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# A note on norm inequalities for positive operators

Journal of Inequalities and Applications20142014:171

https://doi.org/10.1186/1029-242X-2014-171

• Accepted: 23 April 2014
• Published:

## Abstract

In this short note, we present a generalization of a norm inequality due to Bhatia and Kittaneh (Lett. Math. Phys. 43:225-231, 1998), which is also a refinement and a generalization of a result obtained by Kittaneh (Commun. Math. Phys. 104:307-310, 1986).

MSC:15A42, 47A63.

## Keywords

• unitarily invariant norms
• singular values
• weak majorization

## 1 Introduction

Let ${M}_{n}$ be the space of $n×n$ complex matrices. Let $\parallel \cdot \parallel$ denote any unitarily invariant norm on ${M}_{n}$. We shall always denote the singular values of A by ${s}_{1}\left(A\right)\ge \cdots \ge {s}_{n}\left(A\right)\ge 0$, that is, the eigenvalues of the positive semidefinite matrix $|A|={\left(A{A}^{\ast }\right)}^{1/2}$, arranged in decreasing order and repeated according to multiplicity. Let $A,B\in {M}_{n}$ be Hermitian; the order relation $A\ge B$ means, as usual, that $A-B$ is positive semidefinite.

Let $A,B\in {M}_{n}$ be positive semidefinite. Bhatia and Kittaneh [, Theorem 2.2] proved that for any positive integer m,
$\parallel {A}^{m}+{B}^{m}\parallel \le \parallel {\left(A+B\right)}^{m}\parallel .$
(1.1)
Let $A,B\in {M}_{n}$. Kittaneh [, Theorem 2.2] proved that
${\parallel A+B\parallel }_{F}^{2}\le \frac{1}{2}\left({\parallel |A|+|B|\parallel }_{F}^{2}+{\parallel |{A}^{\ast }|+|{B}^{\ast }|\parallel }_{F}^{2}\right),$
(1.2)

where ${\parallel X\parallel }_{F}$ is the Frobenius norm of X. For more information on the Schatten p-norm and its applications, the reader is referred to .

In this note, we present a generalization of inequality (1.1), which is also a refinement and a generalization of (1.2).

## 2 Main results

Now, we show the generalization of inequality (1.1).

Theorem 2.1 Let $A,B\in {M}_{n}$ and suppose that p, q be real numbers with $p>1$ and $\frac{1}{p}+\frac{1}{q}=1$. Then for any positive integer m,
$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {\left(|A{|}^{m}+|B{|}^{m}\right)}^{p/2}\parallel }^{1/p}{\parallel {\left(|{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\right)}^{q/2}\parallel }^{1/q}.$
(2.1)
Proof Let $A,B\in {M}_{n}$ with polar decompositions $A=U|A|$ and $B=V|B|$. It is known [, p.15] that
$\left[\begin{array}{cc}|A|& {A}^{\ast }\\ A& |{A}^{\ast }|\end{array}\right]=\left[\begin{array}{cc}I& 0\\ 0& U\end{array}\right]\left[\begin{array}{cc}|A|& |A|\\ |A|& |A|\end{array}\right]\left[\begin{array}{cc}I& 0\\ 0& {U}^{\ast }\end{array}\right]\ge 0.$
It follows that
${\left[\begin{array}{cc}|A|& {A}^{\ast }\\ A& |{A}^{\ast }|\end{array}\right]}^{m}+{\left[\begin{array}{cc}|B|& {B}^{\ast }\\ B& |{B}^{\ast }|\end{array}\right]}^{m}={2}^{m-1}\left[\begin{array}{cc}|A{|}^{m}+|B{|}^{m}& |A{|}^{m}{U}^{\ast }+|B{|}^{m}{V}^{\ast }\\ U|A{|}^{m}+V|B{|}^{m}& |{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\end{array}\right]\ge 0.$
So, by Proposition 1.3.2 of [, p.13], we have
$U|A{|}^{m}+V|B{|}^{m}={\left(|A{|}^{m}+|B{|}^{m}\right)}^{1/2}K{\left(|{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\right)}^{1/2}$
for some contraction K. For $k=1,\dots ,n$, by using Horn’s inequality [, p.72], we know that
$\prod _{j=1}^{k}{s}_{j}\left(U|A{|}^{m}+V|B{|}^{m}\right)\le \prod _{j=1}^{k}{s}_{j}^{1/2}\left(|A{|}^{m}+|B{|}^{m}\right){s}_{j}^{1/2}\left(|{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\right).$
(2.2)
Let
$X=diag\left({s}_{1}^{1/2}\left(|A{|}^{m}+|B{|}^{m}\right),\dots ,{s}_{n}^{1/2}\left(|A{|}^{m}+|B{|}^{m}\right)\right)$
and
$Y=diag\left({s}_{1}^{1/2}\left(|{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\right),\dots ,{s}_{n}^{1/2}\left(|{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\right)\right).$
Then inequality (2.2) is equivalent to
$\prod _{j=1}^{k}{s}_{j}\left(U|A{|}^{m}+V|B{|}^{m}\right)\le \prod _{j=1}^{k}{s}_{j}\left(XY\right).$
Since weak log-majorization implies weak majorization, we get
$\sum _{j=1}^{k}{s}_{j}\left(U|A{|}^{m}+V|B{|}^{m}\right)\le \sum _{j=1}^{k}{s}_{j}\left(XY\right).$
(2.3)
Thanks to the Fan dominance principle [, p.93], we know that inequality (2.3) is equivalent to
$\parallel U|A{|}^{m}+V|B{|}^{m}\parallel \le \parallel XY\parallel .$
(2.4)
By Hölder’s inequality for unitarily invariant norms [, p.95] (see also [, Theorem 2.4]), we obtain
$\parallel XY\parallel \le {\parallel {X}^{p}\parallel }^{1/p}{\parallel {Y}^{q}\parallel }^{1/q}.$
(2.5)
It follows from (2.4) and (2.5) that
$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {\left(|A{|}^{m}+|B{|}^{m}\right)}^{p/2}\parallel }^{1/p}{\parallel {\left(|{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\right)}^{q/2}\parallel }^{1/q}.$

