Applications of Kato’s inequality for n-tuples of operators in Hilbert spaces, (II)

Abstract

In this paper, by the use of famous Kato’s inequality for bounded linear operators, we establish some new inequalities for n-tuples of operators and apply them to functions of normal operators defined by power series as well as to some norms and numerical radii that arise in multivariate operator theory. They provide a natural continuation of the results in previous paper with (I) in the title.

MSC: 47A50, 47A63.

1 Introduction

In 1952, Kato [1] proved the following generalization of the Schwarz inequality:

$$|〈Tx,y〉{|}^2\le 〈{\left(T^{\ast }T\right)}^{\alpha }x,x〉〈{\left(T{T}^{\ast }\right)}^{1-\alpha }y,y〉$$

(1.1)

for any $x,y\in H$, $\alpha \in [0,1]$, and T is a bounded linear operator on H.

Utilizing the operator modulus notation, we can write (1.1) as follows:

$$|〈Tx,y〉{|}^2\le 〈|T{|}^{2\alpha }x,x〉〈|T^{\ast }{|}^{2(1-\alpha )}y,y〉.$$

(1.2)

For results related to Kato’s inequality, see [1–17] and [18].

In the recent paper [19], by employing Kato’s inequality (1.2), Dragomir, Cho and Kim established the following results for sequences of bounded linear operators on complex Hilbert spaces.

Theorem 1 Let $(T_1,\dots ,T_n)\in \mathcal{B}(H)\times \cdots \times \mathcal{B}(H):={\mathcal{B}}^{(n)}(H)$ be an n-tuple of bounded linear operators on the Hilbert space $(H;〈\cdot ,\cdot 〉)$ and $(p_1,\dots ,p_n)\in {\mathbb{R}}_{+}^{\ast n}$ be an n-tuple of nonnegative weights not all of them equal to zero, then

$$\begin{array}{rl}\sum _{j=1}^n{p}_j|〈T_{j}x,y〉|\le & {〈\sum _{j=1}^n{p}_j\left(\frac{|T_j{|}^{2\alpha }+|T_j{|}^{2(1-\alpha )}}{2}\right)x,x〉}^{1/2}\\ \times {〈\sum _{j=1}^n{p}_j\left(\frac{|T_j^{\ast }{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right)y,y〉}^{1/2}\end{array}$$

(1.3)

for any $\alpha \in [0,1]$ and any $x,y\in H$.

Theorem 2 With the assumptions in Theorem 1, we have

$$\begin{array}{r}\sum _{j=1}^n{p}_j|〈T_{j}x,y〉{|}^2\\ \le \frac{1}{2}\sum _{j=1}^n{p}_j({\parallel T_{j}x\parallel }^{2\alpha }{\parallel T_j^{\ast }y\parallel }^{2(1-\alpha )}+{\parallel T_j^{\ast }y\parallel }^{2\alpha }{\parallel T_{j}x\parallel }^{2(1-\alpha )})\\ \le \frac{1}{2}[{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }y\parallel }^2\right)}^{1-\alpha }+{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2\right)}^{1-\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }y\parallel }^2\right)}^{\alpha }]\\ \le \frac{1}{2}\sum _{j=1}^n{p}_j({\parallel T_{j}x\parallel }^2+{\parallel T_j^{\ast }y\parallel }^2)\end{array}$$

(1.4)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

For various related results, see the papers [20–22] and [23–27].

Motivated by the above results, we establish in this paper more inequalities for n-tuples of bounded linear operators that can be obtained from Kato’s result (1.2) and apply them to functions of normal operators defined by power series as well as to some norms and numerical radii that can be associated with these n-tuples of bounded linear operators on Hilbert spaces. The paper is a natural continuation of [19].

2 Some inequalities for n-tuples of operators

The following result holds.

Theorem 3 Let $(T_1,\dots ,T_n)\in {\mathcal{B}}^{(n)}(H)$ be an n-tuple of bounded linear operators on the Hilbert space $(H;〈\cdot ,\cdot 〉)$ and $(p_1,\dots ,p_n)\in {\mathbb{R}}_{+}^{\ast n}$ be an n-tuple of nonnegative weights not all of them equal to zero, then

$$\begin{array}{rl}\left|〈\sum _{j=1}^n{p}_j\left(\frac{T_j+T_j^{\ast }}{2}\right)x,y〉\right|\le & \sum _{j=1}^n{p}_j\left|〈\frac{T_j+T_j^{\ast }}{2}x,y〉\right|\\ \le & \sum _{j=1}^n{p}_j\left[\frac{|〈T_{j}x,y〉|+|〈T_j^{\ast }x,y〉|}{2}\right]\\ \le & {〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]x,x〉}^{1/2}\\ \times {〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]y,y〉}^{1/2}\end{array}$$

(2.1)

for any $\alpha \in [0,1]$ and, in particular, for $\alpha =\frac{1}{2}$

$$\begin{array}{rl}\left|〈\sum _{j=1}^n{p}_j\left(\frac{T_j+T_j^{\ast }}{2}\right)x,y〉\right|\le & \sum _{j=1}^n{p}_j\left|〈\frac{T_j+T_j^{\ast }}{2}x,y〉\right|\\ \le & \sum _{j=1}^n{p}_j\left[\frac{|〈T_{j}x,y〉|+|〈T_j^{\ast }x,y〉|}{2}\right]\\ \le & {〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]x,x〉}^{1/2}\\ \times {〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]y,y〉}^{1/2}\end{array}$$

(2.2)

for any $x,y\in H$.

Proof The first two inequalities are obvious by the properties of the modulus.

Utilizing Kato’s inequality, we have

$$|〈T_{j}x,y〉|\le {〈|T_j{|}^{2\alpha }x,x〉}^{1/2}{〈|T_j^{\ast }{|}^{2(1-\alpha )}y,y〉}^{1/2}$$

(2.3)

and, by replacing x with y, we have

$$|〈T_{j}y,x〉|\le {〈|T_j{|}^{2\alpha }y,y〉}^{1/2}{〈|T_j^{\ast }{|}^{2(1-\alpha )}x,x〉}^{1/2},$$

i.e.,

$$\left|〈T_j^{\ast }x,y〉\right|\le {〈|T_j^{\ast }{|}^{2(1-\alpha )}x,x〉}^{1/2}{〈|T_j{|}^{2\alpha }y,y〉}^{1/2}$$

(2.4)

for any $j\in \{1,\dots ,n\}$ and $x,y\in H$.

Adding inequalities (2.3) and (2.4) and utilizing the elementary inequality

$$ab+cd\le {(a^2+c^2)}^{1/2}{(b^2+d^2)}^{1/2},a,b,c,d\ge 0,$$

we get

$$\begin{array}{rl}|〈T_{j}x,y〉|+\left|〈T_j^{\ast }x,y〉\right|\le & {〈|T_j{|}^{2\alpha }x,x〉}^{1/2}{〈|T_j^{\ast }{|}^{2(1-\alpha )}y,y〉}^{1/2}\\ +{〈|T_j^{\ast }{|}^{2(1-\alpha )}x,x〉}^{1/2}{〈|T_j{|}^{2\alpha }y,y〉}^{1/2}\\ \le & {〈[|T_j{|}^{2\alpha }+\left|T_j^{\ast }{|}^{2(1-\alpha )}\right]x,x〉}^{1/2}\\ \times {〈[|T_j{|}^{2\alpha }+\left|T_j^{\ast }{|}^{2(1-\alpha )}\right]y,y〉}^{1/2}\end{array}$$

(2.5)

for any $j\in \{1,\dots ,n\}$ and $x,y\in H$.

