Improved Heinz inequality and its application

Abstract

We obtain an improved Heinz inequality for scalars and we use it to establish an inequality for the Hilbert-Schmidt norm of matrices, which is a refinement of a result due to Kittaneh.

Mathematical Subject Classification 2010: 26D07; 26D15; 15A18.

1. Introduction

Let M _n be the space of n × n complex matrices and ||·|| stand for any unitarily invariant norm on M _n . So, ||UAV|| = ||A|| for all A ∈ M _n and for all unitary matrices U, V ∈ M _n . If A = [a _ij ] ∈ M _n , then

$${∥A∥}_2={\left(\sum _{i,j=1}^n|a_{ij}{|}^2\right)}^{1/2}$$

is the Hilbert-Schmidt norm of matrix A. It is known that the Hilbert-Schmidt norm is unitarily invariant.

The classical Young's inequality for nonnegative real numbers says that if a, b ≥ 0 and 0 ≤ v ≤ 1, then

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

(1.1)

with equality if and only if a = b. Young's inequality for scalars is not only interesting in itself but also very useful. If $v=\frac{1}{2}$, by (1.1), we obtain the arithmetic-geometric mean inequality

$$2\sqrt{ab}\le a+b.$$

(1.2)

Kittaneh and Manasrah [1] obtained a refinement of Young's inequality as follows:

$$a^v{b}^{1-v}+r_0{\left(\sqrt{a}-\sqrt{b}\right)}^2\le va+\left(1-v\right)b,$$

(1.3)

where r₀ = min {v, 1 − v}.

Let a, b ≥ 0 and 0 ≤ v ≤ 1. The Heinz means are defined as follows:

$$H_v\left(a,b\right)=\frac{a^v{b}^{1-v}+a^{1-v}{b}^v}{2}.$$

It follows from the inequalities (1.1) and (1.2) that the Heinz means interpolate between the geometric mean and the arithmetic mean:

$$\sqrt{ab}\le H_v\left(a,b\right)\le \frac{a+b}{2}.$$

(1.4)

The second inequality of (1.4) is known as Heinz inequality for nonnegative real numbers.

As a direct consequence of the inequality (1.3), Kittaneh and Manasrah [1] obtained a refinement of the Heinz inequality as follows:

$$H_v\left(a,b\right)+r_0{\left(\sqrt{a}-\sqrt{b}\right)}^2\le \frac{a+b}{2},$$

(1.5)

where r₀ = min {v, 1 − v}.

Bhatia and Davis [2] proved that if A, B, X ∈ M _n such that A and B are positive semidefinite and if 0 ≤ v ≤ 1, then

$$2∥A^{1/2}X{B}^{1/2}∥\le ∥A^{v}X{B}^{1-v}+A^{1-v}X{B}^v∥\le ∥AX+XB∥.$$

(1.6)

This is a matrix version of the inequality (1.4). Kittaneh [3] proved that if A, B, X ∈ M _n such that A and B are positive semidefinite and if 0 ≤ v ≤ 1, then

$$∥A^{v}X{B}^{1-v}+A^{1-v}X{B}^v∥\le 4r_0∥A^{1/2}X{B}^{1/2}∥+\left(1-2r_0\right)∥AX+XB∥,$$

(1.7)

where r₀ = min {v, 1 − v}. This is a refinement of the second inequality in (1.6).

In this article, we first present a refinement of the inequality (1.5). After that, we use it to establish a refinement of the inequality (1.7) for the Hilbert-Schmidt norm.

2. A refinement of the inequality (1.5)

In this section, we give a refinement of the inequality (1.5). To do this, we need the following lemma.

