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

## Abstract

In this paper, by the use of the famous Kato’s inequality for bounded linear operators, we establish some inequalities for n-tuples of operators and apply them for functions of normal operators defined by power series as well as for some norms and numerical radii that arise in multivariate operator theory.

MSC:47A63, 47A99.

## 1 Introduction

The ‘square root’ of a positive bounded self-adjoint operator on H can be defined as follows (see, for instance, [[1], p.240]).

If the operator $A\in \mathcal{B}\left(H\right)$ is self-adjoint and positive, then there exists a unique positive self-adjoint operator $B:=\sqrt{A}\in \mathcal{B}\left(H\right)$ such that ${B}^{2}=A$. If A is invertible, then so is B.

If $A\in \mathcal{B}\left(H\right)$, then the operator ${A}^{\ast }A$ is self-adjoint and positive. Define the ‘absolute value’ operator by $|A|:=\sqrt{{A}^{\ast }A}$.

In 1952, Kato [2] proved the following generalization of 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 $x,y\in H$, $\alpha \in \left[0,1\right]$ and T is a bounded linear operator on H.

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

$|〈Tx,y〉{|}^{2}\le 〈|T{|}^{2\alpha }x,x〉〈|{T}^{\ast }{|}^{2\left(1-\alpha \right)}y,y〉.$
(1.2)

For results related to the Kato’s inequality, see [218] and [19].

In the recent paper [20], by employing Kato’s inequality (1.2), Dragomir established the following results for sequences of bonded linear operators on complex Hilbert spaces.

Theorem 1.1 Let $\left({T}_{1},\dots ,{T}_{n}\right)\in \mathcal{B}\left(H\right)×\cdots ×\mathcal{B}\left(H\right):={\mathcal{B}}^{\left(n\right)}\left(H\right)$ be an n-tuple of bounded linear operators on the Hilbert space $\left(H;〈\cdot ,\cdot 〉\right)$ and $\left({p}_{1},\dots ,{p}_{n}\right)\in {\mathbb{R}}_{+}^{\ast n}$ be an n-tuple of nonnegative weights not all of them equal to zero. Then we have

$\sum _{j=1}^{n}{p}_{j}|〈{T}_{j}x,y〉{|}^{2}\le {〈\sum _{j=1}^{n}{p}_{j}|{T}_{j}{|}^{2}x,x〉}^{\alpha }{〈\sum _{j=1}^{n}{p}_{j}|{T}_{j}^{\ast }{|}^{2}y,y〉}^{1-\alpha }$
(1.3)

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

He also obtained the following result.

Theorem 1.2 With the assumptions in Theorem  1.1, we have

$\sum _{j=1}^{n}{p}_{j}|〈{T}_{j}x,y〉|\le {〈\sum _{j=1}^{n}{p}_{j}|{T}_{j}{|}^{2\alpha }x,x〉}^{1/2}{〈\sum _{j=1}^{n}{p}_{j}|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}y,y〉}^{1/2}$
(1.4)

for any $x,y\in H$.

For various related results, see the papers [2131].

Motivated by the above results, we establish in this paper other similar 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 bonded linear operators on Hilbert spaces.

## 2 Some inequalities for an n-tuple of linear operators

Employing Kato’s inequality (1.2), we can state the following new result.

Theorem 2.1 Let $\left({T}_{1},\dots ,{T}_{n}\right)\in {\mathcal{B}}^{\left(n\right)}\left(H\right)$ be an n-tuple of bounded linear operators on the Hilbert space $\left(H;〈\cdot ,\cdot 〉\right)$ and $\left({p}_{1},\dots ,{p}_{n}\right)\in {\mathbb{R}}_{+}^{\ast n}$ be an n-tuple of nonnegative weights, not all of them equal to zero. Then we have

$\begin{array}{rcl}\sum _{j=1}^{n}{p}_{j}|〈{T}_{j}x,y〉|& \le & {〈\sum _{j=1}^{n}{p}_{j}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}\\ ×{〈\sum _{j=1}^{n}{p}_{j}\left(\frac{|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)y,y〉}^{1/2}\end{array}$
(2.1)

for any $x,y\in H$, $\alpha \in \left[0,1\right]$ and, in particular, for $\alpha =\frac{1}{2}$

$\sum _{j=1}^{n}{p}_{j}|〈{T}_{j}x,y〉|\le {〈\sum _{j=1}^{n}{p}_{j}|{T}_{j}|x,x〉}^{1/2}{〈\sum _{j=1}^{n}{p}_{j}|{T}_{j}^{\ast }|y,y〉}^{1/2}$
(2.2)

for any $x,y\in H$.