Multiplying inequalities (2.5) by $p_j\ge 0$ and then summing over j from 1 to n and utilizing the weighted Cauchy-Buniakowski-Schwarz inequality, we have

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\right|〈T_{j}x,y〉|+|〈T_j^{\ast }x,y〉\left|\right]\\ \le \sum _{j=1}^n{p}_j{〈[|T_j{|}^{2\alpha }+\left|T_j^{\ast }{|}^{2(1-\alpha )}\right]x,x〉}^{1/2}{〈[|T_j{|}^{2\alpha }+\left|T_j^{\ast }{|}^{2(1-\alpha )}\right]y,y〉}^{1/2}\\ \le {〈\sum _{j=1}^n{p}_j[|T_j{|}^{2\alpha }+\left|T_j^{\ast }{|}^{2(1-\alpha )}\right]x,x〉}^{1/2}{〈\sum _{j=1}^n{p}_j[|T_j{|}^{2\alpha }+\left|T_j^{\ast }{|}^{2(1-\alpha )}\right]y,y〉}^{1/2}\end{array}$$

(2.6)

for $x,y\in H$, which is equivalent to the third inequality in (2.1). □

Remark 1 The particular case $y=x$ is of interest for providing numerical radii inequalities and can be stated as follows:

$$\begin{array}{rcl}\left|〈\sum _{j=1}^n{p}_j\left(\frac{T_j+T_j^{\ast }}{2}\right)x,x〉\right|& \le & \sum _{j=1}^n{p}_j\left|〈\frac{T_j+T_j^{\ast }}{2}x,x〉\right|\\ \le & \sum _{j=1}^n{p}_j|〈T_{j}x,x〉|\\ \le & 〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]x,x〉\end{array}$$

(2.7)

for any $\alpha \in [0,1]$ and, for $\alpha =\frac{1}{2}$,

$$\begin{array}{rcl}\left|〈\sum _{j=1}^n{p}_j\left(\frac{T_j+T_j^{\ast }}{2}\right)x,x〉\right|& \le & \sum _{j=1}^n{p}_j\left|〈\frac{T_j+T_j^{\ast }}{2}x,x〉\right|\\ \le & \sum _{j=1}^n{p}_j|〈T_{j}x,x〉|\\ \le & 〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]x,x〉\end{array}$$

(2.8)

for any $x\in H$.

The case of unitary vectors provides more refinements as follows.

Remark 2 With the assumptions in Theorem 3, we have

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\frac{|〈T_{j}x,y〉|+|〈T_j^{\ast }x,y〉|}{2}\right]\\ \le {〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]x,x〉}^{1/2}{〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]y,y〉}^{1/2}\\ \le 〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]\right)}^{1/2}x,x〉\\ \times 〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]\right)}^{1/2}y,y〉\\ \le \frac{1}{2}[{〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]\right)}^{1/2}x,x〉}^2\\ +{〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]\right)}^{1/2}y,y〉}^2]\\ \le \frac{1}{2}[〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]x,x〉+〈\sum _{j=1}^n{p}_j\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]y,y〉]\end{array}$$

(2.9)

for any $\alpha \in [0,1]$ and, in particular,

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\frac{|〈T_{j}x,y〉|+|〈T_j^{\ast }x,y〉|}{2}\right]\\ \le {〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]x,x〉}^{1/2}{〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]y,y〉}^{1/2}\\ \le 〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]\right)}^{1/2}x,x〉〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]\right)}^{1/2}y,y〉\\ \le \frac{1}{2}[{〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]\right)}^{1/2}x,x〉}^2+{〈{\left(\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]\right)}^{1/2}y,y〉}^2]\\ \le \frac{1}{2}[〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]x,x〉+〈\sum _{j=1}^n{p}_j\left[\frac{|T_j|+|T_j^{\ast }|}{2}\right]y,y〉]\end{array}$$

(2.10)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$.

The proofs follow by utilizing the Hölder-McCarthy inequalities $〈P^{r}x,x〉\le {〈Px,x〉}^r$ and ${〈Px,x〉}^s\le 〈P^{s}x,x〉$ that hold for the positive operator P, for $r\in (0,1)$, $s\in [1,\mathrm{\infty })$ and $x\in H$ with $\parallel x\parallel =1$. The details are omitted.

In order to employ the above result in obtaining some inequalities for functions of normal operators defined by power series, we need the following version of (2.1).

Remark 3 If we write inequality (2.1) for the normal operators $N_j$, $j\in \{1,\dots ,n\}$, then we get

$$\begin{array}{rl}\left|〈\sum _{j=1}^n{p}_j\left(\frac{N_j+N_j^{\ast }}{2}\right)x,y〉\right|\le & \sum _{j=1}^n{p}_j\left|〈\frac{N_j+N_j^{\ast }}{2}x,y〉\right|\\ \le & \sum _{j=1}^n{p}_j\left[\frac{|〈N_{j}x,y〉|+|〈N_j^{\ast }x,y〉|}{2}\right]\\ \le & {〈\sum _{j=1}^n{p}_j\left[\frac{|N_j{|}^{2\alpha }+|N_j{|}^{2(1-\alpha )}}{2}\right]x,x〉}^{1/2}\\ \times {〈\sum _{j=1}^n{p}_j\left[\frac{|N_j{|}^{2\alpha }+|N_j{|}^{2(1-\alpha )}}{2}\right]y,y〉}^{1/2}\end{array}$$

(2.11)

for any $\alpha \in [0,1]$ and, in particular, for $\alpha =\frac{1}{2}$

$$\begin{array}{rl}\left|〈\sum _{j=1}^n{p}_j\left(\frac{N_j+N_j^{\ast }}{2}\right)x,y〉\right|& \le \sum _{j=1}^n{p}_j\left|〈\frac{N_j+N_j^{\ast }}{2}x,y〉\right|\le \sum _{j=1}^n{p}_j\left[\frac{|〈N_{j}x,y〉|+|〈N_j^{\ast }x,y〉|}{2}\right]\\ \le {〈\sum _{j=1}^n{p}_j|N_j|x,x〉}^{1/2}{〈\sum _{j=1}^n{p}_j|N_j|y,y〉}^{1/2}\end{array}$$

(2.12)

for any $x,y\in H$.

The following results involving quadratics also hold.

