\tau }\left ( z\right ) ,{k}_{\tau}\left ( z\right ) }\right \rangle \right |^{2} \\ =&\frac{1}{\|k_{\tau}\|^{4}}\left |\left \langle {U(1, \overline{\tau}, \overline{\tau}^{2},\ldots ),(1, \overline{\tau}, \overline{\tau}^{2},\ldots ) }\right \rangle \right |^{2} \\ =&\frac{1}{\|k_{\tau}\|^{4}}\left |\left \langle {(0, 1, \overline{\tau}, \overline{\tau}^{2},\ldots ),(1, \overline{\tau}, \overline{\tau}^{2},\ldots ) }\right \rangle \right |^{2} \\ =&\frac{1}{\|k_{\tau}\|^{4}}\left |\sum _{j=0}^{\infty} \overline{\tau}^{j}\tau ^{j+1} \right |^{2}= \frac{1}{\|k_{\tau}\|^{4}}\left (\frac{1}{1-|\tau |^{2}}|\tau | \right )^{2}=|\tau |^{2}, \ \end{aligned}$$

and due to the fact that $U$ is an isometry, we conclude that

$$ \Vert U\widehat{k}_{\tau }\Vert ^{2}=\Vert \widehat{k}_{\tau }\Vert ^{2}=1. $$

Then, we obtain that

$$ |\widetilde{U}\left ( \tau \right )|^{2}\leq |\tau |^{2}< 1=\Vert U \widehat{k}_{\tau }\Vert ^{2}, $$

(2.1)

and by the monotonicity of ${\sigma _{t}}$, we have

$$\begin{aligned} 1=\sup _{\tau \in \mathbb{D} }|\tau |=\sup _{\tau \in \mathbb{D} } \left \vert \widetilde{U}\left ( \tau \right ) \right | \leq \underset{\tau \in \mathbb{D} }{\sup}\left \{\sqrt{\big\vert \widetilde{U}(\tau )\big\vert ^{2}\,\, \sigma _{t}\,\,\big\Vert U\widehat{k}_{\tau}\big\Vert ^{2} }\right \} \leq \sup _{\tau \in \mathbb{D} }\big\Vert {U\widehat{k}_{\tau}} \big\Vert =1 \end{aligned}$$

(2.2)

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

In conclusion, we have that $\mathbf{ber}(U)\leqslant \Vert U\Vert _{\sigma _{t}}\leqslant \Vert U\Vert _{\mathbf{ber}}$, and in particular, $\Vert U\Vert _{\sigma _{t}}=1$ for any $t\in [0,1]$.

Following the ideas from the previous example and given that $\left \vert \widetilde{X}\left ( \tau \right ) \right \vert \leq \big\Vert {X\widehat{k}_{\tau}}\big\Vert$, from the Cauchy-Schwartz inequality, we have that

$$\begin{aligned} \mathbf{ber}(X)\leqslant \Vert X\Vert _{\sigma _{t}}\leqslant \Vert X \Vert _{\mathbf{ber}} \end{aligned}$$

for $t\in [0,1]$ and $X\in{\mathscr{L}}({\mathscr{H}})$. It is easy to see that for the special case $\sigma _{t}=\nabla _{t}\,\,(0\leqslant t \leqslant 1)$, we have $\Vert \cdot \Vert _{\sigma _{t}}=\Vert \cdot \Vert _{ (1-t)- \textbf{ber}}$.

The next proposition shows some properties of $\Vert \cdot \Vert _{\sigma _{t}}$.

Proposition 2.3

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $\sigma _{t}$, $\tau _{\mu}$ be interpolational paths of symmetric means $\sigma$ and $\tau$. Then

• (1) $\Vert X \Vert _{\sigma _{0}}=\mathbf{ber}(X)$ and $\Vert X \Vert _{\sigma _{1}}=\Vert X\Vert _{\mathbf{ber}}$;

• (2) $\Vert X \Vert _{\sigma _{t}}\leqslant \sqrt{\mathbf{ber}^{2}(X)\,\, \sigma _{t}\,\,\Vert X\Vert _{\mathbf{ber}}^{2}}$ for $t\in [0,1]$;

• (3) $\Vert \tau X \Vert _{\sigma _{t}}=\vert \tau \vert \Vert X \Vert _{ \sigma _{t}}$ for all $\tau \in \mathbb{C}$;

• (4) If the functional Hilbert space has the Ber property and $t\in [0,1)$, then $\Vert X \Vert _{\sigma _{t}}=0$ if and only if $X=0$;

• (5) If $\sigma _{t}\leqslant \tau _{s}$, then $\Vert X \Vert _{\sigma _{t}}\leqslant \Vert X \Vert _{\tau _{s}}$ for $s,t\in [0,1]$.

Remark 2.4

If $X\in{\mathscr{L}}({\mathscr{H}})$, then

$$\begin{aligned} \Vert \,\vert X\vert \,\Vert ^{2}_{\mathbf{ber}}&= \underset{\tau \in \Theta }{\sup}\big\Vert \vert X\vert \hat{k}_{\tau}\big\Vert ^{2}= \underset{\tau \in \Theta }{\sup}\langle \vert X\vert \hat{k}_{\tau}, \vert X\vert \hat{k}_{ \tau}\rangle \\ & =\underset{\tau \in \Theta }{\sup}\langle X^{*}X \hat{k}_{\tau}, \hat{k}_{\tau}\rangle = \underset{\tau \in \Theta }{\sup}\langle X \hat{k}_{\tau}, X\hat{k}_{\tau}\rangle = \underset{\tau \in \Theta }{\sup}\big\Vert X \hat{k}_{\tau}\big\Vert ^{2}=\Vert X\Vert ^{2}_{ \mathbf{ber}} \end{aligned}$$

and for a semi-hyponormal operator $X$, i.e. $\vert X^{*} \vert \leqslant \vert X \vert$, the mixed Cauchy-Schwarz inequality $\big\vert \langle X\hat{k}_{\tau}, \hat{k}_{\tau}\rangle \big\vert ^{2} \leqslant \langle \vert X\vert \hat{k}_{\tau}, \hat{k}_{\tau}\rangle \langle \vert X^{*}\vert \hat{k}_{\tau}, \hat{k}_{\tau}\rangle$ implies that $\big\vert \langle X\hat{k}_{\tau}, \hat{k}_{\tau}\rangle \big\vert ^{2} \leqslant \langle \vert X\vert \hat{k}_{\tau}, \hat{k}_{\tau}\rangle ^{2}$ for all $\hat{k}_{\tau}\in{\mathscr{H}}$, and then

$$\begin{aligned} \mathbf{ber}(X)=\underset{\tau \in \Theta }{\sup}\big\vert \langle X\hat{k}_{\tau}, \hat{k}_{\tau}\rangle \big\vert \leqslant \underset{\tau \in \Theta }{\sup}\big\langle \vert X\vert \hat{k}_{\tau}, \hat{k}_{\tau} \big\rangle =\mathbf{ber}(\vert X\vert ). \end{aligned}$$

Using the definition of $\Vert \cdot \Vert _{\sigma _{t}}$ and the monotonicity of $\sigma _{t}$, we have the next result.

Theorem 2.5

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $\sigma _{t}$ be an interpolational path of a symmetric mean $\sigma$ for all $t \in [0,1]$. Then

• (1) If $X$ is hyponormal, i.e., $XX^{*} \leqslant X^{*}X$, then $\Vert X^{*} \Vert _{\sigma _{t}}\leqslant \Vert X \Vert _{\sigma _{t}}$.

• (2) If $X$ is co-hyponormal, i.e., $X^{*}X \leqslant XX^{*}$, then $\Vert X \Vert _{\sigma _{t}}\leqslant \Vert X^{*} \Vert _{\sigma _{t}}$.

• (3) If $X$ is semi-hyponormal, i.e., $\vert X^{*}\vert \leqslant \vert X\vert$, then $\Vert X \Vert _{\sigma _{t}}\leqslant \Vert \,\vert X\vert \, \Vert _{\sigma _{t}}$.

• (4) If $X$ is $(\alpha ,\beta )$-normal, i.e., $\alpha X^{*}X\leqslant XX^{*} \leqslant \beta X^{*}X$ for some positive real numbers $\alpha$ and $\beta$ with $\alpha \leqslant 1\leqslant \beta$, then

  $$\begin{aligned} \alpha \Vert X \Vert _{\sigma _{t}}\leqslant \Vert X^{*} \Vert _{ \sigma _{t}}\leqslant \beta \Vert X \Vert _{\sigma _{t}}. \end{aligned}$$

• (5) If $X$ is normal, then $\Vert X^{*} \Vert _{\sigma _{t}}=\Vert X \Vert _{\sigma _{t}}$.

