A family of refinements of Heinz inequalities of matrices

Abstract

For any unitarily invariant norm $⦀\cdot ⦀$, the Heinz inequalities for operators assert that $2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀\le ⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀\le ⦀AX+XB⦀$, for A, B, and X any operators on a complex separable Hilbert space such that A, B are positive and $\nu \in [0,1]$. In this paper, we obtain a family of refinements of these norm inequalities by using the convexity of the function $f(\nu )=⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀$ and the Hermite-Hadamard inequality.

1 Introduction

Let $M_n(\mathbf{C})$ be the algebra of $n\times n$ complex matrices. We denote by $H_n(\mathbf{C})$ the set of all Hermitian matrices in $M_n(\mathbf{C})$. The set of all positive semi-definite matrices in $M_n(\mathbf{C})$ shall be denoted by $H_n^{+}(\mathbf{C})$. A norm $⦀\cdot ⦀$ on $M_n(\mathbf{C})$ is called unitarily invariant or symmetric if

$$⦀UAV⦀=⦀A⦀$$

for all $A\in M_n(\mathbf{C})$ and for all unitaries $U,V\in M_n(\mathbf{C})$.

The arithmetic-geometric mean inequality for two nonnegative real numbers a and b is

$$\sqrt{ab}\le \frac{a+b}{2},$$

which has been generalized to the context of matrices as follows:

$$2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀\le ⦀AX+XB⦀,$$

where $A,B\in H_n^{+}(\mathbf{C})$, $X\in M_n$, and $⦀\cdot ⦀$ is a unitarily invariant norm on $M_n(\mathbf{C})$.

For $\nu \in [0,1]$ and two nonnegative numbers a and b, the Heinz mean is defined as

$$H_{\nu }(a,b)=\frac{a^{\nu }{b}^{1-\nu }+a^{1-\nu }{b}^{\nu }}{2}.$$

Clearly the Heinz mean interpolates between the geometric mean and the arithmetic mean:

$$\sqrt{ab}\le H_{\nu }(a,b)\le \frac{a+b}{2}.$$

The function $H_{\nu }(a,b)$ has the following properties: it is convex, attains its minimum at $\nu =\frac{1}{2}$, its maximum at $\nu =0$ and $\nu =1$, and $H_{\nu }(a,b)=H_{1-\nu }(a,b)$ for $0\le \nu \le 1$. The generalization of the above inequalities to matrices is due to Bhatia and Davis [1] as follows:

$$2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀\le ⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀\le ⦀AX+XB⦀,$$

(1.1)

where $A,B\in H_n^{+}(\mathbf{C})$, $X\in M_n(\mathbf{C})$, and $\nu \in [0,1]$. For a historical background and proofs of these norm inequalities as well as their refinements, and diverse applications, we refer the reader to the [2–8], and the references therein. Indeed, it has been proved, in [1], that $f(\nu )=⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀$ is a convex function of ν on $[0,1]$ with symmetry about $\nu =\frac{1}{2}$, and attains its minimum there and it has a maximum at $\nu =0$ and $\nu =1$. Moreover, it increases on $[0,\frac{1}{2}]$ and decreases on $[\frac{1}{2},1]$.

In [4, 5], (1.1) is refined by using the so-called Hermite-Hadamard inequality:

$$g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}g(t)dt\le \frac{g(a)+g(b)}{2},$$

where g is a convex function on $[a,b]$.

Recently, in [3] and [7], respectively, the following inequalities were used to get new refinements of (1.1):

$$\begin{array}{r}g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}g(t)dt\le \frac{1}{4}(g(a)+2g\left(\frac{a+b}{2}\right)+g(b))\le \frac{g(a)+g(b)}{2},\\ g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}g(t)dt\le \frac{1}{32}(15g(a)+2g\left(\frac{a+b}{2}\right)+15g(b))\le \frac{g(a)+g(b)}{2}.\end{array}$$

The purpose of this note is to obtain a family of new refinements of Heinz inequalities for matrices. Also the two refinements, given in [3] and [7], are two special cases of this new family.

2 Main results

We start by the following key lemma which plays a central role in our investigation to obtain a further series of refinements of the Heinz inequalities.