Lemma 2.1. [4, 5] Let f(x) be a real valued convex function on an interval [a, b]. For any x₁, x₂ ∈ [a, b], we have

$$f\left(x\right)\le \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}x-\frac{x_{1}f\left(x_2\right)-x_{2}f\left(x_1\right)}{x_2-x_1},x\in \left(x_1,x_2\right).$$

Theorem 2.1. Let a, b ≥ 0 and 0 ≤ v ≤ 1. If r₀ = min {v, 1 − v}, then

$$2H_v\left(a,b\right)\le \left\{\begin{array}{c}\left(1-4r_0\right)\left(a+b\right)+4r_0\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right),v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right],\\ 2\left(4r_0-1\right)\sqrt{ab}+2\left(1-2r_0\right)\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right),v\in \left[\frac{1}{4},\frac{3}{4}\right].\end{array}\right.$$

(2.1)

Proof. It is known that as a function of v, H _v (a, b) is convex and attains its minimum at $v=\frac{1}{2}$. Let

$$f\left(v\right)=2H_v\left(a,b\right)=a^v{b}^{1-v}+a^{1-v}{b}^v,0\le v\le 1.$$

Obviously, f(v) is convex. For $0\le v\le \frac{1}{4}$, since f(v) is convex on 0[1], by Lemma 2.1, we have

$$f\left(v\right)\le \frac{f\left(\frac{1}{4}\right)-f\left(0\right)}{\frac{1}{4}-0}v-\frac{0f\left(\frac{1}{4}\right)-\frac{1}{4}f\left(0\right)}{\frac{1}{4}-0},$$

which is equivalent to

$$f\left(v\right)\le 4\left(f\left(\frac{1}{4}\right)-f\left(0\right)\right)v+f\left(0\right).$$

That is,

$$f\left(v\right)\le \left(\mathsf{\text{1}}-\mathsf{\text{4}}v\right)f\left(0\right)+\mathsf{\text{4}}vf\left(\frac{1}{4}\right).$$

So,

$$a^v{b}^{1-v}+a^{1-v}{b}^v\le \left(\mathsf{\text{1}}-\mathsf{\text{4}}{r}_0\right)\left(a+b\right)+\mathsf{\text{4}}{r}_0\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right).$$

For $\frac{3}{4}\le v\le 1$, similarly, we have

$$f\left(v\right)\le \frac{f\left(1\right)-f\left(\frac{3}{4}\right)}{1-\frac{3}{4}}v-\frac{\frac{3}{4}f\left(1\right)-f\left(\frac{3}{4}\right)}{1-\frac{3}{4}},$$

which is equivalent to

$$f\left(v\right)\le \mathsf{\text{4}}\left(f\left(\mathsf{\text{1}}\right)-f\left(\frac{3}{4}\right)\right)v-\mathsf{\text{3}}f\left(\mathsf{\text{1}}\right)+\mathsf{\text{4}}f\left(\frac{3}{4}\right).$$

That is,

$$f\left(v\right)\le \left(\mathsf{\text{4}}v-\mathsf{\text{3}}\right)f\left(\mathsf{\text{1}}\right)+\mathsf{\text{4}}\left(\mathsf{\text{1}}-v\right)f\left(\frac{3}{4}\right).$$

So,

$$a^v{b}^{1-v}+a^{1-v}{b}^v\le \left(\mathsf{\text{1}}-\mathsf{\text{4}}{r}_0\right)\left(a+b\right)+\mathsf{\text{4}}{r}_0\left(a^{\mathsf{\text{1}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{3}}/\mathsf{\text{4}}}+a^{\mathsf{\text{3}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{1}}/\mathsf{\text{4}}}\right).$$

If $\frac{1}{4}\le v\le \frac{1}{2}$, then by Lemma 2.1, we have

$$f\left(v\right)\le \frac{f\left(\frac{1}{2}\right)-f\left(\frac{1}{4}\right)}{\frac{1}{2}-\frac{1}{4}}v-\frac{\frac{1}{4}f\left(\frac{1}{2}\right)-\frac{1}{2}f\left(\frac{1}{4}\right)}{\frac{1}{2}-\frac{1}{4}},$$

and so

$$f\left(v\right)\le \left(\mathsf{\text{4}}v-\mathsf{\text{1}}\right)f\left(\frac{1}{2}\right)+2\left(1-2v\right)f\left(\frac{1}{4}\right),$$

which is equivalent to

$$a^v{b}^{1-v}+a^{1-v}{b}^v\le 2\left(\mathsf{\text{4}}{r}_0-1\right)\sqrt{ab}+2\left(1-2r_0\right)\left(a^{\mathsf{\text{1}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{3}}/\mathsf{\text{4}}}+a^{3/\mathsf{\text{4}}}{b}^{1/\mathsf{\text{4}}}\right).$$