Proof

• (1) It follows from the hyponormality of $X$ that $\Vert X^{*} \widehat{k}_{\tau}\Vert \leqslant \Vert X \widehat{k}_{ \tau}\Vert$ for all $\tau \in \Theta$. Moreover, $\vert \langle X \widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \vert = \vert \langle X^{*} \widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \vert$ for all $\tau \in \Theta$. Hence, by the monotonicity of $\sigma _{t}$, we get

  $$\begin{aligned} \big\vert \langle X^{*} \widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \big\vert ^{2} \sigma _{t}\Vert X^{*} \widehat{k}_{\tau}\Vert ^{2} \leqslant \big\vert \langle X \widehat{k}_{\tau},\widehat{k}_{\tau} \rangle \big\vert ^{2} \sigma _{t} \Vert X \widehat{k}_{\tau}\Vert ^{2} \quad \textrm{for all}\,\,\tau \in \Theta . \end{aligned}$$

  Then, by the definition of ${\Vert \cdot \Vert _{\sigma}}_{t}$, we have ${\Vert X^{*}\Vert _{\sigma}}_{t}\leqslant {\Vert X\Vert _{\sigma}}_{t}$.

• (2) The proof is similar to that of part (1).

• (3) The condition of semi-hyponormality and the mixed Cauchy-Schwarz inequality imply that $\big\vert \langle X\hat{k}_{\tau}, \hat{k}_{\tau}\rangle \big\vert ^{2} \leqslant \langle \vert X\vert \hat{k}_{\tau}, \hat{k}_{\tau}\rangle ^{2}$ for all $\hat{k}_{\tau}\in{\mathscr{H}}$. Also, $\Vert X\hat{k}_{\tau }\Vert =\Vert \,\vert X\vert \,\hat{k}_{\tau } \Vert$ for all $\hat{k}_{\tau}\in{\mathscr{H}}$. Therefore,

  $$\begin{aligned} \big\vert \langle X \widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \big\vert ^{2} \sigma _{t}\Vert X \widehat{k}_{\tau}\Vert ^{2} \leqslant \langle \vert X \vert \widehat{k}_{\tau},\widehat{k}_{\tau} \rangle ^{2} \sigma _{t} \Vert \,\vert X\vert \, \widehat{k}_{\tau} \Vert ^{2}\quad \textrm{for all}\,\,\tau \in \Theta . \end{aligned}$$

  By taking the supremum over all $\tau \in \Theta$, we get ${\Vert X\Vert _{\sigma}}_{t}\leqslant {\Vert \,\vert X\vert \, \Vert _{\sigma}}_{t}$.

• (4) Since $X$ is $(\alpha ,\beta )$-normal, we have $\alpha \Vert X\hat{k}_{\tau }\Vert \leqslant \Vert X^{*}\hat{k}_{ \tau }\Vert \leqslant \beta \Vert X\hat{k}_{\tau }\Vert$ for all $\tau \in \Theta$. It follows from the fact that $\sigma _{t}$ is increasing in its both variables that

  $$\begin{aligned} \alpha ^{2}\big\vert \langle X \widehat{k}_{\tau},\widehat{k}_{\tau} \rangle \big\vert ^{2} \sigma _{t} \alpha ^{2}\Vert X \widehat{k}_{ \tau}\Vert ^{2} \leqslant \big\vert \langle X^{*} \widehat{k}_{\tau}, \widehat{k}_{\tau}\rangle \big\vert ^{2} \sigma _{t} \Vert X^{*} \widehat{k}_{\tau}\Vert ^{2} \leqslant \beta ^{2} \big\vert \langle X \widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \big\vert ^{2} \sigma _{t} \beta ^{2}\Vert X \widehat{k}_{\tau}\Vert ^{2} \end{aligned}$$

  for all $\tau \in \Theta$. Hence, $\alpha{\Vert X\Vert _{\sigma}}_{t}\leqslant {\Vert X^{*}\Vert _{ \sigma}}_{t}\leqslant \beta{\Vert X\Vert _{\sigma}}_{t}$.

• (5) It follows from the normality of $X$ that $X$ is both hyponormanl and co-hyponormanl, and then by the parts (1) and (2) we have the desired result.

 □

Theorem 2.6

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $t\in [0,1]$. Then the following conditions are equivalent.

• (1) $\Vert X\Vert ^{2}_{\sigma _{t}}=\mathbf{ber}^{2}(X)\,\sigma _{t} \Vert X\Vert _{\mathbf{ber}}^{2}$.

• (2) There exists a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

  $$\begin{aligned} \lim _{n\rightarrow \infty} \big\vert \widetilde{X}({\tau _{n}}) \big\vert =\mathbf{ber}(X)\quad \textrm{and }\quad \lim _{n \rightarrow \infty}\big\Vert X\widehat{k}_{\tau _{n}} \big\Vert = \Vert X\Vert _{\mathbf{ber}}. \end{aligned}$$

Proof

We first prove that (1) implies (2). By the definition of the supremum, there exists a sequence $\big\{\widehat{k}_{\tau _{n}}\big\}$ in ℋ such that

$$ \Vert X\Vert ^{2}_{\sigma _{t}}=\lim _{n\rightarrow \infty} \big\vert \widetilde{X}({\tau _{n}})\big\vert ^{2}\,\sigma _{t} \big\Vert {X}\widehat{k}_{\tau _{n}}\big\Vert ^{2}. $$

It follows from the boundedness of the sequences $\big\{\big\vert \widetilde{X}({\tau _{n}})\big\vert \big\}$ and $\big\{ \big\Vert {X}\widehat{k}_{\tau _{n}}\big\Vert \big\}$ that there exists a subsequence $\big\{\widehat{k}_{{\tau _{n}}_{k}}\big\}$ such that $\big\{\big\vert \widetilde{X}({\tau _{n}}_{k})\big\vert \big\}$ and $\big\{ \big\Vert {X}\widehat{k}_{{\tau _{n}}_{k}}\big\Vert \big\}$ are convergent. Then, we have

$$\begin{aligned} \mathbf{ber}^{2}(X)\,\sigma _{t}\Vert X\Vert _{\mathbf{ber}}^{2} &= \Vert X\Vert ^{2}_{\sigma _{t}} \\ &=\lim _{n\rightarrow \infty}\left (\vert \widetilde{X}({{\tau _{n}}_{k}}) \vert ^{2}\,\sigma _{t}\Vert {X}\widehat{k}_{{\tau _{n}}_{k}}\Vert ^{2} \right ) \\ &\leqslant \mathbf{ber}^{2}(X)\sigma _{t} \Vert X\Vert _{ \mathbf{ber}}^{2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{n\rightarrow \infty}\vert \widetilde{X}({{\tau _{n}}_{k}}) \vert =\mathbf{ber}(X)\quad \textrm{and }\quad \lim _{n\rightarrow \infty}\Vert {X}\widehat{k}_{{\tau _{n}}_{k}}\Vert =\Vert X\Vert _{ \mathbf{ber}}. \end{aligned}$$

Now, we prove that (2) implies (1). We have

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma _{t}}&=\underset{\tau \in \Theta }{\sup}\left \{{\vert \widetilde{X}(\tau )\vert ^{2}\,\, \sigma _{t} \,\,\Vert X\widehat{k}_{\tau}\Vert ^{2} }\right \} \\ &\geqslant \underset{n\rightarrow \infty }{\lim}\left \{{\vert \widetilde{X}{(\tau _{n})}\vert ^{2}\,\, \sigma _{t}\,\,\Vert X\widehat{k}_{\tau _{n}}\Vert ^{2} }\right \} \\ &=\mathbf{ber}^{2}(X)\,\, \sigma _{t}\,\,\Vert X\Vert _{\mathbf{ber}}^{2}. \end{aligned}$$

Hence, $\Vert X\Vert ^{2}_{\sigma _{t}}=\mathbf{ber}^{2}(X)\,\sigma _{t} \Vert X\Vert _{\mathbf{ber}}^{2}$. □

We have seen in Proposition 2.3 that $\Vert \cdot \Vert _{\sigma _{t}}\,\,(0\leqslant t\leqslant 1)$ fulfills the semi-norm properties, except possibly for the triangle inequality. In particular, when $\sigma =\nabla _{t}\,\,(0\leqslant t\leqslant 1)$, we have the next proposition.