Proof Utilizing Kato’s inequality, we have

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

and by replacing α with $1-\alpha$,

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

which by summation gives

$\begin{array}{rcl}|〈{T}_{j}x,y〉|& \le & \frac{1}{2}\left[{〈|{T}_{j}{|}^{2\alpha }x,x〉}^{1/2}{〈|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}y,y〉}^{1/2}\\ +{〈|{T}_{j}{|}^{2\left(1-\alpha \right)}x,x〉}^{1/2}{〈|{T}_{j}^{\ast }{|}^{2\alpha }y,y〉}^{1/2}\right]\end{array}$
(2.3)

for any $j\in \left\{1,\dots ,n\right\}$ and $x,y\in H$. By the elementary inequality

$ab+cd\le {\left({a}^{2}+{c}^{2}\right)}^{1/2}{\left({b}^{2}+{d}^{2}\right)}^{1/2},\phantom{\rule{1em}{0ex}}a,b,c,d\ge 0,$
(2.4)

we have

which by (2.3) produces

$|〈{T}_{j}x,y〉|\le {〈\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}{〈\left(\frac{|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)y,y〉}^{1/2}$
(2.5)

for any $j\in \left\{1,\dots ,n\right\}$ and $x,y\in H$. Multiplying the inequalities (2.5) with the positive weights ${p}_{j}$, summing over j from 1 to n and utilizing the weighted Cauchy-Buniakowski-Schwarz inequality

$\sum _{j=1}^{n}{p}_{j}{a}_{j}{b}_{j}\le {\left(\sum _{j=1}^{n}{p}_{j}{a}_{j}^{2}\right)}^{1/2}{\left(\sum _{j=1}^{n}{p}_{j}{b}_{j}^{2}\right)}^{1/2},$

where $\left({a}_{1},\dots ,{a}_{n}\right),\left({b}_{1},\dots ,{b}_{n}\right)\in {\mathbb{R}}_{+}^{n}$, we have

(2.6)

for any $x,y\in H$, and the inequality in (2.1) is proved. □

Remark 2.1 In order to provide some applications for functions of normal operators defined by power series, we need to state the inequality (2.1) for normal operators ${N}_{j}$, $j\in \left\{1,\dots ,n\right\}$, namely,

$\begin{array}{rcl}\sum _{j=1}^{n}{p}_{j}|〈{N}_{j}x,y〉|& \le & {〈\sum _{j=1}^{n}{p}_{j}\left(\frac{|{N}_{j}{|}^{2\alpha }+|{N}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}\\ ×{〈\sum _{j=1}^{n}{p}_{j}\left(\frac{|{N}_{j}{|}^{2\alpha }+|{N}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)y,y〉}^{1/2}\end{array}$
(2.7)

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

From a different perspective that involves quadratics, we can state the following result as well.

Theorem 2.2 Let $\left({T}_{1},\dots ,{T}_{n}\right)\in {\mathcal{B}}^{\left(n\right)}\left(H\right)$ be an n-tuple of bounded linear operators on the Hilbert space $\left(H;〈\cdot ,\cdot 〉\right)$ and $\left({p}_{1},\dots ,{p}_{n}\right)\in {\mathbb{R}}_{+}^{\ast n}$ be an n-tuple of nonnegative weights, not all of them equal to zero. Then we have

(2.8)

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

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

Utilizing Kato’s inequality for the operator ${T}_{j}$, $j\in \left\{1,\dots ,n\right\}$, we have

$|〈{T}_{j}x,y〉{|}^{2}\le 〈|{T}_{j}{|}^{2\alpha }x,x〉〈|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}y,y〉$
(2.9)

and, by replacing α with $1-\alpha$,

$|〈{T}_{j}x,y〉{|}^{2}\le 〈|{T}_{j}{|}^{2\left(1-\alpha \right)}x,x〉〈|{T}_{j}^{\ast }{|}^{2\alpha }y,y〉$
(2.10)

for any $x,y\in H$.