Theorem 4 Let $(T_1,\dots ,T_n)\in {\mathcal{B}}^{(n)}(H)$ be an n-tuple of bounded linear operators on the Hilbert space $(H;〈\cdot ,\cdot 〉)$ and $(p_1,\dots ,p_n)\in {\mathbb{R}}_{+}^{\ast n}$ be an n-tuple of nonnegative weights not all of them equal to zero, then

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\right|〈T_{j}x,y〉{|}^2+\left|〈T_j^{\ast }x,y〉{|}^2\right]\\ \le \sum _{j=1}^n{p}_j[{\parallel T_{j}x\parallel }^{2\alpha }{\parallel T_j^{\ast }y\parallel }^{2(1-\alpha )}+{\parallel T_{j}y\parallel }^{2\alpha }{\parallel T_j^{\ast }x\parallel }^{2(1-\alpha )}]\\ \le {\left(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }y\parallel }^2\right)}^{1-\alpha }+{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}y\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }x\parallel }^2\right)}^{1-\alpha }\\ \le {\left(\sum _{j=1}^n{p}_j[{\parallel T_{j}x\parallel }^2+{\parallel T_{j}y\parallel }^2]\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j[{\parallel T_j^{\ast }y\parallel }^2+{\parallel T_j^{\ast }x\parallel }^2]\right)}^{1-\alpha }\end{array}$$

(2.13)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

Proof We must prove the inequalities only in the case $\alpha \in (0,1)$, since the case $\alpha =0$ or $\alpha =1$ follows directly from the corresponding case of Kato’s inequality.

Utilizing Kato’s inequality, we have

$$|〈T_{j}x,y〉{|}^2\le 〈|T_j{|}^{2\alpha }x,x〉〈|T_j^{\ast }{|}^{2(1-\alpha )}y,y〉$$

(2.14)

and, by replacing x with y, we have

$$|〈T_j^{\ast }x,y〉{|}^2\le 〈|T_j^{\ast }{|}^{2(1-\alpha )}x,x〉〈|T_j{|}^{2\alpha }y,y〉$$

(2.15)

for any $j\in \{1,\dots ,n\}$ and $x,y\in H$.

By the Hölder-McCarthy inequality $〈P^{r}x,x〉\le {〈Px,x〉}^r$ for $r\in (0,1)$ and $x\in H$ with $\parallel x\parallel =1$, we also have

$$〈|T_j{|}^{2\alpha }x,x〉〈|T_j^{\ast }{|}^{2(1-\alpha )}y,y〉\le {〈|T_j{|}^{2}x,x〉}^{\alpha }{〈|T_j^{\ast }{|}^{2}y,y〉}^{1-\alpha }$$

(2.16)

and

$$〈|T_j^{\ast }{|}^{2(1-\alpha )}x,x〉〈|T_j{|}^{2\alpha }y,y〉\le {〈|T_j{|}^{2}y,y〉}^{\alpha }{〈|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }$$

(2.17)

for any $j\in \{1,\dots ,n\}$ and $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$.

We then obtain by summation

$$\begin{array}{r}|〈T_{j}x,y〉{|}^2+|〈T_j^{\ast }x,y〉{|}^2\\ \le {〈|T_j{|}^{2}x,x〉}^{\alpha }{〈|T_j^{\ast }{|}^{2}y,y〉}^{1-\alpha }+{〈|T_j{|}^{2}y,y〉}^{\alpha }{〈|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }\end{array}$$

(2.18)

for any $j\in \{1,\dots ,n\}$ and $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$.

Now, if we multiply (2.18) with $p_j\ge 0$, sum over j from 1 to n, we get

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\right|〈T_{j}x,y〉{|}^2+\left|〈T_j^{\ast }x,y〉{|}^2\right]\\ \le \sum _{j=1}^n{p}_j{〈|T_j{|}^{2}x,x〉}^{\alpha }{〈|T_j^{\ast }{|}^{2}y,y〉}^{1-\alpha }+\sum _{j=1}^n{p}_j{〈|T_j{|}^{2}y,y〉}^{\alpha }{〈|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }\end{array}$$

(2.19)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in (0,1)$.

Since $〈|T_j{|}^{2}x,x〉={\parallel T_{j}x\parallel }^2$, $〈|T_j^{\ast }{|}^{2}y,y〉={\parallel T_j^{\ast }y\parallel }^2$, $〈|T_j{|}^{2}y,y〉={\parallel T_{j}y\parallel }^2$ and $〈|T_j^{\ast }{|}^{2}x,x〉={\parallel T_j^{\ast }x\parallel }^2$ $j\in \{1,\dots ,n\}$, then we get from (2.19) the first part of (2.13).

Now, on making use of the weighted Hölder discrete inequality

$$\sum _{j=1}^n{p}_j{a}_j{b}_j\le {\left(\sum _{j=1}^n{p}_j{a}_j^p\right)}^{1/p}{\left(\sum _{j=1}^n{p}_j{b}_j^q\right)}^{1/q},p,q>1,\frac{1}{p}+\frac{1}{q}=1,$$

where $(a_1,\dots ,a_n),(b_1,\dots ,b_n)\in {\mathbb{R}}_{+}^n$, we also have

$$\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^{2\alpha }{\parallel T_j^{\ast }y\parallel }^{2(1-\alpha )}\le {\left(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }y\parallel }^{2(1-\alpha )}\right)}^{1-\alpha }$$

and

$$\sum _{j=1}^n{p}_j{\parallel T_{j}y\parallel }^{2\alpha }{\parallel T_j^{\ast }x\parallel }^{2(1-\alpha )}\le {\left(\sum _{j=1}^n{p}_j{\parallel T_{j}y\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }x\parallel }^2\right)}^{1-\alpha }.$$

Summing these two inequalities, we deduce the second inequality in (2.13).

Finally, on utilizing the Hölder inequality

$$ab+cd\le {(a^p+c^p)}^{1/p}{(b^q+d^q)}^{1/q},a,b,c,d\ge 0,$$

where $p>1$ and $\frac{1}{p}+\frac{1}{q}=1$, we have

$$\begin{array}{r}{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }y\parallel }^2\right)}^{1-\alpha }+{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}y\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }x\parallel }^2\right)}^{1-\alpha }\\ \le {(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2+\sum _{j=1}^n{p}_j{\parallel T_{j}y\parallel }^2)}^{\alpha }{(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }y\parallel }^2+\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }x\parallel }^2)}^{1-\alpha }\end{array}$$

and the proof is completed. □

Remark 4 Utilizing the elementary inequality for complex numbers

$$|\frac{z+w}{2}{|}^2\le \frac{|z{|}^2+|w{|}^2}{2},z,w\in \mathbb{C},$$

we have

$$\sum _{j=1}^n{p}_j\left[\right|〈\left(\frac{T_j+T_j^{\ast }}{2}\right)x,y〉{|}^2]\le \sum _{j=1}^n{p}_j\left[\frac{|〈T_{j}x,y〉{|}^2+|〈T_j^{\ast }x,y〉{|}^2}{2}\right]$$