Proposition 2.7

Let $X, Y\in{\mathscr{L}}({\mathscr{H}})$ and $0\leqslant t\leqslant 1$. Then

$$\begin{aligned} \Vert X+Y\Vert _{\nabla _{t}}\leqslant {\Vert X\Vert _{\nabla _{t}}}+ \Vert Y\Vert _{\nabla _{t}}. \end{aligned}$$

Proof

Let $\tau \in \Theta$ be a unit vector. Then

$$\begin{aligned} & t \vert \widetilde{(X+Y)}(\tau )\vert ^{2}+(1- t)\Vert (X+Y) \widehat{k}_{\tau}\Vert ^{2} \\ &\leqslant t \left (\vert \widetilde{X}(\tau )\vert +\vert \widetilde{Y}(\tau )\vert \right )^{2}+(1- t)\left (\Vert X \widehat{k}_{\tau}\Vert +\Vert Y\widehat{k}_{\tau}\Vert \right )^{2} \\ &= t \left (\vert \widetilde{X}(\tau )\vert ^{2}+\vert \widetilde{Y}( \tau )\vert ^{2}+2\vert \widetilde{X}(\tau )\vert \vert \widetilde{Y}( \tau )\vert \right ) \\ &\qquad \qquad \qquad +(1- t)\left (\Vert X\widehat{k}_{\tau}\Vert ^{2}+ \Vert Y\widehat{k}_{\tau}\Vert ^{2}+2\Vert X\widehat{k}_{\tau}\Vert \Vert Y\widehat{k}_{\tau}\Vert \right ) \\ &= t\vert \widetilde{X}(\tau )\vert ^{2}+(1- t)\Vert X\widehat{k}_{ \tau}\Vert ^{2}+ t\vert \widetilde{Y}(\tau )\vert ^{2}+(1- t)\Vert Y \widehat{k}_{\tau}\Vert ^{2} \\ &\qquad \qquad \qquad +2\left ( t\vert \widetilde{X}(\tau )\vert \vert \widetilde{Y}(\tau )\vert +(1- t)\Vert X\widehat{k}_{\tau} \Vert \Vert Y\widehat{k}_{\tau}\Vert \right ). \end{aligned}$$

Moreover, the Cauchy-Schwarz inequality implies that

$$\begin{aligned} t\vert \widetilde{X}(\tau )\vert \vert \widetilde{Y}(\tau )\vert +&(1- t)\Vert X\widehat{k}_{\tau}\Vert \Vert Y\widehat{k}_{\tau}\Vert \\ &\leqslant \sqrt{ t\vert \widetilde{X}(\tau )\vert ^{2}+(1- t)\Vert X \widehat{k}_{\tau}\Vert ^{2}} \sqrt{ t\vert \widetilde{Y}(\tau ) \vert ^{2}+(1- t)\Vert Y\widehat{k}_{\tau}\Vert ^{2}}. \end{aligned}$$

Combining the above inequalities, we have

$$\begin{aligned} & t \vert \widetilde{(X+Y)}(\tau )\vert ^{2}+(1- t)\Vert (X+Y) \widehat{k}_{\tau}\Vert ^{2} \\ &\leqslant t\vert \widetilde{X}(\tau )\vert ^{2}+(1- t)\Vert X \widehat{k}_{\tau}\Vert ^{2}+ t\vert \widetilde{Y}(\tau )\vert ^{2}+(1- t)\Vert Y\widehat{k}_{\tau}\Vert ^{2} \\ &\qquad +2\sqrt{ t\vert \widetilde{X}(\tau )\vert ^{2}+(1- t)\Vert X \widehat{k}_{\tau}\Vert ^{2}} \sqrt{ t\vert \widetilde{Y}(\tau ) \vert ^{2}+(1- t)\Vert Y\widehat{k}_{\tau}\Vert ^{2}} \\ &\leqslant {\Vert X\Vert ^{2}_{\nabla _{t}}}+ {\Vert Y\Vert ^{2}_{ \nabla _{t}}}+2 {\Vert X\Vert _{\nabla _{t}}} {\Vert X\Vert _{ \nabla _{t}}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert X+Y\Vert ^{2}_{\nabla _{t}}= \underset{\tau \in \Theta }{\sup}\left \{ t \vert \widetilde{(X+Y)}(\tau )\vert ^{2}+(1- t) \Vert (X+Y)\widehat{k}_{\tau}\Vert ^{2}\right \} \leqslant \left ({ \Vert X\Vert _{\nabla _{t}}}+\Vert Y\Vert _{\nabla _{t}}\right )^{2}. \end{aligned}$$

 □

In the following theorem, we give an equivalent condition that ${\Vert X+Y\Vert _{\nabla _{t}}}= {\Vert X\Vert _{\nabla _{t}}}+{ \Vert Y\Vert _{\nabla _{t}}}$ for all $X,Y\in{\mathscr{L}}({\mathscr{H}})$.

Theorem 2.8

Let $X, Y\in{\mathscr{L}}({\mathscr{H}})$ and $0< t<1$. Then the following conditions are equivalent.

• (1) ${\Vert X+Y\Vert _{\nabla _{t}}}= {\Vert X\Vert _{\nabla _{t}}}+{ \Vert Y\Vert _{\nabla _{t}}}$.

• (2) There exists a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

  $$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\Re \left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+(1- t) \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right )={\Vert X\Vert _{\nabla _{t}}} {\Vert Y\Vert _{\nabla _{t}}}, \end{aligned}$$

  where $\Re (z)$ denotes the real part of a complex number $z$.

Proof

(1) ⇒ (2) Using the definition of the supremum and the hypothesis, there exists a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\left ( t \vert \widetilde{(X+Y)}(\tau _{n})\vert ^{2}+(1- t) \Vert (X+Y)\widehat{k}_{\tau _{n}}\Vert ^{2}\right )=\left ({\Vert X \Vert _{\nabla _{t}}}+{\Vert Y\Vert _{\nabla _{t}}} \right )^{2}. \end{aligned}$$

Hence,

$$\begin{aligned} & t \vert \widetilde{(X+Y)}(\tau _{n})\vert ^{2}+(1- t)\Vert (X+Y) \widehat{k}_{\tau _{n}}\Vert ^{2} \\ &= t \left (\vert \widetilde{X}(\tau _{n})\vert ^{2}+\vert \widetilde{Y}(\tau _{n})\vert ^{2}+2\Re \left ( \widetilde{X^{*}}( \tau _{n})\widetilde{Y}(\tau _{n})\right )\right ) \\ &\qquad \qquad \qquad +(1- t)\left (\Vert X\widehat{k}_{\tau _{n}} \Vert ^{2}+\Vert Y\widehat{k}_{\tau _{n}}\Vert ^{2}+2\Re \left ( \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right )\right ) \\ &= t\vert \widetilde{X}(\tau _{n})\vert ^{2}+(1- t)\Vert X\widehat{k}_{ \tau _{n}}\Vert ^{2}+ t\vert \widetilde{Y}(\tau _{n})\vert ^{2}+(1- t) \Vert Y\widehat{k}_{\tau _{n}}\Vert ^{2} \\ &\qquad \qquad \qquad +2\Re \left ( t\widetilde{X^{*}}(\tau _{n}) \widetilde{Y}(\tau _{n})+ (1- t)\langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right ) \\ &\leqslant {\Vert X\Vert ^{2}_{\nabla _{t}}}+ {\Vert Y\Vert ^{2}_{ \nabla _{t}}}+2\Re \left ( t \widetilde{X^{*}}(\tau _{n}) \widetilde{Y}(\tau _{n})+(1- t) \langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right ) \\ &\leqslant {\Vert X\Vert ^{2}_{\nabla _{t}}}+ {\Vert Y\Vert ^{2}_{ \nabla _{t}}}+2\left ( t\vert \widetilde{X}(\tau _{n})\vert \vert \widetilde{Y}(\tau _{n})\vert + (1- t)\vert \langle X\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \vert \langle Y \widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \right ) \\ &\leqslant {\Vert X\Vert ^{2}_{\nabla _{t}}}+ {\Vert Y\Vert ^{2}_{ \nabla _{t}}}+2\left ( t\Vert X\widehat{k}_{\tau _{n}}\Vert \Vert Y \widehat{k}_{\tau _{n}}\Vert + (1- t)\vert \langle X\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \vert \langle Y \widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \right ) \\ &\leqslant {\Vert X\Vert ^{2}_{\nabla _{t}}}+ {\Vert Y\Vert ^{2}_{ \nabla _{t}}} +2\sqrt{ t\vert \widetilde{X}(\tau _{n})\vert ^{2}+(1- t) \Vert X\widehat{k}_{\tau _{n}}\Vert ^{2}} \sqrt{ t\vert \widetilde{Y}( \tau _{n})\vert ^{2}+(1- t)\Vert Y\widehat{k}_{\tau _{n}}\Vert ^{2}} \\ &\qquad \qquad \qquad \qquad \textrm{(by the Cauchy-Schwarz inequality)} \\ &\leqslant \left ({\Vert X\Vert _{\nabla _{t}}}+ {\Vert Y\Vert _{ \nabla _{t}}}\right )^{2}. \end{aligned}$$