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

$〈|{T}_{j}{|}^{2\alpha }x,x〉〈|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}y,y〉\le {〈|{T}_{j}{|}^{2}x,x〉}^{\alpha }{〈|{T}_{j}^{\ast }{|}^{2}y,y〉}^{1-\alpha }$
(2.11)

and

$〈|{T}_{j}{|}^{2\left(1-\alpha \right)}x,x〉〈|{T}_{j}^{\ast }{|}^{2\alpha }y,y〉\le {〈|{T}_{j}{|}^{2}x,x〉}^{1-\alpha }{〈|{T}_{j}^{\ast }{|}^{2}y,y〉}^{\alpha }$
(2.12)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$, $j\in \left\{1,\dots ,n\right\}$ and $\alpha \in \left(0,1\right)$.

If we add (2.9) with (2.10) and make use of (2.11) and (2.12), we deduce

$2|〈{T}_{j}x,y〉{|}^{2}\le {〈|{T}_{j}{|}^{2}x,x〉}^{\alpha }{〈|{T}_{j}^{\ast }{|}^{2}y,y〉}^{1-\alpha }+{〈|{T}_{j}^{\ast }{|}^{2}y,y〉}^{\alpha }{〈|{T}_{j}{|}^{2}x,x〉}^{1-\alpha }$
(2.13)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$, $j\in \left\{1,\dots ,n\right\}$ and $\alpha \in \left(0,1\right)$.

Now, if we multiply (2.13) with ${p}_{j}\ge 0$, sum over j from 1 to n, we get

$\begin{array}{rcl}2\sum _{j=1}^{n}{p}_{j}|〈{T}_{j}x,y〉{|}^{2}& \le & \sum _{j=1}^{n}{p}_{j}{〈|{T}_{j}{|}^{2}x,x〉}^{\alpha }{〈|{T}_{j}^{\ast }{|}^{2}y,y〉}^{1-\alpha }\\ +\sum _{j=1}^{n}{p}_{j}{〈|{T}_{j}^{\ast }{|}^{2}y,y〉}^{\alpha }{〈|{T}_{j}{|}^{2}x,x〉}^{1-\alpha }\end{array}$
(2.14)

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

Since $〈|{T}_{j}{|}^{2}x,x〉={\parallel {T}_{j}x\parallel }^{2}$ and $〈|{T}_{j}^{\ast }{|}^{2}y,y〉={\parallel {T}_{j}^{\ast }y\parallel }^{2}$, $j\in \left\{1,\dots ,n\right\}$, then we get from (2.14) the first inequality in (2.8).

Now, on making use of the weighted Hölder discrete inequality

$\sum _{j=1}^{n}{p}_{j}{a}_{j}{b}_{j}\le {\left(\sum _{j=1}^{n}{p}_{j}{a}_{j}^{p}\right)}^{1/p}{\left(\sum _{j=1}^{n}{p}_{j}{b}_{j}^{q}\right)}^{1/q},\phantom{\rule{1em}{0ex}}p,q>1,\frac{1}{p}+\frac{1}{q}=1,$

where $\left({a}_{1},\dots ,{a}_{n}\right),\left({b}_{1},\dots ,{b}_{n}\right)\in {\mathbb{R}}_{+}^{n}$, we also have

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

and

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

Summing these two inequalities, we deduce the second inequality in (2.8).