(2.20)

and by the weighted arithmetic mean-geometric mean inequality

$$a^{\alpha }{b}^{1-\alpha }\le \alpha a+(1-\alpha )b,a,b\ge 0,\alpha \in [0,1],$$

we also have

$$\begin{array}{r}{\left(\sum _{j=1}^n{p}_j[{\parallel T_{j}x\parallel }^2+{\parallel T_{j}y\parallel }^2]\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j[{\parallel T_j^{\ast }y\parallel }^2+{\parallel T_j^{\ast }x\parallel }^2]\right)}^{1-\alpha }\\ \le \alpha \sum _{j=1}^n{p}_j[{\parallel T_{j}x\parallel }^2+{\parallel T_{j}y\parallel }^2]+(1-\alpha )\sum _{j=1}^n{p}_j[{\parallel T_j^{\ast }y\parallel }^2+{\parallel T_j^{\ast }x\parallel }^2].\end{array}$$

(2.21)

If we choose $\alpha =\frac{1}{2}$ and use (2.4), (2.20) and (2.22), we derive

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\right|〈\left(\frac{T_j+T_j^{\ast }}{2}\right)x,y〉{|}^2]\\ \le \sum _{j=1}^n{p}_j\left[\frac{|〈T_{j}x,y〉{|}^2+|〈T_j^{\ast }x,y〉{|}^2}{2}\right]\le \frac{1}{2}\sum _{j=1}^n{p}_j[\parallel T_{j}x\parallel \parallel T_j^{\ast }y\parallel +\parallel T_{j}y\parallel \parallel T_j^{\ast }x\parallel ]\\ \le \frac{1}{2}{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}x\parallel }^2\right)}^{1/2}{\left(\sum _{j=1}^n{p}_j\parallel T_j^{\ast }y\parallel \right)}^{1/2}+\frac{1}{2}{\left(\sum _{j=1}^n{p}_j{\parallel T_{j}y\parallel }^2\right)}^{1/2}{\left(\sum _{j=1}^n{p}_j{\parallel T_j^{\ast }x\parallel }^2\right)}^{1/2}\\ \le {\left(\sum _{j=1}^n{p}_j\left[\frac{{\parallel T_{j}x\parallel }^2+{\parallel T_{j}y\parallel }^2}{2}\right]\right)}^{1/2}{\left(\sum _{j=1}^n{p}_j\left[\frac{{\parallel T_j^{\ast }y\parallel }^2+{\parallel T_j^{\ast }x\parallel }^2}{2}\right]\right)}^{1/2}\\ \le \sum _{j=1}^n{p}_j\left[\frac{{\parallel T_{j}x\parallel }^2+{\parallel T_{j}y\parallel }^2+{\parallel T_j^{\ast }y\parallel }^2+{\parallel T_j^{\ast }x\parallel }^2}{4}\right]\\ =\frac{1}{2}[\sum _{j=1}^n{p}_j〈\frac{|T_j{|}^2+|T_j^{\ast }{|}^2}{2}x,x〉+\sum _{j=1}^n{p}_j〈\frac{|T_j{|}^2+|T_j^{\ast }{|}^2}{2}y,y〉]\end{array}$$

(2.22)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$.

Remark 5 The case of normal operators $N_j$, $j\in \{1,\dots ,n\}$, is of interest for functions of operators and may be stated as follows:

$$\begin{array}{r}\sum _{j=1}^n{p}_j\left[\right|〈\left(\frac{N_j+N_j^{\ast }}{2}\right)x,y〉{|}^2]\\ \le \sum _{j=1}^n{p}_j\left[\frac{|〈N_{j}x,y〉{|}^2+|〈N_j^{\ast }x,y〉{|}^2}{2}\right]\\ \le \frac{1}{2}\sum _{j=1}^n{p}_j[{\parallel N_{j}x\parallel }^{2\alpha }{\parallel N_{j}y\parallel }^{2(1-\alpha )}+{\parallel N_{j}y\parallel }^{2\alpha }{\parallel N_{j}x\parallel }^{2(1-\alpha )}]\\ \le \frac{1}{2}{\left(\sum _{j=1}^n{p}_j{\parallel N_{j}x\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel N_{j}y\parallel }^2\right)}^{1-\alpha }+\frac{1}{2}{\left(\sum _{j=1}^n{p}_j{\parallel N_{j}y\parallel }^2\right)}^{\alpha }{\left(\sum _{j=1}^n{p}_j{\parallel N_{j}x\parallel }^2\right)}^{1-\alpha }\\ \le \frac{1}{2}\sum _{j=1}^n{p}_j[{\parallel N_{j}x\parallel }^2+{\parallel N_{j}y\parallel }^2]\end{array}$$

(2.23)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

3 Inequalities for functions of normal operators

Now, by the help of power series $f(z)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{z}^n$, we can naturally construct another power series which will have as coefficients the absolute values of the coefficient of the original series, namely $f_A(z):={\sum }_{n=0}^{\mathrm{\infty }}|a_n|z^n$. It is obvious that this new power series will have the same radius of convergence as the original series. We also notice that if all coefficients $a_n\ge 0$, then $f_A=f$.

As some natural examples that are useful for applications, we can point out that if

$$\begin{array}{r}f(z)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{z}^n=ln\frac{1}{1+z},z\in D(0,1);\\ g(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n)!}{z}^{2n}=cosz,z\in \mathbb{C};\\ h(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!}{z}^{2n+1}=sinz,z\in \mathbb{C};\\ l(z)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{z}^n=\frac{1}{1+z},z\in D(0,1);\end{array}$$

(3.1)

then the corresponding functions constructed by the use of absolute values of the coefficients are

$$\begin{array}{r}{f}_A(z)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{z}^n=ln\frac{1}{1-z},z\in D(0,1);\\ g_A(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n)!}{z}^{2n}=coshz,z\in \mathbb{C};\\ h_A(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)!}{z}^{2n+1}=sinhz,z\in \mathbb{C};\\ l_A(z)=\sum _{n=0}^{\mathrm{\infty }}{z}^n=\frac{1}{1-z},z\in D(0,1).\end{array}$$

(3.2)

The following result is a functional inequality for normal operators that can be obtained from (2.1).

Theorem 5 Let $f(z)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{z}^n$ be a function defined by power series with complex coefficients and convergent on the open disk $D(0,R)\subset \mathbb{C}$, $R>0$. If N is a normal operator on the Hilbert space H and for $\alpha \in [0,1]$, we have that ${\parallel N\parallel }^{2\alpha },{\parallel N\parallel }^{2(1-\alpha )}<R$, then we have the inequalities

$$\begin{array}{rl}\left|〈\left(\frac{f(N)+f(N^{\ast })}{2}\right)x,y〉\right|\le & {〈\left(\frac{f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )})}{2}\right)x,x〉}^{1/2}\\ \times {〈\left(\frac{f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )})}{2}\right)y,y〉}^{1/2}\end{array}$$

(3.3)

for any $x,y\in H$.