Now, if we let $n\rightarrow \infty$, we get

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\Re \left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+(1- t) \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right )={\Vert X\Vert _{\nabla _{t}}} {\Vert Y\Vert _{\nabla _{t}}}. \end{aligned}$$

(2) ⇒ (1) Suppose that there exists a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\Re \left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+ (1- t)\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \right )={\Vert X\Vert _{\nabla _{t}}} {\Vert Y\Vert _{ \nabla _{t}}}. \end{aligned}$$

Then, for every $n\in \mathbb{N}$, we have

$$\begin{aligned} \Re ^{2}&\left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+ (1- t)\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \right ) \\ &\leqslant \big\vert (1- t)\widetilde{X}(\tau _{n}) \widetilde{Y}( \tau _{n})+ t\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \big\vert ^{2} \\ &\qquad -\Im ^{2}\left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}( \tau _{n})+ (1- t)\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \right ) \\ &\leqslant \big\vert (1- t)\widetilde{X}(\tau _{n}) \widetilde{Y}( \tau _{n})+ t\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \big\vert ^{2}, \end{aligned}$$

where $\Im (z)$ denotes the imaginary part of a complex number $z$. Hence,

$$\begin{aligned} \Re &\left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+ (1- t)\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right ) \\ &\leqslant \big\vert (1- t)\widetilde{X}(\tau _{n}) \widetilde{Y}( \tau _{n})+ t\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \big\vert \\ &\leqslant (1- t)\vert \widetilde{X}(\tau _{n})\vert \vert \widetilde{Y}(\tau _{n})\vert + t\vert \langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \vert \vert \langle Y\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \\ &\leqslant \left ( (1- t)\Vert X\widehat{k}_{\tau _{n}}\Vert \Vert Y \widehat{k}_{\tau _{n}}\Vert + t\vert \langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \vert \vert \langle Y\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \right ) \\ &\leqslant \sqrt{ t\vert \widetilde{X}(\tau _{n})\vert ^{2}+(1- t) \Vert X\widehat{k}_{\tau _{n}}\Vert ^{2}} \sqrt{ t\vert \widetilde{Y}( \tau _{n})\vert ^{2}+(1- t)\Vert Y\widehat{k}_{\tau _{n}}\Vert ^{2}} \\ &\qquad \qquad \qquad \qquad \textrm{(by the Cauchy-Schwarz inequality)} \\ &\leqslant{\Vert X\Vert _{\nabla _{t}}} {\Vert Y\Vert _{\nabla _{t}}}. \end{aligned}$$

(2.3)

It follows from ${ t\vert \widetilde{X}(\tau _{n})\vert ^{2}+(1- t)\Vert X\widehat{k}_{ \tau _{n}}\Vert ^{2}}\leqslant {\Vert X\Vert ^{2}_{\nabla _{t}}}$ and ${ t\vert \widetilde{Y}(\tau _{n})\vert ^{2}+(1- t)\Vert Y\widehat{k}_{ \tau _{n}}\Vert ^{2}}\leqslant {\Vert Y\Vert ^{2}_{\nabla _{t}}}$ that

$$ \underset{n\rightarrow \infty }{\lim}\bigg({ t\vert \widetilde{X}(\tau _{n})\vert ^{2}+(1- t)\Vert X \widehat{k}_{\tau _{n}}\Vert ^{2}}\bigg)= {\Vert X\Vert ^{2}_{\nabla _{t}}} $$

and

$$ \underset{n\rightarrow \infty }{\lim}\bigg({ t\vert \widetilde{Y}(\tau _{n})\vert ^{2}+(1- t)\Vert Y \widehat{k}_{\tau _{n}}\Vert ^{2}}\bigg)= {\Vert Y\Vert ^{2}_{\nabla _{t}}}. $$

Therefore,

$$\begin{aligned} \left ({\Vert X\Vert _{\nabla _{t}}} +{\Vert Y\Vert _{\nabla _{t}}} \right )^{2} &={\Vert X\Vert ^{2}_{\nabla _{t}}} +{\Vert Y\Vert ^{2}_{ \nabla _{t}}}+2{\Vert X\Vert _{\nabla _{t}}} {\Vert Y\Vert _{ \nabla _{t}}} \\ &=\underset{n\rightarrow \infty }{\lim}\bigg( t\vert \widetilde{X}(\tau _{n})\vert ^{2}+(1- t)\Vert X \widehat{k}_{\tau _{n}}\Vert ^{2}\bigg)\\ &\qquad \qquad + \underset{n\rightarrow \infty }{\lim}\bigg( t\vert \widetilde{Y}(\tau _{n})\vert ^{2}+(1- t)\Vert Y \widehat{k}_{\tau _{n}}\Vert ^{2}\bigg) \\ &\qquad \qquad +2\underset{n\rightarrow \infty }{\lim}\Re \left ( t\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+ (1- t)\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \right ) \\ &=\underset{n\rightarrow \infty }{\lim}\bigg[ t \left (\vert \widetilde{X}(\tau _{n})\vert ^{2}+ \vert \widetilde{Y}(\tau _{n})\vert ^{2}+2\Re \left ( \widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})\right )\right ) \\ &\qquad \quad +(1- t)\left (\Vert X\widehat{k}_{\tau _{n}}\Vert ^{2}+ \Vert Y\widehat{k}_{\tau _{n}}\Vert ^{2}+2\Re \left ( \langle X \widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y \widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right ) \right )\bigg] \\ &=\underset{n\rightarrow \infty }{\lim}\bigg( t \vert \widetilde{(X+Y)}(\tau _{n})\vert ^{2}+(1- t) \Vert (X+Y)\widehat{k}_{\tau _{n}}\Vert ^{2}\bigg) \\ &={\Vert X+Y\Vert ^{2}_{\nabla _{t}}} \\ &\leqslant \left ({\Vert X\Vert _{\nabla _{t}}}+{\Vert Y\Vert _{ \nabla _{t}}}\right )^{2} \\ &\qquad \qquad \qquad \qquad (\textrm{by Proposition } 2.7). \end{aligned}$$

Hence, ${\Vert X+Y\Vert _{\nabla _{t}}}={\Vert X\Vert _{\nabla _{t}}}+{ \Vert Y\Vert _{\nabla _{t}}}$. □

Recently, Altwaijry et al. in [1], introduced the following generalization of $\Vert \cdot \Vert _{ t-\textbf{ber}}$. Given non-negative real scalars $\alpha$ and $\beta$ such that $(\alpha , \beta )\neq (0, 0)$ and $X\in{\mathscr{L}}({\mathscr{H}})$, let

$$\begin{aligned} \Vert X\Vert _{\alpha , \beta}^{\textbf{ber}}= \underset{\tau \in \Theta }{\sup}\left \{\sqrt{\beta \big\vert \widetilde{X}(\tau )\big\vert ^{2}+ \alpha \big\Vert X\widehat{k}_{\tau}\big\Vert ^{2} }\right \}. \end{aligned}$$

(2.4)

Then, we have

$$\begin{aligned} \frac{1}{\sqrt{\alpha +\beta}}\Vert X\Vert _{\alpha , \beta}^{ \textbf{ber}}&=\underset{\tau \in \Theta }{\sup}\left \{\sqrt{ \frac{\beta \big\vert \widetilde{X}(\tau )\big\vert ^{2}+\alpha \big\Vert X\widehat{k}_{\tau}\big\Vert ^{2} }{\alpha +\beta}} \right \} \\ &= \underset{\tau \in \Theta }{\sup}\left \{\sqrt{ \frac{\alpha}{\alpha +\beta}\big\Vert X\widehat{k}_{\tau}\big\Vert ^{2}+ \frac{\beta}{\alpha +\beta} \big\vert \widetilde{X}(\tau )\big\vert ^{2}} \right \}=\Vert X\Vert _{\nabla _{t_{0}}}=\Vert X\Vert _{\nabla _{t_{1}}}, \ \end{aligned}$$

(2.5)

where $t_{0}=\frac{\alpha}{\alpha +\beta}$ and $t_{1}=\frac{\beta}{\alpha +\beta}$.