Finally, on utilizing the Hölder inequality

$ab+cd\le {\left({a}^{p}+{c}^{p}\right)}^{1/p}{\left({b}^{q}+{d}^{q}\right)}^{1/q},\phantom{\rule{1em}{0ex}}a,b,c,d\ge 0,$

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

and the proof is concluded. □

Remark 2.2 For $\alpha =\frac{1}{2}$, we get from (2.8) that

(2.15)

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

## 3 Inequalities for functions of normal operators

Now, by the help of power series $f\left(z\right)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n}{z}^{n}$, we can naturally construct another power series which will have as coefficients the absolute values of the coefficient of the original series, namely, ${f}_{A}\left(z\right):={\sum }_{n=0}^{\mathrm{\infty }}|{a}_{n}|{z}^{n}$. 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

$\begin{array}{r}f\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{n}{z}^{n}=ln\frac{1}{1+z},\phantom{\rule{1em}{0ex}}z\in D\left(0,1\right);\\ g\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{\left(2n\right)!}{z}^{2n}=cosz,\phantom{\rule{1em}{0ex}}z\in \mathbb{C};\\ h\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{\left(2n+1\right)!}{z}^{2n+1}=sinz,\phantom{\rule{1em}{0ex}}z\in \mathbb{C};\\ l\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{z}^{n}=\frac{1}{1+z},\phantom{\rule{1em}{0ex}}z\in D\left(0,1\right),\end{array}$
(3.1)

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

$\begin{array}{r}{f}_{A}\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{z}^{n}=ln\frac{1}{1-z},\phantom{\rule{1em}{0ex}}z\in D\left(0,1\right);\\ {g}_{A}\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\left(2n\right)!}{z}^{2n}=coshz,\phantom{\rule{1em}{0ex}}z\in \mathbb{C};\\ {h}_{A}\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\left(2n+1\right)!}{z}^{2n+1}=sinhz,\phantom{\rule{1em}{0ex}}z\in \mathbb{C};\\ {l}_{A}\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{z}^{n}=\frac{1}{1-z},\phantom{\rule{1em}{0ex}}z\in D\left(0,1\right).\end{array}$
(3.2)

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

Theorem 3.1 Let $f\left(z\right)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be a function defined by power series with complex coefficients and convergent on the open disk $D\left(0,R\right)\subset \mathbb{C}$, $R>0$. If N is a normal operator on the Hilbert space H, for $\alpha \in \left(0,1\right)$, we have that ${\parallel N\parallel }^{2\alpha },{\parallel N\parallel }^{2\left(1-\alpha \right)}, then we have the inequality

$\begin{array}{rcl}|〈f\left(N\right)x,y〉|& \le & \frac{1}{2}{〈\left[{f}_{A}\left(|N{|}^{2\alpha }\right)+{f}_{A}\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]x,x〉}^{1/2}\\ ×{〈\left[{f}_{A}\left(|N{|}^{2\alpha }\right)+{f}_{A}\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]y,y〉}^{1/2}\end{array}$
(3.3)

for any $x,y\in H$. In particular, if $\parallel N\parallel , then

$|〈f\left(N\right)x,y〉|\le {〈{f}_{A}\left(|N|\right)x,x〉}^{1/2}{〈{f}_{A}\left(|N|\right)y,y〉}^{1/2}$
(3.4)

for any $x,y\in H$.

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

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

Now, utilizing the inequality (2.9), we can write

$\begin{array}{rcl}|〈\sum _{j=0}^{n}{a}_{j}{N}^{j}x,y〉|& \le & \sum _{j=0}^{n}|{a}_{j}||〈{N}^{j}x,y〉|\\ \le & {〈\sum _{j=0}^{n}|{a}_{j}|\left(\frac{|N{|}^{2j\alpha }+|N{|}^{2j\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}\\ ×{〈\sum _{j=0}^{n}|{a}_{j}|\left(\frac{|N{|}^{2j\alpha }+|N{|}^{2j\left(1-\alpha \right)}}{2}\right)y,y〉}^{1/2}\end{array}$
(3.5)

for any $x,y\in H$ and $n\in \mathbb{N}$. Since ${\parallel N\parallel }^{2\alpha },{\parallel N\parallel }^{2\left(1-\alpha \right)}, then it follows that the series ${\sum }_{j=0}^{\mathrm{\infty }}|{a}_{j}|{\left(|N{|}^{2\alpha }\right)}^{j}$ and ${\sum }_{j=0}^{\mathrm{\infty }}|{a}_{j}|{\left(|N{|}^{2\left(1-\alpha \right)}\right)}^{j}$ are absolute convergent in $\mathcal{B}\left(H\right)$, and by taking the limit over $n\to \mathrm{\infty }$ in (3.5), we deduce the desired result (3.3). □