Lemma 1 Let g be a convex function on the interval $[a,b]$. Then for any positive integer n, we have

$$\begin{array}{rcl}g\left(\frac{a+b}{2}\right)& \le & \frac{1}{b-a}{\int }_a^{b}g(t)dt\le \frac{1}{4n}[(2n-1)g(a)+2g\left(\frac{a+b}{2}\right)+(2n-1)g(b)]\\ \le & \frac{g(a)+g(b)}{2}.\end{array}$$

Proof Since g is convex on $[a,b]$, we have

$$g\left(\frac{a+b}{2}\right)\le \frac{g(a)+g(b)}{2}.$$

Thus

$$(2n-1)g(a)+2g\left(\frac{a+b}{2}\right)+(2n-1)g(b)\le 2ng(a)+2ng(b),$$

whence

$$\frac{1}{4n}((2n-1)g(a)+2g\left(\frac{a+b}{2}\right)+(2n-1)g(b))\le \frac{g(a)+g(b)}{2}.$$

To prove the middle inequality, we start by

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}g(t)dt& =& \frac{1}{b-a}[{\int }_a^{\frac{a+b}{2}}g(t)dt+{\int }_{\frac{a+b}{2}}^{b}g(t)dt]\\ \le & \frac{1}{b-a}[\frac{g(\frac{a+b}{2})+g(a)}{2}\cdot \frac{b-a}{2}+\frac{g(b)+g(\frac{a+b}{2})}{2}\cdot \frac{b-a}{2}]\\ =& \frac{1}{4}[g(a)+2g\left(\frac{a+b}{2}\right)+g(b)]\\ =& \frac{1}{4n}[ng(a)+2ng\left(\frac{a+b}{2}\right)+ng(b)]\\ =& \frac{1}{4n}[ng(a)+2g\left(\frac{a+b}{2}\right)+(2n-2)g\left(\frac{a+b}{2}\right)+ng(b)]\\ \le & \frac{1}{4n}[ng(a)+2g\left(\frac{a+b}{2}\right)+(2n-2)\left[\frac{g(a)+g(b)}{2}\right]+ng(b)]\\ =& \frac{1}{4n}[ng(a)+2g\left(\frac{a+b}{2}\right)+(n-1)g(a)+(n-1)g(b)+ng(b)]\\ =& \frac{1}{4n}[(2n-1)g(a)+2g\left(\frac{a+b}{2}\right)+(2n-1)g(b)].\end{array}$$

 □

Applying the previous lemma on the convex function defined earlier

$$f(\nu )=⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀$$

on the interval $[\mu ,1-\mu ]$ when $0\le \mu \le \frac{1}{2}$ and on the interval $[1-\mu ,\mu ]$ when $\frac{1}{2}\le \mu \le 1$, we obtain the following refinement of the first inequality (1.1) which is a kind of refinements of Theorem 1 in a paper Kittaneh [5] and Theorem 1 in a paper of Feng [3].

Theorem 1 Let $A,B\in H_n^{+}(\mathbf{C})$, and $X\in M_n(\mathbf{C})$. Let n be any positive integer. Then for any $\mu \in [0,1]$, and for every unitarily invariant norm $⦀\cdot ⦀$ on $M_n(\mathbf{C})$, we have

$$\begin{array}{rl}2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀& \le \frac{1}{|1-2\mu |}|{\int }_{\mu }^{1-\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu |\\ \le \frac{1}{2n}[(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀+2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀]\\ \le ⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀.\end{array}$$

(2.1)

Proof First assume that $0\le \mu \le \frac{1}{2}$. Then it follows from Lemma 1 that

$$\begin{array}{rcl}f\left(\frac{\mu +1-\mu }{2}\right)& \le & \frac{1}{1-2\mu }{\int }_{\mu }^{1-\mu }f(t)dt\\ \le & \frac{1}{4n}[(2n-1)f(\mu )+2f\left(\frac{1-\mu +\mu }{2}\right)+(2n-1)f(1-\mu )]\\ \le & \frac{f(\mu )+f(1-\mu )}{2}=f(\mu ).\end{array}$$

Since $f(\mu )=f(1-\mu )$, we have

$$\begin{array}{rcl}f\left(\frac{1}{2}\right)& \le & \frac{1}{1-2\mu }{\int }_{\mu }^{1-\mu }f(t)dt\le \frac{1}{4n}[(4n-2)f(\mu )+2f\left(\frac{1}{2}\right)]\\ \le & \frac{1}{2n}[(2n-1)f(\mu )+f\left(\frac{1}{2}\right)]\le f(\mu ).\end{array}$$

Thus,

$$\begin{array}{rl}2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀& \le \frac{1}{1-2\mu }{\int }_{\mu }^{1-\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ \le \frac{1}{2n}[(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀+2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀]\\ \le ⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀.\end{array}$$