As a consequence of Theorem 2.8 and the previous identity, we derive the following characterization of the equality in the triangle inequality for the norm $\Vert \cdot \Vert _{\alpha , \beta}^{\textbf{ber}}$.

Corollary 2.9

[1, Theorem 10] Let $X, Y\in{\mathscr{L}}({\mathscr{H}})$ and non-negative real scalars $\alpha$ and $\beta$ such that $(\alpha , \beta )\neq (0, 0)$. Then the following conditions are equivalent.

• (1) ${\Vert X+Y\Vert _{\alpha , \beta}^{\textbf{ber}}}= {\Vert X\Vert _{ \alpha , \beta}^{\textbf{ber}}}+{\Vert Y\Vert _{\alpha , \beta}^{ \textbf{ber}}}$.

• (2) There exists a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

  $$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\left ( t_{0}\widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \right )=\Vert X\Vert _{\nabla _{t_{0}}}\Vert Y\Vert _{ \nabla _{t_{0}}}, \end{aligned}$$

  (2.6)

  where $t_{0}=\frac{\alpha}{\alpha +\beta}$.

Proof

We note that from the equality (2.5), the condition (1) is equivalent to

$$\begin{aligned} \Vert X+Y\Vert _{\nabla _{t_{0}}}= {\Vert X\Vert _{\nabla _{t_{0}}}}+{ \Vert Y\Vert _{\nabla _{t_{0}}}}, \end{aligned}$$

(2.7)

with $t_{0}=\frac{\alpha}{\alpha +\beta}$. Now, by Theorem 2.8, the condition (2.7) is equivalent to the existence of a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\Re \left ( t_{0} \widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right )={\Vert X\Vert _{\nabla _{t_{0}}}} {\Vert Y\Vert _{\nabla _{t_{0}}}}. \end{aligned}$$

(2.8)

To finish the proof, it is enough to show that (2.8) is equivalent to

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\left ( t_{0}\widetilde{X^{*}}( \tau _{n})\widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right )={\Vert X\Vert _{\nabla _{t_{0}}}} {\Vert Y\Vert _{\nabla _{t_{0}}}}. \end{aligned}$$

(2.9)

Indeed, if (2.9) holds, then

$$\begin{aligned} &\underset{n\rightarrow \infty }{\lim}\left ( t_{0}\widetilde{X^{*}}( \tau _{n})\widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right ) \\ &={\Vert X\Vert _{\nabla _{t_{0}}}} {\Vert Y\Vert _{\nabla _{t_{0}}}}= \frac{{\Vert X\Vert _{\nabla _{t_{0}}}} {\Vert Y\Vert _{\nabla _{t_{0}}}}+\overline{{\Vert X\Vert _{\nabla _{t_{0}}}} {\Vert Y\Vert _{\nabla _{t_{0}}}}}}{2} \\ &=\underset{n\rightarrow \infty }{\lim}\Re \left ( t_{0} \widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right ). \end{aligned}$$

(2.10)

On the other hand, by (2.3) for any $n\in \mathbb{N}$, we have

$$ \begin{aligned} &\Re ^{2}\left ( t_{0}\widetilde{X}(\tau _{n})\widetilde{Y}(\tau _{n})+ (1-t_{0})\langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \right ) \\ &\leq \left (t_{0} \vert \widetilde{X}(\tau _{n})\vert \vert \widetilde{Y}(\tau _{n})\vert \left |+ (1-t_{0}) \right | \vert \langle X\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \vert \vert \langle Y\widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}} \rangle \vert \right )^{2} \\ &\leqslant{\Vert X\Vert ^{2}_{\nabla _{t_{0}}}} {\Vert Y\Vert ^{2}_{ \nabla _{t_{0}}}}. \end{aligned} $$

Thus, if we assume that condition (2.8) is fulfilled, then we can conclude that

$$ \lim \limits _{n\to \infty}\Im \left ( t_{0}\widetilde{X}(\tau _{n}) \widetilde{Y}(\tau _{n})+ (1-t_{0})\langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right )=0, $$

and

$$ \underset{n\rightarrow \infty }{\lim}\left ( t_{0}\widetilde{X^{*}}( \tau _{n})\widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{ \tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right )={\Vert X\Vert _{\nabla _{t_{0}}}} {\Vert Y\Vert _{\nabla _{t_{0}}}}, $$

and this completes the proof. □

Remark 2.10

We note that condition (2.6) is equivalent to

(2’) There exists a sequence $\{\widehat{k}_{\tau _{n}}\}$ in ℋ such that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim}\left ( \alpha \widetilde{X^{*}}( \tau _{n})\widetilde{Y}(\tau _{n})+\beta \langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right )={\Vert X\Vert _{\alpha , \beta}^{\textbf{ber}}} {\Vert Y\Vert _{\alpha , \beta}^{\textbf{ber}}}. \end{aligned}$$

Indeed, if we denote by $t_{0}=\frac{\alpha}{\alpha +\beta}$, then by (2.6)

$$\begin{aligned} {\Vert X\Vert _{\alpha , \beta}^{\textbf{ber}}} {\Vert Y\Vert _{ \alpha , \beta}^{\textbf{ber}}}&=\sqrt{\alpha +\beta}{\Vert X\Vert _{ \nabla _{t_{0}}}}\sqrt{\alpha +\beta} {\Vert Y\Vert _{\nabla _{t_{0}}}} \\ &=\underset{n\rightarrow \infty }{\lim}(\alpha +\beta )\left ( t_{0}\widetilde{X^{*}}(\tau _{n}) \widetilde{Y}(\tau _{n})+(1- t_{0}) \langle X\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \langle Y\widehat{k}_{\tau _{n}}, \widehat{k}_{\tau _{n}}\rangle \right )= \\ &= \underset{n\rightarrow \infty }{\lim}\left ( \alpha \widetilde{X^{*}}(\tau _{n})\widetilde{Y}(\tau _{n})+\beta \langle X \widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \langle Y \widehat{k}_{\tau _{n}},\widehat{k}_{\tau _{n}}\rangle \right ). \end{aligned}$$

Finally, we remark that in [1, Theorem 10], the authors obtained a similar characterization of the equality ${\Vert X+Y\Vert _{\alpha , \beta}^{\textbf{ber}}}= {\Vert X\Vert _{ \alpha , \beta}^{\textbf{ber}}}+{\Vert Y\Vert _{\alpha , \beta}^{ \textbf{ber}}}$.

3 Some Estimations for $\Vert \cdot \Vert _{\sigma _{t}}$

In this section, we present some upper and lower bounds for $\Vert \cdot \Vert _{\sigma _{t}}$. The following well-known lemmas will be essential to prove our results.

Lemma 3.1

[25] Let $X\in{\mathscr{L}}({\mathscr{H}})$ be a self-adjoint operator with spectrum in an interval $J$ and $\tau \in \Theta$.

• (1) If $f : J \rightarrow \mathbb{R}$ is convex, then $f \big(\langle X\widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \big) \leqslant \langle f(X)\widehat{k}_{\tau}, \widehat{k}_{\tau}\rangle$.

• (2) If $f : J \rightarrow \mathbb{R}$ is concave, then $f \big(\langle X\widehat{k}_{\tau},\widehat{k}_{\tau}\rangle \big) \geqslant \langle f(X)\widehat{k}_{\tau}, \widehat{k}_{\tau}\rangle$.