Proof If N is a normal operator, then for any $j\in \mathbb{N}$, we have that

$$|N^j{|}^2={\left(N^{\ast }N\right)}^j=|N{|}^{2j}.$$

Utilizing inequality (2.11), we have

$$\begin{array}{r}\left|〈\sum _{j=0}^n{a}_j\left(\frac{N^j+{(N^{\ast })}^j}{2}\right)x,y〉\right|\\ \le \sum _{j=0}^n|a{|}_j|〈\frac{N^j+{(N^{\ast })}^j}{2}x,y〉|\\ \le \sum _{j=0}^n|a{|}_j\left[\frac{|〈N^{j}x,y〉|+|〈{(N^{\ast })}^{j}x,y〉|}{2}\right]\\ \le {〈\sum _{j=0}^n|a{|}_j\left[\frac{{(|N{|}^{2\alpha })}^j+{(|N{|}^{2(1-\alpha )})}^j}{2}\right]x,x〉}^{1/2}\\ \times {〈\sum _{j=1}^n|a{|}_0\left[\frac{{(|N{|}^{2\alpha })}^j+{(|N{|}^{2(1-\alpha )})}^j}{2}\right]y,y〉}^{1/2}\end{array}$$

(3.4)

for any $\alpha \in [0,1]$, $n\in \mathbb{N}$ and any $x,y\in H$. Since ${\parallel N\parallel }^{2\alpha },{\parallel N\parallel }^{2(1-\alpha )}<R$, then it follows that the series ${\sum }_{j=0}^{\mathrm{\infty }}|a_j|{(|N{|}^{2\alpha })}^j$ and ${\sum }_{j=0}^{\mathrm{\infty }}|a_j|{(|N{|}^{2(1-\alpha )})}^j$ are absolute convergent in $\mathcal{B}(H)$, and by taking the limit over $n\to \mathrm{\infty }$ in (3.4), we deduce the desired result (3.3). □

Remark 6 With the assumptions in Theorem 5, if we take the supremum over $y\in H$, $\parallel y\parallel =1$, then we get the vector inequality

$$\begin{array}{rl}\parallel \left(\frac{f(N)+f(N^{\ast })}{2}\right)x\parallel \le & \frac{1}{2}{〈(f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )}))x,x〉}^{1/2}\\ \times \parallel f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )})\parallel \end{array}$$

(3.5)

for any $x\in H$, which in its turn produces the norm inequality

$$\parallel \frac{f(N)+f(N^{\ast })}{2}\parallel \le \frac{1}{2}\parallel f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )})\parallel$$

(3.6)

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

Moreover, if we take $y=x$ in (3.3), then we have

$$\left|〈\frac{f(N)+f(N^{\ast })}{2}x,x〉\right|\le \frac{1}{2}〈[f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )})]x,x〉$$

(3.7)

for any $x\in H$, which, by taking the supremum over $x\in H$, $\parallel x\parallel =1$, generates the numerical radius inequality

$$w\left(\frac{f(N)+f(N^{\ast })}{2}\right)\le \frac{1}{2}w[f_A(|N{|}^{2\alpha })+f(|N{|}^{2(1-\alpha )})]$$

(3.8)

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

Making use of the examples in (3.1) and (3.2), we can state the vector inequalities

$$\begin{array}{r}\left|〈\left[\frac{ln{(1_H+N)}^{-1}+ln{(1_H+N^{\ast })}^{-1}}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[ln{(1_H-|N{|}^{2\alpha })}^{-1}+ln{(1_H-|N{|}^{2(1-\alpha )})}^{-1}]x,x〉}^{1/2}\\ \times {〈[ln{(1_H-|N{|}^{2\alpha })}^{-1}+ln{(1_H-|N{|}^{2(1-\alpha )})}^{-1}]y,y〉}^{1/2}\end{array}$$

(3.9)

and

$$\begin{array}{r}\left|〈\left[\frac{{(1_H+N)}^{-1}+{(1_H+N^{\ast })}^{-1}}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[{(1_H-|N{|}^{2\alpha })}^{-1}+{(1_H-|N{|}^{2(1-\alpha )})}^{-1}]x,x〉}^{1/2}\\ \times {〈[ln{(1_H-|N{|}^{2\alpha })}^{-1}+ln{(1_H-|N{|}^{2(1-\alpha )})}^{-1}]y,y〉}^{1/2}\end{array}$$

(3.10)

for any $x,y\in H$ and $\parallel N\parallel <1$.

We also have the inequalities

$$\begin{array}{r}\left|〈\left[\frac{sin(N)+sin(N^{\ast })}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[sinh(|N{|}^{2\alpha })+sinh(|N{|}^{2(1-\alpha )})]x,x〉}^{1/2}\\ \times {〈[sinh(|N{|}^{2\alpha })+sinh(|N{|}^{2(1-\alpha )})]y,y〉}^{1/2}\end{array}$$

(3.11)

and

$$\begin{array}{r}\left|〈\left[\frac{cos(N)+cos(N^{\ast })}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[cosh(|N{|}^{2\alpha })+cosh(|N{|}^{2(1-\alpha )})]x,x〉}^{1/2}\\ \times {〈[cosh(|N{|}^{2\alpha })+cosh(|N{|}^{2(1-\alpha )})]y,y〉}^{1/2}\end{array}$$

(3.12)

for any $x,y\in H$ and N, a normal operator.

If we utilize the following function as power series representations with nonnegative coefficients:

$$\begin{array}{r}\frac{1}{2}ln\left(\frac{1+z}{1-z}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n-1}{z}^{2n-1},z\in D(0,1);\\ {sin}^{-1}(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+\frac{1}{2})}{\sqrt{\pi }(2n+1)n!}{z}^{2n+1},z\in D(0,1);\\ {tanh}^{-1}(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2n+1}{z}^{2n-1},z\in D(0,1);\\ {}_{2}F_1(\alpha ,\beta ,\gamma ,z)=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+\alpha )\mathrm{\Gamma }(n+\beta )\mathrm{\Gamma }(\gamma )}{n!\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\beta )\mathrm{\Gamma }(n+\gamma )}{z}^n,\alpha ,\beta ,\gamma >0,z\in D(0,1);\end{array}$$

(3.13)

where Γ is the gamma function, then we can state the following vector inequalities:

$$\begin{array}{r}\left|〈\left[\frac{exp(N)+exp(N^{\ast })}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[exp(|N{|}^{2\alpha })+exp(|N{|}^{2(1-\alpha )})]x,x〉}^{1/2}\\ \times {〈[exp(|N{|}^{2\alpha })+exp(|N{|}^{2(1-\alpha )})]y,y〉}^{1/2}\end{array}$$

(3.14)

for any $x,y\in H$ and N, a normal operator.