Remark 3.1 With the assumptions in Theorem 3.1, if we take the supremum over $y\in H$, $\parallel y\parallel =1$, then we get the vector inequality

$\begin{array}{rcl}\parallel f\left(N\right)x\parallel & \le & \frac{1}{2}{〈\left[{f}_{A}\left(|N{|}^{2\alpha }\right)+{f}_{A}\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]x,x〉}^{1/2}\\ ×{\parallel {f}_{A}\left(|N{|}^{2\alpha }\right)+{f}_{A}\left(|N{|}^{2\left(1-\alpha \right)}\right)\parallel }^{1/2}\end{array}$
(3.6)

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

$\parallel f\left(N\right)\parallel \le \frac{1}{2}\parallel {f}_{A}\left(|N{|}^{2\alpha }\right)+{f}_{A}\left(|N{|}^{2\left(1-\alpha \right)}\right)\parallel$
(3.7)

for any $\alpha \in \left[0,1\right]$. Making use of the examples in (3.1) and (3.2), we can state the vector inequalities

(3.8)

and

(3.9)

for any $x,y\in H$ and $\parallel N\parallel <1$. We also have the inequalities

$\begin{array}{rcl}|〈sin\left(N\right)x,y〉|& \le & \frac{1}{2}{〈\left[sinh\left(|N{|}^{2\alpha }\right)+sinh\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]x,x〉}^{1/2}\\ ×{〈\left[sinh\left(|N{|}^{2\alpha }\right)+sinh\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]y,y〉}^{1/2}\end{array}$
(3.10)

and

$\begin{array}{rcl}|〈cos\left(N\right)x,y〉|& \le & \frac{1}{2}{〈\left[cosh\left(|N{|}^{2\alpha }\right)+cosh\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]x,x〉}^{1/2}\\ ×{〈\left[cosh\left(|N{|}^{2\alpha }\right)+cosh\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]y,y〉}^{1/2}\end{array}$
(3.11)

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

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

(3.12)

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

$\begin{array}{rcl}|〈exp\left(N\right)x,y〉|& \le & \frac{1}{2}{〈\left[exp\left(|N{|}^{2\alpha }\right)+exp\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]x,x〉}^{1/2}\\ ×{〈\left[exp\left(|N{|}^{2\alpha }\right)+exp\left(|N{|}^{2\left(1-\alpha \right)}\right)\right]y,y〉}^{1/2}\end{array}$
(3.13)

for any $x,y\in H$ and N a normal operator. If $\parallel N\parallel <1$, then we also have the inequalities

(3.14)
(3.15)

and

(3.16)

for any $x,y\in H$. From a different perspective, we also have

Theorem 3.2 With the assumption of Theorem  3.1 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}, then we have the inequalities

$\begin{array}{rcl}|〈f\left(zN\right)x,y〉{|}^{2}& \le & \frac{1}{2}{f}_{A}\left(|z{|}^{2}\right)\left[{〈{f}_{A}\left(|N{|}^{2}\right)x,x〉}^{\alpha }{〈{f}_{A}\left(|N{|}^{2}\right)y,y〉}^{1-\alpha }\\ +{〈{f}_{A}\left(|N{|}^{2}\right)x,x〉}^{1-\alpha }{〈{f}_{A}\left(|N{|}^{2}\right)y,y〉}^{\alpha }\right]\\ \le & \frac{1}{2}{f}_{A}\left(|z{|}^{2}\right)\left(〈{f}_{A}\left(|N{|}^{2}\right)x,x〉+〈{f}_{A}\left(|N{|}^{2}\right)y,y〉\right)\end{array}$
(3.17)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in \left[0,1\right]$. In particular, for $\alpha =\frac{1}{2}$, we have