Lemma 3.2

[24] Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $f$, $g$ be two non-negative continuous functions defined on $[0,\infty )$ such that $f(t)g(t) = t$ for $t\geqslant 0$. Then

$$\begin{aligned} \vert \widetilde{X}(\tau ) \vert \leqslant \vert \widetilde{f^{2}(\vert X\vert )}(\tau )\vert \vert \widetilde{g^{2}(\vert X^{*}\vert )}(\tau ) \vert \end{aligned}$$

for all $\tau \in \Theta$.

For an operator $X\in {\mathscr{L}}({\mathscr{H}})$, the Crawford Berezin number $\widetilde{c}(X)$ is defined as $\widetilde{c}(X)=\underset{\tau \in \Theta }{\inf}\vert \widetilde{X}(\tau )\vert$. In the following theorem, we obtain a lower bound for $\Vert \cdot \Vert _{\sigma _{t}}$ in the terms of $\widetilde{c}(\cdot )$.

Theorem 3.3

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $0\leqslant t\leqslant 1$. Then,

$$\begin{aligned} \Vert X\Vert _{\sigma _{t}}\geqslant \max \left \{\sqrt{\textbf{ber}^{2}(X) {\sigma _{t}}\widetilde{c}^{2}(X^{*}X)}\,,\,\sqrt{\widetilde{c}^{2}(X) {\sigma _{t}}\Vert X\Vert ^{2}_{\textbf{ber}} } \right \}. \end{aligned}$$

Proof

Let ${\tau}\in \Theta$ and $0\leqslant t\leqslant 1$. Then

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma _{t}}&=\underset{\tau \in \Theta }{\sup}\{ \vert \widetilde{X}(\tau ) \vert ^{2} \sigma _{t} \Vert X( \widehat{k}_{\tau})\Vert ^{2} \} \\ &\geqslant \vert \widetilde{X}(\tau ) \vert ^{2} \sigma _{t} \Vert X( \widehat{k}_{\tau})\Vert ^{2} \\ &= \vert \widetilde{X}(\tau ) \vert ^{2} \sigma _{t} \vert \widetilde{X^{*}X}(\tau )\vert ^{2} \\ &\geqslant \vert \widetilde{X}(\tau ) \vert ^{2} \sigma _{t} \widetilde{c}^{2}(X^{*}X). \end{aligned}$$

Taking the supremum over all vectors $\tau \in \Theta$, we get $\Vert X\Vert ^{2}_{\sigma _{t}}\geqslant \textbf{ber}^{2}(X) \sigma _{t} \widetilde{c}^{2}(X^{*}X)$. Similarly, we have

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma _{t}}=\underset{\tau \in \Theta }{\sup}\{ \vert \widetilde{X}(\tau ) \vert ^{2} \sigma _{t} \Vert X( \widehat{k}_{\tau})\Vert ^{2} \} \geqslant \vert \widetilde{X}(\tau ) \vert ^{2} \sigma _{t} \Vert X(\widehat{k}_{\tau})\Vert ^{2} \geqslant \widetilde{c}^{2}(X) \sigma _{t} \Vert X(\widehat{k}_{\tau}) \Vert ^{2}, \end{aligned}$$

whence $\Vert X\Vert ^{2}_{\sigma _{t}}\geqslant \widetilde{c}^{2}(X) \sigma _{t}\Vert X\Vert ^{2}_{\textbf{ber}}$. Combining the above inequalities, we get the desired result. □

In the next result, we get some special case of Theorem 3.3.

Corollary 3.4

Let $X\in{\mathscr{L}}({\mathscr{H}})$, $r\in [-1,1]$ and $0\leqslant \mu \leqslant 1$. Then

$$\begin{aligned} \Vert X\Vert _{m_{r,\mu}}\geqslant \max \left \{ \sqrt[2r]{\frac{(1-\mu )\textbf{ber}^{r}(X)+\mu \widetilde{c}^{r}(X^{*}X)}{2}} \,,\, \sqrt[2r]{\frac{(1-\mu )\widetilde{c}^{r}(X) +\mu \Vert X\Vert ^{r}_{\textbf{ber}} }{2}} \right \}. \end{aligned}$$

In particular,

$$\begin{aligned} \Vert X\Vert _{\nabla _{\mu}}\geqslant \max \left \{\sqrt{{(1-\mu ) \textbf{ber}^{2}(X)+\mu \widetilde{c}^{2}(X^{*}X)}}\,,\,\sqrt{{(1-\mu ) \widetilde{c}^{2}(X) +\mu \Vert X\Vert ^{2}_{\textbf{ber}}} } \right \} \end{aligned}$$

and

$$\begin{aligned} \Vert X\Vert _{\sharp _{\mu}}\geqslant \max \left \{{\textbf{ber}^{(1- \mu )}(X) \widetilde{c}^{\mu}(X^{*}X)}\,,\,{\widetilde{c}^{(1-\mu )}(X) \Vert X\Vert ^{\mu}_{\textbf{ber}} } \right \}. \end{aligned}$$

Proof

Letting $\sigma _{t}$ be the interpolational paths of the power means $m_{r,\mu}$ for $r\in [-1,1]$ and $0\leqslant \mu \leqslant 1$ in Theorem 3.3, we have the first inequality. If we take the weighted arithmetic mean $\nabla _{\mu}$ and the weighted geometric mean $\sharp _{\mu}$, $(0\leqslant \mu \leqslant 1)$ in Theorem 3.3, then we have the second and the third inequalities, respectively. □

Theorem 3.5

Let $X\in{\mathscr{L}}({\mathscr{H}})$, and let $f$, $g$ be two non-negative continuous functions defined on $[0,\infty )$ such that $f(t)g(t) = t$ for $t\geqslant 0$. If $\sigma$ is a mean dominated by the arithmetic mean $\nabla$, then

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \textbf{ber}\left (\frac{1}{4}(f^{4}( \vert X\vert )+g^{4}(\vert X^{*}\vert ))+\frac{1}{2}\vert X\vert ^{2} \right ) \end{aligned}$$

and

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \frac{1}{2}\sqrt{\textbf{ber} \left (f^{4}(\vert X\vert )+g^{2}(\vert X\vert ^{2})\right ) \textbf{ber}\left (f^{2}(\vert X\vert ^{2})+g^{4}(\vert X^{*}\vert ) \right )}. \end{aligned}$$

Proof

Let $\tau \in \Theta$. Then

$$\begin{aligned} \vert \widetilde{X}(\tau )\vert ^{2}\sigma \Vert X\widehat{k}_{\tau} \Vert ^{2} &= \vert \widetilde{X}(\tau )\vert ^{2}\sigma \vert \widetilde{X^{*}X}(\tau )\vert \\ & \leqslant \vert \widetilde{f^{2}(\vert X\vert})(\tau )\vert \vert \widetilde{g^{2}(\vert X^{*}\vert})(\tau )\vert \sigma \vert \widetilde{X^{*}X}(\tau )\vert \\ &\qquad \qquad \qquad (\textrm{by Lemma }3.2) \\ & \leqslant \frac{1}{2}\left (\vert \widetilde{f^{2}(\vert X\vert })( \tau )\vert ^{2}+\vert \widetilde{g^{2}(\vert X^{*}\vert})(\tau ) \vert ^{2}\right )\sigma \vert \widetilde{X^{*}X}(\tau )\vert \\ & \leqslant \frac{1}{2}\left (\vert \widetilde{f^{4}(\vert X\vert})( \tau )\vert +\vert \widetilde{g^{4}(\vert X^{*}\vert})(\tau )\vert \right )\sigma \vert \widetilde{X^{*}X}(\tau )\vert \\ &\qquad \qquad \qquad (\textrm{by Lemma }3.1) \\ &= \frac{1}{2}\left (\vert \widetilde{f^{4}(\vert X\vert )+g^{4}(\vert X^{*}\vert})(\tau )\vert \right )\sigma \vert \widetilde{\vert X\vert ^{2}}(\tau )\vert . \end{aligned}$$