If $\parallel N\parallel <1$, then we also have the inequalities

$$\begin{array}{r}\left|〈\left[\frac{ln(\frac{1_H+N}{1_H-N})+ln(\frac{1_H+N^{\ast }}{1_H-N^{\ast }})}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[ln\left(\frac{1_H+|N{|}^{2\alpha }}{1_H-|N{|}^{2\alpha }}\right)+ln\left(\frac{1_H+|N{|}^{2(1-\alpha )}}{1_H-|N{|}^{2(1-\alpha )}}\right)]x,x〉}^{1/2}\\ \times {〈[ln\left(\frac{1_H+|N{|}^{2\alpha }}{1_H-|N{|}^{2\alpha }}\right)+ln\left(\frac{1_H+|N{|}^{2(1-\alpha )}}{1_H-|N{|}^{2(1-\alpha )}}\right)]y,y〉}^{1/2},\end{array}$$

(3.15)

$$\begin{array}{r}\left|〈\left[\frac{{tanh}^{-1}(N)+{tanh}^{-1}(N^{\ast })}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈[{tanh}^{-1}(|N{|}^{2\alpha })+{tanh}^{-1}(|N{|}^{2(1-\alpha )})]x,x〉}^{1/2}\\ \times {〈[{tanh}^{-1}(|N{|}^{2\alpha })+{tanh}^{-1}(|N{|}^{2(1-\alpha )})]y,y〉}^{1/2}\end{array}$$

(3.16)

and

$$\begin{array}{r}\left|〈\left[\frac{{}_{2}F_1(\alpha ,\beta ,\gamma ,N){+}_2{F}_1(\alpha ,\beta ,\gamma ,N^{\ast })}{2}\right]x,y〉\right|\\ \le \frac{1}{2}{〈{[}_2{F}_1(\alpha ,\beta ,\gamma ,|N{|}^{2\alpha }){+}_2{F}_1(\alpha ,\beta ,\gamma ,|N{|}^{2(1-\alpha )})]x,x〉}^{1/2}\\ \times {〈{[}_2{F}_1(\alpha ,\beta ,\gamma ,|N{|}^{2\alpha }){+}_2{F}_1(\alpha ,\beta ,\gamma ,|N{|}^{2(1-\alpha )})]y,y〉}^{1/2}\end{array}$$

(3.17)

for any $x,y\in H$.

From a different perspective, we also have the following.

Theorem 6 With the assumption of Theorem 5 and if N is a normal operator on the Hilbert space H and $z\in \mathbb{C}$ such that ${\parallel N\parallel }^2,|z{|}^2<R$, then we have the inequalities

$$\begin{array}{r}|〈\left(\frac{f(zN)+f(z{N}^{\ast })}{2}\right)x,y〉{|}^2\\ \le \frac{1}{2}{f}_A(|z{|}^2)[{〈f_A(|N{|}^2)x,x〉}^{\alpha }{〈f_A(|N{|}^2)y,y〉}^{1-\alpha }+{〈f_A(|N{|}^2)y,y〉}^{\alpha }{〈f_A(|N{|}^2)x,x〉}^{1-\alpha }]\\ \le \frac{1}{2}{f}_A(|z{|}^2)[〈f_A(|N{|}^2)x,x〉+〈f_A(|N{|}^2)y,y〉]\end{array}$$

(3.18)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

In particular, for $\alpha =\frac{1}{2}$, we have

$$\begin{array}{rl}|〈\left(\frac{f(zN)+f(z{N}^{\ast })}{2}\right)x,y〉{|}^2& \le f_A(|z{|}^2){〈f_A(|N{|}^2)x,x〉}^{1/2}{〈f_A(|N{|}^2)y,y〉}^{1/2}\\ \le \frac{1}{2}{f}_A(|z{|}^2)[〈f_A(|N{|}^2)x,x〉+〈f_A(|N{|}^2)y,y〉]\end{array}$$

(3.19)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$.

Proof If we use the third and fourth inequalities in (2.23), we have

$$\begin{array}{r}\sum _{j=0}^n|a_j|\left[\right|〈\left(\frac{N^j+{(N^{\ast })}^j}{2}\right)x,y〉{|}^2]\\ \le \frac{1}{2}{(\sum _{j=0}^n|a_j|{\parallel N^{j}x\parallel }^2)}^{\alpha }{(\sum _{j=0}^n|a_j|{\parallel N^{j}y\parallel }^2)}^{1-\alpha }\\ +\frac{1}{2}{(\sum _{j=0}^n|a_j|{\parallel N^{j}y\parallel }^2)}^{\alpha }{(\sum _{j=0}^n|a_j|{\parallel N^{j}x\parallel }^2)}^{1-\alpha }\\ \le \frac{1}{2}\sum _{j=0}^n|a_j|[{\parallel N^{j}x\parallel }^2+{\parallel N^{j}y\parallel }^2]\end{array}$$

(3.20)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

Since N is a normal operator on the Hilbert space H, then

$${\parallel N^{j}x\parallel }^2=〈|N^j{|}^{2}x,x〉=〈|N{|}^{2j}x,x〉$$

for any $j\in \{0,\dots ,n\}$ and for any $x\in H$ with $\parallel x\parallel =1$.

Then from (3.20) we get

$$\begin{array}{r}\sum _{j=0}^n|a_j|\left[\right|〈\left(\frac{N^j+{(N^{\ast })}^j}{2}\right)x,y〉{|}^2]\\ \le \frac{1}{2}{\left(〈\sum _{j=0}^n|a_j||N{|}^{2j}x,x〉\right)}^{\alpha }{\left(〈\sum _{j=0}^n|a_j||N{|}^{2j}y,y〉\right)}^{1-\alpha }\\ +\frac{1}{2}{\left(〈\sum _{j=0}^n|a_j||N{|}^{2j}y,y〉\right)}^{\alpha }{\left(〈\sum _{j=0}^n|a_j||N{|}^{2j}x,x〉\right)}^{1-\alpha }\\ \le \frac{1}{2}[〈\sum _{j=0}^n|a_j||N{|}^{2j}x,x〉+〈\sum _{j=0}^n|a_j||N{|}^{2j}y,y〉]\end{array}$$

(3.21)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

By the weighted Cauchy-Buniakowski-Schwarz inequality, we also have

$$\begin{array}{r}|〈\sum _{j=0}^n{a}_j{z}^j\left(\frac{N^j+{(N^{\ast })}^j}{2}\right)x,y〉{|}^2\\ \le \sum _{j=0}^n|a_j||z{|}^{2j}\sum _{j=0}^n|a_j|\left[\right|〈\left(\frac{N^j+{(N^{\ast })}^j}{2}\right)x,y〉{|}^2]\end{array}$$

(3.22)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$.

Now, since the series ${\sum }_{j=0}^{\mathrm{\infty }}{a}_j{z}^j{N}^j$, ${\sum }_{j=0}^{\mathrm{\infty }}{a}_j{z}^j{(N^{\ast })}^j$, ${\sum }_{j=0}^{\mathrm{\infty }}|a_j||z{|}^{2j}$, ${\sum }_{j=0}^{\mathrm{\infty }}|a_j||N{|}^{2j}$ are convergent, then by (3.21) and (3.22) on letting $n\to \mathrm{\infty }$, we deduce the desired result (3.18). □

Similar inequalities for some particular functions of interest can be stated. However, the details are left to the interested reader.

4 Applications to the Euclidian norm

In [28], the author introduced the following norm on the Cartesian product ${\mathcal{B}}^{(n)}(H):=\mathcal{B}(H)\times \cdots \times \mathcal{B}(H)$, where $\mathcal{B}(H)$ denotes the Banach algebra of all bounded linear operators defined on the complex Hilbert space H:

$${\parallel (T_1,\dots ,T_n)\parallel }_e:=\underset{({\lambda }_1,\dots ,{\lambda }_n)\in {\mathbb{B}}_n}{sup}\parallel {\lambda }_1{T}_1+\cdots +{\lambda }_n{T}_n\parallel ,$$

(4.1)

where $(T_1,\dots ,T_n)\in {\mathcal{B}}^{(n)}(H)$ and ${\mathbb{B}}_n:=\{({\lambda }_1,\dots ,{\lambda }_n)\in {\mathbb{C}}^n|{\sum }_{j=1}^n|{\lambda }_j{|}^2\le 1\}$ is the Euclidean closed ball in ${\mathbb{C}}^n$.