$\begin{array}{rl}|〈f\left(zN\right)x,y〉{|}^{2}& \le {f}_{A}\left(|z{|}^{2}\right){〈{f}_{A}\left(|N{|}^{2}\right)x,x〉}^{1/2}{〈{f}_{A}\left(|N{|}^{2}\right)y,y〉}^{1/2}\\ \le \frac{1}{2}{f}_{A}\left(|z{|}^{2}\right)\left(〈{f}_{A}\left(|N{|}^{2}\right)x,x〉+〈{f}_{A}\left(|N{|}^{2}\right)y,y〉\right)\end{array}$
(3.18)

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

Proof If we use the second and third inequality from (2.8) for powers of operators, we have

(3.19)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in \left[0,1\right]$. Since N is a normal operator on the Hilbert space H, then

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

and

${\parallel {\left({N}^{\ast }\right)}^{j}y\parallel }^{2}=〈|{\left({N}^{\ast }\right)}^{j}{|}^{2}y,y〉=〈|{N}^{\ast }{|}^{2j}y,y〉=〈|N{|}^{2j}y,y〉$

for any $j\in \left\{0,\dots ,n\right\}$ and for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$. Then from (3.19), we have

(3.20)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in \left[0,1\right]$. By the weighted Cauchy-Buniakowski-Schwarz inequality, we also have

$|〈\sum _{j=0}^{n}{a}_{j}{z}^{j}{N}^{j}x,y〉{|}^{2}\le \sum _{j=0}^{n}|{a}_{j}||z{|}^{2j}\sum _{j=0}^{n}|{a}_{j}||〈{N}^{j}x,y〉{|}^{2}$
(3.21)

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

Now, since the series ${\sum }_{j=0}^{\mathrm{\infty }}{a}_{j}{z}^{j}{N}^{j}$, ${\sum }_{j=0}^{\mathrm{\infty }}|{a}_{j}||z{|}^{2j}$, ${\sum }_{j=0}^{\mathrm{\infty }}|{a}_{j}||N{|}^{2j}$ are convergent, then by (3.20) and (3.21), on letting $n\to \mathrm{\infty }$, we deduce the desired result (3.17). □

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

## 4 Applications for the Euclidean norm

In [29], the author has introduced the following norm on the Cartesian product ${\mathcal{B}}^{\left(n\right)}\left(H\right):=\mathcal{B}\left(H\right)×\cdots ×\mathcal{B}\left(H\right)$, where $\mathcal{B}\left(H\right)$ denotes the Banach algebra of all bounded linear operators defined on the complex Hilbert space H:

${\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}:=\underset{\left({\lambda }_{1},\dots ,{\lambda }_{n}\right)\in {\mathbb{B}}_{n}}{sup}\parallel {\lambda }_{1}{T}_{1}+\cdots +{\lambda }_{n}{T}_{n}\parallel ,$
(4.1)

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

It is clear that ${\parallel \cdot \parallel }_{e}$ is a norm on ${B}^{\left(n\right)}\left(H\right)$ and, for any $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$, we have

${\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}={\parallel \left({T}_{1}^{\ast },\dots ,{T}_{n}^{\ast }\right)\parallel }_{e},$

where ${T}_{j}^{\ast }$ is the adjoint operator of ${T}_{j}$, $j\in \left\{1,\dots ,n\right\}$. We call this the Euclidean norm of an n-tuple of operators $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$.

It has been shown in [29] that the following basic inequality for theEuclidean norm holds true:

$\frac{1}{\sqrt{n}}{\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel }^{\frac{1}{2}}\le {\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}\le {\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel }^{\frac{1}{2}}$
(4.2)

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

In the same paper [29], the author has introduced the Euclidean operator radius of an n-tuple of operators $\left({T}_{1},\dots ,{T}_{n}\right)$ by

${w}_{e}\left({T}_{1},\dots ,{T}_{n}\right):=\underset{\parallel x\parallel =1}{sup}{\left(\sum _{j=1}^{n}|〈{T}_{j}x,x〉{|}^{2}\right)}^{\frac{1}{2}}$
(4.3)

and proved that ${w}_{e}\left(\cdot \right)$ is a norm on ${B}^{\left(n\right)}\left(H\right)$ and satisfies the double inequality

$\frac{1}{2}{\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}\le {w}_{e}\left({T}_{1},\dots ,{T}_{n}\right)\le {\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}$
(4.4)

for each n-tuple $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$.