(2.2)

Now, assume that $\frac{1}{2}\le \mu \le 1$. Then, by applying (2.2) to $1-\mu$, it follows that

$$\begin{array}{rl}2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀& \le \frac{1}{2\mu -1}{\int }_{1-\mu }^{\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ \le \frac{1}{2n}[(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀+2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀]\\ \le ⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀.\end{array}$$

(2.3)

Since

$$\begin{array}{r}\underset{\mu \to \frac{1}{2}}{lim}\frac{1}{|1-2\mu |}|{\int }_{\mu }^{1-\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu |\\ =\underset{\mu \to \frac{1}{2}}{lim}\frac{1}{4n}[(4n-2)f(\mu )+f\left(\frac{1}{2}\right)]=2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀,\end{array}$$

the inequalities in (2.1) follow by combining (2.2) and (2.3) and so the required result is proved. □

Applying Lemma 1 to the function $f(\nu )=⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀$ in the interval $[\mu ,\frac{1}{2}]$ on $0\le \mu \le \frac{1}{2}$, and in the interval $[\frac{1}{2},\mu ]$ for $\frac{1}{2}\le \mu \le 1$, we obtain the following, which is a kind of refinements of Theorem 2 in a paper Kittaneh [5] and Theorem 2 in a paper of Feng [3].

Theorem 2 Let $A,B\in H_n^{+}(\mathbf{C})$, and $X\in M_n(\mathbf{C})$. Then, for any positive integer n, any $\mu \in [0,1]$, and for every unitarily invariant norm $⦀\cdot ⦀$ on $M_n(\mathbf{C})$, we have

$$\begin{array}{r}⦀A^{\frac{1+2\mu }{4}}X{B}^{\frac{3-2\mu }{4}}+A^{\frac{3-2\mu }{4}}X{B}^{\frac{1+2\mu }{4}}⦀\\ \le \frac{2}{|1-2\mu |}|{\int }_{\mu }^{\frac{1}{2}}⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu |\\ \le \frac{1}{4n}[(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀+2⦀A^{\frac{1+2\mu }{4}}X{B}^{\frac{3-2\mu }{4}}+A^{\frac{3-2\mu }{4}}X{B}^{\frac{1+2\mu }{4}}⦀\\ +2(2n-1)⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀]\\ \le \frac{1}{2}[⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀+2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀].\end{array}$$

(2.4)

Inequalities (2.4) and the first inequality in (1.1) yield the following refinements of the first inequality in (1.1).

Corollary 1 Let $A,B\in H_n^{+}(\mathbf{C})$, and $X\in M_n(\mathbf{C})$. Then, for any positive integer n, any $\mu \in [0,1]$, and for every unitarily invariant norm $⦀\cdot ⦀$ on $X\in M_n(\mathbf{C})$, we have

$$\begin{array}{rl}2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀\le & ⦀A^{\frac{1+2\mu }{4}}X{B}^{\frac{3-2\mu }{4}}+A^{\frac{3-2\mu }{4}}X{B}^{\frac{1+2\mu }{4}}⦀\\ \le & \frac{2}{|1-2\mu |}|{\int }_{\mu }^{\frac{1}{2}}⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu |\\ \le & \frac{1}{4n}[(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀\\ +2⦀A^{\frac{1+2\mu }{4}}X{B}^{\frac{3-2\mu }{4}}+A^{\frac{3-2\mu }{4}}X{B}^{\frac{1+2\mu }{4}}⦀+2(2n-1)⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀]\\ \le & \frac{1}{2}[⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀+2⦀A^{\frac{1}{2}}X{B}^{\frac{1}{2}}⦀]\\ \le & ⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀.\end{array}$$

(2.5)

Applying the Lemma 1 to the function $f(\nu )=⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀$ on the interval $[\mu ,\frac{1}{2}]$ when $0\le \mu \le \frac{1}{2}$, and on the interval $[\frac{1}{2},\mu ]$ when $\frac{1}{2}\le \mu \le 1$, we obtain the following theorem, which is a kind of refinements of Theorem 3 in a paper Kittaneh [5] and Theorem 3 in a paper of Feng [3].