It is clear that ${\parallel \cdot \parallel }_e$ is a norm on $B^{(n)}(H)$ and for any $(T_1,\dots ,T_n)\in B^{(n)}(H)$, we have

$${\parallel (T_1,\dots ,T_n)\parallel }_e={\parallel (T_1^{\ast },\dots ,T_n^{\ast })\parallel }_e,$$

where $T_j^{\ast }$ is the adjoint operator of $T_j$, $j\in \{1,\dots ,n\}$. We call this the Euclidian norm of an n-tuple of operators $(T_1,\dots ,T_n)\in B^{(n)}(H)$.

It has been shown in [28] that the following basic inequality for the Euclidian norm holds true:

$$\frac{1}{\sqrt{n}}{\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{\frac{1}{2}}\le {\parallel (T_1,\dots ,T_n)\parallel }_e\le {\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{\frac{1}{2}}$$

(4.2)

for any n-tuple $(T_1,\dots ,T_n)\in B^{(n)}(H)$ and the constants $\frac{1}{\sqrt{n}}$ and 1 are best possible.

In the same paper [28], the author introduced the Euclidean operator radius of an n-tuple of operators $(T_1,\dots ,T_n)$ by

$$w_e(T_1,\dots ,T_n):=\underset{\parallel x\parallel =1}{sup}{\left(\sum _{j=1}^n\right|〈T_{j}x,x〉{|}^2)}^{\frac{1}{2}}$$

(4.3)

and proved that $w_e(\cdot )$ is a norm on $B^{(n)}(H)$ and satisfies the double inequality

$$\frac{1}{2}{\parallel (T_1,\dots ,T_n)\parallel }_e\le w_e(T_1,\dots ,T_n)\le {\parallel (T_1,\dots ,T_n)\parallel }_e$$

(4.4)

for each n-tuple $(T_1,\dots ,T_n)\in B^{(n)}(H)$.

As pointed out in [28], the Euclidean numerical radius also satisfies the double inequality

$$\frac{1}{2\sqrt{n}}{\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{\frac{1}{2}}\le w_e(T_1,\dots ,T_n)\le {\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{\frac{1}{2}}$$

(4.5)

for any $(T_1,\dots ,T_n)\in B^{(n)}(H)$ and the constants $\frac{1}{2\sqrt{n}}$ and 1 are best possible.

In [29], by utilizing the concept of hypo-Euclidean norm on $H^n$, we obtained the following representation for the Euclidian norm.

Proposition 1 For any $(T_1,\dots ,T_n)\in B^{(n)}(H)$, we have

$${\parallel (T_1,\dots ,T_n)\parallel }_e=\underset{\parallel y\parallel =1,\parallel x\parallel =1}{sup}{\left(\sum _{j=1}^n\right|〈T_{j}y,x〉{|}^2)}^{\frac{1}{2}}.$$

(4.6)

Theorem 7 For any $(T_1,\dots ,T_n)\in B^{(n)}(H)$, we have

$$\begin{array}{rl}{\parallel (\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\parallel }_e^2& \le {\parallel \sum _{j=1}^n|T_j{|}^2\parallel }^{\alpha }{\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{1-\alpha }\\ \le \alpha \parallel \sum _{j=1}^n|T_j{|}^2\parallel +(1-\alpha )\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel \end{array}$$

(4.7)

and

$$\begin{array}{r}{w}_e^2(\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\\ \le \underset{\parallel x\parallel =1}{sup}\left[{〈\sum _{j=1}^n|T_j{|}^{2}x,x〉}^{\alpha }{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }\right]\\ \le \{\begin{array}{l}{\parallel {\sum }_{j=1}^n|T_j{|}^2\parallel }^{\alpha }{\parallel {\sum }_{j=1}^n|T_j^{\ast }{|}^2\parallel }^{1-\alpha },\\ \parallel \alpha {\sum }_{j=1}^n|T_j{|}^2+(1-\alpha ){\sum }_{j=1}^n|T_j^{\ast }{|}^2\parallel \end{array}\\ \le \alpha \parallel \sum _{j=1}^n|T_j{|}^2\parallel +(1-\alpha )\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel \end{array}$$

(4.8)

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

Proof Making use of inequalities (2.13) and (2.20), we have

$$\begin{array}{r}\sum _{j=1}^n\left[\right|〈\left(\frac{T_j+T_j^{\ast }}{2}\right)x,y〉{|}^2]\\ \le \frac{1}{2}{〈\sum _{j=1}^n|T_j{|}^{2}x,x〉}^{\alpha }{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}y,y〉}^{1-\alpha }\\ +\frac{1}{2}{〈\sum _{j=1}^n|T_j{|}^{2}y,y〉}^{\alpha }{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }\end{array}$$

(4.9)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in [0,1]$.

Taking the supremum over $\parallel x\parallel =\parallel y\parallel =1$ in (4.9), we get

$$\begin{array}{r}{\parallel (\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\parallel }_e^2\\ \le \frac{1}{2}\underset{\parallel x\parallel =1}{sup}{〈\sum _{j=1}^n|T_j{|}^{2}x,x〉}^{\alpha }\underset{\parallel y\parallel =1}{sup}{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}y,y〉}^{1-\alpha }\\ +\frac{1}{2}\underset{\parallel y\parallel =1}{sup}{〈\sum _{j=1}^n|T_j{|}^{2}y,y〉}^{\alpha }\underset{\parallel x\parallel =1}{sup}{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }\\ ={\parallel \sum _{j=1}^n|T_j{|}^2\parallel }^{\alpha }{\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{1-\alpha }\end{array}$$

and inequality (4.7) is proved.

Now, if we take $y=x$ in (4.9), we get

$$\begin{array}{r}\sum _{j=1}^n\left[\right|〈\left(\frac{T_j+T_j^{\ast }}{2}\right)x,x〉{|}^2]\\ \le {〈\sum _{j=1}^n|T_j{|}^{2}x,x〉}^{\alpha }{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}x,x〉}^{1-\alpha }\\ \le 〈[\alpha \sum _{j=1}^n|T_j{|}^2+(1-\alpha )\sum _{j=1}^n\left|T_j^{\ast }{|}^2\right]x,x〉\end{array}$$

(4.10)

for any $x\in H$ with $\parallel x\parallel =1$ and $\alpha \in [0,1]$.