As pointed out in [29], the Euclidean numerical radius also satisfies the double inequality

$\frac{1}{2\sqrt{n}}{\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel }^{\frac{1}{2}}\le {w}_{e}\left({T}_{1},\dots ,{T}_{n}\right)\le {\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel }^{\frac{1}{2}}$
(4.5)

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

In [30], by utilizing the concept of hypo-Euclidean norm on ${H}^{n}$, we obtained the following representation for the Euclidean norm.

Proposition 4.1 For any $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$, we have

${\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}=\underset{\parallel y\parallel =1,\parallel x\parallel =1}{sup}{\left(\sum _{j=1}^{n}|〈{T}_{j}y,x〉{|}^{2}\right)}^{\frac{1}{2}}.$
(4.6)

We can state now the following result.

Theorem 4.1 For any $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$, we have

$\begin{array}{rcl}{\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}^{2}& \le & \frac{1}{2}\left[{\left(\parallel \sum _{j=1}^{n}|{T}_{j}{|}^{2}\parallel \right)}^{\alpha }{\left(\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel \right)}^{1-\alpha }\\ +{\left(\parallel \sum _{j=1}^{n}|{T}_{j}{|}^{2}\parallel \right)}^{1-\alpha }{\left(\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel \right)}^{\alpha }\right]\\ \le & \frac{1}{2}\left[\parallel \sum _{j=1}^{n}|{T}_{j}{|}^{2}\parallel +\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}\parallel \right]\end{array}$
(4.7)

and

(4.8)

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

Proof We have from the second inequality in (2.8)

$\begin{array}{rcl}\sum _{j=1}^{n}|〈{T}_{j}x,y〉{|}^{2}& \le & \frac{1}{2}\left[{\left(〈\sum _{j=1}^{n}|{T}_{j}{|}^{2}x,x〉\right)}^{\alpha }{\left(〈\sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}y,y〉\right)}^{1-\alpha }\\ +{\left(〈\sum _{j=1}^{n}|{T}_{j}{|}^{2}x,x〉\right)}^{1-\alpha }{\left(〈\sum _{j=1}^{n}|{T}_{j}^{\ast }{|}^{2}y,y〉\right)}^{\alpha }\right]\end{array}$
(4.9)

for any $x,y\in H$ with $\parallel x\parallel =\parallel y\parallel =1$ and $\alpha \in \left[0,1\right]$. Taking the supremum over $\parallel x\parallel =\parallel y\parallel =1$, we have

which proves the first part of (4.7). The second part follows by the elementary inequality

${a}^{\alpha }{b}^{1-\alpha }\le \alpha a+\left(1-\alpha \right)b$

for $a,b\ge 0$ and $\alpha \in \left[0,1\right]$. The inequality (4.8) follows from (4.9) by taking $y=x$ and then the supremum over $\parallel x\parallel =1$. □

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

Following [20], we consider the s-p-norm of the n-tuple of operators $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$ by

${\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{s,p}:=\underset{\parallel y\parallel =1,\parallel x\parallel =1}{sup}\left[{\left(\sum _{j=1}^{n}|〈{T}_{j}y,x〉{|}^{p}\right)}^{\frac{1}{p}}\right].$
(5.1)

For $p=2$, we get

${\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{s,2}={\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{e}.$

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

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

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

$\parallel \sum _{j=1}^{n}{T}_{j}\parallel \le {\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{s,1}\le \sum _{j=1}^{n}\parallel {T}_{j}\parallel .$
(5.2)

Similarly, we can also consider the s-p-numerical radius of the n-tuple of operators $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$ by [20]

