A note on norm inequalities for positive operators

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.

1 Introduction

Let $M_n$ be the space of $n\times n$ complex matrices. Let $\parallel \cdot \parallel$ denote any unitarily invariant norm on $M_n$. We shall always denote the singular values of A by $s_1(A)\ge \cdots \ge s_n(A)\ge 0$, that is, the eigenvalues of the positive semidefinite matrix $|A|={(A{A}^{\ast })}^{1/2}$, 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 [[1], Theorem 2.2] proved that for any positive integer m,

$$\parallel A^m+B^m\parallel \le \parallel {(A+B)}^m\parallel .$$

(1.1)

Let $A,B\in M_n$. Kittaneh [[2], Theorem 2.2] proved that

$${\parallel A+B\parallel }_F^2\le \frac{1}{2}({\parallel |A|+|B|\parallel }_F^2+{\parallel \left|A^{\ast }\right|+\left|B^{\ast }\right|\parallel }_F^2),$$

(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 [3].

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,

$$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {(|A{|}^m+|B{|}^m)}^{p/2}\parallel }^{1/p}{\parallel {\left(\right|A^{\ast }{|}^m+\left|B^{\ast }{|}^m\right)}^{q/2}\parallel }^{1/q}.$$

(2.1)

Proof Let $A,B\in M_n$ with polar decompositions $A=U|A|$ and $B=V|B|$. It is known [[4], 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 [[4], p.13], we have

$$U|A{|}^m+V|B{|}^m={(|A{|}^m+|B{|}^m)}^{1/2}K{\left(\right|A^{\ast }{|}^m+\left|B^{\ast }{|}^m\right)}^{1/2}$$

for some contraction K. For $k=1,\dots ,n$, by using Horn’s inequality [[5], p.72], we know that

$$\prod _{j=1}^k{s}_j(U|A{|}^m+V|B{|}^m)\le \prod _{j=1}^k{s}_j^{1/2}(|A{|}^m+|B{|}^m)s_j^{1/2}\left(\right|A^{\ast }{|}^m+\left|B^{\ast }{|}^m\right).$$

(2.2)

Let

$$X=diag(s_1^{1/2}(|A{|}^m+|B{|}^m),\dots ,s_n^{1/2}(|A{|}^m+|B{|}^m))$$

and

$$Y=diag(s_1^{1/2}\left(\right|A^{\ast }{|}^m+\left|B^{\ast }{|}^m\right),\dots ,s_n^{1/2}\left(\right|A^{\ast }{|}^m+\left|B^{\ast }{|}^m\right)).$$

Then inequality (2.2) is equivalent to

$$\prod _{j=1}^k{s}_j(U|A{|}^m+V|B{|}^m)\le \prod _{j=1}^k{s}_j(XY).$$

Since weak log-majorization implies weak majorization, we get

$$\sum _{j=1}^k{s}_j(U|A{|}^m+V|B{|}^m)\le \sum _{j=1}^k{s}_j(XY).$$

(2.3)

Thanks to the Fan dominance principle [[5], 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 [[5], p.95] (see also [[6], 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

$$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {(|A{|}^m+|B{|}^m)}^{p/2}\parallel }^{1/p}{\parallel {\left(\right|A^{\ast }{|}^m+\left|B^{\ast }{|}^m\right)}^{q/2}\parallel }^{1/q}.$$

This completes the proof. □

Remark 2.2 Inequality (2.1) is a norm version of the following scalar triangle inequality:

$$|z_1|z_1{|}^{m-1}+z_2|z_2{|}^{m-1}|\le \left|{(|z_1{|}^m+|z_2{|}^m)}^{p/2}{|}^{1/p}\right|{\left(\right|z_1^{\ast }{|}^m+\left|z_2^{\ast }{|}^m\right)}^{q/2}{|}^{1/q}.$$

Corollary 2.3 Let $A,B\in M_n$. Then for any positive integer m,

$$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {(|A|+|B|)}^m\parallel }^{1/2}{\parallel {\left(\right|A^{\ast }|+|B^{\ast }\left|\right)}^m\parallel }^{1/2}.$$

(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

$$\parallel A|A{|}^{m-1}+B|B{|}^{m-1}\parallel \le {\parallel {(|A|+|B|)}^m\parallel }^{1/2}{\parallel {\left(\right|A^{\ast }|+|B^{\ast }\left|\right)}^m\parallel }^{1/2}.$$

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

$$\parallel A+B\parallel \le {\parallel {(|A|+|B|)}^{p/2}\parallel }^{1/p}{\parallel {\left(\right|A^{\ast }|+|B^{\ast }\left|\right)}^{q/2}\parallel }^{1/q}.$$

In particular,

$$\parallel A+B\parallel \le {\parallel |A|+|B|\parallel }^{1/2}{\parallel \left|A^{\ast }\right|+\left|B^{\ast }\right|\parallel }^{1/2},$$

(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

$$\parallel \left|A^{\ast }\right|+\left|B^{\ast }\right|\parallel \le \parallel \left|A^{\ast }\right|\parallel +\parallel \left|B^{\ast }\right|\parallel \le 2\parallel |A|+|B|\parallel ,$$

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

Remark 2.6 Recently, Zou [[7], Theorem 2.2] and Zou and He [[8], Theorem 2.2] gave some generalizations of inequality (1.1) for normal matrices. Our result is different from the ones obtained by these authors.