Taking the supremum over $\parallel x\parallel =1$ in (4.10), we get the desired result. □

Remark 7 In the particular case $\alpha =\frac{1}{2}$, we get

$$\begin{array}{rl}{\parallel (\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\parallel }_e^2& \le {\parallel \sum _{j=1}^n|T_j{|}^2\parallel }^{1/2}{\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel }^{1/2}\\ \le \frac{1}{2}[\parallel \sum _{j=1}^n|T_j{|}^2\parallel +\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel ]\end{array}$$

(4.11)

and

$$\begin{array}{r}{w}_e^2(\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\\ \le \underset{\parallel x\parallel =1}{sup}\left[{〈\sum _{j=1}^n|T_j{|}^{2}x,x〉}^{1/2}{〈\sum _{j=1}^n|T_j^{\ast }{|}^{2}x,x〉}^{1/2}\right]\end{array}$$

(4.12)

$$\begin{array}{r}\le \{\begin{array}{l}{\parallel {\sum }_{j=1}^n|T_j{|}^2\parallel }^{1/2}{\parallel {\sum }_{j=1}^n|T_j^{\ast }{|}^2\parallel }^{1/2},\\ \parallel {\sum }_{j=1}^n\frac{|T_j{|}^2+|T_j^{\ast }{|}^2}{2}\parallel \end{array}\\ \le \frac{1}{2}[\parallel \sum _{j=1}^n|T_j{|}^2\parallel +\parallel \sum _{j=1}^n|T_j^{\ast }{|}^2\parallel ].\end{array}$$

(4.13)

5 Applications for s-1-norm and s-1-numerical radius

Following [30], we consider the s-p-norm of the n-tuple of operators $(T_1,\dots ,T_n)\in B^{(n)}(H)$ given by

$${\parallel (T_1,\dots ,T_n)\parallel }_{s,p}:=\underset{\parallel y\parallel =1,\parallel x\parallel =1}{sup}\left[{\left(\sum _{j=1}^n\right|〈T_{j}y,x〉{|}^p)}^{\frac{1}{p}}\right].$$

(5.1)

For $p=2$ we get

$${\parallel (T_1,\dots ,T_n)\parallel }_{s,2}={\parallel (T_1,\dots ,T_n)\parallel }_e.$$

We are interested in this section in the case $p=1$, namely on the s-1-norm defined by

$${\parallel (T_1,\dots ,T_n)\parallel }_{s,1}:=\underset{\parallel y\parallel =1,\parallel x\parallel =1}{sup}\sum _{j=1}^n|〈T_{j}y,x〉|.$$

Since for any $x,y\in H$ we have ${\sum }_{j=1}^n|〈T_{j}y,x〉|\ge |〈{\sum }_{j=1}^n{T}_{j}y,x〉|$, then by the properties of the supremum, we get the basic inequality

$$\parallel \sum _{j=1}^n{T}_j\parallel \le {\parallel (T_1,\dots ,T_n)\parallel }_{s,1}\le \sum _{j=1}^n\parallel T_j\parallel .$$

(5.2)

Similarly, we can also consider the s-p-numerical radius of the n-tuple of operators $(T_1,\dots ,T_n)\in B^{(n)}(H)$ defined by [30]

$$w_{s,p}(T_1,\dots ,T_n):=\underset{\parallel x\parallel =1}{sup}\left[{\left(\sum _{j=1}^n\right|〈T_{j}x,x〉{|}^p)}^{\frac{1}{p}}\right],$$

(5.3)

which for $p=2$ reduces to the Euclidean operator radius introduced previously. We observe that the s-p-numerical radius is also a norm on $B^{(n)}(H)$ for $p\ge 1$, and for $p=1$ it satisfies the basic inequality

$$w\left(\sum _{j=1}^n{T}_j\right)\le w_{s,1}(T_1,\dots ,T_n)\le \sum _{j=1}^{n}w(T_j).$$

(5.4)

Theorem 8 For any $(T_1,\dots ,T_n)\in B^{(n)}(H)$, we have

$$\begin{array}{rl}{\parallel (\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\parallel }_{s,1}& \le \parallel \sum _{j=1}^n\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]\parallel \\ \le \frac{1}{2}[\parallel \sum _{j=1}^n|T_j{|}^{2\alpha }\parallel +\parallel \sum _{j=1}^n|T_j^{\ast }{|}^{2(1-\alpha )}\parallel ]\end{array}$$

(5.5)

and

$$\begin{array}{rl}{w}_{s,1}(\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})& \le w_{s,1}(T_1,\dots ,T_n)\\ \le \parallel \sum _{j=1}^n\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]\parallel \\ \le \frac{1}{2}[\parallel \sum _{j=1}^n|T_j{|}^{2\alpha }\parallel +\parallel \sum _{j=1}^n|T_j^{\ast }{|}^{2(1-\alpha )}\parallel ]\end{array}$$

(5.6)

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

Proof Utilizing inequality (2.1), we have

$$\begin{array}{r}\sum _{j=1}^n\left|〈\frac{T_j+T_j^{\ast }}{2}x,y〉\right|\\ \le {〈\sum _{j=1}^n\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]x,x〉}^{1/2}\times {〈\sum _{j=1}^n\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]y,y〉}^{1/2}\end{array}$$

(5.7)

for any $x,y\in H$ and $\alpha \in [0,1]$.

Taking the supremum in (5.7) over $\parallel x\parallel =\parallel y\parallel =1$, we get the first inequality in (5.5).

The second part follows by the triangle inequality.

By inequality (2.7) we have

$$\sum _{j=1}^n\left|〈\frac{T_j+T_j^{\ast }}{2}x,x〉\right|\le \sum _{j=1}^n|〈T_{j}x,x〉|\le 〈\sum _{j=1}^n\left[\frac{|T_j{|}^{2\alpha }+|T_j^{\ast }{|}^{2(1-\alpha )}}{2}\right]x,x〉$$

for any $x\in H$.

Taking the supremum over $\parallel x\parallel =1$, we deduce the desired result (5.6). □

Remark 8 The case $\alpha =\frac{1}{2}$ produces the following chains of inequalities:

$$\begin{array}{rl}\parallel \sum _{j=1}^n\left(\frac{T_j+T_j^{\ast }}{2}\right)\parallel & \le {\parallel (\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\parallel }_{s,1}\\ \le \parallel \sum _{j=1}^n\left(\frac{|T_j|+|T_j^{\ast }|}{2}\right)\parallel \le \frac{1}{2}[\parallel \sum _{j=1}^n|T_j|\parallel +\parallel \sum _{j=1}^n\left|T_j^{\ast }\right|\parallel ]\end{array}$$

(5.8)

and

$$\begin{array}{rl}w\left(\sum _{j=1}^n\left(\frac{T_j+T_j^{\ast }}{2}\right)\right)& \le w_{s,1}(\frac{T_1+T_1^{\ast }}{2},\dots ,\frac{T_n+T_n^{\ast }}{2})\\ \le w_{s,1}(T_1,\dots ,T_n)\\ \le \parallel \sum _{j=1}^n\left(\frac{|T_j|+|T_j^{\ast }|}{2}\right)\parallel \\ \le \frac{1}{2}[\parallel \sum _{j=1}^n|T_j|\parallel +\parallel \sum _{j=1}^n\left|T_j^{\ast }\right|\parallel ].\end{array}$$

(5.9)