${w}_{s,p}\left({T}_{1},\dots ,{T}_{n}\right):=\underset{\parallel x\parallel =1}{sup}\left[{\left(\sum _{j=1}^{n}|〈{T}_{j}x,x〉{|}^{p}\right)}^{\frac{1}{p}}\right],$
(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 ${B}^{\left(n\right)}\left(H\right)$ for $p\ge 1$, and for $p=1$ it satisfies the basic inequality

$w\left(\sum _{j=1}^{n}{T}_{j}\right)\le {w}_{s,1}\left({T}_{1},\dots ,{T}_{n}\right)\le \sum _{j=1}^{n}w\left({T}_{j}\right).$
(5.4)

We can state the following result.

Theorem 5.1 For any $\left({T}_{1},\dots ,{T}_{n}\right)\in {B}^{\left(n\right)}\left(H\right)$, we have

(5.5)

and

${w}_{s,1}\left({T}_{1},\dots ,{T}_{n}\right)\le \parallel \sum _{j=1}^{n}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}+|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{4}\right)\parallel .$
(5.6)

Proof From (2.1) we have

$\begin{array}{rcl}\sum _{j=1}^{n}|〈{T}_{j}x,y〉|& \le & {〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}\\ ×{〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)y,y〉}^{1/2}\end{array}$
(5.7)

for any $x,y\in H$.

Taking the supremum over $\parallel y\parallel =1$, $\parallel x\parallel =1$ in (5.7), we have

$\begin{array}{rcl}{\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{s,1}& \le & {\left[\underset{\parallel x\parallel =1}{sup}〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉\right]}^{1/2}\\ ×{\left[\underset{\parallel y\parallel =1}{sup}〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)y,y〉\right]}^{1/2}\\ =& {\parallel \sum _{j=1}^{n}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)\parallel }^{1/2}\\ ×{\parallel \sum _{j=1}^{n}\left(\frac{|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)\parallel }^{1/2}\end{array}$

and the first inequality in (5.5) is proved. The second part follows by the arithmetic mean-geometric mean inequality.

Now, if we take $y=x$ in (5.7), then we get

$\begin{array}{rcl}\sum _{j=1}^{n}|〈{T}_{j}x,x〉|& \le & {〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}\\ ×{〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉}^{1/2}\\ \le & \frac{1}{2}〈\sum _{j=1}^{n}\left(\frac{|{T}_{j}{|}^{2\alpha }+|{T}_{j}{|}^{2\left(1-\alpha \right)}+|{T}_{j}^{\ast }{|}^{2\alpha }+|{T}_{j}^{\ast }{|}^{2\left(1-\alpha \right)}}{2}\right)x,x〉.\end{array}$

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

Remark 5.1 If we take $\alpha =\frac{1}{2}$ in the first inequality in (5.5), then we deduce

${\parallel \left({T}_{1},\dots ,{T}_{n}\right)\parallel }_{s,1}\le {\parallel \sum _{j=1}^{n}|{T}_{j}|\parallel }^{1/2}{\parallel \sum _{j=1}^{n}|{T}_{j}^{\ast }|\parallel }^{1/2},$
(5.8)

and then we get the following refinement of the generalized triangle inequality:

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

From (5.6) we also have, for $\alpha =\frac{1}{2}$,

${w}_{s,1}\left({T}_{1},\dots ,{T}_{n}\right)\le \parallel \sum _{j=1}^{n}\left(\frac{|{T}_{j}|+|{T}_{j}^{\ast }|}{2}\right)\parallel .$
(5.9)

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## Acknowledgements

The authors wish to thank the anonymous referees for their valuable comments. Also, this research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2012-0008474).

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Correspondence to Yeol Je Cho or Young-Ho Kim.

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The authors declare that they have no competing interests.

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Dragomir, S.S., Cho, Y.J. & Kim, YH. Applications of Kato’s inequality for n-tuples of operators in Hilbert spaces, (I). J Inequal Appl 2013, 21 (2013). https://doi.org/10.1186/1029-242X-2013-21