Theorem 3 Let $A,B\in H_n^{+}(\mathbf{C})$, and $X\in M_n(\mathbf{C})$ and let n be a positive integer. Then:

• (1) for any $0\le \mu \le \frac{1}{2}$ and for every unitarily invariant norm $⦀\cdot ⦀$, we have

  $$\begin{array}{r}⦀A^{\frac{\mu }{2}}X{B}^{1-\frac{\mu }{2}}+A^{1-\frac{\mu }{2}}X{B}^{\frac{\mu }{2}}⦀\\ \le \frac{1}{\mu }{\int }_0^{\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ \le \frac{1}{4n}[(2n-1)⦀AX+XB⦀+2⦀A^{\frac{\mu }{2}}X{B}^{1-\frac{\mu }{2}}+A^{1-\frac{\mu }{2}}X{B}^{\frac{\mu }{2}}⦀\\ +(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀]\\ \le \frac{1}{2}[⦀AX+XB⦀+⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀];\end{array}$$

  (2.6)

• (2) for any $\frac{1}{2}\le \mu \le 1$ and for every unitarily invariant norm $⦀\cdot ⦀$, we have

  $$\begin{array}{r}⦀A^{\frac{1+\mu }{2}}X{B}^{\frac{1-\mu }{2}}+A^{\frac{1-\mu }{2}}X{B}^{\frac{1+\mu }{2}}⦀\\ \le \frac{1}{1-\mu }{\int }_{\mu }^1⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ \le \frac{1}{4n}[(2n-1)⦀AX+XB⦀+2⦀A^{\frac{1+\mu }{2}}X{B}^{\frac{1-\mu }{2}}+A^{\frac{1-\mu }{2}}X{B}^{\frac{1+\mu }{2}}⦀]\\ \le \frac{1}{2}[⦀AX+XB⦀+⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀].\end{array}$$

  (2.7)

Since the function $f(\nu )=⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀$ is decreasing on the interval $[0,\frac{1}{2}]$ and increasing on the interval $[\frac{1}{2},1]$, and using the inequalities (2.6) and (2.7), we obtain a family of refinements of second inequality in (1.1).

Corollary 2 Let $A,B\in H_n^{+}(\mathbf{C})$, and $X\in M_n(\mathbf{C})$ and let n be a positive integer. Then:

• (1) for any $0\le \mu \le \frac{1}{2}$ and for every unitarily invariant norm $⦀\cdot ⦀$, we have

  $$\begin{array}{r}⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀\\ \le ⦀A^{\frac{\mu }{2}}X{B}^{1-\frac{\mu }{2}}+A^{1-\frac{\mu }{2}}X{B}^{\frac{\mu }{2}}⦀\\ \le \frac{1}{\mu }{\int }_0^{\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ \le \frac{1}{4n}[(2n-1)⦀AX+XB⦀+2⦀A^{\frac{\mu }{2}}X{B}^{1-\frac{\mu }{2}}+A^{1-\frac{\mu }{2}}X{B}^{\frac{\mu }{2}}⦀\\ +(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀]\\ \le \frac{1}{2}[⦀AX+XB⦀+⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀]\\ \le ⦀AX+XB⦀;\end{array}$$

  (2.8)

• (2) for any $\frac{1}{2}\le \mu \le 1$ and for every unitarily invariant norm $⦀\cdot ⦀$, we have

  $$\begin{array}{r}⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀\\ \le ⦀A^{\frac{1+\mu }{2}}X{B}^{\frac{1-\mu }{2}}+A^{\frac{1-\mu }{2}}X{B}^{\frac{1+\mu }{2}}⦀\\ \le \frac{1}{1-\mu }{\int }_{\mu }^1⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ \le \frac{1}{4n}[(2n-1)⦀AX+XB⦀+2⦀A^{\frac{1+\mu }{2}}X{B}^{\frac{1-\mu }{2}}+A^{\frac{1-\mu }{2}}X{B}^{\frac{1+\mu }{2}}⦀\\ +(2n-1)⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀]\\ \le \frac{1}{2}[⦀AX+XB⦀+⦀A^{\mu }X{B}^{1-\mu }+A^{1-\mu }X{B}^{\mu }⦀]\\ \le ⦀AX+XB⦀.\end{array}$$

  (2.9)

It should be noted that in inequalities (2.8) and (2.9), we have

$$\begin{array}{r}\underset{\mu \to 0\frac{1}{\mu }}{lim}{\int }_0^{\mu }⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu \\ =\underset{\mu \to 1}{lim}\frac{1}{1-\mu }{\int }_{\mu }^1⦀A^{\nu }X{B}^{1-\nu }+A^{1-\nu }X{B}^{\nu }⦀d\nu =⦀AX+XB⦀.\end{array}$$

Remark 1 The two special values $n=1$ and $n=8$ give the refinements of Heinz inequalities obtained in [3] and [7], respectively.