It follows from $\sigma \leqslant \nabla$ and the above inequalities that

$$\begin{aligned} \vert \widetilde{X}(\tau )\vert ^{2}\sigma \Vert X\widehat{k}_{\tau} \Vert ^{2} &\leqslant \frac{1}{2}\left (\vert \widetilde{f^{4}(\vert X\vert )+g^{4}(\vert X^{*}\vert})(\tau )\vert \right )\sigma \vert \widetilde{\vert X\vert ^{2}}(\tau )\vert \\ &\leqslant \frac{1}{2}\left [\frac{1}{2}\left (\vert \widetilde{f^{4}(\vert X\vert )+g^{4}(\vert X^{*}\vert})(\tau )\vert \right )+ \vert \widetilde{\vert X\vert ^{2}}(\tau )\vert \right ] \\ &\leqslant \bigg\vert \big( \widetilde{\frac{1}{4}\left (f^{4}(\vert X\vert )+g^{4}(\vert X^{*}\vert )\right )+ \frac{1}{2}\vert X\vert ^{2}} \big)(\tau )\bigg\vert \\ &\leqslant \textbf{ber}\left (\frac{1}{4}\left (f^{4}(\vert X\vert )+g^{4}( \vert X^{*}\vert )\right )+ \frac{1}{2}\vert X\vert ^{2}\right ). \end{aligned}$$

Then, by taking the supremum over $\tau \in \Theta$, we get the first result. For the second inequality, we have

$$\begin{aligned} \vert \widetilde{X}(\tau )\vert ^{2}\sigma \Vert X\widehat{k}_{\tau} \Vert ^{2} &= \vert \widetilde{X}(\tau )\vert ^{2}\sigma \vert \vert \widetilde{X}\vert ^{2}(\tau )\vert \\ & \leqslant \vert \widetilde{f^{2}(\vert X\vert})(\tau )\vert \vert \widetilde{g^{2}(\vert X^{*}\vert})(\tau )\vert \sigma \sqrt{\vert \widetilde{f^{2}(\vert X\vert ^{2}})(\tau )\vert \vert \widetilde{g^{2}(\vert X\vert ^{2}})(\tau )\vert} \\ & = \sqrt{\vert \widetilde{f^{2}(\vert X\vert})(\tau )\vert ^{2} \vert \widetilde{g^{2}(\vert X^{*}\vert})(\tau )\vert ^{2}}\sigma \sqrt{\vert \widetilde{f^{2}(\vert X\vert ^{2}})(\tau )\vert \vert \widetilde{g^{2}(\vert X\vert ^{2}})(\tau )\vert} \\ & \leqslant \sqrt{\vert \widetilde{f^{4}(\vert X\vert})(\tau )\vert \vert \widetilde{g^{4}(\vert X^{*}\vert})(\tau )\vert}\sigma \sqrt{ \vert \widetilde{f^{2}(\vert X\vert ^{2}})(\tau )\vert \vert \widetilde{g^{2}(\vert X\vert ^{2}})(\tau )\vert} \end{aligned}$$

$$\begin{aligned} &\leqslant \frac{1}{2}\left (\sqrt{\vert \widetilde{f^{4}(\vert X\vert})(\tau )\vert \vert \widetilde{g^{4}(\vert X^{*}\vert})(\tau )\vert}+ \sqrt{\vert \widetilde{f^{2}(\vert X\vert ^{2}})(\tau )\vert \vert \widetilde{g^{2}(\vert X\vert ^{2}})(\tau )\vert}\right ) \\ & \leqslant \frac{1}{2}\left (\sqrt{\vert \widetilde{f^{4}(\vert X\vert})(\tau )\vert +\vert \widetilde{g^{2}(\vert X\vert ^{2}})(\tau )\vert} \sqrt{\vert \widetilde{f^{2}(\vert X\vert ^{2}}(\tau )\vert +\vert \widetilde{g^{4}(\vert X^{*}\vert})(\tau )\vert})\right ) \\ &\qquad \qquad \qquad (\textrm{by the Cauchy-Schwarz inequality}\,\, \sqrt{ab}+\sqrt{cd}\leqslant \sqrt{a+c}\,\sqrt{b+d}) \\ &\leqslant \frac{1}{2}\left (\sqrt{\vert \big( \widetilde{f^{4}(\vert X\vert )+g^{2}(\vert X\vert ^{2}})\big)(\tau ) \vert \vert \widetilde{f^{2}(\vert X\vert ^{2})+g^{4}(\vert X^{*}\vert})(\tau ) \vert}\right ) \\ &\leqslant \frac{1}{2}\sqrt{\textbf{ber}\left (f^{4}(\vert X\vert )+g^{2}( \vert X\vert ^{2})\right )\textbf{ber}\left (f^{2}(\vert X\vert ^{2})+g^{4}( \vert X^{*}\vert )\right )}. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \frac{1}{2}\sqrt{\textbf{ber} \left (f^{4}(\vert X\vert )+g^{2}(\vert X\vert ^{2})\right ) \textbf{ber}\left (f^{2}(\vert X\vert ^{2})+g^{4}(\vert X^{*}\vert ) \right )}. \end{aligned}$$

 □

For the special case $f(t)=g(t) = \sqrt{t}$, we have the following result.

Corollary 3.6

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and let $\sigma$ is a mean dominated by the arithmetic mean $\nabla$. Then

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \textbf{ber}\left (\frac{3}{4} \vert X\vert ^{2}+\frac{1}{4}\vert X^{*}\vert ^{2} \right ) \end{aligned}$$

(3.1)

and

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \frac{1}{2}\sqrt{2\textbf{ber} \left (\vert X\vert ^{2}\right )\textbf{ber}\left (\vert X\vert ^{2}+ \vert X^{*}\vert ^{2}\right )}. \end{aligned}$$

Remark 3.7

If $X\in{\mathscr{L}}({\mathscr{H}})$ and $\sigma \leqslant \nabla$, then by the inequality (3.1) and also the subadditivity of the Berezin radius, we have

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}&\leqslant \textbf{ber}\left (\frac{3}{4} \vert X\vert ^{2}+\frac{1}{4}\vert X^{*}\vert ^{2} \right ) \\ &\leqslant \frac{3}{4}\textbf{ber}\left (\vert X\vert ^{2}\right )+ \frac{1}{4}\textbf{ber}\left (\vert X^{*}\vert ^{2} \right ). \end{aligned}$$

Now, if $X$ is normal, then by the definition of the Berezin radius, we get $\textbf{ber}(\vert X^{*}\vert )=\textbf{ber}(\vert X\vert )$. Hence, we have

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}&\leqslant \frac{3}{4}\textbf{ber}\left ( \vert X\vert ^{2}\right )+\frac{1}{4}\textbf{ber}\left (\vert X^{*} \vert ^{2} \right ) \\ &\leqslant \textbf{ber}\left (\vert X\vert ^{2}\right ) \\ &\leqslant \textbf{ber}^{2}\left (\vert X\vert \right )\quad ( \textrm{by Lemma }3.1) \end{aligned}$$

(3.2)

Moreover, if $X$ is positive, then the inequality (3.2) and the fact that $\textbf{ber}(X)\leqslant \Vert X\Vert _{\sigma}$ imply that $\Vert X\Vert _{\sigma}=\textbf{ber}(X)$.

Example 3.8

Consider for $\mathbb{C}^{2}$ the standard orthonormal basis $\big\{ e_{1}, e_{1} \big\}$ as a RKHS on the set $\big\{ 1,2 \big\}$. Then for the self-adjoint matrix $X= \begin{bmatrix} 2 & 1 \cr 1& -2 \cr \end{bmatrix}$, which is not positive, we have

$$\begin{aligned} \textbf{ber}(X)=2\lneqq \Vert X\Vert _{\sigma}=\sqrt{5}. \end{aligned}$$

Theorem 3.9

Let $X\in{\mathscr{L}}({\mathscr{H}})$, and let $\sigma$ is a mean dominated by the arithmetic mean $\nabla$ and $0\leqslant \mu \leqslant 1$. Then

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \frac{1}{2} \textbf{ber}\big((1+ \mu )\vert X\vert ^{2}+(1-\mu )\vert X^{*}\vert ^{2}\big). \end{aligned}$$

In particular,

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \frac{1}{2} \textbf{ber}\big( \vert X\vert ^{2}+\vert X^{*}\vert ^{2}\big). \end{aligned}$$