If $\frac{1}{2}\le v\le \frac{3}{4}$, similarly, we have

$$f\left(v\right)\le \frac{f\left(\frac{3}{4}\right)-f\left(\frac{1}{2}\right)}{\frac{3}{4}-\frac{1}{2}}v-\frac{\frac{1}{2}f\left(\frac{3}{4}\right)-\frac{3}{4}f\left(\frac{1}{2}\right)}{\frac{3}{4}-\frac{1}{2}},$$

and so

$$f\left(v\right)\le \left(3-4v\right)f\left(\frac{1}{2}\right)+2\left(2v-1\right)f\left(\frac{3}{4}\right),$$

which is equivalent to

$$f\left(v\right)\le \left(\mathsf{\text{4}}{r}_0-\mathsf{\text{1}}\right)f\left(\frac{1}{2}\right)+2\left(1-2r_0\right)f\left(\frac{3}{4}\right).$$

That is,

$$a^v{b}^{1-v}+a^{1-v}{b}^v\le \mathsf{\text{2}}\left(\mathsf{\text{4}}{r}_0-\mathsf{\text{1}}\right)\sqrt{ab}+\mathsf{\text{2}}\left(\mathsf{\text{1}}-\mathsf{\text{2}}{r}_0\right)\left(a^{\mathsf{\text{1}}/\mathsf{\text{4}}}{b}^{\mathsf{\text{3}}/\mathsf{\text{4}}}+a^{3/\mathsf{\text{4}}}{b}^{1/\mathsf{\text{4}}}\right).$$

This completes the proof. □

Now, we give a simple comparison between the upper bound for a^vb^1-v+ a^1-vb^v in (1.5) and (2.1). If $v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]$, then

$$\begin{array}{ll}\hfill a+b-2r_0{\left(\sqrt{a}-\sqrt{b}\right)}^2& -\left(1-4r_0\right)\left(a+b\right)-4r_0\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right)\\ =2r_0\left(a+b+2\sqrt{ab}-2\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right)\right)\\ \ge 0.\end{array}$$

If $v\in \left[\frac{1}{4},\frac{3}{4}\right]$, then

$$\begin{array}{ll}\hfill a+b-2r_0{\left(\sqrt{a}-\sqrt{b}\right)}^2& -2\left(4r_0-1\right)\sqrt{ab}-2\left(1-2r_0\right)\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right)\\ =\left(1-2r_0\right)\left(a+b+2\sqrt{ab}-2\left(a^{1/4}{b}^{3/4}+a^{3/4}{b}^{1/4}\right)\right)\\ \ge 0.\end{array}$$

So, the inequality (2.1) is a refinement of the inequality (1.5).

3. An application

In this section, we give a refinement of the inequality (1.7) for the Hilbert-Schmidt norm based on the inequality (2.1).

Theorem 3.1. Let A, B, X ∈ M _n such that A and B are positive semidefinite and suppose

that

$$\varphi \left(v\right)=\left|\right|A^{v}X{B}^{1-v}+A^{1-v}X{B}^v|{|}_{\mathsf{\text{2}}},0\le v\le \mathsf{\text{1}}.$$

Then

$$\varphi \left(v\right)\le \left\{\begin{array}{c}\left(1-4r_0\right)\varphi \left(0\right)+4r_0\varphi \left(\frac{1}{4}\right),v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]\\ \left(4r_0-1\right)\varphi \left(\frac{1}{2}\right)+2\left(1-2r_0\right)\varphi \left(\frac{1}{4}\right),v\in \left[\frac{1}{4},\frac{3}{4}\right]\end{array}\right.,$$

(3.1)

where r₀ = min {v, 1 − v}.