This completes the proof. □

Remark 2.2 Inequality (2.1) is a norm version of the following scalar triangle inequality:
$|{z}_{1}|{z}_{1}{|}^{m-1}+{z}_{2}|{z}_{2}{|}^{m-1}|\le |{\left(|{z}_{1}{|}^{m}+|{z}_{2}{|}^{m}\right)}^{p/2}{|}^{1/p}|{\left(|{z}_{1}^{\ast }{|}^{m}+|{z}_{2}^{\ast }{|}^{m}\right)}^{q/2}{|}^{1/q}.$
Corollary 2.3 Let $A,B\in {M}_{n}$. Then for any positive integer m,
$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {\left(|A|+|B|\right)}^{m}\parallel }^{1/2}{\parallel {\left(|{A}^{\ast }|+|{B}^{\ast }|\right)}^{m}\parallel }^{1/2}.$
(2.6)
Proof Putting $p=q=2$ in inequality (2.1), we have
$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel |A{|}^{m}+|B{|}^{m}\parallel }^{1/2}{\parallel |{A}^{\ast }{|}^{m}+|{B}^{\ast }{|}^{m}\parallel }^{1/2}.$
It follows from inequality (1.1) and this last inequality that
$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {\left(|A|+|B|\right)}^{m}\parallel }^{1/2}{\parallel {\left(|{A}^{\ast }|+|{B}^{\ast }|\right)}^{m}\parallel }^{1/2}.$

This completes the proof. □

Remark 2.4 If A and B are positive semidefinite, then by inequality (2.6), we get (1.1).

Remark 2.5 For $m=1$, by inequality (2.1), we get
$\parallel A+B\parallel \le {\parallel {\left(|A|+|B|\right)}^{p/2}\parallel }^{1/p}{\parallel {\left(|{A}^{\ast }|+|{B}^{\ast }|\right)}^{q/2}\parallel }^{1/q}.$
In particular,
$\parallel A+B\parallel \le {\parallel |A|+|B|\parallel }^{1/2}{\parallel |{A}^{\ast }|+|{B}^{\ast }|\parallel }^{1/2},$
(2.7)
which is a generalization and a refinement of inequality (1.2). For the usual operator norm, it is known that
${\parallel A+B\parallel }_{\mathrm{\infty }}\le \sqrt{2}{\parallel |A|+|B|\parallel }_{\mathrm{\infty }},$
(2.8)
which is sharp. Since
$\parallel |{A}^{\ast }|+|{B}^{\ast }|\parallel \le \parallel |{A}^{\ast }|\parallel +\parallel |{B}^{\ast }|\parallel \le 2\parallel |A|+|B|\parallel ,$

we find that inequality (2.7) is a strengthening of inequality (2.8).

Remark 2.6 Recently, Zou [, Theorem 2.2] and Zou and He [, Theorem 2.2] gave some generalizations of inequality (1.1) for normal matrices. Our result is different from the ones obtained by these authors.

## Declarations

### Acknowledgements

The authors wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript. This research was supported by the program of Guizhou Science and technology department and Guizhou Minzu University (No. Qian Science co-J word LKM  09) and the National Natural Science Foundation of China (No. 61263034).

## Authors’ Affiliations

(1)
College of Science, Guizhou Minzu University, Guiyang, Guizhou, 550025, P.R. China
(2)
School of Mathematics and Statistics, Chongqing Three Gorges University, Chongqing, 404100, P.R. China

## References 