Proof

Let $\tau \in \Theta$. Then

$$\begin{aligned} \vert \widetilde{X}(\tau )\vert ^{2}\sigma \,\Vert X\widehat{k}_{\tau} \Vert ^{2} &= \vert \widetilde{X}(\tau )\vert ^{2}\sigma \, \vert \vert \widetilde{X}\vert ^{2}(\tau )\vert \\ & \leqslant \vert \widetilde{\vert X\vert ^{2\mu}}(\tau )\vert \vert \widetilde{\vert X^{*}\vert ^{2(1-\mu )}}(\tau )\vert \sigma \, \vert \vert \widetilde{X}\vert ^{2}(\tau )\vert \\ & \leqslant \vert \widetilde{\vert X\vert ^{2}}(\tau )\vert ^{\mu} \vert \widetilde{\vert X^{*}\vert ^{2}}(\tau )\vert ^{(1-\mu )} \sigma \, \vert \vert \widetilde{X}\vert ^{2}(\tau )\vert \\ &\qquad \qquad \qquad (\textrm{by Lemma }3.1) \\ & \leqslant \big\vert \widetilde{({\mu}\vert X\vert ^{2}+ {(1-\mu )}\vert X^{*}\vert ^{2}})( \tau )\big\vert \sigma \, \vert \vert \widetilde{X}\vert ^{2}(\tau ) \vert \\ &\qquad \qquad \qquad ( \textrm{by the weighted arithmetic-geometric mean inequality}) \\ & \leqslant \frac{1}{2}\left (\big\vert \widetilde{({\mu}\vert X\vert ^{2}+ {(1-\mu )}\vert X^{*}\vert ^{2}})( \tau )\big\vert + \vert \vert \widetilde{X}\vert ^{2}(\tau )\vert \right ) \\ & = \frac{1}{2}\left (\big\vert \widetilde{{((1+\mu )}\vert X\vert ^{2}+ {(1-\mu )}\vert X^{*}\vert ^{2}})( \tau )\big\vert \right ) \\ &\leqslant \frac{1}{2} \textbf{ber}\big((1+\mu )\vert X\vert ^{2}+(1- \mu )\vert X^{*}\vert ^{2}\big). \end{aligned}$$

Taking the supremum over all $\tau \in \Theta$, we get

$$\begin{aligned} \Vert X\Vert ^{2}_{\sigma}\leqslant \frac{1}{2} \textbf{ber}\big((1+ \mu )\vert X\vert ^{2}+(1-\mu )\vert X^{*}\vert ^{2}\big). \end{aligned}$$

If we put $\mu =0$, then we have the second inequality. □

Theorem 3.10

Let $X\in{\mathscr{L}}({\mathscr{H}})$ and $0\leqslant t \leqslant 1$. Then

$$\begin{aligned} \Vert X\Vert _{\nabla _{t}}\leqslant \underset{\lambda \in [0, 1] }{\inf}\sqrt{\lambda \left \Vert X\right \Vert _{\mathrm{ber}}^{2} +(1- \lambda ) \left \Vert X\right \Vert _{\mathrm{ber}}\Bigl((1-t)\left \Vert X\right \Vert _{\mathrm{ber}}+t\: \textbf{ber}\left ( X\right ) \Bigr)}. \end{aligned}$$

Proof

Given $u, v \in \mathscr{H}$ and $\lambda \in [0, 1]$, we have the following refinement of the classical Cauchy-Schwarz inequality:

$$\begin{aligned} \vert \langle u, v \rangle \vert ^{2}&=[(1-\lambda )+\lambda ]\vert \langle u, v \rangle \vert ^{2} \\ &\leq (1-\lambda )\vert \langle u, v \rangle \vert ^{2}+\lambda \Vert u\Vert ^{2}\Vert v\Vert ^{2} \\ &\leq (1-\lambda )\Vert u\Vert \Vert v\Vert \:\vert \langle u, v \rangle \vert + \lambda \Vert u\Vert ^{2}\Vert v\Vert ^{2}. \ \end{aligned}$$

(3.3)

Utilizing the inequality (3.3), yields

$$\begin{aligned} (1-t)\vert \langle u, v \rangle \vert ^{2}&\leq (1-t)(1-\lambda ) \Vert u\Vert \Vert v\Vert \:\vert \langle u, v \rangle \vert + (1-t) \lambda \Vert u\Vert ^{2}\Vert v\Vert ^{2}, \ \end{aligned}$$

(3.4)

and

$$\begin{aligned} t\vert \langle u, w \rangle \vert ^{2}&\leq t(1-\lambda )\Vert u \Vert \Vert w\Vert \:\vert \langle u, w \rangle \vert + t\lambda \Vert u\Vert ^{2}\Vert w\Vert ^{2}, \ \end{aligned}$$

(3.5)

for all $u, v, w \in \mathscr{H}$ and $t\in [0, 1]$. Adding the relations (3.4), (3.5) and replacing $v$ with $\frac{u}{\Vert v\Vert }$, we obtain

$$\begin{aligned} \begin{aligned} (1-t)\Vert u \Vert ^{2}+t\vert \langle u, w \rangle \vert ^{2}&\leq \lambda \Vert u\Vert ^{2}\left ((1-t)+t\Vert w\Vert ^{2}\right ) \\ &\quad +(1- \lambda ) \Vert u\Vert \left ((1-t)\Vert u\Vert +t \Vert w\Vert \: \vert \langle u, w \rangle \vert \right ). \end{aligned} \end{aligned}$$

(3.6)

By substituting $u$ for $X\hat{k}_{\tau}$ and $w$ for $\hat{k}_{\tau}$, we have

$$\begin{aligned} \begin{aligned} (1-t)\Vert X\hat{k}_{\tau}\Vert ^{2}+t\vert \langle X\hat{k}_{\tau}, \hat{k}_{\tau }\rangle \vert ^{2}&\leq \lambda \Vert X\hat{k}_{\tau} \Vert ^{2} \\ &\quad +(1-\lambda ) \Vert X\hat{k}_{\tau}\Vert \left ((1-t) \Vert X\hat{k}_{\tau}\Vert +t \vert \langle X\hat{k}_{\tau}, \hat{k}_{ \tau }\rangle \vert \right ). \end{aligned} \end{aligned}$$

(3.7)

Taking the supremum over all $\tau \in \Theta$, we have

$$\begin{aligned} \Vert X\Vert _{\nabla _{t}}^{2}&=\underset{\tau \in \Theta }{\sup} \left \{(1-t)\Vert X\hat{k}_{\tau}\Vert ^{2}+t\vert \langle X\hat{k}_{ \tau}, \hat{k}_{\tau }\rangle \vert ^{2}\right \} \\ &\leq \underset{\tau \in \Theta }{\sup}\left \{\lambda \Vert X\hat{k}_{ \tau}\Vert ^{2} +(1-\lambda ) \Vert X\hat{k}_{\tau}\Vert \left ((1-t) \Vert X\hat{k}_{\tau}\Vert +t \vert \langle X\hat{k}_{\tau}, \hat{k}_{ \tau }\rangle \vert \right )\right \} \\ &\leq \lambda \left \Vert X\right \Vert _{\mathrm{ber}}^{2} +(1- \lambda ) \left \Vert X\right \Vert _{\mathrm{ber}}\Bigl((1-t)\left \Vert X\right \Vert _{\mathrm{ber}}+t\: \textbf{ber}\left ( X\right ) \Bigr), \ \end{aligned}$$

for any $\lambda \in [0,1]$. □

Remark 3.11

Taking $\lambda =\frac {1}{3}$ in Theorem 3.10, we have a refinement of [1, Theorem 6]. Moreover, from the relation (2.5) and Theorem 3.10, we obtain for any pair of non-negative real numbers $\alpha$, $\beta$ such that $(\alpha , \beta )\neq (0, 0)$,

$$\begin{aligned} \Vert X\Vert _{\alpha , \beta}^{\textbf{ber}}&= \sqrt{\alpha +\beta}\: \Vert X\Vert _{\nabla _{t_{1}}} \\ &\leq \sqrt{\alpha +\beta} \underset{\lambda \in [0, 1] }{\inf}\sqrt{\lambda \left \Vert X\right \Vert _{\mathrm{ber}}^{2} +(1- \lambda ) \left \Vert X\right \Vert _{\mathrm{ber}}\Bigl((1-t_{1}) \left \Vert X\right \Vert _{\mathrm{ber}}+t_{1}\: \textbf{ber}\left ( X \right ) \Bigr)} \\ &=\underset{\lambda \in [0, 1] }{\inf}\sqrt{(\alpha +\beta )\lambda \left \Vert X\right \Vert _{ \mathrm{ber}}^{2} +(1-\lambda ) \left \Vert X\right \Vert _{ \mathrm{ber}}\Bigl(\alpha \left \Vert X\right \Vert _{\mathrm{ber}}+ \beta \: \textbf{ber}\left ( X\right ) \Bigr)} \end{aligned}$$

where $t_{1}=\frac{\beta}{\alpha +\beta}$.