Proof. Since every positive semidefinite matrix is unitarily diagonalizable, it follows that there exist unitary matrices U, V ∈ M _n such that A = U Λ₁U* and B = V Λ₂V*, where Λ₁ = diag (λ₁,..., λ _n ), Λ₂ = diag(µ₁,..., µ _n ) and λ _i , µ _i ≥ 0, i = 1,..., n. Let

$$Y=U^{*}XV=\left[y_{ij}\right].$$

If $v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]$, then by (2.1) and the Cauchy-Schwarz inequality, we have

$$\begin{array}{ll}\hfill \left|\right|A^{v}X{B}^{1-v}+A^{1-v}X{B}^v|{|}_2^2& =\sum _{i,j=1}^n{\left({\lambda }_i^v{\mu }_j^{1-v}+{\lambda }_i^{1-v}{\mu }_j^v\right)}^2|y_{ij}{|}^2\\ \le \sum _{i,j=1}^n{\left(\left(1-4r_0\right)\left({\lambda }_i+{\mu }_j\right)+4r_0\left({\lambda }_i^{1/4}{\mu }_j^{3/4}+{\lambda }_i^{3/4}{\mu }_j^{1/4}\right)\right)}^2|y_{ij}{|}^2\\ ={\left(1-4r_0\right)}^2\sum _{i,j=1}^n{\left({\lambda }_i+{\mu }_j\right)}^2|y_{ij}{|}^2\\ +16r_0^2\sum _{i,j=1}^n{\left({\lambda }_i^{1/4}{\mu }_j^{3/4}+{\lambda }_i^{3/4}{\mu }_j^{1/4}\right)}^2|y_{ij}{|}^2\\ +8r_0\left(1-4r_0\right)\sum _{i,j=1}^n\left({\lambda }_i+{\mu }_j\right)\left({\lambda }_i^{1/4}{\mu }_j^{3/4}+{\lambda }_i^{3/4}{\mu }_j^{1/4}\right)|y_{ij}{|}^2\\ \le {\left(1-4r_0\right)}^2{\varphi }^2\left(0\right)+16r_0^2{\varphi }^2\left(\frac{1}{4}\right)+8r_0\left(1-4r_0\right)\varphi \left(0\right)\varphi \left(\frac{1}{4}\right)\\ ={\left(\left(1-4r_0\right)\varphi \left(0\right)+4r_0\varphi \left(\frac{1}{4}\right)\right)}^2.\end{array}$$

If $v\in \left[\frac{1}{4},\frac{3}{4}\right]$, the result follows from the inequality (2.1) and the same method above. This completes the proof. □

Remark. For the Hilbert-Schmidt norm, by the inequality (1.7), we have

$$\varphi \left(v\right)\le 2r_0\varphi \left(\frac{1}{2}\right)+\left(1-2r_0\right)\varphi \left(0\right).$$

So, for $v\in \left[0,\frac{1}{4}\right]\cup \left[\frac{3}{4},1\right]$, we have

$$\begin{array}{ll}\hfill 2r_0\varphi \left(\frac{1}{2}\right)& +\left(1-2r_0\right)\varphi \left(0\right)-\left(1-4r_0\right)\varphi \left(0\right)-4r_0\varphi \left(\frac{1}{4}\right)\\ =2r_0\left(\varphi \left(\frac{1}{2}\right)+\varphi \left(0\right)-2\varphi \left(\frac{1}{4}\right)\right)\ge 0.\end{array}$$

If $v\in \left[\frac{1}{4},\frac{3}{4}\right]$, then

$$\begin{array}{ll}\hfill 2r_0\varphi \left(\frac{1}{2}\right)& +\left(1-2r_0\right)\varphi \left(0\right)-\left(4r_0-1\right)\varphi \left(\frac{1}{2}\right)-2\left(1-2r_0\right)\varphi \left(\frac{1}{4}\right)\\ =\left(1-2r_0\right)\left(\varphi \left(\frac{1}{2}\right)+\varphi \left(0\right)-2\varphi \left(\frac{1}{4}\right)\right)\ge 0.\end{array}$$

So, the inequality (3.1) is a refinement of the inequality (1.7) for the Hilbert-Schmidt norm.